Classical Finite Transformation Semigroups: An Introduction

Classical Finite Transformation Semigroups

Algebra and ApplicationsVolume 9

Managing Editor:

Alain VerschorenUniversity of Antwerp, Belgium

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Eric FriedlanderNorthwestern University, USA

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Idun ReitenNorwegian University of Science and Technology, Norway

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Algebra and Applications aims to publish well written and carefully refereed mono-graphs with up-to-date information about progress in all fields of algebra, itsclassical impact on commutative and noncommutative algebraic and differential ge-ometry, K-theory and algebraic topology, as well as applications in related domains,such as number theory, homotopy and (co)homology theory, physics and discretemathematics.

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Olexandr Ganyushkin • Volodymyr Mazorchuk

Classical FiniteTransformation Semigroups

An Introduction

ABC

Olexandr GanyushkinKyiv Taras Shevchenko UniversityVolodymyrska Street, [email protected]

Volodymyr MazorchukUppsala UniversityDept. MathematicsSE-751 06 [email protected]

ISBN: 978-1-84800-280-7 e-ISBN: 978-1-84800-281-4DOI: 10.1007/978-1-84800-281-4

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Preface

Semigroup theory is a relatively young part of mathematics. As a separatedirection of algebra with its own objects, formulations of problems, andmethods of investigations, semigroup theory was formed about 60 years ago.

One of the main motivations for the existence of some mathematicaltheories are interesting and natural examples. For semigroup theory the ob-vious candidates for such examples are transformation semigroups. Varioustransformations of different sets appear everywhere in mathematics all thetime. As the usual composition of transformations is associative, each set oftransformations, closed with respect to the composition, forms a semigroup.

Among all transformation semigroups one can distinguish three classicalseries of semigroups: the full symmetric semigroup T (M) of all transforma-tions of the set M ; the symmetric inverse semigroup IS(M) of all partial(that is, not necessarily everywhere defined) injective transformations of M ;and, finally, the semigroup PT (M) of all partial transformations of M . IfM = {1, 2, . . . , n}, then the above semigroups are usually denoted by Tn,ISn and PT n, respectively. One of the main evidences for the importance ofthese semigroups is their universality property: every (finite) semigroup is asubsemigroup of some T (M) (resp. Tn); and every (finite) inverse semigroupis a subsemigroup of some IS(M) (resp. ISn). Inverse semigroups form aclass of semigroups which are closest (in some sense) to groups.

An analogous universal object in group theory is the symmetric groupS(M) of all bijective transformations of M . Many books are dedicated to thestudy of S(M) or to the study of transformation groups in general. Transfor-mation semigroups had much less luck. The “naive” search in MathSciNetfor books with the keywords “transformation semigroups” in the title resultsin two titles, one being a conference proceedings, and another one being old50-page-long lecture notes in Russian ([Sc4]). Just for comparison, an anal-ogous search for “transformation groups” results in 75 titles. And this is inspite of the fact that the semigroup T (M) was studied by Suschkewitschalready in the 1930s. The semigroup IS(M) was introduced by Wagnerin 1952, but the first relatively small monograph about it appeared onlyin 1996 ([Li]). The latter monograph considers some basic questions aboutIS(M): how one writes down the elements of IS(M), when two elementsof IS(M) commute, what is the presentation of IS(M), what are the con-

v

vi PREFACE

gruences on IS(M). For example, such basic semigroup-theoretical notionsas ideals and Green’s relations are mentioned only in the Appendix withoutany direct relation to IS(M).

Much more information about the semigroup T (M) can be found in thelast chapter of [Hi1]; however, it is mostly concentrated around the combi-natorial aspects. Otherwise one is left with the options to search throughexamples in the parts of the abstract theory of semigroups using the classi-cal books [CP1, Gri, Ho3, Ho7, Hi1, Law, Pe] or to look at original researchpapers.

The aim of the present book is to partially fill the gaps in the literature.In the book we introduce three classical series of semigroups, and for themwe describe generating systems, ideals, Green’s relations, various classes ofsubsemigroups, congruences, conjugations, endomorphisms, presentations,actions on sets, linear representations and cross-sections. Some of the resultsare very old and classical, some are quite young. In order not to overloadthe reader with too technical and specialized results, we decided to restrictthe area of the present book to the above-mentioned parts of the theory oftransformation semigroups.

The book was thought to be an elementary introduction to the theory oftransformation semigroups, with a strong emphasis on the concrete exam-ples in the form of three classical series of finite transformation semigroups,namely, Tn, ISn and PT n. The book is primarily directed to students, whowould like to make their first steps in semigroup theory. The choice of thesemigroups Tn, ISn and PT n is motivated not only by their role in semi-group theory, but also by our strong belief that a good understanding ofa couple of interesting and pithy examples is more important for the firstacquaintance with some theory than a formal learning of dozens of theorems.

Another good motivation to consider the semigroups Tn, ISn and PT n atthe same time is the observation that many results about these semigroups,which for each of them were obtained independently by different people andin different times, in reality can be obtained in a unified (or almost unified)way.

Several results which will be presented extend in one or another way tothe cases of infinite transformation semigroups. However, we restrict our-selves to the case of finite semigroups to make the exposition as elementaryand accessible for a wide audience as possible. We are not after the biggestpossible generality. Another argument is that we also try to attract thereader’s attention to numerous combinatorial aspects and applications ofthe semigroups we consider.

With our three principal examples of semigroups on the background wealso would like to introduce the reader to the basics of the abstract theory ofsemigroups. So, along the discussion of these examples, we tried to presentmany important basic notions and prove (or at least mention) as manyclassical abstract results as possible.

PREFACE vii

The requirements for the reader’s mathematical background are verylow. To understand most of the content, it is enough to have a minimalmathematical experience on the level of common sense. Perhaps some famil-iarity with basic university courses in algebra and combinatorics would be asubstantial help. We have tried to define all the notions we use in the book.We have also tried to make all proofs very detailed and to avoid complicatedconstructions whenever possible.

The penultimate section of each chapter is called “Addenda and Com-ments.” A part of it consists of historical comments (which are by no meanscomplete). Another part consists of some remarks, facts, and statements,which we did not include in the main text of the book. The reason is usuallythe much less elementary level of these statements or the more complicatedcharacter of the proofs. However, we include them in the Addenda as fromour point of view they deserve attention in spite of the fact that they do notreally fit into the main text. Some statements here are also given with proofs,but these proofs are less detailed than those in the main text. For this partof the book, our requirements for the reader’s mathematical background aredifferent and are closer to the standard mathematical university curriculum.In the Addenda, we sometimes also mention some open problems and try todescribe possible directions for further investigations.

The division of the book into the main text and the Addenda is notvery strict as sometimes the notions and facts mentioned in the Addendaare used in the main text.

The last section of each chapter contains problems. Some problems (notmany) are also included in the main text. The latter ones are mostly simpleand directed to the reader. Sometimes they also ask to repeat a proof givenbefore for a different situation. These problems are in some sense compulsoryfor the successful understanding of the main text (i.e., one should at leastread them). The additional problems of the last section of each chaptercan be quite different. Some of them are easy exercises, while others aremuch more complicated problems, which form an essential supplement tothe material of the chapter. Hints for solutions of the latter ones can befound at the end of the book.

The book was essentially written during the visit of the first author toUppsala University, which was supported by The Royal Swedish Academyof Sciences and The Swedish Foundation for International Cooperation inResearch and Higher Education (STINT). The financial support of TheAcademy and STINT, and the hospitality of Uppsala University are grate-fully acknowledged. We thank Ganna Kudryavtseva, Victor Maltcev, andAbdullahi Umar for their comments on the preliminary version of the book.

Kyiv, Ukraine Olexandr GanyushkinUppsala, Sweden Volodymyr Mazorchuk

Contents

Preface v

1 Ordinary and Partial Transformations 11.1 Basic Definitions . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 Graph of a (Partial) Transformation . . . . . . . . . . . . . . 31.3 Linear Notation for Partial Transformations . . . . . . . . . . 81.4 Addenda and Comments . . . . . . . . . . . . . . . . . . . . . 101.5 Additional Exercises . . . . . . . . . . . . . . . . . . . . . . . 13

2 The Semigroups Tn , PT n , and ISn 152.1 Composition of Transformations . . . . . . . . . . . . . . . . 152.2 Identity Elements . . . . . . . . . . . . . . . . . . . . . . . . . 172.3 Zero Elements . . . . . . . . . . . . . . . . . . . . . . . . . . . 192.4 Isomorphism of Semigroups . . . . . . . . . . . . . . . . . . . 202.5 The Semigroup ISn . . . . . . . . . . . . . . . . . . . . . . . 222.6 Regular and Inverse Elements . . . . . . . . . . . . . . . . . . 232.7 Idempotents . . . . . . . . . . . . . . . . . . . . . . . . . . . . 262.8 Nilpotent Elements . . . . . . . . . . . . . . . . . . . . . . . . 282.9 Addenda and Comments . . . . . . . . . . . . . . . . . . . . . 312.10 Additional Exercises . . . . . . . . . . . . . . . . . . . . . . . 35

3 Generating Systems 393.1 Generating Systems in Tn, PT n, and ISn . . . . . . . . . . . 393.2 Addenda and Comments . . . . . . . . . . . . . . . . . . . . . 423.3 Additional Exercises . . . . . . . . . . . . . . . . . . . . . . . 43

4 Ideals and Green’s Relations 454.1 Ideals of Semigroups . . . . . . . . . . . . . . . . . . . . . . . 454.2 Principal Ideals in Tn, PT n, and ISn . . . . . . . . . . . . . 464.3 Arbitrary Ideals in Tn, PT n, and ISn . . . . . . . . . . . . . 504.4 Green’s Relations . . . . . . . . . . . . . . . . . . . . . . . . . 534.5 Green’s Relations on Tn, PT n, and ISn . . . . . . . . . . . . 584.6 Combinatorics of Green’s Relations

in the Semigroups Tn, PT n, and ISn . . . . . . . . . . . . . . 60

ix

x CONTENTS

4.7 Addenda and Comments . . . . . . . . . . . . . . . . . . . . . 624.8 Additional Exercises . . . . . . . . . . . . . . . . . . . . . . . 65

5 Subgroups and Subsemigroups 695.1 Subgroups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 695.2 Cyclic Subsemigroups . . . . . . . . . . . . . . . . . . . . . . 705.3 Isolated and Completely Isolated

Subsemigroups . . . . . . . . . . . . . . . . . . . . . . . . . . 745.4 Addenda and Comments . . . . . . . . . . . . . . . . . . . . . 835.5 Additional Exercises . . . . . . . . . . . . . . . . . . . . . . . 88

6 Other Relations on Semigroups 916.1 Congruences and Homomorphisms . . . . . . . . . . . . . . . 916.2 Congruences on Groups . . . . . . . . . . . . . . . . . . . . . 946.3 Congruences on Tn, PT n, and ISn . . . . . . . . . . . . . . . 966.4 Conjugate Elements . . . . . . . . . . . . . . . . . . . . . . . 1036.5 Addenda and Comments . . . . . . . . . . . . . . . . . . . . . 1086.6 Additional Exercises . . . . . . . . . . . . . . . . . . . . . . . 110

7 Endomorphisms 1117.1 Automorphisms of Tn, PT n, and ISn . . . . . . . . . . . . . 1117.2 Endomorphisms of Small Ranks . . . . . . . . . . . . . . . . . 1147.3 Exceptional Endomorphism . . . . . . . . . . . . . . . . . . . 1157.4 Classification of Endomorphisms . . . . . . . . . . . . . . . . 1187.5 Combinatorics of Endomorphisms . . . . . . . . . . . . . . . . 1237.6 Addenda and Comments . . . . . . . . . . . . . . . . . . . . . 1277.7 Additional Exercises . . . . . . . . . . . . . . . . . . . . . . . 128

8 Nilpotent Subsemigroups 1318.1 Nilpotent Subsemigroups and Partial Orders . . . . . . . . . 1318.2 Classification of Maximal Nilpotent

Subsemigroups . . . . . . . . . . . . . . . . . . . . . . . . . . 1348.3 Cardinalities of Maximal Nilpotent,

Subsemigroups . . . . . . . . . . . . . . . . . . . . . . . . . . 1388.4 Combinatorics of Nilpotent Elements in ISn . . . . . . . . . 1418.5 Addenda and Comments . . . . . . . . . . . . . . . . . . . . . 1488.6 Additional Exercises . . . . . . . . . . . . . . . . . . . . . . . 151

9 Presentation 1539.1 Defining Relations . . . . . . . . . . . . . . . . . . . . . . . . 1539.2 A presentation for ISn . . . . . . . . . . . . . . . . . . . . . . 1569.3 A Presentation for Tn . . . . . . . . . . . . . . . . . . . . . . 1619.4 A presentation for PT n . . . . . . . . . . . . . . . . . . . . . 1699.5 Addenda and Comments . . . . . . . . . . . . . . . . . . . . . 1729.6 Additional Exercises . . . . . . . . . . . . . . . . . . . . . . . 173

CONTENTS xi

10 Transitive Actions 17510.1 Action of a Semigroup on a Set . . . . . . . . . . . . . . . . . 17510.2 Transitive Actions of Groups . . . . . . . . . . . . . . . . . . 17710.3 Transitive Actions of Tn . . . . . . . . . . . . . . . . . . . . . 17910.4 Actions Associated with L-Classes . . . . . . . . . . . . . . . 18010.5 Transitive Actions of PT n and ISn . . . . . . . . . . . . . . 18210.6 Addenda and Comments . . . . . . . . . . . . . . . . . . . . . 18510.7 Additional Exercises . . . . . . . . . . . . . . . . . . . . . . . 187

11 Linear Representations 18911.1 Representations and Modules . . . . . . . . . . . . . . . . . . 18911.2 L-Induced S-Modules . . . . . . . . . . . . . . . . . . . . . . 19211.3 Simple Modules over ISn and PT n . . . . . . . . . . . . . . . 19511.4 Effective Representations . . . . . . . . . . . . . . . . . . . . 19811.5 Arbitrary ISn-Modules . . . . . . . . . . . . . . . . . . . . . 20011.6 Addenda and Comments . . . . . . . . . . . . . . . . . . . . . 20411.7 Additional Exercises . . . . . . . . . . . . . . . . . . . . . . . 211

12 Cross-Sections 21512.1 Cross-Sections . . . . . . . . . . . . . . . . . . . . . . . . . . . 21512.2 Retracts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21612.3 H-Cross-Sections in Tn, PT n, and ISn . . . . . . . . . . . . . 21912.4 L-Cross-Sections in Tn and PT n . . . . . . . . . . . . . . . . 22212.5 L-Cross-Sections in ISn . . . . . . . . . . . . . . . . . . . . . 22612.6 R-Cross-Sections in ISn . . . . . . . . . . . . . . . . . . . . . 23112.7 Addenda and Comments . . . . . . . . . . . . . . . . . . . . . 23312.8 Additional Exercises . . . . . . . . . . . . . . . . . . . . . . . 235

13 Variants 23713.1 Variants of Semigroups . . . . . . . . . . . . . . . . . . . . . . 23713.2 Classification of Variants for ISn, Tn,

and PT n . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24013.3 Idempotents and Maximal Subgroups . . . . . . . . . . . . . 24313.4 Principal Ideals and Green’s Relations . . . . . . . . . . . . . 24513.5 Addenda and Comments . . . . . . . . . . . . . . . . . . . . . 24613.6 Additional Exercises . . . . . . . . . . . . . . . . . . . . . . . 249

14 Order-Related Subsemigroups 25114.1 Subsemigroups, Related to the Natural

Order . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25114.2 Cardinalities . . . . . . . . . . . . . . . . . . . . . . . . . . . 25314.3 Idempotents . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25714.4 Generating Systems . . . . . . . . . . . . . . . . . . . . . . . 26014.5 Addenda and Comments . . . . . . . . . . . . . . . . . . . . . 26714.6 Additional Exercises . . . . . . . . . . . . . . . . . . . . . . . 273

xii CONTENTS

Answers and Hints to Exercises 277

Bibliography 283

List of Notation 297

Index 307

Chapter 1

Ordinary and PartialTransformations

1.1 Basic Definitions

The principal objects of interest in the present volume are finite sets andtransformations of finite sets. Let M be a finite set, say M={m1, m2, . . . , mn},where n is a nonnegative integer. Transformation of M is an array of thefollowing form:

α =(

m1 m2 · · · mn

k1 k2 · · · kn

), (1.1)

where all ki ∈ M . If x ∈ M , say x = mi, the element ki will be called thevalue of the transformation α at the element x and will be denoted by α(x).The fact that α is a transformation of M is usually written as α : M → M .As the nature of elements of M is not important for us, instead of M weshall usually consider the set N = Nn = {1, 2, . . . , n}.

Apart from the transformations of M we shall also consider the so-calledpartial transformations of M , that is, transformations of the form α : A→M ,where A = {l1, l2, . . . , lk} is a subset of M . Note that the set A can be empty.Again, the element α can be written in the following tabular form:

α =(

l1 l2 · · · lkα(l1) α(l2) · · · α(lk)

). (1.2)

Abusing notation, we may also write α : M → M for a partial transforma-tion, having in mind that such α is only defined on some elements from M .Note that the order of elements in the first row of arrays (1.1) and (1.2) isnot important.

With each (partial) transformation α as above we associate the followingstandard notions:

• The domain of α: dom(α) = A

O. Ganyushkin, V. Mazorchuk, Classical Finite Transformation Semigroups, Algebra 1

and Applications 9, DOI: 10.1007/978-1-84800-281-4 1,c© Springer-Verlag London Limited 2009

2 CHAPTER 1. ORDINARY AND PARTIAL TRANSFORMATIONS

• The codomain of α: dom(α) = M\A

• The image of α: im(α) = {α(x) : x ∈ A}

The word range, which is also frequently used in the literature, is a synonymof the word image. If dom(α) = M , the transformation α is called full ortotal .

The set of all total transformations of M is denoted by T (M), and theset of all partial transformations of M is denoted by PT (M). Obviously,T (M) ⊂ PT (M). To simplify our notation we set Tn = T (N) and PT n =PT (N).

Sometimes it is convenient to use a slightly modified version of (1.2) forsome α ∈ PT n. In the case of PT n it is natural to form the first row ofthe array for α by simply listing all the elements from N in their naturalorder. Then, to define α completely, one needs a special symbol to indicatethat some element x belongs to dom(α). We shall use the symbol ∅. Inother words, α(x) = ∅ means that x ∈ dom(α). Thus the element α can bewritten in the following form:

α =(

1 2 . . . nk1 k2 . . . kn

), (1.3)

where ki = α(i) if i ∈ dom(α) and ki = ∅ if i ∈ dom(α).

Example 1.1.1 Here is the list of all elements of PT 2:(1 21 1

),

(1 21 2

),

(1 22 1

),

(1 22 2

),

(1 21 ∅

),

(1 22 ∅

),

(1 2∅ 1

),

(1 2∅ 2

),

(1 2∅ ∅

).

The first row of this list consists of total transformations and hence lists allelements in T2.

Proposition 1.1.2 The set Tn contains nn elements and the set PT n con-tains (n + 1)n elements.

Proof. Each element α ∈ Tn is uniquely defined by array (1.3), where eachki ∈ N. Since the choices of kis are independent, we have |Tn| = nn

by the product rule. In the case of PT n, the elements ki can be inde-pendently chosen from the set N∪{∅}. Hence the product rule implies|PT n|=(n + 1)n.

The cardinality |im(α)| of the image of a partial transformation α ∈ PT n

is called the rank of this partial transformation and is denoted by rank(α).Thus rank(α) equals the number of different elements in the second row ofarray (1.2). The number def(α) = n − rank(α) is called the defect of thepartial transformation α.

1.2. GRAPH OF A (PARTIAL) TRANSFORMATION 3

A partial transformation α ∈ PT n is called

• Surjective if im(α) = N

• Injective if x �= y implies α(x) �= α(y) for all x, y ∈ dom(α)

• Bijective if α is both surjective and injective

If α is given by (1.2), then surjectivity means that the second row of array(1.2) contains all elements of N; and injectivity means that all elements inthe second row of array (1.2) are different. Bijective transformations on Nare also called permutations of N.

Proposition 1.1.3 Let α ∈ Tn. Then the following conditions are equiva-lent:

(a) α is surjective

(b) α is injective

(c) α is bijective

Proof. By the definition of a bijective transformation, it is enough to showthat the conditions (a) and (b) are equivalent. We start by proving thatinjectivity implies surjectivity. Let α ∈ Tn be injective and given by (1.3).Because of the injectivity of α, the second row of (1.3) gives n differentelements of the set N, namely, α(1), α(2), . . . , α(n). But N contains exactlyn elements. Hence N = {α(1), α(2), . . . , α(n)}, and thus α is surjective.

Conversely, let α ∈ Tn be surjective and given by (1.3). Then the secondrow of (1.3) contains all n elements of the set N. But this row containsexactly n elements. Hence they all must be different. This implies that α isinjective.

1.2 Graph of a (Partial) Transformation

With each partial transformation α on N one naturally associates a directedgraph Γα. A directed graph (or a digraph) is a pair, Γ = (V, E), where V isa set and E ⊂ V × V . The elements of V are called vertices of Γ and theelements of E are called directed edges or arrows of Γ . If (a, b) ∈ E, thenthe vertex a is called the tail of (a, b) and the vertex b is called the head of(a, b).

The graph Γα = (Vα, Eα) is called the graph of the transformation α andis constructed in the following way: The set Vα of vertices coincides with N;for x, y ∈ N the element (x, y) belongs to Eα if and only if x ∈ dom(α) andα(x) = y.


Example 1.2.1 For the transformation

α =(

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 168 9 7 13 7 16 ∅ 1 13 11 16 4 12 9 14 13

)

the graph Γα has the following form:

10 �� 11

��

��

� 6

��

15

��

2

��

16

��

4

��

14 �� 9 �� 13 �� 12

��

8

��1

�� 3

��

5

��

7

(1.4)

It is obvious that a directed graph Γ = (N, E) will be the graph of sometotal (partial) transformation of N if and only if each vertex is the tail ofexactly one (at most one) arrow. The graph Γα decomposes into a disjointunion of connected components. Intuitively, this is an obvious notion, for ex-ample, graph (1.4) has three connected components. The rigorous definitionis as follows.

First we define a subgraph. If Γ = (V, E) is a directed graph, a subgraphof Γ is a directed graph Γ ′ = (V ′, E′) such that V ′ ⊂ V and E′ ⊂ E. A di-rected graph Γ = (V, E) is called connected if for each partition of V into adisjoint union of nonempty subsets V1 and V2 there exists a ∈ V1 and b ∈ V2

such that either (a, b), or (b, a) is an arrow. If Γ is a directed graph, then theconnected components of Γ are simply the maximal connected subgraphs ofΓ , that is, those connected subgraphs of Γ which are not proper subgraphsof any other connected subgraph of Γ .

Exercise 1.2.2 Prove that two different connected components of a directedgraph Γ do not have common arrows.

To understand the structure of Γα it is of course enough to understandthe structure of its connected components. For this we shall need some moregraph-theoretical notions. Let Γ (V, E) be a directed graph and a, b ∈ V . Anoriented path from a to b in Γ is a sequence x0 = a, x1, . . . , xk = b of verticessuch that (xi, xi+1) ∈ E for each i = 0, 1, . . . , k − 1. Vertex a is called thetail of the path and vertex b is called the head of the path. If we have anoriented path such that a = b and xi �= xj for all 0 ≤ i < j < k, then suchpath is called an (oriented) cycle and is denoted by (x0, x1, . . . , xk−1). If Γdoes not contain any arrow with tail b we will say that our path breaks at b.Analogously one defines infinite paths. Such paths may be without tails,without heads, or without both tails and heads.

Let Γ be a directed graph and v be a vertex of Γ . We define a trajectoryof v as any longest possible path with the tail v. There are two possibilities:


either such trajectory is finite and hence breaks at some point, or this tra-jectory is infinite. For example, 5, 7 is a trajectory of vertex 5 in graph (1.4),which breaks at vertex 7; and 2, 9, 13, 12, 4, 13, 12, 4, 13, . . . is a trajectoryof point 2. In general, a point can have many different trajectories. However,we have the following obvious statement.

Lemma 1.2.3 Let Γ be a directed graph. Then the following conditions areequivalent:

(a) Each vertex of Γ has a unique trajectory

(b) Each vertex of Γ is the tail of at most one arrow

We will say that the infinite trajectory x0 = v, x1, . . . terminates atthe cycle (xk, xk+1, . . . , xk+m−1) if the path xk, xk+1, . . . , xk+m−1 is a cycle,xi = xi+m for all i ≥ k and xk−1 �= xk+m−1. Thus, the trajectory of vertex 2in graph (1.4) terminates at the cycle (13, 12, 4); and the trajectory of vertex4 terminates at the cycle (4, 13, 12).

Proposition 1.2.4 Let α ∈ PT n.

(i) Every vertex of Γα has a unique trajectory.

(ii) The trajectory of each x ∈ N in Γα either breaks at some vertex orterminates at some cycle.

(iii) α is total if and only if the trajectory of each x ∈ N in Γα terminatesat some cycle.

(iv) Let x, y ∈ N. If y occurs in the trajectory of x in Γα, then the trajectoryof y is a subsequence of the trajectory of x.

Proof. Since each vertex of Γα is a tail of at most one arrow, the state-ment (i) follows immediately from Lemma 1.2.3. The statement (iv) followsimmediately from (i).

Assume that the trajectory x = x0, x1, . . . of x does not break. SinceΓα is finite, this trajectory must contain repetitions of some vertices. Letk be minimal for which there exists a repetition of xk and let xk+m be thefirst repetition of xk. Since each vertex of Γα is a tail of at most one arrow,the condition xk = xk+m implies xk+1 = xk+m+1, which, in turn, impliesxk+2 = xk+m+2, and so on. Hence our trajectory terminates at the cycle(xk, xk+1, . . . , xk+m−1). This proves (ii).

If α is total, the trajectory of each vertex cannot break. Hence (iii) followsfrom (ii).

For α ∈ PT n and x ∈ N we denote by trα(x) the trajectory of x in Γα.This is well-defined because of Proposition 1.2.4(i). Define now the binaryrelation ωα on N in the following way: For x, y ∈ N set x ωα y if trα(x) andtrα(y) have at least one common vertex.


Lemma 1.2.5 The relation ωα is an equivalence relation.

Proof. That ωα is reflexive and symmetric is obvious. To prove the tran-sitivity of ωα consider x, y, z ∈ N such that x ωα y and y ωα z. Let a be acommon vertex of trα(x) and trα(y) and b be a common vertex of trα(y)and trα(z). Without loss of generality we can assume that the first occur-rence of a in trα(y) is not later than the first occurrence of b in trα(y). Butthis means that b occurs in trα(a) by Proposition 1.2.4(iv). Another appli-cation of Proposition 1.2.4(iv) implies that b occurs in trα(x). Hence x ωα z,completing the proof.

The equivalence classes of ωα are called the orbits of α. For x ∈ N theorbit of x in Γα will be denoted by oα(x). From the definition of ωα it followsthat for any x ∈ dom(α) we have x ωα α(x). Hence all vertices which occurin trα(x) belong to oα(x). Furthermore, for each x ∈ N we can restrict thepartial transformation α to the orbit K = oα(x), obtaining a new partialtransformation, α(K) ∈ PT (oα(x)). Certainly, α(K) does not depend on thechoice of the vertex in K.

Proposition 1.2.6 For each x ∈ N the graph Γα(K) is a connected compo-nent of Γα.

Proof. From the definition of ωα it follows that the graph Γα(K) is connectedand contains all those arrows of Γα, for which both the heads and the tailsbelong to K. Assume now that (x, y) is an arrow of Γα such that x ∈ K.Then y ∈ trα(x) and hence x ωα y, that is, y ∈ K. If (x, y) is an arrow of Γα

such that y ∈ K, then again y ∈ trα(x) and hence x ωα y, that is, x ∈ K.This means that Γα(K) is not properly contained in any connected subgraphof Γα, which proves our statement.

As an immediate corollary of Proposition 1.2.6 we have:

Corollary 1.2.7 The mapping K → Γα(K) is a bijection between the orbitsof α and the connected components of Γα.

A directed graph Γ=Γ (V, E) is called a tree with the sink a∈V providedthat for each x ∈ V the trajectory of x in Γ is unique and breaks at a. Forinstance, in the example (1.4) if K = oα(3), the connected component Γα(K)

is a tree with the sink 7. A (nonempty) disjoint union of several trees withsinks is called a forest of trees with sinks.

Exercise 1.2.8 Let Γ be a tree with the sink a. Show that Γ is connected;that each b �= a is a tail of exactly one arrow; and that Γ contains neitheroriented nor unoriented cycles.

A directed graph Γ = (V, E) is called a cycle provided that we canenumerate V = {a1, . . . , ak} such that E = {(a1, a2), (a2, a3), . . . , (ak−1, ak),


(ak, a1)}. If Γi = (Vi, Ei), i ∈ I, are directed graphs, then their union Γ =∪i∈IΓi is defined as follows Γ = (V, E), where V = ∪i∈IVi and E = ∪i∈IEi.For example, each directed graph is a union of its connected components.

Theorem 1.2.9 Each connected component of Γα, α ∈ PT n, is either

(i) a tree with a sink, or

(ii) a union of a forest of trees with sinks with a cycle on the set of alltheir sinks.

We note that the forest in Theorem 1.2.9(ii) may contain only one treewith a sink. The union of this tree with a sink with the cycle on its sink is nota tree with a sink anymore. We also note that the forest in Theorem 1.2.9(ii)may also have some trivial trees with sinks, that is, trees consisting only ofsinks. If all trees in this forest are trivial, Theorem 1.2.9(ii) simply describesa cycle.

Proof. Let α ∈ PT n and Γα(K) = (K, EK) be a connected component of Γα,and x ∈ K. Then Lemma 1.2.3 and Proposition 1.2.4(i) imply that x has aunique trajectory in both Γα and Γα(K) . Moreover, since K is a connectedcomponent of Γα we also have that the trajectories of x in Γα and Γα(K)

coincide (and hence they both are equal to trα(x)).Assume first that trα(x) breaks at some vertex, say a. Let y ∈ K be

arbitrary. Then, by definition, trα(x) and trα(y) have a common vertex,say z. Hence trα(z) is a subsequence of both trα(x) and trα(y). But trα(x)breaks at a, which means that trα(z) must break at a as well. This impliesthat trα(y) breaks at a. This means, by definition, that Γα(K) is a tree withthe sink a, that is, of the type Theorem 1.2.9(i).

Now we assume that trα(x) terminates at some cycle, say (a1, a2, . . . , ak).For each ai, i = 1, . . . , k, the trajectory of ai is unique by Proposition 1.2.4(i)and hence is ai, ai+1, . . . , ak, a1, . . . . For every y ∈ K, the trajectory trα(y)has a common subsequence with trα(x) and thus must contain some ai. Fori = 1, . . . , k we denote by Ki the set of all those vertices y from K suchthat the first vertex from the cycle (a1, a2, . . . , ak) in trα(y) is ai. Note thatKi ∩ Kj = ∅ for i �= j by definition.

Assume that i �= j and let (x, y) be an arrow in Γα(K) such that x ∈ Ki

and y ∈ Kj . Then trα(x) has the form x, y, y1, . . . , where y, y1, . . . is justtrα(y). However, the first element from the cycle (a1, a2, . . . , ak) in trα(x) isai, whereas in trα(y) it is aj �= ai. This is possible only in the case of x = ai

and y = aj . This means that every arrow in Γα(K) from some element in Ki

to some element in Kj in fact belongs to the cycle (a1, a2, . . . , ak).For each i = 1, . . . , k, consider the graph Γi = (Ki, Ei), where

Ei = ((Ki × Ki) ∩ E)\{(ai, ai)},


(this means that Ei consists of all arrows from E, which has both tails andheads in Ki, with the exception of the arrow (ai, ai)). Let Γ0 be the cycle(a1, a2, . . . , ak). From the previous paragraph, we have that Γα(K) = ∪k

i=0Γi.Furthermore, Kis are disjoint for i = 1, . . . , k. To show that Γα(K) is ofthe form described in Theorem 1.2.9(ii) it remains to show that each Γi,i = 1, . . . , k, is a tree with the sink ai.

By definition, Γi does not contain any arrow with the tail ai. Let y ∈ Ki.Then trα(y) has the form y = y0, y1, . . . , ym = ai, ai+1, . . . , where ym = ai

is the first occurrence of ai in trα(y). By the definition of Ki, all verticesy1, . . . , ym−1 do not belong to Γ0. Hence the definition of Ei implies thaty = y0, y1, . . . , ym = ai is the trajectory of y in Γi, and it breaks at ai. Inother words, the trajectory of each vertex in Γi breaks at ai and thus Γi isa tree with the sink ai. This completes the proof.

Example 1.2.10 The graph Γα from Example 1.2.1 is given by (1.4) andhas three connected components. The third component is a tree with thesink 7. The second component is a cycle (that is, the union of the cycle(1, 8) with the corresponding forest of trivial trees with sinks). The firstcomponent is the union of the cycle (4, 13, 12) with the following forest oftrees with sinks:

15 �� 14

��

��

� 11

��

10��

2 �� 9

��

��

� 16

��

6��

13

12 4

An immediate corollary of Theorem 1.2.9 is the following:

Corollary 1.2.11 Different cycles of Γα belong to different connected com-ponents of Γα.

1.3 Linear Notation for Partial Transformations

The graphical presentation of a transformation α ∈ PT n via Γα is verytransparent, but also rather space consuming. For a plain mathematicaltext, it would be very useful to have some space-saving alternative. Forpermutations this is known as the cyclic notation and can be easily describedby the following example:

Example 1.3.1 For the permutation

α =(

1 2 3 4 5 6 7 8 9 10 11 12 13 14 159 8 15 2 10 1 14 4 7 5 6 11 13 3 12

)

1.3. LINEAR NOTATION FOR PARTIAL TRANSFORMATIONS 9

the graph Γα has the following form:

1 �� 9 �� 7 �� 14 �� 3

��

6

��

11�� 12�� 15��

2 �� 8

��

4

�� 5

��10

�� 13

Using the notation for cycles, introduced on page 4 in the paragraph afterExercise 1.2.2, we may write

α = (1, 9, 7, 14, 3, 15, 12, 11, 6)(2, 8, 4)(5, 10)(13).

Clearly the above notation is not uniquely defined. Writing a cycle wecan start from each of its vertices. Moreover, the order of cycles in the cyclicnotation can also be chosen in an arbitrary way. In this subsection, we wouldlike to generalize this notation to be able to use it for all elements of PT n.A very good hint how to do this is given by Theorem 1.2.9, which roughlysays that we only have to find a nice notation for trees with sinks. We callour notation linear and will define it recursively.

Assume for the moment that the graph Γα, where α ∈ PT n, is a treewith the sink a. If a is the only vertex of Γ , we shall write Γα = [a] (orsimply α = [a]). If Γα contains some other vertices, then it has to have thefollowing form:

��

. . .

��

��

. . .

��

Γ1 . . . . . . . . . Γk

a1

��

ak

��

a

(1.5)

For i = 1, . . . , k, the subgraph Γi of the graph (1.5) above is a tree with thesink ai and has strictly less vertices than Γα. Assume that we already havethe linear notation Γi for Γi, i = 1, . . . , k. Then the linear notation for Γα

(and α) is defined recursively as follows

Γα = [Γ1, Γ2, . . . , Γk; a].

This defines the notation for the elements given by Theorem 1.2.9(i).Assume now that Γα is connected and given by Theorem 1.2.9(ii). Then

Γα is the union of some cycle, say (a1, . . . , ak), with certain disjoint trees Γi

with sinks ai, i = 1, . . . , k. Let Γi, i = 1, . . . , k, be the corresponding linearnotation. In this case, we define the linear notation for Γα (and α) as follows

Γα = (Γ1, Γ2, . . . , Γk).


Finally, for any α ∈ PT n, we define the linear notation for Γα (and α)to be the product of linear notation over all connected components of Γα,written in an arbitrary order. In the same way as the classical cycle notationfor permutations, the linear notation for (partial) transformartions is uniqueonly up to permutation of certain components of the notation. Namely, theconnected components can be written in an arbitrary order, and on eachstep of the recursive procedure the order of the components Γ1, . . . , Γk canalso be chosen arbitrarily.

Note that, by the above definition, the ordinary cycle (a1, a2, . . . , ak) isdenoted by ([a1], [a2], . . . , [ak]). This is of course not very practical, so toavoid this unnecessary complication inside the notation for cycles (but notfor trees with sinks!) we shall usually skip the brackets “[ ]” surroundingtrivial trees with sinks. Sometimes, if n is fixed, one can also skip all loops(i.e., the elements of the form (x) = ([x])). This just means that all x ∈ N,which do not appear in the notation, correspond to loops. It is clear thatthis does not give rise to any confusion, moreover, it restores the originalnotation for usual cycles and permutations.

Example 1.3.2 For the transformation α from Example 1.2.1 we have

α = ([[[[15]; 14], [2]; 9], [[[10]; 11], [6]; 16]; 13], 12, 4)(1, 8)[[3], [5]; 7].

1.4 Addenda and Comments

1.4.1 To use graphs for presentation of transformations was proposed bySuschkewitsch in [Su1].

1.4.2 Let α ∈ PT n. The element x ∈ N satisfying α(x) = x is usually calleda fixed point of α. If α is a permutation, then all the connected componentsof Γα are cycles. Fixed points of α correspond to cycles of length 1. Inthe cyclic notation for α such cycles are usually omitted (for the identityelement one thus has to use a special notation, for example ε). If after suchsimplification the cyclic notation for α contains only one cycle, say of lengthk, the α is usually called a cycle of length k. Cycles of length 2 are calledtranspositions.

1.4.3 An alternative “linear” notation for total transformations was pro-posed in [AAH]. Although [AAH] works only with total transformationsit is fairly straightforward to generalize their notation to cover all partialtransformations. Roughly speaking the [AAH]-notation reduces to listingthe trajectories of all vertices of the graph Γα. If we know the trajectoryx0 = x, x1, x2, . . . of the vertex x, then of course we know the trajectories ofeach of the vertices x1, x2, . . . , so we can omit the latter ones. A trajectoryis denoted by [a1, a2, . . . , ak|aj ], where 1 ≤ j ≤ k. This means the following:

1.4. ADDENDA AND COMMENTS 11

• If j < k, the trajectory a1, a2, . . . of the vertex a1 terminates at thecycle (aj , aj+1, . . . , ak)

• If j = k and ak does not occur previously, then the trajectory a1, a2, . . .of the vertex a1 terminates at the cycle (ak)

• If j = k and ak does occur previously, it means that we already knowthe trajectory of ak, in this case the trajectory of the vertex a1 isobtained by attaching a1, a2, . . . , ak to the known trajectory of ak

On each step we choose any vertex, say a, whose trajectory is not yet writtendown, and we write down the trajectory of a until we either reach a vertex,whose trajectory is already written down, or we terminate the trajectory ofa in some cycle. To make the notation as short as possible, on each step oneshould try to choose a new vertex, which is not a head of any arrow in Γα.

For example, for the transformation

α =(

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 168 9 7 13 7 16 7 1 13 11 16 4 12 9 14 13

)

the [AAH]-notation for α will have the following form:

[15, 14, 9, 13, 12, 4|13][10, 11, 16, 13|13][6, 16|16][2, 9|9][8, 1|8][3, 7|7][5, 7|7].

For comparison, our notation for α looks as follows:

([[[[15]; 14], [2]; 9], [[[10]; 11], [6]; 16]; 13], 12, 4)(1, 8)([[3], [5]; 7]).

From our point of view the [AAH]-notation has some disadvantages,namely,

• The cyclic notation for permutations is not a partial case of the [AAH]-notation

• The [AAH]-notation is long, that is, it always contains elements oc-curring more than one time

• The [AAH]-notation is by far not unique, even up to permutations ofcertain components of this notation

Another disadvantage of the [AAH]-notation, related to the compositionof transformations, will be discussed in 2.9.3. An advantage of the [AAH]-notation in comparison to our notation is that it contains less brackets.

1.4.4 In [Ka] it was shown that the number of those α ∈ Tn for which Γα

is connected equals

n!n−1∑k=0

nk−1

k!


This can be proved, for example, in the following way:From Theorem 1.2.9 it follows that Γα is connected if and only if it is

a union of a forest of trees with sinks with a cycle on the set of all sinks.For each such α and for each i ≥ 0 we define the set N

(i)α as the set of all

x ∈ N such that the first element from the cycle occurs in trα(x) on step i.Obviously, N

(0)α consists of the elements of our cycle and N = ∪i≥0N

(i)α is

a disjoint union. Furthermore, α(N (i)α ) ⊂ N

(i−1)α for i > 0. The number of

those α, for which |N (i)α | = ni, 0 ≤ i ≤ t, and such that

∑ti=0 ni = n equals

(n

n0, n1, . . . , nt

)(n0 − 1)!nn1

0 nn21 . . . nnt

t−1 (1.6)

(here the polynomial coefficient(

nn0,n1,...,nt

)gives the number of ordered

partitions of N into blocks with cardinalities n0,. . . , nt, respectively; (n0−1)!is the number of ways to form a cycle out of n0 elements; and n

ni+1

i is thenumber of maps from the block with ni+1 elements to the block with ni

elements). We can rewrite (1.6) as follows

n!n0

· nn10

n1!· nn2

1

n2!. . .

nntt−1

nt!. (1.7)

Now to find the number Xn0 of those α ∈ Tn for which Γα is connected andcontains a cycle of length n0 one has to add up all summands of the form(1.7) for all t ≤ n − n0 and all decompositions n1 + · · · + nt = n − n0. Letn − n0 = k. Then

nk−1

k!=

(n0 + k)k−1

k!=

k−1∑i=0

1k!

· (k − 1)!i!(k − 1 − i)!

· ni0 · kk−1−i =

=k∑

n1=1

nn1−10

(n1 − 1)!· kk−n1

(k − n1)!,

where for the last equality we substituted i by n1 − 1. Continuing in thesame way we get

nk−1

k!=

=k∑

n1=1

nn1−10

(n1 − 1)!·

k−n1∑n2=1

nn2−11

(n2 − 1)!· · ·

k−n1−···−nt−1∑nt=1

nnt−1t−1

(nt − 1)!· n−1

t =

=1n0

k∑n1=1

nn10

n1!

k−n1∑n2=1

nn21

n2!· · ·

k−n1−···−nt−1∑nt=1

nntt−1

nt!=

=∑

n1+···+nt=n−n0

1n0

· nn10

n1!· nn2

1

n2!. . .

nntt−1

nt!. (1.8)

1.5. ADDITIONAL EXERCISES 13

Hence Xn0 = n! · nk−1

k! , where k = n−n0. Since 1 ≤ n0 ≤ n, the final answeris now computed as follows

n∑n0=1

Xn0 = n!n−1∑k=0

nk−1

k!.

1.5 Additional Exercises

1.5.1 Let N denote the set of all positive integers. Give an example of atransformation α : N → N such that

(a) α is injective but not surjective.

(b) α is surjective but not injective.

1.5.2 Prove that limn→∞|PT n||Tn| = e.

1.5.3 Directed graphs Γi = (Vi, Ei), i = 1, 2, are called isomorphic providedthat there exists a bijection ϕ : V1 → V2 which induces a bijection from E1

to E2. Compute the number of pairwise nonisomorphic graphs Γα, where

(a) α ∈ T3.

(b) α ∈ T4.

(c) α ∈ PT 2.

(d) α ∈ PT 3.

(e) α ∈ PT 4.

1.5.4 Find the number of those partial transformations α ∈ PT 8, whosegraphs are isomorphic to the following graph:

•

��•

•

��

•�� •��

• �� •

��

•��

1.5.5 For α ∈ PT n characterize dom(α), im(α), dom(α), rank(α), anddef(α) in terms of Γα.

1.5.6 Compute the number of those α ∈ Tn (resp. α ∈ PT n) for whichim(α)

(a) Does not contain given elements a1, a2,. . . , ak


(b) Contains given elements a1, a2,. . . , ak

(c) Coincides with the given set {a1, a2, . . . , ak}1.5.7 Prove that the number of those α ∈ PT n, for which Γα is a tree witha sink, equals nn−1.

1.5.8 Prove that the number of those α ∈ PT n, for which Γα does notcontain cycles, equals

∑nk=1

(n−1k−1

)nn−k.

1.5.9 (a) Find the number of those α ∈ Tn which fix at least one (resp.exactly one) element (that is, α(x) = x for at least one or exactly oneelement x ∈ N, respectively).

(b) The same problem for PT n.

1.5.10 Let Γ = (V, E) be a directed graph. Consider the set X , whichconsists of all possible unordered partitions of V into disjoint unions ofnonempty subsets Vis such that for each i �= j the graph Γ does not con-tain any arrow from Vi to Vj . The set X is partially ordered in the naturalway with respect to inclusions of components of partitions. Prove that thepartition of V , which corresponds to the partition of Γ into connected com-ponents, is the minimum of X .

1.5.11 Let Γ = (V, E) be a directed graph. Consider the set Y, whichconsists of all possible unordered partitions of V into disjoint unions ofnonempty subsets Vis such that for each i we have that the subgraph(Vi, (Vi ×Vi)∩E) is connected. The set Y is partially ordered in the naturalway with respect to inclusions of components of partitions. Prove that thepartition of V , which corresponds to the partition of Γ into connectedcomponents, is the maximum of Y.

1.5.12 For α ∈ Tn, let tk(α) denote the number of those x ∈ N for which|{y ∈ N : α(y) = x}| = k. Prove that

(a)n∑

k=0

tk(α) = n,

(b)n∑

k=0

ktk(α) = n.

1.5.13 For α ∈ PT n let tk(α) denote the number of those x ∈ N for which|{y ∈ N : α(y) = x}| = k. Prove that

(a)n∑

k=0

tk(α) = n,

(b)n∑

k=0

ktk(α) ≤ n.

Chapter 2

The Semigroups Tn , PT n ,and ISn

2.1 Composition of Transformations

Let X and Y be two sets. A mapping from X to Y is an array of the form

f =(

xf(x)

)x∈X

,

where all f(x) ∈ Y . This is usually denoted by f : X → Y . The element f(x)is called the value of the mapping f at the element x. A transformation, asdefined in Sect. 1.1, is just a mapping from a set to itself. Let now X, Y, Y ′, Zbe sets such that Y ⊂ Y ′ and let f : X → Y and g : Y ′ → Z be twomappings. In this situation, we can define the product or the composition gfof f and g by the following rule: The composition gf is the mapping fromX to Z such that for all x ∈ X we have (gf)(x) = g(f(x)). In particular,we can always compose two total transformations of the same set and theresult will be a total transformation of this set.

The above definition admits a straightforward generalization to partialmappings. A partial mapping from X to Y is a mapping α : X ′ → Y , whereX ′ ⊂ X. In this case, we say that the partial mapping α is defined onelements from X ′. Again, a partial transformation, as defined in Sect. 1.1, isa partial mapping from a set to itself. One usually abuses notation and writesα : X → Y just emphasizing that α is a partial mapping. Let α : X → Y andβ : Y → Z be two partial mappings. We define their product or compositionβα as the partial mapping, defined on all those x ∈ X for which α and βare defined on the elements x and α(x), respectively; on such x the valueof βα is given by (βα)(x) = β(α(x)). In particular, we can always composetwo partial transformations of the same set and the result will be another



16 CHAPTER 2. THE SEMIGROUPS Tn, PT n, AND ISn

partial transformation of this set. We also note that the definition of thecomposition of total transformations is just a special case of that of partialtransformations.

Proposition 2.1.1 The composition of (partial) mappings is associative,that is, if α, β, and γ are partial mappings, then the composition γ(βα) isdefined if and only if the composition (γβ)α is defined, and if they both aredefined, we have γ(βα) = (γβ)α.

Proof. Follows immediately from the following picture:

X Y Z V

•x

α��

βα

��

γ(βα)=(γβ)α

��•y β ��

γβ

�� •z γ �� •

v

Associativity of the composition of partial transformations naturallyleads to the notion of a semigroup. Let S be a nonempty set, and let · :S × S → S be a binary operation on S. Then (S, ·) is called a semigroupprovided that · is associative, that is, a · (b · c) = (a · b) · c for all a, b, c ∈ S.To simplify the notation, in the case when the operation · is clear from thecontext one usually writes S for (S, ·). Furthermore, one usually writes abinstead of a · b.

Exercise 2.1.2 Let (S, ·) be a semigroup. Show that the value of the producta1a2 · · · an, where all ai ∈ S, does not depend on the way of computing it(that is, of putting brackets into this product).

Let (S, ·) be a semigroup. From Exercise 2.1.2 it follows that for everya ∈ S we have a well-defined element ak = a · a · · · a︸︷︷︸

ktimes

. The number of elements

in S is called the cardinality of S and is denoted by |S|.By Proposition 2.1.1, in both Tn and PT n the composition of (partial)

transformations is an associative operation. Hence we have:

Proposition 2.1.3 Both Tn and PT n are semigroups with respect to thecomposition of (partial) transformations.

The semigroup Tn is called the full transformation semigroup on the setN or the symmetric semigroup of all transformations of N. The semigroupPT n is called the semigroup of all partial transformations on N.

2.2. IDENTITY ELEMENTS 17

A nonempty subset T of a semigroup (S, ·) is called a subsemigroup ofS provided that T is closed with respect to · (that is, a · b ∈ T as soon asa, b ∈ T ). Obviously, in this case, T itself is a semigroup with respect to therestriction of the operation · to T . The fact that T is a subsemigroup of Sis usually denoted by T < S.

Exercise 2.1.4 Show that for arbitrary α, β ∈ PT n the following is true:

(a) dom(βα) ⊂ dom(α)

(b) im(βα) ⊂ im(β)

(c) rank(βα) ≤ min(rank(α), rank(β))

2.2 Identity Elements

An element e of a semigroup S is called a left or a right identity providedthat ea = a, or ae = a, respectively, for all a ∈ S. An element e, which is aleft and a right identity at the same time, is called a two-sided identity orsimply an identity.

If S contains some left identity el and some right identity er we haveel = el · er = er and hence these two elements coincide. Hence in this case Scontains a unique identity element, which is, moreover, a two-sided identity.However, a semigroup may contain many different left identities or manydifferent right identities (see Exercise 2.10.2). It is possible for a semigroupto contain neither left nor right identities. An example of such a semigroupis the semigroup (N, +). Another example is the semigroup {2, 3, 4, . . . } withrespect to the ordinary multiplication.

A semigroup which contains a two-sided identity element is called amonoid. The absence of an identity element can be easily repaired in thefollowing way.

Proposition 2.2.1 Each semigroup can be extended to a monoid by addingat most one element.

Proof. Let (S, ·) be a semigroup. If S contains an identity, we have nothingto prove. If S does not contain any identity element, consider the set S1 =S ∪ {1}, where 1 �∈ S. Define the binary operation ∗ on S1 as follows: Fora, b ∈ S1 set

a ∗ b =

⎧⎪⎨⎪⎩

a · b, a, b ∈ S;a, b = 1;b, a = 1.

A direct calculation shows that ∗ is associative, hence S1 is a semigroup.Furthermore, from the definition of ∗ we have that 1 is the identity elementin S1. Moreover, the restriction of the operation ∗ to S coincides with theoriginal operation ·. Hence S is a subsemigroup of S1.


Denote by εn : N → N the identity transformation

εn =(

1 2 · · · n1 2 · · · n

).

If n is clear from the context we shall sometimes write ε instead of εn. Thefollowing statement is obvious.

Proposition 2.2.2 The transformation εn is the (two-sided) identity ele-ment in both Tn and PT n. In particular, both, Tn and PT n, are monoids.

Let S be a monoid with the identity element 1. An element a ∈ S is calledinvertible or a unit provided that there exists b ∈ S such that ab = ba = 1.Such an element b, if it exists, is unique. Indeed, assume that b1 and b2 aredifferent such elements, then

b1 = b1 · 1 = b1(ab2) = (b1a)b2 = 1 · b2 = b2.

The element b is called the inverse of a and is denoted by a−1. Note that ifb is the inverse of a, then a is the inverse of b. In other words, (a−1)−1 = a.The set of all invertible elements of the monoid S is denoted by S∗. Notethat 1 ∈ S∗ since 1 · 1 = 1. In particular, S∗ is not empty.

The above terminology and notation deserve some explanation. Usuallythe operation in an abstract semigroup is thought of as a multiplication.Since the element 1 is the identity element in such multiplicative semigroupsas N, Z, R, C, it is natural to denote the identity element of an abstractsemigroup by the same symbol 1. This also justifies the notions “unit” and“inverse.” However, there are many semigroups where the operation is notthe multiplication, for example the semigroup (Z, +). The identity elementin this semigroup is the number 0 and not the number 1. And the inverse ofthe number n ∈ Z is the number −n and not the number n−1 (note that thelatter one is not always defined, and when it is defined, it is not an integerin general).

A monoid in which each element has an inverse is called a group.

Proposition 2.2.3 Let S be a monoid with the identity element 1. ThenS∗ is a group.

Proof. Obviously, if a ∈ S∗, then a−1 ∈ S∗ as well. If a, b ∈ S∗, then we have

ab · b−1a−1 = a · bb−1 · a−1 = a · 1 · a−1 = aa−1 = 1.

Analogously one shows that b−1a−1 · ab = 1 and hence b−1a−1 = (ab)−1. Inparticular, ab ∈ S∗. Thus S∗ is a submonoid of S and each element of S∗

has an inverse in S∗. The claim follows.

Proposition 2.2.4 Let α ∈ Tn, or α ∈ PT n. Then α is invertible if andonly if α is a permutation on N.

2.3. ZERO ELEMENTS 19

Proof. Assume that α is invertible and β is a (partial) transformation suchthat αβ = βα = ε. Note that dom(ε) = N. Hence Exercise 2.1.4(a) impliesdom(α) = N. Further, if x, y ∈ N are such that x �= y, then ε(x) �= ε(y).If α(x) = α(y), we would get ε(x) = β(α(x)) = β(α(y)) = ε(y), a contra-diction. This means that α(x) �= α(y). Hence α is everywhere defined andinjective and thus is a permutation by Proposition 1.1.3.

Conversely, if

α =(

1 2 · · · ni1 i2 · · · in

)

is a permutation, the element

α =(

i1 i2 · · · in1 2 · · · n

)

is a permutation as well and a direct computation shows that αβ = βα = ε,that is, α is invertible.

The group T ∗n = PT ∗

n of all permutations on N is called the symmetricgroup on N and is denoted by Sn.

2.3 Zero Elements

An element 0 of a semigroup S is called a left or a right zero provided that0a = 0, or a0 = 0, respectively, for all a ∈ S. An element 0 which at thesame time is a left and a right zero, is called a two-sided zero or simply azero.

If S contains some left zero 0l and some right zero 0r, we have 0l =0l · 0r = 0r and hence these two elements coincide. Hence in this case Scontains a unique zero element, which is, moreover, a two-sided zero. Theanalog of Proposition 2.2.1 is the following statement.

Proposition 2.3.1 Each semigroup can be extended to a semigroup withzero by adding at most one element.

Proof. Let (S, ·) be a semigroup. If S contains a zero, we have nothing toprove. Otherwise, consider the set S0 = S ∪ {0}, where 0 �∈ S. Define thebinary operation ∗ on S0 as follows: For a, b ∈ S0 set

a ∗ b =

⎧⎪⎨⎪⎩

a · b, a, b ∈ S;0, b = 0;0, a = 0.

A direct calculation shows that ∗ is associative, hence S0 is a semigroup.Furthermore, from the definition of ∗ we have that 0 is the zero element in S0

and that the restriction of ∗ to S coincides with ·. Hence S is a subsemigroupof S0.


Denote by 0n the partial transformation 0n : ∅ → N on N. We havedom(0n) = ∅. If n is clear from the context, we shall usually write simply0 instead of 0n.

Proposition 2.3.2 0n is the zero element of the semigroup PT n.

Proof. The equalities α0n = 0nα = 0n, α ∈ PT n, are obvious.

For a ∈ N define the total transformation 0a : N → N via 0a(x) = a forall x ∈ N. The transformation 0a is called the constant transformation.

Proposition 2.3.3 For n > 1 the semigroup Tn does not contain any rightzeros. α ∈ Tn is a left zero if and only if α = 0a for some a ∈ N. Inparticular, the semigroup Tn contains exactly n left zeros.

Proof. From the obvious equality 0aβ = 0a for any β ∈ Tn we obtain thateach constant transformation on Tn is a left zero. Hence Tn contains at leastn different left zeros. In particular, for n > 1 the semigroup Tn cannotcontain any right zeros.

Note that constant transformations are the only transformations ofrank 1. Let α ∈ Tn be a transformation of rank at least 2. Then for any 0a

the rank of α0a is 1 by Exercise 2.1.4(c) and hence the equality α0a = α isnot possible. This implies that α is not a left zero of Tn.

We note that the semigroup T1 consists of the identity element only. Thiselement is the zero element at the same time.

2.4 Isomorphism of Semigroups

Let (S, ·) and (T, ∗) be two semigroups. The semigroups S and T are saidto be isomorphic (denoted by S ∼= T ) provided that there exists a bijectionϕ : S → T such that

ϕ(a) ∗ ϕ(b) = ϕ(a · b) for all a, b ∈ S. (2.1)

The bijection ϕ is called an isomorphism from S to T . Note that if ϕ : S → Tis an isomorphism, then ϕ−1 : T → S is an isomorphism as well.

Exercise 2.4.1 Show that the relation of being isomorphic semigroups isan equivalence relation.

Example 2.4.2 It is rather handy to present small semigroups using theirmultiplication tables, which is also called Cayley tables. Such table is a squarematrix with |S| rows and |S| columns, which are indexed by the elementsof S. At the intersection of the ath row and the bth column, a, b ∈ S, one

2.4. ISOMORPHISM OF SEMIGROUPS 21

writes the product ab. For example, the semigroup T2 = {ε, (12), 01, 02} hasthe following Cayley table:

· ε (12) 01 02

ε ε (12) 01 02

(1, 2) (1, 2) ε 02 01

01 01 01 01 01

02 02 02 02 02

. (2.2)

The following set of 2 × 2 matrices is also a semigroup with respect to theusual matrix multiplication

S ={(

1 00 1

),

(0 11 0

),

(1 10 0

),

(0 01 1

)}.

To simplify the notation we denote

E =(

1 00 1

), A =

(0 11 0

), B =

(1 10 0

), C =

(0 01 1

).

The Cayley table for S has the following form

· E A B C

E E A B C

A A E C B

B B B B B

C C C C C

. (2.3)

It is easy to check that the bijection

ϕ =(

ε (12) 01 02

E A B C

)

transforms the Cayley table (2.2) to the Cayley table (2.3) and hence is anisomorphism from T2 to S.

The primary importance of the semigroup Tn is revealed by the followingstatement.

Theorem 2.4.3 (Cayley’s Theorem) Each finite semigroup S of cardinalityn is isomorphic to a subsemigroup of either Tn or Tn+1.

Proof. First we assume that S contains the identity element 1. Let S ={a1 = 1, a2, . . . , an} and define the mapping ϕ : S → Tn as follows

ϕ(a) =(

1 2 · · · ni1 i2 · · · in

),


where for k = 1, . . . , n, the number ik is uniquely determined by the equalityaak = aik .

Since ak · 1 = ak, we have ϕ(ai)(1) = i for all i = 1, . . . , n. In particular,ϕ(ai) �= ϕ(aj) if i �= j and hence the mapping ϕ is injective. The mappingϕ : S → ϕ(S) is therefore even bijective.

Further, the equality (ab)ak = a(bak) implies the equality ϕ(ab)(k) =ϕ(a)(ϕ(b)(k)) for all k = 1, . . . , n and a, b ∈ S. This means that ϕ(ab) =ϕ(a)ϕ(b) for all a, b ∈ S and thus ϕ is an isomorphism from S to the sub-semigroup ϕ(S) of Tn.

If S does not contain any identity element, we can consider S as a sub-semigroup of the semigroup S1 and |S1| = n + 1. By the above, S1 is iso-morphic to a subsemigroup of Tn+1, and the claim follows.

2.5 The Semigroup ISn

Obviously, the composition αβ of two partial injections α and β is againa partial injection. This observation leads to the definition of another veryimportant subsemigroup of PT n, namely, the semigroup ISn of all partialinjective transformations of N, which is usually called the symmetric inversesemigroup.

Theorem 2.5.1

|ISn| =n∑

k=0

(n

k

)2

· k!.

Proof. Each partial injection α : N → N can be considered as a bijectionα : dom(α) → im(α). Let us count the number of such bijections of rank k.The set A = dom(α) can be chosen in

(nk

)different ways. The set B = im(α)

can be independently chosen in(nk

)different ways. If A and B are fixed,

there are exactly k! different bijections from A to B. Hence we have exactly(nk

)·(nk

)·k! bijections of rank k. Since k can be an arbitrary integer between

0 and n, the statement of the theorem is obtained by applying the sumrule.

Partial injections α : N → N are also called partial bijections or partialpermutations on N.

If α : N → N is a partial injection, then each x ∈ N has at mostone preimage. In particular, the graph Γα is much simpler than for generalpartial transformations. From Theorem 1.2.9 it follows that each connectedcomponent of Γα is either a cycle or has the form

a1 �� a2 �� a3 �� . . . �� ak−1 �� ak . (2.4)

In our linear notation such a graph is denoted by

[[[. . . [[[a1]; a2]; a3]; . . . ]; ak−1]; ak].

2.6. REGULAR AND INVERSE ELEMENTS 23

Since the order of the brackets in this expression is fixed, we can simplify thenotation by omitting all inner brackets. So, in what follows the graph (2.4)(and the corresponding element) will be denoted by [a1, a2, a3, . . . , ak−1, ak].Such element is called a chain. This allows us to use for partial permutationsa simplified version of linear notation, which we call the chain-cycle notation.For example, the partial permutation

α =(

1 2 3 4 5 6 7 8 9 10 11 12 13 14 1514 12 11 13 7 ∅ 3 15 ∅ ∅ 5 2 1 9 10

),

written in the chain-cycle notation has the following form:

α = (5, 7, 3, 11)(2, 12)[4, 13, 1, 14, 9][8, 15, 10][6].

Obviously the chain-cycle notation for a partial permutation α is unique upto a permutation of components and cyclic permutation of elements in eachcycle. Note that no permutations of elements in chains are allowed.

As ISn contains the identity element ε and the zero element 0 of thebigger semigroup PT n, these elements will be the identity element and thezero element of ISn, respectively. Moreover, we have inclusions Sn ⊂ IS∗

n ⊂PT ∗

n = Sn, which imply IS∗n = Sn. The following diagram characterizes the

connection between the principal objects of the present book:

PT n

Tn

� �

��ISn

� �

��

Sn

� �

��

��

Note that all these inclusions are proper for n > 1.

2.6 Regular and Inverse Elements

An element a of a semigroup S is called regular provided that there existsb ∈ S such that aba = a. Elements a, b ∈ S form a pair of inverse elementsprovided that aba = a and bab = b. Set

VS(a) = {b ∈ S : aandbis a pair of inverse elements}.

If a and b is a pair of inverse elements, then one says that a is an inverseof b and b is an inverse of a. This might be slightly confusing, since in suchgenerality the inverse element of a given element is not uniquely defined.Note that this notion extends the notion of an inverse element for invertibleelements: if a ∈ S is invertible, then aa−1a = a and a−1aa−1 = a−1. Hencea and a−1 is a pair of inverse elements.


Exercise 2.6.1 Let a ∈ S be invertible. Show that VS(a) = {a−1}.

If a has at least one inverse (i.e., VS(a) �= ∅), then the element a isobviously regular. The converse is also true.

Proposition 2.6.2 Let a ∈ S be regular and b ∈ S be such that aba = a.Then a and c = bab is a pair of inverse elements.

Proof. Follows from the following computation

aca = a · bab · a = aba · ba = aba = a

cac = bab · a · bab = b · aba · bab = b · aba · b = bab = c.

The semigroup S is called regular provided that every element of S isregular. Obviously, each group is a regular semigroup.

Theorem 2.6.3 The semigroups Tn, PT n, and ISn are regular.

Proof. Let α ∈ PT n. Let us construct the element β ∈ Tn as follows: Foreach x ∈ im(α) take some y ∈ N such that α(y) = x and set β(x) = y. Forx �∈ im(α) define β(x) = 1. A direct calculation shows that αβα = α. Itfollows that both Tn and PT n are regular semigroups.

Let now α ∈ ISn. Define β ∈ ISn as follows: First of all we set dom(β) =im(α), and for each x ∈ im(α) we define β in the same way as before.Since α was a partial bijection, that is, a bijection from dom(α) to im(α),the element β is simply the inverse bijection from im(α) to dom(α). Inparticular, β ∈ ISn. As before, we still have the equality αβα = α andhence the semigroup ISn is regular as well.

Let us study the structure of inverse elements in the semigroups Tn andPT n in more detail.

Theorem 2.6.4 Let α, β ∈ PT n. Then the element β is inverse to the ele-ment α if and only if the following conditions are satisfied:

(a) dom(β) ⊃ im(α).

(b) β(a) ∈ {x ∈ N : α(x) = a} for all a ∈ im(α).

(c) im(β) = β(im(α)).

Proof. Let im(α) = {a1, a2, . . . , ak}. For i = 1, . . . , k set Bi = {x ∈ N :α(x) = ai}. Then the equality αβα = α is equivalent to the following con-dition:

β(ai) ∈ Bi for alli = 1, 2, . . . , k. (2.5)

Assume now that the condition (2.5) is satisfied. Then for all i we have(αβ)(ai) = ai and (βαβ)(ai) = β(ai). Set B = {β(y) : y ∈ im(α)}. From

2.6. REGULAR AND INVERSE ELEMENTS 25

the previous equality we have (βα)(b) = b for all b ∈ B. As im(βαβ) ⊂ B,the equality βαβ = β requires im(β) ⊂ B. However, the latter condition isalso sufficient. Indeed, as β(y) ∈ B for any y ∈ dom(β), we have (βαβ)(y) =(βα)(β(y)) = β(y) and thus βαβ = β. This completes the proof.

Theorem 2.6.4 is also true for the semigroup Tn, just in this case thecondition (a) is superfluous since it is automatically satisfied.

Corollary 2.6.5 Let α ∈ PT n, im(α) = {a1, a2, . . . , ak}, Bi = {x ∈ N :α(x) = ai} and mi = |Bi|. Then

(i) |VPT n(α)| = m1m2 · · ·mk · (k + 1)n−k;

(ii) if α is total, then |VTn(α)| = m1m2 · · ·mk · kn−k.

Proof. For each ai ∈ im(α) we have to choose β(ai) ∈ Bi. As for different ithese choices are independent, we have m1m2 · · ·mk different ways to defineβ on im(α).

After this we have to define β on the set N\im(α). From Theorem 2.6.4it follows that the restriction of β to N\im(α) is an arbitrary (partial in thecase of PT n) mapping from N\im(α) to the set {β(y) : y ∈ im(α)}. Thelatter set contains exactly k elements. Hence the restriction of β to N\im(α)can be chosen in kn−k different ways for Tn and in (k +1)n−k different waysfor PT n.

Exercise 2.6.6 Show that for any a ∈ N we have

(a) |VTn(0a)| = n.

(b) |VPT n(0a)| = n · 2n−1.

A regular semigroup S is called an inverse semigroup provided that eacha ∈ S has a unique inverse element. This inverse element is usually denotedby a−1 (this can now be justified by the requirement that it is unique). FromCorollary 2.6.5 it follows that in the case of n > 1 the semigroups Tn andPT n are not inverse semigroups.

Theorem 2.6.7 The semigroup ISn is an inverse semigroup.

Proof. Let α ∈ ISn and β ∈ VPT n(α). Let us try to analyze in whichcase β is a partial permutation. Since α is injective, the sets Bi from theproof of Theorem 2.6.4 consist of one element each. Hence β is uniquelydefined on im(α) by Theorem 2.6.4(b). At the same time, the injectivity ofβ and Theorem 2.6.4(c) imply that β must be undefined on N\im(α). ThusVPT n(α) ∩ ISn contains a unique element, implying that ISn is an inversesemigroup.


Since the notion of an inverse element is a natural extension of the corre-sponding notion for invertible elements, inverse semigroups are very naturalgeneralizations of groups. In some sense they form the class of semigroups,which is “closest” to groups.

Exercise 2.6.8 Prove the following identities in any inverse semigroup

(a) (a−1)−1 = a,

(b) (ab)−1 = b−1a−1.

2.7 Idempotents

Identities and zero elements in semigroups are special cases of a more generalnotion. An element e of a semigroup S is called an idempotent provided thate2 = e. The set of all idempotents of S is denoted by E(S).

Theorem 2.25 describes idempotents in PT n. To formulate it we needsome notation. If α ∈ PT n and B ⊂ N, one says that B is invariant withrespect to α provided that α(x) ∈ B for all x ∈ B ∩ dom(α). In case B isinvariant with respect to α we can define the restriction α|B of α to B asfollows: α|B is a partial transformation on B, dom(α|B) = B ∩ dom(α), andfor x ∈ dom(α|B) we have α|B(x) = α(x).

Exercise 2.7.1 Show that for each α ∈ PT n the sets N, ∅, and im(α) areinvariant with respect to α.

Theorem 2.7.2 α ∈ PT n is an idempotent if and only if im(α) ⊂ dom(α)and the restriction α = α|im(α) is the identity transformation on im(α).

Proof. If α2 = α, for all x ∈ dom(α) we have α(x) = α2(x) = α(α(x)).Hence im(α) ⊂ dom(α) and for each y ∈ im(α) we have α(y) = y.

Conversely, if α acts as the identity on im(α), we have α2(x) = α(x) forany x ∈ dom(α). Since dom(α2) ⊂ dom(α), it follows that α2 = α.

Corollary 2.7.3 The element α ∈ ISn is an idempotent if and only if αis the identity transformation of some A ⊂ N. In particular, ISn containsexactly 2n idempotents.

Proof. For α ∈ ISn we have |im(α)| = |dom(α)| by the injectivity of α.Hence im(α) ⊂ dom(α) in this case implies im(α) = dom(α). The condi-tion that α|im(α) is the identity transformation means exactly that α is theidentity transformation on im(α). Since N has exactly 2n subsets, the claimfollows.

For A ⊂ N we denote by εA the unique idempotent of ISn such thatdom(εA) = A. We have ε = εN and 0 = ε∅.

2.7. IDEMPOTENTS 27

Corollary 2.7.4 The number un of idempotents in the semigroup Tn equals

un =n∑

k=1

(n

k

)kn−k.

Proof. To define an idempotent α of rank k we have to choose a k-elementset im(α) (this can be done in

(nk

)different ways), and then we have to define

a mapping from N\im(α) to im(α) in an arbitrary way (this can be donein kn−k different ways). Hence Tn contains exactly

(nk

)kn−k idempotents of

rank k. The statement is now obtained applying the sum rule.

Corollary 2.7.5 The number u′n of idempotents in the semigroupPT n equals

u′n =

n∑k=0

(n

k

)(k + 1)n−k.

Proof. The difference with the proof of Corollary 2.7.4 is just the facts thatPT n contains idempotents of rank 0, and that the mapping from N\im(α)to im(α) may be partial.

There is a very close connection between the idempotents and the regularelements. Indeed, each idempotent e is a regular element as e · e · e = e.Conversely, if a is a regular element and aba = a, then the elements ab andba are idempotents: ab · ab = aba · b = ab and ba · ba = b · aba = ba. Interms of idempotents, the inverse semigroups can be detected in the set ofall regular semigroups using the following:

Theorem 2.7.6 A regular semigroup S is inverse if and only if all idem-potents of S commute.

Proof. Let S be a regular semigroup in which all idempotents commute. Letfurther a ∈ S and b1, b2 ∈ S be two inverse elements to a. Then ab1a = a,b1ab1 = b1, ab2a = a, b2ab2 = b2, and as we mentioned above, the elementsab1, ab2, b1a, and b2a are idempotents. Hence we have

b1 = b1ab1 = b1ab2ab1 = b1 · ab2 · ab1 = b1 · ab1 · ab2 = b1ab2 == b1ab2ab2 = b1a · b2a · b2 = b2a · b1a · b2 = b2 · ab1a · b2 = b2ab2 = b2.

This means that each a ∈ S has a unique inverse and thus S is inverse.Conversely, let S be an inverse semigroup and e, f ∈ S be two idempo-

tents. Let us first prove that their product ef is an idempotent. Considera = (ef)−1. We have

ef · a · ef = ef, a · ef · a = a.

Using this we obtain the following equalities:

ef · ae · ef = ef · a · e2f = ef · a · ef = ef,

ae · ef · ae = a · e2f · ae = a · ef · a · e = ae.


Hence ae is also an inverse to ef , which means ae = a. Analogously oneshows that fa = a. Thus

a2 = ae · fa = a · ef · a = a,

and hence a is an idempotent. In particular, a−1 = a, and thus a = ef .Analogously one shows that fe is an idempotent as well. Moreover, we

have

fe · ef · fe = f · e2 · f2 · e = fe · fe = fe,

ef · fe · ef = e · f2 · e2 · f = ef · ef = ef.

This means that (ef)−1 = fe. However, since ef is an idempotent, we alsohave (ef)−1 = ef . Thus fe = ef and we are done.

A semigroup S is called commutative or abelian if ab = ba for all a, b ∈ S.From Theorem 2.7.6 it follows that the set E(S) of idempotents of an inversesemigroup S is a commutative subsemigroup of S. In particular, E(ISn) isa commutative semigroup of order 2n.

Exercise 2.7.7 Prove that εA · εB = εA∩B for all A, B ⊂ N and use thisto show that the semigroup E(ISn) is isomorphic to the semigroup of allsubsets of N with respect to the operation of the intersection of subsets.

Exercise 2.7.8 (a) Show that E(T2) is a noncommutative subsemigroupof T2.

(b) Show that E(Tn) is not a subsemigroup of Tn for all n > 2.

Exercise 2.7.9 Show that E(PT n) is not a subsemigroup of PT n for alln > 1.

We know already that for n > 1 the semigroups Tn and PT n are notinverse. This fact also follows from Theorem 2.7.6 and Exercises 2.7.8 and2.7.9.

2.8 Nilpotent Elements

An element a of a semigroup S with the zero 0 is called nilpotent or a nil-element provided that ak = 0 for some k ∈ N. The minimal k for whichak = 0 is called the nilpotency degree or nilpotency class of the element aand is denoted by nd(a).

Proposition 2.8.1 α ∈ PT n is nilpotent if and only if the graph Γα doesnot contain cycles.

2.8. NILPOTENT ELEMENTS 29

Proof. If Γα contains a cycle, say (a1, a2, . . . , am), then a1 ∈ dom(αk) for allk > 0 (it is easy to see that αk(a1) = a(1+k)mod m). As the zero element 0of PT n satisfies dom(0) = ∅, we have αk �= 0 for all k > 0 and hence theelement α cannot be nilpotent.

If Γα does not contain any cycle, the trajectory of each vertex a mustbreak. More precisely, this trajectory has the form a0 = a, a1, a2, . . . , am,where all ai are pairwise different and Γα does not contain any arrow withthe tail am (i.e., am �∈ dom(α)). This means that αm(a) = am and thata �∈ dom(αm+1) since otherwise we would have αm+1(a) = α(am), whichdoes not make sense since am �∈ dom(α). Since the length of any trajectorydoes not exceed n, we have dom(αn) = ∅ and thus α is nilpotent.

From the proof of Proposition 2.8.1 we, in fact, derive the following:

Proposition 2.8.2 The nilpotency degree of a nilpotent element α ∈ PT n

equals the length of the longest trajectory in Γα.

Corollary 2.8.3 The partial permutation

α = (a1, . . . , ak) · · · (b1, . . . , bl)[c1, . . . , cp] · · · [d1, . . . , dq] ∈ ISn

is nilpotent if and only if it does not contain cycles. If α ∈ ISn is nilpotent,the nilpotency degree of α equals the maximum max(p, . . . , q) of the lengthsof all chains.

Denote by S(n, k) the Stirling number of the second kind, that is, thenumber of partitions N = A1 ∪ A2 ∪ · · · ∪ Ak into an unordered disjointunion of k nonempty blocks.

Theorem 2.8.4 ([LU1]) The semigroup PT n contains(nk

)· S(n, k + 1) · k!

nilpotent elements of rank k.

Proof. We have the unique nilpotent element of rank 0, namely, the element0 itself. On the other hand,

(n0

)· S(n, 1) · 0! = 1. Hence for the rest of the

proof we may assume k ≥ 1.Let α ∈ PT n be a nilpotent element of rank k. Assume that im(α) =

{a1, a2, . . . , ak}. Then the sets Ai = {x ∈ N : α(x) = ai}, i = 1, . . . , k, andAk+1 = dom(α) form a partition N = A1 ∪ A2 ∪ · · · ∪ Ak+1 of the set Ninto k +1 block (note that Ak+1 �= ∅ since α is nilpotent). This observationsuggests the following procedure for constructing all nilpotent elements ofrank k (we construct some nilpotent element α): First we choose im(α) ={a1, . . . , ak}. This is a k-element subset of N and hence it can be chosen in(nk

)different ways. Then we choose a partition, N = B1 ∪ B2 ∪ · · · ∪ Bk+1

into an unordered disjoint union of nonempty blocks. This can be done inS(n, k + 1) different ways. Now for each ai we have to choose its preimage,which is one of the blocks B1, . . . , Bk+1, such that the resulting element is


nilpotent. The choice of the preimage for ai means that we choose somearrows for Γα in such a way that the resulting graph does not contain anycycles.

The preimage of a1 can be any of the blocks B1, . . . , Bk+1, which doesnot contain a1 (since otherwise Γα would contain a loop at a1, which is acycle of length 1). This means that the preimage of a1 can be chosen in kdifferent ways.

We proceed by induction. Assume that we have already chosen the preim-ages for a1, . . . , am (and let these preimages be the blocks Bi1 , . . . , Bim , re-spectively). Denote X = Bi1 ∪ · · · ∪ Bim and consider two different cases.

Case 1: am+1 �∈ X. This means that there is no arrow with the tail am+1

yet. Hence the preimage of am+1 could be any of the remaining blocks, whichdoes not contain am+1. This preimage can be chosen in (k+1−m)−1 = k−mdifferent ways.

Case 2: am+1 ∈ X. Consider the trajectory a(0) = am+1, a(1), . . . , a(t) of

am+1 in that part of Γα, which is already constructed. Since the constructedpart of Γα does not contain cycles, this trajectory breaks at a(t) �∈ X. Inorder to ensure that the choice of the preimage for am+1 does not create acycle in Γα, it is necessary and sufficient to make sure that this preimagedoes not contain a(t). Hence we again have (k + 1−m)− 1 = k−m ways tochoose the preimage for am+1.

From the above we have that for fixed {a1, . . . , ak} and N = B1 ∪ B2 ∪· · ·∪Bk+1 the element α can be constructed in k(k−1) · · · (k− (k−1)) = k!different ways. Hence the total number of nilpotent elements of rank k equals(nk

)· S(n, k + 1) · k!.

Theorem 2.8.5 The number of nilpotent elements of defect k in the semi-group ISn equals the signless Lah number L′

n,k = n!k!

(n−1k−1

).

Proof. By Corollary 2.8.3, a nilpotent partial permutation of defect k hasthe following form:

[i1, i2, . . . , im1 ][im1+1, im1+2, . . . , im2 ] · · · [imk−1+1, imk−1+2, . . . , imk].

To get the above expression from the permutation i1, i2, . . . , in of 1, 2, . . . , n,it is enough to choose the ends im1 , im2 , . . . , imk−1

of the first k − 1 chains(as mk = n automatically). This can be done in

(n−1k−1

)different ways. Going

through all permutations we will get chain-cycle notation for all nilpotentelements of defect k. Since the order of chains in the chain-cycle notationis not important (because all chains in the chain-cycle notation commute),every nilpotent element of defect k will be counted k! times. Indeed, thechains in each chain-cycle notation can be permuted in k! different wayswithout changing the corresponding partial transformation of N; however,each of these permutations of chains corresponds to a different permutation


of N. This means that n! permutations of N give k! repetitions of eachnilpotent element. Hence the number of nilpotent elements of defect k equalsn!k!

(n−1k−1

).

Corollary 2.8.6 The semigroup ISn contains

n∑k=1

n!k!

(n − 1k − 1

)

nilpotent elements.


2.9.1 A subset M ⊂ N is called invariant with respect to S ⊂ PT n providedthat M is invariant with respect to each α ∈ S. Obviously, the union andthe intersection of invariant subsets is an invariant subset. Further, for eachα ∈ PT n and x ∈ N the set of vertices of trα(x) is invariant with respectto α.

Let α ∈ PT n. If there exists a partition N = N1 ∪ N2 of N into adisjoint union of two invariant subsets, then the study of the action of α onN reduces to the study of the action of α on N1 and N2. Continuing thisprocedure, the study of the action of α on N reduces to the study of theaction of α on all its orbits:

Proposition 2.9.1 Each orbit of α is invariant with respect to α. If N =N1 ∪ N2 is a partition into invariant subsets, then each of them is a unionof orbits.

Hence the partition of N into orbits is the finest possible partition of Ninto a disjoint union of nonempty invariant subsets.

2.9.2 Let α ∈ PT n and let N = N1 ∪ N2 be a partition into invariantsubsets. For each i = 1, 2 set

α(i)(x) =

{α(x), x ∈ Ni,

x, otherwise.

In other words, α(i) acts on Ni in the same way as α, and α(i) acts on thecomplement N1−i as the identity transformation. It is easy to see that

α = α(1)α(2) = α(2)α(1).

Analogous transformations α(i) and the corresponding decompositionα = α(1)α(2) · · ·α(m) can be defined for any partition N = N1∪N2∪· · ·∪Nm

of N into a disjoint union of arbitrarily many invariant subsets. In particular,if some Ni is an orbit, the graph of the transformation α(i) has the following


form: one of its components coincides with the corresponding component ofΓα (the one which corresponds to the orbit Ni), and all other componentsconsist of loops. In some sense α(i) can be identified with the correspondingconnected component. Then the decomposition of α described in Sect. 1.3can be interpreted as a decomposition of α into a product of its connectedcomponents. In particular, for permutation we get the usual decompositioninto a product of independent cycles.

