# Thread: Proving G is abelian

1. ## Proving G is abelian

Prove that a group$\displaystyle G$ in which all $\displaystyle m-th$ power elements commute with $\displaystyle m-th$ power elements and all $\displaystyle n-th$ power elements commute with $\displaystyle n-th$ power elements, where $\displaystyle m,n$ are relatively prime is abelian.

2. Prove that for a, b in G, $\displaystyle a^{km+ln}$ commutes with $\displaystyle b^{km+ln}$, for all integers k, l. Then use the fact that m and n are relatively prime to deduce that k, l can be chosen so that km+ln=1.

3. Originally Posted by Opalg
Prove that for a, b in G, $\displaystyle a^{km+ln}$ commutes with $\displaystyle b^{km+ln}$, for all integers k, l. Then use the fact that m and n are relatively prime to deduce that k, l can be chosen so that km+ln=1.
That is what I was thinking but how do you actually prove that? Notice that it sufficies to prove that $\displaystyle a^n b^m = b^m a^n$ for all $\displaystyle a,b\in G$ because of what you just said. I do not see how this follows in any straightforward way.

4. Originally Posted by ThePerfectHacker
That is what I was thinking but how do you actually prove that? Notice that it sufficies to prove that $\displaystyle a^n b^m = b^m a^n$ for all $\displaystyle a,b\in G$ because of what you just said. I do not see how this follows in any straightforward way.
You're right, of course. I was overlooking that.

There is a neat trick for proving that $\displaystyle a^n b^m = b^m a^n$, which I found on another forum. The key to it is the calculation $\displaystyle (a^mb^n)^m = a^m(b^na^m)^m(b^na^m)^{-1}b^n = (b^na^m)^ma^m(b^na^m)^{-1}b^n = (b^na^m)^m$, and similarly $\displaystyle (a^mb^n)^n = (b^na^m)^n$. Since m and n are coprime, it follows that $\displaystyle a^mb^n = b^na^m$.