# Product of Commuting Elements with Inverses

## 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:

$x \circ y \circ x^{-1} \circ y^{-1} = e_S = x^{-1} \circ y^{-1} \circ x \circ y$

if and only if $x$ and $y$ commute.

## Proof

As $\struct {S, \circ}$ is a monoid, it is by definition a semigroup.

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

From Invertible Element of Monoid is Cancellable, we also have that $x, y, x^{-1}, y^{-1}$ are cancellable.

So let :

 $\ds x \circ y$ $=$ $\ds y \circ x$ by hypothesis: $x$ and $y$ commute $\ds \leadstoandfrom \ \$ $\ds x \circ y \circ x^{-1}$ $=$ $\ds y$ Conjugate of Commuting Elements $\ds \leadstoandfrom \ \$ $\ds x \circ y \circ x^{-1} \circ y^{-1}$ $=$ $\ds y \circ y^{-1}$ Monoid Axiom $\text S 0$: Closure $\ds \leadstoandfrom \ \$ $\ds x \circ y \circ x^{-1} \circ y^{-1}$ $=$ $\ds e_S$ Invertibility of $y$

Similarly:

 $\ds x \circ y$ $=$ $\ds y \circ x$ by hypothesis: $x$ and $y$ commute $\ds \leadstoandfrom \ \$ $\ds y^{-1} \circ x \circ y$ $=$ $\ds x$ Conjugate of Commuting Elements $\ds \leadstoandfrom \ \$ $\ds x^{-1} \circ y^{-1} \circ x \circ y$ $=$ $\ds x^{-1} \circ x$ Monoid Axiom $\text S 0$: Closure $\ds \leadstoandfrom \ \$ $\ds x^{-1} \circ y^{-1} \circ x \circ y$ $=$ $\ds e_S$ Invertibility of $x$

$\blacksquare$