Nine Chapters on the Semigroup Art

Home , Aperiodic semigroup, Empty semigroup, Semigroup action

Alan J. Cain

Lecture notes for a tour through semigroups

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This work is licensed under the Creative Commons Attribu- tion–Non-Commercial–NoDerivs 4.0 International Licence. To view a copy of this licence, visit https://creativecommons.org/licenses/by-nc-nd/4.0/ Contents

Preface v Prerequisites vii ◆ Acknowledgements vii

Chapter 1 | Elementary semigroup theory 1 Basic concepts and examples 1 ◆ Generators and subsemigroups 8 ◆ Binary relations 11 ◆ Orders and lattices 15 ◆ Homomorphisms 19 ◆ Congruences and quotients 20 ◆ Generating equivalences and congruences 22 ◆ Subdirect products 28 ◆ Actions 29 ◆ Cayley graphs 30 ◆ Exercises 32 ◆ Notes 34

Chapter 2 | Free semigroups & presentations 36 Alphabets and words 36 ◆ Universal property 37 ◆ Proper- ties of free semigroups 40 ◆ Semigroup presentations 41 ◆ Exercises 50 ◆ Notes 52

Chapter 3 | Structure of semigroups 53 Green’s relations 53 ◆ Simple and 0-simple semigroups 56 ◆ D-class structure 58 ◆ Inverses and D-classes 61 ◆ Schützenberger groups 63 ◆ Exercises 66 ◆ Notes 68

Chapter 4 | Regular semigroups 70 Completely 0-simple semigroups 72 ◆ Ideals and completely 0-simple semigroups 78 ◆ Completely simple semigroups 79 ◆ Completely regular semigroups 81 ◆ Left and right groups 83 ◆ Homomorphisms 85 ◆ Exercises 86 ◆ Notes 88

Chapter 5 | Inverse semigroups 90 Equivalent characterizations 90 ◆ Vagner–Preston theorem 94 ◆ The natural partial order 97 ◆ Clifford semigroups 99 ◆ Free inverse semigroups 103 ◆ Exercises 113 ◆ Notes 116

Chapter 6 | Commutative semigroups 117 Cancellative commutative semigroups 117 ◆ Free commutative semigroups 119 ◆ Rédei’s theorem 121 ◆ Exercises 124 ◆ Notes 125

• iii Chapter 7 | Finite semigroups 126 Green’s relations and ideals 126 ◆ Semidirect and wreath products 128 ◆ Division 130 ◆ Krohn–Rhodes decomposition theorem 135 ◆ Exercises 142 ◆ Notes 143

Chapter 8 | Varieties & pseudovarieties 144 Varieties 144 ◆ Pseudovarieties 152 ◆ Pseudovarieties of semigroups and monoids 154 ◆ Free objects for pseudovarieties 156 ◆ Projective limits 157 ◆ Pro-V semigroups 159 ◆ Pseudoidentities 162 ◆ Semidirect products of pseudovarieties 167 ◆ Exercises 168 ◆ Notes 169

Chapter 9 | Automata & finite semigroups 170 Finite automata and rational languages 170 ◆ Syntactic semigroups and monoids 179 ◆ Eilenberg correspondence 183 ◆ Schützenberger’s theorem 190 ◆ Exercises 196 ◆ Notes 197

Solutions to exercises 198

Bibliography 246

Index 250

List of Tables

Table 8.1 Varieties of semigroups 151 Table 8.2 Varieties of monoids 151 Table 8.3 Varieties of semigroups with a unary operation −1 152 Table 8.4 S-pseudovarieties of semigroups 164 Table 8.5 M-pseudovarieties of monoids 164

Table 9.3 Varieties of rational ∗-languages 188 Table 9.4 Varieties of rational +-languages 188

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• iv Preface

A preface is frequently a superior composition to the work itself ‘ — Isaac D’Israeli,’ ‘Prefaces’. In: Curiosities of Literature.

• This course is a tour through selected areas of semigroup theory. There are essentially three parts: ◆ Chapters 1–3 study general semigroups, including presentations for semigroups and basic structure theory. ◆ Chapters 4–6 examine special classes: namely regular, inverse, and commutative semigroups. ◆ Chapters 7–9 study finite semigroups, their classification using pseudovarieties, and connections with the theory of automata and regular languages. The course is broad rather than deep. It is not intended to be comprehensive: it does not try to study (for instance) structure theory as deeply as Howie, Fundamentals of Semigroup Theory, pseudovarieties as deeply as Almeida, Finite Semigroups and Universal Algebra, or languages as deeply as Pin, Varieties of Formal Languages; rather, it samples highlights from each area. It should be emphasized that there is very little that is original in this course. It is heavily based on the treatments in these and other standard textbooks, as the bibliographic notes in each chapter make clear. The main novelty is in the selection and arrangement of material, the slightly slower pace, and the general policy of avoiding leaving proofs to the reader when the corresponding results are required for later proofs. Figure P.1 shows the dependencies between the chapters. At the end of each chapter, there are a number of exercises, intended to reinforce concepts introduced in the chapter, and also to explore some related topics that are not covered in the main text. The most important exercises are marked with a star ✴ . Solutions are supplied for all exercises. At the end of each chapter are bibliographic notes, which give sources and suggestions for further reading. Warnings against potential misunderstandings are marked (like this) with a ‘dangerous bend’ symbol, as per Bourbaki or Knuth. Important observations that are not potential misunderstandings per se are marked with an ‘exclamation’ symbol (like this). This course was originally delivered to master’s students at the Uni-

• v Chapter 1 Elementary semigroup theory

Chapter 2 Free semigroups & presentations

Chapter 3 Structure of semigroups

Chapter 4 Regular semigroups

Chapter 5 Inverse semigroups

Chapter 6 Commutative semigroups

Chapter 7 Finite semigroups

Chapter 8 FIGURE P.1 Varieties & pseudovarieties Chart of the dependencies between the chapters. Dotted arrows indicate that the Chapter 9 dependency is only in the Automata & finite semigroups exercises, not in the main text. versities of Porto and Santiago de Compostella. The course was covered during 56 hours of classes, which included lectures and discussions of the exercises. Revisions have increased the length of the notes, and about 70 hours of class time would now be required to cover them fully. These notes were heavily revised in 2013–15. Most of the main text is now stable, but Chapter 8 will be further revised, and further exercises will be added. At present, the index is limited to names and ‘named results’ only. There may be minor typesetting problems that arise from the ‘in-development’ status of the LuaLaTEX software and many of the required packages. The author welcomes any corrections, observations, or constructive criticisms; please send them to the email address on the copyright page.

Preface • vi Prerequisites There are few formal prerequisites: general mathematical maturity is the main one. An understanding of the most basic concepts from elementary group theory is assumed, such as the definition of groups, cosets, and factor groups. Some knowledge of linear algebra will help with understanding certain examples, but is not vital. For Chapters 1 and 5, knowledge of the basic definitions of graph theory is assumed. Some basic topology is necessary to appreciate part of Chapter 8 fully (although most of the chapter can be understood without it, and the relevant sections can simply be skipped), and some background in universal algebra is useful, but not essential. For Chapter 9, some experience with formal language theory and automata is useful, but again not essential.

Acknowledgements Attila Egri-Nagy, Darij Grinberg, Akihiko Koga, and Guil- herme Rito made valuable suggestions and indicated various errors. Some exercises were suggested by Victor Maltcev. Typos were pointed out by Nick Ham, Samuel Herman, José Manuel dos Santos dos Santos, and Alexandre Trocado. Many improvements are due to the students who took the first version of this course: Miguel Couto, Xabier García, and Jorge Soares. The imperfections that remain are my responsibility. The title alludes to 九章算術 (Jiuzhāngˇ Suànshù), Nine Chapters on the Mathematical Art. A. J. C. •

Acknowledgements • vii Elementary semigroup theory 1

I use the word “elementary” in the sense ‘ in which professional mathematicians use it — G.H. Hardy, A Mathematician’s’ Apology, § 21.

• A binary operation ∘ on a set 푆 is a map ∘ ∶ 푆 × 푆 → 푆. Binary operation This operation is associative if 푥 ∘ (푦 ∘ 푧) = (푥 ∘ 푦) ∘ 푧 for all elements 푥, 푦, 푧 ∈ 푆.A semigroup is a non-empty set equipped with an associative Semigroup binary operation. Semigroups are therefore one of the most basic types of algebraic structure. We could weaken the definition further by removing the associativity condition and requiring only a binary operation on a set. A structure that satisfies this weaker condition is called a magma or groupoid. (These ‘groupoids’ are different from the category-theoretic notion of groupoid.) On the other hand, we can strengthen the definition by requiring an identity and inverses. Structures satisfying this stronger condition are of course groups. However, there are many more semigroups than groups. For instance, there are 5 essentially different groups with 8 elements (the cyclic group 퐶8, the direct products 퐶4 × 퐶2 and 퐶2 × 퐶2 × 퐶2, the dihedral group 퐷4, and the quaternion group 푄8), but there are 3 684 030 417 different (non-isomorphic) semigroups with 8 elements. Some authors define a semigroup as a (possibly empty) set equipped with ‘Empty semigroup’ an associative binary operation. That is, the empty set forms the ‘empty semigroup’. This has advantages from a category-theoretic viewpoint. Note, however, that other definitions must be adjusted if a semigroup can be empty. In these notes, semigroups are required to be non-empty.

Basic concepts and examples Throughout this chapter, 푆 will denote a semigroup with operation ∘. Formally, we write (푆, ∘) to indicate that we are considering the set 푆 with the operation ∘, but we will only do this when we need to distinguish a particular operation. Unless we need to distinguish between

• 1 different operations, we will often write 푥푦 instead of 푥∘푦 (where 푥, 푦 ∈ 푆) and we will call the operation multiplication and the element 푥푦 (i.e. the Multiplication, product result of applying the operation to 푥 and 푦) the product of the elements 푥 and 푦. In order to compute a product like 푥푦푧푡 (or, equivalently, 푥 ∘ 푦 ∘ 푧 ∘ 푡), where 푥, 푦, 푧, 푡 ∈ 푆, we have to insert balanced pairs of brackets into the product to show in what order we perform the multiplications. We might insert brackets in any of the following five ways:

((푥푦)푧)푡, (푥(푦푧))푡, (푥푦)(푧푡), 푥((푦푧)푡), 푥(푦(푧푡)).

The following result shows that the choice of how to insert balanced pairs of brackets is unimportant:

P r o p o s i t i o n 1 . 1. Let 푠1, … , 푠푛 ∈ 푆. Every way of inserting balanced Associativity pairs of brackets into the product 푠1푠2 ⋯ 푠푛 gives the same result. Proof of 1.1. We will prove that any insertion of brackets into the product gives the same result as 푠1(푠2(푠3 ⋯ 푠푛) ⋯). We proceed by induction on 푛. For 푛 = 1, the result is trivially true, for there is only one way to insert balanced pairs of brackets into the product 푠1. This is the base case of the induction. So assume that the result holds for all 푛 < 푘; we aim to show it is true for 푛 = 푘. Take some bracketing of the product 푠1푠2 ⋯ 푠푘 and let 푡 be the result. This bracketing is a product of some bracketing of 푠1 ⋯ 푠ℓ and some bracketing of 푠ℓ+1 ⋯ 푠푘, for some ℓ with 1 ⩽ ℓ < 푘. Now consider two cases: ◆ Suppose ℓ = 1. By the assumption, the result of inserting brackets into 푠ℓ+1 ⋯ 푠푘 = 푠2 ⋯ 푠푘 is equal to 푠2(푠3(⋯ 푠푘) ⋯). Thus 푡 = 푠1(푠2(푠3(⋯ 푠푘) ⋯)), which is the result with 푛 = 푘. ◆ Suppose ℓ > 1. By the assumption, the result of the bracketing of 푠1 ⋯ 푠ℓ is 푠1(푠2(⋯ 푠ℓ) ⋯) and the result of the bracketing of 푠ℓ+1 ⋯ 푠푘 is 푠ℓ+1(푠ℓ+2(⋯ 푠푘) ⋯). Thus

푡 = (푠1(푠2(⋯ 푠ℓ) ⋯))(푠ℓ+1(푠ℓ+2(⋯ 푠푘) ⋯))

= 푠1(((푠2(⋯ 푠ℓ) ⋯))(푠ℓ+1(푠ℓ+2(⋯ 푠푘) ⋯))) [by associativity]

= 푠1(푠2(푠3 ⋯ 푠푘) ⋯), [by assumption with 푛 = 푘 − 1] which is the result with 푛 = 푘. Hence, by induction, the result holds for all 푛. 1.1 Thus, by Proposition 1.1, there is no ambiguity in writing a product 푠1푠2 ⋯ 푠푛 (where each 푠푖 ∈ 푆): the product is the same regardless of how we insert the brackets. Any group is also a semigroup. The most familiar example of a semigroup that is not a group is the set of natural numbers ℕ = {1, 2, 3, …}

Basic concepts and examples • 2 under the operation of addition. This is not a group since it does not contain inverses. Note that, for our purposes, the set of natural numbers ℕ = {1, 2, 3, …} does not include 0. Let 푒 be an element of 푆. If 푒푥 = 푥 for all 푥 ∈ 푆, the element 푒 is a Identity, monoid left identity. If 푥푒 = 푥 for all 푥 ∈ 푆, the element 푒 is a right identity. If 푒푥 = 푥푒 = 푥 for all 푥 ∈ 푆, then 푒 is a two-sided identity or simply an identity. A semigroup that contains an identity is called a monoid. Let 푧 be an element of 푆. If 푧푥 = 푧 for all 푥 ∈ 푆, the element 푧 is a left Zero zero. If 푥푧 = 푧 for all 푥 ∈ 푆, the element 푧 is a right zero. If 푧푥 = 푥푧 = 푧 for all 푥 ∈ 푆, then 푧 is a two-sided zero or simply a zero.

E x a m p l e 1 . 2. Let us give some examples of semigroups: a) The integers ℤ form a semigroup under two different operations: addition + and multiplication ⋅ . The semigroup (ℤ, +) is a monoid with identity 0; but in (ℤ, ⋅ ), the element 0 is a zero. b) The trivial semigroup contains only one element 푒, with multiplication Trivial semigroup obviously defined by 푒푒 = 푒. Since 푒 is (trivially) an identity, this semigroup is also called the trivial monoid. c) A null semigroup is a semigroup with a zero 푧 in which the product Null semigroup of any two elements is 푧. It is easy to see that this multiplication is associative. Notice that we can define a null semigroup on any non- empty set by choosing some element 푧 and defining all products to be 푧. d) If every element of 푆 is a left zero (that is, 푥푦 = 푥 for all 푥, 푦 ∈ 푆), Right/left zero semigroup then 푆 is a left zero semigroup. If every element of 푆 is a right zero (that is, 푥푦 = 푦 for all 푥, 푦 ∈ 푆), then 푆 is a right zero semigroup. We can define a left zero semigroup on any non-empty set 푋 by defining the multiplication 푥푦 = 푥 for all 푥, 푦 ∈ 푋; it is easy to see that this multiplication is associative. Similarly, we can define a right zero semigroup on any non-empty set 푋 by defining the multiplication 푥푦 = 푦 for all 푥, 푦 ∈ 푋. e) Any ring is a semigroup under multiplication.

P r o p o s i t i o n 1 . 3. If 푒 is a left identity of 푆 and 푒′ is a right identity Uniqueness of an identity of 푆 then 푒 = 푒′. Consequently, a semigroup contains at most one identity.

Proof of 1.3. Since 푒 is a left identity 푒푒′ = 푒′. Since 푒′ is a right identity, 푒 = 푒푒′. Hence 푒 = 푒푒′ = 푒′. 1.3

P r o p o s i t i o n 1 . 4. If 푧 is a left zero of 푆 and 푧′ is a right zero of 푆 then Uniqueness of a zero 푧 = 푧′. Consequently, a semigroup contains at most one zero.

Proof of 1.4. Since 푧 is a left zero, 푧푧′ = 푧. Since 푧′ is a right zero, 푧푧′ = 푧′. Hence 푧 = 푧푧′ = 푧′. 1.4

Basic concepts and examples • 3 It therefore makes sense to use the special notations 0 and 1 for the unique zero and identity of a semigroup. If we need to specify the zero or identity of a particular semigroup 푆, we will use 0푆 and 1푆. Let 1 be a new element not in the semigroup 푆. Extend the multiplic- Adjoining an identity or zero ation on 푆 to 푆 ∪ {1} by 1푥 = 푥1 = 푥 for all 푥 ∈ 푆 and 11 = 1. It is easy to prove that this extended multiplication is associative. Then 푆 ∪ {1} is a monoid with identity 1. Similarly, let 0 be a new element not in 푆 and extend the multiplication on 푆 to 푆 ∪ {0} by 0푥 = 푥0 = 00 = 0 for all 푥 ∈ 푆. Again, this extended multiplication is associative. Then 푆 ∪ {0} is a semigroup with zero 0. For any semigroup 푆, define

푆 if 푆 has an identity, 푆1 = { 푆 ∪ {1} otherwise; 푆 if 푆 has a zero, 푆0 = { 푆 ∪ {0} otherwise.

The semigroups 푆1 and 푆0 are called, respectively, the monoid obtained by adjoining an identity to 푆 if necessary and the semigroup obtained by adjoining a zero to 푆 if necessary. Throughout these notes, maps are written on the right and composed Notation for maps left to right. To clarify: let 휑 ∶ 푋 → 푌 and 휓 ∶ 푌 → 푍 be maps. The result of applying 휑 to an element 푥 of 푋 is denoted 푥휑. The composition of 휑 and 휓 is denoted 휑 ∘ 휓 or simply 휑휓, and is a map from 푋 to 푍 with 푥(휑휓) = (푥휑)휓 for all 푥 ∈ 푋. For 푋′ ⊆ 푋, the restriction of the map 휑 to 푋′ is denoted 휑|푋′. Let X = { 푋푖 ∶ 푖 ∈ 퐼 } (where 퐼 is an index set) be a collection of Cartesian product sets. Informally, the cartesian product ∏푖∈퐼 푋푖 of the sets in X is the set of tuples with |퐼| components, where, for each 푖 ∈ 퐼, the 푖-th component is an element of 푋푖. More formally, the cartesian product ∏푖∈퐼 푋푖 is the set of maps 휎 from 퐼 to ⋃푖∈퐼 푋푖 such that 푖휎 ∈ 푋푖 for each 푖 ∈ 퐼. We think of the map 휎 as a tuple with 푖-th component 푖휎. We will use both map notation and (especially when the index set 퐼 is finite) tuple notation for elements of cartesian products. When 퐼 = {1, … , 푛} is finite, we write

푋1 × … × 푋푛 for ∏푖∈퐼 푋푖, and we say the cartesian product is finitary. Finitary cartesian product When the sets 푋푖 are all equal to a set 푋 (that is, when we consider the 퐼 cartesian product of |퐼| copies of the set 푋), we write 푋 for ∏푖∈퐼 푋푖. Let S = { 푆푖 ∶ 푖 ∈ 퐼 } (where 퐼 is an index set) be a collection of Direct product semigroups. The direct product of the semigroups in S is their cartesian product ∏푖∈퐼 푆푖 with componentwise multiplication: (푖)(푠푡) = (푖)푠(푖)푡, or, using tuple notation,

( … , 푠푖, … )( … , 푡푖, … ) = ( … , 푠푖푡푖, … ).

It is easy to prove that this componentwise multiplication is associative, and so the direct product is itself a semigroup.

Basic concepts and examples • 4 For 푥 ∈ 푆 and 푛 ∈ ℕ, define Exponent

푛 times 푥푛 = ⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞푥푥⋯ 푥 . (1.1)

Notice that, in general, 푥푛 is only defined for positive 푛. If 푆 is a monoid, 0 푛 define 푥 = 1푆. Any element 푥 , where 푛 ∈ ℕ ∪ {0} is a power of 푥; if Power, positive power 푛 > 0, it is a positive power of 푥. Notice that if 푆 is not a monoid, then every power is a positive power. As an immediate consequence of Proposition 1.1, 푥푚푥푛 = 푥푚+푛 for Exponent laws all 푥 ∈ 푆 and 푚, 푛 ∈ ℕ. If 푆 is a monoid, then 푥푚푥푛 = 푥푚+푛 for all 푥 ∈ 푆 and 푚, 푛 ∈ ℕ ∪ {0}. Let 푥 ∈ 푆 and consider the positive powers 푥, 푥2, 푥3,…. There are Periodic element, two possibilities: either all these positive powers of 푥 are distinct or there index, period is some 푘, ℓ ∈ ℕ with 푘 < ℓ such that 푥푘 = 푥ℓ. In the latter case, 푥 is said to be periodic; notice that in a finite semigroup, this latter case must hold. Choose ℓ as small as possible; then 푥ℓ is the first positive power of 푥 that is equal to some earlier positive power. Let 푚 = ℓ − 푘; then 푥푘 = 푥푘+푚. Repeatedly multiplying this equality by 푥푚, one sees that 푥푘 = 푥푘+푟푚 for all 푟 ∈ ℕ ∪ {0}. Let 푛 ∈ ℕ ∪ {0}. Then 푛 = 푚푟 + 푖 for some 푟 ∈ ℕ ∪ {0} and 푖 ∈ {0, … , 푚 − 1}, and so 푥푘+푛 = 푥푘+푚푟+푖 = 푥푘+푖. Therefore every power of 푥 after 푥푘 is equal to one of

푥푘, 푥푘+1, … , 푥푘+푚−1.

Thus, by the minimality of the choice of ℓ, there are 푘 + 푚 − 1 distinct positive powers of 푥, namely

푥, 푥2, … , 푥푘−1, 푥푘, 푥푘+1, … , 푥푘+푚−1.

We call 푘 the index of 푥 and 푚 the period of 푥.A periodic semigroup is Periodic semigroup one in which every element is periodic. Note that all finite semigroups are periodic. An element 푥 of 푆 is an idempotent if 푥2 = 푥. The set of idempotents Idempotent, 퐸(푆), of 푆 is denoted 퐸(푆). If every element of 푆 is an idempotent, then 푆 is semigroup of idempotents a semigroup of idempotents. For example, a right zero semigroup is a semigroup of idempotents. For any subsets 푋 and 푌 of 푆, define 푋푌 = { 푥푦 ∶ 푥 ∈ 푋, 푦 ∈ 푌 }. Write Product of subsets 푥푌 for {푥}푌 and 푋푦 for 푋{푦}. Since multiplication in 푆 is associative, so is this product of subsets: for subsets 푋, 푌, and 푍 of 푆, we have 푋(푌푍) = (푋푌)푍. By analogy with (1.1), for 푋 ⊆ 푆 and 푛 ∈ ℕ, define

푛 times 푋푛 = 푋푋⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞ ⋯ 푋 .

The semigroup 푆 is nilpotent if it contains a zero and there exists some Nilpotent semigroup, 푛 ∈ ℕ such that 푆푛 = {0}. The semigroup 푆 is a nilsemigroup if it contains nilsemigroup a zero and for every 푥 ∈ 푆, there exists some 푛 ∈ ℕ such that 푥푛 = 0.

Basic concepts and examples • 5 Note that this is incompatible with the definition of a ‘nilpotent group’. A non-trivial group is never nilpotent in this semigroup sense. The semigroup 푆 is left-cancellative if Cancellativity

(∀푥, 푦, 푧 ∈ 푆)(푧푥 = 푧푦 ⇒ 푥 = 푦); right-cancellative if

(∀푥, 푦, 푧 ∈ 푆)(푥푧 = 푦푧 ⇒ 푥 = 푦); and cancellative if it is both left- and right-cancellative. Note that a non- trivial semigroup with zero cannot be cancellative. The semigroup 푆 is commutative if 푥푦 = 푦푥 for all 푥, 푦 ∈ 푆. For Commutativity instance, (ℤ, +) and (ℤ, ⋅ ) are both commutative. A non-trivial left zero semigroup is not commutative. Let 푀 be a monoid. Let 푥 ∈ 푀. Suppose that there exists an element 푥′ Left and right inverse such that 푥푥′ = 1. Then 푥′ is a right inverse for 푥, and 푥 is right invertible. Similarly, suppose there exists an element 푥″ such that 푥″푥 = 1. Then 푥″ is a left inverse for 푥, and 푥 is left invertible. If 푥 is both left and right invertible, then 푥 is invertible.

P r o p o s i t i o n 1 . 5. Let 푀 be a monoid, and let 푥 ∈ 푀. Suppose 푥 Right and left is invertible and let 푥′ be a right inverse of 푥 and 푥″ a left inverse. Then inverses coincide 푥′ = 푥″.

Proposition 1.5 says that right and left inverses coincide when they both exist. The existence of one does not imply the existence of the other.

Proof of 1.5. Since 푥′ and 푥″ are, respectively, right and left inverses of 푥, we have 푥푥′ = 1 and 푥″푥 = 1. Hence 푥′ = 1푥′ = 푥″푥푥′ = 푥″1 = 푥″. 1.5

Thus if 푥 is an invertible element of a monoid 푀, denote the unique Group right and left inverse of 푥 by 푥−1. A monoid in which every element is invertible is of course a group. Let 푥 ∈ 푆. If there is an element 푦 ∈ 푆 such that 푥푦푥 = 푥, then the Regular element element 푥 is regular. Notice that in this case, 푥푦 and 푦푥 are idempotent, since (푥푦)2 = 푥푦푥푦 = (푥푦푥)푦 = 푥푦 and (푦푥)2 = 푦푥푦푥 = 푦(푥푦푥) = 푦푥. If every element of 푆 is regular, then 푆 is a regular semigroup. Regular semigroup An element 푥′ ∈ 푆 such that 푥 = 푥푥′푥 and 푥′푥푥′ = 푥′ is an inverse of Inverse 푥. Notice that this is entirely different from the notion of left/right inverses above. We will never use ‘inverse’ (on its own) to refer to a left or right inverse.

P r o p o s i t i o n 1 . 6. Let 푥 ∈ 푆. Then 푥 has an inverse if and only if 푥 is regular.

Basic concepts and examples • 6 Proof of 1.6. Obviously if 푥 has an inverse, then it is regular. So suppose 푥 is regular. Then there exists 푦 ∈ 푆 such that 푥푦푥 = 푥. Let 푥′ = 푦푥푦. Then 푥푥′푥 = 푥(푦푥푦)푥 = (푥푦푥)푦푥 = 푥푦푥 = 푥 and 푥′푥푥′ = (푦푥푦)푥(푦푥푦) = 푦(푥푦푥)푦푥푦 = 푦푥푦푥푦 = 푦(푥푦푥)푦 = 푦푥푦 = 푥′, so 푥′ is an inverse of 푥. 1.6

In the proof of Proposition 1.6, the element 푦 might not be an inverse of 푥: for example, let 푆 be a semigroup with a zero and let 푥 = 0 and 푦 ≠ 0. Then 푥푦푥 = 푥 but 푦푥푦 ≠ 푦. An element 푥 can have more than one inverse; see Example 1.7(e). The Set of inverses 푉(푥) set of inverses of 푥 is denoted 푉(푥). Notice also that a zero 0 of a semigroup has an inverse, namely 0 itself. In general, if 푒 ∈ 푆 is idempotent, then 푒3 = 푒2 = 푒 and so 푒 is an inverse of itself. In particular, every idempotent is regular. E x a m p l e 1 . 7. a) Let 푈 = {0, … , 푘} for some 푘 ⩾ 0. Define an operation △ on 푈 by 푚 △ 푛 = min{푚, 푛}. It is easy to see that △ is associative, and so (푈, △) is a semigroup. Notice that 0△푚 = 푚△0 = 0 and 푘 △ 푚 = 푚 △ 푘 = 푚 for all 푚 ∈ 푈. Hence 푈 has zero 0 and identity 푘. Furthermore, 푚 △ 푚 = 푚 for all 푚 ∈ 푈, so every element of 푈 is an idempotent. Finally, 푚 △ 푛 = 푛 △ 푚 for all 푚, 푛 ∈ 푈 and so 푈 is commutative. b) Similarly, define an associative operation △ on ℕ ∪ {0} = {0, 1, 2, …} by 푚 △ 푛 = min{푚, 푛}. Then (ℕ ∪ {0}, △) has a zero 0 but has no identity. It is commutative and all its elements are idempotents. c) Consider the set of all 2 × 2 integer matrices:

푎 푏 푀 (ℤ) = { [ ] ∶ 푎, 푏, 푐, 푑 ∈ ℤ }. 2 푐 푑

With the usual matrix multiplication, 푀2(ℤ) is a monoid with identity 1 0 0 0 [ ] and zero [ ]. It is easy to see that 푀 (ℤ) is not commut- 0 1 0 0 2 ative, and that not all of its elements are idempotent. Since 푀2(ℤ) contains a zero, it is not cancellative. d) Now let 푉 be the set of all 2 × 2 integer matrices with non-zero determinant:

푎 푏 푎 푏 푉 = { [ ] ∶ 푎, 푏, 푐, 푑 ∈ ℤ ∧ det [ ] ≠ 0 }. 푐 푑 푐 푑

Again, 푉 is a monoid. Let 푃, 푄, 푅 ∈ 푉. Suppose 푅푃 = 푅푄. Since −1 det 푅 ≠ 0, the matrix 푅 has a (left and right) inverse 푅 ∈ 푀2(ℚ). [Note that 푅−1 ∉ 푉 whenever det 푅 ≠ ±1, so 푉 is not a group.] So 푅−1푅푃 = 푅−1푅푄 and so 푃 = 푄. Hence 푉 is left-cancellative. Similarly, it is right-cancellative and therefore cancellative.

Basic concepts and examples • 7 e) Let 퐿 be a left zero semigroup and 푅 a right zero semigroup. Let Rectangular band 퐵 = 퐿 × 푅. The semigroup 퐵 is an |퐿| × |푅| rectangular band, or simply a rectangular band. For (ℓ1, 푟1), (ℓ2, 푟2) ∈ 퐵, we have

(ℓ1, 푟1)(ℓ2, 푟2) = (ℓ1ℓ2, 푟1푟2) = (ℓ1, 푟2),

since (in particular) ℓ1 is a left zero and 푟2 is a right zero. Thus every element of 퐵 is idempotent, since (ℓ, 푟)(ℓ, 푟) = (ℓ, 푟) for all (ℓ, 푟) ∈ 퐵. Furthermore, for any (ℓ1, 푟1), (ℓ2, 푟2) ∈ 퐵, we have

(ℓ1, 푟1)(ℓ2, 푟2)(ℓ1, 푟1) = (ℓ1ℓ2ℓ1, 푟1푟2푟1) = (ℓ1, 푟1)

(ℓ2, 푟2)(ℓ1, 푟1)(ℓ2, 푟2) = (ℓ2ℓ1ℓ2, 푟2푟1푟2) = (ℓ2, 푟2).

Hence (ℓ2, 푟2) is an inverse of (ℓ1, 푟1). Thus every element is an inverse of every element. (ℓ2, 푟2) The name ‘rectangular band’ comes from the following diagrammatic interpretation of the multiplication. The elements of the semigroup correspond to the cells of a grid whose rows are indexed by 퐿 and whose columns are indexed by 푅. So (ℓ1, 푟1) corresponds to (ℓ1, 푟1) (ℓ1, 푟2) the cell in row ℓ and column 푟 . In terms of cells, the product of two 1 1 FIGURE 1.1 elements is the cell in the row of the first multiplicand and the column Diagrammatic interpretation of of the second multiplicand; see Figure 1.1. [The reason for indexing multiplication in a rectangular rows by the first coordinate and columns by the second coordinate band. will become clear in Chapter 4.]

The opposite semigroup 푆opp of 푆 is the semigroup with the same set as Opposite semigroup 푆 but ‘reversed multiplication’.That is, for 푥, 푦 ∈ 푆, the product 푥푦 in 푆opp is equal to the product 푦푥 in 푆. It is easy to check that 푆opp is indeed a semigroup. Notice that if 푆 is commutative, then 푆opp and 푆 are the same semigroup.

Generators and subsemigroups

A non-empty subset 푇 of 푆 is a subsemigroup if it is closed Subsemigroup under multiplication; that is, if 푇푇 ⊆ 푇.A proper subsemigroup is any subsemigroup except 푆 itself. A submonoid is a subsemigroup that happens to be a monoid. A subgroup is a subsemigroup that happens to be a group. P r o p o s i t i o n 1 . 8. The set of invertible elements of a monoid forms a subgroup. Proof of 1.8. Let 푇 be the set of invertible elements of a monoid 푀. Note that 푇 is non-empty since 1 ∈ 푇. Let 푥, 푦 ∈ 푇. Then since 푥 and 푦 are invertible we have 푦−1푥−1푥푦 = 푦−1푦 = 1 and 푥푦푦−1푥−1 = 푥푥−1 = 1.

Generators and subsemigroups • 8 Hence 푥푦 is invertible and so lies in 푇. Hence 푇 is a subsemigroup of 푀. Furthermore, 1 ∈ 푇 is also an identity for 푇 and so 푇 is a submonoid of 푀. Finally, let 푥 ∈ 푇; then 푥푥−1 = 푥−1푥 = 1 and so 푥−1 is invertible and thus 푥−1 ∈ 푇. Hence 푇 is closed under taking inverses. Therefore 푇 is a subgroup of 푀. 1.8

The set of invertible elements of a monoid is called its group of units; Group of units Proposition 1.8 justifies this name. The following result is a useful characterization of subgroups of a semigroup:

L e m m a 1 . 9. Let 퐺 be a non-empty subset of 푆. Then 푔퐺 = 퐺푔 = 퐺 for all 푔 ∈ 퐺 if and only if 퐺 is a subgroup of 푆.

Proof of 1.9. Notice first that if 퐺 is a subgroup and 푔 ∈ 퐺, then 퐺 = 푔푔−1퐺 ⊆ 푔퐺 ⊆ 퐺, so 퐺 = 푔퐺. Similarly, 퐺 = 퐺푔. For the converse, suppose that 푔퐺 = 퐺푔 = 퐺 for all 푔 ∈ 퐺. For any 푔, ℎ ∈ 퐺, the product 푔ℎ lies in 푔퐺 = 퐺. Hence 퐺 is a subsemigroup. Let 푔 ∈ 퐺. Since 퐺 = 퐺푔, it follows that 푔 ∈ 퐺푔, and so there exists 푒 ∈ 퐺 such that 푔 = 푒푔. Let ℎ ∈ 퐺. Since 퐺 = 푔퐺, there exists 푥 ∈ 퐺 such that ℎ = 푔푥. Hence 푒ℎ = 푒푔푥 = 푔푥 = ℎ. Since ℎ ∈ 퐺 was arbitrary, 푒 is a left identity for 퐺. Similarly 퐺 contains a right identity 푓, and so 푒 = 푓 is an identity for 퐺 by Proposition 1.3. So 퐺 is a submonoid with identity 1퐺. Finally, since 1퐺 ∈ 푔퐺 = 퐺푔, the element 푔 is right and left invertible and its right and left inverses coincide by Proposition 1.5. Since 푔 ∈ 퐺 was arbitrary, 퐺 is a subgroup. 1.9

Let 푇 be a non-empty subset of 푆. The subset 푇 is a left ideal of 푆 if it is Ideal closed under left multiplication by any element of 푆; that is, if 푆푇 ⊆ 푇. It is a right ideal of 푆 if it is closed under right multiplication by any element of 푆; that is, if 푇푆 ⊆ 푇. It is a two-sided ideal, or simply an ideal, of 푆 if it is closed under both left and right multiplication by elements of 푆; that is, if 푆푇 ∪ 푇푆 ⊆ 푇. Every ideal, whether left, right, or two-sided, is a subsemigroup. For any 푥 ∈ 푆, define Principal ideal

퐿(푥) = 푆1푥 = {푥} ∪ 푆푥, 푅(푥) = 푥푆1 = {푥} ∪ 푥푆, 퐽(푥) = 푆1푥푆1 = {푥} ∪ 푥푆 ∪ 푆푥 ∪ 푆푥푆.

Then 퐿(푥), 푅(푥), and 퐽(푥) are, respectively, the principal left ideal generated by 푥, the principal right ideal generated by 푥, and the principal ideal generated by 푥. As their names imply, they are, respectively, a left ideal, a right ideal, and a (two-sided) ideal.

Generators and subsemigroups • 9 E x a m p l e 1 . 1 0. a) Consider the semigroup (ℕ, +). Let 푛 ∈ ℕ and let 퐼푛 = { 푚 ∈ ℕ ∶ 푚 ⩾ 푛 }. Then 퐼푛 is an ideal of ℕ; indeed, 퐼푛 = 퐿(푛) = 푅(푛) = 퐽(푛). b) Let 푆 be a right zero semigroup. Let 푇 be a non-empty subset of 푆. Then 푆푇 = 푇 since 푥푦 = 푦 for any 푥 ∈ 푆 and 푦 ∈ 푇. So 푇 is a left ideal of 푆. On the other hand, 푇푆 = 푆 and so 푇 is a right ideal if and only if 푇 = 푆. c) Let 퐺 be a group. Let 푇 be a non-empty subset of 퐺. For any 푥 ∈ 퐺 and 푦 ∈ 푇, we have 푥 = 푥푦−1푦 ∈ 퐺푦; hence 퐺푦 = 퐺. So 푇 is a left ideal if and only if 푇 = 퐺; similarly 푇 is a right ideal if and only if 푇 = 퐺. So the only left ideal or right ideal of 퐺 is 퐺 itself.

Let T = { 푇푖 ∶ 푖 ∈ 퐼 } be a collection of subsemigroups of 푆. It is Generating a subsemigroup easy to see that if their intersection ⋂ T = ⋂푖∈퐼 푇푖 is non-empty, it is also a subsemigroup. So let 푋 be a non-empty subset of 푆 and let T be the collection of subsemigroups of 푆 that contain 푋. The collection T has at least one member, namely the semigroup 푆 itself, and every subsemigroup in T contains 푋, so ⋂ T is non-empty and is thus a subsemigroup. Indeed, it is the smallest subsemigroup of 푆 that contains 푋. This subsemigroup, denoted ⟨푋⟩, is called the subsemigroup generated by 푋. If 푋 ⊆ 푆 is such that ⟨푋⟩ = 푆, then 푋 is a generating set for 푆 and Generating set 푋 generates 푆. If there is a finite generating set for 푆, then 푆 is said to be finitely generated.

P r o p o s i t i o n 1 . 1 1. Let 푋 be a non-empty subset of 푆. Then ⟨푋⟩ = { 푥1푥2 ⋯ 푥푛 ∶ 푛 ∈ ℕ, 푥푖 ∈ 푋 }.

Proof of 1.11. Let 푈 = { 푥1푥2 ⋯ 푥푛 ∶ 푛 ∈ ℕ, 푥푖 ∈ 푋 }. Then 푈 is closed under multiplication and so is a subsemigroup of 푆. Furthermore, 푋 ⊆ 푈. Hence 푈 must be one of the 푇푖 in T, and so ⟨푋⟩ ⊆ 푈. Since 푋 ⊆ ⟨푋⟩ and ⟨푋⟩ is closed under multiplication, 푈 ⊆ ⟨푋⟩. Therefore ⟨푋⟩ = 푈. 1.11

Suppose 푆 is generated by a single element 푥; that is, 푆 = ⟨{푥}⟩ (which Monogenic semigroup we abbreviate to 푆 = ⟨푥⟩). Then 푆 is a monogenic semigroup, and, by Proposition 1.11, 푆 = { 푥푛 ∶ 푛 ∈ ℕ }. If the element 푥 is periodic with index 푘 and period 푚, then 푆 = {푥, 푥2, … , 푥푘+푚−1}. Let

퐾 = {푥푘, 푥푘+1, … , 푥푘+푚−1}.

It is easy to see that 퐾 is an ideal of 푆.

P r o p o s i t i o n 1 . 1 2. The ideal 퐾 is a subgroup of 푆.

Proof of 1.12. Let 퐼 = {푘, 푘 + 1, … , 푘 + 푚 − 1}, so that 퐾 = { 푥푛 ∶ 푛 ∈ 퐼 }. Then 퐼 is a complete set of representatives for congruence classes of the integers modulo 푚. In particular there is some 푝 ∈ 퐼 such that 푝 ≡ 0 mod 푚; note that 푝 = 푟푚 for some 푟 ∈ ℕ. Let 푒 = 푥푝 = 푥푟푚. Then

Generators and subsemigroups • 10 푥푛푒 = 푥푛푥푟푚 = 푥푛+푟푚 = 푥푛 and similarly 푒푥푛 = 푥푛 for any 푛 ∈ 퐼; hence 푒 is an identity for 퐾. Now let 푛 ∈ 퐼. Choose 푞 ∈ 퐼 with 푞 ≡ −푛 mod 푚. Then 푞 + 푛 ≡ 0 mod 푚 and so 푞 + 푛 = 푠푚 for some 푠 ∈ ℕ. Hence 푥푞푥푛 = 푥푞+푛 = 푥푠푚. Since 푠푚 ⩾ 푘 and since 푟푚 is the unique multiple of 푚 in 퐼, it follows that 푠푚 = 푟푚 + 푡푚 for some 푡 ∈ ℕ ∪ {0}. Hence 푥푞푥푛 = 푥푟푚+푡푚 = 푥푟푚 = 푒, and similarly 푥푛푥푞 = 푒. Hence 푥푞 is a right and left inverse for 푥푛; since 푛 ∈ 퐼 was arbitrary, every element of 퐾 has an inverse in 퐾. 1.12

Note that 푥푚 may not be an identity for 푚. It is true that 푥푛푥푚 = 푥푚푥푛 = 푥푛, but if 푘 > 푚, then 푥푚 ∉ 퐾. Given a subset 푋 of a monoid 푀, we can also define the submonoid Generating a submonoid generated by 푋. Let T be the collection of submonoids of 푀 that contain 푋 ∪ {1푀}. The intersection of the submonoids in T is non-empty and thus a submonoid. This is the smallest submonoid of 푀 with identity 1푀 that contains 푋. This submonoid, denoted Mon⟨푋⟩, is called the submonoid generated by 푋. Reasoning similar to the proof of Proposition 1.11 yields the following result:

P r o p o s i t i o n 1 . 1 3. Let 푋 ⊆ 푀. Then Mon⟨푋⟩ = { 1푀푥1푥2 ⋯ 푥푛 ∶ 푛 ∈ ℕ ∪ {0}, 푥푖 ∈ 푋 }. 1.13

Essentially, when we generate a submonoid of a monoid, we always Monoid generator include the identity of the monoid. If 푋 ⊆ 푀 is such that Mon⟨푋⟩ = 푀, then 푋 is a monoid generating set for 푀 and 푋 generates 푀 as a monoid. Notice that if 푋 is a generating set for 푀, then 푋 is also a monoid Generating set and generating set; on the other hand, if 푋 is a monoid generating set for 푀, monoid generating set then 푋 ∪ {1푀} is a generating set for 푀. Thus 푀 is finitely generated if and only if there is a finite monoid generating set for 푀.

Binary relations Recall that a relation 휌 between a set 푋 and a set 푌 is simply a subset of 푋 × 푌, and 푥 휌 푦 is equivalent to (푥, 푦) ∈ 휌. The identity relation on 푋 is the relation Identity relation

id푋 = { (푥, 푥) ∶ 푥 ∈ 푋 }.

The converse 휌−1 of 휌 is the relation Converse of a relation

휌−1 = { (푦, 푥) ∶ (푥, 푦) ∈ 휌 }.

The converse relation 휌−1 is not in general a left or right inverse of 휌, even when 휌 is a map. Let 휌 be a relation between 푋 and 푌 and 휎 be a relation between 푌 Composition of relations

Binary relations • 11 and 푍. Define the composition of 휌 and 휎 to be

휌 ∘ 휎 = { (푥, 푧) ∈ 푋 × 푍 ∶ (∃푦 ∈ 푌)((푥 휌 푦) ∧ (푦 휎 푧)) }. (1.2)

Notice that 휌 ∘ 휎 is a relation between 푋 and 푍. Furthermore, notice that 휌 ∘ id푌 = 휌 and id푌 ∘ 휎 = 휎. For any 푥 ∈ 푋, let 푥휌 = { 푦 ∈ 푌 ∶ 푥 휌 푦 }. Then 휌 is a partial map Partial/full map from 푋 to 푌 if |푥휌| ⩽ 1 for all 푥 ∈ 푋. Furthermore, 휌 is a full map, or simply a map from 푋 to 푌 if |푥휌| = 1 for all 푥 ∈ 푋. Suppose 휌 is a partial map from 푋 to 푌. When 푥휌 is the empty set, we say that 푥휌 is undefined; when 푥휌 is the singleton set {푦}, we say that 푥휌 is defined and write 푥휌 = 푦 instead of 푥휌 = {푦}. The definition of a map given here, and the notation in the last paragraph, agree with the standard concept and notation of a map. Further- more, when 휌 and 휎 are maps, (1.2) simply defines the usual composition of maps. Thus we have recovered the usual notion of maps in a more general setting. For any partial map 휌 from 푋 to 푌, the domain of 휌 is the set Domain, image, preimage

dom 휌 = { 푥 ∈ 푋 ∶ (∃푦 ∈ 푌)((푥, 푦) ∈ 휌) }. (1.3)

That is, dom 휌 is the subset of 푋 on which 휌 is defined. If 휌 is a map, we have dom 휌 = 푋. The image of 휌 is the set

im 휌 = { 푦 ∈ 푌 ∶ (∃푥 ∈ 푋)((푥, 푦) ∈ 휌) }. (1.4)

The preimage under 휌 of 푌′ ⊆ 푌 is the set

푌′휌−1 = { 푥 ∈ 푋 ∶ (∃푦 ∈ 푌′)((푦, 푥) ∈ 휌−1)} = { 푥 ∈ 푋 ∶ (∃푦 ∈ 푌′)((푥, 푦) ∈ 휌) }.

We will be particularly interested in binary relations on 푋; that is, Binary relations, B푋 relations from 푋 to itself. Let B푋 denote the set of all binary relations on 푋. It is easy to show that ∘ is an associative operation on B푋 and so (B푋, ∘) is a semigroup, called the semigroup of binary relations on 푋. Furthermore, id푋 is an identity and so B푋 is a monoid. A partial map from 푋 to itself is called a partial transformation of Partial/full transformation 푋. A map from 푋 to itself is called a full transformation, or simply a transformation of 푋. The set of all partial transformations of 푋 is P푋; the P푋, T푋, S푋 set of all [full] transformations of 푋 is T푋. Finally, S푋 denotes the set of bijections on 푋. This is the well-known symmetric group on 푋. Clearly S푋 ⊆ T푋 ⊆ P푋 ⊆ B푋.

P r o p o s i t i o n 1 . 1 4. a) P푋 is a submonoid of B푋;

b) T푋 is a submonoid of P푋;

c) S푋 is a subgroup of T푋.

Binary relations • 12 Proof of 1.14. a) Let 휌, 휎 ∈ P푋 and suppose 푦, 푦′ ∈ 푥(휌 ∘ 휎). Then by the definition of ∘, there exist 푧, 푧′ ∈ 푋 such that (푥, 푧) ∈ 휌 and (푧, 푦) ∈ 휎, and (푥, 푧′) ∈ 휌 and (푧′, 푦′) ∈ 휎. Since 휌 ∈ P푋, we have |푥휌| ⩽ 1 and so 푧 = 푧′. Since 휎 ∈ P푋, we have |푧휎| ⩽ 1 and so 푦 = 푦′. Hence |푥(휌 ∘ 휎)| ⩽ 1 and so 휌 ∘ 휎 ∈ P푋.

b) Let 휌, 휎 ∈ T푋. Let 푥 ∈ 푋. Since 휌 ∈ T푋, we have |푥휌| = 1. So let 푧 = 푥휌. Since 휎 ∈ T푋, we have |푧휎| = 1. So 푥(휌 ∘ 휎) contains (푥, 푧휎) and so |푥(휌 ∘ 휎)| ⩾ 1. By part a), |푥(휌 ∘ 휎)| = 1. Therefore 휌 ∘ 휎 ∈ T푋. c) This is immediate because the composition of two bijections is a bijection. 1.14

In light of Proposition 1.14, T푋 is called the semigroup of transformations on 푋 and P푋 is called the semigroup of partial transformations on 푋. Any bijection 휌 ∈ S푋 can be denoted by the usual disjoint cycle Two-line notation for transformations notation from group theory. A partial (or full) transformation 휌 ∈ P푋 can be denoted using a 2 × |푋| matrix: the (1, 푥)-th entry is 푥 and the (2, 푥)-th entry is either 푥휌 (when 푥휌 is defined) or ∗ (indicating that 푥휌 is undefined). For example, if 푋 = {1, 2, 3} and 1휌 = 2, and 2휌 is undefined, and 3휌 = 1, then

1 2 3 휌 = ( ). 2 ∗ 1

E x a m p l e 1 . 1 5. Let 푋 = {1, 2}. Then

◆ S푋 consists of two elements: 1 2 1 2 id = ( ) and ( ); 푋 1 2 2 1

◆ T푋 consists of four elements: the two elements in S푋, and the transformations 1 2 1 2 ( ) and ( ); 1 1 2 2

◆ P푋 consists of nine elements: the four elements in T푋, and the partial transformations 1 2 1 2 1 2 1 2 1 2 ( ),( ),( ),( ), and ( ); 1 ∗ 2 ∗ ∗ 1 ∗ 2 ∗ ∗

◆ B푋 consists of all sixteen possible subsets of 푋 × 푋, including the empty set ∅ and 푋 × 푋 itself.

Let us illustrate how elements of the semigroups of partial and full transformations multiply:

Binary relations • 13 E x a m p l e 1 . 1 6. Let 푋 = {1, 2, 3}.

a) Let Multiplication in T푋

1 2 3 1 2 3 휌 = ( ) and 휎 = ( ) 3 1 1 2 2 3

be elements of T푋. Let us compute the product 휌휎. First, 휌 contains the pair (1, 3) and 휎 contains the pair (3, 3), so 휌휎 contains the pair (1, 3). Using our notation for partial and full maps, this says that 1휌 = 3 and 3휎 = 3, and thus 1휌휎 = 3휎 = 3. Similarly, 2휌휎 = 1휎 = 2 and 3휌휎 = 1휎 = 2. Hence

1 2 3 휌휎 = ( ). 3 2 2

b) Let Multiplication in P푋

1 2 3 1 2 3 휌 = ( ) and 휎 = ( ) 3 ∗ 2 1 ∗ 2

be elements of P푋. Let us compute the product 휌휎. First, 1휌휎 = 3휎 = 2; this part of the computation is just like the case of a full map. Next, 2휌 is undefined: that is, 휌 does not contain the pair (2, 푥) for any 푥 ∈ 푋. Hence 휌휎 cannot contain the pair (2, 푦) for any 푦 ∈ 푋. That is, 2휌휎 is undefined. Finally, 3휌 = 2, but 휎 does not contain the pair (2, 푥) for any 푥 ∈ 푋, and hence 휌휎 cannot contain the pair (3, 푥) for any 푥 ∈ 푋. That is, 3휌휎 is undefined. Hence

1 2 3 휌휎 = ( ). 2 ∗ ∗

[During this computation, it may be helpful to think of ‘∗’ as an additional element of 푋 that is mapped to itself by every partial transformation of P푋. Then one can think ‘휌 maps 2 to ∗ and 휎 maps ∗ to ∗, so 휌휎 maps 2 to ∗’ and ‘휌 maps 3 to 2 and 휎 maps 2 to ∗, so 휌휎 maps 3 to ∗’. Remember, however, that ∗ is not an element of 푋, but is simply a notational convenience to indicate where a partial transformation is undefined.]

There are several important properties that a binary relation may have: Reflexive, (anti-)symmetric, transitive a relation 휌 ∈ B푋 is

◆ reflexive if 푥 휌 푥 for all 푥 ∈ 푋, or, equivalently, if id푋 ⊆ 휌; ◆ symmetric if 푥 휌 푦 ⇒ 푦 휌 푥 for all 푥, 푦 ∈ 푋, or, equivalently, if 휌 = 휌−1; ◆ anti-symmetric if (푥 휌 푦) ∧ (푦 휌 푥) ⇒ 푥 = 푦 for all 푥, 푦 ∈ 푋, or, −1 equivalently, if 휌 ∩ 휌 ⊆ id푋;

Binary relations • 14 ◆ transitive if (푥 휌 푦) ∧ (푦 휌 푧) ⇒ 푥 휌 푧 for all 푥, 푦, 푧 ∈ 푋, or, equivalently, if 휌2 ⊆ 휌. Notice that ‘anti-symmetric’ is not the same as ‘not symmetric’: for example, the identity relation id푋 is both symmetric and anti-symmetric. An equivalence relation is a relation that is reflexive, symmetric, and Equivalence relation transitive. An equivalence relation on 푋 partitions the set 푋 into equivalence classes, each made up of related elements.

Orders and lattices

Let 휌 ∈ B푋. The binary relation 휌 is a partial order if it is Partial order reflexive, anti-symmetric, and transitive. We normally use symbols like ⩽, ≼, and ⊑ for partial orders. We write 푥 < 푦 to mean that 푥 ⩽ 푦 and 푥 ≠ 푦; the obvious analogies apply for symbols like ≺ and ⊏.A partially ordered set or poset is a set 푋 equipped with a partial order ⩽, formally denoted (푋, ⩽). If (푋, ⩽) is a partially ordered set and 푌 is a subset of 푋, then 푌 ‘inherits’ the partial order ⩽ from 푋. That is, the restriction of the relation ⩽ to 푌 (that is, ⩽ ∩ (푌 × 푌)) is a partial order on 푌, and so 푌 is also a partially ordered set. We use the same notation for the original partial order on 푋 and for its restriction to 푌. A Hasse diagram of a partial order ⩽ on a set 푋 is a diagrammatic Hasse diagram representation of ⩽. Every element of 푋 is represented by a point on the plane, arranged so that 푥 appears below 푦 whenever 푥 < 푦. If 푥 < 푦 and there is no element 푧 such that 푥 < 푧 < 푦, then a line segment is drawn between 푥 and 푦. Suppose ⩽ is a partial order on 푋. Two elements 푥, 푦 ∈ 푋 are compar- Total order able if 푥 ⩽ 푦 or 푦 ⩽ 푥. The partial order ⩽ is a total order, or simply an order, if all pairs of elements of 푋 are comparable. Suppose ⩽ is a partial order on 푋.A chain is a subset 푌 of 푋 in which Chain, antichain every pair of elements are comparable. An antichain is a subset 푌 of 푋 in which no pair of distinct elements is comparable. Note that it is possible for 푋 itself to be a chain or an antichain. If 푋 is a chain (respectively, antichain), then any subset of 푋 is also a chain (respectively, antichain). E x a m p l e 1 . 1 7. a) For example, the relation ⩽ on the integers ℤ is a partial order: it is reflexive, since 푚 ⩽ 푚 for all 푚; it is anti- symmetric, since 푚 ⩽ 푛 and 푛 ⩽ 푚 imply 푚 = 푛; and it is transitive, {1, 2, 3} since 푚 ⩽ 푛 and 푛 ⩽ 푝 imply 푚 ⩽ 푝. {1, 2} {1, 3} {2, 3} b) Let 푋 be a set. Recall that the power set ℙ푋 is the set of all subsets of {1} {2} {3} 푋. The relation ⊆ on ℙ푋 is a partial order. Figure 1.2 shows the Hasse diagram of ℙ{1, 2, 3}. Notice that ⊆ is not a total order: for instance, {} {1} and {2, 3} are not comparable. FIGURE 1.2 Hasse diagram for ⊆ on ℙ{1, 2, 3}. Orders and lattices • 15 c) Let ∣ be the divisibility relation on ℕ; that is, 푥 ∣ 푦 if and only if there exists 푝 ∈ ℕ such that 푦 = 푝푥. Then ∣ is reflexive, since 푥 ∣ 푥 for all 푥 ∈ ℕ. It is anti-symmetric, since 푥 ∣ 푦 and 푦 ∣ 푥 imply 푦 = 푝푥 and 푥 = 푝′푦 for some 푝, 푝′ ∈ ℕ, which implies 푥 = 푝′푝푥 and so 푝 = 푝′ = 1, which implies 푥 = 푦. It is transitive, since 푥 ∣ 푦 and 푦 ∣ 푧 imply 푦 = 푝푥 and 푧 = 푝′푦 for some 푝, 푝′ ∈ ℕ, which implies 푧 = (푝′푝)푥 and so 푥 ∣ 푧. So ∣ is a partial order on ℕ.

If 푥 ∈ 푋 is such that there is no element 푦 ∈ 푋 with 푦 < 푥 (respect- Minimal/minimum, ively, 푥 < 푦), then 푥 is minimal (respectively, maximal). If 푥 ∈ 푋 is such maximal/maximum that for all elements 푦 ∈ 푋, we have 푥 ⩽ 푦 (respectively 푦 ⩽ 푥), then 푥 is a minimum (respectively, maximum). Therefore, in summary:

푥 is minimal ⇔ (∀푦 ∈ 푋)(푦 ⩽ 푥 ⇒ 푦 = 푥); 푥 is minimum ⇔ (∀푦 ∈ 푋)(푥 ⩽ 푦); 푥 is maximal ⇔ (∀푦 ∈ 푋)(푥 ⩽ 푦 ⇒ 푦 = 푥); 푥 is maximum ⇔ (∀푦 ∈ 푋)(푦 ⩽ 푥).

Notice that a minimum element is also minimal, but that the converse does not hold. A poset does not have to contain minimum or minimal elements. It contains at most one minimum element, for if 푥1 and 푥2 are both minimum, then 푥1 ⩽ 푥2 and 푥2 ⩽ 푥1, and so 푥1 = 푥2 by anti- symmetry. It may contain many distinct minimal elements.

E x a m p l e 1 . 1 8. a) The poset (ℤ, ⩽) does not contain either maximal elements or minimal elements.

b) Let 푋 = {푥, 푦1, 푦2}; define ⩽ on 푋 by

푢 ⩽ 푢 for all 푢 ∈ 푋,

푦1 ⩽ 푥,

푦2 ⩽ 푥.

The Hasse diagram for (푋, ⩽) is as shown in Figure 1.3(a): 푥 is a (necessarily unique) maximum element, and 푦1 and 푦2 are both minimal elements.

c) Let 푋 = {푥, 푦, 푧1, 푧2, …} and define ⩽ by

푢 ⩽ 푢 for all 푢 ∈ 푋, 푦 ⩽ 푥,

푧푖 ⩽ 푥 for all 푖 ∈ ℕ,

푧푖 ⩽ 푧푗 for all 푖, 푗 ∈ ℕ with 푖 ⩾ 푗.

The Hasse diagram for (푋, ⩽) is as shown in Figure 1.3(b): 푥 is a (necessarily unique) maximum element, and 푦 is the unique minimal element, but 푦 is not a minimum.

Orders and lattices • 16 푥 푧1 푦 FIGURE 1.3 푧2 Examples of partial orders, illus- trating minimal/maximal and 푧3 minimum/maximum elements: 푧4 (a) has a maximum 푥 and two minimal elements 푦1 and 푦2; 푥 푧푖 (b) has a unique minimal element 푦 but has no minimum 푦1 푦2 element. (a) (b)

There is a natural partial order of idempotents of a semigroup 푆 that Partial order of idempotents will re-appear in several different settings. Define the relation ≼ on the set of idempotents 퐸(푆) by 푒 ≼ 푓 ⇔ 푒푓 = 푓푒 = 푒.

P r o p o s i t i o n 1 . 1 9. The relation ≼ is a partial order.

Proof of 1.19. Since 푒2 = 푒, we have 푒 ≼ 푒 and so ≼ is reflexive. If 푒 ≼ 푓 and 푓 ≼ 푒, then 푒푓 = 푓푒 = 푒 and 푓푒 = 푒푓 = 푓 and so 푒 = 푓; hence ≼ is anti-symmetric. If 푒 ≼ 푓 and 푓 ≼ 푔, then 푒푓 = 푓푒 = 푒 and 푓푔 = 푔푓 = 푓 and so 푔푒 = 푔푓푒 = 푓푒 = 푒 and 푒푔 = 푒푓푔 = 푒푓 = 푒 and thus 푒 ≼ 푔; hence ≼ is transitive. Therefore ≼ is a partial order. 1.19

Let ⩽ be a partial order on a set 푋. Let 푌 ⊆ 푋.A lower bound for 푌 is Lower bound any element 푧 of 푋 such that 푧 ⩽ 푦 for all 푦 ∈ 푌. Let 퐵 be the set of lower bounds for 푌. If 퐵 is non-empty and has a maximum element 푧, then 푧 is the greatest lower bound or meet or infimumof 푌. The meet of 푌, if it Greatest lower bound, meet exists, is unique and is denoted by ⨅ 푌, or, in the case where 푌 = {푥, 푦}, by 푥 ⊓ 푦. If 푥 ⊓ 푦 exists for all 푥, 푦 ∈ 푋, then 푋 is a meet semilattice or lower Semilattice semilattice. If ⨅ 푌 exists for all 푌 ⊆ 푋, then 푋 is a complete meet semilattice or complete lower semilattice. The obvious definitions apply for upper bound, least upper bound or Upper bound, join join or supremum, ⨆ 푌, 푥 ⊔ 푦, join semilattice or upper semilattice, and complete join semilattice or complete upper semilattice. Most texts use ∧, ∨, ⋀, and ⋁ in place of ⊓, ⊔, ⨅, and ⨆. The square variants are used here to avoid confusion with the symbols for logical conjunction (‘and’) ∧ and disjunction (‘or’) ∨. The partially ordered set (푋, ⩽) is a lattice if it is an upper and lower Lattice semilattice. It is a complete lattice if it is an complete upper semilattice and complete lower semilattice.

E x a m p l e 1 . 2 0. a) In the example of the relation ⊆ on the power set ℙ{1, 2, 3}, we have {1, 2} ⊓ {1, 3} = {1} and {1, 2} ⊓ {3} = {}. Indeed, (ℙ{1, 2, 3}, ⊆) is a complete lattice.

Orders and lattices • 17 b) Let 푋 = {푡, 푥, 푦, 푧1, 푧2, …} and define ⩽ by 푥 ⩽ 푡, 푦 ⩽ 푡, 푡 푧푖 ⩽ 푡 for all 푖 ∈ ℕ, 푥 푦 푧푖 ⩽ 푥 for all 푖 ∈ ℕ,

푧푖 ⩽ 푦 for all 푖 ∈ ℕ, 푧푖 푧푖 ⩽ 푧푗 for all 푖, 푗 ∈ ℕ with 푖 ⩽ 푗. 푧4 Figure 1.4 shows a partial Hasse diagram for (푋, ⩽). Notice that 푥 and 푧3 푦 do not have a meet, but that every pair of elements has a join. So 푧2 (푋, ⩽) is an upper semilattice but not a lower semilattice. However, it is not a complete upper semilattice because the subset { 푧 ∶ 푖 ∈ ℕ } 푖 푧1 does not have a join. FIGURE 1.4 T h e o r e m 1 . 2 1. a) Let (푋, ⩽) be a non-empty lower semilattice. Partial Hasse diagram for the partially ordered set (푋, ⩽). Then (푋, ⊓) is a commutative semigroup of idempotents. Conversely, let (푆, ∘) be a commutative semigroup of idempotents. Semilattice = commutative Define a relation ⩽ on 푆 by 푥 ⩽ 푦 ⇔ 푥 ∘ 푦 = 푥. Then ⩽ is a partial semigroup of idempotents order and (푆, ⩽) is a lower semilattice. b) Let (푋, ⩽) be a non-empty upper semilattice. Then (푋, ⊔) is a commutative semigroup of idempotents. Conversely, let (푆, ∘) be a commutative semigroup of idempotents. Define a relation ⩽ on 푆 by 푥 ⩽ 푦 ⇔ 푥 ∘ 푦 = 푦. Then ⩽ is a partial order and (푆, ⩽) is an upper semilattice. Proof of 1.21. We prove part a); the reasoning for part b) is dual. Suppose (푋, ⩽) is a lower non-empty semilattice. Let 푥, 푦, 푧 ∈ 푋. First, 푥 ⊓ (푦 ⊓ 푧) and (푥 ⊓ 푦) ⊓ 푧 are both the meet of {푥, 푦, 푧} and hence 푥 ⊓ (푦 ⊓ 푧) = (푥 ⊓ 푦) ⊓ 푧. So ⊓ is associative. Next, 푥 ⊓ 푦 and 푦 ⊓ 푥 are both the meet of {푥, 푦}, and so 푥 ⊓ 푦 = 푦 ⊓ 푥. So (푋, ⊓) is commutative. The meet of {푥} is 푥 itself, so 푥 ⊓ 푥 = 푥. Hence every element of (푋, ⊓) is idempotent. So (푋, ⊓) is a commutative semigroup of idempotents. Suppose (푆, ∘) is a commutative semigroup of idempotents and define ⩽ as in the statement of the result. Let 푥, 푦, 푧 ∈ 푆. First, 푥 is idempotent, and so 푥 ∘ 푥 = 푥, and thus 푥 ⩽ 푥. Hence ⩽ is reflexive. Second, suppose that 푥 ⩽ 푦 and 푦 ⩽ 푥. Then 푥 ∘ 푦 = 푥 and 푦 ∘ 푥 = 푦. Since (푆, ∘) is commutative, this shows that 푥 = 푦. Hence ⩽ is anti-symmetric. Third, suppose 푥 ⩽ 푦 and 푦 ⩽ 푧. Then 푥 ∘ 푦 = 푥 and 푦 ∘ 푧 = 푦. So 푥 ∘ 푧 = (푥 ∘ 푦) ∘ 푧 = 푥 ∘ (푦 ∘ 푧) = 푥 ∘ 푦 = 푥, and so 푥 ⩽ 푧. Hence ⩽ is transitive. Finally, we want to show that 푥⊓푦 = 푥∘푦. First of all (푥∘푦)∘푥 = (푥∘푦), so 푥 ∘ 푦 ⩽ 푥 and similarly 푥 ∘ 푦 ⩽ 푦. So 푥 ∘ 푦 is a lower bound for {푥, 푦}. Let 푧 be some lower bound for {푥, 푦}. Then 푧 ⩽ 푥 and 푧 ⩽ 푦. Hence 푧 ∘ 푥 = 푧 and 푧 ∘ 푦 = 푧. So 푧 ∘ (푥 ∘ 푦) = (푧 ∘ 푥) ∘ 푦 = 푧 ∘ 푦 = 푧, and so 푧 ⩽ (푥 ∘ 푦). Hence 푥 ∘ 푦 is the greatest lower bound for {푥, 푦}. Thus (푆, ⩽) is a lower semilattice. 1.21

Orders and lattices • 18 Homomorphisms

Let 푆 and 푇 be semigroups. A map 휑 ∶ 푆 → 푇 is a homo- Homomorphism morphism if (푥푦)휑 = (푥휑)(푦휑) for all 푥, 푦 ∈ 푆. If 푆 and 푇 are monoids, then 휑 is a monoid homomorphism if (푥푦)휑 = (푥휑)(푦휑) for all 푥, 푦 ∈ 푆 and 1푆휑 = 1푇. A monomorphism is an injective homomorphism. If 휑 ∶ 푆 → 푇 Monomorphism, is a surjective homomorphism, then 푇 is a homomorphic image of 푆. isomorphism An isomorphism is a bijective homomorphism. It is easy to prove that a homomorphism 휑 ∶ 푆 → 푇 is an isomorphism if and only if there is a −1 −1 −1 homomorphism 휑 ∶ 푇 → 푆 such that 휑휑 = id푆 and 휑 휑 = id푇. If there is an isomorphism 휑 ∶ 푆 → 푇, then we say 푆 and 푇 are isomorphic and denote this by 푆 ≃ 푇. When two semigroups are isomorphic, we can think of them as the ‘same’ abstract structure in different settings. It is easy to prove that if 휑 ∶ 푆 → 푇 is a homomorphism and 푆′ and 푇′ are subsemigroups of 푆 and 푇 respectively, then 푆′휑 is a subsemigroup of 푇 and 푇′휑−1 is a subsemigroup of 푆 if it is non-empty. In particular, putting 푆′ = 푆 shows that im 휑 is a subsemigroup of 푇. If 휑 is a monomorphism, then 푆 is isomorphic to the subsemigroup im 휑 of 푇. The kernel of a homomorphism 휑 ∶ 푆 → 푇 is the binary relation Kernel of a homomorphism

ker 휑 = { (푥, 푦) ∈ 푆 × 푆 ∶ 푥휑 = 푦휑 }.

Notice that 휑 is a monomorphism if and only if ker 휑 is the identity relation (that is, ker 휑 = id푆). We now give a result showing that every semigroup is isomorphic to a subsemigroup of a semigroup of transformations. This is the analogue of Cayley’s theorem for groups, which states that every group is isomorphic to a subgroup of a symmetric group. For any 푥 ∈ 푆, let 휌푥 ∈ T푆1 be the 휌푥 1 map defined by 푠휌푥 = 푠푥 for all 푠 ∈ 푆 .

T h e o r e m 1 . 2 2. The map 휑 ∶ 푆 → T푆1 given by 푥 ↦ 휌푥 is a mono- Right regular representation morphism.

Proof of 1.22. Let 푥, 푦, 푠 ∈ 푆. Then 푠휌푥휌푦 = (푠푥)휌푦 = (푠푥)푦 = 푠(푥푦) = 푠휌푥푦; hence (푥휑)(푦휑) = 휌푥휌푦 = 휌푥푦 = (푥푦)휑. Therefore 휑 is a homomorphism. Furthermore

푥휑 = 푦휑 ⇒ 휌푥 = 휌푦 ⇒ 1휌푥 = 1휌푦 ⇒ 1푥 = 1푦 ⇒ 푥 = 푦; hence 휑 is injective. 1.22 An endomorphism is a homomorphism from a semigroup to itself. The set of all endomorphisms of 푆 is denoted End(푆) and forms a sub- End(푆) semigroup of T푆. The semigroup 푆 is group-embeddable if there exists a group 퐺 and a Group-embeddability

Homomorphisms • 19 monomorphism 휑 ∶ 푆 → 퐺. In this case, 푆 is isomorphic to the subsemigroup im 휑 of 퐺. Clearly any group-embeddable semigroup is cancellative, but we shall see that there exist cancellative semigroups that are not group- embeddable (see Example 2.14). A map 휑 ∶ 푆 → 푇 is an anti-homomorphism if (푥푦)휑 = (푦휑)(푥휑) for Anti-homomorphism all 푥, 푦 ∈ 푆.

Congruences and quotients

A binary relation 휌 on 푆 is Congruence ◆ left-compatible if (∀푥, 푦, 푧 ∈ 푆)(푥 휌 푦 ⇒ 푧푥 휌 푧푦); ◆ right-compatible if (∀푥, 푦, 푧 ∈ 푆)(푥 휌 푦 ⇒ 푥푧 휌 푦푧); ◆ compatible if (∀푥, 푦, 푧, 푡 ∈ 푆)((푥 휌 푦) ∧ (푧 휌 푡) ⇒ 푥푧 휌 푦푡). A left-compatible equivalence relation is a left congruence; a right-compatible equivalence relation is a right congruence; a compatible equivalence relation is a congruence.

P r o p o s i t i o n 1 . 2 3. A relation 휌 on 푆 is a congruence if and only if it Congruences are is both a left and a right congruence. left/right congruences

Proof of 1.23. Suppose that the relation 휌 is both a left and a right congruence. Let 푥, 푦, 푧, 푡 ∈ 푆 be such that 푥 휌 푦 and 푧 휌 푡. Since 휌 is a right congruence, 푥푧 휌 푦푧. Since 휌 is a left congruence, 푦푧 휌 푦푡. Since 휌 is transitive, 푥푧 휌 푦푡. Hence 휌 is a congruence. Suppose now that 휌 is a congruence. Let 푥, 푦 ∈ 푆 be such that 푥 휌 푦. Let 푧 ∈ 푆. Since 휌 is reflexive, 푧 휌 푧. Since 휌 is a congruence, 푧푥 휌 푧푦 and 푥푧 휌 푦푧. Hence 휌 is both a left and a right congruence. 1.23

Let 휌 be a congruence on 푆. Let 푆/휌 denote the quotient set of 푆 by Factor semigroup 휌 (that is, the set of 휌-classes of 푆). For any 푥 ∈ 푆, let [푥]휌 ∈ 푆/휌 be the 휌-class of 푥; that is, [푥]휌 = { 푦 ∈ 푆 ∶ 푦 휌 푥 }. Define a multiplication on 푆/휌 by

[푥]휌[푦]휌 = [푥푦]휌.

This multiplication is well-defined, in the sense that if we chose different representatives for the 휌-classes [푥]휌 and [푦]휌, we would get the same answer:

([푥]휌 = [푥′]휌) ∧ ([푦]휌 = [푦′]휌) ⇒ (푥 휌 푥′) ∧ (푦 휌 푦′) ⇒ 푥푦 휌 푥′푦′ [since 휌 is a congruence]

⇒ [푥푦]휌 = [푥′푦′]휌.

Congruences and quotients • 20 The factor set 푆/휌, with this multiplication, is a semigroup and is called the ♮ ♮ quotient or factor of 푆 by 휌. The map 휌 ∶ 푆 → 푆/휌, defined by 푥휌 = [푥]휌 is clearly a surjective homomorphism, called the natural map or natural Natural map homomorphism.

T h e o r e m 1 . 2 4. Let 휑 ∶ 푆 → 푇 be a homomorphism. Then ker 휑 is a First isomorphism theorem congruence, and the map 휓 ∶ 푆/ker 휑 → im 휑 with [푥]ker 휑휓 = 푥휑 is an isomorphism, and so 푆/ker 휑 ≃ im 휑.

Proof of 1.24. Let 푥, 푦, 푧, 푡 ∈ 푆. Then

(푥, 푦) ∈ ker 휑 ∧ (푧, 푡) ∈ ker 휑 ⇒ (푥휑 = 푦휑) ∧ (푧휑 = 푡휑) [by definition of ker 휑] ⇒ (푥휑)(푧휑) = (푦휑)(푡휑) ⇒ (푥푧)휑 = (푦푡)휑 [since 휑 is a homomorphism] ⇒ (푥푧, 푦푡) ∈ ker 휑; [by definition of ker 휑] thus ker 휑 is a congruence. Now,

[푥] = [푦] ker 휑 ker 휑 } } ⇔ (푥, 푦) ∈ ker 휑 [by definition of ker 휑-classes] } (1.5) ⇔ 푥휑 = 푦휑 [by definition of ker 휑] } } ⇔ [푥]ker 휑휓 = [푦]ker 휑휓. [by definition of 휓] }

The forward implication of (1.5) shows that 휓 is well-defined. The reverse implication shows that 휓 is injective. The image of 휓 is clearly im 휑. The map 휓 is a homomorphism since 휑 is a homomorphism. Hence 휓 is an isomorphism and so 푆/ker 휑 ≃ im 휑. 1.24

Let 퐼 be an ideal of 푆. Then 휌퐼 = (퐼 × 퐼) ∪ id푆 is a congruence on 푆. Rees factor semigroup The factor semigroup 푆/휌 is also denoted 푆/퐼, and the element [푥] is 퐼 휌퐼 denoted [푥]퐼. The congruence 휌퐼 is called the Rees congruence induced by 퐼, and 푆/퐼 is a Rees factor semigroup. The elements of 푆/퐼 are the 휌퐼-classes, which comprise 퐼 and singleton sets {푥} for each 푥 ∈ 푆 ∖ 퐼. It is easy to 푆 푆/퐼 see that 퐼 is a zero of the factor semigroup 푆/퐼, so it is often convenient to view 푆/퐼 as having elements (푆 ∖ 퐼) ∪ {0}, and to think of forming 푆/퐼 by 푆 ∖ 퐼 푆 ∖ 퐼 starting with 푆 and merging all elements of 퐼 to form a zero; see Figure 1.5.

The following result shows that the ideals of 푆/퐼 are in one-to-one 퐼 0푆/퐼 correspondence with the ideals of 푆 that contain 퐼. FIGURE 1.5 P r o p o s i t i o n 1 . 2 5. Let 퐼 be an ideal of 푆. Let A be the collection of Forming 푆/퐼 from 푆 by merging elements of 퐼 to form a zero. ideals of 푆 that contain 퐼. Let B be the collection of the ideals of 푆/퐼. Then the map 휑 ∶ A → B given by 퐽휑 = 퐽/퐼 is a bijection from A to B that preserves inclusion, in the sense that 퐽 ⊆ 퐽′ ⇒ 퐽휑 ⊆ 퐽′휑. 1.25

Congruences and quotients • 21 A semigroup 퐸 is an ideal extension of 푆 by 푇 if 푆 is an ideal of 퐸 Ideal extension and 퐸/푆 ≃ 푇. Note that for an ideal extension of 푆 by 푇 to exist, 푇 must contain a zero. Note further that there may be many non-isomorphic semigroups that are ideal extensions of 푆 by 푇.

Generating equivalences and congruences In this section, we will study how an equivalence relation or congruence on a semigroup 푆 is generated by a relation on 푆. This section is rather technical, but fundamentally important for future chapters. Throughout this section, let 푋 be a non-empty set. For any 휌 ∈ B푋, Generating equivalences let

R 휌 = ⋂{ 휎 ∈ B푋 ∶ 휌 ⊆ 휎 ∧ 휎 is reflexive }; S 휌 = ⋂{ 휎 ∈ B푋 ∶ 휌 ⊆ 휎 ∧ 휎 is symmetric }; T 휌 = ⋂{ 휎 ∈ B푋 ∶ 휌 ⊆ 휎 ∧ 휎 is transitive }; E 휌 = ⋂{ 휎 ∈ B푋 ∶ 휌 ⊆ 휎 ∧ 휎 is an equivalence relation }.

There is at least one element 휎 ∈ B푋 fulfilling the condition in each of the collections above, namely 휎 = 푋 × 푋. Furthermore, since every element in these collections contains 휌, the intersections 휌R, 휌S, 휌T, and 휌E all contain 휌. It is easy to see that ◆ 휌R, called the reflexive closure of 휌, is the smallest reflexive relation containing 휌; ◆ 휌S, called the symmetric closure of 휌, is the smallest symmetric relation containing 휌; ◆ 휌T, called the transitive closure of 휌, is the smallest transitive relation containing 휌; ◆ 휌E, called the equivalence relation generated by 휌, is the smallest equivalence relation containing 휌.

P r o p o s i t i o n 1 . 2 6. For any 휌 ∈ B푋, R a) 휌 = 휌 ∪ id푋; b) 휌S = 휌 ∪ 휌−1; T ∞ 푛 c) 휌 = ⋃푛=1 휌 ; R S S R −1 d) (휌 ) = (휌 ) = 휌 ∪ 휌 ∪ id푋; R T T R T e) (휌 ) = (휌 ) = 휌 ∪ id푋; E R S T S T R ∞ −1 푛 f) 휌 = ((휌 ) ) = ((휌 ) ) = id푋 ∪ ⋃푛=1(휌 ∪ 휌 ) .

Generating equivalences and congruences • 22 Proof of 1.26. a) Since 휌R is a reflexive relation containing 휌, it is immedi- R ate that 휌 ∪ id푋 ⊆ 휌 . On the other hand, 휌 ∪ id푋 is a reflexive relation containing 휌; since 휌R is the smallest reflexive relation containing 휌, R R we have 휌 ⊆ 휌 ∪ id푋. Hence 휌 = 휌 ∪ id푋. b) Since 휌S is a symmetric relation containing 휌, it is immediate that 휌 ∪ 휌−1 ⊆ 휌S. On the other hand, 휌 ∪ 휌−1 is a symmetric relation containing 휌; since 휌S is the smallest symmetric relation containing 휌, we have 휌S ⊆ 휌 ∪ 휌−1. Hence 휌S = 휌 ∪ 휌−1. c) Since 휌T contains 휌, transitivity implies that it contains 휌2 = 휌 ∘ 휌. Transitivity again implies that 휌T contains 휌3 = 휌 ∘ 휌2. Continuing T 푛 ∞ 푛 inductively, we see that 휌 contains 휌 for all 푛 ∈ ℕ; hence ⋃푛=1 휌 ⊆ 휌T. On the other hand,

∞ (푥, 푦), (푦, 푧) ∈ ⋃ 휌푛 푛=1 ⇒ (∃푘, ℓ ∈ ℕ)((푥, 푦) ∈ 휌푘 ∧ (푦, 푧) ∈ 휌ℓ) ⇒ (∃푘, ℓ ∈ ℕ)((푥, 푧) ∈ 휌푘 ∘ 휌ℓ) ⇒ (∃푘, ℓ ∈ ℕ)((푥, 푧) ∈ 휌푘+ℓ) ∞ ⇒ (푥, 푧) ∈ ⋃ 휌푛. 푛=1

∞ 푛 T So ⋃푛=1 휌 is a transitive relation containing 휌. Since 휌 is the smallest T ∞ 푛 T ∞ 푛 such relation, we have 휌 ⊆ ⋃푛=1 휌 . Hence 휌 = ⋃푛=1 휌 . d) We have

(휌R)S (1.6) = 휌R ∪ (휌R)−1 [by part b)] −1 = 휌 ∪ id푋 ∪ (휌 ∪ id푋) [by part a)] −1 −1 = 휌 ∪ id푋 ∪ 휌 ∪ id푋 [by definition of converse] −1 −1 = 휌 ∪ 휌 ∪ id푋 [since id푋 = id푋] (1.7) S = 휌 ∪ id푋 [by part b)] = (휌S)R. [by part a)] (1.8)

The result is given by lines1.6 ( ), (1.7), and (1.8). R T R T e) Since id푋 ⊆ 휌 , we have id푋 ⊆ (휌 ) . Since id푋 ∘ id푋 = id푋, it follows T R T R from part c) that id푋 = id푋. So id푋 ⊆ (휌 ) . Since 휌 ⊆ 휌 , we have T R T T R T R T 휌 ⊆ (휌 ) . So (휌 ) = 휌 ∪ id푋 ⊆ (휌 ) . R T ∞ Now let (푢, 푣) ∈ (휌 ) . Then by parts a) and c), (푢, 푣) ∈ ⋃푛=1(휌 ∪ 푛 id푋) . So there exists 푛 ∈ ℕ and 푥0, … , 푥푛 ∈ 푋 such that 푢 = 푥0, 푣 = 푥푛, and (푥푖, 푥푖+1) ∈ 휌∪id푋 for 푖 = 0, … , 푛−1. Fix such a sequence with 푛 minimal. Then if 푛 ⩾ 2, no pair (푥푖, 푥푖+1) is in id푋, for this would imply that 푥푖 = 푥푖+1 and so we could shorten the sequence by

Generating equivalences and congruences • 23 deleting one of 푥푖 or 푥푖+1, contradicting the minimality of 푛. So ∞ 푛 T T R (푢, 푣) ∈ id푋 ∪ 휌 ∪ ⋃ 휌 = id푋 ∪ 휌 = (휌 ) . 푛=2 Hence (휌R)T ⊆ (휌T)R and so (휌R)T = (휌T)R. f) Since 휌E is reflexive and contains 휌, it contains 휌R. Since it is symmetric and contains 휌R, it contains (휌R)S. Since it is transitive and contains (휌R)S, it contains ((휌R)S)T. Hence ((휌R)S)T ⊆ 휌E. On the other hand, ((휌R)S)T is transitive by the definition of T. R S T R S R R S T Furthermore, ((휌 ) ) ⊇ (휌 ) ) ⊇ 휌 ⊇ id푋 and so ((휌 ) ) is reflex- R S T ∞ R S 푛 R S 푛 ive. Let (푥, 푦) ∈ ((휌 ) ) = ⋃푛=1((휌 ) ) . Then (푥, 푦) ∈ ((휌 ) ) for some 푛 ∈ ℕ. Hence there exist 푥0, … , 푥푛 ∈ 푋 with 푥0 = 푥, 푥푛 = 푦, R S R S and (푥푖, 푥푖+1) ∈ (휌 ) for 푖 = 0, … , 푛 − 1. Since (휌 ) is symmetric, R S R S 푛 R S T (푥푖+1, 푥푖) ∈ (휌 ) for each 푖, and so (푦, 푥) ∈ ((휌 ) ) ⊆ ((휌 ) ) . So ((휌R)S)T is symmetric. Hence ((휌R)S)T is an equivalence relation containing 휌. Since 휌E is the smallest equivalence relation containing 휌, we have 휌E ⊆ ((휌R)S)T. Hence 휌E = ((휌R)S)T. Finally, notice that

휌E = ((휌R)S)T [by the above reasoning] (1.9) = ((휌S)R)T [by part d)] = ((휌S)T)R [by part e)](1.10) S T = id푋 ∪ (휌 ) [by part a)] −1 T = id푋 ∪ (휌 ∪ 휌 ) [by part b)] ∞ −1 = id푋 ∪ ⋃푛=1(휌 ∪ 휌 ). [by part c)] (1.11)

Lines (1.9), (1.10), and (1.11) give the three required equalities. 1.26

For any 휌 ∈ B푆, let Generating congruences

C 휌 = ⋂{ 휎 ∈ B푆 ∶ 휌 ⊆ 휎 ∧ 휎 is left and right compatible }, # 휌 = ⋂{ 휎 ∈ B푆 ∶ 휌 ⊆ 휎 ∧ 휎 is a congruence }. It is easy to see that ◆ 휌C is the smallest left and right compatible relation containing 휌; ◆ 휌#, called the congruence generated by 휌, is the smallest congruence containing 휌.

C P r o p o s i t i o n 1 . 2 7. For any 휌 ∈ B푆, we have 휌 = { (푝푥푞, 푝푦푞) ∈ 푆 × 푆 ∶ 푝, 푞 ∈ 푆1 ∧ (푥, 푦) ∈ 휌 }. Proof of 1.27. Let 휎 = { (푝푥푞, 푝푦푞) ∈ 푆 × 푆 ∶ 푝, 푞 ∈ 푆1, (푥, 푦) ∈ 휌 }. To prove that 휎 = 휌C, we have to show that 휎 is the smallest left and right compatible relation on 푆 containing 휌. Notice first that if (푥, 푦) ∈ 휌, then

Generating equivalences and congruences • 24 (푥, 푦) = (1푥1, 1푦1) ∈ 휎. Hence 휎 contains 휌. Let (푢, 푣) ∈ 휎 and 푟 ∈ 푆. Then 푢 = 푝푥푞 and 푣 = 푝푦푞 for some (푥, 푦) ∈ 휌. Let 푝′ = 푟푝. Then (푟푢, 푟푣) = (푝′푥푞, 푝′푦푞) ∈ 휎. Hence 휎 is left-compatible. Similarly, 휎 is right compatible. Now let 휏 be some left and right compatible relation that contains 휌. Let (푝푥푞, 푝푦푞) ∈ 휎, where (푥, 푦) ∈ 휌 and 푝, 푞 ∈ 푆1. Then (푥, 푦) ∈ 휏 since 휌 ⊆ 휏. Hence (푝푥푞, 푝푦푞) ∈ 휏 since 휏 is left and right compatible. Thus 휎 ⊆ 휏. Therefore 휎 is the smallest left and right compatible relation containing 휌. 1.27

P r o p o s i t i o n 1 . 2 8. For any 휌, 휎 ∈ B푆, a) (휌 ∪ 휎)C = 휌C ∪ 휎C; b) (휌−1)C = (휌C)−1. Proof of 1.28. a) For 푢, 푣 ∈ 푆, (푢, 푣) ∈ (휌 ∪ 휎)C ⇔ (∃푝, 푞 ∈ 푆1, (푥, 푦) ∈ 휌 ∪ 휎)(푢 = 푝푥푞 ∧ 푣 = 푝푦푞) [by Proposition 1.27] ⇔ (∃푝, 푞 ∈ 푆1, (푥, 푦) ∈ 휌)(푢 = 푝푥푞 ∧ 푣 = 푝푦푞) ∨ (∃푝, 푞 ∈ 푆1, (푥, 푦) ∈ 휎)(푢 = 푝푥푞 ∧ 푣 = 푝푦푞) ⇔ (푢, 푣) ∈ 휌C ∨ (푢, 푣) ∈ 휎C [by Proposition 1.27] ⇔ (푢, 푣) ∈ 휌C ∪ 휎C. b) For 푢, 푣 ∈ 푆, (푢, 푣) ∈ (휌−1)C ⇔ (∃푝, 푞 ∈ 푆1, (푥, 푦) ∈ 휌−1)(푢 = 푝푥푞 ∧ 푣 = 푝푦푞) [by Proposition 1.27] ⇔ (∃푝, 푞 ∈ 푆1, (푦, 푥) ∈ 휌)(푣 = 푝푦푞 ∧ 푢 = 푝푥푞) ⇔ (푣, 푢) ∈ 휌C [by Proposition 1.27] C −1 ⇔ (푢, 푣) ∈ (휌 ) . 1.28

P r o p o s i t i o n 1 . 2 9. For any 휌 ∈ B푆, Characterizing generated congruences ∞ # C E C C −1 푛 휌 = (휌 ) = id푆 ∪ ⋃(휌 ∪ (휌 ) ) . 푛=1 Proof of 1.29. By Proposition 1.26(f), ∞ C E C C −1 푛 (휌 ) = id푆 ∪ ⋃(휌 ∪ (휌 ) ) , 푛=1 so we must prove that 휌# = (휌C)E. That is, we must show that (휌C)E is the smallest congruence containing 휌. By definition, (휌C)E is an equivalence relation containing 휌C, which in turn contains 휌. So 휌 ⊆ (휌C)E.

Generating equivalences and congruences • 25 C E Now let 푥, 푦, 푧 ∈ 푆 and suppose that (푥, 푦) ∈ (휌 ) . If (푥, 푦) ∈ id푆, C E then 푥 = 푦, and so 푧푥 = 푧푦, and thus (푧푥, 푧푦) ∈ id푆 ⊆ (휌 ) . Further- more,

∞ C C −1 푛 (푥, 푦) ∈ ⋃푛=1(휌 ∪ (휌 ) ) ∞ −1 C 푛 ⇒ (푥, 푦) ∈ ⋃푛=1((휌 ∪ 휌 ) ) [by Proposition 1.28] ⇒ (∃푛 ∈ ℕ)(∃푥0, 푥1, … , 푥푛 ∈ 푆)

[(푥 = 푥0) ∧ (푥푛 = 푦) −1 C ∧ (∀푖)((푥푖, 푥푖+1) ∈ (휌 ∪ 휌 ) )] [by definition of ∘]

⇒ (∃푛 ∈ ℕ)(∃푥0, 푥1, … , 푥푛 ∈ 푆)

[(푧푥 = 푧푥0) ∧ (푧푥푛 = 푧푦) −1 C ∧ (∀푖)((푧푥푖, 푧푥푖+1) ∈ (휌 ∪ 휌 ) )] [since (휌 ∪ 휌−1)C is left and right compatible] ⇒ (∃푛 ∈ ℕ)((푧푥, 푧푦) ∈ ((휌 ∪ 휌−1)C)푛) [by definition of ∘] ∞ C C −1 푛 ⇒ (푧푥, 푧푦) ∈ ⋃푛=1(휌 ∪ (휌 ) ) [by Proposition 1.28] ⇒ (푧푥, 푧푦) ∈ (휌C)E.

Hence (푥, 푦) ∈ (휌C)E implies (푧푥, 푧푦) ∈ (휌C)E. Therefore (휌C)E is left- compatible. Similarly, (휌C)E is right-compatible. Hence (휌C)E is a congruence containing 휌. Now suppose that 휏 is a congruence containing 휌. Then 휏 is left and right compatible and so must contain 휌C, which is the smallest left and right compatible relation containing 휌. Furthermore, 휏 is an equivalence relation, and so it must contain (휌C)E, which is the smallest equivalence relation containing 휌C. Hence (휌C)E ⊆ 휏. Therefore (휌C)E is the smallest congruence containing 휌. 1.29

Let E푆 be the set of equivalence relations on 푆 and let C푆 be the set of Lattice of congruences congruences on 푆. Then E푆 and C푆 both admit ⊆ as a partial order. It is easy to see that both (E푆, ⊆) and (C푆, ⊆) are actually lattices: E ◆ for any 휌, 휎 ∈ E푆, we have 휌 ⊓ 휎 = 휌 ∩ 휎 and 휌 ⊔ 휎 = (휌 ∪ 휎) ; # ◆ for any 휌, 휎 ∈ C푆, we have 휌 ⊓ 휎 = 휌 ∩ 휎 and 휌 ⊔ 휎 = (휌 ∪ 휎) .

Suppose 휌, 휎 ∈ C푆. There seems to be an ambiguity in writing 휌 ⊔ 휎: do E we mean the join (휌 ∪ 휎) in the lattice of equivalence relations E푆, or the # join (휌 ∪ 휎) in the lattice of congruences C푆? However,

(휌 ∪ 휎)# = ((휌 ∪ 휎)C)E [by Proposition 1.29] = (휌C ∪ 휎C)E [by Proposition 1.28(a)] = (휌 ∪ 휎)E. [since 휌 and 휎 are compatible]

So there is really no ambiguity in writing 휌 ⊔ 휎.

Generating equivalences and congruences • 26 T P r o p o s i t i o n 1 . 3 0. Let 휌, 휎 ∈ E푆. Then 휌 ⊔ 휎 = (휌 ∘ 휎) . Characterizing join of equivalence relations Proof of 1.30. Since 휌 ∪ 휎 contains both 휌 and 휎, it follows that

휌 ∘ 휎 ⊆ (휌 ∪ 휎) ∘ (휌 ∪ 휎) = (휌 ∪ 휎)2, and more generally that (휌 ∘ 휎)푛 ⊆ (휌 ∪ 휎)2푛. Thus

∞ ∞ (휌 ∘ 휎)T = ⋃(휌 ∘ 휎)푛 ⊆ ⋃(휌 ∪ 휎)푛 = (휌 ∪ 휎)T. (1.12) 푛=1 푛=1

On the other hand, 휌 ∘ 휎 contains 휌 ∘ id푆 = 휌 (since 휎 is reflexive) and contains id푆 ∘ 휎 = 휎 (since 휌 is reflexive), and thus 휌 ∪ 휎 ⊆ 휌 ∘ 휎. Hence (휌 ∪ 휎)T ⊆ (휌 ∘ 휎)T. Combine this with (1.12) to see that

(휌 ∪ 휎)T = (휌 ∘ 휎)T. (1.13)

Then

휌 ⊔ 휎 = (휌 ∪ 휎)E = (((휌 ∪ 휎)R)S)T [by Proposition 1.26(f)] −1 T = ((휌 ∪ 휎) ∪ (휌 ∪ 휎) ∪ id푆) [by Proposition 1.26(d)] −1 −1 T = (휌 ∪ 휎 ∪ 휌 ∪ 휎 ∪ id푆) = (휌 ∪ 휎)T [since 휌 and 휎 are reflexive and symmetric] = (휌 ∘ 휎)T. [by (1.13)]

This completes the proof. 1.30

P r o p o s i t i o n 1 . 3 1. Let 휌, 휎 ∈ E푆. If 휌 ∘ 휎 = 휎 ∘ 휌, then 휌 ⊔ 휎 = 휌 ∘ 휎. Join of commuting equivalence relations Proof of 1.31. Suppose 휌 ∘ 휎 = 휎 ∘ 휌. Then

(휌 ∘ 휎)2 =휌∘휎∘휌∘휎=휌∘휌∘휎∘휎=휌2 ∘ 휎2. (1.14)

But 휌2 ⊆ 휌 and 휎2 ⊆ 휎 since 휌 and 휎 are transitive. Furthermore, 휌 = 2 2 2 휌 ∘ id푆 ⊆ 휌 since 휌 is reflexive, and similarly 휎 ⊆ 휎 . Hence 휌 = 휌 and 휎2 = 휎 and so (휌 ∘ 휎)2 = 휌 ∘ 휎 by (1.14). Hence (휌 ∘ 휎)푛 = 휌 ∘ 휎 for all 푛 ∈ ℕ, and thus

휌 ⊔ 휎 = (휌 ∘ 휎)T [by Proposition 1.30] ∞ 푛 = ⋃푛=1(휌 ∘ 휎) [by Proposition 1.26(c)] = 휌 ∘ 휎. 1.31

Generating equivalences and congruences • 27 Subdirect products

Let S = { 푆푖 ∶ 푖 ∈ 퐼 } be a collection of semigroups. For each 푗 ∈ 퐼, there is a projection map from the direct product ∏푖∈퐼 푆푖 to 푆푗, taking an element of the direct product to its 푗-th component:

휋푗 ∶ ∏ 푆푖 → 푆푗, 푥휋푗 = (푗)푥. 푖∈퐼

Notice that every 휋푗 is a surjective homomorphism. A subdirect product of S is [a semigroup isomorphic to] a subsemi- Subdirect product group 푃 of the direct product ∏푖∈퐼 푆푖 such that 푃휋푗 = 푆푗 for all 푗 ∈ 퐼. Let 푆 be a semigroup. A collection of surjective homomorphisms Separation by surjective homomorphisms 훷 = { 휑푖 ∶ 푆 → 푆푖 ∶ 푖 ∈ 퐼 } is said to separate the elements of 푆 if they have the property that

(∀푖 ∈ 퐼)(푥휑푖 = 푦휑푖) ⇒ 푥 = 푦. P r o p o s i t i o n 1 . 3 2. A semigroup 푆 is a subdirect product of a collection of semigroups S = { 푆푖 ∶ 푖 ∈ 퐼 } if and only if there exists a collection of surjective homomorphisms 훷 = { 휑푖 ∶ 푆 → 푆푖 ∶ 푖 ∈ 퐼 } that separate the elements of 푆. Proof of 1.32. If 푆 is a subdirect product of S, then the collection of projection maps restricted to 푆 (that is, the collection { 휋푖|푆 ∶ 푆 → 푆푖 ∶ 푖 ∈ 퐼 }) separates the elements of 푆. On the other hand, suppose the collection 훷 separates the elements of 푆. Define 휓 ∶ 푆 → ∏푖∈퐼 푆푖 by letting the 푖-th component of 푠휓 be 푠휑푖; that is, (푖)(푠휓) = 푠휑푖. Then 휓 is a homomorphism since each 휑푖 is a homomorphism. Furthermore, 푠휓 = 푡휓 implies that 푠휑푖 = 푡휑푖 for all 푖 ∈ 퐼, which implies 푠 = 푡 since 훷 separates the elements of 푆. Hence 휓 is injective. So 푆 is isomorphic to the subsemigroup im 휓 of ∏푖∈퐼 푆푖. Finally, the projection maps 휋푖 are all surjective since each 휑푖 is surjective. So im 휑 is a subdirect product of S. 1.32

P r o p o s i t i o n 1 . 3 3. Let 푆 be a semigroup and let 훴 = { 휎푖 ∶ 푖 ∈ 퐼 } be a collection of congruences on 푆. Let 휎 = ⋂ 훴. Then 푆/휎 is a subdirect product of { 푆/휎푖 ∶ 푖 ∈ 퐼 }.

Proof of 1.33. For each 푖 there is a homomorphism 휑푖 ∶ 푆/휎 → 푆/휎푖 with [푥] 휑 = [푥] . (These maps are well-defined since 휎 ⊆ 휎 .) Clearly, the 휎 푖 휎푖 푖 homomorphisms 휑푖 are surjective. Furthermore, the collection 훷 = { 휑푖 ∶ 푖 ∈ 퐼 } separates the elements of 푆/휎, since if [푥]휎휑푖 = [푦]휎휑푖 for all 푖 ∈ 퐼, then [푥] = [푦] and thus (푥, 푦) ∈ 휎 for all 푖 ∈ 퐼, which implies 휎푖 휎푖 푖 (푥, 푦) ∈ 휎 = ⋂ 훴 and so [푥]휎 = [푦]휎. Therefore 푆/휎 is a subdirect product of { 푆/휎푖 ∶ 푖 ∈ 퐼 } by Proposition 1.32. 1.33 P r o p o s i t i o n 1 . 3 4. Let 푀 be a monoid and let 퐸 be an ideal ex- Ideal extensions of monoids tension of 푀 by a semigroup 푇. Then 퐸 is a subdirect product of 푀 and are subdirect products 푇.

Subdirect products • 28 Proof of 1.34. By definition, 푀 is an ideal of 퐸 and 푇 is the Rees factor semigroup 퐸/푀. Let 휑 ∶ 퐸 → 푇 be the natural homomorphism 푥휑 = [푥]푀. Let 휓 ∶ 퐸 → 푀 be given by 푥휓 = 푥1푀. Then

(푥휓)(푦휓) = 푥1푀푦1푀

= 푥푦1푀 [since 푦1푀 lies in the ideal 푀 of 퐸] = (푥푦)휓.

Thus 휓 is a homomorphism. Both 휑 and 휓 are clearly surjective. Further- more, if 푥휑 = 푦휑 and 푥휓 = 푦휓, then either 푥, 푦 ∈ 퐸∖푀 and [푥]푀 = [푦]푀 and so 푥 = 푦, or 푥, 푦 ∈ 푀 and 푥1푀 = 푦1푀 and so 푥 = 푦. Thus the collection of surjective homomorphisms {휑, 휓} separates elements of 퐸 and so 퐸 is a subdirect product of 푀 and 푇. 1.34

Actions

A semigroup action of a semigroup 푆 on a set 퐴 is an oper- Semigroup action ation ⋅ ∶ 퐴 × 푆 → 퐴 that is compatible with the semigroup multiplication, in the sense that

(푎 ⋅ 푥) ⋅ 푦 = 푎 ⋅ (푥푦) (1.15) for all 푎 ∈ 퐴 and 푥, 푦 ∈ 푆. We call such a semigroup action an action of 푆 on 퐴, or an 푆-action on 퐴, and say that 푆 acts on 퐴.

E x a m p l e 1 . 3 5. a) Any subsemigroup 푆 of T퐴 acts on 퐴 by 푎 ⋅ 휌 = 푎휌 (where 휌 ∈ T퐴). b) Let 푆 be a subsemigroup of a semigroup 푇. Then 푆 acts on 푇 via 푡 ⋅ 푥 = 푡푥 for all 푡 ∈ 푇 and 푥 ∈ 푆. In particular, this holds when 푇 = 푆 or when 푇 = 푆1.

Given an action ⋅, we can define a map 휑 ∶ 푆 → T퐴, where the transformation 푠휑 is such that 푎(푠휑) = 푎 ⋅ 푠. The condition (1.15) implies that 휑 is a homomorphism. Conversely, given a homomorphism 휑 ∶ 푆 → T퐴, we can define an action ⋅ by 푎 ⋅ 푠 = 푎(푠휑), which satisfies (1.15) since 휑 is a homomorphism. There is thus a one-to-one correspondence between actions of a semigroup 푆 on 퐴 and homomorphisms 휑 ∶ 푆 → T퐴. An action of 푆 on 퐴 is free if distinct elements of 푆 act differently on Free, transitive, every element of 퐴, or, equivalently, regular actions

(∀푥, 푦 ∈ 푆)((∃푎 ∈ 퐴)(푎 ⋅ 푥 = 푎 ⋅ 푦) ⇒ 푥 = 푦).

An action of 푆 on 퐴 is transitive if 퐴 is non-empty and for all 푎, 푏 ∈ 퐴, there is some element 푠 ∈ 푆 such that 푎 ⋅ 푠 = 푏. That is, the action is transitive if one can reach start at any element of 퐴 and reach any element

Actions • 29 (possibly the same one) by acting by some element of 푆. An action is regular if it is both free and transitive. It is easy to see that if 푆 has a regular action on 퐴, then |푆| = |퐴|. Suppose 퐴 is also a semigroup. An action of 푆 on 퐴 is by endomorph- Action by endomorphisms isms if 푠휑 ∈ End 퐴 for each 푠 ∈ 푆; in this case, 푎푏 ⋅ 푥 = (푎 ⋅ 푥)(푏 ⋅ 푥) for all 푎, 푏 ∈ 퐴 and 푥 ∈ 푆. The above discussions concern right semigroup actions. There is a Left action dual notion of a left semigroup action of 푆 on 퐴, which is an operation ⋅ ∶ 푆 × 퐴 → 퐴 satisfying 푠 ⋅ (푡 ⋅ 푎) = (푠푡) ⋅ 푎; this corresponds to a map 휑 ∶ 푆 → T퐴, where 푎(푠휑) = 푠 ⋅ 푎. This map 휑 is an anti-homomorphism since 푎(푡휑)(푠휑) = (푡 ⋅ 푎)(푠휑) = 푠 ⋅ (푡 ⋅ 푎) = 푠푡 ⋅ 푎 = 푎((푠푡)휑). The definitions of actions being free, transitive, regular, and byendo- morphisms also apply to left actions. The correspondence of right actions with homomorphisms and left actions with anti-homomorphisms depends on writing maps on the right and composing them left-to-right. When maps are written on the left and composed right-to-left, right actions correspond to anti- homomorphisms and left actions to homomorphisms.

Cayley graphs Let 푆 be a semigroup or monoid with a generating set 퐴. The right (respectively, left) Cayley graph 훤(푆, 퐴) (respectively, 훤′(푆, 퐴)) of 푆 with respect to 퐴 is the directed graph with vertex set 푆 and, for every 푥 ∈ 푆 and 푎 ∈ 퐴, an edge from 푥 to 푥푎 (respectively, 푎푥) labelled by 푎. By default ‘Cayley graph’ means ‘right Cayley graph’. E x a m p l e 1 . 3 6. a) Let 푀 be the monoid (ℕ ∪ {0}) × (ℕ ∪ {0}). Let 푎 = (1, 0) and 푏 = (0, 1) and let 퐴 = {푎, 푏}. The Cayley graph 훤(푀, 퐴) is an infinite grid; part of it is shown in Figure 1.6.

b) Let 푋 = {1, 2}. Let 퐴 = {휎, 휋, 휐} ⊆ P푋, where 1 2 1 2 1 2 휎 = ( ) , 휋 = ( ), and 휐 = ( ). 2 1 1 1 1 ∗

Then 퐴 generates P푋. The Cayley graph 훤(P푋, 퐴) is shown in Figure 1 2 1.7. Note the subgroup S and the subsemigroup T , and that ( ) 푋 푋 ∗ ∗ is a zero and a sink vertex of the graph.

Cayley graphs • 30 (0, 3) 푎 (1, 3) 푎 (2, 3) 푎 (3, 3)

푏 푏 푏 푏

(0, 2) 푎 (1, 2) 푎 (2, 2) 푎 (3, 2)

푏 푏 푏 푏

(0, 1) 푎 (1, 1) 푎 (2, 1) 푎 (3, 1)

푏 푏 푏 푏 FIGURE 1.6 푎 푎 푎 Cayley graph of 푀 = (ℕ ∪ (0, 0) (1, 0) (2, 0) (3, 0) {0}) × (ℕ ∪ {0}).

1 2 휎 1 2 ( ) ( ) 1 2 휎 2 1 휋 휋 휐 1 2 휎 1 2 휐 휋, 휐 ( ) ( ) 1 1 휎, 휋 2 2

1 2 휎 1 2 휐 1 2 휎 1 2 휋, 휐 ( ) ( ) 휋, 휐 ( ) ( ) 1 ∗ 휎, 휋 2 ∗ ∗ 1 휎, 휋 ∗ 2

휐 1 2 휐 휎, 휋, 휐 ( ) ∗ ∗ FIGURE 1.7 Cayley graph of P{1,2}.

c) Let 푆 = {푥, 푦} be a two-element right zero semigroup and let 퐴 = 푆. The i) right and ii) left Cayley graphs 훤(푆, 퐴) and 훤′(푆, 퐴) are shown in Figure 1.8.

For groups, Cayley graphs have special properties. First, the left and right Cayley graphs are isomorphic under the map sending each vertex and each edge label to its inverse. Second, the Cayley graphs are connected, and indeed strongly connected. Third, the Cayley graphs are homogeneous, which essentially means that a neighbourhood of any vertex ‘looks like’ the corresponding neighbourhood of any other vertex. The graphs in Example 1.36(c) show that the left and right Cayley graphs of a semigroup need not be isomorphic; the second graph shows that the Cayley graph of a semigroup need not be connected. All the graphs in Example 1.36 except (c)(ii) show that Cayley graphs of semigroups need not be homogeneous.

Cayley graphs • 31 푦 푦 푦 푥, 푦

푥 푦 FIGURE 1.8 Right (i) and left (ii) Cayley 푥 푥 푥 푥, 푦 graphs of a two-element right zero semigroup {푥, 푦}. (i) (ii) Exercises [See pages 198–204 for the solutions.] 1.1 Prove that if 푆 is a semigroup and 푒 ∈ 푆 is both a right zero and a right identity, then 푆 is trivial. 1.2 Prove the following: a) If 푆 is a monoid with identity 1, the semigroup 푆0 obtained by adjoining a zero if necessary is also a monoid with identity 1. b) If 푆 is a semigroup with zero 0, the monoid 푆1 obtained by adjoining an identity if necessary also has zero 0. ✴1.3 Let 푆 be a left-cancellative semigroup. Suppose that 푒 ∈ 푆 is an idempotent. Prove that 푒 is a left identity. Deduce that a cancellative semigroup can contain at most one idempotent, which must be an identity. ✴1.4 Prove that a right zero semigroup is left-cancellative. ✴1.5 Prove that a finite cancellative semigroup is a group.

1.6 Prove from the definition that id푋 is an identity for B푋. Does B푋 contain a zero? 1.7 Does there exist a non-trivial semigroup that does not contain any proper subsemigroups? ✴1.8 Give an [easy] example of an infinite periodic semigroup.

1.9 Does either T푋 or P푋 contain a zero? A left zero? A right zero? [Note that the answer may depend on |푋|.] 1.10 The power semigroup of a semigroup 푆 is the set ℙ푆 of all subsets of Power semigroup 푆 under the operation 푋푌 = { 푥푦 ∶ 푥 ∈ 푋, 푦 ∈ 푌 } for 푋, 푌 ∈ ℙ푆. (Recall from page 5 that 푋(푌푍) = (푋푌)푍 for all 푋, 푌, 푍 ∈ ℙ푆.) a) Prove that ℙ푆 contains a subsemigroup isomorphic to 푆. b) Prove that ∅ is a zero of ℙ푆. Prove that (ℙ푆) ∖ {∅} is a subsemigroup of ℙ푆. c) Let 푀 be a monoid. Prove that (ℙ푀) ∖ {∅} is cancellative if and only if 푀 is trivial. d) Prove that (ℙ푆) ∖ {∅} is a right zero semigroup if and only if 푆 is a right zero semigroup. ✴1.11 Let 푋 = {1, … , 푛} with 푛 ⩾ 2. Recalling the cycle notation for per- mutations from group theory, let 휏 = (1 2) and 휁 = (1 2 … 푛 − 1 푛). Note that 휏, 휁 ∈ S푋; indeed, from elementary group theory, we know

Exercises • 32 that S푋 = ⟨휏, 휁⟩. For any 푖, 푗 ∈ 푋 with 푖 ≠ 푗, let |푖 푗| denote the transformation 휑푖,푗 ∈ T푋 such that 푖휑푖,푗 = 푗휑푖,푗 = 푗, and 푥휑푖,푗 = 푥 for 푥 ∉ {푖, 푗}. a) Prove the following four identities when 푛 ⩾ 3, and only the last identity for 푛 ⩾ 2; note that the elements appearing in first three identities all lie in T푋 only when 푛 ⩾ 3: (1 푖)|1 2|(1 푖) = |푖 2| for 푖 ⩾ 3; (2 푗)|1 2|(2 푗) = |1 푗| for 푗 ⩾ 3; (1 푖)(2 푗)|1 2|(2 푗)(1 푖) = |푖 푗| for 푖, 푗 ⩾ 3 and 푖 ≠ 푗; (푖 푗)|푖 푗|(푖 푗) = |푗 푖| for 푖, 푗 ⩾ 1 and 푖 ≠ 푗.

b) Let 휑 ∈ T푋. Suppose | im 휑| = 푟 < 푛. Let 푖, 푗 ∈ 푋 with 푖 ≠ 푗 be such that 푖휑 = 푗휑. Let 푘 ∈ 푋 ∖ im 휑. Show that 휑 = |푖 푗|휑′, where 푖휑′ = 푘 and 푥휑′ = 푥휑 for 푥 ≠ 푖.

c) Deduce that T푋 = ⟨휏, 휁, |1 2|⟩. 1.12 Let 푆 be a finite monoid. Prove that 푥 ∈ 푆 is right-invertible if and only if it is left-invertible. [Hint: use the fact that 푥 is periodic.]

✴1.13 Prove than an element of T푋 is a) left-invertible if and only if it is surjective; b) right-invertible if and only if it is injective. 1.14 Let (푆, ⩽) be a lattice. a) Prove that (푥 ⊓ 푦) ⊔ 푥 = 푥 and (푥 ⊔ 푦) ⊓ 푥 = 푥 for any 푥, 푦 ∈ 푆. b) Deduce that (∀푥, 푦, 푧 ∈ 푆)(푥 ⊓ (푦 ⊔ 푧) = (푥 ⊓ 푦) ⊔ (푥 ⊓ 푧)) ⇔ (∀푥, 푦, 푧 ∈ 푆)(푥 ⊔ (푦 ⊓ 푧) = (푥 ⊔ 푦) ⊓ (푥 ⊔ 푧)). [Equivalently: ⊓ distributes over ⊔ if and only if ⊔ distributes over ⊓.] ✴1.15 Give an example of a map 휑 from a monoid 푆 to a monoid 푇 that is a homomorphism but not a monoid homomorphism. ✴1.16 Let 푆 and 푇 be semigroups and let 휑 ∶ 푆 → 푇 be a homomorphism. The homomorphism 휑 is a categorical monomorphism if, for any semigroup 푈 and homomorphisms 휓1, 휓2 ∶ 푈 → 푆,

휓1 ∘ 휑 = 휓2 ∘ 휑 ⇒ 휓1 = 휓2, (1.16) and a categorical epimorphism if, for any semigroup 푈 and homomorphisms 휓1, 휓2 ∶ 푇 → 푈,

휑 ∘ 휓1 = 휑 ∘ 휓2 ⇒ 휓1 = 휓2. (1.17) [These are the definitions of ‘monomorphism’ and ‘epimorphism’ used in category theory; the word ‘categorical’ is simply being used to avoid ambiguity here.]

Exercises • 33 a) Prove that 휑 is a monomorphism (as defined on page 19) if and only if it is a categorical monomorphism. [Therefore, for semigroups, monomorphisms and categorical monomorphisms coincide and there is no risk of confusion in using the term ‘monomorphism’.] b) i) Prove that a surjective homomorphism is a categorical epimorphism. ii) Prove that the inclusion map 휄 ∶ ℕ → ℤ is a categorical epimorphism. [Hint: prove the contrapositive of (1.17) with 휑 = 휄.] [Therefore, for semigroups, categorical epimorphisms are not necessarily surjective. For groups, ‘surjective homomorphism’ and ‘categorical epimorphism’ are equivalent. Some authors define ‘epimorphism’ as ‘surjective homomorphism’ for semigroups, but this risks confusion.]

1.17 Prove that if we restrict the maps 휌푥 in Theorem 1.22 to 푆 (instead of 1 푆 ), then the map 푥 ↦ 휌푥 may or may not be injective. [Hint: show that this map is injective if 푆 is a right zero semigroup but not if it is a left zero semigroup.] 1.18 Let 푌 be a semilattice. Prove that 푌 is a subdirect product of copies of the two-element semilattice 푇 = {푒, 푧}, where 푒 > 푧. 1.19 Let 퐼 and 퐽 be ideals of 푆 such that 퐼 ⊆ 퐽. Prove that 푆/퐽 ≃ (푆/퐼)/(퐽/퐼). 1.20 Let 퐼 and 퐽 be ideals of 푆. Prove that 퐼∩퐽 and 퐼∪퐽 are ideals. [Remember to prove that 퐼 ∩ 퐽 ≠ ∅.] Prove that (퐼 ∪ 퐽)/퐽 ≃ 퐼/(퐼 ∩ 퐽). 1.21 Let 푆 be a semigroup with a zero and let 푇 be a subset of 푆 that contains 0푆 and at least one other element. Prove that 푇 = 퐺 ∪ {0푆} for some subgroup 퐺 of 푆 if and only if 푡푇 = 푇푡 = 푇 for all 푡 ∈ 푇 ∖ {0푆}. [This is an analogue of Lemma 1.9 for groups with a zero adjoined.]

Notes

Most of the definitions and results in this chapter are ‘folklore’. ◆ The exposition owes much to the standard accounts in Clifford & Preston, The Algebraic Theory of Semigroups, ch. 1 and Howie, Fundamentals of Semigroup Theory, ch. 1, which are probably the ne plus ultra of how to explain this material, and to a lesser extent Grillet, Semigroups, ch. i and Higgins, Techniques of Semi- group Theory, ch. 1. ◆ The number of non-isomorphic semigroups of order 8 is from Distler, ‘Classification and Enumeration of Finite Semigroups’, Table A.16. Exercise 1.11 appears as Howie, Fundamentals of Semigroup Theory, Exercise 1.6, but contains a minor error in the original. ◆ For an alternative approach to basic semigroup theory, Ljapin, Semigroups covers fundamental topics in much greater detail. For an account of structure theory that allows a semigroup to be

Notes • 34 empty, see Grillet, Semigroups. For further reading on the issues discussed in Exercise 1.16, the standard text on category theory remains Mac Lane, Categories for the Working Mathematician. For the situation for groups, see Linderholm, ‘A group epimorphism is surjective’. •

Notes • 35 Free semigroups & presentations 2

how can we think both of presentations as conforming to objects, and ‘ objects as conforming to presentations? is, not the first, but the highest task of transcendental philosophy. — Friedrich Wilhelm Joseph’ von Schelling, System of Transcendental Philosophy, § 3.

• Informally, a free semigroup on a set 퐴 is the unique biggest, most ‘general’ semigroup generated by [any set in bijection with] 퐴, in the sense that all other semigroups generated by 퐴 are homomorphic images (and thus factor semigroups) of the free semigroup on 퐴. This chapter studies some of the interesting properties of free semigroups and then explains their role in semigroup presentations, which can be used to define and manipulate semigroups as factor semigroups of free semigroups.

Alphabets and words

An alphabet is an abstract set of elements called letters Alphabet, letter, word or symbols. Let 퐴 be an alphabet. A word over 퐴 is a finite sequence (푎1, 푎2, … , 푎푚), where each term 푎푖 of the sequence is a letter from 퐴. The length of this word is 푚. There is also a word of length 0, which is the empty sequence (). This is called the empty word. The set of all 퐴+, 퐴∗ words (including the empty word) over 퐴 is denoted 퐴∗. The set of all non-empty words (that is, of length 1 or more) over 퐴 is denoted 퐴+. Multiplication of words is simply concatenation: that is, for all words Multiplication of words ∗ (푎1, 푎2, … , 푎푚), (푏1, 푏2, … , 푏푛) ∈ 퐴 ,

(푎1, 푎2, … , 푎푚)(푏1, 푏2, … , 푏푛) = (푎1, 푎2, … , 푎푚, 푏1, 푏2, … , 푏푛) It is easy to see that this multiplication is associative and so 퐴∗ is a semigroup; furthermore, the empty word () is an identity and so 퐴∗ is a monoid. Since the product of two words of non-zero length must itself have non-zero length, 퐴+ is a subsemigroup of 퐴∗; indeed, 퐴∗ is [isomorphic to] (퐴+)1. Because of associativity, we simply write 푎1푎2 ⋯ 푎푛 for (푎1, 푎2, … , 푎푛) Notation for words

• 36 푏3 푏 푎 푏2 푏2푎

푏 푏푎푏 푏 푎 푎 푏 푏푎 푏푎2

푎푏2 푏 푎 푏 푎푏 푎푏푎

푏 푎2푏 푏 FIGURE 2.1 Part of the Cayley graph 휀 푎 푎 푎 2 푎 3 푎 푎 훤(퐴∗, 퐴), where 퐴 = {푎, 푏}. and write 휀 for the empty word. For any word 푢 ∈ 퐴∗, denote the length of 푢 by |푢|, and notice that |푢| = 0 if and only if 푢 = 휀. Note further that |푢푣| = |푢| + |푣| for any 푢, 푣 ∈ 퐴∗. A subword of a word 푎1푎2 ⋯ 푎푛 (where 푎푖 ∈ 퐴) is any word of the Subword form 푎푖 ⋯ 푎푗, where 1 ⩽ 푖 ⩽ 푗 ⩽ 푛.A prefix of 푎1푎2 ⋯ 푎푛 is a subword 푎1 ⋯ 푎푖, where 1 ⩽ 푖 ⩽ 푛. The Cayley graph 훤(퐴∗, 퐴) is an infinite tree; an example for 퐴 = {푎, 푏} is shown in Figure 2.1. This is obvious, because if we start at 휀 and follow the path labelled by 푢 ∈ 퐴∗, then we end up at the vertex 푢. Thus a path uniquely determines a vertex and so the graph must be a tree.

Universal property

Let 퐹 be a semigroup and let 퐴 be an alphabet. Let 휄 ∶ Free semigroup 퐴 ,→ 퐹 be an embedding of 퐴 into 퐹. Then (퐹, 휄) is a free semigroup on 퐴 if, for any semigroup 푆 and map 휑 ∶ 퐴 → 푆, there is a unique homomorphism 휑+ ∶ 퐹 → 푆 that extends 휑 (that is, with 휄휑+ = 휑). Using

Universal property • 37 diagrams, this definition says that (퐹, 휄) is a free semigroup on 퐴 if 휄 퐴 퐹 } } for all , there exists a unique } } 휑 } 휄 푆 퐴 퐹 } (2.1) } + + } homomorphism 휑 such that 휑 . } 휑 } 푆 } Usually, we just write ‘퐹 is a free semigroup on 퐴’ instead of the precisely correct ‘(퐹, 휄) is a free semigroup on 퐴’. P r o p o s i t i o n 2 . 1. Let 퐴 be an alphabet and let 퐹 be a semigroup. Uniqueness of the Then 퐹 is a free semigroup on 퐴 if and only if 퐹 is isomorphic to 퐴+. free semigroup on 퐴 Proof of 2.1. Part 1. Let us first show that 퐴+ is a free semigroup on 퐴. Let 휄 ∶ 퐴 ,→ 퐴+ be the natural embedding map. Let 푆 be a semigroup and 휑 ∶ 퐴 → 푆 be a map. Define 휑+ ∶ 퐴+ → 푆 by + (푎1푎2 ⋯ 푎푛)휑 = (푎1휑)(푎2휑) ⋯ (푎푛휑). (2.2) It is easy to see that 휑+ is a homomorphism and that 휄휑+ = 휑. We now have to prove that 휑+ is unique. So let 휓 ∶ 퐴+ → 푆 be an arbitrary + homomorphism with 휄휓 = 휑. For any 푎1푎2 ⋯ 푎푛 ∈ 퐴 ,

(푎1푎2 ⋯ 푎푛)휓

= (푎1휓)(푎2휓) ⋯ (푎푛휓) [since 휓 is a homomorphism] + + + + = (푎1휑 )(푎2휑 ) ⋯ (푎푛휑 ) [since 휄휓 = 휑 = 휄휑 ] + + = (푎1푎2 ⋯ 푎푛)휑 . [since 휑 is a homomorphism] and so 휓 = 휑+. Hence 휑+ is the unique homomorphism from 퐴+ to 푆 with 휄휑+ = 휑, and so 퐴+ is free on 퐴. Now suppose that 퐹 is isomorphic to 퐴+ via an isomorphism 휗 ∶ 퐴+ → 퐹. The embedding map is 휗휄 ∶ 퐴 ,→ 퐹. Let 휑 ∶ 퐴 → 푆 be a map. Let 휏 = 휗휑+ (where 휑+ is the homomorphism defined in (2.2)); then 휏 ∶ 퐹 → 푆 is a homomorphism extending 휑. To see that it is unique, let 휎 ∶ 퐹 → 푆 be an arbitrary homomorphism extending 휑. Then 휗−1휎 ∶ 퐴+ → 푆 is a homomorphism extending 휑. Since 퐴+ is a free semigroup, −1 + −1 + 휗 휎 = 휑 , and so 휎 = id퐹휎 = 휗휗 휎 = 휗휑 = 휏. So 휏 ∶ 퐹 → 푆 is the unique homomorphism extending 휑 and so 퐹 is a free semigroup on 퐴. Part 2. Suppose that 퐹 is a free semigroup on 퐴; the aim is to show that 퐹 + + is isomorphic to 퐴 . Let 휄1 ∶ 퐴 ,→ 퐴 and 휄2 ∶ 퐴 ,→ 퐹 be the embedding + + maps. Since 퐴 is free on 퐴, we can put 휄1, 퐴 , 휄2 and 퐹 in the places of 휄, + 퐹, 휑 and 푆 in (2.1) to see that there is a homomorphism 휄2 such that 휄 퐴 1 퐴+ + . (2.3) 휄2 휄2 퐹

Universal property • 38 + Similarly, since 퐹 is free on 퐴, we can put 휄2, 퐹, 휄1 and 퐴 in the places of + 휄, 퐹, 휑 and 푆 in (2.1) to see that there is a homomorphism 휄1 such that 휄 퐴 2 퐹 휄+ . (2.4) 휄1 1 퐴+ Combining (2.3) and (2.4) in two ways, we get the following diagrams: 휄 휄 퐴 1 퐴+ 퐴 2 퐹

휄2 + 휄1 + 휄2 휄1 퐹 and 퐴+ . (2.5) 휄1 휄2 + + 휄1 휄2 퐴+ 퐹

+ + + + Therefore 휄1 = 휄1휄2 휄1 and 휄2 = 휄2휄1 휄2 . In diagrammatic terms, this corresponds to simplifying the diagrams in (2.5) to give 휄 휄 퐴 1 퐴+ 퐴 2 퐹 + + and + + . (2.6) 휄2 휄1 휄 휄 휄1 휄2 1 2 퐴+ 퐹 Clearly the following diagrams commute: 휄 휄 퐴 1 퐴+ 퐴 2 퐹

id퐴+ and id퐹 . (2.7) 휄1 휄2 퐴+ 퐹

+ Therefore, by the left-hand diagrams in (2.6) and (2.7), if we put 휄1, 퐴 , + 휄1 and 퐴 in place of 휄, 퐹, 휑, and 푆 in (2.1), then the homomorphisms + + + 휄2 휄1 and id퐴+ are both possibilities for 휑 . But (2.1) requires that there is + + + a unique such homomorphism 휑 , so 휄2 휄1 = id퐴+ . Similarly, using the + + + right-hand diagrams in (2.6) and (2.7), we obtain 휄1 휄2 = id퐹. Hence 휄1 + and 휄2 are mutually inverse isomorphisms, and so 퐹 is isomorphic to + 퐴 . 2.1

We could repeat the discussion above, but for monoids instead of Free monoids semigroups. Let 퐹 be a monoid and let 퐴 be an alphabet, and let 휄 ∶ 퐴 ,→ 퐹 be an embedding of 퐴 into 퐹. Then (퐹, 휄) is a free monoid on 퐴 if, for any monoid 푆 and map 휑 ∶ 퐴 → 푆, there is a unique monoid homomorphism 휑∗ ∶ 퐹 → 푆 extending 휑; that is, with 휄휑∗ = 휑. One can prove an analogy of Proposition 2.1 for monoids, showing that a monoid 퐹 is a free on 퐴 if and only if 퐹 ≃ 퐴∗. As with free semigroups, we usually write ‘퐹 is the free monoid on 퐴’ instead of ‘(퐹, 휄) is the free monoid on 퐴’.

Universal property • 39 Properties of free semigroups In preparation for our study of presentations, we begin by examining the structure of free semigroups and monoids.

P r o p o s i t i o n 2 . 2. Let 푀 be a submonoid of 퐴∗. Let 푁 = 푀 ∖ {휀}. Then 푁 ∖ 푁2 is the unique minimal (monoid) generating set for 푀.

Proof of 2.2. Clearly, any generating set for 푀 must contain 푁∖푁2. So we must show that Mon⟨푁∖푁2⟩ = 푀. Clearly Mon⟨푁∖푁2⟩ ⊆ 푀; we have to prove that 푀 ⊆ Mon⟨푁∖푁2⟩. Wealready know that 휀 ∈ Mon⟨푁∖푁2⟩, so it remains to show that 푁 ⊆ Mon⟨푁 ∖ 푁2⟩. Assume that all words of length less than ℓ in 푁 lie in Mon⟨푁 ∖ 푁2⟩. Let 푢 ∈ 푁 with |푢| = ℓ. If 푢 ∈ 푁 ∖ 푁2, then 푢 ∈ Mon⟨푁 ∖ 푁2⟩. On other other hand, if 푢 ∉ 푁 ∖ 푁2, then 푢 ∈ 푁2 and so 푢 = 푢′푢″ for 푢′, 푢″ ∈ 푁. Hence |푢′| = |푢| − |푢″| and |푢″| = |푢| − |푢′|. Since neither 푢′ nor 푢″ is the empty word, this gives |푢′|, |푢″| < |푢| = ℓ. So, by assumption, 푢′, 푢″ ∈ Mon⟨푁 ∖ 푁2⟩ and so 푢 ∈ Mon⟨푁 ∖ 푁2⟩. Hence, by induction, 2 푁 ⊆ Mon⟨푁 ∖ 푁 ⟩. 2.2

The base of a submonoid or subsemigroup 푀 of 퐴∗ is defined to be Base 푁∖푁2, where 푁 = 푀∖{휀}. Thus the base is the unique minimal monoid generating set for 푀 if 푀 is a submonoid, and is the unique minimal generating set for 푀 if 푀 is a subsemigroup that is not a submonoid. As an immediate application of Proposition 2.2, we see that 퐴 is the base of 퐴∗ and 퐴+.

P r o p o s i t i o n 2 . 3. A semigroup 푆 is free if and only if every element of 푆 has a unique representative as a product of elements of 푆 ∖ 푆2.

Proof of 2.3. Clearly every element of 퐴+ has a unique representative as a product of elements of 퐴 = 퐴+ ∖ (퐴+)2. So assume that every element of 푆 has a unique representative as a product of elements of 퐴 = 푆 ∖ 푆2. We will show that 푆 satisfies the definition of freedom. Let 푇 be a semigroup and 휑 ∶ 퐴 → 푇 a map. Define a + + map 휑 ∶ 푆 → 푇 by letting 푠휑 = (푎1휑)(푎2휑) ⋯ (푎푛휑), where 푎1푎2 ⋯ 푎푛 is the unique representative of 푠 as a product of elements 푎푖 ∈ 퐴. Notice that if 푡 ∈ 푆 is uniquely represented 푏1 ⋯ 푏푚 where 푏푖 ∈ 퐴, then 푠푡 has + unique representative 푎1 ⋯ 푎푛푏1 ⋯ 푏푚. Hence 휑 is a homomorphism. It is clear that 휑+ is the unique homomorphism extending 휑 and so 푆 is free on 퐴. 2.3

∗ P r o p o s i t i o n 2 . 4. Let 퐴 = {푥, 푦}. Let 퐵 = { 푏푖 ∶ 푖 ∈ ℕ }. Then 퐴 Free monoid of rank 2 contains a submonoid isomorphic to 퐵∗. contains a free monoid of countably infinite rank ∗ 푖 ∗ Proof of 2.4. Define a map 휑 ∶ 퐵 → 퐴 by 푏푖휑 = 푥푦 푥. Since 퐵 is free on 퐵, this map 휑 extends to a unique homomorphism, which we also denote 휑, from 퐵∗ to 퐴∗.

Properties of free semigroups • 40 Suppose, with the aim of obtaining a contradiction, that 휑 is not injective. Then there exist 푢, 푣 ∈ 퐵∗ with 푢휑 = 푣휑. Suppose 푢 and 푣 begin with the same symbol 푏; that is, 푢 = 푏푢′ and 푣 = 푏푣′. Then (푏휑)(푢′휑) = (푏휑)(푣′휑) and so 푢′휑 = 푣′휑 by cancellativity in 퐴∗. So we can replace 푢 by 푢′ and 푣 by 푣′ and repeat this process until we have words 푢 and 푣 beginning with different symbols. Therefore assume that 푢 and 푣 begin with symbols 푏푖 and 푏푗 respectively, where 푖 ≠ 푗; that is, 푢 = 푏푖푢′ and 푣 = 푏푗푣′. 푖 푗 Then 푥푦 푥(푢′휑) = (푏푖휑)(푢′휑) = (푏푗휑)(푣′휑) = 푥푦 푥(푣′휑). Assume 푖 > 푗; the other case is similar. By cancellativity in 퐴∗, we have 푦푖−푗푥(푢′휑) = 푥(푣′휑), which is a contradiction since 푖 − 푗 > 0. Therefore 휑 is injective ∗ and so 퐵 is isomorphic to im 휑. 2.4 As a consequence of Proposition 2.4, we see that the free monoid on {푥, 푦} contains submonoids isomorphic to all free monoids on countable sets. A similar result holds for free semigroups.

E x a m p l e 2 . 5. Let 퐴 = {푥} and let 푆 = ⟨푥2, 푥3⟩. Then 푆∖푆2 = {푥2, 푥3}. Free semigroups can contain But 푥5 ∈ 푆 and 푥5 = 푥2푥3 = 푥3푥2, so 푥5 has two distinct representatives non-free subsemigroups as a product of elements of {푥2, 푥3}. Hence 푆 is not a free semigroup by Proposition 2.3.

Example 2.5 shows that a free semigroup contains subsemigroups that are not themselves free. In contrast, every subgroup of a free group is itself a free group by the famous Nielsen–Schreier theorem.

Semigroup presentations

The reason why free semigroups are interesting is that Every semigroup is a every semigroup is isomorphic to a quotient of a free semigroup. To see quotient of a free semigroup this, let 휑 ∶ 퐴 → 푆 be such that im 휑 generates 푆. (We could, for instance, choose 퐴 to be a set of the same cardinality as 푆 and 휑 to be a bijection.) Then, 휑 extends to a homomorphism 휑+ ∶ 퐴+ → 푆. Since im 휑 generates 푆, we have im 휑+ = 푆. By Theorem 1.24, 퐴+/ker 휑+ ≃ im 휑+ = 푆. That is, 푆 is isomorphic to the quotient 퐴+/ker 휑+. This is slightly interesting, but its real importance is when we turn it around. Instead of starting with a semigroup and knowing that it is a quotient of a free semigroup, we can specify a free semigroup 퐴+ and a congruence 휎 and so define the corresponding quotient semigroup 퐴+/휎. This is the idea of a semigroup presentation. It allows us to specify and reason about a semigroup as a quotient of a free semigroup: that is, as a quotient 퐴+/휎 for some congruence 휎 on the free semigroup 퐴+. By Proposition 2.1, in order to specify the free semigroup, it is sufficient to

Semigroup presentations • 41 specify the alphabet 퐴. In order to specify the congruence 휎, it is sufficient to specify some binary relation 휌 that generates 휎. A semigroup presentation is a pair Sg⟨퐴 | 휌⟩, where 퐴 is an alphabet Presentations and 휌 is a binary relation on 퐴+. The elements of 퐴 are called generating symbols, and the elements of 휌 (which are pairs of words in 퐴+) are called defining relations. The presentation Sg⟨퐴 | 휌⟩ defines, or presents, any semigroup isomorphic to 퐴+/휌#. Let 푆 be a semigroup presented by Sg⟨퐴 | 휌⟩. Then 푆 is isomorphic to 퐴+/휌# and so there is a one-to-one correspondence between elements of 푆 and 휌#-classes. Thus we can think of a word 푤 ∈ 퐴+ as representing + the element of 푆 corresponding to [푤]휌# . If 푢, 푣 ∈ 퐴 represent the same # element of 푆 (that is, if (푢, 푣) ∈ 휌 , or, equivalently, if [푢]휌# = [푣]휌# ), we say that 푢 and 푣 are equal in 푆 and write 푢 =푆 푣. Let 푇 be a semigroup. Let 휑 ∶ 퐴 → 푇 be a map such that 퐴휑 generates Assignment of generators 푇; such a map is called an assignment of generators. In this case, the unique homomorphism 휑+ ∶ 퐴+ → 푇 extending 휑 is surjective. The semigroup 푇 satisfies a defining relation (푢, 푣) ∈ 휌 with respect Satisfying a defining relation to an assignment of generators 휑 ∶ 퐴 → 푇 if 푢휑+ = 푣휑+. Notice that 푇 satisfies all defining relations in 휌 with respect to 휑 ∶ 퐴 → 푇 if and only if 휌 ⊆ ker 휑+. By definition, any semigroup defined by the presentation Sg⟨퐴 | 휌⟩ satisfies the defining relations 휌 with respect to the assignment # ♮ + # of generators (휌 ) |퐴 ∶ 퐴 → 퐴 /휌 . P r o p o s i t i o n 2 . 6. Let 푇 be a semigroup, and suppose 푇 satisfies the defining relations in 휌 with respect to an assignment of generators 휑 ∶ 퐴 → 푇. Then 푇 is a homomorphic image of the semigroup presented by Sg⟨퐴 | 휌⟩.

Proof of 2.6. Since 푇 satisfies the defining relations 휌 with respect to 휑, we have 휌 ⊆ ker 휑+. Since ker 휑+ is a congruence by Theorem 1.24, and 휌# is the smallest congruence containing 휌, it follows that 휌# ⊆ ker 휑+. + # + So the map 휓 ∶ 퐴 /휌 → 푇 defined by [푢]휌# 휓 = 푢휑 is a well-defined + homomorphism, and is clearly surjective since 휑 is surjective. 2.6 By Proposition 2.6, we can think of semigroup presented by Sg⟨퐴 | 휌⟩ as the largest semigroup generated by 퐴 and satisfying the defining relations in 휌. An elementary 휌-transition is a pair (푤, 푤′) ∈ (휌C)S = 휌C ∪ (휌C)−1, Elementary transition which we denote 푤 ↔휌 푤′. By Proposition 1.27, 푤 ↔휌 푤′ if and only if 푤′ can be obtained from 푤 by substituting a subword 푦 for a subword 푥 of 푤, where (푥, 푦) ∈ 휌 or (푦, 푥) ∈ 휌. In this situation, we say that we apply the defining relation (푥, 푦) or (푦, 푥) to the word 푤 and obtain 푤′. Let 푢, 푣 ∈ 퐴+. If there is a sequence of elementary 휌-transitions 푢 = 푤0 ↔휌 … ↔휌 푤푛 = 푣, then we say (푢, 푣) is a consequence of 휌, or (푢, 푣) can be deduced from 휌. The following result shows the connection between this notion and presentations:

Semigroup presentations • 42 P r o p o s i t i o n 2 . 7. Let 푆 be presented by Sg⟨퐴 | 휌⟩ and let 푢, 푣 ∈ 퐴+. Then 푢 =푆 푣 if and only if (푢, 푣) is a consequence of 휌; that is, if and only if there is a sequence of elementary 휌-transitions

푢 = 푤0 ↔휌 푤1 ↔휌 … ↔휌 푤푛 = 푣. Proof of 2.7. First of all, note that

푢 =푆 푣 ⇔ (푢, 푣) ∈ 휌# ⇔ (푢, 푣) ∈ (휌C)E [by Proposition 1.29] ∞ C C −1 푛 ⇔ (푢, 푣) ∈ id퐴+ ∪ ⋃푛=1(휌 ∪ (휌 ) ) [by Proposition 1.26(f)] ⇔ (푢 = 푣) ∨ (∃푛 ∈ ℕ)((푢, 푣) ∈ (휌C ∪ (휌C)−1)푛) + ⇔ (∃푛 ∈ ℕ ∪ {0})(∃푤0, … , 푤푛 ∈ 퐴 )

[(푢 = 푤0) ∧ (푤푛 = 푣) C C −1 ∧ (∀푖)((푤푖, 푤푖+1) ∈ 휌 ∪ (휌 ) )] + ⇔ (∃푛 ∈ ℕ ∪ {0})(∃푤0, … , 푤푛 ∈ 퐴 )

[(푢 = 푤0) ∧ (푤푛 = 푣) ∧ (∀푖)(푤푖 ↔휌 푤푖+1))].

Hence 푢 =푆 푣 if and only if there is there is a sequence of elementary 휌-transitions from 푢 to 푣. 2.7 The next result gives a usable condition for when a given presentation defines a particular semigroup. Afterwards, we will see how this result yields a practical proof method.

P r o p o s i t i o n 2 . 8. Let 푆 be a semigroup. Then Sg⟨퐴 | 휌⟩ presents 푆 if Condition for and only if there is an assignment of generators 휑 ∶ 퐴 → 푆 such that Sg⟨퐴 | 휌⟩ to define 푆 a) 푆 satisfies the defining relations in 휌 with respect to 휑, and b) if 푢, 푣 ∈ 퐴+ are such that 푢휑+ = 푣휑+, then (푢, 푣) is a consequence of 휌. Proof of 2.8. Suppose first that Sg⟨퐴 | 휌⟩ presents 푆. Then 푆 is isomorphic to 퐴+/휌#, so we can let 휑 be the restriction of natural homomorphism # ♮ + + # (휌 ) |퐴 ∶ 퐴 → 퐴 /휌 . Then condition a) holds from the definition and condition b) holds from Proposition 2.7. Now suppose that conditions a) and b) hold. Since 푆 satisfies the defining relations in 휌, we have 휌 ⊆ ker 휑+ and so 휌# ⊆ ker 휑+. If (푢, 푣) ∈ ker 휑+, then 푢휑+ = 푣휑+ and so (푢, 푣) is a consequence of 휌 and hence (푢, 푣) ∈ 휌# by Proposition 2.7. Hence 휌# = ker 휑+ and therefore 푆 ≃ + + + # 퐴 /ker 휑 ≃ 퐴 /휌 ; thus Sg⟨퐴 | 휌⟩ presents 푆. 2.8 There is a standard three-step method for directly proving that a presentation defines a particular semigroup:

M e t h o d 2 . 9. To prove that a presentation Sg⟨퐴 | 휌⟩ defines a particu- Method for proving lar semigroup 푆: Sg⟨퐴 | 휌⟩ defines 푆

Semigroup presentations • 43 1) Define an assignment of generators 휑 ∶ 퐴 → 푆, and prove that 푆 satisfies the defining relations in 휌 with respect to 휑. 2) Find a set of words 푁 ⊆ 퐴+ such that for every word 푤 ∈ 퐴+ there is a word ̂푤∈ 푁 such that (푤, ̂푤) is a consequence of 휌. + 3) Prove that 휑 |푁 is injective. In Method 2.9, step 1 establishes that condition a) of Proposition 2.8 holds. Now let 푢, 푣 ∈ 퐴+ be such that 푢휑+ = 푣휑+. Step 2 shows that (푢, ̂푢) and (푣, ̂푣) are consequences of 휌; that is, there are sequences of elementary 휌-transitions 푢 ↔휌 … ↔휌 ̂푢 and 푣 ↔휌 … ↔휌 ̂푣. Since 푆 satisfies the relations in 휌, this implies that ̂푢휑+ = ̂푣휑+, so step 3 shows that ̂푢= ̂푣, and thus there is a sequence of elementary 휌-transitions 푢 ↔휌 … ↔휌 ̂푢= ̂푣↔휌 … ↔휌 푣; that is, (푢, 푣) is a consequence of 휌. This establishes condition b) of Proposition 2.8 and so proves that Sg⟨퐴 | 휌⟩ presents 푆. Before giving some examples to illustrate the theory described above, we introduce a convention to simplify notation. When we explicitly list generating symbols and defining relations in a presentation, we do not write the braces {} enclosing the list of elements in the two sets. So instead of Sg⟨{푎1, 푎2, …} | {(푢1, 푣1), (푢2, 푣2), …}⟩, we write Sg⟨푎1, 푎2, … | (푢1, 푣1), (푢2, 푣2), …⟩. E x a m p l e 2 . 1 0. a) Let us prove that the presentation Sg⟨퐴 | ⟩ (with no defining relations) defines the free semigroup 퐴+. To see this, it # + + suffices to notice that ∅ = id퐴+ , and 퐴 /id퐴+ ≃ 퐴 . [Following Method 2.9 for the sake of illustration, let 휑 be the embedding map 휄 ∶ 퐴 ,→ 퐴+. Clearly 퐴+ trivially satisfies all defining relations with respect to 휑; this is step 1. Let 푁 = 퐴+; then every word in 퐴+ itself in 푁 and so step 2 is immediately proved. Finally, step 3 is trivial since 휑+ is the identity map and so injective.] b) Now we prove that the presentation Sg⟨푎 | (푎2, 푎)⟩ defines the trivial semigroup {푒}. To see this, it suffices to notice that {(푎2, 푎)}# = {푎}+ × {푎}+. [Following Method 2.9 for the sake of illustration, let 휑 ∶ {푎} → {푒} be given by 푎휑 = 푒. Then 푎2휑+ = (푎휑+)2 = 푒2 = 푒 = 푎휑+ and so the semigroup {푒} satisfies the defining relation (step 1). Let 푁 = {푎}. Then any word in {푎}+ can be transformed to the unique word 푎 ∈ 푁 by repeatedly applying the defining relation (step 2). Finally, 푁 contains + only a single element and hence 휑 |푁 is trivially injective (step 3).] c) Less trivially, we now prove that the presentation Sg⟨퐴 | (푎푏, 푎) ∶ 푎, 푏 ∈ 퐴⟩ defines a left zero semigroup on a set of size |퐴|. Following Method 2.9, let 푆 be the left zero semigroup with |퐴| elements and let 휑 ∶ 퐴 → 푆 be a bijection. Then (푎푏)휑+ = (푎휑+)(푏휑+) = 푎휑+ since 푆 is a left zero semigroup, and so 푆 satisfies the defining relations (step 1). Let 푁 = 퐴; then any word in 퐴+ can be transformed to one in 푁 by

Semigroup presentations • 44 applying defining relations to replace a subword 푎푏 by 푎; this yields a shorter word and so ends with a word in 퐴 (step 2). Finally, if 푎, 푏 ∈ 푁 are such that 푎휑+ = 푏휑+, then 푎휑 = 푎휑+ = 푏휑+ = 푏휑, which implies + 푎 = 푏 since 휑 is a bijection; thus 휑 |푁 is injective (step 3).

d) Consider the set 푀3(ℤ) of all 3 × 3 integer matrices. Let

1 0 0 1 1 0 1 0 1 푃 = [0 1 1] , 푄 = [0 1 0] , 푅 = [0 1 0] , [0 0 1] [0 0 1] [0 0 1]

and let 푆 be the subsemigroup of 푀3(ℤ) generated by {푃, 푄, 푅}. Let us prove that 푆 is presented by

Sg⟨푎, 푏, 푐 | (푏푎, 푎푏푐), (푐푎, 푎푐), (푐푏, 푏푐)⟩.

First, let 휑 ∶ {푎, 푏, 푐} → 푆 be given by 푎휑 = 푃, 푏휑 = 푄, and 푐휑 = 푅. Straightforward calculations show that 푆 satisfies the defining relations with respect to 휑+ (step 1). Let

푁 = { 푎푖푏푗푐푘 ∶ 푖, 푗, 푘 ∈ ℕ ∪ {0} ∧ 푖, 푗, 푘 not all 0 }.

Every word in {푎, 푏, 푐}+ can be transformed to one in 푁 as follows: First, by applying the second and third defining relations from left to right, we move all symbols 푐 to the right of the word. Then, if there is some symbol 푏 to the left of a symbol 푎, we apply the first defining relation, and move the ‘new’ symbol 푐 to the right of the word. We repeat this step until there is no symbol 푏 to the left of a symbol 푎. This process must terminate because no application of a relation changes the number of symbols 푎 or 푏 in the word. At the end of the process, we are left with a word in 푁 (step 2). Finally, a simple calculation shows that 1 푗 푘 (푎푖푏푗푐푘)휑+ = [0 1 푖] , [0 0 1]

푖 푗 푘 푖′ 푗′ 푘′ and so if 푎 푏 푐 =푆 푎 푏 푐 , then 푖 = 푖′, 푗 = 푗′, and 푘 = 푘′; hence + 휑 |푁 is injective (step 3). [Note that the subgroup of 푀3(ℤ) generated by {푃, 푄, 푅} is the famous discrete Heisenberg group 퐻3(ℤ).]

We could repeat the discussion of presentations above, but reasoning Monoid presentations about monoids instead of semigroups. Every monoid is a quotient of a free monoid. In a monoid presentation Mon⟨퐴 | 휌⟩ the defining relations in 휌 are of the form (푢, 푣) for 푢, 푣 ∈ 퐴∗. In particular, they can be of the form (푢, 휀) or (휀, 푢) or (휀, 휀). The presentation Mon⟨퐴 | 휌⟩ presents the monoid 퐴∗/휌#. The notion of an assignment of generators carries over to monoids. The analogies of Propositions 2.6, 2.7, and 2.8 all hold for

Semigroup presentations • 45 monoids, using 퐴∗ instead of 퐴+ and 휑∗ instead of 휑+ as appropriate. Thus the monoid presented by Mon⟨퐴 | 휌⟩ is the largest monoid generated by 퐴 and satisfying the defining relations in 휌. If 푀 is presented by Mon⟨퐴 | 휌⟩, ∗ then for 푢, 푣 ∈ 퐴 , we have 푢 =푀 푣 if and only if there is a sequence of elementary 휌-transitions from 푢 to 푣. Finally, Method 2.9 works for monoids, again with 휑∗ instead of 휑+.

E x a m p l e 2 . 1 1. a) Let us prove that the monoid (ℕ ∪ {0}) × (ℕ ∪ {0}) is presented by Mon⟨푎, 푏 | (푎푏, 푏푎)⟩. Following the monoid version of Method 2.9, let 휑 ∶ {푎, 푏} → (ℕ ∪ {0}) × (ℕ ∪ {0}) be defined by 푎휑 = (1, 0) and 푏휑 = (0, 1). Then (푎푏)휑∗ = (1, 0)(0, 1) = (1, 1) = (0, 1)(1, 0) = (푏푎)휑∗, so (ℕ ∪ {0}) × (ℕ ∪ {0}) satisfies the defining relations with respect to 휑 (step 1). Let 푁 = { 푎푖푏푗 ∶ 푖, 푗 ∈ ℕ ∪ {0} } Then every word in {푎, 푏}∗ can be transformed to one in 푁 by applying the defining relation to move symbols 푎 to the left of symbols 푏 (step 2). Finally, note that if 푎푖푏푗휑∗ = 푎푖′푏푗′휑∗, then (푖, 푗) = (푖′, 푗′) ∗ and so 푖 = 푖′ and 푗 = 푗′; thus 휑 |푁 is injective (step 3).

b) Let 휏, 휎 ∈ Tℕ be given by

1 2 3 4 … 1 2 3 4 … 휏 = ( ) , 휎 = ( ), 2 3 4 5 … 1 1 2 3 …

and let 퐵 be the submonoid generated by {휏, 휎}. Let us prove that 퐵 is presented by Mon⟨푏, 푐 | (푏푐, 휀)⟩. Define 휑 ∶ {푏, 푐} → 퐵 by 푏휑 = 휏 and ∗ ∗ 푐휑 = 휎. Then (푏푐)휑 = 휏휎 = idℕ = 휀휑 , so 퐵 satisfies the defining relation with respect to 휑 (step 1). Let 푁 = { 푐푖푏푗 ∶ 푖, 푗 ∈ ℕ ∪ {0} }; then every word in {푏, 푐}∗ can be transformed to one in 푁 by using the defining relations to replace subwords 푏푐 by 휀 (effectively ‘deleting’ the subword 푏푐) and ultimately yielding one in 푁 (step 2). Finally,

(푐푖푏푗)휑∗ 1 2 … 푖 + 1 푖 + 2 … = ( ) 1 1 … 1 2 … 1 2 3 4 … ( ) 푗+1 푗+2 푗+3 푗+4 … 1 2 … 푖 + 1 푖 + 2 … = ( ), 푗+1 푗+1 … 푗+1 푗+2 …

and so in the image of 푐푖푏푗, the image of 1 is 푗 + 1 and the maximum element of the domain with image 푗 + 1 is 푖 + 1. That is, the image ∗ determines 푖 and 푗. Hence 휑 |푁 is injective (step 3). This monoid 퐵 defined by Mon⟨푏, 푐 | (푏푐, 휀)⟩ is the bicyclic monoid. Bicyclic monoid Every element of the bicyclic monoid is represented by a unique word of the form 푐푖푏푗, where 푖, 푗 ∈ ℕ ∪ {0}, and we normally work with these representatives of elements of 퐵. Multiplying using these

Semigroup presentations • 46 representatives is concatenation followed by deletion of subwords 푏푐. That is,

푐푖+푘−푗푏ℓ if 푘 ⩾ 푗, 푐푖푏푗 푐푘푏ℓ = { 퐵 푐푖푏푗−푘+ℓ if 푘 ⩽ 푗.

A presentation is finite if both 퐴 and 휌 are finite. The semigroup is Finite presentation finitely presented if is defined by some finite presentation.

P r o p o s i t i o n 2 . 1 2. Suppose 푆 is finitely presented and let 휑 ∶ 퐴 → 푆 Finite presentability be an assignment of generators (with 퐴 possibly being infinite). Then there is independent of exists a finite subset 퐵 of 퐴 and 휌 ⊆ 퐵+ × 퐵+ such that Sg⟨퐵 | 휌⟩ is a finite the generating set presentation defining 푆. Proof of 2.12. Since 푆 is finitely presented, it is defined by a finite presentation Sg⟨퐶 | 휏⟩. For brevity, let 휓 ∶ 퐶 → 푆 be the natural assignment of # ♮ generators (휏 ) |퐶, so that 푐휓 = [푐]휏# . For each 푐 ∈ 퐶, there exists a word 푐휁 ∈ 퐴+ such that 푐 and 푐휁 represent the same element of 푆. (We can choose 푐휁 to be any word in (푐휓)(휑+)−1.) Thus we have a map 휁 ∶ 퐶 → 퐴+ such that 휁+휑+ = 휓+. Let

퐵 = { 푏 ∈ 퐴 ∶ (∃푐 ∈ 퐶)(푏 appears in 푐휁) };

+ + + + thus 퐶휁 ⊆ 퐵 and so 휁 휑|퐵 = 휓 . Notice that 퐵 is finite since 퐶 is finite. Notice that ⟨퐵휑|퐵⟩ ⊇ ⟨퐶휓⟩ = 푆, so 휑|퐵 ∶ 퐵 → 푆 is an assignment of generators. 퐵 휂 휁 퐶 Similarly, for every 푏 ∈ 퐵, there exists a word 푏휂 ∈ 퐶+ such that 푏 and 푏휂 represent the same element of 푆. (We can choose 푏휂 to be any word 휂+ + −1 + + + + + + in (푏휑|퐵)(휓 ) .) Thus we have a map 휂 ∶ 퐵 → 퐶 such that 휂 휓 = 휑|퐵 . 퐵 퐶 + + + + + (Figure 2.2 shows the relationship between 휑|퐵 , 휓 , 휁 , and 휂 .) 휁 + + Let 휑|퐵 휓 푆 + + + + 휌 = { (푝휁 , 푞휁 ) ∶ (푝, 푞) ∈ 휏 } ∪ { (푏, 푏휂 휁 ) ∶ 푏 ∈ 퐵 }. FIGURE 2.2 + + Maps used in the proof of Pro- Note first that 휌 ⊆ 퐵 ×퐵 . Now, if (푝, 푞) ∈ 휏, then 푆 satisfies this defining position 2.12 + + + + + + relation with respect to 휓, so 푝휓 = 푞휓 , and hence 푝휁 휑|퐵 = 푞휁 휑|퐵 . + + + + + + Furthermore, if 푏 ∈ 퐵, then 푏휑|퐵 = 푏휂 휓 = 푏휂 휁 휑|퐵 . So 푆 satisfies every defining relation in 휌 with respect to 휑|퐵. + + + It remains to prove that if 푢, 푣 ∈ 퐵 are such that 푢휑|퐵 = 푣휑|퐵 , then + + + + (푢, 푣) is a consequence of 휌. So suppose 푢휑|퐵 = 푣휑|퐵 . Then 푢휂 휓 = 푣휂+휓+. So (푢휂+, 푣휂+) is a consequence of 휏. That is, there is a sequence of elementary 휏-transitions

+ + 푢휂 = 푤0 ↔휏 푤1 ↔휏 … ↔휏 푤푛 = 푣휂 . So, applying 휁+ to this sequence, we see that by the definition of 휌, there is a sequence of elementary 휌-transitions

+ + + + + + + 푢휂 휁 = 푤0휁 ↔휌 푤1휁 ↔휌 … ↔휌 푤푛휁 = 푣휂 휁 . (2.8)

Semigroup presentations • 47 Suppose 푢 = 푢1푢2 ⋯ 푢푘 and 푣 = 푣1푣2 ⋯ 푣ℓ, where 푢푖, 푣푖 ∈ 퐵. By the definition of 휌, there are also sequences of elementary 휌-transitions + + 푢 = 푢1푢2 ⋯ 푢푘 ↔휌 (푢1휂 휓 )푢2 ⋯ 푢푘 ↔휌 … + + + + + + + + } (2.9) ↔휌 (푢1휂 휓 )(푢2휂 휓 ) ⋯ (푢푘휂 휓 ) = 푢휂 휁 and + + + + + + + + 푣휂 휁 = (푣1휂 휓 )(푣2휂 휓 ) ⋯ (푣ℓ휂 휓 ) ↔휌 … + + } (2.10) ↔휌 (푣1휂 휓 )푣2 ⋯ 푣ℓ ↔휌 푣1푣2 ⋯ 푣ℓ = 푣. Concatenating the sequences (2.8), (2.9), and (2.10) shows that (푢, 푣) is a consequence of 휌 and so completes the proof. 2.12 We now give two more important examples. Example 2.13 shows that a semigroup can be finitely generated but not finitely presented. Example 2.14 then shows that cancellativity is not a sufficient condition for group-embeddability.

E x a m p l e 2 . 1 3. Let 푋 = {푥푦푧, 푦푧, 푦푡, 푥푦, 푧푦, 푧푦푡} ⊆ {푥, 푦, 푧, 푡}+. Let Finitely generated but 푆 be the subsemigroup of {푥, 푦, 푧, 푡}+ generated by 푋. not finitely presented Suppose, with the aim of obtaining a contradiction, that 푆 is finitely presented. Let 퐴 = {푎, 푏, 푐, 푑, 푒, 푓} and let 휑 ∶ 퐴 → 푆 be given by 푎휑 = 푥푦푧, 푏휑 = 푦푧, 푐휑 = 푦푡, 푑휑 = 푥푦, 푒휑 = 푧푦, 푓휑 = 푧푦푡. Clearly 푆 is presented by Sg⟨퐴 | ker 휑+⟩, since 퐴+/ker 휑+ ≃ 푆 by Theo- rem 1.24. Thus, by Proposition 2.12, 푆 is defined by a finite presentation Sg⟨퐴 | 휎⟩. Assume without loss of generality that 휎 contains no defining relations of the form (푢, 푢). Let 훼 be greater than the maximum length of a side of a defining relation in 휎. Now, (푎푏훼푐)휑+ = 푥(푦푧)훼+1푦푡 = (푑푒훼푓)휑+. 훼 훼 That is, 푎푏 푐 =푆 푑푒 푓. By Proposition 2.7, there is a sequence of elementary 휎-transitions 훼 훼 푎푏 푐 ↔휎 … ↔휎 푑푒 푓. (2.11) For any 훽 ∈ ℕ∪{0}, the word 푎푏훽 is the unique word over 퐴 representing (푎푏훽)휑 = 푥(푦푧)훽+1, the word 푏훽푐 is the unique word over 퐴 representing (푏훽푐)휑 = (푦푧)훽푦푡, and for 훽 ≠ 0 the word 푏훽 is the unique word over 퐴 representing (푏훽)휑 = (푦푧)훽. Hence 휎 cannot contain any defining relation of the form (푎푏훽, 푢) or (푏훽푐, 푣) or (푏훽, 푤). Thus in the sequence of elementary 휎-transitions (2.11), the first step must involve applying a defining relation of which one side is 푎푏훼푐. This contradicts the fact that 훼 is greater than the maximum length of a side of a defining relation in 휎. Therefore 푆 is not finitely presented.

Semigroup presentations • 48 Example 2.5 showed that it is possible for a free semigroup to contain subsemigroups that are not themselves free. By showing that a free semigroup can contain finitely generated subsemigroups that are not even finitely presented, Example 2.13 provides an even stronger contrast to the Nielsen–Schreier theorem.

E x a m p l e 2 . 1 4. Let 푆 be the semigroup presented by Sg⟨퐴 | 휌⟩, where Cancellative but not 퐴 = {푎, 푏, 푐, 푑, 푒, 푓, 푔, ℎ} and let 휌 = {(푎푒, 푏푓), (푐푓, 푑푒), (푑푔, 푐ℎ)}. We will group-embeddable prove that 푆 is cancellative but not group-embeddable. Proving that 푆 is cancellative involves many cases, so we prove the left-cancellativity condition for the generator represented by 푐; the other cases are similar. Suppose that 푐푢 =푆 푐푣; we aim to prove that 푢 =푆 푣. Then there is a sequence of elementary 휌-transitions

푐푢 = 푤0 ↔휌 … ↔휌 푤푛 = 푐푣. (2.12)

Without loss of generality, assume that 푛 is minimal among all such sequences. Suppose, with the aim of obtaining a contradiction, that at some step in this sequence the initial symbol 푐 is altered. This must involve applying one of the defining relations (푐푓, 푑푒) or (푑푔, 푐ℎ). Assume the former; the latter case is similar. Thus (2.12) is of the form

푐푢 = 푤0 ↔휌 … ↔휌 푐푓푤′ ↔휌 푑푒푤′ ↔휌 … ↔휌 푤푛 = 푐푣.

Now, no defining relation has one side starting with a symbol 푒, so the symbol 푒 must remain in the terms of the sequence until the defining relation (푐푓, 푑푒) is applied again to alter the initial symbol 푑. (We know that this relation must be applied because the sequence of elementary 휌-transitions ends with 푤푛 = 푐푣.) Thus2.12 ( ) is of the form

푐푢 = 푤0 ↔휌 … ↔휌 푐푓푤′ ↔휌 푑푒푤′ ↔휌 …

↔휌 푑푒푤″ ↔휌 푐푓푤″ ↔휌 … ↔휌 푤푛 = 푐푣.

Since the distinguished symbol 푒 is present throughout the subsequence 푑푒푤′ ↔휌 … ↔휌 푑푒푤″, so does the symbol 푑. Because the symbols 푑푒 are not involved in any of the intermediate steps, there is no need to include the two elementary 휌-transitions 푐푓푤′ ↔휌 푑푒푤′ and 푑푒푤″ ↔휌 푐푓푤″. That is, we can remove the elementary 휌-transitions 푐푓푤′ ↔휌 푑푒푤′ and 푑푒푤″ ↔휌 푐푓푤″ and replace the prefixes 푑푒 by 푐푓 in the subsequence 푑푒푤′ ↔휌 … ↔휌 푑푒푤″ and obtain a strictly shorter sequence of elementary 휌-transitions from 푐푢 to 푐푣. This contradicts the minimality of 푛. Therefore the initial symbol 푐 is never altered. Thus we can delete the initial symbol 푐 from each step in (2.12) to obtain a sequence of elementary 휌-transitions from 푢 to 푣. Hence 푢 =푆 푣. This argument proves that the left-cancellativity condition holds for the generator 푐. Reasoning similarly for the symbols in 퐴 ∖ {푐} shows

Semigroup presentations • 49 that 푆 is left-cancellative; symmetrical arguments show that 푆 is right cancellative. Thus 푆 is cancellative. Suppose 푆 is group-embeddable. Then there is a monomorphism 휑 ∶ 푆 → 퐺, where 퐺 is a group. Then

(푎푔)휑 = (푎휑)(푔휑) = (푎휑)(푒휑)(푒휑)−1(푔휑) −1 = (푏휑)(푓휑)(푒휑) (푔휑) [since 푎푒 =푆 푏푓] = (푏휑)(푐휑)−1(푐휑)(푓휑)(푒휑)−1(푔휑) −1 = (푏휑)(푐휑) (푑휑)(푒휑)(푒휑)(푔휑) [since 푐푓 =푆 푑푒] = (푏휑)(푐휑)−1(푑휑)(푔휑) −1 = (푏휑)(푐휑) (푐휑)(ℎ휑) [since 푑푔 =푆 푐ℎ] = (푏휑)(ℎ휑) = (푏ℎ)휑.

But 푎푔 ≠푆 푏ℎ, since there is no sequence of elementary 휌-transitions from 푎푔 to 푏ℎ because 푎푔 does not contain a subword that forms one side of a defining relation in 휌. This contradicts 휑 being a monomorphism and so 푆 is not group-embeddable.

Several of the syntactic arguments used in this chapter and in the exercises could be simplified by using the tools of string-rewriting. However, presenting the necessary theory is beyond the scope of this course.

Exercises [See pages 204–210 for the solutions.] ✴2.1 A semigroup 푆 is equidivisible if for all 푥, 푦, 푧, 푡 ∈ 푆, the following Equidivisibility holds:

푥푦 = 푧푡 ⇒ (∃푝 ∈ 푆)(푥 = 푧푝 ∧ 푡 = 푝푦) ∨ (∃푞 ∈ 푆)(푧 = 푥푞 ∧ 푦 = 푞푡).

a) Prove that groups are equidivisible. b) Prove that free monoids are equidivisible. ✴2.2 Let 푢, 푣 ∈ 퐴+. Prove that

푢푣 = 푣푢 ⇔ (∃푤 ∈ 퐴+)(∃푖, 푗 ∈ ℕ)(푢 = 푤푖 ∧ 푣 = 푤푗).

[Hint: to prove the left-hand side implies the right-hand side, use induction on |푢푣|.] 2.3 Let 푢, 푣, 푤 ∈ 퐴+ be such that 푢푣 = 푣푤.

Exercises • 50 a) Using induction on |푣|, prove that there exist 푠, 푡 ∈ 퐴∗ and 푘 ∈ ℕ ∪ {0} such that 푢 = 푠푡, 푣 = (푠푡)푘푠, and 푤 = 푡푠. b) Prove part a) in a different way by letting 푘 be maximal (possibly 푘 = 0) such that 푣 = 푢푘푠 for some 푠 ∈ 퐴∗. 2.4 Let 푢, 푣 ∈ 퐴+. Show that the subsemigroup ⟨푢, 푣⟩ is free if and only if 푢푣 ≠ 푣푢. 2.5 Let 푆 be a semigroup and let 푋 be a generating set for 푆, with |푋| ⩾ 2. Suppose that for all 푥푖, 푦푖 ∈ 푋 and 푛 ∈ ℕ, we have 푥1 ⋯ 푥푛 = 푦1 ⋯ 푦푛 ⇒ (∀푖 ∈ {1, … , 푛})(푥푖 = 푦푖). Prove that 푆 is free with basis 푋.

✴2.6 Let 푛 ∈ ℕ. Let 푋 = {푥1, 푥2, … , 푥푛}. Let 푀 be the set ℙ푋 under the operation of union; then 푀 is a monoid with identity ∅. The aim of this exercise is to use Method 2.9 to prove that 푀 is defined by 2 Mon⟨퐴 | 휌⟩, where 퐴 = {푎1, … , 푎푛} and 휌 = { (푎푖 , 푎푖), (푎푖푎푗, 푎푗푎푖) ∶ 푖, 푗 ∈ {1, … , 푛} }. a) Do step 1 of Method 2.9: define an assignment of generators 휑 ∶ 퐴 → 푀 and show that 푀 satisfies the defining relations in 휌 with respect to 휑. [Hint: the monoid 푀 is generated by elements {푥1}, {푥2}, … , {푥푛}.] 푒1 푒2 푒푛 b) Do step 2 of Method 2.9: let 푁 = { 푎1 푎2 ⋯ 푎푛 ∶ 푒푖 ⩽ 1 } and prove that for every 푤 ∈ 퐴∗ there is a word ̂푤∈ 푁 such that (푤, ̂푤) is a consequence of 휌. ∗ c) Do step 3 of Method 2.9: prove that 휑 |푁 is injective. ✴2.7 Prove that Mon⟨푎, 푏 | (푎푏푎, 휀)⟩ defines (ℤ, +). 2.8 Let 푀 be defined by Mon⟨퐴 | 휌⟩, where 퐴 = {푎, 푏, 푐} and 휌 = {(푎푏푐, 휀)}. Let 푁 = 퐴∗ ∖ 퐴∗푎푏푐퐴∗, so that 푁 consists of all words over 퐴 that do not contain a subword 푎푏푐. Prove that every element of 푀 has a unique representative in 푁, and that this representative can be obtained by taking any word representing that element and iteratively deleting subwords 푎푏푐.

✴2.9 Let 퐵2 be the semigroup consisting of the following five matrices: 0 0 0 1 0 0 1 0 0 0 [ ],[ ],[ ],[ ],[ ]. 0 0 0 0 1 0 0 0 0 1

Show that 퐵2 is presented by Sg⟨퐴 | 휎 ∪ 휁⟩, where 퐴 = {푎, 푏, 푧} and 휎 = {(푎2, 푧), (푏2, 푧), (푎푏푎, 푎), (푏푎푏, 푏)}, 휁 = {(푧푎, 푧), (푎푧, 푧), (푧푏, 푧), (푏푧, 푧), (푧2, 푧)}.

[Hint: note that the defining relations in 휁 imply that 푧 is mapped to the zero of 퐵2.] 2.10 Let 퐵 be the bicyclic monoid Mon⟨푏, 푐 | (푏푐, 휀)⟩. a) Prove that 푐훾푏훽 is idempotent if and only if 훽 = 훾.

Exercises • 51 b) Prove that 푐훾푏훽 is right-invertible if and only if 훾 = 0. [Dual reasoning will show that 푐훾푏훽 is left-invertible if and only if 훽 = 0.] ✴2.11 Let 퐵 be the bicyclic monoid Mon⟨푏, 푐 | (푏푐, 휀)⟩. Draw a part of the Cayley graph 훤(퐵, {푏, 푐}) including all elements 푐훾푏훽 with 훾, 훽 ⩽ 4. ✴2.12 Let 푆 be a semigroup and let 푒, 푥, 푦 ∈ 푆 be such that 푒푥 = 푥푒 = 푥, 푒푦 = 푦푒 = 푦, 푥푦 = 푒, and 푦푥 ≠ 푒. a) Prove that all powers of 푥 and all powers of 푦 are distinct. (That is, 푥 and 푦 are not periodic elements.) b) Prove that if 푥푘 = 푦ℓ for some 푘, ℓ ∈ ℕ ∪ {0}, then 푘 = ℓ = 0. c) Prove that if 푦푘푥ℓ = 푒 for some 푘, ℓ ∈ ℕ ∪ {0}, then 푘 = ℓ = 0. d) Prove that if 푦푘푥ℓ = 푦푚푥푛 for some 푘, ℓ, 푚, 푛 ∈ ℕ ∪ {0}, then 푘 = 푚 and ℓ = 푛. e) Deduce that the subsemigroup ⟨푥, 푦⟩ of 푆 is isomorphic to the bicyclic monoid. ✴2.13 Let 퐵 be the bicyclic monoid and 휑 ∶ 퐵 → 푆 a surjective homomorphism. Prove that 푆 is either isomorphic to the bicyclic monoid or a group.

Notes

The section on properties of free semigroups and monoids is largely based on Howie, Fundamentals of Semigroup Theory, ch. 7. ◆ The discussion of semigroup presentations is partly based on Ruškuc, ‘Semigroup Presentations’, chs 1 & 3. ◆ For further reading on free semigroups and monoids see Harju, ‘Lecture Notes on Semigroups’, § 4.1–2 and Howie, Fundamentals of Semigroup Theory, § 7.2 on submonoids of free monoids and connections to coding theory. Lothaire, Combinatorics on Words is a broad study of words and contains a great deal of relevant material. For further reading on semigroup presentations, Ruškuc, ‘Semigroup Presentations’ is an essential text, but see also Higgins, Techniques of Semigroup Theory, § 1.7 & ch. 5 for an introduction to using diagrams to reason about semigroup presentations. ◆ For string-rewriting and its application to semigroup theory, see Book & Otto, String Rewriting Systems; for rewriting more generally, see Baader & Nipkow, Term Rewriting and All That. ◆ Example 2.14 is derived from the criterion for group-embeddability in Malcev, ‘On the immersion of an algebraic ring into a field’. ◆ Exercise 2.5 is adapted from Gallagher, ‘On the Finite Generation and Presentability of Diagonal Acts…’, Proof of Proposition 3.1.12. •

Notes • 52 Structure of semigroups 3 structure can be considered as a complex of ‘ relations, and ultimately as multi-dimensional order. —’ Alfred Korzybski Science and Sanity, bk I, pt. I, ch. 2.

• The aim of this chapter is to understand better the structure of semigroups. We want to divide the semigroup into sections in such a way that we can understand the semigroup in terms of those parts and their interaction. One goal is to understand the semigroup in terms of groups; then we assume that our work is done and we hand on the problem to a group theorist.

Green’s relations

The most fundamental tools in understanding a semi- Green’s relations group are its Green’s relations. These relate elements depending on the ideals they generate, and, as we shall see, give a lot of information about the structure of a semigroup and how its elements interact. On a semigroup, there are five Green’s relations: H, L, R, D, and J. We start by defining L, R, and J: for a semigroup 푆, define L, R, and J

1 1 푥 L 푦 ⇔ 푆 푥 = 푆 푦, } 1 1 } 푥 R 푦 ⇔ 푥푆 = 푦푆 , } (3.1) 1 1 1 1 } 푥 J 푦 ⇔ 푆 푥푆 = 푆 푦푆 . } It is easy to see that L, R, and J are all equivalence relations. Useful characterizations of these relations, which we will use at least as often as the definitions in3.1 ( ), are given by the following result:

P r o p o s i t i o n 3 . 1. The relations L, R, and J on a semigroup 푆 satisfy Characterization of L, R, J the following:

푥 L 푦 ⇔ (∃푝, 푞 ∈ 푆1)((푝푥 = 푦) ∧ (푞푦 = 푥)); 푥 R 푦 ⇔ (∃푝, 푞 ∈ 푆1)((푥푝 = 푦) ∧ (푦푞 = 푥)); 푥 J 푦 ⇔ (∃푝, 푞, 푟, 푠 ∈ 푆1)((푝푥푟 = 푦) ∧ (푞푦푠 = 푥)).

• 53 Proof of 3.1. We prove the result for L; similar reasoning applies for R and J. Suppose 푥 L 푦. Then by (3.1), 푆1푥 = 푆1푦. Since 푦 ∈ 푆1푦, it follows that 푦 ∈ 푆1푥 and so there exists 푝 ∈ 푆1 such that 푝푥 = 푦. Similarly, there exists 푞 ∈ 푆1 such that 푞푦 = 푥. Now suppose that there exist 푝, 푞 ∈ 푆1 such that 푝푥 = 푦 and 푞푦 = 푥. Then 푆1푥 = 푆1푞푦 ⊆ 푆1푦, and similarly 푆1푦 = 푆1푝푥 ⊆ 푆1푥. Hence 1 1 푆 푥 = 푆 푦 and so 푥 L 푦. 3.1

P r o p o s i t i o n 3 . 2. L ∘ R = R ∘ L. L and R commute

Proof of 3.2. Let (푥, 푦) ∈ L ∘ R. Then there exists 푧 ∈ 푆 such that 푥 L 푧 and 푧 R 푦. By Proposition 3.1, there exist 푝, 푞, 푟, 푠 ∈ 푆1 such that 푝푥 = 푧, 푞푧 = 푥, 푧푟 = 푦, and 푦푠 = 푧. Let 푧′ = 푞푧푟. Then 푥푟 = 푞푧푟 = 푧′ and 푧′푠 = 푞푧푟푠 = 푞푦푠 = 푞푧 = 푥, so 푥 R 푧′, and 푞푦 = 푞푧푟 = 푧′ and 푝푧′ = 푝푞푧푟 = 푝푥푟 = 푧푟 = 푦, so 푧′ L 푦. Hence (푥, 푦) ∈ R ∘ L. Thus L ∘ R ⊆ R ∘ L. Similarly R ∘ L ⊆ L ∘ R and so L ∘ R = R ∘ L. 3.2 As a consequence of Propositions 1.31 and 3.2, we see that L ⊔ R = L ∘ R. Recall from page 26 that the meet of two equivalence relations is their intersection, so L ⊓ R = L ∩ R. The meet and join of L and R play an important role, so they are also counted as Green’s relations and have particular notations: H and D

H = L ⊓ R = L ∘ R, D = L ⊔ R = L ∩ R.

From either (3.1) or Proposition 3.1, one sees that L ⊆ J and R ⊆ J. So J is an upper bound for {L, R} and so D = L ⊔ R ⊆ J. Furthermore, J it is immediate that H ⊆ L and H ⊆ R. In fact, all of these inclusions are in general strict by Exercises 3.5, 3.6, and 3.7, or by Exercise 3.11; see D Figure 3.1. However, in some special classes of semigroups we do have equality of some of the relations. L R For instance, let 퐺 be a group. Then in 퐺, all of Green’s relations are equal to the universal relation 퐺 × 퐺. That is, all elements of 퐺 are H-, L-, H R-, D-, and J-related. FIGURE 3.1 Hasse diagram of Green’s rela- P r o p o s i t i o n 3 . 3. In a periodic semigroup, the Green’s relations D tions in a general semigroup and J coincide. D = J for periodic semigroups Proof of 3.3. Suppose 푆 is periodic. We already know D ⊆ J, so we have to prove the opposite inclusion. Let 푥 J 푦. Then there exist 푝, 푞, 푟, 푠 ∈ 푆1 such that 푝푥푟 = 푦 and 푞푦푠 = 푥. So 푥 = 푞푝푥푟푠 and so 푥 = (푞푝)푛푥(푟푠)푛 for all 푛 ∈ ℕ, and

Green’s relations • 54 similarly 푦 = (푝푞)푛푦(푠푟)푛 for all 푛 ∈ ℕ. Since 푆 is periodic, there exist 푘, ℓ ∈ ℕ such that (푞푝)푘 and (푠푟)ℓ are idempotent. Let 푧 = 푝푥. Then

푥 = (푞푝)푘푥(푟푠)푘 = (푞푝)2푘푥(푟푠)푘 = (푞푝)푘((푞푝)푘푥(푟푠)푘) = (푞푝)푘푥 = ((푞푝)푘−1푞)푧.

Hence 푥 L 푧. Furthermore, 푧푟 = 푝푥푟 = 푦 and

푧 = 푝푥 = 푝(푞푝)ℓ+1푥(푟푠)ℓ+1 = (푝푞)ℓ+1푝푥푟(푠푟)ℓ푠 = (푝푞)ℓ+1푝푥푟(푠푟)2ℓ푠 = (푝푞)ℓ+1푦(푠푟)2ℓ푠 = (푝푞)ℓ+1푦(푠푟)ℓ+1(푠푟)ℓ−1푠 = 푦((푠푟)ℓ−1푠).

Hence 푧 R 푦. Therefore 푥 D 푦. Thus J ⊆ D and so D = J. 3.3

P r o p o s i t i o n 3 . 4. a) The relation L is a right congruence. b) The relation R is a left congruence.

Proof of 3.4. For any 푥, 푦, 푧 ∈ 푆,

푥 L 푦 ⇒ 푆1푥 = 푆1푦 ⇒ 푆1푥푧 = 푆1푦푧 ⇒ 푥푧 L 푦푧, and so L is a right congruence. Dual reasoning shows that R is a left congruence. 3.4 In general, L is not a left congruence and R is not a right congruence; see Exercise 3.4.

For 푎 ∈ 푆, denote by 퐻푎, 퐿푎, 푅푎, 퐷푎, and 퐽푎 the H-, L-, R-, D, and 퐻푎, 퐿푎, 푅푎, 퐷푎, and 퐽푎 J-classes of 푎, respectively. By the containment between Green’s relations described above,

퐻푎 ⊆ 퐿푎, 퐻푎 ⊆ 푅푎, 퐿푎 ⊆ 퐷푎, 푅푎 ⊆ 퐷푎, and 퐷푎 ⊆ 퐽푎.

There are natural partial orders on the collection of L-classes 푆/L, the Partial order of collection of R-classes 푆/R, and the collection of J-classes 푆/J induced 푆/L, 푆/R, and 푆/J by inclusion order of ideals:

1 1 퐿푥 ⩽ 퐿푦 ⇔ 푆 푥 ⊆ 푆 푦, } 푅 ⩽ 푅 ⇔ 푥푆1 ⊆ 푦푆1, (3.2) 푥 푦 } 1 1 1 1 퐽푥 ⩽ 퐽푦 ⇔ 푆 푥푆 ⊆ 푆 푦푆 . }

It follows immediate from (3.2) that for all 푥 ∈ 푆 and 푝, 푞 ∈ 푆1,

퐿푝푥 ⩽ 퐿푥, 푅푥푞 ⩽ 푅푥, 퐽푝푥푞 ⩽ 퐽푥.

Green’s relations • 55 Simple and 0-simple semigroups

A semigroup is simple if it contains no proper ideals; thus Simple/0-simple 푆 is simple if its only ideal is 푆 itself. A semigroup 푆 with a zero is 0-simple if it is not a null semigroup and its only proper ideal is {0}; thus 푆 is 0-simple if 푆2 ≠ ∅ and 푆 and {0} are the only ideals of 푆. The notion of a simple semigroup is not a generalization of a ‘simple group’,in the sense of a group that contains no proper non-trivial normal subgroups. Groups never contains proper ideals, so groups are always simple semigroups. Let 푆 be a semigroup and let 퐼 be an ideal (respectively, left ideal, Minimal/0-minimal ideal right ideal) of 푆. Then 퐼 is minimal if there is no ideal (respectively, left ideal, right ideal) 퐽 of 푆 that is strictly contained in 퐼. Suppose now that 푆 contains a zero. Then 퐼 is 0-minimal if 퐼 ≠ {0} and there is no ideal (respectively, left ideal, right ideal) 퐽 ≠ {0} of 푆 that is strictly contained in 퐼.

P r o p o s i t i o n 3 . 5. A semigroup contains at most one minimal ideal. Uniqueness of minimal ideals Proof of 3.5. Suppose 퐼 and 퐽 are minimal ideals of a semigroup 푆. Then 퐼퐽 is an ideal of 푆 and 퐼퐽 ⊆ 퐼푆 ⊆ 퐼 and 퐼퐽 ⊆ 푆퐽 ⊆ 퐽. Hence, by the minimality of 퐼 and 퐽, we have 퐼 = 퐼퐽 and 퐽 = 퐼퐽 and hence 퐼 = 퐽. 3.5

A semigroup 푆 might not contain a minimal ideal. For example, the Kernel of a semigroup ideals 퐼푛 of (ℕ, +) defined in Example 1.10(a) form an infinite descending chain: ℕ = 퐼1 ⊇ 퐼2 ⊇ 퐼3 ⊇ …. But Proposition 3.5 shows that if a semigroup 푆 contains a minimal ideal, it is unique. Such a unique minimal ideal is called the kernel of 푆 and is denoted 퐾(푆). Notice that if 푆 is a 퐾(푆) semigroup with zero, 퐾(푆) = {0}.

L e m m a 3 . 6. If a semigroup 푆 is 0-simple, then 푆2 = 푆. 푆2 = 푆 for 0-simple semigroups Proof of 3.6. Note that 푆2 is an ideal of 푆, since 푆푆2 ⊆ 푆2 and 푆2푆 ⊆ 푆2. Now, 푆2 ≠ {0} since 푆 is not null (by the definition of 0-simple). Hence 2 푆 = 푆 since 푆 is 0-simple. 3.6

L e m m a 3 . 7. A semigroup 푆 is 0-simple if and only if 푆푥푆 = 푆 for all 푥 ∈ 푆 ∖ {0}.

Proof of 3.7. Suppose 푆 is 0-simple. Then 푆2 = 푆 by Lemma 3.6 and so 푆3 = 푆2푆 = 푆푆 = 푆. For any 푥 ∈ 푆, the principal ideal 푆푥푆 is either {0} or 푆 since 푆 is 0-simple. Let 푇 = { 푥 ∈ 푆 ∶ 푆푥푆 = {0} }. It is easy to prove that 푇 is an ideal of 푆. Since 푆 is 0-simple, it follows that 푇 = 푆 or 푇 = {0}. Suppose that 푇 = 푆. Then 푆푥푆 = {0} for all 푥 ∈ 푆, which implies 푆3 = {0}, which is a contradiction since 푆3 = 푆 by the previous paragraph. Hence 푇 = {0}, and so 푆푥푆 = 푆 for all 푥 ∈ 푆 ∖ {0}.

Simple and 0-simple semigroups • 56 For the converse, suppose 푆푥푆 = 푆 for all 푥 ∈ 푆 ∖ {0}. Note first that 푆 cannot be null. Let 퐼 be some ideal of 푆. Suppose 퐼 ≠ {0}. Then there exists some 푦 ∈ 퐼 ∖ {0}, and 푆푦푆 = 푆. Hence 푆 = 푆푦푆 ⊆ 퐼 ⊆ 푆 and so 퐼 = 푆. So for any ideal 퐼 of 푆, either 퐼 = {0} or 퐼 = 푆, and so 푆 is 0-simple. 3.7

P r o p o s i t i o n 3 . 8. a) A 0-minimal ideal of a semigroup with a 0-minimal ideals zero is either null or 0-simple. are 0-simple or null b) A minimal ideal of a semigroup is simple.

Proof of 3.8. a) Let 푆 be a semigroup with a zero, and let 퐼 be a 0-minimal ideal of 푆. Suppose 퐼 is not null. Then 퐼2 ≠ {0}. Hence, since 퐼2 ⊆ 퐼 is an ideal of 푆 and 퐼 is 0-minimal, we have 퐼2 = 퐼 and so 퐼3 = 퐼. Let 푥 ∈ 퐼 ∖ {0}. Then 푆1푥푆1 is an ideal of 푆 contained in 퐼. Since 푥 ∈ 푆1푥푆1, we have 푆1푥푆1 ≠ {0}; hence 푆1푥푆1 = 퐼 since 퐼 is 0-minimal. Thus 퐼 = 퐼3 = 퐼푆1푥푆1퐼 ⊆ 퐼푥퐼 ⊆ 퐼. Therefore 퐼푥퐼 = 퐼 for all 푥 ∈ 퐼 ∖ {0} and so 퐼 is 0-simple by Lemma 3.7. So 퐼 is either null or 0-simple. b) First, note that if 푆 has a zero 0, its unique minimal ideal is {0}, which is simple. So suppose that 퐼 is a minimal ideal of a semigroup 푆 that does not contain a zero. Then 퐼2 is an ideal of 푆 and 퐼2 ⊆ 퐼. So 퐼2 = 퐼 since 퐼 is minimal. Hence 퐼3 = 퐼. Suppose 퐽 is an ideal of 퐼. Let 푥 ∈ 퐽. Then 퐼푥퐼 ⊆ 퐽 since 퐽 is an ideal of 퐼. Then 푆1푥푆1 is an ideal of 푆 and 푆1푥푆1 ⊆ 퐼; hence 푆1푥푆1 = 퐼 since 퐼 is minimal. Therefore 퐽 ⊆ 퐼 = 퐼3 = 퐼푆1푥푆1퐼 ⊆ 퐼푥퐼 ⊆ 퐽 and so 퐽 = 퐼. So 퐼 is simple. 3.8 For any 푥 ∈ 푆, recall that 퐽(푥) = 푆1푥푆1, and that the J-class of 푥, denoted 퐽푥, is the set of all elements of the semigroup that generate (as a principal ideal) 퐽(푥). Let 퐼(푥) = 퐽(푥) ∖ 퐽푥. Notice that 퐼(푥) = { 푦 ∈ 푆 ∶ 퐼(푆) 퐽푦 < 퐽푥 }. L e m m a 3 . 9. Let 푆 be a semigroup and 푥 ∈ 푆. Then 퐼(푥) is either empty or an ideal of 푆.

Proof of 3.9. Suppose 퐼(푥) ≠ ∅. Let 푦 ∈ 퐼(푥) and 푧 ∈ 푆. Then 푦푧 ∈ 퐽(푥) since 퐽(푥) is an ideal. But 퐽(푦푧) ⊆ 퐽(푦) ⊊ 퐽(푥) (since 퐽(푦) = 퐽(푥) would imply 푦 ∈ 퐽푥). Hence 푦푧 ∈ 퐼(푥). Similarly 푧푦 ∈ 퐼(푥). Hence 퐼(푥) is an ideal. 3.9

The factor semigroups 퐽(푥)/퐼(푥) (where 푥 is such that 퐼(푥) ≠ 0) and Principal factors the kernel 퐾(푆) are called the principal factors of 푆.

P r o p o s i t i o n 3 . 1 0. Let 푆 be a semigroup. If the kernel 퐾(푆) exists, it Principal factors are is simple. All other principal factors of 푆 are either null or 0-simple. null or 0-simple

Proof of 3.10. By Proposition 3.8(b), if 퐾(푆) exists, it is simple. The principal factor 퐽(푥)/퐼(푥) is a 0-minimal ideal of 푆/퐼(푥) and so is 0-simple by Proposition 3.8(a). 3.10

Simple and 0-simple semigroups • 57 A principal series of a semigroup 푆 is a finite chain of ideals Principal series

퐾(푆) = 푆1 ⊊ 푆2 ⊊ … ⊊ 푆푛 = 푆 (3.3) that is maximal in the sense that there is no ideal 퐼 such that 푆푖 ⊊ 퐼 ⊊ 푆푖+1. Not all semigroups admit principal series. Indeed, even if a semigroup has a kernel, it may not admit a principal series: for example, let 푆 be the semigroup (ℕ, +). Then 푆0 has a minimal ideal {0} but no principal series. We now have an analogy for semigroups of the Jordan–Hölder theorem for groups, which states that any composition series for a group contains the the same composition factors in some order.

T h e o r e m 3 . 1 1. Let 푆 be a semigroup admitting a principal series ‘Jordan–Hölder theorem’ for semigroups (3.3). Then the factors 푆푖+1/푆푖 are, in some order, isomorphic to the principal factors of 푆.

Proof of 3.11. [Not especially difficult, but technical and omitted.] 3.11

D-class structure Since L ⊆ D and R ⊆ D, every D-class must be both a union of L-classes and a union of R-classes. On other hand, suppose that an L-class 퐿푥 and a R-class 푅푦 intersect. Then there is some element 푧 ∈ 퐿푥 ∩ 푅푦. So 푥 L 푧 R 푦 and so 푥 D 푦. Hence 퐿푥 and 푅푦 are both contained within the same D-class. Therefore an L-class and an R-class intersect if and only if they are contained within the same D-class. Thus we can visualize a D-class in the following useful way: Imagine the elements of this D-class arranged in a rectangular pattern. This pattern is divided into a grid of cells. Each column of cells is an L-class; each row is an R-class, and every cell is the H-class that is the intersection of the L- and R-class forming the column and row that contain that cell. This visualization is called an egg-box diagram; see Figure 3.2. A useful mnemonic for remembering the arrangement of an egg-box diagram is: R-classes are Rows and L-classes are coLumns. For a concrete example of an egg-box diagram, see Figure 3.7 on page 67, which is drawn using the result in Exercise 3.3.

G r e e n ’ s L e m m a 3 . 1 2. a) Let 푥, 푦 ∈ 푆 be such that 푥 L 푦 and Green’s lemma let 푝, 푞 ∈ 푆1 be such that 푝푥 = 푦 and 푞푦 = 푥. Then the ‘left multiplication’ maps 휆 | and 휆 | (where 푡휆 = 푧푡) are mutually inverse 푝 푅푥 푞 푅푦 푧 bijections between 푅푥 and 푅푦. Furthermore, both of these maps preserve L-classes, in the sense that 푡휆 | L 푡 and 푡휆 | L 푡, and so 휆 | 푝 푅푥 푞 푅푦 푝 퐻푥 and 휆 | are mutually inverse bijections between 퐻 and 퐻 . (See 푞 퐻푦 푥 푦 Figure 3.3.)

D-class structure • 58 FIGURE 3.2 An egg-box diagram for the D- 퐻푥 푅푥 class 퐷푥. The R-class 푅푥 and 푥 the L-class 퐿푥 are represented by the row and column that intersect in the box represent- 퐿 ing the H-class 퐻푥, which con- 푥 tains the element 푥.

휆 | 휆 | = id 푝 푅푥 푞 푅푦 푅푥

푥 푅푥

FIGURE 3.3 Green’s lemma: if 푝 and 푞 re- 휆푞 휆푝 휆푞|푅 휆 | 휆푞|푅 휆 | 푦 푝 푅푥 푦 푝 푅푥 spectively left-multiply 푥 to give 푦 and 푦 to give 푥, then the left multiplication maps 휆푝 and 휆푞 restrict to mutually inverse bijections between the 푦 푅푦 R-classes 푅푥 and 푅푦 (the rows containing 푥 and 푦), and both 퐿푥 = 퐿푦 휆 | 휆 | = id of these restricted maps pre- 푞 푅푦 푝 푅푥 푅푦 serve L-classes (columns).

b) Let 푥, 푦 ∈ 푆 be such that 푥 R 푦 and let 푝, 푞 ∈ 푆1 be such that 푥푝 = 푦 and 푦푞 = 푥. Then the ‘right multiplication’ maps 휌 | and 휌 | (where 푝 퐿푥 푞 퐿푦 푡휌푧 = 푡푧) are mutually inverse bijections between 퐿푥 and 퐿푦, and both of these maps preserve R-classes.

Proof of 3.12. We prove only part a); the other part is proved by a dual argument. First, notice that

P r o p o s i t i o n 3 . 1 3. Let 푥, 푦 ∈ 푆 be such that 푥 D 푦. Then |퐻푥| = H-classes in the same D-class have |퐻푦|. the same cardinality Proof of 3.13. Assume 푥 D 푦. So there exists 푧 such that 푥 L 푧 and 푧 R 푦. Let 푝, 푞, 푟, 푠 ∈ 푆1 be such that 푝푥 = 푧, 푞푧 = 푥, 푧푟 = 푦, and 푦푠 = 푧. By Lemma 3.12, 휆 | ∶ 퐻 → 퐻 is a bijection, and 휌 | ∶ 퐻 → 퐻 is 푝 퐻푥 푥 푧 푟 퐻푧 푧 푦 a bijection. So 휆 | 휌 | ∶ 퐻 → 퐻 is a bijection, and hence |퐻 | = 푝 퐻푥 푟 퐻푧 푥 푦 푥 |퐻푦|. 3.13 P r o p o s i t i o n 3 . 1 4. Let 퐻 be an H-class of 푆. Then either: Two types of H-class a) 퐻2 ∩ 퐻 = ∅, or b) the following equivalent statements hold: i) 퐻2 ∩ 퐻 ≠ ∅; ii) 퐻 contains an idempotent; iii) 퐻2 = 퐻; iv) 퐻 is a subsemigroup of 푆; v) 퐻 is a subgroup of 푆. Proof of 3.14. If 퐻2 ∩퐻 = ∅ there is nothing further to prove. So suppose that 퐻2 ∩ 퐻 ≠ ∅. Then there exist 푠, 푡 ∈ 퐻 such that 푠푡 ∈ 퐻. Then 푠 H 푠푡. In particular, 푠 R 푠푡. So by Lemma 3.12(b), 휌푡|퐻 is a bijection from 퐻 to itself. Similarly 푡 L 푠푡, and thus, by Lemma 3.12(a), 휆푠|퐻 is a bijection from 퐻 to itself. Now let 푧 ∈ 퐻. Then 푠푧 = 푧휆푠|퐻 and 푧푡 = 푧휌푡|퐻 are both in 퐻. Again by Lemma 3.12, 휌푧|퐻 and 휆푧|퐻 are bijections from 퐻 to itself. Since 푧 ∈ 퐻 was arbitrary, it follows that 푧퐻 = 퐻푧 = 퐻 for all 푧 ∈ 퐻. Therefore 퐻 is a subgroup by Lemma 1.9. We have shown that statement i) implies statement v). Statement v) clearly implies statements ii), iii), and iv), and each of these implies statement i). So all five statements are equivalent. 3.14 A maximal subgroup is a subgroup that does not lie inside any larger subgroup.

P r o p o s i t i o n 3 . 1 5. The maximal subgroups of 푆 are precisely the Maximal subgroup = H-class H-classes of 푆 that contain idempotents. containing an idempotent Proof of 3.15. Since every element of a subgroup is H-related, it follows that any subgroup is contained within a single H-class. So a maximal subgroup 퐺 is contained within a single H-class 퐻. But 퐻 therefore contains an idempotent 1퐺 and so is itself a subgroup by Proposition 3.14. Hence 퐻 = 퐺. 3.15

D-class structure • 60 C o r o l l a ry 3 . 1 6. An H-class contains at most one idempotent. 3.16

P r o p o s i t i o n 3 . 1 7. Let 푒 ∈ 푆 be idempotent. Then 푒푥 = 푥 for all Idempotents are ‘left/right identities’ 푥 ∈ 푅푒 and 푦푒 = 푦 for all 푦 ∈ 퐿푒. for their R/L-classes 1 Proof of 3.17. Suppose 푥 ∈ 푅푒. Then there exists 푝 ∈ 푆 such that 푒푝 = 푥. Hence 푒푥 = 푒푒푝 = 푒푝 = 푥. Hence 푒 is a left identity for 푅푒. Similarly 푒 is a right identity for 퐿푒. 3.17

P r o p o s i t i o n 3 . 1 8. Let 푥, 푦 ∈ 푆 with 푥 D 푦. Then 푥푦 ∈ 퐿푦 ∩ 푅푥 if Products located by idempotents and only if 퐿푥 ∩ 푅푦 contains an idempotent. (See Figure 3.4.)

푒 푦 푅푦 Proof of 3.18. Suppose that 푥푦 ∈ 퐿푦 ∩ 푅푥. In particular 푥푦 R 푥. Hence there exists 푞 ∈ 푆1 such that 푥푦푞 = 푥. By Lemma 3.12, 휌 | ∶ 퐿 → 퐿 푦 퐿푥 푥 푥푦 and 휌 | ∶ 퐿 → 퐿 are mutually inverse R-class preserving bijections 푞 퐿푥푦 푥푦 푥 between 퐿푥 and 퐿푥푦. Since 푥푦 L 푦, these maps are in fact mutually inverse R-class preserving bijections between 퐿푥 and 퐿푦 Hence (푦푞)2 = 푦푞푦푞 = 푦휌 | 휌 | 휌 | = 푦휌 | = 푦푞. Hence 푦푞 is 푞 퐿푦 푦 퐿푥 푞 퐿푦 푞 퐿푦 푥 푥푦 푅푥 idempotent. Furthermore, 푦푞 = 푦휌 | ∈ 퐿 ∩ 푅 . 푞 퐿푦 푥 푦 퐿푥 퐿푦 Now suppose that 퐿푥 ∩ 푅푦 contains an idempotent 푒. Then 푒푦 = 푦 by Proposition 3.17. Since 푒 R 푦, the map 휌 | ∶ 퐿 → 퐿 is an R-class FIGURE 3.4 푦 퐿푒 푒 푦 Products are located by idem- preserving bijection by Lemma 3.12. Hence 푥푦 ∈ 푅푥 ∩ 퐿푦. 3.18 potents: 푥푦 ∈ 퐿푦 ∩ 푅푥 if and only if 퐿푥 ∩ 푅푦 contains 푒 ∈ 퐸(푆).

Inverses and D-classes Proposition 3.18 shows a close relationship between the product of two elements of a D-class and idempotents in that D-class. It is thus not surprising that idempotents and inverses in a D-class are also connected.

P r o p o s i t i o n 3 . 1 9. If 푥 ∈ 푆 is regular, then every element of 퐷푥 is Either every element of regular. 퐷푥 is regular or none are Proof of 3.19. Suppose 푥 is regular. Then there exists 푦 ∈ 푆 such that 푥푦푥 = 푥. Suppose 푧 L 푥. Then there exist 푝, 푞 ∈ 푆1 such that 푝푧 = 푥 and 푞푥 = 푧. Hence 푧 = 푞푥 = 푞푥푦푥 = 푧푦푝푧 and so 푧 is regular. So every element of 퐿푥 is regular. A dual argument shows that if 푡 ∈ 푆 is regular, every element of 푅푡 is regular. Combining these, we see that if 푥 is regular, every element of 퐷푥 is regular. 3.19

A D-class is regular if all its elements are regular, and otherwise is Regular/irregular D-class irregular.

P r o p o s i t i o n 3 . 2 0. In a regular D-class, every L-class and every Idempotents in a R-class contains an idempotent. regular D-class

Inverses and D-classes • 61 Proof of 3.20. Let 푥 ∈ 푆 be such that 퐷푥 is regular. In particular, 푥 is regular and so 푥푦푥 = 푥 for some 푦 ∈ 푆. Now, 푦푥 L 푥 and (푦푥)2 = 푦푥푦푥 = 푦푥. So 푦푥 is an idempotent in 퐿푥. Similarly 푥푦 is an idempotent in 푅푥. Thus every L-class and R-class contains an idempotent. 3.20 Recall that 푉(푥) denotes the set of inverses of 푥. P r o p o s i t i o n 3 . 2 1. If 푥 lies in a regular D-class, then: a) if 푥′ ∈ 푉(푥), then 푥 R 푥푥′ L 푥′ and 푥 L 푥′푥 R 푥′ and so 푥 D 푥′;

b) if 푧 ∈ 퐷푥 is such that 퐿푧 ∩ 푅푥 contains an idempotent 푒 and 푅푧 ∩ 퐿푥 contains an idempotent 푓, then 퐻푧 contains some 푡 ∈ 푉(푥) with 푥푡 = 푒 and 푡푥 = 푓; 푥′푥 푥′ 푅푥′ c) an H-class contains at most one member of 푉(푥). Proof of 3.21. a) Let 푥′ ∈ 푉(푥). Then 푥푥′푥 = 푥 and 푥′푥푥′ = 푥′. Then 푥 R 푥푥′ L 푥′ and so 푥 D 푥′; furthermore 푥 L 푥′푥 R 푥′. (See Figure 3.5.) 1 b) Since 푥 R 푒, there exists 푝, 푞 ∈ 푆 with 푥푝 = 푒 and 푒푞 = 푥. Let 푥 푥푥′ 푅푥 푡 = 푓푝푒. Then 퐿푥 퐿푥′ 푥푡푥 = 푥푓푝푒푥 [by definition of 푡] FIGURE 3.5 = 푥푝푥 [since 푥푓 = 푥 and 푒푥 = 푥 by Proposition 3.17] 푥 and 푥′ ∈ 푉(푥) in a regular D-class = 푒푥 [since 푥푝 = 푒] = 푥 [since 푒푥 = 푥 by Proposition 3.17]

and

푡푥푡 = 푓푝푒푥푓푝푒 [by choice of 푡] = 푓푝푥푝푒 [since 푥푓 = 푥 and 푒푥 = 푥 by Proposition 3.17] = 푓푝푒2 [since 푥푝 = 푒] = 푓푝푒 [since 푒 is idempotent] = 푡. [by definition of 푡]

Hence 푡 ∈ 푉(푥). Furthermore, 푥푡 = 푥푓푝푒 = 푥푝푒 = 푒2 = 푒. Finally, note that 휌 | ∶ 퐿 → 퐿 and 휌 | ∶ 퐿 → 퐿 are mutually inverse 푝 퐿푥 푥 푒 푞 퐿푒 푒 푥 R-class preserving bijections by Lemma 3.12(b). Hence

Inverses and D-classes • 62 Now combine some of the facts we have established: from 푡 = 푓푝푒 and 푒 = 푥푡, we see that 푡 L 푒; from 푡 = 푓푝푒 and 푓 = 푡푥, we see that 푓 푧, 푡 푅푧 푡 R 푓. Hence 푡 ∈ 퐿푒 ∩ 푅푓 = 퐻푧. (See Figure 3.6.) c) Suppose 푥′, 푥″ ∈ 푉(푥) and 푥′ H 푥″; we aim to show 푥′ = 푥″. Then 푥푥′ and 푥푥″ are idempotents lying inside 퐿푥′ ∩ 푅푥 = 퐿푥″ ∩ 푅푥. Hence 푥푥′ = 푥푥″ by Corollary 3.16. Similarly 푥′푥 = 푥″푥. Therefore 푥′ = 푥′푥푥′ = 푥′푥푥″ = 푥″푥푥″ = 푥″. 3.21 푥 푒 푅푥 Example 1.7 noted that every element of a rectangular band is an 퐿푥 퐿푧 inverse of every element. Exercise 3.5 shows that the H-classes of a rectangular band are the singleton sets. Thus it is possible for an element 푥 FIGURE 3.6 Inverse 푡 corresponding to to have an inverse in every H-class. Exercise 3.5 also notes that a rectan- idempotents 푒 and 푓 in a gular band consists of a single D-class (which must be regular, since all regular D-class elements of a rectangular band are idempotent), so all these inverses of 푥 are D-related to 푥, which fits with Proposition 3.21(a).

C o r o l l a ry 3 . 2 2. Let 푒, 푓 ∈ 푆 be idempotents. Then 푒 D 푓 if and only if there exist 푥 ∈ 푆 and 푥′ ∈ 푉(푥) such that 푥푥′ = 푒 and 푥′푥 = 푓.

Proof of 3.22. Suppose 푒 D 푓. Then 퐷푒 = 퐷푓 is a regular D-class since it contains the regular elements 푒 and 푓. Choose 푥 ∈ 푅푒 ∩퐿푓 and 푧 ∈ 퐿푒 ∩푅푓. Then by Proposition 3.21(b), 퐻푧 contains some 푥′ ∈ 푉(푥) such that 푥푥′ = 푒 and 푥′푥 = 푓. Suppose now that 푥 ∈ 푆 and 푥′ ∈ 푉(푥) are such that 푥푥′ = 푒 and 푥′푥 = 푓. Since 푒 = 푥푥′ and 푒푥 = 푥푥′푥 = 푥, it follows that 푥 R 푒. A dual argument shows that 푥 L 푓. Thus 푒 R 푥 L 푓 and so 푒 D 푓. 3.22

Schützenberger groups Let 푆 be a semigroup and let 퐻 be an H-class of 푆. Let Stab(퐻) = { 푥 ∈ 푆1 ∶ 퐻푥 = 퐻 }. Clearly, the adjoined identity 1 lies in Stab(퐻) Stab(퐻). If 푥, 푦 ∈ Stab(퐻), then 퐻푥푦 = 퐻푦 = 퐻 and so 푥푦 ∈ Stab(퐻); 1 thus Stab(퐻) is a submonoid of 푆 . Define a relation 휎퐻 on Stab(퐻) by 휎퐻

푥 휎퐻 푦 ⇔ (∀ℎ ∈ 퐻)(ℎ푥 = ℎ푦).

Let 푥 휎퐻 푦 and 푧 휎퐻 푡. Let ℎ ∈ 퐻. Then ℎ푥 = ℎ푦 by the definition of 휎퐻. Since 푥, 푦 ∈ Stab(퐻), we have ℎ푥 = ℎ푦 = ℎ′ ∈ 퐻. Thus ℎ′푧 = ℎ′푡, again by the definition of 휎퐻, and so ℎ(푥푧) = ℎ(푦푡). Since ℎ ∈ 퐻 was arbitrary, 푥푧 휎퐻 푦푡. Therefore 휎퐻 is a congruence on Stab(퐻). Let 훤(퐻) denote 훤(퐻) the factor semigroup Stab(퐻)/휎퐻. P r o p o s i t i o n 3 . 2 3. Let 퐻 be an H-class of a semigroup. Then 훤(퐻) is a group.

Schützenberger groups • 63 Proof of 3.23. First of all note that 훤(퐻) is a monoid with identity [1] , 휎퐻 since it is a quotient of the monoid Stab(퐻). Let 푥 ∈ Stab(퐻) and let ℎ ∈ 퐻. Then ℎ푥 ∈ 퐻. In particular, ℎ푥 R ℎ 1 and so there exists 푞 ∈ 푆 such that ℎ푥푞 = ℎ. Hence by Lemma 3.12, 휌푥|퐻 and 휌푞|퐻 are mutually inverse bijections. In particular, 퐻푞 = 퐻휌푞 = 퐻, and so 푞 ∈ Stab(퐻). Thus for any ℎ′ ∈ 퐻, we have

ℎ′푥푞 = ℎ′휌푥|퐻휌푞|퐻 = ℎ′ = ℎ′1,

ℎ′푞푥 = ℎ′휌푞|퐻휌푥|퐻 = ℎ′ = ℎ′1.

Hence 푥푞 휎 1 and 푞푥 휎 1, and so [푥] [푞] = [1] and [푞] [푥] = 퐻 퐻 휎퐻 휎퐻 휎퐻 휎퐻 휎퐻 [1] . Since 푥 ∈ Stab(퐻) was arbitrary, this proves that 훤(퐻) is a group. 휎퐻 3.23

The group 훤(퐻) is called the Schützenberger group of 퐻. This notion Schützenberger group associates a group to every H-class, not just those for which 퐻2 ∩ 퐻 ≠ ∅ (see Proposition 3.14). We will see that when 퐻 is a group H-class, 훤(퐻) is actually isomorphic to 퐻.

P r o p o s i t i o n 3 . 2 4. Let 퐻 be an H-class of a semigroup. Then the 훤(퐻) acts regularly on 퐻 Schützenberger group 훤(퐻) acts regularly on 퐻 via ℎ ⋅ [푥] = ℎ푥. 휎퐻 Proof of 3.24. First of all, note that the action ℎ⋅[푥] = ℎ푥 is well-defined, 휎퐻 since if [푥] = [푦] , then 푥 휎 푦 and so ℎ푥 = ℎ푦 by the definition of 휎퐻 휎퐻 퐻 휎퐻. Let ℎ, ℎ′ ∈ 퐻. Since in particular ℎ R ℎ′, there exists 푝 ∈ 푆1 such that ℎ푝 = ℎ′. By Lemma 3.12, 휌푝|퐻 is a bijection from 퐻 to itself, and so 푝 ∈ Stab(퐻), and hence [푝] ∈ 훤(퐻). Furthermore, ℎ ⋅ [푝] = ℎ푝 = ℎ′. 휎퐻 휎퐻 So 훤(퐻) acts transitively on 퐻. To show that 훤(퐻) acts freely on 퐻, we have to show that [푝] is the 휎퐻 unique element that acts on ℎ to give ℎ′. So suppose ℎ ⋅ [푦] = ℎ′. Let 휎퐻 푔 ∈ 퐻. Since 푔 L ℎ, there exists 푞 ∈ 푆1 such that 푞ℎ = 푔. Then

푔푦 = 푞ℎ푦 = 푞ℎ ⋅ [푦] = 푞ℎ′ = 푞ℎ ⋅ [푝] = 푞ℎ푝 = 푔푝. 휎퐻 휎퐻 Since this holds for all 푔 ∈ 퐻, it follows that 푦 휎 푝 and so [푦] = [푝] . 퐻 휎퐻 휎퐻 Hence 훤(퐻) acts freely on 퐻. Thus the action of 훤(퐻) on 퐻 is regular. 3.24

C o r o l l a ry 3 . 2 5. Let 퐻 be an H-class of a semigroup. Then |훤(퐻)| = An H-class and its |퐻|. Schützenberger group have the same size Proof of 3.25. Since 훤(퐻) acts regularly on 퐻, there is a one-to-one correspondence between the elements of 퐻 and the elements of 훤(퐻) and so |퐻| = |훤(퐻)|. 3.25

Schützenberger groups • 64 Strictly speaking, 훤(퐻) is the right Schützenberger group of 퐻, be- Left Schützenberger group cause the definitions of Stab(퐻) and 휎퐻 are in terms of right multiplication of elements of 퐻. This seems arbitrary, because we could make similar definitions using left multiplication:

Stab′(퐻) = { 푥 ∈ 푆1 ∶ 푥퐻 = 퐻 };

푥 휎′퐻 푦 ⇔ (∀ℎ ∈ 퐻)(푥ℎ = 푦ℎ);

훤′(퐻) = Stab′(퐻)/휎′퐻. Clearly, reasoning dual to the proofs of Propositions 3.23 and 3.24 shows that 훤′(퐻) is a group that acts regularly on 퐻 on the left via [푥] ⋅ℎ = 푥ℎ. 휎퐻′ The group 훤′(퐻) is called the left Schützenberger group of 퐻.

P r o p o s i t i o n 3 . 2 6. 훤(퐻) ≃ 훤′(퐻). Right and left Schützenberger groups Proof of 3.26. Fix some ℎ ∈ 퐻. Define a map 휑 ∶ 훤(퐻) → 훤′(퐻) as are isomorphic follows. For any 푠 ∈ 훤(퐻), since 훤′(퐻) acts regularly on 퐻, there is a unique 푠′ ∈ 훤′(퐻) such that ℎ ⋅ 푠 = 푠′ ⋅ ℎ. Define 푠휑 to be this 푠′. Similarly, since 훤(퐻) acts regularly on 퐻, we can define a map 휓 ∶ 훤′(퐻) → 훤(퐻) by letting 푡휓 be the unique element of 훤(퐻) such that 푡 ⋅ ℎ = ℎ ⋅ (푡휓). Clearly 휑 and 휓 are mutually inverse and thus are bijections. Let [푥] ∈ 훤(퐻) and [푦] ∈ 훤′(퐻). Let 푔 ∈ 퐻. Then 휎퐻 휎퐻′

[푦] ⋅ (푔 ⋅ [푥] ) = [푦] ⋅ (푔푥) 휎퐻′ 휎퐻 휎퐻′ } } = 푦푔푥 } } (3.4) = (푦푔) ⋅ [푥]휎 } 퐻 } = ([푦] ⋅ 푔) ⋅ [푥] . 휎퐻′ 휎퐻 } Let 푠, 푡 ∈ 훤(퐻). Then

(푠휑)(푡휑) ⋅ ℎ = (푠휑) ⋅ (ℎ ⋅ 푡) [by definition of 휑] = ((푠휑) ⋅ ℎ) ⋅ 푡 [by (3.4)] = (ℎ ⋅ 푠) ⋅ 푡 [by definition of 휑] = ℎ ⋅ (푠푡) [by definition of an action; see (1.15)] = ((푠푡)휑) ⋅ ℎ. [by definition of 휑]

Since 훤′(퐻) acts regularly on 퐻, it follows that (푠휑)(푡휑) = (푠푡)휑. Therefore 휑 is an isomorphism. 3.26

P r o p o s i t i o n 3 . 2 7. Let 푆 be a semigroup and let 푥, 푦 ∈ 푆. If 푥 D 푦, Schützenberger groups are the same then 훤(퐻푥) ≃ 훤(퐻푦). throughout a D-class Proof of 3.27. Suppose first that 푥 L 푦. Then there exist 푝, 푞 ∈ 푆1 such that 푝푥 = 푦 and 푞푦 = 푥. So by Lemma 3.12, 휆 | ∶ 퐻 → 퐻 and 휆 | ∶ 푝 퐻푥 푥 푦 푞 퐻푦 퐻푦 → 퐻푥 are mutually inverse bijections. Hence 푝퐻푥 = 퐻푦 and 푞퐻푦 =

Schützenberger groups • 65 퐻푥. Suppose that 푧 ∈ Stab(퐻푥). Then 퐻푦푧 = 푝퐻푥푧 = 푝퐻푥 = 퐻푦 and so 푧 ∈ Stab(퐻푦). Thus Stab(퐻푥) ⊆ Stab(퐻푦) and similarly Stab(퐻푦) ⊆ Stab(퐻푥). So Stab(퐻푥) = Stab(퐻푦). Now let 푧, 푡 ∈ Stab(퐻 ). Suppose 푧 휎 푡. Then 푥푧 = 푥푡. Let 푦′ ∈ 퐻 . 푥 퐻푥 푦 Since 푥 L 푦′, there exists 푝′ ∈ 푆1 such that 푦′ = 푝′푥 and so 푦′푧 = 푝′푥푧 = 푝′푥푡 = 푦′푡. Since 푦′ ∈ 퐻 was arbitrary, 푧 휎 푡. Hence 휎 ⊆ 휎 . Sim- 푦 퐻푦 퐻푥 퐻푦 ilarly 휎 ⊆ 휎 and so 휎 = 휎 . Therefore the Schützenberger groups 퐻푦 퐻푥 퐻푥 퐻푦 훤(퐻 ) = Stab(퐻 )/휎 and 훤(퐻 ) = Stab(퐻 )/휎 are isomorphic. 푥 푥 퐻푥 푦 푦 퐻푦 On the other hand, if 푥 R 푦, dual reasoning shows that the left Schützenberger groups 훤′(퐻푥) and 훤′(퐻푦) are isomorphic. The result follows from Proposition 3.26. 3.27 Notice that from Corollary 3.25 and Proposition 3.27 we immediately recover the result that if 푥, 푦 ∈ 푆 are such that 푥 D 푦, then |퐻푥| = |퐻푦| (Proposition 3.13).

P r o p o s i t i o n 3 . 2 8. Let 푆 be a semigroup and let 퐻 be an H-class of 푆. If 퐻 is a subgroup of 푆, then 훤(퐻) ≃ 퐻.

Proof of 3.28. Suppose 퐻 is a group. Then 퐻 ⊆ Stab(퐻). The restriction of the natural map 휎♮ | ∶ 퐻 → Stab(퐻)/휎 , which maps ℎ to [ℎ] , is a 퐻 퐻 퐻 휎퐻 homomorphism. Let 푠 ∈ 훤(퐻). Let ℎ = 1 ⋅ 푠. Then since 1 ⋅ [ℎ] = ℎ and 훤(퐻) acts 퐻 퐻 휎퐻 freely on 퐻, we have 푠 = [ℎ] . Hence 휎♮ | is surjective. 휎퐻 퐻 퐻 Let 푔, ℎ ∈ 퐻 with 푔휎♮ | = ℎ휎♮ | . Then [푔] = [ℎ] and so 퐻 퐻 퐻 퐻 휎퐻 휎퐻 푔 = 1 ⋅ [푔] = 1 ⋅ [ℎ] = ℎ. Hence 휎♮ | is injective. 퐻 휎퐻 퐻 휎퐻 퐻 퐻 ♮ So 휎퐻|퐻 is an isomorphism from 퐻 to 훤(퐻). Hence 훤(퐻) ≃ 퐻. 3.28 Propositions 3.27 and 3.28 have the following consequence:

C o r o l l a ry 3 . 2 9. If 퐻 and 퐻′ are H-classes that are subgroups within the same D-class, then 퐻 ≃ 퐻′. 3.29

Exercises [See pages 210–214 for the solutions.] ✴3.1 Prove that any two elements of a subgroup of a semigroup are H- related. ∗ 3.2 Prove that in a free monoid 퐴 , we have H = L = R = D = J = id퐴∗ .

✴3.3 Let 푋 be a set and let 휎, 휏 ∈ T푋. Prove the following: a) 휎 L 휏 ⇔ im 휎 = im 휏; b) 휎 R 휏 ⇔ ker 휎 = ker 휏;

Exercises • 66 im 휎 = {1, 2, 3}

⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞ 1 2 3 1 2 3 1 2 3 ( ) ( ) ( ) Kernel classes 1 2 3 2 3 1 3 1 2 |im 휎| = 3 {1}, {2}, {3} 1 2 3 1 2 3 1 2 3 ( ) ( ) ( ) 1 3 2 2 1 3 3 2 1 ⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞

im 휎 = {1, 2} im 휎 = {1, 3} im 휎 = {2, 3}

⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞ ⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞ ⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞ 1 2 3 1 2 3 1 2 3 ( ) ( ) ( ) Kernel classes 1 1 2 1 1 3 2 2 3 {1, 2}, {3} 1 2 3 1 2 3 1 2 3 ( ) ( ) ( ) 2 2 1 3 3 1 3 3 2 ⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞ 1 2 3 1 2 3 1 2 3 ( ) ( ) ( ) Kernel classes 1 2 1 1 3 1 2 3 2 |im 휎| = 2 {1, 3}, {2} 1 2 3 1 2 3 1 2 3 ( ) ( ) ( ) 2 1 2 3 1 3 3 2 3 ⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞ 1 2 3 1 2 3 1 2 3 ( ) ( ) ( ) Kernel classes 1 2 2 1 3 3 2 3 3 {1}, {2, 3} 1 2 3 1 2 3 1 2 3 ( ) ( ) ( ) 2 1 1 3 1 1 3 2 2 ⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞

im 휎 = {1} im 휎 = {2} im 휎 = {3} FIGURE 3.7 ⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞ ⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞ ⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞ Kernel class 1 2 3 1 2 3 1 2 3 The egg-box diagrams of the ( ) ( ) ( ) |im 휎| = 1 {1, 2, 3} 1 1 1 2 2 2 3 3 3 three D-classes of T{1,2,3}. ⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞ Idempotents are shaded.

c) 휎 D 휏 ⇔ 휎 J 휏 ⇔ |im 휎| = |im 휏|.

Note that the eggbox diagrams for the D-classes of T{1,2,3} are as shown in Figure 3.7. ✴3.4 Give examples to show that L is not in general a left congruence and R is not in general a right congruence. [Hint: Use Exercise 3.3.] ✴3.5 Let 퐵 = 퐿 × 푅 be a rectangular band. Prove that the R-classes of 퐵 are the sets {ℓ} × 푅 where ℓ ∈ 퐿, that the L-classes of 퐵 are the sets 퐿 × {푟} where 푟 ∈ 푅, that 퐵 consists of a single D-class, and that H is the identity relation. ✴3.6 Prove that if 푆 is a cancellative semigroup and does not contain an identity element, then H = L = R = D = id푆. ✴3.7 Let

푎 푏 푆 = { [ ] ∶ 푎, 푏 ∈ ℝ, 푎, 푏 > 0 } ⊆ 푀 (ℝ). 0 1 2

Exercises • 67 Prove that 푆 is a subsemigroup of 푀2(ℝ). Prove that 푆 is cancellative and has no identity, so that H = L = R = D = id푆 by Exercise 3.6. Prove that 푆 is simple, so that J = 푆 × 푆.

3.8 Let 푋 = {1, … , 푛} for some 푛 ∈ ℕ. In the semigroup T푋, prove that the H-class containing 휏 is a subgroup if and only if |im 휏| = |im(휏2)|. 3.9 Recall that the bicyclic monoid 퐵 is presented by Mon⟨푏, 푐 | (푏푐, 휀)⟩ and that every element of 퐵 has a unique representative of the form 푐훾푏훽. Prove that the R-classes of 퐵 are sets { 푐훾푏훽 ∶ 훽 ∈ ℕ ∪ {0} } (where 훾 ∈ ℕ ∪ {0} is fixed) and L-classes of 퐵 are sets { 푐훾푏훽 ∶ 훾 ∈ ℕ ∪ {0} } (where 훽 ∈ ℕ ∪ {0} is fixed). Deduce that 퐵 has a single D-class. 3.10 Let 푅 be an R-class and 퐿 an L-class of a semigroup 푆 and suppose 퐿 ∩ 푅 contains an idempotent. Let 퐷 be the D-class containing 퐿 and 푅. Prove that 퐿푅 = 퐷. 3.11 Let 푀 be defined by Mon⟨퐴 | 휌⟩, where 퐴 = {푎, 푏, 푐} and 휌 = (푎푏푐, 휀). Prove that for any 푤 ∈ 푀,

푤 H 휀 ⇔ 푤 =푀 휀; 푤 L 휀 ⇔ 푤 ∈ Mon⟨푏푐, 푐⟩; 푤 R 휀 ⇔ 푤 ∈ Mon⟨푎, 푎푏⟩; 푤 D 휀 ⇔ 푤 ∈ Mon⟨푏푐, 푐⟩Mon⟨푎, 푎푏⟩; 푤 J 휀 ⇔ 푤 ∈ Mon⟨푏푐, 푐⟩Mon⟨푎, 푎푏⟩ ∪ Mon⟨푏푐, 푐⟩푏Mon⟨푎, 푎푏⟩.

[Hint: Recall from Exercise 2.8 that every element of 푀 has a unique representative in 푁 = 퐴∗ ∖ 퐴∗푎푏푐퐴∗, and that such a representative can be obtained by iteratively deleting subwords 푎푏푐.] Note that, consequently, all Green’s relations are distinct in 푀. 3.12 Let 푆 be a regular semigroup containing a unique idempotent. Prove that 푆 is a group. 3.13 Let 푀 be a group-embeddable monoid. a) Prove that an element of 푥 is right- and left-invertible if and only if it is R-related to 1푀. b) Prove that 푀 either has one R-class or infinitely many R-classes.

Notes

The exposition of Green’s relations in this chapter owes much to Clifford & Preston, The Algebraic Theory of Semigroups, §§ 2.1–2.4 and Howie, Fundamentals of Semigroup Theory, §§ 2.1–2.4. The discussions of Schützen- berger groups in Clifford & Preston, The Algebraic Theory of Semigroups, § 2.6

Notes • 68 and Grillet, Semigroups, § ii.3 use a different, but equivalent, definition. ◆The example in Exercise 3.7 is due to Andersen, ‘Ein bericht über die Struktur abstrakter Halbgruppen’. ◆ The definition of the relations L, R, and J, the results on principal series, Green’s lemma, and the basic structure of D-classes are all from Green, ‘On the structure of semigroups’. The interaction of inverses and D-classes is due to Miller & Clifford, ‘Regular D-classes in semigroups’. Schützenberger groups first appear, in a rather different form, in Schützenberger, ‘D̅ représentation des demi-groupes’. ◆ For a proof of Theorem 3.11, see Clifford & Preston, The Algebraic Theory of Semigroups, § 2.6. For background reading on the Jordan–Hölder theorem for groups, see Robinson, A Course in the Theory of Groups, § 3.1. •

Notes • 69 Regular semigroups 4 It looks just a little more mathematical and regular ‘ than it is; its exactitude is obvious, but its inexactitude is hidden; its wildness lies in wait. — G.K. Chesterton,’ Orthodoxy, ch. vi.

• Groups are semigroups that have many of the prop- Properties of groups erties we have encountered in previous chapters. For example, groups are cancellative, regular, simple, and all of their elements have unique inverses. In this chapter, we begin to study regular semigroups, because within the class of regular semigroups there is a very interesting hierarchy of classes of semigroups that are more or less ‘group-like’, some of which have very neat structure theorems (in the sense that there is a neat description of the structure of a semigroup in this class). Figure 4.1 outlines the relationship between these classes, and it is useful to refer back to this chart to see how new definitions and results fit into the general setting. (Note that we have not yet defined many of the terms in this figure.) Recall two basic properties of a group 퐺: every element 푥 ∈ 퐺 has a −1 unique inverse 푥 ; and the identity 1퐺 is the unique idempotent, and this idempotent commutes with every element of 퐺. A semigroup is regular if and only if every element has an inverse (Proposition 1.6), but there is no requirement that these inverses are unique: in a rectangular band, every element is an inverse of every element (Example 1.7(e)). Furthermore, every element of a rectangular band is idempotent, but they do not commute. As we shall see, the class of semigroups in which every element has a unique inverse, which are called ‘inverse semigroups’,is very important, and this turns out to be the class of regular semigroups in which idempotents commute (Theorem 5.1). If we impose another condition and require that the idempotents are central (that is, they commute with every element), we obtain the class of Clifford semigroups, which turn out to to have a very neat characterization as ‘strong semilattices of groups’. If we restrict further, and require that there is only one idempotent, we arrive at the class of groups. This is just an example; Figure 4.1 shows other ways in which we can impose properties that groups satisfy and obtain classes of more ‘group-like’ semigroups. However, we are going to begin by defining some of these classes of semigroups in terms of properties that the inverse operation satisfies; this helps prepare the way for the study of varieties in Chapter 8. Later, we

• 70 Semigroup

All elements 0-simple Simple and regular and has has primitive Unique inverses primitive Regular idempotents idempotents semigroup Idempotents commute

Idempotents All elements in subgroups central Completely Inverse regular

simple semigroup semigroup completely semigroups Semilattice of Idempotents Simple central togsemilattice Strong Idempotents Completely Completely groups of central Clifford FIGURE 4.1 0-simple simple semigroup Chart of the classes of semigroup semigroup semigroup semigroup -Rees matrix Rees matrix

over a group over a group semigroup considered in this 0 Only one Only one chapter and the following

L-class R-class product Direct one. Labels on arrows indicate fgopand group of semigroup ih zero right a possible extra condition Left Right (there may be others) that group group left zero restricts the larger class to the semigroup of group and Direct product Only one Only one Unique smaller. Grey text summarizes R-class L-class idempotent the structure theorem for the adjacent class. (Some of these classes have not yet been Group defined.) will show how these classes fit into the chart in Figure 4.1. Let 푆 be a semigroup. If 푆 is a group, the map 푥 ↦ 푥−1 that sends an element to its inverse is a unary operation on 푆 that satisfies certain −1 −1 properties. For instance, by definition 푥푥 = 푥 푥 = 1푆 for all 푥 ∈ 푆. But −1 also satisfies other properties: for all 푥, 푦 ∈ 푆,

(푥−1)−1 = 푥, (푥푦)−1 = 푦−1푥−1, 푥−1푥 = 푥푥−1, 푥푥−1푥 = 푥, 푥푥−1푦푦−1 = 푦푦−1푥푥−1, 푥푥−1 = 푦푦−1.

If we require that a semigroup with an operation −1 satisfies only some of these properties, we may no longer have a group. Instead, we obtain different types of semigroup depending on which conditions are required. Let 푆 be a semigroup equipped with an operation −1. If 푆 satisfies the Regular semigroup condition that for all 푥 ∈ 푆,

푥푥−1푥 = 푥, then 푆 is clearly regular, as defined on page 6. [Note that in a regular semigroup, an element may have many different inverses. However, we can always define an operation −1 by choosing a particular inverse for

Regular semigroups • 71 each element.] If 푆 satisfies the two conditions that for all 푥 ∈ 푆, (푥−1)−1 = 푥, 푥푥−1푥 = 푥, (4.1) then again 푆 is regular and for any 푦 ∈ 푆, we have 푦푦−1푦 = 푦 and 푦−1푦푦−1 = 푦−1(푦−1)−1푦−1 = 푦−1 and so 푦−1 is an inverse of 푦. If 푆 Completely satisfies the three conditions that for all 푥 ∈ 푆, regular semigroup (푥−1)−1 = 푥, 푥−1푥 = 푥푥−1, 푥푥−1푥 = 푥, (4.2) it is a completely regular semigroup. We will look at regular and completely regular semigroups in this chapter. If 푆 satisfies the four conditions that Inverse semigroup for all 푥, 푦 ∈ 푆, (푥−1)−1 = 푥, (푥푦)−1 = 푦−1푥−1, } (4.3) 푥푥−1푥 = 푥, 푥푥−1푦푦−1 = 푦푦−1푥푥−1, it is an inverse semigroup. Finally, if 푆 satisfies the four conditions that for Clifford semigroup all 푥, 푦 ∈ 푆, (푥−1)−1 = 푥, 푥−1푥 = 푥푥−1, } (4.4) 푥푥−1푥 = 푥, 푥푥−1푦푦−1 = 푦푦−1푥푥−1, it is a Clifford semigroup. We will look at inverse semigroups and Clifford semigroups in Chapter 5.

Completely 0-simple semigroups The aim of this section is to introduce the concept of a completely 0-simple semigroup and to present a classification result for such semigroups, the Rees–Suschkewitsch theorem, which was one of the most important results in the early development of semigroup theory. We study completely 0-simple semigroups for two reasons. First, we saw in Proposition 3.10 that the principal factors of a semigroup are either 0-simple or null, and completely 0-simple semigroup are an important subclass of 0-simple semigroups. Furthermore, studying completely 0-simple semigroups will lead naturally to studying completely simple semigroups, and we will see that both completely 0-simple and completely simple semigroups are regular (Lemma 4.6(b) and Proposition 4.13), and that a simple semigroup is completely simple if and only if it is completely regular (Theorem 4.16). Recall that the set of idempotents 퐸(푆) of a semigroup 푆 admits a Primitive idempotent natural partial order given by 푒 ≼ 푓 ⇔ 푒푓 = 푓푒 = 푒. (See Proposition 1.19.) In a semigroup with a zero, 0 is the unique minimal idempotent; in such a semigroup, an idempotent is primitive if it is minimal within the set of non-zero idempotents of the semigroup. A semigroup is completely Completely 0-simple if it is 0-simple and contains at least one primitive idempotent. 0-simple semigroup

Completely 0-simple semigroups • 72 P r o p o s i t i o n 4 . 1. A finite 0-simple semigroup is completely 0-simple. Finite 0-simple ⇒ completely 0-simple Proof of 4.1. Let 푆 be a finite 0-simple semigroup. Now, if there are non- zero idempotents in 푆, there must be a primitive idempotent in 푆, since otherwise there would be infinite descending chains of idempotents, and this is impossible since 푆 is finite. So we must simply rule out the possibility that 0 is the only idempotent. So suppose, with the aim of obtaining a contradiction, that the only idempotent in 푆 is 0. Let 푥 ∈ 푆 ∖ {0}. Then by Lemma 3.7 there exist 푝, 푞 ∈ 푆 with 푝푥푞 = 푥. Hence 푝푛푥푞푛 = 푥 for all 푛 ∈ ℕ. Since 푆 is finite and thus periodic, 푝푚 is idempotent for some 푚 ∈ ℕ. Thus 푝푚 = 0 since 0 is the only idempotent in 푆. Therefore 푥 = 푝푚푥푞푚 = 0푥푞푚 = 0, which contradicts the choice of 푥. So it is impossible for 0 to be the only idempotent of 푆. This completes the proof. 4.1

We are now going to show how to construct examples of completely Rees matrix semigroup 0-simple semigroups. Let 퐺 be a group, let 퐼 and 훬 be abstract index sets, and let 푃 be a regular 훬 × 퐼 matrix with entries from 퐺0. (Recall that a matrix is regular if every row and every column contains at least one non-zero entry. By a ‘훬 × 퐼 matrix’ we mean simply a matrix whose rows are indexed by 훬 and whose columns are indexed by 퐼.) Let 푝휆푖 be the (휆, 푖)-th entry of 푃. Let 푆 be the set 퐼 × 퐺0 × 훬. Define a multiplication on 푆 by

(푖, 푥, 휆)(푗, 푦, 휇) = (푖, 푥푝휆푗푦, 휇). This multiplication is associative since

(푖, 푥, 휆)((푗, 푦, 휇)(푘, 푧, 휈)) = (푖, 푥, 휆)(푗, 푦푝휇푘푧, 휈)

= (푖, 푥푝휆푗푦푝휇푘푧, 휈)

= (푖, 푥푝휆푗푦, 휇)(푘, 푧, 휈) = ((푖, 푥, 휆)(푗, 푦, 휇))(푘, 푧, 휈), and so 푆 is a semigroup. Let 푇 = 퐼 × {0} × 훬. It is easy to see that 푇 is an ideal of 푆. Clearly, 푆∖푇 = 퐼×퐺×훬. Notice that if (푖, 푥, 휆), (푗, 푦, 휇) ∈ 푆∖푇, then (푖, 푥, 휆)(푗, 푦, 휇) ∈ 푇 if and only if 푝휆푗 = 0. Let M0[퐺; 퐼, 훬; 푃] be the Rees factor semigroup 푆/푇. Then the semigroup M0[퐺; 퐼, 훬; 푃] can be viewed as the set (푆 ∖ 푇) ∪ {0}: that is, M0[퐺; 퐼, 훬; 푃] can be viewed as the set (퐼 × 퐺 × 훬) ∪ {0} under the multiplication

(푖, 푥푝 푦, 휇) if 푝 ≠ 0, (푖, 푥, 휆)(푗, 푦, 휇) = { 휆푗 휆푗 0 if 푝휆푗 = 0, 0(푖, 푥, 휆) = (푖, 푥, 휆)0 = 00 = 0.

The semigroup M0[퐺; 퐼, 훬; 푃] is called the 퐼 × 훬 Rees matrix semigroup over 퐺0 with regular sandwich matrix 푃.

Completely 0-simple semigroups • 73 (1, 푦, 2) (1, 푥푝23푦, 5) if 푝23 ≠ 0 FIGURE 4.2 Multiplication in a Rees matrix semigroup M [퐺; 퐼, 훬; 푃]. 푝11 푝12 푝13 0 The product (1, 푦, 2)(3, 푥, 5) is [푝 푝 푝 ] [ 21 22 23] either (1, 푦푝 푥, 5) or 0, de- 푃 = [푝 푝 푝 ] 23 [ 31 32 33] pending on the value of 푝23. (3, 푥, 5) [ ] T 푝41 푝42 푝43 The shape of 푃 is the same as the shape of the grid, and [푝51 푝52 푝53] the cells containing the multi- 0 if 푝23 = 0 푝11 푝21 푝31 푝41 푝51 plicands, the cell correspond- T [ ] 푃 = 푝12 푝22 푝32 푝42 푝52 ing to 푝23, and cell containing the product (if it is non-zero) 푝13 푝23 푝33 푝43 푝53 [ ] form the corners of a rectangle.

Diagrammatically, we can place the non-zero elements of this Rees matrix semigroup in a rectangular pattern, divided into a grid of cells indexed by the sets 퐼 and 훬, so that the (푖, 휆)-th cell contains all elements of the form (푖, 푔, 휆), where 푔 ∈ 퐺. Figure 4.2 illustrates how the multiplication works in terms of this diagram. Compare this with Figure 1.1. This is of course reminiscent of an egg-box diagram, and we will see that it actually is an egg-box diagram: the columns, rows, and cells of this grid are the non-zero L-, R-, and H-classes of the Rees matrix semigroup.

P r o p o s i t i o n 4 . 2. For any group 퐺, index sets 퐼 and 훬, and matrix Rees matrix ⇒ 0 completely 0-simple 푃 over 퐺 , the semigroup M0[퐺; 퐼, 훬; 푃] is completely 0-simple.

Proof of 4.2. For brevity, let 푆 = M0[퐺; 퐼, 훬; 푃]. Let (푖, 푥, 휆) ∈ 푆 ∖ {0}. Let (푗, 푦, 휇) ∈ 푆 ∖ {0}. Since 푃 is regular, we can choose 휈 ∈ 훬 and 푘 ∈ 퐼 such that 푝휈푖 ≠ 0 and 푝휆푘 ≠ 0. Then −1 −1 −1 (푗, 1퐺, 휈)(푖, 푥, 휆)(푘, 푝휆푘 푥 푝휈푖 푦, 휇) −1 −1 −1 = (푗, 1퐺푝휈푖푥푝휆푘푝휆푘 푥 푝휈푖 푦, 휇) = (푗, 푦, 휇). Hence, since (푗, 푦, 휇) ∈ 푆 ∖ {0} was arbitrary, and since 0 = 0(푖, 푥, 휆)0, we have 푆 ⊆ 푆(푖, 푥, 휆)푆. Since (푖, 푥, 휆) ∈ 푆 ∖ {0} was arbitrary, 푆 is 0-simple by Lemma 3.7. Now, (푖, 푥, 휆) ∈ 푆∖{0} is an idempotent if and only if (푖, 푥, 휆)(푖, 푥, 휆) = −1 (푖, 푥푝휆푖푥, 휆) = (푖, 푥, 휆), which is true if and only if 푝휆푖 ≠ 0 and 푥 = 푝휆푖 . −1 Hence the idempotents in 푆 ∖ {0} are elements of the form (푖, 푝휆푖 , 휆). Furthermore,

−1 −1 (푖, 푝휆푖 , 휆) ≼ (푗, 푝휇푗 , 휇) −1 −1 −1 −1 −1 ⇔ (푖, 푝휆푖 , 휆)(푗, 푝휇푗 , 휇) = (푗, 푝휇푗 , 휇)(푖, 푝휆푖 , 휆) = (푖, 푝휆푖 , 휆) −1 −1 −1 −1 −1 ⇔ (푖, 푝휆푖 푝휆푗푝휇푗 , 휇) = (푗, 푝휇푗 푝휇푖푝휆푖 , 휆) = (푖, 푝휆푖 , 휆) ⇔ (푖 = 푗) ∧ (휆 = 휇) −1 −1 ⇔ (푖, 푝휆푖 , 휆) = (푗, 푝휇푗 , 휇).

Completely 0-simple semigroups • 74 Hence every idempotent in 푆 ∖ {0} is primitive. Thus 푆 certainly contains primitive idempotents and so is completely 0-simple. 4.2 Proposition 4.2 gives a method for constructing completely 0-simple semigroups. In fact, all completely 0-simple semigroups arise in this way:

P r o p o s i t i o n 4 . 3. Let 푆 be a completely 0-simple semigroup. Then Completely 0-simple ⇒ Rees matrix 푆 ≃ M0[퐺; 퐼, 훬; 푃] for some group 퐺, index sets 퐼 and 훬, and regular sandwich matrix 푃.

Proof of 4.3. Let 푆 be completely 0-simple. We have to define a Rees matrix semigroup M0[퐺; 퐼, 훬; 푃] and show it is isomorphic to 푆. Since 푆 is completely 0-simple, it contains a primitive idempotent. We first describe the R-classes and L-classes of primitive idempotents: L e m m a 4 . 4. For any primitive idempotent 푒 of 푆,

a) 푅푒 = 푒푆 ∖ {0},

b) 퐿푒 = 푆푒 ∖ {0}. Proof of 4.4. We prove part a); a dual argument gives part b). Note that by definition 푒 ≠ 0. Every element of 푅푒 is a right multiple of 푒 and cannot be 0. Hence 푅푒 ⊆ 푒푆 ∖ {0}. Let 푥 ∈ 푒푆∖{0}. So 푥 = 푒푠 for some 푠 ∈ 푆∖{0}. Hence 푒푥 = 푒푒푠 = 푒푠 = 푥. Since 푆 is 0-simple, by Lemma 3.7 there exist 푝, 푞 ∈ 푆 with 푝푥푞 = 푒. Let 푝′ = 푒푝푒. Then 푝′푥푞 = 푒푝푒푥푞 = 푒푝푥푞 = 푒푒 = 푒. Let 푓 = 푥푞푝′. Then 푓2 = 푥푞푝′푥푞푝′ = 푥푞푒푝′ = 푥푞푒푒푝푒 = 푥푞푒푝푒 = 푥푞푝′ = 푓. So 푓 is idempotent. Furthermore, 푒푓 = 푒푥푞푝′ = 푥푞푝′ = 푓 and 푓푒 = 푥푞푝′푒 = 푥푞푒푝푒푒 = 푥푞푒푝푒 = 푥푞푝′ = 푓. So 푒푓 = 푓푒 = 푓 and hence 푓 ≼ 푒. Suppose that 푓 = 0; then 푒 = 푒2 = 푝′푥푞푝′푥푞 = 푝′푓푥푞 = 0, which is a contradiction. Hence 푓 ≠ 0. But 푒 is primitive and therefore ≼- minimal among non-zero idempotents; thus 푒 = 푓 = 푥푞푝′. Since 푥 = 푒푠, it follows that 푥 R 푒 and so 푥 ∈ 푅푒. Hence 푒푆 ∖ {0} ⊆ 푅푒. 4.4 L e m m a 4 . 5. For any 푥 ∈ 푆 ∖ {0},

a) 푅푥 = 푥푆 ∖ {0},

b) 퐿푥 = 푆푥 ∖ {0}. Proof of 4.5. We prove part a); a dual argument gives part b). As in the proof of Lemma 4.4(a), 푅푥 ⊆ 푥푆 ∖ {0}. So let 푦 ∈ 푥푆 ∖ {0}. Then 푦 = 푥푠 for some 푠 ∈ 푆 ∖ {0}. Let 푒 be a primitive idempotent of 푆. Since 푆 is 0- simple, by Lemma 3.7 there exist 푝, 푞 ∈ 푆 with 푝푒푞 = 푥. So 푦 = 푝푒푞푠. By Lemma 4.4(a), 푒푞푠, 푒푞 ∈ 푅푒. Since R is a left congruence by Propostion 3.4, 푦 = 푝푒푞푠 R 푝푒푞 = 푥. So 푦 ∈ 푅푥 and hence 푥푆 ∖ {0} ⊆ 푅푥. 4.5 We can now deduce information about the D-class structure of 푆: L e m m a 4 . 6. a) The D-classes of 푆 are 0 and 푆 ∖ {0}. b) The semigroup 푆 is regular.

Completely 0-simple semigroups • 75 c) For all 푥, 푦 ∈ 푆 ∖ {0}, if 퐿푥 ∩ 푅푦 contains an idempotent, then 푥푦 ∈ 푅푥 ∩ 퐿푦; otherwise, 푥푦 = 0. Proof of 4.6. a) Let 푥, 푦 ∈ 푆 ∖ {0}. Suppose 푥푆푦 = {0}. Then

푆2 = 푆푥푆푆푦푆 ⊆ 푆(푥푆푦)푆 = 푆{0}푆 = {0},

which contradicts the fact that 0-simple semigroups are (by definition) not null. Hence 푥푆푦 contains some non-zero element 푡. Now, 푡 ∈ 푥푆 ∖ {0} and 푡 ∈ 푆푦 ∖ {0}. Hence 푡 ∈ 푅푥 and 푡 ∈ 퐿푦 by Lemma 4.5. Thus 푥 R 푡 L 푦 and so 푥 D 푦. So the D-classes of 푆 must be 푆 ∖ {0} and {0}. b) The primitive idempotent 푒 lies in 푆∖{0} and so every element of 푆∖{0} is regular by Proposition 3.19. Since 0 is also regular, the semigroup 푆 is regular.

c) Let 푥, 푦 ∈ 푆 ∖ {0}. By part b), 푥 D 푦. Suppose 퐿푥 ∩ 푅푦 contains an idempotent. Then 푥푦 ∈ 푅푥 ∩ 퐿푦 by Proposition 3.18. Suppose 퐿푥 ∩ 푅푦 does not contain an idempotent. Then 푥푦 ∉ 푅푥 ∩ 퐿푦, and so 푥푦 = 0, since if 푥푦 = 0, then by Lemma 4.6 푥푦 ∈ 푥푆 ∖ {0} = 푅푥 and 푥푦 ∈ 푆푦 ∖ {0} = 퐿푦, contradicting 푥푦 ∉ 푅푥 ∩ 퐿푦. 4.6 Let 퐻 be an H-class of 푆 contained in the D-class 푆∖{0}. Let 푥, 푦 ∈ 퐻. Then either 푥푦 = 0 or 푥푦 ∈ 푅푥 ∩ 퐿푦 = 퐻 by Lemma 4.6(c). Suppose first that 푥푦 = 0. Let 푧, 푡 ∈ 퐻. Since 푧 L 푥 and 푡 R 푦, we have 푧 = 푝푥 and 푡 = 푦푠 for some 푝, 푠 ∈ 푆1. Then 푧푡 = 푝푥푦푠 = 푝0푟 = 0. Since 푧, 푡 ∈ 퐻 were arbitrary, 퐻2 = {0}. On the other hand, suppose that 푥푦 ∈ 퐻. Then 퐻 is a subgroup by Proposition 3.14. So we can divide the H-classes in the D-class 푆 ∖ {0} into zero H-classes and group H-classes. Let 퐼 be the set of R-classes and let 훬 be the set of L-classes in 푆 ∖ {0}. Write the R- and L-classes as 푅(푖) and 퐿(휆) for 푖 ∈ 퐼 and 휆 ∈ 훬, and write 퐻(푖휆) for 푅(푖) ∩ 퐿(휆). We will treat 퐼 and 훬 as abstract index sets, and these will ultimately be the index sets for the Rees matrix semigroup we are constructing. Since 푆 ∖ {0} is a regular D-class by Lemma 4.6(b), every R-class and every L-class contains an idempotent by Proposition 3.20 and thus contains some group H-class. Therefore assume without loss that there is some element 1 ∈ 퐼 ∩ 훬 such that 퐻(11) is a group H-class. For brevity, write 퐻 for 퐻(11). (1휆) (1) 푟 휌 | (1) 휆 For each 푖 ∈ 퐼 and 휆 ∈ 훬, fix arbitrary elements 푟휆 ∈ 퐻 ⊆ 푅 and 푟휆 퐿 (1) (푖1) (1) (1) 푥 푅 푞푖 ∈ 퐻 ⊆ 퐿 . Since 1퐻 is idempotent, it is a left identity for 푅 and (1) a right identity for 퐿 by Proposition 3.17. So 1퐻푟휆 = 푟휆 and 푞푖1퐻 = 푞푖. 휆 | (1) (1) (휆) 푞푖 푅 Therefore, by Lemma 3.12, 휌 | (1) ∶ 퐿 → 퐿 restricts to a bijection 푟휆 퐿 (푖1) (1) (1) between 퐻 and 퐻 and 휆 | (1) ∶ 푅 → 푅 restricts to a bijection 푞푖 푅 between 퐻(푖1) and 퐻(푖휆) for each 휆 ∈ 훬. Thus there is a unique expression (푖휆) 푞푖푥푟휆 (푖) 푞푖푥푟휆, where 푥 ∈ 퐻, for every element of 퐻 . 푞푖 푅 퐿(1) 퐿(휆) FIGURE 4.3 Completely 0-simple semigroups • 76 (1휆) Choosing 푟휆 ∈ 퐻 and 푞푖 ∈ (푖1) 퐻 gives bijections 휌 | (1) 푟휆 퐿 and 휆 | (1) and so a unique 푞푖 푅 expression 푞푖푥푟휆 for each element of 퐻(푖휆0. Therefore the map 휑 ∶ (퐼 × 퐻 × 훬) ∪ {0} → 푆 defined by (푖, 푥, 휆)휑 = 푞푖푥푟휆 and 0휑 = 0 is a bijection. To turn (퐼 × 퐻 × 훬) ∪ {0} into a 퐼 × 훬 Rees matrix semigroup over 퐻0, it remains to define a sandwich matrix 푃. For each 푖 ∈ 퐼 and 휆 ∈ 훬, let 푝휆푖 = 푟휆푞푖, and let 푃 be the 훬 × 퐼 matrix whose (휆, 푖)-th entry is 푝휆푖. By Lemma 4.6(c), 푝 = 푟 푞 ∈ 푅 ∩ 퐿 = 푅(1) ∩ 퐿(1) = 퐻 if and only if 휆푖 휆 푖 푟휆 푞푖 푅 ∩ 퐿 contains an idempotent and is thus a group H-class; otherwise 푞푖 푟휆 0 푝휆푖 = 0. Hence each 푝휆푖 lies in 퐻 . Furthermore, since every R-class and every L-class contains an idempotent, for every 푖 ∈ 퐼 there exists 휆 ∈ 훬 such that 푅 ∩ 퐿 contains an idempotent and so 푝 ∈ 퐻, and 푞푖 푟휆 휆푖 thus 푝휆푖 ≠ 0. Thus every column of 푃 contains a non-zero entry. Similarly every row of 푂 contains a non-zero entry. Therefore 푃 is a regular matrix. So 휑 is now a bijection from M0[퐻; 퐼, 훬; 푃] to 푆. For any elements (푖, 푥, 휆), (푗, 푦, 휇) ∈ M0[퐻; 퐼, 훬; 푃] ∖ {0},

((푖, 푥, 휆)휑)((푗, 푦, 휇)휑) = (푞푖푥푟휆)(푞푗푦푟휇)

= 푞푖푥(푟휆푞푗)푦푟휇

= 푞푖푥푝휆푗푦푟휇 (푖, 푥푝 푦, 휇)휑 if 푝 ≠ 0 = { 휆푗 휆푗 0휑 if 푝휆푗 = 0 = ((푖, 푥, 휆)(푗, 푦, 휇))휑.

Furthermore, ((푖, 푥, 휆)휑)(0휑) = 푞푖푥푟휆0 = 0 = ((푖, 푥, 휆)0)휑 and similarly for other multiplications involving 0. Hence the map 휑 is a homomorphism and hence an isomorphism between M0[퐻; 퐼, 훬; 푃] and 푆. 4.6 Combining Propositions 4.2 and 4.3, we get the following characterization of completely 0-simple semigroups:

Rees–Suschkewitsch Theorem 4.7. A semigroup 푆 is com- Rees–Suschkewitsch pletely 0-simple if and only if there exist a group 퐺, index sets, 퐼 and theorem 훬, and a regular 훬 × 퐼 matrix 푃 with entries from 퐺0 such that 푆 ≃ M0[퐺; 퐼, 훬; 푃]. 4.7 One of the advantages of this characterization is that it gives us a neat description of the H, L, R, D, and J-classes:

P r o p o s i t i o n 4 . 8. Let 푆 ≃ M0[퐺; 퐼, 훬, 푃] be a completely 0-simple Green’s relations semigroup. in completely 0-simple semigroups a) In 푆, the relations D and J coincide, and 푆 has two D-classes {0} and 푆 ∖ {0} = 퐼 × 퐺 × 훬. b) The L-classes of 푆 are {0} and sets of the form 퐼 × 퐺 × {휆}. c) The R-classes of 푆 are {0} and sets of the form {푖} × 퐺 × 훬. d) The H-classes of 푆 are {0} and sets of the form {푖} × 퐺 × {휆}.

Completely 0-simple semigroups • 77 Proof of 4.8. a) By Lemma 4.6, 푆 has two D-classes, {0} and 푆∖{0}. Since 푆 is 0-simple, it has only two ideals: 푆 and {0}. If 푥 ∈ 푆 ∖ {0}, then 푥 ∈ 푆1푥푆1, so 푆1푥푆1 = 푆. On the other hand, 푆10푆1 = {0}. So 푆 and {0} are also the J-classes of 푆. b) Since {0} is the D-class of 0, it is alsoe the L-class of 0. Let (푖, 푥, 휆)∖{0}. By Lemma 4.5(b), we have 퐿(푖,푥,휆) = 푆(푖, 푥, 휆)∖{0}. First, note that 푆(푖, 푥, 휆) ∖ {0} ⊆ 퐼 × 퐺 × {휆} ∖ {0} by the definition of multiplication in M0[퐺; 퐼, 훬; 푃]. On the other hand, let (푗, 푦, 휆) ∈ 퐼 × 퐺 × {휆} ∖ {0}. Let 푠 = −1 −1 (푗, 푦푥 푝휇푖 , 휇), where 휇 is such that 푝휇푖 ≠ 0; such a 휇 exists because −1 푃 is regular. Note that 푝휇푖 ∈ 퐺, so 푝휇푖 exists. Then

−1 푠(푖, 푥, 휆) = (푗, 푦푥푝휇푖 , 휇)(푖, 푥, 휆) −1 −1 = (푗, 푦푥 푝휇푖 푝휇푖푥, 휆) = (푗, 푦, 휆).

So 퐼 × 퐺 × {휆} ∖ {0} ⊆ 푆(푖, 푥, 휆) ∖ {0}. Thus 퐿(푖,푥,휆) = 퐼 × 퐺 × {휆} ∖ {0}. c) The reasoning is dual to part b).

d) First, 퐻0 = 퐿0 ∩ 푅0 = {0}. For (푖, 푥, 휆) ∈ 푆 ∖ {0}, we have 퐻(푖,푥,휆) = 퐿(푖,푥,휆) ∩ 푅(푖,푥,휆) = (퐼 × 퐺 × {휆}) ∩ ({푖} × 퐺 × 훬) = {푖} × 퐺 × {휆}. 4.8

Ideals and completely 0-simple semigroups This section characterizes the 0-simple semigroups that are also completely 0-simple. We require some definitions. A semigroup 푆 is group-bound if every 푥 ∈ 푆 has some power 푥푛 lying in a subgroup Group-bound semigroup of 푆. A semigroup satisfies the condition minL (respectively, minR) if any minL, minR subset of the partial order 푆/L (respectively, 푆/R) has a minimal element.

T h e o r e m 4 . 9. Let 푆 be 0-simple. The following are equivalent: Characterization of a) 푆 is completely 0-simple; 0-simple semigroups that are completely 0-simple b) 푆 is group-bound;

c) 푆 satisfies the conditions minL and minR. Proof of 4.9. Part 1 [a) ⇒ b)]. Suppose 푆 is completely 0-simple. Let 푥 ∈ 푆. 2 2 Then either 퐻푥 is a subgroup and 푥 ∈ 퐻푥, or else 푥 = 0. In either case, 푥2 lies in a subgroup. Thus 푆 is group-bound. Part 2 [b) ⇒ c)]. Suppose 푆 is group-bound. Let 푥, 푦 ∈ 푆 ∖ {0} be such that 퐿푥 ⩽ 퐿푦. We are going to show that 퐿푥 = 퐿푦. First, notice that 푥 = 푝푦 for some 푝 ∈ 푆1. Furthermore, 푦 = 푞푥푟 for some

Ideals and completely 0-simple semigroups • 78 푞, 푟 ∈ 푆 by Lemma 3.7, since 푆 is 0-simple. Then 푦 = 푞푥푟 = 푞푝푦푟 and so 푦 = (푞푝)푛푦푟푛 for all 푛 ∈ ℕ. Fix 푛 so that 푔 = (푞푝)푛 lies in a subgroup 푛 푛 −1 −1 푛 퐺. Then 1퐺푦 = 1퐺푔푦푟 = 푔푦푟 = 푦 and so 푦 = 푔 푔푦 = 푔 (푞푝) 푦 = −1 푛−1 −1 푛−1 푔 (푞푝) 푞푝푦 = 푔 (푞푝) 푞푥. Hence 퐿푦 ⩽ 퐿푥 and so 퐿푥 = 퐿푦. There- fore 퐿푥 ⩽ 퐿푦 ⇒ 퐿푥 = 퐿푦, and this certainly implies that any subset of 푆/L is has a minimal element. So 푆 satisfies minL. Similarly, 푆/R satisfies minR.

Part 3 [c) ⇒ a)]. Suppose 푆 satisfies minL and minR. Suppose, with the aim of obtaining a contradiction, that 푆 does not contain a primitive idempotent. Then 푆 contains an infinite descending chain of non-zero idempotents

푒1 ≻ 푒2 ≻ 푒3 ≻ … . Notice that for 푒, 푓 ∈ 퐸(푆), by the definition of the partial order ≼ on 퐸(푆) and the relation R, we have

푒 ≼ 푓 ⇒ 푒 = 푓푒 ⇒ 푒푆 = 푓푒푆 ⊆ 푓푆 ⇒ 푅푒 ⩽ 푅푓 and similarly 푒 ≼ 푓 ⇒ 퐿푒 ⩽ 퐿푓. Hence 퐿 ⩾ 퐿 ⩾ 퐿 ⩾ … and 푅 ⩾ 푅 ⩾ 푅 ⩾ … , 푒1 푒2 푒3 푒1 푒2 푒3 Since 푆 satisfies min and min , the set of L-classes { 퐿 ∶ 푖 ∈ ℕ } L R 푒푖 contains a minimal element 퐿 and the set of R-classes { 푅 ∶ 푖 ∈ ℕ } 푒푗 푒푖 contains a minimal element 푅 . Let ℓ = max{푗, 푘}; then 퐿 = 퐿 and 푒푘 푒ℓ 푒ℓ+1 푅 = 푅 , and so 퐻 = 퐻 . By Corollary 3.16, 푒 = 푒 , which 푒ℓ 푒ℓ+1 푒ℓ 푒ℓ+1 ℓ ℓ+1 is a contradiction. Thus 푆 contains a primitive idempotent and so is completely 0-simple. 4.9

Completely simple semigroups

An idempotent of a semigroup without zero is primitive if Primitive idempotent it is minimal. A semigroup without zero is completely simple if it is simple Completely and contains a primitive idempotent. simple semigroup ‘Primitive idempotent’ has different meanings for semigroups with and without zero: in a semigroup with a zero, a primitive idempotent is a minimal non-0 idempotent; in a semigroup without zero, a primitive idempotent is a minimal idempotent.

P r o p o s i t i o n 4 . 1 0. A finite simple semigroup is completely simple. Finite simple ⇒ completely simple Proof of 4.10. Let 푆 be a finite simple semigroup. Since 푆 is finite, every element has a power which is an idempotent. So 푆 contains at least one idempotent. Furthermore, there must be a primitive idempotent in 푆, since otherwise there would be an infinite descending chain of idempotents, and this is impossible since 푆 is finite. 4.10

Completely simple semigroups • 79 Define a new version of the Rees matrix construction as follows. Let 퐺 be a group, let 퐼 and 훬 be abstract index sets, and let 푃 be a 훬 × 퐼 matrix with entries from 퐺, with the (휆, 푖)-th entry of 푃 being 푝휆푖. Let M[퐺; 퐼, 훬; 푃] be the set 퐼 × 퐺 × 훬 with multiplication

(푖, 푥, 휆)(푗, 푦, 휇) = (푖, 푥푝휆푗푦, 휇).

Then we have the following characterization of completely simple semigroups, paralleling the Rees–Suschkewitsch theorem:

T h e o r e m 4 . 1 1. A semigroup 푆 is completely simple if and only if there Characterization exist a group 퐺, index sets 퐼 and 훬, and a 훬 × 퐼 matrix 푃 with entries from of completely simple semigroups 퐺 such that 푆 ≃ M[퐺; 퐼, 훬; 푃]. 4.11

Theorem 4.11, and many other properties of completely simple semigroups, are consequences of the following observations: ◆ 푆 is simple if and only if 푆0 is 0-simple; ◆ an idempotent is primitive in 푆 if and only if it is primitive in 푆0’; ◆ for any group 퐺, index sets 퐼 and 훬, and 훬 × 퐼 matrix 푃 with entries 0 from 퐺, we have (M[퐺; 퐼, 훬; 푃]) = M0[퐺; 퐼, 훬; 푃]. Notice that in the second condition above, ‘primitive’ means ‘minimal’ in 푆 and ‘minimal non-0’ in 푆0. The proof of the following characterization of Green’s relations in completely simple semigroups is similar to the proof of Proposition 4.8.

P r o p o s i t i o n 4 . 1 2. Let 푆 ≃ M[퐺; 퐼, 훬, 푃] be a completely simple Green’s relations semigroup. in completely simple semigroups a) In 푆, the relations D and J coincide, and 푆 consists of a single D-class. b) The L-classes of 푆 are sets of the form 퐼 × 퐺 × {휆}. c) The R-classes of 푆 are sets of the form {푖} × 퐺 × 훬. d) The H-classes of 푆 are sets of the form {푖} × 퐺 × {휆}. 4.12

P r o p o s i t i o n 4 . 1 3. A semigroup is completely simple if and only if it Characterization of is regular and every idempotent is primitive. regular semigroups that are completely simple Proof of 4.13. Suppose 푆 is completely simple. Then 푆 ≃ M[퐺; 퐼, 훬; 푃]. So 푆 consists of a single D-class. Furthermore, 푆 contains idempotents, which are regular elements. Hence 푆 is regular by Proposition 3.19. By following the reasoning in the proof of Proposition 4.2 (and ignoring mentions of the zero), every idempotent in 푆 is primitive. Now suppose that 푆 is regular and every idempotent is primitive. We have to show that 푆 is simple. Since 푆 is regular, every D-class contains an idempotent. So every J-class contains an idempotent. Let 퐽푒 be a J-class

Completely simple semigroups • 80 and let 퐽푓 ⩽ 퐽푒, where 푒 and 푓 are idempotents. Then 푓 = 푝푒푞 for some 푝, 푞 ∈ 푆1. Let 푔 = 푒푞푓푝푒. Then

푔2 = 푒푞푓푝푒푒푞푓푝푒 [by definition of 푔] = 푒푞푓푝푒푞푓푝푒 [since 푒 is idempotent] = 푒푞푓푓푓푝푒 [since 푓 = 푝푒푞] = 푒푞푓푝푒 [since 푓 is idempotent] = 푔; [by definition of 푔] thus 푔 is idempotent. Furthermore 푔푒 = 푒푔 = 푔 and so 푔 ≼ 푒. Since 푒 is primitive (since all idempotents are primitive), it follows that 푔 = 푒. Therefore 푓 = 푝푒푞 and 푒 = 푔 = 푒푞푓푝푒. Hence 퐽푒 = 퐽푓. Since all J-classes contain idempotents, 푆 contains only one J-class. Hence all elements 푥 ∈ 푆 generates the same principal ideal, 푆1푥푆1, and so 푆 = 푆1푥푆1 for all 푥 ∈ 푆. Thus 푆 is the only ideal of 푆 and so 푆 is simple. 4.13

P r o p o s i t i o n 4 . 1 4. A completely simple semigroup is a group if and Characterization of only if it contains exactly one idempotent. completely simple semigroups that are groups Proof of 4.14. One direction of this result is obvious: a group contains exactly one idempotent. Suppose 푆 is completely simple and contains exactly one idempotent. By Proposition 4.13, 푆 is regular. Hence, by Proposition 3.20, every L- and every R-class of 푆 contains an idempotent. Since 푆 contains only one idempotent, it contains only one L-class and only one R-class and so consists of a single H-class, which is a group by Proposition 3.14. 4.14

Completely regular semigroups Recall that a semigroup 푆 is completely regular if it is equipped with a unary operation −1 satisfying the conditions in (4.2); namely that for all 푥 ∈ 푆,

(푥−1)−1 = 푥, 푥−1푥 = 푥푥−1, 푥푥−1푥 = 푥.

T h e o r e m 4 . 1 5. Let 푆 be a semigroup. Then the following are equivalent: Characterization a) 푆 is completely regular; of completely regular semigroups b) every element of 푆 lies in a subgroup of 푆; c) every H-class of 푆 is a subgroup. Proof of 4.15. Part 1 [a) ⇒ b)]. Suppose 푆 is completely regular. Let 푥 ∈ 푆. Then 푒 = 푥푥−1 = 푥−1푥 is an idempotent, since 푒2 = (푥푥−1)(푥푥−1) = −1 −1 −1 (푥푥 푥)푥 = 푥푥 = 푒. By Proposition 3.18, 푥 ∈ 푅푒 ∩ 퐿푒 = 퐻푒, which is a subgroup. So every element of 푆 lies in a subgroup.

Completely regular semigroups • 81 Part 2 [b) ⇒ c)]. Suppose every element of 푆 lies in a subgroup. Let 푥 ∈ 푆. Then 푥 ∈ 퐺 for some subgroup 퐺 of 푆. Then 푥 H 1 and so 퐻 = 퐻 , 퐺 푥 1퐺 which contains an idempotent and is thus a subgroup. So every H-class of 푆 is a subgroup. Part 3 [c) ⇒ a)]. Suppose every H-class of 푆 is a subgroup. Define −1 by −1 letting 푥 (where 푥 ∈ 푆) be the unique inverse of 푥 in the subgroup 퐻푥. It is clear that −1 satisfies the conditions (4.2) and 푆 is thus completely regular. 4.15 The next result is analogous to Theorem 4.9:

T h e o r e m 4 . 1 6. Let 푆 be simple. The following are equivalent: Characterization of a) 푆 is completely simple; simple semigroups that are completely simple b) 푆 is completely regular;

c) 푆 satisfies the conditions minL and minR; Proof of 4.16. Part 1 [a) ⇒ b)]. Suppose 푆 is completely simple. Then by Theorem 4.11, every element of 푆 lies in a subgroup of 푆. So 푆 is completely regular by Theorem 4.15. Part 2 [b) ⇒ c)]. Suppose 푆 is completely regular. Then every element of 푆 lies in a subgroup of 푆 by Theorem 4.15. So every element of 푆0 lies in 0 a subgroup and so 푆 is group-bound and therefore satisfies minL and minR by Theorem 4.9. 0 Part 3 [c) ⇒ a)]. Suppose 푆 satisfies minL and minR. Then so does 푆 and so 푆0 is completely 0-simple by Theorem 4.9. Hence 푆 is completely simple. 4.16

A semilattice of semigroups is a semigroup 푆 for which there exists a Semilattice of semigroups semilattice 푌 and a collection of disjoint subsemigroups 푆훼 of 푆, where 훼 ∈ 푌, such that 푆 = ⋃훼∈푌 푆훼 and 푆훼푆훽 ⊆ 푆훼⊓훽 (see Figure 4.4). A semilattice of completely simple semigroups is one in which every 푆훼 is completely simple; a semilattice of groups is one in which every 푆훼 is a group.

T h e o r e m 4 . 1 7. Every completely regular semigroup is a semilattice of Characterization completely simple semigroups. of completely regular semigroups Proof of 4.17. Let 푆 be a completely regular semigroup. By Theorem 4.15, each H-class of 푆 is a subgroup. So for any 푥 ∈ 푆, we have 푥2 H 푥 and hence 푥2 J 푥. Hence for any 푥, 푦 ∈ 푆, we have 퐽푥푦 = 퐽(푥푦)2 = 퐽푥(푦푥)푦 ⩽ 퐽푦푥. By symmetry, 퐽푦푥 ⩽ 퐽푥푦 and so 퐽푥푦 = 퐽푦푥. Let 푥 J 푦. Then there exist 푟, 푠 ∈ 푆1 with 푟푦푠 = 푥. Let 푧 ∈ 푆. Then

퐽푧푥 = 퐽푧푟푦푠 ⩽ 퐽푧푟푦 = 퐽푟푦푧 ⩽ 퐽푦푧 = 퐽푧푦.

By symmetry 퐽푧푦 ⩽ 퐽푧푥 and hence 퐽푧푥 = 퐽푧푦. So 푧푥 J 푧푦. Similarly 푥푧 J 푦푧. Therefore J is a congruence. The factor semigroup 푆/J is a

Completely regular semigroups • 82 푆훼 푆훽 푥 푦

FIGURE 4.4 푥푦 Multiplying in a semilattice of semigroups: the product of 푥 ∈ 푆 and 푦 ∈ 푆 lies in the sub- 푆훼⊓훽 훼 훽 semigroup 푆훼⊓훽. commutative semigroup of idempotents since 푥2 J 푥 and 푥푦 J 푦푥 for all 푥, 푦 ∈ 푆. Hence 푆/J is a semilattice by Theorem 1.21. Since J is a congruence, 퐽푥퐽푦 ⊆ 퐽푥푦. In particular, 퐽푥퐽푥 ⊆ 퐽푥2 = 퐽푥, so each 퐽푥 is a subsemigroup. The aim is to show 퐽푥푦퐽푥 = 퐽푥 for all 푦 ∈ 퐽푥, and so deduce that 1 퐽푥 is simple. Let 푧 ∈ 퐽푥. Since 푦 J 푧, there exist 푝, 푞, 푟, 푠 ∈ 푆 such that 푝푦푞 = 푧 and 푟푧푠 = 푦. [We cannot immediately deduce that 푧 ∈ 퐽푥푦퐽푥, because 푝 and 푞 may not lie in 퐽푥.] Write 1푦 for the identity of 퐻푦 and 1푧 for the identity of 퐻푧. Since 푦, 푧 ∈ 퐽푥, it follows that 1푦, 1푧 ∈ 퐽푥. Then (1푧푝)푦(푞1푧) = 1푧푧1푧 = 푧 and (1푦푟)푧(푠1푦) = 1푦푦1푦 = 푦. Furthermore, 퐽 ⩾ 퐽 = 퐽 = 퐽 and 퐽 ⩽ 퐽 = 퐽 . Hence 1 푝 ∈ 퐽 . 1푧푝 (1푧푝)푦(푞1푧) 푧 푥 1푧푝 1푧 푥 푧 푥 Similarly 푞1푧 ∈ 퐽푥. Hence 푧 = (1푧푝)푦(푞1푧) ∈ 퐽푥푦퐽푥. Since 푧 ∈ 퐽푥 was arbitrary, 퐽푥 ⊆ 퐽푥푦퐽푥. Clearly 퐽푥푦퐽푥 ⊆ 퐽푥 and so 퐽푥 = 퐽푥푦퐽푥. Since 푦 ∈ 퐽푥 was arbitrary, 퐽푥 is simple. Thus, since 퐽푥 is completely regular, it is completely simple by Theorem 4.16. To see 푆 is a semilattice of completely simple semigroups, let 푌 be the semilattice 푆/J and write 푆훼 instead of 훼 ∈ 푆/J. 4.17

Left and right groups This section discusses left and right groups, which are semigroups that are very close to being groups, and which have a very easy characterization, which we will deduce from our results on completely 0-simple semigroups. A semigroup is left simple if it contains no proper left ideals, and right Left/right simple semigroup simple if it contains no proper right ideals. P r o p o s i t i o n 4 . 1 8. Let 푆 be left or right simple. Then 푆 is simple. Proof of 4.18. Suppose 푆 is left simple; the reasoning for the right simple case is parallel. Let 푥 ∈ 푆. Then 푆푥 = 푆 since 푆 is left simple. So 푆2 = 푆푆 ⊇

Left and right groups • 83 푆푥 = 푆 and so 푆 = 푆2 = 푆푥. Hence 푆푥푆 = 푆3 = 푆2 = 푆 and so the only ideal of 푆 is 푆 itself. Thus 푆 is simple. 4.18

Note that Proposition 4.18 shows that being left simple (or right simple) is a stronger condition than being simple. This contrasts (for example) cancellativity: being left-cancellative is a weaker condition than being cancellative. A semigroup is a left group if is left simple and right cancellative, and Left/right group a right group if it is right simple and left cancellative.

T h e o r e m 4 . 1 9. Let 푆 be a semigroup. The following are equivalent: Characterization a) 푆 is a left group; of left groups b) 푆 is left simple and contains an idempotent; c) 푆 is completely simple semigroup and has only one L-class; d) 푆 ≃ 푍 × 퐺, where 푍 is a left zero semigroup and 퐺 is a group. There is a natural analogue of Theorem 4.19 for right groups. Note that taking 퐺 trivial in part d) shows that a left zero semigroup is a left group. Proof of 4.19. Part 1 [a) ⇒ b)] Suppose 푆 is a left group. By definition, 푆 is left simple. Let 푥 ∈ 푆. Since 푆 is left simple, 푆푥 = 푆. So there exists 푒 ∈ 푆 such that 푒푥 = 푥. Thus 푒2푥 = 푒푥. Since 푆 is right-cancellative, 푒2 = 푒. Part 2 [b) ⇒ c)] Suppose 푆 is left simple and 퐸(푆) ≠ ∅. Since 푆 is left simple, 푆1푥 = 푆 for all 푥 ∈ 푆, and so 푆 consists of a single L-class and thus a single D-class. Since 퐸(푆) ≠ ∅, some H-class in this L-class contains an idempotent, which is a regular element. By Proposition 3.19, all elements of 푆 are regular. By Proposition 3.20, every R-class of 푆 contains an idempotent. Since 푆 has only one L-class, this means that every H- class of 푆 contains an idempotent and so is a group by Proposition 3.14. Thus 푆 is completely regular by Theorem 4.15. Since 푆 is left simple and thus simple by Proposition 4.18, 푆 is completely simple by Theorem 4.16. Part 3 [c) ⇒ d)] Since 푆 is completely simple, 푆 = M[퐺; 퐼, 훬; 푃] for some group 퐺, index sets 퐼 and 훬, and matrix 푃. Since 푆 has only one L-class, 훬 = {1} by Proposition 4.12. Make 퐼 a left zero semigroup by defining 푖푗 = 푖 for all 푖, 푗 ∈ 퐼. Define a map

−1 휑 ∶ 퐼 × 퐺 → 푆; (푖, 푔)휑 = (푖, 푝1푖 푔, 1). Note that in this definition, the pair (푖, 푔) is in the direct product 퐼×퐺, and −1 the triple (푖, 푝1푖 푔, 1) is in the Rees matrix semigroup M[퐺; 퐼, 훬; 푃] = 푆. Then

((푖, 푔)휑)((푗, ℎ)휑) −1 −1 = (푖, 푝1푖 푔, 1)(푗, 푝1푗 ℎ, 1) [by definition of 휑]

Left and right groups • 84 −1 −1 = (푖, 푝1푖 푔푝1푗푝1푗 ℎ, 1) [by multiplication in M[퐺; 퐼, 훬; 푃]] −1 = (푖, 푝푖1 푔ℎ, 1) [by multiplication in 퐺] = (푖, 푔ℎ)휑 [by definition of 휑] = ((푖, 푔)(푗, ℎ))휑, [by multiplication in 퐼 × 퐺] so 휑 is a homomorphism. Furthermore,

−1 −1 (푖, 푔)휑 = (푗, ℎ)휑 ⇒ (푖, 푝1푖 푔, 1) = (푗, 푝1푗 ℎ, 1) [by definition of 휑] −1 −1 ⇒ 푖 = 푗 ∧ 푝1푖 푔 = 푝1푗 ℎ −1 −1 ⇒ 푖 = 푗 ∧ 푝1푖 푔 = 푝1푖 ℎ [using 푖 = 푗] ⇒ 푖 = 푗 ∧ 푔 = ℎ [by cancellation in 퐺] ⇒ (푖, 푔) = (푗, ℎ); thus 휑 is injective. Finally, for any (푖, 푔, 1) ∈ 푆 = M[퐺; 퐼, 훬; 푃], we have (푖, 푝1푖푔)휑 = −1 (푖, 푝1푖 푝1푖푔, 1) = (푖, 푔, 1), so 휑 is surjective. So 휑 is an isomorphism, and 푆 ≃ 퐼 × 퐺. Since 퐼 is a left zero semigroup and 퐺 is a group, this completes this part of the proof. Part 4 [d) ⇒ a)] Let (푥, 푔), (푦, ℎ), (푧, 푖) ∈ 푍 × 퐺. Then

(푥, 푔)(푦, ℎ) = (푥, 푔)(푧, 푖) ⇒ (푦, 푔ℎ) = (푧, 푔푖) [since 푍 is a right zero semigroup] ⇒ (푦 = 푧) ∧ (푔ℎ = 푔푖) ⇒ (푦 = 푧) ∧ (ℎ = 푖) [since 퐺 is a group] ⇒ (푦, ℎ) = (푧, 푖).

So 푍 × 퐺 is right-cancellative. Furthermore, (푥, 푔)(푦, 푔−1ℎ) = (푦, ℎ) and so (푔, 푥)(푍 × 퐺) = 푍 × 퐺 for all (푥, 푔) ∈ 푍 × 퐺. Hence 푍 × 퐺 is left simple, and so 푍 × 퐺 is a left group. 4.19

Homomorphisms We close this chapter with the following result, showing that homomorphisms preserve regularity and that, within regular semigroups, the preimage of an idempotent must contain an idempotent.

P r o p o s i t i o n 4 . 2 0. Let 푆 be a regular semigroup, 푇 a semigroup (not necessarily regular), and let 휑 ∶ 푆 → 푇 be a homomorphism. a) The subsemigroup im 휑 of 푇 is a regular semigroup and that if 푥′ ∈ 푆 is an inverse of 푥 ∈ 푆, then 푥′휑 is an inverse of 푥휑.

Homomorphisms • 85 b) If 푒 ∈ im 휑 is idempotent, 푓휑 = 푒, and 푧 ∈ 푆 is an inverse of 푓2, then 푓푧푓 is idempotent and (푓푧푓)휑 = 푒.

Proof of 4.20. a) Clearly im 휑 is a semigroup; we have to show it is inverse. Let 푦 ∈ im 휑. Then there exists 푥 ∈ 푆 with 푥휑 = 푦. Since 푆 is regular, there exists an inverse 푥′ for 푥. Let 푦′ = 푥′휑. Then 푦푦′푦 = (푥휑)(푥′휑)(푥휑) = (푥푥′푥)휑 = 푥휑 = 푦 and similarly 푦′푦푦′ = 푦. So 푦′ is an inverse for 푦. Since 푦 ∈ im 휑 was arbirary, im 휑 is regular. b) Let 푔 = 푓푧푓. Since 푧 is an inverse of 푓2, we have 푧푓2푧 = 푧 and 푓2푧푓2 = 푓2. Then 푔2 = 푓(푧푓2푧)푓 = 푓푧푓 = 푔 and so 푔 is idempotent. Furthermore

푔휑 = (푓푧푓)휑 [by choice of 푔] = (푓휑)(푧휑)(푓휑) [since 휑 is a homomorphism] = 푒(푧휑)푒 [since 푒 = 푓휑] = 푒2(푧휑)푒2 [since 푒 is idempotent] = (푓휑)2(푧휑)(푓휑)2 [since 푒 = 푓휑] = (푓2푧푓2)휑 [since 휑 is a homomorphism] = (푓2휑) [since 푓2푧푓2 = 푓2] = (푓휑)2 [since 휑 is a homomorphism] = 푒2 [since 푓휑 = 푒] = 푒. [since 푒 is idempotent]

This completes the proof. 4.20

Exercises [See pages 215–220 for the solutions.] 4.1 Let 퐺 be a group, let 퐼 = {1} and 훬 = {1} be index sets (each containing only one element), and 푃 a 훬 × 퐼 matrix over 퐺. a) By defining a suitable isomorphism, prove that M[퐺; 퐼, 훬; 푃] ≃ 퐺. b) Give an example to show that if we replace 퐺 by a monoid 푀 and construct M[푀; 퐼, 훬; 푃] using the same multiplication, we can have M[푀; 퐼, 훬; 푃] ≄ 푀. 4.2 Prove that every completely simple semigroup is equidivisible. 4.3 Let 푆 be a completely simple semigroup. Prove that a) L, R, and H are congruences on 푆; b) 푆/L is a right zero semigroup and 푆/R is a left zero semigroup; c) 푆/H is isomorphic to the rectangular band 푆/R × 푆/L. 4.4 Let 푆 be a completely simple semigroup.

Exercises • 86 a) Suppose |푆| = 푝, where 푝 is a prime. Prove that 푆 is [isomorphic to] either a right zero semigroup, a left zero semigroup, or a group. b) Suppose |푆| = 푝푞, where 푝 and 푞 are primes. Prove that 푆 is [isomorphic to] either a rectangular band, a right group, or a left group. ✴4.5 a) Let 푆 and 푇 be completely regular semigroups and 휑 ∶ 푆 → 푇 a homomorphism. Show that (푧휑)−1 = 푧−1휑 for all 푧 ∈ 푆. b) Give an example of regular semigroups 푆 and 푇 that have operations −1 satisfying (4.1), and a homomorphism 휑 ∶ 푆 → 푇 such that (푧휑)−1 ≠ 푧−1휑 for some 푧 ∈ 푆. ✴4.6 Let 퐺 and 퐻 be groups, 퐼, 퐽, 훬, and 훭 be index sets, 푃 be a 훬 × 퐼 regular matrix over 퐺0, and 푄 be a 퐽 × 훭 regular matrix over 퐻0.

a) Suppose 휑 ∶ M0[퐺; 퐼, 훬; 푃] → M0[퐻; 퐽, 훭; 푄] is an isomorphism. i) Prove that there exist bijections 훼 ∶ 퐼 → 퐽 and 훽 ∶ 훬 → 훭 such that (푖, 푎, 휆)휑 ∈ {푖훼} × 퐻 × {휆훽} and such that 푝휆푖 = 0 ⇔ 푞(휆훽)(푖훼) = 0. ii) Assume without loss that 1 ∈ 퐼 ∩ 훬. Define an isomorphism 훾 ∶ 퐺 → {1} × 퐺 × {1} ⊆ M0[퐺; 퐼, 훬; 푃] and an isomorphism 휂 ∶ 퐻 → {1훼} × 퐻 × {1훽} ⊆ M0[퐻; 퐽, 훭; 푄]. Deduce that 휗 = 훾휑휂−1 is an isomorphism from 퐺 to 퐻. −1 −1 iii) Check that (푖, 푥, 휆) = (푖, 1퐺, 1)(1, 푝11 푥, 1)(1, 푝11 , 휆). Now let 푢푖, 푣휆 ∈ 퐻 be such that

(푖, 1퐺, 1)휑 = (푖훼, 푢푖, 1훽) and

−1 −1 (1, 푝11 , 휆)휑 = (1훼, 푞(1훼)(1훽)푣휆, 휆훽).

Using the fact that (푖, 푝휆푖, 휆) = (푖, 1퐺, 휆)(푖, 1퐺, 휆), prove that

푝휆푖휗 = 푣휆푞(휆훽)(푖훼)푢푖 (4.5) for all 푖 ∈ 퐼 and 휆 ∈ 훬. b) Suppose that there exists an isomorphism 휗 ∶ 퐺 → 퐻, bijections 훼 ∶ 퐼 → 퐽 and 훽 ∶ 훬 → 훭 and elements 푢푖 and 푣휆 such that (4.5) holds for all 푖 ∈ 퐼 and 휆 ∈ 훬. Show that M0[퐺; 퐼, 훬; 푃] ≃ M0[퐻; 퐽, 훭; 푄]. ✴4.7 Let 퐺 be a group, 퐼 and 훬 index sets, and let 푃 be a 훬 × 퐼 matrix over 퐺0 that is not necessarily regular (that is, 푃 can have rows or columns where all the entries are 0). Let 푆 = M0[퐺; 퐼, 훬; 푃], where the multiplication is the same as in the usual Rees matrix semigroup. Prove that 푆 is regular (as a semigroup) if and only if 푃 is regular (as a matrix).

Exercises • 87 4.8 Let 푆 be a 0-simple semigroup.

a) Suppose 푆 satisfies minL. By applying this condition to the set of L- classes that are not equal to {0}, show that 푆 contains a 0-minimal left ideal. [Dually, if 푆 satisfies minR, it contains a 0-minimal right ideal.] b) Now suppose 푆 that 푆 contains a 0-minimal left ideal and a 0- minimal right ideal. i) Prove that if 퐾 is a 0-minimal left ideal of 푆 with 퐾2 ≠ {0}, then 퐾 = 푆푥 for any 푥 ∈ 퐾 ∖ {0}. ii) Let 퐿 be a 0-minimal left ideal of 푆, and suppose 푥 ∈ 푆 is such that 퐿푥 ≠ {0}. Prove that 퐿푥 is a 0-minimal left ideal of 푆. [Hint: to prove 퐿푥 is 0-minimal, consider the set 퐽 = { 푦 ∈ 퐿 ∶ 푦푥 ∈ 퐾 }.] iii) By considering the subset 퐿푆 of 푆, prove that there exists 푥 ∈ 푆 with 퐿푥 ≠ {0}. iv) Let 푀 = ⋃{ 퐿푥 ∶ 푥 ∈ 푆, 퐿푥 ≠ {0} }. Prove that 푀 = 푆 and deduce that 푆 is the union of its 0-minimal left ideals. [Dually, 푆 is the union of its 0-minimal right ideals.] v) Using part iii), the dual version of part iv), and Exercise 1.21, prove that there exists a 0-minimal right ideal 푅 such that 퐿푅 = 푆 and 푅퐿 is a group with a zero adjoined. vi) Let 푒 be the identity of the group 푅퐿 ∖ {0}. Prove that 푒 is a primitive idempotent. This proves that 푆 is completely 0-simple.

Since a completely 0-simple semigroup satisfies minL and minR by Theorem 4.9, this exercise has shown that a 0-simple semigroup is completely 0-simple if and only it has a 0-minimal left ideal and a 0-minimal right ideal. 4.9 Let 푆 be a left-cancellative semigroup. Let 퐺 be a subgroup of 푆. Sup- pose 퐺 is also a left ideal of 푆. Prove that 푆 is a right group.

Notes

The exposition here is based on Howie, Fundamentals of Semi- group Theory, ch. 3 and Clifford & Preston, The Algebraic Theory of Semigroups, § 2.5. ◆ The Rees–Suschkewitsch theorem (Theorem 4.7) was originally proved in Rees & Hall, ‘On semi-groups’; the analogue for completely simple semigroups (Theorem 4.11) is the earlier version, having been essentially proved in Suschkewitsch, ‘Über die endlichen Gruppen…’ ◆ The results on the structure of completely regular semigroups are due to Clifford, ‘Semigroups admitting relative inverses’. ◆ The analogy of Theorem 4.19 for right groups is due to Susch- kewitsch, ‘Über die endlichen Gruppen…’; for a more accessible proof (which

Notes • 88 does not use Green’s relations or the Rees–Suschkewitsch theorem), see Clifford & Preston, The Algebraic Theory of Semigroups, Theorem 1.27 ◆ Exercise 4.9 is from Cain, Robertson & Ruškuc, ‘Cancellative and Malcev presentations for finite Rees index subsemigroups and extensions’, Proposition 8.3. ◆ For further reading, Petrich, Completely Regular Semigroups seems to be the most recent monograph in the area. •

Notes • 89 Inverse semigroups 5 The sensibility of man to trifles, and his insensibility ‘ to great things, indicates a strange inversion. — Blaise’ Pascal, Pensées, § iii.198.

• Recall that an inverse semigroup is one equipped with Inverse semigroup an operation −1 satisfying the four conditions in (4.3): namely, that for all 푥, 푦 ∈ 푆,

(푥−1)−1 = 푥, (5.1) (푥푦)−1 = 푦−1푥−1, (5.2) 푥푥−1푥 = 푥, (5.3) 푥푥−1푦푦−1 = 푦푦−1푥푥−1. (5.4)

Clifford & Preston’s 1961 view that ‘[i]nverse semigroups constitute probably the most promising class of semigroups for study’ has proved accurate, and the field has grown into a vast and active one. We can only survey a minuscule part of it here.

Equivalent characterizations We being by giving alternative characterizations of inverse semigroups. Some texts define inverse semigroups using one of these alternative characterizations.

T h e o r e m 5 . 1. The following are equivalent: Characterizations of a) 푆 is an inverse semigroup; inverse semigroups b) every element of 푆 has a unique inverse; c) 푆 is regular and its idempotents commute; d) every L-class and every R-class of 푆 contains exactly one idempotent.

Proof of 5.1. The plan is as follows: parts 1–3 of this proof show that b), c), and d) are equivalent. Then parts 4 and 5 show, respectively, that a) implies c) and that b) implies a).

• 90 Part 1 [b) ⇒ c)]. Suppose every element of 푆 has a unique inverse. Then 푆 is clearly regular. Let 푒, 푓 ∈ 퐸(푆). Then

(푒푓)(푓(푒푓)−1푒)(푒푓) = 푒푓2(푒푓)−1푒2푓 [rearranging brackets] = 푒푓(푒푓)−1푒푓 [since 푒 and 푓 are idempotent] = 푒푓 [by definition of inverse] and

(푓(푒푓)−1푒)(푒푓)(푓(푒푓)−1푒) = 푓(푒푓)−1푒2푓2(푒푓)−1푒 [rearranging brackets] = 푓(푒푓)−1푒푓(푒푓)−1푒 [since 푒 and 푓 are idempotent] = 푓(푒푓)−1푒 [by definition of inverse] and so 푓(푒푓)−1푒 is an inverse of 푒푓. Since inverses are unique, (푒푓)−1 = 푓(푒푓)−1푒. Hence

((푒푓)−1)2 = 푓(푒푓)−1푒푓(푒푓)−1푒 [since (푒푓)−1 = 푓(푒푓)−1푒] = 푓(푒푓)−1푒 [by definition of inverse] = (푒푓)−1 [since (푒푓)−1 = 푓(푒푓)−1푒] and so (푒푓)−1 is idempotent. Thus (푒푓)−1(푒푓)−1(푒푓)−1 = (푒푓)−1 and so the uniqueness of inverses implies that 푒푓 = ((푒푓)−1)−1 = (푒푓)−1 and so 푒푓 is idempotent. A similar argument shows that 푓푒 is idempotent. Hence

(푒푓)(푓푒)(푒푓) = 푒푓2푒2푓 = 푒푓푒푓 = 푒푓 and

(푓푒)(푒푓)(푓푒) = 푓푒2푓2푒 = 푓푒푓푒 = 푓푒.

Hence 푓푒 = (푒푓)−1 = 푒푓. Thus idempotents of 푆 commute. Part 2 [c) ⇒ d)]. Suppose that 푆 is regular and that its idempotents commute. Since 푆 is regular, every L-class contains at least one idempotent by Proposition 3.20. So suppose a particular L-class contains idempotents 푒 and 푓. Then both 푒 and 푓 are right identities for this L-class by Proposi- tion 3.17. So 푒푓 = 푒 and 푓푒 = 푓. Since idempotents commute, 푒푓 = 푓푒 and so 푒 = 푓. So each L-class contains a unique idempotent. Similarly each R-class contains a unique idempotent. Part 3 [d) ⇒ b)]. Suppose every L-class and every R-class of 푆 contains a unique idempotent. Let 푥 ∈ 푆. By Proposition 3.21, the inverses of 푥 are in one-to-one correspondence with pairs of idempotents (푒, 푓) ∈ 푅푥 × 퐿푥.

Equivalent characterizations • 91 Since 푅푥 and 퐿푥 each contain a unique idempotent, 푥 therefore has a unique inverse. So every element of 푆 has a unique inverse. Part 4 [a) ⇒ c)]. Suppose 푆 is an inverse semigroup. Let 푥 ∈ 푆. Then 푥푥−1푥 = 푥 by (5.3) and so 푆 is regular. Let 푒 ∈ 퐸(푆). Then

푒−1 = 푒−1(푒−1)−1푒−1 [by (5.3)] = 푒−1푒푒−1 [by (5.1)] = 푒−1푒푒푒−1 [since 푒 is idempotent] = 푒−1(푒−1)−1푒푒−1 [by (5.1)] = 푒푒−1푒−1(푒−1)−1 [by (5.4)] = 푒푒−1푒−1푒 [by (5.1)] = 푒(푒푒)−1푒 [by (5.2)] = 푒푒−1푒 [since 푒 is idempotent] = 푒. [by (5.3)]

Hence 푒푒−1 = 푒2 = 푒 and 푒−1푒 = 푒2 = 푒 for any 푒 ∈ 퐸(푆). Now let 푒, 푓 ∈ 퐸(푆). Then 푒푓 = 푒푒−1푓푓−1 = 푓푓−1푒푒−1 = 푓푒 by (5.4). Thus idempotents of 푆 commute. Part 5 [b) ⇒ a)]. Suppose every element of 푆 has a unique inverse. Then for any 푥 ∈ 푆, we have 푥푥−1푥 = 푥; thus (5.3) holds. By the uniqueness of inverses, (푥−1)−1 = 푥; thus (5.1) holds. Let 푥, 푦 ∈ 푆. Then 푥푥−1 and 푦푦−1 are idempotents and so commute by parts 1 and 2 of this proof; thus (5.4) holds. Therefore

푥푦(푦−1푥−1)푥푦 = 푥(푦푦−1)(푥−1푥)푦 [rearranging brackets] = 푥푥−1푥푦푦−1푦 [by (5.4), which holds] = 푥푦 [by definition of inverse] and

(푦−1푥−1)푥푦(푦−1푥−1) = 푦−1(푥−1푥)(푦푦−1)푥−1 [rearranging brackets] = 푦−1푦푦−1푥−1푥푥−1 [by (5.4), which holds] = 푦−1푥−1. [by definition of inverse]

Hence, by the uniqueness of inverses, (푥푦)−1 = 푦−1푥−1; thus (5.2) holds. Thus 푆 is an inverse semigroup. 5.1 We now prove some consequences of these alternative characterizations.

P r o p o s i t i o n 5 . 2. Let 푆 be an inverse semigroup. Then 퐸(푆) is a 퐸(푆) is a semilattice subsemigroup of 푆 and forms a semilattice. when 푆 is inverse

Equivalent characterizations • 92 Proof of 5.2. By Theorem 5.1, all elements of 퐸(푆) commute. Hence, if 푒, 푓 ∈ 퐸(푆), then (푒푓)2 = 푒푓푒푓 = 푒2푓2 = 푒푓 and so 푒푓 ∈ 퐸(푆). So 퐸(푆) is a subsemigroup of 푆. Furthermore, 퐸(푆) is a commutative semigroup of idempotents and hence a semilattice by Theorem 1.21. 5.2

P r o p o s i t i o n 5 . 3. Let 푆 be an inverse semigroup. Then 푆 is a group if Characterization of inverse and only if 푆 contains exactly one idempotent. semigroups that are groups Proof of 5.3. In one direction, this result is obvious: if 푆 is a group, then it is an inverse semigroup and 1푆 is the unique idempotent in 푆. So suppose 푆 is an inverse semigroup and 푒 is the unique idempotent in 푆. Let 푥 ∈ 푆. Then 푥푥−1 and 푥−1푥 are idempotents and so 푒 = 푥푥−1 = 푥−1푥. Thus 푒푥 = 푥푥−1푥 = 푥 and 푥푒 = 푥푥−1푥 = 푥. So 푒 is an identity for 푆. Furthemore, since 푒 = 푥푥−1 = 푥−1푥 for all 푥 ∈ 푆, every element of 푥 is right- and left-invertible and so 푆 is a group. 5.3

Let 푆 and 푇 be inverse semigroups. A homomorphism 휑 ∶ 푆 → 푇 is Inverse semigroup an inverse semigroup homomorphism if 푥−1휑 = (푥휑)−1 for all 푥 ∈ 푆. homomorphism In Chapter 1, we saw the distinction between a [semigroup] homomorphism and a monoid homomorphism: a homomorphism between two monoids may preserve multiplication, but not preserve the identity (see Exercise 1.15). Thus it is conceivable that there exists a homomorphism between two inverse semigroups that is not an inverse semigroup homomorphism. However, the following result shows that the two notions coincide:

P r o p o s i t i o n 5 . 4. Let 푆 be an inverse semigroup and let 푇 a semi- Homomorphism from group (not necessarily inverse), and let 휑 ∶ 푆 → 푇 be a homomorphism. an inverse semigroup is Then im 휑 is an inverse semigroup, and 휑 is an inverse semigroup homo- an inverse semigroup homomorphism morphism. Proof of 5.4. By Proposition 4.20(a), im 휑 is regular and 푥−1휑 is an inverse of 푥휑 for any 푥 ∈ 푆. Let 푒, 푓 ∈ im 휑 be idempotents. By Proposition 4.20(b), there are idempotents 푔, ℎ ∈ 푆 with 푔휑 = 푒 and ℎ휑 = 푓. Since 푆 is an inverse semigroup, 푔ℎ = ℎ푔 by Theorem 5.1. Thus 푒푓 = (푔휑)(ℎ휑) = (푔ℎ)휑 = (ℎ푔)휑 = (ℎ휑)(푔휑) = 푓푒. Hence idempotents commute in im 휑 and so im 휑 is an inverse semigroup by Theorem 5.1. Since inverses are unique in inverse semigroups, it follows that (푥휑)−1 = 푥−1휑 for all 푥 ∈ 푆, so 휑 is an inverse semigroup homomorphism. 5.4

C o r o l l a ry 5 . 5. Let 퐺 be a group and let 푇 a semigroup (not necessar- Homomorphism from ily a group or inverse), and let 휑 ∶ 퐺 → 푇 be a homomorphism. Then im 휑 a group preserves is a group, and 휑 ∶ 퐺 → im 휑 is an inverse semigroup homomorphism and identity and inverses −1 −1 a monoid homomorphism. (That is, 푥 휑 = (푥휑) for all 푥 ∈ 퐺 and 1퐺휑 is an identity for im 휑.) Proof of 5.5. Proposition 5.4 shows that im 휑 is an inverse semigroup and 휑 is an inverse semigroup homomorphism. Let 푦 ∈ im 휑 and let 푥 ∈ 퐺 be

Equivalent characterizations • 93 −1 −1 such that 푥휑 = 푦. Then (1퐺휑)푦 = ((푥푥 )휑)(푥휑) = (푥푥 푥)휑 = 푥휑 = 푦, and similarly 푦(1퐺휑) = 푦. Hence 1퐺휑 is an identity for im 휑. 5.5 The last consequence we prove is more technical, but we will make use of it in the next section.

L e m m a 5 . 6. Let 푆 be an inverse semigroup. a) For any 푒, 푓 ∈ 퐸(푆), we have 푆푒 = 푆푓 ⇒ 푒 = 푓. b) For any 푒, 푓 ∈ 퐸(푆), we have 푆푒 ∩ 푆푓 = 푆푒푓. c) For any 푥 ∈ 푆, we have 푆푥 = 푆푥−1푥. d) For 푥 ∈ 푆 and 푒 ∈ 퐸(푆), the element 푓 = 푥−1푒푥 is idempotent and 푒푥 = 푥푓.

Proof of 5.6. a) Since 푒 = 푒푒 ∈ 푆푒 = 푆푓, we deduce that 푒 = 푥푓 for some 푥 ∈ 푆. Then 푒푓 = 푥푓2 = 푥푓 = 푒. Similarly 푓푒 = 푓. Since idempotents commute by Theorem 5.1, 푒 = 푓. b) Obviously 푆푒푓 ⊆ 푆푓 and, since idempotents commute, 푆푒푓 = 푆푓푒 ⊆ 푆푒. So 푆푒푓 ⊆ 푆푒 ∩ 푆푓. Let 푥 ∈ 푆푒 ∩ 푆푓. Then 푥 = 푦푒 and 푥 = 푧푓 for some 푦, 푧 ∈ 푆. Then 푥 = 푧푓 = 푧푓2 = 푥푓 = 푦푒푓 ∈ 푆푒푓. So 푆푒 ∩ 푆푓 ⊆ 푆푒푓 and hence 푆푒 ∩ 푆푓 = 푆푒푓. c) Obviously 푆푥−1푥 ⊆ 푆푥. But 푆푥 = 푆푥푥−1푥 ⊆ 푆푥−1푥 and so 푆푥 = 푆푥−1푥. d) Since 푥푥−1 is an idempotent, and idempotents commute in 푆, 푓2 = 푥−1푒푥푥−1푒푥 = 푥−1푥푥−1푒푒푥 = 푥−1푒푥 = 푓, so 푓 is idempotent. Fur- −1 −1 thermore, 푒푥 = 푒푥푥 푥 = 푥푥 푒푥 = 푥푓. 5.6

Vagner–Preston theorem Theorem 1.22 showed that every semigroup embeds into T푋 for some 푋. Cayley’s theorem shows that every group embeds into S푋 for some 푋. The Vagner–Preston theorem, to which this section is devoted, is an analogue of these results for inverse semigroups. Let 휏 ∈ P푋. Recall from (1.3) that the domain of 휏, denoted dom 휏, is Partial bijection the subset of 푋 on which 휏 is defined. If 휏 ∶ dom 휏 → im 휏 is a bijection, then 휏 is a partial bijection. The set of partial bijections on 푋 is denoted

I푋. (The symbol I stands for ‘injection’.) Notice that if 휏, 휎 ∈ I푋, then I푋

푥 ∈ dom(휏휎) ⇔ (∃푦 ∈ 푋)((푥, 푦) ∈ 휏휎) ⇔ (∃푦 ∈ 푋)(∃푧 ∈ 푋)((푥, 푧) ∈ 휏 ∧ (푧, 푦) ∈ 휎) ⇔ (∃푧 ∈ 푋)((푥, 푧) ∈ 휏 ∧ 푧 ∈ dom 휎) ⇔ (푥 ∈ dom 휏) ∧ (푥휏 ∈ dom 휎) ⇔ (푥휏 ∈ im 휏) ∧ (푥휏 ∈ dom 휎)

Vagner–Preston theorem • 94 dom 휏 im 휏

푧 푋 푥 휏 휎 푦 FIGURE 5.1 The domain of the composition dom 휎 im 휎 of two partial bijections 휏 and 휎; the shaded area is dom(휏휎).

⇔ (푥휏 ∈ im 휏 ∩ dom 휎) ⇔ 푥 ∈ (im 휏 ∩ dom 휎)휏−1.

That is,

dom(휏휎) = (im 휏 ∩ dom 휎)휏−1. (5.5)

(See Figure 5.1.) For 푥, 푦 ∈ dom 휏휎, we have 푥, 푦 ∈ dom 휏 and 푥휏, 푦휏 ∈ dom 휎 and so

푥휏휎 = 푦휏휎 ⇒ 푥휏 = 푦휏 [since 휎 is injective] ⇒ 푥 = 푦. [since 휏 is injective]

Hence 휏휎 is a bijection from dom(휏휎) to im(휏휎) and so 휏휎 ∈ I푋. Thus I푋 is a subsemigroup of P푋. −1 For 휏 ∈ I푋, let 휏 be the partial bijection with domain im 휏 and Inverse of a partial bijection image dom 휏 defined by (푥휏)휏−1 = 푥. (That is, 휏 is defined by inverting the bijection 휏 ∶ dom 휏 → im 휏.) Note that

−1 −1 휏휏 = iddom 휏 and 휏 휏 = idim 휏. (5.6)

P r o p o s i t i o n 5 . 7. For any set 푋, the semigroup of partial bijections I푋 is an inverse semigroup I푋 is an inverse semigroup. −1 −1 Proof of 5.7. Let 휏 ∈ I푋. Since 휏휏 = iddom 휏 and 휏 휏 = idim 휏 by (5.6), we have 휏휏−1휏 = 휏 and 휏−1휏휏−1 = 휏−1. Hence 휏−1 is an inverse of 휏. Thus I푋 is regular. Suppose 휎 ∈ I푋 is an inverse of 휏. Then 휏휎휏 = 휏 and 휎휏휎 = 휎. Suppose, with the aim of obtaining a contradiction, that im 휏 ⊈ dom(휎휏). So there exists 푡 ∈ im 휏 ∖ dom(휎휏); thus 푡 ∈ im 휏 but 푡 ∉ im 휏 ∩ dom(휎휏). Hence im 휏 ∩ dom(휎휏) ⊊ im 휏. Therefore

dom 휏 = dom(휏휎휏) = (im 휏 ∩ dom(휎휏))휏−1 [by (5.5)] ⊊ (im 휏)휏−1 = dom 휏.

Vagner–Preston theorem • 95 The strict inclusion is a contradiction; hence im 휏 ⊆ dom(휎휏) ⊆ dom 휎. Similarly, from 휎휏휎 = 휎 we obtain dom 휎 ⊆ im 휏. Therefore dom 휎 = im 휏 = dom 휏−1. For any 푥 ∈ dom 휎, we have 푥 ∈ im 휏 and so 푥 = 푦휏 for some 푦 ∈ 푋. Hence 푥휎 = 푦휏휎 = 푦휏휎휏휏−1 = 푦휏휏−1 = 푥휏−1. Hence 휎 = 휏−1. So 휏−1 is the unique inverse of 휏. Since each element of I푋 has a unique inverse, I푋 is an inverse semigroup by Theorem 5.1. 5.7

Let 푆 be an inverse semigroup and let 푇 be a subsemigroup of 푆. Then Inverse subsemigroup 푇 is an inverse subsemigroup of 푆 if it is also an inverse semigroup, or, equivalently, if it is closed under taking inverses in 푆.

Vagner–Preston Theorem 5.8. For any inverse semigroup 푆, Vagner–Preston theorem there exists a set 푋 and a monomorphism 휑 ∶ 푆 → I푋. Hence every inverse semigroup is isomorphic to some inverse subsemigroup of I푋.

Proof of 5.8. Let 푋 = 푆. For each 푥 ∈ 푆, let 휏푥 be the partial transformation −1 with domain 푆푥 and defined by 푦휏푥 = 푦푥. Thus 휏푥 is simply 휌푥 (as −1 defined on page 19) restricted to 푆푥 . Note that im 휏푥 = (dom 휏푥)휏푥 = 푆푥−1푥 = 푆푥, by Lemma 5.6(c). −1 −1 Let us prove that 휏푥 ∈ I푋. Let 푦, 푧 ∈ 푆푥 , with 푦 = 푝푥 and 푧 = 푞푥−1. Then

푦휏푥 = 푧휏푥 ⇒ 푦푥 = 푧푥 ⇒ 푝푥−1푥 = 푞푥−1푥 ⇒ 푝푥−1푥푥−1 = 푞푥−1푥푥−1 ⇒ 푝푥−1 = 푞푥−1 ⇒ 푦 = 푧.

So 휏푥 is a partial bijection and so 휏푥 ∈ I푋. −1 −1 Let us now prove that (휏푥) = 휏푥−1 . If 푧 ∈ dom 휏푥 = 푆푥 , then −1 푧휏푥휏푥−1 휏푥 = 푧푥푥 푥 = 푧푥 = 푧휏푥. If 푧 ∈ dom 휏푥−1 = 푆푥, then 푧휏푥−1 휏푥휏푥−1 = −1 −1 −1 푧푥 푥푥 = 푧푥 = 푧휏푥−1 . Furthermore, dom 휏푥−1 = 푆푥 = im 휏푥 and −1 −1 im 휏푥−1 = 푆푥 = dom 휏푥. Hence (휏푥) = 휏푥−1 . Define 휑 ∶ 푆 → I푋 by 푥휑 = 휏푥. We first prove that 휑 is injective. Let 푥, 푦 ∈ 푆. Then

푥휑 = 푦휑 ⇒ 휏푥 = 휏푦 [by definition of 휑]

⇒ dom 휏푥 = dom 휏푦 −1 −1 ⇒ 푆푥 = 푆푦 [by definition of 휏푥 and 휏푦] ⇒ 푆푥푥−1 = 푆푦푦−1 [by Lemma 5.6(c)] ⇒ 푥푥−1 = 푦푦−1 [by Lemma 5.6(a)] −1 −1 ⇒ 푥푥 휏푥 = 푦푦 휏푦 [since 휏푥 = 휏푦] −1 −1 ⇒ 푥푥 푥 = 푦푦 푦 [by definition of 휏푥 and 휏푦] ⇒ 푥 = 푦;

Vagner–Preston theorem • 96 thus 휑 is injective. Let 푥, 푦 ∈ 푆. Then

−1 dom(휏푥휏푦) = (im 휏푥 ∩ dom 휏푦)휏푥 [by (5.5)] −1 −1 −1 = (푆푥 푥 ∩ 푆푦 )휏푥 [by definition of 휏푥 and 휏푦] −1 −1 −1 = (푆푥 푥 ∩ 푆푦푦 )휏푥 [by Lemma 5.6(c)] −1 −1 −1 = (푆푥 푥푦푦 )휏푥 [by Lemma 5.6(b)] −1 −1 −1 = (푆푥 푥푦푦 )휏푥−1 [since 휏푥 = 휏푥−1 ] −1 −1 −1 = 푆푥 푥푦푦 푥 [by definition of 휏푥−1 ] = 푆푥푥−1푥푦푦−1푥−1 [by Lemma 5.6(c)] = 푆푥푦푦−1푥−1푥푥−1 [since idempotents commute] = 푆푥푦푦−1푥−1 = 푆(푥푦)(푥푦)−1 = 푆(푥푦)−1 [by Lemma 5.6(c)]

= dom 휏푥푦, [by definition of 휏푥푦] and for all 푧 ∈ dom 휏푥푦, we have 푧휏푥휏푦 = 푧푥푦 = 푧휏푥푦. Hence (푥휑)(푦휑) = 휏푥휏푦 = 휏푥푦 = (푥푦휑). Thus 휑 is a monomorphism. 5.8

Notice that the image of 푆 in I푋 is an inverse subsemigroup of I푋 by Proposition 5.4. However, some subsemigroups of I푋 are not inverse; see Exercise 5.1.

The natural partial order

Elements of I푋 are maps, and thus relations, and thus simply subsets of 푋×푋. So we can apply the partial order ⊆ to I푋. However, ⊆ can be characterized using the algebraic structure of I푋, since

휎 ⊆ 휏 ⇔ 휎 = 휏|dom 휎

⇔ 휎 = iddom 휎휏 ⇔ 휎 = 휎휎−1휏.

Since every inverse monoid embeds into I푋 for some 푋 by Theorem 5.8, we can transfer this algebraic definition to arbitrary inverse semigroups by defining 푥 ≼ 푦 ⇔ 푥 = 푥푥−1푦.

L e m m a 5 . 9. For 푥, 푦 ∈ 푆, the following are equivalent: Characterizing the relation ≼ a) 푥 ≼ 푦; b) 푥 = 푒푦 for some 푒 ∈ 퐸(푆); c) 푥 = 푦푓 for some 푓 ∈ 퐸(푆);

The natural partial order • 97 d) 푥 = 푦푥−1푥. Proof of 5.9. Part 1 [a) ⇒ b)]. Suppose 푥 ≼ 푦. Then 푥 = 푥푥−1푦, and 푒 = 푥푥−1 is an idempotent. Part 2 [b) ⇒ c)]. Suppose 푥 = 푒푦. Let 푓 = 푦−1푒푦. Then 푓2 = 푦−1푒푦푦−1푒푦 = 푦−1푦푦−1푒2푦 = 푦−1푒푦 = 푓; thus 푓 is idempotent. Furthermore, 푦푓 = 푦푦−1푒푦 = 푒푦푦−1푦 = 푒푦 = 푥. Part 3 [c) ⇒ d)]. Suppose 푥 = 푦푓. Then 푥푓 = 푦푓2 = 푦푓 = 푥 and so 푦푥−1푥 = 푦푥−1푥푓 = 푦푓푥−1푥 = 푥푥−1푥 = 푥. Part 4 [d) ⇒ a)]. Suppose 푥 = 푦푥−1푥. Then 푥 = 푦푦−1푦푥−1푥 = 푦푥−1푥푦−1푦. Let 푒 = 푦푥−1푥푦−1, so that 푥 = 푒푦. Then 푒2 = 푦푥−1푥푦−1푦푥−1푥푦−1 = 푦푥−1푥푥−1푥푦−1푦푦−1 = 푦푥−1푥푦−1 = 푒, so 푒 is idempotent. Hence 푒푥 = 푒2푦 = 푒푦 = 푥, and so 푒푥푥−1 = 푥푥−1. Thus 푥푥−1푦 = 푒푥푥−1푦 = 푥푥−1푒푦 = 푥푥−1푥 = 푥, and so 푥 ≼ 푦 by definition. 5.9 P r o p o s i t i o n 5 . 1 0. The relation ≼ is a partial order. ≼ is a partial order Proof of 5.10. Since 푥 = 푥푥−1푥, we have 푥 ≼ 푥 for any 푥 ∈ 푆; thus 푥 is reflexive. If 푥 ≼ 푦 and 푦 ≼ 푥, then by 푥 = 푥푥−1푦 and 푦 = 푦푦−1푥. Hence 푥 = 푥푥−1푦푦−1푥 = 푦푦−1푥푥−1푥 = 푦푦−1푥 = 푦; thus ≼ is anti-symmetric. If 푥 ≼ 푦 and 푦 ≼ 푧, then 푥 = 푒푦 and 푦 = 푓푧 for some 푒, 푓 ∈ 퐸(푆), and so 푥 = (푒푓)푧, and hence 푥 ≼ 푧 (since 푒푓 is in the subsemigroup 퐸(푆)); thus ≼ is transitive. 5.10

Proposition 5.10 justifies the choice of the symbol ≼ for this relation, Natural partial order which is called the natural partial order on an inverse semigroup. Notice that if 푥 and 푦 are idempotents, then by the commutativity of idempotents this agrees with the definition of the natural partial order for idempotents (see Proposition 1.19). We are therefore justified in using the same symbol ≼ for both relations. P r o p o s i t i o n 5 . 1 1. a) The relation ≼ is compatible (with multiplication); that is, 푥 ≼ 푦 ∧ 푧 ≼ 푡 ⇒ 푥푧 ≼ 푦푡 for all 푥, 푦, 푧, 푡 ∈ 푆. b) The relation ≼ is compatible with inversion; that is, 푥 ≼ 푦 ⇒ 푥−1 ≼ 푦−1 for all 푥, 푦 ∈ 푆. Proof of 5.11. a) Let 푥, 푦, 푧, 푡 ∈ 푆. Then (푥 ≼ 푦) ∧ (푧 ≼ 푡) ⇒ (∃푒, 푓 ∈ 퐸(푆))((푥 = 푒푦) ∧ (푧 = 푡푓)) [by Lemma 5.9] ⇒ (∃푒, 푓 ∈ 퐸(푆))((푥푧 = 푒푦푧) ∧ (푦푧 = 푦푡푓)) ⇒ (푥푧 ≼ 푦푧) ∧ (푦푧 ≼ 푦푡) [by Lemma 5.9] ⇒ 푥푧 ≼ 푦푡. [since ≼ is transitive]

The natural partial order • 98 b) Let 푥, 푦 ∈ 푆. Then

푥 ≼ 푦 ⇒ (∃푒 ∈ 퐸(푆))(푥 = 푒푦) [by Lemma 5.9] ⇒ (∃푒 ∈ 퐸(푆))(푥−1 = 푦−1푒) [by (5.2) and 푒−1 = 푒] ⇒ (∃푒 ∈ 퐸(푆))(푥−1 = 푦−1푦푦−1푒) ⇒ (∃푒 ∈ 퐸(푆))(푥−1 = 푦−1푒푦푦−1) ⇒ (∃푓 ∈ 퐸(푆))(푥−1 = 푓푦−1) [since 푦−1푒푦 ∈ 퐸(푆) by Lemma 5.6(d)] −1 −1 ⇒ 푥 ≼ 푦 . [by Lemma 5.9] 5.11

The natural partial order can serve as a measure of how ‘close’ an inverse semigroup is to being a group:

P r o p o s i t i o n 5 . 1 2. Let 푆 be an inverse semigroup. Then 푆 is a group Characterizing inverse if and only if ≼ is the identity relation on 푆. semigroups that are groups using ≼ Proof of 5.12. Suppose 푆 is a group. Then

−1 푥 ≼ 푦 ⇔ 푥 = 푥푥 푦 ⇔ 푥 = 1푆푦 ⇔ 푥 = 푦; thus ≼ is the identity relation. Now suppose that ≼ is the identity relation. Let 푒, 푓 ∈ 퐸(푆). Then 푒푓 ≼ 푒 and 푒푓 ≼ 푓; hence 푒 = 푒푓 = 푓. Thus 푆 contains a unique idempotent and so 푆 is a group by Proposition 5.3. 5.12

Clifford semigroups

Recall that a semigroup 푆 is a Clifford semigroup if it Clifford semigroup satisfies the conditions in (4.4). Thus 푆 is a Clifford semigroup if it is completely regular and, for all 푥, 푦 ∈ 푆,

푥푥−1푦푦−1 = 푦푦−1푥푥−1. (5.7)

We are going to prove a structure theorem for Clifford semigroups, but first we need to a stronger version of the notion of a semilattice of semigroups, which we introduced in the previous chapter. If we know that 푆 is a semilattice of semigroups 푆훼, we know something of the coarse structure of 푆: we know that if 푥 ∈ 푆훼 and 푦 ∈ 푆훽, then 푥푦 ∈ 푆훼⊓훽 (see Figure 4.4). The new version is stronger in that it describes precisely what products are, rather than simply where they are in the semilattice. Suppose that we

Clifford semigroups • 99 have a semilattice 푌, disjoint semigroups 푆훼 for each 훼 ∈ 푌, and, for all 훼 ⩾ 훽, homomorphisms 휑훼,훽 ∶ 푆훼 → 푆훽 satisfying the conditions

(∀훼 ∈ 푌)(휑훼,훼 = id훼) (5.8) 푆훼 푆훽 (∀훼, 훽, 훾 ∈ 푌)((훼 ⩾ 훽 ⩾ 훾) ⇒ (휑훼,훽휑훽,훾 = 휑훼,훾)) (5.9) 푥 푦

Then we can define a multiplication on 푆 = ⋃훼∈푌 푆훼 as follows: for each 휑훼,훼⊓훽 휑훽,훼⊓훽 푥 ∈ 푆훼 and 푦 ∈ 푆훽, the product 푥푦 is defined to be (푥휑훼,훼⊓훽)(푦휑훽,훼⊓훽). That is, we use the homomorphisms to map 푥 and 푦 ‘down’ into 푆훼⊓훽 and multiply them there; see Figure 5.2. For any 푥 ∈ 푆훼, 푦 ∈ 푆훽, 푧 ∈ 푆훾, 푥휑훼,훼⊓훽 푦휑훽,훼⊓훽 푥푦 푥(푦푧)

= 푥((푦휑훽,훽⊓훾)(푧휑훾,훽⊓훾)) [by definition of multiplication] 푆훼⊓훽

= (푥휑훼,훼⊓훽⊓훾)((푦휑훽,훽⊓훾)(푧휑훾,훽⊓훾))휑훽⊓훾,훼⊓훽⊓훾 FIGURE 5.2 [by definition of multiplication] Multiplying in a strong semilattice of semigroups = (푥휑훼,훼⊓훽⊓훾)(푦휑훽,훽⊓훾휑훽⊓훾,훼⊓훽⊓훾)(푧휑훾,훽⊓훾휑훽⊓훾,훼⊓훽⊓훾)

[since 휑훽⊓훾,훼⊓훽⊓훾 is a homomorphism]

= (푥휑훼,훼⊓훽⊓훾)((푦휑훽,훼⊓훽⊓훾)(푧휑훾,훼⊓훽⊓훾)) [by (5.9)]

= ((푥휑훼,훼⊓훽⊓훾)(푦휑훽,훼⊓훽⊓훾))(푧휑훾,훼⊓훽⊓훾) [by associativity in 푆훼⊓훽⊓훾] = (푥푦)푧, [by similar reasoning] and so this multiplication is associative. This semigroup 푆 is a strong Strong semilattice of semigroups/groups semilattice of semigroups and is denoted S[푌; 푆훼; 휑훼,훽]. If every 푆훼 is a group, it is a strong semilattice of groups. An element 푥 of a semigroup 푆 is central if 푥푦 = 푦푥 for all 푦 ∈ 푆. Central element

T h e o r e m 5 . 1 3. The following are equivalent: Characterization of a) 푆 is a Clifford semigroup; Clifford semigroups b) 푆 is a semilattice of groups; c) 푆 is a strong semilattice of groups; d) 푆 is regular, and the idempotents of 푆 are central; e) 푆 is regular, and every D-class of 푆 contains a unique idempotent.

Proof of 5.13. Part 1 [a) ⇒ b)]. Let 푆 be a Clifford semigroup. Then 푆 is completely regular and so is a semilattice of completely simple semigroups −1 −1 푆훼 by Theorem 4.17. Let 푒, 푓 be idempotents. Then 푒 = 푒푒 푒 = 푒푒푒 = 푒푒−1 by (4.2) and similarly 푓 = 푓푓−1 and so 푒푓 = 푓푒 by (5.7). So all idempotents of 푆 commute. Now, 푆훼 is completely simple and so 푆훼 ≃ M[퐺; 퐼, 훬; 푃] for some group 퐺, index sets 퐼 and 훬, and matrix 푃 over −1 −1 퐺. Let 푒, 푓 ∈ 푆훼 be idempotents. Then 푒 = (푖, 푝휆푖 , 휆) and 푓 = (푗, 푝휇푗 , 휇), −1 −1 −1 −1 and (푖, 푝휆푖 푝휆푗푝휇푗 , 휇) = 푒푓 = 푓푒 = (푗, 푝휇푗 푝휇푖푝휆푖 , 휆). Hence 푖 = 푗 and 휆 = 휇 and so 푒 = 푓. So each 푆훼 contains only one idempotent. Thus, by Proposition 4.14, 푆훼 is a group. Therefore 푆 is a semilattice of groups.

Clifford semigroups • 100 Part 2 [b) ⇒ c)]. Let 푆 be a semilattice of groups 푆훼, where 훼 ∈ 푌. To prove that 푆 is a strong semilattice of groups, we have to define homomorphisms 휑훼,훽 for all 훼, 훽 ∈ 푌, prove that (5.8) and (5.9) hold, and show that the strong semilattice of groups S[푌; 푆훼; 휑훼,훽] is isomorphic to 푆. Write 1훼 for the identity of the group 푆훼. Then for 훼 ⩾ 훽 and 푥 ∈ 푆훼, we have 1훽푥 ∈ 푆훽. Hence we can define a map 휑훼,훽 ∶ 푆훼 → 푆훽 by 푥휑훼,훽 = 1훽푥. For 푥, 푦 ∈ 푆훼,

(푥휑훼,훽)(푦휑훼,훽)

= 1훽푥1훽푦 [by definition of 휑훼,훽]

= 1훽푥푦 [since 1훽푥 ∈ 푆훽 and thus (1훽푥)1훽 = 1훽푥]

= (푥푦)휑훼,훽; [by definition of 휑훼,훽] hence 휑 is a homomorphism. Clearly 휑 = id , so (5.8) holds. For 훼,훽 훼,훼 푆훼 훼 ⩾ 훽 ⩾ 훾, for any 푥 ∈ 푆훼

푥휑훼,훽휑훽,훾

= (1훽푥)휑훽,훾 [by definition of 휑훼,훽]

= 1훾1훽푥 [by definition of 휑훽,훾]

= (1훽휑훽,훾)푥 [by definition of 휑훽,훾]

= 1훾푥 [by Corollary 5.5]

= 푥휑훼,훾; [by definition of 휑훼,훾] hence (5.9) holds. Finally, for any 푥 ∈ 푆훼 and 푦 ∈ 푆훽,

푥푦 = 1훼⊓훽푥푦 [since 푥푦 ∈ 푆훼⊓훽]

= 1훼⊓훽푥1훼⊓훽푦 [since 1훼⊓훽푥 ∈ 푆훼⊓훽]

= (푥휑훼,훼⊓훽)(푦휑훽,훼⊓훽). [by definition of 휑훼,훼⊓훽 and 휑훼,훼⊓훽]

Therefore 푆 is isomorphic to S[푌; 푆훼; 휑훼,훽].

Part 3 [c) ⇒ d)]. A strong semilattice of groups 푆 = S[푌; 푆훼; 휑훼,훽] is −1 certainly regular: for each 푥 ∈ 푆훼, let 푥 be the inverse of 푥 in the group 푆훼. The idempotents of 푆 are the identities of the groups 푆훼. Write 1훼 for the identity of 푆훼. Then for any 훽 ∈ 푌 and 푥 ∈ 푆훽,

1훼푥 = (1훼휑훼,훼⊓훽)(푥휑훽,훼⊓훽) = 1훼⊓훽(푥휑훽,훼⊓훽)

= (푥휑훽,훼⊓훽) = (푥휑훽,훼⊓훽)1훼⊓훽 = (푥휑훽,훼⊓훽)(1훼휑훼,훼⊓훽) = 푥1훼.

Thus every idempotent of 푆 is central.

Part 4 [d) ⇒ e)]. Each D-class 퐷푥 must contain at least one idempotent, namely 푥푥−1. Suppose 푒 and 푓 are idempotent and 푒 D 푓. Then by

Clifford semigroups • 101 Proposition 3.21(b) there exists an element 푥 and inverse 푥′ such that 푥푥′ = 푒 and 푥′푥 = 푓. Therefore 푒 = 푒2 = 푥푥′푥푥′ [since 푥푥′ = 푒] = 푥푓푥′ [since 푥′푥 = 푓] = 푥푥′푓 [since 푓 is central] = 푥푥′푥′푥 [since 푓 = 푥′푥] = 푒푥′푥 [since 푥푥′ = 푒] = 푥′푒푥 [since 푒 is central] = 푥′푥푥′푥 [since 푒 = 푥푥′] = 푓2 = 푓. [since 푓 = 푥′푥] Hence every D-class of 푆 contains a unique idempotent. Part 5 [e) ⇒ a)]. Since every D-class contains a unique idempotent, every D-class consists of a single H-class by Proposition 3.20, and so D = H. Furthermore, each of these H-classes is a group by Proposition 3.14, and so every element of 푆 lies in a subgroup and thus 푆 is completely regular by Theorem 4.15. Thus, by Theorem 4.17, 푆 is a semilattice of completely simple semigroups 푆훼. Every element of a completely simple semigroup is D-related, and so every 푆훼 is contained within a single D-class and is thus a group. So 푆 is a semilattice of groups and thus, by the second part of this proof, a strong semilattice of groups S[푌; 푆훼; 휑훼,훽]. Hence for 푥 ∈ 푆훼 and −1 −1 −1 −1 푦 ∈ 푆훽, we have 푥푥 푦푦 = 1훼1훽 = 1훼⊓훽 = 1훽1훼 = 푦푦 푥푥 . 5.13 In particular, Theorem 5.13(d) implies that in a Clifford semigroup, idempotents commute; hence, by Theorem 5.1, Clifford semigroups are inverse semigroups. Notice that this is not obvious from the conditions (4.3) and (4.4). Let 푆 be a Clifford semigroup. By Theorem 5.13, 푆 is isomorphic to a Natural partial order on Clifford semigroups strong semilattice of groups S[푌; 퐺훼; 휑훼,훽]. Let 푥 ∈ 퐺훼 and 푦 ∈ 퐺훽. Then 푥 ≼ 푦 ⇔ 푥 = (푥푥−1)푦

⇔ 푥 = 1훼푦

⇔ 푥 = (1훼휑훼,훼⊓훽)(푦휑훽,훼⊓훽)

⇔ (푥 = (1훼휑훼,훼)(푦휑훽,훼)) ∧ (훼 ⊓ 훽 = 훼)

⇔ (푥 = 푦휑훽,훼⊓훽) ∧ (훼 ⩽ 훽). Thus the natural partial order ≼ precisely corresponds to the homomorphisms 휑훼,훽 and the order of the semilattice (푌, ⩽). In particular, we have

1훼 ≼ 1훽 ⇔ 1훼 = 1훽휑훽,훼 ∧ (훼 ⩽ 훽) ⇔ 훼 ⩽ 훽.

Since the identities of the groups 퐺훼 are precisely the idempotents of 푆, we see that (퐸(푆), ≼) and (푌, ⩽) are isomorphic. In particular, every

Clifford semigroups • 102 semilattice (푌, ⩽) is a Clifford semigroup S[푌; 퐺훼, 휑훼,훽] where the groups 퐺훼 are all trivial.

Free inverse semigroups Let 퐴 be an alphabet. Let 퐴−1 be a set of new symbols bijection with 퐴 under the map 푎 ↦ 푎−1. Extend this map to an involution −1 −1 −1 of 퐴 ∪ 퐴 by defining (푎 ) = 푎. For any word 푎1푎2 ⋯ 푎푛 ∈ (퐴 ∪ −1 ∗ −1 −1 −1 −1 퐴 ) , define (푎1푎2 ⋯ 푎푛) = 푎푛 ⋯ 푎2 푎1 . Let FInvS(퐴) be semigroup presented by Sg⟨퐴 ∪ 퐴−1 | 휌⟩, where

휌 = { (푢푢−1푢, 푢) ∶ 푢 ∈ (퐴 ∪ 퐴−1)+ } ∪ { (푢푢−1푣푣−1, 푣푣−1푢푢−1) ∶ 푢, 푣 ∈ (퐴 ∪ 퐴−1)+ }.

P r o p o s i t i o n 5 . 1 4. The semigroup FInvS(퐴) is an inverse semigroup, −1 where the inverse of [푢]휌# ∈ FInvS(퐴) is [푢 ]휌# .

−1 −1 Proof of 5.14. Define ([푢]휌# ) = [푢 ]휌# . We aim to prove that the conditions (5.1)–(5.4) are satisfied. First of all, it is necessary to check that −1 the operation is well-defined on FInvS(퐴). Suppose [푢]휌# = [푣]휌# . Then there is a sequence of elementary 휌-transitions 푢 = 푤0 ↔휌 푤1 ↔휌 −1 −1 + … ↔휌 푤푛 = 푣. Apply (as an operation on (퐴 ∪ 퐴 ) ) to every term in this sequence. This yields a sequence of elementary 휌-transitions −1 −1 −1 −1 −1 −1 −1 푢 = 푤0 ↔휌 푤1 ↔휌 … ↔휌 푤푛 = 푣 ; hence [푢 ]휌# = [푣 ]휌# . Now let 푢, 푣 ∈ FInvS(퐴). It is immediate from the definition of −1 that

−1 −1 −1 −1 ([푢]휌# ) = [(푢 ) ]휌# = [푢]휌# and

−1 −1 −1 −1 [푢푣]휌# = [(푢푣) ]휌# = [푣 푢 ]휌# −1 −1 −1 −1 = [푣 ]휌# [푢 ]휌# = [푣]휌# [푢]휌# ; thus (5.1) and (5.2) hold. Furthermore,

−1 −1 [푢]휌# [푢]휌# [푢]휌# = [푢]휌# [푢 ]휌# [푢]휌# −1 = [푢푢 푢]휌# [by definition of 휌]

= [푢]휌# and

−1 −1 −1 −1 [푢]휌# [푢]휌# [푣]휌# [푣]휌# = [푢]휌# [푢 ]휌# [푣]휌# [푣 ]휌# −1 −1 = [푢푢 푣푣 ]휌#

Free inverse semigroups • 103 −1 −1 = [푣푣 푢푢 ]휌# [by definition of 휌] −1 −1 = [푣]휌# [푣 ]휌# [푢]휌# [푢 ]휌# −1 −1 = [푣]휌# [푣]휌# [푢]휌# [푢]휌# ; thus (5.3) and (5.4) hold. Hence FInvS(퐴) is an inverse semigroup. 5.14

Let 퐹 be an inverse semigroup, let 퐴 be an alphabet, and let 휄 ∶ 퐴 → 퐹 Free inverse semigroup be an embedding of 퐴 into 퐹. Then the inverse semigroup 퐹 is a free inverse semigroup on 퐴 if, for any inverse semigroup 푆 and map 휑 ∶ 퐴 → 푆, there is a unique homomorphism 휑 ∶ 퐹 → 푆 that extends 휑 (that is, with 휄휑 = 휑). Using diagrams, this definition says that 퐹 is a free inverse semigroup on 퐴 if 휄 퐴 퐹 } } for all with 푆 inverse, there exists } } 휑 } 휄 푆 퐴 퐹 } (5.10) } } a unique homomorphism 휑 such that 휑 . } 휑 } 푆 } This definition is analogous to the definition of the free semigroup on 퐴 (see pages 37 f.). In Chapter 8, we will see definitions of ‘free objects’ in a much more general setting. Like the free semigroup on 퐴, the free inverse semigroup on 퐴 is unique up to isomorphism: P r o p o s i t i o n 5 . 1 5. Let 퐴 be an alphabet and let 퐹 be an inverse Uniqueness of the free semigroup. Then 퐹 is a free inverse semigroup on 퐴 if and only if 퐹 ≃ inverse semigroup on 퐴 FInvS(퐴).

Proof of 5.15. Let 휄 ∶ 퐴 → FInvS(퐴) be the natural map 푎휄 = [푎]휌# . Let 푆 be a inverse semigroup and 휑 ∶ 퐴 → 푆 a map. Extend 휑 to a map 휑′ ∶ 퐴∪퐴−1 → 푆 by defining 푎−1휑′ = (푎휑)−1 for 푎−1 ∈ 퐴−1. Since (퐴∪퐴−1)+ is the free semigroup on 퐴, the map 휑′ extends to a unique homomorphism −1 + 휑″ ∶ (퐴 ∪ 퐴 ) → 푆 with (푎1푎2 ⋯ 푎푛)휑″ = (푎1휑′)(푎2휑′) ⋯ (푎푛휑′), −1 where 푎푖 ∈ 퐴 ∪ 퐴 . Since 푆 is an inverse semigroup, (푢푢−1푢)휑″ = (푢휑″)(푢−1휑″)(푢휑″) [since 휑″ is a homomorphism] = (푢휑″)(푢휑″)−1(푢휑″) [by definition of 휑′] = 푢휑″ and (푢푢−1푣푣−1)휑″ = (푢휑″)(푢−1휑″)(푣휑″)(푣−1휑″) = (푢휑″)(푢휑″)−1(푣휑″)(푣휑″)−1 = (푣휑″)(푣휑″)−1(푢휑″)(푢휑″)−1 [since 푆 is inverse] = (푢푢−1푣푣−1)휑″

Free inverse semigroups • 104 for all 푢, 푣 ∈ (퐴 ∪ 퐴−1)+. Thus 휌 ⊆ ker 휑″ and so there is a well-defined homomorphism 휑 ∶ FInvS(퐴) → 푆 with [푢]휌# 휑 = 푢휑″. That is, the following diagram commutes:

퐴 휄 FInvS(퐴)

(휌#)♮ 퐴 ∪ 퐴−1 (퐴 ∪ 퐴−1)+ 휑 휑″ 휑′ 휑 푆

It remains to prove that 휑 is the unique homomorphism such that 휄휑 = 휑. So let 휓 ∶ FInvS(퐴) → 푆 be such that 휄휓 = 휑. Then for all 푎 ∈ 퐴, we have [푎]휌# 휓 = 푎휄휓 = 푎휑 and

−1 −1 −1 [푎 ]휌# 휓 = ([푎]휌# ) 휓 [by definition of in FInvS(퐴)] −1 = ([푎]휌# 휓) [by Proposition 5.4] = (푎휄휓)−1 = (푎휑)−1 = 푎−1휑. [by Proposition 5.4]

−1 Hence for any 푎푖 ∈ 퐴 ∪ 퐴 ,

([푎1푎2 ⋯ 푎푛]휌# )휓 = ([푎1]휌# [푎2]휌# ⋯ [푎푛]휌# )휓

= ([푎1]휌# 휓)([푎2]휌# 휓) ⋯ ([푎푛]휌# )휓)

= (푎1휑)(푎2휑) ⋯ (푎푛휑)

= ([푎1]휌# 휑)([푎2]휌# 휑) ⋯ ([푎푛]휌# )휑)

= ([푎1푎2 ⋯ 푎푛]휌# )휑. Thus 휓 = 휑. Therefore FInvS(퐴) is a free inverse semigroup on 퐴. Now let 퐹 be a free inverse semigroup on 퐴. Let 휄1 ∶ 퐴 → FInvS(퐴) and 휄2 ∶ 퐴 → 퐹 be the embedding maps. Following the same argument as for free semigroups on 퐴 (see the proof of Proposition 2.1), this leads to 휄2 ∶ FInvS(퐴) → 퐹 and 휄1 ∶ 퐹 → FInvS(퐴) being mutually inverse isomorphisms. 5.15

We could repeat the discussion above for monoids instead of sem- Free inverse monoid igroups. The monoid FInvM(퐴) is presented by Mon⟨퐴 ∪ 퐴−1 | 휌⟩.A monoid 퐹 is a free inverse monoid on 퐴 if, for any inverse monoid 푆 and map 휑 ∶ 퐴 → 푆, there is a unique monoid homomorphism 휑 ∶ 퐹 → 푆 extending 휑; that is, with 휄휑 = 휑. One can prove an analogy of Proposi- tion 5.15 for monoids, showing that an inverse monoid 퐹 is a free inverse monoid on 퐴 if and only if 퐹 ≃ FInvM(퐴). Notice that because there is

Free inverse semigroups • 105 no defining relation in 휌 that has the empty word 휀 as one of its two sides, there is no non-empty word that is equal to 휀 in FInvM(퐴). Therefore FInvM(퐴) ≃ (FInvS(퐴))1. Since free inverse semigroups and monoids are such fundamental objects, we would like to be able to solve the word problem: given two words in (퐴∪퐴−1)+ (respectively, (퐴∪퐴−1)∗), do they represent the same element of FInvS(퐴) (respectively, FInvM(퐴))? This appears difficult: for example, 푎푎푎−1푎−1푎−1푎푏푏−1푎푏−1푏푐푎푎−1푐푐−1 −1 −1 −1 −1 −1 −1 −1 −1 } (5.11) =FInvS(퐴) 푎 푎푏푏 푎푎푎 푐푎푎 푐푐 푐 푏 푏푎 푎푐, but this not obvious. However, we now introduce a representation of elements of FInvM(퐴) that makes it easy to answer this question. Let 푇 be a finite non-empty directed tree with edges labelled by sym- Tree with edge bols in 퐴. Extend the set of labels to 퐴 ∪ 퐴−1 by adopting the following labels from 퐴 ∪ 퐴−1 convention: for all 푎 ∈ 퐴 and vertices 훽 and 훾,

−1 푎 means the same as 푎 (5.12) 훽 훾 훽 훾

Denote the set of vertices of 푇 by 푉(푇). By definition, |푉(푇)| ⩾ 1. Let 훽, 훾 ∈ 푉(푇). If 훽 and 훾 are adjacent, then 훽훾 will denote the edge from 훽 to 훾.A (훽, 훾)-walk on 푇 is a sequence 훽 = 훿0, 훿1, … , 훿푛 = 훾 such that 훿푖−1 and 훿푖 are adjacent for 푖 = 1, … , 푛.A (훽, 훾)-walk 훽 = 훿0, 훿1, … , 훿푛 = 훾 spans 푇 if every vertex of 푇 appears at least once among the 훿푖. The (훽, 훾)- path on 푇, denoted 휋(훽, 훾), is the unique (훽, 훾)-walk 훽 = 훿0, 훿1, … , 훿푛 = 훾 such that no vertex of 푇 occurs more than once among the 훿푖; the integer 푛 is the length of 휋(훽, 훾). Notice that there is a trivial path at 훽, namely 휋(훽, 훽), which has length 0. For a (훽, 훾)-walk 휎 = (훽 = 훿0, … , 훿푚 = 훾), define w(휎) = 푥1푥2 ⋯ 푥푚, −1 where 푥푖 ∈ 퐴∪퐴 is the label on the edge 훿푖−1훿푖 for 푖 = 1, … , 푚 (recalling the convention (5.12)). Note that w(휋(훽, 훽)) = 휀. A word tree over 퐴 is a finite non-empty directed tree 푇 with edges Word tree, Munn tree labelled elements of 퐴 (using the convention (5.12)), and where there is no vertex that has two distinct incoming edges with the same label or two distinct outgoing edges with the same label. That is, a word tree does not contain subgraphs } 푎 푎 7 2 13 } (5.13) or } 푏 푎 푎 푎 푎 } 훼 푎 푐 휔 푇 1 푇 푎 푐 A Munn tree over 퐴 is a word tree 푇 with two distinguished vertices 훼푇 푏 and 휔푇 (not necessarily distinct). 5 10 15 Figure 5.3 gives an example of a Munn tree. Notice it satisfies the con- FIGURE 5.3 dition (5.13). Furthermore, both words in (5.11) label spanning (훼 , 휔 )- Munn tree 푇 for the words 푇 푇 푎2푎−3푎푏푏−1푎푏−1푏푐푎푎−1푐푐−1 walks in this Munn tree. This is how Munn trees allow us to solve the and 푎−1푎푏푏−1푎2푎−1푐푎푎−1 푐푐−2푏−1푏푎−1푎푐.

Free inverse semigroups • 106 word problem for FInvM(퐴): we will prove that two words represent the same element of FInvM(퐴) if and only if they have isomorphic Munn trees. To be precise, an isomorphism between two word trees 푇1 and 푇2 is a bijection 휑 ∶ 푉(푇1) → 푉(푇2) such that there is an edge 푣푣′ labelled by 푎 in 푇1 if and only if there is an edge (푣휑)(푣′휑) labelled by 푎 in 푇2. If 푇1 and 푇2 are Munn trees, then such a map is an isomorphism if, in addition, 훼 휑 = 훼 and 휔 휑 = 휔 . 푇1 푇2 푇1 푇2 −1 Suppose we have a word 푢 = 푥1푥2 ⋯ 푥푛, where 푥푖 ∈ 퐴 ∪ 퐴 . Let us Constructing a Munn tree from a word describe how to construct a Munn tree 푇 with a spanning (훼푇, 휔푇)-walk 휎 such that w(휎) = 푢. We will initially construct a tree 푇 with distinguished vertices 훼푇 and 휔푇 such that there is a spanning (훼푇, 휔푇)-walk 휎 on 푇 such that w(휎) = 푢. This tree may not satisfy (5.13). We will then modify 푇 to turn it into a Munn tree. To begin, let 푇 be the graph with 푛 + 1 vertices 훿0, 훿1, 훿2, … , 훿푛−1, 훿푛 with edges 훿푖−1훿푖 having label 푥푖 for 푖 = 1, … , 푛 (recall the convention (5.12)). Notice that this tree is simply a path. Let 훼푇 = 훿0 and 휔푇 = 훿푛. Note that 푇 is a tree with distinguished vertices 훼푇 and 휔푇. Let 휎 be unique path from 훼푇 to 휔푇. Then w(휎) = 푢. (The graph at the top of Figure 5.4 is the result of this construction for 푢 = 푎2푎−3푎푏푏−1푎푏−1푏푐푎푎−1푐푐−1.) Note that 푇 satisfies all the conditions we want except possibly (5.13). Now let us modify 푇. If 푇 satisfies (5.13), then it is a word tree and so a Munn tree and we are finished. So suppose 푇 does not satisfy (5.13). Then by the convention (5.12), 푇 contains a subgraph

훿푗 푥 훿ℓ 푥 훿푘 for some 푥 ∈ 퐴 ∪ 퐴−1. Fix such a subgraph. Modify 푇 by folding the (identically-labelled) edges 훿푗훿ℓ and 훿푗훿ℓ together and merging the vertices 훿푗 and 훿푘. If we merge 훼푇 (respectively, 휔푇) with some vertex, the resulting merged vertex is still 훼푇 (respectively, 휔푇). Then 푇 is still a tree and the walk 휎 (which is, after all, simply a sequence of vertices) is still a spanning (훼푇, 휔푇)-walk for 푇. However, 푇 now contains one vertex fewer than before. Repeat this process. Since each such modification reduces the number of vertices of 푇, then process must halt with a tree 푇 satisfying (5.13), which is the desired Munn tree. (Figure 5.4 illustrates this process for the word 푢 = 푎2푎−3푎푏푏−1푎푏−1푏푐푎푎−1푐푐−1.) We now establish four lemmata that lead up to the main result. For brevity, we write 퐹 for FInvM(퐴). First, we must make some more definitions. Let 휎 = (훽 = 훿0, … , 훿푚 = 훾) and 휏 = (훾 = 휂0, … , 휂푛 = 휁) be,

Free inverse semigroups • 107 훼 푎 푎 푎 푎 푎 푎 푏 푏 푎 푏 10 1 2 3 4 5 6 7 8 9 푏 merging 1 & 3; 4 & 6; 6 & 8; 9 & 11; 12 & 14; 14 & 휔 11 7 13 푐 푏 푎 12 훼 푎 1 푎 푎 9 푐 휔 푎 4 푎 푎 푏 푐 13 2 5 10 15 푎 14 merging 훼 & 4 푐 7 13 15 푏 푎 푐 1 푎 푎 9 푐 훼 휔 휔 푎 푎 푏 푐 2 5 10 15

merging 1 & 9

7 2 13 푏 푎 푎 훼 푎 푐 휔 FIGURE 5.4 1 Folding a linear graph to pro- 푎 푐 푏 duce a Munn tree for the word 5 10 15 푎2푎−3푎푏푏−1푎푏−1푏푐푎푎−1푐푐−1. respectively, a (훽, 훾)- and a (훾, 휁)-walk on 푇. Define a (훽, 휁)-walk 휎휏 by

휎휏 = (훽 = 훿0, … , 훿푚−1, 훾, 휂1, … , 휂푛 = 휁). Clearly one can extend this to products of three or more walks and this product is associative (whenever it is defined). We also define 휎−1 to be 푘 the (훾, 훽)-walk (훾 = 훿푚, … , 훿0 = 훽). If 휎 is a (훽, 훽)-walk, then 휎 has the obvious meaning for all 푘 ∈ ℕ. L e m m a 5 . 1 6. Let 휎 be a (훽, 훾)-walk and 휏 a (훾, 휁)-walk on a word tree 푇. Then: a) w(휎휏) = w(휎) w(휏); b) 휏 = 휎−1 if and only if w(휏) = (w(휎))−1. Proof of 5.16. Part a) and the forward implication in part b) are immediate from the definition. It remains to prove the reverse implication in part b). −1 −1 So suppose w(휏) = (w(휎)) = 푥1 ⋯ 푥푚 (where 푥푖 ∈ 퐴 ∪ 퐴 ). Then 휎 and 휏 both contain 푚 + 1 vertices, with 휎 = (훽 = 훿0, … , 훿푚 = 훾) and 휏 = (훾 = 휂0, … , 휂푚 = 휁). We will prove that 훿푚−푗 = 휂푗 by induction on 푗. We already know that 훿푚 = 훾 = 휂0; this is the base of the induction. For the induction step, suppose that 훿푚−푗 = 휂푗. Now, 훿푚−푗훿푚−푗−1 and 휂푗휂푗+1 both have label 푥푗. So, since 푇 satisfies (5.13), 훿푚−푗−1 = 휂푗+1. This proves −1 the induction step. So 훿푚−푗 = 휂푗 for all 푗 = 1, … , 푚. Hence 휏 = 휎 . 5.16

Free inverse semigroups • 108 The next lemma essentially says that each Munn tree is associated with a unique element of FInvM(퐴):

L e m m a 5 . 1 7. If 휎 and 휏 are spanning (훼푇, 휔푇)-walks on a Munn tree 푇, then w(휎) =퐹 w(휏). Proof of 5.17. If |푉(푇)| = 1, then 휎 and 휏 consist of the single vertex in 푉(푇) and so w(휎) = 휀 = w(휏). In the remaining cases, we use induction on |푉(푇)| ⩾ 2 to prove the following statement: If 휎 and 휏 are spanning (훽, 훾)-walks on a word tree 푇, then w(휎) =퐹 w(휏). For the base of the induction, let |푉(푇)| = 2. Let 휁 be the unique vertex in 푇 ∖ {훽}, let 휋 = 휋(훽, 휁) and let 푥 = w(휋). Note that 푥 ∈ 퐴 ∪ 퐴−1 since 휋 has length 1, because there are only two vertices in 푇. We now consider the cases 훾 = 훽 and 훾 = 휁 separately: ◆ 훾 = 훽. Then 휎 = (휋휋−1)푘 and 휏 = (휋휋−1)ℓ for some 푘, ℓ ∈ ℕ ∪ {0} and so, by Lemma 5.16 and the defining relations in 휌,

−1 푘 −1 −1 ℓ w(휎) = (푥푥 ) =퐹 푥푥 =퐹 (푥푥 ) = w(휏).

◆ 훾 = 휁. Then 휎 = (휋휋−1)푘휋 and 휏 = (휋휋−1)ℓ휋 for some 푘, ℓ ∈ ℕ ∪ {0} and so, by Lemma 5.16 and the defining relations in 휌,

−1 푘 −1 ℓ w(휎) = (푥푥 ) 푥 =퐹 푥 =퐹 (푥푥 ) 푥 = w(휏).

In either case, the result holds for |푉(푇)| = 2. For the inductive step, let 푛 > 2. Suppose that if 휎 and 휏 are spanning (훽, 훾) walks on a tree 푇 such that |푉(푇)| < 푛, then w(휎) =퐹 w(휏).

C l a i m . If 휎0 is a (휉, 휉)-walk on a subtree 푇 of 푇 such that |푉(푇)| < 푛, 2 then (w(휎0)) =퐹 w(휎0).

Proof of Claim. Let 푇0 be the subtree of 푇 spanned by 휎0. Then we have 2 |푉(푇0)| ⩽ |푉(푇)| < 푛 and both 휎0 and 휎0 are spanning (휉, 휉)-walks on 푇0. 2 2 Thus, by the induction hypothesis, w(휎0) =퐹 w(휎0 ) = (w(휎0)) . Claim Now let 휎 and 휏 be spanning (훽, 훾)-walks on 푇. We consider separately the case where 훽 is an endpoint of 푇 and the case where 훽 is not an endpoint of 푇. ◆ 훽 is an endpoint (or leaf vertex) of 푇. Let 휉 be the unique vertex of 푇 adjacent to 훽 and let 푇 be the subtree of 푇 obtained by deleting 훽 and the edge 훽휉. Let 휋 = 휋(훽, 휉) and let 푥 = w(휋). Now we consider the sub-cases 훽 = 훾 and 훽 ≠ 훾 separately.

■ 훾 = 훽. Then for some (휉, 휉)-walks 휎1, 휎2, … , 휎ℎ on 푇 and some 푘푖 ∈ ℕ ∪ {0} (where 푖 = 0, … , ℎ),

−1 푘0 −1 푘1 −1 푘ℎ −1 휎 = 휋(휋 휋) 휎1(휋 휋) 휎2 ⋯ 휎ℎ(휋 휋) 휋 .

Free inverse semigroups • 109 Let 푢푖 = w(휎푖) for 푖 = 1, … , ℎ. By Lemma 5.16,

−1 푘0 −1 푘1 −1 푘ℎ −1 w(휎) = 푥(푥 푥) 푢1(푥 푥) 푢2 ⋯ 푢ℎ(푥 푥) 푥

2 2 However, 푢푖 = (w(휎푖)) =퐹 w(휎푖) = 푢푖 by the Claim. That is, each −1 푢푖 is idempotent. Hence, since 푥 푥 is also an idempotent, and idempotents commute,

−1 푘0+푘1+⋯+푘ℎ −1 −1 w(휎) =퐹 푥(푥 푥) 푢1푢2 ⋯ 푢ℎ푥 =퐹 푥푢푥 ,

where 푢 = 푢1푢2 ⋯ 푢ℎ. Furthermore, we have 푢 = w(휎), where 휎 = 휎1휎2 ⋯ 휎ℎ. Note that 휎 is a (휉, 휉)-walk. Moreover, since 휎 spans 푇, it follows that 휎 spans 푇. −1 Similarly, w(휏) =퐹 푥푣푥 , where 푣 = w(휏) for some spanning (휉, 휉)-walk 휏 of 푇. But |푉(푇)| = 푛−1 and so 푢 =퐹 푣 by the inductive hypothesis. Hence

−1 −1 w(휎) =퐹 푥푢푥 =퐹 푥푣푥 =퐹 w(휏).

■ 훾 ≠ 훽. Then 훾 is a vertex of 푇. Therefore, for some (휉, 휉)-walks 휎1, 휎2, … , 휎ℎ and a (휉, 훾)-walk 휎∞ on 푇 and some 푘푖 ∈ ℕ ∪ {0} (where 푖 = 0, … , ℎ),

−1 푘0 −1 푘1 −1 푘ℎ 휎 = 휋(휋 휋) 휎1(휋 휋) 휎2 ⋯ 휎ℎ(휋 휋) 휎∞.

Let 푢푖 = w(휎푖) for 푖 = 1, … , ℎ and 푢∞ = w(휎∞). By Lemma 5.16,

−1 푘0 −1 푘1 −1 푘ℎ w(휎) = 푥(푥 푥) 푢1(푥 푥) 푢2 ⋯ 푢ℎ(푥 푥) 푢∞

2 By the Claim, 푢푖 =퐹 푢푖 is idempotent for 푖 = 1, … , ℎ. Hence

−1 푘0+푘1+⋯+푘ℎ w(휎) =퐹 푥(푥 푥) 푢1푢2 ⋯ 푢ℎ푢∞ = 푥푢,

where 푢 = 푢1푢2 ⋯ 푢ℎ푢∞. Furthermore, we have 푢 = w(휎), where 휎 = 휎1휎2 ⋯ 휎ℎ휎∞ is a (휉, 훾)-walk that spans 푇 since 휎 spans 푇. Similarly, w(휏) =퐹 푥푣, where 푣 = w(휏) for some spanning (휉, 휉)-walk 휏 of 푇. By the inductive hypothesis, 푢 =퐹 푣 and so w(휎) = w(휏).

◆ 훽 is not an endpoint of 푇. Then we can split 푇 into two subtrees 푇1 and 푇2 such that |푉(푇1)| < 푛 and |푉(푇2)| < 푛, and 푉(푇1) ∩ 푉(푇2) = {훽}. Interchanging 푇1 and 푇2 if necessary, assume that 훾 ∈ 푉(푇2). Then

휎 = 휎1휎2 ⋯ 휎ℎ,

where ℎ is even, the 휎1, 휎3, 휎5, … , 휎ℎ−1 are (훽, 훽)-walks (possibly trivial) on the subtree 푇1, the 휎2, 휎4, 휎6, … , 휎ℎ−2 are (훽, 훽)-walks (possibly trivial) on the subtree 푇2, and 휎ℎ is a (훽, 훾)-walk (possibly trivial) on

Free inverse semigroups • 110 the subtree 푇2. Let 푢푖 = w(휎푖) for 푖 = 1, … , ℎ. Then for 푖 = 1, … , ℎ−1, 2 by the Claim we have 푢푖 =퐹 푢푖 and so 푢푖 is an idempotent. Hence

w(휎) = 푢1푢2푢3푢4 ⋯ 푢ℎ−1푢ℎ

=퐹 푢1푢3 ⋯ 푢ℎ−1푢2푢4 ⋯ 푢ℎ

= 푢1푢2,

where 푢1 = w(휎1) and 푢2 = w(휎2), and 휎푘 = 휎1휎3 ⋯ 휎ℎ−1 and 휎2 = 휎2휎4 ⋯ 휎ℎ. Note that 휎1 is a (훽, 훽)-walk on 푇1 and 휎2 is a (훽, 훾)-walk on 푇2. Since 휎 spans 푇, it follows that 휎1 spans 푇1 and 휎2 spans 푇2. Similarly, we can show that w(휏) =퐹 푣1푣2, where 푣1 = w(휏1) and 푣1 = w(휏1) for some spanning (훽, 훽)-walk 휏1 of 푇1 and spanning (훽, 훾)-walk 휏2 of 푇2, respectively. Thus, by the inductive hypothesis, w(휎1) =퐹 w(휏1) and w(휎2) =퐹 w(휏2). Hence w(휎) = 푢1푢2 =퐹 푣1푣2 =퐹 w(휏). This completes the inductive step and so the result holds. Claim Now we want to show each element of FInvM(퐴) is associated to a unique Munn tree. As a first step, the next lemma shows that each element of (퐴 ∪ 퐴−1)∗ is associated to a unique Munn tree.

L e m m a 5 . 1 8. Let 푇 and 푇 be Munn trees. Let 휎 be a spanning (훼푇, 휔푇)- walk in 푇 and 휏 a spanning (훼푇, 휔푇)-walk in 푇 such that w(휎) = w(휏). Then 푇 and 푇 are isomorphic.

Proof of 5.18. Let 푥1푥2 ⋯ 푥푚 = w(휎) = w(휏) and suppose

휎 = (훼푇 = 훿0, … , 훿푚 = 휔푇) and 휏 = (훼푇 = 휂0, … , 휂푚 = 휔푇), where 푥푖 is the label on 훿푖−1훿푖 and 휂푖−1휂푖 for 푖 = 1, … , 푚. Let 푇푖 and 푇푖 be the subtrees of 푇 and 푇 spanned by the walks (훿0, … , 훿푖) and (휂0, … , 휂푖), respectively, for 푖 = 0, … , 푚. Notice that 푇 = 푇푚 and 푇 = 푇푚 since 휎 and 휏 span 푇 and 푇, respectively. Clearly the map 휑0 ∶ 푇0 → 푇0 defined by 훿0휑0 = 휂0 is trivially an isomorphism of word trees. Suppose that we have an isomorphism 휑푖−1 ∶ 푇푖−1 → 푇푖−1 such that 훿푗휑푖−1 = 휂푗 for 푗 = 0, … , 푖 − 1. We show that this can be extended to an isomorphism 휑푖 ∶ 푇푖 → 푇푖. We consider the cases 훿푖 ∈ 푉(푇푖−1) and 훿푖 ∉ 푉(푇푖−1) separately.

◆ 훿푖 ∈ 푉(푇푖−1). Then 푇푖 = 푇푖−1. Since 훿푖 is adjacent to 훿푖−1 and 훿푖−1훿푖 has label 푥푖, there exists 휁 ∈ 푉(푇푖−1) such that 휁 = 훿푖휑푖−1 and 휁 is adjacent to 휂푖−1 with the edge 휂푖−1휁 having label 푥푖. However, 휂푖 is adjacent to 휂푖−1 in 푇 and 휂푖−1휂푖 has label 푥푖. Since 푇 satisfies (5.13), 휂푖 = 휁 = 훿푖휑푖−1. Thus 푇푖 = 푇푖−1. So define 휑푖 ∶ 푇푖 → 푇푖 by 휑푖 = 휑푖−1; then 훿푗휑푖 = 휂푗 for 푗 = 0, … , 푖 and so 휑푖 is an isomorphism of word trees.

Free inverse semigroups • 111 ◆ 훿푖 ∉ 푉(푇푖−1). Suppose with the aim of obtaining a contradiction that 휂푖 ∈ 푇푖−1. Since 휂푖 is adjacent to 휂푖−1 and 휂푖−1휂푖 has label 푥푖, there exists −1 휉 ∈ 푉(푇푖−1) such that 휉 = 휂푖휑푖−1 and 휉 is adjacent to 훿푖−1 with the edge 훿푖−1휉 having label 푥푖. Since 훿푖 is adjacent to 훿푖−1 in 푇 and 훿푖−1훿푖 −1 has label 푥푖, and 푇 satisfies (5.13), we have 훿푖 = 휉 ∈ (푉(푇푖−1))휑푖−1 = 푉(푇푖−1), which is a contradiction. Hence 휂푖 ∉ 푇푖−1. Thus we can extend 휑푖−1 to an isomorphism of word trees 휑푖 ∶ 푇푖 → 푇푖 by defining 훿푖휑푖 = 휂푖. By induction on 푖, there exists an isomorphism 휑푛 ∶ 푇푛 → 푇푛. Note that 훼푇휑푛 = 훿0휑푛 = 휂0휑푛 = 훼푇휑푛 and similarly 휔푇휑푛 = 휔푇휑푛. So 휑푛 is an isomorphism of Munn trees. 5.18 The next result strengthens the previous one, showing that each element of FInvM(퐴) is associated to a unique Munn tree.

L e m m a 5 . 1 9. Let 푇 and 푇 be Munn trees. Let 휎 be a spanning (훼푇, 휔푇)- walk in 푇 and 휏 a spanning (훼푇, 휔푇)-walk in 푇 such that w(휎) =퐹 w(휏). Then 푇 and 푇 are isomorphic. Proof of 5.19. It is sufficient to prove the result when w(휎) and w(휏) differ by a single elementary 휌-transition. ◆ w(휎) = 푝푢푞 and w(휏) = 푝푢푢−1푢푞 for 푝, 푢, 푞 ∈ (퐴 ∪ 퐴−1)∗ with 푢 ≠ 휀. So there exist (훼푇, 훽)-, (훽, 훾)-, and (훾, 휔푇)-walks 휎1, 휎2, and 휎3 on 푇 such that 휎 = 휎1휎2휎3, where w(휎1) = 푝, w(휎2) = 푢, and w(휎3) = 푞. −1 Let 휐 be the (훼푇, 휔푇)-walk 휎1휎2휎2 휎2휎3. Since 휎 spans 푇, so does 휐. By Lemma 5.16, w(휐) = 푝푢푢−1푢푞 = w(휏). Therefore, by Lemma 5.18, the Munn trees 푇 and 푇 are isomorphic. ◆ w(휎) = 푝푢푢−1푣푣−1푞 and w(휏) = 푝푣푣−1푢푢−1 for 푝, 푢, 푣, 푞 ∈ (퐴∪퐴−1)∗ with 푢, 푣 ≠ 휀. So there exist (훼푇, 훽)-, (훽, 훾)-, (훽, 훿)- and (훽, 휔푇)-walks −1 −1 휎1, 휎2, 휎3, and 휎4 on 푇 such that 휎 = 휎1휎2휎2 휎3휎3 휎4, where w(휎1) = 푝, w(휎2) = 푢, w(휎3) = 푣 and w(휎4) = 푞. Let 휐 be the (훼푇, 휔푇)-walk −1 −1 휎1휎3휎3 휎2휎2 휎4. Since 휎 spans 푇, so does 휐. By Lemma 5.16, w(휐) = 푝푣푣−1푢푢−1푞 = w(휏). Therefore, by Lemma 5.18, the Munn trees 푇 and 푇 are isomorphic. 5.19 T h e o r e m 5 . 2 0. Let 푇 and 푇 be Munn trees, and let 휎 be a spanning Equal in FInvM(퐴) ⇔ isomorphic Munn trees (훼푇, 휔푇)-walk on 푇, and let 휏 be a spanning (훼푇, 휔푇) walk on 푇. Then w(휎) =퐹 w(휏) if and only if 푇 and 푇 are isomorphic. Proof of 5.20. If 푤(휎) = w(휏), then 푇 and 푇 are isomorphic by Lemma 5.19. On the other hand, suppose 휑 ∶ 푇 → 푇 is an isomorphism. Then 휑 maps 휎 to a spanning (훼푇, 휔푇)-walk 휎 of 푇. Note that w(휎) = w(휎). Then by Lemma 5.17, w(휎) = w(휏) and so w(휎) = w(휏). 5.20 We can use Munn trees to compute multiplications in FInvM(퐴). Suppose we have two Munn trees 푇 and 푇 . Pick a spanning (훼 , 휔 )- 1 2 푇1 푇1 walk 휎 on 푇 and a spanning (훼 , 휔 )-walk 휎 with elements. Merge the 1 1 푇2 푇2 1

Free inverse semigroups • 112 2 3 8 5 푏 푎 푎 푎 훼 푎 휔 훼 푐 휔 푇1 푇1 푇2 푇2 푎 푏 푏 푐 1 4 7 6 FIGURE 5.5 merging Multiplying using Munn

휔 & 훼 7 trees: from Munn trees for 푇1 푇2 & −1 −1 2 −1 −1 and 4 푎 푎푏푏 푎 푎 푏 푏

; −1 −1 −1 −1

1 3 8 5 8 2 3 5 푐푎푎 푐푐 푏 푏푎 푎푐, we

푎 & 푏 푎 푏 푎 푎 compute a Munn tree for the

푎 merging 훼 훼 푎 푐 휔 훼 푎 푐 휔 product by merging the ‘휔’ 푇 ∞ 푇 푇 ∞ 푇 푏 of the first tree with the ‘훼’ of 푎 푐 푎 푏 푐 푏 the second and then folding 2 4 7 6 1 4 6 edges.

the vertices 휔 and 훼 to obtain a tree 푇, and let 훼 = 훼 and 휔 = 휔 . 푇1 푇2 푇 푇1 푇 푇2 Let 휎 = 휎1휎2. Then 휎 is a spanning (훼, 휔)-walk on 푇. It remains to fold edges together until (5.13) is satisfied, as we did to construct Munn trees initially. Figure 5.5 illustrates the process.

Exercises [See pages 220–229 for the solutions.]

✴5.1 Let 푋 = {1, 2}. Find a subsemigroup of I푋 that contains only two elements and which is not an inverse subsemigroup. 5.2 Let 퐺 be a group and let 푆 be the set of isomorphisms between subgroups of 퐺. Prove that 푆 is an inverse subsemigroup of I퐺. 5.3 Let 푋 be a set and let 휎, 휏 ∈ I푋. Prove the following: a) 휎 L 휏 ⇔ im 휎 = im 휏; b) 휎 R 휏 ⇔ dom 휎 = dom 휏; c) 휎 D 휏 ⇔ 휎 J 휏 ⇔ |dom 휎| = |dom 휏|. ✴5.4 Let 푋 = {1, … , 푛} with 푛 ⩾ 3. Let 휏 = (1 2) and 휁 = (1 2 … 푛 − 1 푛). As remarked in Exercise 1.11, from elementary group theory, we know that S푋 = ⟨휏, 휁⟩. For 푘 = 1, … , 푛, let

퐽푘 = { 휎 ∈ I푋 ∶ |dom 휏| = 푘 }.

(Notice that 퐽푛 = S푋.) Fix an element 훽 of 퐽푛−1. a) Let 훾 ∈ 퐽푛−1 and let 휋 ∶ dom 훾 → dom 훽 be a bijection. Prove that there exists 휌 ∈ S푋 such that 휋훽휌 = 훾. Deduce that 퐽푛−1 ⊆ ⟨휏, 휁, 훽⟩.

b) Prove that 퐽푘 ⊆ 퐽푘+1퐽푛−1 for 푘 = 0, 1, … , 푛 − 2. Deduce that I푋 = ⟨휏, 휁, 훽⟩.

Exercises • 113 5.5 Let 푋 be an infinite set.

a) Let 휏 ∈ I푋 be such that dom 휏 = 푋 and im 휏 ⊊ 푋. (So 휏 is a bijection from 푋 to a proper subset of itself.) Prove that ⟨휏, 휏−1⟩ is isomorphic to the bicyclic monoid.

b) Let 퐼 be an abstract index set with |퐼| ⩾ 2 and let { 휏푖 ∶ 푖 ∈ 퐼 } ⊆ I푋 be a collection of partial bijections such that dom 휏푖 = 푋 and all the images im 휏푖 are disjoint. (So the 휏푖 are bijections from 푋 to −1 disjoint subsets of 푋.) Prove that ⟨{ 휏푖, 휏푖 ∶ 푖 ∈ 퐼 }⟩ is an inverse monoid isomorphic to the monoid

Mon⟨푧, 푏 , 푐 for 푖 ∈ 퐼 ∣ (푏 푐 , 휀), (푏 푐 , 푧), 푖 푖 푖 푖 푖 푗 } } (푏 푧, 푧), (푧푏 , 푧), } 푖 푖 (5.14) } (푐푖푧, 푧), (푧푐푖, 푧), (푧푧, 푧) } for 푖, 푗 ∈ 퐼 with 푖 ≠ 푗⟩. }

[These monoids are called the polycyclic monoids.] Polycyclic monoid ✴5.6 A semigroup is orthodox if it is regular and its set of idempotents form Orthodox a subsemigroup. a) Prove that a Clifford semigroup is orthodox. b) Prove that a semigroup is completely simple and orthodox if and only if it is isomorphic to the direct product of a rectangular band and a group. ✴5.7 Prove that a completely 0-simple semigroup is inverse if and only if it is isomorphic to M0[퐺; 퐼, 퐼; 푃] where 푃 is a diagonal 퐼 × 퐼 matrix.

✴5.8 Let 푆 be a cancellative semigroup. An element 휏 of I푆1 is a partial right Partial right translation translation if dom 휏 is a left ideal of 푆1 and for any 푥 ∈ dom 휏 and 푦 ∈ 푆1, we have (푦푥)휏 = 푦(푥휏).

a) Prove that if 휏 ∈ I푆1 is a partial right translation, then im 휏 is a left ideal of 푆1. 1 1 b) Note that for each 푥 ∈ 푆, the map 휌푥 ∶ 푆 → 푆 (where 푡휌푥 = 푡푥) is injective and so lies in I푆1 . Let 휑 ∶ 푆 → I푆 be the homomorphism defined by 푥 ↦ 휌푥. Let 푇 be the inverse subsemigroup of I푆1 generated by im 휑. Prove that the set of partial right translations in I푆1 is an inverse subsemigroup of of I푆1 and contains 푇. ✴5.9 Prove that the bicyclic monoid 퐵 = Mon⟨푏, 푐 | (푏푐, 휀)⟩ is an inverse semigroup. [Hint: use the characterization of idempotents in Exercise 2.10(a).] ✴5.10 Let 푆 be an inverse semigroup, and let 푥 ∈ 푆 and 푒 ∈ 퐸(푆). Prove that 푥 ≼ 푒 ⇒ 푥 ∈ 퐸(푆). 5.11 Prove that the FInvM({푎}) is isomorphic to the set

퐾 = { (푝, 푞, 푟) ∶ 푝, 푞, 푟 ∈ ℤ, 푝 ⩽ 0, 푟 ⩾ 0, 푝 ⩽ 푞 ⩽ 푟 }

Exercises • 114 with the operation

(푝, 푞, 푟)(푝′, 푞′, 푟′) = (min{푝, 푝′ + 푞}, 푞 + 푞′, max{푟, 푞 + 푟′}).

[Hint: each element of FInvM({푎}) corresponds to a Munn tree 푇 of the form 푎 푎 푎 푎 푎 푎 훼푇 휔푇

View vertices along this path as having an ‘푥-coordinate’ relative to 훼푇. Let 푝, 푞, and 푟 be, respectively, the 푥-coordinates of the leftmost endpoint, the vertex 휔푇, and the rightmost endpoint.] 5.12 Using Exercise 5.11 and the map

휑 ∶ 퐾 → 퐵 × 퐵; (푝, 푞, 푟)휑 = (푐−푝푏−푝+푞, 푐푟푏−푞+푟).

prove that FInvM({푎}) is a subdirect product of two copies of the bicyclic monoid. 5.13 Let 푀 be a monoid presented by Mon⟨퐴 | 휌⟩. Let 휑 ∶ 푀 → 푀 be an Bruck–Reilly extensions endomorphism. The Bruck–Reilly extension of 푀 with respect to 휑, denoted BR(푀, 휑), is the monoid presented by

Mon⟨퐴 ∪ {푏, 푐} ∣ } (5.15) 휌 ∪ { (푏푐, 휀), (푏푎, (푎휑)푏), (푎푐, 푐(푎휑)) ∶ 푎 ∈ 퐴 }⟩,

where we view 푎휑 in the defining relations as some word in 퐴∗ representing that element of 푀. a) Prove that every element of BR(푀휑) is represented by a word of the form 푐훾푤푏훽, where 훾, 훽 ∈ ℕ ∪ {0} and 푤 ∈ 퐴∗.

b) i) Prove that if 훾 = 훾′, and 훽 = 훽′, and 푤 =푀 푤′, then we have 훾 훽 훾′ 훽′ 푐 푤푏 =BR(푀,휑) 푐 푤′푏 . ii) Let

푋 = (ℕ ∪ {0}) × 푀 × (ℕ ∪ {0}) = { (훾, 푤, 훽) ∶ 훾, 훽 ∈ ℕ ∪ {0}, 푤 ∈ 푀 }.

Define

훽 (훾, 푤, 훽)휏푎 = (훾, 푤(푎휑 ), 훽) for each 푎 ∈ 퐴;

(훾, 푤, 훽)휏푏 = (훾, 푤, 훽 + 1); (훾 + 1, 푤휑, 0) if 훽 = 0, (훾, 푤, 훽)휏 = { 푐 (훾, 푤, 훽 − 1) if 훽 > 0.

Prove that the map 휓 ∶ 퐴 → T푋 given by 푥휓 = 휏푥 for all 푥 ∈ 퐴 ∪ {푏, 푐} extends to a well-defined homomorphism 휓 ∶

Exercises • 115 BR(푀, 휑) → T푋. Prove that the homomorphism 휓 is injective. 훾 훽 훾′ 훽′ Deduce that if 푐 푤푏 =BR(푀,휑) 푐 푤′푏 , then 훾 = 훾′, 훽 = 훽′, and 푤 =푀 푤′. [Note that, since 휓 is injective, BR(푀, 휑) is isomorphic to a particular subsemigroup of T푋. Since this subsemigroup is independent of the choice of the presentation Mon⟨퐴 | 휌⟩ for 푀, the Bruck–Reilly extension BR(푀, 휑) is also is independent of the choice of the presentation for 푀.] c) Deduce that 푀 embeds into BR(푀, 휑). 5.14 Let 푀 be a monoid and let 휑 ∶ 푀 → 푀 be defined by 푥휑 = 1 for all 푥 ∈ 푀. Prove that BR(푀, 휑) is simple. [Thus, as a consequence of Exercise 5.13, every semigroup 푆 embeds into a simple semigroup BR(푆1, 휑).]

Notes

The exposition of the Vagner–Preston representation theorem is based on Clifford & Preston, The Algebraic Theory of Semigroups, § 1.9 and Howie, Fundamentals of Semigroup Theory, § 5.1. The discussion of Clifford semigroups is based on Howie, Fundamentals of Semigroup Theory, §§ 4.1–2. ◆ The introduction of free inverse semigroups follows Lawson, Inverse Semi- groups, ch. 6; the explanation of Munn trees follows closely Munn, ‘Free Inverse Semigroups’ (which is a model of clarity) except that we consider free inverse monoids rather than free inverse semigroups. ◆ See Clifford & Preston, The Algebraic Theory of Semigroups, p. 28 for the quotation in the introduction. ◆ The Vagner–Preston theorem (Theorem 5.8), and much of the basic theory of inverse semigroups, is found in Vagner, ‘Generalized groups’ and Preston, ‘Inverse semi-groups with minimal right ideals’; Preston, ‘Representations of inverse semi-groups’. The structure theorem for Clifford semigroups is due to Clifford, ‘Semigroups admitting relative inverses’, though the terminology is later. ◆ For further reading, the standard text on inverse semigroups remains Petrich, In- verse Semigroups, but Lawson, Inverse Semigroups provides a geometric and topological perspective. •

Notes • 116 Commutative semigroups 6 The two operations, suicide and going to MIT, didn’t commute ‘ — Murray Gell-Mann,’ ‘The Making of a Physicist’. In: Edge.org.

• Abelian groups (that is, commutative groups) have a simpler structure and are better understood than general groups, especially in the finitely generated case. It is therefore unsurprising that commutative semigroups also have a well-developed theory. However, there are still many more commutative semigroups than abelian groups. For instance, there are three essentially different (non-isomorphic) abelian groups with 8 elements (the cyclic group 퐶8 and the direct products 퐶4 × 퐶2 and 퐶2 × 퐶2 × 퐶2), but there are 221 805 non-isomorphic commutative semigroups with 8 elements. A large theory of commutative semigroups has developed, and we will sample only two areas: first, in structure theory, how cancellative commutative semigroups are group-embeddable; second, free commutative semigroups and their congruences, leading to the result that finitely generated semigroups are always finitely presented.

Cancellative commutative semigroups Example 2.14 showed that a cancellative semigroup is not necessarily group-embeddable. However, in this section we will see that a cancellative commutative semigroup is always group-embeddable. The method used to construct the group from the cancellative semigroup is essentially the same as that used to construct a field from an integral domain (for example, to construct (ℚ, +, ⋅ ) from (ℤ, +, ⋅ )).

T h e o r e m 6 . 1. Let 푆 be a cancellative commutative semigroup. Then Cancellative commutative 푆 embeds into a group 퐺 via a monomorphism 휑 ∶ 푆 → 퐺 such that semigroups are 퐺 = (푆휑)(푆휑)−1 = { 푥푦−1 ∶ 푥, 푦 ∈ 푆 }. group-embeddable.

Proof of 6.1. First of all, note that 푆 embeds into 푆1 and that if 퐺 = 푆푆−1, then 퐺 = 푆1(푆1)−1, and so we assume without loss of generality that 푆 is a monoid.

• 117 Define a relation 휎 on 푆 × 푆 by (푥, 푦) 휎 (푧, 푡) ⇔ 푥푡 = 푧푦. It is trivial to prove 휎 is reflexive and symmetric since 푆 is commutative, and 휎 is transitive since (푥, 푦) 휎 (푧, 푡) ∧ (푧, 푡) 휎 (푝, 푞) ⇒ 푥푡 = 푧푦 ∧ 푧푞 = 푝푡 ⇒ 푥푡푧푞 = 푧푦푝푡 ⇒ 푥푞 = 푝푦 [since 푆 is cancellative and commutative] ⇒ (푥, 푦) 휎 (푝, 푞). Thus 휎 is an equivalence relation. Furthermore, (푥, 푦) 휎 (푧, 푡) ∧ (푥′, 푦′) 휎 (푧′, 푡′) ⇒ 푥푡 = 푧푦 ∧ 푥′푡′ = 푧′푦′ ⇒ 푥푡푥′푡′ = 푧푦푧′푦′ ⇒ 푥푥′푡푡′ = 푧푧′푦푦′ [since 푆 is commutative] ⇒ (푥푥′, 푦푦′) 휎 (푧푧′, 푡푡′). Thus 휎 is a congruence. Let 퐺 = (푆 × 푆)/휎. Let [(푥, 푦)]휎 ∈ 퐺; then (1푆푥)푦 = (1푆푦)푥 since 푆 is commutative. Hence (1푆푥, 1푆푦) 휎 (푥, 푦) and thus [(1푆, 1푆)]휎[(푥, 푦)]휎 = [(1푆푥, 1푆푦)]휎 = [(푥, 푦)]휎. Similarly, [(푥, 푦)]휎[(1푆, 1푆)]휎 = [(푥, 푦)]휎. So 퐺 is a monoid with identity [(1푆, 1푆)]휎. Furthermore, 1푆(푥푦) = (푦푥)1푆, since 푆 is commutative, and therefore (푥푦, 푦푥) 휎 (1푆, 1푆). Hence [(푥, 푦)]휎[(푦, 푥)]휎 = [(푥푦, 푦푥)]휎 = [(1푆, 1푆)]휎 and similarly [(푦, 푥)]휎[(푥, 푦)]휎 = [(1푆, 1푆)]휎. Thus [(푦, 푥)]휎 is a left and right inverse for [(푥, 푦)]휎. So 퐺 is a group. Let 휑 ∶ 푆 → 퐺 be defined by 푠휑 = [(푠, 1푆)]휎. It is clear that 휑 is a homomorphism. Furthermore, 휑 is injective since

푥휑 = 푦휑 ⇒ [(푥, 1푆)]휎 = [(푦, 1푆)]휎

⇒ 푠1푆 = 푡1푆 ⇒ 푠 = 푡. Therefore 푆 embeds into 퐺. Finally, note that

−1 −1 [(푥, 푦)]휎 = [(푥, 1푆)]휎[(1푆, 푦)]휎 = (푥휑)(푦휑) ∈ (푆휑)(푆휑) ; −1 hence 퐺 = (푆휑)(푆휑) . 6.1 Let 푆 be a commutative cancellative semigroup. By Theorem 6.1, there is a monomorphism 휑 from 푆 into a group 퐺 such that 퐺 = (푆휑)(푆휑)−1. We can therefore identify 푆 with a subsemigroup of 퐺 such that 퐺 = 푆푆−1. For any commutative cancellative semigroup 푆, denote by 퐺(푆) some fixed group containing 푆 as a subsemigroup, such that 퐺(푆) = 푆푆−1. (Actually, one can prove that 퐺(푆) is unique up to isomorphism, but we will not need this result.)

Cancellative commutative semigroups • 118 Free commutative semigroups Let 퐴 be an alphabet. Let FCommS(퐴) be semigroup presented by Sg⟨퐴 | 휌⟩, where

휌 = { (푎푏, 푏푎) ∶ 푎, 푏 ∈ 퐴 }.

The following result is essentially immediate:

P r o p o s i t i o n 6 . 2. FCommS(퐴) is a commutative semigroup. 6.2

Let 퐹 be a commutative semigroup, let 퐴 be an alphabet, and let 휄 ∶ Free commutative 퐴 → 퐹 be an embedding of 퐴 into 퐹. Then the commutative semigroup 퐹 semigroups is the free commutative semigroup on 퐴 if, for any commutative semigroup 푆 and map 휑 ∶ 퐴 → 푆, there is a unique homomorphism 휑 ∶ 퐹 → 푆 that extends 휑 (that is, with 휄휑 = 휑). Using diagrams, this definition says that 퐹 is a free commutative semigroup on 퐴 if

휄 퐴 퐹 } } for all with 푆 commutative, there exists } } 휑 } 휄 푆 퐴 퐹 } (6.1) } } a unique homomorphism 휑 such that 휑 . } 휑 } 푆 }

This definition is analogous to the definition of the free semigroup on 퐴 (see pages 37 f.) and free inverse semigroup on 퐴 (see page 104). As already noted, in Chapter 8, we will see definitions of ‘free objects’ in a much more general setting. Reasoning analogous to the proof of Proposition 5.15 establishes the following result:

P r o p o s i t i o n 6 . 3. Let 퐴 be an alphabet and let 퐹 be a commutative Uniqueness of the semigroup. Then 퐹 is a free commutative semigroup on 퐴 if and only if free commutative semigroup on 퐴 퐹 ≃ FCommS(퐴). 6.3

As in the discussions of ‘free’ and ‘free inverse’, we could repeat the Free commutative monoids reasoning above, but for monoids instead of semigroups. The monoid FCommM(퐴) is presented by Mon⟨퐴 ∪ 퐴−1 | 휌⟩. A monoid 퐹 is a free commutative monoid on 퐴 if, for any commutative monoid 푆 and map 휑 ∶ 퐴 → 푆, there is a unique monoid homomorphism 휑 ∶ 퐹 → 푆 extending 휑, with 휄휑 = 휑. A commutative monoid 퐹 is a free commutative monoid on 퐴 if and only if 퐹 ≃ FCommM(퐴). We have FCommM(퐴) ≃ (FCommS(퐴))1.

P r o p o s i t i o n 6 . 4. Let 퐴 be a finite alphabet. Then FCommM(퐴) ≃ (ℕ ∪ {0})퐴.

Free commutative semigroups • 119 (Recall the notation for cartesian and direct products from page 4.) Proof of 6.4. Following Method 2.9, we aim to prove that Mon⟨퐴 | 휌⟩ presents (ℕ ∪ {0})퐴. Define a map 휑 ∶ 퐴 → (ℕ ∪ {0})퐴, where 푎휑 is such that (푎)(푎휑) = 1 and (푥)(푎휑) = 0 for 푥 ≠ 푎. (That is, 푎휑 is the tuple whose 푎-th component is 1 and all other components 0.) Clearly (ℕ ∪ {0})퐴 satisfies the defining relations in 휌 with respect to 휑. Suppose 퐴 = {푎1, … , 푎푛} and let

푘1 푘2 푘푛 푁 = { 푎1 푎2 ⋯ 푎푛 ∶ 푘1, 푘2, … , 푘푛 ∈ ℕ ∪ {0} }.

It is obvious that every word in 퐴∗ can be transformed to one in 푁 by 푘1 푘2 푘푛 applying defining relations from 휌. Finally, since (푎푖)((푎1 푎2 ⋯ 푎푛 )휑), 퐴 we see that 휑|푁 is injective. So Mon⟨퐴 | 휌⟩ presents (ℕ ∪ {0}) . 6.4

Proposition 6.4 does not hold if 퐴 is infinite. The tuples 푎휑 as defined in the proof do not generate (ℕ ∪ {0})퐴 when 퐴 is infinite, because no (finite) product of these tuples is equal to (for example) the tuple with all components 1.

P r o p o s i t i o n 6 . 5. Let 푆 be a finite generated commutative semigroup (respectively, commutative monoid), let 휑 ∶ 퐴 → 푆 be an assignment of generators (with 퐴 finite), and let 휑 ∶ FCommS(퐴) → 푆 (respectively, 휑 ∶ FCommM(퐴) → 푆) be the homomorphism extending 휑. Suppose there is a finite set 휎 ⊆ ker 휑 such that 휎# = ker 휑. Then 푆 is finitely presented.

Proof of 6.5. We prove the result for semigroups; the reasoning for monoids is similar. Let 휑+ ∶ 퐴+ → 푆 be the homomorphism extending 휑. For brevity, let 휓 be the natural homomorphism (휌#)♮ ∶ 퐴+ → FCommS(퐴), # ♮ # ♮ # where 푎휓 = 푎(휌 ) = [푎]휌# . Note that ker 휓 = ker(휌 ) 휌 . The following diagram commutes:

휓 = (휌#)♮ 퐴 휄 퐴+ FCommS(퐴) 휑+ 휑 휑 푆

To show that 푆 is finitely presented, we must find a finite subset of B퐴+ that generates ker 휑+. # ♮ −1 # ♮ −1 For each (푥, 푦) ∈ 휎, fix 푤푥 ∈ 푥((휌 ) ) and 푤푦 ∈ 푦((휌 ) ) and let 휎̂ = { (푤푥, 푤푦) ∶ (푥, 푦) ∈ 휎) }. Note that 휎̂ is finite since 휎 is finite. We will prove that (휎̂ ∪ 휌)# = ker(휓휑) = ker 휑+. −1 Make the following definition: for any 휏 ∈ B푇, let 휏휓 = { (푠, 푡) ∈ 푆 × 푆 ∶ (푠휓, 푡휓) ∈ 휏 }. Let (푢, 푣) ∈ (휎C)휓−1. So (푢휓, 푣휓) ∈ 휎C. So by Proposition 1.27, there exist 푝, 푞 ∈ FCommS(퐴) and (푥, 푦) ∈ 휎 such that 푢휓 = 푝푥푞 and

Free commutative semigroups • 120 푣휓 = 푝푦푞. Let 푝′, 푞′ ∈ 푆 be such that 푝′휓 = 푝 and 푞′휓 = 푞. Then (푝′푤푥푞′)휓 = 푢휓 and (푝′푤푦푞′)휓 = 푣휓. Therefore (푢, 푝′푤푥푞′) ∈ ker 휓 = # # # 휌 and (푝′푤푦푞, 푣) ∈ ker 휓 = 휌 . Since (푝′푤푥푞′, 푝′푤푦푞′) ∈ 휎̂ , we have

# # # 푢 휌 푝′푤푥푞′ 휎̂ 푝′푤푦푞′ 휌 푣, and so (푢, 푣) ∈ (휎̂ ∪ 휌)#. Thus (휎C)휓−1 ⊆ (휎̂ ∪ 휌)#. Since (휎̂ ∪ 휌)# is symmetric, (휎C)휓−1 ∪ (휎C)−1휓−1 ⊆ (휎̂ ∪ 휌)#. Now let 푢, 푣 ∈ 퐴+. Then

(푢, 푣) ∈ ker(휓휑) ⇒ (푢휓, 푣휓) ∈ 휎# ∞ C C −1 푛 ⇒ (푢휓 = 푣휓) ∨ (푢휓, 푣휓) ∈ ⋃푛=1(휎 ∪ (휎 ) ) [by Proposition 1.26(f)]

⇒ (∃푛 ∈ ℕ ∪ {0})(∃푤0, … , 푤푛 ∈ FCommS(퐴))

[(푢휓 = 푤0) ∧ (푤푛 = 푣휓) C C −1 ∧ (∀푖)((푤푖, 푤푖+1) ∈ 휎 ∪ (휎 ) )] + ⇒ (∃푛 ∈ ℕ ∪ {0})(∃푤′0, … , 푤′푛 ∈ 퐴 )

[(푢휓 = 푤′0휓) ∧ (푤′푛휓 = 푣휓) C −1 C −1 −1 ∧ (∀푖)((푤′푖, 푤′푖+1) ∈ 휎 휓 ∪ (휎 ) 휓 )] [since 휓 is surjective] + ⇒ (∃푛 ∈ ℕ ∪ {0})(∃푤′0, … , 푤′푛 ∈ 퐴 )

[(푢휓 = 푤′0휓) ∧ (푤′푛휓 = 푣휓) # ∧ (∀푖)((푤′푖휓, 푤′푖+1) ∈ (휎′ ∪ 휌) )] [since (휎C)휓−1 ∪ (휎C)−1휓−1 ⊆ (휎̂ ∪ 휌)#] ⇒ (푢, 푣) ∈ (휎′ ∪ 휌)#)].

Therefore ker(휓휑) ⊆ (휎′ ∪ 휌)#. On the other hand, if (푢, 푣) ∈ 휎′, then (푢휓, 푣휓) ∈ 휎 ⊆ ker 휑, so 푢휓휑 = 푣휓휑 and so (푢, 푣) ∈ ker(휓휑). If (푢, 푣) ∈ 휌 ⊆ ker 휓, then 푢휓 = 푣휓, so 푢휓휑 = 푣휓휑 and so (푢, 푣) ∈ ker(휓휑). Thus 휎′ ∪ 휌 ⊆ ker(휓휑) and hence (휎′ ∪ 휌)# ⊆ ker(휓휑) since ker(휓휑) is a congruence. Therefore (휎′ ∪ 휌)# = ker(휓휑) = ker 휑+ and so 푆 is defined by the finite presentation Sg⟨퐴 | 휎′ ∪ 휌⟩. 6.5

Rédei’s theorem

When the alphabet 퐴 has 푛 elements, we write write 퐹푛 = FCommM(퐴) for brevity and (by Proposition 6.4) we view elements of

Rédei’s theorem • 121 FCommM(퐴) as 푛-tuples of non-negative integers from ℕ ∪ {0}. Define a relation ⩽ on 퐹푛 by

(푥1, … , 푥푛) ⩽ (푦1, … , 푦푛) ⇔ (∀푖 ∈ {1, … , 푛})(푥푖 ⩽ 푦푖).

It is easy to see that ⩽ is a partial order on 퐹푛. Notice that there are no infinite ⩽-decreasing sequences in 퐹푛.

T h e o r e m 6 . 6. Every antichain in 퐹푛 is finite. Dickson’s theorem Proof of 6.6. The proof is by induction on 푛. For the base of the induction, let 푌 be an antichain in ℕ ∪ {0} ≃ 퐹1. Since ⩽ is a total order on ℕ ∪ {0}, every pair of elements is comparable and thus 푌 contains at most one element. Thus the result holds for 푛 = 1. Now suppose the result holds for all 푛 < 푘; we aim to prove it for 푘 푛 = 푘. Let 푌 ⊆ (ℕ ∪ {0}) ≃ 퐹푘 be an antichain. For each 푡 ∈ ℕ ∪ {0} and for each 푖 ∈ {1, … , 푘}, let

푌푖,푡 = { (푥1, … , 푥푘) ∈ 푌 ∶ 푥푖 = 푡 }. The set

{ (푥1, … , 푥푖−1, 푥푖+1, … , 푥푘) ∶ (푥1, … , 푥푖−1, 푡, 푥푖+1, 푥푘) ∈ 푌푖,푡 }. is an antichain (possibly empty) of 퐹푘−1 and therefore finite by the induction hypothesis. Hence each set 푌푖,푡 is finite. Fix some 푦 ∈ 푌, with 푦 = (푦1, … , 푦푘). Let 푧 = (푧1, … , 푧푘) ∈ 푌 ∖ {푦}. Since 푌 is an antichain, 푦 ≰ 푧. Hence 푦푖 > 푧푖 for some 푗 ∈ {푖, … , 푘}. Hence 푛

푌 = {푦} ∪ ⋃{ (푧1, … , 푧푘) ∈ 푌 ∶ 푧푖 < 푦푖 } 푖=1 푛 푦푗−1

= {푦} ∪ ⋃ ⋃ { (푧1, … , 푧푘) ∈ 푌 ∶ 푧푖 = 푡 } 푖=1 푡=0 푛 푦푗−1

= {푦} ∪ ⋃ ⋃ 푌푖,푡. 푖=1 푡=0

Each set 푌푖,푡 is finite, and so 푌 itself is finite. Since 푌 was an arbitrary antichain in 퐹푘, this establishes the induction step and so proves the result. 6.6

P r o p o s i t i o n 6 . 7. Let 휎 be a congruence on 퐹푛. Then there is a finite set 휌 ⊆ 휎 such that 휌# = 휎.

Proof of 6.7. Define the lexicographic order ⊑ on 퐹푛 by

(푥1, … , 푥푛) ⊏ (푦1, … , 푦푛) ⇔ (∃푘 ∈ {1, … , 푛})

(푥푘 < 푦푘 ∧ (∀푗 < 푘)(푥푗 = 푦푘)).

Rédei’s theorem • 122 Then ⊑ is a total order on 퐹푛 and is compatible. (That is, 푥 ⊑ 푦 ⇒ 푥푧 ⊑ 푦푧 for all 푥, 푦, 푧 ∈ 퐹푛.) Furthermore, ⊑ is a well-order (that is, every non- empty subset of 퐹푛 has a ⊑-minimum element). In particular, every 휎-class [푥]휎 has a ⊑-minimum element 푞푥. Let

푄 = { 푞푥 ∶ 푥 ∈ 퐹푛 } = { 푦 ∈ 퐹푛 ∶ (∀푥 ∈ 퐹푛)(푦 휎 푥 ⇒ 푦 ⊑ 푥) }. So 푄 consists of the ⊑-minimal elements of all the 휎-classes. Let 푅 be the complement of 푄 in 퐹푛; that is,

푅 = { 푥 ∶ 푞푥 ⊏ 푥 } = { 푥 ∈ 퐹푛 ∶ (∃푦 ∈ 퐹푛)(푦 휎 푥 ∧ 푦 ⊏ 푥) }. Then 푅 consists of the non-⊑-minimal elements of all the 휎-classes. Fur- thermore,

(푥 ∈ 푅) ∧ (푧 ∈ 퐹푛) ⇒ (푞푥 ⊏ 푥) ∧ (푧 ∈ 푆) ⇒ 푞푥푧 ⊏ 푥푧 ⇒ 푥푧 ∈ 푅; hence 푅 is an ideal of 퐹푛. Let 푀 be the set of ⩽-minimal elements of 푅. Then 푀 is an antichain and so finite by Theorem 6.6. Let

휌 = { (푞푚, 푚) ∶ 푚 ∈ 푀 }. Notice that 휌 is finite because 푀 is finite. # The aim is now to show that 휌 = 휎. Since 푞푚 휎 푚 for each 푚 ∈ 푀, it is immediate that 휌 ⊆ 휎 and so 휌# ⊆ 휎. # # To prove that 휎 ⊆ 휌 , the first step is to prove that 푞푥 휌 푥 for all 푥 ∈ 퐹푛. # Suppose, with the aim of obtaining a contradiction, that (푞푥, 푥) ∉ 휌 for some 푥 ∈ 퐹푛. Then, since ⊑ is a well-order, there is a ⊑-minimum 푠 ∈ 퐹푛 # such that (푞푠, 푠) ∉ 휌 . This element 푠 cannot be in 푄, since otherwise # 푞푠 = 푠 and so (푞푠, 푠) ∈ 휌 . Furthermore, 푠 cannot be in 푀, since otherwise # (푞푠, 푠) ∈ 휌 by definition and hence (푞푠, 푠) ∈ 휌 . Thus 푠 ∈ 푅 ∖ 푀 and so 푠 > 푚 for some 푚 ∈ 푀. Therefore 푠 = 푚푡 for some 푡 ∈ 퐹푛. # Let 푢 = 푞푚푡. Since (푞푚, 푚) ∈ 휌 and (푞푚, 푚) ∈ 휎, we have (푢, 푠) = # (푞푚푡, 푚푡) ∈ 휌 and so (푢, 푠) ∈ 휎. Notice that (푢, 푠) ∈ 휎 implies 푞푢 = 푞푠. Furthermore, 푢 ⊏ 푠 since 푞푚 ⊏ 푚 and ⊑ is compatible. Since 푠 is # # ⊑-minimum with (푞푠, 푠) ∉ 휌 , it follows that (푞푢, 푢) ∈ 휌 . Therefore # # # 푠 휌 푢 휌 푞푥 = 푞푠. Thus (푞푠, 푠) ∈ 휌 , which is a contradiction, and hence # (푞푥, 푥) ∈ 휌 for all 푥 ∈ 퐹푛. Finally, let (푥, 푦) ∈ 휎. Then 푞푥 = 푞푦. By the previous paragraph, # # # # (푞푥, 푥) and (푞푦, 푦) are in 휌 . Thus 푥 휌 푞푥 = 푞푦 휌 푦; hence (푥, 푦) ∈ 휌 . # # That is, 휎 ⊆ 휌 , and therefore 휎 = 휌 . 6.7 The following result is immediate from Propositions 6.5 and Proposi- tion 6.7:

R é d e i ’ s T h e o r e m 6 . 8. Every finitely generated commutative mon- Rédei’s theorem oid is finitely presented. 6.8

Rédei’s theorem • 123 Exercises [See pages 229–232 for the solutions.] ✴6.1 Let 푆 be a commutative semigroup. Let 퐺 and 퐺′ be abelian groups such that 퐺 = 푆푆−1 and 퐺′ = 푆푆−1. Prove that there is an isomorphism 휓 ∶ 퐺 → 퐺′ such that 휓|푆 maps 푆 ⊆ 퐺 to 푆 ⊆ 퐺′. ✴6.2 Let 푆 be a commutative semigroup. Let 퐼 be an ideal of 푆, and let 퐺 be an abelian group. Let 휑 ∶ 퐼 → 퐺 be a homomorphism. Prove that there is a unique extension of 휑 to a homomorphism 휑̂ ∶ 푆 → 퐺. 6.3 Let 푆 be a non-trivial subsemigroup of (ℕ ∪ {0}, +). Prove that there exists 푑 ∈ ℕ ∪ {0} such that 푆 ⊆ 푑ℕ and 푑ℕ ∖ 푆 is finite. 6.4 Let 푆 be a subsemigroup of (ℤ, +). Prove that either every element of 푆 is non-negative, or every element of 푆 is non-positive, or 푆 is a subgroup. ✴6.5 A semigroup 푆 is right-reversible (respectively, left-reversible) if every Right- and left-reversibility two elements of 푆 have a common left (respectively, right) multiple; that is, if for all 푥, 푦 ∈ 푆, there exist 푧, 푡 ∈ 푆1 such that 푧푥 = 푡푦 (respectively, 푥푧 = 푦푡). Let 푆 be a cancellative right-reversible semigroup; the aim of this Ore’s theorem exercise is to prove that 푆 is group-embeddable; this is Ore’s theorem, and generalizes Theorem 6.1. Let 휑 ∶ 푆 → I푆 be the homomorphism defined by 푥 ↦ 휌푥. Let 푇 be the inverse subsemigroup of I푆 generated by im 휑. By Exercise 5.8(b), every element of 푇 is a partial right translation. Define a relation ∼ on 푇 by

훼 ∼ 훽 ⇔ (∃훿 ∈ 푇)((훿 ⊆ 훼) ∧ (훿 ⊆ 훽))

for all 훼, 훽 ∈ 푇. Notice that

훿 ⊆ 훼 ⇔ ((dom 훿 ⊆ dom 훼) ∧ (∀푥 ∈ dom 훿)(푥훿 = 푥훼)).

a) Prove that ∼ is an congruence. b) Let 퐺 = 푇/∼. Prove that 퐺 is a group. c) Let 훼, 훽 ∈ 푇. Prove that 훼 ∘ 훽 is not the empty relation, and so deduce that 푇 does not contain the empty relation. ♮ d) Let 휓 = 휑 ∘ ∼ (that is, 푥휓 = [푥휑]∼). Prove that 휓 is a monomorphism and so deduce that 푆 is group-embeddable. ✴6.6 Let 푆 = (ℕ ∪ {0}) × (ℕ ∪ {0}) and define a multiplication on 푆 by

(푚, 푛)(푝, 푞) = (푚 + 푝, 2푝푛 + 푞)

Check that this multiplication is associative, so that 푆 is a semigroup. Prove that 푆 is left reversible but not right reversible.

Exercises • 124 Notes

The number of commutative semigroups of with 8 elements is from Grillet, Commutative Semigroups, p. 1. ◆ Rédei’s theorem (Theorem 6.8) was first proved by Rédei, The Theory of Finitely Generated Commutative Semigroups; see also Clifford & Preston, The Algebraic Theory of Semigroups, § 9.3. The proof given here is from Grillet, ‘A short proof of Rédei’s theorem’. ◆ Ore’s theorem (Exercise 6.5) is contained in a theorem about rings proved, using different terminology, in Ore, ‘Linear equations in non-commutative fields’; the proof here is due to Rees, ‘On the group of a set of partial transformations’. ◆ For further reading, Grillet, Commutative Semigroups is a comprehensive monograph, but with a very terse style, and Rosales & García-Sánchez, Finitely Generated Commutative Monoids is an accessible introduction to structural and computational aspects. •

Notes • 125 Finite semigroups 7 The known is finite, the unknown infinite; intellectually we standon ‘ an islet in the midst of an illimitable ocean of inexplicability. — T.H.’ Huxley, On the Reception of the Origin of Species.

• In this chapter, we begin the detailed study of finite semigroups. Although Green’s relations will play a role, other techniques are used to understand finite semigroups. In particular we will introduce the notion of divisibility, where one semigroup is a homomorphic image of a subsemigroup of another. The goal of this chapter is to prove the Krohn–Rhodes theorem, which says that every finite semigroup divides a wreath product of finite groups and finite aperiodic semigroups, which, as we shall see, are finite semigroups where all subgroups are trivial. This leads naturally into the classification of finite semigroups by means of pseudovarieties, which is the topic of next chapter.

Green’s relations and ideals As a consequence of Proposition 3.3, we know that the Green’s relations D and J coincide for finite semigroups. P r o p o s i t i o n 7 . 1. Let 푀 be a finite monoid with identity 1. Then 퐻1 = 퐿1 = 푅1 = 퐷1 = 퐽1. Furthermore, 퐻1 is the group of units of 푀, and 푀 ∖ 퐻1 is either empty or an ideal of 푀. 1 Proof of 7.1. Let 푥 ∈ 푅1. Then there exists 푦 ∈ 푀 = 푀 such that 푥푦 = 1. Since 푀 is finite, 푥푚+푘 = 푥푚 for some 푚, 푘 ∈ ℕ. Then 푥푘 = 푥푚+푘푦푚 = 푥푚푦푚 = 1, and so 푦푥 = 1푦푥 = 푥푘푦푥 = 푥푘−1푥 = 푥푘 = 1 Hence 푥 H 1. Therefore 푅1 ⊆ 퐻1. The opposite inclusion is obvious, so 푅1 = 퐻1. Similarly 퐿1 = 퐻1. So 퐷1 contains only one L-class and only one R-class and so 퐷1 = 퐻1. Finally, 퐽1 = 퐻1 since D = J. This reasoning also shows that 퐻1 is contained in the group of units of 푀. On the other hand, all elements of group of units of 푀 are H-related to 1, so the group of units of 푆 is 퐻1. For any 푦 ∈ 푀 ∖ 퐻1 = 푀 ∖ 퐽1, we have 퐽푦 < 퐽1 by (3.2). So 푀 ∖ 퐻1 = 퐼(1) = { 푦 ∈ 푆 ∶ 퐽푦 < 퐽1 }, which is either empty or an ideal by Lemma 3.9. 7.1

• 126 P r o p o s i t i o n 7 . 2. Let 푆 and 푆′ be finite semigroups and let 휑 ∶ 푆 → 푆′ be a surjective homomorphism. Let 퐺′ be a maximal subgroup of 푆′. Then there is a maximal subgroup 퐺 of 푆 such that 퐺휑 = 퐺′.

Proof of 7.2. Let 퐺′ be a maximal subgroup of 푆′. Then 푇 = 퐺′휑−1 is a subsemigroup of 푆 and 푇휑 = 퐺′. Since 푇 is finite, it has a kernel; let 퐾 = 퐾(푆), which is a simple ideal of 푇 by Proposition 3.10. Since 휑 is surjective, 퐾휑 is an ideal of the group 퐺′ and so 퐾휑 = 퐺′. Since 퐾 is finite it is also completely simple by Proposition 4.10. So, by Theorem 4.11, 퐾 ≃ M[퐺; 퐼, 훬; 푃] for some group 퐺, index sets 퐼 and 훬, and matrix 푃 over 퐺. View 휑|퐾 as a surjective homomorphism from M[퐺; 퐼, 훬; 푃] to 퐺′. For each 푖 ∈ 퐼 and 휆 ∈ 훬, let 퐺푖휆 be the subset {푖} × 퐺 × {휆} of M[퐺; 퐼, 훬; 푃]. Then M[퐺; 퐼, 훬; 푃] is the union of the various 퐺푖휆, and 퐺푖휆퐺푗휇 ⊆ 퐺푖휇. In particular, every 퐺푖휆 is a subgroup of 푇. Let 퐺′푖휆 = 퐺푖휆휑. Then each 퐺′푖휆 is a subgroup of 퐺′, and 퐺′푖휆퐺′푗휇 ⊆ 퐺′푖휇. In particular, 퐺′푖휆퐺′푗휆 ⊆ 퐺′푖휆, which implies 퐺′푗휆 = 1퐺′퐺′푗휆 ⊆ 퐺′푖휆. Similarly 퐺′푖휆 ⊆ 퐺′푖휇. Thus all the 퐺′푖휆 are equal. Since 휑 is surjective, 퐺′ is the union of the 퐺′푖휆 and thus equal to any one of the 퐺′푖휆. Hence 퐺′ = 퐺푖휆휑 for any 푖 ∈ 퐼 and 휆 ∈ 훬. 7.2

P r o p o s i t i o n 7 . 3. Let 푆 be a finite semigroup and let 푥, 푦 ∈ 푆. If 푥 H 푦, then 푦 ∈ 푥퐺 for some subgroup 퐺 of 푆.

Proof of 7.3. Let 퐻 be an H-class of 푆. Apply Proposition 7.2 to the natural ♮ ♮ surjective homomorphism 휎퐻 ∶ Stab(퐻) → 훤(퐻) to see that 퐻 = 퐺휎퐻 for some subgroup 퐺 of Stab(퐻). By Proposition 3.24, we see that 푦 ∈ 푥 ⋅ 훤(퐻) = 푥퐺 for all 푥, 푦 ∈ 퐻. 7.3

A semigroup 푆 is aperiodic if for every 푥 ∈ 푆, there exists 푛 ∈ ℕ such Aperiodic semigroups that 푥푛 = 푥푛+1. Notice that aperiodic semigroup are actually periodic. For example, any semigroup of idempotents (such as a semilattice) satisfies 푥 = 푥2 and so is aperiodic.

P r o p o s i t i o n 7 . 4. Let 푆 be a finite semigroup. The following are Characterization of equivalent: aperiodic finite semigroups a) 푆 is aperiodic. b) all subgroups of 푆 are trivial;

c) H = id푆; Proof of 7.4. Part 1 [a) ⇒ b)]. Suppose 푆 is aperiodic. Let 퐺 be a subgroup 푚 푚+1 of 푆 and let 푥 ∈ 퐺. Then 푥 = 푥 for some 푚 ∈ ℕ. Hence 푥 = 1퐺 by cancellativity in 퐺. So 퐺 is trivial. Thus all subgroups of 푆 are trivial. Part 2 [b) ⇒ c)]. Suppose that all subgroups of 푆 are trivial. Let 퐻 be an H-class of 푆. Then 훤(퐻), which is a homomorphic image of a subgroup

Green’s relations and ideals • 127 of Stab(퐻), is trivial. Since |퐻| = |훤(퐻)|, it follows that 퐻 is trivial by Proposition 7.3. Hence H = id푆.

Part 3 [c) ⇒ a)]. Suppose that H = id푆. Let 푥 ∈ 푆. Since 푆 is finite, 푥푚 = 푥푚+푘 for some 푚, 푘 ∈ ℕ. The set of elements {푥푚, 푥푚+1, … , 푥푚+푘−1} is a subgroup and so all its elements are H-related. Since H = id푆, this set thus contains only one element, which implies 푘 = 1. Hence 푥푚 = 푥푚+1. Thus 푆 is aperiodic. 7.4 We end this section by proving the following two results, although we will not use them until Chapter 9:

L e m m a 7 . 5. Let 푆 be a finite semigroup and let 푛 ⩾ |푆|. Then 푆푛 = 푆퐸(푆)푆.

Proof of 7.5. Let 푒 ∈ 퐸(푆). Then 푆푒푆 = 푆푒푛−2푆 ⊆ 푆푛. Thus 푆퐸(푆)푆 ⊆ 푆푛. 푛 Let 푥 ∈ 푆 . Then 푥 = 푥1 ⋯ 푥푛, where 푥푖 ∈ 푆. Suppose first that all the products 푥1 ⋯ 푥푘 for 푘 ⩽ 푛 are distinct. Then every element of 푆 is equal to some product 푥1 ⋯ 푥푘. Hence, since 푆, being finite, contains at least one idempotent, some product 푒 = 푥1 ⋯ 푥푘 is idempotent. Hence 3 푥 = 푒푥푘+1 ⋯ 푥푛 = 푒 푥푘+1 ⋯ 푥푛 ∈ 푆푒푆 ⊆ 푆퐸(푆)푆. Now suppose that 푖 푥1 ⋯ 푥푘 = 푥1 ⋯ 푥ℓ for some 푘 < ℓ. Then 푥1 ⋯ 푥푘 = 푥1 ⋯ 푥푘(푥푘+1 ⋯ 푥ℓ) 푖 for all 푖 ∈ ℕ. Let 푖 be such that 푒 = (푥푘+1 ⋯ 푥ℓ) is idempotent. Then 3 푛 푥1 ⋯ 푥푛 = 푥1 ⋯ 푥푘푒푥ℓ+1 ⋯ 푥푛 = 푥1 ⋯ 푥푘푒 ∈ 푆푒푆 ⊆ 푆퐸(푆)푆. Thus 푆 ⊆ 푆퐸(푆)푆. 7.5

L e m m a 7 . 6. Let 푆 be a finite semigroup and let 푥, 푦 ∈ 푆. a) If 푥 D 푥푦, then 푥 R 푥푦. b) If 푥 D 푦푥, then 푥 L 푦푥.

Proof of 7.6. We prove only part a); dual reasoning gives part b). Suppose 푥 D 푥푦. Since D = J by Proposition 3.3, there exist 푝, 푞 ∈ 푆1 such that 푝푥푦푞 = 푥. Thus 푝푛푥(푦푞)푛 for all 푛 ∈ ℕ. Since 푆 is finite, there exists 푘 ∈ ℕ such that 푒 = 푝푘 is idempotent. Thus, in particular, 푒푥(푦푞)푘 = 푥, and also 푒푥 = 푒푒푥(푦푞)푘 = 푒푥(푦푞)푘 = 푥. Combining these 푘 gives 푥(푦푞) = 푥. So 푥 R 푥푦. 7.6

Semidirect and wreath products

Let 푆 and 푇 be semigroups and let 푇 act on 푆 from the Semidirect product left by endomorphisms; let 휑 ∶ 푇 → End(푆) be the anti-homomorphism corresponding to this left action. To avoid having to write extra brackets, we will write 푡푠 instead of 푡 ⋅ 푠. The semidirect product of 푆 and 푇 with

Semidirect and wreath products • 128 respect to 휑 is denoted 푆 ⋊휑 푇 and is the cartesian product 푆 × 푇 with multiplication defined by

푡1 (푠1, 푡1)(푠2, 푡2) = (푠1 푠2 , 푡1푡2). (7.1)

This multiplication is associative (see Exercise 7.6) and so 푆 ⋊휑 푇 is a semigroup. Notice that 푆 ⋊휑 푇 has cardinality |푆||푇|. Notice that for any semigroups 푆 and 푇, we can take the trivial left action, where 푡 ⋅ 푠 =푡 푠 = 푠 for 푡 ∈ 푇 and 푠 ∈ 푆; this corresponds to the trivial anti-homomorphism 휑 ∶ 푇 → End(푆), with 푦휑 = id푆 for all 푦 ∈ 푇. In this case, (푠1, 푡1)(푠2, 푡2) = (푠1푠2, 푡1푡2). Thus the direct product is a special case of the semidirect product. Recall from page 4 that 푆푇 is the direct product of copies of the set 푆 Wreath product indexed by 푇, or formally the set of maps from 푇 to 푆. Define a left action of 푇 on 푆푇 by letting 푦⋅푓 =푦 푓 be such that (푥)푦 푓 = (푥푦)푓. This satisfies the definition of a left action: for all 푦, 푧 ∈ 푇, we have 푧 ⋅ (푦 ⋅ 푓) = 푧푦 ⋅ 푓 since, for all 푥 ∈ 푇, (푥)(푧⋅(푦⋅푓)) = (푥)푧 (푦 푓) = (푥푧)푦 푓 = (푥푧푦)푓 = (푥)푧푦 푓 = 푧푦⋅푓. Let 휑 be the anti-homomorphism that corresponds to this action. The 푇 wreath product of 푆 and 푇, denoted 푆≀푇, is the semidirect product 푆 ⋊휑 푇. Thus the product in 푆 ≀ 푇 is

푡1 (푓1, 푡1)(푓2, 푡2) = (푓1 푓2 , 푡1푡2). Since this multiplication is derived from the multiplication in direct and semidirect products, we know it is associative. Hence 푆 ≀ 푇 is a semigroup. Notice that 푆 ≀ 푇 has cardinality |푆||푇||푇|. Let 푆, 푇, 푈 be finite semigroups. Then

|(푆 ≀ 푇) ≀ 푈| = |푆 ≀ 푇||푈||푈| = (|푆||푇||푇|)|푈||푈| = |푆||푇||푈||푇||푈||푈| and

|푈| |푆 ≀ (푇 ≀ 푈)| = |푆||푇≀푈||푇 ≀ 푈| = |푆||푇| |푈||푇||푈||푈|. Therefore the wreath product, as an operation on semigroups, is not associative. P r o p o s i t i o n 7 . 7. If 푀 and 푁 are monoids, then 푀 ≀ 푁 is a monoid with identity (푒, 1푁), where 푒 ∶ 푁 → 푀 is the constant map with (푥)푒 = 1푀 for all 푥 ∈ 푁. Proof of 7.7. Suppose 푀 and 푁 are monoids. Let . Then for any (푓, 푛) ∈ 푀 ≀ 푁, we have

(푒, 1푁)(푓, 푛) 1푁 = (푒 푓, 1푁푛) 1푁 = (푓, 푛) [since (푥)푒 푓 = (푥)푒(푥1푁)푓 = 1푀(푥)푓 = (푥)푓]

Semidirect and wreath products • 129 and

(푓, 푛)(푒, 1푁) 푛 = (푓 푒, 푛1푁) 푛 푛 = (푓, 푛); [since (푥)푓 푒 = (푥)푓(푥푛) 푒 = (푥)푓1푀 = (푥)푓] hence (푒, 1푁) is an identity for the monoid 푀 ≀ 푁. 7.7

Division

A semigroup 푆 divides a semigroup 푇, denoted 푆 ≼ 푇, Division if 푆 is a homomorphic image of a subsemigroup of 푇. Notice that the divisibility relation ≼ is reflexive. Although the divisibility relation is reflexive, most texts use the notation 푆 ≺ 푇 instead of 푆 ≼ 푇.

P r o p o s i t i o n 7 . 8. The divisibility relation ≼ is transitive. Proof of 7.8. Let 푆, 푇, 푈 be semigroups with 푆 ≼ 푇 and 푇 ≼ 푈. Then there are subsemigroups 푇′ of 푇 and 푈′ of 푈 and surjective homomorphisms 휑 ∶ 푇′ → 푆 and 휓 ∶ 푈′ → 푇. Let 푈″ = 푇′휓−1. Since 푇′ is a subsemigroup of 푇, it follows that 푈″ is a subsemigroup of 푈′ and thus of 푈. Furthermore, 휓|푈′ ∘ 휑 ∶ 푈″ → 푆 is a surjective homomorphism. So 푆 ≼ 푈. 7.8 The relation of divisibility seems rather ‘artificial’ here, but it is arises very naturally through the connection between semigroups and finite automata, which we will study in Chapter 9. P r o p o s i t i o n 7 . 9. Let 푆 and 푇 be semigroups. Then 푆 and 푇 and their direct product 푆 × 푇 divide their wreath product 푆 ≀ 푇. Proof of 7.9. Since 푆 and 푇 are homomorphic images of 푆 × 푇 under the projection maps 휋푆 ∶ 푆 × 푇 → 푆 and 휋푇 ∶ 푆 × 푇 → 푇, we have 푆 ≼ 푆 × 푇 and 푇 ≼ 푆 × 푇. Since ≼ is transitive (by Proposition 7.8), it suffices to prove that 푆 × 푇 ≼ 푆 ≀ 푇. 푇 For each 푠 ∈ 푆, let 푓푠 ∈ 푆 have all components equal to 푠. Define a map 휓 ∶ 푆 × 푇 → 푆 ≀ 푇 by (푠, 푡)휓 = (푓푠, 푡). Then ((푠, 푡)휓)((푠′, 푡′)휓)휓

= (푓푠, 푡)(푓푠′, 푡′) 푡 = (푓푠 푓푠′ , 푡푡′)

= (푓푠푠′, 푡푡′) 푡 [since (푥)(푓푠 푓푠′ ) = (푥)푓푠(푥푡)푓푠′ = 푠푠′ = (푥)푓푠푠′ for all 푥 ∈ 푇] = (푠푠′, 푡푡′)휓 = ((푠, 푡)(푠′, 푡′))휓.

Division • 130 So 휓 is a homomorphism. Furthermore,

(푠, 푡)휓 = (푠′, 푡′)휓 ⇒ (푓푠, 푡) = (푓푠′, 푡′) ⇒ 푠 = 푠′ ∧ 푡 = 푡′ ⇒ (푠, 푡) = (푠′, 푡′); thus 휓 is injective. Thus 휓 ∶ 푆 × 푇 → im 휓 ⊆ 푆 ≀ 푇 is an isomorphism, and so 휓−1 is an surjective homomorphism from the subsemigroup im 휓 of 푆 ≀ 푇 to the semigroup 푆 × 푇. So 푆 × 푇 ≼ 푆 ≀ 푇. 7.9 P r o p o s i t i o n 7 . 1 0. Let 푀 be a monoid and let 퐸 be an ideal extension of 푀 by 푇. Then 퐸 ≼ 푇 ≀ 푀. Proof of 7.10. By Proposition 1.34, 퐸 is a subdirect product of 푇 and 푀. That is, 퐸 is a subsemigroup of 푇 × 푀 and hence 퐸 divides 푇 × 푀. The result follows from Propositions 7.8 and 7.9. 7.10 P r o p o s i t i o n 7 . 1 1. If 푆′ ≼ 푆 and 푇′ ≼ 푇, then 푆′ ≀ 푇′ ≼ 푆 ≀ 푇. Proof of 7.11. The strategy is to prove this in two cases: when 푆′ and 푇′ are subsemigroups of 푆 and 푇, and when 푆′ and 푇′ are homomorphic images of 푆 and 푇. The general result follows immediately. a) Suppose 푆′ and 푇′ are subsemigroups of 푆 and 푇. Let

푈 = { (푓, 푡) ∈ 푆 ≀ 푇 ∶ 푇′푓 ⊆ 푆′ ∧ 푡 ∈ 푇′ }.

The immediate aim is to prove that 푈 is a subsemigroup of 푆 ≀ 푇. Let 푡1 (푓1, 푡1), (푓2, 푡2) ∈ 푈. So (푓1, 푡1)(푓2, 푡2) = (푓1 푓2 , 푡1푡2). First, 푡1푡2 ∈ 푇′ since 푇′ is a subsemigroup of 푇. Furthermore, for all 푥 ∈ 푇′,

푡1 (푥)(푓1 푓2 ) = ((푥)푓1)((푥푡1)푓2) ∈ (푇′푓)(푇′푓) ⊆ 푆′

푡1 since 푆′ is a subsemigroup of 푆. Hence (푓1 푓2 , 푡1푡2) ∈ 푈. Thus 푈 is a subsemigroup of 푆 ≀ 푇. Define 휑 ∶ 푈 → 푆′ ≀ 푇′ by (푓, 푡)휑 = (푓|푇′, 푡). It is clear that 휑 is a surjective homomorphism and so 푆′ ≀ 푇′ ≼ 푆 ≀ 푇. b) Suppose 휑 ∶ 푆 → 푆′ and 휓 ∶ 푇 → 푇′ are surjective homomorphisms. Let

푈 = { (푓, 푡) ∈ 푆 ≀ 푇 ∶ ker 휓 ⊆ ker(푓휑) }. (7.2)

As in part a), the first task is to prove that 푈 is a subsemigroup of 푆 ≀ 푇. First, note that 푈 is non-empty, because any map 푓 ∈ 푆푇 with ker 휓 ⊆ ker 푓 satisfies the condition in (7.2). Now let (푓1, 푡1), (푓2, 푡2) ∈ 푈. Let 푥, 푦 ∈ 푇 with 푥휓 = 푦휓. Then (푥)푓2휑 = (푦)푓2휑 since ker 휓 ⊆ ker(푓2휑). Furthermore, (푥푡1)휓 = (푥휓)(푡1휓) = (푦휓)(푡1휓) = (푦푡1)휓 and so (푥푡1)푓2휑 = (푦푡2)푓2휑 since ker 휓 ⊆ ker(푓2휑). Hence

푡1 푡1 (푥)푓1 푓2 휑 = (푥)푓1휑(푥푡1)푓2휑 = (푦)푓1휑(푦푡1)푓2휑 = (푦)푓1 푓2 휑.

Division • 131 푡1 푡1 Thus ker 휓 ⊆ ker 푓1 푓2 휑, and so (푓1, 푡1)(푓2, 푡2) = (푓1 푓2 , 푡1푡2) ∈ 푈. So 푈 is a subsemigroup of 푆 ≀ 푇. For any map 푓 ∶ 푇 → 푆 such that ker 휓 ⊆ ker(푓휑), there is a unique map 푓′ ∶ 푇′ → 푆′ such that 휓푓′ = 푓휑. Define 휗 ∶ 푈 → 푆′ ≀ 푇′ by (푓, 푡)휗 = (푓′, 푡휓). Notice that since 휓 is surjective, for any map 푓′ ∈ 푆′푇′ there is a map 푓 ∈ 푆푇 with 휓푓′ = 푓휑; hence 휗 is surjective. 푡1 Let (푓1, 푡1), (푓2, 푡2) ∈ 푈, then (푓1, 푡1)(푓2, 푡2) = (푓1 푓2 , 푡1푡2). Further, 푡1휓 (푓1, 푡1)휗(푓2, 푡2)휗 = (푓1′, 푡1휓)(푓2′, 푡2휓) = (푓1′ 푓2′, (푡1푡2)휓). Now

푡1휓 (푦휓)푓1′ 푓2′

= (푦휓)푓1′((푦휓)(푡1휓))푓2′ [by def. of the product and action]

= (푦휓)푓1′(푦푡1)휓푓2′ [since 휓 is a homomorphism]

= (푦)푓1휑(푦푡1)푓2휑 [by definition of 푓1′ and 푓2′]

= ((푦)푓1(푦푡1)푓2)휑 [since 휑 is a homomorphism] 푡1 = ((푦)푓1 푓2 )휑, [by def. of the product and action]

and so

푡1 푡1휓 (푓1 푓2 , 푡1푡2)휗 = (푓1′ 푓2′, (푡1푡2)휓). (7.3)

Hence

((푓1, 푡1)(푓2, 푡2))휗 푡1 = (푓1 푓2 , 푡1푡2)휗 [multiplication in 푆 ≀ 푇] 푡1휓 = (푓1′ 푓2′, (푡1푡2)휓) [by (7.3)]

= (푓′1, 푡1휓)(푓′2, 푡2휓) [factoring in 푆′ ≀ 푇′]

= (푓1, 푡1)휗(푓2, 푡2)휗,

and thus 휗 is a homomorphism. Therefore 푆′ ≀ 푇′ ≼ 푆 ≀ 푇. 7.11

Let 푆 be a semigroup. Let 푆′ be a set in bijection with 푆 under 푥 ↦ 푥′. Constant extension Define a multiplication on 푆∪푆′ as follows: multiplication in 푆 is as before (so that 푆 is a subsemigroup of 푆 ∪ 푆′), and for all 푥, 푦 ∈ 푆,

푥푦′ = 푥′푦′ = 푦′, 푥′푦 = (푥푦)′. (7.4)

It is easy but tedious to prove that this multiplication is associative (see Ex- ercise 7.9). The set 푆 ∪ 푆′ is thus a semigroup, called the constant extension of 푆 and denoted 퐶(푆).

P r o p o s i t i o n 7 . 1 2. If 푆 ≼ 푇, then 퐶(푆) ≼ 퐶(푇).

Proof of 7.12. If 푆 is a subsemigroup of 푇, then 퐶(푆) is a subsemigroup of 퐶(푇). Suppose 푆 is a homomorphic image of 푇. Then there exists some surjective homomorphism 휑 ∶ 푇 → 푆. Define 휑̂ ∶ 퐶(푇) → 퐶(푆) by

Division • 132 푥휑̂ = 푥휑 and 푥′휑̂ = (푥휑)′. Checking the various cases in (7.4) shows that 휑̂ is a homomorphism. It is clearly surjective. So 퐶(푆) is a homomorphic image of 퐶(푇). Hence 푆 ≼ 푇 implies 퐶(푆) ≼ 퐶(푇). 7.12 P r o p o s i t i o n 7 . 1 3. Let 푀 be a monoid and 푆 a semigroup. Then 퐶(푆 ≀ 푀) ≼ 퐶(푆)푀 ≀ 퐶(푀).

In place of this result, several textbooks claim incorrectly that 퐶(푆≀푀) ≼ 퐶(푆) ≀ 퐶(푀).

Proof of 7.13. Define a map 휓 ∶ 퐶(푆 ≀ 푀) → 퐶(푆)푀 ≀ 퐶(푀) by

where (푦)푓 ∈ 퐶(푆)푀 is defined by { ext (푓, 푚)휓 = (푓 , 푚), (푥)[(푦)푓 ] = (푥푦)푓 ext { ext {for all 푥 ∈ 푀 and 푦 ∈ 퐶(푀); where (푦)푓 ∈ 퐶(푆)푀 is defined by { con (푓, 푚)′휓 = (푓 , 푚′), (푥)[(푦)푓 ] = ((푥)푓)′ con { con {for all 푥 ∈ 푀 and 푦 ∈ 퐶(푀).

Notice that (푥)[(푦)푓con] does not depend on 푦. That is, 푓con is a constant map from 퐶(푀) to 퐶(푆)푀. 푀 The maps 푓ext and 푓con are maps from 퐶(푀) to 퐶(푆) . That is, (푦)푓ext and (푦)푓con are maps from 푀 to 퐶(푆) for all 푦 ∈ 퐶(푀). Notice in 푀 particular that 푓con is a constant map from 퐶(푀) to 퐶(푆) , but for 푦 ∈ 퐶(푀), the map (푦)푓con is in general not constant. We are going to prove that 휓 is a monomorphism. Let us first prove that 휓 is injective. To begin, observe that (푓, 푚)휓 = (푔, 푛)′휓 implies (푓ext, 푚) = (ℎcon, 푛′), which can never happen since 푚 ∈ 푀 and 푛′ ∈ 푀′. Thus to prove that 휓 is injective, we only have to check that no two distinct elements of 푆≀푀 are mapped to the same element and that no two distinct elements (푆 ≀ 푀)′ are mapped to the same element:

a) Suppose (푓, 푚)휓 = (푔, 푚)휓. Then 푓ext = 푔ext and 푚 = 푛, and hence (푦)푓ext = (푦)푓ext for all 푦 ∈ 푀, or, equivalently, (푥푦)푓 = (푥푦)푔 for all 푥, 푦 ∈ 푀. In particular, putting 푥 = 1 shows that (푦)푓 = (푦)푔 for all 푦 ∈ 푀 and so 푓 = 푔; hence (푓, 푚) = (푔, 푚).

b) Suppose (푓, 푚)′휓 = (푔, 푛)′휓. Then (푓con, 푚′) = (푔con, 푛′) and so 푓con = 푔con and 푚′ = 푛′. So ((푥)푓)′ = (푥)[(푦)푓con] = (푥)[(푦)푔con] = ((푥)푔)′, and thus (푥)푓 = (푥)푔 for all 푥 ∈ 푀. Hence 푓 = 푔 and so (푓, 푚)′ = (푔, 푛)′. Therefore 휓 is injective. Next, we have to prove that 휓 is a homomorphism. There are four cases to consider, depending on whether each multiplicand lies in 푆 ≀ 푀 or (푆 ≀ 푀)′. We explain one case in full here and outline the others; the details are left to Exercise 7.11.

Division • 133 a) Let (푓, 푚), (푔, 푛) ∈ 푆 ≀ 푀. We first have to prove: 푚 푚 (푓 푔)ext = 푓ext 푔ext . (7.5) Since both sides of (7.5) are maps from 퐶(푀) to 퐶(푆)푀, we must 푚 푚 prove that (푦)(푓 푔)ext = (푦)푓ext 푔ext for all 푦 ∈ 퐶(푀); since both sides of this equality are maps from 푀 to 퐶(푆), we must prove that 푚 푚 (푥)[(푦)(푓 푔)ext] = (푥)[(푦)푓ext 푔ext ] for all 푥 ∈ 푀 and 푦 ∈ 퐶(푀). We proceed as follows: 푚 (푥)[(푦)(푓 푔)ext] 푚 = (푥푦)푓 푔 [by definition of ext] = (푥푦)푓(푥푦푚)푔 [by def. of the product and action] 푚 = (푥)[(푦)푓ext](푥)[(푦) 푔ext ] [by definition of ext] 푚 푀 = (푥)[(푦)푓ext(푦) 푔ext ], [by multiplication in 퐶(푆) ] 푚 푀 퐶(푀) = (푥)[(푦)푓ext 푔ext ]; [by multiplication in (퐶(푆) ) ] this proves (7.5). Now we have: ((푓, 푚)(푔, 푛))휓 = (푓푚 푔, 푚푛)휓 푚 = ((푓 푔)ext, 푚푛) 푚 = (푓ext 푔ext , 푚푛) [by (7.5)]

= (푓ext, 푚)(푔ext, 푛) = (푓, 푚)휓(푔, 푛)휓. b) Let (푓, 푚)′ ∈ (푆 ≀ 푀)′ and (푔, 푛) ∈ 푆 ≀ 푀. By Exercise 7.11, 푚 푚′ (푓 푔)con = 푓con 푔ext . (7.6) Now we have: ((푓, 푚)′(푔, 푛))휓 = (푓푚 푔, 푚푛)′휓 푚 = ((푓 푔)con, (푚푛)′) 푚′ = (푓con 푔ext , 푚′푛) [by (7.6)]

= (푓con, 푚′)(푔′, 푛) = (푓, 푚)′휓(푔, 푛)휓. c) Let (푓, 푚) ∈ 푆 ≀ 푀 and (푔, 푛)′ ∈ (푆 ≀ 푀)′. By Exercise 7.11, 푚 푔con = 푓ext 푔con . (7.7) Now we have: ((푓, 푚)(푔, 푛)′)휓 = (푔, 푛)′휓

= (푔con, 푛′) 푚 = (푓ext 푔con , 푚푛′) [by (7.7)]

= (푓ext, 푚)(푔con, 푛′) = (푓, 푚)휓(푔, 푛)′휓.

Division • 134 d) Let (푓, 푚)′, (푔, 푛)′ ∈ (푆 ≀ 푀)′. By Exercise 7.11,

푚 푔con = 푓con 푔con . (7.8)

Now we have:

((푓, 푚)′(푔, 푛)′)휓 = (푔, 푛)′휓

= (푔con, 푛′) 푚 = (푓con 푔con , 푚′푛′) [by (7.8)]

= (푓con, 푚′)(푔con, 푛′) = (푓, 푚)′휓(푔, 푛)′휓.

Hence 휓 is a homomorphism and thus a monomorphism. Therefore 휓−1 is a surjective homomorphism from the subsemigroup im 휓 of 퐶(푆)푀≀퐶(푀) 푀 to 퐶(푆 ≀ 푀), and so 퐶(푆 ≀ 푀) ≼ 퐶(푆) ≀ 퐶(푀). 7.13

C o r o l l a ry 7 . 1 4. Let 푀 be a finite monoid and 푆 a semigroup. Then 퐶(푆 ≀ 푀) divides a wreath product of copies of 퐶(푆) and 퐶(푀).

Proof of 7.14. By Proposition 7.13, let 푀 be a monoid and 푆 a semigroup. Then 퐶(푆 ≀ 푀) ≼ 퐶(푆)푀 ≀ 퐶(푀). But 퐶(푆)푀 is a direct product of |푀| copies of 퐶(푆) and so 퐶(푆)푀 ≀ 퐶(푀) divides a wreath product of copies of 퐶(푆) and 퐶(푀) by Propositions 7.9 and 7.11. The result follows by the transitivity of ≼. 7.14

Krohn–Rhodes decomposition theorem Let 푈 be the monoid obtained by adjoining an identity 3 1 푎 푏 to a two-element right zero semigroup {푎, 푏}. So 푈 has elements {1, 푎, 푏} 3 1 1 푎 푏 and is multiplication table is as shown in Table 7.1. Notice that 푥2 = 푥 for 푎 푎 푎 푏 all 푥 ∈ 푈3, and so 푈3 is aperiodic. 푏 푏 푎 푏 The Krohn–Rhodes theorem is often stated in the form ‘every finite semigroup divides a wreath product of finite groups and finite aperiodic TABLE 7.1 semigroups’. We will prove a stronger form by showing that every finite Multiplication table of 푈3. semigroup divides a wreath product of its own subgroups and copies of 푈3. We first of all note that it suffices to prove the theorem for monoids since 푆 ≼ 푆1. The proof is by induction on the number of elements in the monoid. The core of the induction is Lemma 7.16, which shows that a monoid 푆 is either a group, a left simple semigroup with an identity adjoined, monogenic, or can be decomposed as 푆 = 퐿 ∪ 푇, where 퐿 is a left ideal and 푇 is a submonoid and 퐿1 and 푇 have fewer elements than 푆. The theorem is trivial for groups, and we will prove it for left simple semigroups with identities adjoined (Lemma 7.18) and for monogenic

Krohn–Rhodes decomposition theorem • 135 semigroups (Lemma 7.19); these cases form the base of the induction. The fourth possibility, of decomposition as 푆 = 퐿 ∪ 푇, supplies the induction step. The reader may wish to look ahead to Figure 7.1 on page 142 to keep track of the roles of the various lemmata. We need the following auxiliary result before we prove Lemma 7.16:

L e m m a 7 . 1 5. Let 푆 be a finite semigroup. Then at least one of the following is true: a) 푆 is trivial; b) 푆 is left simple; c) 푆 is monogenic; d) 푆 = 퐿 ∪ 푇, where 퐿 is a proper left ideal of 푆 and 푇 is a proper subsemigroup of 푆.

Proof of 7.15. Suppose that none of the properties a), b), or c) is true; we aim to prove d). Since 푆 is not left simple, it contains proper left ideals. Since it is finite, it has a maximal proper left ideal 퐾. Let 푥 ∈ 푆 ∖ 퐾. Then 퐾 ∪ 푆1푥 is a left ideal that strictly contains 퐾. Since 퐾 is maximal, 퐾 ∪ 푆1푥 = 푆. If 푆1푥 ≠ 푆, then let 퐿 = 퐾 and 푇 = 푆1푥 and the proof is complete. So assume 푆1푥 = 푆. Then 푆 == 푆푥 ∪ {푥} ⊆ 푆푥 ∪ ⟨푥⟩. If 푆 ≠ 푆푥, then let 퐿 = 푆푥 and 푇 = ⟨푥⟩ and the proof is complete since 푇 ≠ 푆 because 푆 is not monogenic. So assume 푆 = 푆푥. Let 푀 = { 푦 ∈ 푆 ∶ 푦푥 ∈ 퐾 }. Then 푀 non-empty (since 푀푥 = 퐾) and is a left ideal of 푆. Furthermore, it is a proper left ideal because 퐾 is a proper left ideal and 푆푥 = 푆. If 푀 ⊈ 퐾, then 푀 ∪ 퐾 is a left ideal of 푆 strictly containing the maximal left ideal 퐾 and so 푀 ∪ 퐾 = 푆; set 퐿 = 푀 and 푇 = 퐾 and the proof is complete. So assume 푀 ⊆ 퐾; that is,

푦푥 ∈ 퐾 ⇒ 푦 ∈ 퐾. (7.9)

Repeat the reasoning above for all 푥 ∈ 푆 ∖ 퐾. Either some such 푥 allows us to complete the proof, or (7.9) holds for all 푥 ∈ 푆 ∖ 퐾. In the former case, the proof is finished. In the latter case, take the contrapositive to see that 푦 ∈ 푆 ∖ 퐾 ⇒ 푦푥 ∈ 푆 ∖ 퐾 for all 푥 ∈ 푆 ∖ 퐾. Therefore 푆 ∖ 퐾 is a subsemigroup. So let 퐿 = 퐾 and 푇 = 푆 ∖ 퐾; the proof is complete. 7.15

L e m m a 7 . 1 6. Let 푆 be a finite monoid. Then at least one of the following is true: a) 푆 is a group; b) 푆 is a left simple with an identity adjoined; c) 푆 is monogenic; d) 푆 = 퐿 ∪ 푇, where 퐿 is a left ideal of 푆 and 푇 is a submonoid of 푆, and 퐿1 and 푇 both have fewer elements than 푆.

Krohn–Rhodes decomposition theorem • 136 Proof of 7.16. Suppose that none of the properties a), b), and c) is true; we aim to prove d). Let 퐺 be the group of units of 푆. Consider two cases: a) 퐺 is trivial. Then 푆 ∖ 퐺 = 푆 ∖ {1} is an ideal by Proposition 7.1 and thus a subsemigroup of 푆. Since 푆 is not left simple with an identity adjoined, we know that 푆 ∖ {1} is not left simple. Apply Lemma 7.15 to 푆 ∖ {1} to see that 푆 ∖ {1} = 퐿 ∪ 푄, where 퐿 is a proper left ideal of 푆∖{1} and 푄 is a proper subsemigroup of 푆∖{1}. Since 퐿 ≠ 푆∖{1}, we know that 퐿 ∪ {1} ≠ 푆. Let 푇 = 푄 ∪ {1}; then 푇 is a proper submonoid of 푆 and 푆 = 퐿 ∪ 푇. b) 퐺 is non-trivial. Then let 퐿 = 푆 ∖ 퐺 and let 푇 = 퐺. Then 푆 = 퐿 ∪ 푇. Since 퐺 is non-trivial, 퐿 ∪ {1} ≠ 푆, and since 푆 is not a group, 푇 ≠ 푆. In both cases, 푆 = 퐿 ∪ 푇, where 퐿 is a left ideal of 푆 and 푇 is a submonoid of 푆. Furthermore, in both cases 퐿1 and 푇 both have fewer elements than the original monoid 푆. 7.16 Now we turn to proving the cases forming the base of the induction; that is, monoids consisting of a left simple semigroup with an identity adjoined, and monogenic monoids. To prove the former case in Lemma 7.18, we will need the following lemma, which essentially shows that the theorem holds for left zero semigroups: L e m m a 7 . 1 7. Every finite left zero semigroup divides a wreath product of copies of 푈3.

Proof of 7.17. Let 퐿푛 be a left zero semigroup with 푛 elements. The strategy 1 1 1 is to proceed by induction and show that 퐿푛 ≼ 퐿푛−1 ≀ 퐿1. The base of 1 the induction is proved by observing that the semigroup 퐿1 = {0, 1} is a homomorphic image of 푈3. 1 1 1 For each 푥 ∈ 퐿푛−1, define 푓푥 ∶ 퐿1 → 퐿푛−1 by (1)푓푥 = 푥 and (0)푓푥 = 1. Let

1 퐾 = { (푓푥, 0) ∶ 푥 ∈ 퐿푛−1 }.

0 1 Furthermore, (푓푥, 0)(푓푦, 0) = (푓푥 푓푦 , 0) = (푓푥, 0), since for all 푧 ∈ 퐿1,

0 (푧)푓푥 푓푦 = (푧)푓푥(푧0)푓푦 = (푧)푓푥(0)푓푦 = (푧)푓푥1 = (푧)푓푥.

1 1 1 Hence 퐾 is a left zero subsemigroup of 퐿푛−1 ≀ 퐿1. Notice that 푛 = |퐿푛−1| = 1 1 |퐾|. Furthermore, the wreath product 퐿푛−1 ≀퐿1 is a monoid by Proposition 1 1 1 7.7. Therefore 퐾 ∪ {1} is a subsemigroup of 퐿푛−1 ≀ 퐿1 isomorphic to 퐿푛. 1 1 1 Hence 퐿푛 ≼ 퐿푛−1 ≀ 퐿1. 1 By Proposition 7.11, 퐿푛 divides a wreath product of 푛 copies of 푈3; thus 퐿푛 also divides a wreath product of 푛 copies of 푈3 by the transitivity of divisibility. 7.17 L e m m a 7 . 1 8. Let 푆 be a finite left simple semigroup. Then 푆1 divides the wreath product of a subgroup of 푆 and copies of 푈3.

Krohn–Rhodes decomposition theorem • 137 Proof of 7.18. Since 푆 is finite, it contains an idempotent. Thus by Theorem 4.19, we see that 푆 is isomorphic to 푍×퐺, where 푍 is a left zero semigroup and 퐺 is a subgroup of 푆. By Lemma 7.17, 푍 divides a wreath product of copies of 푈3. So by Propositions 7.9 and 7.11, 푍 × 퐺 divides a wreath product of 퐺 and copies of 푈3. 7.18 We now turn to proving the remaining case in the base of the induction, namely monogenic monoids: L e m m a 7 . 1 9. Let 푆 be a finite monogenic monoid. Then 푆 divides the wreath product of a subgroup of 푆 and copies of 푈3. Proof of 7.19. The monogenic monoid 푆 = {1, 푥, … , 푥푘, … , 푥푘+푚−1} (with 푥푘+푚 = 푥푘) is an ideal extension of the subgroup 퐺 = {푥푘, … , 푥푘+푚−1} 푘 푘 푘 by the monogenic monoid 퐶푘 = {1, 푥, … , 푥 } (with 푥 = 푥 + 1). We proceed by induction on 푘 and show that 퐶푘 divides a subsemigroup of 퐶푘−1 ≀ 퐶1. The base case of the induction is proven by observing 2 that 퐶1 = {1, 푥} (with 푥 = 푥) divides 푈3, since it is isomorphic to the subsemigroup {1, 푎} of 푈3. 푖−1 푖 For 푖 ∈ ℕ, define 푓푖 ∶ 퐶1 → 퐶푘−1 by (1)푓푖 = 푥 and (0)푓푖 = 푥 . Let

푈 = {1} ∪ { (푓푖, 0) ∶ 푖 ∈ ℕ } ⊆ 퐶푘−1 ≀ 퐶1.

0 Let (푓푖, 0)(푓푗, 0) ∈ 푈. Then (푓푖, 0)(푓푗, 0) = (푓푖 푓푗 , 0) = (푓푖+푗, 0) since

0 푖 푗 푖+푗 (0)푓푖 푓푗 = (0)푓푖(0)푓푗 = 푥 푥 = 푥 = (0)푓푖+푗, 0 푖−1 푗 (1)푓푖 푓푗 = (1)푓푖(0)푓푗 = 푥 푥 = (1)푓푖+푗.

푖 Hence 푈 is a submonoid of 퐶푘−1 ≀ 퐶1. In particular, (푓푖, 0) = (푓1, 0) for all 푖 ∈ ℕ, and so 푈 is the monogenic submonoid of 퐶푘−1 ≀ 퐶1 generated by (푓1, 0). Finally, note that

푘+1 푘 (푓1, 0) = (푓푘+1, 0) = (푓푘, 0) = (푓1, 0) , 푘 푘−1 (푓1, 0) = (푓푘, 0) ≠ (푓푘−1, 0) = (푓1, 0) ; and so

2 푘 푈 = {1, (푓1, 0), (푓1, 0) , … , (푓1, 0) }.

Hence 푈 is isomorphic to 퐶푘, and therefore 퐶푘 ≼ 퐶푘−1 ≀ 퐶1. Thus every 퐶푘 divides a wreath product of 푈3 by Propositions 7.9 and 7.11. So 푆, being an ideal extension of 퐺 and 퐶푘, divides a wreath product of 퐺 and copies of 푈3 by Proposition 7.10. 7.19 Finally, we are ready to being proving the induction step, in the case where the monoid has been decomposed as the union of a left ideal and a subsemigroup. We require the following four lemmata, and then we can quickly prove the theorem.

Krohn–Rhodes decomposition theorem • 138 L e m m a 7 . 2 0. Let 푆 be a semigroup and suppose 푆 = 퐿 ∪ 푇, where 퐿 is a left ideal of 푆 and 푇 is a subsemigroup of 푆. Then 푆 ≼ 퐿1 ≀ 퐶(푇1).

Proof of 7.20. Let 푖 ∶ 퐶(푇1) → 퐿1 be the constant map defined by (푡)푖 = 1 1 1 (푡′)푖 = 1 for all 푡 ∈ 푇 . For each 푥 ∈ 퐿, let 푓푥 ∶ 퐶(푇 ) → 퐿 be the right 1 translation defined by (푡)푓푥 = (푡′)푓푥 = 푡푥 for all 푡 ∈ 푇 . Notice that 푡푥 ∈ 퐿 since 퐿 is a left ideal. Let

1 1 푉 = { (푖, 푡) ∶ 푡 ∈ 푇 } ∪ { (푓푥, 푡′) ∶ 푥 ∈ 퐿, 푡 ∈ 푇 }.

We aim to show 푉 is a subsemigroup of 퐿1 ≀ 퐶(푇1) and that 푆 is a homomorphic image of 푉. We have four cases to consider: a) Let (푖, 푡), (푖, 푢) ∈ 푉. Then (푖, 푡)(푖, 푢) = (푖푡 푖, 푡푢) = (푖, 푡푢) since

(푠)푖푡 푖 = (푠)푖(푠푡)푖 = 1 = (푠)푖, (푠′)푖푡 푖 = (푠′)푖(푠′푡)푖 = 1 = (푠′)푖

for all 푠 ∈ 푇1. 푡 b) Let (푖, 푡), (푓푦, 푢′) ∈ 푉. Then (푖, 푡)(푓푦, 푢′) = (푖 푓푦 , 푡푢′) = (푓푡푦, 푢′), since

푡 (푠)푖 푓푦 = (푠)푖(푠푡)푓푦 = 1푠푡푦 = (푠)푓푡푦, 푡 (푠′)푖 푓푦 = (푠′)푖(푠′푡)푓푦 = 1푠′푡푦 = (푠′)푓푡푦

for all 푠 ∈ 푇1. 푡′ c) Let (푓푥, 푡′), (푖, 푢) ∈ 푉. Then (푓푥, 푡′)(푖, 푢) = (푓푥 푖, 푡′푢) = (푓푥, (푡푢)′) since

푡′ (푠)푓푥 푖 = (푠)푓푥(푠푡′)푖 = (푠)푓푥1 = (푠)푓푥, 푡′ (푠′)푓푥 푖 = (푠′)푓푥(푠′푡′)푖 = (푠′)푓푥1 = (푠′)푓푥

for all 푠 ∈ 푇1.

d) Let (푓푥, 푡′), (푓푦, 푢′) ∈ 푉. Then

푡′ (푓푥, 푡′)(푓푦, 푢′) = (푓푥 푓푦 , 푡′푢′) = (푓푥푡푦, 푢′),

since

푡′ (푠)푓푥 푓푦 = (푠)푓푥(푠푡′)푓푦 = (푠)푓푥(푡′)푓푦 = 푠푥푡푦 = (푠)푓푥푡푦, 푡′ (푠′)푓푥 푓푦 = (푠′)푓푥(푠′푡′)푓푦 = (푠′)푓푥(푡′)푓푦 = 푠푥푡푦 = (푠′)푓푥푡푦

for all 푠 ∈ 푇1.

Krohn–Rhodes decomposition theorem • 139 Hence 푉 is a subsemigroup of 퐿1 ≀ 퐶(푇1). Define a map 휑 ∶ 푉 → 푆 by (푖, 푡)휑 = 푡 and (푓푥, 푡′)휑 = 푥푡. Then, using the four cases above, ((푖, 푡)(푖, 푢))휑 = (푖, 푡푢)휑 = 푡푢 = (푖, 푡)휑(푖, 푢)휑,

((푖, 푡)(푓푦, 푢′))휑 = (푓푡푦, 푢′)휑 = 푡푦푢 = (푖, 푡)휑(푓푦, 푢′)휑,

((푓푥, 푡′)(푖, 푢))휑 = (푓푥, (푡푢)′)휑 = 푥푡푢 = (푓푥, 푡′)휑(푖, 푢)휑,

((푓푥, 푡′)(푓푦, 푢′))휑 = (푓푥푡푦, 푢′)휑 = 푥푡푦푢 = (푓푥, 푡′)휑(푓푦, 푢′)휑; hence 휑 is a homomorphism. Finally, note that 푇 ⊆ im 휑 since (푖, 푡)휑 = 푡 for all 푡 ∈ 푇 and 퐿 ⊆ im 휑 since (푓푥, 1)휑 = 푥 for all 푥 ∈ 퐿. Hence 푆 = 퐿 ∪ 푇 = im 휑 and so 휑 is a surjective homomorphism. Thus 푆 ≼ 1 1 퐿 ≀ 퐶(푇 ). 7.20 The following result is essentially a more precise version of Lemma 7.20 that holds when we decompose a monoid into the union of its group of units and the set of remaining elements: L e m m a 7 . 2 1. Let 푆 be a monoid and let 퐺 be its group of units. Then 퐼 = 푆 ∖ 퐺 is an ideal of 푆 and 푆 ≼ 퐼1 ≀ 퐺. Proof of 7.21. First, notice that 푆∖퐺 is an ideal by Proposition 7.1. For each 1 1 −1 푥 ∈ 퐼 , define a map 푓푥 ∶ 퐺 → 퐼 by (푔)푓푥 = 푔푥푔 for all 푔 ∈ 퐺. Notice −1 1 that (푔)푓1 = 푔푔 = 1 for all 푔 ∈ 퐺. Let 푉 = { (푓푥, 푔) ∶ 푥 ∈ 퐼 , 푔 ∈ 퐺 }. Let (푓푥, 푔), (푓푦, ℎ) ∈ 푉. Then for any 푘 ∈ 퐺,

푔 (푘)푓푥 푓푦 = (푘)푓푥(푘푔)푓푦 = 푘푥푘−1푘푔푦푔−1푘−1 = 푘(푥푔푦푔−1)푘−1

= (푘)푓푥푔푦푔−1 . Therefore

푔 (푓푥, 푔)(푓푦, ℎ) = (푓푥 푓푦 , 푔ℎ) = (푓푥푔푦푔−1 , 푔ℎ). (7.10)

Notice that 푥푔푦푔−1 ∈ 퐼1 since 푥 is in 퐼1 and 퐼 is an ideal. Thus 푉 is a subsemigroup of 퐼1 ≀ 퐺. Define a map 휑 ∶ 푉 → 푆 by (푓푥, 푔)휑 = 푥푔. This map 휑 is well-defined since 푓푥 = 푓푦 ⇒ (1)푓푥 = (1)푓푦 ⇒ 푥 = 푦. Furthermore,

((푓푥, 푔)(푓푦, ℎ))휑 = (푓푥푔푦푔−1 , 푔ℎ)휑 [by (7.10)] = 푥푔푦푔−1푔ℎ = 푥푔푦ℎ

= (푓푥, 푔)휑(푓푦, ℎ)휑.

So 휑 is a homomorphism. Finally, 퐺 ⊆ im 휑 since (푓1, 푔)휑 = 푔 for all 1 푔 ∈ 퐺 and 퐼 ⊆ im 휑 since (푓푥, 1)휑 = 푥 for all 푥 ∈ 푆. So 휑 is surjective 1 and so 푆 ≼ 퐼 ≀ 퐺. 7.21

Krohn–Rhodes decomposition theorem • 140 The following lemma shows that the result holds for right zero semigroups, but we only use it to prove the next lemma. L e m m a 7 . 2 2. Every finite right zero semigroup, and every finite right zero semigroup with an identity adjoined, divides a wreath product of copies of 푈3.

Proof of 7.22. Let 푅푘 denote the right zero semigroup with 푘 elements. 1 1 Notice that 푅푘 is a subsemigroup of 푅ℓ and 푅푘 is a submonoid of 푅ℓ for 푘 ⩽ ℓ. The direct product of 푛 copies of 푈3 contains subsemigroups 1 isomorphic to 푅2푛 and 푅2푛 , so for any 푘 the direct product of sufficiently 1 1 many copies of 푈3 contains 푅푘 and 푅푘. So 푅푘 and 푅푘 divide a wreath product of copies of 푈3 by Proposition 7.9. 7.22 L e m m a 7 . 2 3. Let 푆 be a finite semigroup. If 푆 divides a wreath product of groups and copies of 푈3, then 퐶(푆) divides a wreath product of copies of those same groups and copies of 푈3.

Proof of 7.23. Supppose that 푆 divides a wreath product of groups 퐺푖 and copies of 푈3. By Propositions 7.11, 7.12, and 7.14, 퐶(푆) divides a wreath product of the monoids 퐶(퐺푖) and copies of 퐶(푈3). Now, the semigroup 퐶(푈3) is a right zero semigroup with an identity adjoined, which divides a wreath product of copies of 푈3 by Lemma 7.22. The group of units of 퐶(퐺푖) is 퐺푖 and 퐿 = 퐶(퐺푖) ∖ 퐺푖 is an ideal of 퐶(퐺푖) by Proposition 7.1. Furthermore, 퐿 is a right zero semigroup by (7.4). So 퐶(퐺푖) divides 1 1 퐿 ≀ 퐺푖 by Lemma 7.21. Since 퐿 is a right zero semigroup with an identity adjoined, it divides the wreath product of copies of 푈3 by Lemma 7.22. So 1 퐿 ≀ 퐺푖 divides a wreath product of 퐺푖 and copies of 푈3 by Proposition 7.11. So 푆 divides a wreath product of the groups 퐺푖 and copies of 푈3. 7.23 Finally, using these lemmata, we can proof the Krohn–Rhodes Theo- rem. To keep track of the roles of the various lemmata, see Figure 7.1.

Krohn–Rhodes Theorem 7.24. Let 푆 be a finite semigroup. Krohn–Rhodes theorem Then 푆 divides a wreath product of subgroups of 푆 and copies of 푈3. Proof of 7.24. Let 푆 be a semigroup; we will show that 푆 divides a wreath 1 product of its subgroups and copies of 푈3. Since 푆 ≼ 푆 , we can assume 푆 is a monoid. The strategy is induction on the number of elements in 푆. The base case of the induction is when 푆 has one element. In this case, 푆 is trivial, and so 푆 is a group and the result holds immediately. So assume the result holds for all monoids with fewer elements than 푆. As already noted, the result clearly holds if 푆 is a group and in particular if 푆 is trivial. It also holds by Lemma 7.18 if 푆 is a left simple semigroup with an identity adjoined, and if 푆 is monogenic by Lemma 7.19. So assume 푆 is not trivial, not a group, not a left simple semigroup with an identity adjoined, and not monogenic. By Lemma 7.16, 푆 = 퐿 ∪ 푇,

Krohn–Rhodes decomposition theorem • 141 Wreath prod. of 푈3 and subgroups of 푆1

Wreath prod. Wreath prod. of 푈3 and of 푈3 and subgroups of 푆1 subgroups of 푆1 Pr. 7.11 ≼ Induction, Induction ≼ ≼ Lem. 7.23 Wreath prod. Wreath prod. 1 1 of 푈3 and of 푈3 and 퐿 ≀ 퐶(푇 ) subgroups of 푆1 subgroups of 푆1 ≼ Lem. 7.20 ≼ Lem. 7.18 ≼ Lem. 7.19 푆1 = 퐿 ∪ 푇 푆1 left simple 퐿 a left ideal, 푆1 a group with identity 푆1 monogenic 푇 a submonoid, adjoined 1 ⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟|퐿 |, |푇| < |푆| 1 푆 FIGURE 7.1 ≼ Diagram showing how the various lemmata are used to prove 푆 the Krohn–Rhodes theorem.

where 퐿 is a left ideal and 푇 is a submonoid of 푆 and both 퐿1 and 푇 have fewer elements than 푆. So by the induction hypothesis, 퐿1 divides a wreath product of subgroups of 퐿1 (which are also subgroups of 푆) and copies of 푈3, and similarly 푇 divides a wreath product of subgroups of 푇 (which are also subgroups of 푆) and copies of 푈3. By Lemma 7.23, 퐶(푇) also divides a wreath product of subgroups of 푆 and copies of 푈3. By Proposition 7.11, 푆 thus divides a wreath product of subgroups of 푆 and copies of 푈3. Thus, by induction, the result holds for all monoids 푆. 7.24

Exercises [See pages 232–236 for the solutions.] 7.1 Let 푀 be a finite monoid. Prove that 푀 is a group if and only if 푀푥푀 = 푀 for all 푥 ∈ 푀.

7.2 Let 푆 be a finite semigroup. Let 퐽푥 be a nontrivial J-class of 푆. Prove that there is a regular J-class 퐽푦 such that 퐽푥 ⩽ 퐽푦. ✴7.3 a) Prove that a finite nilsemigroup is nilpotent. b) Give an example of an infinite nilsemigroup that is not nilpotent.

Exercises • 142 7.4 Let 푆 and 푆′ be finite semigroups and let 휑 ∶ 푆 → 푆′ be a surjective homomorphism. a) Let 퐽 be a J-class of 푆. Prove that there is a J-class 퐽′ of 푆′ such that 퐽휑 ⊆ 퐽′. b) Let 퐽′ be a J-class of 푆′. Prove that there is a J-class 퐽 of 푆 such that 퐽휑 ⊆ 퐽′. If 퐽 is minimal such that 퐽휑 ⊆ 퐽′, then 퐽휑 = 퐽′. ✴7.5 Prove that if 푆 is a finite semigroup in which H is the equality relation and 푇 ≼ 푆, then in 푇 the relation H is also the equality relation. Give an example to show that this is may not be true when 푆 is infinite. ✴7.6 Prove that the multiplication defined for semidirect products (7.1) is associative. 7.7 Prove that if 푀 and 푁 are groups, 푀 ≀ 푁 is a group. 7.8 Suppose that 푆 and 푇 are cancellative semigroups. Must 푆 ≀ 푇 be cancellative? ✴7.9 Prove that the product defined by (7.4) for the constant extension is associative.

7.10 Let 푀 be a non-trivial monoid. For each 푥 ∈ 푀, let 휌푥 ∈ T푀 and 휏푥 ∈ T푀 be defined by 푦휌푥 = 푦푥 and 푦휏푥 = 푥. Prove that 퐶(푀) is isomorphic to the subset { 휌푥, 휏푥 ∶ 푥 ∈ 푀 } of T푀. ✴7.11 Using a technique similar to the proof of (7.5), prove (7.6), (7.7), (7.8)

Notes

The Krohn–Rhodes theorem was first stated and proved for automata in Krohn & Rhodes, ‘Algebraic theory of machines I’. ◆ The proof in this chapter is due to Lallement, Semigroups and Combinatorial Applications, and incorporates the correction published in Lallement, ‘Augmentations and wreath products of monoids’. ◆ Rhodes & Steinberg, The 픮-theory of Finite Semigroups is the most comprehensive monograph on finite semigroup theory, but is decidedly non-elementary. •

Notes • 143 Varieties & pseudovarieties 8

Variety of opinion is necessary for objective knowledge. ‘ And a method that encourages variety is also the only method that is compatible with a humanitarian outlook. — Paul Feyeraband, Against’ Method, § 3.

• The aim of this chapter is to introduce varieties and pseudovarieties of semigroups and monoids. These are classes of that are well-behaved and can, in particular, be defined using sets of equations. For instance, the class of all commutative semigroups forms a variety, and the class of all finite commutative semigroups forms a pseudovariety, and both are defined by the equation 푥푦 = 푦푥. We will formalize these notions later in the chapter, and we will see how varieties and pseudovarieties can be defined and manipulated in different ways. Pseudovarieties are important in the study of finite semigroups for the following reason: there are many finite semigroups of a given size, but most of them are boring. Of the 3 684 030 417 non-isomorphic semigroups with 8 elements, 3 661 522 792 of them are nilpotent semigroups 푆 satisfying 푆3 = {0}. (Essentially, the reason there are so many of these semigroups is that any multiplication in which all products of length 3 are equal to 0 is trivially associative.) Pseudovarieties allow us to isolate the more interesting classes. The concepts of varieties and pseudovarieties are actually broader than semigroups: varieties make sense for any type of algebraic structure, and pseudovarieties make sense for any type of finite algebraic structure. Therefore we will begin by discussing varieties in terms of universal algebra.

Varieties

An algebra is a set 푆 equipped with some operations { 푓푖 ∶ Algebras and operations 푓푖훼 푖 ∈ 퐼 }. An operation 푓푖 on 푆 is simply a map 푓푖 ∶ 푆 → 푆 for some 푓푖훼 ∈ ℕ ∪ {0}. This 푓푖훼 is called the arity of 푓푖. For instance, if 푆 is a Arity of an operation semigroup, the multiplication operation ∘ is a map ∘ ∶ 푆2 → 푆 and so has arity 2. If 푆 is a inverse semigroup, the inverse operation −1 is a map

• 144 −1 ∶ 푆 → 푆 and so has arity 1. If 푆 is a monoid, we can view the identity 0 0 element 1푆 as an operation, or as a map 1푆 ∶ 푆 → 푆 (where 푆 = ∅); the operation 1푆 has arity 0. Operations of arity 1 are called unary; operations of arity 2 are called binary; operations of arity 0 are called constants. Notice that we have a map 훼 ∶ { 푓푖 ∶ 푖 ∈ 퐼 } → ℕ ∪ {0}. A type T of an algebra is a set of operations symbols { 푓푖 ∶ 푖 ∈ 퐼 } and a Types map 훼 ∶ { 푓푖 ∶ 푖 ∈ 퐼 } → ℕ ∪ {0} determining the arity of each operations. We can write the type simply by listing the pairs in the map 훼 (viewed as a set). A semigroup has type {(∘, 2)}, a monoid has type {(∘, 2), (1, 0)} (the identity operation 1 is constant), and a lattice has type {(⊓, 2), (⊔, 2)}. An algebra of type T is called a T-algebra. Notice that some structures can be viewed as algebras in more than one way, and thus have more than one type. Let 퐺 be a group. Then −1 퐺, viewed as a group, is a {(∘, 2), (1퐺, 0), ( , 1)}-algebra; 퐺, viewed as a monoid, is a {(∘, 2), (1퐺, 0)}-algebra; 퐺, viewed as a semigroup, is a {(∘, 2)}- algebra. Strictly speaking, we should distinguish a symbol 푓푖 from the operation 푓푖: for instance, we use the same symbol ∘ to refer to the different multiplications in different semigroups. We will want to use the same symbol to discuss operations of the same arity in different structures. Let T = { (푓푖, 푓푖훼) ∶ 푖 ∈ 퐼 } be a type. We are now going to give the definition of subalgebras, homomorphisms, congruences, and direct products of T-algebras. These definitions are straightforward generaliza- tions of the definitions for semigroups. Let 푆 be a T-algebra. A subset 푆′ of 푆 is a subalgebra of 푆 if 푆′ is closed Subalgebras under all the operations in T: that is, for each 푖 ∈ 퐼, we have

푥 , … , 푥 ∈ 푆′ ⇒ (푥 , … , 푥 )푓 ∈ 푆′. (8.1) 1 푓푖훼 1 푓푖훼 푖

In particular, this means that 푓푖 ∈ 푆′ whenever 푓푖훼 = 0. Let 푋 ⊆ 푆. The subalgebra generated by 푋 is defined to be the intersection of all subalgebras that contain 푋. It is easy to prove (cf. Proposition 1.11) that the subalgebra generated by 푋 consists of all elements that can be obtained by starting from 푋 and applying the operations 푓푖. Let 푆 and 푇 be T-algebras. Then 휑 ∶ 푆 → 푇 is a homomorphism if for Homomorphisms each 푖 ∈ 퐼, we have

((푥 , … , 푥 )푓)휑 = (푥 휑, … , 푥 휑)푓. (8.2) 1 푓푖훼 푖 1 푓푖훼 푖

(Notice that on the left-hand side of (8.2), 푓푖 is an operation on 푆, while on the right-hand side, it is an operation on 푇.) An injective homomorphism is a monomorphism, and a bijective homomorphism is an isomorphism. If 휑 ∶ 푆 → 푇 is a surjective homomorphism, 푇 is a homomorphic image of 푆. Let 푆 be a T-algebra. A binary relation 휌 on 푆 is a congruence if for Congruences

Varieties • 145 each 푖 ∈ 퐼,

(∀푥 , 푦 , … , 푥 , 푦 ) 1 1 푓푖훼 푓푖훼 (푥 휌 푦 ∧ … ∧ 푥 휌 푥 1 1 푓푖훼 푓푖훼 ⇒ (푥 , … , 푥 )푓 휌 (푦 , … , 푦 )푓). 1 푓푖훼 푖 1 푓푖훼 푖

Let S = { 푆푗 ∶ 푗 ∈ 퐽 } be a collection of T-algebras. The direct product of Direct products the T-algebras in S is their cartesian product ∏푗∈퐽 푆푗 with the operations performed componentwise:

(푗)(푠 , … , 푠 )푓 = ((푗)푠 , … , (푗)푠 )푓. 1 푓푖훼 푖 1 푓푖 푖

Let 퐴 be a non-empty set and let 퐹T(퐴) be the smallest set of all Free T-algebras formal expressions (that is, words) over 퐴 ∪ { 푓푖 ∶ 푖 ∈ 퐼 } ∪ {(} ∪ {)} ∪ {, } satisfying the following two conditions:

퐴 ⊆ 퐹T(퐴); 푢 , … , 푢 ∈ 퐹 (퐴) ⇒ (푢 , … , 푢 )푓 ∈ 퐹 (퐴). 1 푓푖훼 T 1 푓푖훼 푖 T For instance, if T is {(푓, 2), ( ′ , 1)} and 퐴 = {푎, 푏, 푐}, then the words (푎, (((푐, 푏)푓)′, 푐)푓)푓 and (((푏)′, ((푏, 푎)푓, (푐)′)푓)푓 are elements of 퐹T(퐴). The set 퐹T(퐴) is obviously a T-algebra and is called the free T-algebra or absolutely free T-algebra. Notice that 퐹T(퐴) is generated by 퐴. Let 휄 ∶ 퐴 → 퐹T(퐴) be the inclusion map. For any T-algebra 푆 and map 휑 ∶ 퐴 → 푆, there is a unique extension of 휑 to a homomorphism 휑̂ ∶ 퐹T(퐴) → 푆. That is, 휄휑̂ = 휑, or, equivalently, the following diagram commutes: 휄 퐴 퐹T(퐴) 휑 휑̂ (8.3) 푆

This property is reminiscent of some definitions we have already seen: free semigroups and monoids (see pages 37 f.), free inverse semigroups and monoids (see page 104), and free commutative semigroups (see page 119), and we shall say more about it later. Let X be a non-empty class of T-algebras. Let ℍX denote the class ℍ, 핊, ℙ of all T-algebras that are homomorphic images of the algebras in X. Let 핊X denote the class of all T-algebras that are subalgebras of algebras in X. Let ℙX denote the class of all T-algebras that are direct products of the T-algebras in X. That is,

ℍX = { 푆 ∶ (∃푇 ∈ X)(푆 is a homomorphic image of 푇) }; 핊X = { 푆 ∶ (∃푇 ∈ X)(푆 is a subalgebra of 푇) };

ℙX = { 푆 ∶ (∃{ 푇푖 ∶ 푖 ∈ 퐼 } ⊆ X)(푆 = ∏푖∈퐼푇푖) }.

Varieties • 146 Thus ℍ, 핊, and ℙ are unary operators on classes of algebras. Notice that X is contained in ℍX, 핊X, and ℙX. A non-empty class of T-algebras is a variety of T-algebras if it is closed Variety under the operations ℍ, 핊, and ℙ. That is, X is a variety if ℍX∪핊X∪ℙX ⊆ X.

E x a m p l e 8 . 1. a) Let 1 be the class containing only the trivial semigroup 퐸 = {푒}. Then 1 is a variety, since the only subsemigroup of 퐸 is 퐸 itself, the only homomorphic image of 퐸 is 퐸 itself, and any direct product of copies of 퐸 is isomorphic to 퐸. b) Let S be the class of all semigroups (viewed as {(∘, 2)}-algebras). Any homomorphic image of a semigroup is itself a semigroup, so S is closed under ℍ. Subalgebras are subsemigroups, and so S is closed under 핊. A direct product of semigroups is a semigroup, so S is closed under ℙ. Therefore S is a variety. c) Let M be the class of all monoids (viewed as {(∘, 2), (1, 0)}-algebras). A homomorphic image of a monoid is again a monoid, so M is closed under ℍ. Subalgebras are submonoids because the subalgebra must be closed under the ‘operation’ 1: that is, they must contain the constant 1. So M is closed under 핊. A direct product of monoids is itself a monoid. Therefore M is a variety. d) Let Com be the class of all commutative semigroups (viewed, like members of S, as {(∘, 2)}-algebras). Since any subalgebra or homomorphic image of a commutative semigroup is itself a commutative semigroup, and a direct product of commutative semigroups is commutative, Com is a variety. −1 e) Let G be the class of all groups, viewed as {(∘, 2), (1퐺, 0), ( , 1)}-algebras. Then subalgebras are closed under multiplication and taking inverses; thus subalgebras are subgroups. Since any subalgebra or homomorphic image of a group is also a group, and any direct product of groups is also a group, G is a variety. Notice that the class of all groups G viewed as {(∘, 2)}-algebras is not a variety, because in this case subalgebras are subsemigroups and so G is not closed under taking subalgebras; for example, G contains ℤ but not its subsemigroup ℕ. f) Let Inv be the class of inverse semigroups, viewed as {(∘, 2), (−1, 1)}- algebras. Then Inv is a variety.

Let V be a variety of T-algebras and let 퐴 be a non-empty set. Let 푆 ∈ V. Let 휑 ∈ 푆퐴. We know there is a unique extension of 휑 to a homomorphism 휑̂ ∶ 퐹T(퐴) → 푆. Now, im 휑̂ is a subalgebra of 푆 and so im 휑̂ ∈ V since V is closed under forming subalgebras. Let

휌 = ⋂{ ker 휑̂ ∶ 휑 ∈ 푆퐴, 푆 ∈ V }.

Varieties • 147 Then 휌, being an intersection of congruences on 퐹T(퐴), is also a congruence. Furthermore, 휌 ⊆ ker 휑̂ for any 휑 ∈ 푆퐴. Hence for each 푆 ∈ V and 퐴 휑 ∈ 푆 , there exists a unique homomorphism 휑 ∶ 퐹T(퐴)/휌 → 푆 such ♮ that 휌 휑 = 휑̂. Thus 휑 ∶ 퐹T(퐴)/휌 → 푆 is the unique homomorphism such that 휑 = 휄휑̂ = 휄휌♮휑, and the following diagram commutes:

♮ 휄 휌 퐴 퐹T(퐴) 퐹T(퐴)/휌 휑 휑̂ 휑 푆

By a result for T-algebras analogous to Proposition 1.33, 퐹T(퐴)/휌 is a 퐴 subdirect product of { 퐹T(퐴)/ ker 휑̂ ∶ 휑 ∈ 푆 , 푆 ∈ V }. Now, each algebra 퐹T(퐴)/ ker 휑̂ is a subalgebra of an element of V and therefore is itself a member of V (since 핊V = V). Hence 퐹T(퐴)/휌 ∈ 핊ℙV = V. The T-algebra 퐹T(퐴)/휌 is called the V-free algebra, and is denoted V-free algebras ♮ 퐹V(퐴). Notice that there is a map 휗 ∶ 퐴 → 퐹V(퐴) given by 푥휗 = 푥휄휌 = [푥]휌 such that the universal property holds: for any 푆 ∈ V and map 휑 ∶ 퐴 → 푆, there is a unique homomorphism 휑 ∶ 퐹V(퐴) → 푆 such that 휗휑 = 휑, or, in diagrammatic terms:

휗 퐴 퐹V(퐴) 휑 휑 (8.4) 푆

Notice that if V is the variety of all T-algebras, 퐹V(퐴) = 퐹T(퐴) and we recover diagram (8.3). Let us apply this definition to some concrete varieties. Let V be the variety of all semigroups S. Then the definition ofa S-free algebra co- incides with the definition of a free semigroup, and the diagram (8.4) becomes identical to the second diagram in (2.1). Since 퐹S(퐴) ∈ S, we see + that 퐹S(퐴) ≃ 퐴 by Proposition 2.1. Similarly, if we apply the definition to the variety of inverse semigroups Inv, the diagram (8.4) becomes identical to the second diagram in (5.10) and we see that 퐹Inv(퐴) ≃ FInvS(퐴) by Proposition 5.15. With the variety of commutative semigroups Com, the diagram (8.4) becomes identical to the second diagram in (6.1) and we see that 퐹Com(퐴) ≃ FCommS(퐴) by Proposition 6.3. Thus for any variety V of T-algebras we have (for each set 퐴) a T- algebra 퐹V(퐴) ∈ V and a map 휗 ∶ 퐴 → 퐹V(퐴) with the universal property. This indicates that varieties have some of the nice properties of the classes of semigroups, inverse semigroups, and commutative semigroups. But varieties also have another very useful property: they are precisely those collections of algebras that can be defined using sets of equations called laws.

Varieties • 148 For T-algebras, a law over an alphabet 퐴 is a pair of elements 푢 and 푣 of Laws 퐹T(퐴), normally written as a formal equality 푢 = 푣.A T-algebra 푆 satisfies the law 푢 = 푣 if, for every map 휑 ∶ 퐴 → 푆, we have 푢휑̂ = 푣휑̂ (where 휑̂ is the homomorphism in diagram (8.3)). Informally, 푆 satisfies 푢 = 푣 if we every possible substitution of elements of 푆 for letters of 퐴 in the words 푢 and 푣 gives elements that are equal. For instance, commutative semigroups, viewed as {(∘, 2)}-algebras, satisfy the law 푥 ∘ 푦 = 푦 ∘ 푥. Semigroups of idempotents satisfy the law 푥 ∘ 푥 = 푥. All semigroups satisfy the law 푥 ∘ (푦 ∘ 푧) = (푥 ∘ 푦) ∘ 푧, and all monoids satisfy the laws 푥 ∘ 1 = 푥 and 1 ∘ 푥 = 푥. A law over 퐴 is sometimes called an identity or identical relation over 퐴, but we will avoid this potentially confusing terminology. Let E be a class of T-algebras. Suppose there is a set 퐿 of laws over an Equational classes alphabet 퐴 such that 푆 ∈ E if and only if 푆 satisfies every law in 퐿. Then E is the equational class defined by 퐿.

Birkhoff’s Theorem 8.2. Let T be a type. Then a class of T- Birkhoff’s theorem algebras is a variety if and only if it is an equational class.

Proof of 8.2. Part 1. Suppose X is an equational class. Then there is a set of laws 퐿 over an alphabet 퐴 such that 푆 ∈ X if and only if 푆 satisfies every law in 퐿. To prove that X is a variety, we must show that it is closed under ℍ, 핊, and ℙ. Let 푆 ∈ X, and let 푇 be a T-algebra and 휓 ∶ 푆 → 푇 a surjective homomorphism. Let 푢 = 푣 be a law in 퐿. Let 휑 ∶ 퐴 → 푇 be a map. Define a map 휗 ∶ 퐴 → 푆 by letting 푎휗 ∈ 푆 be such that 푎휗휓 = 푎휑 (such an 푎휗 exists because 휓 is surjective). Notice that 휗휓̂ and 휑̂ are homomorphisms from 퐹T(퐴) to 푆 extending 휗휓 = 휑 and so, by the uniqueness of such homomorphisms, 휗휓̂ = 휑̂. Since 푆 satisfies 퐿, we have 푢휗̂ = 푣휗̂; hence 푢휑̂ = 푢휗휓̂ = 푣휗휓̂ = 푣휑̂. So 푇 satisfies 푢 = 푣. Hence 푇 satisfies every law in 퐿 and so 푇 ∈ X. Thus X is closed under ℍ. Let 푆 ∈ X and let 푇 be a subalgebra of 푆. Let 푢 = 푣 be a law in 퐿. Then if 휑 ∶ 퐴 → 푇, then 휑 is also a map from 퐴 to 푆 and so 푢휑̂ = 푣휑̂ since 푆 satisfies 푢 = 푣. Hence 푇 also satisfies 푢 = 푣. So 푇 satisfies every law in 퐿 and so 푇 ∈ X. Thus X is closed under 핊. Let { 푆푗 ∶ 푗 ∈ 퐽 } ⊆ X and suppose 푇 is the direct product of { 푆푗 ∶ 푗 ∈ 퐽 }. Let 푢 = 푣 be a law in 퐿. Let 휑 ∶ 퐴 → 푇 be a map. For each 푗 ∈ 퐽, let 휑푗 = 휑휋푗, where 휋푗 ∶ 푇 → 푆푗 is the projection homomorphism. So 휑푗 is a map from 퐴 to 푆푗, and 푆푗 satisfies 푢 = 푣, and thus 푢휑̂푗 = 푣휑̂푗. The map 휓 ∶ 퐹T(퐴) → 푇 with (푗)(푥휓) = 푥휑̂푗 is a homomorphism extending 휑, so, by the uniqueness condition, 휓 = 휑̂. Since 푢휑̂푗 = 푣휑̂푗 for each 푗, we have 푢휑̂ = 푢휓 = 푣휓 = 푣휑̂; hence 푇 satisfies 푢 = 푣. So 푇 satisfies every law in 퐿 and so 푇 ∈ X. Thus X is closed under ℙ. So X is closed under ℍ, 핊, and ℙ, and so is a variety. Part 2. Suppose now that V is a variety. Let 퐴 be an infinite alphabet.

Varieties • 149 Recall that 퐹V(퐴) = 퐹T(퐴)/휌, where 휌 = ⋂{ ker 휑̂ ∶ 휑 ∈ 푆퐴, 푆 ∈ V }.

We aim to show that V is the equational class defined by 휌, viewing the set of pairs 휌 ⊆ 퐹T(퐴) × 퐹T(퐴) as a set of laws. Let 푆 ∈ V. Let (푢, 푣) ∈ 휌; notice that 푢, 푣 ∈ 퐹T(퐴). Then (푢, 푣) ∈ ker 휑̂ and thus 푢휑̂ = 푣휑̂ for any 휑 ∈ 푆퐴. So 푆 satisfies the law 푢 = 푣. Thus every 푆 ∈ V satisfies the law 푢 = 푣 for any (푢, 푣) ∈ 휌. Conversely, suppose that 푆 satisfies the law 푢 = 푣 for every (푢, 푣) ∈ 휌. Let 퐵 be an alphabet with cardinality greater than or equal to both 푆 and 퐴. Let 퐹V(퐵) be the V-free algebra generated by 퐵; then 퐹V(퐵) = 퐹T(퐵)/휋 for some congruence 휋 on 퐹T(퐵). Since 퐵 has cardinality greater than or equal to 푆, there is a surjective homomorphism 휓 ∶ 퐹T(퐵) → 푆. We are going to prove that 휋 ⊆ ker 휓, which will imply that we have a well-defined surjective homomorphism 휗 ∶ 퐹V(퐵) → 푆 with [푥]휋휗 = 푥휓, which will in turn imply 푆 ∈ V. So let (푢, 푣) ∈ 휋. Let 퐵0 be the subset of 퐵 containing the letters that appear in 푢 or 푣; notice that 퐵0 is finite. Let 퐴0 be a finite subset of 퐴 such that there is a bijection 휉0 ∶ 퐴0 → 퐵0. Since 퐵 has cardinality greater than or equal to 퐴, there is an injection 휉 ∶ 퐴 → 퐵 extending 휉0. Since 휉 is injective, there is a right inverse 휂 ∶ 퐵 → 퐴 of 휉 (that is, 휉휂 = id퐴). ̂ Then 휉 extends to a monomorphism 휉 ∶ 퐹T(퐴) → 퐹T(퐵), and 휂 extends ̂ to a homomorphism ̂휂∶ 퐹T(퐵) → 퐹T(퐴). Since 휉 is injective, there are ̂ ̂ uniquely determined 푢0, 푣0 ∈ 퐹T(퐴) such that 푢0휉 = 푢 and 푣0휉 = 푣. Notice that 푢휂 = 푢0 and 푣휂 = 푣0. ♮ Consider the map 휂휌 ∶ 퐹T(퐵) → 퐹V(퐴). Since 퐹V(퐵) lies in the variety V, we must have 휋 ⊆ 휂휌♮. In particular, 푢휂휌♮ = 푣휂휌♮ and so (푢0, 푣0) ∈ 휌. Thus 푆 satisfies the law 푢0 = 푣0. Therefore, since 휂휓 ∶ 퐹T(퐵) → 푆 is a homomorphism, 푢0휂휓 = 푣0휂휓 and so 푢휓 = 푣휓. Hence 휋 ⊆ ker 휓 and so we have a well-defined homomorphism 휗 ∶ 퐹V(퐵) → 푆 with [푥]휋휗 = 푥휓. Therefore 푆 is a homomorphic image of 퐹V(퐵) ∈ V and so lies in the variety V. Thus we have proved that 푆 ∈ V if and only if 푆 satisfies every law 푢 = 푣 in 휌. Therefore V is an equational class. 8.2

Theorem 8.2 shows that every variety can be defined by a set of laws. Finitely based variety However, in general an infinite set of laws is required. This is true even for varieties of semigroups. However, in some cases, a finite set of laws suffice. Such varieties are said tobe finitely based. Let T be the type {(∘, 2)}. The variety consisting of all semigroups S is defined by the law 푥 ∘ (푦 ∘ 푧) = (푥 ∘ 푦) ∘ 푧. We use this type when working with varieties of semigroups, and we will always implicitly assume this law and write 푥푦 for 푥 ∘ 푦. Examples are summarized in Table 8.1. Let T be the type {(∘, 2), (1, 0)}. The variety consisting of all monoids M is defined by the laws 푥(푦푧) = (푥푦)푧, 1푥 = 푥, and 푥1 = 푥. When

Varieties • 150 Variety Symbol Defining laws

Semigroups S — TABLE 8.1 Null semigroups Z 푥푦 = 푧푡 Varieties of semigroups. The Left zero semigroups LZ 푥푦 = 푥 law 푥(푦푧) = (푥푦)푧 is implicitly Right zero semigroups RZ 푥푦 = 푦 assumed.

Variety Symbol Defining laws Monoids M — TABLE 8.2 Trivial monoid 1 푥 = 1 Varieties of monoids, viewed as {(∘, 2), (1, 0)}-algebras. The Commutative monoids Com 푥푦 = 푦푥 laws 푥(푦푧) = (푥푦)푧, 푥1 = 푥, 2 푥 = 푥, and 1푥 = 푥 are implicitly as- Semilattices with identities Sl { 푥푦 = 푦푥 sumed. working with varieties of monoids we will use this type and implicitly assume these laws. Examples are summarized in Table 8.2. Finally let T be the type {(∘, 2), (−1, 1)}. We will use this type when working with semigroups equipped with an inverse operation −1, such as regular and inverse semigroup. In this context, we will assume the laws (푥푦)푧 = 푥(푦푧), 푥푥−1푥 = 푥 and (푥−1)−1 = 푥. Examples are summarized in Table 8.3. Another way to define a variety of T-algebras is to use a specified Variety generated by X set of T-algebras to generate a variety. Let X be a set of T-algebras. The intersection of all varieties of T-algebras containing X is itself a variety, called the variety of T-algebras generated by X, or simply the variety generated by X. It is easy to prove that the variety generated by X consists of all X-algebras that can be obtained from X by repeatedly forming subsemigroups, homomorphic images, and direct products. That is, the variety generated by X is

{ 핆1핆2 ⋯ 핆푛X ∶ 푛 ∈ ℕ, 핆푖 ∈ {ℍ, 핊, ℙ} }. (8.5)

L e m m a 8 . 3. For any non-empty class of T-algebras X, we have

핊ℍX ⊆ ℍ핊X; ℙℍX ⊆ ℍℙX; ℙ핊X ⊆ 핊ℙX.

Proof of 8.3. Let 푆 ∈ 핊ℍX. Then there is T-algebra 푇 ∈ X and a surjective homomorphism 휑 ∶ 푇 → 푈 such that 푆 is a subalgebra of 푈. Let 푇′ = −1 푆휑 = { 푡 ∈ 푇 ∶ 푡휑 ∈ 푆 }. Then 푇′ is a subalgebra of 푇 and 휑|푇′ ∶ 푇′ → 푆 is a surjective homomorphism. So 푆 ∈ ℍ핊X. Let 푆 ∈ ℙℍX. Then there is a collection of T-algebras { 푇푖 ∶ 푖 ∈ 퐼 } ⊆ X and a collection of surjective homomorphisms 훷 = { 휑푖 ∶ 푇푖 → 푈푖 ∶ 푖 ∈ 퐼 } such that 푆 = ∏푖∈퐼 푈푖. Define a homomorphism 휓 ∶ ∏푖∈퐼 푇푖 → 푆 by

Varieties • 151 Variety Symbol Defining laws Completely regular sgrps CR 푥푥−1 = 푥−1푥 (푥푦)−1 = 푦−1푥−1, Inverse semigroups Inv { TABLE 8.3 푥푥−1푦푦−1 = 푦푦−1푥푥−1 Varieties of semigroups with a unary operation −1. The laws −1 −1 푥푥 = 푥 푥, −1 Clifford semigroups Cl { 푥(푦푧) = (푥푦)푧, 푥푥 푥 = 푥, 푥푥−1푦푦−1 = 푦푦−1푥푥−1 and (푥−1)−1 = 푥 are implicitly Groups G 푥푥−1 = 푦푦−1 assumed.

(푖)(푥휓) = ((푖)푥)휑푖. Then 휓 is surjective since each 휑푖 is surjective. So 푆 ∈ ℍℙX. Let 푆 ∈ ℙ핊X. Then there is a collection of T-algebras { 푇푖 ∶ 푖 ∈ 퐼 } ⊆ X and a subalgebras 푈푖 of 푇푖 such that 푆 = ∏푖∈퐼 푈푖. Then 푆 is a subalgebra of ∏푖∈퐼 푇푖. So 푆 ∈ 핊ℙX. 8.3 As an immediate consequence of Lemma 8.3 and (8.5), and the fact that the operators ℍ, 핊, and ℙ are idempotent, we obtain the following result:

P r o p o s i t i o n 8 . 4. Let X be a class of T-algebras. The variety generated by X is ℍ핊ℙX. 8.4

Pseudovarieties Varieties are not useful for studying and classifying finite algebras, for the simple reason that every non-trivial variety contains infinite algebras: if a variety contains an algebra 푆 with two elements, then it contains the direct product of infinitely many copies of 푆, which is of course infinite. Clearly, if we take a class X of finite T-algebras, then ℍX and 핊X also contain only finite T-algebras. The problem, therefore, is the operator ℙ. To modify the notion of variety in order to study finite algebras, we therefore introduce a new operator on classes of T-algebras.

Let ℙfinX denote the class of all T-algebras that are finitary direct ℙfin products of the algebras in X. That is,

ℙfinX = { 푆 ∶ (∃{푇1, … , 푇푛} ⊆ X)(푆 = 푇1 × 푇2 × … × 푇푛) }.

A non-empty class of finite T-algebras is a pseudovariety of T-algebras if it Pseudovariety is closed under the operations ℍ, 핊, and ℙfin. That is, X is a pseudovariety if ℍX ∪ 핊X ∪ ℙfinX ⊆ X. E x a m p l e 8 . 5. a) Let 1 be the class containing only the trivial semigroup (or monoid) 퐸 = {푒}. Then 1 is a pseudovariety both when

Pseudovarieties • 152 we view 퐸 as a {(∘, 2)}-algebra and when we view 퐸 as a {(∘, 2), (1, 0)}- algebra, since the only subalgebra of 퐸 is 퐸 itself, the only homomorphic image of 퐸 is 퐸 itself, and any finitary direct product of copies of 퐸 is isomorphic to 퐸. b) Let S be the class of all finite semigroups. Then S is a pseudovariety. c) Let M be the class of all finite monoids (viewed as {(∘, 2), (1, 0)}-algebras. Then S is a pseudovariety. d) Let Com be the class of all finite commutative monoids (viewed as {(∘, 2), (1, 0)}-algebras). Then Com is a pseudovariety.

e) Let G be the class of all finite groups, which we view as {(∘, 2), (1퐺, 0), (−1, 1)}-algebras; then G is a pseudovariety. In contrast with varieties, that the class of all finite groups viewed as {(∘, 2)}-algebras is a pseudovariety, because in this case subalgebras are subgroups (since for elements 푥 of a group with 푛 elements, 푥푛 = 1 and 푥−1 = 푥푛−1). f) Let Inv be the class of all finite inverse semigroups, which we view as {(∘, 2), (−1, 1)}-algebras. Then Inv is a pseudovariety. g) Let N be the class of all finite nilpotent semigroups. Then N is a pseudovariety. (See Exercise 8.2(a).) h) Let A be the class of all finite aperiodic monoids, viewed as {(∘, 2), (1, 0)}-algebras. Then A is a pseudovariety. Notice that N ⊆ A.

Notice that we are using the same symbols for certain varieties and pseudovarieties: for instance, Com is used to denote both the variety of commutative monoids and the pseudovariety of finite commutative monoids. This will not cause confusion, because from now on we will only use them to denote pseudovarieties. Just as with varieties, we have the idea of generating a pseudovariety Pseudovariety of finite T-algebras. Let X be a set of finite T-algebras. The intersection generated by X of all pseudovarieties of T-algebras containing X is itself a pseudovariety, called the pseudovariety of finite T-algebras generated by X, or simply the pseudovariety generated by X, and is denoted VT(X) It is easy to prove that VT(X) consists of all (necessarily finite) X-algebras that can be obtained from X by repeatedly forming subalgebras, homomorphic images, and finitary direct products. That is,

VT(X) = { 핆1핆2 ⋯ 핆푛X ∶ 푛 ∈ ℕ, 핆푖 ∈ {ℍ, 핊, ℙfin} }. (8.6)

We have the following analogue of Proposition 8.4:

P r o p o s i t i o n 8 . 6. Let X be a class of finite T-algebras. Then VT(X) = ℍ핊ℙfinX.

Pseudovarieties • 153 Proof of 8.6. For any non-empty class of finite T-algebras X, we have 핊ℍX ⊆ ℍ핊X;

ℙfinℍX ⊆ ℍℙfinX;

ℙfin핊X ⊆ 핊ℙfinX; to see this, follow the reasoning in the proof of Lemma 8.3, restricting the index sets 퐼 in the direct products to be finite. The result follows immediately. 8.6

Let V and W be pseudovarieties of T-algebras. The class of pseudova- Join and meet of rieties of T-algebras is ordered by the usual inclusion order ⊆. Then it is pseudovarieties easy to see that

V ⊔ W = VT(V ∪ W), and, since V ∩ W is a variety, V ⊓ W = V ∩ W. So the class of T-algebras is a lattice. Furthermore, if we consider only subpseudovarieties of a fixed pseudovariety V (such as S or M), then the class of such subpseudovarieties forms a sublattice.

Pseudovarieties of semigroups and monoids From this point onwards, we will consider only pseudovarieties of semigroups and pseudovarieties of monoids. These pseudovarieties have different types: pseudovarieties of semigroups have type S = {(∘, 2)} and pseudovarieties of monoids have type M = {(∘, 2), (1, 0)}. In pseudovarieties of semigroups, the homomorphisms are the usual semigroup homomorphisms and the subalgebras are subsemigroups. In pseudovarieties of monoids, the homomorphisms are monoid homomorphisms, and the subalgebras are submonoids that contain the identity of the original monoid. In previous chapters, by ‘submonoid’ we meant ‘any subsemigroup that forms a monoid’.But such a submonoid may not be a subalgebra: For example, let 푆 = {1, 0} be the two-element semilattice with 1 > 0. Then 푆 is a monoid with identity 1, and contains the submonoid 푇 = {0}. However, 푇 is not an M-subalgebra of 푆, because an M- subalgebra must include the constant 1. For brevity, we call a submonoid that contains the identity of the original monoid an M-submonoid. We will use the term S-pseudovarieties for pseudovarieties of semi- S-pseudovarieties, groups, and M-pseudovarieties for pseudovarieties of monoids. The reas- M-pseudovarieties oning for the two types often runs in parallel, but there are important differences.

Pseudovarieties of semigroups and monoids • 154 Notice that S-pseudovarieties are closed under division: if V is an S- pseudovariety, 푇 ∈ V, and 푆 ≼ 푇, then by definition there is an surjective homomorphism 휑 ∶ 푇′ → 푆, where 푇′ is a subsemigroup of 푇; hence 푆 ∈ ℍ핊V = V. In fact, M-pseudovarieties are also closed under division, as a consequence of the following result:

P r o p o s i t i o n 8 . 7. Let 푆 be a semigroup and 푀 a monoid and suppose Division in M- 푆 ≼ 푀. Then there is an M-submonoid 푀′ of 푀 and a surjective monoid pseudovarieties homomorphism 휑 ∶ 푀′ → 푆1. Consequently, if 푁 is a monoid, then 푁 ≼ 푀 if and only if there is an M-submonoid 푀′ of 푀 and a surjective monoid homomorphism 휑 ∶ 푀′ → 푁.

Proof of 8.7. Suppose 푆 ≼ 푀. Then there is a subsemigroup 푇 of 푀 and a surjective homomorphism 휓 ∶ 푇 → 푆. Let 푀′ = 푇 ∪ {1푀} and extend 휓 to a monoid homomorphism 휑 ∶ 푀′ → 푆 by defining 1푀휑 = 1푆1 and 푥휑 = 푥휓 for all 푥 ∈ 푇. (Notice that 휑 is well-defined, since if 1푀 ∈ 푇, then (푧휓)(1푀휓) = (푧1푀)휓 = 푧휓 and (1푀휓)(푧휓) = (1푀푧)휓 = 푧휓 and so 푆 is a monoid with identity 1푀휓 because 휓 is surjective.) 8.7 We now introduce two operators that allow us to connect S-pseudovarieties of semigroups and M-pseudovarieties of monoids.

For any M-pseudovariety of monoids V, let VSg

VSg = VS(V).

So to obtain VSg from V we simply treat the monoids in V as semigroups, form all finite direct products, then all subsemigroups, and then all (semigroup) homomorphic images. From Proposition 8.7, we see that

1 푆 ∈ VSg ⇔ 푆 ∈ V. (8.7)

For any S-pseudovariety of semigroups, let VMon

VMon = { 푆 ∈ V ∶ 푆 is a monoid }.

That is VMon consists of the monoids that, when viewed as semigroups, belong to V. It is easy to see that VMon is an M-pseudovariety of monoids.

L e m m a 8 . 8. For any M-pseudovariety of monoids V, we have (VSg)Mon = V.

Proof of 8.8. Let 푆 be a finite monoid. Then

푆 ∈ (VSg)Mon

⇔ 푆 is a monoid that belongs to VSg ⇔ 푆1 ∈ V [by (8.7)] 1 ⇔ 푆 ∈ V. [since 푆 = 푆 ] 8.8

Pseudovarieties of semigroups and monoids • 155 P r o p o s i t i o n 8 . 9. The operator Sg is an embedding of the lattice of M-pseudovarieties of monoids into the lattice of of the S-pseudovarieties of semigroups.

Proof of 8.9. It is immediate from (8.7) that Sg is a lattice homomorphism. By Lemma 8.8,

VSg = WSg ⇒ (VSg)Mon = (WSg)Mon ⇒ V = W, so Sg is injective. 8.9

An S-pseudovariety of semigroups W is monoidal if W = VSg for some Monoidal pseudovariety M-pseudovariety of monoids.

E x a m p l e 8 . 1 0. a) The S-pseudovariety of all finite semigroups S is monoidal, because S = MSg by (8.7), where M is the M-pseudovariety of all finite monoids. b) The S-pseudovariety of all finite nilpotent semigroups N is not monoidal. To see this, suppose, with the aim of obtaining a contradiction, that N = VSg for some M-pseudovariety of all finite monoids V. Then 푛 NMon = V by (8.8). Let 푀 ∈ N be a monoid. Then 1푀 = 0푀 for some 푛 ∈ ℕ, since 푀 is nilpotent, which implies that 푀 is trivial. Hence NMon = 1, and so N = VSg = (NMon)Sg = 1Sg = 1, which is a contradiction.

Free objects for pseudovarieties If we want to follow the same path for pseudovarieties as for varieties, our next step should be to construct a ‘free V-semigroup’ for each S-pseudovariety V and a ‘free W-monoid’ for each M-pseudovariety W, and then to devise an analogue of laws and prove an analogue of Birk- hoff’s theorem. However, this is much more difficult for pseudovarieties than for varieties. We will outline the problems and describe the solution in this section and the next two sections. To simplify the explanation, we will only discuss S-pseudovarieties, but every result and construction in these sections has a parallel for M-pseudovarieties, replacing semigroups with monoids, homomorphisms with monoid homomorphisms, and subsemigroups with M-submonoids as appropriate. The basic problem in finding free objects for pseudovarieties is very simple: free objects are usually infinite, and members of a pseudovariety are always finite. Consider the S-pseudovariety N of finite nilpotent semi- + groups. For any finite alphabet 퐴 and 푛 ∈ ℕ, let 퐼푛 = { 푤 ∈ 퐴 ∶ |푤| ⩾ 푛 }. + + Then 퐴 /퐼푛 is a nilpotent semigroup; thus 퐴 /퐼푛 ∈ N. The semigroup + 푛 퐴 /퐼푛 contains at least 푛 elements (and indeed contains |퐴| elements if

Free objects for pseudovarieties • 156 |퐴| ⩾ 2). Thus, by taking 푛 to be arbitrarily large, we see that N contains arbitrarily large 퐴-generated semigroups. Since an 퐴-generated free object for N must map surjectively to each of these semigroups, it is clear that no semigroup in N is free. If we try to approach the idea of a free object through laws, we en- + counter another problem. It is clear that 퐴 /퐼푛 satisfies no law in at most |퐴| variables where the two sides of the law have length less than 푛. So if we try to base our free objects on laws, all S-pseudovarieties containing N will have the same free object. Let us look at free objects from another direction. The idea is that a free 퐴-generated object for a class X should be just general enough to be more general than any 퐴-generated object in X. Suppose we take two semigroups 푆1 and 푆2 in an S-pseudovariety V. Let 휑1 ∶ 퐴 → 푆1 and 휑2 ∶ 퐴 → 푆2 be functions such that im 휑1 generates 푆1 and im 휑2 generates 푆2. Let 푇 be the subsemigroup of 푆1 × 푆2 generated by { (푎휑1, 푎휑2) ∶ 푎 ∈ 퐴 }. Then 푇 is 퐴-generated and lies in V, since V is closed under ℙfin and 핊. Furthermore, the following diagram commutes:

퐴 휑1 휑2

푆1 푇 푆2 휋1 휋2

Thus 푇 is more general than both 푆1 and 푆2 as an 퐴-generated member of V. Furthermore, 푇 is the smallest such member of V. We could iterate this process, but, as our discussion of N shows, we will never find an element of V that is more general than all other members of V. A limiting process is needed.

Projective limits

A partially ordered set (퐼, ⩽) is a directed set if every pair Directed set of elements of 퐼 have an upper bound. [Notice that a directed set is not necessarily a join semilattice, because some pairs of elements might not have least upper bounds.] A topological semigroup is a semigroup equipped with a topology Topological semigroup such that the multiplication operation is a continuous mapping. Any semigroup can be equipped with the discrete topology and thus becomes a topological semigroup. Notice that finite semigroups are compact. For any 퐴-generated alphabet 퐴, an 퐴-generated topological semigroup is a pair (푆, 휑), where 푆 is topological semigroup a topological semigroup and 휑 ∶ 퐴 → 푆 is a map such that im 휑 generates a dense subsemigroup of 푆. We will often denote such an 퐴-generated topological semigroup by the map 휑 ∶ 퐴 → 푆. A homomorphism between Homomorphisms between 퐴-generated topological semigroups

Projective limits • 157 퐴-generated topological semigroups 휑1 ∶ 퐴 → 푆1 and 휑2 ∶ 퐴 → 푆2 is a continuous homomorphism 휓 ∶ 푆1 → 푆2 such that 휑1휓 = 휑2. A projective system is a collection of 퐴-generated topological sem- Projective system igroups { 휑푖 ∶ 퐴 → 푆푖 ∶ 푖 ∈ 퐼 }, where 퐼 is a directed set, such that for all 푖, 푗 ∈ 퐼 with 푖 ⩾ 푗 there is a connecting homomorphism 휆푖,푗 from 휑푖 ∶ 퐴 → 푆푖 to 휑푗 ∶ 퐴 → 푆푗 satisfying the following properties: for each 푖 ∈ 퐼, the homomorphism 휆푖,푖 is the identity map; for all 푖, 푗, 푘 ∈ 퐼 with 푖 ⩾ 푗 ⩾ 푘, we have 휆푖,푗휆푗,푘 = 휆푖,푘. The projective limit of this projective system is an 퐴-generated topo- Projective limit logical semigroup 훷 ∶ 퐴 → 푆 equipped with homomorphisms 훷푖 from 훷 ∶ 퐴 → 푆 to 휑푖 ∶ 퐴 → 푆푖. such that the following properties hold: 퐴 1) For all 푖, 푗 ∈ 퐼 with 푖 ⩾ 푗, we have 훷푖휆푖,푗 = 훷푗. 훹 2) If there is another 퐴-generated topological semigroup 훹 ∶ 퐴 → 푇 푇 훷 휑푖 휑푗 and homomorphisms 훹푖 from 훹 ∶ 퐴 → 푇 to 휑푖 ∶ 퐴 → 푆 such that for all 푖, 푗 ∈ 퐼 with 푖 ⩾ 푗, we have 훹휆 = 훹 , then there exists 훩 푖 푖,푗 푗 훹푖 훹푗 a homomorphism 훩 from 훹 ∶ 퐴 → 푇 to 훷 ∶ 퐴 → 푆 such that 푆 훩훷푖 = 훹푖. That is, the diagram in Figure 8.1 commutes. 훷푖 훷푗 Let us first show that the projective limit is unique (up to isomor- 푆 푆 푖 휆 푗 phism); we will then show that it exists. Suppose 훷 ∶ 퐴 → 푆 and 훷′ ∶ 퐴 → 푖,푗 푆′ are both projective limits of the projective system { 휑 ∶ 퐴 → 푆 ∶ 푖 ∈ 퐼 }. FIGURE 8.1 푖 푖 Property 2) of the projective By property 2) above, there are homomorphisms 훩 from 훷 ∶ 퐴 → 푆 to limit of { 휑푖 ∶ 퐴 → 푆푖 ∶ 푖 ∈ 훷′ ∶ 퐴 → 푆′ and 훩′ from 훷′ ∶ 퐴 → 푆′ to 훷 ∶ 퐴 → 푆. Thus we have 퐼 } with connecting homomor- 훷훩훩′ = 훷 and 훷′훩′훩 = 훷′; hence 훩훩′|퐴훷 = id퐴훷 and 훩′훩|퐴훷′ = id퐴훷′. phisms 휆푖,푗. Hence 훩훩′ restricted to the subsemigroup generated by 퐴훷 is the identity Uniqueness of the map; since this subsemigroup is dense in 푆 and 훩 and 훩′ are continuous, projective limit we have 훩훩′ = id푆. Similarly 훩′훩 = id푆′. So 훩 and 훩′ are mutually inverse isomorphisms between 훷 ∶ 퐴 → 푆 and 훷′ ∶ 퐴 → 푆′. In order to construct the projective limit, we proceed as follows. Let Construction of the projective limit 푆 = { 푠 ∈ ∏ 푆푖 ∶ (∀푖, 푗 ∈ 퐼)(푖 ⩾ 푗 ⇒ ((푖)푠)휆푖,푗 = (푗)푠) }. 푖∈퐼 Notice that

푠, 푡 ∈ 푆

⇒ (∀푖, 푗 ∈ 퐼)(푖 ⩾ 푗 ⇒ ((푖)푠)휆푖,푗 = (푗)푠 ∧ ((푖)푡)휆푖,푗 = (푗)푡)

⇒ (∀푖, 푗 ∈ 퐼)(푖 ⩾ 푗 ⇒ ((푖)푠)휆푖,푗((푖)푡)휆푖,푗 = (푗)푠(푗)푡)

⇒ (∀푖, 푗 ∈ 퐼)(푖 ⩾ 푗 ⇒ ((푖)(푠푡))휆푖,푗 = (푗)(푠푡)) ⇒ 푠푡 ∈ 푆; thus 푆 is a subsemigroup of ∏푖∈퐼 푆푖. Furthermore, 푆 is equipped with the induced topology from the product topology on ∏푖∈퐼 푆푖. Let 훷 ∶ 퐴 → 푆 be defined by (푖)(푎훷) = 푎휑푖. For each 푖 ∈ 퐼, let 훷푖 be the projection homomorphism from 푆 to 푆푖.

Projective limits • 158 P r o p o s i t i o n 8 . 1 1. 훷 ∶ 퐴 → 푆 is an 퐴-generated topological semigroup and satisfies the properties 1) and 2) above. Hence 훷 ∶ 퐴 → 푆 is a projective limit. Proof of 8.11. We first have to show that 퐴훷 generates a dense subsemigroup of 푆. Since the topology of 푆 is induced by the product topology on ∏푖∈퐼 푆푖, we can work with the product topology instead. Let 푠 ∈ 푆. Let 퐾 be a neighbourhood of 푠. Assume without loss that 퐾 is an open set

(in the product topology). Thus 퐾 = ∏푖∈퐼 퐾푖, where each 퐾푖 ⊆ 푆푖 is open and 퐾푖 = 푆푖 for all but finitely many 푖 ∈ 퐼. Let 푖푗 ∈ 퐼 (where 푗 = 1, … , 푛) be the indices for which 퐾 ≠ 푆 . 푖푗 푖푗 Let ℎ be an upper bound for { 푖푗 ∶ 푗 = 1, … , 푛 }; such an ℎ exists because 퐼 is a directed set. Let 퐿 = ⋂푛 퐾 휆−1 ⊆ 푆 . Notice that 푗=1 푖푗 ℎ,푖푗 ℎ (ℎ)푠 ∈ ((푖 )푠)휆−1 for all 푗 = 1, … , 푛, so 퐿 contains (ℎ)푠 and is thus non- 푗 ℎ,푖푗 empty. Furthermore, 퐿 is an intersection of open sets because each 휆푖,푗 is continuous and each 퐾 is open; hence 퐿 is itself open. Since 퐴휑 푖푗 ℎ + generates a dense subset of 푆ℎ, the set there is a word 푤 ∈ 퐴 such that 푤휑ℎ ∈ 퐿. Let 푡 = 푤훷. Thus (푖)푡 = 푤휑푖 for all 푖 ∈ 퐼. For 푗 = 1, … , 푛, we have

(푖 )푡 = 푤휑 = 푤휑 휆 ∈ 퐿휆 ⊆ 퐾 . 푗 푖푗 푘 푘,푖푗 푘,푖푗 푖푗 Hence 푤훷 = 푡 ∈ 퐾. Therefore 퐴훷 generates a dense subset of 푆.

Since 푆 consists of elements 푠 ∈ ∏푖∈퐼 푆푖 with ((푖)푠)휆푖,푗 = (푗)푠, it is immediate that 훷푖휆푖,푗 = 훷푗; hence 훷 ∶ 퐴 → 푆 satisfies property 1). Now let 훹 ∶ 퐴 → 푇 be an 퐴-generated topological semigroup as described in property 2). Define 훩 ∶ 푇 → 푆 by (푖)(푡훩) = 푡훹푖. (Note that 푡훩 ∈ 푆 since 훹푖휆푖,푗 = 훹푗.) Then this map is a continuous homomorphism since each 훹푖 is continuous. Finally, 훩훷푖 = 훩휋푖 = 훹푖. So 훷 ∶ 퐴 → 푆 satisfies property 2). 8.11

Notice that if all of the 푆푖 are compact, so is ∏푖∈퐼 푆푖 by Tychonoff’s theorem. Furthermore, since each 휆푖,푗 is continuous and the condition ((푖)푠)휆푖,푗 = (푗)푠 involves only two components of the product, 푆 is closed in the ∏푖∈퐼 푆푖. Hence if all the 푆푖 are compact, 푆 is also compact. A profinite semigroup is a projective limit of a projective system of Profinite semigroup finite semigroups for some suitable choice of generators. Notice that any finite semigroup is [isomorphic to] a profinite semigroup. To see this, let 푆 be a finite semigroup, and take 퐼 = {1} and 푆1 = 푆. It is easy to see that the projective limit of this projective system is isomorphic to 푆.

Pro-V semigroups

Let V be an S-pseudovariety. A profinite semigroup 푆 is Pro-V semigroup

Pro-V semigroups • 159 pro-V if it is a projective limit of a projective system containing only semigroups from V. Let us return of the problem of finding a free object for V. For a generating set 퐴, the idea is to take the projective limit of the projective system containing every 퐴-generated semigroup in V. Strictly speaking, we take one semigroup from every isomorphism class in V, and let the connecting homomorphisms be the unique homomorphisms that respect ̅ the generating set 퐴. The projective limit of this system is denoted Ω퐴V. If the 퐴-generated semigroups in V are { 휑푖 ∶ 퐴 → 푆푖 ∶ 푖 ∈ 퐼 }, then there ̅ is a natural map 휄 ∶ 퐴 → Ω퐴V given by (푖)(푎휄) = 푎휑푖. (This is the map 훷 in the discussion of the projective limit above.) Denote by Ω퐴V the ̅ [dense] subsemigroup of Ω퐴V generated by 퐴휄. ̅ The following result essentially says that the profinite semigroup Ω퐴V is a free object for 퐴-generated pro-V semigroups:

̅ P r o p o s i t i o n 8 . 1 2. For any pro-V semigroup 푆 and map 휗 ∶ 퐴 → 푆, Ω퐴V is a free object for V ̂ ̅ there is a unique continuous homomorphism 휗 ∶ Ω퐴V → 푆 such that 휄휗̂ = 휗; that is, such that the following diagram commutes:

휄 ̅ 퐴 Ω퐴V 휗 휗̂ 푆

Proof of 8.12. Since pro-V semigroups are subdirect products of members of V, it is sufficient to consider the case when 푆 lies in V. Without loss of generality, assume 푆 is generated by 퐴휗. Then 푆 is [isomorphic to] an 퐴-generated semigroup in V; that is, 푆 is [isomorphic to] one of the 퐴-generated semigroups 휑푗 ∶ 퐴 → 푆푗 in the projective system whose ̅ projective limit is Ω퐴V. ̂ ̅ Let 휗 be the projection 휋푗 ∶ Ω퐴V → 푆푗 ≃ 푆. Finally, we have to show that 휗̂ is the unique continuous homomorphism with this property. Let ̅ 휓 ∶ Ω퐴V → 푆 be some continuous homomorphism with 휄휓 = 휗. Since ̂ ̂ 휄휗 = 휗, we see that 휓|퐴휄 = 휗|퐴휄 and hence, since 퐴휄 generates Ω퐴V, we have 휓| = 휗|̂ . Since Ω V is dense in Ω̅ V and 휓 is continuous, we Ω퐴V Ω퐴V 퐴 퐴 have 휓 = 휗̂. 8.12 ̅ In light of Proposition 8.12, for any S-pseudovariety V, we call Ω퐴V Free pro-V semigroup the free pro-V semigroup on 퐴. P r o p o s i t i o n 8 . 1 3. Let V be an S-pseudovariety that is not the trivial ̅ S-pseudovariety 1. Then the map 휄 ∶ 퐴 → Ω퐴V is injective. Proof of 8.13. Since V ≠ 1, there are arbitrarily large semigroups in V. Hence for any 푎, 푏 ∈ 퐴 with 푎 ≠ 푏, there is some 휑푖 ∶ 퐴 → 푆푖 such that 푎휑푖 ≠ 푏휑푖. Therefore, (푖)(푎휄) = 푎휑푖 ≠ 푏휑푖 = (푖)(푏휄), and so 푎휄 ≠ 푏휄. Thus 휄 is injective. 8.13

Pro-V semigroups • 160 Proposition 8.13 means that, when we consider any non-trivial S- ̅ pseudovariety V, we can identify 퐴 with the subset 퐴휄 of Ω퐴V. From now on, assume that V is a non-trivial S-pseudovariety. L e m m a 8 . 1 4. Let 푆 be a pro-V semigroup and let 퐾 ⊆ 푆. Then the following conditions are equivalent: a) there exists a continuous homomorphism 휑 ∶ 푆 → 퐹 such that 퐹 ∈ V and 퐾 = 퐾휑휑−1; b) 퐾 is a clopen subset of 푆. Proof of 8.14. Suppose that condition a) holds. Then since 퐹 is finite and has the discrete topology, 퐾휑 is clopen in 퐹. Since 휑 is a continuous homomorphism, 퐾 is clopen since it is the pre-image under 휑 of the 퐾휑. Thus condition b) holds. Now suppose that condition b) holds and 퐾 is a clopen subset of 푆. Now, 푆 be a subdirect product of semigroups in V. That is, 푆 is a subsemigroup of ∏푖∈퐼 푇푖 for some 푇푖 ∈ V. Then 퐾 = 푆 ∩ (퐾1 ∪ … ∪ 퐾푛), where each 퐾ℓ is a product of the form ∏푖∈퐼 푋ℓ,푖 with 푋ℓ,푖 ⊆ 푆푖 and 푋ℓ,푖 = 푆푖 for all but finitely many indices. Let

퐽 = { 푖 ∈ 퐼 ∶ (∃ℓ ∈ {1, … , 푛})(푇ℓ,푖 ≠ 푆푖) }; notice that 퐽 is finite. Let 휑 ∶ 푆 → ∏푖∈퐽 푆푖 be the natural projection. −1 Then 휑 is continuous, ∏푖∈퐽 푆푖 is finite, and 퐾 = 퐾휑휑 . Thus condition a) holds. 8.14 P r o p o s i t i o n 8 . 1 5. Let 푆 be pro-V and let 푇 be profinite. Let 휑 ∶ 푆 → 푇 be a continuous homomorphism. Then im 휑 is pro-V and belongs to V if it is finite. Proof of 8.15. Since 푇 is a subdirect product of finite semigroups, it is sufficient to consider the case where 푇 is finite and 휑 is surjective and show that 푇 ∈ V. −1 For each 푡 ∈ 푇, let 퐾푡 = 푡휑 . Then every 퐾푡 is a pre-image of a clopen set under the continuous homomorphism 휑 and so is clopen. By Lemma 8.14, there is, for each 푡 ∈ 푇, a continuous homomorphism −1 휓푡 ∶ 푆 → 퐹푡 with 퐹푡 ∈ V such that 퐾푡휓푡휓푡 = 퐾푡. Let 퐹 = ∏푡∈푇 퐹푡; notice that 퐹 ∈ V since 푇 is finite. Let 휓 ∶ 푆 → 퐹 be defined by (푡)(푥휓) = 푥휓푡. Then ker 휓 ⊆ ker 휑. Hence there is a homomorphism 휗 ∶ im 휓 → 푇 given by (푥휓)휗 = 푥휑. Since 휑 is surjective, 휗 is a surjective homomorphism from the subsemigroup im 휓 of 퐹 to the semigroup 푇. Hence 푇 ≼ 퐹 and so 푇 ∈ V. 8.15 ̅ Propositions 8.12, 8.13, and 8.15 together show that Ω퐴V is a very good analogue for pseudovarieties of free algebras for varieties: maps ̅ from 퐴 can be extended to homomorphisms from Ω퐴V, the ‘basis’ 퐴 ̅ (usually) embeds in Ω퐴V, and, finally, the only finite semigroups that are ̅ homomorphic images of Ω퐴V are the semigroups in V.

Pro-V semigroups • 161 Pseudoidentities Earlier in this chapter, we saw how varieties of T-algebras, and in particular varieties of semigroups, can be defined using laws. Recall that a law in a variety V of T-algebras is a pair 푢, 푣 ∈ 퐹T(퐴), usually written as a formal equality 푢 = 푣, and that a T-algebra 푆 satisfies this law if 푢휑̂ = 푣휑̂ for all homomorphisms 휑̂ ∶ 퐹T(퐴) → 푆 extending maps 휑 ∶ 퐴 → 푆. Now that we have free objects for S-pseudovarieties available, we can the study the analogue of laws for finite semigroups. ̅ Let V be an S-pseudovariety. A V-pseudoidentity is a pair 푢, 푣 ∈ Ω퐴V, Pseudoidentities usually written as a formal equality 푢 = 푣. A pro-V semigroup 푆 satisfies ̅ this pseudoidentity if, for every continuous homomorphism 휗 ∶ Ω퐴V → 푆 we have 푢휗 = 푣휗. ̅ Now let V be an M-pseudovariety. Then Ω퐴V also exists, with the corresponding properties, and is a monoid. So we also have V-pseudo- ̅ identities 푢 = 푣 in this case, where 푢, 푣 ∈ Ω퐴V, and here 푢 or 푣 may be the ̅ identity of Ω퐴V. In this context, a pro-V monoid 푀 satisfies this pseudo- ̅ identity if, for every continuous monoid homomorphism 휗 ∶ Ω퐴V → 푀 we have 푢휗 = 푣휗.

L e m m a 8 . 1 6. Let V and W be S-pseudovarieties (respectively, M-pseu- ̅ ̅ dovarieties) with W ⊆ V and let 휋 ∶ Ω퐴V → Ω퐴W be the natural projection homomorphism (respectively, monoid homomorphism). Then for ̅ any 푢, 푣 ∈ Ω퐴V, every semigroup in W satisfies 푢 = 푣 if and only if 푢휋 = 푣휋. 8.16

Let 훴 be a set of V-pseudoidentities. Let ⟦훴⟧V denote the class of all 푆 ∈ V that satisfy all the V-pseudoidentities in 훴.

Reiterman’s Theorem 8.17. Let W be a subclass of a S-pseudo- Reiterman’s theorem variety (respectively, M-pseudovariety) V. Then W is an S-pseudovariety (respectively, M-pseudovariety) if and only if W = ⟦훴⟧V for some set 훴 of V-pseudoidentities.

Proof of 8.17. We prove the result for S-pseudovarieties; the same reasoning works for M-pseudovarieties with the standard modifications.

Part 1. Suppose W = ⟦훴⟧V. By reasoning parallel to the proof of Theo- rem 8.2, we see that W is closed under ℍ, 핊, and ℙfin and is thus an S- pseudovariety. Part 2. Suppose W is an S-pseudovariety. Fix a countably infinite alphabet 퐴. Let 훴 be the set of all V-pseudoidentities 푢 = 푣 satisfied by all ̅ semigroups in W, where 푢, 푣 ∈ Ω퐵V and 퐵 ⊆ 퐴. Clearly W ⊆ ⟦훴⟧V; we aim to prove equality. Let X = ⟦훴⟧V and let 푆 ∈ X. Then since 퐴 is infinite and 푆 is finite, there exists some 퐵 ⊆ 퐴 and a surjective continuous homomorphism ̅ ̅ ̅ 휑 ∶ Ω퐵X → 푆. Let 휋 ∶ Ω퐵X → Ω퐵W be the natural projection.

Pseudoidentities • 162 ̅ Suppose 푢, 푣 ∈ Ω퐵X are such that 푢휋 = 푣휋. Then by Lemma 8.16, every semigroup in W satisfies 푢 = 푣. Thus 푢 = 푣 is a V-pseudoidentity in 훴; and thus 푆 satisfies 푢 = 푣. In particular, 푢휑 = 푣휑. This shows that ker 휋 ⊆ ker 휑. ̅ Therefore the map 휓 ∶ Ω퐵W → 푆 defined by (푥휋)휓 = 푥휑 is a well- defined surjective homomorphism. −1 ̅ For any subset 퐾 of 푆, the subset 퐾휑 of Ω퐵X is closed because 휑 is continuous. The map 휋 maps closed sets to closed sets because it is a projection of compact spaces. Hence 퐾휓−1 = 퐾휑−1휋 is closed. Thus 휓 is continuous. By Proposition 8.15, 푆 ∈ W. Therefore ⟦훴⟧V = X ⊆ W. 8.17 If V is an S-pseudovariety (respectively M-pseudovariety) and 훴 is a Bases of pseudoidentities set of S-pseudoidentities (respectively, M-pseudoidentities) such that V = ⟦훴⟧S (respectively, V = ⟦훴⟧M), then 훴 is called a basis of pseudoidentities for V. If there is a finite set of pseudoidentities 훴 such that V = ⟦훴⟧S (respectively, ⟦훴⟧M), then V is finitely based. In order to actually write down useful pseudoidentities, we introduce Notation for some new concepts and notation. Let 푇 be a finite semigroup, 푥 ∈ 푇, pseudoidentities 푛!+푖 and 푖 ∈ ℤ. Consider the sequence (푥 )푛. This sequence is eventually constant: for all 푛 > max{|푖|, |푇|}, all terms 푥푛!+푖 are equal. More generally, 푛!+푖 let 푇 be a profinite semigroup. Then the sequence (푥 )푛 converges to a 휔+푖 ̅ limit, which we denote 푥 . In particular, this holds when 푇 is Ω퐴S and 푥 ∈ 퐴. ̅ Let 푆 be finite and let 휗 ∶ Ω퐴S → 푆 be a continuous homomorphism, the powers of 푥휗 are not all distinct: we have (푥휗)푚+푘 = (푥휗)푚 for some 푚, 푘 ∈ ℕ. Let (푥휗)푛 be the identity of the cyclic group 퐶 = {(푥휗)푚, … , (푥휗)푚+푘−1}. Since (푥휗)푛 = (푥휗)푚! = 푥푚!휗 for all 푚 ⩾ 푛, we have (푥휔)휗 = (푥휗)푛. That is, (푥휔)휗 is the unique idempotent power of 푥휗. Furthermore, 푥휔−1휗 is the inverse of 푥휔+1휗 in 퐶. We can interpret this notation in 푆 by define new operations 휔+푖 on finite semigroups. For any finite semigroup 푆, the operation 휔 ∶ 푆 → 푆 takes any element 푦 to its unique idempotent power 푦휔. For any 푘 ∈ ℕ, the operation 휔+푘 ∶ 푆 → 푆 takes any element 푦 to 푦휔푦푘, and 휔−푘 ∶ 푆 → 푆 takes 푦 to the inverse of 푦휔+푘 in the [finite] cyclic subgroup {푦휔, 푦휔+1, …}. We can now give explicit examples of pseudoidentities defining particular pseudovarieties of finite semigroups and monoids. See Tables 8.4 and 8.5 for a summary. E x a m p l e 8 . 1 8. a) The S-pseudovariety of all finite aperiodic semigroups A is defined by the pseudoidentity 푥휔+1 = 푥휔. b) The S-pseudovariety of finite nilpotent semigroups N is defined by the pseudoidentities 푦푥휔 = 푥휔 and 푥휔푦 = 푥휔. These pseudoidentities essentially say that 푥휔휗 is a zero, and so we abbreviate them by 푥휔 = 0. c) The S-pseudovariety of finite groups G is defined by the S-pseudoidentities 푦푥휔 = 푦 and 푥휔푦 = 푦. Since these S-pseudoidentities

Pseudoidentities • 163 Pseudovariety Symbol Pseudoidentities See also Semigroups S — Trivial semigroup 1 푥 = 푦 Null semigroups Z 푥푦 = 푧푡 Nilpotent semigroups N 푥휔 = 0 Exa. 8.18(b) Left zero semigroups LZ 푥푦 = 푥 Right zero semigroups RZ 푥푦 = 푦 Rectangular bands RB 푥푦푥 = 푥 Exer. 8.4 Comp. simple sgrps CS 푥휔+1 = 푥 Exer. 8.8 휔 Comp. regular sgrps CR (푥푦) 푥 = 푥 Exer. 8.9 TABLE 8.4 휔 Left simple sgrps LS 푥푦 = 푥 Exer. 8.10 S-pseudovarieties of semi- Right simple sgrps RS 푦휔푥 = 푥 Exer. 8.10 groups. The pseudoidentity Left-trivial sgrps K 푥휔푦 = 푥휔 pp. 188–189 푥(푦푧) = (푥푦)푧 is implicitly Right-trivial sgrps D 푦푥휔 = 푥휔 pp. 188–189 assumed in every case.

Pseudovariety Symbol Pseudoidentities See also Monoids M — Trivial monoid 1 푥 = 1 Commutative monoids Com 푥푦 = 푦푥 푥2 = 푥, Semilattices with ident. Sl { 푥푦 = 푦푥 Aperiodic monoids A 푥휔+1 = 푥휔 Exa. 8.18(a) L-trivial monoids L 푦(푥푦)휔 = (푥푦)휔 Pr. 8.20(a) R-trivial monoids R (푥푦)휔푥 = (푥푦)휔 Pr. 8.20(b) (푥푦)휔푥 = (푥푦)휔, J-trivial monoids J { Pr. 8.20(c) 푦(푥푦)휔 = (푥푦)휔 TABLE 8.5 Comp. regular monoids CR (푥푦)휔푥 = 푥 Exer. 8.9 M-pseudovarieties of mon- 휔 oids. The pseudoidentities Groups G 푥 = 1 Exa. 8.18(c) 푥(푦푧) = (푥푦)푧, 푥1 = 1, and 푥푦 = 푦푥, 1푥 = 푥 are implicitly assumed Abelian groups Ab { 푥휔 = 1 in every case.

휔 ̅ essentially say that 푥 휗 (where 휗 ∶ Ω퐴S → 푆 is a homomorphism) is an identity, we abbreviate them by 푥휔 = 1. If we consider the M- pseudovariety of finite groups instead, then 푥휔 = 1 is a genuine M- pseudoidentity.

We now have two different ways to define pseudovarieties: wecan specify a set of 푆- or 푀-pseudoidentities and consider the S-or M-pseudovarieties of semigroups or monoids they define, or we can specify a set of finite semigroups or monoids and consider the S- or M-pseudovariety they generate. These ways of defining pseudovarieties interact with the lattices of pseudovarieties in different but complementary ways.

Pseudoidentities • 164 If 훴 and 훵 are sets of 푆-pseudoidentities, then

⟦훴⟧S ⊓ ⟦훵⟧S = ⟦훴 ∪ 훵⟧S, (8.8) and similarly for 푀-pseudoidentities. For example, the M-pseudovariety of finite Abelian groups is

휔 휔 Ab = ⟦푥푦 = 푦푥⟧M ⊓ ⟦푥 = 1⟧M = ⟦푥푦 = 푦푥, 푥 = 1⟧M. On the other hand, for any classes X and Y of finite semigroups,

VS(X) ⊔ VS(Y) = VS(X ∪ Y), (8.9) and similarly for finite monoids. Furthermore, the operators Sg and Mon interact with pseudoidentities in a pleasant way. Let 훴 be a set of M-pseudoidentities. Let 훴Sg be the set of S-pseudoidentities that can be obtained from 훴 as follows: 1) by substituting 1 for some of the variables; 2) replacing pseudoidentities of the form 푢 = 1, where 푢 is not the identity, by 푢푥 = 푥 and 푥푢 = 푥, where 푥 is a new symbol not in 푢; 3) deleting the pseudoidentity 1 = 1 if it is present.

P r o p o s i t i o n 8 . 1 9. a) Let 훴 be a set of S-pseudoidentities. Then

⟦훴⟧M = (⟦훴⟧S)Mon. b) Let 훴 be a set of M-pseudoidentities. Then

⟦훴 ⟧ = (⟦훴⟧ ) . Sg S M Sg Proof of 8.19. a) Let 푆 be a finite monoid. Then

푆 ∈ ⟦훴⟧M ⇔ 푆 is a monoid that satisfies all the S-pseudoidentities in 훴

⇔ 푆 is a monoid that belongs to ⟦훴⟧S

⇔ 푆 ∈ (⟦훴⟧S)Mon. b) For brevity, let V = ⟦훴⟧ . It is clear that V = ⟦훴 ⟧ , and so V ⊆ M Sg M Sg ⟦훴′⟧S. Conversely, if 푆 satisfies all the S-pseudoidentities in 훴Sg, then 푆1 satisfies all the M-pseudoidentities in 훴 . Thus 푆 ∈ ⟦훴 ⟧ implies Sg Sg S 1 푆 ∈ V, which implies 푆 ∈ SSg by (8.7). Hence ⟦훴′⟧S ⊆ VSg. 8.19 Proposition 8.19 allows us to switch from M-pseudoidentities for an M-pseudovariety of monoids V to S-pseudoidentities for corresponding monoidal S-pseudovariety of semigroups VSg.

Pseudoidentities • 165 A semigroup 푆 is H-trivial (respectively, L-trivial, R-trivial, D-trivial, J-trivial) if H (respectively, L, R, D, J) is the identity relation id푆. A finite semigroup is H-trivial if and only if it is aperiodic by Proposition 7.4, and is D-trivial if and only if it is J-trivial by Proposition 3.3. In particular, therefore, the class of H-trivial finite monoids is the M-pseudovariety A. Let L, R, and J be, respectively, the classes of L-, R-, and J-trivial monoids. P r o p o s i t i o n 8 . 2 0. a) The class L is an M-pseudovariety of monoids, and

휔 휔 L = ⟦푦(푥푦) = (푥푦) ⟧M. b) The class R is an M-pseudovariety of monoids, and

휔 휔 R = ⟦(푥푦) 푥 = (푥푦) ⟧M. c) The class J is an M-pseudovariety of monoids, and

휔 휔 휔 휔 J = L ⊓ R = ⟦푦(푥푦) = (푥푦) , (푥푦) 푥 = (푥푦) ⟧M 휔 휔 휔 휔+1 = ⟦(푥푦) = (푦푥) , 푥 = 푥 ⟧M. Proof of 8.20. a) Let 푆 ∈ L; that is, 푆 is a finite L-trivial monoid. Then 푆 is H-trivial and so aperiodic, and therefore 푧휔 = 푧휔+1 for all 푧 ∈ 푆. Let 푥, 푦 ∈ 푆. Then 푥(푦(푥푦)휔) = (푥푦)휔 and so 푦(푥푦)휔 L (푥푦)휔 and 휔 휔 so 푦(푥푦) = (푥푦) since L = id푆. So every finite L-trivial monoid satisfies the pseudoidentity 푦(푥푦)휔 = (푥푦)휔. On the other hand, let 푆 be a finite monoid satisfying the pseudoidentity 푦(푥푦)휔 = (푥푦)휔. Let 푧, 푡 ∈ 푆 be such that 푧 L 푡. Then there exist 푝, 푞 ∈ 푆 such that 푝푧 = 푡 and 푞푡 = 푧. Then 푡 = (푝푞)푡 = (푝푞)2푡 = … and so 푡 = (푝푞)휔푡. Similarly 푧 = 푞푡 = 푞(푝푞)푡 = 푞(푝푞)2푡 = … and so 푧 = 푞(푝푞)휔푡. Substitute 푝 for 푥 and 푞 for 푦 in the pseudoidentity to 휔 휔 see that 푡 = (푝푞) 푡 = 푞(푝푞) 푡 = 푧. So L = id푆. Thus 푆 is L-trivial and so 푆 ∈ L. Thus the class of finite L-trivial monoids L is a pseudovariety, and 휔 휔 L = ⟦푦(푥푦) = (푥푦) ⟧S. b) The reasoning is dual to part a). c) A monoid is D-trivial if and only if it is both L-trivial and R-trivial, and a finite monoid is J-trivial if and only if it is D-trivial. Thus 휔 휔 휔 휔 J = L ⊓ R, and L ⊓ R = ⟦푦(푥푦) = (푥푦) , (푥푦) 푥 = (푥푦) ⟧M by (8.8). 휔 휔 휔 휔 Suppose 푆 ∈ ⟦푦(푥푦) = (푥푦) , (푥푦) 푥 = (푥푦) ⟧M. Putting 푥 = 푦 in the first pseudoidentity shows that 푥(푥2)휔 = (푥2)휔. Since 푥휔 = (푥2)휔, it follows that 푥휔 = 푥휔+1 for all 푥 ∈ 푆. Let 푥, 푦 ∈ 푆 and let 푛 be large enough that (푥푦)휔 = (푥푦)푛 and (푦푥)휔 = (푦푥)푛. Then (푦푥)휔푦 = (푦푥)푛푦 = 푦(푥푦)푛 = 푦(푥푦)휔. So 푆 satisfies the pseudoidentity (푥푦)휔 = 휔 휔 휔 휔 휔+1 (푦푥) . Hence 푆 ∈ ⟦(푥푦) = (푦푥) , 푥 = 푥 ⟧M. 휔 휔 휔 휔+1 Now suppose 푆 ∈ ⟦(푥푦) = (푦푥) , 푥 = 푥 ⟧M. Let 푥, 푦 ∈ 푆. Us- ing both pseudoidentities, we see that (푥푦)휔 = (푥푦)휔+1 = (푦푥)휔+1 =

Pseudoidentities • 166 푦(푥푦)휔푥. Hence (푥푦)휔 = 푦2(푥푦)휔푥2 = 푦3(푥푦)휔푥3 = … and so (푥푦)휔 = 푦휔(푥푦)휔푥휔 = 푦휔+1(푥푦)휔푥휔 = 푦(푥푦)휔. Similarly, (푥푦)휔 = 휔 휔 휔 휔 휔 (푥푦) 푥. Therefore 푆 ∈ ⟦푦(푥푦) = (푥푦) , (푥푦) 푥 = (푥푦) ⟧S. Thus

휔 휔 휔 휔 ⟦푦(푥푦) = (푥푦) , (푥푦) 푥 = (푥푦) ⟧M 휔 휔 휔 휔+1 = ⟦(푥푦) = (푦푥) , 푥 = 푥 ⟧M. 8.20 We now define another operator that connects M-pseudovarieties of monoids with S-pseudovarieties of semigroups. For any M-pseudovariety of monoids V, let

핃V = { 푆 ∈ S ∶ (∀푒 ∈ 퐸(푆))(푒푆푒 ∈ V) }.

For any semigroup 푆 and 푒 ∈ 퐸(푆), the subset 푒푆푒 forms a submonoid Local submonoid, locally V whose identity is 푒, called the local submonoid of 푆 at 푒. Thus a semigroup in 핃V is said to be locally V. Let 훴 be a basis of M-pseudoidentities for a M-pseudovariety of monoids V. Let 푧 be a new symbol that does not appear in 훴. Let 훴′ be the set of S-pseudovarieties obtained by substituting 푧휔푥푧휔 for 푥 in every M- pseudoidentity in 훴, for every symbol 푥 that appears in 훴, and substitut- 휔 ing 푧 for 1 in every M-pseudoidentity in 훴. Then 핃V = ⟦훴′⟧S, and so 핃V is an S-pseudovariety of semigroups. In particular, the variety of locally trivial semigroups is

휔 휔 휔 핃1 = ⟦푧 푥푧 = 푧 ⟧S; we will study these further in the next chapter.

Semidirect product of pseudovarieties

The semidirect product of two S-pseudovarieties V and Semidirect product W, denoted V ⋊ W, is the S-pseudovariety generated by all semidirect of pseudovarieties products 푆 ⋊휑 푇, where 푆 ∈ V, 푇 ∈ W, and 휑 ∶ 푇 → End(푆) is an anti-homomorphism. We will not explore semidirect products of pseudovarieties in detail; we mention only the following result, which allows us to re-state the Krohn–Rhodes theorem in a more elegant form:

P r o p o s i t i o n 8 . 2 1. The semidirect product of S-pseudovarieties is associative.

Proof of 8.21. [Technical, and omitted.] 8.21

Notice that it is the semidirect product of S-pseudovarieties that is associative. There is no natural definition for the associativity of the semidirect

Semidirect products of pseudovarieties • 167 product of semigroups: by the definition of semidirect products (see pages 128–129), the expression (푆 ⋊휑 푇) ⋊휓 푈 only makes sense if the map 휓 is an anti-homomorphism from 푈 to End(푆 ⋊ 휑푇), whereas the expression 푆 ⋊휑 (푇 ⋊휓 푈) only makes sense if the map 휓 is an anti- homomorphism from 푈 to End(푇). The Krohn–Rhodes theorem shows that every finite semigroup isa wreath product of its subgroups and copies of the aperiodic semigroup 푈3. Now, if V and W are S-pseudovarieties and 푆 ∈ V and 푇 ∈ W, then 푆푇 ∈ V (since V is closed under finitary direct products); hence 푆 ≀ 푇 ∈ V ⋊ W. Notice furthermore that V ⊆ V⋊W since every pseudovariety contains the trivial semigroup. Therefore the Krohn–Rhodes theorem can be restated in terms of S-pseudovarieties as

‘⋊ A ⋊ G’ appears 푘 times S = ⋃ G ⋊⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞A ⋊ G ⋊ ⋯ ⋊ A ⋊ G. 푘∈ℕ∪{0}

Exercises [See pages 236–243 for the solutions.] 8.1 Let 푆 be cancellative semigroup that satisfies a law 푢 = 푣 where 푢, 푣 ∈ 퐴+ and 푢 and 푣 are not equal words. Without loss of generality, assume |푢| ⩽ |푣|. Let 푤 ∈ 퐴∗ be the longest common suffix of 푢 and 푣. (That is, 푢 = 푢′푤 and 푣 = 푣′푤, where 푢′ and 푣′ do not end with the same letter.) Prove that a) if 푢 = 푤, then 푆 is a group; b) if 푢 ≠ 푤 then 푆 is group-embeddable. ✴8.2 a) Prove, directly from the definition, that the class of finite nilpotent semigroups is a pseudovariety. b) Prove that the class all nilpotent semigroups is not a variety. ✴8.3 Recall that a semigroup is orthodox if it is regular and its idempotents form a subsemigroup. Prove that the class of orthodox completely regular semigroups forms a variety and that it is defined by the laws 푥푥−1 = 푥−1푥 and 푥푦푦−1푥−1푥푦 = 푥푦. 8.4 Let RB be the class of rectangular bands. a) Prove, directly from the definition, that RB is a variety. b) Prove that RB is defined by the law 푥푦푥 = 푥. c) Prove that RB is also defined by the laws 푥2 = 푥 and 푥푦푧 = 푥푧. d) Give an example of a semigroup that satisfies 푥푦푧 = 푥푧 but is not a rectangular band.

Exercises • 168 8.5 Let X be the class of semigroups isomorphic to a direct product of a group and a rectangular band. Prove that X is a variety and is defined by the laws 푥푥−1 = 푥−1푥 and 푥−1푦푦−1푥 = 푥−1푥.

8.6 Let T be a type, and let { V푖 ∶ 푖 ∈ 퐼 } be a collection of pseudovarieties of T-algebras. Prove that ⋂푖∈퐼 V푖 is a pseudovariety. 8.7 Prove that (VMon)Sg ⊆ V for any S-pseudovariety of semigroups V. Give an example to show that the inclusion may be strict. ✴8.8 Prove that the pseudovariety of finite completely regular semigroups 휔+1 CR is ⟦푥 = 푥⟧S. ✴8.9 Prove that the pseudovariety of finite completely simple semigroups 휔 CS is ⟦(푥푦) 푥 = 푥⟧S. 휔 8.10 Prove that ⟦푥푦 = 푥⟧S is the class of finite left simple semigroups. 휔 [Dual reasoning shows that ⟦푦 푥 = 푥⟧S is the class of finite right simple semigroups.]

Notes

The number of non-isomorphic nilpotent semigroups of order 8 is from Distler, ‘Classification and Enumeration of Finite Semigroups’, Table A.4. ◆ The section on varieties follows Howie, Fundamentals of Semigroup Theory, § 4.3 in outline, but in a universal algebraic context instead of the restricted context of {(∘, 2), (−1, 1)}-algebras. ◆ The exposition of pseudovarieties contains elements from Almeida, Finite Semigroups and Universal Algebra, §§ 3.1, 5.1, & 7.1. ◆ The discussion of free objects for pseudovarieties, profinite semigroups, pro-V semigroups, and pseudoidentities is based on Almeida, ‘Profinite semigroups and applications’ and Almeida, Finite Semigroups and Universal Algebra, ch. 3. ◆ For a proof of Proposition 8.21, see Almeida, Finite Semigroups and Universal Algebra, § 10.1. ◆ For further reading, Almeida, Finite Semigroups and Universal Algebra is the more accessible text, and Rhodes & Steinberg, The 픮-theory of Finite Semigroups the more recent and comprehensive monograph. Pin, Varieties of Formal Languages, ch. 2 and Eilenberg, Automata, Languages, and Machines (Vol. B), ch. v give rather different treatments. •

Notes • 169 Automata & finite semigroups 9

We do not praise automata for accurately producing ‘ all the movements they were designed to perform, because the production of these movements occurs necessarily. It is the designer who is praised ’ — René Descartes, Principles of Philosophy, Part One, § 37 (trans. John Cottingham).

• This chapter explores the connection between finite semigroups and rational languages. Rational languages are sets of words that are recognized by finite automata, which are mathematical models of simple computers. After discussing the necessary background on the theory of languages and automata, we will explore its connection to the theory of finite semigroups. The goal is the Eilenberg correspondence, which associates pseudovarieties of finite semigroups to certain classes of rational languages. We will then study some consequences of this correspondence.

Finite automata and rational languages

Let 퐴 be an alphabet. A language over 퐴 is a subset of Language 퐴∗. So a language over 퐴 is a set of words with letters in 퐴. We will be interested in a particular class of languages over 퐴 called the rational languages. To motivate the definition of this class, we first introduce finite automata. A finite automaton is a mathematical model of a computer with a very simple form of operation: it reads an input word (a sequence of symbols over an alphabet) one symbol at a time, and either accepts or rejects this input. The automaton can be in one of a finite number of internal states at any point. As it reads a symbol, it changes its state to a new one that is dependent on its current state and the symbol it reads. If can start in one of a given set of initial states, read an input word symbol-by-symbol, and end up in one of a given set of accept states, it accepts this input.

• 170 푏 푏 휀 푎 푞0 푞1 FIGURE 9.1 An example of a finite auto- 푎, 푏 maton.

It is easier to start with an example of a finite automaton rather than a formal definition. The directed graph in Figure 9.1 represents a finite automaton. The vertices of the graph represent he states of the automaton. The state 푞0 is marked with an incoming arrow ‘from nowhere’: this indicates that it is an initial state. The state 푞1 has a double outline: this indicates that it is an accept state. (In this example, there is only one initial state and one accept state; generally there may be more than one of each, and it is possible for a state to be both an initial and an accept state.) The edges and their labels indicate how the automaton behaves: for example,

◆ the edge from 푞0 to 푞1 labelled by 푎 says that if the automaton is in state 푞0 and reads the symbol 푎, it can change to state 푞1;

◆ the edge from 푞1 to 푞0 labelled by the empty word 휀 says that if the automaton is in state 푞1, it can change to state 푞0 without reading any symbols (that is, it can ‘spontaneously’ change from state 푞1 to state 푞0).

Notice that if the automaton is in state 푞1 and reads a symbol 푏, it can either change to state 푞0 or return to state 푞1. That is, the automaton is non-deterministic: there is an element of choice in how it functions. Thus the automaton is said to accept a word 푤 ∈ {푎, 푏}∗ if there is some sequence of choices it can make so that it starts in the initial state 푞0, reads 푤, and finishes in the accept state 푞1. In terms of the graph, this is equivalent to saying that the automaton accepts 푤 if there is a directed path in the graph starting at 푞0 and ending at 푞1, labelled by 푤. (The label on a path is the concatenation of the labels on its edges.) Hence this automaton accepts 푏푎푎, since this word labels the path

푏 푎 휀 푎 푞0 푞0 푞1 푞0 푞1 ⏟⏟⏟⏟⏟⏟⏟⏟⏟ ⏟⏟⏟⏟⏟⏟⏟⏟⏟ ∈ 퐼 ∈ 퐹

On the other hand, it does not accept the word 푏, because the only path with label 푏 starting at 푞0 is the path

푏 푞0 푞0 ⏟⏟⏟⏟⏟⏟⏟⏟⏟ ⏟⏟⏟⏟⏟⏟⏟⏟⏟ ∈ 퐼 ∉ 퐹 which does not end at 푞1 Formally, a finite automaton, or simply an automaton, A is formally Finite automaton a quintuple (푄, 퐴, 훿, 퐼, 퐹), where 푄 is a finite set of states, 퐴 is a finite

Finite automata and rational languages • 171 alphabet, 훿 ∶ 푄 × (퐴 ∪ {휀}) → ℙ푄 is a map called the transition function, 퐼 ⊆ 푄 is a set of distinguished states called the initial states or start states, and 퐹 ⊆ 푄 is a distinguished set of states called accept states or final states. We think of an automaton as a directed graph with labelled edges, with vertices being the states, and, for each 푞 ∈ 푄 and 푎 ∈ 퐴, and for each 푞′ ∈ (푞, 푎)훿, an edge labelled by 푎 from 푞 to 푞′. We can thus represent an automaton in a diagrammatic form, with the states being nodes connected by arrows. Initial states are marked with an incoming arrow ‘from nowhere’. Accept states have double borders. For each 푞 ∈ 푄 and 푎 ∈ 퐴, there is an arrow labelled by 푎 ∈ 퐴 from 푞 to each element of 휀 푎 푏 (푞, 푎)훿. The label on a path in such a graph is the product of the labels on 푞0 ∅ {푞1} {푞0} the edges in that path. 푞1 {푞0} {푞0} 푄 For example, let A be automaton in Figure 9.1. Then A has state set TABLE 9.1 Values of (푞, 푎)훿 푄 = {푞0, 푞1}. The set of initial states is 퐼 = {푞0}, the set of final states is 퐹 = {푞1}, and the transition function 훿 ∶ 푄 × (퐴 ∪ {휀}) → ℙ푄 is as given in Table 9.1. We say that an automaton A = (푄, 퐴, 훿, 퐼, 퐹) accepts a word 푤 ∈ 퐴∗ Accepted word if there is a directed path in the diagram starting at an initial state in 퐼 and ending at an accept state in 퐹, and labelled by 푤. The idea is that the automaton is a model of a computer that can start in any state in 퐼. While in state 푞, it can read a letter 푎 from an input tape and change to any state in (푞, 푎)훿, or it can change to any state in (푞, 휀)훿 without reading any input. The automaton accepts its input if, when it has finished reading all the input letters, it is in a state in 퐹. The set of all words accepted by an automaton A is denoted 퐿(A), and Language recognized is called the language recognized by A. If a language 퐿 ⊆ 퐴∗ is recognized by an automaton by some finite automaton, it is called a recognizable language. Recognizable language Our description of an automaton reading input involves an element of choice. The automaton is non-deterministic: First, the automaton can start in any state in 퐼. Second, the action it takes when it is in a particular state with a particular input letter to read is not fixed: the automaton can change to one of several other states on reading that letter, and may indeed change to another state without reading any input. An automaton where there is no such choice is called deterministic. Deterministic automaton More formally, an automaton A = (푄, 퐴, 훿, 퐼, 퐹) is deterministic if 퐼 contains exactly one state, 훿(푞, 휀) = ∅ for all 푞 ∈ 푄, and 훿(푞, 푎) contains a single state for all 푞 ∈ 푄 and 푎 ∈ 퐴. In terms of the diagram, A is deterministic if there is only one state with an incoming edge ‘from nowhere’,no edge is labelled by 휀, and for every state 푞 ∈ 푄 and 푎 ∈ 퐴, there is at most one edge starting at 푞 and labelled by 푎. So in a deterministic automaton, there is at most one path starting at a given state and labelled by a given word. (Such a path may not exist, since there might not an edge with the required label present at some point.) However, although deterministic automata seem to be much more re- strictive than non-deterministic ones, the class of deterministic automata

Finite automata and rational languages • 172 actually has the same ‘recognizing power’ as the class of all automata, in a sense made precise by the following result:

T h e o r e m 9 . 1. Let 퐿 be a recognizable language. Then there is a de- Recognizable languages terministic automaton that recognizes 퐿. are recognized by deterministic automata Proof of 9.1. Let A = (푄, 퐴, 훿, 퐼, 퐹) be an automaton, possibly non-deterministic, that recognizes 퐿. For the purposes of this proof, we define the 휀-closure of a set 푃 ⊆ 푄 to be the set

퐶휀(푃) = { 푟 ∈ 푄 ∶ (∃푝 ∈ 푃)(there is a path in A from 푝 to 푞 labelled by 휀) }.

We are going to define a new automaton 퐷(A) = (ℙ푄, 퐴, 휂, 퐽, 퐺). Note that the state set of 퐷(A) is the power set of the state set of A. The idea is that each state of 퐷(A) is a set that records every possible state that A could be at a given time. The following definitions formalize this idea. The set of initial states 퐽 is the singleton set {퐶휀(퐼)}. (Note that before reading any symbol, A could be in any state in 퐶휀(퐼).) The transition function 휂 has domain 푄 × (퐴 ∪ {휀}) and codomain the power set of the power set of 푄. The function 휂 is defined by

(푆, 휀)휂 = ∅,

(푆, 푎)휂 = {퐶휀({ 푝 ∈ 푄 ∶ (∃푞 ∈ 푆)((푞, 푎)훿 = 푝) })}.

We emphasize that (푆, 푎)휂 contains a single element 퐶휀(…). Note that, starting in a state in 푆 and reading a symbol 푎, the automaton A could be in any state in (푆, 푎)휂. Finally, the set of accept states is

퐺 = { 푈 ∈ ℙ푄 ∶ 푈 ∩ 퐹 ≠ ∅ }.

Note that 퐷(A) is deterministic. We now have to prove that 퐿(A) = 퐿(퐷(A)). Suppose 푤 = 푤1 ⋯ 푤푛 ∈ 퐿(A). Then there is a path in the graph of A from a state in 퐼 to a state in 퐹. We are going to prove that there is a path in 퐷(A) from the (unique) initial state to an accept state with the same label. Let 푞0, … , 푞푘 be the states on a path in A labelled by 푤, with 푞0 ∈ 퐼 and 푞푘 ∈ 퐹. For 푗 = 0, … , 푛 − 1, let 푖푗 the the subscript of the state immediately before the edge labelled by the symbol 푤푗+1 on this path. So there is a path in A from 푞 ∈ 퐼 to 푞 labelled by 휀, so 푞 ∈ 퐶 (퐼). 0 푖0 푖0 휀 So 푞 is in the unique initial state of 퐷(A). Let 푄 = 퐶 (퐼); then we have 푖0 푖0 휀 푞 ∈ 푄 . 푖0 푖0 Proceed by induction. For any 푗, we have 푞 ∈ (푞 , 푤 )훿, and we 푖푗+1 푖푗 푗 know that there is a path in A from 푞 to 푞 labelled by 휀. Therefore 푖푗+1 푖푗+1

Finite automata and rational languages • 173 푎, 푏 ∅ {푞1}

푎 푏

푎 {푞0} 푄 푎, 푏 FIGURE 9.2 The deterministic automaton equivalent to the one in Figure 푏 9.1. we have 푞 ∈ 퐶 ({푞 }) ⊆ (푄 , 푤 )휂. Let 푄 = (푄 , 푤 )휂; then we 푖푗+1 휀 푖푗+1 푖푗 푗 푖푗+1 푖푗 푗 have 푞 ∈ 푄 . 푖푗+1 푖푗+1 Finally, there is a path from 푞 to 푞 ∈ 퐹 labelled by 휀, so (푄 , 푤 )휂 푖푛−1+1 푘 푖푛 푛 must contain 푞 , and hence (푄 , 푤 )휂 ∩ 퐹 ≠ ∅. Let 푄 = (푄 , 푤 )휂. 푘 푖푛−1 푛 푖푛 푖푛−1 푛 Thus the (unique) path in 퐷(A) labelled by 푤1 ⋯ 푤푛 and starting from the (unique) initial state 푄 ∈ 퐽 visits states 푄 , 푄 , … , 푄 , 푄 and 푖0 푖0 푖0 푖푛−1 푖푛 ends at an accept state. So 푤 ∈ 퐿(퐷(A)). Now suppose that 푤 = 푤1 … , 푤푛 ∈ 퐿(퐷(A). Then there is a sequence of states 푄0, … , 푄푛, with 푄0 ∈ 퐽 and 푄푛 ∈ 퐺, and (푄푖−1, 푤푖)휂 = 푄푖 for 푖 = 1, … , 푛. By the definition of 퐺, there is some 푞′푛 ∈ 푄푛 ∩ 퐹. Proceed by induction. For any 푗 = 1, … , 푛, by the definition of 휂, there exist 푞푗−1 ∈ 푄푗−1 and 푞푗 ∈ 푄푗 such that 푞푗 ∈ (푞푗−1, 푤푗)훿 and there with a path from 푞푗 to 푞′푗 in A labelled by 휀. Finally, since 푄0 ∈ 퐽, we have 푄0 = 퐶휀(퐼) and so there is a path in A labelled by 휀 from some 푞0 ∈ 퐼 to 푞′0. Hence there is a path in A from 푞0 ∈ 퐼 to 푞′푛 ∈ 퐹 passing through 푞′0, 푞1, 푞′1 … , 푞푛−1, 푞′푛−1, 푞푛 (and other intermediate states) and labelled by 푤 = 푤1 ⋯ 푤푛. So 푤 ∈ 퐿(A). 휀 푎 푏 Thus the recognizable language 퐿 is recognized by the deterministic ∅ ∅ {∅} {∅} automaton 퐷(A). 9.1 {푞0} ∅ {푄} {{푞0}} {푞1} ∅ {{푞0}} 푄 Applying the construction in the proof of Theorem 9.1 to the ex- 푄 ∅ {푄} 푄 ample automaton A above, the resulting deterministic automaton 퐷(A) TABLE 9.2 recognizing 퐿(A) has set of initial states 퐽 = {{푞0}}, set of accept states Values of (푆, 푎)휂 퐺 = {{푞1}, 푄}, and transition function 휂 ∶ ℙ푄 × (퐴 ∪ {휀}) → ℙ(ℙ푄) as shown in Table 9.2. Diagrammatically, 퐷(A) is shown in Figure 9.2. We will need to make a distinction between two classes of languages. ∗-languages, +-languages A ∗-language is a subset of 퐴∗; that is, it may include the empty word. A +-language is a subset of 퐴+; that is, it does not contain the empty word. Of course, every +-language can also be viewed as a ∗-language. But the distinction is important when we perform operations on languages, and when we develop the correspondence of classes of languages and pseudovarieties. Let 퐴 be an alphabet. We are going to define some operations on the Boolean operations, classes of languages over 퐴. Let 퐿 and 퐾 be ∗-languages over 퐴. Then Boolean algebra

Finite automata and rational languages • 174 퐾 ∪ 퐿 and 퐾 ∩ 퐿 are, respectively, the union and intersection of 퐾 and 퐿. The language 퐴∗ ∖ 퐿 is the complement of 퐿 in 퐴∗. Notice that the class of ∗-languages is closed under union, intersection, and complement. These are the Boolean operations on the class of ∗-languages. We will say that a class of ∗-languages that is closed under the Boolean operations is a Boolean algebra. [The notion of a Boolean algebra is more general than this, but this definition will suffice for us.] For +-languages, we have the same union and intersection operations. However, the complement operation is different: 퐴+ ∖퐿 is the complement of 퐿 in 퐴+, and is also a +-language. The class of +-languages is closed under the operations of union, intersection, and this new complement operation; these are the Boolean operations on the class of +-languages. Again, we say that a class of +-languages that is closed under the Boolean operations is a Boolean algebra. The concatenation 퐾퐿 of the ∗-languages or +-languages 퐾 and 퐿 is Kleene star ∗, Kleene plus + the set of words of the form 푢푣, where 푢 ∈ 퐾 and 푣 ∈ 퐿. The submonoid of 퐴∗ generated by 퐾 is 퐾∗; the subsemigroup generated by 퐾 is 퐾+. Note that when 퐾 = 퐴, this agrees with the notation for the free monoid and free semigroup. However, when 퐾 ≠ 퐴, the sets 퐾∗ and 퐾+ are in general not the free monoid and free semigroup on 퐾. For example, if 퐾 = {푎2, 푎3}, then 푎2푎3 = 푎3푎2, so 퐾∗ and 퐾+ are not the free monoid and free semigroup on 퐾. The operations ∗ and + are called the Kleene star and Kleene plus. Notice that the class of ∗-languages is closed under the operations ∗ and +, and the class of +-languages is closed under the operation +. A language over 퐴 = {푎1, … , 푎푛} is rational or regular if it can be Rational/regular language obtained from the languages ∅, {휀}, {푎1}, {푎2}, …, {푎푛}, by applying (zero or more times) the operations of union, concatenation, and Kleene star.

K l e e n e ’ s T h e o r e m 9 . 2. A language over a finite alphabet is ra- Kleene’s theorem tional if and only if it is recognizable.

Proof of 9.2. To show that any rational language is recognizable, it suffices to show that the languages ∅, {휀}, and {푎} (for 푎 ∈ 퐴) are recognizable, and then to prove that the class of recognizable languages is closed under concatenation, union, and Kleene star. In our various constructions, we will simply draw automata. First, notice that

푞0 푞1 recognizes ∅;

푞0 recognizes {휀}; and

푎 푞0 푞1 recognizes {푎} (for 푎 ∈ 퐴).

So ∅, {휀}, and {푎} (for 푎 ∈ 퐴) are recognizable.

Finite automata and rational languages • 175 Now suppose that 퐾 and 퐿 are recognizable languages. So there are automata A = (푄A, 퐴, 훿A, 퐼A, 퐹A) and B = (푄B, 퐵, 훿B, 퐼B, 퐹B) recognizing 퐾 and 퐿 respectively, which we will sketch as

⋮퐼A A 퐹A⋮ and ⋮퐼B B 퐹B⋮ respectively. Then

휀 휀 ⋮퐼A A 퐹A⋮ ⋮퐼 B 퐹 ⋮ recognizes 퐾퐿; 휀 B B 휀

} } 퐼 퐹 } ⋮ A A A⋮ } } } recognizes 퐾 ∪ 퐿; } } } } ⋮퐼B B 퐹B⋮ } } } 휀

+ and ⋮퐼A A 퐹A⋮ recognizes 퐾 .

Thus 퐾퐿, 퐾∪퐿, and 퐾+ are recognizable languages. Hence 퐾∗ = 퐾+ ∪{휀} is also recognizable. This proves that every rational language is recognizable. So let 퐿 be a recognizable language. Then by Theorem 9.1, there is a deterministic automaton A = (푄, 퐴, 훿, 퐼, 퐹) recognizing 퐿. Suppose that 푄 = {푞1, … , 푞푛} and 퐼 = {푞1}. For each 푖, 푗 ∈ {1, … , 푛} and 푘 ∈ {0, … , 푛}, let 푅[푖, 푗; 푘] be the set of labels on paths that start at 푞푖, end at 푞푗, and visit only intermediate vertices in {푞1, … , 푞푘}. (So 푅[푖, 푗; 0] is the set of labels on paths that start at 푞푖, end at 푞푗, and do not visit any intermediate vertices.) Note that

퐿 = 퐿(A) = ⋃ 푅[1, 푗; 푛]. (9.1) 1⩽푗⩽푛 푞푗∈퐹

Thus we aim to prove that 푅[푖, 푗; 푘] is rational for all 푖, 푗 ∈ {1, … , 푛} and 푘 ∈ {0, … , 푛}; this will suffice to prove 퐿 is rational. We proceed by induction on 푘.

Finite automata and rational languages • 176 First, consider 푘 = 0. The words in 푅[푖, 푗; 0] label paths from 푞푖 to 푞푗 that visit no intermediate vertices, and so can have length at most 1 (length 0 is possible if 푖 = 푗). So 푅[푖, 푗; 0] is either ∅ or a union of sets {휀} and {푎} (for 푎 ∈ 퐴). This is the base of the induction. Now let 푘 > 0 and assume that 푅[푖, 푗; 푘 − 1] is rational for all 푖, 푗. Consider a path 훼 from 푞푖 to 푞푗 that only visits intermediate vertices from {푞1, … , 푞푘}. Now, if 훼 does not visit 푞푘, then its label is in 푅[푖, 푗; 푘 − 1]. Otherwise we can decompose 훼 into subpaths between visits to 푞푘: that is, let 훼 be the concatenation of subpaths 훼0, 훼1, … , 훼푚 where 훼0 is the subpath from 푞푖 up to the first visit to 푞푘, the 훼ℓ (for ℓ = 1, … , 푚 − 1) are the subpaths between visits to 푞푘, and 훼푚 is the subpath from the last visit to 푞푘 to the state 푞푗. Then the label on 훼0 is in 푅[푖, 푘; 푘 − 1], the labels on the 훼ℓ (for ℓ = 1, … , 푛 − 1) are in 푅[푘, 푘; 푘 − 1], and the label on 훼푛 is in 푅[푘, 푗; 푘 − 1]. Since 훼 was arbitrary, this shows that

푅[푖, 푗; 푘] = 푅[푖, 푗; 푘−1]∪푅[푖, 푘; 푘−1](푅[푘, 푘; 푘−1])∗푅[푘, 푗; 푘−1].

By the induction hypothesis, each of the sets 푅[푖, 푗; 푘−1] are rational. Thus 푅[푖, 푗; 푘] is rational, since it obtained from these sets using concatenation, union, and Kleene star. Hence, by induction, all the sets 푅[푖, 푗; 푘] are rational. Therefore 퐿 is rational by (9.1). 9.2 As a consequence of Theorem 9.2, the class of rational languages is closed under complementation. Hence we may view the rational languages over 퐴 as the languages obtainable from {푎} (for 푎 ∈ 퐴) and {휀} by applying the operations of union, concatenation, and Kleene star, and also intersection, complement, and Kleene plus. A ∗-language 퐿 over 퐴 is star-free if it can be obtained from the Star-free/plus-free languages languages {푎}, where 푎 ∈ 퐴, and {휀} using only the operations of union, intersection, complementation, and concatenation. For any ∗-language 퐿 over an alphabet 퐴 and word 푢 ∈ 퐴∗, define Left-/right-quotients the languages of a language

푢−1퐿 = { 푤 ∈ 퐴∗ ∶ 푢푤 ∈ 퐿 } 퐿푢−1 = { 푤 ∈ 퐴∗ ∶ 푤푢 ∈ 퐿 }; the languages 푢−1퐿 and 퐿푢−1 are, respectively, the left and right quotients of 퐿 with respect to 푢. Similarly, for any +-language 퐿 over an alphabet 퐴 and word 푢 ∈ 퐴∗, define

푢−1퐿 = { 푤 ∈ 퐴+ ∶ 푢푤 ∈ 퐿 } 퐿푢−1 = { 푤 ∈ 퐴+ ∶ 푤푢 ∈ 퐿 }.

Notice that the class of +-languages is closed under forming left and right quotients. The following result shows that the classes of rational ∗- and +-languages are also closed under forming left and right quotients:

Finite automata and rational languages • 177 P r o p o s i t i o n 9 . 3. If 퐿 is a rational ∗-language (respectively, +-language), then 퐿 has only finitely many distinct left and right quotients, all of which are rational ∗-languages (respectively, +-language). Proof of 9.3. We will prove the result for ∗-language, the proof for +- languages is identical, with the additional observation the class of +- languages is closed under forming left and right quotients. Let A = (푄, 퐴, 훿, 퐼, 퐹) be an automaton with 퐿 = 퐿(A). Let 푢 ∈ 퐴∗. Let 퐽 ⊆ 푄 consist of all states A can reach starting at a state in 퐼 and reading 푢. Let 퐽A = (푄, 퐴, 훿, 퐽, 퐹). Then 푤 ∈ 퐿(퐽A) if and only if −1 −1 −1 푢푤 ∈ 퐿(A). That is, 푢 퐿 = 푢 퐿(A) = 퐿(퐽A). So 푢 퐿 is rational. Since there are only finitely many possibilities for 퐽, there are only finitely many distinct languages 푢−1퐿. Similarly, let 퐺 ⊆ 푄 consist of all states in which A can start and reach a state in 퐹 after reading 푢. Let A퐺 = (푄, 퐴, 훿, 퐼, 퐺). Then 푤 ∈ 퐿(A퐺) if −1 −1 −1 −1 and only if 푤푢 ∈ 퐿(A). That is, 퐿푢 = 퐿(A)푢 = 퐿(A퐺)푢 . So 퐿푢 is rational. Since there are only finitely many possibilities for 퐺, there are −1 only finitely many distinct languages 퐿푢 . 9.3

Let A = (푄, 퐴, 훿, {푞0}, 퐹) be a deterministic automaton. Note that for Complete automaton each 푞 ∈ 푄 and 푎 ∈ 퐴, the set (푞, 푎)훿 is either empty or contains a single element. If 훿 is such that (푞, 푎)훿 is never empty (that is, there is exactly one element in each (푞, 푎)훿), then the automaton A is complete. In terms of the graph, a complete deterministic automaton has exactly one edge starting at each vertex with each label in 퐴. Let A = (푄, 퐴, 훿, {푞0}, 퐹) be a complete deterministic automaton. For Transition monoid of an automaton each 푎 ∈ 퐴, there is a map 휏푎 ∶ 푄 → 푄 with 푞휏푎 given by (푞, 푎)훿 = {푞휏푎}. Notice that 휏푎 ∈ T푄 for each 푎 ∈ 퐴. So we have a homomorphism 휑 ∶ ∗ 퐴 → T푄 extending the map 푎 ↦ 휏푎. The set im 휑 is a submonoid of T푄 called the transition monoid of A. For any word 푤 ∈ 퐴∗, the element 푤휑 is a transformation of 푄. For any 푞 ∈ 푄, the state 푞(푤휑) is the state that A will reach if it starts in 푞 and reads 푤. Let 푌 = { 휎 ∈ im 휑 ∶ 푞0휎 ∈ 퐹 } ⊆ T푄. Then 푤휑 ∈ 푌 if and only if 푤 ∈ 퐿(A). That is, we have a monoid T푄 with ∗ a subset 푌 and a homomorphism 휑 ∶ 퐴 → T푄 that describes 퐿(A) as the inverse image of 푌 under 휑. This motivates the following definition. A ∗-language 퐿 over 퐴 is recognized by a homomorphism into a monoid Language recognized 푀, or more simply recognized by a monoid 푀, if there exists a monoid by a homomorphism homomorphism 휑 ∶ 퐴∗ → 푀, where 푀 is a monoid with a subset 푀′ of 푀 such that 퐿 = 푀′휑−1, or, equivalently, with 퐿 = 퐿휑휑−1. Similarly, a +- language 퐿 over 퐴 is recognized by a homomorphism into a semigroup 푆, or more simply recognized by a semigroup 푆, if there exists a homomorphism 휑 ∶ 퐴+ → 푆, where 푆 is a semigroup, with 퐿 = 퐿휑휑−1. Notice that if 퐿 is a ∗-language (respectively, a +-language) recognized by 휑 ∶ 퐴∗ → 푀 + −1 −1 (respectively, 휑 ∶ 퐴 → 푆), then 퐿 = 휑휑 and so 퐿 = ⋃푥∈퐿휑 푥휑 . Each set 푥휑−1 consists of words that map to 푥 and so are related by ker 휑. That is, each 푥휑−1 is a ker 휑-class, so 퐿 is a union of ker 휑-classes. Notice that

Finite automata and rational languages • 178 in the discussion in the previous paragraph, T푄 is a finite monoid. So any recognizable language is recognized by a finite monoid. Furthermore, any recognizable +-language 퐿 is recognized by a finite semigroup, since in this case the initial state of an automaton recognizing 퐿 cannot also be an accepting state, and hence 푌 in the discussion above does not + contain id푄, so that we can use 휑|퐴+ ∶ 퐴 → (T푄 ∖ {id푄}), noting that −1 −1 퐿 = 푌휑|퐴+ = 퐿휑|퐴+ 휑|퐴+ . On the other hand, suppose 퐿 is a ∗-language over 퐴 recognized by a finite monoid 푀. Let 휑 ∶ 퐴∗ → 푀 be a homomorphism recognizing 퐿, so that by 퐿 = 퐿휑휑−1 Then we can construct an automaton A recognizing 퐿 as follows. The state set is 푀. The set of initial states is {1푀}, the set of accept states is 퐿휑, and the transition function 훿 ∶ 푀 × (퐴 ∪ {휀}) → 푀 is given by (푚, 푎)훿 = {푚(푎휑)}. It is easy to see that 퐿(A) = 퐿휑휑−1 = 퐿, since the unique path starting at 1푀 and labelled by 푤 = 푤1 ⋯ 푤푛 (where 푤푖 ∈ 퐴) is

푤1 푤2 푤3 푤푛 1푀 푤1휑 (푤1푤2)휑 … (푤1 ⋯ 푤푛)휑

This path ends in 퐿휑 if and only if 푤 ∈ 퐿. Similarly, if a +-language 퐿 is recognized by a finite semigroup 푆, we can construct an automaton recognizing 푆 with state set 푆1. Thus we have proven the following result: T h e o r e m 9 . 4. A ∗-language is recognizable if and only if it is recognized by a finite monoid. A +-language is recognizable if and only if it is recognized by a finite semigroup. 9.4

Let V be an M-pseudovariety of monoids (respectively, an S-pseudo- V-recognizability variety of semigroups). A ∗-language (respectively, +-language) over 퐴 is V-recognizable if it is recognized by some monoid (respectively, semigroup) in V. Thus Theorem 9.4 says that a ∗-language (respectively +- language) is recognizable if and only if it is M-recognizable (respectively, S-recognizable). At this point, our goal is to describe classes of languages that are V-recognizable for a given pseudovariety V.

Syntactic semigroups and monoids We are now going to study particular semigroups and monoids associated to languages, known as syntactic monoids and semigroups. These will be of fundamental importance in establishing a connection between pseudovarieties and classes of recognizable languages. ∗ For any ∗-language 퐿 over 퐴, define a relation 휎퐿 on 퐴 as follows: 휎퐿 for 푢, 푣 ∈ 퐴∗,

∗ 푢 휎퐿 푣 ⇔ (∀푝, 푞 ∈ 퐴 )(푝푢푞 ∈ 퐿 ⇔ 푝푣푞 ∈ 퐿). (9.2)

Syntactic semigroups and monoids • 179 + For any +-language 퐿, define 휎퐿 on 퐴 in exactly the same way, using (9.2); note in particular that 푝 and 푞 still range over 퐴∗, not 퐴+. P r o p o s i t i o n 9 . 5. Let 퐿 be a ∗-language (respectively, +-language) over 퐴. Then: ∗ + a) The relation 휎퐿 is a congruence on 퐴 (respectively, 퐴 ). b) The language 퐿 is a union of 휎퐿-classes. c) If 휌 is a congruence on 퐴∗ (respectively, 퐴+) with the property that 퐿 is a union of 휌-classes, then 휌 ⊆ 휎퐿. Proof of 9.5. We prove the result for ∗-languages; the reasoning for +- languages is parallel.

a) It is immediate from the definition that 휎퐿 is reflexive, symmetric, and transitive. So 휎퐿 is an equivalence relation. Let 푢 휎퐿 푣 and let 푠 ∈ 퐴∗. Then 푝푢푞 ∈ 퐿 ⇔ 푝푣푞 ∈ 퐿 for all 푝, 푞 ∈ 퐴∗. In particular, this holds for all 푝 of the form 푝′푠; hence 푝′푠푢푞 ∈ 퐿 ⇔ 푝′푠푣푞 ∈ 퐿 for all ∗ 푝′, 푞 ∈ 퐴 . Hence 푠푢 휎퐿 푠푣. So 휎퐿 is left-compatible; similarly it is right-compatible and is thus a congruence.

b) Let 푢 ∈ 퐿 and let 푣 휎퐿 푢. Put 푝 = 푞 = 휀 in the definition of 휎퐿 to see that 푣 ∈ 퐿. Thus if any 휎퐿-class intersects 퐿, it is contained in 퐿. Therefore 퐿 is a union of 휎퐿-classes. c) Let 휌 be a congruence on 퐴∗ such that 퐿 is a union of 휌-classes. Then (푢, 푣) ∈ 휌 ⇒ (∀푝, 푞 ∈ 퐴∗)((푝푢푞, 푝푣푞) ∈ 휌) [since 휌 is a congruence] ⇒ (∀푝, 푞 ∈ 퐴∗)((푝푢푞, 푝푣푞) ∈ 퐿 ∨ (푝푢푞, 푝푣푞) ∉ 퐿) [since 퐿 is a union of 휌-classes] ⇒ (∀푝, 푞 ∈ 퐴∗)(푝푢푞 ∈ 퐿 ⇔ 푝푣푞 ∈ 퐿)

⇒ (푢, 푣) ∈ 휎퐿;

thus 휌 ⊆ 휎퐿. 9.5

For any language 퐿 over an alphabet 퐴, the congruence 휎퐿 is called Syntactic congruence the syntactic congruence of 퐿. ∗ For any ∗-language 퐿, the factor monoid 퐴 /휎퐿 is called the syntactic Syntactic monoid monoid of 퐿 and is denoted SynM 퐿, and the natural monoid homomor- ♮ ∗ ∗ phism 휎퐿 ∶ 퐴 → 퐴 /휎퐿 = SynM 퐿 is the syntactic monoid homomorphism of 퐿. + For any +-language 퐿, the factor semigroup 퐴 /휎퐿 is called the syn- Syntactic semigroup tactic semigroup of 퐿 and is denoted SynS 퐿, and the natural homomor- ♮ + + phism 휎퐿 ∶ 퐴 → 퐴 /휎퐿 = SynS 퐿 is the syntactic homomorphism of 퐿. The importance of syntactic monoids and semigroups is the following result. Essentially, it says that the syntactic monoid of a ∗-language is the smallest monoid that recognizes that language, and similarly for +- languages and semigroups:

Syntactic semigroups and monoids • 180 P r o p o s i t i o n 9 . 6. a) Let 퐿 be a ∗-language. Then 퐿 is recognized by a monoid 푀 if and only if SynM 퐿 ≼ 푀. b) Let 퐿 be a +-language. Then 퐿 is recognized by a semigroup 푆 if and only if SynS 퐿 ≼ 푆. Proof of 9.6. We prove only part a); the proof for part b) follows by replacing ‘퐴∗’ by ‘퐴+’, ‘submonoid’ by ‘subsemigroup’, ‘SynM’ by ‘SynS’, and ‘monoid homomorphism’ by ‘homomorphism’ throughout. Let 휑 ∶ 퐴∗ → 푀 recognize 퐿. So 퐿 = 퐿휑휑−1. Then ker 휑 is a congru- ∗ ence on 퐴 and 퐿 is a union of ker 휑-classes. Hence ker 휑 ⊆ 휎퐿. Define a map 휓 ∶ im 휑 → SynM 퐿 by (푢휑)휓 = [푢] ; this map is a well-defined 휎퐿 monoid homomorphism since ker 휑 ⊆ 휎퐿. It is clearly surjective. Since im 휑 is an M-submonoid of 푀 and 휓 ∶ im 휑 → SynM 퐿 is a surjective homomorphism, SynM 퐿 ≼ 푀. For the other direction, we first prove that SynM 퐿 recognizes 퐿. Since 퐿 is a union of 휎 -classes, it follows that 퐿 = ⋃ [푢] = ⋃ 푥(휎♮)−1 = 퐿 푢∈퐿 휎퐿 푥∈퐿 퐿 ♮ ♮ −1 ♮ ∗ 퐿휎퐿(휎퐿) . Thus 휎퐿 ∶ 퐴 → SynM 퐿 recognizes 퐿. Let SynM 퐿 ≼ 푀. So there is an M-submonoid 푁 of 푀 and a surjective M-homomorphism 휓 ∶ 푁 → SynM 퐿. For each 푎 ∈ 퐴, define a map ♮ ∗ 휑 ∶ 퐴 → 푁 by choosing 푎휑 such that (푎휑)휓 = 푎휎퐿. Since 퐴 is free on 퐴, there is a unique extension of 휑 to a homomorphism 휑̂ ∶ 퐴∗ → 푁; ♮ ∗ ♮ notice that (푢휑)휓̂ = 푢휎퐿 for all 푢 ∈ 퐴 since 휓 and 휎퐿 are monoid ho- ♮ −1 momorphisms. Let 푁′ = 퐿휎퐿휓 . Then, viewing 휑̂ as a homomorphism from 퐴∗ to 푀, we have ♮ ♮ −1 ♮ −1 −1 −1 퐿 = 퐿휎퐿(휎퐿) = 퐿휎퐿휓 휑̂ = 푁′휑̂ , and so 푀 recognizes 퐿. 9.6 Proposition 9.6 is actually the original motivation behind the concept of division. P r o p o s i t i o n 9 . 7. Let 퐴 and 퐵 be alphabets. For all ∗-languages 퐿 Properties of and 퐾 over 퐴, for all 푎 ∈ 퐴, and for all monoid homomorphisms 휑 ∶ 퐵∗ → syntactic monoids 퐴∗: a) SynM 퐿 = SynM(퐴∗ ∖ 퐿); b) SynM(퐿 ∩ 퐾) ≼ (SynM 퐿) × (SynM 퐾); c) SynM(퐿 ∪ 퐾) ≼ (SynM 퐿) × (SynM 퐾); d) SynM(푎−1퐿) ≼ SynM 퐿; e) SynM(퐿푎−1) ≼ SynM 퐿; f) SynM(퐿휑−1) ≼ SynM 퐿. Proof of 9.7. a) For any 푢, 푣 ∈ 퐴∗, we have ∗ 푢 휎퐿 푣 ⇔ (∀푝, 푞 ∈ 퐴 )(푝푢푞 ∈ 퐿 ⇔ 푝푣푞 ∈ 퐿) ⇔ (∀푝, 푞 ∈ 퐴∗)(푝푢푞 ∈ 퐴∗ ∖ 퐿 ⇔ 푝푣푞 ∈ 퐴∗ ∖ 퐿)

⇔ 푢 휎퐴∗∖퐿 푣;

Syntactic semigroups and monoids • 181 ∗ Hence 휎퐿 = 휎퐴∗∖퐿 and so SynM 퐿 = SynM(퐴 ∖ 퐿). b) Define a monoid homomorphism 휑 ∶ 퐴∗ → (SynM 퐿) × (SynM 퐾) ♮ ♮ ♮ ♮ by 푢휑 = (푢휎퐿, 푢휎퐾). Let 푆 = 퐿휎퐿 × 퐾휎퐾 ⊆ (SynM 퐿) × (SynM 퐾). Then

푢 ∈ 푆휑−1 ♮ ♮ ⇒ 푢휑 ∈ 퐿휎퐿 × 퐾휎퐾 ♮ ♮ ♮ ♮ ⇒ (∃푣 ∈ 퐿, 푤 ∈ 퐾)((푣휎퐿, 푤휎퐾) = (푢휎퐿, 푢휎퐾)) ♮ ♮ ♮ ♮ ⇒ (∃푣 ∈ 퐿, 푤 ∈ 퐾)((푣휎퐿 = 푢휎퐿) ∧ (푤휎퐾 = 푢휎퐾)) ♮ ♮ −1 ♮ ♮ −1 ⇒ (푢 ∈ 퐿휎퐿(휎퐿) ) ∧ (푢 ∈ 퐾휎퐾(휎퐾) ) ⇒ (푢 ∈ 퐿) ∧ (푢 ∈ 퐾) ⇒ 푢 ∈ 퐿 ∩ 퐾.

Hence 푆휑−1 ⊆ 퐿 ∩ 퐾. On the other hand,

푢 ∈ 퐿 ∩ 퐾 ⇒ 푢 ∈ 퐿 ∧ 푢 ∈ 퐾 ♮ ♮ ♮ ♮ ⇒ 푢휑 = (푢휎퐿, 푢휎퐾) ∈ 퐿휎퐿 × 퐾휎퐾 = 푆 ⇒ 푢 ∈ 푆휑−1,

so 푆휑−1 ⊆ 퐿 ∩ 퐾. Hence 푆휑−1 = 퐿 ∩ 퐾. Thus 휑 ∶ 퐴∗ → (SynM 퐿) × (SynM 퐾) recognizes 퐿 ∩ 퐾, and so SynM(퐿 ∩ 퐾) ≼ (SynM 퐿) × (SynM 퐾) by Proposition 9.6(a). c) Define a monoid homomorphism 휑 ∶ 퐴∗ → (SynM 퐿) × (SynM 퐾) ♮ ♮ ♮ ♮ by 푢휑 = (푢휎퐿, 푢휎퐾). Let 푆 = (퐿휎퐿 × SynM 퐾) ∪ (SynM 퐿 × 퐾휎퐾) ⊆ (SynM 퐿) × (SynM 퐾). Then

푢 ∈ 푆휑−1 ♮ ♮ ⇒ 푢휑 ∈ (퐿휎퐿 × SynM 퐾) ∪ (SynM 퐿 × 퐾휎퐾) + ♮ ♮ ♮ ♮ ⇒ (∃푣 ∈ 퐿, 푤 ∈ 퐴 )((푣휎퐿, 푤휎퐾) = (푢휎퐿, 푢휎퐾)) + ♮ ♮ ♮ ♮ ∨ (∃푣 ∈ 퐴 , 푤 ∈ 퐾)((푣휎퐿, 푤휎퐾) = (푢휎퐿, 푢휎퐾)) ♮ ♮ ♮ ♮ ⇒ (∃푣 ∈ 퐿)(푣휎퐿 = 푢휎퐿) ∨ (∃푤 ∈ 퐾)(푤휎퐾 = 푢휎퐾) ♮ ♮ −1 ♮ ♮ −1 ⇒ (푢 ∈ 퐿휎퐿(휎퐿) ) ∨ (푢 ∈ 퐾휎퐾(휎퐾) ) ⇒ (푢 ∈ 퐿) ∨ (푢 ∈ 퐾) ⇒ 푢 ∈ 퐿 ∪ 퐾.

Syntactic semigroups and monoids • 182 Hence 푆휑−1 ⊆ 퐿 ∪ 퐾. On the other hand,

푢 ∈ 퐿 ∪ 퐾 ⇒ 푢 ∈ 퐿 ∨ 푢 ∈ 퐾 ♮ ♮ ⇒ 푢휑 = (푢휎퐿, 푢휎퐾) ∈ ♮ ♮ (퐿휎퐿 × SynM 퐾) ∪ (SynM 퐿 × 퐾휎퐾) = 푆 ⇒ 푢 ∈ 푆휑−1,

so 푆휑−1 ⊆ 퐿 ∩ 퐾. Hence 푆휑−1 = 퐿 ∪ 퐾. Thus 휑 ∶ 퐴+ → (SynM 퐿) × (SynM 퐾) recognizes 퐿 ∪ 퐾, and so SynM(퐿 ∪ 퐾) ≼ (SynM 퐿) × (SynM 퐾) by Proposition 9.6(a). ♮ ♮ d) Let 푆 = 퐿휎퐿 ⊆ SynM 퐿. Let 푠 = 푎휎퐿. Define

푠−1푆 = { 푥 ∈ SynM 퐿 ∶ 푠푥 ∈ 푆 }.

Then 푎−1퐿 = (푠−1푆)휑−1 and so 휑 ∶ 퐴+ → SynM 퐿 recognizes 푎−1퐿. Hence SynM(푎−1퐿) ≼ SynM 퐿 by Proposition 9.6(a). e) This is similar to part d). ♮ + −1 f) The homomorphism 휑휎퐿 ∶ 퐵 → SynM 퐿 recognizes 퐿휑 since

−1 ♮ ♮ −1 −1 ♮ ♮ −1 −1 ♮ ♮ −1 −1 −1 퐿휑 휑휎퐿(휑휎퐿) = 퐿휑 휑휎퐿(휎퐿) 휑 = 퐿휎퐿(휎퐿) 휑 = 퐿휑 .

−1 Hence SynM(퐿휑 ) ≼ SynM 퐿 by Proposition 9.6(a). 9.7 Essentially the same proofs yield the corresponding results for syntactic semigroups:

P r o p o s i t i o n 9 . 8. Let 퐴 and 퐵 be alphabets. For all +-languages 퐾 Properties of and 퐿 over 퐴, for all 푎 ∈ 퐴, and for all homomorphisms 휑 ∶ 퐵+ → 퐴+: syntactic semigroups a) SynS 퐿 = SynS(퐴+ ∖ 퐿); b) SynS(퐿 ∩ 퐾) ≼ (SynS 퐿) × (SynS 퐾); c) SynS(퐿 ∪ 퐾) ≼ (SynS 퐿) × (SynS 퐾); d) SynS(푎−1퐿) ≼ SynS 퐿; e) SynS(퐿푎−1) ≼ SynS 퐿; −1 f) SynS(퐿휑 ) ≼ SynS 퐿. 9.8

Eilenberg correspondence The classes of languages that correspond to pseudovarieties are called ‘varieties of rational languages’. However, the class of languages that are V-recognizable for some pseudovariety V is also dependent on the finite alphabet 퐴 as well. Thus we do not formally define

Eilenberg correspondence • 183 a ‘variety of rational languages’ as a class, but rather as a correspondence that associates finite alphabets to classes of languages. We also need to distinguish between ∗-languages and +-languages. To be precise, a variety Variety of rational of rational ∗-languages is formally defined to be a correspondence V that ∗-languages associates to each finite alphabet 퐴 a class of rational ∗-languages V(퐴∗) with the following properties: 1) The class V(퐴∗) is a Boolean algebra. (That is, it is closed under union, intersection, and complement in 퐴∗.) 2) For all 퐿 ∈ V(퐴∗) and 푎 ∈ 퐴, the right and left quotient languages 푎−1퐿 = { 푤 ∈ 퐴∗ ∶ 푎푤 ∈ 퐿 } and 퐿푎−1 = { 푤 ∈ 퐴∗ ∶ 푤푎 ∈ 퐿 } are also in V(퐴∗) 3) For all finite alphabets 퐵, for all ∗-languages 퐿 ∈ V(퐵∗), and for all monoid homomorphisms 휑 ∶ 퐴∗ → 퐵∗, we have 퐿휑−1 ∈ V(퐴∗). Similarly, a variety of rational +-languages is a correspondence V that Variety of rational associates to each finite alphabet 퐴 a class of rational languages V(퐴+) +-languages with the following properties: 1) The class V(퐴+) is a Boolean algebra. (That is, it is closed under union, intersection, and complement in 퐴+.) 2) For all 퐿 ∈ V(퐴+) and 푎 ∈ 퐴, the right and left quotient languages 푎−1퐿 = { 푤 ∈ 퐴+ ∶ 푎푤 ∈ 퐿 } and 퐿푎−1 = { 푤 ∈ 퐴+ ∶ 푤푎 ∈ 퐿 } are also in V(퐴+). 3) For all finite alphabets 퐵, for all +-languages 퐿 ∈ V(퐵+), and for all homomorphisms 휑 ∶ 퐴+ → 퐵+, we have 퐿휑−1 ∈ V(퐴+). E x a m p l e 9 . 9. a) The correspondence E such that E(퐴+) = {∅, 퐴+} is a variety of rational +-languages. To see this, first note that each E(퐴+) is clearly closed under union, intersection, and complement. Next, for any 푎 ∈ 퐴, we have 푎−1∅ = ∅푎−1 = ∅ ∈ E(퐴+) and 푎−1퐴+ = 퐴+푎−1 = 퐴+ ∈ E(퐴+), so E(퐴+). Finally, for any homomorphism 휑 ∶ 퐵+ → 퐴+, we have ∅휑−1 = ∅ ∈ E(퐵+) and 퐴+휑−1 = 퐵+ ∈ E(퐵+). b) Let M be the correspondence that associates to each finite alphabet 퐴 the class of all ∗-languages over 퐴. It is easy to see that M is a variety of rational ∗-languages. c) A +-language 퐿 over an alphabet 퐴 is said to be cofinite if 퐴+ ∖ 퐿 is finite. Let N be the correspondence that associates to each finite alphabet 퐴 the class of all finite or cofinite languages over 퐴. Then N is a variety of rational +-languages (see Exercise 9.4). There is a natural correspondence, known as the Eilenberg correspond- Eilenberg correspondence ence, between varieties of rational ∗-languages and M-pseudovarieties of monoids, and between varieties of rational +-languages and S-pseudovarieties of semigroups. For varieties of rational ∗-languages and M- pseudovarieties of monoids, the correspondence is defined as follows:

Eilenberg correspondence • 184 ◆ Let V be a variety of rational ∗-languages. The corresponding M- pseudovariety of monoids V is generated by all monoids SynM 퐿 such that 퐿 ∈ V(퐴∗) for some finite alphabet 퐴. That is, we have a map

{ SynM 퐿 ∶ 퐿 ∈ V(퐴∗) V ↦ V ( ). } (9.3) M for some finite alphabet 퐴 }

◆ Let V be an M-pseudovariety of monoids. The corresponding variety of rational ∗-languages V associates to each finite alphabet 퐴 the class of languages 퐿 such that SynM 퐿 ∈ V, or, equivalently, the class of languages 퐿 such that 퐿 is recognized by some monoid in V. That is, we have a map

V ↦ V, where V(퐴∗) = { 퐿 ⊆ 퐴∗ ∶ SynM 퐿 ∈ V } } (9.4) for each finite alphabet 퐴.

The correspondence for +-varieties of rational languages and S-pseudovarieties of semigroups is defined similarly: ◆ Let V be a variety of rational +-languages. The corresponding S- pseudovariety of semigroups V is generated by all semigroups SynS 퐿 such that 퐿 ∈ V(퐴+) for some finite alphabet 퐴. That is, we have a map

{ SynS 퐿 ∶ 퐿 ∈ V(퐴∗) V ↦ V ( ). } (9.5) S for some finite alphabet 퐴 }

◆ Let V be an S-pseudovariety of semigroups. The corresponding variety of rational +-languages V associates to each finite alphabet 퐴 the class of languages 퐿 such that SynS 퐿 ∈ V, or, equivalently, the class of languages 퐿 such that 퐿 is recognized by some semigroup in V. That is, we have a map

V ↦ V, where V(퐴+) = { 퐿 ⊆ 퐴+ ∶ SynS 퐿 ∈ V } } (9.6) for each finite alphabet 퐴.

Eilenberg’s Theorem 9.10. The maps (9.3) and (9.4) are mutu- Eilenberg’s theorem ally inverse, and the maps (9.5) and (9.6) are mutually inverse. Proof of 9.10. We will prove that (9.3) and (9.4) are mutually inverse; the other case is similar. Let V be a variety of rational ∗-languages. Let V be the M-pseudovariety of monoids associated to it by (9.3). Let W be the variety of rational ∗- languages associated to V by (9.4). We aim to show that V(퐴∗) = W(퐴∗). for each finite alphabet 퐴. First, we prove that V(퐴∗) ⊆ W(퐴∗). Let 퐿 ∈ V(퐴∗). Then SynM 퐿 ∈ V by the definition of (9.3), and so 퐿 ∈ W(퐴∗) by the definition of (9.4). Hence V(퐴∗) ⊆ W(퐴∗).

Eilenberg correspondence • 185 Next we prove that W(퐴∗) ⊆ V(퐴∗). This part of the proof is more complicated. Let 퐿 ∈ W(퐴∗). Then SynM 퐿 ∈ V by the definition of (9.4). Now, V is generated by

X = { SynM 퐾 ∶ 퐾 ∈ V(퐴∗) for some finite alphabet 퐴 }; that is, V = ℍ핊ℙfinX. Hence there exist alphabets 퐴푖 and ∗-languages ∗ 퐾푖 ∈ V(퐴푖 ) for 푖 = 1, … , 푛 such that 푛

SynM 퐿 ≼ ∏ SynM 퐾푖. 푖=1 푛 ∗ 푛 Let 푈 = ∏푖=1 퐴푖 and 푇 = ∏푖=1 SynM 퐾푖. Define a map

훾 ∶ 푈 → 푇, (푤 , … , 푤 )훾 = (푤 휎♮ , … , 푤 휎♮ ); 1 푛 1 퐾1 푛 퐾푛 then 훾 is a surjective homomorphism because all of the maps 휎♮ ∶ 퐴∗ → 퐾푖 푖 SynM 퐾푖 are surjective homomorphisms. Since SynM 퐿 ≼ 푇, the monoid 푇 recognizes 퐿; that is, there is a homomorphism 휑 ∶ 퐴∗ → 푇 and a subset 푀 of 푇 such that 퐿 = 푀휑−1. Define 휓 ∶ 퐴 → 푈 by letting 푎휓 be such that 푎휓훾 = 푎휑; since 퐴∗ is free on 퐴, this map extends to a unique monoid homomorphism 휓̂ ∶ 퐴∗ → 푈. Notice that 푢휓훾̂ = 푢휑 for 푢 ∈ 퐴∗ since 휑 and 훾 are monoid ∗ ∗ homomorphisms. For each 푖 = 1, … , 푛, let 휓푖 ∶ 퐴 → 퐴푖 be such that

푢휓̂ = (푢휓1, … , 푢휓푛)

∗ and 휑푖 ∶ 퐴 → SynM 퐾푖 be such that

푢휑 = (푢휑1, … , 푢휑푛).

Then 휑 = 휓 휎♮ . 푖 푖 퐾푖 We have

퐿 = 푀휑−1 = ⋃ 푚휑−1. 푚∈푀 Since V(퐴∗) is a Boolean algebra, it is sufficient to show that 푚휑−1 ∈ ∗ V(퐴 ) for all 푚 ∈ 푀. If 푚 = (푠1, … , 푠푛) ∈ 푀 ⊆ 푇, where 푠푖 ∈ SynM 퐾푖, then 푛 −1 −1 푚휑 = ⋂ 푠푖휑푖 . 푖=1 Again, since V(퐴∗) is a Boolean algebra, it is sufficient to show that −1 ∗ 푠푖휑푖 ∈ V(퐴 ) for all 푠푖 ∈ SynM 퐾푖 and 푖 = 1, … , 푛. Since 푠 휑−1 = 푠 (휎♮ )−1휓−1, the closure of V under homomorphism 푖 푖 푖 퐾푖 푖 pre-images shows that it is sufficient to prove that 푠 (휎♮ )−1 ∈ V(퐴∗). 푖 퐾푖 푖

Eilenberg correspondence • 186 ∗ For 푤 ∈ 퐴푖 , define

∗ 푅푤 = { (푝, 푞) ∶ 푝, 푞 ∈ 퐴푖 , 푝푤푞 ∈ 퐾푖 } ∗ −1 −1 = { (푝, 푞) ∶ 푝, 푞 ∈ 퐴푖 , 푤 ∈ 푝 퐾푖푞 }; then for any 푢, 푣 ∈ 퐴∗ we have 푢 휎 푣 if and only if 푅 = 푅 . Hence 퐾푖 푢 푣 푢휎♮ (휎♮ )−1, which is the 휎 -class of 푢 ∈ 퐴∗, is given by 퐾푖 퐾푖 퐾푖 푖

푢휎♮ (휎♮ )−1 = ⋂ 푝−1퐾 푞−1 ∖ ⋃ 푝−1퐾 푞−1. (9.7) 퐾푖 퐾푖 푖 푖 (푝,푞)∈푅푢 (푝,푞)∉푅푢

−1 −1 By Proposition 9.3, there are only finitely many distinct sets 푝 퐾푖푞 . Therefore the intersections and unions in (9.7) are finite. By repeated ∗ application of the closure of V(퐴푖 ) under left and right quotients, each −1 −1 ∗ ∗ of the sets 푝 퐾푖푞 lies in V(퐴푖 ). Since V(퐴 ) is closed under unions, intersections, and complements, 푢휎♮ (휎♮ )−1 lies in V(퐴∗). 퐾푖 퐾푖 푖 Finally, let 푢 be such that 푠 = 푢휎♮ . Then 푠 (휎♮ )−1 ∈ V(퐴∗). This 푖 퐾푖 푖 퐾푖 푖 completes the proof. 9.10 It is important to notice that although Eilenberg’s theorem shows that there is a one-to-one correspondence between M-pseudovarieties of semigroups and varieties of rational ∗-languages, and between S-pseudovarieties of monoids and varieties of rational +-languages, it does not actually give a concrete method for describing a variety of rational languages if we know a pseudovariety, or vice versa. The following result is therefore an instance of the Eilenberg correspondence, but it is not a consequence of Theorem 9.10. In general, finding and proving instances of Eilenberg’s correspondence can be difficult, although this particular result is straightforward.

T h e o r e m 9 . 1 1. The Eilenberg correspondence associates the S-pseudovariety of nilpotent semigroups N with the variety of finite or cofinite rational +-languages N.

Proof of 9.11. Let 푆 ∈ N, with 푆푛 = 0 for all 푥 ∈ 푆. Let 퐴 be a finite alphabet and suppose 휑 ∶ 퐴+ → 푆 recognizes a +-language 퐿. ∗ Suppose that 퐿휑 contains 0푆. Then if 푤 ∈ 퐴 with |푤| ⩾ 푛, then −1 푤휑 = 0푆 and so 푤 ∈ 퐿휑휑 = 퐿. Hence 퐿 contains all words with at least 푛 letters and so is cofinite. Thus 퐿 ∈ N(퐴+) ∗ Suppose that 퐿휑 does not contain 0푆. Then if 푤 ∈ 퐴 with |푤| ⩾ 푛, then 푤 ∉ 퐿, since otherwise 퐿휑 ∋ 푤휑 ∈ 푆푛 = {0}. Hence 퐿 contains only words with fewer than 푛 symbols and so 퐿 is finite. Thus 퐿 ∈ N(퐴+). Thus if 퐿 is a +-language over 퐴 recognized by a semigroup in N, then 퐿 ∈ N(퐴+). On the other hand, let 퐿 be a +-language in N(퐴+). So 퐿 is either finite or cofinite. Then there exists some 푛 ∈ ℕ such that either 퐿 ∩ 퐼푛 = ∅ or

Eilenberg correspondence • 187 Variety of rational ∗-languages M-pseudo- (class associated to 퐴∗) Symbol variety See also TABLE 9.3 Varieties of rational ∗- {∅, 퐴∗} E 1 languages and M-pseudovari- Rational ∗-languages over 퐴 M M eties related by the Eilenberg Star-free ∗-languages over 퐴 SF A Th. 9.19 correspondence.

Variety of rational +-languages S-pseudo- (class associated to 퐴+) Symbol variety See also {∅, 퐴+} E 1 Rational +-languages over 퐴 S S Finite/cofinite +-langs over 퐴 N N Th. 9.11 푋퐴∗ ∪ 푌, where K K Th. 9.12(a) 푋, 푌 are finite +-languages 퐴∗푋 ∪ 푌, where D D Th. 9.12(b) 푋, 푌 are finite +-languages 푘 ∗ 푍 ∪ ⋃푖=1 푥푖퐴 푦푖, L1 핃1 Th. 9.13 where 푍 is a finite +-language, { + 푘 ∈ ℕ ∪ 0, and 푥푖, 푦푖 ∈ 퐴 ∗ 푘 ∗ TABLE 9.4 푍 ∪ ⋃푎∈퐴 푎퐴 푎 ∪ ⋃푖=1 푎푖퐴 푎′푖, RB RB Exer. 9.5 where Varieties of rational +-languages and S-pseudovarieties 푍 ⊆ 퐴, 푘 ∈ ℕ ∪ 0 and related by the Eilenberg { correspondence. 푎푖, 푎′푖 ∈ 퐴 with 푎푖 ≠ 푎′푖

∗ + 퐿 ⊇ 퐼푛, where 퐼푛 = { 푤 ∈ 퐴 ∶ |푤| ⩾ 푛 }. Notice that 퐼푛 is an ideal of 퐴 , + 푛 and that 푆 = 퐴 /퐼푛 is a nilpotent semigroup with 푆 = 0푆; thus 푆 ∈ N. ♮ + Then the natural homomorphism 휌 ∶ 퐴 → 푆 recognizes 퐿. 9.11 퐼푛 The remainder of this chapter is devoted to more involved results that, like Theorem 9.11, are instances of the Eilenberg correspondence. All such from this chapter are summarized in Tables 9.3 and 9.4. A semigroup 푆 is left-trivial if 푒푠 = 푒 for all 푒 ∈ 퐸(푆) and 푠 ∈ 푆, and right-trivial if 푠푒 = 푒 for all 푒 ∈ 퐸(푆) and 푠 ∈ 푆. The finite left-trivial 휔 휔 semigroups form the S-pseudovariety K = ⟦푥 푦 = 푥 ⟧S, and the finite 휔 휔 right-trivial semigroups form the S-pseudovariety D = ⟦푦푥 = 푥 ⟧S Let K be the correspondence where K(퐴+) is the class of all +-languages of the form 푋퐴∗ ∪푌, where 푋 and 푌 are finite +-languages over 퐴; and let D be correspondence where D(퐴+) is the class of all +-languages of the form 퐴∗푋 ∪ 푌, where 푋 and 푌 are finite +-languages over 퐴. T h e o r e m 9 . 1 2. a) K is a variety of +-languages and is associated by the Eilenberg correspondence to the S-pseudovariety K; b) D is a variety of +-languages and is associated by the Eilenberg correspondence to the S-pseudovariety D.

Eilenberg correspondence • 188 Proof of 9.12. We will prove part a); dual reasoning gives part b). Let 퐿 = 푋퐴∗ ∪ 푌, where 푋 and 푌 are finite +-languages over 퐴. Now, 푌 is finite and so lies in N(퐴+). Hence SynS 푌 lies in N by Theorem 9.11. Hence SynS 푌 satisfies the S-pseudoidentity 푥휔푦 = 푥휔 and so SynS 푌 ∈ K. Let 푛 be the length of the longest word in 푋. Let 푤 ∈ 퐴+ be such that |푤| = 푛 and let 푡 ∈ 퐴∗. Then 푢푤푣 ∈ 푋퐴∗ ⇔ 푢푤푡푣 ∈ 푋퐴∗, so ∗ 푛 ∗ 푤 휎푋퐴∗ 푤푡. So 푠푡 = 푠 for all 푠 ∈ (SynS 푋퐴 ) and 푡 ∈ SynS 푋퐴 . In particular 푒푡 = 푒 for all idempotents 푒, since 푒푛 = 푒. Thus SynS 푋퐴∗ satisfies the S-pseudoidentity 푥휔푦 = 푥휔 and so SynS 푋퐴∗ ∈ K. Finally, note that SynS 퐿 ≼ (SynS 푋퐴∗) × (SynS 푌) by by Proposition 9.6(b). Hence SynS 퐿 ∈ ℍ핊ℙK = K. Now suppose that 퐿 is recognized by a semigroup 푆 ∈ K. Then there is a homomorphism 휑 ∶ 퐴+ → 푆 such that 퐿 = 퐿휑휑−1. Let 푛 = |푆|. Then, by Lemma 7.5, 푆푛 = 푆퐸(푆)푆 = 푆퐸(푆) since 푒푥 = 푒 for all 푒 ∈ 퐸(푆) and 푥 ∈ 푆. Suppose that 푤푡 ∈ 퐿 with |푤| = 푛. Then 푤휑 ∈ 푆푛 = 푆퐸(푆) and so 푤휑 = 푠푒 for some 푠 ∈ 푆 and 푒 ∈ 퐸(푆). It follows that (푤푡)휑 = 푠푒(푡휑) = 푠푒 = 푤휑 since 푒푥 = 푒 for all 푥 ∈ 푆. Hence 푤 ∈ 퐿 and 푤퐴∗ ⊆ (푠푒푆)휑−1 = (푠푒)휑−1 = 푤휑휑−1. Thus if 푤푡 ∈ 퐿, where |푤| = 푛, then 푤퐴∗ ⊆ 퐿. Hence 퐿 = 푋퐴∗ ∪ 푌, where 푋 ⊆ 퐴푛 and 푌 is a set of words of length less than 푛, + so 퐿 ∈ K(퐴 ). 9.12 Notice that a left- or right-trivial semigroup is also locally trivial: if 푆 is left-trivial, then 푒푠 = 푒 for all 푠 ∈ 푆 and 푒 ∈ 퐸(푆), and hence 푒푠푒 = 푒2 = 푒, which shows that 푆 is locally trivial. Hence K ∪ D ⊆ 핃1. Let L1 be the correspondence where L1(퐴+) is the class of languages of the form

푘 ∗ 푍 ∪ ⋃ 푥푖퐴 푦푖, (9.8) 푖=1

+ where 푥푖 and 푦푖 are words 퐴 and 푍 is a finite +-language. Some texts define L1(퐴+) to be the class of languages of the form 푍 ∪ 푋퐴∗푌, where 푋, 푌, and 푍 are finite +-languages, and claim that this is equivalent to (9.8). This is incorrect, because (for example) the language 푎{푎, 푏}∗푎 ∪ 푏{푎, 푏}∗푏 cannot be expressed in the form 푍 ∪ 푋{푎, 푏}∗푌.

T h e o r e m 9 . 1 3. The Eilenberg correspondence associates the S-pseudovariety 핃1 with the variety of +-languages L1.

Proof of 9.13. We will first of all show that the syntactic semigroups of 푤퐴∗ and 퐴∗푤 are in 핃1 for all 푤 ∈ 퐴+. Since 푤퐴∗ ∈ K(퐴+), it follows by Theorem 9.12(a) that SynS(푤퐴∗) ∈ K. Hence SynS(푤퐴∗) is left-trivial and so locally trivial, and so SynS(푤퐴∗) ∈ 핃1. Similarly SynS(퐴∗푤) ∈ 핃1. Let 푍 be a finite +-languages over 퐴. Then

푍 = ⋃{푤} = ⋃(푤퐴∗ ∖ ⋃ 푤푎퐴∗). 푤∈푍 푤∈퐾 푎∈퐴

Eilenberg correspondence • 189 So SynS 푍 ∈ 핃1 by Proposition 9.8 and since 핃1 is closed under finitary direct products and division. Furthermore, for any 푥, 푦 ∈ 퐴+, we have

|푥|+|푦| 푥퐴∗푦 = (푥퐴∗ ∩ 퐴∗푦) ∖ ⋃ 퐴푖. 푖=1 So SynS(푥퐴∗푦) ∈ 핃1 by Proposition 9.8 again. Therefore by Proposition 9.8,

푘 ∗ SynS(푍 ∪ ⋃ 푥푖퐴 푦푖) ∈ 핃1. 푖=1 On the other hand, let 퐿 be recognized by some semigroup 푆 in 핃1. Then 푆 is locally trivial and there is a homomorphism 휑 ∶ 퐴+ → 푆 such that 퐿휑휑−1 = 퐿. Let 푛 = |푆|. Let 푒 ∈ 퐸(푆) and 푠 ∈ 푆. Then (푒푠)2 = 푒푠푒푠 = 푒푠 since 푒푠푒 = 푒 because 푆 is locally trivial. Hence 푒푠 ∈ 퐸(푆). Similarly 푠푒 ∈ 퐸(푆). Thus 퐸(푆) is an ideal, and so, by Lemma 7.5, 푆푛 = 푆퐸(푆)푆 = 퐸(푆). Let 푤 ∈ 퐿 be such that |푤| ⩾ 2푛. Then 푤 = 푥푣푦 with |푥| = |푦| = 푛. Now, 푆푛 = 퐸(푆) by the previous paragraph, so 푥휑, 푦휑 ∈ 푆푛 = 퐸(푆). Hence 푤휑 = (푥휑)(푣휑)(푦휑) ∈ (푥휑)푆(푦휑). Hence 푥퐴∗푦 ⊆ 푤휑휑−1 ⊆ 퐿. Thus 퐿 is a finite union of languages of the form 푥퐴∗푦 (where |푥| = |푦| = 푛) and a + finite set of words of length at most 2푛. Thus 퐿 ∈ L1(퐴 ). 9.13

C o r o l l a ry 9 . 1 4. 핃1 = K ⊔ D.

Proof of 9.14. Since K ∪ D ⊆ 핃1, it remains to show that 핃1 ⊆ K ⊔ D. Let V be the +-variety of rational languages associated to K ⊔ D by the Eilenberg correspondence. Then V(퐴+) contains the languages 푤퐴∗ and 퐴∗푤 and hence the Boolean algebra generated by these languages, + namely L1(퐴 ). Therefore 핃1 ⊆ K ⊔ D. 9.14 Notice that the proof of Corollary 9.14 essentially involves using the Eilenberg correspondence to convert a question about S-pseudovarieties of semigroups into one about varieties of +-rational languages and back again. Although it would be possible to give a pure pseudovariety-theoretic proof of this result, the proof via the Eilenberg correspondence is much more straightforward.

Schützenberger’s theorem The aim of this final section, and the capstone of the entire course, is Schützenberger’s theorem, which shows that the star-free

Schützenberger’s theorem • 190 rational languages are precisely those languages recognized by aperiodic monoids. Before embarking on the proof, we need to introduce a new concept. A relational morphism between two semigroups 푆 and 푇 is a relation Relational morphism 휑 ⊆ 푆 × 푇 such that 1) 푥휑 ≠ ∅ for all 푥 ∈ 푆; 2) (푥휑)(푦휑) ⊆ (푥푦)휑 for all 푥, 푦 ∈ 푆. Notice that any homomorphism is also a relational morphism. We need to establish some basic properties of relational morphisms that we will use to prove Schützenberger’s theorem. The first three lemmata follow immediately from this definition: L e m m a 9 . 1 5. A relational morphism 휑 ⊆ 푆 × 푇 between semigroups 푆 and 푇 is a subsemigroup of 푆 × 푇. The projection homomorphisms from 푆 × 푇 to 푆 and 푇 restricts to homomorphisms 훼 ∶ 휑 → 푆 and 훽 ∶ 휑 → 푇 −1 such that 훼 is surjective and 휏 = 훼 훽. 9.15 L e m m a 9 . 1 6. If 휑 ∶ 푆 → 푇 is a surjective homomorphism from a semigroup 푆 to a semigroup 푇, then 휑−1 ⊆ 푇 × 푆 is a relational morphism between 푇 and 푆. 9.16 L e m m a 9 . 1 7. If 휑 ⊆ 푆 × 푇 and 휓 ⊆ 푇 × 푈 are relational morphisms between semigroups 푆 and 푇, and between semigroups 푇 and 푈, respectively, then 휑휓 ⊆ 푆 × 푈 is a relational morphism between 푇 and 푆. 9.17 L e m m a 9 . 1 8. Let 휑 ⊆ 푆 × 푇 be a relational morphism between finite semigroups 푆 and 푇. Suppose that 푇 is aperiodic and that for all 푒 ∈ 퐸(푇), the subsemigroup 푒휑−1 is aperiodic. Then 푆 is aperiodic. Proof of 9.18. Let 푥 ∈ 푆. Since 푆 is finite, 푥푘+푚 = 푥푘 for some 푘, 푚 ∈ ℕ, and 퐻 = {푥푘, 푥푘+1, … , 푥푘+푚−1} is a subgroup of 푆. Let 훼 ∶ 휑 → 푆 and 훽 ∶ 휑 → 푇 be as in Lemma 9.15, so that 휑 = 훼−1훽. Then 퐻훼−1 is a subgroup of 휑, and so 퐻훼−1훽 = 퐻휑 is a subgroup of 푇. Since 푇 is aperiodic 퐻휑 is trivial by Proposition 7.4, 퐻휑 = 푒 for some idempotent 푒 of 푇. By the hypothesis, 푒휑−1 ⊇ 퐻 is aperiodic, and so 푚 = 1 and 푘+1 푘 푥 = 푥 . Since 푥 ∈ 푆 was arbitrary, this proves that 푆 is aperiodic. 9.18

Schützenberger’s Theorem 9.19. The Eilenberg correspond- Schützenberger’s theorem ence associates the variety of star-free rational ∗-languages SF and the pseudovariety A of aperiodic monoids. Proof of 9.19. Let A be the ∗-variety of rational languages associated to A. We have to prove that SF(퐴∗) = A(퐴∗) for all finite alphabets 퐴. So fix a finite alphabet 퐴. Part 1 [SF(퐴∗) ⊆ A(퐴∗)]. The class of ∗-languages SF(퐴∗) consists of the languages that can be obtained from the languages {푎} (for 푎 ∈ 퐴) and {휀} using the Boolean operations and concatenation.

Schützenberger’s theorem • 191 Let us therefore begin by showing that A(퐴∗) contains the languages {푎} (for 푎 ∈ 퐴) and {휀}. Let 푎 ∈ 퐴. Let 푆 = {푥, 0} be a two-element null semigroup with all products equal to 0. Let 휓 ∶ 퐴 → 푆1 be the map with 푎휓 = 푥 and 푏휓 = 0 for all 푏 ∈ 퐴 ∖ {푎}, and let 휓∗ ∶ 퐴∗ → 푆1 be the unique extension of 휓 to a monoid homomorphism. It is easy to see that {푎} = {푥}휓−1 = {푎}휓휓−1, thus {푎} is recognized by the monoid 푆1. Clearly 푆1 is an aperiodic monoid; thus 푆1 ∈ A. Hence {푎} ∈ A(퐴∗) by (9.3). Finally, {휀} = 푎−1{푎} ∈ A(퐴∗) by the definition of a ∗-variety of rational languages. Further, by the definition of a ∗-variety of rational languages, A(퐴∗) is a Boolean algebra and thus closed under the Boolean operations. It therefore remains to show that A(퐴∗) is closed under concatenation. So let 퐾, 퐿 ∈ A(퐴∗); we aim to prove that 퐾퐿 ∈ A(퐴∗). Then both SynM 퐾 and SynM 퐿 belong to A by (9.4). That is, SynM 퐾 and SynM 퐿 are aperiodic. ♮ ∗ Consider the three syntactic monoid homomorphisms 휎퐾 ∶ 퐴 → ♮ ∗ ♮ ∗ SynM 퐾, 휎퐿 ∶ 퐴 → SynM 퐿, and 휎퐾퐿 ∶ 퐴 → SynM(퐾퐿). Let 휂 = ♮ −1 ♮ ∗ (휎퐾퐿) . Since 휎퐾퐿 is surjective, 휂 ⊆ SynM(퐾퐿)×퐴 is a relational morphism by Lemma 9.16. Let 휁 ∶ 퐴∗ → SynM 퐾 × SynM 퐿 be defined by 푢휁 = ♮ ♮ (푢휎퐾, 푢휎퐿); clearly 휁 is a homomorphism and thus a relational morphism. Let 휑 = 휂휁; then 휑 is a relational morphism between SynM(퐾퐿) and (SynM 퐾) × (SynM 퐿) by Lemma 9.17. We want to use Lemma 9.18 and the relational morphism 휑 to show that SynM(퐾퐿) is aperiodic. Let (푒1, 푒2) ∈ 퐸((SynM 퐾) × (SynM 퐿)). Let −1 −1 2 푚 ∈ (푒1, 푒2)휑 . Then 푚 = 푔휂 for some 푔 ∈ (푒1, 푒2)휁 . Then 푔 휁 = 2 2 2 ♮ (푒1 , 푒2 ) = (푒1, 푒2) = 푔휁. Thus (푔 , 푔) ∈ ker 휁 ⊆ ker 휎퐾 = 휎퐾. Similarly 2 (푔 , 푔) ∈ 휎퐿. Suppose 푢푔2푣 ∈ 퐾퐿 for some 푢, 푣 ∈ 퐴∗. Then 푢푔2푣 = 푥푦, for 푥 ∈ 퐾 and 푦 ∈ 퐿. Then, by equidivisibility, either there exists 푝 ∈ 퐴∗ such that 푥 = 푢푔푝 and 푝푦 = 푔푣, or there exists 푞 ∈ 퐴∗ such that 푥푞 = 푢푔 and 2 푦 = 푞푔푣. Assume the former case; the latter is similar. Since (푔 , 푔) ∈ 휎퐾, we have 푢푔2푝 ∈ 퐾, and so 푢푔2푝푦 = 푢푔3푣 ∈ 퐾퐿. This shows that 푢푔2푣 ∈ 퐾퐿 implies 푢푔3푣 ∈ 퐾퐿. Now suppose 푢푔3푣 ∈ 퐾퐿 for some 푢, 푣 ∈ 퐴∗. Then 푢푔3푣 = 푥푦, for 푥 ∈ 퐾 and 푦 ∈ 퐿. Then, by equidivisibility, either there exists 푝 ∈ 퐴∗ such that 푥 = 푢푔2푝 and 푝푦 = 푔푣, or there exists 푞 ∈ 퐴∗ such that 푥푞 = 푢푔2 and 푦 = 푞푔푣. Assume the former case; the latter is similar. 2 2 Since (푔 , 푔) ∈ 휎퐾, we have 푢푔푝 ∈ 퐾, and so 푢푔푝푦 = 푢푔 푣 ∈ 퐾퐿. This shows that 푢푔3푣 ∈ 퐾퐿 implies 푢푔2푣 ∈ 퐾퐿. 3 2 Combining the last two paragraphs shows that (푔 , 푔 ) ∈ 휎퐾퐿, and so 3 3 ♮ 2 ♮ 2 −1 푚 = 푔 휎퐾퐿 = 푔 휎퐾퐿 = 푚 . Since 푚 was an arbitrary element of 푒휑 , it follows that the subsemigroup 푒휑−1 is aperiodic. Since both SynM 퐾 and SynM 퐿 are aperiodic, (SynM 퐾) × (SynM 퐿) is aperiodic. Hence, by Lemma 9.18, SynM(퐾퐿) is aperiodic. Thus SynM(퐾퐿) ∈ A, and so

Schützenberger’s theorem • 192 퐾퐿 ∈ A(퐴∗) by (9.3). So A(퐴∗) is closed under concatenation. Thus A(퐴∗) contains every language in SF(퐴∗). Part 2 [A(퐴∗) ⊆ SF(퐴∗)]. The aim is to prove that any ∗-language recognized by an aperiodic monoid 푀 (and thus belonging to A(퐴∗)) lies in SF(퐴∗). The strategy is to proceed by induction on |푀|. For brevity, let 훥푀 = { 푁 ∶ 푁 ≼ 푀 ∧ 푁 ≠ 푀 }. That is, 훥푀 is the class of monoids that strictly divide 푀. Let 퐴∗훥푀 be the class of ∗-languages over 퐴 recognized by some monoid in 훥푀. The base of the induction consists of the cases |푀| = 1 and |푀| = 2. First, suppose |푀| = 1. Let 퐿 be a ∗-language over 퐴 recognized by 휑 ∶ 퐴∗ → 푀. Then either 퐿 = ∅휑−1 = ∅ or 퐿 = 푀휑−1 = 퐴∗, and both these languages are in SF(퐴∗) by definition of a ∗-variety of rational languages. Now suppose |푀| = 2. Then 푀 is the two-element semilattice {1, 0} with 1 > 0. [To see this, let {1, 푧} be an aperiodic monoid. Then 11 = 1, 1푧 = 푧1 = 푧, and either 푧푧 = 1 or 푧푧 = 푧. But in the former case, we have a cyclic group, which is not aperiodic. Hence 푧푧 = 푧 and we have a commutative semigroup of idempotents.] Let 퐿 be a ∗-language recognized by 휑 ∶ 퐴∗ → 푀. Let 퐵 = { 푎 ∈ 퐴 ∶ 푎휑 = 0 }. Then

0휑−1 = ⋃ 퐴∗푏퐴∗, 푏∈퐵 1휑−1 = 퐴∗ ∖ ⋃ 퐴∗푏퐴∗. 푏∈퐵

Then 0휑−1 ∈ SF(퐴∗), since SF(퐴∗) contains the languages 퐴∗ and {푏} for any 푏 ∈ 퐵 and is by definition closed under concatenation and union, and 1휑−1 ∈ SF(퐴∗), since SF(퐴∗) is by definition closed under complementation. Since one of the four cases 퐿 = ∅, 퐿 = 0휑−1, 퐿 = 1휑−1, and 퐿 = 푀휑−1 = 0휑−1 ∪ 1휑−1 holds, and since SF(퐴∗) is closed under union, it follows that 퐿 ∈ SF(퐴∗). We have completed the base of the induction; we turn now to the induction step. Let |푀| ⩾ 3 and suppose that every language in 퐴∗훥푀 lies in SF(퐴∗); that is, 퐴∗훥푀 ⊆ SF(퐴∗). We must prove that every language recognized by 푀 lies in SF(퐴∗). Let 퐿 be a ∗-language over 퐴 recognized by 푀. Then there exists a homomorphism 휑 ∶ 퐴∗ → 푀 and a subset 푃 of 푀 such that 퐿 = 푃휑−1. If 휑 is not surjective, then 퐿 is recognized by the proper submonoid im 휑 of 푀 and so, since im 휑 ∈ 훥푀, by induction 퐿 ∈ 퐴∗훥푀 ⊆ SF(퐴∗). So assume that 휑 is surjective. Furthermore, since

퐿 = 푃휑−1 = ⋃ 푚휑−1 푚∈푃 and SF(퐴∗) is by definition closed under union, it suffices to prove the case where 퐿 = 푚휑−1.

Schützenberger’s theorem • 193 Let

퐾 = ⋂{ 퐼 ∶ 퐼 is an ideal of 푀 and |퐼| ⩾ 2 }. (9.9)

Then 퐾 is an ideal of 푀. For use later in the proof, we now establish some properties of 퐾, considering separately the cases where 푀 has a zero and where 푀 does not have a zero: ◆ 푀 has a zero. Let 퐷 = 퐾 ∖ {0}. Suppose 퐷 is non-empty. Then for any 푥 ∈ 퐷, we have {0, 푥} ≠ 푀푥푀 ⊆ 퐾 (since 퐾 is an ideal and 푥 ≠ 0), and thus 푀푥푀 = 퐾 (since 푀푥푀 is one of the ideals 퐼 in the intersection (9.9) and so 퐾 ⊆ 푀푥푀). So 퐷 is a single J-class of 푀 and so a single D-class of 푀 by Proposition 3.3. Furthermore, 퐾 is a 0-minimal ideal and so either 0-simple or null by Proposition 3.8(a). So either 퐷 is empty, or else 퐷 is a single D-class and 퐾 is 0-simple or null. ◆ 푀 does not have a zero. Then 퐾 is the kernel of 푀 and so simple by Proposition 3.8(b). (Since if there were an ideal with only one element 푧, then 푆푧 = {푧} and 푧푆 = {푧} and so 푧 would be a zero.) Furthermore, 퐾2 = 퐾. (Since if 퐾2 ⊊ 퐾, then 퐾2 would be an ideal of 푆 strictly contained in 퐾.) We now consider separately the three cases where 푚 ∉ 퐾, where 푚 is the zero of 푀, and where 푚 is not a zero of 푀 (but 푀 may or may not contain a zero): a) 푚 ∉ 퐾. Then there exists an ideal 퐼 of 푀 with |퐼| ⩾ 2 such that 푚 ∉ 퐼. Let 휌퐼 = (퐼 × 퐼) ∪ id푀 be the Rees congruence. Then 푚 = ♮ ♮ −1 −1 ♮ ♮ −1 −1 ♮ ♮ −1 휌퐼 (휌퐼 ) and hence 푚휑 = 푚휌퐼 (휌퐼 ) 휑 = (푚휌퐼 )(휑휌퐼 ) . Thus −1 ♮ ∗ 푚휑 is recognized by the homomorphism 휑휌퐼 ∶ 퐴 → 푀/퐼 and so is recognized by 푀/퐼. Since |푀/퐼| = |푀| − |퐼| + 1 < |푀| (since |퐼| ⩾ 2|), we have 푀/퐼 ∈ 훥푀 and so by induction 푚휑−1 ∈ SF(퐴∗).

b) 푀 has a zero and 푚 = 0푀. Let 퐶 = { 푎 ∈ 퐴 ∶ 푎휑 = 0 }. The first step is to prove that

0휑−1 = 퐴∗퐶퐴∗ ∪ ⋃ 퐴∗푎(푛휑−1)푎′퐴∗, (9.10) (푎,푛,푎′)∈퐸 where

퐸 = { (푎, 푛, 푎′) ∈ (퐴 ∖ 퐶) × (푀 ∖ 퐾) × (퐴 ∖ 퐶) ∶ (푎휑)푛(푎′휑) = 0 ∧ (푎휑)푛 ≠ 0 ∧ 푛(푎′휑) ≠ 0 }.

First, notice that (퐴∗퐶퐴∗)휑 = 푀(퐶휑)푀 ⊆ {0}. If (푎, 푛, 푎′) ∈ 퐸, then (퐴∗푎(푛휑−1)푎′퐴∗)휑 = 푀(푎휑)푛(푎′휑)푀 = 푀0푀 = {0}. This shows that the right-hand side of (9.10) is contained in the left-hand side. Let 푓 ∈ 0휑−1 ∖ 퐴∗퐶퐴∗ = 0휑−1 ∩ (퐴 ∖ 퐶)∗. Since 푀 has at least two elements, 1 ≠ 0 and so 푓 ≠ 휀. Let 푔 be the longest left factor

Schützenberger’s theorem • 194 of 푓 such that 푔휑 ≠ 0. (Such a left factor exists since 휀휑 = 1 and 푓휑 = 0.) Then 푓 = 푔푎′푔′, where 푔휑 ≠ 0 and (푔푎′)휑 = 0. Note that since 푓 ∈ (퐴 ∖ 퐶)∗, we have 푔 ∈ (퐴 ∖ 퐶)∗ and 푎′ ∈ 퐴 ∖ 퐶. Let ℎ be the longest right factor of 푔 such that (ℎ푎′)휑 ≠ 0. (Such a right factor of 푔 exists because 푎′휑 ≠ 0 and (푔푎′)휑 = 0.) Then 푔 = ℎ′푎ℎ, where (ℎ푎′)휑 ≠ 0 and (푎ℎ푎′)휑 = 0. Note that since 푔 ∈ (퐴 ∖ 퐶)∗, we have ℎ ∈ (퐴 ∖ 퐶)∗ and 푎 ∈ 퐴 ∖ 퐶. Furthermore, since 푔휑 ≠ 0, we have (푎ℎ)휑 ≠ 0. Let 푛 = ℎ휑. Suppose, with the aim of obtaining a contradiction, that 푛 ∈ 퐾. Since 퐾 is an ideal, 푛(푎′휑) ∈ 퐾. Since 푛(푎′휑) = (ℎ푎′)휑 ≠ 0, we have 푛(푎′휑) ∈ 퐷. Thus 푛 D 푛(푎′휑), and so, by Lemma 7.6(a), 푛 R 푛(푎′휑). Similarly, since (푎휑)푛 = (푎ℎ)휑 ≠ 0, we have (푎휑)푛 L 푛. Hence, by Lemma 3.12, (푎휑)푛(푎′휑) L 푛(푎′휑) and therefore we have (푎휑)푛(푎′휑) D 푛. Thus (푎휑)푛(푎′휑) lies in the D-class 퐷, which contradicts the fact that (푎ℎ푎′)휑 = 0. Hence 푛 ∉ 퐾, and thus 푓 = ℎ′푎ℎ푎′푔′ ∈ 퐴∗푎(푛휑−1)푎′퐴∗ with (푎, 푛, 푎′) ∈ 퐸. This shows that the left-hand side of (9.10) is contained in the right-hand side. Since 푛 ∉ 퐾, the reasoning in case a) shows that 푛휑−1 ∈ 퐴∗훥푀 and thus 푛휑−1 ∈ SF(퐴∗). Since SF(퐴∗) is closed under Boolean operations and concatenation, it follows from (9.10) that 푚휑−1 = 0휑−1 ∈ SF(퐴∗). c) 푚 ∈ 퐾 ∖ {0} (where 퐾 ∖ {0} = 퐾 if 푀 does not contain a zero). Now, 푚푀 ⊆ 퐾 since 퐾 is an ideal. Hence all elements of 푚푀 ∖ {0} are D- related and hence R-related by Lemma 7.6. Thus 푅푚 = 푚푀 ∖ {0}. Similarly, 퐿푚 = 푚푀 ∖ {0}.

{푚} = 퐻푚 [by Proposition 7.4]

= 푅푚 ∩ 퐿푚 = (푚푀 ∖ {0}) ∩ (푀푚 ∖ {0}) = (푚푀 ∩ 푀푚) ∖ {0}.

(When 푀 does not contain a zero, this becomes {푚} = 푚푀 ∩ 푀푚.) Thus, since by case b) we already know that 0휑−1 is in SF(퐴∗), it is sufficient to prove that (푚푀)휑−1 and (푀푚)휑−1 are in SF(퐴∗). We will prove (푚푀)휑−1 ∈ SF(퐴∗); the other case is similar. The first step is to prove that

(푚푀)휑−1 = 0휑−1 ∪ ⋃(푛휑−1)푎퐴∗. (9.11) (푛,푎)∈퐹 where

퐹 = { (푛, 푎) ∈ (푀 − 퐾) × 퐴 ∶ 푛(푎휑) ∈ 푅푚 }.

(We formally let 0휑−1 = ∅ if 푀 does not contain a zero.)

Schützenberger’s theorem • 195 −1 ∗ −1 −1 If (푛, 푎) ∈ 퐹, then (푛휑 )푎퐴 = 푅푚휑 ⊆ (푚푀)휑 . Trivially, 0휑−1 ⊆ (푚푀)휑−1. Thus the right-hand side of (9.10) is contained in the left-hand side. Let 푓 ∈ (푚푀)휑−1. If 푓휑 = 0, then 푓 ∈ 0휑−1. So assume 푓휑 ≠ 0. Then 푓휑 ∈ 푚푀 ∖ {0} = 푅푚. Since 1 ∉ 퐾 (since 푀 has at least two elements), 휀휑 = 1 ∉ 푅푚. Let 푔 be the longest left factor of 푓 such that 푔휑 ∉ 푅푚. (Such a longest left factor exists since 휀휑 ∉ 푅푚 and 푓휑 ∈ 푅푚.) Hence 푓 = 푔푎푔′ where 푔휑 ∉ 푅푚 and (푔푎)휑 ∈ 푅푚, where 푔, 푔′ ∈ 퐴∗ and 푎 ∈ 퐴. Let 푛 = 푔휑. Suppose, with the aim of obtaining a contradiction, that 푛 ∈ 퐾. Then 푅푛 = 푛푀 ∖ {0} and so 푛(푎휑) ∈ 푅푛. But 푛(푎휑) ∈ 푅푚, so 푅푛 = 푅푚 and so 푛 ∈ 푅푚, which contradicts 푛 = 푔휑 ∉ 푅푚. Thus 푛 ∉ 퐾, and so (푛, 푎) ∈ 퐹. Therefore 푓 ∈ (푛휑−1)푎퐴∗ This shows that the left-hand side of (9.11) is contained in the right-hand side. Thus, for any (푛, 푎) ∈ 퐹, we have 푛 ∉ 퐾 and so 푛휑−1 ∈ 퐴∗훥푀 by case a), and thus 푛휑−1 ∈ SF(퐴∗). Hence (푀푚)휑−1 is in SF(퐴∗) by (9.11). This completes the induction step and thus the proof. 9.19 Finis.

Exercises [See pages 243–245 for the solutions.] ✴9.1 Prove that a language 퐿 ⊆ 퐴∗ is rational if and only if SynM 퐿 is finite. [Hint: this is an easy consequence of results in this chapter.] 9.2 Let 퐴 = {푎, 푏}. Let 퐿 be the language of words over 퐴 that contain at least one symbol 푎 and at least one symbol 푏. (That is, 퐿 = 퐴+ ∖ ({푎}+ ∪ {푏}+).) Find a homomorphism 휑 ∶ 퐴+ → 푆 that recognizes 퐿. [Hint: 푆 can be taken to have 3 elements.] ✴9.3 A Dyck word is a string of balanced parentheses: that is, a word in Dyck word { ( , ) }∗ where every opening parenthesis ( matches a corresponding closing parentheses ) to its right, and vice versa. For instance,

( ) ( ( ( ) ( ) ) ( ) ) ( ( ) ) is a Dyck word, but

( ) ) ( ( ) ( ) ) ( ) ( ( ( ) ( is not a Dyck word.

An equivalent characterization of Dyck words is the following: define ∗ a map 퐶 ∶ { ( , ) } × ℕ ∪ {0} → ℤ as follows: let 퐶(푤1 ⋯ 푤푛, 푖) be the number of symbols ( minus the number of symbols ) in the prefix 푤1 ⋯ 푤푖. A word 푤 is a Dyck word if and only if 퐶(푤, |푤|) = 0 and

Exercises • 196 퐶(푤, 푖) ⩾ 0 for all 푖 = 1, … , |푤|. Consider the two example words above, with 퐶(푤, 푖) plotted graphically:

If 푤 = ( ) ( ( ( ) ( ) ) ( ) ) ( ( ) ) , then 퐶(푤, |푤|) = 0 and 퐶(푤, 푖) ⩾ 0 for all 푖.

If 푤 = ( ) ) ( ( ) ( ) ) ( ) ( ( ( ) ( , then 퐶(푤, |푤|) ≠ 0 and 퐶(푤, 푖) < 0 for some 푖.

Let 퐷 be the language of Dyck words. Prove that SynM 퐷 is isomorphic to the bicyclic monoid. ✴9.4 Without using the Eilenberg correspondence, prove that the correspondence N in Example 9.9(c) (with N(퐴+) being the class of finite or cofinite languages over 퐴) is a variety of rational languages. ✴9.5 Let RB be the S-pseudovariety of rectangular bands. Let RB be the variety of rational +-languages associated to RB by the Eilenberg correspondence. Prove that RB(퐴+) is the class of all +-languages of the form 푛 ∗ ∗ 푍 ∪ ⋃ 푎퐴 푎 ∪ ⋃ 푎푖퐴 푎′푖, (9.12) 푎∈푍 푖=1

where 푍 ⊆ 퐴 and 푎푖, 푎′푖 ∈ 퐴 with 푎푖 ≠ 푎′푖 for each 푖.

Notes

For further reading on automata and rational languages, see Hopcroft & Ullman, Introduction to Automata Theory, Languages, and Compu- tation, ch. 2, Lawson, Finite Automata, or Howie, Automata and Languages. ◆ Theorem 9.1 is due to Rabin & Scott, ‘Finite automata and their decision problems’. ◆ Theorem 9.2 was first stated, in rather different terminology, in Kleene, ‘Representation of events in nerve nets and finite automata’. ◆ The discussion of syntactic monoids and semigroups and Eilenberg’s correspondence is based on Eilenberg, Automata, Languages, and Machines (Vol. B), ch. vii and Pin, ‘Syn- tactic semigroups’, §§ 2.2–3. ◆ The proof of Schützenberger’s theorem is a blend of the original proof by Schützenberger, ‘On finite monoids having only trivial subgroups’ and its exposition in Pin, Varieties of Formal Languages, § 4.2 ◆ For further reading on the connection between semigroups and languages, see Pin, Varieties of Formal Languages or Pin, ‘Mathematical Foundations of Automata Theory’. •

Notes • 197 Solutions to exercises

A solution which does not prepare for the next round ‘ with some increased insight is hardly a solution at all. —’ R.W. Hamming, The Art of Doing Science and Engineering, p. 200.

Exercises for chapter 1 [See pages 32–34 for the exercises.] 1.1 Let 푥 ∈ 푆. Then 푥 = 푥푒 since 푒 is a right identity, and 푒 = 푥푒 since 푒 is a right zero. Hence 푥 = 푥푒 = 푒. Thus 푒 is the only element of 푆. 1.2 a) If 푆 contains a zero, then 푆0 = 푆 and there is nothing to prove. Otherwise 푆0 = 푆 ∪ {0}. Then 푥1 = 푥1 = 푥 for all 푥 ∈ 푆 since 1 is an identity for 푆, and 01 = 10 = 0 by the definition of 푆0. Hence 1 is an identity for 푆0. b) The reasoning is similar to part a). 1.3 Let 푆 be left-cancellative and let 푒 ∈ 푆 be an idempotent. Let 푥 ∈ 푆. Since 푒 is idempotent, 푒푒푥 = 푒푥. Since 푆 is left-cancellative, 푒푥 = 푥. Since 푥 ∈ 푆 was arbitrary, this proves that 푒 is a left identity for 푥. Suppose now that 푆 is cancellative and that 푒, 푓 ∈ 푆 are idempotents. By the preceding paragraph and the symmetric result for right-cancellativity, 푒 and 푓 are left and right identities for 푆. By Pro- position 1.3, 푒 = 푓. 1.4 Let 푆 be a right zero semigroup. Suppose 푥, 푦, 푧 ∈ 푆 are such that 푧푥 = 푧푦. Since 푆 is a right zero semigroup, 푧푥 = 푥 and 푧푦 = 푦. Hence 푥 = 푧푥 = 푧푦 = 푦. That is, 푥 = 푦. So 푧푥 = 푧푦 ⇒ 푥 = 푦 for all 푥, 푦, 푧 ∈ 푆 and thus 푆 is left-cancellative. 1.5 Let 푆 be a finite cancellative semigroup. Let 푥 ∈ 푆. Then 푥 is periodic and so some power of 푥 is an idempotent. By Exercise 1.3, this idem- 푛 potent is an identity 1푆 for 푆. Now let 푦 ∈ 푆 be arbitrary. Then 푦 is 푛 idempotent for some 푛 ∈ ℕ. Again by Exercise 1.3, 푦 = 1푆 and so 푦푛−1 is a left and right inverse for 푦. Since 푦 ∈ 푆 was arbitrary, 푆 is a group.

1.6 Let 휌 ∈ B푋. Let 푥, 푦 ∈ 푋. Then

(푥, 푦) ∈ 휌 ∘ id푋

⇔ (∃푧 ∈ 푋)((푥, 푧) ∈ 휌 ∧ (푧, 푦) ∈ id푋) [by definition of ∘]

• 198 ⇔ (∃푧 ∈ 푋)((푥, 푧) ∈ 휌 ∧ (푧 = 푦)) [by definition of id푋] ⇔ (푥, 푦) ∈ 휌.

So 휌 ∘ id푋 = 휌 and similarly id푋 ∘ 휌 = 휌. So id푋 is the identity of B푋. The zero of B푋 is the empty relation ∅. So see this, we must prove that 휌 ∘ ∅ = ∅ ∘ 휌 = ∅. So suppose, with the aim of obtaining a contradiction, that 휌 ∘ ∅ ≠ ∅. Then (푥, 푦) ∈ 휌 ∘ ∅ for some 푥, 푦 ∈ 푋. Then there exists 푧 ∈ 푋 such that (푥, 푧) ∈ 휌 and (푧, 푦) ∈ ∅. But (푧, 푦) ∈ ∅ is a contradiction. So 휌 ∘ ∅ = ∅ and similarly ∅ ∘ 휌 = ∅. 1.7 No. Let 푆 be a non-trivial semigroup. Choose some element 푥 ∈ 푆 and let 푇 = { 푥푛 ∶ 푛 ∈ ℕ } be the subsemigroup generated by 푥. If 푇 is finite (that is, if 푥 is periodic), then some 푥푛 is idempotent and so {푥푛} is a subsemigroup of 푆; furthermore, it must be a proper subsemigroup since 푆 is non-trivial. If, on the other hand, 푇 is infinite, then { 푥2푛 ∶ 푛 ∈ ℕ } is a proper subsemigroup of 푇 and hence of 푆. 1.8 The easiest examples are infinite right or left zero semigroups, and the semigroups (ℕ, △) and (ℤ, △) from Example 1.7(a)–(b).

1.9 The empty relation ∅ is a partial transformation. It is a zero for B푋, so it is certainly a zero for P푋. By Proposition 1.4, this is the unique left and right zero in P푋. Let us prove that the semigroup of transformations T푋 contains exactly |푋| right zeros, namely the constant maps 휏푥 ∶ 푋 → 푋 defined by 푦휏푥 = 푥 for all 푦 ∈ 푋. Each map 휏푥 is a right zero because for any 휎 ∈ T푋, we have 푦휎휏푥 = 푥 for all 푦 ∈ 푋, and so 휎휏푥 = 휏푥. Suppose 휏 ∈ T푋 is a right zero. Then 휎휏 = 휏 for all 휎 ∈ T푋. In particular, this is true for all 휎 ∈ S푋. Choose some 푦 ∈ 푋 and let 푥 = 푦휏. Now let 푧 ∈ 푋. Choose 휎 ∈ S푋 with 푧휎 = 푦. Then 푧휏 = 푧휎휏 = 푦휏 = 푥. Since 푧 ∈ 푋 was arbitrary, we have 휏 = 휏푥. Thus the right zeros in T푋 are precisely the constant maps 휏푥. Suppose 휌 ∈ T푋 is a left zero. Then for all 푥 ∈ 푋, we have 휌 = 휌휏푥 = 휏푥 since 휌 is a left zero and 휏푥 is a right zero. Hence |푋| = 1 and so T푋 is trivial (and so contains a zero). Hence if |푋| ⩾ 2, then T푋 cannot contain a left zero. 1.10 a) Define 휑 ∶ 푆 → ℙ푆 by 푧휑 = {푧}. Then

(푧휑)(푡휑) = {푧}{푡} = {푧푡} = (푧푡)휑.

and

푧휑 = 푡휑 ⇒ {푧} = {푡} ⇒ 푧 = 푡.

So 휑 is a monomorphism and so 휑 ∶ 푆 → im 푆 ⊆ ℙ푆 is an isomorphism. b) For any 푋 ∈ ℙ푆, we have

푋∅ = { 푥푦 ∶ 푥 ∈ 푋 ∧ 푦 ∈ ∅ } = ∅

Solutions to exercises • 199 (1 푖)|1 2|(1 푖) |푖 2| (1 푖)(2 푗)|1 2|(2 푗)(1 푖) |푖 푗| 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ = = 푖 푖 푖 푖 푖 푖 푖 푖 ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ 푗 푗 푗 푗 푗 푗 푗 푗 FIGURE S.3 Generating (a) |2 푗| and (b) |푖 푗| (a) (b) using transpositions and |1 2|.

and similarly ∅푋 = ∅. So ∅ is a zero for ℙ푆. If 푋, 푌 ∈ (ℙ푆) ∖ {∅} then there exist 푥 ∈ 푋 and 푦 ∈ 푌 and so 푥푦 ∈ 푋푌 and hence 푋푌 ≠ ∅. So (ℙ푆) ∖ {∅} is a subsemigroup of ℙ푆.

c) Suppose 푀 is non-trivial. Let 푥 ∈ 푀 ∖ {1푀}. Let 푁 = 푀 ∖ {푥}. Then 푀 ≠ 푁 ≠ ∅ but 푀푀 = 푀 and 푀푁 = 푀 since both 푀 and 푁 contain 1푀. Hence (ℙ푀) ∖ {∅} is not cancellative. On the other hand, suppose 푀 is trivial. Then ℙ푀 = {∅, 푀}. Hence (ℙ푀) ∖ {∅} is trivial and thus cancellative. d) Let 푆 be a right zero semigroup. Let 푋, 푌 ∈ (ℙ푆) ∖ {∅}. Then 푋푌 = { 푥푦 ∶ 푥 ∈ 푋 ∧ 푦 ∈ 푌 } ={푦∶푥∈푋∧푦∈푌} [since 푦 is a right zero] = { 푦 ∶ 푦 ∈ 푌 } = 푌. So (ℙ푆) ∖ {∅} is a right zero semigroup. On the other hand, if ℙ푆 is a right zero semigroup, so is its subsemigroup im 휑 ≃ 푆, where 휑 is the monomorphism from part a). So 푆 is a right zero semigroup. 1.11 a) To prove the four identities, we have to show that the transformations on each side act the same way on every element of 푋. For the first identity, let 푖 ⩾ 3. Then: 1(1 푖)|1 2|(1 푖) = 푖|1 2|(1 푖) = 푖(1 푖) = 1 = 1|푖 2|; 2(1 푖)|1 2|(1 푖) = 2|1 2|(1 푖) = 2(1 푖) = 2 = 2|푖 2|; 푖(1 푖)|1 2|(1 푖) = 1|1 2|(1 푖) = 2(1 푖) = 2 = 2|푖 2|; 푥(1 푖)|1 2|(1 푖) = 푥|1 2|(1 푖) = 푥(1 푖) = 푥 = 푥|푖 2| for 푥 ∈ 푋 ∖ {1, 2, 푖}. (Figure S.3(a) illustrates the first identity diagrammatically.) The second identity is proved similarly. For the third identity, let 푖, 푗 ⩾ 3 with 푖 ≠ 푗. Then: 1(1 푖)(2 푗)|1 2|(2 푗)(1 푖) = 푖(2 푗)|1 2|(2 푗)(1 푖) = 푖|1 2|(2 푗)(1 푖) = 푖(2 푗)(1 푖) = 푖(1 푖) = 1 = 1|푖 푗|;

Solutions to exercises • 200 2(1 푖)(2 푗)|1 2|(2 푗)(1 푖) = 2(2 푗)|1 2|(2 푗)(1 푖) = 푗|1 2|(2 푗)(1 푖) = 푗(2 푗)(1 푖) = 2(1 푖) = 2 = 2|푖 푗|; 푖(1 푖)(2 푗)|1 2|(2 푗)(1 푖) = 1(2 푗)|1 2|(2 푗)(1 푖) = 1|1 2|(2 푗)(1 푖) = 2(2 푗)(1 푖) = 푗(1 푖) = 푗 = 푖|푖 푗|; 푗(1 푖)(2 푗)|1 2|(2 푗)(1 푖) = 푗(2 푗)|1 2|(2 푗)(1 푖) = 2|1 2|(2 푗)(1 푖) = 2(2 푗)(1 푖) = 푗(1 푖) = 푗 = 푗|푖 푗|; 푥(1 푖)(2 푗)|1 2|(2 푗)(1 푖) = 푥(2 푗)|1 2|(2 푗)(1 푖) = 푥|1 2|(2 푗)(1 푖) = 푥(2 푗)(1 푖) = 푥(1 푖) = 푥 = 푥|푖 푗| for 푥 ∈ 푋 ∖ {1, 2, 푖, 푗}.

(Figure S.3(b) illustrates the third identity diagrammatically.) For the fourth identity, let 푖 ≠ 푗. Then:

푖(푖 푗)|푖 푗|(푖 푗) = 푗|푖 푗|(푖 푗) = 푗(푖 푗) = 푖 = 푖|푗 푖|; 푗(푖 푗)|푖 푗|(푖 푗) = 푖|푖 푗|(푖 푗) = 푗(푖 푗) = 푖 = 푗|푗 푖|; 푥(푖 푗)|푖 푗|(푖 푗) = 푥|푖 푗|(푖 푗) = 푥(푖 푗) = 푖 = 푥|푗 푖| for 푥 ∈ 푋 ∖ {푖, 푗}.

b) To prove that |푖 푗|휑′ = 휑, we must show that both sides act the same way on every element of 푋. By the definition of 휑′,

푖|푖 푗|휑′ = 푗휑′ = 푗휑, 푥|푖 푗|휑′ = 푥휑′ = 푥휑 for 푥 ≠ 푖.

c) Since ⟨휏, 휁⟩ = S푋, we have (푖 푗) ∈ ⟨휏, 휁, |1 2|⟩ for all 푖, 푗 ∈ 푋. From part a), the first two identities show that |푖 2| and |1 푗| are in ⟨휏, 휁, |1 2|⟩ for all 푖, 푗 ∈ 푋 ∖ {1, 2}. Combining this with the fourth identity shows that |2 푗| and |푖 1| are in ⟨휏, 휁, |1 2|⟩. Together with the third identity, this shows that |푖 푗| ∈ ⟨휏, 휁, |1 2|⟩ for all 푖, 푗 ∈ 푋. Now proceed by induction on |푋 ∖ im 휑|. If |푋 ∖ im 휑| = 0, then im 휑 = 푋 and so 휑 is surjective and so (since 푋 is finite) injective. Hence 휑 ∈ S푋 = ⟨휏, 휁⟩ ⊆ ⟨휏, 휁, |1 2|⟩. So assume that 휓 ∈ ⟨휏, 휁, |1 2|⟩ is true for all 휓 ∈ T푋 with |푋 ∖ im 휓| = 푘 − 1 < 푛. Let 휑 be such that |푋 ∖ im 휑| = 푘. Then by parts a) and b), we have 휑 = |푖 푗|휑′ = (1 푖)(2 푗)|1 2|(2 푗)(1 푖)휑′, where im 휑′ ⊊ im 휑. Hence |푋 ∖ im 휑′| = 푘 − 1 and so 휑′ ∈ ⟨휏, 휁, |1 2|⟩. Hence 휑 ∈ ⟨휏, 휁, |1 2|⟩. By induction, T푋 = ⟨휏, 휁, |1 2|⟩. 1.12 Suppose 푥 is right invertible. Then there exists 푦 ∈ 푆 such that 푥푦 = 1. Since 푆 is finite, 푥푘 = 푥푘+푚 for some 푘, 푚 ∈ ℕ. So 1 = 푥푘푦푘 = 푥푘+푚푦푘 = 푥푚 = 푥푚−1푥 and so 푥푚−1 is a left inverse for 푥. Similarly, if 푥 is left invertible, it is right invertible.

1.13 a) Let 휌 ∈ T푋 be left-invertible. Then there exists 휎 ∈ T푋 such that 휎 ∘ 휌 = id푋. Let 푥 ∈ 푋. Then 푥(휎 ∘ 휌) = 푥. So (푥휎)휌 = 푥. So 휌 is surjective.

Solutions to exercises • 201 Now let 휌 ∈ T푋 be surjective. Define 휎 ∈ T푋 as follows. For each 푥 ∈ 푋, choose 푦 ∈ 푋 such that 푦휌 = 푥. (Such a 푦 exists because 휌 is surjective.) Define 푥휎 = 푦. Clearly 휎 ∘ 휌 = id푋 and so 휌 is left-invertible.

b) Let 휌 ∈ T푋 be right-invertible. Then there exists 휎 ∈ T푋 such that 휌 ∘ 휎 = id푋. Then 푥휌 = 푦휌 ⇒ (푥휌)휎 = (푦휌)휎 ⇒ 푥 = 푦 and so 휌 is injective. Now let 휌 ∈ T푋 be injective. Define 휎 ∈ T푋 as follows. For 푥 ∈ im 휌, let 푦 ∈ 푋 be the unique element such that 푦휌 = 푥. Define 푥휎 = 푦. For 푥 ∈ 푋 ∖ im 휌, define 푥휎 arbitrarily. Clearly 휌 ∘ 휎 = id푋 and so 휌 is right-invertible. 1.14 a) By definition, 푥 ⊓ 푦 ⩽ 푥. So the least upper bound of 푥 ⊓ 푦 and 푥 (which is the definition of (푥 ⊓ 푦) ⊔ 푥) must be 푥 itself. Dual reasoning gives (푥 ⊔ 푦) ⊓ 푥 = 푥. b) Assume that for all 푝, 푞, 푟 ∈ 푆, we have 푝⊓(푞⊔푟) = (푝⊓푞)⊔(푝⊓푟). (We have re-labelled variables to avoid confusion.) Then (푥 ⊔ 푦) ⊓ (푥 ⊔ 푧) = ((푥 ⊔ 푦) ⊓ 푥) ⊔ ((푥 ⊔ 푦) ⊓ 푧) [by assumption, with 푝 = (푥 ⊔ 푦), 푞 = 푥, 푟 = 푧] = 푥 ⊔ ((푥 ⊔ 푦) ⊓ 푧) [by part a)] = 푥 ⊔ ((푥 ⊓ 푧) ⊔ (푦 ⊓ 푧)) [by assumption, with 푝 = 푧, 푞 = 푥, 푟 = 푦] = (푥 ⊔ (푥 ⊓ 푧)) ⊔ (푦 ⊓ 푧) [by associativity of ⊔] = 푥 ⊔ (푦 ⊓ 푧). [by part a)] The other direction is similar. 1.15 There are many examples. For instance, let 푆 be any non-trivial monoid, let 푇 = 푆0, and define 휑 ∶ 푆 → 푇 by 푥휑 = 0 for all 푥 ∈ 푆. It is easy to see that 휑 is a homomorphism, but 1푆휑 = 0 ≠ 1푆0 . 1.16 a) Let 휑 be a monomorphism (that is, an injective homomorphism), and let 휓1, 휓2 ∶ 푈 → 푆 be such that 휓1 ∘ 휑 = 휓2 ∘ 휑. Let 푥 ∈ 푈. Then 푥휓1휑 = 푥휓2휑 and so 푥휓1 = 푥휓2 since 휑 is injective. Since this is true for all 푥 ∈ 푈, it follows that 휓1 = 휓2. This proves that 휑 is a categorical monomorphism. Now let 휑 be a categorical monomorphism. Suppose, with the aim of obtaining a contradiction, that 휑 is not injective. Then there exist 푥, 푦 ∈ 푆 with 푥 ≠ 푦 such that 푥휑 = 푦휑. Define maps 푛 푛 휓1, 휓2 ∶ ℕ → 푆 by 푛휓1 = 푥 and 푛휓2 = 푦 . It is easy to see that 휓1 and 휓2 are homomorphisms. Then for any 푛 ∈ ℕ, 푛 푛 푛 푛 푛휓1휑 = 푥 휑 = (푥휑) = (푦휑) = 푦 휑 = 푛휓2휑,

and so 휓1 ∘ 휑 = 휓2 ∘ 휑. Hence 휓1 = 휓2 by (1.16), which contradicts 1휓1 = 푥 ≠ 푦 = 1휓2 and so proves that 휑 is a monomorphism.

Solutions to exercises • 202 b) i) Let 휑 be a surjective homomorphism. Suppose 휓1, 휓2 ∶ 푇 → 푈 are such that 휑 ∘ 휓1 = 휑 ∘ 휓2. Let 푥 ∈ 푇. Then since 휑 is surjective, there exists 푦 ∈ 푆 with 푦휑 = 푥. Thus

푥휓1 = 푦휑휓1 = 푦휑휓2 = 푥휓2.

Since this holds for all 푥 ∈ 푇, it follows that 휓1 = 휓2. This proves that 휑 is a categorical epimorphism.

ii) Let 휓1, 휓2 ∶ ℤ → 푈 be such that 휓1 ≠ 휓2 (which is the negation of the right-hand side of (1.17)). Then there exists 푛 ∈ ℤ such that 푛휓1 ≠ 푛휓2. Either 푛 or −푛 lies in im 휄, and so either 푛휄휓1 ≠ 푛휄휓2 or (−푛)휄휓1 ≠ (−푛)휄휓2, and thus 휄∘휓1 ≠ 휄∘휓2 (which is the negation of the left-hand side of (1.17) with 휑 = 휄). Thus 휄 is a categorical epimorphism.

1.17 Suppose 푆 is a right zero semigroup. Let 푥, 푦 ∈ 푆. Then 휌푥 = 휌푦 ⇒ 푧휌푥 = 푧휌푦 ⇒ 푧푥 = 푧푦 ⇒ 푥 = 푦 and so the map 푥 ↦ 휌푥 is injective. Suppose now that 푆 is a left zero semigroup. Let 푥, 푦 ∈ 푆 with 푥 ≠ 푦. Then 푧푥 = 푧푦 for all 푧 ∈ 푆. Hence 푧휌푥 = 푧휌푦 for all 푧 ∈ 푆, and so 휌푥 = 휌푦. Thus 푥 ↦ 휌푥 is not injective. 1.18 For each 푦 ∈ 푌, let 푇푦 be a copy of 푇, and define a map 휑푦 ∶ 푌 → 푇푦 푒 if 푥 ⩾ 푦, 푥휑 = { 푦 푧 otherwise. Let 푥, 푥′, 푦 ∈ 푌. Then

(푥휑푦) ⊓ (푥′휑푦) = 푒 ⊓ 푒 = 푒 = (푥 ⊓ 푥′)휑푦 if 푥, 푥′ ⩾ 푦;

(푥휑푦) ⊓ (푥′휑푦) = 푒 ⊓ 푧 = 푧 = (푥 ⊓ 푥′)휑푦 if 푥 ⩾ 푦, 푥′ ≱ 푦;

(푥휑푦) ⊓ (푥′휑푦) = 푧 ⊓ 푒 = 푧 = (푥 ⊓ 푥′)휑푦 if 푥 ≱ 푦, 푥′ ⩾ 푦;

(푥휑푦) ⊓ (푥′휑푦) = 푧 ⊓ 푧 = 푧 = (푥 ⊓ 푥′)휑푦 if 푥, 푥′ ≱ 푦.

So 휑푦 is a homomorphism. It is clearly surjective. Now,

(∀푦 ∈ 푌)(푥휑푦 = 푥′휑푦)

⇒ (푥휑푥 = 푥′휑푥) ∧ (푥휑푥′ = 푥′휑푥′)

⇒ (푒 = 푥′휑푥) ∧ (푥휑푥′ = 푒) ⇒ (푥′ ⩾ 푥) ∧ (푥 ⩾ 푥′) ⇒ 푥 = 푥′.

So the collection of surjective homomorphisms { 휑푦 ∶ 푌 → 푇푦 ∶ 푦 ∈ 푌 } separates elements of 푌, and so 푌 is a subdirect product of { 푇푦 ∶ 푦 ∈ 푌 }. 1.19 Define a homomorphism 휑 ∶ 푆/퐼 → 푆/퐽 by [푥]퐼휑 = [푥]퐽. Since 퐼 ⊆ 퐽, the homomorphism 휑 is well defined. Its image is clearly 푆/퐽. Now, ([푥]퐼, [푦]퐼) ∈ ker 휑 ⇔ [푥]퐽 = [푦]퐽 ⇔ 푥, 푦 ∈ 퐽 ⇔ [푥]퐼, [푦]퐼 ∈ 퐽/퐼. Hence, by Theorem 1.24, 푆/퐽 ≃ (푆/퐼)/ker 휑 ≃ (푆/퐼)/(퐽/퐼).

Solutions to exercises • 203 1.20 Notice that 퐼퐽 ⊆ 퐼푆∩푆퐽 ⊆ 퐼∩퐽, so 퐼∩퐽 ≠ ∅. Furthermore, 푆(퐼∩퐽)푆 ⊆ 푆퐼푆∩푆퐽푆 ⊆ 퐼∩퐽, since 퐼 and 퐽 are ideals; thus 퐼∩퐽 is an ideal. Similarly, 푆(퐼 ∪ 퐽)푆 ⊆ 푆퐼푆 ∪ 푆퐽푆 ⊆ 퐼 ∪ 퐽 and so 퐼 ∪ 퐽 is an ideal. Define a homomorphism 휑 ∶ 퐼 → (퐼 ∪ 퐽)/퐽 by 푥휑 = [푥]퐽. Let [푦]퐽 ∈ (퐼 ∪ 퐽)/퐽. If 푦 ∈ 퐽 then let 푧 ∈ 퐼 ∩ 퐽 and notice that 푧휑 = [푧]퐽 = [푦]퐽; if 푦 ∉ 퐽 then 푦 ∈ 퐼 and 푦휑 = [푦]퐽. Hence im 휑 is (퐼 ∪ 퐽)/퐽. Now for any 푥, 푦 ∈ 퐼, we have (푥, 푦) ∈ ker 휑 ⇔ [푥]퐽 = [푦]퐽 ⇔ 푥, 푦 ∈ 퐽. Hence (퐼 ∪ 퐽)/퐽 ≃ 퐼/(퐼 ∩ 퐽) by Theorem 1.24.

1.21 Suppose first that 푇 = 퐺 ∪ {0푆} and let 푡 ∈ 푇 ∖ {0푆} = 퐺. Then 푡퐺 = 퐺푡 = 퐺 by Lemma 1.9 and 푡0푆 = 0푆푡 = 0푆, so 푡푇 = 푇푡 = 푇. Conversely, suppose that 푡푇 = 푇푡 = 푇 for all 푡 ∈ 푇 ∖ {0푆}. Let 퐺 = 푇 ∖ {0푆}. By assumption, 푇 contains at least one element other than 0푆, so 퐺 ≠ ∅. For any 푠, 푡 ∈ 푇, we have 푠, 푡 ∈ 푠푇 = 푇, so 푇 is a subsemigroup. Suppose, with the aim of obtaining a contradiction, that there exist 푔, ℎ ∈ 퐺 with 푔ℎ = 0푆. Then

푇 = 푔푇 ⊆ 푇푇 = (푇푔)(ℎ푇) = 푇(푔ℎ)푇 = 푇0푆푇 = {0푆}, contradicting 퐺 ≠ ∅. So for all 푔, ℎ ∈ 퐺, we have 푔ℎ ∈ 퐺. Since 푔0푆 = 0푆푔 = 0푆, it follows that 푔퐺 = 퐺푔 = 퐺 for all 푔 ∈ 퐺. Hence, by Lemma 1.9, 퐺 is a subgroup of 푆.

Exercises for chapter 2 [See pages 50–52 for the exercises.] 2.1 a) Let 퐺 be a group and suppose that 푥, 푦, 푧, 푡 ∈ 퐺 are such that 푥푦 = 푧푡. Then we can take 푝 = 푧−1푥 = 푡푦−1, and it follows that 푥 = 푧푝 and 푡 = 푝푦. b) Suppose 푥, 푦, 푧, 푡 ∈ 퐴∗ are such that 푥푦 = 푧푡. Let 푥푦 = 푧푡 = 푎1 ⋯ 푎푛, where 푎푖 ∈ 퐴. Then, by the definition of multiplication in 퐴∗, we have

푥 = 푎1 ⋯ 푎푘, 푦 = 푎푘+1 ⋯ 푎푛, 푧 = 푎1 ⋯ 푎ℓ, 푡 = 푎ℓ+1 ⋯ 푎푛, for some 0 ⩽ 푘, ℓ ⩽ 푛 + 1. (We allow 푘 and ℓ to take the values 0 and 푛 + 1 and formally take subwords 푎푖 ⋯ 푎푗 where 푗 < 푖 to mean the empty word 휀.) If 푘 ⩽ ℓ, then the situation is as follows: 푥 푦 푥푦 = 푧푡 = ⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞푎1 ⋯ 푎푘 ⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞푎푘+1 ⋯ 푎ℓ ⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟푎ℓ+1 ⋯ 푎푛 푧 푡

and thus we let 푞 = 푎푘+1 ⋯ 푎ℓ; then 푧 = 푥푞 and 푦 = 푞푡. On the other hand, if 푘 ⩾ ℓ, let 푝 = 푎ℓ+1 ⋯ 푎푘; then 푥 = 푧푝 and 푡 = 푝푦.

Solutions to exercises • 204 2.2 If 푢 = 푤푖 and 푣 = 푤푗, then 푢푣 = 푤푖+푗 = 푣푢. In the other direction, suppose that 푢푣 = 푣푢. First note that if 푢 = 휀, then we can take 푤 = 푣, so that 푢 = 푤0 and 푣 = 푤1; similar reasoning holds when 푣 = 휀. So assume henceforth that neither 푢 nor 푣 is the empty word. Now proceed by induction on |푢푣|. If |푢푣| ⩽ 2 and 푢푣 = 푣푢, then since neither 푢 nor 푣 is 휀, it follows that 푢 and 푣 both have length 1, so 푢 = 푣. So assume the result holds for |푢푣| < 푘 and suppose 푢푣 = 푣푢. By Exercise 2.1, there exists 푝 ∈ 퐴∗ such that 푢 = 푣푝 and 푢 = 푝푣 (or there exists 푞 ∈ 퐴∗ such that 푣 = 푢푞 and 푣 = 푞푢, which says the same thing). If 푝 = 휀 then 푢 = 푣. Otherwise, since 푣푝 = 푝푣, the induction hypothesis shows that 푣 = 푤푗 and 푝 = 푤푖 for some 푤 ∈ 퐴∗ and 푖, 푗 ∈ ℕ. Thus 푢 = 푤푖+푗. Hence, by induction, the result holds for all 푢, 푣 ∈ 퐴∗. 2.3 a) Suppose 푢푣 = 푣푤. If |푣| = 0, then let 푠 = 휀, 푡 = 푢, 푘 = 0. Since 푣 = 휀 and 푢 = 푤, we have 푢 = 푠푡, 푣 = (푠푡)푘푠, 푤 = 푡푠. So suppose the result holds for |푣| < 푘. Then if |푣| = 푘, by equidivisibility we have either 푢 = 푣푝 and 푝푣 = 푤 for some 푝 ∈ 퐴∗ or 푢푞 = 푣 and 푣 = 푞푤 for some 푞 ∈ 퐴∗. In the former case, let 푠 = 푣, 푡 = 푝, and 푘 = 0; then 푢 = 푠푡, 푣 = (푠푡)푘푠, and 푤 = 푡푠. In the latter case, first note that if |푞| = 0 we have 푢푞 = 푞푤, with |푞| < |푣|. By the induction hypothesis, 푢 = 푠푡, 푡 = (푠푡)푘푠, and 푤 = 푡푠 for some 푠, 푡 ∈ 퐴∗ and 푘 ∈ ℕ ∪ {0}. Then 푣 = 푢푞 = (푠푡)푘+1푠. This proves the induction step. b) Let 푘 be maximal such that 푣 = 푢푘푠 for some 푠 ∈ 퐴∗. Then 푢푘+1푠 = 푢푣 = 푣푤 = 푢푘푠푤 and so by cancellativity 푢푠 = 푠푤. So by equidivisibility, either 푠 is a left factor of 푢 or 푢 is left factor of 푠. But the latter contradicts the maximality of 푘. Hence 푢 = 푠푡 for some 푡 ∈ 퐴∗. Hence 푣 = (푠푡)푘푠 and so (푠푡)푘+1푠 = 푢푣 = 푣푤 = (푠푡)푘푠푤 and so by cancellativity 푤 = 푡푠. 2.4 First, notice that if ⟨푢, 푣⟩ is free, then every element of ⟨푢, 푣⟩ has a unique representation as a a product of elements of {푢, 푣}; hence 푢푣 ≠ 푣푢. So suppose ⟨푢, 푣⟩ is not free. Without loss of generality, assume |푢| ⩾ |푣| and let 푢 = 푣푘푧, where 푘 ∈ ℕ ∪ {0} is maximal and 푧 ∈ 퐴∗. Then there are two distinct products 푥1 ⋯ 푥푚 and 푦1 ⋯ 푦푛 (where 푥푖, 푦푖 ∈ {푢, 푣}) such that 푥1 ⋯ 푥푚 = 푦1 ⋯ 푦푛. By cancellativity, assume 푥1 ≠ 푦1. Interchanging the two products if necessary, assume 푥1 = 푢 and 푦1 = 푣. Let ℓ ∈ ℕ be maximal such that 푦1 = 푦2 = … = 푦ℓ = 푘 ℓ 푣. Then 푣 푧푥2 ⋯ 푥푚 = 푣 푦ℓ+1 ⋯ 푦푛. By cancellativity, 푧푥2 ⋯ 푥푚 = ℓ−푘 푣 푦ℓ+1 ⋯ 푦푛. By equidivisibility, either 푧 = 푣푝 and 푝푥2 ⋯ 푥푚 = ℓ−푘 ∗ ℓ−푘 푣 푦ℓ+1 ⋯ 푦푛 for some 푝 ∈ 퐴 , or 푣 = 푧푞 and 푞푣 푦ℓ+1 ⋯ 푦푛. The former case is impossible since 푘 is maximal; thus the latter case holds. 푘 So 푢 = (푧푞) 푧. Repeat this reasoning but focusing on 푢푚 and 푣푛 shows that 푣 is a right factor of 푢. But since 푢 = (푧푞)푘푧 and |푣| =

Solutions to exercises • 205 |푧| + |푞| = |푞푧|, we conclude that 푣 = 푞푧. Hence 푧푞 = 푣 = 푞푧, and so 푢푣 = (푞푧)푘푧푞푧 = (푧푞)푘푧푞푧 = 푧푞(푧푞)푘푧 = 푧푞(푞푧)푘푧 = 푣푢.

2.5 Suppose that 푝1 … 푝푘 = 푞1 … 푞ℓ, where 푝푖, 푞푖 ∈ 푋. Suppose, with the aim of obtaining a contradiction, that 푘 ≠ ℓ. Without loss of generality, assume 푘 < ℓ. Let 푟 ∈ 푋 ∖ {푞푘+1}; such an element 푟 exists since |푋| ⩾ 2. Now, 푝1 ⋯ 푝푘푟푞1 ⋯ 푞ℓ = 푞1 ⋯ 푞ℓ푟푝1 ⋯ 푝푘. Both products have length 푘 + ℓ + 1 and so their corresponding terms are equal by the supposition. In particular, 푟 = 푞푘+1, which contradicts the choice of 푟. Hence 푘 = ℓ, and so by the supposition 푝푖 = 푞푖 for all 푖. Since 푆 is generated by 푋, this proves that 푆 is free with basis 푋.

2.6 a) Define 휑 ∶ 퐴 → 푀 by 푎푖 ↦ {푥푖}. Since (푎푖푎푗)휑 = {푥푖} ∪ {푥푗} = 2 {푥푖, 푥푗} = {푥푗}∪{푥푖} = (푎푗푎푖)휑 and (푎푖 )휑 = {푥푖}∪{푥푖} = {푥푖} = 푎푖휑, the monoid 푀 satisfies the defining relations in 휌 with respect to 휑. b) Let 푤 ∈ 퐴∗. We can find a sequence of elementary transition from 푒1 푒2 푒푛 푤 to a word 푎1 푎2 ⋯ 푎푛 ∈ 푁, where each 푒푖 ⩽ 1 as follows. First we use the defining relations (푎푖푎푗, 푎푗푎푖) to find a sequence from 푒1 푒2 푒푛 푤 to a word 푎1 푎2 ⋯ 푎푛 , where each 푒푖 ∈ ℕ ∪ {0}. Then we use 2 the defining relations (푎푖 , 푎푖) to find a sequence from this word to one where each 푒푖 ⩽ 1. 푒1 푒2 푒푛 푐1 푐2 푐푛 c) Let 푎1 푎2 ⋯ 푎푛 , 푎1 푎2 ⋯ 푎푛 ∈ 푁 so that 푒푖, 푐푖 ⩽ 1. Then:

푒1 푒2 푒푛 ∗ 푐1 푐2 푐푛 ∗ (푎1 푎2 ⋯ 푎푛 )휑 = (푎1 푎2 ⋯ 푎푛 )휑

⇒ { 푥푖 ∶ 푒푖 = 1 } = { 푥푖 ∶ 푐푖 = 1 }

⇒ (∀푖)(푒푖 = 푐푖) 푒1 푒2 푒푛 푐1 푐2 푐푛 ⇒ 푎1 푎2 ⋯ 푎푛 = 푎1 푎2 ⋯ 푎푛 .

∗ Hence 휑 |푁 is injective. 2.7 We apply Method 2.9. For brevity, let 퐴 = {푎, 푏} and 휌 = {(푎푏푎, 휀)}. Let 휑 ∶ 퐴 → ℤ be defined by 푎휑 = 1 and 푏휑 = −2. Then ℤ satisfies the defining relation in 휌 since (푎푏푎)휑∗ = 1 − 2 + 1 = 0 = 휀휑∗. [Recall that 0 is the identity of ℤ under addition.] Let

푁 = { 푎푖 ∶ 푖 ∈ ℕ ∪ {0} } ∪ { 푏푖 ∶ 푖 ∈ ℕ } ∪ { 푎푏푖 ∶ 푖 ∈ ℕ }.

Now, there are sequences of elementary transitions

푎푏 ↔휌 푎푏푎푏푎 ↔휌 푏푎

and

푎푎푏 ↔휌 푎푎푏푎푏푎 ↔휌 푎푏푎 ↔휌 휀.

Thus we can first of all transform any word in 퐴+ by applying defining relations to replace subwords 푏푎 by 푎푏, which ultimately yields a word

Solutions to exercises • 206 of the form 푎푖푏푗. Then we can replace subwords 푎푎푏 by 휀, which must ultimately yield a word consisting either entirely of symbols 푎, entirely of symbols 푏, or by a single symbol 푎 followed by symbols 푏; that is, a word in 푁. Finally note that

푎푖휑∗ = 푖 for 푖 ∈ ℕ ∪ {0}, 푏푖휑∗ = −2푖 for 푖 ∈ ℕ, (푎푏푖)휑∗ = −2푖 + 1 for 푖 ∈ ℕ.

∗ It is now easy to see that 휑 |푁 is injective. Hence Mon⟨퐴 | 휌⟩ defines (ℤ, +) 2.8 Deleting a subword 푎푏푐 is an elementary 휌-transition, and so does not alter the element represented. Thus given any word 푤 ∈ 퐴∗, one can obtain a word ̂푤∈ 푁 with 푤 =푀 ̂푤 by deleting subwords 푎푏푐. Thus every element of 푀 has at least one representative in 푁; it remains to prove uniqueness. So suppose some element of 푀 has two representatives 푢, 푣 ∈ 푁 with 푢 ≠ 푣. Since 푢 =푀 푣, there is a sequence of elementary 휌- transitions

푢 = 푤0 ↔휌 푤1 ↔휌 … ↔휌 푤푛 = 푣.

Consider the collection of such sequences with the maximum length of an intermediate word 푤푖 being minimal, and choose and fix such a sequence where the fewest words 푤푖 have this maximum length. Note that 푛 > 0 since 푢 ≠ 푣. Consider some intermediate word 푤푖 of this maximum length. Note that 푖 ≠ 0 and 푖 ≠ 푛, since the words 푤0 and 푤푛 do not contain subwords 푎푏푐, and so the words 푤1 and 푤푛 must be obtained by inserting subwords 푎푏푐 into 푤0 and 푤푛 respectively, and so |푤1| > |푤0| and |푤푛−1| > |푤푛|. So there are words 푤푖−1 and 푤푖+1, and these are obtained from 푤 by applying the defining relation (푎푏푐, 휀). Because 푤푖 has maximum length among the intermediate words, 푤푖−1 and 푤푖+1 must both be obtained by deleting subwords 푎푏푐 from 푤푖. Now, they cannot be obtained by deleting the same subword 푎푏푐, for then

푢 = 푤0 ↔휌 푤1 ↔휌 … ↔휌 푤푖−1 = 푤푖+1 ↔휌 … ↔휌 푤푛 = 푣.

would be a sequence of elementary 휌-transitions from 푢 to 푣 where the number of intermediate words of maximum length is smaller, or (if no other intermediate word had length |푤푖|) a smaller maximum length of intermediate words; in either case, this is a contradiction. Hence 푤푖−1 and 푤푖+1 are obtained by deleting different subwords 푎푏푐 from 푤푖. ∗ Thus 푤푖 = 푝푎푏푐푞푎푏푐푟 for some 푝, 푞, 푟 ∈ 퐴 , and either 푤푖−1 = 푝푞푎푏푐푟 and 푤푖+1 = 푝푎푏푐푞푟, or 푤푖−1 = 푝푎푏푐푞푟 and 푤푖+1 = 푝푞푎푏푐푟. Assume the

Solutions to exercises • 207 former case; the latter is similar. Then there is a sequence of elementary 휌-transitions

푢 = 푤0 ↔휌 푤1 ↔휌 … ↔휌 푤푖−1 = 푝푞푎푏푐푟

↔휌 푝푞푟 ↔ 푝푎푏푐푞푟 = 푤푖+1 ↔휌 … ↔휌 푤푛 = 푣.

Since |푝푞푟| < |푤푖|, this is a sequence where the number of intermediate words of maximum length is smaller, or (if no other intermediate word had length |푤푖|) a smaller maximum length of intermediate words; in either case, this is a contradiction. Hence every element of 푀 has a unique representative in 푁.

2.9 To define an assignment of generators 휑 ∶ 퐴 → 퐵2, proceed as 0 0 follows. As noted in the question, 푧휑 must be the zero matrix [ ]. 0 0 Furthermore, (푎휑)2 and (푏휑)2 must be the zero matrix. Calculating the squares of the available matrices shows that 푎휑 and 푏휑 must be 0 1 0 0 in {[ ],[ ]}. Since 푎 and 푏 can be swapped in 휎 ∪ 휁 to give 0 0 1 0 the same set of defining relations, it does not matter which matrix we assign to each of 푎휑 and 푏휑. So define 0 1 0 0 0 0 푎휑 = [ ] , 푏휑 = [ ] , 푧휑 = [ ]. 0 0 1 0 0 0

Straightforward calculations show that 퐵2 satisfies all the defining relations in 휎 ∪ 휁 with respect to 휑. Let 푁 = {푧, 푎, 푏, 푎푏, 푏푎}. Let 푤 ∈ 퐴+. If 푤 contains a symbol 푧, then applying defining relations from 휁 shows that (푤, 푧) is a consequence of 휎 ∪ 휁. If 푤 contains 푎2 or 푏2, then applying a single relation (푎2, 푧) or (푏2, 푧) introduces a symbol 푧, and so by the previous sentence (푤, 푧) is a consequence of 휎 ∪ 휁. Finally, if 푤 contains no 푎2 or 푏2 or 푧, then it consists of alternating symbols 푎 and 푏, and so applying relations (푎푏푎, 푎) or (푏푎푏, 푏) transforms it to a word 푢 ∈ {푎, 푏, 푎푏, 푏푎}, and (푤, 푢) is a consequence of 휎 ∪ 휁. + Lastly, 휑 |푁 is injective since the five words in 푁 = {푧, 푎, 푏, 푎푏, 푏푎} correspond to the five matrices in 퐵2 (in the order listed in the question). 2.10 a) Suppose 푐훾푏훽 is idempotent. If 훾 > 훽, then

훾 훽 2 훾 훽 훾 훽 훾+훾−훽 훽 훾 훽 (푐 푏 ) =퐵 푐 푏 푐 푏 =퐵 푐 푏 ≠퐵 푐 푏 . If 훾 < 훽, then

훾 훽 2 훾 훽 훾 훽 훾 훽+훽−훾 훾 훽 (푐 푏 ) =퐵 푐 푏 푐 푏 =퐵 푐 푏 ≠퐵 푐 푏 . 훾 훾 2 훾 훾 훾 훾 훾 훾 Hence 훾 = 훽. On the other hand, (푐 푏 ) = 푐 푏 푐 푏 =퐵 푐 푏 and so 푐훾푏훾 is idempotent.

Solutions to exercises • 208 푏 푏 푏 푏 푐4 푐4푏 푐4푏2 푐4푏3 푐4푏4 푐 푐 푐 푐 푐 푏 푏 푏 푏 푐3 푐3푏 푐3푏2 푐3푏3 푐3푏4 푐 푐 푐 푐 푐 푏 푏 푏 푏 푐2 푐2푏 푐2푏2 푐2푏3 푐2푏4 푐 푐 푐 푐 푐 푏 푏 푏 푏 푐 푐푏 푐푏2 푐푏3 푐푏4 푐 푐 푐 푐 푐 푏 푏 푏 푏 FIGURE S.4 휀 푏 푏2 푏3 푏4 Part of the Cayley graph of the 푐 푐 푐 푐 bicyclic monoid.

b) Suppose first that 푐 is right-invertible. Then there exists 푐휁푏휂 such 휁 휂 1+휁 휂 that 푐푐 푏 =퐵 휀. But this is impossible, since 푐 푏 ≠퐵 휀 since 1 + 휁 > 0. Now suppose that 푐훾푏훽, where 훾 ⩾ 1, has a right inverse 훾−1 훽 푥. Then 푐푐 푏 푥 =퐵 휀 and so 푐 is right-invertible, which is a contradiction. Hence if 푐훾푏훽 is right-invertible, then 훾 = 0. On 훽 훽 훽 훽 the other hand, 푏 푐 =퐵 휀 and so 푐 is a right inverse for 푏 . 2.11 The Cayley graph 훤(퐵, {푏, 푐}) is shown in Figure S.4. 2.12 a) Suppose, with the aim of obtaining a contradiction, that 푥푘 = 푥푘+푚 for some 푘, 푚 ∈ ℕ. Then 푒 = 푥푘푦푘 = 푥푘+푚푦푘 = 푥푚. Then 푦 = 푒푦 = 푥푚푦 = 푥푚−1푒 = 푥푚−1 and so 푦푥 = 푥푚 = 푒, which is a contradiction. So 푥 is not periodic. Similarly 푦 is not periodic. b) Suppose 푥푘 = 푦ℓ. Then 푥푘+ℓ+1 = 푥ℓ+1푦ℓ = 푥. Since 푥 is not periodic, this forces 푘 = ℓ = 0. c) Suppose 푦푘푥ℓ = 푒. Suppose, with the aim of obtaining a contradiction, that ℓ > 0. Then 푦푥 = 푒푦푥 = 푦푘푥ℓ푦푥 = 푦푘푥ℓ−1푥 = 푦푘푥ℓ = 푒, which is a contradiction. Thus ℓ = 0, and so 푦푘+1 = 푒푦 = 푦 and so 푘 = 0 since 푦 is not periodic. d) Suppose, with the aim of obtaining a contradiction, that 푦푘푥ℓ = 푦푚푥푛 with either 푘 ≠ 푚 or ℓ ≠ 푛. Assume 푘 ≠ 푚; the other case is similar. Interchanging the two products if necessary, assume that 푘 < 푚. Then 푥ℓ = 푒푥ℓ = 푥푘푦푘푥ℓ = 푥푘푦푚푥푛 = 푒푦푚−푘푥푛 = 푦푚−푘푥푛. If ℓ ⩾ 푛, then 푦푚−푘 = 푦푚−푘푥푛푦푛 = 푥ℓ푦푛 = 푥ℓ−푛, which contradicts part b). If ℓ ⩽ 푛, then 푒 = 푥ℓ푦ℓ = 푦푚−푘푥푛푦ℓ = 푦푚−푘푥푛−ℓ, which contradicts part c). e) Define 휑 ∶ 퐵 → ⟨푥, 푦⟩ by 푏휑 = 푥 and 푦휑. The given properties of 푒, 푥, and 푦 show that 휑 is a well-defined homomorphism; it is clearly surjective; part d) shows that it is injective.

Solutions to exercises • 209 2.13 Let 푒 = 휀휑, 푥 = 푏휑, and 푦 = 푐휑. Since 휑 is a homomorphism, 푒, 푥, and 푦 satisfy the conditions 푒푥 = 푥푒 = 푥, 푒푦 = 푦푒 = 푦, and 푥푦 = 푒. Note further that 푆 = ⟨푥, 푦⟩ since 휑 is surjective. If the condition 푦푥 ≠ 푒 is also satisfied, then by Exercise 2.12, 푆 is isomorphic to the bicyclic monoid. On the other hand, if 푦푥 = 푒, then every element of 푆 is left- and right-invertible and so 푆 is a group.

Exercises for chapter 3 [See pages 66–68 for the exercises.] 3.1 Let 퐺 be a subgroup of a semigroup. Let 푥, 푦 ∈ 퐺. Let 푝 = 푥−1푦 and 푞 = 푦−1푥. Then 푥푝 = 푦 and 푦푞 = 푥. So 푥 R 푦. Similarly 푥 L 푦. Hence 푥 H 푦. 3.2 Suppose 푢, 푣 ∈ 퐴∗ are such that 푢 R 푣. Then there exist 푝, 푞 ∈ 퐴∗ such that 푢푝 = 푣 and 푣푞 = 푝. Then 푢푝푞 = 푝, so |푢| + |푝| + |푞| = |푢|, and so |푝| = |푞| = 0. Thus 푝 = 푞 = 휀 and so 푢 = 푣. That is, R is the identity relation id퐴∗ . Similarly, the Green’s relations L, and J are the identity relation. Hence H = R ⊓ L and D = R ⊔ L are the identity relation.

3.3 a) Suppose 휎 L 휏. Then there exist 휋, 휌 ∈ T푋 such that 휋휎 = 휏 and 휌휏 = 휎. Therefore

im 휎 = 푋휎 ⊇ (푋휋)휎 = im(휋휎) = im 휏,

and similarly im 휏 ⊇ im(휌휏) = im 휎. Hence im 휎 = im 휏. Now suppose im 휎 = im 휏. For each 푥 ∈ 푋, we have 푥휏 ∈ im 휏 = im 휎 and so we can define 푥휋 to be some element of 푋 such that (푥휋)휎 = 푥휏. Then 휋휎 = 휏. Similarly we can define 휌 ∈ T푋 so that 휌휏 = 휎. Hence 휎 L 휏. b) Suppose 휎 R 휏. Then there exist 휋, 휌 ∈ T푋 such that 휎휋 = 휏 and 휏휌 = 휎. Therefore (푥, 푦) ∈ ker 휎 ⇒ 푥휎 = 푦휎 ⇒ 푥휎휋 = 푦휎휋 ⇒ 푥휏 = 푦휏 ⇒ (푥, 푦) ∈ ker 휏. Thus ker 휎 ⊆ ker 휏. Similarly, ker 휏 ⊆ ker 휎. Hence ker 휎 = ker 휏. Now suppose ker 휎 = ker 휏. We aim to define 휋 ∈ T푋 such that 휎휋 = 휏. For each 푥 ∈ im 휎, choose 푦푥 ∈ 푋 such that 푦푥휎 = 푥. Note that each 푧 ∈ 푋 is related by ker 휎 to 푦푧휎 and to no other 푦푥. Since ker 휎 = ker 휏, we have (푧, 푦푧휎) ∈ ker 휏 and so 푧휏 = 푦푧휎휏. For each 푥 ∈ im 휎, define 푥휋 = 푦푥휏. For 푥 ∉ im 휎, let 푥휋 be arbitrary. Then for all 푧 ∈ 푋, we have 푧휎 ∈ im 휎 and so 푧휎휋 = 푦푧휎휏 = 푧휏; hence 휎휋 = 휏. Similarly, we can define 휌 ∈ T푋 so that 휏휌 = 휎. Hence 휎 R 휏.

c) Suppose 휎 D 휏. Then there exists 휐 ∈ T푋 such that 휎 L 휐 R 휏. Since 휐 R 휏, there exist 휋, 휌 ∈ T푋 such that 휐휋 = 휏 and 휏휌 = 휐.

Solutions to exercises • 210 Hence 휏휌휋 = 휏 and 휐휋휌 = 휐. Therefore 휌|im 휏 ∶ im 휏 → im 휐 and 휋|im 휐 ∶ im 휐 → im 휏 are mutually inverse bijections. So |im 휐| = |im 휏|. Since 휎 L 휐, we have im 휎 = im 휐 and thus |im 휎| = |im 휐| = |im 휏|. Now suppose |im 휎| = |im 휏|. Then there is a bijection 휇 ∶ im 휎 → im 휏. Extend 휇 to a map 휋 ∈ T푋 by defining 푥휋 arbitrarily −1 for 푥 ∈ 푋∖im 휎. Similarly extend 휇 to a map 휌 ∈ T푋. Let 휐 = 휎휋. Then 휐휌 = 휎, so 휐 R 휎. Furthermore im 휐 = im(휎휋) = im(휎휇) = im 휏, so 휐 L 휏 by part a). Hence 휎 D 휏. Suppose 휎 J 휏. Then there exist 휋, 휌, 휋′, 휌′ ∈ T푋 such that 휎 = 휋휏휌 and 휏 = 휋′휎휌′. Therefore |im 휎| = |푋휎| = |푋휋휏휌| ⩽ |푋휏휌| ⩽ |푋휏| = |im 휏|; similarly |im 휏| ⩽ |im 휎|. So |im 휎| = |im 휏|. Hence 휎 D 휏. There- fore J ⊆ D. Since D ⊆ J in general, it follows that D = J.

3.4 Consider the following elements of T{1,2,3}: 1 2 3 1 2 3 1 2 3 휌 = ( ) ; 휎 = ( ) ; 휏 = ( ). 1 2 2 1 3 3 1 1 3 Notice that 1 2 3 1 2 3 1 2 3 휌휎 = ( ) ; 휌휏 = ( ) ; 휎휏 = ( ). 1 3 3 1 1 1 1 3 3 Thus im 휎 = im 휏 = {1, 3}, but im 휌휎 = {1, 3} ≠ {1} = im 휌휏 and so (휎, 휏) ∈ L but (휌휎, 휌휏) ∉ L by Exercise 3.3(a). So L is not a left congruence in T{1,2,3}. Similarly, ker 휌 = ker 휎 but ker 휌휏 ≠ ker 휎휏 and so (휌, 휎) ∈ R but (휌휏, 휎휏) ∉ R by Exercise 3.3(b). So R is not a right congruence in T{1,2,3}. 3.5 Let (ℓ1, 푟1), (ℓ2, 푟2) ∈ 퐵. Then

(ℓ1, 푟1) R (ℓ2, 푟2) ⇒ (∃(푘, 푠) ∈ 퐵)((ℓ1, 푟1)(푘, 푠) = (ℓ2, 푟2))

⇒ (∃(푘, 푠) ∈ 퐵)((ℓ1, 푠) = (ℓ2, 푟2))

⇒ ℓ1 = ℓ2.

On the other hand, if (ℓ, 푟1), (ℓ, 푟2) ∈ {ℓ}×푅, then (ℓ, 푟1)(ℓ, 푟2) = (ℓ, 푟2) and (ℓ, 푟2)(ℓ, 푟1) = (ℓ, 푟1) and so (ℓ, 푟1) R (ℓ, 푟2). So the R-classes of 퐵 are the sets {ℓ} × 푅. The result for L-classes is proved similarly. Therefore

(ℓ1, 푟1) H (ℓ2, 푟2) ⇔ ((ℓ1, 푟1) L (ℓ2, 푟2)) ∧ ((ℓ1, 푟1) L (ℓ2, 푟2))

⇔ (푟1 = 푟2) ∧ (ℓ1 = ℓ2), and so H is the identity relation. Finally, let (ℓ1, 푟1), (ℓ2, 푟2) ∈ 퐵. Then (ℓ1, 푟1) R (ℓ1, 푟2) L (ℓ2, 푟2) and so (ℓ1, 푟1) D (ℓ2, 푟2). Hence 퐵 has consists of a single D-class.

Solutions to exercises • 211 3.6 If 푥 R 푦, then there exist 푝, 푞 ∈ 푆1 with 푥푝 = 푦 and 푦푞 = 푥. So 푥푝푞 = 푥. Suppose that 푝푞 ∈ 푆. Then for any 푧 ∈ 푆, we have 푥푝푞푧 = 푥푧 and so 푝푞푧 = 푧 by cancellativity. So 푝푞 is a left identity for 푆 and in particular an idempotent. By Exercise 1.3, 푝푞 is an identity, which is a contradiction. So 푝푞 ∉ 푆 and thus 푝푞 = 1, the adjoined identity of 푆1. Hence 푝 = 푞 = 1 and so 푥 = 푦. Thus R = id푆. Similarly L = id푆, and so H = R ⊓ L = id푆 and D = R ⊔ L = id푆. 3.7 Let 푎, 푏, 푐, 푑, 푒, 푓 ∈ ℝ with 푎, 푏, 푐, 푑, 푒, 푓 > 0. Then

푎 푏 푐 푑 푎푐 푎푑 + 푏 [ ][ ] = [ ]; 0 1 0 1 0 1

since 푎푐 > 0 and 푎푑+푏 > 0, we see that 푆 is a subsemigroup of 푀2(ℝ). Furthermore,

푎 푏 det [ ] = 푎 > 0; 0 1

thus every matrix in 푆 is invertible; hence 푆 is a subsemigroup of the general linear group GL2(ℝ) and so cancellative. Furthermore,

푒 푓 푎 푏 푎 푏 [ ][ ] = [ ] 0 1 0 1 0 1 푒푎 푒푏 + 푓 푎 푏 ⇒ [ ] = [ ] 0 1 0 1 ⇒ 푒푎 = 푎 ∧ 푒푏 + 푓 = 푏 ⇒ 푒 = 1 ∧ 푒푏 + 푓 = 푏 ⇒ 푒 = 1 ∧ 푓 = 0,

which shows that 푆 does not contain a left identity; thus 푆 does not contain an identity. Finally, let 푔, ℎ ∈ ℝ with 푔, ℎ > 0. Choose 푓 = 1, 푑 = 0, 푐 = ℎ/(푎 + 푏), and 푒 = 푔/푐푎. Then 푐, 푑, 푒, 푓 > 0 and

푐 푑 푎 푏 푒 푓 [ ][ ][ ] 0 1 0 1 0 1 푐푎푒 푐푎푓 + 푐푏 + 푑 = [ ] 0 1 푐푎(푔/푐푎) (ℎ/(푎 + 푏))푎 + (ℎ/(푎 + 푏))푏 = [ ] 0 1 푔 ℎ = [ ]. 0 1

Thus for any 푥 ∈ 푆, we have 푆푥푆 = 푆 and so 푆 is simple. Hence J = 푆 × 푆.

Solutions to exercises • 212 2 2 3.8 Suppose 퐻휏 is a subgroup. Then 휏 ∈ 퐻휏. In particular, 휏 D 휏 and so |im 휏| = |im 휏2|. Now suppose that |im 휏| = |im(휏2)|. First, notice that im(휏2) = 푋휏2 ⊆ 푋휏 = im 휏. Since |im(휏2)| = |im 휏|, we have im 휏2 = im 휏 since im(휏2) and im 휏 are finite (because 푋 is finite). Also, (푥, 푦) ∈ ker 휏 ⇒ 푥휏 = 푦휏 ⇒ 푥휏2 = 푦휏2 ⇒ (푥, 푦) ∈ ker(휏2), and so ker 휏 ⊆ ker(휏2). So each ker(휏2)-class is a union of ker 휏-classes. Suppose, with the aim of obtaining a contradiction, that ker(휏2) − ker 휏 ≠ ∅. Then some ker(휏2)-class is a union of at least two distinct ker 휏-classes. So im(휏2) ⊊ im 휏. Hence, since im(휏2) and im 휏 are finite, |im(휏2)| < |im 휏|, which is a contradiction. So ker(휏2) = ker 휏. Hence 휏 L 휏2 2 2 and 휏 R 휏 and so 휏 H 휏 . Therefore 퐻휏 is a subgroup. 3.9 First, let 푥, 푦 ∈ { 푐훾푏훽 ∶ 훽 ∈ ℕ ∪ {0} }. Interchanging 푥 and 푦 if necessary, suppose 푥 = 푐훾푏훽 and 푦 = 푐훾푏훿 where 훽 ⩽ 훿. Then 푥푏훿−훽 = 푦 and 푦푐훿−훽 = 푥. Hence 푥 R 푦. Now suppose 푐훾푏훽 R 푐훾+휂푏훿 for some 휂 > 0. Then since R is a left 훽 훾 훾 훽 훾 훾+휂 훿 휂 훿 congruence, we have 푏 =퐵 푏 푐 푏 R 푏 푐 푏 =퐵 푐 푏 . Therefore 휂 훿 훽 휂 훿 훽 there exists 푝 ∈ 퐵 such that 푐 푏 푝 =퐵 푏 . Hence 푐 푏 푝푐 =퐵 휀 and so 푐휂 is right-invertible, which contradicts Exercise 2.10(b). Hence { 푐훾푏훽 ∶ 훽 ∈ ℕ ∪ {0} } is an R-class. Similarly, L-classes are of the form { 푐훾푏훽 ∶ 훾 ∈ ℕ ∪ {0} }. Finally, note that 푐훾푏훽 R 푐훾푏훿 L 푐휂푏훿 and so 푐훾푏훽 D 푐휂푏훿. Thus 퐵 consists of a single D-class. 3.10 Let 푒 ∈ 퐿 ∩ 푅 be idempotent. Then 푒 is a right identity for 퐿 and a left identity for 푅. For any 푦 ∈ 푅, we have 푒푦 = 푦 and so 휌푦|퐿 is a bijection from 퐿 to 퐿푦. Let 푧 ∈ 퐷. Choose 푦 ∈ 푅 ∩ 퐿푧. Since 휌푦|퐿 is a bijection, there exists 푥 ∈ 퐿 such that 푧 = 푥휌푦|퐿 = 푥푦 ∈ 퐿푅. Hence 퐷 ⊆ 퐿푅. Let 푥 ∈ 퐿 and 푦 ∈ 푅. Since 퐿 ∩ 푅 contains the idempotent 푒, we have 푥푦 ∈ 퐿푦 ∩ 푅푥 ⊆ 퐷 by Proposition 3.18. Hence 퐿푅 ⊆ 퐷.

3.11 Let 푤 ∈ Mon⟨푏푐, 푐⟩. Then 푤 = 훽1 ⋯ 훽푛, where each 훽푖 is either 푏푐 or 푐. Let

푎 if 훽푖 = 푏푐, 훼푖 = { 푎푏 if 훽푖 = 푐. Then

훼푛 ⋯ 훼1푤 = 훼푛 ⋯ 훼1훽1 ⋯ 훽푛

= 훼푛 ⋯ 훼2푎푏푐훽2 ⋯ 훽푛

=푀 훼푛 ⋯ 훼2훽2 ⋯ 훽푛 ⋮

=푀 휀.

Clearly, 푤휀 =푀 푤, so if 푤 ∈ Mon⟨푏푐, 푐⟩, then 푤 L 휀.

Solutions to exercises • 213 Now suppose 푤 ∈ 푁 with 푤 L 휀. Then there is a word 푢 ∈ 푁 such that 푢푤 =푀 휀. By Exercise 2.8, 휀 can be obtained from 푢푤 by deleting subwords 푎푏푐. Neither 푢 nor 푣 contain subwords 푎푏푐, so 푢푤 must have a subword 푎푏푐 across the ‘boundary’ of 푢 and 푤. That is, we have either:

◆ 푤 = 푏푐푤′ and 푢 = 푢′푎, with 푢′푤′ =푀 푢푤 =푀 휀; or

◆ 푤 = 푐푤′ and 푢 = 푢′푎푏, with 푢′푤′ =푀 푢푤 =푀 휀. Again, 푢′ and 푤′, being subwords of 푢 and 푤, do not contain subwords 푎푏푐, so the same reasoning applies. Proceeding by induction, we see that 푤 ∈ Mon⟨푏푐, 푐⟩ (and 푢 ∈ Mon⟨푎, 푎푏⟩, although that is not important). Hence if 푤 L 휀, then 푤 ∈ Mon⟨푏푐, 푐⟩. Symmetrical reasoning shows that 푤 R 휀 if and only if 푤 ∈ Mon⟨푎, 푎푏⟩. Since H = L ∩ R, it follows that 푤 H 휀 if and only if 푤 ∈ Mon⟨푏푐, 푐⟩ ∩ Mon⟨푎, 푎푏⟩ = {휀}. Since 휀 is an idempotent, Exercise 3.10 shows that the D-class of 휀 is Mon⟨푏푐, 푐⟩Mon⟨푎, 푎푏⟩. If 푤 ∈ Mon⟨푏푐, 푐⟩Mon⟨푎푏, 푎⟩, then 푤 D 휀 and so 푤 J 휀. If 푤 ∈ Mon⟨푏푐, 푐⟩푏Mon⟨푎푏, 푎⟩, then by the results for L and R, there exist 푝, 푞 ∈ 푀 such that 푝푤푞 =푀 푏, so 푎푝푤푞푐 =푀 푎푏푐 =푀 휀. Now suppose 푤 ∈ 푁 with 푤 J 휀. Then there exist 푢, 푣 ∈ 푁 such that 푢푤푣 =푀 휀. So 휀 can be obtained from 푢푤푣 by deleting subwords 푎푏푐. Any subwords 푎푏푐 in 푢푤푣 must be across the boundaries of 푢 and 푤 and of 푣 and 푤. Proceeding by induction as in the L case, we see that 푤 = Mon⟨푏푐, 푐⟩푥Mon⟨푎, 푎푏⟩, where 푥 is either 휀 or a single letter 푏. 3.12 Since 푆 is regular, L-class and every R-class of 푆 contains an idempotent. Since there is only one idempotent in 푆, there is only one R-class and only one L-class in 푆. Hence R = L = H = 푆 × 푆. So 푆 consists of a single 퐻-class, which contains an idempotent and is thus a subgroup.

3.13 a) Let 푥 ∈ 푅1. Then there exists 푞 ∈ 푀 such that 푥푞 = 1. Since 푀 is group-embeddable, 푞푥 = 1. Thus any element of 푅1 is right- and left-invertible. On the other hand, if 푥 ∈ 푀 is right-invertible, then 푥 ∈ 푅1. So 푥 ∈ 푅1 if and only if 푥 is right- and left-invertible.

b) Suppose 푀 has at least two R-classes. Then 푀 ∖ 푅1 is non-empty. Let 푥 ∈ 푀 ∖ 퐻1. Then 푥 is not right or left invertible. Suppose that 푥푘 R 푥ℓ for some 푘 < ℓ. Then there exists 푝 ∈ 푀 such that 푥푘 = 푥ℓ푝. Hence 푥ℓ−푘푝 = 1 and so 푥 has a right inverse 푥ℓ−푘−1푝. This is a contradiction. So all of the powers of 푥 lie in different R-classes.

Solutions to exercises • 214 Exercises for chapter 4 [See pages 86–88 for the exercises.] −1 4.1 a) Define 휑 ∶ 퐺 → M[퐺; 퐼, 훬; 푃] by 푥 ↦ (1, 푥푝11 , 1). Then

−1 −1 (푥휑)(푦휑) = (1, 푥푝11 , 1)(1, 푦푝11 , 1) −1 −1 = (1, 푥푝11 푝11푦푝11 , 1) −1 = (1, 푥푦푝11 , 1) = (푥푦)휑.

So 휑 is a homomorphism. Furthermore,

−1 −1 푥휑 = 푦휑 ⇒ (1, 푥푝11 , 1) = (1, 푦푝11 , 1) −1 −1 ⇒ 푥푝11 = 푦푝11 ⇒ 푥 = 푦.

−1 So 휑 is injective. Finally, since 퐺 is a group, (1, 푥푝11 , 1) will range over M[퐺; 퐼, 훬, 푃] = {1} × 퐺 × {1} as 푥 ranges over 퐺. So 휑 is surjective. Hence 휑 is an isomorphism.

b) Let 푀 = {푒, 푧} be a semilattice with 푒 > 푧. Let 푝휆푖 = 푧. Let (푖, 푥, 휆) and (푖, 푦, 휆) be arbitrary elements of M[푀; 퐼, 훬; 푃]. Then

(푖, 푥, 휆)(푖, 푦, 휆) = (푖, 푥푝휆푖푦, 휆) = (푖, 푥푧푦, 휆) = (푖, 푧, 휆).

So M[푀; 퐼, 훬; 푃] is a null semigroup and so not isomorphic to 푀. 4.2 A completely simple semigroup is isomorphic to M[퐺; 퐼, 훬; 푃] for some group 퐺, index sets 퐼 and 훬, and matrix 푃. Suppose that we have (푖1, 푔1, 휆1)(푖2, 푔2, 휆2) = (푗1, ℎ1, 휇1)(푗2, ℎ2, 휇2). The, by the definition of the product in M[퐺; 퐼, 훬; 푃], we have (푖 , 푔 푝 푔 , 휆 ) = 1 1 휆1푖2 2 2 (푗 , ℎ 푝 ℎ , 휇 ), and so 1 1 휇1푗2 2 2

푖1 = 푗1, (S.13)

휆2 = 휇2, (S.14) 푔 푝 푔 = ℎ 푝 ℎ . (S.15) 1 휆1푖2 2 1 휇1푗2 2

Let 푞 = (푗 , 푝−1 ℎ−1푔 , 휆 ). Then 2 휇1푗2 1 1 1

(푗1, ℎ1, 휇1)푞 = (푗 , ℎ , 휇 )(푗 , 푝−1 ℎ−1푔 , 휆 ) [by definition of 푞] 1 1 1 2 휇1푗2 1 1 1 = (푗 , ℎ 푝 푝−1 ℎ−1푔 , 휆 ) 1 1 휇1푗2 휇1푗2 1 1 1 = (푗1, 푔1, 휆1)

= (푖1, 푔1, 휆1) [by (S.13)]

Solutions to exercises • 215 and

푞(푖2, 푔2, 휆2) = (푗 , 푝−1 ℎ−1푔 , 휆 )(푖 , 푔 , 휆 ) [by definition of 푞] 2 휇1푗2 1 1 1 2 2 2 = (푗 , 푝−1 ℎ−1푔 푝 푔 , 휆 ) 2 휇1푗2 1 1 휆1푖2 2 2 = (푗 , 푝−1 ℎ−1ℎ 푝 ℎ , 휇 ) [by (S.14) and (S.15)] 2 휇1푗2 1 1 휇1푗2 2 2 = (푗2, ℎ2, 휇2).

Hence M[퐺; 퐼, 훬; 푃] is equidivisible. 4.3 a) Let 푥, 푦 ∈ 푆 ≃ M[퐺; 퐼, 훬; 푃] with 푥 L 푦. Then by Proposition 4.12, 푥 = (푖, 푔, 휆) and 푦 = (푗, ℎ, 휆) for some 푖, 푗 ∈ 퐼, 푔, ℎ ∈ 퐺, and 휆 ∈ 훬. Let 푧 = (푘, 푓, 휇) ∈ 푆. Then 푧푥 = (푘, 푓푝휇푖푔, 휆) and 푧푦 = (푘, 푓푝휇푗ℎ, 휆). Since 푧푥, 푧푦 ∈ 퐼 × 퐺 × {휆}, we have 푧푥 L 푧푦. Hence L is left compatible. We already know L is a right congruence by Proposition 3.4(a). So L is a congruence. Similarly, R is a congruence and so H = L ∩ R is a congruence.

b) Let [(푖, 푔, 휆)]L, [(푗, ℎ, 휇)]L ∈ 푆/L. Then [(푖, 푔, 휆)]L[(푗, ℎ, 휇)]L = [(푖, 푔푝휆푗ℎ, 휇)]L = [(푗, ℎ, 휇)]L (since (푖, 푔푝휆푗ℎ, 휇) and (푗, ℎ, 휇) are L-related). Hence 푆/L is a right zero semigroup. Similarly 푆/R is a left zero semigroup. c) Define a map 휑 ∶ 푆/H → 푆/R × 푆/L by

[(푖, 푔, 휆)]H휑 = ([(푖, 푔, 휆)]R, [(푖, 푔, 휆)]L).

Using the fact that H = L ∩ R, it is easy to show that this map is well-defined and injective. It is clearly surjective, and is a homomorphism since R and L are congruences. So 푆/H ≃ 푆/R × 푆/L. 4.4 Since 푆 is completely simple, 푆 ≃ M[퐺; 퐼, 훬, 푃]. Hence |푆| = |퐼|×|퐺|× |훬|. a) Since 푝 = |퐼| × |퐺| × |훬|, one of the following three cases must hold: i) |퐼| = 푝, |퐺| = 1, and |훬| = 1. Since 퐺 is trivial and |훬| = 1, the R-classes of 푆 are single elements by Proposition 4.12(c). Thus 푆 ≃ 푆/R is a left zero semigroup by Exercise 4.3(b). ii) |퐼| = 1, |퐺| = 1, and |훬| = 푝. This is similar to case i), and shows that 푆 is a right zero semigroup. iii) |퐼| = 1, |퐺| = 푝, and |훬| = 1. Then 푆 is a group by Exercise 4.1. b) Since 푝푞 = |퐼| × |퐺| × |훬|, one of the following cases must hold (interchanging 푝 and 푞 if necessary): i) |퐼| = 푝푞, |퐺| = 1, and |훬| = 1. As in part a)i), 푆 ≃ 푆/R is a left zero semigroup and so a left group by Theorem 4.19. [We could also use the fact that 푆 has only one L-class and apply Theorem 4.19.]

Solutions to exercises • 216 ii) |퐼| = 푝, |퐺| = 푞, and |훬| = 1. Then 푆 = 퐼 × 퐺 × {휆}. Thus 푆 has only one L-class and so is a left group by Theorem 4.19. iii) |퐼| = 푝, |퐺| = 1, and |훬| = 푞. Then the H-classes of 푆 are single elements by Proposition 4.12(d). So 푆 ≃ 푆/H is a rectangular band by Exercise 4.3(c) iv) |퐼| = 1, |퐺| = 푝푞, and |훬| = 1. As in part a)iii), 푆 is a group (and thus both a left and a right group). v) |퐼| = 1, |퐺| = 푝, and |훬| = 푞. This is dual to case ii), and shows that 푆 is a right group. vi) |퐼| = 1, |퐺| = 1, and |훬| = 푝푞. This is dual to case i), 푆 ≃ 푆/L is a right zero semigroup and so a right group. 4.5 a) Let 푧 ∈ 푆. Then 푧푧−1 R 푧 and 푧−1푧 L 푧. So 푧푧−1 = 푧−1푧 H 푧. Similarly 푧푧−1 = 푧−1푧 H 푧−1. So 푧 H 푧−1. Since every H-class of 푆 is a subgroup, 푧−1 is the unique group inverse of 푧 in this subgroup. The H-class of 푧휑 is also a subgroup and (푧휑)−1 is the unique group inverse of 푧휑 in this subgroup. Then 휑| is a group 퐻푧 −1 −1 homomorphism into the subgroup 퐻푧휑 and so 푧 휑 = (푧휑) .

b) There are many possible examples. Let 푆 = {푠1, 푠2} and 푇 = {푡1, 푡2} −1 −1 −1 be left zero semigroups. Define on 푆 by 푠1 = 푠2 and 푠2 = 푠1. −1 −1 −1 −1 Define on 푇 by 푡1 = 푡1 and 푡2 = 푡2. In both cases, satisfies −1 −1 −1 (푥 ) = 푥 and 푥푥 푥 = 푥. Define 휑 ∶ 푆 → 푇 by 푠1휑 = 푡1 and −1 −1 −1 푠2휑 = 푡2. Then (푠1휑) = 푡1 = 푡1 but 푠1 휑 = 푠2휑 = 푡2. 4.6 a) i) The isomorphism 휑 maps non-zero R-classes bijectively to non-zero R-classes. Since the R-classes of M0[퐺; 퐼, 훬; 푃] are sets of the form {푖}×퐺×훬 and the R-classes of M0[퐻; 퐽, 훭; 푄] are sets of the form {푗} × 퐺 × 훭, there must be a bijection 훼 ∶ 퐼 → 퐽 such that (푖, 푎, 휆)휑 ∈ {푖훼} × 퐻 × 훭. Similarly there is a bijection 훽 ∶ 훬 → 훭 such that (푖, 푎, 휆)휑 ∈ 퐼 × 퐻 × {휆훽}. Combining these statements shows that (푖, 푎, 휆)휑 ∈ {푖훼} × 퐻 × {휆훽}. Since 휑 must map group H-classes to group H-classes, we have 푝휆푖 ≠ 0 if and only if 푝(휆훽)(푖훼) ≠ 0. −1 ii) Let 훾 ∶ 퐺 → {1}×퐺×{1} be defined by 푥훾 = (1, 푝11 푥, 1). Then −1 −1 −1 −1 (푥훾)(푦훾) = (1, 푝11 푥, 1)(1, 푝11 푦, 1) = (1, 푝11 푥푝11푝11 푦, 1) = −1 (1, 푝11 푥푦, 1) = (푥푦)훾, so 훾 is a homomorphism. Furthermore, −1 −1 훾 is injective since 푥훾 = 푦훾 ⇒ (1, 푝11 푥, 1) = (1, 푝11 푦, 1) ⇒ −1 −1 푝11 푥 = 푝11 푦 ⇒ 푥 = 푦. Finally, 훾 is surjective since for any (1, 푥, 1) ∈ {1} × 퐺 × {1}, we have (푝11푥)훾 = (1, 푥, 1). So 훾 is an isomorphism. Similarly, the map 휂 ∶ 퐻 → {1훼} × 퐻 × {1훽} defined by −1 푥휂 = (1훼, 푞(1훽)(1훼)푥, 1훽) is an isomorphism. By part i), 휑|{1}×퐺×{1} ∶ {1} × 퐺 × {1} → {1훼} × 퐻 × {1훽} is an isomorphism, so the composition 휗 = 훾휑휂−1 = −1 훾휑|{1}×퐺×{1}휂 is an isomorphism from 퐺 to 퐻.

Solutions to exercises • 217 iii) First,

−1 −1 −1 −1 (푖, 1퐺, 1)(1, 푝11 푥, 1)(1, 푝11 , 1) = (푖, 1퐺푝11푝11 푥푝11푝11 ) = (푖, 푥, 휆).

Now, for all 푥 ∈ 퐺,

−1 −1 (1, 푝11 푥, 1)휑 = 푥훾휑 = 푥휗휂 = (1훼, 푞(1훽)(1훼)(푥휗), 1훽). Therefore for any 푥 ∈ 퐺,

(푖, 푥, 휆)휑 −1 −1 = ((푖, 1퐺, 1)(1, 푝11 푥, 1)(1, 푝11 , 휆))휑 −1 −1 = (푖, 1퐺, 1)휑(1, 푝11 푥, 1)휑(1, 푝11 , 휆)휑 −1 −1 = (푖훼, 푢푖, 1훽)(1훼, 푞(1훽)(1훼)(푥휗), 1훽)(1훼, 푞(1훽)(1훼)푣휆, 휆훽) −1 −1 = (푖훼, 푢푖푞(1훽)(1훼)푞(1훽)(1훼)(푥휗)푞(1훽)(1훼)푞(1훽)(1훼)푣휆, 휆훽)

= (푖훼, 푢푖(푥휗)푣휆, 휆훽)휑. Hence

(푖훼, 푢푖(푝휆푖휗)푣휆, 휆훽)

= (푖, 푝휆푖, 휆)휑

= ((푖, 1퐺, 휆)(푖, 1퐺, 휆))휑

= (푖, 1퐺, 휆)휑(푖, 1퐺, 휆)휑

= (푖훼, 푢푖푣휆, 휆훽)(푖훼, 푢푖푣휆, 휆훽)

= (푖훼, 푢푖푣휆푞(휆훽)(푖훼)푢푖푣휆, 휆훽);

thus 푝휆푖휗 = 푣휆푞(휆훽)(푖훼)푢푖 by cancellativity in 퐻. b) Define a map 휑 ∶ M0[퐺; 퐼, 훬; 푃] → M0[퐻; 퐽, 훭; 푄] by

(푖, 푥, 휆)휑 = (푖훼, 푢푖(푥휗)푣휆, 휆훽), and 0휑 = 0. Then 휑 is a homomorphism since

(푖, 푥, 휆)휑(푖′, 푦, 휆′)휑

= (푖훼, 푢푖(푥휗)푣휆, 휆훽)(푖′훼, 푢푖′(푦휗)푣휆′, 휆′)

= (푖훼, 푢푖(푥휗)푣휆푞(휆훽)(푖′훼)푢푖′(푦휗)푣휆′, 휆′)

= (푖훼, 푢푖(푥휗)(푝휆푖′휗)(푦휗)푣휆′, 휆′)

= (푖훼, 푢푖((푥푝휆푖′푦)휗)푣휆′, 휆′)

= (푖, 푥푝휆푖′푦, 휆′)휑 = ((푖, 푥, 휆)(푖′, 푦, 휆′))휑.

Furthermore, 휑 is a bijection since 훼, 훽, and 휗 are all bijections. So 휑 is an isomorphism from M0[퐺; 퐼, 훬; 푃] to M0[퐻; 퐽, 훭; 푄].

Solutions to exercises • 218 4.7 Suppose 푃 is regular. Then 푆 = M0[퐺; 퐼, 훬; 푃] is completely simple and so regular by the Lemma 4.6(b). [Alternatively: Since 푃 contains −1 some non-zero element 푝휆푖, the element (푖, 푝휆푖 , 휆) is idempotent and thus regular. Thus the D-class 퐼×퐺×훬 is regular by Proposition 3.19.] Suppose 푃 is not regular. Then 푃 has a row or a column all of whose entries are 0. Suppose all the entries in the row indexed by 휆 are 0; the reasoning for columns is similar. Let (푖, 푥, 휆) ∈ 퐼 × 퐺 × {휆}. Then for (푗, 푦, 휇) ∈ M0[퐺; 퐼, 훬; 푃] ∖ {0}, we have (푖, 푥, 휆)(푗, 푦, 휇) = 0 since 푝휆푗 = 0. Hence there is no element 푧 ∈ M0[퐺; 퐼, 훬; 푃] with (푖, 푥, 휆)푧(푖, 푥, 휆) = (푖, 푥, 휆). Thus 푆 is not regular.

4.8 a) Since 푆 satisfies minL, the set of L-classes that are not equal to {0} there is a minimal element. Let 퐿푥 be such a minimal L-class not equal to {0}. Then 푆푥 is a left ideal not equal to {0}. Suppose 퐿 is some left ideal contained in 푆푥 and not equal to {0}. Pick 푦 ∈ 퐿 ∖ {0}. Then 푆푦 ⊆ 푆푥 and so 퐿푦 ⊆ 퐿푥. Since 퐿푥 is minimal among non-{0} L-classes, 퐿푥 = 퐿푦 and so 푆푥 = 푆푦. So 푆푥 must be a 0-minimal left ideal. b) i) Let 푥 ∈ 퐾 ∖ {0}. Then 푆푥 is a left ideal of 푆 and is contained in 퐾. Since 퐾 is 0-minimal, either 푆푥 = 퐾 or 푆푥 = {0}. Suppose, with the aim of obtaining a contradiction, that 푆푥 = {0}. Then {0, 푥} is a left ideal of 푆 contained in 퐾 and not equal to {0}. Since 퐾 is 0-minimal, 퐾 = {푥, 0}. But then 퐾2 = {0}, which is a contradiction. So 퐾 = 푆푥. ii) It is immediate that 퐿푥 is a left ideal. Suppose 퐾 ≠ {0} is a left ideal contained in 퐿푥. Let 퐽 = { 푦 ∈ 퐿 ∶ 푦푥 ∈ 퐾 }. Then 퐽 ⊆ 퐿 and 퐽 is a left ideal, since

푦 ∈ 퐽 ∧ 푠 ∈ 푆 ⇒ 푦 ∈ 퐿 ∧ 푦푥 ∈ 퐾 ∧ 푠 ∈ 푆 [by definition of 퐽] ⇒ 푠푦 ∈ 퐿 ∧ 푠푦푥 ∈ 퐾 [since 퐿 and 퐾 are left ideals] ⇒ 푠푦 ∈ 퐽. [by definition of 퐽]

Since 퐿 is 0-minimal and 퐽 ≠ {0}, we have 퐽 = 퐿 and so 퐽푥 = 퐿푥. Furthermore, 퐽푥 ⊆ 퐾 by the definition of 퐽 and 퐾 ⊆ 퐿푥 by the definition of 퐾, and so 퐾 = 퐽푥 = 퐿푥. Hence 퐿푥 is 0-minimal. iii) Note that 퐿푆 is an ideal since 푆퐿푆푆 ⊆ (푆퐿)(푆2) ⊆ 퐿푆. So, since 푆 is 0-simple, either 퐿푆 = {0} or 퐿푆 = 푆. Suppose, with the aim of obtaining a contradiction, that 퐿푆 = {0}. Then 퐿푆 ⊆ 퐿 and so 퐿 is an ideal. Since 퐿 ≠ {0}, we have 퐿 = 푆. Hence 푆2 = 퐿푆 = {0} and so 푆 is null, which contradicts 푆 being 0-simple. Therefore 퐿푆 = 푆. So there exists 푥 ∈ 푆 with 퐿푥 ≠ {0}. iv) The set 푀 is a union of 0-minimal left ideals and is thus itself a left ideal. By part iii), 푀 ≠ {0}. Let 푚 ∈ 푀 and 푡 ∈ 푆. Then

Solutions to exercises • 219 푚 ∈ 퐿푥 for some 푥 ∈ 푆 and so 푚푡 ∈ 퐿푥푡 ⊆ 푀. Hence 푀푆 ⊆ 푀 and so 푀 is also a right ideal. So 푀 is an ideal and not equal to {0}. Since 푆 is 0-simple, we have 푀 = 푆. v) Let 퐿 be a 0-minimal left ideal. For any 0-minimal right ideal 푅, the set 퐿푅 is an ideal and hence, since 푆 is 0-simple, either 퐿푅 = {0} or 퐿푅 = 푆. By part iii), there exists some 푥 ∈ 푆 with 퐿푥 ≠ {0}. By the dual version of part iv), 푥 lies in some 0- minimal right ideal. Fix a 0-minimal right ideal 푅 containing 푥. Then 퐿푅 ≠ {0} and so 퐿푅 = 푆. Notice that since 푅 is a right ideal, 푅퐿 ⊆ 푅. Similarly, 푅퐿 ⊆ 퐿. Let 푥 ∈ 푅퐿 ∖ {0} ⊆ 푅 ∖ {0}. Then 푅 = 푥푆 by the dual version of part i). Since 푆 = 퐿푅 = 퐿푥푆, we have 퐿푥 ≠ {0} and so 퐿푥 is a 0-minimal left ideal by part ii). However, 퐿푥 ⊆ 퐿 since 푥 ∈ 푅퐿 ⊆ 퐿. Therefore, since 퐿 is 0-minimal and 퐿푥 ≠ {0}, we have 퐿푥 = 퐿 and so 푅퐿푥 = 푅퐿. Similarly 푥푅퐿 = 푅퐿. Hence 푅퐿 is a group with a zero adjoined by Exercise 1.21. vi) Let 푓 be a non-zero idempotent in 푆 with 푓 ≼ 푒. Then 푒푓 = 푓푒 = 푓. Since 푒 ∈ 푅퐿 ⊆ 푅 ∩ 퐿, it follows from part i) and its dual version that 푅 = 푒푆 and 퐿 = 푆푒. Hence 푓 = 푒푓푒 ∈ 푒푆푒 = 푒푆2푒 = (푒푆)(푆푒) = 푅퐿, since 푆2 = 푆 by Lemma 3.6. Since 푅퐿 ∖ {0} is a group, 푒 = 푓. So 푒 is a primitive idempotent. Hence 푆 is completely 0-simple. 4.9 Let 푅 be a right ideal of 푆. Let 푟 ∈ 푅 and ℓ ∈ 퐺. Then 푟ℓ ∈ 푅 ∩ 퐺 since 푅 is a right ideal and 퐺 is a left ideal. So 푅 ∩ 퐺 ≠ ∅. Then 푅 ∩ 퐺 is a right ideal of 퐺, since 푅 is a right ideal and 퐺 is a subgroup. But 퐺 is a group, and thus its only right ideal is 퐺 itself. Hence 푅∩퐺 = 퐺, and so 퐺 ⊆ 푅. In particular, 1퐺 ∈ 푅. Let 푥 ∈ 푆. Then 1퐺푥 = 1퐺1퐺푥 since 1퐺 is idempotent, and so 푥 = 1퐺푥 since 푆 is left-cancellative. Therefore 푥 = 1퐺푥 ∈ 1퐺푆 ⊆ 푅푆 ⊆ 푅. Hence 푆 ⊆ 푅 and so 푆 = 푅. Therefore 푆 does not contain any proper right ideals and so is right simple. Since it is also left-cancellative, 푆 is a right group.

Exercises for chapter 5 [See pages 113–116 for the exercises.] 1 2 1 2 5.1 Let 휏 = ( ) and 휁 = ( ). Then 휏휏 = 휏휁 = 휁휏 = 휁휁 = 휁. So 2 ∗ ∗ ∗ 푇 = {휏, 휁} is a null semigroup and 휏 does not have an inverse in 푇. 1 2 [Of course, 휏 does have an inverse in I ; indeed 휏−1 = ( ).] 푋 ∗ 1

5.2 Let 휎1, 휎2 ∈ 푆. Then there exist subgroups 퐻1, 퐻′1, 퐻2, and 퐻′2 of 퐺 such that 휎1 ∶ 퐻1 → 퐻′1 and 휎2 ∶ 퐻2 → 퐻′2 are isomorphisms.

Solutions to exercises • 220 −1 −1 Then dom(휎1휎2) = (im 휎1 ∩ dom 휎2)휎1 = (퐻′1 ∩ 퐻2)휎1 . Now, 퐻′1 ∩ 퐻2 is a subgroup of 퐺. (In particular, it contains 1퐺 and so is non-empty.) Thus dom(휎1휎2) is a subgroup of 퐺 and so im(휎1휎2) is also a subgroup of 퐺. So 휎1휎2 ∈ 푆. Thus 푆 is a subsemigroup of I퐺. −1 −1 Furthermore, 휎1 ∶ 퐻′1 → 퐻1 is also an isomorphism; thus 휎1 ∈ 푆. Thus 푆 is an inverse subsemigroup of I퐺. 5.3 a) Suppose that 휎 L 휏. Then there exist 휋, 휌 ∈ I푋 such that 휋휎 = 휏 and 휌휏 = 휎. Therefore

im 휎 = 푋휎 ⊇ (푋휋)휎 = im(휋휎) = im 휏,

and similarly im 휏 ⊇ im(휌휏) = im 휎. Hence im 휎 = im 휏. Now suppose that im 휎 = im 휏. Let 휋 = 휏휎−1. Then

−1 휋휎 = 휏휎 휎 = 휏idim 휎 = 휏idim 휏 = 휏. Similarly, let 휌 = 휎휏−1; then 휌휏 = 휎. Hence 휎 L 휏.

b) Suppose that 휎 R 휏. Then there exist 휋, 휌 ∈ I푋 such that 휎휋 = 휏 and 휏휌 = 휎. Therefore

dom 휏 = dom 휎휋 = (im 휎 ∩ dom 휋)휎−1 ⊆ (im 휎)휎−1 = dom 휎

and similarly dom 휎 ⊆ (im 휏)휏−1 = dom 휏. Thus dom 휎 = dom 휏. Now suppose that dom 휎 = dom 휏. Let 휋 = 휎−1휏. Then

−1 휎휋 = 휎휎 휏 = iddom 휎휏 = iddom 휏휏 = 휏. Similarly, let 휌 = 휏−1휎; then 휏휌 = 휎. Hence 휎 R 휏.

c) Suppose that 휎 D 휏. Then there exists 휐 ∈ I푋 such that 휎 L 휐 R 휏, and so,

Now suppose that |dom 휎| = |dom 휏|. Then there is a bijection −1 휋 ∶ dom 휎 → dom 휏. Note that 휋 ∈ I푋. Let 휐 = 휋 휎. Then 휎 = 휋휐, and so 휎 L 휐. Furthermore,

dom 휐 = dom(휋−1휎) = (im 휋−1 ∩ dom 휎)휋 = (dom 휋 ∩ dom 휎)휋 = (dom 휎)휋 = dom 휏,

Solutions to exercises • 221 and so 휐 R 휏 by part b). Hence 휎 D 휏. Suppose 휎 J 휏. Then there exist 휋, 휌, 휋′, 휌′ ∈ I푋 such that 휎 = 휋휏휌 and 휏 = 휋′휎휌′. Therefore

|dom 휎| = |im 휎| = |푋휎| = |푋휋휏휌| ⩽ |푋휏휌| = |푋휏| = |im 휏| = |dom 휏|;

dom(휋훽) = (im 휋 ∩ dom 훽)휋−1 = (im 휋)휋−1 = dom 휋 = dom 훾.

Hence 휋훽 R 훾 by Exercise 5.3(b). Thus there exists 휌′ ∈ I푋 such that 휋훽휌′ = 훾. Since |im 훽| = |dom 훽| = 푛 − 1 = |dom 훾| = |im 훾|, it follows that dom 휌′ ⊇ im 훽. Extend 휌′ to a permutation 휌 ∈ S푋. Then 휌 and 휌′ agree on im 훽. Hence 휋훽휌 = 휋훽휌′ = 훾. Since 휋, 휌 ∈ S푋 = ⟨휏, 휁⟩, it follows from the previous paragraph that 퐽푛−1 ⊆ S푋훽S푋 ⊆ ⟨휏, 휁, 훽⟩.

b) Let 휎 ∈ 퐽푘. Pick 푥 ∈ 푋∖dom 휎 and 푦 ∈ 푋∖im 휎 and extend 휎 to 휎′ by defining 푥휎′ = 푦. Then 휎′ ∈ 퐽푘+1, and 휎 = 휎′id푋−{푥} ∈ 퐽푘+1퐽푛−1. Hence 퐽푘 ⊆ 퐽푘+1퐽푛−1. 푛−푘 By induction on 푘, we see that 퐽푘 ⊆ 퐽푛−1 ⊆ ⟨휏, 휁, 훽⟩. Since this holds for 푘 = 0, … , 푛 − 1, and since obviously 퐽푛 = S푋 = ⟨휏, 휁⟩ ⊆ 푛 ⟨휏, 휁, 훽⟩, it follows that I푋 = ⋃푘=0 퐽푘 ⊆ ⟨휏, 휁, 훽⟩. −1 −1 5.5 a) Let 푀 = ⟨휏, 휏 ⟩. Note that 휏휏 = id푋, so 푀 is a monoid. We will use Method 2.9 to prove that 푀 is defined by Mon⟨푏, 푐 | (푏푐, 휀)⟩. Define 휑 ∶ {푏, 푐} → 푀 by 푏휑 = 휏 and 푐휑 = 휏−1. Then 푀 satisfies the defining relation with respect to 휑 since (푏푐)휑∗ = −1 ∗ 푖 푗 휏휏 = id푋 = 휀휑 . Let 푁 = { 푐 푏 ∶ 푖 ∈ ℕ ∪ {0} }; any word in {푏, 푐}∗ can be transformed to one in 푁 by applying the defining relation to delete subwords 푏푐. Finally, let 푥 ∈ 푋 ∖ 푋휏 (note that such an 푥 exists since im 휏 ⊊ 푋). Then for 푘 ∈ ℕ ∪ {0}, we have 푥휏푘 ∈ 푋휏푘 ∖푋휏푘+1. In particular, the 푥휏푘 are all distinct. Note that 푥 ∉ im 휏 = dom 휏−1. Thus (푥휏푘)(푐푖푏푗)휑∗ = 푥휏푘휏−푖휏푗 is defined if and only if 푘 ⩾ 푖, in which case it is equal to 푥휏푘−푖+푗. So the minimum 푘 for which (푥휏푘)(푐푖푏푗)휑∗ is defined is 푖, and the image of 푥휏푖 under (푐푖푏푗)휑∗ is 푥휏푗. So (푐푖푏푗)휑∗ determines 푖 and 푗, and ∗ so 휑 |푁 is injective. This completes the proof. [This proof is essentially just Example 2.11(b) rephrased in terms of partial bijections.]

Solutions to exercises • 222 −1 −1 b) Let 푀 = ⟨{ 휏푖, 휏푖 ∶ 푖 ∈ 퐼 }⟩. Note that 휏푖휏푖 = id푋, so 푀 is a monoid. For any 푖1, … , 푖푘 ∈ 퐼 and 휖1, … , 휖푘 ∈ {1, −1}, 휏휖1 ⋯ 휏휖푘 휏−휖푘 ⋯ 휏−휖1 휏휖1 ⋯ 휏휖푘 푖1 푖푘 푖푘 푖1 푖1 푖푘 = 휏휖1 ⋯ 휏휖푘−1 id 휏−휖푘−1 ⋯ 휏−휖1 휏휖1 ⋯ 휏휖푘 푖1 푖푘−1 푋 푖푘−1 푖1 푖1 푖푘 = 휏휖1 ⋯ 휏휖푘−1 휏−휖푘−1 ⋯ 휏−휖1 휏휖1 ⋯ 휏휖푘 푖1 푖푘−1 푖푘−1 푖1 푖1 푖푘 ⋮= id 휏휖1 ⋯ 휏휖푘 푋 푖1 푖푘 = 휏휖1 ⋯ 휏휖푘 . 푖1 푖푘 So (휏휖1 ⋯ 휏휖푘 )−1 = 휏−휖푘 ⋯ 휏−휖1 ∈ 푀. Hence 푀 is an inverse 푖1 푖푘 푖푘 푖1 monoid. Furthermore, for 푖, 푗 ∈ 퐼 with 푖 ≠ 푗 since im 휏푖 and −1 −1 im 휏푗 = dom 휏푗 are disjoint, 휏푖휏푗 = ∅, and ∅ is a zero for B푋 and thus for 푀. We will use Method 2.9 to prove that 푀 is defined by (5.14). Let 휑 ∶ { 푏푖, 푐푖 ∶ 푖 ∈ 퐼 } ∪ {푧} → 푀 be given by 푏푖휑 = 휏푖 and −1 푐푖휑 = 휏푖 for each 푖 ∈ 퐼, and 푧휑 = ∅. Then for 푖, 푗 ∈ 퐼 with 푖 ≠ 푗,

∗ −1 ∗ (푏푖푐푖)휑 = 휏푖휏푖 = id푋 = 휀휑 ; (S.16) ∗ −1 ∗ (푏푖푐푗)휑 = 휏푖휏푗 = ∅ = 푧휑 ; (S.17) ∗ ∗ (푏푖푧)휑 = 휏푖∅ = ∅ = 푧휑 ; (S.18) ∗ ∗ (푧푏푖)휑 = ∅휏푖 = ∅ = 푧휑 ; (S.19) ∗ −1 ∗ (푐푖푧)휑 = 휏푖 ∅ = ∅ = 푧휑 ; (S.20) ∗ −1 ∗ (푧푐푖)휑 = ∅휏푖 = ∅ = 푧휑 ; (S.21) (푧푧)휑∗ = ∅∅ = ∅ = 푧휑∗. (S.22)

Thus 푀 satisfies the defining relations in (5.14) with respect to 휑. Let

∗ ∗ 푁 = { 푐푖 ∶ 푖 ∈ 퐼 } { 푏푖 ∶ 푖 ∈ 퐼 } ∪ {푧}.

∗ Any word in { 푧, 푏푖, 푐푖 ∶ 푖 ∈ 퐼 } can be transformed to one in 푁 by applying defining relations to remove any subwords 푏푖푐푗 (for any 푖, 푗 ∈ 퐼, replacing them with 푧 if 푖 ≠ 푗), and then to replacing any two-symbol subword that contains a 푧 into 푧 alone. ∗ The remaining step is to show that 휑 |푁 is injective. Now, −1 −1 푥휏푖휏푗 is defined if and only if 푖 = 푗, and 푥휏푖 is defined if and only if 푥 ∈ im 휏푖. So 푥휏 휏 ⋯ 휏 휏−1휏−1 ⋯ 휏−1 푖ℓ 푖ℓ−1 푖1 푗1 푗2 푗푚

is defined for all 푥 ∈ 푋 if and only if ℓ ⩾ 푚 and 푖ℎ = 푗ℎ for ℎ = 1, … , 푚. So suppose

(푐 ⋯ 푐 푏 ⋯ 푏 )휑∗ = (푐 ⋯ 푐 푏 ⋯ 푏 )휑∗. 푗1 푗푚 푖1 푖푛 푗1′ 푗푚′ ′ 푖1′ 푖푛′′

Solutions to exercises • 223 Interchanging the two sides if necessary, assume 푚 ⩽ 푚′. Now,

푥휏 휏 ⋯ 휏 (푐 ⋯ 푐 푏 ⋯ 푏 )휑∗ 푗푚 푗푚−1 푗1 푗1 푗푚 푖1 푖푛 = 푥휏 휏 ⋯ 휏 휏−1 ⋯ 휏−1휏 ⋯ 휏 푗푚 푗푚−1 푗1 푗1 푗푚 푖1 푖푛 is defined for all 푥 ∈ 푋. So

푥휏 휏 ⋯ 휏 (푐 ⋯ 푐 푏 ⋯ 푏 )휑∗ 푗푚 푗푚−1 푗1 푗1′ 푗푚′ 푖1′ 푖푛′′ = 푥휏 휏 ⋯ 휏 휏−1 ⋯ 휏−1휏 ⋯ 휏 푗푚 푗푚−1 푗1 푗1 푗푚 푖1 푖푛 is defined for all 푥 ∈ 푋. So 푚 ⩾ 푚′, and thus 푚 = 푚′, and 푗 = 푗′ for ℎ = 1, … , 푚. Now, 푥 = 푥휏 휏 ⋯ 휏 (푐 ⋯ 푐 )휑∗, ℎ ℎ 푗푚 푗푚−1 푗1 푗1 푗푚 so 푥(푏 ⋯ 푏 )휑∗ = 푥(푏 ⋯ 푏 )휑∗ for all 푥 ∈ 푋. Interchanging 푖1 푖푛 푖1′ 푖푛′′ the two sides if necessary, assume 푛 ⩾ 푛′. Then

푥(푏 ⋯ 푏 )휑∗휏−1 ⋯ 휏−1 = 푥휏 ⋯ 휏 휏−1 ⋯ 휏−1 푖1 푖푛 푖푛 푖1 푖1 푖푛 푖푛 푖1 is defined for all 푥 ∈ 푋. So

푥(푏 ⋯ 푏 )휑∗휏−1 ⋯ 휏−1 = 푥휏 ⋯ 휏 휏−1 ⋯ 휏−1 푖1′ 푖푛′′ 푖푛 푖1 푖1′ 푖푛′′ 푖푛 푖1

is defined for all 푥 ∈ 푋. So 푛 ⩽ 푛′, and thus 푛 = 푛′, and 푖ℎ = 푖′ℎ for ℎ = 1, … , 푛. Hence

푐 ⋯ 푐 푏 ⋯ 푏 = 푐 ⋯ 푐 푏 ⋯ 푏 . 푗1 푗푚 푖1 푖푛 푗1′ 푗푚′ ′ 푖1′ 푖푛′′ Finally, note that we have shown that 푥(푐 ⋯ 푐 푏 ⋯ 푏 )휑∗ is 푗1 푗푚 푖1 푖푛 always defined for some element 푥 ∈ 푋. Hence 푧휑∗ = ∅ ≠ (푐 ⋯ 푐 푏 ⋯ 푏 )휑∗. Thus 휑∗| is injective. 푗1 푗푚 푖1 푖푛 푁 5.6 a) Let 푆 be a Clifford semigroup. Then 푆 ≃ S[푌; 퐺훼; 휑훼,훽], for some semilattice 푌, groups 퐺훼, and homomorphisms 휑훼,훽 ∶ 퐺훼 → 퐺훽. Let 푒 and 푓 be idempotents in 푆. Then 푒 ∈ 퐺훼 and 푓 ∈ 퐺훽 for some 훼, 훽 ∈ 푌. Thus 푒 = 1훼 and 푓 = 1훽, where 1훼 and 1훽 are the identities of 퐺훼 and 퐺훽. So 푒푓 = 1훼1훽 = (1훼휑훼,훼⊓훽)(1훽휑훽,훼⊓훽) = 1훼⊓훽1훼⊓훽 = 1훼⊓훽. So the idempotents of 푆 form a subsemigroup. Since 푆 is regular by Theorem 5.13, 푆 is orthodox. b) Let 푆 be completely simple and orthodox. By Theorem 4.11, 푆 ≃ M[퐺; 퐼, 훬; 푃] for some group 퐺, index sets 퐼 and 훬 and regular matrix 푃 over 퐺. View 퐼×훬 as a rectangular band. Without loss of generality, assume that there is a symbol 1 in 퐼 ∩ 훬. The elements −1 −1 (1, 푝휆1 , 휆) and (푗, 푝1푗 , 1) are idempotents of 푆, and so, since 푆 is −1 −1 −1 −1 orthodox, their product (1, 푝휆1 휆)(푗, 푝1푗 , 1) = (1, 푝휆1 푝휆푗푝1푗 , 1) −1 −1 −1 is also an idempotent; hence 푝휆1 푝휆푗푝1푗 = 푝11 . Define a map −1 −1 휑 ∶ 퐺 × (퐼 × 훬) → 푆 by (푔, (푖, 휆))휑 = (푖, 푝1푖 푔푝11푝휆1 , 휆). Then (푔, (푖, 휆))휑(ℎ, (푗, 휇))휑 −1 −1 −1 −1 = (푖, 푝1푖 푔푝11푝휆1 , 휆)(푗, 푝1푗 ℎ푝11푝휇1 , 휇)

Solutions to exercises • 224 −1 −1 −1 −1 = (푖, 푝1푖 푔푝11푝휆1 푝휆푗푝1푗 ℎ푝11푝휇1 , 휇) −1 −1 −1 = (푖, 푝1푖 푔푝11푝11 ℎ푝11푝휇1 , 휇) −1 −1 = (푖, 푝1푖 푔ℎ푝11푝휇1 , 휇) = (푔ℎ, (푖, 휇))휑; thus 휑 is a homomorphism. It is clearly injective and surjective and thus an isomorphism. For the converse, let 퐺 be a group and let 퐼×훬 be a rectangular band. Let 푃 be the 훬 × 퐼 matrix all of whose entries are 1퐺. It is straightforward to see that 휑 ∶ 퐺 × (퐼 × 훬) → M[퐺; 퐼, 훬; 푃], (푔, (푖, 휆)) ↦ (푖, 푔, 휆) is an isomorphism; thus 퐺 × (퐼 × 훬) is completely simple. The only idempotent in 퐺 is 1퐺 and every element of 퐼 × 훬 is idempotent. Hence the set of idempotents of 퐺×(퐼×훬) is {1퐺}×(퐼×훬), which is clearly a subsemigroup. Since 퐺 × (퐼 × 훬) is completely simple, it is regular by Proposition 4.13, and hence is orthodox. 5.7 Let 푆 be a completely 0-simple inverse semigroup. By Theorem 4.7, 푆 ≃ M0[퐺; 퐼, 훬; 푃] for some group 퐺, index sets 퐼 and 훬, and regular matrix 푃 over 퐺0. Since 푆 is inverse, every L-class and every R-class contains exactly one idempotent. Now, the non-zero idempotents of −1 M0[퐺; 퐼, 훬; 푃] are elements of the form (푖, 푝휆푖 , 휆), where 푖 ∈ 퐼 and 휆 ∈ 훬 are such that 푝휆푖 ≠ 0. The non-zero R-classes of M0[퐺; 퐼, 훬; 푃] are the sets {푖} × 퐺 × 훬; the non-zero L-classes of M0[퐺; 퐼, 훬; 푃] are the sets 퐼 × 퐺 × {휆}. So for each 푖, there is a unique 휆 such that 푝휆푖 is non-zero, and vice versa. Hence there is a bijection 휓 ∶ 퐼 → 훬 so that 푖휓 is the unique element of 훬 with 푝(푖휓)푖 ≠ 0. Hence |퐼| = |훬| Since 훬 an abstract index set, we can reorder it and the rows of 푃 so that 푃 becomes diagonal. Now we can simply replace the index set 훬 with 퐼. Now suppose that 푆 ≃ M0[퐺; 퐼, 퐼; 푃], where 푃 is diagonal. Then 푆 is completely 0-simple and therefore regular. The idempotents of −1 M0[퐺; 퐼, 퐼; 푃] are the elements (푖, 푝푖푖 , 푖). If 푖 ≠ 푗, then 푝푖푗 = 0 (since −1 −1 푃 is diagonal) and so (푖, 푝푖푖 , 푖)(푗, 푝푗푗 , 푗) = 0. So the idempotents of 푆 commute and so 푆 is inverse. 5.8 a) Let 푥 ∈ im 휏. Then 푥 = 푧휏 for some 푧 ∈ 푆1. Let 푦 ∈ 푆1. Since 휏 is a partial right translation, dom 휏 is a left ideal and so 푦푧 ∈ dom 휏; furthermore, (푦푧)휏 = 푦(푧휏) = 푦푥 and so 푦푥 ∈ im 휏. Thus im 휏 is a left ideal of 푆1. 1 b) Let 휏, 휎 ∈ I푆1 be partial right translations. Let 푥, 푦 ∈ 푆 . Suppose 푥휏휎 is defined. Then both 푥 ∈ dom 휏 and 푥휏 ∈ dom 휎. Since dom 휏 is a left ideal, 푦푥 ∈ dom 휏 and (푦푥)휏 = 푦(푥휏). Since dom 휎 is a left ideal, 푦(푥휏) ∈ dom 휎 and (푦(푥휏))휎 = 푦(푥휏휎). Hence 푦푥 ∈ dom(휏휎) and (푦푥)휏휎 = 푦(푥휏휎). So 휏휎 is a partial right translation.

Solutions to exercises • 225 Suppose 푥휏−1 is defined. Let 푧 = 푥휏−1. Then 푧 ∈ dom 휏 and 푧휏 = 푥. Since dom 휏 is a left ideal, 푦푧 ∈ dom 휏 and (푦푧)휏 = 푦(푧휏) = 푦푥. So 푦푥 ∈ dom 휏−1 and (푦푥)휏−1 = 푦푧 = 푦(푥휏−1). So 휏−1 is a partial right translation. Hence the set of partial right translations forms an inverse subsemigroup of I푆1 . Since every 휌푥 is a partial right translation, 푇 is a subsemigroup of the set of partial right translation. 5.9 Let 푥 = 푐훾푏훽 ∈ 퐵 be arbitrary. Let 푦 = 푐훽푏훾. Then

훾 훽 훽 훾 훾 훽 훾 훽 훾 훽 푥푦푥 = 푐 푏 푐 푏 푐 푏 =퐵 푐 푏 = 푐 푏 = 푥. So 푥 is regular. The idempotents of 퐵 are elements of the form 푐훾푏훾 by Exercise 2.10(a). Thus, given two idempotents 푒 = 푐훾푏훾 and 푓 = 푐훽푏훽, we see that if 훾 ⩾ 훽,

훾 훾 훽 훽 훾 훾−훽 훽 훾 훾 푒푓 = 푐 푏 푐 푏 =퐵 푐 푏 푏 =퐵 푐 푐 훽 훾−훽 훾 훽 훽 훾 훾 =퐵 푐 푐 푏 =퐵 푐 푏 푐 푏 = 푓푒 and similarly 푒푓 = 푓푒 if 훾 ⩽ 훽. So 퐵 is a regular semigroup whose idempotents commute and so is inverse by Theorem 5.1. 5.10 For 푥 ∈ 푆 and 푒 ∈ 퐸(푆),

푥 ≼ 푒 ⇒ 푥 = 푥푥−1푒 [by definition of ≼] ⇒ 푥2 = 푥푥−1푒푥푥−1푒 ⇒ 푥2 = 푥푥−1푥푥−1푒푒 [since idempotents commute in 푆] ⇒ 푥2 = 푥푥−1푒 [since 푥푥−1 and 푒 are idempotents] ⇒ 푥2 = 푥 [since 푥 = 푥푥−1푒] ⇒ 푥 ∈ 퐸(푆).

5.11 Consider an element 푢 of FInvM({훼}). Let 푇1 be the (unique) Munn tree corresponding to 푢. Let 푝, 푞, and 푟 be, respectively, the ‘푥-coordinates’ of the leftmost endpoint, the vertex 휔푇, and the rightmost endpoint. Notice that 푝 ⩽ 0, 푟 ⩾ 0, and 푝 ⩽ 푞 ⩽ 푟, so that (푝, 푞, 푟) ∈ 퐾. In this way, we determine a map 휑 ∶ FInvM({푎}) → 퐾. Clearly, a unique Munn tree of the given form can be reconstructed from any triple (푝, 푞, 푟) ∈ 퐾, so 휑 is injective and surjective. Let 푣 be another element of FInvM({푎}) and let 푇2 be the corresponding Munn tree. Let the triple (푝′, 푞′, 푟′) ∈ 퐾 correspond to 푇2. Consider multiplying 푢 and 푣 using the corresponding Munn trees 푇1 and 푇2 to get a Munn tree 푇 corresponding 푢푣. The process is illustrated in Figure S.5. First we merge the vertices 휔 and 훼 to 푇1 푇2 form a vertex that we call ∞, and let 훼 = 훼 and 휔 = 휔 . Then 푇 푇1 푇 푇2 we fold edges together until we get the Munn tree 푇. It is easy to see from the diagram that the coordinate of 휔푇 relative to 훼푇 is 푞 + 푞′, that the coordinate of the leftmost endpoint of 푇 relative to 훼푇 is the

Solutions to exercises • 226 푝 푞 푟 ⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞푎 푎 ⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞푎 푎 ⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞푎 푎 훼 휔 푇1 푇1 푎 푎 푎 푎 푎 푎 훼 휔 ⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟푇⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟2 푇⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟2 푝′ 푞′ 푟′ merge 휔 & 훼 , 푇1 푇2 let 훼 = 훼 and 휔 = 휔 푇 푇1 푇 푇2

푎 푎 푎 푎 푎 푎 훼푇 ∞ FIGURE S.5 푎 푎 푎 푎 푎 푎 휔 Multiplication of elements of 푇 FInvM({푎}) using Munn trees. The numbers 푝, 푞, 푟 are, re- folding arrows and merging vertices spectively, the ‘푥-coordinates’ relative to 훼 of the left end- 푇1 푎 푎 푎 푎 푎 푎 point of 푇1, the vertex 휔푇 , and 훼 휔 1 ⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟푇⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟푇 the right endpoint of 푇1; the min{푝, 푝′ + 푞} 푞 + 푞′ max{푟, 푞 + 푟′} numbers 푝′, 푞′, 푟′ play a similar role for 푇2.

smaller of 푝 and 푞 + 푝′, and the coordinate of the rightmost endpoint of 푇 relative to 훼푇 is the greater of 푟 and 푞 + 푟′. That is, the triple (min{푝, 푞 + 푝′}, 푞 + 푞′, max{푟, 푞 + 푟′}) corresponds to 푇. Thus the map 휑 is a homomorphism and thus an isomorphism. 5.12 By Exercise 5.11, the monoid 퐾 is isomorphic to FInvM({푎}). Thus it is sufficient to prove that 퐾 is a subdirect product of 퐵 × 퐵. The map 휑 is a homomorphism since

(푝, 푞, 푟)휑(푝′, 푞′, 푟′)휑 = (푐−푝푏−푝+푞, 푐푟푏−푞+푟)(푐−푝′푏−푝′+푞′, 푐푟′푏−푞′+푟′) = (푐−푝+푝−푞+max{−푝+푞,−푝′}푏−푝′+푞′+푝′+max{−푝+푞,−푝′}, 푐푟+푞−푟+max{−푞+푟,푟′}푏푟′−푞′−푟′+max{−푞+푟,푟′}) = (푐max{−푝,−푝′−푞}푏max{−푝+푞+푞′,−푝′+푞′}, 푐max{푟,푞+푟′}푏max{푟−푞−푞′,−푞′+푟′}) = (푐max{−푝,−푝′−푞}푏푞+푞′+max{−푝,−푝′−푞}, 푐max{푟,푞+푟′}푏max{푟,푞+푟′}−푞−푞′) = (− max{−푝, −푝′ − 푞}, 푞 + 푞′, max{푟, 푞 + 푟′})휑 = (min{푝, 푝′ + 푞}, 푞 + 푞′, max{푟, 푞 + 푟′})휑 = ((푝, 푞, 푟)(푝′, 푞′, 푟′))휑.

Furthermore, 휑 is injective since

(푝, 푞, 푟)휑 = (푝′, 푞′, 푟′)휑 ⇒ (푐−푝푏−푝+푞, 푐푟푏−푞+푟) = (푐−푝′푏−푝′+푞′, 푐푟′푏−푞′+푟′)

Solutions to exercises • 227 ⇒ (−푝 = −푝′) ∧ (−푝 + 푞 = −푝′ + 푞′) ∧ (푟 = 푟′) ⇒ (푝, 푞, 푟) = (푝′, 푞′, 푟′). So 휑 embeds 퐾 into 퐵×퐵. Finally, as 푝 and 푞 range over ℕ∪{0}, clearly −푝 −푝+푞 (푝, 푞, 푞)휑휋1 = 푐 푏 ranges over 퐵, and as 푞 and 푟 range over 푟 −푞+푟 ℕ ∪ {0}, clearly (푞, 푞, 푟)휑휋2 = 푐 푏 ranges over 퐵. So im 휑 projects surjectively to both copies of 퐵, and so 퐾 is a subdirect product of two copies of 퐵. 5.13 a) Since BR(푀, 휑) is generated by 퐴 ∪ {푏, 푐}, every element is represented by some word 푢 ∈ (퐴 ∪ {푏, 푐})∗. Using the defining relations (푏푐, 휀), we can delete any subword 푏푐. Then, using defining relations of the form (푏푎, (푎휑)푏), we can replace any subword 푏푎 by (푎휑)푏 and any subword 푎푐 by 푐(푎휑). Iterating this process, we eventually find a word 푣 containing no subwords 푏푐, 푏푎 or 푎푐: that is, 푣 = 푐훾푤푏훽 for some 훾, 훽 ∈ ℕ ∪ {0} and 푤 ∈ 퐴∗.

b) i) Suppose that 훾 = 훾′, 훽 = 훽′, and 푤 =푀 푤′. Then there is a sequence of elementary 휌-transitions from 푤 to 푤′. Since 휌 is a subset of the defining relations in (5.15), 푤 and 푤′ represent the same element of BR(푀, 휑). Hence 푐훾푤푏훽 and 푐훾푤′푏훽 represent the same element of BR(푀, 휑). ii) It is easy to prove that for all defining relations (푢, 푣) in (5.15), we have 푢휓 = 푣휓 and so 휓 is well-defined. Suppose now that 푐훾푤푏훽 and 푐훾′푤′푏훽′ represent the same element of BR(푀, 휑). Then (푐훾푤푏훽)휓 = (푐훾′푤′푏훽′)휓. Thus 훾 훽 (훾, 푤, 훽) = (0, 1푀, 0)((푐 푤푏 )휓) 훾′ 훽′ = (0, 1푀, 0)((푐 푤′푏 )휓) = (훾, 푤′, 훽), and so 훾 = 훾′ and 훽 = 훽′. c) Define a map 휗 ∶ 푀 → BR(푀, 휑) by 푤휗 = 푤. This is clearly a homomorphism, and 0 0 0 0 푤휑 =BR(푀,휑) 푤′휑 ⇒ 푐 푤푏 =BR(푀,휑) 푐 푤′푏 ⇒ 푤 =푀 푤′ by parts a) and b). Hence 휗 is injective and so 푀 embeds into BR(푀, 휑). 5.14 Let 푆 = BR(푀, 휑). We aim to show that 푆푥푆 = 푆 for all 푥 ∈ 푆. Suppose 푥 = 푐훾푤푏훽, where 푤 ∈ 푀. Let 푐훿푢푏휁 be an arbitrary element of 푆. Let 푝 = 푐훿푢푏훾+1 and 푞 = 푐훽+1푏휁. Then 푝푥푞 = 푐훿푢푏훾+1푐훾푤푏훽푐훽+1푏휁 훿 휁 =푆 푐 푢푏푤푐푏 훿 휁 =푆 푐 푢푏푐(푤휑)푏 훿 휁 =푆 푐 푢푏푐푏 훿 휁 =푆 푐 푢푏 .

Solutions to exercises • 228 So 푐훿푢푏휁 = 푝푥푞 ∈ 푆푥푆. Since 푐훿푢푏휁 ∈ 푆 was arbitrary, 푆 = 푆푥푆. Hence any ideal of 푆 must be 푆 itself. So 푆 is simple.

Exercises for chapter 6 [See page 124 for the exercises.] 6.1 For clarity, let 휄 ∶ 푆 → 퐺 and 휄′ ∶ 푆 → 퐺′ be the embedding maps. Define 휓 ∶ 퐺 → 퐻 by (푥휄)(푦휄)−1휑 = (푥휄′)(푦휄′)−1 for 푥, 푦 ∈ 푆. Let 푥1, 푥2, 푦1, 푦2 ∈ 푆. Then −1 −1 (푥1휄)(푦1휄) = (푥2휄)(푦2휄)

⇔ (푥1휄)(푦2휄) = (푥2휄)(푦1휄)

⇔ 푥1푦2 = 푥2푦1 [since 휄 is an injective homomorphism]

⇔ (푥1휄′)(푦2휄′) = (푥2휄′)(푦1휄′) [since 휄′ is an injective homomorphism] −1 −1 ⇔ (푥1휄′)(푦1휄′) = (푥2휄′)(푦2휄′) −1 −1 ⇔ ((푥1휄)(푦1휄) )휓 = ((푥2휄)(푦2휄) )휓. The forward implication shows 휓 is well-defined; the reverse implication shows it is injective. Furthermore

−1 −1 ((푥1휄)(푦1휄) )휓((푥2휄)(푦2휄) )휓 −1 −1 = (푥1휄′)(푦1휄′) (푥2휄′)(푦2휄′) [by definition of 휓] −1 −1 = (푥1휄′)(푥2휄′)(푦1휄′) (푦2휄′) [by commutativity] −1 = (푥1푥2)휄′((푦2푦1)휄′) [by inverses in 퐻] −1 = (((푥1푥2)휄)(푦2푦1휄) )휓 [by definition of 휓] −1 −1 = ((푥1휄)(푥2휄)(푦1휄) (푦2휄) )휓 [by inverses in 퐺] −1 −1 = (((푥1휄)(푦1휄) )((푥2휄)(푦2휄) ))휓, [by commutativity] so 휓 is a homomorphism. Finally, let 푠휄 ∈ 푆휄. Then for arbitrary 푧 ∈ 푆, (푠휄)휓 = ((푠푧휄)(푧휄)−1)휓 = ((푠푧휄′)(푧휄′)−1) = 푠휄′, so 휓 is clearly maps 푆휄 surjectively to 푆휄′. 6.2 Fix 푥 ∈ 퐼. For 푠 ∈ 푆 ∖ 퐼. Define 푠휑̂ to be (푥휑)−1((푥푠)휑); notice that 푥푠 ∈ 퐼 since 퐼 is an ideal. Now, for 푠′ ∈ 푆 and 푦 ∈ 퐼, (푠휑)(푦̂ 휑)̂ = (푥휑)−1((푥푠)휑)(푦휑) [by definition of 휑̂] = (푥휑)−1((푥푠푦)휑) [since 휑 is a homomorphism] = (푠푦)휑;̂ [by definition of 휑̂]

Solutions to exercises • 229 furthermore, (푠휑)(푦̂ 휑)̂ = (푠푦)휑̂ by commutativity of 푆 and 퐺. For 푠, 푠′ ∈ 푆,

(푠휑)(푠′̂ 휑)̂ = (푥휑)−1((푥푠)휑)(푥휑)−1((푥푠′)휑) [by definition of 휑̂] = (푥휑)−1(푥휑)−1((푥푠)휑)((푥푠′)휑) [since 퐺 is abelian] = (푥휑)−1(푥휑)−1((푥푠푥푠′)휑) [since 휑 is a homomorphism] = (푥휑)−1(푥휑)−1((푥푥푠푠′)휑) [since 푆 is commutative] = (푥휑)−1(푥휑)−1(푥휑)((푥푠푠′)휑) [since 휑 is a homomorphism and 푥, 푥푠푠′ ∈ 퐼] −1 −1 = (푥휑) ((푥푠푠′)휑) [since (푥휑) (푥휑) = 1퐺] = (푠푠′)휑.̂ [by definition of 휑̂]

Together with the fact that 휑 is a homomorphism, this shows that 휑̂ is a homomorphism. Finally, suppose 휓 ∶ 푆 → 퐺 is a homomorphism extending 휑. Then (푥푠)휓 = (푥휓)(푠휓) for any 푠 ∈ 푆 ∖ 퐼. Hence (푥푠)휑 = (푥휑)(푠휓) since 푥, 푥푠 ∈ 퐼, and so 푠휓 = (푥휑)−1((푥푠)휑) = 푠휑̂. Hence 휓 = 휑̂ and so 휑̂ is the unique extension of 휑 to 푆. 6.3 Let 푑 = gcd(푆); this is well-defined since 푆 ≠ {0}. Then if 푥 ∈ 푆, then 푥 = 푑푘 ∈ 푑ℕ, so 푆 ⊆ 푑ℕ. Furthermore, there exist 푧1, … , 푧푛 ∈ 푆 and 푘1, … , 푘푛 ∈ ℤ such that 푘1푧1 + 푘2푧2 + ⋯ + 푘푛푧푛 = 푑, hence moving all the terms where 푘푖 is negative to the right of the equality, we get 푠 + 푑 = 푠′ for two elements 푠 and 푠′ of 푆. Suppose 푠 = 푑푡 and 푠′ = 푑푡′. Now let 푛 ∈ ℕ with 푛 ⩾ (푡 − 1)푡 + (푡 − 1); we aim to prove that 푑푛 ∈ 푆. Let 푛 = 푞푡 + 푟, where 푞 ∈ ℕ and 0 ⩽ 푟 < 푡. Then 푞 ⩾ (푡 − 1) ⩾ 푟 and so 푞 − 푟 ⩾ 0. Now, 푛 = (푟푡 + 푟) + (푞 − 푟)푡 = 푟(푡 + 1) + (푞 − 푟)푡, and so 푑푛 = 푟(푑푡 + 푑) + (푞 − 푟)푡푑 = 푟(푠 + 푑) + (푞 − 푟)푠 = 푟푠′ + (푞 − 푟)푠 ∈ 푆. Thus, if 푥 ∈ 푑ℕ ∖ 푆, then 푥 = 푑푛 for 푛 < (푡 − 1)푡 + (푡 − 1). Thus 푑ℕ ∖ 푆 is finite. 6.4 If 푆 = {0}; then all three conditions hold. So assume 푆 ≠ {0} and suppose 푆 contains both a positive integer 푝 and a negative integer 푛. Let 푆+ = { 푠 ∈ 푆 ∶ 푠 > 0 } and 푆− = { 푠 ∈ 푆 ∶ 푠 < 0 }; clearly 푆+ and 푆− are subsemigroups of 푆. Let 푑 = gcd(푆), 푑+ = gcd(푆+), and 푑− = gcd(푆−). Clearly, 푑 ⩽ 푑+ and 푑 ⩽ 푑−. Since 푑 = 푠 − 푠′ for some 푠, 푠′ ∈ 푆, we have 푑 = (푠 + 푘푝) − (푠′ + 푘푝) = (푠 + 푘푛) − (푠′ + 푘푛) for all 푘 ∈ ℕ. Thus 푑 is both the difference between two elements of 푆+ and the difference between two elements of 푆−. Hence 푑 ⩾ 푑+ and 푑 ⩾ 푑− and hence 푑 = 푑+ = 푑−. Thus 푆+ ⊆ 푑ℕ and 푆− ⊆ −푑ℕ, and 푑ℕ ∖ 푆+ and −푑ℕ ∖ 푆− are finite. Hence 푑푘, 푑(푘 + 2) ∈ 푆+ ⊆ 푆 and −푑(푘 + 1) ∈ 푆− ⊆ 푆 for large 푘. Hence 푑 = 푑(푘 + 2) − 푑(푘 + 1) ∈ 푆 and −푑 = 푑푘−푑(푘+1) ∈ 푆. So 푑ℤ ⊆ 푆 ⊆ 푆− ∪{0}∪푆+ ⊆ −푑ℕ∪{0}∪푑ℕ = 푑ℤ. Hence 푆 = 푑ℤ is a subgroup of ℤ.

Solutions to exercises • 230 6.5 a) From the definition, ∼ is clearly reflexive and symmetric. Suppose 훼 ∼ 훽 and 훽 ∼ 훾. Then there exist 훿 and 휁 with 훿 ⊆ 훼, 훿 ⊆ 훽, 휁 ⊆ 훼, and 휁 ⊆ 훽. Let 휂 = 휁휁−1훿. Then dom 휂 ⊆ dom 휁 and for any 푥 ∈ dom 휂, we have 푥휂 = 푥휁휁−1훿 = 푥훿 = 푥훽 = 푥휁 and so 휂 ⊆ 훿 ⊆ 훼 and 휂 ⊆ 휁 ⊆ 훾. Hence 훼 ∼ 훾. Therefore ∼ is transitive. Suppose 훼1 ∼ 훽1 and 훼2 ∼ 훽2. Then there exist 훿1 and 훿2 with 훿1 ⊆ 훼1, 훿1 ⊆ 훽1, 훿2 ⊆ 훼2 and 훿2 ⊆ 훽2. Hence 훿1훿2 ⊆ 훼1훼2 and 훿1훿2 ⊆ 훽1훽2. Hence 훼1훼2 ∼ 훽1훽2. Therefore ∼ is a congruence. b) Let 훼, 훽 ∈ 푇. Let 휁 = 훼−1훽 and 휂 = 훽훼−1. Then 훼휁 = 훼훼−1훽 ⊆ 훽 and so 훼휁 ∼ 훽; similarly 휂훼 = 훽훼−1훼 ⊆ 훽 and so 휂훼 ∼ 훽. Thus for any [훼]∼, [훽]∼ ∈ 퐺, there exist [휁]∼, [휂]∼ ∈ 퐺 with [훼]∼[휁]∼ = [휂]∼[훼]∼ = [훽]∼; hence [훼]∼퐺 = 퐺[훼]∼ = 퐺 for any [훼]∼ ∈ 퐺. Thus 퐺 is a group. c) Let 훼, 훽 ∈ 푇. Then im 훼 is a left ideal of 푆 by Exercise 5.8(a) and dom 훽 is a left ideal of 푆 since 훽 is a partial right transformation. Since 푆 is right-reversible, im 훼 ∩ dom 훽 ≠ ∅. Hence 훼훽 ≠ ∅. −1 Since 푇 is generated by the non-empty elements 휌푥 and 휌푥 , we see that 푇 does not contain the empty relation.

d) Suppose 푥휓 = 푦휓; then [휌푥]∼ = [휌푦]∼ and so 휌푥 ∼ 휌푦. Then there exists 훿 ∈ 푇 such that 훿 ⊆ 휌푥 and 훿 ⊆ 휌푦. By the previous paragraph, 훿 is not the empty relation. So let 푧 ∈ dom 훿. Then 푧휌푥 = 푧휌푦. Thus 푧푥 = 푧푦 and so 푥 = 푦 by cancellativity. Hence 휓 ∶ 푆 → 퐺 is a monomorphism and so 푆 is group-embeddable. 6.6 Let (푚, 푛), (푝, 푞), (푟, 푠) ∈ 푆. Then (푚, 푛)((푝, 푞)(푟, 푠)) = (푚, 푛)(푝 + 푟, 2푟푞 + 푠) = (푚 + 푝 + 푟, 2푝+푟푛 + 2푟푞 + 푠) = (푚 + 푝 + 푟, 2푟(2푝푛 + 푞) + 푠) = (푚 + 푝, 2푝푛 + 푞)(푟, 푠) = ((푚, 푛)(푝, 푞))(푟, 푠); thus the multiplication is associative. 푚1 Let (푚1, 푛1), (푚2, 푛2) ∈ 푆. Let 푝1 = 푚2, 푞1 = 2 푛2, 푝2 = 푚1, and 푚2 푞2 = 2 푛2. Then

푝1 (푚1, 푛1)(푝1, 푞1) = (푚1 + 푝1, 2 푛1 + 푞1) 푚2 푚1 = (푚1 + 푚2, 2 푛2 + 2 푛2) and

푝2 (푚2, 푛2)(푝2, 푞2) = (푚2 + 푝2, 2 푛2 + 푞2) 푚1 푚2 = (푚2 + 푚1, 2 푛2 + 2 푛2);

Solutions to exercises • 231 so (푚1, 푛1)(푝1, 푞1) = (푚2, 푛2)(푝2, 푞2). Since (푚1, 푛1) and (푚2, 푛2) were arbitrary, 푆 is left-reversible. Suppose 푆 is right-reversible. Then (1, 0) and (1, 1) have a common left multiple. Hence there exist elements (푝1, 푞1) and (푝2, 푞2) such that (푝1, 푞1)(1, 0) = (푝2, 푞2)(1, 1). Thus (푝1 + 1, 2푞1) = (푝2 + 1, 2푞2 + 1), which is a contradiction, since 2푞1 is even and 2푞2 +1 is odd. Therefore 푆 is not right-reversible.

Exercises for chapter 7 [See pages 142–143 for the exercises.] 7.1 Let 푀 be a group. Then 푀 is simple and so 푀푥푀 = 푀 for all 푥 ∈ 푀. Now suppose 푀푥푀 = 푀 for all 푥 ∈ 푀. Then for each 푥 ∈ 푀, there exists 푝, 푞 ∈ 푀 such that 푝푥푞 = 1푀. Hence 푥 J 1푀 and so 푥 H 1푀 by Proposition 7.1. Thus 푥 lies in the group of units of 푀. So all elements of 푀 are invertible and so 푀 is a group.

7.2 In finite semigroups, J = D, so 퐽푥 = 퐷푥. Since 퐷푥 is non-trivial, it contains some element 푧 ≠ 푥 such that 푧 R 푥. That is, there exist 푝, 푞 ∈ 푆1 such that 푥푝 = 푧 and 푧푞 = 푥; notice that 푝, 푞 ∈ 푆 since 푥 ≠ 푧. Hence 푥푝푞 = 푥, and so 푥(푝푞)푘 = 푥 for all 푘 ∈ ℕ. Since 푆 is finite, there is some ℓ ∈ ℕ such that (푝푞)ℓ is idempotent. Let 푦 = (푝푞)ℓ; 2 then 푦 = 푦 and 푥푦 = 푥. By the ordering of J-classes, 퐽푥 = 퐽푥푦 ⩽ 퐽푦. Since 푦 is idempotent and thus regular, every element of 퐷푦 = 퐽푦 is regular by Proposition 3.19.

7.3 a) Let 푆 be a finite nilsemigroup. Let 푛 = |푆|. Let 푥1, … , 푥푛+1 ∈ 푆. Consider the 푛 + 1 products

푥1, 푥1푥2, … , 푥1 ⋯ 푥푛, 푥1 ⋯ 푥푛+1. Since |푆| = 푛, at least two of these 푛 + 1 products must be equal: that is, 푥1 ⋯ 푥푘 = 푥1 ⋯ 푥푘+ℓ for some 푘 ∈ {1, … , 푛} and ℓ ∈ {1, … , 푛 + 1 − 푘}. Hence 푚 푥1 ⋯ 푥푘 = 푥1 ⋯ 푥푘푥푘+1 ⋯ 푥푘+ℓ = 푥1 ⋯ 푥푘(푥푘+1 ⋯ 푥푘+ℓ) for all 푚 ∈ ℕ. Since 푆 is a nilsemigroup, there is some 푚 ∈ ℕ with 푚 푚 (푥푘+1 ⋯ 푥푘+ℓ) = 0. Thus 푥1 ⋯ 푥푘 = 푥1 ⋯ 푥푘(푥푘+1 ⋯ 푥푘+ℓ) = 0 푛 and so 푥1 ⋯ 푥푛 = 0 (since 푘 ⩽ 푛). Therefore 푆 = {0} and so 푆 is nilpotent.

b) Let 푆 = {0}∪{ 푥푖,푗 ∶ 푖 ∈ ℕ, 푗 ⩽ 푖 }. Define a product on 푆 as follows: 푥 if 푖 = 푘 and 푗 + ℓ ⩽ 푖, 푥 푥 = { 푖,푗+ℓ 푖,푗 푘,ℓ 0 otherwise,

푥푖,푗0 = 0푥푖,푗 = 00 = 0.

Solutions to exercises • 232 It is easy to check that this operation is associative. For any 푥푖,푗 ∈ 푆, 푖+1 we have 푥푖,푗 = 0 since 푗(푖 + 1) > 푖. Thus 푆 is a nilsemigroup. 푛 푛 However, for any 푛 ∈ ℕ, we have 푥푛,1 = 푥푛,푛 ≠ 0, so 푆 ≠ {0}. Thus 푆 is not nilpotent. 7.4 a) Let 푥′, 푦′ ∈ 퐽휑. Then 푥′ = 푥휑 and 푦′ = 푦휑 for some 푥, 푦 ∈ 퐽. Thus there exist 푝, 푞, 푟, 푠 ∈ 푆1 such that 푝푥푞 = 푦 and 푟푦푠 = 푥. Then (푝휑)푥′(푞휑) = 푦′ and (푟휑)푦′(푠휑) = 푥′ (where we view 1휑 as the identity of (푆′)1) and so 푥′ J 푦′. So all elements of 퐽휑 are contained within a single J-class 퐽′ of 푆′.

b) Let 푥′ ∈ 퐽′. Then 푥′ = 푥휑 for some 푥 ∈ 푆. Let 퐽 = 퐽푥. Since all elements of 퐽휑 are J-related by part a), we see that 퐽휑 ⊆ 퐽′. Let 퐽 be minimal such that 퐽휑 ⊆ 퐽′. Let 퐼 = 푆1퐽푆1. Then 퐼 = 푆1푥푆1 for any 푥 ∈ 퐽, by the definition of J. Let 푦′ ∈ 퐽′. Then 푦′ J 푥휑 and so there exist 푝′, 푞′ ∈ (푆′)1 such that 푦′ = 푝′(푥휑)푞′. Therefore 푦′ ∈ (푆′)1(푥휑)(푆′)1 = (푆1푥푆1)휑 = 퐼휑 since 휑 is surjective. So 퐽′ ⊆ 퐼휑. Let 푦 ∈ 퐼 and let 퐾 = 퐽푦. By part a), there exists some J- class 퐾′ of 푆′ such that 퐾휑 ⊆ 퐾′. We now want to prove that 푦 ∉ 퐽 implies 푦휑 ∉ 퐽′. So suppose that 푦 ∉ 퐽. Then 퐾 = 퐽푦 < 퐽. Therefore 퐾휑 ⊈ 퐽′ since 퐽 was chosen to be minimal such that 퐽휑 ⊆ 퐽′. Hence 퐾′ ≠ 퐽′. Suppose, with the aim of obtaining a contradiction, that 푦휑 ∈ 퐽′. Then there exists 푝′, 푞′, 푟′, 푠′ ∈ (푆′)1 with 푝′(푦휑)푞′ = 푥휑 and 푟′(푥휑)푠′ = 푦휑 for some 푥 ∈ 퐽. Since 휑 is surjective, this shows that 푦 J 푥 and so 푦 ∈ 퐽, which is a contradiction. Therefore 푦휑 ∉ 퐽′. Thus for any 푦 ∈ 퐼, we have 푦 ∉ 퐽 implies 푦휑 ∉ 퐽′. Hence 푦휑 ∈ 퐽′ implies 푦 ∈ 퐽, which implies 푦휑 ∈ 퐽휑. Since 퐽′ ⊆ 퐼휑, this shows that 퐽′ ⊆ 퐽휑. Thus 퐽휑 = 퐽′. 7.5 It suffices to prove this when 푇 is a subsemigroup of 푆 and when 푇 is a homomorphic image of 푆. In both cases, 푇 is finite because 푆 is, and thus for both 푆 and 푇 the property of having H being the equality relation is equivalent to aperiodic. Let 푇 be a subsemigroup of 푆. Let 푥 ∈ 푇. Since 푥 ∈ 푆 and 푆 is aperiodic, there exists 푘 ∈ ℕ such that 푥푘 = 푥푘+1. Since this is true for all 푥 ∈ 푇, the subsemigroup 푇 is aperiodic. Now let 휑 ∶ 푆 → 푇 be a surjective homomorphism. Let 푦 ∈ 푇. Then there exists 푥 ∈ 푆 such that 푥휑 = 푦. Since 푆 is aperiodic, 푥푘 = 푥푘+1 for some 푘 ∈ ℕ. Hence 푦푘 = (푥휑)푘 = 푥푘휑 = 푥푘+1휑 = (푥휑)푘+1 = 푦푘+1. Since this is true for all 푦 ∈ 푇, the semigroup 푇 is aperiodic. This completes the proof. In the free semigroup {푎}+, the relation H is the equality relation, but any finite non-trivial cyclic group is a homomorphic image of {푎}+, and in groups all elements are H-related.

Solutions to exercises • 233 7.6 Let (푠1, 푡1), (푠2, 푡2), (푠3, 푡3) ∈ 푆 ⋊휑 푇. Then

((푠1, 푡1)(푠2, 푡2))(푠3, 푡3) 푡1 = (푠1 푠2 , 푡1푡2)(푠3, 푡3) [by (7.1)] 푡1 푡1푡2 = (푠1 푠2 푠3 , 푡1푡2푡3) [by (7.1)]

푡1 푡1 푡2 = (푠1 푠2 ( 푠3 ), 푡1푡2푡3) [by the definition of a left action]

푡1 푡2 = (푠1 (푠2 푠3 ), 푡1푡2푡3) [since the action is by endomorphisms]

푡2 = (푠1, 푡1)(푠2 푠3 , 푡2푡3) [by (7.1)]

= (푠1, 푡1)((푠2, 푡2)(푠3, 푡3)); [by (7.1)] thus the multiplication (7.1) is associative. 7.7 Suppose 푀 and 푁 are groups. Then 푀 ≀ 푁 is a monoid with identity (푒, 1푁) by Proposition 7.7. Let (푓, 푛) ∈ 푀 ≀ 푁. Define 푓′ ∈ 푁 → 푀 by (푥)푓′ = ((푥푛−1)푓)−1. Then

(푓, 푛)(푓′, 푛−1) = (푓푛 푓′, 푛푛−1)

= (푒, 1푁), since

(푥)푓푛 푓′ = (푥)푓(푥푛)푓′ = (푥)푓((푥푛푛−1)푓)−1 −1 = (푥)푓((푥)푓) = 1푀, and

(푓′, 푛−1)(푓, 푛) −1 = (푓′푛 푓, 푛−1푛)

= (푒, 1푁), since

푛−1 −1 −1 (푥)푓′ 푓 = (푥)푓′(푥푛 )푓 = (푥)푓′((푥)푓′) = 1푀; thus (푓′, 푛−1) is a right and left inverse for (푓, 푛). Hence 푀 ≀ 푁 is a group. 7.8 The wreath product 푆 ≀ 푇 must be right-cancellative but is not necessarily left-cancellative. For (푓, 푠), (푔, 푡), (ℎ, 푢) ∈ 푆 ≀ 푇,

(푓, 푠)(ℎ, 푢) = (푔, 푡)(ℎ, 푢) ⇒ (푓푠 ℎ, 푠푢) = (푔푡 ℎ, 푡푢) ⇒ 푓푠 ℎ = 푔푡 ℎ ∧ 푠푢 = 푡푢 ⇒ (∀푥 ∈ 푇)((푥)푓(푥푠)ℎ = (푥)푔(푥푡)ℎ) ∧ 푠 = 푡 [since 푇 is cancellative]

Solutions to exercises • 234 ⇒ (∀푥 ∈ 푇)((푥)푓(푥푠)ℎ = (푥)푔(푥푠)ℎ) ∧ 푠 = 푡 [substituting 푠 = 푡] ⇒ (∀푥 ∈ 푇)((푥)푓 = (푥)푔) ∧ 푠 = 푡 [since 푆 is cancellative] ⇒ 푓 = 푔 ∧ 푠 = 푡.

Now let 푆 = 푇 = ℕ ∪ {0} (under +) and define a map 푓 ∶ 푆 → 푇 by (0)푓 = 1 and (푥)푓 = 0 for all 푥 ∈ 푇 ∖ {0} and a map 푔 ∶ 푆 → 푇 by (푥)푔 = 0 for all 푥 ∈ 푇. Then

(푔, 1)(푓, 1) = (푔1 푓, 2) = (푔1 푔, 2) = (푔, 1)(푔, 1)

since (푥)푔1 푓 = (푥)푔 + (푥 + 1)푓 = 0 + 0 = (푥)푔 + (푥 + 1)푔 = (푥)푔1 푔 for all 푥 ∈ 푇. Hence 푆 ≀ 푇 is not left-cancellative. 7.9 This is a tedious analysis of products of three elements in 퐶(푆). Each element is either in 푆 or 푆′; there are thus eight cases. Let 푥, 푦, 푧 ∈ 푆. Then: ◆ (푥푦)푧 = 푥(푦푧), since 푆 is a subsemigroup of 푆 ∪ 푆′; ◆ (푥푦)푧′ = 푧′ = 푥푧′ = 푥(푦푧′); ◆ (푥푦′)푧 = 푦′푧 = (푦푧)′ = 푥(푦푧)′ = 푥(푦′푧); ◆ (푥푦′)푧′ = 푧′ = 푥푧′ = 푥(푦′푧′); ◆ (푥′푦)푧 = (푥푦)′푧 = ((푥푦)푧)′ = (푥(푦푧))′ = 푥′(푦푧), using associativity in 푆 for the third equality; ◆ (푥′푦)푧′ = 푧′ = 푥′푧′ = 푥′(푦푧′); ◆ (푥′푦′)푧 = 푦′푧 = (푦푧)′ = 푥′(푦푧)′ = 푥′(푦′푧); ◆ (푥′푦′)푧′ = 푧′ = 푥′푧′ = 푥′(푦′푧′). Therefore the product defined7.4 by( ) is associative.

7.10 Define a map 휓 ∶ 퐶(푀) → T푀 by 푥휓 = 휌푥 and 푥′휓 = 휏푥 for 푥 ∈ 푀. Clearly im 휑 = { 휌푥, 휏푥 ∶ 푥 ∈ 푀 }. We cannot have 푥휓 = 푦′휓, for 푥휓 is a non-constant map and 푦′휓 is a constant map. So to check injectivity, we simply check that 휓|푀 and 휓푀′ are injective:

푥휓 = 푦휓 ⇒ 휌푥 = 휌푦 ⇒ 1휌푥 = 1휌푦 ⇒ 푥 = 푦,

푥′휓 = 푦′휓 ⇒ 휏푥 = 휏푦 ⇒ 1휏푥 = 1휏푦 ⇒ 푥 = 푦.

Finally, to check that 휓 is a homomorphism, we must check the various cases of multiplication in the definition of 퐶(푀):

(푥휑)(푦′휑) = 휌푥휏푦 = 휏푦 = 푦′휑 = (푥푦′)휑

(푥′휑)(푦′휑) = 휏푥휏푦 = 휏푦 = 푦′휑 = (푥′푦′)휑

(푥′휑)(푦휑) = 휏푥휌푦 = 휏푥푦 = (푥푦)′휑.

So 휓 is an isomorphism.

Solutions to exercises • 235 7.11 Let 푥 ∈ 푀 and 푦 ∈ 퐶(푀). Then

푚 (푥)[(푦)(푓 푔)con] 푚 = ((푥)푓 푔)′ [by definition of con] = ((푥)푓)′(푥푚)푔 [by def. of the product and action]

= (푥)[(푦)푓con](푥)[(푚′)푔ext] [by definition of ext and con]

= (푥)[(푦)푓con](푥)[(푦푚′)푔ext] [by def. of the product in 퐶(푀)] 푚′ 푀 = (푥)[(푦)푓con(푦) 푔ext ] [by multiplication in 퐶(푆) ] 푚′ 푀 퐶(푀) = (푥)[(푦)푓con 푔ext ]; [by multiplication in (퐶(푆) ) ]

this proves (7.6). Next,

(푥)[(푦)푔con]

= ((푥)푔)′ [by definition of con] = (푥푦)푓((푥)푔)′ [by def. of the product in 퐶(푆)]

= (푥)[(푦)푓ext](푥)[(푦푚)푔con] [by definition of ext and con] 푚 푀 = (푥)[(푦)푓ext(푦) 푔con ] [by multiplication in 퐶(푆) ] 푚 푀 퐶(푀) = (푥)[(푦)푓ext 푔con ]; [by multiplication in (퐶(푆) ) ]

this proves (7.7). Finally,

(푥)[(푦)푔con]

= ((푥)푔)′ [by definition of con] = ((푥)푓)′((푥)푔)′ [by def. of the product in 퐶(푆)]

= (푥)[(푦)푓con](푥)[(푦푚)푔con] [by definition of con] 푚 푀 = (푥)[(푦)푓con(푦) 푔con ] [by multiplication in 퐶(푆) ] 푚 푀 퐶(푀) = (푥)[(푦)푓con 푔con ]; [by multiplication in (퐶(푆) ) ]

this proves (7.8).

Exercises for chapter 8 [See pages 168–169 for the exercises.] 8.1 a) Suppose 푤 = 푢. Then for any homomorphism 휗 ∶ 퐴+ → 푆 we have 푤휗 = 푢휗 = 푣휗 = (푣′휗)(푤휗). Then for any 푞 ∈ 푆, we have 푞(푤휗) = 푞(푣′휗)(푤휗) and so by cancellativity 푞 = 푞(푣′휗). So 푣′휗 is a right identity for 푆 and thus (by cancellativity) an identity. Let 푎 and 푏 be the first and last letters of 푣′ (which may or may not be distinct). For 푠 ∈ 푆, put 푎휗 = 푏휗 = 푠 to see that 푠 is right and left invertible. Thus 푆 is a group.

Solutions to exercises • 236 b) Suppose 푤 ≠ 푢. Since 푤 is the longest common suffix of 푢 and 푣, we know that 푢′ and 푣′ end with different letters 푎 and 푏 of 퐴. That is, 푢′ = 푢″푎 and 푣′ = 푣″푎. Let 푠, 푡 ∈ 푆. Let 휗 ∶ 퐴 → 푆 be such that 푎휗 = 푠 and 푏휗 = 푡. Then (푢″휗)푠(푤휗) = 푢휗 = 푣휗 = (푣″휗)푡(푤휗) and so (푢″휗)푠 = (푣″휗)푡 by cancellativity. Hence 푠 and 푡 have a common left multiple. Since this holds for all 푠, 푡 ∈ 푆, the semigroup 푆 is group-embeddable by Exercise 6.5. 8.2 a) Let N be the class of finite nilpotent semigroups. Let 푆 ∈ N. So 푆푛 = {0} for some 푛 ∈ ℕ. First, let 푇 be a subsemigroup of 푆. Then 푇푛 ⊆ 푆푛 = {0}; hence 푇 ∈ N. So N is closed under 핊. Second, let 휑 ∶ 푆 → 푈 be a surjective homomorphism. Then 푛 푛 푛 푈 = (푆휑) = 푆 휑 ⊆ {0푆}휑 = {0푈}. So 푈 ∈ N. Thus N is closed under ℍ. Third, let 푆 , … , 푆 be nilpotent; then 푆푛푖 = {0 } for 1 푘 푖 푆푖 some 푛푖 ∈ ℕ for each 푖 = 1, … , 푘. Let 푛 be the maximum of the various 푛푖. Then

(푆 ×…×푆 )푛 ⊆ 푆푛 ×… 푆푛 = {0 }×… {0 } = {(0 , … , 0 )}; 1 푘 1 푘 푆1 푆푘 푆1 푆푘

hence 푆1 × … × 푆푘 ∈ N. Thus N is closed under ℙfin. Therefore N is a pseudovariety. + b) Let 퐴 = {푎}. For each 푘 ∈ ℕ, let 퐼푘 = { 푤 ∈ 퐴 ∶ |푤| ⩾ 푘 }. Then + + 푘 퐼푘 is an ideal of 퐴 . Let 푆푘 = 퐴 /퐼푘; then 푆푘 = {0푆 }. So each 푆푘 ∞ 푘 is nilpotent. Let 푆 = ∏푖=1 푆푘. Let 푠 ∈ 푆 be such that (푘)푠 = 푎 ∈ 푆푘 푛 푛 for all 푘 ∈ ℕ. Then for any 푛 ∈ ℕ, we have (푛 + 1)푠 = 푎 ∈ 푆푛+1; hence (푛 + 1)푠푛 ≠ 0 , and so 푠푛 ≠ 0 for any 푛 ∈ ℕ. Thus 푆푛+1 푆 푛 푆 ≠ {0푆} for any 푛 ∈ ℕ. Hence 푆 is not nilpotent. Therefore the class of nilpotent semigroups is not closed under ℙ and so is not a variety. 8.3 Note first that we are working with algebras of type {(∘, 2), (−1, 1)}. Let 푆 be an orthodox completely regular semigroup. Let 휑 ∶ 푆 → 푇 be a surjective homomorphism. Then 푇 is regular by Proposition 4.20, and furthermore (푥휑)−1 = (푥−1휑) since homomorphisms for algebras of this type must also preserve −1. Therefore since 푆 is completely regular and thus satisfies the laws (4.2), 푇 also satisfies these laws, so 푇 is completely regular. Finally, if 푒, 푓 ∈ 푇 are idempotents, then 푒 = 푥푥−1 and 푓 = 푦푦−1 for some 푥, 푦 ∈ 푇 by Theorem 4.15. Let 푝, 푞 ∈ 푆 be such that 푝휑 = 푥 and 푞휑 = 푦. Then 푝푝−1 and 푞푞−1 are idempotent. So 푝푝−1푞푞−1 is idempotent (since 푆 is orthodox) and so (푝푝−1푞푞−1)휑 = 푥푥−1푦푦−1 = 푒푓 is idempotent. So the idempotents of 푇 form a subsemigroup and so 푇 is orthodox. Now let 푇 be a subalgebra of 푆. Then 푇 also satisfies the laws (4.2) and is thus completely regular. Finally, the set of idempotents of 푇 is the intersection of the set of idempotents of 푆, which is a subsemigroup, and 푇, which is also a subsemigroup. Hence the set of idempotents of 푇 is a subsemigroup.

Solutions to exercises • 237 Finally, let { 푆푖 ∶ 푖 ∈ 퐼 } be a collection of orthodox completely regular semigroups. Then each 푆푖 satisfies the laws (4.2) and so their product ∏푖∈퐼 푆푖 does also. The set of idempotents in ∏푖∈퐼 푆푖 is the product of the sets of idempotents in each 푆푖 and hence forms a subsemigroup. Now let 푆 be an orthodox completely regular semigroup. Then 푆 satisfies the laws (4.2). Let 푥, 푦 ∈ 푆. Note that 푥−1푥 and 푦푦−1 are idempotents, and so their product 푥−1푥푦푦−1 is idempotent since 푆 is orthodox. Thus

푥푦푦−1푥−1푥푦 = 푥푥−1푥푦푦−1푥−1푥푦푦−1푦 = 푥푥−1푥푦푦−1푦 [since 푥−1푥푦푦−1 is idempotent] = 푥푦.

Therefore 푆 satisfies the law 푥푦푦−1푥−1푥푦 = 푥푦. Now suppose 푆 satisfies the laws (4.2) and 푥푦푦−1푥−1푥푦 = 푥푦. Then 푆 is completely regular. Let 푒, 푓 ∈ 푆 be idempotents; then 푒 = 푥−1푥 and 푓 = 푦푦−1 for some 푥, 푦 ∈ 푆 by Theorem 4.15. Then

(푒푓)2 = (푥−1푥푦푦−1)2 = 푥−1푥푦푦−1푥−1푥푦푦−1 = 푥−1푥푦푦−1; [since 푥푦푦−1푥−1푥푦 = 푥푦] = 푒푓.

Hence the idempotents of 푆 form a subsemigroup and so 푆 is orthodox. 8.4 a) Let 푆 = 퐿 × 푅 be a rectangular band, where 퐿 is a left zero semigroup and 푅 is a right zero semigroup. Let 휑 ∶ 푆 → 푇 be a surjective homomorphism. Fix (ℓ, 푟) ∈ 푆. Let 퐿푇 = (퐿 × {푟})휑 and 푅푇 = ({ℓ} × 푅)휑. Notice that 퐿푇 is a left zero semigroup and 푅푇 is a right zero semigroup; hence 퐿푇 ×푅푇 is a rectangular band. Define 휓 ∶ 퐿푇 ×푅푇 → 푇 by (ℓ푡, 푟푡)휓 = ℓ푡푟푡. Let (1) (1) (2) (2) (1) (2) (1) (2) (ℓ푡 , 푟푡 ), (ℓ푡 , 푟푡 ) ∈ 퐿푇 × 푅푇. Let ℓ , ℓ ∈ 퐿 and 푟 , 푟 ∈ 푅 (푖) (푖) (푖) (푖) be such that (ℓ , 푟)휑 = ℓ푡 and (ℓ, 푟 )휑 = 푟푡 for 푖 = 1, 2. Then

(1) (1) (2) (2) (ℓ푡 , 푟푡 )휓(ℓ푡 , 푟푡 )휓 (1) (1) (2) (2) = ℓ푡 푟푡 ℓ푡 푟푡 = (ℓ(1), 푟)휑(ℓ, 푟(1))휑(ℓ(2), 푟)휑(ℓ, 푟(2))휑 = ((ℓ(1), 푟)(ℓ, 푟(1))(ℓ(2), 푟)(ℓ, 푟(2)))휑 = (ℓ(1), 푟(2))휑 = ((ℓ(1), 푟)(ℓ, 푟(2)))휑 = (ℓ(1), 푟)휑(ℓ, 푟(2))휑

Solutions to exercises • 238 (1) (2) = ℓ푡 푟푡 (1) (2) = (ℓ푡 , 푟푡 )휓 (1) (1) (2) (2) = ((ℓ푡 , 푟푡 )(ℓ푡 , 푟푡 ))휓; thus 휓 is a homomorphism. Furthermore,

(1) (1) (2) (2) (ℓ푡 , 푟푡 )휓 = (ℓ푡 , 푟푡 )휓 (1) (1) (2) (2) ⇒ ℓ푡 푟푡 = ℓ푡 푟푡 ⇒ (ℓ(1), 푟)휑(ℓ, 푟(1))휑 = (ℓ(2), 푟)휑(ℓ, 푟(2))휑 ⇒ ((ℓ(1), 푟)(ℓ, 푟(1)))휑 = ((ℓ(2), 푟)(ℓ, 푟(2)))휑 ⇒ (ℓ(1), 푟(1))휑 = (ℓ(2), 푟(2))휑 ⇒ (ℓ(1), 푟(1))휑(ℓ, 푟)휑 = (ℓ(2), 푟(2))휑(ℓ, 푟)휑 ∧ (ℓ, 푟)휑(ℓ(1), 푟(1))휑 = (ℓ, 푟)휑(ℓ(2), 푟(2))휑 ⇒ ((ℓ(1), 푟(1))(ℓ, 푟))휑 = ((ℓ(2), 푟(2))(ℓ, 푟))휑 ∧ ((ℓ, 푟)(ℓ(1), 푟(1)))휑 = ((ℓ, 푟)(ℓ(2), 푟(2)))휑 ⇒ (ℓ(1), 푟)휑 = (ℓ(2), 푟)휑 ∧ (ℓ, 푟(1))휑 = (ℓ, 푟(2))휑 (1) (2) (1) (2) ⇒ ℓ푡 = ℓ푡 ∧ 푟푡 = 푟푡 (1) (1) (2) (2) ⇒ (ℓ푡 , 푟푡 ) = (ℓ푡 , 푟푡 ), so 휓 is injective. Finally, 휓 is surjective since

im 휓 = 퐿푇푅푇 = (퐿 × {푟})휑({ℓ} × 푅)휑 = ((퐿 × {푟})({ℓ} × 푅))휑 = (퐿 × 푅)휑 = 푇.

Hence 푇 is isomorphic to the rectangular band 퐿푇 × 푅푇; thus 푇 ∈ RB. So RB is closed under forming homomorphic images. Now let 푇 be a subsemigroup of 푆. Let 퐿푇 = { ℓ ∈ 퐿 ∶ (∃푟 ∈ 푅)((ℓ, 푟) ∈ 푇) } and 푅푇 = { 푟 ∈ 푅 ∶ (∃ℓ ∈ 퐿)((ℓ, 푟) ∈ 푇) }. Notice that 퐿푇 is also a left zero semigroup and 푅푇 is also a right zero semigroup. Clearly 푇 ⊆ 퐿푇 × 푅푇; we now establish the opposite inclusion. Let (ℓ푡, 푟푡) ∈ 퐿푇 × 푅푇. Then there exist 푟 ∈ 푅 and ℓ ∈ 퐿 such that (ℓ푡, 푟) ∈ 푇 and (ℓ, 푟푡) ∈ 푇. Thus (ℓ푡, 푟푡) ∈ (ℓ푡, 푟)(ℓ, 푟푡) ∈ 푇. So 푇 = 퐿푇 ×푅푇 is a rectangular band. So RB is closed under taking subsemigroups. Finally, let { 푆푖 ∶ 푖 ∈ 퐼 } be a collection of rectangular bands. Then 푆푖 ≃ 퐿푖 × 푅푖 for some left zero semigroup 퐿푖 and right zero semigroup 푅푖, for each 푖 ∈ 퐼. Then

∏ 푆푖 = ∏(퐿푖 × 푅푖) ≃ (∏ 퐿푖) × (∏ 푅푖). 푖∈퐼 푖∈퐼 푖∈퐼 푖∈퐼

Solutions to exercises • 239 Since ∏푖∈퐼 퐿푖 is a left zero semigroup and ∏푖∈퐼 푅푖 is a right zero semigroup, ∏푖∈퐼 푆푖 ∈ RB. Hence RB is closed under forming direct products. Thus RB is a variety. b) Let 푆 = 퐿×푅 be a rectangular band. Let 푥 = (푙1, 푟1) and 푦 = (푙2, 푟2). Then 푥푦푥 = (푙1, 푟1)(푙2, 푟2)(푙1, 푟1) = (푙1, 푟1) = 푥. So 푆 satisfies this law. Suppose 푆 satisfies the law 푥푦푥 = 푥. Fix some 푡 ∈ 푆. Let 퐿 = 푆푡 and 푅 = 푡푆. Then for any 푝푡, 푝′푡 ∈ 퐿, we have 푝푡푝′푡 = 푝푡 by the law (with 푥 = 푡 and 푦 = 푝′). So 퐿 is a left zero semigroup and similarly 푅 is a right zero semigroup. Furthermore, for any 푝, 푞.푟 ∈ 푆,

푝푟 = 푝푞푝푟 [by the law with 푥 = 푝 and 푦 = 푞] } = 푝푞푟푞푝푟 [by the law with 푥 = 푞 and 푦 = 푟] (S.23) } = 푝푞푟. [by the law with 푥 = 푟 and 푦 = 푞푝] } Define 휓 ∶ 푆 → 퐿 × 푅 by 푝휓 = (푝푡, 푡푝). Then

(푝휓)(푞휓) = (푝푡, 푡푝)(푞푡, 푡푞) = (푝푡, 푡푞) = (푝푞푡, 푡푝푞) [using (S.23) in both components] = (푝푞)휓,

so 휓 is a homomorphism. Notice that this also shows that for any 푝푡 ∈ 퐿, 푡푞 ∈ 푅, we have (푝푞)휓 = (푝푡, 푡푞); thus 휓 is surjective. Finally, for any 푝, 푞 ∈ 푆,

푝휓 = 푞휓 ⇒ (푝푡, 푡푝) = (푞푡, 푡푞) ⇒ 푝푡 = 푞푡 ∧ 푡푝 = 푡푞 ⇒ 푝푡푝 = 푞푡푝 ∧ 푞푡푝 = 푞푡푞 ⇒ 푝푡푝 = 푞푡푞 ⇒ 푝 = 푞, [applying the law on both sides]

so 휓 is injective. So 푆 is [isomorphic to] a rectangular band and so 푆 ∈ RB. c) Any rectangular band satisfies the law 푥푦푧 = 푥푧 by (S.23). Every element of a rectangular band is idempotent, so 푥2 = 푥 is also satisfied. Let 푆 satisfy the laws 푥2 = 푥 and 푥푦푧 = 푥푧. To prove that 푆 is a rectangular band, follow the reasoning in part b) with the following minor differences: First, 퐿 is a left zero semigroup since 푝푡푝′푡 = 푝푡푡 = 푝푧 by applying first 푥푦푧 = 푥푧 and then 푥2 = 푥;

Solutions to exercises • 240 similarly 푅 is a right zero semigroup. Second, to prove 휓 is a homomorphism, apply 푥푦푧 = 푥푧 to both components. Finally, the last step in proving 휓 is injective becomes 푝푡푝 = 푞푡푞 ⇒ 푝2 = 푞2 ⇒ 푝 = 푞, by applying first 푥푦푧 = 푥푧 and then 푥2 = 푥. d) Let 푆 be a non-trivial null semigroup. Then for any 푥, 푦, 푧 ∈ 푆, we have 푥푦푧 = 0푆 and 푥푧 = 0푆. However, 푆 is not a rectangular band 2 because 푥 ≠ 푥 for any 푥 ≠ 0푆. 8.5 Let 푆 = 퐺 × 퐿 × 푅, where 퐺 is a group, 퐿 is a left zero semigroup, and 푅 is a right zero semigroup. Let 휑 ∶ 푆 → 푇 be a homomorphism. Fix (1퐺, ℓ, 푟) ∈ 푆. Let 퐻 = (퐺 × {ℓ} × {푟})휑, 퐿푇 = ({1퐺} × 퐿 × {푟})휑 and 푅푇 = ({1퐺} × {ℓ} × 푅)휑. Reasoning parallel to Example 8.4 shows that 푇 ≃ 퐻 × 퐿푇 × 푅푇. Notice that (푔, ℓ, 푟)−1 = (푔−1, ℓ, 푟). Let 푇 be a subalgebra of 푆. Let 퐻 = { 푔 ∈ 퐺 ∶ (∃(ℓ, 푟) ∈ 퐿 × 푅)((푔, ℓ, 푟) ∈ 푇 }. We first prove that if (푔, ℓ, 푟) ∈ 푇, then 퐻 × {(ℓ, 푟)} ⊆ 푇. Let ℎ ∈ 퐻; then (ℎ, ℓ′, 푟′) ∈ 푇 for some ℓ′ ∈ 퐿, 푟′ ∈ 푅. Hence 푇 contains

(푔, ℓ, 푟)(푔, ℓ, 푟)−1(ℎ, ℓ′, 푟′)(푔, ℓ, 푟)(푔, ℓ, 푟)−1 = (푔푔−1ℎ푔푔−1, ℓ, 푟) = (ℎ, ℓ, 푟),

and thus 퐻 × {(ℓ, 푟)} ⊆ 푇. Now reason as in Example 8.4 to see that 푇 = 퐻 × 퐿푇 × 푅푇 and thus 푇 ∈ X. Let { 푆푖 ∶ 푖 ∈ 퐼 } be a collection of semigroups in X. Then for all 푖 ∈ 퐼, we have 푆푖 ≃ 퐺푖 × 퐿푖 × 푅푖 for some group 퐺푖, left zero semigroup 퐿푖 and right zero semigroup 푅푖. Hence

∏ 푆푖 ≃ ∏(퐺푖 × 퐿푖 × 푅푖) ≃ (∏ 퐺푖) × (∏ 퐿푖) × (∏ 푅푖); 푖∈퐼 푖∈퐼 푖∈퐼 푖∈퐼 푖∈퐼

since ∏푖∈퐼 퐺푖 is a group, ∏푖∈퐼 퐿푖 is a left zero semigroup, and ∏푖∈퐼 푅푖 is a right zero semigroup, we see that ∏푖∈퐼 푆푖 is [isomorphic to] the direct product of a group and a rectangular band. So ∏푖∈퐼 푆푖 ∈ X. Let 푆 = 퐺 × 퐿 × 푅, where 퐺 is a group, 퐿 is a left zero semigroup, and 푅 is a right zero semigroup. Let 푥 = (푔, ℓ, 푟) and 푦 = (푔′, ℓ′, 푟′). Then

푥푥−1 = (푔, ℓ, 푟)(푔−1, ℓ, 푟)

= (1퐺, ℓ, 푟) = (푔−1, ℓ, 푟)(푔, ℓ, 푟) = 푥−1푥

and

푥−1푦푦−1푥 = (푔−1, ℓ, 푟)(ℎ, ℓ′, 푟′)(ℎ−1, ℓ′, 푟′)(푔, ℓ, 푟) = (푔−1ℎℎ−1푔, ℓ, 푟)

Solutions to exercises • 241 = (1퐺, ℓ, 푟) = (푔−1, ℓ, 푟)(푔, ℓ, 푟) = 푥−1푥.

So 푆 satisfies these laws. Now suppose that 푆 satisfies the given laws. For any 푥, 푦 ∈ 푆, we have 푥 = 푥푥−1푥 = 푥푥−1푦푦−1푥 ∈ 푆푦푆. So 푆 is simple by the analogue of Lemma 3.7 for simple semigroups. Let 푒, 푓 ∈ 푆 be idempotents; then 푒 = 푥푥−1 and 푓 = 푦푦−1 for some 푥, 푦 ∈ 푆. Then 푒푓푒 = 푥푥−1푦푦−1푥푥−1 = 푥푥−1푥푥−1 = 푥푥−1 = 푒. So the idempotents of 푆 form a rectangular band by Example 8.4. Since rectangular bands are completely simple, they contain primitive idempotents. Hence 푆 contains a primitive idempotent. So 푆 is completely simple. Since the idempotents of 푆 form a subsemigroup, 푆 is orthodox. Hence 푆 is a direct product of a rectangular band and a group by Exercise 5.6(b).

8.6 Let 푆 ∈ ⋂푖∈퐼 V푖. Then 푆 ∈ V푖 for all 푖 ∈ 퐼. Let 푇 be a homomorphic image (respectively, subalgebra) of 푆. Since each V푖 is a pseudovariety, 푇 ∈ V푖 for all 푖 ∈ 퐼. Hence 푇 ∈ ⋂푖∈퐼 V푖. So ⋂푖∈퐼 V푖 is closed under forming homomorphic images and subalgebras. Now let 푆1, … , 푆푛 ∈ ⋂푖∈퐼 V푖. Then 푆푗 ∈ V푖 for each 푖 ∈ 퐼 and 푗 = 1, … , 푛. So 푆1 ×…×푆푛 ∈ V푖 for each 푖 ∈ 퐼 and so 푆1 × … × 푆푛 ∈ ⋂푖∈퐼 V푖. So ⋂푖∈퐼 V푖 is closed under forming finitary direct products. Therefore ⋂푖∈퐼 V푖 is a pseudovariety. 8.7 Let V be an S-pseudovariety of semigroups. Then

푆 ∈ (VMon)Sg 1 ⇒ 푆 ∈ VMon [by (8.7)] ⇒ 푆1 is a monoid in V ⇒ 푆 ∈ V. [since 푆 is closed under taking subsemigroups]

Let V be the S-pseudovariety of rectangular bands. Then VMon = 1, since the only monoid that is a rectangular band is the trivial monoid, and so (VMon)Sg = VS(1) = 1 ≠ V. 8.8 Let 푆 be a completely regular semigroup. Let 푠 ∈ 푆. By Theorem 4.15, ̅ 푠 lies in a subgroup 퐺 of 푆. If 휗 ∶ Ω{푥}S → 푆 is such that 푥휗 = 푠, then 푥휔휗 is the idempotent power of 푆, which much be the identity of 퐺. So 푥휔+1휗 = (푥휔휗)(푥휗) = 1푠 = 푠 = 푥휗, so 푆 satisfies the pseudoidentity 푥휔+1 = 푥. Now suppose that 푆 satisfies 푥휔+1 = 푥. Let 푠 ∈ 푆 and choose ̅ 휔 푘 휗 ∶ Ω{푥}S → 푆 with 푥휗 = 푠. Then 푥 휗 = 푠 for some 푘 ∈ ℕ. So 푠푘 = 푥휔휗 = 푥휗 = 푠. Thus 푠 lies in the cyclic group {푠, 푠2, … , 푠푘−1}. Hence every element of 푆 lies in a subgroup and so 푆 is completely regular by Theorem 4.15. 8.9 Let 푆 be a completely simple semigroup; thus 푆 = M[퐺; 퐼, 훬; 푃] for some group 퐺, index sets 퐼 and 훬, and matrix 푃 over 퐺. Let (푖, 푔, 휆)

Solutions to exercises • 242 ̅ and (푗, ℎ, 휇) be elements of 푆. If 휗 ∶ Ω{푥}S → 푆 is such that 푥휗 = (푖, 푔, 휆) and 푦휗 = (푗, ℎ, 휇), then we have (푥푦)휗 = (푖, 푔, 휆)(푗, ℎ, 휇) = 푘 푘−1 (푖, 푔푝휆푗ℎ, 휇). Now, (푖, 푔푝휆푗ℎ, 휇) = (푖, (푔푝휆푗ℎ푝휇푖) 푔푝휆푗ℎ, 휇) for all 휔 푘−1 푘 ∈ ℕ. Thus (푥푦) 휗 is (푖, (푔푝휆푗ℎ푝휇푖) 푔푝휆푗ℎ, 휇) for some 푘. Since 휔 휔 −1 (푥푦) 휗 is always an idempotent, we have (푥푦) 휗 = (푖, 푝휇푖 , 휇). There- 휔 −1 −1 fore we have ((푥푦) 푥)휗 = (푖, 푝휇푖 , 휇)(푖, 푔, ℎ) = (푖, 푝휇푖 , 휇)(푖, 푔, 휆) = −1 (푖, 푝휇푖 푝휇푖푔, 휆) = (푖, 푔, 휆) = 푥휗. Thus 푆 satisfies the pseudoidentity (푥푦)휔푥 = 푥. Now suppose that 푆 satisfies (푥푦)휔+1 = 푥. Let 푠, 푡 ∈ 푆 and choose ̅ 휔 푘 휗 ∶ Ω{푥}S → 푆 with 푥휗 = 푠 and 푦휗 = 푡. Then (푥푦) 푥휗 = (푠푡) 푠 for some 푘 ∈ ℕ. Hence 푠 = (푠푡)푘푠 ∈ 푆푡푆 and so 푆 is simple by the analogue of Lemma 3.7 for simple semigroups. Arguing as in Exercise 8.8 but with 푥휗 = 푦휗 = 푠, we see that 푠 lies in the {푠, 푠2, … , 푠2푘}. Hence every element of 푆 lies in a subgroup and so 푆 is completely regular by Theorem 4.15. Since 푆 is completely regular and simple, it is completely simple by Theorem 4.16. 8.10 Let 푆 be left simple. Let 푒 be an idempotent of 푆. Then 푆푒 = 푆 since 푆 is left simple. Let 푠 ∈ 푆; then 푠 = 푠′푒 for some 푠′ ∈ 푆. Therefore 푠푒 = 푠′푒푒 = 푠′푒 = 푠, and so 푒 is a right identity for 푆. For any homomorphism ̅ 휔 휔 휗 ∶ Ω{푥,푦}(푆), the element 푦 휗 is an idempotent of 푆. Hence (푥푦 )휗 = (푥휗)(푦휔휗) = 푥휗. Thus 푆 satisfies the pseudoidentity 푥푦휔 = 푥. 휔 ̅ Now suppose 푆 satisfies 푥푦 = 푥. Let 푠, 푡 ∈ 푆. Let 휗 ∶ Ω{푥,푦}S be such that 푥휗 = 푠 and 푦휗 = 푡. Then (푦휔)휗 will be some idempotent power of 푡, say 푡푘 for some 푘 ∈ ℕ. Then 푠푡푘 = (푥푦휔)휗 = 푥휗 = 푠. Hence 푠 ∈ 푆푡. Thus 푆 = 푆푡 for all 푡 ∈ 푆 and so 푆 is left simple.

Exercises for chapter 9 [See pages 196–197 for the exercises.] 9.1 Suppose 퐿 is rational. Then it is recognized by a finite semigroup 푆 by Theorem 9.4. By Proposition 9.6, SynM 퐿 divides 푆. Hence SynM 퐿 is finite. Suppose SynM 퐿 is finite. The monoid SynM 퐿 recognizes 퐿 by Proposition 9.6. Since 퐿 is recognized by a finite monoid, it is rational by Theorem 9.4. 9.2 Let 푆 be the three element semilattice {0, 푥, 푦} with 푥 > 0 and 푦 > 0. Let 푎휑 = 푥 and 푏휑 = 푦. Then {푥}휑−1 = {푎}+ and {푦}휑−1 = {푏}+; hence {0}휓−1 = 퐿. ∗ ∗ 9.3 By definition, SynM 퐷 = { ( , ) } /휎퐷. Let 푤1 ⋯ 푤푛 ∈ { ( , ) } . Then

Solutions to exercises • 243 for any 푖, 푗,

{퐶(푤1 ⋯ 푤푛, 푖) if 푖 ⩽ 푗, { {퐶(푤1 ⋯ 푤푛, 푗) + 1 if 푖 = 푗 + 1, 퐶(푤 ⋯ 푤 () 푤 ⋯ 푤 , 푖) = 1 푗 푗+1 푛 {퐶(푤 ⋯ 푤 , 푗) if 푖 = 푗 + 2, { 1 푛 {퐶(푤1 ⋯ 푤푛, 푖) if 푖 ⩾ 푗 + 2. In particular,

퐶(푤1 ⋯ 푤푗 () 푤푗+1 ⋯ 푤푛, 푛 + 2) = 0 ⇔ 퐶(푤1 ⋯ 푤푛, 푛) = 0,

퐶(푤1 ⋯ 푤푗 () 푤푗+1 ⋯ 푤푛, 푖) ⩾ 0 for all 푖

⇔ 퐶(푤1 ⋯ 푤푛, 푖) ⩾ 0 for all 푖.

Hence for any words 푝, 푞 ∈ { ( , ) }∗, we have 푝 () 푞 ∈ 퐷 if and only if 푝푞 ∈ 퐷. Hence () 휎 휀. That is, [ ( ] [ ) ] = [휀] . Furthermore, )( 퐷 휎퐷 휎퐷 휎퐷 is not a Dyck word, so )( is not 휎 -related to 휀. That is [ ) ] [ ( ] ≠ 퐷 휎퐷 휎퐷 [휀] . Hence, by Exercise 2.12 with 푥 = [ ( ] , 푦 = [ ) ] , and 푒 = 휎퐷 휎퐷 휎퐷 [휀] , and noting that [ ( ] and 푦 = [ ) ] generate SynM 퐷, we see 휎퐷 휎퐷 휎퐷 that SynM 퐷 is isomorphic to the bicyclic monoid. 9.4 Let 퐾, 퐿 ∈ N(퐴+). If both 퐾 and 퐿 are finite, then 퐾 ∪ 퐿 and 퐾 ∩ 퐿 are finite and so 퐾 ∪ 퐿, 퐾 ∩ 퐿 ∈ N(퐴+). If one of 퐾 or 퐿 is finite and the other cofinite, then 퐾 ∪ 퐿 is cofinite and 퐾 ∩ 퐿 is finite and so 퐾 ∪ 퐿, 퐾 ∩ 퐿 ∈ N(퐴+). If both 퐾 and 퐿 are cofinite, then 퐾 ∪ 퐿 and 퐾 ∩ 퐿 are cofinite and so 퐾 ∪ 퐿, 퐾 ∩ 퐿 ∈ N(퐴+). So N(퐴+) is closed under union and intersection. If 퐾 is finite, 퐴+ ∖ 퐾 is cofinite and so 퐴+ ∖ 퐾 ∈ N(퐴+); if 퐾 is cofinite, 퐴+ ∖ 퐾 is finite and so 퐴+ ∖ 퐾 ∈ N(퐴+). So N(퐴+) is closed under complementation. Let 퐿 ∈ N(퐴+) and 푎 ∈ 퐴. If 퐿 is finite, it contains only word of length less than 푛 for some fixed 푛 ∈ ℕ. So 푎−1퐿 and 퐿푎−1 contain only words of length less than 푛 − 1. So 푎−1퐿 and 퐿푎−1 are finite and so 푎−1퐿, 퐿푎−1 ∈ N(퐴+). On the other hand, if 퐿 is cofinite, it contains all words in 퐴+ of length greater than 푛 for some fixed 푛 ∈ ℕ. So 푎−1퐿 and 퐿푎−1 contain all words in 퐴+ of length greater than 푛 − 1. So 푎−1퐿 and 퐿푎−1 are cofinite and so 푎−1퐿, 퐿푎−1 ∈ N(퐴+). Let 퐿 ∈ N(퐵+) and let 휑 ∶ 퐴+ → 퐵+ be a homomorphism. If 퐿 is finite, it contains only word of length less than 푛 for some fixed 푛 ∈ ℕ. Let 푤 ∈ 퐴+ have length greater than 푛. Then 푤휑 has length greater than 푛 and so 푤휑 ∉ 퐿. So 퐿휑−1 contains only words of length less than 푛; thus 퐿휑−1 is finite and so 퐿휑−1 ∈ N(퐴+). On the other hand, if 퐿 is cofinite, it contains all words in 퐴+ of length greater than 푛 for some fixed 푛 ∈ ℕ. Let 푤 ∈ 퐴+ have length greater than 푛. Then 푤휑 has length greater than 푛 and so 푤휑 ∈ 퐿. So 퐿휑−1 contains all words in 퐴+ of length greater than 푛; thus 퐿휑−1 is cofinite and so 퐿휑−1 ∈ N(퐴+). 9.5 Suppose that 퐾 is a +-language over 퐴 recognized by some finite rectangular band 푆. Then there is a homomorphism 휑 ∶ 퐴+ → 푆

Solutions to exercises • 244 such that 퐾 = 퐾휑휑−1. Recall from Exercise 8.4(c) that 푆 satisfies the pseudoidentities 푥2 = 푥 and 푥푦푧 = 푥푧. Thus, for 푎, 푎′ ∈ 퐴 and 푤 ∈ 퐴∗, we have (푎푤푎′)휑 ∈ 퐾휑 if and only if (푎푎′)휑 ∈ 퐾휑, or equivalently 푎푤푎′ ∈ 퐾 if and only if 푎푎′ ∈ 퐾. Therefore

푎퐴∗푎′ ∩ 퐾 ≠ ∅ ⇒ (∃푢 ∈ 퐴∗)(푎푢푎′ ∈ 퐾) ⇒ 푎푎′ ∈ 퐾 ⇒ (∀푤 ∈ 퐴∗)(푎푤푎′ ∈ 퐾) ⇒ 푎퐴∗푎′ ⊆ 퐾.

On the other hand, if 푎퐴∗푎′ ⊆ 퐾, then obviously 푎퐴∗푎′ ∩ 퐾 ≠ ∅. Therefore:

푎퐴∗푎′ ⊆ 퐾 ⇔ 푎퐴∗푎′ ∩ 퐾 ≠ ∅. (S.24)

Reasoning similar to the above and also using (푎푎)휑 = 푎휑 proves that

푎퐴∗푎 ⊆ 퐾 ⇔ 푎 ∈ 퐾. (S.25)

∗ Let 푍 be the subset of 퐴 that lies in 퐾 and let 퐾1 = 푍 ∪ ⋃푎∈푍 푎퐴 푎. Then by (S.25), 퐾1 ⊆ 퐾. Again by (S.25), 퐾1 must be precisely the words in 퐾 that start and end with the same letter. Let 퐾2 be the set of words in 퐾 that start and end with different letters. By (S.24), if there a ∗ word in 퐾2 that starts with 푎 and ends with 푎′, then all words in 푎퐴 푎′ lie in 퐾2. There are only finitely many possible choices for 푎 and 푎′, 푛 ∗ so 퐾2 = ⋃푖=1 푎푖퐴 푎′푖 for suitable 푎푖 and 푎′푖. Hence 퐾 = 퐾1 ∪ 퐾2 is a language of the form (9.12). Now suppose that 퐾 has the form (9.12). Then whether a word in 퐴+ lies in 퐾 depends only on its first and last letters. Let 푠, 푡 ∈ 퐴+. Then for any 푝, 푞 ∈ 퐴∗, the first letters of 푝푠푡푠푞 and 푝푠푞 are either both the first letter of 푝, and thus equal, or (when 푝 = 휀) both the first letter of 푠, and thus equal. Similarly, the last letters of 푝푠푡푠푞 and 푝푠푞 are equal. So 푝푠푡푠푞 ∈ 퐾 if and only if 푝푠푞 ∈ 퐾. Hence 푠푡푠 휎퐾 푠, or [푠] [푡] [푠] = [푠] . Since 푠, 푡 ∈ 퐴+ were arbitrary, this proves 휎퐾 휎퐾 휎퐾 휎퐾 that SynS 퐾 satisfies the pseudoidentity 푥푦푥 = 푥. Hence SynS 퐾 is a rectangular band and SynS 퐾 ∈ RB.

•

Solutions to exercises • 245 Bibliography

The bibliographical references at the end of this book do not ‘ make up a bibliography, they are only a legal device aimed at avoiding accusations of having omitted the names of persons from whom I took direct quotations. —’ Umberto Eco, Kant and the Platypus, Introduction (trans. William Weaver).

Almeida, J. Finite Semigroups and Universal Algebra. Series in Algebra 3. World Scientific, 1994. isbn: 9789810218959. ———‘Profinite semigroups and applications’.In: Structural Theory of Automata, Semigroups, and Universal Algebra. NATO Science Series II: Mathemat- ics, Physics and Chemistry 207. Dordrecht: Springer, 2005, pp. 1–45. isbn: 9781402038150. doi: 10.1007/1-4020-3817-8_1. Andersen, O. ‘Ein bericht über die Struktur abstrakter Halbgruppen’.Staat- sexamensarbeit. Hamburg, 1952.

Baader, F. & Nipkow, T. Term Rewriting and All That. Cambridge Univer- sity Press, 1999. isbn: 9780521455206. Book, R. V. & Otto, F. String Rewriting Systems. Texts and Monographs in Computer Science. Springer, 1993. isbn: 9780387979656.

Cain, A. J., Robertson, E. F. & Ruškuc, N. ‘Cancellative and Malcev presentations for finite Rees index subsemigroups and extensions’.In: Journal of the Australian Mathematical Society 84, no. 1 (2008), pp. 39–61. doi: 10.10 17/s1446788708000086. Clifford, A. H. & Preston, G. B. The Algebraic Theory of Semigroups. Vol. 1. Mathematical Surveys 7. Providence, RI: American Mathematical Society, 1961. ——— The Algebraic Theory of Semigroups. Vol. 2. Mathematical Surveys and Monographs 7. Providence, RI: American Mathematical Society, 1967. Clifford, A.H. ‘Semigroups admitting relative inverses’.In: Annals of Math- ematics. 2nd ser. 42 (1941), pp. 1037–1049. doi: 10.2307/1968781.

Distler, A. ‘Classification and Enumeration of Finite Semigroups’. Ph.D. thesis. University of St. Andrews, 2010. url: http://hdl.handle.net/10023/94 5.

• 246 Eilenberg, S. Automata, Languages, and Machines. Vol. B. Pure and Applied Mathematics 59. New York: Academic Press, 1976.

Gallagher, P. ‘On the Finite Generation and Presentability of Diagonal Acts, Finitary Power Semigroups and Schützenberger Products’.Ph.D. thesis. University of St Andrews, 2005. Green, J. A. ‘On the structure of semigroups’. In: Annals of Mathematics. 2nd ser. 54 (1951), pp. 163–172. doi: 10.2307/1969317. Grillet, P. A. Semigroups: An Introduction to the Structure Theory. Mono- graphs and Textbooks in Pure and Applied Mathematics 193. New York: Marcel Dekker, 1995. isbn: 9780824796624. ——— Commutative Semigroups. Advances in Mathematics 2. Springer, 2011. isbn: 9781441948571. Grillet, P. A. ‘A short proof of Rédei’s theorem’.In: Semigroup Forum 46, no. 1 (Dec. 1993), pp. 126–127. doi: 10.1007/BF02573555.

Harju, T. ‘Lecture Notes on Semigroups’.Unpublished lecture notes, Univer- sity of Turku. 1996. url: http://users.utu.fi/harju/semigroups/semigroups96 .pdf. Higgins, P. M. Techniques of Semigroup Theory. With a forew. by G. B. Pre- ston. Oxford Science Publications. Oxford: Clarendon Press, 1992. isbn: 9780198535775. Hopcroft, J. E. & Ullman, J. D. Introduction to Automata Theory, Lan- guages, and Computation. 1st ed. Reading, MA: Addison–Wesley, 1979. isbn: 9780201029888. Howie, J. M. Automata and Languages. Oxford Science Publications. Oxford: Clarendon Press, 1991. isbn: 9780198534242. ——— Fundamentals of Semigroup Theory. London Mathematical Society Mono- graphs: New Series 12. Oxford: Clarendon Press, 1995. isbn: 9780198511946.

Kleene, S.C. ‘Representation of events in nerve nets and finite automata’.In: Automata Studies. Ed. by C. E. Shannon & J. McCarthy. Annals of Mathem- atics Studies 34. Princeton, NJ: Princeton University Press, 1956, pp. 3–41. isbn: 978-0-691-07916-5. Krohn, K. & Rhodes, J. ‘Algebraic theory of machines I: Prime decomposition theorem for finite semigroups and machines’. In: Transactions of the American Mathematical Society 116 (1965), pp. 450–464. doi: 10.2307/1994127.

Lallement, G. Semigroups and Combinatorial Applications. New York: John Wiley & Sons, 1979. isbn: 9780471043799. ———‘Augmentations and wreath products of monoids’.In: Semigroup Forum 21, no. 1 (Dec. 1980), pp. 89–90. doi: 10.1007/BF02572539. Lawson, M. V. Inverse Semigroups: The Theory of Partial Symmetries. River Edge, NJ: World Scientific, 1998. isbn: 9789810233167. doi: 10.1142/978981281 6689.

Bibliography • 247 Lawson, M. V. Finite Automata. Boca Raton, FL: Chapman & Hall/CRC, 2004. isbn: 9781584882558. Linderholm, C. E. ‘A group epimorphism is surjective’.In: The American Mathematical Monthly 77, no. 2 (Feb. 1970), p. 176. url: https://www.jstor.or g/stable/2317336. Ljapin, E. S. Semigroups. Trans. by A. A. Brown, J. M. Danskin, D. Foley, S. H. Gould, E. Hewitt, S. A. Walker & J. A. Zilber. 3rd ed. Translations of Mathematical Monographs 3. Providence, RI: American Mathematical Society, 1974. Lothaire, M. Combinatorics on Words. With a forew. by R. Lyndon. Correc- ted edition. Encyclopedia of Mathematics and its Applications 17. Cambridge University Press, 1997. isbn: 9780521599245.

Mac Lane, S. Categories for the Working Mathematician. 2nd ed. Graduate Texts in Mathematics 5. Springer, 1998. isbn: 9780387984032. Malcev, A.I. ‘On the immersion of an algebraic ring into a field’.In: Math- ematische Annalen 113 (1937), pp. 686–691. doi: 10.1007/BF01571659. Miller, D.D. & Clifford, A.H. ‘Regular D-classes in semigroups’. In: Transactions of the American Mathematical Society 82 (1956), pp. 270–280. doi: 10.2307/1992989. Munn, W. D. ‘Free Inverse Semigroups’.In: Proceedings of the London Math- ematical Society 29, no. 3 (1st Nov. 1974), pp. 385–404. doi: 10.1112/plms/s3-2 9.3.385.

Ore, Ø. ‘Linear equations in non-commutative fields’. In: Annals of Mathemat- ics. 2nd ser. 32, no. 3 (July 1931), pp. 463–477. doi: 10.2307/1968245.

Petrich, M. Inverse Semigroups. New York: John Wiley & Sons, 1984. isbn: 9780471875451. ——— Completely Regular Semigroups. Canadian Mathematical Society Series of Monographs and Advanced Texts 23. New York: John Wiley & Sons, 1999. isbn: 9780471195719. Pin, J. E. Varieties of Formal Languages. Trans. by A. Howie. With a forew. by M. P. Schützenberger. Foundations of Computer Science. New York: Plenum Publishing, 1986. isbn: 9781461293002. doi: 10.1007/978-1-4613-2215-3. Pin, J. ‘Mathematical Foundations of Automata Theory’.Unpublished draft. 13th Mar. 2019. Pin, J. ‘Syntactic semigroups’.In: Handbook of Formal Languages. Vol. 1: Word, Language, Grammar. Ed. by G. Rozenberg & A. Salomaa. Berlin: Springer, 1997, pp. 679–746. isbn: 9783642638633. doi: 10.1007/978-3-642-59136-5_10. Preston, G.B. ‘Inverse semi-groups with minimal right ideals’.In: Journal of the London Mathematical Society. 1st ser. 29 (1954), pp. 404–411. doi: 10.1112/jlms/s1-29.4.404. ———‘Representations of inverse semi-groups’.In: Journal of the London Math- ematical Society. 1st ser. 29 (1954), pp. 411–419. doi: 10.1112/jlms/s1-29.4.411.

Bibliography • 248 Rabin, M.O. & Scott, D. ‘Finite automata and their decision problems’. In: International Business Machines Journal of Research and Development 3 (1959), pp. 114–125. doi: 10.1147/rd.32.0114. Rédei, L. The Theory of Finitely Generated Commutative Semigroups. Ed. by N. Reilly. International Series of Monographs in Pure and Applied Mathematics. Oxford: Pergamon Press, 1965. Rees, D. & Hall, P. ‘On semi-groups’.In: Mathematical Proceedings of the Cambridge Philosophical Society 36, no. 04 (Oct. 1940), p. 387. doi: 10.1017 /S0305004100017436. Rees, D. ‘On the group of a set of partial transformations’.In: Journal of the London Mathematical Society. 1st ser. 22, no. 4 (1947), pp. 281–284. doi: 10.1112/jlms/s1-22.4.281. Rhodes, J. & Steinberg, B. The 픮-theory of Finite Semigroups. Springer Monographs in Mathematics. New York: Springer, 2009. isbn: 9780387097800. doi: 10.1007/b104443. Robinson, D. J. S. A Course in the Theory of Groups. Graduate Texts in Mathematics 80. Springer, 1995. isbn: 9780387944616. doi: 10.1007/978-1-44 19-8594-1. Rosales, J. C. & García-Sánchez, P. A. Finitely Generated Com- mutative Monoids. Commack, NY: Nova Science Publishers, 1999. isbn: 9781560726708. Ruškuc, N. ‘Semigroup Presentations’.Ph.D. thesis. University of St Andrews, 1995. url: https://hdl.handle.net/10023/2821.

Schützenberger, M. P. ‘On finite monoids having only trivial subgroups’. In: Information and Control 8, no. 2 (Apr. 1965), pp. 190–194. doi: 10.1016/S0 019-9958(65)90108-7. Schützenberger, M.-P. ‘D̅ représentation des demi-groupes’.In: Comptes rendus hebdomadaires des séances de l’Académie des Sciences 244, no. 2 (Apr.– June 1957), pp. 1994–1996. url: https://gallica.bnf.fr/ark:/12148/bpt6k7215 /f154.image. ———‘Sur la représentation monomiale des demi-groupes’.In: Comptes rendus hebdomadaires des séances de l’Académie des Sciences 246, no. 1 (Jan.–Mar. 1958), pp. 865–867. url: https://gallica.bnf.fr/ark:/12148/bpt6k3198s/f871 .image. Suschkewitsch, A. ‘Über die endlichen Gruppen ohne das Gesetz der eindeutigen Umkehrbarkeit’.In: Mathematische Annalen 99 (1928), pp. 30–51. doi: 10.1007/BF01459084.

Vagner, V. V. ‘Generalized groups’.In: Doklady Akademii Nauk SSSR 84 (1952), pp. 1119–1122.

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Bibliography • 249 Index

Never index your own book. ‘ ’ — Kurt Vonnegut, Cat’s Cradle, ch. 55.

• In this index, the ordering of entries is strictly lexicographic, ignoring punctuation and spacing. Symbols outside the Latin alphabet are collected at the start of the index, even if ‘auxiliary’ Latin symbols are used: thus 푆1 is included in this set, since it is the notation ‘ 1’ that is being defined. Brief definitions are given for notation. This index currently covers only Chapter 1 and part of Chapter 3, plus names and ‘named results’. It will gradually be expanded to a full index.

∧: logical conjunction; ‘and’ automaton: see ‘finite state automaton’; 0-simple, 56–58 vi–vii ∨: logical disjunction; ‘or’ 1, 1푆: identity of a semigroup 푆 Baader, Franz, 52, 246 0, 0푆: zero of a semigroup 푆 biber, 256 푆1: monoid obtained by adjoining an BibLaTEX, 256 identity to 푆 if necessary; 4 bijection, 13 푆0: semigroup obtained by adjoining a binary operation, 1 zero to 푆 if necessary; 4 associative, 1–2 ⟨푋⟩: subsemigroup generated by 푋; 10 binary relation, 11–15, 20, 22 ⨅ 푌, 푥 ⊓ 푦: meet; 17 Birkhoff’s theorem, 149 ⨆ 푌, 푥 ⊔ 푦: join; 17 Book, Ronald V., 52, 246 휌R: reflexive closure of 휌; 22 Bourbaki, Nicolas, v 휌S: symmetric closure of 휌; 22 bracket, 2 휌T: transitive closure of 휌; 22 Brown, Arthur A., 248 휌E: equivalence relation generated by 휌; B푋: set of binary relations on 푋; 12 22 휌C: smallest left and right compatible Cain, Alan James, i–ii, vi–vii, 89, 246 relation containing 휌; 24 cancellative semigroup, 6, 7, 20, 32 휌#: congruence generated by 휌; 24 finite implies group, 32 cartesian product, 4 휌푥: transformation that right-multiplies by 푥; 19 finitary, 4 category theory, 1, 33, 35 action: see ‘semigroup action’ Cayley graph, 30–31 Almeida, Jorge, v, 169, 246 of a group, 31 Andersen, Olaf, 69, 246 right/left, 30–31 antichain, 15 Cayley’s theorem, 19 anti-homomorphism, 20, 30 chain, 15, 56 anti-symmetric binary relation, 15 Chesterton, Gilbert Keith, 70 associativity: see ‘binary operation, Clifford, A. H., 34, 68–69, 88–89, 116, 125, associative’ 246

• 250 Clifford, Alfred Hoblitzelle, 69, 88, 116, epimorphism, 34 246, 248 categorical, 33–34 commutative semigroup, v–vi, 6, 7–8 of groups, 35 of idempotents, 18 equivalence class, 15, 20 comparable elements, 15 equivalence relation, 15, 22–27, 53 compatible binary relation, 20, 24–26 characterization of join, 27 complete lattice: see ‘lattice, complete’ commuting complete lower semilattice: see characterization of join, 27 ‘semilattice, complete’ generated by a binary relation complete upper semilattice: see characterization of, 22 ‘semilattice, complete’ lattice of equivalence relations, 26–27 composition of binary relations, 11 exponent, 5 congruence, 20–21, 28 laws, 5 lattice of congruences, 26 congruence generated by a binary factor group, vii relation, 24–26 factor semigroup, 20–22, 57–58 characterization of, 25 Feyeraband, Paul Karl, 144 converse of a relation, 11 finitely generated, 10 correction, vi finite semigroup, v–vi, 5, 33 coset, vii cancellative implies group, 32 Costa, Alfredo Manuel Gouveia da, 246 finite state automaton, v Cottingham, John, 170 Foley, D., 248 Couto, Miguel Ângelo Marques Lourenço ‘folklore’, 34 do, vii free semigroup, vi Creative Commons, ii full map: see ‘map’ cyclic group, 1 full transformation: see ‘transformation’

D: see also ‘Green’s relation’; 53–55 Gallagher, Peter Timothy, 52, 247 퐷푎: D-class of 푎; 55 García Martinez, Xabier, vii Danskin, John Moffatt, 248 García-Sánchez, Pedro A., 125, 249 Descartes, René, 170 Gell-Mann, Murray, 117 determinant: see ‘matrix, determinant of generating set, 10 a’ Gould, Sydney Henry, 248 dihedral group, 1 graph, vii direct product, 1, 4, 8, 28, 30 greatest lower bound: see ‘meet’ D’Israeli, Isaac, v Green, James Alexander: see also ‘Green’s Distler, Andreas, 34, 169, 246 lemma’, ‘Green’s relations’; 69, 247 distributivity, 33 Green’s lemma, 58 dom 휌: domain of 휌; 12 Green’s relation: see also ‘H, L, R, D, J’; Dyck, Walther Franz Anton von: see 53–55 ‘Dyck word’ inclusion of, 54–55 Dyck word, 196–197 partial order from L, R, J, 55 Grillet, Pierre Antoine, 34–35, 69, 125, 247 퐸(푆): set of idempotents of 푆; 5 Grinberg, Darij, vii Eco, Umberto, 246 group, vii, 1–2, 6, 10, 32, 53–54, 56, 58 Egri-Nagy, Attila, vii composition series, 58 Eilenberg correspondence, 184–196 group-embeddable semigroup, 19 Eilenberg, Samuel: see also ‘Eilenberg group of units, 9 correspondence’, ‘Eilenberg’s groupoid, 1 theorem’; 169, 197, 247 Eilenberg’s theorem, 185 H: see also ‘Green’s relation’; 53, 54 ‘empty semigroup’, 1, 35 퐻푎: H-class of 푎; 55 End(푆): endomorphisms of 푆; 19 Hall, P., 88, 249 endomorphism, 19 Hamming, Richard Wesley, 198

Index • 251 Ham, Nick, vii join semilattice: see ‘semilattice’ Hardy, Godfrey Harold, 1 ‘Jordan–Hölder theorem’ for semigroups, Harju, Tero Juhani, 52, 247 58 Hasse diagram, 15–16, 18, 54 Herman, Samuel, vii 퐾(푆): kernel of a semigroup; 56 Hewitt, Edwin, 248 ker 휑: kernel of the homomorphism 휑; 19 Higgins, Peter Michael, 34, 52, 247 kernel, 56, 57 homomorphic image, 19 of a homomorphism: see homomorphism, 19–20, 21, 28–30, 33–34 ‘homomorphism, kernel of a’ kernel of a, 19, 21 Kleene, Stephen Cole: see also ‘Kleene’s monoid, 19, 33 theorem’; 197, 247 Hopcroft, John Edward, 197, 247 Kleene’s theorem, 175 Howie, A., 248 Knuth, Donald Ervin, v Howie, John Mackintosh, v, 34, 52, 68, 88, Koga, Akihiko (古賀明彦), vii 116, 169, 197, 247 Korzybski, Alfred Habdank Skarbek, 53 Huxley, Thomas Henry, 126 Krohn, Kenneth Bruce: see also ‘Krohn–Rhodes theorem’; 143, 247 퐼(푥): the set 퐽(푥) ∖ 퐽푥; 57 Krohn–Rhodes theorem, 141 id푋: identity relation on 푋; 11 ideal, 9–10, 34, 53–58 L: see also ‘Green’s relation’; 53, 55 left: see ‘left ideal’ commutes with R, 54 minimal 퐿푎: L-class of 푎; 55 uniqueness of, 56 퐿(푥): principal left ideal generated by 푥; 9 principal, 9 Lallement, Gérard, 143, 247 right: see ‘right ideal’ language, vii two-sided: see ‘ideal’ regular: see ‘regular language’ ideal extension, 22, 28–29 lattice, 17, 33 idempotent, 5, 7–8, 32 complete, 17 partial order of, 17 of congruences: see ‘congruence, idempotents lattice of congruences’ semigroup of: see ‘semigroup of of equivalence relation: see also idempotents’ ‘equivalence relation, lattice of identity: see also ‘monoid’; 1, 3, 7, 9, 12, 32 equivalence relations’ adjoining, 4, 32 Lawson, Mark Verus, 116, 197, 247–248 left: see ‘left identity’ least upper bound: see ‘join’ right: see ‘right identity’ left-cancellative semigroup, 6, 32 two-sided: see ‘identity’ left-compatible binary relation, 20 uniqueness of, 3 left congruence, 20, 55 identity relation, 11, 12, 19, 32 left ideal, 9–10 im 휌: image of 휌; 12 0-minimal, 56 index of an element, 5 minimal, 56 infimum: see ‘meet’ principal, 9 integers left identity, 3, 32 as a partially ordered set, 15 left inverse, 6 as a semigroup, 3 left-invertible element, 6, 33 inverse, 1, 6–7, 8 left zero, 3, 32 inverse semigroup, v–vi left zero semigroup, 3, 6, 8, 34 invertible element, 6, 8–9, 33 Linderholm, C. E., 35, 248 isomorphism, 19, 21 linear algebra, vii Lisbon, i J: see also ‘Green’s relation’; 53, 54–55 Ljapin, Evgeniĭ Sergeevich (Ляпин, 퐽푎: J-class of 푎; 55 Евгений Сергеевич), 34, 248 퐽(푥): principal ideal generated by 푥; 9 Lothaire, M., 52, 248 join, 17, 54 lower bound, 17

Index • 252 lower semilattice: see ‘semilattice’ Otto, Friedrich, 52, 246 LuaLaTEX, vi, 256 Lyndon, Roger Conant, 248 ℙ푋: power set; set of all subsets of 푋 P푋: set of partial transformations on 푋; Mac Lane, Saunders, 35, 248 12 magma, 1 partially ordered set, 15, 16 Malcev, Anatoly Ivanovich (Мальцев, subset of, 15 Анатолий Иванович), 52, 248 partial map, 12 Maltcev, Victor, vii partial order, 15–18, 17 map, 12 partial transformation, 12–13 domain of, 12 Pascal, Blaise, 90 image of, 12 periodic element, 5 notation for, 4, 30 periodic semigroup, 5, 33, 54–55 preimage under, 12 infinite, 32 matrix period of an element, 5 determinant of a, 7 Petrich, Mario, 89, 116, 248 matrix semigroup, 7 PGF/TikZ, 256 maximal element, 16 Pin, Jean-Éric, v, 169, 197, 248 maximum element, 16 Porto, i McCarthy, John, 247 Porto, University of, vi meet, 17, 54 poset: see ‘partially ordered set’ meet semilattice: see ‘semilattice’ power, 5 Miller, Don Dalzell, 69, 248 positive, 5 minimal element, 16 power semigroup, 32 minimum element, 16 power set, 15, 17 Mon⟨푋⟩: submonoid generated by 푋; 11 presentation, v–vi monogenic semigroup, 10 monoid: see ‘monoid presentation’ monoid: see also ‘identity’; 3, 7, 11–12, semigroup: see ‘semigroup 28–29, 32–33 presentation’ presentation of: see ‘monoid Preston, Gordon Bamford: see also presentation’ ‘Vagner–Preston theorem’; 34, trivial: see ‘trivial semigroup’ 68–69, 88–89, 116, 125, 246–248 monomorphism, 19, 34 principal factor, 57–58 categorical, 33–34 principal series, 58 multiplication, 2 product of elements, 2 Munn, William Douglas, 116, 248 product of subsets, 5 pseudovariety, v–vi natural homomorphism: see ‘natural map’ natural map, 21, 29 quaternion group, 1 natural numbers quotient semigroup: see ‘factor as a semigroup, 2, 10, 30, 58 semigroup’ nilpotent group, 6 nilpotent semigroup, 5 R: see also ‘Green’s relation’; 53, 55 nilsemigroup, 5 commutes with L, 54 Nine Chapters on the Mathematical Art 푅푎: R-class of 푎; 55 (九章算術; Jiuzhāngˇ Suànshù), vii 푅(푥): principal right ideal generated by 푥; Nipkow, Tobias, 52, 246 9 ,197 ,( ןיִּבַררזועלֵאָכיִמ ) null semigroup, 3, 56–57 Rabin, Michael Oser 249 opposite semigroup, 8 rectangular band, 8 order, 15 Rédei, László: see also ‘Rédei’s theorem’; Ore, Øystein: see also ‘Ore’s theorem’; 125, 125, 249 248 Rédei’s theorem, 123 Ore’s theorem, 124 Rees congruence, 21

Index • 253 Rees, David: see also ‘Rees–Suschkewitsch cancellative: see ‘cancellative theorem’; 88, 125, 249 semigroup’ Rees factor semigroup, 21–22 commutative: see ‘commutative Rees–Suschkewitsch theorem, 77, 80 semigroup’ reflexive binary relation, 15, 20, 22 finite: see ‘finite semigroup’ reflexive closure of a binary relation, free: see ‘free semigroup’ 22–24 inverse: see ‘inverse semigroup’ characterization of, 22 left zero: see ‘left zero semigroup’ regular element, 6–7 matrix: see ‘matrix semigroup’ regular language, v null: see ‘null semigroup’ regular semigroup, v–vi, 6 periodic: see ‘periodic semigroup’ Reilly, Norman R., 249 presentation of: see ‘semigroup Reiterman’s theorem, 162 presentation’ relation: see ‘binary relation’ regular: see ‘regular semigroup’ Rhodes, John Lewis: see also right zero: see ‘right zero semigroup’ ‘Krohn–Rhodes theorem’; 143, 169, simple: see ‘simple semigroup’ 247, 249 trivial: see ‘trivial semigroup’ right-cancellative semigroup, 6 zero-simple: see ‘0-simple semigroup’ right-compatible binary relation, 20 semigroup action, 29–30 right congruence, 20, 55 by endomorphisms, 30 right ideal, 9–10 free, 30 0-minimal, 56 left, 30 minimal, 56 regular, 30 principal, 9 right, 30 right identity, 3, 32 transitive, 30 right inverse, 6 semigroup of binary relations, 12, 13, 32 right-invertible element, 6, 33 semigroup of idempotents, 5 right regular representation, 19, 34 semigroup of partial transformations, 13, right zero, 3, 32 30, 32 right zero semigroup, 3, 5, 8, 10, 31–32, 34 computation(, 14 ring, 3 computation), 14 Rito, Guilherme Miguel Teixeira, vii semigroup of transformations, 13, 19, Robertson, Edmund Frederick, 89, 246 29–30, 32–33 Robinson, Derek J. S., 69, 249 computation(, 14 Rosales, José Carlos, 125, 249 computation), 14 Rozenberg, Grzegorz, 248 semilattice, 17–18, 34 Ruškuc, Nikola, 52, 89, 246, 249 as a commutative semigroup of idempotents, 18 S푋: set of bijections on 푋; 12 complete, 17 Salomaa, A., 248 Shannon, Claude Elwood, 247 Santiago de Compostella, University of, vi simple group, 56 Santos, José Manuel dos Santos dos, vii simple semigroup, 56–58 Schelling, Friedrich Wilhelm Joseph von, Soares, Jorge Fernando Valentim, vii 36 Steinberg, Benjamin, 143, 169, 249 Schützenberger group, 63–66 structure of a semigroup, v–vi right and left, 65 subdirect product, 28–29, 34 Schützenberger, Marcel-Paul: see also subgroup, 8–9, 12, 19 ‘Schützenberger group’, submonoid: see also ‘subsemigroup’; 8, 12 ‘Schützenberger’s theorem’; 69, 197, generating, 11 248–249 subsemigroup: see also ‘submonoid’; 8–10, Schützenberger’s theorem, 191 19, 29, 32 Scott, Dana, 197, 249 generating, 10 semigroup, v–vi, 1–15, 19–22, 24–35 proper, 8, 32 0-simple: see ‘0-simple semigroup’ supremum: see ‘join’

Index • 254 Suschkewitsch, Anton Kazimirovich: see Ullman, Jeffrey David, 197, 247 also ‘Rees–Suschkewitsch theorem’ universal algebra, vii Suschkewitsch, Anton Kazimirovich upper bound, 17 (Сушкевич, Антон Казимирович), upper semilattice: see ‘semilattice’ 88, 249 symmetric binary relation, 14–15, 22 푉(푥): set of inverses of 푥; 7 symmetric closure of a binary relation, Vagner–Preston theorem, 96 22–24 Vagner, Viktor Vladimirovich: see also characterization of, 22 ‘Vagner–Preston theorem’; 116, 249 symmetric group, 12, 13, 19, 30, 32 variety: see also ‘pseudovariety’; vi Vonnegut, Kurt, 250 T푋: set of transformations on 푋; 12 Tilson, Bret Ransom, 247 Walker, Sue Ann, 248 topology, vii Weaver, William Fense, 246 total order: see ‘order’ transformation, 12–13 xindy, 256 two-line notation, 13 transitive binary relation, 15, 20, 22 zero, 3, 6–7, 32, 56 transitive closure of a binary relation, adjoining, 4, 32 22–24 left: see ‘left zero’ characterization of, 22 right: see ‘right zero’ trivial monoid: see also ‘trivial two-sided: see ‘zero’ semigroup’; 3 uniqueness of, 3 trivial semigroup, 3, 32 zero-simple semigroup: see ‘0-zero Trocado, Alexandre, vii semigroup’ tuple, 4 Zilber, Joseph Abraham, 248

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Index • 255 Colophon

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