2.9.3 Let us interpret [a1, a2, . . . , ak|ai] as the notation for the element fromTn, which acts as the element

(a1 a2 · · · ak−1 ak

a2 a3 · · · ak ai

)

on the set A = {a1, a2, . . . , ak}, and as the identity on the set N\A. Thenthe linear notation

μ = [a1, a2, . . . , ak|ai][b1, b2, . . . , bl|bj ] · · · [c1, c2, . . . , cm|ch],

described in 1.4.3 can be considered as the following decomposition of μ:

μ = [c1, c2, . . . , cm|ch] · · · [b1, b2, . . . , bl|bj ][a1, a2, . . . , ak|ai]

(note that the factors should be taken in the reverse order). In the gen-eral case, the factors of this decomposition do not correspond to connectedcomponents, the factors themselves (and even the number of these factors)depend on the choice of certain elements along the way to write down thisdecomposition, and these factors do not commute. This is another disadvan-tage of the notation from [AAH].

2.9.4 Let α ∈ PT n. As each orbit of α is invariant with respect to αand the intersection of invariant sets is invariant, each minimal (with re-spect to inclusions) nonempty invariant set is contained in some orbit. FromTheorem 1.2.9 it follows that the trajectory of each vertex from a fixed con-nected component either contains a cycle (if it exists in this component) orterminates at components sink. On the other hand, the set of all verticesin a cycle, and the set consisting of the sink are obviously invariant. Henceonly these sets are minimal (with respect to inclusions) nonempty invariantsets.

Let now K be an orbit of α. If the corresponding connected componentof Γα contains a cycle, the set of all vertices of this cycle is called the kernelof the orbit K. If K does not contain any cycle, we say that the kernel ofthis orbit is empty.

Proposition 2.9.2 Let α ∈ PT n, R be an orbit of α containing a cycle andK be the kernel of R. Assume |K| = k > 0. Then


(i) αk(x) = x for all x ∈ K.

(ii) If y ∈ R and m > 0 are such that αm(y) = y, then y ∈ K, k|m andK = {y, α(y), . . . , αk−1(y)}.

2.9.5 Let α ∈ PT n. The stable image of α is the set

stim(α) =⋂k∈N

im(αk).

Proposition 2.9.3 For each α ∈ PT n the set stim(α) is invariant withrespect to α. The restriction of α to stim(α) is a permutation, moreover,stim(α) is the maximum subset of N (with respect to inclusions) such thatthe restriction of α to this subset is defined and is a permutation.

It is easy to see that stim(α) is the union of kernels of all orbits of α.The restriction of α to stim(α) is called the permutational part of α.

2.9.6 Let α ∈ PT n. The stable rank of α is defined as the following number:strank(α) = mink∈N rank(αk).

Proposition 2.9.4 strank(α) = |stim(α)|.

2.9.7 The notation for the elements of ISn, introduced in Sect. 2.5, is closeto the notation used in [Li].

2.9.8 The role of ISn in the theory of inverse semigroups is analogous to therole of Sn in group theory, and to the role of Tn in semigroup theory. Namely,we have the following analog of Cayley’s Theorem for inverse semigroups:

Theorem 2.9.5 (Preston–Wagner) Any inverse semigroup T is isomorphicto a subsemigroup of the semigroup IS(T ) of all partial injective transfor-mations of the set T .

The idea of the proof is as follows: for each a ∈ T we consider thetransformation ρa : a−1T → aT , defined via ρa(x) = ax. This transfor-mation turns out to be bijective. Hence we can consider ρa as a partialpermutation on T . For the mapping ϕ : T → IS(T ), ϕ(a) = ρa, one eas-ily checks that ϕ(ab) = ϕ(a)ϕ(b) and that ϕ is injective. For details, see[CP1, Theorem 1.20].

2.9.9 If (S, ·) is a semigroup, we can consider the new semigroup (S, ∗),where a ∗ b = b · a. The semigroup (S, ∗) is called the opposite semigroup

or the dual semigroup or simply the dual of S. It is usually denoted by←S .

Obviously, the second dual of S is isomorphic to S. If←S ∼= S, the semigroup

S is called self-dual.


If S is an inverse semigroup, from Exercise 2.6.8 it follows that the map-

ping a → a−1 is an isomorphism from S to←S . Hence all inverse semigroups

are self-dual, in particular, the semigroup ISn is self-dual.However, not any semigroup is self-dual. For example, for n > 1 the

semigroup Tn is not self-dual. Indeed, we know that Tn contains left zeros

but not right zeros (see Proposition 2.3.3). Hence←Tn contains right zeros

but no left zeros. At the same time each isomorphism must map left zerosto left zeros, and right zeros to right zeros.

2.9.10 The number of nilpotent elements in PT n is given by the followingstatement.

Theorem 2.9.6 ([LU1]) PT n contains exactly (n + 1)n−1 nilpotent ele-ments.

Proof. For α ∈ PT n and k ≥ 0 let Nk denote the set of all x ∈ N forwhich the trajectory of x breaks at the kth step, that is, has the formx0 = x, x1, . . . , xk. Assume that the nilpotency degree of α equals t. In thiscase from the proof of Proposition 2.8.1 it follows that the sets N0, N1, . . . , Nt

are not empty and form a partition of N. Moreover, α(Ni) ⊂ Ni−1 for all i.Hence each nilpotent element of nilpotency degree t can be obtained in thefollowing way: First we choose an ordered partition N = N0∪N2∪· · ·∪Nt ofN into t+1 nonempty blocks (this can be done in

(n

n0,n1,...,nt

)different ways,

where ni = |Ni|, i = 0, . . . , t). Then for each i > 0 we define some mappingfrom Ni to Ni−1 (altogether such mappings can be defined in nn1

0 nn21 · · ·nnt

t−1

different ways).The total number of nilpotent elements in PT n is denoted by n(PT n).

By the above, we have

n(PT n) =∑

1≤t<nn0+n1+···+nt=n

(n

n0, n1, . . . , nt

)nn1

0 nn21 · · ·nnt

t−1 =

=∑

1≤t<nn0+n1+···+nt=n

n!n0!

· nn10

n1!· nn2

1

n2!· · ·

nntt−1

nt!.

Using (1.8) we have

n(PT n) =n∑

n0=1

n!(n0 − 1)!

∑1≤t<n−n0

n1+···+nt=n−n0

1n0

· nn10

n1!· nn2

1

n2!· · ·

nntt−1

nt!=

=n∑

n0=1

n!(n0 − 1)!

· nk−1

k!=

n−1∑k=0

(n − 1)!(n − 1 − k)!k!

· nk = (n + 1)n−1.


In [LU1] Theorem 2.9.6 is proved in a different way. One more way toprove it is to use [Hi1, Lemma 6.1.4]. Yet another way to prove this theoremis to use Cayley’s theorem on the number of labeled trees: If we have a tree,say Γ on the set {1, 2, . . . , n + 1} of vertices, we can consider it as a rootedtree with the root (n + 1). In particular, we have a well-defined notion of adistance of a vertex to the root. Now we can define the nilpotent elementα from PT n as follows: If i is a vertex, let j denote the unique vertex ofΓ such that (i, j) is an edge and whose distance to the root is smaller thanthat of i. Define α(i) = j if j is not the root and α(i) = ∅ otherwise. Oneverifies that the correspondence Γ → α is a bijection from the set of alllabeled trees on {1, 2, . . . , n+1} to the set of all nilpotent elements in PT n.

From Theorem 2.9.6 it follows that

n(PT n)|PT n|

=(n + 1)n−1

(n + 1)n=

1n + 1

.

From Theorems 2.8.4 and 2.9.6 we get the following identity.

Corollary 2.9.7

n−1∑k=0

(n

k

)S(n, k + 1)k! = (n + 1)n−1.


2.10.1 Let S be a nonempty set such that 0 ∈ S. For all a, b ∈ S seta · b = 0. Show that (S, ·) is a semigroup (it is called a semigroup with zeromultiplication or a null semigroup).

2.10.2 Let S be a nonempty set. For a, b ∈ S define a · b = b. Show that(S, ·) is a semigroup, each element of which is at the same time a left identityand a right zero (such S is usually called a semigroup of left identities ora right zero semigroup; analogously, using a · b = a one defines a left zerosemigroup or a semigroup of right identities).

2.10.3 Show that Tn+1 contains a subsemigroup, isomorphic to the semi-group PT n.

2.10.4 Let X be a nonempty set and B(X) be the set of all subsets of X (theBoolean of X). Show that both (B(X),∪) and (B(X),∩) are commutativesemigroups. Furthermore, show that these two semigroups are isomorphic.

2.10.5 Let K be an orbit of some transformation α ∈ PT n. Prove that thekernel of K coincides with the set

∩x∈K{αm(x) : m ≥ 0}.


2.10.6 Let α ∈ Tn. Prove that x ∈ N belongs to the kernel of an orbit K ifand only if {y : αk(y) = xfor somek > 0} = K.

2.10.7 Let α ∈ PT n. Characterize stim(α) and strank(α) in terms of Γα.

2.10.8 Let α ∈ PT n.

(a) Prove that stim(α) is invariant with respect to αk for each k > 0.

(b) Let β ∈ PT n be such that α = βk for some k > 0. Prove that stim(α)is invariant with respect to β.

2.10.9 ([HS]) Denote by un the number of idempotents in Tn. For a primep show that

(a) up+1 ≡ 2 mod p;

(b) up+2 ≡ 7 mod p.

2.10.10 ([HS]) Let un be as in 2.10.9 and set u0 = 1. Show that

un+1 =n∑

k=0

(n

k

)(k + 1)un−k.

2.10.11 Show that for each α ∈ PT n the set VPT n(α) · α is a left zerosemigroup and the set α · VPT n(α) is a right zero semigroup.

2.10.12 Prove that each regular semigroup with only one idempotent is agroup.

2.10.13 Prove that the operations of extending a semigroup with the el-ement 1 (as in Proposition 2.2.1) and with the element 0 (as in Proposi-tion 2.3.1) commute, that is, (S1)0 = (S0)1 (note the equality and not theisomorphism sign).

2.10.14 Prove that the set A × B with respect to the operation (a1, b1) ∗(a2, b2) = (a1, b2) is a semigroup in which each pair of elements is a pair ofinverse elements (such a semigroup is called a rectangular band).

2.10.15 Show that the elements

α = (a1, . . . , ak) · · · (b1, . . . , bl)[c1, . . . , cp] · · · [d1, . . . , dq],β = [dq, . . . , d1] · · · [cp, . . . , c1](bl, . . . , b1) · · · (ak, . . . , a1)

of the semigroup ISn form a pair of inverse elements.

2.10.16 Prove that for n > 1 the set of left zeros in the semigroup Tn formsa noncommutative semigroup.


2.10.17 Prove or disprove the following statements:

(a) If the idempotents a and b commute, their product ab is an idempotent.

(b) If the product ab of two idempotents, a and b, is an idempotent, then aand b commute.

2.10.18 ([GH1]) Prove that

(a) ISn contains n! nilpotent elements of defect 1,

(b) PT n contains n! nilpotent elements of defect 1.

2.10.19 ([LU1]) Let Nn denote the total number of nilpotent elements inthe semigroup ISn. Prove that Nn = |ISn| − n|ISn−1|, n > 1.

2.10.20 ([BRR]) Prove the following recursive relation (for n > 2):

|ISn| = 2n|ISn−1| − (n − 1)2|ISn−2|.

2.10.21 ([GM, Theorem 9]) For Nn as in 2.10.19 show that

limn→∞

Nn

|ISn|= 0.

2.10.22 For α ∈ PT n denote by c(α) the number of connected componentsof the graph Γα. Prove the following:

(a)1

|Sn|∑

α∈Sn

c(α) = 1 +12

+ · · · + 1n

.

(b) ([Hi1, Lemma 6.1.12])

1|Tn|

∑α∈Tn

c(α) =n∑

k=1

n!k(n − k)!nk

.

(c) ([GM5, Corollary 1])

∑α∈ISn

c(α) =n∑

k=1

(1 +

1k

)|ISn−k|n(n − 1) · · · (n − k + 1).

2.10.23 Prove that the semigroup PT n is not self-dual for n > 1.

2.10.24 (a) Let α, β ∈ Tn. Show that we either have SnαSn = SnβSn, orSnαSn ∩ SnβSn = ∅.


(b) For α ∈ Tn let tk(α) be as in Exercise 1.5.12. Set

t(α) = (t0(α), t1(α), . . . , tn(α))

and call this vector the type of α. Show that SnαSn = SnβSn if and onlyif t(α) = t(β).

2.10.25 (a) Let α, β ∈ PT n. Show that we either have SnαSn = SnβSn, orSnαSn ∩ SnβSn = ∅.

(b) For α ∈ PT n let tk(α) be as in Exercise 1.5.13. Set

t(α) = (t0(α), t1(α), . . . , tn(α))

and call this vector the type of α. Show that SnαSn = SnβSn if and onlyif t(α) = t(β).

Chapter 3

Generating Systems

3.1 Generating Systems in T n , PT n , and ISn

Let S be a semigroup and A ⊂ S be a set. An element s ∈ S is said tobe generated by A provided that s can be written as a finite product ofelements from A. The set of all elements from S generated by A is usuallydenoted by 〈A〉. If S is a monoid, we will understand the identity element ofthis monoid as a trivial finite product of elements from A (of length 0). Thus1 ∈ 〈A〉 for any A ⊂ S. A subset A of the semigroup S is called a generatingsystem or a generating set for S provided that 〈A〉 = S. A generating systemis called irreducible provided that each proper subset of A is no longer agenerating system.

One of the first most natural questions in the study of some semigroupis to describe some (irreducible) generating system and to classify all (irre-ducible) generating systems.

Until the end of this section we let S denote one of the semigroups Tn,PT n, or ISn. Let us recall that each of these semigroups contains Sn as thegroup of units (see Proposition 2.2.4).

Lemma 3.1.1 (i) Each generating system of S must contain a generatingsystem of Sn.

(ii) If A is an irreducible generating system of S, then A ∩ Sn is an irre-ducible generating system of Sn.

Proof. Sn coincides with the set of all elements of S of rank n. ByExercise 2.1.4(c), an element of rank n can be a product of elements ofrank n only. This implies both (i) and (ii).

Lemma 3.1.2 Each generating system of S must contain at least one ele-ment of rank (n − 1).

Proof. The semigroup S contains elements of rank (n − 1). The set of ele-ments of rank n coincides with Sn and is closed under composition. Hence



40 CHAPTER 3. GENERATING SYSTEMS

from the inequality of Exercise 2.1.4(c) it follows that an element of rank(n − 1) can be written as a product of elements of rank (n − 1) and n only.Moreover, an element of rank (n − 1) cannot be written as a product ofelements of rank n. The claim follows.

Theorem 3.1.3 Let A be a generating system of Tn. Then this system is ir-reducible if and only if A = A1 ∪ {α}, where A1 is an irreducible generatingsystem of Sn and rank(α) = n − 1.

Proof. Taking Lemmas 3.1.1 and 3.1.2 into account, it is enough to showthat any set of the form A = A1∪{α}, where A1 is an irreducible generatingsystem of Sn and rank(α) = n − 1 is a generating system of Tn.

As A1 is an irreducible generating system of Sn, we have Sn = 〈A1〉 ⊂〈A〉. Let us first show that 〈A〉 contains all elements of Tn of rank (n − 1).Assume that

α =(

1 2 · · · i − 1 i i + 1 · · · j − 1 j j + 1 · · · na1 a2 · · · ai−1 a ai+1 · · · aj−1 a aj+1 · · · an

)

and let

β =(

1 2 · · · k − 1 k k + 1 · · · l − 1 l l + 1 · · · nb1 b2 · · · bk−1 b bk+1 · · · bl−1 b bl+1 · · · bn

)

be an arbitrary element from Tn of rank (n− 1). Let π ∈ Sn be an arbitrarypermutation satisfying π(k) = i and π(l) = j. Then

απ =(

1 2 · · · k − 1 k k + 1 · · · l − 1 l l + 1 · · · nc1 c2 · · · ck−1 a ck+1 · · · cl−1 a cl+1 · · · cn

)

is an element of rank (n − 1). Consider now the permutation

τ =(

c1 c2 · · · ck−1 ck+1 · · · cl−1 cl+1 · · · cn a xb1 b2 · · · bk−1 bk+1 · · · bl−1 bl+1 · · · bn b y

),

where {x} = N\im(α) and {y} = N\im(β). A direct calculation shows thatβ = ταπ and hence β ∈ 〈A〉.

So, 〈A〉 contains all elements of Tn of defects 0 and 1. Let us prove byinduction on k that 〈A〉 contains all elements of Tn of defect k. We onlynow have to prove the induction step. Let γ ∈ Tn be an element of defectk, k > 1. Since def(γ) > 0, there exists a ∈ im(γ) such that its preimageB = {x ∈ N : γ(x) = a} contains more than one element. Let b1, b2 ∈ B besuch that b1 < b2. Finally, let a′ ∈ N\im(γ). Consider the idempotent

μ =(

1 2 · · · b1 − 1 b1 b1 + 1 · · · b2 − 1 b2 b2 + 1 · · · n1 2 · · · b1 − 1 b1 b1 + 1 · · · b2 − 1 b1 b2 + 1 · · · n

)

of rank (n−1) and the transformation δ, which coincides with γ on N\{b2},and such that δ(b2) = a′. Then def(δ) = (k − 1) and hence δ ∈ 〈A〉 by

3.1. GENERATING SYSTEMS IN Tn, PT n, AND ISn 41

induction. We also know that μ ∈ 〈A〉 since def(μ) = 1. At the same time,we have γ = δμ and hence γ ∈ 〈A〉. This completes the induction step andthe proof.

Theorem 3.1.4 Let A be a generating system of ISn. Then this system isirreducible if and only if A = A1∪{α}, where A1 is an irreducible generatingsystem of Sn and rank(α) = (n − 1).

Proof. Repeat Theorem 3.1.3 with the following changes: For the partialpermutations

α =(

1 2 · · · i − 1 i i + 1 · · · na1 a2 · · · ai−1 ∅ ai+1 · · · an

)

and

β =(

1 2 · · · j − 1 j i + 1 · · · nb1 b2 · · · bj−1 ∅ bj+1 · · · bn

)

the permutations π and τ should be chosen as follows: π(j) = i;

τ =(

c1 c2 · · · cj−1 cj+1 · · · cn xb1 b2 · · · bj−1 bj+1 · · · bn y

).

For the partial permutation

γ =(

i1 i2 · · · ik ik+1 · · · inj1 j2 · · · jk ∅ · · · ∅

)

the partial permutations δ and μ should be chosen as follows:

δ =(

i1 i2 · · · ik ik+1 ik+2 · · · inj1 j2 · · · jk jk+1 ∅ · · · ∅

)

μ =(

i1 i2 · · · ik ik+1 ik+2 · · · ini1 i2 · · · ik ∅ ik+2 · · · in

).

Theorem 3.1.5 Let A be a generating system of PT n. Then this system isirreducible if and only if A = A1 ∪ {α, β}, where A1 is an irreducible gener-ating system of Sn, α is a total transformation of rank (n − 1) and β is apartial permutation of rank (n − 1).

Proof. A composition of total transformations is a total transformation. Onthe other hand, a composition which consists of nontotal maps only, resultsin an element which is not total either. Hence from the proof of Lemma 3.1.2,it follows that each generating system of the semigroup PT n must contain atleast one total transformation, say α, of rank (n−1) and at least one partialtransformation, say β, of rank (n−1), which is not total. We have |dom(β)| ≤

42 CHAPTER 3. GENERATING SYSTEMS

(n − 1) and |im(β)| = (n − 1). Moreover, the mapping β : dom(β) → im(β)is surjective. This implies |dom(β)| = n − 1 and β : dom(β) → im(β) isbijective.

Taking Lemma 3.1.1 and the previous paragraph into account, we haveto only show that any A = A1∪{α, β}, where A1 is an irreducible generatingsystem of Sn, α is a total transformation of rank (n − 1), and β is a partialpermutation of rank (n − 1), generates PT n.

By Theorems 3.1.3 and 3.1.4 the sets A1 ∪ {α} and A1 ∪ {β} generateTn and ISn, respectively. Each γ ∈ PT n can be extended to a total trans-formation γ′ ∈ Tn. Then we have γ = γ′εdom(γ) and εdom(γ) ∈ ISn. Henceγ ∈ 〈A〉. This completes the proof.


3.2.1 Theorem 3.1.3 was proved in [Vo2].

3.2.2 Although all groups are semigroups, group theory and semigroup the-ory are two completely different directions of modern algebra. They differboth in formulations of the principal problems and in methods of their solu-tions. In particular, a problem in semigroup theory is usually considered assolved if it is reduced to a problem in group theory. Theorems 3.1.3–3.1.5 aregood examples of such reduction. They reduce the problem of classificationof all irreducible generating systems for the semigroups PT n, Tn, and ISn

to the analogous problem for the group Sn. Irreducible generating systemsin Sn have very complicated structure. Two classical irreducible generatingsystems are: the Coxeter system {(1, 2), (2, 3), . . . , (n−1, n)} and the system{(1, 2), (1, 2, . . . , n)} consisting of two generators. However, the problem toclassify all irreducible generating systems in Sn is still open. At the sametime Theorems 3.1.3–3.1.5 give a satisfactory classification of irreduciblegenerating systems for PT n, Tn, and ISn.

3.2.3 In each of the semigroups PT n, Tn, and ISn, the set of all noninvert-ible elements forms a subsemigroup (called the singular part of the originalsemigroup). Hence we have a new natural problem: to study generating sys-tems for the singular parts of our semigroups PT n, Tn, and ISn. Here, apartfrom irreducibility, one could also impose other conditions, for example toconsist of idempotents, to consist of elements of defect 1, to consists of ele-ments with prescribed structure of connected components, and so on. Oneof the first works in this direction was [Ho1] in which it was shown thatthe singular part of Tn is generated by idempotents of defect 1. One of themost recent works is [AAH], where the authors reprove some known results(in particular, the above-mentioned result from [Ho1]) and give several newexamples of generating systems for the singular part of Tn.



3.3.1 Prove that each element of defect k from Tn can be written as aproduct of a permutation and k idempotents of defect 1.

3.3.2 Let A ⊂ Sn be a subset consisting of transpositions. Define the (un-oriented) graph ΓA in the following way: the vertices are {1, 2, . . . , n}, andfor i, j ∈ {1, 2, . . . , n} the graph ΓA contains the edge (i, j) if and only if(i, j) ∈ A. Prove the following:

(a) A is a generating system of Sn if and only if ΓA is connected.

(b) A is an irreducible generating system of Sn if and only if ΓA is a tree.

(c) Sn has exactly nn−2 irreducible generating systems, consisting of trans-positions.

3.3.3 (a) Show that for n > 2 the group Sn does not contain any one-element generating system.

(b) Show that (1, 2), (2, 3),. . . , (n − 1, n) generate Sn.

(c) Show that (1, 2) and (1, 2, . . . , n) generate Sn.

3.3.4 ([Ho1]) Prove that the semigroup Tn\Sn is generated by the set of allidempotents of defect 1.

Chapter 4

Ideals and Green’s Relations

4.1 Ideals of Semigroups

Let S be a semigroup. If A, B ⊂ S, we set AB = {ab : a ∈ A; b ∈ B}.A subset I ⊂ S is called a left ideal provided that for all a ∈ S and b ∈ I

we have ab ∈ I (in other words, SI ⊂ I). Analogously, I is called a rightideal provided that IS ⊂ I. Left and right ideals are also called one-sidedideals. A subset I ⊂ S is called a two-sided ideal or simply an ideal providedthat it is both a left and a right ideal.

Let I be a (one-sided) ideal of S. Then we have I · I ⊂ I by definitionand hence I is a subsemigroup of S. The converse is not true in the generalcase. For example, the group Sn is a subsemigroup of each of the semigroupsTn, PT n, and ISn. However, for each (partial) transformation α �∈ Sn bothsets αSn and Snα do not have any common elements with Sn.

It is easy to see that both the intersection and the union of an arbitraryfamily of left (right, two-sided) ideals of S are in turn a left (resp. right,two-sided) ideal of S.

Exercise 4.1.1 Show that for any A ⊂ S the set AS is a right ideal of S;the set SA is a left ideal of S; and the set SAS is a two-sided ideal of S.

For each semigroup S we denote the semigroup S by S1 providedthat S contains an identity element, and the semigroup constructed inProposition 2.2.1 provided that S does not contain any identity element.Analogously we define S0. We also note that for k > 1 the notation Sk meanssomething completely different, namely, Sk = {a1a2 · · · ak : ai ∈ S. i =1, . . . , k}.

A left (resp. right or two-sided) ideal I of S is called principal providedthat there exists a ∈ S such that I = S1a (resp. I = aS1, I = S1aS1). Theelement a is called the generator of the ideal I. Note that a ∈ S1a, a ∈ aS1,and a ∈ S1aS1 by definition.



46 CHAPTER 4. IDEALS AND GREEN’S RELATIONS

Proposition 4.1.2 Each left (right or two-sided) ideal is a union of prin-cipal left (resp. right or two-sided) ideals.

Proof. Since a ∈ S1a, if I ⊂ S is a left ideal, we have I = ∪a∈IS1a. For

other ideals the argument is similar.

4.2 Principal Ideals in T n , PT n , and ISn

Since the semigroups Tn, PT n, and ISn are in fact monoids, we have S = S1

for these semigroups.

Theorem 4.2.1 Let S denote one of the semigroups Tn, PT n, or ISn. Thenfor each α ∈ S the right principal ideal generated by α has the following form

αS = {β ∈ S : im(β) ⊂ im(α)}.

Proof. Denote X = {β ∈ S : im(β) ⊂ im(α)}. If γ ∈ S, we have αγ(x) =α(γ(x)) by definition. Hence im(αγ) ⊂ im(α) and thus αS ⊂ X.

To prove the inverse inclusion consider an arbitrary β ∈ X. We haveim(β) ⊂ im(α). For each b ∈ im(β) choose some ab such that α(ab) = b.Consider the transformation γ for which dom(γ) = dom(β) and such thatfor x ∈ dom(β) we have γ(x) = ab if and only if β(x) = b. A direct calculationthen shows that αγ = β. Note that in the case S = ISn the injectivity of αand β gives the following implications for x1, x2 ∈ dom(γ):

x1 �= x2 ⇒ b1 = β(x1) �= β(x2) = b2 ⇒ ab1 �= ab2 ⇒ γ(x1) �= γ(x2).

Hence γ ∈ ISn. Thus β ∈ αS. This implies that X ⊂ αS and completes theproof.

Corollary 4.2.2 (i) Each of the semigroups PT n and ISn has exactly2n different principal right ideals.

(ii) The semigroup Tn has exactly 2n − 1 different principal right ideals.

Proof. From Theorem 4.2.1 we have that a principal right ideal is uniquelydetermined by im(α), that is, by a subset of N, which must be nonempty inthe case of Tn. The claim follows.

Corollary 4.2.3 Let S denote one of the semigroups Tn, PT n, or ISn andα ∈ S be such that rank(α) = k. Then

|αS| =

⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

kn, S = Tn;(k + 1)n, S = PT n;

k∑i=0

(n

i

)(k

i

)i!, S = ISn.

4.2. PRINCIPAL IDEALS IN Tn, PT n, AND ISn 47

Proof. Theorem 4.2.1 says that in the case of Tn the ideal αS is the setof all mappings from N to im(α), where |N| = n and |im(α)| = k. Hence|αS| = kn.

In the case of PT n, Theorem 4.2.1 says that αS is the set of all partialmappings from N to im(α), that is, the set of all mappings from N toim(α) ∪ {∅}. Hence |αS| = (k + 1)n.

Finally, for ISn, Theorem 4.2.1 says that αS is the set of all partial injec-tions from N to im(α). Such a partial injection can have rank i = 0, 1, . . . , k.If i is fixed, to define such a partial injection, say β, we have to choose itsdomain, that is, an i-element subset of N, which can be done in

(ni

)different

ways; then we have to choose im(β), that is, an i-element subset of im(α),which can be done in

(ki

)different ways; finally we have to define a bijection

from dom(β) to im(β), which can be done in i! different ways. The statementnow follows by applying the product rule and the summing up over all i.

With each α ∈ PT n, we associate the binary relation πα on N in thefollowing way

x πα y ⇔((

x, y ∈ dom(α)andα(x) = α(y))or(x, y �∈ dom(α)

)). (4.1)

In other words, x πα y if either α is not defined on both x and y, or α isdefined on both x and y and α(x) = α(y). It is clear that πα is an equivalencerelation and the equivalence classes of πα are dom(α) and all full preimagesof elements from im(α).

Theorem 4.2.4 Let S denote one of the semigroups Tn, PT n, or ISn. Thenfor each α ∈ S the left principal ideal generated by α has the following form

Sα = {β ∈ S : dom(β) ⊂ dom(α) and πα ⊂ πβ}. (4.2)

Proof. Denote the right-hand side of (4.2) by X. Let β ∈ Sα. Then β = γαfor some γ ∈ S. For arbitrary x ∈ dom(β) we have β(x) = γ(α(x)). Hencex ∈ dom(α) and thus dom(β) ⊂ dom(α). Furthermore, if x, y ∈ dom(β) andα(x) = α(y), we have β(x) = γ(α(x)) = γ(α(y)) = β(y). Thus πα ⊂ πβ andwe have Sα ⊂ X.

Assume now that β ∈ X. Set M = α(dom(β)) and define the mappingγ : M → N as follows:

γ(α(y)) = β(y). (4.3)

The mapping γ is well-defined since if x = α(y1) = α(y2), we have y1 πα y2,which implies y1 πβ y2 and thus β(y1) = β(y2). Hence the value γ(x) isuniquely defined.

If S = Tn, we have dom(β) = N and M = α(N) = im(α). In this case, wecan extend γ to a total transformation on N in an arbitrary way. Abusingnotation we denote the resulting total transformation also by γ. If S �=Tn, we just consider γ as a partial transformation on N with domain M .In particular, if S = ISn and x1 = α(y1), x2 = α(y2) are two different


elements in M , we have y1 �= y2 and from the injectivity of β it follows thatγ(x1) = β(y1) and γ(x2) = β(y2) are also different. This means that in thiscase γ ∈ ISn as well.

From the definition of γ we have that y �∈ dom(β) implies either y �∈dom(α), or α(y) �∈ M . Indeed, assume that y ∈ dom(α) and α(y) ∈ M .Then there exists y1 ∈ dom(β) such that α(y) = α(y1). Hence y πα y1 andthus y πβ y1, implying y ∈ dom(β), a contradiction. This means that fory �∈ dom(β) the equality (γα)(y) = γ(α(y)) implies that y �∈ dom(γα).

If y ∈ dom(β), we have α(y) ∈ M and the equality (4.3) implies that(γα)(y) = γ(α(y)) = β(y). This means that β = γα and thus β ∈ Sα. ThusX ⊂ Sα completing the proof.

If S = Tn, then dom(α) = dom(β) = N and hence the first condition in(4.2) can be omitted. If S = ISn, then the restriction of πα to dom(α) be-comes the equality relation. Hence dom(β) ⊂ dom(α) automatically impliesπα ⊂ πβ . This means that for the semigroup S = ISn the second conditionin (4.2) can be omitted.

The number of all unordered partitions N = N1 ∪ N2 ∪ · · · ∪ Nk of theset N into disjoint unions of nonempty blocks is called the nth Bell numberand is denoted by Bn. From the definition of the Stirling numbers S(n, k) ofthe second kind we immediately have the equality Bn =

∑nk=1 S(n, k).

Corollary 4.2.5 (i) The number of different principal left ideals in thesemigroup Tn equals Bn.

(ii) The number of different principal left ideals in the semigroup PT n

equals Bn+1.

(iii) The number of different principal left ideals in the semigroup ISn

equals 2n.

Proof. From Theorem 4.2.4 we have that a principal left ideal of Tn isuniquely determined by the partition of N into the equivalence classes ofthe relation πα. Since all partitions obviously give rise to some ideals, weobtain that the number of ideals equals Bn. This proves (i).

To prove (ii) we take a new symbol, say n + 1. Let α ∈ PT n. With πα

we associate the following equivalence relation π′α on the set N ∪ {n + 1}:

If dom(α) �= N, we adjoin n + 1 to the block dom(α); if dom(α) = N, thenwe say that {n + 1} is a new equivalence class. Obviously, π′

α is uniquelydetermined by πα. Conversely, any π′

α uniquely determines πα and dom(α).

By Theorem 4.2.4 each principal left ideal of PT n is uniquely deter-mined by the pair (πα, dom(α)). Hence the previous paragraph says thatthe mapping (πα, dom(α)) → π′

α is a bijection from the set of all principalleft ideals of PT n to the set of all unordered partitions of N ∪ {n + 1}, thelatter containing n + 1 element. Thus the number of different principal leftideals of PT n equals Bn+1. This proves (ii).

4.2. PRINCIPAL IDEALS IN Tn, PT n, AND ISn 49

Finally, to prove (iii) we note that by Theorem 4.2.4 each principal leftideal of ISn is uniquely determined by the set dom(α), which is just a subsetof N. Hence the number of such ideals is 2n.

Remark 4.2.6 Corollary 4.2.5(iii) also follows from Corollary 4.2.2(ii) and2.9.9.

Corollary 4.2.7 Let S denote one of the semigroups Tn, PT n, or ISn, andα ∈ S be an element of rank k. Then we have the following:

|Sα| =

⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

nk, S = Tn;(n + 1)k, S = PT n;

k∑i=0

(n

i

)(k

i

)i!, S = ISn.

Proof. We start with the cases S = Tn and S = PT n. As πα ⊂ πβ , eachtransformation β ∈ Sα is uniquely determined by its values on the set ofthose equivalence classes of πα which are contained in dom(α). We have ksuch classes and for each of them we have to choose the value of β on thisclass, which is an element from N (or N ∪ {∅} in the case of PT n). Nowfor Tn and PT n the statements are obtained by applying the product rule.

In the case S = ISn the proof is similar to that of the correspondingpart of Corollary 4.2.3. The only difference is that now one has to considerpartial injections from dom(α) to N.

Theorem 4.2.8 Let S denote one of the semigroups Tn, PT n, or ISn andα ∈ S. Then the principal ideal generated by α has the following form:

SαS = {β ∈ S : rank(β) ≤ rank(α)}. (4.4)

Proof. Let X denote the right-hand side of (4.4). The inclusion SαS ⊂ Xfollows from Exercise 2.1.4(c), so we have only to prove the inclusion X ⊂SαS.

Let im(α) = {a1, a2, . . . , ak}, and β ∈ X be such that rank(β) = m, andim(β) = {b1, b2, . . . , bm}. Then m ≤ k and for each i = 1, . . . , k we choosesome element ci in the set Ai = {x ∈ N : α(x) = ai}. Define γ ∈ PT n

in the following way: dom(γ) = dom(β), and for all y ∈ Bj = {z ∈ N :β(z) = bj}, j = 1, . . . , m, we set γ(y) = cj . Note that β ∈ Tn implies γ ∈ Tn

and β ∈ ISn implies γ ∈ ISn by construction. Let δ be any permutationsatisfying δ(ai) = bi, i = 1, . . . , m. Then a direct calculation gives δαγ = β,which implies β ∈ SαS. Thus X ⊂ SαS and the proof is complete.

Corollary 4.2.9 Each of the semigroups PT n and ISn contains (n + 1)different principal ideals. The semigroup Tn contains n different principalideals.


Proof. By Theorem 4.2.8 a principal ideal is uniquely determined by therank of the generator. For PT n and ISn the rank varies between 0 and n,for Tn it varies between 1 and n.

Corollary 4.2.10 Let S denote one of the semigroups Tn, PT n, or ISn,and α ∈ S be an element of rank k. Then we have the following:

|SαS| =

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

k∑i=1

S(n, i)n!

(n − i)!, S = Tn;

k∑i=0

S(n + 1, i + 1)n!

(n − i)!, S = PT n;

k∑i=0

(n

i

)2

i!, S = ISn.

Proof. According to Theorem 4.2.8, to determine |SαS| we have to find thenumber ti of elements of rank i ∈ N and then form the sum

∑i≤k ti.

If S = Tn, then an element of rank i is uniquely determined by anunordered partition of N into i nonempty blocks (which can be done inS(n, i) different ways), and then an injective mapping from the blocks ofthis partition into N (this can be done in n(n − 1) · · · (n − i + 1) = n!

(n−i)!

different ways). Hence in this case ti = S(n, i) n!(n−i)! .

For S = PT n the argument is almost the same with the difference thatwe should consider partitions of N∪{n+1} instead of those of N. The blockcontaining n + 1 corresponds to dom(β). Hence ti = S(n + 1, i + 1) n!

(n−i)! inthis case.

For S = ISn the number of partial permutations of rank k was computedin Theorem 2.5.1.

4.3 Arbitrary Ideals in T n , PT n , and ISn

Theorem 4.3.1 Let S denote one of the semigroups Tn, PT n, or ISn. Thenall two-sided ideals in S are principal and are generated by any element ofthe ideal, which has the maximal possible rank.

Proof. Let I ⊂ S be a two-sided ideal. Choose α ∈ I of maximal possiblerank. As SIS ⊂ I, we have SαS ⊂ I. On the other hand, by Theorem 4.2.8the set SαS contains all elements whose rank does not exceed that of α.Hence I ⊂ SαS and thus I = SαS.

Let S denote one of the semigroups Tn, PT n, or ISn. For k ≤ n setIk = {α ∈ S : rank(α) ≤ k}. By Theorem 4.3.1, each Ik is an ideal of S,and each ideal in S is of the form Ik for some k. In particular, the set of all

4.3. ARBITRARY IDEALS IN Tn, PT n, AND ISn 51

ideals of S forms a chain with respect to inclusions. For Tn this chain hasthe form

I1 ⊂ I2 ⊂ · · · ⊂ In = Tn.

For PT n and ISn this chain has the form

{0} = I0 ⊂ I1 ⊂ I2 ⊂ · · · ⊂ In = PT n (orISn).

Note that In\In−1 = Sn in all cases.The Boolean B(X) of a set X is partially ordered with respect to inclu-

sions. Recall that an antichain of a partially ordered set is a subset suchthat each two elements in this subset are not comparable. In particular, ifA1, A2, . . . , Am ⊂ N, the set {A1, . . . , Am} is an antichain of B(N) if andonly if for all i �= j we have both Ai �⊂ Aj and Aj �⊂ Ai.

Let L ⊂ B(N) be an antichain. Set

IL = {α ∈ S : there existsA ∈ Lsuch thatim(α) ⊂ A}.

Theorem 4.3.2 Let S denote one of the semigroups Tn, PT n, or ISn.

(i) For each antichain L ⊂ B(N) the set IL is a right ideal of S.

(ii) Let L1 and L2 be two antichains in B(N). Then L1 �= L2 impliesIL1 �= IL2.

(iii) For each right ideal I of S there exists an antichain L ⊂ B(N) suchthat I = IL.

Proof. For each α ∈ IL and μ ∈ S from im(αμ) ⊂ im(α) (see Exercise 2.1.4(b))it follows that im(αμ) ∈ IL. This proves (i).

Let L1 �= L2. Without loss of generality we may assume L1\L2 �= ∅. LetA ∈ L1\L2. We have to consider two possible cases.

Case 1: A is not comparable with any element of L2. In this case eachα ∈ S for which im(α) = A is contained in IL1 but not in IL2 . HenceIL1 �= IL2 .

Case 2: There exists B ∈ L2 such that A and B are comparable. ThenB ⊂ L2\L1. Without loss of generality we may assume B ⊂ A. As B �= A,each element α ∈ S for which im(α) = A is contained in IL1\IL2 . HenceIL1 �= IL2 in this case as well.

The above proves (ii).To prove (iii) we first observe that the maximal (with respect to inclu-

sions) elements of the set {im(α) : α ∈ I} obviously form an antichain, sayL, of B(N). By the definition of L, for each α ∈ I there exists A ∈ L suchthat im(α) ⊂ A. Hence α ∈ IL and I ⊂ IL.

On the other hand, for each β ∈ IL there exists A ∈ L and α ∈ I suchthat im(β) ⊂ A and im(α) = A. By Theorem 4.2.1 we have β ∈ αS. ButαS ⊂ I since α ∈ I and I is a right ideal. Hence β ∈ I. This means thatIL ⊂ I and thus IL = I. The proof is complete.


The identity mapping from S to the dual semigroup←S maps left ide-

als to right ideals and vice versa. For the inverse semigroup ISn we have

ISn∼=

←ISn via the mapping a → a−1. Hence Theorem 4.3.2 can be used

to describe all left ideals in ISn. The only change one has to make isto substitute im by dom. Indeed, the equalities im(α−1) = dom(α) andim(α) = dom(α−1) follow from the proof of Theorem 2.6.7. Hence we getthe following statement:

Theorem 4.3.3 For each antichain L of B(X) the set

LI = {α ∈ S : there existsA ∈ Lsuch thatdom(α) ⊂ A}

is a left ideal of ISn and the mapping L → LI is a bijection from the set ofall antichains of B(X) to the set of all left ideals of ISn.

As Tn and PT n are not self-dual (see 2.9.9 and Exercise 2.10.23), thedescription of all left ideals in these semigroups requires more efforts.

Let Partn denote the set of all (unordered) partitions of N. Define apartial order on Partn in the following way: Let ρ = A1 ∪ · · · ∪ Ak andτ = B1 ∪ · · · ∪Bm be two partitions. We will write ρ � τ provided that eachblock Ai is a union of some blocks of the partition τ (i.e., of some Bjs). Inparticular, the maximum partition is the one into one element blocks andthe minimum partition is the one with one block.

Let ρ be a partition of N. Denote by πρ the corresponding equivalencerelation (x πρ y if and only if x and y belong to the same block of ρ). It iseasy to see that ρ � τ if and only if πρ ⊃ πτ .

Example 4.3.4 If we have A1 = {a1, . . . , ap}, A2 = {b1, . . . , bq}, . . . , Ak ={c1, . . . , cr}, the partition A1 ∪ A2 ∪ · · · ∪ Ak can be written for exampleas follows:

a1, . . . , ap|b1, . . . , bq| · · · |c1, . . . , cr.

Using this notation the Hasse diagram of the partially ordered set of allpartitions of {1, 2, 3, 4} looks as follows:

1|2|3|4

��

��

��

!!!!!!!!

!!!!!!

1|23|4

��

14|2|3

��

��

1|24|3

��

��

13|2|4

��

��

12|3|4

��

��

"""""""""""""""""""

1|2|34

��

######

######

######

######

######

14|23

!!!!!!!!

!!!!!!!!

1|234

124|3

13|24 123|4

��

134|2

12|34

��

��

1234

4.4. GREEN’S RELATIONS 53

To each element α ∈ Tn we associate the partition ρα of N into equiv-alence classes with respect to the relation πα. Then ρα � ρβ if and only ifπα ⊃ πβ .

Theorem 4.3.5 For each antichain L in Partn the set

LI = {α ∈ Tn : there existsρ ∈ Lsuch thatρα � ρ}

is a left ideal of Tn and the mapping L → LI is a bijection from the set ofall antichains of Partn to the set of all left ideals of Tn.

Proof. For any μ, ν ∈ Tn the equality μ(x) = μ(y) implies the equality(νμ)(x) = (νμ)(y). Hence πμ ⊂ πνμ and ρνμ � ρμ. Thus for arbitrary α ∈ LIand β ∈ Tn, from ρα � ρ it follows that ρβα � ρα � ρ. Hence βα ∈ LI andLI is a left ideal.

It is clear that for each partition ρ ∈ Partn there exists α ∈ Tn such thatρα = ρ. Hence using the same arguments as in the proof of Theorem 4.3.2(ii)(with the obvious substitution of im(α) by ρα) we obtain that for allantichains L1 and L2 of Partn such that L1 �= L2 the ideals L1I and L2Iare different.

Finally, to prove that each left ideal of Tn has the form LI for someantichain L in Partn one should just follow the proof of Theorem 4.3.2(iii):If I is a left ideal of Tn, the antichain L is defined as the set of maximal(with respect to �) elements of the set {ρα : α ∈ I}. One should also useTheorem 4.2.4 instead of Theorem 4.2.1.

With each element α ∈ PT n we associate the partition ρα of N∪{n+1}as it was done in the proof of Corollary 4.2.5(ii).

Theorem 4.3.6 For each antichain L in Partn+1 the set

LI = {α ∈ PT n : there existsρ ∈ Lsuch thatρα � ρ}

is a left ideal of PT n and the mapping L → LI is a bijection from the setof all antichains of Partn+1 to the set of all left ideals of PT n.

Proof. The proof repeats that of Theorem 4.3.5.

4.4 Green’s Relations

In this section, we introduce several equivalence relations on semigroups,which play a central role in the structure theory. Let S be a semigroup.Elements a, b ∈ S are called L-equivalent provided that they generate thesame principal left ideal. In other words, aLb if and only if S1a = S1b.Equivalence classes of the relation L are called L-classes. For a ∈ S, theL-class containing a will be denoted by L(a). In other words, aLb if andonly if a ∈ L(b). The following easy but very useful fact follows immediatelyfrom the definition.


Proposition 4.4.1 aLb if and only if there exist x, y ∈ S1 such that a = xband b = ya.

In the dual way we define the relation R: Elements a, b ∈ S are calledR-equivalent provided that they generate the same principal right ideal. Inother words, aRb if and only if aS1 = bS1. Equivalence classes of the relationR are called R-classes. For a ∈ S the R-class containing a will be denotedby R(a).

Proposition 4.4.2 aRb if and only if there exist x, y ∈ S1 such that a = bxand b = ay.

Note that for every x ∈ S1 the equality S1a = S1b implies the equalityS1ax = S1bx; and the equality aS1 = bS1 implies the equality xaS1 = xbS1.In other words, the relation L is compatible with the right multiplication byelements of S1 (this is also called right compatible), that is, aLb impliesaxLbx. Dually, the relation R is left compatible, that is, aRb implies xaRxb.A relation which is both left and right compatible is simply called compatible.

If X is a set and ξ, η are two binary relations on X, then the productξ ◦ η is defined in the following way:

ξ ◦ η = {(a, b) : there existsc ∈ Xsuch that(a, c) ∈ ξand(c, b) ∈ η}.

Exercise 4.4.3 Show that the set of all binary relations on X with respectto the product defined above is a monoid.

Lemma 4.4.4 The relations L and R commute, that is, L ◦ R = R ◦ L.

Proof. Let (a, b) ∈ L ◦ R. Then there exists c ∈ S such that aLc and cRb.By Propositions 4.4.1 and 4.4.2, there exist x, y ∈ S1 such that a = xc,b = cy. Moreover, cRb implies xcRxb, and aLc implies ayLcy. But xc = a,xb = xcy = ay, and cy = b. Hence aRxcy and xcyLb, which implies (a, b) ∈R ◦ L, and thus L ◦ R ⊂ R ◦ L.

Analogously one shows that R◦L ⊂ L◦R and hence L◦R = R◦L.

Exercise 4.4.5 Let ξ and η be two equivalence relations on X. Show thatξ ◦ η = η ◦ ξ implies that ξ ◦ η is again an equivalence relation. Moreover,show that this product is the minimum equivalence relation which containsboth ξ and η.

The minimum equivalence relation on S which contains both R and Lis denoted by D and is called the D-relation. From Lemma 4.4.4 it followsthat D = L◦R = R◦L. All other notions and notation for D are similar tothe ones used for L. In particular, D(a) denotes the D-class of an element a.

The intersection of two equivalence relations is always an equivalencerelation. We define the H-relation as the intersection of R and L. All other


notions and notation for H are similar to the ones used for L. In particular,H(a) denotes the H-class of an element a.

Finally, we will say that the elements a and b are J -equivalent providedthat they generate the same principal two-sided ideal, that is, S1aS1 =S1bS1. All other notions and notation for J are similar to the ones usedfor L. In particular, J (a) denotes the J -class of an element a.

It is obvious that R ⊂ J and L ⊂ J . In particular, it follows thatD ⊂ J . Hence we have the following diagram depicting the introducedrelations on S:

L �

��

��

�

H �

��

��

��

� D � � �� J

R �

��

The relations L, R, H, D, and J on the semigroup S are called Green’srelations after J. A. Green, who introduced them in 1951 in [Gr].

Lemma 4.4.6 The following conditions are equivalent:

(a) aDb.

(b) R(a) ∩ L(b) �= ∅.

(c) L(a) ∩R(b) �= ∅.

Proof. Since the relation D is symmetric, it is enough to show that (a)⇔(b).Let aDb. As D = R◦L, there exists c ∈ S1 such that aRc and cLb. But

this means that c ∈ R(a) and c ∈ L(b). Thus R(a) ∩ L(b) �= ∅, proving theimplication (a)⇒(b).

Assume that c ∈ R(a)∩L(b). Then c ∈ R(a) and c ∈ L(b), which meansthat aRc and cLb. By definition, this implies that (a, b) ∈ R ◦ L = D.

Lemma 4.4.6 says that every L-class and every R-class which belong tothe same D-class have a nonempty intersection. Hence it is convenient tothink of a D-class as a rectangular table in which the rows correspond to,say, R-classes and the columns correspond to L-classes:

R(a1)R(a2)

...R(ak)

. . .

. . ....

......

. . .

L(b1) L(b2) . . . L(bm)


The cells in this table give all H-classes, contained in our D-class. FromLemma 4.4.6 it follows that all cells are nonempty. The rectangular tableabove is called the egg-box diagram of the D-class.

Lemma 4.4.7 (Green) Let S be a semigroup and a, b ∈ S be such that aRb.Let further u, v ∈ S1 be such that au = b and bv = a.

(i) The mapping Λu : x → xu maps L(a) to L(b), and the mappingλv : y → yv maps L(b) to L(a).

(ii) λu : L(a) → L(b) and λv : L(b) → L(a) are mutually inverse bijections.

(iii) Both λu and λv preserve R-classes, that is, xRλu(x) and yRλv(y) forall x ∈ L(a) and y ∈ L(b).

Here is an illustration of the statement of Lemma 4.4.7:

R(a) = R(b) a = bv b = au

xλu �� xu

yv yλv��

L(a) L(b)

Proof. The statement (i) follows from the fact that L is right compatible.Let x ∈ L(a). Then there exists p ∈ S1 such that x = pa. Hence

(λvλu)(x) = λv(λu(x)) = xuv = pauv = pa = x.

This means that λvλu is the identity transformation of L(a). Analogouslyone shows that λuλv is the identity transformation of L(b). This proves thestatement (ii).

Let x ∈ L(a). Using (ii) we have

λu(x) = xu and x = λv(xu) = xu · v = λu(x) · v.

Using Proposition 4.4.2 we have xRλu(x). Hence λu preserves R. Since λv

is inverse to λu, it must preserve R as well. This completes the proof.

From Green’s Lemma it follows that the mappings λu and λv inducemutually inverse bijections between the sets of H-classes in L(a) and L(b).Moreover, λu and λv also induce mutually inverse bijections between thecorresponding H-classes.

Applying Green’s Lemma to the opposite semigroup←S we obtain the

following dual version of this lemma.


Lemma 4.4.8 Let S be a semigroup and a, b ∈ S be such that aLb. Letu, v ∈ S1 be such that ua = b and vb = a.

(i) The mapping μu : x → ux maps R(a) to R(b), and the mapping μv :y → vy maps R(b) to R(a).

(ii) μu : R(a) → R(b) and μv : R(b) → R(a) are mutually inverse bijec-tions.

(iii) Both μu and μv preserve L-classes, that is, xLμu(x) and yLμv(y) forall x ∈ R(a) and y ∈ R(b).

We note that the existence of the elements u and v, which are essentiallyused in Lemmas 4.4.7 and 4.4.8, follows from Propositions 4.4.1 and 4.4.2.

Corollary 4.4.9 All H-classes inside the same D-class are of the samecardinality.

Proof. Let aDb. As D = L ◦ R, there exists c ∈ S1 such that aRc andcLb. By Propositions 4.4.1 and 4.4.2, there exist u, v ∈ S such that c = au,b = vc. By Green’s Lemma, the mapping λu is a bijection from H(a) toH(c). By the dual of Green’s Lemma the mapping μv is a bijection fromH(c) to H(b).

H(a)

��

H(c)

��

R(a) aλu

�� c

μv

��

b H(b)��

L(b)

Hence the composition μvλu is a bijection from H(a) to H(b).

Exercise 4.4.10 Let S be a semigroup such that for each a ∈ S we haveaS = Sa = S. Show that S is a group.

Theorem 4.4.11 Let H be an H-class of S. Then the following conditionsare equivalent:

(a) H is a group.

(b) H contains an idempotent.

(c) There exist a, b ∈ H such that ab ∈ H.


Proof. Assume that H is a group. Let e ∈ H be the identity element. Thene · e = e and hence (a)⇒(b). The implication (b)⇒(c) is obvious.

To complete the proof we have to prove the implication (c)⇒(a). Leta, b, ab ∈ H. As abLb and aRab, by Green’s Lemma we have

aH = H = Hb. (4.5)

Let now x ∈ H be arbitrary. From (4.5) we have ax ∈ H and xb ∈ H.Analogously to the proof of (4.5) we have Hx = xH = H. As x is arbitrary,H · H = H and thus H is a semigroup. The statement now follows fromExercise 4.4.10.

4.5 Green’s Relations on T n , PT n , and ISn

The description of principal ideals in Tn, PT n, and ISn obtained in Sect. 4.2allows us to describe Green’s relations in these semigroups. As before for a(partial) transformation α we denote by ρα the partition of N (in the case ofTn) or N∪{n+1} (in the case of PT n, or ISn) generated by the relation πα.

Theorem 4.5.1 Let S be one of the semigroups Tn, PT n, or ISn, andα, β ∈ S. Then

(i) αRβ if and only if im(α) = im(β)

(ii) αLβ if and only if ρα = ρβ

(iii) αHβ if and only if im(α) = im(β) and ρα = ρβ

(iv) αDβ if and only if rank(α) = rank(β)

(v) αJ β if and only if rank(α) = rank(β)

Proof. The statement (i) follows from Theorem 4.2.1.The statement (ii) follows from Theorem 4.2.4.The statement (iii) follows from (i) and (ii) since H = R∩ L.The statement (v) follows from Theorem 4.2.8.As D ⊂ J , from (v) it follows that αDβ implies rank(α) = rank(β).

Conversely, let rank(α) = rank(β). Set im(α) = {a1, a2, . . . , ak} and im(β) ={b1, b2, . . . , bk} and consider the element γ, defined in the following way:

γ(x) = bj if and only ifα(x) = aj , j = 1, . . . , k.

Obviously ργ = ρα and hence αLγ by (ii). On the other hand, im(γ) =im(β) and hence γLβ by (i). This means that (α, β) ∈ L ◦ R = D andproves (iv).

4.5. GREEN’S RELATIONS ON Tn, PT n, AND ISn 59

Corollary 4.5.2 In the semigroups Tn, PT n, and ISn we have D = J .

As we have already mentioned after Theorem 4.2.4, for the semigroupISn the condition ρα = ρβ is equivalent to the condition dom(α) = dom(β).

Proposition 4.5.3 Let S be one of the semigroups Tn, PT n, or ISn, andα, β ∈ S. Then αLβ if and only if there exists μ ∈ Sn such that α = μβ.

Proof. Let α = μβ, where μ ∈ Sn. Since μ is invertible, we have β = μ−1α.Hence αLβ by Proposition 4.4.1.

Conversely, let αLβ. Then ρα = ρβ . Let A1, . . . , Ak be those blocks ofρα which are contained in dom(α) and let α(x) = ai and β(x) = bi for allx ∈ Ai, i = 1, . . . , k. Take any permutation μ ∈ Sn such that μ(bi) = ai,i = 1, . . . , k. Then a direct calculation shows that α = μβ.

Proposition 4.5.4 Let α, β ∈ ISn. Then

(i) αRβ if and only if there exists μ ∈ Sn such that α = βμ.

(ii) αDβ if and only if there exists μ, ν ∈ Sn such that α = νβμ.

Proof. The statement (i) follows from Proposition 4.5.3 and the fact that

ISn∼=

←ISn.

If α = νβμ for some μ, ν ∈ Sn, then β = ν−1αμ−1 and hence the principalideals generated by α and β coincide. This means αJ β. The latter impliesαDβ using Theorem 4.5.1(iv) and (v).

Conversely, if αDβ, then there exists γ such that αLγ and γLβ. ByProposition 4.5.3 we have α = νγ for some ν ∈ Sn, and by (i) we haveγ = βμ for some μ ∈ Sn. Thus α = νβμ.

Example 4.5.5 Here we present the egg-box diagrams for all D-classes ofthe semigroup PT 3. This semigroup has four D-classes Di, i = 0, 1, 2, 3,indexed by the rank of elements inside the D-class. The transformation(

1 2 3a1 a2 a3

)from PT 3 is given by the second row (a1a2a3) (do not mix

this with our notation for cycles). To accommodate the picture to the pagesize we are forced to transpose it. So, in our picture rows are L-classes andcolumns are R-classes. For each L-class L we show on the left of the diagramthe partition ρα of the domain of α ∈ L, and for each R-class R we showabove the diagram the set im(α), where α ∈ R. The elements marked with∗ are idempotents.


D3 : 1|2|3{1, 2, 3}

(123)∗, (132), (213), (231), (312), (321)

D2 : {1, 2} {1, 3} {2, 3}12|313|21|231|21|32|3

(112), (221) (113)∗, (331) (223)∗, (332)(121)∗, (212) (131), (313) (232), (323)∗

(122)∗, (211) (133)∗, (311) (233), (322)(12∅)∗, (21∅) (13∅), (31∅) (23∅), (32∅)(1∅2), (2∅1) (1∅3)∗, (3∅1) (2∅3), (3∅2)(∅12), (∅21) (∅13), (∅31) (∅23)∗, (∅32)

D1 : {1} {2} {3}123121323123

(111)∗ (222)∗ (333)∗

(11∅)∗ (22∅)∗ (33∅)(1∅1)∗ (2∅2) (3∅3)∗

(∅11) (∅22)∗ (∅33)∗

(1∅∅)∗ (2∅∅) (3∅∅)(∅1∅) (∅2∅)∗ (∅3∅)(∅∅1) (∅∅2) (∅∅3)∗

D0 :∅

(∅∅∅)∗

If from the above tables we delete all elements containing the symbol ∅, weobtain the egg-box diagram for the semigroup T3. If from the above tableswe delete all elements containing some repetitions of 1, 2, or 3, we obtainthe egg-box diagram for the semigroup IS3.

4.6 Combinatorics of Green’s Relationsin the Semigroups T n , PT n , and ISn

From the definition of Green’s relations we have that the number of L-, R-and J -classes in a semigroup coincides with the number of principal left,right and two-sided ideals, respectively. Hence from Corollaries 4.2.2, 4.2.5,and 4.2.9 we get:

Proposition 4.6.1 (i) The semigroup Tn contains Bn different L-classes,(2n − 1) different R-classes and n different J -classes.

(ii) The semigroup PT n contains Bn+1 different L-classes, 2n different R-classes and n + 1 different J -classes.

(iii) The semigroup ISn contains 2n different L-classes, 2n different R-classes and n + 1 different J -classes.

4.6. COMBINATORICS OF GREEN’S RELATIONS 61

Since in Tn, PT n, and ISn we have J = D by Theorem 4.5.1, in whatfollows we will not consider J -classes.

For each semigroup Tn, PT n, and ISn and each k = 0, . . . , n we denoteby Dk the D-class which consists of elements of rank k. Then Tn has D-classesDk, k = 1, . . . , n, and PT n and ISn have an additional D-class D0. Notethat in each of these semigroups the ideal Ik has the form Ik = ∪i≤kDk andhence Dk = Ik\Ik−1.

The cardinality of Dk was already determined during the proof ofCorollary 4.2.10:

Proposition 4.6.2 Let S be one of the semigroups Tn, PT n, or ISn. Wehave

|Dk| =

⎧⎪⎪⎨⎪⎪⎩S(n, k) n!

(n−k)! , S = Tn;

S(n + 1, k + 1) n!(n−k)! , S = PT n;(

nk

)2k!, S = ISn.

Proposition 4.6.3 For each of the semigroups Tn, PT n, and ISn there is anatural bijection between the k-element subsets of N and the R-classes insidethe D-class Dk. In particular, Dk contains exactly

(nk

)different R-classes.

Proof. By Theorem 4.5.1 each R-class is uniquely defined by the imageof elements from this class. For R-classes inside Dk the image can be anarbitrary k-element subset of N. The statement follows.

Proposition 4.6.4 (i) For the semigroup Tn there is a natural bijectionbetween partitions of N into k blocks and the L-classes inside the D-class Dk. In particular, Dk contains S(n, k) different L-classes.

(ii) For the semigroup PT n there is a natural bijection between partitionsof N ∪ {n + 1} into k + 1 blocks and the L-classes inside the D-classDk. In particular, Dk contains S(n + 1, k + 1) different L-classes.

(iii) For the semigroup ISn there is a natural bijection between k-elementsubsets of N and the L-classes inside the D-class Dk. In particular,Dk contains exactly

(nk

)different L-classes.

Proof. By Theorem 4.5.1 each L-class L of Tn is uniquely determined bysome partition ρα of the set N. This partition is the same for all elementsof L. For L-classes inside the Dk such partition can be an arbitrary partitionof N into k blocks. The statement (i) follows.

The proof of (ii) is analogous to that of (i) with substitution of N byN ∪ {n + 1}.

The proof of (iii) repeats that of Proposition 4.6.3 considering domainsinstead of images.


Proposition 4.6.5 (i) In each of the semigroups Tn, PT n and ISn everyL-class inside the Dk contains exactly

(nk

)different H-classes.

(ii) In semigroups Tn, PT n and ISn every R-class inside the Dk containsS(n, k), S(n + 1, k + 1) and

(nk

)different H-classes, respectively.

(iii) In semigroups Tn, PT n and ISn every D-class Dk contains S(n, k)(nk

),

S(n + 1, k + 1)(nk

)and

(nk

)2 different H-classes, respectively.

Proof. Let D be a D-class. By Lemma 4.4.6, there is a bijection betweenthe H-classes inside D and pairs (R, L), where L and R are an L-class andan R-class inside D, respectively. The bijection is given by taking R ∩ L.Hence the number of H-classes inside R equals the number of L-classes insideD, and vice versa. The statement now follows from Propositions 4.6.3 and4.6.4.

Theorem 4.6.6 For each of the semigroups Tn, PT n, and ISn the cardi-nality of any H-class inside the D-class Dk equals k!.

Proof. Let H be an H-class inside the D-class Dk. By Theorem 4.5.1(iii) itconsists of all elements α for which the image im(α) and the partition ρα

are fixed. Let im(α) = {a1, . . . , ak} and B1, . . . , Bk be those blocks of ρα

which are contained in dom(α). Then each element from H is completelydetermined by a surjective function from the set {B1, . . . , Bk} to the set{a1, . . . , ak}. The number of such functions is obviously k!.

From Theorem 4.6.6 and Proposition 4.6.5 we obtain:

Corollary 4.6.7 (i) In each of the semigroups Tn, PT n, and ISn eachL-class inside the D-class Dk contains exactly

(nk

)k! elements.

(ii) In the semigroups Tn, PT n, and ISn each R-class inside the D-classDk contains S(n, k)k!, S(n+1, k+1)k! and

(nk

)k! elements, respectively.


4.7.1 Two-sided ideals of Tn were independently described by Mal’cev in[Ma1] and Vorob’ev in [Vo1]. However, principal one-sided ideals of Tn hadbeen described already by A. Suschkewitsch in [Su2]. Two-sided ideals of ISn

and PT n were described by Liber in [Lib] and Sutov in [Sut2], respectively.The cardinalities of all principal one-sided ideals of Tn were determined in[HM]. We did not manage to find any description of all one-sided ideals forTn, PT n, or ISn in the literature.

4.7.2 Green’s relations were introduced in [Gr]. They play a very importantrole in the study of the structure of semigroups. One of the main reasonsis that Green’s Lemma 4.4.7 in some sense allows us to define a kind ofCartesian coordinate system on each D-class.


4.7.3 Green’s relations for Tn were described by Miller and Doss, see [Do]. Anice presentation of this description can be found in [CP1, Sect. 2.2]. Green’srelations for ISn and PT n were described by Reilly in [Re1] and FitzGeraldand Preston in [FP], respectively.

4.7.4 In Sect. 2.7 we already mentioned a very close connection betweenregular elements and idempotents. Another aspect of this connection is thefollowing:

Proposition 4.7.1 Let S be a semigroup and a ∈ S. Then the followingconditions are equivalent:

(a) a is regular.

(b) There exists an idempotent e ∈ S such that aRe.

(c) There exists an idempotent e ∈ S such that aLe.

Proof. Assume that a is regular and b is an inverse of a. Then ab = a · b anda = ab · a. Hence aRab by Proposition 4.4.2. At the same time e = ab =(aba)b = (ab)2 is an idempotent. This proves the implication (a)⇒(b).

Let aRe for some idempotent e. Then, by Proposition 4.4.2, there existx, y ∈ S1 such that ax = e and ey = a. If x = 1, then a = e is an idempotent,hence regular. If x �= 1, we have x ∈ S and axa = ea = e · ey = ey = a.Hence a is regular, which proves the implication (b)⇒(a).

The equivalence (a)⇔(c) is proved similarly.

Corollary 4.7.2 Let S be a semigroup and s ∈ S. If the element a is regular,then every element from D(a) is regular as well.

A D-class containing a regular element is called a regular D-class. As thesemigroups Tn, PT n, and ISn are regular, we have the following:

Corollary 4.7.3 (i) In the semigroups Tn, PT n, and ISn all principal(right, left, two-sided) ideals are generated by idempotents.

(ii) In the semigroups Tn, PT n, and ISn each L-class and each R-classcontain an idempotent.

4.7.5 The D-class Dk of the semigroup Tn contains(nk

)kn−k idempotents.

This is shown in the proof of Corollary 2.7.4. The D-class Dk of the semi-group PT n contains

(nk

)(k + 1)n−k idempotents. This is shown in the proof

of Corollary 2.7.5.

4.7.6 Let G be a group and a, b ∈ G. Then each of the equations ax = b andya = b has a (unique) solution. Hence if G is a subgroup of some semigroupS, then any two elements from G are both L- and R-equivalent in S. Hencethey are H-equivalent. This means that any subgroup of S is contained in


some H-class, say H. In particular, H must contain an idempotent: theidentity element of this group. By Theorem 4.4.11, in this case H itself is agroup.

Theorem 4.7.4 Let H be an H-class of Tn, PT n, or ISn, which containsan idempotent of rank k. Then H is isomorphic to the symmetric group Sk.

Proof. Let ε ∈ H be an idempotent. Let further im(ε) = {a1, . . . , ak} andfor i = 1, . . . , k let Bi be the full preimage of ai. From ε2 = ε we have thatai ∈ Bi for all i. Then, by Theorem 4.5.1(iii) the elements from H are givenby all possible injections from the set {B1, . . . , Bk} to the set {a1, . . . , ak}. Inparticular, the restriction to {a1, . . . , ak} defines a bijective mapping from Hto the symmetric group on {a1, . . . , ak}. It is easy to see that this mappingis compatible with the composition and hence is an isomorphism. The imageof this mapping is obviously isomorphic to Sk.

4.7.7 Let S be one of the semigroups Tn, PT n, or ISn. Let further H1 andH2 be two different H-classes of S inside the same D-class. If both H1 andH2 are groups, then from Theorem 4.7.4 it follows that H1

∼= H2. This is aspecial case of the following more general statement.

Theorem 4.7.5 Let S be a semigroup and H1 and H2 be two differentH-classes of S inside the same D-class. If both H1 and H2 are groups, thenH1

∼= H2 as groups.

Proof. Denote the idempotents of H1 and H2 by e and f , respectively. Thene and f are the identities in the groups H1 and H2, respectively. Let a ∈R(e) ∩ L(f). As eRa, there exists b ∈ S1 such that a = eb. Hence ea =e · eb = eb = a. Analogously one shows that af = a.

H(e)

��

R(e) eλa

��

μa′

��

a

μa′

��

R(f) a′λa �� f H(f)��

L(e) L(f)

By Green’s Lemma the mapping λa : x → xa is a bijection of the L-classL(e) onto the L-class L(a) = L(f). Let a′ ∈ L(e) be such that λa(a′) = f .


As a′e = a′, the dual Green’s lemma says that the mapping μa′ : y → a′y isa bijection from R(e) to R(f). In particular, μa′(a) = a′a = λa(a′) = f .

The composition ϕ = μa′λa : x → a′xa is a bijection from H(e) to H(f).Since f = a′a and a = af , by Green’s Lemma the mapping μa : R(f) →R(e) is inverse to the mapping μa′ . Analogously λa′ : L(f) → L(e) is inverseto λa. Hence (λa′μa)(f) = e. On the other hand,

(λa′μa)(f) = afa′ = af · a′ = aa′.

This shows that aa′ = e, which implies the following:

ϕ(xy) = a′xya = a′xeya = a′xa · a′ya = ϕ(x)ϕ(y).

Hence ϕ is an isomorphism from H1 to H2.

4.7.8 If S is a semigroup with the zero element 0, then for each nonemptysubset A ⊂ S we can define the left annihilator of A as follows:

Annl(A) = {x ∈ S : xa = 0for alla ∈ A}.

Analogously one defines the right annihilator of A as follows:

Annr(A) = {x ∈ S : ax = 0for alla ∈ A}.

It is easy to see that Annl(A) is a left ideal of S and Annr(A) is a right idealof S for each A.

If A = {a}, one usually uses the notation Annl(a), or Annr(a) instead ofAnnl({a}), or Annr({a}), respectively.


4.8.1 Prove the identity

Bn+1 =n∑

k=1

(k + 1)S(n, k) =n∑

k=0

(n

k

)Bk.

4.8.2 Let I be an ideal of one of the semigroups Tn, PT n, or ISn. Provethat I2 = I.

4.8.3 Prove that in each of the semigroups Tn, PT n, or ISn the numberaR of right ideals satisfies the inequality

22n> aR >

n∑k=1

2(nk) − n.


4.8.4 Find the number of left ideals in ISn for

(a) n = 2,

(b) n = 3,

(c) n = 4.

4.8.5 Let ξ be an equivalence relation on some set X. Prove that ξ ◦ ξ = ξ.

4.8.6 (a) Let S = Tn, or S = PT n. Prove that αRαη for each η ∈ Sn andeach α ∈ S.

(b) Let S = Tn, n > 3, or S = PT n, n > 1. Show that for α, β ∈ S therelation αRβ in general does not imply the existence of η ∈ Sn suchthat α = βη.

4.8.7 ([HM]) Let n > 3, 1 < k < n, and α ∈ Tn be an element of rank k.Prove that |αTn| ≥ |Tnα|, moreover, the equality is possible only if n = 4and k = 2.

4.8.8 Compute the number of elements in each D-class of the semigroups

(a) IS5,

(b) T5,

(c) PT 5.

4.8.9 Prove that a regular semigroup is inverse if and only if for each prin-cipal one-sided ideal I there exists a unique idempotent which generates I.

4.8.10 An ideal I ⊂ S is called prime provided that ab ∈ I implies a ∈ I,or b ∈ I for all a, b ∈ S. Find all prime ideals in Tn, PT n, and ISn.

4.8.11 An ideal I ⊂ S is called semiprime provided that a2 ∈ I impliesa ∈ I for all a ∈ S. Find all semiprime ideals in Tn, PT n, and ISn.

4.8.12 An ideal I ⊂ S is called reflexive provided that ab ∈ I implies ba ∈ Ifor all a, b ∈ S. Find all reflexive ideals in Tn, PT n, and ISn.

4.8.13 Find the number of idempotents in each R-class inside Dk for thesemigroup

(a) Tn

(b) PT n

4.8.14 (a) Let α ∈ Tn be a transformation of rank k and n1, . . . , nk bethe cardinalities of the full preimages of elements from im(α). Find thenumber of idempotents in R(α).

(b) The same problem for PT n.


4.8.15 Prove that the class H(α) of the semigroup Tn contains an idempo-tent if and only if for each block B of the partition ρα we have B∩im(α) �= ∅.

4.8.16 (a) Let R be an R-class of some semigroup S and e ∈ R be anidempotent. Show that e is a left identity on R, that is, ex = x for allx ∈ R.

(b) Let L be an L-class of some semigroup S and e ∈ L be an idempotent.Show that e is a right identity on L, that is, xe = x for all x ∈ L.

4.8.17 For each α ∈ ISn prove that |Annl(α)| = |Annr(α)|.

4.8.18 Let α ∈ PT n. Determine:

(a) |Annl(α)|

(b) |Annr(α)|

4.8.19 Prove that in the semigroups PT n and ISn none of the ideals Ik,0 < k < n, is a (left or right) annihilator of some set.

Chapter 5

Subgroupsand Subsemigroups

5.1 Subgroups

Let S be a semigroup and G be a subgroup of S. Then G contains theunique idempotent e (the identity element of G). Conversely, if e ∈ S is anidempotent, then {e} forms a trivial subgroup of S. This shows that thereis a close connection between the subgroups of S and idempotents of S. Inthis section we would like to illustrate this connection. We start from thefollowing two obvious statements:

Lemma 5.1.1 Let e ∈ E(S). Then eSe is a submonoid of S with the identityelement e.

Lemma 5.1.2 Let G be a subgroup of S with the identity element e ∈ E(S).Then G is a subgroup of (eSe)∗.

A subgroup G of S is called maximal provided that G is not properlycontained in any other subgroup of S.

Theorem 5.1.3 (i) For each e ∈ E(S) there is a unique maximal sub-group Ge of S in which e is the identity element.

(ii) If e, f ∈ E(S) and e �= f , then Ge ∩ Gf = ∅.

Proof. From Lemma 5.1.2 it follows that any subgroup of S in which eis the identity element is contained in (eSe)∗. The latter is a group byProposition 2.2.3. Hence Ge = (eSe)∗. This proves (i).

Let a ∈ Ge ∩ Gf and let b and c denote the inverses to a in Ge andGf , respectively. We have ab = e and ca = f . From this and the facts thatfa = a (since a ∈ Gf ) and ae = e (since a ∈ Ge) we have

e = ab = fab = fe = cae = ca = f,

a contradiction.



70 CHAPTER 5. SUBGROUPS AND SUBSEMIGROUPS

In 4.7.6 it is shown that maximal subgroups of S are exactly theH-classes H(e), e ∈ E(S). As each H-class contains at most one idempotentand different H-classes do not have common elements, we get an alternativeproof of Theorem 5.1.3. In what follows for e ∈ E(S) we shall denote thecorresponding maximal subgroup by Ge. For the semigroups Tn, PT n, andISn the following statement is a direct consequence of Theorems 4.5.1and 4.7.4:

Theorem 5.1.4 Let S denote one of the semigroups Tn, PT n, or ISn. Forε ∈ E(S) we have:

(i)Gε = {α ∈ S : im(α) = im(ε), ρα = ρε}.

(ii) If rank(ε) = k, then Gε∼= Sk.

An element g ∈ S is called a group element provided that g ∈ Ge forsome e ∈ E(S).

Exercise 5.1.5 Let S denote one of the semigroups Tn, PT n, or ISn. Showthat an element α ∈ S is a group element if and only if the full subgraphof Γα with the vertex set N\stim(α) is an empty graph, that is, it does notcontain any arrow.

Proposition 5.1.6 (i) The semigroup Tn contains∑n

k=1

(nk

)kn−kk! group

elements.

(ii) The semigroup PT n contains∑n

k=1

(nk

)(k + 1)n−kk! group elements.

(iii) The semigroup ISn contains∑n

k=1

(nk

)k! group elements.

Proof. In the proofs of Corollaries 2.7.4, 2.7.5, and 2.7.3, it was shown thatthe semigroups Tn, PT n, and ISn contain

(n

k

)kn−k,

(n

k

)(k + 1)n−k, and

(n

k

)

idempotents of rank k, respectively. After this the claim follows directly fromTheorems 5.1.3(ii) and 5.1.4(ii).

5.2 Cyclic Subsemigroups

A semigroup S is called cyclic provided that it has a generating systemconsisting of one element. If this element is a, one writes S = 〈a〉.

Let S = 〈a〉 be a finite cyclic semigroup. Then the sequence of elementsa1 = a, a2, a3, . . . must contain repeating elements. Assume that the ele-ments a, a2, . . . , al are pairwise different and al+1 = ak, where k ≤ l. The

5.2. CYCLIC SUBSEMIGROUPS 71

number l is called the order of a and is denoted by |a|, the number k is calledthe index of a and the number m = (l + 1)− k is called the period of a. Thepair (k, m) is called the type of a.

From ak+m = ak we get that the different elements in the sequence a = a,a2, a3, . . . are

〈a〉 = {a, a2, . . . , ak+m−1}.

Lemma 5.2.1 If a and b are of the same type, the semigroups 〈a〉 and 〈b〉are isomorphic.

To prove Lemma 5.2.1 we have to recall some notation. Recall that forx ∈ Z and y ∈ N the expression x mod y denotes the unique number 0 ≤z < y such that x − z is divisible by y. This number is called the residue ofx modulo y. The set {x+ sy : s ∈ Z} is called the residue class of x moduloy and is denoted by x. The set Zy of all residue classes modulo y forms agroup with respect to the addition + of residues, defined as follows:

a+b = a + b. (5.1)

This group is usually referred to as the group of residue classes modulo y.

Exercise 5.2.2 Show that the addition of residue classes, defined in (5.1),is well defined, that is, a = a′ and b = b′ imply a + b = a′ + b′.

Exercise 5.2.3 Check that (Zy, +) is indeed a group.

Proof of Lemma 5.2.1. If a is an element of type (k, m), then

as · at =

{as+t, s + t < k + m

ak+(s+t−k)mod m s + t ≥ k + m.(5.2)

Hence in the semigroup 〈a〉 the multiplication is completely determined bythe type of a. Thus if b has the same type as a, the mapping ai → bi,1 ≤ i ≤ k + m − 1, is an isomorphism of semigroups.

Lemma 5.2.4 For each pair (k, m) ∈ N2 there exists a cyclic semigroup,

generated by an element of type (k, m).

Proof. A direct calculation shows that the element

α = [1, 2, . . . , k](k + 1, k + 2, . . . , k + m) ∈ ISk+m

has type (k, m) and hence generates the necessary cyclic semigroup.

Lemma 5.2.5 Let 〈a〉 be a cyclic semigroup, generated by the element aof type (k, m). Then the set P = {ak, . . . , ak+m−1} is a subgroup of 〈a〉.Moreover, this subgroup is isomorphic to Zm.


Proof. That P is a subsemigroup of 〈a〉 is obvious. Consider the mapping

ϕ : P → Zm

ak+i → k + i,

where i = 0, 1, . . . , m − 1. The mapping ϕ is obviously a bijection. On theother hand, we have

ϕ(ak+i ·ak+j) = ϕ(a2k+i+j) = 2k + i + j = k + i+k + j = ϕ(ak+i)+ϕ(ak+j).

Hence ϕ is an isomorphism.

Corollary 5.2.6 Let S be a semigroup and a ∈ S be an element of finiteorder. Then the semigroup 〈a〉 contains an idempotent. In particular, eachfinite semigroup contains an idempotent.

Proof. To prove the first statement we just have to note that the identityelement of the group P from Lemma 5.2.5 is an idempotent. The proof isthen completed by observing that in a finite semigroup all elements havefinite order.

The identity element of P is the ak+i for which ϕ(ak+i) = 0, that is, i ≡−k(modm). It is worth noting that ak+(−k mod m) is the unique idempotentof 〈a〉, as for each b ∈ 〈a〉\P we have bk ∈ P and hence b cannot be anidempotent.

The cyclic semigroup 〈a〉, generated by an element of type (k, m), canbe depicted via the following diagram.

ak+1 ak+2 . . .

a1 a2 a3 . . . ak

��

ak+m−1

��

ak+m−2 . . .

The results obtained above can be summarized in the following statement.

Theorem 5.2.7 (Structure theorem for finite cyclic semigroups)

(i) Let S be a finite cyclic semigroup, generated by the element a. Thenthere exist k, m ∈ N such that S = 〈a〉 = {a1, a2, . . . , ak+m−1} andak+m = ak.

(ii) For each k, m ∈ N there exists a finite cyclic semigroup, generated byan element of type (k, m).

5.2. CYCLIC SUBSEMIGROUPS 73

(iii) Two finite cyclic semigroups are isomorphic if and only if they aregenerated by elements of the same type.

(iv) If a has type (k, m), then the set {ak, ak+1, . . . , ak+m−1} is a subgroupof 〈a〉, which is isomorphic to the group (Zm, +).

(v) Every finite cyclic semigroup contains a unique idempotent.

Proof. Almost everything was already proved in Lemmas 5.2.1, 5.2.4, 5.2.5and Corollary 5.2.6. The only thing which we have to show is that if thesemigroups 〈a〉 and 〈b〉 are isomorphic, then the types of a and b coincide.Let (k, m) be the type of a and (k′, m′) be the type of b. The fact that 〈a〉and 〈b〉 are isomorphic implies k + m = k′ + m′. From the observation afterCorollary 5.2.6 we have that an element of 〈a〉 is a group element if andonly if it belongs to P . In particular, 〈a〉 contains exactly m group elements.As 〈a〉 and 〈b〉 are isomorphic, we get that 〈b〉 contains exactly m groupelements as well. Hence m = m′ and thus k = k′ as well, since we alreadyknow that k + m = k′ + m′. This completes the proof.

Proposition 5.2.8 Let α ∈ PT n. Then the index of α equals the minimumk for which im(αk) = stim(α), and the period of α equals the order of thepermutation α|stim(α).

Proof. For every i > 0 the set im(αi) is invariant with respect to α. Hencewe have the following chain of invariant sets:

im(α) ⊃ im(α2) ⊃ im(α3) ⊃ · · · . (5.3)

As im(αi) = im(αi+1) implies im(αi) = im(αi+j) = stim(α) for all j > 0,the chain (5.3) has the form

im(α) � im(α2) � · · · � im(αk) = im(αk+1) = · · · .

