let $a\sim b$ iff for some integer $k$, $a^k = b^k$ Let $G$ be a group, 
 Let $a\sim b$ iff  for some integer $m$, $a^m = b^m$.

I am having a problem trying to figure out how to prove that the transitive property. I know that you start off by Assuming $a\sim b$ and $b\sim c$, such that $a^m = b^m$ and $b^n = c^n$ for some $m$, $n$ that exist in the integers. I am stuck and not sure where to go from here. I just need a nudge in the right direction.
Note that $a^{mn}=b^{mn}$ and $b^{nm}=c^{nm}$.

Remark: The relation is uninteresting if we allow exponents to be $0$.
Here's another approach. If $a^k = b^k$, then $a^k/b^k = (a/b)^k = 1$. Thus, $a\sim b$ iff $(a/b)^k$. Of course if an element has order dividing $k$, then so does its inverse, so $(b/a)^k = ((a/b)^{-1})^k = ((a/b)^k)^{-1} = 1^{-1} = 1$.