All Elements Self-Inverse then Abelian

Theorem
If every element in a group is self-inverse, then that group is abelian.

Proof
If every element in a group $\left({G, \circ}\right)$ is self-inverse, then $\forall x \in G: x \circ x = e$.

Also, for all $\forall x, y \in G: \left({x \circ y}\right) \circ \left({x \circ y}\right) = e$, that is, $x \circ y$ is also self-inverse.

From Self-Inverse Elements that Commute, it follows that $y \circ x = x \circ y$, i.e. that $x$ and $y$ commute.

As this is true for all elements of $\left({G, \circ}\right)$, it follows that all elements of $\left({G, \circ}\right)$ commute with all other elements of $\left({G, \circ}\right)$.

Thus, by definition, $\left({G, \circ}\right)$ is abelian.