Inverse Completion of Commutative Semigroup is Inverse Completion of Itself

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Theorem

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

Let $\struct {C, \circ} \subseteq \struct {S, \circ}$ be the subsemigroup of cancellable elements of $\struct {S, \circ}$.

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


Then $\struct {T, \circ'}$ is its own inverse completion.


Proof

Let $x \circ' y^{-1}$ be cancellable for $\circ'$, where $x \in S$ and $y \in C$.

We have that $y$ is invertible for $\circ'$.

So by Invertible Element of Associative Structure is Cancellable:

$y$ is cancellable for $\circ'$.

Now by definition of inverse element:

$x = \paren {x \circ' y^{-1} } \circ' y$

Thus $x$ is also cancellable for $\circ'$.

By Cancellable Elements of Semigroup form Subsemigroup, $x$ is cancellable for $\circ$.

So:

$x \in S \implies x \in C$

Thus $x$ is invertible for $\circ'$.

Hence by Inverse of Product in Associative Structure:

$x \circ' y$ is invertible for $\circ'$.

So every cancellable element of $\struct {T, \circ'}$ is invertible.

So, by definition, $\struct {T, \circ'}$ is its own inverse completion.

$\blacksquare$


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