Classical Finite Transformation Semigroups: An Introduction

Classical Finite Transformation Semigroups

Algebra and Applications Volume 9 Managing Editor: Alain Verschoren University of Antwerp, Belgium Series Editors: Alice Fialowski Eötvös Loránd University, Hungary Eric Friedlander Northwestern University, USA John Greenlees Sheffield University, UK Gerhard Hiss Aachen University, Germany Ieke Moerdijk Utrecht University, The Netherlands Idun Reiten Norwegian University of Science and Technology, Norway Christoph Schweigert Hamburg University, Germany Mina Teicher Bar-llan University, Israel

Algebra and Applications aims to publish well written and carefully refereed monographs with up-to-date information about progress in all fields of algebra, its classical impact on commutative and noncommutative algebraic and differential geometry, K-theory and algebraic topology, as well as applications in related domains, such as number theory, homotopy and (co)homology theory, physics and discrete mathematics. Particular emphasis will be put on state-of-the-art topics such as rings of differential operators, Lie algebras and super-algebras, group rings and algebras, C*algebras, Kac-Moody theory, arithmetic algebraic geometry, Hopf algebras and quantum groups, as well as their applications. In addition, Algebra and Applications will also publish monographs dedicated to computational aspects of these topics as well as algebraic and geometric methods in computer science.

Olexandr Ganyushkin

•

Volodymyr Mazorchuk

Classical Finite Transformation Semigroups An Introduction

ABC

Olexandr Ganyushkin Kyiv Taras Shevchenko University Volodymyrska Street, 64 Kyiv Ukraine [email protected]

Volodymyr Mazorchuk Uppsala University Dept. Mathematics SE-751 06 Uppsala Sweden [email protected]

ISBN: 978-1-84800-280-7 e-ISBN: 978-1-84800-281-4 DOI: 10.1007/978-1-84800-281-4 British Library Cataloguing in Publication Data A catalogue record for this book is available from the British Library Library of Congress Control Number: 2008941019 c Springer-Verlag London Limited 2009  Apart from any fair dealing for the purposes of research or private study, or criticism or review, as permitted under the Copyright, Designs and Patents Act 1988, this publication may only be reproduced, stored or transmitted, in any form or by any means, with the prior permission in writing of the publishers, or in the case of reprographic reproduction in accordance with the terms of licenses issued by the Copyright Licensing Agency. Enquiries concerning reproduction outside those terms should be sent to the publishers. The use of registered names, trademarks, etc., in this publication does not imply, even in the absence of a specific statement, that such names are exempt from the relevant laws and regulations and therefore free for general use. The publisher makes no representation, express or implied, with regard to the accuracy of the information contained in this book and cannot accept any legal responsibility or liability for any errors or omissions that may be made. Printed on acid-free paper Springer Science+Business Media springer.com

Preface Semigroup theory is a relatively young part of mathematics. As a separate direction of algebra with its own objects, formulations of problems, and methods of investigations, semigroup theory was formed about 60 years ago. One of the main motivations for the existence of some mathematical theories are interesting and natural examples. For semigroup theory the obvious candidates for such examples are transformation semigroups. Various transformations of different sets appear everywhere in mathematics all the time. As the usual composition of transformations is associative, each set of transformations, closed with respect to the composition, forms a semigroup. Among all transformation semigroups one can distinguish three classical series of semigroups: the full symmetric semigroup T (M ) of all transformations 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 . If M = {1, 2, . . . , n}, then the above semigroups are usually denoted by Tn , IS n and PT n , respectively. One of the main evidences for the importance of these semigroups is their universality property: every (finite) semigroup is a subsemigroup of some T (M ) (resp. Tn ); and every (finite) inverse semigroup is a subsemigroup of some IS(M ) (resp. IS n ). Inverse semigroups form a class of semigroups which are closest (in some sense) to groups. An analogous universal object in group theory is the symmetric group S(M ) of all bijective transformations of M . Many books are dedicated to the study of S(M ) or to the study of transformation groups in general. Transformation semigroups had much less luck. The “naive” search in MathSciNet for books with the keywords “transformation semigroups” in the title results in two titles, one being a conference proceedings, and another one being old 50-page-long lecture notes in Russian ([Sc4]). Just for comparison, an analogous search for “transformation groups” results in 75 titles. And this is in spite of the fact that the semigroup T (M ) was studied by Suschkewitsch already in the 1930s. The semigroup IS(M ) was introduced by Wagner in 1952, but the first relatively small monograph about it appeared only in 1996 ([Li]). The latter monograph considers some basic questions about IS(M ): how one writes down the elements of IS(M ), when two elements of IS(M ) commute, what is the presentation of IS(M ), what are the con-

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PREFACE

gruences on IS(M ). For example, such basic semigroup-theoretical notions as ideals and Green’s relations are mentioned only in the Appendix without any direct relation to IS(M ). Much more information about the semigroup T (M ) can be found in the last chapter of [Hi1]; however, it is mostly concentrated around the combinatorial aspects. Otherwise one is left with the options to search through examples in the parts of the abstract theory of semigroups using the classical books [CP1, Gri, Ho3, Ho7, Hi1, Law, Pe] or to look at original research papers. 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 them we describe generating systems, ideals, Green’s relations, various classes of subsemigroups, congruences, conjugations, endomorphisms, presentations, actions on sets, linear representations and cross-sections. Some of the results are very old and classical, some are quite young. In order not to overload the reader with too technical and specialized results, we decided to restrict the area of the present book to the above-mentioned parts of the theory of transformation semigroups. The book was thought to be an elementary introduction to the theory of transformation semigroups, with a strong emphasis on the concrete examples in the form of three classical series of finite transformation semigroups, namely, Tn , IS n and PT n . The book is primarily directed to students, who would like to make their first steps in semigroup theory. The choice of the semigroups Tn , IS n and PT n is motivated not only by their role in semigroup theory, but also by our strong belief that a good understanding of a couple of interesting and pithy examples is more important for the first acquaintance with some theory than a formal learning of dozens of theorems. Another good motivation to consider the semigroups Tn , IS n and PT n at the same time is the observation that many results about these semigroups, which for each of them were obtained independently by different people and in 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 to the cases of infinite transformation semigroups. However, we restrict ourselves to the case of finite semigroups to make the exposition as elementary and accessible for a wide audience as possible. We are not after the biggest possible generality. Another argument is that we also try to attract the reader’s attention to numerous combinatorial aspects and applications of the semigroups we consider. With our three principal examples of semigroups on the background we also would like to introduce the reader to the basics of the abstract theory of semigroups. So, along the discussion of these examples, we tried to present many important basic notions and prove (or at least mention) as many classical abstract results as possible.

PREFACE

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The requirements for the reader’s mathematical background are very low. To understand most of the content, it is enough to have a minimal mathematical experience on the level of common sense. Perhaps some familiarity with basic university courses in algebra and combinatorics would be a substantial 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 complicated constructions whenever possible. The penultimate section of each chapter is called “Addenda and Comments.” A part of it consists of historical comments (which are by no means complete). Another part consists of some remarks, facts, and statements, which we did not include in the main text of the book. The reason is usually the much less elementary level of these statements or the more complicated character of the proofs. However, we include them in the Addenda as from our point of view they deserve attention in spite of the fact that they do not really 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 part of the book, our requirements for the reader’s mathematical background are different and are closer to the standard mathematical university curriculum. In the Addenda, we sometimes also mention some open problems and try to describe possible directions for further investigations. The division of the book into the main text and the Addenda is not very strict as sometimes the notions and facts mentioned in the Addenda are used in the main text. The last section of each chapter contains problems. Some problems (not many) are also included in the main text. The latter ones are mostly simple and directed to the reader. Sometimes they also ask to repeat a proof given before for a different situation. These problems are in some sense compulsory for the successful understanding of the main text (i.e., one should at least read them). The additional problems of the last section of each chapter can be quite different. Some of them are easy exercises, while others are much more complicated problems, which form an essential supplement to the material of the chapter. Hints for solutions of the latter ones can be found at the end of the book. The book was essentially written during the visit of the first author to Uppsala University, which was supported by The Royal Swedish Academy of Sciences and The Swedish Foundation for International Cooperation in Research and Higher Education (STINT). The financial support of The Academy and STINT, and the hospitality of Uppsala University are gratefully acknowledged. We thank Ganna Kudryavtseva, Victor Maltcev, and Abdullahi Umar for their comments on the preliminary version of the book.

Kyiv, Ukraine Uppsala, Sweden

Olexandr Ganyushkin Volodymyr Mazorchuk

Contents Preface

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1 Ordinary and Partial Transformations 1.1 Basic Definitions . . . . . . . . . . . . . . . 1.2 Graph of a (Partial) Transformation . . . . 1.3 Linear Notation for Partial Transformations 1.4 Addenda and Comments . . . . . . . . . . . 1.5 Additional Exercises . . . . . . . . . . . . . 2 The 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 2.10

Semigroups Tn , PT n , and IS n Composition of Transformations . Identity Elements . . . . . . . . . . Zero Elements . . . . . . . . . . . . Isomorphism of Semigroups . . . . The Semigroup IS n . . . . . . . . Regular and Inverse Elements . . . Idempotents . . . . . . . . . . . . . Nilpotent Elements . . . . . . . . . Addenda and Comments . . . . . . Additional Exercises . . . . . . . .

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3 Generating Systems 39 3.1 Generating Systems in Tn , PT n , and IS n . . . . . . . . . . . 39 3.2 Addenda and Comments . . . . . . . . . . . . . . . . . . . . . 42 3.3 Additional Exercises . . . . . . . . . . . . . . . . . . . . . . . 43 4 Ideals and Green’s Relations 4.1 Ideals of Semigroups . . . . . . . . . . . 4.2 Principal Ideals in Tn , PT n , and IS n . 4.3 Arbitrary Ideals in Tn , PT n , and IS n . 4.4 Green’s Relations . . . . . . . . . . . . . 4.5 Green’s Relations on Tn , PT n , and IS n 4.6 Combinatorics of Green’s Relations in the Semigroups Tn , PT n , and IS n . .

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Addenda and Comments . . . . . . . . . . . . . . . . . . . . . Additional Exercises . . . . . . . . . . . . . . . . . . . . . . .

5 Subgroups and Subsemigroups 5.1 Subgroups . . . . . . . . . . . . . 5.2 Cyclic Subsemigroups . . . . . . 5.3 Isolated and Completely Isolated Subsemigroups . . . . . . . . . . 5.4 Addenda and Comments . . . . . 5.5 Additional Exercises . . . . . . .

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74 83 88

6 Other Relations on Semigroups 6.1 Congruences and Homomorphisms 6.2 Congruences on Groups . . . . . . 6.3 Congruences on Tn , PT n , and IS n 6.4 Conjugate Elements . . . . . . . . 6.5 Addenda and Comments . . . . . . 6.6 Additional Exercises . . . . . . . .

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7 Endomorphisms 7.1 Automorphisms of Tn , PT n , and IS n 7.2 Endomorphisms of Small Ranks . . . . 7.3 Exceptional Endomorphism . . . . . . 7.4 Classification of Endomorphisms . . . 7.5 Combinatorics of Endomorphisms . . . 7.6 Addenda and Comments . . . . . . . . 7.7 Additional Exercises . . . . . . . . . .

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8 Nilpotent Subsemigroups 8.1 Nilpotent Subsemigroups and Partial 8.2 Classification of Maximal Nilpotent Subsemigroups . . . . . . . . . . . . 8.3 Cardinalities of Maximal Nilpotent, Subsemigroups . . . . . . . . . . . . 8.4 Combinatorics of Nilpotent Elements 8.5 Addenda and Comments . . . . . . . 8.6 Additional Exercises . . . . . . . . . 9 Presentation 9.1 Defining Relations . . . 9.2 A presentation for IS n . 9.3 A Presentation for Tn . 9.4 A presentation for PT n 9.5 Addenda and Comments 9.6 Additional Exercises . .

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Orders

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CONTENTS

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10 Transitive Actions 10.1 Action of a Semigroup on a Set . . . 10.2 Transitive Actions of Groups . . . . 10.3 Transitive Actions of Tn . . . . . . . 10.4 Actions Associated with L-Classes . 10.5 Transitive Actions of PT n and IS n 10.6 Addenda and Comments . . . . . . . 10.7 Additional Exercises . . . . . . . . .

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11 Linear Representations 11.1 Representations and Modules 11.2 L-Induced S-Modules . . . . 11.3 Simple Modules over IS n and 11.4 Effective Representations . . 11.5 Arbitrary IS n -Modules . . . 11.6 Addenda and Comments . . . 11.7 Additional Exercises . . . . .

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12 Cross-Sections 12.1 Cross-Sections . . . . . . . . . . . . . . 12.2 Retracts . . . . . . . . . . . . . . . . . 12.3 H-Cross-Sections in Tn , PT n , and IS n 12.4 L-Cross-Sections in Tn and PT n . . . 12.5 L-Cross-Sections in IS n . . . . . . . . 12.6 R-Cross-Sections in IS n . . . . . . . . 12.7 Addenda and Comments . . . . . . . . 12.8 Additional Exercises . . . . . . . . . .

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13 Variants 13.1 Variants of Semigroups . . . . . . . . . 13.2 Classification of Variants for IS n , Tn , and PT n . . . . . . . . . . . . . . . . 13.3 Idempotents and Maximal Subgroups 13.4 Principal Ideals and Green’s Relations 13.5 Addenda and Comments . . . . . . . . 13.6 Additional Exercises . . . . . . . . . . 14 Order-Related Subsemigroups 14.1 Subsemigroups, Related to the Natural Order . . . . . . . . . . . . . . . . . . 14.2 Cardinalities . . . . . . . . . . . . . . 14.3 Idempotents . . . . . . . . . . . . . . . 14.4 Generating Systems . . . . . . . . . . 14.5 Addenda and Comments . . . . . . . . 14.6 Additional Exercises . . . . . . . . . .

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Answers and Hints to Exercises

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Bibliography

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List of Notation

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Index

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Chapter 1

Ordinary and Partial Transformations 1.1

Basic Definitions

The principal objects of interest in the present volume are finite sets and transformations 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 the following form:   m1 m2 · · · mn , (1.1) α= k1 k2 · · · kn where all ki ∈ M . If x ∈ M , say x = mi , the element ki will be called the value 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 we shall usually consider the set N = Nn = {1, 2, . . . , n}. Apart from the transformations of M we shall also consider the so-called partial 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:   l2 ··· lk l1 . (1.2) α= α(l1 ) α(l2 ) · · · α(lk ) Abusing notation, we may also write α : M → M for a partial transformation, 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) is not important. With each (partial) transformation α as above we associate the following standard notions: • The domain of α: dom(α) = A O. Ganyushkin, V. Mazorchuk, Classical Finite Transformation Semigroups, Algebra and Applications 9, DOI: 10.1007/978-1-84800-281-4 1, c Springer-Verlag London Limited 2009 

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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 synonym of the word image. If dom(α) = M , the transformation α is called full or total . The set of all total transformations of M is denoted by T (M ), and the set 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) for some α ∈ PT n . In the case of PT n it is natural to form the first row of the array for α by simply listing all the elements from N in their natural order. Then, to define α completely, one needs a special symbol to indicate that some element x belongs to dom(α). We shall use the symbol ∅. In other words, α(x) = ∅ means that x ∈ dom(α). Thus the element α can be written in the following form:   1 2 ... n , (1.3) α= k1 k2 . . . kn where ki = α(i) if i ∈ dom(α) and ki = ∅ if i ∈ dom(α). Example 1.1.1  1 1   1 2 , 1 ∅

Here is the list of all elements of PT 2 :        2 1 2 1 2 1 2 , , , , 1 1 2 2 1 2 2         1 2 1 2 1 2 1 2 , , , . 2 ∅ ∅ 1 ∅ 2 ∅ ∅

The first row of this list consists of total transformations and hence lists all elements in T2 . Proposition 1.1.2 The set Tn contains nn elements and the set PT n contains (n + 1)n elements. Proof. Each element α ∈ Tn is uniquely defined by array (1.3), where each ki ∈ N. Since the choices of ki s are independent, we have |Tn | = nn by the product rule. In the case of PT n , the elements ki can be independently 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 of array (1.2). The number def(α) = n − rank(α) is called the defect of the partial transformation α.

1.2. GRAPH OF A (PARTIAL) TRANSFORMATION

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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 in the second row of array (1.2) are different. Bijective transformations on N are also called permutations of N. Proposition 1.1.3 Let α ∈ Tn . Then the following conditions are equivalent: (a) α is surjective (b) α is injective (c) α is bijective Proof. By the definition of a bijective transformation, it is enough to show that the conditions (a) and (b) are equivalent. We start by proving that injectivity 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 different elements of the set N, namely, α(1), α(2), . . . , α(n). But N contains exactly n 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 second row of (1.3) contains all n elements of the set N. But this row contains exactly n elements. Hence they all must be different. This implies that α is injective.

1.2

Graph of a (Partial) Transformation

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

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CHAPTER 1. ORDINARY AND PARTIAL TRANSFORMATIONS

Example 1.2.1 For the transformation   1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 α= 8 9 7 13 7 16 ∅ 1 13 11 16 4 12 9 14 13 the graph Γα has the following form: 10

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BB BB BB BB !

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7

It is obvious that a directed graph Γ = (N, E) will be the graph of some total (partial) transformation of N if and only if each vertex is the tail of exactly one (at most one) arrow. The graph Γα decomposes into a disjoint union of connected components. Intuitively, this is an obvious notion, for example, graph (1.4) has three connected components. The rigorous definition is as follows. First we define a subgraph. If Γ = (V, E) is a directed graph, a subgraph of Γ is a directed graph Γ  = (V  , E  ) such that V  ⊂ V and E  ⊂ E. A directed graph Γ = (V, E) is called connected if for each partition of V into a disjoint 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 the connected components of Γ are simply the maximal connected subgraphs of Γ , that is, those connected subgraphs of Γ which are not proper subgraphs of any other connected subgraph of Γ . Exercise 1.2.2 Prove that two different connected components of a directed graph Γ do not have common arrows. To understand the structure of Γα it is of course enough to understand the structure of its connected components. For this we shall need some more graph-theoretical notions. Let Γ (V, E) be a directed graph and a, b ∈ V . An oriented path from a to b in Γ is a sequence x0 = a, x1 , . . . , xk = b of vertices such that (xi , xi+1 ) ∈ E for each i = 0, 1, . . . , k − 1. Vertex a is called the tail of the path and vertex b is called the head of the path. If we have an oriented path such that a = b and xi = xj for all 0 ≤ i < j < k, then such path is called an (oriented) cycle and is denoted by (x0 , x1 , . . . , xk−1 ). If Γ does 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 trajectory of v as any longest possible path with the tail v. There are two possibilities:

1.2. GRAPH OF A (PARTIAL) TRANSFORMATION

5

either such trajectory is finite and hence breaks at some point, or this trajectory 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 trajectory of 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 are equivalent: (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 at the 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 2 in graph (1.4) terminates at the cycle (13, 12, 4); and the trajectory of vertex 4 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 or terminates at some cycle. (iii) α is total if and only if the trajectory of each x ∈ N in Γα terminates at some cycle. (iv) Let x, y ∈ N. If y occurs in the trajectory of x in Γα , then the trajectory of y is a subsequence of the trajectory of x. Proof. Since each vertex of Γα is a tail of at most one arrow, the statement (i) follows immediately from Lemma 1.2.3. The statement (iv) follows immediately from (i). Assume that the trajectory x = x0 , x1 , . . . of x does not break. Since Γα is finite, this trajectory must contain repetitions of some vertices. Let k be minimal for which there exists a repetition of xk and let xk+m be the first 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, implies xk+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) follows from (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 binary relation ωα on N in the following way: For x, y ∈ N set x ωα y if trα (x) and trα (y) have at least one common vertex.

6

CHAPTER 1. ORDINARY AND PARTIAL TRANSFORMATIONS

Lemma 1.2.5 The relation ωα is an equivalence relation. Proof. That ωα is reflexive and symmetric is obvious. To prove the transitivity of ωα consider x, y, z ∈ N such that x ωα y and y ωα z. Let a be a common 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 occurrence of a in trα (y) is not later than the first occurrence of b in trα (y). But this means that b occurs in trα (a) by Proposition 1.2.4(iv). Another application 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 the orbit of x in Γα will be denoted by oα (x). From the definition of ωα it follows that for any x ∈ dom(α) we have x ωα α(x). Hence all vertices which occur in trα (x) belong to oα (x). Furthermore, for each x ∈ N we can restrict the partial transformation α to the orbit K = oα (x), obtaining a new partial transformation, α(K) ∈ PT (oα (x)). Certainly, α(K) does not depend on the choice of the vertex in K. Proposition 1.2.6 For each x ∈ N the graph Γα(K) is a connected component of Γα . Proof. From the definition of ωα it follows that the graph Γα(K) is connected and contains all those arrows of Γα , for which both the heads and the tails belong 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 subgraph of 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 orbits of α and the connected components of Γα . A directed graph Γ =Γ (V, E) is called a tree with the sink a ∈ V provided that for each x ∈ V the trajectory of x in Γ is unique and breaks at a. For instance, 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 with sinks 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 neither oriented nor unoriented cycles. A directed graph Γ = (V, E) is called a cycle provided that we can enumerate V = {a1 , . . . , ak } such that E = {(a1 , a2 ), (a2 , a3 ), . . . , (ak−1 , ak ),

1.2. GRAPH OF A (PARTIAL) TRANSFORMATION

7

(ak , a1 )}. If Γi = (Vi , Ei ), i ∈ I, are directed graphs, then their union Γ = ∪i∈I Γi is defined as follows Γ = (V, E), where V = ∪i∈I Vi and E = ∪i∈I Ei . 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 all their sinks. We note that the forest in Theorem 1.2.9(ii) may contain only one tree with a sink. The union of this tree with a sink with the cycle on its sink is not a 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 of sinks. If all trees in this forest are trivial, Theorem 1.2.9(ii) simply describes a 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 a unique trajectory in both Γα and Γα(K) . Moreover, since K is a connected component 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 implies that trα (y) breaks at a. This means, by definition, that Γα(K) is a tree with the 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 . For i = 1, . . . , k we denote by Ki the set of all those vertices y from K such that the first vertex from the cycle (a1 , a2 , . . . , ak ) in trα (y) is ai . Note that Ki ∩ 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 just trα (y). However, the first element from the cycle (a1 , a2 , . . . , ak ) in trα (x) is ai , 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 )},

8

CHAPTER 1. ORDINARY AND PARTIAL TRANSFORMATIONS

(this means that Ei consists of all arrows from E, which has both tails and heads 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) = ∪ki=0 Γi . Furthermore, Ki s are disjoint for i = 1, . . . , k. To show that Γα(K) is of the 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 vertices y1 , . . . , ym−1 do not belong to Γ0 . Hence the definition of Ei implies that y = y0 , y1 , . . . , ym = ai is the trajectory of y in Γi , and it breaks at ai . In other words, the trajectory of each vertex in Γi breaks at ai and thus Γi is a 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) and has three connected components. The third component is a tree with the sink 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 first component is the union of the cycle (4, 13, 12) with the following forest of trees with sinks: 15

/ 14

@@ @@ @@ @@  /9 2 @@ @@ @@ @@ 

13

11 o

| || || | }| |

| || || | }| | 16 o 6

10

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 components of Γα .

1.3

Linear Notation for Partial Transformations

The graphical presentation of a transformation α ∈ PT n via Γα is very transparent, but also rather space consuming. For a plain mathematical text, it would be very useful to have some space-saving alternative. For permutations this is known as the cyclic notation and can be easily described by the following example: Example 1.3.1 For the permutation   1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 α= 9 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: 1O

/9

/7

/ 14

6o

11 o

12 o

15

/3 ~ ~ ~~ ~~ ~ ~

2O 4

/8

H5

13Y 

10

Using the notation for cycles, introduced on page 4 in the paragraph after Exercise 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 we can start from each of its vertices. Moreover, the order of cycles in the cyclic notation can also be chosen in an arbitrary way. In this subsection, we would like 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 roughly says that we only have to find a nice notation for trees with sinks. We call our notation linear and will define it recursively. Assume for the moment that the graph Γα , where α ∈ PT n , is a tree with the sink a. If a is the only vertex of Γ , we shall write Γα = [a] (or simply α = [a]). If Γα contains some other vertices, then it has to have the following form:

Γ1

... ... // // // //    // //    // //   ... // . . .  // . . .  // //   //  //  /  /  a a1 OO o k OOO ooo OOO o o OOO oo OOO ooo ' wooo

Γk (1.5)

a For i = 1, . . . , k, the subgraph Γi of the graph (1.5) above is a tree with the sink ai and has strictly less vertices than Γα . Assume that we already have the linear notation Γ˜i for Γi , i = 1, . . . , k. Then the linear notation for Γα (and α) is defined recursively as follows Γα = [Γ˜1 , Γ˜2 , . . . , Γ˜k ; 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 Γ˜i , i = 1, . . . , k, be the corresponding linear notation. In this case, we define the linear notation for Γα (and α) as follows Γα = (Γ˜1 , Γ˜2 , . . . , Γ˜k ).

10

CHAPTER 1. ORDINARY AND PARTIAL TRANSFORMATIONS

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 notation for permutations, the linear notation for (partial) transformartions is unique only up to permutation of certain components of the notation. Namely, the connected components can be written in an arbitrary order, and on each step of the recursive procedure the order of the components Γ˜1 , . . . , Γ˜k can also be chosen arbitrarily. Note that, by the above definition, the ordinary cycle (a1 , a2 , . . . , ak ) is denoted by ([a1 ], [a2 ], . . . , [ak ]). This is of course not very practical, so to avoid this unnecessary complication inside the notation for cycles (but not for trees with sinks!) we shall usually skip the brackets “[ ]” surrounding trivial 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 that this does not give rise to any confusion, moreover, it restores the original notation 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 by Suschkewitsch in [Su1]. 1.4.2 Let α ∈ PT n . The element x ∈ N satisfying α(x) = x is usually called a fixed point of α. If α is a permutation, then all the connected components of Γα are cycles. Fixed points of α correspond to cycles of length 1. In the cyclic notation for α such cycles are usually omitted (for the identity element one thus has to use a special notation, for example ε). If after such simplification the cyclic notation for α contains only one cycle, say of length k, the α is usually called a cycle of length k. Cycles of length 2 are called transpositions. 1.4.3 An alternative “linear” notation for total transformations was proposed in [AAH]. Although [AAH] works only with total transformations it is fairly straightforward to generalize their notation to cover all partial transformations. Roughly speaking the [AAH]-notation reduces to listing the trajectories of all vertices of the graph Γα . If we know the trajectory x0 = x, x1 , x2 , . . . of the vertex x, then of course we know the trajectories of each of the vertices x1 , x2 , . . . , so we can omit the latter ones. A trajectory is 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 the cycle (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 know the trajectory of ak , in this case the trajectory of the vertex a1 is obtained 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 written down, 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 of a in some cycle. To make the notation as short as possible, on each step one should try to choose a new vertex, which is not a head of any arrow in Γα . For example, for the transformation   1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 α= 8 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 occurring more than one time • The [AAH]-notation is by far not unique, even up to permutations of certain components of this notation Another disadvantage of the [AAH]-notation, related to the composition of 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−1  nk−1 n! k! k=0

12

CHAPTER 1. ORDINARY AND PARTIAL TRANSFORMATIONS

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. (i) For each such α and for each i ≥ 0 we define the set Nα as the set of all x ∈ N such that the first element from the cycle occurs in trα (x) on step i. (0) Obviously, Nα consists of the elements of our cycle and N = ∪i≥0 Nα(i) is (i) (i−1) a disjoint union. Furthermore, α(Nα ) ⊂ Nα for i > 0. The number of  (i) those α, for which |Nα | = ni , 0 ≤ i ≤ t, and such that ti=0 ni = n equals   n t (n0 − 1)!nn0 1 nn1 2 . . . nnt−1 (1.6) n0 , n1 , . . . , nt   (here the polynomial coefficient n0 ,n1n,...,nt gives the number of ordered partitions of N into blocks with cardinalities n0 ,. . . , nt , respectively; (n0 −1)! n is the number of ways to form a cycle out of n0 elements; and ni i+1 is the number of maps from the block with ni+1 elements to the block with ni elements). We can rewrite (1.6) as follows n

n t n! nn0 1 nn1 2 · . . . t−1 . · n0 n1 ! n2 ! nt !

(1.7)

Now to find the number Xn0 of those α ∈ Tn for which Γα is connected and contains 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 . Let n − n0 = k. Then (n0 + k)k−1  1 (k − 1)! nk−1 = = · · ni · k k−1−i = k! k! k! i!(k − 1 − i)! 0 k−1 i=0

=

k  n1 =1

k k−n1 nn0 1 −1 · , (n1 − 1)! (k − n1 )!

where for the last equality we substituted i by n1 − 1. Continuing in the same way we get nk−1 = k! k  = n1 =1

k−n1 −···−nt−1 k−n t −1 1 nn2 −1  nnt−1 nn0 1 −1 1 · ··· · n−1 = (n1 − 1)! (n2 − 1)! (nt − 1)! t n2 =1

=

1 n0

k  n1 =1

nt =1

k−n nn0 1 1

n1 !

n2 =1

=

nn1 2 ··· n2 !

k−n1 −···−nt−1



n1 +···+nt =n−n0



nt =1

t nnt−1 = nt !

n

n t 1 nn0 1 nn1 2 · . . . t−1 . (1.8) · n0 n1 ! n2 ! nt !

1.5. ADDITIONAL EXERCISES

13

k−1

Hence Xn0 = n! · n k! , where k = n − n0 . Since 1 ≤ n0 ≤ n, the final answer is now computed as follows n 

Xn0 = n!

n0 =1

1.5

n−1  k=0

nk−1 . k!

Additional Exercises

1.5.1 Let N denote the set of all positive integers. Give an example of a transformation 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 provided that 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 , whose graphs are isomorphic to the following graph: • 

•

•o 

•

•O o

•

/•o

•

1.5.5 For α ∈ PT n characterize dom(α), im(α), dom(α), rank(α), and def(α) in terms of Γα . 1.5.6 Compute the number of those α ∈ Tn (resp. α ∈ PT n ) for which im(α) (a) Does not contain given elements a1 , a2 ,. . . , ak

14

CHAPTER 1. ORDINARY AND PARTIAL TRANSFORMATIONS

(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 with a sink, equals nn−1 . 1.5.8 Prove that the  number α ∈ PT n , for which Γα does not  of those n−k . contain cycles, equals nk=1 n−1 n k−1 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 one element 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 , which consists of all possible unordered partitions of V into disjoint unions of nonempty subsets Vi s such that for each i = j the graph Γ does not contain any arrow from Vi to Vj . The set X is partially ordered in the natural way with respect to inclusions of components of partitions. Prove that the partition of V , which corresponds to the partition of Γ into connected components, is the minimum of X . 1.5.11 Let Γ = (V, E) be a directed graph. Consider the set Y, which consists of all possible unordered partitions of V into disjoint unions of nonempty subsets Vi s such that for each i we have that the subgraph (Vi , (Vi × Vi ) ∩ E) is connected. The set Y is partially ordered in the natural way with respect to inclusions of components of partitions. Prove that the partition of V , which corresponds to the partition of Γ into connected components, 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 n  tk (α) = n, (a) k=0

(b)

n 

ktk (α) = n.

k=0

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 n  tk (α) = n, (a) k=0

(b)

n  k=0

ktk (α) ≤ n.

Chapter 2

The Semigroups Tn , PT n , and IS n 2.1

Composition of Transformations

Let X and Y be two sets. A mapping from X to Y is an array of the form   x , f= f (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, as defined in Sect. 1.1, is just a mapping from a set to itself. Let now X, Y, Y  , Z be sets such that Y ⊂ Y  and let f : X → Y and g : Y  → Z be two mappings. In this situation, we can define the product or the composition gf of f and g by the following rule: The composition gf is the mapping from X 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 the result will be a total transformation of this set. The above definition admits a straightforward generalization to partial mappings. A partial mapping from X to Y is a mapping α : X  → Y , where X  ⊂ X. In this case, we say that the partial mapping α is defined on elements from X  . Again, a partial transformation, as defined in Sect. 1.1, is a partial mapping from a set to itself. One usually abuses notation and writes α : X → Y just emphasizing that α is a partial mapping. Let α : X → Y and β : Y → Z be two partial mappings. We define their product or composition βα as the partial mapping, defined on all those x ∈ X for which α and β are defined on the elements x and α(x), respectively; on such x the value of βα is given by (βα)(x) = β(α(x)). In particular, we can always compose two partial transformations of the same set and the result will be another

O. Ganyushkin, V. Mazorchuk, Classical Finite Transformation Semigroups, Algebra and Applications 9, DOI: 10.1007/978-1-84800-281-4 2, c Springer-Verlag London Limited 2009 

15

CHAPTER 2. THE SEMIGROUPS Tn , PT n , AND IS n

16

partial transformation of this set. We also note that the definition of the composition of total transformations is just a special case of that of partial transformations. 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) is defined if and only if the composition (γβ)α is defined, and if they both are defined, we have γ(βα) = (γβ)α. Proof. Follows immediately from the following picture:

X

Y

Z

V

a _ ] [ Y W g e c γβ U S i k α Q( ,• m , , • • γ β 6z 9 v• y x Q S m k U W Y [ ] _ a c e g i βα

γ(βα)=(γβ)α

Associativity of the composition of partial transformations naturally leads 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 semigroup provided 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 the context one usually writes S for (S, ·). Furthermore, one usually writes ab instead of a · b. Exercise 2.1.2 Let (S, ·) be a semigroup. Show that the value of the product a1 a2 · · · 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 every a ∈ S we have a well-defined element ak = a  · a · · · a. The number of elements ktimes

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 the composition of (partial) transformations. The semigroup Tn is called the full transformation semigroup on the set N or the symmetric semigroup of all transformations of N. The semigroup PT 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 of S provided that T is closed with respect to · (that is, a · b ∈ T as soon as a, b ∈ T ). Obviously, in this case, T itself is a semigroup with respect to the restriction of the operation · to T . The fact that T is a subsemigroup of S is 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 provided that ea = a, or ae = a, respectively, for all a ∈ S. An element e, which is a left and a right identity at the same time, is called a two-sided identity or simply an identity. If S contains some left identity el and some right identity er we have el = el · er = er and hence these two elements coincide. Hence in this case S contains a unique identity element, which is, moreover, a two-sided identity. However, a semigroup may contain many different left identities or many different right identities (see Exercise 2.10.2). It is possible for a semigroup to contain neither left nor right identities. An example of such a semigroup is the semigroup (N, +). Another example is the semigroup {2, 3, 4, . . . } with respect to the ordinary multiplication. A semigroup which contains a two-sided identity element is called a monoid. The absence of an identity element can be easily repaired in the following way. Proposition 2.2.1 Each semigroup can be extended to a monoid by adding at most one element. Proof. Let (S, ·) be a semigroup. If S contains an identity, we have nothing to prove. If S does not contain any identity element, consider the set S 1 = S ∪ {1}, where 1 ∈ S. Define the binary operation ∗ on S 1 as follows: For a, b ∈ S 1 set ⎧ ⎪a · b, a, b ∈ S; ⎨ a ∗ b = a, b = 1; ⎪ ⎩ b, a = 1. A direct calculation shows that ∗ is associative, hence S 1 is a semigroup. Furthermore, from the definition of ∗ we have that 1 is the identity element in S 1 . Moreover, the restriction of the operation ∗ to S coincides with the original operation ·. Hence S is a subsemigroup of S 1 .

18

CHAPTER 2. THE SEMIGROUPS Tn , PT n , AND IS n Denote by εn : N → N the identity transformation   1 2 ··· n . εn = 1 2 ··· n

If n is clear from the context we shall sometimes write ε instead of εn . The following statement is obvious. Proposition 2.2.2 The transformation εn is the (two-sided) identity element 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 called invertible 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 are different such elements, then b1 = b1 · 1 = b1 (ab2 ) = (b1 a)b2 = 1 · b2 = b2 . The element b is called the inverse of a and is denoted by a−1 . Note that if b 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 ∗ . Note that 1 ∈ S ∗ since 1 · 1 = 1. In particular, S ∗ is not empty. The above terminology and notation deserve some explanation. Usually the operation in an abstract semigroup is thought of as a multiplication. Since the element 1 is the identity element in such multiplicative semigroups as N, Z, R, C, it is natural to denote the identity element of an abstract semigroup by the same symbol 1. This also justifies the notions “unit” and “inverse.” However, there are many semigroups where the operation is not the multiplication, for example the semigroup (Z, +). The identity element in this semigroup is the number 0 and not the number 1. And the inverse of the number n ∈ Z is the number −n and not the number n−1 (note that the latter one is not always defined, and when it is defined, it is not an integer in 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. Then S ∗ is a group. Proof. Obviously, if a ∈ S ∗ , then a−1 ∈ S ∗ as well. If a, b ∈ S ∗ , then we have ab · b−1 a−1 = a · bb−1 · a−1 = a · 1 · a−1 = aa−1 = 1. Analogously one shows that b−1 a−1 · ab = 1 and hence b−1 a−1 = (ab)−1 . In particular, 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 and only if α is a permutation on N.

2.3. ZERO ELEMENTS

19

Proof. Assume that α is invertible and β is a (partial) transformation such that αβ = βα = ε. Note that dom(ε) = N. Hence Exercise 2.1.4(a) implies dom(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 contradiction. This means that α(x) = α(y). Hence α is everywhere defined and injective and thus is a permutation by Proposition 1.1.3. Conversely, if   1 2 ··· n α= i1 i2 · · · in is a permutation, the element α=



i1 i2 · · · 1 2 ···

in n



is a permutation as well and a direct computation shows that αβ = βα = ε, that is, α is invertible. The group Tn∗ = PT ∗n of all permutations on N is called the symmetric group 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 that 0a = 0, or a0 = 0, respectively, for all a ∈ S. An element 0 which at the same time is a left and a right zero, is called a two-sided zero or simply a zero. 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 S contains a unique zero element, which is, moreover, a two-sided zero. The analog of Proposition 2.2.1 is the following statement. Proposition 2.3.1 Each semigroup can be extended to a semigroup with zero by adding at most one element. Proof. Let (S, ·) be a semigroup. If S contains a zero, we have nothing to prove. Otherwise, consider the set S 0 = S ∪ {0}, where 0 ∈ S. Define the binary operation ∗ on S 0 as follows: For a, b ∈ S 0 set ⎧ ⎪ ⎨a · b, a, b ∈ S; a ∗ b = 0, b = 0; ⎪ ⎩ 0, a = 0. A direct calculation shows that ∗ is associative, hence S 0 is a semigroup. Furthermore, from the definition of ∗ we have that 0 is the zero element in S 0 and that the restriction of ∗ to S coincides with ·. Hence S is a subsemigroup of S 0 .

20

CHAPTER 2. THE SEMIGROUPS Tn , PT n , AND IS n

Denote by 0n the partial transformation 0n : ∅ → N on N. We have dom(0n ) = ∅. If n is clear from the context, we shall usually write simply 0 instead of 0n . Proposition 2.3.2 0n is the zero element of the semigroup PT n . Proof. The equalities α0n = 0n α = 0n , α ∈ PT n , are obvious. For a ∈ N define the total transformation 0a : N → N via 0a (x) = a for all x ∈ N. The transformation 0a is called the constant transformation. Proposition 2.3.3 For n > 1 the semigroup Tn does not contain any right zeros. α ∈ Tn is a left zero if and only if α = 0a for some a ∈ N. In particular, the semigroup Tn contains exactly n left zeros. Proof. From the obvious equality 0a β = 0a for any β ∈ Tn we obtain that each constant transformation on Tn is a left zero. Hence Tn contains at least n different left zeros. In particular, for n > 1 the semigroup Tn cannot contain any right zeros. Note that constant transformations are the only transformations of rank 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 is not possible. This implies that α is not a left zero of Tn . We note that the semigroup T1 consists of the identity element only. This element 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 said to be isomorphic (denoted by S ∼ = T ) provided that there exists a bijection ϕ : 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 → T is an isomorphism, then ϕ−1 : T → S is an isomorphism as well. Exercise 2.4.1 Show that the relation of being isomorphic semigroups is an equivalence relation. Example 2.4.2 It is rather handy to present small semigroups using their multiplication tables, which is also called Cayley tables. Such table is a square matrix with |S| rows and |S| columns, which are indexed by the elements of 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 } has the following Cayley table: · ε (1, 2) 01 02

ε (12) 01 ε (12) 01 (1, 2) ε 02 01 01 01 02 02 02

02 02 01 . 01 02

(2.2)

The following set of 2 × 2 matrices is also a semigroup with respect to the usual matrix multiplication         1 0 0 1 1 1 0 0 S= , , , . 0 1 1 0 0 0 1 1 To simplify the notation we denote         1 0 0 1 1 1 0 0 E= , A= , B= , C= . 0 1 1 0 0 0 1 1 The Cayley table for S has the following form · E A B C

E E A B C

A A E B C

B B C B C

C C B B C

.

(2.3)

It is easy to check that the bijection   ε (12) 01 02 ϕ= E A B C transforms the Cayley table (2.2) to the Cayley table (2.3) and hence is an isomorphism from T2 to S. The primary importance of the semigroup Tn is revealed by the following statement. Theorem 2.4.3 (Cayley’s Theorem) Each finite semigroup S of cardinality n 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   1 2 ··· n , ϕ(a) = i1 i2 · · · in

CHAPTER 2. THE SEMIGROUPS Tn , PT n , AND IS n

22

where for k = 1, . . . , n, the number ik is uniquely determined by the equality aak = 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 mapping ϕ : 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 subsemigroup ϕ(S) of Tn . If S does not contain any identity element, we can consider S as a subsemigroup of the semigroup S 1 and |S 1 | = n + 1. By the above, S 1 is isomorphic to a subsemigroup of Tn+1 , and the claim follows.

2.5

The Semigroup IS n

Obviously, the composition αβ of two partial injections α and β is again a partial injection. This observation leads to the definition of another very important subsemigroup of PT n , namely, the semigroup IS n of all partial injective transformations of N, which is usually called the symmetric inverse semigroup. Theorem 2.5.1 |IS n | =

n  2  n k=0

k

· k!.

Proof. Each partial injection α : N → N can be considered as a bijection α : dom(α) → im(α). Let us count the number of such bijections of rank k.  n different ways. The set B = im(α) The set A = dom(α) can be chosen in   k can be independently chosen in nk different ways. If A and B are fixed, there n nare  exactly k! different bijections from A to B. Hence we have exactly k · k · 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 sum rule. Partial injections α : N → N are also called partial bijections or partial permutations on N. If α : N → N is a partial injection, then each x ∈ N has at most one preimage. In particular, the graph Γα is much simpler than for general partial transformations. From Theorem 1.2.9 it follows that each connected component of Γα is either a cycle or has the form a1

/ a2

/ a3

/ ...

/ ak−1

In our linear notation such a graph is denoted by [[[. . . [[[a1 ]; a2 ]; a3 ]; . . . ]; ak−1 ]; ak ].

/ ak .

(2.4)

2.6. REGULAR AND INVERSE ELEMENTS

23

Since the order of the brackets in this expression is fixed, we can simplify the notation 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 permutations a simplified version of linear notation, which we call the chain-cycle notation. For example, the partial permutation   1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 α= , 14 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 up to a permutation of components and cyclic permutation of elements in each cycle. Note that no permutations of elements in chains are allowed. As IS n contains the identity element ε and the zero element 0 of the bigger semigroup PT n , these elements will be the identity element and the zero element of IS n , 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:

.

Tn

PT < ncGG GG zz z GG z GG zz z 1Q z IS bEE ;w n w EE ww EE EE wwww 1Q -w 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 exists b ∈ S such that aba = a. Elements a, b ∈ S form a pair of inverse elements provided 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 inverse of b and b is an inverse of a. This might be slightly confusing, since in such generality the inverse element of a given element is not uniquely defined. Note that this notion extends the notion of an inverse element for invertible elements: if a ∈ S is invertible, then aa−1 a = a and a−1 aa−1 = a−1 . Hence a and a−1 is a pair of inverse elements.

24

CHAPTER 2. THE SEMIGROUPS Tn , PT n , AND IS n

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 is obviously 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 is regular. Obviously, each group is a regular semigroup. Theorem 2.6.3 The semigroups Tn , PT n , and IS n are regular. Proof. Let α ∈ PT n . Let us construct the element β ∈ Tn as follows: For each x ∈ im(α) take some y ∈ N such that α(y) = x and set β(x) = y. For x ∈ im(α) define β(x) = 1. A direct calculation shows that αβα = α. It follows that both Tn and PT n are regular semigroups. Let now α ∈ IS n . Define β ∈ IS n 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). In particular, β ∈ IS n . As before, we still have the equality αβα = α and hence the semigroup IS n is regular as well. Let us study the structure of inverse elements in the semigroups Tn and PT n in more detail. Theorem 2.6.4 Let α, β ∈ PT n . Then the element β is inverse to the element 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 condition: (2.5) β(ai ) ∈ Bi for alli = 1, 2, . . . , k. 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 is also 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 the condition (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)| = m1 m2 · · · mk · (k + 1)n−k ; (ii) if α is total, then |VTn (α)| = m1 m2 · · · mk · k n−k . Proof. For each ai ∈ im(α) we have to choose β(ai ) ∈ Bi . As for different i these choices are independent, we have m1 m2 · · · mk different ways to define β on im(α). After this we have to define β on the set N\im(α). From Theorem 2.6.4 it follows that the restriction of β to N\im(α) is an arbitrary (partial in the case of PT n ) mapping from N\im(α) to the set {β(y) : y ∈ im(α)}. The latter set contains exactly k elements. Hence the restriction of β to N\im(α) can be chosen in k n−k different ways for Tn and in (k + 1)n−k different ways for 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 each a ∈ S has a unique inverse element. This inverse element is usually denoted by a−1 (this can now be justified by the requirement that it is unique). From Corollary 2.6.5 it follows that in the case of n > 1 the semigroups Tn and PT n are not inverse semigroups. Theorem 2.6.7 The semigroup IS n is an inverse semigroup. Proof. Let α ∈ IS n and β ∈ VPT n (α). Let us try to analyze in which case β is a partial permutation. Since α is injective, the sets Bi from the proof of Theorem 2.6.4 consist of one element each. Hence β is uniquely defined on im(α) by Theorem 2.6.4(b). At the same time, the injectivity of β and Theorem 2.6.4(c) imply that β must be undefined on N\im(α). Thus VPT n (α) ∩ IS n contains a unique element, implying that IS n is an inverse semigroup.

CHAPTER 2. THE SEMIGROUPS Tn , PT n , AND IS n

26

Since the notion of an inverse element is a natural extension of the corresponding notion for invertible elements, inverse semigroups are very natural generalizations 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−1 a−1 .

2.7

Idempotents

Identities and zero elements in semigroups are special cases of a more general notion. An element e of a semigroup S is called an idempotent provided that e2 = 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 need some notation. If α ∈ PT n and B ⊂ N, one says that B is invariant with respect to α provided that α(x) ∈ B for all x ∈ B ∩ dom(α). In case B is invariant with respect to α we can define the restriction α|B of α to B as follows: α|B is a partial transformation on B, dom(α|B ) = B ∩ dom(α), and for 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) are invariant 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) for any x ∈ dom(α). Since dom(α2 ) ⊂ dom(α), it follows that α2 = α. Corollary 2.7.3 The element α ∈ IS n is an idempotent if and only if α is the identity transformation of some A ⊂ N. In particular, IS n contains exactly 2n idempotents. Proof. For α ∈ IS n 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 condition that α|im(α) is the identity transformation means exactly that α is the identity transformation on im(α). Since N has exactly 2n subsets, the claim follows. For A ⊂ N we denote by εA the unique idempotent of IS n such that dom(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 n    n n−k k . un = k k=1

Proof. To define an idempotent  α of rank k we have to choose a k-element set 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 arbitraryway  (this can be done in k n−k different ways). Hence Tn contains exactly nk k n−k idempotents of rank k. The statement is now obtained applying the sum rule. Corollary 2.7.5 The number un of idempotents in the semigroup PT n equals n    n un = (k + 1)n−k . k k=0

Proof. The difference with the proof of Corollary 2.7.4 is just the facts that PT 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 regular elements. 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 and ba are idempotents: ab · ab = aba · b = ab and ba · ba = b · aba = ba. In terms of idempotents, the inverse semigroups can be detected in the set of all regular semigroups using the following: Theorem 2.7.6 A regular semigroup S is inverse if and only if all idempotents of S commute. Proof. Let S be a regular semigroup in which all idempotents commute. Let further a ∈ S and b1 , b2 ∈ S be two inverse elements to a. Then ab1 a = a, b1 ab1 = b1 , ab2 a = a, b2 ab2 = b2 , and as we mentioned above, the elements ab1 , ab2 , b1 a, and b2 a are idempotents. Hence we have b1 = b1 ab1 = b1 ab2 ab1 = b1 · ab2 · ab1 = b1 · ab1 · ab2 = b1 ab2 = = b1 ab2 ab2 = b1 a · b2 a · b2 = b2 a · b1 a · b2 = b2 · ab1 a · b2 = b2 ab2 = 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 idempotents. Let us first prove that their product ef is an idempotent. Consider a = (ef )−1 . We have ef · a · ef = ef, a · ef · a = a. Using this we obtain the following equalities: ef · ae · ef = ef · a · e2 f = ef · a · ef = ef, ae · ef · ae = a · e2 f · ae = a · ef · a · e = ae.

28

CHAPTER 2. THE SEMIGROUPS Tn , PT n , AND IS n

Hence ae is also an inverse to ef , which means ae = a. Analogously one shows that f a = a. Thus a2 = ae · f a = a · ef · a = a, and hence a is an idempotent. In particular, a−1 = a, and thus a = ef . Analogously one shows that f e is an idempotent as well. Moreover, we have f e · ef · f e = f · e2 · f 2 · e = f e · f e = f e, ef · f e · ef = e · f 2 · e2 · f = ef · ef = ef. This means that (ef )−1 = f e. However, since ef is an idempotent, we also have (ef )−1 = ef . Thus f e = 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 inverse semigroup S is a commutative subsemigroup of S. In particular, E(IS n ) is a commutative semigroup of order 2n . Exercise 2.7.7 Prove that εA · εB = εA∩B for all A, B ⊂ N and use this to show that the semigroup E(IS n ) is isomorphic to the semigroup of all subsets of N with respect to the operation of the intersection of subsets. Exercise 2.7.8 (a) Show that E(T2 ) is a noncommutative subsemigroup of 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 all n > 1. We know already that for n > 1 the semigroups Tn and PT n are not inverse. This fact also follows from Theorem 2.7.6 and Exercises 2.7.8 and 2.7.9.

2.8

Nilpotent Elements

An element a of a semigroup S with the zero 0 is called nilpotent or a nilelement provided that ak = 0 for some k ∈ N. The minimal k for which ak = 0 is called the nilpotency degree or nilpotency class of the element a and is denoted by nd(a). Proposition 2.8.1 α ∈ PT n is nilpotent if and only if the graph Γα does not contain cycles.

2.8. NILPOTENT ELEMENTS

29

Proof. If Γα contains a cycle, say (a1 , a2 , . . . , am ), then a1 ∈ dom(αk ) for all k > 0 (it is easy to see that αk (a1 ) = a(1+k) mod m ). As the zero element 0 of PT n satisfies dom(0) = ∅, we have αk = 0 for all k > 0 and hence the element α cannot be nilpotent. If Γα does not contain any cycle, the trajectory of each vertex a must break. 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 with the tail am (i.e., am ∈ dom(α)). This means that αm (a) = am and that a ∈ dom(αm+1 ) since otherwise we would have αm+1 (a) = α(am ), which does not make sense since am ∈ dom(α). Since the length of any trajectory does 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 ] ∈ IS n is nilpotent if and only if it does not contain cycles. If α ∈ IS n is nilpotent, the nilpotency degree of α equals the maximum max(p, . . . , q) of the lengths of all chains. Denote by S(n, k) the Stirling number of the second kind, that is, the number of partitions N = A1 ∪ A2 ∪ · · · ∪ Ak into an unordered disjoint union 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 element   0 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, and Ak+1 = dom(α) form a partition N = A1 ∪ A2 ∪ · · · ∪ Ak+1 of the set N into k + 1 block (note that Ak+1 = ∅ since α is nilpotent). This observation suggests the following procedure for constructing all nilpotent elements of rank k (we construct some nilpotent element α): First we choose im(α) = {a 1 , . . . , ak }. This is a k-element subset of N and hence it can be chosen in n k different ways. Then we choose a partition, N = B1 ∪ B2 ∪ · · · ∪ Bk+1 into an unordered disjoint union of nonempty blocks. This can be done in S(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

30

CHAPTER 2. THE SEMIGROUPS Tn , PT n , AND IS n

nilpotent. The choice of the preimage for ai means that we choose some arrows for Γα in such a way that the resulting graph does not contain any cycles. The preimage of a1 can be any of the blocks B1 , . . . , Bk+1 , which does not contain a1 (since otherwise Γα would contain a loop at a1 , which is a cycle of length 1). This means that the preimage of a1 can be chosen in k different ways. We proceed by induction. Assume that we have already chosen the preimages for a1 , . . . , am (and let these preimages be the blocks Bi1 , . . . , Bim , respectively). 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, which does not contain am+1 . This preimage can be chosen in (k+1−m)−1 = k−m different 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 constructed part of Γα does not contain cycles, this trajectory breaks at a(t) ∈ X. In order to ensure that the choice of the preimage for am+1 does not create a cycle in Γα , it is necessary and sufficient to make sure that this preimage does not contain a(t) . Hence we again have (k + 1 − m) − 1 = k − m ways to choose 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 n k · S(n, k + 1) · k!. Theorem 2.8.5 The number of nilpotent elements of defect k in the semin−1  . group IS n equals the signless Lah number Ln,k = n! k! k−1 Proof. By Corollary 2.8.3, a nilpotent partial permutation of defect k has the 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−1 k−1 different ways. Going through all permutations we will get chain-cycle notation for all nilpotent elements of defect k. Since the order of chains in the chain-cycle notation is not important (because all chains in the chain-cycle notation commute), every nilpotent element of defect k will be counted k! times. Indeed, the chains in each chain-cycle notation can be permuted in k! different ways without changing the corresponding partial transformation of N; however, each of these permutations of chains corresponds to a different permutation

2.9. ADDENDA AND COMMENTS

31

of N. This means that n! permutations of N give k! repetitions of each nilpotent   element. Hence the number of nilpotent elements of defect k equals n! n−1 k! k−1 . Corollary 2.8.6 The semigroup IS n contains   n  n! n − 1 k! k − 1 k=1

nilpotent elements.

2.9

Addenda and Comments

2.9.1 A subset M ⊂ N is called invariant with respect to S ⊂ PT n provided that M is invariant with respect to each α ∈ S. Obviously, the union and the intersection of invariant subsets is an invariant subset. Further, for each α ∈ PT n and x ∈ N the set of vertices of trα (x) is invariant with respect to α. Let α ∈ PT n . If there exists a partition N = N1 ∪ N2 of N into a disjoint union of two invariant subsets, then the study of the action of α on N reduces to the study of the action of α on N1 and N2 . Continuing this procedure, the study of the action of α on N reduces to the study of the action 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 union of orbits. Hence the partition of N into orbits is the finest possible partition of N into a disjoint union of nonempty invariant subsets. 2.9.2 Let α ∈ PT n and let N = N1 ∪ N2 be a partition into invariant subsets. For each i = 1, 2 set  α(x), x ∈ Ni , α(i) (x) = x, otherwise. In other words, α(i) acts on Ni in the same way as α, and α(i) acts on the complement 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 decomposition α = 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

32

CHAPTER 2. THE SEMIGROUPS Tn , PT n , AND IS n

form: one of its components coincides with the corresponding component of Γα (the one which corresponds to the orbit Ni ), and all other components consist of loops. In some sense α(i) can be identified with the corresponding connected component. Then the decomposition of α described in Sect. 1.3 can be interpreted as a decomposition of α into a product of its connected components. In particular, for permutation we get the usual decomposition into a product of independent cycles. 2.9.3 Let us interpret [a1 , a2 , . . . , ak |ai ] as the notation for the element from Tn , 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. Then the 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 general case, the factors of this decomposition do not correspond to connected components, the factors themselves (and even the number of these factors) depend on the choice of certain elements along the way to write down this decomposition, and these factors do not commute. This is another disadvantage of the notation from [AAH]. 2.9.4 Let α ∈ PT n . As each orbit of α is invariant with respect to α and the intersection of invariant sets is invariant, each minimal (with respect to inclusions) nonempty invariant set is contained in some orbit. From Theorem 1.2.9 it follows that the trajectory of each vertex from a fixed connected component either contains a cycle (if it exists in this component) or terminates at components sink. On the other hand, the set of all vertices in a cycle, and the set consisting of the sink are obviously invariant. Hence only these sets are minimal (with respect to inclusions) nonempty invariant sets. Let now K be an orbit of α. If the corresponding connected component of Γα contains a cycle, the set of all vertices of this cycle is called the kernel of the orbit K. If K does not contain any cycle, we say that the kernel of this orbit is empty. Proposition 2.9.2 Let α ∈ PT n , R be an orbit of α containing a cycle and K be the kernel of R. Assume |K| = k > 0. Then

2.9. ADDENDA AND COMMENTS

33

(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 and K = {y, α(y), . . . , αk−1 (y)}. 2.9.5 Let α ∈ PT n . The stable image of α is the set  stim(α) = im(αk ). k∈N

Proposition 2.9.3 For each α ∈ PT n the set stim(α) is invariant with respect 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 that the 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 IS n , introduced in Sect. 2.5, is close to the notation used in [Li]. 2.9.8 The role of IS n in the theory of inverse semigroups is analogous to the role 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 isomorphic to a subsemigroup of the semigroup IS(T ) of all partial injective transformations of the set T . The idea of the proof is as follows: for each a ∈ T we consider the transformation ρa : a−1 T → aT , defined via ρa (x) = ax. This transformation turns out to be bijective. Hence we can consider ρa as a partial permutation on T . For the mapping ϕ : T → IS(T ), ϕ(a) = ρa , one easily 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.

CHAPTER 2. THE SEMIGROUPS Tn , PT n , AND IS n

34

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 IS n 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 zeros to left zeros, and right zeros to right zeros. 2.9.10 The number of nilpotent elements in PT n is given by the following statement. Theorem 2.9.6 ([LU1]) PT n contains exactly (n + 1)n−1 nilpotent elements. Proof. For α ∈ PT n and k ≥ 0 let Nk denote the set of all x ∈ N for which the trajectory of x breaks at the kth step, that is, has the form x0 = x, x1 , . . . , xk . Assume that the nilpotency degree of α equals t. In this case 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 the following way: First we choose an ordered partition N = N0 ∪N  2 ∪· · ·∪Nt of N into t + 1 nonempty blocks (this can be done in n0 ,n1n,...,nt different ways, where ni = |Ni |, i = 0, . . . , t). Then for each i > 0 we define some mapping t from Ni to Ni−1 (altogether such mappings can be defined in nn0 1 nn1 2 · · · nnt−1 different ways). The total number of nilpotent elements in PT n is denoted by n(PT n ). By the above, we have    n t n(PT n ) = nn1 nn2 · · · nnt−1 = n0 , n1 , . . . , nt 0 1 1≤t


=

1≤t
n

n t n! nn0 1 nn1 2 · · · · · t−1 . n0 ! n1 ! n2 ! nt !

Using (1.8) we have

n(PT n ) =

n  n0 =1

=

n  n0 =1

n! (n0 − 1)!



n

1≤t
n t 1 nn0 1 nn1 2 · · · · t−1 = · n0 n1 ! n2 ! nt !

nk−1  n! (n − 1)! · = · nk = (n + 1)n−1 . (n0 − 1)! k! (n − 1 − k)!k! n−1 k=0

2.10. ADDITIONAL EXERCISES

35

In [LU1] Theorem 2.9.6 is proved in a different way. One more way to prove it is to use [Hi1, Lemma 6.1.4]. Yet another way to prove this theorem is 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 rooted tree with the root (n + 1). In particular, we have a well-defined notion of a distance of a vertex to the root. Now we can define the nilpotent element α from PT n as follows: If i is a vertex, let j denote the unique vertex of Γ such that (i, j) is an edge and whose distance to the root is smaller than that of i. Define α(i) = j if j is not the root and α(i) = ∅ otherwise. One verifies that the correspondence Γ → α is a bijection from the set of all labeled 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 ) (n + 1)n−1 1 = . = n |PT n | (n + 1) n+1 From Theorems 2.8.4 and 2.9.6 we get the following identity. Corollary 2.9.7 n−1  k=0

2.10

 n S(n, k + 1)k! = (n + 1)n−1 . k

Additional Exercises

2.10.1 Let S be a nonempty set such that 0 ∈ S. For all a, b ∈ S set a · b = 0. Show that (S, ·) is a semigroup (it is called a semigroup with zero multiplication 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 identity and a right zero (such S is usually called a semigroup of left identities or a right zero semigroup; analogously, using a · b = a one defines a left zero semigroup or a semigroup of right identities). 2.10.3 Show that Tn+1 contains a subsemigroup, isomorphic to the semigroup PT n . 2.10.4 Let X be a nonempty set and B(X) be the set of all subsets of X (the Boolean of X). Show that both (B(X), ∪) and (B(X), ∩) are commutative semigroups. Furthermore, show that these two semigroups are isomorphic. 2.10.5 Let K be an orbit of some transformation α ∈ PT n . Prove that the kernel of K coincides with the set ∩x∈K {αm (x) : m ≥ 0}.

36

CHAPTER 2. THE SEMIGROUPS Tn , PT n , AND IS n

2.10.6 Let α ∈ Tn . Prove that x ∈ N belongs to the kernel of an orbit K if and 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 = β k 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 prime p 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    n (k + 1)un−k . = k k=0

2.10.11 Show that for each α ∈ PT n the set VPT n (α) · α is a left zero semigroup 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 a group. 2.10.13 Prove that the operations of extending a semigroup with the element 1 (as in Proposition 2.2.1) and with the element 0 (as in Proposition 2.3.1) commute, that is, (S 1 )0 = (S 0 )1 (note the equality and not the isomorphism 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 of inverse 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 IS n form a pair of inverse elements. 2.10.16 Prove that for n > 1 the set of left zeros in the semigroup Tn forms a noncommutative semigroup.

2.10. ADDITIONAL EXERCISES

37

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 a and b commute. 2.10.18 ([GH1]) Prove that (a) IS n 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 in the semigroup IS n . Prove that Nn = |IS n | − n|IS n−1 |, n > 1. 2.10.20 ([BRR]) Prove the following recursive relation (for n > 2): |IS n | = 2n|IS n−1 | − (n − 1)2 |IS n−2 |. 2.10.21 ([GM, Theorem 9]) For Nn as in 2.10.19 show that Nn = 0. n→∞ |IS n | lim

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

1 1 1  c(α) = 1 + + · · · + . |Sn | 2 n α∈Sn

(b) ([Hi1, Lemma 6.1.12]) n  n! 1  c(α) = . |Tn | k(n − k)!nk α∈Tn

k=1

(c) ([GM5, Corollary 1])  α∈IS n

c(α) =

 n   1 1+ |IS n−k |n(n − 1) · · · (n − k + 1). k 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 Sn αSn = Sn βSn , or Sn αSn ∩ Sn βSn = ∅.

38

CHAPTER 2. THE SEMIGROUPS Tn , PT n , AND IS n

(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 Sn αSn = Sn βSn if and only if t(α) = t(β). 2.10.25 (a) Let α, β ∈ PT n . Show that we either have Sn αSn = Sn βSn , or Sn αSn ∩ Sn βSn = ∅. (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 Sn αSn = Sn βSn if and only if t(α) = t(β).

Chapter 3

Generating Systems 3.1

Generating Systems in T n , PT n , and IS n

Let S be a semigroup and A ⊂ S be a set. An element s ∈ S is said to be generated by A provided that s can be written as a finite product of elements from A. The set of all elements from S generated by A is usually denoted by A. If S is a monoid, we will understand the identity element of this monoid as a trivial finite product of elements from A (of length 0). Thus 1 ∈ A for any A ⊂ S. A subset A of the semigroup S is called a generating system or a generating set for S provided that A = S. A generating system is called irreducible provided that each proper subset of A is no longer a generating system. One of the first most natural questions in the study of some semigroup is to describe some (irreducible) generating system and to classify all (irreducible) generating systems. Until the end of this section we let S denote one of the semigroups Tn , PT n , or IS n . Let us recall that each of these semigroups contains Sn as the group of units (see Proposition 2.2.4). Lemma 3.1.1 (i) Each generating system of S must contain a generating system of Sn . (ii) If A is an irreducible generating system of S, then A ∩ Sn is an irreducible generating system of Sn . Proof. Sn coincides with the set of all elements of S of rank n. By Exercise 2.1.4(c), an element of rank n can be a product of elements of rank n only. This implies both (i) and (ii). Lemma 3.1.2 Each generating system of S must contain at least one element of rank (n − 1). Proof. The semigroup S contains elements of rank (n − 1). The set of elements of rank n coincides with Sn and is closed under composition. Hence O. Ganyushkin, V. Mazorchuk, Classical Finite Transformation Semigroups, Algebra and Applications 9, DOI: 10.1007/978-1-84800-281-4 3, c Springer-Verlag London Limited 2009 

39

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 of elements of rank n. The claim follows. Theorem 3.1.3 Let A be a generating system of Tn . Then this system is irreducible if and only if A = A1 ∪ {α}, where A1 is an irreducible generating system of Sn and rank(α) = n − 1. Proof. Taking Lemmas 3.1.1 and 3.1.2 into account, it is enough to show that any set of the form A = A1 ∪ {α}, where A1 is an irreducible generating system 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   1 2 ··· i − 1 i i + 1 ··· j − 1 j j + 1 ··· n α= a1 a2 · · · ai−1 a ai+1 · · · aj−1 a aj+1 · · · an and let β=



1 2 ··· b1 b2 · · ·

k − 1 k k + 1 ··· bk−1 b bk+1 · · ·

l − 1 l l + 1 ··· bl−1 b bl+1 · · ·

n bn



be an arbitrary element from Tn of rank (n − 1). Let π ∈ Sn be an arbitrary permutation satisfying π(k) = i and π(l) = j. Then   1 2 ··· k − 1 k k + 1 ··· l − 1 l l + 1 ··· n απ = c1 c2 · · · ck−1 a ck+1 · · · cl−1 a cl+1 · · · cn is an element of rank (n − 1). Consider now the permutation   c1 c2 · · · ck−1 ck+1 · · · cl−1 cl+1 · · · cn a x , τ= b1 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 that β = τ απ and hence β ∈ A. So, A contains all elements of Tn of defects 0 and 1. Let us prove by induction on k that A contains all elements of Tn of defect k. We only now have to prove the induction step. Let γ ∈ Tn be an element of defect k, k > 1. Since def(γ) > 0, there exists a ∈ im(γ) such that its preimage B = {x ∈ N : γ(x) = a} contains more than one element. Let b1 , b2 ∈ B be such that b1 < b2 . Finally, let a ∈ N\im(γ). Consider the idempotent   1 2 · · · b1 − 1 b1 b1 + 1 · · · b2 − 1 b2 b2 + 1 · · · n μ= 1 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 IS n

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 and the proof. Theorem 3.1.4 Let A be a generating system of IS n . Then this system is irreducible if and only if A = A1 ∪{α}, where A1 is an irreducible generating system of Sn and rank(α) = (n − 1). Proof. Repeat Theorem 3.1.3 with the following changes: For the partial permutations   1 2 ··· i − 1 i i + 1 ··· n α= a1 a2 · · · ai−1 ∅ ai+1 · · · an 

and β=

1 2 ··· b1 b2 · · ·

j − 1 j i + 1 ··· bj−1 ∅ bj+1 · · ·

n bn



the permutations π and τ should be chosen as follows: π(j) = i;   c1 c2 · · · cj−1 cj+1 · · · cn x . τ= b1 b2 · · · bj−1 bj+1 · · · bn y For the partial permutation  i1 i2 · · · γ= j1 j2 · · ·

ik ik+1 · · · jk ∅ · · ·

in ∅



the partial permutations δ and μ should be chosen as follows:   i1 i2 · · · ik ik+1 ik+2 · · · in δ= j1 j2 · · · jk jk+1 ∅ · · · ∅  μ=

i1 i2 · · · i1 i2 · · ·

ik ik+1 ik+2 · · · ik ∅ ik+2 · · ·

in in

 .

Theorem 3.1.5 Let A be a generating system of PT n . Then this system is irreducible if and only if A = A1 ∪ {α, β}, where A1 is an irreducible generating system of Sn , α is a total transformation of rank (n − 1) and β is a partial permutation of rank (n − 1). Proof. A composition of total transformations is a total transformation. On the other hand, a composition which consists of nontotal maps only, results in 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 at least one total transformation, say α, of rank (n − 1) and at least one partial transformation, say β, of rank (n−1), which is not total. We have |dom(β)| ≤

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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) is bijective. Taking Lemma 3.1.1 and the previous paragraph into account, we have to only show that any A = A1 ∪{α, β}, where A1 is an irreducible generating system of Sn , α is a total transformation of rank (n − 1), and β is a partial permutation of rank (n − 1), generates PT n . By Theorems 3.1.3 and 3.1.4 the sets A1 ∪ {α} and A1 ∪ {β} generate Tn and IS n , respectively. Each γ ∈ PT n can be extended to a total transformation γ  ∈ Tn . Then we have γ = γ  εdom(γ) and εdom(γ) ∈ IS n . Hence γ ∈ A. This completes the proof.

3.2

Addenda and Comments

3.2.1 Theorem 3.1.3 was proved in [Vo2]. 3.2.2 Although all groups are semigroups, group theory and semigroup theory are two completely different directions of modern algebra. They differ both in formulations of the principal problems and in methods of their solutions. In particular, a problem in semigroup theory is usually considered as solved if it is reduced to a problem in group theory. Theorems 3.1.3–3.1.5 are good examples of such reduction. They reduce the problem of classification of all irreducible generating systems for the semigroups PT n , Tn , and IS n to the analogous problem for the group Sn . Irreducible generating systems in Sn have very complicated structure. Two classical irreducible generating systems 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 to classify all irreducible generating systems in Sn is still open. At the same time Theorems 3.1.3–3.1.5 give a satisfactory classification of irreducible generating systems for PT n , Tn , and IS n . 3.2.3 In each of the semigroups PT n , Tn , and IS n , the set of all noninvertible elements forms a subsemigroup (called the singular part of the original semigroup). Hence we have a new natural problem: to study generating systems for the singular parts of our semigroups PT n , Tn , and IS n . Here, apart from irreducibility, one could also impose other conditions, for example to consist of idempotents, to consist of elements of defect 1, to consists of elements with prescribed structure of connected components, and so on. One of the first works in this direction was [Ho1] in which it was shown that the singular part of Tn is generated by idempotents of defect 1. One of the most recent works is [AAH], where the authors reprove some known results (in particular, the above-mentioned result from [Ho1]) and give several new examples of generating systems for the singular part of Tn .

3.3. ADDITIONAL EXERCISES

3.3

43

Additional Exercises

3.3.1 Prove that each element of defect k from Tn can be written as a product of a permutation and k idempotents of defect 1. 3.3.2 Let A ⊂ Sn be a subset consisting of transpositions. Define the (unoriented) graph ΓA in the following way: the vertices are {1, 2, . . . , n}, and for 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 transpositions. 3.3.3 (a) Show that for n > 2 the group Sn does not contain any oneelement 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 all idempotents 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 right ideal provided that IS ⊂ I. Left and right ideals are also called one-sided ideals. A subset I ⊂ S is called a two-sided ideal or simply an ideal provided that 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 definition and hence I is a subsemigroup of S. The converse is not true in the general case. For example, the group Sn is a subsemigroup of each of the semigroups Tn , PT n , and IS n . However, for each (partial) transformation α ∈ Sn both sets αSn and Sn α do not have any common elements with Sn . It is easy to see that both the intersection and the union of an arbitrary family 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 S 1 provided that S contains an identity element, and the semigroup constructed in Proposition 2.2.1 provided that S does not contain any identity element. Analogously we define S 0 . We also note that for k > 1 the notation S k means something completely different, namely, S k = {a1 a2 · · · ak : ai ∈ S. i = 1, . . . , k}. A left (resp. right or two-sided) ideal I of S is called principal provided that there exists a ∈ S such that I = S 1 a (resp. I = aS 1 , I = S 1 aS 1 ). The element a is called the generator of the ideal I. Note that a ∈ S 1 a, a ∈ aS 1 , and a ∈ S 1 aS 1 by definition. O. Ganyushkin, V. Mazorchuk, Classical Finite Transformation Semigroups, Algebra and Applications 9, DOI: 10.1007/978-1-84800-281-4 4, c Springer-Verlag London Limited 2009 

45

46

CHAPTER 4. IDEALS AND GREEN’S RELATIONS

Proposition 4.1.2 Each left (right or two-sided) ideal is a union of principal left (resp. right or two-sided) ideals. Proof. Since a ∈ S 1 a, if I ⊂ S is a left ideal, we have I = ∪a∈I S 1 a. For other ideals the argument is similar.

4.2

Principal Ideals in T n , PT n , and IS n

Since the semigroups Tn , PT n , and IS n are in fact monoids, we have S = S 1 for these semigroups. Theorem 4.2.1 Let S denote one of the semigroups Tn , PT n , or IS n . Then for 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 have im(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 that for x ∈ dom(β) we have γ(x) = ab if and only if β(x) = b. A direct calculation then shows that αγ = β. Note that in the case S = IS n the injectivity of α and 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 ∈ IS n . Thus β ∈ αS. This implies that X ⊂ αS and completes the proof. Corollary 4.2.2 (i) Each of the semigroups PT n and IS n has exactly 2n 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 uniquely determined by im(α), that is, by a subset of N, which must be nonempty in the case of Tn . The claim follows. Corollary 4.2.3 Let S denote one of the semigroups Tn , PT n , or IS n and α ∈ S be such that rank(α) = k. Then ⎧ ⎪ S = Tn ; kn , ⎪ ⎪ ⎪ ⎨(k + 1)n , S = PT n ; |αS| =    k  n k ⎪ ⎪ ⎪ i!, S = IS n . ⎪ ⎩ i i i=0

4.2. PRINCIPAL IDEALS IN Tn , PT n , AND IS n

47

Proof. Theorem 4.2.1 says that in the case of Tn the ideal αS is the set of all mappings from N to im(α), where |N| = n and |im(α)| = k. Hence |αS| = k n . In the case of PT n , Theorem 4.2.1 says that αS is the set of all partial mappings from N to im(α), that is, the set of all mappings from N to im(α) ∪ {∅}. Hence |αS| = (k + 1)n . Finally, for IS n , Theorem 4.2.1 says that αS is the set of all partial injections 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 its domain, 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 statement now 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 the following 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 is defined on both x and y and α(x) = α(y). It is clear that πα is an equivalence relation and the equivalence classes of πα are dom(α) and all full preimages of elements from im(α). Theorem 4.2.4 Let S denote one of the semigroups Tn , PT n , or IS n . Then for 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 = γα for some γ ∈ S. For arbitrary x ∈ dom(β) we have β(x) = γ(α(x)). Hence x ∈ dom(α) and thus dom(β) ⊂ dom(α). Furthermore, if x, y ∈ dom(β) and α(x) = α(y), we have β(x) = γ(α(x)) = γ(α(y)) = β(y). Thus πα ⊂ πβ and we have Sα ⊂ X. Assume now that β ∈ X. Set M = α(dom(β)) and define the mapping γ : 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) is uniquely defined. If S = Tn , we have dom(β) = N and M = α(N) = im(α). In this case, we can extend γ to a total transformation on N in an arbitrary way. Abusing notation 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 = IS n and x1 = α(y1 ), x2 = α(y2 ) are two different

48

CHAPTER 4. IDEALS AND GREEN’S RELATIONS

elements in M , we have y1 = y2 and from the injectivity of β it follows that γ(x1 ) = β(y1 ) and γ(x2 ) = β(y2 ) are also different. This means that in this case γ ∈ IS n 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 and thus y πβ y1 , implying y ∈ dom(β), a contradiction. This means that for y ∈ 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. Thus X ⊂ 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 = IS n , then the restriction of πα to dom(α) becomes the equality relation. Hence dom(β) ⊂ dom(α) automatically implies πα ⊂ πβ . This means that for the semigroup S = IS n the second condition in (4.2) can be omitted. The number of all unordered partitions N = N1 ∪ N2 ∪ · · · ∪ Nk of the set N into disjoint unions of nonempty blocks is called the nth Bell number numbers S(n, k) of and is denoted by Bn . From the definition of the Stirling  the 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 the semigroup 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 IS n equals 2n . Proof. From Theorem 4.2.4 we have that a principal left ideal of Tn is uniquely determined by the partition of N into the equivalence classes of the relation πα . Since all partitions obviously give rise to some ideals, we obtain 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 πα on the set N ∪ {n + 1}: If dom(α) = N, we adjoin n + 1 to the block dom(α); if dom(α) = N, then we say that {n + 1} is a new equivalence class. Obviously, πα is uniquely determined by πα . Conversely, any πα uniquely determines πα and dom(α). By Theorem 4.2.4 each principal left ideal of PT n is uniquely determined by the pair (πα , dom(α)). Hence the previous paragraph says that the mapping (πα , dom(α)) → πα is a bijection from the set of all principal left ideals of PT n to the set of all unordered partitions of N ∪ {n + 1}, the latter containing n + 1 element. Thus the number of different principal left ideals of PT n equals Bn+1 . This proves (ii).

4.2. PRINCIPAL IDEALS IN Tn , PT n , AND IS n

49

Finally, to prove (iii) we note that by Theorem 4.2.4 each principal left ideal of IS n is uniquely determined by the set dom(α), which is just a subset of 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) and 2.9.9. Corollary 4.2.7 Let S denote one of the semigroups Tn , PT n , or IS n , and α ∈ S be an element of rank k. Then we have the following: ⎧ ⎪ S = Tn ; nk , ⎪ ⎪ ⎪ ⎨(n + 1)k , S = PT n ; |Sα| = k     ⎪ n k ⎪ ⎪ i!, S = IS n . ⎪ ⎩ i i i=0

Proof. We start with the cases S = Tn and S = PT n . As πα ⊂ πβ , each transformation β ∈ Sα is uniquely determined by its values on the set of those equivalence classes of πα which are contained in dom(α). We have k such classes and for each of them we have to choose the value of β on this class, which is an element from N (or N ∪ {∅} in the case of PT n ). Now for Tn and PT n the statements are obtained by applying the product rule. In the case S = IS n the proof is similar to that of the corresponding part of Corollary 4.2.3. The only difference is that now one has to consider partial injections from dom(α) to N. Theorem 4.2.8 Let S denote one of the semigroups Tn , PT n , or IS n and α ∈ 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 ⊂ X follows 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, and im(β) = {b1 , b2 , . . . , bm }. Then m ≤ k and for each i = 1, . . . , k we choose some 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 ∈ IS n implies γ ∈ IS n by construction. Let δ be any permutation satisfying 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 IS n contains (n + 1) different principal ideals. The semigroup Tn contains n different principal ideals.

50

CHAPTER 4. IDEALS AND GREEN’S RELATIONS

Proof. By Theorem 4.2.8 a principal ideal is uniquely determined by the rank of the generator. For PT n and IS n 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 IS n , and α ∈ S be an element of rank k. Then we have the following: ⎧ k ⎪  ⎪ n! ⎪ ⎪ , S = Tn ; S(n, i) ⎪ ⎪ (n − i)! ⎪ i=1 ⎪ ⎪ ⎪ k ⎨ n! |SαS| = , S = PT n ; S(n + 1, i + 1) ⎪ (n − i)! ⎪ i=0 ⎪ ⎪ ⎪ k  2 ⎪  ⎪ n ⎪ ⎪ i!, S = IS n . ⎪ ⎩ i i=0

Proof. According to Theorem 4.2.8, to determine |SαS| we have  to find the number 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 an unordered partition of N into i nonempty blocks (which can be done in S(n, i) different ways), and then an injective mapping from the blocks of n! this partition into N (this can be done in n(n − 1) · · · (n − i + 1) = (n−i)! n! different ways). Hence in this case ti = S(n, i) (n−i)! . For S = PT n the argument is almost the same with the difference that we should consider partitions of N∪{n+1} instead of those of N. The block n! in containing n + 1 corresponds to dom(β). Hence ti = S(n + 1, i + 1) (n−i)! this case. For S = IS n the number of partial permutations of rank k was computed in Theorem 2.5.1.

4.3

Arbitrary Ideals in T n , PT n , and IS n

Theorem 4.3.1 Let S denote one of the semigroups Tn , PT n , or IS n . Then all two-sided ideals in S are principal and are generated by any element of the ideal, which has the maximal possible rank. Proof. Let I ⊂ S be a two-sided ideal. Choose α ∈ I of maximal possible rank. As SIS ⊂ I, we have SαS ⊂ I. On the other hand, by Theorem 4.2.8 the 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 IS n . For k ≤ n set Ik = {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 IS n

51

ideals of S forms a chain with respect to inclusions. For Tn this chain has the form I1 ⊂ I2 ⊂ · · · ⊂ In = Tn . For PT n and IS n this chain has the form {0} = I0 ⊂ I1 ⊂ I2 ⊂ · · · ⊂ In = PT n (orIS n ). Note that In \In−1 = Sn in all cases. The Boolean B(X) of a set X is partially ordered with respect to inclusions. Recall that an antichain of a partially ordered set is a subset such that each two elements in this subset are not comparable. In particular, if A1 , A2 , . . . , Am ⊂ N, the set {A1 , . . . , Am } is an antichain of B(N) if and only 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 IS n . (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 implies IL1 = IL2 . (iii) For each right ideal I of S there exists an antichain L ⊂ B(N) such that 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 = ∅. Let A ∈ L1 \L2 . We have to consider two possible cases. Case 1: A is not comparable with any element of L2 . In this case each α ∈ S for which im(α) = A is contained in IL1 but not in IL2 . Hence IL1 = IL2 . Case 2: There exists B ∈ L2 such that A and B are comparable. Then B ⊂ 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 . Hence IL1 = IL2 in this case as well. The above proves (ii). To prove (iii) we first observe that the maximal (with respect to inclusions) elements of the set {im(α) : α ∈ I} obviously form an antichain, say L, of B(N). By the definition of L, for each α ∈ I there exists A ∈ L such that im(α) ⊂ A. Hence α ∈ IL and I ⊂ IL . On the other hand, for each β ∈ IL there exists A ∈ L and α ∈ I such that im(β) ⊂ A and im(α) = A. By Theorem 4.2.1 we have β ∈ αS. But αS ⊂ I since α ∈ I and I is a right ideal. Hence β ∈ I. This means that IL ⊂ I and thus IL = I. The proof is complete.

52

CHAPTER 4. IDEALS AND GREEN’S RELATIONS ←

The identity mapping from S to the dual semigroup S maps left ideals to right ideals and vice versa. For the inverse semigroup IS n we have ← IS n ∼ = IS n via the mapping a → a−1 . Hence Theorem 4.3.2 can be used to describe all left ideals in IS n . The only change one has to make is to substitute im by dom. Indeed, the equalities im(α−1 ) = dom(α) and im(α) = dom(α−1 ) follow from the proof of Theorem 2.6.7. Hence we get the 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 IS n and the mapping L → L I is a bijection from the set of all antichains of B(X) to the set of all left ideals of IS n . As Tn and PT n are not self-dual (see 2.9.9 and Exercise 2.10.23), the description of all left ideals in these semigroups requires more efforts. Let Partn denote the set of all (unordered) partitions of N. Define a partial order on Partn in the following way: Let ρ = A1 ∪ · · · ∪ Ak and τ = B1 ∪ · · · ∪ Bm be two partitions. We will write ρ  τ provided that each block Ai is a union of some blocks of the partition τ (i.e., of some Bj s). In particular, the maximum partition is the one into one element blocks and the minimum partition is the one with one block. Let ρ be a partition of N. Denote by πρ the corresponding equivalence relation (x πρ y if and only if x and y belong to the same block of ρ). It is easy 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 example as follows: a1 , . . . , ap |b1 , . . . , bq | · · · |c1 , . . . , cr . Using this notation the Hasse diagram of the partially ordered set of all partitions of {1, 2, 3, 4} looks as follows: 1|2|3|4 TT

jjjojo  jjjoojooo  j j j  o jjj

1|23|4 14|2|3 1|24|3

??OOOTTT ?? OOOTTTT OO TTTT ? T

13|2|4 12|3|4 1|2|34

// OOO // OOO //  // tt lll//ll  //  OOO/O/  OOO/O/   tttl/tl//lll ///  /O/ OO //  /O/ OO tll /  / / / OOOO // OOOlOtltltltl /// /// OOO/ lllttO OO   ///   ///   l  //  // lllllt/t/O/tOtOOO OOOOO ///  ///   llll tt  O  O  14|23 TT1|234 124|3 13|24 123|4 134|2 j 12|34 TTTT OOO ??  oooo jjjjjj TTTT OOO ??  j  o TTTTOOO?? ooojjjjj TT  ojjj

1234

4.4. GREEN’S RELATIONS

53

To each element α ∈ Tn we associate the partition ρα of N into equivalence classes with respect to the relation πα . Then ρα  ρβ if and only if π α ⊃ 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 → L I is a bijection from the set of all 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 ∈ L I and β ∈ Tn , from ρα  ρ it follows that ρβα  ρα  ρ. Hence βα ∈ L I and L I is a left ideal. It is clear that for each partition ρ ∈ Partn there exists α ∈ Tn such that ρα = 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 all antichains L1 and L2 of Partn such that L1 = L2 the ideals L1 I and L2 I are different. Finally, to prove that each left ideal of Tn has the form L I for some antichain 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 use Theorem 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 → L I is a bijection from the set of 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 the same principal left ideal. In other words, aLb if and only if S 1 a = S 1 b. Equivalence classes of the relation L are called L-classes. For a ∈ S, the L-class containing a will be denoted by L(a). In other words, aLb if and only if a ∈ L(b). The following easy but very useful fact follows immediately from the definition.

54

CHAPTER 4. IDEALS AND GREEN’S RELATIONS

Proposition 4.4.1 aLb if and only if there exist x, y ∈ S 1 such that a = xb and b = ya. In the dual way we define the relation R: Elements a, b ∈ S are called R-equivalent provided that they generate the same principal right ideal. In other words, aRb if and only if aS 1 = bS 1 . Equivalence classes of the relation R are called R-classes. For a ∈ S the R-class containing a will be denoted by R(a). Proposition 4.4.2 aRb if and only if there exist x, y ∈ S 1 such that a = bx and b = ay. Note that for every x ∈ S 1 the equality S 1 a = S 1 b implies the equality = S 1 bx; and the equality aS 1 = bS 1 implies the equality xaS 1 = xbS 1 . In other words, the relation L is compatible with the right multiplication by elements of S 1 (this is also called right compatible), that is, aLb implies axLbx. 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 product ξ ◦ η is defined in the following way: S 1 ax

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 respect to 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 ∈ S 1 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 that ξ ◦ η = η ◦ ξ implies that ξ ◦ η is again an equivalence relation. Moreover, show that this product is the minimum equivalence relation which contains both ξ and η. The minimum equivalence relation on S which contains both R and L is denoted by D and is called the D-relation. From Lemma 4.4.4 it follows that D = L ◦ R = R ◦ L. All other notions and notation for D are similar to the 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 equivalence relation. We define the H-relation as the intersection of R and L. All other

4.4. GREEN’S RELATIONS

55

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 provided that they generate the same principal two-sided ideal, that is, S 1 aS 1 = S 1 bS 1 . All other notions and notation for J are similar to the ones used for 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 that D ⊂ J . Hence we have the following diagram depicting the introduced relations on S: > L @o @@ ~ @@ ~~ @@ ~~ ~ / ~~ @  H @o D ? @@ ~ @@ ~~ @@ ~~ ~ @ / ~~

/J

R The relations L, R, H, D, and J on the semigroup S are called Green’s relations 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 ∈ S 1 such that aRc and cLb. But this means that c ∈ R(a) and c ∈ L(b). Thus R(a) ∩ L(b) = ∅, proving the implication (a)⇒(b). Assume that c ∈ R(a) ∩ L(b). Then c ∈ R(a) and c ∈ L(b), which means that 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 to the same D-class have a nonempty intersection. Hence it is convenient to think 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 )

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CHAPTER 4. IDEALS AND GREEN’S RELATIONS

The cells in this table give all H-classes, contained in our D-class. From Lemma 4.4.6 it follows that all cells are nonempty. The rectangular table above 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 ∈ S 1 be such that au = b and bv = a. (i) The mapping Λu : x → xu maps L(a) to L(b), and the mapping λv : 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) for all 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

x yv o

L(a)

b = au

λu

/ xu

λv

y

L(b)

Proof. The statement (i) follows from the fact that L is right compatible. Let x ∈ L(a). Then there exists p ∈ S 1 such that x = pa. Hence (λv λu )(x) = λv (λu (x)) = xuv = pauv = pa = x. This means that λv λu is the identity transformation of L(a). Analogously one shows that λu λv is the identity transformation of L(b). This proves the statement (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 induce mutually inverse bijections between the sets of H-classes in L(a) and L(b). Moreover, λu and λv also induce mutually inverse bijections between the corresponding H-classes. ←

Applying Green’s Lemma to the opposite semigroup S we obtain the following dual version of this lemma.

4.4. GREEN’S RELATIONS

57

Lemma 4.4.8 Let S be a semigroup and a, b ∈ S be such that aLb. Let u, v ∈ S 1 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 bijections. (iii) Both μu and μv preserve L-classes, that is, xLμu (x) and yLμv (y) for all x ∈ R(a) and y ∈ R(b). We note that the existence of the elements u and v, which are essentially used 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 same cardinality. Proof. Let aDb. As D = L ◦ R, there exists c ∈ S 1 such that aRc and cLb. 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) to H(c). By the dual of Green’s Lemma the mapping μv is a bijection from H(c) to H(b). H(c)

H(a)

 

R(a)

a

λu

/

c

μv



b o

H(b)

L(b) Hence the composition μv λu 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 have aS = Sa = S. Show that S is a group. Theorem 4.4.11 Let H be an H-class of S. Then the following conditions are equivalent: (a) H is a group. (b) H contains an idempotent. (c) There exist a, b ∈ H such that ab ∈ H.

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CHAPTER 4. IDEALS AND GREEN’S RELATIONS

Proof. Assume that H is a group. Let e ∈ H be the identity element. Then e · e = e and hence (a)⇒(b). The implication (b)⇒(c) is obvious. To complete the proof we have to prove the implication (c)⇒(a). Let a, 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 from Exercise 4.4.10.

4.5

Green’s Relations on T n , PT n , and IS n

The description of principal ideals in Tn , PT n , and IS n obtained in Sect. 4.2 allows 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 of Tn ) or N∪{n+1} (in the case of PT n , or IS n ) generated by the relation πα . Theorem 4.5.1 Let S be one of the semigroups Tn , PT n , or IS n , and α, 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 and proves (iv).

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

59

Corollary 4.5.2 In the semigroups Tn , PT n , and IS n we have D = J . As we have already mentioned after Theorem 4.2.4, for the semigroup IS n 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 IS n , and α, 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−1 α. Hence αLβ by Proposition 4.4.1. Conversely, let αLβ. Then ρα = ρβ . Let A1 , . . . , Ak be those blocks of ρα which are contained in dom(α) and let α(x) = ai and β(x) = bi for all x ∈ 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 ∈ IS n . 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 ← IS n ∼ = IS n .

If α = νβμ for some μ, ν ∈ Sn , then β = ν −1 αμ−1 and hence the principal ideals generated by α and β coincide. This means αJ β. The latter implies αDβ using Theorem 4.5.1(iv) and (v). Conversely, if αDβ, then there exists γ such that αLγ and γLβ. By Proposition 4.5.3 we have α = νγ for some ν ∈ Sn , and by (i) we have γ = βμ for some μ ∈ Sn . Thus α = νβμ. Example 4.5.5 Here we present the egg-box diagrams for all D-classes of the 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 3 from PT 3 is given by the second row (a1 a2 a3 ) (do not mix a1 a2 a3 this with our notation for cycles). To accommodate the picture to the page size we are forced to transpose it. So, in our picture rows are L-classes and columns are R-classes. For each L-class L we show on the left of the diagram the partition ρα of the domain of α ∈ L, and for each R-class R we show above the diagram the set im(α), where α ∈ R. The elements marked with ∗ are idempotents.

60

CHAPTER 4. IDEALS AND GREEN’S RELATIONS {1, 2, 3} D3 :

1|2|3

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

{1, 2}

D2 : 12|3 13|2 1|23 1|2 1|3 2|3

{1, 3}

{2, 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 : 123 12 13 23 1 2 3

{1}

{2}

{3}

(111)∗

(222)∗

(11∅)∗

(22∅)∗

(333)∗ (33∅) (3∅3)∗ (∅33)∗ (3∅∅) (∅3∅) (∅∅3)∗

(1∅1)∗ (2∅2) (∅11) (∅22)∗ (1∅∅)∗ (2∅∅) (∅1∅) (∅2∅)∗ (∅∅1) (∅∅2) D0 :

∅ (∅∅∅)∗

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

4.6

Combinatorics of Green’s Relations in the Semigroups T n , PT n , and IS n

From the definition of Green’s relations we have that the number of L-, Rand 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 Rclasses and n + 1 different J -classes. (iii) The semigroup IS n contains 2n different L-classes, 2n different Rclasses and n + 1 different J -classes.

4.6. COMBINATORICS OF GREEN’S RELATIONS

61

Since in Tn , PT n , and IS n we have J = D by Theorem 4.5.1, in what follows we will not consider J -classes. For each semigroup Tn , PT n , and IS n and each k = 0, . . . , n we denote by Dk the D-class which consists of elements of rank k. Then Tn has D-classes Dk , k = 1, . . . , n, and PT n and IS n have an additional D-class D0 . Note that in each of these semigroups the ideal Ik has the form Ik = ∪i≤k Dk and hence Dk = Ik \Ik−1 . The cardinality of Dk was already determined during the proof of Corollary 4.2.10: Proposition 4.6.2 Let S be one of the semigroups Tn , PT n , or IS n . We have ⎧ ⎪S(n, k) n! , S = Tn ; ⎪ (n−k)! ⎨ n! |Dk | = S(n + 1, k + 1) (n−k)! , S = PT n ; ⎪ ⎪ ⎩n2 k!, S = IS . k

n

Proposition 4.6.3 For each of the semigroups Tn , PT n , and IS n there is a natural bijection between the k-element subsets of Nand  the R-classes inside the 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 image of elements from this class. For R-classes inside Dk the image can be an arbitrary k-element subset of N. The statement follows. Proposition 4.6.4 (i) For the semigroup Tn there is a natural bijection between partitions of N into k blocks and the L-classes inside the Dclass Dk . In particular, Dk contains S(n, k) different L-classes. (ii) For the semigroup PT n there is a natural bijection between partitions of N ∪ {n + 1} into k + 1 blocks and the L-classes inside the D-class Dk . In particular, Dk contains S(n + 1, k + 1) different L-classes. (iii) For the semigroup IS n there is a natural bijection between k-element subsets 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 by some partition ρα of the set N. This partition is the same for all elements of L. For L-classes inside the Dk such partition can be an arbitrary partition of N into k blocks. The statement (i) follows. The proof of (ii) is analogous to that of (i) with substitution of N by N ∪ {n + 1}. The proof of (iii) repeats that of Proposition 4.6.3 considering domains instead of images.

62

CHAPTER 4. IDEALS AND GREEN’S RELATIONS

Proposition 4.6.5 (i) In each of the semigroups Tn , PT n and IS n every   L-class inside the Dk contains exactly nk different H-classes. (ii) In semigroups Tn , PT n and IS  n every R-class inside the Dk contains S(n, k), S(n + 1, k + 1) and nk different H-classes, respectively.   (iii) In semigroups Tn , PT n and IS n every D-class Dk contains S(n, k) nk ,    2 S(n + 1, k + 1) nk and nk different H-classes, respectively. Proof. Let D be a D-class. By Lemma 4.4.6, there is a bijection between the H-classes inside D and pairs (R, L), where L and R are an L-class and an 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 inside D, and vice versa. The statement now follows from Propositions 4.6.3 and 4.6.4. Theorem 4.6.6 For each of the semigroups Tn , PT n , and IS n the cardinality 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) it consists 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 completely determined 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 IS n each L-class inside the D-class Dk contains exactly nk k! elements. inside the D-class (ii) In the semigroups Tn , PT n , and IS n each R-class  Dk contains S(n, k)k!, S(n+1, k+1)k! and nk k! elements, respectively.

4.7

Addenda and Comments

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 had been described already by A. Suschkewitsch in [Su2]. Two-sided ideals of IS n 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 for Tn , PT n , or IS n in the literature. 4.7.2 Green’s relations were introduced in [Gr]. They play a very important role in the study of the structure of semigroups. One of the main reasons is that Green’s Lemma 4.4.7 in some sense allows us to define a kind of Cartesian coordinate system on each D-class.

4.7. ADDENDA AND COMMENTS

63

4.7.3 Green’s relations for Tn were described by Miller and Doss, see [Do]. A nice presentation of this description can be found in [CP1, Sect. 2.2]. Green’s relations for IS n and PT n were described by Reilly in [Re1] and FitzGerald and Preston in [FP], respectively. 4.7.4 In Sect. 2.7 we already mentioned a very close connection between regular elements and idempotents. Another aspect of this connection is the following: Proposition 4.7.1 Let S be a semigroup and a ∈ S. Then the following conditions 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 and a = 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 exist x, y ∈ S 1 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 the semigroups Tn , PT n , and IS n are regular, we have the following: Corollary 4.7.3 (i) In the semigroups Tn , PT n , and IS n all principal (right, left, two-sided) ideals are generated by idempotents. (ii) In the semigroups Tn , PT n , and IS n each L-class and each R-class contain an idempotent.   4.7.5 The D-class Dk of the semigroup Tn contains nk k n−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 and ya = b has a (unique) solution. Hence if G is a subgroup of some semigroup S, then any two elements from G are both L- and R-equivalent in S. Hence they are H-equivalent. This means that any subgroup of S is contained in

64

CHAPTER 4. IDEALS AND GREEN’S RELATIONS

some H-class, say H. In particular, H must contain an idempotent: the identity element of this group. By Theorem 4.4.11, in this case H itself is a group. Theorem 4.7.4 Let H be an H-class of Tn , PT n , or IS n , which contains an idempotent of rank k. Then H is isomorphic to the symmetric group Sk . Proof. Let  ∈ H be an idempotent. Let further im() = {a1 , . . . , ak } and for i = 1, . . . , k let Bi be the full preimage of ai . From 2 =  we have that ai ∈ Bi for all i. Then, by Theorem 4.5.1(iii) the elements from H are given by all possible injections from the set {B1 , . . . , Bk } to the set {a1 , . . . , ak }. In particular, the restriction to {a1 , . . . , ak } defines a bijective mapping from H to the symmetric group on {a1 , . . . , ak }. It is easy to see that this mapping is compatible with the composition and hence is an isomorphism. The image of this mapping is obviously isomorphic to Sk . 4.7.7 Let S be one of the semigroups Tn , PT n , or IS n . Let further H1 and H2 be two different H-classes of S inside the same D-class. If both H1 and H2 are groups, then from Theorem 4.7.4 it follows that H1 ∼ = H2 . This is a special case of the following more general statement. Theorem 4.7.5 Let S be a semigroup and H1 and H2 be two different H-classes of S inside the same D-class. If both H1 and H2 are groups, then H1 ∼ = H2 as groups. Proof. Denote the idempotents of H1 and H2 by e and f , respectively. Then e and f are the identities in the groups H1 and H2 , respectively. Let a ∈ R(e) ∩ L(f ). As eRa, there exists b ∈ S 1 such that a = eb. Hence ea = e · eb = eb = a. Analogously one shows that af = a. H(e) 

R(e)

e

λa

μa

R(f )



a L(e)

/

a

μa

λa

 / f o

H(f )

L(f )

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

4.8. ADDITIONAL EXERCISES

65

As a e = a , the dual Green’s lemma says that the mapping μa : y → a y is a 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 inverse to λa . Hence (λa μa )(f ) = e. On the other hand, (λa μa )(f ) = af a = 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 nonempty subset 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 ideal of S for each A. If A = {a}, one usually uses the notation Annl (a), or Annr (a) instead of Annl ({a}), or Annr ({a}), respectively.

4.8

Additional Exercises

4.8.1 Prove the identity Bn+1 =

n  k=1

n    n Bk . (k + 1)S(n, k) = k k=0

4.8.2 Let I be an ideal of one of the semigroups Tn , PT n , or IS n . Prove that I 2 = I. 4.8.3 Prove that in each of the semigroups Tn , PT n , or IS n the number aR of right ideals satisfies the inequality n

22 > aR >

n  k=1

n 2(k ) − n.

66

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4.8.4 Find the number of left ideals in IS n 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 and each α ∈ S. (b) Let S = Tn , n > 3, or S = PT n , n > 1. Show that for α, β ∈ S the relation αRβ in general does not imply the existence of η ∈ Sn such that α = βη. 4.8.7 ([HM]) Let n > 3, 1 < k < n, and α ∈ Tn be an element of rank k. Prove that |αTn | ≥ |Tn α|, moreover, the equality is possible only if n = 4 and k = 2. 4.8.8 Compute the number of elements in each D-class of the semigroups (a) IS 5 , (b) T5 , (c) PT 5 . 4.8.9 Prove that a regular semigroup is inverse if and only if for each principal 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 IS n . 4.8.11 An ideal I ⊂ S is called semiprime provided that a2 ∈ I implies a ∈ I for all a ∈ S. Find all semiprime ideals in Tn , PT n , and IS n . 4.8.12 An ideal I ⊂ S is called reflexive provided that ab ∈ I implies ba ∈ I for all a, b ∈ S. Find all reflexive ideals in Tn , PT n , and IS n . 4.8.13 Find the number of idempotents in each R-class inside Dk for the semigroup (a) Tn (b) PT n 4.8.14 (a) Let α ∈ Tn be a transformation of rank k and n1 , . . . , nk be the cardinalities of the full preimages of elements from im(α). Find the number of idempotents in R(α). (b) The same problem for PT n .

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67

4.8.15 Prove that the class H(α) of the semigroup Tn contains an idempotent 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 an idempotent. Show that e is a left identity on R, that is, ex = x for all x ∈ 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 ∈ IS n 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 IS n none of the ideals Ik , 0 < k < n, is a (left or right) annihilator of some set.

Chapter 5

Subgroups and Subsemigroups 5.1

Subgroups

Let S be a semigroup and G be a subgroup of S. Then G contains the unique idempotent e (the identity element of G). Conversely, if e ∈ S is an idempotent, then {e} forms a trivial subgroup of S. This shows that there is a close connection between the subgroups of S and idempotents of S. In this section we would like to illustrate this connection. We start from the following two obvious statements: Lemma 5.1.1 Let e ∈ E(S). Then eSe is a submonoid of S with the identity element 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 properly contained in any other subgroup of S. Theorem 5.1.3 (i) For each e ∈ E(S) there is a unique maximal subgroup 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 e is the identity element is contained in (eSe)∗ . The latter is a group by Proposition 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 and Gf , respectively. We have ab = e and ca = f . From this and the facts that f a = a (since a ∈ Gf ) and ae = e (since a ∈ Ge ) we have e = ab = f ab = f e = cae = ca = f, a contradiction. O. Ganyushkin, V. Mazorchuk, Classical Finite Transformation Semigroups, Algebra and Applications 9, DOI: 10.1007/978-1-84800-281-4 5, c Springer-Verlag London Limited 2009 

69

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In 4.7.6 it is shown that maximal subgroups of S are exactly the H-classes H(e), e ∈ E(S). As each H-class contains at most one idempotent and different H-classes do not have common elements, we get an alternative proof of Theorem 5.1.3. In what follows for e ∈ E(S) we shall denote the corresponding maximal subgroup by Ge . For the semigroups Tn , PT n , and IS n the following statement is a direct consequence of Theorems 4.5.1 and 4.7.4: Theorem 5.1.4 Let S denote one of the semigroups Tn , PT n , or IS n . For  ∈ E(S) we have: (i) G = {α ∈ S : im(α) = im(), ρα = ρ }. (ii) If rank() = k, then G ∼ = Sk . An element g ∈ S is called a group element provided that g ∈ Ge for some e ∈ E(S). Exercise 5.1.5 Let S denote one of the semigroups Tn , PT n , or IS n . Show that an element α ∈ S is a group element if and only if the full subgraph of Γα with the vertex set N\stim(α) is an empty graph, that is, it does not contain any arrow.    Proposition 5.1.6 (i) The semigroup Tn contains nk=1 nk k n−k k! group elements.    (ii) The semigroup PT n contains nk=1 nk (k + 1)n−k k! group elements.    (iii) The semigroup IS n contains nk=1 nk k! group elements. Proof. In the proofs of Corollaries 2.7.4, 2.7.5, and 2.7.3, it was shown that the semigroups Tn , PT n , and IS n contain       n n−k n n n−k k , (k + 1) , and k k k idempotents of rank k, respectively. After this the claim follows directly from Theorems 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 system consisting of one element. If this element is a, one writes S = a. Let S = a be a finite cyclic semigroup. Then the sequence of elements a1 = a, a2 , a3 , . . . must contain repeating elements. Assume that the elements a, a2 , . . . , al are pairwise different and al+1 = ak , where k ≤ l. The

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71

number l is called the order of a and is denoted by |a|, the number k is called the index of a and the number m = (l + 1) − k is called the period of a. The pair (k, m) is called the type of a. From ak+m = ak we get that the different elements in the sequence a = a, 2 a , 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 for x ∈ 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 of x modulo y. The set {x + sy : s ∈ Z} is called the residue class of x modulo y and is denoted by x. The set Zy of all residue classes modulo y forms a group 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  s+t
(5.2)

Hence in the semigroup a the multiplication is completely determined by the 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) ∈ IS k+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 a of type (k, m). Then the set P = {ak , . . . , ak+m−1 } is a subgroup of a. Moreover, this subgroup is isomorphic to Zm .

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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 the other 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 finite order. Then the semigroup a contains an idempotent. In particular, each finite semigroup contains an idempotent. Proof. To prove the first statement we just have to note that the identity element of the group P from Lemma 5.2.5 is an idempotent. The proof is then completed by observing that in a finite semigroup all elements have finite order. The identity element of P is the ak+i for which ϕ(ak+i ) = 0, that is, i ≡ −k(mod m). It is worth noting that ak+(−k mod m) is the unique idempotent of a, as for each b ∈ a\P we have bk ∈ P and hence b cannot be an idempotent. The cyclic semigroup a, generated by an element of type (k, m), can be depicted via the following diagram. ak+1

a1

a2

a3

...

w ww ww w w ww

ak+2

...

ak+m−2

...

ak GG

GG GG GG G

ak+m−1

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. Then there exist k, m ∈ N such that S = a = {a1 , a2 , . . . , ak+m−1 } and ak+m = ak . (ii) For each k, m ∈ N there exists a finite cyclic semigroup, generated by an element of type (k, m).

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73

(iii) Two finite cyclic semigroups are isomorphic if and only if they are generated by elements of the same type. (iv) If a has type (k, m), then the set {ak , ak+1 , . . . , ak+m−1 } is a subgroup of 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, 5.2.5 and Corollary 5.2.6. The only thing which we have to show is that if the semigroups 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 after Corollary 5.2.6 we have that an element of a is a group element if and only 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 group elements as well. Hence m = m and thus k = k  as well, since we already know that k + m = k  + m . This completes the proof. Proposition 5.2.8 Let α ∈ PT n . Then the index of α equals the minimum k for which im(αk ) = stim(α), and the period of α equals the order of the permutation α|stim(α) . Proof. For every i > 0 the set im(αi ) is invariant with respect to α. Hence we 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. Let m 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 m elements have the same image, but their restrictions to this image are permutations αk |stim(α) , αk+1 |stim(α) ,. . . , αk+m−1 |stim(α) , which are all different since the order of α|stim(α) is m. It is obvious that αk+m cannot be equal to any of αi , i < k, because the image of αk+m is different from that of each such αi . At the same time, we claim that αk+m = αk . Indeed, for every x ∈ 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 has index k and period m, as claimed. The statement is proved.

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Exercise 5.2.9 For each α ∈ Tn the period of α equals the least common multiple 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 all cardinalities of trajectories for the vertices of the full subgraph of Γα with the vertex set N\stim(α).

5.3

Isolated and Completely Isolated Subsemigroups

A subsemigroup T of a semigroup S is called completely isolated provided that 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 isolated 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 that an ∈ T implies a ∈ T for all a ∈ S and n ∈ N. It is obvious that each completely isolated subsemigroup is isolated. Each semigroup is a completely isolated subsemigroup of itself. Exercise 5.3.2 Show that a proper subsemigroup T  S is isolated if and only 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 an isolated subsemigroup. In particular, for each a ∈ S there exists a minimum isolated subsemigroup of S, containing a. (ii) If a union of some family of (completely) isolated subsemigroups of S is 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 unique √ idempotent, say e, by Theorem 5.2.7. This means that a ∈ e and thus √ T ⊂ ∪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 maximal subgroup, corresponding to e. √ Remark 5.3.6 For the semigroups Tn , PT n , and IS n we have ε = Sn . √ Exercise 5.3.7 Show that 0 is not a subsemigroup of IS n if n > 1. To describe all (completely) isolated subsemigroups of Tn , PT n , and IS n we will need a number of auxiliary statements. For simplicity we shall call noninvertible elements of semigroups singular. Also for simplicity, we shall use the following notation for the elements from PT n :   A1 A2 . . . Am , α= a1 a2 . . . am which means that α(a) = ai for all a ∈ Ai , i = 1, . . . , m, and that on the complement B = N\(A1 ∪ · · · ∪ Am ) the element α acts as the identity on all elements, on which it is defined. If we consider several elements at the same 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 IS n , and T be a proper isolated subsemigroup of S. If T contains all singular idempotents of 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, contains √ √ a singular idempotent, say e. But a ∈ e by definition, and hence e ⊂ T by Lemma 5.3.4(i). Hence In−1 ⊂ T . On the other hand, if T would contain some invertible element, T would contain the identity transformation as a power of this element, and hence the whole Sn by Remark 5.3.6 and Lemma 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 all left zeros of Tn , then T contains all singular idempotents of Tn . Proof. Since in T2 each singular idempotent is a left zero, we may assume n > 2. Let A ⊂ N and let a ∈ A, b, c ∈ A, b = c. We have 

b c A c a a



2 =

b c A a a a



 =

b c A a b a

2 ,

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CHAPTER 5. SUBGROUPS AND SUBSEMIGROUPS 

b c A a b a



b c A c a a



 =

b c A b a a

 . 

b c A a a a

 ,

Hence if an isolated subsemigroup T contains the element   b c A then T contains the element as well. Applying the last stateb a a ment inductively startingfromleft zeros, we get that T contains all idemA potents of the form A = , where A ⊂ N, |A| > 1, and a ∈ A. For the a element A we have that the corresponding partition ρ A contains a unique block A of cardinality |A| ≥ 2, and each of the remaining blocks of this partition consists of one element. Let now   A1 A2 · · · Ak = a1 a2 · · · ak be a singular idempotent, where A1 , . . . , Ak are all blocks of ρ , which consists of more than one element. The proof is now completed by observing that  = A1 A2 · · · Ak . Let  ∈ E(Tn ) be a singular idempotent and A ⊂ N. We will say that  is singular on A provided that A is invariant with respect to  and  acts as the identity transformation on the complement N\A. If A is the unique block of ρ , which contains more than one element, we will say that the idempotent ρ 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 for some A ⊂ N, |A| ≥ 3, the semigroup T contains all idempotents, singular on 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 contains all constants on A ∪ {k}. Let a ∈ A. Fix a partition A\{a} = A1 ∪ A2 and ai ∈ Ai , i = 1, 2. The element   A1 a A2 k ωa = a a a k is a constant on A and hence belongs to T by our assumptions. For the elements     A1 a A2 k A1 a A2 k β= , , γ= a1 a1 k a2 a1 k k a   A1 a A2 k δ= k k k a

5.3. ISOLATED SUBSEMIGROUPS

77

we have that the elements     A1 a A2 k A1 a A2 k , γ2 = , δ 2 = ωa β2 = a1 a1 a2 k a1 a a k are singular idempotents on A. Hence β, γ, δ ∈ T . The latter implies that the following elements belong to T as well:     A1 a A2 k A1 a A2 k , μk = δμa = . μa = ωa βγ = a a a a k k k k The statement follows. Corollary 5.3.11 Let T be an isolated subsemigroup of Tn . Assume that for some A ⊂ N, |A| ≥ 3, the semigroup T contains all idempotents, singular on A. Then T contains In−1 . Proof. Inductively applying Lemma 5.3.10 we obtain that T contains all singular 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 of rank (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 thatT contains all constants on the set {1, 2, 3}. For the element 1 2 3 α= we have α2 = ε2,1 and hence α ∈ T . Then 3 3 2  αε2,3 =

1 2 3 3 3 3



 ∈ T, ε2,3 α =

1 2 3 2 2 2

 ∈ T.



 1 2 3 Analogously one also shows that ∈ T . By Lemma 5.3.9 we 1 1 1 thus have that T contains all idempotents, singular on A. The statement now follows from Corollary 5.3.11. Lemma 5.3.13 Let T be an isolated subsemigroup of Tn . Assume that T contains εm1 ,k1 and εm2 ,k2 for some pairwise different k1 , k2 , m1 , m2 . Then T 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 enough to show that T contains all constants on {1, 2, 3, 4}. For the element α =  1 2 3 4 we have α3 = ε3,4 and hence α ∈ T . Thus β = ε1,2 αε1,2 ∈ T 2 3 1 1

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and a direct commutation shows that β is the constant on {1, 2, 3, 4} with image {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 are also in T . Lemma 5.3.14 Let T be an isolated subsemigroup of Tn . Assume that T contains some idempotent  of rank r < n − 1. Then T contains an idempotent of rank (r + 1). Proof. If the partition ρ has some block B of cardinality |B| ≥ 3, then an idempotent of rank (r + 1) in T can be easily obtained using the same construction as in the proof of Lemma 5.3.9. If all blocks of ρ contain at most two elements, there should be more than one two-element block. Hence without loss of generality we may assume that   1 2 3 4 5 ··· n . = 1 1 3 3 a5 · · · an For the elements  1 2 3 4 5 ··· α= 3 4 1 1 a5 · · ·

n an



 ,

β=

1 2 3 4 5 ··· 3 3 1 2 a5 · · ·

n an



we have α2 = β 2 =  and hence α, β ∈ T . Thus   1 2 3 4 5 ··· n βα = ∈ T, 1 2 3 3 a4 · · · an 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 T contains some idempotent  of rank (n − 2). Then T contains In−1 .   1 2 3 Proof. Without loss of generality we may assume that  = , or 1 1 1   1 2 3 4 = . 1 1 3 3   1 2 3 In the case  = we may use the same construction as in 1 1 1 the proof of Lemma 5.3.9 to get ε1,2 , ε1,3 ∈ T . Thus T contains In−1 by Lemma 5.3.12.   1 2 3 4 In the case  = we may use the same construction as 1 1 3 3 in the proof of Lemma 5.3.14 to get ε1,2 , ε3,4 ∈ T . Thus T contains In−1 by Lemma 5.3.13. √ Lemma 5.3.16 Let  ∈ E(Tn ) be of rank (n − 1). Then  = G .

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79

√ √ Proof. The inclusion G ⊂  is obvious. Let now α ∈ . Then αm =  for some m > 0. From the inequalities n > rank(α) ≥ rank() = n − 1 we have rank(α) = n − 1. As im(α) is invariant with respect to α, and αm = , we have stim(α) = im(α). From Proposition 5.2.8 it now follows that 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 ∈ G . This proves the √ inclusion  ⊂ G and completes the proof. Theorem 5.3.17 Let n > 2. A subsemigroup T of Tn is isolated if and only if it belongs to the following set:   {Tn , Sn , In−1 } ∪ Gεm,k ∪ Gεk,m : m, k ∈ N, m = k ∪   ∪ ∪m∈M Gεm,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 subsemigroup of Tn . Let T be an isolated subsemigroup of Tn such that T ∩ Sn = ∅. As Sn is √ a 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 T contains an idempotent of rank (n−1). Thus T contains a generating system for 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 . From Lemmas 5.3.14 and 5.3.15 it follows that in this case T contains only idempotents 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 some i, j, then T ⊃ In−1 by Lemma 5.3.13, which is not possible. Hence for all i, j we have {mi , ki } ∩ {mj , kj } = ∅. Moreover,   k l m εl,m εm,k = l l l is an idempotent of rank (n − 2) as well. For i = j the possibility mi = mj is prohibited by Lemma 5.3.12. Hence we either have ki = kj , or mi = kj and mj = ki . In the last case the only possibility is E(T ) = {εm,k , εk,m } as for any 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 of the sets P = Gεm,k ∪ Gεk,m and Q = Gεm1 ,k ∪ · · · ∪ Gεmt ,k (5.4)

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is a subsemigroup of Tn . In the first case of (5.4) for arbitrary α ∈ Gεm,k and β ∈ Gεk,m we have ραβ = ρβ = ρα and im(αβ) = im(α). Hence αβ ∈ Gεm,k by 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 Theorem 4.5.1(iii) we have αβ ∈ H(β) ⊂ Q. Hence Q is a subsemigroup as well. This completes the proof. Remark 5.3.18 The only isolated subsemigroup of T1 is T1 itself. The semigroup T2 has five isolated subsemigroups: T2 , S2 , I1 , {ε1,2 }, {ε2,1 }. Corollary 5.3.19 For n > 1 the only completely isolated subsemigroups of Tn are Tn , Sn , and In−1 . Proof. Since each completely isolated subsemigroup is isolated, we have just to check which of the subsemigroups, given by Theorem 5.3.17, are completely isolated. It is obvious that Tn is completely isolated. Further, both Sn and In−1 are completely isolated as each of these semigroups is the complement 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 of rank (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 , a contradiction. Analogously one shows that {ε2,1 } is not completely isolated. Assume n > 2. Then the complement T = Tn \T contains Sn and the element   1 2 3 4 ··· n α= , 1 1 2 3 ··· n − 1 which is not a group element by Exercise 5.1.5. By Theorem 3.1.3 the set Sn ∪ {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 of IS n . The elements of N which will be omitted in the notation of some transformation α will be assumed to be fixed points of α. Lemma 5.3.20 If an isolated subsemigroup T ⊂ IS n contains some idempotent  of rank at most (n − 2), then T contains at least two different idempotents of rank (n − 1).

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81

Proof. Let rank() = n − k, k > 1, and dom() = {a1 , a2 , . . . , ak }. Then the elements α = [a1 , a2 , . . . , ak ] and β = [ak , a2 , . . . , a1 ] satisfy αk = β k = . Hence α, β ∈ T and thus βα, αβ ∈ T . But both βα = [ak ] and αβ = [a1 ] are idempotents of rank (n − 1), and they are obviously different. Lemma 5.3.21 If an isolated subsemigroup T ⊂ IS n contains two different idempotents 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 means that 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] and hence [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 β 3 = γ 3 = α. Hence the idempotent [z] = γβ is also contained in T . In particular, T contains all singular idempotents of rank (n − 1). As every singular idempotent  = [a1 ][a2 ] . . . [ak ] of IS n is a product of singular idempotents [ai ], i = 1, . . . , k, of rank (n − 1), we obtain that T contains all singular idempotents. Hence T contains In−1 by Lemma 5.3.8. Theorem 5.3.22 The only isolated subsemigroups of IS n are IS n , Sn , In−1 , and G[x] , x ∈ N. Proof. Let T be an isolated subsemigroup of IS n . Repeating the arguments of the first part of the proof of Theorem 5.3.17 one shows that either T ∈ {IS n , 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 IS n are IS n , Sn , and In−1 . Proof. Each completely isolated subsemigroup is isolated. So, we have just to go through the list given by Theorem 5.3.22. The semigroups IS n , 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 = IS 1 . We shall need the following obvious result: Lemma 5.3.24 (i) Let S be a semigroup and A < B < S. If A is completely isolated in S, then A is completely isolated in B.

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(ii) Let S be a semigroup, A < S completely isolated and B < S. Then A ∩ B is either empty or a completely isolated subsemigroup of B. Theorem 5.3.25 For n > 1 the semigroup PT n has seven completely isolated 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. Hence the 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 , they both are completely isolated by Exercise 5.3.1. Let T be a completely isolated subsemigroup of PT n . The intersection T ∩ Tn is either a empty or a completely isolated subsemigroup of Tn because of Lemma 5.3.24(ii). Hence T ∩ Tn is either Tn , or Sn , or Tn \Sn by Corollary 5.3.19. The product of two noninvertible transformations of N is not 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   A ··· B C = ∈ PT n \Tn . a ··· b ∅ Let x ∈ im() and c ∈ C. Set x = (x, c)(x, c). We claim that x ∈ T . Indeed, this is obvious if (x, c) ∈ T . If (x, c) ∈ T , then (x, c)x (x, c) =  ∈ T and hence x ∈ T since T is completely isolated. Thus T contains the element  = x , which is an idempotent and im( ) = im()\{x}. Continuing inductively we get that T contains the element 0n . From 0n = [1][2] . . . [n] we have that T  [k] for some k ∈ N. By the same arguments as above we get that for 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, all singular idempotents from IS n . If n = 2, the above gives us that PT 2 contains three minimal completely isolated subsemigroups: S2 , T2 \S2 , and PT 2 \T2 = IS 2 \S2 . Any union of these subsemigroups is again a semigroup, hence completely isolated by Proposition 5.3.3(ii). Assume now that n ≥ 3. We continue the consideration of the case T ∩ (PT n \Tn ) = ∅ and recall that we already know that T contains all singular idempotents from IS n . Consider the following elements in PT n :       1 2 3 1 2 3 1 2 3 α= , β= , αβ = . 1 1 ∅ ∅ 3 3 ∅ ∅ ∅ αβ is a singular idempotent of IS n . Hence α ∈ T , or β ∈ T . Without loss of generality we may assume α ∈ T . As in the previous paragraph one shows that πατ ∈ T for all π, τ ∈ Sn , which implies that T contains all idempotents

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83

from PT n \Tn of rank (n − 2), that is, all elements of the form εz,y [x], where x, y, z ∈ N are arbitrary different elements. Now let   A1 · · · Ak B ∈ PT n \Tn δ= a1 · · · ak ∅ be any idempotent, written such that all blocks Ai , i = 1, . . . , k, consist of more than one element. For each i = 1, . . . , k consider the idempotent   Ai B . δi = ai ∅ 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 = δ1 δ2 · · · δk and hence δ ∈ T . Therefore T contains all idempotents from PT n \Tn and thus T contains PT 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 , and PT n \Tn . Any union of these subsemigroups is again a semigroup, hence completely isolated by Proposition 5.3.3(ii). This completes the proof.

5.4

Addenda and Comments

5.4.1 For each semigroup S the Boolean B(S) has the following natural structure 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 is associative). The semigroup B(S) is called the global semigroup or the power semigroup of S. Theorem 5.2.7 was proved by Frobenius in 1895 for the elements of the semigroup B(S), where S is a finite group (this restriction did not play an essential role in the proof). Since that time Theorem 5.2.7 has 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 to show that J ⊂ D. Let aJ b, that is, S 1 aS 1 = S 1 bS 1 . Then there exist u, v, x, y ∈ S 1 such that uav = b and xby = a. Thus (xu)a(vy) = a and even (xu)k a(vy)k = a for all k ∈ N. As S 1 is finite, there exist k and l such that e = (xu)k and f = (vy)l are idempotents. We have a = (xu)kl a(vy)kl = eaf,

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which implies ea = a and af = a. From the first equality we have (xu)k−1 x · 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 for any 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 IS n , and G is a subgroup 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 IS n . Then the only maximal subsemigroups of S are Sn ∪ In−2 and In−1 ∪ G, where G is a maximal subgroup of Sn . Proof. Let T be a maximal subsemigroup of S. If Sn ⊂ T , then T ∪ In−1 is a subsemigroup and T ⊂ T ∪ In−1  S. Hence T = T ∪ In−1 . On the other hand, G = T ∩ Sn is a subsemigroup of Sn and hence is a subgroup since Sn is finite. Furthermore, G has to be a maximal subgroup for otherwise there would exist a subgroup G of Sn such that G  G  Sn . In this case G ∪ 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) for otherwise T = S by Theorems 3.1.3 or 3.1.4. Hence T ⊂ Sn ∪ In−2 . On the other 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 ∈ IS n : rank(α) = n − 1}. Proof. The proof is similar to that of Theorem 5.4.2. One has only to note that each element of rank (n − 1) from PT n is either a total transformation or a partial injection. By the above theorems, the classification of maximal subsemigroups in PT n , Tn , and IS n reduces to the classification of maximal subgroups of the symmetric group Sn . The latter problem is very difficult. The only known solution so far, which can be found in [LPS], is based on the so-called classification of finite simple groups. The correctness of the latter result is still questioned by many specialists.

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5.4.4 From Theorem 5.4.2 it is easy to see that each maximal subsemigroup of IS n is inverse. Thus as a bonus we get a description of all maximal inverse subsemigroups of IS n , see [X.Ya1]. Analogous questions can be asked for both PT n and Tn , especially taking into account that every α ∈ Tn is contained in some inverse subsemigroup of Tn , see [Sc5]. These questions were 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 the computation of the order of α can be easily reduced to the computation of the order of the permutation α|stim(α) . If π = (a1 , . . . , ak ) · · · (b1 , . . . , bl ), then the order of π is just the least common multiple (lcm) of the lengths k, . . . , l of 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 determination of 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 max

n1 +···+nk =n

lcm(n1 , . . . , nk ),

which is also very difficult. Analogous problems for the semigroups PT n , Tn , and IS n are of the same level of difficulty. 5.4.6 On the other hand, the asymptotic behavior of Pmax (Sn ) is much easier to describe. Already in 1903 E. Landau has proved in [La] the following formula:  (5.5) ln(Pmax (Sn )) ∼ n ln(n), n → ∞. A very elementary proof of this statement can be found in [Mi]. Some additional 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 the asymptotic behavior of the orders of elements in the semigroups Tn , PT n , and IS n . Theorem 5.4.4 ln(Pmax (Tn )) ∼ ln(Pmax (PT n )) ∼ ln(Pmax (IS n )) ∼



n ln(n), n → ∞.

5.4.7 From the proof of Theorem 2.4.3 it follows that each finite semigroup S 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 the identity transformation in Tn . The group S ∗ becomes, under such identification, a subgroup of Sn . As a consequence of this and Lemma 5.3.24(ii) we obtain: Proposition 5.4.5 If S is a finite semigroup, then both S ∗ and S\S ∗ are completely isolated subsemigroups of S.

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The statement of Proposition 5.4.5 is wrong in the general case as the group of units of an infinite semigroup does not have to be completely isolated (a bijective transformation can be a composition of two transformations, each of which is not bijective). 5.4.8 Let S be a semigroup. There is a natural partial order on the set of all J -classes of S, given by the inclusion of the principal two-sided ideals: J (a) ≤ J (b)

⇔

S 1 aS 1 ⊂ S 1 bS 1 .

If S is finite, then J = D by Theorem 5.4.1 and hence the above partial order 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 all D-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 either belong 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 IS n satisfy all conditions of Theorem 5.4.6. Proof. Consider S as a subsemigroup of some Tn . Then aRb in S implies aRb in Tn by Proposition 4.4.2. Hence aRb implies im(a) = im(b) by Theorem 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 the fact 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 ∪ · · · ∪ Gek is closed with respect to multiplication. Let e = ei and f = ej for i = j and a ∈ Ge , b ∈ Gf . Assume first that all ei s belong to the same R-class. We claim that in this case ab ∈ Gf . As S is a subsemigroup of Tn , we have im(a) = im(b) by Theorem 4.5.1(i). Since both a and b are group elements, different elements of 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 ei s 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 technical than that for Tn . Hence we just present here the result and an idea of the proof. To work out all the details of the proof is left to the reader.

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Theorem 5.4.7 A subsemigroup T of PT n is isolated if and only if T belongs 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 the list 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 of T (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})) N\{x}.



∪m∈M Gεm,x , where x ∈ N and ∅ = M ⊂

Sketch of the Proof. By Theorem 5.3.25, the semigroup PT n has three minimal completely isolated subsemigroups, namely, Sn , Tn \Sn , and PT n \Tn . If T ⊂ PT n is an isolated subsemigroup and S is a minimal completely isolated subsemigroup of PT n , then S ∩ T is an isolated subsemigroup of S. Let us first determine all possible intersections of T with the minimal completely isolated subsemigroups of PT n . The semigroup T ∩ Sn is obviously either empty or equals Sn . From Theorem 5.3.17 the semigroup T ∩ (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 subsemigroup of PT n \Tn , then it is given by (c). So, let T be a proper isolated subsemigroup of PT n \Tn . First we claim that T contains an idempotent e such that |dom(e)| = 1. This can be proved analogously to the proof of Lemma 5.3.20. Then one shows that such T cannot contain 0n . For this one shows that if T contains 0n , then T contains all idempotents of PT n \Tn and hence must coincide with the latter. This again can be done generalizing the arguments of Lemma 5.3.20. The next step is to show that T contains an idempotent e such that all blocks of ρe on which e is defined contain at most two elements. This is similar to the proof of Lemma 5.3.9. Generalizing the arguments of the proof of Lemma 5.3.21 one shows that all idempotents of T must satisfy |dom(e)| = 1. From this it follows easily that there should exist x ∈ N such that dom(α) = N\{x} for all α ∈ T . In particular, it follows that x ∈ im(α) for all α ∈ T and hence the set N\{x} is invariant with respect to all elements from T . The restriction to N\{x} defines a homomorphism from T to T (N\{x}), the image of which is 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 ) is isolated (this also follows from Exercise 5.5.13(a)).

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It is left to see whether we can form a union of some of the semigroups given by (a), (b), and (c). A case-by-case analysis shows that the only possible 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 from different components of our union it is prohibited to get elements of smaller ranks. We leave the details to the reader. 5.4.10 There are of course many other natural classes of subsemigroups in the semigroups Tn , PT n , and IS n , defined using various conditions. For instance, 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 all α such that x ≤ y implies α(x) ≤ α(y) for all x, y ∈ N). Some of these classes 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

Additional Exercises

5.5.1 Let G be a maximal subgroup of Tn , or IS n . Show that G is a maximal subgroup of PT n . 5.5.2 Let S be a finite semigroup and a ∈ S. Prove that the following conditions 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 that the element α is a group element if and only if α and αm form a pair of inverse elements. 5.5.5 Let S be a semigroup and e ∈ E(S). Show that eSe is the maximum submonoid of S, which has e as the identity element.

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5.5.6 Let α ∈ PT n be a group element. Show that the restriction α|im(α) is a 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 S has a unique irreducible generating system. (b) Show that for each k ∈ N there exists a finite cyclic group and an irreducible 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 the number 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  n k nn2 1 nn3 2 · · · nk k−1 nnk+1 P (nk+1 , m). n1 +···+nk +nk+1 =n

(b) Show that the number of elements of type (k, m) in the semigroup PT n equals  n k nn2 1 nn3 2 · · · nk k−1 nnk+1 P (nk+1 , m). n1 +···+nk +nk+1 ≤n

(c) Show that the number of elements of type (k, m) in the semigroup IS n equals  [n2 ]n1 [n3 ]n2 · · · [nk ]nk−1 P (nk+1 , m). n1 +···+nk +nk+1 =n

5.5.10 Prove that a (one-sided) ideal I of a semigroup S is an isolated subsemigroup of S if and only if I is semiprime. 5.5.11 Construct a semigroup S and a subsemigroup T such that a2 ∈ T implies a ∈ T for any a ∈ S, but at the same time T is not an isolated subsemigroup 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 isolated subsemigroup 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 completely isolated subsemigroup of T . Is it true that X is a completely isolated subsemigroup of S? 5.5.14 Show that the intersection of completely isolated subsemigroups is not completely isolated in general.

Chapter 6

Other Relations on Semigroups 6.1

Congruences and Homomorphisms

Recall that a binary relation ρ on a semigroup S is called left compatible provided that a ρ b implies ca ρ cb for all a, b, c ∈ S; ρ is called right compatible provided that a ρ b implies ac ρ bc for all a, b, c ∈ S; and ρ is called compatible provided that it is both left and right compatible. A left compatible equivalence relation is called a left congruence. A right compatible equivalence relation is called a right congruence. A compatible equivalence relation is called a congruence. Example 6.1.1 (a) From Sect. 4.4 we know that Green’s relation L is a right 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 × S consisting of just one equivalence class S. If S is a set and ρ is an equivalence relation on S, there are several ways to show that some a, b ∈ S belong to the same class of ρ. This can be written as follows: a ρ b, or (a, b) ∈ ρ, or a ≡ b(ρ) or simply a ≡ b if there is no confusion about which congruence we are considering. In what follows we may use any of the above notation. Exercise 6.1.2 Prove that the intersection of an arbitrary family of congruences on S is a congruence on S. Lemma 6.1.3 Let S be a semigroup and ρ be a congruence on S. Let further K be an equivalence class of ρ. If K contains some ideal of S, then K is an ideal of S itself. In addition, at most one class of ρ can be an ideal. O. Ganyushkin, V. Mazorchuk, Classical Finite Transformation Semigroups, Algebra and Applications 9, DOI: 10.1007/978-1-84800-281-4 6, c Springer-Verlag London Limited 2009 

91

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Proof. Let I be the ideal of S, contained in K. Fix some b ∈ I and let a ∈ K and x ∈ S be arbitrary. As a ≡ b, we have xa ≡ xb and ax ≡ bx. But xb, 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 classes of an equivalence relation on S. Let S be a semigroup and I ⊂ S an ideal. Define the equivalence relation ρI on S as follows: ρI = (I × I) ∪ {(a, a) : a ∈ S\I}. In other words, all elements from I form one equivalence class, and all other equivalence classes are trivial (each consists of a single element). The fact 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 the ideal I. If ρ is a congruence on S and a ∈ S, the equivalence class of a in ρ is denoted 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 all equivalence classes. For a, b ∈ S set a · b := ab.

(6.1)

Lemma 6.1.4 The formula (6.1) defines a binary associative operation on S/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 operation on S/ρ. Let a ∈ a and b ∈ b. Since ρ is compatible we have ab ≡ 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 multiplication in S: (a · b) · c = ab · c = abc = a · bc = a · (b · c). The semigroup (S/ρ, ·) constructed above is called the quotient or the factor of S modulo the congruence ρ. If ρ = ρI , one usually simply writes S/I instead of S/ρI . The semigroup S/I is called the Rees quotient of S modulo the ideal I. If S is a monoid with the identity element 1, then from (6.1) we get that 1 will be the identity element in S/ρ. The quotient S/ιS is naturally identified with S. The quotient S/ωS is the one-element semigroup. Let now (S, ·) and (T, ∗) be two semigroups. A mapping ϕ : S → T is called 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 of the notion of a homomorphism. An isomorphism is just a bijective homomorphism. Other special cases of homomorphisms are (1) injective homomorphisms, which are called monomorphisms and (2) surjective homomorphisms, which are called epimorphisms. Monomorphisms are usually denoted by the symbol →, and epimorphisms by the symbol . A homomorphism ϕ : S → S is called an endomorphism of S. A bijective endomorphism is called an automorphism. The set of all endomorphisms of S is denoted by End(S) and the set of all automorphisms of S is denoted by Aut(S). Both these sets are not empty since they both contain the identity transformation on S. Proposition 6.1.5 The set End(S) is a monoid with respect to the composition 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. Since this 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 composition of bijections is a bijection, Aut(S) is a subsemigroup of End(S). Let ϕ ∈ 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 bijection. This means that ϕ−1 ∈ Aut(S) and thus all elements of Aut(S) are invertible. 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 canonical projection or the canonical epimorphism πρ : S  S/ρ via a → a. The mapping 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)}

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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 compatible is proved analogously. On the other hand, if ρ is a congruence on S, then we have the homomorphism πρ : S  S/ρ and Ker(πρ ) = ρ. This means that each congruence is the kernel of some homomorphism. If ϕ : S  T is an epimorphism, T is called a homomorphic image of S. It turns out that all homomorphic images of a semigroup can be characterized up to isomorphism. Theorem 6.1.6 Let ϕ : S → T be a homomorphism of semigroups, and π 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: ϕ / S II u: T II u II u I u π III$ , u ψ $ S/Ker(ϕ)

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 have ϕ(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 completes the proof. From Theorem 6.1.6 it follows that, up to isomorphism, all homomorphic images 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 congruences on the semigroups Tn , PT n , and IS n . This will be done in Sect. 6.3. In the present 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. Then the 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 implies a ≡ ab. Hence ab ∈ 1, proving (i). Assume that S is a group and let a ∈ 1. Since ρ is left compatible we have that 1 ≡ a implies a−1 ≡ a−1 a = 1. Hence a−1 ∈ 1, proving (ii). Lemma 6.2.2 Let G be a group, ρ a congruence on G and H = 1. Then for any g ∈ G and h ∈ H we have g −1 hg ∈ H. Proof. Since ρ is left compatible, we have that 1 ≡ h implies g −1 ≡ g −1 h. Since ρ is right compatible, we have that g −1 ≡ g −1 h implies 1 = g −1 g ≡ g −1 hg. Hence g −1 hg ∈ H. Recall that a subgroup H of a group G is called normal provided that g −1 hg ∈ H for all g ∈ G and h ∈ H. The fact that H is a normal subgroup of 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 cosets of G modulo H is denoted by G/H. Given G and H as above define the binary 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 an equivalence relation on G. Proof. H is a subgroup, in particular, 1 ∈ H. Hence a = a · 1 ∈ aH, which means that ρH is reflexive. Let (a, b) ∈ ρH . Then a = bh for some h ∈ H by definition. Since H is a 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 ∈ cH implying (a, c) ∈ ρH . This means that ρH is transitive and completes the proof. 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 of a group.

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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 subgroup H  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 equivalence relation on G. Let (a, b) ∈ ρH and g ∈ G. Then a = bh for some h ∈ H and 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 the right, we have ag = bhg = bg(g −1 hg). As H is normal, we have g −1 hg ∈ H and 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 is left compatible we get 1 = a−1 a ≡ a−1 b. Hence a−1 b ∈ H and multiplying with a from the left we get b ∈ aH, that is, (a, b) ∈ ρH . This shows that ρ ⊂ ρH . On the other hand, let (a, b) ∈ ρH . Then (1, a−1 b) ∈ ρH by the left stability of ρH . But then a−1 b ∈ H = 1 and hence (1, a−1 b) ∈ ρ. Multiplying the latter with a from the left we get (a, b) ∈ ρ as ρ is left compatible. Hence ρH ⊂ ρ and thus ρ = ρH , proving (ii). We already know that G/ρ is a monoid with the identity element 1. To prove (iii) one just has to observe that a · a−1 = 1 and hence all elements of G/ρ are invertible. This completes the proof.

6.3

Congruences on T n , PT n , and IS n

Let S denote one of the semigroups Tn , PT n , or IS n . 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 following form:   A1 A2 · · · Ak Ak+1 , α= a1 a2 · · · ak ∅ (6.4)   A1 A2 · · · Ak Ak+1 β= , aμ(1) aμ(2) · · · aμ(k) ∅ where the sets A1 , . . . , Ak , Ak+1 are pairwise disjoint, μ ∈ Sk , and the set Ak+1 may be empty. If S = Tn , we always have Ak+1 = ∅. If S = IS n , we have |A1 | = · · · = |Ak | = 1. Sometimes we shall omit the set Ak+1 in our notation. For every k = 1, 2, . . . , n and for every normal subgroup R  Sk we define an equivalence relation ≡R on S as follows:

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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 are given by (6.4), then μ ∈ R. Lemma 6.3.1 The relation ≡R is well defined, that is, it does not depend on the order of the blocks A1 , . . . , Ak+1 in the presentation of α in the form (6.4). Proof. Assume that we permute the Ai s in some way, say B1 = Aη(1) , B2 = Aη(2) , . . . , Bk = Aη(k) . In this numeration, we have     B1 B2 · · · Bk B1 B2 · · · Bk , β= , α= bτ (1) bτ (2) · · · bτ (k) b1 b2 · · · bk 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−1 μη)(i) . Hence τ = η −1 μη. As R is a normal subgroup of Sk , we have τ ∈ R if and only 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 of rank = k is also obvious. Let us show that R is symmetric and transitive for elements of rank k. Let   A1 A2 · · · Ak , α= a1 a2 · · · ak     A1 A2 · · · Ak A1 A2 · · · Ak β= = , aμ(1) aμ(2) · · · aμ(k) b1 b2 · · · bk   A2 · · · Ak A1 . γ= bτ (1) bτ (2) · · · bτ (k) Then

 α=  γ=

A1

A2

bμ−1 (1) bμ−1 (2)

··· ···

A2 ··· A1 aμ(τ (1)) aμ(τ (2)) · · ·

Ak



bμ−1 (k) Ak aμ(τ (k))

,  .

If α ≡R β and β ≡R γ, then μ, τ ∈ R and hence both μ−1 and μτ are elements of R as well, since R is a group. Thus β ≡R α and α ≡R γ, which means that ≡R is indeed an equivalence relation.

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Let us now show that ≡R is both left and right compatible. If we have rank(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 case rank(α) = rank(β) = k. Let α and β be given by (6.4) and α ≡R β. For any   a1 a2 · · · ak · · · γ= c1 c2 · · · c k · · · we have γα =



A1 A2 · · · c1 c2 · · ·

Ak ck



 , γβ =

A2 · · · A1 cμ(1) cμ(2) · · ·

Ak cμ(k)

 .

If some of the symbols c1 , c2 , . . . , ck are equal to ∅ or coincide, then we have rank(γα), rank(γβ) < k and hence γα ≡R γβ. In the other case we have γαHγβ. Hence we can consider γα and γβ as two elements, written in the form (6.4). The corresponding permutation of indices is μ ∈ R. Hence γα ≡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 hence αγ ≡R βγ. In the other case we have     B1 B2 · · · Bk B1 B2 · · · Bk , βγ = αγ = cμ(1) cμ(2) · · · cμ(k) c1 c2 · · · ck and the same argument as in the previous paragraph shows that αγ ≡R βγ again. 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 S is either the uniform congruence or has the form ≡R for some R. To prove this we will need a series of auxiliary lemmas.  Recall  from Sect. 2.3 that for N a ∈ N we denote by 0a the element 0a = ∈ Tn . Recall also from a Sect. 2.7 that for A ⊂ N the element εA is the idempotent of IS n such that dom(εA ) = A. Lemma 6.3.4 Let ρ be a congruence on Tn . If ρ = ιS , then the ideal I1 is contained in some class of ρ. Proof. By assumption, there exist α = β ∈ Tn such that α ≡ β. This means that for some x ∈ N we have a = α(x) = β(x) = b. Take arbitrary y, z ∈ N and let   a b ··· γ= . y z ···

6.3. CONGRUENCES ON Tn , PT n , AND IS n

99

Then γα0x = 0y and γβ0x = 0z , implying 0y ≡ 0z . Hence all elements from I1 belong to the same class of ρ. Lemma 6.3.5 Let ρ be a congruence on S, and α, β ∈ S be such that rank(α) = k, rank(β) = m, k > m, and α ≡ β. Then Ik ⊂ α. Proof. First, we consider the case S = PT n or S = IS n . As α ∈ Ik , it is enough to show that Ik ⊂ 0. For this we show that Il ⊂ 0 for all l ≤ k by induction 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 some a0 ∈ im(α)\im(β) and complete it to some (l + 1)-element subset A = {a0 , a1 , . . . , al } ⊂ im(α). From α ≡ β we have εA α ≡ εA β. However, im(εA β) ⊂ {a1 , . . . , al } and thus rank(εA β) ≤ l. This implies εA β ∈ 0 and thus εA α ∈ 0 as well. But im(εA α) = A and thus rank(εA α) = l + 1. As 0 contains the ideal I0 , by Lemma 6.1.3 the set 0 must be an ideal itself. Hence 0 contains the principal ideal SεA αS, which coincides with Il+1 by Theorem 4.2.8. Let now S = Tn . By Lemma 6.3.4 one of the classes of ρ contains I1 . Now the proof is analogous to the one given above. The only difference is that instead of the idempotent εA one should consider the following idempotent:   a0 X1 · · · Xl  εA = , 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 by Lemmas 6.3.4 and 6.1.3 it always contains a unique congruence class which is 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 congruence. 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 the idempotent , uniquely defined by the following conditions: ⎧ ⎪ ⎨x, x = a; (x) = b, x = aandS = Tn ; ⎪ ⎩ ∅, x = aandS = Tn . Then we have α ≡ β = β. Hence α ≡ α. On the other hand, im(α) = im(α)\{a} and hence rank(α) < rank(α). Applying Lemma 6.3.5 we get α ∈ Iρ , a contradiction. Hence im(α) = im(β).

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Let im(α) = {a1 , . . . , ak }. Then we may assume that    B1 B2 · · · A1 A2 · · · Ak Ak+1 , β= α= a1 a2 · · · ak ∅ a1 a2 · · ·

Bk Bk+1 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)) and there exist elements x, y ∈ dom(α), x = y, which belong to the same block of ρα but to different blocks of ρβ . Consider the second case first. In this case k ≥ 2 and without loss of generality we may assume that there exist b1 ∈ B1 and b2 ∈ B2 such that b1 , b2 ∈ Al , l < k + 1. Consider the element   a1 a2 · · · ak · · · . γ= b1 b2 · · · bk · · · From γα ≡ γβ we have (γα)2 ≡ (γβ)2 . But  A1 A2 A3 · · · 2 (γα) = bl bl c3 · · ·  B1 B2 B3 · · · (γβ)2 = b1 b2 b3 · · ·

Ak Ak+1 ck ∅ Bk Bk+1 bk ∅

 ,  .

Hence rank((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, a contradiction. Therefore ρα = ρβ in this case. Consider now the case Ak+1 = Bk+1 . Without loss of generality we may assume that there exists d1 ∈ Ak+1 ∩ B1 . Choose any di ∈ Bi , i = 2, . . . , k, and consider the element   a1 a2 a3 · · · ak · · · . δ= d1 d2 d3 · · · dk · · · From δα ≡ δβ we have (δα)2 ≡ (δβ)2 . But  A1 A2 A3 · · · 2 (δα) = ∅ s2 s3 · · ·  B1 B2 B3 · · · (δβ)2 = d1 d2 d3 · · ·

Ak Ak+1 sk ∅ Bk Bk+1 dk ∅

 ,  .

Using the same arguments as in the previous paragraph, we conclude that α ∈ Iρ , a contradiction. Therefore ρα = ρβ in this case as well. From Theorem 4.5.1(iii) it now follows that αHβ. This completes the proof.

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101

Lemma 6.3.7 Let ρ be a congruence on S, and α ∈ S be an element of rank k 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, generated by α). So, we may now assume that α ∈ Iρ . By Lemma 6.3.6, we may assume that     Aμ(1) Aμ(2) · · · Aμ(k) A1 A2 · · · Ak , β= , α= a1 a2 · · · ak 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 } ∈ S and since μ = (1, 2) we have     A2 A1 = = ε{a1 } β = γ2 . γ1 = ε{a1 } α = a1 a1 Moreover, even dom(γ1 ) = A1 = A2 = dom(γ2 ), that is, γ1 ∈ H(γ2 ). However, γ1 ≡ γ2 and as rank(γ1 ) = 1, the inclusion I1 ⊂ Iρ follows now from Lemma 6.3.6. If k > 2, then, since μ is different from the identity transformation, it is easy 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   a1 · · · at at+1 at+2 · · · ak · · · . γ= a1 · · · at at at+2 · · · ak · · · Then γα ≡ γβ. The partition ργα is obtained from ρα uniting the blocks At 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 get At ∪ At+1 = Aμ(t) ∪ Aμ(t+1) . Hence ργα = ργβ . From Lemma 6.3.6 it thus follows that γα ∈ Iρ . As rank(γα) = k − 1, it follows that Iρ ⊃ Ik−1 . In the case S = IS n the proof is similar, one has just to choose t such that μ(t) = t and consider the element   a1 . . . at−1 at at+1 · · · ak · · · . γ= a1 · · · at−1 ∅ at+1 · · · ak · · · Lemma 6.3.8 Let ρ be a congruence on S and Iρ = Ik . If α, β ∈ S are such that rank(α) > k + 1 and rank(β) > k + 1, then α ≡ β if and only if α = β. Proof. By Lemma 6.3.7, the inequality α = β implies Iρ ⊃ Ik+1 . The claim follows. Lemma 6.3.9 Let ρ be a congruence on S, which is neither identity nor uniform. Assume that Iρ = Ik−1 . Then ρ has the form ≡R for some normal subgroup R of Sk .

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Proof. If Iρ = Ik−1 , all elements of rank at least (k + 1) form one-element equivalence classes by Lemma 6.3.8. Hence we have only to find out the structure of equivalence classes for elements of rank k. By Lemma 6.3.6, for such elements we have that γ ≡ δ implies γHδ. Let γ be an idempotent of rank k. Then H(γ) is a group with the identity element γ by Theorem 4.4.11. Moreover, H(γ) ∼ = Sk by Theorem 5.1.4. The set 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 the set 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 idempotent γ be of the following form:   M1 M 2 · · · M k . γ= m1 m2 · · · mk Then each element δ ∈ H(γ) has the form  M2 · · · M1 δ= mμ(1) mμ(2) · · ·

Mk 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 of rank k in S. Assume that they are given by (6.4). Choose in each block Ai some element ai and consider the elements     a1 · · · ak · · · M 1 M2 · · · Mk , , π = ν= a1 a2 · · · ak m1 · · · mk · · ·     A1 A2 · · · Ak m1 · · · mk · · · , , π1 = ν1 = m1 m2 · · · mk a1 · · · ak · · · where π, π1 ∈ Sn . Then γ = παν, δ = πβν, α = π1 γν1 , β = π1 δν1 . Hence α ≡ β if and only if γ ≡ δ. From the previous paragraph we know that the latter is the case if and only if μ ∈ R. Hence ρ =≡R , completing the proof. 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 IS n . (i) For each k, 1 ≤ k ≤ n, and for each normal subgroup R of Sk the relation ≡R is a congruence on S. The congruence ≡R is uniform if and only if n = 1 and S = Tn , and is the identity congruence if and only if k = 1.

6.4. CONJUGATE ELEMENTS

103

(ii) Let ρ be a congruence on S, which is not uniform. Then there exists k ∈ {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) follows 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 element b ∈ G provided that there exists c ∈ G such that a = c−1 bc. Exercise 6.4.1 Show that the binary relation “a is conjugate to b” is an equivalence relation on G. Conjugation plays a very important role in group theory. Hence it is very natural to study various generalizations of this notion for certain classes of semigroups, in the best case, for all semigroups. There are several ways to generalize this notion. The most direct one works for monoids and is defined as follows: Let S be a monoid and G = S ∗ be the group of units of S. An element a ∈ S is said to be G-conjugate to an element b ∈ S provided that there exists c ∈ S ∗ such that a = c−1 bc. Obviously, if S is a group, then the relations 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 an equivalence relation on S. Recall that two graphs Γ1 = (V1 , E1 ) and Γ2 = (V2 , E2 ) are said to be isomorphic provided that there exists a bijection ϕ : V1 → V2 such that for all a, b ∈ V1 we have (a, b) ∈ E1 if and only if (ϕ(a), ϕ(b)) ∈ E2 . In other words, two graphs are isomorphic if they can be obtained from each other via a renumeration of vertices. Proposition 6.4.3 Let S be one of the semigroups PT n , IS n , or Tn . Two elements α, β ∈ S are G-conjugate if and only if the graphs Γα and Γβ are isomorphic. Proof. In our case we have S ∗ = Sn . Let α, β ∈ S and π ∈ Sn be such that α = π −1 βπ. 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 −1 βπ as 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 α



α(x)

π

/ π(x) β

π

 / π(α(x)) = β(π(x))

(6.6)

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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 from Γβ 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 from Γα to Γβ , then from (6.6) it follows that α = π −1 βπ. A more interesting semigroup generalization of the notion of group conjugation is the following one: Let S be a semigroup. Elements a, b ∈ S are said to be primarily S-conjugate provided that there exist u, v ∈ S 1 such that a = uv and b = vu. The fact that a and b are primarily S-conjugate is denoted 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 from IS 3 . 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 of the 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 that a ∼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−1 bc, then for u = c−1 and v = bc we have a = uv and vu = bcc−1 = b. Hence a ∼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 y i for all i ∈ N. (ii) If e = xi and f = y j are idempotents and x ∼pS y, then e ∼pS f . Proof. Letx = ab and y = ba for some a, b ∈ S. Then xi = (ab)i =   i−1 i i i−1 a (ba) b and y = (ba) = (ba) b a implying xi ∼pS y i and proving (i). To prove (ii) we observe that e = xi = xij and f = y j = y ij and that xij ∼pS y ij by (i). Lemma 6.4.7 Let S be a semigroup and e, f, g be three idempotents from S 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. Since e is an idempotent we also have e = e2 = xyxy = xf y and analogously f = yex, e = ugv, g = veu. This implies g = vxf yu and f = yugvx. Therefore g = g 2 = (vxf )(yug) and f = f 2 = (yug)(vxf ) and thus g ∼pS f .

6.4. CONJUGATE ELEMENTS

105

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 that x ∼S y. Let i, j ∈ N be such that e = xi and f = y j are idempotents. Then e ∼pS f . Proof. Let x = x1 , x2 , . . . , xl = y be a sequence of elements from S such that xi ∼pS xi+1 for i = 1, . . . , l−1. Since S is finite, for every i = 2, . . . , l−1 there i exists mi ∈ N such that yi = xm is an idempotent. Let y1 = e and yl = f . i 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 the relation ∼pS and ∼S to E(S) coincide. Proof. For e, f ∈ E(S) that e ∼pS f implies e ∼S f is obvious. The converse implication 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 ∈ S 1 be such that uv = e and vu = f . Then e = e2 = uvuv = uf v and f = f 2 = vuvu = veu. Hence S 1 eS 1 = S 1 f S 1 and we have eJ 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 and f = 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 η

μ

η

μ

μ

η

μ

η

μ

a1 → b1 → a2 → b2 → . . . → ak−1 → bk−1 → ak → bk → a1 . It follows that (b1 , b2 , . . . , bk ) = (η(a1 ), . . . , η(ak )) is a cycle for β. Obviously b1 , . . . , bk ∈ stim(β) and hence (b1 , b2 , . . . , bk ) is a cycle of the permutation 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 bijections between cycles of α and cycles of β, which preserve lengths of cycles. The claim follows.

106

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Theorem 6.4.13 Let S denote one of the semigroups PT n , Tn , or IS n , and α, β ∈ S. Then α ∼S β if and only if strank(α) = strank(β) and ct(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: M0 (γ) = stim(γ), M0 (γ) = im(γ)\dom(γ), M0 (γ) = M0 (γ) ∪ M0 (γ), M1 (γ) = {x ∈ M0 (γ) : α(x) ∈ M0 (γ)}, Mi (γ) = {x : α(x) ∈ Mi−1 (γ)}, i ≥ 2.

(6.7)

Note that M0 (γ) = ∅ for γ ∈ Tn . From Theorem 1.2.9, which describes the structure of connected components of Γγ it follows that the sets M0 (γ), M0 (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 implies Mj (γ) = ∅ for all j > i. Of course it makes sense to consider only those sets from (6.7), which are nonempty, say these are the sets M0 (γ), . . . , Mk (γ). Then we have the following picture: M0 (γ)

γ

gOOO OOO γ OOO OOO γ O o rr r r γ rr rrr r r x r M1 (γ) < r

... o

γ

o

Mk−1 (γ)

γ

Mk (γ)

M0 (γ) Now let us apply the above to our element α. If S = PT n , or S = IS n and k > 0, or if S = Tn and k > 1, we consider an idempotent , which satisfies the following conditions: • im() = M0 (α) ∪ M1 (α) ∪ · · · ∪ Mk−1 (α); • dom() = im() if S = PT n , or S = IS n ; and (Mk (α)) ⊂ Mk−1 (α) if S = Tn . In the case S = PT n , or S = IS n the idempotent  is uniquely determined, while in the case S = Tn we might have a choice. For such  we obviously have α = α. On the other hand, for the element α1 = α the nonempty part of the sequence (6.7) will be one step shorter. Further, for all i = 0, . . . , k − 2

6.4. CONJUGATE ELEMENTS

107

we obviously have Mi (α1 ) = Mi (α). If S = PT n , or S = IS n , we also have Mk−1 (α1 ) = Mk−1 (α) if k > 1 and M0 (α1 ) = M0 (α) if k = 1. If S = 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 = IS n . Let α = αk if k > 0, and α = α in the opposite case. Then M0 (α ) = M0 (α ) = stim(α) = im(α). Analogously we obtain the element β  . We have α ∼S α , β ∼S β  and using the assumptions of our theorem we also have α|stim(α) = α |stim(α) = (a1 , . . . , ap )(b1 , . . . , bq ) . . . (c1 , . . . , cr ), β|stim(β) = β  |stim(β) = (a1 , . . . , ap )(b1 , . . . , bq ) . . . (c1 , . . . , cr ). Consider the elements  a1 · · · μ=  a  1 · · · a1 · · · η= a1 · · ·

ap ap ap ap

b1 b1 b1 b1

··· ··· ··· ···

bq bq bq bq

··· ··· ··· ···

c1 c1 c1 c1

··· ··· ··· ···

 cr , cr  cr , cr

(6.8)

(6.9)

where dom(μ) = dom(α ) = im(α ) and dom(η) = dom(β  ) = im(β  ). We have 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 opposite case. For this element the sequence (6.7) has at most two nonempty sets: M0 (α ) = M0 (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 the theorem we also have the decompositions (6.8). Consider some permutation ˜−1 . μ ˜, which coincides with the element μ from (6.9) on im(α ). Let η˜ = μ  Then η˜ coincides with the element η from (6.9) on im(β ) and for the element δ = μ ˜α · η˜ we have α ∼pS δ as α = η˜ · μ ˜α . Moreover, we also have   stim(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 as follows:   D1 · · · Dm d1 ··· dm  β = d1 · · · dm π(d1 ) · · · π(dm ) (some Di s may be empty). At the same time we can write δ as follows:   F1 · · · Fm d1 ··· dm δ= . d1 · · · dm π(d1 ) · · · π(dm ) For the element τ=



··· F1 −1 π (d1 ) · · ·

Fm d1 · · · −1 π (dm ) d1 · · ·

dm dm



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CHAPTER 6. OTHER RELATIONS ON SEMIGROUPS

we then have τ β  = 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 only if 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 of PT 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 α and β are conjugate.

6.5

Addenda and Comments

6.5.1 Congruences on Tn were described by Mal’cev in [Ma1]. This description was used by Liber in [Lib] to describe congruences on IS n . Congruences on 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 for all α ∈ T and σ ∈ Sn we have σ −1 ασ ∈ T . For example, the semigroups PT n , Tn and IS n are Sn -normal subsemigroups of PT n . Moreover, any ideal in 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 almost all 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 of the same type. The proof of this very general result is technically slightly more difficult as one has to consider the cases of an arbitrary collection of operations with different numbers of arguments. 6.5.4 Let Sem denote the category, whose objects are semigroups, and morphisms are all possible homomorphisms between semigroups. Categorical monomorphisms in Sem are exactly the injective homomorphisms of semigroups, while the non-surjective embedding (N, +) → (Z, +) is a categorical epimorphism 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 alternating group), and the subgroup Ek consisting of the identity transformation. For n = 4 this was proved by Galois in about 1830. A short and easy argument for 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 Klein 4-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:

6.5. ADDENDA AND COMMENTS

109

Corollary 6.5.1 (i) For n > 1 all congruences on each of the semigroups PT n , Tn and IS n 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 IS n , or Tn is a restriction of some congruence on PT n . Moreover, for n > 1 different congruences on PT n restrict to different congruences on both IS n and Tn . 6.5.6 The criterion for S-conjugacy of two elements in IS n given in Theorem 6.4.13 was obtained in [GK1]. For the semigroups PT n and Tn an analogous result is obtained in [KuMa1]. Several abstract results about 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 : αβ

if and only if Φα ⊂ Φβ .

Note that the restriction of the natural partial order to Tn coincides with the equality relation. It is easy to see that the natural partial order is a compatible relation. The restriction of the natural partial order to IS n has a transparent algebraic interpretation: Lemma 6.5.2 Let α, β ∈ IS n . Then α  β if and only if αβ −1 = αα−1 . It is hence natural and interesting to try to understand the structure of all compatible partial orders on the semigroups PT n , Tn , and IS n . It turns out that the only compatible partial order on Tn is the equality relation Δ (see [Mo2]). For PT n and IS n 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, the relation Ψk1 ,...,km ,lm ,...,l1 = (Δ ∪ (Ik1 × Il1 ) ∪ · · · ∪ (Ikm × Ilm ))∩ 

(6.10)

is a compatible partial order on PT n . Moreover, different collections give different compatible partial orders. (ii) Each compatible partial order on the semigroup PT n has either the form Ψk1 ,...,km ,lm ,...,l1 or is transposed to some Ψk1 ,...,km ,lm ,...,l1 .

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Theorem 6.5.4 ([Mo1]) (i) For each collection of numbers as follows: 0 ≤ k1 < k2 < · · · < km < lm < · · · < l1 ≤ n, m ≥ 1, the relation Ψk1 ,...,km ,lm ,...,l1 as in (6.10) is a compatible partial order on IS n . Moreover, different collections give different compatible partial orders. (ii) Each compatible partial order on IS n different from Δ has either the form Ψk1 ,...,km ,lm ,...,l1 or is transposed to some Ψk1 ,...,km ,lm ,...,l1 . Corollary 6.5.5 ([Mo1]) orders on PT n .

(i) There are exactly 2n+2 −5 compatible partial

(ii) There are exactly 2n+1 − 1 compatible partial orders on IS n .

6.6

Additional Exercises

6.6.1 Prove that the alternating group An of all even permutations is a normal 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 relations on a group. 6.6.4 Let ρ be a congruence on a semigroup S. Let ϕ : S  S/ρ be an epimorphism such that Ker(ϕ) = ρ. Is it possible to claim that ϕ always coincides 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 the natural partial order coincides with Tn . 6.6.7 Prove that for the semigroup IS n the following conditions are equivalent: (a) α  β (b) α−1  β −1 (c) βα−1 = αα−1 (d) α−1 β = α−1 α (e) αβ −1 = αα−1 (f) β −1 α = α−1 α (g) αβ −1 α = α (h) α−1 βα−1 = α−1 6.6.8 Prove Corollary 6.5.5.

Chapter 7

Endomorphisms 7.1

Automorphisms of T n , PT n , and IS n

Recall from Sect. 6.1 that an automorphism of a semigroup S is a bijective mapping 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−1 xa, x ∈ S, is an automorphism of S. Proof. For x, y ∈ S we have Λa (xy) = a−1 xya = a−1 xa · a−1 ya = Λa (x)Λa (y) and hence Λa is an endomorphism of S. Further for all x ∈ S we have Λa−1 Λa (x) = aa−1 xaa−1 = x, Λa Λa−1 (x) = a−1 axa−1 a = x. Hence Λa is a bijection with the inverse Λa−1 . The claim follows. Automorphisms of the form Λa , a ∈ S ∗ , are called inner automorphisms of 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−1 a−1 = Λa−1 (bxb−1 ) = Λa−1 Λb−1 (x). Hence Λa−1 Λb−1 = Λ(ab)−1 . This proves (i). In the proof of Proposition 7.1.1 we have shown that Λ−1 a = Λa−1 for all a ∈ S ∗ . Together with the above proof of (i) we hence have that Inn(S) O. Ganyushkin, V. Mazorchuk, Classical Finite Transformation Semigroups, Algebra and Applications 9, DOI: 10.1007/978-1-84800-281-4 7, c Springer-Verlag London Limited 2009 

111

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CHAPTER 7. ENDOMORPHISMS

is closed with respect to both composition of maps and taking the inverse map. This means that Inn(S) is a subgroup of Aut(S). Let now a ∈ S ∗ , ψ ∈ Aut(S) and x ∈ S. Then ψ −1 Λa ψ(x) = ψ −1 (a−1 ψ(x)a) = ψ −1 (a−1 )xψ −1 (a) = Λψ−1 (a) (x). This means that ψ −1 Λa ψ = 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 IS n . Then Inn(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 elements Λσ and Λτ of Inn(S) are also different. Proof. As σ = τ , there exists x ∈ N such that y = σ(x) = τ (x) = z. If S = Tn or S = PT n , then Λσ (0x ) = 0y = 0z = Λτ (0x ). Hence Λσ = Λτ . If S = IS n , then Λσ (ε{x} ) = ε{y} = ε{z} = Λτ (ε{x} ). Hence Λσ = Λτ again. Lemma 7.1.4 says that the epimorphism which was described in Proposition 7.1.2(i) is injective. In particular, this mapping induces an isomorphism Inn(S) ∼ = S∗ ∼ = Sn . To complete the proof we only need to show that any automorphism of S is inner. Since we already have n! different (inner) automorphisms of S, it is in fact enough to show that |Aut(S)| ≤ n!. We shall do it via a case-by-case analysis. Let ψ ∈ Aut(S). Since S is finite and all two-sided ideals of S form a chain (by Theorem 4.3.1), ψ induces the identity mapping on the set of all two-sided ideals. In particular, for all i = 0, 1, . . . , n we have that the restriction map ψ|Ii : Ii → Ii is bijective. Case 1. S = Tn . In this case we have I1 = {0a : a ∈ N}. Since the restriction 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 a homomorphism 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 it is 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. Take now some α ∈ Tn and let ψ(α) = β. Applying ψ to the equality α0b = 0a

7.1. AUTOMORPHISMS OF Tn , PT n , AND IS n

113

we get that this equality is equivalent to the equality β0b = 0a . This means that M (α) = M (β), in particular, α(x) = β(x) for all x ∈ N. Therefore β = α. Hence ψ(α) = α for all α ∈ Tn and the proof in the case S = Tn is completed. Case 2. S = IS n . The set X = {ε{a} : a ∈ N} is the set of all idempotents of rank one, that is, the set of all idempotents in I1 \I0 . As an automorphism, ψ sends idempotents to idempotents. Since the restriction of ψ to both I1 and I0 is a bijection, ψ induces a permutation on X, in particular, 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 mapping ψ → ψ is a homomorphism of groups. To complete the proof it is enough to show that the kernel of this mapping is trivial. In other words, let ψ ∈ Aut(IS n ) be such that ψ(ε{a} ) = ε{a} for all a ∈ N. We have to show that ψ(α) = α for all α ∈ IS n . For α ∈ IS n 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 ∈ IS n and let ψ(α) = β. Applying ψ to the inequality ε{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 ∈ IS n and the proof in the case S = IS n is completed as well. Case 3. S = PT n . We could apply exactly the same arguments as in the proof given in Case 2 provided that we could prove that ψ preserves the set X. 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, if im(β) ⊂ dom(α). Hence the right annihilator Annr (α) = {β : αβ = 0} of the element α contains exactly (|dom(α)| + 1)n elements. Moreover, the equality ψ(0) = 0 implies that αβ = 0 if and only if ψ(α)ψ(β) = 0. Hence ψ(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) = ∅, that is, 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. Analogously to 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.

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7.2

CHAPTER 7. ENDOMORPHISMS

Endomorphisms of Small Ranks

Recall from Sect. 6.1 that an endomorphism of a semigroup S is a mapping ϕ : S → S such that ϕ(x · y) = ϕ(x) · ϕ(y) for all x, y ∈ S. Our main goal for the rest of this chapter is to describe all endomorphisms of the semigroups Tn , IS n , and PT n . The final classification will be done basically using the case-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 of small ranks. Let S denote one of the semigroups Tn , IS n , or PT n . Let  ∈ S be an idempotent. Define the mapping Φ : S → S via Φ (α) =  for all α ∈ S. The following statement is obvious: Lemma 7.2.1 (i) For any idempotent  the mapping Φ is an endomorphism of S of rank one. (ii) If  =  , then Φ = Φ  . (iii) Every endomorphism of S or rank one has the form Φ for some idempotent  ∈ S. Let , δ ∈ S be different idempotents such that δ = δ = δ. Define the mapping Ψ ,δ : S → S as follows:  , α ∈ Sn ; Ψ ,δ (α) = δ, otherwise. Lemma 7.2.2 (i) For any pair , δ of different idempotents of S satisfying δ = δ = δ the mapping Ψ ,δ is an endomorphism of S of rank two. (ii) If the pairs , δ and  , δ  are different, then Ψ ,δ = Ψ  ,δ . (iii) Every endomorphism of S or rank two has the form Φ ,δ for some unequal idempotents , δ ∈ S satisfying δ = δ = 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 congruence Ker(ϕ) contains exactly two congruence classes. Going through the list of all congruences on S given by Theorem 6.3.10, we see that the only congruence on S with exactly two congruence classes is ≡Sn . The congruence classes are S ∗ and In−1 . As the representatives of the two congruence classes of this congruence we can take for example the identity transformation εn and any (left) zero element γ of S. These are idempotents and obviously satisfy εn γ = γεn = γ. Since they belong to different congruence classes, we have  = ϕ(εn ) = ϕ(γ) = δ. Moreover, δ = δ = δ. By definition, in this notation we have ϕ = Φ ,δ . This completes the proof of (iii).

7.3. EXCEPTIONAL ENDOMORPHISM

115

Let , δ ∈ S be different idempotents such that δ = δ = δ. Let further τ ∈ G be an element of order two (i.e., τ 2 = , τ = ) such that τ δ = δτ = δ. τ : S → S as follows: Define the mapping Θ ,δ ⎧ ⎪ ⎨, α ∈ An ; τ (α) = τ, α ∈ Sn \An ; Θ ,δ ⎪ ⎩ δ, otherwise. τ is an Lemma 7.2.3 (i) For any triple , δ, τ as above the mapping Θ ,δ endomorphism of S of rank three. 

τ = Θ τ . (ii) If the triples , δ, τ and  , δ  , τ  are different, then Θ ,δ  ,δ  τ for some , (iii) Every endomorphism of S or rank three has the form Θ ,δ δ, and τ as above.

Proof. The statement (i) is proved by a direct calculation and the statement (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 congruence Ker(ϕ) contains exactly three congruence classes. Going through the list of all congruences on S given by Theorem 6.3.10, we see that the only congruence on S with exactly three congruence classes is ≡An , n > 1. The congruence classes are An , Sn \An , and In−1 . As the representatives of the three congruence classes of this congruence we can take for example the identity transformation εn , the transposition (1, 2) and any (left) zero element γ of S. These elements obviously satisfy εn γ = γεn = γ, γ(1, 2) = γ, and (1, 2) ∈ Gεn . Further, either (1, 2)γ = γ or (1, 2)γ is another left zero, and hence belongs to In−1 . Since our representatives belong to different congruence classes, we have  = ϕ(εn ) = ϕ(γ) = δ and ϕ((1, 2)) = τ is an element of order two in G . Further, from the above we also get that δ = δ = δ τ . This and τ δ = δτ = δ. By definition, in this notation we have ϕ = Θ ,δ proves (iii).

7.3

Exceptional Endomorphism

Let ϕ ∈ Aut(Sn ) and S = IS n or S = PT n . Define the mapping Ωϕ : S → S as follows:  ϕ(α), α ∈ Sn ; Ωϕ (α) = 0, otherwise. Lemma 7.3.1 Ωϕ is an endomorphism of both IS n and PT n . Proof. Follows by a direct calculation.

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To proceed we need some additional notation. Let α ∈ PT n be an element of rank (n − 1). We would like to associate with α some element α ˆ of rank one. There are two possibilities. The first one is dom(α) = N. Then we define α ˆ as the unique element of rank one such that α) = N\im(α). dom(ˆ α) = dom(α), im(ˆ Here is an example of such α and α ˆ for n = 5:     1 2 3 4 5 1 2 3 4 5 α= , α ˆ= . 2 5 ∅ 1 3 ∅ ∅ 4 ∅ ∅ The second possibility is that there exists a unique pair a, b ∈ N such that α(a) = α(b). Then we define α ˆ as the unique element of rank one such that dom(ˆ α) = {a, b}, im(ˆ α) = N\im(α). Here is an example of such α and α ˆ for n = 5:     1 2 3 4 5 1 2 3 4 5 α= , α ˆ= . 2 5 2 1 3 4 ∅ 4 ∅ ∅ Define the mapping Ξ : PT n → PT n , n > 1, as follows: ⎧ ⎪ ⎨α, rank(α) = n; Ξ(α) = α ˆ , rank(α) = n − 1; ⎪ ⎩ 0, otherwise. Note that Ξ maps IS n to IS n . Abusing notation we denote the induced mapping on IS n also by Ξ. Lemma 7.3.2 Let S = IS n 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 are permutations or if one of them has rank at most (n − 2). There are three cases left. Case 1. α ∈ Sn , rank(β) = n−1. In this case rank(αβ) = n−1 and hence  Since α is a permutation, we have (7.1) reduces to the equality αβˆ = αβ.  dom(αβ) = dom(β) and im(αβ) = α(im(β)). Hence the equality αβˆ = αβ follows 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  Since β is a permutation, we have (7.1) reduces to the equality αβ ˆ = αβ. −1  ˆ β = αβ dom(αβ) = β (dom(α)) and im(αβ) = im(α). Hence the equality α follows 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) contains two different a and b such that α(a) = α(b) or if im(β) ∩ dom(α) = ∅. In both cases the definition of Ξ gives α ˆ βˆ = 0. The second possibility is that rank(αβ) = n − 1. This is possible if and only if im(β) ⊂ dom(α) and α is injective on im(β). In particular, from the definition of Ξ we have that the ˆ ⊂ dom(ˆ unique element in N\im(β) belongs to dom(ˆ α), that is, im(β) α). We  = dom(β) ˆ and im(αβ)  = im(ˆ  thus get dom(αβ) α), which implies α ˆ βˆ = αβ since both elements in the last equality have rank one. This proves (i). To prove (ii) one can just observe that ϕΞ and ψΞ act differently on idempotents of rank (n − 1). Now let n = 4. In this case we have an additional normal subgroup of S4 , namely, V4 . The cosets of S4 modulo V4 are the following sets: 

X1 =  X2 =  X3 =  X4 =  X5 =  X6 =

1 2 3 4 1 2 3 4 1 2 3 4 2 1 3 4 1 2 3 4 1 3 2 4 1 2 3 4 3 2 1 4 1 2 3 4 2 3 1 4 1 2 3 4 3 1 2 4

       1 2 3 4 1 2 3 4 1 2 3 4 , , , , 2 1 4 3 3 4 1 2 4 3 2 1   1 , 1   1 , 2   1 , 4   1 , 1   1 , 4

2 3 4 2 4 3 2 3 4 4 1 3 2 3 4 1 2 3 2 3 4 4 2 3 2 3 4 2 1 3

  1 , 4   1 , 3   1 , 1   1 , 4   1 , 1

2 3 4 3 1 2 2 3 4 1 4 2 2 3 4 4 3 2 2 3 4 1 3 2 2 3 4 3 4 2

  1 , 3   1 , 4   1 , 2   1 , 3   1 , 2

2 3 4 4 2 1 2 3 4 2 3 1 2 3 4 3 4 1 2 3 4 2 4 1 2 3 4 4 3 1

 ,  ,  ,  ,  .

We see that each of these cosets contains a unique element α satisfying α(4) = 4 (the first element of each coset). Mapping the whole coset to this element defines the endomorphism ψ : S4 → S4 with kernel V4 . Define the following transformations of PT 4 :   ψ(α), α ∈ S4 ; ψ(α), α ∈ S4 ; Υ1 (α) = Υ2 (α) = 0, otherwise; ε{4} , otherwise;  ψ(α), Υ3 (α) = 04 ,

α ∈ S4 ; Υ4 (α) = otherwise;



ψ(α)ε{1,2,3} , α ∈ S4 ; 0, otherwise;

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  ψ(α)ε{1,4} , α ∈ S4 ; ψ(α)ε{2,4} , α ∈ S4 ; Υ6 (α) = Υ5 (α) = 0, otherwise; 0, otherwise;  ψ(α)ε{3,4} , α ∈ S4 ; Υ7 (α) = 0, otherwise. Note that Υ1 , Υ2 , and Υ4 preserve IS 4 and Υ3 preserves T4 . Abusing notation we shall denote the induced mappings on IS 4 and T4 by the same symbols. Exercise 7.3.3 Let i ∈ {1, 2, . . . , 7} and ϕ, ψ ∈ Aut(PT 4 ) be two different automorphisms. Show that the restrictions of ϕΥi and ψΥi to S4 are different. 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 IS 4 and T4 in appropriate cases. Proof. We prove the statement (i) for Υ1 . All other cases are similar. We have 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 the fact 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 is a 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 one of the following forms: (a) Λπ , where π ∈ Sn . (b) Φ , where  ∈ Tn is an idempotent. (c) Ψ ,δ , where  = δ ∈ Tn are idempotents such that δ = δ = δ. τ , where  = δ ∈ T are idempotents such that δ = δ = δ and (d) Θ ,δ n τ ∈ G is an element of order two such that τ δ = δτ = δ.

(ii) In addition to the above endomorphisms, the semigroup T4 has also endomorphisms Λπ Υ3 , 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 α such that πα = α for all π ∈ Sn . Proof. Let α ∈ Tn , x ∈ N and y = α(x). Let further z ∈ N be such that z = 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 , IS n , or PT n . Let ϕ ∈ End(S). Assume that all elements of Sn form one-element congruence classes 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 then contained in some maximal subgroup. But, by Theorem 5.1.4, all maximal subgroups of S are isomorphic to Sm , m ≤ n, and there is only one maximal subgroup isomorphic to Sn , namely, Sn itself. Hence ϕ maps Sn to Sn and thus 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 all elements of Sn form one-element congruence classes. From Lemma 7.4.3 we have 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 to some 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 the restriction of ϕ to Sn is bijective by the previous paragraph, we have πγ = γ for all π ∈ Sn . But Tn , n > 1, does not contain any γ with such property by Lemma 7.4.2. This means that the case 1 < k < n is impossible, which completes the proof. Let n = 4 and ϕ ∈ End(Tn ). If Ker(ϕ) is a uniform congruence, then ϕ has rank one and thus coincides with some Φ by Lemma 7.2.1(iii). If Ker(ϕ) is the identity congruence, then ϕ is an automorphism and thus coincides with some Λπ by Theorem 7.1.3. In all other cases Ker(ϕ) =≡R , where R is En , An , or Sn by Lemma 7.4.4. If R = Sn , then ϕ has rank two and thus coincides with some Ψ ,δ by Lemma 7.2.2(iii). If R = An , then ϕ has rank τ by Lemma 7.2.3(iii). three and thus coincides with some Θ ,δ 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 shows that ϕ induces an automorphism of Sn . The ideal In−1 is a congruence class and should be sent by ϕ to some idempotent γ. As ϕ|Sn ∈ Aut(Sn ), as in the

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proof of Lemma 7.4.4 we obtain πγ = γ for all π ∈ Sn . But such element γ does 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 above is the fact that R can be equal to V4 . For such R we have ϕ(S4 ) ∼ = S3 , which S , where γ is an idempotent may be a subgroup of either S4 or some Gγ ∼ = 3 of 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, isomorphic to 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 now the endomorphism  ϕ, i = 4;  ϕ = (7.3) Λ(i,4) ϕ, i = 4. 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 there exists π ∈ S4 such that Λπ ϕ restricts to the identity map on H4 . The ideal I3 is a congruence class and thus should be mapped by Λπ ϕ to some noninvertible idempotent α of T4 . As in the proof of Lemma 7.4.4 we get βα = α for all β ∈ H4 , in particular, for β = (1, 2, 3). Hence for any x ∈ im(α) we should have β(x) = x, which means that im(α) = {4}, and thus α = 04 . Therefore Λπ ϕ = Υ3 and hence ϕ = Λπ−1 Υ3 . 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 that δα = αδ = δ for all α ∈ Gγ . From γδ = δ we have im(δ) ⊂ im(γ). In the proof of Theorem 4.7.4 it was shown that the restriction to im(γ) defines an isomorphism from Gγ to S(im(γ)). Take any α ∈ Gγ , whose image in S(im(γ)) under this isomorphism does not have any fixed point. Then for any x ∈ N we have α(δ(x)) = δ(x) since δ(x) ∈ im(γ). This means that the equality 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 IS n has one of the following forms: (a) Λπ , where π ∈ Sn . (b) Φ , where  ∈ IS n is an idempotent. (c) Ψ ,δ , where  = δ ∈ IS n are idempotents such that δ = δ = δ. τ , where  = δ ∈ IS are idempotents such that δ = δ = δ (d) Θ ,δ n and τ ∈ G is an element of order two such that τ δ = δτ = δ.

7.4. CLASSIFICATION OF ENDOMORPHISMS

121

(e) Ωψ , ψ ∈ Aut(Sn ). (f ) Λπ Ξ, π ∈ Sn . (ii) In addition to the above endomorphisms, the semigroup IS 4 also has endomorphisms Λπ Υi , i = 1, 2, 4, π ∈ Sn . Proof. The statement is obvious for n = 1 so until the end of the proof we assume n > 1. Let ϕ ∈ End(IS n ). Assume first that all elements of Sn form separate congruence classes of Ker(ϕ). By Lemma 7.4.3, ϕ preserves Sn and induces an automorphism on it. Consider first the case n = 6. Then this automorphism is inner (see 7.6.2) and hence there exists π ∈ Sn such that ϕ = Λπ ϕ induces on Sn the identity mapping. Set  = εN\{1} . The element ϕ () must be an idempotent, say ϕ () = εB for some B ⊂ N. To determine B more precisely we shall need an auxiliary notion and statement. For α ∈ IS n the set {π ∈ Sn : απ = πα} is called the centralizer of α in 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 = εA π is equivalent to εA = πεA π −1 . Obviously πεA π −1 = επ(A) . Hence εA = πεA π −1 is equivalent to εA = επ(A) , that is, A = π(A). This proves the first claim and the second claim follows immediately from the first one. Applying ϕ to π = π we get πεB = εB π as ϕ (π) = π for all π ∈ Sn . This implies that CSn () ⊂ 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 of S. Let f : A → T be any mapping. Then there exists at most one homomorphism F : S → T such that F |A = f . Proof. As S = A, each x ∈ S can be written in the form x = a1 a2 · · · ak for some ai ∈ A. Then F (x) = F (a1 a2 · · · ak ) = F (a1 )F (a2 ) · · · F (ak ) = f (a1 )f (a2 ) · · · f (ak ), which means that F (x) is uniquely determined by f .

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By Theorem 3.1.4, the semigroup IS n is generated by Sn and . By Lemma 7.4.9 this means that any endomorphism of IS n is uniquely determined by its values on Sn and . In the situation we have ϕ (π) = π for all π ∈ Sn and we have three possibilities for ϕ (), namely, ϕ () = , ϕ () = ε{1} and ϕ () = 0. The first possibility is realized by the identity automorphism, the second one is realized by the endomorphism Ξ, and the third one is realized by the endomorphism Ωθ , where θ is the identity automorphism of Sn . This means that the original automorphism ϕ = Λπ−1 ϕ is of the form (ia), (if), or (ie). Let now n = 6. In addition to the cases considered above we have to consider the case when ϕ induces on Sn an automorphism, which is not inner. Such automorphisms are known explicitly, see 7.6.2. In particular, each such automorphism maps transpositions from the subgroup CS6 () = {π ∈ S6 : π(1) = 1} < S6 to permutations, each of which is a product of three different commuting transpositions. 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 7.4.8 implies that the only possibility for ϕ() in this case is 0. This means that ϕ is of type (ie). Now we have to consider all cases when ϕ does not induce an automorphism of Sn . If Ker(ϕ) = ωIS n , 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 is a nontrivial normal subgroup of Sn . If R = Sn , then ϕ is of type (ic) by Lemma 7.2.2(iii). If R = An , then ϕ is of type (id) by Lemma 7.2.3(iii). It remains to consider the case R = V4 . We leave it to the reader to verify that in this case ϕ is of type (ii). Theorem 7.4.10 (i) Let n = 4. Then each endomorphism of PT n has one of the following forms: (a) Λπ , where π ∈ Sn (b) Φ , where  ∈ PT n is an idempotent (c) Ψ ,δ , where  = δ ∈ PT n are idempotents such that δ = δ = δ τ , where  = δ ∈ PT (d) Θ ,δ n are idempotents such that δ = δ = δ and τ ∈ G is an element of order two such that τ δ = δτ = δ

(e) Ωϕ , ϕ ∈ Aut(Sn ) (f ) Λπ Ξ, π ∈ Sn (ii) In addition to the above endomorphisms, the semigroup PT 4 also has endomorphisms Λπ Υi , i = 1, 2, . . . , 7, π ∈ Sn . Exercise 7.4.11 Prove Theorem 7.4.10.

7.5. COMBINATORICS OF ENDOMORPHISMS

7.5

123

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 ⎞ ⎛ m−1

m−2r n  2 n−m m−r−k   m ·k ⎠ n! ⎝1 + 2r · (n − m)! · (m − 2r − k)! · k! · r! m=1 r=0

(7.5)

k=1

endomorphisms. Proof. The statement (i) is obvious. The statement (ii) follows from the formula of (iii) and the note that T4 has 24 additional endomorphisms described in Theorem 7.4.1(ii). Hence we have to prove (iii). The semigroup Tn has n! automorphisms by Theorem 7.1.3. By Theorem 7.4.1, if we forget about the additional endomorphisms T4 , we have only to count the number of endomorphisms of Tn of ranks 1, 2, and 3. By Lemma 7.2.1(iii) the endomorphisms 1 correspond bijectively  n−m  of rank n m endomorphisms of to idempotents of Tn . Hence we have nm=1 m rank 1 by Corollary 2.7.4. By Lemma 7.2.2(iii) the endomorphisms of rank 2 correspond to pairs of idempotents (, δ) of Tn satisfying the condition δ = δ = δ. Lemma 7.5.2 Let , δ ∈ Tn be two idempotents. Then the following conditions are equivalent: (a) δ = δ = δ (b) im(δ) ⊂ im() and ρδ  ρ Proof. If δ = δ, then im(δ) ⊂ im() by Exercise 2.1.4(b). If δ = δ, then ρδ  ρ by Theorem 4.2.4. Hence (a) implies (b). Conversely, if im(δ) ⊂ im() and ρδ  ρ , then a direct calculation shows that δ = δ = δ. Hence (b) implies (a). Now we can use Lemma 7.5.2 to count the number of all pairs (, δ) of different idempotents of Tn satisfying the condition δ = δ = δ. Let m = rank() and k = rank(δ). Since  = δ and im(δ)  n  ⊂ im(), we have 1 ≤ k < m ≤ n. The image of  can be chosen in m different ways. By Lemma 7.5.2, the  image of δ should be chosen inside the im() and this can be done in m k different ways. As  is an idempotent, (x) = x for all x ∈ im(). To define  on other elements, we should map each x ∈ N\im() to some element of im(). This can be done in mn−m different ways. For x ∈ im()\im(δ) let Ax = {y ∈ N : (y) = x}. From Lemma 7.5.2, to define δ we should map each Ax , x ∈ im()\im(δ) to some element of im(δ). This

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CHAPTER 7. ENDOMORPHISMS

can be done in k m−k different ways. Using the multiplication rule and then adding everything up for all possible m and k we get n m−1    n m mn−m k m−k m k

(7.6)

m=2 k=1

endomorphisms of rank 2. By Lemma 7.2.3(iii) the endomorphisms of rank 3 correspond to triples (, δ, τ ) of Tn , where  and δ are different idempotents of Tn satisfying the condition δ = δ = δ, and τ ∈ G is an element of order 2 such that τ δ = δτ = δ. The restrictions on  and δ are given by Lemma 7.5.2. Lemma 7.5.3 Let , δ ∈ Tn be two idempotents such that δ = δ = δ. Let further τ ∈ G . Then the following conditions are equivalent: (a) τ δ = δτ = δ (b) τ (x) = x for all x ∈ im(δ); and δ(x) = δ(y) for all x, y ∈ im() such that x = y and τ (x) = y Proof. Assume (a). Let x ∈ im(δ). Then x = δ(x) = τ δ(x) = τ (x). Let x, y ∈ im() 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 τ δ = δ. By assumptions we have τ  = τ and δ = δ. Hence the equality δτ = δ is equivalent to δτ  = δ, which means that the equality δτ = δ should be checked only for x ∈ im(). For those x ∈ im() for which τ (x) = x, the equality δτ (x) = δ(x) is obvious. If τ (x) = y = x for some x, y ∈ im(), 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 r m commuting transpositions, where 0 < r ≤  m 2 . There are 2 ways to choose m−2 the first transposition, 2 ways to choose the second transposition, and so on. As the order of transpositions is not important, for each r as above we have that Sm contains exactly m m−2   · · · · · m−2r+2 m! 2 · 2 2 = (7.7) r! (m − 2r)! · r! · 2r elements of order 2, which are products of r commuting transpositions. Let us now count the number of triples (, δ, τ ) as above. The rank  m of  n can have values m = 2, . . . , n. The image of  can be chosen in m different ways and to define  completely we should assign some element from im() to each 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 G ∼ = Sm and hence it can

7.5. COMBINATORICS OF ENDOMORPHISMS

125

be a product of r commuting transpositions, where 0 < r ≤  m 2 . However, , that is, 0 < r ≤  m−1 τ (x) = x for all x ∈ im(δ) = ∅. Hence r = m 2 2 . For m! each such r by (7.7) we have (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 different ways to choose im(δ). For x ∈ im() set Ax = k {y ∈ N : (y) = x}. As ρδ  ρ by Lemma 7.5.2(b) and δ(x) = δ(y) for all x, y ∈ im() such that x = y and τ (x) = y by Lemma 7.5.3(b), to define δ we should assign to each Ax , x ∈ im()\im(δ), an element of im(δ), such that the same element is assigned to Ax and Ay as soon as τ (x) = y. Such assignment can be done in k m−r−k different ways. Adding everything up, we get m−1

m−2r    n  2   n m − 2r m! · mn−m · k m−r−k (7.8) m k (m − 2r)! · r! · 2r m=2 r=1

k=1

endomorphisms of order 3. If for r = 0 in the formula (7.8) we allow k to vary from 1 to (m − 1), we will obtain the formula (7.6). Further, the formula (7.6) would count all endomorphisms of rank one for Tn as well if we allow k = m and the value 1 for m. Taking this into account, adding n! automorphisms, and rewriting binomial coefficients in terms of factorials gives the formula (7.5). Corollary 7.5.4

(i) The semigroup IS 1 has three endomorphisms.

(ii) The semigroup IS 2 has 14 endomorphisms. (iii) The semigroup IS 4 has 282 endomorphisms. (iv) The semigroup IS 6 has 5,244 endomorphisms. (v) The semigroup IS n , n = 1, 2, 4, 6, has ⎞ ⎛ m

n  2 m−3r  2 ⎠ n! ⎝3 + (n − m)! · (m − 2r)! · r!

(7.9)

m=0 r=0

endomorphisms. Proof. The semigroup IS 1 has the trivial automorphism and two endomorphisms 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 additional endomorphisms of IS 4 given by Theorem 7.4.7(ii) and 6! additional endomorphisms of IS 6 corresponding to the outer automorphisms of S6 in Theorem 7.4.7(ie). The statement (ii) follows from (v) taking into account that for n = 2 the only endomorphism of type Theorem 7.4.7(ie) has rank 3 (and hence is of type Theorem 7.4.7(id) as well), and all endomorphisms of type Theorem 7.4.7(if) are in fact automorphisms. So, we have to prove only the statement (v).

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For n > 1 the semigroup IS n has n! automorphisms by Theorem 7.1.3, n! endomorphisms of type (ie), which correspond to inner automorphisms of Sn , and n! endomorphisms of type (if). So again we are left to count the number 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 (, δ, τ ) where  and δ are idempotents satisfying δ = δ = δ (not necessarily different) and τ ∈ G is an element  n of order at most 2 satisfying τ δ = δτ = δ. Let m = rank(). Then we choices for im() = A. Assume now that τ is a product of exactly have m m! r transpositions, 0 ≤ r ≤  m 2 . Then by (7.7) we have exactly (m−2r)!r!2r possibilities to choose τ . To proceed we need to know the restrictions on δ imposed by the condition τ δ = δτ = δ. Lemma 7.5.5 For α ∈ IS n and B ⊂ N we have αεB = εB α = εB if and only 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 for all x ∈ B, then αεB = εB is obvious, which also reduces εB α = εB to εB αεB = εB , that is, εB α = εB should be checked only for x ∈ B. However, for such x the equality εB α = 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 to choose δ. Summing everything up gives

 n  2   n m

m=0 r=0

m! · 2m−2r m (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 ⎞ ⎛ m

n  2 m−2r n−m · (k + 1)m−r−k   (m + 1) ⎠ n! ⎝3 + 2r · (n − m)! · (m − 2r − k)! · r! · k! m=0 r=0 k=0

endomorphisms. Exercise 7.5.7 Prove Corollary 7.5.6.

(7.10)

7.6. ADDENDA AND COMMENTS

7.6

127

Addenda and Comments

7.6.1 Automorphisms of Tn were described by Schreier in [Sch]. A different argument was later proposed by Mal’cev in [Ma1]. Automorphisms of IS n 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 . The semigroup 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 reason that all automorphisms of Sn , n = 6, are inner is purely numerical. Any automorphism maps conjugate elements to conjugate elements and preserves the order of an element. Let X be the set of all transpositions in Sn . This is a conjugacy class of elements of order two. A straightforward combinatorial computation shows that for n = 6 all other conjugacy classes of elements of order two have cardinalities different from |X|. Hence for n = 6 any automorphism maps transpositions to transpositions. From here it is fairly straightforward to deduce that any automorphism is inner. Details can be found for example in [KM]. Each outer (that is, not inner) automorphism of S6 maps a transposition to a product of three commuting transpositions. 7.6.3 Endomorphisms of IS n , Tn , and PT n were described and their numbers were counted in [ST1], [ST2], and [ST3], respectively. Although the generic statement is correct, there are some typos in the exceptional cases of [ST1] and [ST3]. In particular, for n = 6 the authors missed the endomorphisms of the form Ωϕ , where ϕ is an outer automorphism of S6 . 7.6.4 Here is the table for |End(S)|, where S = IS n , Tn , and PT n for small n: n |End(IS n )| |End(Tn )| |End(PT n )|

1 3 1 3

2 14 7 18

3 54 40 138

4 282 345 1,374

5 918 3,226 13,178

6 5,244 38,503 170,772

7 25,560 529,614 2,507,690

8 168,828 8,219,025 41,387,036

7.6.5 The problem to describe all homomorphisms (monomorphisms, epimorphisms) from S to T , where S, T ∈ {IS n , Tn , PT n : n ≥ 1}, is open in general. 7.6.6 In [ST1], [ST2], and [ST3] for S = IS n , Tn , and PT n Schein and → 0, n → ∞. More generally they ask to Teclezghi asked whether |End(S)| |S| find an estimate for the asymptotic of |End(S)|. For the semigroup IS n both questions were answered in [JM]: Theorem 7.6.1 ([JM]) (ii)

|End(IS n )| |IS n |

(i) |End(IS n )| ∼ 3n!, n → ∞.

