Group theory, order of a product of two elements

[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.

Homework Statement

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.

Homework Equations

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.

 

[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.

 

[Barre]

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.

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.