Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
This might be relevant:… – 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).

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.