Inverse Completion of Commutative Semigroup is Abelian Group

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

Let all the elements of $\struct {S, \circ}$ be cancellable.

Then an inverse completion of $\struct {S, \circ}$ is an abelian group.

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

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

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

has been shown to be a commutative monoid.

Taking the group axioms in turn:

As $\struct {T, \circ'}$ is a commutative semigroup, it is by definition closed.

As $\struct {T, \circ'}$ is a commutative semigroup, it is by definition associative.

Let $x \in S$.

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

As $\struct {T, \circ'} = \struct {S \circ' S^{-1}, \circ'}$ is closed:
 * $x \circ' x^{-1} \in T$

This holds for all $x \in S$.

As $\struct {T, \circ'}$ is a commutative semigroup:
 * $\exists e \in T: \forall x \in S: x \circ' x^{-1} = e^{-1} = x \circ' x$

Thus $\struct {T, \circ'}$ has an identity element.

Every element of $S$ has an inverse in $S^{-1}$.

Therefore every element of $S$ and $S^{-1}$ is invertible.

From Inverse of Product in Associative Structure, 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 $\struct {T, \circ'}$ is a group.

Commutativity of $\circ$ has been demonstrated.

Hence the result, by definition of abelian group.