Orbit-Stabilizer Theorem

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

Proof
Let us define the mapping:
 * $\phi: G \to G * x$

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

where $*$ denotes a group action.

It is clear that $\phi$ is surjective, because from the definition all elements of $X$ are 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:
 * $g \equiv h \left({\bmod\, \operatorname{Stab} \left({x}\right)}\right)$

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

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

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

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

The result follows.

Alternate Proof
Lemma: For $K$ subgroup of $G$,
 * $g * x = h * x \iff gK = hK$, where $gK$ and $hK$ are cosets of $K$

Notice that $g * x$ and $h * x$ are elements of of the orbit

We conclude that there is a bijection between the elements of the orbit and the cosets of any subgroup $K$.

Therefore:
 * $|G:K|=|Orb(x)|$

Recalling,
 * $|G|=|K|*|H:K|$

And because $Stab(x)$ is a subset of $G$, setting $Stab(x)=K$ yields:
 * $|G|=|Stab(x)|*|Orb(x)|$