Subtotal semigroups

In this post, we will investigate when it is that every subset of a semigroup is a subalgebra, and when it is that every subsemigroup is a radical subsemigroup. A semigroup is called subtotal if every subset is a subsemigroup. The structure of subtotal semigroups will be described. It will be shown that in a semigroup $S$ if every subset of $S$ is a subsemigroup then every subset is also radical.

Theorem 1. let $A \subseteq B \subseteq C$ be a chain of semigroups. Then the radical of $A$ in $B$ is less then the radical of $A$ in $C$: $\sqrt(A)_B \subseteq \sqrt(A)_C$.

Proof. let $A \in B$ then $\sqrt(A)_B$ is equal to $\{ x \in B : \exists n : x^n \in A \}$. On the other hand, $\sqrt(A)_C$ is equal to $\{ x \in C : \exists n : x^n \in A \}$. Then by $B \subseteq C$ we have: \[ x \in B \implies x \in C \] Therefore, we have the following implication for all $x \in B$: \[ x \in B \wedge (\exists n : x^n \in A) \implies x \in C \wedge (\exists n : x^n \in A) \] It follows that the set formed by comprehesion by the first proposition is included in the set so $\sqrt(A)_B \subseteq \sqrt(A)_C$. $\square$

This easily demonstrates that the property of being a radical subsemigroup is hereditary, that is if a semigroup $A$ is a radical subsemigroup of $C$ then it is radical in every intermediate subsemigroup.

Corollary 1. let $A \subseteq B \subseteq C$ be a chain of semigroups. If $A$ is radical in $C$ then it is radical in $B$.

By this corollary, we know that if $(x)$ is a principal subsemigroup that is radical, then it also must be radical in any monogenic subsemigroup that contains it. Therefore, it is sufficient to investigate the radical subsemigroups of a monogenic semigroup.

Theorem 2. let $S$ be a monogenic semigroup then the only radical subsemigroups of $S$ are $\emptyset$ and $S$.

Proof. (1) let $S$ be finite. Then $S$ contains a unique idempotent $e$, which is contained in every non-empty subsemigroup. As every element generates $e$ we have $\sqrt(e) = S$. Therefore, if $X$ is non-empty we have $e \in X$ which implies $\sqrt(e) \subseteq X$ which implies $S \subseteq X \subseteq S$ which implies $X = S$.

(2) let $S$ be infinite. Then every infinite monogenic semigroup is a well ordered semigroup in a natural way. Let $X$ be a non-empty subsemigroup then by well ordering $X$ has a least element $n$. Additionally, $1$ is a generator for this element by iterating it $n$ times. Therefore, $1 \in X$ but since $\sqrt(\{1\}) = S$ we have that $S \subseteq X \subseteq S$ which implies $X = S$. $\square$

It follows that the lattice ordered family of radical subsemigroups of a monogenic subsemigroup is equal to an indiscrete topology. Furthermore, if every subsemigroup of a monogenic semigroup is radical then it is trivial.

Corollary 2. let $S$ be a monogenic semigroup. If every subsemigroup of $S$ is radical, then $S$ is a trivial monoid on one element.

By the combination of these two collaries we can complete our classification of semigroups for which every subsemigroup is radical.

Theorem 3. let $S$ be a semigroup. Every subsemigroup of $S$ is radical iff $S$ is a band.

Proof. (1) if $S$ is a band then the iteration preorder on $S$ is trivial, therefore every subsemigroup must be radical (2) if every subsemigroup of $S$ is radical, then by corollary one every subsemigroup of a monogenic subsemigroup of $S$ is radical. By corollary two the only monogenic subsemigroups for which every subsemigroup is radical are trivial monoids. Therefore, every monogenic subsemigroup of $S$ must be a trivial monoid, so that every element of $S$ is idempotent which means it is a band. $\square$

A moore family is what is formed by relaxing the union closure condition of a cotopological space. The lattice of Moore families on a set is typically a more broad and useful setting for applications then the lattice of topological spaces. Within this lattice of lattices, we have the following chain of inclusions:

Radical subsemigroups $\subseteq$ Subsemigroups $\subseteq$ Subsets

Subsemigroups and radical subsemigroups are rank two Moore families while the family of all subsets is really rank zero rather then rank two, because it is defined by a nullary closure operation rather then a binary one. The fact that the lattice of subsemigroups is rank two follows from basic universal algebra.

Corollary. let $S$ be a semigroup then $Sub(S)$ is a rank two Moore family

The interesting thing about this is that not all set systems, or even all Moore families, emerge as the families of subsemigroups of a semigroup. In particular, it is enough to get that $S$ is a 1-total and 2-total for it to be completely total.

