Let $P$ be a finite upper bounded partially ordered set. Then there exists a family of commutative semigroup isomorphism types which have $P$ in common as their factorisation order. We will study these factorisation order equal families of isomorphism types by first considering the special case in which $P$ is a total order.
Structure theory:
We will clasify finite commutative tosetal semigroups by first classifying finite commutative unipotent tosetal semigroups. We will start by locating the unique idempotent in such a semigroup and examining the idempotents in .
Proposition. The maximal element in a finite commutative partially ordered semigroup $S$ is idempotent
Proof. Let $x$ be the maximal element in $S$. Suppose that $x$ is not idempotent, then there exists an element $y$ such that $x^2 = y$ which implies $x \subseteq y$ which contradicts the supposition that $x$ is maximal. Therefore, $x$ must idempotent. $\square$
Proposition. All finite commutative unipotent totally ordered semigroups $S$ are monogenic
Proof. Let $x$ be the the maximal element of $S$ and $y$ the minimal element. Then by the previous proposition, $x$ is idempotent. The minimal element $y$ generates a monogenic subsemigroup $(y)$. Suppose that $(y)$ is not equal to the entire semigroup $S$ then $(y)$ has a maximal element which is idempotent, and since it doesn't generate the entire semigroup it is distinct from $x$. This implies that there are two idempotents in the semigroup which contradicts the condition that it is unipotent. Therefore, $y$ generates the entire semigroup, which means that $S$ is monogenic. $\square$
This should make the structure theory of finite commutative tosetal semigroups clear. By semilattice decompositions all finite commutative posetal semigroups can be decomposed into semilattices of finite commutative unipotent semigroups.
Proposition. Finite commutative tosetal semigroups are semilattices of monogenic semigroups
Proof. By semilattice decompositions every finite commutative semigroup can be decomposed into a semilattice of unipotent semigroups. All of these unipotent semigroups are monogenic. Therefore, finite commutative tosetal semigroups are semilattices of monogenic semigroups. $\square$
With this classification in place, we can introduce numerics to measure the number of commutative semigroups associated to a finite total order.
Theorem. There are $2^{n-1}$ finite commutative tosetal semigroups on $n$ elements.
Proof. Let $F$ be the family of commutative semigroups on the total order $T_n$. We can prove this constructively by constructing a bijection $f : \mathcal{P}(1..n-1) \to F$.
(1) Let $S$ a member of $\mathcal{P}(1..n-1)$ and construe its elements as idempotents in the semigroup. Then we can take each of these elements of $\mathcal{P}(1..n-1)$ and add to them all the non-idempotents before them until we get to an idempotent to get a family of intervals with cardinalities $N_1, ... N_{i-1}$. Then take the ordinal sum of semigroups of monogenic semigroups with orders $N_1, ... N_{i-1}$ to get a semigroup isomorphism type in $F$.
(2) Let $S$ be a semigroup isomorphism type and a member of $F$. Then we can map $S$ to $\mathcal{P}(1..n-1)$ by taking the set of all proper idempotents of $S$. By extending the set of idempotents to a family of intervals, we can get a different semilattice of monogenic subsemigroups for each set of idempotents. By the the classification of finite commutative tosetal semigroups as semilattices of monogenic subsemigroups, this determines finite commutative tosetal semigroups up to isomorphism. $\square$
Boolean algebra of isomorphism types:
The fact that there are $2^{n-1}$ isomorphism types of commutative semigroups on a finite total order suggests that there is an underlying boolean algebra structure of isomorphism types. Recall from basic lattice theory that partitions of a finite total order into intervals forms a boolean algebra.
Proposition. the lattice of all partitions of a finite total order $T_n$ into intervals forms a boolean algebra with $T_{n-1}$ elements.
You can take a finite commutative tosetal semigroup and get a partition into intervals from it by taking its semilattice decomposition, which is natural way to get a connection to boolean algebras.
Proposition. semilattice decompositions of finite commutative tosetal semigroups are partitions of total orders into intervals
This makes makes the family of commutative semigroups on a total order into a boolean algebra. This is similar to the boolean algebra of lattice congruences on a finite total order, which is also constructed from partitions into intervals.
The minimal element of $Part(S)$ is the finest partition of $S$. As a semilattice has the finest semilattice decomposition, it is the minimal element of the boolean algebra of isomorphism types. It makes sense to have semilattices be the minimal, because they are the least upper bound function so every greater element is a greater upper bound.
Addendum:
Let $P$ be an upper bounded lower common point free tree. Then the set of non-maximal elements forms a disjoint union of total orders, and so every semigroup principal ideal is a tosetal semigroup. Then a semigroup $S$ with this order type is fully determined by its maximal principal ideals, which are all tosetal. In particular, this fully classifies all semigroups with order type [n, 1].
Proposition. let $P$ be a partial order of order $n$ with maximum height two then there are $n$ isomorphism types on $P$ and they are totally ordered by number of idempotents.
We see that for the simplest order types: those that are $T_n$ or $[n, 1]$ their families of commutative semigroups are lattice ordered. This is true for any upper bounded lower common point free interval ordered tree, in which case the poset of isomorphism types is a power multiset lattice. In the case of $[\{[1,1],[1,1]\}, 1]$ which is not an interval order, however, the ten isomorphism types do not form a lattice because the set orbit poset of 2+2 poset does not.
Proposition. Let $P$ be the diamond partial order with type [1,2,1]. Then the minimal element of $P$ is idempotent.
Proof. Let $x$ be the minimal element of $P$ and suppose that $x$ is not idempotent, then $x^2$ is another element in $S$. Let $y$ be one of the two middle elements not generated by $x$ then the fact that $x$ is less then $y$ implies that $xy=y$ which means that $x^2y=y$, and so $x^2 \subseteq y$ which contradicts the fact that the only predecessor of $y$ is $x$. Therefore, $x$ must be idempotent. $\square$
Therefore, the total order of semigroups on [1,2,1] is the same as those on [2,1]. The minimal idempotent has no effect. It would be wrong to assume that this is the case for any element whose order ideal is a diamond because we must remember that order ideals and semigroup ideals don't necessarily need to coincide. Recall that the factorisation preorder of a subsemigroup is a subpreorder of the preorder of the semigroup.
Proposition. the factorisation preorder of a subsemigroup $S \subset T$ is a subpreorder of the subpreorder of the factorisation preorder of $T$ induced by $S$.
The relationship between action preorders and subalgebras is monotone. If the factorisation preorder of a subsemigroup is reduced that means there is external actions by elements outside the semigroup principal ideal that determine what elements an element is a factor of.
In general, we study commutative semigroups using three things (1) the classification of commutative groups which are the H classes of a commutative semigroup (2) the partially ordered commutative semigroups that emerge from quotient by the H congruence which determines how H classes are related to one another and (3) the action of predecessor Schutzenberger groups on sucessor H classes.
References:
Commutative semigroups by Grillet
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