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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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If $H$ is a finite group of order $r$ and $x\in H$, we can see that $x^r=1$. This is because we have the set $\langle x\rangle:=\{1,x,x^2,\ldots,x^r\}\subset H$, and if $x^r\neq 1$ then $\langle x\rangle$ has more elements than $H$, a contradiction. We say that $\min\{t\in\mathbb{N}:x^t=1\}$ is the order of $x$.

If $t$ is the order of $x$, then $r=st+b$ for some $s$ and $0\leq b

Let's look at the problem now. We can see that $AB=\{ab:a\in A,b\in B\}$ has order less than or equal to $mn$ (one reason is that the function $A\times B\to AB$ that sends $(a,b)\mapsto ab$ is surjective and $|A\times B|=mn$).

When exactly is $ab=a'b'$? This happens if and only if $a(a')^{-1}=b(b)^{-1}$ (because $G$ is abelian). We see that $(a(a')^{-1})$ is in $A$, and so $1=(a(a')^{-1})^{m}=(b(b')^{-1})$. Similarly, we have that $1=(b(b')^{-1})^n=(a(a')^{-1})^m$. Because of what we said earlier we have that the order of $b(b')^{-1}$ divides $m$ and $n$ (and the same with the order of $a(a')^{-1}$). Since $m$ and $n$ are relatively prime, these orders must be equal to 1, and so $b=b'$ and $a=a'$. Thus every element of $AB$ is uniquely written as $ab$ with $a\in A$ and $b\in B$, and so we see that the function $A\times B\to AB$ we described earlier is a bijection, and so we're done.

Of course in this proof I proved Lagrange's theorem....

Another proof would consist of showing that the function $A\times B\to AB$ is an isomorphism (although I'm sure you would use Lagrange too).

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    I kind of assumed Lagrange at first when I said $x^r=1$... It really should be $x^r=x^s$ for some $s$... but the proof I gave below that the order of an element divides the order of the group works here too.2011-10-26