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.
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