Introduction to Abstract Algebra 9782759829163

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Current Natural Sciences

Libin LI and Kaiming ZHAO

Introduction to Abstract Algebra

Printed in France

EDP Sciences – ISBN(print): 978-2-7598-2915-6 – ISBN(ebook): 978-2-7598-2916-3 DOI: 10.1051/978-2-7598-2915-6 All rights relative to translation, adaptation and reproduction by any means whatsoever are reserved, worldwide. In accordance with the terms of paragraphs 2 and 3 of Article 41 of the French Act dated March 11, 1957, “copies or reproductions reserved strictly for private use and not intended for collective use” and, on the other hand, analyses and short quotations for example or illustrative purposes, are allowed. Otherwise, “any representation or reproduction – whether in full or in part – without the consent of the author or of his successors or assigns, is unlawful” (Article 40, paragraph 1). Any representation or reproduction, by any means whatsoever, will therefore be deemed an infringement of copyright punishable under Articles 425 and following of the French Penal Code. The printed edition is not for sale in Chinese mainland. Customers in Chinese mainland please order the print book from Science Press. ISBN of the China edition: Science Press 978-7-03-067958-1 Ó Science Press, EDP Sciences, 2022

Preface Topics included are: Basic properties of groups; subgroups, symmetric groups, alternating groups, cyclic groups, cosets; direct products, finitely generated abelian groups, quotient groups and homomorphisms; basic properties of rings; subrings, integral domains, quotient rings and ring homomorphisms, quaternions; ideal theory, isomorphism theorems; unique factorization domains, Euclidean domains and Gaussian integers; polynomial rings, fields, field extensions, algebraic closure, finite fields. Overview and Approach: In addition to being an important branch of mathematics in its own right, abstract algebra is now an essential tool in number theory, geometry, topology, and, to a lesser extent, analysis. Thus, it is a core requirement for all mathematics majors. Outside of mathematics, abstract algebra also has many applications in cryptography, coding theory, quantum chemistry, and physics. Abstract algebra is the field of mathematics that studies algebraic structures such as groups, rings, fields, and modules; we will primarily study groups, rings and fields in this course. The power of abstract algebra is embedded in its name: it gives us an arena in which we may study disparate mathematical objects together and abstractly, without considering a particular instance or occurrence. For example, the multiplication of nonzero real numbers, symmetries of a molecule, roots

ii

Preface

of the unity, actions of a Rubik's cube, and loops on surfaces all form groups. By exploring groups abstractly, we can derive properties and structures that apply to all examples that we currently know or may discover in the future. With this in mind it should come as no surprise that abstract algebra builds a language that is used in nearly every field of mathematics. The applications of abstract algebra within and beyond mathematics are not the only reasons to study abstract algebra. First, learning abstract algebra is one of the best ways to practice working through complex concepts and to develop your abstract reasoning abilities. Second, studying abstract algebra provides a window into what it is like to do research in mathematics. Perhaps most importantly, you will experience the intrinsic beauty of mathematics during this course. While the aesthetic nature of abstract algebra is difficult to describe, it is obvious to any of its practitioners.

About this book: This book is intended as a textbook for a oneterm senior undergraduate (or graduate) course in abstract algebra, to prepare students for further readings, for example, Group Theory, Galois Theory. K. Zhao used the earlier drafts of this book as the textbook when he taught MA323 (and MA523) at Wilfrid Laurier University since 2015. L. Li used the earlier drafts as his textbook in his Abstract Algebra course since 2018. When they prepared these lecture notes they mainly took [F] as their reference. The book contains 168 exercises of varying difficulty with sample solutions. Besides standard ones, many of the exercises are very interesting. Some are rather hard. It is not a surprise if the reader cannot solve some of the exercises, particularly for first learners.

Acknowledgments: K. Zhao likes to thank Mr. Julian J. Park for his kindness of letting K. Zhao use the notes he took from Zhao's lectures in the fall of 2014. Both authors are grateful to many people, including, Junchao Wei,

Preface

iii

Yang Chen, Ying Zheng, Yan Wang, Liufeng Cao, Xinyu Zhou, for their comments and suggestions on earlier drafts of the book. The authors wish all instructors and students who use this book a happy mathematical journey they will undertake into this delightful and beautiful realm of abstract algebra.

Libin Li, Kaiming Zhao Octoher, 2020

Contents Chapter 1

Groups and Generating Sets · · · · · · · · · · · · · · · · · · · · 1

1.1

Binary operations · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · ·1

1.2

Isomorphic binary structures · · · · · · · · · · · · · · · · · · · · · · · · · · · · · 6

1.3 Groups · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · 9 1.4 Subgroups · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · 15 1.5 1.6

Cyclic groups · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · ·19 Generating sets · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · 25

1.7 Exercises · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · 27 Chapter 2 Permutation Groups and Alternating 2.1

Groups · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · ·31 Permutation groups · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · ·31

2.2

Alternating groups · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · 38

2. 3 Exercises · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · 43 Chapter 3 Finitely Generated Abelian Groups and

Quotient Groups · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · 45 3.1 3.2

The theorem of Lagrange · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · 45 Finitely generated abelian groups · · · · · · · · · · · · · · · · · · · · · · · 48

3.3 3.4

Properties of homomorphisms · · · · · · · · · · · · · · · · · · · · · · · · · · ·57 Quotient groups and isomorphism theorems · · · · · · · · · · · · · 60

3.5

Automorphism groups · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · 67

3.6

Simple groups · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · 69

3. 7

Exercises · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · 74

Contents

vi

Chapter 4 4.1 4.2

Rings, Quotient Rings and Ideal Theory · · · · · · 78

Basic definitions · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · 78 Integral domains · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · ·84

4.3 Noncommutative rings · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · 88 4.4 Quaternions · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · 95 4.5 Isomorphism theorems · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · ·101 4.6 Euler's theorem · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · 107 4. 7 Ideal theory · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · 109 4.8 Exercises · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · 115 Chapter 5 Unique Factorization Domains · · · · · · · · · · · · · · · ·119 5.1 Basic definitions .. · .... · .... · .. · .... · .... · .... · .. · .... ·119 5.2 Principal ideal domains · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · ·122 5.3 Euclidean domains · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · 125 5.4 Polynomial rings over UFDs · · · · · · · · · · · · · · · · · · · · · · · · · · · 129 5.5 Multiplicative norms · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · ·134 5.6 Exercises · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · 138 Chapter 6 Extension Fields · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · 141 6.1 Prime fields and extension fields · · · · · · · · · · · · · · · · · · · · · · · 141 6.2 Algebraic and transcendental elements · · · · · · · · · · · · · · · · · 145 6.3 Algebraic extensions and algebraic closure · · · · · · · · · · · · · ·152 6.4 Finite fields · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · 157 6.5 Exercises · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · 162 Appendix A Equivalence Relations and Quotient Set · · ·165

