Group Actions

Prelim: Group Isomorphisms and Homomorphisms
Needed for Burnsides Lemma

Group Actions Introduction

Definition:

Let $G$ be a group and $X$ be a set. Then we can define a group action as:

G \times X \to X

(g, x) = g x

Additionally, we have it so $e x = x$ and $g_{1} (g_{2} x) = (g_{1} g_{2}) x$ for all $g \in G$ and $x \in X$ .
Essentially we have Identity and associativity property.

Example:

Take the Dihedral group $D_{n}$ acting on a regular n-gon. This is a group action where the elements of $D_{n}$ act on the vertices of the n-gon (which is a set of n elements).

For example, we can take a look at $D_{4}$ acting on the vertices of a square; $X = {1, 2, 3, 4}$ . Then we know $r = (1234)$ and this acts on the vertex $1$ by sending it to $2$ . i.e. $(1234) 1 = 2$ . Clearly:

D_{4} \times X \to X s . t X = {1, 2, 3, 4}

Which is a group action.

Acting by conjugation

A group can act on itself by conjugation through the following definition:
Let $G$ be a group and let $X = G$ . Then a group acts on itself via conjugation via:

G \times X \to X s . t . (g, x) = g x g^{- 1}

This works as we can prove identity and associativity through the following:

Identity:
1. $e^{- 1} = e$ so $(e, x) = e x e = x$ .
Associativity
1. Let $g_{1}, g_{2} \in G$ .
2. Then $g_{1} (g_{2} x) = g_{1} (g_{2} x g_{2}^{- 1}) = g_{1} g_{2} x g_{2}^{- 1} g_{1}^{- 1} = g_{1} g_{2} x (g_{1} g_{2})^{- 1} = (g_{1} g_{2}) x$

Orbits and Stabilizers

Orbits

Two elements in a set are G-equivalent if an element of $G$ can take the first element to the second. Rigorously, x is G-equivalent to y if $\exists g \in G$ such that $x = g y$ . This creates an equivalence relation where all the classes are called orbits, denoted $O$ and the specific orbit of x (all the places where G takes x) is denoted $O_{x}$ .

To reiterate, you can think of $O_{x}$ as all the places $G$ takes $x$

Lets prove G-equvilance is a equivalence relation:

Reflexive: x~x
1. Clearly any group contains the identity, and we have defined group actions such that ex=x. Thus x~x
Symmetry:
1. Assume x~y. There there exists g s.t. $x = g y$ . However, this means $y = g^{- 1} x$ and as G is a group and has inverses clearly the inverse of g is in G. Thus y~x
Transitivity:
1. Assume x ~ y and y ~ g. Then we have it so $x = g_{1} y$ and $y = g_{2} z$ .
2. From here, we can show $x = g_{1} y = g_{1} g_{2} z$ .
3. Then $x = g_{1} g_{2} z$ and $g_{1} g_{2} \in G$ as it is a group. thus x~z

Stabilizer

Stabilizers are the subgroup of elements $g \in G$ that do not move an element $x \in X$ . i.e. collection of $g$ where $g x = x$ . This is denoted as $G_{x}$ for all elements in $G$ which fix $x$ . This creates a subgroup in $G$ .

Proving that $G_{x}$ is a subgroup of $G$ :

Identity:
1. By definition $e$ fixes every element, thus $e \in G_{x}$
Closure:
1. Let $g_{1}, g_{2} \in G_{x}$ . Then, $(g_{1} g_{2}) x = g_{1} (g_{2} x) = g_{1} x = x$ as both elements fix x. Thus $g_{1} g_{2} \in G_{x}$
Inverses:
1. Suppose $g \in G$ . Then $g x = x$ , so $x = g^{- 1} x$ and we have it that $g^{- 1}$ is also in $G_{x}$ .

Example

Lets consider $D_{4}$ acting on the vertices of a square. What are the orbits and stabilizers?
First note: $$D_4 = {e,r^1,r^2,r^3,s,sr^1,sr^2,sr^3}$$
Which I drew out and represented as permutations in $S_{4}$ here

e = (1), r^{1} = (1234), r^{2} = (13) (24), r^{3} = (1432)

s = (21) (34), s r = (13), s r^{2} = (23) (14), s r^{3} = (42)

and $$X = {1,2,3,4}$$
for the $O_{1}$ we see that 1 can go anywhere through just the rotations. So we only have one orbit and all the vertices are G-equivalent. $$\mathcal{O}_1 = \mathcal{O}_2 = \mathcal{O}_3 = \mathcal{O}_4 = {1,2,3,4}$$ Then, for the stabilizer of 1, or $G_{1} = {e, s r^{3}}$ . Clearly these elements fix r, and $(s r^{3})^{2} = e$ so this is a subgroup isomorphic to $Z_{2}$ . I put the stabilizer groups for every element bellow:

G_{1} = {e, s r^{3}}

G_{2} = {e, s r}

G_{3} = {e, s r^{3}}

G_{4} = {e, s r}

You can observe all these stabilizer groups are isomorphic to $Z_{2}$ and are all subgroups of $D_{4}$ .

Connection $∣ O_{x} ∣= [G : G_{x}]$

The size of the orbit of x is the same as the number of left cosets of the stabilizer group in G. We can show this here:
#review proof if you have time.
For $x = 1$

we saw that $O_{1} = {1, 2, 3, 4}$ and so $∣ O_{1} ∣= 4$ .
The size of $∣ D_{4} ∣= 2 (4) = 8$
Lastly, size of $∣ G_{1} ∣= 2$ . So we know there are 4 cosets of $G_{1}$ in $D_{4}$
Thus, $∣ O_{1} ∣= [G : G_{1}] = 4$