Past exam of the mathematics course of the University of Cambridge / 2016 / ia / Paper 3 / 5D / i / Solution

No such nonabelian group exists. Every element, including the identity, satisfies $g^2=1$, so $g^{-1}=g$. For any $a,b$,Therefore the group is an abelian group. This proves that a group of exponent two is abelian without any finiteness assumption.