Commutation of Inverses in Monoid

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Let $\struct {S, \circ}$ 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 $y$ if and only if $x^{-1}$ commutes with $y^{-1}$.


Necessary Condition

Let $x$ commute with $y$.


\(\ds x^{-1} \circ y^{-1}\) \(=\) \(\ds \paren {y \circ x}^{-1}\) Inverse of Product
\(\ds \) \(=\) \(\ds \paren {x \circ y}^{-1}\) $x$ commutes with $y$
\(\ds \) \(=\) \(\ds y^{-1} \circ x^{-1}\) Inverse of Product

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


Sufficient Condition

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

From the above, $\paren {x^{-1} }^{-1}$ commutes with $\paren {y^{-1} }^{-1}$.

From Inverse of Inverse in Monoid, $\paren {x^{-1} }^{-1} = x$ and $\paren {y^{-1} }^{-1} = y$.

Thus $x$ commutes with $y$.