Commutation of Inverses in Monoid

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

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

Then:

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

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

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

From Inverse of Inverse in Associative Structure, $\left({x^{-1}}\right)^{-1} = x$ and $\left({y^{-1}}\right)^{-1} = y$.

Thus $x$ commutes with $y$.