AATA Groups Acting on Sets

Section 14.1 Groups Acting on Sets

Let

X

be a set and

G

be a group. A (left) action of

G

X

is a map

G \times X \to X

given by

(g, x) \mapsto g x,

where

$e x = x$ for all $x \in X;$
$(g_{1} g_{2}) x = g_{1} (g_{2} x)$ for all $x \in X$ and all $g_{1}, g_{2} \in G .$

🔗

Under these considerations

X

is called a

G

-set. Notice that we are not requiring

X

to be related to

G

in any way. It is true that every group

G

acts on every set

X

by the trivial action

(g, x) \mapsto x;

however, group actions are more interesting if the set

X

is somehow related to the group

G .

🔗

Example 14.1.

🔗

Let

G = G L_{2} (R)

and

X = R^{2} .

Then

G

acts on

X

by left multiplication. If

v \in R^{2}

and

I

is the identity matrix, then

I v = v .

A

and

B

are

2 \times 2

invertible matrices, then

(A B) v = A (B v)

since matrix multiplication is associative.

🔗

Example 14.2.

🔗

Let

G = D_{4}

be the symmetry group of a square. If

X = {1, 2, 3, 4}

is the set of vertices of the square, then we can consider

D_{4}

to consist of the following permutations:

{(1), (1 3), (2 4), (1 4 3 2), (1 2 3 4), (1 2) (3 4), (1 4) (2 3), (1 3) (2 4)} .

🔗

The elements of

D_{4}

act on

X

as functions. The permutation

(1 3) (2 4)

acts on vertex

1

by sending it to vertex

3,

on vertex

2

by sending it to vertex

4,

and so on. It is easy to see that the axioms of a group action are satisfied.

🔗

In general, if

X

is any set and

G

is a subgroup of

S_{X},

the group of all permutations acting on

X,

then

X

is a

G

-set under the group action

(σ, x) \mapsto σ (x)

🔗

for

σ \in G

and

x \in X .

🔗

Example 14.3.

🔗

If we let

X = G,

then every group

G

acts on itself by the left regular representation; that is,

(g, x) \mapsto λ_{g} (x) = g x,

where

λ_{g}

is left multiplication:

\begin{matrix} e \cdot x = λ_{e} x = e x = x \\ (g h) \cdot x = λ_{g h} x = λ_{g} λ_{h} x = λ_{g} (h x) = g \cdot (h \cdot x) . \end{matrix}

🔗

H

is a subgroup of

G,

then

G

is an

H

-set under left multiplication by elements of

H .

🔗

Example 14.4.

🔗

Let

G

be a group and suppose that

X = G .

H

is a subgroup of

G,

then

G

is an

H

-set under conjugation; that is, we can define an action of

H

G,

H \times G \to G,

🔗

via

(h, g) \mapsto h g h^{- 1}

🔗

for

h \in H

and

g \in G .

Clearly, the first axiom for a group action holds. Observing that

\begin{aligned} (h_{1} h_{2}, g) & = h_{1} h_{2} g (h_{1} h_{2})^{- 1} \\ = h_{1} (h_{2} g h_{2}^{- 1}) h_{1}^{- 1} \\ = (h_{1}, (h_{2}, g)), \end{aligned}

🔗

we see that the second condition is also satisfied.

🔗

Example 14.5.

🔗

Let

H

be a subgroup of

G

and

L_{H}

the set of left cosets of

H .

The set

L_{H}

is a

G

-set under the action

(g, x H) \mapsto g x H .

🔗

Again, it is easy to see that the first axiom is true. Since

(g g^{'}) x H = g (g^{'} x H),

the second axiom is also true.

🔗

G

acts on a set

X

and

x, y \in X,

then

x

is said to be

G

-equivalent to

y

if there exists a

g \in G

such that

g x = y .

We write

x \sim_{G} y

x \sim y

if two elements are

G

-equivalent.

🔗

Proposition 14.6.

🔗

Let

X

be a

G

-set. Then

G

-equivalence is an equivalence relation on

X .

Proof.

The relation

\sim

is reflexive since

e x = x .

Suppose that

x \sim y

for

x, y \in X .

Then there exists a

g

such that

g x = y .

In this case

g^{- 1} y = x;

hence,

y \sim x .

To show that the relation is transitive, suppose that

x \sim y

and

y \sim z .

Then there must exist group elements

g

and

h

such that

g x = y

and

h y = z .

z = h y = (h g) x,

and

x

is equivalent to

z .

