**Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 1.1 Exercise 1.1.25**

Let $G$ be a group. Prove that if $x^2 = 1$ for all $x \in G$, then $G$ is abelian.

Solution: Note that since $x^2 = 1$ for all $x \in G$, we have $x^{-1} = x$. Now let $a,b \in G$. We have $$ab = (ab)^{-1} = b^{-1} a^{-1} = ba.$$ Thus $G$ is abelian.