By Proposition 2.9.3, the restriction of α to stim(α) is a permutation. Letm be the order of this permutation. Then the elements α1, α2, . . . , αk+m−1

are different. Indeed, the first k elements have different images, the last melements have the same image, but their restrictions to this image are per-mutations αk|stim(α), αk+1|stim(α),. . . , αk+m−1|stim(α), which are all differentsince the order of α|stim(α) is m. It is obvious that αk+m cannot be equalto any of αi, i < k, because the image of αk+m is different from that ofeach such αi. At the same time, we claim that αk+m = αk. Indeed, for everyx ∈ dom(αk) we have

αk+m(x) = αm(αk(x)) = αk(x)

as αk(x) ∈ stim(α) and αm acts as the identity transformation on stim(α).Since dom(αk+m) ⊂ dom(αk), we get αk+m = αk. This means that α hasindex k and period m, as claimed. The statement is proved.


Exercise 5.2.9 For each α ∈ Tn the period of α equals the least commonmultiple of the cardinalities of the kernels of all orbits of α.

Exercise 5.2.10 For each α ∈ Tn the index of α is the maximum over allcardinalities of trajectories for the vertices of the full subgraph of Γα withthe vertex set N\stim(α).

5.3 Isolated and Completely IsolatedSubsemigroups

A subsemigroup T of a semigroup S is called completely isolated providedthat ab ∈ T implies a ∈ T , or b ∈ T for all a, b ∈ S.

Exercise 5.3.1 Show that a proper subsemigroup T � S is completely iso-lated if and only if its complement T = S\T is a subsemigroup. In particular,if T is completely isolated, then T is completely isolated as well.

A subsemigroup T of a semigroup S is called isolated provided thatan ∈ T implies a ∈ T for all a ∈ S and n ∈ N. It is obvious that eachcompletely isolated subsemigroup is isolated. Each semigroup is a completelyisolated subsemigroup of itself.

Exercise 5.3.2 Show that a proper subsemigroup T � S is isolated if andonly if T is a union of subsemigroups.

Immediately from the definitions, Exercises 5.3.1 and 5.3.2 we have

Proposition 5.3.3 Let S be a semigroup.

(i) The intersection of any family of isolated subsemigroups of S is anisolated subsemigroup. In particular, for each a ∈ S there exists aminimum isolated subsemigroup of S, containing a.

(ii) If a union of some family of (completely) isolated subsemigroups of Sis a semigroup, then it is a (completely) isolated subsemigroup of S.

For each e ∈ E(S) set√

e = {x ∈ S : xm = efor somem > 0}.

Lemma 5.3.4 Let S be a semigroup.

(i) If T ⊂ S is an isolated subsemigroup, then√

e ⊂ T for all e ∈ T .

(ii) If T ⊂ S is isolated and S is finite, then T = ∪e∈E(T )

√e.

5.3. ISOLATED SUBSEMIGROUPS 75

Proof. The statement (i) is obvious. From (i) it follows that ∪e∈E(T )

√e ⊂ T .

On the other hand, S is finite and hence if a ∈ T , then 〈a〉 contains a uniqueidempotent, say e, by Theorem 5.2.7. This means that a ∈ √

e and thusT ⊂ ∪e∈E(T )

√e. This proves (ii) and completes the proof.

Remark 5.3.5 If S is finite, then√

e always contains Ge, that is, the max-imal subgroup, corresponding to e.

Remark 5.3.6 For the semigroups Tn, PT n, and ISn we have√

ε = Sn.

Exercise 5.3.7 Show that√

0 is not a subsemigroup of ISn if n > 1.

To describe all (completely) isolated subsemigroups of Tn, PT n, and ISn

we will need a number of auxiliary statements. For simplicity we shall callnoninvertible elements of semigroups singular. Also for simplicity, we shalluse the following notation for the elements from PT n:

α =(

A1 A2 . . . Am

a1 a2 . . . am

),

which means that α(a) = ai for all a ∈ Ai, i = 1, . . . , m, and that on thecomplement B = N\(A1 ∪ · · · ∪ Am) the element α acts as the identity onall elements, on which it is defined. If we consider several elements at thesame time, we assume that they act in the same way on the complement B.

Lemma 5.3.8 Let S be one of the semigroups Tn, PT n, or ISn, and T bea proper isolated subsemigroup of S. If T contains all singular idempotentsof S, then T = In−1.

Proof. The ideal In−1 coincides with the set of all singular elements in S.The cyclic subsemigroup of S, generated by a singular element a, containsa singular idempotent, say e. But a ∈ √

e by definition, and hence√

e ⊂ Tby Lemma 5.3.4(i). Hence In−1 ⊂ T . On the other hand, if T would containsome invertible element, T would contain the identity transformation asa power of this element, and hence the whole Sn by Remark 5.3.6 andLemma 5.3.4(i). This would imply T = S, a contradiction. Hence T =In−1.

Lemma 5.3.9 Let T be an isolated subsemigroup of Tn. If T contains allleft zeros of Tn, then T contains all singular idempotents of Tn.

Proof. Since in T2 each singular idempotent is a left zero, we may assumen > 2. Let A ⊂ N and let a ∈ A, b, c �∈ A, b �= c. We have

(b c Ac a a

)2

=(

b c Aa a a

)=(

b c Aa b a

)2

,


(b c Aa b a

)(b c Ac a a

)=(

b c Ab a a

).

Hence if an isolated subsemigroup T contains the element(

b c Aa a a

),

then T contains the element(

b c Ab a a

)as well. Applying the last state-

ment inductively starting from left zeros, we get that T contains all idem-

potents of the form εA =(

Aa

), where A ⊂ N, |A| > 1, and a ∈ A. For the

element εA we have that the corresponding partition ρεA contains a uniqueblock A of cardinality |A| ≥ 2, and each of the remaining blocks of thispartition consists of one element. Let now

ε =(

A1 A2 · · · Ak

a1 a2 · · · ak

)

be a singular idempotent, where A1, . . . , Ak are all blocks of ρε, which con-sists of more than one element. The proof is now completed by observingthat ε = εA1εA2 · · · εAk

.

Let ε ∈ E(Tn) be a singular idempotent and A ⊂ N. We will say that εis singular on A provided that A is invariant with respect to ε and ε acts asthe identity transformation on the complement N\A.

If A is the unique block of ρε, which contains more than one element, wewill say that the idempotent ρε is a constant on A. Clearly, via restriction,all constants on A can be considered as left zeros of the semigroup T (A).

Lemma 5.3.10 Let T be an isolated subsemigroup of Tn. Assume that forsome A ⊂ N, |A| ≥ 3, the semigroup T contains all idempotents, singularon A. Then for each k ∈ N\A the semigroup T contains all idempotents,singular on A ∪ {k}.

Proof. Taking Lemma 5.3.9 into account, it is enough to show that T con-tains all constants on A ∪ {k}. Let a ∈ A. Fix a partition A\{a} = A1 ∪ A2

and ai ∈ Ai, i = 1, 2. The element

ωa =(

A1 a A2 ka a a k

)

is a constant on A and hence belongs to T by our assumptions. For theelements

β =(

A1 a A2 ka1 a1 k a2

), γ =

(A1 a A2 ka1 k k a

),

δ =(

A1 a A2 kk k k a

)


we have that the elements

β2 =(

A1 a A2 ka1 a1 a2 k

), γ2 =

(A1 a A2 ka1 a a k

), δ2 = ωa

are singular idempotents on A. Hence β, γ, δ ∈ T . The latter implies thatthe following elements belong to T as well:

μa = ωaβγ =(

A1 a A2 ka a a a

), μk = δμa =

(A1 a A2 kk k k k

).

The statement follows.

Corollary 5.3.11 Let T be an isolated subsemigroup of Tn. Assume that forsome A ⊂ N, |A| ≥ 3, the semigroup T contains all idempotents, singularon A. Then T contains In−1.

Proof. Inductively applying Lemma 5.3.10 we obtain that T contains allsingular idempotents and the statement follows from Lemma 5.3.8.

For k, m ∈ N, m �= k, let εm,k denote the unique idempotent of Tn ofrank (n − 1) satisfying εm,k(m) = εm,k(k) = m.

Lemma 5.3.12 Let T be an isolated subsemigroup of Tn. If for some k1 �= k2

and m the semigroup T contains εm,k1 and εm,k2 , then T contains In−1.

Proof. Without loss of generality we may assume k1 = 1, m = 2, and k2 = 3.Let us show that T contains all constants on the set {1, 2, 3}. For the element

α =(

1 2 33 3 2

)we have α2 = ε2,1 and hence α ∈ T . Then

αε2,3 =(

1 2 33 3 3

)∈ T, ε2,3α =

(1 2 32 2 2

)∈ T.

Analogously one also shows that(

1 2 31 1 1

)∈ T . By Lemma 5.3.9 we

thus have that T contains all idempotents, singular on A. The statementnow follows from Corollary 5.3.11.

Lemma 5.3.13 Let T be an isolated subsemigroup of Tn. Assume that Tcontains εm1,k1 and εm2,k2 for some pairwise different k1, k2, m1, m2. ThenT contains In−1.

Proof. Without loss of generality we may assume m1 = 1, k1 = 2, m2 = 3,k2 = 4. By the same arguments as in the proof of Lemma 5.3.12, it is enoughto show that T contains all constants on {1, 2, 3, 4}. For the element α =(

1 2 3 42 3 1 1

)we have α3 = ε3,4 and hence α ∈ T . Thus β = ε1,2αε1,2 ∈ T


and a direct commutation shows that β is the constant on {1, 2, 3, 4} withimage {1}. At the same time αβ ∈ T and αβ is the constant on {1, 2, 3, 4}with image {2}. Analogously one shows that the remaining constants arealso in T .

Lemma 5.3.14 Let T be an isolated subsemigroup of Tn. Assume that Tcontains some idempotent ε of rank r < n− 1. Then T contains an idempo-tent of rank (r + 1).

Proof. If the partition ρε has some block B of cardinality |B| ≥ 3, thenan idempotent of rank (r + 1) in T can be easily obtained using the sameconstruction as in the proof of Lemma 5.3.9. If all blocks of ρε contain atmost two elements, there should be more than one two-element block. Hencewithout loss of generality we may assume that

ε =(

1 2 3 4 5 · · · n1 1 3 3 a5 · · · an

).

For the elements

α =(

1 2 3 4 5 · · · n3 4 1 1 a5 · · · an

), β =

(1 2 3 4 5 · · · n3 3 1 2 a5 · · · an

)

we have α2 = β2 = ε and hence α, β ∈ T . Thus

βα =(

1 2 3 4 5 · · · n1 2 3 3 a4 · · · an

)∈ T,

and the element βα is an idempotent of rank (r + 1).

Lemma 5.3.15 Let T be an isolated subsemigroup of Tn. Assume that Tcontains some idempotent ε of rank (n − 2). Then T contains In−1.

Proof. Without loss of generality we may assume that ε =(

1 2 31 1 1

), or

ε =(

1 2 3 41 1 3 3

).

In the case ε =(

1 2 31 1 1

)we may use the same construction as in

the proof of Lemma 5.3.9 to get ε1,2, ε1,3 ∈ T . Thus T contains In−1 byLemma 5.3.12.

In the case ε =(

1 2 3 41 1 3 3

)we may use the same construction as

in the proof of Lemma 5.3.14 to get ε1,2, ε3,4 ∈ T . Thus T contains In−1 byLemma 5.3.13.

Lemma 5.3.16 Let ε ∈ E(Tn) be of rank (n − 1). Then√

ε = Gε.


Proof. The inclusion Gε ⊂√

ε is obvious. Let now α ∈ √ε. Then αm = ε for

some m > 0. From the inequalities

n > rank(α) ≥ rank(ε) = n − 1

we have rank(α) = n − 1. As im(α) is invariant with respect to α, andαm = ε, we have stim(α) = im(α). From Proposition 5.2.8 it now followsthat the index of α is 1 and hence by Theorem 5.2.7 the semigroup 〈α〉is a group. Hence α is a group element and thus α ∈ Gε. This proves theinclusion

√ε ⊂ Gε and completes the proof.

Theorem 5.3.17 Let n > 2. A subsemigroup T of Tn is isolated if and onlyif it belongs to the following set:

{Tn,Sn, In−1} ∪{Gεm,k

∪ Gεk,m: m, k ∈ N, m �= k

}∪

∪{∪m∈MGεm,k

: k ∈ N, ∅ �= M ⊂ N\{k}}

.

Proof. As both Sn and In−1 are subsemigroups, Sn∩In−1 = ∅ and Tn = Sn∪In−1, each of the semigroups Sn, In−1, and Tn is an isolated subsemigroupof Tn.

Let T be an isolated subsemigroup of Tn such that T ∩Sn �= ∅. As Sn isa finite group, T must contain ε and hence Sn =

√ε ⊂ T by Lemma 5.3.4(i).

If at the same time T ∩ In−1 �= ∅, then Lemma 5.3.14 guarantees that Tcontains an idempotent of rank (n−1). Thus T contains a generating systemfor Tn by Theorem 3.1.3, which implies that T = Tn. If, on the other hand,T ∩ In−1 = ∅, we have T = Sn.

It is left to consider the case when T ∩ Sn = ∅ and T �= In−1. FromLemmas 5.3.14 and 5.3.15 it follows that in this case T contains only idem-potents of rank (n − 1), that is, of the form εm,k.

Let E(T ) = {εmi,ki : i = 1, . . . , t}. If {mi, ki} ∩ {mj , kj} = ∅ for somei, j, then T ⊃ In−1 by Lemma 5.3.13, which is not possible. Hence for alli, j we have {mi, ki} ∩ {mj , kj} �= ∅. Moreover,

εl,mεm,k =(

k l ml l l

)

is an idempotent of rank (n−2) as well. For i �= j the possibility mi = mj isprohibited by Lemma 5.3.12. Hence we either have ki = kj , or mi = kj andmj = ki. In the last case the only possibility is E(T ) = {εm,k, εk,m} as forany other εm′,k′ such that {m′, k′}∩{k, m} �= ∅ we would have k = k′ = m,which is impossible.

So, we have either E(T ) = {εm,k, εk,m}, or E(T ) = {εmi,k : i = 1, . . . , t}.Because of Lemmas 5.3.4(ii) and 5.3.16 it remains to show that each ofthe sets

P = Gεm,k∪ Gεk,m

and Q = Gεm1,k∪ · · · ∪ Gεmt,k

(5.4)


is a subsemigroup of Tn. In the first case of (5.4) for arbitrary α ∈ Gεm,k

and β ∈ Gεk,mwe have ραβ = ρβ = ρα and im(αβ) = im(α). Hence αβ ∈

Gεm,kby Theorem 4.5.1(iii) and because of the symmetry of the situation

we conclude that Gεm,k∪ Gεk,m

is a semigroup.In the second case of (5.4) we have

im(εm1,k) = im(εm2,k) = · · · = im(εmt,k) = N\{k}.

For any α, β ∈ Q we have ραβ = ρβ and im(αβ) = N\{k}. By Theorem4.5.1(iii) we have αβ ∈ H(β) ⊂ Q. Hence Q is a subsemigroup as well. Thiscompletes the proof.

Remark 5.3.18 The only isolated subsemigroup of T1 is T1 itself. Thesemigroup T2 has five isolated subsemigroups: T2, S2, I1, {ε1,2}, {ε2,1}.

Corollary 5.3.19 For n > 1 the only completely isolated subsemigroups ofTn are Tn, Sn, and In−1.

Proof. Since each completely isolated subsemigroup is isolated, we have justto check which of the subsemigroups, given by Theorem 5.3.17, are com-pletely isolated. It is obvious that Tn is completely isolated. Further, bothSn and In−1 are completely isolated as each of these semigroups is the com-plement of the other one.

Let T be an isolated subsemigroup of Tn, different from Tn, Sn, and In−1.Then from Theorem 5.3.17 we get that T consists only of group elements ofrank (n − 1).

If n = 2, then neither {ε1,2} nor {ε2,1} is completely isolated. Indeed,assume {ε1,2} is completely isolated. Then (1, 2) · ε2,1 = ε1,2 ∈ T . As (1, 2)has rank 2, we have (1, 2) �∈ T , implying ε2,1 ∈ T and hence T = In−1, acontradiction. Analogously one shows that {ε2,1} is not completely isolated.

Assume n > 2. Then the complement T = Tn\T contains Sn and theelement

α =(

1 2 3 4 · · · n1 1 2 3 · · · n − 1

),

which is not a group element by Exercise 5.1.5. By Theorem 3.1.3 the setSn ∪ {α} generates the whole Tn and hence T cannot be a subsemigroup.Hence T is not completely isolated by Exercise 5.3.1.

Let us now describe isolated and completely isolated subsemigroups ofISn. The elements of N which will be omitted in the notation of sometransformation α will be assumed to be fixed points of α.

Lemma 5.3.20 If an isolated subsemigroup T ⊂ ISn contains some idem-potent ε of rank at most (n − 2), then T contains at least two differentidempotents of rank (n − 1).


Proof. Let rank(ε) = n− k, k > 1, and dom(ε) = {a1, a2, . . . , ak}. Then theelements α = [a1, a2, . . . , ak] and β = [ak, a2, . . . , a1] satisfy αk = βk = ε.Hence α, β ∈ T and thus βα, αβ ∈ T . But both βα = [ak] and αβ = [a1] areidempotents of rank (n − 1), and they are obviously different.

Lemma 5.3.21 If an isolated subsemigroup T ⊂ ISn contains two differentidempotents of rank (n − 1), then T contains In−1.

Proof. Assume that T contains the idempotents [x] and [y], x �= y. If n = 2,then T contains [x][y] and thus contains all singular idempotents. This meansthat T contains In−1 by Lemma 5.3.8.

Assume now that n > 2 and let z �∈ {x, y}. Then ([x](y, z))2 = [x] andhence [x](y, z) ∈ T . This means that T also contains the element

α = [y] · [x](y, z) · [y] = [x][y][z].

Further, T contains both β = [x, y, z] and γ = [z, y, x] since β3 = γ3 = α.Hence the idempotent [z] = γβ is also contained in T . In particular, Tcontains all singular idempotents of rank (n − 1). As every singular idem-potent ε = [a1][a2] . . . [ak] of ISn is a product of singular idempotents [ai],i = 1, . . . , k, of rank (n− 1), we obtain that T contains all singular idempo-tents. Hence T contains In−1 by Lemma 5.3.8.

Theorem 5.3.22 The only isolated subsemigroups of ISn are ISn, Sn,In−1, and G[x], x ∈ N.

Proof. Let T be an isolated subsemigroup of ISn. Repeating the argumentsof the first part of the proof of Theorem 5.3.17 one shows that either T ∈{ISn,Sn, In−1}, or T � In−1.

If T � In−1, from Lemmas 5.3.20 and 5.3.21 we have that E(T ) = {[x]}for some x ∈ N. But then T =

√[x] = G[x] by Lemma 5.3.4(ii). On the

other hand, the semigroup√

[x] = G[x] is obviously isolated.

Corollary 5.3.23 The only completely isolated subsemigroups of ISn areISn, Sn, and In−1.

Proof. Each completely isolated subsemigroup is isolated. So, we have justto go through the list given by Theorem 5.3.22. The semigroups ISn, Sn,and In−1 are obviously completely isolated. On the other hand, if n > 1,none of the semigroups G[x], x ∈ N, is completely isolated. Indeed, let y �= x.If G[x] were completely isolated, from (x, y)[y](x, y) = [x] and (x, y) �∈ G[x]

it would follow [y] ∈ G[x], which is not the case.

Finally, let us describe all completely isolated subsemigroups of PT n.Note that PT 1 = IS1. We shall need the following obvious result:

Lemma 5.3.24 (i) Let S be a semigroup and A < B < S. If A is com-pletely isolated in S, then A is completely isolated in B.


(ii) Let S be a semigroup, A < S completely isolated and B < S. ThenA ∩ B is either empty or a completely isolated subsemigroup of B.

Theorem 5.3.25 For n > 1 the semigroup PT n has seven completelyisolated subsemigroups, namely, PT n, Tn, Sn, PT n\Sn, Tn\Sn, PT n\Tn,PT n\(Tn\Sn).

Proof. A product of two nontotal transformations is not total either. Hencethe complement PT n\Tn of the subsemigroup Tn is a subsemigroup itself.This means that both PT n\Tn and Tn are completely isolated in PT n.Further, as both Sn and In−1 = PT n\Sn are subsemigroups of PT n, theyboth are completely isolated by Exercise 5.3.1.

Let T be a completely isolated subsemigroup of PT n. The intersectionT ∩ Tn is either a empty or a completely isolated subsemigroup of Tn be-cause of Lemma 5.3.24(ii). Hence T ∩ Tn is either Tn, or Sn, or Tn\Sn byCorollary 5.3.19. The product of two noninvertible transformations of N isnot invertible, and the product of two nontotal transformations is not total.Hence the subsemigroup Tn\Sn is completely isolated.

Assume that T ∩ (PT n\Tn) �= ∅. Then T contains some idempotent

ε =(

A · · · B Ca · · · b ∅

)∈ PT n\Tn.

Let x ∈ im(ε) and c ∈ C. Set εx = (x, c)ε(x, c). We claim that εx ∈ T . Indeed,this is obvious if (x, c) ∈ T . If (x, c) �∈ T , then (x, c)εx(x, c) = ε ∈ T andhence εx ∈ T since T is completely isolated. Thus T contains the element ε′ =εεx, which is an idempotent and im(ε′) = im(ε)\{x}. Continuing inductivelywe get that T contains the element 0n. From 0n = [1][2] . . . [n] we havethat T � [k] for some k ∈ N. By the same arguments as above we get thatfor each l �= k the semigroup T contains the element (k, l)[k](k, l) = [l].In particular, T contains all [k], k ∈ N, and all their products, that is, allsingular idempotents from ISn.

If n = 2, the above gives us that PT 2 contains three minimal completelyisolated subsemigroups: S2, T2\S2, and PT 2\T2 = IS2\S2. Any union ofthese subsemigroups is again a semigroup, hence completely isolated byProposition 5.3.3(ii).

Assume now that n ≥ 3. We continue the consideration of the caseT ∩ (PT n\Tn) �= ∅ and recall that we already know that T contains allsingular idempotents from ISn. Consider the following elements in PT n:

α =(

1 2 31 1 ∅

), β =

(1 2 3∅ 3 3

), αβ =

(1 2 3∅ ∅ ∅

).

αβ is a singular idempotent of ISn. Hence α ∈ T , or β ∈ T . Without loss ofgenerality we may assume α ∈ T . As in the previous paragraph one showsthat πατ ∈ T for all π, τ ∈ Sn, which implies that T contains all idempotents


from PT n\Tn of rank (n− 2), that is, all elements of the form εz,y[x], wherex, y, z ∈ N are arbitrary different elements.

Now let

δ =(

A1 · · · Ak Ba1 · · · ak ∅

)∈ PT n\Tn

be any idempotent, written such that all blocks Ai, i = 1, . . . , k, consist ofmore than one element. For each i = 1, . . . , k consider the idempotent

δi =(

Ai Bai ∅

).

Let Ai = {ai, x1, . . . , xk} and b ∈ B. We have

δi = εN\B · εai,x1 [b] · εai,x2 [b] · · · εai,xk[b]

and hence δi ∈ T . At the same time δ = δ1δ2 · · · δk and hence δ ∈ T .Therefore T contains all idempotents from PT n\Tn and thus T containsPT n\Tn by Lemma 5.3.4(ii).

The above explanation gives us that for n > 2 the semigroup PT n

contains three minimal completely isolated subsemigroups: Sn, Tn\Sn, andPT n\Tn. Any union of these subsemigroups is again a semigroup, hencecompletely isolated by Proposition 5.3.3(ii). This completes the proof.


5.4.1 For each semigroup S the Boolean B(S) has the following naturalstructure of a semigroup with respect to the multiplication: A · B = {a · b :a ∈ A, b ∈ B}, A, B ⊂ S (it is very easy to check that this multiplication isassociative). The semigroup B(S) is called the global semigroup or the powersemigroup of S. Theorem 5.2.7 was proved by Frobenius in 1895 for theelements of the semigroup B(S), where S is a finite group (this restrictiondid not play an essential role in the proof). Since that time Theorem 5.2.7has been rediscovered by many different authors.

5.4.2 Finite semigroups have the following property:

Theorem 5.4.1 Let S be a finite semigroup. Then D = J .

Proof. As we already know that D ⊂ J (see Sect. 4.4), it is enough toshow that J ⊂ D. Let aJ b, that is, S1aS1 = S1bS1. Then there existu, v, x, y ∈ S1 such that uav = b and xby = a. Thus (xu)a(vy) = a and even(xu)ka(vy)k = a for all k ∈ N. As S1 is finite, there exist k and l such thate = (xu)k and f = (vy)l are idempotents. We have

a = (xu)kla(vy)kl = eaf,


which implies ea = a and af = a. From the first equality we have

(xu)k−1x · ua = a and u · a = ua.

Hence aLua. Analogously one shows that aRav, implying uaRuav, that is,uaRb. Thus aDb.

5.4.3 A subsemigroup T ⊂ S is called maximal provided that T �= S and forany subsemigroup X ⊂ S the inclusion T ⊂ X implies T = X, or X = S.

Obviously, if S is one of the semigroups PT n, Tn, or ISn, and G is asubgroup of Sn, then the set In−1 ∪ G is a subsemigroup of S.

Theorem 5.4.2 Let S be one of the semigroups Tn or ISn. Then the onlymaximal subsemigroups of S are Sn ∪ In−2 and In−1 ∪ G, where G is amaximal subgroup of Sn.

Proof. Let T be a maximal subsemigroup of S. If Sn �⊂ T , then T ∪ In−1 isa subsemigroup and T ⊂ T ∪ In−1 � S. Hence T = T ∪ In−1. On the otherhand, G = T ∩ Sn is a subsemigroup of Sn and hence is a subgroup sinceSn is finite. Furthermore, G has to be a maximal subgroup for otherwisethere would exist a subgroup G′ of Sn such that G � G′

� Sn. In this caseG′ ∪ In−1 is a subsemigroup of S and T � G′ ∪ In−1 � Sn, a contradiction.

If Sn ⊂ T , the semigroup T cannot contain any element of rank (n−1) forotherwise T = S by Theorems 3.1.3 or 3.1.4. Hence T ⊂ Sn ∪ In−2. On theother hand, Sn ∪In−2 is a subsemigroup of S and hence T = Sn ∪In−2.

Theorem 5.4.3 The only maximal subsemigroups of PT n are:

(a) In−1 ∪ G, where G is a maximal subgroup of Sn;

(b) PT n\A, where A = {α ∈ Tn : rank(α) = n − 1};

(c) PT n\B, where B = {α ∈ ISn : rank(α) = n − 1}.

Proof. The proof is similar to that of Theorem 5.4.2. One has only to notethat each element of rank (n− 1) from PT n is either a total transformationor a partial injection.

By the above theorems, the classification of maximal subsemigroups inPT n, Tn, and ISn reduces to the classification of maximal subgroups of thesymmetric group Sn. The latter problem is very difficult. The only knownsolution so far, which can be found in [LPS], is based on the so-called clas-sification of finite simple groups. The correctness of the latter result is stillquestioned by many specialists.


5.4.4 From Theorem 5.4.2 it is easy to see that each maximal subsemigroupof ISn is inverse. Thus as a bonus we get a description of all maximalinverse subsemigroups of ISn, see [X.Ya1]. Analogous questions can be askedfor both PT n and Tn, especially taking into account that every α ∈ Tn iscontained in some inverse subsemigroup of Tn, see [Sc5]. These questionswere studied in particular in [Ni, Re2, KS, H.Ya].

5.4.5 Let α ∈ PT n. By Proposition 5.2.8 and Exercises 5.2.9 and 5.2.10 thecomputation of the order of α can be easily reduced to the computation ofthe order of the permutation α|stim(α). If π = (a1, . . . , ak) · · · (b1, . . . , bl), thenthe order of π is just the least common multiple (lcm) of the lengths k, . . . , lof the cycles. However, given some m to find the number of those σ ∈ Sn

which have order m is a difficult problem. It reduces to the determinationof all decompositions n = n1 + · · · + nk such that lcm(n1, . . . , nk) = m.Determination of the maximal possible order Pmax(Sn) for elements in Sn

is equivalent to finding

maxn1+···+nk=n

lcm(n1, . . . , nk),

which is also very difficult. Analogous problems for the semigroups PT n,Tn, and ISn are of the same level of difficulty.

5.4.6 On the other hand, the asymptotic behavior of Pmax(Sn) is mucheasier to describe. Already in 1903 E. Landau has proved in [La] the followingformula:

ln(Pmax(Sn)) ∼√

n ln(n), n → ∞. (5.5)

A very elementary proof of this statement can be found in [Mi]. Some addi-tional comments about the asymptotic behavior of Pmax(Sn) can be found in[Sz]. From the formula (5.5) one easily obtains the following result about theasymptotic behavior of the orders of elements in the semigroups Tn, PT n,and ISn.

Theorem 5.4.4

ln(Pmax(Tn)) ∼ ln(Pmax(PT n)) ∼ ln(Pmax(ISn)) ∼√

n ln(n), n → ∞.

5.4.7 From the proof of Theorem 2.4.3 it follows that each finite semigroupS can be considered as a subsemigroup of some Tn, moreover if S is a monoid,then one can even assume that the identity element of S coincides with theidentity transformation in Tn. The group S∗ becomes, under such identifi-cation, a subgroup of Sn. As a consequence of this and Lemma 5.3.24(ii) weobtain:

Proposition 5.4.5 If S is a finite semigroup, then both S∗ and S\S∗ arecompletely isolated subsemigroups of S.


The statement of Proposition 5.4.5 is wrong in the general case as thegroup of units of an infinite semigroup does not have to be completely iso-lated (a bijective transformation can be a composition of two transforma-tions, each of which is not bijective).

5.4.8 Let S be a semigroup. There is a natural partial order on the set ofall J -classes of S, given by the inclusion of the principal two-sided ideals:

J (a) ≤ J (b) ⇔ S1aS1 ⊂ S1bS1.

If S is finite, then J = D by Theorem 5.4.1 and hence the above partialorder is also a partial order on the set of all D-classes of S.

Theorem 5.4.6 Let S be a finite monoid and assume that in the set of allD-classes of S, different from D(1), there is the maximum D-class D.

(i) For any idempotent e ∈ D we have√

e = Ge.

(ii) Let e1, e2, . . . , ek be a collection of idempotents from D which eitherbelong to the same R-class or to the same L-class. Then Ge1 ∪ Ge2 ∪· · · ∪ Gek

is an isolated subsemigroup of S.

We note that the semigroups Tn, PT n, and ISn satisfy all conditions ofTheorem 5.4.6.

Proof. Consider S as a subsemigroup of some Tn. Then aRb in S impliesaRb in Tn by Proposition 4.4.2. Hence aRb implies im(a) = im(b) by Theo-rem 4.2.1. Analogously, aLb implies ρa = ρb by Theorem 4.2.4. In particular,all elements from the same D-class of S have the same rank.

Obviously D(ak) ≤ D(a). Hence for any idempotent e ∈ D we have√e ⊂ D. Let a ∈ √

e. From the equalities am = e, rank(a) = rank(e), and thefact that im(a) is invariant with respect to a, we get that stim(a) = im(a).Hence, by Proposition 5.2.8, a has index one and thus is a group element.It follows that

√e ⊂ Ge, which proves (i).

To prove (ii) it is enough to show that Ge1 ∪ Ge2 ∪ · · · ∪ Gekis closed

with respect to multiplication. Let e = ei and f = ej for i �= j and a ∈ Ge,b ∈ Gf . Assume first that all eis belong to the same R-class. We claim thatin this case ab ∈ Gf . As S is a subsemigroup of Tn, we have im(a) = im(b) byTheorem 4.5.1(i). Since both a and b are group elements, different elementsof im(a) = im(b) must belong to different classes of the partitions ρa and ρb,respectively. Hence ρab = ρb and im(ab) = im(a). This means that ab ∈ Gf

by Theorem 4.5.1(iii).If all eis belong to the same L-class, then using analogous arguments one

shows that ab ∈ Ge. This completes the proof.

5.4.9 The description of isolated semigroups for PT n is even more technicalthan that for Tn. Hence we just present here the result and an idea of theproof. To work out all the details of the proof is left to the reader.


Theorem 5.4.7 A subsemigroup T of PT n is isolated if and only if Tbelongs to the following list:

(a) Completely isolated subsemigroups of PT n, given by Theorem 5.3.25.

(b) Isolated but not completely isolated subsemigroup of Tn\Sn (given by thelist from Theorem 5.3.17 with {Sn, Tn, In−1} taken away).

(c) Semigroups T (x, P ), where x ∈ N and P is an isolated subsemigroup ofT (N\{x}), defined as follows:

T (x, P ) = {α ∈ PT n : dom(α) = {x}, α|N\{x} ∈ P}.

All possibilities for the semigroup P are given by Theorem 5.3.17.

(d) Semigroups T (x,S(N\{x}))⋃

∪m∈MGεm,x , where x ∈ N and ∅ �= M ⊂N\{x}.

Sketch of the Proof. By Theorem 5.3.25, the semigroup PT n has three min-imal completely isolated subsemigroups, namely, Sn, Tn\Sn, and PT n\Tn.If T ⊂ PT n is an isolated subsemigroup and S is a minimal completelyisolated subsemigroup of PT n, then S ∩T is an isolated subsemigroup of S.

Let us first determine all possible intersections of T with the minimalcompletely isolated subsemigroups of PT n. The semigroup T ∩ Sn is ob-viously either empty or equals Sn. From Theorem 5.3.17 the semigroupT ∩ (Tn\Sn) is either empty, or equals Tn\Sn or is given exactly by (b).The complicated part is to show that if T ∩ (PT n\Tn) is a proper subsemi-group of PT n\Tn, then it is given by (c).

So, let T be a proper isolated subsemigroup of PT n\Tn. First we claimthat T contains an idempotent e such that |dom(e)| = 1. This can be provedanalogously to the proof of Lemma 5.3.20. Then one shows that such Tcannot contain 0n. For this one shows that if T contains 0n, then T containsall idempotents of PT n\Tn and hence must coincide with the latter. Thisagain can be done generalizing the arguments of Lemma 5.3.20. The nextstep is to show that T contains an idempotent e such that all blocks of ρe onwhich e is defined contain at most two elements. This is similar to the proofof Lemma 5.3.9. Generalizing the arguments of the proof of Lemma 5.3.21one shows that all idempotents of T must satisfy |dom(e)| = 1. From this itfollows easily that there should exist x ∈ N such that dom(α) = N\{x} forall α ∈ T . In particular, it follows that x �∈ im(α) for all α ∈ T and hence theset N\{x} is invariant with respect to all elements from T . The restrictionto N\{x} defines a homomorphism from T to T (N\{x}), the image of whichis an isolated subsemigroup of T (N\{x}), call it P . This gives T = T (x, P )as in (c). On the other hand, it is easy to check that every T = T (x, P ) isisolated (this also follows from Exercise 5.5.13(a)).


It is left to see whether we can form a union of some of the semigroupsgiven by (a), (b), and (c). A case-by-case analysis shows that the only pos-sible unions are either the ones already described in (a) or the ones given by(d). The main argument of the analysis is that multiplying elements fromdifferent components of our union it is prohibited to get elements of smallerranks. We leave the details to the reader.

5.4.10 There are of course many other natural classes of subsemigroups inthe semigroups Tn, PT n, and ISn, defined using various conditions. For in-stance, one can consider abelian subsemigroups, or nilpotent subsemigroups,or subsemigroups defined via some combinatorial conditions (for example,consisting of all α such that α(x) ≤ x for all x ∈ N, or consisting of allα such that x ≤ y implies α(x) ≤ α(y) for all x, y ∈ N). Some of theseclasses are of independent interest and are intensively studied. In this book,we shall only briefly discuss nilpotent subsemigroups later on in Chap. 8.


5.5.1 Let G be a maximal subgroup of Tn, or ISn. Show that G is a maximalsubgroup of PT n.

5.5.2 Let S be a finite semigroup and a ∈ S. Prove that the followingconditions are equivalent:

(a) a is a group element.

(b) 〈a〉 is a group.

(c) ak = a for some k > 1.

(d) a is regular and commutes with at least one of its inverses.

5.5.3 Let α ∈ PT n and assume that rank(αk) = rank(αk+1). Prove that

(a) rank(αk+j) = rank(αk) for all j > 0

(b) strank(α) = strank(αk)

(c) im(αk+j) = im(αk) for all j > 0

(d) stim(α) = stim(αk)

5.5.4 Let α ∈ PT n. Assume that for some m > 1 the restriction αm|stim(α)

is the permutation, which is inverse to the permutation α|stim(α). Prove thatthe element α is a group element if and only if α and αm form a pair ofinverse elements.

5.5.5 Let S be a semigroup and e ∈ E(S). Show that eSe is the maximumsubmonoid of S, which has e as the identity element.


5.5.6 Let α ∈ PT n be a group element. Show that the restriction α|im(α) isa permutation.

5.5.7 Let α ∈ PT n. Show that α is a group element if and only if im(α) =im(α2).

5.5.8 (a) Let S be a cyclic semigroup, which is not a group. Show that Shas a unique irreducible generating system.

(b) Show that for each k ∈ N there exists a finite cyclic group and an irre-ducible generating system in this group, containing exactly k elements.

5.5.9 Let [p]q = p(p − 1)(p − 2) · · · (p − q + 1), and let P (n, m) denote thenumber of elements of order m in the symmetric group Sn.

(a) Show that the number of elements of type (k, m) in the semigroup Tn

equals ∑n1+···+nk+nk+1=n

nn12 nn2

3 · · ·nnk−1

k nnkk+1P (nk+1, m).

(b) Show that the number of elements of type (k, m) in the semigroup PT n

equals ∑n1+···+nk+nk+1≤n

nn12 nn2

3 · · ·nnk−1

k nnkk+1P (nk+1, m).

(c) Show that the number of elements of type (k, m) in the semigroup ISn

equals ∑n1+···+nk+nk+1=n

[n2]n1 [n3]n2 · · · [nk]nk−1P (nk+1, m).

5.5.10 Prove that a (one-sided) ideal I of a semigroup S is an isolatedsubsemigroup of S if and only if I is semiprime.

5.5.11 Construct a semigroup S and a subsemigroup T such that a2 ∈ Timplies a ∈ T for any a ∈ S, but at the same time T is not an isolatedsubsemigroup of S.

5.5.12 Find the number of isolated subsemigroups in Tn, n > 2.

5.5.13 (a) Let T be an isolated subsemigroup of S and X be an isolatedsubsemigroup of T . Show that X is an isolated subsemigroup of S.

(b) Let T be a completely isolated subsemigroup of S and X be a completelyisolated subsemigroup of T . Is it true that X is a completely isolatedsubsemigroup of S?

5.5.14 Show that the intersection of completely isolated subsemigroups isnot completely isolated in general.

Chapter 6

Other Relationson Semigroups

6.1 Congruences and Homomorphisms

Recall that a binary relation ρ on a semigroup S is called left compatibleprovided that a ρ b implies ca ρ cb for all a, b, c ∈ S; ρ is called right compatibleprovided that a ρ b implies ac ρ bc for all a, b, c ∈ S; and ρ is called compatibleprovided that it is both left and right compatible.

A left compatible equivalence relation is called a left congruence. A rightcompatible equivalence relation is called a right congruence. A compatibleequivalence relation is called a congruence.

Example 6.1.1 (a) From Sect. 4.4 we know that Green’s relation L is aright congruence and Green’s relation R is a left congruence.

(b) Each semigroup S has two trivial congruences: the equality relation(the identity congruence ιS) and the uniform congruence ωS = S × Sconsisting of just one equivalence class S.

If S is a set and ρ is an equivalence relation on S, there are severalways to show that some a, b ∈ S belong to the same class of ρ. This can bewritten as follows: a ρ b, or (a, b) ∈ ρ, or a ≡ b(ρ) or simply a ≡ b if thereis no confusion about which congruence we are considering. In what followswe may use any of the above notation.

Exercise 6.1.2 Prove that the intersection of an arbitrary family of con-gruences on S is a congruence on S.

Lemma 6.1.3 Let S be a semigroup and ρ be a congruence on S. Let furtherK be an equivalence class of ρ. If K contains some ideal of S, then K is anideal of S itself. In addition, at most one class of ρ can be an ideal.



92 CHAPTER 6. OTHER RELATIONS ON SEMIGROUPS

Proof. Let I be the ideal of S, contained in K. Fix some b ∈ I and leta ∈ K and x ∈ S be arbitrary. As a ≡ b, we have xa ≡ xb and ax ≡ bx. Butxb, bx ∈ I since I is an ideal. This implies xa ∈ K and ax ∈ K.

Let I and J be two ideals of S. Then IJ ⊂ I∩J , in particular, I∩J �= ∅.Hence I and J cannot be at the same time two different equivalence classesof an equivalence relation on S.

Let S be a semigroup and I ⊂ S an ideal. Define the equivalence relationρI on S as follows:

ρI = (I × I) ∪ {(a, a) : a ∈ S\I}.

In other words, all elements from I form one equivalence class, and allother equivalence classes are trivial (each consists of a single element). Thefact that I is an ideal guarantees that ρI is a congruence. Indeed, if a ≡ b(ρI),then either a = b, or a, b ∈ I. In the last case ac, bc, ca, cb ∈ I for any c ∈ S.The congruence ρI is called the Rees congruence on S with respect to theideal I.

If ρ is a congruence on S and a ∈ S, the equivalence class of a in ρ isdenoted by aρ or simply by a if the congruence ρ is clear from the context.

Let ρ be a congruence on a semigroup S. Consider the set S/ρ of allequivalence classes. For a, b ∈ S set

a · b := ab. (6.1)

Lemma 6.1.4 The formula (6.1) defines a binary associative operation onS/ρ. In other words, (S/ρ, ·), where · is defined by (6.1), is a semigroup.

Proof. First we have to show that (6.1) defines a well-defined binary op-eration on S/ρ. Let a′ ∈ a and b′ ∈ b. Since ρ is compatible we haveab ≡ ab′ ≡ a′b′ and hence a ·b = a′ ·b′, that is, the operation · is well-defined.The associativity of · follows from the associativity of the multiplicationin S:

(a · b) · c = ab · c = abc = a · bc = a · (b · c).

The semigroup (S/ρ, ·) constructed above is called the quotient or thefactor of S modulo the congruence ρ. If ρ = ρI , one usually simply writesS/I instead of S/ρI . The semigroup S/I is called the Rees quotient of Smodulo the ideal I.

If S is a monoid with the identity element 1, then from (6.1) we getthat 1 will be the identity element in S/ρ. The quotient S/ιS is naturallyidentified with S. The quotient S/ωS is the one-element semigroup.

Let now (S, ·) and (T, ∗) be two semigroups. A mapping ϕ : S → T iscalled a homomorphism provided that for all a, b ∈ S we have

ϕ(a · b) = ϕ(a) ∗ ϕ(b). (6.2)

6.1. CONGRUENCES AND HOMOMORPHISMS 93

The notion of an isomorphism from Sect. 2.4 is just a special case ofthe notion of a homomorphism. An isomorphism is just a bijective ho-momorphism. Other special cases of homomorphisms are (1) injectivehomomorphisms, which are called monomorphisms and (2) surjectivehomomorphisms, which are called epimorphisms. Monomorphisms are usu-ally denoted by the symbol ↪→, and epimorphisms by the symbol �.

A homomorphism ϕ : S → S is called an endomorphism of S. A bijectiveendomorphism is called an automorphism. The set of all endomorphisms ofS is denoted by End(S) and the set of all automorphisms of S is denoted byAut(S). Both these sets are not empty since they both contain the identitytransformation on S.

Proposition 6.1.5 The set End(S) is a monoid with respect to the compo-sition of mappings. Aut(S) is the group of units in End(S), in particular,Aut(S) is a group.

Proof. Let ϕ, ψ ∈ End(S). For any a, b ∈ S we have

ϕ(ψ(ab)) = ϕ(ψ(a)ψ(b)) = ϕ(ψ(a))ϕ(ψ(b)).

Hence End(S) is closed with respect to the composition of mappings. Sincethis composition is associative (by Proposition 2.1.1), End(S) is a semigroup.The identity element of End(S) is the identity transformation. As the com-position of bijections is a bijection, Aut(S) is a subsemigroup of End(S). Letϕ ∈ Aut(S) and consider the inverse bijection ϕ−1. Applying ϕ−1 to (6.2)for any a, b ∈ S we have

ab = ϕ−1(ϕ(a)ϕ(b)). (6.3)

Take x = ϕ(a) and y = ϕ(b). Then a = ϕ−1(x) and b = ϕ−1(y) and (6.3)becomes

ϕ−1(x)ϕ−1(y) = ϕ−1(xy)

and x, y ∈ S are arbitrary since a, b ∈ S were arbitrary and ϕ−1 is a bijec-tion. This means that ϕ−1 ∈ Aut(S) and thus all elements of Aut(S) areinvertible. On the other hand, each invertible transformation from End(S)is a bijection and hence belongs to Aut(S). This completes the proof.

If S is a semigroup and ρ is a congruence on S, we can define the canonicalprojection or the canonical epimorphism πρ : S � S/ρ via a → a. Themapping πρ is surjective by definition and is a homomorphism by (6.1).

Let S and T be two semigroups and ϕ : S → T be a homomorphism.The equivalence relation

Ker(ϕ) = {(a, b) ∈ S × S : ϕ(a) = ϕ(b)}


is called the kernel of ϕ. This relation is a congruence on S. Indeed,

ϕ(a) = ϕ(b) ⇒ ϕ(ca) = ϕ(c)ϕ(a) = ϕ(c)ϕ(b) = ϕ(cb),

which shows that Ker(ϕ) is left compatible. That Ker(ϕ) is right compatibleis proved analogously. On the other hand, if ρ is a congruence on S, then wehave the homomorphism πρ : S � S/ρ and Ker(πρ) = ρ. This means thateach congruence is the kernel of some homomorphism.

If ϕ : S � T is an epimorphism, T is called a homomorphic image of S. Itturns out that all homomorphic images of a semigroup can be characterizedup to isomorphism.

Theorem 6.1.6 Let ϕ : S → T be a homomorphism of semigroups, andπ be the canonical projection π : S � S/Ker(ϕ). Then the mapping ψ :S/Ker(ϕ) → T defined via ψ(a) = ϕ(a) is a monomorphism and ϕ = ψπ,that is, the following diagram commutes:

Sϕ ��

π �� $$$$

$$$$$$

T

S/Ker(ϕ) � ψ

��%%

%%

%

Moreover, if ϕ is an epimorphism, then ψ is an isomorphism.

Proof. First, we have to check that ψ is well defined. For a′ ∈ a we haveϕ(a′) = ϕ(a) by the definition of Ker(ϕ). Hence ψ(a′) = ϕ(a′) = ϕ(a) =ψ(a).

Now let us check that ψ is injective. Indeed, if ψ(a) = ψ(b), then ϕ(a) =ϕ(b) by definition and hence a ∈ b, that is, a = b.

The next step is to check that ψ is a homomorphism. Let a, b ∈ S/Ker(ϕ).Then

ψ(a · b) = ψ(ab) = ϕ(ab) = ϕ(a)ϕ(b) = ψ(a)ψ(b).

Now for a ∈ S we have ψ(π(a)) = ψ(a) = ϕ(a) and hence ϕ = ψπ.Finally, if ϕ is surjective, then so is ψ since ϕ = ψπ. We already know that

ψ is always injective. Hence ψ is bijective for surjective ϕ. This completesthe proof.

From Theorem 6.1.6 it follows that, up to isomorphism, all homomorphicimages of a semigroup S are exhausted by the quotients of S.

6.2 Congruences on Groups

Our main goal in the first part of this chapter is to describe all congruenceson the semigroups Tn, PT n, and ISn. This will be done in Sect. 6.3. In thepresent section, we briefly recall the description of all congruences on groups.

6.2. CONGRUENCES ON GROUPS 95

Lemma 6.2.1 (i) Let S be a monoid and ρ be a congruence on S. Thenthe class 1 is a subsemigroup of S.

(ii) If S is a group, then 1 is a group as well.

Proof. Let a, b ∈ 1. Since ρ is left compatible we have that 1 ≡ b impliesa ≡ ab. Hence ab ∈ 1, proving (i).

Assume that S is a group and let a ∈ 1. Since ρ is left compatible wehave that 1 ≡ a implies a−1 ≡ a−1a = 1. Hence a−1 ∈ 1, proving (ii).

Lemma 6.2.2 Let G be a group, ρ a congruence on G and H = 1. Thenfor any g ∈ G and h ∈ H we have g−1hg ∈ H.

Proof. Since ρ is left compatible, we have that 1 ≡ h implies g−1 ≡ g−1h.Since ρ is right compatible, we have that g−1 ≡ g−1h implies 1 = g−1g ≡g−1hg. Hence g−1hg ∈ H.

Recall that a subgroup H of a group G is called normal provided thatg−1hg ∈ H for all g ∈ G and h ∈ H. The fact that H is a normal subgroupof G is usually denoted by H � G.

Let G be a group and H be a subgroup of G. For g ∈ G the set gH ={gh : h ∈ H} is called a left coset of G modulo H. The set of all left cosetsof G modulo H is denoted by G/H. Given G and H as above define thebinary relation ρH on G as follows: (a, b) ∈ ρH if and only if a ∈ bH.

Lemma 6.2.3 Let G be a group and H be a subgroup of G. Then ρH is anequivalence relation on G.

Proof. H is a subgroup, in particular, 1 ∈ H. Hence a = a · 1 ∈ aH, whichmeans that ρH is reflexive.

Let (a, b) ∈ ρH . Then a = bh for some h ∈ H by definition. Since H isa group, h−1 ∈ H and hence b = bhh−1 = ah−1, implying b ∈ aH. Hence(b, a) ∈ ρH and ρH is symmetric.

Let (a, b), (b, c) ∈ ρH . Then a = bh and b = ch′ for some h, h′ ∈ H.Hence a = ch′h. Since H is a subgroup, we have h′h ∈ H and thus a ∈ cHimplying (a, c) ∈ ρH . This means that ρH is transitive and completes theproof.

As an immediate corollary from Lemma 6.2.3 we obtain:

Corollary 6.2.4 Let G be a group, H be a subgroup of G, and a, b ∈ G.Then either aH = bH or aH ∩ bH = ∅.

Now we are ready to describe all possible congruences and quotients ofa group.


Theorem 6.2.5 Let G be a group.

(i) If H � G, then ρH is a congruence on G.

(ii) Every congruence on G has the form ρH for some normal subgroupH � G.

(iii) If ρ is a congruence on G, then the semigroup G/ρ is, in fact, a group.

Proof. Let H �G. We already know from Lemma 6.2.3 that ρH is an equiva-lence relation on G. Let (a, b) ∈ ρH and g ∈ G. Then a = bh for some h ∈ Hand multiplying with g from the left we get ga = gbh, that is, (ga, gb) ∈ ρH .Thus ρH is left compatible. On the other hand, multiplying with g from theright, we have ag = bhg = bg(g−1hg). As H is normal, we have g−1hg ∈ Hand hence (ag, bg) ∈ ρH . Thus ρH is right compatible as well. This proves (i).

Let ρ be a congruence on G. Set H = 1. By Lemmas 6.2.1(ii) and 6.2.2,H is a normal subgroup of G. Let a, b ∈ G be such that (a, b) ∈ ρ. Since ρ isleft compatible we get 1 = a−1a ≡ a−1b. Hence a−1b ∈ H and multiplyingwith a from the left we get b ∈ aH, that is, (a, b) ∈ ρH . This shows thatρ ⊂ ρH . On the other hand, let (a, b) ∈ ρH . Then (1, a−1b) ∈ ρH by the leftstability of ρH . But then a−1b ∈ H = 1 and hence (1, a−1b) ∈ ρ. Multiplyingthe latter with a from the left we get (a, b) ∈ ρ as ρ is left compatible. HenceρH ⊂ ρ and thus ρ = ρH , proving (ii).

We already know that G/ρ is a monoid with the identity element 1. Toprove (iii) one just has to observe that a · a−1 = 1 and hence all elements ofG/ρ are invertible. This completes the proof.

6.3 Congruences on T n, PT n, and ISn

Let S denote one of the semigroups Tn, PT n, or ISn.From Theorem 4.5.1(iii) we have that some elements α, β ∈ S belong to

the same H-class of the D-class Dk if and only if they have the followingform:

α =(

A1 A2 · · · Ak Ak+1

a1 a2 · · · ak ∅

),

β =(

A1 A2 · · · Ak Ak+1

aμ(1) aμ(2) · · · aμ(k) ∅

),

(6.4)

where the sets A1, . . . , Ak, Ak+1 are pairwise disjoint, μ ∈ Sk, and the setAk+1 may be empty. If S = Tn, we always have Ak+1 = ∅. If S = ISn, wehave |A1| = · · · = |Ak| = 1. Sometimes we shall omit the set Ak+1 in ournotation.

For every k = 1, 2, . . . , n and for every normal subgroup R�Sk we definean equivalence relation ≡R on S as follows:

6.3. CONGRUENCES ON Tn, PT n, AND ISn 97

• If rank(α) < k, then α ≡R β if and only if rank(β) < k.

• If rank(α) > k, then α ≡R β if and only if α = β.

• If rank(α) = k, then α ≡R β if and only if αHβ and if α and β aregiven by (6.4), then μ ∈ R.

Lemma 6.3.1 The relation ≡R is well defined, that is, it does not dependon the order of the blocks A1, . . . , Ak+1 in the presentation of α in the form(6.4).

Proof. Assume that we permute the Ais in some way, say B1 = Aη(1), B2 =Aη(2), . . . , Bk = Aη(k). In this numeration, we have

α =(

B1 B2 · · · Bk

b1 b2 · · · bk

), β =

(B1 B2 · · · Bk

bτ(1) bτ(2) · · · bτ(k)

),

where bi = aη(i), i = 1, . . . , k. Then for all i we have

bτ(i) = aμ(η(i)) = bη−1(μ(η(i))) = b(η−1μη)(i).

Hence τ = η−1μη. As R is a normal subgroup of Sk, we have τ ∈ R if andonly if μ ∈ R. The claim follows.

Lemma 6.3.2 The relation ≡R is a congruence on S.

Proof. First we check that ≡R is an equivalence relation on S. That ≡R

is reflexive is obvious. That R is symmetric and transitive for elements ofrank �= k is also obvious. Let us show that R is symmetric and transitive forelements of rank k. Let

α =(

A1 A2 · · · Ak

a1 a2 · · · ak

),

β =(

A1 A2 · · · Ak

aμ(1) aμ(2) · · · aμ(k)

)=(

A1 A2 · · · Ak

b1 b2 · · · bk

),

γ =(

A1 A2 · · · Ak

bτ(1) bτ(2) · · · bτ(k)

).

Then

α =(

A1 A2 · · · Ak

bμ−1(1) bμ−1(2) · · · bμ−1(k)

),

γ =(

A1 A2 · · · Ak

aμ(τ(1)) aμ(τ(2)) · · · aμ(τ(k))

).

If α ≡R β and β ≡R γ, then μ, τ ∈ R and hence both μ−1 and μτ areelements of R as well, since R is a group. Thus β ≡R α and α ≡R γ, whichmeans that ≡R is indeed an equivalence relation.


Let us now show that ≡R is both left and right compatible. If we haverank(α), rank(β) < k, then for any γ the ranks of the elements αγ, βγ, γα,and γβ do not exceed (k − 1) as well (by Exercise 2.1.4(c)). Hence

αγ ≡R βγ and γα ≡R γβ. (6.5)

If α = β, then relations (6.5) are obvious. So, we are left with the caserank(α) = rank(β) = k. Let α and β be given by (6.4) and α ≡R β. For any

γ =(

a1 a2 · · · ak · · ·c1 c2 · · · ck · · ·

)

we have

γα =(

A1 A2 · · · Ak

c1 c2 · · · ck

), γβ =

(A1 A2 · · · Ak

cμ(1) cμ(2) · · · cμ(k)

).

If some of the symbols c1, c2, . . . , ck are equal to ∅ or coincide, then wehave rank(γα), rank(γβ) < k and hence γα ≡R γβ. In the other case wehave γαHγβ. Hence we can consider γα and γβ as two elements, written inthe form (6.4). The corresponding permutation of indices is μ ∈ R. Henceγα ≡R γβ again. This proves that ≡R is left compatible.

To prove that ≡R is right compatible set Bi = {x ∈ N : γ(x) ∈ Ai},i = 1, . . . , k. If Bi = ∅ for some i, then rank(αγ), rank(βγ) < k and henceαγ ≡R βγ. In the other case we have

αγ =(

B1 B2 · · · Bk

c1 c2 · · · ck

), βγ =

(B1 B2 · · · Bk

cμ(1) cμ(2) · · · cμ(k)

)

and the same argument as in the previous paragraph shows that αγ ≡R βγagain. This proves that ≡R is right compatible and completes the proof.

Remark 6.3.3 If R = S1, then ≡R is the identity congruence on S.

Our ultimate goal in this section is to show that any congruence on Sis either the uniform congruence or has the form ≡R for some R. To provethis we will need a series of auxiliary lemmas. Recall from Sect. 2.3 that for

a ∈ N we denote by 0a the element 0a =(

Na

)∈ Tn. Recall also from

Sect. 2.7 that for A ⊂ N the element εA is the idempotent of ISn such thatdom(εA) = A.

Lemma 6.3.4 Let ρ be a congruence on Tn. If ρ �= ιS, then the ideal I1 iscontained in some class of ρ.

Proof. By assumption, there exist α �= β ∈ Tn such that α ≡ β. This meansthat for some x ∈ N we have a = α(x) �= β(x) = b. Take arbitrary y, z ∈ Nand let

γ =(

a b · · ·y z · · ·

).


Then γα0x = 0y and γβ0x = 0z, implying 0y ≡ 0z. Hence all elements fromI1 belong to the same class of ρ.

Lemma 6.3.5 Let ρ be a congruence on S, and α, β ∈ S be such thatrank(α) = k, rank(β) = m, k > m, and α ≡ β. Then Ik ⊂ α.

Proof. First, we consider the case S = PT n or S = ISn. As α ∈ Ik, it isenough to show that Ik ⊂ 0. For this we show that Il ⊂ 0 for all l ≤ k byinduction on l. That 0 ∈ 0 is obvious. Assume that l < k and that Il ⊂ 0.Let us show that Il+1 ⊂ 0 as well.

We have im(α)\im(β) �= ∅ because of our assumptions. Take somea0 ∈ im(α)\im(β) and complete it to some (l + 1)-element subset A ={a0, a1, . . . , al} ⊂ im(α). From α ≡ β we have εAα ≡ εAβ. However,im(εAβ) ⊂ {a1, . . . , al} and thus rank(εAβ) ≤ l. This implies εAβ ∈ 0and thus εAα ∈ 0 as well. But im(εAα) = A and thus rank(εAα) = l +1. As0 contains the ideal I0, by Lemma 6.1.3 the set 0 must be an ideal itself.Hence 0 contains the principal ideal SεAαS, which coincides with Il+1 byTheorem 4.2.8.

Let now S = Tn. By Lemma 6.3.4 one of the classes of ρ contains I1. Nowthe proof is analogous to the one given above. The only difference is thatinstead of the idempotent εA one should consider the following idempotent:

ε′A =(

a0 X1 · · · Xl

a0 a1 · · · al

),

where Xi are arbitrary such that ai ∈ Xi, i = 1, . . . , l.

If ρ is a congruence on S, different from the identity congruence, then byLemmas 6.3.4 and 6.1.3 it always contains a unique congruence class whichis an ideal of S. We denote this class by Iρ.

Lemma 6.3.6 Let ρ be a congruence on S, different from the identity con-gruence. If α ≡ β and α �∈ Iρ, then αHβ.

Proof. Without loss of generality we may assume n > 1.Assume that im(α) �= im(β). Without loss of generality we may assume

im(α)\im(β) �= ∅. Let a ∈ im(α)\im(β) and b ∈ im(α)\{a}. Consider theidempotent ε, uniquely defined by the following conditions:

ε(x) =

⎧⎪⎨⎪⎩

x, x �= a;b, x = aandS = Tn;∅, x = aandS �= Tn.

Then we have εα ≡ εβ = β. Hence α ≡ εα. On the other hand, im(εα) =im(α)\{a} and hence rank(εα) < rank(α). Applying Lemma 6.3.5 we getα ∈ Iρ, a contradiction. Hence im(α) = im(β).


Let im(α) = {a1, . . . , ak}. Then we may assume that

α =(

A1 A2 · · · Ak Ak+1

a1 a2 · · · ak ∅

), β =

(B1 B2 · · · Bk Bk+1

a1 a2 · · · ak ∅

),

where the sets Ak+1 and/or Bk+1 may be empty. Assume that ρα �= ρβ.Then either Ak+1 �= Bk+1, or Ak+1 = Bk+1 (that is, dom(α) = dom(β)) andthere exist elements x, y ∈ dom(α), x �= y, which belong to the same blockof ρα but to different blocks of ρβ.

Consider the second case first. In this case k ≥ 2 and without loss ofgenerality we may assume that there exist b1 ∈ B1 and b2 ∈ B2 such thatb1, b2 ∈ Al, l < k + 1. Consider the element

γ =(

a1 a2 · · · ak · · ·b1 b2 · · · bk · · ·

).

From γα ≡ γβ we have (γα)2 ≡ (γβ)2. But

(γα)2 =(

A1 A2 A3 · · · Ak Ak+1

bl bl c3 · · · ck ∅

),

(γβ)2 =(

B1 B2 B3 · · · Bk Bk+1

b1 b2 b3 · · · bk ∅

).

Hencerank((γα)2) ≤ k − 1 < k = rank((γβ)2).

Applying Lemma 6.3.5 we get Ik ⊂ Iρ, and hence α ∈ Iρ as rank(α) = k, acontradiction. Therefore ρα = ρβ in this case.

Consider now the case Ak+1 �= Bk+1. Without loss of generality we mayassume that there exists d1 ∈ Ak+1 ∩ B1. Choose any di ∈ Bi, i = 2, . . . , k,and consider the element

δ =(

a1 a2 a3 · · · ak · · ·d1 d2 d3 · · · dk · · ·

).

From δα ≡ δβ we have (δα)2 ≡ (δβ)2. But

(δα)2 =(

A1 A2 A3 · · · Ak Ak+1

∅ s2 s3 · · · sk ∅

),

(δβ)2 =(

B1 B2 B3 · · · Bk Bk+1

d1 d2 d3 · · · dk ∅

).

Using the same arguments as in the previous paragraph, we conclude thatα ∈ Iρ, a contradiction. Therefore ρα = ρβ in this case as well.

From Theorem 4.5.1(iii) it now follows that αHβ. This completes theproof.


Lemma 6.3.7 Let ρ be a congruence on S, and α ∈ S be an element of rankk such that there exists β �= α with the property α ≡ β. Then Iρ ⊃ Ik−1.

Proof. If α ∈ Iρ, then Iρ contains even Ik (the principal ideal, generatedby α). So, we may now assume that α �∈ Iρ. By Lemma 6.3.6, we may assumethat

α =(

A1 A2 · · · Ak

a1 a2 · · · ak

), β =

(Aμ(1) Aμ(2) · · · Aμ(k)

a1 a2 · · · ak

),

where μ ∈ Sk is different from the identity. In particular, k > 1. Let k = 2.If S = Tn, then I1 ⊂ Iρ follows from Lemma 6.3.4. If S �= Tn, then ε{a1} ∈ Sand since μ = (1, 2) we have

γ1 = ε{a1}α =(

A1

a1

)�=(

A2

a1

)= ε{a1}β = γ2.

Moreover, even dom(γ1) = A1 �= A2 = dom(γ2), that is, γ1 �∈ H(γ2). How-ever, γ1 ≡ γ2 and as rank(γ1) = 1, the inclusion I1 ⊂ Iρ follows now fromLemma 6.3.6.

If k > 2, then, since μ is different from the identity transformation, it iseasy to see that we can choose t, t+1 ∈ N such that {t, t+1} �= {μ(t), μ(t+1)}. Consider first the case S = Tn, or S = PT n. Take any element

γ =(

a1 · · · at at+1 at+2 · · · ak · · ·a1 · · · at at at+2 · · · ak · · ·

).

Then γα ≡ γβ. The partition ργα is obtained from ρα uniting the blocksAt and At+1. On the other hand, the partition ργβ is obtained from ρβ

uniting the blocks Aμ(t) and Aμ(t+1). As {t, t + 1} �= {μ(t), μ(t + 1)}, we getAt ∪ At+1 �= Aμ(t) ∪ Aμ(t+1). Hence ργα �= ργβ . From Lemma 6.3.6 it thusfollows that γα ∈ Iρ. As rank(γα) = k − 1, it follows that Iρ ⊃ Ik−1.

In the case S = ISn the proof is similar, one has just to choose t suchthat μ(t) �= t and consider the element

γ =(

a1 . . . at−1 at at+1 · · · ak · · ·a1 · · · at−1 ∅ at+1 · · · ak · · ·

).

Lemma 6.3.8 Let ρ be a congruence on S and Iρ = Ik. If α, β ∈ S aresuch that rank(α) > k + 1 and rank(β) > k + 1, then α ≡ β if and only ifα = β.

Proof. By Lemma 6.3.7, the inequality α �= β implies Iρ ⊃ Ik+1. The claimfollows.

Lemma 6.3.9 Let ρ be a congruence on S, which is neither identity noruniform. Assume that Iρ = Ik−1. Then ρ has the form ≡R for some normalsubgroup R of Sk.


Proof. If Iρ = Ik−1, all elements of rank at least (k + 1) form one-elementequivalence classes by Lemma 6.3.8. Hence we have only to find out thestructure of equivalence classes for elements of rank k. By Lemma 6.3.6, forsuch elements we have that γ ≡ δ implies γHδ.

Let γ be an idempotent of rank k. Then H(γ) is a group with the identityelement γ by Theorem 4.4.11. Moreover, H(γ) ∼= Sk by Theorem 5.1.4. Theset H(γ) × H(γ) is a union of congruence classes of ρ by Lemma 6.3.6.This also induces a congruence on H(γ), call it ργ . This means that theset G(γ) = {δ ∈ H(γ) : δ ≡ γ} is a normal subgroup of H(γ) and ργ

coincides with the partition of H(γ) into left cosets of H(γ) modulo G(γ)by Theorem 6.2.5.

We shall need an explicit form of the above statement. Let the idempo-tent γ be of the following form:

γ =(

M1 M2 · · · Mk

m1 m2 · · · mk

).

Then each element δ ∈ H(γ) has the form

δ =(

M1 M2 · · · Mk

mμ(1) mμ(2) · · · mμ(k)

).

And an isomorphism H(γ) ∼= Sk is given by using the mapping δ → μ, say.Under this isomorphism G(γ) is mapped to some normal subgroup of Sk,call it R.

Let us now show that ρ =≡R. Let α and β be two H-related elements ofrank k in S. Assume that they are given by (6.4). Choose in each block Ai

some element a′i and consider the elements

ν =(

M1 M2 · · · Mk

a′1 a′2 · · · a′k

), π =

(a1 · · · ak · · ·m1 · · · mk · · ·

),

ν1 =(

A1 A2 · · · Ak

m1 m2 · · · mk

), π1 =

(m1 · · · mk · · ·a1 · · · ak · · ·

),

where π, π1 ∈ Sn. Then γ = παν, δ = πβν, α = π1γν1, β = π1δν1. Henceα ≡ β if and only if γ ≡ δ. From the previous paragraph we know thatthe latter is the case if and only if μ ∈ R. Hence ρ =≡R, completing theproof.

Now we can summarize the above results into the following:

Theorem 6.3.10 Let S be one of the semigroups PT n, Tn, or ISn.

(i) For each k, 1 ≤ k ≤ n, and for each normal subgroup R of Sk therelation ≡R is a congruence on S. The congruence ≡R is uniform ifand only if n = 1 and S = Tn, and is the identity congruence if andonly if k = 1.

6.4. CONJUGATE ELEMENTS 103

(ii) Let ρ be a congruence on S, which is not uniform. Then there existsk ∈ {1, 2, . . . , n} and a normal subgroup R of Sk such that ρ =≡R.

Proof. The statement (i) is proved in Lemma 6.3.2. The statement (ii) fol-lows from Lemmas 6.3.4 to 6.3.9.

6.4 Conjugate Elements

Let G be a group. An element a ∈ G is said to be conjugate to an elementb ∈ G provided that there exists c ∈ G such that a = c−1bc.

Exercise 6.4.1 Show that the binary relation “a is conjugate to b” is anequivalence relation on G.

Conjugation plays a very important role in group theory. Hence it is verynatural to study various generalizations of this notion for certain classes ofsemigroups, in the best case, for all semigroups. There are several ways togeneralize this notion. The most direct one works for monoids and is definedas follows: Let S be a monoid and G = S∗ be the group of units of S. Anelement a ∈ S is said to be G-conjugate to an element b ∈ S provided thatthere exists c ∈ S∗ such that a = c−1bc. Obviously, if S is a group, then therelations of the usual conjugation and of G-conjugation coincide.

Exercise 6.4.2 Show that the binary relation “a is G-conjugate to b” is anequivalence relation on S.

Recall that two graphs Γ1 = (V1, E1) and Γ2 = (V2, E2) are said to beisomorphic provided that there exists a bijection ϕ : V1 → V2 such that forall a, b ∈ V1 we have (a, b) ∈ E1 if and only if (ϕ(a), ϕ(b)) ∈ E2. In otherwords, two graphs are isomorphic if they can be obtained from each othervia a renumeration of vertices.

Proposition 6.4.3 Let S be one of the semigroups PT n, ISn, or Tn. Twoelements α, β ∈ S are G-conjugate if and only if the graphs Γα and Γβ areisomorphic.

Proof. In our case we have S∗ = Sn. Let α, β ∈ S and π ∈ Sn be such thatα = π−1βπ. For x, y ∈ N the arrow (x, y) is an arrow of the graph Γα, or Γβ

if and only if y = α(x), or y = β(x), respectively. We first rewrite α = π−1βπas follows: πα = βπ. Then for any x ∈ dom(α) we have π(α(x)) = β(π(x)),that is, the following diagram is commutative:

x π ��

α

��

π(x)

β

��α(x) π �� π(α(x)) = β(π(x))

(6.6)


The above diagram says that π maps arrows from Γα to arrows from Γβ.Analogously one shows that the inverse permutation π−1 maps arrows fromΓβ to arrows from Γα. Hence π is an isomorphism from Γα to Γβ .

The converse statement is obvious. If π : N → N is an isomorphism fromΓα to Γβ, then from (6.6) it follows that α = π−1βπ.

A more interesting semigroup generalization of the notion of group con-jugation is the following one: Let S be a semigroup. Elements a, b ∈ S aresaid to be primarily S-conjugate provided that there exist u, v ∈ S1 suchthat a = uv and b = vu. The fact that a and b are primarily S-conjugate isdenoted by a ∼pS b. From the definition it is obvious that the relation ∼pS

is reflexive and symmetric. However, it is not transitive in general.

Exercise 6.4.4 Let α = [1, 2, 3], β = [1, 2][3], γ = [1][2][3] be elements fromIS3. Show that α ∼pS β and β ∼pS γ but α �∼pS γ.

Denote by ∼S the minimal transitive binary relation on S, which contains∼pS . Such minimal binary relation is usually called the transitive closure ofthe initial relations.

Exercise 6.4.5 Show that ∼S is an equivalence relation on S.

We will say that the elements a, b ∈ S are S-conjugate provided thata ∼S b. The notion of S-conjugation generalizes that of G-conjugation.Indeed, if a, b ∈ S are G-conjugate and c ∈ S∗ is such that a = c−1bc,then for u = c−1 and v = bc we have a = uv and vu = bcc−1 = b. Hencea ∼S b (and even a ∼pS b).

Proposition 6.4.6 Let S be a semigroup and let x, y ∈ S. Then we have:

(i) If x ∼pS y, then xi ∼pS yi for all i ∈ N.

(ii) If e = xi and f = yj are idempotents and x ∼pS y, then e ∼pS f .

Proof. Let x = ab and y = ba for some a, b ∈ S. Then xi = (ab)i =a((ba)i−1b

)and yi = (ba)i =

((ba)i−1b

)a implying xi ∼pS yi and proving

(i). To prove (ii) we observe that e = xi = xij and f = yj = yij and thatxij ∼pS yij by (i).

Lemma 6.4.7 Let S be a semigroup and e, f, g be three idempotents fromS such that e ∼pS f and e ∼pS g. Then f ∼pS g.

Proof. Let x, y, u, v ∈ S be such that e = xy, f = yx, e = uv, g = vu. Sincee is an idempotent we also have e = e2 = xyxy = xfy and analogously f =yex, e = ugv, g = veu. This implies g = vxfyu and f = yugvx. Thereforeg = g2 = (vxf)(yug) and f = f2 = (yug)(vxf) and thus g ∼pS f .


An immediate corollary from Lemma 6.4.7 is the following:

Corollary 6.4.8 Let S be a semigroup. Then the restriction of ∼pS to E(S)is an equivalence relation.

Corollary 6.4.9 Let S be a finite semigroup and x, y ∈ S be such thatx ∼S y. Let i, j ∈ N be such that e = xi and f = yj are idempotents. Thene ∼pS f .

Proof. Let x = x1, x2, . . . , xl = y be a sequence of elements from S such thatxi ∼pS xi+1 for i = 1, . . . , l−1. Since S is finite, for every i = 2, . . . , l−1 thereexists mi ∈ N such that yi = xmi

i is an idempotent. Let y1 = e and yl = f .From Proposition 6.4.6(ii) we obtain yi ∼pS yi+1 for all i = 1, . . . , l − 1.Applying Lemma 6.4.7 inductively, we get that e ∼pS yi for all i = 2, . . . , l.In particular, e ∼pS f .

Corollary 6.4.10 Let S be a finite semigroup. Then the restrictions of therelation ∼pS and ∼S to E(S) coincide.

Proof. For e, f ∈ E(S) that e ∼pS f implies e ∼S f is obvious. The converseimplication is a special case of Corollary 6.4.9.

Corollary 6.4.11 Let S be a finite semigroup and e, f ∈ E(S). Then e ∼pS

f if and only if eDf .

Proof. Let u, v ∈ S1 be such that uv = e and vu = f . Then e = e2 =uvuv = ufv and f = f2 = vuvu = veu. Hence S1eS1 = S1fS1 and we haveeJ f . By Theorem 5.4.1 we thus also have eDf since S is finite.

Conversely, if eDf , we can take a ∈ R(e) ∩ L(f) and a′ ∈ R(f) ∩ L(e)as in the proof of Theorem 4.7.5. For such a and a′ we have e = aa′ andf = a′a and hence e ∼pS f .

Let π ∈ Sn. The cyclic type of π is the vector ct(π) = (l1, l2, . . . , ln),where li is the number of cycles of length i in the cyclic decomposition of π(or, equivalently, in the graph Γπ). Obviously 1 · l1 + 2 · l2 + · · ·+ n · ln = n.

Lemma 6.4.12 Let α, β ∈ PT n and α ∼pS β. Then we have strank(α) =strank(β) and ct(α|stim(α)) = ct(β|stim(β)).

Proof. Let α = μη and β = ημ. For each a1 ∈ stim(α) consider the cycle(a1, a2, . . . , ak) of the permutation α|stim(α), which contains a1. We have

a1η→ b1

μ→ a2η→ b2

μ→ . . .μ→ ak−1

η→ bk−1μ→ ak

η→ bkμ→ a1.

It follows that (b1, b2, . . . , bk) = (η(a1), . . . , η(ak)) is a cycle for β. Obvi-ously b1, . . . , bk ∈ stim(β) and hence (b1, b2, . . . , bk) is a cycle of the per-mutation β|stim(β). Note that the lengths of the cycles (a1, a2, . . . , ak) and(b1, b2, . . . , bk) coincide. It follows that η and μ induce mutually inverse bi-jections between cycles of α and cycles of β, which preserve lengths of cycles.The claim follows.


Theorem 6.4.13 Let S denote one of the semigroups PT n, Tn, or ISn,and α, β ∈ S. Then α ∼S β if and only if strank(α) = strank(β) andct(α|stim(α)) = ct(β|stim(β)).

Proof. The necessity follows from Lemma 6.4.12. Let us prove the sufficiency.For any γ ∈ S define the following sets:

M ′0(γ) = stim(γ),

M ′′0 (γ) = im(γ)\dom(γ),

M0(γ) = M ′0(γ) ∪ M ′′

0 (γ),M1(γ) = {x �∈ M0(γ) : α(x) ∈ M0(γ)},Mi(γ) = {x : α(x) ∈ Mi−1(γ)}, i ≥ 2.

(6.7)

Note that M ′′0 (γ) = ∅ for γ ∈ Tn. From Theorem 1.2.9, which de-

scribes the structure of connected components of Γγ it follows that thesets M ′

0(γ), M ′′0 (γ), M1(γ), M2(γ), . . . are pairwise disjoint and that

γ(Mi(γ)) ⊂ Mi−1(γ) for all i > 0. Obviously Mi(γ) = ∅ for some i impliesMj(γ) = ∅ for all j > i. Of course it makes sense to consider only those setsfrom (6.7), which are nonempty, say these are the sets M0(γ), . . . , Mk(γ).Then we have the following picture:

M ′′0 (γ)

γ��

γ

&&&&&&

&&&&&&

& . . .γ�� γ�� γ��

γ �� M1(γ) Mk−1(γ) Mk(γ)

M ′0(γ)

Now let us apply the above to our element α. If S = PT n, or S = ISn

and k > 0, or if S = Tn and k > 1, we consider an idempotent ε, whichsatisfies the following conditions:

• im(ε) = M0(α) ∪ M1(α) ∪ · · · ∪ Mk−1(α);

• dom(ε) = im(ε) if S = PT n, or S = ISn; and ε(Mk(α)) ⊂ Mk−1(α) ifS = Tn.

In the case S = PT n, or S = ISn the idempotent ε is uniquely determined,while in the case S = Tn we might have a choice. For such ε we obviously haveα = εα. On the other hand, for the element α1 = αε the nonempty part ofthe sequence (6.7) will be one step shorter. Further, for all i = 0, . . . , k − 2


we obviously have Mi(α1) = Mi(α). If S = PT n, or S = ISn, we alsohave Mk−1(α1) = Mk−1(α) if k > 1 and M0(α1) = M ′

0(α) if k = 1. IfS = Tn, it is easy to see that Mk−1(α1) = Mk−1(α) ∪ Mk(α). Further,strank(α1) = strank(α) and α|stim(α) = α1|stim(α1).

Now we can apply the same procedure to α1, obtaining the element α2,and so on. We obtain the sequence α0 = α, α1,. . . , such that αi ∼pS αi+1

for all i. Now we have to consider two different cases.Case 1. S = PT n, or S = ISn. Let α′ = αk if k > 0, and α′ = α in the

opposite case. Then M0(α′) = M ′0(α

′) = stim(α) = im(α). Analogously weobtain the element β′. We have α ∼S α′, β ∼S β′ and using the assumptionsof our theorem we also have

α|stim(α) = α′|stim(α) = (a1, . . . , ap)(b1, . . . , bq) . . . (c1, . . . , cr),β|stim(β) = β′|stim(β) = (a′1, . . . , a

′p)(b

′1, . . . , b

′q) . . . (c′1, . . . , c

′r).

(6.8)

Consider the elements

μ =(

a1 · · · ap b1 · · · bq · · · c1 · · · cr

a′1 · · · a′p b′1 · · · b′q · · · c′1 · · · c′r

),

η =(

a′1 · · · a′p b′1 · · · b′q · · · c′1 · · · c′ra1 · · · ap b1 · · · bq · · · c1 · · · cr

),

(6.9)

where dom(μ) = dom(α′) = im(α′) and dom(η) = dom(β′) = im(β′). Wehave α′ = ηβ′ · μ and β′ = μ · ηβ′. Hence α′ ∼pS β′ and thus α ∼S β.

Case 2. S = Tn. Let α′ = αk−1 if k > 1, and α′ = α in the oppositecase. For this element the sequence (6.7) has at most two nonempty sets:M0(α′) = M ′

0(α′) and possibly M1(α′) = N\M0(α′). Analogously we get

the element β′. As above, we have stim(α) = stim(α′) = im(α′), stim(β) =stim(β′) = stim(β′), α ∼S α′, β ∼S β′ and by the assumptions of thetheorem we also have the decompositions (6.8). Consider some permutationμ, which coincides with the element μ from (6.9) on im(α′). Let η = μ−1.Then η coincides with the element η from (6.9) on im(β′) and for the elementδ = μα′ · η we have α′ ∼pS δ as α′ = η · μα′. Moreover, we also havestim(δ) = im(β′) and δ|im(β′) = β′|im(β′).

Set π = β′|im(β′). If im(β′) = {d1, d2, . . . , dm}, then β′ can be written asfollows:

β′ =(

D1 · · · Dm d1 · · · dm

d1 · · · dm π(d1) · · · π(dm)

)

(some Dis may be empty). At the same time we can write δ as follows:

δ =(

F1 · · · Fm d1 · · · dm

d1 · · · dm π(d1) · · · π(dm)

).

For the element

τ =(

F1 · · · Fm d1 · · · dm

π−1(d1) · · · π−1(dm) d1 · · · dm

)


we then have τβ′ = β′ and β′τ = δ. Hence β′ ∼pS δ, which implies α ∼S β.This completes the proof.

Corollary 6.4.14 Let α, β ∈ Sn. Then α and β are conjugate if and onlyif ct(α) = ct(β).

Proof. If α and β are conjugate, then α = π−1 · βπ for some π ∈ Sn.As β = βπ · π−1, we get α ∼pS β if we consider α and β as elements ofPT n. Hence ct(α) = ct(β) by Theorem 6.4.13. Conversely, if ct(α) = ct(β),choosing μ and η as in (6.9) we get μ ∈ Sn, η = μ−1 and α = ηβμ. Hence αand β are conjugate.


6.5.1 Congruences on Tn were described by Mal’cev in [Ma1]. This descrip-tion was used by Liber in [Lib] to describe congruences on ISn. Congruenceson PT n were described by Sutov in [Sut2].

6.5.2 A subsemigroup T of PT n is said to be Sn-normal provided that forall α ∈ T and σ ∈ Sn we have σ−1ασ ∈ T . For example, the semigroupsPT n, Tn and ISn are Sn-normal subsemigroups of PT n. Moreover, any idealin each of these semigroups is an Sn-normal subsemigroup of PT n as well.

