Commutation of Inverses in Monoid

Theorem

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 $y^{-1}$.

Let $x$ commute with $y$.

Then:

$\ds x^{-1} \circ y^{-1}$

$=$

$\ds \paren {y \circ x}^{-1}$

$\ds $

$=$

$\ds \paren {x \circ y}^{-1}$

$x$ commutes with $y$

$\ds $

$=$

$\ds y^{-1} \circ x^{-1}$

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

$\Box$

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 $y$.

$\blacksquare$

1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text I$: Algebraic Structures: $\S 4$: Neutral Elements and Inverses: Theorem $4.5: \ 3^\circ$