Appendix B Zorn's Lemma · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · 167 Appendix C Quotient field · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · ·169 Reference · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · 170 Index························································· ·171

Notations You are supposed to be familiar with the following conventional notations throughout this course: • Z : the set of integers • N : the set of positive integers • Q : the set of rational numbers

• IR : the set of real numbers • C : the set of complex numbers

• z+ = {o, 1, 2, ... } • Q+ = { r E Q : r > 0} • JR+

=

{r

E

IR : r > 0}

• Z*

=

Z\ {0}

• Q*

=

Q\ {0}

• IR* = IR\ {0} • C*=C\{0} • Mmxn(F) :the set of all m x n matrices with entries in a field F

Chapter 1 Groups and Generating Sets The early history of group theory dates from the 19th century. Group theory has three main historical sources: number theory, the theory of algebraic equations, and geometry. The number-theoretic strand was begun by Leonhard Euler, and developed by Gauss's work on modular arithmetic groups related to quadratic fields. Galois, in the 1830s, was the first to employ groups to determine the solvability of polynomial equations. Felix Klein initiated the Erlangen Program. Sophus Lie, in 1884, started using groups (now called Lie groups) attached to analytic problems. The theory of groups was unified starting around 1880. In this first chapter we will introduce the most important concept in this course: group. Then establish some basic properties on groups.

1.1

Binary operations

Let's begin with some examples of binary algebraic structures. Example 1.1 Let U = {z E C: lzl = 1} be the unit complex circle where lzl is the norm or absolute value of z. For all z~, z 2 E U

since

then z1 z2 E U, i.e., U is closed under multiplication.

Chapter 1

2

Example 1.2

For n EN, let

Un Let w

=

= {Z E

27r . . 27r cos- + 'l smn n

Un For all Zl, Z2 E

un

Groups and Generating Sets

=

2..-i en

= { wk :

k

C : zn =

=

=

1}

~

U.

( 27ri ) exp . Then n

0, 1, · · · , n -

1} .

since

so z1 z2 E Un, i.e., Un is closed under multiplication. Example 1.3 Given r E JR.+, for rt, r 2 E JR., we define

We know that the relation

f".J

is an equivalence relation on JR., and the

equivalence classes, containing s E JR., is [s] Rr = lR/ ""=

=

s

+ r.Z.

{[s] : s E lR} is the quotient set oflR by

f".J.

We know that The addition

and multiplication below are well defined:

(1) [s]

+ [t] = [s + t],

(2) [s] · [t]

=

[st].

Then 1Rr is closed under the above addition and multiplication. Example 1.4

{r + nk: k

E

Given n EN, forr E Z, we define [r] = r+nZ =

Z}. Note that [n + 1] = [1]. Let Zn = {[r] : r

Zn

=

E

Z}. Then

{[0], [1], · .. , [n- 1]}.

We define:

(1) [r] + [s] = [r + s], (2) [r] · [s] = [rs]. Then Zn is closed under the above addition and multiplication. It

is quite often to denote [r] as ambiguity.

r,

or even simply as r if there is no D

1.1

Binary operations

3

Example 1.5 Describe the rotations and reflections that carry an equilateral 6.ABC onto itself (see Figure 1). Why is a rotation not a reflection? Solution First we have three rotations: (1) The identity map r0 : A 1---t A, B 1---t B, C 1---+ C; (2) r 1 : A 1---t B, B H C, C H A; (3) r 2 : A H C, C H B, B H A. Then we have three reflections with respect to the heights ha, hb, he: (1) SA: A H A,B H c,c H B; (2) SB : B 1---t B, A H C, I-+ A; (3) sc : C H C, A H B, B H A. Then the set of all symmetries of 6.ABC is G = {r0 , r1, r 2 , sA, sB, sc}. We can have compositions of elements in G. For example, if you first apply r 1 and then apply sA, you obtain sB, since

c

A~B~C,

B~C~B, C~A~A.

Later you will see that this G is the dihedral group D 3 .

c

Figure 1

symmetry of equilateral triangle

The above examples involve binary operations. In general:

D

Chapter 1

4

Definition 1.1 a map

Groups and Generating Sets

A binary operation* on a nonempty setS is

S x S--+ S, (a, b)

t-+

a* b.

A setS with a given binary operation* is called a binary algebraic structure, or binary structure, denoted by (S, *). Example 1.6 It is easy to see that (IR, +), (Z,-), (U, ·), (Un, ·), and (Z, ·) are all binary algebraic structures. However, (IR,-;-) is not a binary algebraic structure due to potential division by zero. Definition 1.2 Let* be a binary operation on S and H ~ S. We say that H is closed under * if a * b E H for all a, b E H. This is also called the induced operation * on H. In this case, (H, *) is called a substructure of (S, *), denoted by H Example 1. 7 Let

Is H closed under matrix operations

~

S.

+ and · ?

Solution It is easy to see that H is closed under matrix operation +. Let's do the case for multiplication. Take

Then

X .y

=

(a b) (c d) b a d c

=

be) .

(ac + bd ad+ bc+ad bd+ac

Clearly, this is in H. Therefore, His closed under +and ·.

D

Remark 1.1 For sets A, B, we have the set BA = {all maps from A to B}. Note that IBAI = IBIIAI where IAI is the size or cadinality of

A.

1.1

Binary operations

5

Example 1.8 Consider the set JRR. For f, g E JRR, define (1) (f ± g)(x) = f(x) ± g(x); (2) (f · g)(x) = f(x)g(x); (3) (f o g)(x) = f(g(x)). It is easy to verify that JRIR is closed under the above opemtions. Example 1.9 Say 2 * 5 = 2, 3 * 3 opemtion.

On N, define a* b = min {a, b} for all a, b E N. =

3. It is easy to verify that N is closed under the

Definition 1.3 A binary opemtion * on a set S is commutative if a* b = b *a for all a, b E S. It is associative if (a* b) * c = a* (b *c) for all a, b, c E S. Remark 1.2 a1 *a2*· · •*an is well-defined for any a1, a2, · · · , an E S, if * is associative. Theorem 1.4 Let f,g, hE 8 8 . Then (fog) o h = f o (go h). In other words, the composition of maps is associative. Proof It is easy to verify. D Example 1.10 LetS= {a,b,c} and* be a binary opemtion defined on S. The opemtion can be represented in a table like below:

*

a

b

c

a b

a*a a*b a*c b*a b*b b*C

c

C*a

C*b

C*C

Example 1.11

LetS= {a, b, c, · · · , z} . How many different binary opemtions on S can you define? Solution By definition 1.1, a binary operation* is an element in ssxs. Therefore, we have that

So there are 2626 x 26 different binary operations on S.