Levi has shown in [Le] that for n > 4 all nontrivial congruences on almostall Sn-normal subsemigroups have the form ≡R. The results of Mal’cev,Liber, and Sutov are special cases of this general result of Levi.

6.5.3 Theorem 6.1.6 is true for a very general class of universal algebras ofthe same type. The proof of this very general result is technically slightlymore difficult as one has to consider the cases of an arbitrary collection ofoperations with different numbers of arguments.

6.5.4 Let Sem denote the category, whose objects are semigroups, and mor-phisms are all possible homomorphisms between semigroups. Categoricalmonomorphisms in Sem are exactly the injective homomorphisms of semi-groups, while the non-surjective embedding (N, +) ↪→ (Z, +) is a categoricalepimorphism in Sem. It is a good exercise to prove this.

6.5.5 Normal subgroups of the symmetric group Sk are very well known.If n �= 4, then the only normal subgroups of Sk are: the group Sk itself,the subgroup Ak of all even permutation (also known as the alternatinggroup), and the subgroup Ek consisting of the identity transformation. Forn �= 4 this was proved by Galois in about 1830. A short and easy argumentfor this can be found for example in [KM]. In addition to S4, A4, and E4,the group S4 has one more normal subgroup, namely, the so-called Klein4-group V4 = {ε, (12)(34), (13)(24), (14)(23)}. We note that S1 = A1 = E1

and A2 = E2. Hence from Theorem 6.3.10 we have the following corollary:


Corollary 6.5.1 (i) For n > 1 all congruences on each of the semigroupsPT n, Tn and ISn form the following chain:

ιS =≡S1�≡E2�≡S2�≡E3�≡A3�≡S3�≡E4�≡V4�≡A4�≡S4�

�≡E5�≡A5�≡S5� · · · �≡En�≡An�≡Sn� ωS .

(ii) Each congruence on ISn, or Tn is a restriction of some congruence onPT n. Moreover, for n > 1 different congruences on PT n restrict todifferent congruences on both ISn and Tn.

6.5.6 The criterion for S-conjugacy of two elements in ISn given inTheorem 6.4.13 was obtained in [GK1]. For the semigroups PT n andTn an analogous result is obtained in [KuMa1]. Several abstract resultsabout S-conjugacy can be found in [Ku2, KuMa1, KuMa3, KuMa4].

6.5.7 With each α ∈ PT n one associates the binary relation

Φα = {(x, y) : x ∈ N, y = α(x)}

on N. This can be used to define the natural partial order on PT n:

α � β if and only if Φα ⊂ Φβ .

Note that the restriction of the natural partial order to Tn coincides withthe equality relation.

It is easy to see that the natural partial order is a compatible relation.The restriction of the natural partial order to ISn has a transparent alge-braic interpretation:

Lemma 6.5.2 Let α, β ∈ ISn. Then α � β if and only if αβ−1 = αα−1.

It is hence natural and interesting to try to understand the structure ofall compatible partial orders on the semigroups PT n, Tn, and ISn. It turnsout that the only compatible partial order on Tn is the equality relation Δ(see [Mo2]). For PT n and ISn the answer is much more interesting.

Theorem 6.5.3 ([Mo1]) (i) For each collection of numbers as follows:0 ≤ k1 < k2 < · · · < km ≤ lm < · · · < l1 ≤ n, m ≥ 1, km < n, therelation

Ψk1,...,km,lm,...,l1 = (Δ ∪ (Ik1 × Il1) ∪ · · · ∪ (Ikm × Ilm))∩ � (6.10)

is a compatible partial order on PT n. Moreover, different collectionsgive different compatible partial orders.

(ii) Each compatible partial order on the semigroup PT n has either theform Ψk1,...,km,lm,...,l1 or is transposed to some Ψk1,...,km,lm,...,l1.


Theorem 6.5.4 ([Mo1]) (i) For each collection of numbers as follows:0 ≤ k1 < k2 < · · · < km < lm < · · · < l1 ≤ n, m ≥ 1, the rela-tion Ψk1,...,km,lm,...,l1 as in (6.10) is a compatible partial order on ISn.Moreover, different collections give different compatible partial orders.

(ii) Each compatible partial order on ISn different from Δ has either theform Ψk1,...,km,lm,...,l1 or is transposed to some Ψk1,...,km,lm,...,l1.

Corollary 6.5.5 ([Mo1]) (i) There are exactly 2n+2−5 compatible partialorders on PT n.

(ii) There are exactly 2n+1 − 1 compatible partial orders on ISn.


6.6.1 Prove that the alternating group An of all even permutations is anormal subgroup of Sn.

6.6.2 Prove that the Klein group V4 is normal in S4.

6.6.3 Describe all left compatible and all right compatible equivalence re-lations on a group.

6.6.4 Let ρ be a congruence on a semigroup S. Let ϕ : S � S/ρ be anepimorphism such that Ker(ϕ) = ρ. Is it possible to claim that ϕ alwayscoincides with the canonical epimorphism?

6.6.5 ([Mo2]) Prove that the only compatible partial order on Tn is Δ.

6.6.6 Show that the set of all maximal elements of PT n with respect to thenatural partial order coincides with Tn.

6.6.7 Prove that for the semigroup ISn the following conditions are equiv-alent:

(a) α � β

(b) α−1 � β−1

(c) βα−1 = αα−1

(d) α−1β = α−1α

(e) αβ−1 = αα−1

(f) β−1α = α−1α

(g) αβ−1α = α

(h) α−1βα−1 = α−1

6.6.8 Prove Corollary 6.5.5.

Chapter 7

Endomorphisms

7.1 Automorphisms of T n, PT n, and ISn

Recall from Sect. 6.1 that an automorphism of a semigroup S is a bijectivemapping ϕ : S → S such that ϕ(x · y) = ϕ(x) · ϕ(y) for all x, y ∈ S.

Proposition 7.1.1 Let S be a monoid. For each a ∈ S∗ the mapping Λa :x → a−1xa, x ∈ S, is an automorphism of S.

Proof. For x, y ∈ S we have

Λa(xy) = a−1xya = a−1xa · a−1ya = Λa(x)Λa(y)

and hence Λa is an endomorphism of S. Further for all x ∈ S we have

Λa−1Λa(x) = aa−1xaa−1 = x, ΛaΛa−1(x) = a−1axa−1a = x.

Hence Λa is a bijection with the inverse Λa−1 . The claim follows.

Automorphisms of the form Λa, a ∈ S∗, are called inner automorphismsof S. The set of all inner automorphisms of S is denoted by Inn(S).

Proposition 7.1.2 Let S be a monoid.

(i) The mapping a → Λa−1, a ∈ S∗, is an epimorphism from S∗ to Inn(S).

(ii) Inn(S) is a normal subgroup of Aut(S).

Proof. The mapping a → Λa−1 , a ∈ S∗, is obviously surjective. Let a, b ∈ S∗.For any x ∈ S we have

Λ(ab)−1(x) = abx(ab)−1 = abxb−1a−1 = Λa−1(bxb−1) = Λa−1Λb−1(x).

Hence Λa−1Λb−1 = Λ(ab)−1 . This proves (i).In the proof of Proposition 7.1.1 we have shown that Λ−1

a = Λa−1 forall a ∈ S∗. Together with the above proof of (i) we hence have that Inn(S)



112 CHAPTER 7. ENDOMORPHISMS

is closed with respect to both composition of maps and taking the inversemap. This means that Inn(S) is a subgroup of Aut(S).

Let now a ∈ S∗, ψ ∈ Aut(S) and x ∈ S. Then

ψ−1Λaψ(x) = ψ−1(a−1ψ(x)a) = ψ−1(a−1)xψ−1(a) = Λψ−1(a)(x).

This means that ψ−1Λaψ = Λψ−1(a) ∈ Inn(S) and completes the proof of (ii).

Theorem 7.1.3 Let S be one of the semigroups Tn, PT n, or ISn. ThenInn(S) = Aut(S) ∼= Sn.

Proof. We start our proof with the following lemma:

Lemma 7.1.4 Let σ and τ be different elements of Sn. Then the elementsΛσ and Λτ of Inn(S) are also different.

Proof. As σ �= τ , there exists x ∈ N such that y = σ(x) �= τ(x) = z. IfS = Tn or S = PT n, then Λσ(0x) = 0y �= 0z = Λτ (0x). Hence Λσ �= Λτ .If S = ISn, then Λσ(ε{x}) = ε{y} �= ε{z} = Λτ (ε{x}). Hence Λσ �= Λτ

again.

Lemma7.1.4 says that theepimorphismwhichwasdescribed inProposition7.1.2(i) is injective. In particular, this mapping induces an isomorphismInn(S) ∼= S∗ ∼= Sn. To complete the proof we only need to show that anyautomorphism of S is inner. Since we already have n! different (inner) au-tomorphisms of S, it is in fact enough to show that |Aut(S)| ≤ n!. We shalldo it via a case-by-case analysis.

Let ψ ∈ Aut(S). Since S is finite and all two-sided ideals of S forma chain (by Theorem 4.3.1), ψ induces the identity mapping on the set ofall two-sided ideals. In particular, for all i = 0, 1, . . . , n we have that therestriction map ψ|Ii : Ii → Ii is bijective.

Case 1. S = Tn. In this case we have I1 = {0a : a ∈ N}. Since therestriction of ψ to I1 is a bijection, we can define the permutation ψ ∈ Sn

by the following rule: ψ(0a) = 0ψ(a), a ∈ N. The mapping ψ → ψ is ahomomorphism of groups. Indeed,

0ηψ(a) = ηψ(0a) = η(ψ(0a)) = η(0ψ(a)) = 0ηψ(a)

for all a ∈ N and hence ηψ = ηψ. Since |Sn| = n!, to complete the proof itis enough to show that the kernel of the mapping ψ → ψ is trivial.

In other words, let ψ ∈ Aut(Tn) be such that ψ(0a) = 0a for all a ∈ N.We have to show that ψ(α) = α for all α ∈ Tn. For α ∈ Tn set

M(α) = {(a, b) ∈ N × N : α · 0b = 0a}.

From the definition it is clear that (a, b) ∈ M(α) if and only if α(b) = a. Takenow some α ∈ Tn and let ψ(α) = β. Applying ψ to the equality α0b = 0a

7.1. AUTOMORPHISMS OF Tn, PT n, AND ISn 113

we get that this equality is equivalent to the equality β0b = 0a. This meansthat M(α) = M(β), in particular, α(x) = β(x) for all x ∈ N. Thereforeβ = α. Hence ψ(α) = α for all α ∈ Tn and the proof in the case S = Tn iscompleted.

Case 2. S = ISn. The set X = {ε{a} : a ∈ N} is the set of all idem-potents of rank one, that is, the set of all idempotents in I1\I0. As anautomorphism, ψ sends idempotents to idempotents. Since the restrictionof ψ to both I1 and I0 is a bijection, ψ induces a permutation on X, inparticular, we can define the permutation ψ ∈ Sn by the following rule:ψ(ε{a}) = ε{ψ(a)}, a ∈ N. As in the previous case one checks that the map-ping ψ → ψ is a homomorphism of groups. To complete the proof it isenough to show that the kernel of this mapping is trivial.

In other words, let ψ ∈ Aut(ISn) be such that ψ(ε{a}) = ε{a} for alla ∈ N. We have to show that ψ(α) = α for all α ∈ ISn. For α ∈ ISn set

M(α) = {(a, b) ∈ N × N : ε{a}αε{b} �= 0}.

From the definition it is clear that (a, b) ∈ M(α) if and only if α(b) = a.Now take some α ∈ ISn and let ψ(α) = β. Applying ψ to the inequalityε{a}αε{b} �= 0 we get the inequality ε{a}βε{b} �= 0. This means that M(α) =M(β), and thus α = β. Hence ψ(α) = α for all α ∈ ISn and the proof inthe case S = ISn is completed as well.

Case 3. S = PT n. We could apply exactly the same arguments as in theproof given in Case 2 provided that we could prove that ψ preserves the setX. To prove this we will need the following:

Lemma 7.1.5 For α ∈ PT n we have

|dom(α)| = |dom(ψ(α))| and |im(α)| = |im(ψ(α))|.

Proof. Obviously αβ = 0 if and only if dom(α) ∩ im(β) = ∅, that is, ifim(β) ⊂ dom(α). Hence the right annihilator Annr(α) = {β : αβ = 0}of the element α contains exactly (|dom(α)| + 1)n elements. Moreover, theequality ψ(0) = 0 implies that αβ = 0 if and only if ψ(α)ψ(β) = 0. Henceψ(Annr(α)) = Annr(ψ(α)). As any automorphism is a bijection, we have|Annr(α)| = |Annr(ψ(α))| and hence |dom(α)| = |dom(ψ(α))|.

Analogously, we have βα = 0 if and only if im(α) ∩ dom(β) = ∅, thatis, if dom(β) ⊂ im(α). Hence the left annihilator Annl(α) = {β : βα = 0}of the element α contains exactly (n + 1)n−|im(α)| elements. Analogouslyto the previous paragraph one shows that ψ(Annl(α)) = Annl(ψ(α)) and|Annl(α)| = |Annl(ψ(α))|. Thus |im(α)| = |im(ψ(α))|.

From Lemma 7.1.5 it follows that for any a ∈ N the element α = ψ(ε{a})is an idempotent such that |dom(α)| = 1 and |im(α)| = 1. Hence α ∈ X.This completes the proof.


7.2 Endomorphisms of Small Ranks

Recall from Sect. 6.1 that an endomorphism of a semigroup S is a mappingϕ : S → S such that ϕ(x ·y) = ϕ(x) ·ϕ(y) for all x, y ∈ S. Our main goal forthe rest of this chapter is to describe all endomorphisms of the semigroupsTn, ISn, and PT n. The final classification will be done basically using thecase-by-case analysis. If S is a semigroup and ϕ ∈ End(S), the number |ϕ(S)|is called the rank of ϕ. We start with constructing some endomorphisms ofsmall ranks.

Let S denote one of the semigroups Tn, ISn, or PT n. Let ε ∈ S be anidempotent. Define the mapping Φε : S → S via Φε(α) = ε for all α ∈ S.The following statement is obvious:

Lemma 7.2.1 (i) For any idempotent ε the mapping Φε is an endomor-phism of S of rank one.

(ii) If ε �= ε′, then Φε �= Φε′.

(iii) Every endomorphism of S or rank one has the form Φε for some idem-potent ε ∈ S.

Let ε, δ ∈ S be different idempotents such that εδ = δε = δ. Define themapping Ψε,δ : S → S as follows:

Ψε,δ(α) =

{ε, α ∈ Sn;δ, otherwise.

Lemma 7.2.2 (i) For any pair ε, δ of different idempotents of S satisfyingεδ = δε = δ the mapping Ψε,δ is an endomorphism of S of rank two.

(ii) If the pairs ε, δ and ε′, δ′ are different, then Ψε,δ �= Ψε′,δ′ .

(iii) Every endomorphism of S or rank two has the form Φε,δ for someunequal idempotents ε, δ ∈ S satisfying εδ = δε = δ.

Proof. The statement (i) is proved by a direct calculation and the statement(ii) is obvious. To prove (iii) we should assume that S �= T1 as |T1| = 1.Let ϕ ∈ End(S) be an endomorphism of rank two. Then the congruenceKer(ϕ) contains exactly two congruence classes. Going through the list of allcongruences on S given by Theorem 6.3.10, we see that the only congruenceon S with exactly two congruence classes is ≡Sn . The congruence classesare S∗ and In−1. As the representatives of the two congruence classes ofthis congruence we can take for example the identity transformation εn andany (left) zero element γ of S. These are idempotents and obviously satisfyεnγ = γεn = γ. Since they belong to different congruence classes, we haveε = ϕ(εn) �= ϕ(γ) = δ. Moreover, εδ = δε = δ. By definition, in this notationwe have ϕ = Φε,δ. This completes the proof of (iii).

7.3. EXCEPTIONAL ENDOMORPHISM 115

Let ε, δ ∈ S be different idempotents such that εδ = δε = δ. Let furtherτ ∈ Gε be an element of order two (i.e., τ2 = ε, τ �= ε) such that τδ = δτ = δ.Define the mapping Θτ

ε,δ : S → S as follows:

Θτε,δ(α) =

⎧⎪⎨⎪⎩

ε, α ∈ An;τ, α ∈ Sn\An;δ, otherwise.

Lemma 7.2.3 (i) For any triple ε, δ, τ as above the mapping Θτε,δ is an

endomorphism of S of rank three.

(ii) If the triples ε, δ, τ and ε′, δ′, τ ′ are different, then Θτε,δ �= Θτ ′

ε′,δ′.

(iii) Every endomorphism of S or rank three has the form Θτε,δ for some ε,

δ, and τ as above.

Proof. The statement (i) is proved by a direct calculation and the state-ment (ii) is obvious. To prove (iii) we should assume n > 1 as |PT 1| = 2.Let ϕ ∈ End(S) be an endomorphism of rank three. Then the congruenceKer(ϕ) contains exactly three congruence classes. Going through the list ofall congruences on S given by Theorem 6.3.10, we see that the only con-gruence on S with exactly three congruence classes is ≡An , n > 1. Thecongruence classes are An, Sn\An, and In−1. As the representatives of thethree congruence classes of this congruence we can take for example the iden-tity transformation εn, the transposition (1, 2) and any (left) zero elementγ of S. These elements obviously satisfy εnγ = γεn = γ, γ(1, 2) = γ, and(1, 2) ∈ Gεn . Further, either (1, 2)γ = γ or (1, 2)γ is another left zero, andhence belongs to In−1. Since our representatives belong to different congru-ence classes, we have ε = ϕ(εn) �= ϕ(γ) = δ and ϕ((1, 2)) = τ is an elementof order two in Gε. Further, from the above we also get that εδ = δε = δand τδ = δτ = δ. By definition, in this notation we have ϕ = Θτ

ε,δ. Thisproves (iii).

7.3 Exceptional Endomorphism

Let ϕ ∈ Aut(Sn) and S = ISn or S = PT n. Define the mapping Ωϕ : S → Sas follows:

Ωϕ(α) =

{ϕ(α), α ∈ Sn;0, otherwise.

Lemma 7.3.1 Ωϕ is an endomorphism of both ISn and PT n.

Proof. Follows by a direct calculation.


To proceed we need some additional notation. Let α ∈ PT n be an ele-ment of rank (n − 1). We would like to associate with α some element α ofrank one. There are two possibilities. The first one is dom(α) �= N. Then wedefine α as the unique element of rank one such that

dom(α) = dom(α), im(α) = N\im(α).

Here is an example of such α and α for n = 5:

α =(

1 2 3 4 52 5 ∅ 1 3

), α =

(1 2 3 4 5∅ ∅ 4 ∅ ∅

).

The second possibility is that there exists a unique pair a, b ∈ N such thatα(a) = α(b). Then we define α as the unique element of rank one such that

dom(α) = {a, b}, im(α) = N\im(α).

Here is an example of such α and α for n = 5:

α =(

1 2 3 4 52 5 2 1 3

), α =

(1 2 3 4 54 ∅ 4 ∅ ∅

).

Define the mapping Ξ : PT n → PT n, n > 1, as follows:

Ξ(α) =

⎧⎪⎨⎪⎩

α, rank(α) = n;α, rank(α) = n − 1;0, otherwise.

Note that Ξ maps ISn to ISn. Abusing notation we denote the inducedmapping on ISn also by Ξ.

Lemma 7.3.2 Let S = ISn or S = PT n, n > 1.

(i) Ξ is an endomorphism of S.

(ii) If ϕ, ψ ∈ Aut(PT n) and ϕ �= ψ, then ϕΞ �= ψΞ.

Proof. To prove (i) we have to check that

Ξ(αβ) = Ξ(α)Ξ(β) (7.1)

for all α, β ∈ S. By the definition of Ξ this is obvious if both α and β arepermutations or if one of them has rank at most (n − 2). There are threecases left.

Case 1. α ∈ Sn, rank(β) = n−1. In this case rank(αβ) = n−1 and hence(7.1) reduces to the equality αβ = αβ. Since α is a permutation, we havedom(αβ) = dom(β) and im(αβ) = α(im(β)). Hence the equality αβ = αβfollows from the definition of Ξ.

7.3. EXCEPTIONAL ENDOMORPHISM 117

Case 2. rank(α) = n−1, β ∈ Sn. In this case rank(αβ) = n−1 and hence(7.1) reduces to the equality αβ = αβ. Since β is a permutation, we havedom(αβ) = β−1(dom(α)) and im(αβ) = im(α). Hence the equality αβ = αβfollows from the definition of Ξ.

Case 3. rank(α) = n−1, rank(β) = n−1. Here we have two possibilities.The first one is that rank(αβ) < n − 1. This is possible if im(β) containstwo different a and b such that α(a) = α(b) or if im(β) ∩ dom(α) �= ∅. Inboth cases the definition of Ξ gives αβ = 0. The second possibility is thatrank(αβ) = n − 1. This is possible if and only if im(β) ⊂ dom(α) and α isinjective on im(β). In particular, from the definition of Ξ we have that theunique element in N\im(β) belongs to dom(α), that is, im(β) ⊂ dom(α). Wethus get dom(αβ) = dom(β) and im(αβ) = im(α), which implies αβ = αβsince both elements in the last equality have rank one. This proves (i). Toprove (ii) one can just observe that ϕΞ and ψΞ act differently on idempotentsof rank (n − 1).

Now let n = 4. In this case we have an additional normal subgroup ofS4, namely, V4. The cosets of S4 modulo V4 are the following sets:

X1 ={(

1 2 3 41 2 3 4

),

(1 2 3 42 1 4 3

),

(1 2 3 43 4 1 2

),

(1 2 3 44 3 2 1

)},

X2 ={(

1 2 3 42 1 3 4

),

(1 2 3 41 2 4 3

),

(1 2 3 44 3 1 2

),

(1 2 3 43 4 2 1

)},

X3 ={(

1 2 3 41 3 2 4

),

(1 2 3 42 4 1 3

),

(1 2 3 43 1 4 2

),

(1 2 3 44 2 3 1

)},

X4 ={(

1 2 3 43 2 1 4

),

(1 2 3 44 1 2 3

),

(1 2 3 41 4 3 2

),

(1 2 3 42 3 4 1

)},

X5 ={(

1 2 3 42 3 1 4

),

(1 2 3 41 4 2 3

),

(1 2 3 44 1 3 2

),

(1 2 3 43 2 4 1

)},

X6 ={(

1 2 3 43 1 2 4

),

(1 2 3 44 2 1 3

),

(1 2 3 41 3 4 2

),

(1 2 3 42 4 3 1

)}.

We see that each of these cosets contains a unique element α satisfyingα(4) = 4 (the first element of each coset). Mapping the whole coset to thiselement defines the endomorphism ψ : S4 → S4 with kernel V4. Define thefollowing transformations of PT 4:

Υ1(α) =

{ψ(α), α ∈ S4;0, otherwise;

Υ2(α) =

{ψ(α), α ∈ S4;ε{4}, otherwise;

Υ3(α) =

{ψ(α), α ∈ S4;04, otherwise;

Υ4(α) =

{ψ(α)ε{1,2,3}, α ∈ S4;0, otherwise;


Υ5(α) =

{ψ(α)ε{1,4}, α ∈ S4;0, otherwise;

Υ6(α) =

{ψ(α)ε{2,4}, α ∈ S4;0, otherwise;

Υ7(α) =

{ψ(α)ε{3,4}, α ∈ S4;0, otherwise.

Note that Υ1, Υ2, and Υ4 preserve IS4 and Υ3 preserves T4. Abusing no-tation we shall denote the induced mappings on IS4 and T4 by the samesymbols.

Exercise 7.3.3 Let i ∈ {1, 2, . . . , 7} and ϕ, ψ ∈ Aut(PT 4) be two differ-ent automorphisms. Show that the restrictions of ϕΥi and ψΥi to S4 aredifferent.

Lemma 7.3.4 (i) Υi ∈ End(PT 4) for all i = 1, . . . , 7.

(ii) If ϕ, ψ ∈ Aut(PT 4) and ϕ �= ψ, then ϕΥi �= ψΥi for all i = 1, . . . , 7.

(iii) The statement (ii) holds as well for the restrictions to IS4 and T4 inappropriate cases.

Proof. We prove the statement (i) for Υ1. All other cases are similar. Wehave to check that

Υ1(αβ) = Υ1(α)Υ1(β) (7.2)

for all α, β ∈ PT 4. If both α and β are permutations, (7.2) follows from thefact that ϕ is a homomorphism. If both α and β are not permutations, (7.2)follows from the fact that 0 is an idempotent. If exactly one of α and β isa permutation, then αβ is not a permutation and hence both sides of (7.2)are equal to 0. This proves (i).

Finally, (ii) and (iii) follow from Exercise 7.3.3.

7.4 Classification of Endomorphisms

Theorem 7.4.1 (i) Let n �= 4. Then each endomorphism of Tn has oneof the following forms:

(a) Λπ, where π ∈ Sn.

(b) Φε, where ε ∈ Tn is an idempotent.

(c) Ψε,δ, where ε �= δ ∈ Tn are idempotents such that εδ = δε = δ.

(d) Θτε,δ, where ε �= δ ∈ Tn are idempotents such that εδ = δε = δ and

τ ∈ Gε is an element of order two such that τδ = δτ = δ.

(ii) In addition to the above endomorphisms, the semigroup T4 has alsoendomorphisms ΛπΥ3, π ∈ Sn.

7.4. CLASSIFICATION OF ENDOMORPHISMS 119

Proof. We start with some auxiliary statements:

Lemma 7.4.2 For n > 1 the semigroup Tn does not contain any element αsuch that πα = α for all π ∈ Sn.

Proof. Let α ∈ Tn, x ∈ N and y = α(x). Let further z ∈ N be such thatz �= y. Then (y, z)α(x) = z �= y = α(x) and hence (y, z)α �= α.

Lemma 7.4.3 Let S denote one of the semigroups Tn, ISn, or PT n. Letϕ ∈ End(S). Assume that all elements of Sn form one-element congruenceclasses of Ker(ϕ). Then ϕ(Sn) ⊂ Sn and ϕ|Sn ∈ Aut(Sn).

Proof. By assumptions, ϕ(Sn) ∼= Sn is a subgroup of S. It is of course thencontained in some maximal subgroup. But, by Theorem 5.1.4, all maximalsubgroups of S are isomorphic to Sm, m ≤ n, and there is only one maximalsubgroup isomorphic to Sn, namely, Sn itself. Hence ϕ maps Sn to Sn andthus induces an automorphism of Sn.

Lemma 7.4.4 Let ϕ ∈ End(Tn). Then either Ker(ϕ) = ιTn or Ker(ϕ) = ωTn

or Ker(ϕ) =≡R, where R is a normal subgroup of Sn.

Proof. The statement is obvious for n = 1 so we assume n > 1.Assume that Ker(ϕ) has the form ≡R, where R is a normal subgroup of

Sk and k < n. From the definition of ≡R (see Sect. 6.3) it follows that allelements of Sn form one-element congruence classes. From Lemma 7.4.3 wehave that ϕ maps Sn to Sn and induces an automorphism of Sn.

If k = 1, then ≡R coincides with ιTn . If k > 1, then by the definition of≡R the ideal Ik−1 forms one congruence class and hence is mapped by ϕ tosome idempotent, say γ. But Ik−1 is an ideal and hence satisfies αβ ∈ Ik−1

for all α ∈ Sn and β ∈ Ik−1. Applying ϕ yields ϕ(α)γ = γ. Since therestriction of ϕ to Sn is bijective by the previous paragraph, we have πγ = γfor all π ∈ Sn. But Tn, n > 1, does not contain any γ with such propertyby Lemma 7.4.2. This means that the case 1 < k < n is impossible, whichcompletes the proof.

Let n �= 4 and ϕ ∈ End(Tn). If Ker(ϕ) is a uniform congruence, then ϕhas rank one and thus coincides with some Φε by Lemma 7.2.1(iii). If Ker(ϕ)is the identity congruence, then ϕ is an automorphism and thus coincideswith some Λπ by Theorem 7.1.3. In all other cases Ker(ϕ) =≡R, where Ris En, An, or Sn by Lemma 7.4.4. If R = Sn, then ϕ has rank two and thuscoincides with some Ψε,δ by Lemma 7.2.2(iii). If R = An, then ϕ has rankthree and thus coincides with some Θτ

ε,δ by Lemma 7.2.3(iii).For n �= 4 it remains to consider the case when R = En. In this case all

elements of Sn form separate congruence classes. Hence Lemma 7.4.3 showsthat ϕ induces an automorphism of Sn. The ideal In−1 is a congruence classand should be sent by ϕ to some idempotent γ. As ϕ|Sn ∈ Aut(Sn), as in the


proof of Lemma 7.4.4 we obtain πγ = γ for all π ∈ Sn. But such element γdoes not exist by Lemma 7.4.2. Hence the case when R = En is not possible.

Let now n = 4. In this case the only difference with the arguments aboveis the fact that R can be equal to V4. For such R we have ϕ(S4) ∼= S3, whichmay be a subgroup of either S4 or some Gγ

∼= S3, where γ is an idempotentof rank 3 (see Theorem 5.1.4).

Assume first that ϕ(S4) ⊂ S4. Then ϕ(S4) ∼= S3 is a subgroup of S4.

Exercise 7.4.5 Show that S4 contains exactly four subgroups, isomorphicto S3, namely, the subgroups Hi = {α ∈ S4 : α(i) = i}, i = 1, 2, 3, 4.

From Exercise 7.4.5, ϕ(S4) = Hi for some i ∈ {1, 2, 3, 4}. Consider nowthe endomorphism

ϕ′ =

{ϕ, i = 4;Λ(i,4)ϕ, i �= 4.

(7.3)

We have ϕ′(S4) = H4, in particular, ϕ′ induces an automorphism of H4∼= S3

via restriction.

Exercise 7.4.6 Show that Aut(S3) = Inn(S3) ∼= S3.

Taking into account (7.3) and using Exercise 7.4.6 we obtain that thereexists π ∈ S4 such that Λπϕ restricts to the identity map on H4. The idealI3 is a congruence class and thus should be mapped by Λπϕ to some nonin-vertible idempotent α of T4. As in the proof of Lemma 7.4.4 we get βα = αfor all β ∈ H4, in particular, for β = (1, 2, 3). Hence for any x ∈ im(α) weshould have β(x) = x, which means that im(α) = {4}, and thus α = 04.Therefore Λπϕ = Υ3 and hence ϕ = Λπ−1Υ3.

Finally, assume that ϕ(S4) = Gγ , where γ has rank 3. Let δ = ϕ(I3).Then δ is an idempotent and the fact that I3 is an ideal implies thatδα = αδ = δ for all α ∈ Gγ . From γδ = δ we have im(δ) ⊂ im(γ). Inthe proof of Theorem 4.7.4 it was shown that the restriction to im(γ) de-fines an isomorphism from Gγ to S(im(γ)). Take any α ∈ Gγ , whose imagein S(im(γ)) under this isomorphism does not have any fixed point. Then forany x ∈ N we have α(δ(x)) �= δ(x) since δ(x) ∈ im(γ). This means that theequality αδ = δ is not possible, a contradiction. Hence the case ϕ(S4) = Gγ ,where γ has rank 3, is not possible. This completes the proof.

Theorem 7.4.7 (i) Let n �= 4. Then each endomorphism of ISn has oneof the following forms:

(a) Λπ, where π ∈ Sn.

(b) Φε, where ε ∈ ISn is an idempotent.

(c) Ψε,δ, where ε �= δ ∈ ISn are idempotents such that εδ = δε = δ.

(d) Θτε,δ, where ε �= δ ∈ ISn are idempotents such that εδ = δε = δ

and τ ∈ Gε is an element of order two such that τδ = δτ = δ.

7.4. CLASSIFICATION OF ENDOMORPHISMS 121

(e) Ωψ, ψ ∈ Aut(Sn).

(f) ΛπΞ, π ∈ Sn.

(ii) In addition to the above endomorphisms, the semigroup IS4 also hasendomorphisms ΛπΥi, i = 1, 2, 4, π ∈ Sn.

Proof. The statement is obvious for n = 1 so until the end of the proof weassume n > 1. Let ϕ ∈ End(ISn). Assume first that all elements of Sn formseparate congruence classes of Ker(ϕ). By Lemma 7.4.3, ϕ preserves Sn andinduces an automorphism on it. Consider first the case n �= 6. Then thisautomorphism is inner (see 7.6.2) and hence there exists π ∈ Sn such thatϕ′ = Λπϕ induces on Sn the identity mapping. Set ε = εN\{1}. The elementϕ′(ε) must be an idempotent, say ϕ′(ε) = εB for some B ⊂ N. To determineB more precisely we shall need an auxiliary notion and statement.

For α ∈ ISn the set {π ∈ Sn : απ = πα} is called the centralizer of αin Sn and is denoted by CSn(α).

Lemma 7.4.8 For A ⊂ N we have

CSn(εA) = {π ∈ Sn : π(A) = A}.

In particular, |CSn(εA)| = |A|! · (n − |A|)!.

Proof. Since π is invertible, πεA = εAπ is equivalent to εA = πεAπ−1.Obviously πεAπ−1 = επ(A). Hence εA = πεAπ−1 is equivalent to εA = επ(A),that is, A = π(A). This proves the first claim and the second claim followsimmediately from the first one.

Applying ϕ′ to πε = επ we get πεB = εBπ as ϕ′(π) = π for all π ∈ Sn.This implies that CSn(ε) ⊂ CSn(εB). Using Lemma 7.4.8 we obtain that

{π ∈ Sn : π(N\{1}) = N\{1}} ⊂ {π ∈ Sn : π(B) = B}. (7.4)

This gives the following possibilities for B: B = N\{1}, B = {1} or B = N.

Lemma 7.4.9 Let S, T be semigroups and A ⊂ S be a generating system ofS. Let f : A → T be any mapping. Then there exists at most one homomor-phism F : S → T such that F |A = f .

Proof. As S = 〈A〉, each x ∈ S can be written in the form x = a1a2 · · · ak

for some ai ∈ A. Then

F (x) = F (a1a2 · · · ak) = F (a1)F (a2) · · ·F (ak) = f(a1)f(a2) · · · f(ak),

which means that F (x) is uniquely determined by f .


By Theorem 3.1.4, the semigroup ISn is generated by Sn and ε. ByLemma 7.4.9 this means that any endomorphism of ISn is uniquely de-termined by its values on Sn and ε. In the situation we have ϕ′(π) = πfor all π ∈ Sn and we have three possibilities for ϕ′(ε), namely, ϕ′(ε) = ε,ϕ′(ε) = ε{1} and ϕ′(ε) = 0. The first possibility is realized by the identityautomorphism, the second one is realized by the endomorphism Ξ, and thethird one is realized by the endomorphism Ωθ, where θ is the identity auto-morphism of Sn. This means that the original automorphism ϕ = Λπ−1ϕ′ isof the form (ia), (if), or (ie).

Let now n = 6. In addition to the cases considered above we have toconsider the case when ϕ induces on Sn an automorphism, which is notinner. Such automorphisms are known explicitly, see 7.6.2. In particular,each such automorphism maps transpositions from the subgroup

CS6(ε) = {π ∈ S6 : π(1) = 1} < S6

to permutations, each of which is a product of three different commutingtranspositions. It is easy to see that this implies that the only subset of N,invariant under all such permutations, is the set N itself. Hence Lemma 7.4.8implies that the only possibility for ϕ(ε) in this case is 0. This means thatϕ is of type (ie).

Now we have to consider all cases when ϕ does not induce an automor-phism of Sn. If Ker(ϕ) = ωISn , then ϕ is of type (ib) by Lemma 7.2.1(iii).In all other cases from Theorem 6.3.10 we have Ker(ϕ) =≡R, where R isa nontrivial normal subgroup of Sn. If R = Sn, then ϕ is of type (ic) byLemma 7.2.2(iii). If R = An, then ϕ is of type (id) by Lemma 7.2.3(iii). Itremains to consider the case R = V4. We leave it to the reader to verify thatin this case ϕ is of type (ii).

Theorem 7.4.10 (i) Let n �= 4. Then each endomorphism of PT n hasone of the following forms:

(a) Λπ, where π ∈ Sn

(b) Φε, where ε ∈ PT n is an idempotent

(c) Ψε,δ, where ε �= δ ∈ PT n are idempotents such that εδ = δε = δ

(d) Θτε,δ, where ε �= δ ∈ PT n are idempotents such that εδ = δε = δ

and τ ∈ Gε is an element of order two such that τδ = δτ = δ

(e) Ωϕ, ϕ ∈ Aut(Sn)

(f) ΛπΞ, π ∈ Sn

(ii) In addition to the above endomorphisms, the semigroup PT 4 also hasendomorphisms ΛπΥi, i = 1, 2, . . . , 7, π ∈ Sn.

Exercise 7.4.11 Prove Theorem 7.4.10.

7.5. COMBINATORICS OF ENDOMORPHISMS 123

7.5 Combinatorics of Endomorphisms

Corollary 7.5.1 (i) The semigroup T1 has one endomorphism.

(ii) The semigroup T4 has 345 endomorphisms.

(iii) The semigroup Tn, n �= 1, 4, has

n!

⎛⎝1 +

n∑m=1

�m−12

∑r=0

m−2r∑k=1

mn−m · km−r−k

2r · (n − m)! · (m − 2r − k)! · k! · r!

⎞⎠ (7.5)

endomorphisms.

Proof. The statement (i) is obvious. The statement (ii) follows from the for-mula of (iii) and the note that T4 has 24 additional endomorphisms describedin Theorem 7.4.1(ii). Hence we have to prove (iii).

The semigroup Tn has n! automorphisms by Theorem 7.1.3. By Theorem7.4.1, if we forget about the additional endomorphisms T4, we have only tocount the number of endomorphisms of Tn of ranks 1, 2, and 3.

By Lemma 7.2.1(iii) the endomorphisms of rank 1 correspond bijectivelyto idempotents of Tn. Hence we have

∑nm=1

(nm

)mn−m endomorphisms of

rank 1 by Corollary 2.7.4.By Lemma 7.2.2(iii) the endomorphisms of rank 2 correspond to pairs

of idempotents (ε, δ) of Tn satisfying the condition εδ = δε = δ.

Lemma 7.5.2 Let ε, δ ∈ Tn be two idempotents. Then the following condi-tions are equivalent:

(a) εδ = δε = δ

(b) im(δ) ⊂ im(ε) and ρδ � ρε

Proof. If εδ = δ, then im(δ) ⊂ im(ε) by Exercise 2.1.4(b). If δε = δ, thenρδ � ρε by Theorem 4.2.4. Hence (a) implies (b). Conversely, if im(δ) ⊂ im(ε)and ρδ � ρε, then a direct calculation shows that εδ = δε = δ. Hence (b)implies (a).

Now we can use Lemma 7.5.2 to count the number of all pairs (ε, δ)of different idempotents of Tn satisfying the condition εδ = δε = δ. Letm = rank(ε) and k = rank(δ). Since ε �= δ and im(δ) ⊂ im(ε), we have1 ≤ k < m ≤ n. The image of ε can be chosen in

(nm

)different ways. By

Lemma 7.5.2, the image of δ should be chosen inside the im(ε) and thiscan be done in

(mk

)different ways. As ε is an idempotent, ε(x) = x for all

x ∈ im(ε). To define ε on other elements, we should map each x ∈ N\im(ε)to some element of im(ε). This can be done in mn−m different ways. Forx ∈ im(ε)\im(δ) let Ax = {y ∈ N : ε(y) = x}. From Lemma 7.5.2, to defineδ we should map each Ax, x ∈ im(ε)\im(δ) to some element of im(δ). This


can be done in km−k different ways. Using the multiplication rule and thenadding everything up for all possible m and k we get

n∑m=2

m−1∑k=1

(n

m

)(m

k

)mn−mkm−k (7.6)

endomorphisms of rank 2.By Lemma 7.2.3(iii) the endomorphisms of rank 3 correspond to triples

(ε, δ, τ) of Tn, where ε and δ are different idempotents of Tn satisfying thecondition εδ = δε = δ, and τ ∈ Gε is an element of order 2 such thatτδ = δτ = δ. The restrictions on ε and δ are given by Lemma 7.5.2.

Lemma 7.5.3 Let ε, δ ∈ Tn be two idempotents such that εδ = δε = δ. Letfurther τ ∈ Gε. Then the following conditions are equivalent:

(a) τδ = δτ = δ

(b) τ(x) = x for all x ∈ im(δ); and δ(x) = δ(y) for all x, y ∈ im(ε) suchthat x �= y and τ(x) = y

Proof. Assume (a). Let x ∈ im(δ). Then x = δ(x) = τδ(x) = τ(x). Letx, y ∈ im(ε) be such that x �= y and τ(x) = y. Then δ(x) = δτ(x) = δ(y).Hence we have (b).

Assume (b) and let x ∈ N. As τ(x) = x for all x ∈ im(δ), we have τδ = δ.By assumptions we have τε = τ and δε = δ. Hence the equality δτ = δ isequivalent to δτε = δε, which means that the equality δτ = δ should bechecked only for x ∈ im(ε). For those x ∈ im(ε) for which τ(x) = x, theequality δτ(x) = δ(x) is obvious. If τ(x) = y �= x for some x, y ∈ im(ε),then δτ(x) = δ(x) follows from the second condition of (b). Thus δτ = δ,implying (a). This completes the proof.

An element of order 2 in the symmetric group Sm is a product of rcommuting transpositions, where 0 < r ≤ �m

2 �. There are(m2

)ways to choose

the first transposition,(m−2

2

)ways to choose the second transposition, and

so on. As the order of transpositions is not important, for each r as abovewe have that Sm contains exactly

(m2

)·(m−2

2

)· · · · ·

(m−2r+2

2

)r!

=m!

(m − 2r)! · r! · 2r(7.7)

elements of order 2, which are products of r commuting transpositions.Let us now count the number of triples (ε, δ, τ) as above. The rank m of ε

can have values m = 2, . . . , n. The image of ε can be chosen in(

nm

)different

ways and to define ε completely we should assign some element from im(ε) toeach element outside this image. This can be done in mn−m different ways.Now we choose τ . This is an element of order 2 in Gε

∼= Sm and hence it can

7.5. COMBINATORICS OF ENDOMORPHISMS 125

be a product of r commuting transpositions, where 0 < r ≤ �m2 �. However,

τ(x) = x for all x ∈ im(δ) �= ∅. Hence r �= m2 , that is, 0 < r ≤ �m−1

2 �. Foreach such r by (7.7) we have m!

(m−2r)!r!2r ways to choose τ .After we have chosen τ let k be the rank of δ. We have 1 ≤ k ≤ m− 2r.

There are(m−2r

k

)different ways to choose im(δ). For x ∈ im(ε) set Ax =

{y ∈ N : ε(y) = x}. As ρδ � ρε by Lemma 7.5.2(b) and δ(x) = δ(y) forall x, y ∈ im(ε) such that x �= y and τ(x) = y by Lemma 7.5.3(b), to defineδ we should assign to each Ax, x ∈ im(ε)\im(δ), an element of im(δ), suchthat the same element is assigned to Ax and Ay as soon as τ(x) = y. Suchassignment can be done in km−r−k different ways. Adding everything up,we get

n∑m=2

�m−12

∑r=1

m−2r∑k=1

(n

m

)(m − 2r

k

)m! · mn−m · km−r−k

(m − 2r)! · r! · 2r(7.8)

endomorphisms of order 3. If for r = 0 in the formula (7.8) we allow kto vary from 1 to (m − 1), we will obtain the formula (7.6). Further, theformula (7.6) would count all endomorphisms of rank one for Tn as well ifwe allow k = m and the value 1 for m. Taking this into account, addingn! automorphisms, and rewriting binomial coefficients in terms of factorialsgives the formula (7.5).

Corollary 7.5.4 (i) The semigroup IS1 has three endomorphisms.

(ii) The semigroup IS2 has 14 endomorphisms.

(iii) The semigroup IS4 has 282 endomorphisms.

(iv) The semigroup IS6 has 5,244 endomorphisms.

(v) The semigroup ISn, n �= 1, 2, 4, 6, has

n!

⎛⎝3 +

n∑m=0

�m2 ∑

r=0

2m−3r

(n − m)! · (m − 2r)! · r!

⎞⎠ (7.9)

endomorphisms.

Proof. The semigroup IS1 has the trivial automorphism and two endomor-phisms of type Theorem 7.4.7(ib). This proves (i). The statements (iii) and(iv) follow from the general formula (v) taking into account 72 additionalendomorphisms of IS4 given by Theorem 7.4.7(ii) and 6! additional en-domorphisms of IS6 corresponding to the outer automorphisms of S6 inTheorem 7.4.7(ie). The statement (ii) follows from (v) taking into accountthat for n = 2 the only endomorphism of type Theorem 7.4.7(ie) has rank3 (and hence is of type Theorem 7.4.7(id) as well), and all endomorphismsof type Theorem 7.4.7(if) are in fact automorphisms. So, we have to proveonly the statement (v).


For n > 1 the semigroup ISn has n! automorphisms by Theorem 7.1.3,n! endomorphisms of type (ie), which correspond to inner automorphismsof Sn, and n! endomorphisms of type (if). So again we are left to count thenumber of endomorphisms of ranks 1, 2, and 3.

Let us count endomorphisms of rank at most three. By Lemmas 7.2.1,7.2.2, and 7.2.3 these are given by triples (ε, δ, τ) where ε and δ are idem-potents satisfying εδ = δε = δ (not necessarily different) and τ ∈ Gε is anelement of order at most 2 satisfying τδ = δτ = δ. Let m = rank(ε). Then wehave

(nm

)choices for im(ε) = A. Assume now that τ is a product of exactly

r transpositions, 0 ≤ r ≤ �m2 �. Then by (7.7) we have exactly m!

(m−2r)!r!2r

possibilities to choose τ . To proceed we need to know the restrictions on δimposed by the condition τδ = δτ = δ.

Lemma 7.5.5 For α ∈ ISn and B ⊂ N we have αεB = εBα = εB if andonly if α(x) = x for all x ∈ B.

Proof. We have εB(x) = x for all x ∈ B by definition. Hence already αεB =εB implies α(x) = x for all x ∈ B. On the other hand, if α(x) = x forall x ∈ B, then αεB = εB is obvious, which also reduces εBα = εB toεBαεB = εB, that is, εBα = εB should be checked only for x ∈ B. However,for such x the equality εBα = εB follows immediately from α(x) = x.

Lemma 7.5.5 says that im(δ) can be an arbitrary set of fixed points for τ .By construction τ has (m − 2r) fixed points. Hence we have 2m−2r ways tochoose δ. Summing everything up gives

n∑m=0

�m2 ∑

r=0

(n

m

)m! · 2m−2r

(m − 2r)! · r! · 2r

endomorphisms of rank at most 3. The claim follows.

Corollary 7.5.6 (i) The semigroup PT 1 has three endomorphisms.

(ii) The semigroup PT 2 has 18 endomorphisms.

(iii) The semigroup PT 4 has 1,374 endomorphisms.

(iv) The semigroup PT 6 has 170,772 endomorphisms.

(v) The semigroup PT n, n �= 1, 2, 4, 6, has

n!

⎛⎝3 +

n∑m=0

�m2 ∑

r=0

m−2r∑k=0

(m + 1)n−m · (k + 1)m−r−k

2r · (n − m)! · (m − 2r − k)! · r! · k!

⎞⎠ (7.10)

endomorphisms.

Exercise 7.5.7 Prove Corollary 7.5.6.



7.6.1 Automorphisms of Tn were described by Schreier in [Sch]. A differentargument was later proposed by Mal’cev in [Ma1]. Automorphisms of ISn

and PT n were described by Liber and Sutov in [Lib] and [Sut2], respectively.

7.6.2 For n �= 2, 6 all automorphisms of Sn are inner and Aut(Sn) ∼= Sn. Thesemigroup S2 has only the trivial automorphism (which is of course inner).In contrast to S2, for the semigroup S6 we have |Aut(S6)/Inn(S6)| = 2.A noninner automorphism of S6 was first constructed in [Hoe]. The reasonthat all automorphisms of Sn, n �= 6, are inner is purely numerical. Anyautomorphism maps conjugate elements to conjugate elements and preservesthe order of an element. Let X be the set of all transpositions in Sn. This isa conjugacy class of elements of order two. A straightforward combinatorialcomputation shows that for n �= 6 all other conjugacy classes of elementsof order two have cardinalities different from |X|. Hence for n �= 6 anyautomorphism maps transpositions to transpositions. From here it is fairlystraightforward to deduce that any automorphism is inner. Details can befound for example in [KM]. Each outer (that is, not inner) automorphismof S6 maps a transposition to a product of three commuting transpositions.

7.6.3 Endomorphisms of ISn, Tn, and PT n were described and their num-bers were counted in [ST1], [ST2], and [ST3], respectively. Although thegeneric statement is correct, there are some typos in the exceptional casesof [ST1] and [ST3]. In particular, for n = 6 the authors missed the endo-morphisms of the form Ωϕ, where ϕ is an outer automorphism of S6.

7.6.4 Here is the table for |End(S)|, where S = ISn, Tn, and PT n forsmall n:

n 1 2 3 4 5 6 7 8|End(ISn)| 3 14 54 282 918 5,244 25,560 168,828|End(Tn)| 1 7 40 345 3,226 38,503 529,614 8,219,025

|End(PT n)| 3 18 138 1,374 13,178 170,772 2,507,690 41,387,036

7.6.5 The problem to describe all homomorphisms (monomorphisms, epi-morphisms) from S to T , where S, T ∈ {ISn, Tn,PT n : n ≥ 1}, is open ingeneral.

7.6.6 In [ST1], [ST2], and [ST3] for S = ISn, Tn, and PT n Schein andTeclezghi asked whether |End(S)|

|S| → 0, n → ∞. More generally they ask tofind an estimate for the asymptotic of |End(S)|. For the semigroup ISn bothquestions were answered in [JM]:

Theorem 7.6.1 ([JM]) (i) |End(ISn)| ∼ 3n!, n → ∞.

(ii) |End(ISn)||ISn| → 0, n → ∞.


Proof. Set

Xn = n! ·n∑

m=0

�m2 ∑

r=0

2m−3r

(n − m)! · (m − 2r)! · r! .

Because of Corollary 7.5.4(v) to prove (i) it is enough to show thatXn/n! → 0, n → ∞. We have

Xn =n∑

m=0

�m2 ∑

r=0

(n

m

)(m

2r

)(2r

r

)· 2m−3r · r! (7.11)

All binomial coefficients in (7.11) are smaller than 2n and r ≤ �n/2�. Hence

Xn ≤ 8n · �n/2�! ·n∑

m=0

�m2 ∑

r=0

2m−3r ≤ 16n · (n/2)! · n2. (7.12)

Using the Stirling formula we have

16n · (n/2)! · n2

n!∼16n · n2 · nn/2 · en · √πn

2n/2 · en/2 · nn ·√

2πn=

=1√2· en ln 16+2 ln n+n

2ln n+n−n

2ln 2−n

2−n ln n. (7.13)

Since the exponent is −12n lnn + O(n), we obtain that the right-hand side

of (7.13) approaches 0 for large n. This proves (i).By Theorem 2.5.1 we have |ISn| ≥

(n

n−1

)2(n− 1)! = n · n!. Hence, using(i), we have

0 ≤ |End(ISn)||ISn|

∼ 3 · n!|ISn|

≤ 3 · n!n · n!

→ 0, n → ∞.

The claim (ii) follows.

For S = Tn and PT n the corresponding problems are still open.


7.7.1 Show that the only element α ∈ PT n satisfying the condition πα = αfor all π ∈ Sn is the element α = 0.

7.7.2 Determine all elements α ∈ PT n which satisfy the condition απ = αfor all π ∈ Sn.

7.7.3 Let A ⊂ N. Show that α ∈ ISn satisfies the condition αεA = εAα ifand only if both A and N\A are invariant with respect to α. Use this to showthat the centralizer of εA in ISn can be identified with IS(A)× IS(N\A).


7.7.4 Determine |{α ∈ IS3 : αε{1} = ε{1}α}|.

7.7.5 If S is a semigroup, then the center of S is the set

Z(S) = {α ∈ S : αβ = βαfor allβ ∈ S}.

Determine Z(S), where S = ISn, PT n, and Tn.

7.7.6 Show that Aut(I1) = Sn for the ideal I1 of ISn.

7.7.7 ([ST1, ST2, ST3]) An endomorphism ϕ ∈ End(S) is called a retractionprovided that ϕ · ϕ = ϕ. Classify all retractions of ISn, Tn, and PT n.

7.7.8 Classify all endomorphisms and all retractions of Sn.

7.7.9 Let S be a finite semigroup such that End(S) = Aut(S). Prove that|S| = 1.

7.7.10 Let S = (Q>0, +) be the semigroup of all positive rational numberswith respect to addition. Show that End(S) = Aut(S) ∼= (Q>0, ·).

7.7.11 Classify all endomorphisms of the semigroup (N, +) of all positiveintegers with respect to addition.

7.7.12 Classify all endomorphisms of the finite cyclic semigroup S generatedby an element of type (k, m).

7.7.13 Let ε ∈ PT n be an idempotent of rank one with domain D andimage {x}. Show that π ∈ Sn commutes with ε if and only if π(x) = x andπ(D) = D. Use this to show that the number of permutations commutingwith ε equals (|D| − 1)! · (n − |D|)!.

Chapter 8

Nilpotent Subsemigroups

8.1 Nilpotent Subsemigroups and Partial Orders

A semigroup S with the zero element 0 is said to be nilpotent provided thatthere exists k ∈ N such that Sk = {0}, that is, a1 · a2 · · · ak = 0 for alla1, . . . , ak ∈ S. If S is nilpotent, then the minimal k ∈ N such that Sk = {0}is called the nilpotency degree or nilpotency class of S and is denoted bynd(S).

Remark 8.1.1 Note that here we slightly abuse the notation S1. In thepresent chapter, we will use this notation to denote the first power of S,that is, the semigroup S itself, and not the semigroup S with the adjointidentity element, as we did before.

Proposition 8.1.2 Let S be a finite semigroup with the zero element 0.Then the following conditions are equivalent:

(a) S is nilpotent

(b) Every element a ∈ S is nilpotent

Proof. If S is nilpotent, nd(S) = k and a ∈ S, then ak = 0. Hence a isnilpotent. This proves the implication (a)⇒(b).

Conversely assume that each element of S is nilpotent. Let a1, a2, . . . , ak

be arbitrary elements of S such that a1a2 · · · ak �= 0. For i = 1, . . . , k setbi = a1a2 · · · ai and note that bi �= 0. Assume that bi = bj for some i < jand let x = ai+1ai+2 · · · aj . Then for all m ∈ N we have

bixm = (bix)xm−1 = (bj)xm−1 = bix

m−1 = · · · = bi. (8.1)

Hence (8.1) implies that xm �= 0 for all m. This contradicts (b). Hence allelements bi, i = 1, . . . , k, are different and nonzero and thus k < |S|. Inparticular, S|S| = {0} and thus S is nilpotent. This proves the implication(b)⇒(a).



132 CHAPTER 8. NILPOTENT SUBSEMIGROUPS

Corollary 8.1.3 Let S be a finite nilpotent semigroup. Then nd(S) ≤ |S|.Exercise 8.1.4 Let S be the Rees quotient of the semigroup (Q>0, +) ofall positive rational numbers with respect to the addition modulo the ideal,consisting of all rational number greater than 1. Show that each element ofS is nilpotent and that S is not nilpotent.

The aim of the present chapter is to study nilpotent subsemigroups of thesemigroups Tn, PT n, and ISn. Note that only PT n, and ISn contain a zeroelement, so the notion of nilpotent elements and nilpotent subsemigroups isreally natural only for these two semigroups. In this natural case, one justconsiders those nilpotent subsemigroups of PT n or ISn, which share thezero element with the original semigroup. Of course one can also considernilpotent subsemigroups in which the zero element is different from thezero element of the original semigroup. In such interpretation, it also makessense to study nilpotent subsemigroups of Tn. However, it turns out thatthis study for Tn, PT n, and ISn can be basically reduced to the study ofnilpotent subsemigroups of PT n or ISn with the global zero element. Weshall explain this in the Addenda. Until then we restrict our attention to thesemigroups PT n and ISn and consider only those nilpotent subsemigroupsof these semigroups whose zero element is the global zero element 0.

For a semigroup S with the zero element 0 denote by Nil(S) the set ofall nilpotent subsemigroups of S with the zero element 0. The set Nil(S)is nonempty (it contains the nilpotent subsemigroup {0}) and is partiallyordered with respect to inclusions. The subsemigroup {0} is the minimumelement of Nil(S). If the semigroup S itself is nilpotent, then S is the max-imum element of S. In the case when S is not nilpotent, the set Nil(S)may contain many different maximal elements. They are called the maximalnilpotent subsemigroups of S.

For k ∈ N denote also by Nilk(S) the subset of Nil(S) consisting of allnilpotent subsemigroups of nilpotency degree k. The set Nil(S) decomposesinto a disjoint union of Nilk(S), k ∈ N. Note that Nil1(S) =

{{0}

}. For

a given k the set Nilk(S) may be empty. For example, Nilk(S) is certainlyempty for all k big enough if S is finite. If it is not empty, then it inheritsa partial order with respect to inclusions from Nil(S). In general, the setNilk(S) may have many minimal and many maximal elements.

Our main goal for this chapter is to classify all maximal elements in allNilk(S), where S = PT n or ISn. The combinatorial tool we shall use forthis is the set On of all antireflexive partial orders on the set N. The set On

itself is partially ordered with respect to inclusions.The inclusion ISn ↪→ PT n induces inclusions Nil(ISn) ↪→ Nil(PT n)

and Nilk(ISn) ↪→ Nilk(PT n) for all k ∈ N. Hence we can identify Nil(ISn)and Nilk(ISn) with the corresponding images of the above inclusions.

For T ∈ Nil(PT n) define the binary relation τT on N as follows:

xτT y if and only if there existsσ ∈ T such that σ(y) = x.

8.1. NILPOTENT SUBSEMIGROUPS AND PARTIAL ORDERS 133

Lemma 8.1.5 τT ∈ On.

Proof. Let x, y, z ∈ N be such that xτT y and yτT z. Then there exist α, β ∈ Tsuch that α(y) = x and β(z) = y. This gives (αβ)(z) = α(β(z)) = α(y) = x.Hence xτT z and the relation τT is transitive.

Assume that xτT x for some x ∈ N. Then there exists α ∈ T such thatα(x) = x. But this means that αk(x) = x for all k ∈ N. Hence α is notnilpotent, contradicting the nilpotency of T . Hence τT is antireflexive andthus belongs to On.

By Lemma 8.1.5, we have the mapping T → τT from the set Nil(PT n)to On. It restricts to a mapping from Nil(ISn) to On.

For each τ ∈ On define the subsemigroups Nτ ∈ PT n and N ′τ ∈ ISn as

follows:

Nτ = {α ∈ PT n : α(x)τxfor allx ∈ dom(α)}, N ′τ = Nτ ∩ ISn.

Lemma 8.1.6 (i) Nτ ∈ Nil(PT n).

(ii) N ′τ ∈ Nil(ISn).

Proof. The statement (ii) follows from (i). To prove (i) we consider anyαi ∈ Nτ , i = 1, . . . , n. Let α = α1α2 · · ·αn and assume that x ∈ dom(α). Fori = 1, . . . , n set βi = αiαi+1 · · ·αn. Then x ∈ dom(βi) for all i = 1, . . . , n andwe set xi = βi(x) for all such i. Set xn+1 = x. Then we have xi = αi(xi+1) forall i = 1, . . . , n. Hence xiτxi+1 for all i = 1, . . . , n. As τ is an antireflexivepartial order, we get that all elements x1, x2, . . . , xn+1 must be different,contradicting |N| = n. Hence dom(α) = ∅ and thus α = 0. This impliesthat Nn

τ = {0} and therefore Nτ ∈ Nil(PT n).

By Lemma 8.1.6 we have the mapping τ → Nτ from On to Nil(PT n)and the mapping τ → N ′

τ from On to Nil(ISn). The crucial properties ofthe mappings we have constructed are:

Proposition 8.1.7 (i) If S, T ∈ Nil(PT n) and S ⊂ T , then τS ⊂ τT .

(ii) If τ, σ ∈ On and τ ⊂ σ, then Nτ ⊂ Nσ and N ′τ ⊂ N ′

σ.

(iii) For any σ ∈ On we have τNσ = σ and τN ′σ

= σ.

Proof. The statements (i) and (ii) are obvious from the definitions, so wehave to prove only the statement (iii). Let x, y ∈ N be such that xσy.Consider the element α ∈ ISn of rank one, defined as follows: dom(α) = {y}and α(y) = x. Then α ∈ N ′

σ ⊂ Nσ. From the definition it follows that yτN ′σx

and hence σ ⊂ τN ′σ

and σ ⊂ τNσ .


Conversely, let x, y ∈ N be such that xτNσy (for N ′σ the arguments are

similar). Then there exists α ∈ Nσ such that α(y) = x. But the latterimplies xσy by the definition of Nσ. Hence τNσ ⊂ σ and the statement (iii)follows.

Let τ ∈ On. The maximal possible k ∈ N for which there exist elementsx1, . . . , xk ∈ N such that xiτxi+1 for all i = 1, . . . , k − 1 is called the heightof τ . Thus any linear order on N has height n, while the empty partialorder on N has height 1. For k ∈ N we denote by O

(k)n the subset of On

consisting of all partial orders of height k. Obviously, we have the followingdecomposition into a disjoint union of subsets:

On =n⋃

k=1

O(k)n .

Lemma 8.1.8 (i) If T ∈ Nilk(PT n), then τT ∈ O(k)n .

(ii) If τ ∈ O(k)n , then Nτ ∈ Nilk(PT n) and N ′

τ ∈ Nilk(ISn).

Proof. Let α1, . . . , αk−1 ∈ T be such that α = α1α2 · · ·αk−1 �= 0. Fix somex ∈ dom(α) and for i = 1, . . . , k− 1 set xi = αiαi+1 · · ·αk−1(x). Then underthe convention xk = x we have xiτT xi+1 for all i = 1, . . . , k − 1 and hencethe nilpotency degree of T does not exceed the height of τT .

Let xi, i = 1, . . . , m, be elements of N such that xiτT xi+1 for all i =1, . . . , m − 1. By the definition of τT there exist αi ∈ T , i = 1, . . . , m − 1,such that αi(xi+1) = xi. Then for α = α1α2 · · ·αm we have xm ∈ dom(α)and hence α �= 0. This implies that the height of τT does not exceed nd(T ).This proves (i).

The statement (ii) follows from the statement (i) and the assertion ofProposition 8.1.7(iii).

Corollary 8.1.9

Nil(PT n) =n⋃

k=1

Nilk(PT n), Nil(ISn) =n⋃

k=1

Nilk(ISn).

Proof. Follows from Lemma 8.1.8(i) since, obviously, the maximal possibleheight of a partial order on N is n.

8.2 Classification of Maximal NilpotentSubsemigroups

Let k ∈ N, 1 ≤ k ≤ n, and m = (M1, M2, . . . , Mk) be an ordered partition ofN into a disjoint union of nonempty subsets, that is, N = M1∪M2∪· · ·∪Mk;

8.2. CLASSIFICATION OF NILPOTENT SUBSEMIGROUPS 135

Mi �= ∅, i = 1, . . . , k; and Mi ∩Mj = ∅ if i �= j. Define the partial order τmon N in the following way:

xτmy if and only if x ∈ Mi, y ∈ Mj , and i < j.

Example 8.2.1 For m = {{1, 2}, {3, 4}, {5, 6, 7}} the Hasse diagram of thepartial order τm is as follows:

5

''''

'''

!!!!!!!!

!!!!!!!!

!!!!! 6

��

''''

''' 7

��

��

��

��

3

((((((

((((((

(( 4

))))))

))))))

))

1 2

Proposition 8.2.2 (i) For m as above the order τm is a maximal elementin O

(k)n .

(ii) Every maximal element in O(k)n has the form τm for some m as above.

(iii) If m �= n, then τm �= τn.

Proof. First we note that τm ∈ O(k)n . Indeed, if x1, . . . , xm are such that

xiτmxi+1 for all i = 1, . . . , m − 1, then all xi’s belong to different Mj ’s andhence m ≤ k. At the same time, taking some xi ∈ Mi, i = 1, . . . , k, we havexiτmxi+1 for all i = 1, . . . , k − 1. Hence the height of τm equals k.

Next we claim that τm is a maximal element of O(k)n . Indeed, let σ ∈ On

be such that τm � σ. Then there exist x, y ∈ N such that xσy, x ∈ Mi,y ∈ Mj and j ≤ i. Choose any xs ∈ Ms, s = 1, . . . , i − 1, and any yt ∈ Mt,t = j + 1, . . . , k. As τm ⊂ σ, for the sequence

z1 = x1, . . . , zi−1 = xi−1, zi = x, zi+1 = y, zi+2 = yj+1, . . . , zi+k−j+1 = yk

we have zsσzs+1 for all s = 1, . . . , i+k−j. As j ≤ i, we have i+k−j+1 ≥ k+1and hence the height of σ is at least k + 1. Thus σ �∈ O

(k)n . This shows that

τm is a maximal element of O(k)n and thus proves (i).

Now to prove (ii) we have to show that each σ ∈ O(k)n is contained in

some τm. Let σ ∈ O(k)n . Define M1 as the set of all minimal elements of

N with respect to σ, M2 as the set of all minimal elements of N\M1 withrespect to σ, M3 as the set of all minimal elements of N\(M1 ∪ M2) withrespect to σ and so on.

We claim that Mk �= ∅. Indeed, as σ ∈ O(k)n , there exist xi, i = 1, . . . , k,

such that xiσxi+1, i = 1, . . . , k − 1. Assume xi ∈ Mf(i), i = 1, . . . , k. Thenf(i) < f(j) if i < j and hence f(k) ≥ k. At the same time Mf(k) �= ∅ andthus Mk �= ∅.


Now we claim that Mk+1 = ∅. Assume not and let xk+1 ∈ Mk+1. Then,by construction, there exist xi ∈ Mi, i = 1, . . . , k, such that xiσxi+1 for alli = 1, . . . , k. This contradicts the fact that the height of σ is k.

Finally, we claim that σ ⊂ τm, where m = (M1, M2, . . . , Mk). Let x ∈Mi, y ∈ Mj and i ≤ j. Then yσx would imply that x is not minimal inN\(M1 ∪ · · · ∪ Mi−1), which contradicts the definition of Mi. Hence for allx and y as above we have that yσx does not hold. This means that σ ⊂ τm,which proves the claim (ii).

The claim (iii) is obvious.

Now we are ready to formulate our main result of this chapter:

Theorem 8.2.3 Let k ∈ Z, 1 ≤ k ≤ n.

(i) For each m as above the semigroups Nτm and N ′τm are maximal ele-

ments in Nilk(PT n) and Nilk(ISn), respectively.

(ii) Each maximal element of Nilk(PT n) and Nilk(ISn), has, respectively,the form Nτm and N ′

τm for some m as above.

(iii) If m �= n, then Nτm �= Nτn and N ′τm �= N ′

τn.

Proof. Assume that Nτm ⊂ T for some T ∈ Nilk(PT n). Then τNτm⊂ τT

by Proposition 8.1.7(i). However, τNτm= τm by Proposition 8.1.7(iii). As

τm is a maximal element in O(k)n by Proposition 8.2.2(i) and τT ∈ O

(k)n by

Lemma 8.1.8(i), we get that τm = τT . Hence T ⊂ Nτm by the definition ofNτm . This implies T = Nτm and proves (i) in the case of PT n. In the caseof ISn the proof is just the same.

Let now T ∈ Nilk(PT n) be a maximal element. Then τT ∈ O(k)n and

hence τT ⊂ τm for some m as above by Proposition 8.2.2(ii). Now Propo-sition 8.1.7(ii) implies that T ⊂ Nτm . Moreover, Nτm ∈ Nilk(PT n) byLemma 8.1.8(ii). Hence T = Nτm by the maximality of T . This proves (ii)in the case of PT n. In the case of ISn the proof is just the same.

Finally, let m �= n. Without loss of generality we may assume that inthis case there exist x, y ∈ N such that xτmy holds while xτny does nothold. Consider the element α ∈ ISn of rank one, which is uniquely definedby the property α(y) = x. Then α ∈ Nτm and α ∈ N ′

τm , while α �∈ Nτn andα �∈ N ′

τn . This completes the proof.

Example 8.2.4 If M1 = {1}, M2 = {2, 3} and m = {M1, M2}, then thesemigroup Nτm has the following elements:(

1 2 3∅ ∅ ∅

),

(1 2 3∅ ∅ 1

),

(1 2 3∅ 1 ∅

),

(1 2 3∅ 1 1

).

The first three of these elements form the semigroup N ′τm .

8.2. CLASSIFICATION OF NILPOTENT SUBSEMIGROUPS 137

Corollary 8.2.5 If k < n, then each maximal element of Nilk(PT n) andNilk(ISn) is contained in some maximal element of the sets Nilk+1(PT n)and Nilk+1(ISn), respectively.

Proof. Let m = (M1, M2, . . . , Mk). As k < n, by the Pigeonhole Principlethere exists i ∈ {1, 2, . . . , k} such that |Mi| > 1. Let x ∈ Mi and consider

n = (M1, . . . , Mi−1, {x}, Mi\{x}, Mi+1, . . . , Mk).

Obviously τm ⊂ τn and n ∈ O(k+1)n . Hence, by Proposition 8.1.7(ii) we have

Nτm ⊂ Nτn and N ′τm ⊂ N ′

τn . As by Theorem 8.2.3(ii) each maximal elementof Nilk(PT n) and Nilk(ISn) has, respectively, the form Nτm and N ′

τm forsome m as above, the claim follows.

Corollary 8.2.6 (i) The maximal elements of Nil(PT n) and Niln(PT n)(respectively of Nil(ISn) and Niln(ISn)) coincide.

(ii) Both PT n and ISn contain n! maximal nilpotent subsemigroups (thatis, maximal elements in Nil(PT n) and Nil(ISn), respectively).

(iii) If T1 and T2 are maximal nilpotent subsemigroups of PT n (or ISn),then there exists ϕ ∈ Inn(PT n) (resp. Inn(ISn)) such that ϕ(T1) = T2.In particular, T1

∼= T2.

Proof. The statement (i) follows immediately from Corollaries 8.2.5 and8.1.9.

As |N| = n, any element m = (M1, . . . , Mn) ∈ O(n)n is just a permutation

of the elements of N. Conversely, each permutation defines a unique elementof the form m = (M1, . . . , Mn). Hence O

(n)n contains exactly n! maximal

elements. This proves (ii).To prove (iii) consider m = ({a1}, . . . , {an}) and n = ({b1}, . . . , {bn}).

Let further α ∈ Sn such that α(ai) = bi, i = 1, . . . , n. Then a direct calcula-tion shows that α−1Nτnα = Nτm , which proves (iii) and thus completes theproof.

Remark 8.2.7 Strictly speaking a “maximal nilpotent subsemigroup” isa maximal element of Nil(S). For example, in formulations of theorems wewill use only this meaning. However, in the general discussion (for example,in the title of this section), it is often convenient to use this phrase meaningmaximal elements of some Nilk(S).

Exercise 8.2.8 Prove that |O(n)n | = n!.

Exercise 8.2.9 Let m = (M1, . . . , Mk) and n = (M ′1, . . . , M

′k) be such that

|Mi| = |M ′i | for all i = 1, . . . , k. Show that Nτm

∼= Nτn and N ′τm

∼= N ′τn .


Corollary 8.2.10 The number of maximal elements in both Nilk(PT n) andNilk(ISn) equals

k∑i=0

(−1)i

(k

i

)(k − i)n = k! · S(n, k).

Proof. By Theorem 8.2.3(ii), the number of maximal elements in both setsNilk(PT n) and Nilk(ISn) equals the number X of ordered partitions N =M1∪· · ·∪Mk into a disjoint union of nonempty subsets. From the definition ofS(n, k) we immediately get X = k!S(n, k). In contrast, each ordered partitionM1∪· · ·∪Mk of N as above defines a surjective function f : N → {1, 2, . . . , k}via f(i) = t such that i ∈ Mt, i = 1, . . . , n. This correspondence is obviouslybijective and hence the statement follows by a standard application of theinclusion–exclusion formula.

8.3 Cardinalities of Maximal Nilpotent,Subsemigroups

Let m = (M1, . . . , Mk) be as in the previous section. From Exercise 8.2.9 itfollows that |Nτm | and |N ′

τm | depend only on the cardinalities mi = |Mi|,i = 1, . . . , k. Set also m0 = 1. We fix this notation for this section.

Proposition 8.3.1 (i)

|Nτm | =k∏

i=1

(m0 + m1 + m2 + · · · + mi−1)mi .

(ii) Each maximal nilpotent subsemigroup of PT n has cardinality n!.

Proof. To construct some α ∈ Nτm we have to define α(x) for each x ∈ N.If x ∈ Mi, then the definition of Nτm says that either x �∈ dom(α) or α(x) ∈M1∪M2∪· · ·∪Mi−1. Hence we have m0 +m1 +m2 + · · ·+mi−1 independentchoices for α(x) for each such x. The statement (i) is now obtained applyingthe product rule.

The statement (ii) is a special case of (i) as for a maximal nilpotentsubsemigroup we have k = n and mi = 1 for all i.

In the case of ISn the analogous question is much more interesting andnontrivial. To answer it we will need some preparation. We fix the followingnotation:

X(m1, . . . , mk) = |N ′τm |.

If f(x) = amxm + am−1xm−1 + · · · + a1x + a0 is a polynomial with integer

coefficients, we set

f(B) = amBm + am−1Bm−1 + · · · + a1B1 + a0,

where Bi denotes the i-th Bell number.

8.3. CARDINALITIES OF NILPOTENT SUBSEMIGROUPS 139

Exercise 8.3.2 Let f(x) and g(x) be two polynomials with integer coeffi-cients and r ∈ Z. Show that (f + g)(B) = f(B) + g(B) and (rg)(B) = r(g(B)).

Finally, for m ∈ N we set [x]m = x(x − 1)(x − 2) . . . (x − m + 1) anddefine the polynomial Fm1,...,mk

(x) as follows:

Fm1,...,mk(x) = [x]m1 [x]m2 · · · · · [x]mk

.

In the case of ISn the cardinalities of N ′τm are given by the following quite

amazing result:

Theorem 8.3.3 (i) |N ′τm | = Fm1,...,mk

(B).

(ii) Each maximal nilpotent subsemigroup of ISn has cardinality Bn.

Proof. We start with the statement (ii). By Corollary 8.2.6(i), the maximalnilpotent subsemigroups of ISn correspond to the case mi = 1 for all i. So,we may assume m = ({a1}, . . . , {an}). Under such assumptions let α ∈ N ′

τm .Then the connected components of Γα form an unordered partition of Ninto a disjoint union of subsets. Conversely, let X1, . . . , Xs be an unorderedpartition of N into a disjoint union of subsets. For each j = 1, . . . , s let|Xj | = kj and, further, let us write the elements of Xj in the same order asthey appear in the sequence a1, . . . , an. Let xj

1, xj2, . . . , x

jkj

be the resultingsequence. Now define the element α as follows:

dom(α) =s⋃

j=1

{xj2, x

j3, . . . , x

jkj},

and α(xjt ) = xj

t−1 for all j = 1, . . . , s and t = 2, . . . , kj . In other words, wedefine α as the element with the following graph Γα:

x11 x1

2α�� . . .α�� x1

k1

α��

x21 x2

2α�� . . .α�� x2

k2

α��

......

......

xs1 xs

2α�� . . .α�� xs

ks

α��

From the definition of N ′τm we have α ∈ N ′

τm . Moreover, the connectedcomponents of Γα are exactly X1, . . . , Xs. It is easy to see that α is theonly element of N ′

τm such that the connected components of Γα are exactly


X1, . . . , Xs. Hence we obtain a bijection between the elements of N ′τm and

partitions of N into unordered disjoint unions of nonempty subsets. Thismeans that |N ′

τm | = Bn in this case, proving (ii).To prove the statement (i) we will need one auxiliary lemma.

Lemma 8.3.4

X(m1, . . . , mi−2, mi−1, 1, mi+1, . . . , mk) == X(m1, . . . , mi−2, mi−1, mi+1 + 1, . . . , mk)+

+ mi+1 · X(m1, . . . , mi−2, mi−1, mi+1, . . . , mk).

Proof. Assume that |Mi| = 1, or in other words that

m = (M1, . . . , Mi−1, {a}, Mi+1, . . . , Mk).

We decompose N ′τm into a disjoint union of two sets T1 and T2 as follows:

T1 = {α ∈ N ′τm : a �∈ α(Mi+1)}, T2 = {α ∈ N ′

τm : a ∈ α(Mi+1)}.

By definition, the set T1 coincides with N ′τn , where

n = (M1, . . . , Mi−1, Mi+1 ∪ {a}, Mi+2, . . . , Mk).

Let Mi+1 = {a1, . . . , ami+1}. For j ∈ {1, 2, . . . , mi+1} define

T j2 = {α ∈ T2 : α(aj) = a}.

Then T2 = T 12 ∪ · · · ∪ T

mi+1

2 is a partition of T2 into a disjoint union ofnonempty subsets. Fix now j ∈ {1, 2, . . . , mi+1}. With every α ∈ T j

2 , weassociate the element α′ ∈ N ′

τk, where k = (M1, . . . , Mi−1, Mi+1, . . . , Mk),

in the following way:

α′(y) =

{α(y), y �= aj ;α(a), y = aj .

Obviously, the mapping α → α′ from T j2 to N ′

τkis bijective. Hence |T j

2 | =|N ′

τk|. This gives

X(m1, . . . , mi−1, 1, mi+1, mi+2, . . . , mk) == |N ′

τm | = |T1| + |T2| = |N ′τn | + mi+1|N ′

τk| =

= X(m1, . . . , mi−1, mi+1 + 1, mi+2, . . . , mk)++ mi+1 · X(m1, . . . , mi−1, mi+1, mi+2, . . . , mk)

and the claim follows.

8.4. COMBINATORICS OF NILPOTENT ELEMENTS IN ISn 141

We prove the statement (i) by induction on the sum m′ of all mi’s suchthat mi > 1. If m′ = 0, then mi = 1 for all i. In particular, F1,...,1(x) = xn

and F1,...,1(B) = Bn = |N ′τm | according to the definitions and the statement

(ii), proved above.Assume now that m′ > 0. Choose some i such that mi ≥ 2. Applying

Lemma 8.3.4 and the inductive assumption we get

X(m1, . . . , mi−1, mi, . . . , mk) == X(m1, . . . , mi−1, 1, mi − 1, mi+1, . . . , mk)−

− (mi − 1) · X(m1, . . . , mi−1, mi − 1, mi+1, . . . , mk) == Fm1,...,mi−1,1,mi−1,mi+1,...,mk

(B)−− (mi − 1) · Fm1,...,mi−1,mi−1,mi+1,...,mk

(B). (8.2)

Now we observe that

[x]m = (x − (m − 1))[x]m−1 = [x]m−1[x]1 − (m − 1)[x]m−1.

From this and the definition of the polynomials Fm1,...,mk(x) we get

Fm1,...,mi−1,mi,...,mk(x) = Fm1,...,mi−1,1,mi−1,mi+1,...,mk

(x)−− (mi − 1) · Fm1,...,mi−1,mi−1,mi+1,...,mk

(x). (8.3)

Applying (8.3) to (8.2) and using Exercise 8.3.2 we get

X(m1, . . . , mi−1, mi, . . . , mk) = Fm1,...,mi−1,mi,...,mk(B),

completing the proof.

8.4 Combinatorics of Nilpotent Elements in ISn

In this section, we discover some combinatorial relations between differentnumbers, associated with the semigroup ISn, especially with nilpotent ele-ments of ISn. Define the following numbers:

In = the cardinality of ISn

Nn = the total number of nilpotent elements in ISn

Ln = the total number of chains in the chain decompositions ofall elements of ISn

Mn = the total number of chains in the chain decompositions of allnilpotent elements of ISn,

Pn = the total number of fixed points of all elements of ISn


Example 8.4.1 The semigroup IS1 has 2 elements:(

11

),

(1∅

).

For this semigroup, we have

I1 = 2, N1 = 1, L1 = 1, M1 = 1, P1 = 1.

Example 8.4.2 The semigroup IS2 has 7 elements:(

1 21 2

),

(1 22 1

),

(1 21 ∅

),

(1 22 ∅

),

(1 2∅ 1

),

(1 2∅ 2

),

(1 2∅ ∅

).

For this semigroup, we have

I2 = 7, N2 = 3, L2 = 6, M2 = 4, P2 = 4.

Now we will present some combinatorial relations involving In, Nn, Ln,Mn, and Pn.

Proposition 8.4.3Ln =

∑α∈ISn

strank(α).

Proof. Consider the sets

A = {(α, c, x) : α ∈ ISn, c is a cycle of α, x is a point of c},B = {(β, l) : β ∈ ISn, l is a chain from the chain decomposition of β}.

Obviously, our statement is equivalent to the equality |A| = |B|.Consider the mapping f : A → B, defined as follows:

f((α, (x, a, . . . , b), x)) = (β, [x, a, . . . , b]),

where β is obtained from α substituting the cycle (x, a, . . . , b) with the chain[x, a, . . . , b]. Consider also the mapping g : B → A, which is defined asfollows: g((β, [x, a, . . . , b])) = (α, (x, a, . . . , b), x), where α is obtained fromβ substituting the chain [x, a, . . . , b] with the cycle (x, a, . . . , b). Obviouslyf and g are inverse to each other and thus |A| = |B|.

Theorem 8.4.4

Pn +1nLn = In. (8.4)


Proof. Rewrite the equality (8.4) in the following form:

nPn + Ln = nIn. (8.5)

Consider the following sets:

A = {(α, x) : α ∈ ISn, x ∈ N},B = {(β, l) : β ∈ ISn, l is a chain for the chain decomposition of β},

C = {(γ, y, z) : γ ∈ ISn, y is a fixed point of γ, z ∈ N}.

