Stabilizer is Subgroup

Theorem
Let $\left({G, \circ}\right)$ 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
From the group action axiom $GA\,2$:
 * $e * x = x \implies e \in \operatorname{Stab} \left({x}\right)$

and so $\operatorname{Stab} \left({x}\right)$ cannot be empty.

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

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

Then:
 * $x = \left({g^{-1} \circ g}\right) * x = g^{-1} * \left({g * x}\right) = g^{-1} * x$

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

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