Conjugate of Commuting Elements
Jump to navigation
Jump to search
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:
\(\ds x \circ y\) | \(=\) | \(\ds y \circ x\) | ||||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds x \circ y \circ x^{-1}\) | \(=\) | \(\ds y \circ x \circ x^{-1}\) | |||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds x \circ y \circ x^{-1}\) | \(=\) | \(\ds y\) | Definition of Invertible Element |
$\blacksquare$