Lemma. let $S$ be a semigroup and suppose that every subset of order one or two is a subsemigroup of $S$, then every subset of $S$ is a subsemigroup.

Proof. let $X$ be a subset of $S$ then we have that $\forall a,b \in X : ab \in \{a,b\}$. Then the fact that $\{a,b\} \subseteq X$ implies that $ab \in X$. Therefore, $X$ is composition closed. $\square$

This clearly reduces the problem of subalgebraic totality to one of graph theory. In particular, we can define the totality graph for any semigroup.

Definition. let $S$ be a semigroup. Then the totality graph of $S$ has as vertices all elements of $S$, as loops all idempotents, and as edges all size two subbands.

By the preceding theorem, every clique of the totality graph is a subsemigroup. The resulting clique complex is the subclass closed component of the Moore family of subsemigroups of the semigroup $S$. That this set system is a clique complex follows by the rank of the Moore family.

Lemma. let $S$ be a semilattice then $S$ is subtotal iff $S$ is a total order.

Proof. (1) let $S$ be a total order semilattice then $\forall x,y : x \vee y \in \{x,y\}$ so that $S$ is 2-total, it follows that $S$ is subtotal (2) let $S$ be subtotal, then $\forall x,y :x \vee y \in \{x,y\}$. This means $x \vee y = x$ or $x \vee y = y$. By the definition of the ordering on a semilattice, this then implies either $x \subseteq y$ or $y \subseteq x$ which means that $x,y$ are comparable. It follows by the fact that every element of $S$ is comparable that $S$ is a total order. $\square$

Much like the commuting graph of a semigroup can be determined subalgebraically, so to can the comparability graph of a semilattice. By this lemma, when $S$ is a semilattice the comparability graph and the totality graph coincide.

Corollary. let $S$ be a semilattice then the comparability graph of $S$ is equal to its totality graph.

This solves the problem of totality for semilattices, and it relates subtotal semigroups to total order semilattices. However, in order to complete the theory we need to investigate rectangular bands more.

Lemma. let $S$ be a rectangular band then $S$ is subtotal iff $S$ is pure.

Proof. (1) let $S$ be a pure rectangular band, then $S$ is either left zero or right zero. This means that if it is left zero $ab = a$ or if it is right zero then $ab = b$. In either case, $ab \in \{a,b\}$ which means that $S$ is 2-total.

(2) let $S = L_n \times R_m$ then if $S$ is not a pure rectangular band we have $2 \le n,m$. It follows that we can select elements $(1,2) \circ (2,1) = (1,1)$ so that $S$ is not 2-total. $\square$

A band is a 1-total semigroup, so that every singleton is a subsemigroup. A subtotal semigroup is simply a 1-total semigroup that is also 2-total, therefore every subtotal semigroup is a band. Every band is a semilattice of rectangular bands. By this construction, and the previous two lemmas we have a structure theorem for subtotal semigroups.

Theorem. let $S$ be a subtotal semigroup. Then $S$ is a total order semilattice of pure rectangular bands.

Proof. (1) by the fact that $S$ is subtotal it is a band, and then by the classification theorem for bands this means that it is a semilattice of rectangular bands (2) by the fact that only pure rectangular bands are subtotal it follows that each rectangular band must be pure (3) finally the semilattice must be subtotal which must be a total order. $\square$

The class of pure rectangular bands is the union of the atomic varieties of left zero and right zero bands. It follows that each subtotal semigroup can contain either left zero semigroups, right zero semigroups, or both. A subtotal semigroup is called mixed if it contains one of each of the different types.

• Unmixed subtotal semigroup: all pure rectangular bands are either all left or all right.
• Mixed subtotal semigroup: these contain both left and right zero components

The simplest example of a mixed subtotal semigroup is the semigroup on four elements $[L_2,R_2]$ constructed by an ordered pair of a left zero semigroup and a right zero semigroup, or its anti-isomorphic form $[R_2,L_2]$. Every subtotal band is also clearly normal. This completes our understanding of subtotal semigroups. It is not hard to deal with the dual case, when a semigroup or group is subtrivial. Subtriviality can be considered to be the same as atomicity, so this is equivalent to finding atoms in the lattice of subalgebras.

Proposition. (1) atoms in the lattice of subsemigroups are trivial monoids and (2) atoms in the lattice of subgroups are prime order cyclic groups.

Every atomic subsemigroup is clearly monogenic. In the finite case, we always have an idempotent subsemigroup as an atomic which means $S$ must be a trivial monoid. On the other hand, the infinite monogenic semigroup has no atoms, which means nothing covers the empty set. The empty subsemigroup is an order topological limit point of any maximal chain, because every maximal chain of subsemigroups will terminate at the empty set.