TITLE: The conditions $a^2b^2=b^2a^2$ and $a^3b^3=b^3a^3$ make the group $G$ abelian?
QUESTION [9 upvotes]: Let $G$ be a group. Assume that $a^2b^2=b^2a^2$ and $a^3b^3=b^3a^3$ for all $a, b \in G$. Prove that the group $G$ is abelian.

Also please tell me whether there is any standard approach in proving commutativity of groups like this problem.

REPLY [10 votes]: Let $M \subset G$ be the subgroup generated by all squares and let $N \subset G$ be the subgroup generated by all cubes. These subgroups are clearly abelian normal subgroups. Since 2 and 3 are coprime $G = MN$ (use Bézout's Theorem!) and $M \cap N$ 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 $[M,N] \subseteq (M \cap N)$. Let $x \in M$ and $y \in N$.
Then $[x, y] = x^{−1}y^{−1}xy \in M \cap N$. Hence $[x, y] = z$ with $z \in Z(G)$. Hence $y^{−1}xy = zx$, whence $y^{−1}x^3y=z^3x^3$. Since $x^3 \in N$ it commutes with $y$, so $z^3=1$. Similarly $z^2=1$. Since $2$ and $3$ are relatively prime, we conclude $z=1$.