Subset Product defining Inverse Completion of Commutative Semigroup is Commutative Semigroup
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:
- $S \circ' C^{-1}$ is a commutative semigroup
where $S \circ' C^{-1}$ is the subset product of $S$ with $C^{-1}$ under $\circ'$ in $T$.
Proof
Note that by definition of inverse completion, $\struct {T, \circ'}$ is a semigroup.
Thus $\circ'$ is associative.
First it is demonstrated that $S \circ' C^{-1}$ is a semigroup.
Let $x, z \in S$.
Let $y, w \in C$.
Then:
\(\ds \paren {x \circ' y^{-1} } \circ' \paren {z \circ' w^{-1} }\) | \(=\) | \(\ds x \circ' \paren {y^{-1} \circ' z} \circ' w^{-1}\) | Associativity of $\circ'$ | |||||||||||
\(\ds \) | \(=\) | \(\ds x \circ' \paren {z \circ' y^{-1} } \circ' w^{-1}\) | Commutation with Inverse in Monoid | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {x \circ' z} \circ' \paren {y^{-1} \circ' w^{-1} }\) | Associativity of $\circ'$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {x \circ' z} \circ' \paren {w \circ' y}^{-1}\) | Inverse of Product in Monoid | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {x \circ z} \circ' \paren {w \circ y}^{-1}\) | $\circ'$ extends $\circ$ |
Thus:
- $\paren {x \circ z} \circ' \paren {w \circ y}^{-1} \in S \circ' C^{-1}$
proving that $S \circ' C^{-1}$ is closed.
Therefore by Subsemigroup Closure Test:
- $S \circ' C^{-1}$ is a subsemigroup of $\struct {T, \circ'}$
and thus a semigroup.
$\Box$
It remains to be shown that $\circ'$ is a commutative operation.
Let $\paren {x \circ' y^{-1} }$ and $\paren {z \circ' w^{-1} }$ be two arbitrary elements of $S \circ' C^{-1}$.
By Element Commutes with Product of Commuting Elements, $x, y, z, w$ all commute with each other under $\circ$.
As $\circ'$ is an extension of $\circ$, it follows that $x, y, z, w$ also all commute with each other under $\circ'$.
Then:
\(\ds \paren {x \circ' y^{-1} } \circ' \paren {z \circ' w^{-1} }\) | \(=\) | \(\ds x \circ' \paren {y^{-1} \circ' z} \circ' w^{-1}\) | Associativity of $\circ'$ | |||||||||||
\(\ds \) | \(=\) | \(\ds x \circ' \paren {z \circ' y^{-1} } \circ' w^{-1}\) | Commutation with Inverse in Monoid | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {x \circ' z} \circ' \paren {y^{-1} \circ' w^{-1} }\) | Associativity of $\circ'$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {z \circ' x} \circ' \paren {w^{-1} \circ' y^{-1} }\) | Commutation of Inverses in Monoid | |||||||||||
\(\ds \) | \(=\) | \(\ds z \circ' \paren {x \circ' w^{-1} } \circ' y^{-1}\) | Associativity of $\circ'$ | |||||||||||
\(\ds \) | \(=\) | \(\ds z \circ' \paren {w^{-1} \circ' x} \circ' y^{-1}\) | Commutation with Inverse in Monoid | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {z \circ' w^{-1} } \circ' \paren {x \circ' y^{-1} }\) | Associativity of $\circ'$ |
So $x \circ' y^{-1}$ commutes with $z \circ' w^{-1}$.
It follows by definition that $S \circ' C^{-1}$ is a commutative semigroup.
$\blacksquare$
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {IV}$: Rings and Fields: $\S 20$: The Integers: Theorem $20.1: \ 2^\circ$