Cayley automaton semigroups

Cayley automaton semigroups

Victor Maltcev
Abstract.

In this paper we characterize when a Cayley automaton semigroup is a group, is trivial, is finite, is free, is a left zero semigroup, or is a right zero semigroup.

Mathematical Institute, University of St Andrews

St Andrews, Fife KY16 9SS, Scotland, victor.maltcev@gmail.com

1. Introduction & Main Results

In a bright work [8] Silva & Steinberg introduce the notion of a Cayley automaton of a semigroup: having a finite semigroup , let be the automaton with state set and alphabet set , obtained from the Cayley graph of by letting the output symbol on the arc leading from and labeled by to be :

Every state from can be viewed as a transformation on the set of all infinite sequences . The semigroup generated by all such transformations, associated to the states of , is obviously the automaton semigroup generated by the automaton (in the sense of [2] and [7]). Silva & Steinberg prove the following

Theorem 1.1.

Let be a finite non-trivial group. Then is a free semigroup of rank .

Under slightly other prospective, Cayley automaton semigroups, derived from monoids, appeared in a work by Mintz [6]. In particular, he proves that if is a finite -trivial monoid, then is a finite -trivial semigroup.

The aim of this paper is the following theorems and propositions:

Theorem 1.2.

For a finite semigroup , the following statements are equivalent:

  1. is a group.

  2. is trivial.

  3. is an inflation of a right zero semigroup by null semigroups.

We prove Theorem 1.2 in Section 3.

Theorem 1.3.

Let be a finite semigroup. Then is finite if and only if is -trivial.

The proof of sufficiency of Theorem 1.3 in the case when is a monoid was proved in [6]. The analogous characterization for the so-called dual Cayley automaton semigroups can be found in [5]. The proof of Theorem 1.3 is contained in Section 4.

Making use of Theorem 1.3, in Sections 5 and 6 we will prove the following three propositions:

Proposition 1.4.

Let be a finite semigroup. Then is a free semigroup if and only if the minimal ideal of consists of a single -class, in which every -class is not a singleton, and there exists such that for all .

Proposition 1.5.

Let be a finite semigroup. Then is a right zero semigroup if and only if for all .

Proposition 1.6.

Let be a finite semigroup. Then is a left zero semigroup if and only if is the minimal ideal of and if this ideal forms a right zero semigroup.

In the final Section 7 we will discuss our main results and their corollaries. Before we start proving our statements in the next section we give all necessary notation and lemmas needed for the proofs.

2. Auxiliary Lemmas

Let be a finite semigroup. In order to avoid confusion, we will denote the states in by an overline: is a symbol, and is a state. The sequences from can be vied as paths in the infinite -rooted tree. So, following [7], we will think about in terms of the wreath recursion:

where , defined by , corresponds to the action of on the first level of the -tree. Notice that for all . Hence

Iterating this formula one obtains that when automaton is in state and reads a symbol , it moves to the state

The transformation on the set , correspondent to the action of on the first level of the -tree is obviously .

Lemma 2.1.

Let be a finite semigroup. Then for all , in if and only if .

Proof.

Let . Then if and only if and for all . Recursing this, we obtain that if and only if and for all . It remains to notice that implies for all . ∎

Now we calculate Cayley automaton semigroups of special type semigroups:

Lemma 2.2.

Let be a finite left zero semigroup. Then is a right zero semigroup with elements.

Proof.

Suppose is in state and reads symbol . Then, by the definition of , it outputs and moves to the same state . Thus for all . Hence for any and :

and so . It remains to note that, by Lemma 2.1, if , then . ∎

Lemma 2.3.

Let be a finite semigroup and let be a finite right zero semigroup. Then .

Proof.

Let and . Then it follows from Lemma 2.1 that in . Hence coincides with for any fixed . It is now easy to check that gives rise to an isomorphism from onto . ∎

Corollary 2.4.

Let be a finite right zero semigroup. Then is trivial.

The proof of the last lemma of this section is easy, it can be found in [5].

Lemma 2.5.

Let be a finite semigroup and let . If all , and belong to the same -class of then .

3. Proof of Theorem 1.2

We recall that a semigroup is an inflation of a right zero semigroup by null semigroups if and can be partitioned into disjoint subsets (for each ) such that and for all .

Proof of Theorem 1.2.

