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I want to show that if $g,h\in G$ are group elements with finite coprime orders $m$ and $n$ and $gh=hg$ then the order of $gh$ is $mn$.

We have that

$$ (gh)^{mn}=g^{mn}h^{mn}=1 $$

using $gh=hg$, so $|gh|\leq mn$. On the other hand, $mn=\operatorname{gcd}(m,n)\operatorname{lcm}(m,n)=\operatorname{lcm}(m,n)$, since $m$ and $n$ are coprime. I am not sure how to proceed, I tried assuming that $|gh|<mn$ and arriving at a contradiction, but it did not take me anywhere.

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Let $r$ be such that $(gh)^r = 1$. Then $1 = (gh)^{rm} = g^{rm}h^{rm} = h^{rm}$, and so $n\mid rm$ and thus $n\mid r$. Similarly you can show that $m\mid r$, and combining these results gives $mn\mid r$.

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