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If $A$ and $B$ are finite subgroups, of orders $m$ and $n$, respectively, of the abelian group $G$, prove that $AB$ is a subgroup of order $mn$ if $m$ and $n$ are relatively prime.

Lagrange's theorem has not been introduced in this part of the book, so please refrain from using it. I think I managed to prove it by considering the minimal generating set of the subgroups. But my proof is quite long, while this is supposedly middle-level question of the problem set. So I expect to see some simple proofs.

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    This might be relevant: http://math.stackexchange.com/questions/28332/is-lagranges-theorem-the-most-basic-result-in-finite-group-theory2011-10-26
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    Without the "relatively prime" condition, this is a famous problem from Herstein's *Topics in Algebra*, and an elementary solution was only published a couple of years ago IIRC.2011-10-26
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    @Steve: Herstein's "toughest problem" was "if an abelian group has elements of order $m$ and $n$, then it has an element of order $\mathrm{lcm}(mn)$. It's hard given what the book had covered up until that point.2012-01-24
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    The problem is equivalent to proving that $A \cap B = \{ e \}$.2012-01-24

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