D

Chapter 1

6

1.2

Groups and Generating Sets

Isomorphic binary structures

We will study the following questions: How to compare two binary algebraic structures? When are they essentially the same?

Definition 1.5 Let (8, *) and (8', *') be two algebraic structures. A map cp : 8 ---7 8' is a called homomorphism from 8 to 8' if

cp(x * y) = cp(x) *' cp(y),

Vx, y E 8.

A one-to-one and onto homomorphism from 8 to 8' is called an isomorphism from 8 to 8'. If there exists an isomorphism from (8, *) to (8', *'), we say that 8 and 8' are isomorphic, denoted by 8 ~ 8', or

(8,*) "'(8',*'). Example 1.12

Un

= { wk : k =

For n EN, recall that 0, 1, · · · , n -

We will show that= (Un, ·)

~

1} , Zn = {0, I,··· , n - 1} .

(Zn,+). It is clear that the map

is one-to-one and onto. Moreover, for all k~, k2 E Z, we have

Therefore cp is an isomorphism from Un to Zn· Note that w can be replaced by wr, for any r with gcd(r,n) = 1. D Theorem 1.6 Suppose that Then the following hold.

8~, 8 2

and 83 are binary structures.

(1) Reflexivity: 81 "'81. (2) Symmetry: If 81 "' 82, then 82 "' 81. (3) Transitivity: If 81 "' 82 and 82 "' 83, then 81 "' 83. In other words, "' is an equivalence relation on any collection of binary structures.

1.2

Isomorphic binary structures

7

Proof (1) This is clear. (2) Since 8 1 "" 8 2 , there exists an isomorphism cp : 8 1 ---+ 82 . Therefore cp- 1 : 8 2 ---+ 8 1 is also one-to-one and onto. For all x, y E 82 , we need to verify that cp- 1 (x*y) = cp- 1 (x)*cp- 1 (y). There exist x', y' E 8 1 such that cp(x') = x, cp(y') = y. Then x' = cp- 1 (x), y' = cp-1 (y). Since cp is an isomorphism, so cp(x' * y') = cp(x') * cp(y') = x * y. Thus

Then cp-1 is an isomorphism from 8 2 to 81, i.e., 8 2 "'81.

(3) Let 'P1 : 81---+ 82, 'P2: 82---+ Sa be isomorphisms. Then cp2ocp1: 8 1 ---+ 8 3 is a bijection. The homomorphism property is easy to verify. D Show that (R, +) "'(JR+, ·). Proof Construct a map cp : R ---+ JR+, r t---t er. We can find the inverse cp-1 : JR+---+ IR, x t---t lnx. Example 1.13

So cp is one-to-one and onto. Now we verify the homomorphism property. For rl? r2 E IR, cp(r1 + r2) = er1 +r2 = er1 • er2 cp is an isomorphism from (R, +) to (R+, ·).

Example 1.14 are not isomorphic.

=

cp(r1) · cp(r2). Thus D

Show that the binary structures (Q, +) and (IR, +)

Proof This follows from the fact that

IQI = ~0 =1- c = IIRI.

D

Example 1.15 Show that the binary structures (Z, +) and (z+, +) are not isomorphic.

Proof By Contradiction, for any c E Z, the equation x + y = c E Z has infinitely many solutions. Choose c = 0 and consider the equation x + y = 0 in z+, which has only one solution. If there exists

Chapter 1

8

Groups and Generating Sets

an isomorphism cp: (Z, +) -+ (z+, +),then

cp(x) + cp(y)

=

cp(x + y)

=

cp(c)

would also have infinitely many solutions. Clearly, this is a contradic-

D

tioo.

Example 1.16

Show that the binary structures (Z, ·) and (z+, ·)

are not isomorphic. Proof The equation x · x = 1 has two solutions in (Z, ·). Note that for all c E z+, the equation x · x = c has at most one solution in

(z+, ·).

o

Let (S, *) be a binary structure. An element e of S is an identity element if e * x = x * e = x, for all xES. Theorem 1.8 Any binary structure (S, *) has at most one identity element. That is, if it has one identity element, then it is unique. Proof Suppose that both e and e' are two identities in S. Then Definition 1. 7

we know e = e' * e = e'.

D

Let (S, *), ( S', *') be two isomorphic binary structures. If S has an identity, then so does S'. Proof Since S ,...., S', there exists an isomorphism cp : S -+ S'. Let e be the identity element of S. We claim that cp(e) is the identity in S'. For all x E S', there exists y E S so that cp(y) = x. Then Theorem 1. 9

x

*' cp(e) = cp(y) *' cp(e) = cp(y *e) = cp(y) = x.

Similarly, cp(e)

*' x =

x. Therefore, cp(e) is the identity element for S'. D

Example 1.17

Show that the binary structures

(~,

·) and (C, ·)

are not isomorphic. Proof The equation x · x = c, for all c E C, has solutions inC, while the equation x · x = -1 does not have any solution in ~. D

1.3

Groups

9 X

Example 1.18 Let

: G -+ G' is a homomorphism if it is a homomorphism of the binary structures, i.e., ¢>(ab) = ¢>(a)¢>(b), for all a, bE G.

Two groups (G, ·a) and (H, ·H) are isomorphic if they are isomorphic as binary structures, denoted by G

Example 1.25 and 3.

rv

H.

Determine all nonisomorphic groups of order 2

Solution Let IGI = 2. We know that e E G. Let G = {e, a}. Assume that the multiplication table is written as :

t: a X

If x

= a,

then aa

=a=

ae ::::} a

=

e (by Theorem 1.11). This

is impossible. So X = e. u2 = {1, -1} is a group. If we map 1 t-+ e and -1 t-+ a, we have an isomorphism. Note that (Z 2 , +) (U2 , ·).So l"oJ

there is exactly one group of order 2. From cancellation rule, we know that each row or column in a group product table cannot have equal element. Now, let G be a group with order 3 and G = { e, a, b}.

