Master Thesis
Diagonal Ranks of
Semigroups
Author: Ilia Barkov
Supervisor: prof. Andrei Khrennikov Co-supervisor: prof. Igor Kozhukhov (National Research University of Electronic Technology)
Examiner: Dr. Per-Anders Svensson Date: 2013-11-12
Course Code: 5MA11E Subject:mathematics Level: advanced
Abstract
We introduce the notion of diagonal ranks of semigroups, which are numerical characteristics of semigroups. Some base properties of diagonal ranks are ob-tained. A new criterion for a monoid being a group is obtained using diagonal ranks. For some semigroup classes we investigate whether their diagonal acts are finitely generated or not. For the semigroups of full transformations, partial transformations and binary relations we find the general form of the generating pairs.
Contents
1 Introduction 2
2 Thesis outline 4
3 Preliminary definitions 5
4 Background 7
5 Diagonal ranks of semigroups 11
6 Generating pairs of some diagonal acts 19
7 Some positive and some negative results on the finite generation
of diagonal acts 24 8 Conclusion 28 A Reference information 30 A.1 Algebras . . . 30 A.2 Modules . . . 30 A.3 Epigroups . . . 31
A.4 (0-)simple and completely (0-)simple semigroups . . . 32
A.5 Lattices . . . 34
A.6 P-adic numbers . . . 34
Chapter 1
Introduction
One of the aims of abstract algebra is to derive general properties, which can be applied to a variety of objects. So the natural notion of a vector space was first generalised to modules over rings and then further to acts over semigroups. But it has turned out that these generalisation possesses unique properties and themselves can be used as applications and tools for studying other algebraic objects. This can be said about diagonal acts over semigroups, which a special cases of general acts over semigroups. By restricting ourselves to this case we become able to find out more about the structure of underlying semigroups. Diagonal acts are the case.
First the notion of a module over ring was derived as a generalisation of a vector space. Now scalars do not need to form a field, but only a ring. Then it was generalised further to acts over semigroups. Now we get a semigroup instead of a ring and any set instead of a group. Acts over semigroups are useful not only as generalisation, but also as means to model automata. So acts are another bridge between algebra and discrete mathematics.
The notion of a diagonal act first appeared out of a question about finite generation of the corresponding act. Back then it was defined implicitly. First to define it formally and make use of it was Robertson in 2002, who used it to study finite generation of wreath products. And since wreath products are used in Krohn-Rhodes decompostion, it became clear that diagonal acts are worth to be studied as independent objects.
Such work was carried out first by Gallagher in 2005, then by Gallagher and Ruˇskuc in 2006. Gallagher studied different classes of semigroups in the context of finite generation of corresponding diagonal acts. He found out that many important classes of semigroups, for example, commutative and idempotent do not have a finitely generated act. It was proved that an infinite group has a finitely generated bi-act if and only if it has finitely many conjugacy classes, but so far there is no example of a group satisfying this condition. From the work of Gallagher it is clear that cyclic diagonal acts are rather rare. Slightly more semigroups have finitely generated acts.
2006. It was shown that for any infinite set the semigroups of transformations, partial transformations, binary relations have cyclic diagonal acts, while the semigroups of bijections do not has even a finitely generated act. Then it was shown that the semigroups of partial injections does not have finitely generated left and right diagonal acts, but its bidiagonal act is cyclic. This example shows that bidiagonal acts are in some way independent from left and right diagonal acts.
The Gallagher-Ruˇskuc line of work was continued by Apraksina in 2011. She studied the semigroups of isotone transformations and the semigroup of continuous maps. Conditions on the set having a finite generated diagonal act of isotone transformations were derived. It was also shown that the semigroup of continuous maps of a segment of the real line has cyclic right diagonal act, but its left diagonal act is not finitely generated. As it was with the semigroups of partial injections, this statement shows independence of different diagonal acts. Now it can be said that all three diagonal act are independent.
The purpose of this work is to continue a study of diagonal acts over arbi-trary semigroups. We introduce the notion of (left, right, bi-) diagonal rank of a semigroup and n-order diagonal ranks. For semigroups with a finitely gen-erated right diagonal act the upper bound of an n-order right diagonal rank is obtained. We show that if a semigroup has a finitely generated right (left) act then every irreducible generating set of this act is minimal. We prove that for some semigroups the right diagonal rank of a product is equal to a product of ranks and show that the right diagonal rank of a monoid is equal to its cardi-nality. Finally, we describe all generating sets of diagonal acts over some classic semigroups. Some of the present results were published by the authors in [3].
Chapter 2
Thesis outline
All necessary preliminary definitions are given in 3.
In Section 4 we give detailed background on diagonal acts.
In Section 5 we introduce the notion of diagonal act, which is the main subject of study in this thesis.
In Section 6 we find generating pairs for some diagonal acts.
In Section 7 we diagonal study ranks of various semigroup classes and give generalisations for some already known results.
Since this thesis is devoted not to algebraic objects, but rather to the concrete property of semigroups, we use different kinds of semigroups to study and to illustrate obtained results. All necessary definitions and examples are given in Appendix A. We will make references to that appendix when needed.
In Appendix B we highlight an important connection between acts over semigroups and automata. There we also introduce a possible application of diagonal acts.
Chapter 3
Preliminary definitions
Definition 3.1. The right act over a semigroup S is a set X with a map X × S → X, (x, s) 7→ xs satisfying (xs)s0 = x(ss0) for all x ∈ X, s, s0∈ S (see [17]). The left S-act Y over the semigroup S is defined analogously: S ×Y → Y , (s, y) 7→ sy, s(s0y) = (ss0)y for all s, s0∈ S, y ∈ Y .
Definition 3.2. Let S, T be semigroups. The set Z is called an (S, T )-act (bi-act over S and T), if it is a left S-(bi-act and a right T -(bi-act at the same time and (sz)t = s(zt) for all z ∈ Z, s ∈ S, t ∈ T. The right S-act X, left S-act Y and (S, T )-bi-act Z may be denoted by XS,SY , andSZT.
Definition 3.3. Let S be a semigroup. The Cartesian product S ×S can be con-sidered as an (S, S)-bi-act with operations (a, b)s = (as, bs), s(a, b) = (sa, sb) (s, a, b ∈ S). Then we call S(S × S), (S × S)S and S(S, S)S the left
diago-nal act, the right diagodiago-nal act and the diagodiago-nal bi-act over the semigroup S. The left, right act and the bi-actS(Sn), (Sn)S,S(Sn)S of order n are defined
analogously where Sn= S × . . . × S
| {z }
n
and s(a1, a2, . . . , an) = (sa1, sa2, . . . , san),
(a1, a2, . . . , an)s = (a1s, a2s, . . . , ans) for all a1, a2, . . . , an, s ∈ S.
It is well known that one can add an identity element to any semigroup. Indeed, if S is a semigroup, then let
S1= (
S if S is a monoid,
S ∪ {1} otherwise
Definition 3.4. LetSA be a left S-act. We say that subset G ⊆ A is a
generat-ing set of this act if S1G = A. Generating sets of the left act B
S and the bi-act SCT are defined analogously. If A can be chosen finite then the acts is said to be
finitely generated. If A can be chosen to be a singleton then X is called cyclic. The following example shows how to construct a generating set in a simple case. Consider an additive group Z3. Elements of Z3× Z3 can be partitioned
class iff there is s ∈ Z3 such that (x, y)s = (u, v). For illustration see Figure
3.1. From this figure it is clear that any set of representatives can be chosen as
Figure 3.1: Illustration of a generating set
Chapter 4
Background
The first implicit mention of diagonal acts over semigroups was in [5], where the following problem was posed. Suppose S is a monoid containing elements a and b such that every element of S × S is of the form (au, bu) for some u in S (i.e., S × S, considered as a right S-act is cyclic).
• Show that S must be a singleton if S is any one of the following: finite, commutative, idempotent, or inverse.
• Show that S need not be a singleton in general. The solution can be found in [10].
Further diagonal acts were studied by a number of authors. We present a brief survey of the these works.
The notion was introduced by Robertson ([21]) and Thomson ([24]). They used diagonal acts to study wreath products.
Let S, T be two semigroups. The unrestricted wreath product (see, for ex-ample, [21]) S o T is given as (ST × T, ◦), where ◦ is defined by (f, t) ◦ (g, u) =
(ftg, tu), wheretg ∈ ST is defined by
(x)tg = (xt)g.
Theorem 4.1. ([21], Theorem 4.1) Let S be an infinite semigroup whose diag-onal act is finitely generated, and let T be a finite non-trivial semigroup. Then S o T is finitely generated iff the following conditions are satisfied:
(i) S2= S and T2= T (here S2= {ab | a, b ∈ S});
(ii) S is finitely generated.
Theorem 4.2. ([21], Theorem 4.2) Let S be an infinite semigroup whose di-agonal act is infinitely generated, and let T be a finite non-trivial semigroup. Then S o T is finitely generated iff the following conditions are satisfied:
Property of semigroup S Nontrivial S, cyclic right/left act? Infinite S, f.g. right/left act? Nontrivial S, cyclic bi-act? Infinite S, f.g. bi-act?
