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If $g^2 = e$ for all $g$ in $G$, then $G$ is abelian

Proof. Assume that for all $g$ in $G$ that $g^2 = e$ holds. Let $a,b \in G$. Then since $G$ is a group, we know that $ab \in G$. Further, we know that $a^2 = e$ and that $b^2 = e$. Now take $(ab)^2 = abab$. Because $(ab)^2 = e$, we have that $abab = e$. Now, multiply $b^{-1}a^{-1}$ on both sides on the right side of $abab = e$, then we get $ab = b^{-1}a^{-1}$. As $a^2 = e$, we have that $a = a^{-1}$. The same holds argument for $b$. So we get $ab = b^{-1}a^{-1} = ba$. Wrapping everything together, we can write the proof as