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 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.