Finite No N/A No N/A
Commutatitve No No No No
Inverse No No Yes Yes
Completely reg-ular No No No No Idempotent No No No No Completely sim-ple No No No Yes Completely zero-simple No No No Yes Cancellative No No No Yes
Left cancellative No No ??? Yes
Right cancella-tive No No ??? Yes Bruck-Reilly ex-tension No No No No Locally finite No No ??? ???
Table 4.1: A summary of results from [12]
(ii) S is finitely generated;
(iii) every element of T is contained in the principal right ideal generated by a right identity.
Some properties of diagonal acts over arbitrary semigroups were found in [12]. The main findings are summarized in Table 4.1.
For an arbitrary set X let T (X) be the set of all transformations α : X → X, x 7→ xα with multiplication via x(αβ) = (xα)β. For α ∈ T (X) denote im α = Xα (the image of α). Obviously T (X) is a semigroup. Denote by P (X) the
semigroup of partial transformations X → X, i.e. maps α : X1 → X, where
X1⊆ X. The set X1is called the domain of α and denoted dom α. Multiplication
in P (X) is defined as follows:
x(αβ) = (
(xα)β if x ∈ dom α and xα ∈ dom β,
not defined otherwise.
By B(X) we denote the semigroup of all binary relations on the set X, i.e. B(X) = {σ ⊆ X × X} with the multiplication
(x, y) ∈ στ ⇔ ∃t (x, t) ∈ σ ∧ (t, y) ∈ τ,
where x, y, t ∈ X and σ, τ ∈ B(X). It is well known that T (X) is a subsemigroup of P (X) and P (X) is a subsemigroup of B(X). Further, define a set of finite-to-one transformations (i.e. no infinite subset is mapped to one element) F (X), partially injective transformations I(X) and a set of all bijections S(X). The main results from [13] are summarised in Table 4.2.
A study of diagonal acts of transformations semigroups was continued by Apraksina in [1]. Let X be a partially ordered set. Then by O(X) we denote the semigroup of isotone (i.e. order-preserving) transformations and by P O(X) we denote the semigroup of partial isotone transformation.
Semigroup Right act Left act Bi-act
B(X) cyclic cyclic cyclic
P (X) cyclic cyclic cyclic
T (X) cyclic cyclic cyclic
F (X) cyclic infinite cyclic
I(X) infinite infinite cyclic
S(X) infinite infinite infinite
Table 4.2: A summary of results from [13]
Theorem 4.3. ([1], Theorem 3) Let X be a partially ordered set. Then the diagonal right act (P O(X) × P O(X))P O(X) is cyclic iff there are X1, X2⊆ X
such that
(i) X1∼= X2∼= X;
(ii) x 6≤ y for x ∈ X1, y ∈ X2 and y ∈ X1, x ∈ X2.
Theorem 4.4. ([1], Theorem 4) Let X be a partially ordered set. Then the diagonal right act (O(X) × O(X))O(X) is cyclic iff there are X1, X2⊆ X such
that
(i) X1∼= X2∼= X;
(ii) for any two isotone maps ϕ : X1→ X, ψ : X2→ X there are an extension
δ : X → X, i.e. such a map that δ X 1 = ϕ and δ X 2 = ψ.
Proof. First we prove necessity. Let (O(X) × O(X))O(X) be a cyclic diagonal
act and (α, β) be a generating pair. There is δ ∈ O(X) such that αδ = βδ = 1X.
So α and β are isomorphisms. Indeed, choose x, y ∈ X such that x ≤ y. Then x = xαδ = uδ ≤ vδ = uαδ = y,
where uδ = x, vδ = y.
Let X1= Xα, X2= Xβ. The condition (i) is shown in the above paragraph.
Let ϕ : X1 → X and ψ : X2 → X be isotone maps. Then αφ, βϕ ∈ O(X).
There is γ ∈ O(X) such that αγ = αϕ and βγ = βψ. If x ∈ X1 then
xγ = yαγ = yαϕ = xφ. So δ X 1 = φ. Analogously δ X 2
= ψ. So the condition (ii) holds.
Now we prove sufficiency. Suppose that conditions (i)-(ii) hold. There are
isomorphism α : X → X1 and β : X → X2 due to the condition (i). We prove
Let α1, β1∈ O(X). It is clear that ϕ = α−1α1∈ O(X) and maps X1→ X.
In view of (ii) there is an extension of φ to δ : X → X. Let x ∈ X. Then xα1= xαα−1α1= xαϕ = xαδ.
Hence α1 = αδ. Analogous argument for β shows that (α, β) is a generating
pair of (O(X) × O(X))O(X).
Let X be a topological space. Denote by C(X) the semigroup of all con-tinuous maps α : X → X with the multiplication x(αβ) = (xα)β for x ∈ X, α, β ∈ C(X).
Theorem 4.5. ([1], Theorem 5,6) If X is a segment of the real line then the right diagonal act (C(X) × C(X))C(X) is cyclic but the left diagonal act C(X)(C(X) × C(X)) has exactly 2ℵ0 of generators.
Proof. Let α0 : X → X1, β0 : X → X2 be continuous injections, where X1,
X2⊆ [0, 1] and X1∩ X2= ∅. We show that (α0, β0) is a generating pair of the
act (C(X) × C(X))C(X). Choose some (α, β) ∈ C(X). Define δ as follows:
xδ = (
xα−10 α, if x ∈ X1,
xβ0−1β, if x ∈ X2.
Immediate check gives us
xα0δ = xα0α−10 α = xα,
so α0δ = α. Analogously β0δ = α.
Now consider the left diagonal act C(X)(C(X) × C(X)). Suppose that
{(ai, bi) | i ∈ I} is a generating set of this act Then for every a ∈ (0, 1) we
can find δa and i ∈ I such that δaαi = 1X and δaβi = θa, where θa is a
constant map. Let Ia = im δa. From δaαi = 1X we know that δa is a
contin-uous bijection. Hence Ia is a segment and δa is a homeomorphism. But from
δaβi = θa we can see that Iaβi = {a}. Let Ia0 be the maximal open subset of
Ia. Note that Ia0 ⊆ aβ −1
i . In other words, for every a ∈ X there is i ∈ I such
that aβ−1i contains some interval. It means that there is a map ϕ : X → I, a 7→ i. Let a, b ∈ X be such that aϕ = bϕ. Then aβi−1∩ bβi−1 = ∅ and the set iϕ−1= {a | aβi−1contains some interval} is finite or countable. But Iϕ−1= X and therefore |I| = 2ℵ0.
The detailed overview of results on finite generation of diagonal acts can be found in the Ph.D thesis of Peter Gallagher ([11]). In [14] Andrew Gilmour studied flatness properties of diagonal acts.
Chapter 5
Diagonal ranks of
semigroups
Let S be a semigroup. The right diagonal rank of semigroup S (denoted rdr S) is the least cardinality of a generating set of the diagonal right act (S × S)S:
rdr S = min|A| : A ⊆ S × S ∧ AS1= S × S .
The left diagonal rank ldr S and bidiagonal rank bdr S are defined analogously: ldr S = min|A| : A ⊆ S × S ∧ S1A = S × S ,
bdr S = min|A| : A ⊆ S × S ∧ S1AS1= S × S .
It follows from the definition that the act (S × S)S is cyclic iff rdr S = 1. The same takes place for left diagonal acts and bidiagonal acts.
Now we define diagonal ranks of order n (or n-order diagonal ranks):
rdrnS = min |A| : A ⊆ S × . . . × S | {z } n ∧ AS1= S × . . . × S | {z } n , ldrnS = min |A| : A ⊆ S × . . . × S | {z } n ∧ S1A = S × . . . × S | {z } n , bdrnS = min |A| : A ⊆ S × . . . × S | {z } n ∧ S1AS1= S × . . . × S | {z } n .
Consider some simple properties of diagonal acts.
• rdr(S/ρ) ≤ rdr S for any semigroup S and congruence ρ.
Let {(ai, bi)}i be a generating set of (S × S)S. It is clear that {(¯ai, ¯bi)}i,
where ¯ai is a class representative, is a generating set for the diagonal act
• rdr(S ∪ {0}) ≤ 3 · rdr S.
Let S be a semigroup without zero and {(ai, bi)}i be a generating set of
(S × S)S. Then to get the pair (x, 0) we need the pair (ai, 0), where i is
such that (ai, bi)s = (x, y) for some y ∈ S, s ∈ S1. Analogously for the
pairs of the form (0, x).
• rdr(S ∪ {1}) = ∞ if S is infinite and not a monoid.
Let {(ai, bi)}i be a generating set of (S × S)S. It is clear that there is no
way to get the pair (x, 1) except by adding in into the generating set. • bdr S ≤ ldr S, rdr S. This follows from the definition of a diagonal rank. Diagonal ranks of semigroups are interesting semigroup characteristics. For example, if S is a finite semigroup and |S| = n then
• rdr S = n ⇔ ldr S = n if S is a group.
Each pair (a, b) ∈ G gives us n different pair. So rdr S ≤ n2
n = n.
• rdr S = n2⇔ if S is a left zero semigroup.
Any pair of the form (a, b) can be obtained only from the same pair (a, b). So there should be n2pair in a generating set.
• rdr S = n(n − 1) if S is a right zero semigroup.