14

Chapter 1

e e a b

a

b

e a a x b

b

Groups and Generating Sets

Sox= e or x =b. Suppose that x =e. Then

e a b

e a e a a e

b b

b

b

This is impossible. Sox= b.

e a b

e e a b

a a b e

b b

e a

We know that (Z3 , +) "' (U3 , ·) is a group of order 3. So there is one group of order 3. D

Show that any two of the groups are not isomorphic: (U,·), (R,+), (R*,·). Solution Consider the equation x * x * x * x = e, which has: (1) 4 solutions in U; (2) 1 solution in (R, +) since 4x = 0 {::::::::} x = 0; (3) 2 solutions in (R*, ·). So any two of them cannot be isomorphic. D Example 1.26

Let G be a group of even order. Show that there exists a E G such that a -=/:- e and a 2 = e. Proof It is easy to see that a 2 = e if and only if a = a- 1 . Thus we need to find such an element. Note that the cardinality Example 1.27

Subgroups

1.4

15

Let IGI = 2n. Then let { a1, a1 1 } , · • · , { ar, a;:-1 } be all distinct such subsets of G. Then G is a disjoint union and equal to r

G=

U {ai, ai

1

}.

i=l

We may assume that

We deduce that r

r

i=l

i=2

Thus there exists io E {2, · · · , r} so that I {ai0 , ai;; 1 } I = 1. Then aio = ai;; 1, i.e., a~0 = e and aio # e. Take a = aio and we have the desired result. D

1.4

Subgroups

Assume that (G, ·) is a group. If G is commutative, we generally use "+" to denote the operation, and denote e = 0 :

a·a···a---ta+a+···+a=na

'-.....--' n

n

in (G,+), the additive group. We also have that

(1) -na =(-a)+···+ (-a); n

(2) {0} is the identity group.

16

Chapter 1

Groups and Generating Sets

Definition 1.18 A subset H ~ G is called a subgroup of G if H is a group with the same opemtion as G, i.e.,

(1) H is closed under"·", (2) e E H, (3) a- 1 E H, for all a E H. We denote this by writing H ofGare{e},G.

~

G (or G

~

H). The trivial subgroups

We have the obvious meaning for H < G or G >H.

(Z, +) < (Q, +) < (IR, +); SLn(IR) < GLn(IR);

Example 1.28

(Q+,·) < (Q*,·) < (IR*,"). Definition 1.19 ForK, H

~

G, we define the inverse of H and

the product of two sets as H-1 ={h-1

:

heH} andHK={hk: heH,keK}.

By the definitions of group and subgroup, it is easy to prove the following theorem.

Theorem 1.20 following hold:

Let H

~G.

Then H

~

G if and only if all the

( 1) H H ~ H (i.e., H is closed under its own product); (2) e E H; (3) H- 1 ~H. Corollary 1.21 Let H ~ G and H =J 0. Then H ~ G if and only if HH- 1 ~ H, i.e., ab-1 E H for all a,b E H. Proof (::::}) This direction is clear.

( w E (wk) ¢::::::> :3 l E Z such that wkl = w ¢::::::> wkl- 1 = 1 ¢::::::> nlkl - 1 ¢::::::> :3 q E Z such that kl - 1 = nq D ¢::::::> gcd( k, n) = 1.

Example 1.31 Let On(IR) ={A E Mnxn(IR) : AAt =In}. Show that On(IR) ~ GLn(IR). Proof Since In E On(IR), we know that On(IR) # 0. Let A, B E On(IR). We compute

AAt =I* IAI·IAtl = 1

*

2

IAI = 1 * IAI = ±1,

yielding that A E GLn(IR). So On(IR)

~

GLn(IR). We compute

(AB- 1 )(AB- 1 )t = AB- 1 (B- 1 )t At= AB- 1 (Bt)-1 At = AB- 1 (B- 1)- 1At = AIA- 1 =In. Thus AB- 1 E On(IR). We see that On(IR) ~ GLn(IR).

Definition 1.25 Z( G)

D

The center of a group G is defined as

= {a E G

: ag

= ga,

Vg E G} ~ G.

Let a E G. The centralizer of a in G is Za (a)

= {

x E G : ax

=

xa}

~

G.

The verification of this definition is left as Exercises (17-18). Lemma 1.26 Let Hi ~ G fori E I, where I is an index set.

Then

Proof We have e E Hi for all i. So e E niEI Hi # 0. Now let x, y E niEIHi. Then x, y E Hi for each i. So xy, x- 1 E Hi since Hi ~ G. We deduce that xy, x- 1 E niEIHi. Thus, by definition, niEIHi ~G.

o

1.5

Cyclic groups

1.5

19

Cyclic groups

Let (G, ·,e) be a group and a E G. Recall that l(a)l or ord(a) is the order of a. Lemma 1.27 Let a E G with ord(a) = n. (1) The order ord(a) of a is the least positive integer n such that an= e.

(2) If a"

= e,

then nlk. Proof (1) If for any integers k < l we have ak =/= a', then (a) would be an infinite set, contradicting the assumption that ord(a) = n. There are integers k < l such that a" = al, i.e., ak-l = e. Let m be the least positive integer m such that am= e. For any k E Z, write k = mq + r where 0 :::;;; r < m. Since am = e then So (a)= {e,a,a 2 , • • • ,am-1 }. Thus n = m. (2) Write k = qn + r with 0 ~ r ~ n- 1. Thenar = e, yielding

r

= 0, i.e., nlk.

D

Every cyclic group is abelian. Proof Let G be a cyclic group, say G = (a). Let x, y x = an,y =am for m,n E Z. Then Theorem 1.28

E

So G is abelian.

D

Recall the following. Lemma 1.29 (Division Algorithm)

Let m, n

E

Z with m =/= 0.

Then there exist unique q, r E Z, so that n = mq + r, Theorem 1.30

G, say

0 :::;;; r < lml.

Any subgroup of a cyclic group is cyclic.

20

Chapter 1

Groups and Generating Sets

Proof Let G = (a) be a cyclic group and H ::::;; G. We need to show that H is cyclic. If H = {e} , then clearly H = (e) is cyclic. Now suppose that H # {e} . Then there exists n E N so that an E H. Let m be the smallest positive integer with am E H. We will show that

H =(am). Let an E H. By the division algorithm, there exist q, r E .Z, 0 ::::;;

r < m so that n

and (am)-q

=

=

mq+r. Then

a-mq E H, ar = an(am)-q E H. If r

#

0, this would

contradict the choice of m. So r = 0, i.e., an= (am)q E (am).

We see that H::::;; (am) ::::;; H. Thus H

= (am).

D

Corollary 1.31 The subgroups of (.Z, +) are precisely of the form (n) = (n.Z, +) for some n E .z+. In particular, (0} = {0} and

(1}

=

.z.

Theorem 1.32 Let k1, k2, · · · , kr E .Z. (1) The nonnegative integer d which is the genemtor of the cyclic group {a1k1 + a2k2

+ · · · + arkr: ai E .Z}

is gcd(k1, k2, · · · , kr)·

(2) There are ai E .Z such that d = a1k1 + a2k2 + · · · + arkr. Proof (2) is clear. (1) Since ki E {a1k1 + a2k2 + · · · + arkr: ai E .Z} = (d), we see that dlki for each i. Let d'lki for each i. Since there are ai E .Z such that d = a1k1 + a2k2 + · · · + arkr. D It follows that d'ld. Therefore d = gcd(k1, k 2, · · · , kr)· If gcd(k1, k2 , • • • , kr) = 1, then k1, k 2 , • • • , kr are said to be co-

prime, or relatively prime. Lemma 1.33 Let r, s E .Z. Then gcd(r, s) = 1 if and only if there exist a, b E .Z so that ar + bs = 1.

