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Let $m, n$ be coprime positive integers. Let $G$ be an abelian group of order $mn$. Let

$$ H = \{ g^m : g \in G \} \quad K = \{ g^n : g \in G \} \\ $$

Show that $G = HK \cong H \times K$

So far I've shown that that $HK \cong H \times K$ but I'm stuck figuring out how to show that $G = HK$. I know that $HK \subseteq G $ but I'm unsure how to prove the other direction. Any help would be appreciated.

Note: This is part of a bigger question. In the previous parts I've shown that $H, K$ are subgroups of $G$, that any element of $H$ has order dividing $n$, any element of $K$ has order dividing $m$, and $H \cap K = \{ id \}$. I thought I'd include them just in case they are useful.

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    $\begingroup$ It might be worth noting that for any group $G$, we have $G=HK$. Indeed, since $\gcd(m,n)=1$, there exist integers $a,b$ such that $am+bn=1$. Then for all $g\in G$, we have $$g = g^1 = g^{am+bn} = (g^a)^m (g^b)^n \in HK.$$ However, in general neither $H$ nor $K$ are subgroups when $G$ is not abelian. $\endgroup$ Commented Nov 1, 2023 at 17:05

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You can do this in one fell swoop, given the information you have already.

Use this comment. Then $G=HK$.

But $G\cong H\times K$ as an inner direct product iff all of:

  • $H, K\unlhd G$ (but you have shown $H,K\le G$, and $G$ is abelian);
  • $H\cap K=\{e\}$ (shown); and
  • $G=HK$.
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