Product of Commuting Elements with Inverses

Theorem
Let $\left({S, \circ}\right)$ be a monoid whose identity is $e_S$. Let $x, y \in S$ such that $x$ and $y$ are both invertible.

Then $x \circ y \circ x^{-1} \circ y^{-1} = e_S$ iff $x$ and $y$ commute.

Proof
As $\left({S, \circ}\right)$ is a monoid, it is by definition a semigroup.

Therefore $\circ$ is associative, so we can dispense with parentheses.

From Invertible Elements of Semigroup Also Cancellable, we also have that $x, y, x^{-1}, y^{-1}$ are cancellable.

So: