Exercises in Cellular Automata and Groups [1 ed.] 9783031103902, 9783031103919

Table of contents :
Foreword
Preface
Contents
Partial List of Notations
1 Cellular Automata
1.1 Summary
1.1.1 Configuration Spaces
1.1.2 The Prodiscrete Topology
1.1.3 Periodic Configurations
1.1.4 Cellular Automata
1.1.5 The Curtis-Hedlund-Lyndon Theorem
1.1.6 Induction and Restriction of Cellular Automata
1.1.7 Invertible Cellular Automata
1.2 Exercises
2 Residually Finite Groups
2.1 Summary
2.1.1 Equivalent Definitions of Residual Finiteness
2.1.2 The Class of Residually Finite Groups
2.1.3 Divisible Groups
2.1.4 Hopfian Groups
2.2 Exercises
3 Surjunctive Groups
3.1 Summary
3.1.1 Definition of Surjunctivity
3.1.2 The Class of Surjunctive Groups
3.1.3 Expansive Actions
3.1.4 Compactness of the Space of MarkedSurjunctive Groups
3.2 Exercises
4 Amenable Groups
4.1 Summary
4.1.1 Equivalent Definitions of Amenability
4.1.2 The Class of Amenable Groups
4.1.3 Amenability of Solvable and Nilpotent Groups
4.2 Exercises
5 The Garden of Eden Theorem
5.1 Summary
5.1.1 Interiors, Closures, and Boundaries
5.1.2 Tilings
5.1.3 Pre-injective Maps
5.1.4 Garden of Eden Configurations
5.1.5 Entropy
5.1.6 The Garden of Eden Theorem
5.2 Exercises
6 Finitely Generated Groups
6.1 Summary
6.1.1 The Word Metric
6.1.2 Labeled Graphs
6.1.3 Cayley Graphs
6.1.4 Growth of Finitely Generated Groups
6.1.5 The Grigorchuk Group
6.1.6 The Kesten-Day Characterization of Amenability
6.1.7 Quasi-Isometries
6.2 Exercises
7 Local Embeddability and Sofic Groups
7.1 Summary
7.1.1 Local Embeddability
7.1.2 LEF and LEA-Groups
7.1.3 The Hamming Metric
7.1.4 Sofic Groups
7.1.5 Sofic Groups and Ultraproducts
7.1.6 Geometric Characterization of Finitely Generated Sofic Groups
7.1.7 Surjunctivity of Sofic Groups
7.2 Exercises
8 Linear Cellular Automata
8.1 Summary
8.1.1 Rings
8.1.2 Group Rings
8.1.3 NZD-Groups, Unique-Product Groups, and Orderable Groups
8.1.4 Linear Shift Spaces
8.1.5 Linear Cellular Automata
8.1.6 Restriction and Induction of Linear Cellular Automata
8.1.7 Group Ring Representation of Linear Cellular Automata
8.1.8 Matrix Representation of Linear Cellular Automata
8.1.9 The Closed Image Property for LinearCellular Automata
8.1.10 Invertible Linear Cellular Automata
8.1.11 Mean Dimension
8.1.12 The Garden of Eden Theorem for LinearCellular Automata
8.1.13 The Discrete Laplacian
8.1.14 Linear Surjunctivity
8.2 Exercises
References
Index

Citation preview

Springer Monographs in Mathematics

Tullio Ceccherini-Silberstein Michel Coornaert

Exercises in Cellular Automata and Groups Foreword by Rostislav I. Grigorchuk

Springer Monographs in Mathematics Editors-in-Chief Minhyong Kim, School of Mathematics, Korea Institute for Advanced Study, Seoul, South Korea International Centre for Mathematical Sciences, Edinburgh, UK Katrin Wendland, School of Mathematics, Trinity College Dublin, Dublin, Ireland Series Editors Sheldon Axler, Department of Mathematics, San Francisco State University, San Francisco, CA, USA Mark Braverman, Department of Mathematics, Princeton University, Princeton, NJ, USA Maria Chudnovsky, Department of Mathematics, Princeton University, Princeton, NJ, USA Tadahisa Funaki, Department of Mathematics, University of Tokyo, Tokyo, Japan Isabelle Gallagher, Département de Mathématiques et Applications, Ecole Normale Supérieure, Paris, France Sinan Güntürk, Courant Institute of Mathematical Sciences, New York University, New York, NY, USA Claude Le Bris, CERMICS, Ecole des Ponts ParisTech, Marne la Vallée, France Pascal Massart, Département de Mathématiques, Université de Paris-Sud, Orsay, France Alberto A. Pinto, Department of Mathematics, University of Porto, Porto, Portugal Gabriella Pinzari, Department of Mathematics, University of Padova, Padova, Italy Ken Ribet, Department of Mathematics, University of California, Berkeley, CA, USA René Schilling, Institute for Mathematical Stochastics, Technical University Dresden, Dresden, Germany Panagiotis Souganidis, Department of Mathematics, University of Chicago, Chicago, IL, USA Endre Süli, Mathematical Institute, University of Oxford, Oxford, UK Shmuel Weinberger, Department of Mathematics, University of Chicago, Chicago, IL, USA Boris Zilber, Mathematical Institute, University of Oxford, Oxford, UK

This series publishes advanced monographs giving well-written presentations of the “state-of-the-art” in fields of mathematical research that have acquired the maturity needed for such a treatment. They are sufficiently self-contained to be accessible to more than just the intimate specialists of the subject, and sufficiently comprehensive to remain valuable references for many years. Besides the current state of knowledge in its field, an SMM volume should ideally describe its relevance to and interaction with neighbouring fields of mathematics, and give pointers to future directions of research.

Tullio Ceccherini-Silberstein • Michel Coornaert

Exercises in Cellular Automata and Groups Foreword by Rostislav I. Grigorchuk

Tullio Ceccherini-Silberstein Dipartimento di Ingegneria Università degli Studi del Sannio Benevento, Italy

Michel Coornaert Institut de Recherche Mathématique Avancée Université de Strasbourg Strasbourg, France

ISSN 1439-7382 ISSN 2196-9922 (electronic) Springer Monographs in Mathematics ISBN 978-3-031-10390-2 ISBN 978-3-031-10391-9 (eBook) https://doi.org/10.1007/978-3-031-10391-9 © The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 This work is subject to copyright. All rights are solely and exclusively licensed by the Publisher, whether the whole or part of the material is concerned, specifically the rights of translation, reprinting, reuse of illustrations, recitation, broadcasting, reproduction on microfilms or in any other physical way, and transmission or information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed. The use of general descriptive names, registered names, trademarks, service marks, etc. in this publication does not imply, even in the absence of a specific statement, that such names are exempt from the relevant protective laws and regulations and therefore free for general use. The publisher, the authors, and the editors are safe to assume that the advice and information in this book are believed to be true and accurate at the date of publication. Neither the publisher nor the authors or the editors give a warranty, expressed or implied, with respect to the material contained herein or for any errors or omissions that may have been made. The publisher remains neutral with regard to jurisdictional claims in published maps and institutional affiliations. This Springer imprint is published by the registered company Springer Nature Switzerland AG The registered company address is: Gewerbestrasse 11, 6330 Cham, Switzerland Paper in this product is recyclable.

To Tommaso and Gea To Élise

Foreword

It is a privilege to introduce this book which is a comprehensive complement to another wonderful book of Tullio Ceccherini-Silberstein and Michel Coornaert, namely Cellular Automata and Groups (Springer Monographs in Mathematics, 2010). These two monographs are devoted to the important links between four areas of mathematics: the theory of cellular automata, amenability and soficity, geometric group theory, and dynamical systems. The contents include a number of remarkable applications to other exciting topics such as Conway’s “Game of Life”; the Banach-Tarski paradox; the classical Garden of Eden Theorem of Moore and Myhill and its generalizations in various directions; group growth; subshifts; algorithmic problems; and Kaplansky conjectures on group rings, all presented in an elegant and accessible form. This new book consists of more than 600 fully solved exercises including the 306 exercises already proposed without solution in the previous book. It also contains a reasonable amount of background material as well as a presentation of some significant results that appeared after 2010. I am sure that the book will become popular among specialists in different areas of mathematics and will serve them to teach a variety of courses. It will be accessible not only to graduate students but also to mature students of undergraduate level. The authors are masters in book writing and the current book once more confirms this. I recommend it to the reader with a great pleasure. College Station, TX, USA May 2023

Rostislav I. Grigorchuk

vii

Preface

This book is meant as a complement to our monograph Cellular Automata and Groups, Springer Monographs in Mathematics, 2010 [CAG]. It consists of more than 600 fully solved exercises including the 306 exercises already presented without solution in [CAG]. The material covered by these exercises is connected with many areas of mathematics such as general topology, dynamical systems, symbolic dynamics, group theory, combinatorial and geometric group theory, geometry and topology of low-dimensional manifolds, commutative and non-commutative rings, group rings, module theory, automata theory and theoretical computer science. Roughly speaking, a cellular automaton can be described as follows. One is given a set, called the alphabet or the set of states, and a group, called the universe or the set of cells. A configuration is a map from the universe to the alphabet. The universe has a natural action on the configuration set, called the shift action. A cellular automaton is then a map from the set of all configurations into itself satisfying the following local property: the state of the output configuration at a given cell only depends, in a homogeneous way with respect to the shift action, on the states of the input configuration on a finite neighborhood of the cell. There is a natural uniform structure on the configuration set, and the cellular automata are precisely the uniformly continuous and equivariant maps. As in [CAG], one of the recurrent themes is the search for relations between the properties of certain cellular automata and those of the underlying groups. A famous result in this direction is the theorem due to Bartholdi, Machì, Scarabotti, and the first named author asserting that a group is amenable if and only if every cellular automaton with finite alphabet satisfies the Garden of Eden theorem. Another one is the Gromov-Weiss surjunctivity theorem which says that every injective cellular automaton with finite alphabet over a sofic group is surjective. Our treatment of cellular automata is not limited to the finite alphabet case. We consider, for instance, linear cellular automata for which the alphabet set is a vector space over an arbitrary field. Linear cellular automata over finite-dimensional vector spaces can be represented by matrices with entries in the group ring of the universe with coefficients in the field. This leads to surprising connections between

ix

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Preface

the theory of linear cellular automata and celebrated conjectures of Kaplansky about the structure of group rings. In the present book, we generalize the notion of a cellular automaton considered in [CAG] in two directions. Firstly, we allow the input and output configuration spaces to have distinct alphabets. Secondly, we shall consider cellular automata between subshifts, that is, shift-invariant closed subsets of the configurations spaces. Subshifts play a central role in the theory of dynamical systems and have been intensively used, since the pioneering work of Morse and Hedlund in the first half of the twentieth century, to analyze complex systems coming from geometry such as geodesic flows on Riemannian manifolds and Anosov diffeomorphisms. From a categorical point of view, subshifts constitute the objects of a concrete category whose morphisms are the cellular automata. The Garden of Eden theorem and the Gromov-Weiss surjunctivity theorem can be extended to cellular automata between subshifts provided that suitable dynamical properties such as irreducibility and finite type conditions are satisfied by the subshifts. The Shannon-Kolmogorov notion of entropy plays a crucial role in the proofs of these generalizations. When the universe is the group of integers, there are strong connections between symbolic dynamics, graph theory, and the theory of formal languages and coding in theoretical computer science. A beautiful and thorough exposition of this interplay is offered by the monograph of Doug Lind and Brian Marcus, Introduction to Symbolic Dynamics and Coding, Cambridge University Press, 1995, now also available in its second and expanded 2021 edition [LinM]. For groups other than the integers, the study of subshifts and cellular automata reveals several new phenomena and this constitutes one of the main motivations for these exercises. The book is organized as follows. We have kept the division into eight chapters from [CAG]. Each chapter begins with a summary of the main definitions and results contained in the corresponding chapter of [CAG]. Each solution is detailed and entirely self-contained, in the sense that it only requires a standard undergraduate-level background in abstract algebra and general topology, together with results established in [CAG] and in previous exercises. Comments at the end of exercises provide historical and bibliographical information, an account of related recent developments, and suggestions for further readings. The present book should be suitable either for classroom or individual use and it is intended for both students and mature researchers. We hope that, together with its companion [CAG], now also available in its second edition [CAG2], this book will result in a useful and stimulating introduction to the subject as well as a source of inspiration for further new developments. We are grateful to Laurent Bartholdi, Matteo Cavaleri, Yves de Cornulier, Slava Grigorchuk, Pierre de la Harpe, Xuan Kien Phung, and Ralph Strebel for helpful discussions. We wish also to thank the staff of Springer Verlag, in particular Elena Griniari, Senior Editor, Mathematics, for her unwavering support, encouragement, and guidance throughout the development of this book project, and Francesca Fer-

Preface

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rari, Assistant Editor, for her invaluable assistance with our manuscript finalization and related matters. Rome, Italy Strasbourg, France

Tullio Ceccherini-Silberstein Michel Coornaert

Contents

1

Cellular Automata . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.1 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.1.1 Configuration Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.1.2 The Prodiscrete Topology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.1.3 Periodic Configurations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.1.4 Cellular Automata . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.1.5 The Curtis-Hedlund-Lyndon Theorem . . . . . . . . . . . . . . . . . . . . . . . 1.1.6 Induction and Restriction of Cellular Automata . . . . . . . . . . . . . 1.1.7 Invertible Cellular Automata. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2 Exercises. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

1 1 1 2 2 3 3 4 4 5

2

Residually Finite Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.1 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.1.1 Equivalent Definitions of Residual Finiteness . . . . . . . . . . . . . . . 2.1.2 The Class of Residually Finite Groups . . . . . . . . . . . . . . . . . . . . . . . 2.1.3 Divisible Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.1.4 Hopfian Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2 Exercises. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

115 115 115 116 116 117 117

3

Surjunctive Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1.1 Definition of Surjunctivity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1.2 The Class of Surjunctive Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1.3 Expansive Actions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1.4 Compactness of the Space of Marked Surjunctive Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2 Exercises. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

167 167 167 167 168

Amenable Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.1 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.1.1 Equivalent Definitions of Amenability . . . . . . . . . . . . . . . . . . . . . . . 4.1.2 The Class of Amenable Groups. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

199 199 199 200

4

168 169

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Contents

4.1.3 Amenability of Solvable and Nilpotent Groups. . . . . . . . . . . . . . 201 4.2 Exercises. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 201 5

The Garden of Eden Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.1 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.1.1 Interiors, Closures, and Boundaries . . . . . . . . . . . . . . . . . . . . . . . . . . 5.1.2 Tilings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.1.3 Pre-injective Maps. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.1.4 Garden of Eden Configurations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.1.5 Entropy. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.1.6 The Garden of Eden Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.2 Exercises. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

243 243 243 244 244 245 245 246 246

6

Finitely Generated Groups. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.1 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.1.1 The Word Metric . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.1.2 Labeled Graphs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.1.3 Cayley Graphs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.1.4 Growth of Finitely Generated Groups . . . . . . . . . . . . . . . . . . . . . . . . 6.1.5 The Grigorchuk Group . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.1.6 The Kesten-Day Characterization of Amenability . . . . . . . . . . . 6.1.7 Quasi-Isometries. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.2 Exercises. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

295 296 296 297 298 299 300 301 302 303

7

Local Embeddability and Sofic Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.1 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.1.1 Local Embeddability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.1.2 LEF and LEA-Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.1.3 The Hamming Metric . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.1.4 Sofic Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.1.5 Sofic Groups and Ultraproducts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.1.6 Geometric Characterization of Finitely Generated Sofic Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.1.7 Surjunctivity of Sofic Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.2 Exercises. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

465 465 465 466 467 467 468

Linear Cellular Automata. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.1 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.1.1 Rings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.1.2 Group Rings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.1.3 NZD-Groups, Unique-Product Groups, and Orderable Groups. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.1.4 Linear Shift Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.1.5 Linear Cellular Automata . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.1.6 Restriction and Induction of Linear Cellular Automata . . . . . 8.1.7 Group Ring Representation of Linear Cellular Automata . . . 8.1.8 Matrix Representation of Linear Cellular Automata . . . . . . . .

513 513 513 514

8

469 469 469

515 515 516 516 517 517

Contents

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8.1.9

The Closed Image Property for Linear Cellular Automata . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.1.10 Invertible Linear Cellular Automata . . . . . . . . . . . . . . . . . . . . . . . . . . 8.1.11 Mean Dimension . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.1.12 The Garden of Eden Theorem for Linear Cellular Automata . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.1.13 The Discrete Laplacian. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.1.14 Linear Surjunctivity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.2 Exercises. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

518 518 518 519 520 520 520

References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 603 Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 617

Partial List of Notations

The notations and mathematical conventions used throughout this book are the same as in [CAG]. We recall in particular the following ones. N := {0, 1, 2, . . . } is the set of non-negative integers. Z is the ring of integers. .Q is the field of rational numbers. .R is the field of real numbers. .C is the field of complex numbers. .∅ is the empty set. .A ⊂ B means that every element of A is an element of B. .A  B means that .A ⊂ B and .A = B. .A \ B is the set consisting of all elements of A that are not in B. B .A is the set of all maps from B to A. .|A| is the cardinality (number of elements) of the finite set A. . denotes a disjoint union. All rings and algebras are unital, i.e., the multiplicative operation admits a neutral element. • All fields are commutative. • .Matd (R) is the ring of .d × d matrices with coefficients in the ring R. t is the transpose of the matrix A. • .A

• • • • • • • • • • • • •

. .

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Chapter 1

Cellular Automata

This chapter introduces configuration spaces over general groups, their subshifts, and the cellular automata between them. Finiteness properties for subshifts such as the finite type condition, Gromov’s splicability, and soficity are investigated. The chapter also includes a detailed study of mixing properties of subshifts such as topological transitivity, irreducibility, topological mixing, and strong irreducibility. Group subshifts and the finite Markov property for groups are treated. Nilpotency, limit sets, and stability of cellular automata are discussed in detail. In the important particular case when the underlying group is the group of integers, the properties of subshifts are expressed in terms of those of their associated languages.

1.1 Summary 1.1.1 Configuration Spaces Π Let G be a group and let A be a set. Consider the set .AG := g∈G A = {x : G → A}, consisting of all maps from G to A. We equip .AG with the prodiscrete uniform structure, that is, the product uniform structure obtained by taking the discrete uniform structure on each factor A of .AG , and with the G-shift, that is, the left action of G defined by .gx(h) := x(g −1 h) for all .x ∈ AG and .g, h ∈ G. The set G is then called the space of configurations of the universe G over the alphabet .A A. Given a configuration .x ∈ AG and a subset .E ⊂ G, we shall denote by .x|E the restriction of x to E, that is, the element .x|E ∈ AE defined by .x|E (g) := x(g) for all .g ∈ E. A base of entourages for the prodiscrete uniform structure on .AG is given by the sets .WΩ ⊂ AG × AG defined by WΩ := {(x, y) ∈ AG × AG : x|Ω = y|Ω },

.

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 T. Ceccherini-Silberstein, M. Coornaert, Exercises in Cellular Automata and Groups, Springer Monographs in Mathematics, https://doi.org/10.1007/978-3-031-10391-9_1

(1.1) 1

2

1 Cellular Automata

where .Ω runs over all finite subsets of G. These entourages .WΩ are called the basic entourages of .AG .

1.1.2 The Prodiscrete Topology Let G be a group and let A be a set. The topology on .AG associated with the prodiscrete uniform structure is called the prodiscrete topology on .AG . It is the product topology obtained by taking the discrete topology on each factor A of .AG . A pattern over the group G and the alphabet A, or shortly a .(G, A)-pattern, is a map .p : Ω → A, where .Ω is a finite subset of G. The set .Ω is called the support of the pattern p and it is denoted by .supp(p). The cylinder associated with a .(G, A)-pattern p is the subset .C(p) ⊂ AG given by C(p) := {x ∈ AG : x|supp(p) = p}.

.

(1.2)

Note that .C(p) is both open and closed in .AG . Moreover, the cylinders .C(p), where p runs over all .(G, A)-patterns, form a base of open sets for the prodiscrete topology on .AG . Given a configuration .x ∈ AG , the cylinders V (x, Ω) := C(x|Ω ) = {y ∈ AG : y|Ω = x|Ω },

.

(1.3)

where .Ω runs over all finite subsets of G, form a neighborhood base of x.

1.1.3 Periodic Configurations Let G be a group and let A be a set. Let H be a subgroup of G. One says that a configuration .x ∈ AG is fixed by H , or that x is an H -fixed configuration, if one has .hx = x for all .h ∈ H . This amounts to saying that the stabilizer of x in G, that is, the subgroup .StabG (x) ⊂ G consisting of all .g ∈ G such that .gx = x, satisfies .H ⊂ StabG (x). The set .Fix(H ) consisting of all H -fixed configurations is closed in .AG for the prodiscrete topology [CAG, Proposition 1.3.2]. Consider the set .H \G := {H g : g ∈ G} consisting of all right cosets of H in G and the canonical surjective map .ρ : G → H \G given by .ρ(g) := H g for all .g ∈ G. Given an element H \G , i.e., a map .y : H \G → A, we can form the composite map .y ◦ ρ : G → .y ∈ A A which is an element of .AG . We then have .y ◦ ρ ∈ Fix(H ). Moreover, the map ∗ H \G → Fix(H ) defined by .ρ ∗ (y) := y ◦ ρ for all .y ∈ AH \G is bijective .ρ : A [CAG, Proposition 1.3.3]. Consequently, if the set A is finite and H is a subgroup of finite index of G, then the set .Fix(H ) is finite with cardinality .| Fix(H )| = |A|[G:H ] , where .[G : H ] denotes the index of H in G [CAG, Corollary 1.3.4].

1.1 Summary

3

One says that a configuration .x ∈ AG is periodic if there exists a finite index subgroup H of G such that .x ∈ Fix(H ). This amounts to saying that .StabG (x) is of finite index in G or equivalently that the orbit of x under the shift action of G on .AG is finite. In the case when .G = Z is the additive group of integers, a subset .H ⊂ Z is a finite index subgroup if and only if there is an integer .n ≥ 1 such that Z .H = nZ. A configuration .x ∈ A such that .x ∈ Fix(nZ) for some .n ≥ 1 is called an n-periodic configuration.

1.1.4 Cellular Automata Let G be a group and let A and B be sets. A cellular automaton over the group G and the alphabets A and B is a map .τ : AG → B G satisfying the following property: there exist a finite subset .S ⊂ G and a map .μ : AS → B such that τ (x)(g) = μ((g −1 x)|S )

.

(1.4)

for all .x ∈ AG and .g ∈ G. Such a set S is then called a memory set for .τ and the map .μ : AS → B is called the local defining map for .τ associated with S. Since the intersection of two memory sets for a cellular automaton .τ is also a memory set for .τ , there exists a unique memory set .S0 ⊂ G for .τ with minimal cardinality. This memory set .S0 is called the minimal memory set of .τ . A finite subset .S ⊂ G is a memory set for .τ if and only if .S0 ⊂ S (cf. [CAG, Section 1.5]).

1.1.5 The Curtis-Hedlund-Lyndon Theorem Let G be a group and let A and B be sets. A map .τ : AG → B G is a cellular automaton if and only if it is G-equivariant (that is, .τ (gx) = gτ (x) for all .g ∈ G and .x ∈ AG ) and uniformly continuous with respect to the prodiscrete uniform structures on .AG and .B G (cf. [CAG, Theorem 1.9.1]). When the alphabet A is finite, the above characterization may be expressed as follows: a map .τ : AG → B G is a cellular automaton if and only if it is G-equivariant and continuous with respect to the prodiscrete topologies on .AG and .B G . This is the Curtis-Hedlund-Lyndon theorem (cf. [CAG, Theorem 1.8.1]). The composition of two uniformly continuous (resp. G-equivariant) maps is a uniformly continuous (resp. G-equivariant) map. Therefore, the composition of two cellular automata is a cellular automaton. More precisely, given another set C, if G → B G and .σ : B G → C G are cellular automata, then .σ ◦ τ : AG → C G .τ : A is also a cellular automaton. As an immediate consequence, we have that the set G → AG is a monoid for the composition .CA(G; A) of all cellular automata .τ : A of maps, the neutral element being the identity cellular automaton .IdAG defined by G .IdAG (x) = x for all .x ∈ A .

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1 Cellular Automata

1.1.6 Induction and Restriction of Cellular Automata Let now .H ⊂ G be a subgroup. Suppose that .τ : AG → B G is a cellular automaton admitting a memory set S such that .S ⊂ H . Then the local defining map .μ : AS → B associated with S also defines a cellular automaton .τH : AH → B H (over the subgroup H and the same alphabets A and B) by setting τH (x)(h) := μ((h−1 x)|S )

.

(1.5)

for all .x ∈ AH and .h ∈ H . The cellular automaton .τH is called the restriction (cellular automaton) of .τ (from G to H ). Conversely, if .τ : AH → B H is a cellular automaton over the group H and the alphabets A and B, with memory set .S ⊂ H and associated local defining map .μ : AS → B, then (1.4) defines a cellular automaton .τ G : AG → B G over the group G and the alphabets A and B. The cellular automaton .τ G is called the induction (cellular automaton) of .τ (from H to G). We denote by .CA(G, H ; A) the set of all cellular automata .τ : AG → AG admitting a memory set S such that .S ⊂ H . This is a submonoid of .CA(G; A) isomorphic to .CA(H ; A) (cf. [CAG, Proposition 1.7.1 and Proposition 1.7.2]).

1.1.7 Invertible Cellular Automata Let G be a group and let A be a set. One says that a cellular automaton .τ : AG → AG is invertible (or reversible) if .τ is bijective and the inverse map .τ −1 : AG → AG is also a cellular automaton. This is equivalent to the existence of a cellular automaton G → AG such that .τ ◦σ = σ ◦τ = Id . The set of invertible cellular automata .σ : A AG over the group G and the alphabet A is exactly the group .ICA(G; A) consisting of all invertible elements of the monoid .CA(G; A). If .AG is equipped with its prodiscrete uniform structure, a map .τ : AG → AG is an invertible cellular automaton if and only if .τ is a G-equivariant uniform automorphism of .AG [CAG, Theorem 1.10.1]. In the case when the alphabet set A is finite, every bijective cellular automaton G → AG is invertible [CAG, Theorem 1.10.2]. .τ : A Invertibility of cellular automata is preserved under the operations of induction and restriction. More precisely, if .τ : AG → AG is a cellular automaton admitting a memory set contained in a subgroup H of G and .τH : AH → AH denotes the cellular automaton over H obtained by restriction of .τ , then .τ is invertible if and only if .τH is invertible [CAG, Proposition 1.10.4].

1.2 Exercises

5

1.2 Exercises Exercise 1.1 A uniform space X is called uniformly homogeneous if, for all x, y ∈ X, there exists a uniform automorphism φ : X → X such that φ(x) = y. Let G be a group and let A be a set. Show that AG , equipped with its prodiscrete uniform structure, is uniformly homogeneous. Solution Let x, y ∈ AG . For g ∈ G define φg ∈ Sym(A) by setting ⎧ φg :=

.

IdA

if x(g) = y(g)

(x(g) y(g))

otherwise,

where (a b) denotes the transposition exchanging two distinct elements a, b ∈ A. Observe that, if A is equipped with the discrete uniform structure, then φg : A → A is a uniform automorphism. As the product of uniform automorphisms is a uniform automorphism, we deduce that the map φ : AG → AG defined by φ(z)(g) := φg (z(g))

.

for all z ∈ AG and g ∈ G is a uniform automorphism of AG (equipped with its prodiscrete uniform structure). Moreover, φ(x)(g) = φg (x(g)) = y(g) for all g ∈ G, that is, φ(x) = y. This shows that AG is uniformly homogeneous. █ Comment Observe that the uniform automorphism φ we constructed is involutive, i.e., φ ◦ φ is the identity map on AG . Exercise 1.2 Let G be a countable group and let A be a set. Fix a non-decreasing sequence ∅ = E0 ⊂ E1 ⊂ · · · ⊂ En ⊂ · · ·

.

 of finite subsets of G such that n≥0 En = G. For x, y ∈ AG , define v(x, y) ∈ N ∪ {∞} and 0 ≤ d(x, y) ≤ 1 by v(x, y) := sup{n ≥ 0 : x|En = y|En }

.

and d(x, y) := 2−v(x,y) ,

.

with the usual convention 2−∞ = 0.

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1 Cellular Automata

(a) Show that d(x, y) ≤ max{d(x, z), d(z, y)}

.

(1.6)

for all x, y, z ∈ AG . (b) Show that d is a metric on AG . (c) Show that the uniform structure defined by d is the prodiscrete uniform structure on AG . (d) Show that the topology defined by d is the prodiscrete topology on AG . (e) Show that the metric space (AG , d) is complete. Solution (a) Let x, y, z ∈ AG . If x|En = z|En and z|En = y|En , then x|En = y|En . It follows that v(x, y) ≥ min{v(x, z), v(z, y)}

.

and hence d(x, y) = 2−v(x,y) ≤ 2− min{v(x,z),v(z,y)} = max{2−v(x,z) , 2−v(z,y) }

.

= max{d(x, z), d(z, y)}. This gives us (1.6). (b) For all x, y ∈ AG , we have 0 ≤ v(x, y) ≤ ∞ so that 0 ≤ d(x, y) ≤ 1. On the other hand, d(x, y) = 0 if and only if v(x, y) = ∞, that is, x = y. Moreover, we clearly have v(x, y) = v(y, x) and hence d(x, y) = d(y, x). Finally, the triangle inequality follows from (1.6) which gives us d(x, y) ≤ max{d(x, z), d(z, y)} ≤ d(x, z) + d(z, y)

.

for all x, y, z ∈ AG . (c) Denote by U the uniform structure on AG defined by d. Then the sets W (d, r) ⊂ AG × AG defined by W (d, r) := {(x, y) ∈ AG × AG : d(x, y) < r},

.

where r > 0, form a base of entourages of U (cf. [CAG, Example B.1.5]). Recall that the sets WΩ := {(x, y) ∈ AG × AG : x|Ω = y|Ω },

.

where Ω runs over all finite subsets of G, form a base of entourages of the prodiscrete uniform structure on AG . Observe that W (d, 2−n ) = WEn

.

(1.7)

1.2 Exercises

7

for all n ≥ 0. Now, for every r > 0, there exists n ≥ 0 such that 2−n ≤ r. As this implies W (d, 2−n ) ⊂ W (d, r), we deduce that the sets W (d, 2−n ), n ≥ 0, form a base of entourages of U . On the other hand, for every finite subset Ω ⊂ G, there exists n ≥ 0 such that Ω ⊂ En . As this implies WEn ⊂ WΩ , it follows that the sets WEn , n ≥ 0, form a base of entourages for the prodiscrete uniform structure on AG . Thus, we deduce from (1.7) that U coincides with the prodiscrete uniform structure on AG . (d) This immediately follows from (c). (e) Let (xn )n≥0 be a Cauchy sequence in the metric space (AG , d). Given N ≥ 0, there exists n0 = n0 (N ) ≥ 0 such that d(xn , xm ) ≤ 2−N for all n, m ≥ n0 . The latter means that xn (g) = xm (g) for all g ∈ EN and n, m ≥ n0 . It follows that, for every g ∈ G, the A-valued sequence (xn (g))n≥0 is eventually constant. Define x ∈ AG by setting x(g) := xn (g) if g ∈ EN and n ≥ n0 (N ).

.

Let us show that x = limn→∞ xn in AG . Fix ε > 0. Let N ≥ 0 be such that 2−N < ε and set n0 = n0 (N ) as above. If n ≥ n0 , we then have x(g) = xn (g) for all g ∈ EN , that is, x|EN = xn |EN . It follows that, for n ≥ n0 , we have d(x, xn ) ≤ 2−N < ε. Thus x = limn→∞ xn . This shows that the metric space (AG , d) is complete. █ Comment Inequality (1.6) is called the ultrametric inequality. A metric space (X, d) for which d satisfies the ultrametric inequality is called an ultrametric space. A classical example of an ultrametric space is provided by the field Qp of p-adic numbers, equipped with the p-adic metric, for any prime p (see e.g. [Robe]). A topological space that is homeomorphic to a complete metric space is called a completely metrizable space. Exercise 1.2 shows that, if G is a countable group and A is a set, then AG is completely metrizable for the prodiscrete topology. Exercise 1.3 (Ultrauniform Spaces) A uniform structure U on a set X is called a ultrauniform structure if U admits a base of entourages consisting of graphs of equivalence relations. In other words, U is a ultrauniform structure if for every entourage U ∈ U , there exists an entourage V ∈ U which is symmetric (i.e., V = V ∗ := {(y, x) : (x, y) ∈ V }) and transitive (i.e., V ⊃ V ◦ V := {(x, z) : there exists y ∈ X such that (x, y), (y, z) ∈ V }) such that V ⊂ U . A set equipped with a ultrauniform structure is called a ultrauniform space. (a) Show that a set equipped with its discrete uniform structure is a ultrauniform space. (b) Show that the product of a family of ultrauniform spaces, equipped with its product uniform structure, is a ultrauniform space. (c) Show that the prodiscrete uniform structure on the product of a family of sets is ultrauniform. (d) Let X be a ultrauniform space and let Y ⊂ X be a subset. Show that the uniform structure induced on Y by the uniform structure on X is also ultrauniform. (e) Let (X, d) be a metric space and suppose that d satisfies the ultrametric inequality (1.6). Show that the uniform structure defined by d on X is ultrauniform.

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Solution (a) The discrete uniform structure on a set X is the uniform structure Ud whose entourages consist of all subsets of X × X containing the diagonal ΔX := {(x, x) : x ∈ X} ⊂ X × X. As ΔX ∈ Ud is the graph of the equality relation on X, which is an equivalence relation, it follows that Ud is an ultrauniform structure. Π (b) Let (Xλ , Uλ )λ∈Λ be a family of ultrauniform spaces and equip X := λ∈Λ Xλ with the product uniform structure U . Recall that a base of entourages for U is obtained by taking all subsets of X × X which are of the form Π .

λ∈Λ

Vλ ⊂

Π λ∈Λ

( (Xλ × Xλ ) =

Π λ∈Λ

) Xλ ×

(

Π

) Xλ

λ∈Λ

where Vλ ∈ Uλ for all λ ∈ Λ and Vλ = Xλ × Xλ for all but finitely many λ ∈ Λ. Let U ∈ U . Then there exist a finite subset Λ0 ⊂ Λ and Π entourages Uλ ∈ Uλ with Uλ = Xλ × Xλ for all λ ∈ Λ \ Λ0 such that λ∈Λ Uλ ⊂ U . Since Uλ is ultrauniform, we can find, for each λ ∈ Λ0 , a symmetric and transitive entourage Π Vλ ⊂ Uλ . For λ ∈ Λ \ Λ0 , set Vλ := Xλ × Xλ . Then the entourage V := λ∈Λ Vλ ∈ U is clearly symmetric and transitive. As V ⊂ U , we deduce that U is ultrauniform. (c) This follows immediately from (a) and (b). (d) Let UX be a ultrauniform structure on X. Recall that the uniform structure induced on Y by UX is UY := {V ∩ (Y × Y ) : V ∈ UX }. Let UY ∈ UY . Then there exists UX ∈ UX such that UY = UX ∩ (Y × Y ). Since UX is ultrauniform, there exists a symmetric and transitive entourage VX ∈ UX such that VX ⊂ UX . Set VY := VX ∩ (Y × Y ) ∈ UY . Clearly, VY is symmetric and transitive. As VY ⊂ UY , this shows that UY is ultrauniform. (e) If d satisfies the ultrametric inequality, then the basic entourages W (d, ε) := {(x, y) ∈ X × X : d(x, y) < ε}, ε > 0, are clearly both symmetric and transitive. Therefore, the uniform structure defined by d is ultrauniform. █ Comment Ultrauniform spaces are also called non-Archimedean in the literature. For more on ultrauniform spaces, the reader is refered to the paper by Windisch [Win] and the references therein. Exercise 1.4 (First Axiom of Countability) One says that a topological space X satisfies the first axiom of countability if every point x ∈ X admits a countable base of neighborhoods. (a) Show that every metrizable space satisfies the first axiom of countability. (b) Let G be a group and let A be a set having more than one element. Equip AG with its prodiscrete topology. Show that the following conditions are equivalent: (1) G is countable; (2) AG is metrizable; (3) AG satisfies the first axiom of countability. Solution (a) If X is a metric space and x ∈ X, then the open balls centered at x of radius 2−n , where n runs over N, form a countable base of neighborhoods of x.

1.2 Exercises

9

(b) The implication (1) =⇒ (2) follows from Exercise 1.2 while (2) =⇒ (3) follows from (a). To prove that (3) =⇒ (1), we proceed by contradiction. Suppose that G is uncountable and AG satisfies the first axiom of countability. Fix some configuration x ∈ AG . Since AG satisfies the first axiom of countability, there exists a countable neighborhood base (Vn )n∈N of x. Recall that the sets V (x, Ω) := {y ∈ AG : y|Ω = x|Ω }, where Ω runs over all finite subsets of G, also form a neighborhood base of x. Thus, for each n ∈ N, there is a finite subset Ωn ⊂ G such that V (x, Ωn ) ⊂ Vn . Since G is uncountable and  a countable union of finite sets is countable, we can find an element g ∈ G \ n∈N Ωn . As V (x, {g}) is a neighborhood of x, there exists n ∈ N such that Vn ⊂ V (x, {g}). This implies V (x, Ωn ) ⊂ V (x, {g}). Now, as A has more than one element, we can construct a configuration y ∈ AG which coincides with x on G \ {g} and satisfies y(g) /= x(g). Then y is in V (x, Ωn ) but not in V (x, {g}). This gives us a contradiction. Therefore (3) =⇒ (1). █ Exercise 1.5 Let G be a group and let A be a set. Let Pf (G) denote the set of all finite subsets of G. Recall that a (G, A)-pattern is a map p : Ω → A, where Ω = supp(p) ∈ Pf (G) is the support of p. Denote by P(G, A) the set consisting of all (G, A)-patterns. For g ∈ G and p ∈ P(G, A), we denote by gp the (G, A)-pattern defined by supp(gp) := g supp(p) and (gp)(h) := p(g −1 h) for all h ∈ supp(gp). (a) Show that the map α : G × P(G, A) → P(G, A), defined by α(g, p) := gp for all g ∈ G and p ∈ P(G, A), is an action of the group G on the set P(G, A). (b) Equip the Cartesian product AG × Pf (G) with the diagonal action of G, defined by g(x, Ω) := (gx, gΩ) for all x ∈ AG and Ω ∈ Pf (G). Show that the map Φ : AG × Pf (G) → P(G, A), defined by Φ(x, Ω) := x|Ω for all x ∈ AG and Ω ∈ Pf (G), is G-equivariant. (c) Let g ∈ G and let Ω ∈ Pf (G). Show that gAΩ = AgΩ .

.

(1.8)

Solution (a) It is immediate to check that 1G p = p and g1 (g2 p) = (g1 g2 )p for all p ∈ P(G, A) and g1 , g2 ∈ G. This shows that α is an action of G on P(G, A). (b) For all g ∈ G, x ∈ AG , and Ω ∈ Pf (G), we have Φ(g(x, Ω)) = Φ(gx, gΩ) = (gx)|gΩ = g(x|Ω ) = gΦ(x, Ω).

.

This shows that Φ is G-equivariant. (c) Let p ∈ AΩ . Then supp(p) = Ω and supp(gp) = g supp(p) = gΩ so that gp ∈ AgΩ . This shows the inclusion gAΩ ⊂ AgΩ . We then have g −1 AgΩ ⊂ −1 Ag gΩ = AΩ , so that AgΩ ⊂ gAΩ . This shows (1.8). █ Exercise 1.6 (Second Axiom of Countability) One says that at topological space satisfies the second axiom of countability if it admits a countable base of open subsets.

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(a) Show that every topological space that satisfies the second axiom of countability also satisfies the first. (b) Let G be a group and let A be a set having more than one element. Equip AG with its prodiscrete topology. Show that AG satisfies the second axiom of countability if and only if G and A are both countable. (c) Let G be a countable group and let A be an uncountable set. Show that AG , equipped with its prodiscrete topology, satisfies the first axiom of countability but not the second. Solution (a) Let X be a topological space that satisfies the first axiom of countability. If B is a countable base of open subsets of X and x ∈ X, then the set consisting of all U ∈ B such that x ∈ U is a countable base of neighborhoods of x. Therefore X satisfies the first axiom of countability. (b) Suppose first that AG satisfies the second axiom of countability. As the cylinders Ca := {x ∈ AG : x(1G ) = a}, a ∈ A, are non-empty and pairwise disjoint open subsets of AG , the set A must be countable. From (a) and Exercise 1.4(b), we deduce that G is countable. Conversely, suppose now that G and A are both countable. Let P(G, A) denote the set of all patterns over the group G and the alphabet A. Given a finite subset Ω ⊂ G, there are only countably many pattterns p ∈ P(G, A) with support Ω since A is countable. On the other hand, as G is countable, there are only countably many finite subsets Ω ⊂ G. Consequently, the set P(G, A) is countable. As the cylinders C(p), p ∈ P(G, A), form a base of open subsets of AG , we deduce that AG satisfies the second axiom of countability. (c) The space AG does not satisfy the second axiom of countability by (b). It satisfies the first by Exercise 1.4(b) █ Comment A topological space that is completely metrizable and satisfies the second axiom of countability is called a Polish space. Let G be a group and let A be a set with more than one element. Equip AG with its prodiscrete topology. It immediately follows from Exercise 1.6(b) and Exercise 1.2 that AG is a Polish space if and only if G and A are both countable. Exercise 1.7 Let G be a group and let A be a set having more than one element. Let Ω1 and Ω2 be two finite subsets of G. (a) Show that WΩ1 ⊂ WΩ2 if and only if Ω2 ⊂ Ω1 . (b) Show that WΩ1 ∪Ω2 = WΩ1 ∩ WΩ2 . (c) Show that WΩ1 ⊂ WΩ2 if and only if V (x, Ω1 ) ⊂ V (x, Ω2 ) for all x ∈ AG . (d) Let x ∈ AG . Show that V (x, Ω1 ) ⊂ V (x, Ω2 ) if and only if Ω2 ⊂ Ω1 . (e) Show that V (x, Ω1 ∪ Ω2 ) = V (x, Ω1 ) ∩ V (x, Ω2 ) for all x ∈ AG . Solution (a) Let (x, y) ∈ AG × AG . Then ( ) WΩ1 ⊂ WΩ2 ⇔ (x|Ω1 = y|Ω1 ) ⇒ (x|Ω2 = y|Ω2 ) ⇔ Ω2 ⊂ Ω1 .

.

1.2 Exercises

11

(b) Let (x, y) ∈ AG × AG . We have (x, y) ∈ WΩ1 ∩ WΩ2 ⇔ x|Ω1 = y|Ω1 and x|Ω2 = y|Ω2 ⇔ x|Ω1 ∪Ω2 = y|Ω1 ∪Ω2

.

⇔ (x, y) ∈ WΩ1 ∪Ω2 . Note that the inclusion WΩ1 ∪Ω2 ⊂ WΩ1 ∩ WΩ2 also follows from (a). (c) Suppose that WΩ1 ⊂ WΩ2 . Then, for all x ∈ AG , we have V (x, Ω1 ) = {y ∈ G A : (x, y) ∈ WΩ1 } ⊂ {y ∈ AG : (x, y) ∈ WΩ2 } = V (x, Ω2 ). Conversely, suppose that V (x, Ω1 ) ⊂ V (x, Ω2 ) for all x ∈ AG . We then have WΩ1 = {(x, y) ∈ AG × AG : y ∈ V (x, Ω1 )}

.

⊂ {(x, y) ∈ AG × AG : y ∈ V (x, Ω2 )} = WΩ2 , showing that WΩ1 ⊂ WΩ2 . (d) and (e) follow from (a) and (b), respectively, after observing that V (x, Ω) = {y ∈ AG : (x, y) ∈ WΩ } for all x ∈ AG and Ω ⊂ G. █ Exercise 1.8 (Topologically Transitive Actions) An action of a group G on a topological space X is said to be topologically transitive if, for all non-empty open subsets U and V of X, there exists an element g ∈ G such that U ∩ gV /= ∅. (a) Suppose that a group G acts continuously on a topological space X. For Y ⊂ ◦

X, denote by Y (resp. Y ) the interior (resp. closure) of Y in X. Show that if Y is a ◦

G-invariant subset of X then the sets Y and Y are both G-invariant. (b) Suppose that a group G acts continuously on a topological space X. Show that the following conditions are equivalent: (i) the action of G on X is topologically transitive; (ii) every non-empty G-invariant open subset U of X is dense in X; (iii) if F is a G-invariant closed subset of X with non-empty interior then F = X. (c) Suppose that a group G acts on a uniform space X. Show that the action of G on X is topologically transitive if and only if the following holds: (UTT)

for all x1 , x2 ∈ X and every entourage W of X, there exists g ∈ G and x ∈ X such that (x, x1 ) ∈ W and (gx, x2 ) ∈ W.

.

12

1 Cellular Automata

(d) Suppose that a group G acts on a metric space (X, d). Show that the action of G on X is topologically transitive if and only if the following holds: for all x1 , x2 ∈ X and ε > 0, there exists g ∈ G and x ∈ X such that

(MTT)

d(x, x1 ) < ε and d(gx, x2 ) < ε.

.

Solution (a) Suppose that Y is a G-invariant subset of X. Denoting by T the set of open subsets of X, we have

◦

Y =

.

U.

U ∈T and U ⊂Y

Let g ∈ G. Using the fact that the map x |→ gx is a homeomorphism of X, we get ⎛ ◦

gY = g ⎝



.

U ∈T and U ⊂Y

⎞ U⎠ =





gU =

U ∈T and U ⊂Y

U.

U ∈T and U ⊂g −1 Y ◦

◦

◦

As g −1 Y = Y by G-invariance of Y , we deduce that g Y = Y . This shows that Y is G-invariant. Since G \ Y is G-invariant and Y is equal to the complement in X of the interior of G \ Y , we deduce that Y is also G-invariant. (b) If U is a non-empty open G-invariant subset of X and the action of G is topologically transitive, then U meets every non-empty open subset V of X so that U is dense in X. This shows (i) =⇒ (ii). Conversely, suppose (ii) and let U, V be  non-empty open subsets of X. Then V ' := g∈G gV is a non-empty G-invariant open subset of X. By (ii), the set V ' meets U , so that there exists g ∈ G such that gV meets U . This shows that (ii) =⇒ (i). To prove (ii) =⇒ (iii) (resp. (iii) =⇒ (ii)), it suffices to take as U the interior of F (resp. to take as F the closure of U ) and use (a). (c) Suppose first that the action of G on X is topologically transitive. Let x1 , x2 ∈ X and let W be an entourage of X. Then, for each i ∈ {1, 2}, the set W [xi ] := {y ∈ X : (y, xi ) ∈ W } is an neighborhood of xi (cf. [CAG, Section B.1]). Therefore, there exist open sets U and V of X such that x2 ∈ U ⊂ W [x2 ] and x1 ∈ V ⊂ W [x1 ]. By topological transitivity, we can find g ∈ G such that U ∩ gV /= ∅. This implies g −1 U ∩ V = g −1 (U ∩ gV ) /= ∅. As any x ∈ g −1 U ∩ V satisfies that (gx, x2 ) ∈ W and (x, x1 ) ∈ W , this shows that condition (UTT) is satisfied. Conversely, suppose that the action of G on X satisfies (UTT). Let U and V be two non-empty open subsets of X. Let x2 ∈ U and x1 ∈ V . Then, for each i ∈ {1, 2}, we can find an entourage Wi of X such that Wi [xi ] := {y ∈ X : (y, xi ) ∈ Wi }, i = 1, 2, satisfy W1 [x1 ] ⊂ V and W2 [x2 ] ⊂ U . As W := W1 ∩ W2 is an entourage of X, it follows from condition (UTT) that there exists g ∈ G and x ∈ X such that (x, x1 ) ∈ W and (gx, x2 ) ∈ W . Then x ∈ W [x2 ] ∩ g −1 W [x1 ] ⊂ W2 [x2 ] ∩

1.2 Exercises

13

g −1 W1 [x1 ] ⊂ V ∩ g −1 U . We deduce that U ∩ gV = g(V ∩ g −1 U ) /= ∅. This shows that the action of G on X is topologically transitive. (d) This immediately follows from (c) since the sets Wε := {(y, z) ∈ X × X : d(y, z) < ε}, ε > 0, form a base of entourages for the uniform structure on X █ associated with the metric d (cf. [CAG, Example B.1.5]). Exercise 1.9 (Irreducible Actions) An action of a group G on a topological space X is said to be irreducible if, for every finite subset F ⊂ G and all non-empty open subsets U and V of X, there exists g ∈ G \ F such that U ∩ gV /= ∅. (a) Show that a finite group admits no irreducible action. (b) Show that every irreducible action is topologically transitive. (c) Suppose that a group G acts on a uniform space X. Show that the action of G on X is irreducible if and only if the following holds: (UI)

for every finite subset F ⊂ G, for all x1 , x2 ∈ X, and for every entourage W of X, there exists g ∈ G \ F and x ∈ X such that (x, x1 ) ∈ W and (gx, x2 ) ∈ W.

.

(d) Suppose that a group G acts on a metric space (X, d). Show that the action of G on X is irreducible if and only if the following holds: (MI)

for every finite subset F ⊂ G, for all x1 , x2 ∈ X, and for every ε > 0, there exists g ∈ G \ F and x ∈ X such that d(x, x1 ) < ε and d(gx, x2 ) < ε.

.

Solution (a) Suppose that G is a finite group. Then, taking F := G, we have G \ F = G \ G = ∅. Therefore G admits no irreducible action on any topological space X. (b) Suppose that the action of a group G on a topological space X is irreducible. Let U and V be non-empty subsets of X. Taking F := ∅ in the definition of irreduciblity, we see that there exists g ∈ G such that U ∩ gV /= ∅. This shows that the action of G on X is topologically transitive. (c) The proof is, mutatis mutandis, as the one in Exercise 1.8(c). (d) The proof is, mutatis mutandis, as the one in Exercise 1.8(d). █ Exercise 1.10 Let X be a topological space and let f : X → X be a homeomorphism. Recall that the action of Z on X generated by f is the continuous action α : Z × X → X defined by α(n, x) := f n (x) for all (n, x) ∈ Z × X. Show that the action of Z on X generated by f is irreducible if and only if f satisfies the following condition: (HI)

for all non-empty open subsets U and V of X, there exists an integer k ≥ 1 such that U ∩ f k (V ) /= ∅.

14

1 Cellular Automata

Solution By definition, the action generated by f is irreducible if and only if f satisfies the following condition: (C)

for every finite subset F ⊂ Z and all non-empty open subsets U and V of X, there exists n ∈ Z \ F such that U ∩ f n (V ) /= ∅.

Suppose (C). Let U and V be non-empty open subsets of X. By applying (C) with F := ∅, we see that there exists n ∈ Z such that U ∩ f n (V ) /= ∅. If n ≥ 1, then (HI) is satisfied with k := n ≥ 1. Suppose now n ≤ 0. Consider the non-empty open subset W ⊂ X defined by W := U ∩ f n (V ). By applying (C) with U and V both replaced by W and F replaced by the interval {n, n + 1, . . . , −n}, we see that there exists m ∈ Z \ {n, n + 1, . . . , −n} such that W ∩ f m (W ) /= ∅. We have f m (W ) = f m (U ∩ f n (V )) = f m (U ) ∩ f m+n (V ) since f is bijective. Therefore U ∩ f n (V ) ∩ f m (U ) ∩ f m+n (V ) /= ∅.

.

(1.9)

If m ≥ −n+1, then (HI) is satisfied by taking k := m+n ≥ 1 since U ∩f m+n (V ) /= ∅ by (1.9). Otherwise, we have m ≤ n − 1 and we can take k := n − m ≥ 1 since U ∩ f n−m (V ) = f −m (f m (U ) ∩ f n (V )) /= ∅ by (1.9). This shows that (C) implies (HI). Conversely, suppose that (HI) is satisfied. Let U and V be non-empty subsets of X. Let us first show, by induction on n, that, for every n ∈ N, there exists an integer k ≥ n + 1 such that U ∩ f k (V ) /= ∅. For n = 0, this directly follows from (HI). Suppose now that, for some given n ∈ N, there exists k ≥ n + 1 such that U ∩f k (V ) /= ∅. By applying (HI) to the non-empty open sets U and f k (V ), we see ' that there exists an integer k ' ≥ 1 such that U ∩ f k +k (V ) /= ∅. As k ' + k ≥ n + 2, this completes induction. Now let F ⊂ Z be a finite subset. Choose n ∈ N such that F ⊂ (−∞, n]. By the first part of the proof, there exists an integer k ≥ n + 1 such that U ∩ f k (V ) /= ∅. We have k ∈ Z \ F since k ≥ n + 1. This shows that (HI) implies (C). █ Exercise 1.11 (Topologically Mixing Actions) An action of a group G on a topological space X is said to be topologically mixing if, for all non-empty open subsets U and V of X, there exists a finite subset F ⊂ G such that U ∩ gV /= ∅ for all g ∈ G \ F . (a) Suppose that a finite group G acts on a topological space X. Show that the action of G on X is topologically mixing. (b) Suppose that an infinite group G acts on a topological space X. Show that if this action is topologically mixing then it is irreducible and topologically transitive. (c) Suppose that a group G acts on a uniform space X. Show that the action of G on X is topologically mixing if and only if the following holds: (UTM)

for all x1 , x2 ∈ X and for every entourage W of X, there exists a finite subset F ⊂ G such that, for every g ∈ G \ F , there exists x ∈ X such that (x, x1 ) ∈ W and (gx, x2 ) ∈ W.

.

1.2 Exercises

15

(d) Suppose that a group G acts on a metric space (X, d). Show that the action of G on X is topologically mixing if and only if the following holds: (MTM)

for all x1 , x2 ∈ X and for every ε > 0, there exists a finite subset F ⊂ G such that for all g ∈ G \ F there exists x ∈ X such that d(x, x1 ) < ε and d(gx, x2 ) < ε.

.

Solution (a) As G is finite, we can take F := G in the definition. (b) Let F ⊂ G be a finite subset and let U and V be two non-empty open subsets of X. By topological mixing, there exists a finite subset F0 ⊂ G such that U ∩ gV /= ∅ for all g ∈ G \ F0 . As G \ (F ∪ F0 ) ⊂ G \ F0 , this implies that any g ∈ G \ (F ∪ F0 ) satisfies U ∩ gV /= ∅. Note that G \ (F ∪ F0 ) /= ∅ since G is infinite. As G \ (F ∪ F0 ) ⊂ G \ F , we deduce that there exists g ∈ G \ F such that U ∩ gV /= ∅. This shows that the action of G on X is irreducible. It then follows from Exercise 1.9(b) that the action is also topologically transitive. (c) The proof is, mutatis mutandis, as the one in Exercise 1.8(c). (d) The proof is, mutatis mutandis, as the one in Exercise 1.8(d). █ Exercise 1.12 Let G be a group and let A be a set. Equip AG with its prodiscrete topology. (a) Show that the shift action of G on AG is topologically mixing. (b) Suppose that the group G is infinite. Show that the shift action of G on AG is irreducible and topologically transitive. (c) Suppose that the group G is finite and that the set A has more than one element. Show that the shift action of G on AG is neither irreducible nor topologically transitive. Solution (a) Let U1 and U2 be non-empty open subsets of AG . Then, using the notation introduced in (1.2), there exist, for each i ∈ {1, 2}, a finite subset Ωi ⊂ G and a pattern pi ∈ AΩi such that C(pi ) ⊂ Ui .

.

For all g ∈ G, we have that gC(p2 ) = C(gp2 ),

.

where gp2 is the pattern with support gΩ2 defined by gp2 (h) := p2 (g −1 h) for all h ∈ gΩ2 . Consider now the finite subset F ⊂ G given by F := Ω1 Ω2−1 . Observe that if g ∈ G \ F then Ω1 ∩ gΩ2 = ∅. This implies C(p1 ) ∩ C(gp2 ) /= ∅

.

16

1 Cellular Automata

since we can construct a configuration x ∈ C(p1 ) ∩ C(gp2 ) by arbitrarily extending the pattern with support Ω1 ∪ gΩ2 which coincides with p1 on Ω1 and with gp2 on gΩ2 . As C(p1 ) ∩ C(gp2 ) = C(p1 ) ∩ gC(p2 ) ⊂ U1 ∩ gU2 ,

.

we deduce that U1 ∩ gU2 /= ∅ for all g ∈ G \ F . This shows that the shift action of G on AG is topologically mixing. (b) This immediately follows from (a) and Exercise 1.11(b). (c) Since G is finite, the prodiscrete topology on AG is the discrete one. Let a1 and a2 be two distinct elements in the alphabet A. Consider, for each i ∈ {1, 2}, the constant configuration xi ∈ AG defined by xi (g) := ai for all g ∈ G and the open subset Ui ⊂ AG given by Ui := {xi }. As x1 and x2 are fixed by the action of G, there is no g ∈ G such that U1 ∩ gU2 /= ∅. It follows that the action of G on AG is not topologically transitive. It is not irreducible either by Exercise 1.9(c) (or, more directly, by Exercise 1.9(b)). █ Exercise 1.13 Let G be a group acting on two topological spaces X1 and X2 . Suppose that there exists a continuous G-equivariant surjective map f : X1 → X2 . Show that if the action of G on X1 is topologically transitive (resp. irreducible, resp. topologically mixing) then the action of G on X2 is topologically transitive (resp. irreducible, resp. topologically mixing). Solution Let U2 and V2 be any two non-empty open subsets of X2 . Observe that U1 := f −1 (U2 ) and V1 := f −1 (V2 ) are non-empty open subsets of X1 since f is continuous and surjective. Note also that, for all g ∈ G, we have f −1 (gV2 ) = gf −1 (V2 ) by the G-equivariance of f and hence U1 ∩ gV1 = f −1 (U2 ) ∩ gf −1 (V2 ) = f −1 (U2 ) ∩ f −1 (gV2 ) = f −1 (U2 ∩ gV2 ). Using the surjectivity of f , we deduce that if U2 ∩ gV2 /= ∅ then U1 ∩ gV1 /= ∅. Suppose first that the action of G on X1 is topologically transitive. Then there exists g ∈ G such that U1 ∩ gV1 /= ∅. By the above observation, this implies U2 ∩ gV2 /= ∅. This shows that the action of G on X2 is topologically transitive. Suppose now that the action of G on X1 is irreducible and let F ⊂ G be a finite subset. Then there exists g ∈ G \ F such that U1 ∩ gV1 /= ∅. We then deduce as above that U2 ∩ gV2 /= ∅. This shows that the action of G on X2 is irreducible. Finally, suppose that the action of G on X1 is topologically mixing. Then there exists a finite subset F ⊂ G such that U1 ∩ gV1 /= ∅ for all g ∈ G \ F . As above, this implies that U2 ∩ gV2 /= ∅ for all g ∈ G \ F . This shows that the action of G on X2 is topologically mixing. █ Exercise 1.14 (Nilpotent Cellular Automata) Let G be a group and let A be a set. One says that a cellular automaton τ : AG → AG is nilpotent if there exists an integer n ≥ 0 satisfying the following property: there exists a configuration y ∈ AG such that every configuration x ∈ AG satisfies τ n (x) = y. The smallest integer n ≥ 0 satisfying this property is then called the nilpotency class of the nilpotent cellular automaton τ .

1.2 Exercises

17

Let τ : AG → AG be a nilpotent cellular automaton with nilpotency class n. (a) Show that there exists a unique constant configuration cτ ∈ AG such that every x ∈ AG satisfies τ n (x) = cτ . (b) Show that Fix(τ ) = {cτ }. (c) Show that τ k = τ n for every integer k ≥ n. (d) Let σ : AG → AG be a cellular automaton commuting with τ . Show that τ ◦ σ is a nilpotent cellular automaton with nilpotency class ≤ n and that cτ ◦σ = cτ . (e) Let k ≥ 1 be an integer. Show that τ k is a nilpotent cellular automaton with nilpotency class ≤ n and that one has cτ k = cτ . (f) Let σ : AG → AG be a cellular automaton. Show that σ is nilpotent if and only if there is an integer k ≥ 1 such that σ k is nilpotent. Solution (a) Since τ is nilpotent with nilpotency class n, there is a configuration y ∈ AG such that τ n (x) = y for all x ∈ AG . As τ is G-equivariant, we have gy = gτ n (x) = τ n (gx) = y.

.

It follows that the configuration y is fixed by the G-shift action and hence constant. Setting cτ := y, we have τ n (x) = cτ for all x ∈ AG . Uniqueness of cτ is obvious. (b) We have τ (cτ ) = τ (τ n (cτ )) = τ n+1 (cτ ) = τ n (τ (cτ )) = cτ , showing that cτ ∈ Fix(τ ). Conversely, if x ∈ Fix(τ ), then τ n (x) = x, showing that cτ = x. Thus Fix(τ ) = {cτ }. (c) Let k ≥ n. For all x ∈ AG , we have τ k (x) = τ n (τ k−n (x)) = cτ = τ n (x).

.

This shows that τ k = τ n . (d) As τ and σ commute, we have (τ ◦ σ )n = τ n ◦ σ n . It follows that (τ ◦ σ )n (x) = τ n (σ n (x)) = cτ

.

for all x ∈ AG . This shows that τ ◦ σ is nilpotent with nilpotency class ≤ n and that cτ ◦σ = cτ . (e) This immediately follows from (d) after taking σ := τ k−1 so that τ ◦ σ = τ ◦ τ k−1 = τ k . (f) Necessity follows from (e). Conversely, suppose that there is an integer k ≥ 1 such that σ k is nilpotent. Denoting by n the nilpotency class of σ k , there is an integer n ≥ 1 such that σ kn (AG ) = (σ k )n (AG ) is reduced to a single configuration. It follows that σ is nilpotent (with nilpotency class ≤ kn). █ Comment It is known that, for any integer d ≥ 1, the problem of deciding whether a given cellular automaton over Zd is nilpotent or not is undecidable. This result ˇ was first established for d ≥ 2 by Culík, Pachl, and Yu in [CulPY] and then

18

1 Cellular Automata

extended to the case d = 1 by Kari in [Kar1]. The proofs consist in reducing to the undecidability of certain tiling problems of the plane. Exercise 1.15 (Example of a Nilpotent Cellular Automaton of Nilpotency Class n) Let A := {0, 1} and let n be a positive integer. Consider the subset S ⊂ Z given by S := {−1, 0, 1, . . . , n − 2} and the map μ : AS → A defined by ⎧ ⎪ ⎪ ⎨0 .μ(p) := 0 ⎪ ⎪ ⎩p(0)

if p(−1) = p(0) = p(1) = · · · = p(n − 2) = 1, if p(−1) = 0 and p(0) = 1, otherwise,

for all p ∈ AS . Show that the cellular automaton τ : AZ → AZ admitting S as a memory set and μ as the associated local defining map is nilpotent of class n. Solution Let x ∈ AZ . Observe that every chain of 1s appearing in the configuration τ (x) has length at most n − 1. Moreover, every further application of τ has the effect of suppressing the 1 located on the left of each chain of 1s. This implies that τ n (x) = c0 , where c0 ∈ AZ is the constant configuration given by c0 (k) := 0 for all k ∈ Z. This shows that τ is nilpotent with nilpotency class at most n. To prove that the nilpotency class of τ is exactly n, consider, for each integer m such that 0 ≤ m ≤ n the configuration xm ∈ AZ defined by xm (k) := 1 if m ≤ k ≤ n and xm (k) := 0 otherwise. Clearly τ (x0 ) = x2 and τ (xm ) = xm+1 for all 2 ≤ m ≤ n − 1. Therefore τ n−1 (x0 ) = xn /= c0 . It follows that the cellular automaton τ is nilpotent of nilpotency class n. █ Exercise 1.16 Let G be a group and let A be a set. Let H be a subgroup of G and suppose that τ : AG → AG is a cellular automaton admitting a memory set contained in H . Show that τ is nilpotent of nilpotency class n if and only if the restriction cellular automaton τH : AH → AH is nilpotent of nilpotency class n. Solution Consider the monoid CA(G, H ; A) consisting of all cellular automata σ : AG → AG admitting a memory set contained in H . We know from [CAG, Proposition 1.7.2] that the restriction map σ |→ σH is a monoid isomorphism from CA(G, H ; A) onto CA(H ; A). Observe also that it immediately follows from the definition of the restriction map that σ is constant if and only if σH is constant. More precisely, given a ∈ A, we have σ = ca , where ca ∈ CA(G, H ; A) is the constant cellular automaton given by ca (x)(g) := a for all x ∈ AG and g ∈ G, if and only if σH = da , where da ∈ CA(H ; A) is the constant cellular automaton given by da (y)(h) := a for all y ∈ AH and h ∈ H . Suppose now that τ is nilpotent of nilpotency class at most n. Then the cellular automaton τ n is constant, so that there exists a ∈ A such that τ n = ca . This implies that (τH )n = (τ n )H = da . Thus τH is nilpotent with nilpotency class at most n. Conversely, suppose that τH is nilpotent with nilpotency class at most n. Then there exists a ∈ A such that (τ n )H = (τH )n = da . This implies τ n = ca . Therefore τ is nilpotent with nilpotency class at most n.

1.2 Exercises

19

We deduce from the above that τ is nilpotent with nilpotency class n if and only if τH is nilpotent with nilpotency class n. █ Exercise 1.17 Let G be a group. Let A and B be two sets. Let τA : AG → AG and τB : B G → B G be cellular automata. For x ∈ (A × B)G , let xA ∈ AG and xB ∈ B G be the configurations defined by x(g) := (xA (g), xB (g)) for all g ∈ G. Show that the map τ : (A × B)G → (A × B)G given by τ (x)(g) := (τA (xA )(g), τB (xB )(g)) for all g ∈ G is a cellular automaton. Solution Let SA ⊂ G (resp. SB ⊂ G) be a memory set for τA (resp. τB ). Then S := SA ∪ SB is a memory set for both τA and τB . Denote by μA : AS → A (resp. μB : B S → B) the associated local defining map for τA (resp. τB ). For y ∈ (A × B)S , let yA ∈ AS and yB ∈ B S be the patterns defined by y(s) := (yA (s), yB (s)) for all s ∈ S and consider the map μ : (A × B)S → A × B defined by μ(y) := (μA (yA ), μB (yB ))

.

for all y ∈ (A × B)S . Given x ∈ (A × B)G and g ∈ G we have gx = (gxA , gxB ) and x|S = (xA |S , xB |S ) so that τ (x)(g) = (τA (xA )(g), τB (xB )(g)) .

= (μA ((g −1 xA )|S ), μB ((g −1 xB )|S ) = μ((g −1 x)|S ).

This shows that τ is a cellular automaton admitting S as a memory set and μ as the associated local defining map. █ Exercise 1.18 Let G be a group and let S be a finite subset of G of cardinality k. Let A and B be two finite sets with cardinality m := |A| and n := |B|. Show that k there are exactly nm cellular automata τ : AG → B G admitting S as a memory set. Solution Let us denote by CA(G; A, B; S) the set of all cellular automata S τ : AG → B G admitting S as a memory set. The map Φ : B A → CA(G; A, B; S), S which associates with μ ∈ B A the unique cellular automaton τμ admitting S as a memory set and μ as the associated local defining map, is surjective by definition. S Suppose that μ, ν ∈ B A satisfy that τμ = τν . Given y ∈ AS , let x ∈ AG be a configuration extending y. As μ(y) = μ(x|S ) = τμ (x)(1G ) = τν (x)(1G ) = ν(x|S ) = ν(y),

.

20

1 Cellular Automata

we deduce that μ = ν. Thus Φ is also injective, and therefore bijective. As a consequence, we have | CA(G; A, B; S)| = |B A | = |B||A| S

|S|

.

k

= nm . █

Exercise 1.19 Let G be a countable group and let A and B be two finite sets. Show that the set of all cellular automata τ : AG → B G is countable. Solution This follows immediately from Exercise 1.18 and the fact that, G being countable, there are only countably many finite subsets S ⊂ G. █ Exercise 1.20 Let G := Z2 and A := {0, 1}. Let τ : AG → AG denote the cellular automaton associated with the Game of Life (see [CAG, Example 1.4.3.(a)]). Let y ∈ AG be the constant configuration defined by y(g) := 1 for all g ∈ G (all cells are alive). Find a configuration x ∈ AG such that y = τ (x). Solution Consider the configuration x ∈ AG defined by ⎧ x(m, n) :=

.

1 if (m, n) ∈ (3Z × 3Z) ∪ ((Z \ 3Z) × (3Z + 2)) 0 otherwise,

for all (m, n) ∈ G. Observe that, in the configuration x, every cell g ∈ G admits exactly 3 live cells (possibly including itself) in its neighborhood. It follows that τ (x)(g) = 1, as a result of either a birth (if x(g) = 0) or a survival (if x(g) = 1). This shows that τ (x) = y. █ Exercise 1.21 Let A be a set and suppose that G is a trivial group. Show that the monoid CA(G; A) is canonically isomorphic to the monoid Map(A) consisting of all maps from A to A (with composition of maps as the monoid operation). Also show that the group ICA(G; A) is canonically isomorphic to the symmetric group Sym(A) of A. Solution As G = {1G }, every cellular automaton τ ∈ CA(G; A) admits S := {1G } as a memory set. We identify AG = AS and A via the map y |→ y(1G ). From the solution to Exercise 1.18 and the notation therein, we have that CA(G; A) = CA(G; A; {1G }) and the map Φ : Map(A) = AA → CA(G; A) is a bijection. Let us show that Φ is a monoid morphism. Given μ, ν ∈ Map(A) and x ∈ A we have τμ◦ν (x)(1G ) = (μ ◦ ν)(x) = μ(ν(x)) .

= μ(τν (x)) = [τμ (τν (x))](1G ) = (τμ ◦ τν )(x)(1G ).

1.2 Exercises

21

This shows that τμ◦ν (x) = (τμ ◦τν )(x). We deduce that Φ(μ◦ν) = τμ◦ν = τμ ◦τν = Φ(μ) ◦ Φ(ν). We conclude that Φ is a monoid isomorphism. The second statement follows from the first one since ICA(G; A) (respectively Sym(A)) is the subgroup of invertible elements of the monoid CA(G; A) (respectively Map(A)). █ Exercise 1.22 (Constant Cellular Automata) Let G be a group and let A be a set. A configuration x : G → A (resp. a cellular automaton τ : AG → AG ) is said to be constant if it is a constant map. An element u in a monoid M is called left-absorbing (resp. right-absorbing) if every v ∈ M satisfies uv = u (resp. vu = u). (a) Show that a configuration x ∈ AG is constant if and only if it is fixed by the shift action of G on AG . (b) Let τ : AG → AG be a cellular automaton. Show that the following conditions are equivalent: (i) τ is constant; (ii) there exists a constant configuration c ∈ AG such that every x ∈ AG satisfies τ (x) = c; (iii) the minimal memory set of τ is the empty set; (iv) τ is left-absorbing in the monoid CA(G; A). (c) Suppose that A has more than one element. Show that the monoid CA(G; A) contains no right-absorbing elements. Solution (a) If x ∈ AG is constant, then there is a ∈ A such that x(h) = a for all h ∈ G. This implies that (gx)(h) = x(g −1 h) = a = x(h) for all g, h ∈ G and hence gx = x for all g ∈ G. Conversely, suppose that x ∈ AG is fixed by the shift action of G on AG . Then for all g ∈ G, we have x(g) = (g −1 x)(1G ) = x(1G ), showing that the configuration x is constant. (b) Suppose that τ is constant. This means that there exists a configuration y ∈ AG such that every configuration x ∈ AG satisfies τ (x) = y. Using the fact that τ is G-equivariant, we get gy = gτ (x) = τ (gx) = y for all g ∈ G. This shows that y is fixed by the shift action of G on AG . Therefore the configuration y is constant by (a). This proves (i) =⇒ (ii). The converse implication is trivial. Suppose now that there exists a ∈ A such that τ (x)(g) = a for all x ∈ AG and g ∈ G. Then τ is the cellular automaton admitting ∅ as a memory set and local defining map μ : A∅ → A defined by setting μ(p) := a for the unique element p ∈ A∅ . Conversely, suppose that τ admits ∅ as a memory set. Denote by μ : A∅ → A the corresponding local defining map. Then there exists a ∈ A such that μ(p) = a, where p is the unique element in A∅ . Then one has τ (x)(g) = μ((g −1 x)|∅ ) = a for all x ∈ AG and g ∈ G. This proves (ii) ⇐⇒ (iii). Finally, let us show (i) ⇐⇒ (iv). Suppose first that the cellular automaton τ : AG → AG is constant. This means that there is a configuration y ∈ AG such that every configuration x ∈ AG satisfies τ (x) = y. If σ : AG → AG is a cellular automaton, we have τ ◦ σ (x) = τ (σ (x)) = y = τ (x) for all x ∈ AG . Therefore τ ◦ σ = τ . This shows that τ is left-absorbing in CA(G; A). Conversely, suppose

22

1 Cellular Automata

that τ ∈ CA(G; A) is left-absorbing. Choose a constant configuration c ∈ AG and consider the constant cellular automaton σ ∈ CA(G; A) defined by σ (x) := c for all x ∈ AG . As τ is left-absorbing, we have τ = τ ◦ σ . This implies that τ (x) = (τ ◦ σ )(x) = τ (σ (x)) = τ (c) for all x ∈ AG . Therefore τ is constant. (c) Suppose that τ ∈ CA(G; A) is right-absorbing. Let a and b be distinct elements in A. Consider the constant cellular automata σa , σb ∈ CA(G; A) defined, for all x ∈ AG and g ∈ G, by σa (x)(g) := a and σb (x)(g) := b, respectively. As τ is right-absorbing, we have σa ◦ τ = σb ◦ τ = τ . This yields a contradiction since σa ◦ τ = σa and σb ◦ τ = σb , as σa and σb are left-absorbing by (b), and σa /= σb . █ Exercise 1.23 Let G be a group and let A be a set. Let τ : AG → AG be a cellular automaton. Show that τ admits a memory set which is reduced to a single element if and only if there exist an element s ∈ G and a map f : A → A such that one has τ (x)(g) = f (x(gs)) for all x ∈ AG and g ∈ G. Solution Suppose first that τ admits a memory set which is reduced to a single element, say S = {s}, and denote by μ : AS = A → A the associated local defining map. Then for x ∈ AG and g ∈ G we have τ (x)(g) = μ((g −1 x)|S ) = μ(x(gs)),

.

and one can take f := μ. Conversely, suppose that there exists an element s ∈ G and a map f : A → A such that one has τ (x)(g) = f (x(gs)) for all x ∈ AG and g ∈ G. Then, setting S := {s}, we have τ (x)(g) = f (x(gs)) = f ((g −1 x)(s)) = f ((g −1 x)|S ) for all x ∈ AG and g ∈ G. This shows that τ admits S as a memory set (with associated local defining map f ). █ Exercise 1.24 Let G be a group and let A be a set. Let τ : AG → AG be a cellular automaton. Let S be a memory set for τ and denote by μ : AS → A the associated local defining map. Show that the following conditions are equivalent: (C1) S is the minimal memory set of τ ; (C2) for every s ∈ S, there exist two patterns p1 , p2 ∈ AS such that μ(p1 ) /= μ(p2 ) and p1 (t) = p2 (t) for all t ∈ S \ {s}; (C3) for every s ∈ S, there exist two configurations x1 , x2 ∈ AG such that τ (x1 )(1G ) /= τ (x2 )(1G ) and x1 (g) = x2 (g) for all g ∈ G \ {s}. Solution If (C2) is not satisfied, then there exists an element s ∈ S such that the image under μ of a pattern p ∈ AS only depends on the restriction of p to S \ {s}. This implies that there exists a map μ' : S ' := S \ {s} → A such that τ (x)(1G ) = μ' (x|S ' ) for all x ∈ AG . Using the G-equivariance of τ , we get τ (x)(g) = (g −1 τ (x))(1G ) = τ (g −1 x)(1G ) = μ' ((g −1 x)|S ' )

.

for all x ∈ AG and g ∈ G. Thus S ' is a memory set for τ and hence the memory set S is not minimal. Therefore (C1) implies (C2).

1.2 Exercises

23

Suppose now that (C2) is satisfied and let s ∈ S. Then there exist two patterns p1 , p2 ∈ AS such that μ(p1 ) /= μ(p2 ) and p1 (t) = p2 (t) for all t ∈ S \ {s}. Consider two configurations x1 , x2 ∈ AG that coincide on G \ S and such that x1 |S = p1 and x2 |S = p2 . Then x1 (g) = x2 (g) for all g ∈ G \ {s}. On the other hand, we have τ (x1 )(1G ) = μ(p1 ) /= μ(p2 ) = τ (x2 )(1G ). Thus (C3) is satisfied. This shows that (C2) implies (C3). Suppose now that (C1) is not satisfied. Then the minimal memory set S0 of τ is a proper subset of S. Choose an element s ∈ S \ S0 . If two configurations x1 , x2 ∈ AG coincide on G \ {s}, then they coincide on S0 so that τ (x1 )(1G ) = τ (x2 )(1G ). Therefore (C3) is not satisfied. This shows that (C3) implies (C1) and completes the proof of the equivalence of the three conditions. █ Exercise 1.25 Prove that there are exactly 218 cellular automata τ : {0, 1}Z → {0, 1}Z whose minimal memory set is {−1, 0, 1}. Solution From Exercise 1.18, we have that there are exactly: |S|

• |{0, 1}||{0,1}| = 22 = 28 = 256 cellular automata admitting a given set S with |S| = 3 as a memory set, |T | 2 • |{0, 1}||{0,1}| = 22 = 24 = 16 cellular automata admitting a given set T with |T | = 2 as a memory set, |U | • |{0, 1}||{0,1}| = 22 = 4 cellular automata admitting a given set U with |U | = 1 as a memory set, |∅| 0 • |{0, 1}||{0,1}| = 22 = 2 cellular automata admitting the empty set ∅ as a memory set (these are the two constant cellular automata). 3

As a consequence, there are exactly: • 4 − 2 = 2 cellular automata admitting a given set U with |U | = 1 as a minimal memory set, • 16 − 2 − 2 × 4 = 10 cellular automata admitting a given set T with |T | = 2 as a minimal memory set, and • 256 − 2 − 3 × 2 − 3 × 10 = 256 − 38 = 218 cellular automata admitting a given set S with |S| = 3 as a minimal memory set. █ Exercise 1.26 Let τ ∈ CA(Z2 ; {0, 1}) denote the cellular automaton associated with the Game of Life (see [CAG, Example 1.4.3.(a)]). Show that the minimal memory set of τ is the set {−1, 0, 1}2 . Solution It is clear from the definition of the cellular automaton associated with the Game of Life that its minimal memory set S is contained in {−1, 0, 1}2 . Let 2 us denote by μ : {0, 1}{−1,0,1} → {0, 1} the local defining map associated with the memory set {−1, 0, 1}2 . Let g ∈ {−1, 0, 1}2 . Suppose first that g = (0, 0) an

24

1 Cellular Automata

consider the patterns p0 , p1 : {−1, 0, 1}2 → {0, 1} defined by ⎧ ⎪ ⎪ ⎨i .pi (h) := 1 ⎪ ⎪ ⎩0

if h = g if h ∈ {(−1, −1), (1, 1)} otherwise,

for all h ∈ {−1, 0, 1}2 and i = 0, 1. We have μ(p0 ) = 0 (the cell g was dead and remains dead) and μ(p1 ) = 1 (the cell g was alive and survives). This shows that g = (0, 0) ∈ S. Suppose now that g /= (0, 0) an consider the patterns q0 , q1 : {−1, 0, 1}2 → {0, 1} defined by

qi (h) :=

.

⎧ ⎪ ⎪ ⎨i 1 ⎪ ⎪ ⎩0

if h = g if h ∈ {(0, 0), −g} otherwise

for all h ∈ {−1, 0, 1}2 and i = 0, 1. We have μ(q0 ) = 0 (the cell (0, 0) was alive and dies) and μ(q1 ) = 1 (the cell (0, 0) was alive and survives). This shows that g ∈ S. It follows that {−1, 0, 1}2 ⊂ S, and equality follows. █ Exercise 1.27 Let G be a group and let A be a set. Let σ, τ ∈ CA(G; A). Let S0 (resp. T0 , resp. C0 ) denote the minimal memory set of σ (resp. τ , resp. σ ◦ τ ). Prove that C0 ⊂ S0 T0 . Give an example showing that this inclusion may be strict. Solution By the definition of a cellular automaton, a finite subset S ⊂ G is a memory set for σ (resp. τ , resp. σ ◦ τ ) if and only if given any configuration x ∈ AG and g ∈ G, the value σ (x)(g) (resp. τ (x)(g), resp. [σ ◦ τ ](x)(g)) only depends on the restriction x|gS of x to gS = {gs : s ∈ S}. As a consequence, for g ∈ G, [σ ◦ τ ](x)(g) = σ (τ (x))(g) depends only on the restriction of τ (x) to gS0 . By repeating the same argument again, we deduce that, for every s ∈ S0 , the element τ (x)(gs) depends only on the restriction of x to gsT0 . Therefore, [σ ◦ τ ](x)(g) depends only on the restriction of x to gS0 T0 . This shows that S0 T0 is a memory set for σ ◦ τ , showing that C0 ⊂ S0 T0 , by minimality. Let us show that, in general, we may have a strict inclusion. Let A := {0, 1} and G := Z. Let also S0 = T0 := {0, 1} ⊂ Z. Consider the cellular automata σ, τ : AG → AG with memory sets S0 and T0 and associated local defining maps μ : AS0 → A and ν : AT0 → A defined, respectively, by ⎧ μ(y) :=

0

if y(0)y(1) = 00

1

otherwise,

.

1.2 Exercises

25

for all y ∈ AS0 , and ⎧ ν(z) :=

0

if z(0)z(1) = 01

1

otherwise,

.

for all z ∈ AT0 . It is clear that S0 (resp. T0 ) is a minimal memory set for σ (resp. τ ). It is also clear that τ (AG ) ⊂ {x ∈ AG : x(n)x(n + 1) /= 00, for all n ∈ Z}. As a consequence, we have σ ◦ τ is the constant cellular automaton mapping each configuration x ∈ AG to the constant configuration x1 where x1 (g) := 1 for all g ∈ G. It follows that C0 = ∅, while (we use additive notation for Z) S0 + T0 = {0, 1, 2}. █ Exercise 1.28 Let G be a group and let A be a set. Let H be a subgroup of G. (a) Let τ ∈ CA(G, H ; A). Show that τ and τH ∈ CA(H ; A) have the same minimal memory set. (b) Let σ ∈ CA(H ; A). Show that σ and σ G ∈ CA(G, H ; A) have the same minimal memory set. Solution Let T ⊂ G (resp. S ⊂ H ) and TH ⊂ H (resp. S G ⊂ G) denote the minimal memory set of τ (resp. σ ) and τH (resp. σ G ), respectively. It follows from the definitions that TH ⊂ T and S G ⊂ S. (a) Taking σ = τH (and S = TH ) we deduce that TH ⊂ T = (TH )G ⊂ TH . (b) Taking τ = σ G (and T = S G ) we deduce that S G ⊂ S = (S G )H ⊂ S G . █ Exercise 1.29 A map f : X → Y between sets X and Y is said to be countableto-one if the preimage set f −1 (y) := {x ∈ X : f (x) = y} is countable (finite or infinite) for each y ∈ X. Let A be a set and let G be an uncountable group. Show that every countable-to-one cellular automaton τ : AG → AG is injective. Solution Let τ : AG → AG be a countable-to-one cellular automaton. Suppose by contradiction that τ is not injective. Let S ⊂ G be a memory set for τ with associated local defining map μ : AS → A and denote by H the subgroup of G generated by S. Let R ⊂ G be a complete set of representatives for the left cosets of H in G, so that every element g ∈ G can be uniquely written in the form g = rh with r ∈ R and h ∈ H . Observe that R must be infinite since G is assumed to be uncountable while H is finitely generated and hence countable. Let σ : AH → AH denote the cellular automaton over H obtained by restriction of τ to H , so that σ (rh) = μ((h−1 x)|S ) for all x ∈ AH , r ∈ R, and h ∈ H . As τ is not injective by our hypothesis, σ is not injective either (see [CAG, Proposition 1.7.4.(i)]). Thus there are distinct configurations x0 , x1 ∈ AH such that σ (x0 ) = σ (x1 ). For each subset E ⊂ R, denote by χE : R → {0, 1} the characteristic map of E and consider the configuration yE ∈ AG defined by yE (rh) := xχE (r) (h) for all r ∈ R and h ∈ H . Clearly the configurations yE are all distinct and have the same image under τ , namely the configuration z ∈ AG given by z(rh) := σ (x0 )(h) = σ (x1 )(h) for all r ∈ R and h ∈ H . As R is infinite, this implies that τ −1 (z) is uncountable, a contradiction since τ is assumed to be countable-to-one. █

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Exercise 1.30 Let G be a group and let A be a set. Let F be a non-empty finite subset of G and set B := AF . Equip the sets AG and B G with their prodiscrete uniform structure and the G-shift action. Show that the map Φ : AG → B G , defined by Φ(x)(g) := (g −1 x)|F for all x ∈ AG and g ∈ G, is a G-equivariant uniform embedding. Solution For x ∈ AG and g, h ∈ G, we have Φ(gx)(h) = (h−1 (gx))|F = ((g −1 h)−1 x)|F .

= Φ(x)(g −1 h) = (gΦ(x))(h),

showing that Φ(gx) = gΦ(x). We deduce that Φ is G-equivariant. We claim that Φ is injective. To see this, let x, y ∈ AG and suppose that Φ(x) = Φ(y). Let us show that x = y. Since F is not empty, given h ∈ G, we can find g ∈ G such that f := g −1 h ∈ F . Then x(h) = x(gf ) = (g −1 x)(f ) = (g −1 x)|F (f ) = Φ(x)(g)(f )

.

= Φ(y)(g)(f ) = (g −1 y)|F (f ) = y(gf ) = y(h). As h was arbitrary, this shows that x = y. The claim follows. Let us show that Φ B := {(y, y ' ) ∈ is uniformly continuous. Let Ω ⊂ G be a finite subset and set WΩ A G G ' := B ×B : y|Ω = y |Ω }. We claim that the entourage WΩF {(x, x ' ) ∈ AG ×AG : A B A and set ' G x|ΩF = x |ΩF } of A satisfies (Φ × Φ)(WΩF ) ⊂ WΩ . Let (x, x ' ) ∈ WΩF ' ' := := y Φ(x) (resp. y Φ(x )). Then, for g ∈ Ω we have y(g)(f ) = (g −1 x)(f ) = x(gf ) = x ' (gf ) = (g −1 x ' )(f ) = y ' (g)(f ),

.

B. for all f ∈ F . This shows that y|Ω = y ' |Ω , that is, (Φ × Φ)(x, x ' ) = (y, y ' ) ∈ WΩ B with finite Ω ⊂ G constitute a base The claim follows. Since the entourages WΩ for the prodiscrete uniform structure on B G , this shows that the map Φ is uniformly continuous. After recalling (cf. [CAG, Section B.2]) that given uniform spaces X and Z a map f : X → Z is a uniform embedding if f is injective and both f and g : Y → X, where Y := f (X) and g(y) := f −1 (y) for all y ∈ Y , are uniformly continuous, we are left to show that setting Y := Φ(AG ) ⊂ B G , the map Ψ : Y → AG defined by setting Ψ (y) := Φ −1 (y) for all y ∈ Y is uniformly continuous. Let Ω ⊂ G A := {(x, x ' ) ∈ AG × AG : x| ' be a finite subset and set WΩ Ω = x |Ω }. Let us B ' ) ∈ Y × Y : y| := = y| show that the entourage WΩF {(y, y −1 ΩF −1 ΩF −1 } satisfies B A B ' (Ψ ×Ψ )(WΩF −1 ) ⊂ WΩ . Let (y, y ) ∈ WΩF −1 and set x := Ψ (y) = Φ −1 (y) (resp. B −1 ) = y ' (gf −1 ) x ' := Ψ (y ' ) = Φ −1 (y ' )). Since (y, y ' ) ∈ WΩF −1 , we have y(gf

1.2 Exercises

27

for all g ∈ Ω and f ∈ F . Fix f ∈ F , then x(g) = x(gf −1 f ) = ((gf −1 )−1 x)(f ) = y(gf −1 )(f )

.

= y ' (gf −1 )(f ) = ((gf −1 )−1 x ' )(f ) = x ' (gf −1 f ) = x ' (g). A. for all g ∈ Ω. This shows that x|Ω = x ' |Ω , that is, (Ψ × Ψ )(y, y ' ) = (x, x ' ) ∈ WΩ A Since the entourages WΩ with finite Ω ⊂ G constitute a base for the prodiscrete uniform structure on AG , we deduce that the map Ψ is uniformly continuous. This shows that Φ is a G-equivariant uniform embedding. █

Comment Suppose that A is finite. Then B is finite as well and, moreover, AG and B G are compact Hausdorff spaces. After recalling that every injective continuous map f : X → Y , with X (resp. Y ) a compact (resp. Hausdorff) uniform space, is a uniform embedding (cf. [CAG, Proposition B.2.5]), in order to show that the injective map Φ is a uniform embedding it suffices to show that Φ is continuous. Exercise 1.31 Let G be a group and let A be a set. A configuration x ∈ AG is called almost constant if there exists an element a ∈ A such that x(g) = a for all but finitely many elements g ∈ G. (a) Show that the set of almost constant configurations is dense in AG for the prodiscrete topology. (b) Let τ : AG → AG be a cellular automaton and let x ∈ AG be an almost constant configuration. Show that the configuration τ (x) is almost constant. Solution (a) Let x ∈ AG and let U be a neighborhood of x. Then, using the notation introduced in (1.3), there exists a finite subset Ω ⊂ G such that V (x, Ω) ⊂ U . Choose an arbitrary element a ∈ A and consider the configuration y ∈ AG that coincides with x on Ω and satisfies y(g) = a for all g ∈ G \ Ω. Then y is almost constant. Moreover, we have y ∈ V (x, Ω) ⊂ U . This shows that the set of almost constant configurations is dense in AG . (b) Since the configuration x is almost constant, there is an element a ∈ A such that the set Ω := {g ∈ G : x(g) /= a} is finite. Let S ⊂ G be a memory set for τ and let μ : AS → A denote the associated local defining map. Consider the pattern p ∈ AS defined by p(s) := a for all s ∈ S and let b := μ(p) ∈ A. Then, for all g ∈ G \ ΩS −1 , the restriction to S of the configuration g −1 x is equal to p, so that τ (x)(g) = b by (1.4). As the set ΩS −1 is finite, this shows that the configuration τ (x) is almost constant. █ Exercise 1.32 Let G be a group and let H be a finite index subgroup of G. Let T ⊂ G be a complete set of representatives for the right cosets of H in G, so that G = ⨅t∈T H t. Let A be a set and define B := AT . Equip AG (resp. B H ) with its prodiscrete uniform structure and with the G-shift (resp. H -shift) action. Show that the map Ψ : AG → B H defined by Ψ (x)(h)(t) := x(ht) for all x ∈ AG , h ∈ H , and t ∈ T is an H -equivariant uniform isomorphism.

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1 Cellular Automata

Solution Let x, y ∈ AG and h, k ∈ H . We have, for all t ∈ T , Ψ (kx)(h)(t) = kx(ht) = x(k −1 ht) = Ψ (x)(k −1 h)(t) = kΨ (x)(h)(t)

.

showing that Ψ (kx) = kΨ (x). Thus Ψ is H -equivariant. Suppose now that Ψ (x) = Ψ (y). Let g ∈ G and let h ∈ H and t ∈ T be the unique elements such that g = ht. We have x(g) = x(ht) = Ψ (x)(h)(t) = Ψ (y)(h)(t) = y(ht) = y(g),

.

so that x = y. This shows that Ψ is injective. Let z ∈ B H and define x ∈ AG by setting x(g) := z(h)(t) for all g ∈ G, where h ∈ H and t ∈ T are the unique elements such that g = ht. We then have Ψ (x)(h)(t) = x(ht) = z(h)(t),

.

showing that Ψ (x) = z. It follows that Ψ is surjective. Finally, let ΩH ⊂ H be a finite set and consider the entourage W B (ΩH ) := {(z, z' ) ∈ B H × B H : z|ΩH = z' |ΩH }.

.

(1.10)

After noticing that T and therefore ΩG := ΩH T ⊂ G are finite since [G : H ] < ∞, we claim that the entourage W A (ΩG ) := {(x, x ' ) ∈ AG × AG : x|ΩG = x ' |ΩG }

.

(1.11)

satisfies (Ψ × Ψ )(W A (ΩG )) ⊂ W B (ΩH ). Indeed, if (x, x ' ) ∈ W A (ΩG ) and h ∈ ΩH we have Ψ (x)(h)(t) = x(ht) = x ' (ht) = Ψ (x ' )(h)(t)

.

for all t ∈ T , showing that (Ψ × Ψ )(x, x ' ) = (Ψ (x), Ψ (x ' )) ∈ W B (ΩH ). We deduce that Ψ is uniformly continuous for the prodiscrete topologies on AG and B H . Let us show that the inverse map Ψ −1 : B H → AG is also uniformly continuous. It is clear that given y ∈ B H we have Ψ −1 (g) = y(h)(t) for all g = ht ∈ G. Given a finite subset ΩG ⊂ G we can find a finite subset ΩH ⊂ H such that ΩG ⊂ ΩH T . Let us show that the entourages (1.10) and (1.11) satisfy (Ψ −1 × Ψ −1 )(W B (ΩH )) ⊂ W A (ΩG ). Indeed, if (y, y ' ) ∈ W B (ΩH ) and h ∈ ΩH we have Ψ −1 (y)(ht) = y(h)(t) = y ' (h)(t) = Ψ −1 (y ' )(ht) for all t ∈ T . Since ΩG ⊂ ΩH T , we deduce that (Ψ −1 × Ψ −1 )(y, y ' ) = (Ψ −1 (y), Ψ −1 (y ' )) ∈ W A (ΩG ), showing that (Ψ −1 ×Ψ −1 )(W B (ΩH )) ⊂ W A (ΩG ), and uniform continuity of Ψ −1 follows as well. This shows that Ψ is an H -equivariant uniform isomorphism. █

1.2 Exercises

29

Exercise 1.33 Let G be a group and let A be a set. For each s ∈ G, let τs : AG → AG be the cellular automaton defined by τs (x)(g) := x(gs) for all x ∈ AG and g ∈ G (cf. [CAG, Example 1.4.3.(e)]). (a) Show that τs ∈ ICA(G; A) for every s ∈ G. (b) Show that the map Φ : G → ICA(G; A) defined by φ(s) := τs for all s ∈ G is a group homomorphism. (c) Show that if A has more than one element, then Φ is injective but not surjective. Solution (a) Let s ∈ S. For all x ∈ AG and g ∈ G, we have [τs ◦ τs −1 ](x)(g) = τs (τs −1 (x))(g) = τs −1 (x)(gs) = x(gss −1 ) = x(g)

.

and [τs −1 ◦ τs ](x)(g) = τs −1 (τs (x))(g) = τs (x)(gs −1 ) = x(gs −1 s) = x(g),

.

showing that τs ◦ τs −1 = IdAG = τs −1 ◦ τs . Therefore, the map τs : AG → AG is bijective and its inverse map is the cellular automaton τs −1 . This shows that the cellular automaton τs is reversible, that is, τs ∈ ICA(G; A). (b) Let s1 , s2 ∈ G. Then, for all g ∈ G and x ∈ AG , [τs1 ◦ τs2 ](x)(g) = τs1 (τs2 )(x)(g) = (τs2 )(x)(gs1 ) = x(gs1 s2 ) = τs1 s2 (x)(g),

.

showing that Φ(s1 ) ◦ Φ(s2 ) = τs1 ◦ τs2 = τs1 s2 = Φ(s1 s2 ). We deduce that Φ : G → ICA(G; A) is a group homomorphism. (c) Let a and b be distinct elements in A. Let s ∈ G and define the configuration xs ∈ AG by setting ⎧ xs (g) :=

a

if g = s −1

b

otherwise,

.

for all g ∈ G. Suppose that Φ(s) = IdAG . Then xs (s −1 ) = IdAG (xs )(s −1 ) = Φ(s)(xs )(s −1 ) = xs (s −1 s) = xs (1G )

.

and the definition of the configuration xs shows that s = 1G . We deduce that Φ is injective. In order to show that Φ is not surjective, we first observe that for s ∈ G the cellular automaton τs has minimal memory set {s} and that the unique cellular automaton with minimal memory set {1G } is τ1G = IdAG . Consider now the transposition f := (a, b) ∈ Sym(A) and let τ : AG → AG denote the cellular automaton with (minimal) memory set S := {1G } and local defining map f . We

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have τ ∈ ICA(G; A) (indeed τ = τ −1 ) and τ /= IdAG . We deduce that there is no s ∈ G such that Φ(s) = τs equals τ . As a consequence, Φ is not surjective. █ Exercise 1.34 Let G be a group and let A be a set with more than one element. Prove that the set I consisting of all invertible cellular automata τ : AG → AG admitting a memory set which is reduced to a single element is a subgroup of ICA(G; A) isomorphic to the direct product G × Sym(A). Solution We first observe that a memory set for τ ∈ I which is reduced to a single element is in fact the minimal memory set of τ : indeed a cellular automaton with memory set equal to the empty set is constant and therefore is not invertible. Let τ ∈ I and denote by S = {s} its memory set and by μ : AS = A → A the associated local defining map. We claim that μ is invertible. Suppose, by contradiction, that there exist distinct elements a, b ∈ A such that μ(a) = μ(b). Consider the constant configurations xa , xb ∈ AG defined by xa (g) := a and xb (g) := b for all g ∈ G. We have τ (xa )(g) = μ((g −1 xa )|S ) = μ(xa (s)) = μ(a)

.

= μ(b) = μ(xb (s)) = μ((g −1 xb )|S ) = τ (xb )(g) for all g ∈ G, so that τ (xa ) = τ (xb ). Since xa /= xb , this shows that τ is not injective, a contradiction. Thus μ ∈ Sym(A), proving the claim. Let τ1 , τ2 ∈ I with memory sets S1 = {s1 }, S2 = {s2 } and associated local defining maps μ1 : AS1 = A → A and μ2 : AS2 = A → A, respectively. Let us show that τ1 ◦ τ2 has memory set S = S1 S2 = {s1 s2 } and associated local defining map μ = μ1 ◦ μ2 . For x ∈ AG and g ∈ G we have (τ1 ◦ τ2 )(x)(g) = τ1 (τ2 (x))(g) = μ1 ((g −1 τ2 (x))|S1 ) = μ1 (τ2 (x(gs1 ))) .

= μ1 (μ2 (((gs1 )−1 x)|S2 )) = μ1 (μ2 (x(gs1 s2 ))) = μ((g −1 x)|S ).

As a consequence, τ1 ◦ τ2 ∈ I . Moreover, if τ ∈ I has memory set S = {s} and associated local defining map μ ∈ Sym(A), the cellular automaton with memory set S −1 = {s −1 } and associated local defining map μ−1 is the inverse of τ , showing that τ −1 ∈ I . From this we deduce that I is a subgroup of ICA(G; A). Consider the map Φ : I → G × Sym(A) defined by Φ(τ ) := (s, μ), where S = {s} is the (minimal) memory set of τ and μ : AS = A → A is the associated local defining map (which is invertible by the above claim), for all τ ∈ I . It is clear that Φ is a bijection (since A has more than one element!). The above computations show that

1.2 Exercises

31

Φ(τ1 ◦ τ2 ) = (s1 s2 , μ1 ◦ μ2 ) = Φ(τ1 )Φ(τ2 ), for all τ1 , τ2 ∈ I . We conclude that Φ is a group isomorphism. █ Exercise 1.35 (Elementary Cellular Automata) Let A := {0, 1}. One says that a cellular automaton τ : AZ → AZ is an elementary cellular automaton if the set S := {−1, 0, 1} is a memory set for τ . This amounts to saying that the minimal memory set of τ is contained in S. Denote by E the set consisting of all elementary cellular automata. For n ∈ N, let In := {0, 1, . . . , n − 1} denote the set consisting of all non-negative integers that are smaller than n. (a) Show that the map ϕ : AS → I8 , defined by ϕ(p) := 4p(−1) + 2p(0) + p(1) for all p ∈ AS , is bijective. (b) For τ ∈ E , denote by μτ : AS → A the local defining map of τ associated with S. Show that the map W : E → I256 , defined by W (τ ) :=

∑

.

μτ (p)2ϕ(p)

p∈AS

for all τ ∈ E , is bijective. (c) Compute W (τ ) for each of the following elementary cellular automata τ : AZ → AZ : (i) the constant cellular automaton given by τ (x)(n) := 0 for all x ∈ AZ and n ∈ Z; (ii) the constant cellular automaton given by τ (x)(n) := 1 for all x ∈ AZ and n ∈ Z; (iii) the identity cellular automaton given by τ (x) := x for all x ∈ AZ ; (iv) the right-shift given by τ (x)(n) := x(n − 1) for all x ∈ AZ and n ∈ Z; (v) the left-shift given by τ (x)(n) := x(n + 1) for all x ∈ AZ and n ∈ Z; (vi) the cellular automaton given by τ (x)(n) := min(x(n), x(n + 1)) for all x ∈ AZ and n ∈ Z; (vii) the cellular automaton given by τ (x)(n) := x(n) + x(n + 1) mod 2 for all x ∈ AZ and n ∈ Z; (viii) the cellular automaton given by τ (x)(n) := x(n − 1) + x(n + 1) mod 2 for all x ∈ AZ and n ∈ Z; (ix) the cellular automaton given by τ (x)(n) := x(n−1)+x(n)+x(n+1) mod 2 for all x ∈ AZ and n ∈ Z; (x) the cellular automaton given by τ (x)(n) := x(n−1)×x(n+1) for all x ∈ AZ and n ∈ Z; (xi) the majority rule cellular automaton given by τ (x)(n) := 0 if x(n1 ) + x(n) + x(n + 1) ≤ 1 and τ (x)(n) := 1 otherwise for all x ∈ AZ and n ∈ Z. Solution (a) It suffices to observe that if (a, b, c) ∈ A3 , then 4a + 2b + c is the integer whose expansion in base 2 is abc. (b) First observe that the map τ |→ μτ is a bijection from E onto the set consisting of all maps from AS to A. It follows that E has cardinality 28 = 256. Note

32

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also that if (ai )i∈I8 ∈ A8 , then base 2 is

∑ i∈I8

ai 2i is the integer in I256 whose expansion in

a8 a7 a6 a5 a4 a3 a2 a1 .

.

This implies that W is bijective. (c) By applying the formula defining W (τ ), we successively get: (i) (ii) (iii) (iv) (v) (vi) (vii) (vii) (ix) (x) (xi)

W (τ ) = 0; W (τ ) = 20 + 21 + 22 + 23 + 24 + 25 + 26 + 27 = 255; W (τ ) = 22 + 23 + 26 + 27 = 204; W (τ ) = 24 + 25 + 26 + 27 = 240; W (τ ) = 21 + 23 + 25 + 27 = 170; W (τ ) = 23 + 27 = 136; W (τ ) = 21 + 22 + 25 + 26 = 102; W (τ ) = 21 + 23 + 24 + 26 = 90; W (τ ) = 21 + 22 + 24 + 27 = 150; W (τ ) = 25 + 27 = 160; W (τ ) = 23 + 25 + 26 + 27 = 232. █

Comment If τ : {0, 1}Z → {0, 1}Z is an elementary cellular automaton, the integer W (τ ) is called the Wolfram number of τ (cf. [Kar3, Section 2.5]). It follows from (b) that an elementary cellular automaton is entirely determined by its Wolfram’s number which can be any integer from 0 to 255. When τ is an elementary cellular automaton with Wolfram’s number n, one also says that τ is Rule n. Exercise 1.36 Let G := Z and A := {0, 1}. Fix an integer n ≥ 3 and let S := {−1, 0, 1, . . . , n}. Consider the element α ∈ AS (resp. β ∈ AS ) defined by α(−1) = α(n) := 0 and α(k) := 1 for 0 ≤ k ≤ n − 1 (resp. β(−1) = β(0) = β(n) := 0 and β(k) := 1 for 1 ≤ k ≤ n − 1) and the map μ : AS → A defined by μ(α) := 0, μ(β) := 1 and μ(y) := y(0) for y ∈ AS \ {α, β}. Let τ : AG → AG be the cellular automaton with memory set S and local defining map μ. (a) Show that S is the minimal memory set of τ . (b) Show that τ is an invertible cellular automaton and that τ −1 = τ . Solution (a) By virtue of Exercise 1.24, we only need to show that for every s ∈ S there exist patterns p1 , p2 ∈ AS satisfying that μ(p1 ) /= μ(p2 ) and p1 (t) = p2 (t) for all t ∈ S \ {s}. • For s = −1, then we can take p1 := α and p2 := α ' ∈ AS defined by α ' (n) := 0 and α ' (k) := 1 for −1 ≤ k ≤ n − 1; • if s = 0, then we can take p1 := α and p2 := β; • if s = k, with 1 ≤ k ≤ n − 1, we can take p1 := α and p2 := α '' ∈ AS defined by α '' (−1) = α '' (k) = α '' (n) := 0 and α '' (h) := 1 for 0 ≤ h ≤ n − 1 and h /= k;

1.2 Exercises

33

• if s = n, then we can take p1 := β and p2 := β ' ∈ AS defined by β(−1) = β(0) := 0 and β(k) := 1 for 1 ≤ k ≤ n. This shows that S is the minimal memory set of τ . (b) Let x ∈ AG and set y := τ (x). From the definitions, after identifying the pattern α (resp. β) with the word α(−1)α(0)α(1) · · · α(n) ∈ An+2 (resp. β(−1)β(0)β(1) · · · β(n) ∈ An+2 ), and denoting by x|{k−1,k,...,k+n} (resp. y|{k−1,k,...,k+n} ) the word x(k −1)x(k)x(k +1) · · · x(k +n) (resp. y(k −1)y(k)y(k + 1) · · · y(k + n)), for all k ∈ Z, it follows that

y|{k−1,k,...,k+n} =

.

⎧ ⎪ ⎪ ⎨β

if x|{k−1,k,...,k+n} = α

α ⎪ ⎪ ⎩x| {k−1,k,...,k+n}

otherwise,

if x|{k−1,k,...,k+n} = β

for all k ∈ G. It is then clear that τ 2 = IdAG , in other words, τ is bijective (and █ therefore invertible) with inverse τ −1 = τ . Comment The class of involutive cellular automata presented in Exercise 1.36 is taken from [AmoP]. Exercise 1.37 Let G := Z and let K be a field. Let A := K[[t]] denote the ring of all formal power series in one indeterminate t with coefficients in K. Consider the map τ : AG → AG defined by τ (x)(n) := x(n) − tx(n + 1)

.

for all x ∈ AG and n ∈ Z. Then τ is a bijective cellular automaton whose inverse map σ : AG → AG is given by ⎛ ⎞ i ∑ ∑ ⎝ .σ (y)(n) := yn+j,i−j ⎠ t i . i∈N

j =0

∑ i for all y ∈ AG , where y(n) := i∈N yn,i t for all n ∈ G (cf. [CAG, Example 1.10.3]). Show that σ is discontinuous, with respect to the prodiscrete topology on AZ , at every configuration y ∈ AZ . Solution Let y ∈ AZ and set x := σ (y). Consider the finite set Ω := {0} ⊂ Z and the neighborhood V (x, Ω) := {x ' ∈ AZ : x ' (0) = x(0)}. We claim that there is no finite subset Ω ' ⊂ Z such that σ (V (y, Ω ' )) ⊂ V (x, Ω). Suppose, by contradiction, that such a finite subset Ω ' exists and let M > 0 be such that Ω ' ⊂ (−∞, M].

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Consider the configuration y ' ∈ AZ defined by y ' (n) := y(n) if n ≤ M and y ' (n) := y(n) + 1 if n ≤ M + 1. As y ' |Ω ' = y|Ω ' we have y ' ∈ V (y, Ω ' ). Moreover, setting x ' := σ (y ' ) we have x ' (0) = y ' (0) + ty ' (1) + t 2 y ' (2) + · · · + t M y ' (M) + t M+1 y ' (M + 1) + . . . = y(0) + ty(1) + t 2 y(2) + · · · + t M y(M) + t M+1 (y(M + 1) + 1) + . . . .

= x(0) + t M+1 + . . . /= x(0), showing that σ (V (y, Ω ' )) /⊂ V (x, Ω). This proves the claim. We deduce that the map σ is not continuous at y. █ Exercise 1.38 Let G be a group and let H ⊂ G be a subgroup. Let A and B be two sets. Suppose that τ : AG → B G is a cellular automaton over the group G admitting a memory set S ⊂ H and let τH : AH → B H denote the cellular automaton over the group H obtained by restriction of τ to H . Show that τ (AG ) is closed in B G for the prodiscrete topology on B G if and only if τH (AH ) is closed in B H for the prodiscrete topology on B H . Solution Choose a complete set of representatives R ⊂ G for the left cosets of H in G. As every g ∈ G can be uniquely writtenΠin the form g = rh with r ∈ R and h ∈ H , we have a bijective map ϕ : AG → r∈R AH given by ϕ(x) = (xr )r∈R , where xr ∈ AH is defined by xr (h) := x(rh) for all x ∈ AG , r ∈ R, and h ∈ H . Clearly ϕ isΠa homeomorphism if we equip AG and AH with their prodiscrete G H topologies and r∈R AH with the product Π one.HSimilarly, if we equip B and B with their prodiscrete topologies Π and r∈R B with the product one, we have a homeomorphism ψ : B G → r∈R B H given by ψ(y) = (yr )r∈R , where yr ∈ B H G is defined by yr (h) now that Π := y(rh) for all y ∈ B , r ∈ R, and h ∈ H . Observe −1 G ) is closed ψ ◦ τ ◦ ϕ = r∈R τ (cf. [CAG, Section 1.7]). It follows that τ (A H Π Π in B G if and only if r∈R τH (AH ) is closed in r∈R B H . As this last condition is clearly equivalent to τH (AH ) being closed in B H (cf. [CAG, Proposition A.4.3]), this completes the proof. █ Comment This result is taken from [CecC1]. Exercise 1.39 (Subshifts) Let G be a group and let A be a set. A subset X ⊂ AG is called a subshift of AG if X is G-invariant i.e., gx ∈ X for all x ∈ X and g ∈ G, and closed in AG with respect to the prodiscrete topology. (a) Show that ∅ and AG are subshifts of AG . The subshift AG is called the full subshift. (b) Let Y ⊂ AG be a G-invariant subset and let Y denote the closure of Y in AG . Show that Y is a subshift of AG . (c) Let x ∈ AG . Show that the closure Gx of the G-orbit of x in AG is a subshift of AG . This subshift is called the orbit closure of the configuration x.

1.2 Exercises

35

∩ (d) Let (Xi )i∈I be a family of subshifts of AG . Show that i∈I Xi is a subshift of AG .  (e) Let (Xi )i∈I be a finite family of subshifts of AG . Show that i∈I Xi is a subshift of AG . (f) Let B be a set and let τ : AG → B G be a G-equivariant continuous map (e.g. a cellular automaton). Let Y ⊂ B G be a subshift. Show that τ −1 (Y ) is a subshift of AG . (g) Suppose now that A is finite and let B be a set. Let X be a subshift of AG and let τ : AG → B G be a cellular automaton. Show that τ (X) is a subshift of B G . Solution (a) In any topological space T , both the empty set ∅ and the whole space T are closed in T . Since ∅ and AG are clearly G-invariant, we deduce that ∅ and AG are both subshifts of AG . (b) In order to show that Y is a subshift, we only need to check for its Ginvariance. But this immediately follows from Exercise 1.8(a) since Y is assumed to be G-invariant. (c) This follows from (b) since the orbit Gx of x is a G-invariant subset of AG . (d) As an arbitrary intersection of G-invariant (resp. closed) subsets of AG is G∩ invariant (resp. closed), we deduce that i∈I Xi is G-invariant and closed, that is, it is a subshift of AG . (e) As an arbitrary union of G-invariant subsets of AG is G-invariant and a finite  union of closed subsets of AG is itself closed, we deduce that i∈I Xi is G-invariant and closed, that is, it is a subshift of AG . (f) The set τ −1 (Y ) is G-invariant since Y is G-invariant and τ is G-equivariant. On the other hand, τ −1 (Y ) is closed in AG since Y is closed in B G and τ is continuous. This shows that τ −1 (Y ) is a subshift of AG . (g) We first observe that τ (X) is G-invariant since X is G-invariant and τ is Gequivariant. On the other hand, AG is compact since A is finite. As X is closed in AG , it follows that X is itself compact. Since τ is continuous, we deduce that τ (X) is compact. As B G is Hausdorff, it follows that τ (X) is closed in B G . This shows that τ (X) is a subshift of B G . █ Comment The statement in (f) becomes false when A is infinite (see Exercise 1.45(c) or [CAG, Example 3.3.3]). Exercise 1.40 (Cellular Automata Between Subshifts) Let G be a group. Let A and B be sets. Let X ⊂ AG and Y ⊂ B G be two subshifts. One says that a map σ : X → Y is a cellular automaton if there exists a cellular automaton τ : AG → B G such that σ (x) = τ (x) for all x ∈ X. Let σ : X → Y be a map. Show that the following conditions are equivalent: (CA1) (CA2)

σ is a cellular automaton; σ is G-equivariant (with respect to the actions of G on X and Y induced by the shift actions of G on AG and B G ) and uniformly continuous (with respect to the uniform structures induced on X and Y by the prodiscrete uniform structures on AG and B G );

36

(CA3)

1 Cellular Automata

there exists a finite subset S ⊂ G and a map λ : XS → B, where XS := {x|S : x ∈ X}, such that σ (x)(g) = λ((g −1 x)|S )

.

for all x ∈ X and g ∈ G. Solution Suppose (CA1). Thus there exists a cellular automaton τ : AG → B G such that σ (x) = τ (x) for all x ∈ X. Since τ : AG → B G is a cellular automaton, we know that τ is uniformly continuous and G-equivariant. As the restriction of any uniformly continuous (resp. G-equivariant) map to a G-invariant subset is uniformly continuous (resp. G-equivariant), we deduce that σ is uniformly continuous and Gequivariant. This shows the implication (CA1) =⇒ (CA2). Suppose now (CA2). Since σ is uniformly continuous, we can find a finite subset S ⊂ G such that B (σ ×σ )(WSA ∩(X×X)) ⊂ W{1 ∩(Y ×Y ) = {(y1 , y2 ) ∈ Y ×Y | y1 (1G ) = y2 (1G )}, G}

.

where WSA := {(x1 , x2 ) ∈ AG × AG : x1 |S = x2 |S }. This shows that for x ∈ X the value σ (x)(1G ) only depends on x|S . Thus there exists a map λ : XS → B such that σ (x)(1G ) = λ(x|S ) for all x ∈ X. As σ is G-equivariant, we deduce that if x ∈ X and g ∈ G then σ (x)(g) = (g −1 σ (x))(1G ) = σ (g −1 x)(1G ) = λ(g −1 x)|S .

.

This proves the implication (CA2) =⇒ (CA3). Finally, suppose that σ satisfies (CA3). Extend arbitrarily λ to a map μ : AS → B and let τ : AG → B G denote the cellular automaton with memory set S and local defining map μ. Then, for all x ∈ X and g ∈ G, we have σ (x)(g) = λ((g −1 x)|S ) = μ((g −1 x)|S ) = τ (x)(g)

.

and hence σ (x) = τ (x). This shows the implication (CA3) =⇒ (CA1).

█

Exercise 1.41 Let G be a group. Let A and B be two sets with A finite. Suppose that X ⊂ AG and Y ⊂ B G are subshifts and let σ : X → Y be a map. Show that the following conditions are equivalent: (C1) σ is a cellular automaton; (C2) σ is G-equivariant and continuous (with respect to the topologies induced on X and Y by the prodiscrete topologies on AG and B G ). Solution If σ is a cellular automaton, then σ is G-equivariant and uniformly continuous by Exercise 1.40. As uniform continuity implies continuity, this shows the implication (C1) =⇒ (C2).

1.2 Exercises

37

Conversely, suppose that σ is G-equivariant and continuous. As A is finite and X is closed in AG , the subshift X is compact. It follows that every continuous map from X to Y is uniformly continuous (cf. [CAG, Theorem B.2.3]). The map σ : X → Y being G-equivariant and uniformly continuous, it is a cellular automaton by Exercise 1.40. This shows the implication (C2) =⇒ (C1). █ Exercise 1.42 Let G be an infinite group and let A be a set. Show that if X ⊂ AG is a subshift with non-empty interior then X = AG . Solution As the action of G on AG is topologically transitive by Exercise 1.12(b), it suffices to apply implication (i) =⇒ (iii) from Exercise 1.8(b). █ Exercise 1.43 Let G be an infinite group and let A be a set. Let τ : AG → AG be a cellular automaton. Show that the following conditions are equivalent: (i) τ is nilpotent; (ii) there exists a configuration y ∈ AG such that for every x ∈ AG , there is an integer nx ≥ 1 such that τ n (x) = y for all n ≥ nx . (iii) there exists a constant configuration y ∈ AG such that for every x ∈ AG , there is an integer nx ≥ 1 such that τ nx (x) = y. Solution The implication (i) =⇒ (ii) is obvious. Suppose (ii). Given g ∈ G and taking n ≥ max(ny , ngy ), we get gy = gτ n (y) = n τ (gy) = y. This shows that the configuration y is fixed by G and therefore constant. Therefore (ii) implies (iii). Suppose (iii). We consider first the case when G is countably infinite. In this case, the configuration space AG is a Baire space since it admits a complete metric compatible with its topology by Exercise 1.2(e). By (iii), for each integer n ≥ 1, the set Xn := (τ n )−1 (y) = {x ∈ AG : τ n (x) = y}

.

 is a subshift of AG (cf. Exercise 1.39(f)) and we have that AG = n≥1 Xn by our hypothesis on τ . Since AG is a Baire space, there is an integer n0 ≥ 1 such that Xn0 has a non-empty interior. By applying Exercise 1.42, we deduce that Xn0 = AG , that is, τ n0 (AG ) = {y}. Thus τ is nilpotent. This shows (iii) =⇒ (i) when G is countably infinite. Consider now the general case. Suppose that G is a (possibly uncountable) infinite group. Let M be a memory set for τ . Since G is infinite, there exists a countably infinite subset S ⊂ G. Then the subgroup H ⊂ G generated by M ∪ S is countably infinite and satisfies M ⊂ H . The restriction cellular automaton τH : AH → AH satisfies condition (iii) as well, namely there exists nx ' ≥ 1 n ' such that τHx (x ' ) = y ' . By the first part of the proof, τH is nilpotent. We then deduce from Exercise 1.16 that τ is also nilpotent. This completes the proof of (iii) =⇒ (i). █

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Exercise 1.44 (Patterns Appearing in a Subshift) Let G be a group and let A be a set. One says that a pattern p ∈ P(G, A) appears in a configuration x ∈ AG if p = x|supp(p) . Given a subset X ⊂ AG , one says that a pattern p ∈ P(G, A) appears in X if there exists a configuration x ∈ X such that p appears in x. Let P(X) ⊂ P(G, A) denote the set of all patterns appearing in a subset X. (a) Let X, Y ⊂ AG and suppose that Y is a subshift. Show that one has X ⊂ Y if and only if P(X) ⊂ P(Y ). (b) Let X ⊂ AG be a subshift and let x ∈ AG . Show that one has x ∈ X if and only if every pattern appearing in x is in P(X). (c) Let X, Y ⊂ AG be subshifts. Show that one has X = Y if and only if P(X) = P(Y ). (d) Let X ⊂ AG be a subshift and let P := P (X). Show that P satisfies the following properties: (P1) (P2) (P3)

P is a G-invariant subset of P(G, A); if p ∈ P and Ω ⊂ supp(p), then p|Ω ∈ P ; if p ∈ P and Ω is a finite subset of G such that supp(p) ⊂ Ω, then there exists a pattern q ∈ P with support Ω such that q|supp(p) = p.

(e) Suppose that the group G is countable. Show that if P ⊂ P(G, A) satisfies conditions (P1), (P2), and (P3) in (d), then there exists a unique subshift X ⊂ AG such that P = P (X). Solution (a) The fact that X ⊂ Y implies P(X) ⊂ P(Y ) is immediate. To prove the converse, suppose that P(X) ⊂ P(Y ) and let x ∈ X. Then, for every finite subset Ω ⊂ G, we have x|Ω ∈ P(X) and hence x|Ω ∈ P(Y ). It follows that there exists y ∈ Y such that x|Ω = y|Ω . This shows that x is in the closure of Y for the prodiscrete topology on AG . As Y is a subshift and hence closed in AG , we deduce that x ∈ Y . This shows that X ⊂ Y . (b) This immediately follows from (a) after replacing X by {x} and Y by X. (c) This is immediate from (a) by using a double inclusion. (d) We have the following. (P1)

(P2) (P3)

Let g ∈ G and p ∈ P . Then there exists x ∈ X such that p = x|supp(p) . Then gp = (gx)|g supp(p) (cf. Exercise 1.5(b)). As gx ∈ X since X is G-invariant, we deduce that gp ∈ P . This shows that P is G-invariant. Let p ∈ P . Then there exists x ∈ X such that x|supp(p) = p. Suppose that Ω ⊂ supp(p). Then p|Ω = x|Ω so that p|Ω ∈ P . Let p ∈ P . Then there exists x ∈ X such that x|supp(p) = p. Suppose now that Ω is a finite subset of G such that supp(p) ⊂ Ω. Then q := x|Ω ∈ P and q|supp(p) = x|supp(p) = p.

(e) Suppose that P ⊂ P(G, A) satisfies conditions (P1), (P2), and (P3) in (d). Let X ⊂ AG denote the set consisting of all the configurations x ∈ AG such that the pattern x|Ω is in P for every finite subset Ω of G. Let us shows that X is a subshift satisfying P (X) = P .

1.2 Exercises

39

The fact that X is G-invariant follows from (P1). Suppose that (xi )i∈I is a net of configurations in X converging to a configuration x ∈ X. If Ω is a finite subset of G, then we have x|Ω = xi |Ω for i large enough. As xi |Ω ∈ P since xi ∈ X, we deduce that x|Ω ∈ P for every finite subset Ω of G. Therefore x ∈ X. This shows that X is closed in AG . Thus X is a subshift of AG . Let us shows that P (X) = P . If p ∈ P (X), then there exists x ∈ X such that x|supp(p) = p. This implies p ∈ P by definition of X. Thus P (X) ⊂ P . Conversely, suppose that p ∈ P . Since  G is countable, there is a sequence (Ωn )n∈N of finite subsets of G such that G = n∈N Ωn . We can assume Ω0 = supp(p). Let p0 := p. By induction, it follows from (P3) that there exists a sequence (pn )n∈N of patterns in P such that supp(pn ) = Ωn and pn+1 |Ωn = pn for all n ∈ N. Consider the unique configuration x ∈ AG such that x|Ωn = pn for all n ∈ N. For every finite subset Ω ⊂ G, there exists n ∈ N such that Ω ⊂ Ωn . As x|Ω = pn |Ω and pn ∈ P , it follows from (P2) that x|Ω ∈ P . Therefore x ∈ X. As p = p0 = x|Ω0 , we deduce that p ∈ P (X). Thus P ⊂ P (X). This shows that P = P (X). Uniqueness of X follows from (c). █ Exercise 1.45 Let G be a group. Let A and B be sets. Let f : A → B be a map and consider the map f∗ : AG → B G defined by f∗ (x) = f ◦ x for all x ∈ AG . (a) Show that f∗ : AG → B G is a cellular automaton. (b) Show that if A is finite and X is a subshift of AG then f∗ (X) is a subshift of B G . (c) Let G := Z, A := Z, B := {0, 1}, and let f : A → B be defined by f (n) := 0 if n < 0 and f (n) := 1 otherwise. Let X := {xn : n ∈ Z} ⊂ AZ , where xn (m) := n + m for all n, m ∈ Z. Show that X is a subshift of AZ but f∗ (X) is not a subshift of B Z . Solution (a) For all x ∈ AG and g ∈ G, we have f∗ (x)(g) = f (x(g)) = f ((g −1 x)(1G )).

.

Thus f∗ is a cellular automaton with memory set {1G } and local defining map f : A{1G } = A → B. (b) This immediately follows from (a) and Exercise 1.39(f). (c) The subset X, being the Z-orbit of the configuration x0 , is clearly Z-invariant. Let us show that the complement Y := AZ \ X of X is open. Let y ∈ Y and set n := y(0). Observe that y(0) /= xm (0) for all m ∈ Z such that m /= n. Moreover, since y /= xn , we can find m ∈ Z such that y(m) /= xn . As a consequence, if Ω := {0, m}, then the neighborhood V (y, Ω) = C(y|Ω ) := {z ∈ AZ : z|Ω = y|Ω } satisfies that V (y, Ω) ⊂ Y . This shows that Y is open in AZ . We deduce that X = AZ \ Y is closed in AZ . Observe that f∗ (X) = {zn : n ∈ Z}, where zn := f∗ (xn ) ∈ B Z is the configuration given by zn (m) = 0 if m < −n and zn (m) = 1 otherwise. Now the constant configuration z ∈ B Z defined by z(n) = 1 for all n ∈ Z is the limit of

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zn as n → ∞. However, z does not belong to f∗ (X). Therefore f∗ (X) is not closed in B Z for the prodiscrete topology and hence f∗ (X) is not a subshift of B Z . █ Exercise 1.46 Let G be a group. Let A and B be two sets. Equip AG and B G with their prodiscrete topologies and the shift actions of G. Suppose that the group G acts continuously on a topological space Y . (a) Let Z ⊂ Y be a closed G-invariant subset and let f : AG → Y be a Gequivariant continuous map. Show that f −1 (Z) := {x ∈ AG : f (x) ∈ Z} is a subshift of AG . (b) Suppose that Y is Hausdorff and let f, g : AG → Y be G-equivariant continuous maps. Show that Eq(f, g) := {x ∈ AG : f (x) = g(x)} is a subshift of AG . (c) Let f : AG → B G be a G-equivariant continuous map (e.g. a cellular automaton over the group G and the alphabets A and B) and let Z ⊂ B G be a subshift. Show that f −1 (Z) is a subshift of AG . (d) Let f, g : AG → B G be G-equivariant continuous maps (e.g. cellular automata over the group G and the alphabets A and B). Show that Eq(f, g) := {x ∈ AG : f (x) = g(x)} is a subshift of AG . (e) Let f : AG → AG be a G-equivariant continuous map (e.g. a cellular automaton over the group G and the alphabet A). Show that Fix(f ) := {x ∈ AG : f (x) = x} is a subshift of AG . Solution (a) This follows from the fact that the preimage of a closed subset under a continuous map is itself a closed subset and the preimage of a G-invariant subset under a G-equivariant map is itself G-invariant. (b) Consider the map F : AG → Y × Y defined by F (x) := (f (x), g(x)) for all x ∈ AG . Equip Y × Y with the product topology and the diagonal action of G, i.e., the action given by g(y1 , y2 ) := (gy1 , gy2 ) for all g ∈ G and y1 , y2 ∈ Y . Then Eq(f, g) = F −1 (Δ), where Δ := {(y, y) : y ∈ Y } is the diagonal in Y × Y . The diagonal Δ is clearly G-invariant. Moreover, Δ is closed in Y × Y since Y is Hausdorff. Thus Eq(f, g) is a subshift of AG by (a). (c) This follows from (a) by taking Y = B G . (d) This follows from (b) since B G is Hausdorff. (e) This follows from (d) since Fix(f ) = Eq(f, g) with g : AG → AG being the identity map. █ Exercise 1.47 Let G be a group and let A be a set. Given a set F ⊂ P(G, A) of (G, A)-patterns, define the subset XF ⊂ AG as being the set consisting of all configurations x ∈ AG satisfying the following condition: for every g ∈ G, there is no pattern in F appearing in gx, i.e., XF := {x ∈ AG : (gx)|supp(p) /= p for all g ∈ G and p ∈ F }.

.

(a) Let F ⊂ P(G, A). Show that XF is a subshift of AG .

1.2 Exercises

41

(b) Let X ⊂ AG be a subshift and let P (X) ⊂ P(G, A) denote the set consisting of all (G, A)-patterns appearing in X. One says that a set F ⊂ AG of (G, A)patterns is a defining set of forbidden patterns for X if X = XF . Show that the set F (X) := P(G, A) \ P (X) is a defining set of forbidden patterns for X and that every set of forbidden patterns for X is contained in F (X). (c) Let (Xi )i∈I ∩ be a family of subshifts of AG and consider the subshift X ⊂ AG defined by X := i∈I Xi (cf. Exercise 1.39(d)). Suppose that (Fi )i∈I is a family a defining set of forbidden patterns for Xi for of subsets of P(G, A) such that Fi is each i ∈ I . Show that the set F := i∈I Fi is a defining set of forbidden patterns for X. Solution (a) By definition, we have XF =

∩ ∩

.

g −1 (AG \ C(p)).

g∈G p∈F

As C(p) is open in AG and the G-shift action is continuous, the set g −1 (AG \ C(p)) is closed in AG for all g ∈ G and p ∈ F . Since the intersection of any family of closed subsets is itself closed, we deduce that XF is closed in AG . On the other hand, for every h ∈ G, we have g(hx) = (gh)x for all x ∈ AG and g ∈ G. We deduce that if x ∈ XF , then hx ∈ XF for all h ∈ G. Thus XF is G-invariant. This shows that XF is a subshift. (b) If x ∈ X, then gx ∈ X for all g ∈ G since X is invariant under the Gshift action. Therefore all patterns appearing in gx are in P (X), that is, x ∈ XF (X) . This shows that X ⊂ XF (X) . To prove the converse inclusion, suppose now that x ∈ XF (X) . Let Ω be a finite subset of G. Then x|Ω ∈ / F (X) and hence x|Ω ∈ P (X). Thus every pattern appearing in x is in P (X). We deduce that x ∈ X by Exercise 1.44(b). This shows that XF (X) ⊂ X. Therefore X = XF (X) . Finally, suppose that F ⊂ P(G, A) is a defining set of forbidden patterns for X. If p ∈ F , then p ∈ / P (X) since otherwise there would be x ∈ X such that p appears in X. This shows that F ⊂ F (X). (c) Let XF denote the subshift of AG admitting F as a defining set of forbidden patterns. Since Fi ⊂ F , we clearly have XF ⊂ Xi for every i ∈ I . Therefore XF ⊂ X. To prove the reverse inclusion, suppose that x ∈ X. Let g ∈ G. Since x ∈ Xi , we have (gx)|supp(p) /= p for all i ∈ I and p ∈ Fi . We deduce that (gx)|supp(p) /= p for all p ∈ F . As g ∈ G was arbitrary, this shows that x ∈ XF . Therefore X = XF . █ Exercise 1.48 (Subshifts of Finite Type) Let G be a group and let A be a finite set. One says that a subshift X ⊂ AG is of finite type if it admits a finite defining set of forbidden patterns. (a) Let X ⊂ AG be a subshift of finite type. Show that X admits a finite defining set of forbidden patterns having all the same support.

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1 Cellular Automata

(b) Let Ω ⊂ G be a finite subset and let A ⊂ AΩ . Define X(A ) ⊂ AG by X(A ) := {x ∈ AG : (gx)|Ω ∈ A for all g ∈ G}.

.

(1.12)

Show that X(A ) is a subshift of finite type of AG . (c) Let X ⊂ AG be a subshift of finite type. Show that there exist a finite subset Ω ⊂ G and a subset A ⊂ AΩ such that X = X(A ). Such a set A is called a defining set of admissible patterns and the subset Ω is called a memory set for the subshift of finite type X. (d) Let X ⊂ AG be a subshift of finite type and let Ω ⊂ G be a memory set for X. Let g ∈ G and let Λ be a finite subset of G such that Ω ⊂ Λ. Show that gΩ and Λ are also memory sets for X. (e) Show that the full subshift AG is a subshift of finite type of itself. (f) Let X1 , X2 ⊂ AG be two subshifts of finite type. Show that X1 ∩ X2 is a subshift of finite type. (g) Show that the union of two subshifts of finite type may fail to be of finite type. (h) Fix a ∈ A and consider the constant configuration xa ∈ AG defined by xa (g) := a for all g ∈ G. Show that the singleton X := {xa } is a subshift of finite type of AG . Solution (a)  Let P ⊂ P(G, A) be a finite defining set of forbidden patterns for X. Then Ω := p∈P supp(p) is a finite subset of G. Let Q denote the set consisting of all q ∈ AΩ satisfying the following property: there exists p ∈ P such that q|supp(p) = p. Observe that Q is finite since AΩ is finite. Clearly Q is also a defining set of forbidden patterns for X. (b) Setting P := AΩ \ A , we have X(A ) = {x ∈ AG : (gx)|Ω ∈ / P for all g ∈ G}.

.

Thus X(A ) is the subshift of AG admitting P as a defining set of forbidden patterns. This subshift is of finite type since the set P is finite. (c) Let X ⊂ AG be a subshift of finite type. By (a), there exists a finite subset Ω ⊂ G and a subset P ⊂ AΩ that is a defining set of forbidden patterns for X. Let A := AΩ \ P . Then X = {x ∈ AG : (gx)|Ω ∈ / P for all g ∈ G}

.

= {x ∈ AG : (gx)|Ω ∈ A for all g ∈ G}, showing that X = X(A ).

1.2 Exercises

43

(d) Let A ⊂ AΩ be a defining set of admissible patterns for X. Using the action of G on the set of all (G, A)-patterns introduced in Exercise 1.5, we have gA ⊂ AgΩ and X = X(A )

.

= {x ∈ AG : (hx)|Ω ∈ A for all h ∈ G} = {x ∈ AG : (ghx)|gΩ ∈ gA for all h ∈ G} = {x ∈ AG : (kx)|gΩ ∈ gA for all k ∈ G} = X(gA ), showing that gΩ is a memory set for X. Let B ⊂ AΛ denote the set consisting of all patterns q ∈ AΛ such that there exists p ∈ A satisfying p = q|Ω . We claim that B is a defining set of admissible patterns for X. Indeed, X(B) := {x ∈ AG : (gx)|Λ ∈ B for all g ∈ G} so that if x ∈ X(B) and g ∈ G then q := (gx)|Λ ∈ B so that p := q|Ω = ((gx)|Λ )|Ω = (gx)|Ω ∈ A . As g ∈ G was arbitrary, this shows that x ∈ X(A ) = X. We deduce that X(B) ⊂ X. Conversely, if x ∈ X = X(A ) and g ∈ G then p := (gx)|Ω ∈ A and q := (gx)|Λ ∈ AΛ satisfies that q|Ω = p ∈ A so that (gx)|Λ = q ∈ B. As g ∈ G was arbitrary, this shows that x ∈ X(B) and the inclusion X ⊂ X(B) follows. We deduce that X = X(B) and prove the claim. Thus Λ is also a memory set for X. (e) The empty set ∅ is a defining set of forbidden patterns for AG . As the set ∅ is finite, this shows that AG is a subshift of finite type of itself. (f) Choose, for each i ∈ {1, 2}, a finite defining set of forbidden patterns Pi for Xi . Clearly P1 ∪ P2 is a finite defining set of forbidden patterns for X1 ∩ X2 . This shows that X1 ∩ X2 is a subshift of finite type of AG . (g) Let A := {0, 1, 2} and G := Z. Let p1 , p2 : {0} → Z denote the patterns defined by p1 (0) := 1 and p2 (0) := 2. Consider the subshift of finite type X1 ⊂ AZ (resp. X2 ⊂ AZ ) admitting {p1 } (resp. {p2 }) as a defining set of forbidden paterns. Thus, X1 (resp. X2 ) consists of the configurations x ∈ AZ that do not take the value 1 (resp. 2). Suppose by contradiction that the subshift X1 ∪ X2 is of finite type. Let Ω ⊂ Z be a memory set for X1 ∪ X2 . Choose an integer N ≥ 0 such that Ω ⊂ [−N, N] and consider the configuration x ∈ AZ defined by

x(k) :=

.

⎧ ⎪ ⎪ ⎨0

if k ∈ [−N, N ],

1 ⎪ ⎪ ⎩2

if k > N,

if k < −N,

for all k ∈ Z. We clearly have x /∈ X1 ∪ X2 . However, for all n ∈ Z, there exists y ∈ X1 ∪ X2 such that x and y coincide on n + [−N, N ] and hence on n + Ω. This

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1 Cellular Automata

contradicts the fact that Ω is a memory set for X1 ∪ X2 . Therefore X1 ∪ X2 is not of finite type. (h) Taking Ω := {1G } and A := {a} ⊂ A = AΩ , we have X = X(A ). This shows that X is a subshift of finite type. █ Exercise 1.49 Let G be a group and let A be a finite set. (a) Let X ⊂ AG be a subshift. Let Ω ⊂ G be a finite subset and let A ⊂ Ω A denote the set of all patterns with support Ω appearing in X. Let X(A ) ⊂ AG denote the subshift of finite type admitting A as a defining set of admissible patterns. Show that X ⊂ X(A ). (b) With the notation in (a), find an example showing that the inclusion X ⊂ X(A ) may be strict. (c) Find an example of a subshift of finite type X ⊂ AG with memory set Ω ⊂ G admitting two distinct defining sets A , A ' ⊂ AΩ of admissible patterns. (d) Let X ⊂ AG be a subshift of finite type. Let Ω ⊂ G be a memory set for X and let A1 , A2 ⊂ AΩ be two defining sets of admissible patterns for X. Show that A := A1 ∩ A2 is a set of admissible patterns for X. (e) Let X ⊂ AG be a subshift of finite type and let Ω ⊂ G be a memory set for X. Let A1 , A2 , . . . , An ⊂ AΩ denote ∩ all the defining sets of admissible patterns with support Ω for X and set A := ni=1 Ai . Show that A is a set of admissible patterns for X and that it equals the set of all patterns with support Ω appearing in X. Solution (a) Let x ∈ X. Then for all g ∈ G one has gx ∈ X and therefore (gx)|Ω ∈ A . By (1.12), we have x ∈ X(A ). This shows that X ⊂ X(A ). (b) Let A := {0, 1} and let X ⊂ AZ denote the subshift consisting of the two constant configurations. Then, taking Ω := {0} ⊂ Z, we have A = A so that X ⊊ X(A ) = AZ . (c) Let A := {0, 1} and let X ⊂ AZ denote the subshift reduced to the the constant configuration x0 ∈ AZ defined by x0 (n) = 0 for all n ∈ Z. Let Ω := {0, 1} and consider the subsets A , A ' ⊂ AΩ given by A := {p0 } and A ' := {p0 , p1 }, where p0 , p1 ∈ AΩ are defined by p0 (0) = p1 (0) = 0, p0 (1) = 0, and p1 (1) = 1. Clearly X = X(A ) = X(A ' ) but A /= A ' . (d) For i = 1, 2 we have A ⊂ Ai so that X(A ) ⊂ X(Ai ) = X. To prove the reverse inclusion, suppose that x ∈ X. Then for all g ∈ G and i = 1, 2 we have (gx)|Ω ∈ Ai , since Ai is a defining set of admissible patterns for X. We deduce that (gx)|Ω ∈ A for all g ∈ G, so that x ∈ X(A ). This shows that X ⊂ X(A ). We deduce that X = X(A ), that is, A is a defining set of admissible patterns for X. (e) It follows from (d) that A is a set of admissible patterns for X. Let us denote by A ' ⊂ AΩ the set of all patterns with support Ω appearing in X. Let p ∈ A ' . Then there exists x ∈ X such that x|Ω = p. Since A is a defining set of admissible patterns for X and x ∈ X, we have that (gx)|Ω ∈ A for all g ∈ G. Taking g = 1G we deduce that p ∈ A . This shows that A ' ⊂ A . As a consequence, if X' ⊂ AG denotes the subshift of finite type admitting A ' as a defining set of admissible patterns, we have X' ⊂ X. On the other hand, it follows from (a) that X ⊂ X' . Thus

1.2 Exercises

45

X = X' . Therefore A ' is a defining set of admissible ∩ patterns for X, that is, there exists 1 ≤ i0 ≤ n such that A ' = Ai0 . Thus A = ni=1 Ai ⊂ Ai0 = A ' . It follows █ that A = A ' . Exercise 1.50 Let G be a group and let A be a finite set. Let X ⊂ AG be a subset. Let x0 ∈ {0, 1}G denote the constant 0-configuration. Show that the following conditions are equivalent: (i) X is a subshift of finite type of AG ; (ii) there exists a cellular automaton τ : AG → {0, 1}G such that X = τ −1 (x0 ). Solution Suppose first that X is a subshift of finite type. Let M ⊂ G and A ⊂ AM be a memory set and a defining set of admissible patterns for X. Consider the cellular automaton τ : AG → {0, 1}G with memory set M and local defining map μ : AM → {0, 1} defined by ⎧ μ(y) :=

0

if y ∈ A

1

otherwise.

.

Let us show that τ −1 (x0 ) = X. Let x ∈ X. Then we have (g −1 x)|M ∈ A and therefore τ (x)(g) = μ((g −1 x)|M ) = 0 for all g ∈ G. Thus, τ (x) = x0 . This shows that X ⊂ τ −1 (x0 ). On the other hand, if x ∈ / X, then we can find g ∈ G such that (g −1 x)|M ∈ / A . We deduce that τ (x)(g) = 1, showing that x ∈ / τ −1 (x0 ). This proves the equality X = τ −1 (x0 ). Conversely, suppose that there exists a cellular automaton τ : AG → {0, 1}G such that X = τ −1 (x0 ). Let M ⊂ G be a memory set for τ and denote by μ : AM → {0, 1} the corresponding local defining map. Set A := μ−1 (0) ⊂ AM . Let us show that X = X(A ), where X(A ) ⊂ AG is the subshift of finite type defined by (1.12). Let x ∈ X. Since τ (x) = x0 , we have μ((g −1 x)|M ) = τ (x)(g) = 0 and hence (g −1 x)|M ∈ A for every g ∈ G. This shows that x ∈ X(A ), and the inclusion X ⊂ X(A ) follows. Conversely, suppose that x ∈ X(A ). Then, for every g ∈ G we have (g −1 x)|M ∈ A and therefore τ (x)(g) = μ((g −1 x)|M ) = 0. Thus τ (x) = x0 , and therefore x ∈ τ −1 (x0 ) = X. Thus X(A ) ⊂ X. We conclude that X = X(A ) is of finite type. █ Exercise 1.51 Let G be a group and let A and B be two finite sets. Let τ : B G → AG be a cellular automaton and suppose that X ⊂ AG is a subshift of finite type. Show that Y := τ −1 (X) ⊂ B G is a subshift of finite type. Solution Since X is of finite type, it follows from the implication (i) =⇒ (ii) in Exercise 1.50 that there exists a cellular automaton σ : AG → {0, 1}G such that X = σ −1 (x0 ), where x0 ∈ {0, 1}G denotes the constant 0-configuration. Consider the composite cellular automaton η := σ ◦ τ : B G → {0, 1}G . We have Y = τ −1 (X) = τ −1 (σ −1 (x0 )) = η−1 (x0 ). It then follows from the implication (ii) =⇒ (i) in Exercise 1.50 that Y ⊂ B G is a subshift of finite type. █

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Exercise 1.52 Let G be a group and let A be a finite set. Let X ⊂ AG be a subshift. Denote by F the set of all finite subsets of G. For Ω ∈ F , let πΩ : AG → AΩ denote the projection map and set XΩ := {x ∈ AG : πΩ (gx) ∈ πΩ (X) for all g ∈ G}.

.

(a) Let Ω ∈ F . Show that XΩ is a subshift of finite type of AG . (b) Show that one has X=

∩

.

XΩ .

(1.13)

Ω∈F

(c) Show that X is of finite type if and only if there exists Ω ∈ F such that X = XΩ . (d) Let Ω, Ω ' ∈ F such that Ω ⊂ Ω ' . Show that XΩ ' ⊂ XΩ and deduce that if X = XΩ then X = XΩ ' . Solution (a) Using the notation introduced in (1.12), we have XΩ = X(A ), where A := πΩ (X). Therefore X is a subshift of finite type by Exercise 1.48(b). (b) Let E denote the right-hand side of (1.13). If Ω ∈ F and x ∈ X, then x ∈ XΩ since gx ∈ X for all g ∈ G by G-invariance of X. Therefore X ⊂ XΩ . This shows the inclusion X ⊂ E. Conversely, suppose that x ∈ E. Given Ω ∈ F , we have x ∈ XΩ . Therefore πΩ (gx) ∈ πΩ (X) for all g ∈ G. Taking g := 1G , we deduce that there exists xΩ ∈ X such that x and xΩ coincide on Ω. It follows that x is in the closure of X. As X is closed in AG , this implies x ∈ X. This shows the inclusion E ⊂ X, completing the proof of (1.13). (c) Sufficiency follows from (a). To prove necessity, suppose that X is of finite type. Let Ω be a memory set for X. Then A := πΩ (X) is a defining set of admissible patterns for X. Therefore X = X(A ) = XΩ . (d) Let x ' ∈ XΩ ' . Then for every g ∈ G, there exists x ∈ X such that πΩ ' (gx ' ) = πΩ ' (x). As Ω ⊂ Ω ' , we have πΩ (gx ' ) = πΩ (x). This shows that x ' ∈ XΩ and the inclusion XΩ ' ⊂ XΩ follows. If in addition X = XΩ , also keeping in mind (b), we have X ⊂ XΩ ' ⊂ XΩ = X. We deduce that XΩ ' = X. █ Exercise 1.53 Let G be a group. Let A and B be two finite sets and let X ⊂ AG be a subshift. We keep the notation from Exercise 1.52. Let τ : AG → B G be a cellular automaton and let T ⊂ G be a memory set for τ . Show that τ (X) is a subshift of B G and that one has τ (XΩT ) ⊂ (τ (X))Ω

.

(1.14)

for all Ω ∈ F . Solution The fact that τ (X) is a subshift of B G follows from Exercise 1.39(g).

1.2 Exercises

47

Let Ω ∈ F . As T is a memory set for τ , if two configurations x, x ' ∈ AG satisfy x|ΩT = x ' |ΩT then one has τ (x)|Ω = τ (x ' )|Ω (cf. [CAG, Lemma 1.4.7]). Suppose now x ' ∈ XΩT . This means that, given g ∈ G, we can find x ∈ X such that (gx ' )|ΩT = x|ΩT . Then from the above observation and after recalling that τ is G-equivariant, we deduce that (gτ (x ' ))|Ω = τ (gx ' )|Ω = τ (x)|Ω ∈ πΩ (τ (X)) (here πΩ denotes the projection map B G → B Ω ). As g ∈ G was arbitrary, this shows that τ (x ' ) ∈ τ (X)Ω , and (1.14) follows. █ Exercise 1.54 Let G be a group. Let A (resp. B) be a finite set and let X ⊂ AG (resp. Y ⊂ B G ) be a subshift. Suppose that the subshifts X and Y are topologically conjugate, i.e., there exists a G-equivariant homeomorphism ϕ : X → Y . Show that X is of finite type if and only if Y is of finite type. Solution Topological conjugacy is a symmetric relation on subshifts. Indeed, if ϕ : X → Y is a G-equivariant homeomorphism then ϕ −1 : Y → X is also a Gequivariant homeomorphism. Therefore, it is enough to show that if X is of finite type then Y is also of finite type. So, let us assume that X is of finite type. By Exercise 1.52(c), there exists a finite subset ΩX ⊂ G such that X = XΩX . By Exercise 1.52(d), up to enlarging ΩX if necessary, we may suppose that 1G ∈ ΩX . By Exercise 1.41, the maps ϕ and ϕ −1 are cellular automata between subshifts. Therefore, there exist cellular automata σ : AG → B G and τ : B G → AG such that σ |X = ϕ : X → Y and τ |Y = ϕ −1 : Y → X. Let S ⊂ G (resp. T ⊂ G) be a memory set for σ (resp. τ ). Up to enlarging S (resp. T ) if necessary (cf. [CAG, Proposition 1.5.2]), it is not restrictive to suppose that 1G ∈ S (resp. 1G ∈ T ). By [CAG, Proposition 1.4.9], the set ST is a memory set for the composite cellular automaton ψ := σ ◦ τ : B G → B G. Set ΩY := ΩX ST ⊂ G and let us show that Y = YΩY . Since 1G ∈ S, we have ΩX ⊂ ΩX S, so that X = XΩX = XΩX S by Exercise 1.52(d). Thus, by applying the result of Exercise 1.53, we get τ (YΩY ) = τ (YΩx ST ) ⊂ (τ (Y ))ΩX S = XΩX S = XΩX = X,

.

so that ψ(YΩY ) = σ (τ (YΩY )) ⊂ σ (X) = ϕ(X) = Y.

.

(1.15)

Let μ : B ST → B denote the local defining map of ψ with respect to its memory set ST . Since ψ|Y = (σ ◦ τ )|Y = ϕ ◦ ϕ −1 = IdY , we have μ((g −1 y)|ST ) = ψ(y)(g) = y(g)

.

(1.16)

for all y ∈ Y and g ∈ G. Let now y ' ∈ YΩY and let g ∈ G. Then there exists y ∈ Y such that (g −1 y ' )|ΩY = y|ΩY . Note that, since 1G ∈ ΩX , we have ST ⊂ ΩX ST = ΩY and therefore

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1 Cellular Automata

(g −1 y ' )|ST = y|ST . As 1G ∈ ST , this implies in particular y ' (g) = (g −1 y ' )(1G ) = y(1G ). Thus, keeping in mind that ψ|Y = IdY , formula (1.16) gives us ψ(y ' )(g) = μ((g −1 y ' )|ST ) = μ(y|ST ) = ψ(y)(1G ) = y(1G ) = y ' (g).

.

Therefore ψ|YΩY = IdYΩY . Using (1.15), we deduce that YΩY = ψ(YΩY ) ⊂ Y . As Y ⊂ YΩY by Exercise 1.52(b), this implies that Y = YΩY . We then deduce from █ Exercise 1.52(c) that Y is of finite type. Comment This result was established in [LinM, Theorem 2.1.10] for G = Z and then extended to G = Zd , d ≥ 1, in [RadS, Theorem 2]. Exercise 1.55 Let G be a countable group, let A be a finite set, and let X ⊂ AG be a subshift. Show that the following conditions are equivalent: (i) X is of finite type; (ii) every descending sequence of subshifts of AG X0 ⊃ X1 ⊃ X2 ⊃ · · · ⊃ Xn ⊃ Xn+1 ⊃ · · ·

.

(1.17)

∩ such that X = n∈N Xn eventually stabilizes, that is, there exists n0 ∈ N such that Xn = Xn0 for all n ≥ n0 ; (iii) there is no infinite strictly descending sequence of subshifts of AG X0 ⊋ X1 ⊋ X2 ⊋ · · · ⊋ Xn ⊋ Xn+1 ⊋ · · ·

.

such that X =

∩

n∈N Xn .

Solution Suppose that X is of finite type. Let Ω ⊂ G be a memory set for X and let A ⊂ AΩ be the set of all patterns with support Ω appearing in X. ∩ Let X0 ⊃ X1 ⊃ · · · be a descending sequence of subshifts in AG such that X = n∈N Xn . For each n ∈ N, let An denote the set of patterns with support Ω appearing in Xn . Then, denoting by Xn' the subshift of finite type admitting An as a defining set of admissible patterns, we have Xn ⊂ Xn' (cf. Exercise 1.49(a)) for all n ∈ N. Since A0 ⊃ A1 ⊃ · · · is a descending sequence of finite sets, it stabilizes, that is, there exists n0 ∈ N such that An = An0 for all n ≥ n0 . It follows that Xn' = Xn' 0 for all n ≥ n0 . Now, since X ⊂ Xn ⊂ Xn' for all n ∈ N, we have X ⊂ Xn' 0 . Let us prove the reverse inclusion. Let p ∈ An0 , so that p ∈ An for all n ∈ N. By definition of An , we can find xn ∈ Xn such that p = xn |Ω , for all n ∈ N. Since A is finite and G is countable, the prodiscrete topology on AG is compact and metrizable (cf. Exercise 1.2). Consequently, we can find a subsequence of (xn )n∈N G converging ∩to some x ∈ A . Since Xn is closed and xn ∈ Xn for all n ∈ N, we have x ∈ X = n∈N Xn . As p = x|Ω ∈ A , we deduce that An0 ⊂ A , so that, recalling that A is a defining set of admissible patterns for X (cf. Exercise 1.49(e)), Xn' 0 ⊂ X. As X ⊂ Xn0 ⊂ Xn' 0 ⊂ X, we have X = Xn0 , showing that the descending sequence (1.17) stabilizes. The implication (i) ⇒ (ii) follows.

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49

The equivalence (ii) ⇐⇒ (iii) is trivial. Suppose now that X is not of finite type. Then every defining set F ⊂ P(G, A) of forbidden patterns for X is infinite. Pick one of these defining sets, say F . Note that since G is countable and A is finite, the set P(G, A) and hence its subset F are countable (cf. Exercise 1.6). It follows that we can chose an increasing sequence  F0 ⊂ F1 ⊂ · · · of finite subsets of P(G, A) such that F = n∈N Fn . Denoting by Xn the subshift of finite type of AG admitting Fn as a defining set of forbidden patterns, we obtain a descending sequence X0 ⊃ X1 ⊃ · · · of subshifts. As F =  ∩ F , we have X = X n n∈N n∈N n by Exercise 1.47(c). Moreover, the sequence X0 ⊃ X1 ⊃ · · · does not stabilize: indeed if there were n0 ∈ N such that Xn = Xn0 for all n ≥ n0 , then we would have X = Xn0 , a contradiction since X is not of finite type. This shows (ii) =⇒ (i). █ Exercise 1.56 Let G be a group and let A be a set. Let X ⊂ AG be a subshift. One says that the subshift X is topologically transitive (resp. irreducible, resp. topologically mixing) if the shift action of G on X is topologically transitive (resp. irreducible, resp. topologically mixing). (a) Show that X is topologically transitive if and only if it satisfies the following condition: (TT)

for all configurations x1 , x2 ∈ X and for every finite subset Ω of G, there exist a configuration x ∈ X and an element g ∈ G such that x|Ω = x1 |Ω and (gx)|Ω = x2 |Ω .

(b) Show that X is irreducible if and only if it satisfies the following condition: (I)

for every finite subset F ⊂ G, for all configurations x1 , x2 ∈ X, and for every finite subset Ω of G, there exist an element g ∈ G \ F and a configuration x ∈ X such that x|Ω = x1 |Ω and (gx)|Ω = x2 |Ω .

(c) Show that X is topologically mixing if and only if it satisfies the following condition: (TM)

for all configurations x1 , x2 ∈ X and for every finite subset Ω of G, there exists a finite subset F ⊂ G such that, for every element g ∈ G \ F , there exists a configuration x ∈ X satisfying x|Ω = x1 |Ω and (gx)|Ω = x2 |Ω .

(d) Show that every irreducible subshift of AG is topologically transitive. (e) Show that if G is infinite, then every topologically mixing subshift of AG is irreducible and topologically transitive. Solution (a) This follows from Exercise 1.8(c) since the set WΩ := {(y, z) ∈ X × X : y|Ω = z|Ω }, where Ω runs over all finite subsets of G, form a base of entourages for the prodiscrete uniform structure on X. (b) The proof is as in (a) by using Exercise 1.9(c) instead of Exercise 1.8(c). (c) The proof is as in (a) by using Exercise 1.11(c) instead of Exercise 1.8(c). (d) This follows immediately from Exercise 1.9(b). (e) This follows immediately from Exercise 1.11(b). █

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Exercise 1.57 Let G be a group and let A be a set. Let X ⊂ AG denote the subset consisting of all constant configurations. (a) Show that X is a subshift of AG . (b) Find a defining set of forbidden patterns for X. (c) Suppose that A contains at least two distinct elements. Show that X is not topologically transitive. (d) Suppose that G is infinite and that A contains at least two distinct elements. Show that X is not topologically mixing. (e) Suppose that A is finite and contains more than one element. Show that X is of finite type if and only if G is finitely generated. Solution (a) The set X is the set of fixed points of the shift action of G on AG (cf. Exercise 1.22(a)). As this action is continuous and the space AG is Hausdorff, we deduce that X is a closed G-invariant subset of AG , i.e., a subshift. (b) For each g ∈ G, set Ωg := {1G , g} ⊂ G and Fg :={p ∈ AΩg : p(1G ) /= p(g)}. It is then clear that X is the subshift admitting F := g∈G Fg ⊂ P(G, A) as a defining set of forbidden patterns. (c) The subshift X is not topologically transitive since it does not satisfy condition (TT) in Exercise 1.56. Indeed, let a and b be two distinct elements in A, and consider the configurations x1 , x2 ∈ X defined by x1 (g) := a and x2 (g) := b for all g ∈ G. Then, taking Ω := {1G }, there are no x ∈ X and g ∈ G such that x|Ω = x1 |Ω and (gx)|Ω = x2 |Ω . (d) This follows from (c) and the fact that, when G is infinite, every topologically mixing subshift is topologically transitive (cf. Exercise 1.11(b)). (e) Suppose first that G is finitely generated and let S ⊂ G be a finite generating subset of G. After replacing S by S ∪ S −1 , we can assume that S = S −1 . Then, given any element g ∈ G, there exist n ∈ N and s1 , s2 , . . . , sn ∈ S such that g = s1 s2 · · · sn . The minimal n ∈ N in such an expression of g is the S-length of g and is denoted by 𝓁S(g). With the notation from (b), let X' denote the subshift of AG admitting F ' := s∈S Fs ⊂ F as a defining set of forbidden patterns. Let us show that X = X' . By definition, we have X' = {x ∈ AG : (gx)(1G ) = (gx)(s) for all g ∈ G and s ∈ S}.

.

We clearly have X ⊂ X' since every configuration in X is constant and therefore fixed by G. Conversely, suppose that x ∈ X' . To prove that x ∈ X, it suffices to show that x(g) = x(1G ) for all g ∈ G. We proceed by induction on the Slength of g. If 𝓁S (g) = 0, then g = 1G and there is nothing to prove. Suppose now that x(g) = x(1G ) for all g ∈ G such that 𝓁S (g) ≤ n. Let g ' ∈ G such that 𝓁S (g ' ) = n + 1. Then we can find g ∈ G with 𝓁S (g) ≤ n and s ∈ S such that g ' = gs. We have x(g ' ) = x(gs) = (g −1 x)(s). As (g −1 x)(s) = (g −1 x)(1G ) since x ∈ X' , it follows that x(g ' ) = (g −1 x)(1G ) = x(g). As x(g) = x(1G ) by our induction hypothesis, we deduce that x(g ' ) = x(1G ). Thus x ∈ X. This completes the proof that X = X' . Since the set F ' is finite, this shows that X is of finite type.

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51

To prove the reverse implication, suppose that X is of finite type. Then there exists a finite subset S ⊂ G such that X admits a defining set of forbidden patterns F ⊂ AS (cf. Exercise 1.48(a)). Let H ⊂ G denote the subgroup generated by S. Let a and b be two distinct elements of A and consider the two constant configurations x1 , x2 ∈ X defined by x1 (g) := a and x2 (g) := b for all g ∈ G. Consider also the configuration x ∈ AG defined by x(g) := a if g ∈ H and x(g) := b otherwise. Let g ∈ G and s ∈ S. If g ∈ H , then we have (gx)(s) = x(g −1 s) = a = x1 (s) since g −1 s ∈ H . On the other hand, if g ∈ G \ H , then we have (gx)(s) = x(g −1 s) = b = x2 (s) since g −1 s ∈ / H . Thus, for all g ∈ G, we have (gx)|S ∈ / F , so that x ∈ X, that is, x is a constant configuration. We deduce that x(g) = x(1G ) = a for all g ∈ G, that is, x = x1 . This implies H = G. This shows that G is finitely generated. █ Exercise 1.58 Let G be a finitely generated group and let A be a finite set. Show that every finite subshift X ⊂ AG is of finite type. Solution Let S ⊂ G be a finite generating subset of G. After replacing S by S ∪ S −1 ∪ {1G }, we can assume that S = S −1 and 1G ∈ S. Then, given any element g ∈ G, there exist n ∈ N and s1 , s2 , . . . , sn ∈ S such that g = s1 s2 · · · sn . The minimal n ∈ N in such an expression of g is the S-length of g, denoted by 𝓁S (g). Suppose that X ⊂ AG is a finite subshift. For all distinct x, y ∈ X, we can find g = gx,y ∈ G such that x(g) /= y(g). Then the subset Δ := {gx,y : x, y ∈ X and x /= y} ⊂ G is finite and the restriction map X |→ AΔ , given by x |→ x|Δ , is injective. Consider the subshift of finite type X' ⊂ AG admitting Ω := {1G } ∪ SΔ as a memory set and P := {x|Ω : x ∈ X} as a defining set of admissible patterns. Let us show that X = X' . We clearly have X ⊂ X' since X is G-invariant. Conversely, suppose that x ' ∈ ' X . By definition of X' , for every g ∈ G, there exists xg ∈ X such that (g −1 x ' )|Ω = (xg )|Ω . Observe that, given g ∈ G, such an xg is unique since Δ ⊂ Ω. Let us show, by induction on the S-length of g, that xg = g −1 x1G

.

(1.18)

for all g ∈ G. If 𝓁S (g) = 0, then g = 1G and there is nothing to prove. Suppose now that (1.18) is satisfied for all g ∈ G such that 𝓁S (g) = n and let h ∈ G such that 𝓁S (h) = n + 1. Then h = gs for some g ∈ G with 𝓁S (g) = n and s ∈ S. For all d ∈ Δ, we have xh (d) = (h−1 x ' )(d)

.

(since Δ ⊂ Ω)

= (g −1 x ' )(sd)

(since h = gs)

= xg (sd)

(since SΔ ⊂ Ω)

= (g −1 x1G )(sd)

(by our induction hypothesis)

= (h−1 x1G )(d)

(since h = gs).

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1 Cellular Automata

Thus xh and h−1 x1G coincide on Δ. As xh , h−1 x1G ∈ X, this implies that xh = h−1 x1G . By induction, we conclude that (1.18) holds for all g ∈ G. Since 1G ∈ Ω, we deduce that x ' (g) = (g −1 x ' )(1G ) = xg (1G ) = (g −1 x1G )(1G ) = x1G (g)

.

for all g ∈ G. This shows that x ' = x1G ∈ X. In conclusion, X = X' is a subshift of finite type. █ Comment The hypothesis that G is finitely generated cannot be removed. Indeed, it follows from Exercise 1.57(e) that if the set A is finite with more than one element and the group G is not finitely generated, then the finite subshift X ⊂ AG consisting of all constant configurations is not of finite type. Exercise 1.59 Let G be a finitely generated group and let A be a finite set. Let B and C be disjoint subsets of A. Suppose that X ⊂ B G and Y ⊂ C G are two subshifts of finite type. Show that the subset Z ⊂ AG defined by Z := X ∪ Y ⊂ B G ∪ C G ⊂ (B ∪ C)G ⊂ AG

.

is a subshift of finite type of AG . Solution Let Ω be a finite subset of G. By taking Ω large enough, we can assume that Ω is a finite symmetric generating subset of G such that 1G ∈ Ω and that Ω is a memory set for both X and Y . Consider the subsets XΩ ⊂ B Ω and YΩ ⊂ C Ω defined by XΩ := {x|Ω : x ∈ X} and YΩ := {y|Ω : y ∈ Y }. As Ω is a memory set for X (resp. Y ), the set X (resp. Y ) consists of all configurations x ∈ B G (resp. y ∈ C G ) such that (gx)|Ω ∈ XΩ (resp. (gy)|Ω ∈ YΩ ) for all g ∈ G. Let P := XΩ ∪ YΩ ⊂ B Ω ∪ C Ω ⊂ (B ∪ C)Ω ⊂ AΩ

.

and consider the subshift of finite type T ⊂ AG consisting of all t ∈ AG such that (gt)|Ω ∈ P for all g ∈ G. Let us show that Z = T . Let z ∈ Z and let g ∈ G. We have z ∈ X or z ∈ Y . If z ∈ X (resp. z ∈ Y ), then (gz)|Ω ∈ XΩ (resp. (gz)|Ω ∈ YΩ ). In either case, (gz)|Ω ∈ P . As g ∈ G was arbitrary, we have z ∈ T . This shows that Z ⊂ T . To prove the converse inclusion, suppose that t ∈ T . Then, for all g ∈ G, we have (gt)|Ω ∈ XΩ or (gt)|Ω ∈ YΩ . Taking g := 1G , we deduce that t (Ω) ⊂ B or t (Ω) ⊂ C. Assume that t (Ω) ⊂ B. Let us show by induction that t (Ω n ) ⊂ B for all n ≥ 1. For n = 1, this is our assumption. Suppose that t (Ω n ) ⊂ B for some n ≥ 1 and let h ∈ Ω n+1 . Then h = gω, where g ∈ Ω n and ω ∈ Ω. We have (g −1 t)(1G ) = t (g) ∈ B by our induction hypothesis. As 1G ∈ Ω and (g −1 t)|Ω ∈ XΩ ∪ YΩ , we deduce that (g −1 t)|Ω ∈ XΩ . This implies t (h) = (g −1 t)(ω) ∈ B and completes induction. As Ω is a symmetric generating subset of G, we deduce that t (G) ⊂ B. It follows that (gt)|Ω ∈ XΩ for all g ∈ G, so that t ∈ X. As X

1.2 Exercises

53

and Y play symmetric roles, we deduce that if t (Ω) ⊂ C then t ∈ Y . Therefore t ∈ X ∪ Y = Z. This shows the inclusion T ⊂ Z. We conclude that Z = T . It follows that Z is a subshift of finite type of AG . █ Comment The condition that B and C are disjoint cannot be removed (see the counterexample in Exercise 1.48(e) for which G = Z, A = {0, 1, 2}, B = {0, 1}, and C = {0, 2}). Note that the condition that G is finitely generated cannot be removed either. Indeed, for any group G, if B := {0} and C := {1}, so that A := B ∪ C = {0, 1}, then it follows from Exercise 1.48(e) that X := B G and Y := C G are subshifts of finite type. However, if G is not finitely generated, it follows from Exercise 1.57(e) that Z := X ∪ Y (this is the subshift consisting of the two constant configurations in AG ) is not of finite type. Exercise 1.60 Let G be a countable group and let A be a finite set. Show that the set of all subshifts of finite type X ⊂ AG is countable. Solution This follows immediately from the fact that there are only countably many finite subsets Ω ⊂ G (resp. A ⊂ AΩ ) which may serve as memory sets (resp. defining sets of admissible patterns) for a subshift of finite type. █ Exercise 1.61 Let G be a group and let A be a finite set. (a) Let τ1 , τ2 : AG → AG be two cellular automata. Show that the set Eq(τ1 , τ2 ) := {x ∈ AG : τ1 (x) = τ2 (x)} ⊂ AG is a subshift of finite type. (b) Let X ⊂ AG . Show that the following conditions are equivalent: (i) X is a subshift of finite type; (ii) there exists a cellular automaton τ : AG → AG such that X = Fix(τ ). Here Fix(τ ) := {x ∈ AG : τ (x) = x} ⊂ AG denotes the fixed-point set of τ . (c) Let τ : AG → AG be a cellular automaton that is idempotent, i.e., such that 2 τ = τ , and set X := τ (AG ). Show that X = Fix(τ ) and deduce that X is a subshift of finite type of AG . Solution (a) Let S1 ⊂ G (resp. S2 ⊂ G) be a memory set for τ1 (resp. τ2 ). Then Ω := S1 ∪ S2 is a memory set for both τ1 and τ2 . Let μ1 : AΩ → A (resp. μ2 : AΩ → A) denote the local defining map for τ1 (resp. τ2 ) associated with Ω. Consider the set of patterns A ⊂ AΩ defined by A := {p ∈ AΩ : μ1 (p) = μ2 (p)},

.

and let us show that .

Eq(τ1 , τ2 ) = X(A ).

(1.19)

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Let x ∈ AG . We have τ1 (x) = τ2 (x) ⇐⇒ τ1 (x)(g) = τ2 (x)(g) for all g ∈ G ⇐⇒ μ1 ((g −1 x)|Ω ) = μ2 ((g −1 x)|Ω ) for all g ∈ G .

⇐⇒ (g −1 x)|Ω ∈ A for all g ∈ G ⇐⇒ x ∈ X(A ) (by (1.12)).

This shows (1.19). We deduce that Eq(τ1 , τ2 ) is a subshift of finite type with memory set Ω and defining set of admissible patterns A . (b) The implication (ii) =⇒ (i) follows from (a) after taking τ1 := τ and τ2 := IdAG . Conversely, suppose that X ⊂ AG is a subshift of finite type. Let us show that Condition (ii) is satisfied. We can assume that the alphabet A has more than one element. Let Ω ⊂ G and A ⊂ AΩ be a memory set and a defining set of admissible patterns for X. Up to enlarging Ω, we can assume 1G ∈ Ω. Consider the cellular automaton τ : AG → AG with memory set Ω and local defining map μ : AΩ → A given by μ(p) :=

.

⎧ p(1G ) ap

if p ∈ A otherwise,

where ap is an arbitrary element in A \ {p(1G )}. Observe that p ∈ A if and only if μ(p) = p(1G ). On the other hand, for all x ∈ AG and g ∈ G, we have τ (x)(g) = μ((g −1 x)|Ω ) and x(g) = (g −1 x)(1G ). We deduce that x ∈ Fix(τ ) if and only if (g −1 x)|Ω ∈ A for all g ∈ G, i.e., if and only if x ∈ X. This shows that Fix(τ ) = X. (c) Let x ∈ X. Then there exists y ∈ AG such that x = τ (y). As τ is an idempotent, we have τ (x) = τ (τ (y)) = τ (y) = x, showing that x ∈ Fix(τ ). Thus X ⊂ Fix(τ ). Conversely, suppose that x ∈ Fix(τ ). Then x = τ (x) ∈ τ (AG ) = X. This shows that Fix(τ ) ⊂ X. We deduce that X = Fix(τ ). Consequently, X is a subshift of finite type by (b). █ Comment The characterization of subshifts of finite type in (b) is used as a definition in [Gro1, Section 5.1]. Exercise 1.62 Let G be a group and let A be a set. Let B ⊂ A. Consider the subsets X, Y ⊂ AG defined by X := {b : b ∈ B} ⊂ AG , where b denotes the constant configuration given by b(g) := b for all g ∈ G, and Y := {y ∈ AG : y(g) ∈ B for all g ∈ G}. (a) Show that X and Y are subshifts of AG . (b) Show that if G is infinite then Y is topologically transitive. (c) Suppose that B has more than one element. Show that X is not topologically transitive.

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55

Solution (a) Let P1 :=

.

{p ∈ AΩ : p(g1 ) /= p(g2 )}, Ω

where Ω = {g1 , g2 } runs over all 2-elements subsets of G, and P2 :=

.



{q ∈ A{g} : q(g) ∈ A \ B},

g∈G

and set P = P1 ∪ P2 . Then XP = XP1 ∩ XP2 . Let us show that X = XP . For a configuration x ∈ AG one has x ∈ / X if and only if (i) there exist distinct elements g1 , g2 ∈ G such that x(g1 ) /= x(g2 ) or (ii) there exists g ∈ G such that x(g) ∈ A \ B. Setting Ω := {g1 , g2 } and p := x|Ω (resp. q := x|{g} ) we see that (i) holds exactly if x ∈ AG \ XP1 , while (ii) holds exactly if x ∈ AG \ XP2 . Thus X = XP is a subshift. Similarly, Y = XP2 is a subshift. (b) Suppose that G is infinite. Let Ω ⊂ G be a finite subset and let y1 , y2 ∈ Y . As G is infinite, we can find g ∈ G such that the sets Ω and g −1 Ω are disjoint. Fix an element b ∈ B and define a configuration y ∈ AG by setting, for all h ∈ G, ⎧ ⎪ ⎪ ⎨y1 (h) .y(h) := y2 (gh) ⎪ ⎪ ⎩b

if h ∈ Ω if h ∈ g −1 Ω otherwise.

Then y ∈ Y and, moreover, y|Ω = y1 |Ω and (gy)|Ω = y2 |Ω . This shows that Y is topologically transitive. (c) Let b1 , b2 be two distinct elements in B and consider the configurations xi := bi ∈ X, i = 1, 2. Let Ω := {1G }. Then, for every g ∈ G \ {1G }, we have Ω ∩ g −1 Ω = ∅. However, there is no b ∈ B such that the configuration x := b ∈ X satisfies that x|Ω = x1 |Ω (that is, b = x(1G ) = b1 ) and (gx)|Ω = x2 |Ω (that is, b = x(g −1 ) = b2 ). This shows that if B has more than one element then X is not topologically transitive. █ Exercise 1.63 We recall that a topological space X is called a Baire space if the intersection of every countable family of open dense subsets of X is dense in X. We also recall that the Baire category theorem asserts that every completely metrizable space is a Baire space (see e.g. [CAG, Theorem I.1.1 and Remark I.1.2]). (a) Let G be a group acting continuously on a topological space X. Show that if there exists a point in X whose G-orbit is dense in X then the action of G on X is topologically transitive. (b) Let G be a group acting continuously on a Baire space X. Suppose that X is non-empty and satisfies the second axiom of countability. Show that the following conditions are equivalent: (1) the action of G on X is topologically transitive; (2)

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there is a point x ∈ X whose G-orbit is dense in X; (3) there exists a non-empty G-invariant Gδ subset D ⊂ X such that the G-orbit of each point x ∈ D is dense in X. (c) Let G be a countable group and let A be a countable (e.g. finite) set. Let X ⊂ AG be a non-empty subshift. Show that the following conditions are equivalent: (1) the subshift X is topologically transitive; (2) there is a configuration x ∈ X whose G-orbit is dense in X; (3) there exists a non-empty G-invariant Gδ subset D ⊂ X such that the G-orbit of each configuration x ∈ D is dense in X. (d) Let G be an infinite countable group and let A be a countable (e.g. finite) non-empty set. Show that there exists a configuration x ∈ AG whose orbit under the G-shift action is dense in AG . (e) Let G be an infinite countable group and let A be an uncountable set. Show that the G-shift action on AG is topologically transitive but that there is no configuration x ∈ AG whose orbit under the G-shift action is dense in AG . Solution (a) Let x ∈ X be a point whose G-orbit is dense in X. Let U and V be non-empty open subsets of X. Since the orbit of x is dense, there are elements g1 , g2 ∈ G such that g1 x ∈ U and g2 x ∈ V . Setting g := g1 g2−1 , we then have g1 x ∈ U ∩ gV so that U ∩ gV /= ∅. This shows that the action of G on X is topologically transitive. (b) The implication (3) =⇒ (2) is trivial. The implication (2) implies (1) follows from (a). Therefore it is enough to prove that (1) implies (3). Suppose that the action of G on X is topologically transitive. As X satisfies the second axiom of countability, there exists a sequence (Un )n∈N of non-empty open subsets of X which form a base of the topology of X. Denote by Ωn the set of points x ∈ X whose G-orbit meets Un . We claim that each Ωn is open and dense in X. As the  action of G is continuous, we have that g −1 Un is open for all g ∈ G. Thus, Ωn = g∈G g −1 Un , being a union of open sets, is itself open. Suppose now that y ∈ X and let V be a neighborhood of y. By topological transitivity, there exists h ∈ G such that Un ∩ hV /= ∅. Choosing some z ∈ Un ∩ hV , we then have in X and h−1 z ∈ Ωn ∩ V , showing that Ωn ∩ V /= ∅. This shows that Ωn is dense∩ completes the proof of our claim. Since X is a Baire space, the set D := n∈N Ωn is a dense subset of X. In particular, D is not empty. The set D is a Gδ subset of X because it is a countable intersection of open subsets. On the other hand, Ωn is clearly G-invariant for every n so that D is also G-invariant. Let us show now that the G-orbit of every point in D is dense in X. Fix a point x ∈ D. Suppose that y ∈ X and let V be a neighborhood of y. Then we can find n ∈ N such that y ∈ Un ⊂ V . As x ∈ Ωn , there exists g ∈ G such that gx ∈ Un ⊂ V . We conclude that the G-orbit of x is dense in X. This shows the implication (1) =⇒ (3). (c) We know from Exercise 1.2 that AG is completely metrizable. As every closed subset of a completely metrizable space is itself completely metrizable, we deduce that X is completely metrizable. Thus, by the Baire category theorem, X is a Baire space. On the other hand, X satisfies the second axiom of countability since AG satisfies it by Exercise 1.6. We then deduce from (b) that conditions (1), (2), and (3) are equivalent.

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(d) This immediately follows from (c) since the action of G on AG is topologically transitive by Exercise 1.12(b). (e) The action of G on AG is topologically transitive by Exercise 1.12(b). For a ∈ A, denote by Ua the set of configurations in AG that take the value a at 1G . Observe that the sets Ua are non-empty pairwise disjoint open subsets of AG . Let x ∈ AG . As gx(1G ) = x(g −1 ) for all g ∈ G, we see that there can be only countably many a ∈ A such that the G-orbit of x meets Ua . █ Exercise 1.64 (Strongly Irreducible Subshifts) Let G be a group and let A be a set. Let Δ ⊂ G be a finite subset. A subshift X ⊂ AG is said to be Δ-irreducible if it satisfies the following condition: if Ω1 and Ω2 are two finite subsets of G such that Ω1 ∩ Ω2 Δ = ∅, then, given any two configurations x1 , x2 ∈ X, there exists a configuration x ∈ X which satisfies x|Ω1 = x1 |Ω1 and x|Ω2 = x2 |Ω2 . A subshift X ⊂ AG is said to be strongly irreducible if there exists a finite subset Δ ⊂ G such that X is Δ-irreducible. (a) Show that the subshift Y ⊂ AG described in Exercise 1.62 is {1G }-irreducible and therefore strongly irreducible. (b) Show that every strongly irreducible subshift X ⊂ AG is topologically mixing. Solution (a) Recall that Y := {y ∈ AG : y(g) ∈ B for all g ∈ G}, where B is a fixed subset of A. Let Ω1 , Ω2 be two finite disjoint subsets of G, and let x1 , x2 ∈ Y . Fix an element b ∈ B and define a configuration y ∈ AG by setting, for all g ∈ G, ⎧ ⎪ ⎪ ⎨y1 (g) .y(g) := y2 (g) ⎪ ⎪ ⎩b

if g ∈ Ω1 if g ∈ Ω2 otherwise.

Then y ∈ Y satisfies that y|Ω1 = y1 |Ω1 and y|Ω2 = y2 |Ω2 . This shows that Y is {1G }-irreducible. (b) Suppose that X ⊂ AG is a strongly irreducible subshift. Choose a finite subset Δ ⊂ G such that X is Δ-irreducible. Let Ω ⊂ G be a finite subset and let x1 , x2 ∈ X. Consider the subset F ⊂ G defined by F := ΩΔΩ −1 . Observe that F is finite and that Ω ∩ g −1 ΩΔ = ∅ for all g ∈ G \ F . Let g ∈ G \ F . By Δ-irreducibility applied to the configurations x1 and g −1 x2 and the finite subsets Ω and g −1 Ω, there exists a configuration x ∈ X satisfying x|Ω = x1 |Ω and x|g −1 Ω = (g −1 x2 )|g −1 Ω . Now, the latter means (gx)(ω) = x(g −1 ω) = (g −1 x2 )(g −1 ω) = x2 (ω) for all ω ∈ Ω, that is, (gx)|Ω = x2 |Ω . Thus X satisfies condition (TM) in Exercise 1.56(c) and therefore it is topologically mixing. █

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Comment Combining the result established in (b) with those of Exercise 1.56(d), (e), we deduce that, given an infinite group G, a set A, and a subshift X ⊂ AG , the properties of X are related by the following implications strongly irreducible =⇒ topologically mixing

.

=⇒ irreducible =⇒ topologically transitive. Each of the reverse implications is false in general. Exercise 1.65 Let G be a group and let A be a set. Let X ⊂ AG be a subshift. Show that the following conditions are equivalent: (i) X is strongly irreducible, i.e., there exists a finite subset Δ ⊂ G such that X is Δ-irreducible; (ii) there exists a finite symmetric subset Δ' ⊂ G such that the following holds: if Ω1 and Ω2 are two finite subsets of G such that Ω1 ∩ Ω2 Δ' = ∅, then, given any two configurations x1 , x2 ∈ X, there exists a configuration x ∈ X which satisfies x|Ω1 = x1 |Ω1 and x|Ω2 = x2 |Ω2 . (iii) there exists a finite subset Δ'' ⊂ G such that the following holds: if Ω1 and Ω2 are two finite subsets of G such that Ω1 Δ'' ∩ Ω2 Δ'' = ∅, then, given any two configurations x1 , x2 ∈ X, there exists a configuration x ∈ X which satisfies x|Ω1 = x1 |Ω1 and x|Ω2 = x2 |Ω2 . (iv) there exists a finite symmetric subset Δ''' ⊂ G such that the following holds: if Ω1 and Ω2 are two finite subsets of G such that Ω1 Δ''' ∩ Ω2 Δ''' = ∅, then, given any two configurations x1 , x2 ∈ X, there exists a configuration x ∈ X which satisfies x|Ω1 = x1 |Ω1 and x|Ω2 = x2 |Ω2 . Solution We first observe that if X satisfies condition (i) (resp. (ii), resp. (iii), resp. (iv)) then it also satisfies condition (i) (resp. (ii), resp. (iii), resp. (iv)) with Δ (resp. Δ' , resp. Δ'' , resp. Δ''' ) replaced by any larger finite subset of G. The implications (ii) =⇒ (i) and (iv) =⇒ (iii) are obvious. The reciprocals are proved by taking Δ' := Δ ∪ Δ−1 and Δ''' := Δ'' ∪ (Δ'' )−1 , respectively, and using the observation above. Suppose (iii). Then setting Δ' := Δ'' (Δ'' )−1 , we have (Δ' )−1 = Δ' and, for two finite subsets Ω1 and Ω2 of G, condition Ω1 Δ'' ∩Ω2 Δ'' = ∅ is equivalent to Ω1 ∩ Ω2 Δ' = ∅. This proves the implication (iii) =⇒ (ii). Conversely, suppose (ii). By the observation above, setting Δ'' := Δ' ∪ {1G } we have that X satisfies condition (ii) with respect to the finite symmetric set Δ := Δ'' (Δ'' )−1 . Then for two finite subsets Ω1 and Ω2 of G, condition Ω1 ∩ Ω2 Δ = ∅ is equivalent to Ω1 Δ'' ∩ Ω2 Δ'' = ∅. This proves the implication (ii) =⇒ (iii). Thus all the above conditions are equivalent. █ Exercise 1.66 (Weak Specification Property) Let X be a uniform space equipped with a uniformly continuous action of a group G. One says that the action of G on X has the weak specification property if it satisfies the following condition: (WSP)

for every entourage U of X, there exists a finite subset Λ = Λ(U ) ⊂ G such that the following holds: for any finite family (Ωi )i∈I of finite subsets

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of G such that Ωj ∩ ΛΩk = ∅ for all distinct j, k ∈ I and for any family of points (xi )i∈I in X, there exists a point x ∈ X such that (gx, gxi ) ∈ U for all i ∈ I and g ∈ Ωi . Such a subset Λ ⊂ G is then called a specification subset for (X, G, U ). Suppose that the action of G on X has the weak specification property. Let Y be a uniform space equipped with a uniformly continuous action of G and suppose there exists a G-equivariant uniformly continuous surjective map f : X → Y . Show that the action of G on Y has the weak specification property. Solution Let V be an entourage of Y . Since f is uniformly continuous, we can find an entourage U of X such that (f × f )(U ) ⊂ V .

.

(1.20)

Let Λ ⊂ G be a specification subset for (X, G, U ). Let (Ωi )i∈I be a finite family of finite subsets of G such that Ωj ∩ ΛΩk = ∅ for all distinct j, k ∈ I , and let (yi )i∈I be a family of points in Y . Since f is surjective, we can find, for each i ∈ I , a point xi ∈ X such that yi = f (xi ). On the other hand, as Λ is a specification subset for (X, G, U ), there exists a point x ∈ X such that (gx, gxi ) ∈ U

.

for all i ∈ I and g ∈ Ωi .

(1.21)

Setting y := f (x) ∈ Y , we then have, for all i ∈ I and g ∈ Ωi , (gy, gyi ) = (gf (x), gf (xi ))

.

= (f (gx), f (gxi ))

(since f is G-equivariant)

= (f × f )(gx, gxi ) ∈ (f × f )(U )

(by (1.21))

⊂V

(by (1.20)).

This shows that Λ is a specification subset for (Y, G, V ) and hence that the action of G on Y has the weak specification property. █ Comment The concept of specification, a property of dynamical systems allowing to approximate sufficiently time-separated pieces of orbits by a single (sometimes required to be periodic) orbit, was introduced by Rufus Bowen in [Bowe, Section 2.9]. Several variants and extensions of Bowen’s original definition of specification appear in the literature (see [ChuL, Definition 6.1], [DenGS, Chapter 21], [KwiLO], [LinS, Definition 5.1], [Rue]). Connections of specification with chaos for iterates of uniformly continuous maps on uniform spaces was investigated in [DasD]. The definition of weak specification presented above is taken from [CecC7] and is equivalent to the one given in [ChuL] and [Li] when restricted to continuous group actions on compact metrizable spaces.

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Exercise 1.67 Let (Xk )k∈K be a (possibly infinite) family of uniform spaces and let G be a group. Suppose that each Xk , k ∈ K, is equipped with a uniformly continuous action of G having the weak specification Π property. Show that the diagonal action of G on the uniform product X := k∈K Xk has the weak specification property. Solution For each k ∈ K, denote by pk : X → Xk the projection map onto Xk . Let U be an entourage of X. Then we can find a finite set L ⊂ K and, for every k ∈ L, and entourage Uk of Xk such that the entourage V of X defined by V := {(x, y) ∈ X × X : (pk (x), pk (y)) ∈ Uk for all k ∈ L}

.

satisfies V ⊂ U . As the action of G on Xk has the weak specification property, we can find, for every k ∈ L, a finite subsetΛk ⊂ G such that Λk is a specification subset for (Xk , G, Uk ). Clearly Λ := k∈L Λk is a specification subset for (X, G, V ) and hence for (X, G, U ). This shows that the diagonal action of G on X has the weak specification property. █ Exercise 1.68 (Weak Specification Property for Ultra-Uniform Spaces) Let X be a ultrauniform space (cf. Exercise 1.3) equipped with a uniformly continuous action of a group G. Show that the action of G on X has the weak specification property, that is, it satisfies condition (WSP) in Exercise 1.66, if and only if the following condition is satisfied: (WSP’)

for every entourage U of X, there exists a finite subset Λ ⊂ G such that the following holds: if Ω1 and Ω2 are two finite subsets of G such that Ω1 ∩ ΛΩ2 = ∅, then, given any two points x1 , x2 ∈ X, there exists a point x ∈ X such that (gx, gxk ) ∈ U for all k ∈ {1, 2} and g ∈ Ωk .

Solution As (WSP) trivially implies (WSP’), we only need to prove the converse implication. Suppose (WSP’). In order to show (WSP), we proceed by induction on |I |. The base case corresponds to |I | = 2 and this is indeed our assumption. Suppose that (WSP) holds whenever |I | ≤ n and let Λ = Λ(W, n) ⊂ G denote a finite subset guaranteeing (WSP) for any equivalence entourage W of X and any index set I with |I | ≤ n. Fix an entourage U of X and let I ' be a finite index set with |I ' | = n + 1. Since X is ultra-uniform, we can find an entourage W ⊂ U which is the graph of an equivalence relation on X. Let us show that Λ := Λ(W, n) also satisfies (WSP) for I ' . Let (Ωi )i∈I ' be a family of finite subsets of G such that Ωj ∩ ΛΩk = ∅ for all distinct j, k ∈ I ' , and let (xi )i∈I ' be a family of points in X. Fix i ' ∈ I ' and set I := I ' \ {i ' } so that |I | = n. Then, by the inductive hypothesis, we can find a point x ' ∈ X such that (gx ' , gxi ) ∈ W

.

for all i ∈ I and g ∈ Ωi .

(1.22)

Set Ω1 := ∪i∈I Ωi and Ω2 := Ωi ' as well as x1 := x ' and x2 := xi ' . Then Ω1 ∩ ΛΩ2 = ∅. By (WSP’), we can find x ∈ X such that (gx, gxk ) ∈ W for all k ∈ {1, 2}

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and g ∈ Ωk , that is, (gx, gx ' ) ∈ W

.

for all i ∈ I and g ∈ Ωi

(1.23)

and (gx, gxi ' ) ∈ W

.

for all g ∈ Ωi ' .

(1.24)

Since the entourage W is the graph of an equivalence relation, we deduce from (1.22) and (1.23) that (gx, gxi ) ∈ W for all i ∈ I and g ∈ Ωi . This, together with (1.24), yields (gx, gxi ) ∈ W ⊂ U for all i ∈ I ' and g ∈ Ωi . This completes the inductive argument and shows the implication (WSP’) =⇒ (WSP). █ Exercise 1.69 Let G be a group and let A be a set. Equip AG with its prodiscrete uniform structure and the shift action of G. Let X ⊂ AG be a subshift. Show that the following conditions are equivalent: (i) X is strongly irreducible; (ii) the shift action of G on X has the weak specification property. Solution Suppose first that X is strongly irreducible. Let Δ ⊂ G be a finite subset such that X is Δ-irreducible. By Exercise 1.3(c), (d), the space X ⊂ AG is ultrauniform. Thus, by virtue of Exercise 1.68, in order to show that the G-shift action on X has the weak specification property, it suffices to check that condition (WSP’) is satisfied. Fix an entourage U of X. Then we can find a finite subset Ω ⊂ G such that the entourage W of X, defined by W = W (Ω) := {(x, y) ∈ X × X : x(ω) = y(ω) for all ω ∈ Ω},

.

is contained in U . Consider the finite set Λ = Λ(U ) := ΩΔ−1 Ω −1 ⊂ G and suppose that Ω1 and Ω2 are two finite subsets of G such that Ω1 ∩ ΛΩ2 = ∅. Let also x1 , x2 ∈ X. Since Ω1−1 Ω ∩ Ω2−1 ΩΔ = ∅ and X is Δ-irreducible, we can find a configuration x ∈ X such that x|Ω −1 Ω = x1 |Ω −1 Ω and x|Ω −1 Ω = x2 |Ω −1 Ω . This 1 1 2 2 implies that (gx)(ω) = (gxk )(ω) for all ω ∈ Ω, k ∈ {1, 2} and g ∈ Ωk . It follows that (gx, gxk ) ∈ W ⊂ U for all k ∈ {1, 2} and g ∈ Ωk . This shows that the G-shift action on X has the weak specification property and proves the implication (i) =⇒ (ii). Conversely, suppose (ii). Let U = W (1G ) := {(x, y) ∈ X ×X : x(1G ) = y(1G )} and let Λ = Λ(U ) ⊂ G be a specification subset for (X, G, U ). Let us show that X is Δ-irreducible with Δ := Λ−1 ⊂ G. Let Ω1 and Ω2 be finite subsets of G such that Ω1 ∩ Ω2 Δ = ∅ and let x1 , x2 ∈ X. As Ω1−1 ∩ ΛΩ2−1 = ∅ and Λ is a specification subset for (X, G, U ), we can find a configuration x ∈ X such that (gx, gxk ) ∈ U for all k ∈ {1, 2} and g ∈ Ωk−1 . This implies x(g) = (g −1 x)(1G ) = (g −1 xk )(1G ) = xk (g) for all k ∈ {1, 2} and g ∈ Ωk . It follows that x|Ωk = xk |Ωk for all k ∈ {1, 2}. This shows that X is Δ-irreducible. The implication (ii) =⇒ (i) follows. █

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Comment In the case when the group G is countable and the set A is finite, this result was established by Li in [Li, Proposition A.1]. The general case is contained in [CecC7, Proposition 6.7]. Exercise 1.70 Let G be a group. Let A and B be sets. Let X ⊂ AG and Y ⊂ B G be subshifts. Suppose that there exists a surjective cellular automaton τ : X → Y . Show that if X is topologically transitive (resp. irreducible, resp. topologically mixing, resp. strongly irreducible) then Y is topologically transitive (resp. irreducible, resp. topologically mixing, resp. strongly irreducible). Solution As the map τ : X → Y is a cellular automaton, it is continuous and G-equivariant by Exercise 1.40. By applying Exercise 1.13, we deduce that if X is topologically transitive (resp. irreducible, resp. topologically mixing) then Y is topologically transitive (resp. irreducible, resp. topologically mixing). Suppose now that X is strongly irreducible. This means that there exists a finite subset Δ ⊂ G such that X satisfies Condition (i) in Exercise 1.65. Let S ⊂ G be a memory set for τ . Consider the finite subset Λ ⊂ G defined by Λ := SΔS −1 . Let Ω1 and Ω2 be two finite subsets of G such that Ω1 ∩ Ω2 Λ = ∅. Then Ω1 S ∩Ω2 SΔ = ∅. Let now y1 , y2 ∈ Y and choose x1 , x2 ∈ X such that τ (x1 ) = y1 and τ (x2 ) = y2 . Since X satisfies Condition (i), there exists x ∈ X such that x|Ω1 S = x1 |Ω1 S and x|Ω2 S = x2 |Ω2 S . As S is a memory set for τ , this implies that the configuration y := τ (x) ∈ Y is such that y|Ω1 = y1 |Ω1 and y|Ω2 = y2 |Ω2 . This shows that τ (X) is strongly irreducible. █ Comment The fact that the strong irreduciblity of X implies that of Y can also be deduced from Exercise 1.66 by using the dynamical characterization of strongly irreducible subshifts established in Exercise 1.69. Exercise 1.71 (The At-Most-One-One Subshift) Let G be a group and let A := {0, 1}. Consider the set X ⊂ AG consisting of all configurations x ∈ AG such that 1 appears at most once in x (i.e., such that the set {g ∈ G : x(g) = 1} is either empty or reduced to a single element). (a) Show that X is a subshift of AG . It is called the at-most-one-one subshift. (b) Describe the set of all cellular automata τ : X → X. (c) Show that X is topologically transitive if and only if G is infinite. (d) Show that X is not irreducible. (e) Show that if G is infinite then X is neither topologically mixing, nor strongly irreducible, nor of finite type. Solution (a) Let F ⊂ P(G, A) denote the set of all patterns p : Ω → A such that Ω ⊂ G has cardinality |Ω| = 2 and p(ω) = 1 for all ω ∈ Ω. It is then clear that X is the subshift of AG admitting F as a defining set of forbidden patterns. (b) For g ∈ G, let xg ∈ X denote the configuration defined by xg (h) := 1 if h = g and xg (h) := 0 if h ∈ G \ {g}. Let also c0 ∈ X denote the constant configuration defined by c0 (g) := 0 for all g ∈ G. Note that hxg = xhg for all g, h ∈ G and that c0 is fixed by G. Clearly X = {c0 } ∪ {xg : g ∈ G}. Let now τ : X → X be a cellular

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automaton. Using the G-equivariance of τ , we get gτ (c0 ) = τ (gc0 ) = τ (c0 ) for all g ∈ G. As c0 is the unique configuration in X that is fixed by G, we deduce that τ (c0 ) = c0 . We have two possibilities for τ (x1G ) ∈ X, namely, either τ (x1G ) = c0 or τ (x1G ) = xg0 for some g0 ∈ G. Using again the G-equivariance of τ , in the first case we have τ (xg ) = τ (gx1G ) = g(τ (x1G )) = gc0 = c0

.

for all g ∈ G. Keeping in mind that τ (c0 ) = c0 , we deduce that τ is the constant cellular automaton given by τ (x) := c0 for all x ∈ X. In the second case, τ (xg ) = τ (gx1G ) = g(τ (x1G )) = gxg0 = xgg0 = τg0 (xg ),

.

for all g ∈ G, where τg0 is the restriction to X of the cellular automaton on AG with memory set S := {g0 } and local defining map μ : AS → A given by μ(y) := y(g0 ) (cf. [CAG, Example 1.4.3.(e)]). (c) Suppose first that G is infinite. In order to show that X is topologically transitive, it suffices to show that there exists a configuration x ∈ X whose Gorbit Gx is dense in X (cf. Exercise 1.63(a)). Taking x := x1G we have Gx = {xg : g ∈ G}. Moreover, if Ω ⊂ G is a finite subset, then taking g ∈ G \ Ω (here we use the fact that G is infinite) we have that (gx)|Ω = xg |Ω = c0 |Ω . This shows that Gx is dense in X and therefore that X is topologically transitive. Suppose now that the group G is finite. Then the topology on AG is discrete. As U := {c0 } and V := {xg : g ∈ G} are disjoint non-empty G-invariant open subsets of X, we deduce that X is not topologically transitive. (d) We can assume that the group G is not trivial. Let us show that X does not satisfy condition (I) in Exercise 1.56(b). Let g0 ∈ G such that g0 /= 1G and set F := {g0−1 } ⊂ G and Ω := {1G , g0 }. Consider the configurations x1 , x2 ∈ X defined by setting x1 (g0 ) := 1 and x1 (h) := 0 for all h ∈ G \ {g0 }, and x2 (1G ) := 1 and x2 (h) := 0 for all h ∈ G \ {1G }. Suppose that g ∈ G \ F and x ∈ AG satisfy that x|Ω = x1 |Ω and (gx)|Ω = x2 |Ω . We then have x(g0 ) = x1 (g0 ) = 1 and x(g −1 ) = (gx)(1G ) = x2 (1G ) = 1. Since g0 /= g −1 , we deduce that x ∈ / X. (e) Suppose that G is infinite. Let g0 ∈ G such that g0 /= 1G . Observe that U := {xg0 } and V := {x1G } are two non-empty subsets of X. As gV = {xg } for all g ∈ G, we deduce that U ∩ gV = ∅ for all g ∈ G \ {g0 }. Since G is infinite, this shows that X is not topologically mixing. It follows from Exercise 1.64(b) that X is not strongly irreducible either. Suppose by contradiction that X is of finite type. Then there exists a finite subset Ω ⊂ G which is a memory set for X. Thus, a configuration y ∈ AG is in X if and only if (gy)|Ω ∈ XΩ for all g ∈ G, where XΩ := {x|Ω : x ∈ X}. Observe that XΩ consists of all patterns p : Ω → A in which 1 appears at most once (i.e., such that the set {ω ∈ Ω : p(ω) = 1} is either empty or reduced to a single element). Take g0 ∈ G such that g0 /= 1G and g0 ∈ / Ω −1 Ω (this is always possible since G is −1 infinite and Ω and therefore Ω Ω are finite). Consider the configuration y ∈ AG defined by y(1G ) = y(g0 ) = 1 and y(g) = 0 if g ∈ G \ {1G , g0 }. We have y ∈ /X

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since 1 appears twice in y. However, for all g ∈ G, we have {1G , g0 } /⊂ gΩ and hence (gy)|Ω ∈ XΩ . This shows that X is not of finite type. █ Comment The at-most-one-one subshift is also called the sunny-side-up subshift by some authors. Exercise 1.72 Let G be a group, let H be a finite index subgroup of G, and let A be a finite set. Let T ⊂ G be a complete set of representatives for the right cosets of H in G and set B := AT . Consider the map Ψ : AG → B H defined in Exercise 1.32. Let X ⊂ AG be a subshift and set X(H,T ) := Ψ (X). (a) Show that X(H,T ) is a subshift of B H . (b) Show that if X(H,T ) is of finite type then X is of finite type. Conversely, show that if G is abelian and X is of finite type then X(H,T ) is of finite type. Solution (a) It follows from Exercise 1.32 that the map Ψ is an H -equivariant uniform isomorphism. Since X is H -invariant and Ψ is H -equivariant, we deduce that X(H,T ) = Ψ (X) is H -invariant. As A is finite, X ⊂ AG is compact for the prodiscrete topology on AG . As Ψ is (uniformly) continuous, we deduce that X(H,T ) = Ψ (X) is a compact subset in B H for the prodiscrete topology on B H . Since the prodiscrete topology on B H is Hausdorff, we deduce that X(H,T ) is closed in B H . This shows that X(H,T ) is a subshift of B H . (b) Set Y := X(H,T ) and suppose that Y is of finite type. It follows from Exercise 1.52(c) that there exists a finite subset ΩH ⊂ H such that, with the notation therein, Y = YΩH . Set ΩG := ΩH T ⊂ G and let us show that X = XΩG . Since X ⊂ XΩG , we only need to check for the reverse inclusion. Recall (cf. the solution to Exercise 1.32) that the entourages W A (ΩG ) := {(x, x ' ) ∈ AG × AG : x|ΩG = x ' |ΩG } and W B (ΩH ) := {(z, z' ) ∈ B H × B H : z|ΩH = z' |ΩH } satisfy (Ψ × Ψ )(W A (ΩG )) ⊂ W B (ΩH ).

.

(1.25)

Let then x ' ∈ XΩG and set y ' := Ψ (x ' ) ∈ B H . Given h ∈ H we can find x ∈ X such that πΩG (hx ' ) = πΩG (x), that is, (hx ' , x) ∈ W A (ΩG ). Setting y := Ψ (x) ∈ Y , from (1.25) and H -equivariance of Ψ (cf. Exercise 1.32) we deduce that (hy ' , y) = (hΨ (x ' ), Ψ (x)) = (Ψ (hx ' ), Ψ (x)) ∈ W B (ΩH ), equivalently, πΩH (hy ' ) = πΩH (y). As h was arbitrary, this shows that y ' ∈ YΩH = Y . As Ψ is bijective, we have x ' = Ψ −1 (y ' ) ∈ Ψ −1 (Y ) = X. Thus XΩG ⊂ X, and equality follows. We deduce from Exercise 1.52(c) that X is of finite type. Suppose now that G is abelian. Suppose that X is of finite type and let Ω ⊂ G be a finite set serving as a memory set for X. Then we can find a finite set ΩH ⊂ H such that Ω ⊂ ΩH T . This way, the finite set ΩG := ΩH T ⊂ G also serves as a memory set for X and we can take AG := XΩG ⊂ AΩG as an associated defining set of admissible patterns for X. Let s, t ∈ T . Then there exist unique h = h(s, t) ∈ H and σt (s) ∈ T such that t −1 s = hσt (s). Clearly, the map σt : T → T is a permutation of T . Given a pattern

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65

p ∈ B ΩH and t ∈ T , we denote by pt the pattern in B ΩH defined by setting pt (h)(s) := p(h)(σt (s))

.

for all h ∈ ΩH and s ∈ T . Also, for p ∈ B ΩH denote by pG ∈ AΩG the pattern defined by setting G p (ht) := p(h)(t) ∈ A for all h ∈ ΩH and t ∈ T . Note that pG is well defined since G = ⨅t∈T H t. Let us show that ΩH serves as a memory set for Y with associated defining set of admissible patterns AH := {p ∈ B ΩH : (pt )G ∈ AG for all t ∈ T } ⊂ B ΩH ,

.

in formulæ, Y = X(AH ). Let y ∈ B H . Suppose first that y ∈ Y . Then there exists x ∈ X such that y = Ψ (x) so that, y(h)(s) = x(hs) for all h ∈ H and s ∈ T . Let k ∈ H and let us show that p := (ky)|ΩH ∈ B ΩH belongs to AH . Let t, s ∈ T and h' ∈ ΩH , and set g := h(s, t)kt ∈ G. Then we have (pt )G (h' s) = pt (h' )(s) = p(h' )(σt (s)) = (ky)(h' )(σt (s))

.

= y(k −1 h' )(h(s, t)−1 t −1 s) = x(k −1 h(s, t)−1 t −1 h' s) = x(g −1 h' s) = (gx)(h' s) = (gx)|ΩG (h' s). Thus, (pt )G = (gx)|ΩG ∈ AG . As t ∈ T was arbitrary, we deduce that (ky)|ΩH = p ∈ AH . Since k ∈ H was arbitrary, this shows that y ∈ X(AH ). Conversely, suppose that y ∈ X(AH ). Setting x := Ψ −1 (y) ∈ AG , we want to show that x ∈ X = Ψ −1 (Y ). Let then g ∈ G and let us show that the pattern (gx)|ΩG ∈ AΩG belongs to AG . Let h ∈ H and t ∈ T such that g = ht. Let also h' ∈ ΩH and s ∈ T . Keeping in mind that G is abelian, recalling that t −1 s = h(s, t)σt (s), and setting k := h(s, t)−1 h ∈ H , we have (gx)(h' s) = x(g −1 h' s) = x(h−1 h' t −1 s) = x(h−1 h' h(s, t)σt (s)) = y(h−1 h' h(s, t))(σt (s)) = y(k −1 h' )(σt (s)) = (ky)(h' )(σt (s)). It follows that setting p := (ky)|ΩH , we have (pt )G (h' s) := p(h' )(σt (s)) = (ky)(h' )(σt (s)) = (gx)(h' s) = (gx)|ΩG (h' s). Since y ∈ X(AH ), we have (ky)|ΩH = p ∈ AH . As a consequence, (gx)|ΩG = (pt )G ∈ AG . As g ∈ G was arbitrary, and AG is a defining set of admissible patterns for X, this shows that x ∈ X so that y = Ψ (x) ∈ Ψ (X) = Y . We deduce that Y = X(AH ) is of finite type. █ Comment When G = Z and H = NZ for some integer N ≥ 1, one can take T := {0, 1, . . . , N − 1} ⊂ Z and one has B = AN . The subshift X(H,T ) ⊂ B Z is then called the Nth higher power subshift of X (cf. [LinM, Definition 1.4.4]. See also Exercise 6.114).

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Exercise 1.73 Let G be a group and let H be a finite index subgroup of G. Let T ⊂ G be a complete set of representatives for the right cosets of H in G, so that G G = ⨅t∈T H t. Let A1 , A2 be two sets and let τ : AG 1 → A2 be a cellular automaton. T For i = 1, 2, set Bi := Ai and consider the H -equivariant uniform isomorphism G H Ψi : AG i → Bi defined by setting Ψi (x)(h)(t) := x(ht) for all x ∈ Ai , h ∈ H , and t ∈ T (cf. Exercise 1.32). Define a map τ (T ) : B1H → B2H by setting τ (T ) := Ψ2 ◦ τ ◦ Ψ1−1 .

.

(a) Show that τ (T ) is a cellular automaton. G (b) Let X ⊂ AG 1 be a subshift and let Y := τ (X) ⊂ A2 denote the corresponding image subshift under τ . Consider the subshifts X(H,T ) := Ψ1 (X) ⊂ B1H and Y (H,T ) := Ψ2 (Y ) ⊂ B2H (cf. Exercise 1.72). Show that τ (T ) (X(H,T ) ) = Y (H,T ) . Solution (a) The map Ψ1−1 (resp. Ψ2 ) is uniformly continuous and H -equivariant, since Ψ1 (resp. Ψ2 ) is an H -equivariant uniform isomorphism (cf. Exercise 1.32). Moreover (cf. [CAG, Theorem 1.9.1]), τ is uniformly continuous and G-equivariant, and a fortiori, H -equivariant. As the composition of uniformly continuous (resp. H equivariant) maps is itself uniformly continuous (resp. H -equivariant), it follows that τ (T ) is uniformly continuous and H -equivariant. We deduce from the uniform version of the Curtis-Hedlund-Lyndon theorem [CAG, Theorem 1.9.1] that τ (T ) is a cellular automaton. (b) We have τ (T ) (X(H,T ) ) = Ψ2 (τ (Ψ1−1 (X(H,T ) ))) = Ψ2 (τ (X)) = Ψ2 (Y ) = Y (H,T ) .

.

█ Exercise 1.74 (Language of a Subshift Over Z) Let A be a set. Let A∗ denote the free monoid based on A, that is, the set consisting of all words on the alphabet A with the concatenation of words as the monoid operation (cf. [CAG, Section D.1]). One says that a word w ∈ A∗ appears in a configuration x ∈ AZ if there exist integers n, m ∈ Z with n ≤ m such that w = x(n)x(n + 1) · · · x(m).

.

By convention, the empty word ϵ ∈ A∗ appears in any configuration x ∈ AZ . The set consisting of all words appearing in a configuration x ∈ AZ is called the language of x and denoted by L(x). More generally, one says that a word w ∈ A∗ appears in

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a subset X ⊂ A∗ if there exists a configuration x ∈ X such that w appears in x. The set

.L(X) := L(x) ⊂ A∗ , (1.26) x∈X

consisting of all words appearing in X, is called the language of X. (a) Let X, Y ⊂ AZ and suppose that Y is a subshift. Show that one has X ⊂ Y if and only if L(X) ⊂ L(Y ). (b) Let X ⊂ AZ be a subshift and let x ∈ AZ . Show that one has x ∈ X if and only if L(x) ⊂ L(X). (c) Let X ⊂ AZ be a subshift and let x ∈ AZ . Show that one has L(x) = L(X) if and only if X is the orbit closure of x. (d) Let X and Y be two subshifts of AZ . Show that one has X = Y if and only if L(X) = L(Y ). (e) One says that a word u ∈ A∗ is a subword of a word w ∈ A∗ if there exist v1 , v2 ∈ A∗ such that w = v1 uv2 . Let X ⊂ AZ be a subshift and let L := L(X) denote the language of X. Show that L satisfies the following conditions: (i) if w ∈ L, then u ∈ L for every subword u of w; (ii) if w ∈ L, then there exist a, b ∈ A such that awb ∈ L. (f) Conversely, show that if a subset L ⊂ A∗ satisfies conditions (i) and (ii) in (e), then there exists a unique subshift X ⊂ AZ such that L = L(X). Solution (a) The fact that X ⊂ Y implies L(X) ⊂ L(Y ) is obvious. Conversely, suppose that L(X) ⊂ L(Y ). Let x ∈ X and let Ω be a finite subset of Z. There exist n, m ∈ Z with n ≤ m such that Ω ⊂ {n, n+1, . . . , m}. We then have x(n)x(n+1) · · · x(m) ∈ L(X) ⊂ L(Y ). Thus, there exist y ∈ Y and integers n' , m' ∈ Z with m' −n' = m−n such that x(n)x(n + 1) · · · x(m) = y(n' )y(n' + 1) · · · y(m' ). By suitably shifting y, we obtain a configuration z ∈ Y which coincides with x on {n, n + 1, . . . , m} and hence on Ω. Therefore x is in the closure of Y . As Y is a subshift and hence closed in AZ , this implies x ∈ Y . This shows that X ⊂ Y . (b) This immediately follows from (a) after replacing X by {x} and Y by X. (c) Suppose that X is the orbit closure of x. Then x ∈ X and hence L(x) ⊂ L(X) by (b). To prove the converse inclusion, let w ∈ L(X). This means that there exists y ∈ AZ and n, m ∈ Z such that w = y(n)y(n + 1) · · · y(m). As the orbit of x is dense in X, there exists k ∈ Z such that y(n)y(n + 1) · · · y(m) = x(k + n)x(k + n + 1) · · · x(k + m).

.

Therefore w appears in x. This shows that L(x) = L(X). Conversely, suppose that L(x) = L(X). We have x ∈ X by (b). On the other hand, suppose that y ∈ X. Let V be a neighborhood of y in X. By definition of the prodiscrete topology, there is an interval [n, m] ⊂ Z such that V contains all

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configurations in X which coincide with y on [n, m]. As L(y) ⊂ L(X) = L(x), there exists k ∈ Z such that y(n) · · · y(m) = x(k + n) · · · x(k + m). Thus V meets the orbit of x. As X is closed in AZ , we deduce that X is equal to the orbit closure of x. (d) This immediately follows from (a) by double inclusion. (e) Suppose that w ∈ L. This means that there exist x ∈ X and integers n, m ∈ Z with n ≤ m such that w = x(n)x(n + 1) · · · x(m). If u is a subword of w, then there are integers k1 , k2 ∈ Z with n ≤ k1 ≤ k2 ≤ m such that u = x(k1 )x(k1 + 1) · · · x(k2 ). Therefore u appears in x and hence in X. This shows (i). Taking a := x(n − 1) and b := x(m + 1), we have awb = x(n − 1)x(n) · · · x(m)x(m + 1). Therefore awb appears in x and hence in X. This shows (ii). (f) Suppose that L ⊂ A∗ satisfies (i) and (ii) in (e). Let X ⊂ AZ be the set consisting of all configurations x ∈ AZ satisfying the following condition: all words w ∈ A∗ appearing in x are in L. Let us show that X is a subshift of AZ and that L(X) = L. Suppose that x ∈ X and that y ∈ AZ is in the orbit of x under the Z-shift. This means that there exists k ∈ Z such that y(i) = x(i − k) for all i ∈ Z. Let n, m ∈ Z with n ≤ m. We then have y(n)y(n + 1) · · · y(m) = x(n − k)x(n − k + 1) · · · x(m − k) ∈ L. Therefore y ∈ X. This shows that X is Z-invariant. Suppose now that z ∈ AZ is in the closure of X. Let n, m ∈ Z with n ≤ m. Since the set Ω := {n, n + 1, . . . , m} is finite and x is in the closure of X for the prodiscrete topology on AZ , there exists a configuration x ∈ X which coincides with z on Ω. This implies z(n)z(n + 1) · · · z(m) = x(n)x(n + 1) · · · x(m) ∈ L. Consequently, z ∈ X. This shows that X is closed in AZ . The inclusion L(X) ⊂ L is obvious. Conversely, suppose that w is a word in L. If n is the length of w, we have w = a1 a2 · · · an with ai ∈ A for 1 ≤ i ≤ n. By induction on (ii), we can find a sequence (wk )k∈N of words in L with w0 = w such that, for all k ∈ N, there exist bk , ck ∈ A such that wk+1 = bk wk ck . Consider the configuration x ∈ AZ defined by

x(i) :=

.

⎧ ⎪ ⎪ ⎨ai b−i+1

⎪ ⎪ ⎩c

i−n

if 1 ≤ i ≤ n if i ≤ 0 if n + 1 ≤ i,

for all i ∈ Z. It then follows from (ii) that x ∈ X. As w = x(1)x(2) · · · x(n) appears in x, we deduce that w ∈ L(X). This shows that X is a subshift of AZ such that L(X) = L. Uniqueness of the subshift X follows from (d). █ Exercise 1.75 Let A be a set and let X, Y ⊂ AZ be subshifts. (a) Show that X∪Y is a subshift of AZ and that one has L(X∪Y ) = L(X)∪L(Y ). (b) Show that X∩Y is a subshift of AZ and that one has L(X∩Y ) ⊂ L(X)∩L(Y ). (c) Give an example showing that the inclusion in (b) may be strict.

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Solution (a) The fact that X ∪ Y is a subshift of AZ follows from Exercise 1.39(e). Since X ⊂ X ∪ Y and Y ⊂ X ∪ Y , we have L(X) ⊂ L(X ∪ Y ) and L(Y ) ⊂ L(X ∪ Y ) (cf. Exercise 1.74(a)). Therefore, L(X) ∪ L(Y ) ⊂ L(X ∪ Y ). To prove the converse inclusion, suppose that u ∈ L(X ∪ Y ). This means that there exists a configuration x ∈ X ∪ Y such that u appears in x. We must have x ∈ X or x ∈ Y . Therefore, u ∈ L(X) or u ∈ L(Y ), that is, u ∈ L(X) ∪ L(Y ). Thus, L(X ∪ Y ) ⊂ L(X) ∪ L(Y ). This shows that L(X ∪ Y ) = L(X) ∪ L(Y ). (b) The fact that X ∩ Y is a subshift of AZ follows from Exercise 1.39(d). Since X ∩ Y ⊂ X and X ∩ Y ⊂ Y , we have L(X ∩ Y ) ⊂ L(X) and L(X ∩ Y ) ⊂ L(Y ) (cf. Exercise 1.74(a)). Therefore, L(X ∩ Y ) ⊂ L(X) ∩ L(Y ). (c) Let A := {0, 1}. Consider the subshifts X, Y ⊂ AZ defined by X := {x} and Y := {y, z}, where x ∈ AZ is the constant configuration given by x(n) := 0 for all n ∈ Z and y, z ∈ AZ are the periodic configurations given by y(2n) = z(2n + 1) := 0 and y(2n + 1) = z(2n) := 1 for all n ∈ Z. We have X ∩ Y = ∅ so that L(X ∩ Y ) = ∅. On the other hand 0 ∈ L(X) ∩ L(Y ) so that L(X) ∩ L(Y ) /= ∅. This shows that L(X ∩ Y ) ⊊ L(X) ∩ L(Y ). █ Exercise 1.76 Let A be a set and let X ⊂ AZ be a subshift. Let L(X) ⊂ A∗ denote the language of X. (a) Show that X is topologically transitive if and only if the following holds: for all words u and v in L(X), there exists a word w ∈ L(X) such that both u and v are subwords of w. (b) Show that X is irreducible if and only if the following holds: for all words u and v in L(X), there exists w ∈ A∗ such that uwv ∈ L(X). (c) Show that X is topologically mixing if and only if the following holds: for all words u and v in L(X), there exists an integer n0 = n0 (u, v) ≥ 0 such that for any integer n ≥ n0 there exists a word w ∈ A∗ of length 𝓁(w) = n satisfying uwv ∈ L(X). (d) Show that X is strongly irreducible if and only if the following holds: there exists an integer n0 ≥ 0 such that, for all words u and v in L(X), and for any integer n ≥ n0 , there exists a word w ∈ A∗ of length 𝓁(w) = n such that uwv ∈ L(X). Solution (a) Suppose first that X is topologically transitive. Let u and v be two words in L(X). Then there exist two configurations x, y ∈ X such that u = x(1)x(2) · · · x(𝓁(u)) and v = y(1)y(2) · · · y(𝓁(v)). Consider the open subsets U and V of X defined by U := {z ∈ X : z(1)z(2) · · · z(𝓁(u)) = u} and

.

V := {z ∈ X : z(1)z(2) · · · z(𝓁(v)) = v}.

(1.27)

Note that U and V are non-empty since x ∈ U and y ∈ V . By topological transitivity of X, we can find g ∈ Z such that U ∩gV /= ∅. Let z ∈ U ∩gV . We have z(1)z(2) · · · z(𝓁(u)) = u, since z ∈ U , and z(1 + g)z(2 + g) · · · z(𝓁(v) + g) = v,

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since (−g)z ∈ V . Setting m := min{1, 1 + g}, n := max{𝓁(u), 𝓁(v) + g}, we have w := z(m)z(m + 1) · · · z(n) ∈ L(X). As both u and v are subwords of w, this shows that the condition is necessary. Conversely, suppose that for all words u and v in L(X), there exists a word w ∈ L(X) such that both u and v are subwords of w. Let U and V be non-empty open subsets of X. By definition of the prodiscrete topology, there are finite subsets FU , FV ⊂ Z and x, y ∈ X such that {z ∈ X : z|FU = x|FU } ⊂ U and {z ∈ X : z|FV = y|FV } ⊂ V .

.

Choose intervals [mU , nU ], [mV , nV ] ⊂ Z such that FU ⊂ [mU , nU ] and FV ⊂ [mV , nV ], and consider the words u and v in L(X) defined by u := x(mU )x(mU + 1) · · · x(nU ) and v := y(mV )y(mV + 1) · · · y(nV ).

.

(1.28)

Observe that if z ∈ X satisfies that z(mU )z(mU + 1) · · · z(nU ) = u (resp. z(mV )z(mV + 1) · · · z(nV ) = v), then z ∈ U (resp. z ∈ V ). As u, v ∈ L(X), by our assumptions there exists w ∈ L(X) such that both u and v are subwords of w. Since w ∈ L(X), we can find z' ∈ X such that w appears in z' . This implies that both u and v appear in z' , so that there are integers h, k ∈ Z such that z' (h + mU )z' (h + mU +1) · · · z' (h+nU ) = u and z' (k+mV )z' (k+mV +1) · · · z' (k+nV ) = v. Setting z := (−h)z' ∈ X and g := k − h ∈ Z, we then have z(mU )z(mU + 1) · · · z(nU ) = u and z(mV + g)z(mV + 1 + g) · · · z(nV + g) = v, showing that z ∈ U ∩ gV . This shows that X is topologically transitive. (b) Suppose that X is irreducible. Let u, v ∈ L(X) and let us show that there exists w ∈ A∗ such that uwv ∈ L(X). Let 𝓁 := max(|u|, |v|) and set Ω := {1, 2, . . . , 𝓁} ⊂ Z (resp. F := {−𝓁, −𝓁 + 1, . . . , 𝓁} ⊂ Z). Since u, v ∈ L(X), we can find two configurations x1 , x2 ∈ X such that x1 (1)x1 (2) · · · x1 (|u|) = u and x2 (1)x2 (2) · · · x2 (|v|) = v. By condition (I) in Exercise 1.56(b), we can find g ∈ G \ F and x ∈ X such that x|Ω = x1 |Ω and (gx)|Ω = x2 |Ω . We thus have x(1)x(2) · · · x(|u|) = x1 (1)x1 (2) · · · x1 (|u|) = u and x(1 − g)x(2 − g) · · · x(|v| − g) = (gx)(1)(gx)(2) · · · (gx)(|v|) = x2 (1)x2 (2) · · · x2 (|v|) = v. Now, if g < 0, we have −g ≥ 𝓁+1 ≥ |u|+1. Then, setting w := x(|u|+1)x(|u|+2) · · · x(−g) we have uwv ∈ L(X), and we are done. If g > 0, we have −g ≤ −𝓁 − 1 ≤ −|v| − 1. Then setting z := x(|v| − g + 1)x(|v| − g + 2) · · · x(0) we have vzu ∈ L(X). Repeating the previous argument with u and v both replaced by vzu and Ω (resp. F ) replaced by Ω ' := {1, 2, . . . , 𝓁' } ⊂ Z (resp. F ' := {−𝓁' , −𝓁' + 1, . . . , 𝓁' } ⊂ Z), where 𝓁' := |vzu|, we can find g ' ∈ Z \ F ' and x ' ∈ X such that x ' (1)x ' (2) · · · x ' (𝓁' ) = vzu and x ' (1−g ' )x ' (2−g ' ) · · · x ' (𝓁' −g ' ) = (g ' x ' )(1)(g ' x ' )(2) · · · (g ' x ' )(𝓁' ) = vzu. If g ' < 0, we have −g ' ≥ 𝓁' +1 = |vzu|+1. Then, setting w := x(𝓁' +1)x(𝓁' +2) · · · x(−g ' ) we have (vzu)w(vwu) ∈ L(X), so that, keeping in mind Exercise 1.74(e), uwv ∈ L(X), and we are done. If g ' > 0, we have −g ' ≤ −𝓁' − 1 = −|vzu| − 1. Then setting w := x(𝓁' − g + 1)x(𝓁' − g + 2) · · · x(0) we have (vzu)w(vzu) ∈ L(X) so that, again, uwv ∈ L(X).

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Conversely, suppose that for all u, v ∈ L(X) there exists w ∈ A∗ such that uwv ∈ L(X). Let Ω ⊂ Z and F ⊂ Z be two finite subsets and let x1 , x2 ∈ X. In order to show that X is irreducible, it is enough to show, by virtue of Exercise 1.56(b), that there exist g ∈ G \ F and a configuration x ∈ X such that x|Ω = x1 |Ω and (gx)|Ω = x2 |Ω . Let us set F ' := F ∪ Ω and let m, n ∈ Z such that F ' ⊂ [m, n]. Set u := x1 (m)x1 (m+1) · · · x1 (n) and v := x2 (m)x2 (m+1) · · · x1 (n). Note that |u| = |v| = n−m+1. Moreover, u, v ∈ L(X) and, by our assumptions, we can find w ∈ A∗ such that uwv ∈ L(X). This means that there exists a configuration x ' ∈ X such that x ' (1)x ' (2) · · · x ' (𝓁) = uwv, where 𝓁 := |uwv| = |u| + |w| + |v|. Setting g ' := m − 1 ∈ Z and x := g ' x ' ∈ X, we have x(m)x(m + 1) · · · x(m + 𝓁 − 1) = (g ' x ' )(m)(g ' x ' )(m+1) · · · (g ' x ' )(m+𝓁−1) = x ' (1)x ' (2) · · · x ' (𝓁) = uwv. In particular, x(m)x(m + 1) · · · x(n) = u and, keeping in mind that Ω ⊂ F ' ⊂ [m, n], we also have x|Ω = x1 |Ω . Setting g := |v| − 𝓁 ∈ Z we have n + g = n + |v| − 𝓁 = n+|v|−(|u|+|w|+|v|) = n−|u|−|w| = m−1−|w| < m so that g ∈ Z\F ' ⊂ Z\F . Moreover, (gx)(m)(gx)(m +1) · · · (gx)(n) = x(m −g)x(m +1−g) · · · x(n−g) = x(m + 𝓁 − |v|) · · · x(m + 𝓁 − 2)x(m + 𝓁 − 1) = v. In particular, keeping in mind that Ω ⊂ [m, n], we also have (gx)|Ω = x2 |Ω . This shows that X satisfies condition (I) in Exercise 1.56(b). We deduce that X is irreducible. (c) Suppose that X is topologically mixing and let u, v ∈ L(X). Then there exist two configurations x, y ∈ X such that u = x(1)x(2) · · · x(𝓁(u)) and v = y(1)y(2) · · · y(𝓁(v)). Consider the two non-empty open subsets U and V of X defined by (1.27). Since X is topologically mixing, there exists a finite subset F ⊂ Z such that U ∩ gV /= ∅ for all g ∈ Z \ F . Let M := max F and choose an integer n0 ≥ 0 such that n0 ≥ M + 1 − 𝓁(u). If n is an integer such that n ≥ n0 , then n + 𝓁(u) ∈ / F and hence U ∩ (𝓁(u) + n)V /= ∅. It follows that there exists a configuration z ∈ X satisfying z(1)z(2) · · · z(𝓁(u)) = u and z(𝓁(u)+n+1)z(𝓁(u)+n+2) · · · z(𝓁(u)+n+𝓁(v)) = v.

.

Consider the word w := z(𝓁(u) + 1)z(𝓁(u) + 2) · · · z(𝓁(u) + n). We have 𝓁(w) = n. Moreover, the word uwv is in L(X) since uwv appears in z ∈ X. Therefore the integer n0 has the required properties. Conversely, suppose that for all words u and v in L(X), there exists an integer n0 (u, v) ≥ 0 such that for every integer n ≥ n0 (u, v) there exists a word w ∈ A∗ of length 𝓁(w) = n satisfying uwv ∈ L(X). Let U and V be two non-empty open subsets of X. As in the second part of (a), chose configurations x ∈ U , y ∈ V , and consider the words u and v in L(X) defined by (1.28). Consider the (possibly empty) interval F := [mU − nV − n0 (v, u), nU − mV + n0 (u, v)] ⊂ Z. Suppose that g ∈ Z \ F . We distinguish two cases. If g > nU − mV + n0 (u, v), then mV + g − nU − 1 ≥ n0 (u, v) so that there exists w ∈ A∗ with 𝓁(w) = mV + g − nU − 1 such that uwv ∈ L(X). Consequently, there is a configuration z ∈ X such that z(mU ) · · · z(nU ) · z(nU + 1) · · · z(mV + g − 1) · z(mV + g) · · · z(nV + g) = uwv.

.

This clearly implies z ∈ U ∩ gV .

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If g < mU − nV − n0 (v, u), then mU − nV − g − 1 ≥ n0 (v, u) so that there exists w ∈ A∗ with 𝓁(w) = mU − nV − g − 1 such that vwu ∈ L(X). Consequently, there is a configuration z ∈ X such that z(mV + g) · · · z(nV + g) · z(nV + g + 1) · · · z(mU − 1) · z(mU ) · · · z(nU ) = vwu.

.

This clearly implies z ∈ U ∩ gV . We deduce that U ∩ gV /= ∅ for all g ∈ Z \ F . As the interval F is finite, this shows that X is topologically mixing. (d) Suppose first that X is strongly irreducible and let Δ be a finite subset of Z such that X is Δ-irreducible. Choose an integer n0 ≥ 0 such that Δ ⊂ [−n0 , ∞). Let u, v ∈ L(X) and let n ≥ n0 be an integer. Then the sets Ω1 := {1, 2, . . . , 𝓁(u)} and Ω2 := {𝓁(u)+n+1, 𝓁(u)+n+2, . . . , 𝓁(u)+n+𝓁(v)} satisfy Ω1 ∩(Ω2 +Δ) = ∅. On the other hand, since u, v ∈ L(X), we can find configurations x1 , x2 ∈ X such that x1 (1)x1 (2) · · · x1 (𝓁(u)) = u and

.

x2 (𝓁(u) + n + 1)x2 (𝓁(u) + n + 2) · · · x2 (𝓁(u) + n + 𝓁(v)) = v. By Δ-irreducibility of X, there exists x ∈ X such that x|Ω1 = x1 |Ω1 and x|Ω2 = x2 |Ω2 . Then the word w := x(𝓁(u) + 1)x(𝓁(u) + 2) · · · x(𝓁(u) + n) has length n and satisfies uwv = x(1)x(2) · · · x(𝓁(u) + n + 𝓁(v)) ∈ L(X).

.

This shows that the condition is necessary. Conversely, suppose that there exists an integer n0 ≥ 0 such that, for all words u, v ∈ L(X), and for any integer n ≥ n0 , there exists a word w ∈ A∗ of length 𝓁(w) = n such that uwv ∈ L(X). Let us show that X is Δ-irreducible for Δ := {−n0 , −n0 + 1, . . . , n0 }. Let Ω1 and Ω2 be two finite subsets of Z such that Ω1 ∩ (Ω2 +Δ) = ∅, and let x1 , x2 ∈ X. We want to show that there exists a configuration x ∈ X such that x|Ω1 = x1 |Ω1 and x|Ω2 = x2 |Ω2 .

.

(1.29)

First observe that Ω1 ∩ Ω2 = ∅, since 0 ∈ Δ. Note also that (Ω1 + Δ) ∩ Ω2 = ∅ since Δ = −Δ. Thus, after possibly exchanging Ω1 and Ω2 , we can assume that min Ω1 < min Ω2 . On the other hand, after enlarging Ω2 if necessary, we can also assume that max Ω1 < max Ω2 . Now observe that, by an immediate induction on k, the integer n0 satisfies the following: (∗) for every sequence of words u1 , u2 , . . . , uk ∈ L(X), k ≥ 2, and any sequence of integers n1 , n2 , . . . , nk−1 ≥ n0 , there exist words w1 , w2 , . . . , wk−1 ∈ A∗ , with wi of length ni for 1 ≤ i ≤ k − 1, satisfying u1 w1 u2 w2 · · · uk−1 wk−1 uk ∈ L(X).

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Let us introduce the following equivalence relation ∼1 on Ω1 (resp. ∼2 on Ω2 ). Given ω1 , ω1' ∈ Ω1 (resp. ω2 , ω2' ∈ Ω2 ) we write ω1 ∼1 ω1' (resp. ω2 ∼2 ω2' ) if and only if there is no element of Ω2 (resp. of Ω1 ) between ω1 and ω1' (resp. between ω2 and ω2' ). Note that the conditions min Ω1 < min Ω2 and max Ω1 < max Ω2 imply that ∼1 and ∼2 have the same number of equivalence classes, say s. Let Ω1 =

s

.

Ω1,i and Ω2 =

i=1

s

Ω2,i

i=1

be the corresponding partitions of Ω1 and Ω2 into equivalence classes. Let us set mi := min Ω1,i , m'i := max Ω1,i , pi := min Ω2,i and pi' := max Ω2,i for i = 1, 2, . . . , s. We then have, after renumbering the equivalence classes if necessary, m1 ≤ m'1 < p1 ≤ p1' < m2 ≤ m'2 < p2 ≤ p2' < · · · < ms ≤ m's < ps ≤ ps' .

.

It follows from the mutual properties of Ω1 and Ω2 that we have ni := pi −m'i > n0 ' for all i = 1, 2, . . . , s, and n'i := mi − pi−1 > n0 for all i = 2, 3, . . . , s. Consider the words ui , vi ∈ L(X) defined by ui = x1 (mi )x1 (mi + 1) · · · x1 (m'i )

.

and

vi = x2 (pi )x2 (pi + 1) · · · x2 (pi' )

for i = 1, 2, . . . , s. By applying (∗) to the sequence of words u1 , v1 , u2 , v2 , . . . , us , vs ∈ L(X) and to the sequence of integers n1 , n'1 , n2 , n'2 , . . . , ns−1 , n's−1 , ns we deduce that we can find words w1 , z1 , w2 , z2 , . . . , ws−1 , zs−1 , ws ∈ A∗ with wi of length ni , for i = 1, 2, . . . , s, and zi of length n'i , for i = 1, 2, . . . , s − 1, such that the word w := u1 w1 v1 z1 u2 w2 v2 z2 · · · us−1 ws−1 vs−1 zs−1 us ws vs

.

belongs to L(X). This implies that there exists x ∈ X such that x(m1 )x(m1 + 1) · · · x(ps' ) = w.

.

Then x satisfies (1.29). This shows that X is strongly irreducible.

█

Exercise 1.77 Let A be a set. Given a set F ⊂ A∗ of words on the alphabet A, define the subset XF ⊂ AZ as being the set consisting of all configurations x ∈ AZ such that there is no word in F appearing in x. (a) Let F ⊂ A∗ . Show that XF is a subshift of AZ . (b) Let X ⊂ AZ be a subshift and let L(X) ⊂ A∗ denote the language of X. One says that a set F ⊂ A∗ is a defining set of forbidden words for X if X = XF . Show that the set F (X) := A∗ \ L(X) is a defining set of forbidden words for X and that every set of forbidden words for X is contained in F (X).

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Solution (a) Given a word w ∈ A∗ with length n, denote by pw the (Z, A)-pattern with support {0, 1, . . . , n − 1} such that p(0)p(1) · · · p(n − 1) = w. To see that XF is a subshift of AZ , it suffices to observe that XF is the subshift of AZ admitting the set {pw : w ∈ L} as a defining set of forbidden patterns (cf. Exercise 1.47). (b) If x ∈ X, then all words appearing in x are contained in L(X). Therefore X ⊂ XF (X) . To prove the converse inclusion, suppose now that x ∈ XF (X) . Then every word appearing in x is in A∗ \F (X) = L(X). Thus x ∈ X by Exercise 1.74(b). Therefore XF (X) ⊂ X. This shows that X = XF (X) . Finally, suppose that F ⊂ A∗ is a defining set of forbidden words for X. If w ∈ F , then there is no x ∈ X such that w appears in x. Therefore w ∈ A∗ \L(X) = F (X). This shows that F ⊂ F (X). █ Exercise 1.78 Let A be a finite set. (a) Show that a subshift X ⊂ AZ is of finite type if and only if it admits a finite defining set of forbidden words. (b) Let n ≥ 0 be an integer and let W ⊂ A∗ be a set of words of length n. Show that the subset X(W ) ⊂ AZ , defined by X(W ) := {x ∈ AZ : x(i)x(i + 1) · · · x(i + n − 1) ∈ W for all i ∈ Z},

.

(1.30)

is a subshift of finite type. (c) Let X ⊂ AZ be a subshift of finite type. Show that there exists an integer n ≥ 0 and a set W ⊂ A∗ of words of length n such that X = X(W ). One then says that the subshift of finite type X has memory length n and that W is a defining set of admissible words for X. Solution (a) Suppose first that X ⊂ AZ is a subshift of finite type. This means that there exists a finite set P of (Z, A)-patterns that is a defining set of forbidden patterns for X. By Exercise 1.48(d), we can assume that all patterns in P have the same support {0, 1, . . . , n−1} for some integer n ≥ 0. Then F := {p(0)p(1) · · · p(n−1) : p ∈ P } is a finite defining set of forbidden words for X. Conversely, suppose that F ⊂ A∗ is a finite defining set of forbidden words for X. For w ∈ F , denote by pw the (Z, A)-pattern with support {0, 1, . . . , 𝓁(w) − 1} such that p(0)p(1) · · · p(𝓁(w) − 1) = w. Clearly {pw : w ∈ F } is a finite defining set of forbidden patterns for X. This shows that X is of finite type. (b) It suffices to observe that, denoting by F the set of words of length n that do not belong to W , the set X is the subshift admitting F as a defining set of forbidden words. As the set F is finite, X is of finite type. (c) By the first part in the proof of (a), there exist an integer n ≥ 0 and a subset F ⊂ A∗ consisting of words of length n that is a defining set of forbidden words for X. Denoting by W ⊂ A∗ the set consisting of all words of length n that are not in F , we clearly have X = X(W ). █

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Exercise 1.79 Let A be a finite set and let X ⊂ AZ be a subshift. Show that X is of finite type if and only if the following holds: there exists an integer n0 ≥ 0 such that if the words u, v, w ∈ A∗ satisfy uw, wv ∈ L(X) and w has length 𝓁(w) ≥ n0 , then one has uwv ∈ L(X). Solution Suppose first that X is of finite type and choose an integer n0 ≥ 0 that is a memory length for X. Then there is a set W ⊂ A∗ of words of length n0 such that a configuration x ∈ AZ is in X if and only if every word of length n0 appearing in x belongs to W . Let u, v, w ∈ A∗ such that uw, wv ∈ L(X) and n := 𝓁(w) ≥ n0 . Let m := 𝓁(u) and p := 𝓁(v). Since uw ∈ L(X), we can find a configuration x ∈ X such that x(−m) · · · x(−1) · x(0) · · · x(n − 1) = uw.

.

Similarly, since wv ∈ L(X), we can find a configuration y ∈ X such that y(0) · · · y(n − 1) · y(n) · · · y(n + p − 1) = wv.

.

Consider then the unique configuration z ∈ A∗ such that z(k) = x(k) for all k ≤ n − 1 and z(k) = y(k) for all k ≥ 0. As n ≥ n0 , every word of length n0 appearing in z must appears either in x or in y and hence belongs to W . It follows that z ∈ X. As z(−m) · · · z(n + p − 1) = uwv, we deduce that uvw ∈ L(X). Conversely, suppose that there exists an integer n0 ≥ 0 such that if the words u, v, w ∈ A∗ satisfy uw, wv ∈ L(X) and 𝓁(w) ≥ n0 , then uwv ∈ L(X). By an immediate induction on the integer k ≥ 3, this implies the following: if u1 , u2 , . . . , uk ∈ A∗ satisfy ui ui+1 ∈ L(X) for all 1 ≤ i ≤ k .

and 𝓁(ui ) ≥ n0 for all 2 ≤ i ≤ k − 1, then u1 u2 · · · uk ∈ L(X).

(1.31)

Let W ⊂ A∗ denote the set consisting of all words of length 2n0 + 1 appearing in X and consider the subshift of finite type X(W ) ⊂ AZ defined by X(W ) := {x ∈ AZ : x(i)x(i + 1) · · · x(i + 2n0 ) ∈ W for all i ∈ Z},

.

(cf. Exercise 1.78). Let us shows that X = X(W ). We clearly have X ⊂ X(W ). To prove the converse inclusion, it is enough to show that L(X(W )) ⊂ L(X) (cf. Exercise 1.74). So suppose that w ∈ L(X(W )). Write w as a product of words w = u1 u2 · · · uk , where k ≥ 0, ui ∈ A∗ for all 1 ≤ i ≤ k, 𝓁(ui ui+1 ) ≤ 2n0 + 1 for all 1 ≤ i ≤ k − 1, and 𝓁(ui ) ≥ n0 for all 2 ≤ i ≤ k − 1. Then ui ui+1 ∈ L(X) for all 1 ≤ i ≤ k − 1 so that w ∈ L(X) by (1.31). This shows that L(X(W )) ⊂ L(X). We deduce that the subshift X = X(W ) is of finite type. █ Exercise 1.80 Let A be a finite set and let X ⊂ AZ be a subshift of finite type. Show that X is topologically mixing if and only if X is strongly irreducible.

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Solution If X is strongly irreducible then X is topologically mixing by Exercise 1.64(b) (the hypothesis that X is of finite type is not needed here). Conversely, suppose that X is topologically mixing. Since X is of finite type, there exists an integer m ≥ 0 such that if the words u, v, w ∈ A∗ satisfy uw, wv ∈ L(X) and 𝓁(w) ≥ m, then uwv ∈ L(X) (cf. Exercise 1.79). On the other hand, since X is topologically mixing, given u, v ∈ L(X), we can find an integer n0 (u, v) ≥ 0 such that for every integer n ≥ n0 (u, v) there exists a word w ∈ A∗ of length 𝓁(w) = n satisfying uwv ∈ L(X) (cf. Exercise1.76(b)). Let us set n0 := max{n0 (u, v) : u, v ∈ L(X) and 𝓁(u), 𝓁(v) ≤ m}.

.

Let u and v in L(X) and let n ≥ n0 . Let us show that there exists a word w ∈ A∗ of length 𝓁(w) = n such that uwv ∈ L(X). If 𝓁(u) ≥ m + 1, we write u = u1 u2 , where u1 , u2 ∈ A∗ and 𝓁(u2 ) = m. Analogously, if 𝓁(v) ≥ m+1, we write v = v1 v2 , where v1 , v2 ∈ A∗ and 𝓁(v1 ) = m. We then distinguish four cases: (1) If 𝓁(u), 𝓁(v) ≤ m, then n ≥ n0 (u, v) and the existence of w is ensured. (2) If 𝓁(u) ≤ m and 𝓁(v) ≥ m + 1, then n ≥ n0 (u, v1 ) and we can find w such that 𝓁(w) = n and uwv1 ∈ L(X). Since the words uwv1 and v = v1 v2 satisfy that 𝓁(v1 ) = m, we deduce that uwv = uwv1 v2 ∈ L(X). (3) If 𝓁(u) ≥ m + 1 and 𝓁(v) ≤ m, then n ≥ n0 (u2 , v) and we can find w such that 𝓁(w) = n and u2 wv ∈ L(X). Since the words u1 u2 and u2 wv satisfy that 𝓁(u2 ) = m, we deduce that uwv = u1 u2 wv ∈ L(X). (4) If 𝓁(u) ≥ m + 1 and 𝓁(v) ≥ m + 1, then n ≥ n0 (u2 , v1 ) and we can find w such that 𝓁(w) = n and u2 wv1 ∈ L(X). Since the words u1 u2 and u2 wv1 satisfy that 𝓁(u2 ) = m, we deduce that uwv1 = u1 u2 wv1 ∈ L(X). Since the words uwv1 and v1 v2 satisfy that 𝓁(v1 ) = m, we deduce that uwv = uwv1 v2 ∈ L(X). In either case, we find w ∈ A∗ with 𝓁(w) = n and uwv ∈ L(X). By Exercise 1.76(c), this shows that X is strongly irreducible. █ Exercise 1.81 Let A := {0, 1}. Consider the subshift of finite type X ⊂ AZ admitting {00, 11} as a defining set of forbidden words. Show that the subshift X is irreducible and topologically transitive but neither topologically mixing nor strongly irreducible. Solution The subshift X consists of the two 2-periodic configurations x0 and x1 defined by xi (k) := k + i mod 2 for all k ∈ Z and i ∈ {0, 1}. Observe that given two words u, v ∈ L(X), we have that either uv, or u0v, or u1v is in L(X). Thus X is irreducible. By Exercise 1.56(d), the subshift X is also topologically transitive. However, X is not topologically mixing. Indeed, 0 ∈ L(X) and any word w ∈ A∗ such that 0w0 ∈ L(X) must be of the form w = 1(01)k for some integer k ≥ 0 and hence has necessarily odd length. This implies that if n0 ≥ 0 is an integer and we choose an even integer n ≥ n0 , there is no w ∈ A∗ of length n such that 0w0 ∈ L(X). Thus X is not topologically mixing by Exercise 1.76(c). It is not strongly irreducible by virtue of Exercise 1.64(b) █

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Exercise 1.82 Let A be a set and let X ⊂ AZ be an irreducible subshift. Suppose that Y, Z ⊂ AZ are subshifts such that Y ∪ Z = X. Show that one has Y = X or Z = X. Solution Suppose by contradiction that Y /= X and Z /= X. Then L(Y ) ⊊ L(X) and L(Z) ⊊ L(X) by Exercise 1.74(a). Therefore, there exist words u, v ∈ L(X) such that u /∈ L(Y ) and v /∈ L(Z). As X is irreducible, there exists a word w ∈ A∗ such that uwv ∈ L(X) (cf. Exercise 1.76(b)). By Exercise 1.75(a), we have L(X) = L(Y ∪ Z) = L(Y ) ∪ L(Z). Therefore, uwv ∈ L(Y ) or uvw ∈ L(Z). If uwv ∈ L(Y ) (resp. uwv ∈ L(Z)), then, since L(Y ) (resp. L(Z)) is closed under taking subwords by Exercise 1.74(e), we have u ∈ L(Y ) (resp. v ∈ L(Z)), a contradiction. █ Comment This is Exercise 2.1.7(a) in [LinM]. Exercise 1.83 Let A := {0, 1} and consider the subset X ⊂ AZ consisting of all x ∈ AZ such that (x(n), x(n + 1)) /= (0, 1) for all n ∈ Z. (a) Show that X is a subshift of finite type. (b) Show that X is topologically transitive. (c) Show that X is not irreducible. (d) Show that X is neither topologically mixing nor strongly irreducible. Solution (a) A configuration x ∈ AZ is in X if and only if the word 01 does not appear in x. Therefore X is the subshift of finite type of AZ admitting {01} as a defining set of forbidden words (cf. Exercise 1.48). (b) The configurations in X are the two constant configurations x0 and x1 and the configurations yn , n ∈ Z, respectively defined by xa (k) := a for all a ∈ A and k ∈ Z, and yn (k) := 1 for all k ≤ n and yn (k) := 0 for all k ≥ n + 1. Observe that yn → x1 as n → ∞ and that yn → x0 as n → −∞. As the configurations yn , n ∈ Z, are in the same orbit under the shift, we deduce that the orbit of y0 is dense in X. This implies that Y is topologically transitive (cf. Exercise 1.63(a)). (c) Let L(X) ⊂ A∗ denote the language of X. We have 0, 1 ∈ L(X) since both 0 and 1 appears in y0 . However, it is clear from the description of the configurations in X given in (b) that there is no word u ∈ A∗ such that 0u1 ∈ L(X). Therefore X is not irreducible (cf. Exercise 1.76(b)). As X is not irreducible, it is neither topologically mixing nor strongly irreducible by Exercise 1.56(e). █ Exercise 1.84 (The Golden Mean Subshift) Let A := {0, 1} and consider the subset X ⊂ AZ consisting of all configurations x ∈ AZ such that (x(n), x(n+1)) /= (1, 1) for all n ∈ Z. (a) Show that X is a subshift of finite type. (b) Show that X is strongly irreducible. (c) Show that X is topologically mixing, topologically transitive, and irreducible. The subshift X is called the golden mean subshift.

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Solution (a) A configuration x ∈ AZ is in X if and only if the word 11 does not appear in x. Therefore X is the subshift of finite type of AZ admitting {11} as a defining set of forbidden words (cf. Exercise 1.48). (b) Let L(X) ⊂ A∗ denote the language of X. Clearly a word w ∈ A∗ is in L(X) if and only if 11 is not a subword of w. Thus, if u, v ∈ L(X) and n ≥ n0 := 1, then u0n v ∈ L(X). This shows that X is strongly irreducible by Exercise 1.76(d). (c) As the subshift X is strongly irreducible, it is also topologically mixing by Exercise 1.64(b). Since X is topologically mixing, it is topologically transitive and irreducible by Exercise 1.56(e). █ Exercise 1.85 (The Even Subshift) Let A := {0, 1} and consider the set X ⊂ AZ consisting of all configurations x ∈ AZ such that the following holds: if x(m) = x(n) = 1 for some integers m, n ∈ Z with m < n, then the number of k ∈ Z such that x(k) = 0 and m < k < n is even. (a) Show that X is a subshift of AZ . (b) Show that X is not of finite type. (c) Show that X is strongly irreducible. (d) Show that X is topologically mixing, topologically transitive, and irreducible. The subshift X is called the even subshift. Solution (a) A configuration x ∈ AZ is in X if and only if there is no n ∈ N such that the word 102n+1 1 appears in X. Therefore X is the subshift of AZ admitting {102n+1 1 : n ∈ N} as a defining set of forbidden words. (b) Suppose by contradiction that X is of finite type. Then there is an integer n ∈ N and a set L of words of length n such that L is a defining set of forbidden words for X. As the word w := 102n+1 1 is a forbidden word for X although no subword of w of length n is in L, this yields a contradiction. (c) Let L(X) denote the language of X and let u, v ∈ L(X). If n ≥ n0 := 3, then one of the words u1n v, u1n−1 0v, u01n−1 v, u01n−2 0v belongs to L(X). It follows that X is strongly irreducible by Exercise 1.76(d). (d) As the subshift X is strongly irreducible, it is also topologically mixing by Exercise 1.64(b). Since X is topologically mixing, it is topologically transitive and irreducible by Exercise 1.56(e). █ Exercise 1.86 Let A := {0, 1} and let X ⊂ AZ be the subshift with defining set of forbidden words {01k 0h 1 : h, k ∈ N and 1 ≤ h ≤ k}.

.

Show that X is topologically mixing, topologically transitive, and irreducible but not strongly irreducible. Solution Let L(X) denote the language of X and let u, v ∈ L(X). Let us show that there exists n0 ∈ N such that for all n ≥ n0 there exists a word w ∈ A∗ of length 𝓁(w) = n such that uwv ∈ L(X). We may express u in the form u = u' 1k 0h ,

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with u' ∈ A∗ not ending with 1 and k, h ≥ 0. Let n0 := k + 1. Then, setting ' w := 0n for all n ≥ n0 we have uwv = u' 1k 0h v, where h' := h + n > k, so that uwv ∈ L(X). It follows from Exercise 1.76(c) that X is topologically mixing. Since every topologically mixing subshift of AZ is topologically transitive and irreducible (cf. Exercise 1.56(e)), we deduce that X is also topologically transitive and irreducible. For k ≥ 1 let uk := 01k ∈ L(X). Let also v := 01 ∈ L(X). Given k ≥ 1, a ' ' word w ∈ A∗ such that uk wv ∈ L(X) is necessarily of the form w = 1k 0h w ' with k ' ≥ 0, h' ≥ k + k ' ≥ k, and w ' ∈ A∗ not starting with 0. In particular, 𝓁(w) ≥ h' ≥ k. As k was arbitrary, this shows that there is no integer n0 ≥ 0 such that, for each k ≥ n0 + 1, there exists a word w ∈ A∗ of length 𝓁(w) = n0 such that uk wv ∈ L(X). It follows from Exercise 1.76(d) that X is not strongly irreducible. █ Exercise 1.87 Let G be a group and let A be a set. Let H be a subgroup of G. Consider the subset X ⊂ AG consisting of all configurations x ∈ AG that are constant on each left coset of H in G, that is, such that x(gh1 ) = x(gh2 ) for all g ∈ G and h1 , h2 ∈ H . Show that X is a subshift of AG . Solution Consider the set F ⊂ P(G, A) consisting of all non-constant (G, A)patterns whose support is a two-elements subset of H . We then have X = {x ∈ AG : (gx)|supp(p) /= p for all g ∈ G and p ∈ F }.

.

Therefore, X is a subshift of AG by Exercise 1.47(a).

█

Exercise 1.88 Let G be a group and let A be a set. The reverse of a configuration x ∈ AG is the configuration x R ∈ AG defined by x R (g) := x(g −1 ) for all g ∈ G. The reverse of a subset X ⊂ AG is the subset XR := {x R : x ∈ X} ⊂ AG . The reverse of a word w = a1 a2 · · · an ∈ A∗ is the word w R := an an−1 · · · a1 ∈ A∗ . The reverse of a language L ⊂ A∗ is the language LR := {w R : w ∈ L} ⊂ A∗ . (a) Show that a subset X ⊂ AG is closed (in the prodiscrete topology) if and only if XR is closed. (b) Give an example of a subshift X ⊂ AG such that XR fails to be a subshift. (c) Suppose that G is abelian. Let x ∈ AG and let g ∈ G. Show that g(x R ) = (g −1 x)R . (d) Suppose that G is abelian and let X ⊂ AG . Show that X is G-invariant if and only if XR is G-invariant. (e) Suppose that G is abelian and let X ⊂ AG . Show that X is a subshift if and only if XR is a subshift. (f) Suppose that G is abelian. Show that a subshift X ⊂ AG is topologically transitive (resp. irreducible, resp. topologically mixing, resp. strongly irreducible) if and only if the subshift XR is topologically transitive (resp. irreducible, resp. topologically mixing, resp. strongly irreducible). (g) Suppose that G is abelian and that the set A is finite. Show that a subshift X ⊂ AG is of finite type if and only if the subshift XR is of finite type.

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(h) Suppose that G = Z and let X ⊂ AZ be a subshift. Let L(X) ⊂ A∗ be the language of X. Show that L(XR ) = L(X)R and give another proof of the fact that X is irreducible if and only if XR is irreducible. Solution (a) Suppose that X ⊂ AG is closed. Let x ∈ AG be in the closure of XR . This means that, given any finite subset Ω ⊂ G, there exists y ∈ XR such that x|Ω = y|Ω . This implies x R (g −1 ) = x(g) = y(g) = y R (g −1 ) for all g ∈ Ω, that is, x R |Ω −1 = y R |Ω −1 . As y R ∈ X, the subset X is closed, and Ω was arbitrary, this shows that x R ∈ X and hence x ∈ XR . We deduce that XR is closed in AG . The converse implication follows after exchanging the roles of XR and X = (XR )R . (b) Let G be a group and let H be a non-normal subgroup ( )of G (e.g., G := Sym3 and H is the subgroup generated by the transposition 1 2 ). Suppose that A has more than one element (e.g., A := {0, 1}) and set X := {x ∈ AG : x(gh1 ) = x(gh2 ) for all g ∈ G and h1 , h2 ∈ H }.

.

Thus, X is the set consisting of the configurations that are constant on each left coset of H in G. This is a subshift of AG by virtue of Exercise 1.87. On the other hand, observe that XR is the set consisting of the configurations that are constant on each right coset of H in G. Let us show that XR is not G-invariant. As H is not normal in G, there exist g0 ∈ G and h0 ∈ H such that g0−1 h0 g0 ∈ / H . Take a, b ∈ A distinct and define y ∈ AG by setting y(g) := a for all g ∈ H and y(g) := b for all g ∈ G \ H . Clearly, y ∈ X ∩ XR . However, g0 y ∈ / XR . Indeed, −1 (g0 y)(g0 ) = y(1G ) = a while (g0 y)(h0 g0 ) = y(g0 h0 g0 ) = b /= a, so that g0 y is not constant on the right coset H g0 . This shows that XR is not G-invariant and hence that XR is not a subshift. (c) For all h ∈ G, we have (g(x R ))(h) = x R (g −1 h) = x(h−1 g) = x(gh−1 ) = −1 (g x)(h−1 ) = (g −1 x)R (h). This shows that g(x R ) = (g −1 x)R . (d) Suppose that X is G-invariant and let x ∈ X. As X is G-invariant, we have g −1 x ∈ X. It follows from (c) that g(x R ) = (g −1 x)R ∈ XR . This shows that XR is G-invariant. The converse implication follows after exchanging the roles of XR and X = (XR )R . (e) This follows immediately from (a) and (d). (f) Recall that a subshift X ⊂ AG is topologically transitive if and only if it satisfies condition (TT) in Exercise 1.56(a), namely, for any finite subset Ω of G and any two configurations x1 , x2 ∈ X, there exist a configuration x = x(Ω; x1 , x2 ) ∈ X and an element g = g(Ω; x1 , x2 ) ∈ G such that (∗)

x|Ω = x1 |Ω and (gx)|Ω = x2 |Ω .

Using (d) we have that (∗) holds if an only if x R |Ω −1 = x1R |Ω −1 and (g −1 x R )|Ω −1 = x2R |Ω −1 . In other words, x(Ω −1 ; x1R , x2R ) = x(Ω; x1 , x2 )R and g(Ω −1 ; x1R , x2R ) = g(Ω; x1 , x2 )−1 . We deduce that X is topologically transitive if and only if XR is. Recall that a subshift X ⊂ AG is irreducible if and only if it satisfies condition (I) in Exercise 1.56(b), namely, for every finite subset F ⊂ G, for all configurations

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x1 , x2 ∈ X, and for every finite subset Ω of G, there exist an element g = g(Ω, F ; x1 , x2 ) ∈ G \ F and a configuration x = x(Ω, F ; x1 , x2 ) ∈ X satisfying condition (∗) above. Using (d) and arguing as for topological transitivity, we have x(Ω −1 , F −1 ; x1R , x2R ) = x(Ω, F −1 ; x1 , x2 )R and g(Ω −1 , F −1 ; x1R , x2R ) = g(Ω, F ; x1 , x2 )−1 . We deduce that X is irreducible if and only if XR is. Recall that a subshift X ⊂ AG is topologically mixing if and only if it satisfies condition (TM) in Exercise 1.56(c), namely, for any finite subset Ω of G and any two configurations x1 , x2 ∈ X, there exists a finite subset F = F (Ω; x1 , x2 ) ⊂ G such that, for all g ∈ G \ F , there exists a configuration x = x(Ω; g; x1 , x2 ) ∈ X such that condition (∗) above is satisfied. Using (d) and the same arguments as for topological transitivity, we have F (Ω −1 ; x1R , x2R ) = F (Ω; x1 , x2 )−1 and x(Ω −1 ; g −1 ; x1R , x2R ) = x(Ω; g; x1 , x2 )R . We deduce that X is topologically mixing if and only if XR is. Let Δ ⊂ G be a finite subset. Recall (cf. Exercise 1.64) that a subshift X ⊂ AG is Δ-irreducible if it satisfies the following condition: if Ω1 and Ω2 are two finite subsets of G such that Ω1 ∩ Ω2 Δ = ∅, then, given any two configurations x1 , x2 ∈ X, there exists a configuration x = x(Ω1 , Ω2 ; x1 , x2 ) ∈ X which satisfies x|Ω1 = x1 |Ω1 and x|Ω2 = x2 |Ω2 .

(∗∗)

Keeping in mind that G is abelian, we have Ω1 ∩ Ω2 Δ = ∅ if and only if Ω1−1 ∩ Ω2−1 Δ−1 = ∅. Moreover, using (d) we have that if Ω1 and Ω2 are finite and Ω1 ∩ Ω2 Δ = ∅, then x(Ω1−1 , Ω2−1 ; x1R , x2R ) = x(Ω1 , Ω2 ; x1 , x2 )R . This shows that X is Δ-irreducible if and only if XR is Δ−1 -irreducible. Since a subshift is strongly irreducible if it is Δ-irreducible for some finite subset Δ ⊂ G, we deduce that X is strongly irreducible if and only if XR is strongly irreducible. (g) Suppose that X ⊂ AG is a subshift of finite type and let Ω ⊂ G and A ⊂ AΩ be a memory set and a defining set of admissible patterns for X, respectively. Thus, for a configuration x ∈ AG one has x ∈ X if and only if (gx)|Ω ∈ A for all g ∈ G. −1 Given a pattern p ∈ AΩ we define pR ∈ AΩ by setting pR (g −1 ) := p(g) for −1 all g ∈ Ω and set A R := {pR : p ∈ A } ⊂ AΩ . Keeping in mind that G is abelian, it is straightforward that for g ∈ G one has (gx)|Ω ∈ A if and only if (g −1 x R )|Ω −1 ∈ A R . This shows that XR is of finite type with Ω −1 and A R as a memory set and a defining set of admissible patterns, respectively. The converse implication follows after exchanging the roles of XR and X = (XR )R . (h) Let w ∈ L(X). Then we can find a configuration x ∈ AZ such that w = x(1)x(2) · · · x(n). Setting y := (n + 1)(x R ) we have y ∈ XR by (d) and, moreover, y(1)y(2) · · · y(n) = ((n + 1)x R )(1)((n + 1)x R )(2) · · · ((n + 1)x R )(n) .

= x R (−n)x R (−n + 1) · · · x R (−1) = x(n)x(n − 1) · · · x(1) = wR .

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This shows that w R ∈ L(XR ) and gives the inclusion L(X)R ⊂ L(XR ). Exchanging the roles of XR and X = (XR )R we deduce that L(XR )R ⊂ L(X), equivalently, L(XR ) ⊂ L(X)R . This shows that indeed L(X)R = L(XR ). Suppose now that X is irreducible. Given u, v ∈ L(XR ) it follows from the previous result that u' := uR and v ' := v R both belong to L(X). As X is irreducible, there exists w ' ∈ A∗ such that v ' w ' u' ∈ L(X). Setting w := (w ' )R we have uwv = (v ' w ' u' )R ∈ L(X)R = L(XR ), showing that XR is irreducible. The converse █ implication follows after exchanging the roles of XR and X = (XR )R . Exercise 1.89 Let A be a countable set. The forward-orbit (resp. backward-orbit) of a configuration x ∈ AZ is the subset {nx : n ∈ N} ⊂ AZ (resp. {(−n)x : n ∈ N} ⊂ AZ ). Let X ⊂ AZ be a non-empty subshift. Show that the following conditions are equivalent: (i) X is irreducible; (ii) there exists a configuration x ∈ X whose backward and forward orbits are both dense in X; (iii) there exists a configuration x ∈ X whose backward-orbit is dense in X; (iv) there exists a configuration x ∈ X whose forward-orbit is dense in X. Solution Let L = L(X) \ {ϵ} ⊂ A∗ denote the set consisting of all non-empty words that appear in X. Suppose that X is irreducible. Since A is countable, the set L is itself countable. Thus, there exists a sequence (wn )n∈N of non-empty words in A∗ such that L = {wn : n ∈ N}. Since w0 ∈ L, we can find a configuration x0 ∈ X such that, setting 𝓁0 := |w0 |, one has x0 (1)x0 (2) · · · x0 (𝓁0 ) = w0 . Since X is irreducible, there exist u0 , v0 ∈ A∗ such that z1 := w1 u0 w0 v0 w1 ∈ L. Thus we can find a configuration x1 ∈ X such that, setting m1 := |w1 | + |u0 | and n1 := |w0 | + |v0 | + |w1 |, one has x1 (−m1 + 1)x1 (−m1 + 2) · · · x1 (n1 ) = z1 . Note that x1 (1)x1 (2) · · · x1 (𝓁0 ) = w0 = x0 (1)x0 (2) · · · x0 (𝓁0 ). Since X is irreducible, there exist u1 , v1 ∈ A∗ such that z2 := w2 u1 z1 v1 w2 ∈ L. Thus we can find a configuration x2 ∈ X such that, setting m2 := |w2 | + |u1 | + |w1 | + |u0 | and n1 := |w0 | + |v0 | + |w1 | + |v1 | + |w2 |, one has x2 (−m2 + 1)x2 (−m2 + 2) · · · x2 (n2 ) = z2 . Note that x2 (−m1 + 1)x2 (−m1 + 2) · · · x2 (n1 ) = z1 = x1 (−m1 +1)x1 (−m1 +2) · · · x1 (n1 ). It is clear that, continuing this way, the sequence (xn )n∈N in X converges in the prodiscrete topology to a unique configuration x := limn→∞ xn ∈ X and that both the forward and backward orbits of x are dense in X. This shows (i) =⇒ (ii). The implication (ii) =⇒ (iii) is obvious. Suppose now that there exists x ∈ X whose backward-orbit is dense in X. Let u, v ∈ L. Since the backward-orbit of x is dense in X, there exists n ∈ N such that x(n + 1)x(n + 2) · · · x(n + |u|) = ((−n)x)(1)((−n)x)(2) · · · ((−n)x)(|u|) = u. Since v ∈ L, there exists y ∈ X such that y(1)y(2) · · · y(|v|) = v. Let 𝓁 ≥ n+|u|+1 and set v ' := y(−𝓁 + 1)y(−𝓁 + 2) · · · y(|v|). Since the backward-orbit of x is dense in X, there exists m ∈ N such that x(m + 1)x(m + 2) · · · x(m + 𝓁 + |v|) = ((−m)x)(1)((−m)x)(2) · · · ((−m)x)(|v ' |) = v ' . Note that x(m + 𝓁 + 1)x(m + 𝓁 + 2) · · · x(m+𝓁+|v|) = v. Then setting w := x(n+|u|+1)x(n+|u|+2) · · · x(m+𝓁),

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it is clear that uwv = x(n + 1)x(n + 2) · · · x(n + |u|) · x(n + |u| + 1)x(n + |u| + 2) · · · x(m + 𝓁) · x(m + 𝓁 + 1)x(m + 𝓁 + 2) · · · x(m + 𝓁 + |v|) ∈ L. We deduce that X is irreducible. This shows the implication (iii) =⇒ (i). With the notation and results from Exercise 1.88, the reverse set XR ⊂ AZ of the subshift X is itself a subshift. Moreover, XR is irreducible if and only if X is irreducible. Since the forward orbit of a configuration x ∈ AZ is equal to the reverse of the backward orbit of the reverse configuration x R (cf. Exercise 1.88(c)), the equivalence (i) ⇐⇒ (iv) follows after exchanging the roles of X and XR . This completes the proof of the equivalence of the four conditions. █ Exercise 1.90 (Sofic Subshifts) Let G be a group and let A be a finite set. One says that a subshift X ⊂ AG is a sofic subshift if there exist a finite set B, a subshift of finite type Y ⊂ B G , and a cellular automaton τ : B G → AG such that X = τ (Y ). (a) Show that every subshift of finite type X ⊂ AG is a sofic subshift. One says that a subshift X ⊂ AG is strictly sofic if it is sofic but not of finite type. (b) Let A := {0, 1} and consider the at-most-one-one subshift X ⊂ AZ , i.e., the subshift consisting of all configurations x ∈ AZ such that 1 appears at most once in x (cf. Exercise 1.71). Show that the subshift X is strictly sofic. Solution (a) If X ⊂ AG is a subshift of finite type, then X is sofic since the identity map τ := IdAG : AG → AG is a cellular automaton and X = τ (X). (b) Consider the subshift of finite type Y ⊂ AZ admiting the singleton {01} as a defining set of forbidden words (this is the subshift studied in Exercise 1.83). The configurations in Y are the two constant configurations ca , a ∈ A, and the configurations yn , n ∈ Z, respectively defined by ca (k) := a for all a ∈ A and k ∈ Z, and yn (k) := 1 for all k ≤ n and yn (k) := 0 for all k ≥ n + 1. Consider the cellular automaton τ : AZ → AZ with memory set S := {0, 1} ⊂ Z and local defining map μ : AS = A2 → A given by μ(1, 0) := 1 and μ(0, 0) = μ(0, 1) = μ(1, 1) := 0. We have τ (c0 ) = τ (c1 ) = c0 and τ (yn ) = xn for all n ∈ Z. Therefore τ (Y ) = {c0 } ∪ {xn : n ∈ Z} = X. This shows that X is sofic. The fact that X is not of finite type follows from Exercise 1.71(d). Alternatively, the following argument may be used. Denoting by L(X) ⊂ A∗ the language of X, we have 10n , 0n 1 ∈ L(X) for every n ∈ N. As 10n 1 ∈ / L(X) for all n ∈ N, this shows that X is not of finite type by Exercise 1.76(a). █ Comment The notion of a sofic subshift was introduced by Weiss in [Weis1]. Exercise 1.91 Let G be a countable group and let A be a finite set. Show that the set of all sofic subshifts X ⊂ AG is countable. Solution Recall that, given a sofic subshift X ⊂ AG , we can find a finite set B, a subshift of finite type Y ⊂ B G , and a cellular automaton τ : B G → AG such that τ (Y ) = X. Without loss of generality, we may suppose that B = {1, 2, . . . , n} for some integer n ≥ 0. The statement then follows from the fact that for every integer n ≥ 0, there are countably many subshifts of finite type Y ⊂ {1, 2, . . . , n}G (cf.

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Exercise 1.60) and countably many cellular automata τ : {1, 2, . . . , n}G → AG (cf. █ Exercise 1.19). Exercise 1.92 (S-Gap Subshifts) Let A := {0, 1} and let S ⊂ N. Consider the subset X(S) ⊂ AZ consisting of all the configurations x ∈ AZ satisfying the following condition: if x(n) = 1, x(n + 1) = x(n + 2) = · · · = x(n + h) = 0, and x(n + h + 1) = 1 for some n ∈ Z and h ∈ N, then h ∈ S. (a) Show that X(S) is a subshift of AZ . It is called the S-gap subshift. (b) Determine X(S) in each of the following cases: (i) (ii) (iii) (iv) (v)

S S S S S

:= ∅; := {0}; := N; := N \ {0}; := 2N.

(c) Suppose that S1 , S2 ⊂ N are distinct. Show that X(S1 ) /= X(S2 ). (d) Let B be a set with more than one element. Deduce from (c) that there are uncountably many subshifts of B Z . Solution (a) For h ∈ N let us set wh := 10h 1 ∈ A∗ . It is then clear that X(S) is the subshift of AZ admitting the set W (S) := {wh : h ∈ N \ S} ⊂ A∗ as a defining set of forbidden words (cf. Exercise 1.77). (b) (i) For S = ∅, a configuration x ∈ AZ is in X(S) if and only if 1 appears at most once. We deduce that X(S) is the at-most-one-one subshift over Z (cf. Exercise 1.71). (ii) For S = {0}, the subshift X(S) consists of all configurations x ∈ AZ such that {n ∈ Z : x(n) = 1} is an interval of Z. This subshift is called the at-most-onechain-of-ones subshift. (iii) For S = N, we have W (S) = ∅. We deduce that X(S) is the full subshift AZ . (iv) For S = N \ {0}, we have W (S) = {w0 } = {11}. We deduce that X(S) is the golden mean subshift (cf. Exercise 1.84). (v) For S = 2N, we have W (S) = {102h+1 1 : h ∈ N}. We deduce that X(S) is the even subshift (cf. Exercise 1.85). (c) By symmetry, we can assume that S1 is not contained in S2 . Let h ∈ S1 \ S2 . Consider the configuration x ∈ AZ defined by x(0) = x(h + 1) := 1 and x(n) := 0 for all n ∈ Z \ {0, h + 1}. Clearly x is in X(S1 ) but not in X(S2 ). Therefore X(S1 ) /= X(S2 ). (d) There are uncountably many subsets S ⊂ N. Up to renaming the elements in B, we may suppose that A ⊂ B. Using (c), we deduce that the subshifts X(S) ⊂ AZ ⊂ B Z , S ⊂ N, constitute a family of uncountably many pairwise distinct subsihfts of B Z . █

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Comment For more on S-gap subshifts of {0, 1}Z , see [LinM]. Note that the definition of an S-gap subshift in [LinM] differs from the one presented here in the case when S is finite. Exercise 1.93 Let A be a finite set with more than one element. Show that there are uncountably many subshifts X ⊂ AZ that are non-sofic. Solution By Exercise 1.92(d), the set of all subshifts of AZ is uncountable. As there are only countably many sofic subshifts of AZ by Exercise 1.91, we deduce that there are uncountably many subshifts of AZ that are not sofic. █ Exercise 1.94 Let G be a group. Let A and B be finite sets. (a) Suppose that τ : AG → B G is a cellular automaton and that X is a sofic subshift of AG . Show that τ (X) is a sofic subshift of B G . (b) Suppose that X ⊂ AG and Y ⊂ B G are topologically conjugate subshifts (cf. Exercise 1.54). Show that X is sofic if and only if Y is sofic. Solution (a) Since X is a sofic subshift of AG , there exist a finite set C, a cellular automaton σ : C G → AG , and a subshift of finite type Y ⊂ C G such that X = σ (Y ). As ρ := τ ◦ σ : C G → B G is a cellular automaton and ρ(Y ) = τ (σ (Y )) = τ (X), we deduce that τ (X) is a sofic subshift of B G . (b) Let f : X → Y be a G-equivariant homeomorphism. Then f is a restriction of a cellular automaton τ : AG → B G . Since Y = f (X) = τ (X), we deduce from (a) that if X is sofic then Y is sofic. Conversely, Y sofic implies X sofic by symmetry. █ Exercise 1.95 Let G be a group and let A be a finite set. Let F be a non-empty finite subset of G and let B := AF . Consider the map Φ : AG → B G defined in Exercise 1.30. Let X ⊂ AG be a subshift and set X[F ] := Φ(X). (a) Show that X[F ] is a subshift of B G . (b) Show that the subshifts X and X[F ] are topologically conjugate. (c) Show that X is topologically transitive (resp. irreducible, resp. topologically mixing, resp. strongly irreducible, resp. of finite type, resp. sofic) if and only if X[F ] is topologically transitive (resp. irreducible, resp. topologically mixing, resp. strongly irreducible, resp. of finite type, resp. sofic). Solution (a) It follows from Exercise 1.30 that the map Φ : AG → B G is a G-equivariant uniform embedding, so that, in particular, it is a cellular automaton. As A is finite, X[F ] = Φ(X) is a subshift of B G by Exercise 1.39(g). (b) This follows from the fact that Φ : AG → B G is an injective cellular automaton such that X[F ] = Φ(X). (c) Since X and X[F ] are topologically conjugate by (b), this follows from █ Exercises 1.70, 1.54, and 1.94(b).

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Comment When G = Z and F = {0, 1, . . . , N − 1} for some integer N ≥ 1, one has B = AN and the subshift X[F ] ⊂ B Z , denoted X[N ] , is called the Nth higher block shift of X (cf. [LinM, Definition 1.4.1]. See also Exercises 6.99 and 6.100). Exercise 1.96 Let A be a finite set. (a) Show that every non-empty subshift of finite type X ⊂ AZ contains a periodic configuration. (b) Deduce from (a) that every non-empty sofic subshift X ⊂ AZ contains a periodic configuration. Solution (a) Let X ⊂ AZ be a non-empty subshift of finite type. Then there exist an integer n ≥ 1 and a subset L ⊂ A∗ , consisting of words of length n, which is a defining set of admissible words for X, i.e, such that a configuration x ∈ AZ is in X if and only if x(i)x(i + 1) · · · x(i + n − 1) ∈ L for all i ∈ Z. Choose a configuration y ∈ X and consider the sequence (wi )i∈Z of words in L defined by wi := y(i)y(i + 1) · · · y(i + n − 1) for all i ∈ Z. Since L is finite, there are integers j < k in Z such that wj = wk . Then the unique periodic configuration z ∈ AZ that coincides with y on the interval [j, k] and admits k − j as a period is in X. (b) Suppose now that X ⊂ AZ is a non-empty sofic subshift. Then we can find a finite set B, a subshift of finite type Y ⊂ B Z , and a cellular automaton τ : B Z → AZ such that X = τ (Y ). By (a), there exists a periodic configuration y ∈ Y . By Zequivariance of τ , the configuration x := τ (y) ∈ X is also periodic. █ Exercise 1.97 Let A := {0, 1}. Denote by X ⊂ AZ and Y ⊂ AZ the golden mean subshift and the even subshift, respectively. Consider the cellular automata σ : AZ → AZ and τ : AZ → AZ with memory set S := {0, 1} and local defining maps ν : AS → A and μ : AS → A given respectively by ν(p) :=

⎧ 1

if (p(0), p(1)) = (1, 0)

.

0

otherwise,

and ⎧ μ(p) :=

1

if p(0) = p(1) = 0

0

otherwise,

.

for all p ∈ AS . (a) Show that σ is idempotent, i.e., σ ◦ σ = σ . (b) Show that X = σ (AZ ) = Fix(σ ). (c) Show that τ (X) = Y . (d) Show that Y is strictly sofic. (e) Show that the cellular automaton ϕ := τ ◦ σ : AZ → AZ satisfies Y = ϕ(AZ ) and give a memory set and the associated local defining map for ϕ.

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Solution (a) Let x ∈ AZ and set y := σ (x). Let n ∈ Z. For all p ∈ AS , we have ν(p) = 0 whenever p(0) = 0. Thus, if y(n) = 0 then σ (y)(n) = 0 = y(n). On the other hand, if y(n) = 1, then we must have x(n) = 1 and x(n + 1) = 0 so that y(n + 1) = 0. This implies σ (y)(n) = 1 = y(n). We conclude that σ (y)(n) = y(n) for all n ∈ Z. Therefore σ (y) = y. This shows that σ ◦ σ = σ . (b) As σ is idempotent, we have σ (AZ ) = Fix(σ ) by Exercise 1.61(c). On the other hand, if x ∈ AZ , then y := σ (x) ∈ X. Indeed, otherwise, we would have (y(n), y(n + 1)) = (1, 1) for some n ∈ Z, a contradiction since it would imply (x(n), x(n + 1)) = (x(n + 1), x(n + 2)) = (1, 0). Thus σ (AZ ) ⊂ X. On the other hand, if x ∈ X, then (x(n), x(n + 1)) /= (1, 1) and hence σ (x)(n) = x(n) for all n ∈ Z. Therefore X ⊂ Fix(σ ) = σ (AZ ). This shows that X = σ (AZ ) = Fix(σ ). (c) Let x ∈ X and set y := τ (x). For n ∈ Z, we have y(n) = 0 if and only if (x(n), x(n + 1)) ∈ {(0, 1), (1, 0)}. Suppose that y(n)y(n + 1) · · · y(n + k + 1) = 10k 1

.

for some n ∈ Z and k ∈ N. Since x ∈ X, this implies x(n)x(n + 1) · · · x(n + k + 2) = 0(01)h 00

.

for some h ∈ N, showing that k = 2h must be even. This shows that τ (X) ⊂ Y . Conversely, suppose now that we are given some y ∈ Y . We construct a configuration x ∈ AZ in the following way. If y(n)y(n + 1) · · · y(n + 2k + 1) = 102k 1

.

for some n ∈ Z and k ∈ N, then we take x(n)x(n + 1) · · · x(n + 2k + 2) := 0(01)k 00.

.

If y(n) = 1 for some n ∈ Z and y(k) = 0 for all k ∈ Z with k > n, then we take x(n) := 0 and x(k) := 0 (resp. x(k) := 1) if k > n and k − n odd (resp. even). If y(n) = 1 for some n ∈ Z and y(k) = 0 for all k ∈ Z with k < n, then we take x(k) := 0 (resp. x(k) := 1) if k < n with n − k even (resp. odd). This defines entirely the configuration x except in the particular case when y is the identically 0 configuration. In that case, we define x(k) for all k ∈ Z by setting x(k) := k mod 2. It is easy to check that x ∈ X and τ (x) = y for all y ∈ Y . Thus Y ⊂ τ (X). This shows that τ (X) = Y . (d) Since X is of finite type, it follows from (c) that Y is sofic. On the other hand, we have seen in Exercise 1.85(b) that Y is not of finite type. As a consequence, Y is strictly sofic. (e) It follows from (b) and (c) that ϕ(AZ ) = τ (σ (AZ )) = τ (X) = Y .

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An immediate computation from the formulae defining ν and μ shows that a memory set for ϕ is T := {0, 1, 2} and that the associated local defining map is the map λ : AT → A given by ⎧ λ(p) :=

1

if (p(0), p(1), p(2)) ∈ {(0, 0, 0), (0, 0, 1), (0, 1, 1), (1, 1, 1)}

0

otherwise,

.

for all p ∈ AT .

█

Comment The cellular automata σ and τ are elementary cellular automata. In Wolfram’s terminology, σ is Rule 68 and τ is Rule 17. The cellular automaton ϕ is not an elementary cellular automaton since its minimal memory set is {0, 1, 2}. However, the map ω : AZ → AZ , defined by ω(x)(n) := ϕ(x)(n − 1) for all x ∈ AZ and n ∈ Z, is an elementary cellular automaton, namely Rule 139. Observe that ω(AZ ) = ϕ(AZ ) = Y . Exercise 1.98 (Splicable Subshifts) Let G be a group and let A be a finite set. Given two configurations x, y ∈ AG and a subset Ω ⊂ G, the splice of x and y over Ω is the configuration z = spliceΩ (x, y) ∈ AG defined by z|Ω := x|Ω and z|G\Ω := y|G\Ω . Let Δ ⊂ G be a finite subset containing 1G . One says that a subshift X ⊂ AG is Δ-splicable if the following holds: (Splic-Δ)

for every finite subset Ω ⊂ G and any two configurations x, y ∈ X such that x|ΩΔ\Ω = y|ΩΔ\Ω , one has spliceΩ (x, y) ∈ X.

One says that a subshift X ⊂ AG is splicable if there exists a finite subset Δ ⊂ G containing 1G such that X is Δ-splicable. (a) Show that every subshift of finite type X ⊂ AG is splicable. (b) Suppose that G is an infinite group and let A := {0, 1}. Show that the atmost-one-one subshift X ⊂ AG is not splicable. (c) Let A := {0, 1}. Show that the even subshift X ⊂ AZ is not splicable. Solution (a) Let X ⊂ AG be a subshift of finite type. Let M ⊂ G be a memory set for X and let A ⊂ AM be a defining set of admissible patterns for X. Recall that this means that X consists of all configurations x ∈ AG such that (gx)|M ∈ A for all g ∈ G. Up to replacing M by {1G } ∪ M ∪ M −1 , we may suppose that 1G ∈ M and M = M −1 . Set Δ := M 2 and let us show that Condition (Splic-Δ) is satisfied. Let Ω ⊂ G be a finite set and let x, y ∈ X be two configurations such that x|ΩΔ\Ω = y|ΩΔ\Ω . Consider the configuration z := spliceΩ (x, y) ∈ AG . Note that z|ΩΔ = x|ΩΔ since z|Ω = x|Ω and z|ΩΔ\Ω = y|ΩΔ\Ω = x|ΩΔ\Ω . We claim that (gz)|M ∈ A for all g ∈ G. Indeed, if g −1 ∈ ΩM, then g −1 M ⊂ ΩM 2 = ΩΔ so that, for all m ∈ M, (gz)(m) = z(g −1 m) = z|ΩΔ (g −1 m) = x|ΩΔ (g −1 m) = x(g −1 m) = (gx)(m).

.

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89

This implies (gz)|M = (gx)|M ∈ A since x ∈ X. On the other hand, if g −1 ∈ G \ ΩM, then g −1 M ⊂ G \ Ω so that, for all m ∈ M, (gz)(m) = z(g −1 m) = z|G\Ω (g −1 m) = y|G\Ω (g −1 m) = y(g −1 m) = (gy)(m).

.

This implies (gz)|M = (gy)|M ∈ A since y ∈ X. This proves the claim, showing that z ∈ X. We deduce that X is splicable. (b) Suppose by contradiction that the at-most-one-one subshift X is splicable. This means that there exists a finite subset Δ ⊂ G containing 1G such that X is Δ-splicable. Take Ω := {1G } and consider the configuration x ∈ X defined by x(1G ) := 1 and x(g) := 0 for all g ∈ G \ {1G }. Since G is infinite, there exists an element g0 ∈ G \ Δ. Consider the configuration y ∈ X defined by y(g0 ) := 1 and y(g) := 0 for all g ∈ G \ {g0 }. Then x and y coincide on ΩΔ \ Ω = Δ \ {1G }. However, the configuration z ∈ AG defined by z := spliceΩ (x, y) is not in X since z(1G ) = x(1G ) = 1 and z(g0 ) = y(g0 ) = 1 with 1G /= g0 . This yields a contradiction. Therefore X is not splicable. (c) Suppose by contradiction that the even subshift X is splicable. This means that there exists a finite subset Δ ⊂ Z containing 0 such that X is Δ-splicable. Up to enlarging Δ, we can assume that Δ = {−n, −n + 1, . . . , n} for some odd integer n ≥ 1. Take Ω := {0} and consider the configurations x, y ∈ X defined by x(0) = y(n + 1) := 1, x(k) := 0 for all k ∈ Z \ {0}, and y(k) := 0 for all k ∈ Z \ {n + 1}. Then x and y coincide on (Ω + Δ) \ Ω = {−n, . . . , −1, 1, . . . , n}. However, the configuration z ∈ AZ defined by z := spliceΩ (x, y) is not in X since z(0) = z(n + 1) = 1 and z(1) = z(2) = · · · = z(n) = 0 with n odd. This yields a contradiction. Therefore X is not splicable. █ Comment The notion of a splicable subshift was introduced by Gromov in [Gro3, Section 8.C] (see also [CecFS, Section 3.8]). Splicable subshifts are called topological Markov fields by some authors (see [Chan1], [Chan2], [BarbGL]). The even subshift is sofic by Exercise 1.97(d). As all subshifts of finite type are splicable by (a) and the even subshift is not splicable by (c), this shows that the image under a cellular automaton of a splicable subshift may fail to be splicable. Exercise 1.99 (Characterization of Splicable Subshifts Over Z) Let A be a finite set and let X ⊂ AZ be a subshift. Let L(X) ⊂ A∗ denote the language of X. Given an integer n ≥ 0, one says that X is n-splicable if the following holds: (Splic-n)

for all u, u' , v, v ' , w1 , w2 ∈ A∗ such that (i) 𝓁(v) = 𝓁(v ' ) = n and 𝓁(w1 ) = 𝓁(w2 ), (ii) uvw1 v ' u' ∈ L(X) and vw2 v ' ∈ L(X), then one has uvw2 v ' u' ∈ L(X).

Show that X is splicable if and only if there exists an integer n ≥ 0 such that X is n-splicable.

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Solution In the following, given h, k ∈ Z with h ≤ k, we set [h, k] := {h, h + 1, . . . , k} ⊂ Z. Suppose first that X is splicable and let Δ ⊂ Z be a finite subset with 0 ∈ Δ such that X is Δ-splicable. Up to enlarging Δ, if necessary, we may suppose that Δ = [−n, n] for some n ∈ N. Let us show that X satisfies Condition (Splic-n). Let u, u' , v, v ' , w1 , w2 ∈ A∗ satisfying (i) and (ii). Let m := 𝓁(w1 ) = 𝓁(w2 ). We may assume that m ≥ 1 since otherwise w1 and w2 are equal to the empty word and there is nothing to prove. Set Ω := [1, m]. Then Ω + Δ = [−n + 1, m + n] and (Ω + Δ) \ Ω = [−n + 1, 0] ∪ [m + 1, m + n]. Let k := 𝓁(u) and k ' := 𝓁(u' ). Since uvw1 v ' u' ∈ L(X), there exists y ∈ X such that y(−n−k+1)y(−n−k+2) · · · y(m+ n + k ' ) = uvw1 v ' u' . On the other hand, since vw2 v ' ∈ L(X), there exists x ∈ X such that x(−n + 1)x(−n + 2) · · · x(m + n) = vw2 v ' . Since x(−n + 1)x(−n + 2) · · · x(0) = v = y(−n+1)y(−n+2) · · · y(0) and x(m+1)x(m+2) · · · x(m+n) = v ' = y(m + 1)y(m + 2) · · · y(m + n), we have x|(Ω+Δ)\Ω = y|(Ω+Δ)\Ω and Condition (Splic-Δ) ensures that the configuration z := spliceΩ (x, y) belongs to X. We deduce that uvw2 v ' u' = z(−n − k + 1)z(−n − k + 2) · · · z(m + n + k ' ) ∈ L(X). This shows that X satisfies Condition (Splic-n). Conversely, suppose that X satisfies Condition (Splic-n) for some integer n ≥ 0. Set Δ := [−n, n] and let us show that X is Δ-splicable. Let Ω ⊂ Z be a finite subset. Then there is a unique partition Ω = ⨅𝓁i=1 Ωi , where Ωi = [hi , ki ], hi , ki ∈ Z with hi ≤ ki for all i = 1, 2, . . . , 𝓁, and hi+1 − ki ≥ 2 for all i = 1, 2, . . . , n − 1. We shall refer to the Ωi s as to the “connected components” of Ω. Let x, y ∈ X be two configurations such that x|(Ω+Δ)\Ω = y|(Ω+Δ)\Ω , and let us show that spliceΩ (x, y) belongs to X. We proceed by induction on the number 𝓁 of connected components of Ω. Suppose first that 𝓁 = 1 so that Ω = [h, k], where h, k ∈ Z with h ≤ k. Then Ω + Δ = [−n + h, k + n] and (Ω + Δ) \ Ω = [−n + h, h − 1] ∪ [k + 1, k + n]. Consider the words v, w1 , w2 , v ' ∈ A∗ defined by v := x(−n + h)x(−n + h + 1) · · · x(h − 1) = y(−n + h)y(−n + h + 1) · · · y(h − 1), w1 := y(h)y(h + 1) · · · y(k), w2 := x(h)x(h + 1) · · · x(k), and v ' := x(k + 1)x(k + 2) · · · x(k + n) = y(k + 1)y(k + 2) · · · y(k + n). Also, for every i ∈ N, define ui , u'i ∈ A∗ by ui := y(−n + h − i)y(−n + h − i + 1) · · · y(−n + h − 1) and u'i := y(k + n + 1)y(k + n + 2) · · · y(k + n + i). Let i ∈ N. We have 𝓁(v) = 𝓁(v ' ) = n, 𝓁(w1 ) = 𝓁(w2 ) = k −h+1, ui vw1 v ' u'i = y(−n+h−i)y(−n+h−i +1) · · · y(k + n + i) ∈ L(X) and vw2 v ' = x(−n + h)x(−n + h + 1) · · · x(k + n) ∈ L(X). Thus, Condition (Splic-n) ensures us that ui vw2 v ' u'i ∈ L(X), that is, there exists zi ∈ X such that zi (−n + h − i)zi (−n + h − i + 1) · · · zi (k + n + i) = ui vw2 v ' u'i . Since zi |[−n+h−i,k+n+i] = zj |[−n+h−i,k+n+i] for all 0 ≤ i ≤ j and X is closed in AZ , the configuration z := limi→∞ zi is well defined and belongs to X. For every i ∈ N, we have z(h)z(h+1) · · · z(k) = zi (h)zi (h+1) · · · zi (k) = w2 = x(h)x(h+1) · · · x(k), that is, z|Ω = x|Ω . Moreover, z(−n+h−i)z(−n+h−i +1) · · · z(h−1) = zi (−n+ h−i)zi (−n+h−i+1) · · · zi (h−1) = ui v = y(−n+h−i)y(−n+h−i+1) · · · y(h− 1) and z(k + 1)z(k + 2) · · · z(k + n + i) = zi (k + 1)zi (k + 2) · · · zi (k + n + i) = v ' u'i = y(k + 1)y(k + 2) · · · y(k + n + i), so that z|Z\Ω = y|Z\Ω . This shows

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91

that spliceΩ (x, y) = z ∈ X. This completes the argument when Ω has only one connected component. Suppose now that the splice over Ω of two configurations x and y such that x|(Ω+Δ)\Ω = y|(Ω+Δ)\Ω belongs to X whenever Ω ⊂ Z has 𝓁 ≥ 1 connected components. Let Ω ⊂ Z be a finite subset with 𝓁 + 1 connected components, say 𝓁+1 Ω = ⨅i=1 Ωi with Ωi = [hi , ki ] for all 1 ≤ i ≤ 𝓁 + 1. Let x and y be two configurations in X such that x|(Ω+Δ)\Ω = y|(Ω+Δ)\Ω . Set Ω ' := ⨅𝓁i=1 Ωi and Ω '' := Ω𝓁+1 , so that Ω = Ω ' ⨅ Ω '' . Let us set h' := h1 (resp. k ' := k𝓁 ) and h'' := h𝓁+1 (resp. k '' := k𝓁+1 ), so that Ω ' ⊂ [h' , k ' ] and Ω '' = [h'' , k '' ]. Note that h'' − k ' = h𝓁+1 − k𝓁 ≥ 2. We distinguish two cases. Suppose first that h'' − k ' ≥ n. Then (Ω + Δ) \ Ω = ((Ω ' + Δ) \ Ω ' ) ∪ (Ω '' + Δ) \ Ω ''

.

and, from the hypothesis x|(Ω+Δ)\Ω = y|(Ω+Δ)\Ω , we deduce that x|(Ω ' +Δ)\Ω ' = y|(Ω ' +Δ)\Ω ' and x|(Ω '' +Δ)\Ω '' = y|(Ω '' +Δ)\Ω '' . By induction, z' := spliceΩ ' (x, y) (resp. z'' := spliceΩ '' (x, y)) belongs to X. Since k ' + n < h'' , we have z' |(Ω ' +Δ)\Ω ' = y|(Ω ' +Δ)\Ω ' = z'' |(Ω ' +Δ)\Ω '

.

so that z := spliceΩ ' (z' , z'' ) belongs to X. Now, on the one hand, z|Ω ' = z' |Ω ' = x|Ω ' (resp. z|Ω '' = z'' |Ω '' = x|Ω '' ), so that z|Ω = x|Ω and, on the other hand, z|Z\Ω = z'' |Z\Ω = y|Z\Ω . This shows that spliceΩ (x, y) = z ∈ X. Suppose now that 2 ≤ h'' − k ' < n. Consider the set Ω := Ω ∪ [k ' , h'' ]. The decomposition of Ω into connected components is Ω = ⨅𝓁i=1 Ω i , where Ω i := Ωi for i = 1, 2, . . . , 𝓁 − 1 and Ω 𝓁 := Ω𝓁−1 ∪ [k ' , h'' ] ∪ Ω𝓁+1 = [h𝓁 , k '' ]. We have (Ω + Δ) \ Ω ⊂ (Ω + Δ) \ Ω so that x|(Ω+Δ)\Ω = y|(Ω+Δ)\Ω . By induction, spliceΩ (x, y) belongs to X. Moreover, since [k ' + 1, h'' − 1] ⊂ (Ω + Δ) \ Ω, we have x|[k ' +1,h'' −1] = y|[k ' +1,h'' −1] . Therefore spliceΩ (x, y) = spliceΩ (x, y) ∈ X. In either case, we get spliceΩ (x, y) ∈ X, which completes the inductive argument. This shows that X is Δ-splicable. █ Exercise 1.100 Let A := {0, 1}. For a ∈ A, let ca ∈ AZ be the constant configuration defined by ca (k) := a for all k ∈ Z. For n ∈ Z, let xn , yn ∈ AZ be the configurations defined, for all k ∈ Z, by xn (k) := 0 and yn (k) := 1 if k < n, xn (k) := 1 and yn (k) := 0 if k ≥ n. Consider the subset X ⊂ AZ defined by X := {c0 , c1 } ∪ {xn , yn : n ∈ Z}.

.

(a) Show that X is a sofic subshift of AZ . (b) Show that the subshift X is not of finite type. (c) Show that the subshift X is splicable.

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Solution (a) Let B := {0, 1, 2, 3} and let Y ⊂ B Z be the subshift of finite type admitting {00, 01, 11, 22, 23, 33} as a defining set of admissible words. Observe that Y consists of the four constant configurations db ∈ B Z (b ∈ B) defined by db (k) := b for all k ∈ Z, together with the configurations en , fn ∈ B Z (n ∈ Z) defined, for all k ∈ Z, by en (k) := 0 and fn (k) := 2 if k < n, en (k) := 1 and fn (k) := 3 if k ≥ n. Consider the cellular automaton τ : B Z → AZ defined by

(τ (x))(n) :=

.

⎧ ⎪ ⎪ ⎨1 0

⎪ ⎪ ⎩x(n)

if x(n) = 2, if x(n) = 3, otherwise

for all x ∈ B Z and n ∈ Z. Observe that c0 = τ (d0 ) = τ (d3 ), c1 = τ (d1 ) = τ (d2 ), xn = τ (en ) and yn = τ (fn ) for all n ∈ Z. Therefore X = τ (Y ). As Y is a subshift of finite type of B Z , this shows that X is a sofic subshift of AZ . (b) Let L(X) ⊂ A∗ denote the language of X. For every n ∈ N, the word 01n (resp. 1n 0) appears in x0 (resp. y0 ). Therefore 01n , 1n 0 ∈ L(X). As 01n 0 ∈ / L(X), it follows from Exercise 1.79 that X is not of finite type. (c) Let us show that X is 1-splicable. Let u, u' , w1 , w2 ∈ A∗ and a, a ' ∈ A such that 𝓁(w1 ) = 𝓁(w2 ), uaw1 a ' u' ∈ L(X), and aw2 a ' ∈ L(X). If a = a ' , then w1 = w2 so that uaw2 a ' u' = uaw1 a ' u' ∈ L(X). If a /= a ' , then wi = a hi (a ' )ki , where ' 0 ≤ hi , ki and hi + ki = 𝓁(wi ), i = 1, 2, and u = a h (resp. u' = (a ' )h ), where h := ' 𝓁(u) (resp. h' := 𝓁(u' )). This implies that uaw2 a ' u' = a h+1+h2 (a ' )k2 +1+h ∈ L(X). Thus, in either case, we have uaw2 a ' u' ∈ L(X). This shows that X is 1-splicable. By Exercise 1.99, this implies that X is splicable. █ Comment Another example of a splicable subshift over Z that is not of finite type (with a 3-letters alphabet instead of a 2-letters one) is given in [Chan1, Proposition 3.6]. Exercise 1.101 Let G be a finitely generated group and let A be a finite set. Let X, Y ⊂ AG be sofic subshifts. Show that X ∪ Y is a sofic subshift of AG . Solution Since X and Y are sofic, there exist finite sets B and C, cellular automata σ : B G → AG and τ : C G → AG , and subshifts of finite type Z ⊂ B G and T ⊂ C G such that X = σ (Z) and Y = τ (T ). Without loss of generality, we can assume that the sets B and C are disjoint. Let D := B ∪ C. Regarding B G and C G as subsets of D G , the set Z ∪ T is a subshift of finite type of D G by Exercise 1.59. Moreover, the cellular automata σ and τ can be extended to a cellular automaton ρ : D G → AG in the following way. Let S be a finite subset of G that is a memory set for both σ and τ . Let μ : B S → A and ν : C S → A denote the associated local defining maps for σ and τ , respectively. As B S and C S are disjoint subsets of D S , we can extend μ and ν to a map η : D S → A. Let now ρ : D G → AG denote the cellular automaton with memory set S and associated local defining map η. Then ρ extends both σ and

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τ . As ρ(Z ∪ T ) = ρ(Z) ∪ ρ(T ) = σ (Y ) ∪ τ (Z) = X ∪ Y , this shows that X ∪ Y is a sofic subshift of AG . █ Exercise 1.102 Let A := {0, 1}. Let X ⊂ AZ (resp. Y ⊂ AZ ) denote the subshift of finite type admitting the singleton {11} (resp. {00}) as a defining set of forbidden words. Show that X ∪ Y is a strictly sofic subshift. Solution The fact that X ∪ Y is a sofic subshift follows from Exercise 1.101. To prove that X ∪ Y is not of finite type, let L(X ∪ Y ) ⊂ A∗ denote the language of X ∪ Y and consider, for a given n ∈ N, the words u, v, w ∈ A∗ given by u := 00, v := 11, and w := (01)n . Clearly uw, wv ∈ L(X ∪ Y ) but uwv ∈ / L(X ∪ Y ). As the length 2n of w can be made arbitrarily large, it follows from Exercise 1.79 that X ∪ Y is not of finite type. This shows that X ∪ Y is strictly sofic. █ Comment Observe that X and Y are two copies of the golden mean subshift. More precisely, X is exactly the golden mean subshhift as defined in Exercise 1.84 while Y can be obtained from X by exchanging the symbols 0 and 1 in each configuration of X, that is, Y = τ (X), where τ : AZ → AZ is the invertible cellular automaton given by τ (x)(n) := 1 − x(n) for all x ∈ AZ and n ∈ Z (in Wolfram’s notation, τ is Rule 51). The subshift X ∩ Y is of finite type by Exercise 1.48(f). Actually X ∩ Y has cardinality 2, it consists of the two 2-periodic non-constant configurations and admits {00, 11} as a defining set of forbidden words. Exercise 1.103 Let G be a group and let A be a set. Let H be a subgroup of G and let X ⊂ AH be a subshift. For x ∈ AG and g ∈ G, denote by xgH ∈ AH the configuration defined by xgH := (gx)|H . (a) Show that the set X(G) defined by X(G) := {x ∈ AG : xgH ∈ X for all g ∈ G}

.

is a subshift of AG . H = hx H for all x ∈ AG , h ∈ H , and g ∈ G. (b) Check that xhg g (c) From now on, we fix a complete set T ⊂ G of representatives for the right cosets of H in G, so that G = ⨅t∈T H t. We suppose that 1G ∈ T . Let x ∈ AG . H Show that one has x ∈ X(G) if and only Π if xt H∈ X for all t ∈ T . ( H ) G (d) Show that the map ϕ : A → t∈T A defined by ϕ(x) := xt t∈T for all x ∈ AG is a uniform isomorphism with respectΠ to the prodiscrete uniform structure on AG and the product uniform structure on t∈T AH , where each copy of AH is equipped with the prodiscrete uniform Π structure. Π Deduce that ϕ|X(G) yields a uniform isomorphism from X(G) onto t∈T X ⊂ t∈T AH . (e) Let x ∈ AH . Define x T ∈ AG by setting x T (t −1 h) := x(h) for all t ∈ T and h ∈ H . Show that x T ∈ X(G) if and only if x ∈ X. Finally observe that x T |H = x.

.

(1.32)

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(f) Given g ∈ G, there exist unique η(g) ∈ H and θ (g) ∈ T such that g = η(g)θ (g).

.

(1.33)

Show that for every t ∈ T fixed, the map g |→ η(tg) is a surjection from G onto H,

.

(1.34)

while, for every g ∈ G fixed, the map t |→ θ (tg) is a permutation of T .

.

(1.35)

(g) Show that if x ∈ AG and g ∈ G then, for all t ∈ T , H (gx)H t = η(tg)xθ(tg) .

.

(1.36)

(h) Show that if the index of H in G is infinite, then the subshift X(G) ⊂ AG is topologically transitive. (i) Suppose A is finite. Show that X(G) is of finite type if and only if X is of finite type. (j) Show that if X(G) is topologically mixing then X is topologically mixing. (k) Show that X(G) is strongly irreducible if and only if X is strongly irreducible. (l) Let B be another alphabet set. Suppose that σ : B H → AH is a cellular G G G automaton )H σ : B H → A be theGinduced cellular automaton. Check that ( Gand let one has σ (x) g = σ (xg ) for all x ∈ B and g ∈ G, and deduce that .

⎛ ⎞ (σ (X))(G) = σ G X(G) .

(1.37)

(m) Suppose that A is finite. Show that if X is sofic then X(G) is sofic. Solution (a) Let x ∈ X(G) and let g ∈ G. Then for all g ' ∈ G we have (gx)H g' = ' (G) (G) is G-invariant. Moreover, (g gx)|H ∈ X, so that gx ∈ X . This shows that X the projection map πH : AG → AH and the maps λg : AG → AG defined by λg (x) := gx for all g ∈ G and x ∈ AG , are continuous. Since X is closed in AH , it follows that for every g ∈ G the set X(g) := {x ∈∩AG : (gx)|H ∈ X} = (πH ◦ λg )−1 (X) is closed in AG . We deduce that X(G) = g∈G X(g) is closed in AG for the prodiscrete topology. This shows that X(G) ⊂ AG is a subshift. (b) Let x ∈ AG , h, k ∈ H , and g ∈ G. We have H xhg (k) = (hgx)|H (k) = (hgx)(k) = (gx)(h−1 k)

.

= (gx)|H (h−1 k) = xgH (h−1 k) = (hxgH )(k). H = hx H . This shows that xhg g

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(c) Necessity is obvious since T ⊂ G. Conversely, suppose that xtH ∈ X for all t ∈ T and let g ∈ G. As T is a complete set of representatives for the right cosets of H in G, there exist h ∈ H and t ∈ T such that g = ht. We then have H = hx H by using (a). Since X is H -invariant and x H ∈ X, this shows that xgH = xht t t xgH ∈ X. Thus, the condition is also sufficient. Π −1 (d) Since G = ⨅t∈T H t = ⨅t∈T t −1 H , the map ψ : AG → t∈T At H , defined by setting ψ(x)t := x|t −1 H for all x ∈ AG and t ∈ T , is, clearly, a uniform −1 isomorphism. Also, for every t ∈ T , the map φt : At H → AH , defined by setting −1 φt (y)(h) := y(t −1 h) for all y ∈ At H and h ∈ H , is a uniform isomorphism. It Π Π Π −1 follows that φ := t∈T φt : t∈T At H → t∈T AH is a uniform isomorphism. Note that for all x ∈ AG , t ∈ T , and h ∈ H , ( ) φt x|t −1 H (h) = x|t −1 H (t −1 h) = x(t −1 h) = (tx)(h) = (tx)|H (h) = xtH (h)

.

that is, ( ) φt x|t −1 H = xtH .

.

Π H We deduce that ϕ = φ ◦( ψ : A)G → t∈T A is a uniform isomorphism. Finally, it Π (G) follows from (c) that ϕ X = t∈T X. (e) For t ∈ T and h ∈ H we have .

⎛ ⎞H xT (h) = (tx T )|H (h) = (tx T )(h) = x T (t −1 h) = x(h), t

( )H showing that x T t = x. Using (c) we deduce that x T ∈ X(G) if and only if x ∈ X. Moreover, keeping in mind that 1G ∈ T , given h ∈ H we have x T (h) = x T (1G h) = x(h), and (1.32) follows. (f) Fix t ∈ T and let h ∈ H . Then setting g := t −1 h ∈ G we have η(tg)θ (tg) = tg = t (t −1 h) = h so that, in particular, η(tg) = h. This proves (1.34). Fix now g ∈ G and let s ∈ T . Then, setting t := θ (sg −1 ) ∈ T and h := η(sg −1 ) ∈ H , we have sg −1 = η(sg −1 )θ (sg −1 ) = ht, so that tg = h−1 s. We deduce, in particular, that θ (tg) = s. This shows that the map t |→ θ (tg) is surjective. Let us show that it is also injective. Let t1 , t2 ∈ T and suppose that θ (t1 g) = θ (t2 g) := t ∈ T . For i = 1, 2 we set hi := η(ti g) ∈ H so that ti g = hi t, equivalently, ti = hi tg −1 . We deduce that H t1 = H (h1 tg −1 ) = H tg −1 = H (h2 tg −1 ) = H t2 , showing that t1 = t2 . This proves (1.35). (g) Let x ∈ AG , g ∈ G, and t ∈ T . Then setting s := θ (tg) ∈ T and h := η(tg) ∈ H , from (1.33) we deduce (tgx)(k) = (hsx)(k) = (sx)(h−1 k) = (sx)|H (h−1 k) = xsH (h−1 k) = (hxsH )(k),

.

H for all k ∈ H . Thus, (gx)H t = (tgx)|H = hxs , and (1.36) follows.

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(h) Suppose that the index of H in G is infinite, that is, T is infinite. Let Ω ⊂ G be a finite subset and let x1 , x2 ∈ X(G) . Let T0 ⊂ T and H0 ⊂ H be two finite subsets such that Ω ⊂ T0−1 H0 . Since T0−1 is finite, T is infinite, and the action of G on G/H is transitive, we can find g0 ∈ G such that .

g0−1 T0−1 H ∩ T0−1 H = ∅,

(1.38)

θ (tg0 ) ∈ / T0 for all t ∈ T0 .

(1.39)

so that (cf. (1.33)), .

( ) Consider the configuration x := ϕ −1 (xsH )s∈T ∈ AG defined by setting H .xs

:=

⎧ η(tg0 )−1 (x2 )H t (x1 )H s

if s = θ (tg0 ) for some t ∈ T0 otherwise.

Note that, by virtue of (1.35), the above configuration is well defined. Since xsH ∈ X for all s ∈ T , we have that x ∈ X(G) . Moreover, keeping in mind (1.39), ⎞ ⎛ x|T −1 H = (xsH )s∈T0 = (x1 )H s

.

s∈T0

0

= x1 |T −1 H 0

so that, in particular (as Ω ⊂ T0−1 H ), x|Ω = x1 |Ω , and, keeping in mind (1.36), ⎞ ⎛ (gx)|T −1 H = (gx)H t

.

0

t∈T0

⎞ ⎛ H = η(tg)xθ(tg)

t∈T0

⎞ ⎛ = (x2 )H t

t∈T0

= x2 |T −1 H 0

so that (gx)|Ω = x2 |Ω . This shows that when the index of H in G is infinite, the subshift X(G) is topologically transitive. (i) Suppose first that X is of finite type and let Ω ⊂ H and A ⊂ AΩ denote a memory set and a defining set of admissible patterns for X, respectively. Given x ∈ AG we have, keeping in mind (a), x ∈ X(G) ⇔ xgH ∈ X for all g ∈ G ⇔ (hxgH )|Ω ∈ A for all g ∈ G and h ∈ H .

H ⇔ (xhg )|Ω ∈ A for all g ∈ G and h ∈ H

⇔ (xgH )|Ω ∈ A for all g ∈ G ⇔ ((gx)|H ) |Ω ∈ A for all g ∈ G ⇔ (gx)|Ω ∈ A for all g ∈ G,

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97

where the last equivalence follows from the inclusion Ω ⊂ H . This shows that Ω and A ⊂ AΩ serve as a memory set and a defining set of admissible patterns also for X(G) . Since by assumption, Ω is finite, this shows that the subshift X(G) is of finite type. Conversely, suppose that X(G) is of finite type and let Ω ⊂ G denote a memory set for X(G) . Up to enlarging Ω, if necessary, we may suppose that Ω is of the form Ω = T0−1 Ω0 ⊂ T0−1 H for some finite subsets T0 ⊂ T and Ω0 ⊂ H . Let us take A := {x|Ω : x ∈ X(G) } ⊂ AΩ as a defining set of admissible patterns for X(G) , and let us show that A0 := {p|Ω0 : p ∈ A } = {x|Ω0 : x ∈ X} ⊂ AΩ0

.

serves as a defining set of admissible patterns for X. We first observe that we have the equality A0 = {x|Ω0 : x ∈ X(G) }.

(1.40)

.

Indeed, given x ∈ X(G) we have x|Ω0 = (x|H )|Ω0 = x1HG |Ω0 ∈ A0 , showing the inclusion A0 ⊃ {x|Ω0 : x ∈ X(G) }. Conversely, given x ∈ AH ,(let x T) ∈ AG as in (g). Then, keeping in mind (1.32), if x ∈ X we have x T |Ω0 = x T |H Ω = x|Ω0 , 0

and the inclusion A0 ⊂ {x|Ω0 : x ∈ X(G) } follows as well. With the above notation, given x ∈ AH we then have x ∈ X ⇔ x T ∈ X(G) ⇔ (gx T )|Ω ∈ A for all g ∈ G ⎞ ⎛ ⇔ (gx T )(t −1 k) t∈T0 ∈ A for all g ∈ G k∈Ω0

⎞ ⎛ ⇔ x T (g −1 t −1 k) t∈T0 ∈ A for all g ∈ G k∈Ω0

.

⎛ ⎞ (using (1.33)) ⇔ x T (θ −1 (tg)η−1 (tg)k) t∈T0 ∈ A for all g ∈ G k∈Ω0

⎞ ⎛ ⇔ x(η−1 (tg)k) t∈T0 ∈ A for all g ∈ G k∈Ω0

⇔ ((η(tg)x)(k)) t∈T0 ∈ A for all g ∈ G k∈Ω0

(using (1.40) and (1.34)) ⇔ ((hx)(k))k∈Ω0 ∈ A0 for all h ∈ H ⇔ (hx)|Ω0 ∈ A0 for all h ∈ H. This shows that Ω0 and A0 serve as a memory set and a defining set of admissible patterns for X, respectively. Since Ω0 is finite, we deduce that X is of finite type.

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(j) Suppose that X(G) is topologically mixing. Let Ω ⊂ H be a finite subset and let x1 , x2 ∈ X. Consider the configurations x1T , x2T ∈ X(G) as defined in (g). Since X(G) is topologically mixing, we can find a finite subset F ⊂ G such that for every g ∈ G \ F there exists x ' ∈ X(G) such that x ' |Ω = x1T |Ω and (gx ' )|Ω = x2T |Ω . Setting x := x ' |H and keeping in mind (1.32), we have x ∈ X and, moreover, x|Ω = (x ' |H )|Ω = x ' |Ω = x1T |Ω = (x1T |H )|Ω = x1 |Ω and, analogously, (gx)|Ω = x2 |Ω . This shows that X is topologically mixing. (k) Suppose that X is strongly irreducible and let Δ ⊂ H be a finite set such that X is Δ-irreducible. Let us show that X(G) is Δ-irreducible. Let Ω1 , Ω2 ⊂ G be two finite subsets such that Ω1 Δ ∩ Ω2 = ∅ and let x1 , x2 ∈ X(G) . Let T1 , T2 ⊂ T be two finite subsets such that Ωi ⊂ Ti−1 H and write Ωi = ⨅t∈Ti t −1 Ωt,i , where Ωt,i ⊂ H , i = 1, 2. For t ∈ T1 ∩ T2 , since Ωt,1 Δ ∩ Ωt,2 = ∅, we can find x t ∈ X such that x t |Ωt,i = (xi )H t |Ωt,i for i = 1, 2. It follows that the configuration x := ϕ −1 ((xt )t∈T ) ∈ AG defined by setting

xt :=

.

⎧ ⎪ ⎪(x1 )H t if t ∈ T1 \ T2 ⎨ x t if t ∈ T1 ∩ T2 ⎪ ⎪ ⎩(x )H otherwise 2 t

satisfies that x ∈ X(G) and x|Ωi = xi |Ωi , i = 1, 2. This shows that X(G) is strongly irreducible. Conversely, suppose that X(G) is strongly irreducible and let Δ ⊂ G be a finite set such that X(G) is Δ-irreducible. Up to enlarging Δ if necessary, we may suppose that there exist two finite sets T0 ⊂ T and Δ0 ⊂ H such that Δ = Δ0 T0 . We claim that X is Δ0 -irreducible. Let Ω1 , Ω2 ⊂ H be two finite subsets such that Ω1 Δ0 ∩ Ω2 = ∅ and let x1 , x2 ∈ X. Let x1T , x2T ∈ X(G) denote the configurations defined in (g). Since Ω1 Δ ∩ Ω2 = ∅, we can find a configuration x ' ∈ X(G) such that x ' |Ωi = xiT |Ωi for i = 1, 2. Setting x := x ' |H and keeping in mind (1.32), we have x ∈ X and x|Ωi = (x ' |H )|Ωi = x ' |Ωi = xiT |Ωi = (xiT |H )|Ωi = xi |Ωi

.

for i = 1, 2. This proves the claim and shows that X is strongly irreducible. (𝓁) Let S ⊂ H be a memory set for σ and let μ : B S → A denote the corresponding local defining map. For x ∈ X(G) , g ∈ G, and h ∈ H we have .

⎛ ⎞H ⎛ ⎞ ⎛ ⎞ σ G (x) (h) = gσ G (x) |H (h) = σ G (gx) (h) = μ((h−1 (gx))|S ) g

= μ((h−1 ((gx)|H )|S )) = μ((h−1 (xgH )|S )) = σ (xgH )(h), ( )H showing that σ G (x) g = σ (xgH ).

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99

Let x ∈ AG . Suppose that x ∈ (σ (X))(G) . Then for every t ∈ T we have xtH ∈ σ (X), that is, there exists yt ∈ X such that σ (yt ) = xtH . Set y := ϕ −1 ((yt )t∈T ) ∈ X(G) ⊂ B G so that (ty)|H = ytH = yt for all t ∈ T . We have σ G (y)(t −1 h) = μ((h−1 ty)|S ) = μ((h−1 yt )|S ) = σ (yt )(h) = xtH (h) = x(t −1 h)

.

G for (all t ∈ x. )This shows that (σ (X))(G) ⊂ ) T and h ∈ H , that is, σ (y)G (= (G) G (G) . Conversely, suppose that x ∈ σ X . Then we can find y ∈ X(G) ⊂ σ X G G B such that x = σ (y). Let t ∈ T and h ∈ H . Then

xtH (h) = (tx)|H (h) = (tx)(h) = σ G (ty)(h)

.

= μ((h−1 ty)|S ) = μ((h−1 ytH )|S ) = σ (ytH )(h) that(is, xtH) = σ (ytH ) ∈ σ (X). This shows that x ∈ (σ (X))(G) . We deduce that σ G X(G) ⊂ (σ (X))(G) , and (1.37) follows. (m) Suppose that X is sofic. Then, we can find a finite alphabet B, a subshift of finite type Y ⊂ B G , and a cellular automaton σ : B H → AH such that σ (Y ) = X. By applying (1.37) we deduce that ⎛ ⎞ X(G) = (σ (Y ))(G) = σ G Y (G) .

.

Since the subshift Y (G) ⊂ B G is of finite type by (i), we deduce that X(G) is sofic. █ Exercise 1.104 (Limit Set of a Cellular Automaton) Let G be a group and let A be a non-empty finite set. Let τ : AG → AG be a cellular automaton. The subset Ω(τ ) ⊂ AG defined by Ω(τ ) :=

∩

.

τ n (AG )

(1.41)

n∈N

is called the limit set of τ . (a) Show that Ω(τ ) is a non-empty subshift of AG . (b) Show that every subset X ⊂ AG such that X ⊂ τ (X) satisfies X ⊂ Ω(τ ). (c) Show that Fix(τ ) ⊂ Ω(τ ). (d) Show that τ (Ω(τ )) = Ω(τ ). (e) One says that x ∈ AG admits a backward orbit if there exists a sequence (xn )n∈N of elements of AG such that x0 = x and xn = τ (xn+1 ) for all n ∈ N. Show that Ω(τ ) is the set of configurations in AG admitting a backward orbit. (f) Show that Ω(τ k ) = Ω(τ ) for every integer k ≥ 1. (g) Show that Ω(τ ) contains at least one constant configuration. (h) Show that τ is surjective if and only if Ω(τ ) = AG . (i) Show that τ is nilpotent if and only if Ω(τ ) is a singleton.

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1 Cellular Automata

(j) Let A := {0, 1}. Determine Ω(τ ) when τ : AZ → AZ is the cellular automaton defined, for all x ∈ AZ and n ∈ Z, by τ (x)(n) := 1 if (x(n), x(n + 1)) = (1, 0) and τ (x)(n) := 0 otherwise. (k) Let A := {0, 1}. Determine Ω(τ ) when τ : AZ → AZ is the cellular automaton defined, for all x ∈ AZ and n ∈ Z, by τ (x)(n) := 1 if x(n) = x(n + 1) = 1 and τ (x)(n) := 0 otherwise. Solution (a) For every n ∈ N, the set τ n (AG ) is a subshift of AG by Exercise 1.39(f). As the intersection of any family of subshifts of AG is a subshift of AG by Exercise 1.39(d), we deduce that Ω(τ ) is a subshift of AG . As Ω(τ ) is the intersection of a decreasing sequence of non-empty closed subsets of AG , it is not empty by compactness of AG . We shall give an alternative proof of the fact that Ω(τ ) is non-empty in (f) below. (b) If X ⊂ AG satisfies X ⊂ τ (X), then X ⊂ τ (X) ⊂ τ 2 (X) ⊂ τ 3 (X) ⊂ . . .

.

and hence X⊂

∩

.

τ n (X) ⊂

n∈N

∩

τ n (AG ) = Ω(τ ).

n∈N

(c) This follows from (b) since τ fixes every element of Fix(τ ) and hence τ (Fix(τ )) = Fix(τ ). (d) Suppose first that x ∈ Ω(τ ). Then, for every n ∈ N, we have x ∈ τ n (AG ) and hence τ (x) ∈ τ n+1 (AG ) ⊂ τ n (AG ). Therefore τ (x) ∈ Ω(τ ) by(1.41). This shows that τ (Ω(τ )) ⊂ Ω(τ ). Conversely, suppose now that x ∈ Ω(τ ). Then, for every n ∈ N, we have x ∈ τ n+1 (AG ), so that there exists yn ∈ AG such that x = τ n+1 (yn ) = τ (τ n (yn )). As τ n (yn ) ∈ τ −1 (x) ∩ τ n (AG ), it follows that τ −1 (x) ∩ τ n (AG ) /= ∅ for every n ∈ N. Now, since ( ) ∩ ∩ −1 −1 n G .τ (x) ∩ Ω(τ ) = τ (x) ∩ τ (A ) = (τ −1 (x) ∩ τ n (AG )) n∈N

n∈N

is the intersection of a decreasing sequence of closed non-empty subsets of the compact space AG , we deduce that τ −1 (x) ∩ Ω(τ ) /= ∅. Thus there exists a configuration y ∈ Ω(τ ) such that x = τ (y), that is, x ∈ τ (Ω(τ )). This shows that Ω(τ ) ⊂ τ (Ω(τ )). (e) Suppose that x ∈ AG admits a backward orbit. This means that there exists a sequence (xn )n∈N in AG such that x0 = x and xn = τ (xn+1 ) for all n ∈ N. This implies x = τ n (xn ) for all n ∈ N. Therefore x ∈ Ω(τ ). The fact that every configuration in Ω(τ ) admits a backward orbit immediately follows from (d).

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(f) Given an integer k ≥ 1, we have kn → ∞ as n → ∞. As AG ⊃ τ (AG ) ⊃ ⊃ · · · τ n (AG ) ⊃ τ n+1 (AG ) ⊃ · · · , we deduce that

τ 2 (AG )

Ω(τ k ) =

∩

.

τ kn (AG ) =

n∈N

∩

τ n (AG ) = Ω(τ ).

n∈N

(g) Let C ⊂ AG denote the set of constant configurations. Since τ is a cellular automaton, we have τ (C) ⊂ C. On the other hand, the set C is in bijection with A and therefore finite. It follows that there is an integer k ≥ 1 and a configuration c ∈ C such that τ k (c) = c. This implies c ∈ Fix(τ k ) ⊂ Ω(τ k ) by (b). As Ω(τ k ) = Ω(τ ) by (d), we deduce that c ∈ Ω(τ ). (h) If τ is surjective, then τ n (AG ) = AG for every n ∈ N, so that Ω(τ ) = AG . Conversely, if Ω(τ ) = AG then τ is surjective since Ω(τ ) ⊂ τ (AG ). (i) Suppose first that τ is nilpotent. This means that there is an integer k ≥ 1 and a constant configuration c ∈ AG such that every x ∈ AG satisfies τ k (x) = c (cf. Exercise 1.14(a)). We then have τ n (AG ) = {c} for every n ≥ k. This implies Ω(τ ) = {c}. To prove the converse implication, we proceed by contradiction. Suppose that τ is not nilpotent and that Ω(τ ) is reduced to a single configuration. By (e), we have Ω(τ ) = {c}, where c ∈ AG is a constant configuration. Denote by a the element of A such that c(g) = a for all g ∈ G. Consider the closed subset K ⊂ AG defined by K := {x ∈ AG : x(1G ) /= a}. Since τ is not nilpotent, for every n ∈ N, there exists xn ∈ AG such that τ n (xn ) /= c. This implies that there exists gn ∈ N such that τ n (xn )(gn ) /= a. Then the configuration yn := gn−1 τ n (xn ) = τ n (gn−1 xn ) is in τ n (AG ) and satisfies yn (1G ) /= a. Therefore yn ∈ K ∩ τ n (AG ), showing that K ∩ τ n (AG ) is non-empty for every n ∈ N. As ( K ∩ Ω(τ ) = K ∩

∩

.

n∈N

) τ (A ) = n

G

∩

(K ∩ τ n (AG ))

n∈N

is the intersection of a decreasing sequence of non-empty closed subsets of AG , we have K ∩ Ω(τ ) /= ∅ by compactness of AG . This is in contradiction with Ω(τ ) = {c} and c ∈ / K. (j) It follows from (c) and (1.41) that Fix(τ ) ⊂ Ω(τ ) ⊂ τ (AZ ). On the other hand, denoting by X ⊂ AZ the golden mean subshift, we have Fix(τ ) = τ (AZ ) = X by Exercise 1.97(b). Therefore Ω(τ ) = X. (k) First observe that, for every integer n ≥ 1, the set τ n (AZ ) is the subshift of finite type of AZ admitting the set {10k 1 : 1 ≤ k ≤ n} as a defining set of forbidden words. We deduce that Ω(τ ) is the subshift of AZ admitting the set {10n 1 : n ≥ 1} as a defining set of forbidden words. This means that Ω(τ ) is the subshift consisting of all bi-infinite sequences of 0s and 1s having at most one chain of 1s. Note that Ω(τ ) is not of finite type. █

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Comment Limit sets of cellular automata were introduced by Wolfram [Wolfr]. The study of limit sets of cellular automata was subsequently pursued by several authors (see [Mil2], [CulPY], [Hur], [Maa], [GuilR], [BalGK] and the references therein). The characterization of nilpotent cellular automata given in (h) is due to ˇ Culík et al. [CulPY, Theorem 3.5]. In [Kar2], Kari proved that, given any integer d ≥ 1, all non-trivial problems about limit sets of finite-alphabet cellular automata over Zd are undecidable (see also [Kar3, Section 7]). The proof consists in reducing the problem to the nilpotency problem, which is known to be undecidable (see the comments to Exercise 1.14). In Wolfram’s terminology, the elementary cellular automaton appearing in (j) is Rule 68 while the one in (k) is Rule 136. Exercise 1.105 (Stable Cellular Automata) Let G be a group and let A be a nonempty finite set. Suppose that τ : AG → AG is a cellular automaton and let Ω(τ ) denote the limit set of τ . One says that τ is stable if there exists k ∈ N such that τ k (AG ) = τ k+1 (AG ). One says that τ is unstable if it is not stable. (a) Show that if τ is surjective or nilpotent then it is stable. (b) Show that if τ is stable then Ω(τ ) is sofic. (c) Show that if Ω(τ ) is of finite type then τ is stable. (d) Show that if Ω(τ ) is of finite type then it is topologically mixing. (e) Let A := {0, 1} and consider the cellular automaton τ : AZ → AZ defined, for all x ∈ AZ and n ∈ Z, by τ (x)(n) := 1 if x(n) = x(n + 1) = 1 and τ (x)(n) := 0 otherwise. Show that τ is unstable and that Ω(τ ) is strictly sofic and not topologically transitive. Solution (a) This is immediate from the definitions. (b) Suppose that τ is stable. This means that there exists k ∈ N such that τ k (AG ) = τ k+1 (AG ). This implies τ n (AG ) = τ k (AG ) for all integers n ≥ k. Therefore Ω(τ ) = τ k (AG ), showing that Ω(τ ) is sofic. (c) Suppose that τ is∩unstable. Then the sequence (τ n (AG ))n∈N is strictly decreasing. As Ω(τ ) = n∈N τ n (AG ), we deduce from the implication (i) =⇒ (iii) in Exercise 1.55 that Ω(τ ) is not of finite type. (d) Suppose that Ω(τ ) is of finite type. By (c), there exists an integer n ≥ 1 such that Ω(τ ) = τ n (AG ). As AG is topologically mixing by Exercise 1.12(a), we deduce from Exercise 1.13 that Ω(τ ) is topologically mixing. (e) As already observed in the solution of Exercise 1.104(k), for every integer n ≥ 1, the set τ n (AZ ) is the subshift of finite type of AZ admitting the set {10k 1 : 1 ≤ k ≤ n} as a defining set of forbidden words. This clearly implies that the sequence τ n (AZ ), n ≥ 1, is strictly decreasing. Thus τ is unstable. As τ is unstable, it follows from (c) that Ω(τ ) is not of finite type. This can be seen directly since we know that Ω(τ ) consists of all sequences of 0s and 1s with at most one chain of 1s. On the other hand, we have Ω(τ ) = σ (X), where X ⊂ B Z is the subshift of finite type over the alphabet B := {0, 1, 2} admitting {10, 20, 21} as a defining set of forbidden words and σ : B Z → AZ is the cellular automaton with memory set S := {0} and local defining map μ : B S = B → A

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given by μ(0) = μ(2) := 0 and μ(1) := 1. Thus Ω(τ ) is strictly sofic. The limit set Ω(τ ) is not topologically transitive since the words defined by u := 010 and v := 110 belong to the language of Ω(τ ) but do not satisfy the condition stated in Exercise 1.76(a). █ Comment This is taken from [Hur] (see also [Maa], [BalGK]). Exercise 1.106 Let G be an infinite group and let A be a non-empty finite set. Let τ : AG → AG be a cellular automaton. Show that the following conditions are equivalent: (ii) τ is nilpotent; (ii) the limit set Ω(τ ) of τ is a singleton; (iii) the limit set Ω(τ ) of τ is finite. Solution The implications (i) =⇒ (ii) and (ii) =⇒ (iii) are obvious. Suppose that Ω(τ ) is finite. Let M be a memory set of τ and let H denote the subgroup of G generated by M. Consider the associated restriction cellular automaton τH : AH → AH . If ρ : AG → AH is the restriction map, then ρ ◦ τ = τH ◦ ρ. Therefore ρ induces a surjective map Ω(τ ) → Ω(τH ). It follows that Ω(τH ) is finite. We claim that actually Ω(τH ) is a singleton. Let us distinguish two cases. Suppose first that H is infinite. As H is finitely generated, we deduce from Exercise 1.58 that Ω(τH ) is a subshift of finite type of AH . By Exercise 1.105(d), it follows that Ω(τH ) is topologically mixing. Since Ω(τH ) is a discrete non-empty finite space, this clearly implies that Ω(τH ) is a singleton. Suppose now that H Π is finite. Recall that, using the identification AG = c∈G/H Ac associated Π with the partition of G into the right cosets of H in G, we get a factorization τ = c∈G/H τc , where each factor map τc : Ac → Ac is conjugate to τH (see [CAG, Section 1.7]). Consequently, there is a bijective map Ω(τ ) → (Ω(τH ))G/H . As H is of infinite index in G and Ω(τ ) is finite, we deduce that Ω(τH ) is also a singleton in that case. This proves our claim. The fact that Ω(τH ) is a singleton implies that τH is nilpotent by Exercise 1.104(i). Finally, the nilpotency of τH implies that of τ by Exercise 1.16. Thus (iii) =⇒ (i). █ Exercise 1.107 Let G be a group and let A be a set. Let N ⊂ G be a normal subgroup and denote by ρ : G → K := G/N the canonical quotient homomorphism. Consider the map ρ ∗ : AK → AG defined by setting ρ ∗ (y) := y ◦ρ for all y ∈ AK . (a) Show that ρ ∗ is uniformly continuous with respect to the prodiscrete uniform structures on AK and AG . (b) Let g ∈ G and y ∈ AK . Show that one has gρ ∗ (y) = ρ ∗ (ρ(g)y).

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Solution (a) Let W ⊂ AG × AG be an entourage of AG . Then there exists a finite subset Ω ⊂ G such that the basic entourage WΩ := {(x1 , x2 ) ∈ AG × AG : x1 |Ω = x2 |Ω } satisfies WΩ ⊂ W . Consider the basic entourage V of AK defined by V := {(y1 , y2 ) ∈ AK × AK : y1 |ρ(Ω) = y2 |ρ(Ω) }. We have (ρ ∗ × ρ ∗ )(V ) = {(ρ ∗ (y1 ), ρ ∗ (y2 )) ∈ AG × AG : y1 |ρ(Ω) = y2 |ρ(Ω) }

.

= {(y1 ◦ ρ, y2 ◦ ρ) ∈ AG × AG : (y1 ◦ ρ)|Ω = (y2 ◦ ρ)|Ω } ⊂ WΩ ⊂ W. This shows that ρ ∗ is uniformly continuous with respect to the prodiscrete uniform structures on AK and AG . (b) For all g ' ∈ G, we have gρ ∗ (y)(g ' ) = ρ ∗ (y)(g −1 g ' ) = y(ρ(g −1 g ' )) = y(ρ(g −1 )ρ(g ' ))

.

= y(ρ(g)−1 ρ(g ' )) = (ρ(g)y)(ρ(g ' )) = ρ ∗ (ρ(g)y)(g ' ). This shows that gρ ∗ (y) = ρ ∗ (ρ(g)y).

█

Exercise 1.108 Let G be a group and let A be a finite set. Let N ⊂ G be a normal subgroup and denote by ρ : G → K := G/N the canonical quotient homomorphism. Let ρ ∗ : AK → AG be the map defined by setting ρ ∗ (y) := y ◦ ρ for all y ∈ AK . Let Y ⊂ AK be a subshift and set X := ρ ∗ (Y ) ⊂ AG . (a) Show that X is a subshift of AG . (b) Suppose that X is of finite type. Show that Y is of finite type. Solution (a) The subshift Y ⊂ AK is closed and therefore compact. Since ρ ∗ is continuous by Exercise 1.107(a), we deduce that X = ρ ∗ (Y ) is closed in AG . Let x ∈ X and g ∈ G. Then there exists y ∈ Y such that x = ρ ∗ (y). Note that ρ(g)y ∈ Y since Y is K-invariant. The formula established in Exercise 1.107(b) gives us gx = gρ ∗ (y) = ρ ∗ (ρ(g)y) ∈ ρ ∗ (Y ) = X. We deduce that X is G-invariant. This shows that X is a subshift. (b) Let Ω ⊂ G be a memory set for X, so that A := {x|Ω : x ∈ X} ⊂ AΩ is an associated defining set of admissible patterns for X (cf. Exercise 1.48). Let us show that Y is a subshift of finite type admitting ρ(Ω) ⊂ K as a memory set and B := {y|ρ(Ω) : y ∈ Y } ⊂ Aρ(Ω) as an associated defining set of admissible patterns. Let y ∈ AK . As ρ ∗ is injective (cf. [CAG, Proposition 1.3.3]), we have y ∈ Y if and only if ρ ∗ (y) ∈ X. Therefore y|ρ(Ω) ∈ B ⇐⇒ ρ ∗ (y)|Ω = (y ◦ ρ)|Ω ∈ A .

.

(1.42)

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It follows that y ∈ Y ⇐⇒ ρ ∗ (y) ∈ X

.

⇐⇒ (gρ ∗ (y))|Ω ∈ A for all g ∈ G ⇐⇒ (ρ ∗ (ρ(g)y))|Ω ∈ A for all g ∈ G ∗

(by Exercise 1.107(b))

⇐⇒ (ρ (ky))|Ω ∈ A for all k ∈ K

(since K = ρ(G))

⇐⇒ (ky)|ρ(Ω) ∈ B for all k ∈ K

(by (1.42)).

This shows that Y is of finite type with memory set ρ(Ω) and associated defining set of admissible patterns B. █ Exercise 1.109 (Group Subshifts) Let G be a group and let A be a finite group. Equip the configuration space AG with the product group structure (thus, given two configurations x, y ∈ AG , their product is defined as the configuration xy ∈ AG given by (xy)(g) := x(g)y(g) for all g ∈ G). A subshift X ⊂ AG which is also a subgroup of AG is called a group subshift. (a) Let X ⊂ AG be a group subshift of finite type. Let Ω ⊂ G be a memory set for X and equip AΩ with the product group structure. Show that there exists a subgroup A of AΩ which is a defining set of admissible patterns for X. (b) Let H ⊂ G be a subgroup. Let Y ⊂ AH be a group subshift. Show that the set X := Y (G) = {x ∈ AG : (gx)|H ∈ Y for all g ∈ G}

.

(cf. Exercise 1.103) is a group subshift of AG . (c) Let N ⊂ G be a normal subgroup and denote by ρ : G → K := G/N the canonical quotient homomorphism. Show that the map ρ ∗ : AK → AG defined by setting ρ ∗ (y) := y ◦ ρ for all y ∈ AK is a group homomorphism. (d) Show that if Y ⊂ AK is a group subshift, then the set X := ρ ∗ (Y ) is a group subshift of AG . Solution (a) Observe that the projection map πΩ : AG → AΩ is a group homomorphism. Since X is a subgroup of AG , we deduce that πΩ (X) is a subgroup of AΩ . As πΩ (X) is a defining set of admissible patterns for X (cf. Exercise 1.48), we can take A := πΩ (X). (b) We already know that X is a subshift of AG by Exercise 1.103(a). On the : AG → AH . Clearly πH is a group other hand, consider the projection ∩ map−1πH−1 homomorphism. We have X = g∈G g πH (Y ). As Y is a subgroup of AH and the shift action of G on AG is by group automorphisms, we deduce that X is a subgroup of AG . This shows that X is a group subshift of AG .

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(c) Let y1 , y2 ∈ AK . Given g ∈ G we have ρ ∗ (y1 y2 )(g) = (y1 y2 )(ρ(g)) = y1 (ρ(g))y2 (ρ(g))

.

= ρ ∗ (y1 )(g)ρ ∗ (y2 )(g) = (ρ ∗ (y1 )ρ ∗ (y2 ))(g). We deduce that ρ ∗ (y1 y2 ) = ρ ∗ (y1 )ρ ∗ (y2 ). This shows that ρ ∗ is a group homomorphism. (d) We already know that X is a subshift of AG by Exercise 1.108(a). On the other hand, Y is a subgroup of AK and ρ ∗ is a group homomorphism by (c). Therefore X = ρ ∗ (Y ) is a subgroup of AG . This shows that X is a group subshift of AG . █ Comment Group subshifts X ⊂ AZ are called Markov subgroups in [Kit2, Section 6.3]. They were studied and classified up to topological conjugacy by Kitchens in [Kit1] (see also [Kit2, Theorem 6.3.3]). Exercise 1.110 Let G, A, and B be groups. Equip AG and B G with their product group structure and the shift action of G. Show that the following conditions are equivalent: (i) there exists a G-equivariant group isomorphism Φ : AG → B G ; (ii) the groups A and B are isomorphic. Solution Suppose (i). The map ϕA : A → AG (resp. ϕB : B → B G ) defined by setting ϕA (a) := xa (resp. ϕB (b) := yb ) where xa ∈ AG (resp. yb ∈ B G ) denotes the constant configuration defined by xa (g) := a (resp. yb (g) := b) for all g ∈ G, is an injective group homomorphism whose image is the subgroup CA ⊂ AG (resp. CB ⊂ B G ) consisting of all constant configurations. Since Φ is a G-equivariant group isomorphism and CA (resp. CB ) is the fixed point set of the shift action, Φ induces by restriction a group isomorphism CA → CB . This gives us group isomorphisms A → CA → CB → B. Thus, the groups A and B are isomorphic. The converse implication is obvious. █ Comment See [Kit1, Proposition 7]. Exercise 1.111 (Groups of Finite Markov Type) One says that a group G is of finite Markov type if for any finite group A every group subshift X ⊂ AG is of finite type. (a) Show that every finite group is of finite Markov type. (b) Show that every group of finite Markov type is finitely generated. (c) Show that every group of finite Markov type is countable. (d) Show that every subgroup of a group of finite Markov type is of finite Markov type. (e) Show that every subgroup of a group of finite Markov type is finitely generated. (f) Let F be a nonabelian free group. Show that F is not of finite Markov type. (g) Let G be a group containing a non-abelian free subgroup. Show that G is not of finite Markov type.

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(h) Show that every quotient of a group of finite Markov type is of finite Markov type. (i) Show that every group containing a finite index subgroup of finite Markov type is itself of finite Markov type. Solution (a) Let G be a finite group. Let A be a finite group and let X ⊂ AG be any subshift. Clearly X is a subshift of finite type admitting Ω := G as a memory set and A := X as an associated defining set of admissible patterns. Therefore G is of finite Markov type. (b) Let G be a group and suppose that G is not finitely generated. Let A be a nontrivial finite group (e.g. A := Z/2Z) and consider the subset X ⊂ AG consisting of all constant configurations. It is clear that X is a subgroup of AG . On the other hand, X is a subshift of AG (cf. Exercise 1.57(a)). Thus, X is a group subshift. As G is not finitely generated and A has more than one element, the subshift X is not of finite type by Exercise 1.57(e). This shows that G is not of finite Markov type. (c) This follows from (b) since every finitely generated group is countable. (d) Let G be a group of finite Markov type and let H be a subgroup of G. Let Y ⊂ AH be a group subshift. The subset X ⊂ AG defined by X := {x ∈ AG : (gx)|H ∈ Y for all g ∈ G}

.

is a subshift of AG by Exercise 1.103(a). It is clear that X is a subgroup of AG . As G is of finite Markov type, the group subshift X is of finite type. We deduce that Y is of finite type by Exercise 1.103(i). This shows that H is of finite Markov type. (e) This immediately follows from (b) and (d). (f) Pick distinct elements a, b from a free base for F . It follows from [CAG, Corollary D.5.3] that for every n ≥ 2 the subgroup H (n) of F generated by the subset {a i ba −i : i = 0, 1, . . . , n} is free of rank n. We thus have an increasing sequence H (2) ⊂ H (3) ⊂ · · · ⊂ H (n) ⊂H (n + 1) ⊂ · · · of subgroups of F . We claim that the union subgroup H := n≥2 H (n) is not finitely generated. Otherwise, there would exist a finite generating subset S ⊂ H and then for each s ∈ S we could find an integer ns ≥ 2 such that s ∈ H (ns ). Setting n := maxs∈S ns we would get S ⊂ H (n) and therefore H (n + 1) ⊂ H ⊂ H (n), a contradiction. This proves the claim. It follows from (e) that F is not of finite Markov type. (g) This is an immediate consequence of (d) and (f). (h) Let G be a group of finite Markov type, let N ⊂ G be a normal subgroup, and set K := G/N. Let A be a finite group and let Y ⊂ AK be a group subshift. Let ρ : G → K denote the canonical quotient homomorphism and consider the map ρ ∗ : AK → AG defined by setting ρ ∗ (y) := y ◦ ρ for all y ∈ AK . It follows from Exercise 1.109(d) that X := ρ ∗ (Y ) is a group subshift of AG . Since G is of finite Markov type, the group subshift X is of finite type. It then follows from Exercise 1.108(b) that Y itself is of finite type. This shows that the quotient group K = G/N is of finite Markov type.

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(i) Let G be a group and suppose that H ⊂ G is a finite index subgroup of finite Markov type. Let A be a finite group and let X ⊂ AG be a group subshift. Let T ⊂ G be a complete set of representatives for the right cosets of H in G. Then B := AT , equipped with the product group structure, is a finite group. It follows from Exercise 1.32 that, when equipping AG (resp. B H ) with its prodiscrete uniform structure and with the G-shift (resp. H -shift) action, the map Ψ : AG → B H defined by Ψ (x)(h)(t) := x(ht) for all x ∈ AG , h ∈ H , and t ∈ T is an H -equivariant uniform isomorphism. Moreover, given x1 , x2 ∈ AG , for all h ∈ H and t ∈ T we have Ψ (x1 x2 )(n)(t) = (x1 x2 )(ht) = x1 (ht)x2 (ht) = Ψ (x1 )(h)(t)Ψ (x2 )(h)(t)

.

= (Ψ (x1 )(h)Ψ (x2 )(h))(t) = (Ψ (x1 )Ψ (x2 ))(h)(t), so that Ψ (x1 x2 ) = Ψ (x1 )Ψ (x2 ). This shows that Ψ is a group homomorphism (and therefore a group isomorphism from AG onto B H ). It follows that Y := Ψ (X) ⊂ B H is a group subshift. Since H is of finite Markov type and B is a finite group, the subshift Y is of finite type. It then follows from Exercise 1.72(b), where Y is denoted X(H,T ) , that X itself is of finite type. This shows that G is of finite Markov type. █ Comment The finite Markov type property is a weakening of the Markov type property introduced by Schmidt in [Schm, Definition 4.1]. Exercise 1.112 Let G be a group. Show that the following conditions are equivalent: (i) G is of finite Markov type; (ii) G is countable and, for every finite group A, any descending sequence of group subshifts of AG X0 ⊃ X1 ⊃ X2 ⊃ · · · ⊃ Xn ⊃ Xn+1 ⊃ · · ·

.

(1.43)

eventually stabilizes, that is, there exists n0 ∈ N such that Xn = Xn0 for all n ≥ n0 . Solution Suppose first that G is of finite Markov type. Then G is countable by Exercise 1.111(c). Let A be a finite group ∩ and let (Xn )n∈N be a descending sequence of group subshifts of AG . Then X := n∈N Xn is a group subshift of AG . As G is of finite Markov type, the subshift X is of finite type. It then follows from Exercise 1.55 that the sequence (Xn )n∈N stabilizes. This shows that (i) implies (ii). Conversely, suppose (ii). Let A be a finite group and let X ⊂ AG be a group subshift. As G is countable, there exists a sequence (Ωn )n∈N of finite subsets of G such that Ωn ⊂ Ωn+1 for all n ∈ N and G = n∈N Ωn . For every n ∈ N, let πn : AG → AΩn denote the projection map and consider the subset Xn ⊂ AG consisting of all configurations x ∈ AG satisfying πn (gx) ∈ πn (X) for all g ∈ G. Clearly, Xn ⊂ AG is a group subshift of finite type admitting Ωn as a memory set

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and πn (X) ⊂ AΩn as an associated defining set of admissible patterns. Moreover, we have X ⊃ Xn ⊃ Xn+1 for all n ∈ N. This implies that (X ∩n )n∈N is a descending sequence of group subshifts of finite type and that X ⊂ n∈N Xn . On the other ∩ hand, given x ∈ n∈N Xn , there exists, for every n ∈ N, a configuration xn ∈ X such that πn (x) = πn (xn ). Since the sequence ∩ (Ωn )n∈N exhausts G and X is closed in AG , we deduce that x ∈ X. Thus, X = n∈N Xn . By (ii), the sequence (Xn )n∈N stabilizes. Therefore, there exists n0 ∈ N such that X = Xn0 . It follows that X is of finite type. This shows that (ii) implies (i). █ Exercise 1.113 Let G be a group. Suppose that there exists a normal subgroup H ⊂ G which is of finite Markov type and such that K := G/H is infinite cyclic. The purpose of this exercise is to show that G is itself of finite Markov type. Let A be a finite group and let X ⊂ AG be a group subshift. Let ρ : G → K denote the canonical quotient homomorphism and let t ∈ G such that ρ(t) generates K. (a) Show that for every n ∈ N the set Yn := {x|H : x ∈ X such that x(ht i ) = 1A for all h ∈ H and − n ≤ i ≤ −1} ⊂ AH (1.44)

.

is a group subshift of AH . (b) Show that Y0 ⊃ Y1 ⊃ · · · ⊃ Yn ⊃ Yn+1 ⊃ · · · and that the sequence (Yn )n∈N stabilizes, that is, there exists n0 ∈ N such that Yn = Yn0 for all n ≥ n0 . j (c) For i, j ∈ Z with i ≤ j , set Si := {t n : i ≤ n ≤ j } ⊂ G. Show that j +1

j

tH Si = H Si+1 .

.

(1.45) j

(d) Given i, j ∈ Z with i ≤ j , let XH S j := {x|H S j : x ∈ X} ⊂ AH Si . Let i i n0 ∈ N be the integer provided by (b) and, for m ∈ N such that m ≥ n0 , set m m X(m) := {x ∈ AG : (gx)|H S−n ∈ XH S−n for all g ∈ G} ⊂ AG .

.

0

0

(1.46)

Let m ∈ N such that m ≥ n0 and let z ∈ X(m). Prove successively the following results: m m ; (1) there exists x0 ∈ X such that x0 |H S−n = z|H S−n 0

0

m m ; (2) there exists x1 ∈ X such that x1 |H S−n = (t −1 z)|H S−n 0 0 (3) the configuration x2 := tx1 is in X and satisfies that x2 |H S m+1

−n0 +1

= z|H S m+1 ; −n0 +1

(4) the configuration x3 := x0−1 x2 is in X and satisfies that x3 (g) = 1A for all m g ∈ H S−n ; 0 +1 (5) the configuration y := (t −(m+1) x3 )|H ∈ AH is in Yn0 ;

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(6) the set L(y) := {x ∈ X : x|H = y and x(ht i ) = 1A for all h ∈ H and i ≤ −1} ⊂ X

.

is nonempty; (7) given x4 ∈ L(y), the configuration x5 := x0 · (t m+1 x4 )−1 is in X and satisfies that x5 |H S m+1 = z|H S m+1 ; −n0

(8) X(m) = X(m + 1).

−n0

(e) Let z0 ∈ X(n0 ). Show that, given a finite subset F ⊂ G, there exists x ∈ X such that x|F = z0 |F . (f) Show that X = X(n0 ). n0 (g) Set S := S−n and consider the finite group B := AS . Show that the map 0 Θ : AG → B H defined by setting Θ(x)(h)(s) := x(hs) for all x ∈ AG , h ∈ H , and s ∈ S is an H -equivariant continuous group homomorphism. (h) Show that Z := Θ(X) ⊂ B H is a group subshift and that for x ∈ AG one has x ∈ X if and only if Θ(gx) ∈ Z for all g ∈ G. (i) Show that X is of finite type. (j) Show that the group G is of finite Markov type. Solution (a) Let n ∈ N. Consider the set Xn := {x ∈ X : x(ht i ) = 1A for all h ∈ H and − n ≤ i ≤ −1} ⊂ AG .

.

Given x ∈ Xn and k ∈ H , we have (kx)(ht i ) = x(k −1 ht i ) = 1A for all h ∈ H and −n ≤ i ≤ −1, showing that kx ∈ Xn . We deduce that Xn is H -invariant. Since the projection maps πg : AG → A, defined by setting πg (x) := x(g) for all x ∈ AG and g ∈ G, are continuous group homomorphisms and X is a closed subgroup of AG , we have that Xn = X

∩

(

−1 ∩ ∩

.

) −1 πht i ({1A })

h∈H i=−n

is a closed subgroup of AG . Since the projection map π : AG → AH is an H equivariant continuous group homomorphism and AG is compact, we deduce that Yn = π(Xn ) is a group subshift of AH . (b) We have that Xn+1 = {x ∈ Xn : x(ht −(n+1) ) = 1A for all h ∈ H } ⊂ Xn so that Yn+1 = π(Xn+1 ) ⊂ π(Xn ) = Yn for all n ∈ N. Using (a), we deduce that (Yn )n∈N is a decreasing sequence of group subshifts of AH . Since H is of finite Markov type, it follows from Exercise 1.112 that the sequence (Yn )n∈N eventually stabilizes. Thus there exists n0 ∈ N such that Yn0 =

∩

.

n∈N

Yn .

(1.47)

1.2 Exercises

111 j

(c) Let i, j ∈ Z. As H is normal in G, we have tH = H t, so that tH Si = tH {t n : i ≤ n ≤ j } = H t{t n : i ≤ n ≤ j } = H {t n+1 : i ≤ n ≤ j } = H {t n : j +1 i + 1 ≤ n ≤ j + 1} = H Si+1 . (d) m m (1) The existence of x0 ∈ X such that x0 |H S−n = z|H S−n follows from (1.46) 0 0 after taking g := 1G . m m (2) The existence of x1 ∈ X such that x1 |H S−n = (t −1 z)|H S−n follows from 0

0

(1.46) after taking g := t −1 . (3) We have x2 = tx1 ∈ X, since X is G-invariant. Moreover, x2 (tht n ) = −1 (t x2 )(ht n ) = x1 (ht n ) = (t −1 z)(ht n ) = z(tht n ) for all h ∈ H and −n0 ≤ n ≤ m, m m . It follows from (1.45), with i := −n0 and j := m, so that x2 |tH S−n = z|tH S−n 0 0 that x2 |H S m+1 = z|H S m+1 . −n0 +1

−n0 +1

(4) The configuration x3 = x0−1 x2 is in X since X is a subgroup of AG . Moreover, by using (1) and (3), we deduce that x3 (g) = x0 (g)−1 x2 (g) = z(g)−1 z(g) = 1A for m ∩ H S m+1 = H S m all g ∈ H S−n −n0 +1 . −n0 +1 0 −1 (5) If g ∈ H S−n0 , then, keeping in mind (1.45), we have that g ' := m m m+1 g ∈ H S−n ⊂ H S−n . Therefore, by using (4), we deduce that t 0 +m+1 0 +1 −(m+1) m+1 x3 )(g) = x3 (t g) = x3 (g ' ) = 1A . This shows that the configuration (t −(m+1) y := (t x3 )|H is in Yn0 (cf. (1.44)). (6) For every n ≥ n0 , we have y ∈ Yn0 = Yn . Therefore, we can find zn ∈ X such that zn |H = y and zn (ht i ) = 1A for all h ∈ H and −n ≤ i ≤ −1. Since X is compact, the sequence (zn )n≥n0 admits a cluster point in X. Any cluster point x4 ∈ X of the sequence (zn )n≥n0 satisfies x4 ∈ L(y). (7) The configuration x5 := x0 · (t m+1 x4 )−1 is in X since X is a group subshift of AG . Moreover, let h ∈ H and set h' := t −(m+1) ht m+1 ∈ H . Then, for −n0 ≤ n ≤ m, we have x5 (ht n ) = x0 (ht n )((t m+1 x4 )(ht n ))−1 = x0 (ht n )x4 (h' t n−m−1 )−1

.

= x0 (ht n ) = z(ht n ),

(1.48)

where we used the fact that, on the one hand, n − m − 1 ≤ −1 and x4 (h' t i ) = 1A m m for all i ≤ −1, and, on the other hand, x0 |H S−n = z|H S−n (cf. (1)). Finally, with 0 0 the previous notation, x5 (ht m+1 ) = x5 (t m+1 h' )

.

= x0 (t m+1 h' )((t m+1 x4 )(t m+1 h' ))−1 = x0 (t m+1 h' )x4 (h' )−1 = x0 (t m+1 h' )((t −(m+1) x3 )(h' ))−1 = x0 (t m+1 h' )x3 (t m+1 h' )−1

(since x4 |H = y = (t −(m+1) x3 )|H )

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1 Cellular Automata

= x2 (t m+1 h' )

(since x3 = x0−1 x2 )

= x2 (ht m+1 ) = z(ht m+1 )

(by (2)).

Keeping in mind (1.48) for −n0 ≤ n ≤ m, this shows that x5 |H S m+1 = z|H S m+1 . −n0

−n0

m ⊂ H S m+1 . For the (8) The inclusion X(m + 1) ⊂ X(m) is obvious since H S−n −n0 0 reverse inclusion, let x ∈ X(m) and let g ∈ G. Setting z := gx ∈ X(m), it follows from (7) that we can find x5 ∈ X such that

(gx)|H S m+1 = z|H S m+1 = x5 |H S m+1 ∈ XH S m+1 .

.

−n0

−n0

−n0

−n0

As g ∈ G was arbitrary, this shows that x ∈ X(m + 1). Thus, X(m) ⊂ X(m + 1) and the equality X(m) = X(m + 1) follows. (e) First observe that by (d)(8) we have X(n0 ) = X(n0 +1) = · · ·  = X(m−1) = X(m) for all m ≥ n0 . Let now F ⊂ G be a finite subset. Since G = n∈Z H t n , we j j can find i, j ∈ Z, with i ≤ j , such that F ⊂ n=i H t n = H Si . Up to replacing j by j + 2n0 , if necessary, we may suppose that j − i ≥ 2n0 . Thus, setting m := m . Since X(n ) j −i −n0 , we have m ≥ n0 and, by virtue of (1.45), t −n0 −i F ⊂ H S−n 0 0 is G-invariant, we have z := t i−n0 z0 ∈ X(n0 ) = X(m). It follows from (d)(1) that m m . Setting x := t n0 −i x0 ∈ X and we can find x0 ∈ X such that x0 |H S−n = z|H S−n 0

0

h' := t −n0 +i ht n0 −i ∈ H , we have, for all i ≤ n ≤ j ,

x(ht n ) = (t n0 −i x0 )(ht n ) = x0 (t −n0 +i ht n ) = x0 (h' t n−n0 +i )

.

= z(h' t n−n0 +i ) = z(t −n0 +i ht n ) = (t n0 −i z)(ht n ) = z0 (ht n ). We deduce that x|H S j = z0 |H S j so that, in particular, x|F = z0 |F . i i (f) We clearly have X ⊂ X(n0 ). On the other hand, it follows from (e) that given z0 ∈ X(n0 ) and any finite set F ⊂ G, there exists x ∈ X such that x|F = z0 |F . Since X is closed for the prodiscrete topology, this shows that z0 ∈ X, and proves the inclusion X(n0 ) ⊂ X. We deduce that X = X(n0 ). (g) Let x ∈ X, let h, k ∈ H , and let s ∈ S. We have Θ(hx)(k)(s) = (hx)(ks) = x(h−1 ks) = Θ(x)(h−1 k)(s) = (hΘ(x))(k)(s), so that Θ(hx) = hΘ(x). This shows that Θ is H -equivariant. The fact that Θ is a group homomorphism is clear. Finally, let x1 ∈ AG and set y1 := Θ(x1 ) ∈ B H . Let W ⊂ B H be a neighborhood of y1 . Then we can find a finite subset Ω ⊂ H such that the basic neighborhood WΩ := {y ∈ B H : y|Ω = y1 |Ω } satisfies WΩ ⊂ W . Consider the basic neighborhood V of x1 in AG defined by V := {x ∈ AG : x|ΩS = x1 |ΩS }. If x ∈ V we have Θ(x)(h)(s) = x(hs) = x1 (hs) = Θ(x1 )(h)(s) for all h ∈ Ω and s ∈ S, so that Θ(x)|Ω = Θ(x1 )|Ω = y1 |Ω . This shows that Θ(V ) ⊂ WΩ ⊂ W . We deduce that Θ is continuous with respect to the prodiscrete topologies on AG and B H . This shows that Θ is an H -equivariant continuous group homomorphism.

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(h) The fact that Z := Θ(X) ⊂ B H is a group subshift follows immediately from (g) and compactness of X. On the other hand, let x ∈ AG . Since X is Ginvariant, we have Θ(gx) ∈ Θ(X) = Z for all g ∈ G. Conversely, suppose that Θ(gx) ∈ Z for all g ∈ G. This means that given g ∈ G there exists z ∈ X such that Θ(gx) = Θ(z). Thus, (gx)(hs) = Θ(gx)(h)(s) = Θ(z)(h)(s) = z(hs) for all h ∈ H and s ∈ S, that is, (gx)|H S = z|H S ∈ XH S . As g ∈ G was arbitrary and n0 recalling that S = S−n , it follows from (f) that x ∈ X(n0 ) = X. 0 (i) Since H is of finite Markov type and B is a finite group, there exist a finite subset ΩH ⊂ H and a subgroup B ⊂ B ΩH such that Z ⊂ B H is the group subshift of finite type with memory set ΩH and with B as an associated set of admissible patterns. Let then x ∈ AG . By using (h), we have that x ∈ X if and only if zg := Θ(gx) ∈ Z for all g ∈ G. Keeping in mind that Θ is H -equivariant, the latter is then satisfied if and only if zg |ΩH ∈ B for all g ∈ G. Set Ω := ΩH S ⊂ G and A := ψ −1 (B) ⊂ AΩ , where ψ : AΩ → B ΩH is the group isomorphism defined by setting ψ(p)(h)(s) := p(hs) for all p ∈ AΩ , h ∈ ΩH , and s ∈ S. Then, arguing verbatim as in the solution to Exercise 1.72(b) with the subshift X(H,T ) ⊂ B H replaced by Z ⊂ B H , we deduce that x ∈ X if and only if (gx)|Ω = ψ −1 (zg |ΩH ) ∈ ψ −1 (B) = A for all g ∈ G. This shows that X is the group subshift of finite type with memory set Ω and A as a finite set of admissible patterns. (j) This follows from (i). █ Comment This is inspired by [CecCP2, Lemma 6.3]. See also [Schm, Lemma 4.4] for an alternative proof. Exercise 1.114 Show that every finitely generated abelian group is of finite Markov type. Solution Let G be a finitely generated abelian group and let us show that G is of finite Markov type. By the structure theorem for finitely generated abelian groups, there exist d ∈ N and a finite abelian group F such that G is isomorphic to Zd × F . This implies that G contains a finite index subgroup isomorphic to Zd . Thus, by Exercise 1.111(i), we can assume G = Zd . The result then easily follows from Exercise 1.113(j) by induction on d. Indeed, for d = 0, the group Zd is trivial and hence of finite Markov type by Exercise 1.111(a). On the other hand, if Zd is of finite Markov type for some d ∈ N, then Zd+1 is itself of finite Markov type by (j) since Zd+1 ∼ █ = Zd × Z. Comment The fact that infinite cyclic groups are of finite Markov type was established by Kitchens in [Kit1, Proposition 4] (see also [Kit2, Lemma 6.3.5] and [LinM, Exercise 2.1.11]). It was subsequently observed that this property also holds for all finitely generated abelian groups by Kitchens and Schmidt [KitS, Remark 3.10.(2)]. See Exercise 4.37 for a more general result.

Chapter 2

Residually Finite Groups

This chapter is mainly devoted to residual finiteness of groups, monoids, and rings, and to the Hopf property for groups. It includes a discussion of the profinite, pronilpotent, prosolvable, and proamenable completions of a group. It also provides an entirely self-contained proof of the failure of the Hopf property for the BaumslagSolitar group .BS(2, 3). The stability of residual finiteness and Hopficity under group constructions such as semi-direct products and wreath products is discussed. FCgroups, ICC-groups, and the transfer homomorphism are also investigated.

2.1 Summary 2.1.1 Equivalent Definitions of Residual Finiteness A group G is called residually finite if for each element g in G with .g /= 1G , there exist a finite group F and a group homomorphism .φ : G → F such that .φ(g) /= 1F . The following conditions on a group G are equivalent (cf. Propositions 2.1.2, 2.1.11, Corollary 2.2.6, and Theorem 2.7.1 in [CAG]): (RF1) (RF2) (RF3)

(RF4) (RF5)

G is residually finite; for all .g, h in G with .g /= h, there exist a finite group F and a group homomorphism .φ : G → F such that .φ(g) /= φ(h); for each finite subset .K ⊂ G, there exist a finite group F and a group homomorphism .φ : G → F such that the restriction .φ|K of .φ to K is injective; the intersection of all finite index subgroups of G is reduced to the identity element of G; the intersection of all finite index normal subgroups of G is reduced to the identity element of G;

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 T. Ceccherini-Silberstein, M. Coornaert, Exercises in Cellular Automata and Groups, Springer Monographs in Mathematics, https://doi.org/10.1007/978-3-031-10391-9_2

115

116

(RF6) (RF7) (RF8)

(RF9)

2 Residually Finite Groups

there exists a family .(Fi )i∈I of finite groups such Π that the group G is isomorphic to a subgroup of the direct product group . i∈I Fi ; for every set A, the set of configurations in .AG which have a finite orbit under the G-shift is dense in .AG for the prodiscrete topology; there exists a set A having at least two distinct elements such that the set of configurations in .AG which have a finite orbit under the G-shift is dense in G for the prodiscrete topology; .A there exists a Hausdorff topological space X equipped with a continuous and faithful action of G such that the set of points of X which have a finite G-orbit is dense in X.

2.1.2 The Class of Residually Finite Groups The class of residually finite groups is closed under taking subgroups, direct products, direct sums, and projective limits (cf. Proposition 2.2.1, Proposition 2.2.2, Corollary 2.2.3, and Proposition 2.2.7 in [CAG]). Moreover, every virtually residually finite group, i.e., every group containing a finite index subgroup which is residually finite, is itself residually finite [CAG, Proposition 2.2.12]. All finite groups, all finitely generated abelian groups, all profinite groups (i.e., limits of projective systems of finite groups), and all free groups are residually finite (cf. Proposition 2.1.3, Corollary 2.2.4, Corollary 2.2.8, and Theorem 2.3.1 in [CAG]). By a theorem of Gilbert Baumslag, the automorphism group of a finitely generated residually finite group is itself residually finite (cf. [CAG, Theorem 2.5.1]).

2.1.3 Divisible Groups A group G is called divisible if for each .g ∈ G and each integer .n ≥ 1, there is an element .h ∈ G such that .hn = g. For example, the additive groups .Q, .R, and .C are divisible. More generally, every .Q-vector space, with its additive underlying group structure, is divisible. In particular, every field of characteristic 0, with its underlying additive group structure, is divisible. A non-trivial divisible group cannot be residually finite ([CAG, Proposition 2.1.8]).

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2.1.4 Hopfian Groups A group G is called Hopfian if every surjective endomorphism of G is injective. Every simple group is Hopfian. The additive group .Q is Hopfian. By a theorem of Anatoly Mal’cev, every finitely generated residually finite group is Hopfian (cf. [CAG, Theorem 2.4.3]).

2.2 Exercises Exercise 2.1 (The Residual Subgroup) Let G be a group and let G0 ⊂ G denote its residual subgroup. Recall that G0 is defined as being the intersection of all the finite index subgroups of G and that G0 is a normal subgroup of G since it is also equal to the intersection of all the normal subgroups of finite index of G (cf. [CAG, Proposition 2.1.11.(ii)]). (a) Show that the group G/G0 is residually finite. (b) Let φ : G → G be a group homomorphism from G to a residually finite group G . Show that G0 ⊂ ker(φ). (c) Suppose that an element x ∈ G satisfies the following condition: for every integer n ≥ 1, there exists y ∈ G such that x = y n . Show that x belongs to the residual subgroup of G. (d) Show that if G is a divisible group then G0 = G. (e) Show that the following conditions are equivalent: (1) G0 = G, (2) G contains no proper subgroup of finite index, (3) G contains no proper normal subgroup of finite index, (4) all finite quotients of G are trivial, (5) all residually finite quotients of G are trivial. Solution (a) Suppose that c ∈ G/G0 \{1G/G0 }. Denoting by π : G → G/G0 the canonical group homomorphism, this means that there exists g ∈ G \ G0 such that π(g) = c. Since G0 is the intersection of all the finite index normal subgroups of G (see [CAG, Proposition 2.1.11.(i)]), we can find a finite index normal subgroup K of G such that g ∈ / K. Denote by ψ : G → G/K the canonical group homomorphism. As G0 ⊂ K = ker(ψ), there is a group homomorphism φ : G/G0 → G/K such that / K. As G/K is finite, this ψ = φ ◦ π . We then have φ(c) = ψ(g) /= 1G/K since g ∈ shows that G/G0 is residually finite. (b) Let F denote the set consisting of all finite index subgroups of G . The preimage φ −1 (H ) of every H ∈ F is a finite index subgroup of G. It follows that G0 ⊂

∩

.

H ∈F

φ −1 (H ) = φ −1 (

∩ H ∈F

H ).

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2 Residually Finite Groups

As ∩ .

H = {1G }

H ∈F

since G is residually finite, we deduce that G0 ⊂ φ −1 (1G ) = ker(φ). (c) Let K be a finite index normal subgroup of G and let n := [G : K] = |G/K|. By hypothesis, there exists y ∈ G such that x = y n . Consider the canonical group homomorphism π : G → G/K. We then have π(x) = π(y n ) = π(y)n = 1G/K . Therefore x ∈ K. As K was arbitrary, this shows that x ∈ G0 . (d) This immediately follows from (c). (e) The equivalence (1) ⇐⇒ (2) is immediate from the definition of G0 . The equivalence (1) ⇐⇒ (3) follows from the fact that G0 is equal to the intersection of the normal subgroups of finite index of G. The equivalence (3) ⇐⇒ (4) as well as the implication (5) =⇒ (4) are obvious. Finally, the implication (1) =⇒ (5) follows from (b). ■ Exercise 2.2 (Maximal Normal Subgroups) One says that a subgroup H of a group G is a maximal normal subgroup of G if H is a proper normal subgroup of G and every proper normal subgroup K of G such that H ⊂ K satisfies H = K. (a) Let G be a group and let H be a maximal normal subgroup of G. Show that the quotient G/H is a non-trivial simple group. (b) Show that every non-trivial finitely generated group admits a maximal normal subgroup. (c) Let G be a non-trivial finitely generated group admitting no proper subgroups of finite index. Show that G admits an infinite simple quotient group. (d) Show that the additive group Q has no maximal normal subgroups. Solution (a) By the third isomorphism theorem, every normal subgroup of G/H is of the form K/H where K is a normal subgroup of G such that H ⊂ K. As H is a maximal normal subgroup of G, this implies K = H or K = G. Therefore G/H is a simple group. It is not trivial since H is a proper subgroup of G. (b) Suppose that G is a non-trivial finitely generated group and let {g1 , . . . , gn } be a finite generating subset of G. Consider the set S consisting of all proper normal subgroups of G partially ordered by inclusion. We first observe that S is not empty since {1 G } ∈ S . Suppose now that C ⊂ S is a totally ordered subset of S . Then K := H ∈C H is a normal subgroup of G. We claim that K is a proper subgroup of G. Indeed, suppose by contradiction that K = G. Then, for each 1 ≤ i ≤ n, we have gi ∈ K so that there exists Hi ∈ C such that gi ∈ Hi . As C is a chain, there exists H ∈ {H1 , . . . , Hn } such that Hi ⊂ H for all 1 ≤ i ≤ n. We then have gi ∈ H for all 1 ≤ i ≤ n and hence H = G, a contradiction since H is a proper subgroup of G. Therefore K ∈ S is an upper bound for C . By applying Zorn’s lemma, we deduce that S admits a maximal element. Such a maximal element is a maximal normal subgroup of G.

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(c) As G is a non-trivial finitely generated group, it admits a maximal normal subgroup H by (b). The quotient G/H is then a simple group by (a). It is infinite since, by hypothesis, G admits no non-trivial finite quotients. (d) Suppose by contradiction that Q has a maximal normal subgroup H . Then the quotient Q/H is a non-trivial simple group by (a). As Q/H is abelian, this implies that Q/H is a finite cyclic group. This yields a contradiction since all finite quotients of Q are trivial. ■ Comment In 1951, Higman [Hig3] proved that the finitely presented group G := is not trivial and has no proper finite index subgroups (see also Proposition 6 in Section 1.4 of [Ser3]). Using (c), Higman deduced that there is a infinite simple group that is a quotient of G, thus proving the existence of infinite finitely generated simple groups. The Thompson groups V and T provide examples of infinite finitely presented simple groups (see [CanFP] and [Hig4]). In [BurgM], Burger and Mozes constructed torsion-free infinite finitely presented simple groups. Exercise 2.3 Show that every quotient of a divisible group is a divisible group. Solution Let φ : G → G be a surjective group homomorphism of a divisible group G onto a group G . Let g  ∈ G and let n ≥ 1 be an integer. Since φ is surjective, we can find an element g ∈ G such that φ(g) = g  . Since G is divisible, we can find an element h ∈ G such that hn = g. Then the element h := φ(h) ∈ G satisfies that ■ (h )n = φ(h)n = φ(hn ) = φ(g) = g  . This shows that G is divisible. Exercise 2.4 Recall that a group G is said to be torsion-free if 1G is the unique element of finite order in G. Show that every torsion-free divisible abelian group G is isomorphic to a direct sum of copies of Q. Solution Let G be a torsion-free abelian group. We shall use additive notation for the group operation on G. Let g ∈ G. Since G is divisible, for every integer n ≥ 1, there exists h ∈ G such that nh = g. Such an h is unique since G is torsion-free. 1 Thus we can write g := h. Clearly the map Q × G → G given by (q, g) |→ qg, n 1 m where qg := m( g) for q = with m, n ∈ Z and n ≥ 1 is well defined and yields, n n together with the additive group operation on G, a Q-vector space structure on G. If B ⊂ G is a basis for the Q-vector space G, then the group G is isomorphic to the direct sum ⊕B Q. ■ Exercise 2.5 (Profinite Topology) Let G be a group. (a) Show that it is possible to define a topology on G by taking as open sets the subsets Ω ⊂ G satisfying the following property: for each g ∈ Ω there is a subgroup of finite index H ⊂ G such that gH ⊂ Ω. This topology is called the profinite topology on G. (b) Let g ∈ G and let H be a finite index subgroup of G. Show that the set gH is both open and closed in G for the profinite topology. (c) Show that G with its profinite topology is a topological group.

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(d) Show that the group G is finite if and only if the profinite topology on G is discrete. (e) Show that G is residually finite if and only if the profinite topology on G is Hausdorff. Solution (a) Let T denote the set of all subsets Ω ⊂ G satisfying the given property. We have ∅ ∈ T since the corresponding condition is void (there is no g ∈ ∅). On the other hand, G ∈ T since for every g ∈ G one may take H = G. Let now (Ωi )i∈I be a (finite or infinite) family of elements in T and set Ω := ∪i∈I Ωi . Let g ∈ Ω. Then there exist an element i0 ∈ I such that g ∈ Ωi0 and a subgroup of finite index H0 ⊂ G such that gH0 ⊂ Ωi0 . As Ωi0 ⊂ Ω, it follows that gH0 ⊂ Ω. This shows that Ω ∈ T . Finally, let Ω1 , Ω2 ∈ T and set Ω := Ω1 ∩ Ω2 . Let g ∈ Ω. Then for i = 1, 2 there exists a subgroup of finite index Hi ⊂ G such that gHi ⊂ Ωi . The subgroup H := H1 ∩ H2 is also of finite index in G and satisfies that gH = g(H1 ∩ H2 ) = gH1 ∩ gH2 ⊂ Ω. We deduce that Ω ∈ T . This shows that T is a topology on G. (b) Let x ∈ gH . This means that there exists h ∈ H such that x = gh. We then have xH = ghH = gH . As H is of finite index in G, this shows that gH is open for the profinite topology on G. As G \ gH is an union of left cosets of H and hence open in G by the previous result, we deduce that gH is also closed in G. (c) It suffices to show that the map ϕ : G × G → G, defined by ϕ(g, h) = gh−1 for all g, h ∈ G, is continuous when G is equipped with its profinite topology and G×G with the product topology. Let U be a open subset of G. Let (g, h) ∈ ϕ −1 (U ). This means that gh−1 ∈ U . As U is open in G, there is a finite index subgroup H ⊂ G such that gh−1 H ⊂ U . Let K be a finite index normal subgroup of G such that K ⊂ H (cf. [CAG, Lemma 2.1.10]). Then gK and hK are open subsets of G by (b), so that (gK) × (hK) is an open neighborhood of (g, h) in G × G. Moreover, we have ϕ((gK) × (hK)) = gK(hK)−1 = gKK −1 h−1 = gKh−1 = gh−1 K ⊂ gh−1 H ⊂ U . This shows that ϕ −1 (U ) is open in G × G. Thus ϕ is continuous. (d) Suppose first that G is finite. Then the trivial subgroup {1G } is of finite index in G. We deduce from (b) that, for all g ∈ G, the singleton {g} is both open and closed in G for the profinite topology. This shows that the profinite topology on G is discrete. Conversely, suppose that the profinite topology on G is discrete. This implies that {1G } is open in G. By definition of the open sets in the profinite topology, it follows that there is a subgroup of finite index H of G such that H ⊂ {1G }. This implies that {1G } is of finite index in G, that is, G is finite. (e) Suppose first that G is residually finite. Let g1 , g2 ∈ G be two distinct elements. Since G is residually finite, we can find a finite group F and a group homomorphism φ : G → F such that φ(g1 ) /= φ(g2 ). As F is finite, the kernel H of φ is of finite index subgroup of G. Setting Ωi := gi H for i = 1, 2, we have Ωi ∈ T , gi ∈ Ωi , and Ω1 ∩ Ω2 = ∅. Thus Ω1 and Ω2 are disjoint neighborhoods of g1 and g2 , respectively. This shows that the topology T is Hausdorff.

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Conversely, suppose that T is Hausdorff. Let g1 , g2 ∈ G with g1 /= g2 . Since T is Hausdorff, we can find two open subsets Ω1 , Ω2 ∈ T such that gi ∈ Ωi and Ω1 ∩ Ω2 = ∅. By definition of the profinite topology, there exist finite index subgroups Hi ⊂ G such that gi Hi ⊂ Ωi for i = 1, 2. Then the subgroup H1 ∩ H2 is also of finite index in G. Therefore there exists a finite index normal subgroup K of G such that K ⊂ H1 ∩ H2 (cf. [CAG, Lemma 2.1.10]). The quotient group F := G/K is finite since K is of finite index in G. On the other hand, as gi K ⊂ gi Hi ⊂ Ωi and Ω1 ∩ Ω2 = ∅, the canonical group homomorphism φ : G → F satisfies that φ(g1 ) /= φ(g2 ). This shows that G is residually finite. ■ Exercise 2.6 Let α, β ∈ C such that |α|, |β| ≥ 2. Consider the matrices A, B ∈ SL2 (C) defined by ⎛

1α .A := 01

⎞

⎛

and

⎞ 10 B := . β1

Show that A and B generate a free subgroup of rank 2 of SL2 (C). Solution The group SL2 (C) naturally acts on the complex projective line P1 (C) = C ∪ {∞}. This action is given by the formula Mz =

.

az + b cz + d

⎛ ⎞ ab ∈ SL2 (C) and z ∈ P1 (C) (with the usual calculus conventions cd regarding ∞). Consider the non-empty disjoint subsets Y, Z ⊂ P1 (C) defined by

for M =

Y := {z ∈ C : |z| < 1}

.

and

Z := {z ∈ P1 (C) : |z| > 1} = {z ∈ C : |z| > 1}∪{∞}.

Let k ∈ Z \ {0}. For all z ∈ P1 (C), we have Ak z = z + kα and hence |Ak z| ≥ |k||α| − |z|. As |α| ≥ 2, we deduce that Ak Y ⊂ Z. On the other hand, we have B k z = 1/(kβz +1) for all z ∈ P1 (C). As |kβz +1| ≥ |k||β||z| − 1 and |β| ≥ 2, we deduce that B k Z ⊂ Y . By the Klein Ping-Pong theorem [CAG, Theorem D.5.1], this shows that the matrices A and B generate a free subgroup of rank 2 of SL2 (Z). ■ Comment By taking α = β = 2, we recover Lemma 2.3.2 in [CAG]. Exercise 2.7 (Residual Finiteness of Free Groups) The goal of this exercise is to present an alternative proof of the fact that all free groups are residually finite (cf. [CAG, Theorem 2.3.1]). Let X be a set and let G denote the free group based on X. Let g ∈ G with g /= 1G and let g = an · · · a2 a1

.

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denote the reduced form of G. Recall that one has n ≥ 1, ai ∈ X ∪ X−1 for all i ∈ {1, . . . , n}, and ai ai−1 /= 1G for all i ∈ {2, . . . , n}. For each x ∈ X, define the subsets Ex+ , Ex− ⊂ {1, . . . , n} by Ex+ := {i : ai = x} and Ex− := {i : ai = x −1 }.

.

Define also the subsets Fx+ , Fx− ⊂ {2, . . . , n + 1} by Fx+ := {i + 1 : i ∈ Ex+ } and Fx− := {i + 1 : i ∈ Ex− }.

.

(a) Show that Ex+ ∩ Fx− = Fx+ ∩ Ex− = ∅.

.

for all x ∈ X. (b) Let x ∈ X. Show that there exists a permutation σx ∈ Symn+1 such that σx (i) = i + 1 for all i ∈ Ex+ and σx−1 (i) = i + 1 for all i ∈ Ex− . (c) Show that there exists a group homomorphism φ : G → Symn+1 such that φ(x) = σx for all x ∈ X. (d) Compute φ(g)(1) and conclude. Solution (a) We have i ∈ Ex+ ∩ Fx− if and only if ai = x and ai−1 = x −1 . Similarly, we have i ∈ Fx+ ∩ Ex− if and only if ai−1 = x and ai = x −1 . As ai ai−1 /= 1G for all i ∈ {2, . . . , n}, we deduce that Ex+ ∩ Fx− = Fx+ ∩ Ex− = ∅. (b) By (a), there exists σx ∈ Symn+1 (in general not unique) such that σx (i) = i + 1 for all i ∈ Ex+ and σx (i + 1) = i, or equivalently σx−1 (i) = i + 1, for all x ∈ Ex− . (c) This follows from the universal property of free groups (cf. [CAG, Definition D.2.1]). (d) For each i ∈ {1, . . . , n}, there is a unique x ∈ X such that ai ∈ {x, x −1 }. If i ∈ Ex+ , then ai = x and hence φ(ai )(i) = σx (i) = i + 1. Otherwise, i ∈ Ex− and hence φ(ai )(i) = σx−1 (i) = i + 1. Consequently, we have φ(ai )(i) = i + 1 for all i ∈ {1, . . . , n}. As φ(g) = φ(an · · · a2 a1 ) = φ(an ) · · · φ(a2 )φ(a1 ),

.

we deduce that φ(g)(1) = n + 1. In particular, φ(g) /= 1Symn+1 . As the group Symn+1 is finite, this shows that the group G is residually finite. ■ Exercise 2.8 Show that the class of residually finite groups is not closed under taking quotients. Solution Let G be a group that is not residually finite (e.g. the additive group Q). Then there exist a free group F and a surjective group homomorphism F → G

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(cf. [CAG, Corollary D.4.2]). As every free group is residually finite, this shows that the class of residually finite groups is not closed under taking quotients. ■ Exercise 2.9 Let X be an infinite set. Show that the symmetric group Sym(X) is not residually finite. Solution As X is infinite, it contains an infinite countable subset. Therefore Sym(X) contains a subgroup isomorphic to Sym(Q). On the other hand, by Cayley’s theorem (cf. [CAG, Theorem C.1.2]), the group Sym(Q) contains a subgroup isomorphic to the addtive group Q. As Q is not residually finite and the class of residually finite groups is closed under taking subgroups, this shows that Sym(X) is not residually finite. ■ Comment As an alternative proof, one can use the fact that the subgroup Sym0 (X) ⊂ Sym(X) consisting of all permutations of X with finite support is not residually finite (cf. Lemma 2.6.3 in [CAG]). Exercise 2.10 Let G be a residually finite group and let A be a finite set with more than one element. Let ICA(G; A) denote the group consisting of all reversible cellular automata τ : AG → AG (cf. [CAG, Section 1.10]). Show that the group ICA(G; A) is residually finite if and only if G is residually finite. Solution Suppose first that G is residually finite. Denote by S Π the set consisting of all finite index subgroups of G and consider the group P := H ∈S Sym(Fix(H )), where Fix(H ) ⊂ AG denotes the set of all configurations that are fixed by H . Let τ ∈ ICA(G; A). For every H ∈ S , the set Fix(H ) is finite since it is in bijection with AH \G (cf. [CAG, Corollary 1.3.4]). As τ (Fix(H )) ⊂ Fix(H ) by the G-equivariance of τ , we deduce that τ induces by restriction an element τH ∈ Sym(Fix(H )). We get a group homomorphism ϕ : ICA(G; A) → P by setting ϕ(τ ) := (τH )H ∈S for all τ ∈ ICA(G; A). If a configuration x ∈ AG has a finite G-orbit, then Stab(x) ∈ S . Thus a cellular automaton in the kernel of ϕ must fix all configurations in AG with finite G-orbit. Since G is residually finite, the set of configurations with finite G-orbit is dense in AG for the prodiscrete topology. Using the continuity of cellular automata with respect to the prodiscrete topology, we deduce that ϕ is injective. On the other hand, the group P is a direct product of finite groups and therefore residually finite. As every subgroup of a residually finite group is itself residually finite, it follows that ICA(G; A) is residually finite. Conversely, suppose that the group ICA(G; A) is residually finite. As the set A has more than one element, it follows from Exercise 1.33 that there is an injective group homomorphism Φ : G → ICA(G; A). As the class of residually finite groups is closed under taking subgroups, we deduce that G is residually finite. ■ Comment A monoid M is said to be residually finite if for all m1 , m2 ∈ M with m1 /= m2 , there exist a finite monoid F and a monoid homomorphism φ : M → F such that φ(m1 ) /= φ(m2 ). Let M be a monoid and let A be a finite set with |A| ≥ 2. Equip A with the discrete topology. Then the space AM of all maps from M to A is a compact Hausdorff totally disconnected space with respect to the product topology. The right

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multiplication on M induces an action of M on AM , which is called the M-shift. A cellular automaton over the monoid M and the alphabet A is a continuous selfmapping of AM that commutes with the M-shift. The set CA(M; A) of all cellular automata over M and A is a monoid for the composition of maps. In [CecC6], it is shown that the monoid CA(M; A) is residually finite if and only if the monoid M is residually finite. In the particular case when M is a group G, this gives the result of Exercise 2.10. Exercise 2.11 (Residually Finite Rings) A ring R is said to be residually finite if for every non-zero element r ∈ R, there exist a finite ring F and a ring homomorphism φ : R → F such that φ(r) /= 0F . (a) Show that every finite ring is residually finite. (b) Show that the ring Z is residually finite. (c) Show that any subring of a residually finite ring is itself residually finite. (d) Show that every product of residually finite rings is itself residually finite. (e) Show that every ring which is the limit of a projective system of residually finite rings is residually finite. (f) Let p be a prime number. Show that the ring Zp of p-adic integers is residually finite. (g) Let R be a ring. Show that the following conditions are equivalent: (1) the ring R is residually finite; (2) given any two distinct elements r, s ∈ R, there exist a finite ring F and a ring homomorphism φ : R → F such that φ(r) /= φ(s); (3) given any finite subset Ω ⊂ R, there exist a finite ring F and a ring homomorphism φ : R → F such that the restriction of φ to Ω is injective; (4) the intersection of all the finite index two-sided ideals of R is reduced to the zero element of R. (h) Let R be a residually finite ring. Let U (R) denote the set consisting of all invertible elements of R. Show that the additive group R and the multiplicative group U (R) are both residually finite. (i) Let R be a residually finite ring and let n ≥ 1 be an integer. Show that the matrix ring Matn (R) is residually finite. (j) Let R be a residually finite ring and let n ≥ 1 be an integer. Show that the group GLn (R) is residually finite. Solution (a) We can always take F := R and φ := IdR when R is a finite ring. (b) Let n ∈ Z with n /= 0. Choose an integer m such that m > |n|. Then the ring homomorphism φ : Z → Z/mZ given by reduction modulo m satisfies φ(n) /= 0. This shows that Z is residually finite. (c) Let R be a residually finite ring and let S be a subring of R. Let s be a nonzero element in S. Since the ring R is residually finite, there exist a finite ring F and a ring homomorphism φ : R → F such that φ(r) /= 0F . By restricting φ to S, we get a ring homomorphism φ|S : S → F such that φ|S (s) = φ(s) /= 0F . This shows that S is a residually finite ring. Π (d) Let (Ri )i∈I be a family of residually finite rings and let R := i∈I Ri denote their direct product. Let r = (ri )i∈I ∈ R such that r /= 0R . Then there exists

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j ∈ I such that rj /= 0Rj . Since the ring Rj is residually finite, there exist a finite ring F and a ring homomorphism φ : Rj → F such that φ(rj ) /= 0F . The ring homomorphism ψ : R → F , defined by ψ := φ ◦ πj , where πj : R → Rj is the canonical projection, satisfies ψ(r) = φ(rj ) /= 0F . This shows that R is a residually finite ring. (e) Let (Ri )i∈I be a projective system of residually finite rings and Π let R := lim Ri . By construction of a projective limit, R is a subring of the ring i∈I Ri . By ← − using (c) and (d), we deduce that the ring R is residually finite. (f) This immediately follows from (a) and (e) since Zp is the limit of the projective sequence of finite rings (Z/pn Z)n≥0 . (g) If R is residually finite and r, s ∈ R are distinct, then r − s /= 0R so that there exist a finite ring F and a ring homomorphism φ : R → S such that φ(r − s) /= 0F . As φ(r − s) = φ(r) − φ(s), this implies φ(r) /= φ(s) and hence shows that (1) implies (2). If (2) is satisfied and Ω ⊂ R is a finite set, then for all distinct r, s ∈ Ω, there exist a finite ring Frs and a ring homomorphism φrs : R → Frs such that φrs (r) /= Π := φrs (s). Consider the product ring F r,s∈Ω:r/=s Frs . Then the restriction to Ω Π of the ring homomorphism r,s∈Ω:r/=s φrs : R → F is injective. As F is finite, this shows that (2) implies (3). Clearly (3) implies (1) by taking Ω = {r, 0R }. If we suppose (4) and r ∈ R is a non-zero element, then there exists a finite index two-sided ideal I ⊂ R such that r ∈ / I . Then the quotient ring F := R/I is finite and r is not sent to 0F under the canonical ring morphism R → F . This shows that (4) implies (1). Finally, if we suppose the ring R residually finite and r is a non-zero element in R, then there exist a finite ring F and a ring homomorphism φ : R → F such that φ(r) /= 0F . This implies that r ∈ / ker(φ). As ker(φ) is a finite index ideal of R, we conclude that (4) is satisfied. Therefore (1) implies (4). (h) Let r ∈ R such that r /= 0R . Since the ring R is residually finite, there exist a finite ring F and a ring homomorphism φ : R → F such that φ(r) /= 0F . As φ is a ring homomorphism from the additive group R into the additive group F , this shows that the additive group R is residually finite. Let now r ∈ U (R) such that r /= 1R . Since the ring R is residually finite, there exist a finite ring F and a ring homomorphism φ : R → F such that φ(r) /= φ(1R ). By restriction, φ induces a group homomorphism ρ : U (R) → U (F ). As U (F ) is finite and ρ(r) = φ(r) /= φ(1R ) = ρ(1R ), this shows that U (R) is a residually finite group. (i) Let A = (Aij )1≤i,j ≤n ∈ Matn (R) such that A /= 0. Then there exist k, l ∈ {1, . . . , n} such that Akl /= 0R . Since R is residually finite, there exist a finite ring F and a ring homomorphism φ : R → F such that φ(Akl ) /= 0F . Observe that φ induces a ring homomorphism ψ : Matn (R) → Matn (F ) given by (ρ(M))ij = φ(Mij ) for all M ∈ Matn (R). We have (ψ(A))kl = φ(Akl ) /= 0F and hence ψ(A) /= 0. As Matn (F ) is finite, this shows that Matn (R) is a residually finite ring. (j) This immediately follows from (f) and (g) since GLn (R) = U (Matn (R)). ■

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Exercise 2.12 (Residual Finiteness of the Ring of n-Adic Rationals) Let n ≥ 2 be an integer. Let Z[1/n] denote the ring of n-adic rationals, i.e., the subring of Q consisting of all r ∈ Q that can be written in the form r = knm for some k, m ∈ Z. (a) Show that the ring Z[1/n] is residually finite. (b) Show that the additive group Z[1/n] is residually finite and Hopfian but not finitely generated. Solution (a) Observe that Z[1/n] is the smallest subring of Q in which the integer n is invertible. Let r ∈ Z[1/n] such that r /= 0. Then r = knm for some k, m ∈ Z with k /= 0. Choose a prime number p such that p divides neither n nor k. Then there is a unique ring homomorphism φ : Z[1/n] → Z/pZ. As n and k have non-zero images under φ and Z/pZ is a field, we have φ(r) = φ(knm ) = φ(k)φ(n)m /= 0. This shows that Z[1/n] is a residually finite ring. (b) The fact that the ring Z[1/n] is residually finite implies that its underlying additive group is residually finite (cf. Exercise 2.11(f)). Let f be a group endomorphism of Z[1/n]. Put a := f (1). If r ∈ Z[1/n], then k r = m for some k, m ∈ Z with m ≥ 0. We then have nm f (r) = f (k) = kf (1) = n ka and hence f (r) = ra. If f is surjective, we must have a /= 0 so that f is injective. This shows that the group Z[1/n] is Hopfian. For each integer m ≥ 0, denote by Hm the subgroup of Z[1/n] consisting of k all r ∈ Z[1/n] that can be written in the form r = m for some k ∈ Z. Clearly n (Hm )m≥0 is a strictly increasing sequence of subgroups of Z[1/n] whose union is Z[1/n]. This shows that Z[1/n] is not finitely generated. ■ Comment If p is a prime number not dividing n then the class of n is invertible in Z/pk Z for every k ≥ 0. It follows that the ring Z[1/n] embeds in the ring of p-adic integers Zp = lim Z/pk Z. As the ring Zp is residually finite and every subring of ← −k a residually finite ring is residually finite (cf. Exercise 2.11(c), (f)), this yields an alternative proof of the fact that the ring Z[1/n] is residually finite. The ring Z[1/2] is called the ring of dyadic or binary rationals. The ring Z[1/10] is the ring of decimal rationals. Exercise 2.13 (The Baumslag-Solitar Group BS(1, m)) Let m be an integer such that |m| ≥ 2. Consider the group G given by the presentation G := . The group G is the Baumslag-Solitar group BS(1, m). (a) Show that every element g ∈ G may be written in the form g = a i bj a k for some i, j, k ∈ Z such that i ≤ 0 and k ≥ 0. (b) Show that there is a unique group homomorphism ρ : G → GL2 (Q) that satisfies ⎛ ⎞ ⎛ ⎞ m0 11 .ρ(a) = and ρ(b) = . 0 1 01

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(c) Show that ρ is injective and that the image of ρ is the subgroup H ⊂ GL2 (Q) given by  ⎧⎛ n ⎞ m r : n ∈ Z, r ∈ Z[1/m] , 0 1

H :=

.

(2.1)

where Z[1/m] denotes the ring of m-adic rationals, i.e., the subring of Q consisting of all rationals r ∈ Q which can be written in the form r = umv for some u, v ∈ Z (cf. Exercise 2.12). (d) Show that G is not abelian. (e) Show that G is torsion-free. (f) Show that there is a short exact sequence of groups 0 → Z[1/m] → G → Z → 0. (g) Show that G is residually finite. (h) Show that G is Hopfian. (i) Show that G is not of finite Markov type. Solution (a) As a and b generate G, every element g ∈ G can be written in the form g = a α1 bβ1 a α2 bβ2 · · · a αn bβn

.

(2.2)

for some integer n ≥ 0 and αi , βi ∈ Z for 1 ≤ i ≤ n. From the defining relation aba −1 = bm , we deduce (i) ab = bm a, (ii) ab−1 = b−m a, (iii) b−1 a −1 = a −1 b−m , m and (iv) ba −1 = a −1 bm . Using (i) and (ii), we get by induction a α bβ = bαβ a α for all α, β ∈ Z with α ≥ 1. This allows us to move any positive power of a in the right-hand side of (2.2) to the top right. Similarly, relations (iii) and (iv) yield m bβ a −α = a −α bαβ , which can be used to move any negative power of a to the top left. At the end of this process, this gives us an expression of g of the form g = a i bj a k for some i, j, k ∈ Z such that i ≤ 0 and k ≥ 0. (b) Consider the matrices A=

.

⎛ ⎞ m0 0 1

⎛ and

B=

⎞ 11 . 01

Clearly A, B ∈ GL2 (Q). As ABA−1 =

.

⎛ ⎞⎛ ⎞⎛ 1 ⎞ ⎛ ⎞ m0 11 m 0 = 1 m = B m, 0 1 0 1 01 0 1

there is a unique group homomorphism ρ : G → GL2 (Q) satisfying ρ(a) = A and ρ(b) = B.

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(c) Let g ∈ G. By (a), we can write g = a i bj a k for some i, j, k ∈ Z. Then ⎛ i ⎞ ⎛ ⎞ ⎛ k ⎞ ⎛ i+k i ⎞ mj m 0 1j m m 0 .ρ(g) = A B A = = . 0 1 0 1 0 1 01 i

j

k

This shows that the image of ρ is the subgroup H ⊂ GL2 (Q) defined by (2.1). We also deduce that if g ∈ ker(ρ), then mi+k = 1 and mi j = 0, so that k = −i, j = 0, and hence g = 1G . This shows that ρ is injective. (d) The matrices A and B do not commute. As ρ is injective, we deduce that a and b do not commute in G either. In particular, the group G is not abelian. (e) Suppose that M=

.

⎛ n ⎞ m r ∈H 0 1

⎛ ⎞ 10 satisfies = 1H = for some integer k ≥ 1. Taking determinants, we get 01 mnk = 1 and hence n = 0. Thus Mk

⎛

1r .M = 01 k

⎞k

⎛ ⎞ 1 kr = 0 1

so that r = 0. In conclusion, we have M = 1H . This shows that H is torsion-free. As the group G is isomorphic to H by (c), we deduce that G is itself torsion-free. (f) The map H → Z, defined by ⎛ n ⎞ m r . | n, → 0 1 is a surjective homomorphism of groups whose kernel is isomorphic to Z[1/m]. As G is isomorphic to H , this shows that there is a short exact sequence of groups 0 → Z[1/m] → G → Z → 0. (g) By Exercise 2.12(a), the ring Z[1/m] is residually finite. It follows from Exercise 2.11(h) that the group GL2 (Z[1/m]) is residually finite. As H is a subgroup of GL2 (Z[1/m]) and the group G is isomorphic to H , we conclude that G is residually finite. (h) Since the group G is finitely generated and residually finite, it is Hopfian by Mal’cev’s theorem (cf. [CAG, Theorem 2.4.3]). (i) By (f), the group G contains a subgroup isomorphic to the additive group Z[1/m]. As the group Z[1/m] is not finitely generated, we deduce from Exercise 1.111(e) that G is not of finite Markov type. ■

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Comment Baumslag-Solitar groups were introduced in [BauS]. The fact that H (and hence G) is residually finite can also be deduced from the theorem of Mal’cev which asserts that every finitely generated linear group is residually finite. Exercise 2.14 (The Baumslag-Solitar Group BS(2, 3)) Consider the group G given by the presentation G := . Thus, G is the quotient group G = F /N, where F is the free group based on two generators x, y and N is the normal closure of xy 2 x −1 y −3 in F . Let ρ : F → G denote the quotient group homomorphism. The group G is the Baumslag-Solitar group BS(2, 3). (a) Consider the elements a, b ∈ G defined by a := ρ(x) and b := ρ(y). Show that every element g ∈ G may be uniquely written in the form g = bη0 a ε1 bη1 a ε2 bη2 · · · a εn bηn ,

.

(2.3)

where n ∈ N, η0 ∈ Z, εi ∈ {−1, 1} and ηi ∈ {−1, 0, 1}, i = 1, 2, . . . , n, are subject to the conditions if εi = 1 for some 1 ≤ i ≤ n, then ηi /= −1; if ηi = 0 for some 1 ≤ i ≤ n − 1, then εi /= −εi+1 . (b) Show that there is a unique group endomorphism φ : G → G such that φ(a) = a and φ(b) = b2 . (c) Show that φ is surjective. (d) Show that the element g ∈ G given by g := b−1 (aba −1 b−1 )2 is in the kernel of φ. (e) Show that G is not Hopfian. (f) Show that G is not residually finite. Solution (a) We have 1G = ρ(xy 2 x −1 y −3 ) = ρ(x)ρ(y)2 ρ(x)−1 ρ(y)−3 = ab2 a −1 b−3 , so that ab2 = b3 a and a −1 b3 = b2 a −1 . We deduce that for all k ∈ Z, ε ∈ {0, 1}, and η ∈ {−1, 0, 1}, we have ab2k+ε = ab2k bε = ab2k a −1 abε = b3k abε

.

(2.4)

and a −1 b3k+η = a −1 b3k bη = a −1 b3k aa −1 bη = b2k a −1 bη .

.

(2.5)

Since a and b generate G, every element g ∈ G can be written in the form g = s1 s2 · · · sk for some k ≥ 0 and s1 , s2 , . . . , sk ∈ {a, b, a −1 , b−1 }. By repeated uses of (2.4) and (2.5), we can move every even (respectively, multiple of three) power of b to the left past an a (resp. past an a −1 ) and finally obtain an expression for g as in (2.3).

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Let us show that such a form is indeed unique. Let W denote the set of all w ∈ F whose reduced form is w = y η0 x ε1 y η1 x ε2 y η2 · · · x εn y ηn ,

.

(2.6)

where n ∈ N, η0 ∈ Z, εi ∈ {−1, 1} and ηi ∈ {−1, 0, 1}, i = 1, 2, . . . , n, are subject to conditions (i) and (ii). We can make G act on W essentially by multiplying on the left and then reducing. More precisely, denoting by Sym(W ) the symmetric group of W , we shall define a group homomorphism Ψ : F → Sym(W ) such that Ψ (xy 2 x −1 y −3 ) = 1Sym(W ) , equivalently, Ψ (xy 2 x −1 ) = Ψ (y 3 ).

.

(2.7)

The kernel of such a Ψ contains N so that it induces a group homomorphism Ψ : G = F /N → Sym(W ), yielding the required action. In order to define Ψ , we first introduce four maps σx , τx , σy , τy : W → W in the following way. Let w ∈ W as in (2.6) and set w  := x ε2 y η2 · · · x εn y ηn ∈ W so that w = y η0 x ε1 y η1 w  .

.

In order to define σx (w) observe that there exist unique η0 ∈ Z and η ∈ {0, 1} such that η0 = 2η0 + η. We then set ⎧ σx (w) :=



y 3η0 +η1 w 

.

y

3η0

xy η x ε1 y η1 w 

if (η, ε1 ) = (0, −1) otherwise.

Note that σx (w) ∈ W . In order to define τx (w) observe that there exist unique η0 ∈ Z and η ∈ {−1, 0, 1} such that η0 = 3η0 + η . We then set ⎧ .τx (w) :=



y 2η0 +η1 w  y

2η0

η

x −1 y x ε1 y η1 w 

if (η , ε1 ) = (0, 1) otherwise.

Note that τx (w) ∈ W . It also follows by a case-by-case check that τx (σx (w)) = w = σx (τx (w)) for all w ∈ W , that is, τx ◦ σx = IdW = σx ◦ τx , so that σx ∈ Sym(W ) with σx−1 = τx . We also set σy (w) := y η0 +1 x ε1 y η1 w 

.

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and τy (w) := y η0 −1 x ε1 y η1 w  .

.

Note that both σy (w) and τy (w) belong to W . Moreover, it is straightforward that τy (σy (w)) = w = σy (τy (w)) for all w ∈ W , that is, τy ◦ σy = IdW = σy ◦ τy , so that σy ∈ Sym(W ) with σy−1 = τy . Since {x, y} is a base for F , the map {x, y} → F such that x |→ σx and y |→ σy extends to a unique group homomorphism Ψ : F → Sym(W ) satisfying Ψ (x) = σx , Ψ (x −1 ) = σx−1 = τx , Ψ (y) = σy , and Ψ (y −1 ) = σy−1 = τy . Let us now verify (2.7). For (η0 , ε1 ) = (3η0 , 1) we have (η1 , ε2 ) /= (0, −1) and 

Ψ (xy 2 x −1 )(w) = σx (σy2 (τx (y 3η0 xy η1 w  ))) 

= σx (σy2 (y 2η0 +η1 w  )) 

= σx (y 2(η0 +1)+η1 w  ) .



= y 3(η0 +1) xy η1 w  

= σy3 (y 3η0 xy η1 w  ) = Ψ (y 3 )(w) and, for η0 = 3η0 + η , with η ∈ {0, 1} such that (η , ε1 ) /= (0, 1) we have 



Ψ (xy 2 x −1 )(w) = σx (σy2 (τx (y 3η0 +η x ε1 y η1 w  ))) 



= σx (σy2 (y 2η0 x −1 y η x ε1 y η1 w  )) 



= σx (y 2(η0 +1) x −1 y η x ε1 y η1 w  ) .





= y 3(η0 +1)+η x ε1 y η1 w  



= σx3 (y 3η0 +η x ε1 y η1 w  ) = Ψ (y 3 )(w). This completes the argument defining the group homomorphism Ψ : G → Sym(W ).

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Let now g ∈ G. We have already observed at the beginning that there exists an element w ∈ W such that g = ρ(w). We now show that such a word w is indeed unique. If (2.6) is the expression of w, then applying Ψ (g) ∈ Sym(W ) to y 0 ∈ W we obtain Ψ (g)(y 0 ) = Ψ (w)(y 0 ) = Ψ (y η0 x ε1 y η1 x ε2 y η2 · · · y ηn−1 x εn y ηn )(y 0 ) = Ψ (y η0 x ε1 y η1 x ε2 y η2 · · · y ηn−1 x εn )(Ψ (y ηn )(y 0 )) = Ψ (y η0 x ε1 y η1 x ε2 y η2 · · · y ηn−1 x εn )(y ηn ) .

= Ψ (y η0 x ε1 y η1 x ε2 y η2 · · · y ηn−1 )(x εn y ηn ) ··· = Ψ (y η0 x ε1 )(y η1 x ε2 y η2 · · · y ηn−1 x εn y ηn ) = Ψ (y η0 )(x ε1 y η1 x ε2 y η2 · · · y ηn−1 x εn y ηn ) = y η0 x ε1 y η1 x ε2 y η2 · · · y ηn−1 x εn y ηn = w.

This shows that there is a unique expression of g in the form (2.3). (b) Uniqueness follows from the fact that a and b generate G. To prove existence, we first observe that, by the universal property of free groups, there exists a group homomorphism ψ : F → G such that ψ(x) = a and ψ(y) = b2 . Since ψ(xy 2 x −1 y −3 ) = ψ(a)ψ(b)2 ψ(a)−1 ψ(b)−3

.

= ab4 a −1 b−6 = (ab2 a −1 )2 b6 = b6 b−6 = 1G we have ker(ψ) ⊂ N. Therefore there exists a group homomorphism φ : G = F /N → G such that φ ◦ ρ = ψ. We then have φ(a) = φ(ρ(x)) = ψ(x) = a and φ(b) = φ(ρ(y)) = ψ(y) = b2 , showing that φ has the desired properties. (c) The elements a and b are in the image of φ since a = φ(a) and b = b3 b−2 = ab2 a −1 b−2 = φ(a)φ(b)φ(a)−1 φ(b)−1 = φ(aba −1 b−1 ).

.

As a and b generate G, this shows that φ is surjective. (d) Using (2.8), we get φ(g) = φ(b−1 (aba −1 b−1 )2 )

.

= φ(b)−1 φ(aba −1 b−1 )2 = b−2 b2 = 1G . This shows that g is in the kernel of φ.

(2.8)

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(e) We have g = b−1 aba −1 b−1 aba −1 b−1 /= 1G by (a). Since g is in the kernel of φ by (d), we deduce that φ is not injective. As φ is surjective by (c), this shows that G is not Hopfian. (f) Since the group G is finitely generated but not Hopfian, it is not residually finite by Mal’cev’s theorem. ■ Comment The first example of a finitely presented non-Hopfian group was discovered by Higman [Hig2] in 1951. It is the group with presentation . The Baumslag-Solitar group BS(2, 3) was the first example of a finitely generated one-relator non-Hopfian group (see [BauS]). An expression as in (2.3) is called a normal form of g. More generally, if G := is a presentation for a group G, i.e., R is a subset of the free group F based on X and G = F /N, where N is the normal closure of R, a normal form for G is an injective map ν : G → F such that, denoting by ρ : F → G the quotient group homomorphism, the composite ρ ◦ ν is the identity map on G. The proof of (a) presented here is based on a general technique developed by van der Waerden [Wae]. Let Γ be a group with presentation Γ = and let α : H → K be a group isomorphism between two subgroups H and K of Γ . Then the group with presentation Γ = is called the Higman-NeumanNeuman extension (briefly, the HNN extension) of Γ via α. The new generating symbol t is called the stable letter. Thus, BS(2, 3) is the HNN extension of Γ := with respect to the isomorphism α : H → K, where H := , K := , and α(b2 ) := b3 (the stable letter is t = a). The following result, called Britton’s lemma, provides a normal form for elements in an HNN-extension. Let w = w0 t ε1 w1 t ε2 w2 · · · t εn wn , where w0 and wi are words over X and εi ∈ {−1, 1} for all i = 1, 2, . . . , n. Suppose that either n = 0 and w0 /= 1Γ in Γ , or n ≥ 1 and if εi = 1 = −εi+1 (resp. εi = −1 = −εi+1 ) then wi ∈ / H (resp. wi ∈ / K) for all i = 1, 2, . . . , n − 1. Then w /= 1Γ in Γ . Britton’s lemma can be used to give an alternative proof of the fact that a group element g ∈ G as in (2.3) (e.g., the element g := b−1 (aba −1 b−1 )2 in (d)) is different from the identity element 1G for (n, η0 ) /= (0, 0). Exercise 2.15 Let G be a group and let H be a normal subgroup of G. (a) Suppose that there exists a surjective group homomorphism ϕ : G → G such that ϕ(H ) ⊂ H and (G \ H ) ∩ ϕ −1 (H ) /= ∅. Show that the group G/H is not Hopfian. (b) Suppose that there exists a group automorphism α ∈ Aut(G) such that α(H )  H . Show that the group G/H is not Hopfian. Solution (a) Let ρ : G → G/H denote the canonical group homomorphism. As ϕ(H ) ⊂ H , there is a unique group homomorphism ψ : G/H → G/H such that ψ ◦ ρ = ρ ◦ ϕ. Observe that ψ is surjective since ρ and ϕ are. On the other hand, we have H = ker(ρ). Thus, if we take g ∈ (G \ H ) ∩ ϕ −1 (H ), then ρ(g) /= 1G/H and

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ψ(ρ(g)) = ρ(ϕ(g)) = 1G/H , showing that ρ(g) is a non-trivial element in the kernel of ψ. Therefore, ψ is not injective. This shows that G/H is not Hopfian. (b) This follows from (a) by taking ϕ := α. Indeed, α(H )  H implies H  α −1 (H ) since α is bijective. ■ Comment See [Cor, Lemma 2.3]. Exercise 2.16 Let G be a Hopfian group and let H be a non-trivial group. Show that the groups G and G × H are not isomorphic. Solution Suppose that there exists a group isomorphism ϕ : G → G × H . Let π : G × H → G denote the projection onto the first factor. Then the composite group homomorphism π ◦ ϕ : G → G is surjective but not injective. This is in contradiction with the fact that G is Hopfian. ■ Comment This shows in particular that if G is a non-trivial Hopfian group then G is not⊕isomorphic to Π its square G × G. Given any non-trivial group K, the K and groups n∈N n∈N K provide examples of non-trivial groups that are isomorphic to their square. Note that these examples are not finitely generated. In 1974, Jones [Jon2] constructed the first example of a non-trivial finitely generated group isomorphic to its square. Meyer [Mei] provided further examples of nontrivial finitely generated groups G such that G ∼ = G × S, where S = G × G or G ∼ is a simple group. The question of the existence of a non-trivial finitely presented group isomorphic to its square reamins open. Exercise 2.17 Show that all elements of the additive group Q/Z have finite order but that Q/Z is not residually finite. Solution Given m, n ∈ Z with n ≥ 1, we denote by [m/n] := m/n + Z the class of the rational number m/n mod Z. Since n[m/n] = [m] = 0, we deduce that each element of the additive group Q/Z has finite order. On the other hand, given an element g = [m/n] ∈ Q/Z and an integer n ≥ 1, the element h := [m/nn ] ∈ Q/Z satisfies that n h = g. This shows that Q/Z is divisible. As a consequence, Q/Z is not residually finite. ■ Comment We can also use the fact that Q is divisible and the result of Exercise 2.3 to show that Q/Z is divisible. Exercise 2.18 Show that the multiplicative group Q∗ of nonzero rational numbers is residually finite. Solution Let P denote the set of prime numbers. Every element q ∈ Q∗ can be uniquely written in the form q=ε

Π

.

p∈P

pαp

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where ε ∈ {1, −1}, αp ∈ Z for all p ∈ P, and αp = 0 for all but finitely many p ∈ P. The map q |→ (ε, (αp )p∈P ) yields a group isomorphism from Q∗ onto the group ⎛ Z/2Z ⊕ ⎝

⊕

.

⎞ Z⎠ .

p∈P

As any direct sum of residually finite groups is itself residually finite, this shows that Q∗ is residualy finite. ■ Exercise 2.19 Show that the automorphism group Aut(Q) of the additive group Q is isomorphic to the multiplicative group Q∗ . Solution Given q ∈ Q, the map αq : Q → Q defined by αq (q  ) = qq  for all q  ∈ Q is an automorphism of the additive group Q and αq (1) = q. Consider now the map Φ : Aut(Q) → Q∗ defined by Φ(α) := α(1) for all α ∈ Aut(Q). For α ∈ Aut(Q) and m ∈ Z we have α(m) = α(1)m. By unique divisibility of Q, we have, more generally, α(q) = α(1)q for all q ∈ Q.

.

(2.9)

As a consequence, Φ(α ◦ β) = α(β(1)) = α(1)β(1) = Φ(α)Φ(β), for all α, β ∈ Aut(Q), showing that Φ is a group homomorphism. From (2.9), we immediately deduce that Φ is injective. Finally, since Φ(αq ) = q for every q ∈ Q, we have that Φ is also surjective, and therefore bijective. ■ Comment Combining the result of Exercise 2.18 with the result of Exercise 2.19, we see that the additive group Q provides an example of a group that is not residually finite although its automorphism group is residually finite. Exercise 2.20 Let G be a group with trivial center and suppose that the automorphism group of G is residually finite. Show that G is residually finite. Solution Let Aut(G) denote the automorphism group of G. The group homomorphism ρ : G → Aut(G), defined by (ρ(g))(x) := gxg −1 for all g, x ∈ G, is injective since G ha trivial center. This implies that G is isomorphic to a subgroup of Aut(G). As Aut(G) is residually finite, it follows that G is itself residually finite. ■ Exercise 2.21 Let E be an infinite-dimensional vector space over a field K. Show that the additive group E is not Hopfian. Solution Let B ⊂ E be a base for the vector space E. Since B is infinite, there exists a surjective but not injective map f : B → B. Extend f to an endomorphism g of the vector space E by K-linearity. Then g is K-linear and hence a group endomorphism of the additive group E. As g is surjective but not injective, this shows that the additive group E is not Hopfian. ■

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Exercise 2.22 Let K be a field of characteristic p > 0. (a) Show that the additive group K is residually finite. (b) Show that the additive group K is not Hopfian unless K is finite. Solution (a) If B ⊂ K is a basis for K, viewed as a vector space over ⊕ its prime field Z/pZ, then the additive group K is isomorphic to the group B Z/pZ. As any direct sum of residually finite groups is itself residually finite, we deduce that the additive group K is residually finite. (b) If K is infinite then it is an infinite-dimensional vector space over its prime field. This implies that the additive group K is not Hopfian by the result of Exercise 2.21. ■ Exercise 2.23 Show that the additive group R is neither residually finite nor Hopfian. Solution The group R is not residually finite since it is divisible. On the other hand, since R is uncountable while Q is countable, R is an infinitedimensional vector space over its subfield Q. Thus, by the result of Exercise 2.21, the additive group R is not Hopfian. ■ Exercise 2.24 Show that the multiplicative group R∗ of nonzero real numbers is neither residually finite nor Hopfian. Solution The set of all positive real numbers is a subgroup of the multiplicative group R∗ isomorphic to the additive group R via the logarithmic map. As R is not residually finite by Exercise 2.23 and the class of residually finite groups is closed under taking subgroups, we deduce that R∗ is not residually finite. Consider now the additive group G = R ⊕ Z/2Z. The map G → R∗ given by (x, y) |→ exp(x +iy) is clearly a group isomorphism. By the result of Exercise 2.23, the additive group R is not Hopfian so that there exists a group endomorphism f of R which is surjective but not injective. Then the group endomorphism of G given by (x, y) |→ (f (x), y) is also surjective but not injective. This shows that G and hence R∗ are not Hopfian. ■ Exercise 2.25 Let p be a prime number. Show that the additive group Zp of p-adic integers is Hopfian. Solution Recall (cf. [CAG, Example 2.2.9]) that the group Zp of p-adic integers is the limit of the projective system of groups (Z/pn Z, φn,m ), where φn,m : Z/pm Z → Z/pn Z is the group homomorphism induced by the inclusion pm Z ⊂ pn Z for all integers 0 ≤ n ≤ m. Thus, Zp is the subgroup of the product group P := Π n n≥0 Z/p Z consisting of all x = (xn )n≥0 ∈ P such that φn,n+1 (xn+1 ) = xn for all n ≥ 0. Observe that, for each n ≥ 0, the projection map πn : Zp → Z/pn Z, defined by πn (x) = xn , is a surjective group homomorphism whose kernel is the subgroup Hn := pn Zp ⊂ Zp . Let now f be a surjective group endomorphism of Zp and let us show that f is injective. We have f (Hn ) = f (pn Zp ) = pn f (Zp ) = pn Zp = Hn for each n ≥ 0. Thus f induces a surjective quotient group endomorphism fn of

2.2 Exercises

137

Zp /Hn . Using the fact that Zp /Hn ∼ = Z/pn Z is finite, we deduce that fn is injective. It ∩ follows that if x ∈ Zp is in the kernel of f , then x ∈ Hn for all n ≥ 0. As n≥0 Hn = {0}, we conclude that f is injective. This shows that the additive group ■ Zp is Hopfian. Exercise 2.26 Let F be a free group of finite rank n. Suppose that S is a finite generating subset of F of cardinality |S| ≤ n. Use the fact that F is Hopfian to prove that |S| = n and that S is a base for F . Solution Let B ⊂ F be a base for F , so that |B| = n (cf. [CAG, Section D.2]). Since |S| ≤ |B|, there exists a surjective map f : B → S. By the universal property of free groups, f extends to a group homomorphism φ : F → F . This homomorphism is surjective since S generates F . As F is Hopfian, φ is a group automorphism of F . The image of a base of F under an automorphism is clearly a base of F . We deduce that S = f (B) = φ(B) is a base of F with cardinality n. ■ Exercise 2.27 Let F be a free group of finite rank n. Suppose that S is a finite generating subset of F . Show that S has cardinality at least n. Solution By Exercise 2.26, we cannot have |S| < n.

■

Exercise 2.28 Let G be a finitely generated group and let n ≥ 1 be an integer. Show that G contains only a finite number of subgroups of index n. Solution Let Hn denote the set consisting of all subgroups of G of index n. Let H ∈ Hn . Then the set G/H := {gH : g ∈ G} of left cosets of H in G has cardinality n. The group G naturally acts on G/H by left multiplication. Observe that the stabilizer of the coset H ∈ G/H for this action is precisely the subgroup H . Choose a bijection tH : {1, . . . , n} → G/H such that tH (1) = H . Using tH , we can pullback the action of G on G/H to get an action αH of G on {1, . . . , n}. Consider now the group homomorphism ρH : G → Sym(n) associated with αH . As the stabilizer of 1 for αH is H , the map Hn → Hom(G, Sym(n)) given by H |→ ρH is injective. On the other hand, the set Hom(G, Sym(n)) is finite since G is finitely generated (cf. [CAG, Lemma 2.4.4]). It follows that Hn is finite. ■ Comment The result is due to Marshall Hall (see [Hal1, p. 189]) who also gave an explicit recurrence formula for computing the number of subgroups of index n in a free group of finite rank r. The study of the assymptotic behavior of the number of subgroups of finite index n in a finitely generated group is a subject of active current research and there is a vast literature devoted to it (see in particular the monograph [LubS] and the references therein). Exercise 2.29 (Characteristic Subgroups) Let G be a group. A subgroup C ⊂ G is said to be characteristic if it is invariant under all automorphisms of G, that is, α(C) = C for all α ∈ Aut(G). (a) Show that every characteristic subgroup of G is normal in G. (b) Find an example of a group G containing a subgroup H that is normal but not characteristic in G.

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(c) Let G be a group. Show that the center, the derived subgroup, and the residual subgroup of G are all characteristic subgroups of G. (d) Let G be a finitely generated group and let H ⊂ G be a finite index subgroup. Show that there exists a finite index characteristic subgroup C of G such that C ⊂ H. (e) Find an example of a countable group G containing a finite index subgroup H such that there is no finite index characteristic subgroup C of G satisfying C ⊂ H . Solution (a) Let G be a group and let H be a characteristic subgroup of G. Consider, for each g ∈ G, the associated inner automorphism αg ∈ Aut(G) defined by αg (x) = gxg −1 for all x ∈ G. Since H is characteristic, we have αg (H ) = H for all g ∈ G. Therefore ghg −1 ∈ H for all g ∈ G and h ∈ H . This shows that H is a normal subgroup of G. (b) Take any non-trivial abelian group A (e.g. A = Z/2Z or A = Z) and consider the group G := A × A. Then the subgroup H := A × {0A } is normal in G since every subgroup of an abelian group is normal. However, H is not characteristic in G since it is not invariant under the automorphism α ∈ Aut(G) defined by α((x, y)) = (y, x) for all x, y ∈ A. (c) This is clear since all these subgroups are defined in terms of the group structure only. (d) Let n := [G : H ]. By Exercise 2.28, ∩ the set Hn of all subgroups of G of index n is finite. Therefore the subgroup C := K∈Hn K is of finite index in G. Moreover, C ⊂ H since H ∈ Hn , and C is characteristic in G since every automorphism of G preserves the index of its subgroups and therefore permutes the elements of Hn . Thus C has the required properties. (e) Let A := Z/2Z. Let G := ⊕n∈N A be the direct sum of a family of copies of A indexed by N. For each n ∈ N, let πn : G → A denote the projection on the n-th factor. The subgroup H := ker(π0 ) is of index 2 in G. Suppose that C is a characteristic subgroup of G such that C ⊂ H . We claim that C = {0G }. To see this, denote by αn the automorphism of G that exchanges the 0-th and the n-th coordinates of each element of G. We have π0 ◦ αn = πn so that .

ker(πn ) = αn−1 (ker(π0 )) = αn−1 (H ) ⊃ αn−1 (C) = C.

∩ We deduce that C ⊂ n∈N ker(πn ) and the claim follows. In particular, C is of infinite index in G since G is infinite. ■ Exercise 2.30 (External Semidirect Product) Let H and K be two groups and let α : K → Aut(H ) be a group homomorphism from K into the automorphism group of H . Equip the set G := H × K with the binary operation defined by (h, k) · (h , k  ) := (hα(k)(h ), kk  )

.

for all h, h ∈ H and k, k  ∈ K.

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(a) Show that G is a group. (b) Show that the maps ι : H → G and t : K → G, given by ι(h) := (h, 1K ) for all h ∈ H and t (k) := (1H , k) for all k ∈ K, are group monomorphisms and that the map π : G → K, given by π(h, k) := k for all h ∈ H and k ∈ K, is a group epimorphism. (c) Show that ι

π

1 → H→ G → K → 1

.

(2.10)

is a short exact sequence of groups and that the group homomorphism t : K → G satisfies π ◦ t = IdK . (d) Show that the following holds: (i) ι(H ) is a normal subgroup of G and the quotient group G/ι(H ) is canonically isomorphic to K, (ii) G = ι(H )t (K), (iii) ι(H ) ∩ t (K) = {1G }. (e) Check that t (k)ι(h)t (k)−1 = ι(α(k)(h)) for all h ∈ H and k ∈ K. Solution (a) Let h, h , h ∈ H and k, k  , k  ∈ K. We have ) ( (h, k) · (h , k  ) · (h , k  ) = (hα(k)(h ), kk  ) · (h , k  )

.

= (hα(k)(h )α(kk  )(h ), kk  k  ), and ) ( ) ( (h, k) · (h , k  ) · (h , k  ) = (h, k) · h α(k  )(h ), k  k 

.

= (hα(k)(h α(k  )(h )), kk  k  ) = (hα(k)(h α(k  )(h )), kk  k  ) = (hα(k)(h )α(kk  )(h ), kk  k  ), ( ) ( ) so that (h, k) · (h , k  ) · (h , k  ) = (h, k) · (h , k  ) · (h , k  ) . This shows associativity. On the other hand, we have (1H , 1K ) · (h, k) = (1H α(1K )(h), 1K k) = (1H IdH (h), K) = (h, k) and (h, k) · (1H , 1K ) = (hα(k)(1H ), k1K ) = element. (h1H , k) = (h, k), showing that 1G := (1H , 1K() is an identity ) Moreover, (h, k) · (α(k −1 )(h−1 ), k −1 ) = (hα(k) α(k −1 )(h−1 ) , kk −1 ) = (hα(kk −1 )(h−1 ), 1K ) = (hα(1K )(h−1 ), 1K ) = (hh−1 , 1K ) = (1H , 1K ) = 1G and (α(k −1 )(h−1 ), k −1 ) · (h, k) = (α(k −1 )(h−1 )α(k −1 )(k), k −1 k) = (α(k −1 (h−1 h), 1K ) = (α(k −1 )(1H ), 1K ) = (1H , 1K ) = 1G , so that (α(k −1 )(h−1 ), k −1 ) is an inverse for (h, k). This shows that G is a group.

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(b) Let h, h ∈ H and k, k  ∈ K. We have ι(h)ι(h ) = (h, 1K ) · (h , 1K ) = (hα(1K )(h ), 1K 1K ) = (hh , 1K ) = ι(hh ),

.

t (k)t (k  ) = (1H , k) · (1H , k  ) = (1H α(k)(1H ), kk  )

.

= (1H 1H , kk  ) = (1H , kk  ) = t (kk  ) and ( ) π (h, k) · (h , k  ) = π(hα(k)(h ), kk  ) = kk  = π(h, k)π(h , k  ).

.

This shows that ι, t, and π are group homomorphisms. The fact that ι and t are injective (resp. π is surjective) is obvious. (c) We have seen in (b) that ι is a group monomorphism and that π is a group epimorphism. As Im(ι) = {(h, 1K ) : h ∈ H } = ker(π ), it follows that (2.10) is a short exact sequence of groups. Moreover, (π ◦ t)(k) = π(t (k)) = π(1H , k) = k = IdK (k) for all k ∈ K, showing that π ◦ t = IdK . (d) From the exactness of (2.10) and the first isomorphism theorem for groups, we deduce that ι(H ) = ker(π ) is normal in G and that π induces a group isomorphism from G/ι(H ) onto K. We have (h, k) = (h, 1K ) · (1H , k) = ι(h)t (k) for all h ∈ H and k ∈ K, showing that G = ι(H )t (K). Finally, we have ι(H ) ∩ t (K) = {(h, 1K ) : h ∈ H } ∩ {(1H , k) : k ∈ K} = {(1H , 1K )} = {1G }. (d) Let h ∈ H and k ∈ K. We have t (k)ι(h)t (k)−1 = (1H , k) · (h, 1K ) · (1H , k)−1 = (α(k)(h), k) · (1H , k −1 ) .

= (α(k)(h), 1K ) = ι(α(k)(h)). ■

Comment The group G is called the semidirect product of the groups H and K with respect to the group homomorphism α : K → Aut(H ) and is denoted by H α K. Note that H α K is the direct product group H ×K in the case when α is trivial. Exercise 2.31 (Split Exact Sequences) An exact sequence of groups ι

π

1 → H→ G → K → 1

.

is said to be split provided that there exists a group homomorphism t : K → G such that π ◦ t = IdK .

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Two exact sequences of groups ι

π

ι

π

1 → H→ G → K → 1

.

(2.11)

and 1 → H → G → K → 1

.

(2.12)

are said to be equivalent provided that there exists a group isomorphism ϕ : G → G such that ι = ϕ ◦ ι

and

.

π = π  ◦ ϕ,

(2.13)

that is, the diagram

.

is commutative. (a) Suppose that two short exact sequences of groups are equivalent. Show that if one is split so is the other. (b) Let H α K be the semidirect product of two groups H and K with respect to a group homomorphism α : K → Aut(H ). Consider the maps ια : H → H α K defined by ια (h) := (h, 1K ) for all h ∈ H and the map πα : H α K → K defined by πα (h, k) := k for all (h, k) ∈ H α K. Show that ια

πα

1 → H → H α K → K → 1

.

(2.14)

is a split exact sequence of groups. (c) Let 1 → H → G → K → 1 be an exact sequence of groups. Show that the following conditions are equivalent: (i) the exact sequence 1 → H → G → K → 1 is split; (ii) there exists a group homomorphism α : K → Aut(H ) such that the exact sequence 1 → H → G → K → 1 is equivalent to the exact sequence (2.14). (d) Let K be a free group. Show that every exact sequence of groups 1 → H → G → K → 1 is split.

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Solution (a) Suppose that the exact sequences of groups (2.11) and (2.12) are equivalent and let ϕ : G → G be a group isomorphism such that (2.13) holds. Assume that (2.11) is split and let t : K → G be a group homomorphism such that π ◦ t = IdK . Then t  := ϕ ◦t : K → G is a group homomorphism such that π  ◦t  = π  ◦(ϕ ◦t) = (π  ◦ ϕ) ◦ t = π ◦ t = IdK . This shows that (2.12) is split. (b) The fact that (2.14) is a split exact sequence of groups follows from Exercise 2.30(c). (c) The fact that (ii) implies (i) follows from (a) and (b). Conversely, suppose that ι

π

1 → H→ G → K → 1

.

is a split exact sequence of groups and let t : K → G be a group homomorphism such that π ◦ t = IdK . Given h ∈ H and k ∈ K we set ⎛ ⎞ −1 t (k)ι(h)t (k)−1 . .α(k)(h) := ι We claim that α(k) ∈ Aut(H ) for all k ∈ K and that α : K → Aut(H ) is a group homomorphism such that G ∼ = H α K. First of all, given k ∈ K, we have, for all h1 , h2 ∈ H , ⎛ ⎞ α(k)(h1 h2 ) = ι−1 t (k)ι(h1 h2 )t (k)−1 ⎛ ⎞ = ι−1 t (k)ι(h1 )ι(h2 )t (k)−1 ⎛ ⎞ . = ι−1 t (k)ι(h1 )t (k)−1 · t (k)ι(h2 )t (k)−1 ⎛ ⎞ ⎛ ⎞ = ι−1 t (k)ι(h1 )t (k)−1 · ι−1 t (k)ι(h2 )t (k)−1 = α(k)(h1 ) · α(k)(h2 ), showing that α(k) : H → H is a group homomorphism. Moreover, since, for all h ∈ H, ⎛ ⎞ α(k −1 ) ◦ α(k) (h) = α(k −1 ) (α(k)(h)) ⎛ ⎛ ⎞ ⎞ = ι−1 t (k −1 )ι(ι−1 t (k)ι(h)t (k)−1 )t (k −1 )−1 ⎛ ⎞ . = ι−1 t (k −1 )(t (k)ι(h)t (k)−1 t (k −1 )−1 = ι−1 (ι(h))) = h,

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we deduce that α(k) is bijective, with inverse α(k)−1 = α(k −1 ). Moreover, for all k1 , k2 ∈ K and h ∈ H , we have ⎛ ⎞ α(k1 k2 )(h) = ι−1 t (k1 k2 )ι(h)t (k1 k2 )−1 ⎛ ⎞ = ι−1 t (k1 )t (k2 )ι(h)t (k2 )−1 t (k1 )−1 ⎛ ⎛ ⎞ ⎞ = ι−1 t (k1 )ι(ι−1 t (k2 )ι(h)t (k2 )−1 )t (k1 )−1 . ⎛ ⎞ = ι−1 t (k1 )ι(α(k2 )(h))t (k1 )−1 = α(k1 ) (α(k2 )(h)) = (α(k1 ) ◦ α(k2 ))(h), showing that α : K → Aut(H ) is a group homomorphism. This proves the claim. Consider the map ϕ : G → H α K defined by setting ϕ(g) := (ι−1 (gt (π(g))−1 ), π(g)) for all g ∈ G. Note that ϕ is well defined since gt (π(g))−1 ∈ ker(π ) = ι(H ) and π(g) ∈ K for all g ∈ G. For g1 , g2 ∈ G we have ϕ(g1 g2 ) = (ι−1 (g1 g2 t (π(g1 g2 ))−1 ), π(g1 g2 )) = (ι−1 (g1 g2 t (π(g2 ))−1 t (π(g1 ))−1 ), π(g1 )π(g2 )) = (ι−1 (g1 t (π(g1 ))−1 · t (π(g1 ))g2 t (π(g2 ))−1 t (π(g1 ))−1 ), π(g1 )π(g2 )) = (ι−1 (g1 t (π(g1 ))−1 ) .

· ι−1 (t (π(g1 ))ι(ι−1 (g2 t (π(g2 ))−1 ))t (π(g1 ))−1 ), π(g1 )π(g2 )) = (ι−1 (g1 t (π(g1 ))−1 ) · α(π(g1 ))(ι−1 (g2 t (π(g2 ))−1 )), π(g1 )π(g2 )) = (ι−1 (g1 t (π(g1 ))−1 ), π(g1 )) · (ι−1 (g2 t (π(g2 ))−1 ), π(g2 )) = ϕ(g1 ) · ϕ(g2 ).

This shows that ϕ is a group homomorphism. Let h ∈ H and k ∈ K. Then the element g := ι(h)t (k) ∈ G satisfies π(g) = k so that gt (π(g))−1 = ι(h)t (k)t (k)−1 = ι(h) and therefore ϕ(g) = (h, k), showing that ϕ is surjective. Finally let g ∈ G such that ϕ(g) = (1H , 1K ). Since π(g) = 1K , we deduce that g ∈ ker(π ) = ι(H ). As 1H = ι−1 (gt (π(g))−1 ) = ι−1 (gt (1K )−1 ) = ι−1 (g1G ) = ι−1 (g), we deduce that g = ι(1H ) = 1G , and injectivity of ϕ follows. We deduce that ϕ is a group isomorphism. We are only left to show that ϕ yields the required equivalence of the short exact sequences (2.11) and (2.14). This follows immediately since ϕ(ι(h)) = (h, 1K ) = ια (h) for all h ∈ H and πα (ϕ(g)) = πα (ι−1 (gt (π(g))−1 ), π(g)) = π(g) for all g ∈ G, showing that ια = ϕ ◦ ι and π = πι ◦ ϕ.

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(d) Let 1 → H → G → K → 1 be a short exact sequence of groups. Let X ⊂ K be a free base for K. Since π : G → K is surjective, we can find, for each x ∈ X, an element s(x) ∈ G such that π(s(x)) = x. This defines a map s : X → G. Since K is free with base X, the map s uniquely extends to a group homomorphism t : K → G. We have π(t (x)) = π(s(x)) = x for all x ∈ X. As X generates K, we deduce that π ◦ t = IdK . This shows that the exact sequence π ■ 1 → H → G → K → 1 is split. Exercise 2.32 (Internal Semidirect Product) Let G be a group. Let H and K be subgroups of G and denote by ι : H → G and t : K → G the inclusion maps. (a) Show that the following conditions are equivalent. (i) H is normal in G, G = H K, and H ∩ K = {1G }; (ii) H is normal in G and every element g ∈ G can be uniquely written in the form g = hk with h ∈ H and k ∈ K; (iii) there is a short exact sequence of groups ι

π

1 → H → G → K → 1,

.

(2.15)

such that π ◦ t = IdK . (b) Suppose that one (and hence any) of the equivalent conditions in (a) is satisfied. Let α : K → Aut(H ) denote the group homomorphism associated with the action of K on H through conjugation, i.e., α(k)(h) = khk −1 for all k ∈ K and h ∈ H (this is well defined since H is normal in G). Show that the map ϕ : H α K → G defined by ϕ((h, k)) := hk for all h ∈ H and k ∈ K, is a group isomorphism. Solution (a) Suppose (i). We claim that every element g ∈ G can be uniquely written as a product g = hk with h ∈ H and k ∈ K. Indeed, such elements h and k exist since G = H K. Suppose that h1 k1 = h2 k2 , where h1 , h2 ∈ H and k1 , k2 ∈ K. −1 Then h−1 2 h1 = k2 k1 ∈ H ∩ K = {1G }, so that h1 = h2 and k1 = k2 . This proves the claim. This shows the implication (i) =⇒ (ii). Suppose (ii). Define a map π : G → K by setting π(hk) = k for all h ∈ H and k ∈ K. Using the fact that H is normal in G, we have, for all h1 , h2 ∈ H and k1 , k2 ∈ K, π(h1 k1 h2 k2 ) = π(h1 (k1 h2 k1−1 ) · k1 k2 ) = k1 k2 = π(h1 k1 )π(h2 k2 ),

.

showing that π is a group homomorphism. We have π(k) = k for all k ∈ K. Therefore π is surjective and satisfies π ◦ t = IdK . It is clear that ker(π ) = H = ι(H ), so that (2.15) is a short exact sequence of groups. This shows that (ii) =⇒ (iii). Finally, suppose (iii). As (2.15) is an exact sequence, we have ι(H ) = ker(π ) so that H = ι(H ) is normal in G. On the other hand, since π ◦ t = IdK , we have π(k) = k for all k ∈ K. As H = ker(π ), we deduce that H ∩ K = {1G }. Let g ∈ G

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and set k := π(g) ∈ K. We have π(gk −1 ) = π(g)π(k)−1 = kk −1 = 1G , so that g = hk with h := gk −1 ∈ ker(π ) = H . Therefore G = H K. This shows (iii) =⇒ (i). (b) Let h1 , h2 ∈ H and k1 , k2 ∈ K. We have ϕ(h1 , k1 )ϕ(h2 , k2 ) = h1 k1 h2 k2

.

= h1 (k1 h2 k1−1 )k1 k2 = ϕ(h1 (k1 h2 k1−1 ), k1 k2 ) = ϕ((h1 , k1 ) · (h2 , k2 )), showing that ϕ is a group homomorphism. As ϕ is bijective by (ii), we conclude that ϕ is a group isomorphism. ■ Exercise 2.33 (Residual Finiteness of Semidirect Products) Let H and K be two groups and let α : K → Aut(H ) be a group homomorphism from H into the automorphism group of K. Consider the semidirect product G := H α K. One can regard H and K as subgroups of G via the group embeddings ι : H → G and t : K → G introduced in Exercise 2.30. Every element g ∈ G can be uniquely written in the form g = hk with h ∈ H and k ∈ K, and the map π : G → K defined by π(g) := k is a group epimorphism with kernel H . Moreover, one has khk −1 = α(k)(h) for all h ∈ H and k ∈ K. Suppose that the groups H and K are residually finite and that the group H is finitely generated. The goal of the exercise is to show that G is residually finite. (a) Let g ∈ G\H . Show that there is a finite group F and a group homomorphism φ : G → F such that φ(g) /= 1F . (b) Let L be a finite index subgroup of H . Show that there exists a characteristic subgroup C of H which is of finite index in H such that C ⊂ L. (c) Show that α induces a group homomorphism ^ α : K → Aut(H /C). (d) Let N denote the kernel of ^ α . Show that N is a finite index normal subgroup of K and that ^ α induces a group homomorphism β : K/N → Aut(H /C). (e) Consider the semidirect product F := (H /C) β (K/N). Show that the group F is finite and that the map φ : G → F , defined by φ(g) := (hC, kN) for all g = hk ∈ G, h ∈ H and k ∈ K, is a group homomorphism. (f) Use (a)–(e) to show that the group G is residually finite. Solution (a) As g ∈ / H = ker(π ), the group homomorphism π : G → K satisfies that k := π(g) /= 1K . Since K is residually finite, there is a finite group F and a homomorphism ψ : K → F such that ψ(k) /= 1F . Then the composite homomorphism φ := ψ ◦ π : G → F satisfies that φ(g) = ψ(k) /= 1F . (b) This follows from Exercise 2.29(d) since H is finitely generated. (c) For all k ∈ K, we have α(k)(C) = C, since C is a characteristic subgroup of H . Therefore α(k) induces a group automorphism ^ α (k) of H /C satisfying

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2 Residually Finite Groups

^ α (k)(hC) := α(k)(h)C for all h ∈ H . The map ^ α : K → Aut(H /C) is clearly a group homomorphism since α : K → Aut(H ) is a group homomorphism. (d) The group Aut(H /C) is finite since C is of finite index in H . As N is the kernel of the group homomorphism ^ α : K → Aut(H /C), we deduce that N is a normal subgroup of finite index in K. Moreover, ^ α induces a quotient group homomorphism β : K/N → Aut(H /C). (e) The group F is finite since H /C and K/N are both finite. Let g, g  ∈ G and let h, h ∈ H and k, k  ∈ K such that g = hk and g  = h k  . We then have φ(gg  ) = φ(hk · h k  )

.

= φ(hα(k)(h ) · kk  )

(by definition of α)

= (hα(k)(h )C, kk  N) 



(by definition of φ)

= (hβ(kN )(h C), kk N)

(by definition of β)

= (hC, kN )(h C, k  N)

(by definition of the multiplication in F )

 

= φ(hk)φ(h k )

(by definition of φ)

= φ(g)φ(g  ). This shows that φ is a group homomorphism. (f) Let g ∈ G \ {1G }. Let us show that there exist a finite group F and a group homomorphism φ : G → F such that φ(g) /= 1F . If g ∈ / H then by (a) we are done. Now suppose that g ∈ H . Since H is residually finite, there is a finite index subgroup L ⊂ H with g ∈ / L. Choose C ⊂ L as in (b) and keep the notations introduced in (c), (d), and (e). Then the group homomorphism φ : G → F = (H /C) β (K/N) satisfies φ(g) = φ(g · 1K ) = (gC, N ) /= (C, N ) = (1H /C , 1K/N ) = 1F

.

since g ∈ / C. As F is finite by (e), this completes the proof that G is residually finite. ■ Comment The final result of Exercise 2.33, namely that the semidirect product of a finitely generated residually finite group with a residually finite group is itself residually finite, is due to Mal’cev [Mal2]. Note that there are semidirect products of residually finite groups that are not residually finite (cf. [CAG, Proposition 2.6.5] and Exercise 2.36(i)). Exercise 2.34 Let G be a group. Suppose that G contains a finitely generated normal subgroup H such that H is residually finite and G/H is free. Show that G is residually finite. Solution Since G/H is free, the short exact sequence 1 → H → G → G/H → 1 splits (cf. Exercise 2.31(d)), so that G is isomorphic to a semidirect product H α G/H for some group homomorphism α : G/H → Aut(H ) (cf. Exercise 2.31). As

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every free group is residually finite, it follows from Exercise 2.33 that G is residually finite. ■ Exercise 2.35 Let G be a group. Suppose that G contains a finitely generated normal subgroup H such that H is residually finite and G/H is (finite or infinite) cyclic. Show that G is residually finite. Solution In the case when G/H is finite, this follows from the fact that every virtually residually finite group is itself residually finite. When G/H is infinite and hence isomorphic to Z, this is a particular case of Exercise 2.34. ■ Exercise ⊕ 2.36 (Wreath Product) Let K and H be two groups. Consider the group D := h∈H Kh , the direct sum of a family (Kh )h∈H of copies Π of K indexed by H . In other words, D is the subgroup of the product group h∈H K = K H = {x : H → K} consisting of all d : H → K such that d(h) = 1K for all but finitely many h ∈ H . Let σ : H → Aut(D) denote the group homomorphism associated with the restriction to D ⊂ K H of the shift action of H on K H . This means that, given h ∈ H and d ∈ D, the element d  := σ (h)(d) ∈ D is given by the formula d  (h ) = d(h−1 h ) for all h ∈ H . The semidirect product G := D σ H associated with σ (cf. Exercise 2.33) is called the wreath product of the groups K and H and is denoted by K  H . Since G is a semidirect product of D and H , the groups D and H can be regarded as subgroups of G. We also regard K as a subgroup of D (and hence of G) by using the identification of K1H with K. (a) Show that hKh−1 = Kh for all h ∈ H . (b) Let d ∈ D. Show that d can be uniquely written in the form d=

Π

.

hkh h−1 ,

h∈S

where S ⊂ H is a finite subset and 1K /= kh ∈ K for all h ∈ S. (c) Show that G is generated by K ∪ H . (d) Show that G is finitely generated whenever K and H are both finitely generated. (e) Suppose that K is residually finite and that H is finite. Show that G is residually finite. (f) Suppose that K is abelian. Let H  be a group and let ϕ : H → H  be a group homomorphism. Let G := K  H  . Show that ϕ uniquely extends to a group homomorphism Φ : G → G fixing every element of K. (g) Suppose that K and H are residually finite and that K is abelian. Show that G is residually finite. (h) Suppose that K is non-abelian and that H is infinite. Show that G is not residually finite. (i) Show that the group G is residually finite if and only if the following two conditions are both satisfied: (1) the groups K and H are both residually finite, (2) the group K is abelian or the group H is finite.

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(j) Suppose that H is infinite. Show that if the group K is non-trivial then G is not of finite Markov type. Solution (a) We have ⊕ hKh−1 = hK1H h−1 = σ (h)(K1H ) = Kh . (b) As D = h∈H Kh , the element d can be uniquely written in the form d=

Π

.

kh

h∈S

with S ⊂ H finite and 1K /= kh ∈ Kh for all h ∈ S. To complete the proof, it suffices to observe that every kh can be uniquely written in the form kh = hkh h−1 , with kh ∈ K, by (a). (c) This immediately follows from (b) since G = DH . (d) This immediately follows from (c). (e) It suffices to show that for every g ∈ G \ {1G }, there is a finite index subgroup L of G such that g ∈ / L. If g ∈ / D, then we can take L := D since G/D = H is finite. Suppose now that 1G /= g ∈ D. The group D is residually finite since K is residually finite and the class of residually finite groups is closed under taking direct sums. Therefore there exists a subgroup of finite index N ⊂ D such that g ∈ / N. As D is of finite index in G, we deduce that N is also of finite index in G. Thus we can take L := N. (f) Uniqueness ⊕ follows from the fact that K ∪ H generates G by (c). Let D  := h ∈H  Kh ⊂ G denote the direct sum of a family of copies of K indexed by H  . Then the canonical isomorphisms Kh → Kϕ(h) (h ∈ H ), given by hkh−1 |→ ϕ(h)kϕ(h)−1 for all k ∈ K, extend to a group homomorphism ϕ : D → D  (here we use the hypothesis that K is abelian). Observe that ϕ(hdh−1 ) = ϕ(h)ϕ(d)ϕ(h)−1

.

(2.16)

for all h ∈ H and d ∈ D. Recall that every g ∈ G can be uniquely written in the form g = dh with d ∈ D and h ∈ H . We define a map Φ : G → G by setting for all g = dh, with d ∈ D and h ∈ H , Φ(g) := ϕ(d)ϕ(h).

.

(2.17)

Clearly Φ satisfies Φ(h) = ϕ(h) for all h ∈ H and Φ(k) = ϕ(k) = k for all k ∈ K. Let now g1 , g2 ∈ G and write gi = di hi , where di ∈ D and hi ∈ H , for each i ∈ {1, 2}. We then have Φ(g1 g2 ) = Φ(d1 h1 d2 h2 )

.

= Φ(d1 (h1 d2 h−1 1 )h1 h2 )

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149

= ϕ(d1 (h1 d2 h−1 1 ))ϕ(h1 h2 )

(by (2.17))

= ϕ(d1 )ϕ(h1 d2 h−1 1 )ϕ(h1 )ϕ(h2 ) = ϕ(d1 )ϕ(h1 )ϕ(d2 )ϕ(h1 )−1 ϕ(h1 )ϕ(h2 )

(by (2.16))

= ϕ(d1 )ϕ(h1 )ϕ(d2 )ϕ(h2 ) = Φ(g1 )Φ(g2 )

(by (2.17)).

This shows that Φ is a group homomorphism. (g) Let g ∈ G \ {1G }. If g ∈ / D, then the canonical quotient homomorphism π : G → G/D = H satisfies that h := π(g) /= 1H . Since H is residually finite, we can find a finite group F and a group homomorphism ψ : H → F such that ψ(h) /= 1F . Then the composite homomorphism φ := ψ ◦ π : G → F satisfies that φ(g) = ψ(h) /= 1F . Suppose now that 1G /= g ∈ D. By (b), we can write g=

Π

.

hkh h−1 ,

h∈S

for some non-empty finite subset S ⊂ H and 1K /= kh ∈ K for all h ∈ S. Since H is residually finite, we can find, by the characterization (RF3) of residual finiteness, a finite group H  and a group homomorphism ϕ : H → H  whose restriction to S is injective. Let G := K  H  . By (f), there is a group homomorphism Φ : G → G extending ϕ that fixes every element of K. Then g  := Φ(g) satisfies g =

Π

.

ϕ(h)kh ϕ(h)−1 .

h∈S

As the restriction of ϕ to S is injective, we deduce that g  /= 1G . Since H  is a finite group and K is residually finite, the group G is residually finite by (d). Therefore we can find a finite group F and a group homomorphism ψ : G → F such that ψ(g  ) /= 1F . It follows that the composite homomorphism φ := ψ ◦ Φ : G → F satisfies that φ(g) = ψ(g  ) /= 1F . This shows that G is residually finite. (h) Suppose by contradiction that G is residually finite. Since K is non-abelian, there are elements k1 , k2 ∈ K such that k1 k2 /= k2 k1 . As G is residually finite, there exist a finite group F and a group homomorphism φ : G → F such that φ(k1 k2 ) /= φ(k2 k1 ). Since φ is a group homomorphism, it follows that the elements φ(k1 ) and φ(k2 ) do not commute. On the other hand, as F is finite and H is infinite, we can find a non-trivial element h ∈ H in the kernel of φ. The element hk1 h−1 ∈ Kh commutes with k2 since h /= 1H . Since φ(hk1 h−1 ) = φ(h)φ(k1 )φ(h)−1 = φ(k1 ), we deduce that φ(k1 ) commutes with φ(k2 ), a contradiction.

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(i) Suppose first that G is residually finite. Then H and K are residually finite since the class of residually finite groups is closed under taking subgroups. If K is non-abelian, then H is finite by (h). Thus conditions (1) and (2) are both satisfied. Conversely, suppose that K and H are residually finite. If K is abelian, then G is residually finite by (g). On the other hand, G is residually finite if H is finite by (e). This shows that G is residually finite if conditions (1) and (2) are both satisfied. (j) If K is not trivial then the group D is not finitely generated. As D is a subgroup of G, this implies that G is not of finite Markov type by Exercise 1.111(e). ■ Comment The characterization of groups whose wreath product is residually finite given in (i) is due to Gruenberg [Gru]. Given a group K, it follows from (d) and (i) that the wreath product K  Z is finitely generated whenever K is finitely generated and that K  Z is residually finite if and only if K is abelian and residually finite. For example, the groups (Z/2Z)  Z and Z  Z are finitely generated and residually finite, while the group Sym(n)  Z is finitely generated but not residually finite as soon as n ≥ 3 (cf. [CAG, Proposition 2.6.5]). The group (Z/2Z)  Z is commonly known as the lamplighter group (see Exercise 4.44). The wreath product K  H as defined in this exercise is sometimes called the restricted wreath product of the groups K and H while the unrestricted wreath product of the groups K and H is defined as being the semidirect product K u H := (K H ) η H , where η : H → Aut(K H ) is the group homomorphism associated with the shift action of H on K H . Clearly K  H = K u H if the group H is finite. Exercise 2.37 Let n ≥ 1 be an integer. All matrices considered here are n × n matrices with entries in Z. An elementary row operation on matrices is a transformation of one of the following types: Type 1: Type 2: Type 3: Type 4:

multiplication of row i by −1, operation denoted by Ri ← −Ri ; interchange of rows i and j , operation denoted by Ri ↔ Rj ; addition of row j to row i, operation denoted by Ri ← Ri + Rj ; subtraction of row j to row i, operation denoted by Ri ← Ri − Rj .

Here 1 ≤ i, j ≤ n and i /= j . The elementary column operations Ci ← −Ci , Ci ↔ Cj , Ci ← Ci + Cj , and Ci ← Ci − Cj are defined analogously. A matrix is called an elementary matrix if it can be obtained from the identity matrix In by applying an elementary row operation. One says that a matrix A is equivalent to a matrix B, and one writes A ∼ B, if B can be obtained from A by performing a finite sequence of elementary row and column operations. (a) Show that the transpose of an elementary matrix is an elementary matrix. (b) Show that every elementary matrix is in GLn (Z) and that its inverse is an elmentary matrix. (c) Show that the matrix obtained from a matrix A by applying an elementary row (resp. column) operation T is of the form T (A) = EA (resp. T (A) = AE), where E is an elementary matrix. (d) Show that ∼ is an equivalence relation.

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151

(e) Let A ∈ GLn (Z). Show that every matrix that is equivalent to A is in GLn (Z). (f) Let A ∈ GLn (Z) and let (a1 a2 · · · an ) be the first row of A. Show that gcd(a1 , a2 , . . . , an ) = 1. (g) Let A ∈ GLn (Z). Show that A is equivalent to a matrix whose first row is (1 0 0 · · · 0). (h) Let A be an n×n matrix with entries in Z. Show that the following conditions are equivalent: (C1) (C2) (C3)

A ∈ GLn (Z); A is equivalent to In ; A is a finite product of elementary matrices.

(i) Show that the group GLn (Z) is finitely generated. (j) Show that the group SLn (Z) is finitely generated. Solution (a) For 1 ≤ i, j ≤ n and i /= j , denote by Di (resp. Sij , resp. Tij , resp. Uij ) the elementary matrix obtained from In by applying the elementary row operation Ri ← −Ri (resp. Ri ↔ Rj , resp. Ri ← Ri + Rj , resp. Ri ← Ri − Rj ). Denoting by t A the transpose of a matrix A, one easily checks that t Di = Di , t S = S , t T = T , and t U = U . This shows that the transpose of every ij ij ij ji ij ji elementary matrix is an elementary matrix. (b) The determinant of a square matrix is a multilinear and antisymmetric function of its rows. It follows that each elementary row operation on a square matrix preserves the determinant up to sign. Thus every elementary matrix E satisfies det(E) = ± det(In ) = ±1 so that E ∈ GLn (Z). In fact, with the notation introduced in the solution of (a), we have Di−1 = Di , Sij−1 = Sij , Tij−1 = Uij , and Uij−1 = Tij . Thus the inverse of every elementary matrix is an elementary matrix. (c) When applying an elementary row operation T to a matrix A, each row of T (A) is a linear combination of the rows of A. Thus there is a matrix E such that T (A) = EA. We have T (In ) = EIn = E, showing that E is an elementary matrix. If now T is an elementary column operation, then the transpose of T (A) is obtained by applying an elementary row operation to the transpose of A. Therefore t (T (A)) = E tA for some elementary matrix E. This gives us T (A) = A tE. As the transpose of an elementary matrix is itself an elementary matrix by (a), this completes the proof. (d) Clearly ∼ is reflexive and transitive. To prove symmetry, it is enough to observe that if B is obtained from A by applying an elementary row or column operation, then B = EA or B = AE for some elementary matrix E by (c), so that A = E −1 B or A = BE −1 . As E −1 is an elementary matrix by (b), we deduce that A can be obtained from B by applying an elementary row or column operation. (e) If B is equivalent to A, then it follows from (c) that there are elementary matrices E1 , . . . , Ep and F1 . . . Fq such that B = Ep . . . E1 AF1 . . . Fq .

.

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As every elementary matrix is in GLn (Z) by (b), this implies that B ∈ GLn (Z). (f) If d := gcd(a1 , a2 , . . . , an ), then d can be factor out of det(A). As det(A) = ±1 since A ∈ GLn (Z), we must have d = 1. (g) By applying elementary column operations of Type 1, we can transform A into a matrix A whose first row contains only non-negative entries. Let M denote the largest integer appearing in the first row of A . Note that M ≥ 1 since det(A ) /= 0. Suppose that M appears at column i in the first row of A . By applying elementary column operations of the form Cj ← Cj − Ci with j /= i, we can transform A into a matrix A such that all entries in the first row are in the interval [0, M] with M appearing only once in the first row (at column i). If there is a non-zero entry in the first row of A , say at column j , other than M, then by applying to A the elementary column operation Ci ← Ci −Cj , we arrive to a matrix A whose entries in the first row are all in [0, M  ] for some 0 < M  < M. Continuing this way, the process must stop after a finite number of steps when arriving to a matrix B with only one non-zero entry in the first row. This non-zero entry must be 1 since the entries in the first row of B are setwise coprime by (f). After possibly applying to B an elementary column operation of Type 2, we arrive to a matric C equivalent to A whose first row is (1 0 0 · · · 0) as desired. (h) Let us show, by induction on n, that every matrix A ∈ GLn (Z) is equivalent to In . If n = 1, then A = I1 or A = −I1 and the result is obvious. Suppose now that the statement is true for all matrices in GLn−1 (Z) and that A ∈ GLn (Z). By (d), we can find a finite sequence of elementary row operations transforming the first row of A into (1 0 0 · · · 0). Then, by a finite sequence of elementary row operations of the form Ri ← Ri ± R1 , where 2 ≤ i ≤ n, we can transform the first column of the matrix into the column t (1 0 0 · · · 0) without modifying the first row. This way, A is equivalent to a matrix of the form I1 ⊕ A , where A ∈ GLn−1 (Z). Now, elementary row/column operations with 2 ≤ i, j ≤ n will not affect the first row and first column of I1 ⊕ A . By induction, we can find a finite sequence of such elementary row/column operations transforming A into In−1 . The resulting form of the full matrix will then be I1 ⊕ In−1 = In , as desired. This shows the implication (C1) ⇒ (C2). If the matrices A and In are equivalent, then we deduce from (c) that there are elementary matrices E1 , . . . , Ep and F1 . . . Fq such that A = Ep · · · E1 In F1 · · · Fq = Ep · · · E1 F1 · · · Fq .

.

This shows that (C2) ⇒ (C3). Finally, (C3) ⇒ (C1) follows from (b). (i) The implication (C1) ⇒ (C3) in (h) shows that the set of elementary matrices generates GLn (Z). Since this set is finite, we deduce that the group GLn (Z) is finitely generated. (j) This immediately follows from (i). Indeed, SLn (Z) is a subgroup of index 2 of GLn (Z) and every finite index subgroup of a finitely generated group is itself finitely generated [CAG, Proposition 6.6.6]. ■

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153

Comment The implication (C1) ⇒ (C3) in (h) actually shows that GLn (Z) is generated as a monoid by the n(5n − 3)/2 elementary matrices Di , Sij , Tij , and Uij (1 ≤ i, j ≤ n, i /= j ). Note that the n2 elmentary matrices Di , Tij suffice to generate GLn (Z) as a monoid since Uij = Dj Tij Dj and Sij = Di Tij Dj Tj i Di Tij . For instance, as ) ( 0 ) GL2 (Z) (is1 generated ( a )monoid by the ( 4) elementary matrices 0 , T := 1 1 , and T := 1 0 . The two elementary := , D D1 := −1 2 1 2 01 11 0 1 ( 0 )−1 matrices T1 and S := 01 10 suffice to generate GL2 (Z) as a group since T2 = ST1 S, D1 = T1−1 ST1 ST1−1 S, and D2 = ST1−1 ST1 ST1−1 . It is known that the group GLn (Z) can be generated by two suitably chosen matrices for any n. For instance, the matrices ⎛ ⎞ 100 . ⎝ 1 1 0⎠ 001

⎛ and

⎞ 0 10 ⎝ 0 0 1⎠ −1 0 0

generate GL3 (Z) (see [Tro], [CoxM, p. 85]). Exercise 2.38 Let n ≥ 1 be an integer. Show that the group GLn (Z) is Hopfian. Solution The group GLn (Z) is finitely generated by Exercise 2.37 and residually finite (cf. [CAG, Proposition 2.1.5]). Therefore GLn (Z) is Hopfian by Mal’cev’s theorem (cf. [CAG, Theorem 2.4.3]). ■ Exercise 2.39 Let A be a finitely generated abelian group and let G := Aut(A) denote its automorphism group. (a) Prove that G is residually finite. (b) Prove that G is finitely generated. (c) Prove that G is Hopfian. Solution (a) This follows from Baumslag’s theorem since every finitely generated abelian group is residually finite (b) By the structure theorem for finitely generated abelian groups, we know that A = T ⊕ F where T is a finite subgroup of A and F is a free abelian subgroup of A with finite rank r ≥ 0. Observe that every automorphism α of A satisfies α(T ) = T and therefore induces an automorphism of A/T . As A/T ∼ = Zr , we deduce =F ∼ that there is a short exact sequence of groups 1 → Aut(T ) → G → GLr (Z) → 1.

.

The group Aut(T ) is finite since T is finite and the group GLr (Z) is finitely generated by Exercise 2.37. As the class of finitely generated groups is closed under taking extensions, we deduce that G is finitely generated. (c) This immediately follows from (a) and (b) by applying Mal’cev’s theorem (cf. [CAG, Theorem 2.4.3]).. ■

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Comment There exist finitely generated groups, and even finitely presented groups, whose automorphism groups are not finitely generated. For instance, it is known that the automorphism group of the Baumslag-Solitar group BS(2, 4) := is not finitely generated (see [ColL]). Exercise 2.40 Let G denote the subgroup of Sym(Z) generated by the translation T : n |→ n + 1 and the transposition S = (0 1) (thus G is the group G1 considered in Section 2.6 of [CAG]). Let H be a non-trivial normal subgroup of G. Show that the group H is not residually finite. Solution Denote, as in [CAG, Section 2.6], by Sym0 (Z) the subgroup of Sym(Z) consisting of all permutations of Z with finite support and by Sym+ 0 (Z) the subgroup of Sym0 (Z) consisting of all permutations of Z with finite support and signature 1. Recall that every g ∈ G can be uniquely written in the form g = T k σ with σ ∈ Sym0 (Z) and k ∈ Z, and that the map u : G → Z, defined by u(g) := k for all g ∈ G, is a surjective group homomorphism with kernel Sym0 (Z) (cf. [CAG, + Proof of Proposition 2.6.1]). The subgroup H ∩ Sym+ 0 (Z) is normal in Sym0 (Z). + As Sym0 (Z) is a simple group (cf. [CAG, Theorem C.4.3]) and H is non-trivial, it + + follows that H ∩ Sym+ 0 (Z) = {1} or H ∩ Sym0 (Z) = Sym0 (Z). + + + If H ∩ Sym+ 0 (Z) = Sym0 (Z), then Sym0 (Z) is a subgroup of H . As Sym0 (Z) is a non-trivial simple group, it is not residually finite. Using the fact that the class of residually finite groups is closed under taking subgroups, we then deduce that H is not residually finite in this case. + Suppose now that H ∩ Sym+ 0 (Z) = {1}. Since Sym0 (Z) is a subgroup of Sym0 (Z) of index 2, this implies that H ∩ Sym0 (Z) is either trivial or has order 2. Suppose first that H ∩ Sym0 (Z) = {1}. Then the restriction of u to H is injective so that H is infinite cyclic. Let α be a generator of H . Since H is normal in G, for every g ∈ G, there exists n ∈ Z such that gαg −1 = α n . By applying the homomorphsm u to each side of this equality, we get n = 1. Thus every element of G commutes with α. This would imply that the conjugacy class of α in G is reduced to α, which is impossible since the conjugacy class of every non-trivial element in G is infinite. Suppose now that H ∩ Sym0 (Z) has order 2. Then the non-trivial element β ∈ H ∩ Sym0 (Z) is a product of an odd number of transpositions with disjoint supports, say β = (i1 i2 )(i3 i4 ) · · · (ik−1 ik )

.

with i1 , i2 , . . . , ik ∈ Z distinct. Let j ∈ Z \ {i1 , i2 , . . . , ik } and consider the element γ ∈ H given by γ := τβτ −1 , where τ := (i1 j ). We then have γ = (j i2 )(i3 i4 ) · · · (ik−1 ik )

.

and βγ = (i1 i2 )(j i2 ) = (i1 i2 j ).

.

2.2 Exercises

155

As βγ ∈ H , we deduce that H contains a 3-cycle. This is a contradiction since any 3-cycle has signature 1 and we are assuming H ∩ Sym+ ■ 0 (Z) = {1}. Comment By taking H = G, we recover the fact that G is not residually finite (cf. [CAG, Proposition 2.6.1]). Exercise 2.41 Show that the wreath product G := Sym+ 5 Z is a Hopfian group. Solution By definition of the wreath product (see Exercise 2.36), the group G is ⊕ the semidirect product G = H ψ Z, where H := H i∈Z i is the direct sum of a family (Hi )i∈Z of copies of the alternating group Sym+ 5 and ψ : Z → Aut(H ) is the group homomorphism associated with the shift action of Z on H . Every h ∈ H can be uniquely written in the form h=

∑

.

(2.18)

hi ,

i∈Z

where hi ∈ Hi for all i ∈ Z and hi = 1Hi for all but finitely many i ∈ Z. The subgroup H ⊂ G is the kernel of a ∑ surjective homomorphism ρ : G → Z and there is t ∈ ρ −1 (1) such that tht −1 = i∈Z hi−1 for all h ∈ H written as in (2.18). Every g ∈ G can be uniquely written in the form g = t n h, where n = ρ(g) ∈ Z and h ∈ H . Let φ : G → G be a surjective group endomorphism of G. Note that H is the set of torsion elements of G. As the image under φ of every torsion element is also a torsion element, it follows that φ(H ) ⊂ H . This implies in particular that ρ(φ(G)) = ρ(G) = Z is generated by ρ(φ(t)), so that ρ(φ(t)) ∈ {−1, 1}. In the sequel, we set ε := ρ(φ(t)) ∈ {−1, 1}. On the other hand, for every i ∈ Z, we have Hi = t −i H0 t i . We deduce that the group φ(H0 ) is not trivial. Indeed, otherwise we would have φ(Hi ) = φ(t −i H0 t i ) = φ(t)−i φ(H0 )φ(t)i = {1G } for all i ∈ Z and hence φ(H ) = {1G }. This would imply that φ(G) is an infinite cyclic group, contradicting the surjectivity of φ. Consider now, for each i ∈ Z, the projection map πi : H → Hi and the group homomorphism ψi : H0 → Hi defined by ψi (h0 ) := πi (φ(h0 )) for all h0 ∈ H0 . Since Sym+ 5 is a simple group (cf. [CAG, Theorem C.4.3]), ψi is either the trivial homomorphism or an isomorphism. As φ(H0 ) is a non-trivial finite group, the integers i1 := min{i ∈ Z : ψi is an isomorphism} and i2 := max{i ∈ Z : ψi is an isomorphism} are well defined, and we have φ(H0 ) ⊂

⊕

.

Hi .

(2.19)

i1 ≤i≤i2

Let k ∈ Z. Then ρ(φ(t −k )) = −kρ(φ(t)) = −kε and hence φ(t −k ) = t −kε uk for some uk ∈ H . Since Hk = t −k H0 t k , it follows that kε φ(Hk ) = φ(t −k H0 t k ) = φ(t −k )φ(H0 )φ(t k ) = t −kε uk φ(H0 )u−1 k t

.

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2 Residually Finite Groups

for all k ∈ Z. This gives us ⎛ φ(Hk ) ⊂ t −kε uk ⎝

⊕

.

=

⊕

⎞ kε Hi ⎠ u−1 k t

(by (2.19))

i1 ≤i≤i2 kε (t −kε uk Hi u−1 k t )

i1 ≤i≤i2

=

⊕

(t −kε Hi t kε )

(since Hi is normal in H )

i1 ≤i≤i2

=

⊕

Hi+kε ,

i1 ≤i≤i2

and hence, by shifting the summation index, ⊕

φ(Hk ) ⊂

.

(2.20)

Hi

i1 +kε≤i≤i2 +kε

for all k ∈ Z. Suppose now that φ is not injective. Therefore there exists a non-trivial element h in the kernel of φ. As ρ(φ(g)) = ερ(g) for all g ∈ G, we have h ∈ H . After conjugating h by t i for some i ∈ Z, we can assume that h = h0 + h1 + · · · + hn , where n ≥ 0, hk ∈ Hk for all 0 ≤ k ≤ n, and h0 /= 1H0 , hn /= 1Hn . We then have φ(h) = φ(

∑

.

0≤k≤n

hk ) =

∑ 0≤k≤n

φ(hk ) ∈

∑

φ(Hk ).

0≤k≤n

We deduce from (2.20) that ψi1 (h) = ψi1 (h0 ) if ε = 1 and that ψi2 (h) = ψi2 (hn ) if ε = −1. This gives a contradiction since, on the one hand, h is in the kernel of φ and hence in the kernels of ψi1 and ψi2 , while, on the other, h0 and hn are non-trivial elements, and ψi1 and ψi2 are isomorphisms. We deduce that φ is injective. This shows that the group G is Hopfian. ■ Comment The group G, which is the group G2 considered in Section 2.6 of [CAG], is finitely generated but not residually finite (cf. [CAG, Proposition 2.6.5] or Exercise 2.36(d), (h)). Exercise 2.42 (Profinite Completion of a Group) Let G be a group. Denote by Nf q the set of all normal subgroups of finite index of G, partially ordered by reverse inclusion. (a) Show that Nf q is a directed set. (b) For H, K ∈ Nf q with H ⊂ K, let ϕK,H : G/H → G/K denote the canonical homomorphism. Show that the directed set Nf q together with the homomorphisms ϕK,H forms a projective system of groups. The limit of this

2.2 Exercises

157

projective system is called the profinite completion of the group G and is denoted ^ by G. ^ and that the kernel (c) Show that there is a canonical homomorphism η : G → G of η is the residual subgroup of G. (d) Prove that G is residually finite if and only if the canonical homomorphism ^ is injective. η: G → G Solution (a) Let H, K ∈ Nf q . Then N := H ∩ K satisfies that N ∈ Nf q and H, K  N. This shows that the set Nf q is directed. (b) Let H, K, N ∈ Nf q such that H ⊂ K ⊂ N. We have ϕH,H : G/H → G/H is the identity map and ϕN,H = ϕN,K ◦ ϕK,H . This shows that the directed set Nf q together with the homomorphisms ϕK,H form a projective system of groups. (c) We have ^ = {(gH H )H ∈N : gH ∈ G and gK K = gH K for all H, K ∈ Nf q s.t. H ⊂ K}. G fq

.

This is a group with multiplication defined by   (gH H )H ∈Nf q · (gH H )H ∈Nf q := (gH gH H )H ∈Nf q .

.

^ given by Note that the identity element is 1G ^ := (H )H ∈Nf q . The map η : G → G η(g) := (gH )H ∈Nf q is a well defined group homomorphism. Moreover, for g ∈ G we have η(g) = 1G only if gH = H , equivalently, g ∈ H , for all H ∈ Nf q . ^ if and∩ This shows that ker(η) = H ∈Nf q H . (d) Since G is residually finite if and only if the residual subgroup of G is trivial (cf. (RF1) and (RF5)), from (c) we deduce that G is residually finite if and only if ^ is injective. ker(η) = {1G }, equivalently, η : G → G ■ Exercise 2.43 (The Transfer Homomorphism) Let G be a group and let H := Z(G) denote its center. Suppose that H has finite index n in G. Let T ⊂ G be a complete set of representatives for the cosets of H in G. Thus, for each g ∈ G, there exists a unique pair (θ (g), η(g)) ∈ T × H such that g = θ (g)η(g). (a) Let g ∈ G. Show that the map σg : T → T , defined by σg (t) := θ (gt) for all t ∈ T , is bijective. (b) Define a map ϕ : G → H by setting ϕ(g) :=

Π

.

η(gt)

(2.21)

t∈T

for all g ∈ G (observe that ϕ(g) does not depend on the order of the factors in the right-hand side of (2.21) since H is abelian). Show that ϕ is a group homomorphism. (c) Show that ϕ(g) = g n for all g ∈ G. (d) Show that (xy)n = x n y n for all x, y ∈ G.

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2 Residually Finite Groups

Solution (a) The group G acts on G/H by left multiplication. Therefore σg is bijective with inverse σg −1 . (b) Let g1 , g2 ∈ G. Then, given t ∈ T , we have g1 g2 t = g1 θ (g2 t)η(g2 t) = θ (g1 θ (g2 t))η(g1 θ (g2 t))η(g2 t).

.

We deduce that η(g1 g2 t) = η(g1 θ (g2 t))η(g2 t) = η(g1 σg2 (t))η(g2 t). Thus, keeping in mind that H is abelian, we get ϕ(g1 g2 ) =

Π

.

η(g1 g2 t)

t∈T

=

Π

η(g1 σg2 (t))η(g2 t)

t∈T

=

Π

η(g1 σg2 (t))

t∈T

=

Π

η(g1 t)

t∈T

Π

Π

η(g2 t)

t∈T

η(g2 t)

(since σg2 ∈ Sym(T ) by (a))

t∈T

= ϕ(g1 )ϕ(g2 ). This shows that ϕ is a group homomorphism. (c) Let g ∈ G. Using the cycle decomposition of the permutation σg ∈ Sym(T ), we get a partition (Ti )1≤i≤k of T such that, for each 1 ≤ i ≤ k, the set Ti has cardinality ri ≥ 1 and, choosing an arbitrary element ti ∈ Ti , we have Ti = {σgri −1 (ti ), σgri −2 (ti ), . . . , σg (ti ), ti } and σgri (ti ) = ti . As η(gt) = (σg (t))−1 gt for all t ∈ T , we deduce that η(gσgr (ti )) = (σgr+1 (ti ))−1 gσgr (ti )

.

for all 1 ≤ i ≤ k and 0 ≤ r ≤ ri − 1, so that Π .

η(gt) = η(gσgri −1 (ti ))η(gσgri −2 (t)) · · · η(gσg (ti ))η(gti )

t∈Ti

= ti−1 gσgri −1 (ti )(σgri −1 (ti ))−1 gσgri −2 (ti ) · · · (σg2 (ti ))−1 gσg (ti )(σg (ti ))−1 gti = ti−1 g ri ti .

2.2 Exercises

Since

Π t∈Ti

159

η(gt) is in the center of G, this gives us ⎛ g ri = ti ⎝

Π

.

⎞ η(gt)⎠ ti−1 =

t∈Ti

Π

η(gt).

t∈Ti

Consequently, we finally get ϕ(g) =

Π

.

η(gt) =

t∈T

k Π Π

η(gt) =

i=1 t∈Ti

k Π

g ri = g |T | = g n .

i=1

(d) This is an immediate consequence of (b) and (c).

■

Comment The construction of the homomorphism ϕ admits the following generalization. Let G be a group and let H be a finite index subgroup (not necessarily normal or abelian) of G. Fix a complete set of representatives T ⊂ G for the left cosets of H in G. Thus, for each g ∈ G, there exists a unique pair (θ (g), η(g)) ∈ T × H such that g = θ (g)η(g). Suppose we are also given a group homomorphism ψ : H → A from H into an abelian group A. Then the map ψ ∗ : G → A, defined by setting ψ ∗ (g) :=

Π

.

ψ(η(gt))

t∈T

for all g ∈ G, is a group homomorphism which does not depend on the choice of T (cf. [Robi, 10.1.1]). It is called the transfer of ψ. In the case when A = H is the center Z(G) of G, the transfer of ψ = IdH : H → H is the homomorphism ϕ : G → H defined in (2.21), i.e., ϕ = (IdH )∗ . The transfer was introduced by Schur [Schu] and rediscovered by Artin [Art]. The transfer map is an important tool for the study of finite groups and it is used for example in one of the steps of the proof of the classification theorem of finite simple groups. Artin applied transfer to Galois groups in his work on reciprocity laws and principalization in algebraic number theory. If A = H /H  is the abelianization of H (here H  = D(H ) = [H, H ] denotes the derived subgroup of H ), then the transfer of the abelianization map ψ : H → H /H  is called the transfer homomorphism of G into H . The following description of ψ ∗ : G → H /H  is given in [Pas, Lemma 5.1.10]. Let K be a field. Define the projection map π : K[G] → K[H ] by setting π(α) := α|H for all α ∈ K[G] (cf. Exercise 8.85). Then the map fT : K[G] → Matn (K[H ]), defined by setting fT (α) := (π(tαs −1 ))t,s∈T

.

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2 Residually Finite Groups

for all α ∈ K[G] is a K-algebra embedding. Moreover, the abelianization map H → H /H  naturally extends to a K-algebra homomorphism Φ : Matn (K[H ]) → Matn (K[H /H  ]). Then, for all g ∈ G, we have ψ ∗ (g) = ε(g) det(Φ(fT (g))),

.

where ε(g) ∈ {1, −1} is the signature of the permutation σg ∈ Sym(T ) and .

det : Matn (K[H /H  ]) → K[H /H  ]

is the determinant map. Since H /H  is abelian, we have G ⊂ ker(ψ ∗ ) and ψ ∗ induces a homomorphism ver : G/G → H /H 

.

(the notation “ver” comes for the German word Verlagerung (= transfer), a term which was coined by Hasse) [Pas, Chapter 5, Section 1]. There are also various homological interpretations for the definition of transfer (see [Hat, Section 2.G] and [Ser2, Section 7.8]). Exercise 2.44 (FC-Groups) Let G be a group. Denote by Z(G) its center and by D(G) its derived subgroup. One says that G is an FC-group if every conjugacy class of G is finite. (a) Show that if D(G) is finite then G is an FC-group. (b) Show that if Z(G) has finite index in G then G is an FC-group. (c) Show that if G is an FC-group then G/Z(G) is residually finite. (d) Show that if G is a finitely generated FC-group then G/Z(G) is finite. (e) Show that if G is a finitely generated FC-group then there exists an integer d ≥ 0 such that G contains a finite index subgroup isomorphic to Zd . Solution (a) Suppose that D(G) is finite. Let x ∈ G. For every g ∈ G, we have gxg −1 = x(x −1 gxg −1 ) = x[x −1 , g] ∈ xD(G). Thus the conjugacy class of x is finite. This shows that G is an FC-group. (b) The group G acts on itself by conjugation via the map G × G → G given by (g, x) |→ gxg −1 . The orbits of this action are precisely the conjugacy classes of G. As every g ∈ Z(G) satisfies gxg −1 = x for all x ∈ G, this action induces an action of the group G/Z(G) on G with the same orbits. Thus G is an FC-group whenever G/Z(G) is finite. (c) Suppose that G is an FC-group. Let X denote the discrete topological space whose underlying set is G. Let us make G act on X by conjugation as in (b). The kernel of this action is precisely Z(G) so that the induced action of G/Z(G) on X is faithful. This action of G/Z(G) is continuous since the topology on X is the discrete one. Moreover, all orbits are finite since G is an FC-group. Thus the group G/Z(G) satisfies (RF9) and is therefore residually finite.

2.2 Exercises

161

(d) Suppose that G is a finitely generated FC-group and let S ⊂ G be a finite generating subset for G. Denote by C(s) the Πconjugacy class of s ∈ S. Consider the group homomorphism φ : G → F := s∈S Sym(C(s)) obtained by making G act by conjugation on the conjugacy classes of the elements of S. An element g ∈ G is in the kernel of φ if and only if it commutes with each element in S. As S generates G, we deduce that the kernel of φ is Z(G). It follows that the group G/Z(G) embeds in F . As F is finite, this shows that G/Z(G) is itself finite. (e) Suppose that G is a finitely generated FC-group. Then Z(G) is of finite index in G by (d). As G is finitely generated, this implies that Z(G) is itself finitely generated (cf. [CAG, Proposition 6.6.2]). From the classification of finitely generated abelian groups, there exist an integer d ≥ 0 and a finite group F such that Z(G) ∼ = Zd × F . It follows that G contains a finite index subgroup isomorphic to d Z . ■ Comment FC-groups were investigated by Baer in [Bae]. Being an FC-group is a finiteness condition for groups in the sense that all finite groups are FC-groups and there exist infinite groups that are not FC. All finite groups, all abelian groups, and, more generally, all finite-by-abelian groups are FC-groups by (a). The class of FC-groups is clearly closed under taking subgroups, quotients, and direct sums. The infinite dihedral group D∞ is not an FC-group since it contains an infinite conjugacy class consisting of all elements of order 2. As there is a short exact sequence of groups 0 → Z → D∞ → Z/2Z → 0,

.

this shows in particular that the class of FC-groups is not closed under taking extensions and that an abelian-by-finite group may fail to be a FC-group. However, central-by-finite groups are FC by (b). Conversely, finitely generated FC-groups are central-by-finite by (d). It turns out that every central-by-finite group is finite-byabelian (see [Neu3, Theorem 5.3]). The monographs [Gor] and [Tom] are entirely devoted to FC-groups. Exercise 2.45 Let G be an FC-group. (a) Let x, y ∈ G. Let H denote the subgroup of G generated by x and y. Show that the center of H has finite index in H . (b) Deduce from (a) and Exercise 2.43 that every torsion-free FC-group is abelian. (c) Show that the subset T ⊂ G, consisting of all elements of G with finite order, is a characteristic (and therefore normal) subgroup of G. (d) Show that G/T is a torsion-free abelian group. Solution (a) For h ∈ H , let ZH (h) := {k ∈ H : hk = kh} denote the centralizer of h in H . Since x and y generate H , the center Z(H ) of H satisfies Z(H ) = ZH (x) ∩ ZH (y). As every subgroup of an FC-group is an FC-group, the group H is itself an FC-

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2 Residually Finite Groups

group. Therefore ZH (x) and ZH (y) have both finite index in H . It follows that Z(H ) also has finite index in H . (b) Suppose that G is torsion-free. Keeping the notation introduced in (a) and denoting by n the index of Z(H ) in H , we deduce from Exercise 2.43 that the map ϕ : h |→ hn is a group homomorphism from H into Z(H ). Thus, since Z(H ) is abelian, we obtain 1H = [ϕ(x), ϕ(y)] = ϕ([x, y]) = [x, y]n .

.

As H ⊂ G is torsion-free, this implies [x, y] = 1H . Consequently, x and y commute. As x and y were arbitrary, this shows that G is abelian. (c) Let x, y ∈ T with respective orders p and q. Keeping the notation introduced in (a) and using the fact that ϕ(x), ϕ(y)−1 ∈ Z(H ) commute, we get (xy −1 )npq = (ϕ(xy −1 ))pq = (ϕ(x)ϕ(y)−1 ))pq = (ϕ(x))pq (ϕ(y)−1 )pq

.

= (x n )pq (y −n )pq = (x p )nq (y q )−np = 1G . We deduce that xy −1 ∈ T . As 1G ∈ T , this shows that T is a subgroup of G. The subgroup T ⊂ G is characteristic since every automorphism of G is orderpreserving. (d) Suppose that x ∈ G/T has finite order n. Let g ∈ G representing x. Then g n ∈ T . Denoting by N the order of g n , we have g N n = (g N )n = 1G , showing that g ∈ T . Therefore x = 1G/T . This shows that G/T is torsion-free. As every quotient of an FC-group is an FC-group, the group G/T is an FCgroup. Since every torsion-free FC-group is abelian by (b), we deduce that G/T is abelian. ■ Comment All these results are due to Bernhard Neumann (see in particular [Neu3, Theorem 5.1]). Exercise 2.46 Let G be a group. Denote by Δ the subset of G consisting of all elements whose conjugacy class is finite. (a) Show that Δ is a characteristic (and hence normal) subgroup of G. (b) Show that Δ is an FC-group. (c) Show that if G is torsion-free then Δ is abelian. Solution (a) For g ∈ G, denote by CG (g) := {hgh−1 : h ∈ G} the conjugacy class of g in G. We have 1G ∈ Δ since CG (1G ) = {1G }. Let x, y ∈ Δ. We have CG (xy −1 ) = {gxy −1 g −1 : g ∈ G} = {(gxg −1 ) · (gyg −1 ) : g ∈ G} ⊂ CG (x)CG (y)−1 . Since CG (x), CG (y) ⊂ G are finite subsets, we deduce that CG (xy −1 ) is also finite, that is, xy −1 ∈ Δ. This shows that Δ is a subgroup of G. If α ∈ Aut(G), we have CG (α(g)) = α(CG (g)), showing that α(Δ) ⊂ Δ. Thus, Δ is a characteristic subgroup of G.

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(b) Let x ∈ Δ and denote by CΔ (x) the conjugacy class of x in Δ. We clearly have CΔ (x) ⊂ CG (x). Therefore CΔ (x) is finite for all x ∈ Δ. This shows that Δ is an FC-group. (c) This immediately follows from Exercise 2.45(b) since Δ is torsion-free (as a subgroup of the torsion-free group G) and an FC-group by (b). ■ Exercise 2.47 (ICC-Groups) One says that a group G has the infinite conjugacy class property (ICC-property for short) or that G is an ICC-group for short, if every conjugacy class of G except {1G } is infinite. (a) Let G be an ICC-group. Show that the center of G is trivial. (b) Let G be an ICC-group. Show that every non-trivial normal subgroup of G is infinite. (c) Let X be an infinite set and let Sym0 (X) denote the group consisting of all permutations of X with finite support. Show that Sym0 (X) is an ICC-group. (d) Show that every non-abelian free group is an ICC-group. Solution (a) If g is in the center of G, then the conjugacy class of G is reduced to g. Therefore g = 1G since G has the ICC property. (b) Let H be a normal subgroup of G such that H /= {1G }. Let h be an element of H such that h /= 1G . Since H is normal, the conjugacy class of h is entirely contained in H . On the other hand, this conjugacy class is infinite since G has the ICC-property. Therefore H is infinite. (c) Let G := Sym0 (X). Consider an element g /= 1G in G. Then the support Y := {x ∈ X : g(x) /= x} of g is a non-empty finite subset of X. As X is infinite, there exists a sequence (Xn )n∈N of finite subsets of X such that |Xn | = |Y | for all n ∈ N and Xn /= Xm for all distinct n, m ∈ N. Choose, for each n ∈ N, an element gn ∈ G such that gn (Y ) = Xn . Observe that the support of gn ggn−1 is Xn . Consequently, the elements gn ggn−1 , n ∈ N, are all distinc. Thus the conjugacy class of g is infinite. This shows that G is an ICC-group. (d) Let G be a non-abelian free group with base X. Consider an element g ∈ G with g /= 1G and its reduced form g = x1n1 x2n2 · · · xknk ,

.

with k ≥ 1, xi ∈ X, ni ∈ Z \ {0} for all 1 ≤ i ≤ k, and xi /= xi+1 for all 1 ≤ i ≤ k − 1 (cf. [CAG, Corollary D.3.2]). Since the free group G is non-abelian, the set X has more than one element. Therefore there exists y ∈ X such that y /= x1 . Then, for each integer n ≥ 1, the reduced form of y n gy −n begins with y n . Thus the elements y n gy −n , n ≥ 1, are all distinct. This implies that the conjugacy class of g is infinite. Therefore G is an ICC-group. ■ Exercise 2.48 Let G be a finitely generated group and let φ : G → G be a group endomorphism of G. Let F be a finite group and, for n ∈ N, denote by Rn the set consisting of all group homomorphisms f : φ n (G) → F .

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(a) Show that the set Rn is finite. (b) Let ψn : φ n (G) → φ n+1 (G) denote the group homomorphism obtained by restriction of φ. Show that the map ρn : Rn+1 → Rn , defined by ρn (f ) := f ◦ ψn for all f ∈ Rn+1 , is injective. (c) Show that the sequence (|Rn |)n∈N is eventually constant. (d) Show that ρn is bijective for n large enough. Solution (a) The group φ n (G) is finitely generated since G is finitely generated and φ n is a group homomorphism. As the group F is finite, we deduce that Rn is finite (cf. [CAG, Lemma 2.4.4]). (b) The injectivity of ρn immediately follows from the surjectivity of ψn . (c) Since ρn : Rn+1 → Rn is injective by (b), we have |Rn+1 | ≤ |Rn | for all n ∈ N. Consequently, the sequence (|Rn |)n∈N is eventually constant. (d) By (c), there exists n0 ∈ N such that |Rn+1 | = |Rn | for all n ≥ n0 . As ρn is injective by (b), we deduce that ρn is bijective for all n ≥ n0 . ■ Exercise 2.49 Let G be a finitely generated residually finite group and let φ : G → G be a group endomorphism such that φ(G) has finite index in G. (a) For n ∈ N,∩ let Hn := φ n (G). Show that Hn+1 has finite index in Hn . (b) Let Nn := g∈Hn gHn+1 g −1 . Show that Nn is a finite index normal subgroup of Hn . (c) Let ψn : Hn → Hn+1 denote the group homomorphism obtained by restriction of α. Show that ψn (Nn ) ⊂ Nn+1 . (d) Show that ψn induces a surjective quotient group homomorphism θn : Hn /Nn → Hn+1 /Nn+1 . (e) Show that there exists n0 ∈ N such that θn is a group isomorphism for all n ≥ n0 . (f) Let Kn := Hn ∩ ker(φ). Show that Kn = Kn0 for all n ≥ n0 . (g) Show that the restriction of φ to Hn0 is injective. (h) Show that the kernel of φ is finite. (i) Suppose that G contains no finite non-trivial normal subgroups (e.g., G is torsion-free). Show that φ is injective. Solution (a) We proceed by induction. For n = 0, we have [Hn : Hn+1 ] = [G : φ(G)] < ∞ by our assumption. Suppose now that [Hn : Hn+1 ] < ∞ for some n ∈ N. Let R ⊂ G be a complete set of representatives for the left cosets of Hn+1 in Hn . If g ∈ Hn+1 , then there exists h ∈ Hn such that g = φ(h). Writing h = rk with r ∈ R and k ∈ Hn+1 , we get g = φ(rk) = φ(r)φ(k) ∈ φ(r)Hn+2 . This shows that [Hn+1 : Hn+2 ] ≤ |φ(R)| ≤ |R| = [Hn : Hn+1 ] < ∞ and completes induction.

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(b) This follows from (a) and [CAG, Lemma 2.1.10]. (c) We have ⎛ ψn (Nn ) = ψn ⎝

∩

.

⊂

∩

⎞ gHn+1 g −1 ⎠

g∈Hn

ψn (g)ψn (Hn+1 )ψn (g)−1

g∈Hn

=

∩

ψn (g)Hn+2 ψn (g)−1

g∈Hn

=

∩

gHn+2 g −1

g∈Hn+1

= Nn+1 , where the last but one equality follows from the surjectivity of ψn : Hn → Hn+1 . This shows that Nn ⊂ Nn+1 . (d) By (c), the group homomorphism ψn : Hn → Hn+1 induces a quotient group homomorphism θn : Hn /Nn → Hn+1 /Nn+1 . The homomorphism θn is surjective since ψn is surjective. (e) By (d), we have |Hn /Nn | ≥ |Hn+1 /Nn+1 | for all n ∈ N. It follows that the sequence (|Hn /Nn |)n∈N is eventually constant, i.e., there exists n0 ∈ N such that |Hn /Nn | = |Hn+1 /Nn+1 | for all n ≥ N0 . As θn is surjective, we deduce that θn is a group isomorphism for n ≥ n0 . (f) Since Hn+1 ⊂ Hn , we have Kn+1 = Hn+1 ∩ ker(φ) ⊂ Hn ∩ ker(φ) = Kn for all n. Suppose now n ≥ n0 . Since Kn ⊂ ker(φ), we have ψn (Kn ) = {1Hn+1 }. As θn : Hn /Nn → ∩ Hn+1 /Nn+1 is a group isomorphism by (e), this implies Kn ⊂ Nn . Since Nn = g∈Hn gHn+1 g −1 ⊂ Hn+1 , this gives us Kn ⊂ Hn+1 ∩ker(φ) = Kn+1 . We conclude that Kn = Kn+1 for all n ≥ n0 . By induction, it follows that Kn = Kn0 for all n ≥ n0 . (g) Suppose by contradiction that the restriction of φ to Hn0 is not injective, i.e., Kn0 /= {1G }. Let g ∈ Kn0 such that g /= 1G . Since G is residually finite, there exists a finite group F and a group homomorphism η ∈ Hom(G, F ) such that η(g) /= 1F . For n ∈ N, let ηn ∈ Hom(Hn , F ) denote the restriction of η to Hn . By Exercise 2.48(d), there exist n ≥ n0 and f ∈ Hom(Hn+1 , F ) such that ηn = f ◦ ψn . As g ∈ Kn = Hn ∩ ker(φ) by (f), it follows that η(g) = ηn (g) = f (ψn (g)) = f (φ(g)) = f (1G ) = 1F . This is a contradiction. (h) By (g), we have ker(φ) ∩ Hn0 = {1G }. This implies that the restriction to ker(φ) of the canonical map from G onto the set of left cosets of Hn0 in G is injective. As Hn0 is of finite index in G by (a), we deduce that ker(φ) is finite. (i) This immediately follows from (h) since ker(φ) is a normal subgroup of G. ■

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Comment These results are all due to Hirshon [Hirsh] (see also [BriHM, Proposition 4.4] and [BriR, Proposition 3.3]). Note that (g) implies that φ is injective if it is surjective. Thus, it is an extension of Mal’cev’s theorem on the Hopficity of finitely generated residually finite groups. Wise [Wis] gave an example of a finitely generated residually finite group G admitting a group endomorphism φ such that there is no n ∈ N such that the restriction of φ to φ n (G) is injective. This answered in the negative a question raised in [Hirsh].

Chapter 3

Surjunctive Groups

This chapter is devoted to surjunctivity of maps, groups, dynamical systems, and subshifts. It includes a study of minimal subshifts, almost periodic configurations, strongly aperiodic subshifts, Toeplitz subshifts, the Thue-Morse sequence, and the Morse subshift.

3.1 Summary 3.1.1 Definition of Surjunctivity A group G is said to be surjunctive if it satisfies the following condition: for any finite set A, every injective cellular automaton .τ : AG → AG is surjective (and hence bijective).

3.1.2 The Class of Surjunctive Groups Every subgroup of a surjunctive group is itself surjunctive [CAG, Proposition 3.2.1]. If .P is a property of groups, a group G is said to be locally .P if all finitely generated subgroups of G satisfy .P. A group is surjunctive if and only if it is locally surjunctive [CAG, Proposition 3.2.2]. Every locally residually finite group is surjunctive [CAG, Corollary 3.3.6]. In particular, all finite groups, all locally finite groups, all residually finite groups, all abelian groups, and all free groups are surjunctive.

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 T. Ceccherini-Silberstein, M. Coornaert, Exercises in Cellular Automata and Groups, Springer Monographs in Mathematics, https://doi.org/10.1007/978-3-031-10391-9_3

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3.1.3 Expansive Actions An action of a group G on a uniform space X is called expansive if there exists an entourage .U0 of X satisfying the following property: for all points .x, y ∈ X with .x /= y, there exists .g ∈ G such that .(gx, gy) ∈ / U0 (equivalently, if .x, y ∈ X satisfy that .(gx, gy) ∈ U0 for all .g ∈ G, then .x = y). Such an entourage .U0 is then called an expansiveness entourage for the action (see [CAG, Section 3.5]). Suppose that a group G acts on a metric space .(X, d). Then the action is expansive if and only if there exists a real number .c > 0 satisfying the following property: for all .x, y ∈ X with .x /= y, there exists .g ∈ G such that .d(gx, gy) ≥ c (equivalently, if .x, y ∈ X satisfy that .d(gx, gy) < c for all .g ∈ G, then .x = y). Such a real number c is then called an expansiveness constant for the action (cf. [CAG, Remark 3.5.1.(c)]). Let G be a group, let A be a set and equip .AG with its prodiscrete uniform structure. Then the shift action of G on .AG is expansive (indeed, the set .W ({1G }) := {(x, y) ∈ AG × AG : x(1G ) = y(1G )} is an expansiveness entourage for the G-shift action (cf. [CAG, Proposition 3.5.2])).

3.1.4 Compactness of the Space of Marked Surjunctive Groups Let .Γ be a group. A .Γ -quotient is a pair .(G, ρ), where G is a group and .ρ : Γ → G is a surjective group homomorphism. We introduce an equivalence relation on the set of .Γ -quotients by declaring that two .Γ -quotients .(G1 , ρ1 ) and .(G2 , ρ2 ) are equivalent provided there exists a group isomorphism .φ : G2 → G1 such that .ρ1 = φ ◦ ρ2 . An equivalence class of .Γ -quotients is called a .Γ -marked group. Two .Γ quotients .(G1 , ρ1 ) and .(G2 , ρ2 ) are equivalent if and only if .ker(ρ1 ) = ker(ρ2 ). Thus, we may identify the set of .Γ -marked groups with the set .N (Γ ) of all normal subgroups of .Γ . Equip .{0, 1}Γ with its prodiscrete uniform structure and identify .P(Γ ), the set of all subsets of .Γ , with .{0, 1}Γ via the map .A |→ χA , where .χA : Γ → {0, 1} denotes the characteristic map of .A ⊂ Γ . A base of entourages for the induced uniform structure on .N (Γ ) ⊂ P(Γ ) is given by the sets VF := {(N1 , N2 ) ∈ N (Γ ) × N (Γ ) : N1 ∩ F = N2 ∩ F },

.

where F ranges over all finite subsets of .Γ . The space .N (Γ ) of all .Γ -marked groups is a totally disconnected compact Hausdorff topological space ([CAG, Proposition 3.4.1]). The subspace .S (Γ ) := {N ∈ N (Γ ) : Γ /N is surjunctive} is closed (and therefore compact) in .N (Γ ) ([CAG, Theorem 3.7.1]).

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3.2 Exercises Exercise 3.1 Let X be a set and let f : X → X be an injective map. Suppose that there exists n ∈ N such that f n (X) = f n+1 (X). Show that f is surjective. Solution Let x ∈ X. As f n (X) = f n+1 (X), there exists y ∈ X such that f n (x) = f n+1 (y). This implies that f n (x) = f n (f (y)). As f n is injective since f is injective, we deduce that x = f (y). This shows that f is surjective. ■ Exercise 3.2 Let K be an algebraically closed field. Show that every injective polynomial map f : K → K is surjective. Solution Let f : K → K be an injective polynomial map and let a ∈ K. By definition of a polynomial map, there exists a polynomial P ∈ K[x] such that f (k) = P (k) for all k ∈ K. Observe that deg(P ) ≥ 1 since f is injective. Consider the polynomial Q := P − a. Since K is algebraically closed and deg(Q) = deg(P ) ≥ 1, there exists b ∈ K such that Q(b) = 0. We then have f (b) = P (b) = Q(b) + a = a. This shows that f is surjective. ■ Exercise 3.3 Show that every injective polynomial map f : R → R is surjective. Solution Let f : R → R be an injective polynomial map. Since f is injective and continuous, it follows from the intermediate value theorem that f is strictly monotone and that its image is an interval. On the other hand, |f (x)| tends to infinity as x tends to ∞ or −∞ since f is a non-constant polynomial map. This implies that f (R) = R. Thus f is surjective. ■ Comment More generally, every injective polynomial map f : Rn → Rn , n ≥ 1, is surjective (see [BiaR]). Exercise 3.4 Show that the polynomial map f : Q → Q defined by f (x) = x 3 is injective but not surjective. Solution The map f is injective since it is strictly increasing. We claim that 2 is not in the image of f . Indeed, suppose by contradiction that there exists q ∈ Q such that f (q) = 2. Then, writing q = a/b with a, b ∈ Z and b /= 0, we would get a 3 = 2b3 , which is clearly impossible since the number of 2s appearing in the prime decomposition of a 3 is a multiple of 3 while the number of 2s appearing in the prime decomposition of 3b3 is congruent to 1 modulo 3. This proves our claim that 2 is not in the image of f . Thus f is not surjective. ■ Exercise 3.5 Show that every injective holomorphic map f : C → C is surjective. Solution Let f : C → C be an injective holomorphic map. Suppose by contradiction that f is not surjective. Since f is not constant, it follows from Picard’s little theorem that there exists a ∈ C such that f (C) = C \ {a}. On the other hand, the injectivity of f implies that f ' never vanishes. This implies that C is biholomorphically equivalent to f (C), a contradiction since C is simply-connected while C \ {a} is not. ■

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Comment In complex analysis, an injective holomorphic function defined on an open subset of C is called a univalent function. For n ≥ 2, Picard’s little theorem becomes false and there are injective holomorphic maps Cn → Cn that are not surjective. In fact, for every integer n ≥ 2, there exist Fatou-Bieberbach domains in Cn , i.e., proper open subsets of Cn that are biholomorphically equivalent to Cn . Examples of Fatou-Bieberbach domains in C2 were first given by Fatou [Fat] and Bieberbach [Bie]. Exercise 3.6 Give an example of a real analytic map f : R → R which is injective but not surjective. Solution The arctangent function f : R → R does the job. Indeed, it is injective since it is strictly increasing. It is not surjective since its image is the open interval (−π/2, π/2). Finally, it is real analytic since its derivative is the map x |→ 1/(1 + x 2 ), which is a rational function with non-vanishing denominator. ■ Comment The exponential function f : R → R also does the job. Exercise 3.7 Let K be a field and let V be a vector space over K. Show that V is finite-dimensional if and only if every injective endomorphism f : V → V is surjective. Solution Suppose first that V is finite-dimensional and let f : V → V be an endomorphism. By the rank-nullity theorem, we have dim(Im(f ))+dim(ker(f )) = dim(V ). We deduce that Im(f ) = V if and only if ker(f ) = {0}. Thus f is surjective if and only if f is injective. Suppose now that V is infinite-dimensional and let B ⊂ V be a base for V . Since B is an infinite set, there exists an injective map g : B → B that is not surjective. As B is a base for V , the map g (uniquely) extends to an endomorphism f of V . The image of f is the vector subspace of V generated by g(B). As g(B)  B, we deduce that f is not surjective. ■ Exercise 3.8 (Artinian Modules) Let R be a ring and let M be a left (or right) R-module. One says that M is Artinian if every descending chain of submodules of M N0 ⊃ N1 ⊃ N2 ⊃ . . .

.

eventually stabilizes (i.e., there is an integer n0 ≥ 0 such that Nn = Nn+1 for all n ≥ n0 ). Show that if M is Artinian then every injective endomorphism f : M → M is surjective. Solution Let M be an Artinian left (resp. right) R-module and let f : M → M be an injective endomorphism. Consider the descending sequence of submodules of M defined by Nn := Im(f n ) for n ≥ 0. Since M is Artinian, there exists an integer n0 ≥ 0 such that Nn0 = Nn0 +1 . This means that for every y ∈ M there exists x ∈ M such that f n0 (y) = f n0 +1 (x) = f n0 (f (x)). Since f is injective, so is f n0 . Thus y = f (x). This shows that f is surjective. ■

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Exercise 3.9 Let U := {z ∈ C : |z| = 1}. Show that every injective continuous map f : U → U is surjective. Solution Let f : U → U be an injective continuous map. Since U is compact and Hausdorff, f induces a homeomorphism from U onto f (U ). Suppose by contradiction that f is not surjective. Then there is a point a ∈ U such that f (U ) ⊂ U \ {a}. As f (U ) is closed in U , there exist distinct points z1 , z2 ∈ U such that z1 , z2 ∈ f (U ) and f (U ) is entirely contained in the arc I = [z1 , z2 ] ⊂ U going from z1 to z2 in the positive direction. Observe that f (U ) = I since f (U ) must be connected. This is a contradiction since U remains connected after removing an arbitrary point from it while it is clear that the arc I does not have this property. ■ Comment This result also follows from Exercise 3.10 since U is a compact onedimensional topological manifold. Exercise 3.10 Let n ≥ 0 be an integer. An n-dimensional topological manifold is a non-empty Hausdorff topological space X such that each point in X admits a neighborhood homeomorphic to Rn . Show that if X is a compact n-dimensional topological manifold, then every injective continuous map f : X → X is surjective. Solution Let X be a compact n-dimensional manifold and let f : X → X be an injective continuous map. By compactness, X has finitely many connected components X1 , . . . , Xk and each Xi , 1 ≤ i ≤ k, is a connected compact n-dimensional manifold. Each f (Xi ) is a closed subset of Xj for some j = j (i). On the other hand, by the Brower invariance of domain theorem, f (Xi ) is open in Xj . As Xj is connected, we deduce that f (Xi ) = Xj . Finally, the injectivity of f implies that f permutes the connected components Xi of X. This shows that f is surjective. ■ Exercise 3.11 Let P be a property of groups. Show that every subgroup of a locally P group is itself locally P. Solution Let G be a locally P group and let H be a subgroup of G. If K is a finitely generated subgroup of H , then K is a finitely generated subgroup of G and hence K has property P. This shows that H is a locally P group. ■ Exercise 3.12 Let P be a property of groups. Let G be a group. Show that G is locally P if and only if all its finitely generated subgroups are locally P. Solution Necessity follows from the previous exercise. The converse follows from the observation that a finitely generated group which is locally P in fact has property P. ■ Exercise 3.13 Show that the additive group Q is locally cyclic. Solution Let H be a finitely generated subgroup of Q. Thus there are finitely many rational numbers q1 , . . . , qn ∈ Q such that H = {α1 q1 + · · · + αn qn : αi ∈ Z for all i = 1, . . . , n}.

.

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Write each qi in the form qi = ai /bi , with ai , bi ∈ Z and bi ≥ 1, and let d := lcm(b1 , . . . , bn ) denote the least common multiple of the integers b1 , . . . , bn . Clearly every q ∈ H can be written in the form q = m/d for some m ∈ Z. Moreover, the set I consisting of all m ∈ Z such that m/d ∈ H is an ideal of Z. As the ring Z is principal, there exists c ∈ Z such that I consists of all integer multiples of c. It follows that the group H is generated by c/d and hence it is cyclic. This shows that Q is locally cyclic. ■ Exercise 3.14 Show that every elment of a locally finite group has finite order. Solution Let G be a locally finite group and let g ∈ G. Since G is locally finite, the subgroup H ⊂ G generated by g is finite. We deduce that g has order |H | < ∞. ■ Exercise 3.15 Let G be an abelian group. Show that G is locally finite if and only if every element of G has finite order. Solution The condition is necessary by Exercise 3.14. Conversely, let G be an abelian group in which every element has finite order and let H be a finitely generated subgroup of G. By the structure theorem for finitely generated abelian groups, there are subgroups T and F of H , with T finite and F isomorphic to Zr for some integer r ≥ 0, such that H = T ⊕ F . As every element of G has finite order, we must have r = 0 so that H = T is finite. This shows that G is locally finite. ■ Exercise 3.16 Show that every subgroup and every quotient of a locally finite group is a locally finite group. Solution Let G be a locally finite group. The fact that every subgroup of G is locally finite follows from Exercise 3.11. Let now G' be a quotient of G and let ϕ : G → G' be a surjective homomorphism. Let H ' ⊂ G' be a finitely generated subgroup and let us show that it is finite. Let {h'1 , . . . , h'n } ⊂ H ' be a finite generating subset for H ' . Let h1 , . . . , hn ∈ G such that ϕ(hi ) = h'i for all i = 1, . . . , n, and denote by H ⊂ G the subgroup they generate. Since H is finitely generated and G is locally finite, we deduce that H and therefore its homomorphic image H ' = ϕ(H ) are finite. ■ Exercise 3.17 Let G be a group. Suppose that G contains a normal subgroup N such that both N and G/N are locally finite. Show that G is locally finite. Solution Let H be a finitely generated subgroup of G. Denote by π : G → G/N the canonical group homomorphism. Since H is finitely generated, the subgroup π(H ) ⊂ G/N is also finitely generated. As G/N is locally finite, we deduce that π(H ) is finite. Therefore K := H ∩N is a finite index subgroup of H . As every finite index subgroup of a finitely generated group is itself finitely generated, it follows that K is finitely generated. Since K is a subgroup of N and N is locally finite, this implies that K is finite. Consequently, being an extension of the finite group π(H ) by the finite group K, the subgroup H is itself finite. This shows that G is locally finite. ■

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Exercise 3.18 Let G be a group which is the limit of an inductive system of locally finite groups. Show that G is locally finite. Solution Let (Gi , ψj,i ) be an inductive system of locally finite groups such that G = lim Gi . We denote by ψi : Gi → G the canonical group homomorphism. − → Let H ⊂ G be a finitely generated subgroup and let us show that H is finite. Let {h1 , . . . , hn } ⊂ H be a finite generating subset for H . Then for each k = 1, . . . , n we can find i(k) ∈ I and gk ∈ Gi(k) such that hk = ψi(k) (gk ). As I is a directed set, there exists i ∗ ∈ I such that i(k) ≤ i ∗ for all k = 1, . . . , n. Then the elements gk∗ := ψi ∗ ,i(k) (gk ) ∈ Gi ∗ satisfy that hk = ψi ∗ (gk∗ ). Since Gi ∗ is locally finite, the subgroup H ∗ ⊂ Gi ∗ generated by the elements g1∗ , . . . , gn∗ is finite. It follows that H = ψi ∗ (H ∗ ) is also finite. ■ Exercise 3.19 Show that the direct sum of any family of locally finite groups is a locally finite group. ⊕ Solution Let (Gi )i∈I be a familyΠof locally finite groups and let G := i∈I Gi . Consider the product group P := i∈I Gi and denote, for each i ∈ I , by πi : G → Gi the projection group homomorphism. Then G is the subgroup of P consisting of all g ∈ P such that πi (g) = 1Gi for all but finitely many i ∈ I . Let now S be a finite subset of G and let H denote the subgroup of G generated by S. For each i ∈ I , since the group Gi is locally finite, there is a finite subgroup Hi of Gi such that πi (S) ⊂ Hi . On the other hand, there is a finite subset J ⊂ I such that i ∈ I \ J . We deduce that H embeds via projection into the πi (S) = {1Gi } for allΠ product group PJ := i∈J Hi . As PJ is finite, it follows that H is finite. This shows that G is locally finite. ■ Exercise 3.20 Let G be a group. Let S denote the set consisting of all normal locally finite subgroups of G. (a) Show that if H ∈U S and K ∈ S then H K ∈ S . (b) Show that M := H ∈S H is a normal locally finite subgroup of G and that every normal locally finite subgroup of G is contained in M. Solution (a) Let H ∈ S and K ∈ S . Since H and K are normal subgroups, their product H K is also a normal subgroup of G. Indeed, 1G = 1G · 1G ∈ H K and, for all h1 , h2 , h ∈ H , k1 , k2 , k ∈ K, and g ∈ G we have −1 (h1 k1 ) · (h2 k2 )−1 = h1 k1 k2−1 h2 = (h1 h2 ) · (h−1 2 (k1 k2 )h2 ) ∈ H K

.

and g(hk)g −1 = ghg −1 · gkg −1 ∈ H K.

.

Let us show that H K is locally finite. Let S = {g1 , . . . , gn } be a finite subset of H K. Then, for all 1 ≤ i ≤ n, there are elements hi ∈ H and ki ∈ K such that gi = hi ki . Note that (hi ki ) · (hj kj ) = (hi hj ) · (h−1 j ki hj )kj . Let us denote by

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H ' ⊂ H the subgroup generated by the hi ’s. Since H is locally finite, we deduce that H ' is finite. Let then K ' ⊂ K denote the subgroup generated by all the H ' conjugates of the ki ’s. Again, since K is locally finite and K ' is finitely generated, K ' is finite. It follows that the subgroup generated by S is contained in the finite set H ' K ' and is therefore finite. (b) By (a), the set M is an inductive union of normal locally finite subgroups of G. Using the result of Exercise 3.18, we deduce that M is a normal locally finite subgroup of G. The fact that M contains every normal locally finite subgroup of G is immediate from the definition of M. ■ Exercise 3.21 Let G be an infinite locally finite group and let A be a finite set. Let τ : AG → AG be a cellular automaton. Show that the following conditions are all equivalent: (i) (ii) (iii) (iv) (v)

τ τ τ τ τ

is surjective; is injective; is finite-to-one; is countable-to-one; is bijective.

Solution Let S ⊂ G be a memory set for τ and let μ : AS → A denote the associated local defining map. Since G is locally finite, the subgroup H ⊂ G generated by S is finite. Consider the restriction τH : AH → AH of τ . Since the set AH is finite, τH is surjective if and only if it is injective. As τ is surjective (resp. injective) if and only if τH is surjective (resp. injective) by [CAG, Proposition 1.7.4]), it follows that τ is surjective if and only if it is injective. This proves the equivalences (i) ⇐⇒ (ii) ⇐⇒ (v). The implications (ii) =⇒ (iii) and (iii) =⇒ (iv) are obvious. To complete the proof it suffices to show that (iv) =⇒ (ii). Suppose that τ is not injective and let us show that τ is not countable-to-one. As τ is not injective, the restriction cellular automaton τH is not injective either (cf. [CAG, Proposition 1.7.4]). Thus there are distinct configurations x0 , x1 ∈ AH such that τH (x0 ) = τH (x1 ). Denote by R ⊂ G a complete set of representatives for the left cosets of H in G, so that every element g ∈ G can be uniquely written in the form g = rh with r ∈ R and h ∈ H . Observe that R is infinite since G is infinite while H is finite. For each subset E ⊂ R, denote by χE : R → {0, 1} the characteristic map of E and consider the configuration yE ∈ AG defined by yE (rh) := xχE (r) (h) for all r ∈ R and h ∈ H . Clearly the configurations yE are all distinct and have the same image under τ , namely the configuration z ∈ AG given by z(rh) := (τH (x0 ))(h) = (τH (x1 ))(h) ■ for all r ∈ R and h ∈ H . This shows that τ −1 (z) is uncountable. Exercise 3.22 Let G be a locally finite group and let A be a set. Show that every bijective cellular automaton τ : AG → AG is invertible. Solution Let τ : AG → AG be a bijective cellular automaton with memory set S ⊂ G. Consider the subgroup H ⊂ G generated by S. As S is finite and G is locally finite, the subgroup H is finite. Let τH : AH → AH denote the cellular automaton

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over the group H obtained by restriction of τ to H (see [CAG, Section 1.7]). As τ is bijective, τH is also bijective (cf. [CAG, Proposition 1.7.4]). Since the group H is finite, every map AH → AH is a cellular automaton over H . Consequently, the cellular automaton τH is invertible. As every cellular automaton obtained by induction from an invertible cellular automaton is itself invertible (cf. [CAG, Proposition 1.10.4]), we conclude that τ is itself invertible. ■ Exercise 3.23 Let G be a locally finite group and let A be a set. Let τ : AG → AG be a cellular automaton. Show that τ (AG ) is closed in AG with respect to the prodiscrete topology. Solution Let S ⊂ G be a memory set for τ and let H denote the subgroup of G generated by S. Consider the cellular automaton τH : AH → AH over the group H obtained from τ by restriction to H . Since G is locally finite, the group H is finite. This implies that the prodiscrete topology on AH is the discrete one. It follows that τH (AH ) is closed in AH . By applying the result of Exercise 1.38, we deduce that τ (AG ) is closed in AG for the prodiscrete topology. ■ Exercise 3.24 Let G and A be two groups. Suppose that there is an element s0 ∈ G of infinite order and that the group A is non-trivial. Show that the map τ : AG → AG defined by τ (x)(g) := x(g)−1 x(gs0 ), for all x ∈ AG and g ∈ G, is a cellular automaton over the group G which is surjective but not injective. Solution We first observe that τ is indeed a cellular automaton over the group G, admitting S := {1G , s0 } ⊂ G as a memory set and the map μ : AS → AS defined by μ(p) := p(1G )−1 p(s0 ) for all p ∈ AS as the associated local defining map. Denote by H the infinite cyclic subgroup of G generated by s0 . As S ⊂ H , we can consider the cellular automaton τH : AH → AH over the group H obtained by restriction of τ . Let z ∈ AH . Then the configuration y ∈ AH defined by

y(s0n ) :=

.

