Does every commutative idempotent semigroup have a representation as a union-closed family of sets

Consider a finite semigroup $$S$$ whose semigroup operation $$\times$$ is commutative and whose elements are idempotent.

Does there exist a finite union-closed family of finite sets $$\mathcal{M}$$ such that there is a bijection $$\tau: S \rightarrow \mathcal{M}$$ where $$\forall a,b \in S, \tau(a \times b) = \tau(a) \cup \tau(b)$$?

I feel that this may not be the case necessarily since I haven't encoded any information about "subsets,containment..." into the axioms i've stipulated on the semi group, but from a purely algebraic standpoint it would seem that this characterizes the union very well.

• Retag your question, it is not set theory. Apr 4 '19 at 2:00
• Re tagged to combinatorics since family of union closed sets originates from there Apr 4 '19 at 2:01

Yes. Given $$a\in S$$, consider the set $$I(a)=\{b\in S:ab=b\}$$. I claim that that $$I(ab)=I(a)\cap I(b)$$. Indeed, if $$c\in I(a)\cap I(b)$$ then $$(ab)c=a(bc)=ac=c$$ so $$c\in I(ab)$$. Conversely, if $$c\in I(ab)$$, then $$ac=a(abc)=a^2bc=abc=c$$ so $$c\in I(a)$$, and similarly $$c\in I(b)$$. Also, if $$I(a)=I(b)$$, then $$a\in I(b)$$ and $$b\in I(a)$$ so $$a=ab=b$$.
So, this is almost what you wanted, but with intersections instead of unions. To get unions, you can just take complements and define $$\tau(a)=S\setminus I(a)$$ and let $$\mathcal{M}$$ be the image of $$\tau$$.
Semigroups of this sort are known as (unbounded) semilattices and are a basic object of study in order theory. They are often thought of as ordered sets via the ordering $$a\leq b$$ if $$ab=b$$ (or $$ab=a$$, depending on the context); the algebraic structure can then be defined in terms of the ordering.
• @M.Vinay: I can see it even while it's deleted. Your approach is basically the same as defining $\tau(a)=\{b\in S:ab=a\}$. That typically won't satisfy $\tau(ab)=\tau(a)\cup\tau(b)$. Indeed, suppose the operation on $S$ really is $\cup$. Then $\tau(ab)=\tau(a)\cup\tau(b)$ would be saying $c\subseteq a\cup b$ iff $c\subseteq a$ or $c\subseteq b$, which isn't true in general. Apr 4 '19 at 2:57
• The corresponding fact for intersections is true, though: $c\subseteq a\cap b$ iff $c\subseteq a$ and $c\subseteq b$. That's why it's more natural to first get a representation where the operation becomes intersection, as in my answer. Apr 4 '19 at 2:59