Inverse Completion Commutative Semigroup
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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 $T = S \circ' C^{-1}$, and is a commutative semigroup.
Proof
Proof of Subsemigroup
Let $x, z \in S, y, w \in C$.
Then by Associativity and Commutativity Properties, $x, y, z, w$ all commute with each other under $\circ$.
So:
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \left({x \circ' y^{-1} }\right) \circ' \left({z \circ' w^{-1} }\right)\) | \(=\) | \(\displaystyle x \circ' \left({y^{-1} \circ' z}\right) \circ' w^{-1}\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | Associativity of $\circ'$ | ||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle x \circ' \left({z \circ' y^{-1} }\right) \circ' w^{-1}\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | Commutation with Inverse | ||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \left({x \circ' z}\right) \circ' \left({y^{-1} \circ' w^{-1} }\right)\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | Associativity of $\circ'$ | ||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \left({x \circ' z}\right) \circ' \left({w \circ' y}\right)^{-1}\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | Inverse of Product | ||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \left({x \circ z}\right) \circ' \left({w \circ y}\right)^{-1}\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | $\circ'$ extends $\circ$ |
Thus $\left({x \circ z}\right)\circ' \left({w \circ y}\right)^{-1} \in S \circ' C^{-1}$, proving that $S \circ' C^{-1}$ is closed, therefore a subsemigroup of $\left({T, \circ'}\right)$.
$\Box$
Proof of Commutative Subsemigroup
Let $\left({x \circ' y^{-1} }\right)$ and $\left({z \circ' w^{-1} }\right)$ be two arbitrary elements of $S \circ' C^{-1}$.
By Associativity and Commutativity Properties, $x, y, z, w$ all commute with each other under $\circ$.
Then:
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \left({x \circ' y^{-1} }\right) \circ' \left({z \circ' w^{-1} }\right)\) | \(=\) | \(\displaystyle x \circ' \left({y^{-1} \circ' z}\right) \circ' w^{-1}\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | Associativity of $\circ'$ | ||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle x \circ' \left({z \circ' y^{-1} }\right) \circ' w^{-1}\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | Commutation with Inverse | ||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \left({x \circ' z}\right) \circ' \left({y^{-1} \circ' w^{-1} }\right)\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | Associativity of $\circ'$ | ||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \left({z \circ' x}\right) \circ' \left({w^{-1} \circ' y^{-1} }\right)\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | Commutation of Inverses | ||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle z \circ' \left({x \circ' w^{-1} }\right) \circ' y^{-1}\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | Associativity of $\circ'$ | ||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle z \circ' \left({w^{-1} \circ' x}\right) \circ' y^{-1}\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | Commutation with Inverse | ||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \left({z \circ' w^{-1} }\right) \circ' \left({x \circ' y^{-1} }\right)\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | 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 subsemigroup of $\left({T, \circ'}\right)$.
$\Box$
Proof of Equality
Let $a \in C$.
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle x\) | \(\in\) | \(\displaystyle S\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \implies\) | \(\displaystyle \) | \(\displaystyle x\) | \(=\) | \(\displaystyle x \circ \left({a \circ' a^{-1} }\right)\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | as $a$ is invertible in $C$ | ||
| \(\displaystyle \) | \(\displaystyle \implies\) | \(\displaystyle \) | \(\displaystyle x\) | \(=\) | \(\displaystyle \left({x \circ a}\right) \circ' a^{-1}\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | Associativity | ||
| \(\displaystyle \) | \(\displaystyle \implies\) | \(\displaystyle \) | \(\displaystyle x\) | \(\in\) | \(\displaystyle S \circ' C^{-1}\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | Definition of Subset Product | ||
| \(\displaystyle \) | \(\displaystyle \implies\) | \(\displaystyle \) | \(\displaystyle S\) | \(\subseteq\) | \(\displaystyle S \circ' C^{-1}\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | Definition of Subset |
Then:
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle y\) | \(\in\) | \(\displaystyle C\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \implies\) | \(\displaystyle \) | \(\displaystyle y^{-1}\) | \(=\) | \(\displaystyle a \circ' a^{-1} \circ' y^{-1}\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | as $a$ is invertible in $C$ | ||
| \(\displaystyle \) | \(\displaystyle \implies\) | \(\displaystyle \) | \(\displaystyle y^{-1}\) | \(=\) | \(\displaystyle a \circ' \left({y \circ' a}\right)^{-1}\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | Inverse of Product | ||
| \(\displaystyle \) | \(\displaystyle \implies\) | \(\displaystyle \) | \(\displaystyle y^{-1}\) | \(=\) | \(\displaystyle a \circ' \left({y \circ a}\right)^{-1}\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | $\circ'$ extends $\circ$ | ||
| \(\displaystyle \) | \(\displaystyle \implies\) | \(\displaystyle \) | \(\displaystyle y^{-1}\) | \(\in\) | \(\displaystyle S \circ' C^{-1}\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | Definition of Subset Product | ||
| \(\displaystyle \) | \(\displaystyle \implies\) | \(\displaystyle \) | \(\displaystyle C^{-1}\) | \(\subseteq\) | \(\displaystyle S \circ' C^{-1}\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | Definition of Subset |
Thus, as $C^{-1} \subseteq S \circ' C^{-1}$ and $S \subseteq S \circ' C^{-1}$, we have $S \cup C^{-1} \subseteq S \circ' C^{-1}$ by Union Smallest.
By the definition of the inverse completion, we see that $T = S \circ' C^{-1}$.
$\blacksquare$
Sources
- Seth Warner: Modern Algebra (1965)... (previous)... (next): $\S 20$: Theorem $20.1: \ 1^\circ, \ 2^\circ$
