# $|AB|=mn$ if $\gcd(m,n)=1$

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: math.stackexchange.com/questions/28332/… –  joriki Oct 26 '11 at 10:12
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. –  user641 Oct 26 '11 at 11:36
@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. –  Arturo Magidin Jan 24 '12 at 16:37
The problem is equivalent to proving that $A \cap B = \{ e \}$. –  N. S. Jan 24 '12 at 16:52

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 $0\leq b<t$. We have that $1=x^r=(x^{t})^sx^b=1^sx^b=x^b=1$, and because of the minimality of $t$, we must have that $b=0$. This shows that $t\mid r$. In fact the same proof shows that if $k$ is any number such that $x^k=1$, then $t\mid k$.

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. –  rfauffar Oct 26 '11 at 15:16