The equality (8.5) is equivalent to the equality |C|+ |B| = |A|. To prove thelatter equality, we decompose A into a disjoint union A = A1 ∪ A2, where

A1 = {(α, x) ∈ A : x belongs to some chain of α},A2 = {(α, x) ∈ A : x belongs to some cycle of α}.

Consider the transformation which maps the cycle (x, a, . . . , b) to the chain[x, a, . . . , b]. As in the proof of Proposition 8.4.3 this transformation inducesa bijection A2 → B. Hence |A2| = |B|.

To prove |A1| = |C|, we construct mutually inverse bijections f : A1 → Cand g : C → A1. Consider any element (α, x) ∈ A1. If x is the starting pointof some chain [x, a, . . . , b] of length at least 2 from the chain decompositionof α, we define f((α, x)) = (γ, x, a), where γ is obtained from α substitutingthe chain [x, a, . . . , b] with the cycle (x) and the cycle (a, . . . , b). If x is theonly point of the chain [x], we define f((α, x)) = (γ, x, x), where γ is obtainedfrom α substituting the chain [x] with the cycle (x). Finally, if x is containedin some chain [a, . . . , b, x, c, . . . , d] and is different from the starting point aof this chain, we define f((α, x)) = (γ, x, b), where γ is obtained from αsubstituting the chain [a, . . . , b, x, c, . . . , d] with the cycle (x) and the chain[a, . . . , b, c, . . . , d].

Let now (γ, y, z) ∈ C. If y = z, we define g((γ, y, z)) = (α, y), whereα is obtained from γ substituting the cycle (y) with the chain [y]. If zis a point of some chain [a1, . . . , as, z, b1, . . . , bt] in the chain decomposi-tion of γ, we set g((γ, y, z)) = (α, y), where α is obtained from γ substi-tuting the cycle (y) and the chain [a1, . . . , as, z, b1, . . . , bt] with the chain[a1, . . . , as, z, y, b1, . . . , bt]. Finally, if z is a point of some cycle (z, a1, . . . , as)of γ, we set g((γ, y, z)) = (α, y), where α is obtained from γ substituting thecycles (y) and (z, a1, . . . , as) with the chain [y, z, a1, . . . , as].

Obviously, f and g are inverse to each other implying |A1| = |C|, andthe statement follows.

Let Nn denote the set of all nilpotent elements of ISn. We have |Nn|=Nn.

Proposition 8.4.5In = Nn + Pn.


Proof. The element α ∈ ISn can have some fixed points only in the casewhen α is not nilpotent. For every α �∈ Nn let Aα = stim(α) and Aα =N\Aα. Consider the set

Mα = {β ∈ ISn : stim(β) = Aαand α|Aα= β|Aα

}.

Obviously, |Mα| = strank(α)! and either Mα = Mβ or Mα ∩ Mβ = ∅ forarbitrary α, β ∈ ISn. Hence the sets Mα form a partition of ISn\Nn into adisjoint union of subsets.

Lemma 8.4.6 The total number of fixed points of all elements of Sn

equals n!.

Proof. Consider the following sets:

A = {(α, x, y) : α ∈ Sn, x, y ∈ N, α(x) = x},B = {(β, z) : β ∈ Sn, z ∈ N}.

The statement of lemma is obviously equivalent to the equality |A| = |B|.Define the function f : A → B as follows: f((α, x, y)) = (β, y), where

β = α if x = y, and β is obtained from α substituting the cycles (x) and(y, a1, . . . , ak) with the cycle (x, y, a1, . . . , ak) if x �= y. Further define thefunction g : B → A as follows: g((β, z)) = (α, x, z), where α = β and x = zif β(z) = z, and α is obtained from β substituting the cycle (x, z, a1, . . . , ak)with the cycles (x) and (z, a1, . . . , ak) in the case β(z) �= z. Obviously f andg are mutually inverse bijections and hence |A| = |B|.

By Lemma 8.4.6 the total number of fixed points for all elements in Mα

equals strank(α)! and hence equals the cardinality of Mα. Hence the totalnumber Pn of all fixed points for all elements of ISn equals the total numberof those elements of ISn which are not nilpotent. The claim follows.

Theorem 8.4.7 (i) Nn = 1nLn.

(ii) In = 1n+1Mn+1.

Proof. Using Theorem 8.4.4 and Proposition 8.4.5 we have

1nLn = In − Pn = In − (In − Nn) = Nn.

This proves (i).Consider the following sets:

A = {(x, α) : x ∈ {1, 2, . . . , n + 1}, α ∈ IS({1, 2, . . . , n + 1}\{x})}B = {(β, l) : β ∈ Nn+1, l is a chain of β}

To prove (ii) it is enough to show that |A| = |B|.


Define the mapping f : A → B in the following way. Let (x, α) ∈ A andassume that (

a1 . . . ak

ai1 . . . aik

)

is the permutational part of α, where a1 < a2 < · · · < ak. Set f((x, α)) =(β, l) ∈ B, where l = [ai1 , . . . , aik , x] and β is obtained from α substitutingthe permutational part with l.

Define the mapping g : B → A, g : (β, l) → (x, α) in the following way: ifl = [a1, . . . , ak, ak+1], we set x = ak+1 and α is obtained from β substitutingl with the permutational part

(ai1 . . . aik

a1 . . . ak

),

where ai1 < ai2 < · · · < aik are elements a1, . . . , ak, written with respectto the natural increasing order. Obviously, f and g are mutually inversebijections, which implies |A| = |B| and completes the proof.

Theorem 8.4.8 (i) Nn = In−1 + Ln−1.

(ii) In = Nn + Mn.

Proof. Consider the following sets:

A = {α ∈ Nn : n �∈ im(α)},B = {α ∈ Nn : n ∈ im(α)},

C = {(β, l) : β ∈ ISn−1, lis a chain ofβ}.

We have A∪B = Nn, A∩B = ∅, |C| = Ln−1. Hence the equality (i) wouldfollow if we showed that |A| = In−1 and |B| = |C|.

Lemma 8.4.9 |A| = In−1.

Proof. This proof is rather similar to that of Theorem 8.4.7(ii). Define thefunction f : A → ISn−1 in the following way: Let α ∈ A and [n, a1, . . . , ak]be the chain starting at n (it exists since n �∈ im(α)). We set f(α) = β,where β(y) = α(y) for all y �∈ {a1, . . . , ak} and on {a1, . . . , ak} the elementβ is defined to be the following permutation:

(ai1 . . . aik

a1 . . . ak

),

where ai1 < ai2 < · · · < aik are elements a1, . . . , ak, written with respect tothe natural increasing order.


Define further the function g : ISn−1 → A as follows: For β ∈ ISn−1 letstim(β) = {a1, . . . , ak}, where a1 < a2 < · · · < ak. Assume further that onstim(β) the element β acts as the following permutation:

(a1 . . . ak

ai1 . . . aik

).

We set g(β) = α, where

α(y) =

⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

β(y), y �∈ stim(β);ai1 , y = n;aij+1 , y = aijandj < k;∅, y = aik .

Obviously, f and g are mutually inverse bijections, and hence |A| = In−1.

Lemma 8.4.10 |B| = |C|.

Proof. Define the function f : B → C in the following way: Let α ∈ B and[a1, . . . , ak, n, b1, . . . , bm] be the chain containing n (note that k ≥ 1 sincen ∈ im(α)). We set f(α) = (β, l), where β is obtained from α substituting thechain [a1, . . . , ak, n, b1, . . . , bm] by the chain [a1, . . . , ak] and the permutation:

(bi1 . . . bim

b1 . . . bm

),

where bi1 < bi2 < · · · < bim are elements b1, . . . , bm, written with respect tothe natural increasing order. Further, we set l = [a1, . . . , ak].

Define the function g : C → B as follows: For (β, l) ∈ C let l =[a1, . . . , ak], stim(β) = {b1, . . . , bm}, where b1 < b2 < · · · < bm and assumethat on stim(β) the element β acts as the following permutation:

(b1 . . . bm

bi1 . . . bim

).

We set g(β, l) = α, where α is obtained from β substituting l and the abovepermutation by the chain [a1, . . . , ak, n, bi1 , . . . , bim ]. Obviously, f and g aremutually inverse bijections, and hence |B| = |C|.

The statement (i) now follows from Lemmas 8.4.9 and 8.4.10.Using Theorem 8.4.7(ii) and Proposition 8.4.5 we can rewrite the equality

of (ii) as follows:Pn = nIn−1. (8.6)


To prove (8.6) consider the following two sets:

D = {(α, x) : α ∈ ISn, x ∈ N, α(x) = x},E = {(β, y) : y ∈ N, β ∈ IS(N\{y})}.

The equality (8.6) is equivalent to the equality |D| = |E|.Define the mapping f : D → E in the following way: f((α, x)) = (β, x),

where β is the restriction of α to N\{x}. Define the mapping g : E → D inthe following way: g((β, y)) = (α, y), where

α(z) =

{β(z), z �= y;y, z = y.

Obviously, f and g are mutually inverse bijections and hence |D| = |E|. Thisproves (8.6) and completes the proof.

Corollary 8.4.11 Pn = Mn.

Proof. Follows from Proposition 8.4.5 and Theorem 8.4.8(ii).

Theorem 8.4.12 Nn = |{α ∈ ISn : 1 �∈ dom(α)}|.

Proof. Set A = {α ∈ ISn : 1 �∈ dom(α)} and consider the following decom-position of A into a disjoint union of subsets:

A =⋃

X⊂{2,3,...,n}AX ,

where AX = {α ∈ A : stim(α) = X}.Consider also the following decomposition of Nn into a disjoint union of

subsets:Nn =

⋃X⊂{2,3,...,n}

BX ,

where

BX = {β ∈ Nn : β contains a chain [b1, . . . , bl, 1, a1, . . . , ak]and {a1, . . . , ak} = X}.

For a fixed X = {a1, . . . , ak}, a1 < a2 < · · · < ak, define the mapping f :AX → BX as follows: f(α) = β, where β is obtained from α substituting thechain [b1, . . . , bl, 1] of α (which exists as 1 �∈ dom(α)) and the permutationalpart (

a1 . . . ak

ai1 . . . aik

)(8.7)

of α by the chain [b1, . . . , bl, 1, ai1 , . . . , aik ].


Define the mapping g : BX → AX as follows: g(β) = α, where α isobtained from β substituting the chain [b1, . . . , bl, 1, ai1 , . . . , aik ] by the chain[b1, . . . , bl, 1] and the permutational part given by (8.7). The mappings f andg are obviously mutually inverse bijections and hence |AX | = |BX | for allX. The statement of our theorem follows.


8.5.1 In the context of ring theory, nilpotent semigroups were studied al-ready in [Ko, Lev]. The oldest purely semigroup theoretical paper, dedicatedto the study of nilpotent semigroups, which we managed to find, is [She].

8.5.2 An alternative notion of nilpotent semigroups was proposed in [Ma2].This notion generalizes the notion of nilpotent groups. Finite nilpotent semi-groups in our sense are sometimes referred to as nil-semigroups.

8.5.3 From the results of [KRS] it follows that almost all semigroups arenilpotent in the following sense: Let An denote the total number of semigroupstructures on N and Bn denote the total number of nilpotent semigroupstructures on N, then Bn/An → 1, n → ∞. The proof of the main statementof [KRS] (which implies the above claim) is only outlined and no completeargument was ever published, so some specialists put this proof in question.Nevertheless, the experimental data obtained so far strongly suggest thatthe above statement is true. Thus in [SYT] it is shown that 99% of almost2×109 semigroups consisting of 8 elements are nilpotent.

8.5.4 The idea to use partial orders for description of maximal nilpotentsubsemigroups appears first in the papers [GK2, GK3, GK4]. In particular,the results of Sect. 8.1 and 8.2 are essentially proved in these papers. Thisidea has proved to be very useful and applicable to many different situations,see for example [GM1, GM2, KuMa3, Sh1, Sh2, St1, Ts4]. Furthermore, thisidea led to the development of a very general and abstract approach to thestudy of nilpotent subsemigroups proposed in [GM6].

8.5.5 Theorem 8.3.3 is due to M. Pavlov. It appeared in [GP] for the firsttime, and can also be found in [GM4]. In [GM2] analogous ideas were suc-cessfully applied to determine the cardinalities of some maximal nilpotentsubsemigroups of the semigroup IOn of all partial order-preserving injec-tions on the chain 1 < 2 < · · · < n.

8.5.6 The results of Sect. 8.4 are obtained in [GM5]. Some of these resultswere reproved in [JM] using generating functions.

8.5.7 Let A ⊂ N and T ⊂ ISn be a nilpotent semigroup with the zeroelement εA.


Lemma 8.5.1 α(x) = x for all α ∈ T and x ∈ A.

Proof. Since εA is the zero element in T and εA(x) = x for all x ∈ A, wehave

α(x) = α(εA(x)) = (αεA)(x) = εA(x) = x.

Lemma 8.5.2 Let α ∈ T and x ∈ dom(α) ∩ (N\A). Then α(x) ∈ N\A.

Proof. Assume that α(x) = y ∈ A. Then α(x) = α(y) by Lemma 8.5.1,which contradicts the fact that α is a partial injection. The claim follows.

By Lemma 8.5.2 we can consider the restriction map

T → IS(N\A)α → α|N\A.

By Lemma 8.5.1 this mapping is even injective. Since εA|N\A is the zeroelement of IS(N\A), the image of T under this restriction is a nilpotentsubsemigroup of IS(N\A), now with the usual zero element. This reducesthe study of all nilpotent subsemigroups of ISn to the study of nilpotentsubsemigroups with the usual zero element.

8.5.8 Define the mapping in : PT n → Tn+1 in the following way: for α ∈PT n and x ∈ {1, 2, . . . , n + 1} set

in(α)(x) =

{α(x), x ∈ dom(α);n + 1, otherwise.

A direct calculation shows that in is a homomorphism. Furthermore, in isobviously injective. The mapping in is called the canonical inclusion of PT n

to Tn+1.

Proposition 8.5.3 (i) Let T ⊂ PT n be a nilpotent subsemigroup withthe usual zero. Then in(T ) is a nilpotent subsemigroup of Tn+1 withthe zero element 0n+1.

(ii) Let T ⊂ Tn+1 be a nilpotent subsemigroup with the zero element 0n+1.Then T = in(S) for some nilpotent subsemigroup S ⊂ PT n with theusual zero.

Proof. The statement (i) follows from the facts that in is a homomorphismand in(0) = 0n+1.

To prove (ii) consider any nilpotent subsemigroup T ⊂ Tn+1 with thezero element 0n+1. For each α ∈ T we have

α(n + 1) = α(0n+1(n + 1)) = (α0n+1)(n + 1) = 0n+1(n + 1) = n + 1.


In particular, for α, β ∈ T we have α �= β if and only if there exists x ∈{1, 2, . . . , n} such that α(x) �= β(x). Let S be the set of all γ ∈ PT n such thatin(γ) ∈ T . Then the previous argument implies that in induces a bijectionbetween S and T . A direct calculation shows that S is in fact a subsemigroupof PT n. This completes the proof.

Proposition 8.5.3 reduces the study of nilpotent subsemigroups of Tn+1

with the zero element 0n+1 to the study of nilpotent semigroups of PT n withthe usual zero. Since all left zeros of Tn+1 can be mapped to 0n+1 using someinner automorphisms, we can reduce the study of all nilpotent subsemigroupsof Tn+1 with zero elements of the form 0x, x ∈ {1, 2, . . . , n+1}, to the studyof nilpotent semigroups of PT n with the usual zero.

8.5.9 Let now ε ∈ PT n be any idempotent and T ⊂ PT n be any nilpotentsubsemigroup of PT n with the zero element ε. Note that if ε ∈ Tn, thenfrom α(n+1)n

= ε for all α ∈ T it follows that dom(α) = N for all α ∈ T , inparticular, T is a nilpotent subsemigroup of Tn. So the study of all nilpotentsubsemigroups of Tn reduces to the study of all nilpotent subsemigroups ofPT n.

Let A = dom(ε), {a1, . . . , ak} = im(ε) and A = A1 ∪ · · · ∪ Ak be thepartition such that ε(x) = ai for all x ∈ Ai.

Lemma 8.5.4 Each α ∈ T has the following properties:

(a) α(x) ∈ Ai for all i and all x ∈ Ai.

(b) α(ai) = ai for all i.

(c) α(x) ∈ N\A for all x ∈ (N\A) ∩ dom(α).

Proof. Let x ∈ Ai. Then for α(x) we have ε(α(x)) = (εα)(x) = ε(x) = ai.Hence α(x) ∈ Ai, which proves (a). The statement (c) is proved analogously.

For any i = 1, . . . , k we have α(ai) = α(ε(ai)) = (αε)(ai) = ε(ai) = ai.This proves (b) and completes the proof.

By Lemma 8.5.4 we can restrict each α ∈ T to each of the sets A1, . . . , Ak,N\A. These restrictions in fact define homomorphisms from T to T (Ai),i = 1, . . . , k, and PT (N\A). If α, β ∈ T and α �= β, then the images of αand β with respect to at least one of these homomorphisms are different.The images of these homomorphisms are nilpotent subsemigroups of T (Ai),whose zero elements are some left zeros of T (Ai), and nilpotent subsemi-groups of PT (N\A) with the usual zero. This reduces the study of nilpotentsubsemigroups of PT n with the zero element ε to the study of nilpotent sub-semigroups of PT n with the usual zero element and the study of nilpotentsubsemigroups of Tn, whose zero is a left zero of Tn (see 8.5.8). It might bea good exercise for the reader to write down all necessary details.


8.5.10 Let M be a finite set and G be a subgroup of S(M). Define thebinary relation ∼ on M as follows: x ∼ y if and only if there exists g ∈ Gsuch that x = g(y). It is easy to check that ∼ is an equivalence relation.The equivalence classes of this relation are called the G-orbits of M . Let OG

denote the number of G-orbits. For g ∈ G let fi(g) denote the number offixed points of g. In the case G = S(M) the number of G-orbits on M isobviously 1. Hence Lemma 8.4.6 is a special case of the following famousstatement, known as Burnside’s Lemma, the Cauchy-Frobenius Lemma orthe Orbit-counting Theorem:

Theorem 8.5.5OG =

1|G|

∑g∈G

fi(g).

The proof of this statement can be found in many general books oncombinatorics and group theory. We refer the reader to [Ne].

8.5.11 From Exercise 2.10.21 and Proposition 8.4.5 it follows that Pn/In → 1,n → ∞. This is an asymptotic analogue of Lemma 8.4.6 for ISn.

8.5.12 If m = (M1, . . . , Mk) and n = (M ′1, . . . , M

′k), then the semigroups

N ′τ(m) and N ′

τ(n) are isomorphic if and only if |Mi| = |M ′i | for all i. This is

proved in [GK4].


8.6.1 For each k ∈ N construct a nilpotent semigroup S such that |S| =nd(S) = k.

8.6.2 Show that every nilpotent semigroup contains exactly one idempotent.

8.6.3 Let S and T be two nilpotent semigroups. Consider the set S × T ={(s, t) : s ∈ S, t ∈ T} with the operation (s, t) · (s′, t′) = (ss′, tt′). Show thatS × T is a nilpotent semigroup.

8.6.4 Let ϕ : S � T be an epimorphism of semigroups.

(a) Show that the semigroup T is nilpotent provided that S is nilpotent.

(b) Construct an example of S, T , and ϕ as above such that T is nilpotentand S is not nilpotent.

8.6.5 Let ϕ : S � T be an epimorphism of semigroups. Assume that T isnilpotent and nd(T ) = k. Let 0 be the zero element of T and assume thatX = ϕ−1(0) = {s ∈ S : ϕ(s) = 0} is a nilpotent semigroup and nd(X) = l.Show that S is a nilpotent semigroup and nd(S) ≤ l · k.


8.6.6 ([Ko, Lev]) Let Matn(C) denote the semigroup of all n × n complexmatrices with respect to the usual matrix multiplication. Let further T ⊂Matn(C) be some nilpotent subsemigroup. Show that nd(T ) ≤ n.

8.6.7 ([Ko, Lev, Fa, KuMa3]) Classify all maximal nilpotent subsemigroupsof the semigroup Matn(C).

8.6.8 Let T ⊂ PT n (or T ⊂ ISn) be a subsemigroup, containing I1. Showthat for each k > 0 the mapping X → T ∩X defines a bijection between theset of maximal elements in Nilk(PT n) (resp. Nilk(ISn)) and the set Nilk(T ).

8.6.9 Construct a subsemigroup T ⊂ ISn, which contains I1 and for whichthere exist two nonisomorphic maximal nilpotent subsemigroups of nilpo-tency degree n.

8.6.10 For each m as in Sect. 8.2 determine the cardinality of the subsemi-group N ′

τm ∩ I1 of ISn.

8.6.11 Prove that Nn equals the number of all partial injections from N to{1, 2, . . . , n − 1}.

8.6.12 ([GM5]) Prove that

n−1∑k=0

(n − k)(

n

k

)2

k! =n∑

k=1

[n]kIn−k.

8.6.13 ([GM5]) Prove that the following two sets have the same cardinality:

A = {(α, l) : α ∈ ISn, l is a chain of α},B = {(β, x) : β ∈ ISn, 1 ∈ dom(β), x = βk(1) for some k ≥ 0}.

8.6.14 ([GM2, GM4]) Prove that the number of those α ∈ ISn, whichsatisfy the following two conditions:

(a) α(x) < α(y) for all x, y ∈ N, x < y

(b) α(x) < x for all x ∈ N

equals the n-th Catalan number Cn = 1n+1

(2nn

).

Chapter 9

Presentation

9.1 Defining Relations

Let A be a nonempty set, which we will call the alphabet. Elements of Awill be called letters. Any finite nonempty sequence a1a2 · · · ak of elementsfrom A will be called a word or a word over A. The set of all words over A isdenoted A+. If u = a1a2 · · · ak and v = b1b2 · · · bm are two words, we definetheir product or concatenation or juxtaposition as follows:

uv = a1a2 · · · akb1b2 · · · bm.

It is obvious that this binary operation is associative, which turns A+ intoa semigroup. This semigroup is called the free semigroup with base A. Ob-viously, A is a generating system of A+.

Let now S be a semigroup and A be a generating system for S. On the onehand, then every element s ∈ S can be written as a product s = a1a2 · · · ak ofsome elements from A, however not uniquely in general. On the other hand,every product a1a2 · · · ak can be considered as an element of A+. If thereexist two words u = a1a2 · · · ak and v = b1b2 · · · bm in A+, which determinethe same element s ∈ S, we say that the relation u = v holds in S.

Proposition 9.1.1 The binary relation

ρ(S, A) = {(u, v) ∈ A+ × A+ : u = v is a relation in S}

is a congruence on A+, moreover A+/ρ(S, A) ∼= S canonically.

Proof. The mapping ϕ : A+ → S, which sends the word a1a2 · · · ak to theproduct a1a2 · · · ak, is obviously a homomorphism of semigroups. It is sur-jective since A generates S. We further have ρ(S, A) = Ker(ϕ) by definition.Hence ρ(S, A) is a congruence on A+. The claim A+/ρ(S, A) ∼= S followsfrom Theorem 6.1.6.



154 CHAPTER 9. PRESENTATION

Fixing some representatives in all classes of ρ(S, A) determines a canon-ical form for each element of S with respect to the generating system A.Sometimes, it is convenient to identify the elements from S with their canon-ical forms. Such identification might be useful if the following two naturalconditions are satisfied:

• There exists a constructable description for all canonical forms

• If the canonical forms of some elements g, h ∈ S are known, then onecan determine the canonical form of the element gh

Let u = v be a relation in S = 〈A〉. We will say that some congruenceρ on A+ contains the relation u = v provided that (u, v) ∈ ρ. In particular,the uniform congruence ωA+ = A+ × A+ contains all possible relations. Asthe intersection of an arbitrary family of congruences is a congruence, foreach set Σ of relations, there exists an unique minimal congruence ρΣ onA+, which contains all relations from Σ. If ρΣ = ρ(S, A), the set Σ is calleda set or a system of defining relations for S with respect to A. In this case,one says that S is a semigroup generated by A with the system Σ of definingrelations. This is denoted S = 〈A|Σ〉. The pair 〈A|Σ〉 is called a presentationof S. Note that a system of defining relations for S with respect to A is notunique in general.

Let Σ be a subset of A+ × A+. A pair (u, v) ∈ A+ × A+ will be calleda Σ-pair provided that u = v or there exist decompositions u = su1t andv = sv1t (where s, t, u1, v1 ∈ A+ or either s or t or both are empty), suchthat either (u1, v1) or (v1, u1) belongs to Σ.

Proposition 9.1.2 Congruence ρΣ coincides with the set of all pairs (u, v)for which there exists a finite collection w1, w2, . . . , wk ∈ A+ such that eachof the pairs (u, w1), (w1, w2), . . . , (wk−1, wk) and (wk, v) is a Σ-pair.

Proof. Let Ω denote the set of all pairs (u, v), which satisfy the condition ofthe proposition. Obviously, each Σ-pair is contained in ρΣ. As each congru-ence is an equivalence relation, in particular, is transitive, we get Ω ⊂ ρΣ.To complete the proof it is enough to show that Ω is a congruence.

Let (u, v) ∈ Ω and (u, w1), (w1, w2), . . . , (wk−1, wk), (wk, v) be the cor-responding collection of Σ-pairs. Then for every w ∈ A+ we have that thepairs (wu, ww1), (ww1, ww2), . . . , (wwk−1, wwk), (wwk, wv) are Σ-pairs aswell. Hence (wu, wv) ∈ Ω and Ω is left compatible. Analogously one showsthat Ω is right compatible. This completes the proof.

A relation u = v is said to follow from Σ provided that (u, v) ∈ ρΣ.

Remark 9.1.3 Assume that both u = v and v = w follow from Σ. Then,obviously, u = w follows from Σ as well.

9.1. DEFINING RELATIONS 155

If S contains the unit element 1, we will always assume that every gen-erating system A of S contains 1 and every system Σ of defining relationscontains all relations of the form a · 1 = a and 1 · a = a, a ∈ A. Such rela-tions will be called trivial. As trivial relations for monoids should always bepresent, we always assume that they are present, and normally omit them.

Let now A and B be two generating systems for S. Then for every a ∈ Athere exists v(a) ∈ B+ such that a = v(a), and for every b ∈ B there existsw(b) ∈ A+ such that b = w(b). We fix such v(a), a ∈ A, and w(b), b ∈ B.For arbitrary x = a1a2 · · · ak ∈ A+ and any y = b1b2 · · · bm ∈ B+ set

v(x) = v(a1)v(a2) · · · v(ak) ∈ B+, w(y) = w(b1)w(b2) · · ·w(bm) ∈ A+.

Let Σ be an arbitrary system of relations for the generating system A.Define a new system ΣB

A of relations for the generating system B in thefollowing way:

• If Σ contains the relation a1 · · · ak = a′1 · · · a′m, then ΣBA contains the

relation v(a1 · · · ak) = v(a′1 · · · a′m)

• If b ∈ B and w(b) = a1 · · · ak, then ΣBA contains the relation b =

v(a1 · · · ak) (in the proof of Theorem 9.1.4 it will be convenient todenote the right-hand side of this relation by f(b))

• ΣBA does not contain any other nontrivial relations

Theorem 9.1.4 Let Σ be a system of defining relations of S with respectto the generating system A. Then ΣB

A is a system of defining relations of Swith respect to the generating system B.

Proof. From the definitions we have ΣBA ⊂ ρ(S, B). Hence, it is enough to

show that every relation in ρ(S, B) follows from ΣBA . Let

b1b2 · · · bk = b′1b′2 · · · b′m (9.1)

be some relation in ρ(S, B). For every i, 1 ≤ i ≤ k, and every j, 1 ≤ j ≤ m,the system ΣB

A contains the relations bi = f(bi) and b′j = f(b′j). This yieldsthat the relations

b1b2 · · · bk = f(b1)f(b2) · · · f(bk) and b′1b′2 · · · b′m = f(b′1)f(b′2) · · · f(b′m)

follow from ΣBA (even belong to ΣB

A).From (9.1) it follows that ρ(S, A) contains the relation w(b1b2 · · · bk) =

w(b′1b′2 · · · b′m). As ρ(S, A) = ρΣ, by Proposition 9.1.2 there exists a finite

collection w1, . . . , wr ∈ A+ such that all pairs

(w(b1b2 · · · bk), w1), (w1, w2), . . . , (wr−1, wr), (wr, w(b′1b′2 · · · b′m))


are Σ-pairs. However, if (x, y) is a Σ-pair, then the pair (v(x), v(y)) is aΣB

A-pair by definition. This means that the relation

v(w(b1b2 · · · bk)) = v(w(b′1b′2 · · · b′m))

follows from ΣBA .

Observe now that the word v(w(b1b2 · · · bk)) coincides with the wordf(b1)f(b2) · · · f(bk), and at the same time the word v(w(b′1b

′2 · · · b′m)) coin-

cides with the word f(b′1)f(b′2) · · · f(b′m). Hence, the relation (9.1) followsfrom ΣB

A by Remark 9.1.3. As every relation from ρ(S, B) follows from ΣBA ,

we obtain that ΣBA is a system of defining relations of S with respect to the

generating system B and the proof is complete.

9.2 A presentation for ISn

According to Lemma 3.1.1 and Theorem 3.1.4, a subset A ⊂ ISn is a gen-erating system for ISn if and only if A contains some generating system A1

of the symmetric group Sn and an element of rank (n − 1).

Lemma 9.2.1 Transpositions (1, 2), (1, 3), . . . , (1, n) generate Sn.

Proof. This follows from the equality (i, j) = (1, i)(1, j)(1, i) and the stan-dard fact that the set of all transpositions generates Sn (see Exercise 3.3.2).

The above implies that we can choose for example the following generat-ing system A for ISn: we take (n−1) transpositions πk = (1, k), k = 2, . . . , n,and the idempotent ε(1) = ε{2,3,...,n}. Let Σ1 denote an arbitrary system ofdefining relations for the group Sn with respect to the generating systemA1 = {π2, . . . , πn}. We also set ε(k) = πkε(1)πk for all k = 2, . . . , n.

Theorem 9.2.2 Let n ≥ 4 and Σ denote the union of Σ1 with the set of thefollowing relations:

(a) ε2(1) = ε(1)

(b) ε(1)ε(2) = ε(2)ε(1)

(c) ε(2)πk = πkε(2), k = 3, 4, . . . , n

(d) ε(k)π2 = π2ε(k), k = 3, 4, . . . , n

(e) π2ε(1)ε(2) = ε(1)ε(2)

Then Σ is a system of defining relations for the semigroup ISn with respectto the generating system A.

9.2. A PRESENTATION FOR ISn 157

To prove this theorem we will need several lemmas. Note that, since Σ1

is a system of defining relations for Sn, we may use arbitrary relations fromSn in the proof.

Lemma 9.2.3 The relation ε2(i) = ε(i) follows from Σ for any i.

Proof. As π2i = ε and ε2

(1) = ε(1) belongs to Σ, we have

ε2(i) = πiε(1)πi · πiε(1)πi

= πiε(1) · ε(1)πi

= πiε(1)πi

= ε(i).

Lemma 9.2.4 For any k �= 1 and k �= i the relation πiε(k) = ε(k)πi followsfrom Σ.

Proof. If k = 2 or i = 2, the claim follows from the relations (c) and (d),respectively. If both k �= 2 and i �= 2, then we have

πiε(k) = πiπkπ2ε(2)π2πk

= πkπ2πkπiπkε(2)π2πk

(by (c)) = πkπ2ε(2)πkπiπkπ2πk

= πkπ2ε(2)π2πkπi

= ε(k)πi

using the relations πiπkπ2 = πkπ2πkπiπk and π2πkπi = πkπiπkπ2πk in Sn.

Lemma 9.2.5 Let k �= 1. If π ∈ Sn is a cyclic permutation such thatπ(i) = k, then π can be written as π = τ(1, k)(1, i), where τ(k) = k.

Proof. For i �= 1 we consider all possible cases:

(1) if i = k, then π = π(1, k)(1, i)

(2) if π = (i, k), then π = (1, i)(1, k)(1, i)

(3) if π = (i, k, a, . . . , b), a �= 1, b �= 1, then π = (1, a, . . . , b, i)(1, k)(1, i)

(4) if π = (1, i, k, a, . . . , b), then π = (1, a, . . . , b)(1, k)(1, i)

(5) if π = (i, k, 1, a, . . . , b), then π = (i, a, . . . , b)(1, k)(1, i)

If i = 1, then the only possible cases are (2) and (3). The case (2) is trivial.In the case (3) we have π = (1, a, . . . , b)(1, k).


Lemma 9.2.6 Let k �= 1 and π ∈ Sn be such that π(k) = k. Then π maybe written as π = πj1πj2 · · ·πjm such that k �∈ {j1, j2, . . . , jm}.

Proof. This statement is obvious.

Lemma 9.2.7 Let π ∈ Sn be such that π(i) = k. Then the relation πε(i) =ε(k)π follows from Σ.

Proof. We first consider the case k, i �= 1. From Lemmas 9.2.5 and 9.2.6 itfollows that π can be written as a product π = πj1πj2 · · ·πjmπkπi such thatk �∈ {j1, . . . , jm}. Then

πε(i) = πj1πj2 · · ·πjmπkπi · πiε(1)πi

= πj1πj2 · · ·πjmπkε(1)πi

= πj1πj2 · · ·πjmπkε(1)πk · πkπi

= πj1πj2 · · ·πjmε(k)πkπi

(by Lemma 9.2.4) = ε(k)πj1πj2 · · ·πjmπkπi

= ε(k)π.

If k = 1 and i �= 1, for the permutation τ = π2π we have τ(i) = 2 and henceτε(i) = ε(2)τ by the arguments above. From this we obtain

πε(i) = π2π2πε(i)

= π2τε(i)

= π2ε(2)τ

= π2ε(2)π2π

= ε(1)π.

The case i = 1 now follows by symmetry.

Lemma 9.2.8 The relation ε(i)ε(k) = ε(k)ε(i) follows from Σ.

Proof. The statement is obvious in the case i = k. We split the rest intothree different cases:

First we consider the case when i = 2. Because of the relation (b) wemay assume k > 2. We have

ε(2)ε(k) = ε(2)πkε(1)πk

(by Lemma 9.2.4) = πkε(2)ε(1)πk

(by (b)) = πkε(1)ε(2)πk

= πkε(1)πk · πkε(2)πk

= ε(k)πkε(2)πk

(by (c)) = ε(k)ε(2)πkπk

= ε(k)ε(2).

9.2. A PRESENTATION FOR ISn 159

Now we consider the case i = 1. Because of the relation (b) we mayassume k > 2. We have

ε(1)ε(k) = π2ε(2)π2ε(k)

(by Lemma 9.2.4) = π2ε(2)ε(k)π2

(using the case i = 2) = π2ε(k)ε(2)π2

(by Lemma 9.2.4) = ε(k)π2ε(2)π2

= ε(k)ε(1).

Finally we consider the case i, k > 2. We have

ε(i)ε(k) = ε(i)πkε(1)πk

(by Lemma 9.2.4) = πkε(i)ε(1)πk

(using the case i = 1) = πkε(1)ε(i)πk

(by Lemma 9.2.4) = πkε(1)πkε(i)

= ε(k)ε(i).

Lemma 9.2.9 The relation πiε(1)ε(i) = ε(1)ε(i) follows from Σ.

Proof. For i = 2 the statement coincides with the relation (e). For i > 2 wefirst observe that

ε(1)ε(2)π2 = ε(2)ε(1)π2

= π2ε(1)π2 · π2ε(2)π2 · π2

= π2ε(1)ε(2)

(by (e)) = ε(1)ε(2).

We further have:

πiε(1)ε(i) = πiε(1)πiε(1)πi

= πiπ2ε(2)π2πiε(1)π2π2πi

(using π2πi = πiπ2πiπ2) = πiπ2ε(2)πiπ2πiπ2ε(1)π2π2πi

(by Lemma 9.2.4) = πiπ2πiε(2)π2πiε(2)π2πi

(by Lemma 9.2.4) = πiπ2πiπ2π2ε(2)π2ε(2)πiπ2πi

(using π2πi = πiπ2πiπ2) = π2πiε(1)ε(2)π2πiπ2

(using the observation above) = π2πiε(1)ε(2)πiπ2

(by Lemma 9.2.4) = π2πiε(1)πiε(2)π2

= π2ε(i)ε(2)π2

(by Lemma 9.2.4) = ε(i)π2ε(2)π2

= ε(i)ε(1).


Lemma 9.2.10 For any transposition (i, j) = πjπiπj we have that the re-lation (i, j)ε(i)ε(j) = ε(i)ε(j) follows from Σ.

Proof. Using Lemma 9.2.4 we have

πjπiπjε(i)ε(j) = πjπiε(i)πjε(j)πjπj

= πjπiε(i)ε(1)πj

(by Lemmas 9.2.8 and 9.2.9) = πjε(i)ε(1)πj

(by Lemma 9.2.4) = ε(i)πjε(1)πj

= ε(i)ε(j).

Lemma 9.2.11 Let {1, 2, . . . , n} = {i1, . . . , im} ∪ {j1, . . . , jn−m} and as-sume m ≥ 2. Let further π, τ ∈ Sn be such that π(jk) = τ(jk) for allk = 1, . . . , n − m. Then the relation

πε(i1) · · · ε(im) = τε(i1) · · · ε(im) (9.2)

follows from Σ.

Proof. For π and τ as above we have π = τμ, where μ acts on the set{j1, . . . , jn−m} as the identity transformation. Hence we can write μ =μr · · ·μ1, where each factor is a transposition on {i1, . . . , im}. Let μ1 =(a, b). As ε(i1), . . . , ε(im) commute by Lemma 9.2.8 and are idempotents byLemma 9.2.3, we have

ε(i1) · · · ε(im) = ε(a)ε(b)ε(i1) · · · ε(im). (9.3)

Now we compute:

πε(i1) · · · ε(im) = τμr · · ·μ1ε(i1) · · · ε(im)

(by (9.3)) = τμr · · ·μ1ε(a)ε(b)ε(i1) · · · ε(im)

(by Lemma 9.2.10) = τμr · · ·μ2ε(a)ε(b)ε(i1) · · · ε(im)

(by (9.3)) = τμr · · ·μ2ε(i1) · · · ε(im).

Now the statement follows by induction.

Proof of Theorem 9.2.2. A straightforward calculation shows that the rela-tions (a)–(e) hold in ISn.

From Lemmas 9.2.4 and 9.2.7, it follows that every word over the alpha-bet A can be written in the form

πε(j1)ε(j2) · · · ε(jm), (9.4)

9.3. A PRESENTATION FOR Tn 161

where m ≥ 0 and π ∈ Sn. Using Lemmas 9.2.3 and 9.2.8, the word (9.4) canbe written in the form

πε(i1)ε(i2) · · · ε(ik),

where k ≥ 0 and 1 ≤ i1 < i2 < · · · < ik ≤ n.It is left to show that every relation in ISn of the form

πε(i1)ε(i2) · · · ε(ik) = τε(j1)ε(j2) · · · ε(jm), (9.5)

where π, τ ∈ Sn, k ≥ 0, m ≥ 0, 1 ≤ i1 < i2 < · · · < ik ≤ n and 1 ≤ j1 <j2 < · · · < jm ≤ n, follows from Σ.

For α = πε(i1)ε(i2) · · · ε(ik) we have dom(α) = {i1, . . . , ik}. Analogouslyfor β = τε(j1)ε(j2) · · · ε(jm) we have dom(β) = {j1, . . . , jm}. Hence (9.5)implies k = m, i1 = j1, . . . , ik = jk and also π(l) = τ(l) for any elementl ∈ N\{i1, . . . , ik}. Hence the relation (9.5) has the form

πε(i1)ε(i2) · · · ε(ik) = τε(i1)ε(i2) · · · ε(ik). (9.6)

If k = 0 or k = 1, the permutations π and τ coincide and thus π = τ followsfrom Σ1, implying that (9.6) follows from Σ. If k ≥ 2, the relation (9.6)follows from Σ by Lemma 9.2.11. This completes the proof.

9.3 A Presentation for T n

Recall that in Sect. 5.3 we denoted by εm,k the idempotent of rank (n−1) inTn such that εm,k(m) = εm,k(k) = m. From Lemma 3.1.1 and Theorem 3.1.3it follows that the set A, which consists of ε, all transpositions (i, j) ∈ Sn

and all idempotents εm,k of rank (n − 1), is a generating system of Tn.Let Σ1 denote an arbitrary system of defining relations for the group Sn

with respect to the generating system A1, consisting of the identity elementand all transpositions. Denote by Σ the system of relations obtained byadding to Σ1 the following relations where the letters i, j, k, l denote differentelements of N:

(a) (k, l)εi,j(k, l) = εi,j

(b) (j, k)εi,j(j, k) = εi,k

(c) (k, i)εi,j(k, i) = εk,j

(d) (i, j)εi,j(i, j) = εj,i

(e) εi,jεl,k = εl,kεi,j

(f) εi,jεi,k = εi,kεi,j = εi,jεj,k

(g) εi,jεk,i = εk,j(i, j)


(h) εi,jεk,j = εk,j

(i) εi,jεi,j = εi,j

(j) εi,jεj,i = εi,j

(k) εi,j(i, j) = εi,j

Theorem 9.3.1 The system Σ is a system of defining relations for Tn withrespect to the generating system A.

As in the previous section, to prove this theorem we will need severaltechnical lemmas. By the same arguments as in the previous section we willfreely use all relations in the symmetric group.

Lemma 9.3.2 The relation πεi,jπ−1 = επ(i),π(j) follows from Σ for every

π ∈ Sn.

Proof. If π is the identity element, the claim is obvious. If π is a trans-position, the claim follows from the relations (a)–(d). In the general casethe claim follows by inductions since every permutation can be written as aproduct of transpositions.

Lemma 9.3.3 Let k > 1 and f = εik,jkεik−1,jk−1

· · · εi1,j1 be such that

{ik, jk} ∩ {j1, . . . , jk−1} �= ∅ and {im, jm} ∩ {j1, . . . , jm−1} = ∅

for all m < k. Then the relation

f =

{εik−1,jk−1

εik−2,jk−2· · · εi1,j1 , jk ∈ {j1, . . . , jk−1},

(ik, jk)εik−1,jk−1· · · εi1,j1 , otherwise

follows from Σ.

Proof. We prove the statement by induction on k. For k = 2 the claimfollows from the relations (h) and (i) if j2 = j1 or the relations (g), (b) andLemma 9.3.2 if j2 �= j1.

Assume now that k ≥ 3 and that the statement is proved if the numberof factors is less than k. Consider first the case jk ∈ {j1, . . . , jk−1}. Letjk = jm for some m < k. If m > 1, then from the inductive assumption wehave

εik,jkεik−1,jk−1

· · · εim,jm = εik−1,jk−1· · · εim,jm ,

which implies f = εik−1,jk−1εik−2,jk−2

· · · εi1,j1 as required.Let now m = 1. If εi2,j2 and εi1,j1 commute, we have


· · · εi2,j2εi1,j1 = εik,jkεik−1,jk−1

· · · εi3,j3εi1,j1εi2,j2

(by the inductive assumption) = εik−1,jk−1· · · εi3,j3εi1,j1εi2,j2

= εik−1,jk−1· · · εi3,j3εi2,j2εi1,j1 .


Finally, assume that εi2,j2 and εi1,j1 do not commute. Because of (e),this is possible only in the case {i1, j1} ∩ {i2, j2} �= ∅. We have j1 �∈ {i2, j2}by assumption. The case i1 = i2 is not possible because of the relation (f).Hence i1 = j2. In this case the relation (f) yields

εi2,j2εi1,j1 = εi2,j1εi2,j2 . (9.7)

Using (9.7) we compute:


· · · εi3,j3εi2,j2εi1,j1 = εik,jkεik−1,jk−1

· · · εi3,j3εi2,j1εi2,j2

(by the inductive assumption) = εik−1,jk−1· · · εi3,j3εi2,j1εi2,j2

(by (9.7)) = εik−1,jk−1· · · εi3,j3εi2,j2εi1,j1 .

If jk �∈ {j1, . . . , jk−1}, then ik ∈ {j1, . . . , jk−1}. This case is dealt with byanalogous arguments and is left to the reader.

Let J = {j1, . . . , jk} ⊂ N be nonempty and i ∈ N\J . The relation(f) implies that the elements εi,j1 , . . . , εi,jk

pairwise commute. Hence theproduct εi,j1εi,j2 · · · εi,jk

does not depend on the order of the factors. Letus denote this product by εi,J . Note that from (e) we have εi1,J1εi2,J2 =εi2,J2εi1,J1 provided that (J1 ∪ {i1}) ∩ (J2 ∪ {i2}) = ∅.

Lemma 9.3.4 Let k ≥ 1 and f = εik,jkεik−1,jk−1

· · · εi1,j1 be such that

{im, jm} ∩ {j1, . . . , jm−1} = ∅

for all m ≤ k. Then there exists a partition {j1, . . . , jk} = K1∪· · ·∪Kr intodisjoint nonempty blocks and elements {k1, . . . , kr} ∈ {i1, . . . , ik} such that

(Kp ∪ {kp}) ∩ (Kq ∪ {kq}) = ∅ if p �= q

and the relation f = εkr,Krεkr−1,Kr−1 · · · εk1,K1 follows from Σ.

Proof. For k = 1 the claim is obvious. The rest is proved by induction on k.From the inductive assumption we have

εik−1,jk−1· · · εi1,j1 = εls,Ls · · · εl1,L1 , (9.8)

where L1 ∪ · · · ∪ Ls = {j1, . . . , jk−1}, l1, . . . , ls ∈ {i1, . . . , ik−1} and also(Lp ∪ {lp}) ∩ (Lq ∪ {lq}) = ∅ for p �= q.

We first assume that ik ∈ {l1, . . . , ls}. As all factors on the right-handside of (9.8) commute, we may assume ik = ls. In this case we have

εik,jkεls,Ls = εls,jk

εls,Ls

= εls,L′s,

where L′s = Ls ∪ {jk}. This gives the necessary decomposition for the

product f : f = εls,L′sεls−1,Ls−1 · · · εl1,L1 .


Assume ik �∈ {l1, . . . , ls}. If jk �∈ {l1, . . . , ls}, then we set ls+1 = ik,Ls+1 = {jk}, and have the necessary decomposition for the product f : f =εls+1,Ls+1εls,Ls · · · εl1,L1 . If jk ∈ {l1, . . . , ls}, then as before we may assumejk = ls. Let Ls = {x1, . . . , xt}. Then we compute:

εik,jkεls,Ls = εik,lsεls,x1 · · · εls,xt

(by the relation (f)) = εik,x1εik,lsεls,x2 · · · εls,xt

= · · ·= εik,x1 · · · εik,xtεik,ls .

If we now set l′s = ik and L′s = Ls ∪ {ls}, we again obtain the necessary

decomposition for the product f : f = εl′s,L′sεls−1,Ls−1 · · · εl1,L1 .

As in Exercise 2.10.24 for α ∈ Tn we let t(α) = (t0(α), . . . , tn(α)) denotethe type of α.

Lemma 9.3.5 Let Δ = (t0, t1, . . . , tn) be a vector with nonnegative integercoefficients. Then the following conditions are equivalent:

(i) There exists α ∈ Tn such that Δ = t(α).

(ii)n∑

k=0

tk = n andn∑

k=0

ktk = n.

Proof. The implication (i)⇒(ii) is Exercise 1.5.12. To prove the implication(ii)⇒(i) let Δ be such that the condition (ii) is satisfied. Define the idem-potent transformation πΔ in the following way: Consider the usual orderon N. Partition N into an ordered collection of disjoint nonempty subsetsas follows: first take t1 subsets K1 = {1}, K2 = {2}, . . . , Kt1 = {t1}; thentake t2 subsets L1 = {t1 + 1, t1 + 2}, . . . , Lt2 = {t1 + 2t2 − 1, t1 + 2t2}; thentake t3 subsets M1 = {t1 + 2t2 + 1, t1 + 2t2 + 2, t1 + 2t2 + 3}, . . . , Mt3 ={t1 +2t2 +3t3−2, t1 +2t2 +3t3−1, t1 +2t2 +3t3}; and so on. Let k1, . . . , kt1 ,l1, . . . , lt2 , m1, . . . , mt3 and so on denote the corresponding minimal elementsin these subsets. Set

πΔ =(

K1 · · · Kt1 L1 · · · Lt2 M1 · · · Mt3 · · ·k1 · · · kt1 l1 · · · lt2 m1 · · · mt3 · · ·

). (9.9)

It follows from the construction that t(πΔ) = Δ.

Remark 9.3.6 The transformation πΔ from Lemma 9.3.5 can be con-structed in a different way: For every x ∈ N there exists a unique num-ber k(x) such that

∑i<k(x)

iti < x and∑

i≤k(x)

iti ≥ x. Then a straightforward

calculation shows that the assignmentπΔ(x) =

∑i<k(x)

iti + k(x)

⌊x − 1 −

∑i<k(x) iti

k(x)

⌋+ 1

defines πΔ.


Using (9.9), to each vector Δ, which satisfies Lemma 9.3.5(ii), we asso-ciate the word

εΔ = εl1,L1\{l1} · · · εlt2 ,Lt2\{lt2}εm1,M1\{m1} · · · εmt3 ,Mt3\{mt3} · · · . (9.10)

Lemma 9.3.7 Let μ, η ∈ E(Tn) be such that t(μ) = t(η). Then there existsσ ∈ Sn such that η = σμσ−1.

Proof. SetMμ

k = {x ∈ N : |{y ∈ N : μ(y) = x}| = k} .

Similarly define Mηk As |Mμ

k | = |Mηk | for each k, we can define bijections

τk : Mμk → Mη

k for all k > 0. For every x ∈ Mμk each of the sets P x

k ={y ∈ N : μ(y) = x} and Qx

k = {y ∈ N : η(y) = τk(x)} contains exactly kelements. Moreover, x ∈ P x

k and τk(x) ∈ Qxk. Hence we can define bijections

Θxk : P x

k → Qxk such that θx

k(x) = τk(x).As

⋃k>0

⋃x

P xk = N and P x1

k1∩P x2

k2= ∅ if (k1, x1) �= (k2, x2), the bijections

{θxk} define a permutation σ ∈ Sn. The equality η = σμσ−1 is checked by a

direct calculation.

Lemma 9.3.8 For each word u over A there exist α, β ∈ Sn and a uniquelydefined vector Δ satisfying Lemma 9.3.5(ii) such that the relation u = αεΔβfollows from Σ.

Proof. The word u has the form γ1γ2 · · · γm, where each γ is either an el-ement from Sn or has the form εi,j . By Lemma 9.3.2, the word u can bereduced to the form

u = εik,jk· · · εi1,j1μ, (9.11)

where μ ∈ Sn. Among all possible relations of this form we choose the onewith minimal possible k. Then we claim that {im, jm} ∩ {j1, . . . , jm−1} = ∅

for all m ≤ k. Indeed, if not, then there exists t ≤ k such that {it, jt} ∩{j1, . . . , jt−1} �= ∅ and we may assume that t is minimal having this prop-erty. Then from Lemma 9.3.3 we obtain that

u = εik,jk· · · εit+1,jt+1νεit−1,jt−1εi1,j1μ, (9.12)

where ν ∈ Sn. Applying Lemma 9.3.2 again we reduce the right-hand side of(9.12) to the form (9.11) with k−1 factors of the form εi,j . This contradictsthe minimality of k.

Thus {im, jm} ∩ {j1, . . . , jm−1} = ∅ for all m ≤ k. By Lemma 9.3.4, wecan reduce u to the form

u = εkr,Kr · · · εk1,K1μ,

where (Kp ∪ {kp}) ∩ (Kq ∪ {kq}) = ∅ for p �= q.


The product γ = εkr,Kr · · · εk1,K1 is an idempotent of Tn. Let Δ = t(γ).By Lemma 9.3.7 there exists α ∈ Sn such that γ = αεΔα−1. If we now setβ = α−1μ, the relation u = αεΔβ follows from Σ by Lemma 9.3.2.

From Exercise 2.10.24 we have that u = α1εΔ1β1 = α2εΔ2β2 impliest(εΔ1) = t(εΔ2). The proof is now completed by the observation that εΔ isuniquely determined by its type.

Lemma 9.3.9 Let α ∈ Tn and t(α) = (t0, t1, . . . , tn). Then

|SnαSn| =(n!)2

t0!t1! · · · tn!(1!)t1 · · · (n!)tn.

Proof. From Exercise 2.10.24 we have that β ∈ SnαSn if and only if t(β) =t(α). In particular, rank(β) = t1 + t2 + · · ·+ tn. Each such element β can beobtained in the following way: Take any partition of N into t1 blocks withone element, t2 blocks with two elements and so on. After this we chooseim(β) (which we can do in

(n

rank(β)

)different ways) and define an arbitrary

bijection between our blocks and im(β) (which we can do in (rank(β))!different ways).

To get all necessary partitions of N consider some permutation i1,i2, . . . , in of the elements from N and say that the blocks with one elementare {i1}, {i2}, . . . , {it1}; the blocks with two elements are {it1+1, it1+2}, . . . ,{it1+2t2−1, it1+2t2}; and so on. As the order of blocks of the same cardinalityand the order of elements in each block are not important, each partitionwill be obtained exactly t1! · · · tn!(1!)t1 · · · (n!)tn times. Hence

|SnαSn| =n!

t1! · · · tn!(1!)t1 · · · (n!)tn

(n

t1 + · · · + tn

)(t1 + · · · + tn)!

=(n!)2

t0!t1! · · · tn!(1!)t1 · · · (n!)tn.

Lemma 9.3.10 Let α ∈ Tn and Δ = t(α) = (t0, t1, . . . , tn). Then thenumber of congruence classes in ρΣ, which contain words of the form αεΔβ,α, β ∈ Sn, does not exceed

(n!)2

t0!t1! · · · tn!(1!)t1 · · · (n!)tn.

Proof. Let πΔ be as in (9.9). Denote by PΔ the set of all elements from Sn,for which the sets B1 = {k1, k2, . . . , kt1}, B2 = {l1, . . . , lt2} and so on areinvariant. Set B0 = N\(B1 ∪ · · · ∪Bn). Obviously PΔ is a subgroup of Sn oforder pΔ = t0!t1! · · · tn!.

Let us show that for any ν ∈ PΔ, there exists μ ∈ Sn such that therelation νεΔμ = εΔ follows from Σ. Note that PΔ is generated by those


transpositions (i, j) for which both i and j belong to the same Bk, k =0, . . . , n. Hence, it is enough to consider the case when ν = (i, j) such thati, j ∈ Bk.

Consider the set {x ∈ N : πΔ(x) = i} =: {i1, . . . , ik}. As i ∈ {i1, . . . , ik},we may assume i = i1. Analogously, we define the set {j1 = j, j2, . . . , jk}.Set μ = (i1, j1) · · · (ik, jk) and let us show that νεΔμ = εΔ follows from Σ.

If i, j, k, l are pairwise different, we have

(k, l)εi,j = (k, l)εi,j(k, l)2

(by (a)) = εi,j(k, l).

From this and the relation (9.10) it follows that it is enough to prove therelation

(i1, j1)εi1,i2 · · · εi1,ikεj1,j2 · · · εj1,jk(i1, j1) · · · (ik, jk) =

= εi1,i2 · · · εi1,ikεj1,j2 · · · εj1,jk(9.13)

Using Lemma 9.3.2, the left-hand side of (9.13) can be reduced to thefollowing:

εj1,i2 · · · εj1,ikεi1,j2 · · · εi1,jk(i2, j2) · · · (ik, jk).

Using the relation (e), the latter can be reduced to

εj1,i2εi1,j2εj1,i3εi1,j3 · · · εj1,ikεi1,jk(i2, j2) · · · (ik, jk).

Finally, using (a) the last expression reduces to

εj1,i2εi1,j2(i2, j2)εj1,i3εi1,j3(i3, j3) · · · εj1,ikεi1,jk(ik, jk). (9.14)

As soon as, using (g), we have

εk,jεi,l(l, j) = εk,jεj,lεi,j

(by (f)) = εk,jεk,lεi,j

(by (f)) = εk,lεk,jεi,j

(by (h)) = εk,lεi,j ,

the word (9.14) reduces to the form

εj1,j2εi1,i2εj1,j3εi1,i3 · · · εj1,jkεi1,ik .

The latter reduces to the right-hand side of (9.13) using (e).Now let QΔ denote the set of all permutations from Sn, with respect to

which all sets in the upper row of (9.9) are invariant. Then QΔ is a subgroupof Sn of order qΔ = (1!)t1 · · · (n!)tn . Let us show that for any θ ∈ QΔ therelation εΔθ = εΔ follows from Σ. As above, it is enough to consider the


case θ = (i, j), where both i and j belong to the same set R from the upperrow of (9.9). Let R = {i1, . . . , ik}. From (a) and (9.10) it follows that it isenough to prove the relation

εi1,i2 · · · εi1,ik(i, j) = εi1,i2 · · · εi1,ik . (9.15)

From (f) we have that the εi1,im ’s commute. If i = i1 or j = i1, we canpermute the factors such that the factor εi1,im is such that {i1, im} = {i, j}.In this case (9.15) follows from (k). If i, j �= i1, we have

εi1,i2 · · · εi1,ik(i, j) = · · · εi1,iεi1,j(i, j)(by (f)) = · · · εi1,iεi,j(i, j)(by (k)) = · · · εi1,iεi,j

(by (f)) = · · · εi1,iεi1,j .

For every ν ∈ PΔ let μν be such that the relation νεΔμν = εΔ followsfrom Σ. Then for arbitrary α, β ∈ Sn the congruence class of ρΣ, containingαεΔβ, will contain all ανεΔθμνβ, where ν ∈ PΔ and θ ∈ QΔ. Hence this classcontains at least |PΔ| · |QΔ| = pΔqΔ words from SnεΔSn. This implies thatthe number of congruence classes of ρΣ which contain words from SnεΔSn

does not exceed

|Sn|2pΔqΔ

=(n!)2

t0!t1! · · · tn!(1!)t1 · · · (n!)tn.

Proof of Theorem 9.3.1. The mapping ϕ : A+ → Tn for which the letters aremapped to the corresponding elements of Tn is, obviously, an epimorphism.The kernel of this epimorphism coincides with ρ(Tn,A). By Theorem 6.1.6,we have Tn

∼= A+/Ker(ϕ). It is easy to check that all relations (a)–(k) holdin Tn. Hence Ker(ϕ) ⊃ ρΣ. If Ker(ϕ) �= ρΣ, we would have

|Tn| = |A+/Ker(ϕ)| < |A+/ρΣ|.

However, from Lemmas 9.3.8–9.3.10, it follows that |A+/ρΣ| ≤ |Tn|. HenceKer(ϕ) = ρΣ and the proof is complete.

Remark 9.3.11 From Theorem 3.1.3 it follows that Tn is already generatedby the subset A′ of A, consisting of transpositions (1, 2), . . . , (1, n), and theidempotent ε1,2. Using Theorem 9.1.4 the system Σ of defining relations withrespect to A can be transformed into a system Σ′ of defining relations withrespect to A′.

9.4. A PRESENTATION FOR PT n 169

9.4 A presentation for PT n

For the semigroup PT n we consider the set

A = {(1, 2), (1, 3), . . . , (1, n), ε1,2, ε(1)},

which is a generating system by Theorem 3.1.5. Let Σ denote the system ofrelations, which consists of the system Σ from Theorem 9.2.2, the system Σ′

from Remark 9.3.11 and the following relations:

(a) ε(2)ε1,2 = ε1,2, ε1,2ε(2) = ε(2)

(b) ε(1)ε1,2 = ε1,2ε(1)ε(2)

(c) ε(3)ε1,2 = ε1,2ε(3)

Theorem 9.4.1 For n ≥ 4 the system Σ is a system of defining relationsfor PT n with respect to the generating system A.

We again will need several lemmas. Taking into account that Σ containssystems of defining relations for both ISn and Tn, we will freely use relationsbetween elements of each of these semigroups.

Lemma 9.4.2 The relation εi,jε(i)ε(j) = ε(i)εi,j follows from Σ.

Proof. Let us first show that ε1,jε(1)ε(j) = ε(1)ε1,j follows from Σ. We have

ε1,jε(1)ε(j) =(from Tn) = (2, j)ε1,2(2, j)ε(1)ε(j)

(from ISn) = (2, j)ε1,2ε(1)ε(2)(2, j)(by (b)) = (2, j)ε(1)ε1,2(2, j)

(from ISn) = ε(1)(2, j)ε1,2(2, j)(from Tn) = ε(1)ε1,j .

Using this we have

εi,jε(i)ε(j) =(from Tn) = (1, i)ε1,j(1, i)ε(i)ε(j)

(from ISn) = (1, i)ε1,jε(1)ε(j)(1, i)(previous relation) = (1, i)ε(1)ε1,j(1, i)

(from ISn) = ε(i)(1, i)ε1,j(1, i)(from Tn) = ε(i)εi,j .


Lemma 9.4.3 If k �= i and k �= j, then the relation εi,jε(k) = ε(k)εi,j followsfrom Σ.

Proof. Let us first show that for k > 2 the relation ε1,2ε(k) = ε(k)ε1,2 followsfrom Σ. If k = 3, the relation coincides with (c). For k > 3 we have

ε1,2ε(k) =(from ISn) = ε1,2(1, k)(1, 3)(1, k) · (1, k)ε(3) · (1, 3)(1, k)

(from ISn and Tn) = (1, k)(1, 3)(1, k)ε1,2 · ε(3)(1, k) · (1, 3)(1, k)(by (c)) = (1, k)(1, 3)(1, k) · ε(3)ε1,2 · (1, k)(1, 3)(1, k)

(from ISn and Tn) = (1, k)(1, 3) · ε(3)(1, k) · (1, k)(1, 3)(1, k)ε1,2

(from ISn) = ε(k)ε1,2.

To prove the general case choose t �= 1, 2 and μ ∈ Sn such that μ(1) = i,μ(2) = j and μ(t) = k. Using the previous relation we have

εi,jε(k) = με1,2μ−1ε(k)

(from ISn) = με1,2ε(t)μ−1

(previous relation) = με(t)ε1,2μ−1

(from ISn) = ε(k)με1,2μ−1

= ε(k)εi,j .

Lemma 9.4.4 The relation εi,jε(j) = ε(j) follows from Σ.

Proof. Let μ ∈ Sn be such that μ(1) = i, μ(2) = j. We have

εi,jε(j) = με1,2μ−1ε(j)

(from ISn) = με1,2ε(2)μ−1

(by (a)) = με(2)μ−1

(from ISn) = ε(j).

Lemma 9.4.5 Let N = {i1, . . . , in−m} ∪ {j1, . . . , jm}, where m ≥ 1. Thenfor the element

γ =(

i1 · · · in−m {j1, . . . , jm}i1 · · · in−m i1

)∈ Tn

the relation γε(j1) · · · ε(jm) = ε(j1) · · · ε(jm) follows from Σ.

Proof. Observe that γ = εi1,j1εj1,j2εj2,j3 · · · εjm−1,jm . As the ε(t)’s commute,the statement of the lemma follows from Lemma 9.4.4.

9.4. A PRESENTATION FOR PT n 171

Lemma 9.4.6 Let N = {i1, . . . , in−m} ∪ {j1, . . . , jm}, where m ≥ 1. Ifα, β ∈ Tn are such that α(ik) = β(ik) for all k, 1 ≤ k ≤ n − m, then therelation

αε(j1) · · · ε(jm) = βε(j1) · · · ε(jm) (9.16)

follows from Σ.

Proof. Let γ be as in Lemma 9.4.5. Then the relation (9.16) can be rewrittenas follows:

αγε(j1) · · · ε(jm) = βγε(j1) · · · ε(jm).

For α and β, satisfying our assumptions, we have αγ = βγ in Tn. Henceαγ = βγ follows from Σ, which yields the statement of the lemma.

Proof of Theorem 9.4.1. One easily checks the relations (a)–(c) for the cor-responding elements in PT n.

From the ISn case (Lemmas 9.2.4 and 9.2.7) we know how to commutethe ε(i)’s with transpositions. From Lemmas 9.4.2 and 9.4.3 we know howto commute the ε(i)’s with the εk,j ’s. Using these rules, every word in thealphabet A can be written in the form

αε(j1) · · · ε(jm), m ≥ 0, α ∈ Tn.

Since the ε(i)’s commute, we may assume 1 ≤ j1 < · · · < jm ≤ n.It is left to show that in the semigroup PT n every relation of the form

αε(j1) · · · ε(jm) = βε(i1) · · · ε(ik), (9.17)

where α, β ∈ Tn, m, k ≥ 0, 1 ≤ j1 < · · · < jm ≤ n and 1 ≤ i1 < · · · < ik ≤ n,follows from Σ. From the equalities

dom(αε(j1) · · · ε(jm)) = {j1, . . . , jm},dom(βε(i1) · · · ε(ik)) = {i1, . . . , ik}

it follows that k = m and is = js for all s = 1, . . . , k. Further from (9.17) itfollows that α(l) = β(l) for any l ∈ N\{i1, . . . , ik}. Hence the relation (9.17)has the form

αε(j1) · · · ε(jm) = βε(j1) · · · ε(jm). (9.18)

If m = 0, we have an equality of two elements in Tn, which follows from Σas the latter contains a system of defining relations for Tn. If m > 0, therelation (9.18) follows from Σ by Lemma 9.4.6. This completes the proof.



9.5.1 Theorem 9.1.4 is proved by Aizenshtat in [Ai1]. Theorems 9.2.2 and9.4.1 are taken from Sutov’s announcement [Sut1], where proofs are onlyoutlined. Theorem 9.3.1 is proved in [II].

9.5.2 The group Sn contains lots of generating systems, even irreducibleones (see for example Exercises 3.3.2 and 3.3.3). For many of them somesystems of defining relations are known. In particular, for the generatingsystem {πi : i = 2, . . . , n} from Sect. 9.2 one of the systems of definingrelations is proposed by Carmichael, see [Ca, p. 169]. This system looks asfollows:

π22 = · · · = π2

n = (π2π3)3 = · · · = (πn−1πn)3 = (πnπ2)3 = ε;

(πiπi+1πiπj)2 = ε, 2 ≤ i, j ≤ n, j �∈ {i, i + 1}, πn+1 = π2.

9.5.3 In representation theory one mostly uses the following system ofCoxeter generators of Sn:

τ1 = (1, 2), τ2 = (2, 3), . . . , τn−1 = (n − 1, n).

The name refers to the fact that Sn is an example of a Coxeter group,the latter being defined axiomatically via a fixed presentation. For Sn thecorresponding system of defining relations looks as follows:

τ21 = · · · = τ2

n−1 = ε;

(τiτj)2 = ε, i ≤ j − 2;

(τiτi+1)3 = ε, 1 ≤ i ≤ n − 2.

This system was proposed by Moore in [Moo]. Using the first relation, thetwo last relations can be rewritten as follows:

τiτj = τjτi, i ≤ j − 2; τiτi+1τi = τi+1τiτi+1.

The latter relations are known as braid relations.

9.5.4 Independently of Sutov, Popova in 1961 proposed a system of definingrelations for ISn with respect to the generating system {τ1, . . . , τn−1, ε(1)}(see [Pp]). This system consists of the relations from 9.5.3 and the followingrelations:

ε(1) = ε2(1) = τ2ε(1)τ2 = τn−1τn−2 · · · τ2ε(1)τ2 · · · τn−2τn−1;

(ε(1)τ1)2 = ε(1)τ1ε(1) = (τ1ε(1))2.(9.19)

Another system of defining relations for the same generating system canbe found in [Li, Chap. 9]. It again consists of the relations from 9.5.3 andthe following relations:

ε2(1) = ε(1); (ε(1)τ1)2 = (ε(1)τ1)3; ε(1)τk = τkε(1), 2 ≤ k < n. (9.20)


Solomon in [So] proposes a presentation for ISn with respect to thegenerating system {τ1, . . . , τn−1, ν = [1, 2, . . . , n]}. It is obtained by addingto the relations from 9.5.3 the following relations (here 1 ≤ i ≤ n − 1):

τiνi+1 = νi+1; νn−i+1τi = νn−i+1; ντi = τi+1ν;

ντn−1τn−2 · · · τ1ν = ν.

Delgado and Fernandes show in [DF] that the following collection of rela-tions is a system of defining relations for ISn with respect to the generatingsystem {π = (1, 2), τ = (1, 2, . . . , n), ε(1)}:

π2 = τn = (τπ)n−1 = (πτn−1πτ)3 = ε;

(πτn−kπτk)2 = ε, 2 ≤ k ≤ n − 2;

τn−1πτε(1)τn−1πτ = τπε(1)πτn−1 = ε(1) = ε2

(1);

(ε(1)π)2 = ε(1)πε(1) = (πε(1))2.

9.5.5 Let S be a semigroup. A presentation S = 〈A|Σ〉 of S is called ir-reducible provided that A is an irreducible generating system and Σ is aminimal system of relations in the sense that any proper subset of Σ is nota system of defining relations for S with respect to A. Obviously, if A isirreducible and Σ is finite, then we can always find a subset Σ′ of Σ suchthat the presentation S = 〈A|Σ′〉 is irreducible.

For the semigroups ISn, PT n, and Tn, it is clear that every irreduciblepresentation of each of these semigroups contains an irreducible presentationof the symmetric group Sn.

Irreducible presentations are of course “most effective”. However, ratheroften some other properties, for example simplicity, existence of canonicalforms, existence of effective algorithms for reduction to canonical forms,etc., are more important than irreducibility. With respect to irreduciblepresentations of Tn the following result of Aizenshtat from [Ai1] looks ratherremarkable:

Theorem 9.5.1 For every irreducible generating system of Tn there is anirreducible presentation of Tn with respect to this system, which is obtainedadding not more than seven relations to the corresponding system of definingrelations for Sn.


9.6.1 Prove that A is the only irreducible generating system for A+.

9.6.2 Describe the congruence ρ(S, A) for the cyclic semigroup S = 〈a〉,generated by the element a of type (k, m).


9.6.3 Let S be a semigroup, A = S, and Σ consist of all relations s · t = st,s, t ∈ S. Prove that S = 〈A|Σ〉 is a presentation of S.

9.6.4 Let Σ denote some system of defining relations for the semigroupS with respect to the generating system A. Let Σ′ be another system ofrelations with respect to A. Prove that Σ′ is a system of defining relationsif and only if every relation from Σ follows from Σ′.

9.6.5 ([KuMa2]) Consider for ISn the generating system

A = {τ1, . . . , τn−1, ε(1), . . . , ε(n)}.

Show that a system of defining relations for A can be obtained by addingto the relations from 9.5.3 the following relations:

ε2(i) = ε(i); ε(i)ε(j) = ε(j)ε(i); τiε(i)ε(i+1) = ε(i)ε(i+1);

τiε(i) = ε(i+1)τi; τiε(j) = ε(j)τi, j �= i, i + 1.

9.6.6 Prove that

〈ε, ε(1)|ε2 = ε, ε2(1) = ε(1), εε(1) = ε(1)ε = ε(1)〉

is a presentation of IS1.

9.6.7 Show that the relation (1, 2)2 = ε together with the relations

ε2(1) = ε(1); ε(1)ε(2) = ε(2)ε(1); (1, 2)ε(1)ε(2) = ε(1)ε(2); (1, 2)ε(1) = ε(2)(1, 2)

are defining relations for IS2 with respect to the generating system ε(1),(1, 2).

9.6.8 Write down a system of defining relations of IS3 with respect to thegenerating system {ε, π2, π3, [1, 2]}.

9.6.9 ([Sut1]) Let

α =(

{i1, . . . , im} B1 · · · Bk

i c1 · · · ck

)∈ Tn,

where i �∈ {c1, . . . , ck}. Using Lemmas 9.4.2 and 9.4.3 show that the relationαε(i1) · · · ε(im) = ε(i)α follows from Σ.

9.6.10 Prove that every relation from (9.19) follows from relations (9.20).

Chapter 10

Transitive Actions

10.1 Action of a Semigroup on a Set

Let S be a semigroup and M be a nonempty set. An action of S on M is,roughly speaking, a realization of each element from S as a (partial) trans-formation of M such that the product in S corresponds to the compositionof (partial) transformations. More formally, an action of S on M is a ho-momorphism from S to one of the semigroups T (M), PT (M), or IS(M).Depending on the choice of the latter semigroup, one speaks of the actionof S on M by transformations, by partial transformations, or by partialpermutations, respectively.

If S has a unit element, then it is usually required that it is representedby the identity transformation of M . In particular, if S is a group, then eachelement of S is represented by an invertible transformation of M and thusan action of a group on M is just a homomorphism from this group to thegroup of all invertible transformations of M (the symmetric group on M).

Let ϕ and ψ be actions of S on some sets M and N , respectively (weallow N = M). These actions are called equivalent or similar if they can bereduced to each other by some bijection between elements of M and N , thatis, if there exists a bijection π : M → N such that the following diagramcommutes for each s ∈ S:

Mϕ(s) ��

π

��

M

π

��N

ψ(s) �� N

(that is, πϕ(s) = ψ(s)π for all s ∈ S). If M = N , from Theorem 7.1.3we get that an equivalent requirement is ϕ = Λπψ, where Λπ is the innerautomorphism of Tn, PT n, or ISn, respectively, which is induced by π.

To describe all actions of a given semigroup on a given set, even upto similarity, is usually a difficult problem. It is thus natural to consider



176 CHAPTER 10. TRANSITIVE ACTIONS

some classes of actions, defined by some natural properties. One of the mostnatural properties of an action is transitivity. An action ϕ of S on M iscalled transitive provided that for any x, y ∈ M (note that x = y is allowed)there exists s ∈ S such that ϕ(s) maps x to y.

Let ϕ be an action of S on M and N ⊂ M . The subset N is said to beinvariant with respect to ϕ provided that for every m ∈ N and for everys ∈ S such that ϕ(s)(m) is defined we have ϕ(s)(m) ∈ N . The whole M andthe empty subset are invariant with respect to any action. The action ϕ iscalled quasi-transitive provided that M cannot be written as a disjoint unionof two nonempty subsets, invariant with respect to ϕ. Each transitive actionis obviously quasi-transitive. The converse is true for groups; however, forsemigroups it need not be true in the general case.

Exercise 10.1.1 Let S be a semigroup acting on M . Show that this actionis transitive if and only if the only invariant subsets with respect to thisaction are M and the empty subset.

Exercise 10.1.2 Construct an example of a quasi-transitive but not tran-sitive action of a finite semigroup S on a finite set M .

An action ϕ is called faithful provided that ϕ is injective. By Cayley’sTheorem (Theorem 2.4.3), each finite semigroup admits a faithful action on afinite set by transformations. Analogously, by the Preston-Wagner Theorem(Theorem 2.9.5), each inverse semigroup admits a faithful action on a finiteset by partial permutations.

Example 10.1.3 The identity maps define the actions of Sn, Tn, PT n, andISn on N, which are called natural. These actions are obviously transitiveand faithful. More general, any automorphism of Sn, Tn, PT n, and ISn

defines a faithful action of the respective semigroup on N. For Tn, PT n, andISn such actions are all similar to the natural action by Theorem 7.1.3. ForSn such actions are similar to the natural action unless n = 6, see 7.6.2.

Example 10.1.4 Mapping all elements of a semigroup S to the identitytransformation of {1} defines an action trivS of S, which is called trivial. Thisaction is obviously transitive; however, it is faithful if and only if |S| = 1.

Exercise 10.1.5 Let ϕ be an action of S on M . Show that ϕ is transitive ifand only if for every x, y ∈ M there exists a sequence x = x0, x1, x2, . . . , xk =y (where k may depend on both x and y) of elements from M and a sequences1, s2, . . . , sk of elements from S such that ϕ(si)(xi−1) = xi for all i.

In the present chapter, we will describe all transitive actions of Tn, PT n,and ISn on finite sets by appropriate transformations up to similarity. With-out loss of generality, we may assume that our semigroups act on the set Nk

for some k. To start with we recall the description of transitive actions forgroups.

10.2. TRANSITIVE ACTIONS OF GROUPS 177

10.2 Transitive Actions of Groups

Let G be a group acting on a set M via ϕ. To simplify the notation, ifthe action ϕ is fixed, for g ∈ G and m ∈ M we will write g · m instead ofϕ(g)(m) (and we will use this notation for semigroups as well). The kernelKer(ϕ) has the form ρH with respect to some normal subgroup H � G byTheorem 6.2.5. Abusing both the language and notation we call H the kernelof ϕ and denote it by Ker(ϕ).

Let m ∈ M . Then the set

StG(m) = {g ∈ G : g · m = m}

is called the stabilizer of m with respect to the action ϕ. Note that StG(m)is nonempty as it contains the identity element of G. From the definition wealso have that

Ker(ϕ) =⋂

m∈M

StG(m). (10.1)

Example 10.2.1 Consider the natural action of the symmetric group Sn

on N. Then the stabilizer of the element n ∈ N consists by definition of allπ ∈ Sn such that π(n) = n. This stabilizer can be canonically identified withSn−1 via the restriction to the subset {1, 2, . . . , n − 1}, which is invariantwith respect to all π ∈ StSn(n).

Lemma 10.2.2 Let G be a group acting on a set M .

(i) For every m ∈ M the set StG(m) is a subgroup of G.

(ii) For every m ∈ M and g ∈ G we have StG(m) = g−1StG(g · m)g.

Proof. As we have already noted, the set StG(m) contains the identity el-ement 1 of G. If g, h ∈ StG(m), then g · (h · m) = g · m = m and hencegh ∈ StG(m). Finally, if g ∈ StG(m), then applying g−1 to m = g ·m we get

g−1 · m = g−1 · (g · m) = g−1g · m = 1 · m = m (10.2)

and hence g−1 ∈ StG(m). The statement (i) follows.Let h ∈ StG(g · m). Then, using (10.2) we get

g−1hg · m = g−1 · (h · (g · m)) = g−1 · (g · m) = m,

hence g−1hg ∈ StG(m) and thus g−1StG(g · m)g ⊂ StG(m). Analogously,we have ghg−1 ∈ StG(g · m) provided that h ∈ StG(m), which impliesgStG(m)g−1 ⊂ StG(g · m). Multiplying the latter inclusion with g−1 fromthe left and with g from the right, we get StG(m) ⊂ g−1StG(g ·m)g and thestatement (ii) follows.


Let G be a group and H be a subgroup of G. Consider the set G/H ={gH : g ∈ G} of all left cosets of G modulo H (see Sect. 6.2).

Proposition 10.2.3 (i) For x, g ∈ G the assignment x ·H gH := xgHdefines a transitive action of G on G/H.

(ii) The kernel of this action is the normal subgroup⋂g∈G

g−1Hg of G.

Proof. Let 1 denote the identity element of G. Then 1 ·H gH := 1gH = gH,and hence 1 acts as the identity transformation on G/H. Further, if g, x, y ∈G, we have

x ·H (y ·H gH) = x ·H ygH = xygH = xy ·H gH,

which proves that ·H is indeed an action. As for any x, y ∈ G we have(xy−1) ·H yH = xy−1yH = xH, it follows that ·H is transitive. This provesthe statement (i).

As 1 ·H 1H = 1H and StG(1H) = H, the statement (ii) follows from(10.1) and Lemma 10.2.2(ii).

Let G act transitively on M and let m ∈ M be fixed. Set H = StG(m)and for each k ∈ M choose some gk ∈ G such that gk(m) = k. Using theseconventions, we have the following complete description of transitive groupactions up to similarity:

Theorem 10.2.4 (i) The action of G on M is similar to the action ·Hof G on G/H, defined in Proposition 10.2.3, via the mapping f : M →G/H defined as follows: f(k) = gkH.

(ii) If H and F are subgroups of G, then the actions ·H and ·F are similarif and only if there exists g ∈ G such that H = g−1Fg (in other wordsif and only if H and F are conjugate).

Proof. Let g ∈ G and k ∈ M . On the one hand, we have g ·H f(k) =g ·H gkH = ggkH. On the other hand, let g · k = l for some l ∈ M . Thenf(g · k) = f(l) = glH. At the same time,

g−1l ggk · m = g−1

l · (g · (gk · m)) = g−1l · (g · k) = g−1

l · l = m

by (10.2). This implies g−1l ggk ∈ H and thus ggkH = glH. Therefore g ·H

f(k) = f(g · k), which proves the statement (i).We first prove the sufficiency part of the statement (ii). Assume that

H = g−1Fg for some g ∈ G. With respect to the action ·F we have F =StG(F ) by definition. Applying Lemma 10.2.2(ii) for the same action weobtain StG(g−1F ) = g−1StG(F )g = H. Now the fact that ·H and ·F aresimilar follows from (i).

10.3. TRANSITIVE ACTIONS OF Tn 179

To prove the necessity part of (ii) assume that ·H and ·F are similar viasome mapping α : G/H → G/F . Let α(H) = gF for some g ∈ G. Then forany x ∈ H we have

x ·F (gF ) = x ·F (α(H)) = α(x ·H H) = α(xH) = α(H) = gF.

In particular, H ⊂ StG(gF ). If x �∈ H, then xH �= H and thus α(xH) �= gFsince α is bijective, which implies x �∈ StG(gF ). This means that H =StG(gF ). From Lemma 10.2.2(ii) it follows that H and F are conjugate.This completes the proof.

From Theorem 10.2.4, we see that all groups, in particular, the sym-metric group Sn, admit a lot of different transitive actions on finite sets.Moreover, from Proposition 10.2.3(ii) and 6.5.5 it follows that, apart fromfew exceptions, all such actions of Sn are faithful. It will be interesting tocompare this result with the corresponding results for the semigroups Tn,PT n, and ISn, which we will obtain in the next sections.

10.3 Transitive Actions of T n

In this section, we determine, up to similarity, all transitive actions of Tn.As Tn is finite, it is enough to consider finite sets. The answer is given bythe following:

Theorem 10.3.1 Let ϕ be a transitive action of Tn on Nm for some m.Then either m = 1 and ϕ is the trivial action or m = n and ϕ is similar tothe natural action.

To prove this theorem we will need the following lemma:

Lemma 10.3.2 Let ψ be an action of Tn on Nm.

(i) For every a ∈ N each block of the partition ρψ(0a) of Nm is invariantwith respect to ψ.

