Condensation of commutative cancellative semigroups

The class of commutative cancellative semigroups is subalgebra closed, but it is not quotient closed. The free commutative monoid $F(x)$ is commutative and cancellative but $\frac{F(x)}{x^2=x}$ is not. If it is in the case that the condensation $\frac{S}{H}$ is cancellative, it is therefore not an inherent property of the class of commutative cancellative semigroups.

Theorem. let $S$ be a commutative cancellative semigroup then $\frac{S}{H}$ is cancellative.

Proof. suppose $a =H a'$ and $ax =_H a'y$ then $a = ba'$ and $a' = ca$ for some $b,c$. Furthermore, $ax \lambda = a'y$ ad $ax = a'y \mu$. By plugging in we get \[ (ba')\lambda x = a'y \Rightarrow a'(b\lambda x) = a'y \] \[ ax = (ca)\mu y \Rightarrow a(x) = a(c \mu y) \] By the cancellative property this yields \[ ax = a(c\mu y) \Rightarrow x = (c\mu)y \] \[ a'(b\lambda x) = a'y \Rightarrow y=(b \lambda)x \] Thusly, $x =_H y$. We can conclude that $a =H a'$ and $ax =_H a'y$ implies that $x =H y$. This is our first step. Let $H_1,H_2,H_3$ be three $H$ classes and suppose that $H_1 H_2 = H_1 H_3$ then we have: \[ \exists h_1,h_1' \in H_1, h_2 \in H_2,h_3 \in H_3 : h_1 h_2 =_H h_1' h_3 \] We have that $h_1 =_H h_1'$ as well ast that $h_1 h_2 =_H h_1' h_3$ but as we previously concluded that this means that $h_2 =_H h_3$. If $h_2 \in H_2$ and $h_3 \in H_3$ then $H_2 = H_3$ because $H$ classes are disjoint.

Then $H_1 H_2 = H_1 H_3 \Rightarrow H_2 = H_3$ implies that the condensation semigroup $\frac{S}{H}$ is a commutative cancellative J-trivial semigroup. $\square$

This is an infinite source of commutative cancellative J-trivial semigroups. For example, it follows from this theorem that the ideal multiplication semigroup of a principal ideal domain is a commutative cancellative J-trivial semigroup with zero.