Showing that $x \sim y$ if $x^n=y^n$ defines an equivalence relation on group $G$.

Question

How can we show that given a group $G$ and $x,y \in G$ that $x \sim y$ if $x^n = y^n$, for some $n > 0$, defines an equivalence relation? The only issue is transitivity, because we need to work with $x^n = y^n$ and $y^m = z^m$. I tried to use the division algorithm but that didn't seem to help. Any clues or help is appreciated.

Answer 1

Let $x\sim y$ and $y\sim z$. Then there exist $n,m\in\Bbb N$ such that $x^n=y^n$ and $y^m=z^m$. Consider $x^{mn}$: we have

$$\begin{align} x^{mn}&=(x^n)^m\\ &=(y^n)^m\\ &=(y^m)^n\\ &=(z^m)^n\\ &=z^{mn}. \end{align}$$

Hence $x\sim z$.