Theorem – Orbit-Stabilizer theorem

Contents
1. Theorem
2. Proof
3. Consequences of the theorem.

Theorem

Let $\langle G, * \rangle$ be a finite group and let $H$ be a subgroup of $G$ . Let $X$ be a set. Consider the orbit and the stabiliser of the group action, respectively $orb(x)$ and $stab(x)$ . We have that, for any $x\in X$ ,

$|orb(x)|\times |stab(x)| = |G|$

Proof

We will prove that there exists a bijection $f: orb(x)\rightarrow \{gH ~|~ g\in G\}$ , where $\{gH ~|~ g\in G\}$ is the set of left cosets of $H$ in $G$ . This follows because the number of left cosets of $H$ in $G$ is equal to $\frac{|G|}{|stab(x)|}$ , so a bijection would prove that

$|orb(x)| = \frac{|G|}{|stab(x)|}$

First, we define $f(y) = send_x(y)$ . We now prove that this function is both surjective and injective (and hence bijective). To prove injectivity, let

$\begin{array}{ccc} f(y_1) = f(y_2) & \implies & send_x(y_1) = send_x(y_2) \\\\\ &\implies& \{g\in G ~|~ h(g,x)=y_1\} = \{g\in G ~|~ h(g,x)=y_2\} \\\\ &\implies& \{g\in G ~|~ x=h^{-1}(g,y_1)\} = \{g\in G ~|~ x=h^{-1}(g,y_2)\} \\\\ &\implies&x=h^{-1}(g,y_1) =h^{-1}(g,y_2) \\\\ &\implies& g_1 = g_2 \end{array}$ .

To prove surjectivity, let $gH$ be a left coset of $H$ in $G$ . Then,

$gH = send_x(g*x)$

But $g*x \in orb(X)$ , so for every left coset, there is a corresponding element $g*x \in orb(X)$ . This proves surjectivity. $\square$