1.1 Review of Important Basics

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In this short section we gather together some of the basics of elementary

group theory, and at the same time establish a bit of the notation which will

be used in these notes. The following terms should be well-understood by

the reader (if in doubt, consult any elementary treatment of group theory):

1 group, abelian group, subgroup, coset, normal subgroup, quotient group,

order of a group, homomorphism, kernel of a homomorphism, isomorphism,

normalizer of a subgroup, centralizer of a subgroup, conjugacy, index of a

subgroup, subgroup generated by a set of elements Denote the identity element

of the group G by e, and set G# = G − {e}. If G is a group and if H

is a subgroup of G, we shall usually simply write H _ G. Homomorphisms

are usually written as left operators: thus if _ : G ! G0 is a homomorphism

of groups, and if g 2 G, write the image of g in G0 as _(g).

The following is basic in the theory of finite groups.

Theorem 1.1.1 (Lagrange’s Theorem) Let G be a finite group, and

let H be a subgroup of G. Then |H| divides |G|.

The reader should be quite familiar with both the statement, as well as

the proof, of the following.

Theorem 1.1.2 (The Fundamental Homomorphism Theorem) Let

G, G0 be groups, and assume that _ : G ! G0 is a surjective homomorphism.

1Many, if not most of these terms will be defined below.

1

2 CHAPTER 1. GROUP THEORY

Then

G/ker_ _= G0

via gker_ 7! _(g). Furthermore, the mapping

_−1 : {subgroups of G0} ! {subgroups of G which contain ker _}

is a bijection, as is the mapping

_−1 : {normal subgroups of G0} ! { normal subgroups of G which contain ker _}

Let G be a group, and let x 2 G. Define the order of x, denoted by o(x),

as the least positive integer n with xn = e. If no such integer exists, say

that x has infinite order, and write o(x) = 1. The following simple fact

comes directly from the division algoritheorem in the ring of integers.

Lemma 1.1.3 Let G be a group, and let x 2 G, with o(x) = n < 1. If k is

any integer with xk = e, then n|k.

The following fundamental result, known as Cauchy’s theorem , is very

useful.

Theorem 1.1.4 (Cauchy’s Theorem) Let G be a finite group, and let p

be a prime number with p dividing the order of G. Then G has an element

of order p.

The most commonly quoted proof involves distinguishing two cases: G

is abelian, and G is not; this proof is very instructive and is worth knowing.

Let G be a group and let X _ G be a subset of G. Denote by hXi

the smallest subgroup of G containing X; thus hXi can be realized as the

intersection of all subgroups H _ G with X _ H. Alternatively, hXi can

be represented as the set of all elements of the form xe1

1 xe2

2 · · · xer

r where

x1, x2, . . . xr 2 X, and where e1, e2, . . . , er 2 Z. If X = {x}, it is customary

to write hxi in place of h{x}i. If G is a group such that for some x 2 G,

G = hxi, then G is said to be a cyclic group with generator x. Note that, in

general, a cyclic group can have many generators.

The following classifies cyclic groups, up to isomorphism:

1.1. REVIEW OF IMPORTANT BASICS 3

Lemma 1.1.5 Let G be a group and let x 2 G. Then

hxi _=

(

(Z/(n), +) if o(x) = n,

(Z, +) if o(x) = 1.

Let X be a set, and recall that the symmetric group SX is the group of

bijections X ! X. When X = {1, 2, . . . , n}, it is customary to write SX

simply as Sn. If X1 and X2 are sets and if _ : X1 ! X2 is a bijection, there

is a naturally defined group isomorphism __ : SX1 ! SX2 . (A “naturally”

defined homomorphism is, roughly speaking, one that practically defines

itself. Given this, the reader should determine the appropriate definition of

__.)

If G is a group and if H is a subgroup, denote by G/H the set of left

cosets of H in G. Thus,

G/H = {gH| g 2 G}.

In this situation, there is always a natural homomorphism G ! SG/H,

defined by

g 7! (xH 7! gxH),

where g, x 2 G. The above might look complicated, but it really just

means that there is a homomorphism _ : G ! SG/H, defined by setting

_(g)(xH) = (gx)H. That _ really is a homomorphism is routine, but

should be checked! The point of the above is that for every subgroup of

a group, there is automatically a homomorphism into a corresponding symmetric

group. Note further that if G is a group with H _ G, [G : H] = n,

then there exists a homomorphism G ! Sn. Of course this is established

via the sequence of homomorphisms G ! SG/H ! Sn, where the last map

is the isomorphism SG/H

_= Sn of the above paragraph.

Exercises 1.1

1. Let G be a group and let x 2 G be an element of finite order n. If

k 2 Z, show that o(xk) = n/(n, k), where (n, k) is the greatest common

divisor of n and k. Conclude that xk is a generator of hxi if and only

if (n, k) = 1.

2. Let H,K be subgroups of G, both of finite index in G. Prove that

H\K also has finite index. In fact, [G : H\K] = [G : H][H : H\K].

4 CHAPTER 1. GROUP THEORY

3. Let G be a group and let H _ G. Define the normalizer of H in G by

setting NG(H) = {x 2 G| xHx−1 = H}.

(a) Prove that NG(H) is a subgroup of G.

(b) If T _ G with T _ NG(H), prove that KT _ G.

4. Let H _ G, and let _ : G T ! SG/H be as above. Prove that ker_ =

xHx−1, where the intersection is taken over the elements x 2 G.

5. Let _ : G ! SG/H exactly as above. If [G : H] = n, prove that

n||_(G)|, where _(G) is the image of G in SG/H.

6. Let G be a group of order 15, and let x 2 G be an element of order

5, which exists by Cauchy’s theorem. If H = hxi, show that H / G.

(Hint: We have G ! S3, and |S3| = 6. So what?)

7. Let G be a group, and let K and N be subgroups of G, with N normal

in G. If G = NK, prove that there is a 1 − 1 correspondence between

the subgroups X of G satisfying K _ X _ G, and the subgroups T

normalized by K and satisfying N \ K _ T _ N.

8. The group G is said to be a dihedral group if G is generated by two elements

of order two. Show that any dihedral group contains a subgroup

of index 2 (necessarily normal).

9. Let G be a finite group and let C× be the multiplicative group of

complex numbers. If _ : G ! C× is a non-trivial homomorphism,

prove that

X

x2G

_(x) = 0.

10. Let G be a group of even order. Prove that G has an odd number of

involutions. (An involution is an element of order 2.)

1.2. THE CONCEPT OF A GROUP ACTION 5

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