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This may be a straightforward question, but if I have a group $G$ acting on a set $A$, and two elements $a,b\in A$ belong to the same orbit, how do I show that their stabilizers are conjugate.

So far I know that $a=gb$ for some $g\in G$. Do I just need to show that $g$ times some element of the stabilizer of $b$ is equal to a stabilizer of $a$?

Shaun
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Jesse Jahn
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2 Answers2

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The stabilizer $A$ of $a$ is the $g$ conjugate of the stabilizer $B$ of $b$ : $$h\in A \iff ha= a \iff hg^{-1}bg=g^{-1}bg \iff ghg^{-1}\in B$$ In other words, $B=gAg^{-1}$

Alexander Gruber
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Thomas
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    Writing $g^{-1}bg$ does not make much sense when the group acts from the left, or? Do you mean $$ h \in A \Leftrightarrow ha = a \Leftrightarrow hgb = gb \Leftrightarrow g^{-1}hgb = b \Leftrightarrow g^{-1}hg \in B? $$ – StefanH Mar 06 '18 at 13:42
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It is $b \in O(a) \Leftrightarrow \exists \bar g \in G \mid b= \bar g \cdot a$. Then:

\begin{alignat}{1} \operatorname{Stab}(b) &= \{g \in G \mid g \cdot b = b\} \\ &= \{g \in G \mid g \cdot (\bar g \cdot a) = \bar g \cdot a\} \\ &= \{g \in G \mid (g \bar g) \cdot a = \bar g \cdot a\} \\ &= \{g \in G \mid \bar g^{-1}\cdot((g \bar g) \cdot a) = \bar g^{-1}\cdot(\bar g \cdot a)\} \\ &= \{g \in G \mid (\bar g^{-1}g \bar g) \cdot a = a\} \\ \tag 1 \end{alignat}

Now, call $g':=\bar g^{-1}g \bar g$; then, $g=\bar gg'\bar g^{-1}$ and $(1)$ reads:

\begin{alignat}{1} \operatorname{Stab}(b) &= \{\bar gg'\bar g^{-1} \in G \mid g' \cdot a = a\} \\ &= \{\bar gg'\bar g^{-1} \in G \mid g' \in \operatorname{Stab}(a)\} \\ &= \bar g \operatorname{Stab}(a)\bar g^{-1} \end{alignat}