Conjugate of Commuting Elements

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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:

\(\displaystyle x \circ y\) \(=\) \(\displaystyle y \circ x\)
\(\displaystyle \leadstoandfrom \ \ \) \(\displaystyle x \circ y \circ x^{-1}\) \(=\) \(\displaystyle y \circ x \circ x^{-1}\)
\(\displaystyle \leadstoandfrom \ \ \) \(\displaystyle x \circ y \circ x^{-1}\) \(=\) \(\displaystyle y\) Definition of Invertible Element

$\blacksquare$