2
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ Which examples did you try? $\endgroup$ Commented Sep 5, 2020 at 19:46
  • 1
    $\begingroup$ @Shaun $ \, \mathbb{N} \, $, $ \, \mathbb{N} \setminus \{ 0 \} \, $, the free semigroups and the multiplicative $ \ S = \{ 0,1,x \} \ $, with $ \ |S|=3 \ $ and $ \ x^2=x \ $. $\endgroup$ Commented Sep 5, 2020 at 19:53

3 Answers 3

Reset to default
2
$\begingroup$

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 $$\{2\}\ne 2(\Bbb{N}\setminus\{0\}).$$

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
-1
$\begingroup$

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

$\endgroup$

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.