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) $


HINT: Note that $\operatorname{stab}_G(g\cdot x)\subseteq g\operatorname{stab}_G(x)g^{-1}$ iff $g^{-1}\operatorname{stab}_G(g\cdot x)g\subseteq\operatorname{stab}_G(x)$. To prove the latter, suppose that $a\in\operatorname{stab}_G(x)$, and show that $(g^{-1}ag)\cdot x=x$. 
To show that $g\operatorname{stab}_G(x)g^{-1}\subseteq\operatorname{stab}_G(g\cdot x)$, again let $a\in\operatorname{stab}_G(x)$, and see whether $gag^{-1}$ stabilizes $g\cdot x$. Both are straightforward calculations if you understand the defining properties of group actions.
