Self-Inverse Elements Commute iff Product is Self-Inverse

Theorem
Let $\struct {G, \circ}$ be a group.

Let $x, y \in \struct {G, \circ}$, such that $x$ and $y$ are self-inverse.

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

Proof
Let the identity element of $\struct {G, \circ}$ be $e_G$.

Necessary Condition
Let $x$ and $y$ commute.

Then:

Thus $\paren {x \circ y} \circ \paren {x \circ y} = e_G$, proving that $x \circ y$ is self-inverse.

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

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

Thus:
 * $\paren {x \circ x} \circ \paren {y \circ y} = e_G \circ e_G = e_G$

Because $x \circ y$ is self-inverse, we have:
 * $\paren {x \circ y} \circ \paren {x \circ y} = e_G$

Thus:

So $x$ and $y$ commute.