The proof follows via the chain .

is clear.

. Let be an inflation of a right zero semigroup . Then for all , we have . Hence and so, by Lemma 2.1, for all . It remains to prove that for any fixed , the element is an idempotent. We have and . Thus it suffices to prove that . This holds since for all .

. We will prove by induction on that if is a group, then is an inflation of a right zero semigroup by null semigroups. The base case is trivial. So suppose the implication holds for all semigroups of cardinality and that is a group.

Let be the transformation semigroup on . The subsemigroup in is a homomorphic image of the group and so is a group. This implies that the images and kernels of the mappings must coincide. These conditions can be translated as

  • for all and

  • if and only if , for all .

Notice that the condition that for all is equivalent to . The rest of the proof depends on whether or not.

Case 1: .

Then, by the observations above, for all . So each , acting via left-multiplication, permutes . Then for any , some power of is a left identity for . Then for all , the condition implies . Hence is left cancellative. The condition that for all implies that is right simple.

Therefore is a right group and so for some group and a right zero semigroup , see [1]. By Lemma 2.3, we have . Since the Cayley automaton semigroup over every non-trivial group is a free semigroup, must trivial. Then is a right zero semigroup and so (3) holds.

Case 2: .

The condition of the case means that contains indecomposable elements.

Recall that the kernels of all the mappings coincide. Partition into these kernel classes and notice that, for every , the equality holds if and only if and come from the same class. Furthermore, since the mappings generate a subgroup in , it follows that every kernel class contains an image point, which must of course be an element of (the image of every is ).

The remainder of the proof we will work out in two subcases:

Subcase a: for all there exists with .

Consider an arbitrary and find the corresponding . That means that there exists an element where . Since , there exists some with . Obviously then . Hence . That is, for all . This is equivalent to that for all and . The kernel class that contains is not a singleton, for it must contain an element from and itself is indecomposable. Take an arbitrary . Then for any , a contradiction.

Subcase b: There exists such that for all there holds .

Fix such an . Obviously is a subsemigroup of .

We will show now that is a homomorphic image of and thus that is a group. Let be the minimal ideal in . Then is simple and so, being finite, is completely simple. Hence is a Rees matrix semigroup. Take now an arbitrary . Since is a Rees matrix semigroup, there exists such that . Then and

for all . Thus in . Since is a group, we derive now that . Since , the element must lie in . Hence in . Restricting the action of the states from to yields the automaton . Therefore is a homomorphic image of , and as so is a group.

So, by the induction hypothesis, is an inflation of a right zero semigroup by null semigroups. Suppose without loss of generality that . Then are the correspondent null semigroups from . For each , let be the right zero in . Then . In particular, since is indecomposable. Take now and . Recall that as soon as and are from the same kernel class.

  1. If and , then .

  2. If and , then .

  3. Let and . Let for some . Then . Since , it follows that and so . Hence .

  4. If , then .

Thus is an inflation of a right zero semigroup and the induction step is established. ∎

4. Proof of Theorem 1.3

Proof of Theorem 1.3.

. Suppose that is finite. Take any -class in . With the seek of a contradiction, suppose that . Let . Then for every , by [1, Lemma 2.21], the mapping , , is a bijection of onto itself. The set of all these bijections forms the so-called dual Schützenberger group of . By [1, Theorem 2.22] we have . Let be the dual group of : that is has the same underlying set as but in we have for all ..

Take arbitrary . Then for all :

We also have .

Take now . Then

and .

Take and consider the restriction of to . From the very definition of , it now follows that the mapping gives rise to a well-defined homomorphism from onto . It means that has a free semigroup on points, as a homomorphic image, and so is infinite. Thus is infinite, a contradiction.

. We will prove by induction on that if is -trivial then is finite. The base case is obvious. Assume that we have proved this for all -trivial semigroups of size . Take now any -trivial semigroup with . Let be the set of all maximal -classes from and let be the complement of all these -classes in . If is empty then consists only of one -class and then is simple. Since it is -trivial we have that is a rectangular band, where is some left zero semigroup and is some right zero semigroup. Combining Lemmas 2.2 and 2.3, we have that is a right zero semigroup on points and so is finite. So in the remainder of the proof we may assume that . Notice that is an ideal in .

Step 1: is finite.

