Inverse Completion of Commutative Semigroup is Abelian Group

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

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

From Inverse Completion Commutative Semigroup, $T$ is commutative.

The product of two invertible elements is invertible.

Every element of $S \circ' S^{-1}$ is invertible in $T$, from the definition of inverse completion and the fact that every element of $S$ is cancellable.

Thus every element of $T$ is invertible.

As Invertible Elements of Monoid form Subgroup, $T$ is therefore a group.

So $\left({T, \circ'}\right)$ is commutative, and a group, and therefore an abelian group.