Pairs of the form (s, s) are obtained from any pair: (x, y)s = (s, s). All other pairs should be in a generating set.
• rdr S = n2− 1 if S is a zero semigroup.
Only (0, 0) can be obtained without using an external identity element. So the remaining n2− 1 elements should be in a generating set.
Further we will show that if S is a monoid and rdr S = |S| then S is a group. It was proved in [12] (Lemma 7.9) that if rdr S < ∞, then rdrnS < ∞ for
any n ≥ 2. It is obvious that rdrnS < ∞ ⇒ rdr S < ∞ for any n ≥ 2. In
this master thesis we prove that rdrnS ≤ (rdr S)n for all n, if S is an infinite
semigroup and its right diagonal act (S × S)S is finitely generated.
From Table 4.2 and Theorem 4.4 one can see that the conditions rdr S < ∞ and ldr S < ∞ are independent. Moreover, Table 4.2 shows that there are semigroups with infinite right and left diagonal ranks, but with a cyclic bidiagonal rank.
Let G be a generating set of the diagonal act (S × S)S. Then GS1= S × S.
Under some quite natural conditions this equality can be turned into a more strong one: GS = S × S. The same holds for the act
S × . . . × S | {z } n S .
We say that a generating set G for the act S × . . . × S | {z } n S is irreducible if none of its proper subsets G0⊂ G is a generating set for this act. Obviously any
finite generating set can be reduced to irreducible one. We say that a generating set is minimal if it consists of the least possible number of elements.
Proposition 5.1. Let S be an infinite semigroup with finitely generated right diagonal act. Let G =n(a(i)1 , . . . , a(i)n )|i = 1, 2, . . . , n1
o
be an irreducible gener-ating set of the act (Sn)S. Then for any u1, . . . , un ∈ S there exist i ≤ n1 and
s ∈ S such that (a(i)1 , . . . , a(i)n )s = (u1, . . . , un).
Proof. The existence of such s ∈ S1that (a(i) 1 , . . . , a
(i)
n )s = (u1, . . . , un) for some
i follows from the definition of a generating set. We have to prove that there exists s ∈ S with the same property. It is sufficient to show that for every i ≤ n1
there exists such s ∈ S that
(a(i)1 , . . . , a(i)n )s = (a(i)1 , . . . , a(i)n ). As the act (Sn)
S is finitely generated and n ≥ 2, the act (S × S)S is finitely
generated too. Let H = {(ai, bi)|i = 1, 2, . . . , n2} be a generating set of the
act (S × S)S. Denote A = {a
(α)
β |α ≤ n1, β ≤ n}, B = {bα|α ≤ n2}. Since
the semigroup S is infinite, there is c ∈ S such that c 6∈ A ∪ AB. Since G is a generating set, for some p, q ∈ S1 and j, k ≤ n
1 we have
(a(j)1 , . . . , a(j)n−1, a(j)n )p = (a(i)1 , . . . , a(i)n−1, c) and
(a(k)1 , . . . , a(k)n−1, a(k)n )q = (a(i)1 , . . . , c, a(i)n ).
As c 6∈ A, then p, q ∈ S. We may find (am, bm) ∈ H and t ∈ S1 such that
(am, bm)t = (p, q). We have (a(j)1 am, . . . , a (j) n−1am, a(k)n bm)t = (a (i) 1 , . . . , a (i) n−1, a (i) n ).
If t = 1, then q = bm, and therefore c = a (k) n−1q = a
(k)
n−1bm ∈ AB, which is a
contradiction with the choice of c. Hence t 6= 1. If (a(j)1 am, . . . , a (j) n−1am, a(k)n bm) = (a (i) 1 , . . . , a (i) n−1, a (i) n ) then
(a(i)1 , . . . , a(i)n )t = (a(i)1 , . . . , a(i)n ),
and t 6= 1, which was required. Otherwise take l ≤ n1 and s ∈ S1 such that
(a(l)1 , . . . , a(l)n )s = (a(j)1 am, . . . , a (j) n−1am, a(k)n bm). Then we obtain (a(l)1 , . . . , a(l)n )st = (a (i) 1 , . . . , a (i) n ).
If l 6= i then (a(l)1 , . . . , a(l)n ) can be removed from the generating set G, which
Note that the above statement does not hold for a finite semigroup. An example is a right zero semigroup.
The next theorem shows the connection between n-order diagonal ranks and diagonal ranks of order 2.
Theorem 5.2. If S is an infinite semigroup and rdr S < ∞, then rdrnS ≤
(rdr S)n for odd n and rdrnS ≤ (rdr S) n−1
for even n. An analogous statement holds for ldr S and ldrnS.
Proof. We shall prove by induction on n. We may consider only right diagonal ranks. Let m = rdr S and {(g1, h1), . . . , (gm, hm)} be a generating set of the
act (S × S)S. Assume that this set is irreducible. Then by Proposition 5.1 for any triple (a1, a2, a3) ∈ S3 there are s, t ∈ S and i, j ≤ m such that (a1, a2) =
(gi, hi)s, (a3, a3) = (gj, hj)t. Next by applying Proposition 5.1 we fix k ≤ m
and u ∈ S such that (gk, hk)u = (s, t). So (gigk, higk, gjhk)u = (a1, a2, a3).
Therefore the act (S × S × S)S has a generating set consisting of at most m3 elements.
Suppose that the statement holds for all p < n. Let (a1, a2, . . . , an) be an
arbitrary element from Sn and k < n, m1 = rdrkS, m2= rdrn−kS. Let A be
an irreducible generating set consisting of m1elements of the act Sk, and B be
a generating set consisting of m2elements of the act Sn−k. Then (u1, . . . , uk)s =
(a1, . . . , ak) for some (u1, . . . , uk) ∈ A, s ∈ S and (uk+1, . . . , un)t =(ak+1, . . . , an)
for an appropriate (uk+1, . . . , un) ∈ Sn−k and t ∈ S. We choose i ≤ m
and v ∈ S such that (gi, hi)v = (s, t). So (u1gi, . . . , ukgi, ukhi, . . . , unhi)v =
(a1, . . . , ak, ak+1, . . . , an). Therefore rdrnS ≤ min1<k<p{rdrkS · rdrn−kS} ·
rdr S.
Let i = k, j = n − k. If n is even then we may think that i, j are even too. Then by the induction basis we have: rdriS ≤ mi−1 and rdrjS ≤ mj−1, and
hence rdrnS ≤ m · mi−1· mj−1= mn−1. If n is odd, then assume i is odd and
j is even. Then rdrnS ≤ m · mi· mj−1= mn.
Remark. This theorem does not give any information about tightness of this bound so it is still an open question.
Note that as a corollary for an infinite semigroup S we obtain already proven in [1] statement: if the act (S × S)S [S(S × S)] is cyclic then the act (Sn)S
[S(Sn)] is also cyclic for any n > 2 ([1], Theorem 1).
Further we remark that the diagonal right/left acts and the diagonal bi-acts are unary algebras, i.e. algebras whose all operations are unary. We recall a result of A.V.Kartashov.
Theorem 5.3. (Kartashov, [15], Th. 1). Let A be an algebra (see Appendix A.1 for definition and examples) with signature Σ = {ϕi|i ∈ I}, where all operations
ϕi are unary. If A is finitely generated, then every irreducible generating set of
A is minimal.
The authors had proved this theorem before they were pointed to this result, so we present an independent proof.
Proof. Operation symbols of the algebra A we write on the right side of the argument. For a, b ∈ A put a ≤ b ⇔ a = b or a = bϕi1. . . ϕik for some
ϕi1, . . . , ϕik ∈ Σ. The binary relation ≤ is reflexive and transitive by definition,
so it is a quaisorder. Let G = {g1, . . . , gn} be a minimal generating set of
the algebra A, and B = {b1, . . . , bm} be some irreducible generating set of this
algebra. We have to show that m = n.
Let bi ∈ B. Then bi ≤ gj for some gj ∈ G. Besides gj ≤ bk for some
bk ∈ B. Hence bi ≤ bk. Since the set B is irreducible, then i = k, and therefore
gj ≤ bi. The fact that gk ≤ bl ⇒ bl ≤ gk can be proved in a similar way. So
bi≤ gj ⇔ gj≤ bi. For all bi∈ B there is gj ∈ G such that bi≤ gj and gj ≤ bi.
This gj is unique, since if bi ≤ gj, gj0 then gj ≤ bi ≤ gj0, from which j = j0.
Therefore, we have a map f : B → G such that gj= f (bi) ⇔ bi≤ gj⇔ gj ≤ bi.
If bi, bi0 ≤ gj, then bi ≤ gj ≤ bi0, by the irreducibility of the set B we have
bi= bi0. Therefore the map f is an injection. If gj – arbitrary element from G,
then gj ≤ bifor some bi∈ B, and therefore f (bi) = gj. It shows that f : B → G
is a bijection and therefore m = n. This implies
Corollary 5.4. For any act (and any bi-act) over a semigroup (over the semi-groups) any irreducible generating set is minimal.
Note that for universal algebras whose operations are not all unary, the
above statement does not hold in general. An example is the group Z6 =
{0, 1, 2, 3, 4, 5}, where the set {2, 3} is irreducible, but not a minimal generating set.
