Naturally Ordered Semigroup Axioms imply Commutativity

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Theorem

Consider the naturally ordered semigroup axioms:

A naturally ordered semigroup is a (totally) ordered commutative semigroup $\struct {S, \circ, \preceq}$ satisfying:

\((\text {NO} 1)\)   $:$   $S$ is well-ordered by $\preceq$      \(\ds \forall T \subseteq S:\) \(\ds T = \O \lor \exists m \in T: \forall n \in T: m \preceq n \)      
\((\text {NO} 2)\)   $:$   $\circ$ is cancellable in $S$      \(\ds \forall m, n, p \in S:\) \(\ds m \circ p = n \circ p \implies m = n \)      
\(\ds p \circ m = p \circ n \implies m = n \)      
\((\text {NO} 3)\)   $:$   Existence of product      \(\ds \forall m, n \in S:\) \(\ds m \preceq n \implies \exists p \in S: m \circ p = n \)      
\((\text {NO} 4)\)   $:$   $S$ has at least two distinct elements      \(\ds \exists m, n \in S:\) \(\ds m \ne n \)      


Axioms $\text {NO} 1$, $\text {NO} 2$ and $\text {NO} 3$ together imply the commutativity of the naturally ordered semigroup $\struct {S, \circ, \preceq}$.


Proof

From Naturally Ordered Semigroup Axiom $\text {NO} 1$: Well-Ordered, $\struct {S, \circ, \preceq}$ has a smallest element.

This is identified as zero: $0$.

From Zero is Identity in Naturally Ordered Semigroup, $0$ is the identity element of $\struct {S, \circ, \preceq}$,


It may be the case that $S$ is a singleton such that $S = \set 0$ .

Then $\struct {S, \circ}$ degenerates to the trivial group.

From Trivial Group is Abelian, it follows that $\circ$ is commutative.


Let $S^* := S \setminus \set 0$ denote the complement of $\set 0$ in $S$.

From Naturally Ordered Semigroup Axiom $\text {NO} 1$: Well-Ordered, $S^*$ also has a smallest element.

This we will call $1$.

It will be shown that $1$ commutes with every element of $S$.


Let $T \subseteq S$ be the set of element of $S$ which commute with $1$.

We have a priori that $0$ is the identity element of $\struct {S, \circ, \preceq}$.

Hence:

$1 \circ 0 = 1 = 0 \circ 1$

and it is seen that $0 \in T$.

Now suppose $n \in T$.

That is:

$1 \circ n = n \circ 1$

We have:

\(\ds 1 \circ \paren {n \circ 1}\) \(=\) \(\ds \paren {1 \circ n} \circ 1\) Semigroup Axiom $\text S 1$: Associativity
\(\ds \) \(=\) \(\ds \paren {n \circ 1} \circ 1\) by hypothesis

So

$n \in T \implies n \circ 1 \in T$.

It follows from Principle of Mathematical Induction for Naturally Ordered Semigroup that:

$T = S$

That is, all the element of $S$ commute with $1$.

Hence the result.

$\blacksquare$


Sources

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