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Math Help - Proving G is abelian

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    Proving G is abelian

    Prove that a group G in which all m-th power elements commute with m-th power elements and all n-th power elements commute with n-th power elements, where m,n are relatively prime is abelian.
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    Prove that for a, b in G, a^{km+ln} commutes with 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, a^{km+ln} commutes with 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 a^n b^m = b^m a^n for all 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 a^n b^m = b^m a^n for all 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 a^n b^m = b^m a^n, which I found on another forum. The key to it is the calculation (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 (a^mb^n)^n = (b^na^m)^n. Since m and n are coprime, it follows that a^mb^n = b^na^m.
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