Structure of Inverse Completion of Commutative Semigroup

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Theorem

Let $\struct {S, \circ}$ be a commutative semigroup.

Let $\struct {C, \circ} \subseteq \struct {S, \circ}$ be the subsemigroup of cancellable elements of $\struct {S, \circ}$.

Let $\struct {T, \circ'}$ be an inverse completion of $\struct {S, \circ}$.


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 $a \in C$.

\(\ds x\) \(\in\) \(\ds S\)
\(\ds \leadsto \ \ \) \(\ds x\) \(=\) \(\ds x \circ \paren {a \circ' a^{-1} }\) Definition of Invertible Element
\(\ds \leadsto \ \ \) \(\ds x\) \(=\) \(\ds \paren {x \circ a} \circ' a^{-1}\) Definition of Associative Operation
\(\ds \leadsto \ \ \) \(\ds x\) \(\in\) \(\ds S \circ' C^{-1}\) Definition of Subset Product
\(\ds \leadsto \ \ \) \(\ds S\) \(\subseteq\) \(\ds S \circ' C^{-1}\) Definition of Subset


Then:

\(\ds y\) \(\in\) \(\ds C\)
\(\ds \leadsto \ \ \) \(\ds y^{-1}\) \(\in\) \(\ds C^{-1}\) Definition of Inverse of Subset of Monoid
\(\ds \leadsto \ \ \) \(\ds y^{-1}\) \(=\) \(\ds a \circ' a^{-1} \circ' y^{-1}\) Definition of Invertible Element
\(\ds \leadsto \ \ \) \(\ds y^{-1}\) \(=\) \(\ds a \circ' \paren {y \circ' a}^{-1}\) Inverse of Product
\(\ds \leadsto \ \ \) \(\ds y^{-1}\) \(=\) \(\ds a \circ' \paren {y \circ a}^{-1}\) Definition of Extension of Operation
\(\ds \leadsto \ \ \) \(\ds y^{-1}\) \(\in\) \(\ds S \circ' C^{-1}\) Definition of Subset Product
\(\ds \leadsto \ \ \) \(\ds C^{-1}\) \(\subseteq\) \(\ds S \circ' C^{-1}\) 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:

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


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:

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


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:

$x \circ' y^{-1} \in \gen {S \cup C^{-1} }$

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


By definition of set equality:

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


and so by definition of the inverse completion:

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

$\blacksquare$


Sources

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