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SCUAVWS OUTLINE OF
THEORY AND PltOBLEMS
OP
GROUP THEORY
ВТ
BENJAMIN BAUMSLAC, Ph.D.
BRUCE CHANDLER, Ph.D.
Department of Mathematics
New York Uniwrsity
SCHAUM'S OUTLINE SERIES
McGraw-Hill
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Copyright © 1968 by McGraw-Hill, Inc. All Rights Reserved. Printed in the
United States of America. No part of this publication may be reproduced,
stored in a retrieval system, or transmitted, in any form or by any means,
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prior written permission of the publisher.
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Typography by Signs and Symbols, Inc., New York, N. Y.

Preface
The study of groups arose early in the nineteenth century in connection with the
solution of equations. Originally a group was a set of permutations with the property that
the combination of any two permutations again belongs to the set. Subsequently this
definition was generalized to the concept of an abstract group, which was defined to be a
set, not necessarily of permutations, together with a method of combining its elements
that is subject to a few simple laws.
The theory of abstract groups plays an important part in present day mathematics
and science. Groups arise in a bewildering number of apparently unconnected subjects.
Thus they appear in crystallography and quantum mechanics, in geometry and topology, in
analysis and algebra, in physics, chemistry and even in biology.
One of the most important intuitive ideas in mathematics and science is symmetry.
Groups can describe symmetry; indeed many of the groups that arose in mathematics and
science were encountered in the study of symmetry. This explains to some extent why
groups arise so frequently.
Although groups arose in connection with other disciplines, the study of groups is in
itself exciting. Currently there is vigorous research in the subject, and it attracts the
energies and imagination of a great many mathematicians.
This book is designed for a first course in group theory. It is mainly intended for
college and first year graduate students. It is complete in itself and can be used for
self-study or as a text for a formal course. Moreover, it could with advantage be used
as a supplement to courses in group theory and modern algebra. Little prior knowledge is
assumed. The reader should know the beginnings of elementary number theory, a summary
of which appears in Appendix A. An acquaintance with complex numbers is needed for
some problems. In short, a knowledge of high school mathematics should be a sufficient
prerequisite, and highly motivated and bright high school students will be able to
understand much of this book.
The aim of this book is to make the study of group theory easier. Each chapter begins
with a preview and ends with a summary, so that the reader may see the ideas as a whole.
Each main idea appears in a section of its own, is motivated, is explained in great detail,
and is made concrete by solved problems.
Chapter 1 presents the rudiments of set theory and the concept of binary operation,
which are fundamental to the whole subject. Chapter 2, on groupoids, further explores
the concept of binary operation. In most courses on group theory the concept of groupoid
is usually treated briefly if at all. We have chosen to treat it more fully for the following
reasons: (a) A thorough understanding of binary compositions is thereby obtained, (b) The
important ideas of homomorphism, isomorphism and Cayley's theorem occur both in the
chapter on groupoids and in the chapters on groups, and the repetition ensures familiarity.
Chapter 3 shows that the concept of group is natural by producing a large number of
examples of groups that arise in different fields. Here are discussed groups of real and
complex numbers, the symmetric groups, symmetry groups, dihedral groups, the group of
Mobius transformations, automorphism groups of groupoids and fields, groups of matrices,
and the full linear group.
Chapter 4 is concerned with the homomorphism theorems and cyclic groups. The
concept of homomorphism is fundamental, and thus the theorems of this chapter are
indispensable for further study.

Chapter 5 is on finite groups. The Sylow theorems are proved, the concept of external
direct product is introduced, and groups up to order 15 are classified. The chapter
concludes with the Jordan-Holder theorem and a proof that most alternating groups are simple.
Chapter 6 is on abelian groups. Two important classes of abelian groups are treated:
finitely generated and divisible groups. Undergraduates will probably find their needs
are met by the material through Section 6.3. Graduate students will certainly want to
continue.
Chapter 7 is on permutational representations and extensions. Chapter 8 is on free
groups and presentations. Those who would like to study the theory of groups more
deeply will find a guide to the literature at the end of the book.
Chapters 1.-4 must be read in order, although, if desired, only the first three sections of
Chapter 3 need be read at first (the other sections of Chapter 3 may be studied when they
are needed). The order of reading Chapters 5-8 can be varied, although part of Chapter 7
is required for the last sections of Chapter 8.
The reader need not work all the solved problems; he should decide for himself how
much practice he needs. Some of the problems are designed to clarify the immediately
preceding text, and the reader will find that the solutions may overcome some of his
obstacles. On the whole, however, it is advisable to attempt the problems before reading
their solutions. The numerous supplementary problems, some of which are very difficult,
serve as a review of the material of each chapter.
We thank Prof. Gilbert Baumslag for giving us access to several chapters of
unpublished notes and for many useful suggestions. We thank Sister Weiss for reading two
chapters of an early draft, Harold Brown for much helpful advice, Henry Hayden for
typographical arrangement and art work, and Louise Baggot for the typing. Finally we
express our appreciation to Daniel Schaum and Nicola Monti for their unfailing editorial
cooperation.
B. BAUMSLAG
B. CHANDLER
June 1968

CONTENTS
Chapter 1 SETS, MAPPINGS AND BINARY OPERATIONS
Preview of Chapter 1
1.1 SETS
a. Basic notions, b. Union and intersection.
1.2 CARTESIAN PRODUCTS
a. Definition, b. Equivalence relations, с Partitions and equivalence relations,
d. The division notation.
1.3 MAPPINGS
a. Definition of mapping, .b. Formal definition of mapping, с Types of
mappings.
1.4 COMPOSITION OP MAPPINGS
1.5 BINARY OPERATIONS
a. Definition, b. The multiplication table.
A look back at Chapter 1
Page
1
1
11
17
19
24
Chapter 2 GROUPOIDS
Preview of Chapter 2
2.1 GROUPOIDS
a. Definition of a groupoid. b. Equality of groupoids.
COMMUTATIVE AND ASSOCIATIVE GROUPOIDS
IDENTITIES AND INVERSES IN GROUPOIDS
a. The identity of a groupoid. b. Inverses in a groupoid,
SEMIGROUPS WITH AN IDENTITY ELEMENT
a. Uniqueness of inverses, b. The semigroup of mappings of a set into itself,
c. Notation for a mapping, d. The order in a product.
HOMOMORPHISMS OP GROUPOIDS AND CAYLEY'S THEOREM
a. Definition of a homomorphism. b. Epimorphism, monomorphism, and
isomorphism, c. Properties of epimorphisms. d. Naming and isomorphisms, e. Mx
and semigroups.
A look back at Chapter 2
2.2
2.3
2.4
2.5
26
26
29
30
33
40
47
Chapter 3 GROUPS AND SUBGROUPS
Preview of Chapter 3
3.1
3.2
3.3
3.4
3.5
GROUPS
Definition. Examples of groups of numbers.
SUBGROUPS
THE SYMMETRIC AND ALTERNATING GROUPS
a. The symmetric group on X. b. Even and odd permutations, с The
alternating groups, d. The order of An.
GROUPS OP ISOMETRIES
a. Isometries of the line. b. Two points determine an isometry. с Isometries
of the plane, d. Isometries are products of reflections, translations and
rotations, e. Symmetry groups, f. The dihedral groups.
THE GROUP OF MOBIUS TRANSFORMATIONS
a. Defining the group, b. 2 X 2 matrices.
3.6 SYMMETRIES OP AN ALGEBRAIC STRUCTURE
a. Automorphisms of groupoids. b. Fields of complex numbers, c.
Automorphisms of fields, d. Vector spaces, e. Linear transformations. The full linear
group.
A look back at Chapter 3
50
50
54
56
64
77
83
91

CONTENTS
Chapter 4 ISOMORPHISM THEOREMS Page
Preview of Chapter 4 94
4.1 FUNDAMENTALS 94
a. Preliminary remarks, b. More about subgroups, с Exponents.
4.2 CYCLIC GROUPS 101
a. Fundamentals of cyclic groups, b. Subgroups of cyclic groups.
4.3 COSETS 107
a. Introduction to the idea of coset. b. Cosets form a partition. Lagrange's
theorem, с Normal subgroups, d. Commutator subgroups, centralizers,
normalizes, e. Factor groups.
4.4 HOMOMORPHISM THEOREMS 117
a. Homomorphisms and factor groups: The homomorphism theorem, b.
Correspondence theorem. Factor of a factor theorem, с The subgroup isomorphism
theorem, d. Homomorphisms of cyclic groups.
A look back at Chapter 4 127
Chapter 5 FINITE GROUPS
Preview of Chapter 5 130
5.1 THE SYLOW THEOREMS ISO
a. Statements of the Sylow theorems, b. Two lemmas used in the proof of the
Sylow theorems, с Proofs of the Sylow theorems.
5.2 THEORY OF p-GROUPS 139
a. The importance of p-groups in finite groups, b. The center of a p-group.
с The upper central series.
5.3 DIRECT PRODUCTS AND GROUPS OF LOW ORDER 143
a. Direct products of groups, b. Groups of small order: orders ρ and 2p.
c. Groups of small order: orders 8 and 9. d. Groups of small order: orders 12
and 15.
5.4 SOLVABLE GROUPS 158
a. Definition of solvable groups, b. Properties of, and alternative definition
for, solvable groups.
5.5 COMPOSITION SERIES AND SIMPLE GROUPS 163
a. The Jordan-Holder theorem, b. Proof of Jordan-Holder theorem, c. Cycles
and products of cycles, d. Transpositions, and even and odd permutations.
e. The simplicity of An, η — 5.
A look back at Chapter 5 174
Chapter 6 ABELIAN GROUPS
Preview of Chapter 6 177
6.1 PRELIMINARIES 178
a. Additive notation and finite direct sums. b. Infinite direct sums, с The
homomorphic property of direct sums and free abelian groups.
6.2 SIMPLE CLASSIFICATION OF ABELIAN GROUPS,
AND STRUCTURE OF TORSION GROUPS 188
a. Tentative classifications: torsion, torsion-free, and mixed, b. The torsion
subgroup, c. Structure of torsion groups. Prufer groups, d. Independence
and rank.
6.3 FINITELY GENERATED ABELIAN GROUPS 196
a. Lemmas for finitely generated free abelian groups, b. Fundamental theorem
of abelian groups, с The type of a finitely generated abelian group, d.
Subgroups of finitely generated abelian groups.
6.4 DIVISIBLE GROUPS 205
a. p-Prufer groups. Divisible subgroups, b. Decomposition theorem for
divisible groups.
A look back at Chapter 6 211

CONTENTS
Chapter 7 PERMUTATIONAL REPRESENTATIONS Page
Preview of Chapter 7 214
7.1 CAYLEY'S THEOREM 214
a. Another proof of Cayley's theorem, b. Cayley's theorem and examples of
groups.
7.2 PERMUTATIONAL REPRESENTATIONS 216
7.3 DEGREE OF A REPRESENTATION AND FAITHFUL REPRESENTATIONS.. 217
a. Degree of a representation, b. Faithful representations.
7.4 PERMUTATIONAL REPRESENTATIONS ON COSETS 218
7.5 FROBENIUS' VARIATION OF CAYLEY'S THEOREM 222
a. The kernel of a coset representation, b. Frobenius' theorem.
7.6 APPLICATIONS TO FINITELY GENERATED GROUPS 227
a. Subgroups of finite index, b. Remarks about the proof of Theorem 7.4.
с Marshall Hall's theorem, d. One consequence of Theorem 7.5.
7.7 EXTENSIONS 232
a. General extension, b. The splitting extension, с An analysis of splitting
extensions, d. Direct product.
7.8 THE TRANSFER 240
a. Definition, b. Proof that τ is a homomorphism. с Proof that τ is independent
of the choice of transversal, d. A theorem of Schur.
A look back at Chapter 7 243
Chapter 8 FREE GROUPS AND PRESENTATIONS
Preview of Chapter 8 245
8.1 ELEMENTARY NOTIONS 245
a. Definition of a free group, b. Length of an element. Alternative description
of a free group, с Existence of free groups, d. Homomorphisms of free
groups.
8.2 PRESENTATIONS OF GROUPS 253
a, Definitions, b. Illustrations of presentations.
8J THE SUBGROUP THEOREM FOR FREE GROUPS; AN EXAMPLE 259
8.4 PROOF OF THE SUBGROUP THEOREM FOR FREE GROUPS 260
a. Plan of the proof, b. Schreier transversals, c. A look at the elements aXiS.
d. The proof of the subgroup theorem, e. Subgroups of finite index, f.
Intersection of finitely generated subgroups.
A look back at Chapter 8 266
Appendix A NUMBER THEORY 269
Appendix В A GUIDE TO THE LITERATURE 270
IN D EX 274
SYMBOLS AND NOTATIONS
278

Chapter 1
Sets, Mappings and Binary Operations
Preview of Chapter 1
This chapter begins with a few remarks about sets. A set is a collection of objects.
For example, the real numbers form a set, the objects being the numbers.
The real numbers have an operation called addition. Addition essentially involves two
numbers, for the addition of a single number is meaningless, while the addition of three
or more numbers is repeated addition of two numbers. Because addition involves two
numbers it is called a binary operation.
The main object of this chapter is to define precisely the notion of a binary operation.
The concept of binary operation is required to define the concept of group.
We introduce the important ideas of cartesian product and mapping. Welding them
together gives rise to an explicit definition of a binary operation. Another important idea
is that of equivalence relation, which is a generalization of the idea of equality. The reader
will also pick up much useful notation.
1.1 SETS
a. Basic notions
Set is synonymous with collection. The objects in a set are termed the elements of
the set. Usually we denote sets by capital Latin letters, for example, B, G, T. We shall
denote
(i) the set of positive integers 1,2,3, . .. by Ρ
(ii) the set of nonnegative integers 0,1,2, . .. by N
(iii) the set of all integers by Ζ
(iv) the set of rational numbers by Q
(v) the set of real numbers by R
(vi) the set of complex numbers by C.
The elements of a set will usually be denoted by small Latin letters such as s, t, u, etc.
By s Ε S we mean "s is an element of S" or "s belongs to S". In particular, 2 e P. If s
is not an element of S, we write s £ S and read this as "s is not an element of S" or "s does
not belong to S" or "s is not in S". For example, —1 £ P.
In dealing with sets it is advantageous to abbreviate the phrase "the set whose elements
are" by using braces. Thus, for example, we write {1,2} for the set whose elements are 1
and 2 and similarly we write {-1, 0,1,2, ...} for the set whose elements are -1, 0,1,2, ... .
A variation of this notation is useful to describe a set in terms of a property which singles
out its elements. Thus we write {χ \ χ has the property Ψ) for the set of all those elements
χ which have the property Ψ. Here ψ stands for some "understandable" property; to
illustrate: {χ \ χ is a real number} is R, the set of real numbers. (Ψ here is the property of
being a real number.) Notice that we read {χ \ χ is a real number} as the set of all those
elements χ which have the property that ж is a real number, or the set of all those elements
χ such that ж is a real number.
1

2
SETS, MAPPINGS AND BINAEY OPERATIONS
[CHAP. 1
В
We can now introduce some useful notation.
(i) If a and b are real numbers and a <b, then the open interval (a, b) is defined by
(a, b) — {x | χ e R and α < χ < Ъ).
у
(ii) Again if α and b are real numbers and α < b, then the closed interval [a, b] is defined
by [а,Ъ] = {χ Ι χ G R and a^x — b).
(iii) In coordinate geometry (0,0) denotes the origin,
(0,1) the point A, and (1,0) the point B. In
general, if a, b are any two elements of a set,
(a, b) is called the ordered pair formed from α
and b. We will say (a, b) = (c, e?) if and only
if a — с and b = d. (a, b) is called an ordered
pair because the order of α and b matters; (a, b)
is not the same as (b,a) if α ¥■ Ъ.
This idea enables us to define the Euclidean plane as {ρ \ ρ — (ж, y) where x, у G R},
where the distance between (x,y) and (xi,yi) is defined by λ/{χ — X\f + {y — y\f. We
write i?2= {p\ p — (x,y) with х,у €i?}. Rz is not the Euclidean plane we normally
think of. Rather it can be interpreted as the set of coordinates of the Euclidean plane.
Having defined the Euclidean plane it is easy to define, for example, circles and discs.
(iv) A circle with radius r and center at the origin is defined by {p\ ρ — (x,y) G R2- and
31 + ^ = 7»}.
(v) The disc with radius r and center at the origin is defined by {p \ ρ = (χ, у) G R2 and
Xi + yZ^r1}.
We say that two sets S and Τ are equal, and write S = T, if every element of S belongs
to Τ and every element of Τ belongs to S. Thus {2,2,3,3} = {2,3} = {3,2}.
If every element of the set S is also an element of the set T, we say S is a subset of Τ and
express this briefly by writing ScT. ScT means S С Τ but S ¥> T. Thus Ρ С N and
Ρ с N.
We can use the notion of subset to give a criterion for the equality of sets.
Proposition 1.1: Set S is equal to set Τ if and only if S С Τ and Τ QS.
Proof: SqT and Τ С S expresses in symbols "every element of S belongs to Τ and
every element of Τ belongs to S".
Problems
1.1. Are the following statements true?
(i) 2 € {2}
(ii) 3€{2,4}
(iii) ζ Й {a, b}, ζ ¥= a and ζ ¥= Ь
Solution:
(i) True (iv) False (vii) False, since л/2 is not a rational number.
(ii) False (v) True (viii) True, for {■/-1)2 = —1 and —1 is an integer.
(iii) True (vi) False (ix) False, since —1 € P.
1.2. Check the truth of the following assertions.
(i) {2} is a subset of {2}. (v) PcQ (ix) Q С Q
(ii) If S is any set, ScS. (vi) Ζ QQ (x) {3,a, b, c} = {3,a, b, 3,c, 6}
(iii) {а} с {a, b}, α Φ Ъ (vii) QcR
(iv) {2,3} - {3,4} (viii) ficC
Solution:
(i) True. The only element of {2} is 2, and 2 € {2}.
(ii) True. Any element in S is an element in S.
(iv) α С {a, 6}
(ν) 5 e ρ
(vi) fep
(vii)
(viii)
(ix)
л/2 е Q
(V^i )2<=z
dFi^ep

Sec. 1.1]
SETS
3
(iii) True, a is the only element of {a} and α € {a, b}. But b e {a, b} and Ь S {a}.
Consequently {a} ¥= {a, b}.
(iv) False. 2 €{2,3} and 2 €{3,4}.
(v) True. Any positive integer is a rational number but not all rational numbers are positive
integers.
(vi) True, All integers are rational numbers.
(vii) True. All rational numbers are real numbers but R ¥= Q.
(viii) True. For if α £ β, then a - a + Ог е C. But V^l £ R implies R ¥* C.
(ix) False. Q = Q.
(x) True. 3, a, b and с are the only elements of {3, a, b, 3, c, b}. Therefore the sets are equal.
1.3. Are the following statements true?
(i) Ζ =■ {x
(ii) N = {x
(iii) Q = {x
(iv) Ρ = {χ
(ν) С = {χ
(vi) Ζ = {χ
χ is real and x > 0}
a; £ Q and ar — 0}
χ — α/Ь, where 6^0 and a,b €. Z}
x^N and ar2 = l}
χ = μ + ir, where »,«£й and гв = —1}
ar is real and a;2 £ P}
Solution:
(i) False, {a; | χ is real and χ > 0} has no negative elements.
(ii) False. |€{a;| ar £ Q and ar — 0} but f € JV. Hence the sets are not equal.
(iii) True. A rational number is defined as the set of all numbers of the form a/b where b ·Φ· 0
and a,b £ Z,
(iv) True. For if χ € Ρ, then χ € N and a;2 = 1. Thus χ € {a; | χ € AT and a;2 ^ 1}. Now if
ar £ JV and a;2 — 1, i.e. a; £ {a; | ar G N and a;2 — 1}, then χ ¥■■ 0. Hence ar £ P, as the only
element in N which is not in Ρ is zero.
(y) True. The property that χ — и + iv where u,v £ β and г2 — —1 is the defining property
for complex numbers.
(vi) False, a;2 G Ρ implies χ ¥■ 0. But OeZ.
b. Union and intersection
Let S and Τ be sets. Then the union of S and T, written SUT and read "S union T", is
defined as the set whose elements are either in S or in Τ (or in both S and T). For example,
{1,2,3} U {2,5,6} ={1,2,3,5,6} and PU{0} = JV. Clearly, SqSUT and TcSUT,
Indeed it follows from the definition of SUT that any set containing both S and Τ contains
SUT, so we say SUT is the smallest set containing S and T.
Similarly if {S, T, U .. .} is any set of sets, we define SU TU £/U ■ · ·, the union of S and
Τ and £7 and ..., to be the set whose elements are the elements that belong to at least one of
the sets S,T,U, ... .
SuTuUU ... is said to be the smallest set containing the sets S, T,U,. ... To illustrate,
{1,2}U{3,4}U{5,6}U··· = P.
If S and Τ are sets, we may consider the common part or intersection of S and T. The
intersection is denoted by SnT and read as "S intersection T", For example, suppose
S= {1,2,3} and T= {2,5,6}. Then SnT ^{2}. Repeating the definition, SnT is the
set of those elements which belong simultaneously to S and to T. Here the possibility arises
that there are no elements of S which belong also to T. We shall agree to the convention
that there is a set, which we denote by 0, with no elements. Again we shall agree to the
convention that the empty set 0 is a subset of every set. Two sets are termed disjoint if
they have an empty intersection. Thus {1,2} and {3,4} are disjoint. This notion of
intersection can be generalized to any number of sets in the same way that the notion of union
was generalized from two to any number. To be precise, the intersection of sets S,T,U, ...,
written SnTnUn ■ ■ ·, is the set of all those elements which belong simultaneously to S,
toT,toU, Notice that S Л Τ can be thought of as the largest subset of S which is also
a subset of T. Similarly SnTnUn- ■ · is the largest subset of S which is contained in
Τ and in U and in ... .

4
SETS, MAPPINGS AND BINARY OPERATIONS
[CHAP. 1
If SqT we define Τ — S, the difference between Τ and S, by
Τ - S = {x j χ G Τ and χ & S]
Thus if Γ = {1,2,3,4} and S = {1,2}, then T-S= {3,4}. For any sets Γ and S such
that TdS,
T-(T-S) = S (1.1)
We prove equation (1.1) by showing that
right side of equation (1.1) С left side of equation (1.1)
and left side of equation (1.1) С right side of equation (1.1)
To do this suppose χ G S. Then clearly χ G Τ but x<£T-S. So χ S Τ - (T - S); in
other words, the right side of equation (1.1) is contained in the left side of equation (1.1).
The reverse inclusion is obtained similarly. Suppose χ e Τ - (Τ — S). Then χ Ε. Τ and
χ & Τ — S. Therefore χ G Τ and χ G S, i.e. χ G S. So the left side is contained in the
right side and we have proved equation (1.1) by virtue of Proposition 1.1.
Problems
1.4. Check that the following statements are correct:
(Ϊ) {1,2} U {1,2,3,4} = {1,2,3,4}
(ii) {a, <?} и {<?, /} и {fit, h) ~ {a, e, f, g, h)
(iii) {...,-2,-1,0} U {0,1,2,...} - Ζ
(iv) If a < b, а, Ь ей, then [а,Ь] = (а,Ь) U {a} U {b}.
(v) {p | Ρ = (ж, J/) e В2, хг + j/2 = 72} U {ρ Ι ρ = (ж, j/) e Л», ж2 + j/2 < 72} = {ρ j ρ = (ж, j/) € В2,
а;2+ у2 ί= 72}. (In other words, the union of the circle of radius 7 and the disc of radius 7
without its boundary, is the disc of radius 7 itself.)
Solution:
(iv) Let χ £ [a,b]. By definition, α — χ — b. If a < χ < b, we have a; €Ξ (a,b). If a; = α or
a; = b, then a; e {a} or a; €Ξ {6} respectively. Therefore [а, Ь] с (a, b) U {a} U {b}. Now for
any a; € (а, Ь) и {a} U {b}, either χ € (α, Ь), а; € {a.}, or !te {b}; and in each case a — χ — b.
Hence (a,b) U {a} U {b} с [a,b]. The equality follows from Proposition 1.1.
(v) If ρ = (χ, у) is any element of the disc, then x2 + j/2 — 72. a;2 + yz < 72 implies ρ €Ξ disc
without its boundary; and хг + у2 = 72 implies ρ €Ξ boundary. Thus the disc С boundary U
disc without its boundary. The reverse inclusion can be checked similarly. Proposition 1.1
then implies the sets are equal.
1.5. Check the following statements:
(i) {1,2} Π {1,2,3,4} = {1,2,3}
(ii) {a, e} η {e, /} Π {g,h} — 0 where a,e,f,g,h are distinct.
(iii) {...,-2,-1,0} Π {0,1,2, ...} = {0}
(iv) If a < b, a, b € R, then [a, b] η {a, b} = {a} U {b}.
Solution:
(i) False. 3€ {1,2,3,4} but 3€{1,2}.
(ii) True. The three sets {a, e}, {«j, /}, {fir, h) have no elements in common.
(iii) True, since 0 is the only element in the intersection and 0 is the only element in {0}.
(iv) True, α and b are the only elements of {a, b] η {a, b} and a, b € {a} U {b}. Furthermore,
a and b are the only elements of {a}u{b}.
1.6. Check that the following statements are correct:
(i) {1,2,3,4} -{1,2} = {3,4}
(ii) {1,2,3,4} -{1,2,3,4} = 0
(iii) If a,b € R and a < b, then [a, b] — (a, b) = {a, b}.

1]
SETS
5
(νΐΐΐ)
(ix)
W
(xi)
(xii)
SnS = S
SU(TuV) - SUTUU
Sn(TnU) = (SnT)nU
Su(TnU) = (SuT)n(SuU)
S-S=0
If S, Τ and U are any three sets, prove the following-:
(i) SuT=TuS
(ii) SnT=TnS
(iii) ScSuT and ScTuS
(iv) SnTcS and TnS CS
(v) SU0=S
(vi) Sn 0 = 0 (xiii) S-{S~S) = S
(vii) SuS-S
Solution:
(i) Let a;GSur. By the definition of union of sets, ж € S or к € Г. Hence a; £ Г US and
SuTqTuS. Similarly if xSTuS, it follows that TuScSuT. Consequently SuT =
TuS by Proposition 1.1.
(ii) If χ £ Sn T, then, by the definition of intersection, χ £ S and χ €. T. a; is therefore an
element of TnS and SnT С TnS. The reverse inclusion, TnS Q SnT, is established in the
same manner. The equality follows from Proposition 1.1.
(iii) By the definition of union, SnT contains all elements of S and of Г. So χ €. S implies
χ e SuT and S Q SuT. Using (i) above, we also have S Q TuS.
(iv) x£.SnT implies χ e S and χ € Т. In particular, χ € S. Thus SnTcS. Part (ii) allows
us to write TnS С S.
(v) SCSU0 by (iii). If a;eSU0, then i£S, for a; € 0 by definition. Therefore SU0CS.
Hence by Proposition 1.1, SU0 = S.
(vi) By (iv), Sn0 С 0. But we know that 0 is a subset of any set and, in particular, 0 С Sn$.
Hence Sn0 = 0.
(vii) ScSuS, using (iii). Now, x€.SuS implies χ € S. Hence SuScS and the equality
follows.
(viii) By (iv), SnS Q S. But ж € S implies a;£SnS. Thus ScSnS and the equality follows.
(ix) Let χ e SU(Tu U). Then ж £ S or areTuu. Thus χ € S or a; € Γ or iGl/.
Consequently ΐεδυΓυϋ, Hence SU(TuU) С SuTuU. If areSuTut/, then χ £ S or a; € Γ
or are 17. If a;eS, ж <= SU(Tu 17). If ж <= Г or 17, a; <= Su (Ги 17). Hence SuiUtJc
SU(TuU). The result follows.
(χ) χ €. Sn(TnU) implies i£S and a; £ (Γη 17), which in turn implies a; € Γ and a; £ 17.
From χ e S and a; € Γ it follows that a; € (SnT) and, as x&U, χ S (SnT)n 17.
Therefore Sn (Γη t/) с (SnT) η U. Similarly (Sn Τ) η ϋ с Sn (Τη U).
(xi) SU(TnU) С (SuT)n(SuU). For, if a; <= SU(Tn t/), then к € S or а;<=ГпС/. Now, χ <= S
implies aieSuT, a;eSUl7, and consequently к € (SU Г)П (Su 17). If x€.TnU, then a; € Г
and iet/; hence x€.SuT, X£.SUU and, as before, χ € (SU T)n(SuU). (SuT)n(SuU) с
Sn(TuU) is established in a similar manner.
(xii) If S — S contains an element x, then i£S and χ g S which is impossible. Hence S — S
must be empty.
(xiii) S-S= 0 from (xii). Thus S-(S-S)~S-0 and clearly S - 0 = S.
Prove the following statements:
(i) If S С Τ and U is any set, then SuU С TuU.
(ii) If S С Τ and С/ is any set, then Sn U с Γη С/.
(iii) If S С Г and Г С С/, then S С U.
(iv) S С Г if and only if SnT = S.
(v) TcS if and only if S = TuS.
(vi) If TcS, then (S ~ Г) U Г = S.
Solution:
(i) Let iceSuC/. Then a; € S or net/, If a; € S then a; € Г since S С Г, and
consequently xGTuU. If геи, then x^TuU. Thus SUUcTuU.

6
SETS, MAPPINGS AND BINARY OPERATIONS
[CHAP. 1
(ii) xGSnU implies χ €. S and χ £ U. As ScT, we also have χ € Т. Therefore x£.Tc\U
and Snt/ С Tr\U.
(iii) ScT means that if χ € S, then г£Г; and since Τ q U, this in turn implies χ Ε. U.
Hence Set/.
(iv) First, assume SnT = S. The equality implies any element ж in S belongs to Sr\T. But
xGSnT implies χ € Т. Hence a; € Г and S С Г. Secondly, let ScT. If χ e S, then
ϊ£Γ and so zeSnr. Therefore ScSnT. The reverse inclusion SdTqS is true by
Problem 1.7(iv). By Proposition 1.1, S - SnT.
(v) Assume S-TuS. If геГ, then «eTuS and, since S=TuS, χ e S. Hence Г С S.
On the other hand if we assume Τ qS, then ж € TuS implies that «eS. Consequently
TUS CjS. By Problem 1.7(iii) we have S С TuS. Thus TuS = S.
(vi) It follows from the definition that (S- T) с S. Using (i) above, (S- T)UT Q SUT. But by
(v), TcS implies S-TuS, and-we have (S-T)UT cS. To show S c(S~T)uT, let
χ e S. Either xGT or ϊίΓ. If «GT, then χ £ (S - T) U Γ. If ΐίΓ, then ж £ S - Τ
and we also have χ £ (S - Γ) U T. Thus S c(S-T)UT. The equality follows by Proposition 1.1.
1.2 CARTESIAN PRODUCTS
a. Definition
The plane Й2 (see Section 1.1a) consists of all pairs (x, у) of real numbers χ and y. We
shall also denote R2 by i? x i?; thus i? x R is defined as the set of all ordered pairs (x, y) with
x ε R and у &R.
There is a natural extension of this notation to any two sets, S and T:
SxT = {p\p=(8,t),se S, t G T)
For example,
{1,2} χ {1,2,3} - {(1,1), (1,2), (1,3), (2,1), (2, 2), (2,3)}
In words, SxT is defined to be the set of all ordered pairs of elements (s, t), the first
member of each pair always belonging to S and the second member always belonging to T. We
term SxT the cartesian product of S and T. It is worth pointing out that, just as in R2,
two elements of SxT are equal iff (if and only if) they are identical, i.e. (s, t) = (s', t/) if and
only if s = s' and t = t/. If either S = 0 or Τ = 0, we interpret S x Τ as 0.
One defines similarly the cartesian product Si x S2 x · ■ · X S„ of the та sets Si, S2, .. .,Sn
(n < ю) as the set of all it-tuples ($i, s2,..., s*) with si G Si, s2 G As with
the cartesian product of two sets, (si, s2, ..., sn) = (si, si, ..., sn) iff Si = si, S2 = si, .. ., sn = si.
For example,
{1,2} x {2,3} x {4,5} = {(1,2,4), (1,2,5), (1,3,4), (1,3, 5),
(2,2,4), (2,2,5), (2,3,4), (2,3,5)}
If any one of the sets S,, S2, ..., S„ = 0, then we shall interpret Si x S2 x ■ ■ · x Sn to be 0.
If each St = S, then Si x S2 x ■ · ■ x Sn is denoted by S".
We often are interested in certain subsets of SxT. For example, in elementary analytic
geometry one investigates lines, circles (see Section 1.1a), ellipses and other figures in the
plane; these are subsets of R2.
Problems
1.9. Let S= {1,2,3}, Τ = {1,5}. Verify the following statements:
(i) S Χ Τ = {(1,1), (1, 5), (2,1), (2, 5), (3,1), (3, 5)}
(ii) TXS = {(1,1), (5,1), (1, 2), (5, 2), (1, 3), (5, 3)}
(iii) SXT?6 TXS

Sec. 1.2]
CARTESIAN PRODUCTS
7
(iv) S2 Φ Γ2 (S^ = SXS and T2 = TXT)
(v) (SXT)XS^SX(TXS)
Solution:
(i) Clearly these are the only ordered pairs (ж, y) with ж £ S and у €■ T.
(ii) As in (i), these are the only ordered pairs (a;, y) with χ £ Γ and у €■ S.
(iii) Looking at (i) and (ii), we find (5,1) e Г X S and (5, l)gSxT. Hence S Χ Γ Φ Τ Χ S.
(iv) (3,3) e S2 but (3, 3) € T2.
(v) An element (ж, у) of (S X Г) X S has ж £ S Χ Γ and у <= S. Thus ((1,1),5) <= (S Χ Γ) Χ S. But
((1,1), 5) eSX(TXS), since (1,1) € S.
1.10. Let S, Г and U be any three sets. Prove the following:
(i) S Χ Τ = Τ X S iff either S = Τ or at least one of the two is empty.
(ii) If (x, y) e S2, then (y, x) <= S2.
(iii) SXT = SXU iff either Γ = 17 or S = 0.
(iv) (SXT)XU = SX(TXU) iff at least one of the sets S, T, U is empty.
Solution:
(i) S = Τ implies S Χ Τ = Τ2 = Τ X S; and S = 0 or Τ = 0 implies, by the definition of the
cartesian product of any set and the empty set, SXT = 0 = rxS. Therefore S Χ Τ = Τ Χ S
whenever S = T, S = 0, or Τ = 0. To prove the converse, assume S Χ Τ = Τ XS. We may
also assume S Φ 0 and Τ Φ 0. Let ί£Τ and β £ S (such elements exist since S Φ 0
and Τ Φ 0). Then (β, ί) € S Χ Τ and, as SXT = TXS, (s,t)eTXS. It follows, from the
definition of Τ X S, that t e S and s £ Γ. Therefore Г С S and ScT, We conclude, using
Proposition 1.1, S = T.
(ii) (x, y) € S2 means χ £ S and у € S. Hence (?/, ж) £ S2.
(iii) Clearly if Τ ~U or S = 0, we have SXT ~ SXU. Conversely, let SXT - SXU and
S Φ 0 (if S = 0 we have nothing to prove). If Τ Φ 0, let t €. T. Then (β, ί) € S Χ Τ for
any «eS; and, as S Χ Τ = S X £/, (<,i)eSXt/. But (s, i) £ S X £/ means ί € С/. Hence
Τ о U. A similar argument gives U oT, and we conclude Τ ~U. If Γ = 0, then
SXT = 0 and SXE/ = 0. 17 = 0 follows from SX£/ = 0; for if U Φ 0, then SxU
would not be empty, by virtue of our assumption S Φ 0. Thus both Τ Φ 0 and Τ = 0 give
Γ = U.
(iv) If either S,T orU is empty, (S Χ Γ) X U = 0 - S X (T X U). Conversely, assume (S Χ Τ) Χ U =
SX(TXU). If S Φ 0, Τ Φ 0 and U Φ 0, there is at least one element (ж, у) in (S Χ Τ) Χ U,
χ € S Χ Τ and у €.U. But (ж,?/) must also be an element of S X (T X £/). Hence ж G S. This
is a contradiction, for ж cannot be both an element of S and an element of S X T. Therefore
the assumption that S Φ 0, Τ Φ 0 and U Φ 0 must be false and so either S, Τ or U is empty.
1.11. (i) If A = {p | ρ = (ж, J/) £ Ρ2 and ж < ?/} (recall that Ρ is the set of positive integers), prove
that:
(a) (ж, ж) ЙА for every ж £ P.
(b) if (ж,?у)£А and if (y,z)€.A, then (ж,г)€Л.
(ii) Let В = {ρ Ι ρ = (χ, у) 6Р2,*- ?/}.
(α) Show that (ж, ж) € β for every ж £ P.
(b) Prove that if (ж, у) е В and if (?/, г) € β, then (ж, г) £ β.
(c) If (ж, у) £ В, when is (у, ж) € β?
(iii) Let Λ = {ρ \ ρ — (χ, у) & Ζ2 with χ — у divisible by 3}. Prove'.
(a) (ж, ж) € Л for all χ GZ.
(b) If (#, i/) € A, then (?/, ж) € Л.
(c) If (ж,г/)€Л and (у, г) £Л, then (ж,г)£Л.
(iv) Let С/ be the points of the plane R2 on or above
the X axis and let L be the points below the X axis.
Put V=U2UL2. Notice that VoR2XR2. Prove:
(a) If (л, a) € Д2 X -R2, then (a, a) £ У.
(b) If (a,/?) e V, then (/3,a) € У.
(c) If (й,0)€У and 08,7)еУ, then (а, у) € У.
(d) y^#2X.R2.
Ж

8
SETS, MAPPINGS AND BINARY OPERATIONS
[CHAP. 1
Solution:
(i) (a) (x, x) Й A for every χ £ Ρ, since χ is not less than x.
(i>) (x, y) GA implies χ < у, and {у, ζ) e A implies у < г. Now χ < у and у < г imply
а; < ζ, since x, у, ζ £ P. By definition of set A, we have (x, z) £ A,
(ii) (a) a; ^ m for all ar £ P. Hence (x, x) £ B.
(6) (ar, y) e В and (y, z) £ В imply χ — у " and J/ — ζ respectively. It follows that χ — ζ
and, by definition of set B, (x, z) £ B.
(c) (V> x) £B iff x ~ y. For if (a;, y) GB, χ — у; and if (j/, a;) £ В, у ^- χ. But a; ^ у
and у — χ iff a; == y.
(iii) (a) For any ϊ£2, a; — ж = 0, and zero is divisible by 3. Consequently (x, x) £ A.
(6) («, 1/) € A means χ — у is divisible by 3, i.e. χ — у == 3qf where qf is some integer. Now
y — x== —(x — y) = —3qr = 3(—qr). Therefore y — χ is divisible by 3 and (y, x) € A.
(c) (a;, j/) e A means x — y — Sq for some integer q, and (y, z) GA means у — ζ = Sr for
some integer r. Thus
ж - г = (х- у) + {у - ζ) = 3q + 3r - 3(q + r)
and so a; — ζ is divisible by 3 and (x, z) G A.
(iv) (a) (a, a) £ Д2 X Д2 implies a £ R2, which means a £ U or «eL. Hence (α, α) £ Ϊ72 or
(«(«)€Д and (a, a) £ Е/аиЬ*= F.
(6) If (a, /3) e V, then (a, «eiP or (α, β) £ L2 and by Problem 1.10(H), (0, a) £ U* or
(β, a) £ L2. Hence (β, a) £ V.
(c) (a, /8) £ V implies either a,/} £[7 or a,j3eL, but not both, since 1/nL = 0. Now
(iJjlEV implies β,у £ 17 or L. If /},γ£17, then α,β£ϋ since /} cannot be an
element of both C7 and L, and hence (α, γ) £ U2. Similarly if β, у £ L, we have (α, γ) £ £A
Therefore (α, γ) £ t/2uL2 = V.
(<i) Let a £ Ϊ7 and /8 £ L. Then (a, |8)ei?X Д2, but (a, /8) Й V.
b. Equivalence relations
Similarity of triangles is an example of an equivalence relation. This means that if
s, t and и are any triangles, the following three conditions hold:
(i) s is similar to s.
(ii) If s is similar to t, then t is similar to s.
(iii) If s is similar to t, and ί is similar to u, then s is similar to u.
Another example of an equivalence relation is congruence of triangles since (i), (ii) and
(iii) above hold also if "similar" is replaced by "congruent". Continuing in this vein, if X
is any non-empty set and R is a "relation on X", i.e. if for any pair of elements x,y € X
either χ is related to у by R (written xRy and read "x is related to у by R") or not, then
R is termed an equivalence relation in X if:
(i) xRx for all χ G X.
(ii) If xRy, then yRx.
(iii) If xRy and yRz, then жйг.
One objection to this definition of equivalence relation is that "relation on X" is vaguely
defined. We shall therefore define the idea of equivalence relation by means of sets and
subsets.
Let us consider again the notion of similarity of triangles. Let Τ be the set of all
triangles. Let S be the subset of Τ χ Τ defined by (e, t)GS iff s is similar to t. If t is a
triangle, t is similar to itself, so (t,t) e S. If (s,t) and (t,u) G S, then s and t are similar
and t and и are similar. Thus s and и are similar. Hence (s, u) £ S. Similarly if (s, t) G S,
it is clear that (i, s) G S.
Consequently, discarding our informal approach, we have the following

Sec. 1.2]
CARTESIAN PRODUCTS
9
Definition: Let X be a non-empty set and let R be a subset of X2. Then R is called an
equivalence (or an equivalence relation) in X if the following conditions are
satisfied.
(i) (ж, χ) G R for all ж G X (reflexive property),
(ii) If (x,y)G.R, then (y,x)GR (symmetric property),
(iii) If (ж, у) £R and (#, z) e i?, then (ж, г) е i? (transitive property).
Problem
1.12. Examine Problem 1.11 for equivalence relations.
Solution:
(i) A is not an equivalence relation in P, as (χ, χ) &.Α for all x.
(ii) В is not an equivalence relation in P, as we know (x, y) € β and (j/, x) €. В occurs only if
χ = y. Thus though (1,2) <= β, (2,1) g β.
(iii) A is an equivalence relation in Z2 as А С 2й and the reflexive, symmetric and transitive
properties hold.
(iv) V QR2XR2 and the reflexive, symmetric and transitive properties hold, so V is an equivalence
in R*.
c. Partitions and equivalence relations
Suppose R is an equivalence relation in X. We say ж is unrelated to y, or ж is related to
у by i?, if (ж, г/) ε i?. If (ж, κ)εβ we shall sometimes write жДг/. То illustrate let
X = {1,2,3,4} and let
R = {(1,1), (2,2), (3,8), (4,4), (1,3), (3,1), (3,4), (1,4), (4,3), (4,1)} (1.2)
Then it is easy to check that R satisfies the three necessary conditions for it to be an
equivalence relation. Now 3i?4 since (3,4) G R, but 2Й4 is an incorrect assertion since
(2,4) gi.
Note that we have used a notation that fits in with the notation of Section 1.2b where
we informally introduced an equivalence relation.
An equivalence relation in X is intimately connected with a partition of X, i.e. a
decomposition of X into disjoint subsets of X such that every element of X belongs to some subset.
Examples of partitions of {1,2,3,4,5} are
{1}, {2,3}, {4,5}
and {1,3,4}, {2}, {5}
On the other hand {1,2}, {2}, {3,4,5} is not a partition of {1,2,3,4,5}.
If R is the equivalence relation (1.2) in {1,2,3,4}, then all the elements of {1,3,4} are
i?-related to 1, i.e.,
LR1, li?3, li?4
This suggests a means of getting a partition of a set X. In order to explain, we need some
additional notation. Let R be an equivalence relation in a set X. If χ € X, we define
xR = {y\ yGX and (ж,у) ei?}. xR is thus a certain subset of X. This subset xR is
called the R-elass of x, or the R-equivalence class of x, or the R-block of x. A subset of X
will be called an R-class or R-Ыоск if it is the й-class or i?-bk>ck of some element ж G X.
To illustrate these terms, consider the equivalence relation R given by (1.2). Here
1R = {1,3,4}, 2R = {2}, and SR = № = \R
Thus the i?-c!asses here are simply {1,3,4} and {2}. Notice that these i?-classes constitute
a partition of {1, 2,3,4}.
More generally, we have the following

10
SETS, MAPPINGS AND BINARY OPERATIONS
[CHAP. 1
Theorem 12: Let X be a non-empty set and let R be an equivalence relation in X. Then
(i) if xRnx'R Φ φ, then xR = x'R,
(ii) χ G xR for every χ GX.
Thus the й-classes constitute a partition of X, for (i) guarantees that
distinct й-classes are disjoint, while (ii) shows that every element of X
appears in at least one of the β-classes.
Proof: First, we verify (i). Suppose xRnx'R¥> 0. Then there is an element у G xR
which lies also in x'R, i.e. {x, y) G R and {x', y) G R. As β is an equivalence, it follows from
the symmetric property that (y, x') G R. But (x, y) G R and (y, x') G R imply, by the
transitive property, (x, x') G R. Now if ζ G x'R, then (x', z) G R; and hence by the
transitive property applied to (x, x') and (a/, z), we find (x, z) G R. This means, by the very
definition of xR, that ζ G xR. Since 2 was any element of x'R, we have proved x'R С ж.К.
The reverse inequality follows by a similar argument. Hence x'R = xR as required.
The verification of (ii) is trivial since (x, x) G R means χ G xR. This completes the
proof of Theorem 1.2.
Problems
1.13. (i) Prove that Ε = {(0, 0), (1,1), (2, 2), (3, 3), (0, 2), (1, 3), (2, 0), (3,1)} is an equivalence relation in
S = {0,1,2,3}.
(ii) Find the ^-equivalence blocks (a) 0E, (b) IE, (c) 2E, (d) BE.
Solution:
(i) The reflexive property holds, i.e. (x, x) € Ε for all i£S, since (0,0), (1,1), (2, 2), (3,3) € E.
To show that Ε is symmetric, let us examine all pairs (x, y) where χ Φ y. There are only four,
namely (0, 2), (1,3), (2,0), (3,1). Clearly if (x, у) is any one of the four, so is (y, x). When
x = y, (x, y) = (y, x). Thus (x,y) e Ε implies (y,x)€.E. Ε is also transitive. Let (x,y)€.E
and (y,z) € E. Suppose χ Фу. Then {x, у) can be (0,2), (1,3), (2,0) or (3,1). If {x,y) = (0,2),
then (y, z) = (2, 0) or (2, 2) and (x, z) = (0, 0) or (0, 2) respectively. Hence (x, z) € E.
Similarly if (x, y) = (1,3), (2,0) or (3,1), it can be shown that (x, z) € E. When χ = у, (y,z)£.E
means (x,z) &E. Therefore for any (x,y)€.E and (y,z) €.E, we have (x,z) €. Ε andJE
is transitive.
(ii) (a) 0E = {0, 2}, (b) IE = {1, 3}, (c) 2E = {2, 0}, (d) BE = {3,1}. Observe that 0E = ZE and
IE = BE.
1.14. Let A = {p \ ρ = {χ, у) € Ζ2 with χ — у divisible by 3}. Prove that A is an equivalence relation
in Ζ and find the i?-equivalence classes.
Solution:
It was shown in Problem l.ll(iii), page 7, that A satisfies the three conditions of an
equivalence relation. The i?-equivalance classes are;
(1) 0A = {3g I q £ Z}; for if (0, x) £ A, 0-x = -x is divisible by 3. Also, (0,3g) € A.
(2) 1A = {l-3g| ?eZ}; for if (1,ж)€Л, 1 - χ = Bq and hence x~l-Bq. If χ = 1 - 3g,
1 — χ is divisible by 3; hence (1, x) & 1.4.
(3) 2A = {2-3g| g £Z}; for if (2,x)£A, 2 - χ = 3g and so χ = 2 - 3g. If a; = 2~3g, 2-я;
is divisible by 3; hence (2, x) € A.
0A,1A,2A are the only i?-blocks, for any integer can be written as 3g, 1 — 3g, or 2 — 3g.
Consequently OAulAUZA = Z.
1.15. Let Z* be the set of nonzero integers and let S = Zx Z*. (Recall that Ζ is the set of all integers.)
Let Ε = {ρ Ι ρ = ((r,s), (t,u)) € S2 with ru = st}. Prove that Ε is an equivalence relation in S.

Sec. 1.3]
MAPPINGS
11
Solution:
Ε is reflexive, for (r,s)€.S implies ((r„s), (r,s)) £ S2; and since rs == sr, ((r,s), (r,s)) € E.
The symmetric property of Ε is established by noticing ((r,s), (i, w)) € Ε means (r,«) and (i, w) £ S2,
and no = st. But no = st can be rewritten as ts ~ wr. Hence ((i, w), {r,s)) € E, To show Ε is
transitive, let ((r,«), (i, w)) £ J? and ((i, u), (v, w)) £ E. Then ru ~ st and iw = vw. Since и ^ 0,
j?£ situ s ff
r~— and rw = = ~tw — ~vu = sv. Thus no = sv and so ((r,s),(v,w))€.E. Hence £/ is
transitive. Therefore E is an equivalence relation.
1.16. Prove that S XS is an equivalence relation in S.
Solution:
SXS is reflexive since (i,i)€SxS for all χ €.S. If (ж,J/) € S X S, then a; and j/ € S
and, by definition of SXS, (y, x) £ SXS. Hence SXS is symmetric. Now (ж, у) £ S and
(j/, «)eS imply ж, j/ and г £ S. But then (ж, ζ) €. S X S and S X S is transitive. Thus S X S is
an equivalence relation on S.
1.17. Prove that a set X is infinite if and only if there are infinitely many equivalences in X. (Hard.)
Solution:
Assume there are an infinite number of equivalences in X. If X is finite, then there are at most
a finite number of distinct subsets of X2. Therefore when X is finite there are at most a finite
number of equivalences in X, which contradicts our hypothesis. Hence X must be infinite.
Conversely, assume X is infinite. We exhibit an infinite number of equivalences in X as follows: For
each pair a,b £ й, аф Ь, we define
Я(а,ь) — ίρ\ Ρ = (X'V) e χ2> where either χ = У, (х, у) = (а, Ь) or {x, у) = (Ь, а)}
Now R(a,b) == R(c,d) if and only if {a, b} = {c, d}. Therefore since X is infinite, we can find an
infinite number of different pairs a,b £ X each of which gives a distinct set R^a,b)· Furthermore,
each -β(α,[>) *B an equivalence. To prove that i?fuj f)) satisfies the three conditions of an equivalence
relation, we first notice (x, x) £й(а,Ь) for all χ £ X, by the very definition of R(a,i,y Secondly,
R(a,b·) is symmetric since (x, y) £ R(a,t>) means (x, y) = (a, b) or (b, a), in which case (y, x) =
(b,a) or (a,b) respectively, or χ = у and then (x, y) = (j/, ж). Thirdly, if (x,y) and (j/, z) £-К(а,Ь)>
then (x, z) ei?(aj[)). To see this, notice that (x, y) can only be (a,b), (b,a) or (x,x). (x,y) ~ (a,b)
implies (y, z) = (b, a) or (b, 6); hence (ж, г) == (α, α) or (а, Ь), which are both elements of R{a,by
Similarly, {x, y) = (b,a) implies (x,z) £ Ria,by Finally, {x, y) = (x,x) means {x,z) =
(y,z) ei2(ajb).
d. The division notation
We find it useful to introduce a notation for the i?-classes of an equivalence relation R
ш a set X, namely X/R.
Problems
1.18. What is S/E in Problem 1.13?
Solution: S/E = {OE, IE},
1.19. What is Z/A in Problem 1.14?
Solution: Z/A = {0A, ΙΑ, 2Α}.
L3 MAPPINGS
a. Definition of mapping
Assign to each even integer the value 1, and to each odd integer the value -1. Let us
give the name a to this assignment; thus a assigns to each even element in Z, the set of all
integers, the unique element +1 in the set {1,-1} and to each odd element in Ζ the unique
element -1 in {1,-1}. In less detailed terms a assigns to each element in Ζ a. unique
element in {1, -1}. Such an assignment is termed α mapping from Ζ into {1, -1} or a map

12
SETS, MAPPINGS AND BINARY OPERATIONS
[CHAP. 1
from Ζ into {1, —1}. More generally, if S and Τ are any two non-empty seta, a mapping or
a map from S into Τ is an assignment of a unique element of Τ to each element of S. For
the most part we shall denote mappings by lower case Greek letters such as α, β, γ. If « is a
mapping from S into T, we shall express this fact more briefly by writing a: S -*■ T; this is
read "« is a mapping from S into T". We call S the domain and Τ the codomain of a.
We find it useful to provide further notation and definitions. Suppose that <*: S-* T.
If a assigns to s in S the element t in T, we write a: s -*■ t and read this as "« sends s into
t". We call t the image of s (under «) if «:s->t. It is convenient to have a number of
notations for the image of an element s in S under a mapping a: S -* T; thus we shall write
sa or s", or even «(s) for the image of s under a. For the most part we use the first notation.
If t S Τ and sa = t, we call s a preimage of t. By S« we mean {sa\ s G. S). We call S«
the range of «.
Problems
1.20. Suppose α : Ρ -* Ρ is defined by
(i) а: п->тр for all η € Ρ (iv) α : η -»1 for all η € Ρ
(ii) α : η -* η + 1 f or all и G Ρ (ν) α : 1 -» 2, η -» 1 f or all η G Ρ, η > 1
(iii) α : η -> 2« for all η G Ρ
Note: In (i)-(v) above, the mapping a is defined by describing its "action" on every element of
P. For example, in (i), la = l2 = 1, 2a = 4, 3a == 9, .... Note that each element of Ρ has a unique
element assigned to it.
In each case determine: (a) 2a, 5a, 6a; (6) a preimage (under a) of 2, 5, 6, 27. (c) Is every element
of Ρ an image of some element of Ρ in (i)-(v)? How many preimages (under a) does 2 have in (v)?
How many preimages does 1 have in (iv)? in (v)?
Solution:
(i) (a) 2a = 4; 5a = 25; 6a = 36.
(b) There is no preimage of 2, 5, 6 or 27.
(c) Not every element of Ρ is an image, e.g. 2 is not an image.
(ii) (o.) 2a = 3; 5a = 6; 6a = 7.
(b) la = 2; 4a = 5; 5a = 6; 26a = 27. Hence 1, 4, 5, 26 are the required preimages.
(c) The only element of Ρ which has no preimage is 1.
(iii) (a) 2a ~ 4; 5a = 10; 6a = 12.
(b) 1 is a preimage of 2; 5 has no preimage; 3 is the preimage of 6; 27 has no preimage.
(c) 1 has no preimage, so not every element of Ρ is an image.
(iv) (a) 2a = 1; 5a = 1; 6a = 1.
(b) 2, 5, 6 and 27 have no preimages.
(c) 1 has every element of Ρ as a preimage and no other element of Ρ has a preimage.
(v) (a) 2a = 1; 5a = 1; 6a = 1.
(b) 1 is a preimage of 2; 5, 6 and 7 have no preimages.
(c) 2 has one preimage, namely 1. 1 has an infinite number of preimages. In fact any
element of Ρ not equal to 1 is mapped onto 1. 1 and 2 are the only elements of Ρ which have
preimages.
1.21. Let S = {1, 2, 3}, Τ = {1, 4, 5}.
(i) Write down a mapping of S into T,
(ii) Let α : S -» Τ be defined by la = 4, 2a = 5, 3a = 4. Is every element of Τ an image of some
element in S under a? Is every element of Τ the image of more than one element in SI
Give preimages of 1 and 4.
Solution:
(i) For example, α : S -> Τ defined by la = 1, 2a = 4, 3a = 5.
(ii) Not every element of Τ is an image, for 1 has no preimage. Only 4 is an image of more than
one element in S. 1 has no preimage, and 4 has preimages 1 and 3.

Sec. 1.3]
MAPPINGS
13
1.22. Suppose α: Ρ -» С is defined by
(i) a : χ -» ж2 (ν) α : я; -»»ж + 1
(ii) a; χ -»2a; + 27 ... is +1
(νι) β: ж-»-——-
(iii) a: χ -» ζ where ζ € С is such that z2 = ж. гж - ι
(iv) α : ж -» ζ where ζ G С is such that z3 = ж -1. (vii) a '· x ~* log™ x
(a) Do all of these descriptions really define mappings of Ρ into C?
(b) Is every element in С a preimage of some element of Ρ in (i), (ii), (v), (vi), (vii)?
Solution:
(a) (i) and (ii) define mappings since every ж G Ρ has a unique image, (iii) does not define a
mapping; for example, either 2 or —2 could be taken as 4a. Similarly (iv) is not a mapping, since
there are three complex cube roots of ж — 1, so that each ж has three different images, (v), (vi)
and (vii) define mappings.
(b) In (i), (ii), (v) and (vii), г has no preimage: for (i) i = жг implies ж = VT? Ρ', (ϋ) 2ж + 27 = г
г — 27
implies ж = —г— £ Ρ; (ν) гж +1 = г gives « = l+i?P; (vii) if log10 ж = ί, then 10{ = ж
and thus «SF. In (vi) 1 has no preimage because -——r = 1 implies 1 = —1.
1.23. What is Pa in each of the cases in Problem 1.20?
Solution:
(i) Pa is the set of squares {1,4,9,16, ...}. (iv) Pa = {1}
(ii) Pa = {2,3,4,6, ...} (ν) Ρα = {1,2}
(iii) Pa = {2,4, 6,8, ...}, i.e. all the even integers.
b. Formal definition of mapping
The reader may ask whether our definition of mapping is precise. After all, it depends
upon an English word, assignment, a word that is used in many different ways.
A comparison with Section 1.2b is valuable. In Section 1.2b we introduced the concept
of equivalence relation in X, but as we felt uneasy about it, we redefined it in terms of a
subset of X2. Here too we feel uneasy about our definition of mapping and so we shall
redefine it in terms of sets.
A subset α of S Χ Τ is called a mapping of S into Τ if (s, £i) and (s, ia) £ a occurs only if
fi = fa, and for each s e S there exists an element (s, t) G <*. S is called the domain and Τ
the codomain of a. If a is a mapping of S into Τ (written briefly as a: S -* T) and (s, t) G a,
we call ί the image of s under α and write a: s -*■ t. We also write t = Sa.
It is easy to see the relationship between the old definition and the new. In the old
definition the elements of S were assigned unique elements of T.
Consider the subset of S χ Τ consisting of the pairs (s, t) where t is assigned to s. The
two conditions of the new definition are satisfied by this subset.
In the sequel we will use the definition of Section 1.3a, being confident that if necessary
we could justify our arguments using the definition of a mapping in terms of a subset.
c. Types of mappings
We have talked of mappings without defining what is meant by the equality of two
mappings. We will now remedy this. Suppose a: S-*T and β: S'-* 2". Then we define
« = β if and only if S = S', Τ = Τ', and, for every element s G S, s<* = sa'. In other words,
two mappings are equal if and only if they have the same domain, the same codomain, and
the same "action" on each element of S. For example, let S= {1,2,3,4}, Τ = {4,5,6}.
Let a:S^T be defined by la = 4, 2a = 5, 3a = 6, 4<* = 4; let 0:S-*T be denned by
Ιβ = 4, 2/3 = 5, 3/3 = 6, 4/3 = 5. Then α Φ β since 4α Φ 4/3.

14
SETS, MAPPINGS AND BINARY OPERATIONS
[CHAP. 1
It is important to distinguish certain types of mappings. Thus suppose <*: S -»■ T. Then
we say α is a mapping from S onto Τ (notice that into has now been replaced by onto) if every
element in Τ has at least one preimage in S, i.e. if for every t G Τ there is at least one
element s G S for which sa = t; in this case we call a an onto mapping.
On the other hand we say a is one-to-one if sa = s'a implies s — s', i.e. distinct elements
of S have distinct images in Τ (under a). Finally, we say α is a matching of S and Τ or that
a matches S with Γ or α is a bisection if a is both onto and one-to-one. Two sets are termed
equipotent or of the same cardinality if there exists a matching of the one with the other.
If S is finite and a matches T, then we say S and Τ have the same number of elements. We
denote the number of elements in a set S by \S\. If S is infinite this definition no longer
makes sense unless one takes from all the sets which match S a single fixed set which we
then term \S\, the cardinality of St. A set which matches Ρ is called enumerable, countable,
or countably infinite. Important results are that the set of rational numbers is enumerable
and the set of real numbers is nott.
We require one further definition. Suppose α: S -* Γ and suppose S' С S. Then we
define a mapping from-5' into Τ by simply restricting the domain of a to S'; this mapping is
denoted by «|S, (read a restricted to S') and is called the restriction of a to S'. To be quite
explicit,
ajS,: S' -* Τ is defined by a[S. : s -*■ sa for all s G S'
Problems
1.24. Which of the mappings denned in Problems 1.20-1.22 are (a) onto, (b) one-to-one, (c) matchings?
Solution:
(a) None of the mappings defined in Problems 1.20 and 1.22 is onto by the solution already given to
these problems. The mapping defined in Problem 1.21(ii) is hot onto, since 1 has no preimage.
(b) Problems 1.20(i), (ii), and (iii) define one-to-one mappings: for in (i), if na == ma, then η2 = τη2
and so η = m, since there is only one positive square root of an element in P; in (ii), na = ma
gives n+ 1 = m+ 1 or η = m; and in (iii), na = m,a implies 2w = 2m or η = τη. Clearly
Problems 1.20(iv) and (v) do not define one-to-one mappings. The mapping a in Problem 1.21(ii)
is not one-to-one since la = 3a. All the mappings defined in Problem 1.22 are one-to-one: for in
(i), xa = x'a means x2 = x'2 and, since x, x' £ Ρ, χ = χ'; in (ii), xa = x'a implies 2.x + 27 =
2a;' + 27 or χ = χ'; in (ν), χα = x'a implies ix + 1 = ix' + 1 or χ = χ'; (vi) жа = x'a means
-——г = . , 1 ■ or 2ix = 2ix' or χ = χ'; in (vii), xa = x'a gives log10 a; = log10 ж'= j/ and
hence 10a — χ — x'. (Note that (iii) and (iv) are not mappings.)
(c) None of the mappings defined in Problems 1.20-1.22 is a matching, since none of them is onto.
1.25. Which mappings in Problems 1.20-1.22 are equal?
Solutiou:
None. Problem 1.20 features α : Ρ -» Ρ; Problem 1.21, α : S -» Γ; and Problem 1.22, α : Ρ -» С.
Hence we need only compare the mappings in each exercise.
1.26. Let a-.P^Z be defined by na = — η for all n&P. Is α onto? One-to-one? A matching?
Solution:
a is neither onto nor a matching, since 1 has no preimage. α is one-to-one, for if na == n'a,
then —n = —n' and hence η = η'.
tFor more details see, for example, G. Birkhoff and S. MacLane, A Survey of Modern Algebra, Macmillan,
1953.

Sec. 1.3]
MAPPINGS
15
1.27. Suppose m is a fixed positive integer. Every integer η can be written uniquely in the form
η == qm + r where the remainder r is an element of the set {0,1, . ..,m—1}. For example, if
m = 4, then every integer can be expressed in precisely one of the following forms:
4fc, 4fc+1, 4fc+2, 4fc +3. Let a: Ζ -» {0,1, .. .,m — 1} be defined by na = r, if n-qm+r where
re{0,l,.,„ffl-l}. Prove:
(а) a is onto {0, 1, ..., m — 1}.
(б) If щ, n2 are any two integers, then (nlnz)a = (пхап-га)а..
Solution;
(а) a is onto because 0,1, ..., m— 1 have preimages 0,1, . .., m — 1 respectively.
(б) Let щ = qxm 4- Г] and w2 = q^wi + ri· Then
«i«2 = (9im + г\)(Чгт + **2> = Ч\Ч%™% + rtf^m + rtqxm + rxr% = (<ϊι<Ϊ2™ + η<Ϊ2 + ntfi)™ + Пг2
Now letting rxr2 = g3m + r3, we obtain (π^)» = rs -■ (rjr2)a = (щап2а)а.
1.28. Check which of the following mappings are onto, one-to-one, bisections:
(i) a: С-> R defined by a: a+ib -> az 4- 62
(ii) α : Ζ -» Ρ defined by a : η -» η2 + 1
τι
(iii) a:P-^Q defined by а: и-»-——-.
v 2« + 1
Solution:
(i) a is neither onto nor one-to-one, because az 4- 62 — 0, for any a,b G Й, and —la = la = 1.
Hence a is not a bijection.
(ίί) As 3 has no preimage, α is neither onto nor a bijection. Also α is not one-to-one, since la = 2
and —la — 2.
72-
(iii) 1 has no preimage, since „ , = 1 implies η — — 1 fi P. Therefore α is not onto and hence
ft УЬ
not a bijection. na — n'a means -—r-x = „ , , . or 2n'n 4- η = 2«'« + «'; hence η — и'.
2« + 1 2тс' + 1
Thus each image has a unique preimage and α is one-to-one.
1.29. Let S be the set of open intervals (a, 6) on the real line and let Τ be the set of closed intervals
[a, b]. Define aiS-^T by (a, 6> a = [a, b]. Is a one-to-one? Onto?
Solution:
The mapping is one-to-one and onto. For, if (a, b) α = (a',b')a then [o, 6] = [a',b'\. But this
equality holds iff о = о' and 6 = 6'. α is therefore one-to-one. α is also onto since a closed interval
[a, 6] has a preimage (a, 6>.
1.30. How many mappings are there from {1,2} into itself? From {1,2,3} into itself? In each case,
how many of these mappings are one-to-one? Onto?
Solution:
There are four mappings of {1, 2} into itself, namely: «г defined by la! = 1 and 2a! = 2; a2
defined by 1<*2 = 1 and 2a2 = 1; a3 defined by la:l = 2 and 2a3 = 1; a4 defined by la4 — 2 and
2a4 — 2. Only аг and a3 are one-to-one and onto.
To find the number of mappings of {1,2,3} into itself, we proceed as follows: 1 may have
any of three images under such a mapping, i.e. 1 -»1 or 1 -> 2 or 1 ->■ 3. Also, 2 may have any of 3
images, either 1, 2 or 3, So we have in all 3X3 possibilities for the actions of mappings on 1 and
2. Then 3 can be sent into 1, 2 or 3, giving 3 X 3 X 3 = 27 possible mappings of {1,2, 3} into itself.
There are 3X2X1 =6 possible one-to-one and onto mappings; for when we once choose an image
for 1 there are only two possible images for 2, and then the image of 3 is uniquely determined.
1.31. Let S = {1, 2,3}, Τ = {3,4, 5}, U = {4, 5,6}, and let a:S^T be defined by а:1->3,ЬЗ,3^5.
Let β and γ be the mappings from Τ into U given by /):3^4,4->6,И4, у:3-»4,4^4,б^4.
Cnmniifo Παϊβ. C2.a\R. (3«ϊβ. flab. (2aW. (3aW.
Compute (la)β, (2α)β, (3α)β, (1α)γ, (2α)γ, (3α)γ.
Solution:
(1α)β = 3β = 4; (2α)β = 3β=4; (3α)β = 5Д = 4; (1α)γ = 3γ = 4; (2α)γ = 3γ = 4; (3α)γ = 5γ = 4.

16
SETS, MAPPINGS AND BINARY OPERATIONS
[CHAP. 1
1.32. Let α, β, у be the mappings of Q into Q defined by α : χ -> —, β : χ-* χ + 1, у. χ ^> χ — 1. Prove that
ж + 1
if я; is any element of Q, then ({χγ)α)β ~ —5— · What is a^p 1 ajz ? Is ajp = a|Z ?
Solution: , . , „ ,
((ζγ)α)/ϊ = {(χ~1)α)β = —g— β = —— + 1 = g = ——. atP is a mapping of P
into Q. a.z is a mapping of Ζ into Q. a ¥· a,z since the domain of a.p is not the same as the
domain of a.z.
1.33. If S is a non-empty set, prove that S is infinite if and only if there are infinitely many mappings
of S into S.
Solution:
First we show that if there are infinitely many mappings of S into S, then S is infinite. Assume,
on the contrary, that S is finite and }S\ = n. Now each of the η elements can be mapped onto at
most η images. Hence there are at most nn different mappings of S into S. This contradicts our
assumption that there is an infinite number of such mappings. Hence S is infinite. Conversely, let
S be infinite. We define, for each s €.S, the mapping a„: S -» S by as: χ -> s for all »eS.
{as 1 s e S} is an infinite set since as = as. if and only if s = s', and we assumed S to be infinite.
Therefore we have found an infinite number of mappings of S into S.
1.34. If S is any non-empty set, verify that S matches S.
Solution:
Define the mapping a: S -> S by a: s -> s for each β e S. a is clearly one-to-one and onto.
Hence a· is a matching.
1.35. If S matches T, prove that Τ matches S.
Solution:
И S matches T, then there is a mapping a: S -> Τ which is one-to-one and onto.
Define &:T^>S as follows. Let t e T. Then there is an s&S such that sa — t. The image
of t under a is defined to be s.
We now show δ is a matching. In the first place, 5 is a mapping. For if ta — в and ts. = β',
for some t e T, then by definition of α, βα = t and e'a = t. But a is one-to-one, so that ва = в'а
implies s = s'. Thus the image of an element under 5 is unique. Secondly, α is one-to-one, because
ta=t'S — s implies sa = t and sa = t\ which in turn implies, since a· is a mapping, t=t'.
Thirdly, if s e S, then sa = t for some i e T. By definition of 5, ta = s. Hence every element
of S has a preimage under α and 5 is onto.
1.36. If S matches Τ and if Γ matches U, prove that S matches U. (Hard.)
Solution:
Let a:S-*T and β: Τ -> U be matchings. Then "5: S -> U, defined by 8Έ= (sa)fi, is a
matching, a is a mapping of S into U; for e&€.U, and if ва = Щ and sa = u2 then (эа)/8 = щ
and (sa)/8 = ω2, which implies щ = ω2 since ва has a unique image under β, a is one-to-one, for
ва = β'5 implies ($α)β = (β'α)β. But α and /8 are one-to-one, so that sa — s'a and β = β', α is also
onto, for if u e U then there is a t G Τ such that i/? = u. Now t has a preimage seS under
at. Thus 8Έ = (sat)yS = ίβ = и.
1.37. Let a be the mapping of S = {1,2, 3,4,.5, 6,7,8,9,10} into itself given by α: 1 -> 3, 2 -> 4, 3 -> 5,
4 -> 7, 5 -> 9, 6 -> 10, 7 -> 1, 8 -> 3, 9 -> 4, 10 -> 5. Then there is a useful alternative way of describing
a: we list the elements of S on one line (in any order) and on the following line we place under
each element of S its image under a, enclosing the entire description in parentheses as follows.
/1 2 3 4 5 6 7 8 9 10\
1,3 4579 10 1345/
Describe the following mappings of S into S, using this notation.

Sec. 1.4]
COMPOSITION OF MAPPINGS
17
(i) β: 1 -* 2, 2 -> 2, 3 -* 2, 4 -> 2, 5 -> 2, 6 -* 3, 7 -» 4, 8 -* 4, 9 -> 4, 10 -> 5.
(ii) γ: 1 -> 6, 2 -> 7, 3 -> 8, 4 -» 9, 5 -* 10, 6 -> 1, 7 -* 1, 8 -» 1, 9 -» 1, 10 -> 1.
Use these descriptions to decide whether (iii) β = γ, (iv) /? is one-to-one, (ν) γ is onto.
Solution:
/1 2 3 4 5 6 7 8 9 10\
V2 222234445/
/1 2 3 4 5 6 7 8 9 10\
^6 789 1011111/
(iii) β Φ у since 1β = 2 and 1γ = 6. It is only necessary to compare the bottom rows.
(iv) β is not one-to-one, e.g. 1β = Ζβ = 2. It is only necessary to find a repetition on the bottom row.
(ν) γ is not onto, e.g. 2 has no preimage. It is only necessary to check whether all the integers
1,2, ..., 10 appear in the bottom row.
1.4 COMPOSITION OF MAPPINGS
Definition
Let a: S -* Τ, β: Τ -* U, Because sn G T, we can compute (δα)β. This suggests
"composing the mappings a and β", i.e. defining a mapping of S into U by performing a
and β in succession on each of the elements in S. More precisely we define α°β, the
composition of a with β (in this order) as a mapping of S into U defined by
8(αοβ) — (3α)β, f ОГ all S in S
(Some authors use exactly the opposite order, so that their α ° β is our β ° α.) For example,
let
S = {1,2}, Τ = {3,4,5}, U = {6,7}
and let a:S-*T, β-.T^U be defined by
a: 1^3,2^5, β: 3^6, 4^7, 5^6
Then l(*°j8) = (1β)/β = 3β = 6
2(α°β) = (2α)|8 = 5/8 = 6
Hence α°/ϊ: {1,2}-» {6,7} is defined by
αοβ: 1^6,2^6
This notion of the composition of two mappings is of tremendous importance; hence we
give the following drill problems.
Problems
1.38. Let a: Ρ -» С be defined by not = in + 1 and let β : С -* Ρ be defined by /? : с + ib -» b2, where
г2 = —1. What do a ° β, (α ° β) ° a, and a ° {β ° a) map «eP to? Why is α°βφβ°α^.
Solution:
Let iteP, Then η(α°β) = (ηα)β = (гте + 1)/? = »A Now η{{α°β)°α) = {η{αοβ))α = η2α =
ma+l and η(α° (β°α)) = {ηά){βοα) - ((ίη+ 1)β)α = η2α = inz + l. Hence (α ο β) ο α = α ο (β ο α).
α ο β: Ρ-> Ρ while β°α: С^> С. Hence a« β Φ β°α.
1.39. Let α : Q -* Q be defined by α : α. -> α2 + 2 and let β: Q-» Q be defined by /? : α -> Ια - 2.
Compute α°β,β° a. Are these mappings equal? Compute (α° β)° α, <*° {β° a).

18
SETS, MAPPINGS AND BINARY OPERATIONS
[CHAP. 1
Solution:
Let α e Q. a{a ° /?) = {αα)β = (aa + 2)0 - ${a? + 2) - 2 ~ \Φ - 1. α{β ° α) = (»/?)<* = (|α- 2)α -
(±α - 2)2 + 2. Clearly αοβΦβοα, Furthermore, φ»^)ο«) = (α{αοβ))α = {±α? - 1)α =
(|α2 „ΐ)2 + 2 and α{α°{βοα)) = (αα)(/?°α) = (β? + Ζ)β ° α = ((α2 + 2)/?)α = ((|(α2 + 2) - 2)α =
(^α2- 1)ζ + 2. Note then that (αο β) ο α = αο (β о а).
1.40. Employing the notation of Problem 1.37, compute the following:
/1 2 3 4 5\ /1 2 3 4 5\ ,...ч /12 3 4 5
(1) (2315 i)°{l 2453; (Ш)(2 4531
/1 2 3 4 5W1 2 3 4 5
yi ^2 14 3 5Д2 И 5 1
Solution:
1 2 3 4 5\ ,..s /1 2 3 4 5\ ,.... /1 2 3 4 5\ .. /1 2 3 4 5\
2 4 1 3 6J (П) (3254О (m) (2З54О (№) (2 б 3 4 l)
1.41. Let a:S->7\ β-.Τ-^ϋ, y.V-*V. Prove that (α°/?) °γ = α °(/?° у). (Hard.)
Solution:
If k£S, then s((a°/?)°y) = (β(α ° /?))γ = ((»α)/?)γ and s(»o(/J»y)) = (βα)(/?°γ) = ((8α)/?)γ.
Consequently (a ° /?) ° γ = α ° (/? ° γ).
1.42. Prove that if a : S -» Γ and β : Τ -^ U and α ° /? is onto, then /? is onto. Is α onto? (Hard.)
Solution:
Let и £ [7. As α ° (3 is onto, we can find a preimage of и under α ° β. Let s £ S be a preimage
of ω under a0/?, i.e. 8(α°β) = и. Thus φο^) = (βα)β = и and sa is a preimage of и under β.
Hence /? is onto, a need not be onto, e.g. let f? ■= {1}, Τ = {1,2}, 17 = {1}. Define a: S-> Γ by
In == 1, and β: Τ -> U by 1/? == 2/? = 1. α° β is onto but α is not.
1.43. Prove that if α: S -» Τ1, β : Τ -» U and α°β is one-to-one, then α is one-to-one. Is β one-to-one?
(Hard.)
Solution:
Let 8lts2GS and sya — s^a. s1a = s2a implies (»ια)/? = (82a)/? and, by definition of a°β,
8λα° β = s2a° β. But α0/? is one-to-one, so that вг = s2. Hence a is one-to-one. /? is not necessarily
one-to-one, e.g. let S = {1}, Г = {1,2} and U =■ {1}. Define α : S -> Г by la = 1, and
β: Τ -*U by 1β = 1 and 2/? = 1. α°β is one-to-one but /? is not one-to-one.
1.44. Prove that β ; Τ -* U (T ¥· φ) is one-to-one if and only if for every set S and every pair of
mappings a : S -» Τ and a': S -» T, α ° β = α' ο β implies a = a'. (Hard.)
Solution:
First assume that for every set S and every pair of mappings a: S -» Τ and a': S -* T,
a ° β = α' ° β implies a = a'. Under this hypothesis suppose β is not one-to-one. Then we can find
i, t' £ Τ (ЬФ t') such that tfi = i'/? = и £ U. bet S = {1,2}, let α : S -> Г be defined by la = t
and 2a = t', and let a': S -* Τ be defined by la' = V and 2a' — t. Now a ° /? = α' ° β, since
la о /? = (la)/? = tfi-U, 2αθβ = (2a)/? = f/ϊ = u, la'° β = (la')/? = t'/? = Щ and 2α' ° β = (2α')β =
t/? = и. But α τ^ a'. Hence the assumption that /? is not one-to-one is false and /? must be one-to-one.
To prove the converse, let β be one-to-one. Say we can find a set S and a pair of mappings a
and a' of S into Τ such that a°β = α' ° β and α ?* a'. So there exists s £ S such that sa ¥· sa'.
a ο β — a' ο β means that s{a ° /?) = s{a' ° /?). Hence (sa)/? = {sa')fi. As /? is one-to-one, sa — sa'.
Therefore we have a contradiction and a must be equal to a'.
1.45. Prove that a: S -* Τ {Τ Φ0) is onto iff for every set U and every pair of mappings /?: Τ -* U
and β':Τ-*ϋ such that a ° /? = α ° /?', it follows that /? = /?'. (Hard.)
Solution:
Let us assume that for every set U and every pair of mappings /? and /?' of Τ into U such that
α о /? = α ° /?', it follows that /? = /?'. Say α is not onto, and *! is an element of Τ which has no
Wl 2 3 4 5N
/ ll 2 4 3 5/
12 3 4 5
12 3 4 5
12 3 4 5
2 5 3 4 1

Sec. 1,5]
BINARY OPERATIONS
19
preimage under a. Let U = {1,2} and define the mappings β and β' of Τ into U as follows; f/? = 1
for all С G F, t/T = 1 for all С Φ Ц in Г and („8 = 2. Now if s e S, 8α°β = (sa)/? = 1 and
sa ° /8' = (sa)/?' = 1, since sa φ tv Hence a° β = α°β' and /? ^ /J'. This contradicts the
assumption that α°β = α°β' implies β = β'. Thus α must be onto.
Conversely, assume a is onto and we can find a set U and two mappings, β and β', of Τ into U
such that α°β = α ο β' and β φ β', β Φ β' means there is a t, £ Γ such that t,/? φ t-φ'.
Furthermore, since α is onto, we can find sGS such that sa = tv But α°β = α°β' implies (sa)fi — (sa)/?'
or fi/? = Cj/?'. Here we have a contradiction because we chose Ьф Φ ίφ'. We therefore conclude
β = β>.
L5 BINARY OPERATIONS
a. Definition
The idea of a binary operation is illustrated by the usual operation of addition in Z,
which may be analyzed in the following way. For every pair of integers (m,n) there is
associated a unique integer m + n. We may therefore think of addition as being a brief
description of a mapping of Ζ χ Ζ into Ζ where the image of (m, n) € ZxZ is denoted by
m+n. Any mapping β of SxS into S, where S is any non-empty set, is called a binary
operation in S. We shall sometimes write instead of (s, ΐ)β (the image of (s, t) under β) one
of sot, s't, st, s + t or sx t. We stress that in all these cases the meaning of the various
expressions s ° t, s · t, st, s +1 and s x t is simply the image of (s, t) under the given mapping
β of S x S into S. These notations suppress the binary operation β, so there is danger of
confusion. However, we will work with binary operations and the various notations so
frequently that the reader will become familiar with the pitfalls. Incidentally, we read
sot as "ess circle tee"
s-t as "ess dot tee" or "ess times tee"
st as "ess tee" or "ess times tee"
s + ί as "ess plus tee"
sxt as "ess times tee".
The notation s о t is called the circle notation, the notations s · t and st are termed
multiplicative, and the notation s +1 is termed additive. We sometimes refer to s · t or st as the
product of s and t, and s + t as the sum of s and t. The following problems will help to
make the various notations clear.
Problems
1.46. Convince yourself that the following mappings are binary operations in P.
(i) a : Ρ ΚΡ-*Ρ defined by а: (г, j) -> г2, where (г, /) G P.
(ii) <*: PxP->P defined by a: (i,j)-*i+j, where (i,j)GP.
(iii) a: Ρ xP -> Ρ defined by a : (г, fl-»ix ; (regular multiplication of integers), where (г, j) G P.
(iv) a:PxP-*P defined by а: (г,У)-»2г + 3/, (i,f)£PxP.
(v) a:PxP-*P defined by a : (i,j) -> г+ j+ 1, (i,])ePxP.
1.47. Which of the following are binary operations in Ρ (throughout (г, j) e P1) ?
(i) a: (i,j)-*i + i (in) a: (i, i)-» г-т-У (ν) a: (i,j)-*i
(ii) а : (г, j) -> г — ; (iv) a : (г, ί) -> г + j + ??
Solution:
In (i), a is clearly a mapping from Ρ Χ Ρ into P. So α is a binary operation in P. (ii) and
(iii) do not define binary operations in Ρ because not every element in Ρ Χ Ρ has an image in P:
e.g. in (ii), a: (1,2) -> i - 2 = -1«P; and in (iii), a: (1,2) -> 1 -r-2 = \ ЙР. (iv) and (v) define
mappings from PxP into P. Hence they define binary operations in P.

20
SETS, MAPPINGS AND BINARY OPERATIONS
[CHAP. 1
1.48. Interpret the following (abbreviated) definitions of binary operations in Z, where i and / denote
arbitrary elements of Z.
(i) i ° j — (i + j)2 (iii) i° j = i+ j (v) i'Kj — ϋ
(ii) i + j = i — } — (ix j) (iv) ij = i — iX j (vi) i · j = i + 27 j
Solution:
Throughout, (i, i) € Z2.
(i) a: (i,j)-> (i+ j)2 (iii) а: (г, j) -» г + j (v) a: (г, j)-> i1
(ii) a : (г, У) -» г - ;' - (г X j) (iv) α : (г, j) -* г - г X j (vi) a : (г, j) -» г + 27 j
1.49. Check that the following are binary operations in the plane R2.
(i) (*,У) ° (я',уО = midpoint of the line joining the point (ж,у) G R2 to the point (ж',у') ей2
if (ж, у) Φ (ж', у'). If (x,y) = {x',v'), define (ж,у) ° (ж',у') = (х,у).
(ii) (ж, у) + (ж', у') = (ж + ж', у + у'), where (ж, у), (ж', у') G Д2.
(iii) (ж, у) -(ж', у') = (жж', уу'), where (x,y),(x',y') €. R2.
(iv) (ж, у) - (ж', у') = (ж-ж', у-у'), where (ж, у), (х',у') € R*.
(ν) (ж, у) о (ж', у') = (ж + ж', ж'у + у'), where (ж, у), (ж',у') G R2.
(Notice here how we have abbreviated the definitions of the binary operations considered.)
Solution:
(i) ° is a binary operation since two points determine a unique line and each line has a unique
midpoint.
(ii)-(v) are binary operations because each has a unique image by virtue of the fact that addition,
multiplication and subtraction are binary operations in R.
1.50. Let S = {1,2,3} and let a and β be the following binary operations in S:
a: (1,1)-* 1, (1,2) -1, (1,8)-» 2, (2,1)-* 2, (2,2) - 3, (2,3) -* 3, (3,1) - 3, (3, 2) -» 2, (3,3)-l;
/?: (1,1) -»1, (1,2)-»1, (1,3)-* 2, (2,1) -* 3, (2,2)-» 8, (2,8)-» 8, (8,1)-» 8, (8,2)-*l, (3,3)-* 2.
(i) Is α = /??
(ii) Compute ((Ι,Ι)α,Ι)/?, ((1,1)/?, l)a.
(iii) Compute ((1,2)α,3)α, (1,(2,8)«)«, ((l,2)/?,3)/?, (1,(2,3)/?)/?.
Solution:
(i) αφβ for (2, l)a = 2 and (2,1)/? = 3.
(ii) ((1,1)«, 1)/? = (1,1)/? = 1; ((1,1)/?, 1)« = (1,1)« = 1.
(iii) ((1,2)α,3)α = (1,3)α = 2; ((l,2)/?,3)/? = (l,3)/? = 2;
(1, (2,8)e)« = (1, 3)« = 2; (1, (2,3)/?)/? = (1,3)/? = 2.
1.51. Let S = {1,2,3} and let X be the set of all mappings of S into S.
(i) Compute \X\.
(ii) Verify that the composition ° of mappings is a binary operation in X.
(iii) Is a° β ~ β°α for all elements α, β € X?
Solution:
(i) |X| = 27 (see Problem 1.30, page 15).
(ii) The composition of mappings a: S -» S and β: S -* S is again a mapping of S into S.
Therefore π:Χζ-*Χ, defined by (α,β)π = α°β, (α,β) £ Ζ2, is a binary operation.
(iii) No. For example, if α: S -* S defined by sa = 1 for all s€ S, and /? : S-* S defined by
s/? = 2 for all 8 €.S, are two mappings of S into S, then s(a°/?) = (sa)/? = 1/? = 2 and
s/? ο α = (β/?)α = 2α = 1. Hence α° β Φ β°α.
1.52. Let Q* be the set of nonzero rational numbers. Make sense of the remark that division (denoted as
usual by -3-) is a binary operation in Q*. Check whether the following statements hold for all
a, b, с е Q*.
(i) a -r- b = b -5- a. (iv) If α -*- b = α + с, then b = с.
(ii) (α-^ Ь) -τ- с = а -т- (Ь -т- с), (v) If Ь -τ- α = с -г- a, then b = с.
(iii) ((α -ϊ- Ь) -f- «) ■*- d = а + (Ь -г- (с-г- <*)).

Sec. 1.5]
BINARY OPERATIONS
21
Solution:
Division is a binary operation in Q*; for if w/x G Q* and y/zGQ* where w,x,y,z are
integers, then w/x ■*· ylz = wz/xy G Q*. Hence -r is a mapping of Q* X Q* -* Q*.
(i) False; e.g. 2 4-3^3-^2.
(ii) False; e.g. (2-1)^2 = 1 and 2 -=- (1 ■+- 2) = 2 + | = 4.
(iii) False; e.g. ((1 + 2) + 2) -=- 2 = £ and H(2-i-(2-f 2)) = f.
(iv) True, a -i- b = a ■+- « implies ас = об and, as a^O, u = J.
(v) True, b -r- a — e -τ- a implies ab = «a. Hence b — c, since a ^ 0.
1.53. Let a be the binary operation in R defined by aab = a+ b + ab. Verify that:
(i) For all a,b,e.€R, (a°b) °c = o» (6 °c).
(ii) For all а, Ь G R, a ° 6 = b ° a.
(iii) Prove that if α ^ —1, then uoi = (ioc iff b = c.
Solution:
(i) (a°J)»c = (a+b + ab)°c = α + 6 + ab + с + (a + Ы- ab)«
= а + b + с + be + ab + ас + abc
α о (6 о β) = а°(Ь+с + Ьс) = а + b + с + be + a(b+ c+be)
= а+Ь + <! + Ьс + аЬ + а<; + abc
(ii) a°b = а + b + ab = b + a + ba = Ь ° а
(iii) If b = «, then a°b = aoe for any a. If a°6 = a°<! and а # — 1, then a+b + ab =
а + с + ас. Therefore Ь+аЬ=(! + ас, 6(1 + а) = c(l + α) and, since α ^ — 1, Ь = c.
1.54. Let ° be the binary operation in R'2 defined by (x, y) ° (x', y') = (xx' — yy', yx' + xy'). Verify that for
all {χ,ν)Λ*',ν')Λχ",ν")^№;
(i) (x,y)o(x',yr) = (x\y')°(x,y)
(ii) ((*,*) о (*',10>°(*",у") = (*,v)°((x',V')0(*",V"))
Solution:
(i) (ж, ί/) ° (ж', У') = («»' ~ УУ', У«' + «У') = (я'« ~ V'V, У'« + x'v) = (*', V') ° («, у)
(ii) ((«, V) ° (*', У')) ° (*", y") = (««' ~ W, V*' + *v0 ° (*", V")
= ((aw' - yy')x" - (yx' + xy')y", (yx' + xy')x" + (xx' - yy')y")
= (xx'x" — yy'x" — yx'y" — xy'y", yx'x" + xy'x" + xx'y" — yy'y")
(X, у) о ((«', у') ο (χ", у")) = (а,, у) о (Х>х» - у'у", y'x" + у"Х')
= (х(х'х" - у'у") - у(у'х" + у"х'), у(х'х" - у'у") + х(у'х" + у"х'))
= (хх'х" ~ ху'у" — уу'х" — уу"х', ух'х" — уу'у" + ху'х" + ху"х')
= ({χ,ν)°(χ',ν'))°(χ",ν")
1.55. Let ° be the binary operation in Q denned by
/\ l i. , l ,·\ i. a + b + ab .. , a + b
(a) aob = a — b + ab, (b) a°b — — , (c) aob =—'—
L· 3
Determine which of the above binary operations satisfy
(i) (аоб) о„ = aо(6 °<0 for all a,b,cSQ
(ii) а о b = b ο α for all a, b G Q
Solution:
(i) (a) (aob)°c^ao(boe) for some a,6,cGQ; e.g. (2°0)°2 = (2-0 + 0)°2 = 2°2 =
2-2 + 4 = 4 and 2°(0°2) = 2o(0-2 + 0) = 2o-2 = 2- (-2) -4=0.
(b) (aob)oc^ а°(Ьос) for some a,b,cGQ; e.g. (1 °0) °0 = |°0 = \ and 1 ° (0 ° 0) =
(«) (a°b) 0(i7^ α o(6 о с) for some а, 6, с G Q; e.g. (1°0)°0 = J°0 = | and 1°(0°0) =
(ii) (a) ao6^b°a since 1°0 = 1 and 0°1=— 1.
/м „ь_а+Ь + а& Ь + а+Ьа ,
(b) aob = = = boa
/\ ι a+b b + a .
(c) aob — —r— = Z-^-— = boa

22
SETS, MAPPINGS AND BINARY OPERATIONS
[CHAP. 1
b. The multiplication table
So far we have introduced a number of definitions and notations and familiarized
ourselves with them. The object of this section is to introduce a "table" as a convenient way
of either defining a binary operation in a finite set S or tabulating the effect of a binary
operation in a set S. To explain this procedure, suppose S - {1,2,3} and let μ be the binary
operation in fif defined by
μ.: (1,1)->1, (1,2)-1, (1,3)-* 2, (2,l)->2, (2,2)->3,
(2,3)->3, (3,1)->1, (3,2)->3, (3,3)->2
Then a table which sums up this description of μ is
12 3
112
2 3 3
,
13 2
2
3
We put the number (2, 3)μ = 3 in the square that is the intersection of the row facing
2 (on the left) and the column below 3 (on the top). More generally, in the (i,;)th square,
i.e. the intersection of the ith row (the row labelled or faced by i) and the jth column (the
column labelled by j), we put {i, ])μ.
A table of this kind is termed a multiplication table because it looks like the usual
multiplication tables. One often calls μ a multiplication in S. Thus when we talk about a
multiplication μ in a set S, we mean that μ is a binary operation in S.
There is a reverse procedure to the one described above. For example, suppose we start
out with a table
12 3
1
2
1
1
3
3
2
3
2
2
3
Then there is a natural way of associating with this table a binary operation μ in {1,2,3},
We simply define (ΐ, ί)μ to be the entry in the (г, /)th place in the table. For example,
(1,1)μ=1, (2,3)/ι = 3, (3,2)μ = 3
We shall usually define multiplications in a finite set by means of such tables.
Problems
1.56.· Write down the multiplication tables for the following binary operations in S = {1,2,3}.
(i) a: (1,1) -» 2, (1, 2) -» 3, (1, 3) -» 1, (2,1) -» 3, (2,2) -»1, (2, 3) -» 2, (3,1) -» 1, (3,2) -» 2, (3,3) -* 3.
(ii) β:82->8 defined by (г, ])β = 1 for all (i,j) G S2.
(iii) y. (1,1)-H, (2,2)^1, (3,3)^1, (1,3)-* 2, (3,1)-» 2, (2, 3) -> 1, (3, 2) -» 1, (2,l)->3, (1,2)-*3.
Solution:
123 123 123
1
2
3
2 3 1
3 12
12 3
(H)
111
111
111
(iii)
13 2
3 11
2 1 1

2sHEC* JLeijj
BINARY OPERATIONS
23
1.57. Does the following table define a binary operation in {1,2,3}? In {1,2,3,4}?
12 3
1
2
3
Solution:
The table does not give a binary operation in {1, 2, 3} since (1,3)-»4ё {1, 2, 3}. The table
also does not define a binary operation in {1,2,3,4}, because (1,4), (2,4), (3,4), etc., have no images.
1.58. Write down explicitly the binary operations in {1,2,3,4} defined by the following tables.
1
2
1
3
1
3
4
3
2
12 3 4 1 2 3 4
1 I 2 3 4 112 3 4
2341 22413
3412 33142
4123 44321
1111
2 2 2 2
3 3 3 3
4 4 4 4
(»)
(ii)
(Hi)
Solution:
(i) (1,3) -* 3 U = 1,2, 3,4); (j, 1) -* i (j = 2, 3,4); (2, 2) -> 3; (2,3) -> 4; (3,2) -> 4; (3,3) -> 1;
(3,4)-» 2; (4,3)-* 2; (4,4)-> 8; (4, 2) - 1; (2,4)-»1.
(H) (1, Л -* / 0' = 1,2,3,4); (j, 1) -» j (j - 2, 3,4); (2, 2) -> 4; (2,3) -> 1; (3,2) -> 1; (3,3) -* 4;
(3,4)->2; (4,3)-* 2; (2,4)-»3; (4,2)-» 8; (4,4)-»1.
(iii) (t, Я -» ί (i = 1, 2,3,4 and j = 1,2,3, 4).
1.59. Rewrite the three binary operations in Problem 1.58, using
(a) circle notation, i.e. write (ί,ί)β as i°i,
(b) additive notation, i.e. write (ί,]')β as i+ j,
(«) multiplicative notation, i.e. write (ί,ϊ)β as i· j.
Solution:
(а) Problem 1.58(i): W = j (/ = 1, 2,3,4); j ° 1 = / (/= 1,2, 3,4); 2o2 = 3; 2°3=3°2 = 4;
3°3 = 1; 3°4 = 4°3 = 2; 4 ° 4 = 3; 4°2 = 2°4 = 1.
Problem 1.58(H): W = j (j = 1,2, 3,4); /= 1 = / (j = 2, 3,4); 2°2=4; 2°3 = 3°2 = 1;
3o3 = 4; 3°4=4°3 = 2; 2°4 = 4°2 = 3; 4°4 = 1.
Problem 1.58(iii): i°j-i (i = 1,2, 3,4 and j = 1,2,3,4).
(б) Problem 1.58(i): 1 + /- j (j = 1,2, 3,4); j + 1 = j {j = 1,2,3,4); 2+2 = 3; 2 + 3=3 + 2 = 4;
3 + 3 = 1; 3 + 4 = 4+3 = 2; 4 + 4 = 3; 4 + 2 = 2 + 4 = 1.
Problem 1.58(ii): 1 + j = j and j + 1 = j (j = 1,2,3,4); 2+2 = 4; 2 + 3 = 3 + 2 = 1;
3 + 3 = 4; 3 + 4 = 4+3 = 2; 2 + 4 = 4+2 = 3; 4 + 4 = 1.
Problem 1.58(ш): i+j = i (i = 1, 2, 3, 4 and / = 1, 2, 3,4).
(e) Problem 1.58(i): 1 · i = / and i · 1 = / (/ = 1, 2,3,4); 2 · 2 = 3; 2 · 3 = 3 · 2 = 4; 3-3 = 1;
3-4 = 4-3 = 2; 4-4 = 3; 4-2 = 2-4 = 1.
Problem 1.58(ii): 1 -j = i and j · 1 = j (j = 1,2,3,4); 2 ■ 2 = 4; 2-3 = 3-2=1; 3-3 = 4;
3-4 = 4-3=2; 2-4 = 4-2 = 3; 4-4 = 1.
Problem 1.58(Hi): i-j = i (i = 1,2,3, 4 and j = 1,2, 3,4).

24 SETS, MAPPINGS AND BINARY OPERATIONS [CHAP. 1
A look back at Chapter 1
We began with a few remarks about sets. We then introduced the idea of cartesian
products. This led to the idea of an equivalence relation on a set. Then the notion of a
mapping was defined, followed by the definition of a binary operation.
In this book we are mainly concerned with binary operations in sets. At this stage
the reader may wonder what one could possibly say about binary operations in a set.
Without some specialization we can say very little. In Chapter 2 we begin to place
restrictions on binary operations.
Supplementary Problems
SETS
1.60. The sets S{ (i ~ 1, 2, .. ,,n) are such that S,-cSi+i (i = 1,2, .. .,η-ί). Find SjnS2n · · ■ nSn and
SjUS2U---uS„.
1.61. Let Ε = {n\ n^Z and η even}, О = {n\ n&Z and η odd}, Τ = {n\ n€.Z and η divisible by
3},and F-{«| n<=Z and η divisible by 4}. Find {i)EnT, (ii)E\jT, (Hi) TnFnEnO, (iv) TuF,
(v) Or\F, (vi) Or\T.
1.62. Given A = {-3, -2, -1, 0, {1,2,3,}} and В = {-3, -1, {1,3}}. Find A UB and Α η Β.
1.63. Prove Sn(TuV) = (SnT)u{SnU).
1.64. Let A = {-5,-4,-3, ...,3,4,5}, В = {-4,-2,0,2,4}, С = {-5,-3,-1,1,3,5}, С = {-4,4},
Ε = {-3,-2,-1,0}, F = (3. Which, if any, of these sets take the place of Ζ if (i) Z-C = (3,
(ii) Xr\B = C, (iii) XcC but Ζ is not a subset of A, (iv) X<zB and X is not a subset of E,
(v)ZnCcA, (vi) ZU(BnD) = A?
1.65. Prove ScT if and only if (TnC)uS = Tn(CuS) for every set C.
CARTESIAN PRODUCTS
1.66. Prove S X (TuTV) = (S Χ Γ)υ(5 X TV) for any sets S, Τ and TV.
1.67. Let Ρ be the set of positive integers and S = P2. Show that Ε — {ρ j ρ = ((r, s), (t, w)) e S2 and
?· + «; = « + *} is an equivalence relation on S. Find the E'-class determined by (4,7).
1.68. Find the equivalence relation, E, on Z: (i) if the equivalence classes of Ε are {n\ η = iq for
q £ Z}, {n | и = 1 + 4? for ? € Z}, {« | и = 2 + 4? for ? e Z}, and {«j и = 3 + 4? for ? £ Z};
(ii) if every equivalence class consists of a single integer; (iii) if the equivalence classes are
{?. ~?} ior each q <= Z.
1.69. If Я and F are equivalence relations on S, is (i) 2JnF, (ii) EuF an equivalence relation on SI
1.70. What is wrong with the following argument: Ε is a non-empty subset of S2 which has the symmetric
and transitive properties. If (a, b) € E, then by the symmetric property, (6, a)&E, But by the
transitive property, (a, b) e Ε and (b, a.) e Ε implies (a,,a)&E. Therefore Ε is also reflexive.
1.71. (a) If Ρ is the set of positive integers, show that Ε = {(a, b) \ (a., b) e P2 and a divides 6} is
reflexive and transitive but not symmetric.
(b) Find an example of a subset Ε of Pa which is both symmetric and transitive but not reflexive.
(c) Find an example of a subset Ε of P2 which is reflexive and symmetric but not transitive.

CHAP. 1]
SUPPLEMENTARY PROBLEMS
25
MAPPINGS
1.72. Show that the following define mappings of Ζ into Z.
ί 0 if ж = 0 . . . , , , . . ficific^O
(ι) α: χ -> < „ , . 'ж is the absolute value of ж, i.e. ar = -i
[x/\x\ if ж^О ' ' ' ' [-ж if χ < 0
Г 0 if ja;| = 0 or 1
\_(—l)r if \x\ ^ 0 or 1, and r is the number of distinct primes dividing χ
Γ-1 if χ < 0
(iii) γ: χ -> \ 0 if χ — 0
1 if a; > 0
(iv) 8 : χ -> sin2 a; + cos2 x
1.73. Which of the mappings α, β, у, δ of the preceding problem are equal?
1.74. Find subsets S(¥'0) and T(¥>0) of the real numbers R such that the mappings (i) a : χ -> cos ж,
(ii) /?: a; -> sin a;, and (iii) γ : a; -> tan a; are one-to-one mappings of 5" onto T. Do α, β, у define
mappings of R into Й?
1.75. Let Ε = {и j и e Ρ and и even}. Define α: Ε -> Ε by w« = и for all ti6B, and β : Ε -* Ε
by и/? = 2и for all η & Ε, Find an infinite number of mappings α, β' of Ρ into Ρ such that
<V = « and /^ = β.
1.76. Suppose α is a mapping of a set S into a set Τ and, for any subset W of S, Wa = {t\ t ε Τ and
ί = sa for some self}. If Λ and β are any subsets of S, show: (i) (AvB)a = AaVBa;
(ii) (AnB)a cAanBa; and (iii) Л с В implies AacBa.
1.77. Let S be a subset of a set T, and a: T -* W. Prove: (i) a one-to-one implies ag is one-to-one;
(ii) a onto implies a is onto.
COMPOSITION OF MAPPINGS
1.78. Suppose a : S -> Τ is onto and β: Τ -> U is onto. Show a° β is an onto mapping.
1.79. Given a: S -> Γ is one-to-one and β : Τ -* U is one-to-one. Prove a°/? is one-to-one.
1.80. (i) a : χ -* sin (a;2), (ii) /?:«-> sin (sin (x)), and (iii) γ : χ -* s/l — χ2 define mappings of non-empty
subsets of the real numbers R into R. First, find an appropriate subset in each case. Secondly,
write α, β and γ as the composition of two mappings, giving in each case the domain and codomain
of each mapping defined.
BINARY OPERATIONS
1.81. Let S be a set and q) the set of all subsets of S, i.e. (3J' = {A j ACS}, Show that intersection and
union define binary operations on e/.
1.82. How many different binary operations can be defined on a set of 3 elements?
1.83. Consider the set, F, of mappings f{ {i = 1, 2, . .., 6) of R - {0,1} into R defined for each χ € R - {0,1}
1 χ — 1 1 χ
by: fi'.x-^x; /2: χ-* γζΓ^'> fs:«-*—^—> f-i'-x~*-'> fs-x^-^ΖΓγ' f6-x-*l-x- Show that
composition of mappings is a binary operation on F and write a multiplication table for the operation.

Chapter 2
Groupoids
Preview of Chapter 2
In this chapter we define a set G together with a fixed binary operation о to be a groupoid.
As we remarked at the end of Chapter 1, there is little one can say about binary operations
without making restrictions,
The first restriction is that of associativity. A groupoid with set G and operation о is
said to be associative if fifi ° (.<7г ° <7з) = (дч ° gz) ° gs for all gh g%, g3 in G, Such a groupoid is
called a semigroup. In order of increasing specialization we have the concepts of groupoid,
semigroup, and group.
To define a group we need the concepts of identity and inverse. Hence we discuss these
ideas.
We introduce the semigroup Mx of mappings of X into X. The importance of Μχ is
that, but for the names of the elements, each semigroup is contained in some Μχ.
Two other important concepts we deal with are homomorphism and isomorphism.
Homomorphism is a more general concept than isomorphism. There is an isomorphism
between two groupoids if they are essentially the same but for the names of their elements.
2.1 GROUPOIDS
a. Definition of a groupoid
Consider the set Ζ of integers. Ζ has two binary operations, addition (+) and
multiplication (X). The set Ζ is one thing, a binary operation in Ζ is another; the two together
constitute a groupoid. Repeating this definition in general terms.
Definition: A groupoid is a pair (G, μ) consisting of a non-empty set G, called the carrier,
and a binary operation μ in G.
We shall mostly use a multiplicative notation when dealing with groupoids. Thus we
write g - h or simply gh for (g, Κ)μ, g,h eG. This notation has been used in Chapter 1
in our consideration of binary operation. As an example, let G = {1,2,3} and let μ be the
binary operation in G defined by the following table.
12 3
1
2
3
3 j 2
1
2
1
3
Then the pair (G, μ) is a groupoid. Suppose we use multiplicative notation ·; then
1-1 = 1, 1-2 = 3, 2-2 = 1, 3-2 = 2, etc.
26

Sec. 2.1]
GKOUPOIDS
27
These products look bizarre unless we recall that the notation employed is a shorthand
version of
(1,1)μ = 1, (1,2> = 3, (2,2)μ = 1, (3,2)μ = 2, etc.
If we use the expression "the groupoid G," where G is a set, it is understood that we
have already been given a binary operation μ in G, and that we have been talking about
the groupoid (G, μ).
Suppose now that (G, μ) is a groupoid. If we use the circle notation for μ, i.e. we write
g о h instead of (g, Κ)μ, we shall sometimes write (G, o) to refer to the groupoid (G, μ).
Similarly we write {G,·), {G, +), (G, x) if we employ g-h,g + h,gxh respectively for (g,ίήμ.
Example 1: Let G = {1,2} and let μ be the binary operation in G defined аз follows:
(1,1)0 = 1, (1,2)^=2, (2,1)0 = 1, (2,2)0 = 2
If we use the circle notation, we have
1°1 = 1, lo2 = 2, 2°1=1, 2o2 = 2
The pair (G, μ) or (G, °) is then a groupoid.
Example 2: Let S be the set of all mappings of {1,2,3} into {1,2,3}. Then (S, °) is a groupoid,
-where ° is interpreted as the usual composition of mappings. The composition
makes sense, for if «,|9eS, then
a : {1, 2, 3} -» {1, 2, 3} and β : {1, 2, 3} -* {1,2, 3}
Therefore a ° β, the composition of a and β, is defined by aa° β = (αα)β, α. e {1, 2,3},
and is once again a mapping of {1,2, 3} into itself. (S, °) is indeed a groupoid.
Example 3: Let ° be the binary operation in Q, the rational numbers, defined by a°b = a +
b + ab. Then (Q, °) is a groupoid, because for every pair of rational numbers
α and b, a°b defines a unique rational number a+b + ab.
Example 4: Let R2 be the plane. Further, let there be a cartesian coordinate system in R2 and
let С be the disc of radius 1 with center at the point (2,0) of the coordinate system.
Consider the region R'2—C, the unshaded area in the diagram. We term any path
beginning and ending at 0, which does not meet any point of С (i.e. it is entirely in
R2 — C), a loop in R2 — C. By a path we mean any line which can be traced out by
a pencil without raising the point from the paper. For example, I and m are
loops in R'1 — C. Let L be the set of all such loops in R2 — C. Then there is a
natural binary operation in L which we denote by ·; thus if lv 1% £ L, then Ιχ*1%
is the loop obtained by first tracing out lx, followed by tracing1 l2. This type of
groupoid (L, ·) is of considerable importance in modern topology.

28
GROUPOIDS
[CHAP. 2
Example 5: Let F be the set of all mappings of R, the real numbers, into R. Consider two
elements α, β £ F. We define the mapping a + β : R -* R by a(a + β) — aa + αβ,
α. 6 R. a + β is clearly a unique mapping of -B into R and hence is an element in F.
Therefore + is a binary operation in F and (F, +) is a groupoid. Notice that (F, +)
is not the groupoid with F as the carrier and the composition of mapping's as the
binary operation.
Problems
2.1. Are the following· groupoids?
(i) (S, o) where S = (1, 2, 3,4} and i ° j = 1 for ΐ and / elements of S.
(ii) (Z,—), the set of integers with the usual subtraction of integers as binary operation,
(iii) (P, —), the set of positive integers with the usual subtraction as binary operation,
(iv) (Q, -r), the set of rational numbers with the usual binary operation of division,
(ν) (Ζ, -ΐ-), the set of integers with the usual binary operation of division.
Solution:
(i) (S, o) is a groupoid since ° is clearly a binary operation in S.
(ii) (Z, —) is a groupoid; for if a,b 6 Z, then a— b is a unique element of Z,
(iii) (Я, -) is not a groupoid since a — b € Ρ for all a,b eP, Therefore - is not a binary
operation in P.
(iv) (Q, -s-) is not a groupoid because a-^0 is not defined for any a.e Q and hence -=- is not a binary
operation in Q.
(v) (Z, -f-) is not a groupoid since a-rbii for all a,b e. Z, e.g. 2 + 3 Й 2. Therefore division
is not a binary operation in Z.
1.1, Is (Z, D) a groupoid if ° is defined as (i) a° Ъ = V»+ b. (ϋ) α·0 Ь = (»+ Ь)2, (iii) a.ofe = a.— b — ab,
(iv) a.° b = 0, (v) a°b = at
Solution:
All but (i) define a binary operation in Z. Therefore (Z, o) is a groupoid in (ii) through (v).
The multiplication ° in (i) does not define a binary operation in Ζ since ya+b is not always an
integer.
2.3. Let S be any non-empty set and Τ the set of all subsets of S. Are (7\ Π) and (T, u) groupoids?
Solution:
Both intersection η and union и are binary operations on T, for the intersection or union of
two subsets of S is again a unique subset of S. Thus (T, n) and (T, u) are groupoids.
b. Equality of groupoids
Two groupoids are equal if and only if they have the same carriers and the same binary
operation. Remember, a binary operation was denned as a mapping and two mappings
are equal if and only if they have the same domain and codomain, and the image of each
element ia the same under both mappings. Thus the groupoids described in Examples 1-5
are all different.
Problems
2.4. Are any two of the groupoids in Problems 2.1-2.3 equal?
Solution: No.
2.5. Which of the following pairs define equal groupoids?
(i) (Z,+) and (Ζ, μ), where (a, b)ji = a + b.
(ii) (Z, —) and (Z, °), where acb = α — Ь.
(iii) (Ζ, ο) where а°Ь —a for all α. and b in Z, and (Z, X) where axb = b for all a. and b in Z.
Solution:
The groupoids in (i) are clearly the same. So too are the groupoids in (ii). In (iii) (Z, °) is not
the same as (Z, X); for if α, τ4 b, a° b τ4 α. Χ b.

