For a semigroup $S,$ its power semigroup $P(S)$ is the semigroup of all non-empty subsets of $S$ with the operation given by $AB=\{ab\,|\,a\in A,b\in B\}.$ I would like to know about the cancellable elements of $P(S)$ given some knowledge of cancellability in $S.$

If $s\in S$ is left-cancellable in $S,$ then $\{s\}$ is also left-cancellable in $P(S)$ because if $sA=sB$ and $a\in A,$ then $sa=sb$ for some $b\in B,$ and $a=b\in B$ follows.

There can be cancellable elements of $P(S)$ with more than one element: for a free semigroup $F,$ any set of the free generators is cancellable from both sides, since from its product with any set $A$ we can recover $A$ by looking at the first letters of each word in the product and removing the letters. This is also an example with $S$ two-sided cancellative.

The first question that comes to mind here is whether

we can find $S$ and a cancellable $A\subseteq S$ such that $A$ has an element not cancellable in $S.$

Another question is what happens for cancellative semigroups. For the commutative ones, we can show that only the singletons are (left-)cancellable in $P(S)$ (let's call this property $(P)$.) That is because then, if $x,y\in A,\,x\neq y,$ we have $A(S\setminus\{xy\})=AS,$ which is easy to check using commutativity and cancellation laws in $S.$ With regard to the first question, I don't see if it can be done without the cancellation laws.

Actually, this is in a sense a "good" method of showing that the only cancellable elements of $P(S)$ are singletons. That is, the following two conditions are equivalent for any semigroup $S$ and $A\subseteq S.$

- $A$ is not left-cancellable.
- There exists $c\in S$ such that $A(S\setminus\{c\})=AS.$

Suppose $A$ is not left-cancellable. Then we have $AB=AC$ for some $B\neq C.$ Without loss of generality, take $c\in C\setminus B.$ Then $A(S\setminus\{c\})=AB\cup A(S\setminus\{c\})=AC\cup A(S\setminus\{c\})=AS.$

So to check that a left-cancellative semigroup has the property $(P),$ we need to find such a $c$ for every $A$.

Can this criterion be simplified further? Can we just look for the $c$ for any two given elements of $A$ as in the commutative case? We can in every example I can think of. Can we just use two-element sets $A$ to check it?

The commutative cancellative semigroups are not the only ones satisfying $(P)$. Groups satisfy it as well. Also, the multiplicative structure of non-zero Lipschitz quaternions does.

We also have that any left-cancellative semigroup satisfying $(P)$ will have to satisfy the right Ore condition. Suppose $S$ is left-cancellative and satisfies $(P).$ Let $x,y\in S.$ Then we have $A\neq B$ such that $\{x,y\}A=\{x,y\}B.$ Without loss of generality, let $b\in B\setminus A.$ Then $yb=xa$ for some $a\in A$ since $yb=ya$ is impossible if $y$ is left-cancellable and $b\not\in A.$

However, this semigroup is both two-sided Ore and two-sided cancellative, but doesn't seem to satisfy $(P).$ $\{x,y\}$ seems to be a cancellable subset.

So what is $(P)$ really?

The class of left-cancellative semigroups satisfying it is clearly globally determined, which is why I started to think about this in the first place.