Inverse of Commuting Pair

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

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


Proof

\(\displaystyle x \circ y\) \(=\) \(\displaystyle y \circ x\)
\(\displaystyle \leadstoandfrom \ \ \) \(\displaystyle \struct {x \circ y}^{-1}\) \(=\) \(\displaystyle \struct {y \circ x}^{-1}\)
\(\displaystyle \leadstoandfrom \ \ \) \(\displaystyle \struct {x \circ y}^{-1}\) \(=\) \(\displaystyle x^{-1} \circ y^{-1}\) 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:

\(\displaystyle \paren {x y}^{-1}\) \(=\) \(\displaystyle \paren {\tuple {1 2 3} \tuple {1 3} }^{-1}\)
\(\displaystyle \) \(=\) \(\displaystyle \tuple {1 2}^{-1}\)
\(\displaystyle \) \(=\) \(\displaystyle \tuple {1 2}\)


However:

\(\displaystyle x^{-1} y^{-1}\) \(=\) \(\displaystyle \tuple {1 2 3}^{-1} \tuple {1 3}^{-1}\)
\(\displaystyle \) \(=\) \(\displaystyle \tuple {1 3 2} \tuple {1 3}\)
\(\displaystyle \) \(=\) \(\displaystyle \tuple {2 3}\)
\(\displaystyle \) \(\ne\) \(\displaystyle \paren {x y}^{-1}\)