Background: I am trying to define rational powers of elements in an arbitrary group. For this, I first define $n$-th powers for $n\in\mathbb Z\setminus\{0\}$: I call $y$ an $n$-th root of $x$ iff $y^n = x$.
Now, while trying to come up with a definition for rational exponentiation, and then then trying to deduct the various laws of exponents, I am stuck on the following problem:
Let $G$ be a group and $x, y\in G$. Let $m, n\in\mathbb Z\setminus\{0\}$ such that $\gcd(m, n) = 1$ and $x^m = y^n$. Do $x$ and $y$ respectively have $n$-th and $m$-th roots?
I don't think it should be true in general, but my knowledge of groups is limited. Can you give some counterexample?
Any insights in special cases (for instance, take $G$ to be abelian and/or ordered) will be appreciated a lot too!
A counterexample is $G = {\rm SL}(2,3)$ with $$x = \left(\begin{array}{rr}0&-1\\1&0\end{array}\right) \quad {\rm and}\quad y = \left(\begin{array}{rr}-1&1\\0&-1\end{array}\right).$$ We have $x^2 = y^3 = -I_2$.
Then $x$ does have a cube root because its order is $4$, which is coprime to $3$, but $y$ has no square root, because a square root of $y$ would have order $12$ and there is no such element in $G$. (Note that $G/Z(G) \cong {\rm PSL}(2,3) \cong A_4$ which has no element of order $6$.)