Prove $\operatorname{stab}_G(g \cdot x) = g\operatorname{stab}_G(x)g^{-1}$ Suppose a group $G$ acts on a set $X$. The stabilizer in $G$ of $x \in X$ is
$$
\operatorname{stab}_G(x) = \{a \in G : a \cdot x = x \}
$$ For each $g \in G$, let 
$$
g\operatorname{stab}_G(x)g^{-1} = \{gag^{-1} : a \in \operatorname{stab}_G(x)\}
$$ I have to prove that $$ \operatorname{stab}_G(g \cdot x) = g\operatorname{stab}_G(x)g^{-1} $$
by showing that 


*

*$\operatorname{stab}_G(g \cdot x) \subseteq g\operatorname{stab}_G(x)g^{-1}$

*$g\operatorname{stab}_G(x)g^{-1} \subseteq \operatorname{stab}_G(g \cdot x) $


Let $a\in\operatorname{stab}_G(g\cdot x)$; then $a\cdot(g\cdot x)=g\cdot x$, that is
$$
(ag)\cdot x=g\cdot x
$$
or else
$$
(g^{-1}ag)\cdot x=x
$$
Therefore $b=g^{-1}ag\in\operatorname{stab}_G(x)$, which means that
$$
a=gbg^{-1}
$$
where $b\in\operatorname{stab}_G(x)$. This proves one of the inclusions; can you show the other one with a similar technique?