Recall that an element a of a semigroup S is called left-invertible if Sa = S. Theorem 5.5. Let S, T be semigroups, each of which satisfies one of the fol-lowing conditions:
(i) infinite with finite diagonal rank, (ii) finite with a left-invertible element. Then rdr(S × T ) = rdr S · rdr T .
Proof. Let G be a generating set of a diagonal right act of a semigroup S. First we show that GS = S × S.
For an infinite semigroup it follows from Proposition 5.1.
Consider S to be finite and a be a left-invertible element, i.e. Sa = S. Suppose that GS 6= S × S. Choose a pair (ai, bi) ∈ G such that (ai, bi) 6∈
(ai, bi)S. There are u, v ∈ S such that (u, v)a = (ai, bi). The pair (u, v) is
not in G since G is minimal. So there are (aj, bj) ∈ G and s ∈ S such that
(aj, bj)s = (u, v) and therefore
(aj, bj)sa = (u, v)a = (ai, bi),
Now we prove the theorem statement. Let x1, x2be arbitrary elements of S
and y1, y2be arbitrary elements of T . We are interested in the generating set of
the diagonal act ((S × T ) × (S × T ))S×T. So we have to find such a1, a2 ∈ S,
b1, b2∈ T and (s, t) ∈ (S × T )1 that ((a1, b1), (a2, b2)) (s, t) = ((x1, y1), (x2, y2)) , or ( (a1, a2)s = (x1, x2), (b1, b2)t = (y1, y2).
We can take such pairs (a1, a2) and (b1, b2) from any minimal generating
sets of the diagonal acts of S and T respectively. Irreducibility of the resulting generating set is equivalent to irreducibility of the used generating sets of (S × S)S and (T × T )T. So in view of Theorem 5.3 the resulting generating set is
minimal and consists of rdr S · rdr T elements.
Remark. Note that this statement does not hold in general. Let R be a right zero semigroup. We know that rdr R = n2− n. The semigroup R × R is a right
zero semigroup too since for any a, b, c, d ∈ R we have (a, b) · (c, d) = (ac, bd) = (c, d). So rdr(R × R) = n4− n2, but (rdr R)2= n4+ n2− 2n3.
Remark. This theorem gives us a tool to construct infinite semigroups of arbi-trary right or left diagonal rank. Indeed, consider any semigroup S such that rdr S = 1 and a group G consisting of n elements. Then rdr S × G = n. Corollary 5.6. Let {Si}ni=1 be a family of semigroups and each Si is either
infinite with a finite right diagonal rank or finite with a left-invertible element. Then rdr n Y i=1 Si ! = n Y i=1 rdr Si.
In some cases Theorem 5.5 allows us to find the diagonal rank of a Rees ma-trix semigroup (for definitions of a Rees mama-trix semigroup and a completely sim-ple semigroup see A.4). Consider a Rees matrix semigroup S = M0(G; I; Λ; P ), where |Λ| = 1. Let (a)iλ, (b)jνbe some elements from S. We can write them as
(a)i, (b)j since |Λ| = 1. So by definition of the Rees matrix multiplication we
get
(i, a) · (j, b) = (i, apjb).
Moreover, consider the map ϕ : (i, a) 7→ (i, pia). It is an isomorphism with the
Rees matrix semigroup with the same G and Λ, but with P consisting only of 1’s. Check that it is an isomorphism:
and
(i, a)ϕ(j, b)ϕ = (i, pia)(j, pjb) = (i, piapjb).
So we can mulitply elements of S in the following way: (i, a)(j, b) = (i, ab).
This shows that S is isomorphic with L × G, where L is a left-zero semigroup such that |L| = |Λ|. As we mentioned above, rdr L = |L|2 and rdr G = |G|. So
rdr S = |L|2|G|.
Lemma 5.7. Let S be a semigroup and |S| = n. Then rdr S ≥ n.
Proof. Let |S| = n, rdr S = r and let G = {(a1, b1), . . . , (ar, br)} be an
ir-reducible generating set of the diagonal right act (S × S)S. By multiplying
these pairs by elements of S1 we can obtain no more than r(n + 1) pairs. But
r(n + 1) ≥ n2 since G is a generating set. So
r ≥ n 2 n + 1= n − 1 + 1 n + 1> n − 1. Therefore r ≥ n.
Theorem 5.8. Let M be a monoid of cardinality n and rdr M = n. Then M is a group.
Proof. Let G be a group of all invertible elements of M . Denote N = M \ G. All elements, which are left- or right-invertible elements are invertible since M is finite. Hence if N is not empty then it’s an ideal of M .
Let |G| = k. As the pair (1, 1) can be obtained only from pairs of the form (g, g), g ∈ G, the generating set of the diagonal act (M × M )M should contain
at least one such pair. Note, that without loss of generality we can assume that if one component of the generating pair is from G then the pair has the form (1, a) or (a, 1), a ∈ M .
Let u ∈ N . For the pair (1, u) we have the following: (g, t)x = (1, u),
where g ∈ G, t ∈ N . In view of the note above we can assume that g = 1. This implies x = 1. Hence all pairs of the form (1, u) and (u, 1), where u ∈ N , are present in the generating set.
As pairs from G × G are generated only by the pairs from G × G, then the irreducible generating set of (G × G)G consists of the pairs (1, g), where g ∈ G.
So the irreducible (and therefore minimal) generating set of the diagonal
right act (M × M )M has no more than 2(n − k) + k elements. As rdr M ≤ n
then we have
n ≥ 2(n − k) + k,
This theorem gives a criterion for a monoid being a group.
Note that this does not hold for semigroups without identity in general. For example, consider a left zero semigroup S, |S| = n. It’s right diagonal rank is equal to n2− n. For n = 2 we have 4 − 2 = 2, but obviously S isn’t a group.
Chapter 6
Generating pairs of some
diagonal acts
As we already mentioned, it was proved in [13] that if a set X is infinite, then the right and the left diagonal acts over semigroups T (X), P (X), B(X) are cyclic. Here we present the general forms of generating elements of these acts.
Recall that ∆ is the minimal equivalence relation on a set X, i.e. ∆ = {(x, x) | x ∈ X}. Also throughout this chapter by the symbol θa we will mean
a constant map, i.e. such a map from X to X that θax = a for every x ∈ X.
Theorem 6.1. Let X be an infinite set, T (X) be the transformation monoid of X. The pair (α, β) ∈ T (X) × T (X) is a generating pair of the act (T (X) × T (X))T (X) iff ker α = ker β = ∆ and im α ∩ im β = ∅.
Proof. The sufficiency was proved in [13]. Here we prove the necessity. Since (α, β) is a generating pair, we have
∀ξ, η ∃γ (αγ = ξ ∧ βγ = η).
Let ξ = η = 1X. Then αγ = βγ = 1X, and therefore α and β are injective, i.e.
ker α = ker β = ∆. Let X1= im α, X2= im β. We will show that X1∩ X2= ∅.
Let x ∈ X1∩ X2. Then x = uα = vβ for some u, v ∈ X. Choose a, b ∈ X
such that a 6= b. Then choose δ ∈ T (X) so that αδ = θa, βδ = θb. Hence
a = uθa = uαδ = vβδ = vθb = b, which is a contradiction with the choice of
elements a and b.
We also present a statement which gives an example of a semigroup without identity with a cyclic diagonal act.
Theorem 6.2. Let X be an infinite set, TX be the set of transformations of
this set and x0 be some element of X. Then the right diagonal act over the
Proof. It can be easily seen that the (α, β) from the previous theorem is a generating pair for this diagonal act.
Theorem 6.3. Let X be an infinite set, P (X) be the monoid of partial trans-formations of X. The pair (α, β) ∈ P (X) × P (X) is a generating pair of the
act (P (X) × P (X))P (X) iff dom α = dom β = X, ker α = ker β = ∆ and
im α ∩ im β = ∅.
Proof. Sufficiency was proved in [13]. Here we prove necessity. Let (α, β) · P (X) = P (X) × P (X). For some γ ∈ P (X) we have: αγ = βγ = 1X, therefore
dom α = dom β = X and ker α = ker β = ∆. Now we have to show that im α ∩ im β = ∅. It can be made as such as in the proof of Theorem 6.1.
Then we want to generalize the notions of dom and im on the binary rela-tions. If α ∈ B(X) then we put
dom α = {x | (∃y)(x, y) ∈ α}, im α = {x | (∃y)(y, x) ∈ α}
Theorem 6.4. Let X be an infinite set and B(X) be the monoid of all binary relations on X. A pair (α, β) ∈ B(X) × B(X) is a generating pair of the act (B(X) × B(X))B(X) iff α = α1∪ α2, β = β1∪ β2, where α1, α2, β1, β2∈ B(X)
and the following conditions hold:
(1) α1, β1 are injections with dom α1= dom β1= X
(2) im α1∩ im α2= ∅, im β1∩ im β2= ∅
(3) im α ∩ im β1= ∅, im α1∩ im β = ∅
Proof. Necessity. Let ϕ, ψ be injections such that im ϕ ∩ im ψ = ∅. Since (α, β) is a generating pair, αγ = ϕ, βγ = ψ for some γ ∈ B(X). Note that dom α = dom β = X since dom ϕ = dom ψ = X.
