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 $x$

$\blacksquare$