For a group $G$, show the relation $x\sim y$ defined by $\exists a(y=axa^{-1})$ is an equivalence relation on $G$.

Question

Let G be a group. For $x,y\in G$, define $x\sim y$ if there exists some element $a\in G$ such that $y=axa^{-1}$. Show that ~ defines an equivalence relation on $G$.

Answer 1

$x \sim x$

$a:=e$ satisfies the condition.

$x \sim y \implies y \sim x$

Suppose $x \sim y$. Then there exists an $a$ such that $y=axa^{-1}$. Rearranging the equation: $$a^{-1}ya=x$$ Recall that for an arbitrary group and an element $x$ in the group $(x^{-1})^{-1}=x$. Thus $a^{-1}$ satisfies the condition.

$x \sim y \wedge y \sim z \implies x \sim z$

Suppose $x \sim y$ and $y \sim z$. Then there exist $a,b$ such that: $$x=aya^{-1} \space \space \space (1)$$ $$y=bzb^{-1}$$ From equation (1) we have: $$y=a^{-1}xa$$ $$a^{-1}xa=bzb^{-1}$$ $$b^{-1}a^{-1}xab=z$$ Recall that in an arbitrary group all elements $x,y$ satisfy $(xy)^{-1}=y^{-1}x^{-1}$

So $ab$ indeed satisfies the condition. This concludes the proof.