 Solution

The hypothesis implies $g^2=e$ for every $g\in G$, including the identity, so $g^{-1}=g$. For arbitrary $a,b\in G$, their product also equals its own inverse. The inverse-of-a-product rule therefore givesThus $G$ is an abelian group. Finiteness is not needed for this implication; it holds for every group in which each element squares to the identity.