# Structure of Inverse Completion of Commutative Semigroup

## Theorem

Let $\left({S, \circ}\right)$ be a commutative semigroup.

Let $\left ({C, \circ}\right) \subseteq \left({S, \circ}\right)$ be the subsemigroup of cancellable elements of $\left({S, \circ}\right)$.

Let $\left({T, \circ'}\right)$ be an inverse completion of $\left({S, \circ}\right)$.

Then:

$T = S \circ' C^{-1}$

where:

$C^{-1}$ is the inverse of $C$ in $T$
$S \circ' C^{-1}$ is the subset product of $S$ with $C^{-1}$.

## Proof

Let $a \in C$.

 $\displaystyle x$ $\in$ $\displaystyle S$ $\displaystyle \implies \ \$ $\displaystyle x$ $=$ $\displaystyle x \circ \left({a \circ' a^{-1} }\right)$ Definition of Invertible Element $\displaystyle \implies \ \$ $\displaystyle x$ $=$ $\displaystyle \left({x \circ a}\right) \circ' a^{-1}$ Definition of Associative $\displaystyle \implies \ \$ $\displaystyle x$ $\in$ $\displaystyle S \circ' C^{-1}$ Definition of Subset Product $\displaystyle \implies \ \$ $\displaystyle S$ $\subseteq$ $\displaystyle S \circ' C^{-1}$ Definition of Subset

Then:

 $\displaystyle y$ $\in$ $\displaystyle C$ $\displaystyle \implies \ \$ $\displaystyle y^{-1}$ $\in$ $\displaystyle C^{-1}$ Definition of Inverse of Subset $\displaystyle \implies \ \$ $\displaystyle y^{-1}$ $=$ $\displaystyle a \circ' a^{-1} \circ' y^{-1}$ Definition of Invertible Element $\displaystyle \implies \ \$ $\displaystyle y^{-1}$ $=$ $\displaystyle a \circ' \left({y \circ' a}\right)^{-1}$ Inverse of Product $\displaystyle \implies \ \$ $\displaystyle y^{-1}$ $=$ $\displaystyle a \circ' \left({y \circ a}\right)^{-1}$ $\circ'$ extends $\circ$ $\displaystyle \implies \ \$ $\displaystyle y^{-1}$ $\in$ $\displaystyle S \circ' C^{-1}$ Definition of Subset Product $\displaystyle \implies \ \$ $\displaystyle C^{-1}$ $\subseteq$ $\displaystyle S \circ' C^{-1}$ Definition of Subset

Thus, as:

$C^{-1} \subseteq S \circ' C^{-1}$

and:

$S \subseteq S \circ' C^{-1}$

by Union is Smallest Superset it follows that:

$S \cup C^{-1} \subseteq S \circ' C^{-1}$
$S \circ' C^{-1}$ is a commutative semigroup.

So $S \circ' C^{-1}$ is a semigroup which contains $S \cup C^{-1}$.

By definition of generator of semigroup, it follows that:

$\left\langle{S \cup C^{-1} }\right\rangle \subseteq S \circ' C^{-1}$

Let $z = x \circ' y^{-1} \in S \circ' C^{-1}$.

Then by definition:

$x \in S$

and:

$y^{-1} \in C^{-1}$

and so by definition of generator of semigroup:

$x \circ' y^{-1} \in \left\langle{S \cup C^{-1} }\right\rangle$

Thus: $S \circ' C^{-1} \subseteq \left\langle{S \cup C^{-1} }\right\rangle$

By definition of set equality:

$S \circ' C^{-1} = \left\langle{S \cup C^{-1} }\right\rangle$

and so by definition of the inverse completion:

$T = S \circ' C^{-1}$

$\blacksquare$