Group Action

An action of a group $G$ on a set $X$ is a map $\varphi: G\times X\to X$ which satisfies the following two conditions:

1. $\forall x\in X, \varphi(e,x) = x$ (identity)
2. $\forall g,h\in G, \forall x\in X, \varphi(g,\varphi(h,x)) = \varphi(gh,x)$ (associativity)
   • for a defined map, we denote $\varphi(g,x) = \varphi_g(x) = g\cdot x = g.x = gx$

If a set $X$ is equipped with an action of a group $G$, then we will denote $G\curvearrowright X$

Let $G\curvearrowright X$, denote such action as $\varphi$, let $x\in X$:

1. $\forall g\in G$ the map $\varphi_g : X \to X$ is a bijection
2. Any action $\varphi$ uniquely defines a homomorphism $\psi : G \to \text{Perm}(X)$. And vice, versa, every such homomorphism defines an action of $G$ on $X$.
3. The orbit of $x$ is a set $Gx:=\{g.x: g\in G\} \subseteq X$
4. The stabilizer of $x$ is a set $Stab_x = G_x := \{g\in G : g.x = x\}\subseteq G$ and is a subgroup of $G$
5. Define relation $\sim$ where $x\sim_G y\iff y = gx, g\in G$. Orbits are the equivalence classes respect to $\sim$
6. $\varphi$ is faithful if the kernel of the respective homomorphism $\varphi: G\to Perm(X)$ is trivial. Equivalently, $\forall x\in X, \exists g\in G, g.x \ne x$
7. $\varphi$ is transitive if $\varphi$ admits only one orbit which coincides with $X$. Equivalently, $\forall x,y \in X, \exists g\in G, s.t., g.x=y$
   • Furthermore, $\forall (x_1,\ldots, x_n)\in X^n, \forall (y_1,\ldots,y_n)\in Y^n,\exists g\in G, (g.x_i)= (y_i)$

Some facts:

• $\varphi_l(g,h) = gh$ is a transitive and faithful action(prove by associativity) which used to prove Cayley's theorem.
• We define an action via conjugation of $G$ to itself: $g.h = ghg^{-1}$. This action is not faithful, nor transitive, but can be use to prove Sylow's theorem. The orbits of this action are called conjugacy classes, $\forall g\in G$ the stabilizer of $g$ is the centralizer $C_G(g)$
• the right multiplication does not define an action of non-abelian group to itself
• $\forall x\in X$, the restriction of the action of its orbit $Gx$ is still an action, as $\forall g\in G, y=h.x \in Gx, g.(h.x) = (gh).x \in Gx$

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Let $X,Y$ be two sets equipped with actions $\varphi_X,\varphi_Y$ of a single group $G$. $\varphi_X: X\to X\land f:X\to Y \land \varphi_Y: Y\to Y \implies f:X\to Y$ is $G-$equivariant. In other words, $\forall x\in X, g\in G$ we have $f((\varphi_X)_g(x)) = (\varphi_Y)_g(f(x))$ or $f(g.x) =g.f(x)$.

• $f$ is a bijective map, then we will say that $f$ is an isomorphism of group actions.

Orbit-Stabilizer Theorem: Let $G\curvearrowright X$, let $x\in X$, the map $f: Gx\to G/G_x$ where $y\to \{g\in G | gx =y\}$ is an isomorphism of group actions via left multiplication of $G$.

• Let $X$ be equipped with an action of a finite group $G$ then $\forall x\in X$ we have $|Gx| = |G|/|G_x|$

Burnside's Lemma:: Let $X$ be a finite set, equipped with an action of a finite group $G$. $\forall g\in G$ let us denote $X^g = \{x\in X: g.x = x\}$. Als, by $X/G$ we will denote the set of orbits w.r.t. the action of $G$. Then $|X/G| = \frac{1}{|G|}\sum_{g\in G}|X^g|$