Suppose that $G$ is a group s.t. $x^2=1$ for all $x \in G$. Prove that $G$ is abelian.
If we assume that $x^2 = 1$ for all $ x \in G$, and suppose that $a,b \in G$ and if $x=ab$ we see that $x^2 = abab =aabb = a^2 b^2$. Simplifying we see that $ab=ba$, and because we know that $x= ab$, we see that $1=1$. Therefore it is abelian. $\square$
Is this sound? I'm struggling with this stuff because I think it's so simple and I'm concerned I'm overthinking it.
I'd like to contest the mods decision that this is an exact duplicate. Though there are similar questions, as has been linked, it is not the same thing. If it is a problem with the abstract algebra tag, then delete it and allow it to be an active thread under other tags.