Inverse Completion is Commutative Monoid
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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 $\left({T, \circ'}\right)$ is a commutative monoid.
Proof
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.
$\blacksquare$