Naturally Ordered Semigroups Isomorphism Unique

Theorem

Let $\left({S, \circ, \preceq}\right)$ and $\left({S\,', \circ', \preceq'}\right)$ be naturally ordered semigroups.

Let:

• $0\,'$ be the smallest element of $S\,'$
• $1'$ be the smallest element of $S\,' \setminus \left\{{0\,'}\right\} = S\,'^*$.

Then the mapping $g: S \to S\,'$ defined as:

$\forall a \in S: g \left({a}\right) = \circ'^a 1'$

is an isomorphism from $\left({S, \circ, \preceq}\right)$ to $\left({S\,', \circ', \preceq'}\right)$.

This isomorphism is unique.

Thus, up to isomorphism, there is only one naturally ordered semigroup.

Proof

Proof that Mapping is Isomorphism

Let $T\,' = \operatorname{Cdm} \left({g}\right)$, that is, the codomain of $g$.

By the definition of zero, $0\,'$ is the identity for $\circ'$.

Thus $g \left({0}\right) = \circ'^0 1' = 0\,' \implies 0\,' \in T\,'$.

Now $x' \in T\,' \implies g \left({n}\right) = x'$. Then:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle x' \circ' 1'\)	\(=\)	\(\displaystyle \circ'^n 1' \circ' 1'\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \left({\circ'^n 1'}\right) \circ' \left({\circ'^1 1'}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \circ'^{\left({n \circ 1}\right)} 1'\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle g \left({n \circ 1}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

So $x' \in T\,' \implies x' \circ' 1' \in T\,'$.

Thus, by the Principle of Finite Induction applied to $S\,'$, $T\,' = S\,'$.

So, $\forall a' \in S\,': \exists a \in S: g \left({a}\right) = a'$, thus $g$ is surjective.

We also have from Recursive Mapping to Semigroup: Index Laws that $g \left({a \circ b}\right) = g \left({a}\right) \circ' g \left({b}\right)$ and therefore $g$ is a homomorphism from $\left({S, \circ}\right)$ to $\left({S\,', \circ'}\right)$.

Now:

	\(\displaystyle \)	\(\displaystyle \forall p \in S:\)	\(\displaystyle \)	\(\displaystyle \circ'^p 1'\)	\(\prec'\)	\(\displaystyle \circ'^p 1' \circ' 1'\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \left({\circ'^p 1'}\right) \circ' \left({\circ'^1 1'}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \circ'^{\left({p \circ 1}\right)} 1'\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle \circ'^p 1'\)	\(\prec'\)	\(\displaystyle \circ'^{\left({p \circ 1}\right)} 1'\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

Let $T = \left\{{p \in S: \forall a \in S_p: \circ'^a 1' \prec' \circ'^p 1'}\right\}$

Now $S_0 = \varnothing \implies 0 \in T$.

Suppose $p \in T$.

Then $a \prec p \circ 1 \implies a \preceq p$.

Either of these is the case:

• $a \prec p: p \in T \implies \circ'^a 1' \prec' \circ'^p 1' \prec' \circ'^{\left({p \circ 1}\right)} 1'$
• $a = p: \circ'^a 1' = \circ'^p 1' \prec' \circ'^{\left({p \circ 1}\right)} 1'$

In either case, we have $p \in T \implies p \circ 1 \in T$, and by induction $T = S$.

So $n \prec p \implies \circ'^n 1' \prec' \circ'^p 1'$.

Thus $g$ is a surjective monomorphism and therefore is an isomorphism from $\left({S, \circ, \preceq}\right)$ to $\left({S\,', \circ', \preceq'}\right)$.

$\blacksquare$

Proof that Isomorphism is Unique

Now we need to show that the isomorphism $g$ is unique.

Let $f: S \to S\,'$ be another isomorphism different from $g$.

Suppose $f \left({1}\right) \ne 1'$.

We show by induction that $1' \notin \operatorname{Cdm} \left({f}\right)$.

... Thus $1' \notin \operatorname{Cdm} \left({f}\right)$ which is a contradiction.

Thus $f \left({1}\right) = 1$ and it follows

that $f = g$.

Thus the isomorphism $g$ is unique.

$\blacksquare$

Sources

• Seth Warner: Modern Algebra (1965)... (previous)... (next): $\S 16$: Theorem $16.15$