Commutation with Inverse in Monoid

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

Then $x$ commutes with $y$ $x$ commutes with $y^{-1}$.

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

Then:

So $x$ commutes with $y^{-1}$.

Sufficient Condition
Now let $x$ commute with $y^{-1}$.

From the above, it follows that $x$ commutes with $\left({y^{-1}}\right)^{-1}$.

From Inverse of Inverse in Monoid:
 * $\left({y^{-1}}\right)^{-1} = y$

Thus $x$ commutes with $y$.