1.5

Cyclic groups

21

Lemma 1.34 Let r, s, t E Z and gcd(r, s) = 1. If rlst, then rlt. Proof Since gcd(r, s) = 1, there are a, bE Z such that ar + bs

So we have art + bst

=

= 1.

t. We see that rlt.

D

Theorem 1.35 Let G =(a) be a cyclic group. (1) If IGI = oo, then G rv (Z, +).

(2) If IGI

n, then G rv (Zn, +). Proof Case 1 Suppose that am=/:- e, for all m E N. We claim that am= an , 0, So m- n = 0. So m = n. The claim follows. Note that G = {ak : k E Z}. Define a map t.p : G --+ Z, ak t---+ k. So t.p is 1-1 and onto. For t.p to be an isomorphism, we verify the homomorphism property. For r, s E Z, =

Thus t.p is an isomorphism from G--+ Z, i.e., G ~ (Z, +).

Case 2 There exists r E N so that ar = e. Let n E N be smallest with an = e. We claim that G = {e, a, a2 , • • • , an-1 }, IGI = n. Let am E G be an arbitrary element. Then m = qn + r, 0 ::::;; r ~ n - 1. If follows

Thus G = {e,a, · ·. ,an- 1} . If ah = ak for 0 ::::;; h, k ::::;; n- 1, we may assume that h ::::;; k. So ah-k =e. Notice that 0 ::::;; h- k ::::;; n- 1, we see that h- k = 0, i.e., h = k. So

IGI = n, the claim follows.

22

Chapter 1

Groups and Generating Sets

Define a map cp : G ~ (Zn, + ), ak 1---t k, k E Z. Then it is easy to show that cp is well-defined and cp is 1-1 and onto. Recall that

k = { k, k ± n, k ± 2n, · · · } = [k] and Zn =

{0, I,··· , n- 1}.

~Z

Now we verify the homomorphism pro-

perty for cp to be an isomorphism:

Thus cp is an isomorphism from G to (Zn, +),i.e., G"' (Zn, + ).

D

Let G =(a) with IGI = n. Then n (1) (as) = (agcd(n,s))' I(as) I = ; gcd(n, s) (2) (ar) = (a 8 ) {::::::::} gcd(n, s) = gcd(n, r); (3) G = (as) {::::::::} gcd( n, s) = 1. Proof (1) We will show that they are both subsets of each other. Let d = gcd(n, s). There exists d = xn + ys for x, y E Z, yielding that Theorem 1.36

(Note that an = e since we have order n.) So (ad) ~ (as). By definition, s = dt for some t E Z. Then as= adt = (ad)t E (ad). So (as) ~ (ad). So (as) = (ad)

= { e, ad'

a2d' ... 'a((n/d)-l)d} ~ I(as) I = ~.

(2) (¢::)If d = gcd(n, r) = gcd(n, s), then (ar) =(ad)= (as). (=}) If (ar) = (as), then

n 8 I(ar) I = I(a ) I = } gcd ( n,r )

-

n d( ) gc n,s

=}

gcd(n, r)

=

gcd(n, s)

as required. (3) G=(a1 )=(ar)

{=?

gcd(1,n)=gcd(n,r)

{=?

gcd(n,r)=l. D

1.5

23

Cyclic groups

The Euler cp-function cp: N--+ N is defined

Definition 1.37 by

cp (n) =

I{k E { 1, · · · , n} : gcd (k, n) = 1} I·

It is clear that cp(1)

= cp(2) = 1, cp(3) = 2.

Let n = p~ 1 p;2 • • • p~•, where Pb · · · , Ps are distinct primes, and T1, • • • ,r8 EN. Then

For example, cp(20) = cp(22 5) = 20(1- 1/2)(1- 1/5) = 8. Question 1.38 Let G = (a), IGI = n. (1) How many subgroups does G have? (2) How many generators does G have? Solution (1) From Theorem 1.36 (1), we know that the number of subgroups of a group G is the number of different factors of n, which is denoted by d(n) in number theory. Let n = PI1P~2 • • • p~·, where Pb · · · ,p8 are distinct primes, and TJ, • • • , T 8 EN. Then we have the formula

d(n) = (r1 + 1)(r2 + 1) · · · (rs + 1). (2) From Theorem 1.36 (3), we know that the number of generators of G is given by the function cp(n). D Let G =(a) with IGI = n. Now we interpret Theorem 1.36 in terms of (Z11 , +) by using the isomorphism

cp : G --+ (Zn, +),a k Remark 1.3

1---t

-

k,

kEZ

(1) All the subgroups of (Z11 , +) are given by

{ ((d), +) : 1 ~ d ~ n, din}, -

where I{d) I =

n

d

in Zn.

Chapter 1

24

Groups and Generating Sets

(2) For s E N, (s) = (d) in Zn, where d = gcd(n, s). Moreover, G = (s) ~ gcd(s, n) = 1 and n (3) I(s)l = gcd(s, n)" Definition 1.39(Exponent of Group) The exponent of a group G is the smallest integer n such that an = e for all a E G. Example 1.32 Let p, q be primes. Find the number of generators & subgroups of (Zpq, +). Solution Case 1: p =/:- q. We have

'P(pq)=pq(1-~) (1-~) =(p-1)(q-1). So, Zpq has (p- 1)(q- 1) generators. For the subgroups, we use

d(pq) = (1 + 1)(1 + 1) = 4, which corresponds top, q,pq, and 1. Thus the subgroups of Zpq are (1), (p), (q), and (0). Case 2: p = q. We have

So, Zp2 has r? - p generators. For the subgroups, we use

which corresponds to p,p2 and 1. Thus (1), (p), and (0) are all subgroups of Zpq

=

ZP2.

D

Example 1.33 Let (G, ·) be a group such that aEGis the only element of order 2. Show that a E Z (G).

1.6

25

Generating sets

Analysis We need to show that xa = ax for any x E G, that is, xax- 1 =a. Now we have to use the property that a is the only element of order 2. Thus we need to show that xax- 1 has order 2, deducing that x ax- 1 =a. Proof We have

If xax- 1 = e, then a= e, which contradicts the fact that ord(a) = 2. So xax- 1 -:/:- e and ord(xax- 1) = 2. Since a is the unique order 2 element, we must have xax- 1 =a, i.e., xa =ax, for all x E G. D Example 1.34 gram).

List all the subgroups of (Z18 , +) (Hasse dia-

Solution All factors of 18 are: 1, 2, 3, 6, 9, 18. The subgroups are

(1), (2), (3),(6), (9), (18).

(1) (2) / "

(6)

" (3) /1

(9)

\ I

(0)

1.6

Generating sets

Let (G, ·) be a group with order IGI. The order of aEGis ord(a)

=

l(a) I· Definition 1.40 LetS ~ G. The smallest subgroup of G containing S is the subgroup generated by S, denoted by (S). If G = (S), then we say that S generates G, or S is a generating set of G. If G is generated by a finite set, then G is said to be finitely generated. In other words, (S) = H.

n

S~H~G

Chapter 1

26

Example 1.35

0

~

Groups and Generating Sets

The empty set generates the identity group, i.e.,

{e} = (0).

Theorem 1.41

(S)

LetS~

= { a~

1

• • •

Proof Let K = { a~1 that K = (S).

G. Then

a~ : r E

• • •

z+, ai

a~ : r E

E S, ki E Z}.

z+, ai E S, ki E Z} . We will show

Clearly K ~G. Since K 2 S, we see that K 2 (S). For all a1, a2, · · · , ar E S, k1, k2, · · · , kr E Z, we see that

So K

c;, (S). Thus K = (S).

Corollary 1.42 Then

D

Let (G, +) be an additive group and S

~

G.

In the following we shall write (Zn, +) = {0, 1, · · · , n- 1}.

Theorem 1.43 Let r1, r 2 , • • • , rk E Z. (1) In (Z,+) we have (rbr2,··· ,rk) = (gcd(r1,r2,··· ,rk)).

(2) In (Zn, +) we have (r1, r2, · · · , rk) Proof Let d = gcd(rb r2, · · · , rk)·

=

(gcd(rb r2, · · · , rk, n)).

(1) It follows from Theorem 1.32. (2) In (Z, +), we have

(r1, r2, · · · , rk)

=

{a1r1 + a2r2 + · · · + akrk: ai E Z}

= (gcd(r1, r2, · · · , rk)). Then, in (Zn, +) we have

(d)= (gcd(d, n)) = (gcd(r1, r2, · · · , rk, n)) by Theorem 1.36.

D

1.7

Exercises

27

Example 1.36 List the elements of the subgroup generoted by (1) the subset {12, 18, 14} in Z 36 ; (2) the subset {12, 45, 39} in Z. Solution (1) Since gcd(12, 18, 14, 36) = 2, we see that

(12, 18, 14) in

= (2) = {0, 2, ...

'34}

z36·

(2) Since gcd(12, 45, 39)

1.7

=

3, we have (12, 45, 39)

=

(3)

=

3Z in Z. D

Exercises

(1) Show that 8(2, lR) is a group with respect to matrix multiplication where

S(2,R) ={A= (:

~) E M2x2(R)

: a+ b = c+ d = l,detA # 0}.

(This group is called the stochastic group and the matrices in 8(2, JR) are called stochastic. We can similarly define the stochastic group 8(n,lR) for any n EN). (2) A nonempty set G has a multiplication that is associative and allows left cancellations. If there exists a E G such that x 3 = axa for all x E G, show that G is an abelian group with this multiplication. (3)Let Q[J5]• ={a+ bJ5 =I 0: a, bE Q}. Show that Q[J5]• is a group with usual multiplication. (4) We know that (U6 , ·) (Z6 , + ). Is there an isomorphism of U6 with z6 mapping exp(7ri/3) to 4? (5) Determine whether the binary operation* defined as follows is commutative and whether *is associative. (a) * defined on Q by letting a* b = ab/2, (b) * defined on N by letting a * b = 2ab, f".J

28

Chapter 1

Groups and Generating Sets

(c) *defined on N by letting a* b = ab. (6) Let H be the subset of M2 (IR) consisting of all matrices of the form ( ab -ab)'

Va,b E IR.

Show that H is closed under both matrix addition and matrix multiplication. (a) Show that (C, +) is isomorphic to (H, + ). (b) Show that (C, ·) is isomorphic to (H, ·). (We say that H is a matrix representation of the complex numbers

c.) (7) Let (G, ·,e) be a group and a, bE G. Suppose ord(a) = ord(b) ord(ab) = 2. Then show that ab = ba. (8) Let (G, ·,e) be a group and a, b E G. Show that ord(ab)

=

=

ord(ba).

(9) Let (G, ·,e) be a group and a, b E G. Suppose that ord(a) = m, ord(b) = n, ab = ba, (m, n) = 1. Show that ord(ab) = mn. (10) Let (G, ·,e) be a finite group and n > 2. Show that the number of elements in G with order n is even. (11) Show that a group G cannot be the union of two proper subgroups, in other words, if G = H U K where H and K are subgroups of G, then H = G or K =G. Equivalently, if Hand K are subgroups of a group G, then H UK cannot be a subgroup unless H ~ K or K~H.

(12) Let A= {1, 2, · · · , 11}. (a) Find the number a of different binary operations on A. (b) Difine a group on A. (13) (a) Find a generator of the subgroup A = (352, 880, 1232) of the additive group (.Z56o, +). (b) Find the order of A.

1.7

29

Exercises

(c) Find a generator of the subgroup B = (616, 440, 280) in the integer group (Z, +). (d) Can you give another generator for the subgroup B of (Z560 , +)? (e) Can you give another generator for the subgroup A of (Z,+)? (14) Show that the following two groups are not isomorphic: (Q, +, 0) and (Q*, ·, 1). (15) List the elements of the subgroup generated by (a) the subset {42, 30, 66} in (Z35, + ), (b) the subset {18,24,39} in (Z,+). (16) Let G be a finite group, A, B be nonempty subsets of G with

IAI + IBI > IGI. Show that AB =G. (17) Let G be a group. Prove: Z(G) ~G. (18) Let (G, ·,e) be a group and let a E G. Show that Za(a) ~G. (19) Describe all the elements in the cyclic subgroup of GL(2, IR) generated by the given 2 x 2 matrix. 0 -1) (a) ( -1 0 '

b (1 1) ( ) 0 1 '

(c)

(3 0) 0 2 '

d ( 0 - 2) ( ) -2 0 ·

(20) Show that a group (G, ·,e) that has only a finite number of subgroups must be a finite group. Solution For any a E G, if (a) is infinite, then (a) "' (Z, +) which has infinitely many subgroups nZ for n E N. So (a) is finite for any a E G. LetS= {(a) :a E G} which is a collection of subgroups of G. From the given conditions we know that S is finite and each subgroup inS is finite. Thus G = UnEsH is finite. D (21) Let (G, ·)be an algebraic binary structure such that G is finite and the multiplication satisfies the associativity and the cancellation rules. Show that G is a group. (22) Let G be a finite group, and suppose that for any two subgroups H and K either H ~ K or K ~ H. Prove that G is cyclic of prime power order pk.

