Stabilizer is Subgroup

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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 $G$.


Corollary

Let $G$ be a group whose identity is $e$.

Let $G$ act on a set $X$.

Let $x \in X$.


Then:

$\forall g, h \in G: g * x = h * x \iff g^{-1} h \in \Stab x$


Proof

From the Group Action Axiom $\text {GA} 2$:

$e * x = x \implies e \in \Stab x$

and so $\Stab x$ cannot be empty.


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

\(\ds g, h\) \(\in\) \(\ds \Stab x\)
\(\ds \leadsto \ \ \) \(\ds g * x\) \(=\) \(\ds x\) Definition of Stabilizer of $x$ by $G$
\(\, \ds \land \, \) \(\ds h * x\) \(=\) \(\ds x\) Definition of Stabilizer of $x$ by $G$
\(\ds \leadsto \ \ \) \(\ds g * \paren {h * x}\) \(=\) \(\ds x\)
\(\ds \leadsto \ \ \) \(\ds \paren {g \circ h} * x\) \(=\) \(\ds x\) Group Action Axiom $\text {GA} 1$
\(\ds \leadsto \ \ \) \(\ds g \circ h\) \(\in\) \(\ds \Stab x\) Definition of Stabilizer of $x$ by $G$


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 $\Stab x \le G$.

$\blacksquare$


Sources

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