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 $$e \wedge x = x \implies e \in \operatorname{Stab} \left({x}\right)$$.


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

$$ $$ $$ $$

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


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