5Group actions

IA Groups

5.3 Important actions

Given any group

G

, there are a few important actions we can define. In particular,

we will define the conjugation action, which is a very important concept on

its own. In fact, the whole of the next chapter will be devoted to studying

conjugation in the symmetric groups.

First, we will study some less important examples of actions.

Lemma

(Left regular action)

.

Any group

G

acts on itself by left multiplication.

This action is faithful and transitive.

Proof. We have

1. (∀g ∈ G)(x ∈ G) g(x) = g · x ∈ G by definition of a group.

2. (∀x ∈ G) e · x = x by definition of a group.

3. g(hx) = (gh)x by associativity.

So it is an action.

To show that it is faithful, we want to know that [(

∀x ∈ X

)

gx

=

x

]

⇒ g

=

e

.

This follows directly from the uniqueness of identity.

To show that it is transitive,

∀x, y ∈ G

, then (

yx

−1

)(

x

) =

y

. So any

x

can

be sent to any y.

Theorem

(Cayley’s theorem)

.

Every group is isomorphic to some subgroup of

some symmetric group.

Proof.

Take the left regular action of

G

on itself. This gives a group homo-

morphism

ϕ

:

G → Sym G

with

ker ϕ

=

{e}

as the action is faithful. By the

isomorphism theorem, G

∼

=

im ϕ ≤ Sym G.

Lemma

(Left coset action)

.

Let

H ≤ G

. Then

G

acts on the left cosets of

H

by left multiplication transitively.

Proof. First show that it is an action:

0. g(aH) = (ga)H is a coset of H.

1. e(aH) = (ea)H = aH.

2. g

1

(g

2

(aH)) = g

1

((g

2

a)H) = (g

1

g

2

a)H = (g

1

g

2

)(aH).

To show that it is transitive, given

aH, bH

, we know that (

ba

−1

)(

aH

) =

bH

.

So any aH can be mapped to bH.

In the boring case where

H

=

{e}

, then this is just the left regular action

since G/{e}

∼

=

G.

Definition

(Conjugation of element)

.

The conjugation of

a ∈ G

by

b ∈ G

is

given by

bab

−1

∈ G

. Given any

a, c

, if there exists some

b

such that

c

=

bab

−1

,

then we say a and c are conjugate.

What is conjugation? This

bab

−1

form looks familiar from Vectors and

Matrices. It is the formula used for changing basis. If

b

is the change-of-basis

matrix and

a

is a matrix, then the matrix in the new basis is given by

bab

−1

. In

this case, bab

−1

is the same matrix viewed from a different basis.

In general, two conjugate elements are “the same” in some sense. For example,

we will later show that in

S

n

, two elements are conjugate if and only if they

have the same cycle type. Conjugate elements in general have many properties

in common, such as their order.

Lemma

(Conjugation action)

.

Any group

G

acts on itself by conjugation (i.e.

g(x) = gxg

−1

).

Proof. To show that this is an action, we have

0. g(x) = gxg

−1

∈ G for all g, x ∈ G.

1. e(x) = exe

−1

= x

2. g(h(x)) = g(hxh

−1

) = ghxh

−1

g

−1

= (gh)x(gh)

−1

= (gh)(x)

Definition

(Conjugacy classes and centralizers)

.

The conjugacy classes are the

orbits of the conjugacy action.

ccl(a) = {b ∈ G : (∃g ∈ g) gag

−1

= b}.

The centralizers are the stabilizers of this action, i.e. elements that commute

with a.

C

G

(a) = {g ∈ G : gag

−1

= a} = {g ∈ G : ga = ag}.

The centralizer is defined as the elements that commute with a particular

element a. For the whole group G, we can define the center.

Definition

(Center of group)

.

The center of

G

is the elements that commute

with all other elements.

Z(G) = {g ∈ G : (∀a) gag

−1

= a} = {g ∈ G : (∀a) ga = ag}.

It is sometimes written as C(G) instead of Z(G).

In many ways, conjugation is related to normal subgroups.

Lemma. Let K C G. Then G acts by conjugation on K.

Proof.

We only have to prove closure as the other properties follow from the

conjugation action. However, by definition of a normal subgroup, for every

g ∈ G, k ∈ K, we have gkg

−1

∈ K. So it is closed.

Proposition.

Normal subgroups are exactly those subgroups which are unions

of conjugacy classes.

Proof.

Let

K C G

. If

k ∈ K

, then by definition for every

g ∈ G

, we get

gkg

−1

∈ K

. So

ccl

(

k

)

⊆ K

. So

K

is the union of the conjugacy classes of all its

elements.

Conversely, if

K

is a union of conjugacy classes and a subgroup of

G

, then

for all k ∈ K, g ∈ G, we have gkg

−1

∈ K. So K is normal.

Lemma.

Let

X

be the set of subgroups of

G

. Then

G

acts by conjugation on

X.

Proof. To show that it is an action, we have

0.

If

H ≤ G

, then we have to show that

gHg

−1

is also a subgroup. We

know that

e ∈ H

and thus

geg

−1

=

e ∈ gHg

−1

, so

gHg

−1

is non-empty.

For any two elements

gag

−1

and

gbg

−1

∈ gHg

−1

, (

gag

−1

)(

gbg

−1

)

−1

=

g(ab

−1

)g

−1

∈ gHg

−1

. So gHg

−1

is a subgroup.

1. eHe

−1

= H.

2. g

1

(g

2

Hg

−1

2

)g

−1

1

= (g

1

g

2

)H(g

1

g

2

)

−1

.

Under this action, normal subgroups have singleton orbits.

Definition

(Normalizer of subgroup)

.

The normalizer of a subgroup is the

stabilizer of the (group) conjugation action.

N

G

(H) = {g ∈ G : gHg

−1

= H}.

We clearly have

H ⊆ N

G

(

H

). It is easy to show that

N

G

(

H

) is the largest

subgroup of G in which H is a normal subgroup, hence the name.

There is a connection between actions in general and conjugation of subgroups.

Lemma.

Stabilizers of the elements in the same orbit are conjugate, i.e. let

G

act on X and let g ∈ G, x ∈ X. Then stab(g(x)) = g stab(x)g

−1

.