(ii) The set ⋃a∈N

im(ψ(0a))

is invariant with respect to ψ.

Proof. Let x ∈ Nm and α ∈ Tn. As 0a is a left zero, we have

0a · (α · x) = (0aα) · x = 0a · x

and the statement (i) follows.Let α ∈ Tn, a ∈ N and x = 0a · y for some y ∈ Nm. Then we have

α · x = α · (0a · y) = (α0a) · y = 0α(a) · y

and the statement (ii) follows.


Proof of Theorem 10.3.1. Assume first that ϕ is faithful. To distinguish theelements of Tn and Tm in this proof, we will use the notation α(m) to em-phasize that α is an element of Tm. Because of Lemma 10.3.2(i), the transi-tivity of ϕ implies that for every a ∈ N the element ϕ(0a) must have rankone, that is, there exists f(a) ∈ Nm such that ϕ(0a) = 0(m)

f(a). Because ofLemma 10.3.2(ii), the transitivity of ϕ implies that

Nm =⋃

a∈N

{f(a)}.

As ϕ is faithful, all f(a)’s are different and we get m = n. Hence ϕ determinesan automorphism of Tn. By Theorem 7.1.3 every automorphism of Tn is aconjugation by some element from Sn. By definition, the means that ϕ issimilar to the natural action.

Now assume that ϕ is not faithful. Then, by Theorem 6.3.10 we haveϕ(0a) = ϕ(0b) for all a, b ∈ N. The same arguments as in the previousparagraph imply first that for all a ∈ N we have ϕ(0a) = 0(m)

z for some fixedz ∈ Nm; and then even that Nm = {z}. Hence ϕ is the trivial action in thiscase. This completes the proof.

The following corollary now shows quite a striking difference with thesymmetric group Sn:

Corollary 10.3.3 Every transitive and faithful action of Tn is similar tothe natural action.

10.4 Actions Associated with L-Classes

In this section, we present one general construction of transitive actions bypartial transformation for arbitrary semigroups.

Let S be a semigroup and e ∈ E(S). From Theorem 4.4.11 we know thatthe H-class H(e) is a group. Let H be a subgroup of H(e). Consider theL-class L(e). It is a disjoint union of H-classes, so we can pick a represen-tative in each such H-class. Let a = {ai : i ∈ I} denote this collection ofrepresentatives. Note that ai ∈ S1e by definition and hence aie = ai for alli ∈ I. As we also have e ∈ S1ai by definition, there exists a′i ∈ S1 such thata′iai = e. In this situation from Green’s lemma we have that the assignmentx → aix defines a bijective mapping from H(e) to H(ai), whose inverse isgiven by y → a′iy. In particular, every element from L(e) admits a uniquepresentation as a product aig, where i ∈ I and g ∈ H(e). Note further that{gH : g ∈ H(e)} = z{gH : g ∈ H(e)} for any z ∈ H(e). This, in particular,means that the following set:

M(e, H) = {aigH : i ∈ I, g ∈ H(e)}does not depend on the choice of a.

10.4. ACTIONS ASSOCIATED WITH L-CLASSES 181

For every s ∈ S we define the partial transformation ξe,H(s) of the setM(e, H) by prescribing its value on aigH, g ∈ H(e), in the following way:

s · aigH =

{ajg

′gH, sai = ajg′for somej ∈ I, g′ ∈ H(e);

∅, sai �∈ L(e).

Theorem 10.4.1 (i) ξe,H is a transitive action of the semigroup S bypartial transformations on the set M(e, H).

(ii) If e ∈ E(S) is such that eDe, then there exists an appropriate subgroupH of H(e) such that the actions ξe,H and ξe,H are similar.

Proof. Let s, t ∈ S and i ∈ I. Then st · aigH is defined if and only ifS1stai = S1ai. As S1stai ⊂ S1tai ⊂ S1ai, we have S1stai = S1ai if andonly if S1stai = S1tai and S1tai = S1ai, in other words if and only if botht · aigH and s · taigH are defined.

Now assume that stai, tai ∈ L(e) and further that tai = ajg, saj = akhand (st)ai = alz for appropriate j, k, l ∈ I and g, h, z ∈ H. We have

alz = (st)ai = s(tai) = s(ajg) = (saj)g = (akh)g = ak(hg), (10.3)

which implies k = l and z = hg because of the uniqueness of the presentationof stai. This proves that ξe,H is indeed an action by partial transformations.The transitivity of this action follows immediately from the definitions andGreen’s lemma. The statement (i) follows.

To prove (ii) let us first assume that eLe. For such e from e ∈ S1e itfollows that ee = e and similarly ee = e. In this case for x, y ∈ H(e) we have

e(xy) = exey = exeey = exey = (ex)(ey),

which means that the mapping x → ex from H(e) to H(e) is a group iso-morphism. As xe = x for all x ∈ L(e), we have

{aigH : g ∈ H(e)} = {aiegeH : g ∈ H(e)} = {ai(eg)(eH) : eg ∈ H(e)}.

This means that M(e, H) = M(e, eH), which proves the statement (ii) inthe case eLe.

If e �∈ L(e), we can pick some element a ∈ D(e) such that aRe and aLe.Let u, v ∈ S be such that au = e and ev = a. Then Green’s lemma says thatthe mapping λv : L(e) → L(e), x → xv is a bijection. Let H(e) = λv(H(ai)).From Theorem 4.7.5 it follows that the composition of λv with x → aixinduces a group isomorphism from H(e) to H(e). Let H denote the imageof H under this isomorphism. Note that λv maps the set akgH to the set

akgHv = akegHv = aka′iaiegHv = aka

′iaigHv = aka

′iaigvaiHv.

As the element g = aigv belongs to H(e), we have g = eg. Hence akgHv =aka

′ie(gH), moreover, the factor aka

′ie = aka

′iaiev = akev = akv belongs


to L(e). Then λv induces a bijection from M(e, H) to M(e, eH), which bydefinition commutes with the left multiplication with elements from S. Thestatement (ii) follows and the proof is complete.

From Theorem 10.4.1 it follows that, up to similarity, the action ξe,H

defined above depends only on the choice of a regular D-class and a subgroupH in the abstract group, which is isomorphic to every maximal subgroup ofS, contained in this D-class.

Remark 10.4.2 If s ∈ S is such that the two-sided ideal S1sS1 does notintersect the L-class Le, then the domain of the transformation ξe,H(s) isempty. In particular, if the set

{s ∈ S : S1sS1 ∩ Le = ∅}

contains more than one element, then the action ξe,H is not faithful.

10.5 Transitive Actions of PT n and ISn

In the previous section, we constructed some natural transitive actions ofsemigroups. In this section we will show that for the semigroups PT n andISn these are all transitive actions.

Theorem 10.5.1 (i) Let S be one of the semigroups PT n or ISn. Theneach transitive action of S is either trivial or similar to ξe,H for somee ∈ E(S) and a subgroup H of H(e).

(ii) For the semigroup ISn each action ξe,H is an action by partial permu-tations.

Proof. Let ϕ be a transitive action of S on some set M . As S is finite andϕ is transitive, we have that M is finite as well. We split the proof of thestatement (i) into a sequence of observation. We start with the followingeasy one:

Lemma 10.5.2 Every x ∈ im(ϕ(0)) is invariant with respect to ϕ.

Proof. Let α ∈ S and x = 0 · y for some x, y ∈ M . Then

α · x = α · (0 · y) = (α0) · y = 0 · y = x

and the claim follows.

Assume now that im(ϕ(0)) �= ∅. In this case from Lemma 10.5.2 and thetransitivity of ϕ we derive M = im(ϕ(0)), moreover, M consists of exactlyone element. Hence ϕ is the trivial action.

From now on we assume that im(ϕ(0)) = ∅. Let k > 0 be minimal forwhich there exists an idempotent e of rank k such that im(ϕ(e)) �= ∅. Ourprincipal observation is the following:

10.5. TRANSITIVE ACTIONS OF PT n AND ISn 183

Lemma 10.5.3 M is a disjoint union of im(ϕ(εA)), A ⊂ N, |A| = k.Moreover, im(ϕ(εA)) �= ∅ for all such A.

Proof. First we claim that im(ϕ(εA)) �= ∅ for each A ⊂ N, |A| = k. Indeed,if we assume that im(ϕ(εA)) = ∅, then im(ϕ(α)) = ∅ for all α ∈ SεAS.However, e ∈ SεAS by Theorem 4.2.8, which contradicts our assumptionim(ϕ(e)) �= ∅.

Now we claim that im(ϕ(εA)) ∩ im(ϕ(εB)) = ∅ for any A �= B ⊂ N,|A| = |B| = k. Indeed, as A �= B and |A| = |B| = k we have |A ∩ B| < kand thus for any x ∈ im(ϕ(εB)) we obtain

εA · (εB · x) = (εAεB) · x = εA∩B · x = ∅

by our assumptions. This implies im(ϕ(εA)) ∩ im(ϕ(εB)) = ∅.Finally we claim that the set

N =⋃

A⊂N, |A|=k

im(ϕ(εA))

is invariant with respect to ϕ. Let α ∈ S and x ∈ im(ϕ(εA)) for some A ⊂ Nsuch that |A| = k. Then α ·x = α · (εA ·x) = αεA ·x. If rank(αεA) < k, thenfrom αεA = εim(αεA)αεA it follows that im(ϕ(αεA)) ⊂ im(ϕ(εim(αεA))) = ∅,which means that α · x = ∅. Hence we may assume rank(αεA) = k. Thenfor B = im(αεA) we have |B| = k, implying αεA = εBαεA. From this weobtain

εB · (α · x) = εB · (αεA · x) = (εBαεA) · x = αεA · x = α · (εA · x) = α · x,

implying α · x ∈ im(ϕ(εB)).The statement of the lemma now follows from the transitivity of ϕ. This

completes the proof.

Our next step is the following:

Lemma 10.5.4 Let A ⊂ N, |A| = k. Then ϕ induces a transitive action ofthe group H(εA) on im(ϕ(εA)).

Proof. By Exercise 10.1.1 we have to prove that the only subsets of the setim(ϕ(εA)) invariant with respect to the restriction of ϕ to H(εA) are thewhole im(ϕ(εA)) and the empty set. Assume that this is not the case andlet X ⊂ im(ϕ(εA)) be a proper invariant subset.

Let α ∈ S be such that rank(αεA) = k. Then either im(αεA) = A, inwhich case αεA ∈ H(εA), or im(αεA) = B �= A for some B ⊂ N, |B| = k. Inthe second case, from αεA = εBαεA it follows that α · im(εA) ⊂ im(εB). AsX is invariant with respect to the action of H(εA), for any x ∈ X we have

{α · x : α ∈ S} ⊂ X ∪⋃

B⊂N, |B|=k, B �=A

im(ϕ(εB)) �= M


by our assumptions and Lemma 10.5.3. This contradicts the transitivity ofthe action ϕ.

Let A = {1, 2, . . . , k} and e = εA. For B ⊂ N, |B| = k, let αB denotethe unique increasing bijection from A to B. Note that all αB are elementsof ISn, in particular, we have their inverse elements α−1

B in ISn. Thena = {αB : B ⊂ N, |B| = k} contains exactly one representative of eachH-class in L(e). Finally, fix some x ∈ im(ϕ(εA)) and set H = StH(εA)(x).Our final step in the proof of the statement (i) is the following:

Lemma 10.5.5 The mapping f : M(e, H) → M , αBβH → αBβ · x (hereαB ∈ a and β ∈ H(e)) produces a similarity between the actions ξe,H and ϕ.

Proof. For α ∈ S we have α · x = α · (e · x) = αe · x. If α · x �= ∅, thenrank(αe) = k, in particular, αe ∈ L(e). This implies M = S · x = L(e) · x.Since each element of L(e) belongs to some αiβH, from the transitivity ofϕ we get that f is surjective.

Assume αBβ · x = αCγ · x. Then B = C follows from Lemma 10.5.3.We further have α−1

B αBβ · x = α−1B αBγ · x. Note that α−1

B αB = εA byconstruction, which yields β · x = γ · x. Hence γ−1β ∈ H and βH = γH.This implies that f is injective, and hence bijective.

Let now α ∈ S. If rank(ααB) < k, then both α · αBβH = ∅ andα · (αBβ · x) = (ααBβ) · x = ∅. If rank(ααB) = k, then ααB ∈ L(e) canbe uniquely written as ααB = αCγ for some γ ∈ H(e). Then from thedefinitions we have

f(α · αBβH) = f(αCγβH) = αCγβ · x = ααBβ · x == α · (αBβ · x) = α · f(αBβH),

which completes the proof of the lemma.

The statement (i) is proved.Now we prove the statement (ii). We retain the notation from the para-

graph before Lemma 10.5.5. By Theorem 10.5.1 we may assume e = εA. Letβ ∈ ISn be such that βαBπH = βαCτH ∈ M(e, H) for some αB, αC ∈ a

and π, τ ∈ H(εA). Then

β−1βαBπH = β−1βαCτH ∈ M(e, H) (10.4)

as well since β−1βαBπH �∈ M(e, H) would imply for any h ∈ H the in-equality rank(β−1βαBπh) < rank(βαBπh) and thus

rank(βαBπh) = rank(ββ−1βαBπh) ≤ rank(β−1βαBπh) < rank(βαBπh),

a contradiction.We have β−1β = εD for some D ⊂ N. As both αB and αC are elements

of rank k, from (10.4) it follows that both εDαB and εDαC have rank k


as well. This yields εDαB = αB and εDαC = αC . Now the equality (10.4)implies αBπH = αCτH, which shows that ξe,H(β) is a partial permutation.This completes the proof.

Corollary 10.5.6 Let S be one of the semigroups PT n or ISn. Then everytransitive and faithful action of S is similar to the natural action.

Proof. By Theorem 10.5.1, any action of S is equivalent to some ξe,H . FromTheorem 4.2.8 and the construction of the action ξe,H it follows that for anyα ∈ S such that rank(α) < rank(e) we have S1αS1 ∩ L(e) = ∅. Hence fromRemark 10.4.2 we derive that ξe,H is not faithful unless rank(e) = 1.

If rank(e) = 1, then H(e) ∼= S1 by Theorem 5.1.4(ii). As S1 containsexactly one subgroup, from Theorem 10.5.1 we obtain that S has, up tosimilarity, at most one transitive and faithful action. Our statement nowfollows from the obvious observation that the natural action is both transi-tive and faithful (see Example 10.1.3).


10.6.1 The notion of similarity admits a categorical definition. Consider thecategory A defined as follows: The objects of A are triples (S, M, ϕ), where Sis a semigroup, M is a nonempty set, and ϕ is an action of S on M by somefixed type of transformations. If (S, M, ϕ) and (S′, M ′, ϕ′) are two objects,then A

((S, M, ϕ), (S′, M ′, ϕ′)

)is the set of all pairs (α, f), where α : S → S′

is a homomorphism and f : M → M ′ is a mapping, such that for all s ∈ Sthe following diagram commutes:

Mϕ(s) ��

f��

M

f��

M ′ ϕ′(α(s)) �� M ′.

The composition of two morphisms (α, f) ∈ A((S, M, ϕ), (S′, M ′, ϕ′)

)and

(α′, f ′) ∈ A((S′, M ′, ϕ′), (S′′, M ′′, ϕ′′)

)is defined to be (α′α, f ′f). For a fixed

S let AS denote the subcategory whose objects are (S, M, ϕ) for all possibleM and ϕ, and morphisms are (id, f) for all possible f . Using the categoricallanguage, similar actions of S are exactly the actions which can be obtainedfrom each other by composing with an isomorphism in AS . Both categoriesA and AS seem to be quite complicated and not much is known about them.

10.6.2 The results presented in Sects. 10.3–10.5 are special cases of a moregeneral result obtained by Ponizovskiy in [Po1, Po2]. A special case ofCorollary 10.5.6 for ISn appears also in [Vl1]. In fact, Ponizovskiy classifiedall transitive actions by partial transformations of all finite semigroups. His


answer of course includes the actions ξe,H constructed in Sect. 10.4; how-ever, in the general case one has to consider technically more complicatedactions on L-classes as well. These actions generalize the action ξe,H andhave the form L(e)/ρ, where ρ is a left compatible equivalence relation onL(e). Combinatorial description of such relations even for the semigroup Tn

is complicated.

10.6.3 The actions considered in this chapter are left actions. One definesright actions in the obvious dual way. Left actions of S are naturally the same

as right actions of←S . As ISn is inverse, description of transitive right ISn-

actions is simply the dual to the description of the transitive left actions,presented in Theorem 10.5.1. For the semigroups Tn and PT n descriptionof right actions by partial transformations is much more complicated andrequires the general approach of [Po1]. At the same time an analogue of

Theorem 10.3.1 for the semigroup←Tn trivializes, see Exercise 10.7.7. This

shows that right actions for transformation semigroups are less natural thanleft actions.

10.6.4 There is a rather advanced and well developed abstract theory ofsemigroups acting on sets, see for example [CP2, Chap. 11]. There is also awell-developed abstract theory of semigroup actions, in particular, transitiveactions, by binary relations. This one was developed by Schein, see [Sc2, Sc3].

10.6.5 Schein also developed in [Sc1] a general theory of actions of inversesemigroups in terms of the so-called closed subsemigroups. Let S be an in-verse semigroup. A subsemigroup H of S is said to be closed provided thatα ∈ H implies β ∈ H for every β ∈ S such that α−1α = β−1α (for ele-ments of ISn the latter condition is equivalent to dom(α) ⊂ dom(β) andα(x) = β(x) for any x ∈ dom(α)). For a closed inverse subsemigroup H of asemigroup S Schein naturally defines cosets of S modulo H and constructsa transitive action of S by partial permutations on the set of all cosets.Applications of this approach to ISn are discussed in [Vl1].

The above notion of closed subsemigroups admits a natural generaliza-tion to PT n: a subsemigroup H of PT n is said to be closed provided thatα ∈ H implies β ∈ H for every β ∈ S such that dom(α) ⊂ dom(β) andα(x) = β(x) for any x ∈ dom(α). However, as far as we know, there isno classification of closed subsemigroups of PT n similar to that of closedinverse subsemigroups of ISn, given in [Vl1, Theorem 1] (see also Exer-cise 10.7.10 below). We also do not know of any description of arbitraryclosed subsemigroups of PT n.

10.6.6 Let ϕ be an action of some semigroup S on some set M . The actionϕ is said to be semitransitive provided that for any x, y ∈ M (we allowx = y) there exists some s ∈ S such that either s · x = y or s · y = x.Each transitive action is obviously semitransitive. The converse is not true


in general. The easiest general example is the following: Let S be a finitemonoid with unit 1 and G = G1 �= S. Let further ϕ be an action of G onsome set M . Consider the set M = M ∪ {∗} (we assume ∗ �∈ M) and definethe action ϕ of S on M as follows:

ϕ(s)(m) =

{ϕ(s)(m), s ∈ G, m ∈ M ;∗, otherwise.

It is straightforward to verify that ϕ is an action. It is further easy to seethat ϕ is semitransitive if and only if ϕ is transitive.

The notion of a semitransitive action is a natural generalization of thatof a transitive action. However, it seems that for the moment not much isknown about such actions, even for classical semigroups. Some first resultsin the study of semitransitive actions of inverse semigroups are obtained in[C-O], where the reader will also find some references to the literature, wheresuch actions naturally appear.


10.7.1 Let S denote one of the semigroups Sn, Tn, PT n, or ISn. For α ∈ Sand X ∈ B(N) define

α · X = {α(x) : x ∈ X ∩ dom(α)}.

Show that this defines an action of S on B(N). Will this action be faithful?transitive? quasi-transitive?

10.7.2 Is it true that a quasi-transitive action of an inverse semigroup bypartial permutations is transitive?

10.7.3 Let S denote one of the semigroups Sn, Tn, PT n, or ISn. The groupAut(S) acts on S in the natural way. Is this action faithful? transitive?quasi-transitive?

10.7.4 Let S denote one of the semigroups Sn, Tn, PT n, or ISn. The semi-group End(S) acts on S in the natural way. Is this action faithful? transitive?quasi-transitive?

10.7.5 Show that the assignment α ·π = απα−1 defines an action of ISn onitself by transformations. Is this action faithful? transitive? quasi-transitive?

10.7.6 Let S be a semigroup. Show that the assignment s ·x = sx defines anaction of S on itself by transformations. Prove that this action is transitiveif and only if S has exactly one L-class.

10.7.7 Describe all transitive actions of←Tn by transformations.


10.7.8 Show that any semitransitive action of a group is transitive.

10.7.9 ([KuMa4]) Let S be a finite inverse monoid. For x ∈ S let ex denotethe idempotent, which has the form xk for some k.

(a) Prove that the assignment

a · x =

{axa−1, a−1aex = ex;∅, otherwise,

where a, x ∈ S, defines an action of S by partial transformations onitself.

(b) Prove that for x, y ∈ S the elements x and y are conjugate (that is,x ∼S y) if and only if there exist a, b, z ∈ S such that a · x = b · y = z.

10.7.10 ([Vl1]) Let A ⊂ N and H be a subgroup of H(εA). Denote byC(A, H) the subsemigroup of ISn, which consists of all α for which thereexists β ∈ H such that α(x) = β(x) for all x ∈ A.

(a) Show that C(A, H) is a closed inverse subsemigroup of ISn.

(b) Show that every closed inverse subsemigroup of the semigroup ISn isof the form C(A, H) for appropriate A and H.

10.7.11 ([Vl2, Vl3]) Describe all closed inverse subsemigroups and all transi-tive actions (up to similarity) for every finite inverse semigroup S satisfyingthe following condition: there exists a ∈ S such that the minimal inversesubsemigroup of S containing a coincides with S.

10.7.12 Classify up to similarity all transitive actions of Tn by partial per-mutations on finite sets.

10.7.13 Classify up to similarity all transitive actions of PT n by partialpermutations on finite sets.

10.7.14 Prove that every transitive action of ISn by partial permutationson a finite set remains transitive when restricted to Sn.

10.7.15 ([Po2]) (a) Let ϕ be an action of a finite inverse semigroup S bypartial permutations on a finite set M . Prove that there exists a decom-position M = M1 ∪ M2 ∪ · · · ∪ Mk into invariant subsets such that forany i ∈ {1, 2, . . . , k} either |Mi| = 1 or for any x, y ∈ Mi there existss ∈ S such that ϕ(s)(x) = y.

(b) Prove that the statement (a) is no longer true in general if instead ofan action by partial permutations one considers an action by partialtransformations.

Chapter 11

Linear Representations

11.1 Representations and Modules

In this chapter, we work over the field C of complex numbers, in particular,all vector spaces are vector spaces over C.

Let S be a semigroup and V a vector space (which we always assumeto be finite-dimensional). A representation of S in V is a homomorphism ϕfrom S to EndC(V ), the semigroup of all linear transformations of V . If S isa monoid, one additionally requires that ϕ maps the identity element fromS to the identity transformation on V .

An alternative but equivalent notion is that of a module. If V is a vectorspace, then one says that V has the structure of an S-module (alternativelyis a module over S) if we are given a mapping S × V → V , (s, v) → s · vsuch that the following conditions are satisfied:

• s · (v + w) = s · v + s · w for all s ∈ S, v, w ∈ V

• s · (λv) = λ(s · v) for all s ∈ S, v ∈ V and λ ∈ C

• s · (t · v) = st · v for all s, t ∈ S and v ∈ V

• 1 · v = v for all v ∈ V if 1 is the identity element of S

Exercise 11.1.1 Show that the notion of a representation is equivalent tothat of a module.

The language of modules usually requires simpler notation and hence wewill mostly use it. Let V be an S-module. A subspace W ⊂ V is called asubmodule of V provided that s ·w ∈ W for all w ∈ W and s ∈ S. Obviously,the whole V and the zero vector space 0 are submodules of any module.An S-module V is called simple provided that it is nonzero and the onlysubmodules of V are V and 0. The corresponding representation is calledirreducible.



190 CHAPTER 11. LINEAR REPRESENTATIONS

If V is an S-module and W is a submodule of V , then the quotient spaceV/W carries the natural structure of an S-module given by

s · (v + W ) = sv + W. (11.1)

This module is called the quotient of V modulo W .

Exercise 11.1.2 Verify that (11.1) indeed defines an S-module structureon V/W .

Example 11.1.3 Let S be any semigroup. Then the assignment s · c = c,c ∈ C, defines on C the structure of an S-module. This module is calledthe trivial S-module and is usually denoted by Ctriv. The trivial module issimple because C, being one-dimensional, has only two subspaces, namely,C and 0. The same argument implies that every one-dimensional S-moduleis simple.

Example 11.1.4 Let S denote one of the semigroups Sn, ISn, Tn, or PT n.For α ∈ S let ϕ(α) = (mi,j)n

i,j=1 denote the n × n-matrix such that

mi,j =

{1, α(j) = i;0, otherwise.

For example, we have

ϕ

((1 2 3 4 5 62 1 1 ∅ 5 3

))=

⎛⎜⎜⎜⎜⎜⎜⎝

0 1 1 0 0 01 0 0 0 0 00 0 0 0 0 10 0 0 0 0 00 0 0 0 1 00 0 0 0 0 0

⎞⎟⎟⎟⎟⎟⎟⎠

.

As α is a (partial) transformation, we obtain that each column of the matrixϕ(α) contains at most one nonzero element. It is easy to check that for v ∈ C

n

the assignment α · v = ϕ(α)v defines on Cn the structure of an S-module.

This module is called the natural S-module.If α ∈ Sn, then every row and every column of ϕ(α) contains exactly

one nonzero element. If α ∈ ISn, then every row and every column of ϕ(α)contains at most one nonzero element. The latter corresponds to placementsof nonattacking rooks on an n×n chessboard. Because of this interpretation,the semigroup ISn is sometimes called the rook monoid.

Exercise 11.1.5 Verify that the natural module is indeed a module, andshow that the natural module is simple if and only if S = ISn or S = PT n.

Exercise 11.1.6 Let M be an S-module. Show that M is simple if and onlyif for any nonzero m ∈ M the linear span of s ·m, s ∈ S, coincides with M .

11.1. REPRESENTATIONS AND MODULES 191

Let V and W be two S-modules. A linear mapping f : V → W is called ahomomorphism or an S-homomorphism provided that it commutes with theaction of all elements from S, that is, for every s ∈ S the following diagramis commutative:

Vs· ��

f

��

V

f

��W

s· �� W.

The trivial example of a homomorphism is the zero mapping. A less trivialexample is the identity homomorphism in the case V = W . The set of allS-homomorphisms from V to W is denoted by HomS(V, W ). This set carriesthe natural structure of a vector space.

Two S-modules V and W are called isomorphic or equivalent providedthat HomS(V, W ) contains an isomorphism. The fact that V and W areisomorphic is denoted by V ∼= W . As usually, for most of the questionsabout S-modules, in particular, classification problems, it makes sense toformulate and study them only up to isomorphism. Some basic facts abouthomomorphisms and isomorphisms are collected in the following statement:

Proposition 11.1.7 Let V and W be S-modules and f ∈ HomS(V, W ).

(i) The kernel Ker(f) of f is a submodule of V .

(ii) The image im(f) of f is a submodule of W .

(iii) If both V and W are simple module, then f is either an isomorphismor zero.

Proof. Let v ∈ V be such that f(v) = 0. Then f(s · v) = s · f(v) = s · 0 = 0and hence s · v ∈ Ker(f). This proves the statement (i).

Let v ∈ V . Then s · f(v) = f(s · v), which implies the statement (ii).If V is simple, then for the vector space Ker(f), which is a submodule

of V by (i), we have only two possibilities: Ker(f) = V or Ker(f) = 0. Inthe first case, f is the zero mapping. In the second case, f is injective. Nowim(f) is a submodule of W by (ii). As W is simple, we have two options:im(f) = 0 or im(f) = W . The first one is not possible as V �= 0 and f isinjective. For the second one, we obtain that f is surjective, hence it is anisomorphism.

Corollary 11.1.8 Let V and W be two simple S-modules. Then

HomS(V, W ) ∼={

C, V ∼= W ;0, V �∼= W.


Proof. If V �∼= W , then the fact that HomS(V, W ) = 0 follows immediatelyfrom Proposition 11.1.7(iii).

If V ∼= W , we have an obvious isomorphism of vector spaces betweenHomS(V, W ) and HomS(V, V ). Let f ∈ HomS(V, V ) be a nonzero homo-morphism. As C is algebraically closed, f has an eigenvalue, λ ∈ C say.Then f − λε ∈ HomS(V, V ) (here ε is the identity transformation of V ). Asf −λε has a nontrivial kernel, we have f −λε = 0 by Proposition 11.1.7(iii).Thus f = λε and we are done.

Example 11.1.9 Let S be a finite semigroup. Consider the vector spaceCS with the formal basis {vt : t ∈ S}. For every s ∈ S define a linearoperator on CS by prescribing its action on basis elements in the followingway: s · vt = vst. We have

r · (s · vt) = r · vst = vrst = rs · vt,

which implies that the above assignment defines on CS the structure of anS-module. This module is called the regular S-module.

On the one hand, for every a ∈ S the linear mapping fa : CS → CSdefined via fa(vt) = vta is an endomorphism of CS since for any s ∈ S wehave

fa(s · vt) = fa(vst) = vsta = s · vta = s · fa(vt).

On the other hand, if S is a monoid with identity 1, then any homomorphismf ∈ HomS(CS, CS) is uniquely determined by its value on v1 since

f(vs) = f(s · v1) = s · f(v1)

for all s ∈ S. Hence if f(v1) =∑

a∈S cava =∑

a∈S cafa(v1) for some ca ∈ C,we have f =

∑a∈S cafa. The homomorphisms fa’s are linearly independent

as the set {va = fa(v1) : a ∈ S} is a basis of CS. This means that we havean isomorphism of vector spaces

HomS(CS, CS) ∼= CS.

Exercise 11.1.10 Prove that the homomorphisms {fg : g ∈ S} form asemigroup with respect to the composition. Further, show that this semi-

group is isomorphic to←S .

11.2 L-Induced S -Modules

In this section, we present a construction of some S-modules using modulesover maximal subgroup. The idea is very similar to the one described inSect. 10.4.

Let S be a semigroup for which all regular L-classes are finite. Fix somee ∈ E(S) and let M denote some H(e)-module. Let a = {ai : i ∈ I}

L-INDUCED S-MODULES 193

be a fixed collection of representatives from all H-classes inside L(e), onefor each H-class. In particular, I is finite and indexes R-classes inside theD-class D(e). Then, analogously to Sect. 10.4, we have that the mappingx → aix is a bijection from H(e) to H(ai) for every i ∈ I, in particular,every y ∈ L(e) can be uniquely written in the form aix for some i ∈ I andx ∈ H(e). For i ∈ I let M (i) denote a copy of M and ηi : M → M (i) denotethe canonical identification. Consider the vector space

Va(M) =⊕i∈I

M (i).

Note that dim(Va(M)) = m · dim(M), where m is the number of H-classesinside L(e) (this number is finite by our assumptions). For s ∈ S, i ∈ I andv ∈ M define

s · ηi(v) =

{ηj(x · v), sai = ajx, j ∈ I, x ∈ H(e);0, otherwise.

Proposition 11.2.1 (i) The above assignment defines on the vector spa-ce Va(M) the structure of an S-module.

(ii) Let a′ be another collection of representatives of H-classes inside L(e).Then Va(M) ∼= Va′(M).

Proof. Let s, t ∈ S, i ∈ I and v ∈ M . Because of the first paragraph of theproof of Theorem 10.4.1 we have that stai ∈ L(e) implies that tai ∈ L(e) aswell. Hence it is enough to consider the case stai, tai ∈ L(e). In this case lettai = ajg, saj = akh and stai = alz. Then from (10.3) we have l = k andz = hg. Hence

st · ηi(v) = ηk(z · v) = ηk(hg · v) = ηk(h · (g · v)) = s · ηj(g · v) = s · (t · ηi(v)).

The statement (i) follows.To prove (ii) we consider a and a′. As a′i ∈ H(ai) for all i ∈ I, we

can find gi ∈ H(e) such that a′i = aigi. Define now the linear mappingf : Va(M) → Va′(M) as follows: For v ∈ M and i ∈ I let fi : M (i) → M (i)

be the linear mapping defined via fi(ηi(v)) = ηi(g−1i ·v), and set f = ⊕i∈Ifi.

We consider f as a linear mapping from Va(M) to Va′(M). As M is an H(e)-module, H(e) is a group and gi ∈ H(e), we obtain that fi is an isomorphism.This yields that f is an isomorphism. Now let s ∈ S, i ∈ I and v ∈ M . Ifsai �∈ L(e), then s · f(ηi(v)) = f(s · ηi(v)) = 0 is obvious. So, we assumesai = ajx for some j ∈ I and x ∈ H(e), in particular,

sa′i = saigi = ajxgi = a′jg−1j xgi. (11.2)


To distinguish the actions on Va(M) and Va′(M) we for the moment will usethe notation � for the action on Va′(M). Using (11.2) we compute

s � f(ηi(v)) = s � fi(ηi(v))= s � ηi(g−1

i · v)= ηj(g−1

j xgi · (g−1i · v))

= ηj(g−1j x · v)

= ηj(g−1j · (x · v))

= fj(ηj(x · v))= f(ηj(x · v))= f(s · ηi(v)).

The statement (ii) follows and the proof is complete.

Because of Proposition 11.2.1(ii) in the isomorphism class we can removethe index a in the notation Va(M). We will use simply the notation V (M) inthe sequel. The module V (M) will be called an S-module L-induced from M .Now we would also like to remove the dependence on the choice of e insidea given D-class.

Proposition 11.2.2 Let e, e ∈ E(S) be such that eDe and let M be an H(e)-module. Then there exists an H(e)-module M such that V (M) ∼= V (M).

Proof. Using the above notation, we consider the module Va(M) and assumee ∈ a. Let l ∈ I be such that alRe. Set u = al. By Green’s Lemma themapping x → ux is a bijection from R(e) to R(u), which preserves L-classes.Hence there exists v ∈ R(e) ∩ L(e) such that uv = e.

For all i ∈ I define ai = aiv, and set a = {ai : i ∈ I}. In particular, ev =v and uv = e belong to a. We also have vu = e by the proof of Theorem 4.7.5.Set further M = M (l). Let x ∈ H(e) and x′ be the inverse to x in H(e).Then (xu)v = xe = x, implying xuRx by Proposition 4.4.2, in particular,xuRu as uRx by our assumptions. At the same time x′(xu) = eu = u. Thisimplies xuLu by Proposition 4.4.1 and thus xuHu. Hence xu = ux′ for somex′ ∈ H(e), which by construction means that M is stable under the actionof H(e), in particular, is an H(e)-module. For i ∈ I let M (i) denote a copy ofM , and ηi : M → M (i) denote the canonical identification. We thus obtainthe module Va(M).

Define the linear mapping ϕ : Va(M) → Va(M) as follows: ϕ(ηi(m)) =ηi(ηl(m)), i ∈ I, m ∈ M . By construction, ϕ is an isomorphism of vectorspaces.

To prove our statement for s ∈ S, i ∈ I and m ∈ M we have to checkthe following:

s · ϕ(ηi(m)) = ϕ(s · ηi(m)). (11.3)

11.3. SIMPLE MODULES OVER ISn AND PT n 195

Note that sai �∈ L(e) if and only if saiv = sai �∈ L(e), in which case bothsides of (11.3) are zero. Hence we may assume sai ∈ L(e). Let sai = ajx forsome j ∈ I and x ∈ H(e). Observe that x = ex = vux and hence

sai = saiv = ajxv = ajvuxv = ajv(uxv) = aj(uxv).

We have uxv · u = uxe = u · x = alx and thus the left-hand side of (11.3)equals (we again use the notation � similarly to the one used in the proof ofProposition 11.2.1)

s � ϕ(ηi(m)) = s � ηi(ηl(m))= ηj(uxv · ηl(m))= ηj(ηl(x · m)).

On the other hand, the right-hand side of (11.3) equals

ϕ(s · ηi(m)) = ϕ(ηj(x · m))= ηj(ηl(x · m)).

This proves that ϕ is an isomorphism, and the statement of the propositionfollows.

Remark 11.2.3 From the construction used in Proposition 11.2.2 it followsthat under the identification of H(e) and H(e), constructed in Theorem 4.7.5,the modules M and M are isomorphic.

11.3 Simple Modules over ISn and PT n

In this section, we describe all simple modules for the semigroups ISn andPT n.

For k ∈ {0, 1, 2, . . . , n} let ε(k)n denote the idempotent of ISn with do-

main {1, 2, . . . , k}. In particular, ε(n)n = ε and ε

(0)n = 0. Using Theorem 5.1.4

we identify H(ε(k)n ) with Sk in the natural way. Our main result in this

section is the following:

Theorem 11.3.1 Let S = ISn or S = PT n.

(i) For any k ∈ {0, 1, 2, . . . , n} and any simple Sk-module M the moduleV (M) is a simple S-module.

(ii) Let k, k ∈ {0, 1, 2, . . . , n}, and M and M be simple Sk- and Sk-modules,respectively. Then V (M) ∼= V (M) if and only if k = k and M ∼= M .

(iii) Every simple S-module is isomorphic to the module V (M) for somek ∈ {0, 1, 2, . . . , n} and some simple Sk-module M .


Proof. The H-classes inside L(ε(k)n ) are indexed by I = {A ⊂ N : |A| = k}

by Theorem 4.5.1. Set for simplicity A = {1, 2, . . . , k}.Let M be a simple Sk-module and v ∈ V (M) be a nonzero element. Then

v =∑

A∈I cAvA, where cA ∈ C and vA ∈ M (A). We may assume vA �= 0for all A ∈ I. Let N be the linear span of s · v, s ∈ S. Obviously N is asubmodule of V (M). As v �= 0, there exists B ∈ I such that cB �= 0. Assumethat a = {αA : A ∈ I} as in Sect. 10.5. Then εAαA = αA and εA · vA = vA

for every A ∈ I. If A ∈ I is such that A �= B, then rank(εBεA) < k andhence

εB ·∑A∈I

cAvA =∑A∈I

cAεBεA · vA = cBvB �= 0.

As cB �= 0, we have vB ∈ N , in particular, vB = ηB(w) for some nonzerow ∈ M . As α−1

B αB = ε(k)n , we have α−1

B · vB = α−1B · ηB(w) = w. Hence

w ∈ N . Since the Sk-module M is simple and w �= 0, the minimal Sk-invariants subspace of M (A) containing w must be the whole M (A), implyingM (A) ⊂ N . By definition, αA ·M (A) = M (A), which means M (A) ⊂ N for allA ∈ I. This finally implies N = V (M). Now the fact that V (M) is simplefollows from Exercise 11.1.6. This proves the statement (i).

If k = k and M ∼= M , then V (M) ∼= V (M) is obvious. Hence we assumethat k, k ∈ {0, 1, 2, . . . , n} and M and M are such that there exists anisomorphism ϕ : V (M) ∼= V (M).

Assume first that k �= k. Without loss of generality we may assumek < k. Then ε

(k)n · V (M) = 0 by definition, whereas ε

(k)n · M (A) = M (A).

At the same time, the action of ε(k)n on V (M) is just a conjugate of the

corresponding action on V (M) by the isomorphism ϕ, a contradiction. Thisproves that k = k.

As V (M) = ⊕A∈IM(A) and ε

(k)n αA �∈ L(ε(k)

n ) if A �= A, we have theequality ε

(k)n · V (M) = M (A). Analogously ε

(k)n · V (M) = M (A). As ϕ com-

mutes with the action of ε(k)n we obtain that ϕ(M (A)) ⊂ M (A). Analogously,

one shows that ϕ−1(M (A)) ⊂ M (A). As both mappings are injective, weconclude that the restriction of ϕ to M (A) is an isomorphism with imageM (A). Since ϕ commutes with the action of S, it, in particular, commuteswith the action of Sk. This proves the statement (ii).

Let L be a simple S-module and k ∈ {0, 1, 2, . . . , n} be minimal suchthat there exists an element α ∈ S of rank k, which acts on L in a nonzeroway. As α ∈ Sε

(k)n S by Theorem 4.2.8, it follows that ε

(k)n acts on L in a

nonzero way. Set M = ε(k)n · L �= 0.

Lemma 11.3.2 M is a simple Sk-module.

Proof. Obviously, M is an Sk-module. Assume that this is not the case andlet N be a proper Sk-submodule of M . For A ∈ I set

LA = αAα−1A · L = εA · L.

11.3. SIMPLE MODULES OVER ISn AND PT n 197

By the same arguments as above we have LA �= 0. Further, as εAεB �∈ L(ε(k)n )

if A �= B, it also follows that the sum of all LA’s inside L is direct.The minimal subspace of L, which is invariant with respect to the action

of S and contains N , is the linear span N of s · v, s ∈ S and v ∈ N . Forany s ∈ S and any v ∈ N we have s · v = sε

(k)n · v. If sε

(k)n �∈ L(ε(k)

n ),then rank(sε(k)

n ) < k and we have s · v = 0 by our assumption on k. Ifsε

(k)n ∈ L(ε(k)

n ), then there exists A ∈ I and β ∈ Sk such that sε(k)n = αAβ.

As εAαA = αA, we obtain

N ⊂ N ⊕⊕A �=A

LA �= L,

which contradicts the fact that L is a simple S-module. This completes theproof.

Define a linear mapping ϕ : V (M) → L as follows: ϕ(ηA(v)) = αA · v forall v ∈ M and A ∈ I. Obviously, ϕ �= 0 as ϕ(v) = v for any v ∈ M .

We claim that ϕ is a homomorphism of S-modules. To prove this wehave to check

s · ϕ(ηA(v)) = ϕ(s · ηA(v)) (11.4)

for all s ∈ S, A ∈ I, and v ∈ M . If rank(sαA) < k, then s · ηA(v) = 0by definition and s · ϕ(ηA(v)) = sαA · v = 0 by our assumption on k. Ifrank(sαA) = k, we can write sαA = αBβ for some B ∈ I and β ∈ H(ε(k)

n ).We compute

s · ϕ(ηA(v)) = s · (αA · v)= sαA · v= αBβ · v= αB · (β · v)= ϕ(ηB(β · v))= ϕ(s · ηA(v)).

This proves (11.4). So, we have constructed a nonzero homomorphism ϕfrom V (M) to L. The module V (M) is simple by Lemma 11.3.2 and thestatement (i), the module L is simple by assumption. Now (iii) follows fromProposition 11.1.7(iii).

Corollary 11.3.3 All simple ISn modules are obtained from simple PT n-modules by restriction, in particular, the corresponding simple ISn- andPT n-modules have the same dimension.

Proof. Follows from Theorem 11.3.1, construction of V (M) and the fact thatthe L-classes of ε

(k)n in ISn and PT n coincide.


11.4 Effective Representations

Let S be a semigroup and ϕ : S → EndC(V ) be a representation of S. Therepresentation ϕ is called effective provided that it is injective. The samenotion is used for the corresponding module.

Theorem 11.4.1 Let S denote one of the semigroups Tn, PT n, or ISn.

(i) If V is an effective S-module, then dim(V ) ≥ n.

(ii) The natural S-module is effective.

(iii) If V is an effective S-module of dimension n, then V is isomorphic tothe natural S-module.

Proof. The statement (ii) is obvious. To prove the statement (i) let us firstassume that S = PT n or S = ISn. Consider the element α = [1, 2, . . . , n].We have αn = 0 and αk �= 0 for any k < n. Note that 0 is an idempotent and0αk = αk0 = 0. If V is now an effective S-module, then α defines a linearoperator on V , whose Jordan decomposition contains either Jordan cells ofsize 1 with eigenvalue 1 or nilpotent Jordan cells. Moreover, it must containat least one nilpotent Jordan cell of dimension n. Hence dim(V ) ≥ n.

For S = Tn we consider

β =(

1 2 3 4 . . . n1 1 2 3 . . . n − 1

).

We have βn−1 = 01 and βk �= 01 for all k < n − 1. Hence if now V isan effective Tn-module, then, by the same arguments as above, β defines alinear operator on V , whose Jordan decomposition must have at least onenilpotent Jordan cell of dimension (n−1). Furthermore, the element 01 actson V in a nonzero way, for otherwise all 0a ∈ Tn01Tn would act as zerosand V could not be effective. This implies that β also contains at least oneJordan cell with eigenvalue 1. Thus dim(V ) ≥ n again and the statement(i) is proved.

To prove (iii) we again first consider the case S = PT n or S = ISn. Ourfirst observation is:

Lemma 11.4.2 Let S = PT n or S = ISn and V be an effective module ofdimension n. Then 0 · V = 0.

Proof. Let W = 0 · V and assume W �= 0. As 0 is an idempotent, for anys ∈ S and w ∈ W we have

s · w = s · (0 · w) = (s0) · w = 0 · w = w.

In particular, W is a submodule of V . As we have seen above, s ·w = w forall w ∈ W . As V is effective, we get that for any s �= t ∈ S there should

11.4. EFFECTIVE REPRESENTATIONS 199

exist v ∈ V such that s · v �= t · v. Assume s · v = t · v + w for some w ∈ W .Applying 0 we obtain 0 · (s · v) = 0 · (t · v +w), which yields 0 · v = 0 · v +w.Hence w = 0. However, w = 0 is impossible since s · v �= t · v. This provesthat s · v + W �= t · v + W . The latter implies that the module V/W isan effective module of dimension strictly smaller than n, which contradictsTheorem 11.4.1(i).

As V is effective and 0 · V = 0 by Lemma 11.4.2, we have ε{1} · V �= 0.Fix any nonzero element v ∈ ε{1} · V . Let M be the trivial H(ε{1})-moduleand m ∈ M be any nonzero element.

Lemma 11.4.3 The assignment s · m → s · v, s ∈ S, defines a nonzerohomomorphism f : V (M) → V .

Proof. As ε{1} · m = m, for s ∈ S we have

s · m = s · (ε{1} · m) = (sε{1}) · m.

We have Sε{1} = {0, α{i} : i ∈ N} in the notation of Sect. 10.5. We furtherhave 0 ·v = 0 by Lemma 11.4.2 and 0 ·m = 0 by definition. Observe that fors, t ∈ S the equality s ·m = t ·m implies sε{1} = tε{1} and thus s ·v = t ·v. Asα{i} · m are linearly independent in V (M), the mapping f is a well-definedlinear mapping. It commutes with the action of S by definition, and henceit is a homomorphism of S-modules. Finally, f is obviously nonzero.

The homomorphism f is nonzero and hence injective since the moduleV (M) is simple by Theorem 11.3.1(i). As dim(V (M)) = dim(V ) = n, weobtain that f is an isomorphism, implying that an effective module of di-mension n is unique up to isomorphism. For S = PT n and S = ISn thestatement (iii) now follows from the statement (ii).

Consider now the case S = Tn. First we note that 01 · V �= 0 as V iseffective. Indeed, as 0i ∈ Tn01, i ∈ N, the equality 01 · V = 0 would imply0i · V = 0 for all i, a contradiction. Now we note that (1, 2)01 = 02, inparticular, the idempotent linear transformations 01 and 02 of V have thesame rank and the same kernel. Since they must be different as V is effective,we conclude that their images are different. As these images have the samedimension, they cannot be included into each other and hence there existsa nonzero v ∈ 01 · V such that v �∈ 02 · V .

Let M be the trivial representation of H(01) and m ∈ M be a nonzeroelement. Analogously to Lemma 11.4.3 one shows that the assignment s·m →s · v, s ∈ Tn, defines a nonzero homomorphism f : V (M) → V . To proceedwe have to analyze the module V (M). Let {v1, . . . , vn} denote the standardbasis of C

n.

Lemma 11.4.4 (i) For i ∈ N the assignment 0i · m → vi defines anisomorphism from V (M) to the natural Tn-module C

n.


(ii) The natural Tn-module Cn has exactly three submodules, namely, 0, C

n

and the submodule

N =

{n∑

i=1

aivi :n∑

i=1

ai = 0

}.

Proof. The statemet (i) follows from the relation α0i = 0α(i) in Tn. To prove(ii) let W be a proper Tn-submodule of C

n, and w = (w1, . . . , wn) ∈ W bea nonzero element.

Assume first that w1 + · · · + wn �= 0. In this case, we have 01 · w =(w1 + · · · + wn, 0, . . . , 0), implying that W contains v1. Then W contains(1, i) · v1 = vi for every i ∈ N and hence W = C

n, a contradiction.Assume now that w1 + · · ·+wn = 0. As the action of Sn simply permutes

the entries of w, without loss of generality we may assume w1 �= 0. Then wehave

(1 2 3 4 . . . n1 2 2 2 . . . 2

)· w = (w1, w2 + · · · + wn, 0, . . . , 0),

which implies that W contains the element z = (1,−1, 0, . . . , 0). Apply-ing the elements from Sn we get that W contains all elements of the form(0, . . . , 0, 1,−1, 0, . . . , 0). As such elements form a basis in N , we obtainN ⊂ W . As the codimension of N in C

n is 1 and W �= Cn, it follows that

W = N . This completes the proof.

The mapping f is nonzero and hence from Lemma 11.4.4(ii) we havethat either Ker(f) = 0 or Ker(f) ∼= N . Assume Ker(f) ∼= N . As im(f) ∼=V (M)/Ker(f), we have that im(f) has dimension 1. At the same time,it contains both v and the element 02 · v, which is linearly independentwith v because of our assumptions. This is a contradiction, which provesKer(f) = 0. As both V (M) and V have dimension n, it follows that f is anisomorphism. Now the proof is completed by applying Lemma 11.4.4(i).

11.5 Arbitrary ISn-Modules

Let S be a semigroup and V and W be S-modules. Then the vector spaceV ⊕ W = {(v, w) : v ∈ V, w ∈ W} carries the natural structure of anS-module via s · (v, w) = (s · v, s · w), s ∈ S, v ∈ V , w ∈ W . The moduleV ⊕W is called the direct sum of the modules V and W . The direct sum iscalled trivial if V = 0 or W = 0.

An S-module U is called indecomposable provided that U cannot bedecomposed into a nontrivial direct sum of two modules. In other words, Uis indecomposable if U ∼= V ⊕ W implies V = 0 or W = 0. Every simplemodule is obviously indecomposable. The converse is not true in general.

11.5. ARBITRARY ISn-MODULES 201

Exercise 11.5.1 Show that for n > 1 the natural Tn-module is indecom-posable but not simple.

An S-module U is called semisimple provided that it is isomorphic toa direct sum of simple modules. The corresponding representation is calledcompletely reducible. Obviously every simple module is semisimple. If U is asemisimple module, then there exists essentially a unique way to write it asa direct sum of simple modules as shown in the following statement.

Proposition 11.5.2 Let U be a semisimple S-module such that

U ∼= V1 ⊕ V2 ⊕ · · · ⊕ Vk∼= W1 ⊕ W2 ⊕ · · · ⊕ Wm,

where all Vi’s and Wj’s are simple modules. Then k = m and there existsα ∈ Sk such that Vi

∼= Wα(i) for all i.

Proof. Let L be some simple S-module. We have

HomS(L, U) = HomS(L, V1 ⊕ · · · ⊕ Vn) = HomS(L, V1) ⊕ · · · ⊕ Hom(L, Vn)

(see Exercise 11.7.2). As both L and Vi are simple, from Corollary 11.1.8 weobtain that

|{i : Vi∼= L}| = dim Hom(L, U).

Analogously one obtains

|{j : Wj∼= L}| = dim Hom(L, U).

The statement of the proposition follows.

The main aim of the present section is to prove the following theorem,which, together with Proposition 11.5.2, gives a complete description of allISn-modules:

Theorem 11.5.3 Every ISn-module is semisimple.

Proof. Let V be an ISn-module. We prove the statement by induction ondim(V ). If dim(V ) = 1, the module V is simple (see Example 11.1.3) andhence semisimple.

Let k be such that ε(k)n ·V �= 0 while ε

(i)n ·V = 0 for all i < k. In particular,

we have α · V = αα−1α · V = α · (α−1α · V ) = 0 for any α ∈ ISn such thatrank(α) < k. Let I = {A ⊂ N : |A| = k} and for A ∈ I set UA = εA · V .Define

W =⋂A∈I

{v ∈ V : εA · v = 0}, U =∑A∈I

UA.

Lemma 11.5.4 (i) U is a direct sum of UA’s.

(ii) V = U ⊕ W .


(iii) Both U and W are ISn-submodules.

Proof. Let vA ∈ UA be nonzero and cA ∈ C be such that∑

A∈I cAvA = 0.As for A �= B ∈ I we have rank(εAεB) < k, we compute

εB ·∑A∈I

cAvA =∑A∈I

cAεB · (εA · vA) =∑A∈I

cAεBεA · vA = cBvB.

This yields cB = 0 for any B ∈ I, and the statement (i) follows.As all εA’s are idempotents, we have U ∩ W = 0 from the definition.

Let v ∈ V . Then u =∑

A∈I εA · v ∈ U and for any B ∈ I, using the samearguments as in the previous paragraph, we have

εB · (v − u) = εB ·(

v −∑A∈I

εA · v)

= εB · v −∑A∈I

εBεA · v

= εB · v − εB · v= 0.

Hence v − u ∈ W , implying V = U + W . The statement (ii) follows.Let A ∈ I, vA ∈ UA and α ∈ ISn. If rank(αεA) < k, then

α · vA = αεA · vA = 0 ∈ U.

If rank(αεA) = k, then

α · vA = αεA · vA = εα(A)αεA · vA = εα(A) · (α · vA).

and hence α · vA ∈ Uα(A). This implies that U is an ISn-submodule of V .Let w ∈ W , α ∈ ISn, and A ∈ I. If rank(εAα) < k, then

εA · (α · w) = εAα · w = 0

by our assumption on k. If rank(εAα) = k, then εAα = αεB, where B =α−1(A). Hence

εA · (α · w) = εAα · w = αεB · w = α · (εB · w) = 0.

This shows that W is an ISn-submodule of V and completes the proof.

By our assumptions, we have U �= 0. If W �= 0, then dim(U) < dim(V )and dim(W ) < dim(V ) and hence applying the inductive assumption toboth U and W we obtain a decomposition of V into a direct sum of simplemodules, as required. Hence in what follows we may assume W = 0.

Let A = {1, 2, . . . , k} and for B ∈ I let αB denote the unique increasingbijection from A to B.

11.5. ARBITRARY ISn-MODULES 203

Lemma 11.5.5 For B ∈ I the action of αB induces a bijection from UA

to UB.

Proof. This follows from the obvious equalities α−1B αB = εA and αBα−1

B =εB.

Let (·, ·)1 be any Hermitian scalar product on UA. For v, w ∈ UA set

(v, w) =∑

g∈H(ε(k)n )

(g · v, g · w)1.

For v ∈ UC and w ∈ UB, C �= B ∈ I we set

(v, w) = (w, v) = 0.

Finally, for v, w ∈ UA and B ∈ I we set

(αB · v, αB · w) = (v, w)

and extend the product (·, ·) to the whole V by skew-linearity.

Lemma 11.5.6 (i) (·, ·) is a Hermitian scalar product on V .

(ii) For all v, w ∈ V and α ∈ ISn we have (α · v, w) = (v, α−1 · w).

Proof. The product (·, ·) is bilinear and skew-symmetric by construction.The restriction of (·, ·) to UA is positive definite since (·, ·)1 is positive defi-nite. Now the fact that (·, ·) is positive definite follows from the constructionand Lemma 11.5.4(i). This proves (i).

To prove (ii) let v ∈ UC , w ∈ UB and β ∈ ISn be chosen such thatwe have (β · v, w) �= 0. This, in particular, means 0 �= β · v ∈ UB and thusrank(εBβεC) = k. Hence rank(εCβ−1εB) = k as well and thus εCβ−1εB ·w ∈UC . By definition we have

(β · v, w) =(α−1

B · (β · v), α−1B · w

)=((α−1

B βαC) · (α−1C · v), α−1

B · w)

(11.5)

and analogously

(v, β−1 · w) =(α−1

C · v, (α−1C β−1αB) · (α−1

B · w)). (11.6)

The element γ = α−1B βαC belongs to G = H(εA), γ−1 = α−1

C β−1αB andboth α−1

C · v and α−1B ·w belong to UA. At the same time for any x, y ∈ UA,


using the fact that G is a group, we have

(γ · x, y) =∑g∈G

(g · (γ · x), g · y

)1

=∑g∈G

(gγ · x, g · y

)1

=∑g∈G

(g · x, gγ−1 · y

)1

=∑g∈G

(g · x, g · (γ−1 · y)

)1

= (x, γ−1 · y).

The latter together with (11.5) and (11.6) implies the statement (ii). Theproof is complete.

Let now X be some proper submodule of V . Then for the orthogonalcomplement

X⊥ = {y ∈ V : (y, x) = 0 for all x ∈ X}

using Lemma 11.5.6(ii) we have

(α · y, x) = (y, α−1 · x) = 0

for any y ∈ X⊥, x ∈ X and α ∈ ISn. Hence X⊥ is a proper submodule of Vand we have V = X ⊕X⊥. Applying now the inductive assumption to bothX and X⊥ we produce a decomposition of V into a direct sum of simplemodules. This completes the proof.


11.6.1 The results, presented in this chapter, mostly follow from more gen-eral results for representation of semigroups, obtained by Munn in [Mu1,Mu2, Mu3, Mu4] and Ponizovskiy [Po3]. These more general results, whichalso are more technically complicated and require more advanced knowledgeabout algebras, can be found in [CP1, Chap. 5]. The only results which wedid not manage to find in the literature are those presented in Sect. 11.4.

There is also a well-developed theory of characters of semigroups. Char-acters of commutative semigroups were studied by Schwarz in [Sw1, Sw2,Sw3] and Hewitt and Zuckerman in [HZ1]. This theory is also presented in[CP1, Chap. 5].

Some recent developments in the character theory for modules, especiallyover transformation semigroups, can be found in the works of Putcha [Pu1,Pu2, Pu3]. For inverse semigroups some new approaches appear in [Ste2].Character tables for ISn were studied in [So]. A modern point of view


on classification of simple modules over finite semigroups is presented inExercises 11.7.19 and 11.7.20. It also recently appeared in [GMS].

11.6.2 All simple Tn-modules were explicitly constructed by Hewitt andZuckerman in [HZ2], and their characters were described in [Pu1, Section 2].This description is more complicated than the corresponding description forPT n and ISn presented in Theorem 11.3.1, and the proof is technicallymuch more complicated. To state the result we will need some new notationfor the symmetric group.

Let σ ∈ Sn be a permutation. A pair (i, j) ∈ N×N is called an inversionfor σ provided that i < j and σ(i) > σ(j). The number of inversions for σis denoted invσ. The permutation σ is called odd or even provided that invσ

is odd or even, respectively. For example, the identity permutation is even,while any transposition is odd.

For c ∈ C the assignment σ · c = (−1)invσc defines on C the structureof an Sn-module, called the sign module. This module is simple as it is one-dimensional. For n > 1 the trivial module and the sign module are the onlyone-dimensional Sn-modules. They are obviously nonisomorphic. For n = 1the definition is awkward, however, it is convenient to use the conventionthat the trivial and the sign S1-modules coincide. Now we can formulate thestatement describing all simple Tn-modules. We refer the reader to [HZ2]and [Pu1] for the proofs.

Theorem 11.6.1 (i) The Tn-module V (M) is simple if and only if k = nor M is not isomorphic to the sign Sk-module.

(ii) If M is isomorphic to the sign Sk-module, then the module V (M) hasa unique simple quotient V (M) of dimension

(n−1k−1

).

(iii) Every simple Tn-module is isomorphic either to V (M), where M is thesign Sk-module, or to V (M), where M is an Sk-module, which is notisomorphic to the sign module.

The statement in Theorem 11.6.1(iii) is proved similarly to the corre-sponding statement in Theorem 11.3.1. The really hard part here is to provethe first two statements of Theorem 11.6.1.

A naive straightforward generalization of the arguments from Sect. 11.3leads to some interesting combinatorial questions. Let us follow the argu-ments from Sect. 11.3 and try to prove that the Tn-module V (M) is simplein the case when M is the trivial module and k > 1. To prove this we haveto show that for any nonzero vector v ∈ V (M) the minimal submodule ofV (M), containing v, coincides with V (M). This is equivalent to showingthat for any nonzero vector v ∈ V (M) there exists some idempotent e ∈ Tn

of rank k such that e · v �= 0. The latter statement has the following com-binatorial formulation: Let X denote the matrix whose rows are indexed by


k-element subsets of N and columns are indexed by unordered partitions ofN into k disjoint nonempty blocks. Note that the same objects index theL-classes and the R-classes of Tn for elements of rank k. In particular, theelements of X correspond naturally to H-classes inside Dk. Let the elementof X be 1 if the corresponding H-class contains an idempotent and 0 oth-erwise. Then, following the above arguments, the irreducibility of V (M) isequivalent to the statement that the rank of X equals the number of rowsin X, that is,

(nk

). This result follows from Theorem 11.6.1(i). However, we

do not know any combinatorial proof.The above matrix X contains a natural square submatrix defined using

the following natural injection from the set of all k-element subsets of Nto the set of all partitions of N into k-blocks. With each k-element subset{i1 < i2 < · · · < ik} one associates the partition

{ik + 1, . . . , n, 1, . . . , i1} ∪ {i1 + 1, . . . , i2} ∪ · · · ∪ {ik−1 + 1, . . . , ik}.

We suspect that the determinant of the corresponding square submatrix ofX is always a power of 2 (up to sign), but we cannot prove that.

11.6.3 There is a beautiful combinatorial description of all simple Sn-modules, see for example [Sa]. A partition of n is a vector λ = (λ1, λ2, . . . , λl)with positive integer coefficients such that λ1 ≥ λ2 ≥ · · · ≥ λl and n =λ1 + λ2 + · · · + λl. That λ is a partition of n is usually denoted λ � n. TheFerrers diagram or Young diagram of shape λ is an array of n boxes havingl left-justified rows with λi boxes in the i-th row for 1 ≤ i ≤ l. For example,the Young diagram of shape λ = (4, 2, 2, 1) � 9 is the following:

With every λ � n one associates a simple Sn-module Sλ, called the Spechtmodule.

Let λ � n and a1 < a2 < · · · < an be a sequence of integers. A standard λ-tableau of content {a1, . . . , an} is obtained if we write the numbers a1, . . . , an

in the boxes of the Young tableau so that the entries in all rows increaseleft-to-right and entries in all columns increase top-to-bottom. Here is anexample of a standard (4, 2, 2, 1)-tableau of content N9:

1 2 6 73 54 89


Theorem 11.6.2 (i) Each simple Sn-module is isomorphic to Sλ forsome λ � n.

(ii) For λ, μ � n we have Sλ∼= Sμ if and only if λ = μ.

(iii) Let λ � n. Then the Specht module Sλ has a natural basis, which isindexed by standard λ-tableaux of content N.

The combinatorial description from Theorem 11.6.2(iii) was generalizedto ISn by Grood in [Grd]. As simple ISn-modules are obtained from simplePT n-modules by restriction (Corollary 11.3.3), the same description appliesto PT n as well. As usually, for k ≤ n we identify Sk with the group H(ε(k)

n )in the natural way. Combining the results of Grood with the results fromSect. 11.3 we have:

Theorem 11.6.3 Let S = ISn or S = PT n.

(i) Each simple S-module is isomorphic to M(Sλ) for some λ � k, k ≤ n.

(ii) For λ � k1 and μ � k2, k1, k2 ≤ n, we have M(Sλ) ∼= M(Sμ) if andonly if λ = μ.

(iii) Let λ � k, k ≤ n. Then the module M(Sλ) has a natural basis, which isindexed by standard λ-tableaux, whose contents is a k-element subsetof N.

Let k ≤ n and I denote the set of all k-element subsets of N. If λ � k,then we have a decomposition

M(Sλ) ∼=⊕A∈I

εA · M(Sλ)

into a direct sum of vector spaces. The space εA ·M(Sλ) is a simple H(εA)-module, which is isomorphic to Sλ after an appropriate identification ofH(εA) with Sk. The bases of Theorems 11.6.2(iii) and 11.6.3(iii) are naturallycompatible with this identification.

Tableaux combinatorics can also be used to describe the decompositionof the restriction of simple ISn-modules to ISn−1 (the so-called Branch-ing rule), see [MR]. It is very similar to the corresponding result for thesymmetric group, see for example [Sa] for the latter one.

11.6.4 Let S be a finite semigroup. The semigroup algebra C[S] is the set ofall formal linear combinations

∑s∈S css with complex coefficients, endowed

with the natural bilinear multiplication induced from the multiplication inthe semigroup S.

Each S-module naturally extends to a C[S]-module, and each C[S]-module restricts to an S-module. This produces an isomorphism of categoriesof all finite-dimensional S-modules and all finite-dimensional C[S]-modules.


Denote by Matn(C) the algebra of all complex n×n matrices. In the languageof algebras Theorem 11.5.3 says that the algebra C[ISn] is semisimple, inparticular, it is isomorphic to a direct sum of some Matni(C).

One can also show that C[ISn] is isomorphic to a direct sum of matrixalgebras over symmetric groups as follows: For k ≤ n the algebra

Ak = Mat(nk)

(C) ⊗C C[Sk]

can be realized as the algebra of all(nk

)×(nk

)matrices with coefficients from

C[Sk].

Theorem 11.6.4 ([Mu3]) C[ISn] ∼=n⊕

k=0

Ak.

The most beautiful part of the proof of this fact, presented for examplein [So, Section 2], is the construction of pairwise orthogonal idempotents,which determine the components Ak’s. This is done as follows: For A ⊂ Ndefine

ηA =∑B⊂A

(−1)|A\B|εB. (11.7)

By the inclusion–exclusion formula we have

εA =∑B⊂A

ηB. (11.8)

Lemma 11.6.5 For A, B ⊂ N we have

ηAηB =

{ηB, A = B;0, A �= B.

Proof. We first claim that for A, B ⊂ N we have

εAηB =

{ηB, B ⊂ A;0, otherwise.

(11.9)

This is obvious if B ⊂ A. Assume now that B\A �= ∅. Then

εAηB =∑C⊂B

(−1)|B\C|εA∩C .

The latter sum is a linear combination of the εY ’s, where Y ⊂ A∩B. WriteC = Y ∪ Z, where Z ⊂ B\A �= ∅. Then the coefficient at εY is

∑Z⊂B\A

(−1)|B\(Y ∪Z)| = 0.


This proves the formula (11.9). From (11.7) and (11.9) we have

ηAηB =∑X⊂A

(−1)|A\X|εXηB =

( ∑B⊂X⊂A

(−1)|A\X|)

ηB

The coefficient on the right-hand side is zero unless A = B, when it is equalto 1. This completes the proof.

For k ≤ n now set ηk =∑

|A|=k ηA. Then Lemma 11.6.5 implies

ηkηr =

{ηk, k = r;0, k �= r.

Moreover, (11.8) even says ε =∑n

k=0 ηk, which gives us a decomposition ofthe identity element in C[ISn] into a sum of pairwise orthogonal idempo-tents.

Exercise 11.6.6 Show that ηnα = αηn for any α ∈ ISn.

From Exercise 11.6.6 it follows that

C[ISn] ∼=n⊕

k=0

C[ISn]ηk =n⊕

k=0

ηkC[ISn]ηk

is a decomposition into a direct sum of subalgebras. It is now not difficultto check that ηkC[ISn]ηk

∼= Ak. Using Mobius function, these argumentsgeneralize to arbitrary finite inverse semigroups, see [Ste1].

11.6.5 Description of all finite-dimensional modules for semigroups Tn andPT n is a very hard problem. It is for example known (see [Po4, Pu2, Ri]) thatthe semigroup Tn has finitely many isomorphism classes of indecomposablemodules if and only if n < 5.

11.6.6 Let S be a semigroup and V and W be two S-modules. Then thevector space V ⊗C W carries the natural structure of an S-module via

s · (v ⊗ w) = (s · v) ⊗ (s · w).

The module V ⊗C W is called the tensor product of the modules V and W .Set V ⊗k = V ⊗C V ⊗C · · · ⊗C V , where the right-hand side contains exactlyk factors.

Let now V be the natural representation of ISn. Then the symmetricgroup Sk acts on the space V ⊗k permuting components of the tensor product.This action obviously commutes with the action of ISn. However, the linearclosure of this action of Sk does not give all linear operators which commutewith the ISn-action. The set of all operators, which commute with some


action, is usually called the centralizer of the action. The centralizer of theISn-action on V ⊗k can be explicitly described in terms of the dual symmetricinverse semigroup I∗

k , see [KuMa5]. Furthermore, the original C[ISn]-actionturns out to give the full centralizer of the C[I∗

k ]-action, which gives riseto a Schur-Weyl duality connecting ISn and I∗

k . Moreover, one can alsoshow that the kernel of the action of C[ISn] on V ⊗n coincides with C0. Inparticular, the module V ⊗n contains as submodules, except for the trivialmodule, representatives from all isomorphism classes of simple ISn-modules.

There is another version of the Schur-Weyl duality for ISn in whichISn occurs on the other side. It was discovered in [So, Sect. 5]. Let V be thenatural n-dimensional representation of the group GLn of all nondegeneratecomplex n × n matrices. Let C be the trivial representation of GLn. SetU = V ⊕ C. Then ISk acts on U⊗k in the following way: Let π : U → C bethe projection with kernel V . For α ∈ ISk and ui ∈ U , i = 1, . . . , k, set

α · (u1 ⊗ u2 ⊗ · · · ⊗ uk) = v1 ⊗ v2 ⊗ · · · ⊗ vk,

where

vj =

{uα(j), j ∈ dom(α);π(uj), j �∈ dom(α).

It turns out that the action of C[ISk] gives the full centralizer of the C[GLn]-action and vice versa.

11.6.7 Let S be a semigroup. An element w ∈ S is called an involutionprovided that w ∈ Ge for some e ∈ E(S) and w2 = e. Let B denote the setof all involutions in ISn and let CB denote the C-linear span of vw, w ∈ B.For α ∈ ISn and an involution w ∈ H(e), e ∈ E(ISn), such that αe ∈ D(e)set

invw(α) = |{(i, j) : i, j ∈ dom(w), i < j, w(i) = jandα(i) > α(j)}|.

Theorem 11.6.7 ([KuMa6]) (i) For α ∈ ISn and an involution w ∈H(e), e ∈ E(ISn), the assignment

α · vw =

{(−1)invw(α)v(αe)w(αe)−1 , αe ∈ D(e);0, otherwise.

defines on CB the structure of an ISn-module.

(ii) CB is a multiplicity-free direct sum of all simple ISn-modules.

Theorem 11.6.7(ii) says that CB is a Gelfand model for C[ISn]. Theorem11.6.7 is based on the corresponding result for Sn, proved in [APR].



11.7.1 Let S be a finite monoid with the unit 1, V be any S-module andv ∈ V .

(a) Show that the assignment v1 → v extends to a homomorphism from theregular S-module CS to V .

(b) Use (a) to show that V is a quotient of some CS ⊕ · · · ⊕ CS.

11.7.2 Let S be a semigroup and U, V, W be S-modules. Show that

(a) HomS(U ⊕ V, W ) ∼= HomS(U, W ) ⊕ HomS(V, W ).

(b) HomS(U, V ⊕ W ) ∼= HomS(U, V ) ⊕ HomS(U, W ).

11.7.3 Describe all modules, in particular, all simple modules for the semi-group B(N) with respect to the operation ∩.

11.7.4 Determine the minimal dimension of an effective representation ofthe semigroup B(N) with respect to the operation ∩.

11.7.5 Construct examples of indecomposable but not simple PT n-modules.

11.7.6 Generalize the arguments from Sect. 11.5 and prove an analogue ofTheorem 11.5.3 for arbitrary finite inverse semigroup.

11.7.7 Let S be a finite semigroup. Show that for every s ∈ S there existssome simple S-module V such that s · V �= 0, while t · V = 0 for any t ∈ Ssuch that s �∈ StS.

11.7.8 Let k ≤ n. Show that the subalgebra of C[ISn], generated by the

ideal Ik, is isomorphic tok⊕

i=0

Ai in the notation of 11.6.4.

11.7.9 (a) Show that every semigroup has at least two nonisomorphic sim-ple modules: the trivial module, and the one-dimensional module withthe zero action of S.

(b) Show that every finite nilpotent semigroup has exactly two nonisomor-phic simple modules.

11.7.10 By Theorem 11.5.3, the regular module CISn is semisimple. Provethat for every simple ISn-module V we have

dim HomISn(V, CISn) = dim(V ).

11.7.11 Let S be a finite monoid and G its maximal subgroup of invertibleelements. Let V denote the one-dimensional S-module on which the elementsfrom G act trivially, and all other elements act as zero. Let now M be anysimple S-module. Show that the module M ⊗C V is semisimple.


11.7.12 Let S be a semigroup acting on some set M . Let CM denote thevector space with the basis vm, m ∈ M . Show that CM becomes an S-module via s · vm = vs·m, s ∈ S, m ∈ M .

11.7.13 ([Ste2]) Let S be a finite inverse semigroup. Show that the regularS-module is isomorphic to the module, obtained from the Preston–Wagnerrepresentation of S via the procedure described in Exercise 11.7.12.

11.7.14 ([KuMa5]) Let V denote the natural ISn-module. Show that themodule V ⊗k is effective for any k.

11.7.15 ([So]) Let V denote the natural ISn-module with the standardbasis v1, . . . , vn. For 0 ≤ k ≤ n consider the exterior power Λk(V ).

(a) Show that for α ∈ ISn the assignment

α · (vi1 ∧ · · · ∧ vik) =

{vα(i1) ∧ · · · ∧ vα(ik), {i1, . . . , ik} ⊂ dom(α);0, otherwise,

where i1, . . . , ik ∈ N (and α · 1 = 1 for k = 0) defines on Λk(V ) thestructure of an ISn-module.

(b) Show that the module Λk(V ) is simple for every k.

(c) Show that the module Λk(V ) is isomorphic to V (M), where M is thesign Sk-module.

11.7.16 ([KuMa4]) Let S denote one of the semigroups Tn, PT n, or ISn,and let α, β ∈ S. Show that α ∼S β if and only if for every representation ϕof S the traces of the linear operators ϕ(α) and ϕ(β) coincide.

11.7.17 ([KuMa4]) Show that the statement of Exercise 11.7.16 is false forfinite semigroups in general.

11.7.18 Classify all simple modules over a finite rectangular band.

11.7.19 ([GMS]) Let S be a finite semigroup, e ∈ E(S), and M be a simpleH(e)-module.

(a) Show that the set

N(M) = {v ∈ V (M) : s · v = 0for alls ∈ D(e)}

is a submodule of V (M).

(b) Prove that any submodule X � V (M) is contained in N(M).

(c) Use (b) to show that V (M) has the unique simple quotient V (M) =V (M)/N(M).


11.7.20 ([GMS]) Let S be a finite monoid.

(a) Prove that every simple S-module is isomorphic to V (M) for appropriatee and M as defined in Exercise 11.7.19.

(b) Use (a) to classify all simple S-modules.

Chapter 12

Cross-Sections

12.1 Cross-Sections

Let X be a set and ρ an equivalence relation on X. A subset Y of X is calleda ρ-cross-section provided that Y contains exactly one representative fromeach equivalence class.

In the study of semigroups, it is natural to expect that prospectivecross-sections capture some semigroup theoretical properties. This can beunderstood in several different ways: one may expect semigroup-theoreticalproperties of either the equivalence relation or the cross-section or both.We will concentrate on the most restrictive case, and require that both theequivalence relation and its cross-section have some semigroup theoreticalmeaning.

Let S be a semigroup. We basically know the following natural equiv-alence relations on S: Green’s relations, conjugacy, and congruences. By across-section with respect to any of these equivalence relations we will meana subsemigroup of S which contains exactly one representative from everyequivalence class. The main problem of course is that there is absolutelyno guarantee that given some of these equivalence relations one can findin S some cross-section for this relation. We will see that in some casescross-sections do not exist. The first negative result is the following:

Theorem 12.1.1 Let S denote one of the semigroups Tn, PT n or ISn, andρ be either the relation of Sn-conjugation or the relation of S-conjugation.

(i) If n = 1, then S contains exactly one cross-section with respect to ρ,namely, S itself.

(ii) If n > 1, then S does not contain any cross-section with respect to ρ.

Proof. The statement (i) is obvious. To prove (ii) we will consider two cases.First we assume n ≥ 3. In this case we observe that conjugacy classes

of Sn form separate ρ-classes regardless whether ρ is the Sn-conjugation or



216 CHAPTER 12. CROSS-SECTIONS

the S-conjugation, see Sect. 6.4. Assume that T is a ρ-cross-section. Then Tmust contain some cycle α = (a, b, c) of length three. As T is a subsemigroup,α2 = (a, c, b) belongs to T as well. However, α2 = (b, c)α(b, c), implying thatα and α2 are Sn-conjugate (and thus also S-conjugate). This contradicts ourassumption that T is a ρ-cross-section.

For n = 2 consider first the case S = IS2 or S = T2. In this case S2

is commutative and thus must be contained in any ρ-cross-section T . FromProposition 6.4.3 and Theorem 6.4.13 we obtain that T must also containone of the two idempotents of rank one. Hence T = S by Theorem 3.1.4or Theorem 3.1.3, respectively. However, S is not a ρ-cross-section since allidempotents of rank one are Sn-conjugate (and hence S-conjugate as well).

Finally, let S = PT 2 and T be a ρ-cross-section. As above, T contains S2

and at least one idempotent e of rank 1. But then T must also contain theS2-conjugated (and hence also S-conjugated) idempotent (1, 2)e(1, 2), whichcontradicts our assumption that T is a ρ-cross-section. This completes theproof.

12.2 Retracts

Let S be a semigroup and ρ be a congruence on S. A cross-section withrespect to ρ is called a retract or ρ-retract. Description of retracts for Sbeing one of the semigroups Tn, PT n, or ISn is an easy application ofTheorem 6.3.10, where all congruences on S were described.

Lemma 12.2.1 Let S be one of the semigroups Tn, PT n, or ISn and ρbe a congruence on S. Then a ρ-retract exists if and only if ρ is one of thefollowing:

(i) The identity congruence

(ii) The uniform congruence

(iii) The congruence ≡R from Sect. 6.3, where R is a subgroup of Sn, more-over, if S = Tn, then, additionally, R �= {e}

Proof. Let 1 < k < n and R be a normal subgroup of Sk. Let further ≡R

be the corresponding congruence from Sect. 6.3 and T be an ≡R-retract.Then, by the construction of ≡R, T contains Sn and all idempotents of rank(n− 1). Hence T = S by Theorems 3.1.3–3.1.5, respectively. Since for k > 1the relation ≡R is not the identity congruence by construction, we get acontradiction with the assumption that T is a cross-section for ≡R.

Assume that S = Tn, k = n, R = {e} and T is an ≡R-retract. Then,by construction, T contains Sn and, apart from that, T must contain aunique noninvertible element e. Let x, y ∈ N be such that x ∈ im(e) whiley �∈ im(e). Then (x, y)e �= e, which means that such T is not closed underthe multiplication, a contradiction.

12.2. RETRACTS 217

Now the statement of the lemma follows from Theorem 6.3.10, whichsays that any ρ which is not of the form (i)–(iii), is equal to ρ =≡R for kand R as in the previous paragraphs.

Using Lemma 12.2.1 we just describe the retracts for each of the semi-groups Tn, PT n, and ISn on a case-by-case basis. Recall from 6.5.5 thatnormal subgroups of Sn are: the Sn itself, the subgroup An of even permu-tations and the identity subgroup {ε} (for n �= 4). The group S4 has anadditional normal subgroup V4.

Proposition 12.2.2 (i) The identity congruence on ISn has the uniqueretract, namely, ISn itself.

(ii) If ρ is the uniform congruence on ISn, then there exist 2n retractsfor ρ, all of the form {e}, where e ∈ E(ISn).

(iii) If ρ =≡Sn, then ISn contains 2n − 1 retracts ρ, all of the form {ε, e},where e ∈ E(ISn)\{ε}.

(iv) If ρ =≡An, then ISn contains

n∑k=2

(n

k

) � k+24 �∑

s=1

k!(k − 2(2s − 1))! · (2s − 1)! · 22s−1

ρ-retracts, all of the form {ε, α, e}, where e ∈ E(ISn)\{ε} and theelement α ∈ Sn\An has order two and is such that α(x) = x for anyx ∈ im(e).

(v) If ρ =≡{ε}, then ISn contains the unique ρ-retract Sn ∪ {0}.

(vi) If ρ =≡V4, then IS4 contains eight ρ-retracts of the form

Q{a,b,c}0 = {ε, (a, b), (a, c), (b, c), (a, b, c), (a, c, b),0};

Q{a,b,c}1 = {ε, (a, b), (a, c), (b, c), (a, b, c), (a, c, b), ε{d}},

where {a, b, c} ⊂ N4, and {d} = N4\{a, b, c}.

Proof. The statement (i) is obvious. If ρ is the uniform congruence, then Tmust contain exactly one element, which is automatically idempotent. Anyidempotent is a retract for the uniform congruence. Hence (ii) follows fromCorollary 2.7.3.

If ρ =≡Sn , then ρ has two equivalence classes: Sn and ISn\Sn. Each ofthem is a subsemigroup of ISn, and hence any retract of ρ must consist ofidempotents. There is a unique idempotent ε in Sn. On the other hand, forany e ∈ E(ISn)\{ε}, the set {ε, e} is a retract for ρ. Hence (iii) follows againfrom Corollary 2.7.3.


If ρ =≡Sn , then ρ has three equivalence classes: An, Sn\An, and ISn\Sn.As both An and ISn\Sn are subsemigroups, any retract T must contain anidempotent from each of these two classes. Hence An must be representedin T by ε and ISn\Sn by some e ∈ E(ISn)\{ε}. As α2 ∈ An for anyα ∈ Sn\An, the latter class must be represented by some α of order two. Asα ∈ Sn\An, α must be a product of an odd number r = 2s − 1 of pairwisecommuting transpositions. The condition which would guarantee that T isclosed under multiplication is αe = eα = e. This is obviously equivalent toα(x) = x for any x ∈ dom(e). Now if we let k be the cardinality of dom(e),the statement (iv) follows easily from the formula (7.7).

