Subset Product defining Inverse Completion of Commutative Semigroup is Commutative Semigroup

From ProofWiki
Jump to navigation Jump to search

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