# Group theory, order of a product of two elements

1. Jan 28, 2012

### Barre

I search for an 'elementary' proof of this, where results about structure of abelian groups are not used. I've tried a standard way of proving this, but hit a wall. I'm mainly interested if my work on a proof can be expanded to a full solution.

1. The problem statement, all variables and given/known data
Let $G$ be an abelian group containing elements $a$ and $b$ of orders $m$ and $n$ respectively. Show that $G$ contains an element whoes order is the least common multiple of $m$ and $n$.

2. Relevant equations

3. The attempt at a solution
I'll try to prove $ab$ has order $l = lcm(m,n)$. Clearly $(ab)^l = e$. So we know that $ord(ab) \vert l$. Assuming $(ab)^k = e$ and
$1 < k \leq l$ I would have to prove that $k = l$.
Define $d = gcd(m,n)$. Then we can write $m= m'd$ and $n = n'd$. It's easy to see that $l = m'n'd$, and $gcd(m',n') = 1$.
If $(ab)^k = e$, then $a^k = b^{-k}$ and orders of these elements must be the same. So we have that $(a^k)^m = a^{km} = e = b^{-km}$ and we see that $n \vert km$ which re-written means $n'd \vert km'd$ and $n' \vert km'$ and since $gcd(m',n') = 1$ we get that $n' \vert k$. Repeating this procedure the other way around we can prove that $m' \vert k$ and finally $m'n' \vert k$.

But this does not do the job, since I need to prove $n'm'd \vert k$, and I can't find a way to do this.

2. Jan 28, 2012

### jbunniii

I think that without further assumptions, all you can conclude about the order of ab is that $ord(ab)|l$.

Consider, for example, $b = a^{-1}$. These two elements have the same order, say m. Thus $l = lcm(m,m) = m$. But ord(ab) = ord(e) = 1.

Last edited: Jan 28, 2012
3. Jan 28, 2012

### Barre

Yes, the textbook is a bit unclear I think. As you point out, if we chose an element and it's inverse, the result does not hold. We could restrict this so that a is not the inverse of b.