Prove that stabilizer subgroups of G are conjugate to each other

Question

Suppose that a group $G$ acts on a set $X$. Show that if $x_1$ and $x_2$ in X are in the same $G$-orbit, then their stabilizer subgroups of $G$ are conjugate to each other.

My proof:

Assume $x_1 = g_1x$ and $x_2 = g_2 x$ for some $g_1, g_2 \in G$. Let $h \in G_{x_1}$. We claim that $g_2g_1^{-1}hg_1g_2^{-1}$ is in $G_{x_2}$, thus proving that the two stabilizer subgroups are conjugate to each other.

Indeed, $$\begin{align} x_1&=g_1x\\ g_2g_1^{-1}x_1&=g_2x\\ g_2g_1^{-1}hx_1&=g_2x\\ g_2g_1^{-1}hg_1x&=g_2x\\ (g_2g_1^{-1}hg_1g_2^{-1})x_2&=x_2\\ \end{align}$$ as desired.

I think it is a bit messy. Can you please comment on my proof and leave your own proof so that I can learn in a better way? Thanks in advance.

The proof is fine, but you might start with $x_2=gx_1$ instead of introducing $x$. — Hagen von Eitzen, Nov 10 '14 at 16:07

score 8 · Accepted Answer · answered Nov 10 '14 at 16:28

8

Suppose $x$ and $y$ are in the same orbit, so $x=gy$ for some $g$. If $h$ stabilizes $y$, then $ghg^{-1}x=ghy=gy=x$. This gives an injective homomorphism from $Stab(y)$ to $Stab(x)$. You can verify that $h\mapsto g^{-1}hg$ is its inverse, so the two groups are conjugate.

answered Nov 10 '14 at 16:28

Matt Samuel

58,164

That's clear! Thanks! – Nighty Nov 11 '14 at 06:20

Prove that stabilizer subgroups of G are conjugate to each other

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