Singleton fulfils Naturally Ordered Semigroup Axioms 1 to 3

Theorem
Let $S$ be a singleton:


 * $S = \set s$

for an arbitrary object $s$.

Let $+$ be the operation on $S$ defined as:
 * $\forall s \in S: s + s = s$

Let $\le$ be the relation defined on $S$ as:
 * $s \le s$

Then the algebraic structure:
 * $\struct {S, +, \le}$

is an ordered semigroup which fulfils the axioms:



but:
 * does not fulfil
 * $\struct {S, +}$ is not isomorphic to $\struct {\N, +}$.

Proof
Recall the axioms:

First it is noted that $\le$ is a reflexive relation.

Hence from Reflexive Relation on Singleton is Well-Ordering:
 * $\le$ is a well-ordering.

Hence holds.

holds for $\struct {S, +}$ trivally.

We have that:
 * $0 \in M$

and:
 * $2 \in M$

and trivially holds.

We have that $S$ is a singleton.

Hence:
 * $\forall a, b \in S: a = b$

and so does not hold.

Lack of Isomorphism
It remains to demonstrate that $S$ and $\N$ are not isomorphic.

there exists a (semigroup) isomorphism $\phi$ from $\struct {\N, +}$ to $\struct {S, +}$.

By definition of isomorphism:
 * $\phi$ is a homomorphism
 * $\phi$ is a bijection.

From Mapping is Constant iff Image is Singleton:
 * $\forall n \in \N: \map \phi n = s$

Thus for example:
 * $\map \phi 0 = s$

and:
 * $\map \phi 1 = s$

and it is immediate that $\phi$ is not an injection.

Hence $\phi$ is not a bijection.

This contradicts our assertion that $\phi$ is an isomorphism.

Hence there can be no such semigroup isomorphism between $\struct {S, +}$ and $\struct {\N, +}$.