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I noted a discussion on groups being abelian under a certain restriction on powers of elements, e.g. Maybe this result (probably not too well-known) concludes it all.

Let $m$ and $n$ be coprime natural numbers. Assume that $G$ is a group such that $m$-th powers commute and $n$-th powers commute (that is for all $g, h$ $\in$ $G$: $g^mh^m=h^mg^m$ and $g^nh^n=h^ng^n$). Then $G$ is abelian.

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marked as duplicate by darij grinberg, William, Daniel W. Farlow, Jeremy Rickard, T. Bongers 2 days ago

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

So what is your question? – Derek Holt Sep 27 '11 at 9:00
@Derek: Let $G$ be a finite group. Assume that there are two coprime integers $m$ and $n$ such that for all $g, h \in G$ holds (1) $g^mh^m = h^mg^m$ and (2) $g^nh^n = h^ng^n$. How do you prove that $G$ is abelian? – Someone Sep 27 '11 at 14:06

$(m,n)=1\implies pn+qm=1$.

$(g^nh^m)^{np}=g^n(((h^mg^n)^p)^n(h^mg^n)^{-1})h^m= (h^mg^n)^{pn}g^n(h^mg^n)^{-1}h^m=(h^mg^n)^{pn}$.

$(g^nh^m)^{mq}=g^n((h^mg^n)^{mq} (h^mg^n)^{-1})h^m=g^n((h^mg^n)^{-1} (h^mg^n)^{mq})h^m=(h^mg^n)^{qm}$.

$(g^nh^m)^{np}(g^nh^m)^{mq}=(h^mg^n)^{np}(h^mg^n)^{qm}\implies g^nh^m=h^mg^n$.


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This is false. In any nonabelian group of exponent $m$, $m$th powers and $n$th powers commute for any $n$, in particular any $n$ coprime to $m$.

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Maybe I was not clear enough: I did not mean the $m$-powers commuting with the $n$-powers, but the $m$-powers among each other and the $n$-powers among each other. – Nicky Hekster Sep 27 '11 at 7:59

$G$ does not have to be finite. Let $M \subset G$ be the subgroup generated by all $m$-th powers and let $N \subset G$ be the subgroup generated by all $n$-th powers. These subgroups are clearly abelian normal subgroups. Since $m$ and $n$ are coprime, $G = MN$, and hence $M \cap N$ is contained in the center $Z(G)$ of $G$. To prove that $G$ is abelian it suffices to show that $M$ and $N$ commute, that is $[M,N]=1$. Note that $[M,N] \subset (M \cap N)$ (since $M$ and $N$ are normal subgroups). Let $a \in M$ and $b \in N$. Then $[a, b] = a^{−1}b^{−1}ab \in M \cap N$. Hence $[a, b] = z$ with $z \in Z(G)$. Hence $b^{−1}ab = za$, whence $b^{−1}a^nb=z^na^n$. Since $a^n \in N$ it commutes with $b$, so $z^n=1$. Similarly $z^m=1$. Since $m$ and $n$ are relatively prime, we conclude $z=1$.

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Why do you tag the question finite-groups, if you don't assume your groups to be finite? – j.p. Sep 27 '11 at 19:40
What question? As far as I can see, no question was asked. – Derek Holt Sep 27 '11 at 22:00
I'm confused - were you looking for help, or just posting a result for all to see? The latter isn't really the intended use of this site. – Alon Amit Sep 27 '11 at 22:37

I am also confused with the question ! Isn`t the term " Abelian " synonymous of " "commutative" ? Meaning that a group G is " Abelian " if $ab=ba$ for all $a,b \in G$

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The hypothesis is not "$ab=ba$ for all $a,b\in G$." Only special cases of elements commuting are assumed. – anon Sep 27 '11 at 11:31

Assuming that $G$ is finite (looking at the tags), you can prove that for any fixed prime $p$ all $p$-elements commute ($p$ does not divide $m$ or $n$).

Then conclude that the $p$-Sylow sugroups are normal and abelian for all $p$.

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This approach has recently been written up: -- note that finitely generated metabelian groups are residually finite (a theorem of Philip Hall), so the result extends from the finite group case to arbitrary groups. – Giles Gardam Jun 28 at 19:07

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