Orbit-Stabilizer Theorem/Proof 1

Theorem
Let $G$ be a group which acts on a finite set $X$.

Let $\operatorname{Orb} \left({x}\right)$ be the orbit of $x$.

Let $\operatorname{Stab} \left({x}\right)$ be the stabilizer of $x$ by $G$.

Let $\left[{G : \operatorname{Stab} \left({x}\right)}\right]$ be the index of $\operatorname{Stab} \left({x}\right)$ in $G$.

Then:
 * $\left|{\operatorname{Orb} \left({x}\right)}\right| = \left[{G : \operatorname{Stab} \left({x}\right)}\right] = \dfrac {\left|{G}\right|} {\left|{\operatorname{Stab} \left({x}\right)}\right|}$

Proof
Let us define the mapping:
 * $\phi: G \to \operatorname{Orb} \left({x}\right)$

such that:
 * $\phi \left({g}\right) = g * x$

where $*$ denotes the group action.

It is clear that $\phi$ is surjective, because from the definition $x$ was acted on by all the elements of $G$.

Next, from Stabilizer is Subgroup: Corollary 2:
 * $\phi \left({g}\right) = \phi \left({h}\right) \iff g^{-1} h \in \operatorname{Stab} \left({x}\right)$

This means:
 * $g \equiv h \pmod{ \operatorname{Stab} \left({x}\right)}$

Thus there is a well-defined bijection:
 * $G \mathbin / \operatorname{Stab} \left({x}\right) \to \operatorname{Orb} \left({x}\right)$

given by:
 * $g \, \operatorname{Stab} \left({x}\right) \mapsto g * x$

So $\operatorname{Orb} \left({x}\right)$ has the same number of elements as $G \mathbin / \operatorname{Stab} \left({x}\right)$.

That is:
 * $\left|{\operatorname{Orb} \left({x}\right)}\right| = \left[{G : \operatorname{Stab} \left({x}\right)}\right]$

The result follows.