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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?

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    $\begingroup$ 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. $\endgroup$ – Qudit Apr 5 '15 at 2:38
  • $\begingroup$ see my new attempt, and tell me how that is $\endgroup$ – All About Groups Apr 6 '15 at 22:54
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$G$ 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 $$ G_x =g^{-1} G_{g\cdot x } g $$

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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 $stab(x)$.

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