Show abelian group has element of order [m,n]

[TopCat]

Let G be an abelian group containing elements a and b of order m and n, respectively. Show that G contains an element of order [m,n] (the LCM of m and n).

This is true when (m,n)=1, because mn(a+b) = e, and if |a+b|=h, then h|mn. Now, hm(a+b) →m|h and similarly I find that n|h. But (m,n)=1, so this implies mn|h and mn = [m,n].

Now assume that (m,n)>1 and |a+b|=h. So h(a+b)=e, which implies 1)ha=hb=e, or 2)ha=-hb. If 1), then we note that [m,n] is the smallest number divisible by both m and n, so that h=[m,n]. Now, if 2) is the case, I'm not sure how to proceed. It's trivial if h=[m,n], so we must assume h≠[m,n] and get a contradiction. This hasn't worked for me.

If instead I try and show that [m,n]|h, the best I can get is that [m,n]/(m,n) | h, since I can only use a similar argument to paragraph one with m/(m,n) and n/(m,n) since they are coprime. If I consider the case where m/(m,n) and n are coprime, the result is easy, but then what if they aren't coprime?

 

[micromass]

Maybe you can do this in folloing steos:

1) If h=[m,n], then h(a+b)=0. This proves that the o(a+b) divides [m,n].
2) Prove that m divides o(a+b) and that n divides o(a+b). This will prove that [m,n] divides o(a+b)

PS: I would appreciate it if you were to follow the homework template next time.

 

[TopCat]

Ah, thanks for the tip. I needed to prove the lemma m|h, n|h imply [m,n]|h.

And my apologies. I'll be sure to use the template next time.