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.(adsbygoogle = window.adsbygoogle || []).push({});

1. The problem statement, all variables and given/known data

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].

2. Relevant equations

3. 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.

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# Group theory, order of a product of two elements

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