⎧ ⎪ ⎪ ⎨1A

z(1G )z(s0 ) · · · z(s0n−1 ) ⎪ ⎪ ⎩z(1 )−1 z(s −1 )−1 · · · z(s n )−1 G 0 0

if n = 0 if n ≥ 1 if n ≤ −1

satisfies that τH (y) = z. This shows that τH is surjective. On the other hand, since the group A is non-trivial, we can find a ∈ A such that a /= 1A . Then the two constant configurations y0 , y1 ∈ AH defined by y0 (s0n ) = 1A and y1 (s0n ) = a for all n ∈ Z satisfy that τH (y0 ) = y0 = τH (y1 ). As y0 /= y1 , this shows that τH is not injective. Since τH is surjective but not injective, it follows that τ itself is surjective but not injective by [CAG, Proposition 1.7.4]. ■ Exercise 3.25 Let X be an infinite set. Show that the symmetric group Sym(X) is not locally residually finite. Solution Since X is infinite, it contains an infinite countable subset. We deduce that Sym(X) contains a subgroup isomorphic to Sym(Z). As the subgroup G1 ⊂

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Sym(Z) generated by the translation T : n |→ n + 1 and the transposition S := (0 1) is not residually finite (cf. [CAG, Proposition 2.6.1]), we conclude that Sym(X) is not locally residually finite. ■ Exercise 3.26 Show that every virtually surjunctive group is surjunctive. Solution Let G be a virtually surjunctive group. This means that there exists a finite index subgroup H ⊂ G which is surjunctive. Let A be a finite set and let τ : AG → AG be an injective cellular automaton. Let us show that τ is surjective. Let T ⊂ G be a complete set of representatives for the right cosets of H in G and set B := AT . By Exercise 1.32, the map Ψ : AG → B H , defined by Ψ (x)(h)(t) := x(ht) for all x ∈ AG , h ∈ H , and t ∈ T , is an H -equivariant homeomorphism. Then the map σ := Ψ ◦ τ ◦ Ψ −1 : B H → B H , obtained by conjugating τ by Ψ , is injective, H -equivariant, and continuous. Therefore, σ : B H → B H is an injective cellular automaton over the group H and the finite alphabet B. As H is surjunctive, we deduce that σ is surjective. It follows that τ = Ψ −1 ◦ σ ◦ Ψ is surjective as well. This shows that G is surjunctive. ■ Exercise 3.27 Let G be a group. Suppose that there exists a family (Ni )i∈I of normal subgroups of G satisfying the following properties: (1) for ∩ all i and j in I , there exists k in I such that Nk ⊂ Ni ∩ Nj ; (2) i∈I Ni = {1G }; (3) the group G/Ni is surjunctive for each i ∈ I . Show that G is surjunctive. Solution Let Ω be a finite subset of G. By virtue of condition (2), for every pair (g, h) ∈ Ω × Ω with g /= h we can find i(g, h) ∈ I such that gh−1 ∈ / N . Since Ω is finite, by condition (1) we can find i ∈ I such that N ⊂ i i(g,h) ∩ (g,h)∈Ω×Ω:g/=h Ni(g,h) . Setting Γ := G/Ni , we have that the restriction to Ω of the canonical quotient group homomorphism φ : G → Γ is injective. Moreover, by condition (3), Γ is surjunctive. We conclude that G itself is surjunctive (cf. [CAG, Lemma 3.3.4]). ■ Exercise 3.28 Let X be a topological space and let f : X → X be a map such that f (X) is closed inUX. Suppose that there is a family (Xi )i∈I of subsets of X such that the set Y := i∈I Xi is dense in X and Xi ⊂ f (Xi ) for all i ∈ I . Show that f is surjective. Solution We have Y =



.

i∈I

Xi ⊂



f (Xi ) = f

i∈I

⎛

⎞ Xi

= f (Y ).

i∈I

This implies Y ⊂ f (X) and hence Y ⊂ f (X). As Y is dense in X and f (X) is closed in X, it follows that X = f (X). Thus f is surjective. ■ Comment Compare with [Gro3, Section 4.B].

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Exercise 3.29 Let X be a set. Let (Ai )i∈I be a net of subsets of X. (a) Show that

∩ ∩

. Aj ⊂ Aj . i j ≥i

(3.1)

i j ≥i

(b) Let B be a subset of X. Show that the net (Ai )i∈I converges to B with respect to the prodiscrete topology on P(X) = {0, 1}X if and only if

∩ ∩

.B = Aj = Aj . (3.2) i j ≥i

i j ≥i

Solution For E ⊂ X, let χE : X → U {0, 1} denote the characteristic map of E. Observe∩ that the characteristic map of i Ai is supi χAi and that the charactristic map of i Ai is infi χAi . Thus the charactristic map of the left-hand side of 3.1 is lim infi χAi while the characteristic map of the right-hand side is lim supi χAi . As lim infi αi ≤ lim supi αi for any net of real numbers (αi ), we deduce that the characteristic map of the left-hand side of (3.1) is bounded above by the characteristic map of the right-hand side. This shows the inclusion (3.1). (b) By definition of the prodiscrete topology, the net (Ai )i∈I converges to B if and only if the net (χAi )i∈I converges to χB for the pointwise convergence on {0, 1}X . We know that a net of real numbers (αi ) converges to a real number λ if and only if λ = lim infi αi = lim supi αi . Consequently, the net (Ai ) converges to B if and only if χB = lim inf χAi = lim sup χAi .

.

i

i

This is equivalent to (3.2) by the observations made in (a).

■

Exercise 3.30 Let G be a group and let A be a set. Let H be a subgroup of G. In Exercise 1.103, we associated with each subshift X ⊂ AH the subshift X(G) ⊂ AG defined by X(G) := {x ∈ AG : xgH ∈ X for all g ∈ G},

.

where xgH := (gx)|H for all x ∈ AG and g ∈ G. Let σ : AH → AH be a cellular automaton and denote by σ G : AG → AG the induced cellular automaton. Show that if X, Y ⊂ AH are two subshifts such that σ (X) ⊂ Y , then the cellular automaton σ G |X(G) : X(G) → Y (G) is injective (resp. surjective) if and only if the cellular automaton σ |X : X → Y is injective (resp. surjective). Solution Let T ⊂ G be a complete set of representatives for the right cosets of H in G with 1G ∈ T , so that G = ⨆t∈T tH . Given g ∈ G, there exist unique t ∈ T and h ∈ H such that g = th so that x(g) = x(th) = (t −1 x)(h) = (t −1 x)|H (h) = xtH−1 (h).

.

(3.3)

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3 Surjunctive Groups

⎞ ⎛ Π It follows that the map ϕ : X(G) → t∈T X defined by ϕ(x) := xtH−1 is a t∈T bijection. (G) the configuration defined by Moreover, given ⎛⎛ x ∈⎞ X ⎞we denote by x ∈ X H −1 xt −1 , where setting x = ϕ t∈T

xtH−1 = (t −1 x)|H := x for all t ∈ T .

(3.4)

.

Suppose that σ |X is injective. Let x, z ∈ X(G) and suppose that σ G (x) = It follows from Exercise 1.103(e) that σ (xtH ) = σ (ztH ) for all t ∈ T . As σ |X is injective, we deduce that xtH = ztH for all t ∈ T . It follows that x = ϕ −1 ((xtH )t∈T ) = ϕ −1 ((ztH )t∈T ) = z. This shows that σ G |X(G) is injective. Conversely, suppose that σ |X is not injective and let x, z ∈ X be two distinct configurations such that σ (x) = σ (z). Denoting by x, z ∈ X(G) the corresponding configurations given by (3.4), we have xtH−1 = x /= z = ztH−1 for all t ∈ T , so that (cf. (3.3)) x /= z. Using Exercise 1.103(e) we obtain σ G (z).

.

⎛ ⎛ ⎞H ⎞H σ G (x) = σ (xtH ) = σ (x) = σ (z) = σ (ztH ) = σ G (z) t

t

for all t ∈ T , showing that σ G (x) = σ G (z). It follows that σ G is not injective either. Suppose that σ |X is surjective and let y ∈ Y (G) . For each t ∈ T , by surjectivity of σ |X we can find x t ∈ X such that σ (x t ) = ytH . Consider the configuration x ∈ X(G) defined by setting xtH := x t for all t ∈ T . It follows from Exercise 1.103(e) that σ G (x) = y. This shows that σ G is surjective. Conversely, suppose that σ G |X(G) is surjective and let y ∈ Y . Let y ∈ Y (G) denote the corresponding configuration given by (3.4). Since σ G |X(G) is surjective, we can find x ∈ X(G) such that σ G (x) = y. Setting x := x1HG ∈ X, we have by Exercise 1.103(e), ⎛ ⎞H y = y1HG = σ G (x) = σ (x1HG ) = σ (x),

.

1G

showing that σ |X is surjective.

■

Exercise 3.31 (Surjunctive Subshifts) Let G be a group and let A be a set. One says that a subshift X ⊂ AG is surjunctive if every injective cellular automaton σ : X → X is surjective. (a) Show that every finite subshift X ⊂ AG is surjunctive. (b) Let A and B be finite sets. Suppose that X ⊂ AG and Y ⊂ B G are topologically conjugate subshifts. Show that X is surjunctive if and only if Y is surjunctive.

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Solution (a) If X ⊂ AG is a finite subshift then every injective map from X into itself, in particular every cellular automaton σ : X → X, is surjective. This implies that X is surjunctive. (b) Since X and Y are topologically conjugate, there exists a bijective cellular automaton τ : X → Y . We know that τ −1 : Y → X is also a cellular automaton and that the composite of two cellular automata is a cellular automaton. It follows that a map σ : X → X is a cellular automaton if and only if the map τ ◦ σ ◦ τ −1 : Y → Y is a cellular automaton. As σ is injective (resp. surjective) if and only if τ ◦ σ ◦ τ −1 is injective (resp. surjective), we deduce that X is surjunctive if and only if Y is surjunctive. ■ Exercise 3.32 Let G be a group and let A be a finite set. Let X ⊂ AG be a subshift and let Per(X) denote the set of all periodic configurations in X. (a) Let H be a subgroup of finite index of G and let XH := Fix(H ) ∩ X denote the set consisting of all configurations in X that are fixed by H . Show that XH is a finite set and that σ (XH ) ⊂ XH for every cellular automaton σ : X → X. (b) Deduce from (a) that if Per(X) is dense in X then X is surjunctive. Solution (a) We know that the set of configurations in AG that are fixed by H is in bijection with AH \G and it is therefore finite (cf. [CAG, Proposition 1.3.3 and Corollary 1.3.4]). This implies in particular that XH is finite. If σ : X → X is a cellular automaton, we have σ (XH ) ⊂ XH since σ is G-equivariant. (b) A configuration x ∈ X is periodic if and only if its stabilizer in G is a subgroup of finite index of G. Thus, denoting by S the set of all subgroups of finite index of G, we have

. Per(X) = XH . H ∈S

Let now σ : X → X be an injective cellular automaton. For every H ∈ S , the set XH is finite and we have σ (XH ) ⊂ XH . As every injective self-mapping of a finite set is surjective, this implies that σ (XH ) = XH . Thus σ (Per(X)) = σ (



.

H ∈S

XH ) =

H ∈S

σ (XH ) =



XH = Per(X)

H ∈S

and hence Per(X) ⊂ σ (X) ⊂ X. As σ (X) is closed in X by the compactness of X and the continuity of σ , we conclude that σ (X) = X, i.e., σ is surjective, if Per(X) ■ is dense in X (cf. Exercise 3.28).

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3 Surjunctive Groups

Exercise 3.33 Let G be a group and let A be a set. Let X ⊂ AG be a subshift. Denote by P(X) ⊂ P(G, A) the set of (G, A)-patterns appearing in X and by Per(X) ⊂ X the set of its periodic configurations. Show that Per(X) is dense in X if and only if the following condition is satisfied: for every pattern p ∈ P(X), there exists a finite index subgroup H of G and an element y ∈ AH \G such that, denoting by ρH : G → H \G the canonical surjection g |→ Hg, the configuration y ◦ ρH is in X and the pattern p appears in y ◦ ρH .

Solution We know from [CAG, Proposition 1.3.3] that a configuration x ∈ AG is periodic if and only if there exists a finite index subgroup H of G and an element y ∈ AH \G such that x = y ◦ ρH . The equivalence in the statement immediately follows since, by definition of the prodiscrete topology, Per(X) is dense in X if and only if every pattern p ∈ P(X) appears in some periodic configuration belonging to X. ■ Exercise 3.34 Let A be a set. Given a word w ∈ A∗ of length p ≥ 1, denote by w ∞ ∈ AZ the unique periodic configuration with period p such that w ∞ (0)w ∞ (1) · · · w ∞ (p − 1) = w.

.

(3.5)

Let X ⊂ AZ be a subshift. Denote by L(X) the language of X and by Per(X) the set of periodic configurations in X. (a) Let w ∈ A∗ . Show that one has w ∞ ∈ X if and only if w n ∈ L(X) for all n ∈ N. (b) Show that Per(X) is dense in X if and only if the following holds: (DP)

for every word u ∈ L(X), there exists a non-empty word w ∈ A∗ such that u is a subword of w and w ∞ ∈ X.

Solution (a) Necessity follows from the fact that w n appears in w ∞ for all n ∈ N. Conversely, suppose that w n ∈ X for all n ∈ N. As X is Z-invariant, we deduce that for each n ∈ N, there exists xn ∈ X such that xn (−n)xn (−n + 1) · · · xn (n − 1) = w 2n .

.

As w ∞ and xn coincide on the interval [−n, n − 1] ⊂ Z, the sequence (xn )n∈N converges to w ∞ . This implies that w ∞ ∈ X since X is closed in AZ . (b) Suppose first that Per(X) is dense in X. Let u ∈ L(X) with length n. Then there exists x ∈ X such that x(0)x(1) · · · x(n − 1) = u.

.

Since Per(X) is dense in X, there exists a periodic configuration y ∈ X such that x and y coincide on {0, 1, . . . , n − 1}. If p ≥ 1 is a period of y such that p ≥ n

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and w := y(0)y(1) · · · y(p − 1), then u is a subword of w and w∞ = y ∈ X. This shows necessity. Conversely, suppose that (DP) is satisfied. Let x ∈ X and let Ω be a finite subset of Z. Choose n ∈ N such that Ω ⊂ [−n, n] and consider the word u := x(−n)x(−n + 1) · · · x(n).

.

By condition (DP), there exists a word w ∈ A∗ such that u is a subword of w and y := w∞ ∈ X. By suitably shifting y we get a periodic configuration in X that coincides with x on [−n, n] and hence on Ω. This shows that Per(X) is dense in X. ■ Exercise 3.35 Let A be a finite set and let X ⊂ AZ be an irreducible subshift of finite type. Let Per(X) denote the set of periodic configurations in X. (a) Show that Per(X) is dense in X. (b) Show that X is surjunctive. (c) Let A := {0, 1} and consider the subshift of finite type Y ⊂ AZ defined by Y := {x ∈ AZ : (x(n), x(n + 1)) /= (0, 1) for all n ∈ Z}. Show that Y is topologically transitive and that Per(Y ) is not dense in Y . (d) Show that the subshift Y is surjunctive. Solution (a) Let L(X) ⊂ A∗ denote the language of X and let m ≥ 1 be an integer such that {0, 1, . . . , m − 1} is a memory of X. Given a word w ∈ A∗ of length p ≥ 1, consider the p-periodic configuration w∞ ∈ AZ defined by (3.5). Let us show that Per(X) is dense in X. By Exercise 3.34, it is enough to prove that for every word u ∈ L(X) of length n ≥ m, there exists a word w ∈ A∗ with w ∞ ∈ X such that u is a subword of w. So consider u ∈ L(X) with length n ≥ m. By irreducibility of X, there exists a word v ∈ A∗ such that uvu ∈ L(X). Then w := vu satisfies w ∞ ∈ X since every word of length m appearing in w ∞ is in L(X). This shows that Per(X) is dense in X. (b) This follows from (a), by Exercise 3.32(b). (c) The elements in Y are the two constant configurations x0 and x1 and the configurations yn , n ∈ Z, respectively defined by xa (k) := a for all a ∈ A and k ∈ Z, and yn (k) := 1 for all k ≤ n and yn (k) := 0 for all k ≥ n + 1. Observe that yn → x1 as n → ∞ and that yn → x0 as n → −∞. As the configurations yn , n ∈ Z, are in the same orbit under the shift, we deduce that the orbit of y0 is dense in Y . This implies that Y is topologically transitive (cf. Exercise 1.63(a)). The set Per(Y ) is reduced to the two constant configurations and is therefore not dense in Y . (d) Let τ : Y → Y be an injective cellular automaton. From the Z-equivariance and the continuity of τ , we deduce that there is some k ∈ Z such that τ (yn ) = yn+k for all n ∈ Z and that τ fixes x0 and x1 . This implies that τ is surjective. Therefore Y is surjunctive. ■ Exercise 3.36 Show that the golden mean subshift is surjunctive.

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Solution This directly follows from Exercise 3.35(b) since the golden mean subshift X ⊂ {0, 1}Z is of finite type and irreducible (cf. Exercise 1.84). ■ Exercise 3.37 Let X ⊂ {0, 1}Z denote the even subshift. (a) Show that X contains a dense set of periodic configurations. (b) Show that X is surjunctive. Solution (a) Given a word w ∈ {0, 1}∗ of length p ≥ 1, we denote by w ∞ ∈ {0, 1}Z the periodic configuration with period p such that w∞ (0)w ∞ (1) · · · w ∞ (p − 1) = w. Let us show that the set Per(X) consisting of all periodic configurations in X is dense in X. By Exercise 3.34, it is enough to prove that for every x ∈ X and any n ∈ N, there exists a word w ∈ {0, 1}∗ with w ∞ ∈ X such that the word u := x(0)x(1) · · · x(n) is a subword of w. We distinguish three cases. Case 1: x(0) = x(n) = 1. Then we can take w := u. Case 2: x(0) = 0 and x(n) = 1 (resp. x(0) = 1 and x(n) = 0). Then we can take either w := 1u or w := 10u (resp. either w := u1 or w := u01). Case 3: x(0) = x(n) = 0. Then we can take either w := 1u1 or w := 10u1, or w := 1u01, or w := 10u01. This shows that Per(X) is dense in X. (b) Since Per(X) is dense in X by (a), we deduce from Exercise 3.32(b) that X is surjunctive. ■ Exercise 3.38 Let A := {0, 1, 2} and consider the subshift of finite type X ⊂ AZ with defining set of forbidden words {02, 10, 20, 21}. (a) Show that X is neither strongly irreducible, nor topologically mixing, nor irreducible, nor topologically transitive. (b) Show that X is not surjunctive. Solution (a) Let us list all the configurations in X. There are the three constant configurations xa , a ∈ A, defined by xa (k) := a for all k ∈ Z, the configurations yn , n ∈ Z, defined by yn (k) := 0 for k < n and yn (k) := 1 for n ≤ k, the configurations zn , n ∈ Z, defined by zn (k) := 1 for k ≤ n and zn (k) := 2 for n < k, and the configurations tm,n , m, n ∈ Z and m < n, defined by tm,n (k) := 0 for k ≤ m, tm,n (k) := 1 for m < k ≤ n, and tm,n (k) := 2 for n < k. It is clear from the list of the configurations in X that there is no configuration in X whose Z-orbit is dense in X. Therefore X is not topologically transitive (cf. Exercise 1.63(b)). Since X is not topologically transitive, it is neither irreducible nor topologically mixing (cf. Exercise 1.56(e)). Since X is not topologically mixing, it is not strongly irreducible either (cf. Exercise 1.64(b)). (b) Consider the cellular automaton τ : AZ → AZ with memory set S := {0, 1} and local defining map μ : AS → A defined by ⎧ μ(p) :=

1

if (p(0), p(1)) = (0, 1)

p(0)

otherwise.

.

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We then have τ (xa ) = xa for all a ∈ A, τ (yn ) = yn−1 and τ (zn ) = zn for all n ∈ Z, and τ (tm,n ) = tm−1,n for all m, n ∈ Z with m < n. It follows that τ (X) ⊂ X and that the cellular automaton σ : X → X defined by σ (x) := τ (x) for all x ∈ X is injective. As σ is not surjective since t0,1 is not in the image of σ , this shows that the subshift X is not surjunctive. ■ Comment This example is due to Weiss [Weis2, p. 358]. Exercise 3.39 Let A := {0, 1} and consider the subshift of finite type X ⊂ AZ with defining set of forbidden words {100, 110}. (a) Show that X is neither topologically transitive, nor irreducible, nor topologically mixing, nor strongly irreducible. (b) Show that X is not surjunctive. Solution (a) Let us list all the configurations in X. There are the two constant configurations xa , a ∈ A, defined by xa (k) := a for all k ∈ Z, the two 2-periodic configurations ya , a ∈ A, defined by ya (k) := a if k is even and ya (k) := 1 − a if k is odd, the configurations zn , n ∈ Z, defined by zn (k) := 0 for k < n and zn (k) := k − n mod 2 for n ≤ k, and the configurations tm,n , m, n ∈ Z and m < n with n − m odd, defined by tm,n (k) := 0 for k ≤ m, tm,n (k) := k − m mod 2 for m < k ≤ n, and tm,n (k) := 1 for n < k. It is clear from the list of the configurations in X given above that there is no configuration in X whose orbit closure is X. Therefore X is not topologically transitive (cf. Exercise 1.63(b)). As X is not topologically transitive, it follows that X is not irreducible, nor topologically mixing, nor strongly irreducible (cf. Exercise 1.56(d), (e) and Exercise 1.64(b)). (b) Consider the cellular automaton τ : AZ → AZ with memory set S := {0, 1, 2} and local defining map μ : AS → A defined by μ(p) :=

⎧ 1

.

p(0)

if (p(0), p(1), p(2)) = (0, 0, 1) otherwise.

We then have τ (xa ) = xa and τ (ya ) = ya for all a ∈ A, τ (zn ) = zn−2 for all n ∈ Z, and τ (tm,n ) = tm−2,n for all m, n ∈ Z with m < n. Thus τ (X) ⊂ X and the cellular automaton σ : X → X defined by σ (x) = τ (x) for all x ∈ X is injective. As σ is not surjective since t0,1 is not in the image of σ , this shows that the subshift X is not surjunctive. ■ Exercise 3.40 Let G be a group and let A be a finite set. Suppose that G contains an element of infinite order and that A has more than one element. (a) Show that there exists a subshift of finite type X ⊂ AG which is not surjunctive. (b) Show that if, in addition, G is not virtually cyclic, then there exists a topologically transitive subshift of finite type X ⊂ AG which is not surjunctive.

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3 Surjunctive Groups

Solution (a) Let H ⊂ G be an infinite cyclic subgroup. Up to renaming the elments of A, we can suppose that 0, 1 ∈ A. It follows from Exercise 3.39(b) that there exists a subshift of finite type Y ⊂ {0, 1}H ⊂ AH which is not surjunctive. By Exercise 1.103(i), the subshift X := Y (G) ⊂ AG is of finite type. On the other hand, it follows from Exercise 3.30 that X is not surjunctive. (b) If, in addition, G is not virtualy cyclic, then the subgroup H has infinite index in G and it follows from Exercise 1.103(h) that X is topologically transitive. ■ Exercise 3.41 (W-Subshifts) Let A be a finite set and let X ⊂ AZ be a subshift. One says that X is a W-subshift if there exists an integer n0 ≥ 0 such that the following holds: for all words u, v ∈ L(X), there exists a word w ∈ A∗ .

with length ℓ(w) ≤ n0 such that uwv ∈ L(X).

(3.6)

(a) Show that every W-subshift X ⊂ AZ is irreducible. (b) Show that every strongly irreducible subshift X ⊂ AZ is a W-subshift. (c) Give an example of a W-subshift X ⊂ AZ that is of finite type but not strongly irreducible. (d) Let X ⊂ AZ be a W-subshift. Show that Per(X) is dense in X. (e) Show that every W-subshift X ⊂ AZ is surjunctive. Solution (a) This is an immediate consequence of the definition of a W-subshift since a subshift X ⊂ AZ is irreducible if and only if for all u, v ∈ L(X), there exists w ∈ L(X) such that uvw ∈ L(X) (cf. Exercise 1.76(b)). (b) This immediately follows from the characterization of strongly irreducible subshifts over Z given in Exercise 1.76(c). Indeed, this characterization implies in particular that if X ⊂ AZ is strongly irreducible, then there exists an integer n0 ≥ 0 such that for all words u, v ∈ L(X), there exists a word w ∈ A∗ with length ℓ(w) = n0 such that uwv ∈ L(X). (c) Consider the subshift of finite type X ⊂ {0, 1}Z admitting {00, 11} as a defining set of forbidden words. Thus X consists of the two non-constant 2-periodic configurations. Clearly X is a W-subshift (we can take n0 = 1). However, X is not strongly irreducible, not even topologically mixing (cf. Exercise 1.81). (d) By Exercise 3.34(b), it suffices to show that every word in L(X) appears in some periodic configuration of X. Let n0 ≥ 0 be an integer satisfying (3.6). For each u ∈ L(X), let F (u) denote the non-empty finite set consisting of all words v ∈ A∗ with length ℓ(v) ≤ n0 such that uvu ∈ L(X). Take u0 ∈ L(X) such that F (u0 ) has minimal cardinality and let v0 ∈ F (u0 ).

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Suppose now that w ∈ A∗ is such that u0 wu0 ∈ L(X). We then have F (u0 wu0 ) ⊂ F (u0 ) and hence F (u0 wu0 ) = F (u0 )

.

by minimality of F (u0 ). In particular, we have v0 ∈ F (u0 wu0 ), that is, u0 wu0 v0 u0 wu0 ∈ L(X). By induction, it follows that the sequence of words (wk )k≥1 defined by w1 := u0 wu0 and wk+1 := wk v0 wk for k ≥ 1, satisfies wk ∈ L(X) for all k ≥ 1. As wk+1 = (u0 wu0 v0 )2

.

k −1

u0 vu0

for all k ≥ 1, we deduce that (u0 wu0 v0 )n ∈ L(X)

.

for all n ≥ 0. By Exercise 3.34(a), this implies that the periodic configuration (u0 wu0 v0 )∞ is in X (cf. (3.5)) and hence that w appears in a periodic configuration of X. Consider now an arbitrary word v ∈ L(X). Since X is irreducible by (a), we can find u1 ∈ A∗ such that u0 u1 v ∈ L(X) and u2 ∈ L(X) such that u0 u1 vu2 u0 = (u0 u1 v)u2 u0 ∈ L(X). Applying our previous argument to w := u1 vu2 , we conclude that u1 vu2 and hence v appears in some periodic configuration of X. This shows that the periodic configurations are dense in X. (e) This follows from (d) by Exercise 3.32(b). ■ Comment This is taken from [CecC4]. The proof of (d) (cf. [CecC4, Theorem 5.1]) was kindly communicated to us by Benjamin Weiss. Exercise 3.42 Let G be a group and let A := {0, 1}. Show that the at-most-one-one subshift X ⊂ AG is surjunctive. Solution This follows from Exercise 1.71(b) since, with the notation therein, the injective cellular automata τ : X → X are exactly the cellular automata τg , g ∈ G, which are in fact all bijective. ■ Exercise 3.43 (Minimal Subshifts) Let G be a group and let A be a set. A subshift X ⊂ AG is called a minimal subshift if X is non-empty and the G-orbit of every configuration x ∈ X is dense in X. (a) Show that a subshift X ⊂ AG is minimal if and only if X /= ∅ and there is no subshift Y of AG such that ∅ /= Y  X. (b) Show that the finite minimal subshifts in AG are precisely the finite G-orbits. (c) Show that every minimal subshift X ⊂ AG is topologically transitive. (d) Give an example of a minimal subshift that is not topologically mixing. (e) Show that if X ⊂ AG is an infinite minimal subshift then the G-orbit of every configuration x ∈ X is infinite.

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3 Surjunctive Groups

(f) Show that if X and Y are minimal subshifts of AG then either X = Y or X ∩ Y = ∅. Solution (a) Let X ⊂ AG be a subshift. Suppose that X is minimal. Then X /= ∅. If Y is a subshift of AG such that ∅ /= Y ⊂ X and y ∈ Y , we have Gy ⊂ Y since Y is G-invariant and closed in AG . As Gy = X by minimality of X, we deduce that Y = X. Conversely, suppose that X /= ∅ and there is no subshift Y of AG such that ∅ /= Y  X. Let x ∈ X. Then Y := Gx is a non-empty subshift in AG so that, by our assumptions, Y = X. This shows that the G-orbit of every element x ∈ X is dense in X. It follows that X is minimal. (b) Let x ∈ AG be a periodic configuration. Then X := Gx is a non-empty finite subshift. Given y ∈ X, there exists g ∈ G such that y = gx so that Gy = Gy = Ggx = Gx = X. This shows that X is minimal. Conversely, suppose that X ⊂ AG is minimal and finite. Pick x ∈ X. Since Gx ⊂ X, we deduce that the configuration x is periodic. By (a), Gx = Gx must equal the whole of X. (c) This follows immediately from Exercise 1.63(a) since every minimal subshift contains a dense orbit. (d) Let A := {0, 1} and consider the subshift X ⊂ AZ consisting of the two nonconstant 2-periodic configurations. Observe that X is a finite Z-orbit. Therefore X is minimal by (b). However, X is not topologically mixing (cf. Exercise 1.81). (e) Let X ⊂ AG be an infinite subshift. Let x ∈ X and suppose that x is periodic. Then the finite non-empty subshift Y := Gx is a proper subshift of the infinite subshift X. We deduce from (a) that X is not minimal. (f) Suppose that X and Y are minimal subshifts in AG . If X ∩ Y /= ∅, then the non-empty subshift Z := X ∩ Y satisfies Z ⊂ X and Z ⊂ Y . By (a), we necessarily have X = Z = Y . ■ Comment More generally, given a dynamical system consisting a topological space X equipped with a continuous action of a group G, a subset M ⊂ X is called a minimal set if M is a non-empty closed invariant subset of X and the orbit of every point m ∈ M is dense in M. The notion of a minimal set was introduced in a 1912 paper by George David Birkhoff [Bir] in his study of recurrence in dynamical systems associated with ordinary differential equations. Exercise 3.44 Let G be a group. Let A and B be sets with A finite. Let X ⊂ AG and Y ⊂ B G be subshifts. Suppose that X is non-empty and Y is minimal. Show that every cellular automaton σ : X → Y is surjective. Solution Let σ : X → Y be a cellular automaton. Then there exists a cellular automaton τ : AG → B G such that σ (x) = τ (x) for all x ∈ X. By Exercise 1.39(g), the set σ (X) = τ (X) ⊂ Y is a non-empty subshift. Therefore σ (X) = Y by minimality of Y . This shows that σ is surjective. ■

3.2 Exercises

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Exercise 3.45 Let G be a group and let A be a finite set. Show that every minimal subshift X ⊂ AG is surjunctive. Solution If X ⊂ AG is a minimal subshift, then every cellular automaton σ : X → X is surjective by Exercise 3.44. This implies in particular that X is surjunctive. ■ Exercise 3.46 Let G be a group and let A be a finite set. Show that every non-empty subshift X ⊂ AG contains a minimal subshift. Solution Let X ⊂ AG be a non-empty subshift. Consider the set S of all nonempty subshifts contained in X, partially ordered by inclusion. Note that S is nonempty ∩ since X ∈ S . Let C ⊂ S be a totally ordered subset of S . We claim that Y0 := Y ∈C Y is a lower bound for C in S . We first observe that Y0 is a subshift contained in X since it is the intersection of a family of subshifts contained in X (cf. Exercise 1.39(d)). On the other hand, C is a set of closed subsets of X with the finite ∩intersection property. Indeed, if F ⊂ C is a non-empty finite subset, then Z0 := Z∈F Z is non-empty since Z0 = minZ∈F Z ∈ S . By compactness of X, it follows that Y0 /= ∅. This shows that Y0 ∈ S . By applying Zorn’s lemma, we deduce that the set S admits a minimal element, that is, a minimal subshift contained in X. ■ Comment More generally, Zorn’s lemma shows that if X is a topological space equipped with a continuous action of a group G, then every non-empty compact invariant subset of X contains a minimal set (this fundamental observation goes back to George David Birkhoff). Exercise 3.47 Let A be a finite set and let X ⊂ AZ be an infinite minimal subshift. Show that X is not sofic. Solution Since X is infinite and minimal, it contains no periodic configurations. As every non-empty sofic subshift of AZ contains a periodic configuration by ■ Exercise 1.96(b), we deduce that X is not sofic. Exercise 3.48 (Almost Periodic Configurations) Let G be a group and let A be a set. A subset R ⊂ G is called syndetic if there exists a finite subset K ⊂ G such that the set Kg meets R for every g ∈ G. A configuration x ∈ AG is called almost periodic (or syndetically recurrent) if for every neighborhood U of x in AG the set Rx (U ) := {g ∈ G : gx ∈ U } is syndetic in G. (a) Show that a configuration x ∈ AG is almost periodic if and only if it satisfies the following condition: for every finite subset Ω ⊂ G the set R(x, Ω) := {g ∈ G : (gx)|Ω = x|Ω } is syndetic in G. (b) Let H be a subsgroup of G. Show that H is syndetic in G if and only if H is of finite index in G. (c) Show that every periodic configuration x ∈ AG is almost periodic. (d) Let x ∈ AG be an almost periodic configuration. Show that its orbit closure X := Gx ⊂ AG is a minimal subshift. (e) Suppose that the set A is finite and let X ⊂ AG be a minimal subshift. Show that every configuration x ∈ X is almost periodic.

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3 Surjunctive Groups

(f) Suppose that the set A is finite and let X ⊂ AG be an infinite minimal subshift. Show that every configuration x ∈ X is almost periodic but not periodic. (g) Suppose that the set A is finite. Show that a non-empty subshift X ⊂ AG is minimal if and only if it satisfies the following two conditions: (1) X is topologically transitive; (2) every configuration x ∈ X is almost periodic. Solution (a) This immediately follows from the fact that the sets V (x, Ω) := {y ∈ AG : y|Ω = x|Ω }, where Ω runs over all finite subsets of G, form a neighborhood base of x and the obvious fact that every subset of G containing a syndetic subset is itself syndetic. (b) Suppose that H is syndetic in G. This means that there is a finite subset K ⊂ G such that Kg meets H for every g ∈ G. Therefore G = K −1 H and hence [G : H ] ≤ |K −1 | = |K| < ∞. This shows that H is of finite index in G. Conversely, suppose that H is of finite index in G and let T ⊂ G be a complete set of representatives of the left cosets of H in G. Then T −1 g meets H for every g ∈ G. As |T −1 | = |T | = [G : H ] < ∞, this shows that H is syndetic in G. (c) Suppose that x ∈ AG is periodic. This means that the G-orbit of x is finite, or equivalently, that the stabilizer of x in G, i.e., the subgroup H := {g ∈ G : gx = x} is of finite index in G. By (a), this implies that H is syndetic in G. As H ⊂ Rx (U ) for every neighborhood U of x, we deduce that x is almost periodic. (d) Suppose by contradiction that X is not minimal. This means that there is a subshift Y of AG such that ∅  Y  X. Since X is the orbit closure of x, we have x∈ / Y . As Y is closed in AG , this implies that there is a finite subset Ω ⊂ G such that V (x, Ω) ⊂ X \ Y , i.e., x|Ω /= y|Ω for all y ∈ Y . Let now K be a finite subset of G. Choose an arbitrary configuration y0 ∈ Y . As the set Λ := K −1 Ω is finite and y0 is in the orbit closure of x, there is an element g0 ∈ G such that (g0 x)|Λ = y0 |Λ . This implies that (kg0 x)|Ω = (ky0 )|Ω /= x|Ω for all k ∈ K. Thus the set Kg0 does not meet R(x, Ω). This shows that x is not almost periodic. (e) Let x ∈ X and let Ω be a finite subset of G. Since X is minimal, X is the orbit closure of x. It follows that the open subsets Vg := {y ∈ X : (gy)|Ω = x|Ω }, g ∈ G, U cover X. By compactness of X, there is a finite subset K ⊂ G such that X = k∈K Vk . We deduce that, for every g ∈ G, there is some k ∈ K such that gx ∈ Vk , i.e., kg ∈ R(x, Ω). This shows that x is almost periodic. (f) This immediately follows from (e) since an infinite minimal subshift cannot contain a periodic configuration by minimality (cf. Exercise 3.43(e)). (g) Suppose first that X is a nonemty minimal subshift. If x ∈ X, then the orbit of x is dense in X by minimality so that X is topologically transitive. On the other hand, x is almost periodic by (e). Conversely, suppose that X ⊂ AG is a non-empty subshift satisfying (1) and (2). By (1), there is a configuration x ∈ X whose orbit is dense in X. Since x is almost periodic by (2), we conclude that X is minimal by (d). ■ Comment More generally, if (X, G) is a dynamical system consisting of a topological space X equipped with a continuous action of a group G, a point x ∈ X

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189

is called almost periodic, or syndetically recurrent, if for every neighborhood U of x, there exists a syndetic subset R ⊂ G such that gx ∈ U for all g ∈ G. If the topological space X is regular (i.e., for any closed subset F ⊂ X and any point x ∈ X \ F , there exists a neighborhood U of x and a neighborhood V of F that are disjoint), then the orbit closure of any almost periodic point is a minimal set. If X is compact and M ⊂ X is a minimal set, then every point of M is almost periodic. Exercise 3.49 Let A be a set. Given a configuration x ∈ AZ (resp. a subshift X ⊂ AZ ), denote by L(x) (resp. L(X)) its language. (a) Show that a non-empty subshift X ⊂ AZ is minimal if and only if one has L(x) = L(X) for all x ∈ X. (b) Let x ∈ AZ . Show that the following conditions are equivalent: (AP1) (AP2) (AP3)

x is almost periodic; for every word u ∈ L(x) of length ℓ(u), the set consisting of all n ∈ Z such that x(n)x(n + 1) · · · x(n + ℓ(u) − 1) = u is a syndetic subset of Z; for every word u ∈ L(x), there exists an integer r = r(u) ≥ 1 such that u is a subword of any word v ∈ L(x) of length r.

Solution (a) The subshift X is minimal if and only if it is the orbit closure of every x ∈ X. This is equivalent to L(x) = L(X) for all x ∈ X by Exercise 1.74(c). (b) Let x ∈ AZ . Suppose first that x is almost periodic and let u ∈ L(x) of length ℓ(u) =: p. Then there exist m ∈ Z such that x(m)x(m + 1) · · · x(m + p − 1) = u. Let Ω := {m, m + 1, · · · , m + p − 1}. Since x is almost periodic, there exists a syndetic subset R ⊂ Z such that (gx)|Ω = x|Ω , that is, x(−g+m)x(−g+m+1) · · · x(−g+m+p−1) = u, for all g ∈ R. As the syndeticity of R clearly implies that of the set −R + m, this shows that (AP1) implies (AP2). Suppose now that (AP2) is satisfied. Then, given u ∈ L(x) with length p, there is a syndetic subset R ⊂ Z such that x(n)x(n + 1) · · · x(n + p − 1) = u for all n ∈ R. As R is syndetic, there is an integer k ≥ 1 such that any subset of Z consisting of k consecutive integers meets R. We deduce that if v ∈ L(x) has length at least r := k + p then u is a subword of v. This shows that (AP2) implies (AP3). Finally, suppose that x satisfies (AP3). Let Ω be a finite subset of Z. Choose an interval [p, q] ⊂ Z such that Ω ⊂ [p, q]. Consider the word u := x(p)x(p + 1) · · · x(q) ∈ L(x) and the integer r = r(u) ≥ 1 provided by (AP3). Then for every k ∈ Z, the word u is a subword of the word x(kr)x(kr + 1) · · · x((k + 1)r − 1). Therefore, there is gk ∈ Z such that kr ≤ −gk + p < (k + 1)r and x(−gk + p)x(−gk + p + 1) · · · x(−gk + q) = u. This implies that (gk x)|Ω = x|Ω . As the set {gk : k ∈ Z} ⊂ Z is clearly syndetic, we deduce that x is almost periodic. This shows that (AP3) implies (AP1). ■ Exercise 3.50 Let G be an infinite group and let A be a finite set. Show that every minimal subshift X ⊂ AG is irreducible.

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3 Surjunctive Groups

Solution Suppose that X ⊂ AG is a minimal subshift. Let U and V be two nonempty open subsets of X. Since X is topologically transitive by Exercise 3.43(c), there exists g0 ∈ G such that U ∩ g0 V /= ∅. Let x0 ∈ U ∩ g0 V . Since the configuration g0−1 x0 ∈ V is almost periodic by Exercise 3.48(e), there exists a syndetic subset R ⊂ G such that rg0−1 x0 ∈ V for all r ∈ R. As G is infinite, the set R is itself infinite. Therefore, for every finite subset F ⊂ G, there exists g1 ∈ R such that g1 g0−1 ∈ / F −1 . Setting g := g0 g1−1 , we then have g ∈ / F and x0 ∈ U ∩ gV . This shows that the subshift X is irreducible. ■ Exercise 3.51 (The Thue-Morse Sequence) Let A := {0, 1}. The Thue-Morse sequence is the sequence T : N → A defined by T (n) := 0 if the number of 1s in the binary expansion of n is even and T (n) := 1 otherwise. (a) Check that T (0)T (1) · · · T (15) = 0110100110010110.

.

(b) Show that T (2n) = T (n) and T (2n + 1) = 1 − T (n) for all n ∈ N. (c) Let k, n ∈ N such that k ≤ 2n − 1. Show that T (k + 2n ) = 1 − T (k). (d) Consider the sequences (an )n∈N and (bn )n∈N of elements of A∗ defined by the initial conditions a0 := 0 and b0 := 1, and the recurrence relations an+1 := an bn and bn+1 := bn an for all n ∈ N. Let ι : A∗ → A∗ denote the unique monoid automorphism such that ι(0) := 1 and ι(1) := 0. Show that one has bn = ι(an )

(3.7)

an = T (0)T (1) · · · T (2n − 1)

(3.8)

.

and .

for all n ∈ N. (e) Let k, n ∈ N. Show that one has ⎧ T (k2 )T (k2 + 1) · · · T (k2 + 2 − 1) =

.

n

n

n

n

an

if T (k) = 0,

bn

if T (k) = 1.

(3.9)

(f) Consider the unique monoid homomorphism ϕ : A∗ → A∗ that satisfies ϕ(0) := 01

and

ϕ(1) := 10.

an+1 = ϕ(an )

and

bn+1 = ϕ(bn )

.

Show that one has .

for all n ∈ N.

(3.10)

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191

(g) Show that an = ϕ n (0) and bn = ϕ n (1) for all n ∈ N. (h) For u ∈ A∗ , let ρ(u) := uR (cf. Exercise 1.88) denote the word in A∗ obtained by reading the letters of u from right to left. Thus, ρ : A∗ → A∗ is the unique antiautomorphism of the monoid A∗ that fixes 0 and 1. Show that one has ⎧ (ρ(an ), ρ(bn )) =

.

(an , bn )

if n is even,

(bn , an )

if n is odd,

(3.11)

for all n ∈ N. Solution (a) The binary expansions of the integers 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15 are respectively 0, 1, 10, 11, 100, 101, 110, 111, 1000, 1001, 1010, 1011, 1100, 1101, 1110, 1111. Therefore T (0) = 0, T (1) = 1, T (2) = 1, T (3) = 0, T (4) = 1, T (5) = 0, T (6) = 0, T (7) = 1, T (8) = 1, T (9) = 0, T (10) = 0, T (11) = 1, T (12) = 0, T (13) = 1, T (14) = 1, T (15) = 0. (b) The digital expansion of 2n is obtained from that of n by adding one 0 on the right. Thus 2n and n have the same number of 1s in their digital expansion. It follows that T (2n) = T (n). On the other hand, the digital expansion of 2n + 1 is obtained from that of n by adding one 1 on the right. Thus the numbers of 1s in the digital expansion of 2n + 1 and n do not have the same parity. It follows that T (2n + 1) = 1 − T (n). (c) The digital expansion of k + 2n is obtained from that of k by adding one 1 on the left at position n and completing with 0s if necessary. Thus the number of 1s in the digital expansion of k + 2n and 2n have different paritiy. It follows that T (k + 2n ) = 1 − T (k). (d) The sequences (ι(an ))n∈N and (ι(bn ))n∈N satisfy the initial conditions ι(b0 ) = 0 and ι(a0 ) = 1, as well as the recurrence relations ι(bn+1 ) = ι(bn an ) = ι(bn )ι(an ) and ι(an+1 ) = ι(an bn ) = ι(an )ι(bn ) for all n ∈ N. We deduce that bn = ι(an ) for all n ∈ N by uniqueness of the sequences (an ) and (bn ). We now prove (3.8) by induction. For n = 0, the equality follows from the fact that a0 = T (0) = 0. Suppose that the formula is true for an . Then we have an+1 = an bn

.

= an ι(an ) = T (0)T (1) · · · T (2n −1)ι(T (0)T (1) · · · T (2n −1))

(by induction)

= T (0)T (1) · · · T (2 − 1)ι(T (0))ι(T (1)) · · · ι(T (2 − 1)) n

n

= T (0)T (1) · · · T (2n − 1)T (2n )T (1 + 2n ) · · · T (2n + 2n − 1) (by (c)) = T (0)T (1) · · · T (2n+1 − 1), showing that the formula is also true for an+1 .

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(e) For k = 0, formula (3.9) reduces to (3.8). Suppose now that k ≥ 1. Observe that, given any integer i such that 0 ≤ i ≤ 2n −1, The digital expansion of k2n +i is obtained by writing the digital expansion of k followed by the digital expansion of i with a (possibly empty) block of 0s between them. We deduce that T (k2n + i) = T (i) if T (k) = 0 while T (i +k2n ) = 1−T (i) if T (k) = 1. By using (3.8) and (3.7), this gives us (3.9). (f) Let us first prove the formula an+1 = ϕ(an ) by induction on n. For n = 0, the formula follows from the fact that a1 = a0 b0 = 01 = ϕ(0) = ϕ(a0 ). Suppose that the formula holds true at some rank n ∈ N, i.e., an+1 = ϕ(an ) for some n ∈ N. Then an+2 = an+1 bn+1

.

= an+1 ι(an+1 ) = ϕ(an )ι(ϕ(an )) = ϕ(an )ϕ(ι(an ))

(since ι and ϕ clearly commute)

= ϕ(an ι(an )) = ϕ(an bn ) = ϕ(an+1 ), showing that the formula is also true at rank n + 1. Therefore an+1 = ϕ(an ) for all n ∈ N. To complete the proof of (3.10), it suffices to observe that bn+1 = ι(an+1 ) = ι(ϕ(an )) = ϕ(ι(an )) = ϕ(bn ).

.

(g) This immediately follows from (3.10) by induction since a0 = 0 and b0 = 1. (h) We prove (3.11) by induction on n. For n = 0, the formula follows from the fact that a0 = 0 and b0 = 1 are fixed by ρ. On the other hand, assuming that the formula holds true at some rank n ∈ N, then, using the fact that ρ(an+1 ) = ρ(an bn ) = ρ(bn )ρ(an ), we get ρ(an+1 ) = bn an = bn+1 and ρ(bn+1 ) = an bn = an+1 if n is even, and ρ(an+1 ) = an bn = an+1 and ρ(bn+1 ) = bn an = bn+1 if n is odd, showing that the formula holds also true at rank n + 1. ■ Comment The Thue-Morse sequence, also called the Prouhet-Thue-Morse sequence, first appeared implicitly in a 1851 paper by Prouhet [Prou] dealing with elementary problems in number theory. It was subsequently rediscovered independently by several authors including Thue [Thue] in 1906 and Morse [Mor] in 1921 (see [AllS1] for a nice historical survey). The Thue-Morse sequence is an example of an automatic sequence (see [AllS2, Sect. 5.1]). Exercise 3.52 (The Morse Subshift) Let A := {0, 1}. Keeping the notation introduced in Exercise 3.51, let x ∈ AZ be the configuration defined by x(n) := T (n) if n ≥ 0 and x(n) := T (−n − 1) if n ≤ −1.

3.2 Exercises

193

(a) Let k, n ∈ Z with n ≥ 0. Let wk,n ∈ A∗ be the word of length 2n defined by wk,n := x(k2n )x(k2n + 1) · · · x(k2n + 2n − 1).

.

Show that wk,n ∈ {an , bn } (here an and bn are the words of length 2n in A∗ introduced in Exercise 3.51). (b) Show that the configuration x is almost periodic. (c) Show that the configuration x is not periodic. (d) Let X ⊂ AZ denote the orbit closure of x. Show that X is an infinite minimal subshift of AZ . (e) Show that the subshift X is surjunctive and irreducible. (f) Show that the subshift X is not sofic. Solution (a) For k ∈ N, this follows from Exercise 3.51(e). Suppose now that k ≤ −1. We then have wk,n = x(k2n )x(k2n + 1) · · · x(k2n + 2n − 1)

.

= T (−k2n − 1)T (−k2n − 2) · · · T (−(k + 1)2n ) = ρ(T (−(k + 1)2n )T (−(k + 1)2n + 1) · · · T (−(k + 1)2n + 2n − 1)) and hence wk,n ∈ {ρ(an ), ρ(bn )} by Exercise 3.51(e). As {ρ(an ), ρ(bn )} = {an , bn } by Exercise 3.51(h), we deduce that wk,n ∈ {an , bn }. (b) Suppose that u ∈ L(x), where L(x) ⊂ A∗ denotes the language of x. This means that there exists i ∈ Z such that u = x(i)x(i + 1) · · · x(i + ℓ(u) − 1).

.

Taking n ∈ N such that 2n+1 ≥ ℓ(u), we deduce from (a) that u is a subword of at least one of the four words an an , an bn , bn an , or bn bn . Each of these four words is a subword of both an+3 = an bn bn an bn an an bn and bn+3 = bn an an bn an bn bn an . As every word in L(X) with length 2n+4 contains an+3 or bn+3 as a subword by (a), we deduce that every word v ∈ L(x) with length 2n+4 contains u as a subword. This shows that x is almost periodic by Exercise 3.49(b). (c) Suppose by contradiction that x is p-periodic for some integer p ≥ 1. This would imply that the Thue-Morse sequence T is p-periodic, i.e., T (n + p) = T n) for all n ∈ N.

.

(3.12)

We distinguish two cases. Suppose first that p is odd. Then 2 is an invertible element in the ring Z/pZ. Denoting by m the cardinality of the group of invertible elements of Z/pZ, we

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3 Surjunctive Groups

deduce that 2m is congruent to 1 modulo p. This implies that 2m+1 − 2 is a multiple of p and hence that T (2)T (3)T (4) = T (2m+1 )T (2m+1 + 1)T (2m+1 + 2).

.

This yields a contradiction, since T (2)T (3)T (4) = 101, while T (2m+1 )T (2m+1 + 1)T (2m+1 + 2) is a prefix of bm+1 (by Exercise 3.51(e)) and hence of b2 = 1001, so that T (2m+1 )T (2m+1 + 1)T (2m+1 + 2) = 100. This shows that (3.12) cannot be satisfied for p odd. Suppose now that p is even. Write p = 2r q, where r, q ∈ N \ {0} and q is odd. Recalling that x is p-periodic, we then observe that the sequence S : N → B, where B := {ar , br }, defined by S(n) := T (n2r )T (n2r + 1) · · · T (n2r + 2r − 1)

.

is q-periodic. This leads again to a contradiction by the first part of the proof since S is a disguised form of the Thue-Morse sequence (obtained by replacing 0 by ar and 1 by br ) by Exercise 3.51(e). (d) The fact that X is a minimal subshift follows from (b) and Exercise 3.48(d). The subshift X is not finite since otherwise the configuration x ∈ X would be periodic, contradicting (c). (e) This immediately follows from (d), Exercises 3.45 and 3.50. (f) This immediately follows from (d) by Exercise 3.47. ■ Comment The subshift X is called the Morse subshift, after Morse who used it in [Mor] to construct non-periodic almost periodic geodesics on certain surfaces with negative curvature. Exercise 3.53 (Toeplitz Subshifts) Let G be a group and let A be a finite set. A configuration x ∈ AG is called a Toeplitz configuration if for every neighborhood U of x in AG , there exists a finite index subgroup H of G such that the set Rx (U ) := {g ∈ G : gx ∈ U } contains H . One says that a subshift X ⊂ AG is a Toeplitz subshift if it is the orbit closure of some Toeplitz configuration x ∈ AG . (a) Show that every periodic configuration x ∈ AG is Toeplitz. (b) Show that every Toeplitz configuration x ∈ AG is almost periodic. (c) Show that every Toeplitz subshift X ⊂ AG is minimal. (d) Show that every Toeplitz subshift X ⊂ AG is surjunctive. (e) Let x ∈ AG . Show that the following conditions are equivalent: (TO1) (TO2) (TO3)

x is Toeplitz; for every finite subset Ω ⊂ G, there exists a finite index subgroup H of G such that (hx)|Ω = x|Ω for all h ∈ H ; for every g ∈ G, there exists a finite index subgroup H of G such that (hx)(g) = x(g) for all h ∈ H .

3.2 Exercises

195

Solution (a) Let x ∈ AG be a periodic configuration. Then the stabilizer H ⊂ G of x is a subgroup of finite index in G. As gx = x for all g ∈ H , this shows that x is Toeplitz. (b) This follows from the fact that every finite index subgroup H ⊂ G is syndetic in G (cf. Exercise 3.48(b)). (c) This follows from (b) and Exercise 3.48(d). (d) This follows from (c) and Exercise 3.45. (e) The equivalence between (TO1) and (TO2) follows from the fact that the sets {y ∈ AG : y|Ω = x|Ω }, where Ω runs over all finite subsets of G, form a neighborhood basis of x. To show that (TO2) implies (TO3), it suffices to take Ω := {g}. Finally, suppose that (TO3) is satisfied. Let Ω be a finite subset of G. By (TO3), for every g ∈ Ω, there exists a finite index subgroup Hg of G such that (hx)(g) = x(g) for all h ∈ Hg . As the intersection of a finite family ∩ of finite index subgroups of G is a finite index subgroup of G, the set H := g∈Ω Hg is a finite index subgroup of G. As (hx)|Ω = x|Ω for all h ∈ H , this shows that (TO3) implies (TO2). ■ Exercise 3.54 Let A be a finite set. (a) Show that every Toeplitz subshift X ⊂ AZ is irreducible. (b) Let x ∈ AZ . Show that x is Toeplitz if and only if it satisfies the following condition: (To)

for every n ∈ Z, there exists an integer r ≥ 1 such that x(n) = x(n + kr) for all k ∈ Z;

(c) Given a Toeplitz configuration x ∈ AZ and n ∈ Z we define the integer p(x, n) ≥ 1 by p(x, n) := min{r ≥ 1 : x(n) = x(n + kr) for all k ∈ Z}

.

and the subset P (x) ⊂ N \ {0} by P (x) := {p(x, n) : n ∈ Z}. Let x ∈ AZ be a Toeplitz configuration. Show that x is periodic if and only if the set P (x) is finite. (d) Let A := {0, 1}. Consider the bijective map ι : Z → N defined by ι(n) := 2n if n ≥ 0 and ι(n) := −(2n + 1) if n ≤ −1. Show that the following inductive procedure yields a well defined non-periodic Toeplitz configuration x ∈ AZ : (1) Step 1 consists in setting x(n) := 1 for all n ∈ 2Z. (2) Suppose that Step k − 1 has been performed for some integer k ≥ 2 and let Jk denote the set consisting of all n ∈ Z for which x(n) has not yet been defined. Denoting by nk the integer in Jk whose image under ι is minimal, Step k consists in setting x(n) := k mod 2 for all n ∈ nk + 2k Z. Solution (a) This follows from Exercises 3.53(c) and 3.50. (b) This follows from the equivalence between conditions (TO1) and (TO3) in Exercise 3.53(e) since the finite index subgroups of Z are the subsets of Z of the form H = rZ, where r ≥ 1.

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3 Surjunctive Groups

(c) Suppose first that P (x) is finite. If m is a common multiple of the elements of P (x), then x(n + m) = x(n) for all n ∈ Z. Therefore x is m-periodic. Conversely, suppose that x is p-periodic for some p ≥ 1. We then have p(x, n) ≤ p for all n ∈ Z. This implies that P (x) is finite. (d) For Step 1, we shall take J1 := Z and n1 := 0. For every k ≥ 1, the set of n ∈ Z for which x(n) is first defined at Step k is nk + 2k Z. As ι(nk ) ≥ k − 1 by an immediate induction, we deduce that the sets nk + 2k Z, k ≥ 1, form a partition of Z. For every 1 ≤ i ≤ k − 1, the set ni + 2i Z is the disjoint union of 2k−1 /2i = k−i−1 2 disjoint equivalence classes modulo 2k−1 . We deduce that the set Z \ Jk , consisting of all n ∈ Z such that x(n) is defined before Step k, is the disjoint union of 2k−2 +2k−3 +· · ·+2+1 = 2k−1 −1 equivalence classes modulo 2k−1 . It follows that Jk is one of the 2k−1 equivalence classes modulo 2k−1 , namely Jk = nk + 2k−1 Z. Consider now an integer n ∈ nk + 2k Z. By construction, we have x(n + 2k i) = k mod 2 for all i ∈ Z and x(n) /= x(nk+1 ) = k + 1 mod 2. As n, nk+1 ∈ Jk = nk + 2k−1 Z, so that nk+1 ≡ n mod 2k−1 , this implies that p(x, n) = 2k (here we use the notation introduced in (c)). Thus, the configuration x is Toeplitz but not periodic. ■ Comment The class of Toeplitz sequences was introduced in 1969 by Jacobs and Keane [JacK]. The term “Toeplitz” was chosen by them because of the analogy with a method proposed by Toeplitz [Toe] for constructing almost periodic functions on the real line. In [GarcH, p. 961], Garcia and Hedlund gave the first example of an infinite Toeplitz subshift over Z. In [Oxt], Oxtoby proposed a method for the explicit construction of non-periodic Toeplitz sequences. Toeplitz sequences were called “regularly almost periodic” sequences at the time (cf. [GarcH, p. 957]). Oxtoby’s method was subsequently extended by Williams in [Wil3]. Exercise 3.55 Let A be a finite set and let X ⊂ AZ be an infinite Toeplitz subshift. Show that X is not sofic. Solution This immediately follows from Exercise 3.47 since X is minimal by Exercise 3.53(c). ■ Exercise 3.56 (Strongly Aperiodic Subshifts) Let G be a group and let A be a set. A configuration x ∈ AG is called strongly aperiodic if the stabilizer of x in G is trivial, i.e., gx = x implies g = 1G for all g ∈ G. A subshift X ⊂ AG is called strongly aperiodic if X is non-empty and every configuration x ∈ X is strongly aperiodic. (a) Let G be a finite group and let A be a set with more than one element. Show that there exists a strongly aperiodic subshift X ⊂ AG . (b) Let A be a finite set. Show that there exists no strongly aperiodic sofic subshift X ⊂ AZ . (c) Let A be a set. Show that every infinite minimal subshift X ⊂ AZ is strongly aperiodic. (d) Show that the Morse subshift X ⊂ {0, 1}Z is strongly aperiodic. (e) Let A be a finite set. Show that every infinite Toeplitz subshift X ⊂ AZ is strongly aperiodic.

3.2 Exercises

197

(f) Let A and B be two sets and let τ : B G → AG be a G-equivariant continuous map (e.g., a cellular automaton). Suppose that X ⊂ AG is a strongly aperiodic subshift. Show that τ −1 (X) is a strongly aperiodic subshift of B G . Solution (a) Let a and b be two distinct elements in A. Consider the configuration x ∈ AG such that x(1G ) = b and x(g) = a for all g ∈ G \ {1G }. Let X ⊂ AG denote the G-orbit of x. Clearly X is a strongly aperiodic subshift of AG . (b) This immediately follows from the fact that every non-empty sofic subshift X ⊂ AZ contains a periodic configuration (cf. Exercise 1.96(b)). (c) First observe that a configuration x ∈ AZ is strongly aperiodic if and only if it is not periodic, i.e., the Z-orbit of x is infinite. If X ⊂ AZ is an infinite minimal subshift, then X contains no periodic configurations. Therefore X is strongly aperiodic. (d) This immediately follows from (c) since the Morse subshift is an infinite minimal subshift by Exercise 3.52(d). (e) This immediately follows from (c) since every Toeplitz subshift X ⊂ AZ is minimal by Exercise 3.53(c). (f) The fact that τ −1 (X) is a non-empty subshift of B G follows from Exercise 1.39(f). Let y ∈ τ −1 (X) and consider the configuration x := τ (y) ∈ X. For all g ∈ G, we have gx = gτ (y) = τ (gy) since τ is G-equivariant. Thus, if gy = y then gx = τ (gy) = τ (y) = x and hence g = 1G since X is strongly aperiodic. This shows that τ −1 (X) is strongly aperiodic. ■ Comment Strongly aperiodic subshifts are also called free subshifts by some authors because of the fact that a non-empty subshift X ⊂ AG is strongly aperiodic if and only if the action of G on X is free. In [GaoJS], Gao, Jackson, and Seward proved that there exists a strongly aperiodic subshift X ⊂ {0, 1}G for every countable group G. A simpler proof of this result, based on probabilistic methods, was subsequently given by Aubrun, Barbieri, and Thomassé in [AubBT]. Since all subshifts of finite type are sofic, it follows from (b) that there are no strongly aperiodic subshifts of finite type with finite alphabet over Z. The study of certain decision problems for tilings of the plane led the logician Hao 2 Wang to conjecture in 1961 that AZ contains no strongly aperiodic subshifts of finite type for A finite. This conjecture was shown to be false by Wang’s student Berger [Ber]. Recently, the question of determining which groups admit strongly aperiodic subshifts of finite type with finite alphabet has led to very interesting results (see [Cohe], [Barb1], [Barb2] and the references therein). In [Cohe], it is shown that finitely generated groups with more than one end (i.e., with 2 or infinitely many ends) do not admit strongly aperiodic subshifts of finite type with finite alphabet. In [CoheGR], it is proved that all one-ended word hyperbolic groups (e.g., fundamental groups of closed surfaces of genus at least 2) admit strongly aperiodic subshifts of finite type with finite alphabet. Barbieri has shown in [Barb1] that strongly aperiodic subshifts with finite alphabet also exist over the Grigorchuk group.

Chapter 4

Amenable Groups

This chapter is devoted to group amenability. It includes a detailed study of nilpotent groups, polycyclic groups, and solvable groups. Følner sequences in countable amenable groups such as the free abelian groups and the Heisenberg group are explicitly described. Paradoxical decompositions of nonamenable groups and Tarski numbers of groups are also investigated.

4.1 Summary 4.1.1 Equivalent Definitions of Amenability Let G be a group. Denote by .P(G) the set of all subsets of G and by .[0, 1] the unit interval in .R. A map .μ : P(G) → [0, 1] is called a left-invariant finitely-additive probability measure on G if it satisfies the following conditions: ● .μ(G) = 1 (normalization); ● .μ(A ∪ B) = μ(A) + μ(B) for all .A, B ∈ P(G) such that .A ∩ B = ∅ (finite additivity); ● .μ(gA) = μ(A) for all .A ∈ P(G) and .g ∈ G (left-invariance). A group G is called amenable if it admits a left-invariant finitely-additive probability measure. Let .ℓ∞ (G) denote the Banach space consisting of all bounded functions .x : G → R with the norm .x∞ := supg∈G |x(g)|. There is a linear isometric action of the group G on .ℓ∞ (G) given by .(gx)(h) := x(g −1 h) for all .g, h ∈ G and .x ∈ ℓ∞ (G). A map .m : ℓ∞ (G) → R is called a left-invariant mean on G if it satisfies the following conditions:

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 T. Ceccherini-Silberstein, M. Coornaert, Exercises in Cellular Automata and Groups, Springer Monographs in Mathematics, https://doi.org/10.1007/978-3-031-10391-9_4

199

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4 Amenable Groups

● .m(1) = 1, where, in the left-hand side, .1 ∈ ℓ∞ (G) is defined by .1(g) = 1 for all .g ∈ G (normalization); ● .m(x) ≥ 0 for all .x ∈ ℓ∞ (G) such that .x(g) ≥ 0 for all .g ∈ G (positivity); ● .m(gx) = m(x) for all .x ∈ ℓ∞ (G) and .g ∈ G (left-invariance). Let I be a directed set. A net .(Fi )i∈I of non-empty finite subsets of G is called a left Følner net for G if it satisfies .

lim i∈I

|gFi \ Fi | =0 |Fi |

for all .g ∈ G (here we use .| · | to denote cardinality of finite sets). A left paradoxical decomposition of the group G is a triple .(K, (Ak )k∈K , (Bk )k∈K ), where .K ⊂ G is a finite subset, .(Ak )k∈K and .(Bk )k∈K are families of subsets of G indexed by K such that G = (⨆k∈K kAk ) ⨆ (⨆k∈K kBk ) = (⨆k∈K Ak ) = (⨆k∈K Bk ) .

.

The following conditions on a group G are all equivalent (cf. [CAG, Proposition 4.4.4 and Theorem 4.9.2]): (A1) (A2) (A3) (A4)

G is amenable; G admits a left-invariant mean; G admits a left Følner net; for every finite subset .K ⊂ G and every .ε > 0, there exists a finite subset .F ⊂ G such that .|kF \ F | < ε|F | for all .k ∈ K; (A5) G does not admit any left paradoxical decomposition. Each of the above definitions has its obvious “right” counterpart and it turns out that left-amenability is equivalent to right-amenability for groups (cf. [CAG, Proposition 4.4.4, Proposition 4.7.1, and Section 4.8]). In other words, a group G is amenable if and only if its opposite group .Gopp is.

4.1.2 The Class of Amenable Groups The class of amenable groups includes all finite groups ([CAG, Proposition 4.4.6]), all abelian groups ([CAG, Theorem 4.6.1]), all locally amenable groups ([CAG, Corollary 4.5.11]), and all virtually amenable groups ([CAG, Corollary 4.5.8]). Moreover, the class of amenable groups is closed under taking subgroups ([CAG, Proposition 4.5.1]), taking quotients ([CAG, Proposition 4.5.4]), taking extensions ([CAG, Proposition 4.5.5]), taking direct limits ([CAG, Proposition 4.5.10]), and taking direct sums ([CAG, Corollary 4.5.15]). The free group .F2 of rank two is non-amenable ([CAG, Theorem 4.4.7]). Therefore, any group containing a subgroup isomorphic to .F2 is itself non-amenable ([CAG, Corollary 4.5.2]).

4.2 Exercises

201

4.1.3 Amenability of Solvable and Nilpotent Groups Let G be a group. The commutator of two elements .x, y ∈ G is the element .[x, y] ∈ G defined by .[x, y] := xyx −1 y −1 . If H and K are subgroups of G, we denote by .[H, K] the subgroup of G generated by all commutators .[h, k], where .h ∈ H and .k ∈ K. The subgroup .D(G) := [G, G] is called the derived subgroup, or commutator subgroup, of G. The subgroup .D(G) is normal in G and the quotient group .G/D(G) is abelian. A group is called metabelian if its derived subgroup is abelian. The derived series of a group G is the decreasing sequence .(D i (G))i≥0 of subgroups of G inductively defined by .D 0 (G) := G and .D i+1 (G) := D(D i (G)) for all .i ≥ 0. The subgroups .D i (G) are all normal in G. The group G is said to be solvable if there is an integer .i ≥ 0 such that .D i (G) = {1G }. The smallest integer .i ≥ 0 such that .D i (G) = {1G } is then called the solvability degree of G. Thus a group is trivial (resp. abelian, resp. metabelian) if and only it is solvable with solvability degree 0 (resp. .≤ 1, resp. .≤ 2). The lower central series of G is the decreasing sequence .(C i (G))i≥0 of subgroups of G defined by .C 0 (G) := G and .C i+1 (G) := [C i (G), G] for all .i ≥ 0. The subgroups .C i (G) are all normal in G. The group G is said to be nilpotent if there is an integer .i ≥ 0 such that .C i (G) = {1G }. The smallest integer .i ≥ 0 such that .C i (G) = {1G } is then called the nilpotency degree of G. Every nilpotent group is solvable but there are solvable groups, such as the symmetric group .Sym3 , that are not nilpotent. All solvable groups are amenable [CAG, Theorem 4.6.3]. In particular, all abelian groups, all metabelian groups, and all nilpotent groups are amenable. The group .Z×Sym5 provides an example of an infinite group that is amenable but not solvable.

4.2 Exercises Exercise 4.1 Let G be an infinite group and let μ : P(G) → [0, 1] be a left (or right) invariant finitely additive probability measure on G. Show that every finite subset A ⊂ G satisfies μ(A) = 0. Solution By left (resp. right) invariance, we have μ({g}) = μ(g{1G }) = μ({1G }) (resp. μ({g}) = μ({1G∑ }g) = μ({1G })). Using the fact that μ is finitely additive, we deduce that μ(A) = g∈A μ({g}) = |A|μ({1G }) for every finite subset A ⊂ G. As G is infinite and 0 ≤ μ(A) ≤ 1, this implies μ({1G }) = 0. Therefore μ(A) = |A|μ({1G }) = 0 for every finite subset A ⊂ G. ■ Exercise 4.2 Let X be an infinite set. Show that the symmetric group Sym(X) is not amenable. Solution Suppose first that X is countably infinite. Let G := F2 be the free group of rank two. It follows from Cayley’s theorem (cf. [CAG, Theorem C.1.2]) that G

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is isomorphic to a subgroup of Sym(G). Since G and X are both countable, and symmetric groups on equipotent sets are isomorphic (cf. [CAG, Remark C.1.1]), we deduce that Sym(X) contains a subgroup isomorphic to F2 . This shows that Sym(X) is not amenable. Suppose now that X is a (possibly uncountable) infinite set. Then we can find a countably infinite subset Y ⊂ X. Note that, by the above argument, Sym(Y ) is not amenable. Given σ ∈ Sym(Y ) we define a permutation σ ∈ Sym(X) by setting ⎧ σ (x) :=

σ (x)

if x ∈ Y

x

otherwise,

.

for all x ∈ X. Clearly, the map σ |→ σ yields an injective homomorphism Sym(Y ) → Sym(X), so that we can regard Sym(Y ) as a subgroup of Sym(X). Since subgroups of amenable groups are theirself amenable, we deduce that Sym(X) is not amenable either. ■ Exercise 4.3 Show that every FC-group is amenable. Solution Let G be an FC-group and let H be a finitely generated subgroup of G. As every subgroup of an FC-group is itself an FC-group, the group H is an FC-group. As H is finitely generated, it follows from Exercise 2.44(d) that the quotient of H by its center Z(H ) is a finite group. Since all abelian groups and all finite groups are amenable and the class of amenable groups is closed under taking extensions, we deduce that the group H is amenable. As every locally amenable group is amenable, this shows that G is an amenable group. ■ Exercise 4.4 Show that the wreath product of two amenable groups is an amenable group. Solution Let G1 and G2 be two amenable groups and let G := G1 G2 denote their wreath product (cf. Exercise 2.36). Recall that G is the semidirect product G = D α G2 , where D is the direct sum of a family of copies of G1 indexed by G2 and α : G2 → Aut(D) is the group homomorphism associated with the shift action of G2 on D. As the class of amenable groups is closed under taking direct sums and extensions, we deduce that G is amenable. ■ Exercise 4.5 Suppose that G is a non-amenable group and that N is an amenable normal subgroup of G. Show that the group G/N is non-amenable. Solution This is an immediate consequence of the fact that the class of amenable ■ groups is closed under taking extensions [CAG, Proposition 4.5.5]. Exercise 4.6 Let G1 (resp. G2 ) denote the finitely generated and non residually finite group described in Exercise 2.40 (resp. Exercise 2.41). Show that the groups G1 and G2 are amenable. Solution As G1 := Sym0 (Z)  Z, we have a short exact sequence of groups 1 → Sym0 (Z) → G1 → Z → 1. The group Sym0 (Z) is amenable since it is

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locally finite and the group Z is amenable since it is abelian. The group G1 , being an extension of an amenable group by an amenable group, is itself amenable. The group G2 := Sym+ 5 Z is amenable by Exercise 4.4 since it is the wreath product of two amenable groups. ■ Exercise 4.7 Let G be a group. (a) Suppose that H and K are normal subgroups of G. Show that H K is a normal subgroup of G and that the groups H K/K and H /(H ∩ K) are isomorphic. (b) Suppose that H and K are normal amenable subgroups of G. Show that H K is itself a normal amenable subgroup of G. (c) Show that the set of all normal amenable subgroups of G has a unique maximal element for inclusion. Solution (a) Since 1G ∈ H and 1G ∈ K, we have that 1G = 1G · 1G ∈ H K. Let now h, h1 , h2 ∈ H , k, k1 k2 ∈ K, and g ∈ G. Since K is normal in G, we have −1 −1 −1 (h1 k1 )(h2 k2 )−1 = h1 k1 k2−1 h−1 2 = h1 h2 · h2 (k1 k2 )h2 ∈ H K.

.

This shows that H K is a subgroup of G. Moreover, since both H and K are normal in G, we have g(hk)g −1 = ghg −1 · gkg −1 ∈ H K,

.

showing that H K is itself normal in G. Consider the map ϕ : H K → H /(H ∩ K) defined by setting ϕ(hk) := h(H ∩ K) for all h ∈ H and k ∈ K. Note that ϕ is well defined. Indeed if h, h' ∈ H and k, k ' ∈ K satisfy that hk = h' k ' then (h' )−1 h = k ' k −1 ∈ H ∩ K so that, in particular, h(H ∩ K) = h' (H ∩ K). We have ϕ(h1 k1 · h2 k2 ) = ϕ(h1 h2 · (h−1 2 k1 h2 )k2 ) = h1 h2 (H ∩ K)

.

= h1 (H ∩ K)h2 (H ∩ K) = ϕ(h1 k1 )ϕ(h2 k2 ) and this shows that ϕ is a group homomorphism. Since ker(ϕ) = K, and ϕ is surjective, the fundamental theorem on group homomorphisms gives H K/K = H K/ ker(ϕ) ∼ = Im(ϕ) = H /(H ∩ K). (b) By (a) we have a short exact sequence 1 → K → H K → H /(H ∩ K) → 1. As the group H is amenable, so is its quotient H /(H ∩K). Since K is also amenable, the group H K, being an extension of an amenable group by an amenable group, is itself amenable. (c) Let us denote by S the set consisting of all normal amenable subgroups of G partially ordered by inclusion. We first observe that S is not empty since {1U G} ∈ S . Suppose now that C ⊂ S is a totally ordered subset of S . Then K := H ∈C H is a normal subgroup of G. Moreover, since the class of amenable groups is closed under direct limits, we deduce that K is amenable. Therefore K ∈ S is an upper

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bound for C . By applying Zorn’s lemma, we deduce that S admits a maximal element. Such a maximal element is unique. Indeed, if M1 and M2 are maximal elements of S , then M := M1 M2 is in S by (b). As M contains both M1 and M2 , ■ we deduce that M1 = M = M2 by maximality. Comment The isomorphism in (a) is a particular case of the second isomorphism theorem for groups (see for example [Rot, Section 2.6]) which states that if H and K are subgroups of a group G with K normal in G then H K is a subgroup of G and H K/K ∼ = H /(H ∩ K). The maximal element in (c) is called the amenable radical of the group G. Exercise 4.8 Let G be a group and let H be a subgroup of G. Show that the following conditions are equivalent: (i) [G, G] ⊂ H ; (ii) H is normal in G and G/H is abelian. Solution Suppose (i). Then, for all h ∈ H and g ∈ G, we have ghg −1 = [g, h]h ∈ H · H = H , showing that H is normal in G. Moreover, for all g1 , g2 ∈ G, g1 H · g2 H = g1 g2 H = g2 g1 [g1−1 , g2−1 ]H = g2 g1 H = g2 H · g1 H.

.

This shows that G/H is abelian. Conversely, suppose (ii). Let ϕ : G → G/H denote the quotient homomorphism. Then ϕ([g1 , g2 ]) = ϕ(g1 g2 g1−1 g2−1 ) = ϕ(g1 )ϕ(g2 )ϕ(g1 )−1 ϕ(g2 )−1

.

= [ϕ(g1 ), ϕ(g2 )] = 1G/H , since G/H is abelian. This shows that [g1 , g2 ] ∈ ker(ϕ) = H . As the comutators [g1 , g2 ], g1 , g2 ∈ G, generate [G, G], this implies (i). ■ Exercise 4.9 Let G be a group. Show that G is metabelian if and only if it contains an abelian normal subgroup N such that the group G/N is abelian. Solution Recall that a group G is called metabelian if its commutator subgroup [G, G] is abelian. Suppose that the group G is metabelian. Then the abelian normal subgroup N := [G, G] satisfies that G/N = G/[G, G] is abelian (the latter is, indeed, the abelianization of G). Conversely, suppose that G contains an abelian normal subgroup N such that the group G/N is abelian. Since G/N is abelian, we have [G, G] ⊂ N (cf. Exercise 4.8). As N is abelian, its subgroup [G, G] is abelian as well. This shows that G is metabelian. ■ Exercise 4.10 Let G be a group. Let x, y, z ∈ G. (a) Show that [x, yz] = [x, y] · [y, [x, z]] · [x, z].

.

(4.1)

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(b) Show that [xy, z] = [y, z] · [[z, y], x] · [x, z].

(4.2)

.

Solution (a) We have [x, yz] = xyzx −1 z−1 y −1

.

= xyx −1 [x, z]zz−1 y −1 = xyx −1 [x, z]y −1 = xyx −1 y −1 [y, [x, z]] · [x, z] = [x, y] · [y, [x, z]] · [x, z]. This shows (4.1). (b) We have [xy, z] = [z, xy]−1

.

= ([z, x] · [x, [z, y]] · [z, y])−1

(by (4.1))

= [z, y]−1 · [x, [z, y]]−1 · [z, x]−1 = [y, z] · [[z, y], x] · [x, z]. ■

This shows (4.2).

Exercise 4.11 Let G be a group and let Z(G) denote the center of G. Suppose that H is a subgroup of G such that [H, G] ⊂ Z(G). (a) Let x, x ' ∈ H and y, y ' ∈ G. Show that one has [xx ' , yy ' ] = [x, y] · [x, y ' ] · [x ' , y] · [x ' , y ' ], .

.

[x, y]−1 = [x −1 , y] = [x, y −1 ], . [x, y] = [x , y] = [x, y ]. n

n

n

(4.3) (4.4) (4.5)

(b) Let x ∈ H and y ∈ G. Show that if x or y has finite order then [x, y] has finite order. (c) Suppose that S is a generating subset of H and that T is a generating subset of G. Show that the set C := {[s, t] : s ∈ S, t ∈ T } is a generating subset of [H, G]. (d) Suppose that the groups H and G are finitely generated. Show that the group [H, G] is finitely generated.

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Solution (a) We have [xx ' , yy ' ] = [xx ' , y] · [y, [xx ' , y ' ]] · [xx ' , y ' ]

.

'

'

'

= [xx , y] · [xx , y ]

(by (4.1)) (as [xx ' , y ' ] ∈ [H, G] ⊂ Z(G))

= [x ' , y] · [[y, x ' ], x] · [x, y] · [x ' , y ' ] · [[y ' , x ' ], x] · [x, y ' ] = [x ' , y] · [x, y] · [x ' , y ' ] · [x, y ' ]

(by (4.2)) (as [x ' , y], [y ' , x ' ] ∈ [H, G] ⊂ Z(G])

= [x, y] · [x, y ' ] · [x ' , y] · [x ' , y ' ]

(as [x, y], [x, y ' ] ∈ [H, G] ⊂ Z(G)).

This shows (4.3). By taking x ' = 1G and y ' = y −1 in (4.3), we get 1G = [x, y] · [x, y −1 ] and hence [x, y]−1 = [x, y −1 ]. Similarly, by taking x ' = x −1 and y ' = 1G in (4.3), we get 1G = [x, y] · [x −1 , y] and hence [x, y]−1 = [x −1 , y]. This shows (4.4). By taking x ' = 1G and y ' = y n in (4.3), we get [x, y n+1 ] = [x, y] · [x, y n ]. Similarly, by taking x ' = x n and y ' = 1G in (4.3), we get [x n+1 , y] = [x, y]·[x n , y]. Using induction on n, we deduce that [x, y]n = [x, y n ] = [x n , y] for all n ≥ 0. By applying (4.4), we obtain [x, y n ] = [x, y −n ]−1 = ([x, y]−n )−1 = [x, y]n and [x n , y] = ([x −n , y])−1 = ([x, y]−n )−1 = [x, y]n for all n < 0. This shows (4.5). (b) Suppose that x or y has finite order n ≥ 1. We then deduce from (4.5) that [x, y]n = 1G . This shows that [x, y] has finite order. (c) This immediately follows from (4.3) and (4.4). (d) This immediately follows from (c). ■ Exercise 4.12 (The Second Center of a Group) Let G be a group and let Z(G) denote its center. Let Z2 (G) denote the subset of G consisting of all x ∈ G such that [x, y] ∈ Z(G) for all y ∈ G. (a) Show that Z(G) ⊂ Z2 (G). (b) Show that Z2 (G) is a normal subgroup of G and that Z2 (G)/Z(G) is the center of G/Z(G). (c) Show that if Z(G) is torsion-free then Z2 (G)/Z(G) is torsion-free. Solution (a) If x ∈ Z(G) then [x, y] = 1G ∈ Z(G) for all y ∈ G. This shows that Z(G) ⊂ Z2 (G). (b) Let ρ : G → G/Z(G) denote the canonical group epimorphism. Let N := ρ −1 (H ), where H ⊂ G/Z(G) denotes the center of G/Z(G). Then N is a normal subgroup of G such that Z(G) ⊂ N and N/Z(G) = H . Let us show that Z2 (G) = N. Suppose first x ∈ Z2 (G). Then, for all y ∈ G, we have [ρ(x), ρ(y)] = ρ([x, y]) = 1G/Z(G) since [x, y] ∈ Z(G). As ρ is surjective, this shows that ρ(x)

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is in the center H of G/Z(G). Thus, Z2 (G) ⊂ ρ −1 (H ) = N. Conversely, suppose now that x ∈ N and let y ∈ G. Then ρ([x, y]) = [ρ(x), ρ(y)] = 1G/Z(G) since ρ(x) ∈ H and H is the center of G/Z(G). It follows that [x, y] ∈ Z(G), showing that N ⊂ Z2 (G). This completes the proof that Z2 (G) = N. (c) Suppose that c ∈ Z2 (G)/Z(G) has finite order n. Let x ∈ Z2 (G) such that ρ(x) = c. We have ρ(x n ) = ρ(x)n = cn = 1Z2 (G)/Z(G) , so that x n ∈ Z(G). On the other hand, since [Z2 (G), G] ⊂ Z(G), we can apply (4.5). This gives us [x, y]n = [x n , y] = 1G for all y ∈ G. Therefore [x, y] ∈ Z(G) has finite order. As Z(G) is torsion-free by hypothesis, it follows that [x, y] = 1G for all y ∈ G, so that x ∈ Z(G). Thus, c = ρ(x) = 1Z2 (G)/Z(G) . This shows that Z2 (G)/Z(G) is torsion-free. ■ Comment The subgroup Z2 (G) is called the second center of the group G. The second center is the second term of the upper central series, which is the increasing sequence (Zi (G))i≥0 of subgroups of G inductively defined by Z0 (G) := {1G } and Zi+1 (G) := {x ∈ G : [x, y] ∈ Zi (G) for all y ∈ G}. The first term of the upper central series is the center Z1 (G) = Z(G) of G. For every i ≥ 0, the quotient Zi+1 (G)/Zi (G) is the center of G/Zi (G). One can show that the group G is nilpotent if and only if there exists i ≥ 0 such that Zi (G) = G. The smallest such i is then equal to the nilpotency degree of G. If G is nilpotent of degree d, then C i (G) ⊂ Zd−i (G) for all 0 ≤ i ≤ d. Exercise 4.13 Let G and H be two groups. Let ϕ : G → H be a group homomorphism. (a) Show that ϕ([x, y]) = [ϕ(x), ϕ(y)] for all x, y ∈ G. (b) Let A and B be subgroups of G. Show that ϕ([A, B]) = [ϕ(A), ϕ(B)].

.

(4.6)

(c) Show that ϕ(D i (G)) ⊂ D i (H ) and ϕ(C i (G)) ⊂ C i (H ) for all i ≥ 0. (d) Suppose that ϕ is surjective. Show that ϕ(D i (G)) = D i (H ) and ϕ(C i (G)) = C i (H )

.

for all i ≥ 0. Solution (a) Let x, y ∈ G. Then we have ϕ([x, y]) = ϕ(xyx −1 y −1 )

.

= ϕ(x)ϕ(y)(ϕ(x))−1 (ϕ(y))−1 = [ϕ(x), ϕ(y)].

(since ϕ is a group homomorphism)

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(b) By (a), for all x ∈ A and y ∈ B, we have ϕ([x, y]) = [ϕ(x), ϕ(y)] ∈ [ϕ(A), ϕ(B)].

.

