Show that for any $g\in G$, the stabiliser of $g.x$ is the subgroup $gG_xg^{-1}$

Question

Let $X$ be a G-set and $x \in X$. Show that for any $g\in G$, the stabiliser of $g.x$ is the subgroup $gG_xg^{-1}$

Now the proof I found on this question is as follows: $$h\in G_{g.x}\Leftrightarrow hg.x=g.x\Leftrightarrow g^{-1}hg.x=x\Leftrightarrow g^{-1}hg\in G_x\Leftrightarrow h\in gG_xg^{-1} $$

However I can't follow the reasoning in this part : $$g^{-1}hg\in G_x\Leftrightarrow h\in gG_xg^{-1} $$

How does this hold?

Answer 1

Operate $g$ from left side and $g^{-1}$ from right side. So there exists $k \in G_x$ such that $g^{-1}hg = k$ if and only if $h = gkg^{-1}$ which implies that $h \in gG_xg^{-1}$.