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Thread: Proving G is abelian

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    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.
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    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.
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    Quote Originally Posted by Opalg View Post
    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.
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    Quote Originally Posted by ThePerfectHacker View Post
    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 $\displaystyle a^mb^n = b^na^m$.
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