If ρ =≡{ε}, then any retract T must contain Sn and one additionalnoninvertible element x. To be closed under multiplication, such x mustsatisfy πx = xπ = x for all π ∈ Sn. The only element of ISn which satisfiesthis condition is the element 0. Indeed, if α ∈ ISn is noninvertible and suchthat α �= 0, then n > 1, im(α) �= ∅ and taking a ∈ im(α) and b ∈ N\{a}we obtain (a, b)α �= α. The statement (v) follows. The statement (vi) is leftas an exercise to the reader.

Proposition 12.2.3 (i) The identity congruence on Tn has the uniqueretract, namely, Tn itself.

(ii) If ρ is the uniform congruence on Tn, then there exist∑n

k=1

(nk

)kn−k

retracts for ρ, all of the form {e}, where e ∈ E(Tn).

(iii) If ρ =≡Sn, then Tn contains∑n−1

k=1

(nk

)kn−k retracts for ρ, all of the

form {ε, e}, where e ∈ E(Tn)\{ε}.

(iv) If ρ =≡An, then Tn contains

n∑k=2

(n

k

) � k+24 �∑

s=1

k!(n − k)k−(2s−1)

(k − 2(2s − 1))! · (2s − 1)! · 22s−1

ρ-retracts, all of the form {ε, α, e}, where e ∈ E(Tn)\{ε} and α ∈Sn\An is an element of order two, which preserves the blocks of ρe

and satisfies α(x) = x for any x ∈ im(e).

(v) If ρ =≡V4, then T4 contains four ρ-retracts of the form

Q{a,b,c} = {ε, (a, b), (a, c), (b, c), (a, b, c), (a, c, b), 0d};

where {a, b, c} ⊂ N4 and {d} = N4\{a, b, c}.

Proof. Analogous to that of Proposition 12.2.2 and is left to the reader.

Proposition 12.2.4 (i) The identity congruence on PT n has the uniqueretract, namely, PT n itself.

12.3. H-CROSS-SECTIONS IN Tn, PT n, AND ISn 219

(ii) If ρ is the uniform congruence on PT n, then there exist∑n

k=0

(nk

)(k +

1)n−k retracts for ρ, all of the form {e}, where e ∈ E(PT n).

(iii) If ρ =≡Sn, then PT n contains∑n−1

k=0

(nk

)(k + 1)n−k retracts ρ, all of

the form {ε, e}, where e ∈ E(PT n)\{ε}.

(iv) If ρ =≡An, then PT n contains

n∑k=2

(n

k

) � k+24 �∑

s=1

k!(n − k + 1)k−(2s−1)

(k − 2(2s − 1))! · (2s − 1)! · 22s−1

ρ-retracts, all of the form {ε, α, e}, where e ∈ E(ISn)\{ε} and α ∈Sn\An is an element of order two, which preserves the blocks of ρe

and satisfies α(x) = x for any x ∈ dom(e).

(v) If ρ =≡{ε}, then PT n contains the unique ρ-retract Sn ∪ {0}.

(vi) If ρ =≡V4, then PT 4 contains twelve ρ-retracts of the form Q{a,b,c}0 ,

Q{a,b,c}1 , Q{a,b,c}, as defined in Propositions 12.2.2(vi) and 12.2.3(v).

Proof. Analogous to that of Proposition 12.2.2 and is left to the reader.

12.3 H-Cross-Sections in T n , PT n , and ISn

In this section, we describe cross-sections in Tn, PT n, and ISn with respectto the relation H. We will call such cross-sections simply H-cross-sections.We start with the following negative result:

Theorem 12.3.1 Let S = Tn or S = PT n.

(i) If n = 1, then S contains the unique H-cross-section, namely, S itself.

(ii) If n = 2, then S contains the unique H-cross-section, namely, thesubsemigroup S\{(1, 2)}.

(iii) If n > 2, then S does not contain any H-cross-sections.

Proof. The statement (i) is obvious. To prove the statement (ii) one justobserves that in this case the only H-class of S containing more than oneelement is S2. As S2 is a subsemigroup of S, any H-cross-section must containan idempotent from S2, which is the identity transformation ε. We haveS2\{ε} = {(1, 2)}. As S\{(1, 2)} is a subsemigroup, it is an H-cross-section.This proves the statement (ii).

To prove (iii) we consider the following elements from S:

α1 =(

1 2 3 · · · n1 3 3 · · · 3

), α2 =

(1 2 3 · · · n3 1 1 · · · 1

).


β1 =(

1 2 3 · · · n2 2 3 · · · 3

), β2 =

(1 2 3 · · · n3 3 2 · · · 2

).

γ1 =(

1 2 3 4 · · · n1 2 1 2 · · · 2

), γ2 =

(1 2 3 4 · · · n2 1 2 1 · · · 1

).

From Theorem 4.5.1 we have H(α1) = {α1, α2}, H(β1) = {β1, β2} andH(γ1) = {γ1, γ2}. Assume that T is an H-cross-section of S. Then the in-tersection of T with each of H(α1), H(β1) and H(γ1) consists of exactly oneelement. Since all these H-classes are subgroups of S, the intersection of Twith each of them must contain the identity elements of the correspondingsubgroup; namely, α1, β1, and γ1 respectively. Then T contains the element

π = γ1β1α1 =(

1 2 3 · · · n2 1 1 · · · 1

).

As π �= π2 and π2 ∈ H(π), we obtain that T must contain two elementsfrom H(π). This contradicts our assumption that T is an H-cross-section,which completes the proof.

Let ≺ be a linear order on N. A partial permutation α ∈ ISn is saidto be ≺-order-preserving provided that α(i) ≺ α(j) for all i, j ∈ dom(α)such that i ≺ j. Denote by IO≺

n the set of all ≺-order-preserving partialpermutations from ISn. If ≺ is the natural order, then one simply saysorder-preserving instead of ≺-order-preserving, and uses the notation IOn

for the corresponding set IO≺n . We denote by

←≺ the linear order opposite to

≺. We will also denote by < the natural order on N.

Exercise 12.3.2 Check that IO≺n is a subsemigroup of ISn.

Theorem 12.3.3 (i) For every linear order ≺ on N the subsemigroupIO≺

n is an H-cross-section of ISn.

(ii) For two linear orders ≺1 and ≺2 we have IO≺1n = IO≺2

n if and only if≺1=≺2 or ≺1=

←≺2.

(iii) If n �= 3, then every H-cross-section of ISn has the form IO≺n for

some linear order ≺ on N.

Proof. From Exercise 12.3.2 we know that IO≺n is a subsemigroup of ISn. By

Theorem 4.5.1(iii), each H-class of ISn is determined by some k, 0 ≤ k ≤ n,and a pair A and B of k-element subsets of N, and consists of all bijectionsα : A → B. Write A = {a1 ≺ a2 ≺ · · · ≺ ak} and B = {b1 ≺ b2 ≺ · · · ≺ bk}.Then the element (

a1 a2 · · · ak

b1 b2 · · · bk

)

is the only representative of the corresponding H-class in IO≺n . Hence IO≺

n

is an H-cross-section, which proves the statement (i).

12.3. H-CROSS-SECTIONS IN Tn, PT n, AND ISn 221

It follows directly from the definitions that IO≺1n = IO≺2

n provided that≺1=≺2 or ≺1=

←≺2. Assume now that ≺1 �=≺2 and ≺1 �=

←≺2. Without loss of

generality we may even assume that ≺1 is the natural order < and set ≺=≺2.Since ≺ is neither the natural order nor the opposite to it, there exists i ∈ Nsuch that either i ± 1 ≺ i or i ≺ i ± 1. Then for the element

α =(

i − 1 ii i + 1

)∈ IO<

n

of rank 2 we have α �∈ IO≺n . The statement (ii) follows.

The really interesting part here is to prove the statement (iii). Note thatfor n = 1 the statement is obvious. If n = 2, then we have a unique H-class,containing more than one element, namely, S2. As in Theorem 12.3.1(ii) weobtain that IS2\{(1, 2)} is the unique H-cross-section, which obviously hasthe necessary form. Hence from now on we may assume n ≥ 4.

Let T be an H-cross-section of ISn, n ≥ 4. Define the binary relation≺=≺T on N as follows: a ≺ b if and only if there exists an element α ∈ Tof rank two such that α(1) = a and α(2) = b. Note that for a �= b we alwayshave that exactly one of the inequalities a ≺ b or b ≺ a holds as therealways exists a unique α ∈ T of rank two such that α({1, 2}) = {a, b}. Inother words, ≺ is antisymmetric.

Lemma 12.3.4 T preserves ≺ in the sense that for any a ≺ b and anyβ ∈ T such that {a, b} ⊂ dom(β) we have β(a) ≺ β(b).

Proof. Let α ∈ T be an element of rank two such that α(1) = a and α(2) = b.Then im(α) ⊂ dom(β), which implies that βα has rank two as well. Wehave βα(1) = β(α(1)) = β(a) and βα(2) = β(α(2)) = β(b). This impliesβ(a) ≺ β(b) by definition.

Lemma 12.3.5 The relation ≺ is transitive in the sense that a ≺ b andb ≺ c implies a ≺ c.

Proof. Let a, b, c ∈ N be such that a ≺ b and b ≺ c. Assume that c ≺ a.As n ≥ 4, we can take some d ∈ N\{a, b, c}. Let α ∈ T be such thatα({a, b, c}) = {b, c, d}.

If α(a) = d, then α({b, c}) = {b, c} and as α preserves ≺ by Lemma 12.3.5,we have α(b) = b and α(c) = c. This implies c ≺ d.

If α(a) = c, then α(b) = b is impossible as α must preserve ≺. Henceα(c) = b and α(b) = d. As α preserves ≺, we again obtain c ≺ d. Analogouslyone shows that c ≺ d in the last case α(a) = b.

On the other hand, let β ∈ T be such that β({a, b, c}) = {a, c, d}. Ifβ(a) = d, then β(b) = c and β(c) = a as α preserves ≺. This implies d ≺ c.Analogous arguments lead to the same conclusion in the case β(a) = a andβ(a) = c. Thus d ≺ c, a contradiction.


Therefore, c ≺ a is not possible. As we already know that either c ≺ aor a ≺ c, we conclude that a ≺ c, which completes our proof.

The relation ≺ is antireflexive by definition. We already observed that≺ is also antisymmetric. By Lemma 12.3.5, it is transitive. Hence ≺ is apartial order. As we already saw, any two elements of N are comparablewith respect to ≺. This implies that ≺ is a linear order. Thus we haveT ⊂ IO≺

n by Lemma 12.3.4. On the other hand, IO≺n is an H-cross-section

of ISn by (i). Hence T and IO≺n have the same cardinalities (the total

number of H-classes in ISn) and we obtain T = IO≺n . This proves (iii) and

completes the proof of the theorem.

Corollary 12.3.6 (i) For n �= 1, 3 the semigroup ISn contains exactly n!2

cross-sections with respect to the relation H.

(ii) An element α ∈ ISn belongs to some H-cross-section if and only if αdoes not have cycles of length greater than one.

Proof. By Theorem 12.3.3(iii), each H-cross-section of ISn, n �= 3, is of theform IO≺

n . For n �= 1, the n! linear orders ≺ on N are divided into n!/2 pairsof the type {≺,

←≺}. Now the statement (i) follows from Theorem 12.3.3(ii).

If α ∈ ISn has a cycle of length greater than one, then it is obvious thatα cannot preserve any linear order on N. On the other hand, if α does nothave such cycles, we can write

α = (a1) · · · (ak)[b1, b2, . . . , bl] · · ·

and α preserves the order a1 ≺ · · · ≺ ak ≺ b1 ≺ b2 ≺ · · · ≺ bl ≺ · · · . Thiscompletes the proof.

Exercise 12.3.7 Prove that for n �= 3 all H-cross-sections of ISn are iso-morphic.

12.4 L-Cross-Sections in T n and PT n

Let S denote one of the semigroups Tn or PT n. Fix a linear order ≺ on N.Let

N = {u1 ≺ u2 ≺ · · · ≺ un}. (12.1)

For a nonempty A ⊂ N denote by min(A,≺) the minimal element of A withrespect to the order ≺. If A, B ⊂ N are nonempty and disjoint, we will writeA ≺ B provided that min(A,≺) ≺ min(B,≺). Let L(≺) denote the set ofall elements of S, which have the form

α =(

A1 A2 · · · Ak

u1 u2 · · · uk

), (12.2)

where A1 ≺ A2 ≺ · · · ≺ Ak.

12.4. L-CROSS-SECTIONS IN Tn AND PT n 223

Theorem 12.4.1 (i) For every linear order ≺ on N the set L(≺) is anL-cross-section of S.

(ii) For two orders ≺1 and ≺2 we have L(≺1) = L(≺2) if and only if≺1=≺2.

(iii) Every L-cross-section of S has the form L(≺) for some linear order ≺on N.

To prove this theorem we will need several lemmas.

Lemma 12.4.2 For every linear order ≺ on N the set L(≺) is a subsemi-group of S.

Proof. Let α, β ∈ L(≺) be such that the element α is given by (12.2), while

β =(

B1 B2 · · · Bm

u1 u2 · · · um

),

where B1 ≺ B2 ≺ · · · ≺ Bm. We have to prove that αβ ∈ L(≺).Let Ai ≺ Aj and assume that uj ∈ im(αβ). This yields that for some

a ∈ Aj we have a ∈ im(β). As min(Ai,≺) ≺ min(Aj ,≺) ≺ a, we concludethat min(Ai,≺) ∈ im(β) as well. Thus ui ∈ im(αβ). Therefore, im(αβ) ={u1, . . . , ul} for some l ≤ k.

Let ui ≺ uj be elements from im(αβ) and us = min(Ai,≺). Then by thedefinition we have

{x ∈ N : αβ(x) = ui} =⋃

uz∈Ai

Bz (12.3)

and{x ∈ N : αβ(x) = uj} =

⋃uz∈Aj

Bz,

where we assume that Bz = ∅ if z > m. Note that us ≺ v for any v ∈ Aj

because Ai ≺ Ai and α ∈ L(≺). Hence for any uz ∈ Aj we have the inequalitymin(Bs,≺) ≺ min(Bz,≺) because β ∈ L(≺). As the set from (12.3) containsBs, we get

{x ∈ N : αβ(x) = ui} ≺ {x ∈ N : αβ(x) = uj},

which finally yields αβ ∈ L(≺).

Lemma 12.4.3 For every linear order ≺ on N the set L(≺) contains exactlyone element from every L-class of S.


Proof. Let X ⊂ N if S = PT n or X = N if S = Tn. Let further X =X1 ∪ X2 ∪ · · · ∪ Xk be an unordered partition of X into a disjoint unionof nonempty subsets. Without loss of generality we may assume that wehave X1 ≺ X2 ≺ · · · ≺ Xk. By Theorem 4.5.1(ii), such partitions bijectivelycorrespond to L-classes of the semigroup S inside the D-class Dk. Then theelement (

X1 X2 · · · Xk

u1 u2 · · · uk

)

is the unique representative of L(≺) in the L-class, defined by the partitionX1 ∪ X2 ∪ · · · ∪ Xk. This completes the proof.

Lemma 12.4.4 Let T be an L-cross-section of S and α, β ∈ Dk ∩ T . Thenim(α) = im(β).

Proof. Assume that this is not the case. Set

C = im(α) ∩ im(β), A = im(α)\C, B = im(β)\C.

As |im(α)| = |im(β)| = k, we have |A| = |B| = m. Let us write

A = {a1, . . . , am}, B = {b1, . . . , bm}, C = {c1, . . . , ck−m}.

Let γ ∈ T be the element which belongs to the L-class given by the followingpartition of N:

{a1, b1}∪· · ·∪{am, bm}∪{c1}∪· · ·∪{ck−m−1}∪({ck−m}∪(N\(A∪B∪C))

).

We have that all the elements α, β, γ, γα and γβ have rank k. This yields

im(γα) = im(γ) = im(γβ). (12.4)

Moreover, we also have ρα = ργα, implying α = γα as T is an L-cross-section. Analogously one shows that β = γβ. The statement of the lemmanow follows from (12.4).

Lemma 12.4.5 Let T be some L-cross-section of S, α ∈ Dk ∩ T and β ∈Dk+1 ∩ T . Then im(α) ⊂ im(β).

Proof. Let A = im(α) = {a1, . . . , ak} and γ denote the representative of Tin the L-class, which corresponds to the partition

{a1} ∪ · · · ∪ {ak} ∪ (N\A).

Then we have ρα = ργα and thus γα = α since T is an L-cross-section. Atthe same time im(γα) ⊂ im(γ), which yields im(α) ⊂ im(γ). As γ ∈ Dk+1,we have im(γ) = im(β) by Lemma 12.4.4. This completes the proof.

12.4. L-CROSS-SECTIONS IN Tn AND PT n 225

Proof of Theorem 12.4.1. The statement (i) follows from Lemmas 12.4.2and 12.4.3.

To prove (ii) let ≺1 and ≺2 be two different linear orders on N. Assumethat

N = {a1 ≺1 a2 ≺1 · · · ≺1 an},N = {b1 ≺2 b2 ≺2 · · · ≺2 bn},

(12.5)

and that k is minimal such that ak �= bk. Consider some fixed L-class insideDk. Let α and β be the representatives from L(≺1) and L(≺2) inside thisL-class, respectively. Then by the definition we have

im(α) = {a1, . . . , ak} �= {b1, . . . , bk} = im(β).

Hence α �= β and thus L(≺1) �= L(≺2). The statement (ii) is proved.It remains to prove the statement (iii). Let T be an L-cross-section of S.

From Lemma 12.4.5 it follows that there exists a linear order ≺ on N givenby (12.1), such that for any α ∈ T of rank k we have im(α) = {u1, . . . , uk}.We claim that T ⊂ L(≺).

Let α ∈ T be an element of rank k such that

α =(

A1 A2 · · · Ak

v1 v2 · · · vk

),

where A1 ≺ A2 ≺ · · · ≺ Ak and {v1, . . . , vk} = {u1, . . . , uk}. Assume thatα �∈ L(≺) and let m be the minimal possible index such that vm �= um, inparticular um ≺ vm. Let ul = min(Am,≺). Then

ul ≺ x for any x ∈ Aj such that j > m. (12.6)

Let β ∈ T be an element of rank l. Then im(β) = {u1, . . . , ul} by ourassumptions. Consider the element αβ ∈ T . The image of αβ is the union ofall the α(Ai)’s such that there exists 1 ≤ j ≤ l for which uj ∈ Ai. If j < l,we have uj ≺ ul and thus uj ∈ Ai for some i < m by (12.6) and our choiceof ul. At the same time α(Am) = vm ∈ im(αβ). Hence im(αβ) has the form{vm} ∪ N for some subset N ⊂ {u1, . . . , um−1}. As vm �= um, the latter setis not of the form {u1, . . . , ui} for any i. Hence αβ �∈ T by our assumptions,a contradiction. This shows that α ∈ L(≺) and hence T ⊂ L(≺).

As both T and L(≺) are L-cross-sections, they have the same cardinality.Hence T ⊂ L(≺) implies T = L(≺). This completes the proof of (iii) and ofthe theorem.

Corollary 12.4.6 (i) S contains exactly n! different L-cross-sections.

(ii) All L-cross-sections of S are isomorphic.

Proof. The statement (i) follows immediately from Theorem 12.4.1 as thereare exactly n! different linear orders on N.


To prove the statement (ii) assume that ≺1 and ≺2 are two differentlinear orders on N given by (12.5). Let π ∈ Sn be such that π(ai) = bi,i = 1, . . . , n. Then from the definition we have L(≺1) = π−1L(≺2)π. Thisimplies that L(≺2) is mapped to L(≺1) by an inner automorphism of S.Hence these two semigroups are isomorphic and the proof is complete.

12.5 L-Cross-Sections in ISn

In this section we classify all L-cross-sections in ISn. This description turnsout to be more complicated than the descriptions presented in the previoussections. Hence we will start with some preparation and some auxiliarylemmas.

We first note that, by Theorem 4.5.1(ii), for α, β ∈ ISn the conditionαLβ is equivalent to the condition dom(α) = dom(β). Hence the equalitiesα = β and dom(α) = dom(β) are equivalent for elements α, β from anarbitrary L-cross-section T of ISn. We will frequently use this fact in thissection.

Lemma 12.5.1 Let T be an L-cross-section of ISn. Then α �∈ 〈T\{α}〉 forevery α ∈ T ∩ Dn−1.

Proof. Let α ∈ T ∩ Dn−1 and β, γ ∈ T be such that α = βγ. Assume thatβ �= α and γ �= α. Then, in particular, β, γ �= ε. As T is an L-cross-section,we thus get β, γ ∈ In−1. The equality α = βγ then implies that both β and γhave rank (n− 1). Thus dom(α) = dom(βγ) = dom(γ), which yields α = γ,a contradiction. Thus we either have β = α or γ = α, and the statement ofthe lemma follows.

From Lemma 12.5.1 it follows that every generating system of eachL–cross-section T of ISn must contain the set T ∩ Dn−1. Now we wouldlike to study the latter set in more details.

Lemma 12.5.2 Let T be an L-cross-section of ISn and α ∈ T ∩ Dn−1.Then the chain decomposition of α contains exactly one chain, [a1, . . . , ak]say, and α(x) = x for all x ∈ N\{a1, . . . , ak}.

Proof. The element α contains exactly one chain as def(α) = 1. If this chainis [a1, . . . , ak], then dom(αk) = dom(αk+1), and hence αk = αk+1 as T is anL-cross-section. By our assumptions, the element α induces a bijection onthe set N\{a1, . . . , ak}, hence αk induces a bijection on this set as well. Letx, y ∈ N\{a1, . . . , ak} be such that αk(y) = x. Then, using αk = αk+1, wehave

α(x) = α(αk(y)) = αk+1(y) = αk(y) = x,

which completes the proof.

12.5. L-CROSS-SECTIONS IN ISn 227

For k = 1, . . . , n set αk = [1, 2, . . . , k](k + 1) · · · (n) and denote by Kn

the subsemigroup of ISn, generated by α1, . . . , αn.

Exercise 12.5.3 ([GM3]) (a) Show that αk · αl = αl · αk−1 for every l andk such that k ≤ l.

(b) Show that αmk = αm−1

k · αk−1 for all m > 1 and k > 1.

Corollary 12.5.4 Every element α ∈ Kn can be written in the form

α = αann α

an−1

n−1 · · ·αa11 ,

where ai ∈ {0, 1} for all i = 1, 2, . . . , n, and a1 + · · · + an > 0.

Proof. From Exercise 12.5.3(a) it follows easily that every element α ∈ Kn

can be written in the form α = αann α

an−1

n−1 · · ·αa11 , where a1, a2, . . . , an ∈

{0, 1, 2, . . . }. Taking this into account, the statement of the corollary followsfrom Exercise 12.5.3(b) and the observation that α2

1 = α1.

Exercise 12.5.5 ([GM3]) Let 1 ≤ i1 < i2 < · · · < ik ≤ n and defineα = αikαik−1

· · ·αi1 . Show that

(a) dom(α) = N\{i1, . . . , ik}.

(b) im(α) = N\{1, . . . , k}.

(c) α coincides with the unique increasing bijection from N\{i1, . . . , ik} toN\{1, . . . , k}.

Corollary 12.5.6 The semigroup Kn = Kn ∪ {ε} is an L-cross-section ofthe semigroup ISn.

Proof. From Exercise 12.5.5(a) and Corollary 12.5.4 it follows that for ev-ery A ⊂ N the semigroup Kn contains exactly one element α with domain A.Hence Kn contains exactly one element from every L-class of ISn by The-orem 4.5.1.

Lemma 12.5.7 (i) In the semigroup Kn the inequality xn−1 �= xn has theunique solution αn.

(ii) The element αn satisfies αnn = αn+1

n .

Proof. The statement (ii) is proved by a direct calculation.We prove the statement (i) by induction on n. If n = 1, then the state-

ment (i) is obvious. Let A ⊂ N be such that |A| = k and assume thatα : A → {n − k + 1, . . . , n} be an increasing bijection. Note that, by Exer-cise 12.5.5, in this way we get all elements from Kn. If n ∈ A, then, obviously,α(n) = n. Applying the inductive assumption and the statement (ii) to therestriction of α to the invariant subset Nn−1 ⊂ N, we obtain αn−1 = αn.


Assume now that n �∈ A and rank(α) < n−1. Then im(α) = {i, . . . , n} ⊂{3, . . . , n} by Exercise 12.5.5(b). As α is an increasing function, im(α) isinvariant with respect to α. Let β be the restriction of α to its image.Because of Exercise 12.5.5, we can apply the inductive assumption and thestatement (ii) to β and get βn−2 = βn−1. Multiplying with α from the rightyields αn−1 = αn.

If n �∈ A and rank(α) = n− 1, then dom(α) = Nn−1 and α = αn followsfrom Corollary 12.5.6. The inequality αn−1

n �= αnn is checked by a direct

calculation. This completes the proof.

Lemma 12.5.8 Let T be an L-cross-section of ISn. Assume that T containsthe element α = αk. Let β ∈ T ∩ Dn−1 be such that β(x) = x for somex ∈ {1, . . . , k}. Then β(y) = y for all y, x ≤ y ≤ k.

Proof. Without loss of generality we may assume that x is the minimalelement such that β(x) = x. Consider the set

M = {y : x ≤ y ≤ kandβ(y) �= y}

and assume that it is not empty. Let p be the maximal element in this set.Assume first that p �∈ dom(β). Then dom(β) = N\{p}, and, taking into

account β(z) = z for all p < z ≤ k, we obtain

dom(αk−pβ) = dom(αk−p+1) = N\{p, p + 1, . . . , k}.

Thus αk−pβ = αk−p+1. At the same time

αk−pβ(x) = x + k − p �= x + k − p + 1 = αk−p+1(x).

Therefore, p �∈ dom(β) is not possible.Assume now that p ∈ dom(β). Set β(p) = q �= p and assume that

the length of the unique chain in the chain decomposition of β equals m.Since β(p) �= p, from Lemma 12.5.2 we deduce that p belongs to the chainof β. This implies that for the idempotent βm we have p, q �∈ dom(βm). SetA = {k + 1, . . . , n}, B = dom(βm), C = A ∩ B and A1 = A\C.

All elements from the set A1 occur in the chain of β and hence we haveim(βm) ∩ A1 = ∅. Moreover, because of our choice of x we have that allelements from {1, 2, . . . , x− 1} belong to the chain of the element β as well.Thus im(βm) ⊂ {x, x + 1, . . . , k} ∪ C. This implies that

im(αp−xβm) ⊂ {p, p + 1, . . . , k} ∪ C ⊂ dom(β).

The latter yields dom(αp−xβm) = dom(βαp−xβm) and hence αp−xβm =βαp−xβm as T is an L-cross-section. But

(αp−xβm)(x) = αp−x(βm(x)) = αp−x(x) = p �= q = β(p) = (βαp−xβm)(x).

Hence p ∈ dom(β) is not possible either. Thus, we obtain a contradictionwith our assumption that M is not empty, and the statement of the lemmafollows.

12.5. L-CROSS-SECTIONS IN ISn 229

Let [a1, a2, . . . , ak] be some chain. For every l ∈ {1, 2, . . . , k} the chain[a1, a2, . . . , al] will be called a prefix of [a1, a2, . . . , ak].

Lemma 12.5.9 Let T be an L-cross-section of ISn and α, β ∈ T ∩ Dn−1.Assume that the chains of α and β have at least one common element. Thenone of these chains is a prefix of the other one.

Proof. Without loss of generality we can assume that α = αk. Let A ={1, 2, . . . , k}, B be the set of all elements from the chain of the element βand C = A∩B. From Lemma 12.5.8 it follows that C = {1, 2, . . . , m}, wherem ≤ k.

Assume that x ∈ N\A is such that x �= βa(x) = y ∈ C for some a. Thenfrom α(x) = x and Lemma 12.5.8 we obtain that α(y) = y, which contradictsthe fact that y ∈ C. Hence β = [i1, i2, . . . , im, j1, . . . , jl](f) · · · (g), wherei1, i2, . . . , im is a permutation of 1, 2, . . . , m.

Suppose that (i1, i2, . . . , im) �= (1, 2, . . . , m) (this is an inequality of vec-tors, not cycles). Then there exist elements u, v, p, q ∈ {1, 2, . . . , m} suchthat ip = u, iq = v, u < v, p > q. This implies αv−u(u) = v, βp−q(v) = uand (βp−qαv−u)(u) = u. For the element γ = βp−qαv−u there exists t suchthat γt is an idempotent. Clearly, u ∈ dom(γt) and γt(u) = u. Moreover,dom(γt) ⊂ dom(γ) ⊂ dom(α). Now γ2t = γt implies that dom(γt) = im(γt).Hence dom(γt) = dom(αγt) and thus γt = αγt. But the last equality isimpossible for γt(u) = u �= u + 1 = (αγt)(u).

From the previous paragraph we have (i1, i2, . . . , im) = (1, 2, . . . , m) andthus β = [1, 2, . . . , m, j1, . . . , jl](f) · · · (g). To complete the proof it is nowenough to show that either k = m or l = 0. Assume that this is not thecase, that is, k > m and l > 0. Note that {m + 1, . . . , k} ∩ {j1, . . . , jl} = ∅,α acts as the identity on all elements from B\C = {j1, . . . , jl} and β acts asthe identity on all elements from A\C = {m + 1, . . . , k}. This implies thatk, jl �∈ dom(βα) and k, jl �∈ dom(αβ). Since def(βα) ≤ 2 and def(αβ) ≤ 2,we have that dom(βα) = dom(αβ) = N\{k, jl} and αβ = βα as T is anL-cross-section. But (βα)(m) = m + 1 �= j1 = (αβ)(m). This contradictioncompletes the proof of the lemma.

Let now N = M1 ∪M2 ∪ · · · ∪Mk be an arbitrary partition of N into anunordered disjoint union of nonempty blocks. For i = 1, 2, . . . , k we set mi =|Mi| and choose some linear order ≺i on Mi. The collection {≺1, . . . ,≺k}uniquely determines the partition M1 ∪ M2 ∪ · · · ∪ Mk. Assume that

Mi = {ui1 ≺i ui

2 ≺i · · · ≺i uimi

}.

For every i ∈ {1, . . . , k} and every j ∈ {1, . . . , mi} let αij denote the unique

element of rank (n − 1), which can be written as follows:

αij = [ui

1, ui2, . . . , u

ij ](a)(b) · · · (c).


Let L(≺1, . . . ,≺k) denote the subsemigroup of ISn, which is generated byε and all αi

j , i ∈ {1, . . . , k}, j ∈ {1, . . . , mi}.

Theorem 12.5.10 (i) For any partition N = M1∪M2∪· · ·∪Mk and anychoice of a linear order ≺i for each block of the partition the semigroupL(≺1, . . . ,≺k) is an L-cross-section of ISn.

(ii) L(≺1, . . . ,≺k) = L(≺′1, . . . ,≺′

m) if and only if we have the equality{≺1, . . . ,≺k} = {≺′

1, . . . ,≺′m}.

(iii) Every L-cross-section of ISn has the form L(≺1, . . . ,≺k) for appro-priate partition N = M1∪M2∪· · ·∪Mk and some choice ≺i of a linearorder for every block of the partition.

Proof. Set L = L(≺1, . . . ,≺k). For X ⊂ N and i = 1, 2, . . . , k set Xi =X ∩Mi and denote by Li the subsemigroup of L, generated by ε and all theαi

j , j ∈ {1, . . . , mi}. Then for all α ∈ Li and x �∈ Mi we have α(x) = x, so wecan consider Li as a subsemigroup of IS(Mi). Let us identify Mi with Nmi

such that the order ≺i corresponds to the natural order on Nmi . Then thesemigroup Li is identified with the semigroup Kmi from Corollary 12.5.6.The latter corollary also says that Li is an L-cross-section of Nmi . Thismeans that there exists a unique element βi ∈ Li such that dom(βi) = N\Xi.

From the definition we have that every element from Li commutes withevery element from Lj if i �= j. As every βi acts as the identity outside Xi,we conclude that

dom(β1β2 · · ·βk) =k⋂

i=1

(N\Xi) = N\X.

Hence L contains a representative in every L-class of ISn.From Corollary 12.5.6 we have |Li| = 2mi . As every element from Li

commutes with every element from Lj if i �= j, from the definition we havethat every β ∈ L can be written as the product β = β1β2 · · ·βk, whereβi ∈ Li for every i. Hence

|L| ≤ |L1| · |L2| · · · |Lk| = 2m1 · 2m2 · · · · · 2mk = 2m1+m2+···+mk = 2n.

As L contains a representative in every L-class of ISn, we have |L| ≥ 2n.This yields |L| = 2n and thus L must contain a unique representative inevery L-class of ISn. This proves the statement (i).

To prove the statement (ii) we consider two collections ≺= {≺1, . . . ,≺k}and ≺′= {≺′

1, . . . ,≺′k}. Set L = L(≺1, . . . ,≺k) and L′ = L(≺′

1, . . . ,≺′m). For

every x ∈ N consider the orbit of x with respect to L:

o≺(x) = {α(x) : α ∈ Landx ∈ dom(α)}.

12.6. R-CROSS-SECTIONS IN ISn 231

It is easy to see that the blocks Mi’s are maximal orbits with respect to in-clusions. Hence the equality L = L′ implies k = m and uniquely determinesthe partition N = M1∪· · ·∪Mk. Let Li and L′

i denote the subsemigroups ofL and L′, defined as in the first part of the proof. By Lemma 12.5.7(i), the in-equality xmi−1 �= xmi has unique solutions in both L and L′, which must co-incide provided that L = L′. The solution has the form [ui

1, . . . , uimi

](a) · · · (b)and uniquely determines the order ≺i. Hence ≺i=≺′

i and the statement (ii)follows.

It remains to prove the statement (iii). Let L be an L-cross-section ofISn. By Lemma 12.5.9 the chains of two arbitrarily chosen elements fromL∩Dn−1 are either disjoint or one of these chains is a prefix of the other one.The chains of the elements from L ∩ Dn−1, which are not proper prefixesfor chains of any other element from L ∩ Dn−1, define a disjoint familyM1, . . . , Mk of subsets of N. Moreover, each such maximal chain defines anatural linear order on the corresponding Mi in the following way: the chain[x1, . . . , xm] defines Mi = {x1, . . . , xm} with the order x1 ≺i x2 ≺i · · · ≺i xm.For every x ∈ N there exists α ∈ L ∩ Dn−1 such that dom(α) = N\{x}.This means that x belongs to the unique chain of α and hence x belongs tosome Mi. Therefore L defines a partition N = M1 ∪ · · · ∪ Mk.

From the construction of this partition and Lemma 12.5.9 it follows that

L ∩ Dn−1 = L(≺1, . . . ,≺k) ∩ Dn−1

and hence L contains the subsemigroup of L(≺1, . . . ,≺k), generated by allelements of rank n and (n − 1). However, L(≺1, . . . ,≺k) is generated byelements of rank n and (n − 1) by definition. Hence L(≺1, . . . ,≺k) ⊂ L.

Recall that L(≺1, . . . ,≺k) is an L-cross-section of ISn by (i), while L isan L-cross-section of ISn by our assumption. Hence |L(≺1, . . . ,≺k)| = |L|,implying L(≺1, . . . ,≺k) = L. This completes the proof of the theorem.

12.6 R-Cross-Sections in ISn

Let N = M1∪M2∪· · ·∪Mk be an arbitrary partition of N into an unordereddisjoint union of nonempty blocks. For i = 1, . . . , k set mi = |Mi| and choosesome linear order ≺i on Mi. Assume that

Mi = {ui1 ≺i ui

2 ≺i · · · ≺i uimi

}.

For every i ∈ {1, . . . , k} and every j ∈ {1, . . . , mi} let βij denote the unique

element of rank (n − 1), which can be written as follows:

βij = [ui

j , uij+1, . . . , u

imi

](a)(b) · · · (c).

Let R(≺1, . . . ,≺k) denote the subsemigroup of ISn, which is generated byε and all βi

j , i ∈ {1, . . . , k}, j ∈ {1, . . . , mi}.


Theorem 12.6.1 (i) For any partition N = M1 ∪M2 ∪ · · · ∪Mk and anychoice of a linear order ≺i for each block of the partition the semigroupR(≺1, . . . ,≺k) is an R-cross-section of ISn.

(ii) R(≺1, . . . ,≺k) = R(≺′1, . . . ,≺′

m) if and only if we have the equality{≺1, . . . ,≺k} = {≺′

1, . . . ,≺′m}.

(iii) Every R-cross-section of ISn has the form R(≺1, . . . ,≺k) for appro-priate partition N = M1∪M2∪· · ·∪Mk and some choice ≺i of a linearorder for every block of the partition.

Proof. The mapping α → α−1 is an antiinvolution on ISn. It swaps L- andR-classes of ISn. It also swaps formulations of Theorems 12.5.10 and 12.6.1.Hence Theorem 12.6.1 follows from Theorem 12.5.10.

Corollary 12.6.2 The semigroup ISn contains exactlyn∑

k=1

n!k!

(n − 1k − 1

)dif-

ferent L-cross-sections, and the same number of different R-cross-sections.

Proof. It is enough to prove the statement for L-cross-sections. Let us countthe number of L-cross-sections, which correspond to partitions of N into kblocks.

For this fixed k consider an arbitrary permutation i1, . . . , in of 1, 2, . . . , n,and a (k−1)-element subset, {j1, . . . , jk−1}, of {1, 2, . . . , n−1}. Assume thatj1 < j2 < · · · < jk−1. This defines a decomposition of N into k blocks M1 ={i1, . . . , ij1}, M2 = {ij1+1, . . . , ij2},. . . , Mk = {ijk−1+1, . . . , in} together withlinear orders on these blocks. Since the order of the blocks is not importantfor L(≺1, . . . ,≺k) we get that each L-cross-section is counted exactly k!times. Therefore we have exactly n!

(n−1k−1

)1k! different L-cross-sections for our

fixed k. Summing this over all k we get the necessary statement.

Corollary 12.6.3 The only L-cross-section of ISn, which is an R-cross-section at the same time, is the semigroup E(ISn) of all idempotents ofISn.

Proof. Let L = L(≺1, . . . ,≺k) be an L-cross-section and N = M1 ∪ · · ·∪Mk

be the corresponding partition of N.Assume that mi > 1 for some i. Let X and Y be two different subsets

of Mi of the same cardinality. Set X ′ = N\X and Y ′ = N\Y . Let α, β ∈ Lbe such that dom(α) = X ′ and dom(β) = Y ′. From |Mi\X| = |Mi\Y |and Exercise 12.5.5 we obtain im(α) = im(β). This means that L containstwo different elements from the same R-class and hence L cannot be anR-cross-section.

On the other hand, if mi = 1 for all i, we obtain that k = n, all ≺i’s aretrivial and αi

1 = εN\Mi. Hence L = E(ISn). As E(ISn) is stable under the

mapping α → α−1, E(ISn) is an R-cross-section as well. This completes theproof.



12.7.1 Propositions 12.2.2–12.2.4 are proved in [ST1], [ST2], and [ST3],respectively. Proposition 12.2.3 can also be found in [Pek1]. Theorems 12.3.1and 12.4.1 are taken from [Pek2]. Theorem 12.3.3 is proved in [CR]. Theresults of Sects. 12.5 and 12.6 are obtained in [GM3]. These results were alsoindependently obtained in [YY] by completely different methods.

12.7.2 Cross-sections of other semigroups, in particular some infinite trans-formation semigroups, were studied in [Pek3, Pek4, KMM].

12.7.3 If G is an abelian group and ρ is the conjugacy relation on G, then Gcontains a unique ρ-cross-section, namely, G itself. In the non-abelian casewe have the following:

Proposition 12.7.1 Let G be a finite non-abelian group and ρ be the con-jugacy relation on G. Then G does not contain any ρ-cross-section.

Proof. Assume that T is a ρ-cross-section. Let H be a conjugacy class in Gof maximal cardinality k. Note that k > 1 as G is not abelian. At the sametime G contains a one-element conjugacy class consisting of the identityelement. Since T is a ρ-cross-section, we get

|T | >|G||H| . (12.7)

The group G acts on H by conjugation. Let a ∈ H ∩ T . As the action ofG on H is transitive, from Theorem 10.2.4(i) we have |G| = |H| · |StG(a)|.In particular, |StG(a)| = |G|/|H|. Now we claim that T ⊂ StG(a). Indeed,if g ∈ T , then g−1ag ∈ H ∩ T and thus g−1ag = a as T is a ρ-cross-section.This yields g ∈ StG(a). Hence

|T | ≤ |StG(a)| =|G||H| . (12.8)

Comparing (12.7) and (12.8) we get a contradiction. This proves the state-ment of our proposition.

After Proposition 12.7.1 a natural question is: Does Proposition 12.7.1generalize to infinite groups?

12.7.4 Description of D-cross-section for semigroups Tn, PT n and ISn

seems to be a very difficult problem, even if one considers some specialclasses of cross-sections.

For example, consider the easiest case of the semigroup ISn. A natu-ral problem is to try to classify all D-cross-sections of ISn, consisting ofidempotents. As E(ISn) ∼= (B(N),∩), the problem can be reformulated as


follows: Classify all collections (X0, X1, . . . , Xn) of subsets of N such thatthe following two conditions are satisfied:

• |Xi| = i for every i = 0, 1, . . . , n.

• For any i, j ∈ {0, 1, . . . , n} there exists l ∈ {0, 1, . . . , n} such thatXi ∩ Xj = Xl.

We do not know how to solve this problem (which appears already in [GM3]).We even do not know how to count the number of such collections. Classi-fication of D-cross-section for both Tn and PT n contains this problem as asubproblem. For PT n the latter statement is obvious as ISn ⊂ PT n andthis inclusion induces a bijection on D-classes. For Tn it follows from theembedding of ISn into Tn+1 via

(a1 a2 · · · ak

b1 b2 · · · bk

)→

(a1 a2 · · · ak Nn+1\{a1, . . . , ak}b1 b2 · · · bk n + 1

).

12.7.5 At the moment there is no description/classification of R-cross-sections for semigroups Tn and PT n. As the inclusion Tn ⊂ PT n induces abijection on R-classes, the problem of description of R-cross-sections for Tn

is a subproblem of the corresponding problem for PT n. From the inclusionISn ⊂ PT n we have that every R-cross-section of ISn is an R-cross-sectionof PT n as well.

A natural family of R-cross-sections for Tn is the following: For anonempty subset X ⊂ N such that X = {i1 < i2 < · · · < ik} define theelement

γX =(

1 2 · · · k − 2 k − 1 k k + 1 · · · ni1 i2 · · · ik−2 ik−1 ik ik · · · ik

).

Let R = {γX : ∅ �= X ⊂ N}.

Lemma 12.7.2 R is an R-cross-section for Tn.

Proof. As im(γX) = X, the set R contains exactly one representative fromeach R-class of Tn. A direct calculation shows that R is closed under themultiplication. Hence R is an R-cross-section for Tn.

Conjugating R with elements from Sn one obtains n! different R-cross-sections for Tn. However, Tn contains many other R-cross-sections, see forexample Exercise 12.8.10. We suspect that R is the only R-cross-sectionfor Tn, which contains the element γ{2,3,...,n}, and that this fact could beused to classify R-cross-sections for Tn along the arguments, dual to thoseof Sect. 12.5.



12.8.1 Prove that the only element x ∈ PT n which satisfies πx = xπ = xfor all π ∈ Sn is the element 0.

12.8.2 Prove the statement of Proposition 12.2.2(vi).

12.8.3 Prove Proposition 12.2.3.

12.8.4 Prove Proposition 12.2.4.

12.8.5 ([CR]) Show that the set

I1 ∪ {ε, ε{1,2}, ε{1,3}, ε{2,3}, [2, 1, 3], [1, 2, 3], [3, 1, 2], [1, 3, 2], [3, 2, 1], [2, 3, 1]}

is an H-cross-section of IS3, which is not equal to πIO3π−1 for any π ∈ S3.

12.8.6 Describe all H-cross-sections of IS3.

12.8.7 ([GM3]) Show that the semigroup Kn has the following presentation:

Kn = 〈α1, . . . , αn|α21 = α1; α2

k = αkαk−1, k = 2, . . . , n;αkαl = αlαk−1, 1 ≤ k < l ≤ n〉.

12.8.8 Let S be a semigroup and T be an L-cross-section of S. Assume thatT contains some element b. Show that T contains all idempotents e ∈ E(S)such that bRe.

12.8.9 Classify all L-cross-sections of a rectangular band.

12.8.10 Show that the following set is an R-cross-section of T4, which isnot conjugate to the R-cross-section R from Lemma 12.7.2:

{ε, ε1,2, ε2,1, ε3,4, ε4,3, ε1,2ε3,4, ε1,2ε4,3, ε2,1ε3,4, ε2,1ε4,3,

(2, 3)ε1,2ε3,4, (2, 3)ε2,1ε4,3, 01, 02, 03, 04, }.

12.8.11 ([GM3]) Let n > 3 and assume that the unordered partition N =M1 ∪ · · · ∪ Mk contains exactly m blocks of cardinality at least two. Showthat for arbitrary linear orders ≺i on Mi, i = 1, . . . , k, the L-cross-sectionL(≺1, . . . ,≺k) of ISn is contained in exactly k! · 2m−1 different H-cross-sections.

12.8.12 ([GM3]) Show that for n > 3 every H-cross-section of ISn containsexactly

1 +�n

2 �∑l=1

(2l ·

n−2l∑m=0

(m + l

l

)(n − m − l − 1

l − 1

))

different L-cross-sections (or R-cross-sections).


12.8.13 ([GM3]) Show that the L-cross-sections

L(≺1, . . . ,≺k) and L(≺′1, . . . ,≺′

k),

corresponding to some choices of linear orders on blocks of the partitionsN = M1 ∪ · · · ∪ Mk and N = M ′

1 ∪ · · · ∪ M ′k, respectively, are isomorphic if

there exists a bijection α ∈ Sk such that |Mi| = |M ′α(i)|.

12.8.14 ([CR]) Let n > 3. Show that an element α ∈ ISn is contained in aunique H-cross-section if and only if α is a nilpotent element of nilpotencydegree n.

12.8.15 Construct a natural bijection between the set of all nilpotent ele-ments of ISn and the set of all L-cross-sections of ISn.

Chapter 13

Variants

13.1 Variants of Semigroups

Let S = (S, ·) be a semigroup and a ∈ S. For x, y ∈ S set x◦ay = x·a·y. Then◦a is an associative binary operation on S and hence (S, ◦a) is a semigroup.The semigroup (S, ◦a) is called a variant of S, or, alternatively, the sandwichsemigroup of S with respect to the sandwich element a. The operation ◦a isusually called the sandwich operation. To simplify notation we will denote(S, ◦a) simply by Sa. If a is the identity element of S, then we obviouslyhave ◦a = ·. However, for other a the sandwich operations normally give riseto new semigroups:

Proposition 13.1.1 (i) We have Sa ∼= S provided that a ∈ S∗.

(ii) If S is a monoid, then Sa ∼= S if and only if a ∈ S∗.

Proof. To prove the statement (i) we consider the mapping ϕ : S → S,defined as follows: x → xa−1. For x, y ∈ S we have

ϕ(xy) = (xy)a−1

= xa−1axa−1

= ϕ(x)aϕ(y)= ϕ(x) ◦a ϕ(y).

Hence ϕ : S → Sa is a homomorphism. Since a is invertible, the mappingy → ya is inverse to ϕ. Therefore, ϕ is bijective and hence an isomorphism.This proves the statement (i).

To prove (ii) we assume that Sa ∼= S. In particular, Sa is a monoid aswell. Let b ∈ Sa be the unit element of this monoid. Then for any s ∈ Sa wehave s ◦a b = b ◦a s = s, which is equivalent to s(ab) = s and (ba)s = s. Thefirst equality means that ab is a right identity in S, and the second equalitymeans that ba is a left identity in S. As S is a monoid with the identity



238 CHAPTER 13. VARIANTS

element 1, we have that ab = ba = 1. This implies that a ∈ S∗ and provesthe statement (ii).

Corollary 13.1.2 Let G be a group. Then every variant of G is isomorphicto G.

Corollary 13.1.2 says that study of variants is a purely semigroup theo-retic phenomenon, which simply does not appear in group theory. The nextnatural question to ask would be: How many nonisomorphic semigroups canone obtain as variants of a given semigroup? Later on we will answer thisquestion for the semigroups Tn, PT n, and ISn completely. For now we justgive the following general sufficient condition:

Proposition 13.1.3 Let S be a monoid, a ∈ S and u, v ∈ S∗. Then Sa ∼=Suav.

Proof. Consider the mapping ϕ : S → S, x → v−1xu−1. As u, v ∈ S∗, themapping ϕ is bijective (with the inverse y → vyu). For x, y ∈ S we have

ϕ(x ◦a y) = ϕ(xay)= v−1xayu−1

= (v−1xu−1)(uav)(v−1yu−1)= ϕ(x) ◦uav ϕ(y).

This shows that ϕ is a homomorphism, and hence an isomorphism. Thestatement of the proposition follows.

Proposition 13.1.4 Let S be a semigroup and a ∈ S. Then (Sa)2 = SaS.

Proof. For x, y ∈ Sa we have x ◦a y = xay ∈ SaS. Conversely, for anyxay ∈ SaS we have xay = x ◦a y. The statement follows.

A variant of a semigroup is usually a special construction over somesubsemigroup, called inflation. This point of view gives some feeling aboutthe structure of a variant. To introduce it we need some preparation.

Proposition 13.1.5 Let S be a semigroup and a ∈ S.

(i) The mapping ϕl : S → aS, x → ax, is an epimorphism from Sa to aS.

(ii) The mapping ϕr : S → Sa, x → xa, is an epimorphism from Sa to Sa.

(iii) Assume that a ∈ E(S), then (aS, ◦a) = aS (note the equality and notthe isomorphism sign!).

(iv) Assume that a ∈ E(S), then (Sa, ◦a) = Sa (note the equality and notthe isomorphism sign!).

13.1. VARIANTS OF SEMIGROUPS 239

Proof. The mapping ϕl is surjective by definition. For x, y ∈ S we have

ϕl(x ◦a y) = ϕl(xay)= axay

= ϕl(x)ϕl(y)

and the statement (i) follows. The statement (ii) is proved similarly.If a ∈ E(S), then for any x = ay ∈ aS we have

ax = a(ay) = a2y = ay = x

and hence the restriction of the mapping ϕl from (i) to aS is the identitymapping. Now the statement (iii) follows from the statement (i). Analo-gously, the statement (iv) follows from the statement (ii).

A semigroup S is called an inflation of its subsemigroup T providedthat there exists an epimorphism θ : S → T such that the following twoconditions are satisfied:

• θ2 = θ

• θ(x)θ(y) = xy for all x, y ∈ S

Example 13.1.6 A semigroup with zero multiplication is an inflation ofits subsemigroup containing the zero element via the obvious projectionmapping.

Using Proposition 13.1.5 we define the equivalence relation h on S asfollows:

h =(Ker(ϕl) ∩ Ker(ϕr) ∩

((Sa\(Sa)2) × (Sa\(Sa)2)

))∪

∪ {(a, a) : a ∈ (Sa)2}.

Proposition 13.1.7 Let S be a semigroup and a ∈ S.

(i) h is a congruence on Sa.

(ii) Any cross-section T of h is a subsemigroup of Sa, moreover, Sa is aninflation of T via the projection mapping πT , defined as follows:

πT (x) = t ∈ T if and only if(x, t) ∈ h.


Proof. Let x, y ∈ Sa\(Sa)2 be such that (x, y) ∈ h and s ∈ Sa. Then (x, y) ∈Ker(ϕl) by definition and hence

s ◦a x = sax = say = s ◦a y.

Thus (s ◦a x, s ◦a y) ∈ h. Analogously, one shows that (x ◦a s, x ◦a s) ∈ h.The statement (i) follows.

To prove (ii) we first observe that, by definition, (Sa)2 ⊂ T . If x, y ∈ T ,then x ◦a y ∈ (Sa)2 and hence x ◦a y ∈ T , in particular, T is a subsemigroupof Sa. If x, y ∈ Sa, we have

πT (x) ◦a πT (y) = πT (x)aπT (y)(as (x, πT (x)) ∈ Ker(ϕr)) = xaπT (y)(as (y, πT (y)) ∈ Ker(ϕl)) = xay

= x ◦a y.

At the same time πT (x ◦a y) = x ◦a y as x ◦a y ∈ (Sa)2. Hence πT is anidempotent epimorphism. This proves the statement (ii).

13.2 Classification of Variants for ISn , T n ,and PT n

In this section, we classify variants of ISn, Tn, and PT n up to isomorphism.Our main result in this section is the following theorem:

Theorem 13.2.1 Let S denote one of the semigroups ISn, Tn, or PT n,and α, β ∈ S. Then the following statements are equivalent:

(a) Sα ∼= Sβ.

(b) There exist τ, σ ∈ Sn such that τασ = β.

(c) t(α) = t(β).

Proof. We first observe that the equivalence (b)⇔(c) is just Exercises 2.10.24and 2.10.25. The implication (b)⇒(a) follows immediately from Proposition13.1.3. Hence we are left to prove either the implication (a)⇒(b) or theimplication (a)⇒(c). We will do this for each of the semigroups separately.We start with the easiest case S = ISn.

Let S = ISn and α, β ∈ ISn be such that ISαn∼= ISβ

n. Then the sets(ISα

n)2 and (ISβn)2 must have the same cardinality. As ISn is a monoid, from

Proposition 13.1.4 we obtain that the principal two-sided ideals, generatedby α and β, respectively, must have the same cardinality. From Theorem4.2.8 we therefore obtain rank(α) = rank(β). Further, Theorem 4.5.1(iv)implies αDβ. Finally, Proposition 4.5.4(ii) implies that the condition (b) issatisfied and we obtain (a)⇒(b). This completes the proof.

13.2. CLASSIFICATION OF VARIANTS FOR ISn, Tn, AND PT n 241

We proceed now with the case S = Tn. Let α ∈ Tn. We will need somemore detailed information about T α

n .

Lemma 13.2.2 The set I1 is the minimum two-sided ideal of T αn in the

sense that every two-sided ideal of T αn contains I1.

Proof. The fact that I1 is an ideal is obvious. Let I be an ideal of T αn and

β ∈ I. Then 0i = 0i ◦α β ∈ I for any i ∈ N and the statement follows.

Lemma 13.2.2 says that T αn contains the unique minimum two-sided

ideal. Define the relation ∼α on I1 as follows: for i, j ∈ N set 0i ∼α 0j ifand only if β ◦α 0i = β ◦α 0j for any β ∈ T α

n . Assume that

α =(

A1 A2 · · · Ak

a1 a2 · · · ak

)

is an element of rank k.

Lemma 13.2.3 For i, j ∈ N we have 0i ∼α 0j if and only if there existsl ∈ {1, . . . , k} such that i, j ∈ Al.

Proof. If i, j ∈ Al, then α0i = α0j = 0aland thus β ◦α 0i = β ◦α 0j = β(al).

On the other hand, if i ∈ Al and j ∈ Am for some l �= m, then

ε ◦α 0i = 0al�= 0am = ε ◦α 0j .

Now let α, β ∈ Tn and assume that ψ : T αn → T β

n is an isomorphism.From Lemma 13.2.2 it follows that ψ induces a bijection from I1 ⊂ T α

n toI1 ⊂ T β

n .

Lemma 13.2.4 For γ1, γ2 ∈ I1 ⊂ T αn we have

γ1 ∼α γ2 if and only if ψ(γ1) ∼β ψ(γ2).

Proof. If γ1 ∼α γ2 and σ ∈ T αn is arbitrary, then

ψ(σ) ◦β ψ(γ1) = ψ(σ ◦α γ1)= ψ(σ ◦α γ2)= ψ(σ) ◦β ψ(γ2).

As ψ(σ) is then an arbitrary element of T βn , we get ψ(γ1) ∼β ψ(γ2). The

statement of the lemma now follows for the same argument applies to theisomorphism ψ−1.


By Lemma 13.2.4 the restriction of ψ to I1 induces an isomorphism be-tween the equivalence relations ∼α and ∼β . In particular, for every i ∈ Nthese relations have the same number of equivalence classes of cardinality i.By Lemma 13.2.3 and the definition, we obtain that t(α) = t(β). This com-pletes the proof.

Finally, in the case S = PT n the proof is rather similar to that from thecase S = Tn. First, we observe that the semigroup PT 0

n is the semigroup withzero multiplication. At the same time for any element α �= 0 let i ∈ dom(α).Then

0i ◦α 0i = 0iα0i = 0i �= 0

and hence PT αn �∼= PT 0

n. So, it is enough to consider the case when α, β �= 0.For α �= 0 in the same way as in Lemma 13.2.2 one shows that I1 is the

unique minimum element in the set of all nonzero ideals contained in the set(PT α

n

)2. As the zero element is obviously preserved by any isomorphism, wederive that any isomorphism from PT α

n to PT βn (where both α and β are

nonzero) must leave the set I1 invariant. The rest is absolutely analogous tothe case of Tn. On I1 we consider the relations ∼α and ∼β and it follows thatany isomorphism induces an isomorphism of these relations. A computationof the sizes of the equivalence classes implies t(α) = t(β). This completesthe proof of the whole theorem.

Let p(n) denote the partition function, that is, the number of nonnegativeinteger solutions to the equation

x1 + 2x2 + 3x3 + · · · + nxn = n. (13.1)

We also set p(0) = 1.

Corollary 13.2.5 (i) The semigroup ISn has n+1 pairwise nonisomor-phic variants.

(ii) The semigroup Tn has p(n) pairwise nonisomorphic variants.

(iii) The semigroup PT n hasn∑

k=0

p(k) pairwise nonisomorphic variants.

Proof. From Theorem 13.2.1 we have that the isomorphism classes of vari-ants of ISn are classified by ranks of elements in ISn. These ranks arenumbers from {0, 1, . . . , n} and the statement (i) follows.

Analogously, from Theorem 13.2.1 and Exercises 2.10.24 and 1.5.12 itfollows that variants of Tn are classified by solutions to (13.1). The statement(ii) follows. To prove the statement (iii) is left as an exercise to the reader.

13.3. IDEMPOTENTS AND MAXIMAL SUBGROUPS 243

Let Δ = (t0, t1, . . . , tn) be a vector with nonnegative integer coefficientssuch that

n∑i=0

ti = n andn∑

i=0

iti = k ≤ n. (13.2)

Exercise 13.2.6 Show that for α ∈ PT n the vector t(α) satisfies (13.2) fork = |dom(α)|.

Let πΔ ∈ PT n be the idempotent, given by (9.9) (with the convention

πΔ(x) = ∅ for all x > k). In particular, πΔ ∈ Tn if and only ifn∑

i=0

iti = n;

and πΔ ∈ ISn if and only if ti = 0 for all i > 1.Let S denote one of the semigroups Tn, PT n, or ISn. For α ∈ S from

Exercise 13.2.6 we have that Δ = t(α) satisfies (13.2). From the proof ofLemma 9.3.5 we obtain t(πΔ) = Δ and hence by Theorem 13.2.1 we haveSα ∼= SπΔ . This means that, up to isomorphism, we may always assume thatthe sandwich element α is an idempotent.

13.3 Idempotents and Maximal Subgroups

Let S denote one of the semigroups Tn, PT n, or ISn. As we have alreadynoted, in the case α = 0, the semigroup S0 is the semigroup with zeromultiplication and hence is not really interesting. So, from now on we assumethat the sandwich element α is given by

α =(

A1 A2 · · · Ak

a1 a2 · · · ak

), (13.3)

has rank k > 0 and is an idempotent, that is, ai ∈ Ai for all i = 1, . . . , k.We also set |Ai| = li.

Lemma 13.3.1 Let

β =(

B1 B2 · · · Bm

b1 b2 · · · bm

)∈ S (13.4)

be an element of rank m. Then β is an idempotent of Sα if and only if thereexists an injection f : {1, . . . , m} → {1, . . . , k} such that bi ∈ Af(i) andaf(i) ∈ Bi for all i = 1, . . . , m.

Proof. Direct computation.

Proposition 13.3.2 (i) If α ∈ ISn, then the semigroup ISαn has 2k

idempotents.


(ii) If α ∈ Tn, then the number of idempotents in the semigroup T αn equals

∑∅�=X⊂{1,...,k}

|X|n−|X| ·∏i∈X

li.

(iii) If α ∈ PT n, then the number of idempotents in the semigroup PT αn

equals1 +

∑∅�=X⊂{1,...,k}

(|X| + 1)n−|X| ·∏i∈X

li.

Proof. If β ∈ ISn is given by (13.4), then |Bi| = 1 for all i and we also haveAi = {ai} for all i. Hence the assertion of Lemma 13.3.1 reads as follows:{b1, . . . , bm} ⊂ {a1, . . . , ak} and Bi = {bi} for all i. The statement (i) follows.

To prove (ii) we count the number of idempotents β of T αn , whose im-

age intersects exactly the Ai’s for which i ∈ X ⊂ {1, . . . , k}, X �= ∅. ByLemma 13.3.1, to determine such β we should choose an image for each ai,i ∈ X, in the corresponding block Ai, which can be done in li different ways;this determines im(β) and after that we should arbitrarily choose the imageof all elements from N\{ai : i ∈ X} inside im(β). Since all our choices areindependent, the statement (ii) now follows applying the product rule.

The statement (iii) is proved similarly to the proof of the statement(ii) with two differences: we should take the zero idempotent into account;and when mapping the elements which are left arbitrarily to im(β), weshould leave for all of them the option to be mapped to ∅. The statementfollows.

Corollary 13.3.3 Let β ∈ Sα be an idempotent given by (13.4). Thenthe corresponding maximal subgroup Gβ of Sα is isomorphic to Sm andcoincides, as a set, with the H-class H(β) of the element β, considered asan element from S.

Proof. By Lemma 5.1.2, Gβ coincides with the group of units in β◦αSα◦αβ.In particular, every γ ∈ Gβ has the form

γ = β ◦α γ′ ◦α β = βαγ′αβ (13.5)

for some γ′ ∈ S. As S is finite, we also have that β must belong to the cyclicsubgroup of Sα, generated by γ. This implies rank(β) = rank(γ), which,together with (13.5), yields ρβ = ργ and im(β) = im(γ), that is, γ ∈ H(β)by Theorem 4.5.1.

On the other hand, without loss of generality we may assume that themapping f from Lemma 13.3.1 is the natural inclusion f(i) = i. For γ ∈H(β), because of the previous paragraph, we have γ(Bi) = bτ(i) for someτ ∈ Sk. Then a direct calculation shows that the mapping γ → τ is anisomorphism from H(β) ⊂ Sα to Sk. This completes the proof.

13.4. PRINCIPAL IDEALS AND GREEN’S RELATIONS 245

13.4 Principal Ideals and Green’s Relations

For the study of ideals in variants, it is worth mentioning one more timethat a variant of a semigroup is not a monoid in general. In this section, Sdenotes one of the semigroups Tn, PT n, or ISn, and α is given by (13.3).The proposition that follows reduces the study of principal ideals in Sα tothat of principal ideals in S, see Sect. 4.2.

Proposition 13.4.1 Assume that β ∈ S is given by (13.4). We have

(i) The principal left ideal of S, generated by β, equals {β} ∪ Sαβ.

(ii) The principal right ideal of S, generated by β, equals {β} ∪ βαS.

(iii) The principal two-sided ideal of S, generated by β, equals

{β} ∪ Sαβ ∪ βαS ∪ SαβαS.

Proof. This follows directly from the definitions.

Theorem 13.4.2 Let β ∈ S be given by (13.4) and γ ∈ S be the followingelement of rank p:

γ =(

C1 C2 · · · Cp

c1 c2 · · · cp

). (13.6)

(i) βLγ if and only if β = γ or the following conditions are satisfied:

(a) ρβ = ργ (in particular p = m).

(b) There exists an injective function f : {1, . . . , m} → {1, . . . , k} suchthat bi ∈ Af(i), i = 1, . . . , m.

(c) There exists an injective function g : {1, . . . , p} → {1, . . . , k} suchthat ci ∈ Ag(i), i = 1, . . . , p.

(ii) βRγ if and only if β = γ or the following conditions are satisfied:

(a) im(β) = im(γ) (in particular p = m).

(b) There exists an injective function f : {1, . . . , m} → {1, . . . , k}such that af(i) ∈ Bi for all i = 1, . . . , m.

(c) There exists an injective function g : {1, . . . , p} → {1, . . . , k} suchthat ag(i) ∈ Ci for all i = 1, . . . , p.

(iii) βHγ if and only if β = γ or all conditions (ia)–(ic) and (iia)–(iic) aresatisfied.

(iv) βDγ if and only if one of the following mutually exclusive conditionsis satisfied:


(a) rank(β) = rank(γ) and the conditions (ib), (ic), (iib), (iic) aresatisfied.

(b) Conditions (ia)–(ic) are satisfied while at least one of the condi-tions (iib) or (iic) is not satisfied.

(c) Conditions (iia)–(iic) are satisfied while at least one of the condi-tions (ib) or (ic) is not satisfied.

(d) β = γ, at least one of the conditions (iib) or (iic) is not satisfied,and at least one of the conditions (ib) or (ic) is not satisfied.

(v) D = J .

Proof. Assume that β �= γ and βLγ. Then from Proposition 13.4.1(i) wehave γ ∈ Sα ◦α β, which yields the condition (ib). Similarly we get (ic).From Proposition 4.4.1 we have β = δαγ and γ = δ′αβ for some δ, δ′ ∈ S. Inparticular, β and γ must be L-connected in the original semigroup S. Hencethe condition (ia) follows from Theorem 4.5.1.

Conversely, if all conditions (ia)-(ic) are satisfied, then Sα ◦α β = Sα ◦α γfollows from Theorem 4.5.1. This proves (i). The proof of the statements (ii)is similar. The statement (v) follows from Theorem 5.4.1. The statements(iii) and (iv) follow from the statements (i) and (ii) and the definitions.


13.5.1 The problem to study variants of semigroups seems to go back atleast to Lyapin’s monograph [Ly]. Various sandwich semigroups were studiedby several authors, see for example some older paper [Sy1, Sy2, Hic1, Hic2,Ch1, Ch2, MS1, MS2, MMT] or some more recent papers [KL, Ji, Ts1, Ts2,Ts3, Ts4, Ku1, KT, MT].

The results of Sect. 13.2 are proved in [Sy1, Ts1, Ts3]. The results ofSect. 13.4 are partially proved in [Ts2]. Proposition 13.1.7 is taken from[Ku1]. The results of Sect. 13.3 are partially taken from [MT].

13.5.2 Theorem 13.2.1 roughly speaking says that the semigroups Tn, PT n,and ISn have lots of isomorphic variants. There exist semigroups, all vari-ants of which are not isomorphic. For example, we think that the followingstatement is noteworthy:

Theorem 13.5.1 ([Ts1]) Let B denote the monoid given by the followingpresentation: B = 〈a, b|ab = 1〉. Then for x, y ∈ B we have Bx ∼= By if andonly if x = y.

The monoid B defined above is called the bicyclic semigroup. The semi-group B is an infinite inverse semigroup, hence the statement of Theorem13.5.1 is rather remarkable, especially because of Exercise 13.6.8.


Proof. From the definition we have that every element of B can be uniquelywritten in the form αi,j = biaj , i, j ≥ 0. We further have

αi,jαk,l =

{αi,l+j−k, j ≥ k;αi+k−j,l, j < k,

in particular, αi,j is an idempotent if and only if i = j.Fix now k, m ≥ 0, let α = αk,m and consider the semigroup Bα.

Lemma 13.5.2 (i) E(Bα) = {αm+l,k+l : l ≥ 0}.

(ii) If for i ≥ 0 we set ek,mi = αm+i,k+i, then ek,m

i ◦α ek,mj = ek,m

max(i,j).

Proof. For i, j ≥ 0 we have that αi,j ∈ E(Bα) if and only if

biaj · bkam · biaj = biaj . (13.7)

If j < k and i < m, then a direct calculation reduces (13.7) to the equationbi+k−jaj+m−i = biaj and hence k = j, m = i, a contradiction.

If j < k and i ≥ m, then a direct calculation reduces (13.7) to theequation bi+k−j+i−maj = biaj . Thus, 0 > k − j = m − i ≤ 0, which is againa contradiction. Analogously, one gets a contradiction in the case j ≥ k andi < m.

In the remaining case j ≥ k and i ≥ m a direct calculation reduces (13.7)to the equation i − m = j − k ≥ 0. The statement (i) follows.

The statement (ii) is now proved by a direct calculation.

For i ≥ 0 consider the sets

P k,mi = {γ ∈ Bα : ek,m

i ◦α γ �= γ},Qk,m

i = {γ ∈ Bα : γ ◦α ek,mi �= γ}.

Lemma 13.5.3 |P k,mi ∩ Qk,m

j | = (k + j)(m + i) for all i, j ≥ 0.

Proof. For x, y ≥ 0 we have αx,y �∈ P k,mi if and only if ek,m

i ◦α αx,y = αx,y,that is,

bm+iak+i · bkam · bxay = bm+iam+i · bxay = bxay.

If m + i > x, then the latter equality is equivalent to m + i = x andm + i − x + y = y. Hence m + i = x, a contradiction. If m + i ≤ x, thenthe equality becomes an identity. This means αx,y �∈ P k,m

i if and only ifm + i ≤ x, and hence αx,y ∈ P k,m

i if and only if x < m + i.Analogously, one shows that αx,y ∈ Qk,m

j if and only if y < k + j. Thestatement of the lemma follows.


Assume now that k, m, u, v ≥ 0, α = αk,m, β = αu,v and ϕ : Bα → Bβ

is an isomorphism. Then ϕ maps idempotents to idempotents and preservesthe multiplication of idempotents. Hence from Lemma 13.5.2 it follows thatϕ(ek,m

i ) = eu,vi for all i ≥ 0. This, in particular, yields that ϕ(P k,m

i ) = P u,vi

and ϕ(Qk,mi ) = Qu,v

i for all i ≥ 0. Thus we have

ϕ(P k,m1 ∩ Qk,m

1 ) = P u,v1 ∩ Qu,v

1 ;

ϕ(P k,m1 ∩ Qk,m

0 ) = P u,v1 ∩ Qu,v

0 ;

ϕ(P k,m0 ∩ Qk,m

1 ) = P u,v0 ∩ Qu,v

1 .

However, from Lemma 13.5.3 we have

|P k,m1 ∩ Qk,m

1 | − |P k,m1 ∩ Qk,m

0 | = m + 1;|P u,v

1 ∩ Qu,v1 | − |P u,v

1 ∩ Qu,v0 | = v + 1;

|P k,m1 ∩ Qk,m

1 | − |P k,m0 ∩ Qk,m

1 | = k + 1;|P u,v

1 ∩ Qu,v1 | − |P u,v

0 ∩ Qu,v1 | = u + 1.

Thus m = v and k = u. This completes the proof.

13.5.3 Proposition 13.1.7 says that a typical variant of a semigroup is aninflation of a certain subsemigroup. Automorphism groups of semigroup in-flations have rather special structure, described in [Ku1]: Assume that weare in the situation of Proposition 13.1.7. Then Aut(S) is a semidirect prod-uct of two subgroups A1 and A2. The normal subgroup A1 of Aut(S) isjust the direct product of symmetric groups on equivalence classes of therelation h. The group A1 acts on S in the natural way. The subgroup A2

consists of all automorphisms of T , which are extendable to S, that is, whichare obtained from an automorphism of S by restriction. For variants ofISn this was proved in [KT]. The approach of [Ku1] unifies the results of[GTS, KT, Sy1, St2].

13.5.4 Variants of transformation semigroups are just a special case of thefollowing more general and rather remarkable construction: If X and Y aretwo nonempty sets and α : Y → X is a fixed mapping, then the set F (X, Y )of all mappings from X to Y (note the direction!) becomes a semigroup withrespect to the operation ◦α defined as follows: For f, g ∈ F (X, Y ) we setf ◦α g = fαg. The definition of this semigroup goes back at least to Lyapin,see [Ly]. Several properties of this semigroup were studied in [Sy1, Sy2].



13.6.1 Let S be a finite monoid. Show that all variants of S are isomorphicif and only if S is a group.

13.6.2 Let S be a finite semigroup with zero. Show that all variants of Sare isomorphic if and only if S3 = 0.

13.6.3 Let S be a semigroup. Assume that the group Aut(S) acts tran-sitively on S (i.e., for any x, y ∈ S there exists ϕ ∈ Aut(S) such thatϕ(x) = y). Show that all variants of S are isomorphic.

13.6.4 Show that for every S from the following list all variants of S areisomorphic:

(a) Semigroup with zero multiplication

(b) Left zero semigroup

(c) Right zero semigroup

(d) Rectangular band

(e) The semigroup of positive rational (or real) numbers with respect to theaddition

13.6.5 For a, b ∈ N define a � b = min(a, b). Show that any two variants of(N, �) with respect to two different elements are not isomorphic.

13.6.6 For a, b ∈ N define a � b = max(a, b). Show that any two variants of(N, �) with respect to two different elements are not isomorphic.

13.6.7 Let S = (N, +). Show that Sa �∼= Sb if a �= b.

13.6.8 Let S be a finite inverse semigroup, which is not a group. Show thatthere exist a, b ∈ S such that (S, ◦a) �∼= (S, ◦b).

13.6.9 Let S be a semigroup and T be a variant of S. Show that everyvariant of T is isomorphic to some variant of S.

13.6.10 (a) For a, b ∈ Z define a � b = min(a, b). Show that all variants of(Z, �) are isomorphic.

(b) Let S be some variant of (Z, �) from (a). Show that all variants of S areisomorphic to S.

13.6.11 An element x of a semigroup S is called a mididentity providedthat uxv = uv for any u, v ∈ S. Let S be a monoid and a ∈ S. Show thatx ∈ Sa is a mididentity if and only if axa = a.


13.6.12 ([Ts2]) Compute the number of one-element R-classes and the num-ber of one-element L-classes in ISα

n.

13.6.13 ([Ts2]) Compute the number of one-element R-classes and the num-ber of one-element L-classes in T α

n .

13.6.14 ([Ts2]) Show that ISαn∼=

←ISα

n for any α ∈ ISn.

13.6.15 ([Ts1]) Show that for α, β ∈ Bn the semigroups Bαn and

←B

βn are

isomorphic if and only if β = α−1.

13.6.16 ([Ts4]) Let α ∈ ISn, and Tα denote the set of all β ∈ ISn suchthat β(im(α)) = dom(α).

(a) Show that Tα is a subsemigroup of ISαn.

(b) Prove that the only completely isolated subsemigroups of ISαn are: ISα

n,Tα and ISα

n\Tα.

13.6.17 ([Ts4]) Let α ∈ ISn, and Tα be as in Exercise 13.6.16. For x ∈im(α) let G(x) denote the set of all β ∈ ISn such that β(x) �∈ dom(α) andβ(y) ∈ dom(α)\α−1(x) for all y ∈ im(α)\{x}.

(a) Show that G(x) is an isolated subsemigroup of ISαn.

(b) Prove that every isolated subsemigroup of ISαn coincides with one of the

following semigroups: ISαn, Tα, ISα

n\Tα or G(x), x ∈ im(α).

13.6.18 ([Ts4]) Let α ∈ ISn. Classify all maximal nilpotent subsemigroupsin ISα

n.

13.6.19 ([Ch1]) Let S be a semigroup and a ∈ S. Show that for any Green’srelation X the fact that elements x, y ∈ Sa are X -related in Sa implies thatx, y are X -related already in S.

Chapter 14

Order-RelatedSubsemigroups

14.1 Subsemigroups, Related to the NaturalOrder

In this chapter, we will study certain subsemigroups of Tn, PT n, and ISn,related to the natural order 1 < 2 < · · · < n on the set N. One of thesesubsemigroups, namely, IO<

n (which we simply denote by IOn in the sequel),already appeared in Sect. 12.3 in the study of H-cross-sections of ISn. Inthis chapter, we will study IOn and some other similarly defined semigroupsin more detail.

A partial transformation α ∈ PT n is called order-preserving providedthat x ≤ y implies α(x) ≤ α(x) for all x, y ∈ dom(α). The set of all order-preserving partial transformations from PT n is denoted by POn.

Exercise 14.1.1 Check that POn is a subsemigroup of PT n.

The semigroup POn is called the semigroup of all order-preserving par-tial transformations of the set N. The subsemigroups On = POn ∩ Tn andIOn = POn∩ISn of POn are called the subsemigroup of all order-preserving(total) transformations and the subsemigroup of all order-preserving partialpermutations of N, respectively.

Example 14.1.2 The semigroup PO2 contains the following eight elements:(1 2∅ ∅

),

(1 21 ∅

),

(1 2∅ 1

),

(1 22 ∅

),

(1 2∅ 2

),

(1 21 1

),

(1 21 2

),

(1 22 2

).

The semigroup O2 contains the following three elements:(1 21 1

),

(1 21 2

),

(1 22 2

).



252 CHAPTER 14. ORDER-RELATED SUBSEMIGROUPS

The semigroup IO2 contains the following six elements:(

1 2∅ ∅

),

(1 21 ∅

),

(1 2∅ 1

),

(1 22 ∅

),

(1 2∅ 2

),

(1 21 2

).

A partial transformation α ∈ PT n is called order-decreasing providedthat α(x) ≤ x for all x ∈ dom(α). The set of all order-decreasing partialtransformations from PT n is denoted by PFn.

Exercise 14.1.3 Check that PFn is a subsemigroup of PT n.

The semigroup PFn is called the semigroup of all order-decreasing par-tial transformations of the set N. The subsemigroups Fn = PFn ∩ Tn andIFn = PFn∩ISn of PFn are called the subsemigroup of all order-decreasing(total) transformations and the subsemigroup of all order-decreasing partialpermutations of N, respectively.

Example 14.1.4 The semigroup PF2 contains the following six elements:(

1 2∅ ∅

),

(1 21 ∅

),

(1 2∅ 1

),

(1 2∅ 2

),

(1 21 1

),

(1 21 2

).

The semigroup F2 contains the following two elements:(

1 21 1

),

(1 21 2

).

The semigroup IF2 contains the following five elements:(

1 2∅ ∅

),

(1 21 ∅

),

(1 2∅ 1

),

(1 2∅ 2

),

(1 21 2

).

We also define the following three subsemigroups, which consist of alltransformations, which are both order-preserving and order-decreasing:

PCn = POn ∩ PFn, Cn = On ∩ Fn, ICn = IOn ∩ IFn.

From Examples 14.1.2 and 14.1.4 we see that PC2 = PF2, C2 = F2, andIC2 = IF2. However, in the general case analogous equalities are no longertrue.

Exercise 14.1.5 Prove that for n ≥ 3 all nine semigroups PCn, POn, PFn,Cn, On, Fn, ICn, IOn, and IFn are different.

We complete this section describing some properties of elements in theintroduced semigroups. Every element α ∈ POn can be written in the form

α =(

A1 A2 · · · Ak

a1 a2 · · · ak

),

14.2. CARDINALITIES 253

where a1 < a2 < · · · < ak. Then for every i < j and every a ∈ Ai and b ∈ Aj

we have a < b. In particular, in the case α ∈ On this means that the setsA1, . . . , Ak form a partition of N into intervals (i.e., if x, y ∈ Ai and x < y,then Ai contains all z such that x < z < y).

For α ∈ IOn all sets Ai contain exactly one element, that is, A1 = {b1},A2 = {b2},. . . , Ak = {bk}. Hence b1 < b2 < · · · < bk, and α coincides withthe unique increasing bijection from {b1, . . . , bk} to {a1, . . . , ak}. In particu-lar, α is uniquely determined by dom(α) and im(α) (see also Theorem 12.3.3and its proof).

Proposition 14.1.6 If α ∈ PT n is either order-preserving or order-decreasing,then the permutational part of α is the identity transformation.

Proof. Assume that the permutational part of the element α contains somecycle (a1, a2, . . . , ak). If α is order-decreasing, we have the inequalities a1 ≥a2 ≥ · · · ≥ ak ≥ a1 and hence a1 = a2 = · · · = ak = a1 and k = 1.

Suppose now that α is order-preserving and k > 1. Consider first thecase a1 < a2. Then a2 < a3, a3 < a4, . . . , ak−1 < ak, which yields a1 < ak.At the same time, as α is order-preserving, ak−1 < ak implies ak < a1, acontradiction. The case a1 > a2 is analogous.

Corollary 14.1.7 Let α be a group element of any of the nine semigroups,defined in this section. Then α is an idempotent.

Proof. It follows from Propositions 14.1.6 and 5.2.8.

A semigroup S is called aperiodic or combinatorial provided that allmaximal subgroups of S have order one. From Corollary 14.1.7 it followsthat all our nine semigroups are aperiodic.

14.2 Cardinalities

In this section, we compute the cardinalities of the semigroups defined inthe previous section. The easiest computation is for the semigroup Fn.

Proposition 14.2.1 |Fn| = n!.

Proof. From the definition of Fn it follows that for an element α ∈ Fn theimage α(x) of some x ∈ N can be chosen in x different ways. Moreover, theimages of different elements from N can be chosen independently. Hence|Fn| = 1 · 2 · · · · · n = n!.

The computation of |PFn| is completely similar. The only differenceis that for every x ∈ PFn there appears a new possibility for the image,namely, ∅.

Proposition 14.2.2 |PFn| = 2 · 3 · · · · · (n + 1) = (n + 1)!.


Proposition 14.2.3 |On| =(2n−1

n

).

Proof. An element

α =(

1 2 · · · na1 a2 · · · an

)∈ On

is uniquely determined by the second row (a1, . . . , an). For i = 1, . . . , n setbi = ai + i − 1. Then the mapping

(a1, . . . , an) → (b1, . . . , bn)

is an obvious bijection between the set of all vectors (a1, . . . , an) such that1 ≤ a1 ≤ a2 ≤ · · · ≤ an ≤ n and the set of all vectors (b1, . . . , bn) suchthat 1 ≤ b1 < b2 < · · · < bn ≤ 2n − 1. The vector (b1, . . . , bn) uniquelydetermines the n-element subset {b1, . . . , bn} of {1, 2, . . . , 2n − 1}. Hence|On| =

(2n−1

n

).

Proposition 14.2.4 |IOn| =(2nn

).

Proof. As we have already seen in the previous section, an element α ∈IOn is a unique increasing bijection from dom(α) to im(α), the latter beingsubsets of N of the same cardinality. Hence

|IOn| =n∑

k=0

(n

k

)2

=(

2n

n

).

