Commutation with Inverse in Monoid

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $\struct {S, \circ}$ be a monoid.

Let $x, y \in S$ such that $y$ is invertible.


Then $x$ commutes with $y$ if and only if $x$ commutes with $y^{-1}$.


Proof

Necessary Condition

Let $x$ commute with $y$.

Then:

\(\ds y^{-1} \circ x\) \(=\) \(\ds y^{-1} \circ \paren {x \circ \paren {y \circ y^{-1} } }\) Invertibility of $y$
\(\ds \) \(=\) \(\ds y^{-1} \circ \paren {\paren {x \circ y} \circ y^{-1} }\) Associativity of $\circ$
\(\ds \) \(=\) \(\ds y^{-1} \circ \paren {\paren {y \circ x} \circ y^{-1} }\) $x$ commutes with $y$
\(\ds \) \(=\) \(\ds \paren {y^{-1} \circ \paren {y \circ x} } \circ y^{-1}\) Associativity of $\circ$
\(\ds \) \(=\) \(\ds \paren {\paren {y^{-1} \circ y} \circ x} \circ y^{-1}\) Associativity of $\circ$
\(\ds \) \(=\) \(\ds x \circ y^{-1}\) Invertibility of $y$


So $x$ commutes with $y^{-1}$.

$\Box$


Sufficient Condition

Now let $x$ commute with $y^{-1}$.

From the above, it follows that $x$ commutes with $\paren {y^{-1} }^{-1}$.

From Inverse of Inverse in Monoid:

$\paren {y^{-1} }^{-1} = y$

Thus $x$ commutes with $y$.

$\blacksquare$


Sources