The question proposes that if: $$|g|=|h|=mn$$ then $$|(gh)^m|=n$$ where $m$ and $n$ are any positive integers. I either have to prove that this is true or provide a counter-example. I have tried some basic examples from the integers modulo $10$ as well as trying $h=g^{-1}$ but nothing I have tried contradicts the statement. Is this true for the general case? How would I go about proving it? It seems like it should be easy but I'm stuck on what the order of a product would be, other than that it divides the product of the orders of each element.
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3Why $h=g^{-1}$ wouldn't work? – Watson Nov 14 '16 at 18:37
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See this question. Thank you, @PaulPlummer – Dietrich Burde Nov 15 '16 at 09:19
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If $h = g^{-1}$, then $gh = e$, so $(gh)^m = e^m = e$, so the order of $(gh)^m$ is $1$ (and if $n\neq 1$, which it will be if $g\neq e$, then this is a contradiction to the statement). In general, the order of $(gh)^m$ need not have any relation to the orders of $g$ and $h$: If you consider the group generated by two elements $g$ and $h$ such that $g^2 = h^2 = e$, but $gh\neq hg$, then $(gh)^2$ to any nonzero power will not be the identity, so $(gh)^2$ has infinite order (and $\infty$ doesn't divide $4$). In a finite abelian group you can say more: see Eric Naslund's answer here.