Corollary 14.2.5 |IOn| = 2|On|.

Proposition 14.2.6 |IFn| = Bn+1.

Proof. The chain-cycle notation for an element from IFn contains onlycycles of length 1 and chains. With every partition ρ of the set N∗ ={∗, 1, 2, . . . , n} we associate an element αρ ∈ IFn by the following rule:each x ∈ N such that xρ∗ corresponds to the cycle (x) of α; each blockA = {a1, . . . , ak}, 1 ≤ a1 ≤ · · · ≤ ak ≤ n, corresponds to the chain[ak, ak−1, . . . , a1]. Obviously, the mapping ρ → αρ is a bijection betweenthe set of all partitions of N∗ and IFn. Hence |IFn| = Bn+1.

The number Cn = 1n+1

(2nn

)is called the n-th Catalan number. Set also

C0 = 1.

Exercise 14.2.7 Show that Catalan numbers are uniquely determined bythe following recursive relation:

C0 = 1, Cn+1 =n∑

i=0

CiCn−i, n ≥ 0.

14.2. CARDINALITIES 255

Theorem 14.2.8 (i) |Cn| = Cn.

(ii) |ICn| = Cn+1.

Proof. We start with the statement (i). Let |Cn| = tn and set t0 = 1. Withevery element

α =(

1 2 · · · n1 α(2) · · · α(n)

)∈ Cn

we associate the element

α =(

1 2 · · · n n + 11 α(2) · · · α(n) n + 1

)∈ Cn+1.

Now we partition Cn into n classes C(2)n ,. . . , C(n+1)

n depending on the minimalk > 1 such that α(k) = k. Then for any element α ∈ C(k)

n we have

α =(

1 2 · · · k − 1 k k + 1 · · · n n + 11 α(2) · · · α(k − 1) k α(k + 1) · · · α(n) n + 1

)

and for all i, 1 < i < k, we have α(i) < i. At the same time for all j > k wehave α(j) ≥ k. We separate the following two parts in α:

α− =(

2 · · · k − 1α(2) · · · α(k − 1)

)α+ =

(k k + 1 · · · nk α(k + 1) · · · α(n)

).

With α− we associate the following element:

α′− =

(1 · · · k − 2

α(2) · · · α(k − 1)

)∈ Ck−2

and with α− we associate the following element:

α′+ =

(1 2 · · · n − k + 11 α(k + 1) − k + 1 · · · α(n) − k + 1

)∈ Cn−k+1.

It is easy to see that the correspondence

C(k)n � α ↔ (α−, α+) ↔ (α′

−, α′+) ∈ Ck−2 × Cn−k+1

is a bijection. Hence

tn = |Cn| =n+1∑k=2

|C(k)n | =

n+1∑k=2

|Ck−2| · |Cn−k+1| =

=n+1∑k=2

tk−2 · tn−k+1 =n−1∑i=0

ti · tn−1−i.

As t0 = 1 = C0, the statement (i) follows now from Exercise 14.2.7.


The proof of the statement (ii) is quite similar. Let |ICn| = fn+1 and setf0 = f1 = 1. With every element

α =(

1 2 · · · nα(1) α(2) · · · α(n)

)∈ ICn

we associate the element

α =(

1 2 · · · n n + 1α(1) α(2) · · · α(n) n + 1

)∈ ICn+1.

Now we partition ICn into n + 1 classes IC(1)n ,. . . , IC(n+1)

n depending onthe minimal k such that α(k) = k. Then for any element α ∈ IC(k)

n andfor all i ∈ dom(α), 1 < i < k, we have α(i) < i. At the same time for allj ∈ dom(α) such that j > k we have α(j) > k. In particular, α(1) = ∅ forall k > 1. Hence we can consider the mapping IC(k)

n → ICk−2 ×ICn−k suchthat α → (α−, α+), where

α− =(

1 2 · · · k − 2α(2) α(3) · · · α(k − 1)

)

andα+ =

(1 2 · · · n − k

α(k + 1) − k α(k + 2) − k · · · α(n) − k

)

(in the case α(j) = ∅ we use the convention α(j)− k = ∅). It is easy to seethat the above correspondence is in fact a bijection (in the degenerate casesk = 1, 2, n, n + 1 we obtain a bijection from IC(k)

n to one of the factors inICk−2 × ICn−k). Hence

fn+1 = |ICn| =n+1∑k=1

|IC(k)n | =

n+1∑k=1

|ICk−2| · |ICn−k| =n∑

k=0

fk · fn−k.

The statement (ii) now also follows from Exercise 14.2.7.

Theorem 14.2.9 |POn| =n∑

m=0

(n

m

)(n + m − 1

m

).

Proof. We partition POn into blocks depending on |dom(α)|, α ∈ POn.If |dom(α)| = m, the set dom(α) can be chosen in

(nm

)different ways. If

dom(α) = {a1, . . . , am} is fixed and a1 < a2 < · · · < am, there is an obviousbijection between the elements

α =(

a1 a2 · · · am

b1 b2 · · · bm

)∈ POn

and vectors (b1, . . . , bm) such that b1 ≤ b2 ≤ · · · ≤ bm. At the same time,there is a bijection between the latter vectors and m-element subsets in

14.3. IDEMPOTENTS 257

{1, . . . , n + m − 1} of the form {b1, b2 + 1, . . . , bm + m − 1}. Hence thenumber of elements α ∈ POn such that dom(α) = {a1, . . . , am} equals(n+m−1

m

). Therefore,

|POn| = 1 +n∑

m=1

(n

m

)(n + m − 1

m

)=

n∑m=0

(n

m

)(n + m − 1

m

).

Exercise 14.2.10 Show that

(1 − x)−n =∑m≥0

(n + m − 1

m

)xm.

Corollary 14.2.11 |POn| coincides with the coefficient of xn in the series(1 + x)n(1 − x)−n.

Proof. This follows directly from Theorem 14.2.9, Exercise 14.2.10 and thebinomial formula.

14.3 Idempotents

Recall from Sect. 2.7 that an element α ∈ ISn is an idempotent if and onlyif α is the identity transformation of some subset A ⊂ N. In particular,every idempotent from ISn is both order-preserving and order-decreasing,by definition. This means the following:

Proposition 14.3.1 We have E(ISn) = E(IOn) = E(IFn) = E(ICn), inparticular, each of these sets contains 2n elements.

Proposition 14.3.2 Every idempotent δ ∈ Cn is uniquely determined byits image im(δ) = {1, a2, . . . , ak}, moreover, the set {a2, . . . , ak} can be anarbitrary subset of {2, . . . , n}. In particular, |E(Cn)| = 2n−1.

Proof. We have δ(1) = 1 for all δ ∈ Cn. Let now δ ∈ E(Cn) and im(δ) ={1, a2, . . . , ak} be such that 1 = a1 < a2 < · · · < ak. As any idempotentacts as the identity on its image, for any x ∈ {ai, ai + 1, . . . , ai+1 − 1} fromai ≤ x < ai+1 we obtain

ai = δ(ai) ≤ δ(x) ≤ x < ai+1.

This yields δ(x) = ai. Hence the element δ is uniquely determined by its im-age. Obviously, the set {a2, . . . , ak} can be an arbitrary subset of {2, . . . , n}.The statement of the proposition follows.

Exercise 14.3.3 Prove that Cn contains(n−1k−1

)idempotents of rank k.


Theorem 14.3.4 The semigroup PCn contains 3n+12 idempotents.

Proof. We count the number of nonzero idempotents δ ∈ PCn for whichthe minimum element in im(δ) equals k. Let rank(δ) = i + 1 and im(δ) ={k, b1, . . . , bi}, where k < b1 < · · · < bi. The subset {b1, . . . , bi} of the set{k + 1, . . . , n} can be chosen in

(n−k

i

)different ways.

Every x < k does not belong to dom(δ); every x, k ≤ x < b1, whichbelongs to dom(δ) must be mapped to k; every x, b1 ≤ x < b2, which belongsto dom(δ) must be mapped to b1 and so on. Hence there are 2n−k−i ways todefine the transformation δ on the set {k + 1, . . . , n}\im(δ). Therefore,

|E(PCn)| = 1 +n∑

k=1

(n−k∑i=0

(n − k

i

)2n−k−i

)=

= 1 +n∑

k=1

(1 + 2)n−k = 1 +3n − 1

2=

3n + 12

.

Proposition 14.3.5 (i) |E(Fn)| = Bn.

(ii) |E(PFn)| = Bn+1.

Proof. Let ρ be a partition of N. Define the transformation αρ as follows: IfX is some block of ρ, the transformation αρ maps all elements from X to theminimum element in X. Obviously, this transformation is idempotent. Onthe other hand, every idempotent δ ∈ E(Fn) has this form as, by definition,every block of ρδ must be mapped to the minimum element of this block.This gives us a bijection between E(Fn) and partitions of N. The statement(i) follows.

Let ρ be a partition of N∗ = N ∪ {∗}. Define the transformation αρ ∈PT n as follows: α is not defined on all elements which belong to the blockof ρ containing ∗; if X is some block of ρ which does not contain ∗, thetransformation αρ maps all elements from X to the minimum element inX. As in the previous paragraph one easily verifies that this correspondencedefines a bijection between E(PFn) and partitions of N∗. The statement (ii)follows and the proof is complete.

To count idempotents in the remaining semigroups we recall that Fi-bonacci numbers fn, n ≥ 1, are defined recursively as follows: f1 = f2 = 1and fn = fn−1 + fn−2, n ≥ 3.

Theorem 14.3.6 |E(On)| = f2n.

14.3. IDEMPOTENTS 259

Proof. Let qn = |E(On)|. Partition the set E(On) into the following threedisjoint subsets:

E1(On) = {δ ∈ E(On) : δ(n) < n};E2(On) = {δ ∈ E(On) : δ(n − 1) < n, δ(n) = n};E3(On) = {δ ∈ E(On) : δ(n − 1) = n}.

For every idempotent δ ∈ E1(On) we have δ(n) = δ(n − 1) as δ is order-preserving, and hence |E1(On)| = qn−1. We also obviously have |E2(On)| =qn−1.

Let us now compute |E3(On)|. If δ ∈ E3(On), then the fact that δ isan idempotent implies n − 1 �∈ im(δ). Consider the transformation δ′ :{1, . . . , n − 1} → {1, . . . , n − 1} defined as follows:

δ′(x) =

{δ(x), δ(x) �= n;n − 1, δ(x) = n.

Obviously, δ′ ∈ E(On−1). In this way we obtain exactly those idempotentsfrom E(On−1), which fix the element (n − 1), that is, exactly the elementsfrom E(On−1)\E1(On−1). This implies |E3(On)| = qn−1 − qn−2 and thus

qn = qn−1 + qn−1 + (qn−1 − qn−2) = 3qn−1 − qn−2.

On the other hand, we obviously have q1 = 1 = f2 and q2 = 3 = f4 (seeExample 14.1.2). For Fibonacci numbers we have

f2n = f2n−1 + f2(n−1) = 2f2(n−1) + f2n−3 = 3f2(n−1) − f2(n−2).

Hence qn satisfies the same recursive relation as the sequence f2n. The state-ment of the theorem follows.

Theorem 14.3.7 |E(POn)| = 1 +n∑

k=1

(n

k

)f2k.

Proof. If δ is an idempotent, we have im(δ) ⊂ dom(δ). Consider all idempo-tents from POn whose domain is the set {a1, . . . , ak}, where a1 < · · · < ak.Then each such idempotent has the form

δ =(

a1 a2 · · · ak

ai1 ai2 · · · aik

).

With this element δ we associate the idempotent

δ′ =(

1 2 · · · ki1 i2 · · · ik

)∈ Ok.

It is obvious that the mapping δ → δ′ is a bijection from the set of allidempotents from POn with domain {a1, . . . , ak} and E(Ok). As the domain


of a nonzero idempotent from POn may be an arbitrary nonempty subsetof N, using Theorem 14.3.6 we have

|E(POn)| = 1 +n∑

k=1

(n

k

)|E(Ok)| = 1 +

n∑k=1

(n

k

)f2k.

14.4 Generating Systems

Each semigroup S from the nine semigroups defined in Sect. 14.1 containsthe identity transformation ε. Since the set S\{ε} is a subsemigroup in S,every generating system of S contains ε. The main result of the presentsection is the following:

Theorem 14.4.1 Each of the semigroups On, Fn, Cn, POn, PFn, PCn,IOn, IFn and ICn is generated by ε and elements of rank (n − 1).

Proof. It is certainly enough to show that in each of these semigroups everyelement of rank r, r < n− 1, can be decomposed into a product of elementsof rank strictly greater than r.

Consider first the semigroup PFn. Let α ∈ PFn be such that rank(α) =r < n − 1. We consider the partition ρα of N ∪ {n + 1}, associated to α, asin Sect. 4.3. For x ∈ N∪ {n + 1} we denote by x the block of ρα, containingx. The block n + 1 will be called the trivial block.

Consider the case in which there exist nontrivial blocks of ρα, containingmore than one element. Let aα ∈ N be the maximum element occurring insuch blocks. Then there exists b ∈ aα such that b < aα, and as α(aα) =α(b) ≤ b, we have α(aα) �= aα. Consider the following two transformations:

β1(x) =

⎧⎪⎨⎪⎩

α(x), x < aα;x, x ≥ aαandx ∈ dom(α);∅, x ≥ aαandx �∈ dom(α);

and

γ1(x) =

⎧⎪⎨⎪⎩

x, x < aα;α(x), x ≥ aαandx ∈ dom(α);x, x ≥ aαandx �∈ dom(α).

Then α = γ1β1, moreover, both β1 and γ1 belong to PFn.From the definition, we see that ρα and ρβ1 share almost all blocks, the

only exceptions the block aα of ρα, which in ρβ1 decomposes into two blocks:{aα} and aα\{aα}. As the rank of an element equals the number of nontrivialblocks, we obtain that rank(β1) = rank(α) + 1 > rank(α).

14.4. GENERATING SYSTEMS 261

We have rank(γ1) ≥ rank(α) by Exercise 2.1.4(c). Consider the setsA1 = {1, 2, . . . , aα − 1} and A2 = N\A1. Note that the restrictions γ1|A1

and γ1|A2 are injective. Hence nontrivial blocks of ργ1 contain at most twoelements, moreover, each two-element block contains one element from A1

and one from A2. Furthermore, dom(γ1) = N by definition. Hence the equal-ity rank(γ1) = r implies that γ1 contains at least two two-element blocks.This yields aγ1 > aα.

Now we repeat the same arguments for γ1, which allows us to writeγ1 = γ2β2, where rank(β2) > r and aγ2 > aγ1 in the case rank(γ2) = r. Inthe latter case, we apply the same arguments to γ2 and so on. The sequenceaα < aγ1 < aγ2 < · · · is finite, so in a finite number of steps, say k steps,we will get that γk−1 = γkβk, where both rank(βk) and rank(γk) are strictlygreater than r. Thus α = γkβk · · ·β1 is a decomposition of α into a productof elements of rank strictly greater than r.

Now we consider the case for which all nontrivial blocks of ρα containexactly one element. Then α ∈ ISn and we can write α in the followingform:

α = [a1, . . . , ak][b1, · · · , bl] · · · [c1, · · · , cm](d1) · · · (dq),

where a1 > · · · > ak,. . . , c1 > · · · > cm. As rank(α) < n − 1, the element αcontains at least two chains. Then α = βγ, where

β = [a1, · · · , ak](b1) · · · (bl) · · · (c1) · · · (cm)(d1) · · · (dq),γ = (a1) · · · (ak)[b1, . . . , bl] · · · [c1, . . . , cm](d1) · · · (dq).

(14.1)

Obviously, rank(β) = n − 1 > r, rank(γ) = r + 1 and β, γ ∈ PFn. Thiscompletes the proof of our theorem for the semigroup PFn.

Note that the first part of our proof for PFn (where we considered thecase in which ρα has a nontrivial block consisting of at least two elements)proves the statement of the theorem for the semigroup Fn. Moreover, thesecond part of the proof for PFn proves the statement of the theorem forthe semigroup IFn.

Let us now consider the case of the semigroup POn. Let α ∈ POn besuch that

α =(

A1 · · · Ar

b1 · · · br

),

where b1 < · · · < br and r < n − 1. Assume that there exists k such that|Ak| > 1 and let Ak = {a1, . . . , am}, where a1 < · · · < am. Consider theset B′ = {b′1, . . . , b′r+2}, b′1 < · · · < b′r+2, which is obtained from the setB = {b1, . . . , br} by adding two new elements b′p and b′q, p < q. Then b′1 =b1,. . . , b′p−1 = bp−1, b′p+1 = bp,. . . , b′q−1 = bq−2, b′q+1 = bq−1, . . . , b

′r+2 = br.

If q > k+1, then α decomposes into the product α = γβ, where the elementβ equals(

A1 · · · Ak−1 Ak\{am} am Ak+1 · · · Aq−2 Aq−1 · · · Ar

b′1 · · · b′k−1 b′k b′k+1 b′k+2 · · · b′q−1 b′q+1 · · · b′r+2

)


and the element γ equals(C1 C2 · · · Ck−1 Ck Ck+1 · · · Cq−2 Cq−1 Cq · · · Cr+1

b1 b2 · · · bk−1 bk bk+1 · · · bq−2 b′q bq−1 · · · br

),

whereC1 = {1, . . . , b′1},C2 = {b′1 + 1, . . . b′2},. . .Ck−1 = {b′k−2 + 1, . . . , b′k−1},Ck = {b′k−1 + 1, . . . , b′k+1},Ck+1 = {b′k+1 + 1, . . . , b′k+2},. . .Cr+1 = {b′r+1 + 1, . . . , n}.

If q ≤ k+1, then p < k+1 and the elements β and γ are constructed play-ing a similar game with the element b′p instead of the element b′q. Obviously,both β and γ are elements of POn of rank r + 1.

Assume now that all blocks A1,. . . , Ar consist of oneelement and wehave Ai = {ai} for all i. Extend the set {a1, . . . , ar} in any way to the setA′ = {a′1, . . . , a′r+1}, a′1 < · · · < a′r+1 and the set {b1, . . . , br} to the set B′ ={b′1, . . . , b′r+1}, b′1 < · · · < b′r+1, by adding one more new element, a′k and b′msay, to each set, respectively. Consider also the set C = {1, 2, . . . , r + 2}. Ifk ≤ m, we let β be the unique increasing bijection from A′ to C\{m + 1},and γ be the unique increasing bijection from C\{k} to B′. If k > m, we letβ be the unique increasing bijection from A′ to C\{m}, and γ be the uniqueincreasing bijection from C\{k + 1} to B′. A direct calculation shows thatα = γβ. This proves our theorem for the semigroup POn. At the same time,the first part of the above proof proves the statement of the theorem for thesemigroup On, and the second part of the proof proves the statement of ourtheorem for the semigroup IOn.

To proceed we will need the following statement:

Lemma 14.4.2 Each transformation α ∈ ICn of rank k can be decomposedinto a product αm · · ·α1 of transformations αi ∈ ICn of rank k such thatx − 1 ≤ αi(x) ≤ x for all x ∈ dom(αi).

Proof. Let

α =(

a1 · · · ak

b1 · · · bk

)∈ ICn,

where a1 < · · · < ak. Then

α =(

b′1 · · · b′kb1 · · · bk

)(a1 · · · ak

b′1 · · · b′k

), (14.2)

where

b′j =

{aj , aj = bj ;aj − 1, aj > bj .


We claim that both factors from (14.2) belong to ICn. Indeed, sincebj ≤ b′j ≤ aj , it is enough to check that b′1 < · · · < b′k. Assume that this isnot the case and b′j ≥ b′j+1 for some j. Then b′j = aj , b′j+1 = aj+1 − 1, whichimplies bj = aj and

bj+1 ≤ b′j+1 = aj+1 − 1 ≤ b′j = aj = bj ,

a contradiction.Both factors in (14.2) have rank k and the second factor obviously sat-

isfies the inequality from the formulation. So, the second factor in (14.2)satisfies the condition from the formulation. For the first factor, which is anelement of ICn, we obtain the following: For every j such that aj > bj wehave

b′j − bj = (aj − bj) − 1 < aj − bj .

Now we proceed by induction on N =k∑

j=1

(aj − bj). If the first factor in

(14.2) does not satisfy the condition from the formulation, we apply to it theinductive assumption and find its decomposition into a product of elementsfrom ICn, which satisfy the necessary condition. This completes the proof.

Now the statement of our theorem is easy to prove for the semigroupICn. Indeed, let α ∈ ICn be an element of rank r < n − 1. Consider somedecomposition α = αm · · ·α1 given by Lemma 14.4.2. Every element αi hasthe form

αi = [u + v, u + v − 1, . . . , u] · · · [t + s, t + s − 1, . . . , t](d1) · · · (dq).

Now, taking βi and γi as given by (14.1), we obtain the decompositionαi = βiγi, where both βi and γi belong to ICn and have rank ≥ r + 1.

Consider now the semigroup PCn. Let

α =(

A1 · · · Ar

b1 · · · br

)∈ PCn,

where b1 < · · · < br and r < n − 1. Let ai denote the minimum element inAi. From the definition of PCn we have that bi ≤ ai for all i, and a < aj forall a ∈ Ai as soon as i < j. Hence for the elements

α∗ =(

A1 · · · Ar

a1 · · · ar

), α =

(a1 · · · ar

b1 · · · br

)

we have α∗ ∈ PCn, α ∈ ICn, and α = αα∗.Since we have already proved the theorem for the semigroup ICn, it is

now enough to show that every α∗ as above decomposes into a product ofelements of rank > r. If α∗ ∈ ICn, this is already proved. If α∗ �∈ ICn, wehave the following three possibilities:


(i) There exists k such that |Ak| ≥ 3.

(ii) There exists i < j such that |Ai| = |Aj | = 2.

(iii) |Ai| = 1 for all i �= k, Ak = {ak, a′} and dom(α) = dom(α∗) �= N.

In the case (i) let Ak = A′k ∪ {a′, a′′}, where ak < a′ < a′′. Then we have

(A′

k a′ a′′

ak ak ak

)=(

A′k a′ a′′

ak ak a′′

)(A′

k a′ a′′

ak a′ a′

),

which implies the existence of the necessary decomposition for α∗.In the case (ii) let Ai = {ai, a

′} and Aj = {aj , a′′}. Then we have

(ai a′ aj a′′

ai ai aj aj

)=(

ai a′ aj a′′

ai a′ aj aj

)(ai a′ aj a′′

ai ai aj a′′

),

which implies the existence of the necessary decomposition for α∗.In the case (iii) let Ak = {ak, a

′} and a ∈ N\dom(α). If we have a �∈{ak + 1, . . . , a′ − 1}, then the necessary decomposition for α∗ follows from

(ak a′ aak ak ∅

)=(

ak a′ aak a′ ∅

)(ak a′ aak ak a

).

If a ∈ {ak + 1, . . . , a′ − 1}, then the necessary decomposition for α∗ followsfrom (

ak a a′

ak ∅ ak

)=(

ak a a′

ak ak a′

)(ak a a′

ak ∅ a

).

This completes the proof of our theorem for the semigroup PCn.Finally, let us prove the statement of the theorem for the semigroup Cn.

Let α ∈ Cn. As Cn ⊂ PCn, we again may write α = αα∗ as above. Notethat α∗ ∈ Cn, and hence the constructed above decomposition for α∗ willcontain only factors from Cn. Hence from the above arguments we have adecomposition

α = γk · · · γ1βm · · ·β1,

where all factors have rank > r, all γi’s belong to ICn and all βj ’s belongto Cn.

Every element

γ =(

a1 · · · ak

b1 · · · bk

)∈ ICn,

where a1 < · · · < ak, can be extended to an element γ from Cn as follows:

γ =(

{1, . . . , a1 − 1} {a1, . . . , a2 − 1} . . . {ak, . . . , n}1 b1 · · · bk

)


in the case 1 < b1, or

γ =(

{1, . . . , a2 − 1} {a2, . . . , a3 − 1} · · · {ak, . . . , n}b1 b2 · · · bk

)

in the case 1 = b1. Note that rank(γ) ≥ rank(γ). A direct calculation showsthat

α = γk · · · γ1βm · · ·β1 = γk · · · γ1βm · · ·β1,

which completes the proof of the theorem.

For six of our semigroups, the statement of Theorem 14.4.1 can bestrengthened.

Theorem 14.4.3 The semigroups Fn and PFn are generated by ε and allidempotents of rank (n − 1).

Proof. Taking Theorem 14.4.1 into account, it is enough to show that everyelement of rank (n − 1) decomposes into a product of idempotents of rank(n − 1).

For the semigroup Fn we use induction on the number of fixed points.Obviously, an element α ∈ Fn is an idempotent if and only if it has (n − 1)fixed points. Let α ∈ Fn be an element of rank (n − 1). Then we can write

α =(

1 · · · k · · · m − 1 m m + 1 · · · na1 · · · ak · · · am−1 ak am+1 · · · an

)

for some k < m. Consider the element

β =(

1 · · · k · · · m − 1 m m + 1 · · · na1 · · · ak · · · am−1 m am+1 · · · an

).

Then α = βεk,m. If rank(β) = n, then β = ε and α = εk,m. If rank(β) = n−1,then β has more fixed points than α and hence decomposes into a productof idempotents by the inductive assumption. This proves the statement ofour theorem for the semigroup Fn.

Let now α be an element of rank (n−1) from PFn\Fn and assume thatN\dom(α) = {k}. If k �∈ im(α), then α acts on N\{k} injectively and order-decreasing, and hence is the identity transformation. This implies that α isan idempotent. If k ∈ im(α), the element α can be written in the followingform:(

1 · · · k − 1 k k + 1 · · · k + m − 1 k + m k + m + 1 · · · n1 · · · k − 1 ∅ k + 1 · · · k + m − 1 k bk+m+1 · · · bn

)

for some m > 0. Consider the element β defined as follows:

β(x) =

{x, x < k + m;α(x), x ≥ k + m.


Then α = βε(k) and β ∈ Fn is an element of rank (n − 1). As we havealready proved the statement of the theorem for the semigroup Fn, β de-composes into a product of idempotents. Hence α decomposes into a productof idempotents as well. This completes the proof.

The next statement follows immediately from the definitions:

Lemma 14.4.4 (i) Every element of rank (n− 1) from the semigroup Cn

has the form(

1 · · · k k + 1 · · · m m + 1 · · · n1 · · · k k · · · m − 1 m + 1 · · · n

),

where 1 ≤ k < m ≤ n.

(ii) Every element of rank (n−1) from the semigroup On either belongs toCn or has the form

(1 · · · m m + 1 · · · k k + 1 · · · n1 · · · m m + 2 · · · k + 1 k + 1 · · · n

),

where 1 ≤ m < k ≤ n.

(iii) Every element of rank (n − 1) from the semigroup IOn has the form(

1 · · · · · · · · · · · · k − 1 k + 1 · · · n1 · · · m − 1 m + 1 · · · · · · · · · · · · n

),

where 1 ≤ m ≤ n and 1 ≤ k ≤ n.

(iv) Every element of rank (n − 1) from the semigroup ICn has the form(

1 · · · k − 1 k + 1 · · · m m + 1 · · · n1 · · · k − 1 k · · · m − 1 m + 1 · · · n

),

where 1 ≤ k ≤ m ≤ n.

Theorem 14.4.5 The semigroups On, Cn, POn, and PCn are generated byε and all idempotents of rank (n − 1).

Proof. It is again enough to show that every element of rank (n−1) decom-poses into a product of idempotents. Let α be an element of rank (n − 1)from one of our semigroups. For the semigroup Cn the statement follows fromthe observation that the element α as in Lemma 14.4.4(i) has the followingdecomposition:

α = εm−1,m · · · εk+1,k+2εk,k+1.

For all elements from the semigroup On as in Lemma 14.4.4(ii) one con-structs a decomposition similarly.


If α ∈ POn\On, then α ∈ IOn and has the form as in Lemma 14.4.4(iii).In the case k ≥ m we have

α = εm+1,m · · · εk,k−1ε(k).

The case k ≤ m is similar. This proves our statement for the semigroupsPOn and PCn simultaneously.

Remark 14.4.6 In the semigroups IOn, ICn, and IFn all idempotentsform a commutative subsemigroup. In Examples 14.1.2 and 14.1.4 we sawthat already IO2, IC2, and IF2 are not commutative. Hence none of thesemigroups IOn, ICn, and IFn is generated by idempotents for n > 1.


14.5.1 The semigroup On, disguised as the semigroup of all endomorphismsof a finite linearly ordered set, appears in the works [Ai2, Ai3, Ai4] of Aizen-shtat. It was further studied by Howie in [Ho2] (some results of the latter pa-per overlap with [Ai2]). However, a really intensive study of order-preservingtransformations started in the 1990s.

The semigroup Fn appears already in [Pi] in connection with the studyof formal languages. In 1990–1992, Howie commented on the importance ofthe study of order-decreasing transformations in [Ho5]. A deeper study ofFn seems to start with the works [Um2, Um3] of Umar.

14.5.2 Instead of order-decreasing transformations one can of course studythe dual notion of order-increasing transformation (that is, α(x) ≥ x for allx ∈ N). In this case instead of the semigroups PFn, Fn, IFn, PCn, Cn andICn we obtain the corresponding semigroups PF+

n , F+n , IF+

n , PC+n , C+

n andIC+

n . Obviously, the semigroups S and S+ (where S is from the above list)are isomorphic.

14.5.3 The semigroups defined in Sect. 14.1 admit many variations. Themost obvious one is to substitute the natural linear order on N by someother linear order. The obtained subsemigroups will be obviously isomorphic(Sn-conjugate) to the corresponding original semigroups.

Another natural generalization is to take some partial order on N in-stead of a linear order. Additionally, one may assume that the partial ordercontains a minimal element. This leads to the definition of a huge variety offinite transformation semigroups. Not much is known about them.

Instead of POn, On, and IOn one could consider slightly bigger “super-versions” of these subsemigroups, consisting of all transformations, which areeither order-preserving or order-reversing (i.e., x ≤ y implies α(x) ≥ α(y)for all x, y ∈ dom(α)). This idea appears, in particular, in [CH].


Another variation is connected with those transformations which pre-serve the cyclic orientation rather than the natural order (see for example[Cat, CH, Fe2]). A (partial) transformation α is called orientation-preservingprovided that we have dom(α) = {a1 < a2 < · · · < am} and there exists ksuch that

α(ak) ≤ α(ak+1) ≤ · · · ≤ α(am) ≤ α(a1) ≤ α(a2) ≤ · · · ≤ α(ak−1).

14.5.4 Catalan numbers, their properties, and various algebraic and combi-natorial interpretations are studied in many papers and even monographs,see for example [Sta1, Chap. 6] and [Sta2].

14.5.5 In the coordinate plane consider the piecewise linear paths from(0, 0) to (n, n), which satisfy the following two conditions:

• From the point (a, b) it is allowed to go directly either to the point(a + 1, b) or to the point (a, b + 1) or to the point (a + 1, b + 1).

• It is not allowed to go over the diagonal y = x.

Such a path is called a Schroder path of order n. The number rn of suchpaths equals

rn =1

n + 1

n∑k=0

(n + 1n − k

)(n + k

k

).

The sequence {rn : n ≥ 0} satisfies the following recursion:

r0 = 1; r1 = 2; (n + 2)rn+1 = 3(2n + 1)rn − (n − 1)rn−1, n > 0.

Theorem 14.5.1 ([LU2]) |PCn| = rn.

Proof. Each Schroder path contains equal numbers of horizontal and verticalsteps. Let l be some Schroder path of order n and k be the number ofhorizontal (vertical) steps in l. For every vertical step (i, j) → (i, j + 1)we write down the number a = j + 1. The k numbers, which we obtain,are pairwise different and we write them in the natural order as follows:a1 < a2 < · · · < ak. After that for every horizontal step (i, j) → (i+1, j) wewrite down the number b = j + 1. Some of the k numbers, which we obtain,may coincide; however, we write them (with multiplicities) in the naturalorder b1 ≤ b2 ≤ · · · ≤ bk. Then

αl =(

a1 a2 . . . ak

b1 b2 . . . bk

)

is an element from PCn. One checks that the mapping l → αl is a bijectionfrom the set of all Schroder paths to PCn. The statement follows.


14.5.6 The results about cardinalities of semigroups described in the pre-sent chapter are taken from the following papers: Propositions 14.2.1 and14.2.2 are taken from [Ho6]; Proposition 14.2.3 is taken from [Ho2]; Propo-sition 14.2.4 is taken from [Ga1]; Proposition 14.2.6 is taken from [BRR];Theorem 14.2.8(i) is taken from [Hi2]; Theorem 14.2.9 and Corollary 14.2.11are taken from [GH2].

14.5.7 The paper [LU3] contains the following more compact formula for|E(OPn)|:

|E(OPn)| = 5n−1

2

((√5 + 12

)n

−(√

5 − 12

)n)+ 1.

14.5.8 The results about idempotents described in Sect. 14.3 are takenfrom the following papers: Proposition 14.3.2 is taken from [Hi2]; Theo-rem 14.3.4 is taken from [LU2]; Proposition 14.3.5(i) is taken from [Um2];Theorem 14.3.6 is taken from [Ho6]; Theorem 14.3.7 is taken from [LU3].

14.5.9 Theorem 14.4.1 for the semigroup IOn was proved in [Fe3]; Theorem14.4.3 for the semigroup Fn was proved in [Um2]; Theorem 14.4.5 for thesemigroup On was proved in [Ai2, Ho2]; Theorem 14.4.5 for the semigroupPOn was proved in [GH2].

14.5.10 Let S be a semigroup. The minimum cardinality of a generatingsystem of S is usually called the rank of S and denoted by rank(S). Ifone restricts attention to generating systems of special kind, for exampleconsisting of idempotent or nilpotent elements, one speaks of the idempotentrank idrank(S) of S and the nilpotent rank nilrank(S) of S etc.

Several papers are dedicated to the study of various ranks for some ofthe semigroups considered in this chapter. For example, Aizenshtat provedin [Ai2] that On has a unique irreducible generating system, consisting ofidempotents, namely, the set of all idempotents of rank n and (n−1). In par-ticular, this implies that idrank(On) = 2n−1. Later on (and independently)this was rediscovered by Howie in [Ho2]. Analogous result about irreduciblegenerating systems for Cn is obtained by Higgins in [Hi2]. In particular,idrank(Cn) = n. In [GH2] it is shown that rank(On) = n + 1, rank(POn) =2n, idrank(POn) = 3n − 1. In [Fe3] it is shown that rank(IOn) = n + 1.

We would like to note that for a monoid S one usually leaves the identityelement out of all generating systems, which explains the difference betweenthe above formulations and the original formulations, which can be foundin the cited papers.

14.5.11 If M is a generating system of the semigroup S, then for every a ∈ Sthere exists a decomposition of a into a product of generating element, whichhas the minimal possible length. The maximum of all such minimal lengthsover all a ∈ S is called the depth of M .


For the semigroups On and Cn the depths of certain generating systemswere studied by Higgins in [Hi3]. In particular, Higgins showed that forboth these semigroups the depth of the generating system consisting of allidempotents equals (n−1); and the depth of the generating system consistingof all idempotents of rank at least (n − 1) equals �n2/4�.14.5.12 Set

IOn(n − 1) = {α ∈ IOn : rank(α) = n − 1}.

Let KN denote the full directed graph with the set N of vertices (i.e., forevery x, y ∈ N the graph KN contains a unique oriented edge from x to y).With every α ∈ IOn(n− 1) we associate the arrow a → b of the graph KN,where a = N\dom(α) and b = N\im(α). Then every subset M ⊂ IOn(n−1)is uniquely determined by some subgraph ΓM of KN (on the same set ofvertices) and vice versa. Recall that a directed graph Γ is called stronglyconnected provided that for any two vertices a, b in Γ there exists an orientedpath in Γ going from a to b.

Theorem 14.5.2 ([GM2]) A generating system of IOn is irreducible if andonly if it has the form M ∪ {ε}, where M ⊂ IOn(n − 1) is such that thegraph ΓM is a minimal strongly connected directed subgraph of KN.

Theorem 14.5.3 ([GM2]) Every irreducible generating system of IOn con-tains at least n + 1 elements. In particular, rank(IOn) = n + 1 and IOn

contains exactly (n− 1)! irreducible generating systems of cardinality n + 1.

14.5.13 Let S denote one of the semigroups Fn, PFn, IFn, Cn, PCn, ICn.Each of these semigroups contains a zero element (for Fn and Cn this zeroelement is the transformation 01, for all other semigroups, this zero elementis the transformation 0). The following statement follows directly from def-initions:

Proposition 14.5.4 (i) An element α ∈ S is nilpotent if and only if

(a) α(x) < xfor allx �= 1 (forS = Fn, Cn), or(b) α(x) < xfor allx ∈ dom(α) (forS = PFn, IFn,PCn, ICn).

(ii) The set Nil(S) of all nilpotent elements of S forms an ideal.

Proposition 14.5.5 ([Um2, LU5]) (i) |Nil(Fn)| = (n − 1)!.

(ii) |Nil(Cn)| = Cn−1.

Proof. Taking Proposition 14.5.4 into account, the statement (i) is obvious.To prove the statement (ii) we observe that there is a natural bijectionbetween Cn−1 and Nil(Cn), constructed in the following way:

Cn−1 �(

1 2 . . . n − 11 a2 . . . an−1

)↔

(1 2 3 . . . n − 1 n1 1 a2 . . . an−2 an−1

)∈ Nil(Cn).

The statement (ii) now follows directly from Theorem 14.2.8(i).


14.5.14 Let S denote one of the semigroups POn or IOn. The semigroupS contains the zero element 0. An element α ∈ S is nilpotent if and onlyif α(x) �= x for all x ∈ dom(α). However, in this case nilpotent elementsdo not form an ideal, they do not even form a subsemigroup. For each ofthese semigroups, the subsemigroup, generated by all nilpotent elements,was studied by Garba in [Ga1, Ga2]. In particular, every element of each ofthese semigroups is a product of at most three nilpotent elements.

14.5.15 Nilpotent subsemigroups of IOn are studied in detail in [GM2].

Theorem 14.5.6 ([GM2]) Let ≺ be an arbitrary linear order on N. Thenthe set

T (≺) = {α ∈ IOn : α(x) ≺ xfor allx ∈ dom(α)}

is a maximal nilpotent subsemigroup of IOn and every maximal nilpotentsubsemigroup of IOn coincides with T (≺) for some linear order ≺ on N. Inparticular, IOn contains exactly n! maximal nilpotent subsemigroups.

Theorem 14.5.7 ([GM2]) If < is the natural order on N, then we have|T (<)| ≥ |T (≺)| for any linear order ≺ on N. Moreover, |T (<)| = Cn.

In connection with the latter theorem, we also refer the reader toExercise 8.6.14.

14.5.16 In the following theorem another order-related subsemigroup ap-pears:

Theorem 14.5.8 ([Ho4]) (i) The semigroup SPOn = POn\On of allstrictly partial order-preserving transformations of N is not generatedby idempotents.

(ii) rank(SPOn) = 2n − 2.

14.5.17 A description of all maximal and maximal inverse subsemigroupsof IOn is given in [GM2]. The semigroup IOn contains 2n − 1 maximalsubsemigroups and 2n−1 maximal inverse subsemigroups. All maximal sub-semigroups in On are described in [X.Ya2]. It turns out that On containsn2−2n+2 maximal subsemigroups. The paper [LY] contains a description ofmaximal subsemigroups of On among the semigroups with some additionalproperties, for example regular subsemigroups (there are 2n − 2 maximalsemigroups among all regular subsemigroups) or subsemigroups generatedby idempotents (there are 2n − 3 maximal subsemigroups among all sub-semigroups which are generated by idempotents).

14.5.18 Presentations are known only for some of the semigroups whichwere studied in this chapter. A presentation for On with respect to theirreducible system of generators, consisting of idempotents of rank at least


(n−1), was given already by Aizenshtat in [Ai2]. A presentation for POn wasfound by Solomon in [Sol]. In [Fe3] Fernandes constructs a system of definingrelations for IOn with respect to the n+1-element generating system, whichcorresponds to the oriented cycle n → n − 1 → . . . → 2 → 1 → n in termsof 14.5.12.

14.5.19 Not much seems to be known about congruences on the semigroupsstudied in this chapter. Recall that a semigroup S is said to be semisimpleprovided that all congruences on S are Rees congruences. In [Ai3] Aizenshtatshows that the semigroup On is semisimple. In [Fe3] Fernandes proves thesame result for IOn. Thus there are exactly n congruences on On and n+1congruences on IOn.

14.5.20 Description of both one-sided and two-sided ideals for the semi-groups On, POn and IOn is the same as the corresponding description forthe semigroups Tn, PT n and ISn, respectively (the latter one is given inTheorems 4.2.1, 4.2.4 and 4.2.8). In particular, all Green’s relations on thesemigroups On, POn and IOn are just restrictions of the correspondingGreen’s relations on the semigroups Tn, PT n and ISn, respectively. For de-tails, we refer the reader to [Ai4] for the case On, to [LU3] for the case POn

and to [Fe1, GM2] for the case IOn.The structure of Green’s relations in Fn is more interesting:

Theorem 14.5.9 ([Pi, Um2]) Let α, β ∈ Fn.

(i) αLβ if and only if α = β.

(ii) αRβ if and only if im(α) = im(β) and for every a ∈ N we havemin{x : α(x) = a} = min{y : β(y) = a}.

Corollary 14.5.10 ([Um2]) In the semigroup Fn we have the equalitiesH = L and R = D = J .

14.5.21 Finally, we present some asymptotic results.

Theorem 14.5.11 ([Hi2]) |E(Cn)||Cn| ∼ (n+1)

√πn

2n+1 .

Theorem 14.5.12 ([LU6]) (i) |E(POn)| ∼ 1√5

(5+

√5

2

)n.

(ii) |POn| = 1π

∫ π0 (2 +

√2 cos t)(3 + 2

√2 cos t)n−1dt ∼ (

√2+1)2n

23/4√

πn.

Theorem 14.5.13 ([LU6])

|PCn| =1

π(n + 1)

∫ π

0

(4 + 3√

2 cos t)(3 + 2√

2 cos t)n−1dt ∼ (√

2 + 1)2n+1

23/4n√

πn.



14.6.1 Prove the formula from 14.5.7 using Theorem 14.3.7.

14.6.2 Prove Lemma 14.4.4.

14.6.3 Compute the number of elements of rank (n − 1) in the semigroups

(a) On

(b) Cn

(c) POn

(d) PCn

14.6.4 ([LU5]) Prove that the number of those elements from On, the setof fixed points for which coincides with {1, n}, equals Cn−1.

14.6.5 ([LU5]) Let

f(n, r, k) = |{α ∈ On : α(n) = kandαhas exactly rfixed points}|.

Show that f(n, r, k) =(n+k−2

k−r

)−(n+k−2k−r−1

).

14.6.6 ([Hi2]) Let

F (n, r) = |{α ∈ On : αhas exactly rfixed points}|.

Show that F (n, r) = rn

(2n

n+r

).

14.6.7 ([LU5]) Prove that

|{α ∈ On : α(n) = k}| =(

n + k − 2k − 1

).


|{α ∈ On : rank(α) = r}| =(

n

r

)(n − 1r − 1

).


|{α ∈ On : rank(α) = r and α(n) = k}| =(

k − 1r − 1

)(n − 1r − 1

).

14.6.10 ([LU2]) Prove that the number of those idempotents δ ∈ PCn, inwhich the maximum element from im(δ) is k, equals 2n−k−1(3k−1 + 1).



|{α ∈ POn : |dom(α)| = r and max(im(α)) = k}| =(

n

r

)(k + r − 2

k − 1

).


|{α ∈ POn : |dom(α)| = r}| =(

n

r

)(n + r − 1

n − 1

).

14.6.13 ([LU3]) Prove that the numbers an = |E(PFn)| satisfy the followingrecursion: an+1 = 1 + 5(an − an−1).

14.6.14 ([Um1]) Let F (α) denote the set of all fixed points of an elementα ∈ PFn. Prove that F (αβ) = F (βα) = F (α) ∩ F (β).

14.6.15 For 1 < k ≤ n set Ik = {α ∈ PFn : rank(α) ≤ k} and letSk = Ik/Ik−1 be the corresponding Rees quotient. Prove that every elementof Sk is either an idempotent or a nilpotent.

14.6.16 ([LU4]) Let

J(n, r, k) = |{α ∈ Cn : rank(α) = r and α(n) = k}|.

Prove that

(a) J(n, k, k) =(n−1k−1

).

(b) J(n, r, k) = n−k+1n−r+1

(n−1r−1

)(k−2r−2

).


(a) |{α ∈ Cn : rank(α) = r}| = 1n−r+1

(n−1r−1

)(nr

).

(b) |{α ∈ Cn : α(n) = k}| = n−k+1n

(n+k−2

n−1

).

14.6.18 ([Hi2]) Let N(n, k) denote the number of those elements from Cn

which have exactly k fixed points. Prove that

(a) N(n, k) = k2n−k

(2n−k

n

).

(b) N(n + 1, k) = N(n, k − 1) + 2N(n, k) + N(n, k + 1).

14.6.19 ([HMR]) Show that there exist two elements α, β ∈ Tn such thatTn = 〈On ∪ {α, β}〉.

14.6.20 Prove that the semigroups On, POn and IOn are regular.

14.6.21 Prove that for n > 2 the semigroups Fn, PFn, IFn, Cn, PCn, andICn are not regular.


14.6.22 ([Ai2, GM2]) Prove that for n > 1 we have

|Aut(On)| = |Aut(IOn)| = |Aut(POn)| = 2.

14.6.23 Prove Theorem 14.5.11.

14.6.24 Let t(n, r) = |{α ∈ Fn : rank(α) = r}|. Prove that

t(n, r) = r · t(n − 1, r) + (n − r + 1)t(n − 1, r − 1).

14.6.25 ([Um1]) Prove that

|{α ∈ Fn : rank(α) = r}| =r−1∑k=0

(−1)k

(n + 1

k

)(r − k)n.

14.6.26 ([Um1]) Recall that Stirling numbers of the first kind s(n, k) aredefined as coefficients

x(x − 1)(x − 2) · · · (x − n + 1) =n∑

k=1

s(n, k)xk.

Prove that

|{α ∈ Fn : αhas exactlykfixed points}| = (−1)n−ks(n, k).

14.6.27 ([LU3]) Let α ∈ POn and rank(α) = k. Prove that in the semigroupPOn we have:

(a) |L(α)| =(nk

).

(b) |R(α)| =n∑

i=k

(n

i

)(i − 1k − 1

).

(c) |R(α)| =n−k+1∑

i=1

(n − i

k − i

)2n−k−i−1.

14.6.28 ([GH2]) Prove that the D-class

Dn−1 = {α ∈ On : rank(α) = n − 1}

of the semigroup On contains exactly (n− 1) different L-classes and exactlyn different R-classes.

14.6.29 Prove Theorem 14.5.9.

Answers and Hintsto Exercises

1.5.3 (a): 7, (b): 19, (c): 6, (d): 16, (e): 45.1.5.4 8!.1.5.6 (a): (n − k)n for Tn and (n + 1 − k)n for PT n,(b):

∑ki=0(−1)i

(ki

)(n − i)n for Tn and

∑ki=0(−1)i

(ki

)(n + 1 − i)n for PT n,

(c):∑k

i=0(−1)i(ki

)(k − i)n for Tn and

∑ki=0(−1)i

(ki

)(k + 1 − i)n for PT n.

Hint: For (b) and (c) use the inclusion–exclusion formula.1.5.7 Hint: Use Cayley’s theorem on the number of labeled trees.1.5.8 Hint: See [Hi1, 6.1.1(b)]1.5.9 (a): nn − (n − 1)n for at least one fixed element and n · (n − 1)n−1 forexactly one fixed element,(b): (n+1)n −nn for at least one fixed element and nn for exactly one fixedelement.

2.10.3 Hint: To each α ∈ PT n associate α ∈ Tn+1 as follows:

α(i) =

{α(i), i ∈ dom(α);n + 1, otherwise.

2.10.9 Hint: Use Corollary 2.7.4.2.10.10 Hint: Use Corollary 2.7.4.2.10.18 Hint: Rewrite the signless Lah number L′(n, n − k) in the formL′(n, n − k) =

(nk

)(n−1

k

)k! and use Theorem 2.5.1 and Corollary 2.8.6.

2.10.20 Hint: Use Theorem 2.5.1.2.10.23 Hint: If α �= 0, the equation α ·x = 0 may have at most nn solutions,whereas the equation x · α = 0 may have at most (n + 1)n−1 solutions. Butnn > (n + 1)n−1.

3.3.2 (a) Hint: Use Cayley’s theorem on the number of labeled trees.3.3.3 (a) Hint: Sn is not commutative. (c) Hint: Show that one can write(i, i + 1), i < n, as a product of (1, 2) and powers of (1, 2, . . . , n).

277

278 ANSWERS AND HINTS TO EXERCISES

4.4.10 Hint: For each a ∈ S the condition of the exercise gives elements e(l)a

and e(r)a such that e

(l)a a = ae

(r)a = a. Show first that e

(l)a b = b and be

(r)a = b

for all a, b and then that e(l)a = e

(r)a = e is the identity element of S.

4.8.1 Hint: To prove the first equality count how many different partitions ofN∪ {n + 1} generate the same partition of N. To prove the second equalitycount the number of those partitions of N ∪ {n + 1} for which the blockcontaining the element n + 1 has cardinality k + 1.4.8.3 Hint: The number of anti-chains in B(N) is smaller than the cardinalityof B(N). On the other hand, each collection of elements from B(N) havingthe same cardinality is an anti-chain.4.8.4 (a): 5; (b): 19; (c): 167.4.8.7 For n > 4 the function f(x) = xn/nx satisfies the condition f(2) > 1and f(n) = 1 and has on [2, n] the unique local extremal point, namely,some local maximum.4.8.8 (a): 1, 25, 200, 600, 600, 120; (b): 5, 300, 1, 500, 1, 200, 120; (c): 1, 155,1, 800, 3, 900, 1, 800, 120.4.8.9 Hint: If S is inverse, e, f ∈ S are idempotents and eS = fS, thenfe = e, ef = f and thus e = f . If every principal one-sided ideal is generatedby a unique idempotent and b, c are inverse to a, then for the idempotentsab, ac, ba and ca we have abS = aS = acS, Sba = Sa = Sca, which impliesab = ac, ba = ca and b = bac = c.4.8.10 In and In−1.4.8.11 In and In−1.4.8.12 In.4.8.13 (a): kn−k; (b): (k + 1)n−k.4.8.14 (a): n1n2 · · ·nk; (b): n1n2 · · ·nk.4.8.18 (a): (n + 1)n−rank(α); (b): (|dom(α)| + 1)n.4.8.19 Hint: Each such ideal contains all idempotents of rank one. If εα = 0(αε = 0) for every idempotent ε of rank one, then α = 0.

5.1.5 Hint: Use the proof of Theorem 4.7.4.5.5.1 Hint: Each maximal subgroup is an H-class, but each H-class of bothTn and ISn is at the same time an H-class of PT n.5.5.2 (d) Hint: If b is an inverse to a and ab = ba, then e = ab is anidempotent and a is invertible in eSe.5.5.8 (a) Hint: Every generating system of 〈a〉 must contain a. (b) Hint:If p1, p2, . . . , pk are pairwise different primes and m = p1p2 · · · pk, then{m/p1, m/p2, . . . ,m/pk} is an irreducible generating system of Zm.5.5.9 (a) Hint: An element of type (k, m) in Tn is uniquely determined byan ordered partition N = N1 ∪ N2 ∪ · · · ∪ Nk+1, a collection of mappingsNi → Ni+1, i = 1, . . . , k, and a permutation of order m on Nk+1. (b) Hint:Additionally to (a) we have a (possibly empty) block N0 = dom(α). (c) Anelement of type (k, m) in ISn is uniquely determined by an ordered partition

ANSWERS AND HINTS TO EXERCISES 279

N = N1 ∪ N2 ∪ · · · ∪ Nk+1, a collection of injective mappings Ni → Ni+1,i = 1, . . . , k, and a permutation of order m on Nk+1.5.5.11 Hint: S = (Z3, +), T = {0}.5.5.12 n · 2n−1 + n(n−1)

2 + 3.5.5.13 (b) Hint: No. Consider the semigroup (N, ·) and its subsemigroups pN

and pqN, where p and q are different primes.5.5.14 Hint: Consider an R-class and an L-class of a rectangular band.

6.6.3 These are all decompositions into left resp. right cosets with respectto some subgroup.6.6.4 No.6.6.8 Hint: We have to count the number of collections k1 < k2 < · · · <l2 < l1 from Theorems 6.5.3 and 6.5.4. Use the fact that the number ofways to write n as a sum n = n1 + · · · + nt of positive integers (here t canvary), equals 2n−1. For ISn all necessary collections can be obtained in thefollowing way: write n + 1 as the sum n + 1 = n1 + · · · + nt, t > 1, if tis odd, delete the last summand, and then consider the collection n1 − 1,n1 + n2 − 1, n1 + n2 + n3 − 1,. . . . For PT n all necessary collections can beobtained in the following way: write n + 1 as the sum n + 1 = n1 + · · ·+ nt,t > 1, and consider the collection n1 − 1, n1 + n2 − 1, n1 + n2 + n3 − 1,. . . .If t is odd we can either delete the last summand or double the middle one.If t is even, we can either take this collection, or delete the last summandand double the middle one of those which are left. One has also to take intoaccount that Δ coincides with its own transpose.

7.4.11 Hint: Follow the proof of Theorems 7.4.1 and 7.4.7.7.5.7 Hint: Follow the proof of Corollaries 7.5.1 and 7.5.6.7.7.2 0 and 0a, a ∈ N.7.7.4 14.7.7.5 {ε,0} for ISn and PT n and {ε} for Tn.7.7.6 Hint: Determine first all automorphism which induce the identity map-ping on the set of all idempotents of I1.7.7.7 Hint: Go through the lists given by Theorems 7.4.1, 7.4.7, and 7.4.10.7.7.9 Hint: Each finite semigroup has an idempotent. At the same timemapping the whole semigroup to an idempotent is always an endomorphism.7.7.11 Each endomorphism of (N, +) has the form x → k · x for some fixedk ∈ N.

8.2.9 Hint: Show that Nτm can be mapped to Nτn using an inner automor-phism of PT n. Analogously for ISn.8.6.1 Hint: Consider the cyclic semigroup of type (k, 1).


8.6.4 (b): For example, S is a nontrivial group, T consists of one elementand ϕ is the unique mapping from S to T .8.6.5 Hint: Take the product a1 · · · alk and consider the factors a1 · · · ak,ak+1 · · · a2k,. . . , a(l−1)k+1 · · · akl of this product. Show that they all belongto X.8.6.6 Hint: Consider all possible linear combinations of the elements of Tand show that they form a nilpotent subalgebra of Matn(C).8.6.7 Hint: Prove that each maximal subsemigroup of Matn(C) is a subalge-bra and use flags of subspaces in C

n instead of partial orders.8.6.9 Hint: Consider the subsemigroup T = In ∪ {[1, 2, 3][4] . . . [n]} of ISn.The maximal nilpotent subsemigroups of T , corresponding to the naturallinear order and the linear order 3 < 2 < 1 < 4 < · · · < n, have differentcardinalities.8.6.10 1 +

∑nk=2(m1 + m2 + · · · + mi−1) · mi, where mi = |Mi| for all i.

8.6.11 Hint: Use Theorem 8.4.12.

9.6.2 {(at, at+lm) : t ≥ k, l ≥ 1} ∪ {(an, an) : n ≥ 1}.9.6.9 Write α as a product α = μν, where μ is a permutation and ν =(

{i1, . . . , im} B1 · · · Bk

im b1 · · · bk

)is an idempotent in Tn.

10.1.1 Hint: For every a ∈ M the set {ϕ(s)(a) : s ∈ S} is invariant.10.1.2 For example, the natural action of the subsemigroup S of ISn, gen-erated by the element [1, 2, . . . , n], on N.10.7.1 This action is always faithful but never transitive. It is quasi-transitiveunless S = Sn, or S = Tn.10.7.2 Yes.10.7.3 The action is faithful. It is neither transitive nor quasi-transitive un-less S = S1 = T1.10.7.4 The action is faithful. It is not transitive unless S = S1 = T1. It isnot quasi-transitive unless S = S1 = T1, or S = IS1 = PT 1.10.7.5 The action is faithful and quasi-transitive but never transitive.10.7.7 Only the trivial action.10.7.12 Only the trivial action and the actions trivially induced from Sn (i.e.,where only the action of invertible elements is different from 0). Hint: If n >1 and ϕ : Tn → ISm is a homomorphism, different from the ones listed above,then by Theorem 6.3.10 the homomorphism ϕ is injective on idempotents ofrank (n − 1). However, such idempotents of Tn do not commute in general.10.7.13 Only the trivial action and the actions trivially induced from Sn

(i.e., where only the action of invertible elements is different from 0). Hint:See the hint to 10.7.12.10.7.14 Hint: It is enough to consider the actions of the form ξe,H and showthat for any π ∈ Sn and α ∈ ISn such that π|dom(α) = α we have α ·x = π ·x.10.7.15 (b) Hint: Consider the action a · x = ax.

ANSWERS AND HINTS TO EXERCISES 281

11.1.5 Hint: for S = Sn and S = Tn the subspace {(c1, . . . , cn) ∈ Cn :

c1 + · · · + cn = 0} is a submodule of the natural module.11.5.1 Hint: Use 11.4.4(ii).11.7.3 If A ⊂ N, then the simple module corresponding to A is CA = C withthe action B · c = c, c ∈ C, B ⊂ N, if A ⊂ B and B ·C = 0 otherwise. Everymodule is a direct sum of simple modules.11.7.4 n.11.7.5 For example, the module V (M), where M is the trivial H(01)-module.11.7.9 (b) Hint: A finite semigroup is nilpotent if and only if it has a zeroelement and this zero element is the only idempotent.11.7.11 Hint: All elements s ∈ S\G annihilate M ⊗C V .11.7.17 Hint: Consider for example the semigroup S = 〈x : x2 = x3〉.11.7.18 A rectangular band has two simple modules: the trivial module,and the one-dimensional module with the zero multiplication. A rectangularband is a monoid if and only if it has only one element. Hence in the case,when a rectangular band has exactly one element, and we consider it as amonoid, there is only one simple module: the trivial one.11.7.19 (b) Hint: Let X be a submodule of V (M), which is not contained inN(M). Show first that there exists v ∈ X such that e ·v �= 0. Show then thate · v ∈ X is a nonzero vector in the H(e)-submodule M of V (M). Finally,use the construction of V (M) to show that such X contains V (M).11.7.20 (a) Hint: Let X be a simple S-module. Show first that there existse ∈ E(S) such that e · X �= 0, while f · X = 0 for all idempotents f ∈ SeSsuch that f �∈ D(e). Let M = e · X and M ′ be a simple submodule of M .Similarly to Lemma 11.4.3 show that there is a nonzero homomorphism fromV (M ′) to X. Finally, use Exercise 11.7.19 to conclude that X ∼= V (M ′).

12.3.7 Hint: Show that any two H-cross-sections can be transferred into eachother by conjugation.12.8.6 There are theree cross-sections of the form IO≺

3 , and an additionalcross-section from Exercise 12.8.5.12.8.8 Hint: Use the Miller-Clifford Lemma ([Hi1, Theorem 1.2.5]), whichstates that for any a, b ∈ S we have ab ∈ R(a)∩L(b) if and only if R(b)∩L(a)contains an idempotent.12.8.9 The R-classes. Hint: Use Exercise 12.8.8.

13.6.3 Hint: Show that ϕ : (S, ◦a) → (S, ◦b) is an isomorphism.13.6.4 Hint: Use Exercise 13.6.3.13.6.5 Hint: Count the number of idempotents.13.6.6 Hint: Count the number of those elements which cannot be writtenas a product of other elements.13.6.10 Hint: Use Exercise 13.6.9.


13.6.12 Answer:n∑

p=0

p∑m=1

(n − k

m

)(k

p − m

)(n

p

)p!.

13.6.13 Answer: nn if k = 1, nn −k∑

m=1

(n

m

)S(k, m)

m∑j=1

S(n − k, j)(

m

j

)j! if

k > 1.13.6.15 Hint: Use the fact that (·)−1 : B → B is an involution.

14.2.7 Reference: See for example [Grm, 10.5].14.2.10 Hint: Use induction on n.14.6.1 Hint: From the recursive relation f2k = 3f2(k−1) − f2(k−2) deducef2k = 1√

5

(((3 +

√5)/2)k + ((3 −

√5)/2)k

), insert this formula into the sum

from Theorem 14.3.7 and use the binomial formula.14.6.3 (a) n(n − 1). (b) n(n − 1)/2. (c) n(2n − 1). (d) n(n + 1)/2.14.6.4 The graph of such transformation has two connected componentswith m and n − m vertices, respectively. Such components are in bijectionwith nilpotent elements from Cn and Cn−m, respectively. Using Proposi-tion 14.5.5(ii), one obtains a recursive relation for the number of elementsin question, and it remains to use Exercise 14.2.7.14.6.6 Hint: Use Exercise 14.6.5.14.6.7 Hint: Use Exercise 14.6.5.14.6.9 Hint: The first factor gives the number of ways to choose im(α), thesecond factor gives the number of partitions of N into r intervals.14.6.15 If α ∈ PFn is not an idempotent (that is, not all x ∈ im(α) are fixedpoints), then rank(α2) < rank(α).14.6.16 (a) Hint: Choose maximal elements in α−1(1),. . . , α−1(k − 1).14.6.23 Hint: Use Theorem 14.2.8(i), Proposition 14.3.1 and Stirling’s for-mula for n!.14.6.25 Hint: Use Exercise 14.6.24.14.6.27 Hint: (b) R(α) = {β : im(β) = im(α)}. Let im(α) = {a1, . . . , ak},a1 < · · · < ak. To determine all β we first choose the cardinality i suchthat |dom(β)| = i, then dom(β), and, finally, the minimum elements fromthe inverse images β−1(a1),. . . , β−1(ak). (c) To determine all β we first con-sequently choose the minimum elements in the inverse images β−1(a1),. . . ,β−1(ak). Then every element x ∈ {i + 1, . . . , n}, which is not yet chosen, iseither included to dom(β) (in which case the value of β on it is uniquelydetermined) or not.14.6.29 Hint: To prove (i) use Proposition 4.4.1. To prove (ii) use Proposi-tion 4.4.2.

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List of Notation

(Γ1, Γ2, . . . , Γk) Linear notation for a connected element with a cycle

(x0, x1, . . . , xk−1) Oriented cycle

[Γ1, Γ2, . . . , Γk; a] Linear notation for a tree with a sink

[x]m The polynomial x(x − 1)(x − 2) . . . (x − m + 1)

α A transformation

α(x) The value of α at x

α(x) = ∅ α is not defined at x

α : M → M α is a transformation of M

α|B The restriction of α to B

α(K) Restriction of α to the orbit K

αij An element of the form [ui

1, ui2, . . . , u

ij ](a)(b) · · · (c)

αk The element [1, 2, . . . , k](k + 1) · · · (n)

βij An element of the form [ui

j , uij+1, . . . , u

imi

](a)(b) · · · (c)

ηi The canonical identification of M and M (i)

Γα The graph of α

ιS The identity congruence on S

λ � n λ is a partition of n

λu The mapping x → xu

〈A〉 The set of all elements generated by A

〈a〉 A cyclic semigroup, generated by a

〈A|Σ〉 Semigroup generated by A with defining relations Σ

297

298 LIST OF NOTATION

C[S] The semigroup algebra of S over C

Ctriv The trivial S-module

N The set of all positive integers

Zy The set of all residue classes modulo y

0 The nowhere defined partial transformation

0n The nowhere defined partial transformation of N

GLn The group of invertible n × n matrices

KN The full directed graph on N

m A partition of N into a disjoint union of nonempty subsets

N The set {1, 2, . . . , n}

Nn The set {1, 2, . . . , n}

Z(S) The center of the semigroup S

Ak The alternating group

B(X) The Boolean of X

CSn(α) The centralizer of α in Sn

D Green’s relation D

D(a) The D-class of a

Dk The set of all elements of rank k in Tn, PT n, or ISn

E(S) The set of idempotents of S

Ek The trivial subgroup of Sk

Fn The set of all order-decreasing total transformations

H Green’s relation H

H(a) The H-class of a

IFn The set of all order-decreasing partial permutations

IOn The set of all order-preserving partial permutations

IOn The set of all order-preserving partial permutations

IO≺n The set of all ≺-order-preserving partial permutations

LIST OF NOTATION 299

ISn The symmetric inverse semigroup on N

Ik The set of all elements of rank at most k

I∗k The dual symmetric inverse semigroup

Iρ The unique ideal of the congruence ρ

J Green’s relation J

J (a) The J -class of a

L Green’s relation L

L(a) The L-class of a

M(e, H, a) The set used to construct transitive actions by partial transfor-mations

Nn The set of all nilpotent elements of ISn

On The set of all order-preserving total transformations

PFn The set of all order-decreasing partial transformations

POn The set of all order-preserving partial transformations

PT (M) The set of all partial transformations of M

PT n The set of all partial transformations of N

R Green’s relation R

R(a) The R-class of a

SPOn Semigroup of strictly partial order-preserving transformations of N

Sn The symmetric group on N

T (M) The set of all total transformations of M

Tn The set of all total transformations of N

V4 The Klein 4-group as a subgroup of S4

B The bicyclic semigroup

h Congruence associated with variants

in The canonical inclusion of PT n to Tn+1

On The set of all partial orders on N


oα(x) The orbit of x in Γα

p(n) The partition function

rn The number of Schroder paths of order n

t(α) The type of α

tk(α) The number of x ∈ N for which |{y ∈ N : α(y) = x}| = k

0a The constant transformation with the image {a}

Annl(A) The left annihilator of the set A

Annl(a) The left annihilator of the element a

Annr(a) The right annihilator of the element a

Annr(A) The right annihilator of the set A

Aut(S) The set of all automorphisms of S

ct(π) The cyclic type of the permutation π

def(α) The defect of α

dom(α) The domain of α

End(S) The set of all endomorphisms of S

fi(g) The number of fixed points of g

HomS(V, W ) The set of all S-homomorphisms from V to W

idrank(S) Idempotent rank of the semigroup S

im(α) The image of α

Inn(S) The set of all inner automorphisms of S

invσ The number of inversions for σ

Ker(ϕ) The kernel of the homomorphism ϕ

lcm The least common multiple

Matn(C) The algebra of all complex n × n matrices

min(A,≺) The minimal element of A with respect to ≺

nd(a) The nilpotency degree of a

nd(S) The nilpotency degree of a nilpotent semigroup S


nilrank(S) Nilpotent rank of the semigroup S

Nil(S) The set of all nilpotent subsemigroups of S

Nilk(S) The set of all nilpotent subsemigroups of S of nilpotency degree k

rank(α) The rank of α

rank(S) Rank of the semigroup S

stim(α) The stable image of α

strank(α) The stable rank of α

StG(m) stabilizer of the point m with respect to the action of G

trivS trivial action of a semigroup S

trα(x) The trajectory of x in Γα

Bn The nth Bell number

Cn The n-th Catalan number

Cn The n-th Catalan number

In The cardinality of ISn

Ln The total number of chains in the chain decompositions of all ele-ments of ISn

Mn The total number of chains in the chain decompositions of all nilpo-tent elements of ISn

n(PT n) The total number of nilpotent elements in PT n

Nn The total number of nilpotent elements in ISn

Nn The total number of nilpotent elements in ISn

OG The number of G-orbits

Pn The total number of fixed points of all elements of ISn

S(n, k) Stirling numbers of the second kind

s(n, k) Stirling numbers of the first kind

ωS The uniform congruence on S

ωα Binary relation describing orbits of α

+ The addition of residue classes


dom(α) The codomain of α

a The equivalence class of a

aρ The equivalence class of a with respect to the equivalence relation ρ

Kn The semigroup, generated by α1, . . . , αn in Section 12.5

x The residue class of x

←≺ The linear order opposite to ≺←S The dual of S

π′α A special binary relation on N ∪ {n + 1}

πk The transposition (1, k)

πα A special binary relation on N

πρ The canonical projection S � S/ρ

� Partial order on the set of all partitions of N

ρ(S, A) Kernel of the natural epimorphism A+ � S

ρI The Rees congruence with respect to the ideal I

ρα A partition of N, associated to α

ρΣ The minimal congruence containing all relations from Σ

ΣBA System of relations for B induced from that for A

∼pS Relation of primary S-conjugation

∼S Relation of S-conjugation√

e The set of all x such that xm = e for some m

τT A binary relation on N associated with a nilpotent subsemigroup Tof PT n

Υi exceptional endomorphisms

Δ The equality relation

ε The identity transformation

εA The unique idempotent of ISn with domain A

εn The identity transformation on N


ε(k) The idempotent ε{1,...,k−1,k,...,n}

εi,J The product εi,j1εi,j2 · εi,jk, where J = {j1, . . . , jk}

εm,k The idempotent of rank (n − 1) satisfying εm,k(m) = εm,k(k) = m

Λa The inner automorphisms corresponding to a ∈ S∗

Ωϕ An exceptional endomorphism

Φα The binary relation associated with α

Φε An endomorphism of rank one

Ψε,δ An endomorphism of rank two

Θτε,δ An endomorphism of rank three

|a| The order of the element a

|S| The cardinality of S

Ξ An exceptional endomorphism

ξ ◦ η Product of binary relations ξ and η

a ρ b The pair (a, b) belongs to the binary relation ρ

a · b The product of a and b

a ≡ b a is equivalent to b

a ≡ b(ρ) a is equivalent to b with respect to the equivalence relation ρ

aDb a and b are D-equivalent

aHb a and b are H-equivalent

aJ b a and b are J -equivalent

aLb a and b are L-equivalent

aRb a and b are R-equivalent

a ∼pS b a and b are primarily S-conjugate

a ∼S b a and b are S-conjugate

A+ The set of all (finite) words over the alphabet A

a−1 The inverse of a in an inverse semigroup

a−1 The inverse of a


AB The product of the sets A and B

ab The product of a and b

Eα Arrows of Γα

F (X, Y ) The set of all mappings from X to Y

f : X → Y A mapping from X to Y

G/H The set of all left cosets of G modulo H

g · m The element ϕ(g)(m) for the action ϕ of G on M

Ge The maximal subgroup corresponding to the idempotent e

H � G H is a normal subgroup of G

Kn The semigroup Kn ∪ {ε}

L(≺1, . . . ,≺k) An L-cross-section of ISn

n A nonnegative integer

N ′τ The nilpotent subsemigroup of ISn associated to the partial order τ

on N

Nτ The nilpotent subsemigroup of PT n associated to the partial orderτ on N

R(≺1, . . . ,≺k) An R-cross-section of ISn

S/ρ The quotient of the semigroup S modulo the congruence ρ

S/I The quotient of the semigroup S modulo the Rees congruence ρI

S ∼= T S is isomorphic to T

S∗ The group of units of the semigroup S

S0 S if 0 ∈ S, and S ∪ {0} otherwise

S1 S if 1 ∈ S, and S ∪ {1} otherwise

Sa The variant of S with respect to the sandwich element a

Sk {a1a2 · · · ak : ai ∈ S, i = 1, . . . , k}, k > 1

T < S T is a subsemigroup of S

V (M) S-module L-induced from M

V ∼= W V is isomorphic to W


VS(a) The set of elements, inverse to a ∈ S

Vα Vertices of Γα

x mod y The residue of x modulo y

Index

G-conjugate elements, 103G-orbit, 151S-conjugate elements, 104

primarily, 104S-homomorphism, 191S-module, 189Σ-pair, 154λ-tableau, 206

standard, 206H-cross-section, 219ρ-cross-section, 215

actioncentralizer of, 210

action of group, 175kernel of, 177stabilizer, 177

action of semigroup, 175by partial permutations, 175by partial transformations, 175by transformations, 175equivalent, 175faithful, 176invariant subset, 176left, 186natural, 176quasi-transitive, 176right, 186semitransitive, 187similar, 175transitive, 176trivial, 176

alphabet, 153word over, 153

alternating group, 108

annihilatorleft, 65right, 65

anti-chain, 51arrow, 3

head of, 3tail of, 3

automorphism, 93, 111extendable, 248inner, 111

bandrectangular, 36

Bell number, 48, 138bicyclic semigroup, 246binary operation, 16

associative, 16Boolean, 35braid relations, 172Branching rule, 207

canonical epimorphism, 93canonical form, 154canonical projection, 93Catalan number, 152, 254Cayley table, 20Cayley’s Theorem, 21center, 129centralizer, 121, 210chain, 23

prefix of, 229chain-cycle notation, 23closure

transitive, 104concatenation, 153

307

308 INDEX

congruence, 91identity, 91left, 91Rees, 92right, 91uniform, 91

conjugate elements, 103connected component, 4coset

left, 95Coxeter group, 172cross-section, 215cycle, 4, 6

oriented, 4cyclic semigroup, 70

diagramegg-box, 56

digraph, 3direct sum of modules, 200

trivial, 200directed edge, 3directed graph, 3

arrow of, 3head of, 3tail of, 3

connected, 4directed edge of, 3isomorphic, 13oriented path of, 4subgraph of, 4union of, 7vertex of, 3

dual, 33duality

Schur-Weyl, 210

effective representation, 198egg-box diagram, 56element

G-conjugate, 103S-conjugate, 104

primarily, 104D-equivalent, 54

H-equivalent, 55J -equivalent, 55L-equivalent, 53R-equivalent, 54conjugate, 103generated by a set, 39group, 70index of, 71inverse, 23

pair, 23inverse of, 18invertible, 18mididentity, 249nil-, 28nilpotency class, 28nilpotency degree, 28nilpotent, 28order of, 71period of, 71regular, 23sandwich, 237singular, 75type of, 71unit, 18

endomorphism, 93, 114rank of, 114

epimorphism, 93canonical, 93

equivalentD-, 54H-, 55J -, 55L-, 53R-, 54

Ferrers diagram, 206shape of, 206

Fibonacci number, 258forest, 6free semigroup, 153

base of, 153full transformation semigroup, 16

INDEX 309

functionpartition, 242

Gelfand model, 210generating set, 39generating system, 39

Coxeter, 42depth of, 269irreducible, 39

generatorCoxeter, 172

global semigroup, 83graph

empty, 70isomorphic, 103strongly connected, 270

Green’s lemma, 56Green’s relations, 55group, 18

action of, 175kernel of, 177stabilizer, 177

alternating, 108element of

conjugate, 103sub-

normal, 95symmetric, 19

group element, 70

homomorphic image, 94homomorphism, 92, 191

kernel of, 94

ideal, 45generator of, 45left, 45one-sided, 45prime, 66principal, 45reflexive, 66right, 45semiprime, 66two-sided, 45

idempotent, 26

constant on a set, 76singular on a set, 76

identity, 17left, 17mid-, 249right, 17two-sided, 17

imagehomomorphic, 94stable, 33

inclusioncanonical, 149

induced module, 194inflation, 239interval, 253invariant, 26, 31inverse element, 23inverse semigroup, 25involution, 210irreducible representation, 189isomorphic graphs, 103isomorphism, 20, 93

juxtaposition, 153

kernel, 32, 94

lemmaBurnside’s, 151Cauchy-Frobenius, 151Green, 56

letter, 153linear notation, 9

mapping, 15composition of, 15partial, 15

composition of, 15defined on, 15product of, 15

product of, 15value of, 15

maximal nilpotent subsemigroup,132

mididentity, 249

310 INDEX

moduleequivalent, 191induced, 194isomorphic, 191regular, 192Specht, 206tensor product of, 209

module over a semigroup, 189L-induced, 194direct sum, 200equivalent, 191homomorphism, 191indecomposable, 200induced, 194isomorphic, 191quotient, 190regular, 192semisimple, 201simple, 189sub-, 189trivial, 190

modulesindecomposable, 200semisimple, 201

monoid, 17rook, 190

monomorphism, 93morphism

auto-, 93endo-, 93epi-, 93homo-, 92iso-, 93mono-, 93

natural module, 190nil-semigroup, 148nilpotency class, 131nilpotency degree, 131nilpotent semigroup, 131normal subgroup, 95notation

chain-cycle, 23cyclic, 8

linear, 9number

Bell, 48, 138Catalan, 152, 254Fibonacci, 258Lah

signless, 30Stirling

first kind, 275second kind, 29

operationsandwich, 237

orbitkernel, 32

orderpartial

natural, 109oriented path, 4

pair of inverse elements, 23partial order

height of, 134natural, 109

partial transformationtype of, 38

partition, 206partition function, 242path

break of, 4head of, 4oriented, 4Schroder, 268tail of, 4

permutation, 3≺-order-preserving, 220cyclic type of, 105even, 205inversion for, 205odd, 205order-preserving, 220partial, 22

permutational part, 33power semigroup, 83

INDEX 311

prefix, 229presentation, 154

irreducible, 173presentation of semigroup, 154

irreducible, 173Preston-Wagner Theorem, 33projection

canonical, 93

quotient module, 190

rankstable, 33

rank of semigroup, 269idempotent, 269nilpotent, 269

rectangular band, 36regular D-class, 63regular element, 23regular module, 192relation

binaryproduct of, 54

relationsbraid, 172

representation, 189completely reducible, 201

representation of a semigroupcompletely reducible, 201

representation of semigroup, 189effective, 198irreducible, 189

residue, 71residue classes, 71

group of, 71restriction, 26retraction, 129rook monoid, 190

sandwichelement, 237operation, 237semigroup, 237

Schroder path, 268Schur-Weyl duality, 210

semigroup, 16D-class, 54

regular, 63H-class, 55H-cross-section of, 219J -class, 55L-class, 53R-class, 54ρ-retract, 216abelian, 28action of, 175

equivalent, 175faithful, 176invariant subset, 176left, 186natural, 176quasi-transitive, 176right, 186semitransitive, 187similar, 175transitive, 176trivial, 176

algebra, 207aperiodic, 253automorphism of, 93, 111

inner, 111bicyclic, 246cardinality, 16Cayley table, 20center of, 129combinatorial, 253commutative, 28congruence on, 91

identity, 91left, 91quotient modulo, 92Rees, 92right, 91uniform, 91

cross-sectionH-, 219

cross-section of, 215cyclic, 70defining relation, 154

312 INDEX

dual, 33element

canonical form, 154element of

G-conjugate, 103S-conjugate, 104D-equivalent, 54H-equivalent, 55J -equivalent, 55L-equivalent, 53R-equivalent, 54group, 70index of, 71order of, 71period of, 71regular, 23singular, 75type of, 71

endomorphism of, 93, 114rank of, 114retraction, 129

epimorphism of, 93canonical, 93

factor, 92free, 153

base of, 153full transformation, 16global, 83Green’s relations, 55homomorphism of, 92

kernel of, 94ideal of, 45

generator of, 45left, 45one-sided, 45prime, 66principal, 45reflexive, 66right, 45semiprime, 66two-sided, 45

idempotent of, 26inflation, 239inverse, 25

involution of, 210isomorphism of, 20left zero, 35module over, 189L-induced, 194direct sum, 200equivalent, 191homomorphism of, 191indecomposable, 200induced, 194isomorphic, 191quotient, 190regular, 192semisimple, 201simple, 189sub-, 189trivial, 190

monomorphism of, 93morphism of

auto-, 93, 111endo-, 93, 114epi-, 93iso-, 20mono-, 93

multiplication table, 20nil-, 148nilpotency class, 131nilpotency degree, 131nilpotent, 131

class of, 131degree of, 131

null, 35of all partial transformations, 16of all transformations, 16of left identities, 35of right identities, 35opposite, 33order-decreasing, 252order-preserving, 251power, 83presentation, 154

irreducible, 173quotient, 92

Rees, 92

INDEX 313

rank of, 269idempotent, 269nilpotent, 269

Rees quotient, 92regular, 24relation, 153

contained in congruence, 154defining, 154follows, 154trivial, 155

relation oncompatible, 54, 91Green’s, 55left compatible, 54, 91right compatible, 54, 91

representation of, 189completely reducible, 201effective, 198irreducible, 189

retract of, 216right zero, 35sandwich, 237self-dual, 33semisimple, 272singular part of, 42sub-, 17Sn-normal, 108closed, 186completely isolated, 74isolated, 74maximal, 84

subgroup ofmaximal, 69

submodule, 189symmetric, 16symmetric inverse, 22

dual, 210variant of, 237with zero multiplication, 35

semigroup algebra, 207semisimple semigroup, 272set

generating, 39sign module, 205

signless Lah number, 30simple module, 189singular part, 42Specht module, 206stable image, 33stable rank, 33subgraph, 4

maximal connected, 4subgroup

coset moduloleft, 95

maximal, 69normal, 95

subsemigroup, 17Sn-normal, 108closed, 186completely isolated, 74isolated, 74

completely, 74maximal, 84maximal nilpotent, 132nilpotent

maximal, 132symmetric group, 19

sign module, 205symmetric inverse semigroup, 22system

generating, 39depth of, 269irreducible, 39

tensor product, 209theorem

Cayley, 21orbit-counting, 151Preston-Wagner, 33

trajectory, 4terminates, 5

transformation, 1Sn-centralizer of, 121≺-order-preserving, 220bijective, 3chain, 23

prefix of, 229

314 INDEX

codomain of, 2constant, 20cycle of length k, 10defect of, 2domain of, 1fixed point of, 10full, 2graph of, 3identity, 18image

stable, 33image of, 2injective, 3invariant set w.r.t., 26orbit of, 6order-decreasing, 252order-increasing, 267order-preserving, 220, 251order-reversing, 267order-turning, 268partial, 1permutational part of, 33preserves relation, 56range of, 2rank

stable, 33rank of, 2restriction of, 26stable image, 33

stable rank, 33surjective, 3tabular form of, 1total, 2type of, 38value of, 1

transitive closure, 104transposition, 10tree, 6

forest of, 6sink of, 6trivial, 7

trivial block, 260trivial module, 190

unit, 18

variant of semigroup, 237

word, 153concatenation of, 153juxtaposition, 153

Young diagram, 206shape of, 206

zero, 19left, 19right, 19two-sided, 19

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Classical Finite Transformation Semigroups: An Introduction