Inverse Completion of Commutative Semigroup is Abelian Group

Theorem
Let $\left({S, \circ}\right)$ be a commutative semigroup, all of whose elements are cancellable.

Then an inverse completion of $\left({S, \circ}\right)$ is an abelian group.

Proof
Let $\left({S, \circ}\right)$ be a commutative semigroup, all of whose elements are cancellable.

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

From Inverse Completion is Commutative Monoid:
 * $\left({T, \circ'}\right) = \left({S \circ' S^{-1}, \circ'}\right)$

has been shown to be a commutative monoid.

Taking the group axioms in turn:

G0: Closure
As $\left({T, \circ'}\right)$ is a commutative semigroup, it is by definition closed.

G1: Associativity
As $\left({T, \circ'}\right)$ is a commutative semigroup, it is by definition associative.

G2: Identity
Let $x \in S$.

Then $x^{-1} \in S^{-1}$ by definition.

As $\left({T, \circ'}\right) = \left({S \circ' S^{-1}, \circ'}\right)$ is closed, $x \circ' x^{-1} \in T$.

This holds for all $x \in S$ and so as $\left({T, \circ'}\right)$ is a commutative semigroup:
 * $\exists e \in T: \forall x \in S: x \circ' x^{-1} = e^{-1} = x \circ' x$

Thus $\left({T, \circ'}\right)$ has an identity element.

G3: Inverses
Every element of $S$ has an inverse in $S^{-1}$, and so every element of $S$ and $S^{-1}$ is invertible.

From Inverse of Product, every element of $S \circ' S^{-1}$ is therefore also invertible.

Thus every element of $T$ has an inverse.

All the group axioms are thus seen to be fulfilled, and so $\left({T, \circ'}\right)$ is a group.

Commutativity of $\circ$ has been demonstrated.

Hence the result, by definition of abelian group.