Structure of Inverse Completion of Commutative Semigroup

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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 $a \in C$.

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


Then:

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


Thus, as:

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

and:

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

by Union is Smallest Superset it follows that:

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


From Subset Product defining Inverse Completion of Commutative Semigroup is Commutative Semigroup:

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

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

$\blacksquare$


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