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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 14 hours 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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