Let $G$ a finite group acting on a set $X$. Prove that the number of elements in any of the orbits is a divisor of $|G|$.

By definition, the $G$-orbit of $x\in X$ in $X$ is the set: $$G(x)=\{g \cdot x \in X : g\in G\}$$

How may I prove this? I have tried to define a mapping from $G$ to $G(x)$ for arbitrary $x$ ($\phi(g)=g \cdot x$) and try to prove it is a bijection, but I can only prove that the map is surjective. So I think this approach may be wrong.

Any help or hints will be appreciated.


This is essentially the orbit-stabilizer lemma.

The following map is surjective: $$\left\{\begin{array}{ccc}G&\rightarrow&G(x)\\g&\rightarrow&g\cdot x\end{array}\right.$$ and one has: $$g\cdot x=g'\cdot x\Leftrightarrow g^{-1}g'\in\textrm{Stab}(x).$$ Hence, one has a bijective mapping from $G/\textrm{Stab}(x)$ to $G(x)$ and they therefore have the same cardinality. To conclude, $\textrm{Stab}(x)$ is a subgroup of $G$ and using Lagrange's theorem $[G:\textrm{Stab}(x)]$ divides $|G|$.


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