Semigroups with the property $S=S^1(S\setminus S^2)$

Asked Jul 26 '23 at 20:23

Active Jul 26 '23 at 20:37

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We have encountered semigroups $S$ which have the property $S=S^1(S\setminus S^2)$ (see Equality of two specific classes of subsets of a group), or equivalently $S\subseteq S^1(S\setminus S^2)$.

Are there any characterizations (or classifications) of such semigroups?

Note. It is obvious that a necessary condition is $S^2\neq S$ (i.e., $S^2\subset S$), thus $S$ must not contain any left or right identity element. Also, the property is equivalent to $S^2=S(S\setminus S^2)$, because $S^1(S\setminus S^2)=S(S\setminus S^2)\dot{\cup} (S\setminus S^2)$ and $S=S^2\dot{\cup} (S\setminus S^2)$, where $\dot{\cup}$ denotes the disjoint union.

Every monoid does not have the property, but the additive semigroups $\mathbb{Z}_+$ and $(M,+\infty)$ ($M>0$) enjoy it.

edited Jul 26 '23 at 20:37

asked Jul 26 '23 at 20:23

M.H.Hooshmand

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It is worth stating the definitions (from the linked question) that $S^1 = S \cup {1}$ and $S^2 = S \cdot S$, because these are both nonstandard and a bit confusingly inconsistent. Also, what is $(M,+\infty)$? – Sean Eberhard Jul 27 '23 at 13:14
Well, they are somewhat standard, but they are also confusingly inconsistent. – Sean Eberhard Jul 27 '23 at 13:20
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It is equivalent to say that $S$ is generated by $S \setminus S^2$. – Sean Eberhard Jul 27 '23 at 13:34
@Sean Eberhard. $(M,+\infty)$ is an interval that is a sub-semigroup of $(\mathbb{R},+)$ if $M\geq 0$. – M.H.Hooshmand Jul 29 '23 at 16:38
@Sean Ebrehard. Can you please clarify why $S=\langle S\setminus S^2\rangle$ is considered equivalent? – M.H.Hooshmand Jul 29 '23 at 16:42

Semigroups with the property $S=S^1(S\setminus S^2)$

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