Sec. 2.2]
COMMUTATIVE AND ASSOCIATIVE GROUPOIDS
29
2.2 COMMUTATIVE AND ASSOCIATIVE GROUPOIDS
Definition of commutative and associative groupoids
Let (Z, +) be the groupoid of integers under the usual operation of addition. Then
a + b = b + a (2.1)
and (a + b) + c = a + (b + c) (2.2)
for all a,b,cG Z.
Similarly if (Z, ·) is the groupoid of integers under multiplication,
a · b = b · a (2.8)
and (a · b) · с = a· (b · c) (24)
for all a, b, с G Z.
The analog of (2.1) and (2.3) in an arbitrary groupoid (G, °) is
aob = boa (2.5)
for all a,b GG. Similarly the analog of (2.2) and (24) in (G,o) is
(aob) о с = a ° (bo с) (2.6)
for all a,b,cG G. We term a groupoid satisfying (2.5) commutative or abelian, and a
groupoid satisfying (2.6) associative or a semigroup. Thus a semigroup is an associative
groupoid. Of course it is not clear that there are non-commutative groupoids, i.e. groupoids
which are definitely not commutative, and similarly it is not clear that there are non-
associative groupoids. We settle the issue now. Let G — {1,2} and let ° be the following
binary operation in G:
1
2
Then (G, o) is a groupoid. Observe that 1 ° 2 = 1 but 2 о 1 ^ 2, so G is not commutative.
Furthermore, (2ol)o2=2o2 = l but 2о(1 °2) = (2° 1) = 2, so G is also non-associative,
i.e. G is not a semigroup.
For the most part we shall use the multiplicative notation for a groupoid (G,p) and
simply talk about the groupoid G. If the groupoid is commutative we will use the additive
notation instead of the multiplicative notation, since we are accustomed to addition as a
commutative binary operation, e.g. in the integers.
The order of a groupoid (G, μ.) is the number of elements in G and is denoted by |G|;
(G,/i) is infinite if \G\ is infinite, and finite if |G| is finite.
Problems
2-β. Which of the groupoids in Examples 1, 2, 3 and 5 are commutative and which are associative?
Solution:
The groupoid of Example 1 is not commutative, since 1 ° 2 = 2 and 2 ° 1 = 1, but is associative.
To show (G, °) is associative we must examine the following 8 cases:
{a.) 1°(1°1) = 1°1 = 1, (1°1)°1 = 1°1 := 1 (e) 2 °(2° 1) = 2 ° 1 = 1, (2°2)°1 = 2°1 = 1
(6) 2°(1°1) =2°1 =1, (2°1)°1=1°1=1 (f) 2°(1°2)=2°2 = 2, (2°1)°2 = 1°2 = 2
(c) 1°(2°1) = 1°1 = 1, (lo2)°l = 2°l=l (fir) 1 °(2° 2) = 1 ° 2 = 2, (1 ° 2) "2 = 2° 2 = 2
(d) 1 о (1 a 2) = ι a 2 = 2, (1°1)°2 = 1°2 = 2 (h) 2 ° (2 ° 2) = 2 ° 2 = 2, (2 ° 2) ° 2 = 2 ° 2 = 2
1 1
2 1

30
GROUPOID S
[CHAP. 2
The groupoid of Example 2 is not commutative; for if α is defined by la = 1, 2a = 3 and 3a = 2,
and β is defined by 1/3=2, 2β = 1 and 3/Ϊ = 1, then 1α°/3 = (Ια)/} = 1/ϊ = 2, 1/3 ° α = (1/3)α =
2α = 3. Hence α ° β ¥* β°α. Since the binary composition of mappings is an associative binary
operation (Problem 1.41, page 18), (S,°) is an associative groupoid.
The groupoid in Example 3 is both commutative and associative. a°b =a+b + ab =b + a +
ba = b ° a, since addition and multiplication are commutative binary operations in Q. Also,
а о (b a c) = a ° (b + с + be) — (a + (b + с + be)) + a(b + c + bc) — a + b+ с + be + ab + ac + abc and
(a°b)°c = (a + b + ab)°c — a + b + ab + с + (a + b +ab)c = a + b + ab + с + ac + be + abc.
Using the associative and commutative properties of addition and multiplication in the rationale,
we see a ° (b ° c) — (a ° 6) D c.
In Example б the groupoid (F, +) is commutative because a(a + β) ~ αα + αβ = αβ + αα —
α(β + α), a,fi&F and a& R (here we use the fact that αα, αβ e R and addition is a commutative
binary operation in R). (F, +) is also a semigroup, for a((a + β) + у) = α(α + /3) + αγ = αα +
(αβ + αγ) = αα + α{β + γ) = α(α + {β + γ)) (here we use the associativity of addition in R).
2.7. Construct an example of a commutative groupoid of order 3.
Solution:
Let S = [a, b, c} and the binary operation ° be defined by the multiplication table
abc
a
b
с
(S, °) is clearly a commutative groupoid.
2.8. Show that the set Q* of nonzero rational numbers with binary operation the usual division of
rational numbers, is a groupoid. Is it commutative? Is it associative?
Solution:
If -r and — are any two elements of Q", then -r -*- -j = т~ is a unique element in Q*
о a . b a be
— ¥*0 since a, bf e, d ¥* 0 ). Therefore division is a binary operation in Q*. However, division
be J
is neither commutative (e.g. \^\ = 2Ф^=\^^) nor associative (e.g. % -τ- (1 -=- |) =
£ + i = t*e = (i-i)-i)·
2.9. Which of the groupoids in Examples 1-3, 5 and in Problems 2.7 and 2.8 are finite?
Solution;
The groupoid of Example 1 is clearly finite of order 3.
In Example 2 the set S of all mappings of {1,2,3} into itself contains 27 elements, and so
is finite.
In Example 3, (Q, °) is not finite as there are an infinite number of rational numbers.
In Example 5 the set F is infinite. To show that F is not finite we construct an infinite number
of mappings of R into R as follows. Let p,-: R -> R (i = 1, 2, 3, ...) be defined by rP1 = r for all
rfiei and ipt = 0. Clearly Pi = p} iff i = /. Therefore we have found an infinite number of
different elements in F. Notice the pt are not all the elements of F.
In Problem 2.7 the groupoid has only three elements and is therefore finite.
In Problem 2.8, since there is an infinite number of nonzero rational numbers, Q* is not finite.
a
b
с
b
a
с
с
с
a
(
2.3 IDENTITIES AND INVERSES IN GROUPOIDS
a. The identity of a groupoid
Let G be a groupoid written multiplicatively. An element e in G is called an identity
element of G if
eg = ge = g

Sec. 2.3]
IDENTITIES AND INVERSES IN GROUPOIDS
31
for every g G G, For example in the multiplicative groupoid Ζ of integers, 1 is an identity
element. A less natural example is the groupoid {1,2, 3} with multiplication table given by
2
3
1
3
1
2
1
2
3
The element 3 is an identity element of G since
3. 1 = 1 = 1-3, 3-2 = 2 = 2-3, 3-3
3 = 3-3
One might ask whether or not a groupoid can have more than one identity element.
The following theorem settles this question.
Theorem 2.1: If a groupoid G has an identity element, it has precisely one identity
element. In other words if e and e' are identity elements of G, then e = e'.
Proof; Since e is an identity element of G, ее' — e'. But e' is also an identity element
of G. Hence ее' = e and so e = e'.
It is instructive to reformulate the proof of Theorem 2.1 in a different notation. Thus
we revert to the notation (G, μ) and instead of the multiplicative notation gh we write
(g, Κ)μ. Suppose e and e' are identity elements. Then as e is an identity element, (e, β')μ = e'.
But e' is also an identity element. Hence (e, β')μ — e; and because the image of any element
under the mapping is unique, e = e'.
Problem
2.10. Which of the following groupoids have identity elements?
(i) The groupoid (Z, +) under the usual operation of addition.
(ii) The groupoid of nonzero rational numbers under division.
(iii) The groupoid of complex numbers under multiplication.
(iv) The groupoid of all mappings of {1,2,3,4} into itself under the composition of mappings.
(v) The groupoid with carrier {1,2} and multiplication table
(a)
1
2
1
1
r-l
1
1
2
1
1
2
2
1
2
2
1
2
1
1
(d)
2
r-l
2
2
1
Solution:
(i) (Z, +) has the identity element 0, since 0 + г
г + 0 = ζ for all z ^ Ζ.
(ii) If this groupoid had an identity element e, then e.-i-q = q for all q in the groupoid. In
particular, e -f- e = e and so e = 1. But 1 -τ- 2 = 1/2 ¥■ 2. Hence there is no identity element.
(iii) Recall that a complex number is any number of the form a+bi where а, Ъ е В and
i = V~l. l+Ог is the identity element of this groupoid, since (a+ Ы)(1 + 0г) = α. + Оси" +
Ойг"2 + Ы - а+ Ы = (1 + 0г)(о.+ Ьг).
(iv) The identity mapping ι, defined by ji = J, j e {1,2, 3,4}, is the identity element of the
groupoid since j(aa ι) = (j(r)i = ja = (?Ί)σ = j(iD σ).
(v) Only (d) has an identity element, namely 1.
b. Inverses in a groupoid
If we use multiplicative notation for groupoids, we shall for the most part reserve the
symbol 1 for the identity element.

32
GROUPOIDS
[CHAP. 2
In the multiplicative groupoid of nonzero rational numbers, one talks about inverses;
for example, £ is the inverse of 2, the determining factor being 2x^=1 = ^x2. In
general if G is any groupoid with an identity 1, we term h an inverse of g (h, g G G) if
gh - 1 = hg
Clearly if h is an inverse of g, then g is an inverse of h. Examples follow.
Example 6: Let G = {а, b, с} be the groupoid with multiplication table
a b с
a
b
Example 7:
a
b
с
Ь
с
α
С
α
b
Then α is the identity element of G, as inspection of the table shows. Furthermore,
be = a = cb and αα — α = αα, so that с is an inverse of 6, b is an inverse of c,
and α is its own inverse, a, b and с have no other inverses.
Let G be the groupoid of mappings of {1, 2, 3} into itself. Then the identity mapping
ι defined by ji = j (j — 1,2,3) is the identity element of the groupoid (see
Problem 2.10(iv)). Now let σ e G be defined by U - 2, 2σ = 3, 3σ = 1. Then τ G G,
defined by lr = 3, 2r = 1, Зт — 2, is an inverse of σ because
1σ о г - (1σ)τ = 2r "- 1, 2σ ° г - (2σ)τ = 3r = 2, 3σ о г = (3σ)τ = lr = 3
which implies στ = ι. Similarly, τσ = ι.
Unlike identities, inverses in groupoids are not always unique. For example, let
G = {1,2,3,4} be the groupoid with multiplication table
12 3 4
2 111
3 114
4 2 3 4
Here 1 is the identity element in G. Moreover, 2 · 2 = 1 = 2 · 2 and 2 · 3 = 1 = 3 · 2; thus
2 has two inverses, 2 and 3. Notice that in this groupoid, 4 is not an inverse of 2 even
though 2 ■ 4 = 1. The definition of an inverse requires both 2 ■ 4 and 4 * 2 to equal 1.
Problems
2.11. Consider the groupoids in Problem 2.10 which have identity elements. What elements in each of
the groupoids have inverses?
Solution:
(i) Any integer г has an additive inverse, namely —г; for г + (—г) = 0 = (—ζ) + я.
(iii) All nonzero elements in the groupoid of complex numbers under multiplication have inverses;
i.e. if α + bi (a and b not both aero) is any element in the groupoid, then
a + bi
f 1 \ia-bi
\a + bij\a — bi
a — Ы
a2+ b2
ai + {,2 a1 _|_ &2
is an element in the groupoid and —--^(a+bi) = 1 = (a+bi)——-r-.. (0 + 0i) has no in-
a + bi a + fri
verse, since (0 + 0i)(a + bi) = (0 + 0ΐ).

Sec. 2.4]
SEMIGROUPS WITH AN IDENTITY ELEMENT
33
(iv) Only mappings τ which are one-to-one and onto have inverses. Say τ is a one-to-one mapping
of {1,2,3,4} onto itself, defined by lr = a, 2T = b, Зт = с and 4r = d, where a,b,c,d are
the elements of {1,2,3,4}. We define the mapping σ which will be an inverse of τ, by ασ = 1,
6σ = 2, ca = 3 and da = 4. σ is a mapping of {1,2,3,4} onto {1,2,3,4}, since {a,b,c,d} =
{1, 2,3,4}. 1τ°σ = (1τ)σ = ασ = 1, 2τ °σ = 6σ = 2, 3τ ° σ = βσ = 3, and 4τ ° σ = <2σ = 4. Hence
τ ο σ = ι. We must also show σ ° τ = <. Now ασ ° τ = lr = а, Ъа ° τ = 2т = Ъ, со ° τ = Зт = с,
йсг от = 4т = d. As {α, b, с, й} is {1, 2, 3,4}, σ ° τ = ι. If τ is not one-to-one then there are at
least two elements, a and b (a ¥= b) in {1,2,3,4} which are mapped onto the same element,
c, by τ, i.e. α,τ = с and Ът = с. Now if σ is an inverse of τ, then α(τ°σ) = a and 6(τ°σ) = b
or (йт)а = ca = a and (6τ)σ = σσ — b; thus σσ = a and ca = b. But under a mapping
each element has a unique image. Hence we have a contradiction. So τ has no inverse.
(v) Both 1 and 2 have inverses since 1 · 1 = 1 = 1 ■ 1 and 2 · 2 = 1 = 2 · 2.
2.12. Find the identity of the groupoid (Q, °) where a°b = a + Ъ + ab (see Example 3, page 27).
What elements have inverses?
Solution:
0 is the identity of (Q, °), since 0°» = Ο + α+Οα = a and αο0 = α+0 + 0α=α. To find
an inverse for an element a e Q, we must find an χ G Q such that a°x — Q = x°a. Now
a° χ = a + χ + ax, so that χ must satisfy the equation a + χ + ax = 0. If a = —1, we obtain
—1 + χ — χ = — 1 = 0; thus if a = —1, there is no χ such that tt°a; = 0. If a ¥= —1, the
equation α + ж + αχ ■=- 0 can be solved, giving χ = — ——; thus α ° τ—:— = 0. Since (Q, °) is
1 -r a L ~r a
commutative, -r—,— ° a = 0 so that 7—;— is an inverse of a. Hence all elements of Q, except —1,
'1 + a 1 + α ■*, ν ,
have inverses.
2.13. Find the identity of the groupoid (F, +) in Example 5 and show that every element has an inverse.
Solution:
The mapping ω: R -> R defined by τω = 0 for all r G R, is the identity of (F, +). For
α(α + ω) = aa + Осе = aa + 0 = aa = 0 + aa = αω + α« = α(ω + α) for all α ε i? implies » + ы = а =
ω + α. If /3 is an inverse of a, then α + β = ω and α(α + /3) = αω for all ι£δ. But α(α + β) —
aa + αβ and αω = 0. Thus aa + αβ = 0 or — (αα) = α/δ. Consequently if β is an inverse of a,
the image of aS R under β must be the negative of the image of a under a. We therefore define
an inverse β for the mapping α by αβ = —(aa), ней. β is a mapping of R into R, since — (aa) is
a unique element of R. Furthermore, a(a + β) — aa + αβ — aa + (—(aa)) = 0 = αω implies
α + /S = ω. (F,+) is a commutative groupoid. Hence α + β = ω = β + α and β is an inverse of a.
2.4 SEMIGROUPS WITH AN IDENTITY ELEMENT
a. Uniqueness of inverses
Suppose G is a groupoid with an identity 1 and suppose h is the inverse of g: gh = 1 = /13.
It is tempting to employ the notation used when dealing with real numbers and write g~l
for an inverse of g. The trouble with this notation is that in a groupoid an element may
have more than one inverse, as we have already seen in Section 2.3b. However, the
associative law on a groupoid prohibits this as we see from
Theorem 2.2: Let G be a semigroup with an identity element 1. If g G G has an inverse,
it has precisely one; i.e. if h and h' are inverses of g, then h = hf.
Proof: h =- hi (since hi = h for all h G G)
= h{gh') (ghf = 1, since h' is an inverse of g)
— (hg)hf (by associativity)
= lh' (hg = 1, since h is an inverse of g)

34
GROUPOIDS
[CHAP. 2
Theorem 2.2 entitles us to denote the inverse of an element g in a semigroup, written
multiplicatively, by g~l. Note that if g and h have inverses, then gh has an inverse, namely
h~ig~1. (Notice the reversed order.) For,
(gh)(h~ig-i) = {{gh)h-l)g~' = (g(hh^))g-' = (gl)g-* = 00 -1 = 1
Similarly, (/г.-1 g~l)(gh) = 1.
Problems
2.14. Let G = {1,2,3,4}. The binary operations on G given by the following tables make G into a
groupoid.
1234 1234 1234
(a)
1
2
3
4
12 3 4
2 4 2 4
3 2 14
4 4 4 4
(b)
12 3 4
2 14 3
3 4 12
4 2 3 1
(«)
1
2
3
4
1111
2 2 2 2
3 3 3 3
4 4 4 4
Which of the groupoids are semigroups? Which have an identity? Which elements have inverses?
Solution:
(a) In order to see that in this case G is a semigroup, we must check associativity, i.e. we must
show a(bc) = (ab)e for every a, b.cGG. Notice that when either a, b or e is 1, a{be) is
clearly equal to (ab)e; e.g. if с = 1, (ab)l = ab — а(Ы). Since Ад = 4 = дА for any g & G,
then (а6)« = 4 = a{be) if either a, b or e is 4. Therefore we need only check the products when
a, b and с have values 2 or 3. If two of the three elements a, b, с are equal to 2 and the other
is equal to 2 or 3, then (ab)e = 4 = a(be) because 2-2=4 and 2·3 = 2 = 3·2. The
following calculations take care of the remaining cases:
3(3 ■ 3) = 3 · 1 = 3 = 1 ■ 3 = (3 ■ 3)3
2(3 · 3) = 2 ■ 1 = 2 = 2 · 3 = (2 · 3)3
3(2 · 3) = 3 · 2 = 2 = 2 · 3 = (3 · 2)3
3(3 ■ 2) = 3 · 2 = 2 = (1 ■ 2) = (3 · 3)2
1 is the identity element of G. The only elements which have inverses are 1 and 3; these inverses
are unique.
(b) The associative law does not hold, since 4(2 ■ 3) = 4 ■ 4 = 1 and (4 ■ 2)3 = 2 ■ 3 = 4. 1 is the
identity, and 1, 2, 3 and 4 have inverses. Notice that the inverses are unique even though G is
not a semigroup.
(c) G is a semigroup. Associativity follows from the fact that ab = a for all a,b e G; hence
(ab)c = a = a(bc) for any a, b, с in G. G has no identity element; therefore no element has an
inverse.
2.15. Let G be the groupoid with carrier Q, the set of rational numbers, and binary operation ° defined by
a°b = a+b — ab. Is the groupoid (Q, °) a semigroup? Is there an identity element in (Q, °)? Which
elements of the groupoid have inverses?
Solution:
Notice a+b — ab is a unique rational number, so that (Q,°) is a groupoid. Now if a,b,c are
any elements in (Q, °),
(a°b)°c = (a+b — ab) °c =■ a+b — ab + c — (a + b — ab)c
= a + b + с — ab — ac — be + abc
and
a°(b°c) = а°(Ъ + в—Ъв) = a + b + с — be. — a(b + в—be)
= a + b + с — be — ab — ac + abe
Hence (Q,°) is a semigroup. The identity of (Q,°) is 0 since a°0 = a+ 0 — аО=а and 0°a =
0 + a — Oa = a. Using an analysis similar to that in Problem 2.12, we find that for α. φ 1 the
inverse of a is ■
and that а = 1 has no inverse. To check that
a-1
is the inverse of a,

Sec. 2.4]
SEMIGROUPS WITH AN IDENTITY ELEMENT
35
a· a ( a \ . — a(a — 1) _
а о j = a + — a ) = a + *—~ = 0
a. — 1 a— 1 \a —1/ a—1
Similarly, =■ ° a = 0.
a— 1
2.16. Let G be the groupoid with carrier Q and binary operation ° defined by a° b = a— b + ab. Is the
groupoid (<J, °) a semigroup? Is there an identity element in (Q, °)? Which elements of the groupoid
have inverses?
Solution:
(Q,°) is clearly a groupoid.
α, ο (6 о с) = a ° (Ь — с + Ьс) = а — (Ь — с + be) + a(b — с + be) = a — b + c — bc+ab — ac+ abc
and
(α ο 6) о с — (a—b + ab) ° с = a — b + ab — c+ (a — b + аЬ)с = α — ί> + ab — « + ас — be + abc
These two expressions are not the same for all values of a, b and c. For example, (0 ° 0) ° 1 =
0ol=—ι while 0°(0°1) = 0°— 1 = 1. Hence (Q,°) is not a semigroup. Furthermore, (Q,°)
has no identity element; for if e were an identity, then e°l =1 and e°0 = 0, since 1,0 £ Q.
But e°i=e — l + e=l implies 2e = 2 or e — 1, and Q=e°o=e —0+0 implies e = 0.
This is clearly impossible. Therefore (Q, °) has no identity element.
2.17. If Q is replaced by Z, the integers, as the carrier for the groupoids in Problem 2-15, are the solutions
the same? (Hard.)
Solution:
(Z,°) with ° defined as in Problem 2.16 is a semigroup with an identity, since the
argument for the associativity depended only upon the associativity and commutativity of addition
and multiplication in Q. These laws also hold in Z. The same is true for the proof that 0 is the
identity of (Z, °). However, the inverse of an element a(¥= 1) e Ζ would be _ 1 £ Z. Our problem
then is: for which integers a ¥· 1 is r an integer?
a— 1
Let —ZT = r» an integer. Then a — r{a— 1). Clearly r = 1 is impossible, (a) Assume first
that a > 1. Then r must be positive, and so r — 2. Hence α^2(α —1) and thus Ъ — а— 2.
Therefore a = 2. If α = 2, then ——- = 2 is an integer, (b) Now assume a — 0. If α = 0, then
α α —1
——τ- =0 is an integer. If a < 0, r must be positive and — 2. Then a = r(cu — 1) — 2(a — 1) and
0 — a — 2, which is impossible. Thus 0 and 2 are the only elements with inverses.
2.18. Let G be the mappings of P, the positive integers, into P. Determine whether 6 is a semigroup
with an identity element if the binary operation in Ρ is (i) the composition of mappings, (ri) the
addition of mappings, where a + β is defined by a(a. + β) = αα+ αβ, α,β £ G and »e P.
Solution:
(i) The composition of mappings is an associative binary operation (see Problem 1.41, Page 18).
Then G with the binary operation of composition of mappings is a semigroup. The mapping
ι defined by i°i~ j for all j GP is the identity of G; for if τ € G, then /((о r) = (jt)r —
jr - (jr)i = j(r ο ι).
(ii) In Problem 2.6 we showed that the addition of mappings in the set F of all mappings of R
into R is an associative binary operation. The argument here is similar. Thus G is a semigroup.
И a were an identity element in G and if (9€G, then a + β = β. Thus if j € P, then
j(a + β) = ja + i/3 = j"£; hence jia = 0. But «eG, a:P~*P and, since 0ЙР, this is a
contradiction. Thus (G, +) has no identity. (Compare with Problem 2.13.)
2J9. Let a, b, с be three elements in a semigroup which have inverses. Prove that a(bc) has an inverse
and that this inverse is (c~1b~1)a~i.
Solution:
We need only verify that
{a(6c)}{(e-»6-i)a-i} = а{(Ъе)[(е-1Ь-*)а-*]} = a{[(bc)(c-lb-i)]a,-1} = alia'1} = aa~l = 1
and similarly that {(с~1Ь~1)а~1}{а(Ьс)} = 1 to prove the result. Note our use of the associative law.

36
GROUPOIDS
[CHAP. 2
b. The semigroup of mappings of a set into itself
A particularly important semigroup is the set Mx of all mappings of a given non-empty
set X into itself, where the binary operation is composition of mappings. We repeat the
definition of the composition of mappings in this special case. Suppose μ: X -*■ X and
γ: X-*■ X, i.e. μ, γ e Μ χ. We define μ ο γ to be the mapping of X into X given by
χ(μογ) = (Χμ)γ for all x^X
It is clear that о is a binary operation in Mx. We shall use the multiplicative notation μ· γ,
or simply μγ, instead of μ ο γ. We now show Mx is a semigroup with an identity.
Theorem 2.3: If X is any non-empty set, Mx is a semigroup with an identity element.
Proof: We begin by proving Mx is a semigroup. Let F,y,Pe -Mx and let χ G X.
Then using the definition of composition of mappings,
яМгр)) = (3μ)(γρ) = ((3μ)γ)Ρ = {χ(μγ))Ρ = Χ((μγ)ρ)
Since χ is any element of Χ, (μγ)ρ and μ(γρ) have the same effect on every element of X.
Thus (μγ)ρ = μ(γρ), and so Μ χ is a semigroup.
To show that Mx has an identity element, let ι: X -*■ X be defined by Χι = χ for all
xGX. Then if μ e Mx,
Χ{ιμ) = (#ι)μ = Χμ - (Χμ)ι = Χ(μι)
Thus ψ = μ = μι for all μ G -Μχ; hence ι is an identity element of Mx. The proof of the
theorem is complete.
Not every element in Mx necessarily has an inverse. For example, if X = {1,2,3} and
σ e Mx is defined by 1σ = 1, 2σ — 1, 3σ = 1, then σ has no inverse; for if σγ = ι, we have
1 = 1ι = 1(σγ) = (1σ)γ — 1γ and 2 = 2t = 2(σγ) = (2σ)γ = 1γ, so that 1 would have two
distinct images under y, which contradicts the assumption that γ is a mapping. The subset of
Mx consisting of all those elements which have inverses is very important. We characterize
these elements in the following theorem.
Theorem 2.4: An element in Mx has an inverse if and only if it is one-to-one and onto.
Proof: Suppose μ has an inverse γ; then μ is onto. For if χ €. X, then χ =Χι = χ(γμ) —
{χ-γ)μ and Χγ €. X is a preimage of x. Moreover, μ is one-to-one. For if % = у μ, then
(Χμ)γ = Χ(μγ) = Χι - Χ and (ΐ/μ)γ = ΐ/(μγ) = ΐ/ι = ΐ/
Therefore, since γ is a mapping, #μ = г/μ implies χ = у; in other words, μ is one-to-one.
To prove the converse, we assume μ is one-to-one and onto. Define γ, which will be
shown to be an inverse of μ, as follows: if iel, we define Xy = у where у is that element
of X which is the preimage of χ under μ. To check that the definition of γ is meaningful,
observe that as μ is onto there certainly is at least one element у €. X such that yμ ~ x.
But μ is one-to-one, i.e. distinct elements have distinct images. So у is the unique element
such that уμ = χ. To conclude we show γ is the inverse of μ. Let χ G X and % = y.
Then χ(μγ) = (Χμ)γ — уу — х by the definition of γ, and so μγ = t.
Since μ is onto, each χ €. X must have a preimage у G X, i.e. у μ = χ. Then
«Ы = ШЫ) = {{νμ>Ύ)μ = (0(μγ))μ = № = ^ = χ
and so у μ = ι. Thus γ is the inverse of μ and the proof of Theorem 2.4 is complete.

Sec. 2.4]
SEMIGROUPS WITH AN IDENTITY ELEMENT
37
с Notation for a mapping
When X is finite, say X = {au . . ,,Οη}, there is a convenient way of denoting any
m e Μχ, namely
_ /«I ■■■ an
σ —
\α%σ ... αησ
That is, we place below each element cu of X (i — 1,2, ..., n) its image under σ; e.g. if
X — {1,2,3} and σ e X is defined by 1σ = 1, 2σ ~ 1 and 3σ = 2, then σ is represented by
1 2 3N
1 1 2
Notice that every element of X has a unique image under an element σ in Mx; therefore the
top row (ai az us " ' a*) will contain all the elements of X and under each element will
appear its unique image under σ. All elements of X need not appear in the bottom row,
as we see from the example above.
rroblems
UL (a) Write the element σ in Mx, X = {1,2,3, 4, 5,6}, in the notation introduced above, when σ is
denned by
(i) 1σ = 1, 2σ = 4, 3σ = 5, 4σ = 6, 5σ = 2, 6σ = 6
(ϋ) 1σ =1, 2σ = 1, 3σ = 1, 4σ = 1, 5σ = 1, 6σ = 1
(ίϋ) 1σ = 6, 2σ = 5, 3σ = 4, 4σ ~ 3, 5σ = 2, 6σ = 1
(6) What elements of Λίχ, Χ = {α, 6, ο,ώ,β}, are represented by the following?
/α Ь с d е\ /α Ь с d e\ /abode
\Ь a e d e/ \a e d с b J \e a e b a
Solution:
/1 2 3 4 5 6\ /1 2 3 4 5 6\ /1 2 3 4 5 6\
(й) (1) (l 4 5 6 2 б; (П) (ι 1 1 1 1 l) (Ш) (б 5 4 3 2 J
(6) The mappings defined by
(i) a -» 6, b -» а, с -» с, d -» d, e -» e
(ii) α-» α, 6 -» e, e -» d, d -» e, e -» 6
(iii) a -» e, 6 -» α, ο->ο, d-»6, e -» a
UL Exhibit all elements of Mx when (i) X = {1}, (ii) X = {1,2}, (iii) X = {aua2}.
Solution:
(i) There is only one element in Af{1), namely t г 1 -» 1, the identity mapping.
(ii) The following are the elements of Af{1>2) : ( „)»(, ,Ι'ίο ι/'ίο 2/
/a, »a\ /»i α,\ /«ι α,Ν /„, „2х ^ rf
\»i «2/ \ni »i/ \«2 4/ U aj <01·^
IS. Write out the multiplication tables of the three semigroups in Problem 2.21(i) and (ii).
(i)
/1 2\
\U) The identity of M{U2y is ι =J ).

38
GR0UP01DS
[CHAP. 2
Let
1 2
1 1
σ2
1 2
2 1
and σ3 —
t σι σ2
Then the multiplication table is
"a
ι
αι
σ2
σ3
σ1
σι
σ1
σ1
σ2
"3
ι
°1
σ3
σ3
σ3
σ3
The multiplication is calculated as follows. Since ι is the identity, ι leaves every element of
M,up. unchanged; hence the first row and first column are easily -written down, Since ΐαλ = 1,
j = 1, 2» and if σ £ ^n,2) > ^en ^(σ<»ι) = (^σ)αχ — 1; thus ασχ = σχ. Hence the second column
consists of σχ.
Similarly σσ3 = σ3, so the last column consists of σ3. We must still calculate αλσ2, σ^
and σ3σ2- Now each element of {1,2} is taken to 1 by σ1( and 1 is taken to 2 by σ2, so агог
takes every element to 2; hence σλσ2 — σ3. Now 1σ2«2 = (1σ2)σ3 — 2σ2 = I and 2σ2σ2 =
(2σ2)σ2 = 1σ2 = 2; hence «2^2 :~ '■ If ? e (1> 2}, then jV3 = 2 and so ί'σ3σ2 = 2σ2 = 1. Thus
»3σ2 " σ1·
2.23. Let X = {1,2, .. .,n} and <r € ΛΓΧ. Show that σ has an inverse if and only if the bottom row in
(1 2 и \
), contains every element of -ЯГ once
_ ., 1σ 2σ ■ · ■ ηα/
Solution:
By Theorem 2.4, if σ has an inverse then σ is one-to-one and onto. Hence {1σ, 2α, 3σ, .. .,гш} =
{1,2,..., η} and the bottom row of ( ''' J contains all elements of X once and only
\1σ 2σ ... ησ/
once. Conversely if the bottom row of the representation has all the elements of X once and only
once, then each element in X has a unique image under σ, namely the entry under that particular
element. Therefore σ is one-to-one. σ is also onto, for if j£X it must be one of the elements in
the bottom row and j is then an image of the element of X appearing above it in the representation.
As σ is a one-to-one and onto mapping it has, by Theorem 2.4, an inverse.
2.24. List all elements of M,l2. which have inverses.
Solution:
Problem 2.23 shows us that we must find all representations of elements of M,lt2\ with bottom
/1 2\ ' /1 2"
row containing 1 and 2. Using the result of Problem 2,21(ii), we have ( _ „ ) and ( „ _ ) as the
,1 2/ \2 I,
only elements of Af{li2} wnich have inverses. In the notation of Problem 2.22, ι and σ2 are the only
elements which have inverses.
2.25.
2.26.
List all elements of ΛΓ„ 2 „ which have inverses.
Solution:
The possible representations of elements of M,l2 3- which have inverses are
<r4 =
2 3
2 3
2 3
σ2 -
2 3 1
"5
/1 2
\з 1
2 3
3 2
2 3
2
σ3 -
σβ
2
1
2
2
What are the inverses of the elements in Problem 2.25?
Solution:
Theorem 2.4 explains how to find the inverse of a mapping which is one-to-one and onto. For
example, to find the inverse of σ3, we note σ3 takes 1 -+ 2. Hence σ™': 2 -* 1. σ3: 3 -* 3, hence
σ""1: 3 -+ 3. But then σ3 is its own inverse. Similarly σ"1 = σχ, σ~ι = σ2, α~ι = <r5, σ"1 = σ4,
σ~ι = σβ.

Sec. 2.4]
SEMIGROUPS WITH AN IDENTITY ELEMENT
39
2.27. Prove that the subset of elements of Mx which have inverses is a semigroup with identity under
the usual compositions of mappings.
Solution:
Let S be the elements of Mx which have inverses. Is °, the usual composition of mappings, a
binary operation in SI In other words, is (α, β)-*α°β a mapping of S X S-> S4 If a ° β £ *S,
the answer is yes- So we ask: if a, β have, inverses, does a° β"! Note that [a ° β) ° (β~ι ο α-1) =
a ° (β ° β~~ι) ° α_ϊ = α° ι°α~ι = ι. Hence α°β has the inverse β~1°α~1. As ι is its own inverse,
ieS. As Mx satisfies the associative law, so does S. Hence S is a semigroup.
d. The order in a product
There is one way in which the associative law makes it easier to work in a semigroup
than in a non-associative groupoid. Suppose S is a semigroup and let ab a%, a3 € S. There
are two ways in which one can multiply αϊ, α·ι and α3 together (in this order): (αια2)α3 and
αι(α.2α3). The point of the associative law is that these products coincide. Suppose now
ui, a-i, ав, «к € S. Then we can multiply ab aa, a3, a4 together (in this order) in the following
five ways:
αι(α2(α3α4)), (αια2)(α3α4), ((αια2)α3)α4, (αι(α2α3))α4, iiH^a^ou)
It is conceivable that some of these products give rise to different elements of S. However,
the associative law, which of course involves products of only three elements, prohibits this.
To see this consider first (aia2)(a3uk). By the associative law, (αιθ2)(α3α·ι) = αι(α2(α3α4));
hence the second product coincides with the first. In fact all of the products equal the
first. As a second illustration consider ((aia2)as)ai. Here we have, as desired,
((αια2)»3)θ4 = (αι(α2α3))α4 = αι((α2α3)α4) = αι(α2(α3α4))
In general, we have
Theorem 2.5: Let S be a semigroup and let ah a2, ..., a„ G S. Then any two products of
ui, a2, ..., an coincide, when the at appears first in each product, oa
second, ..., and dn last.
Proof: Assume the contrary, that not all possible products of αϊ, ..., a„ in that order
are equal. We may assume that η is the first integer for which two different products give
rise to different elements. Let χ and у be these two different products. Now χ = uv and
у = WiVi for some u, ν and some U\, V\. Suppose и is the product of the elements αϊ, . .., ar
and ν the product of the elements ar+\, . . .,α™, while u\ is the product of α,ι, .. .,as and ^i
is the product of as+1, ..., an. Without loss of generality we may suppose that s — r. If
s = r then и = U\ and ν — vt, since η is the first integer for which there exist two unequal
products of the same elements. If s < r, then и = (αι. . .as)(as+1. . .ar) while ^i =
(fls+i.. .ar)(ar+i.. .dr.). Hence χ = uv = {(ai.. .as)(as+i.. .ar)}(ir+i.. -an) while у = UiVi =
(a-ι.. .as){(as+i.. .ar)(ar+i.. .a™)}. By the associative law for the three elements (αι.. .as),
(Os+i.. ,ar) and (ar+i- - .cim), we have x — y, contradicting the assumption that not all
possible products are equal. Hence the result follows.
It follows from Theorem 2.5 that in a semigroup, if we are given the order of a product,
the bracketing is immaterial. Thus we write simply aiCfe. - .йп, without brackets, for the
product of αϊ,a2, . . .,dn in this order. For example, if m is any positive integer, we write
α·α a for the product of m a's; a useful abbreviation for such a product is am. If η
is a second positive integer, then am'an = am+n since α"'·αη is simply the product of m + n
a's. Similarly, (am)n = amn.

40
GROUPOIDS
[CHAP. 2
For example, if S = {1,2,3} is the semigroup with binary operation given by the
following table, then l3 = 3, 23 = 3, 33 = 3.
2 3 1
[ 3 1 2
12 3
2.5 HOMOMORPHISMS OF GROUPOIDS AND CAYLEY'S THEOREM
a. Definition of a homomorphism
Before we give the formal definition of a homomorphism of one groupoid into another,
we will give an example. Let (R, ·) be the groupoid of positive real numbers with the binary
operation of ordinary multiplication. Let (R, +) be the groupoid of real numbers with
binary operation the usual addition inside R. Let us define a mapping 0: R-* R by
«e = logio«. Recall that logio(xy) = logioa; + logioy; hence (ху)в = χθ+ув. θ is an
example of a homomorphism. The formal definition is
Definition: A homomorphism from a groupoid (G,a) into a groupoid (Η,β) is a mapping
θ: G -* Η which satisfies the condition
({9и9г)«)в = (0i0,ffa0)|S
for all 0ι,02 in G.
Usually groupoids are written in multiplicative notation. The definition then takes the
form: A homomorphism of (G, ·) into (H, ·) is a mapping θ : G -* Η such that
(0i0s)<? = 0100*0 {2.7)
for all 0i, 02 in G. (2.7) is often expressed as: "Θ preserves multiplication."
Problems
2.28. Let G be the semigroup of integers under the usual addition of integers and let Η be the semigroup
of even integers under the usual addition. Verify that the mapping θ: G -> Я defined by
θ: g -» 2g for all g £ G is a homomorphism of G into H.
Solution:
First we must check to see if θ is a mapping. Since 2g is a unique integer, в is clearly a
mapping. Let gb gz G G. Then дгв + g%B — 2g1 + 2g% — 2{gt + gz) = (fir, + gs)e. Therefore θ
preserves multiplication and is a homomorphism.
2.29. Let G be the semigroup of integers under the usual multiplication of integers, and let Η be the
semigroup of even integers under the usual multiplication. Is σ: G -> ff defined by σ: g -» 1g
for all j£G a homomorphism of G into H?
Solution:
As in the preceding problem, σ is a mapping. But σ is not a homomorphism; for if glt дг G G,
then (g^g^o — 2gtg2 and gltrg2a = 2дг2д2 = Ag^^,. Hence (β$2)σ Φ (0i<r)(02<O for all ад^б.
2.30. Let (G,+) and (H,+) be the semigroups of Problem 2.28. Find a homomorphism of (G,+) into
{H, +) which is not equal to 9.
Solution:
Define т„: G -» Η by τη: g -* ng where и is a fixed even integer. тп is a mapping of G into H,
since ng G Η for any even integer η and «i/ is unique. For each η, τη is a homomorphism because

Sec. 2.5] HOMOMORPHISMS OF GROUPOIDS AND CAYLEY'S THEOREM 41
ffi, 02 e G implies (дг + д2)тп = n{g1 + g2) = ng1 + ng2 = glTn + g2rn. τη Φ θ, since Iff Φ lrn if
η Φ 2. Therefore there is an infinite number of different homomorphisms between (G, +) and (H, +).
Notice rn is an onto mapping if η = 2, because any element in Η is of the form 2q, q an integer,
and <jr2 = 2?. But when и Ф 2, τη is not onto since 2 has no preimage under т. For if qrn = 2,
?eG, then nq = 2 or g = 2/w (if и ^ 0) and 2/w £ G.
• Η defined by a : g -* g2, for
2.31. Let G and Η be as in Problem 2.28. Verify that the mapping a: G
all g £ G, is not a homomorphism.
Solution:
Let gug^^G. Then (iri+fir2)<r = (0i + fi^)2 = 0, + 02 + 20102 and 9i<r + 02"
Hence σ is not a homomorphism for (g1 + g2)a Φ gxa + g^, for all gu g% € G.
C? + 02-
2.32. Is the mapping σ of G into H, G and Η as in Problem 2.28, defined by да
homomorphism?
0 for all g e G, a
Solution:
If ffi, 02 e Cr, then (flfj + fl^V
0 and fifja + flf^ = 0+0 = 0. Hence σ is a homomorphism.
2.33. Let G be the semigroup of positive integers Ρ under the usual addition, and let Η be the semigroup
of positive integers Ρ under the usual multiplication. Show that the mapping ν : G -» Η defined by
gv = 2s for all g £ G is a homomorphism.
Solution:
j/ is clearly a mapping. Let g\,Ui
a homomorphism.
e G. Then (flfi + ir3h = 29l + fl2 = 2Я1292 = flfl4flf24. Hence ч is
12 3
2 3 1
3 12
с
Ь
с
α
b
а
Ъ
а
Ъ
iM. Let G = ({1,2, 3},α) and Η — ({α, b, с), β) be the groupoids with binary operations a and β defined
by the multiplication tables
12 3 a b с
1
a: 2
3
Which of the following mappings are homomorphisms?
(a) 1 -* a, 2 -» b, 3 -» с (of) 1 -» b, 2 -» с, З -» с
(b) 1 -» a, 2 -» a, 3 -> a (e) 1 -» b, 2 -» b, 3 -» b
(c) 1 -» a, 2 -» b, 3 -» b (/) 1 -» c, 2 -» a, 3 -» b
Solution:
We use <r to indicate the mapping in each case.
(α) 1σ2(τ = ab = a and (1 · 2)<r = 2σ = b. <r is not a homomorphism. (b) l<r2<r = αα = с and
(1 · 2)<r = 2σ = α. σ is not a homomorphism. (c) 1σ2<τ = ab = a and (1 · 2)σ = b. σ is not a
homomorphism. (tit) 1σ3<τ = be = α and (1 ■ 3)σ = 3σ = с. σ is not a homomorphism. (e) <r is a
homomorphism since the image of 1, 2 and 3 is b, so that (ί/)σ = b for any i, j t= {1, 2,3), and iajtr = bb = b.
(/) 1σ2(τ = ca= с and (1 · 2)σ = 2<r = α. <r is not a homomorphism.
ZJ5. Let G be the semigroup of positive integers under the usual addition. Determine which mappings
of G into G are homomorphisms:
(i) a: w-»2w+l, (ii) a: w -» 2w2, (iii) σ:η-*ί.
Solution:
(i) (и, + w2)<r = 2(и1 + иг) + ! and и1°г + И2СТ = 2wj + 1 + 2w2 + 1 = 2(w, + w2) + 2· Hence
(«! + η2)σ Φ ηλσ + η^σ, and so σ is not a homomorphism.
2 2
2Wj + 2w2 + 4пгП2. ηλα + щв
η 2
2wj + 2и2. Then nja + n2tr Φ
(ii) («j + n2)a = 2(Wj + w2)2
(Wj + η2)σ. an(i so σ is not a homomorphism.
(iii) («i + %)<r = 1. Ща + η^σ = 1 + 1 = 2, Thus ηλσ + η2σ ^ («ι + «г)0' ап<^ hence σ is not a
homomorphism.

42
GROUPOIDS
[CHAP. 2
b. Epimorphism, monomorphism, and isomorphism
Three special types of homomorphism arise naturally.
1. A homomorphism of groupoid G into groupoid Я may be an onto mapping.
2. A homomorphism of a groupoid G into a groupoid Я may be a one-to-one mapping.
3. A homomorphism of a groupoid G into a groupoid Я may be both onto and one-to-one.
We give these three types of homomorphisms special names.
Definition: Let θ be a homomorphism of a groupoid G into a groupoid H. Then
1. θ is called an epimorphism if θ maps the carrier of G onto the carrier of
Я, i.e. G6 = Я. (See Section 1.3a, page 12, for the definition of GO.)
2. θ is called a monomorphism if θ is a one-to-one mapping of the carrier of
G into the carrier of Я.
3. θ is called a isomorphism if 0 is both an epimorphism and a monomorphism,
i.e. θ is one-to-one and onto.
If there is an isomorphism from groupoid G onto the groupoid H, then we say G and
Я are isomorphic, or G is isomorphic to Я, and write G = H.
Problems
2.36. Let G be the groupoid of integers with addition as binary operation, and Η the even integers with
addition as the binary operation. Let rn for η an even integer be the homomorphism (Problem 2.30)
defined by дтп = ng, for g £ G. When is тп an isomorphism, monomorphism or epimorphism?
Solution:
If η ι*· 0, rn is one-to-one since дтп = д'т„ implies ng — ng' and во д = gr'. ru is not one-to-
one, so it is not a monomorphism. If тп is onto, there exists g £ G such that grrn = ngr = 2. Then
gf = 2/w and η = ±2. Hence r±2 are the only epimorphisms. Thus r±2 are isomorphisms and rn
is a monomorphism when η Φ 0.
2.37. G and Η are finite groupoids and \G\ ?* |H|. Show that G cannot be isomorphic to H.
Solution:
Let e:G-*H be an isomorphism and let glf ,gn be the (distinct) elements of G. Then
gte, g2e, ..., £rn» are distinct (since θ is one-to-one) and are all the elements of Η (since θ is onto).
Hence \H\ = n, which contradicts \G\ ¥* \H\. Thus there exists no isomorphism в : G -» H.
2.38. Prove that if G,H,K are groupoids, then: (i) G s· G; (ii) if G^H, then if ss G; (iii) if G a if
and if =t K, then G ^ K. In other words "se" is an equivalence relation. (Hard.)
Solution:
(i) Let ι: G -» G be the mapping defined by ffi = g for all gr 6 G. t is a one-to-one epimorphism.
Hence it is an isomorphism, and so G = G.
(ii) Let a = G -» Η be an isomorphism. Then we define a mapping β: Η -* G as follows: Let
he H. As α is one-to-one and onto, there exists a unique g 6 G such that gra = A. Put
A/J == ff. Note that ft/} α — ft.
Now /} is onto G, for if g G G, ga = he Η and Κβ - g by definition. Also /J is one-to-
one; for if кф — кф, then кфа = Α2/}α and во kt — ft2.
Finally β is a homomorphism. Let hi.h^GH. Suppose g^a = ftlF fif2a = ^2- Then
&ιβ = Si, A2;8 = ff2· Note that {g^a = д^дца = кгк2. Hence (ft^)/} = вгг^2 = ΚιβΚ2β.
(iii) Let a: G -* Η and β : Η -* Κ be isomorphisms. Let γ = α/}. We shall prove that γ is an
isomorphism. First, γ : G -» K. Secondly, γ is onto; because if ft G K, there exists к G Η
such that fc/ϊ = ft, and there exists g e G such that gr<* = A, so £f (<*/}) = ft/ϊ = ft. Next γ
is one-to-one; for if £Τ,γ = £Τ2γ, (&ι«)β — {('2α)β> and as β is one-to-one, йГщ = дга, which
implies, since α is one-to-one, gy — g2- Finally, we must show that γ is a homomorphism.
(0\92ίΊ = ((Βιθώ<*)β ~ (9ι«92<*)β = (3\α)β{θ2θ)β = 0ι(α/})«τ2(«/ϊ) = giYffiy. Hence G = K.

Sec. 2.6] HOMOMOEPHISMS OF GROUPOIDS AND CAYLEY'S THEOREM 43
2.39. (a) Is the semigroup of integers under the usual addition isomorphic to the multiplicative groupoid
of nonzero real numbers? (Hint: The integers and the reals are not equipotent.)
(6) Is the groupoid of nonzero rational numbers under division isomorphic to the groupoid oi
nonzero rational numbers under multiplication?
(c) Is the semigroup of integers under the usual addition isomorphic to the semigroup of rational
numbers under the usual addition? (Hint: Show that under any homomorphism ff of the
integers under addition into the rationale under addition, re = r(lff).)
Solution:
(o.) No. Because if the integers were isomorphic to the reals, the isomorphism between them would
constitute a matching and hence the reals and the integers would be equipotent.
(b) Let a'. (Q*, -r-) -* (Q*, ·) be a homomorphism between the nonzero rationale under division and
the nonzero rationals under multiplication. Now 1 = 1 -f-1. Thus la = (1 -r l)a — la · la and
so la = ±1. Now ±1 = la = (2 -*· 2)a = 2a2o and so 2a = ±1. Similarly, 3a = ±1. Hence a
is not one-to-one. In particular, a is not an isomorphism. Thus there is no isomorphism
between the two groupoids.
(β) Let ff : (Ζ, +) -* (Q, +) be any homomorphism. Let Iff = q. We shall show by induction on
r that r$ = rq for all r € N. Now q = Iff = (0 + 1)9 = Off + Iff = Off + q and so Off = 0.
Suppose rff = rq for r = n. Consider r = η + 1. (η + l)ff = ηθ + Iff = nq + q = (n + l)q\ Hence
rff = rq for all r £ N, by induction. If r £ Ζ and r is negative, then — r £ N, Then since
0 = (r + — r)ff = rff + (—r)ff = rff + (—r)q, rff = rq. Hence rff = rq for all r £ Z.
If ? = 0, ff is not onto. If q — m/n, with m, η integers and m, η Φ 0, then l/2n £ Q.
Has l/2n a pre-image? If r were an integer such that rff = l/2w, then rq = l/2w and so
r = 1/2?». But l/2m is not an integer. Thus ff is not an epimorphism and there is no
isomorphism,
2.40.
Let (Z, °) be the groupoid with binary operation ° defined by
α ° 6 = α + b + ab
and let (Z, *) be the groupoid with binary operation * defined by
a* b = a + b — ab
Is {Z,°)s*(Z,*)1
Solution:
Let σ; (Ζ, ο) -* (Ζ, *) be defined by ασ ~ —a for β. £ Ζ. σ is clearly a mapping, (a ° 6)σ —
(о. + b + ab)a = -(β,+ b + ab) and <nr * 6σ = —a*—b — (-β.) + (— 6) — (—a)(— 6) = —(о.+ 6 + о.Ь).
Hence σ is a homomorphism. σ is onto, for if α £ (Ζ, *), then —α £ (Z,°) and (—α)σ — " M-™"
ασ = 6σ implies —a = —6 and o. = b. Therefore σ is one-to-one and (Z, °) s (Z, *).
Now
2.41.
Is Mm^M{li2}l
Is M,
{1.2}
'"{1,2,3} ■
Solution:
|ДГ{1}| = 1 and |M{lj2)| =4 (see Problem 2.21, page 37). Therefore, by Problem 2.37, M{i}
cannot be isomorphic to М^г 2j . From Problem 2,25, the subset of elements in M,lt2mz\ which have
inverses has 6 elements. Thus \M^^ | is less then the order of M,vzz-. , and so M,lzy is not
isomorphic to Λί{1 2 3,.
2.42. Give an example of two groupoids of order two which are not isomorphic.
Solution: a b с d
а. а. а. с
b <t <t d
These two groupoids are not isomorphic, since there are only two one-to-one mappings, namely
ff ; a -* e, b -* d and ψ : a -* d, b -* c. Now ff is not an isomorphism, for (ab)ff = ав = е while
affbff = ed == d. ψ is not an isomorphism, for (αα)ψ == αψ ~ d while αψαψ = dd = c.
с d
d с
2.43. Prove that the mapping ff : a + lb -* a — ib is an isomorphism from the groupoid С of complex
numbers under addition with itself.

44
GROUPOIDS
[CHAP. 2
Solution :
9 is onto; for if χ S C, χ = a+ib. Now (a + i(~b))9 = a + ib — x. Also в is one-to-one, since
(αϊ + гЬг)е — (a2 + ib2)ff implies аг — ibj = a2 — ΐ62 and hence αλ — <ц and ί>ι — ί>2·
Finally θ is a homomorphiam, for
[(a, + ibj) + (u2 + ib2)]9 = {_al + a2) — i(bl + b2) — (e-i + ib{)9 + (a2 + ib^e
2.44. Is the homomorphism e of the groupoid С of complex numbers under the usual multiplication of
complex numbers to the groupoid of real numbers under the usual multiplication defined by
θ : a+ ib -> |a + ib\ = + yja? + b2
an epimorphism or monomorphism?
Solution:
It is well known that if xv x2 are two complex numbers, then \x\X2\ — |a?i| |я?и|. The
calculations are
l«i + »ill«a + rt2l = V(<q+b*)(al + bl)
V(ai(t2- btb^2 + {Ьга2 + Ь2аг)г
\ага2 — ЬгЬ2 + г(Ьга2 + Ь2аг)\
Κο-, + ΛΟΚ + ^ϊ)!
Then (χχΧ^Θ = χλ$χ2$ and so В is a homomorphism. On the other hand (a + ib)e = (— a — ib)e,
so в is not one-to-one. Finally В is not onto, for it is always the case that |a;| — 0, and thus there
exists no χ such that χ в = —1.
2.45. Let (Ρ, ■) be the groupoid Ρ under the usual multiplication of positive integers and (R, +) the
groupoid R under the usual addition of real numbers. Is the mapping в of (P, ·) into (R, +) defined by
9: e-~*logio a an epimorphism, monomorphism or isomorphism?
Solution:
As in Section 2.5a, 9 is a homomorphism. If αθ — be, then logj0 a = log10 Ь and hence a = 6.
Thus 9 is a monomorphism. Since 0 = log10 1 < log10 2 < logI0 3 < · ■ ■ , there is no integer such
that log10 x — —1. Hence $ is not onto. Therefore в is a monomorphism but not an epimorphism
nor an isomorphism.
c. Properties of epimorphisms
We will show in this section that if 0 is an epimorphism from the groupoid G to the
groupoid H, then Η shares some of the properties of G.
Theorem 2.6: Let 0 be an epimorphism from the groupoid G to the groupoid H. Then
(a) if G is a groupoid with an identity 1, so is Η and 10 is the identity of
H. Furthermore if / is an inverse of g in G, then /0 is an inverse of
дв in H.
(b) if G is commutative, so is H.
(c) if G is a semigroup, so is H.
Proof:
(a) Let hE-H. We shall prove 10 is the identity of H, i.e. h ■ 10 = h = 10 · h. As 0 is
an epimorphism, 0 is onto and we can find an element g in G such that дв = h. Then
h -10 = 00-10 = (βτ -1)6» = дв = h
and 10 -h = Ιθ-дв = (1·.ςτ)0 - дв = h
Thus 10 is the identity of Η and Я is a groupoid with an identity.
Now suppose g e G has an inverse /. Then gf = 1 = fg. Therefore
дв-fe - (gf)e = 10 = (fg)e = fe-дв
which means /0 is the inverse of дв in H.

Sec. 2.5] HOMOMOEPHISMS OF GROUPOIDS AND CAYLEY'S THEOREM 45
(b) Suppose G is commutative. We show Η is commutative. To this end let h, h' € H.
Because 0 is onto, we can find fir,fir' 6G for which дв = h and fir'0 = h'. Hence
hh' = дв-д'в = (д-д')в = (д'· д)в = д'в-дВ = h'h
and so Η is commutative as claimed.
(c) To show Я is a semigroup, we must prove that multiplication is associative in H.
Let h,h',h" e H. Then we can find fir,fir',fir" e G such that gr0 = h, gr'0 = h' and
fir"0 = h". Since G is associative we have, as required,
{hh')h" = {дв-д'в)д"в = [(дд')в]д"в = ((дд')д")в
= {9{9'9"))θ = fir0[(fir'fir")0] - д9{д'в-д"в) = fc(fc'fc")
d. Naming and isomorphisms
In our study of groupoids we will take isomorphic groupoids to be essentially the same.
To explain why, we will describe a "naming process," beginning with an example.
Let G be the groupoid with binary operation ·, elements 1 and a, and multiplication table
1 a
1
a,
a,
1
We define a new groupoid G by relabeling the elements of G. Let G consist of the elements
α, β and have multiplication table
α β
a
β
β
a
What we have done is to call the elements of G by different names.
In general if G is any groupoid, we can form a new groupoid G by renaming the elements
of G. Thus for each fir e G we take a new element g, ensuring only that f ¥* g if f¥>g, i.e.
don't use the same name twice. If /fir = h, then we define multiplication of elements of
£ by fog = }i. It is easy to prove that G is a groupoid with this multiplication.
We are not interested in distinguishing between groupoids which differ only because
their elements have different names. Considering groupoids to be the same if they are
isomorphic overcomes this snag. To see this we will show that the G constructed from G
above by renaming is isomorphic with G. We must find a one-to-one onto homomorphism
Θ. Define дв = g, i.e. the image of g under θ is the new name of д. в is one-to-one onto,
as one and only one g corresponds to each g. Also (/fir)6> = Ji where fg = h. But /0 °дв =
f°g = k, by definition of the multiplication of G. Hence (fg)S = fe°ge, and G is
isomorphic with G\ Thus isomorphism gets rid of the difficulty of obtaining a new groupoid
on simply renaming.
We look at the problem from another point of view. Suppose F and G are two
isomorphic groupoids, and that θ is an isomorphism between F and G. Then we will apply
our renaming process to show that G and F, F suitably renamed, cannot be distinguished
either as regards their elements or the way they multiply.
Let us as before rename each element fGF, f. But we shall choose / to be /0. This is
a proper renaming since f = g means that /0 = дв, and, as 0 is one-to-one, / = fir. So we
have not used the same name twice. As before, if fg = h we define / ° g to be 7i. Thus
F becomes a groupoid with respect to the binary operation о.

46
GROUPOIDS
[CHAP. 2
How does Ρ compare with Gl Ρ has the same elements as G. But do these elements
multiply the same way? Suppose A and /2 are two elements of F. Then /1 was previously
called /1, and /2 was called /2. If /1/2 = /3, then we defined /1 ° /2 = /3.
Now /i,/2 are also elements of G; in fact, /1 = /i0, /2 — /26. The product /1/2 in G is
therefore /i0/20 = (/1/2)6 = /30 = /3, since б is a homomorphism. Hence the product /io/2
of two elements in F is the same element as the product /1/2 inside G.
Thus a groupoid isomorphic with a groupoid F is indistinguishable from a suitable
renaming of F as far as the elements and the way they multiply are concerned. For this
reason we do not distinguish between groupoids that are isomorphic.
e. Mx and semigroups
The importance of Mx is explained by the following theorem, which says that an
isomorphic copy of any semigroup S is contained in some Mx.
Theorem 2.7: (Cayley's Theorem): Let S be a semigroup with identity. Then there is
a monomorphism of S into Ms. (The semigroup S is an abbreviation for
the semigroup (S, μ) where μ is a binary operation. Ms is the semigroup of
all mappings of the set S into itself, with binary operation the composition
of mappings.)
Proof: Let s£S and let Ps: S -* S be defined by xPs — xs, χ e S. Here xs is the
product of χ and s in S, i.e. (χ, 8)μ. It is clear that Ps is a mapping of S into S, i.e. Ps Ε Ms.
Let Θ: S^· Ms be defined by s0 = Ps. We shall show that θ is a monomorphism. First
we have to check that θ is a homomorphism. Let s, s' € S.
(ss')e = Pss. and s0s'<? = PsPs,
Now if a;eS, %psps. — (ajps)ps. = («s)Ps, = («s)s' = «pKS,. As this is true for all χ e S, pss, —
PBPB,. Hence (ss')e = sus'u.
Secondly we must show that θ is one-to-one. Suppose s6 = s'6; then Ps = ps,- In
particular, if 1 is the identity of S, lPs = lPs,. But then lPs = 1 ■ s = s = 1 · ps. = 1 · s' = sf
and so s = s' and θ is a monomorphism. This completes the proof of Theorem 2.7.
As an illustration of the proof, let S = {au a%, a3J be the semigroup given by the
following multiplication table:
GE>i Ο/η, GE>3
a1
Щ
α·3
«2
Oj
0·3
at
Oj Я2
Notice that αϊ is the identity element of S and that S is a semigroup. Now
_ /i) 111 d)\ _ /ίι Β2 »l\ _ /αϊ Oa <Хз\
Ραι - у αϊ aa a3/' Pa2 - \^α2 o3 ai/' Раз уа3 cti a2/
The mapping (9 is defined by at0 = Pa , a20 = Pa , a30 = p^. It can be checked directly that
θ is a monomorphism.

CHAP. 2]
SUPPLEMENTARY PROBLEMS
47
Let Se = {s6 | sG S} where б is the monomorphism of Theorem 2.7; then S ^ Se and
hence Ms contains an isomorphic copy of S. In Section 2.5d we pointed out that, but for
naming, S and an isomorphic copy were the same. Thus we see that in a rough way every
semigroup with identity appears in some Μ χ. Hence the importance of Mx.
A look back at Chapter 2
We defined a groupoid, associative groupoid (called a semigroup), and commutative
groupoid.
We showed that in a groupoid the identity is unique, while inverses are unique in a
semigroup.
We defined a mapping Θ: (G, a) -> (G, β) to be a homomorphism if [(дь д2)а]в =
[{9ϊθ,02θ)]β. If θ is one-to-one and onto, we called it an isomorphism of (G, <*) onto (G,/3).
We proved Cayley's theorem, that each semigroup has an isomorphic copy in Mx for some
suitable X.
Supplementary Problems
GROUPOIDS
2.46. Let G == {1,-1}. Is ((?,·) a groupoid if · is the usual multiplication of integers?
2.47. Show that (G, ■) is a groupoid when G = {1,-1, г,-г}, г = if—ϊ, and · is the usual multiplication
of complex numbers.
2.48. Suppose Gn is the set of all integers divisible by the integer n. For which η is (Gnl ·), with the usual
multiplication of integers, a groupoid?
2.49. Let Д. г = 1, 2, 3,4, 5, 6, be the set G of mappings of R - {0,1} into R defined for each * e R - {0,1}
1 * — 1 1 *
by /j :»!->«; ί2:χ->γζττ:' fs-x^-~r~' ^:x^--· /s-*-»—τι'. /6: *-^ 1 - *. Suppose Gi} =
{/{. /j} and that ° is the composition of mappings. Determine which of the following are groupoids:
(i) (Ci.2,°), (ii)«?i,3,°), (iii) (GUi,°), (iv)«?li5,°), (v) (G,,6,o), (vi) (C5.e,*)·
2.50. Let F = i/1,/2,/3} and H= {/1,/2»/4»/«} where /{ are the mappings defined in Problem 2.49.
Prove (F, °) is a groupoid while (H, °) is not a groupoid (° is the composition of mappings).
COMMUTATIVE AND ASSOCIATIVE GROUPOIDS
2.51. Let R* be the set of nonzero real numbers. Define the binary operation ° on R* by a°b — \a\ b for
a,b £ Й*. Prove (Д*, °) is an associative groupoid but not a commutative groupoid. Hint:
Η |Ь| = |ob|.
2.52. Define the binary operation · on С = ЙХЙ as (a,b) · (c,d) = (ac, bc + d). Is (G, ·) a
commutative or an associative groupoid?

48
GROUPOIDS
[CHAP. 2
2.53. The binary operation α on J? is defined by a: (a, b) -* \a — b\ for a,b£R. Show that (J2, a) is
commutative but not associative.
2.54. Suppose we define a binary operation + on R by a+b — the minimum of α and b (a, b £ R). Show
that (R, +) is both associative and commutative.
2.55. Let G = {a | a: R ^ R]. For α,/HeG define the mapping α* β = α°/}— /}°α where ° is the
usual composition of mappings and i;(a^-|8°a) = x(a ° β) — χ(β ° a) for all are Д. Prove:
(i) (G, *) is a groupoid; (ii) (G, *) is neither associative nor commutative; (iii) (a * β) * a = a * (β * a)
for all o^eG; (iv) a. * β — (—β) * a where —β is the element of G defined by —β: x^> — (χβ) for
all ar G Д. (Hard.)
2.56. Let G be the set in Problem 2.55. For α, β £ G define a · /3 = «Qp + ftQ« ν^βΓβ „ jg ^е usuaj
- - , α°β + β°α x(a° β) + χ(β°α) . ,, _ „ _ ... ,„ .
composition of mappings and χ ■—=-* = — CJ—= ' for all χ € R. Prove: (ι) (G, ·)
is a groupoid; (ii) (G, ■) is commutative but not associative; (iii) (α· β)· a — α· (β · a) for all
α, β € G.
INVERSES IN GROUPOIDS
2.57. Let R+ be the set of all non-negative real numbers. Define a*b = уа^ + Ь2 for all a, b Gi? +
(Va·2 + b2 is the positive square root). Find an identity in (JU + ,*). What elements have inverses?
2.58. For all α, β € G = {a | a : Ζ -* Ζ}, let a X /J be the mapping defined by x(a Χ β) — χα·χβ where
χ £ Ζ and * is the usual multiplication of integers. Is (G, X) a groupoid? Does it have an identity?
What elements have inverses?
2.59. Let G = {a | a: Ζ -» Q, Q the rational numbers}. Define a X β as in the preceding problem. Does
(G, X) have an identity? What elements have inverses?
2.60. Which of the groupoids in Problems 2.57, 2.58 and 2.59 are commutative and which are associative?
2.61. Define the following binary operation + in J2 + , the non-negative real numbers: α + b = the maximum
of a. and b, a,b£R*. Does (R +,+) have an identity? What elements have inverses?
2.62. Let (G, *) be the groupoid of Problem 2.56. What is the identity of (G, *)? Find an infinite number
of elements which have inverses.
2.63. Show that (G, *), the groupoid of Problem 2.55 has no identity.
SEMIGROUPS WITH AN IDENTITY
2.64. Which elements of (G, ■), where G = {1, —1, i, — i}, ί — V—ί, and · the usual multiplication of
complex numbers, have inverses?
2.65. Let G = {/i, /2, /3, /4, /5, /a} of Problem 2.49. Prove that G with the binary operation ° of
composition of mappings, is a semigroup with an identity. Find the inverse of each element in (G, °).
2.66. Let G = {(a,b,c,d)\ a,b,c,d£Z}. Define
(a,b,c,d)-(a',b', c',d') = (αα' + cb', ba' + db', ac' + cd', b& + dd')
Show that (G, ■) is a semigroup with an identity. Show that the subsets Η — {(1, 0, 0,1), (—1,0,0,-1)}
and F = {(1,0, c, 1) | с € Ζ}, with the binary operation of (G, ·) restricted to Η and F respectively,
are semigroups with an identity. Find the inverses of the elements of Η and F.
2.67. Let G= {a | a a mapping of {1,2,3,4,5} into {1,2}}. For α,/jeG, let a°β be the usual
composition of mappings. Is (G, °) a semigroup with an identity?

CHAP. 2]
SUPPLEMENTARY PROBLEMS
49
2.68. For «,/?£<? = {a | a : {1,2,3,4,5}-> {1,-1}}, define the mapping α Χ β by n(a Χ β) — Па ■ ηβ,
where и£ {1,2,3,4,5} and ■ the usual multiplication of integers. Is (G, X) a semigroup with an
identity? If so, what elements have inverses?
HOMOMORPHISMS OF GROUPOIDS AND CAYLEY'S THEOREM
2.69. Let Я = {(1,0,0,1), (-1,0,0,-1)} be the groupoid defined in Problem 2.66. Show that Я is
isomorphic to the groupoid (G, ■) where G = {1,-1} and · is the usual multiplication of integers.
2.70. If F - {(1,0, c, 1) | с S Z) is the groupoid defined in Problem 2.66 and (Z, +) the groupoid of integers
under addition, prove the two groupoids are isomorphic.
2.71. Which of the groupoids of Problem 2.49 are isomorphic?
2.72. Let G= {/1,/2./з./4,/5,/б} be the groupoid of Problem 2.65, (H, ·) the groupoid with Я = {1,-1}
and · the usual multiplication of integers. Find all possible homomorphisms of G into H. Show
that there is no homomorphism of (G, °) into (F, °) where F — {fb /2, /3} and ° the usual
composition of mappings. (Hard.)
2.73. Can the groupoid (G, *) of Problem 2.55 be a homomorphic image of the groupoid (MR, °) where °
is the usual composition of mappings?
2.74. Suppose G = {a | a : Ζ -» Ζ}, Χ the binary operation defined in Problem 2.58 and ° the usual
composition of mappings. Show that Φ : (G, X) -» (G, °) defined by Φ : a -> a is not a homomorphism.
(See Theorem 2.6, page 44.)

Chapter 3
Groups and Subgroups
Preview of Chapter 3
We define a group as a semigroup with an identity in which every element has an inverse.
The object of this chapter is to show that the concept of a group is natural. This is done
by providing illustrations of groups which arise in various branches of mathematics, The
most important concepts of this chapter are group and subgroup.
3.1 GROUPS
Definition
As we remarked in the preview, a semigroup with an identity in which every element
has an inverse is termed a group. We repeat the definition in more detail:
Definition: A non-empty set S together with a binary operation in S is called a group if
(i) there exists an identity element (usually denoted by) leS; in other
words, Λ Λ , „ ,_ 0
' a · 1 = a — 1 · a for all a£S
Recall that the identity is unique by Theorem 2.1, page 31.
(ii) for every choice of the elements a, b, с G S,
(a· b) · с — a · (b· c)
Thus (i) and (ii) are the conditions for (S, ·) to be a semigroup with
identity.
(iii) every element a <E S has an inverse in S, i.e. there is an element Ъ Ε S
such that ,
α·ϋ = 1 — b'a
This element b is often denoted by a"1. The inverse is unique by Theorem
2.2, page 33.
Whenever we define a group we shall follow the pattern:
I. Define a set S (¥· 0).
II. Define a binary operation in S. (^
III. Verify that the groupoid (S, ·) contains an identity element.
IV. Verify that the groupoid (S, ·) is a semigroup, i.e. is associative.
V. Verify that every element of S has an inverse.
The number of elements of S, \S\, is called the order of the group. (Compare Section 2.2,
page 29.) We will exhibit groups of infinite and finite order. (See following examples and
problems.)
50

Sec. 3.1]
GROUPS
51
Examples of groups of numbers
Example 1: The additive group of integers.
I. Let Ζ be the set of integers.
II. Let + be the binary operation of addition in Z.
III. n+ 0 — η = 0 + n for every »£Z. Thus (Z, +) has an identity element,
IV. If l,m,n are integers,
(l + m) + η = I + (m + n)
i.e. (Z, +) is a semigroup,
V. If и G Ζ, then —η in Ζ has the property
η + (—η) — О = (—η) + η
i.e. —η is an inverse of η in (Ζ, +).
Thus we have shown that the groupoid (Z, +) is a group. This group is usually
referred to as the additive group of integers.
Example 2:
Example 3:
Example 4:
Example 5:
The additive group of rationale.
I. Let Q be the set of rational numbers.
II. Let + be the binary operation of addition in Q.
III. α.+ 0 = α. = 0 + α· for every s6Q, so 0 is an identity element for (Q, +).
IV. If a,i,ee Q, then (a + b) +c = a + (b + c).
V. If a.€ Q, then —a in Q has the property α + (—a) — 0 = (—a.) + a.
The additive group of complex numbers.
The description of this group is left to the reader.
The multiplicative group of nonzero rationals.
I. Let Q* be the set of nonzero rational numbers.
II. Let · be the binary operation of multiplication, i.e. the usual multiplication of
rational numbers.
III. The rational number 1 is clearly an identity in the groupoid ((?*, ·).
IV. If a,b,c& Q*, then
(a·· b) · с = a ' (b' c)
V- If я £ Q*, so is 1/a. and 1 .
a · — — 1 = — · a
a a
Thus every element of Q* has an inverse.
The multiplicative group of nonzero complex numbers.
This group is very similar to that in Example 4. We shall go through the usual
five stages in setting up and describing the group.
I. Let C* be the set of all nonzero complex numbers. Thus
ί
С* = {x\ x = a + ib where * φ 0 + г'О and a,b £ R}
Recall that г2 = -1.
II. We define multiplication of complex numbers as follows:
(a + ib)(c + id) = (wc — bd) + i(ad + be)
This is a binary operation in C* since (ac- bd) + i(ad + bc) is a unique
element in C* (not both ac — bd and ad + be can be zero).
III. 1 + i'0 = 1 € C* and it is clearly an identity in (C*,·)-

52
GROUPS AND SUBGROUPS
[CHAP. 3
IV. Suppose a. + ib, с + id, e + if e C*. Then
[(a. + ib)(c + id)] (e + if) = [(ac — bd) -f i(bc + ad)] (e + if)
= [(ас - bd)e - (be + ad)f] + i[(bc Jr ad)e + (ac - bd)f]
On the other hand,
(a. + ib)[(c + id)(e + if)] = (a. + ib)[(ce - df) + i(de + cf)]
= [a(ce - df) - b(de + cf)] + i[b(ce - df) + aide + cf)]
It follows from these two computations that
(a.+ ib)l(c+id)(e+if)\ = [(a + ib)(c + id)](e + if)
V. We have to check the existence of inverses. Thus suppose a+ib € C*\ then
not both a and b are zero. Hence a2 + b2 Φ 0 and so
C*
a2 + 62 a·2 + b2
Moreover,
Thus we have proved (C*,·) is a group and we term this group the multiplicative
group of nonzero complex numbers.
Problems
3.1. Is (S, °) a group if
(i) S — Ζ and ° is the usual multiplication of integers?
(ii) S = Q and ° is the usual multiplication in Q?
(iii) S = {g | g G Q and g > 0} and · is the usual multiplication of rational numbers?
(iv) S = {z | ζ £ Ζ and ζ = \/2 } and ° is the usual multiplication in Z?
(v) S = Й and ° is the usual addition of real numbers?
(vi) S — Ζ and ° is defined by a° b = 0 for all a·, 6 in Z?
Solutions:
(i) The identity element is the integer 1. (S,°) is not a group because 5ΞΖ but there is no
integer ζ in Ζ such that z°5 = 5°z = l.
(ii) Again the identity is the number 1. There is no g&Q such that g°0 = 1. Hence (S,°) is
not a group.
(iii) (S, ·) is a group. Clearly S ¥= φ and * is a binary operation onS. g ■ 1 = 1 ■ g = g for all
geS; hence 1 is an identity. Multiplication of rational numbers is associative and every
11 1
element in S has an inverse; for if g£S, then — € S and —· q — I — q·—.
q q g
(iv) S-0 since \/2 6? Z. Therefore (S,°) is not a group.
(v) (S, °) is a group. S ¥· 0 and addition is an associative binary operation onS. r + 0 = 0 + r — r
and r + (-r) = 0 = (-r) + r for all r^S.
(vi) (S, °) is not a group because there is no identity element in S.
3.2. Let S be the set of even integers. Show that S is a group under addition of integers.
Solution:
Let a. = 2a·! and b — 2bl be any two elements in S. a+ b = 2(α·! + bt) is a unique element
in S; thus addition is a binary operation on S. Associativity of addition in S follows from the
associativity of addition in Z. 0 = 2*0 is an identity element in S. If a^S, then —aGS since./
a, — 2al implies —a. = 2{—al). Hence a. has an inverse in S, as α·+ (—α.) = (—α.) + a = 0- \
3.3. Let S be the set of real numbers of the form a. 4- i>\/2 where a,b^Q and are not simultaneously
zero. Show that S becomes a group under the usual multiplication of real numbers.

Sec. 3.1]
GROUPS
53
Solution;
(a.+ bV2)(c + dV2) = (ac + 2bd) + (cb + ап)л/2 . If neither a + b^/2 nor с + d^/2 is zero, i.e.
(Ο + Οτ/Ϊ), then their product cannot be zero. Hence the product of elements in S belongs to S.
z- 1 a. - Ьт/2
1 = 1 + 0у2 is an identity for S. Multiplying —— by , we obtain
a. + 6V2 a. - 6^2
а-Ьл/2 а
a + b-fe a2 ~ 2b'Z ft2 _ 2t>2 (fti! _ 2b2)
V2
Hence £ S. The associativity holds because it is true for multiplication of real numbers.
a + 6У2
3.4. Let S be the set of complex numbers of the form a + b\/—5 where a, b £ Q and are not both
simultaneously zero. Show that S becomes a group under the usual multiplication of complex
numbers.
Solution:
(o. + b\/—5 )(c + rfV—5 ) ~ (e-c— bbd) + (be + ad) τ/—Έ and cannot be zero if its factors are
not zero; hence the product of two elements of S belongs to S. 1 = 1 + Oyf—5 is an identity for S.
—— is certainly an inverse for a 4- Ьт/^Е. Multiplying top and bottom by a, — by—Ε, we get
α + &V-5
a + bv^S ft2 + 5&2 ~ α2 + 5Ь2 2 + 5
and so -pzr £ S. The associativity of multiplication in S follows from associativity of multi-
α + by-5
plication of complex numbers.
3.5. Let m be any fixed positive integer and let S = {0,1,2, ...,m— 1}. Define a binary operation in
S by
a°b = a+b if a + b < m
«."li = r if α + Ь = m + r, 0— r < m
Prove that (S, °) is a group of order m. (Hard.)
Solution:
If a, b € S, then aoj ia uniquely defined and belongs to S. α ° 0 = 0 ° α = α, so 0 is an identity.
Note that a°b — a + b — Sm where δ is 0 or 1, for any a,b eS. So b°c — b + с — 5,m where
S, is 0 or 1. о. о (b о c) — a + (b ° c) - S2m where S2 is 0 or 1. Then a°(b°c) = a + b + c-(S1 + S^)m
where both δτ and 52 could be 0 or 1. Hence
a ° (b °c) = a + b + с — i^m where i/j is 0 or 1 or 2
Similarly (a°b) ° с = o. + 6 + c — i/2m where i/2 is 0 or 1 or 2
Now 0 — a°(b°c) <m and 0 -(а»Ь)ое<и. Suppose i/j > i/2; then
a■ о (Ь о с) — о. 4- Ь + с — (i/2 + l)wi = о- + Ь + с — ij2m — т
because ηι is at least i/2 + l. But 0 — a + b + ο — η^η < τη and the above equation implies that
a°(bo e) < 0; this contradicts 0 — a°(b°c). Hence i;, - ι/2. ι/2 > η1 leads in a similar way to a
contradiction. Thus ηι = i/2 and tt»(60e) = (a°b)°c. If α £ S, then m — a^S and
a°(m — a) — (m — α) ο α = 0; hence «ι — α is an inverse to a. Thus S is a group.
3.6. Let S qC (C the set of complex numbers) be the set of all wth roots of unity, where m is a fixed
positive integer. Prove that under the usual multiplication of complex numbers, S becomes a group
of order m.
Solution:
Recall that a complex number χ is an wth root of unity if xm = 1 and that there are exactly
m distinct roots of unity, viz. ei2nT/m, τ — 1,2, .. .,m; also, ei7rx = cosx+isina;. If a,b^S,
then ab is uniquely defined. Since (ab)m — ambm — 1, ab is an mth root of unity and hence ab £ S.
1· a = a· 1 = a, so 1 is an identity. Associativity is true, since it is true for complex numbers in
general. If a £ S, then (l/a)m = l/am — 1; thus 1/a S S and 1/a is the inverse of a.

54
GROUPS AND SUBGROUPS
[CHAP. 3
3.7. Let S ζ.ϋ (C the set of complex numbers) be the set of all roots of unity. Describe one way of
making S into a group.
Solution:
Use as binary operation the usual multiplication of complex numbers. If a,b £ S and a is an
with root of unity, b an wth root of unity, then ab is an mwth root of unity because (ab)mn — am*bmn —
(am)n(bn)m = i»i«« = i. Hence ab e S and is, of course, uniquely defined. 1 £ S and acts as an
identity. 1/a G S and is the inverse of a. Associativity holds for multiplication of complex numbers.
Thus S is a group with respect to the usual multiplication of complex numbers.
3.8. The following table defines a binary operation. la the resultant groupoid a group?
1 2
1 2
2 1
Solution:
We need to check only (β.) associativity, (b) existence of identity and inverse.
(a) To check associativity, we have the following possible questions:
(a) Does 1·(1·1) equal (1 -1) · 1? (e) Does 2 · (1 · 1) equal (2-1) ■ 1?
(b) Does 1 · (1 · 2) equal (1 · 1) · 2? (/) Does 2 · (1 · 2) equal (2 · 1) · 2?
(c) Does 1·(2·1) equal (1·2)·1? (fir) Does 2 · (2 · 2) equal (2 ·2) ·2?
(d) Does 1 · (2 · 2) equal (1 · 2) · 2? (A) Does 2 · (2 · 1) equal (2 · 2) ■ 1?
Checking all these products, we see that associativity holds.
(b) 1 acts as an identity. The inverse of 1 is 1, the inverse of 2 is 2. Hence the table defines a group.
3.9. Write the multiplication table for the group of Problem 3.5 with m = 3 and m = 4.
Solution:
If m — 3, the multiplication table is
0
1
2
If m = 4, the multiplication table is
0
0
1
2
1
2
0
2
0
1
0
1
2
3
1
2
3
0
2
3
0
1
3
0
1
2
3.2 SUBGROUPS
Definition
Let (G, ·) be a group with binary operation · and let Я be a non-empty subset of G.
Then we say Я is a subgroup of G if the operation · restricted to Я is a binary operation in
Я which makes Я into a group.

Sec. 3.2]
SUBGROUPS
55
For example, if G is the group with m = 4 of Problem 3.9, then the subset Я - {0,2}
is a subgroup of G. For when the operation ° in G, as defined in the multiplication table
for G, is restricted to H, it is a binary operation in H, i.e. 0 о 0 = 0 € Я, 0 ° 2 = 2 e Я,
2°0 = 2 e #, and 2°2 = 0 € Я. Я is a group because: Я ^ 0; 0, the identity, is in Я;
the operation ° restricted to Я is an associative binary operation (since the operation in
G is associative); and every element in Я has an inverse in Я.
The following lemma facilitates proving a subset of a group is a subgroup.
Lemma 3.1: Let (G, ·) be a group. Then a subset Η of G is a subgroup of G iff
(i) Я^0 and (ii) if a, b e Я, then ab x e Я
Proof: If Я satisfies these conditions, then Я is a group with respect to the binary
operation. For if Я ¥- 0, then there exists абЯ. Hence ом^1 = 1 e Я. Also, if ЬеЯ
then lb-1 = Ь-1 е Я. Hence а, Ъ е Я implies а(Ь-1)-1 = ab G Η. Associativity is true
in Я, as it is true in G. Thus · is an associative binary operation опЯ, le Я, and the
inverse of every element of Я is an element of Я, Therefore (Я, ·) is a subgroup.
Conversely if Я is a group with respect to ·, then clearly Я satisfies conditions (i) and
(ii) above.
Problems
3.10. Let (Q,+) be the additive group of rationals. Is Ζ a subgroup of Q? Is Ρ a subgroup of Q?
Solution:
Clearly Ζ ¥· 0, and Ζ cQ. If а, b £ (Q, +) then the binary operation is +, and the inverse
of b is — b. So, if a,b e Z, we ask in accordance with Lemma 3.1, whether а + (—6) = a — b e Z.
It is, and so Ζ is a subgroup of (Q, +). Clearly Ρ ■* 0 and Ρ ς. Q. If a.b&P, is »+(-b) =
а —Ь eP? No, for Ρ does not contain negative numbers; and if о = 1 and 6 = 2, then a — b is
negative.
3.11. Is Q a subgroup of (C,+), the additive group of complex numbers?
Solution:
Q ¥■ 0. QCC; for if а € Q, o = o + OieC If a,fe€Q, then a+ (-6) = а — Ь е Q. Hence
Q is a subgroup of (C, +).
3.12. Is 2 — {0} a subgroup of ({?*, ■)> the multiplicative group of nonzero rational numbers?
Solution:
1 is the identity. 3£Z- {0}, but 3 has no inverse in Ζ — {0}. Therefore Ζ - {0} is not a
subgroup of (Q*, ·).
3.13. Is Q", as above, a subgroup of (C*, ·), the multiplicative group of nonzero complex numbers?
Solution:
Q* 7* 0. a,b e Q* implies ab~i is a nonzero rational. Thus Q* is a subgroup of (C*, ·).
3.14. Is Q* a subgroup of the group of real numbers of the form а + by/2 , a,6eQ and а, Ь not
simultaneously zero, under multiplication?
Solution:
Q* ¥■ 0. If α e Q*, then а = а + 0\/2 € {а + Ь\/2 | а, Ь £ Q and not both zero}, since a¥- 0.
Thus £?*С{а+Ьд/2 | o, b e Q and not both 0}. tifbeQ* implies ab~l is a nonzero rational.
Hence Q* is a subgroup.
3.15. Prove that the intersection of two subgroups Η and К of a group G is a subgroup.
Solution:
1еЯ, for as Η is not empty, there is an element h e H. But then Η contains A.A.-1 = 1, the
identity of G. Similarly, leK. Hence ie#nK and ЯпК#0. If g.h&HnK, then
g,h^H and ffA.-"1 € Я. Also, ghl e K. Thus gh~l eflnK and Я η Κ is a subgroup of G.

56
GROUPS AND SUBGROUPS
[CHAP. 3
3.16. By considering the group of Problem 3.5 with m — 6, show that the union of two subgroups is
not necessarily a subgroup.
Solution:
That {0,3} and {0,2,4} are subgroups is easily verified. But U - {0,2,3,4} = {0, 3}u {0, 2, 4}
is not a subgroup as 3 ° 4 = 1 £ U. Therefore ° is not a binary operation in U.
3.3 THE SYMMETRIC AND ALTERNATING GROUPS
a. The symmetric group on X
Let X be any non-empty set. A very important group arises from the set Sx of all
one-to-one mappings of X onto X, called the symmetric group on X. We describe this group
in the usual five steps.
I. Sx is the set of all matchings of the non-empty set X with itself. Clearly, Sx С Мх
(see Section 2.4b, page 36).
II. If σ,τ£ Sx, then we define στ to be the composition of the mappings σ and т.
Here we must verify that στ is a matching of X with itself. Suppose χ e X; then
as τ is onto, we can find x" e X such that x"T — x. But σ is also onto, so we can
find x' e X such that χ'σ - χ". Consequently
Χ'(στ) — (Χ'σ)τ = Χ"τ = Χ
and hence στ is onto. If χ(στ) = у{от), then (χσ)τ = (уа)т by the definition of the
composition of mappings; this means, since r is one-to-one, that χσ — yo. This in
turn implies, since σ is one-to-one, that x — y. Therefore στ is also one-to-one.
Thus composition of mappings is a binary operation in Sx.
III. Clearly the identity mapping ι: χ -» χ is in Sx and is an identity element of Sx.
IV. The groupoid (Sx, ·) is a semigroup, since composition of mappings is associative.
(Theorem 2.3, page 36.)
V. Let σ e Sx. Since σ is one-to-one and onto, Theorem 2.4, page 36, implies σ has an
inverse, τ, in Μ χ. Now, στ = ι = τσ means σ is an inverse of т. By Theorem 2.4
the only elements in Mx which have inverses are those which are one-to-one onto
mappings. Therefore τ e Sx and τ is the required inverse of σ. The proof that
Sx is a group is complete.
We will call an element of Sx a permutation of X, or, for short, a permutation.
In the particular case where X = {1,2, .. .,rc}, we write Sx = S«. S„ is called the
symmetric group of degree n.
\Sn\ is calculated as follows. If σ € Sn, then 1σ can be one of η elements. 2σ can be
one of те — 1 elements, as 1σ has been chosen and a must be one-to-one; so 2σ ¥· 1σ.
Зог can be one of и — 2 elements, as 1σ and 2σ have been chosen and a must be one-to-one;
so Зог is not equal to 2σ or la. Continuing in this way, we conclude that there are
я·(»-!)(%-2) ·■ ■ -2-1 = и!
elements of Sn, i.e. [S„[ = n\
The elements of S3, for example, are

Sec. 3.3]
THE SYMMETRIC AND ALTERNATING GROUPS
57
12 3
12 3
12 3
2 3 1
12 3
3 1 2
12 3
13 2
12 3
3 2 1
12 3
2 13
Here we are using the notation of Section 2.4c, page 37. To find the multiplication
table, we must compute the products. As an example, we calculate σ^.
12 3
1σ,г, 2σ,τ, Зо-jrj
2i-j 3i"j li-j
To do this calculation mentally, we think as follows:
Write down
1 -* 2 (in ffj, 2^3 (in Γι)
12 3
3
2^3 (in ctj), 3 н> 2 (in 7-j). Put 2 beneath 2 to get
12 3
3 2
Only one other number can appear, namely 1. Thus
1 2 3'
0ΓιΓι \ 3 2 1
The multiplication table for S3 is
Tl
*2
Ta
(
(
<Ί
"2
Tl
r2
rs
«Ί
«Ί
σ2
1
T3
T\
T2
<τ2
"2
I
«Ί
T2
Tg'
η
η
τι
T2
ra
1
»l
"ζ
τ2
τ2
ts
Tl
<T2
I
"Ί
TS
T3
Tl
T2
"1
"2
I
12 3
3 2 1
The reader should check some of the entries. Note that σ^ = τζ and τ1σ1 = тя, so that
o-jTj э* r^. Hence S, is not commutative.
Problems
3.17. Calculate αβ, βα, α~\ β~\ {αβ)-1, and {βα)-1 if
Ί 2 3 4 5 6
,236541
"(ί
j and Д = Л
12 3 4 5 6
3 6 6 2 4
Solution:
αβ —
Ί 2 3 4 6 6
,354261
βα
12 3 4 5 6
2 6 4 13 5
To find α-1, we note that χ(αα~ι) — χ and hence α~χ must carry aw to ж. Now we determine
which χ is taken onto 1. 6a = 1, so we must have ία~ι = 6. Next, since la = 2, 2a-1 = 1.
Proceeding in this way, we obtain
_! _ Λ 2 3 4 5 6V
" "U 12543

58
GROUPS AND SUBGROUPS
[CHAP. 3
An easy method of calculating the answer mechanically follows. Take
/1 2 3 4 5 6\
a (,236541/
Interchange the rows,
2 3 6 5 4 1
12 3 4 5 6
)
Rearrange the columns so that the top row reads 12 3 4 5 6,
1 2 3 4 б 6\
6 12 5 4 3/
This method is conceptually the same as the first. To find β"1, interchange the rows to obtain
13 5 6 2 4
12 3 4 5 6
)
and rearrange to get
Ί 2 3 4 5 6
β-ι =
15 2 6 3 4
)■ («,
β)~ι =
12 3 4 5 6
6 4 13 2 5
\ /12 3 4 5 6\
)· {βθ) 1 = (4 1 5 3 6 2)
3.18. Verify that α{βγ) = (αβ)γ where
/12
a = f
2 3 4 5
2 4 5
β =
Solution:
(«/3)У =
α(βγ) =
12 3 4 5
13 2 5 4
12 3 4 5
3 12 4 5
1 2 3 4 5^
3 2 1 5 4,
1 2 3 4 5\
4 3 1 5 2J
γ =
12 3 4 5
4 3 15 2
Χι
2 3 4 5
3 4 2 5
)■
12 3 4 5
4 13 2 5
12 3 4 5
4 13 2 5
3.19. Is the subset Β — {ι,νχ, <г2} а subgroup of Ss? (For this notation see above.)
Solution:
R Φ 0. From the multiplication table of Ss, page 57, the product of any two elements in В is
again in B. Since σ"1 = <τ2 and ".j"1 = <Ί» it follows that icy'1 G R for any k.j/S/?. Hence
В is a subgroup of S3.
3.20. Find all elements of S1 and S2 and exhibit multiplication tables for these groups.
Solution: ,
Sj contains one element ι = I J. 11 1 is a multiplication table for S\. There are two
elements in S<t'. ι
1 2
1 2
and β
=(;;)■
The multiplication table for S2 is
0
1
β
β
I

Sec. 3.3]
THE SYMMETRIC AND ALTERNATING GROUPS
59
3.21. Find the elements of S4.
Solution:
2 3 4
2 3 4
2
3
2
4
■ - G
-G
·« = (з
„3 = ^
•-G
--G
3
4
3
1
3
2
3
1
3
2
σβ
"7
G
'» = (2
G
"9
Γ1
r2
G
2
4
2
1
2
3
2
3
2
4
3
2
3
4
3
1
3
4
3
2
r3 _ Q
--G
2
2
2
2
3
4
3
1
Γ5 _
'β
г8
= G
■G
2
3
2
1
3
1
3
2
α> - (2
α2
"3
<*4
<*5
G
G
G
2
1
2
2
2
2
2
3
2
4
2
2
We give the multiplication table for future reference.
THE SYMMETRIC GROUP OF DEGREE 4
ι σι σ% σΆ ιτ4 σ5 σβ σ7 σ8 σβ τχ τ2 r3 r4 г5 г6 г7 гз βι α2 α3 α4 α5 α6
1
"1
σ2
"3
'4
σ5
"6
σ7
"8
»Β
Τ\
r2
ГЗ
г4
г5
Гб
г7
Γβ
«1
ο-ϊ
«3
Щ
«5
«6
ι
<Ί
*2
»3
"4
"5
"6
σ7
"8
»В
П
^2
ГЗ
Г4
т5
Г6
г7
Т8
«1
а2
"3
«4
«5
а6
σ1
с2
<^3
1
Г8
«5
г4
Тв
а2
гг
"4
«1
"9
а4
"6
«6
"7
«а
ГЗ
"5
7Ί
5-5
"8
г7
"2
"3
е
"1
«3
»8
а4
а6
σ5
ai
Т8
ТЭ
Г2
Г5
Г4
Г7
Гб
т1
"»
<*5
σ4
"β
α2
σ7
"3
ι
"1
"а
Tl
«2
Γ5
Γ7
«5
rS
«3
"Β
«1
»6
α4
Cr7
"6
σ4
Γ2
σ8
Γ8
τ4
β-8
Γ6
σ4
Γ6
α4
^3
σ5
"6
ί
τι
α3
Γ8
α1
«Ί
α5
Cr7
σ9
α2
»3
«6
Γ4
Τ2
σ2
"8
Γ7
Τ5
σ5
α2
"β
«5
σβ
ι
"4
α\
σ2
"β
ΤΑ
τβ
τι
τ\
ΤΗ
Τ2
тъ
τ%
Ο-,
"ι
α4
«3
"3
"9
<Ί3
τ2
«3
τ-t
ι
"4
Of,
и
α4
ть
О-,
«2
"3
«1
aG
"1
<*5
°9
Γΐ
Гб
"β
σ2
ГЗ
Τ8
σ7
η
«1
τ\
Γ6
«6
Γ7
"β
"9
ι
«2
"6
"4
0-S
σ9
α4
«3
σ1
"»
Γ5
Γ2
Γ3
Γ8
ff2
*8
«5
σ5
«2
α4
°2
α3
«"■»
i
ο7
Γ5
Γ7
Γ6
Γ8
τ\
Γ3
Γ2
Γ4
"6
"3
"Β
"4
«Ί
α1
"9
Γ8
«6
ть
Гз
^1
Г2
t
σ7
»8
<^з
«3
«4
σι
«2
"4
"в
«5
σ2
Γΐ
Γ7
Γβ
Γ4
σ5
ΓΙ
£Γ7
Γ4
«1
α2
Γ5
"3
«3
Γ8
σ4
Γ2
(
"5
Г7
Γ3
»8
"2
Γ6
"6
»9
»1
°'5
«6
«4
Γ2
«3
■η
°α
"9
ГЭ
«1
"\
Γ6
«2
4
rl
Γ5
σ2
σ5
Γ8
Γ4
°8
»3
σ4
"7
αβ
«4
"5
гз
"4
г«
αΑ
α1
Γ2
σ9
σ3
Γ7
«5
"2
Γ8
Γ4
1
"8
Γΐ
Γ5
»5
<Ί
α3
«6
CT-jr
<τβ
«2
Γ4
"1
ΓΙ
σ7
°\
re
*5
at
Γ5
"6
те
"5
ι
Τ3
Γ7
"2
"8
Γ2
σ4
α6
аг
σ3
"Β
^3
Γ5
σΒ
Γ8
α6
σ3
ΓΙ
«2
"6
Γ4
α4
Γ7
"Β
"2
Γ2
Γβ
1
"5
Τ3
^3
σ7
«5
<Ί
«1
"4
Γ6
α4
гз
^4
«6
Γ7
σ7
«2
Γ2
σι
σ5
Τ4
Γ8
»8
1
Γ5
ΓΙ
"2
"5
»6
«1
"9
«3
"3
Γ7
σ6
Γ2
"3
σ7
τβ
«6
«5
Γ3
°3
σ8
η
ΓΙ
°Ь
"2
Τί
Γ8
(
α2
«4
σΒ
а1
"4
σ1
Γ8
«β
Γ5
σ9
«5
Γ4
σ1
»4
ΓΙ
«3
гз
σ2
σ8
ΤΒ
Γ2
σ5
1
τ7
α4
αϊ
"3
«2
"7
"6
<*1
ΓΙ
"7
τ-4
Γ2
σΒ
Γ3
"2
«6
"5
<Ί
«4
»6
'3
<*5
«3
α4
^2
1
Γ8
Γ6
Γ7
Γ5
"8
α2
»8
α5
'5
Γ5
"3
Τ\
τί
σι
Γ6
"Ά
σ7
α6
α3
"4
σΒ
"1
α4
Γ7
ί
Γ4
Γ8
σ2
Гз
«3
Γ7
"6
Γ2
σβ
α4
σ2
Γ8
σ4
ΓΙ
σ9
"3
α2
«•β
«1
«5
<Ί
σ7
Γ5
Гз
1
°Ί>
те
Γ4
«4
Γ3
"4
Γβ
»2
α3
»8
Γ5
»6
Γ4
α6
«5
<Ί
σΒ
"7
»3
"2
«1
Γ8
Γ7
"5
1
Γ2
ΓΙ
ff5
"5
а2
"Β
Γ4
<Ί
Γ8
Гз
"3
Γ7
α4
αβ
σ7
σ4
«3
«1
»8
σ6
τβ
<Τϊ
Γ5
Γι
(
Γ2
«6
Γ5
σ9
τ%
τι
Ο-,
4
"S
α1
σ2
«5
α4
«3
«2
<Ί
σβ
"4
σ3
"8
Γ4
гз
τ%
τ\
'

60
GROUPS AND SUBGROUPS
[CHAP. 3
b. Even and odd permutations
We are interested in a special subgroup of Sn, the alternating group of degree n, usually
denoted by An. This subgroup An is obtained from S„ by singling out certain elements.
(1 2 3 \
2 „ 1 ). Then
2σ — 1σ 3σ — 1σ 3σ — 2σ _ 3 — 2 1—2 1 — 3 _ .
If τ = [ ΐ ? ? ), then
3 2 1,
2г — 1г Зг — It Зг — 2г 2 — 3 1 — 3 1 — 2
= -1
2-1 3-1 3-2 2-13-13-2
We say σ is even and τ is odd.
More generally, let us call a €. Sn even (or an even permutation) if
2σ - 1σ 3σ - 1σ 3σ — 2,σ Па - 1σ Πσ - 2σ _ Τίσ — (η — 1)σ
2-1 " 3-1 "3-2 η-1 " η-2 η-(η-1)
On the other hand, we call a e Sn odd (or an odd 'permutation) if
2σ - 1σ t 3σ — Iff % 3σ - 2σ Πσ - la Πσ - 2a Па — {η — 1)σ
"2^1 З^Т 3-2 ^й^7! п-2~ п- (п-1)
The definition of even or odd 19 written more briefly аз
σ is even if 11 —г—r~ = 1
= -1
σ 19 odd
i<fc /с — г
&σ — ia
« Ητί = -'
We shall show that an element in Sn is either even or odd, i.e. Π -τ—г- = ±1.
><« A; — г
Corresponding to each factor fc — г in the denominator, we will find a factor in the
numerator which is either k — i or — (k-ΐ). As ν is a permutation, there exist unique
integers I,m such that la = к, ma = i. If !>m, the factor Ισ-ma- k-i appears in
the numerator. If Km, the factor ma - la = -(k-i) appears in the numerator. The
quotient —, 1— or —f—A- is thus ±1. Note that distinct factors k-i,k' - i' in the
к — % к — %
denominator give rise to distinct associated factors ±(ίσ - mo), ±(l'a - m'a) in the
numerator. For if 1 = 1', m = m', we have k = k' and % - г'.
Thus regrouping the factors in the numerator, the product becomes the product of
factors ±1 and hence is itself ±1. Therefore every permutation of Sn is either even or odd.
There is an easy way of determining whether a permutation a is even or odd. If we
are given a row of integers, we call the number of integers in the row smaller than the
first integer, the number of inversions. Thus for example the number of inversions in the
row 7,4,3,2,1,6,8 is 5. We will use this concept of inversion to find out if a given
permutation
Ί 2 ... η
.1σ 2σ ... Па,
is even or odd.

Sec. 3.3]
THE SYMMETRIC AND ALTERNATING GROUPS
61
To do this we must calculate 11 (-£ Д ) . Much of the calculation is redundant
*<ъ\ к - г /
аз we have proved that the result is always 1 or —1. We must only determine the sign.
In the denominator we always have positive numbers. In the numerator a negative number
ka — ia arises if ia > ka. For fixed i and varying k, the number of negative factors that
arise is the number of к > i for which ia > ka. But this is the number of inversions in the
row ia, (i + l)a, .. .,ησ. The total number of negative factors is the number of inversions
in la, 2a, ..., ίΐσ + the number of inversions in 2σ, 3σ,..., па + the number of inversions
in (η — 1)σ, Βσ. Let this total be /. The product of / negative numbers is positive if / is
even, and negative if / is odd. So a is even or odd according as / is even or odd.
/1 2 3\
Example 6: Is σ = I I even or odd?
The number of inversions in 3,1, 2 is 2. The number of inversions in 1,2 is 0.
Thus the total number of inversions is 2, and hence σ is even.
Problem
3.22. Is σ even or odd?
Ί 2 3
I)
5 6
6 4
(i) σ =
(ii) σ =
(Hi) σ =
Solution:
(i) Number of inversions in
Total number of inversions
Hence σ is even,
(ii) Number of inversions in
(iv)
(v)
2
3
2
1
3 4 Б 6
6 3 2 5
2 14 3
14 3
4 3
Б 3 2 4 1
3 2 4 1
2 4 1
4 1
1
О
1
2
Total number of inversions
Hence σ is even,
(iii) Odd. (iv) Odd. (v) Even.
c. The alternating groups
We shall show that the set of even permutations forms a subgroup of S„.
1. Let An be the set of even permutations in Sn. Then An ¥> 0, since the identity
permutation ι €E A„: . ,
jjja-j, = nft--
= 1
2. If a and r € Sn, then
ka
ГТ Цат) — ί(στ) _ ГТ к(ат) - ΐ(στ)
i<k к - i ~ кк к — г ка
га
la
ГТ (ка)т — (ίσ)τ _ ГТ ка — ία
t<k ka — ia ><* к — г
(3.1)

62
GROUPS AND SUBGROUPS
[CHAP. 3
We will prove that /f v
Π (fct " ** = Π^^- (3.2)
i<k ka — la m<i I — til v '
To do this we will show that each of the factors -τ corresponds to one and only
ih \ - С \ I- m . _
one of the factors ^—r~—^^- ■ Corresponding to each of the factors A—-f in the
ka — ю «το ι — m
right-hand side of {3.2), there exists, since a is a permutation, unique integers p, q such
that pa = I and qa = m. If ρ > q, then a factor
(ρο)τ - (да)т _ It — mr
pa - qa ~ I - m
appears on the left-hand side of (3.2). If ρ < q, then a factor
(qa)r - (ρα)τ _ (ρσ)τ — (qa)r __ It - fflr
qa - pa ~~ pa - qa ~ I - m
appears on the left-hand side of (3.2). We associate with the factor , _ - the (equal)
factor Mr ~ Ыг if p > q and the (equal) factor ^T ~ ^T if p< q. It is
pa- qa x ' qa — pa
easy to see that each of the factors of the right-hand side of (3.2) corresponds in this
way to one and only one of the (equal) factors of the left-hand side of (3.2). Hence (3£)
holds. From (3.1) we therefore obtain
Д Цат) - г(ат) = ГТ It - Шт _ γ[ for - jg
It follows from (3.3) and the rules for multiplying +1 and -1, that we have
Lemma 3.2: The product of
(i) two even permutations is even
(ii) two odd permutations is even
(iii) an odd permutation and an even permutation is odd
(iv) an even permutation and an odd permutation is odd.
Accordingly if a is even, then σ_1 is even too, since σα~ι = t is even. Thus if α, τ £ An,
στ~ι £ A„. A„ is therefore a subgroup of Sn. It is called the alternating group of degree n.
As an illustration let us find the multiplication table for A,. From the list of elements
of Si given in Problem 3.21 we determine that the even permutations are
= /1 2 3 4\ /1 2 3 4\ /1 2 3 4\ /1 2 3 4\
^12 3 4/ σ* _ \2 1 4 Ъ) T» ~ [ъ 2 4 l/ Γβ ^4 13 2J
_ /1 2 3 4\ _ /1 2 3 4\ _ /1 2 3 4\ /1 2 3 4\
"* ~ [Ζ 4 1 2/ r' " [l 3 4 2J T* ~ \4 2 13/ r' " [2 3 1 4j
/1 2 3 4\ /1 2 3 4\ = /1 2 3 4\ = /1 2 3 4
"5 1^4 3 2 lj *2 ^14 2 3/ Ts ^2 4 3 1/ Г8 \3 1 2 4
The multiplication table is easily written down from the multiplication table of Problem
3.21.

Sec. 3.3]
THE SYMMETRIC AND ALTERNATING GROUPS
63
1
σ2
*5
σ8
rl
T2
T3
U
TS
Te
T7
T8
t
1
σ2
σ5
<^8
η
τ2
τ3
τ4
τ5
Τ6
τ7
Tg
σ2
σ2
ι
σ8
σ5
Τ8
τ3
τ2
Τ5
Τ4
τ7
Τβ
ΤΧ
σ5
σ5
σβ
ι
"2
и
re
Τ7
τ1
Τ8
Τ2
τ3
Τ5
σ8
σΒ
σ5
σ2
ι
r5
τ7
τβ
TS
τ1
Τ3
τ2
τί
τ1
τ1
τ4
τ5
τβ
Γ2
t
σ5
τ7
та
σ8
"2
Τβ
τ2
τ2
Τ7
τ3
те
ι
τ1
τ5
σ2
σ5
τκ
τ4
σ8
та
та
те
τ2
Τ7
σ2
Τ8
η
ι
σ»
τχ
TS
"5
τ4
η
τ1
τ»
τ5
Τβ
σ5
ι
Τ3
τ7
"2
^8
τ2
Τ5
Τ5
τ8
τ1
τ4
Τ7
σ8
σ2
Τ2
Τβ
t
σ5
τ3
τβ
τβ
Τ3
τ7
ΐ"2
σ5
Τ4
τ8
σκ
t
Τ5
τι
σ2
Τ7
Τ7
τ2
τβ
τ.4
"Β
Τ5
τ1
^5
^2
Τ4
rfi
1
τβ
τβ
Τ5
τ4
Τι
τ3
σ2
σ8
τ6
τ2
°5
ι
ΤΊ
Problems
3.23. Write out a multiplication table for (i) Ab (ii) A2, (iii) A3.
Solution:
(i) There is only one element in Si, namely ι = I j , and it is an even permutation. Hence a
multiplication table for At is t ΓΓ1. Notice Αλ is the same group as St.
ί ϊ)·{\ ι2)}· (ί Ι)is
(ii) S.
tion.
for A2,
an even permutation and
Ί 2
(ί ί) *
an odd permuta-
Therefore A2=}[ jL and ι QJ where ' = (. „) is a multiplication table
(iii) 53 contains six elements (see example in Section 3.3a). The elements ι, σχ and σ2 are the even
permutations, and a multiplication table for A3 is
σ2
ι
σ1
σ2
(
σ1
σ2
"t
"2
ί
σ2
ι
σι
3.24. Prove Α„ = 5Β implies η = 1.
Solution:
If n> \, Sn must contain a permutation which interchanges 1 and 2 and leaves everything
else fixed, i.e. lr = 2, 2τ = 1, and £r = t (t = 3, .. ., n), гй А„, since τ is an odd permutation,
and therefore A„ =^ S„. By Problem 3.23(i), Ax = St. Hence An = Sn implies и = 1.
3.25. Show that the set S = {ι, σζ, <τ5,σ8} is a subgroup of A4. (For notation see page 62.)
Solution:
If we look at the multiplication table for A4, we see that the product of any two elements in
S is again in S. It is clear that ajtake.S implies σί-σ~1€5 because σ, = π~ι 0" = 2,5, 8).
Hence S is a subgroup of A4.

64
GROUPS AND SUBGROUPS
[CHAP. 3
d. The order of A„
We now count the elements of An. Suppose that η ~ 2. Let τ be the following element
of S„:
v. 1 -»2, 2 ^ 1, 3 -» 3, ..., n^n
We claim τ is odd because
2r — lr 3r — It 3r — 2г Пт — lr Wr— (И— 1)г _ _
"2- Г 3 - 1 3-2 "η- 1" η - (и - 1) ~~
Now let tj, e2, .. .,ffc be the even permutations (i.e. elements of An). Then
ijr, t2r, ...,v (54)
are all odd; moreover, if ttr — tjT, then
Notice that this means there are at least as many odd permutations as even ones. Indeed
there are exactly the same number of each; for if ω is odd, then
ω = (n)r)r
since г2 = t. Since wr is even by Lemma 3.2, every odd permutation is listed in (ЗА). Thus
there are precisely the same number of even permutations as odd ones.
Consequently, as к is the number of even permutations, the number of odd permutations
is also k, and the number of odd and even permutations is 2k. But every permutation of
Sn is either even or odd, and |S„| — η! Therefore к ~ η 1/2.
Putting our results together, we have proved
Theorem 3.3: If η is any positive integer > 1, then Sn is of order η!, and An is of order и!/2.
Problem
3.26. Check to see whether \Aj\ {j = 2,3,4) agrees with Theorem 3.3.
Solution:
|Aa| =1 by Problem 3.23, and 2!/2 = 1.
|Aa| = 3 by Problem 3.23, and 3!/2 = 3.
|A4| = 12 by Section 3.3c, and 4!/2 = 12.
3.4 GROUPS OP ISOMETKIES
a. Isometries of the line
We begin with certain subgroups of SR, the symmetric group on R, the set of real
numbers. We think of the elements of R arranged as points on the real line. Then if а,5€й
it is clear what we mean by the distance between a and b, namely the absolute value, |a— b\,
of a — b. We denote the distance between a and b by d(a, b).
We define a group I(R), the group of isometries of R, as a subgroup of SR in the
following way.
1. Let I(R) be the set of all elements of SR which preserve distance. The elements of this
set will be called isometries of R. To put this definition more explicitly, an element
σ € SR is termed an isometry if and only if
d(a, b) = d(aa, ba)
for every pair of elements a, b € R, Since the identity mapping ι € I(R), I(R) ·¥■ φ.

Sec. 3.4]
GROUPS OF ISOMETRIES
65
2. Suppose aGl(R), Then of course σ J G Sr. We claim alGl(R). To see this,
suppose а,Ъ ей. Then
d{aa~\ bo-1) = d((aa-l)a, (ba~l)a) = d(a, b)
as σ is an isometry. Hence
d(a,b) = d(aa-\ δσ"1)
which is precisely what we need. Thus σ-1 e/(Й). Suppose σ, f ε ДЙ); then г_1£/(й)
and
^(στ"1), Цат-1)) = ίί((θα)τ-\ (Ьа)^1) = ά(<1σ, Ъа) = <ί(α, Ь)
Thus στ-1 е/(й) and /(Й) is a subgroup of SR. Considered as a group in its own right,
I(R) is called the group of isometries of R.
Problems
3.27. Is σ an isometry? (In the following, ж G β.)
(i) σ: ж -» ж 4- 2. (ii) σ : ж -» nx, η an integer Φ 1 or —1. (iii) о: ж -» ж2, (iv) σ: ж -» —ж.
Solution:
(i) First note that σ £5R. <r is an isometry, since d{x,x') = \x — x'\ and
ά{χσ,χ'σ) = й(ж + 2, я' + 2) = \{х + 2) - (ж' + 2)) = |ж - ж'|
(ii) it is not an isometry, since d(x, ж') = |ж — ж'| and
ά(χσ,χ'σ) = d(nx, nx') = \n(x — x')\
so that, for ж =^ ж', d(a;, ж') =^ ^(жс, x'o), e.g. ж = 2, ж' = 1 implies d(x, x') = 1 and
d(xo,x'o) — \η\.
(ίϋ) σ is not an isometry, since d(x, ж') = |ж — ж'| and
d{xc,x'c) = ^(Ж2, Ж'2) = |Ж2-Ж'2|
so that, for ж э* ж', й(ж, ж') =^ й(жст, ж'<т), e.g. ж = 2, ж' = 0.
(iv) σ is an isometry, since σ £ SR and
d(xa, x'o) = d(-x, -x') = \-x - (-x')\ = \-x + x'\ = |ж - ж'| = й(ж, ж')
3.28. Set Ισ = 2, 2<τ = 1 and же = ж for all ж G β excepting 1 and 2. Is it an isometry?
Solution:
No, since d(l, 3) = |3 - 1| = 2 and d(lir, 3ir) = d(2, 3) = 1.
b. Two points determine an isometry
The following lemma gives us a method of determining whether the two isometries are
the same.
Lemma 3.4: If α, τ € I(R) have the same effect on two distinct real numbers a and b, i.e.
ασ = ατ and Ъа — Ът, then σ = т.
Proof: Let с be any element of R. Then
d(c,a) = d(ca,aa) — \ca — ασ\ — d(cT,aT) = \cr — ar\
and hence ca — aa — ±(c~ - aT)
Assume ca ¥* Ст. Since aa = ατ, ca — ασ = +(ст - α,τ) implies ca — ст = aa — от = 0, i.e.
са = ст, contrary to our assumption. Therefore
Ca - da = — (CT — Or) = — CT + Ο,τ, i.e. Ca + CT — 2(θσ)
Similarly, ca + cT = 2(ba). Hence 2aa = 2ba and aa = 5σ. But σ is a permutation.
Consequently a = b, contrary to the hypothesis that a and b are distinct real numbers. We
conclude that ca = cT and, since с is any element in R, a = T.

66
GROUPS AND SUBGROUPS
[CHAP. 3
"Using this lemma it is possible to describe the elements of ЦК). Let σ G I{R) and let
Οσ = a. Now ά(0σ, la) = d(a, la) -\α- la\ = d(0,1) = 1. Hence
a — la — ±1 or 1σ = α ± 1
(i) 1σ = a + 1 and Οσ = a. Let σ* be the member of I(R) defined by mapping r G R to
r + a. a* is clearly an isometry. Then a* and σ agree on 0 and 1. Hence σ = σ*.
(ii) 1σ = a — 1 and Οσ = a. Let a* be the member of I(R) defined by mapping r G R to
—r + a. Then σ and a* agree on 0 and 1. Hence a — a*.
Thus if σ G I(R), ra — er + Οσ where e is either 1 or — 1.
Geometrically, if ra — r + a, it "moves" the real line α units to the right. If ra= -r + a,
the real line is inverted about the origin and then moved a units to the right.
We come now to an interesting subset of / = I(R) which we will prove is a subgroup.
Let (/: Z) = {a | a G /, ησ G Ζ whenever η G Z}. Let a, T G (/: Z). Let 0<т = а, От = b.
a and b must be integers. The effects of σ and T are
гет = er + a where e = 1 or —1
rT = ψ + b where η = 1 or -1
Let μ■.r^■ψ — ηb. Then μ = г-1, for τ·(τμ) = (??Γ + ί>)μ = ^(^r + Ь) — т?Ь = г and
similarly _τ-(μτ) = г. Hence τ·(στ-1) = (ΐη)τ + rj(a—b). Clearly e^ is ±1, and η(α - b) G Z.
Thus if n is an integer, п(ат~l) G Z. Therefore aT~l G (/: Z) and (/: Z) is a subgroup of /.
Problem
3.29. Determine whether the following sets of mappings, with composition as the binary operation, are
groups.
(i) The set of all mappings of the plane of the form
^a : (x, V) ~* ix + <*■' У + a) wi*h (*>!/) £ ί2. α& R
Notice that τα is defined for each real number a.
(ii) The set of all τα with a G Z.
(iii) The set of all τα with a£Q.
(iv) The set of all mappings of real numbers of the form
μα : χ -> ax with α ^ 0, aeQ
(v) The set of all mappings of the plane R2 of the form
/Ό : (x, У) ~* iax< aV) with α G Q
(vi) The set of mappings of (v) but with a £ Q*, i.e. α φ 0.
Solution:
(i) We show first that τα is a permutation of Д2. τα is a matching of R2 -* Д2, for if (x, у)та =
(ж,, j/t)ra, then (x, y) = («!, j/j). If (ж, J/) G R2, there exists (a; — а, у — a) G Й2 and (a; — a,
у — а)та = (χ, у). Hence та is а one-to-one onto mapping and so та is a permutation of R2.
τ_α = r^1. The set of all τα is not empty, and τατ^1 = тат~ь ~ га_ь since (χ, у)тат^l =
{χ, j/)rar_b = {χ+ a, V+ а)т_ь = (x+ a — b, y+ а— Ь) = (х,у)та^.ъ. Thus the set of all ra is
a subgroup of SRi and so it constitutes a group.
(ii) Using (i), we need only consider whether τατ~ι = ra_b belongs to the given set, which it
clearly does.
(iii) This follows by arguing as in (ii).
(iv) It is easy to verify that each μα is a permutation of R. Also, /ι<ι/α) = (μα)~ι- Now μαβ^1 ~
iWi/ь = /Ό/ь- Hence the set of all μα, α¥Ό, aGQ, constitutes a subgroup of the group of all
permutations of R and itself forms a group.

