Commutation with Inverse in Monoid

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Let $\left({S, \circ}\right)$ 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 $y^{-1}$.


Necessary Condition

Let $x$ commute with $y$.


\(\displaystyle y^{-1} \circ x\) \(=\) \(\displaystyle y^{-1} \circ \left({x \circ \left({y \circ y^{-1} }\right)}\right)\) Invertibility of $y$
\(\displaystyle \) \(=\) \(\displaystyle y^{-1} \circ \left({\left({x \circ y}\right) \circ y^{-1} }\right)\) Associativity of $\circ$
\(\displaystyle \) \(=\) \(\displaystyle y^{-1} \circ \left({\left({y \circ x}\right) \circ y^{-1} }\right)\) $x$ commutes with $y$
\(\displaystyle \) \(=\) \(\displaystyle \left({y^{-1} \circ \left({y \circ x}\right)}\right) \circ y^{-1}\) Associativity of $\circ$
\(\displaystyle \) \(=\) \(\displaystyle \left({\left({y^{-1} \circ y}\right) \circ x}\right) \circ y^{-1}\) Associativity of $\circ$
\(\displaystyle \) \(=\) \(\displaystyle x \circ y^{-1}\) Invertibility of $y$

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


Sufficient Condition

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

From the above, it follows that $x$ commutes with $\left({y^{-1}}\right)^{-1}$.

From Inverse of Inverse in Monoid:

$\left({y^{-1}}\right)^{-1} = y$

Thus $x$ commutes with $y$.