For every x ∈ X we select some elements yx, ux such that (x, yx) ∈ α,
(yx, ux) ∈ γ and elements zx, vx such that (x, zx) ∈ β, (zx, vx) ∈ γ. Let
Y = {yx | x ∈ X}, Z = {zx | x ∈ X}. We will show that Y ∩ Z = ∅.
Suppose the opposite. Then yx= zx0 for some x, x0∈ X. Since (x, yx) ∈ α and
(yx, ux) ∈ γ, (x, ux) ∈ αγ = ϕ. Analogously (x0, vx0) ∈ ψ. Next, (x, yx) ∈ α,
yx= zx0 and (zx0, vx0) ∈ γ imply (x, vx0) ∈ ϕ. Also ux= xϕ = vx0 since ϕ is a
map. Therefore ux= vx0 ∈ im ϕ ∩ im ψ, which is a contradiction with the choice
of ϕ, ψ.
Note that ϕ = {(x, ux) | x ∈ X}, ψ = {(x, vx) | x ∈ X} and therefore
ux 6= ux0 and vx 6= vx0 when x 6= x0. Let α1= {(x, yx) | x ∈ X}, α2 = α\α1,
β1= {(x, zx) | x ∈ X}, β2= β\β1.
Obviously, α1, β1 are injections with dom α1 = dom β1 = X. We show
that im α1∩ im α2 = ∅. If im α1∩ im α2 6= ∅ then there is t ∈ X such that
(t, yx) ∈ α and t 6= x. From (yx, ux) ∈ γ follows (t, ux) ∈ αγ = ϕ. Therefore
Now we prove that im α1∩ im β = ∅. Let im α1∩ im β 6= ∅. Then (t, yx) ∈ β
for some t ∈ X. So we have (x, ux) ∈ αγ = ϕ. Since (t, yx) ∈ β and (yx, ux) ∈ γ,
(t, ux) ∈ βγ = ψ. Therefore ux∈ im ϕ ∩ im ψ, which is a contradiction.
The equalities im β1∩im β2= ∅ and im α ∩ im β1= ∅ are proven analogously.
Sufficiency.
Let α, β ∈ B(X) be such that the conditions (1), (2) and (3) are satisfied. Then for any µ, ν ∈ B(X) we construct γ ∈ B(X) as follows:
γ = α−11 µ ∪ β1−1ν. Now we show that (µ, ν) = (α, β)γ. Indeed,
αγ = α(α−11 µ ∪ β1−1ν) = αα−11 µ ∪ αβ1−1µ = αα−11 µ ∪ ∅ = (α1∪ α2)α1−1µ = α1α−11 µ ∪ ∅ = ∆µ = µ.
Similarly βγ = ν
From [2] we also know the form of generating pairs for the left diagonal acts
P (X)(P (X) × P (X)) andT (X)(T (X) × T (X)).
Theorem 6.5. Let X be an infinite set, T (X) be the transformation monoid
of X. Then a pair (α, β) ∈ T (X) × T (X) is a generating pair of the act
T (X)(T (X) × T (X)) iff xα−1∩ yβ−16= ∅ for any x, y ∈ X.
Proof. Necessity. Let T (X) · (α, β) = T (X) × T (X). Take arbitrary x, y ∈ X. For some γ ∈ T (X) we have γ(α, β) = (θx, θy). If a is an element of X, then
aγα = aθx = x and aγβ = aθy = y and therefore aγ ∈ xα−1∩ yβ−1. Thus,
xα−1∩ yβ−16= ∅
Sufficiency. Let xα−1∩ yβ−1 6= ∅ for all x, y ∈ X. For every pair (x, y) fix
some ux,y ∈ xα−1∩ yβ−1. Let α1, β1 ∈ T (X). Construct γ ∈ T (X) as follows:
xγ = uxα1,xβ1 for x ∈ X. For any x ∈ X we have xγα = uxα1,xβ1α = xα1 and
analogously xγβ = uxα1,xβ1β = xβ1. Hence γα = α1, γβ = β1.
Similar arguments can be used to prove the following theorem from [2]. Theorem 6.6. Let X be an infinite set, P (X) be the partial transformation monoid of X. A pair (α, β) ∈ P (X) × P (X) is a generating pair of the act
P (X)(P (X) × P (X)) iff for any x, y ∈ X the following conditions hold:
(1) xα−1∩ yβ−16= ∅;
(2) xα−16⊆ dom β; (3) xβ−16⊆ dom α.
It is also worth to note the following result from [13].
Lemma 6.7. (Gallagher, Ruˇskuc, [13], Lemma 4.2) There exists c ∈ I(X) such that for all p ∈ I(X) there exists s ∈ I(X) such that p = scs−1
Theorem 6.8. (Gallagher, Ruˇskuc, [13], Theorem 4.3) The diagonal bi-act of I(X), the semigroup of partial injective transformations of an infinite set X is cyclic and generated by the pair (c, 1), where c is from the above lemma.
In [13] there was posed the following problem: is there an infinite associative ring R such that the right, left diagonal act or the diagonal bi-act over its mul-tiplicative semigroup (R, ·) is cyclic? As A.A. Tuganbaev informed the authors, the examples of such rings can be found in his work [25]. We say that a matrix is row-finite if each row has only finitely many nonzero entries. By Lemma 2 from [25] for any associative ring A with an identity the ring R of all row-finite infinite square matrices of the same size over A has the next property: every finitely generated left (right) R-module is cyclic and therefore (a, b)R = R × R and R(a0, b0) = R × R for some a, b, a0, b0 ∈ R. Therefore rdr R = ldr R = 1.
We remind, that a skew field is a ring, where every element has a multi-plicative inverse. Of course, while speaking of a skew field we assume that it is non-commutative, since otherwise we get a field.
If A is a skew field, then the ring of all square row-finite matrices of the same size, obviously, is isomorphic to the ring End V of all endomorphisms of some linear space V over A, where the multiplication is done from left to right: v(r1r2) = (vr1)r2 for v ∈ V, r1, r2 ∈ R. In Theorems 6.9, 6.10 all generating
elements of the appropriate diagonal acts are described.
Theorem 6.9. Let V be an infinitely-dimensional linear space over an arbitrary skew field A and R = End V be the ring of all linear operators on V , consid-ered as a semigroup with multiplication defined by v(r1r2) = (vr1)r2. The pair
(α, β) ∈ R × R is a generating element of the act (R × R)Riff ker α = ker β = 0
and im α ∩ im β = 0.
Proof. Necessity. Let (α, β) be a generating pair. Then ∀α1, β1∈ R ∃γ ∈ R (αγ = α1 ∧ βγ = β1).
Taking α1= β1 = 1V we get αγ = βγ = 1V for some γ, which implies that α
and β are injections, and therefore ker α = ker β = 0. If x ∈ im α ∩ im β then x = uα = vβ for some u, v ∈ V. Select such δ ∈ R that αδ = 0 and βδ = 1V.
Then we have v = vβδ = uαδ = 0 and therefore x = 0.
Sufficiency. Let ker α = ker β = 0 and H1∩H2, where H1= im α, H2= im β.
Obviously, maps α : V → H1and β : V → H2are isomorphisms and H1∩ H2=
0. Let H3 be a complement to the sum H1⊕ H2, i.e. V = H1⊕ H2⊕ H3. For
α1, β1∈ End V we define a map γ : H1∪ H2∪ H3 via
xγ = xα−1α1 if x ∈ H1, xβ−1β1 if x ∈ H2, 0 if x ∈ H3.
This map can be uniquely extended to a linear map γ0 : V → V since it is
The next theorem is dual to 6.9.
Theorem 6.10. Let V be an infinitely-dimensional linear space over an ar-bitrary skew field A and R = End V be the ring of all linear operators on V , considered as a semigroup with multiplication defined by v(r1r2) = (vr1)r2. The
pair (α, β) ∈ R × R is a generating pair of the actR(R × R) iff im α = im β = V
and V = ker α + ker β.
Proof. Necessity. Let (α, β) be a generating pair. Then γ · (α, β) = (1V, 1V) for
some γ ∈ R. Hence im α = im β = V .
Let H1= ker α, H2= ker β. We have to show that (x + H1) ∩ (y + H2) 6= ∅
for all x, y ∈ V . Let xα = u, yβ = v, Obviously, there are such a ∈ V and α1, β1 ∈ End V that aα1 = u, aβ1 = v. Since (α, β) is a generating pair, there
is such δ ∈ R that δα = α1, δβ = β1. We have (aδ)α = a(δα) = aα1= u = xα
and therefore aδ − x ∈ ker α, i.e. aδ ∈ x + H1. Analogously it may be proved
that aδ ∈ y + H2. So (x + H1) ∩ (y + H2) 6= ∅. This is true for all x, y ∈ V and
hence H1+ H2= V.