30

Chapter 1

Groups and Generating Sets

(23) Let (G, ·,e) be a finite abelian group of order n, and G = {a~, a2, · · · , an}· Prove that (a1a2 · · · an) 2 = e.

(24) Let G = Ql \{~}.Define a operation* on G:

for a, b E G. Prove that G is a group. (25) Let G be a group and n E N. Suppose that a E G and (ord(a), n) = 1. Find all x E (a) such that xn =a.

Chapter 2 Permutation Groups and Alternating Groups In this chapter we will introduce the most important classes of groups: symmetric groups and alternating groups. Then we establish some basic properties on them.

2.1

Permutation groups

We have learned a lot of examples of abelian groups: (Q, +),(JR.,+), (C, +), (Zn, +), (Q*, ·),(JR.*,·), and(C*, ·). Some non-abelian groups include GLn(R) and SLn(R) for n > 1. Definition 2.1 A permutation on a nonempty set A is a

bijective map cp : A --+ A. The set of all permutations on A is denoted by SA (or Sym(A)). If cp,'lj; E SA, then both cp-1 , cp o 1/J E SA· We will simply write c.p o 1/J = c.p'lj;. Associativity is well-defined and is given by (f o g) o h = f 0 (g 0 h). We will learn how to compute product of permutations in the following example. Example 2.1 Let A= {1, 2, 3, 4, 5, 6}. For a- E SA, we denote (two-row form)

32

Chapter 2

Permutation Groups and Alternating Groups

1

2

. ..

6 )

( a(1) a(2) · · · a(6)

·

If

and

u=G

2 3 4 5 5 4 2 1

then

23 456)

lTT

=

(

1 6 5 4 2 1 3

= (

6 1 2 5 3 4

1 2 3 4 5

6 1 2 5 3

!) ,

where the first two rows is T and the last two rows is a. Theorem 2.2 Let A #0. Then (SA, o) is a group. Proof The operation o is clearly a binary operation on SA. The identity map idA E SA =I 0. We know that o is associative, and idA is the identity since

For

f

f-

of= idA. So (SA, o) is a group.

1

E SA, we have the inverse map

f- 1

E SA such that

f

o f- 1

=

D

Definition 2.3 Let A= {1, 2, · · · , n}. The group Sn =(SA, o) is called the symmetric group on n letters.

We claim that ISnl = n!. In fact, lT E Sn is determined by lT(l), lT(2), · · · , lT(n). Then lT(1) has n choices, lT(2) has n- 1 choices, and in this manner, the last element lT(n) only has 1 choice. Multiplying them will generate ISn I = n!. For n ~ 3, Sn is not abelian.

2.1

Permutation groups

Example 2.2 permutations :

We know that 183 1= 6 and 83 has all the following

c D· c23) c23) G !) ' c23). G23) c23)

G23) 2 3

and

33

'

2 3 1

x=

2 1 3

y=

2 1

'

2 1 3

'

3 2 1

Then we have that

yx~

3 1

=1-

3 1 2

=

xy.

So 8n is not commutative for n ~ 3. Lemma 2.4 Let ¢ : G --+ G' be a group homomorphism. (1) ¢(e) = e'. (2) ¢(a- 1) = ¢(a)- 1, for all a E G. (3) cp(G) ~ G'. (4) If 'P is injective, then G"" cp(G). Proof (1) Since ¢(e)e' = ¢(ee) =¢(e)¢( e), we see that ¢(e) (2) We know that

=

e'.

Thus ¢(a- 1 ) = ¢(a)-I, for all a E G.

(3) e' = cp(e) for a, b E

cp(G) =/:- 0. Let x, y G. We see that E

E

cp(G), i.e., x

=

cp(a), y = cp(b),

Thus cp( G) ~ G'. (4) Consider a function '1/J : G --+ 'P (G), g f-t 'P (g), for all g E G. So '1/J is a bijective map, and is a group homomorphism. So '1/J is an isomorphism, i.e., G "" cp( G). D

34

Chapter 2

Permutation Groups and Alternating Groups

Theorem 2.5(Cayley's Theorem) phic to a subgroup of (Sa, o).

Every group (G, ·) is isomor-

Proof For all g E G, define a map >.9 : G-+ G, Since >.9 o Ag-1

=

>.9-1 o >.9

where x t--+ gx,

Vx E G.

id we see that >.9

=

E

Sa.

Consider the left regular representation of G, cp : G -+ Sa,

where g t--+ >.9 ,

Vg E G.

We claim that cp is injection. In fact we have

So cp is injective. For any g, h E G, we have cp(gh)(x) = >.9 h(x) = ghx, cp(g)cp(h)(x) So cp(gh)

=

=

>.9 >.h(x)

=

>.9 (hx)

Vx E G, =

ghx,

Vx E G.

cp(g)cp(h) for all g, h E G. Therefore cp is an injective

homomorphism from G to Sa. So G"" cp(G):::;; Sa.

Definition 2.6

D

ForgE G define the map

p9 : G-+ G, x t--+ xg,

Vx E G.

The group homomorphism 'lj; : G -+ Sa, g t--+ p9 -1 : G -+ G

is called the right regular representation of G. Example 2.3 Consider D 4 , the symmetric group of a square. We have 4 reflections and 4 rotations.

D 4 has the 8 elements:

2.1

Permutation groups

35

2 3 3 2 2 3 4 3 1

.

··.

2

01 IIll I.::-"~:: IIll IIll I

...... ~ ...... .• : ·.

4

3

In 86 , let

Example 2.4

u

=

(15 22 4343 15 66) .

Find u 100 .

Solution

Start by computing

u'=G

2 3 4 5 2 3 4 5 :) = e

Then we must have that D

Let A

-:1 0 and a

E SA.

For a, b E A, we define that

a rv b -{:::::::::} b = un(a) for some n

E

.Z.

Chapter 2

36 Then

rv

Permutation Groups and Alternating Groups

is an equivalence relation on A (see Appendix A), i.e., for

a,b,c E A, we have (1) Reflexivity: a ,..., a; (2) Symmetry: a ,..., b => b rv a since b = crn(a) ¢::::::> cr-n(b) = CT-n(crn(a)) =a; (3) Transitivity: a ,..., b and b ,..., c => a rv c. Let b = ern( a), c = crm(b) = crmcrn(a) = crm+n(a). Definition 2.7 The equivalence class [a]=017 (a)={crn(a): n E Z} ~A is the orbit of a under cr. A is the disjoint union of orbits. Example 2.5

In 88 , let cr

0 17 (1) = {crn(1) : n

E

=

1 2 3 4 5 6 7 8) (3 8 6 7 4 1 5 2

.