→ 0, n → ∞.

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CHAPTER 7. ENDOMORPHISMS

Proof. Set

n  2  m

Xn = n! ·

m=0 r=0

2m−3r . (n − m)! · (m − 2r)! · r!

Because of Corollary 7.5.4(v) to prove (i) it is enough to show that Xn /n! → 0, n → ∞. We have

   n  2   n m 2r m

Xn =

m=0 r=0

m

2r

r

· 2m−3r · r!

(7.11)

All binomial coefficients in (7.11) are smaller than 2n and r ≤ n/2. Hence

n  2  m

Xn ≤ 8n · n/2! ·

2m−3r ≤ 16n · (n/2)! · n2 .

(7.12)

m=0 r=0

Using the Stirling formula we have √ 16n · (n/2)! · n2 16n · n2 · nn/2 · en · πn √ = ∼ n! 2n/2 · en/2 · nn · 2πn n n n 1 = √ · en ln 16+2 ln n+ 2 ln n+n− 2 ln 2− 2 −n ln n . (7.13) 2 Since the exponent is − 12 n ln n + O(n), we obtain that the right-hand side of (7.13) approaches 0 for large n. This proves (i).  n 2 By Theorem 2.5.1 we have |IS n | ≥ n−1 (n − 1)! = n · n!. Hence, using (i), we have 0≤

|End(IS n )| 3 · n! 3 · n! ∼ ≤ → 0, n → ∞. |IS n | |IS n | n · n!

The claim (ii) follows. For S = Tn and PT n the corresponding problems are still open.

7.7

Additional Exercises

7.7.1 Show that the only element α ∈ PT n satisfying the condition πα = α for all π ∈ Sn is the element α = 0. 7.7.2 Determine all elements α ∈ PT n which satisfy the condition απ = α for all π ∈ Sn . 7.7.3 Let A ⊂ N. Show that α ∈ IS n satisfies the condition αεA = εA α if and only if both A and N\A are invariant with respect to α. Use this to show that the centralizer of εA in IS n can be identified with IS(A) × IS(N\A).

7.7. ADDITIONAL EXERCISES

129

7.7.4 Determine |{α ∈ IS 3 : 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 = IS n , PT n , and Tn . 7.7.6 Show that Aut(I1 ) = Sn for the ideal I1 of IS n . 7.7.7 ([ST1, ST2, ST3]) An endomorphism ϕ ∈ End(S) is called a retraction provided that ϕ · ϕ = ϕ. Classify all retractions of IS n , 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 numbers with respect to addition. Show that End(S) = Aut(S) ∼ = (Q>0 , ·). 7.7.11 Classify all endomorphisms of the semigroup (N, +) of all positive integers with respect to addition. 7.7.12 Classify all endomorphisms of the finite cyclic semigroup S generated by an element of type (k, m). 7.7.13 Let  ∈ PT n be an idempotent of rank one with domain D and image {x}. Show that π ∈ Sn commutes with  if and only if π(x) = x and π(D) = D. Use this to show that the number of permutations commuting with  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 that there exists k ∈ N such that S k = {0}, that is, a1 · a2 · · · ak = 0 for all a1 , . . . , ak ∈ S. If S is nilpotent, then the minimal k ∈ N such that S k = {0} is called the nilpotency degree or nilpotency class of S and is denoted by nd(S). Remark 8.1.1 Note that here we slightly abuse the notation S 1 . In the present 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 adjoint identity 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 is nilpotent. 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 a1 a2 · · · ak = 0. For i = 1, . . . , k set bi = a1 a2 · · · ai and note that bi = 0. Assume that bi = bj for some i < j and let x = ai+1 ai+2 · · · aj . Then for all m ∈ N we have bi xm = (bi x)xm−1 = (bj )xm−1 = bi xm−1 = · · · = bi .

(8.1)

Hence (8.1) implies that xm = 0 for all m. This contradicts (b). Hence all elements bi , i = 1, . . . , k, are different and nonzero and thus k < |S|. In particular, S |S| = {0} and thus S is nilpotent. This proves the implication (b)⇒(a). O. Ganyushkin, V. Mazorchuk, Classical Finite Transformation Semigroups, Algebra and Applications 9, DOI: 10.1007/978-1-84800-281-4 8, c Springer-Verlag London Limited 2009 

131

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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 , +) of all positive rational numbers with respect to the addition modulo the ideal, consisting of all rational number greater than 1. Show that each element of S is nilpotent and that S is not nilpotent. The aim of the present chapter is to study nilpotent subsemigroups of the semigroups Tn , PT n , and IS n . Note that only PT n , and IS n contain a zero element, so the notion of nilpotent elements and nilpotent subsemigroups is really natural only for these two semigroups. In this natural case, one just considers those nilpotent subsemigroups of PT n or IS n , which share the zero element with the original semigroup. Of course one can also consider nilpotent subsemigroups in which the zero element is different from the zero element of the original semigroup. In such interpretation, it also makes sense to study nilpotent subsemigroups of Tn . However, it turns out that this study for Tn , PT n , and IS n can be basically reduced to the study of nilpotent subsemigroups of PT n or IS n with the global zero element. We shall explain this in the Addenda. Until then we restrict our attention to the semigroups PT n and IS n and consider only those nilpotent subsemigroups of 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 of all nilpotent subsemigroups of S with the zero element 0. The set Nil(S) is nonempty (it contains the nilpotent subsemigroup {0}) and is partially ordered with respect to inclusions. The subsemigroup {0} is the minimum element of Nil(S). If the semigroup S itself is nilpotent, then S is the maximum 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 maximal nilpotent subsemigroups of S. For k ∈ N denote also by Nilk (S) the subset of Nil(S) consisting of all nilpotent subsemigroups of nilpotency degree k. The set Nil(S) decomposes   into 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 certainly empty for all k big enough if S is finite. If it is not empty, then it inherits a partial order with respect to inclusions from Nil(S). In general, the set Nilk (S) may have many minimal and many maximal elements. Our main goal for this chapter is to classify all maximal elements in all Nilk (S), where S = PT n or IS n . The combinatorial tool we shall use for this 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 IS n → PT n induces inclusions Nil(IS n ) → Nil(PT n ) and Nilk (IS n ) → Nilk (PT n ) for all k ∈ N. Hence we can identify Nil(IS n ) and Nilk (IS n ) 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 ∈ T such 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 that α(x) = x. But this means that αk (x) = x for all k ∈ N. Hence α is not nilpotent, contradicting the nilpotency of T . Hence τT is antireflexive and thus 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(IS n ) to On . For each τ ∈ On define the subsemigroups Nτ ∈ PT n and Nτ ∈ IS n as follows: Nτ = {α ∈ PT n : α(x)τ xfor allx ∈ dom(α)},

Lemma 8.1.6

Nτ = Nτ ∩ IS n .

(i) Nτ ∈ Nil(PT n ).

(ii) Nτ ∈ Nil(IS n ). Proof. The statement (ii) follows from (i). To prove (i) we consider any αi ∈ Nτ , i = 1, . . . , n. Let α = α1 α2 · · · αn and assume that x ∈ dom(α). For i = 1, . . . , n set βi = αi αi+1 · · · αn . Then x ∈ dom(βi ) for all i = 1, . . . , n and we set xi = βi (x) for all such i. Set xn+1 = x. Then we have xi = αi (xi+1 ) for all i = 1, . . . , n. Hence xi τ xi+1 for all i = 1, . . . , n. As τ is an antireflexive partial order, we get that all elements x1 , x2 , . . . , xn+1 must be different, contradicting |N| = n. Hence dom(α) = ∅ and thus α = 0. This implies that Nτn = {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 → Nτ from On to Nil(IS n ). The crucial properties of the 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 Nτ ⊂ Nσ . (iii) For any σ ∈ On we have τNσ = σ and τNσ = σ. Proof. The statements (i) and (ii) are obvious from the definitions, so we have to prove only the statement (iii). Let x, y ∈ N be such that xσy. Consider the element α ∈ IS n of rank one, defined as follows: dom(α) = {y} and α(y) = x. Then α ∈ Nσ ⊂ Nσ . From the definition it follows that yτNσ x and hence σ ⊂ τNσ and σ ⊂ τNσ .

134

CHAPTER 8. NILPOTENT SUBSEMIGROUPS

Conversely, let x, y ∈ N be such that xτNσ y (for Nσ the arguments are similar). Then there exists α ∈ Nσ such that α(y) = x. But the latter implies 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 elements x1 , . . . , xk ∈ N such that xi τ xi+1 for all i = 1, . . . , k − 1 is called the height of τ . Thus any linear order on N has height n, while the empty partial (k) order on N has height 1. For k ∈ N we denote by On the subset of On consisting of all partial orders of height k. Obviously, we have the following decomposition into a disjoint union of subsets: On =

n 

O(k) n .

k=1

Lemma 8.1.8

(k)

(i) If T ∈ Nilk (PT n ), then τT ∈ On .

(ii) If τ ∈ On , then Nτ ∈ Nilk (PT n ) and Nτ ∈ Nilk (IS n ). (k)

Proof. Let α1 , . . . , αk−1 ∈ T be such that α = α1 α2 · · · αk−1 = 0. Fix some x ∈ dom(α) and for i = 1, . . . , k − 1 set xi = αi αi+1 · · · αk−1 (x). Then under the convention xk = x we have xi τT xi+1 for all i = 1, . . . , k − 1 and hence the nilpotency degree of T does not exceed the height of τT . Let xi , i = 1, . . . , m, be elements of N such that xi τT 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 = α1 α2 · · · 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 of Proposition 8.1.7(iii). Corollary 8.1.9 Nil(PT n ) =

n  k=1

Nilk (PT n ),

Nil(IS n ) =

n 

Nilk (IS n ).

k=1

Proof. Follows from Lemma 8.1.8(i) since, obviously, the maximal possible height of a partial order on N is n.

8.2

Classification of Maximal Nilpotent Subsemigroups

Let k ∈ N, 1 ≤ k ≤ n, and m = (M1 , M2 , . . . , Mk ) be an ordered partition of N 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 τm on N in the following way: xτm y

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 the partial order τm is as follows: 5 >TTTTT

6> jjj 7 >> TTTT >> jjjj j TTTT j >> > j T jjj> >> jjjTjTTTTT>T> jjjj 3 NNN 4 NNN ppppp NpNpNp ppp NNNNN ppp 1

Proposition 8.2.2 (k) in On .

2

(i) For m as above the order τm is a maximal element (k)

(ii) Every maximal element in On has the form τm for some m as above. (iii) If m = n, then τm = τn . (k)

Proof. First we note that τm ∈ On . Indeed, if x1 , . . . , xm are such that xi τm xi+1 for all i = 1, . . . , m − 1, then all xi ’s belong to different Mj ’s and hence m ≤ k. At the same time, taking some xi ∈ Mi , i = 1, . . . , k, we have xi τm xi+1 for all i = 1, . . . , k − 1. Hence the height of τm equals k. (k) Next we claim that τm is a maximal element of On . 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 zs σzs+1 for all s = 1, . . . , i+k−j. As j ≤ i, we have i+k−j+1 ≥ k+1 (k) and hence the height of σ is at least k + 1. Thus σ ∈ On . This shows that (k) τm is a maximal element of On and thus proves (i). (k) Now to prove (ii) we have to show that each σ ∈ On is contained in (k) some τm . Let σ ∈ On . 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 with respect to σ, M3 as the set of all minimal elements of N\(M1 ∪ M2 ) with respect to σ and so on. (k) We claim that Mk = ∅. Indeed, as σ ∈ On , there exist xi , i = 1, . . . , k, such that xi σxi+1 , i = 1, . . . , k − 1. Assume xi ∈ Mf (i) , i = 1, . . . , k. Then f (i) < f (j) if i < j and hence f (k) ≥ k. At the same time Mf (k) = ∅ and thus Mk = ∅.

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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 xi σxi+1 for all i = 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 in N\(M1 ∪ · · · ∪ Mi−1 ), which contradicts the definition of Mi . Hence for all x 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 Nτm are maximal elements in Nilk (PT n ) and Nilk (IS n ), respectively. (ii) Each maximal element of Nilk (PT n ) and Nilk (IS n ), has, respectively, the form Nτm and Nτm for some m as above. (iii) If m = n, then Nτm = Nτn and Nτm = Nτn . 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 (k)

(k)

τm is a maximal element in On by Proposition 8.2.2(i) and τT ∈ On by Lemma 8.1.8(i), we get that τm = τT . Hence T ⊂ Nτm by the definition of Nτm . This implies T = Nτm and proves (i) in the case of PT n . In the case of IS n the proof is just the same. (k) Let now T ∈ Nilk (PT n ) be a maximal element. Then τT ∈ On and hence τT ⊂ τm for some m as above by Proposition 8.2.2(ii). Now Proposition 8.1.7(ii) implies that T ⊂ Nτm . Moreover, Nτm ∈ Nilk (PT n ) by Lemma 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 IS n the proof is just the same. Finally, let m = n. Without loss of generality we may assume that in this case there exist x, y ∈ N such that xτm y holds while xτn y does not hold. Consider the element α ∈ IS n of rank one, which is uniquely defined by the property α(y) = x. Then α ∈ Nτm and α ∈ Nτm , while α ∈ Nτn and α ∈ Nτn . This completes the proof. Example 8.2.4 If M1 = {1}, M2 = {2, 3} and m = {M1 , M2 }, then the semigroup Nτm has the following elements:         1 2 3 1 2 3 1 2 3 1 2 3 . , , , ∅ 1 1 ∅ 1 ∅ ∅ ∅ 1 ∅ ∅ ∅ The first three of these elements form the semigroup Nτm .

8.2. CLASSIFICATION OF NILPOTENT SUBSEMIGROUPS

137

Corollary 8.2.5 If k < n, then each maximal element of Nilk (PT n ) and Nilk (IS n ) is contained in some maximal element of the sets Nilk+1 (PT n ) and Nilk+1 (IS n ), respectively. Proof. Let m = (M1 , M2 , . . . , Mk ). As k < n, by the Pigeonhole Principle there exists i ∈ {1, 2, . . . , k} such that |Mi | > 1. Let x ∈ Mi and consider n = (M1 , . . . , Mi−1 , {x}, Mi \{x}, Mi+1 , . . . , Mk ). (k+1)

Obviously τm ⊂ τn and n ∈ On . Hence, by Proposition 8.1.7(ii) we have Nτm ⊂ Nτn and Nτm ⊂ Nτn . As by Theorem 8.2.3(ii) each maximal element of Nilk (PT n ) and Nilk (IS n ) has, respectively, the form Nτm and Nτm for some 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(IS n ) and Niln (IS n )) coincide. (ii) Both PT n and IS n contain n! maximal nilpotent subsemigroups (that is, maximal elements in Nil(PT n ) and Nil(IS n ), respectively). (iii) If T1 and T2 are maximal nilpotent subsemigroups of PT n (or IS n ), then there exists ϕ ∈ Inn(PT n ) (resp. Inn(IS n )) such that ϕ(T1 ) = T2 . In particular, T1 ∼ = T2 . Proof. The statement (i) follows immediately from Corollaries 8.2.5 and 8.1.9. (n) As |N| = n, any element m = (M1 , . . . , Mn ) ∈ On is just a permutation of the elements of N. Conversely, each permutation defines a unique element (n) of the form m = (M1 , . . . , Mn ). Hence On 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 calculation shows that α−1 Nτn α = Nτm , which proves (iii) and thus completes the proof. Remark 8.2.7 Strictly speaking a “maximal nilpotent subsemigroup” is a maximal element of Nil(S). For example, in formulations of theorems we will 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 meaning maximal elements of some Nilk (S). (n)

Exercise 8.2.8 Prove that |On | = n!. Exercise 8.2.9 Let m = (M1 , . . . , Mk ) and n = (M1 , . . . , Mk ) be such that |Mi | = |Mi | for all i = 1, . . . , k. Show that Nτm ∼ = Nτn and Nτm ∼ = Nτn .

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Corollary 8.2.10 The number of maximal elements in both Nilk (PT n ) and Nilk (IS n ) equals   k  i k (k − i)n = k! · S(n, k). (−1) i i=0

Proof. By Theorem 8.2.3(ii), the number of maximal elements in both sets Nilk (PT n ) and Nilk (IS n ) equals the number X of ordered partitions N = M1 ∪· · ·∪Mk into a disjoint union of nonempty subsets. From the definition of S(n, k) we immediately get X = k!S(n, k). In contrast, each ordered partition M1 ∪· · ·∪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 obviously bijective and hence the statement follows by a standard application of the inclusion–exclusion formula.

8.3

Cardinalities of Maximal Nilpotent, Subsemigroups

Let m = (M1 , . . . , Mk ) be as in the previous section. From Exercise 8.2.9 it follows that |Nτm | and |Nτm | 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  (m0 + m1 + m2 + · · · + mi−1 )mi . i=1

(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 independent choices for α(x) for each such x. The statement (i) is now obtained applying the product rule. The statement (ii) is a special case of (i) as for a maximal nilpotent subsemigroup we have k = n and mi = 1 for all i. In the case of IS n the analogous question is much more interesting and nontrivial. To answer it we will need some preparation. We fix the following notation: X(m1 , . . . , mk ) = |Nτm |. If f (x) = am xm + am−1 xm−1 + · · · + a1 x + a0 is a polynomial with integer coefficients, we set f (B) = am Bm + am−1 Bm−1 + · · · + a1 B1 + a0 , where Bi denotes the i-th Bell number.

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139

Exercise 8.3.2 Let f (x) and g(x) be two polynomials with integer coefficients 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) and define the polynomial Fm1 ,...,mk (x) as follows: Fm1 ,...,mk (x) = [x]m1 [x]m2 · · · · · [x]mk . In the case of IS n the cardinalities of Nτm are given by the following quite amazing result: Theorem 8.3.3

(i) |Nτm | = Fm1 ,...,mk (B).

(ii) Each maximal nilpotent subsemigroup of IS n has cardinality Bn . Proof. We start with the statement (ii). By Corollary 8.2.6(i), the maximal nilpotent subsemigroups of IS n correspond to the case mi = 1 for all i. So, we may assume m = ({a1 }, . . . , {an }). Under such assumptions let α ∈ Nτm . Then the connected components of Γα form an unordered partition of N into a disjoint union of subsets. Conversely, let X1 , . . . , Xs be an unordered partition 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 as they appear in the sequence a1 , . . . , an . Let xj1 , xj2 , . . . , xjkj be the resulting sequence. Now define the element α as follows: s 

dom(α) =

{xj2 , xj3 , . . . , xjkj },

j=1

and α(xjt ) = xjt−1 for all j = 1, . . . , s and t = 2, . . . , kj . In other words, we define α as the element with the following graph Γα : x11 o

α

x12 o

α

... o

α

x1k1

x21 o

α

x22 o

α

... o

α

x2k2

.. . xs1 o

.. . α

xs2 o

.. . α

... o

.. . α

xsks

From the definition of Nτm we have α ∈ Nτm . Moreover, the connected components of Γα are exactly X1 , . . . , Xs . It is easy to see that α is the only element of Nτm such that the connected components of Γα are exactly

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X1 , . . . , Xs . Hence we obtain a bijection between the elements of Nτm and partitions of N into unordered disjoint unions of nonempty subsets. This means that |Nτm | = 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 Nτm into a disjoint union of two sets T1 and T2 as follows: T1 = {α ∈ Nτm : a ∈ α(Mi+1 )},

T2 = {α ∈ Nτm : a ∈ α(Mi+1 )}.

By definition, the set T1 coincides with Nτn , where n = (M1 , . . . , Mi−1 , Mi+1 ∪ {a}, Mi+2 , . . . , Mk ). Let Mi+1 = {a1 , . . . , ami+1 }. For j ∈ {1, 2, . . . , mi+1 } define T2j = {α ∈ T2 : α(aj ) = a}. m

Then T2 = T21 ∪ · · · ∪ T2 i+1 is a partition of T2 into a disjoint union of nonempty subsets. Fix now j ∈ {1, 2, . . . , mi+1 }. With every α ∈ T2j , we associate the element α ∈ Nτk , where k = (M1 , . . . , Mi−1 , Mi+1 , . . . , Mk ), in the following way:  α(y), y = aj ;  α (y) = α(a), y = aj . Obviously, the mapping α → α from T2j to Nτk is bijective. Hence |T2j | = |Nτk |. This gives X(m1 , . . . , mi−1 , 1, mi+1 , mi+2 , . . . , mk ) = = |Nτm | = |T1 | + |T2 | = |Nτn | + mi+1 |Nτk | = = 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 IS n

141

We prove the statement (i) by induction on the sum m of all mi ’s such that mi > 1. If m = 0, then mi = 1 for all i. In particular, F1,...,1 (x) = xn and F1,...,1 (B) = Bn = |Nτm | 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 IS n

In this section, we discover some combinatorial relations between different numbers, associated with the semigroup IS n , especially with nilpotent elements of IS n . Define the following numbers: In = the cardinality of IS n Nn = the total number of nilpotent elements in IS n Ln = the total number of chains in the chain decompositions of all elements of IS n Mn = the total number of chains in the chain decompositions of all nilpotent elements of IS n , Pn = the total number of fixed points of all elements of IS n

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Example 8.4.1 The semigroup IS 1 has 2 elements:     1 1 . , ∅ 1 For this semigroup, we have I1 = 2,

N1 = 1,

L1 = 1,

Example 8.4.2 The semigroup IS 2     1 2 1 2 , , 1 2 2 1    1 2 1 , ∅ 1 ∅

M1 = 1,

P1 = 1.

has 7 elements:     1 2 1 2 , , 1 ∅ 2 ∅    2 1 2 , . 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 8.4.3 Ln =



strank(α).

α∈IS n

Proof. Consider the sets A = {(α, c, x) : α ∈ IS n , c is a cycle of α, x is a point of c}, B = {(β, l) : β ∈ IS n , 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 as follows: g((β, [x, a, . . . , b])) = (α, (x, a, . . . , b), x), where α is obtained from β substituting the chain [x, a, . . . , b] with the cycle (x, a, . . . , b). Obviously f and g are inverse to each other and thus |A| = |B|. Theorem 8.4.4 Pn +

1 Ln = In . n

(8.4)

8.4. COMBINATORICS OF NILPOTENT ELEMENTS IN IS n

143

Proof. Rewrite the equality (8.4) in the following form: nPn + Ln = nIn .

(8.5)

Consider the following sets: A = {(α, x) : α ∈ IS n , x ∈ N}, B = {(β, l) : β ∈ IS n , l is a chain for the chain decomposition of β}, C = {(γ, y, z) : γ ∈ IS n , y is a fixed point of γ, z ∈ N}. The equality (8.5) is equivalent to the equality |C| + |B| = |A|. To prove the latter 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 induces a bijection A2 → B. Hence |A2 | = |B|. To prove |A1 | = |C|, we construct mutually inverse bijections f : A1 → C and g : C → A1 . Consider any element (α, x) ∈ A1 . If x is the starting point of some chain [x, a, . . . , b] of length at least 2 from the chain decomposition of α, we define f ((α, x)) = (γ, x, a), where γ is obtained from α substituting the chain [x, a, . . . , b] with the cycle (x) and the cycle (a, . . . , b). If x is the only point of the chain [x], we define f ((α, x)) = (γ, x, x), where γ is obtained from α substituting the chain [x] with the cycle (x). Finally, if x is contained in some chain [a, . . . , b, x, c, . . . , d] and is different from the starting point a of this chain, we define f ((α, x)) = (γ, x, b), where γ is obtained from α substituting 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), where α is obtained from γ substituting the cycle (y) with the chain [y]. If z is a point of some chain [a1 , . . . , as , z, b1 , . . . , bt ] in the chain decomposition of γ, we set g((γ, y, z)) = (α, y), where α is obtained from γ substituting 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 the cycles (y) and (z, a1 , . . . , as ) with the chain [y, z, a1 , . . . , as ]. Obviously, f and g are inverse to each other implying |A1 | = |C|, and the statement follows. Let Nn denote the set of all nilpotent elements of IS n . We have |Nn |=Nn . Proposition 8.4.5 In = Nn + Pn .

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Proof. The element α ∈ IS n can have some fixed points only in the case when α is not nilpotent. For every α ∈ Nn let Aα = stim(α) and Aα = N \Aα . Consider the set Mα = {β ∈ IS n : stim(β) = Aα and α|Aα = β|Aα }. Obviously, |Mα | = strank(α)! and either Mα = Mβ or Mα ∩ Mβ = ∅ for arbitrary α, β ∈ IS n . Hence the sets Mα form a partition of IS n \Nn into a disjoint 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 the function g : B → A as follows: g((β, z)) = (α, x, z), where α = β and x = z if 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 and g 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 total number Pn of all fixed points for all elements of IS n equals the total number of those elements of IS n which are not nilpotent. The claim follows. Theorem 8.4.7 (ii) In =

(i) Nn = n1 Ln .

1 n+1 Mn+1 .

Proof. Using Theorem 8.4.4 and Proposition 8.4.5 we have 1 Ln = In − Pn = In − (In − Nn ) = Nn . n 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|.

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145

Define the mapping f : A → B in the following way. Let (x, α) ∈ A and assume 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 substituting the permutational part with l. Define the mapping g : B → A, g : (β, l) → (x, α) in the following way: if l = [a1 , . . . , ak , ak+1 ], we set x = ak+1 and α is obtained from β substituting l with the permutational part 

ai1 a1

... ...

aik ak

 ,

where ai1 < ai2 < · · · < aik are elements a1 , . . . , ak , written with respect to the natural increasing order. Obviously, f and g are mutually inverse bijections, 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 ∈ IS n−1 , lis a chain ofβ}. We have A ∪ B = Nn , A ∩ B = ∅, |C| = Ln−1 . Hence the equality (i) would follow 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 the function f : A → IS n−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 element β is defined to be the following permutation: 

ai1 a1

... ...

aik ak

 ,

where ai1 < ai2 < · · · < aik are elements a1 , . . . , ak , written with respect to the natural increasing order.

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Define further the function g : IS n−1 → A as follows: For β ∈ IS n−1 let stim(β) = {a1 , . . . , ak }, where a1 < a2 < · · · < ak . Assume further that on stim(β) the element β acts as the following permutation: 

a1 ai1

... ...

ak aik

 .

We set g(β) = α, where ⎧ ⎪ β(y), ⎪ ⎪ ⎪ ⎨a , i1 α(y) = ⎪ a ij+1 , ⎪ ⎪ ⎪ ⎩∅,

y y y y

∈ stim(β); = n; = aij andj < k; = 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 since n ∈ im(α)). We set f (α) = (β, l), where β is obtained from α substituting the chain [a1 , . . . , ak , n, b1 , . . . , bm ] by the chain [a1 , . . . , ak ] and the permutation: 

bi1 b1

... ...

bim bm

 ,

where bi1 < bi2 < · · · < bim are elements b1 , . . . , bm , written with respect to the 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 assume that on stim(β) the element β acts as the following permutation: 

b1 bi1

... ...

bm bim

 .

We set g(β, l) = α, where α is obtained from β substituting l and the above permutation by the chain [a1 , . . . , ak , n, bi1 , . . . , bim ]. Obviously, f and g are mutually 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: (8.6) Pn = nIn−1 .

8.4. COMBINATORICS OF NILPOTENT ELEMENTS IN IS n

147

To prove (8.6) consider the following two sets: D = {(α, x) : α ∈ IS n , x ∈ N, α(x) = x}, E = {(β, y) : y ∈ N, β ∈ IS(N\