🔗

X

is a

G

-set, then each partition of

X

associated with

G

-equivalence is called an orbit of

X

under

G .

We will denote the orbit that contains an element

x

X

O_{x} .

🔗

Example 14.7.

🔗

Let

G

be the permutation group defined by

G = {(1), (1 2 3), (1 3 2), (4 5), (1 2 3) (4 5), (1 3 2) (4 5)}

🔗

and

X = {1, 2, 3, 4, 5} .

Then

X

is a

G

-set. The orbits are

O_{1} = O_{2} = O_{3} = {1, 2, 3}

and

O_{4} = O_{5} = {4, 5} .

🔗

Now suppose that

G

is a group acting on a set

X

and let

g

be an element of

G .

The fixed point set of

g

X,

denoted by

X_{g},

is the set of all

x \in X

such that

g x = x .

We can also study the group elements

g

that fix a given

x \in X .

This set is more than a subset of

G,

it is a subgroup. This subgroup is called the stabilizer subgroup or isotropy subgroup of

x .

We will denote the stabilizer subgroup of

x

G_{x} .

🔗

Remark 14.8.

🔗

It is important to remember that

X_{g} \subset X

and

G_{x} \subset G .

🔗

Example 14.9.

🔗

Let

X = {1, 2, 3, 4, 5, 6}

and suppose that

G

is the permutation group given by the permutations

{(1), (1 2) (3 4 5 6), (3 5) (4 6), (1 2) (3 6 5 4)} .

🔗

Then the fixed point sets of

X

under the action of

G

are

\begin{matrix} X_{(1)} = X, \\ X_{(3 5) (4 6)} = {1, 2}, \\ X_{(1 2) (3 4 5 6)} = X_{(1 2) (3 6 5 4)} = \emptyset, \end{matrix}

🔗

and the stabilizer subgroups are

\begin{matrix} G_{1} = G_{2} = {(1), (3 5) (4 6)}, \\ G_{3} = G_{4} = G_{5} = G_{6} = {(1)} . \end{matrix}

🔗

It is easily seen that

G_{x}

is a subgroup of

G

for each

x \in X .

🔗

Proposition 14.10.

🔗

Let

G

be a group acting on a set

X

and

x \in X .

The stabilizer group of

x,

G_{x},

is a subgroup of

G .

Proof.

Clearly,

e \in G_{x}

since the identity fixes every element in the set

X .

Let

g, h \in G_{x} .

Then

g x = x

and

h x = x .

(g h) x = g (h x) = g x = x;

hence, the product of two elements in

G_{x}

is also in

G_{x} .

Finally, if

g \in G_{x},

then

x = e x = (g^{- 1} g) x = (g^{- 1}) g x = g^{- 1} x .

g^{- 1}

is in

G_{x} .

🔗

We will denote the number of elements in the fixed point set of an element

g \in G

| X_{g} |

and denote the number of elements in the orbit of

x \in X

| O_{x} | .

The next theorem demonstrates the relationship between orbits of an element

x \in X

and the left cosets of

G_{x}

G .

🔗

Theorem 14.11.

🔗

Let

G

be a finite group and

X

a finite

G

-set. If

x \in X,

then

| O_{x} | = [G : G_{x}] .

Proof.

We know that

| G | / | G_{x} |

is the number of left cosets of

G_{x}

G

by Lagrange’s Theorem (Theorem 6.10). We will define a bijective map

ϕ

between the orbit

O_{x}

X

and the set of left cosets

L_{G_{x}}

G_{x}

G .

Let

y \in O_{x} .

Then there exists a

g

G

such that

g x = y .

Define

ϕ

ϕ (y) = g G_{x} .

To show that

ϕ

is one-to-one, assume that

ϕ (y_{1}) = ϕ (y_{2}) .

Then

ϕ (y_{1}) = g_{1} G_{x} = g_{2} G_{x} = ϕ (y_{2}),

where

g_{1} x = y_{1}

and

g_{2} x = y_{2} .

Since

g_{1} G_{x} = g_{2} G_{x},

there exists a

g \in G_{x}

such that

g_{2} = g_{1} g,

y_{2} = g_{2} x = g_{1} g x = g_{1} x = y_{1};

consequently, the map

ϕ

is one-to-one. Finally, we must show that the map

ϕ

is onto. Let

g G_{x}

be a left coset. If

g x = y,

then

ϕ (y) = g G_{x} .

Prev Top Next