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

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

**Physics Forums | Science Articles, Homework Help, Discussion**