It can be seen to be dual to E-semigroups, because all rectangular-band free E-semigroups are idempotent commutative. This follows from Clifford's theorem which characterizes all rectangular band free semigroups as semilattices. As a result, the intersection of the class of rectangular band free semigroups and E-semigroups is the class of idempotent commutative semigroups.
Definition. a semigroup is rectangular band free if it contains no subsemigroups of order two or greater that are rectangular bands
The relationship between rectangular band freeness and idempotent commutativity are dealt with in the following ontology of classes of semigroups. Idempotent commutative semigroups, as well as commutative semigroups themselves are rectangular-band free.
While the relationship between idempotent commutativity and rectangular band freeness is interesting. It is also noticeable that rectangular band free semigroups tend to be divisibility commutative (L=R) for the smallest semigroups. Rectangular bands are responsible for different L and R relations in the smallest semigroups.
The weakest generalisation of commutativity is that semigroups be anticommutative semigroup free, because the anticommutative semigroups clearly have the least commutative behaviour (the most different Green's L and R relations, the least commutativity, etc). As a result, it is worth highlighting this as a possible generalisation of commutativity.
Proposition. a semigroup of order four or less is divisibility commutative iff it is rectangular band free
We see that semigroups that are of order four or less are rectangular band free iff they are divisibility commutative, but by no means is this the case in general. If it were, then every inverse semigroup would be Clifford.
Example 1. the inverse semigroup on five elements containing all non-permutation charts on two elements is the smallest non-Clifford inverse semigroup. It is idempotent commutative but not divisibility commutative. In general, non-trivial symmetric inverse semigroups are rectangular band free but not divisibility commutative.
Example 2. the monoid of increasing monotone increasings on a three element total order, is a four element tree ordered J-trivial semiband. As a non-commutative J-trivial semiband it is rectangular band free and not idempotent commutativity. In general, non-commutative J-trivial semibands are rectangular band free and not idempotent commutative.
The existence of non-commutative J-trivial semibands as well as non-Clifford inverse semigroups demonstrates that while rectangular band freeness is bascially the weakest generalisation of commutativity (by forbidding anticommutative components) it doesn't imply any of the other nice conditions like divisibility commutative and idempotent commutativity.
There are a couple of things that can be said about the commuting graphs of rectangular band free semigroups. A graph is idempotent commutative provided that all singleton centralizers form a clique. It is rectangular band free only if no non-trivial centralizers are independent sets. The cyclic graph $C_4$ is an ordered pair of rectangular bands in two different ways, so it cannot be rectangular band free.
In general, a complete bipartite graph $K_{n,m}$ with $n,m \ge 2$ is an ordered pair of rectangular bands, and so it cannot be rectangular band free, for example. The path graph $P_4$ on four elements is not idempotent commutative, but it doesn't have any commutativity necessary rectangular band subsemigroups, so it is an example of the weaker condition.
See also:
[1] Divisibility commutativity
[2] Idempotent commutativity
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