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

Proof
Let us define the mapping $$\phi: G \to G \wedge x$$ such that $$\phi \left({g}\right) = g \wedge x$$.

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 $$G / \operatorname{Stab} \left({x}\right) \to G \wedge x$$ given by $$g \operatorname{Stab} \left({x}\right) \mapsto g \wedge x$$.

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

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

The result follows.