# Characterizing commutative semigroups with a factorization property.

Let $$(N, \times)$$ be a commutative semigroup and assume that a countably infinite subset $$P$$ of $$N$$ algebraically generates $$N$$, and let $${\mathcal F}(P)$$ denote the set of all non-empty finite subsets of $$P$$. Assume that there exists a mapping

$$\quad \quad \mathtt P: N \to {\mathcal F}(P)$$

satisfying the following properties:

$$\tag 1 \forall \, p \in P, \mathtt P (p) = \{p\}$$

$$\tag 2 \forall \, a,b,c \in N, \; \text{If } c = ab \text{ then } \mathtt P(c) = \mathtt P(a) \cup \mathtt P(b)$$

Example: The function that maps every integer in $$(\mathbb N^{\ge 2}, *)$$ to its prime factors.

Question 1: Is every such structure the quotient of a universal one generated by the 'alphabet of letters' in $$P$$ creating 'words'?

Question 2: If the answer is yes how to we create the quotients? Are they all defined by factoring out relations, satisfying some rules, so that the conditions of $$\mathtt P$$ are still guaranteed to hold?

My Work

It looks like $$(\mathbb N^{\ge 2}, *)$$ is the universal structure with free generators the set of prime numbers.

• Doesn't your condition imply unique factorization? Then $N$ would have to be a free commutative monoid (without any relations), and thus $N$ is always isomorphic to $(\Bbb N^{\ge2},\cdot)$. – Berci Feb 2 at 0:16