Sec. 3.4]
GROUPS OF ISOMETRIES
67
(v) The set of μα is not a group, as μ0 has no inverse. For ^ maps all points onto the point (0,0), and
by Theorem 2.4, page 36, the only elements of MRi which have an inverse are the one-to-one
and onto mappings.
(vi) The set of all μα in this case forms a group. The set is non-empty. If μα is an element, then
Μι/α = ?al because {χ,ν)(μαμ1/α) = (αχ,αν)μ1/α = (x,y) and (χ,ν)(μ1/αμα) ~ (x, у). By Theorem
2.4 μα is a one-to-one onto mapping and thus μα is a permutation of R2. Now μαμζι — /%/Н/ь =
μα,,, since (χ,ν)μαμ~χ = (ax,ay)μ'1 =(|ж,% )= (χ,ν)μα/ϋ. Hence the set of all βα is a
subgroup of SRi. \b b I
c. Isometries of the plane
Let Ε be the set R2 — RxR. If (xA, у a) — A, (хв, ув) - В are two elements of E, we
define the distance between A and В as
У(жа - xBf + (yA - ув)2
and denote it by d{A,B). Recall that if we interpret A and В as points of the Euclidean
plane with coordinates (xa, у a) and {хв, ув) relative to a given cartesian coordinate system,
then the formula for d(A,B) is the ordinary distance between A and B. The interpretation
of Ε as the Euclidean plane is not an essential part of our argument. Logically we do
without it and work abstractly with the ordered pairs (ж, у).
σ e Se, the symmetric group on E, is called an isometry if for any А, В points of E,
d(A,B) = ά(Ασ,Βσ).
Theorem 3.5: The set / of all isometries of Ε forms a subgroup of Se.
Proof: I is not empty, since the identity mapping is an isometry. We need to show
only that στ~ι&1 whenever σ,τ&Ι. Consider the effect of г-1. As reSg, there exist,
for each pair of points А, В e E, A', B' e Ε such that A'r = A, B'r = B. Then
d(A', B') = d(A'r,B'r) = d(A,B)
since τ is an isometry. But At1 = А', Вт~г = В'. Hence
d{Ar-\Br-1) = d(A,B)
and τ~ι is an isometry. Consequently
d{Aar\ Βστ-1) = d{Aa, Βσ) = d(A, B)
and so στ~Ύ e /. Hence / is a subgroup of SE.
In the next four paragraphs we shall argue informally.
Let us imagine that the Euclidean plane Ε is covered by an infinite rigid metal lamina
S. If we move <S so that it still covers E, we can define an isometry induced by that
movement. Let the points of Ε be denoted P, Q, R, ... and the points of S be denoted A, B, C,... .
In the initial position suppose that A lies on top of Ρ, Β on top of Q, С on top of 22, ... .
After the sheet S is moved, the point A is on top of another point of E, say Pi; the point В
is on top of, say, the point Qr, С is on top of, say, 22i;... .
ABC <— Metalliimiim___ ABC
» m ^_^ —, · · · m ·—
Ρ Q R ^-Euclidean plane -* Ρ Q R P, Q1 R1
(a) Initial position. (6) After a movement.
Fig. 1. The isometry induced by a movement.

68
GROUPS AND SUBGROUPS
[CHAP. 3
Let Θ-.Ε-+Ε be defined by Ρθ = Pi, Qu = Qu RO = Ri, ... . Then we assert θ is an
isometry. For if P,Q are any two points of E, with, in the initial position, A of S on top
of P, and В of S on top of Q, then we have d(A,B) = d(P,Q). After S is moved, A lies on
top of, say, Pi, and В on top of Qb Hence d(A, B) = d{Pu Qi) and so d{Pu Qi) = d(P, Q).
Using this informal approach, we now describe three particular isometries:
1. Rotation about a point. Let 0 be any point of S. Rotate S about 0 through an angle
Ψ. Then the isometry induced by this movement of S is called the rotation about 0
through an angle Ψ.
(a) Initial position. (b) A rotation through an angle Ψ.
Fig. 2
2. Reflection in a line. Let us choose a line in Ε and turn S over this line and back to E.
The isometry obtained in this way is called the reflection in XY.
(a) Initial position. (b) Rotating about XY. (c) Final position.
Fig. 3. A reflection in line XY.
3. Let us choose a line XY. Let θ be an isometry corresponding to a movement of S for
which ΧΘΥΘ, the line joining ΧΘ and Υθ, is parallel to XY. Then θ is called a translation.
We now describe formally these three types of isometries. As a simplification we
describe only rotations about the origin and reflections about the axis OX.
1, A translation таЬ is the mapping defined by
(х>У)та,ь = {х + а,У + Ъ)
It can be shown that for each a,b, T|l>b is an isometry and that (ταΛ)~Ί =т_(1>_ь. (See
Problem 3.30 for details.)
2. A counterclockwise rotation about the origin through an angle θ is the mapping pe
defined by
(x, y)pe = {x cos θ — у sin θ, χ sin θ + у cos θ)
For each θ, ρβ is an isometry and (pe)-J = p_e. (See Problem 3.31 for proof. Problem
3.32 shows why ρ is called a rotation through Θ.)

Sec. 3.4]
GROUPS OF ISOMETRIES
69
3. Define a reflection in OX to be the map ay where
{χ, y)°y = (χ, -v)
This is readily seen to be an isometry and it is easy to show that (<гу)~л = <ry (Problem
3.33).
Since t is the product of two reflections, we call it a reflection. This is a convenience
which will simplify the statement of Theorem 3.8.
Problems
3.30. Show that та>ь is an isometry and that (·τα^,)_1 = т_а>_ь.
Solution:
First we must show that таА, £ SE, so we must show that it is a one-to-one onto mapping.
(я>У)та,ь = (я'>У')та,ь clearly implies (x,y) = (x',y'). If (x,y)<ZEr then (x - а, у- Ь)та_ь =
{χ,у) and so та,ь is onto. Hence rab G Se- Is та,ь an isometry? If A = (xA,yA) and В = (xB,yB),
then
d{A,B) = t/(xa - xB)2 + (yA - yB)2 = d{Ararb, BTa,b)
and та-Ь is an isometry. {ам/)та,ьт_а,_ь = (x + a, У + Ь)т_ау _b =-- (a;,?/). Hence rajbr_aj _b = i.
Similarly, г-а,-ьта,ь = ' and so (га.ь)"1 = т_а,_ь.
3.31. Show that о is an isometry and that (p ) l = ρ . (Hard.)
Solution:
We must show first that p is an element of SE. (x, y)p == (x'r y')p implies
0 6 0
χ cos β — у sin β = χ' cos θ — у' sin 9
χ sin 9 + з/ cos β = χ' sin θ + у' cos 9
Multiply the first equation by cos θ and the second by sin θ and add to obtain
ic(cos2 β + sin2 β) + (—у cos β sin 9 + у cos 9 sin 9)
= «'(cos2 9 + sin2 9) + {—3/' cos 9 sin 9 + y' cos 9 sin 9)
Since cos2 9 + sin2 9 = 1, χ = x'. Similarly by multiplying the first equation of (S.S) by sin 9 and
the second by cos 9 and subtracting the first from the second, we find у = у'. Hence ρ is one-one.
Is it onto? In other words, can we find (x, y) such that (x,y)p ={a,b) for any (ct, b) £ i?2?
That is, does there exist a solution to
χ cos β — у sin 9 = a
(3-е)
χ sin 9 + 3/ cos 9 = &
for x,y £ R1 We solve these equations for a; and у using the same stratagem as above, i.e. multiply
the top equation by cos 9 and the bottom by sin 9 and add to obtain
«(cos2 9 + sin2 9) = a cos 9 + b sin 9 or a; = ct cos 9 + & sin 9
Multiply the first equation of (3.6) by sin β and the second by cos 9 and subtract the first from the
second to obtain
у = i/(sin2 9 + cos2 Θ) — b cos θ — a, sin 9
(3.5)
On substituting these values of χ and у in (3.S), it is easily seen that they satisfy the equation.
I cos 9 —sin 9
(The reader who knows that the condition for existence of a solution to {3.6) is J ¥· 0,
can verify the existence of a solution to (3.6) immediately.) I sm β cos β
Finally, is ρ an isometry? If A = (ха,Уа) and В = (хв,ув), then

70
GROUPS AND SUBGROUPS
[CHAP. 3
d(Ap , Bp )
e e
= V[{«a cos в - yA sin Θ) - (xB cos 9 — yB sin 9)]2 + [{»„ sin 9 + yA cos 9) - (xB sin 9 + з/в cos 9)]2
= У[(«а-«д) cos 9 - (j/A - yB) sin 9]2 + [{xA - xB) sin 9 + (yA - yB) cos 9]2
= V{a?A - *в)2 [cos2 9 + sin2 9] + (yA- yB)* [cos* 9 + sin2 9]
= V(^-^b)2 + (3/a-3/b)2 = d{A, B)
and thus ρ is an isometry.
(x, y)p a = (x cos 9 — у sin 9, a; sin в + у cos 9)p
θ "" 0 —θ
= [{ж cos 9 — у sin 9) cos (—9) — (χ sin 9 + у cos 9) sin (—9),
(ж cos 9 — у sin 9) sin (—9) + (x sin 9 + у cos 9) cos (—9)]
= [a;(sin2 9 + cos2 9), j/(sin2 9 + cos2 9)] = (ж, з/)
and so ρ ρ_ —ι. Similarly p_ ρ = ι.
9 —0
3.32. Show that (0,0) is equidistant from (x,y) and (x,y)p , and that the smallest angle between the line
Lj joining (я, y)p and (0,0), and the line L2 joining (x,y) and (0,0), is в when 0 — 9 — jr, or
2ir — 9 when тг — 9 — 2ir.
Solution:
As we have shown in Problem 3.31, rf((0,0), (x,y)) = rf((0, 0)pe, (я,з/)р ) = d{(0, 0), (я,з/)р ). We
remind the reader of the formula for calculating the angle between two lines meeting at the origin.
If the one line Lj has end point {a,1, bi) and the other Ъ% has end point (o^, b2), then the cosine of
the angle between L1 and L2 is given by
a1a2 + &i&2
V(*f+ &*)<«* +&*)
Put (a^bj) = (ж, у), (a2,b2) = (ж,з/)р . Then if μ is the smallest angle between the lines Lj and L2,
(iC2 + J/2) COS 9
COS μ = ——'—; ' = COS 9
V(«2 + 3/2)(ir2 + 3/2)
Hence μ = 9 if 0-9-it, and μ — 2π — 9 if ir ^ 9 - 2w.
3.33. Show that oy is an isometry and that o·2 = ι.
Solution:
If A=(xA)yA) and В = (жв,з/В),
rf(Acra, Ясгй) = уГ{ха - *в)2 + {-Уа ~ (~Ув))2 = d(A,B)
and so oy preserves distance. Obviously {x,y)ay = (x\ y')oy implies (x, y) = {«', y'). And σ„ is
clearly onto, for (x,-y)ay = (x,y). Hence σΒ is an isometry. (x, y)ayny = (x, ~y)<ry = (x,y). Thus
3.34. Show that rotations about the origin form a subgroup of /.
Solution:
Po, the rotation through zero degrees, is the identity; hence the set of rotations is not empty.
If Ρβ>Ρφ are two rotations, (p,)"1 — P(_aj· Put ~Φ — *» as ifc is annoying to carry the minus sign.
Is ρ (ρ )_1 ~ ρ ρ, a rotation?
β φ a +
(*> y)p ρψ — {x cos θ — у sin 9, χ sin 9 + 3/ cos 9)p
= ({x cos 9—j/ sin 9) cos Sir — (a; sin 9 + 3/ cos 9) sin Sir,
(ж cos 9 — г/ sin 9) sin Φ + (χ sin 9 + у cos 9) cos Ф))
= (ic(cos 9 cos Φ — sin 9 sin Sir) — 3/(sin 9 cos % + cos 9 sin Sir),
ic(cos 9 sin % + sin 9 cos Sir) + i/(cos 9 cos Φ — sin 9 sin *))
— (a; cos (9 + Φ) — у sin (9 + φ), χ sin (9 + %) +3/ cos (9 + Ф))
Hence ρ ρ -= ρ , and ρ (ρ J-1 is a rotation. Thus the rotations form a subgroup of /.

Sec. 3.4]
GROUPS OF ISOMETRIES
71
3.35. Show that the reflections about OX form a subgroup of /.
Solution:
We have only two elements, for there are only the two reflections ι and σ„. Since we have shown
that ay = σ""1, we quickly verify that the two possible products of a reflection and the inverse of a
reflection is again a reflection. Hence the set of reflections forms a subgroup of / and is of order 2.
3.36. Show that the set of translations forms a subgroup of /.
Solution:
та,ъ(тс,й)~~1 = ЧЬт„с,-й = ra__Cjb^d is easily verified as
(х>У)та,ът~с.~<1 = (x + a> У + Ь)т-С,-Л = (x+a-c,y + b-d) = (я, У)та-СзЬ-а
and thus the set of translations forms a subgroup of /.
3.37. Find an element of SE that is not an isometry.
Solution:
Let v<=SE be defined by (χ,ν)σ = (x,y) except that (0, 0)σ = (1, 0) and (1, 0)σ = (0, 0). It
is easy to verify that σ G SE. Now if A = (0,0), В = (1,0) and С = (0,1), then l = d(A,C).
ά(Ασ, Co) = d(B, C) = y/2 Φ d(A, C). Hence σ is not an isometry.
d. Isometries are products of reflections, translations and rotations
We will prove in this section that an isometry is determined uniquely by its action on
any three points not all on a straight line. This enables us to prove that every isometry is
expressible as the product of a reflection, a translation and a rotation. (Note, however,
that there are isometries that are neither reflections, rotations, nor translations.)
Lemma 3.6: Let a £ /. Let А, В and С be three noncollinear points in E. Let Aa — A',
Βσ = В' and Ca = С Then A'B'C is a triangle congruent to ABC.
Proof: d(A', Bf) = d(A, B), d{A', C) = d(A, C) and d(B', C) = d(B, C). Thus we have
two triangles with corresponding sides equal, and so ΔΑΒΟ is congruent to ΔΑ'ΒΌ'.
Lemma 3.7: If σ and r, elements of /, have the same effect on three points A,B,C which
do not lie on a straight line, then σ = г.
Proof: Let D be any point of E. Let d(D, A) = a, d(D, В) = Ь and d{D, С) = с Let
A' = Ασ = Ατ, Β' = Βσ = Вт and С = Са = Ст. Then the distance of Da from A' is a,
from B' is b, and from С is c. Similarly DT is a distance a from А', Ь from Β', and с from
С Geometrically it is possible to see that Da = Dt, for both Da and Dt lie on the
intersection of three circles: Ολ with center A' and radius a; 02 with center B' and radius b;
and On with center С and radius с
Two circles intersect in two points at most. Hence Oi and 02 determine two points.
It is possible for Da and DT to be these two points. But we shall show that as Da and DT
must lie on 03, they must be the same point. If this is not so, then Oi, 02 and 03 have two
points in common. We shall prove geometrically then that А, В, С must lie on the same
straight line.
Let three circles with centers A', B' and С
intersect in two points Ρ and Q. Without loss of
generality we may assume that B' lies between A' and
C". A'P = A'Q and B'P = B'Q; hence A'B' lies on
the perpendicular bisector of PQ. Similarly B'C
lies on the perpendicular bisector of PQ. Therefore
A'B'C is a straight line. But ΔΑΒΟ is congruent
to ΔΑ'ΒΌ' by Lemma 3.6. Hence ABC lies in a
straight line. But we assumed this was not so.
This contradiction proves Lemma 3.7.

72
GROUPS AND SUBGROUPS
[CHAP. 3
Theorem 3.8:
Every isometry of Ε is expressible as the product of a reflection, a rotation
and a translation.
We present an intuitive proof first. The formal proof below follows exactly the same
steps.
Intuitive proof: Let a S /. Let A = (0,0), В = (1,0) and С = (0,1) as shown in
Fig. 1. Our idea is first to find σ1 as a product, Φ, of a translation, a rotation and a
reflection. It is easy then to prove σ is the product of a reflection, rotation and translation.
So we must find a Ψ which is the product of a translation, rotation and reflection such that
σΨ = t. To check that σΨ = ι, we need only prove that for the three noncollinear points
А, В, С the effect of t and σΨ is the same (Lemma 3.7). We build Ψ in three stages so that
we bring (a) Ασ to A; (b) Βσ to B; (c) Ca to C. Let Ασ = (a, b).
О
Βσ·
cv
Ασ
в Χ
о
Btrr-a,~b
β
/
I*
В'
в
= А
or_ai_b
•Οστ_
о
Bc""-a.~bP-fl
- -Аот_а__ьр_в
C<rr_aj_bp_e — С
Fiff. ί
Fig. 2
Fig. S
Apply the translation τ_α _b which moves each point a distance d(A, Ασ) parallel to the
line joining Ασ and A, so that Αστ
A. Let Β' = Βστ
-о, -Ь"
Then B'A is at an angle
0, say, to (λΧ and, since τ _. and σ are isometries, is of length 1. See Fig. 2.
Apply the rotation through an angle of —0 that takes B' onto B. Let С = Cffr_a-_b p_9.
Then since σ, r_0 _b and p_g are isometries, C" is at a distance 1 from A and a distance
\/2 from B. Hence С is either as in Fig. 3, in which case let μ be the reflection in OX, or
else C' = C, in which case let p. be the identity reflection. Let Φ = р._аз-ьр_9р.. Since
А, В and С are mapped to Α, β and С by σΨ, σ* = ι by Lemma 3.7. Hence, using our
remark on the inverse of a product in Section 2.4, page 34,
•a,b
σ = *-i = д-Чр-.)-1^-.,-,,)-1 = μρ,-
and the theorem is true.
Formal proof: We follow the same three steps and use the same notation. Thus
A = (0,0), β = (1,0), С = (0,1) and σ 6/.
(α) Suppose Ασ-{α, b). Then Аат_а_ь = A.
(Ь) В' = 5στ_α _ь must be a distance 1 from A, since 1 = d{A,B) = d(A«rr_0i_b, Ват^ал .^b) =
<i(A, #'). Hence B' is of the form (cos0, sin 0) for some angle Θ. [This is a well known
fact of coordinate geometry. All we must check is that if B' = (d, e) with d? + e2 = 1,
the equations cos θ = d and sin θ = e can be solved for 0.] Then
B'P-e = (COS^, Sin0)p_e
= (cos 0 (cos-0) - sin 0 (sin-0), cos 0 sin (-Θ) + sin 0 cos (-0))
= (cos2 0 4-sin2 0, 0) = (1, 0) = В
Also, Ap = A.

Sec. 3.4]
GROUPS OF ISOMETRIES
73
(с) С = Сат_а__ъР_в is a distance 1 from A and \/2 from B, since Ааг_ам_ъР_в = A,
Вот_а,_ьр_е = B, and от_а<_ьр_в is an isometry. Thus С = (0,1) or (0,-1). Let μ be
the identity if С = (0,1), and let μ be the reflection σ„ if C" = (0, -1). Put
Φ = τ„αд_ьр_6ц. Then ΛσΦ = Λ, ΒαΨ = В and СаФ = С. Hence σΦ = ι by Lemma
3.7, and σ=φ-ι = μΤο>6ρ#.
e. Symmetry groups
Given a figure in the Euclidean plane, we shall define a group which we will call the
symmetry group of the figure.
The word symmetry is not sufficiently precise for the needs of Chemistry and Physics;
so instead of talking about symmetry, we talk about symmetry groups. In comparing two
figures, it usually turns out that what one would normally think of as the more symmetrical
figure has a symmetry group of greater order than the less symmetrical figure. Also, the
symmetry group of what we would normally call a non-symmetrical figure usually turns
out to be of order 1. More generally, we will be concerned with subsets of the Euclidean
plane. (Clearly, a figure is a subset of the Euclidean plane.)
Theorem 3.9: Let £ be any subset of the Euclidean plane. The set, denoted by h, of all
σ € / such that (i) s € S implies so £ S and (ii) to £ S implies t € S,
forms a subgroup of /, called the symmetry group of S. (An element of h,
therefore, is characterized by its mapping elements of S, and only elements
of S, into S.)
Proof: Is ·* 0, as the identity mapping of the Euclidean plane into itself is in Is. If
σ,τ € h, is στ-1 € /s? We first prove τ-1 G /s. If s € S, (βτ~1)τ = s <= S. Because τ € /s,
(ii) implies st_1 G S. Thus τ"1 satisfies (i), i.e. s G S implies St'1 € S.
To show r_1 also satisfies (ii), let tT~l G S. Then (tr*1^ = t € S, since τ satisfies (i).
Hence τ'1 satisfies (ii), and τ-1 G /s.
Now we show στ"1^^. Let s€S. Then So € S and sot~1GS, since σ and t_1 are
in h. Therefore στ-1 satisfies (i). If iar^GS, since τ-1 satisfies (ii), we have to G S.
Furthermore, since σ G /s, (ii) implies t G S and consequently от-1 also satisfies (ii).
Hence or-1 € Is and /s forms a subgroup of /.
Problems
3.38. Find a plane figure S such that
U — {σ Ι σ ε / and for all s S S, βσ S S}
does not form a subgroup of /. (In other words, in Theorem 3.9 we cannot drop condition (ii).)
Solution:
Let S be the infinite half-line starting: at (1,0), i.e. S = {(x,0) \ x^l). Then ть0 S C7. Now
t~J = r_I0. But t_1-0 moves (1,0) 6S to (0, 0) Й S. Hence U is not a subgroup.
3.39. Find the orders of the symmetry group of
(i)
s \E ίο ρ
I μ
В ii С
—-F4 x

GROUPS AND SUBGROUPS
[CHAP. 3
(Hint: Use Theorem 3.8 and argue intuitively. A useful intuitive approach for this problem is to
cut a cardboard figure corresponding to each figure, label its vertices on both sides, and draw its
perimeter onto a sheet of paper. The isometries are obtained on moving each cardboard figure so
that it lies on the drawn perimeter.)
Solution:
(i) Let S = ABCDEF. Let the images of A,B,C,D,E and F under σ e /s be denoted by
A',B',C',D',E' and F'. By Theorem 3.8 it is easy to see intuitively that the image of the plane
figure S will be the congruent figure A'B'C'O'E'F', since Theorem 3.8 states that every element
of / is a product of a reflection, rotation and translation. But if a rigid body is rotated,
translated or reflected it retains the same shape. Now A'B' must lie along AB, as all other
sides of S are either smaller or larger than d(A',B') — d(A,B), which means F' must lie on
F and hence E' on E, etc. Thus σ must be the identity. Accordingly IB — {<} and is of order 1.
(ii) Let S = IJGH. Now |/s| is at least 8, for we have as members of /s:
st: A reflection in the diagonal GI,
Gst - G, Hst = J, /»! = /.
82: A reflection in the diagonal HJ,
Gs2 = /, Hs2 - H, Isz = G.
S3: A reflection in OX,
Gs$ = J, Hs3 — /, /s3 = H.
s4: A reflection in OY,
Gst = H, HsA = G, Is4 = J.
ss: A clockwise rotation about О of 0°,
G8S m G, Hss m H, hs = /.
s6: A clockwise rotation about О of 90°,
Gs6 = H, Hse = /, /s6 » J.
s7: A clockwise rotation about О of 180°,
Gs7 = J, Hs7 = J, Is7 = G.
s8: A clockwise rotation about О of 270°,
Gss - J, Hss = G, Iss - H.
That all of these are distinct is clear. Could |/s| be greater than 8?
Let s ejs. d(G,I) — d(Gs,Is) — y/2. Only two pairs of points of 5 are a distance V2~
apart, namely G,I and if,·/. Therefore the line Gsls is one of the diagonals of S. Similarly
the line HsJs is a diagonal of S. As s is a permutation of S, distinct points of S are mapped
by s to distinct points. This means that at most the following possibilities arise:
(1) Gsls is GI, HsJs is HJ or JH
(2) GsIsialG, HsJs is HJ or JH
(3) Gsls is JH, HsJs is GI or IG
(4) Gsls is HJ, HsJs is GI or IG
Since these represent eight distinct cases and each involves the movement of three points
not in a straight line, by Lemma 3.7 at most one isometry could correspond to any one case.
Then |/s| — 8. But we have already exhibited eight elements of /s. Hence |/s| = 8.
(iii) Let S be the triangle KLM. Then /s contains the following elements:
(α) σι, the identity mapping of /,
Koy — K, Lct! = L, Маг = Μ.
(6) σ2, the reflection in KN,
Κσ2 — К, Lo% — Μ, Μσ2 ~ L.
Hence |/s| — 2. Let σ e /s. Since d(M,L) = VI and the only two points of KLM which are a
distance τ/2 apart are L and M, then either

Sec. 3.4]
GROUPS OP ISOMETRIES
75
(α) Ma = Μ, La = L. Then as К must be a distance 1 from both Μ and L, Κα ~ Κ.
(b) Ma — L, La — M. Then as К must be a distance 1 from both Ma and La, i.e. from
Μ and L, and К is the only point of S which is a distance 1 from both L and M,
Ka = K. Hence |/s| ^ 2, by Lemma 3.7. Therefore |/s! = 2.
f. The dihedral groups
Let S be a regular w-gon, те > 2, e.g. one of the figures below. We will show that in
any isometry of S, vertices are taken to vertices. This will make it easy to determine the
order of the symmetry group of a regular и-gon, η > 2.
Δ
We will take the following geometrical lemma for granted.
Lemma 3.10: Every regular n-gon can be circumscribed by one and only one circle.
We call the center of the circumscribing circle of an и-gon its center.
Lemma 3.11: The center of a regular w-gon S is taken onto itself by any element of Is.
Proof: Since every point of S is within a distance r, say, from the center O, and σ is
an isometry, then every element of So is within a distance r from Οσ. Also, there are points
of So which are exactly a distance r from Οσ, as there are points of S which are exactly a
distance r from O. But So — S. Hence the circle with radius r and center Οσ is a
circumscribing circle of S. But by the previous lemma there is only one circumscribing circle of S.
Thus Οσ = О.
Lemma 3.12:
If S is a regular n-gon and σ G h, then vertices of S are taken onto
vertices of S by a.
Proof: If A is a vertex of S and О is the center of S, Οσ — Ο by Lemma 3.11.
d(Oa, Aa) - d{0, Ασ) — d(0, A). Hence Ασ is a distance r from O, where r is the radius of
the circumscribing circle C. The only points of S on the circumference of С are vertices.
But Ασ is an element of S on the circumference of С Thus Ασ is a vertex.
The symmetry group of the regular n-gon is called the dihedral group of degree n. We
can now calculate the orders of the dihedral groups.
Let the vertices of a regular n-gon S with center О be Alt ...,A„ (in a clockwise
direction).
Let ffj, 1 ^ j ^ n, rotate S about О in a clockwise direction through an angle -——
radians ( = —(j— 1) degrees ] so that Ατσ} = Ar As an example, the effect of σ3 on the
regular pentagon is shown below.
ι3σ3
σ3
5σ3

76
GROUPS AND SUBGROUPS
[CHAP. 3
Let τ be the reflection about the line through At and O, so that Atr = Av Α2τ = An.
The effect of τ on the regular pentagon is shown.
The following diagram illustrates the effect on a regular pentagon of the reflection T
followed by the rotation σ„.
(Α2τ)σ3
The elements σν .
Alaj ¥* Α ^, j ^ k. If
{ΑΎτ)σΆ
ση, τσν ..., τσα are all distinct- For certainly σ. ¥· ak, j ¥* k, as
then
Αιτσι ~ Ai°i ~ Αισι
к·
Thus τσ} — ak implies j
k.
But та. — aj implies r — σν the identity, contrary to assumption. Finally, τσ} — rak implies
°i = <V
So there are at least 2n possible elements of the dihedral group of degree n. But we can
easily show that there are no more than 2n. For if σ e /s, S the regular *t-gon, then there
are η possibilities for Ata. As vertices are taken to vertices, ΑΎσ is one of Av...,An.
Α2σ has only two possibilities once Ata is determined as ά(Αισ,Α2σ) = d(AvA2), and Α2σ
must also be a vertex. Once Al0 and Α2σ are determined, Ata, i — 3,4, ... ,n are also
determined. Hence there are at most two elements σ e /a which map ΑΎσ to Ay Thus there are
at most 2n elements of /s, and so |/s| = 2n.
Let Dn denote the dihedral group of degree n.
Problems
3.40. Find Ds and its multiplication table.
Solution:
The elements of Ds are the ajt τσ,- above. Note that σ^σ2 =
Also note that τ~ι = r and σ,τ = τ^σ^τ ~ ττσρ. Now τσγτ — σι
as Αιτσ^τ — A^ajT — A%r — А3; and Α^τσ^τ — А$г%т
cordingly the multiplication table is as follows:
σ2
σ3
τ
τσ2
τσ3
Oj +1 if 1 ~ j ~ 2, and σ3σ2 '
since <tl is the identity; тоуг ■
σ$τ.
Α^τ — A]. So σ3τ "— τσ% and τσ3
■= σι.
= σ3,
Ac-
σ1
σΙ
σ2
σ3
τ
τσ2
τσ3
σ2
"2
"3
<Ί
τσ2
Ttrg
τ
"3
"3
σ1
"2
τσ3
τ
τσ2
τ
τ
τσ8
τσ2
<Ί
σ3
σ2
τσ2
τσ2
τ
τσ3
σ2
<Ί
"3
τσ3
τσ3
τσ2
τ
"3
'2
<Ί

Sec. 3.5]
THE GROUP OF MOBIUS TRANSFORMATIONS
77
3.41.
Show that the following are subgroups of D3: (i) {ctj}, (ii) {alt σ2, σ3}, (iii) {τσ2, аг}. (Notation is
the same aa in the preceding problem.)
Solution:
It is only necessary to check in each case that the set is not empty and if g, h belong to the set,
gh~l belongs to the set. It is easy to calculate gh~* from the multiplication table of Problem 3.40.
3.42. Find Dit the symmetry group of the square, and its multiplication table.
Solution:
Notice that we have already found the elements of D4 in Problem 3.39(H). We will, however,
Use the notation of Section 3.4f, i.e. σ3·, j = 1,2,3,4, for the rotations and τ for the reflection.
Accordingly the elements of D4 are σ} and τσ,·, j = 1, 2, 3, 4. Now σ3σ2 — σ3+1 for 1 — j — 3,
σ4σ2 — σν and atOj — OjOt for all i and j. Also τ-1 = r, σ{τ — TratT, and тагт — σ^
We show τσ2τ — σ4. Αιτσ2τ — Αγσχτ '■- Α2τ ~ Αν Α2τσ2τ — Α^σ2τ ~ Αχτ '■- Αι, and Α3τσ2τ =
Аз"?? — Α^τ — Α2. Furthermore Αλσ^ ~ Ait Α2σ4 ~ Au and Α3σ4 = Α2. Since τσ2τ and σ4 have the
same effect on the three points Alf A2 and A3, τσ2τ = σ4 by Lemma 3.7.
The following calculations facilitate the construction of a multiplication table for D4. σ2 ~ σ3
implies τσ3τ — τσ2τ = (τσ2τ)(τσ2τ) ~ σ4 = σ3, an<^ σ4 = σ3σ2 implies τσ4τ - τσ3σ2τ — (τσ3τ)(τσ2τ) =
σ3σ4 = σ2. Hence σ2τ — ττσ2τ ~ τσ4, адт — ττσ3τ — τσ3, an<^ σ4τ ~ "σ^τ ~ τσ2. It is now easy to
construct the table:
(>l σ2 σ3 σ4 τ τσ2 τσ3 τσ4
σ2
σ3
τσ2
τσ3
τσ4
σ1
σ2
σ3
σ4
г
τσ2
τσ3
τσ4
σ2
σ3
σ4
σι
τσ2
τσΆ
τσ4
τ
^3
σ4
σι
σ2
τσ3
τσ4
τ
τσ2
σ4
σ1
σ2
σ3
τσ4
τ
τσ2
τσ3
τ
τσ4
τσ3
τσ2
σ1
σ4
σ3
σ2
τσ2
τ
τ·σ4
τσ3
σ&
σ1
σ4
σ3
τσ3
τσ2
τ
τσ4
σ3
σ2
σ1
η
τσ4
τσ3
τσ2
τ
σ4
σ3
σ2
σ1
3.5 THE GROUP OF MOBIUS TRANSFORMATIONS
a. Defining the group
The complex numbers can be represented as points of the Euclidean plane E, the
complex number z — x + iy corresponds to the point with coordinates (x, y). Instead of inquiring
(as we did in Section 3.4c) what are the permutations of Ε that preserve distance, we inquire
what are the permutations of Ε that preserve both angles and their orientation. These are
called conformal mappings. It can be shown (see Ford, L. R., Automorphic Functions,
Chelsea, 1951) that the mappings
σ = a(a,b,c,d) : z-
az + b
cz + d
where a,b,c,d are fixed complex numbers such that ad — bc¥*0, preserve angles and their
orientation. But σ(α, b, c, d) is not always a mapping of Ε to E. Two things can go wrong.
If c^O, then:
(i) ζσ is not defined if ζ = — die, as then the denominator becomes 0.
(ii) There is no complex number that maps to ale. For suppose za — ale, then az + b =
(cz + d)alc and b — adlc — 0 and hence bc — ad — 0, the very condition we assumed
did not hold.

78
GROUPS AND SUBGROUPS
[CHAP. 3
It seems as if we have been cheated in our efforts to argue analogously to Section 3.4c
in order to prove the σ for various a,b,c,d form a group, because not all the σ are
permutations of E.
However, by adding an extra element « to Ε and forming EU{<*>} — Ё we can
overcome these difficulties. « is any object outside E. It is customary to write « for historical
reasons. The reader is cautioned that just as the symbol χ can have different meanings
(e.g. χ is sometimes a number and sometimes an element of a group or a groupoid, etc.), so
«> has different meanings. The « we introduce should not be confused with the « in such
expressions as lim-= 0; it is logically distinct.
I-»oo X
Our idea is to extend σ to Ё in order to patch up the difficulties (i) and (ii) above so that
σ G Sg, the symmetric group on Ё.
(a) If с = 0, define za = ——-r for any complex number z, and put «σ = oo.
cz + a
b
(b) If с ¥* 0, put zv = ——τ for ζ ¥■ —die, z€C, the complex numbers. Put
(—d/c)a — «о and <*>v = a/c
In (a) we have no real problem. Having had to add an extra element, we just let it map
to Itself. In (b) we neatly get rid of both difficulties (i) and (ii) above, for we have both
denned {—dfc)a and found an element to map to a/c.
Μ will denote the set of all mappings of Ё to Ё defined by {3.7). We will show that Μ
is a subgroup of Sg, leaving most of the checking of details to the problems.
First, each of the σ(α, b, c, d) is a member of S$ (Problem 3.45). Next, the inverse of
v(a, b, c, d) is given by σ(—d, b, c, —a) (Problem 3.44). Finally,
σ(θι,6ι, Ci, <Ζι)σ(α2,&2, C2, da) = σ(θ3, Ь3, С3, ds)
for some choice of a3, Ьз, cs, ds (Problem 3.46). Note that σ(1,0,0,1) is the identity mapping
(denoted by <). Hence the product of an element of Μ and the inverse of an element in Μ
belongs to M. Thus Μ is a subgroup of S^. It is called the group of Mobius transformations.
Problems
3.43. Determine the image of (i) i, (ii) 1 + 2i, (iii) », and (iv) —1/3 under σ(2,1,3,1).
Solutions:
ω * --зТТТ = То~1ог (ш) "σ = т
is /ι_ι_ο·λ 2(1 + 2-Q + 1 9 1 . ,. . Λ ,„
<») (1 + 2*> = 80 + 20ΤΪ = ΐ8~ 26г (1V) ~1/3σ = β
3.44. Show that σ(—d, Ь, с, — α) is the inverse of σ(α, b,e, d), given ad—bc¥-0.
Solution:
Case (a): с = 0.
χ σ(α, b, 0, d) a(—d, b, 0, —a) -
<*> σ[α, b, 0, d) σ(—ά, b, 0, -a) = » σ(—d, b, 0, -a) = »
Case (b): с Ф 0.
(i) χ = -die.
ζ σ(α, b, с, d) a[—d, b, c, —a) = *> σ(—d, b, c, —a) = —d/c
;(^)(--) + ь]/(-.,

Sec. 3.5]
THE GROUP OF MOBIUS TRANSFORMATIONS
79
(ii) ζ Φ —die or «. Then
ζ σ(α, b, с,d) σ(-ά, Ь,с, -а) = (fHjTj)<-«*> + Ь / (^+^1 с ~ α
_ bcs + bd — adz — bd _ (be — aJ)g
~~ azc + be — acz — ad be — ad
ζ
«> σ(α, b, с, й) = а/с and (а/с) a(—dt b, с, —а]
Hence σ(α, Ь, с, d) a(—d, b, c, —a) = ι.
Similarly we can show that σ(—d, b, e, —α) σ(α, b,c,d) — и
Hence <?(—d, b, c, —α) = σ(α, b, e, d)~l.
—a) — <»
3.45. Prove σ(α, b, e, d) e S-g for any choice of a, b, c, d such that ои£ — be ¥■ 0.
Solution.-
In Problem 3.44 we have seen that each σ(α, b, e, d) has an inverse. By Theorem 2.4, page 36,
any mapping of a set into itself which has an inverse is a one-to-one onto mapping. Therefore
σ(α, b,e,d) is one-to-one and onto, and so σ(α, b, c, d) is an element of Sg.
3.46. Show that
ofa,, b„ d, <*!) σ(θ2, Ь2, e2, dj) = ст(оз, Ь3, с3, d3)
where a3 = α^ + bj}Clf b3 — c^bj + b2dj, e3 = a^ + <£2c1( and d3 — Ojc2 + d^. Prove also that
a3ds — bses ¥· 0. (Hint: This is very much an endurance test.)
Solution:
Let σ(βι, Ь„ e„ я^) = σ„ σ(α2, Ь2, e2, rf2) = "ь a"d σ(α3, Ь3, e3,<f3) = σ3. Note that
*з<*з ~ &зсз — (αι«2 + &2ci)(&i"2 + ^1^2) — (bi«2 + &2<*i)(aic2 + <^zci)
- («i^i - &ιβι)«»2<ί2 - (a,d, — ^c^bjjCjj = (a^ - δ,β,χα^ — b2e2) τ* Ο
Hence σ3 is a Mobius transformation.
Now if zeB satisfies
(A) if c, τ* 0, г ι* -d,/c,
(B) if c8*0, Ζσ,Φ-d^/e^
(C) if c3 1* 0, ζ Φ —^3/03
(D) г ч* ю
a^ + b,
tiien ;?<Γι<γ*» ~ —— — ~ » .и — -ι — 2<гз
а,г + Ь, (0,02 + ^0,)*+(b^jj + djdu)
Г-С» + Un
с,г + d, г *
Thus except for restrictions (А), (В), (С) and (D), there is nothing more to prove. Since σ,σ2 and
<r3 are permutations, we may ignore one of these cases, say (D). This obtains because if for all
complex numbers ζ, ζσισ2 — ζσ3, then » can only be mapped to one element of Ε by σ,σ2 and σ3.
Accordingly we shall not consider ζ — » in the following case by case analysis.
Case (a), c, = 0. We have two possibilities: (i) c2 = 0, (ii) c2 ¥· 0.
(i) c2 = 0. Then c3 = ajC2 + ега2 — 0. Thus (A), (B) and (C) do not restrict z, and we can
conclude Ст]ст2 = σ3.
(ii) c2 ¥■ 0. We first show ζ — —d^/e3 if and only if *σ, = — d2/c2. ζ — —а$)Съ implies
_ a^-rfg/cg) + b, -a^dg + 03b, _ — ai(b-lc2 + d^ + α^ζΟ, _ _^
A simple computation shows that ζσ, = (а,г + &i)/di = —d^fc^ implies ζ — -d3/c3. Now if
гст, = -<ii/c2, then ζσ^ = (~<ί2/β2)σ2 = ш and, as г = —аъ/сг, ζσζ = «. г = -^^"з implies
Ζσ! = —яГ2/с2. Thus «σ,σ2 = » = «σ3 in this case, and so ах<г2 — σ3·

80
GROUPS AND SUBGROUPS
[CHAP. 3
Case (b), Cj Φ 0. Again there are two possibilities: (i) c2 = 0, (ii) c2 ·* 0.
(i) c2 = 0. Then c3 = Cjd2 and, as a2dz — e2b2 ■»* 0 implies tfcj ¥* 0, it follows that c3 ¥* 0.
Note also that —dg/c3 = —dldi/cid2 = ~~dl/el. Hence we need consider only the possibility
ζ = — d1/ei. If ζ — —dj/Cj, then, as c2 = 0, ζο·1σ2 = °°σ2 = °° while ζσ3 = «. Hence ata2 = σ3.
(ii) c2 ■»* 0. c3 = я-[С2 + Cjdjj = 0 if and only if Oj/Cj = —d^c^
(a) c3 = 0. If ζ = —<Z1/e1, then ζσ^ =*aj = a2/c2 while
_ (-«y^Kttitbj + Ci&2) + (Ь^з + агЪг) α^-α,ίί, + δ,β,)
Ζσ3 —
blez-\-dld2 Ci(bjC2 + djd2)
From Oj/cj = —d2/e2 we have d2 = ("ij/c^cj and
a2(—ajdj + bjC^ a2
Ζσο — r — —
C2(61C1 ~ Ml) c2
Now zaj = — d2/"2 = Oj/c, only if ζ — », and we need not consider this case. Hence ага2 — σ3.
(β) c3 ¥= 0. Then ttj/cj ■»* —d2/e2. If ζ = —dl/c2, ζσισ2 = =°σ2 — *2/β2-
di _ (oja2 +b2Cj)(—dj/cj) + (a2bj+&2<ίι) α2(—ajdj + bjCj) _ α2
"ι (OjCj + d2Cj)(—dj/e,) + (bjCa+d^g) ~ e2(—Ojdj + bjCj) c2
Finally if ζστ — ~-d3/e2, then ζ ——d^e3, for (c^z + Ь1)/(е1г+ d^ = —d2/ez implies a.jC2z +
bjC2 = —e^z — rf!^. Hence (ajC2+ cxd^z — — (&ιβ2 + djd^ and, as c3 — агс2 +d2cl ¥■ 0,
ζ — —(bjc2+ djd^/iajCa + d2Ci) = —d%fc3. Therefore
d3 _ d2 _ _ d3
—— σ,σ2 - -— σ2 _ » _ - — σ3
c3 c2 c3
3.47. Let σ(α, Ь, с, d) = at be a Mobius transformation and к a nonzero complex number. Show that
σ(α, b, e, d) — a(ka, kb, kc, kd).
Solution:
Denote a(ka, kb, kc, kd) by 5. If ζ ¥■ » or —d/c (if e ■/= 0), then, аз fc ^ 0, ζσ, = ^ , . =
kaz + kb ez + d
-r-—, j, = 25. To treat the special cases ζ = » or —d/c (when с # 0), we first assume с = 0.
Then kc — 0, and »σι == » = ю г by definition. Secondly, if с Φ 0, then fee ^ 0. Therefore
α>σι — — — ~~— a>„ and аг = » = (—fed/fce) σ. Thus for all possible choices of z, we have
2ctj — ζσ and hence σ2 = σ.
3.48. Show that the set of all Mobius transformations = {σ(α, b, c, d) j od— be = 1}.
Solution:
Let Й — {σ(α, b, c,d)\ ad—bc~ 1}. If σ = σ(α, Ь,с,d) is any Mobius transformation, then,
by definition, D — ad — bc^O. From Problem 3.47 above we know that σ = σί~~, —— .-^r.-^V
α d ь ad-bc \Vd Vd Vd VdJ
But —— —z= =: —=r = ~; r~ = 1- Hence σ ε Μ. Furthermore, any element of Μ is ob-
viously a Mobius transformation. Thus Й is the set of all Mobius transformations.
3.49. Suppose ad — be ¥■ 0. Prove σ(α, b,c,d) = ι iff a= d and Ь = с = 0.
Solution:
If σ(α, b, c, d) = ι, then ζ = ζσ(α, Ь, с, d). Hence «> = «>a(a,b,c,d) implies с = 0.
0 = 0σ(α, Ь, c,d) — b/d implies 6 = 0.
1 = 1σ(α, b,c,d) = 1·-; implies a — d.
d
If a = d and с = Ь = 0, then, using the results of Problem 3.47, σ(α, b, c, d) — σ(α, 0,0, a) ~
а(1,0,0,1) = и

Sec. 3.5J
THE GROUP OF MOBIUS TRANSFORMATIONS
81
b. 2X2 matrices
In this section we will define the group of two by two matrices and indicate its
relationship to the group of Mobius transformations.
An array
а с
b d
(3.8)
of complex numbers a, b, c, d is called a two by two (2 X 2) matrix. (Since we will only
deal with 2x2 matrices, we usually omit the adjective 2 X 2.) a, b, с and d are called the
entries of matrix (3.8). Two matrices are equal if and only if their entries are the
same, i.e. ia * ) = ( ?' 5j if and only if a = a',b = b', c=C and d = d'. We define
the product of two matrices as follows:
'a c\/o' c'\ ___ ίαα'+cb' ac' + cd'
b d]\V d'J " \ba' + db' be'+ dd'
The product of two matrices is clearly a matrix. A calculation shows (Problem 3.51) matrix
multiplication is an associative binary operation.
The matrix / = f β .. j is the identity matrix, since
a c\fl θλ _ /α + 0 0 + c\ /at] /l 0\f а с
b d)\0 1/ [b + 0 0 + dj \b dj \0 lj\b dt
In order to determine which matrices have inverses, we define the determinant of the matrix
A = ( , , J to be the complex number D(A) = ad-be. It is easy to show D(A) D(B) =
D(AB) for any two matrices Л and В (see Problem 3.52). If A = (a c) and D(A) ¥° 0,
then , v \° d/
d -с \
D(A) D(A)
—Ь α
\D(A) D(Ay
is the inverse of A (see Problem 3.53(i)). If D(A) = 0, then A has no inverse (see Problem
3.53(ii)).
We claim that the set
m =
\\b d I a>b,c,d complex numbers, ad — be¥° 01
with the operation of matrix multiplication is a group. For if Α,Β&ξΜ, then, as
D(A) D(B) = D(AB), D(AB)^0 and AB £ m. The determinant of / = ( J J J is 1.
Hence the identity / is in (M. Furthermore if A EM, then A~l Ε (Μ since D(A)D(A-1) =
D(/) = 1 implies D(A~*) ¥- 0. Therefore Λί is a group. We call Μ the group of 2x2
matrices over the complex numbers.
The relationship between the group of Mobius transformations and the group of 2 X 2
matrices is now evident. For in Problem 3.46 we found
a(ai,bi,ci,di) ατ(θ2, Ьг, С2, da) = <г(йз, ba, ca, da)

82
GROUPS AND SUBGROUPS
[CHAP. 3
where α3 = ακχ2 + biCit b3 = a2bt + b2dlt c3 =■ <Xic2 + d2ci, and da = bic2 + did2. But by the
definition of multiplication,
/αϊ ci \/σ2 c2\ __ /aia2+cibz atc2 + сА\ _ /α3 c3 \
\bi dij\b2 d2J ~~ \bia2 + dib2 bic2+did2J ~~ \Ьз аз/
This does not mean the group of 2 x 2 matrices is identical with the group of Mobius
transformations. For we have seen in Problem 3.47 that σ(α, b, c, d) — a(ka, kb, kc, kd) for
any complex number k¥*0. But [, , U4 j ,, ,, 1, e.g. if к = — 1. The precise
relationship between the two groups will be given in Problem 4.81, page 120.
Problems
(ii) Find the inverse of («)(J_g)· (&) (l θ)' (c) (θ i)' (<0 (θ d)*
Solution:
/ 8ΐ -3 + 2Λ /1 2c\ /b d\ /c a\
« W(*>-« 10 ) <Ь)(о l) W(. .) »U b)
,.., , ч.14 + 5М 14 + 3M „,/0 П ,ч /-t 0\ ,ж /Ι/α -c/ad\
(и) (α)
3 -7
<b)(i о) (с) ( о-*) (d) (о ш)
,14 + 3* 14 + Зг ;
3.51. Show that matrix multiplication is an associative binary operation.
Solution:
Let A-(J ΐ).Β = (*' Чс = ^ СЛ. Then
\ъх dj \ь2 dj [b% dj
{AB)C -
/<*i «ίΝ/Оз c2\l/a3 c3\ _ /0402 + сфг atc2+ с^Л /a3 c3\
\&i di/\h <*2/J\&3 d3) ~ уЪрг+а^ц btc2+ d^J^ba dz)
t(ща^Л- c^aJog + {atcz+ M2)63 (0^+Cjb2)c3 + (ajc2 + Cj^dj
\ (&i*2 + <*1&г)а3 + (&iC2 + dtd2)b3 (Ъгаъ + агЪ2)с3 + (Ь,^ + d
id2)o3\
\d2)ds)
A(BC)
al cl\ /a2a3 + c2&3 aSc8 + с2<*3\
&i <*i/ ЧМя + ^а btfb+djdz/
/«1(0203 + Сг&з) + ci(&2as + <*a&s) ai(«2c3 + eida) + ci(&2c3 + **я«*а)\
\ &i(«2«3 + <>i&s) + <*1<Мз + <къз) Ъ^ацСз + е^з) + а^ЪцСз + dzd%)J
A check of entries shows (AB)C — A(BC).
3.52. Prove D(A)D(B) - D(AB) = D(B)D(A), for any two matrices A and B.
Solution:
/a c\ /a' c'\ . „ /oa'+cb' ac' + cd'\
Let A = [Ъ d)>B = [ν ι} Then AB = (ba' + Λ' »• + ,«') аМ

Sec. 3.6]
SYMMETRIES OF AN ALGEBRAIC STRUCTURE
83
D(A) D(B) = (ad- bc)(a'd' - b'c') - aa'dd' + bb'ee' - adb'c' - bca'd'
= {aa' + cb'Xbc1 + dd') - (ba' + db'^ac1 + cd')
= D(AB)
Because multiplication of complex numbers is commutative, D(A) D(B) = D(B) D(A).
3.53. Let A
С
be a matrix. Prove:
/-
(i) If D(A) ¥■ 0,
D(A) D(A)
\
is the inverse of A.
\D(A) D(A)j
(ii) If D(A) = 0, A has no inverse.
Solution:
/ d —c \ I ad—be
\ /
(i)
(::)
D(A) D(A) _ D(A)
-b a I I bd-db _~_^_
\D(A) D(A)j \ D(A) D(A)
-ge+ ac\
D(A)
-be + da
1 0\
0 l)
d -с \
D{A) D(A)\
а с
b d
(ii) If A' is an inverse of A, then D(A') D(A) = D(I) where / is the identity matrix. But P(i) = 1
and D{A) — 0. Since zero times any number is zero, A cannot have an inverse.
8.54. Show that the following sets of matrices are subgroups of the group <sM of 2 X 2 matrices.
(i) %
(ii) V
(Hi) Ψ
Solution:
а с
b d
а с
Ь d
а с
0 d
a, b, e, d real numbers, ad — be Φ Ο
a, b, с, d complex numbers, ad — be — 1
a, d, с complex numbers, ad ¥■ 0
(i) %Q*M and / = Λ Λ<Ξ.%- И A = f* л)е<31 then'
as
d
D(A)' D(A)' D(A)' D{A)
are all real numbers, A-1 € % and % is a subgroup of <M, as it is easy to check that % is
closed with respect to products.
(ii) XiCe^f and D{I) = 1, so / e U Let A G U Then P(A)Z?(A~i) = P(i) = 1 implies
D(A~1) = 1. Hence li is a subgroup of *M, as it is easy to check that li is closed with respect
to products.
Ί/α —e/ad\
(iii) ?C* and /£9, The inverse of A
С
is
Ш /
. Hence A-'e? if
Ле?, Thus Τ is a subgroup of *M, as it is easily checked that Ψ is closed with respect to
products.
3.6 SYMMETRIES OF AN ALGEBRAIC STRUCTURE
a. Automorphisms of groupoids
We have discussed isometries of plane figures. The corresponding one-to-one onto
mapping of a groupoid is defined as follows.
Definition: Let G be a groupoid. Then an automorphism a of G is a one-to-one mapping
of G onto G such that (ab)a = aaba for all a, b € G.

84
GROUPS AND SUBGROUPS
[CHAP. 3
Note that isometries preserve distance whereas automorphisms of groupoids preserve
groupoid multiplication. As the analog to Theorem 3.5, page 67, we have
Theorem 3.13: The set A of all automorphisms of a groupoid G is a subgroup of SG, the
symmetric group on G.
Proof:
I. i, the identity mapping, belongs to A; hence Α Φ 0.
II. If α, β G A and a, b G G then
(αΙή(αβ) = (Η)α)/? = [(αα)ψα)]β = [α(αβ)][^αβ)]
Thus the composition of mappings is a binary operation in A.
III. The identity mapping is an automorphism, and so A contains an identity.
IV. (A,·) is associative.
V. If a G A, let a-1 be the inverse of a; since a G Sg, a~r makes sense. Let a, b G G and
choose a', b' € A so that a'a = a, b'a — b. Then (a'b')a - ab. Hence (atya-1 = a'b' =
(αα-1)^"1). Thus A is a group, and hence a subgroup of So.
We call A the automorphism group of the groupoid G and sometimes denote it by
aut(G).
Problems
3.55. Find the automorphism group of (G, ·) where G = {a, b) and ■ is defined by the multiplication table
(a)
a
a
b
b
>!
a
(b)
a
a
b
b
Solution:
(a) i, the identity mapping, is the only automorphism for the only other possibility is the mapping
a defined by αα = b and ba = a. But (bb)a ~ aa — b and baba. — αα = α; hence
{bb)a Φ (ba) · (ba). Thus a is not an automorphism.
(b) Define α by aa = b and ba — a. Note that xy — y. Hence (xy)a ~ ya ~ (xa)(ya). The
automorphism group therefore contains the two elements ι and a. Notice aa = t.
3.56. Find the automorphism group of A3. (For table of A3 see Problem 3.23(iii).)
Solution:
Theorem 2.6, page 44, showed that for any homomorphism α of a groupoid G into a groupoid
G', i.e. a mapping of G into G' such that (ffiff2)a = ffiaffi* f°r aU ffi.ff2e G, the image of an
identity in G is an identity in G' and the image of an inverse of g & G is an inverse of да, i.e.
la = 1' (1 an identity of G and 1' an identity of G') and if gh — 1 = hg, gaha — 1' — haga.
Now an automorphism of a group G is a one-to-one homomorphism of G onto G. Therefore if
a is an automorphism of A3, ш = ι. Also σ^ is either стг or σ^ as α is a one-to-one onto mapping.
Hence there are at most two automorphisms of A3.
Let / be the identity mapping, i.e. J = ι, aj ~ σι, σ2Ι = σ2- Let A be the mapping iA — i,
aYA — σ2, σ%Α — σν On checking the homomorphism property, we see that / and A are
automorphisms. Note that A2 — I. Thus the multiplication table is
I A
1 '
I A
A I

Sec. 3.6]
SYMMETRIES OF AN ALGEBRAIC STRUCTURE
85
3.57. Let a be any element of a group G. Define the mapping pa of G into G by pa: g -» а-1 да. Prove
that Pa is an automorphism of G and that PaPb = Pah where Papb is the usual multiplication of
mappings.
Solution:
Pa is clearly a mapping. If gipa = g2pa, then a~lgla = a~lg2a and hence gi=g2.
Therefore pa is one-to-one. Also Pa is onto, for if g G G, g has as pre-image aga~1.
(ffiffzW = a 1{9ι92)α·
a~1glaa~1g2a
9lPa9iPa
and so pu is a homomorphism. Hence pa is an automorphism. Note that we have used the associative
law, so that our argument would not apply to a groupoid which is not a semigroup. If g £ G,
ff(PaPb)
and thus Ра^ = рарь-
(ffpjpb = (а-1да)Рь = b-^a-^gab = (ab)-ig(ab)
ffPab
3.58. Find the automorphism group of S3. (Hint: Use Problem 3.57 to find six automorphisms. Then
prove that there are no other automorphisms. This problem is difficult.)
Solution:
Refer to the multiplication table of S3 given in Section 3.3(a). By Problem 3.57, ρ , Ρσ , ρα , Ρτ ,
PT„, pT„ are all automorphisms of S3. We use the notation of Section 2.4(c), page 37, to denote the
effect of these mappings. We use the multiplication table of Section 3.3(a) to calculate the images
under the automorphisms.
Po-,
t CFj &2 Tl T2 T3
I Οχ CT2 Tj TZ T3
ι CTj σζ Tj tjj τ3
ι σι σ2 τ2 τ3 rt
Pa,
ι σ, σ2 τ, τ2 τ3
ι σι σ2 τ3 τ2 τ2
ι σι σζ τι τζ τ3
ι σζ σ2 τι τ3 τ2
Ρ τ,
t σ2 σ2 Tj τ2 τ3
ι σ2 σι τ3 τζ tj
ι σι σζ Τι τζ τ3
ι σ2 σ2 τ2 τ1 τ3
If ρ were another automorphism, then φ = ι. Once σ2ρ is given, σ2ρ is known as σ2ρ = (σχσ^ρ =
σιρσ2ρ. Now σ2ρ must be either <j1 or σζ, for if say σιρ = rv then σ^ρ = аграгр = τιτι = ι; but this
contradicts ρ a one-to-one mapping, since tp = и Hence there are two possible choices for σλρ.
Now τ2ρ must be one of тц т2 or т3 for if for example, Tjp = j,, then ψ = (τ2τι)ρ = σισι = σ2 = ι,
a contradiction. Hence there are 3 possible choices for Ttp. But once σιρ and τιΡ are known, the
effect of ρ on all the elements of S3 is known, since
τισι = тз &ηά T\<i<i = τζ
So this means that there are at most six possible automorphisms.
To And the multiplication table we use the result of Problem 3.57, that papb = pab.
Pi
Ρσ,
Ρσ0
Ρτ,
Рто
Ρτ»
Pi
Pa,
Ρτ„
Pt,
Pi
ftr,
Ρσ2
Ρτ1
Pt2
P4
P*l
Ρσ2
Pi
PT3
Pt2
P^2
Ρσ2
Pt
Pffj
Pt2
P-3
P'l
Pt!
P-2
"'3
Pi
ftr,
ρσζ
P-2
Pt3
p-i
Ρσ2
Pi
Ρσ,
*i
%
Pt2
Ρσ,
Ρσ2
Pi

86
GROUPS AND SUBGROUPS
[CHAP. 3
b. Fields of complex numbers
The complex numbers С have customarily two binary operations, addition and
multiplication. Notice that if a,b GC, then a — b G C; and if b ¥· 0, ab~l GC. We are often
interested in a subset of С that satisfies the same conditions. This leads us to a field of
complex numbers.
Definition: A subset F of С is called a field of complex numbers if
(a) 1 e F.
(b) Whenever a,b G F, then also a — bGF,
(c) Whenever a,b&F and b φ 0, then ab~l G F.
(The definition of field can be extended to sets which are not contained in the complex
numbers. See, for example, Birkhoff and MacLane, A Survey of Modern Algebra, Macmillan, 1953.)
Of course the complex numbers themselves form a field. Let F be a field.
Recall that the set of complex numbers is a group under the usual binary operation of
addition, denoted by (C, +), and that C* —C~~- {0} is a group (C*, x) under the usual
multiplication of complex numbers (see Example 5, page 51). Therefore, using Lemma 3.1,
page 55, for a subset of a group to be a subgroup, parts (a) and (b) of the definition of a
field imply (F, +) is a subgroup of (C, +), and parts (a) and (c) imply (F*, x), F* =F- {0},
is a subgroup of (C*, x). In view of these remarks the definition of a field is equivalent to:
Lemma 3.14: A subset F of С is a field of complex numbers if
(1) (F, +) is a subgroup of (C, +),
(2) (F*, x) is a subgroup of (C*, x) where F* = F - {0}, С* = С - {0}.
Problems
3.59. Show that R and Q are fields of complex numbers.
Solution:
lefi. If a, be Д, then a—bGR and ab~l G R whenever b Φ 0. The same argument
applies for Q.
3.60. Which of the following sets are fields?
(i) F = {a + tr/2 | a, beQ}
(ii) F = {a+bi | a,bGQ}, г = V^l
(iii) F = {a+bi \ a,bGZ}, г = v^
Solution:
(i) 1=1 + 0V2 G F. Let a + byfi, and a' + b'y/Z be two elements in F.
(a + by/2) - (a' + b'Vz) = (a - a') + (b - b')y/2 G F
and if a' + bY2>0,
<" + ^><«'+ >'«- - ^|S + |^i^e ,
Therefore F is a field.
(ii) F is a field. 1 + 0г = 1 G F.
(а+Ьг) - (а' + Ь'г) - {a-a') + {b - b')i G F
and, if a' + b'i Φ 0,
, , aa! + bb' a'b - ah' .
(.+*>(»'+*'o-i = >т^ + ^т^г е F
(iii) F is not a field, since 1 + i Φ 0, 1+ief but (1 + ή"1 = 1/2 - 1/2г £ F.

Sec. 3.6]
SYMMETRIES OF AN ALGEBRAIC STRUCTURE
87
с Automorphisms of fields
We have discussed isometries of the plane and automorphisms of groupoids. The
corresponding one-to-one onto mapping of a field is defined as follows.
Definition: A one-to-one mapping α of a field F onto itself is termed an automorphism
if
(i) (a + b)a = aa + ba for all a,b G F.
(ii) (ab)a = (aa){ba) for all a, b G F.
Note that the automorphisms of fields preserve both the operations of
addition and multiplication.
Theorem 3.15: The set A of automorphisms of a field F forms a subgroup of the
symmetric group Sf.
Proof: We must prove
I. A ¥^ (jb\ this is true since the identity mapping ι Ε Α.
II. If α, β G A, then
(a + Ь)(а/3) = ((a + Ь)а)/3 = (aa+ba)β = (αα)β + φα)β = α(ο·/3) + Ь(а/3)
and (<Λ)(αβ) = (И)*)/3 - [(Oa)(ba)]fi = [(α*)β][&α)β] = [а(<ф)][Ъ(<ф)]
for all a, b G F. Thus composition of mappings is a binary operation in A.
III. The identity mapping is in A and is an identity element.
IV. {A, ·) is a semigroup, since composition of mappings is associative.
V. If a Ε A, then a E Sf the symmetric group on F. Let a~l be the inverse of a. We
claim a"1 E A and so a will have an inverse in A as desired. Let a, b GF. Then as
a is onto, we can find a', b' G F such that α = of a, b — b'a. Then ab = (a'b')a and
a + b = (a' + b')a. Consequently (ab)a~l = а'Ъ' = {аа~1)(ЪсГ1) and (a + b)^1 =
a' + b' = da-1 + ba~\ Thus a~l G A as desired.
We have proved that the automorphisms of a field form a group. This group is
extremely useful. For additional pertinent remarks and references, see Section 5.4a, page 158.
Problems
3.61. Find the automorphism group of Q.
Solution:
We will use the fact that {Q, +) is a group and (Q*, X), Q* — Q — {0}, is a group. Notice that
(Q, X)is a groupoid (not a group, since 0 has no inverse) and, because of part (ii) of the definition,
the automorphism of the field Q is also an epimorphism (see Section 2.5b, page 42) of groupoid (Q, X)
onto {Q, X). Hence by Theorem 2.6, page 44, 1, the multiplicative identity of (Q, X), is mapped onto
1 by any automoruhism of Q.
Let a be an auotmorphism of Q; then la = 1. Using mathematical induction we show no. = η
for all positive integers n, la = 1. Assume ka = к for some integer к — 1. Then (k + l)a =
ka + la = к + 1, by the automorphism property of a. We conclude that no. = η for all positive
integers n.
Now {Q, +) is a group and, by definition, any automorphism of Q is an epimorphism of the
group {Q, +). Hence by Theorem 2.6, inverses are mapped onto inverses and the identity, 0, of {Q, +)
is mapped onto 0. Therefore (—n)a = —n for all positive integers n, since wa = η and —n is the
additive inverse of n. Furthermore, On = 0. Hence ra — r for all integers. But the automorphism
α is also an epimorphism of the group (Q*, X) onto itself so that (ir)-1** = — α — — for all
positive integers r, because — is the inverse of r. Collecting these facts we see that if — {ηΦ 0) is any
m r / 1\ 1 1 ύά ^
element in Q, then —α = [ τη· - ] a = ma—a = m- — — —. Therefore α is the identity mapping
η V η J η η η
and is the only possible automorphism of Q. The automorphism group of Q is of order one.

88
GROUPS AND SUBGROUPS
[CHAP. 3
3.62. Find the automorphism group of F = {a + 6^2 | a, b G Q}.
Solution:
Any rational number q is an element of F, since q + ОУ^З — q. If a is an automorphism of F,
then for any q G Q, qa — q arguing as in Problem 3.61. Now λ/2 £ F and (\[2л/2.)а — 2α — 2,
since 2 is an element of Q. But у'гаУ'г'а = (У'гу'гЭа - 2a = 2, so that (\/2a)2-2 or V2a=±V2.
We conclude that V^ has only two possible images under an automorphism of F. Hence
(а + Ьт/2)а — аа + (Ьл/2)а ~ aa + baV^« — a + Ь(л/2 a). There are two possibilities: (1)
(a + Ъл/2 )a = a + Ъл/2 , in which case α is the identity automorphism ι; (2) (a + Ъл/2 )a = a + b{—л/2 ),
in which case we must check to see whether α is an automorphism. If a: a + Ъл/2 -* a — Ъл/2 , then
{(а+Ьт/2)(а' + Ь'т/2)}а - {aa' + 2bb' + {a'b + Ь'а)л/2 }a
= aa' + 2bb' - {a'b + Ь'а)^2
and {a + bV2)a(a' + b'V2)a = (а-Ьу/2)(а'-Ь'тД)
- αα' + 2bb' - (a'b + b'a)y/2
Hence {(а + ЬтД)(а'+Ь'л/2)}а = (α + ЬлД)а(а' + Ь'тД )а
Also, {(а + Ьл/2) + (a' + b'V2)}a - {(α + α') + (b + Ъ')т/2 }а
= α + α' - (b + Ь')\/2
and (α + Ьт/2 )а + (а' + Ь'т/2 )а - (а - Ьл/2 ) + (а' - Ь'лД)
= (а+а') - (Ь + Ь')лД
Hence {{а + Ьл/2) + (а' + Ь'л/2)}а = (а+ Ь\/^ )а + (а' + Ъ'лД )а
Thus a is an automorphism of F. The automorphism group of F has two elements ι and
a : a + Ьл/2 -» а — Ь\/^ . Notice αα — ι.
d. Vector spaces
In Physics we represent a force ж by a straight line pointing in the direction the force
is acting and of length proportional to the magnitude of the force. We shall assume for
the moment that all the forces act on a fixed point О and act in the Euclidean plane E.
It is then possible to represent a force by its endpoint, as we know it begins at O. Any
point of course can be represented by its coordinates, so a force can be represented by the
coordinates of its endpoint.
We can talk of increasing the force χ in magnitude by a factor 3, say. The resultant
force is written as Bx. If x~{fi,fi), then Зх ~ (ЗД, 3/2). Similarly if / is any real
number, we define fx to be the force χ increased in magnitude by a factor / and we can prove
/a = (//i,//a).
The sum of two forces χ = (A,/z) and у ~ {g\,gi) is a third force ζ computed by the
parallelogram law. Again it can be shown that ζ = (fi + gt, /2 + З2). We write ζ — x + y.
The set of all 2-tuples (/1, /2) is called a vector space of dimension 2 over the field of real
numbers (because we can multiply the 2-tuples by real numbers).
We shall generalize the concept of two dimensional physical forces in two ways:
(i) We shall deal with arbitrary dimensions and not only 2 or 3. (We must therefore
relinquish our contact with the real world.)
(ii) We shall consider vectors that involve fields other than the real numbers.
Let F be any field. Let V — Fn be the cartesian product of η copies of F. Then V
consists of the η-tuples (A,/2, .. -,fn) where ft G F.
If χ = (A, ...,/„) and у — (дъ . . .,gn) are two elements of V, and / G F, we define
μ; V x V -* V (i.e. μ is a binary operation in V) and ω: FxV -* V by

Sec. 3.6]
SYMMETRIES OF AN ALGEBRAIC STRUCTURE
89
{χ,ν)μ = {fi + gi, h + g%, ...,fn + gn)
(f,x)m = {ffι, ...,//„)
We denote (x, y)p by χ + у, and (/, х)ш by fx.
V together with μ and ω is called the vector space of dimension η over the field F.
The elements of V are called vectors.
Problems
3.63. Find (i) (1, 2,3) + (6, 7, 8), (ii) 4(6, -2,0,3).
Solution:
(!) (1 + 6, 2 + 7, 3 + 8) = (7, 9,11)
(ii) (4 · 6, 4 · (-2), 4 · 0, 4 · 3) = (24, -8, 0, 12)
3.64. Prove that if x, у and 2 are elements of a vector space of dimension n, then x + y — y + x and
(x + y) + г = χ + (у + г).
Solution:
If χ = (Д, ...,fn), у = (gi,...,ffn) and ζ = (hb...,kn), then x + y - y + x -
(A + ffi. · · ■» Λι + ffn) as addition is commutative in any field.
(x+y) + z = ((/i + ffi) + At (/„ + ffn) + AJ = * + (у + г) by associativity of addition.
3.65. Prove that if V is a vector apace of dimension n, then the elements of V form an abelian group
under the operation μ.
Solution:
V is an abelian groupoid by the preceding problem. (0,0, ...,0) is the identity element. The
inverse of (Д, /2, ..., fn) is (-Д, -/2, ..., -/„).
3.66. Prove that if ex = (1, 0, .. .,0), e2 = (0,1, 0, . .., 0), ..., en = (0, 0, ..., 1), then every element *
of V can be represented uniquely in the form
* = /iei + S%4 + ■ ■ ■ + fnen
Solution:
Suppose x — (fv..., fn). Then indeed * = /jex + fsez -\ + fnen.
If x = gle1 + gie2+ Vgnen, then (Д, ...,/„) = (gu .. .,ffn). Hence fx = gx, /2 = g2, ...,
Λ, = ffn an(l the representation is unique.
e. Linear transformations. The full linear group
Let V be a vector space and a: V -» V. Then α is said to be a linear transformation
of F if
(i) (x + y)a - xa + ya (ii) (fx)a - /(to), for all x,y &V and / € i1
For example, let (ft, /2)« = {U, /i). Then
{(/ьЛ)+(01,л)}а = (fa+gi,fi+gt) = {f2,fi)+{g%,gi) = (Л,/а)« + (01,л)«
Also, {/(/ι,Λ)}« = (/A,//i) = /(Д,А) = /((/i,A)«)
Note that linear transformations preserve both the additive and the multiplicative structures
of V.
Now we have the analog of Theorems 3.13 and 3.15. First let us define Ln(V,F) to
consist of all one-to-one linear transformations of V, the vector space of dimension η over F.
Ln(V,F) QSv, the symmetric group of V, clearly.

90
GROUPS AND SUBGROUPS
[CHAP. 3
Theorem 3.16: Ln{V,F) is a subgroup of Sv.
Proof: ι e Ln{V,F) as £ preserves both addition and multiplication. Hence Ln(V,F) У= ψ.
If a GLn(V,F), we ask whether a'1 GLn(V,F). a'1 is one-to-one onto. Is it a linear
transformation? Let x,у e V and / G F. Since a is onto, there exists χγ and y1 such that
xta — x and yta = г/. Of course xt - Xa~l, yt - ya~l; and (xt + yt)a - xta + yla~x+y.
Hence
(^l + Vi) ~ {X1 + V1)aa~~i ~ (x + y)a~~l
and SO Ха"л +Уа~~1 — (X+y)a~~l. AlSO, {fx^a — /(«ja) = fx; SO ((fxi)a)a~~1 = fxt = (/ж)а-1,
i.e. f{xa-1) = Цх)ссК Accordingly *~l G Ln(V,F). Thus if a^etJF/), ^'eLJ^i1)
and we ask whether a/?"1 £ Ln(V,F). We have
(ж+г^а/Г1 = ({χ + ν)α)β-~ι - (Χα + ν^β-1
= {Χα)β-1 + Ы/^1 = ^(а/Г1) +Κ(α]8-1)
and (/^«/Г1 = ((ίχ)<*)β~* = (/(ζ*))^1 = ЛЫ/?"') = /Ив/Г1))
and thus Ln(V, F) is a subgroup of Sv. Ln(V, F) is called the full linear group of dimension n.
Problems
3.67. Show that if α is a linear transformation of V, a vector space of dimension n, then the effect of a
is uniquely determined by its effect on the elements ev ..., en of Problem 3.66.
Solution:
By Problem 3.66 each element of V is of the form χ = flel + ■ ■■ + fnen. Then xa —
fiie^) + · - · +f„(ena). Hence the effect of a is known once its effect on the elements β, e, is
known.
3.68. Show that if a is any mapping of {eit ..., en} -* V, then there exists a linear transformation
S: V -* V such that e}a — eft, j = 1, ..., n.
Solution:
Each element of V is uniquely of the form /^ + · · · + fnen. Define a: V -» V by
(f^i +··· + /„«») 5 = /i(ei«) + ■ · · + /u(e„a)
Then г is a linear transformation, since
{(/lei + · ■ · + /A) + (ίΤιβ, + h gne«)} a - (/i + g1)(ela) + ■■· + (fn + дл){епа)
- (/iei + Η /„«J* + (0ι«Ί + Η ΑΆ)«
and {/(/Λ + · · · + /neJ}S = (//JM + " · ■ + (ffn)(ena)
= /ί(/ι61+-··+/Λ)α}
3.69. Is a.&Ln(V, F) if α is a linear transformation and eta — e,z, e%a ~ e3, ..., e7l_1a = en and
e„« = et?
Solution:
Yes. All we must prove is that a is one-to-one and onto. An arbitrary element fiel + · · · + fnen
has fze1 + fse2+ · ■ ■ + fnen_l + f\en as a pre-image. Also,
(/i«i + V /»eJa - (ΑΊ^ι + h ίΤ„βη)α
implies /t = jflf /2 = gz, ...,fn — gn by Problem 3.66. Hence a is one-to-one. Thus aGLn(V,F).

CHAP. 3]
SUPPLEMENTARY PROBLEMS
91
A look back at Chapter 3
We have met many important groups, including groups of real and complex numbers,
the symmetric group Sn, symmetry groups, the dihedral groups, the automorphism groups
of groupoids and fields, and the full linear group.
Groups thus arise in many different branches of mathematics, and hence general
theorems about groups can be useful in apparently unrelated topics.
In subsequent chapters we will derive general theorems for groups.
Supplementary Problems
GROUPS
3.70. Let η be any positive integer and let Gn = {a + b^/n | a,beZ} where Ζ is the set of integers.
Prove that with respect to addition Gn is a group. When does G„ — Z1
3.71. Let η be any positive integer. Let Gn = {a + ibyfn \ a, b G Z) where i = y—1 and Ζ is the set of
integers. Is Gn a group with respect to addition? Is G„ a group with respect to multiplication of
complex numbers?
3.72. Let D - ZxZ, Ζ the set of integers. Define (a, b) °(c,d) - (a + c, (-1)^6 + d). Prove that D is
a group with respect to this operation °.
3.73. Prove that the group D of Problem 3.72 is not abelian.
3.74. Let G — Ζ X Q, where Ζ is the set of integers and Q the set of rationale. Define (a, b) * (c, d) ~
(a + c, 2cb + d). Prove G is a gTOup with respect to this operation * .
3.75. Is the group of Problem 3.74 abelian?
3.76. If we define (a, b) ° (c, d) = (a + c, 2~cb + d), is G (of Problem 3.74) a group with respect to»?
Is G a group with respect to the operation · defined by (a, b) · {c,d) — (a + c, 2cb — d)?
3.77. Let Β = {θ Ι β: Z^Z). Let W-ZXB. We define a multiplication on W by (m, θ)(η, φ) ~
(m+ η, φ) where for each ζ G Ζ, ζψ = (ζ — η)θ + ζψ. Prove that Ж is a group. (Hard.)
SUBGROUPS
3.78. Let G be a group and G, Q G2 Q · ■ · be subgroups of G. Show that GiUGgU · ■ · is a subgroup of
G. Find a group G and two subgroups G, and G2 of G such that G!UG2 is not a subgroup of G.
3.79. Let Glt G2, ■ ■. be subgroups of G. Prove G1nG2n ■ · ■ is a subgroup of G.
3.80. Let G be an abelian group. Let Η be a subgroup of G. Let S(H) — {χ | χ G G and xx G ff).
Prove that S(H) is a subgroup of G.
3.81. Let D be the group of Problem 3.72. Determine whether Η - {(a, 0) | a&Z) and Κ - {(Ο,α) | α£Ζ)
are subgroups of D,
3.82. Let G be the group of Problem 3.74. Determine whether Η = {(α, 0) | α G Ζ) and K= {(0, q) \ q G Q)
are subgroups of G.

92
GROUPS AND SUBGROUPS
[CHAP. 3
3.83. Let В be as in Problem 3.77. Let С = {θ j θ : Ζ -» Ζ, ze = ζ for all but a finite number of integers
z). Let B' = {x | x = (0, b), tefi}, С = {x\ x= (0, с), с e C). Prove that B' is a subgroup of
W and С is a subgroup of B'. (Hard.)
л
3.84. Using the notation of the preceding problem, let W = {χ \ χ ~ (m, c), where m£Z and с G С}.
Prove W is a subgroup of W. (Hard.)
SYMMETRIC GROUPS AND ALTERNATING GROUPS
3.85. Let a: Ζ-> Ζ be defined by za-z+1 for all ζ G Z. Let /J: Ζ ■+ Ζ be defined by 2ηβ = 2n,
(2n + 1)/} = 2n + 3 for all integers n. Let γ ; Ζ -» Ζ be defined by 2ny - 2(n + 1), (2w+ 1)γ = 2w+ 1
for all integers n. Prove that α, β,у e Sz and show that aa~ βγ ~ у β.
3.86. Let G = SP, where Ρ is the set of positive integers. Let SP — {θ | seSP and ze = г for all
but a finite number of г G P}. Prove that SP is a subgroup of SP.
3.87. Let G = SP. Let G„ = {β | θ G SP and ге - ζ for all геР such that z>n}. Prove that
G = GiUG2U--· .
3.88. Let Я = {в \ в G S5, le = 1}. Prove that Я is a subgroup of S5. What is its order? Let
К = {θ I e G i?5, le = 1 or le = 2}. Prove that К is not a subgroup of S5.
3.89. Let η and r be positive integers. Let
H= {e | гее {1,2, ...,r} for all г € {1,2, ...,r} and б € Sn}
Prove that Я is a subgroup of Sn and find \H\.
3.90. Let Ζ be a set and Υ a proper subset of X. Let Я = {θ \ в G Sx and ув = у for all у G У}.
Let if = {β] i€Sx and jfGF for all у G У}. Prove that Η and if are subgroups of Sx and
that if ЭЯ. Prove that if |У| & 2, Я τ* if.
3.91. Let Η = {θ \ θ G Л5, le = 1}. Prove that Я is a subgroup of A5 and find its order.
3.92. Let А, В be sets with \A\ = 1. Prove that SA x B a SB. (Hard.)
3.93. Prove that if \X\ = \Y\, Sx = SY.
GROUPS OF ISOMETRIES
3.94. Let S be the even integers and {I: S) = {θ | в G /(Й), se G S for all s G S}. Prove (/: S) is a
subgroup of I(R).
3.95. Let (/: Q) - {θ \ θ G I(R) and gfl e Q for all g G Q}. Prove (/; Q) is a subgroup of /(B).
3.96. Find the symmetry group of the figure W.
3.97. Find the symmetry group of the figure 8.
3.98. What is the symmetry group of the graph of у = sin xl
3.99. Determine the symmetry group of the circle.
3.100. Prove that if S is any subspace of the plane and S' is a congruent figure, i.e. there is an isometry
θ such that Se ~ S', then /s = /s,. (Hard.)
THE GROUP OF MOBIUS TRANSFORMATIONS
3.101. Prove that if Μ is the group of MSbius transformations, then the only element me Μ for which
mn = nm for all nG. Μ is m = ι.
оя + Ь
3.102. Let σ{α, Ь, с, d) be the Mobius transformation defined by <r(a,i>,c, d) : z-*—7—x. Prove that
σ(α, b,c,d) = o{a',b',c',d') if and only if either α=α', b = b', c = c', d=-d' or а——а,', Ъ=—Ъ',
с — -с', d = —d', given ad - be = a'd' — b'e' = 1. (Hard.)

CHAP. 3]
SUPPLEMENTARY PROBLEMS
93
3.103. Let N be tbe set of M6biua transformations σ(β, b, c, d) with 6 = 0. Prove that N is a subgroup of
M, the group of all MSbius transformations.
3.104. Prove that if me N {N defined as in Problem 3.103), then there exists β £ N such that ss == m.
3.105. Let V be the set of all matrices f J, where a,b,c,d are integers such that ad—bc = l.
Find the set of all matrices such that
/a b\/a b\ _ /1 0\
\e d)\e d) ~ \0 1/
SYMMETRIES OF AN ALGEBRAIC STRUCTURE
3.106. Prove that the automorphism group of a finite group is finite.
3.107. Find a finite groupoid G with |G| > 2, whose automorphism groupoid is of order t.
3.108. Let G be a non-abelian group. Prove that the automorphism group of G is not of order 1.
3.109. Prove that the subset К of the symmetric group £f4 defined by
_ f/1 2 3 4\ /1 2 3 4\ /1 2 3 4\ 1
K ~ j\2 1 3 4/'\t 2 4 3/42 1 4 8/''J
is a subgroup of £4. Find the automorphism group of K. (Hard.)
3.110. Let F — {a+ ib^/vi \ a, b rational numbers, where i = V-1 }· Verify that F is a field under the
usual operations of addition and multiplication of complex numbers. Determine the automorphism
group of F.
3.111. Let V be the vector space over the rationale of dimension η +1. Let
S = U\ a(=Ln+l(V,F), (1,0, ...,0)« = (1,0,...,0)}
Prove that ST is a subgroup of Ln + i(V,F) and that S s< Ln(V,F).

Chapter 4
Isomorphism Theorems
Preview of Chapter 4
We say that two groups are isomorphic if they are isomorphic groupoids. Here we shall
prove three theorems which provide a means of determining whether two groups are
isomorphic. The main concepts that arise are those of subgroups generated by a set, cosets,
and normal subgroups. We find a structure theorem for cyclic groups. The contents of
this chapter are indispensable for any further understanding of group theory.
4.1 FUNDAMENTALS
a. Preliminary remarks
We begin by reminding the reader of our previous results. A group is a semigroup in
which every element has an inverse. Consequently we have the following.
(1) The identity is unique. (Theorem 2.1, page 31.)
(2) The inverse of an element is unique (Theorem 2.2, page 33), and if G is any group and
g,hEG, then (gk)-1 = u"1^1.
(3) The product of η elements αϊ, ..., an, in that order, is independent of the bracketing
(Theorem 2.5, page 39).
(4) Homomorphisms, monomorphisms and isomorphisms for groups are defined as they are
for groupoids. (Section 2.5, page 40.)
(5) If G is a group, and θ any homomorphism of G into a groupoid, then
Ge = {x\ x = дв, gGG]
is a group. For by Theorem 2.6, page 44, GO has an identity, is associative, and each
element has an inverse. Note that θ maps the identity of G to the identity of GO and
that (д~1)е = (дв)~1 for each g in G.
(6) If 6:G-*K is a homomorphism from the group G to the group K, and if Я is a subgroup
of G, then Ηθ is a subgroup of K. For ΘΙΗ is a homomorphism of Η into К and, by
(5), Ηθ is a group.
(7) Isomorphic groups are roughly the same except for the names of their elements. (See
Section 2.5d, page 45.)
The following theorem is useful.
Theorem 4.1: If a and b are two elements of a group G, then there exist unique elements
χ and у such that ax = b and ya = b,
Proof: We consider first the solution of the equation ax = b. If we put χ = a~lb,
then a(a~rb) = (aa'^b — b. Hence the equation ax = b has a solution.
Suppose axi = b and ax% = b; then axi — ax2. Multiplying both sides of the equation
on the left by a~l, we have
a~4axi) = а~1(ах2), (α~τα)Χι = (α~Ύα)χτ, or Χι = x%
The argument for solving ya = b is similar; in fact у = ba'1 is a solution. Also, if
yia = b and уяа = b, then yta = уф. Multiplying both sides by a"1 on the right, we get
(yia)a~l = 2/i = {уга)а~1 = y2
94

Sec. 4.1)
FUNDAMENTALS
95
Problems
4.1. Prove that the groups given by the following multiplication tables are isomorphic.
-11 0 1
-1
1
1 -1
-1 1
Η
0
1
0 1
1 0
Solution:
Let e:G-*H be defined by Is = 0, —1θ = 1; then β is a one-to-one onto mapping. If it is
also a homomorphism it will be an isomorphism. We must check that (ffij/2)e = дгвд2в for all
possible choices of g1 and ff2 in G, i.e. we must check
(i) (1 · 1)9 = lile (Hi) (-1 ■ -l)e = (-1β)(-1β)
(ii) (-1 · l)ff = (-l0)(li) (iv) (1 · -1)9 = (le) · (-19)
1 by the multiplication table. Iff = 0. 1β·1β = 0·0 = 0,
1-1
Η.
(i) to (iv) hold. (Thus for (i):
Hence (i) holds.) Therefore G '■
Of course e had to be some mapping of G to H. How did we know which was the right mapping
to choose? Examining the multiplication table for G, it is obvious that 1 is the identity for G,
0 is the identity for H. We remarked that any homomorphism must map an identity to an identity.
Thus the choice for θ was quite clear.
4.2. Prove that S2, the symmetric group of degree 2, is isomorphic to G, where G is the group of Problem
3.5, page 53, with m = 2. Prove S3 sr D3 the dihedral group of degree 3, i.e. the symmetry group
of the equilateral triangle. (Difficult.)
Solution:
The multiplication table for G is
0 1
1 0
while the multiplication table for S2 is (Problem 3.20, page 58):
β
(
β
β
1
Let θ: S2 -*G be defined by ш — 0, βθ — 1. Then it can be checked that β is a homomorphism.
As it is one-to-one and onto, S2 ss G.
The multiplication table for S3 is on page 57, that of D3 is in Problem 3.40, page 76. As we
have used the same Greek symbols for S3 and Ds, we face the risk of not knowing whether <r, for
example, refers to an element of Sa or to an element of B9. To avoid such ambiguities, we will
rename the elements of Da, replacing a <r by an s and а г by a i. The multiplication table then
becomes
»1 »2 s3 * ts2 te3
■i
»2
is,
ts$
»1
»2
»3
t
Щ
Щ
»2
4
»1
ta2
te3
t
4
«1
»2
tsa
t
ts2
t
<ss
ts2
4
»3
«2
ta2
t
tsa
s2
Sl
»3
ts3
is2
ί
«3
»2
«1

ISOMORPHISM THEOREMS
[CHAP. 4
Note that an element τ}, j = 1,2,3, satisfies τ}τ} = ι. If θ is an isomorphism from S3 to Ds, then
rfirfi = {τ}τ})β = ιβ = β!. So 9 can only map the ts, j = 1,2,3, among the elements t, ts2, ts3 since
these are the only non-identity elements of D3 which have the property that their squares are s,.
As в must map ι to s^ it maps σ1( σ2 onto the elements s2, s3.
If we know the effect of θ on au since α1σι = σ2, we know the effect of в on σ2. Also if we know
the effect of в on rlt then, because т2 = г1<Г2 and r3 = τ\οχ, we know the effect of θ on all the
elements of S3-
So we have to experiment. A suitable mapping θ : Ss -> Ds must satisfy σ^ = s2 or s3 while
тув = t, ts2 or is3.
We try the following definition. Let ιβ — β1( σ-β = s2 and j^S = t. Then we must have
"29 = s3, ^9 — tss, rse = is2 if в is to be an isomorphism.
To check whether this mapping is an isomorphism, we must check whether this mapping is a
homomorphism. As a mechanical procedure of doing this we use the following table.
"1 °2 Tl T2 T3
The entry in the second row and third column, for example, is calculated as follows: In the
bottom corner we place σχθ · σ2θ. In the top corner we place {σ^α^θ. If θ is a homomorphism,
(al9)(ase) = (σ^α2)θ. Hence these two entries should be the same in each square of the table.
Checking through this table, we see that the entries in each square are equal. Hence θ is a
homomorphism and as it is one-to-one onto, в is an isomorphism.
Prove that if θ : F -> G and φ: G-* F are two homomorphisms such that θψ = identity mapping
on F and ψθ = identity mapping on G, then θ and φ are isomorphisms of F onto G and of G onto
F respectively.
Solution:
в is one-to-one, for if χθ = ув, then χθψ = увф. But ΘΦ is the identity on F. Hence χ =y.
Similarly ψ is one-to-one.
Next let g 6 G; then дф £ F. дфв = g; hence g is the image of an element of F under θ and
so θ is an onto mapping. Thus θ is an isomorphism. Similarly ψ is an isomorphism.
Prove that if glt g2, да are elements of a group G, then the equation g^xg2 = 03 has a unique
solution.
Solution:
If we put χ = flffVafl^"1» we find g1xg2 = g3. И 01^102 = Я&гвг = ϋ-ь then on multiplying by
g-1 on the left and g~l on the right we have g^i(g1xig2)g~l = д^1{д1Х202>£'2г or χλ = x2.

Sec. 4.1]
FUNDAMENTALS
97
4.5. Prove that if G is a finite group and Я is an infinite group, then G and Я are not isomorphic.
Solution:
If G s« H, there is a one-to-one mapping from G onto Я. But this is not possible since G is
finite and Я is infinite.
4.6. Prove that Sn s Sm if and only if η = т.
Solution:
Sn has order n\ and Sm has order ml Now if S„a Sm, then there is a one-to-one mapping of
Sn onto Sm. So Sn and Sm have the same order, i.e. n! = m!, and this implies « = m. On the other
hand every group G is isomorphic to itself. In fact the identity mapping of G onto G is an
isomorphism. Hence η = m implies S„ ^ Sm.
4.7. Prove that if G^H, then Я a G.
Solution:
Let б be an isomorphism from G onto Я. Then β"1 is an isomorphism from Я to G, and so
Η &G. (See Problem 2.38, page 42.)
4.8. Prove that if G ^ Я and Я зг К, then б^К,
Solution:
Suppose # is an isomorphism from G to Я and 0 an isomorphism from Я to K. Then β0 is an
isomorphism from £? to K, i.e. G as if. (See Problem 2.38, page 42.)
4.9. Prove that there are infinitely many groups, no two of which are isomorphic.
Solution:
Consider the symmetric groups SuS2t ·■■ ■ Then by Problem 4.6, no pair of these groups is
isomorphic.
4.10. Prove that if G is a finite group and Я is a subgroup of G, Ή.Φ G, then G and Η are not isomorphic.
Solution:
We observed in the solution of Problem 4.6 that if two finite groups are isomorphic, they have
the same order. Since the order of Я is less than that of G, it follows that G and Я are not
isomorphic.
b. More about subgroups
Let G = S4 and let X = {σ8, r7} where
/l 2 3 4\ . /l 2 3 4\
σβ ~ 1^2 1 4 Ζ) and T? - \2 3 1 4)
(See Problem 3.21, page 59.) Suppose we wish to refer to a product such as σ8τ7 or τ7σ8σ8 or
<г8т7<г81т7г1т7г1<г8"1· It will be convenient to have some general notation. We will write
xll ''' x*n' where ti = ±1, xt € X
to represent the product of η elements chosen from X or the set of inverses of the elements
of X, where x\ will mean ж., and χΓι will mean the inverse of xv For example, if ^ = % = 1,
«a = «4 = «s = «6 = -1, and xt = x3 = ж6 = σ8, хг = x4 = xs = τ7, then х^х%х£*х1*х1*х\* stands
for σΛτ^τ-ιτ-ιας\
Example 1: If g = x^1 · · ■ χ*ηη, then ff~l - h where h = х~*л · ■ · х~*1.
Proof:
gh - x'1 ■■■ х^х"^...х-Ч - ХЧ ...χ*η-ι.\.χ y-ι...ίτ «ι
* ι в η ι ι «—ι я—ι ι
= ··· = x\ix~'i = 1
Similarly hg — 1. '

98
ISOMORPHISM THEOREMS
[CHAP. 4
We proved in Lemma 3.1, page 55, that if Я is a subgroup of G and xv хг G Я, then
χ,Χζ1 G Я. We now generalize this and prove
Lemma 4.2: If Я is a subgroup of G and XcH, then
Я d {ж'1 · · · ж^1 | x( G X, €t- ±1, na positive integer}
Proof: Recall that as Я is a group, iGH and x,y G Я implies i/"1 e Я, жг1 6 Η,
and жг/ G Я. We prove the lemma by induction on n. Let η — 1. Then ж;^J G Я since
a;t G Я. Hence x\l G Я. Assume, by induction, that G Я for η = к. Let
жь+1 G X. Since a?,»***1 G Я where £fc+1 = ±1,
Hence x\l ■ ■ - o£" G Я for all n. This proves the lemma.
Now if X is "large enough", e.g. X = H, we may have
Я = {x\l · · ■ o£n | ж. G X, e, = ±1, η a positive integer}
We ask what happens if X is not "large enough", i.e. if X is a subset of Я and
S = {ж!1 · · · x*£ | ж{ G X, e, = ±1, η a positive integer}
is S a subgroup of G1
Lemma 4.3: Let G be a group and let Ζ be a non-empty subset of G. Let
-S = {ж*1 · · · ж*п | x{GX, fj = ±1, η a positive integer}
Then S is a subgroup of G. If Я is any subgroup containing X, Я э S.
Proof: We must prove that:
(i) S Φ 0; this is true because there exists ж1 G X as X is non-empty,
(ii) If /, g G S, then /flr-1 G S (Lemma 3.1, page 55).
f,g G S means
/ = x\l ■ ■ ■ <« (s = ±1) and
where ж, and y{ are elements of X. Hence g~l -
where xn+1 = ym,..., xn+n = yx and e„+1 = -,„,.... w = -v Therefore fa"1 G S and
S is a subgroup of G. If Я э X, we use the previous lemma to conclude Я D S.
We denote S by gp(X) and call S the subgroup generated by X.
If a group can be generated by a finite set, we call it a finitely generated group.
Example 2: What is gp({l}) in the group of Problem 3.5, page 53, where m — 3? Recall that
the multiplication table is
0 1 2
0
1
2
S = ffj>({l}) = {x[l · · · ss*n | *j € {1}, е{ = ±l,»i positive integer}
Now leS and 2 € S. Also, 1 · 2 = 0 € S. Hence £ф({1}) is the whole group.
We remind the reader that, for example, in the multiplicative group of nonzero rationale
the inverse of a, which we have denoted in this section by a"1, is 1/a, i.e. in this case a-1
has the meaning usually associated with it when α is a number. But in the additive group
of rationale a'1 is -a.
g = y\l
= 1/ ~J,m . . .
"1 = Xft ·
■■■yyi.Vi =
y~Vl and
• · ж'« а?*"·*1 ■
η it+1
*i)
n + m
0 12
12 0
2 0 1

Sec. 4.1] FUNDAMENTALS 99
Problems
4.11. Let G = Z, the additive group of integers. What is gp({l})1
Solution:
9P(W) 2 1 + 1 + 1+ '■•+1 (и — 1). Hence gp({l}) contains all positive integers.
> , '
η times
gp({l}) contains l"1 + l"1 + h 1~» = -1 + (-1) + · · · + (-1) (n ^ 1). Hence gp({l}) con-
, v , , y ,
. . „ . . n times η times
tains all negative integers.
Also, др({Ц) contains 1·l~i= 1+ (-1) = 0.
Thus gp({l}) = Z.
4.12. Let G — Q the additive group of rationale. Find gp({l}).
Solution:
Exactly as in the last problem, gp({l}) = Z. Since no other elements can arise as sums or
differences of 1, др({1}) Φ Q.
4.13. Find the subgroup of the multiplicative group of rationale generated by {2}.
Solution:
2-1 = £. The elements of gp({2}) are either of the form 2" or 2~n, η a positive integer.
4.14. Determine the subgroup Η of S8 generated by σι and τί of Section 3.3a, page 57.
Solution:
We use the multiplication table for S3 shown in page 57. CTj"1 = σ2; hence Η contains σ2. As
Η contains tj, it contains τ8 = τισι and τ2 = т\аг. Thus Η contains all the elements of 53) and so
4.15. Determine the subgroup of the symmetry group of a square generated by those isometries that leave
two vertices fixed. (Hard.) {Hint: To see what is happening, cut out a square from a piece of
cardboard and label the four vertices. Perform the isometries on the figure.)
Solution:
We refer to Problem 3.39(H), page 73. st leaves G and / fixed; s2 leaves Η and J fixed; s5 leaves
all vertices fixed. Hence we require 5 = ffp{{ss,s2,s1}), and this must contain s7, since 8^82 — s7.
It is easy to prove SjSj = s7, for the effect of s7 and SjSj is the same on three points not on a single
straight line and this is sufficient by Lemma 3.7, page 71. Note that the inverses of s2, s„s7 are
s2,*,, s7 respectively.
Let Τ — {s5, s7, s2j s,}. We assert that Γ is a subgroup. All we must check is that tu t2 e Τ
implies t^1 e T. Since t e Τ implies t~i — t, all we must check is that f^ef for ί,,^εΓ,
As s5 is the identity, tfa G Τ if s5 is either ^ or t2. If tj = t2, then t,^ = s5 G T. Therefore
we need only consider the following cases: sxs2 — SjSj = s7 G T; Sis7 = s^ = s2 G T, Finally,
8287 = sjsu — S\ G T. Then Γ is a subgroup of the symmetry group of the square. But 5 2 T, and
Γ 2 {s5, s^Sy}. Hence Τ 2 gp(su, 8%, Sj) = 5, by Lemma 4.3. Thus Τ = S.
4.16. Find the subgroup of M, the group of Mobius transformations (see Section 3.5a, page 78), generated
by
η : z-* —z {гФ «), η : « -* *>
τ: 2 -* 2 + 1 (ζ Φ ■»), τ: °° -* °°
(In the notation of Section 3.5a, η = σ(-1,0,0,1) and τ = σ(1,1,0,1). This is a difficult problem.)
Solution:
Let σ(η f) be the mapping defined by z-* ег+п for я Φ °°, and «>-»«>, where « = ±1 and
и is any integer. In other words, σ(η e) = σ(«, η,Ο, Ι).

100
ISOMORPHISM THEOREMS
[CHAP. 4
We will show that the subgroup generated by τ/ and τ consists of all σ(η t), n any integer,
ΐ = ±1. Let Σ = {CT(n £) j n any integer, с = ±1}. We claim gp(v,r) = Σ. Any element of Σ is
of the form σ, . . Now гт-1 =г —1. Also, ζτ · · ■ τ = z +w and ζτ_1 ■ · · τ-1 = ζ — π for any
π η
positive integer и. Hence for any arbitrary integer n, σ(η η Ε 9ρ{η,τ)· (We must check what
happens to =°, but this presents no difficulty.) Also, σ(ηл)т; = σ(_η _,> € др(у,т) and so σ(η c), e = ±1,
и any integer, belongs to gp(v,r). Thus gp(v,r) 2 Σ.
Note that -η — σ(0 _n and τ = σ(1 ,> belongs to Σ. So we need only show that Σ is a subgroup
of the group of Mobius transformations to conclude that Σ 2 gp(v,r). To show Σ is a subgroup, we
need only show that σίη>£) σ~£ 6) <= Σ, δ = ±1. But <^,e) = *c-8m,8>> since гстст,8> CT(~Sm,e) ~
(δζ + m)S — 8m = 5sz + Sm-{m = 52г = я. Then и(,|Е) σ^| — (ez -+ и) σ(_6ηι6) = e8z + n8 — 8m,
and we conclude that σ(η>6) σ(^6) = a(n6_6m,t8) € Σ.
с. Exponents
We have seen in the previous section that we often are forced to consider the product of
m a's (m > 0), e.g. a a. (Note that as we are dealing with groups, it is not necessary
to indicate in which order the multiplication is performed. See Section 4.1a.) It is
convenient to introduce the notation am for the product of m a's, m > 0. Then am · an is the
product of m a's followed by n a's, i.e. αη·αη = am + n (Section 2.4d, page 39). Our idea is
to extend the exponent notation in a sensible way to zero and negative exponents. We
would naturally like the law „ „ „ . ,, <4
J am-an = am + n (A.1)
to be true when m and n are arbitrary integers. Now if it were true that a°am = am, then
multiplication by a0 leaves am unchanged. Hence we have only one choice in extending the
exponent notation and retaining the law (i.l), namely putting a0 = 1, the identity. Now
if m = -n where n > 0, m + n = 0. Because we want Ц.1) to be satisfied, we must have
am+n - ao - i> j e we muat put am _ (αη)-ι_ -^0^ ^hat (a")-1 = (or1)" = orn. Thus we have
defined am to be
(i) the product of m a's if m > 0,
(ii) 1 if m = 0,
(iii) the product of -m a~*'s if m < 0,
hoping thus to satisfy (i.l) for all m and n.
(If.l) is true if m, n are both positive. If both are nonnegative, again by running through
the possible cases (i.l) holds. If both m and n are negative, then
aman = (a-,)t-m)-(a-,)(-n) - (a-,)c_m + -n) = (a-,)_Cm + n) = am+n
If m and n are nonpositive, again the result is easily verified. If m > 0 and n < 0, then
by checking the various possibilities m > -n, m = —n and m < -n, we find aman =
Another result which holds for exponents is
(am)n = ctmn (4.#)
We already know that (A.2) holds when n = -1. If m, n are positive, (am)" is the product of
n elements, each of which is the product of m a's. Hence (am)n is the product of mn a's. If
m is negative, n positive,
(am)n - ((or1)-"1)" = (or1)"™" as -m > 0, n > 0
= amn as mn < 0
If now η is negative, (am)n = (a»)t-»t-«> - ((a™)4)-" = (a^m)c-">
_ ас-т). с- η) ^y our previous remarks)
Hence {Jf.2) is proved.
= a"

Sec. 4.2]
CYCLIC GROUPS
101
In the study of groups there are two main notations for the binary operation. One is
the multiplicative notation we have employed up until now. The other is the additive
notation. We denote the binary composition by + in this case. The identity is denoted by zero,
0, and the inverse of a by —a. The result of performing η > 0 compositions of the same
element, i.e. of taking a + a + ■ · ■ + a, we denote by na. The law U.1) becomes
V ( ,
η
na + ma = (n + m)a
while the law {Jf.2) becomes n(ma) = (nm)a
In other words, translation takes place according to the following dictionary:
Multiplicative notation Additive notation
ab a + b
1 0
a"1 -a
an na
It is immaterial which notation one uses. But additive notation is most often used for
a group in which the order of the composition of two elements is irrelevant, i.e. in which
a + b = b + a for all a, b in the group. Such a group is called abelian, after the Norwegian
mathematician Niels Henrik Abel, or commutative (Section 2.2, page 29).
Problems
4.17. Find l3, l-4 where 1 GO, the additive group of rationals.
Solution:
In the additive group of rationala the binary operation ia the usual addition. Then l3 means
1 °1 °1 where ° is the binary operation in Q. Hence l3 = 1 + 1 + 1 = 3. Also l-* means (l"""1)*,
i.e. l-bi-ioi-ioi-i where ° is the binary operation under discussion. Now 1-1 = —1 in
(Q,+). Thus l-* = (-l) + (-l)+(-l) + (-l)=-4.
4.18. Find г2, 2~3 where 2 G Q*, the multiplicative group of nonzero rationals.
Solution:
22 = 2 ·2 = 4 and 2~3 = {2~ψ = φ3 = J.
4.19. Find ff3. с = (J J 4 3) , an element of S*.
Solution:
β3 = ι and so a3 = ισ = σ.
4.20. Find τ™, where τ is as defined in Problem 4.16.
Solution:
τ» = σ(η>1). See Problem 4.16.
4.2 CYCLIC GROUPS
a. Fundamentals of cyclic groups
If gp{X) = H, we say Η is generated by X. To get an understanding of groups, a
good plan is to investigate the simpler groups first. So we begin by considering groups
which can be generated by a single element. We call such groups cyclic. Thus a group Η
is cyclic if we can find an element χ e Η such that Η = gp({x}). We will usually write
gp(x) instead of gp({x}).

102
ISOMORPHISM THEOREMS
[CHAP. 4
Lemma 4.4: gp(x) = {t | t = xr, r an integer}. Cyclic groups are abelian.
Proof: gp(x) = {x\l ■ ■ ■ xfnn \ ж. G {ж}, £. = ±1,и> 0}
= {жп · · · же" J е. - ±1, η > 0}
= ^Vi=1 ;| «. = ±1, n> 0У
= {жг | r any integer}
If a,b G gp(x), then α = xT, b = xs, ab = жгж8 = жг+8, bet = xsxT = жг+5. Hence ab = bet for
any two elements of a cyclic group. Thus we have shown that cyclic groups are abelian.
Suppose now that Η = gp(x) and \H\ — m (m < °°). Then we know that the elements
of Η are of the form xr for various integers r. The xl cannot be distinct for all integers i.
Consider x° = 1, x, . . .,χ1"1 and suppose these are distinct but that xl - xk for some к < I,
к s= 0; then ж^ж")"1 = ж1-" = 1. If к Φ 0, m = l-k<l and жт is equal to ж0. But we
assumed this was not so. Hence к = 0 and xl = ж0 = 1.
We will show that $= {^ж.ж2, ...,ж1-1} is actually H. This is easy. First notice
that as ж1 = 1, every positive power of ж is in S. Furthermore, ж-1 — ж1-1. Hence every
negative power of ж lies in S. But Η = {χ/\ r any integer}. Therefore HqS and so
S = Η as stated.
Thus we have proved
Lemma 4.5: Let G be cyclic of order m generated by the element ж. Then G = {ж0, ж1, ...,
xm~1}. Furthermore xm = 1, and жт is the least positive power of ж that is 1.
We ask a simple question: do cyclic groups of order m exist for all finite integers
m > 0? Yes! Let us consider in the symmetric group Sm of degree m the element
Then
1 2
2 3
1 2
3 4
1 2
1 2
m -1 m
m 1
m-2 m-1 m
m 12
and so the elements ι, am, ..., σ™-1 are distinct and Я = gp(°m) is cyclic of order m. Hence
there exist cyclic groups of order m for each m > 0.
And now we ask another question: are there two essentially different cyclic groups of
order m? Rephrasing the question, we ask: are two cyclic groups of order m isomorphic?
Lemma 4.6: Let G = gp(x), Η = gp(y) be each of order m. Then G ^ H.
Proof: G = {x°,x,x2, ...,χ™-1}, Η = {у0,у1, .. .,2/m_I}. Let Θ-.G^H be defined by
ж'0 = г/* (г = 0,1 m - 1)
Then θ is one-to-one onto H, To prove it is an isomorphism we must show it is a homo-
morphism. Consider
(ж*^)б> = (xi+i)e

Sec. 4.2]
CYCLIC GBOUPS
103
Now 0 — i, j ^ra-1. Then 0 — i + j~ 2(m - 1) = 2m — 2 and so i + / — tm + r where
0 — r — m — 1 and t — 0 or 1. Hence
(χί + *)θ = (ж£т + г)0 = (хетХг)в = (Χτ)θ - Уг
But (ж*0)(ж?0) = г/У = ΐ/ι+ί - г/ет+г = Утут = г/г
Hence ((χι)(χ}))θ = (ж'0)(ж50). Thus β is a homomorphism and, as it is one-to-one onto, it
is an isomorphism.
We now ask the obvious question: are there any infinite cyclic groups and are two
infinite cyclic groups isomorphic?
Consider the element σ in the symmetric group Sz on Z, the set of integers, defined by
Ζσ = Ζ + 1, Ζ G Ζ
As ζση ~ z + n, or"1 = ση implies m-n. Then gp(a)~H, say, has an infinite number of
elements and so Η is an infinite cyclic group.
Recall that G = gp(x) — {ж" \ η any integer}. If there exists an integer m > 0 such
that xm — 1, then G will consist of only a finite number of elements (see the remarks
preceding Lemma 4.5). Consequently if G is infinite, there exists no m^O for which xm = 1.
For we have already shown that there can exist no m > 0 for which xm = 1; while if
xm = 1 for m < 0, then жс"т) = 1 and (-ra) > 0. If xl = xn, n¥-1, then ж"-1 - 1. But
this contradicts the condition that there exists no m such that xm = 1. Hence the elements
of G are simply the powers xn of x, and two such powers xm and xn are equal if and only if
m — n.
Now we can easily prove that two infinite cyclic groups are isomorphic. Let G =■ gp(x),
Η = £fp(y) both be infinite cyclic groups. Then each element of G is uniquely of the form
ж", η an integer, and each element of Η is uniquely of the form yn, η an integer. Define
(χη)θ = yn. 0 is a one-to-one onto mapping. Furthermore, (χηχη)θ = (χη+πι)θ - yn+m and
(χηθ)(χηθ) = 2/Ύ71 = yn+m. Hence χηθχηθ = (xnxm)e. Therefore θ is an isomorphism and
G and Η are isomorphic groups.
Collecting our results, we have proved
Theorem 4.7: There exist cyclic groups of all orders, finite and infinite. Any two cyclic
groups of the same order are isomorphic. (We therefore often talk about
the cyclic group of order m, or the infinite cyclic group, or sometimes the
infinite cycle.)
If χ is an element of a group G, then we define the order of χ as the order of gp(x). Note
that if χ is of order m < », then xm =■ 1 and m is the first positive integer r for which
xr ~ 1. If ж is of infinite order, then xm — 1 implies m ~ 0. If ж is of order m, m < °°,
we say ж is of finite order.
Lemma 4.8: Let ж be of order m < °°. If xr =■ 1, then m divides r.
Proof: Put r = qm + s where 0 — s < m. Then 1 = x/ = x4mxs = ж8. As m is the first
integer greater than 0 for which xm — 1, s — 0. Hence m divides r.
Problems
4.21. Prove that the additive group of integers is infinite cyclic.
Solution:
Ζ — flfp(l). As Ζ is infinite, it is infinite cyclic.
4.22. Prove that the group of Problem 3.5, page 53, is cyclic of order m.
Solution:
The group is ffp{l), and its order is m.

104
ISOMORPHISM THEOREMS
[CHAP. 4
4.23. Prove that (Z, +) and the subgroup of M, the group of Mobius transformations, generated by the
mapping т;: я -» я + 1, °=т/ = » are isomorphic.
Solution :
By Theorem 4.7 all we need prove is that gp{v) is infinite, as we know from Problem 4.21 that
(Z, +) is infinite cyclic. But since ζηη = ζ + η, ηη = т/т implies n — т. Thus gp{v) is infinite and
so gp(v) = (Z, +).
/1 2 3\ /1 2 3 4\
4.24. Find the order of (i) σ = ( ) ε S3, (ϋ) σ = ( £S„ (iii) the map ч of ЛГ
defined by ат/ = —я, «τ; = =ο.
Solution: „ „ _
, „ /ι 2 3\ /ι 2 з\
(ι) σ^ι ami σ = „ ο ) „ ο ) = ι· Hence σ is of order 2.
(ϋ) σ, σζ, σ3 are not ι, but σ* = ι. Thus σ is of order 4.
(iii) τ/2 = ι and so τ/ is of order 2.
4.25. Let G be abelian. Let χ,у ε G be of orders r, s respectively. Show that ая/ is of order rs if r and
s are co-prime, i.e. have no common prime divisors.
Solution:
Note that since G is abelian, {xy)n = xnyn for any integer n. Since (xy)rs ~ xrsyTS = 1, the
order of xy divides rs, by Lemma 4.8. If (xy)m — 1, i.e. «"И/™ = 1, then £m = y_m and 1 — (£m)r =
y~mr. Therefore s, the order of y, divides —mr. Since s does not divide r, s must divide m. Similarly
we can show r divides m. Hence rs divides m. So if m is the order of xy, m is divisible by rs and
also m divides rs. Thus the order of xy is rs.
4.26. Show that if G is a cyclic group of order m < =o and s is co-prime to m, then a.s ~ bs (a·, b G G)
implies a. = 6. Find a group G and a nonzero integer n such that there are two elements a,b ε G
with a." = bn but n5*li.
Solution:
Since G is abelian, so {ab~l)s — a.s(b-1)'* — 1. Since G ~ вгзэ(л;) and the order of G is m, then
α·ί>-1 = xr for some r, and (xr)s = 1. Hence я™ = 1 and m divides rs. But s and »i are co-prime;
then m divides r, say r = qrm. Now ab~l — ж*"1 = 1 and so α = Ь.
In S3, let a. ~ Л 3 V b = f1 2 3\. Then »2 = 62 = , but a. ?* b.
\2 ! 3/ V1 3 2/
4.27. Show that if G = firp(a;) and G is of finite order r and s is co-prime to r, then gp(xs) = G.
Solution:
The distinct elements of gp(xs) are 1, £s, ж&, . .., χ(η~^« where (£s)n — xns = 1 and «. is the
least such positive integer. Since xns = 1 and G is of order r, r divides ws. As r and s are co-prime,
r divides n. Hence there are at least r distinct elements in gp(xs). But as G^>gp(xs) and G itself
has only r elements, gp(xs) = G.
4.28. Find a group which is not abelian. {Hint. Consider S3.)
Solution:
See Section 3.3a, page 57, where we pointed out that σ,τ-, Φ τ\σ\. Hence S3 is not abelian.
4.29. Prove that a subgroup Η of S3 is cyclic if Η Φ Ss.
Solution:
A survey of the subgroups of S3 shows that if Я is a subgroup of S3 and Η Φ S3, then Η is
either cyclic of order 3 or cyclic of order 2 or cyclic of order 1. To obtain all the subgroups of S3,
we refer to the multiplication table for S3 in Section 3.3a, page 57, and list all the subsets of S3.
Then we check which subsets are subgroups. Of course since a subgroup must contain the identity,
there is no need to go through the process of finding all subgroups quite so crudely. Nevertheless
this method will suffice.