Sufficiency. Let im α = im β = V and H1+ H2 = V , where H1 = ker α,
H2 = ker β. Let H0 = H1∩ H2 and select such subspaces H10, H20 that H1 =
H0⊕ H10, H2 = H0 ⊕ H20. Then V = H1+ H2 = H0⊕ H10 ⊕ H20. The map
α : H0
2 → V is an isomorphism since H1 = ker α. Analogously we can prove
that β : H0
1 → V is an isomorphism. Let α0 : V → H20 and β0 : V → H10
be inverse isomorphisms. By definition of α0, β0 we get α0α = β0β = 1V and
α0β = β0α = 0. Let α1, β1∈ End V and γ = α1α0+ β1β0. Note that γ ∈ End V
since α1α0 ∈ End V and β1β0. So γα = (α1α0+ β1β0)α = α1· 1V + β1· 0 = α1
Chapter 7
Some positive and some
negative results on the
finite generation of diagonal
acts
The first class of semigroups to study is matrices over an infinitely ∧-distributive lattices. For definitions and examples see A.5.
Theorem 7.1. Let S = BI(L), where I is an infinite set, L be a complete
infinitely ∧-distributive lattice. Then rdr S = ldr S = 1.
Proof. It is sufficient to prove that rdr S = 1, since the dual equality can be proved analogously. Since I is infinite, one can find such subsets I1, I2⊆ I that
|I1| = |I2| = |I| and I1∩ I2= ∅. Let α, β be bijections α : I → I1, β : I → I2.
Consider α, β as (I × I)-matrices over L:
αij = ( 1 if iα = j, 0 otherwise, βij = ( 1 if iβ = j, 0 otherwise.
We will show that (α, β) is a generating pair of the act (S × S)S. Let γ, δ ∈ S. Consider a matrix ϕ defined as follows:
ϕij = γti if i = tα, δtj if i = tβ, 0 if i 6∈ I1∪ I2,
We show that αϕ = γ and βϕ = δ. Indeed, (αϕ)ij =W
k(αik∧ ϕkj) = αi,iα∧
ϕiα,j = 1 ∧ γij = γij and therefore αϕ = γ. The equality βϕ = δ can be proved
This theorem is a generalization of Theorem 2.1 from [13].
Theorem 7.2 ([13], Theorem 2.1). The diagonal right act of BX, the semigroup
of binary relations on an infinite set X, is cyclic.
Indeed, consider a matrix semigroup BI(L), where I is an infinite set and
L = {0, 1}. Then BI(L) coincides with BI, a semigroup of binary relations on
I. Now the matrix ||aij|| has 1 on (i, j) place iff (i, j) ∈ a, where a is a binary
relation.
The next theorem defines a class of infinite monoids whose right diagonal acts are not finitely generated.
Theorem 7.3. If S is an infinite monoid and the set {x ∈ S|Sx = S} is finite then bdr S = ∞.
Proof. Let bdr S = n < ∞ and {(ai, bi)} be a generating set ofS(S × S)S. Let
G = {x ∈ S|Sx = S}. By condition |G| < ∞.
Note that left or right invertible elements are invertible. Indeed, if it is not true then there are p, q such that pq = 1 and qp 6= 1. By Lemma 1.31 from [7] the subsemigroup hp, qi is bicyclic. Hence all qi, i = 0, 1, . . . are distinctive. As q is left invertible, we have Sq = S. Moreover, Sqi= S. So qi∈ G for all i and G is infinite, which is a contradiction.
Since G is finite and S is infinite, there is such element a ∈ S that a 6∈ Gb1G ∪ . . . ∪ GbnG. So (1, a) = s (ai, bi) t for some i ≤ n and s, t ∈ S. As
1 = sait, then 1 = tsai = aits and s, ai, t ∈ G. But a = sbit, which contradicts
to the choice of a.
An immediate application of the proven theorem is given in the following corollary.
Corollary 7.4. If X is an infinite set and RB(X) is the semigroup of all reflexive binary relations on X then rdr(RB(X)) = ldr(RB(X)) = ∞.
Proof. In view of the existence of involution σ → σ−1the semigroup S = RB(X) is antiisomorphic to itself. Therefore we may prove the statement only for the right diagonal rank. The identity of S is ∆ (the smallest equality relation on X). If a ∈ S and a 6= ∆ then a ⊃ ∆ and therefore for any s ∈ S we have sa ⊇ ∆a = a ⊃ ∆ hence ∆ 6∈ Sa. So the equality Sa = S hold only for a = ∆. The rest is done by applying the theorem 7.3.
It was proved in ([12], Theorem 6.1) that no infinite commutative semigroup does not have a finitely generated diagonal act. In the following theorem we generalize this result by substituting commutativity with a weaker condition of invariance. Another generalization is obtained in Theorem 7.7.
We say that the semigroup S is left invariant (right invariant ) if all of its left (right) ideals are ideals.
Theorem 7.5. If S is an infinite left [right] invariant semigroup then rdr S = ∞ [ldr S = ∞].
Proof. Suppose the opposite and let (a1, b1) , . . . , (an, bn) be a generating set
of the act (S × S)S. Since S is left invariant, all right invertible elements are left invertible (and therefore invertible). Let (ai1, bi1) , . . . , (aik, bik) be the pairs
from the generating set, for which bi is invertible. Since S is infinite, there is
t ∈ S\ai1b
−1
i1 , . . . , aikb
−1
ik . Select a pair (t, 1). There are element s ∈ S and
index i such that (ai, bi) s = (t, 1). So ais = t, bis = 1. The element bi is right
invertible, and therefore the inverse b−1i exists. Then s = b−1i and therefore t = aib−1i , which contradicts to the choice of t.
The last theorem is a generalization of Theorem 6.1 from [12].
Corollary 7.6. ([12], Theorem 6.1) If S is an infinite commutative monoid then rdr S = ldr S = ∞.
Indeed, if a semigroup S is left-invariant, then for all a, s ∈ S there is x ∈ S such that sa = xa. For a commutative semigroup x = s. Moreover, a commutative semigroup is both left and right-invariant. For example of such semigroup consider a semigroup S of integer p-adic numbers (see A.6), where m is not a prime. Obviously it is commutative, so rdr S = ldr S = ∞.
Using the Theorem 5.3 one can prove the following theorem ([2]).
Theorem 7.7. ([2], Theorem 4.8) Let S be an infinite semigroup which satisfies some non-trivial semigroup identity u = v (where u, v are elements of a free semigroup). Then the diagonal right act over S is not finitely generated.
Theorem 7.7 is a very useful result since free semigroups plays crucial role in both semigroup theory and automata theory. Free semigroups can be ob-tained using any set (alphabet) A. The resulting infinite semigroup A∗ can be reduced to a variety of other semigroups using semigroup identities. A commu-tative semigroup with an identity uv = vu, where u, v ∈ S and an idempotent semigroup with an identity ee = e, e ∈ S, just to name a few.
We also remind some results from [2], which answer a question about locally finite semigroups ([12]).
For information about epigroups consult A.3.
Theorem 7.8. ([2], Theorem 4.9) If S is an infinite epigroup, then the diagonal bi-actS(S × S)S is not cyclic.
Proof. Let S be an infinite epigroup such that bdr S = 1. Suppose that (a, b) is a generating element of this bi-act. Obviously, S1aS1= S1bS1= S. Denote
I(a) =x ∈ S | S1xS16= S . Note that S1rxS1⊆ S1xS1 6= S for any r ∈ S.
Analogously for xr. Hence I(a) is an ideal of S. The Rees factor semigroup P (a) = S/I(a) = S1aS1/I(a) is called the principal factor of S (see [7], §2.6).
We know that bdrP (a) = 1 since P (a) is a homomorphic image of the semigroup. By Lemma 2.39 from [7] the semigroup P (a) is one of the following: 0-simple, simple, or a zero semigroup. By Munn’s Theorem ([7], Th. 2.55) any (0-)simple epigroup is completely (0-)simple. By Corollaries 5.3, 5.5 from [12] the semigroup P (a) cannot be simple or 0-simple. Now let P (a) be a zero semigroup. An
infinite zero semigroup has an infinite bidiagonal rank since s(a, b)t = (0, 0) for every s, a, b, t from this semigroup and a finite non-trivial semigroup by Theorem 5.2 from [12] has a bidiagonal rank ≥ 2. So if P (a) is a zero semigroup then |P (a)| = 1, i.e. I(a) = ∅. But this cannot be true since a, b ∈ P (a) and a 6= b. This contradiction completes the proof.
We remind (A.3) that every locally finite semigroup is a periodic semigroup and every periodic semigroup is an epigroups. So we have
Corollary 7.9. ([2], Theorem 4.10) If S is an infinite periodic semigroup then bdr S > 1.
The next corollary gives a negative answer to the question posed in [12]: is there an infinite locally finite semigroup whose diagonal bi-act is cyclic? Corollary 7.10. ([2], Theorem 4.11) If S is an infinite locally finite semigroup then bdr S > 1.
Chapter 8
Conclusion
The purpose of this thesis was to introduce a diagonal rank – a numerical characteristic of semigroups, which can be used to study diagonal acts over semigroups. Diagonal ranks were computed for some semigroups, but as one can see from these computations, it takes considerable efforts. So it is better to derive diagonal ranks for classes of semigroups.
In Table 8.1 we present an extended version of Table 4.1. Recent results are marked with stars. Locally infinite semigroups are now substituted with epigroups. There are still lots of questions concerning diagonal ranks of these classes. We don’t know is there an epigroup with finitely generated diagonal bi-act or not. Also we don’t know anything about bi-bi-acts of right and left invariant semigroups. And, of course, semigroup classes are not limited to these listed in the table.
