Let $G$ be a group acting on a set $X$. Let $x,y \in X$ be from the same orbit. I managed to prove that $\mathrm{Stab}(x)$ and $\mathrm{Stab}(y)$ are conjugates. But how to deduce from this that $\mathrm{Stab}(x)$ and $\mathrm{Stab}(y)$ have the same order?

Could we just rely on the fact that each element from $\mathrm{Stab}(x)$ have a conjugate element with the same order from $\mathrm{Stab}(y)$?

Thanks for your help!


1 Answer 1


So, you've shown that $\mathrm{Stab}(x)$ and $\mathrm{Stab}(y)$ are conjugate subgroups of $G$, i.e., there exists some element $g\in G$ for which $\mathrm{Stab}(x)=g^{-1}\mathrm{Stab}(y)g$.

Now, prove that the function $c_g:G\to G$ defined by $c_g(h)=g^{-1}hg$ is a bijection, so that $\mathrm{Stab}(x)$ and its image under $c_g$, namely $\mathrm{Stab}(y)$, must have the same cardinality. (Note: such a function is known as an inner automorphism of $G$, see on Wikipedia.)


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