Divisibility commutativity in the sense of green's relations

Given a commutative semigroup, then that semigroup has an associated preorder $a \le b$ if there exists a c such that $ac = b$. This is a preorder that defines all the ordering properties of a commutative semigroup in a singular standard way. The situation is not so simple in non-commutative semigroups as seen by the different green's relations. But suppose that L = R then we know that since D and H are defined by the meet and join of L and R in the partition lattice this means that all four are equal. \[ L = R = D = H \] In these cases, we can instead focus on a single preorder. Well there are different preorders on a non-commutative semigroup (the Green's preorders each have L,R,J,and H as partitions) these preorders are also equal in a divisibility commutative semigroup. That is, there is a sense that the semigroup has a singular order just like commutative semigroups. We already discussed J-trivial semigroups which are essentially just semigroups on total orders, well semigroups with a symmetric total preorder are essentially groups (which is proven in the finite case).

• Antisymmetric: J-trivial semigroups
• Symmetric: Groups

It makes sense that groups are divisibility commutative, because everything can divide everything else in both directions. So we see that the two basic constituents of divisibility commutative semigroups are J-trivial semigroups and groups. J-trivial semigroups are aperiodic well groups are periodic respectively. This includes non-commutative groups as well, as their non-commutativity doesn't effect their divisibility. In general, divisibility commutative semigroups allow a restricted amount of non-commutativity so long as it doesn't effect divisibility.