# Inverse Completion of Commutative Semigroup is Inverse Completion of Itself

## 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 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:

- $x = \left({x \circ' y^{-1}}\right) \circ' y$

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 $\left({T, \circ'}\right)$ is invertible.

So, by definition, $\left({T, \circ'}\right)$ is its own inverse completion.

$\blacksquare$

## Sources

- 1965: Seth Warner:
*Modern Algebra*... (previous) ... (next): $\S 20$: Theorem $20.1: \ 3^\circ$