Sec. 4.2]
CYCLIC GROUPS
105
4.30. Prove that (Q, +) is not cyclic.
Solution:
If (Q,+) is cyclic, there exists q = m/n, m and η integers (η Φ 0), such that gp(m/n) — Q.
Of course M5*0. Each element (Φ 0) of Q would then be of the form q + q + ■ ■· + q with some
suitable choice of the positive integer r, or else of the form —q—q-
But l/2w e Q. l/2w = q + · ■ · + q implies l/2w = rm/n, i.e. 1 = 2rm; then 1 — 2rm = 0.
τ
But r and m are integers; hence the equation 1 — 2rm = 0 is not true. If l/2w = — q —q —q — ■ · · —qt
r terms in all, a similar argument leads to a contradiction. Therefore (Q, +) is not cyclic.
4.31. Prove that an abelian group generated by a finite number of elements of finite order is finite.
Solution:
Let G = др({хг, ...,xn}), η < *>, and suppose G is abelian. Then every element fir in G is of
the form
g = Ц1 · · · *I; ц e x, tj = ±1)
Since G is abelian, we can rewrite g in the form
9 = xi1·-· xln (У1.У2. ■■■,Ύη integers) (4.S)
To see this we need only observe that if is — it for s < t in the first expression for p·, then
ff = xh · · · ^ · · · xir
i.e. we can always "collect" all occurrences of any χ in a product. Now if xlr x%, . . .,xn are all of
finite order, then the number of distinct elements given by (4.3) is finite. For if fc; is the order of xlt
t = 1,2, ...,n, the distinct powers of xf are l,x(,x?, ...,xki~\ Thus the number of distinct
elements given by (4.3) is at most k^.. .kn, and so G is finite.
b. Subgroups of cyclic groups
Before beginning the study of a new section it is a good idea to list the natural
questions. If we want to know something about the subgroups of cyclic groups, we might ask:
(i) Are subgroups of cyclic groups cyclic?
(ii) Does there exist a subgroup of any given order?
(iii) How many distinct subgroups of a cyclic group (less than or equal to the order of the
group) are there ?
(iv) How many subgroups of a given order are there?
We tackle each of these questions.
Theorem 4.9: (i) Let Я be a subgroup of G= gp(%). Then Η is cyclic and either
Η = gp(xl) where xl is the least positive power of χ which lies in Η or
else Η = {1}. If the order of G is m < °°, then I \ m and the order of
Η is mil. If the order of G is infinite, Η is infinite or Η = {1}.
(ii) Conversely if I is any positive integer dividing m>J;hen S = gp(xl) is
of order mil. Consequently there is a subgroup of order q for any q
that divides m.
(iii) The number of distinct subgroups of G is the same as the number of
distinct divisors of m= \G\ < <».
(iv) There is at most one subgroup of G of any given order for G finite.

106
ISOMORPHISM THEOREMS
[CHAP. 4
Proof:
(i) If Я Φ {1}, there exists χη¥Ί€Η. Aa Η is a subgroup, г"еЯ, Now one of η,—η
is positive. Hence we can talk meaningfully about the smallest positive power xl e Я.
Clearly, H~DS = gp(xl). Suppose xT G H; then r = ql + s, 0 ^ s < I, and
χ^χ1*)-1 = xs <= Η
But xl is the least positive power of χ that belongs to H. Thus s = 0 and so r = ql
and (ж') = (χ1)* <= S. Hence S = Я.
If the order of G is m < °°, then m = gi 4- s, 0 ^ s < I. Now
1 = x·" = ж«1+9 = жв1ж-' e Я
and so ж9 e Я. Then s = 0, as xl is the least positive power of χ that lies in Я. Hence
ί divides m, and m = Iq. Clearly (x1)" = 1, and q is the least positive integer for which
this occurs. Then, by Lemma 4.5, writing α = xl, we have Я = 0ρ(α) = {а0, а1, ..., α"-1}
and hence )H\~ q ~ m/l. If the order of G is infinite, all the powers of χ are distinct,
and so xl, хгх, ... is an infinite set of distinct elements of Я. Thus Я is infinite.
(ii) Let 11 m, I > 0. Put m/l — q and xl — a. Then S — gp(xl) ~ {1,a, ■ ■ -,α""1}, as
a" = xm is the least positive power of α which is 1 (for if a4' = 1, q' < q, then xlq' — 1
and lq' < m, contradicting the fact that xm is the least positive power of χ which is 1).
Thus S is a subgroup of order q. Consequently if we start out with a positive integer
q which divides m and we put I — m/q, then S is a subgroup of G of order q.
(iii) Let h, U,..., h be the distinct divisors of m. Then put Hi =■ gp(xh),..., Я„ = gpix1*).
We know |H4| = m/ij. These are η distinct subgroups of G (because their orders are
different). Are there any more subgroups? By (i) any subgroup Η of G will have to
be generated by xl where ί is a positive integer dividing m. Hence ί — U, say.
Therefore Η ~ Hi. Thus the subgroups of G are simply Hu Я2, ..., Я„, as desired.
(iv) If Я and К are two subgroups of G with |Я| = \K\, then Я and К are Hi and H} of part
(iii) above, for some ί and j. But |Я{| = m/k, \H}\ = m/Ij. Since |Я| = \K\, U-k and
therefore i = ?* and Η — Hi = H5 = K.
The reader will perceive that our knowledge of the cyclic groups is in some ways quite
comprehensive. We know in the case of finite cyclic groups what the distinct subgroups
are, we know they are cyclic and we know which cyclic subgroups appear. In the case
of infinite cyclic groups we can easily prove there are an infinite number of subgroups.
We will distinguish between them in Theorem 4.24, page 126, using the concept of index
which will be introduced in Section 4.3b.
The reader might naturally be led to consider now groups generated by two elements,
hoping that similar powerful conclusions can be obtained, e.g. that every subgroup of a two
generator group is a two generator group. But in going from one to two generators we lose
control. It has been shown that every countable group is a subgroup of a two generator
group, so we can never hope for a simple account of two generator groups-
Problems
4.37. A subgroup Η of a group G is called proper if Η Φ G and Η ¥■ {1}. Let G be a cyclic group of
order a prime p. Prove that G has no proper subgroups. ,
Solution:
We know from Theorem 4.9 that the number of subgroups of G is the same as the number of
distinct divisors of p, which are ρ and 1. Hence the number of distinct subgroups of G is two. As
{1} and G itself are two distinct subgroups, the number of proper subgroups is zero.

Sec, 4.3]
COSETS
107
4.38. Prove that the only groups which have no proper subgroups are the cyclic groups of order ρ and
the group consisting of the identity alone.
Solution:
Let G be a group with no proper subgroups, G ¥° {1}. Let g € G, g ¥■ 1. Then S = gp(g) is a
subgroup by Lemma 4.3. Since g€. S, S = G аз G has no proper subgroups. Hence G is cyclic.
If G is cyclic of order mn, τη, η ¥■ 1, then, by Theorem 4.9, G has a subgroup of order m. But this
is a proper subgroup. Hence G is cyclic of prime order or else possibly infinite cyclic, say G =
gp(x) = {..., as-2, as-1,as0, as1, as2, ...}. But Η = gp(xz) is a subgroup not equal to {1}, and not
equal to G since as € H, Hence G can only be cyclic of order p, a prime.
4.39. Find a group with two distinct subgroups both of the same order. \Hint. Consider S3.]
Solution;
Let T8 = (2 1 3)' Γ2 = (ί 3 2) ■ Then Ыгз)| = |йфЫ| = 2'
4.40. Find a group which is of infinite order but has a subgroup of finite order. {Hint. Try the group
of Mobius transformations, M, of Section 3.5, page 77.)
Solution:
Let ν '· * -* 1/z, » -» ». Then gp(n) is of order 2, but Μ is infinite.
4.41. Let Η be a subgroup of G. Let jeC. Prove that the set S — {g~lhg | ЛеЯ} is a snbgroup
of G. Prove that β: Η -* S, defined by he = g~lhg, is an isomorphism of Η onto S. К К is a
finite cyclic subgroup of G which contains both Η and S, prove that Η = S. (Hard.)
Solution:
Since H¥°0, S ¥■ 0. Let д-^д.дЩ^/ £ S. Then
(flr^iffXflr1**)-1 = g-4iigg-4i-ig = д~Ч^к^)д G S
because Η is a subgroup implies ^/ij'GE Thus S is a subgroup, в is an onto map, since
he — g~lhg. If kte = h^e, then g~lh^g — g-lh^jg. Pre-multiply by g and post-multiply by
Я"1· g(g~1h1g)g~l — g(g~lh2g)g~1. Hence hl — h2 and so в is one-to-one onto. We need only
check that в is a homomorphism to conclude the proof:
MM = fi^'M/' g~lh-2g — g~lhlh2g = (M2)9
Thus Η and S are isomorphic.
If К is a finite cyclic subgroup containing Η and S, then Η and S are both of finite order; and
since they are isomorphic, \H\ — \S\. But, by Theorem 4.9, К has only one subgroup of any given
order. Hence Η — S.
43 COSETS
a. Introduction to the idea of coset
In this section we propose a natural question which introduces the idea of a coset.
Cosets are important for other reasons: (i) With cosets we can perform useful counting
arguments for finite groups, (ii) Cosets of a subgroup sometimes enable us to construct a
new group from an old. We can also see how a group G is built up from one of its
subgroups Η and the group constructed from the cosets of H. (iii) The fundamental idea
of a homomorphism can be re-interpreted in terms of the idea of a group constructed from
cosets.
What is the natural question we ask? In Section 3.4c, page 67, and Section 3.4e,
page 73, we defined the group / of isometries of the plane Ε and the isometry group h of
a given figure S in E. Recall that an isometry σ of Ε belongs to Is if for each t G E,
ta GS implies t G S, and s G S implies so G S. Suppose aGl-h. We ask: which ele-

108
ISOMORPHISM THEOREMS
[CHAP. 4
ments θ of / are such that Se = Sal In the case
where S is an equilateral triangle we know that Sa
is a congruent equilateral triangle by Lemma 3.6,
page 71. So the question we are asking is this:
which other elements θ of I send the equilateral
triangle S onto the equilateral triangle SOr?
Let Su = Sa. How are θ and σ related ? If θ
were equal to a, then θσ~ι = t. If θ and σ were
widely different, we would expect θσ-1 to be
anything but t. Let φ = θσ~ι. We will show that φ is
an element of Is. To do so observe that φ is an
isometry of the plane E, since it is a product of
isometries of E. Now S9 = S<r implies that for
each s G S there is a t G S such that s9 = ίσ, and
conversely. Therefore for each s G S,
δφ = βθσ'1 = {ta)a~l = t G S
Suppose now that ж G Ε and χφ G S. We must show that χ G S to complete the
proof that φ G /s. Suppose χφ = t G S, i.e. ж(#о·-1) — f. Now there exists an s G S such
that ta = s9. Hence ж# = ((atf)*-1)* - ίσ - sB and so χθ - s9. As β is one-to-one, χ = s.
Hence χ G S. This means φ G Is.
We have of course (9 = φσ. If we write /so- = {Υσ| г G Is}, we may put our deduction in
the form θ G Isa. We have thus shown that every isometry θ of Ε satisfying SO = Sor lies
in the set of isometries Isa. Conversely if θ G Isa, then θ = φσ for some φ G Is; then
ΞΘ = 5ψσ = Sa. This means that Isa consists of all the isometries θ of Ε for which S6 —- -Sor.
Such a subset /go- of the group / of all isometries of Ε associated with the subgroup Is of /
is called a right coset of Is in /. More generally we have the following
Definition: Let G be a group and Η a subgroup of G. Then a right coset of Η in G is a
subset of the form H</ = {ж | χ = Л-дг, Л- G Я) for some </ in G. We define
a left coset of Η in G to be a subset of the form gH ~ {χ \ x — gh, h G H).
Note that a coset is a right or left coset according as the element g is on the right or
the left of H.
In the case where the group G is written additively, i.e. + is used to denote the binary
operation, a right coset is written H + g. Of course, H + g = {x\ x = h + g, hG H}.
Problems
4.42. Let G be the cyclic group of order 4 generated by {a}. Let Η = gp(a2). Find all right cosets of
Я in G. Show that two cosets are either equal or else have no elements in common, and prove that
the union of these cosets is G.
Solution:
Η · 1 = Η = {1, α2} is a right coset. Ha = {a, a?} is a right coset Ha* = {a2, a*} = {1, a2} = H.
Ная = {αΡ,α5} = {a3, a} — Ha. Thus the distinct cosets of Η in G are Η and Ha.
HDHa = 0, and HuHa ~ {1, a*, a, a"} = G.
K~—
4.43. Let Η be the trivial subgroup of a group G, i.e. Η — {!}. Determine the distinct right cosets of
Η in G.
Solution:
If g £ G, Hg = {lg} — {g}. Thus the cosets are the sets consisting of single elements of G.

Sec. 4.3]
COSETS
109
4.44. Find the right cosets of Я = др(тя) in 5Я with the notation of Section 3.3a, page 57. What are
the right cosets of К — βρ(σλ) in S3?
Solution:
Я = (ι, r3), Hi = Я, Ηαλ = {аъ τ2}, Ηα2 — {σ2, тг}. These are all the cosets of Η in Ss.
К = {σν σ2, ι}. The cosets of К are Κι = К and -Кг, = {τν τ2, г3}.
4.45. Let А, В, С be subsets of a group G. If X and Υ are subsets of G, we define XY = {fif | g — xy,
x€.X and у G Γ}. Prove that A(BC) - (AB)C. Hence conclude that if Я is a subset of G,
f.geG, then (i) {fg)H=f(gH), (ii) H(fff) = (Я/fc, (iii) </Я)? - /(Яр).
Solution:
Let χ € A(/?C); then χ = ad where aG A and d G /JC. But d G ВС implies d = be, where
b G В and с G C; hence a: = a(bc) = {ab)c G (AB)C and so A(BQ С (AB)C- Similarly (AB)C С
A(BC). Therefore A{BC) = (AB)C. (i), (ii) and (iii) follow immediately, e.g. (i) is the case where
A = {/}, В = {ff} and C = H.
4.46. Let G be a group with a subset H, Show that /Я = Я/ implies /_1Я = Я/-1.
Solution:
/Я = Я/. Hence
/-Н/Я) =/-Ч»Л. (/-'/)# = (/"'Я)/ and Н=(/-»Я)/
Thus Я/-1 = ((/-W)/)/-i = /-1Я.
Note that we have used the "associative law" proved in Problem 4.45.
b. Cosets form a partition. Lagrange's Theorem
In the problems above it is clear that any two right (or left) cosets are either disjoint or
exactly the same and that the union of all the right (or left) cosets of Я in G is G. We recall
that a family of subsets of a set G is a partition of G if they are disjoint and their union is G.
The examples above point to the following:
Theorem 4.10: Let Я be a subgroup of a group G. Then the right (left) cosets of Я in G
form a partition of Я in G, i.e. the union of all the right (left) cosets of
Я in G is G itself and any pair of distinct cosets has empty intersection.
Proof: It is easy to show that each element of G occurs in at least one right coset.
(The proof for left cosets is similar and is not included here.) For if g GG, then g G Hg
since 1 G Я and 1 · g = g.
Suppose now that Ha and Hb are two cosets of Я in G and that НаПНЪ ¥* 0, i.e. there
exists д^НаПНЪ. Then g = hfa = h"b, h',h" (ΞΗ. Hence α = h'-W'b = h"'b, h'" Gfl,
since the product of two elements of Я belongs to Я. Therefore
Ha = {ha | h e Я} = {h{h'"b) | h<=H) С НЬ as hh'"b = ft,b, Ы е Я
Similarly ЯЬ С Яа. Thus Яа = ЯЬ and any two cosets are either disjbini or identical.
In Section 4.3a we mentioned in (i) that cosets are useful for counting arguments; this
follows from Theorem 4.10. If G is of finite order and Я a subgroup of G, then, since
the cosets of Я in G are disjoint, the order of G is the sum of the number of elements in
each coset. We use this fact in proving
Theorem 4.11 (Lagrange's Theorem): The order of a subgroup Я of a finite group G
divides the order of G.
Proof: Let the distinct cosets of Η in G be Hgi,Hg2, . . .,Hgn. Since these form a
partition of G, |С| = |ВД + |Я*|+".+|ВД («)

110
ISOMORPHISM THEOEEMS
[CHAP. 4
What is \Hg\1 We will show that the mapping ΘΒ:Η-* Hg defined by he = hg is a one-
to-one onto mapping and hence |Я| = \Hg\. Clearly θβ is onto by the definition of right
coset. If hig = h2g, then multiplying by g~l on the right we conclude that hi = h2 and so
hiOg = hiOg implies hi = hi. Therefore θα is one-to-one and onto and |Я| = \Нд\. Thus
for each i, \H\ = \Hgi\. Hence \G\ = n\H\ by {44).
Corollary 4.12: Let G be a finite group and g an element of G of order m. Then m
divides \G\.
Proof: The order of g is the order of gp(g) which is a subgroup of G. Then by Theorem
4.11, \gp(g)\ divides \G\. But m — \gp(g)\. Hence m divides \G\.
Corollary 4.13: If G is of finite order n, and g £ G, then д" = 1.
Proof: Every element of G must be of finite order. Let g £ G be of order m. Then,
by the preceding corollary, m divides \G\ and so \G\ = gm. Hence sr|G| = gQm = (gm)" = 1
and the result follows.
Definition: Let the number of right cosets of Η in G be called the index of Η in G. Denote
it by [G:H].
Note that [G: H] is read as "the index of Η in G", i.e. in the opposite order to which
G and Η appear in [G: H].
Corollary 4.14: If G is a finite group, \G\ = \H\ -[G:H].
Proof: In the proof of Theorem 4.11 we conclude with "Hence \G\ = n\H\ ...". Since
η is the number of cosets of Η in G, i.e. η = [G: H], we have |G| = |Я| [G: H].
Problems
4.47. Show that (i) S7 has no subgroup of order 11; (ii) D4 has no subgroup of order 3; (iii) if g € Аъ
and g1 — ι, then g — ι.
Solution:
(i) \S7\ = 7! = 7 · 6 · 5 · 4 · 3 · 2 = 7 · 5 · 32 · 2«. If Я were a subgroup of S7 of order 11, then, by
Theorem 4.11, 11 divides |S7|. But in the prime decomposition of ]jSj-| there is no 11. Hence
there is no subgroup of order 11.
(ii) |D4| = 8. Since 3 does not divide 8, Theorem 4.11 tells us there is no subgroup of order 3.
(iii) If g1 = ι, then either g = ι or g is of order 7, since gm = 1 implies that the order of g
divides m. As m = 7 and is a prime, the only possibility if g ¥= ι is that the order of g is 7.
Now by Corollary 4.12 it would follow that 7 divides |A5|, which is not true. Hence g = ι.
4.48. Prove that if G is a group of prime order, then G is cyclic.
Solution:
If G = {1}, there is nothing to prove. If l¥=g£ G, gp(g) = Η is a subgroup of G. Hence
its order, by Theorem 4.11, divides the prime \G\. As |Я| ¥= 1, }H\ = \G\ since the only divisors of
]G\ are 1 and \G\. Thus Я = G, as Η QG and Ы and G have the same number of elements.
4.49. Give the right and left cosets of Я = gp(M) where ч = "(0,1,1,0) is an element of M, the group
of Mobius transformations, Section 3.6a, page 77.
Solution: "~\
If σ(α, Ь, с, d) G Μ, then
Ha(a,b,c,d) = {ι ·σ(α,Ъ, c,d), v(a, b,c,d)} = {a(a,b,c,d),a(b,a,d,e)}
by the rule for multiplication which is obtained in Problem 3.46, page 79.
Now a(a,b,e,d)H = {σ(α, b, c,d), a(c, d,a, b)}, as is easily checked. Thus we know what the
right and left cosets are in terms of a,b, c,d.

Sec. 4.3]
COSETS
111
с Normal subgroups
We discussed left and right cosets of a subgroup Я in G. Bach gives rise to a partition
of G. How do these partitions compare? In particular, are they the same? Sometimes yes.
In Problem 4.43, the right cosets are just the elements of G; the left cosets can similarly
be shown to be just the elements of G, Sometimes no. In Problem 4.49 the right coset
containing σ(α, b, c,d) also contains σ(ϊ>, a, d, c). But the left coset containing σ(α, b, c, d)
contains v(c, d, a, b). With a suitable choice of a,b,c,d we can ensure that σ(ϊ>, a, d, c) ¥=
v(c, d, a, b), e.g. if a = b = с = 1, d = 0, then σ(1,1,0,1) ¥- <r(l, 0,1,1). Thus the left and
right cosets of Я in G do not coincide.
We ask: when do the right and left cosets of a subgroup Я in a group G coincide?
Suppose every left coset of Я is also a right coset of Я in G. Let a€.G. aH contains a,
as does Ha. Since the right cosets form a partition, the only right coset containing a is Ha.
But we have assumed that there is some right coset which is the same as aH. Hence it must
be Ha, and so aH = Ha. In other words if every left coset of Я in G is a right coset, then
for every a G Я we must have aH = Ha.
Proposition 4.15: A necessary and sufficient condition for the left cosets of Я in G to
provide the same partition as the right is that for each a G G, aH = Ha.
Proof: We have proved above that if every left coset is a right coset, Ha = aH. Let
Ha — aH for all a€.G. {Ha \ a G G} is the set of all the right cosets of Я in G.
Therefore every right coset of Я in G is a left coset of Я in G. Similarly every left coset of
Я in G is a right coset of Я in G. This completes the proof.
If Ha = aH for all a G G, then for each h€.H, ha=ahi for some fciGff." Hence
a~lha G Я for each h G Я.
Conversely if a~lha = hi, for some hi belonging to Я, ha = ah\. Hence HaQaH.
If ah G aH and assuming x'^-hx G Я for all χ G G and all h G Я, then ah = ah(a~la) =
((а~1)-1/ш~1)а G Ha. Accordingly НаэаН and aH = Ha. Thus we have
Proposition 4.16: aH = Ha for all aGG if and only if a~lhaGH for all hGH and
all aGG.
Definition: A subgroup Я of a group G is normal (also called invariant) in G if g~lhg €.H
for all g eG and all h&H. We write Я < G and read it as: "H is a
normal subgroup of G".
By Proposition 4.16, Я is normal in G if and only if Hg = gH for all g G G (equiva-
lently, д-Щд = Η).
In Section 4.3a we gave in (ii) and (iii) reasons for the importance of some cosets. The
cosets we had in mind are those arising from normal subgroups.
Problems
4.50. Prove that every subgroup of an abelian group is a normal subgroup.
Solution:
Let G be abelian and Я any subgroup of G. Then if g € G and he H, g~rhg = h; for since
G is abelian, gh = hg and hence multiplying by g~l on the left, ^ — g~lhg. Then if hGH,
g~xhg GH, Thus Η is a normal subgroup of G. {
4.51. Prove that An <l Sn for each positive integer n.
Solution:
Let α β Αη, τ £ Sn. Is т_1(гт € An1 Now τ is either odd or even. If τ is even, then τ and
τ-1 € An and so τ~ιατ β An. If τ is odd, then τ-1 is also odd, and hence τ-1* is odd, for a e An.
Since τ-1σ and τ are odd, their product rV is even. Hence rV£Aff Note that we used
Lemma 3.2, page 62.

112
ISOMORPHISM THKORKMS
[CHAP. 4
4.52. Let Η be a finite cyclic subgroup of G, and let Η <3 G. Let К be a proper subgroup of H. Prove
that К <3 G. (Hard.)
Solution:
Let К — gp(y). Note that if у is of order m, and g £ G, then g~lyg is also of order m, since
(g~lyg)m ~ g~H)mg = 1 and (g~1yg)T — 1 implies g~xyTg — \ and therefore з/г = 1. Hence иг
divides r, and the order of g~1yg is m. Since Η <3 G, g~lyg £ H. Therefore gp(g~lyg) is a
subgroup of Я of order m. Then by Theorem 4.9(iv), gp(g~lyg) — K. In particular, g~lkg G К for
any fc S G. Hence К <i G.
d. Commutator subgroups, centralizers, normalizers
We will now introduce some subgroups which are normal.
1. If G is a group, we define the center of G, denoted by Z(G), to be
{z\ ζ EG and for all g EG, gz = zg}
Z(G) turns out to be a normal subgroup of G (see problems below).
2. If G is a group and x,y eG, then x~1y~1xy is called the commutator of ж and |/ or, more
briefly, a commutator. We often write [x,y] for the commutator x~ly~lxy. The
subgroup of G generated by all commutators is called the commutator subgroup (also called
the derived group) of G and is denoted by G'. Again G1 turns out to be normal in G.
Proceeding along somewhat different lines, let A be a subset of a group G.
1. The centralizer C(A) of A (in G) is defined by
C(A) = {c | с e G and for all α Ε A, ca = ac]
C(A) is a subgroup of G (see problems below). If A is an abelian subgroup, A is normal
in С (A) (see problems below).
2. The normalizer N(A) of A in G is defined by
N(A) = (n J η ε G and An = nA)
N(A) is a subgroup of G and, if A is a subgroup of G, A is normal in N(A). Furthermore,
if A is a subgroup of G, A is normal in G if and only if N(A) = G. These facts will be
proved in the problems below.
The details concerning the groups Z{G), G', C(A), N(A) appear in the problems below.
In Chapter 5 we will use the concepts we have just introduced.
Problems
4.53. Prove that the inverse of a commutator is a commutator.
Solution:
[x,y] = x~xy~lxy — as, say. So a-1 = у~гх~^х = [у,х].
4.54. Prove that G' is normal in G.
Solution:
We must show that if g e G and hGG', then g~lhgGG'. If h is a commutator, say
h — x~1y~1xy, then
g~lhg = g~1x~lgg~1y~lgg~1xgg~lyg = х^уГ1*^ =■ [*i,»i]
where xt = g~lxg (consequently x~l — g~lx~lg) and уг — g~lyg.
Now any element h of G' is a product of commutators and their inverses; and as an inverse of
a commutator is a commutator, every element h of G' is a product ег ■ · ■ ckof commutators. Therefore
g~xhg = д-Цв! ·■- c,Jg = g-^gg-^g ·■· g~lr.kg = rf,^ · - ■ dk
where di ~ д~1с$. But we have just shown that dt is a commutator. Hence if hSG', g-ihg&G'.

Sec. 4.3]
COSETS
113
4.55. Show that G is abelian if and only if G' = {1}.
Solution:
Suppose G is abelian and that x,y G G. As χ and у commute (i.e. xy = yx), [x,y] — x~ly~lxy ~
χ-ΐχ = l. Then G' is the subgroup of G generated by 1, and G' = 1. Now if G' = {1}, then in
particular any commutator [x,y] = x~1y~1xy = 1. Hence x(x~1y~1xy) = χ and y(y~1xy) = yx,
i.e. xy =■ xy. Thus G is abelian.
4.56. Show that every element in Ая is a commutator of elements in iS3. Hence show that S3 = A3.
Solution:
We use the table of Section 3.3a, page 57. rt = τ"1. τ~1σ~ιτισι = τι^ι^ι = <ί<ί ~ ο%·
τ~ισ~ίτισ% = r1ff1r1ff2 = σ2σ2 = σι· Τ^1ι~1τιι = ι· Thus every element of A3 is a commutator of
elements in S3 (A3 is listed in Problem 3.23, page 63). If we can show that all commutators belong to
A3, then A3 = S3. This is a matter of trying all possibilities, e.g. т~г τ~λτχτ% = tjTjTjTj = av Hence
the result.
4.57. Show that the commutator subgroup of Μ of Section 3.5a, page 77, is infinite. [Hint. Find an
infinite cyclic subgroup generated by a commutator.]
Solution:
[<j(2,0,0,1), a(l, 1,0,1)] = „(1, -1,0,1) = a, say. Now gp(a) is infinite cyclic, as ση = (1, -η, 0,1)
for each n. Since the commutator subgroup of Μ contains gp{a), it is infinite.
4.58. Prove that if G is any group, Z(G) is a normal subgroup of G.
Solution:
1 G Z(G), since lflf = flfl for all g G G. Consequently Z(G) Φ 0. If gv g2 G Z(G) and
fir GG, then g(g1g~1) = (gg1)g~1 =g1(gg-1) = g1g~1g since gg% = g^g implies gf1g = gg~1. It
follows that Z(G) is a subgroup of G. If g G G and hGZ(G), then gh = hg and so h—g~lhg.
Hence fif-Vifir G Z(G) for all g G G and all /ι G Z(G). Thus Z(G) is normal in G.
4.59. Show that C(A) is a subgroup of G and, if A is an abelian subgroup of G, A < C(A).
Solution:
1 G C(A) and so C(A) ^ 0. If 0i,02eC(A), and α G A, then g2a = ag2 and hence
ctAf1 = 0.Γ1α, i.e. 0"1 £C(A). Now ctfif^1 = 0itt0~1 = 0ι0~1α and so fifjfi·^1 £ C(A) if
ΑΊ. 02 e C(A). Therefore C(A) is a subgroup of G. If A is an abelian subgroup, then each a, G A
belongs to C(A). Now if 0 6 C(A), then for each a G A, 0ct = ct0, i.e. g~xa,g = α G A.
Accordingly A < C(A).
4.60. Show that if A is a subset of G, then N(A) is a subgroup of G. Show that if A is a subgToup
of G, then А О G if and only if N(A) - G.
Solution:
N(A) Φ 0, since 1 G 2V(A). Let /, 0 G 2V(A). Using the results of Problems 4.45 and 4.46,
gA = Ag and /A = A/ implies (fg-l)A = f(g~lA) = f(Ag-*) = (fA)g~* = (Af)g-t = A(/0~i).
Hence /, 0G2V(A) implies fg~l G N(A). Therefore N(A) is a subgToup of G. Clearly, if A is a
subgroup, Α ς N(A) and A < 2V(A).
If A is a subgroup and A < G, then for each 0 G G, 0A = A0. Hence 0 G 2V(A) and so
G с N(A). Therefore G = N(A). If A is a subgroup and N(A) = G, then since A < N(A), A < G.
4.61. Find all normal subgroups of S3.
Solution:
We use the notation and multiplication table in page 57.
Clearly {ι} and S3 are both normal subgroups of S3. There are no normal subgroups of Ss
containing elements of order 2 except S3. The elements of order 2 in S3 are, as we readily check by
using the multiplication table for S3, tj, т2 and т3. Suppose for example that a normal subgroup
N of S3 contains тг; then σ~1τισι = σ2τισ1 = τ2 G N. Similarly τ3 G N. Hence 7ίτ2 = σ2 G Ν
and α\ = σι G 2V, and so it follows in this way that N = S3.

114
ISOMORPHISM THEOREMS
[CHAP. 4
We have shown that if N is a normal subgroup of S3, then if N contains elements of order 2,
N = S3. Therefore if Ν Φ {ι}, then N must contain elements of order 3 (there are only elements
of order 1, 2, or 3 in Ss). Now ffJ and σ2 = "? are the оп1у elements of S3 of order 3. In fact
{1, ctj, σ2} is a normal subgroup of S3. For example, τ~1σχτ1 = tj^tj = <т2 е {'. σι> ^г}- Accordingly
S3 has precisely three distinct normal subgroups.
4.62. Show that if A is a subgTOUp of G and В < G, then AB is a subgroup of G, where
AB = {x\ x = ab, aeA, b<=B}
Solution:
AB Φ 0, as 1 — 1-1 SAB. If gltg2GAB, then g^ = o.^, gr2 = a.2b2 where at G A and
Ь; G R NOW , , ,_, _i , -1 , , , „4
ΑΊ^ = αι°ι°2 α2 αι°ζα·ι (where bs G B)
— a1a~1a2b3a^i = a1a~1(a~1)~lb3(a~1)
= ab, say, /
where a = α,^1 G A, and b = a2bsa2l & В as В < G. Thus ϋ\ϋ~^χ G AB and AB is a
subgroup of G.
4.63. Show that the intersection of two normal subgroups is a normal subgroup.
Solution:
We refer the reader to Problem 3.15, page 55, in which we proved that the intersection of two
subgroups is a subgroup. If H,K are normal subgroups of G, then HnK is a subgroup of G. If
cGffnK and fif G G, then g~lcg G Η as c£H and Η < G, and g~leg&K as e&K and
К < G. Thus g~1cg G HnK, and so FnK is normal in G,
e. Factor groups
In Section 4.3a we mentioned that the concept of a coset sometimes gives rise to a new
group. This occurs when, and only when, the group is normal.
Let G be a group and N <\ G. Let us denote by G/N (read as "G over N", or "G factor
N", or "the factor group of G by N") the set of right cosets of N in G. We turn G/N into a
groupoid by defining a binary operation as follows.
Define a product of two cosets Na and Nb to be the coset Nab. This definition of
multiplication depends on a and b. But it is conceivable that if Na — Ναι and Nb = Nbi, that
Ναώι Φ Nab. In such case, what would we take for the "product" of the two cosets, Naibi
or Nabl What we must show is that the product of two cosets is uniquely defined by the
formula NaNb = Nab when N < G. If Nax — Na and Nbi = Nb, then αϊ = na for some
η GN and f>i = mb for some m G.N. Accordingly,
aibi = namb — n(ama~l)ab = iab
where l — n(ama~1). Since N < G, ama~lGN and hence IG.N. Thus aibiGNab.
Since the cosets form a partition, it follows that Naibi = Nab. But this is just what we
wanted to prove.
Thus G/N with this binary operation is a groupoid. Is it an associative groupoid?
{{Na){Nb))(Nc) = {Nab)Nc = N(ab)c - Na(bc) (as G is associative)
= {Na)(Nbc) = (Na)((Nb)(Nc))
and so G/N is an associative groupoid.
Theorem 4.17: G/N is a group. The mapping ν: G -» G/N defined by grv = Ngr is a homo-
morphism of G onto G/N.
Proof: First, G/N is an associative groupoid. N1 — N is an identity, for Na ■ N1 =
Ν{α·1) = Na and Ν·Να - N(1-a) = Na. Next, each element Να has an inverse, for
NaNa'1 = Ν(αα^) = N while Na~lNa = N(a~la) = N. Thus G/N is a group.
Clearly ν is a mapping of G onto G/N. Since (fl^)»- = Ν^β2) and {д^)(д^) -
NgxNg2 = Ng^g^, (g^v - (д^)(д3у) and hence ν is a homomorphism.

Sec. 4.3]
COSETS
115
ν is called the natural homomorphism of G onto its factor group G/N.
Note that in the case where G is a group whose binary operation is +, as we remarked
at the end of Section 4.3a, the elements of G/N are of the form N + g. Instead of using
the multiplicative notation for G/N, we use additive notation. Our definition of the product
of two cosets is written as the sum: (N + gi) + (N + дг) = N + (gi + дг).
Problems
4.64. Prove that Z/E = C2, where Ε is the set of even integers and C2 is the cyclic group of order 2.
Solution:
As Ζ is abelian, Ε is a normal Subgroup of Ζ and Z/E makes sense. A coset of Ε is of the
form Ε + z, z&Z. £4-0 consists of all the even integers, Ε + 1 of all the odd. Since an integer is
either odd or even, these are all the cosets. gp({(E + 1)}) contains £7 + 1 and (E + 1) + (E + 1) = E,
and so gp({(E + 1)}) = Z/E. Thus Z/E is cyclic of order 2. Hence the result
4.65. Is Q/Z = Q, where Q is the additive group of rationals? (Hint. Examine the order of the
elements of Q/Z.)
Solution:
A coset of Ζ in Q is of the form Ζ + q, q G Q. As q — m/n for some integers m, n, it follows
that nq £2. Then
(Z + q) + (Z + q) + ■ ■ ■ + (Z + q) = Ζ
V v I
η terms in all
and therefore Ζ + q is of finite order in Q/Z. Thus every element of Q/Z is of finite order. On the
other hand no element other than 0 of Q is of finite order, for nq Φ 0 if g/0 and η ¥= 0. It is
this fact that we will utilize to prove that Q/Z is not isomorphic to Q. Suppose it were and that
в : Q/Z -» Q was such an isomorphism. Choose ?£Q (q Φ 0); then there exists an element Ζ + r
of Q/Z such that (Z + r)0 = q. Now Ζ + r is of finite order n, say. Then (n(Z 4- r))9 = nq and now
n(Z + r) = Z. Since Ζ is the identity of Q/Z and в is an isomorphism, в takes Ζ to the identity of
Q, namely 0; hence nq = 0. But q Φ 0, so nq Φ 0. This contradiction proves that there exists
no isomorphism of Q/Z onto Q.
4.66. Find SySa.
Solution:
From Problem 4.56, S's = As. Hence the cosets of Sg/Sa are A3 = Α3ι = {ι, σν <τ2} and
Α3τι = {tj,t2, ts}. Thus Sg/S^ consists of these two cosets. The multiplication table is
"■8 ^3Tl
l3Tl
Ab
Α&ι
^8^1
As
4.67. Prove that if G s Я, then G' = Я' and Z(G) = Z(H).
Solution:
Let ff be an isomorphism from G to H. Then * = 9\q> is a monomorphism into Я, since в is
both one-to-one and a homomorphism.
What is G'*? The elements of G' are products of commutators and their inverses. As the
inverse of a commutator is a commutator, each element of G' is of the form ct · ■ · ck, where each
c{ is a commutator. Then the images of G' under Φ are of the form егФ · · · ск·*. If
с = [α, о] = в-»Ь~»оЬ, then сФ = (αί)-1^*)"1**^* = [tt*>ь*]
Thus the «^ are commutators and so ΰ'Φ С Н'.

116
ISOMORPHISM THEOREMS
[CHAP. 4
To show G'* = Я', we need only show that all possible commutators [hu A2] are images under
*. As в is onto, there exists gltg2 such that gx8 = klt gze = k%- Hence [дъвг]^ ~ [hi> '1г]·
Therefore С'Ф = Я' and consequently G' ss H'.
Let ψ ~ ff|Z(G)· Then 0 is a monomorphism of Z(G) into Я, and we need only show that it maps
onto Z{H) to prove Z(G) s Z(H). Let ζ G 2(G). For each k<=H there exists # G G with #9 = h.
Hence zsA = «ffflfS = (zg)e = (gz)e — geze = hze and so Z(G)4>QZ{H). Let χ e 2(H). As в is onto,
there exists у G G with j/ff = ar. Let g & G. Then (flfj/)9 = 002/9 = увдв — (yg)9. As 9 is one-to-
one, gy = yg. Hence у G 2(G), and every element of Z(H) is an image of an element of Z(G). The
result follows.
4.68. (i) If A is abelian show that \47Д' = A.
(ii) Show that GIG' is abelian.
(iii) Show that GIN abelian implies N э G',
Solution:
(i) A' is the subgroup generated by the commutators x^1y~ixy, x,yGA. As A is abelian,
х~^у~1ху — у^1х~^ху — у^Чу — 1 and A' is the subgroup generated by 1. Since any
product of 1 and its inverse is again 1, A' = {1}. Let ν be the natural homomorphism of A to
ΑΙΑ'. To show с is an isomorphism we need only show that it is one-to-one. Suppose αγν ~ a2v,
i.e. {1}^! = {1}»2, i.e. {α^} — {α.^}. Then of course α,γ = a%, so ν is one-to-one. Thus ν is an
isomorphism.
(ii) Let G'x and G'y be two elements of GIG'. Then G'xG'y = G'xy while G'yG'x = G'yx. Now
(ar-1)~1(3/~1)~1a;^1j/_1 = £j/ar~12/~1 € G', and so G'yx contains the element xyx"^"1 -yx — xy.
But G'xy contains xy; hence G'xy = G'yx. Therefore (G'x)(G'y) = (G'y)(G'x) and GIG' is
abelian.
(iii) We need only show that N contains every commutator. For then N contains the subgroup
generated by the commutators, which is G' by definition.
If GIN is abelian and x,y are any elements of G, (<sy)N = xNyN = yNxN = (yx)N. Hence
xy = yxn where η G N. Multiplying on the left by 3/-1 and then by x~l, we obtain
x~1y~1xy — n, i.e. [x,y]&N.
4.69. Show that if Η is a subgroup of G containing G', then Η < G. Show that if Я is of index 2 in
G, then Η < G and GfH is cyclic of order 2.
Solution:
Let k G Я and consider g~1kg. Now g~1kgk~~1 is a commutator, and hence belongs to G' and
thus to Я. Therefore g-lhgh~l ■ h = g'lhg <= Η and Η < G.
If now Я is of index 2, G-HuHg where Hr\Hg = 0. Let h G Я. If fc G G, then fc = At
or fc = Ajgr (At G Я)· Hence
fc-iftu = hj~1hhl G Я or k~1kk = g~1k^~1kklg = g~lh^g
where h2GH. If д~*Ь,?д G Я, we are through. Otherwise д~11цд^Нд, so that g~lh^g = h'g
(A' G Я) and thus flf-1^ = Ί'. Hence fir = /i2A-1 G Я. But this contradicts the assumption
HnHg = 0. Thus we are forced to conclude that k~lhk G Я for every h G Я and every fc G G,
i.e. that Η < G. The two cosets of Я in G are Я and ffgr for some g &H. Accordingly gp(Hg) —
GfH and so GIH is cyclic of order 2.
4.70. Show by considering a suitable non-normal subgroup Я of S3 that the product of two cosets cannot
be defined as on page 114 without ambiguity.
Solution:
We use the notation of Section 3.3a, page 57, in dealing with the symmetric group S3 of degree
3. Let us take Я = gp(rj = {1,^}. Яа1 — {σ1(τ3} = Ητ3 while Наг - {σ2>τ2} = Нтг. In page 114
we defined the product of the coset Ηαλ and Ha2 as Hala2 = H. However, the product of Hts and
Ят2 would be Hr3T2 = Ησι Φ Я. This means that the product of cosets is not uniquely defined.

Sec. 4.4]
HOMOMORPHISM THEOREMS
117
4.4 HOMOMORPHISM THEOREMS
a. Homomorphisms and factor groups: The homomorphism theorem
We now consider the connection between homomorphisms and factor groups. We have
already established in Theorem 4.17 that corresponding to every factor group GIN there
is a homomorphism ν; G -* GIN such that Gv — GIN. What about the converse? Suppose
now that θ is a homomorphism of G into a group H.
We ask: is there a normal subgroup N of G such that GIN~ <?0 ? Let us define the
kernel of θ (denoted Ker Θ) by Ker0 = [g \ g e G, дв = 1}. We will show that Ker0 will
do the trick.
Theorem 4.18 (Homomorphism Theorem; also called the First Isomorphism Theorem):
If θ: G -* Η is a homomorphism of a group G into a group H, then
N — Ker θ is a normal subgroup of G, and η: g9^> Ng defines an
isomorphism of G6 onto GIN.
Proof: First we will show that N is a subgroup. If #i, #2 e N, then
(W)0 = M)(iC) = (i^Hflfeflr1 =1-1 = 1
and so flriflr^1 eiV. Also l€.N, so N ¥-0, the empty set. Thus iV is a subgroup of G.
To prove that it is a normal subgroup of G, let n G N and g GG. We must show that
g~lng e iV; this will hold if {д~1пд)в — 1. But
(<Г'ад)0 = W^Hsrf?) = (^-η^β) = {дв)~\дв) = 1
Hence g~xng e iV and iV is a normal subgroup of G.
Next we must show that η: дв^> Ng defines a mapping. It is conceivable that there
exist gt ¥* g% with gtf — g%Q. We ask: is Ngi — Ng%1 For if not, η is not a mapping as
it is not uniquely defined. Now gigf1 G.N, for (^i^-1)^ = {9^){9^)~l = 1 as дгв = #20.
Hence gig^1 Ξ iV and #1 = ngz where n e N. Thus iVgi and N,92 have sri in common and,
as the right cosets form a partition, Ngi = Ng%. Consequently η is a well defined mapping.
Is it a homomorphism ?
(дгвдгв^ = {(gig%)e)4 = N(gig%) = NgiNg* - {ΰιθ)η^ιθ)η
and so η is a homomorphism.
Finally, is 7? one-to-one? If (srifl)?? = (#20)τ?, then Ngi = Ng2. Then ngi-g2 for some
nG.N, and gzO = (ngi)6—nOgiO = 1· gi9—giO. Thus 7? is one-to-one and hence is an
isomorphism.
Problems
4.71. Let 9 be the homomorphism of Ζ into the multiplicative group of nonzero rational numbers defined
by xe = 1 if ж is even, and xe = —1 if ж is odd. Find the kernel of 9 and examine the claim
G/(Ker 9) sa Gu.
Solution:
Ker 9 = {ж | ж<? = 1} = {χ I a; is even} and G/(Ker в) = {Ker 9, Ker 9 + 1}, so Ker 9 + 1
generates G/(Ker<?). We have (Ker 9 + 1) + (Ker θ + 1) = Kers; hence Ker 9 + 1 is of order 2
and G/(Ker #) is cyclic of order 2. Now Ge = {1,—1}. Ge = irp({—1}) and, since (—1) · (—1) = 1,
Gb is also cyclic of order 2. Therefore G/(Ker 9) and G9 are isomorphic by Theorem 4.7, page 103.
4.72. Check that the kernel of the natural homomorphism of the additive group of integers (Z, +) onto
Z/2Z is 2Z, where 2Z = {x \ χ — 2z, ζ e Z}, i.e. 2Z is the set of even integers.
Solution:
The natural homomorphism ν is defined by zv — 2Z + z. The identity of ZI2Z is the coset 2Z.
Consequently ζ £ Ker ν if and only if 2Z + ζ = 2Z and hence if and only if ζ G Ш. Thus
Ker ν = 2Z.

118
ISOMORPHISM THEOREMS
[CHAP. 4
4.73. Verify that if G is any group, the subgroup generated by the squares of the elements of G, i.e.
elements of the form gg = g%, is a normal subgroup of G.
Solution:
Let S denote the subgroup generated by the squares of the elements of G. Let χ £ S. Then χ
is a product SjS2 ■ ■ · sk of elements of G each of which is a square or the inverse of a square. Since
the inverse of a square is also a square, we may assume each β], .. .,efc is a square. If g G G, then
g~lxg = д-^дд-Ч^д ··■ g~lskg = *ι*2 —*fc
where ί< = g~lstg. We assert that each t{ is a square. Since e{ = r'f for some rt,
к = д~18гд = fg-^gg-^g = {д~1цд)*
and hence the ij are squares as asserted and g~lxg G S. Thus S <i G.
4.74. Let σ be the homomorphism of Q*, the multiplicative group of nonzero rationals into Q* defined by
χσ = |as|. Find the kernel and image of σ. Verify the homomorphism theorem directly in this case.
Solution:
σ is a homomorphism, since (xy)a = \xy\ = \x\ \y\ = (xa)(ya).
Kera = {χ\χσ = 1} = {* | |*| = 1} = {1,-1}.
Q*o = {a; | χ = yo, у £ Q*} = {a; | χ G Q*, χ is positive}
Let ν be the mapping of Q*/(Ker σ) into ζ>*σ that takes
(Ker o)q -» ?σ
Now (Ker a)q = {q, —q}, and so the only other "representation" for the coset (Ker o)q is (Ker σ)(—g).
Hence the other possibility for ν as far as (Kera)g is concerned is that ((Ker σ)ςτ)»> = (— q)a. But
(—q)a = \q\ = q0. Thus ν is a well defined mapping of G/Ker σ into (?*σ. Since
[(Κβτσ)?1·(ΚβΓσ)?2]» = ((Ker σ)^^)* = \дгд2\ = \дг\\д2\
= «Ker o)q,)r ' ((Ker a)g2)f
ρ is a homomorphism. Is ν one-to-one? If (Ker o)qxv = (Ker o)q2v, then we have that ς^σ = q2a,
i.e. |?!| = |g2|. Thus if ?! ^ <j2, ?! = —g2. Therefore (Kera)g, = (Kera)g2. Hence ρ is one-to-one
and it follows that Q*/(Ker σ) э£ Q*ff.
4.75. Let Cr be the group of mappings of the real line R onto itself of the form aa,b: χ -+ ax + Ь, а ?& 0,
а, Ь real numbers, χ £ R, Prove that the map β: аа,ъ~* аа,о *s a homomorphism of G into G.
Find the kernel and the image of this homomorphism and exhibit the isomorphism described in the
homomorphism theorem.
Solution:
Note that aabac>d = a^,ibc + d, since xaaь = ax + b and (xaaib)acd = c(ax+ b) + d. Then
(<4i,«Cjd)9 = Kc.bc + rf)* = aoc,0' K,i,9)(ac,d») = ao,o«c,o = <V,o> and so 9 is a homomorphism. If
aab ε Ker 9, we have a„ bS = a10 as at 0 = t, the identity mapping. Hence {αλ ь \ b any real
number} = Kers, The image of 9 = {aOj0 | α any nonzero real number}.
A typical coset is
(Ker8)acd = {α1(( | b any real number}aCid
= {"с.ьс+dl b any real number}
= {«ce | e any real number}
The isomorphism 4 between G7(Ker 9) and G9 as given by the homomorphism theorem is the one that
takes the coset (Ker9)«cd to ac0.
4.76. Prove that if G is cyclic of order η and ρ divides n, then there is a homomorphism of G onto a cyclic
group of order p. What is the kernel of this homomorphism?
Solution:
Let G = gp(x) and let η = pm. Let Η be cyclic of order ρ, Η = gp(y). We define 9 to be
the mapping , . ^ . ^
9 : х1-*у\ 0^i-n-l

Sec. 4.4]
HOMOMORPHISM THEOREMS
119
9 is well defined as we established in Lemma 4.6 that the elements xl, 0 — i — и — 1, are all the
distinct elements of G. Now let г and / be less than n. We have {xlx')9 = (xi+^tn)e where « = 0 if
t + j — и — 1 while e = 1 if г + j — n. Then
(xlX')9 = yi + i-en = yiyiytn
Since the order of у divides n, y~*n ~ 1. Hence (χίχί)β = у*у} = (ж<9)(ж*9) and so 9 is a homomor-
phism. The kernel of 9 is the set of all xl such that x'9 = 1, 0 — г — и— 1. Since х*9 = yf and
у1 = 1 if and only if ρ divides i,
Kers = {xl | ρ divides г} = {χ", ж2", ..., ж(т1)р} = др(х»)
4.77. Let (R+, ·) be the multiplicative group of positive real numbers. Prove that the mapping of Й +
into (R,+), the additive group of real numbers defined by 9 : χ -* loglox, is a homomorphism. What
is the kernel of 9? What is the image? Using the homomorphism theorem, prove that
Solution:
9 is certainly a mapping of R + into R· Furthermore,
(xy)9 = \ogw(xy) = log10a; + logl0j/ = хв + y9
and so 9 is a homomorphism.
Kers = {x\ loglox = 0, xGR+} = {1}
We assert that R + 9 = R. To see this observe that if χ is any real number, then 10*6 Й+.
Moreover, log 10* = x. Then if у is any element of R, 10«9 — log 10^ = у and hence R+ 9 — R.
The homomorphism theorem states that й+/(Кег 9) s· R+θ = R. As Ker9 = {l}, all we must
do is show that й+/{1} = R+. We can indeed show this in general: if G is any group, G/{1} s= G.
To do this we exhibit the isomorphism. Let ν be the natural homomorphism of G onto G/{1}. Then
we need only show that ν is one-to-one. Suppose gxv — g2v, i.e. {l}tfi = {1}02- Then {gt} = {g2}
and consequently #, = g%. Hence ν is one-to-one and thus an isomorphism. Accordingly
Й+ as й+/{1} as Д and so R+ = Й.
4.78. Prove that the mapping 9 : χ -* ex defines an isomorphism of {R, +) onto {R+, ·), the multiplicative
group of positive real numbers.
Solution:
(x + y)9 = e*+* = e*e" = (ar9)G/9) and so 9 is a homomorphism. If X9=y9, then e* = e" and
e*"*" = 1, from which x — y — 0 and χ ~ y, so 9 is one-to-one. Is 9 onto? Yes, for if у is any
positive real number, the equation ex = у has a solution χ G R. Thus 9 is an isomorphism between
(R,+) and (Й + ,·).
4.79. Prove that [/flf,a] = g"l[f,a>] g[ff,o] for any /, g and α in a given group. Suppose G'qZ(G), the
center of G. Let α be a fixed element of G. Prove that the mapping 9 : g -* [g, a] is a
homomorphism of G into G. WhatisKer9? (Difficult.)
Solution:
Observe that [fg, a] — g~1f~la~tfga. On the other hand,
g~4f.a]fflff.a] = g~llf~l°>~lfa)g(g-la~lga)
= д~Ч~1о>^Ч{адд~1а,"1)да
= g-if^a-ifga = [fff,a]
If G'CZ(G), then [/,e]6Z(G) and g~l[f,a]g = [f,a]g~ig = [/,»]. Therefore [fg,a] =
[/,α][ϊ,ο]. Hence (/#)9 = /9flf9 and 9 is a homomorphism of G into G. Ker9 = {g \ [g,o] — 1,
j6G}. Thus Ker 9 = C(gp{a)), the centralizer in G of tfp(a).
4.80. Prove that if 9: G -> К is a homomorphism and \G\ < », then \G9\ divides |G|.
Solution:
By the homomorphism theorem, G/(Ker 9) s^ G9. By Lagrange's theorem, |КеГ9| divides the
IGI
order of G. Therefore |Gs| = — divides |G|.

120
ISOMORPHISM THEOREMS
[CHAP. 4
4.81. Show that the group Μ of Mobius transformations is a homomorphic image of the group
- = {(:,
a, b, c, d complex numbers, ad — be = 1
Find the kernel of the homomorphism.
Solution:
In Problem 3.48, page 80, we showed
Μ — {σ(α, 6,e, d) | a,b,c,d complex numbers, ad— 6c = 1}
We define the mapping /u; %i-»■ Μ by ( j μ = σ(α, b,c,d). Clearly μ is a mapping.
Furthermore, using the results of Problem 3.46, page 79,
/ajan+ c,6o αι"2+ "ι^4
— ' 'ft
f
Thus
j&2 &1cj+d1il2
= aft^Oj + 0^21 δι<*ί + dtb2, ахег + с^2) btc2 + d^)
'αλ ΰχ\ (fh. c2
S62 <*2
Γ/α1 "A/eg ο2\Ί _ /α,β^+^ΐ
— σ(αγα^ + сг1
/»! <ίΑ (c
= σ{αν blt сv dijaiatfy 62, с2, d2) = Ι \ μ f
μ is a homomorphism. Kernel μ = л ( „ ι)» ( η —ι / f ' Deeause (a d/'1 ~ (T(a'b»c'fi')
and σ(α, Ь, c,d) = σ(1,0,0,1) (the identity of M) if and only if a = d = 1 and b = c = 0 or
а = d = —1 and 6 = с = 0, by Problem 3.49, page 80. Therefore
-■Via i)-("U°)}
b. Correspondence Theorem. Factor of a factor theorem
Let 9:G^K be a homomorphism of G onto К. If Я is a subgroup of G, then ΗΘ is a
subgroup of K. If Я is normal in G, ΗΘ is normal in K. (See Problem 4.82 below.)
What about the reverse of this procedure? Would the preimage of a subgroup S of K,
i.e. the set {g \ g £ G, дв е S}, be a subgroup of Gt We know that if S = {1}, then the
preimage of S is Kerff, and this is indeed a subgroup of G. We generalize this result in
the following theorem.
Theorem 4.19 (Correspondence Theorem): Let 9:G-> К be a homomorphism of G onto
К. The preimage Η of any subgroup S of If is a subgroup of G containing
Ker Θ. If S < К, then Я <d G. Furthermore if Я1 is any other subgroup
of G containing Ker θ such that Ητθ — S, then Я1 = Я.
Proof: Η = {g\ дв G S}. Since Iff is the identity of K, and S contains the identity
of K, then 1еЯ and so Я * 0, the empty set. Also, if g,h e H, (gh^)e = ge(he)~l. As
дв eS and Й.0 G S, it follows that (gh~l)e G S. Hence gfe"1 G Я and Я is a subgroup of
G. Since Ker0 = {x\ χθ= 1} and 1 G S, Ker Θ С Я.
If S <d Я, we must show that Я <d G. Let Л. G Я, jeG; then (g~1hg)e = {g9)~l(h9){ge).
Now fe0 GS, дв eK, and S <d Κ implies {g~lhg)e G S. Then safes' G# and so Я <d G.
Let Я1 be a subgroup of G containing Ker0 and suppose Нгв = S. We will show that
Hi-H. Let hteHu then MeS. Now Я = (g | g0 G S}. Therefore feiGff and
hence Я1 Q Я. On the other hand if h e Я, then he = s<=S. Choose fti e #! such that
hr9-s. Then Mi_1GKer9Cffi and so /iGH, and HQHi, Thus Я = Я,.

Sec. 4.4]
HOMOMORPHISM THEOREMS
121
Corollary 4.20: Let 0: G -» К be an onto homomorphism. Let S be a subgroup of К of
index η < <χ>. Let Я be the preimage of S. Then Я is of index η in G.
Proof: Let Sku Sk2, ..., Skn, where h G if, be the distinct cosets of S in K. As 0 is
an onto homomorphism, there are elements gi, ..., gn of G such that g$ = fo. We claim that
Hgu...,Hgn (4.5)
are the distinct cosets of Я in G,
Suppose that Нд{ = Нд^ then gigf1 ей. Hence gie(gje)~1 G S, i.e. kik^l€S, from
which jSfei = Sfej and ΐ = /. Thus we have shown that all the cosets in (4.5) are distinct.
Let g GG. Then дв G К and so #0 G Skt for some integer t. Hence дв = sfe{ with
sGS. Consider x~gg7l. хв = дв(д{в)~1 ~ skik^1— s. Consequently ggr1 is in the
preimage of S, so that ggr1 G Я. This means that g G Я#{. We have thus shown that every
element of G is a member of one of the cosets in (4.5). It follows that (4.5) consists of all
the cosets of Η in G.
Of course we can always reformulate results about homomorphisms with the aid of the
homomorphism theorem as results about factor groups. Thus we have
Corollary 4.21: Let N < G. Let L be a subgroup of GIN. Then we can write L = Η IN
where Я is a subgroup of G containing N. If L О GIN, then Η < G.
If HilN — HIN where Hi and Η are subgroups of G containing N, then
Ht = H.
Proof: Let ν be the natural homomorphism of G-» GIN. Let Η = {g | gv e L}. Then
by the correspondence theorem Я is a subgroup of G; and if L < GIN, Я <] G. Also,
Hv~L. Since v.g^Ng, Η ν consists of all cosets Nh, h G Я. Because Hz>N = Kerv
by the correspondence theorem above, Я/N makes sense and consists of all the cosets Nh,
h£H. Hence HIN = Яу = L.
Now if HiDN and Я1/Л? = Я/N, then Я^ = L. It then follows by the correspondence
theorem that Hi = Я.
(Problems 4.82-4.89 below may be studied before reading Theorem 4.22.)
The reader may very well wonder what happens when we take a factor group of a
factor group. For example, if N < G and GIN is a group containing a normal subgroup
MIN, then what is (GIN)l(MfN)! The next theorem tells us that this is isomorphic to a
single factor group, i.e. a factor G by one of its normal subgroups.
Theorem 4,22 (Factor of a Factor Theorem, also called the Third Isomorphism Theorem):
If in the factor group GIN there is a normal subgroup MIN, Μ gW, then
Mo G and
GIM s (GIN)I(MIN)
Proof: Let v : G -» G/N be the natural homomorphism of G^ GIN. Let ρ: G/N -»
(GIN)I(MIN) be the natural homomorphism of G/N^ (GIN)I(MIN). Put 0 = yp. Then 0
is a homomorphism of G -» (GfN)l(MIN); and since ν is onto G/N and ρ is onto (GIN)f(MfN),
vP is onto (GIN)I(MIN). Therefore G/Ker (vp) s* (GIN)I(MIN), by the homomorphism theorem.
If g€G, gv = Ng and (Ng)P = (MIN)(Ng); note that here (MfN)(Ng) is a coset of the
normal subgroup MIN in (GIN), i.e. an element in the group (GIN)I(MIN). Now the elements
of MIN are all the cosets Nm, m(=M. The identity of (GIN)!(MIN) is (MIN). We ask,
what is the kernel of vp? It will be all g e G such that gvp ~ (MIN)Ng - MIN. But in

122
ISOMORPHISM THEOREMS
[CHAP. 4
that case Ng G M/N, i.e. Ng = Nm for some m G M. Hence g — nm where ne N.
But Μ 2 N. Therefore g e Μ and so Ker vp С Μ. Note that if m ε Μ, m(vP) =
(M/N)Nm - M/N. Then Ker vP = M. Thus Ж as a kernel of a homomorphism is normal
in G and G/M ^ (G/N)/(M/N), which is the required result.
Problems
4.82. Prove that if в: G -*■ К is a homomorphism of G onto K, and Η is a subgroup of G, then He is a
subgroup of Я. If # < G, prove that He < K.
Solution:
Since 1е€Нв, НеФф. It x1,x2SHe, xl = h1e, хг = hze for some hl,k2€H. Then
Xi^2l = hiiihrf)-1 = Α,ίΑ^"1» = {h^hz^e = he
where й. = fc,fc2"'e H. Hence He is a subgroup of K. If now ff<G, then g~lhg S Я for all
g £G and all fceff, Now any element of Я, say fc, is of the form ge for some ff £ G; and any
element of He is of the form he. Is (tf0)-1ftitf0 G He! Yes, because (ί»)_1(Ί»)ί?» = (g~lhg)e and
as ff-^ff <= H, (ge)~Hhe)ge <= Hi. Thus Hi < Я.
4.83. Let G = D4 = {ctj, σ2, ir3, nr4, τ, τσ2, τσ3, τσ4}. Let Я = ffp(b) be cyclic of order 2. Then в : G -» К
defined by σβ = 1, (τσ{)θ = b, i = 1, 2,3,4, is a homomorphism of G onto Я. (Take this as a fact.)
Find all the subgroups of Я and all their preimages. Check that the assertions of Theorem 4.19 and
Corollary 4.20 hold. (See Problem 3.42, page 77, for the multiplication table.)
Solution:
The subgroups of К are Я, = К and K2 — {1}. G1 = the preimage of Кг = {g \ ge S K) = G.
G2 = the preimage of K2 — {g \ gθ = 1} = {σν σ2, σ3, σ4>-
(α) Glf G2 are subgroups of G containing Ker ί = G2.
(b) Я1 and К2 are normal in K, and Gl and G2 are normal in G.
(c) The subgroups of G containing Ker 9 are Gt and G2. We note then that G$ = Gse implies
i = j where 1 — i, j — 2.
(d) Я! is of index 1 in Я. Gl is of index 1 in G. Я2 is of index 2 in Я. G2 is of index 2 in G, its
cosets being Gj = {<Ί,σ2, σ3, σ4} and rG2 = {τ, τσ2, τσ3, τσ^.
Thus (α), (6), (с) and (d) agree with Theorem 4.19 and Corollary 4.20.
4.84. Let G = gp(a) be cyclic of order 12 and let К = tfp(b) be cyclic of order 4. Let β : G -» Я be
defined by α'ί = Ь*, i = 0,1, 2,. ..,11. Then в is a homomorphism of G onto Я. (Take this as
fact.) Find all the subgroups of Я and all their preimages. Check that the assertions of Theorem
4.19 and Corollary 4.20 hold.
Solution:
The subgroups of Я are Я, = Я, Я2 = {1,62}, K3 = {1}. G, = the preimage of Κλ =
{χ I xe S ЖГ} = G. G2 = the preimage of K2 — {x | se« = 1 or x9 — W\. Clearly 1 S G2.
ав = Ъ Φ 62, hence α Й G2. α2* = 62, so α2 £ G%. Continuing in this fashion we conclude
that G2 = {1, a2, a\ a9, a8, a10}. Finally, G3 = the preimage of K3 = {ж | ж» = 1} = {1,α4,α8}.
(α) Glf G2 and G3 are all subgroups of G containing Ker 9 = G3.
(b) Kj, Я2 and Я3 are normal in K, and GirG2 and G3 are normal in G (trivially, as G is abelian).
(c) The subgroups of G containing the kernel of в are those containing a4. Hence the subgroups of
G containing Ker ί are Gu G2 and G3. Note Gfi = Gfi implies i = j (1 — i, j — 3).
(d) Я, is of index 1 in K. Gt is of index 1 in G. Я2 is of index 2 in Я, its cosets being K2 = {1, 62}
and K2b = {b,i>3}. G2 is of index 2 in G, its cosets being G2 = {1, α2, α4, ββ, a8, a10}, and
Gtfi = {a, a8, a5, a7, a9, a11}. Я3 is of index 4 in K. G3 is of index 4 in G, the distinct cosets
being G3 = {1, a4, a8}, G3a = {a, as, a»}, G3c»3 = {a*,aB, a"}, G,^ = {a», a7, a"}.
Thus (a)t (Ь), (с) and (d) agree with Theorem 4.19 and Corollary 4.20.

Sec. 4.4]
HOMOMORPHISM THEOREMS
123
4.85. Let в : G -» Я be an onto homomorphism. Let К be cyclic of order 10. Prove that G has normal
subgroups of index 2 and 5 and 10.
Solution:
Let К = gp(k). Then the subgroups Ky = {1}, K2 = gp(ks), K3 = gp(k2) are normal subgroups
of К of index 10, 5 and 2 respectively. Consequently their preimages, by Theorem 4.19 and
Corollary 4.20, are normal of index 10, 5 and 2 respectively. Hence the result.
4.86. Let N < G and suppose G/N is cyclic of order 6. Let GIN — gp{Nx). Find all the subgroups
of G/N and express them in the form of Corollary 4.21. (Hard.)
Solution:
Let G/N = K. Let Κλ = {Ν}, K2 = {N,Nxs}, Ks = {Ν, Νχ2,Νχ*} and Я4 = Я. These are
all the subgroups of K. To find the corresponding subgroups of Corollary 4.21 we let ν: G -» G/N
be the natural homomorphism, i.e. gv = Ng. Then let Gs be the preimage of Klt I = 1,2,3,4.
<?! = {fir I gv = 2V> = {g\ Ng^N} ~ {g\ g e N} = 2V
G2 = {fif | fif» = ЛГ or flf = Ate3} = {fir | Ng = N от Ng = Nx*} = NuNx*
G3 = {fir | fifi- = N or fife = Me2 or fifp = 2W1} = ΝυΝχ2υΝχ*
G4 = {fir! fi> £ Я} = G. Then G{/2V = Я< for i = 1,2, 3,4.
4.87. Use the correspondence theorem to prove that if Η is a subgroup of G containing G' (the derived
group of G), then Η <1 G (i.e. prove Problem 4.69 by another method).
Solution:
Let ρ : G -» G/G' be the natural homomorphism; then с is onto. GIG' is abelian by Problem 4.68.
Any subgroup of GIG' is therefore normal, and thus Hv = S, say, is normal in GIG'. By the
correspondence theorem, Η is the preimage of S. Hence using the correspondence theorem once
more, Η is normal in G.
4.88. Let Η be a subgroup of index η in G. Let e: G-* К be a homomorphism onto K. Prove that Ηθ is
of index re in К if Η Э Ker в.
Solution:
It is only necessary to prove that S = He is of finite index in K, for then the result follows
from Corollary 4.20. If fffif,, ..., Hfif„ are the cosets of Η in G, then we claim that {S^e), .. .,S(fir„ff)}
is the set of all the cosets of S in K. We need only show that if kGK, then к fe Sig^) for some
i = 1, .. .,re. As ί is onto, there is a fif e G such that ge = fc. Let tf = fefir{. Then к = дв —
Ь,в(д{в) е S(fir{i). The result now follows from Corollary 4.20.
Alternatively we can show that S'(fl'iff), .. .,S(gnS) are all distinct. Suppose 5(flf(i) = S(fifji).
Then (0ii)(0y)-1 = (fiTji/j-1)» ε S. Hence flf^-1 £H as if, by the correspondence theorem, is the
preimage of S. Accordingly i = j and the index of S in Я is re.
4.89. Let G he a group and let N be a normal subgroup of G, Suppose further that L and Μ are
subgroups of G/N. Then show that we can write L in the form H/N, and Μ in the form K/N, where
Η and К are subgroups of G containing N. Show also that if LcM, HqK; and if L < Μ, Η < К.
Show that if LcM and [M; L] = re < «, then [K : H] = n.
Solution:
This is just an application of Corollary 4.21 to Corollary 4.20 and Theorem 4.19. That L = H/N
and Μ = K/N follows from Corollary 4.21. If ν is the natural homomorphism, we recall that
Η — {fir | gvSL} and Я = {fir| gv e M}. Hence if LcM, Η cK follows immediately. Now if
L < M, we consider the homomorphism 9 : К -> KfN denned by кв — kv for all к е К, i.e.
9 = Ρ|κ· Clearly Кв = Я/iV and the preimage of L is H, the preimage of Μ is if. We can then
conclude from the correspondence theorem that Η <Χ Κ.
4.90. Let G = D4. Let Μ = {ou σ3, τσ3, г}, N = {ог, σ3}. Then accept N < G and Μ < G. Consider
G/N and M/N. Find (G/N)/(MIN) explicitly and check that it is isomorphic to G/M. In other
words, check agreement with the factor of a factor theorem. (Use table on page 77.)
Solution:
G/N consists of the cosets Ax — N — {oltaa}, A2 = σ2Ν = {σ2,σ^, A3 = TN — {τ, τσΆ},
Α4 = (τσ2)Ν = {τα2, τσ4}- Now Μ = {<*ι,σ3, τσ3, τ}, hence M/N consists of the elements AVA3.

124
ISOMORPHISM THEOREMS
[CHAP. 4
M/N <3 G/N. Therefore we can talk of (G/N)/(M/N). The elements of this group are the cosets of
(M/N) in (G/N). These cosets are Вг = {MIN)Al = {Au A3} and B2 = (M/N)A2 = {ΑλΑ2,A3A2} =
{A2, A4}. (A3A2, for example, is calculated as follows: A3A2 = (rN)(a2N) — (τσ2)Ν — A4.)
Multiplication in the group (G/N)/(M/N) is calculated in the usual way for cosets, e.g.,
B2B2 = (M/N)A2(M/N)AZ = (M/N)(A2A2) = <flt/N)Ax = Вг
as A2A2 = (σ2Ν)(σ2Ν) = σ3Ν = N =AV It is clear then that (G/M)/(M/N) = {S1; S2} is the cyclic
group of order 2 generated by B.2.
Now let us decide what G/M is. The elements of G/M are the cosets Cj = Μ = {σ3, σ3, tw3, γ}
and C2-Ma2 = {a2,ai,TailTa2}. As C2C2 = Λίσ2Μσ2 = Μσ3 = Μ, G/M — {Cl,C2} is the cyclic
group of order 2. Therefore G/M = (G/N)/(M/N).
4.91. Let G be the cyclic group of order 12, say G = gp(a). Let Μ = др(аг), Ν = flfp(a6). Consider G/N
and M/iV. Find (G/N)/(M/N) explicitly and check that it is isomorphic to G/M. In other words,
check agreement with the factor of a factor theorem. (Difficult.)
Solution:
G/N consists of the cosets ΑΛ = Ν = {Ι,α6}, Α2 = Να — {α,α1}, Α3 = Να2 = {α2,α*}, Α4 =
Να? = {α3, α9}, Α5 = Να* = {α4, α30}, Αβ = Να5 - {а$,аЩ.
Now Μ = {1, α2, α*, α6, α8, α10}. Hence M/2V consists of the elements А 1?А3 and A5. M/N <3 G/2V,
and we can talk of (G/N)/(M/N). The elements of this group are the cosets of M/N in G/N. These
COSGtS fl!FG
S, = (ΜΙΝ)ΑΛ = {AVAS,A5} and B2 = (M/2V)A2 = {A2,A4,A6}
Multiplication is calculated in the usual way,
B2B2 = (Μ/Ν)Α2(Μ/Ν)Α2 = (M/Ji)(AjA2)
Now the product of A2 and A2 is also the product of cosets. As A2 = ЛГа, А2-^2 = Ν0,2 = ^з· Hence
В,>В2 = (М/ЛГ)А3 = S3. It is clear then that (G/N)/(M/N) = {SI( S2} = др(Вг) is a cyclic group of
order 2.
Now let us decide what G/M is. The elements of G/M are the cosets
Ct = Μ = {1, α2, α4, ae, a», a10} and C2 = Μα = {α, as, α·>, α1, α9, в11}
Clearly G/M = gp(C2) is cyclic of order 2. Hence G/M s (G/M)/(M/N).
4.92. Let G be any group. Let ^(G) be all possible nonisomorphic factor groups of G. Let Fi+1(G) =
{AH possible factor groups of the groups in Fj(G)}. In the particular case that G is cyclic of
CO
order 2s, find all nonisomorphic groups in U F^G), i.e. Fi(G)i)F2(G)O · · · . (Hard.)
Solution:
It is sufficient to consider FX(G). For if L <= F2(G), L = M/N where Μ & F^G). But then
Μ — G/K, by definition of Ft(G). Thus L is a factor of a factor group. Hence by the factor of a
factor theorem it is isomorphic to a factor group of G.
We must therefore find the number of factor groups of G. All subgroups of G are known. G
has unique subgroups of orders 1, 2, 22, 23, 24 and 25 by Theorem 4.9, page 105. The factor groups
will therefore be of orders 25,24,23,22, 2 and 1. In each case the factor groups will be cyclic. For
if G = gp(a) and N is a subgroup of G, gp(Na) — G/N. Hence the result.
c. The subgroup isomorphism theorem
In the homomorphism theorem we were able to say that the image of a homomorphism
9:G^ К was essentially a factor group of G. What can we say about the effect of θ on
Subgroups? Let Я be a subgroup of G. Let 0i = θίΗ, i.e. 0t is the mapping of Η to if
defined by hei = he, h£H. Then 0i is a homomorphism of H~*K, and so Ηθι — Ηθ^
Я/(Кег0!). Now if Ker0 = N = {χ \ χ G G, χθ = 1}, then Kerfli = {ж|жеЯ and
χθΛ-χθ = 1} = HniV. So H0 =H0t^' H/{HC\N). On the other hand, we know H0 is
a subset of G0 and G0 = G/N.
Our question is: what has H/(HC\N) got to do with GIN! It must be isomorphic to
some subgroup of GIN. But which? This is what the subgroup isomorphism theorem is
about.

Sec. 4.4]
HOMOMORPHISM THEOREMS
125
Theorem 4.23 (Subgroup Isomorphism Theorem, also called the Second Isomorphism
Theorem): Let N <3 G and let Я be a subgroup of G. Then HnN < H,
HN is a subgroup of G, and
Я7{HnN) « ΗΝ/Ν
{ΗΝ = {hn\h<EH, wGN})
Proof: If n^HnN and й G Я, then h'^nh^N as N < G, and h^nh^H as
wG#. Therefore hr^nhe. HnN and HnN < Я.
ffiV is a subgroup, for it is not empty; and if Χι, Жг G HN, then «i = feiiti, #2 = Ыгь and
aria:^"1 = hinin^1^1 — Ь,\ПгЪ,%х = hih^^hznsh^1 = fin*
where w3 = WiW^"1eiV, hihi1=h^H and Ыпгкъ^^п* GiV as N <3 G. Hence xix^1 G HN
and #2V is a subgroup of G.
Let φ:Η-*ΗΝ/Ν be defined by U0 = 2Vu. Then ψ is clearly onto, i.e. Я<£ = ΗΝ/Ν. Also
ψ is a homomorphism: {Ык^ф = N{hikz) = NhiNhs — ЫфЫф. By the homomorphism
theorem, Нф и Я/(Кег ψ). Ker ψ = {ζ | ж е Я, «ψ = 1} = {χ | ж G Я, 2Уж = Ν}. If Nx = N,
then 1ж = ж G 2V; and if χ е Ν, Νχ - N. Therefore Ker,!. = {ж j χ G Я, ж G Ν] = HnN.
Hence HN/N^H/{HnN).
Problems
4.93. Let Q* be the multiplicative group of rationale. Let N — {1, —1}. Let Я be the subgroup generated
by {-J}. Find ΗΝ, HN/N and thereby verify the assertion of the subgroup isomorphism theorem
that HN/N^H/HnN.
Solution:
The elements of Я are all of the form (£)r, r various integers.
HN = {x\ x = hn, h<=H, n£ Ν} = {χ \ χ = h or χ = -h, h G Щ
— {x | χ = —(^)r for all integers r}
A coset of HN/N is of the form Nx = {1, —1}ж = {χ, —χ} where χ ε ΗΝ. Now if χ e HN,
χ = +φ<·. Hence each coset is of the form {($)', -φΓ>· Since JV(£) · JV(J) лГ(£) = #(£)<■ and
(£)' e #(£)»·, each coset of HN/N is a power of ЛГф. Thus *p( {#(£)}) = HN/N; and since
(|)Γ (Ε Ν for r ^ 0, HN/N is the infinite cyclic group.
Now HnN = {a | α = (£)r for some r and ж = ±1} = {1}. Hence H/(Hr\N) a Я. But Я is
infinite cyclic. Thus we have verified that H/(HnN) & HN/N.
4.94. Let σ = ( *"" ) and Η = gp({a}). Prove that HAn/An is cyclic of order 2.
\2 1 3 ... n/
Solution:
HAn/An^ H/(HnAn). Now σ is an odd permutation, hence a g An. Also Η = {a, t}, so
ЯПAn = {(}. Therefore ЯАП/АП = Я/{«} s Я. But Я is cyclic of order 2. Thus HAJAn is cyclic
of order 2.
4.95. Let a group G contain two normal subgroups Μ and N. Let Я be a subgroup of G. Prove that
HM/M ^ HN/N if Hr\M = HnN.
Solution:
By the subgroup isomorphism theorem, HM/M = Н/НпМ = H/HnN & HN/N. Hence
HN/N ξ· HM/M by Problem 4.8, page 97.
4.96. If in the preceding problem we know that G/M has every element of order a power of 2, show that
H/(HnN) has every element a power of 2.
Solution:
H/(Hr\N)=H/(HnM)= HM/M QG/M. Hence the result

126
ISOMORPHISM THEOREMS
[CHAP. 4
4.97. Let Η and К be subgroups of G, Ν О G and HN = KN. Prove that H/(Hr\N) = K/(KnN).
Solution:
H/(Hr\N) ss #2V/2V = КАГ/JV - K/{KnN). Then by Problem 4.8, page 97, H/(HnN) a K/(KnN).
4.98. Let G3G,2{1} and Gt О G. Suppose G/Gt and Gt are abelian and Я is any subgroup of G.
Prove that there exists a subgroup #t of Η such that #t О Я, and Я/Я, and Hl are abelian.
(Hard.)
Solution :
Let Ht= HnGi· Then by the subgroup isomorphism theorem, Нг<ЗН, and H/Ht ss EGJGi.
But HGJGi ς G/Gt and G/Gt is abelian. Therefore Я/Яг is abelian. As #t С Gt and Gt is abelian,
we conclude that Яг is abelian.
4.99. Let G3Gt2 G23 {!}■ Let Gt О G, G2 <1 Gt, and suppose G/Gt, Gj^ and G2 are abelian. Prove
that if Я is any subgroup, then it has subgroups Hx and H2 such that Яг <1 H, Я2 О Ht and
Я/Яъ Ht/H2 and Я2 are abelian. (Hard.)
Solution:
Let Ht — HnGi. Then, as in Problem 4.95, Я, <1 Я and Я/#г is abelian.
Now consider #j as a subgroup of Gj. As G2 О Gi, by the subgroup isomorphism theorem
Hir\Gi< Hj and Hi/{HlnG2) ~ fftG2/G2С Gt/G2. Since Gt/G2 is abelian, so is £?,/(£?,ηG2).
Consequently we put Я2 = H-^riG^.
Finally as Я2 £ G2 and Gz is abelian, so is Я2 and the result follows.
d. Homomorphisms of cyclic groups
We return now to study cyclic groups. In Section 4.2b we could state that in a finite
cyclic group there was at most one subgroup of any given order. An analogy for the infinite
cyclic group would have been awkward to formulate without the concept of index which we
have had at our disposal since Section 4.3a.
Recall that a subgroup Я of a group G is of index η in G if there are exactly η distinct
right cosets of Я in G. In the case of finite cyclic groups we have proved that there is one
and only one subgroup Я of any order m dividing |G|. Since [G- H] = \G\lm, there is one
and only one subgroup of any given index dividing the order of G. This gives the clue to
Theorem 4.24: There is one and only one subgroup of any given finite index η > 0 in
the infinite cyclic group.
Proof: Let G = gp{x) where G is infinite cyclic. Let H„ = gp{xn). Then Я„ = {xnr \
all integers r). Hence the cosets Hn,Hnx,. . .,Hnxn~l are all distinct and are all the cosets
of Hn in G. Hence Hn is of index n.
Next if Я is a subgroup of index n, we already know that Я is generated by the smallest
positive power xT G Я (Theorem 4.9, page 105), and this means Η — Яг. But Яг is of index
r. Hence r = n and HT = Hn. This concludes the proof.
With our knowledge of cyclic groups it is easy to apply the homomorphism theorem to
find all homomorphisms of cyclic groups.
Theorem 4.25: Let θ be a homomorphism of a cyclic group G. Then Ge is cyclic; and
if \G\ < <*, \Ge\ divides \G\.
Furthermore if Я is any cyclic group such that \H\ divides \G\, there
is a homomorphism of G onto Я.
If G is infinite cyclic, there is a homomorphism of G onto any cyclic
group.
Proof: If G - gp{x), then G~{xr\ all integers r}, and
Ge = {xr6 | all r) = {{xe)r \ all r) = др(хв)

CHAP. 4]
SUPPLEMENTARY PROBLEMS
127
Thus Ge is cyclic. If \G\ < °°, then by the homomorphism theorem Ge « GIN for some
normal subgroup N of G. Hence \Ge\ = \G\/\N\ (see Lagrange's theorem) and \Ge\ divides |G|.
Suppose Я is cyclic, Η = gp(y), and the order m of Я divides the order of G = gp{x).
Say, \G\ - rm. Let N = gp{xm). Then N = {l,xm, .. .,{xm)r~1}, \N\ - r, and N is of index
m. Because G is abelian, GIN makes sense and is of order m. But G is cyclic and
consequently so is GIN. Hence GIN « H, and Я is a homomorphic image of G. If G is infinite
cyclic, then, as we saw in the proof of Theorem 4.24, [G: Gn] = n if Gn = gp{xn). As
Gn < G, GIGn is of order n. But G/G„ is the homomorphic image of a cyclic group; hence
it is cyclic. Thus G has as homomorphic image any cyclic group of order η, η > 0.
Obviously it also has the infinite cyclic group as a homomorphic image.
A look back at Chapter 4
In this chapter we have thoroughly investigated the simplest class of groups, the cyclic
groups. We know that there are cyclic groups of all orders, we know their subgroups, we
know that they have as homomorphic images only cyclic groups, and we know whether any
cyclic group G has as homomorphic image a given cyclic group. Furthermore the
subgroups of cyclic groups are again cyclic.
We have also introduced the concept of coset. The cosets form a partition of the group.
Using thia fact we obtained Lagrange's theorem which states that the order of a subgroup
divides the order of a finite group. This enables us to eliminate certain groups as possible
subgroups of a given group. We will see later on that it also enables us to find more quickly
the groups of a given order.
Next we have introduced the idea of a normal subgroup. This gives rise to a new way
of looking at homomorphisms, namely as factor groups (see the homomorphism theorem).
The subgroup isomorphism theorem tells us that the subgroup corresponding to a given
subgroup Я of G in a factor group GIN is isomorphic to a factor group of Я itself, namely
ΗΙ(ΗΠΝ).
The factor of a factor theorem tells us that a factor group of a factor group GIN is just
a factor group of G of the form G/M. Finally, the correspondence theorem associates with
each subgroup of the image of a homomorphism Θ: G-* К a unique subgroup of G itself.
Supplementary Problems
FUNDAMENTALS
4.100. Prove that if α and b are elements of a group G and if а~1Ьга = bar then b = a.
4.101. Suppose α and b are elements of a group G. If a2 = 1 and a~ 1Ьга = 6s, prove bs = 1. (Hard.)
4.102. Suppose α and b are elements of a group G. If α~ιί>2» = b3 and Ь~%2Ь = α3, prove α = 1 = b.
(Very hard.)
4.103. Suppose G and Η are groups. Suppose that G cannot be generated by two elements but that Η can.
Prove G and Η are not isomorphic.
4.104. Let X be a non-empty set and let Υ = {у} be disjoint from X. Prove Sx = SXuy if and only if
X is infinite.

128
ISOMORPHISM THEOREMS
[CHAP. 4
CYCLIC GROUPS
4.105. Let G be a cyclic group. Prove that if N is a subgroup of G such that G/N = G, then N = {1}.
4.106. Let (J = ZxZ where Ζ is the set of integers. Define a binary operation ° in G by
(fc, l)°(m,n) = (fc + m, l+n)
where (fc, I), (m,n) ε G. G is a group with respect to this composition. Prove that G is not cyclic.
4Д07. Let Gv Gz, ... be subgroups of a group G. If G,CGi + i, Gj^Gi+j for i=l,2,,,., prove that
GtUGiU ■ ■ · is not a cyclic group. (Hard.)
4.108. Let G be a group and let θ: G -* F Ъе a homomorphiam. Lat С be a cyclic subgroup of G. Let
ce e С for all с e C. If Η is any subgroup of C, prove that he e Η for all he H.
4.109. Let Q be the additive group of rationale with respect to addition. Prove that every two-generator
subgroup (¥· 0) of Q is infinite cyclic.
COSETS
4.110. Let Η be a subgroup of G. Prove that β : Hg -» g-^H is a matching of the right cosets of
Η in G with the left cosets of Η in G.
4.111. Let Η and К be subgroups of a group G. Show that a coset of Η intersection a coset of К is a coset
offlflK.
4.112. Let D be the group of Problem 3.72, page 91. Let N = {(0, ζ) \ ζ e Z}. Prove that N <i D and
D/JV is infinite cyclic.
4.113. Let G be the group of Problem 3.74, page 91. Let N = {(0,g) | q € Q}. Prove that N <t G and
G/W is infinite cyclic. Show that JV is isomorphic with the additive group of rationale.
4.114. Let W be the group of Problem 3.77, page 91. Let Μ = {(0, Ъ) | (О, b) e TF}. Show that Μ <3 G
and that G/M is infinite cyclic.
4.115. Let G be a group, let Η be a subgroup of G, and let g be an element of G. Prove that if N(H) is the
normalizer of Я and N(g^lHg) the normalizer of g~lHg, then g~lN(H)g = N(g~lHg). g~lHg —
{g-ihg | Λ ε Η).
HOMOMORPHISM THEOREMS
4.116. Let <3 = < ( J a, b, e,d e Ζ >. Prove that Q forms a group with respect to the operation +
defined by
/a b\ /at ЬА _ /a + al Ь + ЬЛ
\c d) \Cl dj \c+ci d + dj
/a b\
Let β : g -*■ Ζ be defined by I , J β = a + d. Prove that θ is a homomorphism of Q onto
the additive group of integers and find its kernel. Consider ^/(Ker β) and prove that in accordance
with the homomorphism theorem it is isomorphic with the additive group of integers.
4.117. Let Q = ■{ (a Ъ
vc d
ad —be Φ 0, α, Ь, с, d real numbers V with operation matrix multiplication. Let
$: g -» R*, the nonzero real numbers, be defined by ( ) в — ad — be. Prove that β is a homo-
morphism from (7 onto the multiplicative group of nonzero real numbers and find its kernel. Prove
that д/(Кет θ) ss R* in accordance with the homomorphism theorem.

CHAP. 4]
SUPPLEMENTARY PROBLEMS
129
4.Щ. Let G be any subgroup of Sn, the symmetric group of degree n. Let θ : G -> {1, —1} be the mapping
defined by χθ == 1 if a; is an even permutation and χθ = — 1 if к is an odd permutation. Prove that
θ is a homomorphism of G into the group {1, —1} with operation multiplication of integers. Using
the homomorphism theorem, prove that the even permutations of G form a normal subgroup of G.
4.119. Let G be a group and N a normal subgroup of G. Suppose that Η = GIN has a sequence of
subgroups H = HjDHjrD ··· dH„ where [ff4: ffi+1] = i+1, for i — 1,2,.. .,n— 1. Prove that
G has a sequence of subgroups G= G1'dG2O "· 2 Gn such that [<%: Gi + 1] = *+ 1, i = 1, .. .,« — 1.
4.120. Let A/ and TV be normal subgroups of G with ΜώΝ. Prove that G/N is finite if G/M and A//7V are
finite.
4.121. Let G be a group and TV a normal subgroup of G. Suppose G/N has a factor group which is infinite
cyclic. Prove that G has a normal subgroup of index η for each positive integer n.
4.122. Let G be a finite group with normal subgroups Μ and N. Let Η be a subgroup of G. Suppose that
the orders of Μ and Η and those of N and Η are co-prime. Prove that HMIM as HN/N.
4.123. Let TV, A/ be normal subgroups of G, ΝώΜ. Suppose GIN is cyclic and \N/M\ =2. Prove that
GlM is abelian.
4.124. Find a group G with normal subgroups N and Af, NoM, GIN cyclic, NIM cyclic but GlM not
abelian.

Chapter 5
Finite Groups
Preview of Chapter 5
The most important result of this chapter is a theorem of Sylow which guarantees the
existence of subgroups of prime power order. We prove two other theorems of Sylow
concerning subgroups of prime power order and then examine groups of prime power order.
One result is that groups of prime power order always have non-trivial centers.
In order to construct a new group from any two groups G and Я, we define a binary
operation on the cartesian product of G and Я, The resultant group is called the direct
product of G and Я. A simple condition enables us to conclude that a group is a direct
product.
The concept of direct product together with general theorems about subgroups, e.g. the
Sylow theorems, help us to classify finite groups. In this chapter we find all groups up to
order 15.
We study a class of groups called solvable groups. Solvable groups are used in Galois
theory to determine whether an equation is solvable in terms of fith roots.
An ambitious plan for studying finite groups is to find all simple groups, i.e. groups
without proper normal subgroups, and then see how groups are built from simple groups.
The Jordan-Holder theorem shows that in a sense a group is built from simple groups in
only one way. As yet the task of finding all simple groups is far from complete. We
conclude the chapter by exhibiting a class of simple groups, namely An, for η ^ 5.
5.1 THE SYLOW THEOREMS
a. Statements of the Sylow Theorems
Lagrange's theorem (Theorem 4.11, page 109) tells us that the order of a subgroup
divides the order of a finite group. Conversely one might ask: if G is a finite group and
η \ \G\, is there always a subgroup of order til
The answer to this question is no: A± is of order 12 but has no subgroup of order 6 (see
Problem 5.1 below). The following important theorem, however, ensures the existence of
subgroups of prime power order. In the following ρ will denote a prime.
Theorem 5.1 (First Sylow Theorem):
Let G be a finite group, ρ a prime, and pr the highest power of φ dividing the
order of G. Then there is a subgroup of G of order pr.
Suppose Я is a subgroup of G of order a power of a prime φ, and \H\ is the highest power
of ρ that divides \G\. Then Я is called a Sylow p-subgroup of G. By Theorem 5.1 every
finite group has a Sylow p-subgroup.
In general a group of order a power of the prime ρ is called a p-group. A Sylow
p-subgroup Η of a group G is a maximal p-group in G, i.e. if HqF <zG where F is a p-group,
then F = Я (see Problem 5.4).
130

Sec. 5.1]
THE SYLOW THEOREMS
131
As an illustration we find the Sylow p-subgroups of the symmetric group Sa on {1,2,3},
for ρ = 2 and 3. The elements of Ss given in Section 3.3a, page 57, are
1 2 3\ /1 2 3\ /12 3
12 3/ °2 ~ [S 1 2/ ч ~ \3 2 1
1 2 3\ _ /1 2 3\ _ /1 2 3
2 3 l) Tl ~ [l 3 2/ Тз " \2 1 3
The order of any Sylow 2-subgroup is 2 and the order of any Sylow 3-subgroup is 3, since
|S3| —6 = 2-3. Now t\— t\— tI = (, so the sets {ι,τ,}, {<,т2} and {<,т3} are all subgroups
of order 2 and therefore, by definition, Sylow 2-subgroups of Sr There are no other Sylow
2-subgroups of S3 because σ2 = σ2, σ2 = στ implies Ss has no other elements of order 2.
{(, σν σ2} is the only subgroup of order three in Sa, so it is the only Sylow 3-subgroup of S3.
Theorem 5.2 (Second Sylow Theorem):
If Η is a subgroup of a finite group G and Η is a p-group, then Η is contained
in a Sylow p-subgroup of G.
Two subgroups S and Γ of a group G are called conjugate if there is a g &G such that
g~lSg = T. Recall g~lSg = {g-*sg \ s £ S).
Theorem 5.3 (Third Sylow Theorem):
Any two Sylow p-subgroups of a finite group G are conjugate. The number
sp of distinct Sylow p-subgroups of G is congruent to 1 modulo ρ and sv
divides |G|. (sv is congruent to 1 modulo ρ if sp = 1 + kp for some integer k.)
Before proving the Sylow theorems, we will use them to show that, up to isomorphism,
there is one and only one group of order 15. If \G\ = 15 then, by Theorem 5.1, G has at
least one subgroup of order 3 and at least one of order 5. Now Theorem 5.3 implies that
there are s3 = 1 + 3& subgroups of order 3 and s3\ \G\. But (1 +3fc) | 15 implies к = 0.
Therefore G has one and only one subgroup of order 3. Similarly G has one and only one
subgroup of order 5. These subgroups must be cyclic (Problem 4.48, page 110). Let
Hi = {1, a, a2} be the subgroup of order 3 and Я2 = {1, b, b2, b3, b4} the subgroup of order 5.
Я1ПЯ2 = {1}, because an element ¥-1 cannot have order 3 and 5 simultaneously. We look
at the order of ab in G which must be either 1, 3, 5 or 15. If the order of ab is 1, then
ab = 1 and a= b_1 which is impossible, for ΗιΠΗ2= {1}. If the order of ab is 3, then
gp(ab) = Hi, since Hi is unique. In this case ab = а1 (г = 0, 1 or 2) and b = ai_1 which is
impossible. If the order of ab is 5, gp(ab) = Я2. Hence ab = bl (i = 0,1, 2, 3 or 4) and
α = bl~l which is impossible. Therefore the order of ab is 15 and G is the cyclic group of
order 15 generated by ab.
Further applications of the Sylow theorems are given in the problems below and in
Section 5.3.
Problems
5.1. Show that the alternating group Л 4 has no subgroup of order 6.
Solution:
The elements of Л4 were given in Section 3.3c, page 62. We repeat them here for convenience:
Ί 2 3 4\ /1 2 3 4\ /1 2 3 4\ /12 3 4
1 2 3 4/ "ά \B 2 4 1/ Г6 " ^4 1 3 2/ "' \3 4 1 2
1 2 3 4\ /1 2 3 4\ /1 2 3 4\ /12 3 4
13 4 2/ V4 2 1 3/ V2 3 1 4/ \4 3 2 1
1 2 3 4\ /1 2 3 4\ /1 2 3 4\ _ /1 2 3 4
T2 = I 1 4 2 3/ rs = I 2 4 3 li Г8 = U 1 2 J "« ' \2 1 4 3,

132
FINITE GROUPS
[CHAP. 5
Suppose that A4 has a subgroup Η of order 6. σ2 = σ5 = σ8 = h a» σ2> σ5 and σ8 are of order 2.
Also τ^ is of order 3 for j = 1,2, .. ., 8. Hence the elements of order 2 are σ2, σ5 and σ8, and the
elements of order 3 are τν τ2, ..., τ8. Now H is of order 6, so it contains a subgroup of order 3,
by Theorem 5.1. Therefore т^Н for some i, say τλ £ Η; then t, τν τλ — τ2 £ Η- Η must also
contain an element of order 2, by Theorem 5.1. Hence Η contains a σ(, say Η contains σ2. Because
σ2τι = T4 and TjCTa = гц, if Η contains σ2 it also contains т4, т^ = т$, та and r8 = т7. This would
mean Η has at least 8 distinct elements, which contradicts the assumption that \H] — 6. Thus
σ2 g H. A similar argument shows σ5 and σ8 £ H. This means that Η does not contain subgroups
of order 2, contradicting Theorem 5.1. Therefore our initial assumption is invalid and A4 does not
contain a subgroup of order 6.
5.2. Find all Sylow p-subgroups of A4 for ρ -= 2 and 3.
Solution:
The elements of A4 are given in Problem 5.1. The order of a Sylow 2-subgroup is 4, since
22 is the highest power of two dividing 12, the order of Av Consequently by Lagrange's theorem
none of the t's can be elements of a Sylow 2-subgroup because they are all of order 3 (see Problem 5.1)
and 3 does not divide 4. Now σ,-σ^ = ak where i, /, к £ {2, 5, 8} and σ? = ι for г = 2, 5, 8. Hence
Ρ — {ι, σ2, σ5, σ8} is a subgroup of A 4 of order 4. Ρ is the only possible Sylow 2-subgroup as there
are pnly four elements having order dividing 4, viz. ι, σ2, σ5, σ8, and these elements are in P. The
order of a Sylow 3-subgroup is 3. The sets {i,tut\}, {ι, r3) τ|}, {ι, τ5, τ\] and {ι, rT, r2} are all
subgroups of order 3. These are all the possible Sylow 3-subgroups, as they include all the elements of
order 3.
Alternately we may use Theorem 5.3: s3 = 1 + 3k must divide 12. Clearly к Ф 0 (we already
have four subgroups); and if к > 1, s3 does not divide 12. Hence к = 1 and there are exactly four
Sylow 3-subgroups.
5.3. If Η is a subset of a group G and g € (r, then Ι^-^Η^Ι = |ff|, where g~lHg = {g~lhg \ h e H}.
Solution:
We define a matching a: H -* g~lHg by а:к-*д~гкд for hSH. α is clearly an onto
mapping. To show a is also one-to-one, we must prove кг-= h2 (hlth2£H) if and only if g~1hlg =
g~1h2g. Let Тц = ft2. Then by multiplying on the left by g~* and on the right by g we get
Sf~1'lii7 = ί?-1/^- Similarly g~1hlg = g~lh2g implies /ц = Ag· Henee α is a matching and
\д~Шд\ = \H\.
5.4. Let }G\ == prm (r — 1 and p/ m) and let Ρ be a Sylow p-subgroup of G. Prove that if Η is a
p-group such that PcHcG, then Η = P.
Solution:
Suppose |H| == pt, t — 0. By Lagrange's theorem, ρ'| prm. Since p/m, t—r. But Ρζ,Η
and ]P| = pr. Hence t = r and |P| == \H\, and so Ρ = H.
5.5. If Η is a Sylow p-subgroup of G, then g~lHg is also a Sylow p-subgroup of G.
Solution:
Suppose |G| = prm (r ^ 0 and ρ/«t); then |H| = pr. But |ί?_1Ηί?[ = |H| by Problem 5.3.
Hence g~~lHg is a Sylow p-subgroup of G if it is a subgroup. To prove g~lHg is a subgroup, observe
that (g~1hlg)(g~1h^)~1 = g~1hlh~1g e g~1Hg. From Lemma 3.1, page 55, д~гНд is therefore a
subgroup.
5.6. Prove that a finite group G is a p-group if and only if every element of G has order a power of p.
Solution:
If \G\ = pr then, as every element of G must have order dividing the order of the group,
every element has order a power of p. To prove the converse let every element of G have order a
power of ρ and assume the order of G is not a power of p. Then there is some prime q, q Φ ρ,
such that q | \G]. But by the first Sylow theorem, G has a subgroup Η of order a nonzero power of
q. So Η contains an element g Φ 1. By Lagrange's theorem, the order of g is a nonzero power of
q and hence the order of g is not a power of p. This Contradicts the assumption that all elements
have order a power of p. Hence |G| = pr for some r—Q.

Sec. 5.1]
THE SYLOW THEOREMS
133
5.7. If G has only one Sylow p-subgroup Я, then Η < G.
Solution:
If j£6, g~lHg is a Sylow p-subgroup by Problem 5.5. But G has only one Sylow
p-subgroup. Thus g~lHg = Я for all g €. G and Я < G.
5.8. If \G\ = pq, where ρ and g primes and ρ < q, then G has one and only one subgroup of order g.
Furthermore if q Φ 1 + kp for any integer it, then G is the cyclic group of order pq.
Solution:
By Theorem 5.3 G has sq = 1 + kq Sylow g-subgroups of order q, with к - 0. Also 1 + Jig
divides pg. There can only be four possibilities for 1 + kq as the expression of pq as a product of
primes is unique: 1 + kq = q or 1 + kq = ρ or 1 + kq = pq or 1 + kq = 1. As g does not divide
1 + kq, we are left with the possibilities that 1 + kq = ρ or 1 + kq = 1. Since q > p, 1 + kq Φ ρ
and hence к = 0. Thus there is only one subgroup of order q, say Я.
There are &p = 1 + kp subgroups of order p. Again we have the possibilities 1 + kp = 1,
1 + kp = p, 1 + kp = q, or l + kp = pq as sp divides \G\. Clearly ρ does not divide ί + kp, so
1 + kp = 1 or 1 + fcjj = q. The last is not true by assumption, so again there is only one subgroup
К of order p.
It follows from Problem 5.7 that Η < G,K < G. Also ЯП if = {1} as the nonunit elements of
Я are of order q and those of Я are of order p. If ft £ Я and fc € Я (ft ¥· 1, fc =s* 1), then
ft-ifc-iftfc = ft-i(^"lftfc) e# as Η < G
= (h~lk-lh)k £ Я as Я < G
Hence ft^'fc^'ftfc = 1 and h and к commute. By Lagrange's theorem, the order of hk is p, q or pq.
But (hk)v = fepfcp as ft and A; commute, so (ftA;)» = № Φ 1. Similarly (ftu)4 = A;" ^ 1. Therefore
hk is of order pq, and so Cr is cyclic.
Instead of the argument of the last paragraph, we note that Я = {1,ft, ft2, ,hq~1}, К =
{1, к, к2, ..., к?*1}. Now hk is an element of order 1, p, q or pq. If ftfc is of order p, since there is
only one subgroup of order p, gp(hk) = K, i.e. hk = fe{ for some i, 0 — i — 30 — 1. But then
ЛЕЯ, which contradicts HnK = {1}. Similarly gp(hk) is not of order g, nor of order 1. Thus
gp(hk) is of order pq, and so G is cyclic.
5.9. Show that if q = 1 + kp in Problem 5.8, G is not necessarily cyclic.
Solution:
Consider S3, the symmetric group on {1, 2,3}. \S3\ =6 = 3*2, 3 = 1 + 1 · Ζ and S3 is not a
cyclic group.
5.10. If |Gr| =■ 2p, ρ an odd prime, then G has one and only one subgroup of order ρ and either G has
exactly ρ subgroups of order 2 or it has exactly one subgroup of order 2.
Solution:
From Problem 5.8 we know G has one and only one Sylow 30-subgroup. Because ρ is itself the
highest power of ρ dividing \G\, the Sylow 30-subgroup of G is of order p. Thus there is precisely
one subgroup of G of order p. The number of Sylow 2-subgroups of G is s2 = 1 + kl for some
integer k. Again 1 + 2fe = 1, 2, ρ or 2p. As 2 does not divide 1 + 2k, then l+2fc = l or l+2k = ρ
and the number of Sylow 2-subgroups is either 1 or p.
b. Two lemmas used in the proof of the Sylow theorems
Lemmas 5.4 and 5.6 will provide the tools for proving the Sylow theorems (see
Section 5.1c).
Throughout this section G will be a fixed group and Η a subgroup of G. As usual we
denote subsets of G by А, В, С, etc.
A generalization of the concept of normalizer as defined in Section 4.3d, page 112, will
be essential. We point out once more that if A is a non-empty subset of a group G and
g €Gt then g~lAg = {g"lag | α G A}.
Definition: Let A be a non-empty subset of a group G. The set {&| h~lAh = A, h G Щ
is called the normalizer of A in Η and is written NH(A).

134
FINITE GROUPS
[CHAP. 5
It is easy to prove that 2νΉ(Α) is a subgroup of Я (Problem 5.11). When Η - G, NG(A)
is the normalizer of A as defined in Chapter 4.
Definition; Let A and В be non-empty subsets of G. В is said to be an Η-conjugate of A
if h"%Ah = В for some h EH, (Note that if H~G, then A and В are
conjugate as defined in Section 5.1a.)
The next lemma gives us a formula for calculating the number of distinct subsets of G
which are Я-conjugates of A.
Lemma 5.4: If G is a finite group with subgroup Я and non-empty subset A, the number of
distinct Я-conjugates of A is the index of NH(A) in H, i.e. [H: NH(A)].
Proof: Since [Я; Nh(A)] is the number of distinct right cosets of Nh(A) in H, we need
only define a one-to-one mapping, <*, of the right cosets of 2Vh(A) in Я onto the distinct
Я-conjugates of A. Let a be defined by
a: NH{A)h^h~lAh {hEH)
To show that α is a one-to-one mapping, we must prove that for hi, hi G Я,
Nu(A)hi = Nn(A)hz if and only if fef'AAi = hilAh%
(i) Let h^Ah = h^Ah%. Then A = к,АГ'ШГ1= (ЛяЛГУ^АаАГ1). Hence йвЛГ1 ε
Nh(A) and so fee 2Vh(A)A,i. Since two right cosets are equal or disjoint, we conclude
2Vfi(A)fe, = NB(A)fb. Thus К1АКг = Л^'АЛа implies ΛΓΗ(Α)Λι = 2VH(A)fe.
(ii) If NH(A)hi = NH(A)ht, then £i G 2Vff(A)/i2, i.e. fei = nft2 for some и G 2VH(A).
h^Ahi = (nhzY^-Anht = fe^'w^Anfe = Ы1АЫ
because it~lAn = A by definition of Nn(A). Hence ΛΓΗ(Α)/&ι = NH{A)ht implies
h\lAh\ — h^Ahi. a is clearly onto, so the proof is complete.
Most of our arguments are concerned with sets whose elements are subsets of G. We
denote such sets by script letters cA, <B, etc.
For example, let G be the cyclic group of order 6, G = {Ι,α,.. .,a5}. Subsets of G are,
for example, A = {Ι,α}, Β = {α2,α3,αβ}, С = {α}. An example of a set whose elements are
subsets of G is the set whose elements are A and B. We write ?A = {A, B}. Another such
set would be <B = {A, B, C].
Proposition 5.5: Let cA be a set of subsets of G. We define for А, В G cAr A ~ В if i5 is
an Я-сопjugate of A (i.e. if there exists an element h Ε Η such that
h~lAh = B). Then ~ is an equivalence relation on cA (see Problem 5.16
for the proof).
We will make a few observations about ~, which follow because it is an equivalence
relation on cA. Recall that if A G oA, A ~ = {X | X G A and X ~ A}, i.e. A ~ is the
equivalence class containing A (see Section 1.2c, page 9). Recall that the distinct
equivalence classes are disjoint and that their union is aA (Theorem 1.2, page 10).
By a set of representatives of the equivalence classes we mean a set % which contains
one and only one element from each of the distinct equivalence classes. It follows that
cA is the disjoint union of the sets R~, R <E %. Hence \cA\ = 2) l^~|- We are now in
a position to prove our main lemma. RG%
Lemma 5.6: Let cA {¥* 0) be a set of subsets of G. Suppose that for each A G oA and
each h G Я, h"lAh G cA. Let ~ denote the equivalence relation defined by
A ~ В if β is an Я-conjugate of A. Let % be a set of representatives of the
equivalence classes. Then
\oA\ - Σ [H:Nh(R)]

Sec. 5.1]
THE SYLOW THEOREMS
135
Proof: We know from the remarks above that
И = Σ \R~\
л е<£
But R ~ = {X j X - h-^Rh for some h G Я} since h~lRh & cA for every h G Я. So
i?~ is the set of Я-conjugates of R. The number of such Η-conjugates is, by Lemma 5.4,
[H:NB{R)\. Hence |сЛ| - Σ \H:NH(R)), as claimed.
л е<£
Corollary 5.7: Let F (^ 0) be a subset of G. Let c/i = {fl-'Pflr | flr G G}. Let %, Я and ~
be as in Lemma 5.6. Then
\cA\ = Σ [H:NB{K)] = \G:No(P)}
Re%
Proof: Clearly \cA\ is the number of G-conjugates of P, and the result follows from
Lemma 5.4.
Corollary 5.8: Let cA — {A | A is a subset of G and A has precisely one element}. Let
~ be the equivalence relation in cA when Я = G, and let % be a set of
representatives of the equivalence classes. Let %* = {R \ RC\Z{G) = φ,
R G %}. Then
\G\ = \Z{G)\ + Σ [G:NG(R)]
(We remind the reader that Z{G) =■ {x\ xg — gx for all g G G})
Proof: Clearly \oA\ — \G\; hence
\G\ = Σ IG-.ЩЩ] {5.1)
л e <£
If 2 G Z(G), then {z} G <yi and the number of G-conjugates of {z} is one, namely {z} itself.
Consequently {z} G ^ for each 2 G Z(G). Note that NG{{z}) = G if ζ G Z(G), Hence
adding first the contribution made by all R G % with RnZ(G) ¥· 0 in (5J), we obtain
\Z(G)\ and the result follows.
Note that as R — {r},
Na(R) = {flr I g G G and fl-Vflr G R}
= {g\ g €G and g"lrg - r]
= C{R)
(For the definition of C(R), the centralizer of R in G, see Section 4.3d, page 112.)
Hence Corollary 5.8 takes the form
|G| = \Z(G)\+ Σ [G:C(R)\ (5.2)
R£%*
(5.2) is called the class equation of G.
Problems
5.11. If A is a non-empty subset and Η a subgroup of a group G, then NH{A) is a subgroup of G.
Solution:
NH(A) is clearly a subset of G. NB(A) Φ 0, since leff and l~M.l = A implies 1 e iV„(4).
Let η e iVB(A). n~1An = A implies nAn~l — A, or (п~1)~1Ап^1 = A. Furthermore n~l £ H,
since Я is a subgroup; hence r'eNH(4 If m, (ieiVH(A), (mw)~M.(mn) = n~1('m~1Am)n =
n~1An = A; hence mn € iVH(.4). Accordingly NH(A) is a subgroup of G.

136
FINITE GROUPS
[CHAP. 5
5.12. Check Lemma 5.4 by direct computation when G = S3 and (using the notation of Section 3.3a,
page 57) A = {tj}, Η = {(jt2}.
Solution:
The Я-conjugates of A are ι-1{τχ}' = {τι)> τ2~4τι}τ2 ~ {тз}- Thus the number of Я-conjugates
of A is 2. Lemma 5.4 requires that 2 = [Я: NH(A)]. But NH(A) — {x\ a G Я and x~lAx =
A) = {t}. Hence [H: NH(A)] = 2, as required.
5.13. Check Corollary 5.7 when G = D8, the dihedral group of degree 8 (G = {bf, abl, 0 — i < 8}, where
a,2 = b& — 1 and а~1Ъа — Ь"1; see page 75, with a, — τ, b — σ2), given Ρ = {a} and
Η = {l,b2,b*,b<>}.
Solution:
cA of Corollary 5.7 is given by cA = {fi^Pi? | s£G} = {Pj,Р2,Р3,PJ where Pi = {a},
P2 = {ab2}, P3 = {a.6*}, P4 = {ab6}; thus \oA\ = 4. Using the equivalence relation ~- of the
corollai-y,
Pt~ = {fc-ipxfc | h£H} = {ΡυΡ3}
P2~ = {fc-iPafc| h&H) = {Р2,Р^
For ^ choose one representative from each of the equivalence classes, e.g. choose ^ = {Ρ&,Ρζ}>
Then Lemma 5.6 claims that Kj = 4 = [H:NH(P3)] + [H: NH(P2)\. Now NH{P3) = {1,6*} =
JVH(P2). Hence [Я: 2VH(P3)] = [H; NH(P2)] = 2 and the required equation of Corollary 5.7 holds.
We must also show that [G; NG(P)\ = 4. As
NG{P) = {x | χ G G and x~lPx = P}
= {ts J χ G G and aj-1ow! = a}
= {1, а, Ъ\ α.ί>4}
the index of NG(P) in G is 4, the required number.
5.14. Check Corollary 5.8 when G = S3. Use the notation of page 57.
Solution:
cA (of Corollary 5.8) = {РъРя,Р3,Р^Р&,Рв} where Pt = «, Pz = {σι}, Ρ3 = {σ2}, Ρ4 = {r^,
^5 = {"""г), ^e ~ {тз)· Let ~ be as in Corollary 5.8. Consequently Px ~ = {Pj}, P2~ = {P2,P3},
P4 ~ = {P4, P5, Pel· To define %, we choose one element from each of these equivalence classes.
Let us take % = {Pi,Pz,P4}. As Z(G) = {.}, PjnZ(G) = 0 except for j - 1. Therefore
%* = {P2, P4}· Now ng(pz) = {',σι,σ2} and JVG(P4) = {ι,τχ}. Hence, as required,
\Z(G)\ + [G:NG(P2)] + [G:NG{PJ\ = 1 + 2+3 = 6 = \S3\
5.15. Show that NH(A) = NG(A)nH for any non-empty subset A and subgroup Я of a group G.
Solution:
Let rc£JVH(A); then η G Η and w-1Arc = A. But HqG, so that nGG and by definition
nGNG(A). Consequently NH(A) С NG(A)r\H. If nG2VG(A)nH, then rc-iArc = A and и G H.
Thus 2ν0(Α)ηΗςΛ/Ή(Α) and the equality follows.
5.16. Prove Proposition 5.5, page 134.
Solution:
As Я is a subgroup of G, Η contains the identity, so A = 1"!A1 and thus A ~ A. If A ~ Б,
then there is an element hGH such that h~1Ah = B. Consequently (Λ"1)-1^^"1) = A and
Б ~ A. Finally, if A~B and Б ~ C, then there exist h,gS.H such that h~-lAh = В and
sr-^sr = C. It follows that hg G Я and (M-1^(M = g~4h~lAh)g = g"lBg = C, and so
A ~ С Hence ~ is an equivalence relation on e/i.
5.17. Let А,В be subsets of G. Suppose Б is an Я-conjugate of A, where Я is a subgroup of G. Prove
that [H:NH{A)] = [H;NH{B)}.
Solution:
Let qA = {X\ λ' = g~1Ag or A' = g"lBg for some gGG}. We use Proposition 5.5.
A ~ = В ~, as Б is an Я-conjugate of A. \A ~\ is therefore the number of Я-conjugates of A,
and also the number of Я-conjugates of B. By Lemma 5.4, [Я; NH{A)] = [H: NH(B)].

