Stabilizer is Subgroup

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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:

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


Proof

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


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


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


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

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


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

$\blacksquare$


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