Another line of research is properties of diagonal ranks. In this master thesis we found some of these properties. It was shown that under some conditions the rank of a product of semigroups is equal to the product of ranks. From the rank of a finite monoid we can find whether it is a group or not. Also the upper boundary for diagonal ranks of order n was found.
There are open problems concerning diagonal ranks properties too. What semigroups can have the diagonal (right, left or bi) rank equal to the cardinality of these semigroups? Or, more specially, consider semigroups consisting of n
elements. Whether for every natural m between n and n2 exists a semigroup
such that rdr S = m?
Open problems discussed above give us two different lines of study. But both this lines should be studied simultaneously, not in isolation from one another since knowledge of diagonal ranks properties gives us means to study diagonal ranks of various classes of semigroups, and vice versa, the study of classes of semigroups can highlight some properties of diagonal ranks.
But there is another line of study, independent from these two. We can generalise the notion of a diagonal act to grupoids (grupoid is a set with a binary operation) and study this more general case. Of course, this object cannot model automata, but it may uncover some properties of binary operations, for which
Property of semigroup S Nontrivial S, cyclic right/left act? Infinite S, f.g. right/left act? Nontrivial S, cyclic bi-act? Infinite S, f.g. bi-act?
Finite No N/A No N/A
Commutatitve No No No No
Inverse No No Yes Yes
Completely reg-ular No No No No Idempotent No No No No Completely sim-ple No No No Yes Completely zero-simple No No No Yes Cancellative No No No Yes
Left cancellative No No Yes* Yes
Right
cancella-tive
No No Yes* Yes
Epigroup No No No* ???
Right invariant No* No* ??? ???
Left invariant No* No* ??? ???
Monoid, finite
{x | Sx = S}
No* No* No* No*
Table 8.1: Summary of results
Appendix A
Reference information
In this appendix we give definitions and examples of some semigroups, which are used in this master thesis.
A.1
Algebras
Definition A.1. The ordered pair (A, Σ), where A is any non-empty set and Σ is a non-empty set of operation symbols is called an algebra. The set A is called a universe and Σ is called a signature. With every symbol of the signature associated an operation of some arity. If all operations are unary, the algebra is called a unary algebra.
There are plenty of examples of algebras. Say, a field Z5 is an algebra with
universe {0, 1, 2, 3, 4} and signature ·, |. The notion of diagonal act provides an example of a unary algebra. Indeed, let S be a semigroup. Then the right diagonal act (S × S)S is a unary algebra with universe S × S and signature
Σ = {ϕa| a ∈ S}, where ϕa is an operation such that sϕa= sa for every s ∈ S.
For more information consult [6].
A.2
Modules
In linear spaces elements, called vectors, are multiplied by scalars, which in principal need not be numbers, but any field. But using a field is a very strict restriction, so the notion of a module is introduced.
Definition A.2. Let A be a ring. An additive Abelian (commutative) group M is called a left A-module if there is a mapping A × M → M , whose value on a pair (a, m) ∈ A × M , written am, satisfies the axioms
1. a(m1+ m2) = am1+ am2;
3. (a1a2)m = a1(a2m).
There are many examples of modules, we will name some of them. Let S be any set and M be a left R-module. Then MS the set of all functions S → M is a left R-module with operations defines as
(f + g)(s) = f (s) + g(s), (f g)(s) = f (g(s)), f, g ∈ MS, s ∈ S.
If R is a ring, then for any n ∈ N the set Rn is a left R-module with an
component-wise multiplication.
We can further generalise the theory of modules by allowing M to be any set and A to be a semigroup. This generalisation might be seen as too abstract, but it turned out to be useful in discrete mathematics. For example concerning automata see Appendix B. For more examples consult [18].
A.3
Epigroups
Definition A.3. A semigroup S is called an epigroup if for all a ∈ S exists n ∈ N such that an ∈ Gr S, where Gr S is a group part of the semigroup S (i.e.
the union of all subgroups).
The notion of an epigroup was introduced by Shevrin in [22, 23]. A quasiperi-odic semigroup and a pseudoinverse semigroup are another names for the same object. Now we show that the semigroup of matrices over Cn×nis an epigroup. Definition A.4. Let S be a semigroup, a ∈ S. If there is n ∈ N such that an∈ Gr S, then we say that n is the index of a. Otherwise we say that the index of a is 0.
From [20] we know that for every element of an epigroup there is a generalised inverse ¯a:
a¯a = ¯aa, a¯a2= a, an+1¯a = an,
where n is an index of a, and vice versa, if there is a generalised inverse for every element of a semigroup, then it is an epigroup. For matrices this properties coincide with the notion of a Drazin inverse.
Definition A.5. Let A ∈ Cn×n. A number k is called an index of A if it is the
least number such that rank(Ak+1) = rank(A). The Drazin inverse Ad is such matrix that
AkAdA = Ak, AdA = AAd, AdAAd= Ad. It was introduced in [8] and studied, for example, in [4].
For example we will compute a Drazin inverse in R2×2, the set of square
matrices with elements from R. Consider a matrix
A =0 −3
0 2
It does not have an inverse since det A = 0. According to [26], the Drazin inverse of A can be computed as follows. Assume that the Jordan decomposition of A is A = XC 0 0 N X−1. Then we have Ad = XC −1 0 0 0 X−1. For the given matrix A we have
X =− 3 2 1 1 0 , C = 2 , N = 0 . So Ad = X 1 2 0 0 0 X−1 =0 − 3 4 0 12 .
By simple multiplication of few matrices on can check that Ad satisfies all properties of a Drazin inverse.
Using a Jordan Normal Form the same way it was used in the above example it easy to prove the following statement.
Theorem A.6 ([26], Theorem 2.1). Let A be a matrix from Cn×n. Then the
Drazin inverse if A exists and it is unique. This means that Cn×n is an epigroup.
Another important for us example if locally finite and periodic semigroups. Definition A.7. A semigroup S is called a periodic semigroup if each mono-genic subsemigroup (generated by one element) of S is finite.
Definition A.8. A semigroup S is called a locally finite semigroup if each finitely generated subsemigroup of S is finite.
It is obvious that any locally finite semigroup is periodic since monogenic subsemigroups form a subset of finitely generated semigroups. Moreover, every periodic semigroups is an epigroup. This is clear from the fact that every finite semigroups has an idempotent and some power of every element from a periodic semigroup is equal to an idempotent.
A.4
(0-)simple and completely (0-)simple
semi-groups
Definition A.9. Let S be a semigroup. A subset I is called a right [left] ideal if IS ⊆ I [SI ⊆ I]. If I is both right and left ideal, then it is called an ideal (two-sided ideal).
Definition A.10. A semigroup S is called simple [left simple, right simple] if it contains no proper two-sided [left, right] ideal.
The theory of simple semigroups becomes trivial for semigroups with a zero element 0. So the notion of a 0-simple semigroup is used instead.
Definition A.11. A semigroup S is called simple [left simple, right 0-simple] if (i) S26= 0 and (ii) 0 is the only proper two-sided [left, right] ideal of
S.
Definition A.12. Let E be the set of idempotents of a semigroup S. If e, f ∈ E, we define e ≤ f to mean ef = f e = e. It can be proved that ≤ is a partial ordering of E. If S contains a zero element 0, then 0 ≤ e for every e ∈ E. An idempotent element f of S is called primitive if f 6= 0 and e ≤ f implies e = 0 of e = f .
Definition A.13. By a completely (0-)simple semigroup we mean a (0-)simple semigroup containing a primitive element.
For example, any finite (0-)simple semigroup is completely (0-)simple. Definition A.14. The Rees I × Λ matrix semigroup over the group with zero G0 with sandwich matrix P (denoted M0(G; I; Λ; P )) is a set {(a)
iλ| a ∈ G0, i ∈
I, λ ∈ Λ} with the following operation:
(a)iλ◦ (b)jµ= (apλjb)iµ (a, b ∈ G0; i, j ∈ I; λ, µ ∈ Λ).
The name “sandwich matrix” comes from the fact that multiplication in the Rees matrix semigroup can be written as A ◦ B = A · P · B, where · is a usual matrix multiplication and the whole construction resembles a sandwich.
Importance of Rees semigroups is clear from the Rees theorem ([7], Theorem 3.6)
Theorem A.15 (Rees). A semigroup is completely 0-simple if and only if it is isomorphic with a regular Rees matrix semigroup over a group with zero.
Consider a semigroup S = L × R, where L is a left zero semigroup and R is a right zero semigroup, i.e. if (x1, y1), (x2, y2) ∈ S, then
(x1, y1) · (x2, y2) = (x1, y2).
Let both L and R be two-element semigroups. Note that S is completely simple. Indeed, a left zero semigroup is simple and every it’s element is a primitive idempotent. The same for a right zero semigroup. So their direct product is completely simple too. To illustrate the theorem statement we will find the corresponding Rees matrix semigroup. Let G = {0, 1}, I = Λ = {1, 2} and
P =1 1
1 1
. So
(a)iλ◦ (b)jµ= (apλjb)iµ= (ab)iµ= (1)iµ.