See. 5.1]
THE SYLOW THEOREMS
137
c. Proofs of the Sylow theorems
First we prove a weak form of the first Sylow theorem.
Proposition 5.9: If G is a finite abelian group and ρ is a prime dividing the order of G,
then G has an element of order p.
Proof: We will prove the proposition by induction on the order of G. If \G\ = 1, there
is nothing to prove. Assume the proposition is true for all groups of order less than n,
the order of G, where η > 1. Recall from Section 4.2b, page 105, that if G is cyclic there
is a subgroup of order any integer that divides \G\. Thus if G is cyclic the theorem holds,
and we may therefore assume G is not cyclic. If η is a prime, G is cyclic; hence η is not a
prime.
Suppose h {¥* 1) G G, h of order m. Clearly m < n. Let Η be the cyclic group
generated by h. Η is a proper subgroup of G. Now if ρ | m, by the induction assumption, Η
has an element of order p. If p/m, form the factor group G/H (every subgroup of an
abelian group is a normal subgroup so HO G). Since \H\ > 1, \G/H\ < \G\. As \G/H\ =
|G|/|ff |, ρ | \G\/\H\. Therefore by the induction assumption, G/H has an element g of order p.
Let ν: G -» G/H be the natural homomorphism of a group onto its factor group (see
Theorem 4.17, page 114) and g be a preimage of g under v. Now {gp)v — g" = the identity
of G/H, so gp ей. As Η is of order m, (gm)" = (g")m = 1. Therefore gm has order ρ or
0m = 1. If gm = 1, then gmv = gm = 1. Since <j has order ρ this implies ρ divides m,
contrary to our assumption. Therefore gm is an element of G of order p.
We are now in a position to prove the Sylow theorems. For convenience we repeat
the statement of each theorem.
The First Sylow Theorem: Let G be a finite group, ρ a prime and pr the highest power
of ρ dividing the order of G. Then there is a subgroup of G
of order pr.
Proof: We will prove the theorem by induction on the order η of G. For |G| = 1 the
theorem is trivial. Assume η > 1 and that the theorem is true for groups of order < n.
Suppose \Z(G)\ — с We have two possibilities: (i) ρ \ с or (ii) р/с.
(i) Suppose ρ \ с Z(G) is an abelian group. By Proposition 5.9, Z(G) has an element of
order p. Let N be a cyclic subgroup of Z(G) generated by an element of order p. N <l G,
since any subgroup of Z(G) is normal in G. Consider G/N. Then |G/2V| = n/p by
Corollary 4.14, page 110. Hence by our induction assumption, G/N has a subgroup Η
of order pT~l.
By Corollary 4.21, page 121, there exists a subgroup Я of G such that H/N = H.
As ρτ~χ = \H\ = \H\/\N\- \H\/p, we conclude that \H\ = pr. Thus in this case, G has
a subgroup of order pr.
(ii) Suppose pjfc. The class equation for G is (see Equation (5.2) of Corollary 5.8, page 135)
\G\ = \Z(G)\ + Σ [G:C(R)]
Since ρ | \G\ and р/с, we have ρ/ Σ [G '■ ^(ff)l- Therefore for at least one R G %*,
R e %*
p/[G:C{R)\. But \G\ = [G:C{R)}\C{R)\ by Corollary 4.14 to Lagrange's theorem,
page 110. Hence pr\\C{R)\, since pT\\G\. Now \C{R)\ * \G\; for if \C(R)\ = |G|,
then C(R) = G and i?n#(G) = i?, contrary to the assumption that #n#(G) = 0.
Thus by the induction assumption, C(R) has a subgroup Η of order pr. Consequently so
does G.
In either case we have found a subgroup Η of order pr. The proof is complete.

138
FINITE GROUPS
[CHAP. 5
The following gives a simple formula for the normalizer of a Sylow p-subgroup Ρ in a
subgroup Я of G, where \H\ is a power of p. It will be used in the proof of the second
Sylow theorem.
Lemma 5.10: If G is a finite group, Ρ a Sylow p-subgroup of G, and Я is a subgroup of G
of order a power of p, then
NH(P) = ЯП Р
Proof: P(1HCNh(P), as conjugation by an element of Ρ sends Ρ to itself. We show
NH(P)QPnH. Nh(P)QNg(P) and Ρ < NG(P) (see Problem 5.15 and Problem 4.60,
page 113), so that by the subgroup isomorphism theorem (Theorem 4.23, page 125) we
have: Nh(P)P is a subgroup of G and
NH(P)P/P & Nh(P)/Nh(P)DP
Consequently [ΝΗ(Ρ)Ρ:Ρ] = [ΝΗ{Ρ): ΝΗ(Ρ)Γ)Ρ]. But NH(P) is a p-group, i.e. a group of
order a power of p, since it is a subgroup of the p-group Я. Thus [Nh(P) : Νη(Ρ)Γ)Ρ] is a
power of p. [2VH(P)P: P] is therefore also a power of ρ and, as Ρ is a p-group, |2Vh(P)P|
is a power of p. Accordingly, NH{P)P is a p-group. But Ρ с NH(P)P and Pisa Sylow
p-subgroup. Hence Ρ — Nh(P)P, for Ρ cannot be a proper subgroup of any other p-sub-
group of G (see Problem 5.4, page 132). Nh(P) is therefore a subgroup of P. As Nh(P) С Н,
we conclude NH(P)QHnP.
The Second Sylow Theorem: Let Я be a subgroup of a finite group G, and let Ρ be a
Sylow p-subgroup of G. If Я is a p-group, then Я is
contained in a G-conjugate of P.
Proof: We apply Corollary 5.7, page 135, to cA = {g~lPg \ g G G) to conclude
И = Σ [H:NH(R)] = [G-.2VG(P)]
Re %
By Lemma 5.10, ΝΗ(β) = HnR for each RG%, Hence
[G:NG(P)} = Σ [Я:ЯПЙ] (5,3)
If HnR ¥>H for all β G 4^, as Я is a p-group, the right-hand side of equation (5.3) is
divisible by p. Hence [G: NG(P)] is divisible by p. But PcNG(P), so that ρ does not divide
[G:NG(P)]. This contradiction implies that Hf\R-H for at least one R G %. But as
R G cA, β is a G-conjugate of P. The result follows.
The Third Sylow Theorem: (i) Any two Sylow p-subgroups of a finite group G are
G-conjugate. (ii) The number sP of distinct Sylow p-subgroups of
G is congruent to 1 modulo p. (iii) sv \ ]G].
Proof:
(i) Let Ρ and P' be two Sylow p-subgroups of G. By the second Sylow theorem, P', as a
p-group, is contained in some G-conjugate β of P. But \P'\ = ]β], by Problem 5.3,
page 132. Hence Ρ' = β and P' is conjugate to Ρ under G.
(ii) Let Ρ be any Sylow p-subgroup of G. Since any other Sylow p-subgroup is conjugate
to Ρ and any conjugate of a Sylow p-subgroup is a Sylow p-subgroup (Problem 5.5,
page 132), we conclude by Lemma 5.4 that
sP = [G:Na(P)}
But on putting Ρ = Я in Equation (5.3), we have
sP = Σ [P:PnR]

Sec. 5.2]
THEOKY OP p-GKOUPS
139
Now for exactly one R G %, R — P; for the only P-conjugate of Ρ is Ρ itself and so
Ρ is the only possible representative of its equivalence class. In all other cases,
Pr\R¥=P. Therefore [Р:РПЙ] is a power of ρ for all RG% except one, and for
this one [P: РПЙ] = 1. Hence
sp = 1 + kp
(in) By Corollary 4.14 to Lagrange's theorem, \G\ — [G: NG(P)] \Nq(P)\. Since sp —
[G:Na(P)]t sP\\G\.
5.2 THEORY OF p-GROUPS
a. The importance of p-groups in finite groups
Suppose that G is a finite group. In Section 5.1a we saw that G has a Sylow p-subgroup
for any prime p. (p will be a prime throughout this section.) One reason why the study
of p-groups (groups of order a power of p) is so important is that the structure of the
Sylow p-subgroups of G partly determines the structure of G. One instance is the
following theorem: If G is a finite group whose Sylow p-subgroups are all cyclic, then G has a
normal subgroup N such that GIN and N are both cyclic. (M. Hall, Jr., The Theory of
Groups, Macmillan, 1959, Theorem 9.4.3, page 146.)
In this section we shall determine some of the elementary properties of p-groups.
b. The center of a p-group
A very important property of finite p-groups is given by
Theorem 5.11: If G ¥= {1} and G is a finite p-group, then Z(G), the center of G, is not of
order 1.
Proof: We make use of the class equation (equation (5.2), page 135)
\G\ = \Z(G)\ + Σ [G:C(R)] {5Jt)
R 6 4?.*
It follows immediately from the definition of C(R) and Z(G) that C(R) - G if and only if
RqZ(G). Because the sum on the right side of (5Л) is taken over all R such that
R(-\Z(G) = 0 and because \G\ =pr, p\[G: C(R)] for all R e %*. Hence ρ ] 2 tG: C(R^-
Since ρ | \G\, we can conclude that p] \Z(G)\, which means Z(G) ¥= {1}. Re%*
Corollary 5.12: If G is a group of order pr, r ^ 1, then G ha3 a normal subgroup of
order pr_1.
Proof: The proof is by induction on r. The statement is clearly true for r = 1.
Suppose the corollary is true for all к < r where r > 1. By Theorem 5.11, Z(G) ¥> {!}.
Because ρ ] \Z(G)\, Proposition 5.9 implies Z(G) has an element g of order p. Let N = gp(g).
N < G, since any subgroup of Z(G) is normal in G. Consider GJN. \G/N\ = pr"4
Therefore by the induction assumption, GIN has a normal subgroup Я of order pr~%. By
Corollary 4.21, page 121, there exists a subgroup Η of G which contains N and such that
H/N = H. Then \H\ = pr~4 Furthermore, again by Corollary 4.21, Η Ο G. Thus G
has a normal subgroup of order pr~l and the proof is complete.
Clearly we could repeat this argument until we obtain a sequence of subgroups of G
{1} = Я»СЙ1 С ■·■ С Hr-ι С Hr = G (5Л)
where Я; < Hi+i (i = 0,1, ..., r-1) and \Щ = ρ' (i = 0,1,2, ..., r).

140
FINITE GEOUPS
[CHAP. 5
Problems
5.18. Suppose G is a group with S a subgroup of the center Z(G). Prove G is abelian if G/S is cyclic.
Solution:
Suppose G/S is cyclic. Then we can find a e G such that every element of G/S is a power of
aS. Then if ff, h e G,
g = ah, h = u>z'
for a suitable choice of the integers i and j, with ζ and z' in S. Then
gh = a'zoV = aWzz' = eJatz'e = cJz'atz = hg
Thus every pair of elements of G commute. Hence G is abelian.
5.19. Prove that a group of order p2 is abelian (p a prime).
Solution:
Let G be of order p2 and let Ζ be the center of G. By Theorem 5,11, Ζ Φ {1>. If Ζ = G, G is
abelian. Suppose Ζ Φ G; then \GIZ\ = p, so GIZ is cyclic. By Problem 5.18, it follows that G
is abelian.
5.20. Let A = {0,1, .. ,,p— 1), where ρ is a prime. Then under addition modulo p, A is an abelian
group. Let
G = {(a, b,c) | a,b,cGA}
be the set of all triples (a, b, c) of elements of A, Define
(a, b, e) · (a', b', c') = (a + a', b + Ь', с + c' - bo,')
Prove that with respect to this binary operation, G is a non-abelian group of order ps.
Solution:
It is clear that \G\ = p3. To prove that G is a group, we check first the associative law:
((a, b, c)(a', b', c')) · (a", b", c") = (a + a',b + b', с + c'- ba')(a'\ Ь", с")
= (a + a' + a", b + b' + b", e + c' + c"- ba' - {b + b')a")
On the other hand,
(a, b, e){{a', b\ e')(a", b", c")) = (a, b, c)(a' + a", b' + b", c' + c" - b'a")
= (a + a' + a", b + b' + b", с + с' + с" - b'a," - Ца' + a"))
We check that
с + с' + с" - ba' - (b + Ъ')а" = с + c' + c" - b'a" - b(a' + a")
which is true. Thus G is a semigroup. Now
(a, b,c)· (0,0,0) = (a,b,c) = (0, 0,0) · (a, b, c)
and so (0,0, 0) is the unit element of G. Finally,
(a, b,c)'(—a, -b, —c— ba) = (0,0,0) = (-a, -b, —c— ba)(a,b,c)
and hence every element of G has an inverse.
Our last task is to prove that G is non-abelian. Now
(1,0,0)(0,1,0) = (1,1,0), (0,1,0)(1,0,0) = (1,1,-1)
and thus (1,0,0)(0,1, 0) Φ (0,1,0)(1, 0,0).
5.21. Let A be the additive group of integers modulo p, A = {0,1, .. ,,p - 1>; and let В be the additive
gTOup of integers modulo ρ2, Β = {0,1, ...,p2-l). Let G be the set of all pairs (г, j), i&A,
j e B. Prove that under the binary operation
(i,3)'(?,]') = (i + V.i+i'+H'p)
G is a non-abelian group of order p3.
Solution:
Clearly G is of order p*. We check that G is a semigroup.

Sec. 5.2]
THEOKY OF p-GKOUPS
141
(ft fW. i'W'. i") = <l*+i'< i + ϊ + n'pW. Л
= (t + i + г", j + j' + j" + ji'p + (j + j' + ji'p)i"p)
(i, iW, )')(i",)")) = (*. /)(*'' + *". J' + J" + }'i"p)
= (i + i' + i", j + f + j" + j'i"p + З'(г' + i")v)
To prove that the binary operation in G is associative, we need check only that
ji'p + (j + j' + ji'p)i"p = j'i"p + j(i' + i")p
Since p2 = 0 in B, this equality is readily verified.
The identity element of G is (0,0). The inverse of (i, j) is (—it-i+]'ip). Thus G is a group.
Fmally' (1,0)(0,1) = (1,1), (0,1)(1,0) = (1,1 + p)
and therefore G is non-abelian.
5.22. Prove that the group G in Problem 5.20 has the property that for all g e G, gp = 1 (i.e. (0, 0, 0)) if
ρ is odd. Is this true if ρ = 27
Solution:
Let (a, b, c) g G. Then
(a, b, c)2 = (a, b, c)(a, b, c) = (2a, 2b, 2c — ba)
Continuing, we find
(a, b, c)3 = (a, b, o)*(a, b, c) = (2a, 2b, 2e - Ьа)(а, Ъ, с)
= (За, ЗЬ, 3c- ba-2ba)
By induction it follows that
(a, b, c)p = (pa, pb, pe—ba— 2ba — · · ■ — (p — l)ba)
But pa =0, pb — 0, pc = 0. Finally,
ba+ 2ba+ ··· + (p-l)ba = ^~^Ьа.
since 1 + 2 + · ■ · + ρ — 1 = ^p(p — 1). If ρ is odd, ρ — 1 is even. Therefore ^p(p — 1) is an integer
divisible by p. Hence ^p(p — l)ba = 0. Thus we have (a, b, c)p — 1.
If ρ = 2, then
(a, b, c)2 = (2a, 2b, 2c - ba) = (0, 0, -ba)
In particular if α = 1, Ь = 1 and с = 0, we have (1,1,0)*= (0,0,-1). Thus not every element
of G is of order 2. This result could have been observed by noting that a group G satisfying g2 = 1
for all g in G is abelian. To see this let g,heG. Then as (gh)2 = 1,
gh = (fifft)-i = A-iff-i = h?h~Wg-i = kg
and so G is abelian. But as G is not abelian, not every element is of order 2.
5.23. If ρ is odd, does the group G of Problem 5.21 satisfy gv = 1 (i.e. (0,0)) for all g € G?
Solution:
No, since (0, l)2 = (0,1)(0,1) = (0,2). Inductively, (0,1)" = (0, ρ) Φ (0,0).
5.24. Prove that if G is a group such that gt> = 1 for all fir £ G, then every homomorphic image Я
has the same property, i.e. hP = 1 for all h £ H,
Solution:
Let ff be a homomorphism of G onto Я. Then if hS H, we can find g £ G such that p-fl = Л.
Therefore h? = (gsY = (g")e = Iff = 1.

142
FINITE GROUPS
[CHAP. 5
5.25. Prove that if ρ is an odd prime, then the groups in Problems 5.20 and 5.21 are not isomorphic.
Solution:
Let G be the group defined in Problem 5.20 and let Η be the group defined in Problem 5.21, for
ρ an odd prime. Then by Problem 5.22, if ff e G, g* = 1. But if G s H, it follows from Problem
5.24 that h" - 1 for all h e H. But by Problem 5.23, (0,1)" Φ 1. Therefore G is not iaomorphic
toff.
5.26. Prove that a non-abelian group G of order pa has a center of order ρ (ρ a prime).
Solution:
Let Ζ be the center of G. By Theorem 5.11, 2 Φ {1}. Also, Ζ Φ G since G ia non-abelian.
Now if \Z\ = p2, then |G/Z| = p. Therefore G/Z would be cyclic and hence, by Problem 5.18, G
would be abelian, a contradiction. Thus \Z\ — p.
c. The upper central series
Suppose G is a group. We shall define a series
{1} = Z0 Q Zt Q ··■
of subgroups Zo,Zi,... of G, called the upper central aeries of G. We begin by defining
Ζ а = {1}, and Zi to be the center of G. Next we define Zi. We look at G/Zi. Since every
subgroup of G/Zi is uniquely of the form Η/Ζλ where Я is a subgroup of G containing Zu
the center of G/Zt is of the form Z-JZi (we are using Corollary 4.21, page 121). Notice that
as the center of a group is a normal subgroup, ZyZi is a normal subgroup of G/Zl
Therefore by Corollary 4.21, Zi is a normal subgroup of G.
In general, once Zi has been defined and proved to be a normal subgroup of G, we define
Zl+l/Zt to be the center of G/Zu By Corollary 4.21 it follows that Zl+t < G.
We shall call a group G nilpotent if its upper central series ascends to G in a finite
number of steps.
Our objective in this section is to prove
Theorem 5.13: A finite p-group G is nilpotent.
Proof'. If G={1), there is nothing to prove. If G¥*{\), then Ζχφ {1} by Theorem
5.11. If G/Zl is not the identity, the center of G/Zl = ZJZi φ ZJZi, again by Theorem 5.11.
Notice that if Zt¥*G, then Z2 Φ Zi. Similarly if G Φ Zi, Z3 Φ Ζ2. By induction we can
show that if Zt¥*G, ZiHT* Zt and thus
1 = Zo С Zt С ··· С Zs С Ζι+ι
Since G is finite, Zk = G for some k. Therefore G is nilpotent.
Problems
5.27. Prove that if a non-abelian group G is of order p3, then Z2 — G.
Solution:
Ζγ Φ {1} by Theorem 5.11. So if Zt Φ G, then G/Z1 is of order ρ or pa. Since G/Zx is cyclic
only if Zt- G (Problem 5.18), -we find G/Zx is of order рг and hence abelian (Problem 5.19).
Therefore ZtIZx - GIZb i.e. Z2 = G.
5.28. Let D„ be the dihedral group of order 2та. Prove that D„ is nilpotent if та is a power of 2.
Solution:
Dn has the property that it contains two elements α and b such that aa = 1, bn = 1, а~1Ьа — b~l
and every element of D„ is uniquely expressible in the form ФЫ where i — 0,1 and j = 0,1,
.. .,та — 1. (See Section 3.4f, page 75, where α = τ and b — σ2·)

Sec. 5.3]
DIKECT PKODUCTS AND GKOUPS OF LOW OKDEK
143
Method 1. Suppose η = 2m. Then b2 is of order 2. Hence
иг7™"1 / -,ι. χϊ"1"1 ,,,.,,ί"1""1 ^г"*"1
ο,-ib-1 а = (ο,-ι&α) = (& ι) = δ*1
So &2 commutes with α; clearly &2 commutes with b. Therefore &2 commutes with every
element of G, and so b2 G^. If m = l, then Z1 - Dn; otherwise a & Zv
What other elements can be in the center? If alb' ε Zlt then clearly a"J(alb')a — &Ь>. On
the other hand, а-ЦаЩа = ala~^a = а>Ь~>. Hence Ы = b~i andjfc*)2 = 1, i.e. Ы = b2'""1. Thus
the only possible element in the center other than b2 is ab2 . But as Ь2 е Zlt this
implies α e Ζγ, which is not so. Consequently ΖΎ = {1, Ь2 }.
Similarly Z2 consists of the powers of b2 , ..., Zi consists of the powers of b2 . Therefore
Zm-i consists of the powers of b2. Now this means \G/Zm^t\ = 4, and so G/Zm^1 is abelian.
Thus Zm = G and G is nilpotent
Method 2. jDn| = some power of 2. Hence we can apply Theorem 5.13.
5.29. Prove that A4 is not nilpotent.
Solution:
Zx = {1} in At, as a direct check shows. Hence Ζτ = Z2= ··-, and thus Zn ¥= A4 for every n.
5.3 DIRECT PRODUCTS AND GROUPS OF LOW ORDER
a. Direct products of groups
In Chapter 1 we defined the cartesian product Я χ Κ of two sets Я and К as the set of
all ordered pairs (h, k), h Ε Η and Α; Ε К. If Я and К are groups, we can define a
multiplication of elements of Я х К as follows. Let (hi, kt), (h2, Щ € Я х К and define
(fci,fci)-(fc2,fe) = (hjh^kjkz) (5.5)
where fcifes and kjc2 are the products in the groups Я and ii respectively. The
multiplication defined in (5.5) is clearly a binary operation. The set HxK with binary operation
(5.5) is a group. To see that Я χ ii is a group, let (hir ki), (h2, k2), (hs, k3) Ε Я х X. Then
[(fei, fci) · (Ы, Щ - (h3, Щ = (hike, ktkz) · (Ы, Щ
= ((hxh^hs, (fcife)fcs)
= (АДЛгЛз), ki(kzk3))
= (hi,kj)'(k2h$,kik3)
- (hu kj) · \(h2, fc2) ' (Й-э, fea)]
Multiplication is therefore an associative binary operation on Я x X. If 1 stands
simultaneously for the identity element of Я and of К and (h,k) Ε HxK,
(l,l)(h,k) = (h,k) = (fc,fc)(l,l)
so that (1,1) is an identity of HxK. It is clear that if (h,k) <EHxK, then (h~\k"1) is
its inverse, for (fe.fe)^-1,^-1) = (fefc-^fefe-1) = (1,1). The group Нхй with binary
operation (5.5) is called the external direct product of the groups Η and K. We often refer to
Я χ Κ as just the direct product. We define the internal direct product after Proposition
5.19. If Я and К are finite groups, then it is clear that
\HxK\ = \H\\K\
If Я ¥- {1} and К ¥= {1} are finite groups, then Я х К is neither isomorphic to Я nor
to X, because |Я x K\ ¥* |Я| and [Я х Кj =* |Z|. Therefore the direct product gives us a
simple way of constructing new finite groups. For example, let C2 be the cyclic group of
order 2 generated by g.

144
FINITE GROUPS
[CHAP. 5
Now \CaX C%\ = 4, so we have (as we shall soon see) two non-isomorphic groups of order
4, namely: the cyclic group of order 4, d = {1,Ъ,Ъ2, bs} where b4 = 1, and the group
C2 X (>2-
The multiplication table for C% χ C2 is
(1,1) (l.fii) (fii,l) (i)i,i)i)
(1,1)
(I,»)
(0,1)
(им?)
(l,ff)
(i.i)
(9,9)
(ff.l)
(ff,l)
(ff.ff)
(1,1)
(1,1/)
(ff.ff)
(ff.l)
(1,1/)
(1,1)
Note that all the elements of C2 x C2 are of order 2. Hence Ca is not isomorphic to C2 x C2.
C2 x C2 is called the Klein four group, or simply the four group.
Theorem 5.14: If G = Η xK is the direct product of the groups Я and K, then the sets
Я = {{h, 1)\ hGH, 1 the identity of Я}
К = {(1, k)\ kGK, 1 the identity of K)
are subgroups of G. Furthermore, H = H,K = K; and if a e. Η and
bGK, ab = Sa. Finally, G = HK and ЯпЯ = {(1,1)}, the identity
subgroup of G.
Proof: If (Ль1),(Л2,1)еН, then {hi, l)(u2, l)"1 = (Λι,1)(ΛΓ',1) = (ΛιΛί'.Ι) G H,
, Л л. A.
since Λ-ιΛ-2 GH. Я is clearly non-empty. Therefore Я is a subgroup of G. Similarly К
is a subgroup of G. The mapping α of Я onto Я defined by
α: Л -* (ЛД), h G Я
is clearly an isomorphism. Similarly Я and X are isomorphic. Now let a-(h,l) ε Я
and S = (1, ft) ε £; then
ab = (M)(l,fc) = (u,fe) = (l,fe)(M) = &α
Now ЙПХ= {(1,1)}. Any element (fc.fc) of G can be written as (fc,l)(l,fc), so GqHK.
Clearly Я& с G. Hence G ^ HK and Theorem 5.14 follows.
Corollary 5.15: Let G = HxK and #,£ be as in Theorem 5.14. Then every g ε G
can be written uniquely as a product hk where h ε Я, к £ К.
Proof: If g — (h, к), then 9 = (Л-, 1)(1, к) is an expression for g as the product of an
element in Я by an element in K, If we also have g = (hul)(l,ki), then clearly hi — h
and fci = fc. Thus the expression is unique.
As a converse we have
Theorem 5.16: Let G be a group with subgroups Я and К such that HnK — {1}, the
elements of Я commute with those of K, and HK — G. Then G = HxK.
Proof: We first show that any element g ε G can be written uniquely in the form
g = hk where h ε Я and k £ К. Since G = ЯХ, g = hk for some hGH and kGK.
Suppose gr = feifci and gr = u2/c2 where hu h2 ε Я and fci, fc2 ε Χ. Λ-ifci = h2k2 implies
h^lhi = k2ki1. But НГ\К={1}, and so Ιιϊ% = 1 and fc^'^l. Hence hi-hi and
fci = fc2.
We define the mapping α: G -* Я хЯ by да = (h,k) where g = hk G G. α is a one-to-
one mapping, for we have shown that there is one and only one way of writing g in the form
g — hk, and the elements of Η χ Κ are of the unique form (h, k). To prove α is a homo-
morphism we must demonstrate that if gi = hikx and g% — hik2 are any two elements in
G, then

Sec. 5.3]
DIRECT PRODUCTS AND GROUPS OF LOW ORDER
145
NOW {hfahjcja = (ЛДВДа = {hfokfa) = (h^kjihrkj = {\\)α{\^)α
Hence α is a homomorphism and the result follows.
Note that if H,K are normal subgroups of a group G with HC\K — {1}, then Я and ii
commute elementwise. For if h G H, к G K,
Н-Че-Ък = (h-lk-lh)kGK = h^l{k^hk)GH
Therefore h~lk-lhk G HnK = {1}, and so Я and К commute elementwise. Clearly
G — HK and Я and ii commute elementwise implies Я and ii are normal in G.
Consequently Theorem 5.16 can be stated as follows:
Corollary 5.17: Let G be a group with normal subgroups Я and K, and suppose ЯПХ= {1},
and HK = G. Then G^HXK.
The hypothesis of Theorem 5.16 asserts G must equal HK. But if G is a finite group
and |ЯК| = |G|, we can conclude, since HK С G, that HK — G. It is useful to be able to
count the number of elements in HK, We therefore prove the following proposition.
Proposition 5.18: If G is a finite group with subgroups Я and K, then
[Я| · \K\
\HK\
[ЯП XI
Proof; Let I = HnK. I is a subgroup of G and, since /CK, / is a subgroup of K,
Let Iki, I fa, , . .,lk„ be the η distinct cosets of / in K. Thus
К = Iki U Ifa U ■ · ■ U /Αν
and, by Corollary 4.14, page 110, n= |X|/|/| = |X|/|HnX|.
We claim now that
ЯХ = Hkt U Яй2 U · · · U ЯА;„
For if hk G HK, then k = Щ for some IG1, j an integer between 1 and n. Hence
hk = (hl)kj = h% where ft' G Я, as both ft, I belong to Я. Thus ЯХ = HfaUHfaU ■ ■ - иHfc,.
Now suppose Hkir\Hkj¥=0 for some integers t and j. Then ftfc = ft'fcj for some
h,h'GH, Consequently ft'_1ft = fejfef1, so ksk^1 G I = HC\K. But k}k7xGl implies
that fcj £ /&». Since two cosets are either equal or disjoint, Ik} = Iki. Hence kt = fa.
Thus HktC\Hk} = <jb for г?*/ and
|HX| = |Hfci| + \Hfa\ + · · · + \Hkn\
Now |Яй*| = |Я|, because hifa — Juki if and only if fti = ft2. Therefore
|ЯЯ| ~ И|Я| - \н^к\
since n=|X|/|HnX|.
To illustrate the use of Proposition 5.18, let G be a group of order 28 and Hi and Яг
subgroups of G of orders 7 and 4 respectively. HiCifa= {!}, because an element in Hi
and also in Я2 must have order dividing 7 and 4. Accordingly,
ι**' = [S =28 = 'Gi

146
FINITE GROUPS
[CHAP. 5
Using Proposition 5.18, we can replace Theorem 5.16 in the case of finite groups by
Theorem 5.16': Let G be a finite group with normal subgroups Η and К where \H\ \K\ = \G\.
If either (i) Η ПК = {1} or (ii) HK = G, then G^HxK.
Proof:
(i) HnK = {1} and \H\ \K\ = \G\ implies, by Proposition 5.18, \HK\ = \H\ |K|/|HnX| = \G\.
Since HKCG, we can conclude HK = G. But then the hypotheses of Corollary 5.17
are fulfilled and G^HxK.
(ii) If HK = G, then \HK\ = \G\. Therefore
\G\ = \HK\ = г^Щ; or LffnlT||G| - \H\\K\
But |S||X|=|G| by hypothesis. Hence НПК = {1} and, by Corollary 5.17, G^HxK.
The concept of direct product can easily be generalized to the direct product of a
finite number of groups, GbG?,.. .,Gn (n^2). Let G = Gi χ G%x · ■ ■ x Gn be the cartesian
product of η groups. Define a multiplication in G by (gi, g»,..., gn)(gi, gL · ■ ■, gn) =
(gigi gtgi ■ ■ ·, gngk) for (gu дг>..., gn), (gi, gi, ..., gn) G G. G is then a group (see Problem
5.30 below) called the (external) direct product of the groups Gi, G2, .. ., G„. We denote G by
In Chapter 6 we will define the direct product of an infinite number of groups
differently. Proposition 5.19 below and Corollary 5.15 will provide a link between the two
definitions.
Proposition 5.19: Let Η and К be subgroups of a group G. If
(i) hk = kh for all h G Η and k GK
and
(ii) every element g G G is a unique product of an element in Η
and an element in K, (i.e. g — hk, hG H, k G K; and if g = hikb
hi G H, h e X, then ft = fti and fc = fct),
then G^HxK.
Proof: We need only prove ЯПК= {1} to fulfill the hypotheses of Theorem 5.16.
Suppose g G Η ПК. Then g = fe· 1 = 1 · A; for some fc G Я and k G K. But condition
(ii) implies h = l and fc = 1. Therefore # = 1 and НПК={1}.
If G is a group with subgroups Η and К satisfying conditions (i) and (ii), G is said to be
the internal direct product of Η and K, and we write G = Η ® Κ. By Proposition 5.19,
H®K^HXK.
Problems
5.30. Let G = θτΧ G2 X · ■ ■ X Gn be the cartesian product of и groups. Define a multiplication in G by
(ffi,ff2 ffn)(ffi.ff2. ■■■,β'η) ~ {9i9'i,9nff-l, ...,gng„)
for (gltg2,..., ffn), (g[,д'ъ ..., g^) e G. Show that G is a group.
Solution:
The multiplication is clearly an associative binary operation in G, since multiplication is
associative in each G4. If 1 stands simultaneously for the identity of Gt, г = 1, 2, .. ., η, then (1,1, ..., 1)
is clearly the identity of G. If g = (gi,g%, .. .,gn) G G, then (ff^1,^""1, .. -.fi^1) is the inverse
of g in G.
5.31. If Η ^ Η and К ^ K, where H, Я, К and К are groups, then HxK= HxK.
Solution: _
If а:Н->й and β : К-* К are isomorphisms, we define y;HxK->HxK by y:(fe,ft)->
(fca, fc/J), ft G H, k€.K, γ is a one-to-one mapping, for (fea, fc/ϊ) = (fc'a, У β) if and only if ha = h'a

Sec. 5.3]
DIRECT PRODUCTS AND GROUPS OF LOW ORDER
147
and kβ = fe'/J. Since a and β are one-to-one mappings, ha = h'a and k$ = k'β if and only if
h — h' and к = к'. To show γ is a homomorphisrn, let (h,k), (h',k') еЯхК. Then
[(fe,fc)-(fe',fe')]7 = {hh',kk')y = ({hh')a, (kk')fi) = {hah'a,kpk'p)
= {ha,kp)-(h'a,k'fi) = {h,k)y(h',k')y
Finally, it is clear that γ is an onto mapping.
5.82. Show that G = Η X К is an abelian group if and only if Η and К are both abelian groups.
Solution:
Suppose Η and К are abelian groups. Letting (h, k), (ft/, «') G G,
{h,k)-(h',k') = (hh',kk') = (h'h,k'k) = (h', k') ■ (h,k)
and so G is abelian.
Conversely, suppose G is abelian. Let h.h'GH. Then if 1 is the identity of K, (ft, l)(fc', 1) =
(ft/, l)(fe, 1) or (hh't 1) = (h'h, 1). But this implies hh' = h'h. Hence Я is abelian. Similarly we
can show К is abelian.
5.83. Consider the groups C2 X K4, C4 X C2, C$ where Cn is the cyclic group of order η and K4 the four
group, i.e. the non-cyclic group of order 4, described above. Are any two of these groups isomorphic?
Is any one non-abelian?
Solution:
Let C2 = gp(a), C4 = gp(b), and (78 = вр{в)· We look at the set of elements of order 2 in
each group. Since every isomorphism maps elements of order 2 onto elements of order 2, if there
are more elements of order 2 in one group than in another, these groups cannot be isomorphic. Every
element (Ф1) of C2 X K4 is of order 2, for (с, к)2 = (са, Щ = (1,1) and (1, A;)2 = (l,fc2) = (1,1) for
any к e K4. C8 on the other hand has only one element of order 2, namely ff4, because (ff{)2 ¥· 1 if
г Φ 4, 0 — ί — 7. Now C4 X C2 has at least one element of order 4, (b, 1), and at least two elements
of order 2, (i>2, a) and (1, a). Therefore no two of the groups are isomorphic. As C2, C4, K4 and C8
are abelian, Problem 5.32 implies C2 X K4, C4 X Сг and C8 are also abelian. Thus we have exhibited
three non-isomorphic abelian groups of order 8.
5.84. If Cn and Cm are the cyclic groups of order η and m respectively and (re, m) = 1, then
On X Cm e Cnm, the cyclic group of order nm.
Solution:
Say Cn = gp(g) and Cm = gp(h). Consider the order of the element (g, h) in Cn X Cm. We
claim that the order of (g, h) is nm. If (fir, h)h = (1,1) for some k, then (fir*, fek) = (1,1) and so
fir* = 1 and fe* = 1. Since the order of fir is re and the order of h is m, m\ к and n\ k. Hence к
is divisible by rem. On the other hand, (fir, h)nm = (fir™"1, fenm) = 1 and so the order is nm. Accordingly,
Cn x Cm = ^((ff. Ό)· Therefore CnX Cm = Cnm, since all cyclic groups of the same order are
isomorphic (Theorem 4.7, page 103).
5.85. Show that Css is not isomorphic to C„ X Cs (where C„ is the cyclic group of order re), for any
integer β > 1.
Solution:
Since Cs2 is a cyclic group it has, by Theorem 4.9, page 105, one and only one subgroup of order
β. But Cs X C„ has two subgroups of order s, namely firp((l,fir)) and firp((fir, 1)), where fir is the generator
of Cs. Since subgroups of a given order are mapped onto subgroups of the same order by any
isomorphism, Cs2 cannot be isomorphic to Cs X C„.
5.36. Show that for any prime ρ there are exactly two non-isomorphic groups of order ps.
Solution:
By Problem 5.19, page 140, we know that any group of order jP is abelian. Cvz, the cyclic group
of order p2, and Cp X Cp, where Cv is the cyclic group of order p, are two non-isomorphic groups of
order p2 (Problem B.3B). To see that these are the only possible groups of order p2, consider a

148
FINITE GROUPS
[CHAP. 5
subgroup Я of order ρ in G, a group of order p7-. Such a subgroup exists by Corollary 5.12. Я is
cyclic, since ρ is a prime. Let иЙН. The order of a is either ρ or p2. If the order of a is p2, G
is a cyclic group generated by a. If |fi7>(a)| = p, then gp(a)nH = {1}, for Ь (^1) G gp(a)r\H
implies Я = ίΡίί») and ^ρ(α) = gp{b), since a group of prime order has no proper subgroups.
Also {gp(a)\\H\ = p2 and, as G is abelian (Problem 5.19), gp{a) О G and Я Ο G. Therefore, using
Theorem 5.16', we conclude G = ^p(a) X Я. But да(а) X Я ^ C„ X С by Problem 5.31. Hence
G s Cp X Cp.
5.37. Show (Hl X Я2) χ Я3 s fft χ ff2 X Я3.
Solution:
Define *: (Я1 X H£ X Я3 -* Hx X Я2 х Я3 by Ψ: ((fclr fe2), h3) -*■ (ht, h^, Λ3) for кг е Я, ft,eif2
and 7i3 e Я3. Clearly Ψ is an onto mapping. If ((htl h^, λ3)Ψ = ((Kt, Я2), из)-*, then (fcj, fea ft3) =
(Я,, ϋ2, ^3) and consequently At = Hv h^ = ft2 and h$ = h3. Therefore Ψ is one-to-one. To show
Φ is a homomorphism, let ((ΛΙ( fe2), Λ3) and ((S1( h£, h3) £ (H1 X Я2) X Я3. Then
[((fciJfc2).fca)((^i.Sg),Ra)l*' = ((VMS,^),^)* = ((^ъ M*)> hh)*
= (hjib hjiz, hjtz) = (fci, fczi'la)(*i»''2»^s)
= ((^Дл), ^зЖ^Д^Дз)*
and so Ψ is an isomorphism.
b. Groups of small order: orders ρ and 2p
As an application of the Sylow theorems and the theorems of Section 5.3a we will find,
up to isomorphism, all groups of order less than 16. We will use C« to denote the cyclic
group of order n, and K* to denote the four group. Recall that Κι is defined to be Cz x Cz-
We refer to the notation of Section 5.3a. Let us put 1 = (1,1), χ = (1, g), у — {g, 1) and
2 = {g, g)· The multiplication table for К± is
1
X
У
ζ
1
1
χ
1y
ζ
χ
as
1
ζ
V
V
У
ζ
1
χ
ζ
ζ
У
χ
1
We note that xy — z, xz = у and yz - x. Notice that the multiplication table is
symmetric in x, у and z. If we put χ — a and у = b, then ζ = ab and we can write the
multiplication table in the form
1 a b ab
1
a
b
ab
a
1
ab
b
b
ab
1
a
ab
b
a
1
1
a
b
ab
There is, up to isomorphism, clearly only one group of order 1.
If ρ is a prime, any group of order ρ is cyclic (Problem 4.48, page 110). Up to
isomorphism, there is one and only one cyclic group of order ρ (see Theorem 4.7, page 103). Thus
there is one and only one group of order ρ, ρ a prime. In particular, the only groups of
order 2, 3, 5, 7,11 and 13 are cyclic.
There are precisely two поп-isomorphic groups of order k, namely Ci and Κ± (Section
5.3a and Problem 5.36).

Sec. 5.3]
DIRECT PRODUCTS AND GROUPS OF LOW ORDER
149
Next we show there are precisely two groups in each сазе of order 6, 10 or 14. Note
that 6 = 2-3, 10 = 2 ■ 5 and 14 = 2 · 7, so these groups are of order 2p for some prime
ρ ¥* 2. Let G be a group of order 2p, ρ an odd prime. By Problem 5.10, G has exactly one
subgroup К of order p, and either
(i) exactly one subgroup Η of order 2
or
(ii) precisely ρ subgroups of order 2.
(i) In this case the group G is a cyclic group of order 2p. To see this notice that Я is a
unique Sylow 2-group and ii is a unique Sylow p-subgroup; so, by Problem 5.7, page
133, Η <J G and К < G. Furthermore Hr\K= {1}, for any element common to Η
and К must have order dividing 2 and ρ and hence is the identity. Clearly \H\ \K\ =
2p = \G\. Therefore by Theorem 5.16', G^ HxK. But Η and К are cyclic groups of
order 2 and ρ respectively. Thus by Problem 5.34, G is cyclic of order 2p.
(ii) Let К = gp(a) where av — 1. Since К is the only subgroup of order p, b g К implies
b2 = 1. Clearly, G — KUbK. Hence G consists of the distinct elements
1, a, a2, ..., a"~L, b, ba, ba2, ..., bet?"1 (5.6)
Now if i = 0,1,..., ρ — 1, then
(baf = 1 and ba' = а»~*Ь (5.7)
since ba1 g К and each element of G outside К is of order 2. Also
(baf = (bal)(bal) = 1 implies ba1 = (ba})'1 = (a')"1^1 = ар~*Ь
Now if <5 is any group of order 2p, then it is either of type (i) or (ii). If G is of type
(i), then by our analysis it must be a cyclic group of order 2p. By Theorem 4.7, page 103,
cyclic groups of the same order are isomorphic. Hence all groups of order 2p having
property (i) are isomorphic.
Suppose G is of type (ii). Then, arguing as above, G has a subgroup К — gp(d) of order
ρ and an element Ъ of order 2 such that
G = {1, a, ..., άρ~Λ δ, δα, .. ., bdp_I}
where for i = 0,1, ..., ρ — 1,
φάψ = 1 and ba1 = д?~1Ъ (5.8)
The mapping α: G-* G defined by
a: a1 -* ά\ a: b-*b, a: ba*-* ba1 (i any integer)
is an isomorphism. First, α is a mapping; for if a* = aJ, ρ divides i — j and hence ά' = a'.
Consequently a is well defined on a1. If ba1 = ba1, then a1 = aj and so ρ divides i — j and
a' = a1. Hence a is well defined on the ba*. As a: a1 -» a1, ba1 -» δα* (i = 0,1, ..., ρ — 1), a
is one-to-one and onto, a is also a homomorphism, for gu gi&G implies g\ — b'a* and
g1 = b"at for some choice of /,sG{0,1} and i, t G {0,1,2, .. ,,p — 1}. Using equations
(5,7) and (5.*), we obtain, when s = 0,
(дгд2)а = [(Ъ}а1)(аь)]а = (6%i+t)a = &W+t = ЬШ* = (Ь'а1)а(а*)а = gLag2a
and when s — 1,
(9^2)" ~ [(ЬМ)(Ьа')]а = (Ь%ар-1а()а = (b^Lap-i + t)a = 6i + 1dp-i + t = &Ъа»~1а*
= δ'ά'δά4 = (Ь'а1)а(Ьаь)а = дхадга
Therefore any two groups of order 2p of type (ii) are isomorphic.

150
FINITE GROUPS
[CHAP. 6
So far we have shown that up to isomorphism there are at most two possible groups of
order 2p, ρ a prime. This does not mean that there exist two non-isomorphic groups of
order 2p, for each prime p. But from Theorem 4.7, page 103, we know that for each positive
integer η there exists a cyclic group of order n, and from Section 3.4f, page 75, we know
that for each η the order of the dihedral group Dn is 2n and D„ is not cyclic for η > 2 (it is
not even abelian). There are therefore, up to isomorphism, exactly two groups of order
2p for each prime ρ Φ 2: one a cyclic group and the other Dp. In particular there are
exactly two non-isomorphic groups of order i, i = 6,10,14.
It is worthwhile summarizing our method of finding all groups of order 2p. We first
showed that if a group had order 2p it had to be isomorphic to one of two possible groups.
The isomorphism in case (ii) was shown by using the fact that the elements of such a group
had to satisfy equations (5.7) and (5.8). (As those equations determine the group up to
isomorphism, they are usually called defining relations for the groups; they will be discussed
in detail in Chapter 8.) After demonstrating the isomorphism, we proved that each of the
possible groups exists by exhibiting a group of each type.
c. Groups of small order: orders 8 and 9
Let G be a group of order 8. There are at least three non-isomorphic abelian groups of
order 8: C8, C2 x C* and C2 x if* (see Problem 5.33). We show that if G is abelian it is
isomorphic to one of these three groups.
If G has an element of order 8, G is cyclic. If G has no element of order 8 but has an
element a of order 4, let Я = gp(a). Let b € G —Я. If b is of order 4, b2 is an element
of order 2 and lies in Я (since the coset decomposition of G is HUbH). As a2 is the only
element of Я of order 2, b2 = a2. Hence (ab)2 = a2b2 = a?a2 = 1. Since ab £ Я, we may
assume that there exists an element χ € G - Я of order 2 (x = b if b is of order 2 or else
x = ab). Let X ~ gp(x). Xr\H = {1}. Therefore by Theorem 5.16', G^XxH. Since
Х^Съ and Я =* d, we conclude G ^ Сг х C4.
If G has no elements of order 4 or 8, all its non-identity elements are of order 2. Let
a, b be distinct elements of order 2 in G. Let A = gp(a), В = gp(b). Then AB is a group
by Problem 4.62, page 114. Now AnB= {1} and jAj |Bj = \AB\. So, by Theorem 5.16',
AB ^AxB and consequently AB = C2 x C2, Let c€G-AB and С = др(с); then
СГ)АВ={1}. Thus еа(С2хС2)хС2 = йхСг.
We conclude that there are, up to isomorphism, exactly three abelian groups of order 8.
Assume G is a non-abelian group of order 8. G has an element of order 4 and no elements
of order 8; for if g € G is of order 8, G ^ Cg. On the other hand if all elements of G are of
order 2, then (ab)2 = 1 for any a,b €G and consequently
ba = a2babz = a(abab)b = ab
contrary to our assumption that G is non-abelian. Let a € G be an element of order 4,
and put Η = gp(a). Then G = HUHb for some b € G. Also Η <Ζ G, as it is of index 2
(Problem 4.69, page 116). b2 € H; for if not, the cosets H,Hb,Hb2 would be distinct, and
this would contradict [G: H] = 2. We have four possibilities for b2: (i) b2 = a, (ii) b2 — a2,
(iii) b2 = as, or (iv) b2 = 1.
If (i) or (iii) occurs, clearly G = gp(b), contrary to our assumption. Thus (ii) and (iv)
are the only possibilities.
(ii) b2 = a2. Since Η Ο G, Ъ^аЪ € Я. As α is of order 4, so is Ъ~1аЪ. Thus b~yab = a
or ая. If b~lab = a, then ab = ba. But every element of G can be written as a4> or a{ for
some integer i, since G = HuHb. Hence ab = ba implies G is abelian, contrary to our
assumption. Thus b~lab = a3 or
ab = Ьая (5.9)

Sec 5.31
DIRECT PRODUCTS AND GROUPS OF LOW ORDER
151
Since G = Hl)Hb, the elements of G can be expressed as 1, a, a2, a3, b, ab, a?b, аяЬ. If а
group of this type actually exists, then equation (5.9), b2 = a2 and a4 = 1 provide us with
enough information to construct its multiplication table.
1 α α2 a-ч b ab аЧ a»b
1
a
a2
a»
&
ab
аЧ
аЧ
a
a2
«3
1
аЧ
b
ab
a?b
a2
a-ч
1
a
a26
аяЪ
b
ab
ая
1
a
a2
a&
a*6
a-46
b
b
ab
аЧ
аЧ
a2
a-ч
1
a
ab
a4
аЧ
Ь
a
a?
ая
1
аЧ
аЧ
Ъ
ab
1
a
a2
a*
0-46
b
ab
a26
a'
1
a
a2
Table 5.1
To calculate the products in the table, we used the fact that ab = ba3 and b2 = a2
imply ba = аяЬ since α3δ = a2(baa) = Ь3ая = Ца2а3) = δα.
If G is another non-abelian group of order 8 with an element й of order 4 and an element
Ъ g gp{a) such that δ2 = δ2, then as in our argument above,
G — {ϊ, ά, δ2, ά-\ δ, α2δ, ά6, abs}
with the elements satisfying the equations
u* = 1, δ2 = δ2, δδ = δά»
from which we find a multiplication table for G which is identical to Table 5.1 except that
a is substituted for a and Ь for b. The mapping a:G^G defined by aia^d and
a: alb-* δ'δ, г = 0,1,2,3, is clearly an isomorphism.
Table 5.1 also shows that a group of order 8 of this type actually does exist, for the
table defines a group. To see this, notice that the product of any two elements is again an
element, i.e. the table defines a binary operation, 1 is an identity element, and every element
has an inverse. The only difficulty is checking that the binary operation is associative.
This involves much calculation (the reader should check some of the calculations himself).
We shall give another description of this group in Problem 5.40. This group is called the
Quaternion group and has the interesting property that all its subgroups are normal and
yet it itself is not abelian (Problem 5.43).
We now move on to a discussion of (iv).
(ίν)δ2 = 1. Let K = gV{b); then НПК={1) and G = HK. Now Η < G so that
b~iab £ Hand, since a is of order 4, we have b~lab = а огая. As in (ii), b~*a,b - a implies
G is abelian. Hence b~1ab = ая, which leads to
ba - a*b (5.10)
The elements 1, а, а2, ая, b, ab, a?b, аяЬ are the distinct elements of G. Equation (5.10),
b2 = 1 and o* = 1 enable us to construct the following multiplication table.

152
FINITE GROUPS
[CHAP. 5
1 а а? а* b ab а?Ъ cflb
1
а
a.2
a3
Ь
ab
a?b
а?Ъ
a
β»
as
1
аЗЬ
b
ab
o^b
a*
»з
1
a
az6
af>b
b
ab
a»
1
a
a«
ab
a.2b
a,4
b
b
ab
a4
a»*
1
a
a?
a?
ab
a4
ФЪ
Ь
a?
1
a
a?
аЧ
аЧ
Ъ
ab
a2
»з
1
a
аЧ
Ъ
ab
аЧ
a
a?
a*
1
Table 5.2
As in part (ii), any non-abelian group of order 8 having property (iv) is isomorphic to G.
Such a group exists, for Table 5.2 also defines a group. This group is isomorphic to the
dihedral group Z>4, the group of symmetries of a square (see Problem 5.38).
The two groups given in Tables 5.1 and 5.2 are not isomorphic because the group of
Table 5.1 has exactly one element a2 of order 2, whereas the group of Table 5.2 has five
elements of order 2: a2, b, ab, a?b and а?Ъ. Thus we have shown that there are exactly two
non-isomorphic non-abelian groups of order 8.
To summarize, there are five non-isomorphic groups of order 8, three abelian and two
non-abelian.
Since 9 = 32, we know by Problem 5.36 that there are two and only two novAsotnorphic
groups of order 9, namely C9 and Ся X Cs.
d. Groups of small order: orders 12 and 15
To complete our list of all groups up to order 15, we must find all possible groups of
order 12 and IS. Because 12 = 3 · 22, we know that a group G of order 12 has at least
one Sylow 2-subgroup of order 22 and at least one Sylow 3-subgroup of order 3. The third
Sylow theorem tells us that the number of Sylow 2-subgroups is congruent to one modulo 2
(i.e. sz = 1 + 2k for some integer A;) and s2 divides |G|. When A; = 0, «2=1; and when
к = 1, s2 = 3. If к > 1 it is clear that 1 + 2fe does not divide 12. We therefore have two
possibilities: G has exactly one Sylow 2-subgroup or G has exactly three Sylow 2-subgroups.
A similar argument shows that G has exactly one Sylow 3-subgroup or G has exactly four
Sylow 3-subgroups. Therefore we have four possibilities:
(i) s% - l and sa = 1
(ii) s2 = 1 and s3 = 4
(iii) Sz = 3 and s3 = 1
(iv) Sz = 3 and s3 = 4
Notice that because the Sylow 2-subgroup has order 4 it must be isomorphic to & or
Kb and the Sylow 3-subgroup must be isomorphic to Cs (Section 5.3b). We treat each case
separately.

Sec. 5.3]
DIRECT PRODUCTS AND GROUPS OF LOW ORDER
153
(i) Let F be the Sylow 2-subgroup and Τ the Sylow 3-subgroup of G. Then F < G and
Τ <J G, since a Sylow p-subgroup is a normal subgroup if it is unique (Problem 5.7,
page 133). Furthermore, F(~)T = {1} since any element in the intersection must have
order dividing 3 and 4 and so must be the identity. Moreover, |F| |T| = 12. Hence by
Theorem 5.16', G = FxT. We have two possibilities for F: (a) F = d or (b) F = Ki.
Thus G s* d x Ся = C12 (by Problem 5.34) or G ^ Κι Χ Cs", both these groups are
abelian.
These are the only abelian groups of order 12, for if G is abelian any two conjugate
subgroups are equal. Hence by Theorem 5.3, page 131, s2 = s8 = 1.
In cases (ii) through (iv) we assume G is a non-abelian group. Furthermore, let F be
any Sylow 2-subgroup of G and Τ any Sylow 3-subgroup of G. Then FHT - {1} and, by
Proposition 5.18, page 145, \FT\ = \F\ \T\I\FnT\ = \F\ \T\ = \G\. Thus G = FT. If ft = tf
for all f G F and * G T, then G is abelian since gu ff2 G G implies 9i = /i*i and s^ = /2ta
for some /b/2 G i1 and *i, *2 G T. Now as i1 and Τ are both abelian groups,
gi9z = /1*1/2*2 = /2*2/1*1 = ff2ffi
Hence we also assume if F is any Sylow 2-subgroup of G and Τ any Sylow 3-subgroup of G,
that it is not the case that /* = */ for all / G F and * G T.
(ii) si = 1 and s3 = 4. Let i1 be the (unique) Sylow 2-subgroup and Γ be a Sylow
3-subgroup. There are two possibilities: (a) F s= d and (£>) F = Я4. We treat each case
separately.
(a) Let F= {1,а,а2,ая} where a4 = 1, and T={l,b,b2} where &" = 1. s2 = 1
implies F <J G. Thus b-iabGF, If b_1ab = a, then ab = ba. But this implies
every element of F commutes with every element of T, contrary to our assumption.
Therefore since a has order 4, b~xab Φ a2 and £>-1a£> Φ 1 so that b~lab = ая or
ab = Ьа3. We show that under these assumptions gp(ba) = G. (ba)2 = b(ab)a =
t»2fti - f,2_ gQ (j,ftyi _ (ba)2ba ■= b2ba = a. Hence gp{ba) contains a and £> and thus
coincides with G. Then G is cyclic, contrary to assumption. There is therefore
no non-abelian group of order 12 with s2 = 1, s3 = 4 and Sylow 2-subgroup
isomorphic to Ci.
(b) Let F= {l,x,y,z} be the four group as given in Section 5.3b, and T= {l,c, c2}
where c3 = 1. As in part (a), F<lG so that c~xfc G F for all / G F. Now by
assumption c_1/c Φ f for at least one / G F. Suppose с_1жс ^ ж (the other cases
are similar). We may assume c~xzc = y. (The other case, c~xxc =■ z, is argued
similarly.) Let us, as in Section 5.3b, put χ = а, у = b and ζ = at». Then ac = cb,
which implies a = cbc~1. Now с~хЪсФа, for c~^c = a implies c-1£>c = cbc-1
or b = (^Ьс 2. Then, as c2 = c_1 and c~2 = c, b = c~xbc. Hence a = b, a
contradiction. Similarly c'bc^b and c_1frc τ4 1. Then c-1bc = ab and c-1(ab)c =
(c_1ac)(c_1bc) = bab = b2a = a. Consequently the equations
ac = cb, be = cab, abc = ca, a2 = b2 = 1, cs =1, ab = ba (5-ίί)
hold in G. Now 1, c, c2 determine distinct cosets of F in G. Therefore the elements
of G are
1, c, c2, a, b, ab, ca, cb, cab, c2a, c2b, c2ab
Using equations {5.11), we can write down the multiplication table for G, as shown
in Table 5.3 below.

154
FINITE GROUPS
[CHAP. 5
1
e
e2
a
b
ah
ca
cb
cab
c*a
c4
c2ab
1
1
e
С2
a
b
ab
ca
cb
cab
e2a
c4
e2ab
e
e
с2
1
cb
cab
ea
<£6
c2ai»
e2a
b
ab
a
e2
e2
1
e
e2ab
c2a
c2b
ab
a
b
cab
ea
eb
a
a
ca
e2a
1
ab
b
e
cab
eb
c2
e2ab
сЧ
b
b
eb
сЧ
ab
1
a
cab
с
ea
c2ab
e2
c2a
a&
ab
cab
c2ab
&
a
1
cb
ca
с
e26
e2a
c2
ca
ca
e2a
a
cab
cb
с
c2ab
c2b
c2
ab
b
1
cb
cb
e26
b
с
ca
cab
c2
e2a
c*ab
1
a
ab
cab
cab
e2ab
ab
ca
с
cb
e2a
c2
сЧ
a
1
6
c2a
(Pa
a
CO,
сЧ
c2
e%b
6
1
ab
cb
с
cab
сЧ
сЧ
b
cb
c2a
c2ab
c2
a
ab
1
ca
cab
с
e2ab
c2ab
ab
cab
c2
e2b
c2a
1
b
a
с
cb
ca
Table 5.3
By a similar argument to that used in the discussion of the non-abelian groups of
order 8, any group of order 12 with s2 = 1 and s3 =■ 3 is isomorphic to G. Moreover,
Table 5.3 defines a group: the table defines a binary operation, the identity is 1, and
every element clearly has an inverse. The associativity of the operation must also be
checked, an even more tedious task than in the case of a group of order 8. The
alternating group A4 is a group of this type (Problem 5.38).
(iii) s2 = 3 and s3=l. Let F be a Sylow 2-subgroup and T= {l,c,c2} (c3 = 1) be the
Sylow 3-subgroup. Again we have two possibilities: (a) F = {l,a,a2,a3} & С4 and
(b) F={l,x,y,z) ^K*.
(a) T, being a unique Sylow 3-subgroup, is normal in G and so а~хса& Т. We may
assume a~lca¥*c, otherwise the group is abelian. Hence a~lca=c2 and ca = ac2.
Also, c?a = cac2 — ас4 = ac. The equations which determine a multiplication table
for this group are
ca = ac2, c2a - ac, c3 = 1, a4 = 1 {5.12)
The distinct elements of G are then
1, a, a2, a3, c, c2, ac, a2c, a?c, ac2, a2c2, asc2
and we obtain Table 5.4 below.
We conclude, by an argument similar to that used in the discussion of the non-
abelian groups of order 8, that any group of order 12 with s2 = 3, s3 = 1 and in
which the Sylow 2-subgroups are cyclic of order 4, is isomorphic to the group G
defined in the table. Again it can be checked that the table defines a group, so that
a group of this type exists. We have as yet not encountered an example of this
type of group, but in Problem 5.41 we show that there is a group of 2 by 2 matrices
which is isomorphic to G.

Sec. 5.3] DIRECT PRODUCTS AND GROUPS OF LOW ORDER 155
1 α a2 a3 с с2 ас a2c asc ас2 a2c2 asc2
1
a
a2
a3
с
с2
ас
а2 с
asc
ас2
а.2с2
asc2
а
а*
а3
1
ac%
ас
a2p2
asc2
с2
агс
а?с
с
а2
а3
1
а
а2с
а2с2
Фс
с
ас
а3 с2
с2
ас2
а?
1
а
а2
а?<&
asc
с2
ас2
а2с2
с
ас
а2с
с
ас
а2с
asc
с2
1
ас2
а2с2
аЧ2
а
а2
а3
с2
ас2
а2 с2
asc2
1
с
а
а2
а3
ас
а2с
а3е
ас
а2с
а3с
с
а
ас2
а2
а3
1
и2с2
а3с2
с2
а2 с
а3с
с
ас
а2с2
а2
а3с2
с2
ас2
а3
1
а
asc
с
ас
а2с
а3
а3е2
1
а
а2
с2
ас2
а2с2
ас2
а2с2
а3с2
с2
ас
а
а2с
а3 с
с
а2
а3
1
а2с2
а3с2
с2
etc2
а2
а2с
а3
1
а
а3с
с
ас
а3 с2
с2
ас2
а2са
а3с
а3
с
ас
а2с
1
а
а2
Table 5.4
(Ь) F= {l,x,y,z} and Γ^μ,ο,ρ3}. Since Τ < G, we have /-»c/e Γ for all / e F.
By assumption, for at least one / €.F, f xcf Φ с. We may therefore assume,
without loss of generality, that χ~xcx = c2. Again, as in Section 5.3b, put χ = а, у = b
and ζ = ab. Then ca = ac2. Note that c2a = c(ca) = c(ac2) = (ca)c2 = ac2c2 — ас,
i.e. c2a — ac.
We claim that S= {l,c,c2,a,ca,c2a} is a subgroup. We leave to the reader
the task of checking that the product of any two elements in S is again in S
(ca = ac2 and c2a = ac make this task easy). The identity 1 is in S, and on
inspection we find every element in S has an inverse in S: с"1 = с2, (с2)-1 = с, а-1 = a,
(ca)"1 ~ ca, (с2а)~* = с?а. Hence S is a subgroup of G. \S\ = 6 for c{a = с3' (г = 1
or 2; and j = 0, 1 or 2) implies neT, a contradiction; cla = a (i = 1 or 2) implies
c{ = 1, a contradiction; and ca = c2a implies с = 1, a contradiction. S is
non-abelian (ac = c2a is not equal to ca) and hence is isomorphic to Ds since
there is only one non-abelian group of order 6 (up to isomorphism). Now
[G:S]=2 implies S < G (by Problem 4.69, page 116). Hence blcbGS.
As b~*cb is an element of order 3, it is either с or c2, as all other elements of
S are of order 2. We shall now choose an element h G F, h& S, such that
h^ch = с If 6-^6 = c, let Л- = b. If on the other hand bxcb = c2, let h = ab.
Recall that α_1οα = с2. Then (abj-^iab) = Ъ~^(а^са)Ъ = Ь~^Ъ - b-4b-b~4b =
с2· с2 = с. Hence there exists an element fegS in F (i.e. b or ab) such that u_1cfr = c.
Consider Η = gp(h). Clearly £пЯ={1}, S and Я commute elementwise, and
|S| \H\ = |G|, and so G = S χ Я by Theorem 5.16. But S ^ Z)3 and Я = С2. We
therefore conclude that any group G with S2 = 3, s3 = 1 and Sylow 2-subgroup
isomorphic to Kt, is isomorphic to Da χ C2. The dihedral group De is a group of this
type (Problem 5.38).
(iv) s2 = 3, S3 = 4. Since distinct cyclic groups of order 3 intersect in the identity element,
the four Sylow 3-subgroups have together 9 distinct elements. A Sylow 2-subgroup
is of order 4. Since the intersection of a group of order 4 and a group of order 3 can
only be the identity, it follows that the number of distinct elements in the 4 3-SyIow
subgroups and a single 2-SyIow subgroup is 12.
But \G\ = 12, so there cannot be another distinct Sylow 2-subgroup. Thus there
is no group of type (iv).

156
FINITE GROUPS
[CHAP. 5
To summarize, we have shown that there are up to isomorphism exactly three non-abelian
groups of order 12. They are the groups
(ii) (b) with s2 = 1, s3 = 4 and the Sylow 2-subgroup ^ Ki\ see Table 5.3. (Such a group
is isomorphic to A4.)
(iii) (a) with s2 = 3, s3 = 1 and the Sylow 2-subgroup «* d; see Table 5.4. (Such a group
is isomorphic to a group of 2 by 2 matrices given in Problem 5.41.)
(b) with s2 = 3, s3 = 1 and the Sylow 2-subgroup ss Kt. (Such a group is isomorphic
to Ds x C2 which is isomorphic to D«. See Problem 5.38.)
Clearly no two of these groups are isomorphic. The abelian groups of order 12 are Κα χ Ca
and Си. Thus including the abelian groups, there are five non4somorphic groups of order 12.
If G is of order 15, we have seen that G is cyclic (Section 5.1a). The following table
gives the number of non-isomorphic groups of order 1 through 15.
Order of group
No. of groups
1
1
2
1
3
1
4
2
5
1
6
2
7
1
8
5
9
2
10
2
11
1
12
5
13
1
14
2
15
1
The reader will agree that finding all groups of a given order is difficult. Indeed there
is not even a general method of determining how many non-isomorphic groups of a given
order there can be.
Problems
5.38. Show that
(i) The dihedral group D4 is a group of order 8 isomorphic to the group given in Table 5.2,
page 152.
(ii) The alternating group A4 is isomorphic to the group given in Table 5.3.
(iii) The dihedral group De is isomorphic to Ds X C2.
Solution:
(i) DA is a non-abelian group of order 8 and as such is isomorphic to one of the groups given in
Section 5.3c, Tables 5.1 and 5.2. A check of Table 5.1 shows that there is only one element of
order 2, namely аг. But D4 is the symmetry group of the square (Section 3.4f, page 75). A
reflection τ is of order 2 and if σ is a rotation of 90°, τσ is of order 2 as can be seen in the
discussion of these groups in Chapter 3. Therefore Z?4 must be isomorphic to the group given
in Table 5.2.
(ii) A4 is a non-abelian group of order 12. Hence it is either isomorphic to the group of Table 5.3
or 5.4 or to Ds X C2 (see page 155). As can be seen from the multiplication table for A4 given
in Chapter 3, page 63, A4 has exactly three elements of order 2, namely σ2, σ5 and σ8. Now the
group of Table 5.4 has only one element a2 of order 2, so that it could not be isomorphic to A,,.
We have shown in Problem 5.1, page 131, that A4 has no subgroup of order 6 and Ds is
isomorphic to a subgroup of D3 X C2 by Theorem 5.14. Hence A4 is not isomorphic to DSX C2.
Thus it must be isomorphic to the group given in Table 5.3.
(iii) D6 is of order 12, and is not abelian. D6 is the symmetry group of the hexagon and therefore
has a subgroup of order 6, namely the rotation of 60° about the center. So it is not the case
that DB sa A4, and hence D& cannot be isomorphic to the group of Table 5.3. Also, a reflection
followed by a rotation is an element of order 2. Since there are six such elements in D$, it
cannot be isomorphic to the group of Table 5.4 as this group has only one element of order 2.
The only other possibility is that D& ss Ds X C2.
5.39. Find a cyclic subgroup of order 6 in D3 X C%.
Solution:
Let a. G Ds be of order 3 and b {¥= 1) G C2. Consider the element (a, b) G Ds X C2. (а, Ь)г =
(β.2,1), (a,6)3 = (1,6), fob)* = (a)1)( (o,)b)5 = (a2rb)! and (o,jb)e = (1д). Hence flfp((o,6)) is a cyclic
subgroup of D3 X C2 of order 6.

Sec. 5.3]
DIRECT PRODUCTS AND GROUPS OF LOW ORDER
157
/0 ΐ\ / 0 1\ ,—
5,40. Consider the matrices A = I . 1 and В = [ where г = ν-1 · A and В have nonzero
determinants and thus are elements in the group of all 2X2 matrices with nonzero determinants.
Show that gp(A,B) is a group of order 8 which is isomorphic to the quaternion group (Table 5.1,
page 151).
Solution:
By direct calculation we find
A2 = { * " ) A& = ( _. * ) A* = f " " ) = i, (identity matrix)
в* = f . Г ) ab = (~l 0Ni aw = (, Л Am =
-1
0
-1
0
0
-1
0
-1
1
0
Ό
1
0
1
-1
0
0 il V1 °/ V° ~~~%
B2A = As, BSA = AB and AW = BA
Let G = {I,A*,A3,B,AB,AW,A3B}. We claim that G = {I,A,A*,A3,B,AB,AW,A3B} =
gp(A,B). Clearly, G Q gp{A,B). To show G = gp(A,B), we need only show G is a group, as
А, В € G. Note G is a subset of the group of 2 X 2 matrices with nonzero determinant. Hence the
elements of G satisfy the associative law. By direct calculation we can show that G is closed under
matrix multiplication; the equation BA — A3B simplifies the calculations, e.g.,
(AW)(A3B) = A*(BA)A*B = A*AWAW = AASBAB = AW2 = A
Furthermore every element of G has an inverse in G, e.g. using B3A = AB we have
(AW)-1 = B~lA-3 = B3A = AB
Checking all these details enables us to conclude that G is a group of order 8 and G = gp(A,B).
G is non-abelian, since AB Φ ΒΑ, so G is either isomorphic to the group of Table 5.1 or Table 5.2.
Because G has only one element of order 2, it cannot be isomorphic to the group of Table 5.2. Thus
G is isomorphic to the quaternion group of Table 5.1.
/0 г"\ /e 0\ ,—
5.41. Consider the matrices A = I . I and В = I 1, where г = v~l and e is a nonreal
complex cube root of 3 (so, in particular, e3 = 1 and t^l). A and В have nonzero determinants and
thus are elements in the group of all 2X2 matrices with nonzero determinants (see Section 3.5b,
page 81). Show that gp(A,B) is a group of order 12 which is isomorphic to the group given in
Table 5.4, page 155.
Solution:
We find by direct calculation that
A2 = (~ _?') B* = l'_ ' ) AW
a* = f_°. ~lQ\ в3 = ι: :) азв =
A* = f1 p AB = ( : " } AB2 =
Let H= {A,AZ,A*,A*,B,BZ,AB,A2B,AW,AB2,AW2,A3BZ}. We claim H = gp(A,B). Clearly
Η Qgp(A,B). To prove Η = gp(A,B) we need only show Я is a group, as Α,Β&Η. Note that
Я is a subset of the 2X2 matrices -with nonzero determinant. Hence the elements of Η satisfy the
associative law. To check that Η is closed under matrix multiplication, first note that A~lBA — B2.
Then, for example,
(AW)(AW) = A*A · (A~WA)AW = AW*AW = A4A-WA)(A~WA)AB
= BW2AB = BAB = AA-WAB = ABW = A
Also the inverse of, for example, (A3B) is given by
(A3B)~l = ВЫ = АА-1ВЫ = A(A~WA)(A~WA) = ABW* = AB <= Η
1
0
0
ίε
0\
ι)
ге2
0
—e
0
0
—г'е
0
te2
-ie2\
ie\
о
AW2 =
A3B2 =
a -:)
/ 0 -u
\-it? 0

158
FINITE GROUPS
[CHAP. 5
Checking all these details enables us to conclude that Η is a group of order 12 and ff = gp{A,B).
The mapping a.; a,-* A, and с -* В is an isomorphism of the group of Table 5.4 and H. This
obtains because Η satisfies the equations
BA = АВг, ВгА = АВ, Ва = I, A* = /
(/ the identity matrix). These are the exact counterparts of equations (5.12), page 154.
Consequently the multiplication table for Η is obtained from that for the group of Table 5.4 by renaming
via a.
5.42. Show that a group G of order 48 has a normal subgroup ¥■ {1} or G. (Very difficult.)
Solution:
By the first Sylow theorem, G has a Sylow 2-subgroup of order 16. By the third Sylow theorem,
s2 = 1 + fc2 for some integer к and s21 48. The only odd divisor of 48 is 3, hence s2 = 1 or 3. If
s2 = 1, then the Sylow 2-subgroup is unique and therefore normal (Problem 5.7, page 133). Suppose
s2 = 3. Let Η and К be two of the Sylow 2-subgroups. As Я ПК is a proper subgroup of H,
\HC\K\ 1 16. Then ]ffnK| =8; for if \HnK\ ^ 4, then, by Proposition 5.18, page 145, \HK\ =
|ff| \K\/\HnK\ ^ 16 X 16/4 = 64, which contradicts our assumption that \G\ = 48. Since both ff
and К are of order 16, НпйГ, as a subgroup of index 2, is normal in both Η and К (Problem 4.69,
page 116). Hence HcNG(Hr\K) and Kc2VG(ffnK). Letting N = JVG(ffnK), we have HKqN.
Thus |N| ^ \HK\ = \H\ \K\f\HC\K\ = 32. As \N\ divides 48 and [N[ ^ 32, \N\ = 48 and so N = G.
Because a group is normal in its normalizer, we have HoK <t G.
5.43. Show that all subgroups of the quaternion gToup G are normal subgroups. (G is given in Table 5.1,
page 151.)
Solution:
It is sufficient to check that the cyclic groups are normal in G. For if S is any subgroup,
sGS, and x€G, then χ 'sa; G gp(s) implies x~1sx€S.
Using the multiplication table we can check that for χ — a or 6 and any s G S, χ '1вх G gp(s).
(We leave this check to the reader.)
This is sufficient to prove the result, for every element of G is of the form af6 or a' for
i = 0,1, 2,3.
5.4 SOLVABLE GROUPS
a. Definition of solvable groups
To introduce our concepts we will begin with an example. If Ρ is a group of order pr
where ρ is a prime, then we showed that Ρ has a series of subgroups Pf with
{1} = Po С Ρ, С ■ ■ ■ С Pr = Ρ (5.13)
where each Pf < Pi+i and [Pi+i: Pi] — ρ for each integer i = 0, .. .,r— 1 (see equation
(5Л), page 139).
Let G be a group and suppose it has a series of subgroups
{1} = Go Q Gi Q ■ · · С Gr = G (5.U)
If each Gi < Gi+i for ΐ = 1,.. .,r — 1, then (5.Ц) is called a subnormal series of (for) G.
With this definition, (5.13) is a subnormal series of P.
If (5.1Jf) is a subnormal series for G and [Gi+i: Gi] is some prime (dependent on i), for
г = 0, ..., r — 1, G is called a solvable group and (5.14) is called a solvable series for G.
Accordingly we conclude that Ρ is solvable and that (5.13) is a solvable series for P.
If (5.1Jf) is a subnormal series for G and Gi+i/Gi is simple, i.e. Gi+i/G( has no normal
subgroups other than Gi+ i/Gt and the identity, for г = 0, ..., r — 1, then (5.14) is said to be a
composition series for G. To see that (5.i#) is a composition series for P, note that Pi+i/Pi
is a cyclic group of order ρ and hence is simple. We call the factor groups GJGi-n of the
subnormal series (5.14) the factors of (5-Z-4).

Sec. 5.4]
SOLVABLE GROUPS
159
We shall discuss composition series in greater detail in Section 5.5. We remark that
not all groups have a composition series but finite groups do (Section 5.5a). Our main
concern in this section is the concept of solvable group.
Historically, solvable groups arose in the attempt to find a formula for the roots of an
wth degree polynomial
f(x) — anxn + On-i*"-1 4- ■ · ■ + atx + a0 (5.15)
in terms of the coefficients сц. The formula sought was one which involved the coefficients
do,..., On of the polynomial, integers, and the operations addition, subtraction,
multiplication and division, and a finite number of extraction of roots. For example, if η = 2, then
χ — 1 is a formula giving the roots of (5.15). If the roots of f(x) can be
obtained by such a formula, we say f(x) — 0 is solvable by radicals.
From the Fundamental Theorem of Algebra we know that an nth degree polynomial
with complex coefficients has η complex roots. Let F be the "smallest" field (see Section
3.6b, page 86, for a definition of field) of complex numbers containing the coefficients ш of
f(x). By saying F is the smallest field we mean that if Η is a field containing the coefficients
(h, then FcH. Let Ε be the smallest field containing F and the roots of f(x). Now the
set of automorphisms of Ε forms a group under the composition of mappings (see Theorem
3.15, page 87). The automorphisms of Ε which map every element f G F onto itself is a
subgroup g of the group of all automorphisms of E. The group Q is called the Galois group
of the polynomial f(x). In the beginning of the 19th century, the French mathematician
E. Galois proved (essentially) the following extraordinary theorem: An equation f(x) = 0
is solvable by radicals if and only if the Galois group of f(x) is solvable. It turns out that
not all equations of degree -5 are solvable by radicals because the symmetric group Sn
is not solvable for η — 5. (For details see Birkhoff and MacLane, A Survey of Modern
Algebra, Macmillan, 1953.) In Section 5.5e we will prove that Sn is not solvable.
Problems
5.44. Show that the symmetric group Sn is solvable for η = 1, 2,3.
Solution:
Sj = {ι}; then Sj has the solvable series {ι} CSt and is therefore solvable.
Ί 2\ /1 2'
Sz ~ '\1 2/'\2 l
then {ι} CS2 is a solvable series for Sz and so Sz is solvable.
Let 53 = {ι, σν σ2, rif r^, r3} where
Ί 2 3\ /1 2 3\ /12 3
12 3/ °2 ~ [s 1 2/ Тг " \3 2 1
1 2 3\ /1 2 3\ /12 3
2 3 \j Tl \1 3 2/ Тз \2 I 3
Now Η — {ι, σ1( σ2} is a cyclic snbgroup of 53. Also, [S3: Щ = 2. Hence by Problem 4.69, page
116, Η <] S3. Thus {ι} С Η QS3 is a solvable series for S3, since [H: {t}] = 3 and [S3: H] = 2,
and so S3 is solvable.
5.45. Show that 54 is solvable.
Solution:
The alternating group AA is a subgroup of order 12 in S4. Then [St: A4] = 2 and A4 <| 54
by Problem 4.69, page 116. We have seen in Problem 5.1, page 131, that A4 has no subgroup of order
6. Now A4 is a group with a unique Sylow 2-subgroup F of order 4 and F =s K4, the four group
(see Problem 5.38(ii), page 156, and Section 5.3d, page 153). Since F is a unique Sylow 2-subgroup,
F <] A4 and [A4: F] - 3. F, being a four group, has a normal subgroup К of order 2. Accordingly,

160
FINITE GROUPS
[CHAP. 5
{1} с К с F с А^ с S4
is a subnormal series for S4. As [ЙГ: {1}] = 2, [F: if] = 2, [A4 : F] = 3 and [S4 : A J = 2, S4 is a
solvable group.
b. Properties of, and alternative definition for, solvable groups
An important property of solvable groups is given in
Theorem 5.20: If G is a finite group and N < G is such that N and G/N are solvable
groups, then G is also solvable.
Proof: Let {N} С Ht С Я2 С · · · С Нк - G/N be a subnormal series for G/N with
[Я4+1: Я4) = Pi, pt a prime. By the correspondence theorem (Theorem 4.19, page 120, and
Corollaries 4.20 and 4.21), there are subgroups Η in G such that Hi <1 Hi + i, HJN — Si and
[Ht+1: H] = [#i+1: #{] = pi (« = 0,1,..., к - 1). Therefore
N = Ho Q Ht Q ·■■ Q Hk = G {5,16)
is a series of subgroups of G with Яе <1 Я4+1 and [Hi+ι: Hi] - p;. Now N is also a solvable
group. Hence N has a series
{1} = Яо с id с Я2 с -■- с Κι = 2V (5.1 Г)
where [Ki+i: Z4] is a prime number (г = 1,2,.. .,1-1). Putting the series {5.16) and {5.17)
together, we obtain
{1} = Яо С Xi С ■ ■ ■ e Χι С #i С Я2 С · ■ · С Як = G
which is a solvable series for G. The proof is complete.
Note. In contrast to Theorem 5.20, it is not always true that a group G has a property
if both a normal subgroup N and G/N have the property; for example, the four group Кц
has a normal cyclic subgroup N of order 2 which is cyclic and KJN is cyclic, but X* itself
is not cyclic.
Corollary 5.21: If G is a finite abelian group, G is solvable.
Proof: We use induction on the order η of G. If |G| = 2, the result holds trivially.
Assume that any abelian group of order less then η is solvable. Suppose ρ | η for some
prime p. Then by Proposition 5.9, page 137, G has an element of order p. Let a be such an
element. If ρ т6 те, then \gp{a)\ < \G\ so that gp{a) is solvable by the induction assumption.
Furthermore, since G is abelian, gp{a) is a normal subgroup of G and |G/gp(a)| < \G[.
Hence G/gp{a) is solvable by our induction assumption, and so, by Theorem 5.20, G is
solvable. If ρ = те, then {1} с G is a solvable series and G is therefore solvable in this
case too.
The following theorem leads to an alternative definition of solvability.
Theorem 522: G is a solvable group if and only if G is finite and has a subnormal series
{1} = Ко С Кг С · · - С Кп = G {5.18)
where Ki+JKi is abelian {i— 0,1, .. .,те—1).
Proof: Let G be a solvable group. Then G has a subnormal series with factors of
prime order and hence cyclic. Since a cyclic group is abelian, the solvable series of G is a
series of type {5.18). Conversely, assume G has a subnormal series {5.18) with Ki+JKi
abelian. We prove that G is solvable by induction on те, the length of the subnormal series
{5.18). If η = 1, then G is abelian since G ^ KJKo which is abelian by assumption.
Hence by Corollary 5.21, G is solvable. Assume that any finite group which has a subnormal
series of length less than те in which the factor groups of consecutive terms of the series are
abelian, is solvable. Let G have subnormal series {5.18) of length те. Then Kn-t has a
subnormal series of length те — 1, namely

Sec. 5.4]
SOLVABLE GROUPS
161
Ко С Κι С - · - С Кп-2. С Кп~\
with Ki+t/K, abelian for i - О,1, ,. .,η — 2. Hence by the induction assumption, Kn\ is
solvable. But G/Kn-x is abelian and hence solvable. Using Theorem 5.20, we conclude G
is solvable.
In the theory of infinite groups one usually defines a group G to be solvable if it has a
subnormal series (5.IS) with Κί + ί/Κι an abelian group (г = 0,1, ..., η — 1). By Theorem 5.22
this is equivalent to our original definition for finite groups. Since this formulation of
solvability is more general, we shall henceforth use it as our definition of solvability.
Note that the infinite cyclic group is an example of a group that does not fit the old
definition but does fit the new.
Using this new definition we prove
Theorem 5.23: Let G be a solvable group. Then (i) any subgroup of G is solvable and
(ii) if N < G then G/N is solvable.
Proof:
(i) Let {1} = HoQHiQ· ■ · QHn~G be a subnormal series of G with Η,+ι/Hi abelian
for г = 0,1, . . ., η — 1. We show that if К is a subgroup of G,
{1} = {KnH0) С {К П Ht) С ■-- С (К П Ни) = К (5.19)
is a subnormal series with KnHi+1/KnHi abelian. First we notice that Kf\Hi —
(KnHi+t)nHi (i = 0, 1, ...,n-l) and that Kf\Hn = K. Now Яi < Я.+ ι, and
KdHi+i is a subgroup of Hi+u Applying the subgroup isomorphism theorem (Theorem
4.23, page 125) inside the group Hi+i with subgroup KnHi+l and normal subgroup
Hi, we obtain
{Kf\Hi+x)nHi < Kf\Hi+x
and {KnHi+l)/{{KnHi + i)r\Hi) « ((КГ\Нг + 1)Щ/Нг
Since {KnHi+l)nHi = KnHu it follows that KnHi < KnHi + i and
{ΚηΗί + ί)/{ΚηΗ>) « {(КПН1+1)Щ/Нг
But (Xnfli+i)fliCff)+i, so that we have
(JJl (Ί Hj + i)Hj Hi+t
m Q m
Now Hi+i/Hi is abelian by assumption and hence so is (ΚηΗι+ι)Ητ/Ηι. Therefore (5.19)
Κ Γ\ Η
is a subnormal series for К with abelian factors „ _ „ ' and consequently ii is
solvable.
KnHi
(ii) Let G have subnormal series
{1} = H0 С Я! С Я2 С - ■ - Q Нп - G
where Hi + i/Hi is abelian. Now N <l G. Consider the natural homomorphism
ν : G -> G/N
Any subgroup of G is mapped by υ onto a subgroup of G/N. In particular let #{ = H.v.
We assert Я4 «d Hl+U But this follows from Problem 4.82, page 122 (with θ = v1Hl+1
and Я^1 — G).
Next we assert that Hi+i/Hi is abelian. Let x-Xv,y = yv (x,y GHi+i) be two
elements of Яг+i. Then since Hi + t/Hi is abelian, xy = yxd where d G Hi. Thus
(a?i/)v = (a;v)(i/v) = (i/v)(.Tv)(dv)

162
FINITE GROUPS
[CHAP. 5
But dv G Hi. Consequently
xHiyHi ■= xyHi - (xy)vHi = (yv){xv)dvHi - (yv)(xv)Hi == yHixili
Therefore Я{+,/Я{ is abelian. Thus
{N} = Ho < St < ■ · ■ < Bn = G/N
is a subnormal series of G/N with abelian factors, and so G/N is solvable.
Now that we have a definition of solvable group that applies to infinite groups, we will
extend Theorem 5,20 to infinite groups.
Theorem 5.24: If N < G and G/N and N are solvable groups, then so is G.
Proof: Let {N} = Я0 CHj С Я2 С · · · С Нк - G/N be a subnormal series for G/N in
which Hi+i/Hi is abelian, г = 0,..., к — 1.
By Corollary 4.21, page 121, there are subgroups Hi in G such that Hi < Я4+1
and HJN =^Ht (i = 0,.. .,k). By the factor of a factor theorem (Theorem 4.22, page
121), Hi + JHt = (Ht+i/N)/{HJN) = Hi+JHu Hence the factors Ht + jHt are abelian. Also,
N has a series
{1} = Ко С Kt С K2 С · ■ ■ С Κι = Ν
with Zi + JК abelian for г = 0,1, ..., Ι— 1. Therefore
{1} = Ко С Κι С - ■ · с Κι = JV - Я0 С Яг с - · ■ С Нк = G
is a subnormal series whose factors are abelian. Thus G is solvable.
Problems
5.46. Show that all groups G of order p2, pq or ρ2ς>, where ρ and ς> are distinct primes, are solvable.
(Hard.)
Solution:
If |G?| = p2» then G is abelian (Problem 5.19, page 140) and G is solvable.
If \G\ —pq then from Problem 5.8, page 133, if ρ < q, G has one and only one subgroup Η
of order q. By Problem 5.7, page 133, Η < G. Now \G/H\ = p, hence G/H is abelian. As Η is of
order q, it is abelian. Therefore we have the subnormal series
{1} С Η ς G
with abelian factors and so G is solvable.
If |6?| = p2q then ep = 1 + kp divides p2q, and so the prime factors of 1 + kp must be ρ or q.
Clearly ρ does not divide 1 + kp. Therefore 1 + kp = 1 or q. If 1 + kp = 1, then the Sylow p-sub-
group Η is normal in G (Problem 5.7). As Η is of order ρ2, Η is abelian (Problem 5.19). Thus we
have a subnormal series , . _, _
{1} С Η С G
with G/H abelian (\GIH\ = ς-) and H/{1} abelian. Hence G is solvable.
Suppose, however, that 1 + kp = q; then q > p. Let ίΓ be a Sylow g-subgroup of G. The
number of such Sylow g-subgroups is 1 + Iq. Again 1 + Iq is not divisible by q, and so 1 + Iq — 1, ρ or pz.
But as q > p, the only possibilities are 1 + Iq = 1 or p2.
Case (i): 1 + Iq = 1. In this case there is only one Sylow g-aubgroup К and (by Problem 5.7)
К О G. |K| = q and |G/K| = p2. Hence Я" is abelian and G/K is abelian (Problem 5.19), and it
follows that G is solvable.
Case (ii): l + lq = p2. We will show that this case does not arise by showing that G would
contain too many elements. We have assumed that G has q Sylow p-subgroups (of order p1) and
p2 Sylow g-subgroups (of order q). Any two distinct subgroups of order q intersect in the identity,
so there are pHq — 1) = p2q — pz distinct elements of order q in G. Also, G has at least 2 Sylow
p-subgroups and hence there are at least p2 distinct elements in G of order ρ or p2. In the above
calculations we have not counted the identity, so in all G has at least p2q — p2 + p2 + 1 = p*q + 1
elements, which is absurd, and we conclude that case (ii) does not arise.

Sec. 5.5]
COMPOSITION SERIES AND SIMPLE GROUPS
163
5.47. Show that every nilpotent group is solvable.
Solution:
Consider the upper central series
{1} = Z0 с Ζγ с Z2 с ·■· с Zn - G
of G, a given nilpotent group. Zl+l is defined by the fact that ZiJrXIZi is the center of G/Z, and
of course this implies Zi + 1/Zi is abelian. Hence G is solvable.
5.48. Prove that the converse of Problem 5.47 is false, i.e. not all solvable groups are nilpotent.
Solution:
The symmetric group S3 is solvable (Problem 5.44). A check of the multiplication table for S3
on page 57 shows that the center of S3 is just the identity {ι}. But this implies that the upper
central series for G never ascends to G. Thus S3 is not nilpotent.
5.49. Let G be any group and let G(i> be defined for all positive integers i by G(1) = G', the derived
group of G, and G»+1> = (G(i>)'. Prove that G is solvable if and only if G("> = {1} for some
integer n.
Solution:
Let G<»> = {1}. Then
{1} = g<»> с ··■ с G«> с G
is a subnormal series for G and G(i)/G(1+1) is abelian. Hence G is solvable.
Now let G be solvable. Then there exists a subnormal series
{1} = Hr С ·■■ с Н0 = G
with Hi/Hi + l abelian.
By Problem 4.68, page 116, #,/Hi+1 abelian implies Hi+12H·. We prove, by induction on i,
that Я4э G«>, i — 1,2, . ,.,r. For i = l this is true since Ηχ 2 НО = G' = Ga>. Suppose our
assertion is true for г = n, i.e. H„2 G<B>. Then H^2 (G<n>)' = G(«+»>. But ЯП+12*С so
H„.+ 12GC"+i). Therefore H^G™ for all г. In particular then, {1} = Hr3 G(r). Accordingly,
G<» = {1} and the result follows.
5.5 COMPOSITION SERIES AND SIMPLE GROUPS
a. The Jordan-Holder Theorem
In Section 5.4a we introduced the idea of a composition series for a finite group G.
We recall that a series of subgroups of G
{1} = Go С Gi С · ■ - С Gu = G (5.20)
is a composition series for G if Gf < Gi+i for г = 0,1, .. .,k-l and Gi+i/Gi is simple, i.e.
has precisely two different normal subgroups. This latter statement carries with it the
implication that Gi Φ Gi+i for г = 0, . .., к — 1.
We observe first that every finite group G has a composition series. The easiest way to
see this is by induction on the order, \G\, of G. If |G| = 1, then G has precisely one
composition series:
{1} = Go = G
Suppose then that \G\ ¥* 1 and that every group of order less than \G\ has a composition
series. Now if G is simple, then
{1} = Go С Gi = G
is the only composition series for G. If G is not simple, let N be a normal subgroup of G,
N ¥> {1}, Ν Φ G. We may suppose that N is the largest normal subgroup of G, that is, if
Μ < G and Μ Φ G, then \M\ ^ \N\. By induction, N has a composition series
{1} = No С Νι С --· С Νι - N

164
FINITE GROUPS
[CHAP. 5
We claim that
{1} = No с Νι Q ■·■ С Νι С G
is a composition series for G and note that to prove this we need only show that G/Nt is
simple. But if G/Νι is not simple, it has a non-trivial normal subgroup. By the corollary
to the correspondence theorem this subgroup is of the form KIN ι where К is a normal
subgroup of G. But as К D Νι, this means that \K\ > \Ni\ which contradicts the choice
of Nl Therefore every finite group has a composition series.
This proof does not suggest that if a group has two composition series then they are
related. Surprisingly they are. In order to explain this relationship we associate with the
composition series (5.20) two notions. First we term к the length of the series. Second
we call the factor groups Gi+i/Gi the composition factors of the series (5.20). The
relationship between composition series is given by
Theorem 5.25 (Jordan-Holder): Every finite group G has at least one composition series.
The lengths of all composition series for G are equal.
Finally if
{1} = Go С ··· С Gk = G
and {1} = Но С ■ · · С Нк = G
are a pair of composition series for G, then their
respective composition factors can be paired off in such a way
that paired factors are isomorphic.
We have already proved the first statement of Theorem 5.25. Before illustrating
Theorem 5.25 we restate its last assertion as follows; There is a permutation π of {1, .... k)
such that
Gi + i/Gi — fl'ci + Dir/fl'a + Dir-l
for г = 0, ..., к — 1.
Example 1: Suppose that S3 is the symmetric group on {1,2,3}. Then the series
Ί 2 3\ /1 2 3\ /1 2 3
Д 2 3/' V2 3 1/' U 1 2
m с HL „ J, (о о J, L . Jl· с ss
is a composition series for Ss. Notice that the composition factors are of orders
3 and 2. This is actually the only composition series for S3, since
Ί 2 3\ /12 3\ /1 2 34
,1 2 3/' U 3 l)' V3 1 2/
is the only normal subgroup of S3 which is neither Ss nor the identity subgroup.
Problem
5.50. Let и be a positive integer. What relevance does the Jordan-Holder theorem have to the factoriza- ·
tion of η into a product of primes? (Hint; Let G be the additive group of integers modulo и
(η > 1).)
Solution:
Let {1} = G0 С G, С ■·■ С G, = G
be a composition series for G. Each composition factor Gi+1/Gt (i — 0, ..., I — 1) is simple. As
G( is abelian, each factor βι + 1/β{ is abelian. Hence if Gi+i/G4 has any proper subgroup, it would
not be simple. So Gi + 1/Gi has no proper subgroups. In particular it has no cyclic proper subgroups,
so it must be cyclic of order a prime. The number I is therefore the total number of primes (allowing
for repetitions) dividing n. By Theorem 5.25, I is uniquely determined. So, as we well know, the
total number of prime divisors of и is a constant. Moreover, the uniqueness of the composition
factors (asserted in Theorem 5.25) simply means that these prime divisors themselves are unique.
Putting these two facts together gives the well-known fact that every integer η > 1 can be written
uniquely as a product of primes, if the order in which it is written is disregarded.

Sec. 5.5]
COMPOSITION SERIES AND SIMPLE GROUPS
165
Example 2: Let S„ be the symmetric group on η letters, и — 5. Then
{1} с Ап С Sn
is a composition series for S„. For we shall show that An is simple (in Section 5.5e,
Theorem 5.34). But An <\ Sn and S,JAn is cyclic of order two. Hence this series
is indeed a composition series.
b. Proof of Jordan-Holder theorem
Suppose G is a finite group and suppose
1 = Go С Gi С · · · С Gk = G {5.21)
and 1 = Но с Hi С ■ · · С Hi = G {5.22)
are two composition series for G. We have to prove that к = I and that the composition
factors Gi+i/Gi of {5.21) can be paired off with the composition factors Hi+i/Hi of {5.22) so
that paired composition factors are isomorphic.
The proof is by induction on the order jGj of G. If \G\ = 1, then both assertions are
clear. Thus we assume that \G\ > 1 and that the theorem holds for all groups of order less
than jGj. It is useful to observe that if к — 1, then G is simple. Hence l~l also, and
again the desired conclusion holds. So we assume, in addition to \G\ > 1, that к > 1 (and
hence I > 1).
There are two cases to consider: Gk-i = Я1-1 and Gk-i ¥= Ηι-ι.
Case 1: Gk-i ~— Я1-1.
It is then clear that 1 = Go С ■ · · С Gk-t {5.23)
and 1 = Но С ■ · ■ С Hi-y ~ Gfc-i {5M)
are composition series for Gk-i. But jGfc-ij < \G\. Hence by our induction assumption,
the composition series {5.23) and {5.2U) have the same length, i.e. k — l — l — 1, and so
к = I. Furthermore the composition factors of {5.23) can be paired with the composition
factors of {5.2k) so that paired factors are isomorphic. But the composition factors of
{5.21) are those of {5.23) together with G/Gk-i, Similarly the composition factors of {5.22)
are those of {5.2k) together with G/Gk-ι- Thus it is clear then that the composition factors
of {5.21) can be paired off with the composition factors of {5.22) so that paired factors are
isomorphic. This concludes the proof of Case 1.
Case 2: Gk-i¥-Hi-i.
Our method of proof is to produce a composition series, {5.26), which has isomorphic
composition factors to those of {5.21) (by Case 1) and a composition series, {5.27), which has
isomorphic composition factors to those of {5.22) (by Case 1). We will then show that
{5.26) and {5.27) have isomorphic factors and this will be sufficient to prove the result.
First we will show that Gk-iHi-i = G. Observe that both Gfc-i and H1-1 are normal
subgroups of G, and so Gk-iHi-i is also a normal subgroup of G. Obviously Gk-ιΗι-ι
contains Gk-i properly, so Gh-tHi-i/Gk-i is a non-trivial normal subgroup of G/Gk-ι by the
correspondence theorem. But G/Gk-i is simple, so Gk-iHi-i/Gk-i — G/Gu-i. This means
that
Gk-ιΗι-χ = G {5.25)
We now put F - Gfc-tnffi-i and note that F <3 G. Let
{1} = Fo С ■ ■ · С Fm = F
be a composition series for F. Then we claim that

166
FINITE GROUPS
[CHAP. 5
{1} = П С · ■ · С Fm С Gk-ι С G {5.26)
and {1} = F0 с · · · с Fm с Hi-! с G (5.27)
are both composition series for G. The only facts that need be verified are that Gk-i/Fm and
Hi-i/Fm are simple. Now Fm = Gk-inHi-г and by (5.25), G = Gk-i#i-i. Therefore by
the subgroup isomorphism theorem (Theorem 4.23, page 125),
G/Gk-i = Gk-iHi-JGb-i « #!-!/(№-!П Gfc-i) = Hi-i/Fm {5.28)
and similarly
G/Ht-y - Gk^H^/Hi-t » Gfc-i/(Hi-inGk-i) = Gk-i/Fm {5.29)
Since both G/Gk-i and G/Hi-t are simple, it follows that Hi-t/Fm and Gk-i/Fm are also both
simple.
Let us compare the composition series (5.£i) and (5.2ff). By Case 1 it follows that they
have the same length and their composition factors can be paired off so that paired factors
are isomorphic. Similarly for the composition series {5.22) and {5.27). Let us now compare
the composition series {5.26) and {5.27). They obviously have the same length, m + 2. Thus
the series {5.21) and {5.22) have the same length. What are the composition factors of the
series {5.26) and {5.27)1 The composition factors of {5.26) are
Fi/Fo, ..., Fm/Fm-l, Gk-l/Fm, G/Gk-l
while those of {5.27) are
Fi/Fo, ..., Fm/Fm-l, Hl-i/Fm, G/Hl-i
Now by {5.28), Hi-i/Fm « G/Gk-ι, and by {5.29), Gk-i/Fm = G/Hi-t. Thus the composition
factors of {5.26) and {5.27) can be paired off so that paired factors are isomorphic. It
follows that the composition factors of {5.21) and {5.22) can be paired off so that paired factors
are isomorphic, since the factors of {5.21) can be so paired off with those of {5.26) and the
factors of {5.22) can similarly be paired off with those of {5.27). This completes the proof
of the Jordan-Holder theorem.
From the Jordan-Holder theorem we know that the length of a composition series and
its factors are uniquely determined for the group. This suggests the following scheme for
studying groups which have a composition series. First, find the structure of all groups
with composition series of length 1. These are all the simple groups. Assuming now that
we know all about groups with a composition series of length n, let G be a group with
composition series of length η + 1. Let
{1} = Go с Gi с ··· с Gn с Gn+i = G
be a composition series for G. If F is a group with a normal subgroup N and F/N ξ* Η,
we say that F is an extension of N by H. In this language then, G is an extension of a
group with a composition series of length n, viz. Gn, by a simple group. Hence what we
must know is how the structure of a group which is an extension of one group by another
is determined.
In the next few sections we shall prove that if η > 5, A„ is simple. The groups A„ are
not all the simple groups and indeed there is no classification of simple groups as yet.
This is one of the basic problems of finite group theory.
The question of how a group G is built from Η and К if G is an extension of Η by K, is
considered in Chapter 7. Here too our knowledge is far from complete.
Problem
5.51. If two groups have the same composition factors, are they isomorphic?
Solution:
No. There is a cyclic group G of order 6 and a non-abelian group Η of order 6. They have the
same composition factors, but they are not isomorphic.

Sec. 5.5]
COMPOSITION SERIES AND SIMPLE GROUPS
167
c. Cycles and products of cycles
We begin with an example of a new notation. Let us consider S6. Let θ S Se be
defined by
12 3 4 5 6
1 2 5 6 4 3;
Note that the effect of θ is to take 3->5, 5->4, 4->6 and 6 ■+ 3, and to leave the elements
1, 2 unaltered, θ is called a cycle. We denote it by (3,5,4, 6).
More generally, consider Sn. If ai,.,.,Om are distinct integers in {1,2, ...,«},
(dj, as, .. ., Om) stands for the permutation that maps each integer in {1,2, .,.,«} —
{αϊ, .. .,Om} to itself, and maps
<Ij -* 0,2, djt ->■ Из, .... <Xm-j -* Om, Om "* <Xl
We call such a permutation a cycle of length m. As a convention take a cycle of length 1 to
denote the identity element. A cycle of length 2 is called a transposition.
The inverse of the cycle a = (au ...,α™) is the cycle β = (α™,Om-i, ■ ■ .,m,αϊ), since
α.{αβ) = {α.α)β = α.+ ιβ = α, if iV m
while α^αβ = (αηα)β = αβ = am
If /G {1,2, ...,η}-{α,, .. .,am},
?(«/?) = (/«)/? = Λ8 = j
Hence αβ = t. Similarly /?a = ι.
It is clear that not every permutation is a cycle. Nor is the product of two cycles
/1 2 4 4 \
necessarily a cycle; for example, in Si, (1,2)(3,4) = L 1 . „ ] which is not a cycle.
An obvious question is: can we express every element of Sn as a product of cycles? The
following theorem answers this question.
Theorem 5.26: Every element of Sn can be written as the product of disjoint cycles.
(Cycles (α1( ...,α™) and (bj, .. .,b%) are disjoint if the Of and b-, are distinct,
i.e. if {αϊ,.. .,Omj П [Ьъ .. .,bk) = 0.)
Proof: Let ж G S„. We say that г G {1, ..., n} is fixed by π if ы — г, and we say it is
moved if ы Φ г.
We shall argue by induction on the number of integers moved by the permutation ж.
If τ moves none of the integers {1,2, ...,«}, then ж is the identity permutation. But then
x = (1), as the 1-cycles are all the identity permutation.
Hence we have a basis for induction. So let ж Φ t and suppose that every element of
S» which moves fewer integers than ж, can be written as the product of disjoint cycles.
Xow suppose a1 G {1, ...,«} and that агжФал. Let us define α2,α3, ... by a2 = a^,
«, = a2ir, ..,, as = as_^, .... Let m be the first integer such that атж = α, for some
integer г with 1 — i < m. (As the images of ж belong to {1,2, ...,«}, the terms of the
sequence a,v a2,... cannot all be different.) We shall prove, using the minimality of m,
tkat i=l. Suppose to the contrary that гф\. Then а. = а._гж and атж — a. = α^π.
As - is a one-to-one mapping, ai_1 = am. But this contradicts the choice of the integer m.
Hence г = 1 and атж — av We consider now the cycle (at, ...,ftj and note that m> 1.
Let
τ — (αϊ, .. .,Om) 'ϊ

168
FINITE GROUPS
[CHAP, 5
τ is a permutation that leaves a.,.,.,am fixed, since a(a,,...,a )"гж = a. ,n = a. for
i¥=l, while at{av .. .,aJ~V = <xmir = ar Also if j e {1,2, .. .,w}, then jT ^ j implies
jV Φ j; for if j e {<V . · ■, am}, jr = j. Hence J Й {a^ ..,, am} and so jV = /(<V .... aj"1"- =
jir ¥· j. It follows that г moves fewer integers than ж. Therefore inductively τ can be written
as the product of disjoint cycles, say
τ = τ, ■ · ■ rt
Now if τ, involves an al, af = a.T = a^r^ · ■ · τ,) = aSTr Hence τ} = (af) = t. Let
4t> \, ...,nk (ii < г2 < ■ ■ ■ < ite)
be those т. which do not involve any of the integers ait ..,, am. Clearly
T = Th Ti2 ' ■ " Пк 0Г Г = '
If τ = t, ir = (ap ..., am) and we have proved that ж is actually a cycle. Otherwise
ж = (av ..., am)Tii ■ ■ · rifc and π has been expressed as the product of disjoint cycles. Hence
the result.
Corollary 5.27: If i6St and ava2,...,am are chosen as in the proof of the theorem
(i.e. а2 = агж, ..., and атж = а1 and av .. .,am distinct), then
ж - (α,, ..., ajr
where aT = аж if α ё {a,,..., am}, while <у = a3. for j = 1,..., m.
Proof: This is precisely what we showed in the proof of Theorem 5.26.
This corollary provides us with a method of computing the decomposition of an element
ж G Sn into the product of disjoint cycles. For example, let us write
/123456789 10 11 12 \
π ~~ \S 4 2 1 8 7 9 11 12 10 5 6/
as the product of disjoint cycles. Since \ж Φ 1, we may take 1 for αι. Then αϊ = 1, α2 = 3,
aa = 2, at = 4, a5 = 1. So m = 4. Hence by the corollary,
,,, ,/l 2 3 4 5 6 7 8 9 10 11 12 \
ж - μ,ό,Δ,*)^ 2 3 4 g ? 911121o5 6/
Here the second factor τ on the right is obtained from ж by letting it leave 1,2,3,4 unchanged,
and letting it act as ж does on the remaining integers. Applying the same technique to τ,
we find r = (5,8,ll)(6,7,9,12). Hence
ж = (1,3,2,4)(5,8,11)(6,7,9,12)
In practice it is not necessary to use the corollary rigorously. We need only find the disjoint
cycles by choosing at such that a^ Φ av and then taking the cycle (alt ..., am) where
ai+1 = afir, amir = av and the аг,.. .,am are distinct. Then we choose 6j G {1,2, .. .,n) —
{at,.. .,am} where b^ ^ b1? and find the cycle (bv .. .,bk) with bl+l = bfir, bfcir = bt and
bv ..., bk distinct, and so on.
Not only is every element of Sn a product of cycles, but each element can also be
expressed as the product of transpositions, as can be seen from the following proposition.
Proposition 5.28: Every cycle is a product of transpositions.
Proof: We assert
(αϊ, α2,. .., ak) = (αϊ, α2)(αι, α3) - · ■ (аь а*)
For if α is an integer not in {ab ..., ak}, both the left-hand side and the right-hand side leave
it unchanged.

Sec. 5.5]
COMPOSITION SERIES AND SIMPLE GROUPS
169
If ii* k, then Oi(ai,.. .,ak) = cti+i while
tti(ai, θ2)(αι, aa) ■ ■ ■ (ai,ak) = αι{αλ, a,i)(ai, α,+ι) ■ · · (alt ak)
— ai+i(abdi+2) · ■ ■ (ai,a,k) = «ti + j
If г — к, then α*(αι, ..., ak) = αχ while
a,k(ai, аг) ■ ■ ■ (αϊ, ak) = α*(αι, аъ) = <ij
Thus the effect of the left-hand side and the right-hand side is the same, and эо the
permutations are equal.
Problems
5.52. In SB compute a ■- (1,2, 3,4)(2, 5) and β = (1,2,3,4) -Ц2, 5).
Solution:
la = (1(1,2,3,4))(2,5) = 2(2,5) = 5.
/1 2 3 4 5 6\
ia -- г + 1 if i = 2, 3; 4a = 1; 5a -- 2; 6a = 6. Hence a - ( , „ , , „ ) .
\^5 о 4 12 6/
(l,2,3,4)-i = (4,3,2,1). Then β = (4, 3,2,1)(2, 5).
Ί 2 3 4 5 6
1/3 =4, 2/3 = 1, 3β =5, 4β = 3, δ/} = 2, 6/8 = 6. Hence z3 = w 1 5 3 2 6
5.53. Express a and /3 of Problem 5.52 as products of disjoint cycles.
Solution:
We note that la = 5, 5a = 2, 2a = 3, 3a = 4, 4a = 1. Hence a = (1, 5, 2,3, 4).
Since 1β = 4, 4β - 3, 3/3 = 5 and 5β - 2, 2/} = 1, /3 = (1,4,3,5,2).
/1 2 3 4 5 6 7 8 9 10 11 12 13 14\
5.54. Write α = (^ 4 e g 10 12 14 ! 3 5 7 9 n 13J as the product of disjoint cycles.
Solution:
la = 2, 2a = 4, 4a = 8, 8a = 1. 3a = 6, 6a = 12, 12a = 9, 9a = 3. 5a = 10, 10a =5. 7a = 14,
14a = 13, 13a = 11, 11a = 7. Hence
a = (1,2,4,8)(3,6,12,9)(5,10)(7,14,13,11)
5.55. If α, β € Sn are such that icn^i implies both (ia)/3 = ia and г/3 = г, and ίβ ¥■ i implies
(ίβ)α = г/3, then α and /3 commute. Hence prove that disjoint cycles commute.
Solution:
Let i € {1,2, .. .,те}. If ia ¥= i, then i(a/3) = (ϊα)β ~ ia while i(/3a) = (i/3)a = ia. If ia = i,
then ia/3 = г/3 while
(i/3)a = i/3 if i/3 ■»* i
= ia if ίβ = г
— г
Hence a/3 = /3a. Now if a and /} are disjoint cycles, then ia ¥· i implies α moves i and ia and hence
/3 does not move either i or ia, and so i/3 ~ г and (i«)/3 = ia. Similarly if ίβ Φ % β moves i and
ίβ, and во (г/3)а = г/3. Thus a and /3 commute.
5.56. Express α of Problem 5.54 as the product of transpositions. Is the expression of an element of Sn
as a product of transpositions unique?
Solution: α = (1, 2,4,8)(3, β, 12,9)(б, 10)(7,14,13,11)
= (1,2)(1,4)(1,8)(8.6)(3,12X8,9)(5,10)(7,14)(7,13)(7,11)
As (1, 2)(1,2) = ι, if α = αχα^ ■ · ■ am is the expression of α as a product of transpositions,
then α = (αια2— а„)(1,2)(1,2) is another expression of а ав a product of transpositions. Thus
the expression as a product of transpositions is not unique.

170
FINITE GROUPS
[CHAP. Б
5.57. Find the order of the cycle (alt .. .,ej. (Hard.)
Solution:
If we write alr..,, a,m in a circle, as shown on the right, it is clear that
a = (a·!, ,. .,dj„) moves each t^ one place clockwise, a2 moves each a( two places
further, and in general a' moves each aj /-places further. Hence am moves each
ai m-places further, i.e. am moves each a{ back to itself. To put this more
formally, define am+l = alf am+2 = a2, ■■·. «m+m = am· We will prove by
induction on j that for 0— j — m, a1 maps щ to Oj+3-, i = 1, 2, ..., m. For
3 = 0, a* = ι and the result is true. If the result is true for j = r, consider j = r + 1 — m. Then
щат+1 = (щаг)а = о.(+тл. If г + r < ш, ai+ru = Oj+r+i by the action of a. If i + r = m·, ama =
% = o„s+1 = сц+г+1. If i + r>m, then ai+r = ai+r_m and г'-гг — w<w (as 1 — i — m and
»■ < m). Hence ai+ra = ai+r_ma = ai + r_m + 1 = ai+r+1.
In particular, am = ι. On the other hand, ατα* = α1 + ί ¥= ax for I — s < m. Therefore m is
the smallest power of a that yields the identity. Accordingly m is the order of a.
d. Transpositions, and even and odd permutations
In this section we will produce another way of deciding whether a permutation i9 even
or odd. (See the definition in Section 3.3b, page 60.) By Proposition 5.28 and Theorem
5.26 every permutation is the product of transpositions. However, this decomposition is
not unique (Problem 5.56). What we shall show is that a permutation π is a product of an
even number of transpositions if тг is even, and the product of an odd number of
transpositions if тг is odd.
Our first task is to show that any transposition is odd. We have already noted that
(1,2) is odd (Section 3.3d, page 64). We will use the following lemma.
Lemma 5.29: Let θ G Sn and let (au ..., am) be a cycle. Then
θ~1(αι, .. .,am)0 = {αιθ, .. .,атв)
Proof: Let χ G {1,2, .. .,n). If χ = сцв for some Oi,
χθ~ι{αι, .. .,ат)в - αιθθ~1{αι, .. .,ат)в
— α»(αι, ..., ат)в
= сц+ιθ if %Фт
-αιθ if г = m
If х^сцв for some at, χθ'1 € {аь ...,am]. Then
хв~1{аи .. .,ат)в = (хв~*)в = χ
Thus θ~ι{αι, .. .,α^θ = {αιθ,.. .,α™0).
Theorem 5.30: All transpositions are odd. If θ is an even permutation and is written as
the product of transpositions, the number of transpositions is even. If θ
is odd and is written as the product of transpositions, the number of
transpositions is odd.
Proof: Let (a, b) be any transposition. We know that (1,2) is odd. Let θ be any
permutation such that 10 = α, 2Θ = b. Then, by Lemma 5.29,
{a,b) = 0-1 (1,2)0
If 0 is even, 0"1 is even, 0~1(1>2) i3 odd, and so 0-1(l,2)0 is odd (Lemma 3.2, page 62). If
0 is odd, 0-1 is odd, 0_1(1,2) is even, and hence 0~1(1,2)0 is odd (again by Lemma 3.2).
Therefore (a, b) i9 odd.
Now let 0 be a permutation and let θ = al- ·■ am be the product of transpositions.
Then by Lemma 3.2, 0 is even if and only if m is even, while 0 is odd if and only if m is odd.
This proves the theorem.

