# Is there any characterization and/or classification of subsemigroups of finite monogenic semigroups?

A semigroup $S$ is called monogenic if $S$ is generated by some element $a$ (which is unique if $S$ is not a group) in the sense that $S=\{a^n:n\in\mathbb N\}$.

Observe that each mongenic group is finite cyclic. It is known that each subsemigroup of a monogenic group is a cyclic group.

On the other hand, a subsemigroup of a finite monogenic semigroup need not be monogenic. The simplest example is the subsemigroup $\{a^2,a^3,a^4=a^5\}$ of the monogenic semigroup $\{a,a^2,a^3,a^4=a^5\}$.

Let us call a semigroup $S$ submonogenic if it is isomorphic to a subsemigroup of a monogenic semigroup.

Question 1. Is there any reasonable characterization of (finite) submonogenic semigroups?

It is clear that each finite submonogenic semigroup $S$ has the following properties:

(1) $S$ is commutative;

(2) $S$ has a unique idempotent;

(3) the minimal ideal $I$ of $S$ is a monogenic group;

(4) for any $a,x,y\in S$ the equality $ax=ay\notin I$ implies $x=y$;

(5) for any $n\in\mathbb N$ and $x,y\in S$ the equality $x^n=y^n\notin I$ implies $x=y$.

Question 2. Is each finite semigroup $S$ satisfying the conditions (1)--(3) submonogenic?

Question 2' (added after appearing Mark Sapir's Counterexample to Question 2). Is each finite semigroup $S$ satisfying the conditions (1)--(5) submonogenic?

Question 3. Is any reasonable classification of finite submonogenic semigroups?

• Rings and tails. The super (finite) semigroup has unique numbers m and p which are the least m and p such that a^m = a^m+p. The tail has length m-1, and the ring has p elements. If the subsemigroup is not also monogenic, then there are more elements in the tail (so it looks like the start of a numerical semigroup), and/or not many elements of the ring are needed to finish off generating the sub. Gerhard "Yet Another Meaning Of Ring" Paseman, 2018.01.21. Jan 21 '18 at 22:30

Q2: No, consider the semigroup consisting of five elements $\{a,b,c,d,0\}$ where $ab=ba=ac=ca=bc=cb=d$, all other products are 0. It is obviously a semigroup (the product of any three elements is $0$) and satisfies your conditions (1),(2),(3). Suppose it is inside a monogenic semigroup $M=\langle x\rangle$. Then $a=x^p, b=x^q, c=x^r$ for some $p<q<r$. Therefore $d$ belongs to the "ring" (see the above comment by Gerhard Paseman). Since the "ring" is a subgroup and contains 0, $d0=0d=0=0*0$, we have that $d=0$, a contradiction.

Q1=Q3. I do not think anybody worked on that. I know that the lattice of subsemigroups of a monogenic semigroup can be quite complicated (Repnitsky).

• Thank you for the counterexample. But at least some conjecture concerning Q1? Jan 22 '18 at 0:32
• The Rees factor over the minimal ideal should be a subsemigroup of a monogenic nil-semigroup which probably could be described in some way. So I would first decide if adding that condition is enough (maybe not).
– user6976
Jan 22 '18 at 0:40
• You can read Grillet's book "Commutative Semigroups".
– user6976
Jan 22 '18 at 0:56
• Thank you for the suggestion. I will try to find that book of Grillet and read. At the moment I added two new cancellativity conditions (4),(5) and modified the Question 2. Do you have some (simple) counterexample to Question 2'? Jan 22 '18 at 7:11
• The question is unmotivated. But take the monogenic semigroup $S_n=\langle x\mid x^n=x^{n+1}\rangle$. It is a nil-semigroup. Take the 0-direct product $T_n$ of $S_n$ with itself (the Rees factor of the direct product over the ideal consisting of pairs where one coordinate is 0). $T_n$ satisfies all your conditions. I suspect that if $n=100$ then $T_n$ is not inside a monogenic semigroup.
– user6976
Jan 22 '18 at 13:17