Ultrafilter invariant semigroups

Question

This may be quite basic but I was unable to figure out the answer on an uncountable set. Suppose that $S$ is an infinite set and $U$ a non-principal ultrafilter on $S$. Does there exist a commutative semigroup structure $(S,+)$ on $S$ such that for every $t\in S$ one has $A\in U$ if and only if $\{s+t\colon s\in A\}\in U$?

Answer 1

Choose, by Zorn's lemma, a maximal chain $\mathcal C\subseteq U$ under subset inclusion. Then $\bigcap \mathcal C$ cannot be in $U$, because every element of a non-principal ultrafilter contains a strictly smaller element of the ultrafilter, which we could have extended the chain with. Now, $$ \mathcal D = \{ A\setminus \bigcap\mathcal C \mid A \in \mathcal C \} $$ is a chain in $U$ whose intersection is the empty set.

Now for $x,y\in S$ say that $x\prec y$ if there is a $B\in\mathcal D$ that contains $y$ but doesn't contain $x$. (If we view $\mathcal D$ as going from larger subsets of $S$ towards smaller ones, $x\prec y$ means that $x$ disappears from the subsets before $y$ does).

$\prec$ is a strict partial order (because $\mathcal D$ is a chain); extend it to a total order in an arbitrary way.

Now let your semigroup operation $x+y$ be $\max(x,y)$ with respect to this total order! (Intuitively $x+y$ is the element among $x$ and $y$ that disappears last in $\mathcal D$, except if they disappear together).

Then for every $t$ there is a $B_t\in\mathcal D$ that does not contain $t$, and we then have, for every $A\subseteq S$, $$ A\cap B_t = (A+t)\cap B_t$$