See. 5^]
COMPOSITION SERIES AND SIMPLE GROUPS
171
Problems
5J>8. Determine whether a and β οι Problem 6.62 are odd or even, using Theorem 6.30.
Solution:
a = (1,5,2,3,4) ~ (1,5)(1, 2)(1,3)(1,4). Since a is expressed as the product of an even number
of transpositions, a is even.
β = (1,4,3,5, 2) = (1,4)(1, 3)(1,5)(1,2), and so β is even.
5.59. Determine whether a of Problem 5.54 is even by using Theorem 5.30.
Solution:
a = (1,2,4)8)(3,6,12;9)(5)10)(Y,14,13,11)
= {(1, 2)(1,4)(1, 8)}{(3,6)(3,12)(3,9)}{(5,10)}{(7, 14)(7, 13)(7, 11)}
Thus a is even.
5.60. Determine whether {av ..., am) is even or odd.
Solution:
(β,, .. .,0m) = (α,,βζΧ»!,^) · ■ · (auam), so (ab...,am) is the product of m — 1 transpositions.
Thus (av .. .,<*„,) is even or odd according as m is odd or even.
5.61. Let (aj, ...,Οη), (6„ ...,bm) be two cycles of S„. Prove that there exists »eS„ such that
«~4<h, --»em)9 = (bi, •■■,bm).
Solution:
Let 9 be defined by:
(1) ate = bj for i — 1, ..,, m;
(2) if {e„ ...,cn_m> = {1,2, . ..,n}-{au...,am} and if {du .. .,<f„_m} = {1,2, ...,«} -
{bi, .. ,,bm}, put ete = dt for i — 1, ...,« — m.
Then »eS„ and by Lemma 5.29, »_,(alF .. .,am)9 = (bi bm).
5.62. If G is a group, the set of all elements conjugate to a: is called the conjugaey class containing x.
Write down the conjugaey classes of S4 using cycle notation. (Hard.)
Solution:
The conjugaey class containing an element 9 is the set of all conjugates of 9. Our first task
is to express each element of 54 as the product of disjoint cycles, and then to take all distinct
conjugates of each element, We obtain the following classes:
(i) {(1)}. (As (l) = i, there are no other conjugates of (1)).
(ii) What conjugates of (1,2) are there? We know that в~1(1т2)в = (19,29). As 9 runs through
Sn, Is, 29 run through all distinct pairs (a, 6). As (a, b) = (bra), the conjugaey class
containing (1,2) is {(1> 2)> (1] sh (lj 4)> (2> 3)> (2j 4)i (3j 4)}
(Hi) What conjugates of (1,2,3) are there? As 9-»(l,2,3)9 = (19,29,39), we will get all possible
3-cycles. Hence the conjugaey class containing (1, 2,3) is
{(1, 2,3), (1, 2,4), (1,3, 2), (1,3,4), (1,4, 2), (1,4,3), (2,3,4), (2,4, 3)}
(iv) What is the conjugaey class containing (1,2)(3, 4)?
r'(l,2)(3,4)i = 9-1(1,2)β9-1(3,4)# = (i9)29)(39,49)
As 9 runs through Sn, we will obtain the product of all pairs of disjoint cycles. We recall that
disjoint cycles commute; for example, (1, 2)(3,4) = (3, 4)(1,2). Thus the distinct elements in
the required conjugaey class are
{(1,2)(3,4), (1,3)(2,4), (1,4)(2,3)>
(v) What is the conjugaey class containing (1,2,3,4)? Again we see that we shall obtain all
4-cycles. Hence the required conjugaey class will be
{(1, 2, 3, 4), (1,2,4, 3), (1, 3,2,4), (1, 3,4,2), (1,4, 2,3), (1, 4,3,2)}

172
FINITE GROUPS
[CHAP. 5
e. The simplicity of An, η - 5.
In this section we aim to prove that An is simple for η — 5. What we must do is to show
that if Я <\ An and Я ¥= {ή, then Я = An. Although the proof involves much calculation,
the ideas are easy.
Lemma 5.31: Every element of An is the product of 3-cycles, n ^ 5.
Proof: Every transposition (αϊ, α2) = [e, a.\)(e, а2)(е, αή where e, a\, a% are distinct. Now
every element of An is the product of an even number of transpositions, and therefore the
product of products of pairs of transpositions. So it is enough to prove that a product of
two transpositions is a product of 3-cycles. Consider (αϊ, α2)(ί>ι, ί>2). We may assume that
the four integers αι,α2, bi, bi are distinct, for otherwise (αϊ, a2)(&i,b2) is either the identity
or is itself a 3-cycle. As в^5, there exists an integer e such that 1 ^ e ^ n and
o>u аг, bi, b2, e are all distinct. Hence
(αϊ, a2)(bi, bz) = {e, ai)(e, a2)(e, a,)(e, bi)(e, b2)(e, bi)
= {(e,ai)(e,a2)}{(e,ai)(e,&i)}[(e, b2)(e, &!)} = (e,ai,a2){e,ai,bi){e,bz,bi)
Lemma 5.32: Let Я О An. If Я contains a 3-cycle, then Η - An.
Proof: Let (a,b,c) Ε Я. Then if x,y,z are distinct elements of {1,2, ...,%} and θ is an
element of Sn that sends a to ж, b to у and с to 2, we have θ ~1(a, b, c)9 = (ж, 2/, г) by Lemma 5.29,
page 170. If θ is even, (x, у, ζ) ЕЯ as Я О А„. If б is odd, then there exists e, / distinct
from a, b, с as n — 5. Therefore * = (e, /)0 is even.
*-1(а,Ъ,с)* = e-*(e,f){a,b,c){e,f)e
= в^(а,Ъ,с)в = (x.y.z)
Hence by normality, (x,y,z) e Я. Thus Я contains all 3-cyeles of Sn; and by Lemma 5.31,
H=An.
Lemma 5.33: Let Я < A„. If Я contains the product of two disjoint transpositions, then
H = An.
Proof: Let a = (at,a2)(as,ai) € H. Since n^5, there exists е€{1, ...,те} distinct
from ab α2,α», α*. Let θ — (α,ι,αζ, e); then 9eA„.
fl-'afl = (аг, е)(а3, а4)
α-ΐ0-ια0 = (αι,α2)(α3,α4)(α2,β)(σ.3,α4)
= (a,, a2)(a2, e) = (ai, е,аг)
Thus Я contains a 3-cycle, and the result follows from Lemma 5.32.
Theorem 5.34: An is simple for n ^ 5.
Proof: Let Я < An, Η ¥- {ι}. Then there exists a E Я, α ^ t. Let а = «,-■· ak where
«lf . . .,ak are disjoint cycles. We may suppose without loss of generality that at, .. .,«fc
are arranged so that the length of a{ — length of «i + 1 for г = 1, ..., к — 1, as the «,,..., ak
commute by Problem 5.55, page 169.
Case 1:
Suppose «, = (ap .. ,,am) with m > 3. Let σ = (alf a2,a^. Clearly aeAt, Also σ
commutes with av...,ak, as they move different integers. As σ-1 Ε An and ff< An,
β = a_1ff~W Ε Я. Note that
σ~ιασ = (α2, α3, a,, a4,.. .,ат)а2 ■ ■ · ak
and SO β = аъ1 " ' «["4«2'α3'αΐ'α4' · · -'атК " " ак
= «ГЧ^^а,.^ •■■»aJ
= (аж, аж_,, · · ·, а,Ж α.» αΐ'а*> ■ ■ ■' О
= (а„а2,а4)
Since /ϊ ЕЯ, the result follows from Lemma 5.32.

Sec. 5.5]
COMPOSITION SERIES AND SIMPLE GROUPS
173
Case 2:
Suppose m = 3, and «2 is also a 3-cycle. Let «t = (av a2, a3) and as — (a4, as, ae). Let
σ — (ctj, α3, α4) £ An. Then Η contains
a~W = (σ-1α1σ)(σ-1α2σ) ■ ■ * (σ~1αΗσ) — (a~1ala)(a~1oc2a)oc3oci * * ' «fc
Thus Я contains
α^'σ^'ασ — αΓ^α,Γ1 · · * <χ~1(σ~~1α1σ)(σ~1οίζσ)αΆ " " " «fc
= (α8'αί'αι)(αβ'ανα4)(αΐ'α8'α*)(αί'αβ'αβ) = (avai'a2'aZ'as)
and the result follows from Сазе 1.
Case 3:
Suppose то = 3 and a2,..,,ak are transpositions. Let аг = [μνα2,ct3). Then
= (^,α,,α^α*···»» = Κ,α2,α3)2 = (ava3,a2)
and the result follows from Lemma 5.32.
Case 4:
Suppose all the «, are of length 2. Then m is even, and ax = (ava2), α2 = (α3,α4). Put
a — (ct2, a3, ctj. Then Η contains
σ~γασ = (t^, α3)((Χ4, α2)α3α4 ■ · · «fc
Thus Η contains σ~1ασα~1 = (αχ, α4)(α2, α3)
and the result follows by Lemma 5.33.
Corollary 5,35: The symmetric group Sn is not solvable for η — 5.
Proof; An < Sn and {t} С Ап С Sn is a composition series of Sn for η — 5 (Example 2,
page 165). By the Jordan-H6Ider theorem this is, up to isomorphism, the only possible
composition series. Now \An\—n\/2, which is even for η ^4. Hence [An: {t}] is not a prime.
But if any subnormal series of Sn with factors of prime order existed, it would be a
composition series. Therefore Sn is not solvable.
Problems
5.63. Prove that if G = An, η — 5, then the derived group G' of G is G.
Solution:
We know that G' < G. Hence G' - G or else G' = {<} as G is simple by Theorem 5.34.
If G' — {(}, G is abelian. But An is not abelian for η — 5; for example,
/1 2 3 4 5 6 ... n\
(1,2,3X3,4,5) = (2 4 j в , β ... я)
/1 2 3 4 5 6 ... n\
bUt (3.4,5)(1,2,3) = (2 3 4 5 1 6 ... η)
Therefore G' = G.
5.64. If G =An and и ^ 5, prove that Z{G) = {ι}.
Solution:
As Z(G) < G, Z(G) = G or Z(G) = {<}. As Z(G) is abelian but G is not (see Problem 5.63),

174
FINITE GROUPS
[CHAP. 5
5.65. Without using Theorem 5.34, prove that G' — G = An for η — 5, where G = Sn.
Solution:
We use Lemma 5.33. Let α = (2, 3,4), σ = (1,2,3). β = a-»ff-W = (1,4)(2,3) € G'. Hence
as G' < G, the result follows from Lemma 5.33.
5.66. Are any of AvA2,A3lAi simple?
Solution :
Both Aj and A2 are of order 1, so they are not simple. A3 is of order 3, so A3 is cyclic of prime
order and therefore simple since groups of prime order are simple. Finally A4 is of order 12. We
have already seen that a group of order 12 is solvable. Indeed we saw in Problem 5.45 that its
composition factors have orders 3,2, 2 respectively. Hence A4 is not simple.
5.67. Prove that Sn has no non-trivial element in its center for ιι-З. (Hard.)
Solution:
Let a £ Sn, α Φ ι. Let a. = α.λ · · · a.k be the decomposition of a. into disjoint non-identity cycles.
Let <*! = (au .. ., am). If m — 3, let β == (ab, a2). Then a-1/Sa = (ага, a^a) == (a2, a3) Φ β. Hence
a & Z(Sn). If m == 2, let β = (щ, a2, a3). Then a-1/Sa = (dja, a^, asa) = (a2, dj, b) where Ъ = a3a
and b is an integer different from a! and a2. No matter what Ъ is, α.~ιβα. Φ β since α2/δ = α3 but
α2α-,/δα = оц. Hence α β Z(Sn). Thus no non-trivial element of S„ belongs to £(Sn).
5.68. Prove that A„, Sn and {«} are the only normal subgroups of S„ for и — 5. (Hard.)
Solution:
Let Η < S„. Then Hn4„ < A„. Hence AnnH = An or else Anr\H — {ι}, as A„ is simple
by Theorem 5.34. If Anr\H = A„, then A„ С H. If Η ^ AnJ as A„ is of index 2 in Sn, Η = S„.
If Л„пЯ = {ι}, suppose Η ^ {ι}. If jeff and β ^ ι, 9 is odd. As 92 is even, 92 £ ffnA,
and so 02 = ι. Let τ £ Η, τ э6 «. Then τ is odd, and as вт is even, τ == 0-1 = в. Hence Η consists
of only two elements, « and в. As Sn has no center (Problem 5.67), there exists μ £ Sn such that
μ~ϊθμ,Φθ. But (i_1S(i e Я, as Я< Sn; and since Η = {ι, 9}, this is impossible. This contradicts
the assumption that Η Φ {ι}.
5.69. Prove that S^ = Απ for и - 5.
Solution:
As iS„/A„ is abelian (cyclic of order 2 in fact), An^S^ (Problem 4.68(iii), page 116). As S„ is
notabelian, S'n Φ {ι}. But S^ < S„. Hence S^ = A„ by Problem 5.68. Alternatively, S^DA^=An
by Problem 5.63 or 5.65.
A look back at Chapter 5.
In this chapter we proved the three Sylow theorems. The first gives the existence of
Sylow p-subgroups; the second states that a subgroup of prime power order is a subset of
one of the Sylow p-subgroups; and the third states that all the Sylow p-subgroups are
conjugate.
The proofs of the Sylow theorems used a standard technique of finite group theory:
induction on the order of the group. It is also worth noting our counting arguments, e.g.
in the proof of the class equation.
Using the class equation we proved that a group of prime power order has a non-trivial
center. Then we proved as a consequence that a group of order pr (where ρ is a prime)
has a normal subgroup of order pr_1. By repeatedly taking the center of factor groups, we
defined the upper central series of a group. A group is nilpotent if a term of the upper
central series is the group itself.
Next we gave a method of constructing a group from the cartesian product of two given
groups Η and K. This group had isomorphic copies H,K of Η and К respectively which
satisfied HK — G and Η ПК— {1}. Reversing this analysis we showed that if a group
G had normal subgroups Я and К with HK = G and HDK-{1}, then G^HxK. Using
this result and the Sylow theorems, we classified groups of orders 1,2, ...,15 up to
isomorphism.

CHAP. 5]
SUPPLEMENTAEY PEOBLEMS
175
Then we defined solvable groups, so called because they led to a criterion for the
solvability of equations. We noted that our first definition involving a subnormal series with
factors of prime order did not extend to infinite groups. So we chose a criterion involving
subnormal series with abelian factors for our definition of solvability. We showed that
subgroups and factor groups of solvable groups were solvable, and an extension of a solvable
group by a solvable group was solvable.
We next considered composition series (subnormal series with simple factors) and
proved that every finite group has a composition series. In the Jordan-Holder theorem we
showed that a composition series has a unique length and unique factors up to isomorphism.
In our final section we proved that the groups A„ for η ^ 5 are simple. To do this we
needed to express permutations as products of disjoint cycles. This led to a method of
determining whether a permutation was even or odd. As a consequence of the fact that
A„ is simple, we concluded that Sn is not solvable for η ^ 5.
Supplementary Problems
SYLOW THEOREMS
5.70. Prove that the Sylow 17-subgroup is normal in a group of order 255 == 3 · 5«17.
5.71. Prove that the Sylow 13-subgroup is normal in a group of order 2 · б · 13.
5.72. Let A = {0,1} be the set of integers modulo 2, В = {0,1, ..., 64} be the set of integers modulo 65,
and G be the set of all pairs (a, b), a € Л and b £ B. Define the multiplication
(au bJiav, b2) = (a, + a^, (-l)"*^ + b2)
for (oj, b{), (a2, 62) e G. Prove that G is a group of order 2 * 5 * 13 with respect to this multiplication.
Is a Sylow 2-subgroup normal?
5.73. Verify the class equation (i.e. equation (5.2), page 135) for the group in Problem 5.72.
5.74. Prove that the Sylow p-subgroup is always normal in a group of order 4p, where ρ is a prime — 5.
Is this true when ρ = 3?
5.75. Prove that the normalizer of a Sylow p-subgroup coincides with its own normalizer,
5.76. Let \G\-n and \H\ = m where Я is a subgroup of G. If Hfxg-^Hg = {1} for all g&G — H,
then there are precisely n/m — 1 elements in G which are not in any conjugate of H. (Hint:
Examine Na(H).)
THEORY OF p-GROUPS
5.77. Show that every subgroup of index ρ in a finite p-group is a normal subgroup.
5.78. Let G be a finite p-group and Η be a proper subgroup of G. Show that NG(H) ¥· H.
5.79. Prove that if G is nilpotent and GIG' is cyclic, then G' = {1}. ((?' is the derived group of G).
5.80. Use Problem 5.79 to show that elements of co-prime order commute in a nilpotent group.
5.81. Find an example of a group G with a normal subgroup W such that G/N and -W are nilpotent but
G is not.

176
FINITE GROUPS
[CHAP. 5
5.82. Let G(rj = G and G(i+1) = gp({[g, x\ \ j£G(i) and χ £ G}). The sequence of subgroups
GfnD G(2;3 ■ ·'Э G(03 ■ ■· is called the lower central series of G. Prove that a group G is
nilpotent if and only if G(n) = {1} for some positive integer n.
5.83, Find the lower central series for D2* f°r positive integers n. (See Problem 5,82 for the definition of
lower central series.)
DIRECT PRODUCTS AND GROUPS OF LOW ORDER
5.84 Prove that a finite nilpotent group is isomorphic to a direct product of any one of its Sylow p-sub-
groups and some other subgroup,
5.85. Employ the results of Problem 5.70 to prove that a group of order 255 is isomorphic to a direct
product of its Sylow 17-subgroup and another of its subgroups. Thereby show that a group of order
255 is cyclic.
5.86. Show that the direct product of two nilpotent groups is a nilpotent group.
5.87. Let G = fifpff n ) » ( ι n ) )' Show that |G| = 8. Which group of order 8 is isomorphic
5.88. Show that Dn is isomorphic to G = gp(( n) > ( n ^i ) ) where
—- fi2Tri/n
5.89. Suppose G is a finite group with all its Sylow p-subgroups normal. Show that G is nilpotent.
SOLVABLE GROUPS
5.90. Prove that a group of order pqr is solvable when p, q, r are primes and r > pq.
5.91. Show that the direct product of two solvable groups is solvable,
5.92. Prove that a group of order 4ρ, ρ a prime, is solvable,
5.93. In Problem 5.49, page 163, we showed that a group G is solvable if and only if G<n) = {1} for some
integer n. Use this fact to give alternate proofs of Theorems 5.23, page 161, and 5.24, page 162.
5.94. Show that Dn is solvable for all positive integers w.
COMPOSITION SERIES AND SIMPLE GROUPS
5.95. Find the composition factors of a finite p-group.
5.96. Show that there is no simple group of order prm, where ρ is a prime and m < p.
5.97. Prove Sn is not solvable for η — 5, without using the fact that An is simple. (Hint. Consider
all 3-eycles of Sn.)
5.98. Find a composition series for the quaternion group.
5.99.
Show that, except for groups of prime order, there are no simple groups of order < 60, (Hint. A
difficult case occurs for order 36, See Problem 5,43, page 158.)

Chapter 6
Abelian Groups
Preview of Chapter 6
A group G with binary operation · is abelian if, for all g,h€.G, g'h-h-g.
It 19 customary to иэе " 4- " for the binary operation in abelian groups. We will begin
this chapter with a preliminary section in which we restate some of our results and
definitions in additive notation.
One of the concepts which we will reformulate additively and generalize is the concept
of "direct product" considered in Chapter 5. In abelian groups it is customary to talk about
"direct sum" instead of "direct product". We note that the direct sum of abelian groups is
again abelian.
We call direct sums of infinite cyclic groups free abelian groups. Direct sums satisfy
an important homomorphism property which gives rise to the important fact that every
abelian group is a homomorphic image of a free abelian group.
We then consider classifying abelian groups according to the orders of their elements.
An abelian group G which has every element of finite order can be expressed as a direct
sum of p-groups, i.e. groups every element of which is of order a power of the prime p.
An important p-group that we shall introduce here is the p-Priifer group.
At this point we have three types of abelian groups:
(a) the cyclic groups, (b) the additive group Q of rationale, (c) the p-Priifer groups.
The rest of the chapter is devoted to showing that many abelian groups are direct sums
of these groups.
Recall that a group G is finitely generated if it contains a finite subset X with G - gp{X).
We show that every finitely generated abelian group is a direct sum of cyclic groups.
Furthermore we associate a set of integers to each finitely generated abelian group. This
set of integers, which we call the type of the abelian group, completely classifies finitely
generated abelian groups; in other words, two finitely generated abelian groups are
isomorphic if and only if they have the same type. This theorem is of great importance
in many branches of mathematics.
The additive group of rationals Q has the property that if g e Q and η is any nonzero
integer, then there exists / G Q such that nf = g. We express this by saying that Q is
divisible. The p-Priif er groups are also divisible. Note that these are not the only divisible
groups, e.g. the additive group of reals is also divisible. We obtain the pleasing result that
if A is any divisible abelian group, then A is a direct sum of p-Priifer groups and groups
isomorphic to the additive group of rationals.
Note. Any reader who would like a briefer account of abelian groups may refer to
Sections 6.1a, 6.1c and 6.3. This will bring him quickly to the fundamental theorem of
abelian groups, i.e. every finitely generated abelian group is the direct sum of cyclic groups.
177

178
ABELIAN GROUPS
[CHAP. Θ
6.1 PRELIMINARIES
Here we will practice expressing our ideas in additive notation.
a. Additive notation and finite direct sums
In this chapter aK groups will be abelian, and we will use additive notation throughout.
In terms of additive notation an abelian group is a non-empty set G together with a binary
operation + such that
(i) (a + b) + с = a + (6 + c) for all a, b, с е G.
(ii) a + b - b + a
(iii) There exists an identity element, denoted by 0, such that a + 0 = a for all aGG. The
identity element 0 is often termed the zero of G.
(iv) Corresponding to each α e G there exists an element b such that a + b — 0. This b
is unique and is denoted by -a. The element —a is often termed the negative of a.
Standard abbreviations are as follows:
(i) g + {—h) is written as g — h.
(ii) If и is a positive integer we write ng for g + - · · + g (n times). If η = 0 we write ng
for 0. If η < 0 we write ng for — g + ■ ■ ■ + — g (—n times).
If G is an abelian group and Я is a subgroup, then automatically Η <d G and we may
talk of the factor group G/H. (Warning: Some authors write G — Η for G/H.) Note that,
in additive notation, a coset is simply a set of the form g + H. Instead of talking of
multiplication of cosets, we talk of addition of cosets. Thus the sum of two cosets gi + H and
g2 + H is, by definition,
(gi+H) + (g2 + H) = (gl+g2) + H
The following table is useful in "translating" multiplicative notation into additive
notation, a and b are elements of a group G, and Η and К are subgroups of G.
Multiplicative
Additive
ab
a+b
a~l
—a
1
0
a"
na
ab-*
a-b
HK
H+K
aH
a+H
In Section 5.3a, page 146, we defined the internal direct product of two groups Η and
K. Now, in additive terminology, we speak of a direct sum rather than a direct product.
Instead of writing Η ® K, we write Η ΘΚ. If G — Η θ Κ, Η is called a direct summand
of G. Here we are interested in extending the concept of direct sum from two subgroups
to a finite number of subgroups.
Definition: An abelian group G is said to be the direct sum of its subgroups Gi, ...,Gn if
each g GG can be expressed uniquely in the form
g = gi+ ■·- +g«
where р(е&, г — 1, ..., n. In this case, we write G — Gi θ - - · θ Gn or
G = Σ Gu
If η — 2 we obtain the same definition of internal direct product that we gave in
Chapter 5. (The condition (i) of Proposition 5-19, page 146, falls away, as all groups
studied here are abelian.)

Sec. 6.1]
PRELIMINARIES
179
If G = Gi θ · · · θ Gn, then Gsn G,- = {0} for г ¥· j. For otherwise if χ G G4n G, and
χ ¥- 0, then
x = 9i+ ··· +9n
with 0! = · · · = д{-! - gff+i = · - - = gn — 0 and flf{ = x, and also with gi = ■ · ■ = 0,-~i =
0i+i = · ■ · = gn — 0 and 0j = ж. But this contradicts the definition of direct sum.
Note that if G = H@K and H = L®M, then G-L®M®K (see Problem 6.15).
The following theorem provides a simple criterion for determining when a group is the
direct sum of its subgroups.
Theorem 6.1: Let Gi, . . .,Gn be subgroups of a group G and suppose each element of G
can be expressed as the sum of elements from the subgroups Gi, ..., G„.
Suppose also that an equation
0 = gi+ ··· +gn
with gt G Gi for i—l,...,n, holds only if gi — дг — · ■ · = gn - 0. Then
G is the direct sum of the subgroups Gi,...,G„.
Proof: If g G G, then
9 = 9i + ' · - + 9*
with g( G G,-, г = 1, .. ,,η. We need only show that this expression is unique. Suppose
9 = g* + · · · + gt
is another such expression. Then
0 = {9i-g*i)+ ••■+(9«-9t)
By our hypothesis, {gi-g*) - {gn-gt) = ■·· = (gn-gt) - 0. Hence gt = gt for
г = 1,.. ,,η and the two expressions for g are identical.
The question arises: if Gu ..,, Gn are abelian groups, does there exist a group G
which is the direct sum of isomorphic copies of the groups Gi? This question is answered
in the following theorem.
Theorem 6.2: Let Gi, ..., G„ be abelian groups. Then there exists a group G which is the
direct sum of isomorphic copies of Gi, ..., G„.
Proof: The proof follows closely the argument of Problem 5.30, page 146, so we will be
brief. Here we will use additive notation. Let G be the cartesian product Gi x Gz x ■ · · χ G„.
If (0i, · ■ ■,9n) and (hi, ...,hn) G G, define (flfi, ...,gn) + {hu ...,hn) = {gi + hu ., ,,gn + hn).
η components
Then G is a group. Furthermore, let Gt - -j (0,..., gi, 0,..., 0) | gi G Gi к Then Gi is a
subgroup of G, and G( sa Gi for г = 1, .. „я. It is clear that every element (gi, . .., gn) of
G is uniquely of the form (gu 0,..., 0) + (0, gz, 0, ..., 0) + «· · + (0,0,..., 0, gn). Hence
G = Gi θ · · · θ Gn and the result follows.
An important result which we will prove in Section 6.1c states that if G is the direct
sum of Gi, ..., G„ and Η is the direct sum of Hi, ..., Hn and Gi » Я4 for г = 1, ..., η, then
G^H.
In Section 6.1b we will define the concept of an indexed family Gi, г G l. The reader
who studies Section 6.1c without reading Section 6.1b may take I - {1, .. .,n} and Gi,
η
i G I as shorthand for Gi,..., G„. Then 2 Gi is simply У Gt.
iei i=i

180
ABELIAN GROUPS
[CHAP. 6
Problems
6.1. Prove that the negative of a + b is —a — b.
Solution:
(a + b) + (-a - b) = {a + b) + [(-a) + (-b)] ~ [(a + b) + (-a)] + (-6)
= [a + (b + (-o))] + (-b) = [a + ((-a) + b)] + (-6)
= [(a + (-a)) + b] + (-b) = (0+ 6) + (-b) = Ь + (-&) = 0
6.2. Prove that if G is abelian and Η is a subgroup of G, then G/H is abelian.
Solution; (/ + H) + (fif + H) = (f+g) + H = (g + f) + Η = (ff + H)+(/-H)
6.3. Let Η be a subset of an abelian group G. Prove that if is a subgroup of G if and only if f,g € Я
implies f — gGH.
Solution:
This is exactly the same argument as in Lemma 3.1, page 55.
6.4. Let re be an integer and G an abelian group. Prove that if g, A G Gt then n(g + A) = ng + «Λ.
Solution:
If re = 0, 7i(sr + A) = 0 by definition. Furthermore ng + nh == 0 + 0 = 0. Thus n(g + A) =
ngr + «A when τι = 0.
If η > 0, let re = w + 1. Then w — 0. Inductively we may assume that m(g + A) = mg + mk.
Keeping this in mind,
7i(fif + A) = (m + l)(sr + A) = w(fif + A) + (fir + A) = Wfif + mA + g + A
= mg + g + mh + A = (m + l)g + (m + 1)A = ng + reA
Finally if τι < 0, re = —m for r« > 0, and so
re(ff + A) — m(— (fir + A)) — m((—g) + (—A)) = m(—g) + τη-(-Α) = ng + nh
6.5. Let re be any integer and G an abelian group. Prove that the mapping в which sends g bo ng for
each g in G is a homomorphism.
Solution:
Now, in additive notation, to say that в is a homomorphism means that (g + h)9 = дв + he
for all ff, A€ G. But, by Problem 6.4, (g + h)9 - n(g + h) = ng+nh = дв + he.
6.6. Let G be any abelian group and let X Q G (ХФ0). Prove that
gp(X) = {y | у = ri*! + r2*2 + · · · + rn*n, 7-j eZ, ж,е Х} (ff J)
In particular then gp{x) = {ra; | r £ £}■
Solution:
Let Η be the right-hand side of (ff.i). Then Η is a subgroup of G, for Η Φ 0. Moreover if
A = ri*! + · ■ ■ + »·„»;„ and к — sly1 + ■ · ■ + spyp belong to Я (riP Sj £ £ and xir ys £ X), then
clearly
A - к - nxt + ··■ + rnxn + (s1)y1 + · · · + (sp)yp e Я
Thus Η is a subgroup of G. It follows from the definition of Я that XqH. Since gp(X) is the
smallest subgroup of G containing X, we have then gp(X)QH. But if x\, ...,*„ ε Χ, then
fi^i + · * · + ί·η*η e flpW for every choice of r< € £. Hence ffp(X) 2 H, and so firp(Z) = Я.
When X = {«}, then gp(X) is, by (ff.i), the set of all multiples of x.
6.7. Let Я and К be subgroups of G, Prove that Я + К is a subgroup of G. If Я П if = {0} and
G = H + K, prove that G = Я φ К. (Я+ К = {А + А; | А £ Я and λ; <= К}.)
Solution:
Я+К^0 since both Η Φ φ and K¥*0, So we have to prove that if ι*,ι;€Η + Κ, then
u-v&H+K. Now ω = A + &, ν = A' + ft' (А,А'€Я, к,к'€. К). Thus и-ν = (А-А') +
(& — fc') е Я+ iC since Я and if are subgroups of G. If HnK == {0} and if we consider two
expressions ft1 + fc1 = h2 + k2 where AlhA2€ff and kuka€ K, then χ = ht —h2== к2 —кг belongs to
both Я and K. Therefore χ = 0 and At = A2, &! = fca. Hence the expression of an element in the
form A+ к is unique. Since G = H+K, it follows that G = Я ф К.

Sec. 6.1]
PRELIMINARIES
181
6.8. Let G have a subgroup Я and suppose that G/H is infinite cyclic. Prove that Я is a direct summand
of G.
Solution:
Let G/Я = grp (fir + Я) where jeG and let S - gp(g). Consider x€.Sr\H. Then % =■ ng
for some η e Z. As *еЯ, n(g + H) = ng + Η = Я. But fir + Я is of infinite order in G/H.
Thus η = 0 and so Sn# = {0}. If i€G, χ £ ng + H, i.e. я = wfir + λ for some re e Ζ and
АеЯ. Hence G = S + H. Therefore, by Problem 6.7, G = S φ Я and Я is a direct summand of G.
6.9. Let G == А φ Б and let Я be a subgroup containing A. Prove that Я = Α φ (Β η Η).
Solution:
As every element of G is uniquely of the form a+ b where a €. A and 6 £ В, A + (BnH) ==
Α φ (BnH). What we must prove then is that A + (BnH) = H. If A € Я, h = a+ b where
a€.A and ieB. Hence 6 = A —a. But, as АсЯ, h — α,^Η. Thus b&BnH and ЯсА +
(ВпЯ). But ЯэА and НэВпН. Hence Я = А ф (ВпЯ).
6.10. Let G be abelian and let »eS be of order к, 6 e G of order m. Show that the order of a + b
divides the least common multiple I of τη and re.
Solution:
Since both m and η divide I, let I = qm = rw for some integers q and r. Then l(a + b) = /a +
Z6 = gma + rnb = 0 + 0 == 0. Thus the order of a + b divides I.
6.11. Let G = Я φ tf. Prove that G/Я = Я.
Solution:
G = Η + К. Then by the subgroup isomorphism theorem (Theorem 4.23, page 125), G/H —
(H + K)/H = K/(KnH). Now HnK = {0}, and so G/Я = Й" as required.
6.12. Prove that if G is an abelian group, then the set S of elements of G of order a power of a fixed
prime ρ is a subgroup. Deduce that a finite abelian group has one Sylow p-subgroup for each
prime ρ dividing |G|.
Solution:
If a and 6 are of order a power of p, then —6 is of order a power of p. So, by Problem 6.10,
a— b is of order a power of ρ since the least common multiple of two powers of a prime ρ is a power
of p. Hence S is a subgroup.
If G is finite, then S is the Sylow p-subgroup of G. For if Ρ is any subgroup of G of order a
power of p, by the definition of S, Ρ С S. So every Sylow p-subgroup of G is contained in S. S
itself is of order a power of ρ (Problem 5.6, page 132). Since the order of a Sylow p-subgroup is
the maximal power of ρ dividing the order of G, every Sylow p-subgroup of G must coincide with
S. Thus, for each prime p, there is precisely one Sylow p-subgroup of G.
6.13. Let G be an abelian group of order 36. Prove that:
(i) If fir £ G, then fir = gl + g% where the order of fitj divides 4 and the order of g% divides 9;
(ii) G = Α φ Β where A is the Sylow 2-subgroup and В is the Sylow 3-subgroup of G.
Solution:
(i) Let fir be of order 2'3S. Put 2* = m and 3s = n. Then, since (m, n) = 1, there exist integers
os, b such that am+ bn — 1. Hence
fir = (am + bn)g — (am)g + (bn)g — gt + fif2, say
Now since ngt — n{am)g — a(nm)g — 0, and similarly mfir2 = 0, (i) is proved.
(ii) Clearly AnB = {0}, so A + Β ~ Α φ Β. Now if fir € G, then, by (i), g=ff1 + g2 where firг
is of order dividing 4 and fir2 is of order dividing 9. By the preceding problem, the set of all
elements of order a power of 2 is the Sylow 2-subgroup A, and so gt £ A. Similarly fir2 S B.
Hence fir € A + В and we conclude that G = Α φ Β.

182
ABELIAN GROUPS
[CHAP. 6
6.14. Show that the group С of complex numbers is with respect to addition the direct sum of the
subgroup consisting of all the reals and the subgroup consisting of all the pure imaginary numbers.
Solution:
Let R = {a+ г'О | α an arbitrary real number], and let / = {0 + ίδ | δ an arbitrary real
number} (here г2 = —1). Then clearly R and / are subgroups of C, Rnl = {0}, and, if a+ib £ C,
a + ib = (a + i0) + (0 + гб) belongs to R + I. Thus С = R ® I.
6.15. Let G = Я θ К and Η = L ® Μ. Prove that G = L ® Μ φ Κ.
Solution:
Every element fir of G can be expressed in the form fir = h + к where h e Η and fceK, But
h = I + m where I £ L and m £ M. Hence fir = I + m + k. Now if g — 1г + m, + kt with 1г € L,
m-i^M and kt€.K, put iI + m1 = fcIeH. Then fir = A + fe = ht + kx and consequently h = Aj
and к — kt. As fe = ί + m = it + Ж), ί = ij and m = m^ Hence the result.
b. Infinite direct sums (See JVoie on page 177.)
It is convenient to label the subsets of a set X with the elements of a second set. We
are already familiar with such a device, e.g. in labeling a collection of sets А\,Аг, ... we
have labeled with I — Z, In general, if / is an arbitrary set we shall denote by Air i £ /,
such a collection of labeled sets. A collection of labeled sets Ai, г е /, is called an indexed
family. More formally, let θ : / -* X be an onto mapping. Then θ is said to be an indexing
with the elements of I of the set X. We will denote the image of г by Xi} i.e. Xt = ъв. The
collection Xi, i Gl, is then called a family of indexed sets.
We generalize our definition of direct sum to apply to the direct sum of an infinite
number of subgroups.
Definition: An abelian group G is said to be the direct sum of its subgroups d, i Gl, if
for each g G G, g ¥> 0, there is a unique expression (but for order) for g
of the form
g - ffi + · - ■ + gk
where g} G Gy, j — 1, ..., k, with 1', 2', . .., k' distinct elements of / and no
gi is zero.
Note that as G is abelian,
flfi + ■ ■ · + flffc = gk + gk-i + ■ ■ · + gi
for example; hence the uniqueness of the expression is understood to be without
regard to the order of the elements </i, ..., gk· We write G = 2 Gi·
lei
If / is finite, it is easy to see that a group which is the direct sum in the sense of the
above definition is also the direct sum in the sense of the definition of Section 6.1a, and
conversely. Usually we will use the definition of Section 6.1a whenever / is finite.
We note that if G = ^Gi and *, j e/,tV j, then GtnGj = {0}. For if χ e G(nG,
t e I
and χ ¥* 0, then χ is expressible as
x — gi where οί G Gi
and x—U2 where g2 & G3
But this contradicts the definition of direct sum.

Sec. 6.1]
INFINITE DIRECT SUMS
183
The analog of Theorem 6.1 is the following:
Theorem 6.1': Let Gi, i Gi be subgroups of the abelian group G. Suppose each element
of G can be expressed as the sum of elements of the subgroups Gu Suppose
also that if 1', 2', ..., k' are distinct elements of / and that the equation
0 = £Ti + · · · + flffc
where g,- G G,- holds if and only if gi — да — · · · — gu = 0. Then G = 2 Gi.
As the proof is similar to that of Theorem 6.1, we omit it.
Again, as in Section 6.1a, the question arises: if Gi, г G I is an indexed family of
abelian groups, does there exist a group G which is the direct sum of isomorphic copies of
the groups Gi? This question is answered in the following theorem:
Theorem 6,3: Let Gi, i G I be an indexed family of abelian groups. Then there exists
an abelian group G which is the direct sum of groups isomorphic to Gi.
Proof: Let G = < ff Ι ff : / -* U Gi, iff G G; for all i€ /, and iff is the zero element of
Gi for all but a finite number of iGll. If ff, «£GG, define Ψ-θ + φ by ίΨ-ίθ+ίφ.
We assert that G is a group. ^
First, if θ, φ G G, then θ + φ is clearly a mapping of / into . U Gi and θ + φ maps all but
a finite number of the elements of / onto zero elements. Hence ff + φ £G. Note that
θ +φ — φ + θ so that G is abelian.
Next, G is associative. For if φνφ2,φ3 G G, and if iei",
*((Ψί + Фа) + Фг) = %Φι + Фа) + 4S = Чу + Ч2 + Чз = 4λ + {Ч* + Чг)
= %φί + г{ф2 + фа) = ϊ(ψ, + (фя + ф3))
Hence (фу + ф2) + ф3 = φί + (фг + ф3).
The mapping η: i-*0eG,· for all i in / is the identity of G. For if θ e G, then for all
i in /, ι(η + θ) = iv + iff = 0 + iff = iff so that η + θ - θ.
Finally, if θ GG, define φ: г -* -(iff). Then i(ff + ф)-гв+гф- ίθ + (-(iff)) = 0 = г')?.
Thus θ + φ = η and ψ is the inverse of G.
Now we prove that if G; = {Θ I ff G G and /ff is the zero of G3- for all j in / with perhaps
■*■-*-
the exception of i0}, then Gj s- G(. Note that G,- is a subgroup of G; for if 9,^e G, then
on putting Ψ = θ-φ we note that if }>г, j'€/, /Φ = з'в-з'ф - 0 -0 = 0. Thus *eG(.
л >s л
Next, let v.: Gi-*Gl be defined by 6vi — iff, θ GGr Clearly v} is a mapping of G. into
Gr To see that v{ is one-to-one, suppose θ,φ G G{ and 0v( = φνν This means that iff = 4·
But j"0 = уф — О for every j'G/, j ^ i and so ff = φ. Next we prove v{ is onto. Let a G G.
and define ff: / -* и G. by ίθ - a and j'0 = 0 if jV i. Then ff e G. and ffv{ = a, and so
vf is one-to-one and onto. vt is a homomorphism; for if θ,φ G G( and Ψ = ff + φ, then
(ff + ψ)ν{ = Φν{ = гФ = iff + ίφ = ffv, + ψν,
Therefore vi defines an isomorphism.
Finally we show that G = ]£ G(. We already noted that G is abelian. We need to show
that if ff G G, θ - θι+ · · ■ +вк where each fff belongs to one of the subgroups G3·. If β G G,
then iff is not the zero of Gi for only a finite number of elements of /, say ii, .. .,ik- Let
0i G G(l be such that ΰθι - iiff, 0z G Giz be such that i2ffz = кв, ..., вк G Gik be such that
ifcfffc = iicff. It is clear that

184
ABELIAN GROUPS
[CHAP. 6
θ = 0i + ■ · · + 0k
Now suppose that η, the zero of G, is of the form
η — βι + · ■ · + вк
where θ, € Gy and V, 2', ..., ft' are distinct elements of /. Then for 1 ~ j — fc,
/', = 0 = ?'(0i + + 0k) = j'fl,
(since V,2', . . .,k' are distinct) from which θ} = η for every / = 1, ..., k, Thus
G = Σ&
(ε r
(Remark: Problem 6.17 shows how the mappings of this proof link up with cartesian
products.)
The existence of direct sums is a very powerful result, and we will use it repeatedly.
We note the important result that: given Gi =s Hi for each i G /, then if G is the
direct sum of its subgroups Gi and Η is the direct sum of its subgroups Hi, we can conclude
G^H (Section 6.1c).
Problems ,
6.16. Let G be an abelian group with subgroups Gb ie/, Suppose that G = gp f υ GA and that
Gj η gpi U G; ) = {0} for each j € /. Prove that G is the direct sum of its subgroups Git ie/.
Vf^i /
Solution:
We need only show that if
0 = gx + · ■ ■ + gk
where fif£ S Gv and 1', 2', ...,k' are distinct elements of /, then
ΑΊ = g% ~ ■·■ = gk = 0
Suppose, if possible, that some fitj is not zero, say дг Ф 0. Then
-9ι = 02 + ' '" + Ok
But — gfj £ Gv, say, and flf2 + · ·' + 9k e OP( U G4 ). By hypothesis then, —fit, = 0. Hence
ΑΊ = 0% — '"" = ffk = 0 a"d the result follows.
6.17. Compare the construction of a direct product involving cartesian products with the construction
involving mappings by showing that an ordered pair can be thought of as a mapping. Hence show
that the two groups obtained are isomorphic.
Solution:
An ordered pair can be thought of as a mapping from {1,2}. The image of 1 gives the entry
in the first position, the image of 2 gives the entry in the second position. Let G = Gx X G% and let
G be the group as constructed in Theorem 6.3 with /=={1,2}. Let »:G^G be defined by
$v — (Is, 2$) for any j€G, Then ν is clearly a one-to-one and onto mapping.
Also ν is a homomorphism. For if Ψ = β + ψ,
(θ+ψ)ν = Ψν == (1Φ,2Φ) = (1β + 1φ, 29 + 2φ)
= (Iff, 2ff) + (1ψ, 2ψ) = βν + φν
6.18. Let π- be the ratio of the circumference of a circle to its diameter. Let G be the subgroup of the
additive group of the real numbers generated by the numbers w, n-2, it3, ... . Let Gj =· др{тг1),
г = 1,2, .., . Prove that G = 2 G,- where Ρ is the set of positive integers. Use the fact that
i ε ρ
π is not the root of any polynomial with integer coefficients.
Solution:
Clearly each element of G is of the form gl + ■ ■ ■ + gn where each gi belongs to one of the
groups Gi, Gz, .. . . Then we need only show that 0 = gt + ■ ■ ■ + gn implies that fifj = g% = ■ ■ · =

Sec. 6.1] HOMOMORPHISMS OF DIRECT SUMS; FREE ABELIAN GROUPS 185
gn — 0 where the gt € Gv, with V, ...,»' distinct elements of {1,2, ...}. Now g{ — z$rv where
z,· € Ζ and i — 1, ..., n. Then
0 = Z^v + z2vs' + ■■■ + Ζηπη'
If not all zt are zero, w is the root of a polynomial with integer coefficients, contrary to the statement
in the problem. Hence the result.
e. The homomorphic property of direct sums and free abelian groups
Let G = Α θ В and let Я be a group which contains isomorphic copies A and β of A
and В respectively. Suppose that Я = A + В (but not necessarily that Я = А Ф Ё),
What connection, if any, is there between G and HI
It turns out that Я is a homomorphic image of G. This follows from Theorem 6.4.
This theorem when applied to particular cases leads also to important results: (1) Theorem
6.5 and (2) the concept of a free abelian group.
Theorem 6.4: Let G = A®B and let Я be any group. Let θ,φ be homomorphisms of
A into Я and В into Я respectively. Then there exists a homomorphism
ζ-.G^H such that ζίΑ=β, ζ ,„ = φ.
Proof: If g GG, then g-a + b uniquely where a&A, b SB. Define дс-ав+ Ьф,
ζ is uniquely defined and so is a mapping of G to Я. Note that if gi = at+ bt where сц G A
and ЫЕВ (г = 1,2), then
(ΑΊ + ^Κ = ((<*l + bi) + Κ + \))ζ = ((a^ + aJ + ibt + bJK
= (a, + a2)0 + (&j + Ъ^ф = at6 + a20 + Ьгф + Ьгф
= агв + \ф + a20 + Ь2ф = (a, + Ъг)С + (a2 + \)ζ = д£ + д£
Hence ζ is the required homomorphism as ζ |д = θ, ζ ,Β = φ.
In exactly the same way we can prove that if G — Σ Gi and if for each ie/, θι: Gi -»■ Я
ίει
is a homomorphism of Gi to Я, then there exists a homomorphism Θ: G -» Я such that
0 |c, = 0i. We shall often say that 0 extends the mappings 0e, or that 0 is an extension of the
mappings 0(.
We use this result to prove:
Theorem 6.5: Let Gf ^ Hit г G /. If G is the direct sum of its subgroups G( and Я is
the direct sum of its subgroups Hi, then G ^ H.
Proof: Let 0i: Gi^> Hi be an isomorphism for all г G/. Then by Theorem 6.4 there
exists a homomorphism 6:G^H such that 0 |Gi = 0i for each г G /. To prove 0 is an
isomorphism, we need only show that its kernel is trivial since 0 is clearly onto. If g G G,
then flf = flfi + · · · + flfn where gs belongs to the subgroup Gy with Γ, 2' n' distinct
elements of /. Thus
дв - 0101- + · · · + дпвп- - hi+ ··· + hn, where hi G Hi·
Then дв = 0 only if each hj = 0, since Я is the direct sum of its subgroups Hi. Since
h) = 0j0j- and 0j. is an isomorphism, we have g, = 0 and g = 0. Thus Ker 0 = {0}, and so
О^Я.
We will now apply Theorem 6.4 when each Gi is infinite cyclic. To begin with, let
С = gp(c) be infinite cyclic and Я any abelian group. Note that if ψ is a mapping of {c}
into Я, then there exists a homomorphism в:С-»Н such that 0 |{e} = φ. The
homomorphism 0 is simply defined by putting (гс)в = г(еф) for each rGZ. It is readily seen
that this does define a homomorphism.

186
ABELIAN GROUPS
[CHAP. 6
Now let each Gi be infinite cyclic, G{ = gp(a), i G /. Let X= {d\iGl}. If Θ-.Χ-+Η
where Η is any abelian group, then there exists a homomorphism 0*: Σ Gi -* Я such that
ίει
θ* ix = 9. For, corresponding to each Gu we know from the remark above that there exists
a homomorphism 0e: Сп-* Я such that сД = c<0. Hence it follows by Theorem 6.4 that
there exists a homomorphism 0*: ^Gt^ Η which agrees with 0i on G{. So we have the
required result. 's'
Corollary 6.6: The direct sum G= Tft satisfies the following condition: for every
mapping θ: X -»Я, Я any abelian group, there exists a
homomorphism U*:G^ Η such that 9* lx = 0.
A group G which contains a subset X such that
(i) G = gp(X),
(ii) for every mapping 0: X -» Я, Я any abelian group, there exists a homomorphism
6*:G^H such that 0*,x = 0,
is called α /ree abelian group. G is said to be freely generated by X and X is called a basis
forG.
We have shown that the direct sum of infinite cyclic groups is a free abelian group.
Conversely we have the following
Theorem 6.7: If G is a free abelian group freely generated by a set X = {xt\ i Gi},
then G is the direct sum of its subgroups Gi = gp{xi) and each G; is
infinite cyclic for all i G I.
Proof: This theorem is proved by showing that G is isomorphic to a direct sum of
infinite cyclic groups. To this end let Η be the direct sum of its subgroups Я,,
Я = Τ Я
where Я( = gp(ht) is an infinite cyclic group generated by hu (We know such a direct sum
exists by Theorem 6.3.) Let Θ-.Χ^-Η be the mapping defined by x$ = ht. Then 0 can
be extended to a homomorphism Θ* of G into H, by the definition of a free abelian group.
On the other hand Я is the direct sum of the infinite cyclic groups Ни Thus by Corollary
6.6, the mapping φ: {hi \ г Ε /} -» X defined by hi<j> = x{ can be extended to a homomorphism
φ* of Я into G. Actually φ* and 0* are inverse isomorphisms. To see this, suppose g GG.
Then g = mxv + · · · + nrXr· where Г, ...,r'ei and иь .. .,Wr eZ. Accordingly,
(00*)ψ* = [ni(a:i'0*) + · ■ · + iv(ay0*)]£* = (ηΛι- + ■ ■ ■ + ητΚ·)φ*
= Μι(Αΐ'ψ) + · · · + пЛЪт'Ф) = и>Ж1' + - ■ · + «rav = 9
and so 0*<£* is the identity mapping on G. Similarly φ*θ* is the identity mapping on Я.
This implies that 0* is a one-to-one mapping of G onto Я. For if g, g' G G, then дв* = дг'0*
implies that (sr6>*)^* = {д'9*)ф*. Since (^0*)ψ* = ff and {д'в*)ф* = g', we have g = flr'.
Furthermore if hGH, then A. = (кф*)в*. Thus 0* is one-to-one and onto.
Note that each d is infinite cyclic, since 9* is an isomorphism and G<0* = Я;. Finally
we show that G is the direct sum of its subgroups Gu If 1', 2',.... r' are distinct elements
of / and gi,.. .,gr are nonzero elements of Gu, . -., Gv respectively, then if
gi + ■ ■ · + gr = 0 (ffJ8)
it follows that #10*+ · · · +0r0* = 0. But then, as д{в* G Ην and gi9* ¥= 0, we have
a contradiction as Я = 2 #;■ Hence (0.2) does not hold. Finally flrp(Gi] i Gi) = G, and
soG = У ft. ,er

Sec. 6.1] HOMOMORPHISMS OF DIRECT SUMS; FREE ABELIAN GROUPS 187
Corollary 6.8: Every abelian group is the homomorphic image of some free abelian group.
Proof: If G is an arbitrary group whose elements are 9i, г Ε /, and gp(Xi), г S I, is
infinite cyclic, then as we have seen, F = 2) 9V{Xi) is free abelian and the mapping
ie r
θ : Χι -> gi extends to a homomorphism of F onto G. ►
Problems
6.19. If \gp(a)\=m, \gp(b)\ = η and (m,n)—l, then G = gp(a) φ gp(b) is cyclic of order «in.
Solution:
We show that G ~ gp(a+b). If I is the order of a+b, then l(a + b) =la + lb = Q implies
la — lb = 0, by the definition of a direct sum. Consequently I is divisible by the order of a and the
order of 6. Since (m, n) = 1, mn | I and we conclude mn — I, so that G = gp(a + b).
6.20. Find \G\ where G ia the direct sum of η cyclic groups of order 3.
Solution:
Let Gu ..., Gn be subgroups of G and suppose G = G, φ · · ■ φ G„, where each G4 is of order
3. Each expression of the form fir, + g2 + ■ · ■ + gn, where gt 6 Gt, gives rise to unique elements
of G. There are 3 different possible choices for gt, 3 for g2, etc. Hence the total number of possible
choices is 3·3 3=3™. Hence |G|=3".
6.21. Prove that A is a direct 3unrmand of G if and only if there exists a homomorphism θ of G onto A
such that θ |Λ is the identity on A. (Hint. Use К — Ker 9. Also consider g — ge to prove
geA+K.)
Solution:
Suppose that G = Α φ Β. Then the identity homomorphism on A and the trivial homomorphism
on В extend to a homomorphism в : G -> A which satisfies the required conditions (Theorem 6.4).
Conversely, if such a homomorphism exists, let К — Ker в and let xGHAoK. Then * = хв,
as 9 |Л = identity on A. But χθ = 0, since ж € Ker i. Hence ΑηίΓ = {0} and so gp(A,K) =
A ©K. Now letting g € G, gi —a for some α € A. Then (g — a)e = ge —αβ = α — α = 0 and
g-aeK. Thus fir € А ф ЯГ and G = Α φ Κ.
6.22. A group G is said to be of exponent η if a; € G implies nas = 0 and η is the smallest positive
integer with this property. Let Η be the direct sum of two cyclic groups of order n, generated by
*, and x2 respectively. Put X — {xux^. Prove that if G is any group of exponent η and в is a
mapping of X" into G, then there exists a homomorphism в* of Я into G such that i* |X = в.
Solution:
Let Д, = gp(x\) and Я2 = gp{x^- There is a homomorphism it: #t -> G satisfying
«j»! = a^i, for др(Х\в) is cyclic and is of order dividing n. Similarly there is a homomorphism
62:H2^G satisfying x%e2 — x2$. Thus there is a homomorphism β* : ΗΙφΗ2-» G such that
e* ih, = *i and β· |н2 — *2> by Theorem 6.4. The result follows.
6.23. Let G be freely generated by a finite set X, \X\ = n. Prove that every element of G is uniquely of
the form m^^ + - ■ · + ■mnxn, where щ € Ζ and xi € X.
Solution:
Let G, = gp(Xi). Now by the theorem on free abelian groups we know that G = Gt® ·· · © Gn,
and each G{ is infinite cyclic. Then each element of G is uniquely of the form gt + ■ ■ · + gn where
fifi € G{. But we know from the theory of infinite cyclic groups that g( = mixi uniquely. The result
follows.
6.24. Let G be the direct sum of cyclic groups G( of order 2 where i € P, the set of positive integers.
Let Ε denote the set of even positive integers. Then Η = gp(Gi \ i € E) is clearly a proper
subgroup of G. Prove that Η эе G.
Solution:
Let ot: G; -> G2i for all iGP be an isomorphism. (Such an isomorphism exists, since all the
G4 are cyclic of order 2.) We apply Theorem 6.5 to obtain the result.

188
ABELIAN GROUPS
[CHAP. 6
6.25. Let 6? = Α φ В where A is cyclic of order 32 and В is cyclic of order 52. Prove that aut (G) as
aut (A) © aut (B) (hard) and hence compute |aut(G)[. (aut(G) is the automorphism group of G
(see Section 3.6a, page 83).)
Solution;
If a € aut (A), then α can be used to form an element of aut (G) by letting it act as the
identity on B. (Theorem 6.4 states that α extends to a homomorphism. We must check that it
is one-to-one and onto.) Similarly for elements β of aut (β). We use the symbols α*, β* to
represent corresponding automorphisms of G. Note that
(а+Ь)а*Д* = (αα+δ)/ϊ* = aa+bfi = (а+Ь)Д*а*
from which α*β* — β*α*. Note that a* = β* implies a* = β* — ι, the identity automorphism.
The mapping a -* a* is an isomorphism of aut (A) into a subgroup of aut (G). The mapping β -* β*
is an isomorphism of aut (δ) into a subgroup of aut(G), As (aut (A))* n (aut (Β))* = {ι}, and as
the elements of (aut (A))* commute with the elements of (aut (B))*, we have (aut (A))* + (aut (B))* =
(aut (Α))* φ (aut (В))*, by Theorem 5.16, page 144. Now let e€aut(G). 9U induces an
automorphism вА on A, for Ae must go into a subgroup of order 9 and by Sylow's theorem there is only
one subgroup of order 9 (as G is abelian). Similarly в\в induces an automorphism 0B on B. Then
(а+Ь)вд9в = (ав+Ь)»в = αθ + be = (a+b)e
from which вдОд = s. Thus aut(G) = (aut (Α))* φ (aut (В))* and, by Theorem 6.5,
aut (G) as aut (Α) φ aut (B)
To compute |aut(G)| we must compute |aut(A)| and |aut(B)|. Let <*€aut(A) and let
A = gp(a). a must take α onto an element of order 9; so if aa — aT, (r, 3) ~ 1. Hence the
possibilities are r = 1,2,4,5, 7,8. Each of these gives rise to an automorphism of A, as can be checked.
Thus |aut(A)| = 6. Similarly |aut(B)| = 20. Accordingly, |aut(G)| - 6 X 20 = 120.
6.2 SIMPLE CLASSIFICATION OF ABELIAN GROUPS,
AND STRUCTURE OF TORSION GROUPS
a. Tentative classifications: torsion, torsion-free, and mixed
Consider the following three examples of abelian groups: Q, the additive group of
rationale; Q/Z the factor group of the additive group of the rationale by the integers;
C, the multiplicative group of complex numbers. Each of these groups is not isomorphic
to the others, but how would we prove that? One way is to examine the orders of the
elements of the groups. Now every element of Q except 0 is of infinite order and every
element of Q/Z is of finite order. For if r G Q, r — m/n where m,n are two integers.
Thus n(r + Z) ~ nr + Ζ ~ m + Ζ = Ζ. Let us, to avoid confusion, continue to use the
multiplicative notation for С We assert that С has elements of infinite order and also elements
of finite order. Recall that the identity of С is 1. Note that (~1)2 = 1 implies that —1 is
of order 2 and 3r = 1 if and only if r = 0. Hence —1 is of finite order and 3 is of infinite
order. Summarizing, we have
(i) Q has every element but the identity of infinite order,
(ii) Q/Z has every element of finite order,
(iii) С has elements of finite order and elements of infinite order.
It is then easy to see that the three groups are not isomorphic.
If G is a group in which every element other than the identity is of infinite order, G is
said to be torsion-free. If G is a group in which every element is of finite order, G is said
to be a torsion group. If G has both an element of infinite order and an element (not equal
to the identity) of finite order, G is said to be mixed. These three concepts provide us with
a rough classification of abelian groups and, as we have seen above, distinguish between
Q, Q/Z and C.

Sec. 6.2] SIMPLE CLASSIFICATION. STEUCTUEE OF TOESlON GEOUPS 189
Problems
6.26. Let G be the direct sum of torsion groups. Prove that G is a torsion group.
Solution:
Let G — 2 G> If 9 ε G, g = xt + ■ ■ · + xn for some integer η and x} belongs to some G}·.
If the order of Xj is Pj, then
(i»i Ρη)β = (Pi Pn)*i + · · · + (Pi Ρη)χη = 0 + · ■ · + 0 = 0
6.27. Prove that Q/Z and G, the direct sum of groups G( (i^-Z) where each G{ is cyclic of order 2, are
not isomorphic.
Solution:
Following the method of Problem 6.26, it is easy to prove that every nonzero element of G
is of order 2. Since ^ + Ζ is of order 3, Q/Z and G are not isomorphic.
6.28- Prove that Q/Z and G, the direct sum of groups Gj (iSZ) where each G{ is cyclic of order 3f, are
not isomorphic.
Solution:
G is easily shown to have every element of order some power of 3 by following the method
of Problem 6.26. Since J + Ζ is of order 4 in Q/Z, G and Q/Z are not isomorphic.
b. The torsion subgroup
In Section 6.2a we introduced a tentative classification of abelian groups into torsion-
free, torsion and mixed groups. In this section we consider the question of whether it
would not be possible to split a mixed group into a torsion-free group and a torsion group.
This would provide the following program for investigating abelian groups:
(1) Investigate torsion-free groups.
(2) Investigate torsion groups.
(3) Investigate how they may be put together to form mixed groups.
Such a program is found to be too difficult to accomplish completely, but it does lead
to some significant results.
Let T(G) be the set of all elements of G of finite order. Then T(G) is a subgroup of G,
as we show in the following
Theorem 6.9: T(G) is a subgroup of G (termed the torsion subgroup of G). G/T(G) is
torsion-free.
Proof: Let a,b € G be of order m, η respectively. Then
mn(a — b) = mna — mnb = 0 — 0 = 0
Thus if a,bG T(G), a-bG T(G) and T(G) is a subgroup of G.
Now consider G/T(G). Assume g + T(G) is of finite order n, i.e. n(g + T(G)) = ng +
T(G) = T(G). It follows that ng e T(G). As T(G) consists of all the elements of G of
finite order, there exists m such that m(ng) = 0. Then g is of finite order and g G T(G);
hence g + T(G) = T(G). Therefore the only element of finite order in G/T(G) is the zero
T(G). Thus G/T(G) is torsion-free.
Problems
6.29. Prove that if G is a group and Η a subgroup of G such that G/H is torsion-free, then Я contains
the torsion subgroup of G.
Solution:
Let g in G be of finite order. Then g + Η is of finite order in G/H. Since G/H is torsion-free,
g + Η = Η. This means g £ H, and so every element of finite order in G is contained in H, i.e.
the torsion subgroup of G is contained in Я.

190
ABELIAN GROUPS
[CHAP. 6
6.30. Is the set consisting of 0 and of all elements of infinite order of a group automatically a subgroup?
Solution;
No. Рог example, let G ~ Η ® Κ where Η = gp (h) is infinite cyclic, and К = gp (k) is of
order 2. Now h + k and —h are both of infinite order. However (h + k) + (—ft) — к is of finite
order.
6.31. Prove that if the set consisting of 0 and the elements of infinite order of a group G constitutes a
subgroup, then G is either a torsion-free group or a torsion group.
Solution:
Suppose G is mixed and Η — {h\ hE G and h is either 0 or of infinite order}. Since G is
mixed, we have ff € G of infinite order and g'(¥=0) € G of finite order. Now g' — g is of infinite
order so that (g' - ff) β ff. But as Η is a subgroup, (ff' — g) + g = ff' £ ff. Therefore ff' is 0 or an
element of infinite order, contradicting the choice of g'. Hence G is not mixed.
6.32. Prove that if Τ is the torsion subgroup of G, then ГпЯ is the torsion subgroup of any given
subgroup Η of G.
Solution:
If «efl is of finite order, then α G Γ. Hence α β Γη Η, and so the torsion subgroup of Η
is contained in TnH. Conversely TnH consists of elements of finite order, and so TnH is contained
in the torsion subgroup of H. Thus we have proved that Γη Η is the torsion subgroup of H.
6.33. Find the torsion subgroup of R/Z where R is the group of real numbers under addition and Ζ is
the subgroup of integers.
Solution:
Suppose r + Ζ is of finite order (r € R). Then for some nonzero integer nr n(r + Z) — Z. But
n(r + Z) = nr + Z, and so nr€Z. This means that r is a rational number. Thus T(R/Z) Q Q/Z,
where Q is the subgroup of rational numbers. On the other hand, if a + Ζ €5 Q/Z, then α = m/n
where m, η € Ζ and и/0. So
n(a + Z) - n(m/n + Z) = n{m/n) + Ζ = m + Ζ — Ζ
Hence a + Ζ is of finite order and Q/Z ς T(R/Z), Thus we have proved that Q/Z = Τ (R/Z).
с Structure of torsion groups. Prufer groups
A group G is called a p-group or a p-primary group for some prime ρ if every element
of G is of order a power of p. (If G is finite, it follows that the order of G is a power of p;
see Problem 5.6, page 132. The definition of p-group given here thus coincides with that
of Chapter 5 when the p-group is finite.) In this section we show that a torsion group is
built out of p-groups. Thus the study of torsion groups becomes essentially the study of
p-groups.
Theorem 6.10: Let G be any torsion group and let Gp = {g \ g has order a power of p},
ρ any prime. Then if Π is the set of all primes,
G = Σ GP
pen
Proof: We let the reader show that each Gv is a subgroup of G.
Suppose that g G G and is of order p^p'j · · · prnH, Pi, ..., pn distinct primes and n, ..., r„
positive integers. Let q — p[r · · ■ p^1; then {q,pr^) = 1. Thus there exist integers a and
b such that aq + bpr^— 1. It follows that g = aqg + bpr*g. Now aqg is of order p^n, and so
aqg e Gpn. bp'fg is of order q. Since q is less then the order of g, we may assume inductively
that bp^g is the sum of elements belonging to Gp„_j ,GPn„2,..., Gpl. Thus every element
of G can be expressed as the sum of elements belonging to the Gp, i.e. G is generated by the
subgroups Gp.

Sec. 6.2] SIMPLE CLASSIFICATION. STRUCTURE OF TORSION GROUPS 191
To show that G — JGP, we must prove that gx + ■ ■ ■ + gn = 0 (where gi G GPi and
pen
Pt, ■ ■ ., Pn are distinct primes) occurs only for д\—дч~ · · · = gn — 0.
We proceed by induction on n. For η — 1 it is certainly true. If true for n, consider
gi + ■ · ■ + gn+i = 0
and let fl'n+i be of order pj; + i- Then
Prn + 19t + Prn + l92+ ■■■ +Prn + 1gn + l = 0
and Ρη + ι9ι + ·■ ■ +Pn+i9n = 0
By the inductive hypothesis, we have
Prn+i9i = Prn + l9z = ·■· = Prn + 19n - 0
Now P*+lgi ■= 0 implies g\ = 0 as 5Ί is of order some power of Pi and P\ ¥= p„+i. Arguing
similarly for g%, . . .,g%, we have g\ — g%— ■ ■ ■ ~ gn~0. Hence also gn+i = 0 and we
conclude that G = X GP.
pen
The subgroups Gv are called the p-components of G.
Example 1: Let us apply this theorem to Q/Z, i.e. the additive group of rationals modulo the
integers. Q/Z is clearly a torsion group.
(Q/Z)p = {x + Ζ | χ + Ζ of order a power of p} = {x + Ζ ! prx e Z}
= {m/pr + Ζ | for various integers r and 0 — m < pr~1}
By the theorem, Q/Z - 2 (Q/Z)„.
pen
(Q/Z)p is called the p-Priifer group (also called a group of type pa). In Section
6.4 the p-Prtifer groups will be fundamental. Note that the p-Prflfer group
(Q/Z)p - U Cr where Cr = gp(l/pr + Z), since (Q/Z)p = {m/p* + Ζ \ f or vari-
r = l os
ous integers r and 0 ^ m < pr~i}. Clearly (Q/Z)pQCr and (Q/Z)pC у Cr. The
result follows. r_1
We now have at our disposal cyclic groups of all orders, the additive group of rationals,
and the p-Priifer groups, together with all their direct sums. These, as we shall prove,
constitute a large class of abelian groups.
Problems
6.34. Use Theorem 6.10 to prove that an abelian group of order pq, where ρ and q are different primes,
is the direct sum of a cyclic group of order ρ and a cyclic group of order q.
Solution:
Let G be of order pq. Then by Theorem 6.10, G = Gp φ Gq; for if r is any prime other than
ρ or q, Gr - {0}. Why does Gr = {0}? If g S Gr, then, as G is of order pq, pqg — 0. Hence r
divides pqt which is not the case. Note that Gp ¥= {0}, Gq Φ {0} by Proposition 5.9, page 137.
JGp| divides pq. Since Gq Φ {0}, \Gp\ = ρ or q. As the elements of Gp are of order a power of p,
it follows that \Gp\ = p. Similarly \Gq\ = q. Hence, as the only group of prime order is cyclic,
we have the result.
6.35. Show that a direct sum of p-groups is again a jbgroup.
Solution:
Let G = 2 Gi where each Gi is a p-group. Let g G G. Then g = g^ + · · ■ + gn where each
> e г
gt ε Gv, 1', 2', ...,η'ε/, Let pT be the maximum order of the g{. Then prg = 0. Thus the order
of g is a power of p, and so G is a p-group.

192
ABELIAN GROUPS
[CHAP. 6
6.36. Show that if G is a p-Prufer group, for each integer к Ф 0 and each element g € G there exists
an element Лев such that kh — g. (We shall call a group with this property divisible.)
Solution:
Let i?G. Then g = m/pr + Ζ where 0 — m < pT. Let к — p"l where ρ and I are co-prime,
and let ft[ = m/pr+s + Z. Then р8Аг = д. As I and pr + s are co-prime, there exist integers a and b
such that al + bpr + s - 1. Therefore
ht - {al + bpr+s)ht = alhl + bpr + shl = alhy
Put h = eftj. Then
fcA = psifc = p4ahl = psfcj = ff
6.37. Let G — 2 G{ where Gf is a cyclic group of order pt and p„p8 ... are the primes in ascending
order of magnitude. Let Η be a p-PrBfer group for any prime p. Prove that G is not isomorphic
toH.
Solution:
Let pj be a prime different from p. Then the pjth component Gp} Φ {0} but HP} — {0}. Thus
G is not isomorphic to H.
d. Independence and rank
We introduce here an important concept in abelian group theory, the concept of rank.
Let G be an abelian group. A subset X of G is called independent if whenever xt, .. .,%„
are distinct elements of X and tu, ..., rw are integers such that
П1Х1 + ■ · ' + ПгХг = 0 (6.3)
then riiXi — ■ · · — nrxr — 0.
Note that if G is torsion-free and X is independent, then equation (6.3) implies m =
ηΆ = ■ ■ ■ = rtr = 0. Suppose X = {ж, j г e /} and ж,- ¥- х-, for г ¥- j. Then if X is an
independent set, it follows readily that gp (X) = У. gp{Xi).
ie ι
We need one further definition. An element ж in a group G is dependent on a subset
X of G if
nx + ntXi + ■ ■ ■ + nrxT = 0
for some choice of xu ..., xr e Χ, η and щ e Ζ and nx ¥· 0. In other words, χ is dependent
on the subset X if there is an integer η with nx Φ 0 and иж ε gp(X). We say YcG is
dependent on ZcG if every element of Υ is dependent on X. Observe also that if G is
torsion-free with subsets Χ, Υ and W and if X is dependent on Υ and Υ is dependent on W,
then X is dependent on W. For if ж ε X, then for some integer η ¥* 0, and integers
Wi,.. .,ttr, иж = Mi2/j + ■ ■ · +ttr#r (#i e Y). Since Γ is dependent on W we can find integers
mi ¥= 0, ..., utr ·?* 0 such that tre^i e дгр(И0 for i—l,...,r. Put m = mi · · · mr. Then
clearly mnlylGgp(W) for i = l, ...,r, and consequently mux € gp(W). Furthermore,
since G is torsion-free, mnx ¥= 0. Thus every element of X depends on W.
The main result that we shall now prove is called the Steinitz Exchange Theorem.
Theorem 6.11: Let G be torsion-free and let A = {alt.. .,α™) be an independent subset
of G. Suppose В = {bl3 .. .,b„) is another subset of G such that A is
dependent on B. Then η ^ m and В depends on A U С where С is a subset
of В and \C\ — n — m.
Proof: We will use induction on m. For m = 1 it is clear that η Ш m. Now aj depends
on В means that there exist integers ζ zn such that ζ ¥* 0,
Zdi + Zibi + · · · + Znbn = 0

Sec. 6.2] SIMPLE CLASSIFICATION. STRUCTURE OF TORSION GROUPS 193
and thus for some integer i, Zi^O, l^i^n. Hence В depends on {αϊ} и (Б - {h}). Thus
the result holds for m — 1, with С — В — {Ы}.
Next we assume that the result holds for m — r, and consider the case m = r + 1. Then
by the inductive hypothesis, n^r. Now {«i, .. .,Or} depends on B, and so inductively В
depends on {«i, .. .,ar} U D where DqB and \D\ = n-r. But {Or+i} depends on Б and
В depends on {au ...,a,}UD. Then by our remarks above, {ar+i} depends on {αϊ, ..., Or} и D,
Thus we can find integers у ¥= 0, yit ...,yr,zi, ..., zs such that
yar+i = г/itti + · · · + угаг + ztdi + · · · + zA*, άΊ, . . .,d, G D
Suppose, if possible, that zx- z2- · ■ · = zs - 0. Then i/Or+i = ϊ/ιαι + · · · + улг implies that
the elements au .. .,ar+i are not independent. So some z% ¥= 0. Let С — D — {άή. Then it
is clear that {αϊ, .. .,Or} ufl depends on {αϊ, .. . ,Or+i} и C. Since Б depends on
{αϊ, .. .,Or} UD, then В depends on {αϊ, .. .,Οτ+ι} U С. Finally С QB and |C| = |D|-1.
\C\ — (n — r) — 1 = η — (r +1) = η — m
as desired, and the proof of Steinitz's theorem is complete.
Let us call a subset S of a torsion-free abelian group G a maximal independent set
(i) if S is independent and
(ii) if g £ G and g g S, then Su{g} is not an independent set.
Suppose now that G, a torsion-free abelian group, has a maximal independent set S
that is finite. Let Τ be any other finite maximal independent set. By the Steinitz exchange
theorem, \S\ ^ [Γ|. Also by the same theorem, |Г| — \S\. Hence we can without ambiguity
define the rank of a torsion-free abelian group G which has a finite maximal independent
set S to be |S|. If G does not have a finite maximal independent set S, we shall say G is of
infinite rank.
It is easy to see that if G and Η are isomorphic groups, then they have equal ranks.
As a consequence of these remarks we obtain a result concerning free abelian groups.
If F is free abelian with a finite set of free generators X, then X is a maximal
independent set of F (see Problem 6.41 below). Hence the rank of F is \X\. Similarly if
F is also freely generated by a finite set Y, then rank of F = \Y\. Hence \Y\ - \X\. Thus
we have
Corollary 6.12: If F is a group freely generated by two finite sets X and У, then \X\ = \Y\.
We have as yet not proved that all abelian groups have maximal independent sets. To
do so we need a result called Zorn's lemma. Before stating the lemma, we consider the
following examples:
(a) Let Ψ = {A, B, C, D), where A = {0,1}, В = {1, 2}, С = {0, 2}, D = {0,1,2,3}.
We inquire: does Ψ have a largest element, i.e. one that contains all the elements of
ΨΊ Clearly it does, for D^A, D^B, DqC and DqD. Thus D is a largest element.
(b) Let Ψ - {A, B, C, E), where А, В, С are as in (a), and Ε - {1, 2,3}. Now there is no
largest element of Ψ. We search for some concept replacing that of a largest element.
Note that although Ε is not a largest element, no element other than itself contains it.
Then Ε is called a maximal element. Similarly С and A are maximal elements, whereas
В is not.
(c) If |<P| is finite, then it is clear that Ψ has maximal elements. For we choose any element
Ai of Ψ. If there is an element of Ψ that contains Ai properly, we call it Аг. If there
is an element of Ψ that contains A% properly, we call it A3. Continuing in this way we
get a chain of elements of Ψ, Ax с А& с · ■ · с As с ■ ■ ·. As Ψ has only a finite number
of elements, this chain ends at A„, say. Clearly A„ is a maximal element.

194
ABELIAN GROUPS
[CHAP. 6
(d) On the other hand, not all sets Ψ of sets have maximal elements. For example, let
An = {0,1, .. .,n) for η = 1,2, ... . Let Ψ ~ {Ai\ г £ Ρ, the positive integers). Then
Ψ has no maximal element. For if 1^9, then X = At for some i, and X С Aj+i, but
X^Ai+i.
Zorn's lemma establishes a criterion for determining whether a set Ψ of sets has a
maximal element. What one needs is some condition for handling an ascending sequence
of sets such as the Αι in (d). To state the criterion we need some definitions.
(1) We define A to be a maximal set in Ψ if for each X £ Ψ, XD A implies X — A.
(2) Let С be a subset of Ψ with the property that if Χ, Υ e Q, then either X QY or
Υ QX. Then С is called a chain in Ψ (in (d) above, Ψ itself is a chain).
We are now in a position to state Zorn's lemma: Let Τ be a set of sets. Suppose that
for every chain Q in Ψ, U X is an element of Ψ. Then Ψ has a maximal element.
We will not prove Zorn's lemma. We take it as an axiom. We could assume a more
innocent sounding axiom instead, namely the axiom of choice, which says that an element
from each set may be chosen from a collection of sets. The proof of Zorn's lemma can be
derived from the axiom of choice (see Problem 6.42 for a sketch of the proof).
Using Zorn's lemma we prove
Theorem 6.13: Let G be any abelian group. Then G has a maximal independent set.
Proof: Let Ψ = {X\ XQG and X an independent set). Let С be any chain in Ψ.
Let £ - {Д< I i e Ц· T° apply Zorn's lemma we must show that U = U X,- G Ψ. Clearly
U QG. Is U an independent set? If not, U is a dependent set. This means that it is
possible to find distinct elements uu .. .,iu G U and integers r\, ...,rn such that
TiUi + · ■ · + rnUn — 0
with at least one nm ¥-0. As U = U Xi, ui G Xv, щ G Хг. where 1', 2' are elements of
г G I
1. Then since С *а a chain, either Xv QXv or X& QXv. Thus ui,ua both belong to some
element of Q. Continuing the argument in this way we find that ui, ..., iu, all belong to
some Xi G Q. But this is a contradiction, since every element of С *а independent. So U
is independent and t/e?. We conclude, using Zorn's lemma, that Ψ has a maximal
element and this is precisely the maximal independent set required. Hence the result follows.
From Theorem 6.13 we conclude that if G is of infinite rank, then G has an infinite
maximal independent subset.
Problems
6.38. Find the rank of the additive group Q of rationals.
Solution:
If mjrii and m^/n^ are elements of Q, тъ m2, nlt n2 integers, then
(m2w1)(m1/w1) + (—mjn2)(mz^n2) ~ "
Thus every set of two elements is dependent. Accordingly the rank of Q is 1.
6.39. Show that the p-Priifer group has no independent set consisting of two elements.
Solution :
Let x, у be elements of G, a p-Priifer group, χ Φ 0, у Φ 0. Then χ, у € Cr, say, for some r
(see Example 1, page 191, for the notation Cr). Let Cr ~ gp{g) and let gp(x) be of order p*. Since
VT~tg is of order p', др{рТ~1д) = gp{x). Thus gp(x) = gp{plg) and flfp(j/) = gp(p>g) for some г, j.
If г — ;, it follows that gp(y) 2 gp(x). (If г — /, we merely reverse the roles of χ and y.)
Consequently χ — ry for some integer 0 < r < order of y. Thus (—r)j/ + 1 · χ = 0 and x, у are not
independent.

Sec, 6.2] SIMPLE CLASSIFICATION. STRUCTURE OF TORSION GROUPS 195
6.40. Prove that if Η and К are torsion-free groups of finite rank m and η respectively, then G = if φ Κ
is of rank m + n. (Difficult.)
Solution:
Let ht,.,., km be a set of independent elements of H, and ku ..., kn a set of independent
elements of K. Then the set hu ..., ft^, klt.. ,,kn is independent. For if r^ + · - · + rmhm + sikl +
·■· + впкп = 0, then _
r^! + · · - + rmftm = -s^ enfen
But HnK = {0}, and so
rtht + ■■■ + rmhm = -e^! - · · · - snkn = 0
Thus, by the independence of hi, ..., Am and къ ..., kn,
П = r2 = ··· = rm = βι = β2 = ·■· = βη = 0
Now suppose that {hlt ..., fc^fej, .. .,&„} is not maximal; say, there exists an element h+k where
АеЯ and к е К such that {hlt ..,, km, kt,,.., kn, h + к) is independent. Since {hlt ..., hm, h} is
not independent, there exist integers tt, ..., tm, t not all zero such that
i^i + · · - + tmhm + th = 0
If t = 0 we have a contradiction to {А^ ...,ftm} being independent, as then at least one of tv .. ,,tm
is nonzero. Hence i?4 0.
Next, {k, klt . . .,kn) is not independent, i.e. there exist β, βΙ( ..., βη, not all zero, such that
sfe + etfei + ■ ■ ■ + enfen = 0. Hence arguing as above, it follows that β ¥* 0. Thus
eij/i! + ■ ■ ■ + «£„,*„, + st{h + k) + ίβ^ι + ■ ■ ■ + tsnkn
= e(iA+iifti+ ■·· +tmhm) + t(ak + 8lkl+ h впкп) = 0
But as βί τ* 0, {hv A2, ..., ftm, klr ..., kn> h + k} is not independent. Thus {ht, h2,..., hm, kt, ..., kn)
is a maximal independent set and rank Η φ Κ — m + n.
6.41. (α) If F is free abelian with a finite set of free generators X, prove that the rank of F is \X\.
(b) Prove that F cannot be generated by fewer than \X\ elements.
Solution:
(a) Let X - {xt, ...,xn}. By Theorem 6.7, page 186,
G = др{хг) φ gp(x2) φ ■ · · Φ gp{xn)
We proceed by induction on n. For η = 1, G is infinite cyclic, and the rank of G is clearly 1.
If true for n — r, suppose η — r +1. Then др(хг) φ · · ■ φ gp(xr) is of rank r, and G is
the direct sum of a torsion-free group of rank r and a group of rank 1. By Problem 6.40, G is
thus of rank r + 1, and the result follows by induction for all n.
(b) Let G = gp(gv .. .,gr). X is an independent set. Then by the Steinitz exchange theorem
(Theorem 6.11), as X is dependent on {git.. .,gr}, n — r. Thus we obtain the result.
6.42. Let X be a set and 'P a collection of subsets of X. Suppose that if Λ£9, all subsets of A belong
to Ψ.
(а) Prove that if Де9 is not a maximal element, there exists a set A* = {A, x) € <P with
x&A.
(б) Assume that A* = A if A is maximal, otherwise that A* has been defined equal to {A,x} with
χ g A as stated in (a). Let С be a chain in Ψ, Suppose that if C{, i G /, is a family of elements
in C, then U Cj G £■ Suppose also that if Aef, A* € £■ Prove that <P has a maximal
element. e'
Solution:
(а) If Лет is not maximal, there exists a set Be? such that В ?* Л and ВэЛ. Hence there
exists χ €. В — A. Since {Л, χ) is a subset of B, {A, x) 6 9*.
(б) Let Μ = и С Then Mef, and so M* e C· But Λί* Э Μ. However, Μ contains every
cec
element of C, in particular, it contains M*. Therefore Μ = Μ* and we conclude that Μ is a
maximal element of Ψ.
Remarke. (1) In assuming that A* can be defined we used implicitly the axiom of choice.
(2) The proof of Zorn's lemma requires converting the theorem into this problem. For details
see P. R. Halmos, Naive Set Theory, Van Nostrand, 1960.

196
ABELIAN GROUPS
[CHAP. 6
6.43. Let G be an arbitrary non-abelian group. Prove that G has a maximal abelian subgroup (i.e. one
that is not properly contained in an abelian subgroup of G).
Solution:
We use multiplicative notation for G since it is not abelian. Let Ψ be the set of all abelian
subgroups of G. Let Γ be a chain in Ψ and let U = U X· Then U is a subgroup of G. For if
g, h £ U and if g belongs to Xt € С and A to ΧΆ £ С. then as either Xl с Хй or X2 Q Xt, it
follows that g, h belong to some element X of C· Hence gh*1 € X as X is a subgroup, and gh~y € U.
Also, fifft. = Λ-fif as X is an abelian subgroup of G. Consequently U is abelian. Hence U e <P. By
Zorn's lemma, Ψ has a maximal element M, say. Μ is the maximal abelian group sought.
6.3 FINITELY GENERATED ABELIAN GROUPS
a. Lemmas for finitely generated free abelian groups
In Section 6.3b we will show that all finitely generated abelian groups are direct sums
of cyclic groups. We will do this by using a lemma (Lemma 6.15) about subgroups of free
abelian groups. The relationship between Lemma 6.15 and finitely generated abelian
groups is easily obtained by noting that all abelian groups are factor groups of free abelian
groups.
Lemma 6.14: Let G = gp(ai) φ · · ■ φ др{ап) be the direct sum of infinite cyclic groups.
If bt = αϊ + г2а2 + * ■ · + rnOn, where r%, . ■., rn are any integers, then
G = flrp(bi) Φ gp(ch) θ ■ ■ · Φ gp(an)
Proof: As gp(bi,ch, .. .,a«) = gp(ah .. .,a«) = G, we must only show that if Si, .. .,sn
are any integers, then
Sibi + S2CI2 + · · · + Sndn = О (6Л)
implies all Bi are 0.
Substituting Eh = ai + r202+ · - * +r„an into {6.b) and collecting terms, we obtain
Sidi + (S2 + Bin)as + · · ■ + (βη + 8ιΤη)θη = 0
As G= {d} Φ · · ■ θ {α„},
Si = S2 + 81Г2. = ■ ■ ■ = S„ + 8irn = 0
Thus 81 = 82= ■ · · = sn = 0 and the proof is complete.
The next lemma is a crucial one. We recall that a basis cu ..., cn for a finitely generated
free abelian group G is a set of elements such that G = gp(ci) Φ · ■ · θ gp(cn) (see Section
6.1c).
Lemma 6.15: Let G be free abelian, the direct sum of η cyclic groups. Let Я be a subgroup
of G. Then there exists a basis ci, ..., c„ of G and integers щ, ..,, и* such
that Я = gp(uicltuzc2, .. .,гисп).
Proof: We use a, b, с to denote basis elements of G, h, k, I to denote elements of Я,
q, r, 8, t, u, ν to denote integers. We prove the result by induction on n. For n = l, G is
cyclic and the result is a consequence of Theorem 4.9, page 105. Assume the result is true
for free abelian groups of rank less than η where η > 1. Let G be free abelian of rank n.
We assume also that Я Φ {0}. For if Я = {0}, we may take an arbitrary basis Ci, ..., cn
for G. Then Я = gp(uicb ..., uncn) where 1*1 = · · · = и» = 0.
To every basis we associate an integer, called its size (with respect to H). Let {αϊ, ..., an}
be a basis for G and let q be the smallest nonnegative integer such that there exists h G Η
h = qai + qzCh + · · · + qnan, q%, . . ., <?» integers (6.5)
Then q is termed the size of the basis {αϊ, α2, ..., α„}.
Assume {ait ...,&*} is a basis of smallest size, i.e. if {bu .. ·, bn] is a basis of G, then the
size of {Eh,..., bn} is not less than q.

Sec 6.3]
FINITELY GENERATED ABELIAN GROUPS
197
Let h be as in equation (6.5). We show that q divides q-ι, ..., qn. From the division
algorithm, if qt is not divisible by q, qt = nq + si where 0 < Si < q. Hence
h = q(di + na-i) + ■ · - + s&i + · · · + qn<in
But if we put bt = сц, b2 = a2, ..., bf = at + па;, ,.., b„ = a», we obtain a basis by Lemma
€.14. Furthermore this basis is of smaller size than the size of {α·ι,..., On}, contrary to our
assumption. Thus se = 0 and q divides qt for i = 2,...,n. Let qi = r{q. Then
h = q(ai + ггОг + l· rnan)
Let cY = α·ι + r-aa2 + ■ · · + rnOn. Then, by Lemma 6.14, {Ci, a2,..., a»} is a basis for G. Also
A = qci (6.6)
If k = ti&i 4 + tna„ G Я, it follows that ii is divisible by q. For if ii = uq + υ with
О ^ г; < q, then 1 = k — uh G Я has г> as its coefficient of αχ. As υ < q, by the minimality
of α, ν = 0. Therefore
i = k-uh G flfp(a2, - -., ctn)
Hence I G gp(a2, .. .,а„)Г\Н = L, say. From this we conclude that if к G H, then
к = uh + l (6.7)
where Ζ G L.
By the inductive hypothesis there exist a basis c2,.. .,c„ and integers щ, .. .,%n such
that L is generated by игс%,.. .,UnCn. Hence by (6.7) every element of Я belongs to
gp(h,uzC2, .. .,UnCn). On the other hand, Я contains h, wac2, ..., ^с*. Thus
Я = до(/г., МгСг, ..., w»c„)
Put Wi = q. By (ff.ff),
Я = gp(uiCi, И2С2, ..., WnCn)
Also, Ci,..., cn is a basis for G. Hence the result follows.
Note that if any г*,· is negative, we can replace d by its inverse — c{. In this manner we
can assume that the im are nonnegative.
Lemma 6.16: Suppose G = Α © В. Let Аь ΒΎ be subgroups with Ai С A, BiQB and
N = Ai +Bi. Then G/tf « Α/Αι Φ B/Bi.
Proo/: Let K = A/Ai®B/Bu and let 0:Ah»A/Ai and φ:Β^Β/Βι be the natural
homomorphisms. θ, φ extend to a homomorphism Φ of G into K. Then Ker Φ D Ker θ = Αι
and Ker Φ D Ker φ = Bu Thus Ker*DAi+Bi. Now let aGKer*. Then x = a + b,
a&A,b&B. x* = (a + Ai) + (b + Bi) and this is the identity element only if a. G Ai
and b G Bi. Hence χ G Ai +BY, and so Ker* = Ai + Blm By the homomorphism theorem
(Theorem 4.18, page 117) GIN <* K and the result follows.
Corollary 6.17: Let G be free abelian with basis cb ...,c„. Let Я = gp(uiCi,..., UnCn)
where Wi, ..., un are nonnegative integers. Then G/H is the direct sum of
cyclic groups of orders u[, ..., ιά, where «,' = m if ги ¥- 0 and wi = °° if
wt = 0.
Proof: The result follows by repeated application of Lemma 6.16.
b. Fundamental theorem of abelian groups
The following theorem is called the fundamental theorem of abelian groups.
Theorem 6.18: Let G be a finitely generated abelian group. Then G is the direct sum of a
finite number of cyclic groups.

198
ABE LI AN GROUPS
[CHAP. 6
Proof: G a« F/H where F is a finitely generated free abelian group (Section 6,1c). By-
Lemma 6.15, F has a basis cu ..., c„ such that Η = gp(uiCi, .. .,UnCn) for some nonnegative
integers uh ..., u*. We now apply Corollary 6.17 to conclude that G ^ F/H is the direct
sum of cyclic groups.
Corollary 6.19: If G is finitely generated, it is the direct sum of a finite number of infinite
cyclic groups and cyclic groups of prime power order.
Proof: It is only necessary to show that a cyclic group of composite order is the direct
sum of cyclic groups of prime power order. This we have already done in Problem 6.19,
page 187.
Corollary 6.20; If G is a group with no elements of finite order and G is finitely generated,
then G is free abelian.
Proof: G is the direct sum of a finite number of cyclic groups each of which must be
infinite cyclic as G has no elements of finite order. Thus the result follows.
Problems
6.44. Prove that every finitely generated torsion group is finite.
Solution:
By the fundamental theorem of finitely generated abelian groups, if G is finitely generated it is
the direct sum of a finite number of cyclic groups. If G is a torsion group, then it is the direct sum
of a finite number of finite cyclic groups. Hence G is finite. (Compare with Problem 4.31, page ЮБ.)
00
6.45, Let G = 2 Gi where G( is a cyclic group of order 2 for i = 1, 2,... . Prove that G is not
finitely generated.
Solution:
Every element in G is of finite order, for if g(*& 1) G G, g = gl + дг + · · · + дл, £Г{ G Gf, (г' G Z),
and 2g = 2дг + ■ ■ ■ + 2gn — 0 + ■ · ■ + 0 = 0. Thus G is a torsion group. If G were finitely
generated, G would be finite by the preceding problem. But G is clearly infinite. Therefore G cannot
be finitely generated.
c. The type of a finitely generated abelian group
In Section 6.3b we proved that a finitely generated abelian group is a direct sum of
cyclic groups. However, such a decomposition is not unique: first, the direct summands
are not unique (see Problem 6.46 below); moreover, the number of direct summands can
vary (see Problem 6.19, page 187).
We say that two decompositions are of the same kind if they have the same number
of summands of each order. For example, two decompositions of a group into the direct sum
of three cyclic groups of order 4 and two cyclic groups of infinite order are said to be of
the same kind. A concrete example of two decompositions of the same kind is given in
Problem 6.46.
As we remarked in Corollary 6.19, every finitely generated group can be decomposed
into the direct sum of a finite number of cyclic groups of prime power or else infinite order.
Our aim is to prove
Theorem 6.21: Any two decompositions of a group G into the direct sum of a finite
number of cyclic groups which are either of prime power order (¥· 1) or of
infinite order, are of the same kind.
Proof: We shall separate the proof into four cases: (1) both decompositions involve only
infinite cyclic groups, (2) both decompositions involve only cyclic groups of order a power
of fixed prime p, (3) both decompositions involve no infinite cyclic groups, and (4) the
general case.

Sec. 6.3]
FINITELY GENERATED ABELIAN GROUPS
199
Case I. G - h © · ■ · © h = Λ © ■ · ■ © h
л.
where I,, It for j = 1, ..., к and г = 1, ..., I respectively, are infinite cyclic groups.
From Corollary 6.12, page 193, we conclude that к = l. (Alternatively we may
proceed as in Problem 6.52.)
Case 2.
Both decompositions involve only cyclic groups of order a power of a fixed prime p.
We shall write for any integer n, nG = {ng \ g G G}. If G is a group, nG is a subgroup
(Problem 6.53). To prove case 2 we will need the following lemma.
Lemma 6.22: Let G = Α θ В. If η is any integer, then nG = nA® nB.
Proof: As пАПпВ CAnB = (0), gp{nA,nB) = nA ΦnB. If gSnG, there exists
hG G such that nh = g. Let h = a+b, a E A and b e B. Then g — nh = na + nb.
Accordingly ftGCnAffinBcnG and so nG = nA θ nB,
Corollary 6.23: Let G = Αι θ · · · Φ Ak. Let η be an integer. Then
nG = пАг θ · · · Φ nAk
Proof: We apply Lemma 6.22 to one direct summand at a time. Then the result
follows.
Corollary 6.24: Let G be expressed as the direct sum of fa cyclic groups of order p\
l^i^r. Then pG is expressible as the direct sum of fa cyclic groups
of order pi_1 where 2 — г — r.
Proof: This is an immediate consequence of Corollary 6.23 and the fact that if A
is cyclic of order φ\ ρ A is cyclic of order pi_1. Hence the corollary follows.
We are now in a position to prove case 2. We proceed by induction on the order of
G. If \G\ = 1 or p, then the result is immediate. If the result is assumed true for all
groups of order less than η that satisfy the conditions of case 2, then let \G\=n.
Suppose G is expressed as the direct sum of fa cyclic groups of order pl for 1 ^ г — r, and
also as ls cyclic groups of order pi for 1 — j — s. Then pG is expressible (by Corollary
6.24) as the direct sum of fa cyclic groups of order pi_1 for 2 ^ i^r on the one hand,
and as the direct sum of U cyclic groups of order p1_1 for 2 — г — s on the other. As
\pG\ < \G\, it follows by the induction assumption that r = s and fa = h for 2 — i — r.
Now we must still prove that Jfci = ib But \G\ = pfci(^)^ · ■ · (ρή*' = pb(p*)l> · · ■ [prfr,
and so U = hi. Thus we have proved both decompositions are of the same kind, as
required.
Case 3.
G is expressed in two ways as the direct sum of a finite number of cyclic groups of
prime power order.
We have dealt with the case where only one prime is involved. We proceed by
induction on the number of primes involved. Let ρ be one of the primes involved. Let
Ai, . .., A„ be all the direct summands of order a power of φ in the one decomposition,
Bi, .. .,Bn the other direct summands involved, so that
G = Αχ Φ · · · Φ Am θ Bi Φ ■ ■ ■ θ Bn
Putting А = Αι Φ · · - φ A™ and β = Βι®···φβη, it follows that G = Α Φ Β.
Let Χι, ..., Хк be all the direct summands of order a power of ρ in the second
decomposition, Yi,..., Yi the remaining direct summands, so that
G = Χι Φ ■ · · Φ Хк Φ Υι θ · ■ · Φ Υι
Put Χ = Хг Φ · · · Φ Хк, Υ = Υι Φ · · ■ θ Υι. Then G = X®Y. We claim that A = X
and Β = Υ.

200
ABELIAN GROUPS
[CHAP. 6
Let g GA. Then g = x+y where χ G X and у G Y. Now the order of any-
nonzero element of Υ is coprime to p. As g is of order a power of p, у = 0 (Problem
6.54). Hence g G X, and so AqX. Similarly XqA and we conclude that A = X,
By a similar argument В — Υ.
Thus Αι Φ · ■ ■ © Am = Χι © ■ ■ · © Zfc and Β,Φ ·■■ ® Д„ = Г,Ф ··· ΦΓ,. By the
induction hypothesis, Αι Φ · ■ ■ Φ Am and Χι Φ · · · Φ Xk on the one hand, and #ι Φ · · · Φ Βη
and Y\ Φ · · · Φ Υι on the other, are of the same kind. Hence the two decompositions are
of the same kind and the result follows.
Case 4.
Let G be expressed as the direct sum of cyclic groups of prime power order or of
infinite order in two ways, say
G = Λ θ ■ · ■ Φ Im Φ Fr Φ ■ · ■ Φ Fn = Λ Φ ■ ■ ■ Φ h Φ Ft Φ ■ - ■ θ Ft
where Ι,,Ιι are infinite cyclic groups and Fj,Fj are groups of prime power order.
Let T(G) be the set of all elements of finite order (see Theorem 6.9, page 189). Then
T(G) is the direct sum of the direct summands of finite order in both cases (Problem
6.55). Thus
T{G) = Fi Φ · ■ ■ Θ Fn = Ft Φ ■ - ■ Φ Fi
Hence by case 3, Fi θ · - · θ Fm and Ρλ φ - · · φ /\ are of the same kind.
Also GIT(G) « /ι Φ · · · Φ /m = /ι Φ · ■ ■ θ Ik (by Problem 6.11, page 181). By
Problem 6.56, /ι Φ · · · Φ /m is the direct sum of к infinite cyclic groups. Then к = т by
case 1. Therefore we have proved that h Φ h Φ · · · θ Im θ Fi θ · · · Φ Fn and /ι Φ % Φ
• · · Φ h Φ Fi Φ · · · θ Fi are of the same kind. This completes the proof of the theorem.
If a finitely generated group G is the direct sum of cyclic groups of orders p\l,.. .,p^
and s infinite cyclic groups, where Pi,...,Pk are primes, pi — pi— · · · — Pk, гг,.. .,rk
positive integers with ri^ri+i if ρί = ρί+ι, then the ordered fe + 1-tuple (p[l,.. .,prkk;s) is
called the type of G. (The definition of type differs slightly from book to book. Usually it
is applied only to p-groups.) By Theorem 6.21 the type of G is uniquely defined. We can
now give a criterion for the isomorphism of two finitely generated abelian groups.
Theorem 6.25: If F and G are two finitely generated groups, then they are isomorphic if
and only if they have the same type.
Proof: Let F = Αι Φ · · · Φ Ak. If φ-.F^G is an isomorphism, then G = ΑιφΦ
··· (ВАкф (Problem 6.56). As Αιφ = Αι, it follows that F and G have the same type.
Conversely, if F and G have the same type they are clearly isomorphic (Theorem 6.5,
page 185).
Problems
6.46. Let G = Α φ Β where A and В are cyclic of order 2. Find С and D such that G - С ф D
where С and D are cyclic of order 2 and С Ф A and С ¥*B.
Solution:
Let A = {Ο,α}, Β = {0,b}. Put С - {0,a+ b}. Then С is cyclic of order 2. Also put D - B.
Then С +D = {О, а+Ъ, Ъ,а+Ь+Ь = a}, and so С +D = G. Also CnD = {0}. Thus G - С ф D.
6.47. If the type of F is (/lt..., fk; f) and that of G is (glt..'.,gp; g) where Д = p^, /2 = рт2г, . .-,fk= PTkk,
дг — q*it ..., gt= q"1 and p* < qv where the p{ and g{ are primes, find the type of F φ G.
Solution :
We have p, - рг - *"' - Pk < Qi ~ '" ~ Qv Hence the type of F φ G is
(/i, .. .,fk>9\, · ■-,δϊ,ί + ο)

Sec. 6.3]
FINITELY GENERATED ABELIAN GROUPS
201
6.48. If F, G and Η are finitely generated abelian groups, show that F © G a F φ Η implies that
GszH.
Solution:
Express F, G and Η as direct sums of cyclic groups of prime power and infinite orders. If the
type of F is (/j fk; /), and that of G is (glt..., #,; g) while that of Я is (fc„ ..., fem; ft), then the
type of F φ G is (av .. .,β^ + ιϊ / + ff) where at,.. .,afc + l is /t,.. .,fkl gb .. .,дг in some order, while
the type of F φ Η is (bt, . ..,bfc+m; /+fc) where bb .. .,bk + m is fv.. .,fk,hb ...,hm in some order.
For two abelian groups to be isomorphic we require that (by Theorem 6.25) their types are the same.
Accordingly the types of G and Η are the same and G a H.
6.49. Find up to isomorphism all abelian groups of order 1800.
Solution:
Observe that 1800 = 2*№&t. So an abelian group of order 1800 is a direet sum of a group of
order 23, a group of order 32 and a group of order 52. The possible types of a group of order 23
are (23; 0), (22, 2; 0), (2,2,2; 0). Thus there are precisely 3 groups of order 8. The possible types
of a group of order 32 are (32; 0) and (3, 3; 0), so there are 2 non-isomorphic groups of order 9.
Similarly there are 2 non-isomorphic groups of order 25. Then the total number of non-isomorphic
groups of order 1800 is 3X2X2 = 12.
Compare the ease with which we solve this problem with the effort required to find all the
groups (non-abelian as well as abelian) of order 8 which we have considered in Chapter 5»
6.50. Let ρ be any prime and let m be any integer. Prove that the number of groups of order pm is equal
to the number of ways of writing m — гг + - · ■ + rk where r„ ..., rk are positive integers and
*4 - r2 - ''' -
resolution:
A group of order pm has all its elements of order a power of p. Hence its type will be of
the form (pri, р'г, .. ., prfc; 0) with »Ί — *"2 - *" * — *Ъ· Since G is of order pm, and
\G\ = pripr2---pTb = ρτι + · ■+rk
we conclude that гг+ · · · +rk = m.
6.51. Prove that a cyclic group of order pn, where ρ is a prime, is not expressible as the direct sum of
nontrivial subgroups by the following two methods: (1) direetly, (2) by using Theorem 6.21.
Solution:
(1) Suppose G = Α φ Β where А, В are nontrivial subgroups of G. Clearly |A| —p""1 and
|В|±гр«-1 as |G| = p». Let G = gp(g). Then g = a+b,aGA, and bGB. As p""1» =
ρ"_1α + ρη-ι& = 0, g is of order less than ρη_1. But gp(g) = G and is of order pn. Thus we
have a contradiction.
(2) Since G is cyclic of order pn, the type of G is (pn; 0). Hence by Theorem 6.21, only one of the
direet summands is nonzero, i.e. G cannot be expressed as a direct sum of more than one non-
trivial group.
4J.52. Prove, by considering the direct sum of cyclic groups of order 2, that if G is the direet sum of к
infinite cyclic groups and also the direct sum of I infinite cyclic groups, then к = I.
Solution:
Let G = gp{xt) φ ·-· ® gp{xk). Let Η = gp{2xv ...,2xk). Then by Corollary 6.17, G/H
is the direct sum of к cyclic groups of order 2. Thus \G/H\ = 24 Clearly Η С 2G. Also if gGG,
g = rxxx + ■ · · + rkxk. Then 2g = τ^2χΎ) + · ■ · +rk{2xk) & H, from which 2GQH. Thus H = 2G.
Now by a similar argument we conclude that if G is the direct sum of I infinite cyclic groups,
|G72G| = 2'. Thus I = k.
6.53. Prove that nG is a subgroup of G where и is a given integer.
Solution:
If А, к e nG, h = nf, k = ng where f.gGG. Hence A - к = n(f - g) e nG, and so nG is
a subgroup.

202
ABELIAN GROUPS
[CHAP. 6
6.54. Let G be an abelian group, G = Χ φ Υ. Let χ GX, yGY. Prove that (1) if * and у are of
finite order, then the order of x + y is the least common multiple (1cm) of the orders of χ and y;
(2) if * is of infinite order, χ + у is of infinite order.
Solution:
(1) Let ί = lcm of the orders of * and y. Then l(x + y) — he + ly — 0. Now if m = order of * + y,
then m(x + y) = mx + my = 0 implies mx = 0 and my = 0. This in turn implies that the
order of χ divides m and the order of у divides m. Thus we have the result.
(2) If χ is of infinite order and m(x + y) = 0, then « + my = 0. But by the uniqueness of such
expressions in direct sums, mx = my — 0. Since * is of infinite order, m = 0.
6.55. Let G = Ix φ ■ · · Φ /m Φ ίΊ Φ · * · Φ Fn where each Ii is torsion-free and each F{ finite. Let
T(G) be the set of all elements of finite order. Prove that
T(G) = F, φ ■ · · φ Fn
Solution:
Clearly T(G) Э F, φ · ·. φ Fn. If g<=T(G), g = h+·■ ■ +im + f1+··· + fn where i„ ..., im
are elements of Ilf...,Im respectively, and Д, ...,/„ are elements of Flt...,Fn respectively. As
g is of finite order r, say,
rg = rt1 +ri2+ ■■■ + rim+ rfr+ ··■ + rfn = 0
By definition of the direct sum, it follows that
rii = «a = " ·' = "m = rf\ = "" = rfn = 0
Since /,,..., /m are torsion-free, we have ч = t2 = · · ■ = V = 0. Thus g e Fr φ · · · φ Fn and
the result follows.
6.56. If F - Α, φ · · · φ Ak and ψ: F -> G is an isomorphism, then G = Αλφ φ ■ · · φ Акф.
Solution:
We must show that every element of G is uniquely of the form αλψ + · · · +акф where alt ..., a^
belong to Аъ...,Ак respectively. Now if g G G, there exists /6F such that f<p~g. But
/ = a, + · · · + Ofc and so g = αλφ + · · - + α^ψ. If αΎψ + ■ ■ ■ + α^ψ — α[ψ + · · · + αΗψ, then
(ttj - α[)ψ + ■ · - + (afc - <4)ψ = 0
Let h = аг — a{ + · · · + (fy — Ofr. h belongs to Ker ψ. Since ψ is an isomorphism, h = 0.
By the uniqueness of expression of direct sums,
ii ι Λ
αϊ — tt] = a2 — «2 — — — afc — Ofc = 0
so that »i — »ίι »2 = aL · · ·. ak — ^ί· Therefore each element of G is expressible in the form
αΎψ + ■ · · + Οχφ in one and only one way.
d. Subgroups of finitely generated abelian groups
The purpose of this section is to decide which groups (up to isomorphism) can appear
as subgroups of finitely generated abelian groups. We begin with
Theorem 6.26: Let G be free abelian of rank n. Then any subgroup Η of G is free abelian
of rank less than or equal to и.
Proof: By Lemma 6.15, page 196, there exist a basis съ .. .,cn of G and integers
Mi,.. .,Un such that Η = др(шсг,.. .,гиса). If иъ .. .,ιμ are nonzero, and Mi+i = г^+г =
·■·-«„- , en gp(UlCu . . .,M„C„) = ffP{UiCi) Φ · · · Φ др{ЩСг)
(See Problem 6.57.) Hence the result.
Corollary 6.27: Let A be a finitely generated abelian group. Then every subgroup of A
is finitely generated.
Proof: As A is a finitely generated abelian group, it is isomorphic to some factor group
of a finitely generated free abelian group G, say A ™ G/N. The subgroups of G/N are of
the form H/N where Я is a subgroup of G. By Theorem 6.26, Я is finitely generated and
therefore so is H/N. Consequently every subgroup of A is finitely generated.

Sec. 6.3]
FINITELY GENERATED ABELIAN GROUPS
203
From this corollary we see that only finitely generated abelian groups can occur as
subgroups of finitely generated abelian groups.
Theorem 6.28: Let G be a finitely generated group and Η a subgroup of G. Let G and Η
be expressed as direct sums of infinite cyclic groups and cyclic groups of
prime power order. If the number of infinite cyclic groups in these
decompositions for G and Я are m and к respectively, then k— m.
Proof: Let
G - h © · ■ ■ © Im © Λ Φ ■ ■ ■ © Fn, Η = /ι Φ ■ · ■ Φ h Φ Α Φ ■ ■ · Φ Α
where the h,Ii are infinite cyclic groups and the F{,Fi are cyclic groups of prime power
order. Let T(G) and T(H) be the torsion subgroups of G and Я respectively, i.e. the
respective sets of elements of finite order (Theorem 6.9, page 189). Then T{H) - HnT{G).
Now G/T{G) ss h Φ · · · Φ Im by Problem 6.11, page 181. Thus the rank of G/T{G) is m.
(See the remarks following Theorem 6.11, page 192.) Since (Я + T{G))/T{G)CG/T{G),
(Я + T{G))/T{G) is free abelian of rank less than m, by Theorem 6.26. But
{H + T{G))/T{G) = H/HnT{G) a H/T{H)
by the subgroup isomorphism theorem (Theorem 4.23, page 125). It follows as before
that H/T{H) = /ι Φ ■ · · Φ h. Thus k^m.
Again let G be finitely generated and Я a subgroup of G. (Recall that if F is any abelian
group and ρ a prime, Fv — {f\ f e F and of order a power of p}.) Gv, as a subgroup of a
finitely generated abelian group, is finitely generated (Corollary 6.27) and so is Яр. Clearly
Gv D Яр. Thus we are led to inquire what groups can occur as subgroups of finitely
generated p-groups. Of course a finitely generated p-group is finite (Problem 6.44).
We first require a lemma.
Lemma 6.29: Let G have type (p*"1, .. .,pr"; 0). Then the number of elements of order ρ
in G is pn — 1.
Proof: Let G = Ci Φ ■ ■ ■ Φ Cn where each Ci — gp{d) and the order of d is p4. If
χ e G is of order p, and χ — Uci + ■ ■ · + i„c„ where U e Z, then pTi~l divides U. Hence
the elements of order ρ are a subset of H, where
Η = j/i-'Ci θ · · ■ Φ p^-'Cn
On the other hand every element {¥* 0) of Я is of order p, so Я — {0} is the set of all elements
of order p. Accordingly, as |Я| = pn, the number of elements of order ρ in G is pn — 1.
Theorem 6.30: Let G be a group with type (pr\pr2, .. .,pTm; 0). Let Η be any subgroup.
If the type of Η is (p'1, ..., p'n; 0), then η — m and 0 < s{ ^ n, г = 1, ..., п.
Proof: We proceed by induction on |G|. If \G\ — lorp, the result is trivial. Hence
assume |G| > p, and the result holds for groups of order less than |G|. Now Η has type
(pSl,p4, .. .,ps*;0), and so the number of elements of order ρ in Я is, by Lemma 6.29,
p" —1. Similarly, the number of elements of order ρ in G is, by Lemma 6.29, pm — 1.
Clearly pn — 1 ^ pm — 1 and consequently η — m.
Now \pG\ < |G|. We can therefore assume the result holds by induction for pG and
its subgroup pH.
At this point we experience a minor notational inconvenience. If for example rm > 1,
pG is of type (pri~\ .. .,pr"»~1;0). However, if rm = 1 and rm-\ > 1, pG is of type
(pri_1, .. .,prm-i_1;0). Therefore we need additional notation. Define m* — m if rm> 1;
otherwise define m* to be an integer such that rm* > 1, but rm*+i =1. As η ~ гг - ■■ ■ ~
rm. > 1 and, if m* Φ m, rm*+\— ■ · · = rm = 1, then pG is of type (p1"1"1, . - .,prm*_1; 0).

204
ABELIAN GROUPS
[CHAP. 6
Similarly, let us define η* = η if s» > 1; otherwise define n* to be an integer such that
sn. > 1 but s„»+1 = 1. Arguing as in the paragraph above, pH is of type (p*1-1,..., j/"*-1; 0).
Then by the inductive hypothesis we have
n* - 1 ^ m* - 1 and sf - 1 ^ η - 1 for i = 1, ..., n*
If n* = и, the result follows immediately. If n* ¥= n, then s„»+i = · ■ · = sn = 1. Hence
tv+i — s„.+i, ..., r„^ s„. Thus Si^n for i = l, ...,».
With Theorems 6.28 and 6.30 it is easy to determine, knowing the type of a given finitely
generated abelian group G, the possible types of subgroups of G. (See Problem 6.60.)
It can be shown (Problems 6.62-65) that every factor group of a finite abelian group G
is isomorphic to a subgroup of G. Therefore we know the types of homomorphic images
of finite abelian groups.
Problems
6.57. Let G = A®B and let C,D be subgroups of А,В respectively. Show that C + D = C®D.
(This can obviously be generalized to the direct sum of any number of groups.)
Solution:
As {0} = AnB^CnD, we have CnD = {0}. Thus C + D = C®O.
6.58. Let G have type (j/i,.. -,Ρ^"' ■)* SuPP°Be that for some i — j
Pi = Pi + i= ··· =Pj, Pi-ι^Ρι and Pj^Pj+i
Put ρ = pj. Show that the type of Gp is (pr*, .. .,prs; 0).
Solution:
Decompose G into the direct sum of infinite cyclic groups and groups of prime power order.
Clearly G = A4 φ · ■ ■ φ A} φ В where Ak is of order }A for i - к - j, and В is the direct sum
of the cyclic groups which are not of order a power of this prime ρ in the given decomposition of
G. Then Gp3 Α{ φ ·■ ■ φ Aj. On the other hand, as any nonzero element of finite order from В
is of order coprime to that of p, Gp С A{ φ ■ ■ ■ φ A}. Thus Gp = At φ · · · φ AJ; and the result
follows.
6.59. Let G be of type (ρΓι,..., pr™; и). Show that G has subgroups of type (p"1,..., p*n; v) where η — m
and 1 — 8j — r{ for 1 — г — η and ν — и.
Solution:
Let G = Aj φ · · ■ φ Am φ Ιχ φ · · ■ φ Iu where A( is cyclic of order pri, and Iu .. -,/u are
infinite cyclic groups. Now each A{ has a subgroup B% of order p**, 1 — i — n. By repeated
application of Problem 6.57,
JSi + - ■ · + Bn + It + ■ · · + /„ = By φ ■ · ■ φ Bn φ h φ ■ ■ · φ /„
This is then a subgroup of G of type (p*1,..., p8™; «), as required.
6.60. Let G be of type (33,32, 52,73; 1). Determine whether G has a subgroup Η of type
(a) (3,3,72,7; 1), (b) (3,3, 5, 7; 2), (β) (3», 3*, 5*, 73; 0), (d) (3, 3,7; 1).
Solution:
(a) No, as then G7 is of type (7s; 0) whereas H7 is of type (72, 7; 0) by Problem 6.68. This is a
contradiction to Theorem 6.30.
(b) No. A direct contradiction to Theorem 6.28.
(c) No. Compare Gs and H$ as in (a).
(d) Yes.
6.61. Give an infinite number of examples of an abelian p-group that contains exactly ρ +1 subgroups of
order p.
Solution:
If G is a group with ρ + 1 subgroups of order p, each of them contributes ρ — 1 distinct elements
(the identity is common to all) of order p. Hence in all there are (p — l)(p + 1) = p2 — 1 elements
of order p.

Sec. 6.4]
DIVISIBLE GROUPS
205
By Lemma 6.29 a p-group with p2 — 1 distinct elements of order ρ contains p2 suramands which
are cyclic groups of order a power of p. Let 1 — rit where t = 1, . .., p2, be integers. Then any
group of type (pri, .. .,prP; u) where и is a nonnegative integer, has exactly p2 — 1 elements of order
ρ and thus exactly ρ + 1 subgroups of order p.
6.62. Let G be a p-group. Suppose G = gp(a) φ В. Prove that G = gp(a + b) φ В where bGB is
of order less than or equal to the order of a.
Solution:
If as e gp{a + b) П B, then ж = r(a+b) = bj where blGB and r is an integer. Thus
ra= bt — rb. Since gp(a)nB = {0}, ra= 0. Then r is divisible by the power of ρ which is the
order of a. Consequently rb = 0. Hence by = 0 and χ = 0. Clearly gp(a +b) + В = G and the
result follows.
6.63. Let G be a unite p-group. Prove that if j6G and g is of order p, g appears as an element of
a cyclic direct summand of G.
Solution:
G is the direct sum of cyclic groups, say G = gp{c\) © ** · © ШКО· И д = 0, the result
follows immediately. Otherwise, without loss of generality, suppose that
9 = rlP lcl + · · ■ + rmV m"m
where (rt,p) = 1, г<р^сх Φ 0 and Wj as w2 — ■ · · — wn. Put гхеу—с[. Clearly gp(c[) = gpicj.
Then g = pwi(c[ + d) where d € flrp(e2, ..., c„). As цг is of order ρ and pttlci ^ 0, the order of d
is less than or equal to the order of c{. By Problem 6.62, on putting e = c[+ d, we obtain
G - gp(°) © gp(<>2· ■■■>"„)
Since g G gp (c), the result follows.
6.64. Let G be a finite p-group. Let N = gp(g) be of order p. Prove that G/iV is isomorphic to a
subgroup of G.
Solution:
By Problem 6.63, G = gp(e) φ Β where flr S srp(c). Then G/N = (gp(c)/N) φ Β (Lemma
6.16, page 197). But clearly gp(pc) = gp(c)/N. Thus flrp(pc, B) sb G/JV.
6.65. Let G be a finite group. Prove by induction on \G\ that if N is a subgroup of G, G/N is isomorphic
to a subgroup of G.
Solution:
Assume the result is true for all groups of order less than r. Let \G\ = r and let N be a
subgroup of G. If N = {0}, G/W 3- G and there is nothing to prove. If N is of order a prime p,
G = Gp φ # where Gp is the p-component of G, and JV с Gp. Then G/W ^ (G„/JV) ф£ by
Lemma 6.16. Now GJN и Η, a subgroup of Gp, by the preceding problem. Hence G/N sflfflS
and Η φ Ε is a subgroup of G. If N is not of order a prime, there is an element щ of N of order a
prime ρ by Proposition 5.9, page 137. Let N0 ~ gp(n(,). Then (G/N0)/(N/N0) = G/N. As
|G/2V0[ < \G\, G/N0 has a subgroup Η/Νϋ ss G/JV. Η Φ G, since otherwise N = N„ which is not
true. Thus [#| < |G[ and by induction it has a subgroup К such that К s #/JVfl =* G/W. The
result follows.
6.4 DIVISIBLE GROUPS
a. p-Prufer groups. Divisible subgroups
A group G is said to be divisible if for each integer η Φ 0 and each element g € G
there exists hGG such that nh — g. Both the additive group of rationale and p-Priifer
groups are divisible in this sense (Problem 6.66 below).
If the groups Gif г € /, are divisible, then Ύ, Gi is divisible. For if η Φ 0 is any
(ei
integ«r and g € 2^> Git then g — gt + * * · + gk, say. So there exist hi,..., Ы such that
fe ι
nhi - gh ..,, «и* = flrk. Then
n(fci Η +hk) = ffi + ■ · · + ffk = fif

206
ABELIAN GROUPS
[CHAP. 6
It follows that direct sums of p-Prufer groups and copies of the additive group of
rationale are also divisible. We show that in fact this exhausts all divisible groups.
To prove this we need several facts.
First we remark that a homomorphic image of a divisible group is divisible. For
suppose G is divisible and Я is a subgroup of G. Let g + Η e G/H (g e G) and let η be a
positive integer. Then there exists g' eG such that ng' = g. Accordingly n{g' + H) =
g + H, and so G/H is divisible.
Secondly we remark that if G = Η ®K and G is divisible, so also are Η and K, since
Η s· G/K (Problem 6.11, page 181) is a homomorphic image of a divisible group. Similarly
К is divisible.
Next we need the following theorem which classifies p-Priifer groups.
Theorem 6.31 (Main Theorem on p-Prtifer Groups): Let ρ be a prime. Let G be a group
which is the union of an ascending sequence of subgroups Ci С С2 С · · ·
where CV is cyclic of order pr for τ — 1,2,... . Then G is isomorphic to the
p-Prufer group.
Proof: We may suppose that Cr = gp{cr) and that pcr+i = cr for r = 1,2,... (see Problem
6.67 for the details). Define Θ-.G-* {Q/Z)P by {mcr)6 = m/pT + Ζ for all integers m. We
must prove that θ is an isomorphism. We are however not even certain that θ is a mapping.
The snag is this:
If g = mcr and if also g = ncs, is дв — m/pr + Ζ or is дв — n/ps + Z7
We will show that m/pT + Ζ — n/ps + Z, thus proving that θ is uniquely defined.
Assume without loss of generality that r s= s. It follows that pr~acr — c». Then
mcT = npr~scr from which (m — npr~s)c.r — 0
As cT is of order pT, m — npr~* = kpr for some integer k. Thus m = nprs + kpT from
which m/pr = np~s + k. We therefore conclude that m/pr + Ζ = n/ps + Z. Thus θ is a
mapping.
Next we show that θ is a homomorphism. If g, h e G, then g,h e CV for some integer
r, so that g — scr, h = tcT> with s,t£Z. Then
{g + h)e = {{s + t)cT)e = {s + t)/pr + Z = {s/pr + Z) + (t/pr + Z) = дв + he
Finally θ is one-to-one as
Ker0 = {flrj ge = Z)
— {sc.r\ s an integer, r a positive integer and {scT)e — s/pr + Ζ = Ζ}
= {scr \ s/pr G Z} - {scr | s divisible by pr) = {0}
Thus the result follows.
In the future we will call any group isomorphic to (Q/Z)p a p-Prufer group.
The following result is not only the main tool in Section 6.4b, but is also of interest in
itself.
Theorem 6.32: Let G contain a divisible subgroup D. Then there exists a subgroup К
of G such that G = D φ Κ, i.e. a divisible subgroup is a direct summand.
Proof: We accomplish this proof by Zorn's lemma. Let Φ be the collection of all
subgroups L of G such that LnD = {0}. (Our idea is to pick a maximal subgroup К which
meets D in {0}. Then D+K = D® К and we need only show that D + K = G which will
turn out to be true because of the maximality of K.) Can we apply Zorn's lemma to Ф?
Suppose {LA г & 1} is a chain in Ψ. Is U L( in 9? We require

Sac. 6.4]
DIVISIBLE GROUPS
207
(i) Dn4UrL(= {0}.
(ii) U L· is a subgroup of G.
Part (i) is true because D η ,^Ζ/ϊ ¥* {0} implies DnL} Φ {0} for some L}.
To prove (ii) we must show that if g, h G U Li, then g — hG U Lt. Now ^Де U Lf
implies g G L} and h G Lk. Either Lfc 2 £j or Lj 2 Lfc, so without loss of generality as-
■nme that Lk 2 Lj. Then since Lk is a subgroup of G, g — hG Lk. Therefore g — h G U Lf
and (ii) holds. So Φ has a maximal element, say K, which satisfies DriK= {0}. Thus
Suppose now that D+K ¥^G. Then G/(D θ Χ) is nonzero. We prove first that
G/(D θ Κ) is a torsion group. Suppose the contrary. Then we can find χ G G such that
9P(x + (D © Κ)) is of infinite order in G/(D θ Я). Now ж g К If we put id = ^(Я, ж),
it consists of all elements of the form nx + k, where η is an integer and к is an element of
К. И nx + kGD, then nxG(K + D). So η = 0 is the only possibility. But XnD={0}.
So fc = 0 follows also. Therefore DnKi= {0}. This contradicts the maximality of K.
Thus we have proved that G/(K Φ D) is a torsion group.
Let ж ε G, ж g Ζ Θ D. Then gp(x + (K® D)) is a subgroup of G/(K θ D) of finite order.
Suppose that ж + (Κ θ D) is of order w. It follows then that wxGKQD, but ηίίίφβ
for 0<r<w. Suppose itw = fc + c£. Since D is divisible, we can find diGD so that
wdi = d. Put Жг = ж — di. Then wxi = wx — wc£i = k + d — d = k. Notice that xi 6? К
but «kci € K. Put ifi = irpiafi, K). Then
ifi = {rai + fc| r — 0,1, .. .,w-l, and all fceX)
We claim that iCinD = {0}. For if nd + fee Л, re {0,1, ...,w-l}, and к G K, then
D θ Κ = rXr + (D®K) = φ, + (Л θ Χ))
Since Ж! + (D θ Κ) = х + (D Θ Κ), we must have τ = 0. So fe € D and thus fc = 0.
Therefore KiC\D — {0}. But К is maximal. This contradiction shows that our original
assumption, i.e. G¥*D®K, is false, thus proving the theorem.
Problems
6.66. Use a proof different from that of Problem 6.36, page 192, to prove that a p-Priifer group is divisible.
Solution:
As Q is divisible, so is Q/Z. But as Q/Z = 2 (Q/^)D. each (Q/Z)„ is itself divisible, since
pel
every direct summand of a divisible group is divisible.
6.67. Let G be as defined in Theorem 6.31. Prove that we can choose elements cr such that Cr = gp(er)
and pcr+1 = c„ r = 1,2, ... .
Solution:
Assume by induction that Cj, ...,c„ have been chosen with per+1 = er for r = 1, ...,n— 1
and C{ = gp(ci), i = 1,.. .,n. Let C„+1 = gp(c). Then gp(pc) is a cyclic group of order p" and,
since cyclic groups have only one subgroup of any given order, gp (pe) = Cn. Hence r(pc) = cn
for some integer r. As cn is of order pn, r and ρ are coprime. Thus gp{rc) = Cn+i. Now put
cn+ J = re. Then рсп+ i~en and it is possible to choose the elements cx, c2, ..., as required,
6.68. A p-group G is a p-Prflfer group if and only if it has the following two properties:
(1) every proper subgroup of G is cyclic,
(2) there is a cyclic subgroup of G of order p» for every i = 1,2, ... . (Hard.)
Solution:
First we shall prove that if G satisfies (1) and (2), it is a p-Prufer group.
Let Ct С С2 С — be a sequence of subgroups of G where each C( is a cyclic group of order pl.
If the sequence is infinite, D = U Cs is a p-Prtifer group (Theorem 6.31). Hence it is divisible and
G — D φ Κ by Theorem 6.32. However, if Κ φ {0}, G has a subgroup which is the direct sum of
two cyclic p-groups, and hence is not cyclic, contrary to the hypothesis. Accordingly, G = D is a
p-Priifer group.

ABELIAN GROUPS [CHAP. 6
Suppose now that Сг С Сг Q · · · С Сп is a sequence of subgroups, each C{ cyclic of order p*,
and there exists no subgroup Сп+1эС„ where Cn + i is of order pn + 1. Let Cn = gp(a). We know
that there exists a subgroup В = др(Ь) of order pn+t. Consider gp(a, b). As a finitely generated
abelian group, it is the direct sum of cyclic groups. If it is the direct sum of two or more cyclic
groups, it is not cyclic (Theorem 6.21). Thus gp(a, b) is cyclic, and as G is a p-group, gp(a, b) is
cyclic of order pm where m — n + l. But then gp(a, b) contains a cyclic subgroup of order pn+1
containing Cn, contrary to the hypothesis. The result follows.
Next we note that the p-Prttfer group satisfies condition (2). As for condition (1), let if be a
subgroup of (Q/Z)p. If Η Φ {Ζ}, then let χ e Η, χ Φ 0 + Ζ. Let x = m/pr + Ζ where m is an
integer between 1 and pT — 1. Clearly gp(x) = gp(l/pr + Z). If there is no integer η for which
l/pr + Ζ <£ Η f or r 2= n, then Я = (Q/Z)v. If there exists such an integer η, Η Q gp(l/pn + Z).
Thus Η is cyclic as required.
6.69. Show that if U is the subgroup of the multiplicative group of the complex numbers consisting of
all the reth roots of unity, then U ss Q/Z.
Solution:
U = 5 U- by Theorem 6.10, page 190. Now U„ is the union of cyclic groups of order p',
pen
namely Up is the union of Cj = gp {{χ \ χ is a p'th root of unity}). But Cf is cyclic of order p*. Then
by Theorem 6.31, Up г* (Q/Z)p. Thus, as Q/Z = 2 (Q/^)P (by Theorem 6,10), Q/Z st U by
Theorem 6.5, page 185. p e n
6.70. Show that the additive group of rationale is the union of an ascending sequence of infinite cyclic
groups.
Solution:
Let Qn — gp(l/nl). Then Qn+i2Q„ and U Q„ = Q the additive group of rationals.
6.71. Show that if G is a p-Priifer group and Η, Κ are subgroups of G, then either H^K or ΚώΗ.
Solution:
If one of Η or К is G, the result is true. Assume both Η and К are proper subgroups. Then
by Problem 6.68, Η and К are both finite cyclic groups. Suppose pr = \H\ — \K\ = p*. Then Η
contains a subgroup of order \K[, say Ht. Now both Ht and К are contained in some cyclic subgroup
of G, as G is the union of cyclic subgroups. But then it follows that Hy = K, as there is one and
only one subgroup of order ps in a cyclic p-group of order exceeding ps. Therefore H~2K.
6.72. Let G be an ascending union of infinite cyclic groups Cf such that Ct — gp(c^ and (i+ l)ef+1 = cit
for г = 1,2, ... . Prove that G is isomorphic to the additive group of rationals. (Hard.)
Solution:
Let Q denote the additive group of rationals and let Qs = gp(l/i\), i = 1,2, ... . Clearly,
Qi+1DQi and U Qi= Q- We shall prove that θ below defines a mapping of G to Q. Define
(2c4)9 = z/i\ where z^Z. We must prove that в is uniquely defined.
Suppose zlci = z2ci, zltz2 and i,j integers. If i — i, then c4 = (}1/И)с3. Hence z,c4 =
2i(j!/i0cj = z2ci- Since Cj is infinite cyclic, zi(j\/i\)Cj — z2Cj implies that z^p./V.) — z2.
To prove that в is well defined, we must show that zji] = z^/jl, i.e. zt(j]/il) = z2. But this is
what we have just shown. Since C{e = Q, = #р(1/г!), it follows that Gi 2 U Qi = Q. Hence ί is
an onto mapping. i_1
let a homomorphism? Let f,g G G. We may as well suppose that f,g € Cf for some integer i.
Hence / = «jCj, fir = ζ^, say. f + Я =■ {гг + z2)cx. Then
(f + ff)» = (*i + z2)^- and /i+09 = ~ + ^ = (2i + 22)^·
г! г] г! г!
Thus β is a homomorphism.
Finally, to show that в is an isomorphism, it is sufficient to show that Ker в — {0}. Suppose /
is such that /β = 0. We have that / = zct for some integers г and i. Then ft — г/г! =0 only if
2 = 0. Hence / = 0 and the result follows.

Sec. 6.4]
DIVISIBLE GROUPS
209
b. Decomposition theorem for divisible groups
The results of Section 6.4a enable us to deduce the following decomposition theorem
for divisible groups.
Theorem 6.33: A divisible group is the direct sum of p-Priifer groups and copies of the
additive group of rationale.
Proof: Let G be divisible and let Τ be the torsion subgroup of G. Now for any integer
η and element t £ T, there exists an element g €LG such that ng = t. Since t is of finite
order, so is g, and hence g £ T. Thus Τ is itself divisible. A divisible subgroup is a direct
summand (Theorem 6.32); so G = T®F. Since Tn{F} = 0r F и (Τ Θ F)/T; hence F is
torsion-free by Theorem 6.9, page 189. Moreover, as F is a direct summand, F is itself
divisible. We now consider F and Τ separately.
(a) F.
We show that F is a direct sum of copies of the additive group of rationals. To this
end let S be a maximal independent set (Theorem 6.13, page 194). For each s £ S we
shall define a subgroup С of F. Let rbs = s. For a given s £ S and positive integer
г, there exists by the divisibility of F an element ri+1,s £ F such that (г* + l)ri+i,s = n,e.
We put C„ = 0p(ri>s [ i = 1,2, .. .)■ I* follows then from Problem 6.72 that CS is
isomorphic to the additive group of rational numbers. Note that if χ £ C3, χ Φ 0, then
there is a nonzero multiple of χ which is also a multiple of s, as &¥■ 0. (This is true for
any two nonzero rational numbers.)
We claim that F is actually the direct sum of these subgroups C„ as s ranges over S.
To prove this, suppose that si, s2, .. ., s„ are distinct elements of S and that ci + c2 + · · ·
+ Cn = 0, where с,· £ CSj, Cj ¥* 0 (j = 1, . . .,w).
As we remarked above, there exists a nonzero multiple fcj of each C) which is then a
multiple of ss. Hence there exists a nonzero integer k, namely fci · - · fc,, such that
kty = IjSi, where Ϊ,- is a nonzero integer. As kct + · · ■ + fee = 0 is a consequence of
Ci + cz + · · ■ + Cn = 0, we find therefore, on substituting for the elements kct the elements
ijSj, that liSi + - ■ · + Ιηβη — 0. But the elements independent. From this
contradiction it follows that the Cs generate the direct sum
С = gv{C„\ seS) = χα
se s
But С is divisible since each summand Cs is divisible. Hence С is a direct summand
(Theorem 6.32), i.e. F = С θ D, say. If D ¥* {0}, then let d £ D {d ¥- 0). Clearly
the set SU{d} is definitely larger than S since d&S (d does not even lie in C) and
Sl){d} is independent. This is a contradiction as S is a maximal independent set. So
F — С is a direct sum of copies of the rationals.
(b) T.
First of all Τ = J) T„ by Theorem 6.10. Since Τ is divisible, so also are the
pen
summands Tv. It is sufficient then to assume that Τ is a divisible p-group and to prove
that Τ is a direct sum of p-Prufer groups.
Let Ρ be the set of elements of Τ of order at most p. Ρ is clearly a subgroup. Let
S be a maximal independent subset of Ρ (Theorem 6.13). For each s £ S define
Ci.s> c2.s. · · · inductively as follows: (a) pct,s = s, (b) pci+1[jl, = ci>s for г = 1,2,... .
This is possible since Τ is divisible. Clearly #p(Ci.s) Qgp{cz,s) с ■ ■ ■. Since др{сЬе) is
cyclic of order pi+1, Cs = gp{cUs, c2jS, ...) is an ascending union of cyclic groups of
order v\ one for each i= 1,2,.... Accordingly by Theorem 6.31, C, is a p-Priifer
group. Then С = gp(C* \ в e S) is divisible. By Theorem 6.32, С is a direct summand
of T, i.e. T~CBD. If D/{0}, there is an element of d £ D of order p. Since
ScC, SU{d) is an independent subset of Ρ larger then S, and so we must have D= {0}
by the maximality of S. Thus Τ = C.

210
ABELIAN GROUPS
[CHAP. 6
It remains only to prove that С = 'Σ Ca. If Si, s2,..., sn are distinct elements of
a ε 5
S, and ci + c2 Η + с = 0 where cs e CH for t = 1,..., n, then, similarly to (a)
above, we arrive at a dependent relation between Si, ..., sn, unless ci = c2 = · · ■ = <%. — 0
(Problem 6.78).
The proof of the theorem is complete.
Problems
6.73. Prove that (a) every free abelian group is isomorphic to a subgroup of a divisible abelian group,
and (b) every abelian group is isomorphic to a subgroup of a divisible group.
Solution:
(a) A free abelian group F is a direct sum of infinite cyclic groups C( (ie/): F = 2 C4. Now
(ε ι
we choose one copy of the rationale Qf for each i G /. Let К = 2 Qi ап(* ^ Λι ^ ° be
iEI
chosen in each Q{. F is clearly isomorphic to «гр({<^ | i e /}). But 2 Qi is divisible since each
Qi is divisible. The result follows. ieI
(b) If G is any group, G e F/JV for some free abelian group F and some subgroup N.
Now in (a) we have proved that there exists a divisible group D containing F. Hence D
contains N, and so D/N contains as a subgroup F/N, i.e. G. Now D/N is divisible since a
homomorphic image of a divisible group is divisible. Thus the result follows.
6.74. Suppose G has the property that if Я is any group such that Η 2 G, then G is a direct summand
of H. Prove that G is divisible.
Solution:
G is a subgroup of some divisible group D by Problem 6.73. Thus D = G φ Τ. But every direct
summand of a divisible group is divisible. Therefore G is divisible.
6.75. Let G be an infinite group whose proper subgroups are all finite. Prove that G is a p-Priifer group
by using the theorem which states: if G is a group such that for some integer η Φ- 0r nG = {0},
then G is a direct sum of cyclic groups (of finite order). (This theorem is not proved in this text.)
Solution:
Consider the subgroups nG for all positive integers n. If nG = G for all such n, then G is
divisible, and so G is the direct sum of p-Priifer groups and copies of the additive group of rationale.
As all the proper subgroups of G are finite and the additive group of rationale has an infinite cyclic
group as a proper subgroup, only p-Priifer groups are involved. Since each p-Priifer group is
infinite, G must in fact be a p-Prflfer group.
If on the other hand nG is a proper subgroup of G for some n, then nG is a finite group of order
m, say, and so mnG = {0}. Using the theorem quoted in the statement of the problem, G is the
direct sum of finite cyclic groups. As G is infinite, it must be the direct sum of an infinite number
of cyclic groups Cj. But then the subgroup generated by all but one of the Ct must be infinite. This
contradiction proves the result.
6.76. (a) Let G be any group and let S be the subgroup generated by all the divisible subgroups of G.
Prove that S is divisible.
(b) Prove that G is the direct sum of a divisible group and a group which has no divisible subgroups
other than the identity subgroup {0}.
Solution:
(a) Let β £ S. Then s = hl + h!l+ ■ ■ ■ + hn where each ftf belongs to a divisible subgroup Ht of G.
Thus if 2 ¥* 0 is any integer, there exist fes € Hi such that zkt = ht. As S 2 Я4 for each
1 e 7* fc„ + kz + ■ ■ ■ + kn e S and «(fcj + - - · + fc J = β
Hence S is divisible.

Sec. 6.4]
DIVISIBLE GROUPS
211
(6) Since S is divisible, we can apply our direct summand theorem for divisible subgroups
(Theorem 6.32) to find that G = S θ Τ. As Τ contains no divisible subgroup other than {0},
the result follows.
6.77. (a) Prove that the additive group of rationals Q has a proper subgroup which is not free abelian.
(6) Let G be a torsion-free group, every proper subgroup of which is free abelian. Prove that G
is free abelian.
Solution:
(a) Consider the subgroup Η generated by 1/2,1/4,1/8, ..., 1/2*, ... in the additive group of
rationals. Η is of rank 1 as О is of rank 1 (Problem 6.38, page 194). So if Η is free abelian, it
must be infinite cyclic. But Η is not infinite cyclic; for if it were, Η = gp(zl2i) for some
integers ζ and i. But then 1/2*+l G Η and 1/2*+г € др(я/21). We have only to prove that Η is
not G. This is obvious since 1/3 £ H.
(b) Suppose that nG = G for all positive integers n. Then G is divisible and is the direct sum
of copies of the additive group of rationals. (As G is torsion-free, no p-Priifer group is involved.)
But by (a) above, G will have a non-free subgroup. Thus for some η Φ 0, nG Φ G, and nG is
free abelian.
Now в : g ~> ng is a homomorphism of G onto nG. Ker в = {g | ng = 0} = {0} as G is
torsion-free. Hence в is an isomorphism, and so G is free abelian.
6.78. Prove the result stated at the end of the proof of Theorem 6.33, i.e. prove that if ct + · - - + c„ = 0,
then Cj = c3 = ■ ■ ■ = cn = 0.
Solution:
Assume the contrary. As the order of cv ..., cn is immaterial, we may assume that ολ Φ 0 and
Cj is of highest order. Say ct is of order y»1. Then р1-1^ = mi8i where щ is an integer, i = 1,2,. . ,,n.
Thus we obtain
pi-l (c, + ■ ■ · + cn) = m1sl + m2s2 + · ■ ■ + m„sn
Since mjSj Φ 0, we have a contradiction to the fact that S is an independent set.
A look back at Chapter 6
This chapter was mainly concerned with the structure of divisible and finitely
generated abelian groups.
Direct sums of groups were discussed. Given a family Ai (i G I) of groups, there is
always a group which is the direct sum of groups isomorphic to each of the groups Αι.
Any homomorphism of the direct summands of a group extends to a homomorphism of the
whole group. From this it follows that if two groups are direct sums of isomorphic
subgroups, they are isomorphic. Direct sums of infinite cyclic groups are all the free abelian
groups. An important fact is that every abelian group is a homomorphic image of a free
abelian group.
The torsion group T(G) of a group G was defined, and it was shown that G/T(G) is
torsion-free. It was proved that if G is a torsion group, it is the direct sum of its p-components.
This led to the definition of the p-Prufer group as the p-component of Q/Z.
An application of Zorn's lemma proves every group has a maximal independent subset.
The rank of a group was defined and proved an invariant of the group by the Steinitz
exchange theorem.
In the fundamental theorem of abelian groups, finitely generated abelian groups were
shown to be expressible as the direct sum of cyclic groups. Two finitely generated abelian
groups were shown to be isomorphic if and only if they have the same type. Finally, the
type of a subgroup of a group was shown to be, roughly speaking, "less than" the type of
the group.

212
ABELIAN GROUPS
[CHAP. 6
Divisible groups were discussed. Any group which is the union of cyclic groups of
order a power of φ turns out to be a p-Prufer group. Any divisible subgroup of a group
is also a direct summand. This led to the proof that every divisible group is the direct sum
of isomorphic copies of the additive group of rationale and p-Prufer groups.
Supplementary Problems
DIRECT SUMS AND FREE ABELIAN GROUPS
6.79. If the mapping α -* a~ *, α € G, is an automorphism of the group G, prove G is abelian.
6.80. Suppose that G is a finite group, a € aut (G), a is of order 2, and ga¥= g for all g (¥* 1) £ G.
Show that G is an abelian group. (Hint; First prove G = {д~~г(да) j g G G} and then use Problem
6.79.)
6.81. Denote the set of all homomorphism of an abelian group G into an abelian group Η by Horn (G, H).
If ψ, Ψ € Horn (G, Я), we define ψ + Φ by
д{ф + *) = дф + д* (I)
for all ff£G. Show that Horn (G,H) with the operation defined by (I) is an abelian group.
6.82. If A is an abelian group and Ζ is the group of integers under addition, prove that Horn (Z, A) s« A.
6.83. Prove that the group of rationale under addition is not the direct sum of cyclic groups.
6.84. If G is the direct sum of cyclic subgroups, show that a factor group of G is not necessarily a direct
sum of cyclic subgroups. (Hint: Use the preceding problem and free abelian groups.)
6.85. Let N be a normal subgroup of G. Prove that if G/N is free abelian, N is a direct summand.
TORSION GROUP AND RANK
6.86. If G is a finite group, show that aut (G) = Π aut(Gp), where w is the set of all primes.
ρ € π
6.87. Prove the first Sylow theorem, page 130, for abelian groups using the p-components.
6.88. Let G denote the group of rotations of the plane (see Section 3.4c, page 68). As G is an abelian group
we may use additive notation. Thus the rotation pe followed by rotation pe is ρθ + ρβ = Ρβ +β ·
Let % — {p0 I в — 2vm/n radians where m,n are integers, η > 0}. Prove that (a) "5^ is a subgroup
of G; (b) every element of "5^ has finite order; (c) if %p denotes the ρ component of % for any prime
p, then %p — {p ) в = 2πτη/ρτ radians where m and r are integers} and ^ρ = the p-Priifer group.
6.89. Let a, b be elements of an abelian group. Let the order of α plus the order of b be n. Prove by
induction on и,that α + b is of order the least common multiple of the orders of α and b,
6.90. Let G be an abelian group. Suppose every element of G is of order less than some fixed integer η
and there are elements of order η in G. Prove that the elements of order η generate G.

CHAP. 6]
SUPPLEMENTAEY PROBLEMS
213
FUNDAMENTAL THEOREM FOR FINITELY GENERATED ABELIAN GROUPS
6.91. Prove that if G is a finitely generated abelian group, then G = It φ · · · φ Im φ Су Φ · · · Φ C„ where
Ij is an infinite cyclic group (j — 1, ..., m), C{ is a finite cyclic group of order v{ (i = 1, ..., n),
and vt divides vi+l for i = 1, .. .,n — 1. (Hint: Use the fundamental theorem and then first look
at the highest power of each different prime in the decomposition.)
fi.92. Prove that the automorphism group of a finitely generated abelian group is finite if and only if
there is at most one infinite cyclic summand in a cyclic decomposition of G.
6.93. Find the type of the additive group of integers modulo m for any integer m > 0.
6.94. Let G be a non-cyclic finite abelian group. Show that G has a subgroup of type (p, p; 0) for some
prime p.
6.95. Prove that the automorphism group of a finite non-cyclic abelian group G is non-abelian. (Hint: Use
Problem 6.91 to find suitable elements a,b S G such that the order of α divides the order of b.
Then look at the mappings <*!: a?b*-* α"+ fb*; «2 : o'b'-* awb~* and as: №"ο*-»»'&» where β and t
are integers.)
6.96. Let G be a finite abelian group. Suppose that for each divisor d of G there are at most d
elements in G of order d. Prove that G is cyclic.
6.97. Let G be a finitely generated abelian group. Prove by induction on the number of generators of G
that every subgroup of G is finitely generated.
DIVISIBLE GROUPS
6.98. Show that a divisible abelian group has no subgroup of finite index.
6.99. If G is a nondivisible abelian group, then G has a subgroup of prime index. (Hint·, Use the
following theorem (not proved in this book): An abelian group G for which nG = {0}, n¥*Q, is the
direct sum of cyclic groups.)
6.100. Prove that the additive group of the real numbers is the direct sum of isomorphic copies of
the additive group of rationals.
6.101. (a) Let G = Horn (A,B) (see Problem 6.81) where В is a torsion-free divisible group. Prove that
G is the direct sum of copies of the additive group of rationals.
(b) Let G be as in (a) but with В a divisible p-group. Prove that if A is finite, G is the direct sum
of p-Prflfer groups.
6.102. A subgroup Η of an abelian group A is a pure subgroup of A if whenever na = h e Η for some
a€ A, then there is an hr ε Η such that nh' = h. Prove that (a) a direct summand of an abelian
group is a pure subgroup, and (b) the torsion subgroup of an abelian group is a pure subgroup.
6.103. Prove that all the subgroups of a group in which every element has square-free order is pure.
6.104. Let Η he a pure subgroup of an abelian group G. Prove that if g + A£ G/A, there is an element
j£G such that g + A — g + A and the order of g is equal to the order of g + A.

Chapter 7
Permutational Representations
Preview of Chapter 7
There are three main divisions of this chapter. In the first we generalize Cayley's
theorem, that every group is isomorphic to a permutation group. As consequences of this
generalization we prove the following theorems for G, a group generated by a finite number
of elements: (1) A subgroup of finite index in G is itself finitely generated. (2) The number
of subgroups of fixed finite index in G is finite. (3) If the subgroups of finite index of G
intersect in the identity, then every homomorphism of G onto G is an automorphism.
The second main division of this chapter appears in Section 7.7. We call a group G
an extension of ji group Я by a group К if there is a normal subgroup Η of G such that
G/H — К and Η — Η. We examine G to see how it is built up from Я and K. The most
general case is complicated and we restrict ourselves to a special extension called "the
splitting extension."
Reversing our analysis, we are able to build a group G that is the splitting extension of
a given group Я by a given group K. A particular example of a splitting extension is the
direct product, used in Chapter 5.
Our third division, which begins in Section 7.8, defines a homomorphism of a
group into one of its abelian subgroups. This homomorphism is called the transfer. We
use it to show that a group G with center of finite index has finite derived group.
7.1 CAYLEY'S THEOREM
a. Another proof of Cayley's theorem
We saw in Chapter 2 that every groupoid is isomorphic to a groupoid of mappings.
In particular, every group is isomorphic to a group of permutations. The consequences
of this theorem are important. We repeat the proof here for the case of groups alone.
Theorem 7.1 (Cayley): Every group is isomorphic to a group of permutations.
Proof: Let G be a group. Let ρ be the mapping which assigns to each element g in G
"the following mapping of G into G: *
χ -» xg for each χ G G
Thus the image of g in G under P is, by definition, gp where
gP: x-*xg (xGG)
The definition of P is unambiguous. To prove that ρ defines an isomorphism of G onto
a subgroup of Sg, we have to check that:
(i) gp is a permutation of G for every g £ G;
(ii) ρ is a homomorphism, i.e. if g, h G G, then (gh)P = gP · hp;
(iii) ρ is also an isomorphism, i.e. ρ is one-to-one.
214

Sec. 7.1]
GAYLEY'S THEOREM
215
We deal first with (i). Thus we must prove gP is a one-to-one mapping of G onto G.
If x(gP) - y(gP), then xg - yg. So, multiplying by g~l on the right, we find χ -y. Next
we prove gP is a mapping of G onto G. Suppose χ €. G; then {xg~l){gP) — [%g~l)g = ж and
so gP is onto.
Secondly we prove (ii). For χ G G,
x({gh)P) - x(gh) = (xg)h - (x(gP))(hP) = %{gPKP)
Since (gh)P and gPhP have precisely the same effect on every element of G, (gh)P = gPhP (by
the definition of equality of mappings).
It remains to prove (iii), i.e. P is one-to-one. Suppose gP — hP; then if 1 is the
identity element of G, g — l(gP) = l{hP) = h, i.e. g = h. Therefore P is one-to-one.
(Note: In the proof of Cayley's theorem ρ is a mapping of G into SG, so that gP is itself
a mapping of G to G. Caution and patience are required to avoid confusion in some of our
subsequent equations.)
b. Cayley's theorem and examples of groups
Cayley's theorem tells us that there is an isomorphic image of every group among the
permutation groups of suitably chosen sets. If one demands that a permutation group satisfy
further conditions, one frequently comes across interesting groups (see Chapter 3).
Historically many important groups arose in precisely this way.
Problem
7.1. Describe in detail the isomorphisms given by Cayley's theorem for (i) a cyclic group of order 2,
(ii) a cyclic group of order η (и — 3), (iii) the symmetric group on three letters.
Solution:
(i) Let G be cyclic of order 2. Then G consists of two elements, 1 and a, where Φ = 1 and
1 · ο = α = α·1. Let ρ be as in Theorem 7.1.
lp is the mapping lp: 1 -* 1, a -* a
and ap is the mapping given by ap ; 1 -» a, a -+ 1
Clearly ρ is one-ta-one, as ар Ф lp.
(ii) Let G be cyclic of order n. Then G consists of η elements Ι,α,α2, -
4.5, page 102). Then
lp: 1-* 1, a-* a, ..., an~l-*an~l
, an~1, say (see Lemma
ap'.
a?p:
a,
a ■
a ■
a»,
an-2.
1
1, an~l
■ an~
1,
a."
.—ι
• an~
(iii) The symmetric group on the set {1,2,3} consists of the permutations
Then
Pi
Pi
Рз
Pip
Pip
Рзр
PiP
PsP
Pup
1-+2, 2-»l, 3->3
1 -+ 3, 2 -» 2, 3 -»1
l->3, 2-»l, 3-»2
1 -* 1, 2 -* 2, 3 ->· 3 Pi
1 -> 1, 2 -> 3, 3 -* 2 ps
1 -+ 2, 2 ·* 3, 3 -»1 pe
Pi-"Pi, P2-*Ps, Ра~*Рз> Pi ~* Pi· Ps ~* Ps> Рб"*Рб
Pi"* Pa. Pi^-Pi, Ря^Рь, Pi~*Pe> Ps "* Рз- Рв "* i>4
Pi -» Рз. Рг -» P4. Рз "* Рб> Р4 -» Ps> Ps -* Pa. Pe "» Pi
Pi "* Pi> P2 "* Рз. Рз -» Рг> Ρ4 ■"* Pi. Ps "* Pe» Pe -+ Ps
Pi "* Ps. Рз -* Pe. Рз -*Pi, Pi-* Ps, Ps ~* Pi. Pe "* P2
Pi -* Pe, Pa ~* Ps> Pa "* Pi. P4 "* Pa. Ps "* P4» Pe ~* Pa
It is worth checking {Pip)(Pjp) = (P{Pj)p for some г and i between 1 and 6.

216
PERMUTATIONAL REPRESENTATIONS
[CHAP. 7
7.2 PERMUTATIONAL REPRESENTATIONS
Definition of a permutational representation
A homomorphism of a group G into the symmetric group on the set X is called a
permutational representation of G on X.
So if ρ is the isomorphism provided by Cayley's theorem for the group G, then P is a
permutational representation of G on G,
Repeating the definition of a permutational representation of a group G in detail, we
say that a mapping μ of G into the symmetric group on some set X is a permutational
representation of G if
for all g and h in G. The permutational representation provided by Cayley's theorem is
called the right-regular representation (the adjective right is used because the
representation is obtained by multiplication on the right).
Example 1; (i) Let G be the symmetric group on {1,2,3}, and let p be as in the solution to
Problem 7.1(iii). Then ρ itself is a representation of G as a permutation group
on six elements.
(ii) There is another representation of the symmetric group G on {1,2,3}, the most
natural one. This is the identity isomorphism, for G is itself a permutation
group on {1,2,3}.
(Hi) Let G be the cyclic group of order 2 and let X be the set of all integers
.. .,—1,0,1,... . Let a be the permutation of X defined by
a: 2i-»2t-f 1, 2i-f l-»2t for г - 0, ±1, ...
Thus a sends an even integer to the succeeding odd integer and an odd integer
to the preceding even integer. Let t denote the identity permutation. Then ι
and a together constitute a subgroup Γ of the symmetric group on X since
a2 — ι, ια = at = a
Now suppose G consists of the elements 1 and a. Then the mapping μ: a-* a,
1 -»ι is actually an isomorphism since both G and Γ are cyclic of order two.
So μ is a permutational representation.
(iv) Suppose η is a positive integer and that G is the cyclic group of order n,
G = {Ι,α,α?, ...,α.»-»}
Let a be the permutation of X = Ζ the integers defined by
a: jn-*jn+l, j'w+1-» jn + 2, ..., j»+(n — 1)-» jn (,?* = 0, ±1, ...)
In particular we have
0a = l, la = 2, ..., (n-l)a = 0
Note that a cyclically permutes the integers taken in blocks of n.
It is not difficult to see that a is of order n. First of all, θα2 = 2,
θα3 = 3, ..., 0αη_1 = η — 1. This means that the permutations ι, a, O2, .. .,an_1
are all distinct since they act differently on 0.
If j is any integer, jan = }. So a" = ι and {(, a,ap, ...,an~~1} is a cyclic
group Γ of order n. Hence the mapping μ of G into Γ defined by
1 -* t, a-* a, ..., αη-ΐ-»α"-ΐ
is an isomorphism and therefore a permutational representation of G.

Sec. 7.3] DEGREE OF A REPRESENTATION AND FAITHFUL REPRESENTATIONS
217
(v) Suppose that G is the dihedral group of degree 4. G is the group of
symmetries of the square
D
В
Let g £ G. By Lemma 3.12, page 75, g takes each vertex of ABCD to a
vertex. Since g is one-to-one, Ag, Bg, Cg and Dg are distinct vertices. If we
define X = {A,B,C, D} and yg by xyg — xg for all χ G Z, then ygG Sx.
Let θ: G-+Sx be defined by £re = γβ. Then if a; G Z, £r, h G G,
x(gh)e = xygh = x(gh) = {xg)h = (a;?g)?h = a;(?g?h) = x(ge)(he)
Thus (£rfr)fl = srflfrfl, and so в is a permutational representation of G. If ge
is the identity permutation of X, then £r leaves every vertex of ABCD
unchanged. But by Lemma 3.7, page 71, g = и Hence Ker θ — {ι} and θ is
actually an isomorphism.
7.3 DEGREE OF A REPRESENTATION AND FAITHFUL REPRESENTATIONS
a. Degree of a representation
Definition: The degree of a permutational representation on X (more briefly referred to
as a representation) is the number of elements in X.
Example 2: We inspect Example l(i)-(v).
(i) This representation is of degree 6.
(ii) This representation is of degree 3.
Notice that in (i) and (ii) we have two representations of the same group,
namely the symmetric group on {1, 2, 3}, of different degrees.
(iii) This representation is of infinite degree. Notice that here G is cyclic of order
2. Hence there are representations of finite groups which are of infinite degree.
(iv) This representation is of infinite degree.
(v) This representation is of degree 4.
Problems
7.2. Find a representation of degree 3 for a cyclic group of order 2.
Solution:
Let G be the cyclic group of order 2, say G = {ί,α}. Let Ζ = {1,2,3}. Then there are several
possible representations of G on Z. First of all there is the rather trivial representation
r : 1 -» t, α -» ι
where t is as usual the identity permutation, r is clearly a representation. For no matter which
elements g, h in G we choose,
(gh)r — ι and (дт)(кт) — 11 = t
from which (gh)r = (дт)(кт). Another representation of G involves choosing a different homomor-
phism. Now observe that if μ is a homomorphism of G into the symmetric group Sx, then either
μ = τ or G/i is of order 2. Note also that ΰμ must be a subgroup. So in deciding on a choice of
μ, we need subgroups of Sx of order 2. There are actually 3 of them. To see this let
αλ: 1 -» 2, 2 -» 1, 3 -> 3 a% : 1 -» 3, 2 -» 2, 3 -> 1 <*3 : 1 -> 1, 2 -* 3, 3 -> 2
for i = 1,2,3. Thus the
Then {ι, αϊ}, {t, α2}, {ι, a3} are subgroups of Sx of order two, since οΆ
mappings
μι : 1 -» t, α -»<*! μ2 : 1 -»ι, α -> α2 Мз : ! ~*
are representations of <7 on Z.
<*з

218
PERMUTATIONAL REPRESENTATIONS
[CHAP. 7
The proof that there are precisely three subgroups of order 2 follows from an inspection of
all the subgroups of Sx. Since we have no real need for such proof here, we leave the details to the
reader.
The upshot of these considerations is that we have produced 4 representations of G of degree 3.
7.3. Find representations of degree 10 and 15 respectively of the cyclic group of order 5.
Solution:
Let G be cyclic of order 5. Then we can find α £ G such that G =■ {Ι,α,α^α3, α*}. Let
X = {1,2,. ..,10} Υ = {1,2, ...,15}
and let ay: 1 -> 2, 2 -» 3, 3 -> 4, 4 -* 5, 5 -> 1, 6 -* 7, 7 -* 8, 8 -> 9, 9 -* 10, 10 -* 6
a2 : 1 -> 2, 2 -> 3, 3 -» 4, 4 -> 5, 5 -> 1, j -> j, for j > 5
It follows from a direct calculation that both at and <*2 are of order 5. Then
Tt = {<, a,, αξ, asv a\} Γ2 = {., az, a\, οξ, α*}
are subgroups of order 5 of Sx and SY respectively. Thus
μι: 1 -»ι, α -» «ι, ..., α4 -» α*
and /ί2: 1-+1, α -» α2, ..., α*-* α*
are both representations of G, The first is of degree 10 and the second is of degree 15.
b. Faithful representations
Definition: A representation is termed faithful if it is one-to-one. Both faithful and
non-faithful representations are useful, as we shall see later.
Problems
7.4. Are the representations in Example 1, page 216, faithful?
Solution:
(i) ρ is faithful.
(ii) The identity isomorphism is one-to-one, so this representation is also faithful. Notice that (i)
and (ii) provide examples of faithful representations of the same group which are of different
degrees.
(iii) μ is faithful.
(iv) μ is faithful.
(v) The representation is faithful.
7.5. Inspect the representations of (a) Problem 7.2 and (b) Problem 7.3 for faithfulness.
Solution:
(a) r is not faithful since α Φ 1 and or = ι, i.e. τ is not one-to-one. However μ±, μ2> /*з are faithful.
(b) цх and pa are both faithful.
7.4 PERMUTATIONAL REPRESENTATIONS ON COSETS
Definition: Suppose that G is a group and that Я is a subgroup of G. Then the right
cosets of Η in G are
Hzi, Hx% .. . (7.1)

fce.7.4]
PERMUTATIONAL REPRESENTATIONS ON COSETS
219
If В consists of the identity element alone, then (7.1) is simply an enumeration of the ele-
■aents of G. Here we will show how to obtain a permutational representation of G using
the cosets (7,1) which coincides with the regular representation when Η = {1}.
To describe our representation, let us choose a complete system X of representatives of
the right cosets Hg of Я in Gr with 1 the representative of Я. In other words, we select
in each coset Hg an element which we term the representative of the coset, with 1 the
representative of Η, Χ is then simply the set of chosen representatives. We call such a set
X a right transversal of Я in G.
Given a right transversal X of Я in G, we denote the representative of the coset Hg by
g. Thus g is an element of X. Since two right cosets are either identical or_disjoint, it
follows that Hg = Hg because g G Hg. Notice that if heH, then hg = g since
Hg = Hg = Hhg = Hh~g.
For example, if G is cyclic of order 4, say G — {1, a, a?, a?}, and if Η = (1, α2} is a
subgroup of order JJ of G, then {l,a} is a right transversal of Η in G. We take X = {l,a}
and note that Ϊ = 1, a = a, ά2 = 1, a3 = a.
With each element g in G we associate a mapping yg of X into X, where ys is defined by
7g: x^xg (xeX) (7.2)
In fact γ is a permutation of X. To prove this we first show that if gi, дг G G, then
#ι</2 = </1</2. For </Γ<7·! is the representative of the coset Нд^дг, while д-^дг is the
representative of the coset Hgxg%. But Hgt = Ядг1, and so
Hg\g~2 = #srisr2 = Я</^
Thus ^ϊ = £ϊ<^ for all gu g^eG (7.3)
We use (7.3) to prove that the mapping yg is a permutation. First we prove yg is one-to-one.
Assume хУд = i/yg (ж,?/ G X). Then xg = yg, and so xgg'1 = ygg~l. Using (7.5), we find
that xgg~l = xgg~x = χ = χ and similarly ygg~l = y, from which x = y. Finally we prove
yg is onto. Suppose χ € X; then xg~l G X and
xg~\ - xg~xg = zsrV = *
Hence every element of X is an image. Thus yg is a permutation.
We now define a mapping π of G into Sx. *■ assigns to g in G the mapping yg, so that g-π
is the permutation of X defined by (7.2). The aim of the discussion in this section is to
prove that this mapping π is a permutational representation of G on X. We have only to
verify that it is a homomorphism, i.e. if 9V92^G, then (ff102)7r = (Sri7r)(Sr27r)· Note that
(дхд^тт is a permutation of X. To prove (дгд^тт = (А^На^О we must show that
the effect of the mapping (дхд^тт is the same as the effect of the mapping (g^ig^). (Note
that (svr)(6Vr) is the product of two mappings, i.e. the result of first performing the mapping
(дгтг) and then (eyr).) If χ G X, using (7.3) we find
%{(9ι92» ~ «(i/ii/a) = «№ = ИбУЖМ = *((M(flV0)
Hence (дгд2)тг = (9г1т)(9г2тг) as claimed.
We shall refer to w as a cosei representation of G (with respect to Я). Of course *■
depends on H, G and X. In a sense π is independent of the choice of the transversal X
(see Problem 7.10).

220
PERMUTATIONAL REPRESENTATIONS
[CHAP. 7
Problems
7.6.
Choose right transversals for (a) the center Ζ in the dihedral gToup Ό of degTee 4, and (b) the
center Ζ in the quaternion group Jf( of order 8 (see Table 5.1, page 161).
Solution:
(a) The dihedral group D = {1, a,, a2, a3, a4, as, a6, a7} is most easily described by its multiplication
table:
»2
«3
as
a6
d7
1
1
«1
a2
a3
a4
as
a6
Ο,η
ai
«1
a2
a3
1
Λη
a4
a5
a6
«2
a2
a3
1
«1
«6
a7
a4
as
B3
a3
1
«I
a2
«s
a6
Ο,η
a4
a4
a4
as
a6
Яу
1
at
a2
a3
as
«5
a6
4η
a4
o3
1
«1
«2
«6
a6
a7
a4
a5
a2
a3
1
«I
a-j
Ο,η
a4
a5
««
«I
a2
a3
1
(Here at corresponds to a rotation of 90°, while a4 corresponds to a reflection.) The center Ζ
of G is given by Ζ = {1,α2}· This can be checked by verifying that a^ commutes with every
element of D (and no element other than 1 and a2 has this property).
Finally Ζ,Ζα,^Ζα,ι,Ζα,*, are the cosets of Ζ in G. Thus
X — {l,alta4, a5}
is a right transversal of Ζ in G.
(b) The quaternion group Jj[ of order 8, with elements 1, av a2, a3, a4, a5, a8, a7, is given by the
following multiplication table:
1
at
a2
α·3
a6
a7
(Comparing with Table 6.1, we have the following correspondence: 1 -* 1, a -* «ь а2 -* (Ц, a3 -* a3,
Ь -* a4, ab -* a5, a2b -» a6, asb -* a7.) The center Ζ of J£ is given by Ζ = {l,ttg}. One has only to
check this from the multiplication table.
The cosets of Ζ in J( are Z, Zalt Zait Za6. This again can be checked directly from the
multiplication table. Thus
X = {1, α„ α4, α5}
is a right transversal of Ζ in Jf[.
1
2
J
4
5
6
7
1
1
«1
a2
«3
я4
«5
a6
a7
«I
«l
a2
«3
1
a7
a4
«5
a6
a2
«2
a3
1
a,
ac
a7
a4
«5
a3
<%
1
αϊ
a.2
«s
a6
a7
a4
a4
a4
«5
«6
a7
a2
a3
1
a.
«5
a5
4
a7
α·4
«1
«2
a3
1
a6
a6
a7
a4
a5
1
«1
a2
<*з
a7
a7
a4
a5
a6
«3
1
al
«2

4]
PERMUTATIONAL REPRESENTATIONS ON COSETS
221
Find a coset representation of (a) D of Problem 7.6(a) with respect to Z, and (b) J{ of Problem
7.6(b) with respect to Z.
Solution:
(a) We work out a coset representation я- using the right transversal X given in Problem 7.6. Thus
we must find the permutations of X that w assigns to each element of D. We will calculate in
detail the permutation о^тг. Now X = {1, α,,α4, α5}. Then 1·α, = a,, and so lt/Xjn-) = av
a^a[ = 1 since Ζαϊ* = Ζ = Ζί, and so aj(atn-) = 1. (Ца1 = a5 since Za4at = Za7 = Zab, and
so a4(atn-) = a5. Finally a5at = a4 since Z(a5aj) — Zait and so ^(о^тг) = «4- Therefore
α,ττ : 1 -» α,, α, -* 1, α4 -» α5, α5 -* α4
Similarly the other permutations are
In-: 1 -* 1, at -* au a4 -» a4, a5 -* a5
o^n-: 1 -» 1, dj -» at, a4-»a4, as -» a5
asir : 1 -* a i, at -* 1, a4 -» a5, a5 -* a4
a4v : 1 -* a4, at -* a5, a* -* 1, a5 -* at
a5v : 1 -* as, at -* a4, a4 -* 04, a5 -* 1
a6n-: 1 -» a4, at -» a5, a4 -» 1, a5 -» at
a7n-: 1 -* a5, at -* a4, a4 -* au a5 -* 1
It is instructive to check directly that this mapping тг is a homomorphism of D into the
permutation group on {1,αΙ,α4, α5}.
(b) Again we must find the permutations that π assigns to each element of J{. The argument
follows closely that of (a) above, using the set X = {1, av a4, a5}. Here we give only the
result which the reader is urged to check.
In-: 1 -» 1, dj -» at, α4 -» α4, as -» a5
агтт : 1 -» at, % -» 1, a4 -» a5, a5 -» a4
a2n-: 1 -» 1, at -» аъ а4 -» a4, a5 -» a5
a3n-: 1 -* alf at -* 1, a4 -» a5, a5 -* a4
a4n-: 1 -* ait at -» a5, a4 -» 1, as -» аг
а5тг : 1 -* a5, аг -* a4, a4 -* au a5 -* 1
aen-: 1 -» a4, ftj -» ft5, a4 -» 1, a5 -» a2
а77г : 1 -* <t5, at -» <t4, a4 -» a,, a5 -» 1
/1 2\ /1 2\
Consider the permutation group S,12,. Its elements are Фг = I _ ) and 02 = ( J . Con-
/а b\ /a b\
eider now the permutation group S.ab, . Its elements are Ψ1 = I ) , *2 = 1 ) · *i and
0i, *2 and 02 are essentially the same except for the elements they act on. Give a definition which
will make this idea of "essentially the same" precise.
Solution:
Let F be a permutation group on a set X and let G be a permutation group on a set Y. We say
that F and G are isomorphic as permutation groups if there exists a one-to-one onto correspondence
a: X -* Υ and an onto mapping в : F -* G such that for all χ in X and /EF, («/)<* = {xa)(fe).
(In this problem a : 1 -* a, 2 -* Ь and в : Фг -* *1( 02 -» *2-)
Prove that if F and G are isomorphic as permutation groups, then they are isomorphic as groups.
(Hard.)
Solution:
The в of the solution of Problem 7.8 provides the isomorphism. First we show that it is a
homomorphism. Let flrfjl^F. For any xGX,
(a:a)((/i/2)0) = (w(/i/2))a = (.(xfi)fs)a (ьУ the definition of composition of mappings)
= (<*/i)«)(/se) = ((*«)(/! ·))(/* ·) = (a»)((/i»)(/2·))

222
PERMUTATIONAL REPRESENTATIONS
[CHAP. 7
Since α is a mapping onto Y, it follows that as χ ranges over X, xa ranges over Y. Hence as (f\f2)e
and (fie)(f2e) are permutations, /ιβ/2β = (f\S){S2e). Thus β is a homomorphisrn.
Next we must show that в is one-to-one. Suppose that fxe = f2e; then if χ G Χ, (χα)(/χβ) —
(x«)(f2e). Thus
(xfja = (Xf2)a
Since a is one-to-one, xfx = xf2 from which /t — /2. Hence β is an isomorphism.
7.10. Let G be any group and Η a subgroup of G. Let X{ be a transversal for Я in б and n-( the
corresponding coset representation, i = 1,2. Prove that Gvi and Gv2 are isomorphic as permutation
groups. (Hard.)
Solution:
Since Xt and X2 contain one and only one element in each coset of Η in G, we define a : Xt -» X2
by sending X\ G Xt to хг^Хг if Нхг = Hxz. a is then a one-to-one correspondence. We define
μ : Gtti -» Gtt2 by (flfn-i)/i = flfi72. It is easy to check that μ is a mapping. Let дтгг = 0, fifn-2 = y. We
need only verify that (χβ)α — (χα)γ for each x€:X. Now by definition of β and a,
Hxg = Η(χβ) - Η((χβ)α)
Also, Hwfif = H(xa)g — H((xa)y)
Hence (жа)у = (χβ)α as they are elements of X2 that belong to the same coset, i.e. Hxg. The result
follows.
7.5 FROBENIUS' VARIATION OF CAYLEY'S THEOREM
a. The kernel of a coset representation
Is a coset representation τ of a group G with respect to a subgroup Η ever faithful?
We know that the answer is yes if Η — {1}. The object now is to find the kernel of π.
Theorem 7.2: Let G be a group and Я a subgroup of G. Let π be a coset representation of
G with respect to H. Then the kernel of π is the largest normal subgroup
of G contained in H, i.e. if Ν Ο G and #DiV, then NcKerir.
Before proving Theorem 7.2, it should be noted that from this theorem it follows that π
is faithful if the only normal subgroup of G which is contained in Η is the identity subgroup.
This implies, for example, that if G is a simple group and Η Φ G, π is automatically
faithful. For then, by definition, the only normal subgroups of G are G and the identity subgroup.
This observation has been useful in the theory of finite groups.
Proof: Let X be the right transversal of Η in G from which π was defined. First we
prove that if К is the kernel of π, i.e. the set of all elements g of G such that g-π = *, the
identity mapping of X onto itself, then К is contained in H.
If a, G К and χ G X, then air = t, i.e. ж = ж(отг) = χα. In particular on putting
χ — 1, 1 = ά. Hence a G Я. This means that ii is contained in H. Of course .K, as the
kernel of the homomorphisrn τ, is a normal subgroup of G. To complete the proof of the
theorem, we must show that К is the largest normal subgroup of G contained in H. To do
this it is sufficient to prove that if N is any normal subgroup of G contained in H, then
Ntt = {t}.
Suppose that a G N. If ж Ε Χ, then
Яжа = Hxax~1x
Since N is normal, хах~г G iV. But N is contained in H, and so жаж"-1 G Я. Accordingly
Hxa = Яж. Thus жа = χ which means
х(атг) = χ for all ж G X
Hence α-π = t and N Q К as required.

Sec. 7.5]
FROBENIUS' VARIATION OF CAYLEY'S THEOREM
223
Problems
7.11. Let G be the symmetric group on {1, 2, 3}. Let a : 1 -» 2, 2 -» 3, 3 -»1, and let τ : 1 -> 2, 2 ->1,
3 -» 3. Then the elements of G are t, α, σ2, τ, στ, αιτ. Let (α) Η = gp(a) = {ι, σ, a2} and (6) Η =
др(т) = {ι, τ}. In each case find a coset representation of G with respect to H. Find also the kernels
of both representations.
Solution:
(a) A right transversal of Η in G is {t, τ}. Notice that t~W = a1 = a2. So Η is normal in G.
Hence by Theorem 7.2, every coset representation has kernel precisely H. The coset
representation π associated with the right transversal {ι, τ} is given by
ITT . 1 -» ί, Τ ~* Τ
απ '. ι -* ι, τ -* τ
ο^π : ι -* ι, τ -» τ
7тг : ι -» τ, τ -» ι
(στ)π : ι -* τ, τ -* ι
(σ2τ)π: ί -» τ, τ -* ι
Clearly Ker π = {t, α, a2}.
(6) Let Χ = {ι, σ,σ2}. Since G == HuHaUHa2, Χ is a right transversal of Η in G. The associated
permutational representation ν is
Τ7Γ I
(στ)ττ I
(σ2τ)ιτ-:
ι -+ ι,
^α\
ι ~* σ,
σ->σ2, ο2
σ -* α, α2
α -* ι, α*
α2π : ι -» σ2, σ -» ι, σ2 -* σ
It follows immediately that тг is faithful, since the only element mapped to the identity
permutation of X is ι. Hence the kernel of ж = {(}■
7.12. Let G be the alternating group of degree 4 and let Η be the subgroup consisting of the permutations
α 2 3 4\ /1 2 3 4\ /l 2 3 4'
^2 1 4 3/' U 4 1 2/' U 3 2 1,
Find an associated coset representation of G with respect to H. Is this representation faithful?
Solution:
Let π
12 3 4
2 3 14
Then if
1 2 3 4\ /1 2 3 4\ /1 2 3 4
2 14 3/' T2 ~ U 4 1 2/' Тз = {A 3 2 1
G consists of the elements
ι, σ, σ2, Tj, Tjff, τλσ2, τ2, τ2σ, τ^σ2, τ3, τ3σ, τ3σ2
It follows immediately that a right transversal of Η in G is X= {ι,α,α2}. Let n- be the associated
coset representation. It is easy to check directly that Η is normal in G. Then Kern- = Η by
Theorem 7.2, and so π is not faithful. Finally we list the permutations g-π with g in G:
iir : t -* ι, α -* α, σ2 ~* a2
air : ι -* a, a -* <r2, σ2 ~> ι
τχττ : ι -* ι, α -* α, σΔ ~> σΔ
(тга)ж : ι -* σ, а -* σ2, а2 -» ι
(Tjff2)n·: ι -* a2, a -* ι, a2 -* a
τ2π
(τ2σ)π
(τ2σ2)π
τ3π
(τ3σ)π
(τ3σ2)ν
' ι -* ι, σ -* α, σζ
ί -» α, α -* α2, σ2
. ι -* σ2, σ -* ι, σ2
ι -* ί, α -* α, σ2
ι -* σ, α -* σ2, α2
ι -* α2, α -» ι, σ2
b. Frobenius' theorem
Let Я be a subgroup of G, X a transversal and it the associated coset representation. Let
ρ denote the right regular representation of G. Our idea is to express P in terms of -л.
If χ e X and g e G, we have
z(S7>) = zflr
Of course xg need not lie in X.

224
PEEMUTATIONAL REPRESENTATIONS
[CHAP. 7
We know, however, that xg belongs to the same coset as xg. Hence xg = axg where
a € Я. Now a is clearly dependent on χ and g, and we denote it by ax,g. Substituting ax.s
for a, we have
xg - ах,вхд (74)
or x{gP) - ах,вх{дтт) {7.5)
Now any element of G can be expressed in the form hx for ft € Я and χ € X. Therefore
(hx)(gP) = hxg = h(x(gP)) - hax.g х(дтт), i.e.
(hx)(gP) = (hax,g)(x(g^)) (7.6)
This equation suggests that the effect of gP on an element hx of G can be explained by
what happens to ft (it goes to the element haXlB of Я) and what happens to ж (it goes to
x(gir)).
We express this formally in the following theorem which is due (essentially) to Frobenius.
Theorem 7.3: Let G be a group and Я a subgroup of G. Let X be a right transversal of
Я in G. Then there is a faithful representation θ of G as a group of
permutations of Hx X (the cartesian product of Я and X) defined by
(к,х)(дв) = (Αα»,β, 3!(ί)τπ))
Proof: The proof is an adaptation of the discussion of the last few paragraphs. First
let 7Г be the coset representation with respect to Я with right-transversal X. For each
g € G the permutation дгтг gives rise to a permutation gk of Hx X which is defined as
follows:
(ft, ζ)(ί/λ) = (ft, ζ(ί/π)), (ft € Я, χ e X) (7.7)
Note that grA is a permutation of HxX. For if
(ft,z)(iA) - {h',x'){gk)
then (Л,ж(аг7г)) = (ft', x'{gir))
Since s^ is a permutation of X, it follows from х(дтт) - x'(g-n) that χ = я'. Hence we have
proved that </λ is one-to-one. But gX is also onto since g-π is onto X. Clearly gk is then a
permutation of HxX.
Now as we saw in (7.5), if g € G, then
ОД = а*,в(ж(9,7Г))« (α*.β € Η)
For each g € G define (h,x)ga — (hax,g,x)
We verify that ^σ is a permutation of Я x X. Suppose (ft, x) € Я X X. Then
(Ьа^в,ж)йга = (Α,α^α*,Β, ж) = (ft, ж)
Thus да is a mapping of HxX onto HxX. It remains to verify that </σ is one-to-one.
Suppose that (ft, x)ga = (ft', x')ga. This means that
(hax,g, x) = {h'ax-ig, x')
and therefore we find χ = χ' and ftax,e = ft'aI>e from which ft = ft' and (ft, ж) = (ft', ж').
Hence да is a permutation of Я χ X,
Finally we compute ^σί/λ.
{h,x)(gagk) = {(h,x)ga)gk = (haXjB,x)gk = (ftax,9, х(дтт)) {7.8)
Thus (g*)(gk) = дв {7.9)

Sec. 7.6]
FROBENIUS' VARIATION OF CAYLEY'S THEOREM
225
As both да and gk are permutations of Я x X, дв is a permutation of Η x X.
Let gft, gr2 £ G. We must show that
(ff,fl)(ffafl) = M»)* (7-J0)
(Note that the left-hand side of {7.10) is the product of two permutations.) To facilitate
the proof of (7.10) we introduce the following notation:
ax ~ &{*' α2 = 02π« as ~ {9гд^, a = дтт (7-11)
where g is an element of G.
Equations (7.8), (7.9) and (7.11) yield
(h,x)ge = (kdx.g.xa) (7.12)
Note that as π is a homomorphism, (дгтт)(д2тт) = (дгд2)тт so that
αια2 = аэ (7.13)
Applying (7.12) twice and (7.IS) once, we have
(Кя){дгвд2в) = ((к,х)дхв)д2в = (hax,ei, хаг)д2в = (Λο*.βιοϊβι.β2, (ж«>2)
= (^αχ,βΙαχα,,β2, «(«^a)) - (hax,giaiax,g2, хая) (7.14)
,Οη the other hand, again on using (7.12), we have
(к,х)(длдъ)в = (йа*,в,в2, ж«3) (7.J5)
To prove that б is a homomorphism, we must show that the right-hand side of (7.14) is
equal to the right-hand side of (7.15), i.e. we must show that
To accomplish this we use equations (7.5) and (7.6) and obtain
Ж(ЗД) = β*·|·2 (Ж«3) (7·17)
Also (х9г)92 = (^'.al(xaj)g2 = αχ.αχαχαι,ϋ1ι((χαι)α2) = ατ,βιαχανΒ2(χα3)
from (7.13). This means (^9ι)92 = а'-«1аха1-«2(Хаз) (7.18)
Since «(flfjflfjj) = (хдг)дг the right-hand sides of (7.i7) and (/.iS) are equal. Thus (7.16)
follows and therefore θ is a homomorphism.
It only remains to show that θ is one-to-one. Assume gfi = д2в. Then (Л., x)gx6 =
(h,x)g26 for all pairs (h,x) efixl; and in particular, if (h,x) = (1,1),
(«i..,»l(M) = (αι.·2. ^M)
from which aUai - aUg2 and 1(дхтт) = 1(δτζπ) (7.ί0)
Using equation (7.5) with ж = 1, we see that
9i = 49ip) = «ι.β1(1(Μ)' 92 = 4ff2p) = Λι.β2(1(ί72τ))
Using (7.i#) we conclude that gx = gr Thus б is one-to-one.
This completes the proof of Theorem 7.3.
We call the homomorphism θ of Theorem 7.3 a Frobenius representation (with respect
toH). Of course θ depends on X as well, but it can be shown that in a sense this dependence
does not matter.

226
PERMUTATIONAL REPRESENTATIONS
[CHAP. 7
Problems
7.13. Describe in detail a Frobenius representation for the symmetric group on {1, 2, 3} relative to the
subgroups given in Problem 7.11(a) and (b).
Solution:
(a) The set X = {t, τ} is a right transversal of Η = {ι, α, a2} in G. Let в be the faithful
representation of G on Η Χ X described above. The elements of G are
ι, α, <r2, τ, στ, σ2τ
We use the formula xg = aXlgxg to calculate ax,a. Note that 1 = ι, α = ι, σ2 = t, τ = τ, στ = τ,
σ2τ = τ.
As an illustration of the procedure we calculate ατ,σ. Now та = aT,a-fa. Since та — σ2τ,
we have fa — τ and ατ,σ — таг — a2. Similar calculations lead to
"Ί,Ι — '
ai,<r = *>
a. „2 = <r2
ατ,στ = »2
«ι,σ2τ
α-„2 — σ
Οτ σ2τ
We use the definition of в given in the statement of Theorem 7.3. The effect of я- on g is
given in the solution of Problem 7.11(a). The results, repeated here for convenience, are
иг = air = (a2)w : ι -* t, τ -» τ
ттг = (στ)π· = (σ2τ)π- : ι -» τ, τ -> ι
We can now calculate the effect of дв for each g in G. In particular,
(λ, ι)σθ = (Ααι>0., i(av)) = (ha, ι)
and (Α, τ)σθ = (haTr<r, τ(σπ)) = (ha2, τ) for all A G ff
We list the effect of the permutations ge for the elements of G.
ιθ : (h, x) -> (А, ж) (/lefl.ie X)
αβ : (h, ι) -* (ha, l), (h, τ) -* (ha2, τ) (h G Я)
ff20: (h, ι) ->■ (ha*, ή, {h,r)^(hafT) (h € H)
τ0: (Μ-»№.τ), (Λ, r)-» (Л, О (AGH)
(<jt)9 : (h, ι) -* (ha, τ), (Α, τ) ^ (A<r2, t) (h G Я)
(<г2т)0 : (Α, ι) -> (A<r2, τ), (Α, τ) ^ (/ur, ι) (h € Я)
One could check that 9 is a homomorphism by inspecting (дхв)(д2в) and (ΡΊί^β, where (g\g2 G G).
The above description of в immediately shows that s is one-to-one.
(b) Here Η = {Ι,τ}; X, a right transversal of Η in G, is given by X = {ι,σ,σ2}. The Frobenius
representation в is then given by
Ш
σθ
α2θ
τθ
(ατ)θ
(Α-)β
(h,x) -*(h,x) (hG H, xGX)
(Μ-»№,*), (Λ, σ) - (Λ, ο3), (Α,^)^(Μ
(М-МЛ,"8), (Μ)-» (Μ, (Α,»2)-ΜΑ,*)
(Α, 0 -* (Ατ, 0, (Λ, <0 "* (Κ a2), (h, σ2) -» (hr, a)
(h,i)-*(hr,a2), (Α,σ)^(Ατ,σ), (h, a2) -* (hr,ι)
(Λ, 0 -+ (hr, a), (h, a) -» (At, ,), (A, a2) -» (At, **)
(AGH)
(AGH)
(AGH)
(agh)
(AGH)
7.14. Describe in detail the Frobenius representations for the alternating group of degree 4 relative to
the subgroup Η given in Problem 7.12.
Solution:
Here G consists of
ι, σ, (Γ2, τ„ Tjff, τ,σ2, τ2, τ2σ, τ2σ2, τ3, τ3σ, τ3σ2
and Η= {ι, TlfT2, τ3}. A right transversal of Я in G is Ζ = {ι, a, a2}. The Frobenius
representation в is then described as follows:

Sec. 7.6]
APPLICATIONS TO FINITELY GENERATED GROUPS
227
iff:
ae:
<fie:
Τχθ 1
(Τισ)β :
(η°2)θ :
т2е:
(т2?)е :
(τ2σ2)β :
тзв:
(τ3σ)β:
(τ3σ2)β :
(А, ж) -* (Α, χ),
(Α, ι) -» (Α, σ),
(Α,Ο-ίΑ,ο»),
(Α,0-*(Ατ],0,
(Α, ι) -»(ΑτΙ,σ),
(Α,Ο-^ίΑτ!,^),
(Α, ι) -> (Ατ2> ι),
(Α, ι) -* (Ατ2, σ),
(A, t) -> (Ατ2, σ2),
(Α, 0 -> (Ατ3, ι),
(Α,.) -»(Ατ3,σ),
(Α,0^(Ατ3,σ2),
(Α € Я, κ€Ι)
(Α, σ)->(Λ,«»),
(Α, σ) -* (A, t),
(Α,σ) -» (Ατ^σ),
(Α, σ) -» (Ατ2, α2),
(Α, σ) -> (Ατ2, 0.
(Α,σ) -» (Ατ3,<0,
(Α, σ) -* (Атд,»2),
(Α, σ) -* (Ат3,е),
(Α, σ) -* (ΑΤι, α),
(Α, σ) -> (Ατ„ α2),
(h,a)-*(hTvi),
(Α,σ2)
(λ,σ2
(Α,σ2
(Α,σ»)
(Α,σ*
(Α,σ2)
(Α, ο»
(Α,σ»
(Α,σ*
(Α,*2)
(Α,σ»
-♦(Α,.)
-»(Α,σ)
-» (Ατ,, σ2)
-* (Ατ3, t)
-* (Ατ3, σ)
"* (Ατϊ, σ«)
) -» (ΑΤι, t)
) -> (ΑτΙ; σ)
"* (At* σ«)
-* (Ατ2, ί)
-» (Ατ2, σ)
(Α€Η)
(he Η)
(he Η)
(he Η)
(he Η)
(he Η)
(he Η)
(he Η)
(he Η)
(he Η)
(he Η)
7.6 APPLICATIONS TO FINITELY GENERATED GROUPS
a. Subgroups of finite index
Frobenius' representation, although only a variation of Cayley's, is very useful. Here
we shall give one application of this representation. First we recall a definition given in
Chapter 4.
Definition: A group G is finitely generated if it can be generated by a finite set, i.e. if there
is a finite subset $ (¥· 0) of G such that for each g G. G there are elements
8Vs2, .. .,snE. S and integers c, «n («t = ±1) such that
g = Sll ■■■ snn
Theorem 7.4 (O. Schreier): A subgroup of finite index in a finitely generated group is
finitely generated.
Proof: Let G be a finitely generated group. Let S be a finite set of generators of G,
with |S| = m. Suppose Я is a subgroup of finite index in G. Choose a right transversal
X — {xv ..., x.} of Я in G, with xt the identity. Notice that j < » by assumption. Let
θ be a Frobenius representation of G with respect to Η given in Theorem 7.3. If h G. H,
then h = 1 and auh — h so that we have (1, l){hd) — (h, 1). But h can be written as
л = <'■■■<" (* = ±i)
with slt ...,8neS. Put ij = s'f, i — Ι,.,.,η; then h = tl---tn. Since θ is a
homomorphism,
Let tn = «., ΐ = 1,.. „,·». By repeated applications of the definition of the action of дв in
Я χ X (see Theorem 7.3) we have
(l,l)(fcfl) = Κ^Ια,Χ^)-·-^) = К^1(11,(2ДМ)(У)-(М)
= (ai.t^la^t^la^.ta * * ■ αιαι...αη_,,(„>- l(«i ' ' ' «„)) = (fe> 1)
Hence fe = ai,tIaiaI,ta ' ·· ttIai...an_Iltn. Since at, ...,«„ e Sx, 1(αι···α^£Χ for each
i, 1^»^и. In other words we have expressed h as the product of elements of the form
ax.t where xGX, and t = s-1 where s G S
As \X\ = j and |S| = m, the number of such elements is at most 2mj. This means that Я
is finitely generated.

228
PERMUTATIONAL REPRESENTATIONS
[CHAP. 7
b. Remarks about the proof of Theorem 7.4
We have actually proved that if G can be generated by m elements and Я is of index
j in G, then Я can be generated by 2jm elements. However, it is not difficult to reduce
this number to jm. To do so observe that we have proved that the elements ax,t generate Я,
where χ is an element of a right transversal X of Я in G, We recall that if g denotes the
representative of the coset Hg, then ax,a is defined by
%9 = &х,я%д
(equation (74), page 224). This means that
ο-χ,β = xg(xu)~l
Therefore the elements xt(M)~l generate H, with t = s or t = s~l, where s € S. Note
that ==_
XS~lS = X8~l8 = X = X
by equation (7.3), page 219. Let у — xsl; then p = χ and
aI)S-i = xs-lsx~l = ys(ys)~l = a„.,
i.e. a,,,-i is the inverse of a„,t. Thus Я is actually generated by the elements aXjS, χ € X
and s € S. The number of these elements is jm. (In fact one can lower this bound and
prove that Я can be generated by 1 + (w — l)j elements. For a proof of this result see
Theorem 8.13, page 264.)
Note that in Section 7.6a we have actually proved that Я is generated by ax,,, χ G X,
s G S, without the assumption that \X\ and |S[ are finite.
Problems
7.15. Let G be the symmetric group of degree 3 and let Я be a subgroup of index 2 in G. Find the set
of generators of Η described above.
Solution:
We use the description of G in Problem 7.11(a). Now a subgroup of index 2 in any group is
normal. Thus if Η is of index 2 in G, Η is normal in G. Moreover, [ff| = 3. Hence Η = {ι, a, a3},
since this is the only subgroup of order 3 in G. A right transversal of Η in G is X = {t, τ}.
Clearly G can be generated by a and r, and so Η is generated by
αι-σ = σσ~ι — σ ατ,σ — τσ(τσ)~ι — τστ-1 = «τ2
θιιΤ — τ?-1 = ι αΤιΤ = ττ(ττ)-1 = τ2 = ι
Thus we find that ι, σ, α2 generate Я. Of course Η is actually generated by π alone.
7.16. Let G be any group and let Η be a subgroup of G of index 2. Prove that if G can be generated by
two elements, then Η can be generated by three elements.
Solution:
Suppose that G is a group generated by с and d and that Я is a subgroup of index 2 in G. If
both с and d are in Я, then HoG and Я is not of index 2 as initially assumed. Without loss of
generality we may suppose that c& H. Then the cosete of Я in G are Я and He. Thus every
element of G can be written in the form he or Л (А е Я). This means that {1, c) is a right transversal
of Я in G. Therefore G is generated by the elements
aUc = lc(lc)-i = cc-i = 1, <Ч,а, Be,e> Bc,d
Hence Я is generated by the three elements a1>(j, oscc and ot^.

Sec. 7.6]
APPLICATIONS TO FINITELY GENERATED GROUPS
229
7.17, Let G be generated by α and b and suppose that N is a normal subgroup of G such that G/N is
generated by Na and G/N is infinite cyclic. Find a set of generators for N in terms of α and b.
(Hard.)
Solution:
It is clear that X = {1, a*1,!!*2, ...} is a right transversal of N in G. If b & N then since
Nb = (Na)w for some integer w, ba~w e N. Put с = ba~w. On the other hand if b SN, put
e = b. Clearly gp{a, e) = G. We therefore take S = {a, c}. The generators of N are the elements
ax,s = aw(«s)-1 (ж ε A', s G 5)
If £ = aJ and s = a, axs — aja(a*a)-1 = а%(а^ + 1)-1 — 1
If χ — α,ι and s = c, aIiS = а*с(а*е)-1 = aJca~>
Hence the elements {..., a-ica, c, ooa-1, ...} are a set of generators for N. Since either с = ba~~w
or с = b, we can restate this set of generators in terms of a and 6 thereby obtaining a set of
generators of G of the desired kind.
7.18. Let G be generated by a and 6. Find generators for all possible subgroups of index 2. Hence show
G has at most three subgroups of index 2.
Solution:
Let Η be a subgroup of index 2 in G. Then we may have
(1) α g H, b € Я, (2) о£Я,И Η, (3) α g Η, 6 g H.
In case (1) take λ' = {1, α} and 5 = {а, Ь}. Then generators of Η are the аХшЯ (χ e λ' and
s € 5). Thus Η is generated by b, a2 and aba-*.
In case (2), proceeding as in (1) with λ' = {1, b}, Η is generated by a, b2 and bab'1.
In case (3), ab = « € H. Take X = {Ι,α}, and 5 = {a,c}. Then Η is generated by с, аса-1
and a2.
So the possible subgroups of index 2 are:
(1) gp(b,aba~l,a% (2) flip (a, bab"1, bz), (3) fir ρ (ab, a*ba~\ a2)
7.19. Let G = firp(a, b,c) and let N be a normal subgroup of G of index 3 with G/N = gp(Na). Suppose
N contains b and c. Find a set of generators for N in terms of a, b and c.
Solution:
Choose S={a,b,c] and Χ = {1, α, α*}. Then the elements ax>s, with ж e X and s € 5,
generate N. Thus N is generated by a3, b, c, aba1, aoa~l, a2ba"2, a2ca~2.
7.20. Prove that if a group G contains a subgroup Η of index 2 which is cyclic, then every subgroup of
G of index 2 can be generated by two elements. (Hard.)
Solution:
Let Η be generated by b and suppose that α g H. It follows that G = fifp(a, 6). From Problem
7.18 the possible subgroups of index 2 are
(1) gp(b,aba-\a2) = Щ, (2) gp(a,bab~\b2) - H2, (3) gp(ab,a?ba~i,a?) = H3
Clearly b g H2 or H3, as then each of them would actually be equal to G. Thus Η = Нг. Since
Η is the cyclic group generated by b, aba"1 = br and a2 = 6s for some integers r and s. We have
ba = α-lb"" (7.20)
and a2 = 6* (7.2i)
The generators of H2 are а, ЬаЬ-1 and Ь2. ЬаЬ~1 = (Ьа)Ь-1 = a-1br_1 from (7.20), and so H2 is
generated by а, ЬГ-1,Ь2. But i?p(br-1,b2) is cyclic generated by c, say, as it is a subgroup of a
cyclic group. Thus H2 can be generated by two elements.
The generators of ff3 are ab, a2ba~* and a2. Hence ab, a2ba~l(ab) and a2 are generators for
H3. Using (7.21) we conclude that ab, bs + z and 6s are generators for Я3. But gp(bs+2, bs) is cyclic
generated by c, say; ao ff3 is generated by two elements and the proof is complete.

230
PERMUTATIONAL REPRESENTATIONS
[CHAP. 7
c. Marshall Hall's theorem
The second application of permutational representations is due to Marshall Hall.
Theorem 7.5: The number of subgroups of finite index j in a finitely generated group is
finite.
This theorem may be restated as follows: Let G be a group which is generated by au ..., an
where η < °°, Suppose j is a fixed positive integer. Then the number of subgroups of G
of index j is finite.
Proof: Let S; be the symmetric group on 1, . .., j. For each subgroup Η of index j in
the finitely generated group G choose a right transversal XH (we emphasize that Xa contains
the identity of G). To avoid confusion between the number 1 and the identity of G, we
shall write the identity (for this proof alone) as e. Thus we have e G XH. Let πΗ be the
coset representation of G with respect to the transversal XH (see Section 7.4). Then irH is a
homomorphism of G into SXh. Since \XH\ = j, it is easy to prove that there exists an
isomorphism ώ · Sx„ -»S. such that θ G Sx„ moves e or leaves it fixed according as to
whether θψΗ moves 1 or leaves it fixed (see Problems 7.21 and 7.22 below).
Note that *H == πΗφΉ: G-> S. is a homomorphism of G into Sj since it is the composition
of two homomorphisms. Note also that if Η and К are two subgroups of index j and
Η ¥- К, then ΨΗ Φ *K, for there exists an element g G Η but g € К (or vice versa). Then
е(дтгн) = e for e(girH) = eg = e as g GH (see Section 7.4). Accordingly 1*H = 1. On
the other hand, as g&K, e(g-rrK) ^ e and hence 1ΨΗ¥Ί. Thus ΨΗ=έΨκ if Η ¥= К.
We have therefore found that the number of subgroups of index j in G is certainly not
greater than the number of homomorphisms of G into S.. This is where the fact that G is
finitely generated comes in. For suppose G is generated by av .. ., av. If ψ, θ are
homomorphisms of G into S. such that а$ = α.θ for i = 1,.. .,n, then φ = θ. To prove this,
observe that if g G G, then
g = all' ' ' αϊ £t = ±X· ''j e (1, · ..,и}
and дф = (сцф)'1 ■ · · (aifc</>f = (tn^)'1 · · ■ {atk0)k = дв
Since φ and # agree on every element of G, φ — θ. This means that the number of
homomorphisms of G into S is finite since the number of possible images of the generators of
G is finite (at most (jl)n).
This completes the proof of the theorem.
d. One consequence of Theorem 7.5
Let G be a finite group and θ a homomorphism of G onto G. It follows that θ is an
isomorphism, for |C?I = |C?0| = |GVKer#| and hence Ker# = {1}. If G is not finite, is it
possible to have a homomorphism θ of G onto G with θ not an isomorphism? For example, if
Ρ is a p-Prufer group (see Section 6.2c, page 191), let θ : Ρ -» Ρ be defined by χθ = px, χ GP.
Then Ρθ = Ρ; but as Ρ has an element of order ρ, θ is not an isomorphism.
In the following theorem we prove a result which tells us that for a special class of
groups every onto homomorphism is an isomorphism.
Theorem 7.6 (A. I. MaFcev): Let G be any finitely generated group whose subgroups of
finite index have intersection 1. Then every
homomorphism φ of G onto G is an automorphism.

Sec. 7.6]
APPLICATIONS TO FINITELY GENERATED GROUPS
231
Proof: Let К be the kernel of φ and let L be any subgroup of finite index in G. If L
is of index /, then the number of subgroups of G of index j is finite. Let these subgroups be
L = L\, L2, ..., L]t
Now, by Theorem 4.18, page 117,
G = ΰφ ε* G/K
and so the number of subgroups of index / in G/K is precisely k, the number of subgroups
of index j in G. Let Mt/K, ...,MJK be these subgroups of index j in G/K. Then
Mh ..., Mk are к distinct subgroups of index j in G, by Corollary 4.20, page 121. Thus the
Mi's are simply a rearrangement of the Li. Therefore every Lf is an Mi and so contains K.
In particular
LD К
This means that every subgroup of finite index contains K. Hence К is contained in the
intersection of the subgroups of finite index. By hypothesis, this intersection is 1. So
К = 1, Accordingly φ is one-to-one, and φ is an automorphism.
This theorem is important in current research in group theory.
Problems
7.21. Prove that there exists an isomorphism μ between S^x iX . and S2 such that if β G ®{xvx } anc*
χφ = Xj, then ϊ{βμ) = j. (Hint: see Problems 7.8 and 7.9, page 221.)
Solution:
Let a : {хъ x2} -» {1, 2} be defined by x,a = j, j = 1,2. Let μ be defined by
(xx x2\ _ /1 2\ (xx x2\ _ /1 2\
Then α, μ define a permutation isomorphism (see Problems 7.8 and 7.9) and therefore μ is an
isomorphism with the required effect.
7.22. Prove that in general there exists an isomorphism μ between S[X .. miX } and Sn such that if
в G S^x yX j and xfi = Xj, then ΐ(θμ) — j. (Use Problems 7.8 and 7.9.)
Solution:
Let a: {xlr ...,xn}-*· {1,2, .... n} be defined by xpt = i. If jeS(l x y, define θμ e Sn
to be the mapping that sends i-» j if x$ = Xj (as θ is a permutation of {xu .. -,x„}, θμ is a
permutation of {1,2, ...,«}). It is clear then that μ is onto S„, and hence α,μ provide an isomorphism
of permutation groups. Thus μ is the required isomorphism.
7.23. Prove that if Η and К are subgroups of G, then each coset of Η intersects a coset of К either in the
empty set or in a coset of HnK. Hence prove that if В and К are of finite index in G, so is HnK.
Solution:
If a coset of Η and a coset of К have an element g in common, then the two eosets are Hg and
Kg. χ G HgnKg if and only if χ = hg = kg, for some hG Η and к G K. But kg = kg if and
only if h=k, i.e. he HnK. Thus χ G HgnKg if and only if χ G (Hc\K)g. Then HgnKg =
(HnK)g and the two eosets meet in a coset of HnK.
If Η is of index η and К is of index m, at most nm eosets can be found as intersections of a
eoset of Η with a coset of K. Furthermore these are all the eosets of HnK, for any coset (HnK)g =
HgnKg and so is the intersection of a coset of Η by a coset of K. Therefore HnK is of finite
index in G if both Η and К are.

232
PEKMUTATIONAL EEPEESENTATIONS
[CHAP. 7
7.24. Let G be finitely generated with a subgroup of index j. Prove that the intersection of all subgroups
of index j in G is a normal subgroup of finite index. (Hard.)
Solution:
By Theorem 7.6, G has only a finite number of subgroups, Mlr.. .,Mn say, of index j. Now if
Μ is any subgroup of index j, it is easy to prove that х~гМх = Μ is also of index j. (If the cosets
of Μ are Mgu .. .,Мд}, then the cosets of Й are Мх~1дгх, Mx~lg^e,.. ,,-fflx~lgjX- For if jeG,
x~1gx £ Mgb for some г, implies that g £ x~lMxx~lg{x = Мх-^д^х.) Hence MlnM2n ■ · · Г\Мп = К
is a normal subgroup of G, for if g G К and χ G G, x~lgx £ х~1Мгх, x~lM2x, ■ ■ -, x~lMnx. But
x~1Mlx, ..., χ~ιΜηχ are η distinct subgroups of index j. Hence they must be all the subgroups of
index j (perhaps in a different order). Thus x~lgx £ К and so К О G. К is of finite index by
repeated application of Problem 7.23.
7.25. Let G and Η be two groups and suppose that G satisfies the conditions of Theorem 7.6. Let
в : G -» Η and ψ: Η -> G be epimorphisms. Prove that G = Я.
Solution:
βφ is an epimorphism of G to itself and so by Theorem 7.6, θψ is an isomorphism. Then if
{(^1, g e G, д(вф) Φ 1 and thus ge Φ 1. Therefore θ is one-to-one and в is an isomorphism of
G to H.
7.26. Let N and Μ be unequal normal subgroups of a finitely generated group G, M^N. Suppose the
intersection of the subgroups of finite index in GIN is the identity. Prove that G/M is not isomorphic
with GIN.
Solution:
Let $ : G/M -» G/N be such an isomorphism. Let μ: GIN -» G/M be defined by Ng -» Mg.
It is easy to verify that μ is an epimorphism. Then μβ is a homomorphism of G/N onto itself. By
Theorem 7.6, μβ is an isomorphism. Now if g £ Μ — Ν, (Ng)^e) = Me is the identity of G/N.
Thus μβ is not an isomorphism. This contradiction yields the required result.
7.7 EXTENSIONS
a. General extension
Suppose G is a group with a normal subgroup Η and that G/H a* K. Then, using the
terminology introduced in Chapter 5, G is an extension of Η by K, It is convenient to
generalize this concept and to say that G is an extension of Η by К if G has a subgroup Я with
Я эг Я and G/Я эг χ. it is our aim to investigate how a group is built as an extension
of one group by another.
In this section let G be a fixed group and Η a normal subgroup of G. Let φ be an
isomorphism of G/H onto K. Let X be a left transversal of Я in G, i.e. a set of elements of
G containing one and only one element from each left coset of Я in G with l£l.
If j£G, g = xh for some χ Ε. Χ and some Л- £ Я. It is easy to see that this
expression for g is unique. Let g £ G, ж e Z; then до belongs to some coset of Я in G, say the
coset #Я where у £ X. Therefore
gx = yh
for some h £ Я. Now Л- is uniquely determined by g and ar, we denote h by m9>1. Thus
gx = 2/rng,* (7.22)
The elements mB,x correspond to the elements α,χ,ο introduced in Section 7.5b. (We use ma,x
instead of ax,g because here we are dealing with left instead of right cosets. We will explain
in Section 7.7c the minor reason why we use left cosets here.)
Note that φ, the isomorphism of G/H onto K, is a one-to-one mapping of the set of left
cosets of Η onto K. Therefore we can unambiguously denote the representative of the coset
gH by хн if {дЩф = к. In particular then, χι = 1. With this notation,
X = {xk\ ktK)

CNBC- Ι* ί Ι
EXTENSIONS
233
Notice that as (хкХк-Н)ф = {(хкН)(Хк-Н)}ф = (хкН)ф(хк-Н)ф = kk' where к, к' G К, the
representative of the coset XkXk'H is xkk·. Then, from {7.22) we have
£fcav = Xkk'frij; χ where mx x G Я
We suppress the x'a and write mk,r for Штк,хк,. Thus,
Ж|сЖ|с· = Хкк"Шк,к' where k,krGK, mk,k'^H (7,23)
Every element g in G can be written uniquely in the form xkh where Xk G X, h G Я.
To express the product of two elements xji and Xk-h' as the product of an element of X
by an element of Я, we proceed as follows:
xJi-Xk-h' = XkXk" x^ hxk'hr = Хкк'Шк.к'Хк' hxk'h' (7.24)
Observe that χκΙιχν&Ή. since Я is a normal subgroup of G. So а^е! and
Шк.к'Хк- hxk'h' G Я. The right-hand-side of (7.24.) looks less complicated if we introduce the
notation hk for x^hXk·; then
Xkh-Xk'h' = Xkk'mk,k-hk'hf (7.25)
It appears from equation (7.25) that the extension G of Я by if that we have been
inspecting is determined by the mk,k· and by the images hk of the elements h obtained by
conjugation by the Xk\ i.e. by forming av hxk·. One may conveniently think of the elements
mk,k- in Я as the images of a function m of two variables (coming from K) with values in Я.
In other words, we may think of m as a mapping from the cartesian product Kx К into Я,
where we use mb,k· to denote the image of (k,kf) GKxK under this mapping m.
Continuing with this analysis, let us turn to the elements hk. For each к Ε Κ we have a
mapping, ka say, of Я into Я, namely the mapping which sends an element h in Я to the
element hK. In a way then the group G is made up of two mappings:
(1) a mapping m from KxK into Я,
(2) a mapping α of if into a set of mappings of Я into Я. (The effect of ka is to map
Л to hk.)
Indeed m and a determine G up to isomorphism (see Problem 7.27). If we add enough
conditions to these mappings, one can reverse the procedure we have been outlining and
construct from Я, К and the mappings m and a an extension G of Я by K, (See A. G. Kurosh,
The Theory of Groups, Vol, II, Chelsea, 1960, translated by K. A. Hirsch, for details.) We
will not tackle the general problem but we will consider only a particular case (in
Section 7.7c).
Problems
7.26. Let G be the dihedral group of degree 3. G is an extension of a cyclic group of order 3 by a cyclic
group of order 2. After choosing a suitable left transversal, find the mappings m and a introduced
above.
Solution:
Using the notation of Section 3.4f, page 75, let Η = {аг,а2, σ3}. Then, as can be easily checked,
HOG, Since \G\ - 6, G/H is of order 2 and thus a cyclic group of order 2. Let {ι,τ} be a left
transversal for Η in G. Then
mi,i — if mi,T — T> mT,i ~ T> mr.r - <
in is the identity mapping of Η onto Я, while та sends σ! to σ2, σ2 to ctj and σ3 to as.

234
PERMUTATIONAL REPRESENTATIONS
[CHAP. 7
7.27. Let G, 0 be two groups, both extensions of Η by K. Assume that Η is actually a subgroup of both
G and G. Let X,X be transversals of Η in G,G respectively. Let m,m and a, a be the mappings
obtained above. Prove that if m = m and a = 5, then GsS.
Solution:
The elements of G are uniquely of the form arfc/i, where xkf=. X, h e H, while the elements of
б are uniquely of the form %A, where xk ε X. Let в : G -* G be defined by (arfcA)9 = xkh. Then
s is a one-to-one onto mapping. To prove в is a homomorphism, we consider the product of two
elements of G.
k'
{xkhxyh')e = (xkvmh,rh h')e - (xkk.mKk.h{k'a)h')e = $kk,:nkik-h(k'a)h>
= *kfc"»ifc((c.A(fc'5)A' = *fcA*fc.A' = (a;fcA)i?(avA')9
Thus G s G.
b. The splitting extension
Suppose G is as in Section 7.7a. Consider the particular case where mx,x· = 1 for all
x, x' € X. By examining equation (7.22) ', i.e. the product of two elements in X
is again in X. Furthermore, we have 1 € X. Let геХ, and let у € X be such that
x~lH = yH; then xy € H. Accordingly
xy = lmxa where ιητ,ν e Η
However mx,a = 1, and so xy = 1. Thus ж has an inverse in X, and so X is a subgroup of
G. Since Η <\ G, XH is a subgroup (Theorem 4.23, page 125). But every element of G
is of the form xh, χ e X, h ей. Hence ХЯ = G. Since distinct elements of -X" lie in
distinct cosets of Η in G and 1 € Я, we have
1ПЙ = {1}
Since X and Я are subgroups of G with XH = G, HnX = {1} and Η <i G, G is said to
epiii over Я. Ζ is called a complement of Я.
Note that if G splits over H, we can choose any complement Ζ of Я as a transversal for
Я in G, since two distinct elements of -X" belong to distinct cosets of Я. Now Ζ is a
subgroup of G, and it follows that if we define mg,x as in Section 7.7a with X as transversal,
then mx,x< = 1 for all x, x' in X.
If G splits over Я and Ζ is a complement of Я, then
GIK - HX/H ^ X/HnX ε* Χ
In other words, G is an extension of Я by X; that is, if G splits over Я, G/H is isomorphic
with any complement of Я.
It is convenient to introduce the following definition. We say that a group G is a
splitting extension of Hi by Χι if there exists a normal subgroup Я of G isomorphic to Яг
such that G splits over Я and G/Я = Χι.
Problems
7.28. Prove that the dihedral group D of order 8 is a splitting extension of a cyclic group of order 4 by
a cyclic group of order 2.
Solution:
We use the multiplication table for D given in Problem 7.6(a), page 220. Let Η = {Ι,^,α^,α^};
then Η is cyclic of order 4 since
of = «a, aj = a3, aj = 1

Sec. 7.7]
EXTENSIONS
235
Moreover Я is a normal subgroup of G. This can be verified either by direct calculation, checking
that if d G D and he Η then d~ ihd e H, or by noting that Η is of index 2 in D. Now letting
X = {1, a4}, a£ = 1 and so X is a subgroup of D. Since a4 g Я, it follows that the coseta Η,α^Η
are disjoint. Therefore, as there are 8 elements in Hua4Ht
С = Ηυβ,Η or D = XH
Finally ЯШ= {1}
So D is a splitting extension of Η by A' as required.
7.29. Prove that neither the dihedral group of order 8 nor the quaternion group of order 8 is a splitting
extension of a group of order 2 by a group of order 4.
Solution:
Let Ε stand for either the dihedral group of order 8 or the quaternion group of order 8. Suppose
Ε is a splitting extension of a subgroup Η of order 2 with complement X of order 4. Then Η is a
normal subgroup of G. Now suppose Η = {1, h}. If eGE, then e"1keGH. Since h ¥· 1,
e-iAe ¥■ 1. Thus e~lhe = h for all e£E. In particular if nei, then x~ifix = h. Now
2? = ZH. Since X is of order 4, Ζ is abelian (see Problem 5.19, page 140).
Now suppose e,f£E; then
e = «'A', / = x"h" (x>, x" G Z, k', h" e H)
Recall that every element of Η commutes with every element of Ζ and that Ζ is abelian; then
ef = ar'A'-a/'A" = x'x"k'h" - χ"χΊι"Κ' - d'hl'tib! - fe
Thus Ε is abelian» But neither the dihedral group of order 8 nor the quaternion group of order 8
is abelian. Hence we have a contradiction to the assumption that either of these groups is a
splitting extension of the type described.
Alternate proof: If £ is a splitting extension of Η by X of order 2 and 4 respectively, since
HnX = {1}, Η <\ Ε and Χ <\ Ε (as X is of index 2), it follows that E-XXH, the direct product
of Ζ and Η by Corollary 5.17, page 145. From this it again follows that Ε is abelian, thus producing
a contradiction.
7.30. Prove that the quaternion group of order 8 is an extension of a group of order 4 by a group of order
2, but is not a splitting extension.
Solution:
We use the multiplication table for ^f[, the quaternion group of order 8 given in Problem 7.6(b),
page 220.
First let К = др(аз). Then К is of order 4 and therefore of index 2. Thus if is a normal
subgroup of Jf[, and it follows that Jfl is an extension of К by Jfl/K. Clearly Jf[/K is cyclic of order 2,
and so JH is an extension of a group of order 4 by a group of order 2.
Now suppose Л splits over any subgroup К of order 4. Then Jf[/K is of order 2 and hence
abelian. Therefore К contains the commutator subgroup of Jj{ by Problem 4.68, page 116. In
particular, К contains a2 since _χ _χ
Now we must check that if χ is any element except 1 and α% of J{, then χ is of order 4. This can be
done directly, using the multiplication table for J}{. Suppose now, if possible, that Jf( is a splitting
extension of К by Z. Then the subgroup Ζ is of order 2, say X = {1, a;}. But as we saw above, A'
is of order 4 since χ ¥· 1, χ ¥· α2. So the subgroup X is not of order 2. This is a contradiction and
so the desired result follows.
Alternate proof: If Л is a splitting extension of a subgroup К of order 4 by a subgroup Ζ of
order 2, then KnX = {1} and К < J^. But every subgroup of the quaternion group is normal
(Problem 5.43, page 158). Therefore Ζ < J[. It follows, by Theorem 5.16', page 146, that J[ -
Κ Χ Χ. This implies Л is abelian, which is a contradiction.
7.31. Is the alternating group of degree 4 a splitting extension of a group of order 6 by a group of
order 2?
Solution:
By Problem 5.1, page 131, the alternating group of degree 4 does not contain a subgroup of order
6. Thus the result follows.

236
PERMUTATIONAL REPRESENTATIONS
[CHAP. 7
c. An analysis of splitting extensions
Suppose now that G is a splitting extension of Η by К with complement X. Then by
Section 7.7b, we may use X as a transversal for Я in G, and mXjX· = 1 for all x,x' G X.
Consequently each element g G G is uniquely expressible in the form
g = ЯкА where к G K, he Η (7.26)
and equation (7.25) becomes xkhxk'h' — xkk-hkhf (7.27)
For each к G К the mapping h -> /ifc (A G H) (7.28)
is an automorphism of H, for Η <i G and ftfc = a*' hxk imply, by Problem 3.57, page 85,
that the mapping h -> hk (h G H) is an automorphism of Я.
Finally we remark that if we let a be the mapping which assigns to each element к Ε Κ
the automorphism
ka: h -> hk (h e H)
of H, then a is itself a homomorphism of К into the group of automorphisms of H. (This
explains why we used a left transversal in Section 7.7(a), namely so that the mapping a be
a homomorphism.) We have only to prove that
(kk')a = kak'a (7.29)
To verify (7.29), let us take an arbitrary element h G Я and apply the automorphism
(fcfe')a to Л:
/i[(fcfc')a] = xkk-hxkk' = (xkXk')~lh(xkXk·) since .тьк- = ЖкЖк' by (7.27)
= (.Tk· .Tk )/i(.Tfc.Tb') = Xk'1^ hxkjXk'
- Xk'1 [h(ka)]Xk· = [h(ka)](k'aj
= h[(ka)(k'oc)] by the definition of the product of two automorphisms
Thus we have (kk')a - (ka)(k'a)
We now replace hk' by h(k'a) in (7.27). Then (7.27) becomes
Xkh-Xk-h' = Xkk·'h(k'a)h' (7.30)
What is the situation we have arrived at? We started from a group G which splits over Я.
Then we chose a subgroup X of G such that G = XH and 1пЯ= {1}. We were given an
implicit isomorphism φ of G/H with К and we denoted the element in X which corresponds
to к in К by Xk. Then we observed that the elements of G were uniquely expressible in the
form xkh with к G K, h G H. The way in which elements of G are multiplied was then
computed by making use of the existence of a homomorphism a of К into the automorphism
group of Η and by applying (7.30). Therefore we may suspect that if we are given
(a) a group H,
(b) a group K,
(c) a homomorphism a of К into the automorphism group of Я,
then we can create a splitting extension of Я by K.
Indeed this is the case. All we have to do is to reverse the process we have described.
To be precise, starting with the data (a), (b) and (c), we let G be the cartesian product of
К and H, i.e.
G = KxH = {(fc,h)\keK, h e Щ

Sec. 7.7]
EXTENSIONS
237
We define a binary operation in G according to the formula
(k,h)(k',h') = (kk',h(k'a)h') (7.31)
The reader will note the strong similarity between (7.30) and (7.31).
Before verifying that G is a group and a splitting extension of Η by K, we will make the
similarity of equations (7.30) and (7.31) even more evident. Let xic = (к, 1) and h = (1, h),
and define ka by h(ka) — (1, h(ka)). Then from (7.31) obtain
Xichxfch' = xicic'h(fc'a)h'
This equation resembles equation (7.30) even more closely than (7.31) does.
We will now prove that G is a group and that it is a splitting extension of Я by K.
(i) We note first that (7.31) defines a binary operation in G.
(ii) The binary operation defined by (7.31) is associative:
((k,h)(k>,h'))(k",k") = (kk',h(k'a)h')(k",h")
= ((kk')k", {[h(k'a)h'](k"a)}h") (7.32)
= ((kk')k", {{(h(k'a))(k"a)]\h'(k"a))}h")
Now we work out
(k,h)((k',h')(k",h")) = (k,h)(k'k",k'(k"a)h")
= (k(k'k"), [h(k'k")a)[h'(k"a)h"})
By (c) above, α is a homomorphism, and so (k'k")a = k'a · k"a. Thus
(k,K)((k',h>)(k",h")) = (k(k'k'% [h((k'a)(k"a))}[h'(k"a)hf'))
= ((kk>)k", [(h(k>a))(k"<x)}[h'(k"<x)h"})
(7.33)
The associative law immediately yields the equality of (7.32) and (7.33). Therefore
(7.31) is an associative operation.
(iii) There exists an identity in G: (1,1). Notice, of course, that the left-hand 1 in (1,1) is
the identity of К while the right-hand 1 of (1,1) is the identity of H. To check that
(1,1) is an identity, let (k, h) e G. Then
(l,l)(k,h) - (k,l(kcc)h) = (k,h)
since ka is an automorphism of H, and so maps the 1 of Η to itself. Similarly
(k,h)(l,l) = (k,h(U)l) = (k,h)
since la is the identity automorphism of H, and so leaves Η identically fixed.
(iv) Finally we must check that every element of G has an inverse. Let (к, К) е G. We
claimthat (fc-s л-чй-М)
is the inverse of (k, h). To prove this, we simply observe that
(k,h)(k-\h-l(k-la)) = (1, ^(fc-^p-1^1**)]) = (1- {hh~l){h~la))
since k~la is an automorphism of H. Thus
(k,h)(k-\h-*(k-la)) = (1,1)
Similarly (k-^h-^k-^ik.h) = (1,1)
We have thus verified that (fc_1, h~l(k~la)) is the inverse of (k, h).

238
PERMUTATIONAL REPRESENTATIONS
[CHAP. 7
(i), (ii), (iii) and (iv) above establish in every detail the proof that G is a group.
Next we verify that G is a splitting extension of Я by K. To accomplish this, put
Η = {(1, h) | k G H} and X = {(fc, 1) | fc G X}
Then it is easy to prove that Я and X are subgroups of G and that
k -> (1, Л) and fc -> (fc, 1)
are isomorphisms of Я onto Я and X onto X respectively.
Now to prove that Я is a normal subgroup of G, observe again that if g G G, then
g = (fc,fc) = (М)(1,Л)
If (1, h') G Я, then
g-l{Xh')g = (й,Л)-Ч1,Л')(й,Л)
= ((fcfl)(l,fe))-l(l,fe')((fc.l)(l.*))
= [(1,Л)-чм)-Ч(1,лт1)(1.л))
Since (fc, l)"1 = (k-\ 1), we have (k, l)"1^, A')(u. 1) = (1, Л'(М). Thus
flr-i(i, лов' = (i. Λ_1)(ΐ, Λ'(Μ)(ΐ, Λ) е я
since the product of elements of Я belongs to Я. Hence Я is normal in G as claimed.
Clearly ЯпХ={(1,1)} and
G = KH
as we saw earlier, since (k,h) = (k,l)(l,h). Consequently we have constructed from the
data (a), (b) and (c) a splitting extension G of Я with complement X. Therefore G is a
splitting extension of Η by K.
The group G that we have constructed is called the splitting extension of Η by К via a.
We emphasize the importance of the above discussion and the related problems which
follow.
Problems
7.32. Construct a non-abelian group of order 6 as a splitting extension of a group of order 3 by a group
of order 2.
Solution:
Recall that if we are given (a) a group H, (ft) a group К, (с) а homomorphisrn a of Ζ into the
automorphism group of H, then we can construct a group from this data as follows. Consider the
set G of all the pairs (fe, h) (k G K, h G H) and define a binary operation in G by
(k, h)(k', h') = (k,k', h(k'a)h')
Then G becomes a group which is a splitting extension of Я by K. So in the case at hand, we have
the group Η (the cyclic group of order 3) and the group К (the cyclic group of order 2). We need
the homomorphisrn a. This means in the first place that we need to know more about the
automorphism group of H, the cyclic group of order 3. Now if Η is generated by h, then
Я = {l,fc,fc*}
The mapping which sends each element of an abelian group into its inverse is an automorphism
(check this). So if η : Я -> Я is this automorphism,
η: 1-> 1, fc-> fc-1 = fc2, h.2^h-2 = h

Sec. 7.7]
EXTENSIONS
239
Now ч2 = h and so др(ц) is cyclic of order 2. Thus the groups К and gpW) are isomorphic, and
accordingly we can take a to be the isomorphism of К onto gp(v). Let G be the splitting extension
of Я by К via a. To see how some of the elements of G combine, let К = {1, к}. Then
(к, A) (M) = (Λ«,Λ(*β)1) = (I.**)
Now (k,l)(k,A) = (к*,1(кв)Л) = (1, A), and so (k,A)(k, 1) Φ (к, l)(k, A). Thus G is non-abelian and
is an extension of Я by K.
7.33. Are there non-abelian groups of order 2 X 777?
Solution:
There are non-abelian groups of order 2 X 777, To produce one such group, let Я be cyclic of
order 777. Then Я has an automorphism τ of order 2, namely the mapping τ which sends every
element of Я to its inverse. So there is a homomorphism a of K, the cyclic group on k, of order 2,
into the automorphism group of Я, namely the one which sends A; to т. Let G be the splitting
extension of Η Ъу К via a. Then G is the required group.
7.34. Construct two non-isomorphic non-abelian groups of order 168.
Solution:
Let Ну = gp(hi) be cyclic of order 84 and Kx = gp(ky) cyclic of order 2. Let a be the
homomorphism of Kj into the automorphism group of Нг defined by kja: Aj -» Aj~f. Then Glt the
splitting extension of Я, by Ku is of order 168. The center Ζλ of Gx is of order 2, consisting of (1,1)
and (1, A*2) which can be easily checked by direct calculation.
Now we construct a second group of order 168. Here we take Яг = gp(h^} to be of order 42
and K3 = gp(k^) to be cyclic of order 4.
As usual, Яг has an automorphism of order 2, i.e. the mapping defined by
τ: Α2-+Α.Γ1, г = 0,1, ...,42
Let β be the homomorphism of K2 onto др(т) defined by
ύβ = τ*, j =0,1, 2,3
2
Thus кг/? = t. Let Gz be the splitting extension of Я2 by K2 via β. Then, as the reader may check
by direct calculation, the center Z2 of G2 consists of (1,1), (1, Α|'), (fc|, 1), (kg, A|') and is therefore
of order 4.
Now both (?! and G2 are non-abelian since [Z1| = 2, ]Z2| ~ 4- Moreover if ΰλ and Gz are
isomorphic groups, they have isomorphic centers. Therefore Gi and Gs are not isomorphic.
7.35. Construct all possible groups of order 30 which are extensions of a cyclic group of order 10 by a
cyclic group of order 3.
Solution:
The solution of this problem requires knowledge of the automorphism group of the cyclic group
Я = ffp(A) of order 10. If τ is an automorphism of Я, then hT is of order 10. The possibilities for
At are therefore , .. , , „ ,
At = A8, At = k7, At = ft9
Let tj be the automorphisms defined by the possibilities listed above (г = 1, 2,3). Now т\ = 1, т\ — 1,
τ\ ■— 1; thus none of the automorphisms of Я is of order 3- So the only possible homomorphism a
of a cyclic group К of order 3 into the automorphism group of Η is the one which sends every
element of К onto the identity automorphism. The resultant splitting extension of Я by К is then
abelian (indeed it is isomorphic to the cyclic group of order 30).
7.36. Construct a non-abelian group of order 222 by using splitting extensions.
Solution:
Form a splitting extension of a cyclic group Η of order 111 by a group К = gp(k) of order 2
via the homomorphism taking k to the automorphism which sends every element of Я into its inverse.

240
PERMUTATIONAL REPRESENTATIONS
[CHAP. 7
7.37, Construct a non-abelian group of order ря for each prime ρ by using splitting extensions. (Hard,)
Solution:
Let Η be the direct product of a cyclic group f?p(a) of order ρ with a cyclic group ffp(b) of order
p. Each element of Η is uniquely of the form
arbs with 0 ~ r < p, 0 ~ s < ρ
Let τ: arbs -> arbs + r. We will show that τ is an automorphism of Η of order p. First let m, and η
be any integers. We claim that
(ambn)r = ambm + n
Let m = m' + qp, η = η' + sp where 0 — m' < p, 0 ~ n' < p. Then
(amb*)T = am'bm' + n' ~ ambm'bn' = a™bmbn ■= ambm + "
To verify that τ is a homomorphism, observe that
(ami&ni · am2&"2)r = ami + m2&mi + m2 + ni + 7l2
= α д b L J- ■ a ^ 6 *■ ^ -= (a 1b 1 )r(a 2 & z )r
Clearly Kerr = {1}, and so τ is one-to-one. It is easy to check that τ is also onto. Thus T is an
automorphism. Note that arP = abp = a, so that τν acts as the identity on α and b, which form a
set of generators of H. Hence τ is of order p.
Now let К ~ ffp(k) be of order p. The mapping a : kl -> τ* is an isomorphism. Then we form
the splitting extension of Η by К via a. This gives a group G of order p3. G is non-abelian since
(1, a)(k, 1) = (/ΐ, α&), but (к, 1)(1, α) = (к, а).
ά. Direct product
Consider the special case of a splitting extension G of a group Я by a group К via
a homomorphism α in which « takes if onto the identity group of automorphisms of
H. In this case G consists of the pairs (h, k) (h G H, к G if) with binary operation given
by {h,k){h',k') — (hh',kk'). So (Ϊ is, in the terminology of Section 5.3a, page 143, simply
the external direct product of Η and K. The obvious usefulness of this construction is that
we do not require any knowledge of the automorphism group of Η to construct the direct
product. Notice that if Η = {(l,h) | h G H) _and К = {(fc,l) | к G К}, then G is the
internal direct product of its subgroups Я and K, again in the sense of Chapter 5. We will
not pursue this concept of direct product here any further.
7.8 THE TRANSFER
a. Definition
Suppose G is a group with an abelian subgroup A of finite index. The transfer is a
special homomorphism of G into A. The use of such transfer homomorphisms has been
important in the theory of finite groups. Here we will examine one application of the
transfer and briefly mention another (at the end of Section 7.8d).
To define the transfer τ of G into A, choose a right transversal X of G in A. We repeat
that A is an abelian subgroup of G of finite index n, say. Therefore \X\ = n. Recall that
if jGG, then g is the element of X in the coset Ag. Let Xi, x%, .. ., xn be the elements of
X; then if ffGG, we define a mapping τ of G into A by
дт = xig{xlg)-1 · хгдСШд)"1 Xngix^g)^1
It is clear that gr G A since, as ж,^ and Щ belong to the same coset of A,
%\д{Щ1У1 G A (i - 1,..., n)
This mapping τ is the homomorphism of G into A mentioned above; it is called the transfer
of G into A. There are two items to be verified: (1) τ is a homomorphism and (2) τ is
independent of the choice of the transversal X.

