Stabilizer is Subgroup

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

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

Then for each $x \in X$, $\operatorname{Stab} \left({x}\right)$ is a subgroup of $G$.

Proof

 * $\operatorname{Stab} \left({x}\right)$ can not be empty, because $e * x = x \implies e \in \operatorname{Stab} \left({x}\right)$.


 * Let $g, h \in \operatorname{Stab} \left({x}\right)$.


 * Let $g \in \operatorname{Stab} \left({x}\right)$.

Suppose $g^{-1} \notin \operatorname{Stab} \left({x}\right)$.

Then $g^{-1} * x = y \ne x$.

This is false, therefore $g^{-1} * x = x$ and so $g^{-1} \in \operatorname{Stab} \left({x}\right)$.


 * Thus the conditions for the Two-step Subgroup Test are fulfilled, and $\operatorname{Stab} \left({x}\right) \le G$.