Inverse of Commuting Pair

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 $x$ commutes with $y$ if and only if:

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

Proof

\(\ds x \circ y\)

\(=\)

\(\ds y \circ x\)

\(\ds \leadstoandfrom \ \ \)

\(\ds \struct {x \circ y}^{-1}\)

\(=\)

\(\ds \struct {y \circ x}^{-1}\)

\(\ds \leadstoandfrom \ \ \)

\(\ds \struct {x \circ y}^{-1}\)

\(=\)

\(\ds x^{-1} \circ y^{-1}\)

Inverse of Product

$\blacksquare$

Examples

Elements of Symmetric Group $S_3$

Consider the Symmetric Group on $3$ Letters $S_3$, whose Cayley table is given as:

$\begin{array}{c|cccccc}

\circ & e & (123) & (132) & (23) & (13) & (12) \\ \hline e & e & (123) & (132) & (23) & (13) & (12) \\ (123) & (123) & (132) & e & (13) & (12) & (23) \\ (132) & (132) & e & (123) & (12) & (23) & (13) \\ (23) & (23) & (12) & (13) & e & (132) & (123) \\ (13) & (13) & (23) & (12) & (123) & e & (132) \\ (12) & (12) & (13) & (23) & (132) & (123) & e \\ \end{array}$

Let $x = \tuple {1 2 3}$ and $y = \tuple {1 3}$.

We have:

\(\ds \paren {x y}^{-1}\)

\(=\)

\(\ds \paren {\tuple {1 2 3} \tuple {1 3} }^{-1}\)

\(\ds \)

\(=\)

\(\ds \tuple {1 2}^{-1}\)

\(\ds \)

\(=\)

\(\ds \tuple {1 2}\)

However:

\(\ds x^{-1} y^{-1}\)

\(=\)

\(\ds \tuple {1 2 3}^{-1} \tuple {1 3}^{-1}\)

\(\ds \)

\(=\)

\(\ds \tuple {1 3 2} \tuple {1 3}\)

\(\ds \)

\(=\)

\(\ds \tuple {2 3}\)

\(\ds \)

\(\ne\)

\(\ds \paren {x y}^{-1}\)