Let $G$ be a finite group with a transitive action on $X,|X|=n$. Prove $\forall x,y\in X\exists g\in G$ s.t. $g^{-1}{\rm Stab}_G(x)g={\rm Stab}_G(y)$

Question

Let $G$ be a finite group with a transitive action on set $X,|X|=n$. Prove $\forall x,y\in X \space \exists g\in G$ such that $g^{-1}{\rm Stab}_G(x)g={\rm Stab}_G(y)$

$\implies$: This is a transitive action, hence $ \exists g\in G$ such that $g\cdot y=x$.

$g^{-1}{\rm Stab}_G(x)g\cdot y=g^{-1}{\rm Stab}_G(x)\cdot x=g^{-1}\cdot x=y\implies g^{-1}{\rm Stab}_G(x)g\subseteq {\rm Stab}_G(y).$

I have a problems with the other side, ${\rm Stab}_G(y)\subseteq g^{-1}{\rm Stab}_G(x)g$.

Any help is welcome!

Thanks!

Answer 1

Everything about groups is symmetric. Replace $x$ by $y$ and $g$ by $g^{-1},$ and live happily ever after.

Answer 2

In general, if $y\in O(x)$ (the orbit of $x\in X$), then there is some $g'\in G$ such that $y=g'\cdot x$. But then: \begin{alignat}{1} \operatorname {Stab}_G(y) &= \{g\in G\mid g\cdot y=y\} \\ &= \{g\in G\mid g\cdot (g'\cdot x)=g'\cdot x\} \\ &= \{g\in G\mid (g'^{-1}gg')\cdot x=x\} \\ \tag 1 \end{alignat} Now, call $g'':=g'^{-1}gg'$; then, $g=g'g''g'^{-1}$, and $(1)$ yields: \begin{alignat}{1} \operatorname {Stab}_G(y) &= \{g'g''g'^{-1}\in G\mid g''\cdot x=x\} \\ &= g'\{g''\in G\mid g''\cdot x=x\}g'^{-1} \\ &= g'\operatorname {Stab}_G(x)g'^{-1} \\ \end{alignat} If the action is transitive, then this holds for every $y\in X$, because the whole $X$ is the only one orbit. Moreover, the finiteness seems not to be relevant.