Prove that $\operatorname{Stab}_G(y) \cong\operatorname{Stab}_G(x).$ > Suppose that $G$ acts on a set $X$. Let $x,y \in X$ and suppose that $y \in G\cdot x$. > > Prove that $$\operatorname{Stab}

Prove that $\operatorname{Stab}_G(y) \cong\operatorname{Stab}_G(x).$ > Suppose that $G$ acts on a set $X$. Let $x,y \in X$ and suppose that $y \in G\cdot x$. > > Prove that $$\operatorname{Stab}_G(y) \cong\operatorname{Stab}_G(x).$$ Can anyone help me with this proof?

It is given that $y\in Gx$ which means that there exists $h\in G$ such that $y=hx$.

Thus, if $g\in \text{Stab}_G(y)$ then $gy=y$ hence $ghx=hx$ and so $h^{-1}ghx=x$.

In particular $g\in \text{Stab}_G(y)$ implies that $h^{-1}g h \in \text{Stab}_G(x)$.

Let $\varphi:\text{Stab}_G(y)\rightarrow \text{Stab}_G(x)$ be such that $\varphi(g)=h^{-1}gh$ we just saw above that $\varphi$ is well defined. It is also easy to check that $\varphi$ a homomorphism.

Furthermore, the inverse of $\varphi$ is given by $\varphi^{-1}(g) = hgh^{-1}$. The reason that $\varphi^{-1}$ is well defined is similar to that of $\varphi$ reversing the roles of $x,y$ .