Let $G$ be a group, we denote the subgroup of it generated by $m$ th power of elements of $G$ by $G(m)$. For integers $m,n$, given $G(m),G(n)$ both being abelian, how to prove that $G(gcd(m,n))$ is also abelian? I know we can assume that $m,n$ are relatively prime, to prove $G$ itself is abelian, but so far I haven’t made any progress. So when $m,n$ are relatively prime, it suffices to prove that any $a,b$ in $G$, $a^{m}$ commutes with $b^{n}$, but proving so by definition became a dead end.
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