Inverse Completion is Commutative Monoid

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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 $\left({T, \circ'}\right)$ is a commutative monoid.


From Inverse Completion is Commutative Semigroup:

$\left({T, \circ'}\right)$ is a commutative semigroup.

By definition of inverse completion:

$\forall x \in C: \exists x^{-1} \in C: x \circ x^{-1} \in T$

Thus by definition of inverse element:

$e := x \circ x^{-1}$ is an identity element of $T$.

By Identity is Unique, $e$ is the only such element of $T$.

Thus $\left({T, \circ'}\right)$ is a semigroup with an identity element.

So, by definition, $\left({T, \circ'}\right)$ is a commutative monoid.