# Showing that stabilizer of group action is conjugate to stabilizer.

Let $G\rightarrow X$. Show that $stab(g\cdot x)$ is conjugate to $stab(x)$.

To make $G$ act on itself by conjugation, take $X = G$ and let $g \times x = gxg^{-1}$ Here $g \in G$ and $x \in G$ Since $e \times x = exe^{-1} = x$ and \begin{align*} g_1 \circ (g_2 \circ x) & = g_1 \circ (g_2xg_2^{-1}) \\ & = g_1(g_2xg_2^{-1})g_1^{-1} \\ & = (g_1g_2)x(g_1g_2)^{-1} \\ & = (g_1g_2) \circ x \\ \end{align*} Conjugation is a group action.

Did I do this correctly?

• No, you just showed part of the proof that conjugation is a group action. Showing that the stabilizers are conjugate is not the same as showing that conjugation is a group action. – Qudit Apr 5 '15 at 2:38
• see my new attempt, and tell me how that is – All About Groups Apr 6 '15 at 22:54

$G$ acts on $X$ : We have a relation between stabilizers : $$z\in G_x \Leftrightarrow z\cdot x=x \Rightarrow (gzg^{-1} )\cdot (g\cdot x ) =g\cdot x \Rightarrow gG_xg^{-1}\subset G_{g\cdot x}$$
So $$(G_x\subset )\ g^{-1} G_{g\cdot x} g \subseteq G_x$$ so that $$G_x =g^{-1} G_{g\cdot x } g$$
First, consider some $z\in stab(x)$. Thus, $z\cdot x = x$, which implies that $(gzg^{-1})\cdot(g\cdot x) = gz(g^{-1}\cdot g)\cdot x) = g(z\cdot x) = g\cdot x$ and therefore $gzg^{-1}\in stab(g\cdot x)$ and $stab(g\cdot x)$ is conjugate to $stab(x)$.