Sec 7.8]
ТНБ TRANSFER
241
b. Proof that τ is a homomorphism
We compute {gh)r, where g,h EG:
(gh)r = (ХгдЦа^дЪ)-*) · {хгдЦх^дЪ)-1) (хпдЦхпдЩ~1)
= {xigh{x~[gh)~i) · (хгдЦЩдЪ)'*) {х*дЦЩ}К)~1)
(since Xigh = xlgh by {7.3), page 219)
= (xigix^g)-1 - х^дЦШдЪу1) · (хгд{й&)~* ' ШдЦхмЩ'*)
' (Xng(xHg)~l ■ х^дЦх^дк)-*)
Now every one of the elements х*д(х1д)~1, хздЦЩдК)'1 lies in A; since A is abelian, they must
commute. So we can rewrite {gh)r in the form
{gh)T = [xigCxlg)"1' Хг9{х1а)~х x*gi&3)~1] ,„ _n
■ [xigh(xigh)~i · x-2.gh{x2gh) ' Xngh(xngh)-1]
But observe that ж; -> ж^ (г = 1,..., η)
is a permutation of X (Section 7.4a). This means that
х1дЦх^дЪ)~* · хЧдЦхЧдЬ)~* хТдЦх^дку1
simply consists of the η elements xih(x~ih)'~i (i — l,...,n) multiplied together in some order.
Since A is abelian, the order of such a product is immaterial. Thus
ШдЩШдЪ)-1 · х^дЩЩдЪ)-1 x~^gh{xngh)~* = xihixji)'1 · χ<Μχ%}ι)~χ - · ■ - хпЬ,(хпЩ~1
= hr, by definition
Then it follows from equation {7.3b) that
(gh)r = {gr){hr)
for all g, h in G. Thus τ is a homomorphism.
c. Proof that τ is independent of the choice of transversal
We now prove that τ is independent of the choice of the transversal X.
The proof depends on an analysis of the product
gr - xig(xlg)~l · х2д{Щд)~1 xng{x^g)~l
where again {xhхг, .. .,xn) - X is a right transversal of A in G. We recall from Section
7.4a that the mapping xt -*■ xlg is a permutation of X. Now every permutation of a finite
set can be written as a product of disjoint cycles (Theorem 5.26, page 167). So, after
relabeling the elements of X if necessary, we can assume that
"xlg - хг, xlg = ж3, ..., Xk-ig = Xk, xTg = x\
XkTTg — Xk + 2r Xk + Ig — Xk + S, ..., Xk+l-J = Жк + l, Xk + ig = Xk + 1
Xn—m + ig — Xn— 771 + 2» Xn—m + zg — Хц—m + S) . . ., Xn—ig — Xn, Xn.g — Xn— 771 + 1
(Note that k +1 + · ■ ■ + m = n, where η is the index of A in G.) Then
gr = {х\д{хЧд)~г · Хгд{х1д)~1 азд(ж^Г')
• {Xk+ig{XkTVg)~i · Хк+гд(Хк+2д)~* Хк+1д(хТТГд)*)
' {xn~m+ig{Xn-m+ig)^1' Хп~т+2д{хп-т+2д)~* х^дхпд)"1)
= {XigXz1 · ХгдХъ1 Ж^Ж"1)
• {xk+igxkiz · Жк+г^Жк+э Xk+igXk') · · · ·
• (Xn-m + igXn-m+ΐ ' Хп-т + гдХп-т + В ■ ■ · · · Жп<7Жтг-771+l)

242
PEEMUTATIONAL KEPKESENTATIONS
[CHAP. 7
and thus дт = (x^gkx^) * (xk+iglXkU) («„-„,+ι^^+ι) (7.^5)
Note that xtgkx^1 = (xig(xig)~l) · (хгд(х2д)~1) · · ■ ■ · (xkg(Xk~g)~l) G A, since it is the
product of factors х<д(Щ/)~~1 which belong to A. Similarly, Жк + l g Xk + l> · . · , Xn—m + lX Xn — m+1 fc -A.
We are now able to prove
Lemma 7.7: Let A be an abelian subgroup of finite index in a group G and let X —
{xi,X2, ■. .,Χη}, Υ = [yi,?/2, .. .,Уп] be two left transversals of A in G.
Furthermore let τ and ~ be mappings of G into A defined respectively by
дт - x\g(xlg)~l · xzg{x^g)~l Xng(x^g)~l (where flr G G)
and g~ = Vig(ylgyi · учд(У2д)~1 Упд(у1д)~1 (where g G G)
where, if h G G, h denotes the element of X in the coset Ah and h denotes the
element of Υ in the coset Ah. Then
τ — τ
Proof: We may assume on suitably reordering Υ that
Vi = cwi (cti G A)
Now if xig — Xj, then Ay; = AdjXj = Axj = Axtg = Асцхлд - Ay&. This means that
ШГ = Vi
It follows therefore as in equation (7.35) that if g G G, then
ff~ = (У19кУГ1) ' (Vk + ifftjkli) (j/n-m + ifl^l/n-m+i) (7.S6)
But ?/{ = OtiCi, and we know that the elements xigkx^1, Xk+iglXk+u ..., Xn-m + igmXn~m+i lie
in the abelian subgroup A. Therefore
yk+iglykli — a,]t+i(Xk+iglXkli)aZ+i = Xk+iglx~k+\
Уп~т + 1ХтУп-т + \ — a,n-m + \(Xn-m + \gmXn-m + \)a>n-m+\ = Xn-m+igmXn~m+l
Thus it follows from (7.35), (7.36) and the above remarks that дт = д~.
This lemma establishes that the transfer homomorphism τ is independent of the choice
of transversal. Accordingly we may speak of the transfer of G into A.
d. A theorem of Schur
Using the transfer, we now prove the following important theorem of I. Schur.
Theorem 7-8: Let G be a group whose center A is of finite index. Then the derived group
G' of G is finite.
Proof: Suppose \G/A\ = n. Let X — {xu ..., xn} be a transversal of A in G and let
τ be the transfer of G into A. Now if g G G, then by equation (7.35)
дт = xtgkx~l · Xk+iglXk+t Xn~m+\gmXn-mi-i
But X\gkx~l G A and A is the center of G. So
xigkxTl = х^1(х\дкх^)х^ = дк
Similarly, Xk+iglXkli = g\ ..., Xn~m+igmx~ni-m+i = g™

Sec. 7.8]
THE TRANSFER
243
Thus, for each g&G, gT = gn.
Notice that the image of G under τ is a subgroup of A and is therefore abelian. Then
by Theorem 4.18, page 117, G/K is abelian, where К is the kernel of т. But by Problem 4.68,
page 116, this implies G' CK.
Now as G/A is finite, so is G'A/A. This means that G'/G'DA is also finite since by
Theorem 4.23, page 125, ^^ ш ^,^ {7Щ
Since G'/G'nA is finite, G' itself is finite if G'f\A is finite. Since g-r = gn for every g € G,
every element of the kernel has finite order. It follows that the elements of G'DA have
finite order. We will show that G'C\A is finitely generated. Assuming this true for the
moment, it follows that G'nA is finite (Problem 6.44, page 198). Thus G' is also finite.
This completes the proof of Schur's theorem but for the verification that G'nA is finitely
generated.
We accomplish this by showing first that G' is finitely generated. If g,h & G, then
g — axi and h — bx}, where a, b G A, xu xf G X. Then since A is the center of G, we have
g~lh~lgh = xl'ia~lx^lb~1axibX} = x^1x^lxtXja~lab~1b = x^lxflx&]
This means that there are at most n2 distinct commutators in G. Therefore G' is finitely
generated since it is generated by commutators.
Finally G'C\A is finitely generated since it is of finite index in a finitely generated group
(Theorem 7.4, page 227). This completes the proof of Schur's theorem.
It is worth noting one fact that emerged in this proof: the transfer into the center is
simply the mapping that takes each element g to g11 where η is the index of the center in G.
We end our discussion of the transfer by mentioning that if all the Sylow subgroups of
a finite group G are cyclic, then G is metacyclic (i.e. an extension of a cyclic group by a cyclic
group). This theorem can be proved by using the transfer. The proof is not too complicated;
however, it is lengthy and will not be given here. Reference to a proof may be found in
Section 5.2a, page 139.
A look back at Chapter 7
We re-proved Cayley's theorem, namely that every group is isomorphic to a group of
permutations. The ideas that arose from Cayley's theorem were generalized. In particular
we called a homomorphism of a group G into the symmetric group on a set Ζ a permutational
representation of G (on X). Permutational representations were roughly classified. We
explained an important permutational representation of a group G called a coset
representation. This representation allowed us to provide a variation of Cayley's theorem due to
Frobenius. Then we used Frobenius' theorem and the coset representation to prove three
theorems; (1) a subgroup of finite index in a finitely generated group is finitely generated;
(2) the number of subgroups of fixed finite index in a finitely generated group is finite; (3) if
G is a finitely generated group whose subgroups of finite index have only the identity
subgroup in common, then every homomorphism of G onto itself is also one-to-one, i.e. an
automorphism.
We called a group G an extension of a group Я by К if there is a normal subgroup Я
of G such that G/H ^ К and Η ^ Η. An analysis of this situation was made simpler
because of our discussion of both coset representations and Frobenius' theorem. This
analysis of extensions was specialized to splitting extensions, where we provided a method
of constructing a splitting extension of two groups Я and K.
Finally we defined a special kind of homomorphism of a group into an abelian
subgroup, called the transfer. We then used the transfer to prove that the derived group
of a group whose center is of finite index is finite.

244
PERMUTATIONAL REPRESENTATIONS
[CHAP. 7
Supplementary Problems
PERMUTATION REPRESENTATIONS, COSET REPRESENTATIONS, FROBENIUS THEOREM
7.38. Let Dn be the dihedral group of order 2w and let Cn be its cyclic normal subgroup of order n. Find
explicitly
(1) the representation of Dn onto itself given by Cayley's theorem,
(2) a coset representation of D„ using Cn as the subgroup,
(3) the representation provided by the Frobenius theorem.
7.39. Give a permutation representation of A n of degree 2re.
7.40. Find a faithful representation of G Χ Η on η + m letters if GcSn, He Sm.
EXTENSIONS
7.41. Construct a non-abelian group which is an extension of an infinite cyclic group by a group of order 2.
7.42. Prove that if G is a non-abelian group with an infinite cyclic normal subgroup of index 2, then G
is a splitting extension of an infinite cyclic group by a group of order 2.
7.43. Prove that there are precisely three non-isomorphic extensions of an infinite cyclic group by a group
of order 2.
7.44. Prove that an extension of a cyclic group of even order by a group of order 3 splits.
7.45. Construct a non-abelian group of order 36.
7.46. Construct five non-isomorphic groups of order 5s X 3.
7.47. Let D be the set of infinite sequences of integers, i.e. D consists of the sequences a= ...,
a.i.ag, ab ... (сц € Ζ). If b = ..., &_2, b0, bv . .., define a+ b = ..., a_2 4- b-i, <% + b0, a.j + bu ... .
D is an abelian group under the operation of addition of sequences. For each integer η define a
mapping «„ of D by putting aan = ..., b~ i, b6, blt . .., where b{ = сЦ-„. Prove that
(1) an is an automorphism of D, (2) amOn — am+n,
(3) A = {щ\ г £ Z} is an infinite cyclic group generated by at.
7.48. Let W be the splitting extension of D of Problem 7.47 by the infinite cyclic group С = др{с) via the
mapping that sends e to α2. Prove that W is a non-abelian group. Find a proper subgroup of W
which is isomorphic to W· Prove that W/W is infinite cyclic. (Hard.)
TRANSFER. MAL'CEV'S THEOREM, MARSHALL HALL'S THEOREM.
7.49. Prove that if A is an abelian subgroup of finite index in a simple group G, then the transfer of G
into A sends G into 1.
7.50. Prove that if G is a finite group whose center Ζ has order co-prime to its index, then the transfer
of G into Ζ is onto. (Hard.)
7.51. Prove that an infinite group with a subgroup of finite index is not simple.
7.52. A group is said to be residually finite if the intersection of all its normal subgroups of finite index
is the identity. Prove that an extension of a residually finite group by a finite group is residually
finite. (Hint: The preceding problem gives a clue.)
7.53. Let G be a cyclic extension of a cyclic group N of order re by a cyclic group of order m. Let
N = ffp(a) and G/N = gp(bN). Prove that Ь~гаЬ = aj, where j* is co-prime to re and bm = aM
(where j and к are integers). Prove that jk = к modulo n.
7.54. Prove that a cyclic extension of a finitely generated residually finite group (defined in Problem 7.52)
is residually finite. (Hint: Use Marshall Hall's theorem. Then use the fact that the automorphism
group of a finite group is finite.) (Hard.)
7.55. Prove that if G is a group and Glr G2» ■ · · are subgroups of G such that Ог ^ G2, GjCG^, G2 =£ G3,
GjCGji, ..., then uGj is a subgroup of G which is not finitely generated.
7.56. Let G be a residually finite group (defined in Problem 7.52) and suppose every subgroup of G is
finitely generated. Prove that if Я is a subgroup of G such that H/N г G for some normal
subgroup N of G, then N = {1}. (Hint: Use Problem 7.55.)
7.57. G is a finitely generated group, every element of which has only a finite number of conjugates.
Prove that \G'\ < ». f Hint: П C(gt) = Z(G) if git...tgn are the generators of G. j
7.58. G is a group in which every element has only a finite number of conjugates. Prove that every
element of G' is of finite order. (Hint: Use Problem 7.57.)

Chapter 8
Free Groups and Presentations
Preview of Chapter 8
We begin with a property of the infinite cyclic group and generalize this property to
define free groups. We ask questions similar to those we asked in Chapter 4 concerning
cyclic groups:
(1) Do free groups exist ?
(2) When are two free groups isomorphic?
(3) What are the homomorphisms of free groups ?
(4) What are the subgroups of free groups ?
In answering (3) we will learn that every group is a homomorphic image of a free
group. This provides a new way of describing a group, i.e. as a factor group of a free
group. Such a description of a group is called a presentation.
8.1 ELEMENTARY NOTIONS
a. Definition of a free group
Recall that if G is a group and X (^ 0) a subset of G,
gp(X) = Κ'-..ίμ,εΐ|£.= ±ΐ}
If xl1 ■ ■ · x^ and y[l · ■ ■ y%" are two products with xl,yi G X and et = ±1, η{ = ±1, then
they are said to be identical if η = m, xi = yt and e, = щ for i = 1, ..., m. Two products
are said to be different if they are not identical.
It is easy to see that two different products of the form x\l ■ ■ ■ xcnn can give rise to the
same element of G. For example, if X = {x,y}, then xy and xx"1xy are different products
of the form ж*1 ■ ■ · xe* but they give rise to the same element of G, i.e. xy. To avoid
redundancy, we introduce the concept of a reduced product:
A product where e. = ±1 and x. £ X, is said to be a reduced X-product
if x. = xi+l implies е; ν4 -ξ,+1.
Synonyms for reduced X-product are reduced product (X being understood) and reduced
product in X.
(Examples of reduced products are easily given. Let X- {x,y}; then xy, x~iyxyx~1
and x~^yyxy~x are reduced products. However, yxyxx~1 and x~1yxyy~1 are not reduced
products.)
Lemma 8.1: да(1) ={w|w = lorii! = a reduced product in X).
Proof: Let {w \ w = 1 or w = a reduced product in X) = R. Clearly R С др{Х). If
и G gp(X), then u= x\* · ■ · a£k where ieX and e. = ±1.
We proceed to show that «ей. If к - l, и is a reduced product in X, and so и G Й.
Assume then that any product х[г ■ ■ ■ a^fc G Д for k^n~ 1. Suppose fe = n. If
μ = ж*1 · ■ · x£fc is a reduced product in X, then ад G i?. If ад is not a reduced product, there
245

246
FREE GROUPS AND PRESENTATIONS
[CHAP. 8
exists an integer i such that x.-x.., and e. = -e.... Suppose η > 2; then we can delete
xixi+il t° obtain и as a product involving n — 2 elements of X. By the inductive hypothesis,
и then belongs to R. If n — 2, then и = ж'1^2, where ж1 = x2 and fj = — «^, from which
u=lGR. Thus flrp(X) с β and accordingly gp(X) = R.
Consider now the infinite cyclic group generated by the element x. The reduced products
in {x} are of two kinds:
χ · · · χ — xT or ж_1ж-1 ■ · · ж-1 — x~T
where r is a positive integer. From what has been said about the infinite cyclic group,
we know that if m and η are integers, xm — xn implies that m = n. Thus different reduced
^-products give rise to different elements. (This is by no means the usual situation. For
example in a cyclic group of order 2 generated by y, we have yyyy - yy.) G is said to be
freely generated by {ж}. More generally we have the following definition:
A group G is said to be freely generated by the set X Q G if X ¥· 0, the empty
set, and
(i) gp(X) = G;
(ii) two different reduced X-products define two different nonunit elements of G.
Notice that it follows from (ii) that if χ G X, ж-1 Й I For if not, there exist χ and
y, both elements of X, with y~l = x. But then χ and y~l are two different reduced
X-products which are equal. It also follows from (ii) that ΙίΧ,
A set X of generators of G satisfying (ii) is often called a free set of generators of G.
A group G is free if it is the identity group or if it possesses a free set of generators. G is
also said to be free on X. If G is a free group freely generated by X, then the study of G is
facilitated by the fact that we know exactly whether two X-products are equal. All we need
to do is to express each of the products in reduced product form. If the reduced products
are distinct the elements are not equal. This process of expressing an element as a reduced
X-product can be carried out in a finite number steps. To illustrate, let F be freely
generated by {ж,у}. Are the products / = xyx~ly~lxxx~1y~lyz and g = xyy~xyz equal? We
convert / to a reduced product by deleting inverse pairs (i.e. two adjacent inverse factors).
Thus
/ = xyx ly lx{xx 1)(у~1у)у2 = хух~~гу 1xyi = xyx xy lxyy
Similarly g = xyy. Hence as / and g are equal to different reduced products, they are not
equal. The fact that we can determine in a finite number of steps whether or not two
elements are equal is often expressed by saying that the word problem is decidable for
free groups. (The interested reader may find more details in, e.g., J. J. Rotman, The
Theory of Groups, Allyn and Bacon, 1965.)
Problems
8.1. Let G be freely generated by the set X — {x, y, z]. (a) Write down three distinct elements of G.
(b) Is xyz(yz)~l a reduced product? (c) Is xyy~1z~1yx equal to xz~1yzz~1x'i (d) Express
х2уа(уах2)~1у~я and (xzy)~1xsyz as reduced products.
Solution:
(a) (1) x, y, 2, (2) x, x2, ж3, (3) x, xy, xz
In fact any three different reduced products in {x, y, z) are distinct elements of G.
(b) No, because it is not written in the required a;,1 ■ ■ ■ жпп form, since it involves (г/г)-1. However,
even if we replace (yz)~l by its equivalent z~ly~l to get xyz2~ly-1, this, though in the form
x ' · ■ · χ * is not a reduced product because of the inverse pair zz~l.

Sec. 8.1]
ELEMENTARY NOTIONS
247
(c) Yes, for on expressing each as a reduced product we get xz lyx.
(d) x2ys(ysx2)~~1y^s = х2уъх~~'1у^ъу~ъ = xxyyyx~lx~~ly~ly~ly~~ly~~1y~~1y~1
(xzy)~~lxzy2 = y~~1z~~1x~^xzy2 — y^1z~~1zy2 = y~ly2 - у
8.2. Prove that if G is a free group, G ¥= {1}, then the infinite cyclic group is a subgroup of G.
Solution:
Suppose G is freely generated by X. If χ £ X, consider gp({x}). This is a cyclic group. Now
χ · χ · · · ■ · χ — xr, with r a positive integer, is a reduced product in X, Hence xr = xs, where r and
s are positive integers implies r ~ s. Thus £rp({#}) is infinite and is infinite cyclic and the result
follows.
8.3. Prove that if G is freely generated by X, where X contains at least two elements, then G is not
abelian.
Solution:
There exist two distinct elements x, у of X. Now xy and yx are two different reduced products,
во xy ¥= yx. But this implies that G is not abelian.
8.4. Prove that a finite group G is not free \i G ¥= {1}.
Solution:
In Problem 8.2 we proved that except for the identity, every free group has as a subgroup the
infinite cyclic group. Consequently if G is free it must have the infinite cyclic group as a subgroup.
This is absurd.
8.5. The direct product of two infinite cyclic groups is not a free group.
Solution:
The direct product of two infinite cyclic groups is abelian. But we have proved in Problem 8.3
that a free group is not abelian if it is freely generated by a set X which contains at least two
elements. Hence if the direct product of two infinite cyclic groups is free, it is freely generated by
some set X — {x}, i.e. it is infinite cyclic. But a free abelian group of rank two is not cyclic. This
follows immediately either from the uniqueness of the type of an abelian group (Theorem 6.21,
page 197) or directly.
8.6. Prove that a free group freely generated by X with \X\ — 2 has no center (i.e. its center consists
of the identity element). (Hard.)
Solution:
Suppose G has an element ζ ¥* 1 in its center. Let ζ be expressed as a reduced product
ζ = x*1 · ■ · x*. Let у £ X, у ¥* хг. Consider
yz ~ yxxl '·■«„"
This is a reduced X-product. On the other hand
zy ~ «j1 ■ ■ ■ xfy
If я™*£ y~l, then x*1 ■ · ■ x™y is a reduced product and so clearly zy ¥= yz as yz begins as a
reduced product with у but zy begins with x^ ¥* y. If жи™ = y~l then for η > 1 (for otherwise
X\ — У contrary to the choice of y),
zy - x? ■ ■ · afc",1
and again it is clear that zy and yz are two different reduced products, so zy Φ yz. But as ζ is in
the center, we must have zy ~ yz. Therefore this contradicts the assumption that G has a center.
8.7. Generalize Problem 8.2 by showing that if F is freely generated by {xlt . . .,xn} with » > 1, then
{xu ..., icj freely generates Ft ~ gp{x\, ■ ■., Χι) for i = 1,2, ..., n.
Solution:
Put Χ( — {xu ..., a;;}. By definition, Χϊ generates Fv We have only to prove that two different
reduced -Χ,-products define different elements of F{. But a reduced Xrproduct is obviously also a
reduced X-product and two different reduced X-products define different elements of F. Hence the
result follows immediately.

248
FREE GROUPS AND PRESENTATIONS
[CHAP. 8
b. Length of an element. Alternative description of a free group
Suppose F is freely generated by a set X. If / G F, f Φ 1, it can be expressed in one
and only one way as a reduced product х[г ■ · · a£". We define the length of f with respect
to this free set of generators X to be n. The length of the identity is defined conventionally
to be 0. For example, if F is freely generated by X = {x,y}, then the length of х2у2х~г
is 5, because , „ _, _1
х2угх x = xxyyx '
A very useful technique in arguments involving free groups is to prove results by
induction on the length of elements (see, for example, Lemma 8.9, page 261).
Lemma 8.2: A group F is freely generated by a set Χ Φ 0, the empty set, if and only if
(a) gp{X) — F, and (b) no reduced Z-product is equal to the identity element.
Proof: Assume first (a) and (b) above. To show F is freely generated by X we must
show that two different reduced products in X are not equal and are not the identity. Let
x\l · ■ ■ o£n and ι/?1 · · · y1^1 (where e< = ±1, щ = ±1, and xi,yi € X) be two different reduced
products. Suppose they are equal. We can assume without loss of generality that a£" Ф y1^1;
for if a£" = #jjj», х[г · · · x^Si and y\l · ■ ■ y^S^ are two different reduced products. Since
xei ... χ'η — y4\... y*m an(j χ*η = ущп jt follows that
1 η " 1 " m η " m '
< · ■ ■ <"_V = у7 ■ ■ ■ ylr:
If again ж^1 = У1^1, we can delete them. But we cannot continue indefinitely this way
as x\l · · · x^1 and y*!1 · · · y*1™ are assumed to be different reduced products. So we may
assume хеппФ y1^. It follows then that x[l · · · aS/-"'" ■ · · y~Vl is a reduced product equal to
1. But by (b) it is not 1. Hence we have a contradiction, and so it follows that F is freely
generated by X.
If on the other hand F is given freely generated by X, do (a) and (b) hold? Of course,
since this is entailed by the definition of a group freely generated by a set X.
Problem
8.8. Let F be freely generated by X — {x, y, z). Determine the length of (i) 1, (ii) xzyz^1, and
(iii) / = x~1yxx~iy2xaz~1.
Solution:
(i) The length of 1 is taken to be 0. (This is just the convention mentioned earlier.)
(ii) The length of xzyz"1 is 4.
(iii) The length of / is calculated by expressing it first in reduced form, that is,
/ = x~1yx~1yyxxxz~1
Hence the length of / is 9.
c. Existence of free groups .
An obvious question is; Do free groups exist? Up to now we have tacitly assumed that
they do.
Theorem 8.3: Let η be any positive integer. There exists a free group freely generated
by a set of η elements.
Proof: Since every group is isomorphic to a subgroup of the symmetric group of some
set, it is natural to look for a suitable subgroup of some symmetric group in order to find
a free group of rank n.

Sec. 8.1]
EXISTENCE OF FREE GROUPS
249
Our plan is as follows: we shall introduce a set Τ consisting of certain ordered 1-tuples
of integers, ordered 2-tuples of integers, and so on. We then choose a set X = {ft, .. .,θη}
of permutations of Τ such that X freely generates gp{X).
The elements of Τ are the ordered m-tuples (n, r2, .. ., rm) with η — 0 and
nonzero integers with η + η+ι¥>0 for i - 1, .. .,m- 1. Thus (0,1,2, -3) e Τ but
(1,2) € Τ and (0,1,2,-2) g T.
We define now for each integer i — ±1, ..., ±n a permutation ft of Τ as follows. If
(n, ..., rm) Ε Τ we define
(i) (гъ .. ., rm)0i - (n, ...,rm)i) if rm ¥> -i.
(ii) (n, .. .,rm)ft = (n, .. .,rm~i) if rm = -*.
It is clear that 0» is a well-defined mapping of Τ into T. Moreover ftft-i = t = e~i9i where
с is the identity permutation of Τ (Problem 8.9). Thus each 9t is a permutation of Τ
(Theorem 2.4, page 36), i.e. ft e ST.
Let G = gp{X), where X = {0i, .. .,ft,}. (As ST is a group and IcST, it makes
sense to talk of gp(X).) We prove that X freely generates G, thereby completing the proof
of the theorem.
We have only to verify that every reduced X-product is not t. Let then
f ~ ί "' " θ™· (where 1', .. ,,m' G {1, . . ,,n) and e. = ±1)
be any reduced X-product (so if r' — (r +1)', er+er + 1 v·4 0). We compute the effect of /
on (0). Now ftf1 = θ-i, and so flf1 = <?±ί. Then
By the definition of the product of permutations, we have
(0)/ = ((0)V)(V ■■■%mm') = ((0,eil')V)(V ■■•^rn'»')
Now if l' = 2', then fl + e2^0 or eil'^-e82'. Therefore
(0,fll')V = (0,eil',e82')
It follows in this way that
(0)/ - (0,fll',e22', ...,emm') ¥· (0)
Thus / τ·4 ι since it does not leave (0) fixed. This completes the proof of the theorem.
Note: It is possible to prove similarly that there exist free groups freely generated by
sets of arbitrary cardinality. Since we have not introduced cardinal numbers, we cannot
prove this more general theorem here. The reader who has a knowledge of cardinal
numbers may read the account in J. J. Eotman's The Theory of Groups, Allyn and Bacon, 1965.
Problems
8.9. Prove that θφ-^ = ι and S-jSj = ι, where θι and t are as defined above.
Solution:
Let (r1? . . ., rm) £ Т. Suppose first that rm Φ —i; then
{rl,...,rm)ete-i = (r1( .. .,rm,i)9_t -- (r1( ...,rm)
If rm = -i, (rlt ...,rm)$ie-i = (r1; ...,rm~i)e~i
Now if rm_! = -(—i) = i, then (rx, ...,%) is of the form (r1( . . .,i, -i). Since such an element
does not belong to T, rm_, ** —(—■i). Therefore

250
FREE GROUPS AND PRESENTATIONS
[CHAP. 8
(П>- ■■,»m-i)0-t = {rv . . .,rm_b—£) = (r, g
and so (rj, ...,гт)0{0^ = (rlf ...,rm)
Thus 949-j leaves all elements of Τ unchanged and hence θβ-ι — ι. Similarly θ~t9i = и
8.10. Prove that there exists a free group freely generated by a set Υ such that there is a one-to-one
mapping of Υ onto the positive integers.
Solution:
We proceed exactly as in the proof of Theorem 8.3 except that we define 0; for all nonzero
integers ΐ and put Υ = {θ^,θ^, ...}. The mapping p: Y^>Z + , the positive integers, defined by
(#{)/> — г is a one-to-one onto mapping, for as none of the ei have the same effect on (0), they must
be distinct. If Η — gp{Y), then Η is freely generated by Y.
d. Homomorphisms of free groups
Now it is a fact (see Problem 8.11 below) that if a group G is generated by a set X,
then a homomorphism θ is uniquely determined by its effect on -X", because each element of
G is a product of elements and inverses of elements from X
Consider conversely what would happen if we had a map ί of I into a group Я.
Could we find a homomorphism of the whole group G into Я whose effect on X was the same
as that of 0? In general the answer is no (see Problem 8.12). However, we have the
following result for free groups.
Theorem 8.4: Let F be freely generated by a set X, let Я be any group, and let θ be a
mapping of X into Я. Then there exists a homomorphism 9 of F into Я
such that θ agrees with 9 on X ^is called an extension of Θ.
Proof: Any nonunit element of F is uniquely expressible as a reduced Xproduct
/ = ж*1 ■ ■ ■ xenn where xi G X, e. = ±1
and χ. — x,,, implies e. ¥° —e. , ,.
Define f9- {xi9)4{x29f2 - ■ ■ (ж„0)с" and 1θ- 1 (the latter 1 being the identity of Я).
Clearly θ is a mapping of F into Я agreeing with θ on X. To conclude the proof we must
prove that θ is a homomorphism.
To do this we shall show that if f — x\x ■ ■ ■ a£n where ж. G X and e4 = ±1, then
fe = {xrfp ■ ■ ■ {Хпвр
whether or not x\* ■ ■ ■ ж^" is a reduced product. If η — 1, this is true by the definition of
Θ. Assume it is true for all positive integers η < к and consider f — x\x · · · ж*п when
n — k. If this is a reduced product then fe = (χ\θγ ■ ■ ■ (xnofn by definition. If it is not a
reduced product, then there exists an integer i such that ж4 = ж4+1 and «;=—ei+1-
Consequently,
(Ж10)'1 ■ · ■ fan0)S = (Ж10)61 ■ ■ · (Xtd)** (Χι + ιθ)4 + 1 ■ ■ ■ (Жпй)£"
= (Хгвр · ■ · (Хг-гв)Ч~*(Х1 + *вр+* ■ ■ ■ (Xn&fn
= (ж;1 ■■■χ·^χ^-·-χ^)θ
by our inductive hypothesis. Therefore
{хгвр ■ ■ · (xnefn = (ж;1 - ■ · ж^Ч^ж^1)^2 ■ ■ · <n)o = fe
So if / - x\x ■ ■ ■ a£ and g = y\l ■ ■ - yy, h = ±1, ,, = ±1, xt, y. G X, then
(f9)θ = (a? ■ · · х*н"у]* · · · уУ)0 = (χιθ)*1 ■ · · (Хпвр(У1вр ■ ■ ■ (ymefm
= [{Хгвр ■ ■ ■ (ж,*)*"] [Μ)'1 ■•■Ы?)"т] = ίθ9θ
Hence θ is a homomorphism and the theorem follows.

Sec. 8.1]
HOMOMORPHISMS OF FREE GROUPS
251
Corollary 8.5: Let G be any finitely generated group (i.e. there exists a finite set Υ such
that gv{Y) — G). Then G is a homomorphic image of some free group.
Proof: Let G be finitely generated by a set {yu .. ,,yn}. There exists a free group F
freely generated by [xt, . . .,xn}, say. Define a map 9 from {xir .,., xn} to G by χφ — г/i
for i = 1, ..., n. By Theorem 8.4 there exists a homomorphism 9 of F into G which agrees
with 9 on each Χι, We know that F9 is a subgroup of G. But as F0 contains yi, ..., 2/π, it
contains flrp(2/i, ..., yn) — G. Therefore F9 — G, and so G is a homomorphic image of a
free group.
Note; The same result applies whether G is finitely generated or not. However, to
prove the more general result we must use some of the ideas involving cardinal numbers.
For this reason we have chosen only to consider the finitely generated case.
Our last theorem reveals the importance of free groups. As every group is a
homomorphic image of a free group, from the knowledge of the properties of a free group we
may achieve some understanding of other groups.
Problems
8.11. Let ffj and θ2 be homomorphisms of G -» H. Let G ~ gp(X) and suppose fflf = eZi - Prove that
Solution:
Let g £ G. Then g = x^ ·■■ x£ where xL £ X and t{ — ±1, and
get = (x^J1 ■ ■ ■ (xn8{f* = (x^P ■ ■ ■ (xnezfn = ge2
Thus ffj and e2 have the same effect on the elements of G, and so et = $%.
8.12. Find an example of a group G generated by a set X, a group Η and a map θ : X -* Η, such that
there exists no homomorphism θ : G -» Η with в \ = в.
Solution:
Let G be cyclic of order 2, G = gp{{x}). Put X = {a;} and let Η = gp({y}) be infinite cyclic.
Define θ : X -» Η by x$ — у. Suppose there exists a homomorphism θ : G -» Я, such that asff = y.
Then G$ is a subgroup of Я which contains £rp(y) and hence G $ = H. But Gff, being the image of
a finite group, must be finite. Since Я is infinite, this is clearly impossible.
8.13. Let Η be the cyclic group of order 2, say Η — {1, α}. Let F be freely generated by {x}. The map
β: χ -»a gives rise to a homomorphism of F-*H. Describe the effect of this homomorphism on
all elements of F. Check directly that it is a homomorphism, and find its kernel.
Solution:
The free group on a single generator is infinite cyclic. Its elements are uniquely of the form
Xя, η an integer. Now θ maps xn -» an. If η is even, say η — 2r, an = 1. Hence θ maps яг" -» 1.
If η is odd, say η = 2r + 1, θ maps xn to a.
To check that в is a homomorphism, consider whether (xnxm) в — хп + тв. Now ж""1""»» is α if
n + m is odd, and is 1 if « + m is even. If « + m is odd, then one of the integers η and m is odd and
the other even. Hence хпвхт в = a, as one of as" в, хт В is α while the other is 1. If η + m is even,
then either η and w are both even, or they are both odd; if η and m are both even, хпе^тв = 1 · 1 = 1;
if both are odd, then xnexm θ = a»a— 1. Hence θ is a homomorphism.
The kernel of Β = {χ2* I all integers r}.
8.14. Find a free group that has as a homomorphic image Sn, η any given positive integer.
Solution:
Let F be free on X where \X\ - n\. Let θ : X -* Sn map each element of X onto a distinct
element of Sn. Then by Theorem 8.4 there is a homomorphism θ of F onto Sn that agrees with 9.

252
FREE GROUPS AND PRESENTATIONS
[CHAP. 8
8.15. Prove that a group F freely generated by η + 1 elements has as a homomorphic image a group G
freely generated by и elements, where и is any given positive integer and G is a subgroup of F.
Solution:
Let F be freely generated by X where \X\ = n+ 1. Let Ζ = XjUia} where \Хг\ = и. Then,
as we have shown before, gp(Xi) = G is freely generated by X^ (Problem 8.7, page 247.) Let
в : X -*■ G be denned as follows: If χ £ Χ, χ ¥= a, put x$ = x. If χ = α, put же = 1, the identity.
Then there exists a homomorphism £ of F onto G by Theorem 8.4.
8.16. Let F be a free group with free generating set {a,b}. Let G be the direct product of two infinite
cyclic groups generated by с and d respectively. The map which takes α to с and b to d extends
to a homomorphism of F onto G (since G = gp{{ctd})). Prove that the kernel of this homomorphism
is F', the derived group of F.
Solution:
First we show that if θ is the homomorphism F -*■ G for which αθ = с, Ьв = d, then F' С Ker 9.
F' is generated by the set of all commutators, so it is sufficient to show that each commutator
belongs to Ker в. Now a commutator is of the form [fi,fi\ = f^fz^hfi, where /ι,/2 ε F. Then
[/ii/a]e = [f\S,fie] =1, as G is abelian. Hence F'cKeri.
Now every element of F that does not belong to F' is of the form arbse where not both r and s
are zero and e & F', as F/F' is abelian and aF', bF' generate it. Under 9 such an element goes
onto crds and Cd" = 1 only if both r and s are 0. Thus only the elements of F' belong to Ker в.
Therefore Ker θ = F'.
8.17. Prove that a free group freely generated by η elements, η any positive integer, has a subgTOUp
of index m for each positive integer m.
Solution:
We will exploit the homomorphism property. Let Cm be the cyclic group of order m generated
by an element a, say. Let F be freely generated by X = {xlt .. ,,xn}· Then there exists a
homomorphism в of F onto Cm for which xfi = a, i = 1, . . .. n. Hence by the homomorphism theorem
(Theorem 4.18, page 117),
F/Ker в = Cm
Since \Cm\ = m, the number of cosets of Ker в in F is m. Hence F has a subgroup of index m.
8.18. Let F be freely generated by X. Let У be a subset of F such that fifjo(F) = F, and ]X| = |У| < «.
Use Theorem 8.4 to find an epiraorphism of F onto itself.
Solution:
Let xlf ..., xn be the elements of X, and уг, .. .,yn the elements of У. Define a homomorphism
$: F -*■ F by хф = j/i where i = 1, . .., n, using Theorem 8.4. Since Fe contains У, F» = F.
Hence 9 is an epimorphism of F onto itself.
8.19. Let в be an isomorphism of F with G. Let F be freely generated by X. Prove that G is freely
generated by Хв.
Solution:
Let (xxe) x ■ ■ ■ (χηβ) η be a reduced product in Xe. Then we must show that this reduced product
is not the identity of G. Clearly
(*;i ■ ■ · x£)e = (χ,θΡ ■ ■ ■ (xnefn
and as F is freely generated by X and θ is an isomorphism, the result follows.
8.20. Let F be freely generated by X and G freely generated by У. If в : X -*■ Υ is a one-to-one
correspondence, prove that F ~ G.
Solution:
Let t be the homomorphism of F into G which is an extension of B. (Theorem 8.4.) Clearly $
is onto G. Let ψ : У -*■ X be the mapping such that βψ is the identity mapping on X and ψβ is the
identity mapping on У. Let φ be the extension of ψ to G -*■ H. Then θ φ is the identity on F and
Φ 9 is the identity on G. It follows readily that в is an isomorphism of F with G.

Sec. 8.2]
PRESENTATIONS OF GROUPS
253
8.21. Let F be freely generated by a, b, c. Let N be the normal subgroup generated by c, i.e. the
intersection of all normal subgroups of F containing c. Prove that F/N is freely generated by aN and
bN. (Hard.)
Solution:
Let G be free on x, y. Let 6 be the homomorphism of F onto G defined by αθ = x, be = у and
ce = 1. (Since F is free on a, b, c, such a homomorphism в exists by Theorem 8.4.) Let К be the
kernel of e. First we shall prove that К = N, the normal subgroup generated by c. Then we shall
prove that aN and bN freely generate F/N.
To see that К = N first observe that as ce — 1, N is contained in К (if is a normal subgroup
containing c). On the other hand, suppose / £ F, f (£ N. Then / can be expressed in the form
/ = /i«i
where Д is a reduced {a, byproduct and щ £ N. But then
fe = Λ β
and /ts ia a reduced {ar, j/}-product. So fxe Φ I, which means fe Φ 1 and so f & K. Thus if
f & N, f&.K which implies if is contained in N. Hence as we observed earlier that AT is contained
in K, we find К = .TV as desired.
Now а, Ь, с generate F. So aN, ЬЛ" and cN generate F/N. Since vN = N, aN and bN generate
F/N. We want to prove that aN and bN freely generate F/N. Suppose (y2N) L · · · (ykN)ek is a
reduced {aN, bN}-product.
Now 9 gives rise to an isomorphism ν of F/K with G, i.e. the mapping denned by (fK)v = fe (the
homomorphism theorem, Theorem 4.18, page 117).
Observe that (aK)v = χ and (bK)v ~ у so that
((yiN)c> · ■ ■ GfcJV)*), = ^ ■ ■ ■ <«=
where b/iN)v = x{ £ {ж, у}. The product ar'1 · ■ · ar*" is a reduced {ar, y}-product. But {x, y} freely
generates G. Thus Wj1 — x£ φ ι. Consequently
(yi^f1 ' · ■ (ykNfk Φ Ν
and the proof ia complete.
Instead of completing the proof this way we could refer to Problem 8.19 using the isomorphism
8.2 PRESENTATIONS OF GROUPS
a. Definitions
We have shown (Theorem 8.4) that if F is freely generated by X, then for every group
G and every mapping θ of X into G there is a homomorphism of F into G which extends Θ,
i.e. which agrees with θ on X. This fact will enable us to "present" a given group in terms
of a free group. This idea of a presentation is especially important in topology and analysis
where groups arise in just this way, as the "groups of certain presentations".
First we need a definition. If S is any subset of a group, then the normal closure of S
is defined to be the intersection of all normal subgroups of G containing S. Clearly the
normal closure of S is a normal subgroup of G containing S. Thus the normal closure of
S is often called the normal subgroup generated by S. It is easy to prove that the normal
closure of S is gv{g~leg \ g G G and s&S) (see Problem 8.22 below).
A presentation is defined to be a pair {X;R) where X is a free set of generators of a
free group F and R is a subset of F. The group of the presentation (X;R) is F/N where N
is the normal subgroup of F generated by R; we usually denote the group of a presentation
(X;R) by \X\R\. Finally a presentation of a group G consists, by definition, of a
presentation (X;R) and an isomorphism θ between \X;R\ and G. A presentation (X;R) is finite if
both X and R are finite, and a group G is termed finitely presented if it has a finite
presentation. Not all groups are finitely generated (a necessary prerequisite for being finitely
presented) and not all finitely generated groups are finitely presented. For a more detailed
discussion of these notions the reader may consult R. H. Crowell and R. H. Fox, An
Introduction to Knot Theory, Blaisdell, 1963.

254
FREE GROUPS AND PRESENTATIONS
[CHAP. 8
b. Illustrations of presentations
We shall work some examples to illustrate the definitions of Section 8.2a,
First of all, suppose F is a free group freely generated by 1= {a-, b}. Then the
following
(i) ({а,Ь};{а,Ь})
(ii) ({a, b}; {аЧг, а3Ь3,а*Ь\ ...})
(Hi) {{a,b);{[a,b}})
are patently presentations (by the very definition). The presentations (i) and (Hi) are
finite, but the presentation (ii) is not.
The obvious question is: what are the groups of these presentations?
Clearly the group of (i) is a group of order 1.
What about (ii)? Well, we are interested in F/N where N is the normal subgroup
generated by агЬг, a3bz, а4Ь4,... . We have, since a2b2e.N, a2N = b~2N. Furthermore
oW = b~3N since аяЬя eM Thus it follows that asN = a4N = b~2aN = b~3N.
Cancelling b~2N from both sides yields aN = b~W. Hence ab e N. Indeed we would like to
prove that N is the normal subgroup generated by ab. Let К be the normal subgroup
generated by ab. Since ab e N, we have KcN. Now observe that aK = Ь~гК. Then
αΨΚ = {аЩ{ЫК) = (Ь~'К)(ЬЩ = К
This implies that αψ Ε. Κ (i = 2,3, ...). Hence N is contained in К and so we have proved
that К = N. Therefore F/N is cyclic (because aN and bN clearly generate F/N; but
aN = b~lN). In fact F/N is infinite. To see this suppose G is an infinite cyclic group
generated by g. Let θ be the mapping of {a, b) into G defined by
αθ = g, be = g-1
Let θ be the homomorphism of F into G defined by θ (Theorem 8.4). Clearly θ is onto
and {ab)6 = 1. So if L is the kernel of t, L^N. But as F/L ^ G, F/L is infinite cyclic.
Therefore F/N is also infinite cyclic since, as we have already noted, F/N is cyclic. (Actually,
N = L; however, we don't need this fact here.)
Finally we come to (iii). In fact \{a, b); {[a, b]}\ is free abelian of rank 2. The reader
may attempt to prove this before we do so in a more general case. At this point we
simplify our notation. Instead of using our set-theoretic notation which encloses the
elements of a given set in braces {}, we shall omit the braces in writing presentations. Thus
we write (a, b; [a, b]) for {{a, b); {[a, b]}), [a, b; [a, b]\ for \{a, b}; {[a, b]}\, etc.
Let F be freely generated by αϊ, ..., a„. We shall prove that
| αϊ, θ2, ..., a„; [alr α,] where 1 — i — j — η \
is free abelian of rank η (from which it follows that \a, b; [a, b]\ is free abelian of rank 2).
To this end let Η be the free abelian group of rank n. Then Η is the direct product of
infinite cyclic groups generated by hi, i = 1,..., n, say. We define a homomorphism
Θ: F -* Η by αιθ = hu Now [au οψ = [α{0, α,θ] = [К hi] = 1. Hence [аь a3] G Ker Θ. Let
N be the normal subgroup generated by the [ait aj\, l^i^j^n. Clearly N Q Ker Θ. Note
that [aiN, a5N] = [cti, ai\N = N. As the generators OiN of F/N commute, the elements / of
F are of the form r r
/ = a/ ■ ■ ■ a„ne where cew

Sec. 8.2]
PRESENTATIONS OF GROUPS
255
and if / € N, at least one of the η ¥= 0. Hence as ев - 1, ίθ = h[l ■ ■ ■ hnn. Now as one of
the η ¥> 0, fe ¥> 1. It follows that / € Ker 0, and so N = Ker 0.
Therefore F/2V is the free abelian group of rank n, i.e.
| αϊ, ..., α„; [ai( a.j] with 1 — г =^ j ^ и (
is the free abelian group of rank n.
Note that as N is generated as a normal subgroup by commutators, N С F'. But as
F/N is abelian, N~D F'. Thus N = F' and we therefore conclude that F/F' is free abelian
of rank n.
We state this fact as
Theorem 8.6: Let F be a free group freely generated by a set of η elements (n < *>).
Then F/F' is a free abelian group of rank n.
Corollary 8.7: Let F be a free group. Suppose X and Υ both freely generate F. If |Z| is
finite, then so is \Y\ and 1-Х"] = |У|.
Proof: By Theorem 8.6 we know that F/F' is a free abelian group of rank η = \X\.
If \Y\< », then F/F' is free abelian of rank \Y\, again by Theorem 8.6. Consequently
\X\ = \Y\ since the rank of an abelian group is unique (see Section 6.2d, page 193). It
remains only to prove that |Y| cannot be infinite. To do this, suppose УиУ% .. . ,ϊ/u+i G Y.
Now in a free abelian group of rank η every η + 1 elements are dependent. Hence there
exist integers mi, . . .,wwi, not all zero, such that
(ViF'f1 ■ ■ ■ (yn+iF'f"*1 = F>
i.e. y\F', ..., yn + iF' are dependent. Let A be the free abelian group on fti, .. .,α„ + ι and
let θ be the homomorphism of F to A defined by
Угв = a,i (г = 1, . . ., и + 1), ув - 1 if 2/ € {#ι, . .., г/„+1} and г/ G Υ
The kernel ii of θ contains F' since F/ii is abelian. Then
1 = {Угвр ■ ■ ■ {y^ef^ = a^---a^tx
But A is free abelian on ait . . ., α»+ι. Hence mx = m%= ■ · ■ = mn+i = 0, a contradiction.
Thus \Y\ < да and the corollary has been proved.
It follows from this corollary that if F is a finitely generated free group, then every pair
of sets which freely generate F have the same number of elements. We define the rank of
F to be this common number, i.e. the number of elements in any set which freely generates
F. Note that free groups of the same rank are isomorphic (Problem 8.20).
We can easily give a presentation of A, a free abelian group on air ..., an, with the
results we now have. Let F be free on xu . .., xn and let θ be the isomorphism defined by
{xsF/)e = aJ (г' = 1 и). Then
(xlr ..., Xn, [Xi, Xj) with l^i^ j ^ n)
together with θ is a presentation of A.
In some of the following problems we will often be dealing with factor groups. A simple
convention makes the arguments simpler to follow.
Let G be a group and N a normal subgroup of G. If we use some phrase such as "let
us calculate modulo N," then by G we mean the factor group G/N. We shall mean by
g = h that Ng = Nh. If we say let Μ be a subgroup of G, what we really mean is "let
M/N be a subgroup of G/N." In other words, we must remember that we are talking of a
factor group and instead of writing the cosets, we will simply write the coset representative.
(See Problem 8.24 for an example.)

256
FREE GROUPS AND PRESENTATIONS
[CHAP. 8
Problems
8.22. If G is a-group, show that the normal closure of a subset R (^ 0) of G is gp({g~1rg j g e G and
г ей)).
Solution:
Clearly iV = gp({g~lrg \ g G. G and r ε Л}) is a subgroup of G containing R. Also N is a
normal subgroup of G. Finally any normal subgroup containing R must contain N. Thus the result
follows.
8.23. Let G be a group with subgroups II and К and let [G: II] = и < °° and [G: K] — и. Prove that
if Я 2 К, then Я = К.
Solution:
Let Ж! = 1 and let the distinct cosets of Η in G be II = Hx1,IIx2, .. .,Hxn. As Яз if, it
follows that К»!, Ka^» . -., Kxn are distinct. As [G: K\ — n, G = Kx1\JKx2\J · ■ ■ U Ka;n. If h G H,
then h G Kxf for some integer 1 If ΐ ^ 1, HnKxi = φ since ΗηΗΐ,·=0 for \ф\. Hence
г = 1 and h G Ka^ = K. Accordingly II С К and thus Η ~ K.
8.24. Rewrite the first part of the argument of presentation (ii), page 254, using the modulo N convention
introduced above. Also calculate modulo К and stop after proving K~N.
Solution:
We are interested in F/N where N is the normal subgroup generated by a2b2, asbs, аАЫ, ....
Let us calculate modulo N. Since a,2b2 = 1, a2 = b~2. Furthermore as = b~3 since a3b3 = 1.
Thus it follows that as = α2α = b~2a = b-3. Cancelling b-2 from both sides yields α = Ь-1, and
во ab = 1.
As we are calculating modulo N, this means that ab ε ΛΓ. Indeed we would like to prove that
N is the normal subgroup generated by ab. Let К be the normal subgroup generated by ab. Since
ab G N, we have К cN.
Now we calculate modulo K. a = b"1, and so аЧ>* = b-ty* = 1 (i = 2,3,...). This means as
we are calculating modulo К that а{Ъ{ 6 К (г — 2, 3, . ..). Therefore we have proved that К = N.
8.25. Let G = | a; a2|. Prove that G is cyclic of order 2.
Solution:
The free group F generated by a is infinite cyclic. Now gp(a2) is normal in F since F is abelian.
Hence gp(a2) is the normal subgroup generated by a2. It can be easily seen that gp(a)/gp(a2) is the
cyclic group of order 2. Thus G is cyclic of order 2.
8.26. Prove that G = |α;αη|, where η is any positive integer, is the cyclic group of order η (again with
F freely generated by a).
Solution:
The free group F on a is infinite cyclic and N = gp(an) is the normal subgroup generated by
an. Therefore F/N — G is cyclic of order η (Theorem 4.9, page 105).
8.27. Prove that \a,b; a2, i>2, [a, b]| is the direct product of two cyclic groups of order 2.
Solution:
Although it has not been stated, a and b are (as usual) free generators of a free group F. Let
N be the normal subgroup generated by a2, b2 and [a, b]. In F/N, Na and Nb commute as [Na, Nb] =
2V[a, b] = 2V, since [os, b] ε ΛΓ. Since F/N is generated by 2Vos and Nb, it is therefore abelian. Also
(Na)2 = ΛΓ and (ΛΓ6)2 — ΛΓ. Let A — {N, Na} and В — {Ν, Nb}. As A is a normal subgroup of
FIN, AB is a subgroup of FIN which contains both Na and Nb. It follows that AB = FIN. Thus
|ί7ΛΓ| — |A!|B| — 2 · 2 = 4. On the other hand there is a homomorphism θ of F onto the direct
product of two cyclic groups of order 2. Clearly Ker в contains a2,b2 and [a,b]. Hence Ker в 2 N.
It follows that N — Ker θ (Problem 8.23). Thus FIN is the direct product of two cyclic groups of
order 2.
8.28. Find a presentation for S3. (Hard.)
Solution:
(1 2 3\ /1 2 3\ /1 2 4\
] and σ = | ] . Then σ2 = ( 1 . So ρ,ρσ,ρσΖ and Ι,σ, σ2
2 13/ \2 3 1/ \3 1 2^
are six distinct elements and hence the whole of S3. Now

Sec. 8.2]
PRESENTATIONS OP GROUPS
257
1 2 3\/l 2 3\/l 2 3\ _ /1 2 3
2 1 ЗД2 3 1Д2 1 3/ ~ \3 1 2
Thus the equation ρ~ισρσ~2 = 1 holds. Also p2 = 1 and σ3 = 1. We use only these equations
and hope that they will give rise to a presentation for S3.
Let F be the free group on alt a^ and let G = |aj, a2; a·2, a2, a^la2a,-ia^2\. Furthermore let N
be the normal subgroup of F generated by a\, a'2, a~1a2a1a2~2, and let в : F -> S3 be defined by
(Xjtf = ρ and a2e = a. Then /V с Ker в.
Now we calculate modulo N. We see that a2 — 1 and so aj"1 = a^ Since a| = 1, a^-1 = a^,
a~1a2aia~2 — 1 from which a.^dj = Ojof. Now i1 is generated by a^ a2. Let Μ = gp(a2); then
Af <l F. It follows from Problem 4.62, page 114, that Λί{1, aj is a subgroup of F and as it contains
αχ and Й2, Af{l, c^} = F. Thus the elements of F are {1, α·2, α2}{1, aj, i.e. 1, a2, a^, alt α2Πχ, a2a1
(although we do not know if they are distinct).
It follows that \F/N\ ^ 6. But |F/Ker θ\ = 6 and Nc Ker в, and so Ker θ = N (Problem
8.23). Then
\alta2; α2, α\, α"1 α,2αλα~2\ = Ss
under the isomorphism ψ : atN -> ρ, α2Ν -> σ (by the homomorphism theorem, Theorem 4.18,
page 117).
8.29. Prove that ία·, b; д2, Ьп, а~гЬаЬ\ is isomorphic to the dihedral group of order 2w.
Solution:
Let G be the dihedral group of order 2w. Then G is the symmetry group of the regular w-gon S
(see Section 3.4f, page 75). Recall that <x2 rotates S in a clockwise direction through an angle of
2ir/n. It follows that σ2 is of order n. Put σ = <χ2. If τ is the reflection about ΑλΟ, where Αλ is
any vertex of S, and О the center of S, then τ2 = ι. Moreover every element of G can be written in
the form тсаа, where e = 0,1 and S = 0,1, .. ., η — 1 (see Section 3.4f and note that <χ* = σ{ for
1 — i — n). Let e be the homomorphism of the free group F, freely generated by α and b, onto G
defined by αθ — τ and be — σ. Then α2θ = 1, 6ηί = 1 and
(a~lbab)0 = τ_1στσ — σ_1σ = ι
Thus α2, bn and a~lbab lie in the kernel К of 0. Then N, the normal subgroup of F generated by
a2,bn and a~1bab, is contained in K. Moreover since θ is onto, F/K s G. If we can show that
N = K, then the proof follows.
We calculate (compare Problem 8.28) modulo N. We shall show that \F/N\ — 2w. Since
!i7K| - 2w and КэЛГ, this will establish that ίί = N by Problem 8.23. Modulo Ν,' α'~4α - Ь"1.
Let Μ ~ #р(Ь). Then M<IF. Thus Af{l,a} is a subgroup of F by Problem 4.62, page 114. As it
contains both α and b, M{1, a} = F. The elements of Μ are 1, b, b2, . .., b"_1. Therefore the elements
of F are
ae&s (e = 0,1, а = 0,1, . ..,n-l)
Since we are calculating modulo N, F really stands for F/N. This implies \F/N\ — In. Thus the
proof is complete.
8.30. Prove that G = \a, b; a2, a~lbab\ is infinite. (This group is called the infinite dihedral group.
Hint: Show that each dihedral group Dn is a homomorphic image of G.)
Solution:
Let F be the free group on α and b and let θη be a homomorphism of F onto Dn (as Dn is a two
generator group (see, for example, Problem 8.29), we know such a homomorphism exists). Also
Ker Sin Э {a2, a~"1ba6}. Thus Ker θη Ώ Ν, the normal subgroup generated by a2, a~"1bab. Therefore
G = F/2V has each Dn as a homomorphic image (Dn s (F/N)/(Ker Θη/Ν)). If G were finite of order
fe, say, we would have a contradiction. For then Gsfc as a homomorphic image of a group of order
fe is of order — k. But Gek — Dfc is of order 2fe. This contradiction proves that G is of infinite
order.

258
FREE GROUPS AND PRESENTATIONS
[CHAP. 8
8.31. Prove that every finitely generated group has a presentation.
Solution:
In Section 8.1 d we proved that every finitely generated group is a factor group of a free group.
We will use this fact to prove that every finitely generated group has a presentation. Let G be
an arbitrary group. Let X be chosen so that X freely generates a free group F and. furthermore,
so that there is a mapping e of X onto one of the finite sets of generators for G. Let ψ be the homo-
morphism of F onto G such that Φ agrees with θ on X and let N = Ker ψ. Then (X; N) together
with the isomorphism μ : fN -> /0 is a presentation of G.
8.32. Let G be a finite group with elements 1 = χ χ, x2, . .., xn. Suppose хгх-} = xti ^ , where %α,ί~> ^
{хъ . . ., xn} (in other words (г, j) is an integer between 1 and n). Let F be the free group freely
generated by ab ...,an, N be the normal subgroup of F generated by а^а^у (1 — i, $ — n), and θ
be the homomorphism from
| аъ . .., arL; оцо^а^ where l—i,j — n\
to G defined by (akN)e = xk. Show that
(a„ .. ., an; а{а}а^ where 1 ^ i, j ^ n)
together with θ is a presentation of G.
Solution:
Since afijN = (i(ij)2V, every product of a's in which there are no negative exponents is equal,
modulo N, to some ak. Now if x^1 = a^, then a^a;,- = *н,я — x\, and so a^ = ax modulo N. But
χ\Χχ — ^(ΐ,ΐ) ~ xv Thus
djii! — a.j modulo N
This means that 04 £ 2V and so α{_1 = α·} modulo N. Therefore every product of the ak involving
both positive and negative exponents can be replaced, modulo N, by a product involving only
positive exponents. Accordingly every product of the ak and their inverses is equal, modulo N,
to an οι. It follows that
|ίΥΛΓ| <* η = \G\
Now let ψ be the homomorphism from F to G defined by щФ = xi (for 1 — i — n). Then
{щарй}п)ф = x&FUl» = 1
and so Ker ψ includes α,-α^α^1^. Therefore Ker ψ Э N. Since
|F/Ker ψ\ - n(< «)
and |F/2V| - n, it follows by Problem 8.23 that N = Ker Φ. Thus the mapping
θ: а^^хк
is indeed an isomorphism (Theorem 4.18, page 117).
8.33. Prove that \x, y; x2y2, y2\ = \x, y; x2, y2\.
Solution:
We have a free group F freely generated by χ and у and two normal subgroups N and Μ
generated, as normal subgroups, respectively by x2y2 and y2, and x2 and г/2. We must prove N = M.
Since N contains x2y2 and y2, it contains a:2j/2(j/2)-1 = χ2. Thus N^M. But Мэ{я;2г/2} since
МэМЛ. Hence M3iV and Μ ~ N.
8.34. Show that a free group of rank η < « cannot be generated by и — 1 elements.
Solution:
Let G be free of rank n. Then by Theorem 8.6, GIG' is free abelian of rank n. If G can be
generated by η — 1 elements, G/G' can be generated by η — 1 elements. But this contradicts Problem
6.41(b), page 195.

Sec. 8.3] THE SUBGROUP THEOREM FOR FKEE GROUPS: AN EXAMPLE 259
8.3 THE SUBGROUP THEOREM FOR FREE GROUPS: AN EXAMPLE
The object of the next sections of this chapter is to prove that every subgroup of a free
group is a free group. This theorem is one of the more difficult in this book. So in order to
give the reader a chance to become accustomed to the ideas involved, we first work an
example.
Example 1:
Let F be the free group freely generated by two elements α and b. Consider the set
Υ = {α~~4α,α~^α2, . . .}. Show that Η = (?p(Y) is freely generated by Y. (Thus
we see that a free group freely generated by two elements has a subgroup which is
freely generated by infinitely many elements.)
Proof: Let α~~^αι = yi [i — 0,1,2, ...). Consider any F-reduced product, where
Υ = {у 11 i — 0,1, . , .}. Say / = у у ■ ■ · yn*, where V, .. .,n' are positive integers,
and у у = Vtf + ιγ implies e- ¥= —e,-+I. (Given, for example, у^У\ХУ\Уг, then 1' = 3,
2' = 4, 3' = 1 and 4' = 2; ^ = 1, e,2 = -1, e3 = 1 and e4 - 1.) We will prove
that f¥=l by induction on n, showing that Η is freely generated by Y.
Our inductive hypothesis (on n, the number of yfs that go into a given reduced
product f) is that / when expressed as a reduced product in {a, b} ends in bnan',
(For example,
2/з2/4 2/12/2
a^3ba3 · a~ib~1a/i * a^xba · а~гЬа2
= а-Ча-Ч-^Ьа-Ча2
as a reduced product in {a, b}, and, as asserted, it ends in ba%.)
If η — 1 this is certainly true.
If it is true for η = к, let η = k + 1. Then yy ■ - - ук> ends in b как' when
expressed as a reduced product in {a,b}, i.e. у J ■■■ yk- = zb a■ , where ζ is a
reduced product in a and b such that zb как' is a reduced product (i.e. ζ does not end
in b~i if efc = 1 or, if е^ = —1, ζ does not end in b). If now fc' = (k+ 1)', then,
as / is a reduced product, efc 7* ~~eje + i- Since efc and efe + 1 are the numbers 1 or —1,
efc — e;c+i- Thus
Vv ■ ■ ■ Vatlr = zbefeafcV<fc+4'befc + 1atfc+i'·
= zbeWfc'befc+Ia(fc + I5' = z?/Hefc + 1a(fc + 1)'
Since efc + 1 = e,c, this last expression is a reduced product in {a, b} and / ends in
hefc + ia(fc + i)'_ if however к'¥=(к + 1)', then
2/1'
Efc+i
'· 2/(fc + i)·
Since fe' — (fc + 1)' ^ 0, the last expression is a reduced product and so / expressed
as a reduced product, ends It follows therefore that in both
situations / ¥= 1. Thus Υ freely generates H. Similar arguments will help to prove in
general that a subgroup of a free group is free.
Problems
8.35. Verify both the inductive assumption of the preceding example and that / ¥= 1 where
/
Solution :
Clearly / ¥= 1. The inductive assumption of the preceding example is that / when expressed as a
reduced product should end in ba3, which it does.
2/2 1У3 ЧгУгУз
8.36. Given the existence of a free group freely generated by two elements, prove the existence of free
groups freely generated by any finite number of elements.
Solution:
In Example 1 we have proved that a free group freely generated by two elements has a subset
Υ such that Υ is infinite and Η = ffP(Y) is freely generated by Y. If и is any positive integer
let ylt ..., yn be η distinct elements of Y. Then др{{у1г ...,уп)) is easily shown to be freely
generated by уъ .. ., yn. Thus there exist free groups of rank η for each positive integer n.

260
FREE GROUPS AND PRESENTATIONS
[CHAP. 8
8.37, Let F be freely generated by α and b. Prove that the subgroup of F generated by aba3 and a2b is
freely generated by aba3 and a2b, (Hard.)
Solution:
Let Υ = {aba3, a2b} and let y1 = aba3, уг = a2b. Consider a reduced Y-product
/ = νΐ··· Vn· (i'e {i,2})
Then }' = (} + !)' implies е}Ф — ei+1.
We will show by induction on η that if n' = 1, then / ends in a3 if cn = 1 and ends in b-1a-1
if en = —1; while if n' — 2, then / ends in b if en = 1 and ends in b~xa~z if cn = —1.
For и = 1 this is certainly true.
If it is true for и = к, we must proceed case by case in proving it true for к + 1.
(i) &' = 1-
(a) ejj, = 1. Then by the inductive hypothesis yy ··· Уь; ends in a3. If now (k+ 1)' = 1, then
as / is a reduced product we must have efc+1 = 1 and / ends in a3 -aba3 = aaaabaaa.
Thus / ends in a3 as required. If (k + 1)' = 2 and cfc + 1 = 1, then / ends in a3 ■ a2b, and
so / ends in b as required. If ek + 1 = —1, then / ends in a? · b~~la' 2 and so / ends in
6_1a-2 as required.
(b) ek = —1. Then by the inductive hypothesis yy · · ■ y^ ends in b~la~1. If now (k+ 1)' = 1,
then as / is a reduced product ek + 1 = —1 and / ends in
b-ia-i'a-Sb-ia-1 - b"1a-4b-1a-'1
Hence / ends in b~~1a~1 as required. If {k+l)' — 2 and tfc+1 = l, then / ends in
b~1a~1a2b = b~lab, and so / ends in b as required. If (k + 1)' — 2 and efc + i = —1,
then / ends in b~ia~~x · b-1a~2 and so / ends in b^a-2 as required.
(ii) k' = 2.
(a) ek = 1. Then by our inductive hypothesis j/]< ■ · · Ук' ends in b. If (fc+l)' = l and
eic + 1 ~ 1, / ends in baba3 and so / ends in a3 as required. If (k+ 1)' = 1 and ek + 1 = —1,
then / ends in 6a_3b_1a_1 and / ends in b^1a~"1 as required. If (fc+1)' — 2, then as /
is a reduced product ck + 1 = 1 and / ends in ba2b, and so / ends in b as required.
(b) ek = — 1. Then by our inductive hypothesis у у ■ · · yk> ends in 6-1a~2. If (fc+l)' = l
and it + i = 1, then / ends in b_1a^2aba3 = b~"1a_1ba3, and so / ends in a3 as required.
If (fc+l)' = l and e|c + 1 = -l, then/ends in b-ia"2 · a^b-ia-1 = 6-»a-5b-ia-i, and
so / ends in b_1a_1 as required. If (k + 1)' = 2, then as / is a reduced product ck+1 — —1.
Thus / ends in b-1a~2· b_1a-2 and so / ends in b-1a~2 as required.
Thus we have proved by induction that any reduced product always ends in one of a3,b_1a_1, b
and b-^a""2. Hence / Φ 1 and Υ freely generates gp(Y).
8.4 PROOF OF THE SUBGROUP THEOREM FOR FREE GROUPS
a. Plan of the proof
The subgroup theorem for free groups (due to J. Nielsen and O. Schreier) may be stated
as follows.
Theorem 8.8: Every subgroup Η of a free group F is free.
Suppose that F is freely generated by S. We know from Section 7.6b, page 228, that
if X is any right transversal of Η in F, then the nonunit elements
ax,s = xs{xs)~~l (x G X, s ε S)
(where, for / G F, /is the unique element of X in the coset H/) generate H. Let
Υ = {ax,s | χ G X, s G S, ax,s Φ 1}
Then we shall prove that Η is actually freely generated by Υ provided X is chosen
appropriately. Thus there are two main steps in the proof:
(i) Choose X appropriately.
(ii) Prove that Υ freely generates H.

Sec. 8.4] PROOF OF THE SUBGROUP THEOREM FOR FREE GROUPS 261
Step (ii) of the proof will be broken into two parts: the first part requires a careful look
at the elements ax,s and the second involves looking at the way in which products of these
ax,s and their inverses interact.
b. Schreier transversals
Suppose that X is a transversal for Η in F, where F is freely generated by a set S.
Every element χ (χ Φ 1) in X may be expressed uniquely as a reduced S-product
χ — <Ζια2 · · · a„ (те —1)
where a,- is an element of S or the inverse of an element of S. Recall that η is termed the
length of χ and that the length of 1 is 0. We shall call the elements
1, ab axa2, ..., αλα2 · · · α„
initial segments of x.
Definition: A right trausversal X is called a right Schreier transversal if every initial
segment of an element in X is also in X.
Notice that it follows that if X is a right Schreier transversal, then 1 G Χ.
The main result of this section is the following.
Lemma 8.9: Suppose that F is a free group freely generated by S and that Я is a subgroup
of F. Then we can always find a right Schreier transversal X for Η in F.
Proof: We say that a right coset Hf is of length η if there is an element in Hf of length
η but no element of length less than n. We shall choose X inductively using the lengths of
the cosets of Η in F.
First if Hf is of length 0, then 1 G Hf and so Hf = H. We choose 1 to be the
representative of H.
Suppose that η > 0 and that for each coset of length less than n, representatives have
already been chosen so that every initial segment of a representative is again a
representative. We choose now representatives for the cosets of length n. Let Hf be a coset of length
η and let a,\a,2 ■ ■ · an be an element in Hf of length n. The element aia2 · * ■ On-i is of length
η - 1. Thus the coset Ha,ia2 · ■ ■ αη-ι is of length at most η - 1 (since aioa · ■ · Οπ-ι G
Haxa2 · ■ · an-i). This means that the representative of Ηαλα2 ■ ■ · α,η-ι has already been
chosen, by our induction assumption. Suppose this representative is ЪхЪ2 · · ■ bm. Now
ЩЬ^Ы · ■ ■ bmdn) = (Hbib2 ■ ■ ■ bm)an = (Ηαλα2 · ■ · αη~-\)θη = Ha\d2 · ■ · an
We select b^bi ■ ■ · bman to be the representative of the coset Haxa2 ■ ■ · an. The initial
segments of bib2 ■ · · bm&n, excluding biba · · · bman, are
1, bh . .., bib2 ■ · ■ bm
which we know have already been chosen as representatives. In the same way we select
suitable representatives for all the cosets of length n.
We have therefore verified the induction hypothesis and so we are able, in this way,
to complete the choice of a right Schreier transversal for Η in F.
c. A look at the elements ax,s
Suppose that we have chosen a right Schreier transversal X for Η in F. Consider a
nonunit element ax,s where χ G X and s G S:
CLx.s = XS(XS)~1

262
FREE GROUPS AND PRESENTATIONS
[CHAP. 8
Now let the reduced S-product for a; be χ = cti ■ · ■ ak where at or its inverse belongs to S.
We will allow к = 0; this will be interpreted as χ = 1. Thus we may write
cLx,s = ata2 · · · aks(a1a2 · ■ ■ ukS)"1
Let at ■ · · afcs = b1 ■ ■ - bi where the right-hand side is a reduced S-product with bt or ЬГ1 G S,
i = 1,.. .,1. We assert that αϊ ■ · · aks and bi ■ · ■ bis"1 are not elements of X. For suppose
«ι * ■ ■ aks G X, then αϊ ■ · · <ад = αϊ ■ ■ ■ aks and so a.TjS = 1; whereas if bi · - · bis-1 G X,
we utilize equation (7.5), page 219. and conclude that
(ax,s)_1 — xss'~'1x'~'1 — xss'~'1(xss~l)~1 = xss~1(xss~1)~l
= bi · ■ ■ b^-^bi · ■ ■ bie-1)-1 = Ъг - ■ ■ bis-^bj · ■ ■ bie-1)-1 = 1
From this it follows that the reduced S-product for ax,, is αϊ ■ · · a^bi^1 ■ ■ ■ bf1. For if
not, s cancels either with afc or with Ы~1. But as every initial segment of an element of X
belongs to X, either αϊ ■ ■ ■ aks G X or bi ■ · * bis-1 G X, which is a contradiction. Note
we have proved a^,s¥-l implies χ does not end in s~1 and xs does not end in s.
Let PF = {w | w G S or w-1 G S}. Then we have the following
Lemma 8.10: Let e = 1 or -1. Then if ax,s¥°l,
<tl.» = Ci · · · CmWdn1 · · ■ di1
where the right-hand side is a reduced product and both C\ " " ■ Cm and di · · ■ dn
are elements of X and both ci ■ · · cmw and di - · · dnw~1 lie outside X.
Proof; The lemma is an immediate consequence of the preceding remarks. We need
only note that
a~l = bi · · ■ ^8~1αζ1 ■ ■ ■ αϊ1
for the case e — —1.
Corollary 8.11: Suppose e = ±1 and
«Is = Ci ■ ■ · CmW<2„ ' ■ ■ ■ (Zi *
where the right-hand side is a reduced product, Ci ■ ■ ■ cm G X and
Ci ■ ■ ■ cmM) ё X. Then
(i) if w G S, then e = 1, χ — C\ · · ■ cm and s = w,
(ii) if w £ S, then <r = -1, a; = d\ · ■ ■ d„ and s = w~1.
Proof; It follows immediately from the preceding argument.
Corollary 8.12: Let e = ±1, v= ±1, x,y G X, and s, ί G S. If
a*,s = Ci · · ■ CmWdn1 · ■ ■ άΤ1 and a^.t = Ci · · · cmiue~' - · · ef1
with the right-hand sides reduced products, and Ci · ■ ■ cm G X but
Ci ■ ■ · cmw £ X, then e = 17, χ = у and ί = s.
Proof; It follows from Corollary 8.11.
d. The proof of the subgroup theorem
Suppose again that F is a free group freely generated by S and that Я is a subgroup of
F, Choose a right Schreier transversal X for Я in F. Then
Υ = {θχ,β J χ G X, s G S, ax,s ?*■ 1}
generates Я. It remains only to prove that Υ freely generates H. For this it suffices to
prove that any reduced Y-product is not the identity (Lemma 8.2, page 248).

Sec. 8.4] PROOF OF THE SUBGROUP THEOREM FOR FREE GROUPS 263
Let g = a'1 s · · - a/s {xb ..., xT G X and «i, ...,sreS) be a reduced Y-product. Let
ax\$r be expressed in the form of Lemma 8.10 as e\ ■ · · emvf~x ■ - · f7\ where ν G W and
βι - * · em G X but βι ■ · · emv g X. We will show that g ends in vfZ1 · ■ ■ f~x by induction on
r, the case r = 1 being of course proved in Lemma 8.10.
If the result is assumed true for r - 1 and al~\ ,. is expressed as ci · · · ckwdTx ■ ■ ■ dT1
in the form of Lemma 8.10 with Ci · ■ ■ Ck G X but Ci - · - ckw ί J (w G PF), then αυ\8 ■ · ·
arr~,1s, , ends in wdr1 ■ · ■ dT1.
Consider the product
wdi · ■ · dt ei ■ · ■ emvfn ■ ■ * Л
We can convert it into a reduced product by successively deleting inverse pairs. If m = l
and wdr1 ■ ■ · ίΖΓ'βι ■ - · emv = 1, then by Corollary 8.12, a^rz\lSr_l = {KrT.s )~\ contrary to
the assumption that g is a reduced Y-product. If βι · ■ ■ emv is removed as a result of deleting
inverse pairs in the product df1 · · · d^1e1 ■ · · emv, then e\ ■ · · emv is an initial segment of
dt · ■ · di and hence an element of X. But this is contrary to Lemma 8.10. Similarly
wdi.1 · ■ · «ΖΓ1 is not removed as a result of deleting inverse pairs in the product
wdr1 · ■ - <ΖΓ1βι · - - em. Thus
wdr1 ■ ■ ■ dilei ■ · ■ emvf~l ••■ft1 = w · ■ · vf~l ■ ■ ■ f7l
when expressed as a reduced product by deleting inverse pairs (the · · ■ between w and ν
represent the factors ЙГ1 ■ - · dtlet - · · em left after deleting inverse pairs). Consequently
g ends in w ■ ■ · vf~l · * - /Г1 and the inductive assertion follows. Therefore g ¥-1, and
the result follows.
e. Subgroups of finite index
In Section 7.6b, page 228, we proved that a subgroup of index и in a group generated by
r elements is generated by nr elements. We shall now find the rank of a subgroup of index
и in a free group. To find the rank of the subgroup we use the result of Section 8.4d, i.e.
the nonunit ax,s freely generate the subgroup.
Let F be freely generated by St, . ..,sr and let X = {xb ...,xn} with χι = 1 be a
Schreier transversal for a subgroup Я of index n. Consider the elements
Ctij, s j = XiSj(XiSjj
i = 1, ...,η and j ~ 1, .. .rr. The number of such elements is nr. To find the rank of H,
we wish to determine how many of the aXi,Sj are unit elements. By line 13, page 262:
(1) If х{ ends in s~l, then ax.,s. ~ 1.
(2) If IcWi ends in s,, then ax.,Sj ~ 1.
We show that (1) and (2) are mutually exclusive. Suppose that xt ends in sr1. Then
Xi = Wf·- WmSf1 (where ин, ..., wm G W) is a reduced product. Consequently wm ^ s,
(for otherwise Wt · ■ ■ wms~l is not a reduced product). As x% is in the Schreier transversal
-X", ин ■ ■ · wm also belong to the Schreier transversal. Thus χϋί = Wx ■ ■ · wm = wr ■ · wm
and xWj does not end in ss. Therefore (1) and (2) are mutually exclusive.
Note that if neither %i ends in s}1 nor xls] ends in ss, then α.^ = ^(ίείΐΐ)""1 ¥-1 as Sj
remains when aXi,S} is expressed as a reduced product.
For fixed j, let a. = number of xi G X for which aXi.Sj= 1. Clearly α5 = number of
x.GX for which x. ends in з~л plus the number of x. for which xJ, ends in sr As xx runs
through X, x~si runs through X (x -»xs~,, χ G X, is a permutation, page 219). Thus the
number of x. such that a^T ends in s. is the number of elements in X which end in s,. We
conclude a. — number of xi that end in si or s~\ So the total number (i.e. with j — 1, ..., r)
of aH,s, = 1 is ax+ · ■ · +ar = number of elements of X that end in s. or sjl, j = 1,.. .,r.
But except for xt = 1, every element of X ends in some sj or sf1. Hence ai+ · · · +aT = n — 1.

264
FREE GROUPS AND PRESENTATIONS
[CHAP. 8
Thus there are exactly η - 1 unit elements among the ax.tSj, and we have proved
Theorem 8.13: Let F be a free group of rank r and let Я be a subgroup of index n. Then
Я is a free group of rank n{r — 1) + 1.
Problems
8.38. Let F be freely generated by slt .. .,sr. Let Η be a subgroup of index 2 such that st £ H, but
ef e Η for i = 2, .. .,r. Find a set of free generators for Я by using the method of Section 8.4d.
Verify that the number of free generators agrees with the number given by Theorem 8.13.
Solution:
We choose {IjSj} for the Schreier transversal. Then the nonunit elements among als and
as,.s (i — li. ■>.*■) freely generate H. Now o,s. = ls^ls^)-1. If j = 1, o^. = 1. If j Φ 1,
αϊ,, = le^lsj)-1 = Sj as «j € Я implies e^ = 1. Now
%SJ = «lejieiej)"1 = «ι^Γ1 if J"5*1 (for then в^еЯв!)
= s? if j = l (as ef=l)
Thus the subgroup Η is freely generated by »*,β2, ..., er and «ιβ^™1, .SjS,^"1. Thus Η is of
rank 2(r- 1) + 1, which agrees with Theorem 8.13.
8.39. Let F be freely generated by χ and y. Find a set of free generators for F', the derived group of F.
Solution:
Let X = {xrys | r and s integers}. Then to show that X is a Schreier transversal we need only
show that (1) xry*, of1!/*1 belong to the same coset of F' only if r = rv s = sr, and (2) X is a set
of coset representatives. Both (1) and (2) follow easily on using the fact that F/F' is free abelian
with basis xF', yF' (by Theorem 8.6, page 255). The free generators of F' are the elements αχΤ , χ
and αχγν which are nonunit. Now
β,ΓΛ, = хгу*х{хгу*х)-г = xryax(xr+1ys)~i = xrysxy-*x-r-i φ 1 for «i*0
On the other hand,
aiV-s = xry^y(xrys + i)-i = ι
Thus a set of free generators for F' are the elements xrysxy~sx~r~ l for all integers r and all
integers s Φ 0.
8.40. Let F be a free group of rank r and Η and К subgroups of index n. Prove that Η = К.
Solution:
By Theorem 8.13, Η and К are free of the same rank. Thus they are isomorphic.
8.41. Let F be a free group on generators χ and y. Suppose that R <\ F, у £ R and FIR — ffp(xR)
is infinite cyclic. Prove that the group R/R' is freely generated as an abelian group by the elements
Xnyx~nR', where nG Z. Then prove that for no integer n, is xnyx~nR' in the center of FIR'.
Solution:
The method of Section 8.4d with X = {xn \n£Z} gives for the free generators of R the
elements a>xn = xnyx~n, n€. Z. By an argument similar to Theorem 8.6 (which deals only with
free groups of finite rank), RIR' is free abelian with basis xnyx~nR' for all integers n.
Now as
(xR')(ai4ix-nR')(xR')-1 = x*+iyx-<-* + UR' Φ xny%~*R'
xnyx-"R' does not belong to the center of FIR',

Sec. 8.4] PROOF OF THE SUBGROUP THEOREM FOR FREE GROUPS 265
f. Intersection of finitely generated subgroups
We prove here that if Η and К are subgroups of finite rank of a free group F, then
HnK is also a subgroup of finite rank. (This result is due to Howson.)
To simplify the problem we note that we may assume that F is finitely generated. For
if not, we could consider gp(H, K), which is certainly finitely generated (as Η and К are),
instead of F.
However, as we have seen in Example 1, page 259, a free group of rank 2 has a subgroup
of infinite rank. Therefore we may as well assume that F is free of rank 2. Let F be
freely generated by a and b.
We say that a coset С is single-ended if every element of С when expressed as a reduced
product has the same last factor.
For example, if all the elements of С end in а, С is said to be single-ended. (We might
have for instance bam G C.) But if С contains for example the elements ababab and
ab"1, then С is not single-ended.
A coset which is not single-ended is called double-ended.
The following lemma is crucial:
Lemma 8.14: A subgroup Η of the group freely generated by a and b is of finite rank if
and only if it has a finite number of double-ended cosets.
Proof: Choose a Schreier transversal X for Η in F. Put S — {a, b}.
(1) Let Η be of finite rank. Then only a finite number of the elements a*,* ¥* 1 (where
χ G X and s G S). But we proved in Section 8.4d that every element of Η ended in the
form wdf1 · ■ · άϊ1 where wd^1 · ■ ■ dT1 is either 8(Щ~г or s_1«_1, where ax,s Φ 1. Thus
we concluded that all the elements of Η end in the form wdr1 · ■ · dT1 where there are
only a finite number of wdr1 ■ · - άϊ1 and di ■ · · diw~l € X although di ■ ■ ■ di G X.
Let χ G X. Then if Hx is double-ended, there exists an element h&H such that
in kx, χ cancels completely. As h ends in some wdf1 ■ ■ ■ άϊ1 and d\ ■ · · diw"1 € X, it
follows that if χ cancels completely, χ is an initial segment of dt · · ■ di (otherwise
xvdi ■ ■ · di is an initial segment of χ and hence in X). It follows that as the number
of di · · · di that appear is finite, the number of initial segments is finite, and thus the
number of double-ended cosets is finite.
(2) Let Η be of infinite rank. If ax,s = Х8{хё)~~1 ¥> 1, it follows that χ does not end in s_1
(line 13, page 262) and, of course, «sjaJs)"1 G H. The coset Hx contains χ and
(xs(xs)~1)~~1x = xss"1. Now s_1 appears in the reduced product for xss"1 (line 13, page
262). As χ does not end in s_1, Hx is double-ended.
Now as there are an infinite number of χ such that Ot,s# 1, there are an infinite
number of double-ended cosets.
Theorem 8.15: The intersection of two subgroups Η and К of finite rank is again of
finite rank.
Proof: By Lemma 8.14, Η has only a finite number of double-ended cosets, say
Hi, .. ., #„, and К has only a finite number of double-ended cosets, say K\, . . ■, Km. Now
the cosets of HnK are intersections of cosets of Η and К (Problem 7.23, page 231). Also
the intersection of a single-ended coset of Η with any coset of К is single-ended (and
vice versa). Thus the double-ended cosets of HnK are among the cosets HidKj, i = 1, ..., w
and j = 1, .. .,m. Therefore HnK has at most mn double-ended cosets and accordingly
HnK is of finite rank by Lemma 8.14.

266
FREE GROUPS AND PRESENTATIONS
[CHAP. 8
A look back at Chapter 8
We defined free groups and gave an alternative definition. The existence of free groups
was established and homomorphisms of free groups investigated. The main result was
that if F is a free group freely generated by X, then for every group Я and every mapping
0 of .Χ" into Η there is a homomorphism ψ of F into Η which coincides on X with Θ. As a
consequence every group is a homomorphic image of a free group.
We next discussed presentations of groups. The important notion of rank of a free
group then arose naturally.
We discussed and proved the subgroup theorem for free groups. Using this, the rank
of a subgroup of finite index was calculated. Finally we proved that the intersection of
two subgroups of finite rank in a free group is of finite rank.
Supplementary Problems
ELEMENTARY NOTIONS
8.42. Suppose F is freely generated by α and b. Find elements u,vGF so that и Ф α and
o-1b-1ab = u^V^v
8.43. Prove that a free group of finite rank has only a finite number of elements of length — η for any
fixed integer n.
8.44. Prove that if N <3 G and GIN is free, then G splits over N.
8.45. Let F be a free group freely generated by α and 6. Let N = gp(.. .,а2Ьа-2, аЪа~г, b, a~lba,
a~4a2, ...). Prove N < F and verify that F splits over N.
8.46. Using the notation of the preceding problem show that if N' is the derived group of N, then
N' < F. Verify that F/N' is a splitting extension of N/N' by an infinite cyclic group. Construct
an isomorphic copy of FIN' directly as an extension of a free abelian group by an infinite cyclic
group.
PRESENTATIONS OF GROUPS
8.47. Prove that the group G = \a,b;a~1ba = 62| is not free. (Hint: Prove that GIG' is cyclic. So if
G is free it must be free cyclic. Then show G' Φ {1} by mapping G into a suitable factor group.)
8.48. Let G = \x,y; x%,y%\. Show that G is infinite. (Hint: Let F be the free group freely generated by
χ and y. Let θ : χ -> а, у -> ab be a homomorphism onto the group of Problem 8.30, page 257.)
8.49. Let G = \a,b,c,d; [a,c\,\a,d\,[b,c},[b,d\\. Prove G is the direct product of two free groups each
of rank 2.
8.50. Prove that if G = |arl5 x2,...; 2ar2 = xb ..., (i+ l)iei+1 = ar„ ...|, then G is isomorphic to the
additive group of rationale. (Hint: Note that G has the additive group of rationale as homomorphic
image. Also it is abelian. Use Problem 6.72, page 208.)
8.51. Prove that if ρ is a prime and G — \xu x2, . -.; xlt яф»"1, ..., «f+i»^1, ■ ·.|, then G is isomorphic
to the p-Prtifer group.

CHAP. 8]
SUPPLEMENTARY PROBLEMS
267
8.52. What is the group \хг, x2, ..., xn; x^^1]?
THE SUBGROUP THEOREM FOR FREE GROUPS
8.53. Prove that if F is free on x, y, χ then the elements xyz, x2y2z2, sfiy^z5, х*у*г*, ... freely generate a
subgroup of F.
PROOF OF THE SUBGROUP THEOREM
8.54, Show that if am = bn where a and 6 are elements of a free group (mn φ 0), then a and 6 generate a
cyclic group.
8.55. Let ΛΓ and Μ be nonidentity normal subgroups of G, a free group of rank greater than 1. Prove
that ΝηΜ Φ {1}.
856. Let F be a free group. Prove that there is no sequence of subgroups FlQFzQ- ■ · of F with
F()*P,H and rank of f( = 2, {Hint: Let G— Ufj. Then G is a free group of infinite rank.
Consider GIG' which is free abelian of infinite rank. Obtain a contradiction by showing that GIG'
is of finite rank.)
8.57. Find generators for F2 where F is a free group of rank two and Fz = ffp(x2 \ xGF). (Note that
F/F* is the Klein 4-group.)
8.58. Let F be a group which can be generated by two elements with a free subgroup of rank 3 which is
of index two. Prove F is free by comparing it with a free group of rank 2, (Hard.)
8.59. Let F, R be as in Problem 8.41, page 264. Prove that FIR' has no element other than 1 in its center.

Appendix A
Number Theory
In this book we assume the reader knows the following:
1. The meaning of a divides b, for which we use the notation a\b. The notation a j( b is
read as "a does not divide b".
2. The definition of a prime, i.e. an integer not equal to 1 which is divisible only by 1 and
itself.
3. If a, b are integers, then there exist an integer g and an integer r such that
a — bq + r
where 0 — r < b.
4. The definitions of greatest common divisor of two numbers a and b (the largest integer
which divides both a and b) and lowest common multiple of a and b (the smallest integer
divisible by both a and b). We write the greatest common divisor of a and b as (a, b),
the lowest common multiple of a and b as l.c.m. (a, b). If (a, b) = 1, then a and b are
said to be co-prime.
5. Given integers a and b, there exist integers ρ and g such that
(a, b) = pa + qb
6. The fundamental theorem of arithmetic which says every integer is expressible in one
and only one way (ignoring order) as the product of primes.
7. a s b modulo η means a — b is divisible by n. Also some of the simpler properties
such as a s b and χ s у implies a + x ^ b + y and ax e by.
The reader who does not know this material can consult
(a) Niven, I., and H. S. Zuckerman, An Introduction to the Theory of Numbers, Wiley, 1966.
(b) Upsensky, J. V., and M. A. Heaslet, Elementary Number Theory, McGraw-Hill, 1939.
(c) Birkhoff, G., and S. MacLane, A Survey of Modern Algebra, Macmillan, 1953.
269

Appendix В
A Guide to the Literature
Note: Numbers in brackets refer to the bibliography on pages 272 and 273.
General
Books which contain much of the material of this text (and frequently more) are, in the
order of complexity, [46], [15], [16]. Useful books in algebra are [47] and [23].
Chapter 1
The theory of sets was established by G. Cantor in the last quarter of the 19th century
[1]. A central notion in his work is that of cardinality — two sets have the same cardinality
if there is a one-to-one mapping from the one onto the other. This leads to the idea of a
transfinite number, and much of Cantor's work was devoted to developing these transfinite
numbers (see [2] and [3]).
In the study of sets a number of paradoxes arose ([4] and [5]). This made it desirable
to put set theory on a firm axiomatic footing. Such axiomatic approaches have led to a
number of interesting developments ([4], [5] and [6]).
Recently a different approach to the foundations of mathematics has been originated
by Lawvere, which is based on the notion of category [44] (see [7] for the axioms of a
category).
Chapter 2
The study of groupoids arose in part from a desire to understand more clearly the
axioms for group theory. The theory of groupoids can be subdivided into systems which
satisfy various "natural" conditions. Thus, for example, among the various classes of
groupoids one has semigroups, loops, groups and quasi-groups. Two major references are
[8] and [9].
Chapter 3
It is clear from this chapter that groups arise in a great many mathematical disciplines.
Groups arose initially from 19th century algebra, analysis and geometry. It was hoped
that much of geometry could be handled by associating a group with each geometrical
object. To some extent this aspect of group theory has been discussed in the section on
isometry groups (see also [10], [11], and [14]). For the applications to topology see [37]
and [39], and for knot theory see [40].
Groups arise also in quantum physics, crystallography [12] and chemistry [13]. This
omnipresence of groups is part of the reason for its importance.
In addition, the study of groups has been carried forward for its own sake (see e.g.
[15], [16], [17], [18], [19], [20], [21], [22], [43]). Finally we refer the reader to the study of
groups with a topology: [41], [42].
270

APPENDIX В]
A GUIDE TO THE LITERATURE
271
Chapter 4
Chapter 4 is concerned mainly with cyclic groups and homomorphisms.
As we have dealt exhaustively with cyclic groups, there is not really any further
information available except perhaps for finding the automorphism group of a cyclic group. If
G is cyclic of order n, then the automorphism group of G is the group of integers co-prime
to n, with multiplication modulo n; see e.g. [43].
The concept of homomorphism is basic. One can define homomorphisms for other
algebraic concepts, such as rings. Factor rings can be defined and very similar theorems
obtained [23], [24], [25]. (Indeed, once the group theoretic ones are known, it is routine
to obtain the others.)
Some work, prompted by algebraic topology, has been done on more complex interaction
of group homomorphisms [26].
Chapter 5
Much of this chapter is "arithmetical" in content, i.e. it deals with the order of a
finite group. Thus for example we have the theorem of Lagrange: The order of a
subgroup of a finite group divides the order of the group. There are a great number of
important theorems of this kind, embodying various generalizations of the Sylow theorems
(see [27], [17]). A very important result of a slightly different kind is the remarkable
theorem of W. Feit and J. Thompson, viz. a finite group of odd order is solvable [45].
The classification of groups of small order has not been too successful. This is due to
the extraordinary complexity of these groups. The reader might consult [21], [17] and
[28] for a discussion of this classification problem.
Today the study of finite groups has proceeded at an extraordinary pace. One of the
main aims of this study is the classification of finite simple groups. This classification is by
no means complete, but much progress has been made (see [29]).
Chapter 6
Our proof of the fundamental theorem of abelian groups is nonconstructive, i.e. we do
not have a definite procedure for finding a basis. Such procedures exist, e.g. [30].
There are also other criteria for deciding if an abelian group is a direct sum of cyclic
groups, e.g. if every element is of bounded order [31]. A more comprehensive result is
that of Kulikov [16].
In this chapter we found invariants for finitely generated abelian groups. The
invariants for countable torsion groups are subtler and appear in Ulm's theorem ([31], [32]).
Some of the theorems of abelian groups extend to wider classes of algebraic structures,
e.g. modules over rings ([31]). There has also been an attempt to prove theorems about
classes of groups that are quite close to abelian groups, e.g. solvable and nilpotent groups
([17], [19]).
Excellent sources for abelian groups appear in [16] and [31]; a more encyclopedic
account appears in [32].
Chapter 7
The aim of this chapter is to show how representing groups as permutation groups
leads to information about the groups themselves. There are other types of representation.
The most important of these is the representation of finite groups as их и matrices.
(These nXn matrices constitute a natural generalization of the 2x2 matrices we have
discussed in Chapter 3.) The reader might consult [33], [34].

272
A GUIDE TO THE LITERATURE
[APPENDIX В
In infinite group theory the permutational representation of Frobenius has led to
significant progress (§2 of Chapter 2, etc., of [35]).
The transfer has been used a great deal in finite group theory. A good source of
information is [27]. See also [17].
Chapter 8
Our proof of the existence of free groups is really motivated by Cayley's theorem, as
a glance at the usual proof will reveal [16].
The property of Theorem 8.4, i.e. the property of being able to extend a mapping of the
free generators to a homomorphism of the free group, is of great importance. Indeed, it is
often used as the definition of a free group. With this approach, the definition of free group
is analogous to the definition of free abelian group (Chapter 6). There are even further
generalizations and we can speak of the free group in a variety, where a variety is a
collection of groups satisfying certain conditions. A detailed account appears in [35].
One can also regard a free group as being the free product of infinite cyclic groups.
The free product of two groups is a way of putting the groups together in, roughly speaking,
the freest way ([16], [17]).
Groups occur frequently as presentations. This is unfortunate as it is always difficult
to say what the group of a presentation is or what its properties are. Indeed, there is not
even a general and effective procedure for deciding whether the group of any given
presentation is of order 1. AH this is connected with the word problem, which is, roughly speaking,
to determine in a finite number of steps if two products in the generators of a given
presentation are equal. The word problem is unsolvable in general (see [15]).
An important concept which enables us to change from one presentation to another is
that of Tietze transformations [36].
There are many proofs of the subgroup theorem, including topological ones (e.g. [37]).
An important one is by Nielsen transformations ([16], [17], [36]). One of the advantages
of the method we used in Chapter 8 is that it extends readily to provide generators and
defining relations for subgroups of a group of a presentation [36].
More properties of free groups, free products and presentations of groups appear in
[36]. See also [38] for presentations.
BIBLIOGRAPHY
[1] Cantor, G., Contributions to the Founding of the Theory of Tranefinite Numbers (translated by
P. Ε. Β. Jourdain), Open Court, 1915.
[2] Катке, Е., Theory of Sets (translated from the 2nd German edition by F. Bagemihl), Dover, 1950.
[3] Halmos, P. R., Naive Set Theory, Van Nostrand, i960.
[4] Kleene, S. C, Introduction to Metamathematics, Van Nostrand, 1952.
[5] Cohen, P. J., and R. Hersh, "Non-Cantorian Set Theory", Scientific American, December 1967.
[6] Suppes, P. C., Axiomatic Set Theory, Van Nostrand, 1960.
[7] Freyd, P., Abelian Categories, An Introduction to the Theory of Functors, Harper and Row, 1964.
[8] Bruck, R. H., A Survey of Binary Systems, Springer, 1958.
[9] Clifford, A. H., and G. B. Preston, "The Algebraic Theory of Semigroups", Mathematical Surveys
7, 1961, American Mathematical Society.

APPENDIX B] A GUIDE TO THE LITERATURE 273
Weyl, H., Symmetry, Princeton University Press, 1952.
Hubert, D., and S. Cohn-Vossen, Geometry and the Imagination, translated by P. Nemenyi, Chelsea,
1956.
Hammermesh, M., Group Theory and its Application to Physical Problems, Addison-Wesley, 1962.
Cotton, F. Α., Chemical Applications of Group Theory, Interscience, 1963.
Yale, P. В., Geometry and Symmetry, Holden Day, 1968.
Rotman, J. J., The Theory of Groups: An Introduction, Allyn and Bacon, 1965.
Kurosh, Α., The Theory of Groups, (translated by K. A, Hirsch), Chelsea, i960.
Hall, Jr., M., The Theory of Groups, Macmillan, 1959.
Wielandt, H., Finite Permutation Groups, Academic Press, 1964.
Schenkman, E., Group Theory, Van Nostrand, 1965.
Scott, W., Group Theory, Prentice-Hall, 1964.
Burnside, W., Theory of Groups of Finite Order, Dover, 1955.
Huppert, В., Endliche Gruppen, Springer-Verlag, 1967.
Herstein, I., Topics in Algebra, Blaisdell, 1964.
Van der Waerden, B. L., Modern Algebra, (translated by F. Blum), Ungar, 1953.
Cohen, P. M., Universal Algebra, Harper and Row, 1965.
Northcott, D. G., An Introduction to Homological Algebra, Cambridge U. P., I960.
Wielandt, H., and B. Huppert, "Arithmetical and Normal Structure of Finite Groups," 1960 Institute
on Finite Groups, Editor, M. Hall, Jr., American Mathematical Society, 1962.
Hall, Jr., M., and J. K. Seniour, The Groups of Order 2n (n — 6), Macmillan, 1964.
Carter, R. W., "Simple Groups and Simple Lie Algebras", Journal of the London Mathematical
Society, April 1965, pp. 193-240.
Schreier, 0., and E. Sperner, Introduction to Modern Algebra and Matrix Theory, (translated by
M. Davis and M. Haussner), Chelsea, 1959.
Kaplansky, I., Infinite Groups, U. of Michigan Press, 1954.
Fuchs, L., Abelian Groups, Akademiai Kiado, 1958.
Burrow, M., Representation Theory of Finite Groups, Academic Press, 1965.
Curtiss, C. W., and I. Reiner, Representation Theory of Finite Groups and Associative Algebras,
Interscience, 1962.
Neumann, H., Varieties of Groups, Springer, 1967.
Magnus, W., A. Karass and D. Solitar, Combinatorial Group Theory: Presentations of Groups in
Terms of Generators and Relations, Interscience, 1966.
Massey, W. S-, Algebraic Topology; An Introduction, Harcourt, Brace, 1967.
Coxeter, H. S. M., and W.O.J. Moser, Generators and Relators for Discrete Groups, Springer-Verlag,
1965.
Wallace, A. H., An Introduction to Algebraic Topology, Pergamon, 1957.
Crowell, R. H., and R. H. Fox, Introduction to Knot Theory, Ginn, 1963.
Pontriagin, L. S., Topological Groups, Gordon and Breach, 1966.
Cohn, P., Lie Groups, Cambridge U. P., 1957=
Zassenhaus, H., The Theory of Groups, Chelsea, 1949.
Lawvere, F. W., "The Category of Categories as a Foundation for Mathematics", Proceedings of the
Conference on Categorical Algebra, La Jolla, 1965, Springer-Verlag, 1966.
Feit, W., and J. G. Thompson, "Solvability of groups of odd order", Pacific Journal of Mathematics,
vol. 3, no. 3, pp. 773-1029, 1963.
Lederman, W., Introduction to the Theory of Finite Groups, Oliver and Boyd, 1957.
Birkhoff, G., and S. MacLane, A Survey of Modern Algebra, Macmillan, 1953.

INDEX
Abelian group, 177-212
divisible, 205, 209
finitely generated, 196
free, 186
of type pM = p-Prufer group, 191, 206
primary = p-group, 190
rank of, 193
Abelian groupoid, 29
Additive notation, 19,178
Alternating group, 62
simplicity, 172
Ascending central series = upper central series, 142
Associative groupoid, 29
Automorphism,
group, 84, 87
inner, = mapping pa, 85
of field, 87
of group = of groupoid, 83
Basis of free abelian group, 186
Basis theorem = fundamental theorem for
finitely generated abelian groups, 197
Bijection, 14
Binary operation, 19
Block, R-, 9
Cardinality, 14
Carrier, 26
Cartesian product, 6
Cayley's theorem, 46, 214
Center, 112
Centralizer of a subset, 112
Chain, 194
Circle, 2
Class, equation, 135
equivalence, 9
R-, 9
Closed with respect to multiplication, i.e. product
of elements in the set belong to the set.
Codomain, 12, 13
Common part, 3
Commutative groupoid, 29
Commutator, 112
Commutator subgroup, 112
Commute: α and b commute if ab = ba.
Complement, 234
Component, p-, 191
Composition factors, 164
Composition of maps, 17
Composition series, 158,163
Congruence, 269
Conjugacy classes, 171
Conjugate, subgroups, 131
Conjugates, H-, 134
Correspondence theorem, 120
Coset, 108
double-ended, 265
representation, 219
representative, 219
Countable, 14
Countably infinite, 14
Cycles, 167
Cyclic group, 101
Degree of permutational representation, 217
Dependent, 192
Derived subgroup, 112
Difference of sets, 4
Different products, 245
Dihedral group, 75
infinite, 257
of degree 4, Table 5.2,152
Dimension of vector space, 89
Direct product, 143
external, 143
internal, 146
Direct summand, 178
Direct sums,
finite, 178
infinite, 182
Divisible group, 205, 209
Element, 1
order of, 103
Epimorphism, 42
Equivalence relation, 8
Extension of groups, 232-234
splitting, 234-238
Extension of mapping to homomorphism, 185, 250
Extension property,
free abelian groups, 185
free groups, 250
Factor groups, 114
Factor of a factor theorem (third isomorphism
theorem), 121
Factors of a subnormal series, 158
Faithful representation, 218
Family of indexed sets, 182
Fields of complex numbers, 86
Finitely generated abelian groups, 197
subgroups of, 202
factor groups of, 204
Finitely generated group, 98, 197, 227
Finitely presented, 253
Finite presentation, 253
Four group, 144, 148
274

INDEX
275
Free abelian group, 186
rank of, 193
Free group, 246
intersections of finitely generated subgroups, 265
length of an element, 248
rank of, 255
subgroups, 260
subgroups of finite index, 264
Free on a set, 246
Free set of generators, 246
Freely generated, 186, 246
Frobenius' representation, 225
Frobenius' theorem, 224
Full linear group, 90
Fundamental theorem of arithmetic, 269
Fundamental theorem of finitely generated
abelian groups, 197
Galois group, 159
Generators, free set of, 246
Greatest common divisor, 269
Group,
abelian, 177-212
alternating, 62
automorphism, 83
commutative, 29
commutator, 112
cyclic, 101
definition of, 50
derived, 112
extension, 234, 238
finitely generated, 98, 197, 227
free, 246
isometry, 64, 67
linear, 90
matrices, 81
mixed, 188
Mobius transformations, 77
nilpotent, 142
number of groups of given order, 156
primary, = p-group, 190
Priifer groups, 191
quaternion, 151
simple, 158, 163
symmetric, 56
table, 22
torsion, 188
torsion-free, 188
Group oid, 26
abelian, 29
associative, 29
commutative, 29
equality of, 28
finite, 29
identity, 30
infinite, 29
order, 29
table of, 22
Groups of order
p, 148
p2, 147
2p, 149
8,160
Groups of order (cont.)
12, 152-156
15, 131
Groups of small order, 148-156
Hamiltonian group = quaternion group, 151
Higher central series = upper central series, 142
Homomorphism, natural, 115
Homomorphism, of a groupoid, 40
of a group, 94
Homomorphism theorem (first isomorphism
theorem), 117
Identical products, 245
Identity, of a groupoid, 30
of a group, 50
Iff = if and only if
Image, 12,13
Independence, independent, 192
Index of a subgroup, 110
Index 2,116
Indexed family, 182
Indexing set, 182
Initial segments, 261
Inner automorphism (see Automorphism.)
Intersection, 3
Intersection, of normal subgroups, 114
of subgroups, 55
Invariant subgroup = normal subgroup, 111
Invariants of finitely generated abelian group =
type of abelian group, 200
Inverse, 32
Inverse pairs, 246
Inversion, 60
Isometries, of line, 64
of plane, 67
Isomorphism,
of group, 94
of groupoid, 42
Isomorphism theorems,
first, 117
second, 125
subgroup, 125
third, 121
Jordan-Holder theorem, 164
Kernel, 117
Klein four group, 144, 148
Lagrange's theorem, 109
Largest set, 3
Least common multiple = lowest common
multiple, 269
Length of,
composition series, 164
cycle, 167
element in free group, 248
Linear fractional transformation = Mobius
transformation, 77
Linear group, 90
Mal'cev's theorem, 280

276
INDEX
Map or mapping,
Injection, 14
codomain, 12, 13
composition, 17
definition, 11, 13
domain, 12, 13
equality, 13
matching, 14
one-to-one, 14
onto, 14
restriction of, 14
Matching, 14
Matrices, 81
Matrices, groups of, 81
Maximal independent set, 193
Maximal set, 194
Metacyclic, 243
Mixed group, 188
Mobius transformations, 77
Monomorphism, 42
Multiplication in a set, 2
Multiplication table, 22
Natural homomorphism, 115
Negative, 178
Nielsen-Schreier theorem, 260
Nilpotent group, 142
Normal closure, 253
Normal subgroup, 111
Normal subgroup generated by a set, 253
Normalizes 112,133
Odd permutation, 60
One-to-one mapping, 14
Onto mapping, 14
Operation, binary, 19
Order
of a group, 50
of a groupoid, 29
of an element, 103
Ordered pair, 2
Partition, 9
p°°, group of type = Prufer group, 191
p-component, 191
Periodic = torsion group, 188
Permutation, 56
even, 60
group = symmetric group, 56
odd, 60
Permutational representation, 216
degree of, 217
p-group, 139, 190
p-primary group = p-group, 190
p-Prfifer group, 191
main theorem, 206
Preimage, 12, 120
Presentation, 253
finite, 253
finitely presented, 253
group of a, 253
of a group, 253
Primary component = p-component, 191
Product,
cartesian, 6
direct, 143
reduced, 245
Products,
different, 245
identical, 245
of subsets of a group, 109
Proper subgroup, 106
Prufer group, 191
Quaternions = Hamiltonian groups, 151
Quotient group = factor group, 114
Range, 12
Rank,
free abelian group, 193
free group, 255
torsion-free abelian group, 198
E-block, 9
K-class, 9
Reduced product, 245
Reflection, 68, 69
Reflexive property, 9
Representation,
coset, 219
degree of, 217
Frobenius, 225
permutational, 216
right-regular, 216
Representative of a coset, 219
Representatives of the equivalence classes, 134
Right-regular representation, 216
Rotation, 68, 69
Schreier's subgroup theorem, 260
Schreier transversal, 261
Schur's theorem, 242
Semigroup, 29
Series,
ascending central, = upper central series, 142
composition, 158
solvable, 158
subnormal, 158
Set, 1
Simple group, 158,163,172
Smallest set, 3
Solvable, group, 158, 161
by radicals, 159
series, 158
word problem, 246
Split, 234
Splitting extension, 234, 2S8
Steinitz exchange theorem, 192
Subgroup,
commutator, 112
conjugate, 131
definition, 54
derived, 112
generated by a set, 98
invariant, = normal subgroup, 111
isomorphism theorem (second isomorphism
theorem), 125
normal, 111

INDEX
277
Subgroup (cont.)
of index 2, 116
proper, 106
Sylow, 130
theorem for cyclic groups, 105, 126
theorem for free groups, 260
torsion, 189
Subnormal series, 158
factors of, 158
Subset, 2
subgroup generated by, 98
product of, 109
Sum, direct, 178
finite, 178
infinite, 182
Sum of vectors, 89
Sylow subgroup, 130
Sylow's theorems,
first, 130
second and third, 131
Symmetric group,
definition, 56
of degree 4, 59
of degree 5 or more, 172
Symmetric property, 9
Symmetry groups, 73
Symmetry groups of an algebraic structure, 83
Table, multiplication, 22
Three-cycle, 172
Torsion group, 188
subgroup, 189
Torsion-free group, 188
Transfer, 240
Transitive property, 9
Translation, 68, 69
Transposition, 167
Transversal, 219
Schreier, 261
Type of a finitely generated abelian group, 200
Union, 3
Unit element = identity element, 30
Upper central series, 142
Vector space of dimension n, 89
Vector sum, 89
Word problem, 246
Zero element, 178
Zorn's lemma, 194

Symbols and Notations
Symbols
ant (G) Automorphism group of G, 84
α*,β «ВШ-1, 224
A„ Alternating group of degree n, 61
С Complex numbers, 1
C* Nonzero complex numbers, 51
C(A) Centralizer of A, 112
C„ Cyclic group of order n, 148
Dn Dihedral group of degree n, 76
gp(X) Subgroup generated by X, 98
hk Conjugation by xk, 233
/ Isometries of plane, 67
1(R) Group of isometries of R, 64
/s Symmetry group of S, 73
(/: Z) Group of isometries of R that move integers to integers, 66
KA Klein four group, 148
Ker в Kernel of в, 117
Lch). (a, b) Lowest common multiple of a and 6, 269
Lniy, F) Pull linear group of dimension n, 89
'"fc.fc' т*к-хк-'233
Μ Group of MSbius transformations, 78
<JH Group of 2 X 2 matrices over complex numbers, 81
Mx Semigroup of mappings of X to X, 36
N Nonnegative integers, 1
N(A) Normalizer of A, 112
Nu (A) Normalizer of A in H, 133
Ρ Positive integers, 1
Q Rationals, 1
Q* Nonzero rationals, 20
R Real numbers, 1
R+ Nonnegative real numbers, 48
R1 Euclidean plane, 2
% Set of representatives of the equivalence classes, 134
%* Representatives whose intersection with the center is the empty se'
ii-class Д-equivalence class, 9
sv Number of distinct Sylow j>-subgroups, 131
Sn Symmetric group of degree n, 66
Sx Symmetry group on X, 66
T(G) Torsion subgroup of G, 189
W Elements of S and their inverses, 262
xk Coset representative, 232
Ζ Integers, 1
Z(G) Center of G, 112
278

SYMBOLS AND NOTATIONS
279
Greek Symbols
ye
Π
Mapping that sends χ -» xg, 219
Frobenius representation, 225
Identity mapping, 36, 56
Natural homomorphism, 114
Mapping that sends д -> уд, 219
Set of all primes, 190
«(a, b,e,d)
та,Ъ
Mapping that sends χ to xg, 214
Rotation through angle в, 68
Mobiua transformation, 77
Reflection, 69
Transfer, 240
Translation, 68
Notations
e
{ }
{ I }
<,)
( , )
с
с
и
η
0
χ
S™
xRy
χΒ
Χ/Β
Βα, я°, a(s)
Sa
\s\
a|S'
s°t
s-i
s + t
8t
sXt )
1
σ-1
Is in, belongs to, 1
Is not in, does not belong to, 1
Set, 1
Set defined by a property, 1
Open interval, 2
Closed interval, 2; also
Commutator, 112
Ordered pair, 2; also Greatest common
divisor, 269
Is a subset, 2
Is a proper subset, 2
Union, 3
Intersection, 3
Empty set, 3
Difference of sets, 4
Cartesian product, 6
Cartesian product of S η times, 6
χ is related to у by B, 8
Д-class of x, 9,11
Set of iB-classes, 11; also X
over B, 114
Mapping from S into T, 12
Image of s under a, 12
Range of a, 12
Number of elements of S, 14
Restriction of a to S', 14
Composition of mappings, 17; also Image
under a binary composition, 19
Image under a binary composition, 19
Groupoid G with binary operation μ, 26
Identity of a groupoid, 31
Inverse of the element g, 34
ΟχΟΓ
1m
am
Hg
XY
[G:H]
H<3 G
G'
GIN
hxk
H®K
η
Π G{
1=1
{alf ...,am)
ΗΦΚ
W
2 G,
2 Gt
iei
Gp
S
(X;R)
\X;R\
(«, h [a, b])
(a,b)
Notation for a mapping, 37
Isomorphic, 42
о to the power m, 100
Right coset, 108
Product of two subsets of a
group,109
Index of Я in G, 110
Η is normal in G, 111
Commutator of χ and y, 112
Commutator subgroup of G, 112
G over JV, 114
Equivalence relation by
conjugation, 134
Equivalence class containing
A, 134
External direct product, 143
Internal direct product, 146
Direct product, 146
Cycle of length m, 167
Direct sum of Η and K, 178
Finite direct sum, 178
Infinite direct sum, 182;
see also 179
p-component of G, 190
Friifer group, 191
Type of a group, 200
Transversal element in same
coset as g, 219
Presentation, 253
Group of a presentation, 253
Presentation, 254
Greatest common divisor of
a and b, 269
Congruent, 269