A.5
Lattices
Definition A.16. A set L is called a lattice if it is partially ordered and for any a, b ∈ L there are a supremum a ∨ b and a infimum a ∧ b. From an algebraic point of view these two operations are
• commutative, • associative,
• suffice the law of absorption:
a ∨ (a ∧ b) = a, a ∧ (a ∨ b) = a.
One example of a lattice is R. Another useful example is a set of all subsets of some set X ordered by inclusion. It is a lattice since for any A, B ⊆ X we have
A ∨ B = A ∪ B, A ∧ B = A ∩ B.
Definition A.17. A partially ordered set L is called a complete lattice if every subset A of L has an infimum and supremum.
From the above examples the set of all subsets is a complete lattice, but R isn’t.
Definition A.18. Let L be a complete lattice, I be an arbitrary set. We say that the lattice L is infinitely ∧-distributive if W
i∈Iai ∧ b = Wi∈I(ai∧ b) for
all ai, b ∈ L.
Definition A.19. By BI(L) we denote the set of all matrices ||aij||i,j∈I, aij ∈ L
with the multiplication defined by the rule: if ||aij|| · ||bij|| = ||cij|| then cij =
W
k∈I(aik∧ bkj). Obviously BI(L) is a semigroup.
An important example of such semigroup is the set of binary relations. Let L = {0, 1}. Then a ∈ BI(L) is a matrix which has 1’s at (i, j)-th place if and
only if (i, j) ∈ a, where a is now considered as a binary relation.
A.6
P-adic numbers
Importance of p-adic numbers is clearly seen already from the fact that they were used by Andrew Wiles to prove Fermat’s Last Theorem ([9]). For more information about p-adic numbers consult [16].
Definition A.20. Let p be a prime number. P-adic norm | · |p is a map Q → R.
Let n = 2ν23ν3. . . pνp. . . be a factorisation of a given n ∈ N
0. Then |n|p= p−νp.
Moreover, |0|p = 0 by definition. If x = n/m, where n, m ∈ N then |x|p =
For example, |2/3|2= 1/2, |2/3|3= 3 and |5|3= 1.
Any p-adic number a has an expansion a =a−f pf + . . . + a−1 p + a0+ a1p + . . . + anp n+ . . . = a −f. . . a0. . . an. . . , where aj= 0, 1, . . . , p − 1.
Definition A.21. Let x be a p-adic number and x−f. . . a0. . . ax. . . be its
ex-pansion. If f = 0 then x is called a p-adic integer. All p-adic integers form a ring denoted Zp.
Appendix B
Acts over semigroups and
automata
To show how acts over semigroups and automata are connected, we first intro-duce some definitions from [19].
Definition B.1. A semiautomaton is a triple S = (Z, A, δ), consisting of two nonempty sets Z (the set of states) and A (the input alphabet), and a function δ : Z × A → Z, called the next-state function of S.
Definition B.2. An automaton is a quintuple A = (Z, A, B, δ, λ), where (Z, A, δ) is a semiautomaton, B is a nonempty set called the output alphabet and λ : Z × A → B is a function, called the output function.
It may be convenient to distinguish the notions of a semiautomaton and an automaton, but they denote one algebraic object. To see that consider an automaton A = (Z, A, B, δ, λ). We can construct an equivalent semiautomaton S = (Z0, A, δ0) as follows:
Z0= Z × B,
δ0: Z0× A → Z0, ((z, b), a) 7→ ((z, a)δ, (z, a)λ).
Considering this, we will not distinguish between an automaton and semiau-tomaton.
A real-life example of a semiautomaton is a traffic light. We may consider the next description of a traffic light. Default state of the light is ”Turned off”. After it had received it will change its state to ”Red”. There is a signal ”Change colour” and a signal ”Turn off”. So we have
z1: ”Turned off”, z2: ”Red”,
z3: ”Yellow”, z4: ”Green”,
a1: ”Turn on”, a2: ”Change colour”,
Now we can define δ: δ a1 a2 a3 z1 z2 z1 z1 z2 z2 z3 z1 z3 z3 z4 z1 z4 z4 z2 z1
Before proceeding to acts we note that for an automaton S = (Z, A, δ) we can define an automaton S0= (Z, A∗, δ0), where A∗is a free semigroup. Indeed, let
δ0: Z × A∗, (z, a1, a2, . . . , an) 7→ (. . . ((z, a1)δ, a2)δ . . . , an)δ.
Note that δ0 is nothing else but a definition of the act ZA. If S is a semigroup,
then the act SS, which coincides with the semigroup itself, defines an automaton,
where the set of states an the input alphabet coincide. We can think about a so-called “chat-bot”, a program simulating an interlocutor. This is an automaton, where the set of states and the input alphabet are free semigroups over the set of all English letters. If we consider a diagonal act (S ×S)S, then the corresponding
automaton is a pair of chat-bots fed with the same words and sentences. The aim of the diagonal ranks theory is to find such set of initial states (if it exists) that allows to “persuade” a set of chat-bots in any statement. This is, of course, not a rigorous definition of a problem since in real life we do not use the whole free semigroup over an English alphabet, but a meaningful subset of it. But this simple example and the example above show that acts over semigroups and even more special case, diagonal acts over semigroups, in principle can be used to model real-life objects. It may be even thought, that diagonal acts can have some applications is sociology (for some semigroups applications in sociology see [19], §32), but now it is too early to make any assumption. And the connection between diagonal acts and formal languages, which are required to model a natural language, is yet to be discovered.
Bibliography
[1] T.V. Apraksina. Diagonal acts over the semigroup of transformations. The Chebyshev Acad. J., 12(1):10–16, 2011. (In Russian).
[2] T.V. Apraksina, I.V. Barkov, and I.B. Kozhukhov. Diagonal ranks of semi-groups. Submitted to Semigroup Forum.
[3] I.V. Barkov and I.B. Kozhukhov. Properties of diagonal acts and bi-acts. Proc. Orel State Univ., 6(2):45–51, 12. (In Russian).
[4] A. Ben-Israel and T. Greville. Generalized Inverses: Theory and Applica-tions. Wiley-Interscience, 2nd edition, 2003.
[5] S. Bullman-Fleming and K. McDowell. Problem e3311. Amer. Math.
Monthly, 96:155, 1989.
[6] S. Burris and H. P. Sankappanavar. A Course in Universal Algebra.
Springer-Verlag, 1981.
[7] A. H. Clifford and G. B. Preston. The Algebraic Theory of Semigroups, volume 1. Mathematical Surveys of the American Mathematical Society, No. 7, Providence, R.I., 1967.
[8] M.P. Drazin. Pseudo inverses in associative rings and semigroups. Amer. Math. Mon., 65:506–514, 1958.
[9] F. Q. Gouvˆea. A marvelous proof. The American Mathematical Monthly, 101(3):203–222, 1994.
[10] J. Ferrer. Problem e3311 solution. Amer. Math. Monthly, 97:617, 1990. [11] P. Gallagher. On The Finite Generation And presentability Of Diagonal
Acts, finitary Power Semigroups And schutzenberger products. PhD thesis, University of St Andrews, 2005.
[12] P. Gallagher. On the finite and non-finite generation of diagonal acts. Comm. Algebra, 34:3123–3137, 2006.
[13] P. Gallagher and N. Ruˇskuc. Finite generation of diagonal acts of some infinite semigroups of transformations and relations. Bull. Austral. Math. Soc., 72:139–146, 2005.
[14] A.J. Gilmour. On diagonal acts of monoids. Master’s thesis, University of Waterloo, 2007.
[15] V.K. Kartashov. Independent systems of generators and the hopf property for unary algebras. Discr. Math and Appl., 6:625–630, 2008.
[16] A.Y. Khrennikov. Non-Archimedean Analysis - Quantum Paradoxes, Dy-namical Systems and Biological Models. Kluwer Academic Publishers, Dor-drecht, 1997.
[17] M. Kilp, U. Knauer, and A.V. Mikhalev. Monoids, acts and categories. W. de Gruyter, Berlin, New York, 2000.
[18] G. Lallement. Semigroups and combinatorial applications. Willey-Intersc. Publ., 1979.
[19] R. Lidl and G. Pilz. Applied abstract algebra. Springer-Verlag, New York Berlin Heidelberg, 2nd edition, 1998.
[20] V.D. Munn. Pseudo-inverses in semigroups. Proc. Camb. Phil. Soc., 7:247– 250, 1961.
[21] E. F. Robertson, N. Ruskuc, and M. R. Thomson. On finite generation and other finiteness conditions for wreath products of semigroups. Comm. Algebra, 30(8):3851–3873, 2002.
[22] L. N. Shevrin. On the theory of epigroups. Math. sb., 85(8):129–160, 1994. [23] L. N. Shevrin. On the theory of epigroups. Math. sb., 85(9):153–176, 1994.
(In Russian).
[24] M.R. Thomson. Finiteness Conditions of Wreath Products of Semigroups and Related Properties of Diagonal Acts. PhD thesis, University of St. Andrews, 2001.
[25] A.A. Tuganbaev. Bezout modules and rings. J. Math. Sci., 163(5):596–597, 2009.
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