Solution
= 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 gives
$$
ab=(ab)^{-1}=b^{-1}a^{-1}=ba.
$$
Thus \b[$G$ is an <abelian group>]. Finiteness is not needed for this implication; it holds for every <group> in which each element squares to the identity.