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 [itex]G[/itex] be an abelian group containing elements [itex]a[/itex] and [itex]b[/itex] of orders [itex]m[/itex] and [itex]n[/itex] respectively. Show that [itex]G[/itex] contains an element whoes order is the least common multiple of [itex]m[/itex] and [itex]n[/itex].

Homework Equations

The Attempt at a Solution

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

But this does not do the job, since I need to prove [itex]n'm'd \vert k[/itex], 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 [itex]ord(ab)|l[/itex].

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

 

[Barre]

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

Consider, for example, [itex]b = a^{-1}[/itex]. These two elements have the same order, say m. Thus [itex]l = lcm(m,m) = m[/itex]. 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.