Divisibility commmutative regular semigroups

Divisibility commutative regular semigroups will be examined. But before doing so, its worth stating Green's theorem and its corollaries which are necessary for the other results.

Green's theorem: the H class of every idempotent forms a subgroup of the semigroup. Since idempotents are unique in groups this means that H classes are idempotent separating.

A proof of this theorem is unnecessary as it is well known. The first theorem that we will prove ourselves is that divisibility commutative regular semigroups are completely regular. The second we will prove is that they are inverse.

Theorem: divisibility commutative regular semigroups are completely regular

Proof: A semigroup is completely regular if every elements is included in some subgroup of the semigroup. But this can be restated as the semigroup is completely regular if every H class contains an idempotent. In that case, by Green's theorem we know that each element is contained in a group which is the group of the idempotent of its H equivalence class. A semigroup is regular if each L class and each R class contain an idempotent. In a divisibility commutative semigroup H = L = R so it follows that H is equal to L and R which both contain idempotents in each of their equivalence classes. It follows that H contains idempotents in each of its equivalence classes. This makes it completely regular.

Theorem: divisibility commutative regular semigroups are inverse

Proof: an inverse semigroup can be stated as a regular semigroup such that L and R are both idempotent separating. Regular semigroups already guarantee that L and R classes have idempotents, but this condition ensures that the idempotent in each L and R class is unique. At the same time, we know from Green's theorem that the H classes of any semigroup are always idempotent separating. It follows that if we have divibisility commutative which means that H = L = R the property of being idempotent separating is shared with the L and R classes of the semigroup. This in turn means that the semigroup is inverse.

Theorem: completely regular inverse semigroups are divisibility commutative

Proof: from the fact that the semigroup is completely regular we know that each H class contains an idempotent. Let S be any L class or R class of the semigroup, then since the semigroup is inverse we know that must contain a unique idempotent. That unique idempotent has an H class containing it. If that H class is equal to S then we are done as that L/R class is equal to its H class which means that L = R = H. Suppose on the contrary, we have an L class or an R class that is not equal to the H class of its unique idempotent. Then there must be some element in the class S that is not in the H class of that idempotent. That element has its own distinct H class that it is contained in and that distinct H class and its distinct idempotent must be included in S. Therefore, we arrive at a class S with two different idempotents but that is not possible for an inverse semigroup so we arrive at a contradiction. As a result, we know that each L class and each R class is equal to a corresponding H class. This implies that L = R = H which means that the semigroup is divisibility commutative.

Conclusion: the three properties of being completely regular, inverse, and divisibility commutative form a perfect trinity where any of the two properties implies the other. The intersection of this triad is the class of Clifford semigroups. Clifford semigroups form a natural generalisation of semilattices and groups and they are one of the most basic classes of divisibility commutative semigroups.