# Commutation with Inverse in Monoid

## 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:

 $\ds y^{-1} \circ x$ $=$ $\ds \paren {y^{-1} \circ x} \circ e$ Monoid Axiom $\text S 2$: Identity $\ds$ $=$ $\ds y^{-1} \circ \paren {x \circ e}$ Monoid Axiom $\text S 1$: Associativity $\ds$ $=$ $\ds y^{-1} \circ \paren {x \circ \paren {y \circ y^{-1} } }$ Invertibility of $y$ $\ds$ $=$ $\ds y^{-1} \circ \paren {\paren {x \circ y} \circ y^{-1} }$ Monoid Axiom $\text S 1$: Associativity $\ds$ $=$ $\ds y^{-1} \circ \paren {\paren {y \circ x} \circ y^{-1} }$ $x$ commutes with $y$ $\ds$ $=$ $\ds \paren {y^{-1} \circ \paren {y \circ x} } \circ y^{-1}$ Monoid Axiom $\text S 1$: Associativity $\ds$ $=$ $\ds \paren {\paren {y^{-1} \circ y} \circ x} \circ y^{-1}$ Monoid Axiom $\text S 1$: Associativity $\ds$ $=$ $\ds \paren {e \circ x} \circ y^{-1}$ Invertibility of $y$ $\ds$ $=$ $\ds x \circ y^{-1}$ Monoid Axiom $\text S 2$: Identity

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

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

Thus $x$ commutes with $y$.

$\blacksquare$