Semigroups are associated to two types of lattices: lattices of subalgebras and congruences. Another aspect of the J class structure theory is therefore how the algebraic preorder is effected by taking subalgebras and quotients. This problem is solved by a couple of lemmas.
Lemma 1. let $S \subseteq T$ be a chain of semigroups then the algebraic preorder $\subseteq_S$ is a subpreorder of $\subseteq_T$.
Proof. let $x \subseteq_S y$ then that means that there exists $a,b \in S^1$ such that $axb = y$. Then because $a,b \in S^1$ and $S^1 \subseteq T^1$ we have $a,b \in T^1$. Therefore, $a,b \in T_1 \wedge axb = y$ which implies that $a \subseteq_T b$. $\square$
Lemma 2. let $S$ be a semigroup with algebraic preorder $\subseteq$ and let $C$ be a congruence on $S$. Then the algebraic preorder of $\frac{S}{C}$ is the factor relation on the preorder $\subseteq_S$ determined by ordinary precedence
Proof. let $S,T$ be subsets of a preorder $\subseteq$ with partition $P$ then ordinary precedence $S \frac{\subseteq}{P} T$ is logically equivalent to $\exists s \in S, t \in T : s \subseteq t$. We will show that this preorder of the classes of a partitioned preorder, determines the J class structure of a quotient.
(1) let $\frac{S}{C}$ be the quotient semigroup and let $C_1, C_2$ be congruence classes in $\frac{S}{C}$. Suppose that $C_1 \subseteq C_2$ then we have that there exists $D$ such that $D C_1 = C_2$. The congruence classes $D,C_1,C_2$ are all non-empty so there exists $d, c_1, c_2 \in D,C_1,C_2$ such that : \[ dc_1 = c_2 \] \[ \Rightarrow c_1 \subseteq c_2 \] Then by the fact that $c_1, c_2 \in C_1, C_2$ and $c_1 \subseteq c_2$ we have sufficient conditions for ordinary precedence of congruence classes.
(2) Let $S$ be a semigroup with a congruence $C$ then there is a projection morphism $\pi$ to the quotient semigroup $\frac{S}{C}$: \[ \pi : S \to \frac{S}{C} \] Let $c_1, c_2 \in C_1, C_2$ and suppose $c_1 \subseteq c_2$ then $\exists d : dc_1 = c_2$. We can apply the projection morphism to both sides of this equation to get: \[ \pi(dc_1) = \pi(d)\pi(c_1) = \pi(c_2) \] We have by supposition that $c_1 ,c_2 \in C_1, C_2$ so $\pi(c_1) = C_1$ and $\pi(c_2) = C_2$. By plugging this into the previous equation we get: \[ \pi(d)C_1 = C_2 \] It follows that there exists $\pi(d) \in C : \pi(d)C_1 = C_2$ which implies that $C_1 \subseteq C_2$. Therefore, ordinary precedence implies quotient precedence.
(3) by the previous two propositions the implications between ordinary precedence and quotient precedence are bidirectional, and so the two concepts are equivalent. $\square$
These two lemmas allow us to reason about the J class structure of subalgebras and quotients. The following corollary is an immediate result.
Corollary.
- J-trivial semigroups are subalgebra closed but not quotient closed
- J-total semigroups are quotient closed but not subalgebra closed
(1) antisymmetry is subclass closed, so by lemma (1) J-triviality is preserved under taking subalgebras.
(2) $(\mathbb{N},+)$ is J-trivial, but its quotient $\frac{(\mathbb{N},+)}{mod2}$ is not. Therefore, J-triviality is not necessarily preserved under quotient.
(3) by lemma (3) quotient precedence is determined by ordinary precedence, so simplicity is preserved under the operation of taking quotients.
(4) the additive group of the integers $(\mathbb{Z},+)$ is J-total, but the additive semigroup of the natural numbers $(\mathbb{N},+)$ which it contains is not. Therefore, J-totality is not necessarily preserved under taking subalgebras. $\square$
In the case that a semigroup is not condensible, then its impediments to condensation are determined by its principal factors, which are all 0-simple semigroups. There are two types of J-trivial semigroups on two elements, therefore the condensible 0-simple semigroups belong to two classes determined by their condensations:
- Null semigroup : (condensation I2)
- Simple semigroup with zero : (condensation T2)
Theorem. let $S$ be a condensible semigroup, then if $S$ is condensible all of its principal factors are.
Proof. suppose that not all principal factors of $S$ are condensible. Let $C$ be a J class whose principal factor is not condensible. Then let $X$ be the set of all elements in $C^2 - C$. Then by the fact that $C^2 \cap C \not= \emptyset$ we have that the the congruence closure of the partition containing $C$ and with every other element a singleton, has a closure which adds $X$ to $C$. Therefore, the J class $C$ cannot be a congruence class of any congruence. It follows that J itself cannot be a congruence, so $S$ cannot be condensible. $\square$
This allows us to determine the impediments to condensibility by all the principal factors of the semigroup. On the level of a single 0-simple semigroup, we can evaluate its condensibility based upon the percentage of its non-zero components that go to one component or the other.
Definition. let $S$ be a finite 0-simple semigroup, then the zero percentage of $S$ is the number of pairs of non-zero elements that compose to zero divided by $n^2$ where $n$ is the number of non-zero elements.
A condensible 0-simple semigroup must have a zero percentage of either 100% (a null semigroup) or 0% (a simple semigroup with zero). The extent to which the zero percentage differs from one of the extremes provides a measure of the extent to which a 0-simple semigroup fails to have J as a congruence. The brandt semigroup on five elements, for example, has a 50% percentage because half of its elements go to zero and the other half don't.
The condensibility of principal factors determines the impediments to condensibility. Additionally, the principal factors themselves, which are well understood by Rees matrix semigroups, provide a measure of understanding of the local algebraic structure of J classes. Together with the principal factors construction this allows us to create a full theory of the J class structure of a semigroup, even when that semigroup is not condensible.
See also:
Factor relations
Condensible semigroups
References:
[1] Fundamentals of semigroup theory [Howie]
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