Example of a cancellative power semigroup

Question

Let $ \, S \, $ be a semigroup such that $ \ |S| \geq 2 \ $. Its power semigroup is the power set $ \, \wp(S) \, $ together with the binary operation $$XY = \{ xy \in S : x \in X, \ y \in Y \} \ \ .$$

I am interested in the semigroup $ \ Q_S = \wp(S) \setminus \{ \varnothing \} \ $, which I also call the power semigroup of $ \, S \, $.

A semigroup $ \, S \, $ is said left cancellative if, and only if, for all $ \ x,y,z \in S \ $, if $ \ xy=xz \ $, then $ \ y=z \ $.

A semigroup $ \, S \, $ is said right cancellative if, and only if, for all $ \ x,y,z \in S \ $, if $ \ yx=zx \ $, then $ \ y=z \ $.

I would like to see an example of a semigroup $ \, S \, $ such that $ \, Q_S \, $ is left cancellative and right cancellative.

I tested the most immediate and the most standard examples of semigroups, but none of them resulted in such a desired example. I don't know where to look anymore.

Answer 1

Since $\Bbb{N}\setminus \{0\}$ under addition does not satisfy your condition (see Edit below) there is no semigroup $S$ that does.

Indeed, if $S$ contains an idempotent $e=e^2$ then $\{e\}S=\{e\} eS$ in $Q_S$, hence $S=eS$, so $e$ is a left identity element. Similarly $e$ is a right identity element, so $e$ is the identity element in $S$. But then $SS=S=\{e\}S$ in $Q_S$, so $S=\{e\}$, a contradiction.

Thus $S$ does not contain idempotents. But then $S$ contains an infinite cyclic subsemigroup $C$ which is isomorphic to the additive semigroup $\Bbb{N}\setminus\{0\}$, a contradiction because $Q_C\le Q_S$.

**Edit: ** The semigroup $Q_{\Bbb{N}\setminus\{0\}}$ does not satisfy the cancelation property as @Simon noticed: $$\{2\}+2(\Bbb{N}\setminus\{0\})=2(\Bbb{N}\setminus\{0\})+2(\Bbb{N}\setminus\{0\})$$ but $$\{2\}\ne 2(\Bbb{N}\setminus\{0\}).$$

Answer 2

Let $a \in S$ and let $a^+ = \{a^n \mid n > 0\}$ be the subsemigroup of $S$ generated by $a$. Since $aa^+ = a^+a^+$, one gets $a = a^+$ by right cancellation. Thus $a = a^2$ and $S$ is an idempotent semigroup. Furthermore, $aS = aaS$ implies $S = aS$ by left cancellation. Consequently, $SS = \bigcup_{x \in S} xS = S = aS$, whence $S = a$ by right cancellation. Thus $S$ is the trivial semigroup.

Answer 3

The set of positive integers under addition forms a cancellative semigroup, this is clearly not a group embeddable semigroup.