Let $G$ is a group and $x,y\in G$ then $y\in Cl_{G}(x)\Leftrightarrow x\in Cl_{G}(y)$

Question

Let $G$ is a group and $x,y\in G$ then $y\in Cl_{G}(x)\Leftrightarrow x\in Cl_{G}(y)$.

Let $Cl_{G}(x)=\left \{ gxg^{-1}:g\in G \right \}$ and $Cl_{G}(y)=\left \{ hyh^{-1}:h\in G \right \}$ ($h,g\in G$).

Suppose $y\in Cl_{G}(x)$ then there exists $ x\in G$ such that $y=gxg^{-1}$. But from here I couldn't get $x=hyh^{-1}$ which means $x\in Cl_{G}(y)$

Any idea will be appreciated.

Answer 1

Building on Aryaman's comment: Take $y \in \text{Cl}_G(x)$. Then there exists $g \in G$ such that $y = gxg^{-1}$. Now multiply this expression from the left with $g^{-1}$ and from the right with $g$. You get $x = g^{-1}yg$. If you let $h := g^{-1}$, you now clearly see that $x \in \text{Cl}_G(y)$, since $h^{-1} = (g^{-1})^{-1} = g$.