Self-Inverse Elements Commute iff Product is Self-Inverse

Theorem
Let $\left({G, \circ}\right)$ be a group.

Let $x, y \in \left({G, \circ}\right)$, such that $x$ and $y$ are self-inverse.

Then $x$ and $y$ commute iff $x \circ y$ is also self-inverse.

Proof
Let the identity element of $\left({G, \circ}\right)$ be $e_G$.


 * Let $x$ and $y$ commute.

Then:

Thus $\left({x \circ y}\right) \circ \left({x \circ y}\right)$, proving that $x \circ y$ is self-inverse.


 * Now, suppose that $x \circ y$ is self-inverse.

We already have that $x$ and $y$ are self-inverse.

Thus $\left({x \circ x}\right) \circ \left({y \circ y}\right) = e_G \circ e_G = e_G$.

Because $x \circ y$ is self-inverse, we have $\left({x \circ y}\right) \circ \left({x \circ y}\right) = e_G$.

Thus:

So $x$ and $y$ commute.

Alternative Proof
See the more general Powers of Commutative Elements in Monoids.