Lattice-ordered suborders and semigroups

Let $L$ be a lattice, and $S$ be a suborder of $L$ that forms a lattice. Then the two lattices can be compared to determine when they are similar and when they are different. In particular, for any two elements $a$ and $b$ they can be either join-preserving or meet-preserving based upon their relation to the parent order:

• Join-preserving: two elements $a,b$ are join preserving if $a \vee_S b = a \vee_L b$
• Meet-preserving: two elements $a,b$ are meet preserving if $a \vee_S b = a \vee_L b$

The join-preservation and meet-preservation both form simple graphs on the set of elements of the suborder. The extrema-preservation graph is the intersection of these two graphs. Clearly, if $a \subseteq b$ or if $b \subseteq a$ then the two elements are extrema-preserving.

Proposition. the comparability graph is a subgraph of the extrema-preserving graph

Proof. let $P$ be a pair of two elements in the comparability graph, as comparable elements we know that there must be some $a,b$ such that $a \subseteq b$ in P, then the join in the parent lattice is still $b$ and the meet is still $a$, because the suborder is always order-preserving (even though it isn't always meet preserving and join preserving).

As the comparability graph is a subgraph of the extrema-preserving graph, a suborder is either join-free and meet-free if its incomparable elements are not join-preserving or meet-preserving respectively. It is in this sense that we say that preorder containment families are union-free: they have no non-trivial unions. A suborder is a meet subsemilattice if it is totally meet preserving, a join subsemilattice if it is totally join preserving, and a sublattice if it is both.

In those cases where the suborder is not meet or join preserving, it is interesting to consider the subset of cases in which it is. In particular, a Moore family is a meet-preserving lattice-ordered suborder in the lattice of sets. In this cases, we can consider the subset of pairs for which the union operation is preserved.

Chains of lattices:
Given a chain of three lattices $A$, $B$, $C$ then it is possible for $A$ to be a sublattice of $C$ even though its intermediate lattice $B$ is not a sublattice. Consider a finite group $G$ which is not a prime power cyclic group and let $Sub(G)$ be its lattice of subgroups, then $Sub(G)$ is not a sublattice of $\mathscr{P}(C)$. Then let $A$ be a chain of groups, such as a composition series, then $A$ is a sublattice of $C$ because it is a chain. Sublattices of the parent lattice come from extrema-preserving cliques that are sublattices.

Lattices of subsemigroups
Let $S$ be a semigroup, and let $Sub(S)$ be its lattice of subsemigroups, then generally the lattice of subsemigroups is not a sublattice. One example where it is, is the lattice of subsemigroups of a total order semilattice, a pure rectangular band, or a totally ordered sum of pure rectangular bands. In fact, in those cases every subset of the semigroup is a subsemigroup. As $Sub(S) = \mathscr{P}(S)$ it forms a sublattice. Another case is the lattice of subsemigroups of a prime power cyclic group in which case $Sub(S)$ forms a chain, which is always a sublattice.

In those cases when the lattice of subsemigroups $Sub(S)$ does not necessarily form a sublattice, then it is potentially interesting to consider the cases when the join is simply the union. An obvious case is a group with zero, in that case the singleton set containing the zero element and the rest of the set form a join-preserving pair of subsemigroups, with the union being the entire semigroup. Another example, is the set of even integers $2\mathbb{Z}$ and the set of odd integers $2\mathbb{Z}+1$ both of which form multiplicative subsemigroups and their union is the entire semigroup, so they form a join-preserving pair. In general, the multiplicative semigroup of a prime ideal and its complement form an example of a join-preserving pair.