A criterion for a group to be abelian I noted a discussion on groups being abelian under a certain restriction on powers of elements, e.g. http://tiny.cc/chs45. 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.
 A: $(m,n)=1\implies pn+qm=1$.
$(g^nh^m)^{np}=g^n(((h^mg^n)^p)^n(h^mg^n)^{-1})h^m= (h^mg^n)^{pn}g^n(h^mg^n)^{-1}h^m=(h^mg^n)^{pn}$.
$(g^nh^m)^{mq}=g^n((h^mg^n)^{mq} (h^mg^n)^{-1})h^m=g^n((h^mg^n)^{-1} (h^mg^n)^{mq})h^m=(h^mg^n)^{qm}$.
$(g^nh^m)^{np}(g^nh^m)^{mq}=(h^mg^n)^{np}(h^mg^n)^{qm}\implies g^nh^m=h^mg^n$.
$gh=(gh)^{pn+mq}=(hg)^{pn+mq}=hg$.
A: $G$ 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 $z=1$.
A: 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 $m$.
A: 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 $p$.
A: 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 $a,b \in G$
