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: $G\times X\to X$ $(g,x)=gx$ Additionally, we have it so $ex=x$ 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: $D_4\times X\to Xs.tX=\{1,2,3,4\}$ 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: $G\times X\to Xs.t.(g,x)=gx{g}^{-1}$ This works as we can prove identity and associativity through the following:

1. Identity:
   1. $e^{-1}=e$ so $(e,x)=exe=x$.
2. 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 $x=gy$. This creates an equivalence relation where all the classes are called orbits, denoted $\mathcal{O}$ and the specific orbit of x (all the places where G takes x) is denoted ${\mathcal{O}}_x$.

To reiterate, you can think of ${\mathcal{O}}_x$ as all the places $G$ takes $x$

Lets prove G-equvilance is a equivalence relation:

1. Reflexive: x~x
   1. Clearly any group contains the identity, and we have defined group actions such that ex=x. Thus x~x
2. Symmetry:
   1. Assume xy. There there exists g s.t. $x=gy$. 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 yx
3. 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 $gx=x$. 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 $G$:

1. Identity:
   1. By definition $e$ fixes every element, thus $e\in G_x$
2. 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 $g_1{g}_2\in G_x$
3. Inverses:
   1. Suppose $g\in G$. Then $gx=x$, 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: $D_4=\{e,r^1,r^2,r^3,s,s{r}^1,s{r}^2,s{r}^3\}$ Which I drew out and represented as permutations in $S_4$ here $e=(1),r^1=(1234),r^2=(13)(24),r^3=(1432)$ $s=(21)(34),sr=(13),s{r}^2=(23)(14),s{r}^3=(42)$ and $X=\{1,2,3,4\}$ for the ${\mathcal{O}}_1$ 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 ${\mathbb{Z}}_2$. I put the stabilizer groups for every element bellow: $G_1=\{e,s{r}^3\}$ $G_2=\{e,sr\}$ $G_3=\{e,s{r}^3\}$ $G_4=\{e,sr\}$ You can observe all these stabilizer groups are isomorphic to ${\mathbb{Z}}_2$ and are all subgroups of $D_4$.

Connection $\mid {\mathcal{O}}_x\mid =[G:G_x]$

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 $x=1$

1. we saw that ${\mathcal{O}}_1=\{1,2,3,4\}$ and so $\mid {\mathcal{O}}_1\mid =4$.
2. The size of $\mid D_4\mid =2(4)=8$
3. Lastly, size of $\mid G_1\mid =2$. So we know there are 4 cosets of $G_1$ in $D_4$
4. Thus, $\mid {\mathcal{O}}_1\mid =[G:G_1]=4$