Stabilizer is Subgroup

Theorem
Let $\struct {G, \circ}$ be a group which acts on a set $X$.

Let $\Stab x$ be the stabilizer of $x$ by $G$.

Then for each $x \in X$, $\Stab x$ is a subgroup of $G$.

Proof
From the :
 * $e * x = x \implies e \in \Stab x$

and so $\Stab x$ cannot be empty.

Let $g, h \in \Stab x$.

Let $g \in \Stab x$.

Then:
 * $x = \paren {g^{-1} \circ g} * x = g^{-1} * \paren {g * x} = g^{-1} * x$

Hence $g^{-1} \in \Stab x$.

Thus the conditions for the Two-Step Subgroup Test are fulfilled, and $\Stab x \le G$.