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