Naturally Ordered Semigroup Axioms imply Commutativity
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
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {III}$: The Natural Numbers: $\S 16$: The Natural Numbers
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {III}$: The Natural Numbers: $\S 16$: The Natural Numbers: Exercise $16.5$