Romanian Mathematical Olympiad

a) For $a\in G$, let $C(a)=\{b\mid b\in G$ and $ab=ba\}$. The hypothesis shows that $C(a^2)\subset C(a)$. Since $C(a)\subset C(a^2)$, we see that $C(a)=C(a^2)$, for every $a\in G$. Consequently, $C(a)=C(a^2)=\cdots =C(a^{2^n})=C(e)=G$, for every $a\in G$, that is $G$ is abelian.

b) An example of a nonabelian group with the given property is the multiplicative group $G$ of the matrices of the form $$\left(\begin{array}{ccc}\hat{1}& a& b\\ \hat{0}& \hat{1}& c\\ \hat{0}& \hat{0}& \hat{1}\end{array}\right),\text{ with }a,b,c\in {\mathbb{Z}}_3.$$ Since $A^3=I_3$ for every $A\in G$, if $A^{2}B=B{A}^2$, then $A^{-1}B=B{A}^{-1}$, so $AB=BA$. In effect, any finite group $G$ of odd order bears the given property, since $C(a^2)\subset C(a^{|G|+1})=C(a)\subset C(a^2)$ (so $C(a)=C(a^2)$), for all $a\in G$. Therefore any such nonabelian group makes an example for b).