It suffices to prove that there are finitely many products for any fixed . We have that and are distinct if and only if the restrictions of and on coincide. Notice that . Obviously and act on in the same way as the correspondent products from do. Now the claim of Step 1 follows from the induction hypothesis.

Step 2: is finite.

Take a typical element . Then for all , we have

Having that is an ideal in , we deduce that .

Step 3: is finite.

We need to prove that there are only finitely many distinct products , where all lie in . Take such . We have that

(4.1)

Obviously, to prove Step 3, it suffices to establish that there only finitely many such expressions (4.1). It follows immediately from Step 2, that there are finitely many such expressions with .

Denote by the restriction of to . Notice that if , for some , then . Indeed, if , then and . But since and , we now have that (all three elements , and lie in maximal -classes). In particular, if then and so .

From the above it follows that it suffices to prove that for a fixed with (which is the same as ), there exist only finitely many expressions . Let be the maximum number such that

(4.2)

Since , by Step 2, it suffices to prove that there are finitely many products with (4.2). Consider such one. We have for all . In particular then we have are all from the same -class (in ). This implies . Then, by Lemma 2.5, . Since is a right congruence, we then have that .

Recall that it is enough to prove that there are only finitely many products with (4.2).

Thus it suffices to prove that there exist only finitely many different products such that all come from the same -class inside a -class from . Obviously it suffices to prove this for a fixed -class in .

So, we need to prove that the set

is finite and then Step 3 is established.

Again we consider for the elements , for . It suffices to prove that there are finitely many such -s.

If , then . So we may assume that is such that . Find the maximum such that for all . Since , as before, it suffices to prove that there are finitely many products .

We have . Now, for all , and so by Lemma 2.5 we have that . By Clifford-Miller Theorem we obtain that contains an idempotent. Since is -trivial and we have that is an idempotent and . It follows that forms a left zero subsemigroup in .

By the token as above, we have that it suffices to show that there are finitely many products of the type with and all of them lying in the same -class and forming a left zero semigroup.

We will prove by induction on that the subset , consisting of those products such that there are precisely idempotents among , is finite. This will then prove Step 3.

Base of induction.

. To prove the base case, it is enough to show that is of finite order for all idempotents .

Let be an arbitrary idempotent from . We have . If , then . Let . Consider . If is not an idempotent then .

Let be the set of all idempotents -equivalent to . Let be the set of all such that is an idempotent in . Notice that for , if and only if is an idempotent.

Since the set is finite, we have that there are only finitely many elements with . On the other hand, and , for all . Thus, by wreath recursions for all elements , , we have that is of finite order for every idempotent . Hence the base case is established.

Induction step.

We will do step . Take an arbitrary product . There are precisely different -classes among . Obviously, it would suffice to prove the step if come from fixed -classes (and for every of these -classes there is at least one representative among ). In particular, in the remainder of the proof all the products from will involve these fixed -classes.

With every such product we associate the correspondent -class . We have for all . Notice that if then and . In addition, for every -class in there exists such that lies in this -class.

Now we split into three disjoint sets:

  • The set of all such that .

  • The set of all such that and there are at most idempotents among .

  • The set of all such that and there are precisely idempotents among .

Notice that if and only if is an idempotent. Thus each of depends only on .

If then .

Let . Take . We have . Let be maximum such that for all . Recall that . So, as before, we have that are idempotents. There are at most such idempotents and so split in at most -classes. We have and so there are at most idempotents among . Thus for all , we have . This implies that there are only finitely many for every and .

Let, finally, . Since lie in exactly -classes, we have that all of are idempotents. In particular and involve the same (fixed) -classes as . We also mention that if we fix some -class in such that for some , then the set of all , where , and , exhausts the whole of .

Let be the total (finite) number of elements in . Let also and be the number of -classes in .

Take now any product . We will prove that equals some element from of length less than . That will complete the induction step and the whole proof of Step 3.

Let be all -classes, which in intersection with the fixed -classes give idempotents. Assume without loss of generality that . Note that for some .

By Pigeonhole Principle, we have that there exist such that and for all and . Notice now that . There exists such that . Analogously, by Pigeonhole Principle, we have that there is a subsequence such that and for all and . Proceeding in this way in total at most times we arrive at two indices , such that

and

for all , .

Finally, we remark that if and , then . Thus, from wreath recursions for elements and for all , it now follows that and so