Then

Z} = {1, 3, 6} = 0 17 (3) = 0 17 (6),

0 17 (2) = {2, 8}, and 0 17 (4) = { 4, 7, 5}. Note that the union is the set

{1,2,3,4,5,6, 7,8}. Definition 2.8 A permutation cr E Sn is a cycle if it has at most 1 orbit containing more than 1 element. The length of a cycle cr is the number of elements in the largest orbit, which is also the order

ofcr: lcrl. 1 2 3 4 ) . This is a cycle 2 3 1 4 since Ou(1) = {1, 2, 3} and Ou(4) = {4}. We denote cr = (1, 2, 3) = ( 1 2 3) , and we know that 2. Show that H = An. [Hint: Show that H contains all squares of elements in Sn.] (11) If a E 8n is an (n- 1)-cycle show that its centralizer is (a). (12) Show that Sn = ((1, 2), (2, 3), · · · , (n- 1, n)). (13) Show that Sn = ((1, 2), (1, 2, 3, · · · , n)). (14) If H is a subgroup of 8 5 that contains a transposition and a 5-cycle, show that H = 85 . Can we change 5 to any n ~ 5 in the statement? (15) Find the number of elements of order p in 8P+1 where pis a prime. (16) What is the maximal order of elements in 88 . (17) What is the largest order of a permutation in 8 15? Find a permutation that has such an order. (Even you can try to determine how many different cyclic subgroups of this order.) (18) What are the possible orders of elements in 87 ?

Chapter 3 Finitely Generated Abelian Groups and Quotient Groups In this chapter we will provide the fundamental theorem for finitely generated abelian groups, introduce factor groups, simple groups and some related results.

3.1

The theorem of Lagrange

Assume that (G, ·,e) is a group, H :s; G. Recall that we can define the product of two subsets in G.

Definition 3.1 Let g E G. We say that gH = {g} Hand Hg = H {g} are the left and right cosets of H containing g respectively. Example 3.1 Fix n E N. We know that n'll ~ (Z, +). For r E Z, the coset of n'll containing r is r

+ n'll = [r] =

{r + nk: k E Z}.

Let a, bE G. Then

Theorem 3.2

aH = bH a- 1b E H; (2) Han Hb =/= 0 Ha = Hb ba- 1 E H· ' (1) aH n bH =/= (3)

0

IHI = laHI = IHal.

Proof We prove the left coset case (1). The proof of (2) is similar to (1).

46 Chapter 3

Finitely Generated Abelian Groups and Quotient Groups

{=}

:3h 1, h 2 E H so that ah1 = bh2 1 - - 1 bh2 h-1 a- 1a h1h2 -a 2 1 1 a- b = h1h"2 E H,

~

::lh1, h2 E H so that ah1

{=}

{ ah1} H = {bh 2} H since hH = H, 'Vh E H

{=}

aH=bH.

aHnbH =1- 0

~ {=}

aHnbH =1- 0

=

bh2

To prove (3), construct a map ) Suppose that G is simple and abelian. Every subgroup of G is normal. Since G is simple, the only subgroups of G are { e} and G, and IGI > 1, so for any x E G\ {e} we have (x) =G. Suppose x has infinite order. Then (x2 ) ~ G but {e} =I (x2 ) =I (x), a contradiction. Sox, and therefore G, has finite order. Suppose x has composite order n so for some p > 1 that divides n, (x11) is a proper non-trivial subgroup of G, so G is not simple. Thus G is a cyclic group of prime order. D We prove the following theorem: Theorem 3.46 The alternating groups An are simple for n (also for n = 3).

~

5

The case in brackets follows from the fact that A 3 "' C3 which is simple. We can also check that V4 = {(1), (1, 2)(3, 4), (1, 3)(2, 4), (1, 4) (2, 3)} is a normal subgroup of A 4 , and hence A 4 is not simple. Proof Let H be a nontrivial normal subgroup of An. We shall

show that H =An. We shall do this by two claims. Claim 1 If H contains a 3-cycle, then H = An. We may assume that (1 2 3) E H.

3.6

Simple groups

71

Clearly, (1, 3, 2) = (1, 2, 3? E H. Then for any k > 3 we have (3, 2, k)(1, 3, 2)(3, 2, k)- 1 From Theorem 3.24 we know that H Claim 2

=

=

(1, 2, k) E H.

An.

H contains a 3-cycle.

We separate this into four cases. (1) Suppose H contains an element which can be written in disjoint cycle notation

u

(1 2 3 · · · r)7,

=

for r ~ 4. We now let 8 = (1 2 3) E An. Then by normality of H, we know 8-1 u8 E H. Then u- 18-1 u8 E H. Also, we notice that 7 does not contain 1, 2, 3. So it commutes with 8, and also trivially with (1 2 3 · · · r). We can expand this mess to obtain

u- 1 8- 1 u8

= (r · · · 2 1)(1 3 2)(1 2 3

· · · r)(1 2 3)

= (2 3 r),

which is a 3-cycle. The same argument goes through if u = ( a 1 a 2

•••

ar )7 for any

ab··· ,an. (2) Suppose H contains an element involving at least two disjoint 3-cycles, say (7 = (1 2 3)(4 5 6)7. Now let 8 = (1 2 4), and again calculate

u- 18-1 u8 = (1 3 2)(4 6 5)(1 4 2)(1 2 3)(4 5 6)(1 2 4) = (1 2 4 3 6). This is a 5-cycle, which is necessarily in H. By the previous case, we get a 3-cycle in H too, and hence H =An· (3) Suppose H contains u

=

(1 2 3)7, with 7 a product of 2-cycles

(if 7 contains anything longer, then it would fit in one of the previous two cases). Then u 2 = (1 2 3) 2

= (1 3 2) is a

3-cycle.

72 Chapter 3

Finitely Generated Abelian Groups and Quotient Groups

(4) Suppose H contains a = (1 2)(3 4)T, where T is a product of 2-cycles. We first let 6 = (1 2 3) and calculate

u

=

a- 16- 1a6 = (1 2)(3 4)(1 3 2)(1 2)(3 4)(1 2 3)

=

(1 4)(2 3) E H.

Now let

v = (1 5 2)u(1 2 5) = (1 3) (4 5)

E H.

Note that we used n ~ 5 again. We have yet again landed in the same case. Notice however, that these are not the same transpositions. We multiply uv = (1 4) (2 3) (1 3) (4 5) = (1 2 3 4 5) E H. This is then covered by the first case, and we are done.

D

Theorem 3.47 Let