# 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 $G$.

### 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 $\Stab x \le G$.

$\blacksquare$