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Let $G$ be an abelian group and let $a,b \in G$ with $a$ having finite order $n$ and $b$ having finite order $m$. Assume that $m,n$ are distinct prime numbers. Prove that if $ab \neq e$, then the order of $ab$ is $mn$.

I got $a^mb^n$ becoming $ab$ to the power of m+n since they both belong to the group G.

Toby
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    Assuming so, then $(ab)^{mn} = a^{mn} b^{mn}$ (your calculation is wrong there), which gives $(a^n)^m (b^m)^n = e$. So we know that the order of $ab$ divides $mn$, and since $m,n$ are coprime we must have that the order of $ab$ is 1 or $mn$. The first case is ruled out by the assumption, and we are done. – mi.f.zh Jan 30 '20 at 04:38
  • @nhmwhhxx Thanks! is there other assumptions you can make to prove that $ab \neq e$ other than assuming so? Or is that the only condition we can base it off as? – Toby Jan 30 '20 at 04:46
  • That the product is not the neutral element is always true since the inverse of an element has the same order as the element itself. – Tobias Kildetoft Jan 30 '20 at 04:49

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