Product of Commuting Elements with Inverses

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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.


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

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


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 $y$


\(\ds x \circ y\) \(=\) \(\ds y \circ x\) 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 $x$