# Conjugate of Commuting Elements

## 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 $x \circ y \circ x^{-1} = y$ if and only if $x$ and $y$ commute.

## Proof

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

Therefore it is taken for granted that $\circ$ is associative, so we can dispense with parentheses.

We also take for granted the fact that $x$ and $y$ are cancellable from Invertible Element of Monoid is Cancellable.

So:

 $\ds x \circ y$ $=$ $\ds y \circ x$ $\ds \leadstoandfrom \ \$ $\ds x \circ y \circ x^{-1}$ $=$ $\ds y \circ x \circ x^{-1}$ $\ds \leadstoandfrom \ \$ $\ds x \circ y \circ x^{-1}$ $=$ $\ds y$ Definition of Invertible Element

$\blacksquare$