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 $$G$$.

Corollary 1
Let $$G$$ be a group whose identity is $$e$$, which acts on a set $$X$$.

Then $$\forall x \in X: e \in \operatorname{Stab} \left({x}\right)$$.

Corollary 2
Let $$G$$ be a group whose identity is $$e$$, which acts on a set $$X$$.

Then $$\forall g, h \in G: g \wedge x = h \wedge x \iff g^{-1} h \in \operatorname{Stab} \left({x}\right)$$.

Proof

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


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

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

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

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

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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 $$\operatorname{Stab} \left({x}\right) \le G$$.

Proof of Corollary 1
Follows directly from the fact that $$\operatorname{Stab} \left({x}\right) \le G$$ and Identity of Subgroup.

Proof of Corollary 2
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