# Commutation with Inverse in Monoid

## Theorem

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 $y^{-1}$.

## Proof

### Necessary Condition

Let $x$ commute with $y$.

Then:

 $\displaystyle y^{-1} \circ x$ $=$ $\displaystyle y^{-1} \circ \left({x \circ \left({y \circ y^{-1} }\right)}\right)$ 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 $y^{-1}$.

$\Box$

### Sufficient Condition

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

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

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

Thus $x$ commutes with $y$.

$\blacksquare$