Let $G$ be a group. Prove that $g^2 = e$ for all $g \in G$, then $G$ is abelian. ($e$ is the identity element.)
My Solution: Let $a,b \in G$. Then $a(ab)b = a^2b^2 = e^2 =e$. Now I tried to reverse $ab$ in the brackets to get the same solution to show that the group is also commutative but I was not able to do so.