As [A, B] is the subgroup of G generated by all [x, y], where x ∈ A and y ∈ B, we deduce that ϕ([A, B]) ⊂ [ϕ(A), ϕ(B)].

.

x'

(4.7)

Let now x ' ∈ ϕ(A) and y ' ∈ ϕ(B). Then there exist x ∈ A and y ∈ B such that = ϕ(x) and y ' = ϕ(y). We then have [x ' , y ' ] = [ϕ(x), ϕ(y)] = ϕ([x, y]) ∈ ϕ([A, B]).

.

As [ϕ(A), ϕ(B)] is the subgroup of H generated by all [x ' , y ' ], where x ' ∈ ϕ(A) and y ' ∈ ϕ(B), we deduce that [ϕ(A), ϕ(B)] ⊂ ϕ([A, B]).

.

(4.8)

By combining (4.7) and (4.8), we get (4.6). (c) As D 0 (G) = C 0 (G) = G, D 0 (H ) = C 0 (H ) = H ⊂ ϕ(G), D i+1 (G) = i [D (G), D i (G)] and C i+1 (G) = [C i (G), G] for all i ≥ 0, we deduce from (4.6), by using induction on i, that ϕ(D i (G)) ⊂ D i (H ) and ϕ(C i (G)) ⊂ C i (H ) for all i ≥ 0. (d) Since ϕ is surjective, we have H = ϕ(G). As D 0 (G) = C 0 (G) = G, 0 D (H ) = C 0 (H ) = H = ϕ(G), D i+1 (G) = [D i (G), D i (G)] and C i+1 (G) = [C i (G), G] for all i ≥ 0, we deduce from (4.6), by using induction on i, that ϕ(D i (G)) = D i (H ) and ϕ(C i (G)) = C i (H ) for all i ≥ 0. ■ Exercise 4.14 Show that every subgroup of a solvable (resp. nilpotent) group of solvability (resp. nilpotency) degree d is solvable (resp. nilpotent) with solvability (resp. nilpotency) degree at most d. Solution Let G be a group and let H be a subgroup of G. It follows from Exercise 4.13(c) applied to the inclusion map ϕ : H → G that D i (H ) ⊂ D i (G) and C i (H ) ⊂ C i (G) for all i ≥ 0. If G is solvable, with solvability degree d (resp. nilpotent with nilpotency degree d), we have D d (G) = {1G } (resp. C d (G) = {1G }) and hence D d (H ) = {1G } = {1H } (resp. C d (H ) = {1G } = {1H }). This implies that H is solvable (resp. nilpotent) with solvability (resp. nilpotency) degree at most d. ■ Exercise 4.15 Show that every quotient of a solvable (resp. nilpotent) group of solvability (resp. nilpotency) degree d is solvable (resp. nilpotent) with solvability (resp. nilpotency) degree at most d. Solution Let G be a group and let Q be a quotient of G. This means that there is a surjective group homomorphism ϕ : G → Q. It follows from Exercise 4.13(d) that

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ϕ(D i (G)) = D i (Q) and ϕ(C i (G)) = C i (Q) for all i ≥ 0. If G is solvable, with solvability degree d (resp. nilpotent with nilpotency degree d), we have D d (G) = {1G } (resp. C d (G) = {1G }) and hence D d (Q) = ϕ({1G }) = {1Q } (resp. C d (Q) = ϕ({1G }) = {1Q }). This implies that the group Q is solvable with solvability degree at most d (resp. nilpotent with nilpotency degree at most d). ■ Exercise 4.16 Let G be a group and let G' = [G, G] denote the derived subgroup of G. (a) Show that if G' and G/G' are Hopfian then G is Hopfian. (b) Show that if G is finitely generated and G' is Hopfian then G is Hopfian. Solution (a) Suppose that the groups G' and G/G' are Hopfian. Let φ : G → G be a surjective endomorphism. It follows from Exercise 4.13(d) that φ(G' ) = G' . Thus, φ induces by restriction a surjective homomorphism φ|G' : G' → G' . As G' is Hopfian, φ|G' is injective. On the other hand, the map φ : G/G' → G/G' given by φ(gG' ) := φ(g)G' for all g ∈ G, is well defined. Since φ is a surjective group endomorphism and G' is normal in G, it is clear that φ is a surjective group endomorphism of G/G' . As G/G' is Hopfian, we deduce that φ is injective. Suppose now that g ∈ ker(φ). Then φ(gG' ) = φ(g)G' = G' , so that g ∈ G' by the injectivity of φ. As φ|G' (g) = φ(g) = 1G = 1G' and φ|G' is injective, we deduce that g = 1G' = 1G . Thus, φ is injective. This shows that G is Hopfian. (b) If G is finitely generated, then G/G' is also finitely generated. As every finitely generated abelian group is Hopfian (cf. Corollary 2.2.4 and Theorem 2.4.3 in [CAG]), this implies that G/G' is Hopfian. Thus, the result follows from (a). ■ Comment Thompson’s group F is finitely generated [CanFP, Theorem 3.4]. Moreover, the derived subgroup of F is simple [CanFP, Theorem 4.5] and therefore Hopfian. Thus, the group F is Hopfian by (b). Thompson’s groups T and V are also Hopfian since they are both simple [CanFP, Theorem 5.8, Corollary 5.9, Lemma 6.1, and Theorem 6.9]. Exercise 4.17 Let G be a group and let i ≥ 0 be an integer. (a) Show that the group G/D i (G) (resp. G/C i (G)) is solvable (resp. nilpotent) and has solvability (resp. nilpotency) degree at most i. (b) Suppose that the group G is solvable (resp. nilpotent) with solvability (resp. nilpotency) degree d. Show that the group G/D i (G) (resp. G/C i (G)) is solvable (resp. nilpotent) with solvability (resp. nilpotency) degree min(d, i). (c) Suppose that the group G is nilpotent with nilpotency degree d ≥ 1 and let Z(G) denote the center of G. Show that the group G/Z(G) is nilpotent with nilpotency degree at most d − 1. Solution (a) Consider the canonical group epimorphisms ϕ : G → G/D i (G) and ψ : G → G/C i (G).

.

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By Exercise 4.13(c), we have D j (G/D i (G)) = ϕ(D j (G)) (resp. C j (G/C i (G)) = ϕ(C j (G)) for every j ≥ i. As D j (G) ⊂ D i (G) (resp. C j (G) ⊂ C i (G)) for j ≥ i, we deduce that the group D j (G/D i (G)) (resp. C j (G/C i (G))) is trivial for j ≥ i. This shows that the group G/D i (G) (resp. G/C i (G)) is solvable (resp. nilpotent) and has solvability (resp. nilpotency) degree at most i. (b) Since G is solvable (resp. nilpotent) with solvability (resp. nilpotency) degree d, the sequence (D j (G))0≤j ≤d (resp. (C j (G))0≤j ≤d ) is strictly decreasing. As D j (G/D i (G)) = ϕ(D j (G)) ∼ = D j (G)/D i (G) (resp. C j (G/C i (G)) = j j i ϕ(C (G) ∼ = C (G)/C (G)), it follows that the sequence (D j (G/D i (G)))0≤j ≤min(d,i) (resp. (C j (G/C i (G)))0≤j ≤min(d,i) )

.

is strictly decreasing. Since the group D j (G/D i (G)) (resp. C j (G/C i (G))) is trivial for j = min(d, i), we conclude that G has solvability (resp. nilpotency) degree min(d, i). (c) Since G is nilpotent with nilpotency degree d, we have [C d−1 (G), G] = d C (G) = {1G }, so that C d−1 (G) ⊂ Z(G). As a consequence, G/Z(G) is isomorphic to a quotient of G/C d−1 (G). Since by (a) the latter is nilpotent with nilpotency degree at most d − 1, it follows from Exercise 4.15 that also the former is nilpotent with nilpotency degree at most d − 1. ■ Exercise 4.18 Show that the class of torsion-free groups is closed under forming extensions. Solution Let G be a group and let N ⊂ G be a normal subgroup. Suppose that N and G/N are both torsion-free and let us show that G itself is torsion-free. Let g ∈ G and suppose that g n = 1G for some integer n ≥ 1. If π : G → G/N denotes the quotient homomorphism, we have (π(g))n = π(g n ) = π(1G ) = 1G/N . Since G/N is torsion-free, we have π(g) = 1G/N . As a consequence, g ∈ N = ker(π ). As N is torsion free and g n = 1G = 1N , we deduce that g = 1N = 1G . This shows that G is torsion-free. ■ Exercise 4.19 Let G be a nilpotent group and let Z(G) denote its center. Show that the following conditions are equivalent (i) the group G is torsion-free; (ii) the group Z(G) is torsion-free; (iii) the groups Z(G) and G/Z(G) are torsion-free. Solution Implication (i) =⇒ (ii) is obvious since Z(G) ⊂ G. Implication (iii) =⇒ (i) follows from the fact that the class of torsion-free groups is closed under forming extensions (cf. Exercise 4.18). Let us show, by induction on the nilpotency degree d of G, that (ii) =⇒ (iii). If d = 0, the group G is trivial and there is nothing to prove. Let us assume now that the nilpotency degree of G is d ≥ 1 and that (ii) =⇒ (iii) has been established for all nilpotent groups of degree at most d − 1. Suppose that Z(G) is torsion-free. Then Z2 (G)/Z(G) is torsion-free by Exercise 4.12(c). As Z2 (G)/Z(G) is the center of G/Z(G) by Exercise 4.12(b)

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and G/Z(G) is nilpotent of degree at most d − 1 by Exercise 4.17(c), it follows from our induction hypothesis that the group (G/Z(G))/(Z2 (G)/Z(G)) is torsionfree. As the group G/Z2 (G) is isomorphic to the group (G/Z(G))/(Z2 (G)/Z(G)) by the third isomorphism theorem, we deduce that G/Z2 (G) is torsion-free. Since G/Z2 (G) and Z2 (G)/Z(G) are torsion-free and the class of torsion-free groups is closed under forming extensions (cf. Exercise 4.18), it follows that G/Z(G) is itself torsion-free. This completes the proof of (ii) =⇒ (iii). ■ Exercise 4.20 Let G be a group and let H be a normal subgroup of G such that G/H is solvable (resp. nilpotent). Let d ≥ 0 denote the solvability (resp. nilpotency) degree of G/H . Show that D d (G) ⊂ H (resp. C d (G) ⊂ H ). Solution Consider the canonical group epimorphism ϕ : G → G/H . By Exercise 4.13(c), we have D i (G/H ) = ϕ(D i (G)) (resp. C i (G/H ) = ϕ(C i (G))) for all i ≥ 0. As G/H is solvable (resp. nilpotent) of solvability (resp. nilpotency) degree d, we have that ϕ(D d (G)) (resp. ϕ(C d (G))) is trivial. This implies D d (G) ⊂ H (resp. C d (G) ⊂ H ). ■ Exercise 4.21 Let G be a group. Show that G is solvable (resp. nilpotent) if and only if there exist an integer n ≥ 0 and a decreasing sequence (Hi )0≤i≤n of subgroups of G such that H0 = G, Hn = {1G } and [Hi , Hi ] ⊂ Hi+1 (resp. [Hi , G] ⊂ Hi+1 ) for all 0 ≤ i ≤ n − 1. Solution The condition is necessary. Indeed, if G is solvable with solvability degree d (resp. nilpotent with nilpotency degree d), then we can take n = d and Hi := D i (G) (resp. Hi := C i (G)) for 0 ≤ i ≤ n. Conversely, if there exists a finite sequence (Hi )0≤i≤n of subgroups of G satisfying the conditions stated above, then we get by induction D i (G) ⊂ Hi (resp. C i (G) ⊂ Hi ) for all 0 ≤ i ≤ n. As Hn = {1G }, it follows that D n (G) = {1G } (resp. C n (G) = {1G }), showing that G is solvable (resp. nilpotent). ■ Comment A decreasing sequence (Hi )0≤i≤n of subgroups of a group G such that H0 = G, Hn = {1G }, and [Hi , G] ⊂ Hi+1 for all 0 ≤ i ≤ n − 1, is called a central series in G. Since Hi+1 ⊂ Hi , the condition [Hi , G] ⊂ Hi+1 amounts to saying that Hi is normal in G and Hi /Hi+1 is contained in the center of G/Hi . As observed in the proof, if G is a nilpotent group of nilpotency degree d, then the sequence (C i (G))0≤i≤d is a central series in G. Moreover, every central series (Hi )0≤i≤n satisfies C i (G) ⊂ Hi for all 0 ≤ i ≤ n. This explains the terminology “lower central series” used for designating the sequence (C i (G))0≤i≤d . Exercise 4.22 Let G be a group. Show that G is solvable if and only if there exist an integer n ≥ 0 and a decreasing sequence (Hi )0≤i≤n of subgroups of G such that H0 = G, Hn = {1G }, Hi+1 is normal in Hi and Hi /Hi+1 is abelian for all 0 ≤ i ≤ n − 1. Solution Let (Hi )0≤i≤n be a decreasing sequence of subgroups of G such that H0 = G and Hn = {1G }. By Exercise 4.8, we have [Hi , Hi ] ⊂ Hi+1 if and

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only if Hi+1 is normal in Hi and Hi /Hi+1 is abelian. Thus the statement follows immediately from Exercise 4.21. ■ Exercise 4.23 Show that a group is nilpotent with nilpotency degree ≤ 2 if and only if its derived subgroup is contained in its center. Solution Let G be a group. Then C 0 (G) = G, C 1 (G) = [G, G], and C 2 (G) = [[G, G], G]. By definition G is nilpotent of nilpotency degree ≤ 2 if and only if C 2 (G) = {1G }. This is equivalent to the fact that every element of [G, G] commutes with every element of G, that is, to the inclusion [G, G] ⊂ Z(G). ■ Exercise 4.24 Let G be a group such that G/Z(G) is cyclic. Show that G is abelian (so that G = Z(G)). Solution Let ϕ : G → G/Z(G) denote the canonical group epimorphism and let a ∈ G such that ϕ(a) generates G/Z(G). Given g, g ' ∈ G, we can find c, c' ∈ Z(G) ' ' and n, n' ∈ Z such that g = ca n and g ' = c' a n . We then have gg ' = ca n c' a n = ' ' ' cc' a n a n = c' ca n a n = c' a n ca n = g ' g. This shows that G is abelian. ■ Exercise 4.25 Let G be a group and let H be a subgroup of G. Suppose that H is contained in the center of G and that the group G/H is nilpotent. Show that G is nilpotent. Solution By the characterization of nilpotent groups given in Exercise 4.21, there exists a finite decreasing sequence (Ki )0≤i≤n of subgroups of G/H such that K0 = G/H , Kn = {1G/H }, and [Ki , G/H ] ⊂ Ki+1 for all 0 ≤ i ≤ n − 1. Let ϕ : G → G/H denote the canonical group epimorphism and consider the subgroups Hi of G defined by Hi := ϕ −1 (Ki ) for all 0 ≤ i ≤ n. Then H0 = G, Hn = H , and [Hi , G] = ϕ −1 ([Ki , G/H ]) ⊂ ϕ −1 (Ki+1 ) = Hi+1 for all 0 ≤ i ≤ n−1. Moreover, Hn+1 := [Hn , G] = {1G } since Hn = H is contained in the center of G. By using again Exercise 4.21, this implies that the group G is nilpotent. ■ Exercise 4.26 One says that a group G is a torsion group if all elements of G have finite order. Show that every finitely generated solvable torsion group is finite. Solution Let G be a finitely generated solvable torsion group. We shall use induction on the solvability degree d of G to prove that G is finite. For d = 0, the group G is trivial and there is nothing to prove. Suppose now that G has positive solvability degree d and that the result has been established for groups with solvability degree d − 1. The derived subgroup [G, G] is solvable of degree d − 1. Every quotient of a torsion (resp. finitely generated) group is itself torsion (resp. finitely generated). Thus the abelian group G/[G, G] is a finitely generated torsion group. By the classification theorem for finitely generated abelian groups, it follows that G/[G, G] is finite. As every finite index subgroup of a finitely generated group is itself finitely generated, we deduce that [G, G] is finitely generated. Every subgroup of a torsion group is itself torsion. Thus it follows from our induction hypothesis that [G, G] is finite. The group G, being an extension of a finite group by a finite group, is itself finite. This completes the inductive step. ■

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Exercise 4.27 Let G be a group containing a normal subgroup H such that both G/H and H are solvable. Let d (resp. d ' ) denote the solvability degree of G/H (resp. H ). Show that G is solvable with solvability degree at most d + d ' . Solution By Exercise 4.20, we have D d (G) ⊂ H . We thus have D d+1 (G) = [D d (G), D d (G)] ⊂ [H, H ] = D 1 (H ) and, by induction, D d+i (G) ⊂ D i (H ) for ' ' all i ≥ 0. We deduce that D d+d (G) ⊂ D d (H ) = {1H } = {1G }. This shows that G is solvable with solvability degree at most d + d ' . ■ Exercise 4.28 Show that the direct product of two nilpotent groups is a nilpotent group. Solution Let G1 and G2 be two groups. Let G := G1 × G2 denote their direct product. First observe that if Hi , Ki are subgroups of Gi for i = 1, 2, and H, K are the subgroups of G defined by H := H1 × H2 and K := K1 × K2 , respectively, then [H, K] = [H1 , K1 ] × [H2 , K2 ]. As C 0 (G) = G = G1 × G2 = C 0 (G1 ) × C 0 (G2 ), we get by induction C i (G) = C i (G1 ) × C i (G2 ) for all i ≥ 0. Suppose now that Gi is nilpotent, with nilpotency degree di for i = 1, 2. Taking d := max(d1 , d2 ), we have C d (G) = C d (G1 ) × C d (G2 ) = {1G1 } × {1G2 } = {1G }, while C d−1 (G) = C d−1 (G1 ) × C d−1 (G2 ) /= {1G1 } × {1G2 } = {1G }. This shows that the group G is nilpotent, with nilpotency degree d. ■ Exercise 4.29 Show that a semidirect product of two nilpotent groups may fail to be nilpotent. Solution Consider the symmetric group G := Sym3 , and the cyclic subgroups H and K of G generated respectively by the 3-cycle (1 2 3) and the transposition (1 2). Then H is normal in G, H K = G, and H ∩ H = {1G }. Thus the group G is the semidirect product of H and K. As H and K are abelian, they are nilpotent. On the other hand, as already mentioned in the introduction to this chapter, the group G is not nilpotent. ■ Exercise 4.30 Let G be a nilpotent group. (a) Let H be a non-trivial normal subgroup of G. Show that H ∩ Z(G) /= {1G }. (b) Show that every non-trivial nilpotent group has a non-trivial center. (c) Let G' be a group and let ϕ : G → G' be a group homomorphism. Let ψ : Z(G) → G' denote the restriction of ϕ to Z(G). Show that ϕ is injective if and only if ψ is injective. Solution (a) Let d ≥ 1 denote the nilpotency degree of G. As the lower central series (C i (G))0≤i≤d is decreasing with C 0 (G) = G and C d (G) = {1G }, there exists 0 ≤ k ≤ d −1 such that H ∩C k (G) /= {1G } and H ∩C k+1 (G) = {1G }. Let h ∈ H ∩ C k (G) such that h /= 1G . For all g ∈ G, we have [h, g] ∈ [C k (G), G] = C k+1 (G) so that [h, g] = 1G . Therefore h ∈ Z(G). This shows that H ∩ Z(G) /= {1G }. (b) This immediately follows from (a) applied to H := G.

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(c) The injectivity of ϕ trivially implies that of ψ since ψ is the restriction of ϕ to Z(G). Conversely, suppose that ϕ is not injective. By applying (a) to H := ker(ϕ), we get that ker(ψ) = H ∩ Z(G) /= {1G }. This shows that ψ is not injective either. ■ Exercise 4.31 Let G be a group and i ≥ 0. Show that the group C i (G)/C i+1 (G) is contained in the center of G/C i+1 (G). Solution Let g ∈ G and gi ∈ C i (G). Then [gi , g] ∈ [C i (G), G] = C i+1 (G), so that (cf. Exercise 4.10) [gi C i+1 (G), gC i+1 (G)] = [gi , g]C i+1 (G) = C i+1 (G) = 1G/C i+1 (G) .

.

This shows that gi C i+1 (G) ∈ Z(G/C i+1 (G)).

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Exercise 4.32 Let G be a finite group whose cardinality is a power of a prime number. Show that G is nilpotent. Solution Let p be a prime number and let n ∈ N. Let us prove by induction on n that every group of cardinality pn is nilpotent. If n = 0, the group is trivial and there is nothing to prove. Suppose now that G is a group with cardinality pn , for some n ≥ 1, and that the result is known to be true for groups with cardinality pk with k ≤ n − 1. Consider the action of G on itself by conjugacy, that is, the map G × G → G given by (g, x) |→ gxg −1 for all (g, x) ∈ G × G. The cardinality of every orbit divides |G| and is therefore a power of p. Moreover, the orbit of an element x ∈ G is a singleton if and only if x belongs to the center Z(G) of G. It follows that pn = |G| ≡ |Z(G)| mod p, so that p divides |Z(G)|. In particular, |Z(G)| /= 1, so that the subgroup Z(G) is not trivial. By our induction hypothesis, the group G/Z(G) is nilpotent since it is of cardinality pk for some k ≤ n − 1. As Z(G) is abelian and therefore nilpotent, we deduce from Exercise 4.25 that G itself is nilpotent. ■ Comment A finite group whose cardinality is a positive power of a prime number p is called a p-group. Exercise 4.33 Let p be a prime number. (a) Show that every group with cardinality p or p2 is abelian (and hence nilpotent). (b) Give an example of a group with cardinality p3 that is not abelian. Solution (a) If G is a group with cardinality p, then G is cyclic and therefore abelian. Suppose now that G is a group with cardinality p2 . Then the center Z(G) of G has cardinality p or p2 (see the solution of Exercise 4.32). If Z(G) has cardinality p2 , then G = Z(G) is abelian. If Z(G) has cardinality p, then G/Z(G) has cardinality p and is therefore cyclic. This implies that G is abelian by Exercise 4.24.

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(b) Consider the field K := Z/pZ and the subgroup G of GL3 (K) consisting of the upper triangular matrices whose entries on the diagonal are all equal to 1. Then |G| = p3 . The matrices A, B ∈ G given by ⎛ ⎞ ⎛ ⎞ 110 100 .A = ⎝0 1 0⎠ and B = ⎝0 1 1⎠ 001 001 do not commute since ⎛ ⎞ ⎛ ⎞ 111 110 .AB = ⎝0 1 1⎠ /= ⎝0 1 1⎠ = BA. 001 001 Therefore the group G is not abelian.

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Exercise 4.34 Let G be a group. Given x, y ∈ G, we write x y := yxy −1 . (a) Let x, y, z ∈ G. Show that [[x −1 , y], z]x · [[z−1 , x], y]z · [[y −1 , z], x]y = 1G .

.

(4.9)

(b) Let X, Y , and Z be subgroups of G. Suppose that [[X, Y ], Z] = [[Y, Z], X] = {1G }. Show that [[Z, X], Y ] = {1G }. (c) Let X, Y , and Z be subgroups of G and let N be a normal subgroup of G. Suppose that [[X, Y ], Z] ⊂ N and [[Y, Z], X] ⊂ N. Show that [[Z, X], Y ] ⊂ N. (d) Deduce from (c) that [C i (G), C j (G)] ⊂ C i+j +1 (G)

.

(4.10)

for all i, j ∈ N. Solution (a) For a, b, c ∈ G we have [[a −1 , b], c]a = a((a −1 bab−1 c)(ba −1 b−1 ac−1 )a −1 = bab−1 cba −1 b−1 ac−1 a −1

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= (bab−1 cb)(aca −1 ba)−1 . Setting a := x, b := y, and c := z in (4.11), we obtain [[x −1 , y], z]x = (yxy −1 zy)(xzx −1 yx)−1 ,

.

(4.11)

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setting a := z, b := x, and c := y in (4.11), we obtain [[z−1 , x], y]z = (xzx −1 yx)(zyz−1 xz)−1 ,

.

and setting a := y, b := z, and c := x in (4.11), we obtain [[y −1 , z], x]y = (zyz−1 xz)(yxy −1 zy)−1 .

.

Substituting, we have [[x −1 , y], z]x · [[z−1 , x], y]z · [[y −1 , z], x]y

.

= (yxy −1 zy)(xzx −1 yx)−1 · (xzx −1 yx)(zyz−1 xz)−1 · (zyz−1 xz)(yxy −1 zy)−1 = 1G . (b) Let x ∈ X, y ∈ Y , and z ∈ Z. We have [[x −1 , y], z] ∈ [[X, Y ], Z] and hence [[x −1 , y], z] = 1G . Similarly, we have [[y −1 , z], x] ∈ [[Y, Z], X] and hence [[y −1 , z], x] = 1G . From (4.9), we then get [[z−1 , x], y]z = 1G , so that [[z−1 , x], y] = 1G . As the commutators [z−1 , x], where z ∈ Z and x ∈ X, generate [Z, X], we deduce that every element of [Z, X] commutes with every element of Y , so that [[Z, X], Y ] = {1G }. (c) Let π : G → G/N denote the canonical quotient group homomorphism. We have π([[X, Y ], Z]) = π([[Y, Z], X]) = {1G/N }. As π([[X, Y ], Z]) = [[π(X), π(Y )], π(Z)] and π([[Y, Z], X]) = [[π(Y ), π(Z)], π(X)] (cf. Exercise 4.13(b)), we deduce from (b) that π([[Z, X], Y ]) = [[π(Z), π(X)], π(Y )] = {1G/N }. This shows [[Z, X], Y ] ⊂ N. (d) We proceed by induction on i. For i = 0, the statement is clear since [C 0 (G), C j (G)] = [G, C j (G)] = C j +1 (G). Let i ∈ N and suppose that (4.10) holds true for all j ∈ N. Let j ∈ N and take X := C i (G), Y := C j (G), Z := G, and N := C i+j +2 (G). We then have [[X, Y ], Z] = [[C i (G), C j (G)], G] ⊂ [C i+j +1 (G), G] = C i+j +2 (G) = N

.

and [[Y, Z], X] = [[C j (G), G], C i (G)] = [C j +1 (G), C i (G)]

.

= [C i (G), C j +1 (G)] ⊂ C i+j +2 (G) = N, where the two inclusions follow from the inductive hypothesis. By applying (c), we deduce that [C i+1 (G), C j (G)] = [[C i (G), G], C j (G)] = [[Z, X], Y ] ⊂ N = C i+j +2 (G).

.

This completes induction.

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Comment Formula (4.9) is known as the Hall-Witt identity after Philip Hall and Witt. It is an analogue of the Jacobi identity in the theory of Lie algebras. The result in (c) is known as the three subgroups lemma. Exercise 4.35 (Polycyclic Groups) A group G is called polycyclic if it admits a finite sequence of subgroups {1G } = G0 ⊂ G1 ⊂ G2 ⊂ · · · ⊂ Gn = G

.

such that Gi is normal in Gi+1 and Gi+1 /Gi is a (finite or infinite) cyclic group for all 0 ≤ i ≤ n − 1. Such a sequence of subgroups is then called a polycyclic series of length n for G. (a) Show that every polycyclic group is solvable. (b) Show that every subgroup of a polycyclic group is polycyclic. (c) Show that every quotient of a polycyclic group is polycyclic. (d) Show that the class of polycyclic groups is closed under forming extensions. (e) Show that every polycyclic group is finitely generated. (f) Show that every polycyclic group is residually finite. (g) Show that an abelian group is polycyclic if and only if it is finitely generated. (h) Show that every subgroup of a polycyclic group is finitely generated. (i) One says that a group G satisfies the ascending condition for subgroups if every increasing sequence (Hn )n∈N of subgroups of G eventually stabilizes, i.e., there exists n0 ∈ N such that Hn0 = Hn for all n ≥ n0 . Show that every polycyclic group satisfies the ascending chain condition for subgroups. (j) Show that a solvable group is polycyclic if and only if it satisfies the ascending chain condition for subgroups. Solution (a) Since every cyclic group is abelian, this immediately follows from the characterization of solvable groups given in Exercise 4.22. (b) Let G be a polycyclic group and let H be a subgroup of G. Let {1G } = G0 ⊂ G1 ⊂ · · · ⊂ Gn = G be a polycyclic series for G. Consider the sequence (Hi )0≤i≤n of subgroups of H defined by Hi := H ∩ Gi . For all 0 ≤ i ≤ n − 1, the subgroup Hi is normal in Hi+1 since Gi is normal in Gi+1 . Moreover, the group Hi+1 /Hi naturally embeds into Gi+1 /Gi . As every subsgroup of a cyclic group is itself cyclic, we conclude that the sequence (Hi )0≤i≤n is a polycyclic series for H . This shows that H is polycyclic. (c) Let G be a polycyclic group and let {1G } = G0 ⊂ G1 ⊂ · · · ⊂ Gn = G be a polycyclic series for G. Suppose that Q is a quotient of G. This means that there is a group epimorphism ϕ : G → Q. Consider the sequence (Qi )0≤i≤n of subgroups of Q defined by Qi := ϕ(Gi ). Clearly Qi is a normal subgroup of Qi+1 for all 0 ≤ i ≤ n − 1. Moreover, ϕ induces a group epimorphism Gi+1 /Gi → Qi+1 /Qi . As Gi+1 /Gi is cyclic, we deduce that Qi+1 /Qi is cyclic. Since Q0 = {1Q } and Qn = Q, we conclude that the sequence (Qi )0≤i≤n is a polycyclic series for Q. This shows that Q is polycyclic.

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(d) Let N be a normal subgroup of a group G such that both N and G/N are polycyclic. We want to show that G is polycyclic. Consider a polycyclic series (Hi )0≤i≤n for N and a polycyclic series (Qj )0≤j ≤p for G/N. Let ϕ : G → G/N denote the canonical group epimorphism. Observe that ϕ −1 (Q0 ) = ϕ −1 ({1Q }) = N = Hn . Write Hn+j := ϕ −1 (Qj ) for 1 ≤ j ≤ p. Then Hn+j is a normal subgroup of Nn+j +1 and Hn+j +1 /Hn+j ∼ = Qj +1 /Qj is cyclic for all 0 ≤ j ≤ p − 1. Therefore the sequence (Hi )0≤i≤n+p is a polycyclic sequence for G. This shows that G is polycyclic. (e) Let G be a polycyclic group and let {1G } = G0 ⊂ G1 ⊂ · · · ⊂ Gn = G be a polycyclic series for G. For each 1 ≤ i ≤ n, let ϕi : Gi → Gi /Gi−1 denote the canonical group epimorphism. Choose, for each 1 ≤ i ≤ n, an element xi ∈ Gi such that ϕi (xi ) generates Gi /Gi−1 . Then, given g ∈ G, we can find kn ∈ Z and gn ∈ Gn−1 such that g = gn xnkn . By induction, we deduce that there exist ki ∈ Z, for 1 ≤ i ≤ n, such that g = x1k1 x2k2 . . . xnkn . Consequently, the set {x1 , x2 , . . . , xn } generates G. This shows in particular that G is finitely generated. (f) Let G be a polycyclic group and suppose that (Gi )0≤i≤n is a polycyclic series for G. Let us show by induction on n that G is residually finite. If n = 0, the group G is trivial and there is nothing to prove. Suppose now that n ≥ 1 and that the result has been established for groups admitting a polycyclic series of length n − 1. Since (Gi )0≤i≤n−1 is a polycyclic series of length n − 1, the group Gn−1 is polycyclic and hence finitely generated by (e). Moreover, Gn−1 is residually finite by our induction hypothesis. As G/Gn−1 is cyclic, it then follows from Exercise 2.35 that G is residually finite. (g) Necessity follows from (e). Conversely, suppose that G is an abelian group admiting a finite generating subset S = {x1 , x2 , . . . , xn } and let us show that G is polycyclic. As the map ϕ : Zn → G defined by ϕ(k1 , k2 , . . . , kn ) := x1k1 x2k2 · · · xnkn is a group epimorphism, it is enough to show that Zn is polycyclic by (c). To see this, observe that the sequence (Hi )0≤i≤n of subgroups of Zn , defined by Hi := Zi × {0Zn−i }

.

for all 0 ≤ i ≤ n, is a polycyclic series for Zn since Hi+1 /Hi ∼ = Z for 0 ≤ i ≤ n−1. (h) This immediately follows from (b) and (e). (i) Let G be a polycyclic group and let (Hn )n∈N be an increasing U sequence of subgroups of G. Consider the subgroup H of G defined by H := n∈N Hn . By (h), the group H is finitely generated. Let S ⊂ H be a finite generating subset for H . Then there exists n0 ∈ N such that S ⊂ Hn0 . This implies Hn = H for all n ≥ n0 . This shows that G satisfies the ascending chain condition for subgroups. (j) Necessity follows from (i). Conversely, suppose that G is a solvable group that satisfies the ascending chain condition for subgroups (“acds" for short). Let us show that G is polycyclic by induction on the solvability degree d of G. If d = 0, the group G is trivial and there is nothing to prove. Suppose now d ≥ 1 and that the result has been established

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for all solvable groups with solvability degree at most d − 1. The group [G, G] is solvable with solvability degree d − 1. As every subgroup of a group that satisfies acds also satisfies acds, we deduce from our induction hypothesis that [G, G] is polycyclic. On the other hand, since every group that satisfies acds is finitely generated and every quotient of a finitely generated group is itself finitely generated, the group G/[G, G] is finitely generated. As every finitely generated abelian group is polycyclic by (g), we deduce that G/[G, G] is polycyclic. Finally, since [G, G] and G/[G, G] are both polycyclic, we conclude that G is itself polycyclic by applying (d). ■ Comment The theory of polycyclic groups grew out from a series of papers [Hir1], [Hir2], [Hir3], [Hir4], [Hir5] on infinite solvable groups published between the late 1930s and the early 1950s by Hirsch. Polycyclic groups appear in various branches of mathematics, including number theory, ring theory, arithmetic groups, topology, dynamical systems, and geometric group theory. A group is polycyclic if and only if it is isomorphic to a solvable subgroup of GLn (Z) for some n ≥ 1 (sufficiency was established by Mal’cev [Mal1] and necessity by Louis Auslander [Aus] and Swan [Swa]). The books [Seg] and [Weh] are research monographs entirely devoted to polycyclic groups. A group G is called supersolvable if it admits a finite sequence of subgroups {1G } = G0 ⊂ G1 ⊂ G2 ⊂ · · · ⊂ Gn = G

.

such that Gi is normal in G and Gi+1 /Gi is a (finite or infinite) cyclic group for all 0 ≤ i ≤ n − 1. Clearly, every supersolvable group is polycyclic and therefore solvable. All finitely generated abelian groups and, more generally, all finitely generated nilpotent groups are supersolvable. The alternating group Sym+ 4 provides an example of a group that is polycyclic but not supersolvable. For a group G the following conditions are all equivalent: (1) the group G satisfies the ascending chain condition for subgroups; (2) every subgroup of G is finitely generated; (3) every non-empty set of subgroups of G admits a maximal element for inclusion. A group satisfying these equivalent conditions is called a Noetherian group because of the analogy with the definition of a Noetherian ring. A polycyclic-by-finite group is a group admitting a normal polycyclic subgroup of finite index. This is equivalent to being virtually polycyclic, that is, admitting a polycyclic subgroup of finite index. The fact that all polycyclic groups are Noetherian implies that all polycyclic-by-finite groups are Noetherian. Tarski monsters, constructed in 1979 by Olshanskii [Ols1], yield the first examples of groups that are Noetherian but not polycyclic-by-finite. By Exercise 1.111(e), every group of finite Markov type is Noetherian. Exercise 4.36 Show that every finite solvable group is polycyclic. Solution Let G be a finite solvable group. Let us show that G is polycyclic by induction on the cardinality of G. For |G| = 1, the group G is trivial and there is nothing to prove. Suppose now that |G| = m ≥ 2 and that the result has been

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established for all finite solvable groups with cardinality less than m. Since G is solvable, the quotient G/[G, G] is non-trivial. As G/[G, G] is finite abelian and hence isomorphic to a direct product of finite cyclic groups, there exist a non-trivial finite cyclic group C and a group epimorphism ψ : G/[G, G] → C. Let ϕ : G → G/[G, G] denote the canonical group epimorphism and consider the kernel H of the composite ψ ◦ ϕ. Then H is a proper normal subgroup of G and G/H ∼ = C. Moreover, H is solvable since the class of solvable groups is closed under taking subgroups by Exercise 4.14. Thus, by our induction hypothesis, H is polycyclic. Let (Hi )0≤i≤n be a polycyclic series for H . Then the sequence (Gi )0≤i≤n+1 , where Gi := Hi for all 0 ≤ i ≤ n and Gn+1 := G, is a polycyclic series for G. This shows that G is polycyclic. ■ Comment By a deep theorem due to Feit and Thompson [FeiT], every finite group of odd cardinality is solvable. The Feit-Thompson theorem implies in particular that every non-abelian finite simple group has even cardinality. Exercise 4.37 Show that every virtually polycyclic group is of finite Markov type. Solution Let G be a virtually polycyclic group. Let us show that G is of finite Markov type. Since every group containing a finite index subgroup which is of finite Markov type is itself of finite Markov type by Exercise 1.111(i), we can assume that G is polycyclic. We then proceed by induction on the length n of a polycyclic series {1G } = G0 ⊂ G1 ⊂⊂ · · · ⊂ Gn−1 ⊂ Gn = G. If n = 0 then G is a trivial group and there is nothing to prove (cf. Exercise 1.111(a)). Suppose now n ≥ 1. The group Gn /Gn−1 is either finite or infinite cyclic. As {1G } = G0 ⊂ G1 ⊂ · · · ⊂ Gn−1 is a polycyclic series of length n − 1 for Gn−1 , it follows inductively that G is of finite Markov type by applying either Exercise 1.111(i) or Exercise 1.113(h). ■ Comment This is a particular case of [Schm, Theorem I.4.2]. The question whether or not every group of finite Markov type is virtually polycyclic seems to remain open. Exercise 4.38 Let G be a group. For each integer i ≥ 0, denote by ρi : C i (G) → C i (G)/C i+1 (G) the canonical group epimorphism. (a) Let i ≥ 1 be an integer. Let x ∈ G. Show that [x, y] ∈ C i (G) for all y ∈ C i−1 (G) and that the map ϕxi : C i−1 (G) → C i (G)/C i+1 (G) defined by ϕix (y) := ρi ([x, y])

.

for all y ∈ C i−1 (G) is a group homomorphism whose kernel contains C i (G). (b) Let y ∈ C i−1 (G). Show that [y, x] ∈ C i (G) for all x ∈ G and that the map y ψi : G → C i (G)/C i+1 (G) defined by y

ψi (x) := ρi ([y, x])

.

for all x ∈ G is a group homomorphism whose kernel contains C 1 (G).

4.2 Exercises

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(c) Suppose that the abelianization G/[G, G] = C 0 (G)/C 1 (G) of the group G is finitely generated. Show that the group C i (G)/C i+1 (G) is finitely generated for every i ≥ 0. Solution (a) For all y ∈ C i−1 (G), we have [x, y] ∈ [G, C i−1 (G)] = [C i−1 (G), G] = i C (G). Let y1 , y2 ∈ C i−1 (G). Observe that [y1 , [x, y2 ]] ∈ [G, [G, C i−1 (G)]] = [G, C i (G)] = C i+1 (G). Thus we have ϕix (y1 y2 ) = ρi ([x, y1 y2 ])

.

= ρi ([x, y1 ] · [y1 , [x, y2 ]] · [x, y2 ])

(by (4.1))

= ρi ([x, y1 ])ρi ([y1 , [x, y2 ]])ρi ([x, y2 ]) = ρi ([x, y1 ])ρi ([x, y2 ])

(since [y1 , [x, y2 ]] ∈ C i+1 (G))

= ϕix (y1 )ϕix (y2 ). This shows that ϕix is a group homomorphism. If y ∈ C i (G), then [x, y] ∈ [G, C i (G)] = C i+1 (G) so that y ∈ ker(ϕix ). This shows that the kernel of ϕix contains C i (G). (b) For all x ∈ G, we have [y, x] ∈ [C i−1 (G), G] = C i (G). Let x1 , x2 ∈ G. Observe that [x1 , [y, x2 ]] ∈ [G, [G, C i−1 (G)]] = [G, C i (G)] = i+1 C (G). Thus we have y

ψi (x1 x2 ) = ρi ([y, x1 x2 ])

.

= ρi ([y, x1 ] · [x1 , [y, x2 ]] · [y, x2 ])

(by (4.1))

= ρi ([y, x1 ])ρi ([x1 , [y, x2 ]])ρi ([y, x2 ]) = ρi ([y, x1 ])ρi ([y, x2 ])

(since [x1 , [y, x2 ]] ∈ C i+1 (G))

= ψyi (x1 )ψyi (x2 ). This shows that ψyi is a group homomorphism. For all x1 , x2 ∈ G, we have [y, [x1 , x2 ]] ∈ [C i−1 (G), C 1 (G)] ⊂ C i+1 (G) by y Exercise 4.34(d). This implies [x1 , x2 ] ∈ ker(ψi ). As the elements [x1 , x2 ], where x1 , x2 run over G, generate C 1 (G), we conclude that C 1 (G) is contained in the y kernel of ψi . (c) Since G/[G, G] = C 0 (G)/C 1 (G) is finitely generated, there exists a finite subset S ⊂ G such that ρ0 (S) generates G/[G, G]. Let us show, by induction on i ≥ 0, that the group C i (G)/C i+1 (G) is finitely generated. For i = 0, this follows from our assymption. Suppose now that C i−1 (G)/C i (G) is finitely generated for a given i ≥ 1. Let T ⊂ C i−1 (G) be a finite subset such that ρi−1 (T ) generates C i−1 (G)/C i (G). By definition, the group C i (G) = [C i−1 (G), G] = [G, C i−1 (G)]

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4 Amenable Groups

is generated by all elements of the form [x, y], where x ∈ G and y ∈ C i−1 (G). Therefore C i (G)/C i+1 (G) is generated by all elements of the form ρi ([x, y]), where x ∈ G and y ∈ C i−1 (G). We have ρi ([x, y]) = ϕix (y). By (a), for every x ∈ G, the map ϕix : C i−1 (G) → C i (G)/C i+1 (G) is a group homomorphism whose kernel contains C i (G). It follows that C i (G)/C i+1 (G) is generated by all elements of the form ϕix (t) = ρi ([x, t]), where x ∈ G and t ∈ T . On the other hand, we have ρi ([x, t]) = ρi ([t, x]−1 ) = (ρi ([t, x]))−1 = (ψit (x))−1 . By (b), y for every y ∈ C i−1 (G), the map ψi : G = C 0 (G) → C i (G)/C i+1 (G) is a group homomorphism whose kernel contains [G, G] = C 1 (G). We deduce that C i (G)/C i+1 (G) is generated by all elements of the form ψit (s) = ρi ([t, s]), where s ∈ S and t ∈ T . This shows that C i (G)/C i+1 (G) is finitely generated and completes induction. ■ Exercise 4.39 Let G be a nilpotent group. Show that the following conditions are equivalent: (i) G is finitely generated; (ii) G/[G, G] is finitely generated; (iii) G is polycyclic. Solution We have (i) =⇒ (ii) since every quotient of a finitely generated group is itself finitely generated. Suppose (ii). Let d be the nilpotency degree of G. As G/[G, G] = C 0 (G)/C 1 (G) is finitely generated, it follows from Exercise 4.38(c) that the groups C i (G)/C i+1 (G), 0 ≤ i ≤ d, are all finitely generated. Since every finitely generated abelian group is polycyclic by Exercise 4.35(g), we deduce that the groups C i (G)/C i+1 (G), 0 ≤ i ≤ d, are all polycyclic. As the class of polycyclic groups is closed under forming extensions by Exercise 4.35(d), we conclude that G is polycyclic. Finally, we have (iii) =⇒ (i) since every polycyclic group is finitely generated by Exercise 4.35(e). ■ Exercise 4.40 (Residual Finiteness of Finitely Generated Nilpotent Groups) Show that every finitely generated nilpotent group is residually finite. Solution Let G be a finitely generated nilpotent group. It follows from Exercise 4.39 that G is polycyclic. As every polycyclic group is residually finite by Exercise 4.35(f), we deduce that G is residually finite. ■ Comment The hypothesis that the group is finitely generated cannot be removed. Indeed, the additive group Q is abelian and hence nilpotent. However, Q is not residually finite since it is divisible [CAG, Proposition 2.1.8].

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223

Exercise 4.41 (A Finitely Generated Solvable Group that is not Residually Finite) Let n ≥ 2 be an integer and let R := Z[1/n] denote the ring of n-adic rationals. Consider the set G ⊂ GL3 (R) consisting of all matrices of the form ⎛

⎞ 1 y z k .M(k, x, y, z) := ⎝0 n x ⎠ 0 0 1 with k ∈ Z and x, y, z ∈ R. (a) Show that G is a subgroup of GL3 (R). (b) Show that the group G is finitely generated, residually finite, and Hopfian. (c) Show that G is a solvable group of solvability degree 3. (d) Let ⎛ ⎞ n00 .D := ⎝ 0 1 0⎠ ∈ GL3 (R) 001 and let α denote the inner automorphism of GL3 (R) associated with D so that α(X) = DXD −1 for all X ∈ GL3 (R). Show that α(G) = G. (e) Let H denote the subgroup of G generated by the matrix M(0, 0, 0, 1). Show that H is normal in G and that α(H )  H . (f) Show that the quotient group G/H is not Hopfian. (g) Show that G/H is a finitely generated solvable group with solvability degree 3 that is not residually finite. Solution (a) Given k, k1 , k2 ∈ Z and x, x1 , x2 , y, y1 , y2 , z, z1 , z2 ∈ R, we have ⎛

M(k, x, y, z)−1

.

⎞−1 1 y z = ⎝0 nk x ⎠ 0 0 1 ⎛ ⎞ 1 −n−k y n−k xy − z = ⎝0 n−k −n−k x ⎠ 0 0 1 = M(−k, −n−k x, −n−k y, n−k xy − z)

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4 Amenable Groups

and ⎛ 1 ⎝ M(k1 , x1 , y1 , z1 )M(k2 , x2 , y2 , z2 ) = 0 0 ⎛ 1 . = ⎝0 0

⎞⎛ ⎞ y1 z1 1 y2 z2 nk1 x1 ⎠ ⎝0 nk2 x2 ⎠ 0 1 0 0 1 ⎞ nk2 y1 + y2 z1 + z1 + y1 x2 nk1 +k2 x1 + nk1 x2 ⎠ 0 1

= M(k1 + k2 , x1 + nk1 x2 , nk2 y1 + y2 , z1 + z2 + y1 x2 ). This shows that G is closed under taking inverses and products. As G is not empty, we deduce that G is a subgroup of GL3 (R). (b) Consider the elements N := M(1, 0, 0, 0), A := M(0, 1, 0, 0), B := M(0, 0, 1, 0), and C := M(0, 0, 0, 1). We have N −k AN k = M(−k, 0, 0, 0)M(0, 1, 0, 0)M(k, 0, 0, 0) = M(0, n−k , 0, 0) for all k ∈ N. Since M(0, x1 , 0, 0)M(0, x2 , 0, 0) =∑M(0, x1 + x2 , 0, 0) and every element x ∈ R can −k with n ∈ Z and n = 0 for all but be expressed as a sum x = k k k∈N nk n finitely many k ∈ N, we deduce that every element of the form M(0, x, 0, 0) with x ∈ R can be expressed as a product of elements of {N, N −1 , A, A−1 }. Analogously, as N k BN −k = M(0, 0, n−k , 0) and M(0, 0, y1 , 0)M(0, 0, y2 , 0) = M(0, 0, y1 + y2 , 0), it follows that every element of the form M(0, 0, y, 0) with y ∈ R can be expressed as a product of elements of {N, N −1 , B, B −1 }. Observe now that [M(0, x, 0, 0), M(0, 0, y, 0)]

.

= M(0, x, 0, 0)M(0, 0, y, 0)M(0, x, 0, 0)−1 M(0, 0, y, 0)−1 = M(0, x, 0, 0)M(0, 0, y, 0)M(0, −x, 0, 0)M(0, 0, −y, 0) = M(0, x, y, 0)M(0, −x, −y, 0) = M(0, 0, 0, −xy). In particular, we have M(0, 0, 0, z) = [M(0, z, 0, 0), M(0, 0, −1, 0)] = [M(0, z, 0, 0), B −1 ] for all z ∈ R. As we know that M(0, z, 0, 0) can be expressed as a product of elements in {N, N −1 , A, A−1 }, this shows that M(0, 0, 0, z) can be expressed as a product of elements in {N, N −1 , A, A−1 , B, B −1 }. Let k ∈ Z and x, y, z ∈ R. Using the fact that M(k, x, y, z) = N k M(0, n−k x, y, z)

.

= N k M(0, n−k x, 0, 0)M(0, 0, y, 0)M(0, 0, 0, z),

4.2 Exercises

225

we deduce from our previous observations that G is generated by the set {N, A, B}. This shows that the group G is finitely generated. As the ring R is residually finite by Exercise 2.12(a), we deduce from Exercise 2.11(j) that the group GL3 (R) is residually finite. This implies that G is residually finite since every subgroup of a residually finite group is itself residually finite. The group G is Hopfian since every finitely generated residually finite group is Hopfian by Mal’cev’s theorem [CAG, Theorem 2.4.3]. (c) The map G → Z given by M(k, x, y, z) |→ k is a group homomorphism whose kernel is the Heisenberg group over the ring R HR := {M(0, x, y, z) : x, y, z ∈ R} ⊂ GL3 (R).

.

As HR is metabelian, it follows from Exercise 4.27 that G is solvable with solvability degree at most 2 + 1 = 3. On the other hand, as the matrices C1 := [N, A] = (NAN −1 )A−1 = An−1 = M(0, n − 1, 0, 0) and C2 := [N, B] = −1 (N BN −1 )B −1 = B n −1 = M(0, 0, n−1 − 1, 0) are in the commutator subgroup of G and satisfy [C1 , C2 ] = C1 C2 C1−1 C2−1 = M(0, 0, 0, n + n−1 − 2) /= M(0, 0, 0, 0) = 1G , the group G is not metabelian. This shows that G has solvability degree exactly 3. (d) For all k ∈ Z and x, y, z ∈ R, we have α(M(k, x, y, z)) = DM(k, x, y, z)D −1 ⎞⎛ ⎛ ⎞ ⎛ −1 ⎞ 1 y z n00 n 00 = ⎝0 1 0⎠ ⎝0 nk x ⎠ ⎝ 0 1 0⎠ 001 0 0 1 0 01 . ⎛ ⎞ 1 ny nz = ⎝0 nk x ⎠ 0 0 1 = M(k, x, ny, nz). As every r ∈ R can be written in the form r = nr ' for some r ' ∈ R, we deduce that α(G) = G. (e) Since M(k, x, y, z)M(0, 0, 0, 1) = M(k, 0, 0, z) = M(0, 0, 0, 1) M(k, x, y, z) for all k ∈ Z and x, y, z ∈ R, the matrix M(0, 0, 0, 1) is in the center Z(G) of G. Therefore H ⊂ Z(G). It follows that H is normal in G. We have H = {M(0, 0, 0, z) : z ∈ Z} and α(H ) = {M(0, 0, 0, nz) : z ∈ Z}. As n ≥ 2, it follows that α(H )  H . (f) This follows from (e) and Exercise 2.15(b). (g) The group G/H is finitely generated since it is a quotient of G which is finitely generated by (b). On the other hand, we deduce from (d) and Exercise 4.15 that G/H is solvable with solvability degree at most 3. Let ρ : G → G/H denote the canonical group epimorphism and consider again the matrices C1 , C2 ∈

226

4 Amenable Groups

G introduced in (d). The elements ρ(C1 ) = ρ([N, A]) = [ρ(N), ρ(A)] and ρ(C2 ) = ρ([N, B]) = [ρ(N), ρ(B)] are in the commutator subgroup of G/H . As [ρ(C1 ), ρ(C2 )] = ρ([C1 , C2 ]) and [C1 , C2 ] = M(0, 0, 0, n + n−1 − 2) ∈ / H, we have [ρ(C1 ), ρ(C2 )] /= 1G/H . This shows that G/H is solvable with solvability degree exactly 3. Finally, the fact that G/H is not residually finite follows from (f) in combination with Mal’cev’s theorem [CAG, Theorem 2.4.3]. ■ Comment It was shown by Philip Hall [Hal2] that every finitely generated solvable group of solvability degree at most two (i.e., every finitely generated metabelian group) is residually finite. Our description of G/H is adapted from [Hal3, Section 3.5] (see also [Abe] and [Cor, Remark 2.6]). Examples of finitely presented solvable groups that are not residually finite were constructed by Abels in [Abe]. Exercise 4.42 Let G be a nilpotent group. Suppose that G admits a finite generating subset S all of whose elements have finite order. Show that G is finite. Solution We proceed by induction on the nilpotency degree d of G. If d ≤ 1, the group G is abelian and the result is obvious. Suppose now that d ≥ 2 and that the result has been established for all groups with nilpotency degree at most d − 1. The group G/C d−1 (G) is nilpotent with nilpotency degree d − 1 by Exercise 4.17(b). On the other hand, the group G/C d−1 (G) is generated by π(S) where π : G → G/C d−1 (G) is the canonical group epimorphism. As all elements of π(S) have finite order, we deduce that G/C d−1 (G) is finite by our induction hypothesis. To complete the proof, it suffices to prove that the group C d−1 (G) is finite. We first observe that C d−1 (G) is finitely generated since it is a finite index subgroup of a finitely generated group (this can also be deduced from the fact that G is polycyclic by Exercise 4.39, since every subgroup of a polycyclic group is itself finitely generated by Exercise 4.35(h)). By definition, C d−1 (G) is generated by all elements of the form [y, x], where y ∈ C d−2 (G) and x ∈ G. Moreover, if we fix y ∈ C d−2 (G), we know from Exercise 4.38(b) that the map x |→ [y, x] is a group homomorphism from G into C d−1 (G). This implies that C d−1 (G) is generated by elements of finite order. As C d−1 (G) is abelian and finitely generated, this implies that C d−1 (G) is finite. This completes the proof that G is finite. ■ Comment The result becomes false for polycyclic groups. For example, the infinite dihedral group is polycyclic since it contains an infinite cyclic group of index 2 but it can be generated by two elements of order 2. Exercise 4.43 Let G be a nilpotent group and let T denote the subset of G consisting of all finite order elements in G. (a) Show that T is a normal subgroup of G. (b) Show that if G is finitely generated then T is finite. Solution (a) We clearly have 1G ∈ T . Suppose now that x and y are in T . Consider the subgroup H of G generated by x and y. Since every subgroup of a nilpotent group is itself nilpotent by Exercise 4.14, It follows from Exercise 4.42 that H is finite. As xy −1 ∈ H , we deduce that xy −1 ∈ T . This shows that T is a subgroup of G.

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227

The subgroup T is normal in G since the elements in a conjugacy class of G have all the same order. (b) Suppose that G is finitely generated. Then G is polycyclic by Exercise 4.39. As every subgroup of a polycyclic group is finitely generated by Exercise 4.35(h), the group T is finitely generated. Since T is nilpotent by Exercise 4.14, it then follows from Exercise 4.42 (or Exercise 4.26 since every nilpotent group is solvable) that T is finite. ■ Comment The example of the infinite dihedral group shows that these results become false for polycyclic groups and solvable groups. Indeed, the elements of finite order in the infinite dihedral group are the identity element and the elements of order 2. They form an infinite set which is not closed under multiplication. Exercise 4.44 (The Lamplighter Group) The lamplighter group is the wreath product L := (Z/2Z) Z. Thus, L is the semidirect product of the groups H := ⊕ A n∈Z n and Z, where An := Z/2Z for all n ∈ Z, and Z acts on H by the Z-shift. (a) Show that the group L is metabelian. (b) Show that the group L is residually finite. (c) Show that L is finitely generated. (d) Show that L is not of finite Markov type. (e) Show that L is not virtually polycyclic. (f) Show that L is solvable but not nilpotent. Solution (a) The group L is a semidirect product of H and Z. Therefore H can be viewed as a normal subgroup of L with H /L ∼ = Z. As the groups H and L/H are both abelian, the group L is metabelian by Exercise 4.9. (b) The subgroup H , being the direct sum of the finite (and therefore residually finite) groups An = Z/2Z, is itself residually finite. Since the group K := L/H ∼ =Z is residually finite and abelian, it follows from Exercise 2.36(g) that the wreath product L = H K is residually finite. (c) By Exercise 2.36(d), the group L is finitely generated since it is the wreath product of two finitely generated groups. (d) This follows from Exercise 2.36(j). (e) This follows from (d) since every virtually polycyclic group is of finite Markov type by Exercise 4.37. (f) The group L is solvable since it is metabelian by (a). Every finitely generated nilpotent group is polycyclic by Exercise 4.39. As L is finitely generated by (c) but not polycyclic by (e), we deduce that L is not nilpotent. ■ Comment By Exercise 4.43(b), a finitely generated nilpotent group cannot contain infinitely many elements of finite order. This yields an alternative proof of the fact that L is not nilpotent. Indeed, the group L is finitely generated by (c) while all elements in the infinite subgroup H = ⊕n∈Z An ⊂ L have order at most 2. Exercise 4.45 Let m be an integer such that |m| ≥ 2. Let G be the group given by the presentation G := .

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4 Amenable Groups

(a) Show that G is solvable with solvability degree 2. (b) Show that G is not polycyclic. (c) Show that G is not nilpotent. Solution (a) By Exercise 2.13(f), there is a short exact sequence of groups 0 → Z[1/m] → G → Z → 0. Therefore G is metabelian (cf. Exercise 4.9). As G is not abelian by Exercise 2.13(d), it follows that G is solvable with solvability degree 2. (b) The additive group Z[1/m] is not finitely generated. As Z[1/m] is isomorphic to a subgroup of G and every subgroup of a polycyclic group is finitely generated by Exercise 4.35(h), it follows that G is not polycyclic. (c) This follows from (b) since every finitely generated nilpotent group is ■ polycyclic by Exercise 4.39. Exercise 4.46 Let G be a locally finite group. Let S denote the directed set consisting of all finitely generated subgroups of G partially ordered by inclusion. Prove that the net (H )H ∈S is a Følner net for G. Solution Let g ∈ G. Consider the subgroup Hg ⊂ G generated by g. Let H ∈ S such that Hg ⊂ H . Since g ∈ Hg ⊂ H , we have gH = H and hence gH \ H = ∅. We deduce that limH ∈S |gH \ H |/|H | = 0. This shows that (H )H ∈S is a Følner net for G. ■ Exercise 4.47 (Følner Sequences in a Finite Cartesian Product) Let G1 (1) (resp. G2 ) be a countable amenable group. Let F1 = (Fn )n∈N (resp. F2 = (2) (Fn )n∈N ) be a left Følner sequence for G1 (resp. for G2 ). (1) (2) (a) Show that F = (Fn )n∈N , where Fn := Fn × Fn for all n ∈ N, is a left Følner sequence for G := G1 × G2 . (b) Let r ≥ 1 be an integer. Show that F (r) := (Fnr )n∈N , where Fnr := {0, 1, . . . , n}r ⊂ Zr for all n ∈ N, is a Følner sequence for Zr . Solution (a) Let g = (g1 , g2 ) ∈ G and let n ∈ N. Then gFn \ Fn = (g1 Fn(1) × g2 Fn(2) ) \ (Fn(1) × Fn(2) ) ⎞ ⎛ ⎞ ⎛ . g1 Fn(1) × (g2 Fn(2) \ Fn(2) ) ⊂ (g1 Fn(1) \ Fn(1) ) × g2 Fn(2) so that |gFn \ Fn | ≤ |(g1 Fn(1) \ Fn(1) ) × g2 Fn(2) | + |g1 Fn(1) × (g2 Fn(2) \ Fn(2) )| .

= |g1 Fn(1) \ Fn(1) | · |g2 Fn(2) | + |g1 Fn(1) | · |g2 Fn(2) \ Fn(2) | = |g1 Fn(1) \ Fn(1) | · |Fn(2) | + |Fn(1) | · |g2 Fn(2) \ Fn(2) |.

4.2 Exercises

229 (1)

(2)

Since |Fn | = |Fn | · |Fn |, and keeping in mind that F1 (resp. F2 ) is a left Følner sequence for G1 (resp. G2 ), we deduce that (1)

.

(1)

(2)

(2)

|g1 Fn \ Fn | |g2 Fn \ Fn | |gFn \ Fn | ≤ lim + = 0. (1) (2) n→∞ n→∞ |Fn | |Fn | |Fn | lim

This shows that F = (Fn )n∈N is a left Følner sequence for G. (b) This follows immediately from (a) by an obvious inductive argument, after recalling that F (1) is a Følner sequence for Z (cf. [CAG, Example 4.7.4.(b)]). ■ Exercise 4.48 Show that the sequence (Fn )n∈N , where Fn consists of all rational numbers of the form k/n! with k ∈ N and k ≤ (n + 1)!, is a Følner sequence for the additive group Q. Solution Observe that |Fn | = (n + 1)! for all n ∈ N. Let now q ∈ Q. Suppose first that q ≥ 0 and let a, b ∈ N with b /= 0 such that q = a/b. For n ≥ b we have qn! ∈ N and q + k/n! = (qn! + k)/n! for all k ∈ N, so that the set (q + Fn ) \ Fn = {k/n! : (n + 1)! < k ≤ qn! + (n + 1)!}.

.

has cardinality qn!. Suppose now that q < 0. We then have |(q + Fn ) \ Fn | = |Fn \ (−q + Fn )| = |(−q + Fn ) \ Fn |.

.

In either case, we deduce that .

lim

n→∞

|(q + Fn ) \ Fn | |q|n! |q| = lim = lim = 0. n→∞ (n + 1)! n→∞ n + 1 |Fn |

This shows that the sequence (Fn )n∈N is a Følner sequence for Q.

■

Exercise 4.49 Let G := HZ denote the integral Heisenberg group (cf. [CAG, Example 4.6.5]). Recall that G is the subgroup of SL3 (Z) consisting of all upper triangular matrices with only 1s on the diagonal. For each integer n ≥ 1, define the subset Fn ⊂ G by ⎧⎛ ⎫ ⎞ ⎨ 1x z ⎬ ⎝0 1 y ⎠ ∈ G : 1 ≤ x ≤ n, 1 ≤ y ≤ n, 1 ≤ z ≤ n2 . .Fn := ⎩ ⎭ 001 Show that the sequence (Fn )n≥1 is a Følner sequence for G.

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4 Amenable Groups

Solution We first observe that |Fn | = n4 for all n ≥ 1. Let now g ∈ G, say, ⎛ ⎞ 1x z .g = ⎝0 1 y ⎠ , 001 and let M ∈ N such that max(|x|, |y|, |z|) ≤ M. Let f ∈ Fn , say ⎛

⎞ 1 x ' z' ' .f = ⎝0 1 y ⎠ . 0 0 1 Then, ⎛ ⎞ 1 x + x ' z + z' + xy ' .gf = ⎝0 1 y + y' ⎠ 0 0 1 where −M + 1 ≤ x + x ' ≤ M + n −M + 1 ≤ y + y ' ≤ M + n

.

−M − Mn + 1 ≤ z + z' + xy ' ≤ M + n2 + Mn. Let ⎛ '' '' ⎞ 1x z '' '' .g = ⎝0 1 y ⎠ ∈ gFn \ Fn . 0 0 1 Then at least one of the following conditions holds: (i) −M + 1 ≤ x '' ≤ 0 or n + 1 ≤ x '' ≤ M + n; (ii) −M + 1 ≤ y '' ≤ 0 or n + 1 ≤ y '' ≤ M + n; (iii) −M − Mn + 1 ≤ z'' ≤ 0 or n2 + 1 ≤ z'' ≤ M + n2 + Mn. Counting all such possibilities, we have at most ⎛ ⎞ p(n) := 2M · (n + 2M) · 2M + n2 + 2Mn

.

possibilities for g '' satisfying (i) (resp. (ii)), and at most q(n) := (2M + n)2 · (2(M + Mn))

.

4.2 Exercises

231

possibilities for g '' satisfying (iii). We deduce that |gFn \ Fn | ≤ 2p(n) + q(n).

.

Since p(n) and q(n) are polynomials of degree 3 in n, we have limn→∞ (2p(n) + q(n))/n4 = 0. It follows that .

|gFn \ Fn | = 0. n→∞ |Fn | lim

This shows that (Fn )n≥1 is a Følner sequence for G.

■

Exercise 4.50 Let G be a group. Show that G is amenable if and only if the following condition holds: for every finite subset K ⊂ G and every ε > 0, there exists a finite subset F ⊂ G such that |KF | < (1 + ε)|F |. Solution Suppose first that G is amenable. Let K ⊂ G be a non-empty finite subset and let ε > 0. Set ε' := ε/|K|. By (A4) applied to the pair (K, ε' ), there exists a finite subset F ⊂ G such that |kF \ F | < ε' |F | for all k ∈ K. Since KF = (KF ∩ F ) ∪ (KF \ F ) and KF \ F =



.

(kF \ F ),

k∈K

we have, by taking cardinalities, |KF | = |KF ∩ F | + |KF \ F | ∑ |kF \ F | ≤ |KF ∩ F | + k∈K

.

'

≤ |F | + |K|ε |F | = (1 + ε)|F |. This shows that the condition in the statement is necessary. Conversely, suppose that this condition is satisfied. Let K ⊂ G be a finite subset and ε > 0. Set K ' := K ∪ {1G } and observe that KF \ F = K ' F \ F and F ⊂ K ' F for all F ⊂ G. Since the condition in the statement holds, there exists a finite subset F ⊂ G such that |K ' F | < (1 + ε)|F |. We deduce that |KF \ F | = |K ' F \ F | = |K ' F | − |F | < ε|F |. Given k ∈ K, we have kF \ F ⊂ KF \ F so that |kF \ F | ≤ |KF \ F | < ε|F |.

.

Thus condition (A4) is satisfied, so that G is amenable.

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4 Amenable Groups

Exercise 4.51 Let G be a finite group. Let I be a directed set and let F = (Fi )i∈I be a net of non-empty finite subsets of G. Show that F is a left Følner net for G if and only if there exists i0 ∈ I such that Fi = G for all i ≥ i0 . Solution Suppose first that there exists i0 ∈ I such that Fi = G for all i ≥ i0 . Then we have gFi \ Fi = G \ G = ∅ and hence |gFi \ Fi | = 0 for all i ≥ i0 and g ∈ G. This implies .

lim i

|gFi \ Fi | = 0 for all g ∈ G, |Fi |

showing that F is a left Følner net for G. Conversely, suppose that F is a left Følner net for G. Since G is finite, we can find i0 ∈ I such that |gFi \Fi | < |Fi |/|G| for all g ∈ G and i ≥ i0 . As |Fi |/|G| ≤ 1, this implies gFi \ Fi = ∅, that is, gFi ⊂ Fi for all g ∈ G and i ≥ i0 . Since |Fi | = |gFi |, we deduce that Fi = gFi for all g ∈ G and i ≥ i0 . Using the fact that Fi /= ∅, we conclude that Fi = G for all i ≥ i0 . ■ Exercise 4.52 Let G be an infinite amenable group and let F = (Fi )i∈I be a Følner net for G. Show that limi |Fi | = ∞. Solution Let M > 0. Since G is infinite, we can find a finite subset K ⊂ G such that |K| > M 2 . As F is a left Følner net for G, there exists i0 ∈ I such that |gFi \ Fi | < |Fi |/2 for all g ∈ K and i ≥ i0 . This implies gFi ∩ Fi /= ∅ and hence g ∈ Fi−1 Fi for all g ∈ K and i ≥ i0 . Thus, for all i ≥ i0 , we have K ⊂ Fi−1 Fi and hence M 2 < |K| ≤ |Fi−1 Fi | ≤ |Fi |2 . It follows that |Fi | > M for all i ≥ i0 . This ■ shows that limi |Fi | = ∞. Exercise 4.53 Let G be a group and let H be a finite index subgroup of G. Suppose that F H = (FjH )j ∈J is a left Følner net for H and that T ⊂ G is a complete set of representatives for the left cosets of H in G, so that G = ⨆t∈T tH . Show that the family F G = (FjG )j ∈J , where FjG := T FjH for all j ∈ J , is a left Følner net for G. Solution Let g ∈ G. Then for every t ∈ T , there exist unique h = h(g, t) ∈ H and s = s(g, t) ∈ T such that gt = sh. Thus, setting K = K(g) := {h(g, t) : t ∈ T } ⊂ H , we have gT = {gt : t ∈ T } = {sh : h = h(g, t), s = s(g, t), t ∈ T } ⊂ T K. We deduce that gFjG = gT FjH ⊂ T KFjH and therefore gFjG \ FjG = gT FjH \ T FjH ⊂ T KFjH \ T FjH ⊂ T (KFjH \ FjH )

.

for all j ∈ J . Taking cardinalities and observing that |FjG | = |T FjH | = |FjH | · |T |, we obtain .

|gFjG \ FjG | |FjG |

≤

|T | · |KFjH \ FjH | |FjH | · |T |

=

|KFjH \ FjH | |FjH |

.

4.2 Exercises

233

Since K is a finite subset of H and F H is a right Følner sequence for H , we have that limj |KFjH \ FjH | / |FjH | = 0. We deduce that limj |gFjG \ FjG | / |FjG | = 0. Since g ∈ G was arbitrary, this shows that F G is a left Følner net for H . ■ Comment Incidentally, the above gives an alternative proof of the fact that a virtually amenable group is itself amenable. Exercise 4.54 Let G be an amenable group and let F = (Fi )i∈I be a left Følner net for G. Let A ⊂ G be a finite subset. Show that the net (Fi ∪A)i∈I is a left Følner net for G. Solution By Exercise 4.51, we only need to consider the case when G is infinite. Let g ∈ G and i ∈ I . Then g(Fi ∪ A) \ (Fi ∪ A) ⊂ (gFi \ (Fi ∪ A)) ∪ (gA \ (Fi ∪ A) ⊂ (gFi \ Fi ) ∪ gA,

.

and hence .

|gFi \ Fi | + |gA| |g(Fi ∪ A) \ (Fi ∪ A)| ≤ |Fi | |Fi | =

|gFi \ Fi | |A| + . |Fi | |Fi |

As F is a left Følner net for G and limi |Fi | = ∞ by Exercise 4.52, we deduce that .

lim i

|g(Fi ∪ A) \ (Fi ∪ A)| = 0. |Fi |

This shows that the net (Fi ∪ A)i∈I is a left Følner net for G.

■

Exercise 4.55 Let G be a countable amenable group. Show that G admits a left Følner sequence (Fn )n∈N which exhausts G, i.e., such that Fn ⊂ Fn+1 for all n ∈ N U and G = n∈N Fn . Solution Let (Fn' )n∈N be a left Følner sequence for G. Since G is countable, U we can find an increasing sequence (An )n∈N of finite subsets of G such that n∈N An = G. We inductively define a sequence (Fn )n∈N of non-empty finite subsets of G as follows. (0) For n = 0, we consider the sequence F0 = (Fk )k∈N by setting (0)

Fk

.

:= Fk' ∪ A0

for all k ∈ N. The fact that F0 is a left Følner sequence for G follows from Exercise 4.54 and implies that there exists k0 ∈ N such that (0)

.

|gFk

(0)

\ Fk | (0)

|Fk |

≤1

(0) for all k ≥ k0 and g ∈ A0 . We set F0 := Fk0 . Note that A0 ⊂ F0 .

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4 Amenable Groups

Suppose now that we have already defined the finite subset Fn−1 ⊂ G for some n ≥ 1. We then consider the sequence Fn = (Fk(n) )k∈N by setting (n)

Fk

.

:= Fk ∪ (Fn−1 ∪ An )

for all k ∈ N. The fact that Fn is a left Følner sequence for G follows from Exercise 4.54 and implies that there exists kn ∈ N such that (n)

.

|gFk

(n)

\ Fk |

(n) |Fk |

≤

1 n+1

for all k ≥ kn and g ∈ An . We then set Fn := Fk(n) . Note that Fn−1 ⊂ Fn and n An ⊂ Fn . In this way, we get by induction an increasing sequence F = (Fn )U n∈N of nonempty finite subsets of G such that An ⊂ Fn for all n ∈ N and therefore n∈N Fn = G. Let us show that F is a left Følner sequence for G. Let g ∈ G. Then there exists n0 ∈ N such that g ∈ An0 . We then have (n)

(n)

|gFkn \ Fkn | |gFn \ Fn | 1 = ≤ . (n) |Fn | n + 1 |Fkn | for all n ≥ n0 since An0 ⊂ An . We deduce that .

|gFn \ Fn | = 0. n→∞ |Fn | lim

This shows that F is a left Følner sequence with the required properties.

■

Exercise 4.56 Let G be a group and let Ω be a subset of G. (a) Let A and B be two subsets of G. Show that AΩ \ BΩ ⊂ (A \ B)Ω.

.

(4.12)

(b) Give an example showing that the inclusion in (4.12) may be strict. (c) Suppose that (Fj )j ∈J is a left Følner net for G and that the subset Ω ⊂ G is finite and non-empty. Show that the family (Fj Ω)j ∈J is a left Følner net for G. Solution (a) Given x ∈ AΩ \ BΩ, there exists a ∈ A and ω ∈ Ω such that x = aω. As x∈ / BΩ, we have a ∈ / B. We deduce that x ∈ (A \ B)Ω. This proves (4.12). (b) Let G be a group generated by an elment g of order 2. Take A = Ω = G and B = {g}. Then AΩ \ BΩ = ∅ while (A \ B)Ω = 1G Ω = Ω = G.

4.2 Exercises

235

(c) Let g ∈ G and j ∈ J . By (a), we have Fj Ω \ gFj Ω ⊂ (Fj \ gFj )Ω and hence |Fj Ω \ gFj Ω| ≤ |(Fj \ gFj )Ω| ≤ |Fj \ gFj | · |Ω|. Moreover, since Ω /= ∅, we have |Fj Ω| ≥ |Fj |. We deduce that .

|(Fj \ gFj )| |Fj Ω \ gFj Ω| ≤ |Ω|. |Fj Ω| |Fj |

As (Fj )j ∈J is a left Følner net for G, taking the limit over J we have that .

lim j

|Fj Ω \ gFj Ω| = 0. |Fj Ω|

Since g ∈ G was arbitrary, this shows that the family (Fj Ω)j ∈J is a left Følner net for G. ■ Exercise 4.57 Let (K, (Ak )k∈K , (Bk )k∈K ) be a left paradoxical decomposition of a group G. Show that |K| ≥ 3. Solution It is clear that |K| ≥ 2. Suppose that |K| = 2. Writing K := {k1 , k2 }, we then have ) ( ) ( .G = k1 Ak1 ⨆ k2 Ak2 ⨆ k1 Bk1 ⨆ k2 Bk2 = Ak1 ⨆ Ak2 = Bk1 ⨆ Bk2 . (4.13) Multiplying on the left by k1−1 the first two terms in (4.13) we get G = Ak1 ⨆ k1−1 k2 Ak2 ⨆ Bk1 ⨆ k1−1 k2 Bk2 .

.

It follows that Ak2 = G \ Ak1 = k1−1 k2 Ak2 ⨆ Bk1 ⨆ k1−1 k2 Bk2 ⊃ Bk1

.

and Bk2 = G \ Bk1 = Ak1 ⨆ k1−1 k2 Ak2 ⨆ k1−1 k2 Bk2 ⊃ Ak1 .

.

Analogously, by multiplying on the left by k2−1 the first two terms in (4.13), we get Ak1 ⊃ Bk2 and Bk1 ⊃ Ak2 . This shows that Ak1 = Bk2 and Ak2 = Bk1 . But then (4.13) yields G = k1 Ak1 ⨆ k2 Ak2 ⨆ k1 Ak2 ⨆ k2 Ak1 = k1 (Ak1 ⨆ Ak2 ) ⨆ k2 (Ak1 ⨆ Ak2 ) .

= k1 G ⨆ k2 G = G ⨆ G,

a contradiction. We deduce that |K| ≥ 3.

■

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4 Amenable Groups

Exercise 4.58 Let G be a group and let H be a subgroup of G. Suppose that (K, (Ak )k∈K , (Bk )k∈K ) is a left paradoxical decomposition of H and let T ⊂ G be a complete set of representatives for the right cosets of H in G. For each k ∈ K, set A'k := ⨆t∈T Ak t and Bk' := ⨆t∈T Bk t. Show that (K, (A'k )k∈K , (Bk' )k∈K ) is a left paradoxical decomposition of G. Solution Since (K, (Ak )k∈K , (Bk )k∈K ) is a left paradoxical decomposition of H , we have H = (⨆k∈K kAk ) ⨆ (⨆k∈K kBk ) = ⨆k∈K Ak = ⨆k∈K Bk .

.

(4.14)

Moreover, since T is a complete set of representatives for the right cosets of H in G, we have G = ⨆t∈T H t = H T . Multiplying each term in (4.14) on the right by t ∈ T , we get H t = (⨆k∈K kAk t) ⨆ (⨆k∈K kBk t) = ⨆k∈K Ak t = ⨆k∈K Bk t.

.

Thus, taking disjoint unions over t ∈ T , and recalling that A'k = Ak T and Bk' = Bk T for all k ∈ K, we get ) ( ) ( G = ⨆k∈K kA'k ⨆ ⨆k∈K kBk' = ⨆k∈K A'k = ⨆k∈K Bk' .

.

This shows that (K, (A'k )k∈K , (Bk' )k∈K ) is a left paradoxical decomposition of G. ■ Comment Since a group is amenable if and only if it does not admit any paradoxical decomposition, this yields another proof of the fact that every subgroup of an amenable group is amenable. Exercise 4.59 Let G be a group and let N be a normal subgroup of G. Suppose that (K, (Ak )k∈K , (Bk )k∈K ) is a left paradoxical decomposition of G/N. Let π : G → G/N denote the canonical quotient group homomorphism and let K ' ⊂ G such that π(K ' ) = K and |K ' | = |K|. For each k ∈ K, denote by k ' the unique element in K ' such that π(k ' ) = k and set Ak ' := π −1 (Ak ) (resp. Bk ' := π −1 (Bk )). Show that (K ' , (Ak ' )k ' ∈K ' , (Bk ' )k ' ∈K ' ) is a paradoxical decomposition of G. Solution We have π −1 (kAk ) = k ' Ak ' (resp. π −1 (kBk ) = k ' Bk ' ) for all k ∈ K. Thus, by applying π −1 to each side of the equalities G/N = (⨆k∈K kAk ) ⨆ (⨆k∈K kBk ) = ⨆k∈K Ak = ⨆k∈K Bk ,

.

we get ) ( ) ( G = ⨆k ' ∈K ' k ' Ak ' ⨆ ⨆k ' ∈K ' k ' Bk ' = ⨆k ' ∈K ' Ak ' = ⨆k ' ∈K ' Bk ' .

.

This shows that (K ' , (Ak ' )k ' ∈K ' , (Bk ' )k ' ∈K ' ) is a paradoxical decomposition of G. ■

4.2 Exercises

237

Comment Since a group is amenable if and only if it does not admit any paradoxical decomposition, this yields another proof of the fact that every quotient of an amenable group is amenable. Exercise 4.60 (The Tarski Number of a Group) Let G be a group. Given a (left) paradoxical decomposition P = (K, (Ak )k∈K , (Bk )k∈K ) of G, the integer c(P) := m + n, where m = m(P) := |{k ∈ K : Ak /= ∅}| and n = n(P) := |{k ∈ K : Bk /= ∅}|, is called the complexity of P. The quantity T (G) := inf c(P), where the infimum is taken over all paradoxical decompositions P of G, is called the Tarski number of G. One uses the convention that T (G) = +∞ if G admits no paradoxical decompositions, that is, if G is amenable. (a) Let H be a subgroup of G. Show that T (G) ≤ T (H ). (b) Let N be a normal subgroup of G. Show that T (G) ≤ T (G/N). (c) Show that there exists a finitely generated subgroup H of G such that T (H ) = T (G). (d) Show that one always has T (G) ≥ 4. (e) Show that T (G) = 4 if and only if G contains a subgroup isomorphic to F2 . (f) Suppose that all elements of G have finite order. Show that T (G) ≥ 6. Solution (a) We can assume T (H ) < +∞. Let PH = (K, (Ak )k∈K , (Bk )k∈K ) be a paradoxical decomposition of H such that T (H ) = c(PH ). Let T ⊂ G be a complete set of representatives of the right cosets of H in G. It follows from Exercise 4.58 that PG := (K, (A'k )k∈K , (Bk' )k∈K ), where A'k := Ak T and Bk' := Bk T for all k ∈ K, is a paradoxical decomposition of G. Since Ak = ∅ if and only if A'k = ∅ (resp. Bk = ∅ if and only if Bk' = ∅) for all k ∈ K, we deduce that m(PG ) = m(PH ) (resp. n(PG ) = n(PH )). Thus, T (G) ≤ c(PG ) = c(PH ) = T (H ). (b) We can assume T (G/N) < +∞. Let PG/N = (K, (Ak )k∈K , (Bk )k∈K ) be a paradoxical decomposition of G/N such that T (G/N) = c(PG/N ). Let π : G → G/N denote the canonical quotient homomorphism and denote by K ' ⊂ G a subset such that π(K ' ) = K and |K ' | = |K|. Also, for every k ∈ K denote by k ' the unique element in K ' such that π(k ' ) = k and set Ak ' := π −1 (Ak ) (resp. Bk ' := π −1 (Bk ). It follows from Exercise 4.59 that PG = (K ' , (Ak ' )k ' ∈K ' , (Bk ' )k ' ∈K ' ) is a paradoxical decomposition of G. Since Ak = ∅ if and only if Ak ' = ∅ (resp. Bk = ∅ if and only if Bk ' = ∅) for all k ∈ K, we deduce that m(PG ) = m(PG/H ) (resp n(PG ) = n(PG/H )). Thus, T (G) ≤ c(PG ) = c(PG/N ) = T (G/N). (c) By (a), we can assume T (G) < +∞. Let PG = (K, (Ak )k∈K , (Bk )k∈K ) be a paradoxical decomposition of G such that T (G) = c(PG ). Let H ⊂ G denote the subgroup generated by K. It follows from (a) that T (G) ≤ T (H ). Let us show the reverse inequality. Set A'k := Ak ∩ H and Bk' := Bk ∩ H for all k ∈ K. Keeping in mind that K ⊂ H , it is straightforward that PH := (K, (A'k )k∈K , (Bk' )k∈K ) is a paradoxical decomposition of H . Moreover, since Ak ' = ∅ if Ak = ∅ (resp. Bk ' = ∅ if Bk = ∅) for all k ∈ K, we have that m(PH ) ≤ m(PG ) (resp. n(PH ) ≤ n(PG )). We deduce that T (H ) ≤ c(PH ) ≤ c(PG ) = T (G). This shows that T (H ) = T (G).

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4 Amenable Groups

(d) We can assume T (G) < +∞. Let P = (K, (Ak )k∈K , (Bk )k∈K ) be a paradoxical decomposition of G. Since G = ⨆k∈K Ak = ⨆k∈K Bk , we have m := m(P) ≥ 1 (resp. n := n(P) ≥ 1). If m = 1, say {k ∈ K : Ak /= ∅} = {k1 }, then G = Ak1 . Since G = (⨆k∈K kAk ) ⨆ (⨆k∈K kBk ) = G ⨆ (⨆k∈K kBk ), this forces n = 0, a contradiction. We deduce that m ≥ 2 and, exchanging the roles of the Ak s and Bk s, we also have n ≥ 2. Thus, c(P) = m + n ≥ 4. As the paradoxical decomposition P of G was arbitrary, this shows that T (G) ≥ 4. (e) Let H be the free group of rank two, freely generated by a and b. We denote by A+ (resp. A− ) the subset of G consisting of all elements whose reduced form starts with a positive (resp. negative) power of a. Let also B + denote the subset of H consisting of all elements either of the form b−n for n = 0, 1, 2, . . ., or whose reduced form starts with a positive power of b. Finally, let B − := H \ (A+ ∪ A− ∪ B + ). In [CAG, Example 4.8.2], it is shown that if K := {1H , a, b} and A1H := A− , Aa := a −1 A+ , Ab := ∅, B1H := B − , Ba := ∅, and Bb := b−1 B + , then P := (K, (Ak )k∈K , (Bk )k∈K ) is a paradoxical decomposition of H . Since m(P) = n(P) = 2, we have c(P) = 4. It then follows from (c) that T (H ) = 4. Suppose now that a group G contains a subgroup isomorphic to H . Then, using (a) and (d), we have 4 ≤ T (G) ≤ T (H ) = 4. We deduce that T (G) = 4. Conversely, suppose that T (G) = 4 and let us show that the group G contains a subgroup isomorphic to H . Let P := (K, (Ak )k∈K , (Bk )k∈K ) be a paradoxical decomposition of G such that c(P) = 4, that is, m(P) = n(P) = 2. We have three possibilities according to |{k ∈ K : Ak /= ∅} ∩ {k ∈ K : Bk /= ∅}| being 0, 1, or 2. In the first case, writing {k ∈ K : Ak /= ∅} = {k1 , k2 } and {k ∈ K : Bk /= ∅} = {k3 , k4 }, we have G = k1 Ak1 ⨆ k2 Ak2 ⨆ k3 Bk3 ⨆ k4 Bk4 = Ak1 ⨆ Ak2 = Bk3 ⨆ Bk4 .

.

Then setting a := k1 k2−1 , b := k3 k4−1 , A1 := k1 Ak1 , A2 := k2 Ak2 , B1 := k3 Bk3 , and B2 := k4 Bk4 we get G = A1 ⨆ A2 ⨆ B1 ⨆ B2 = A1 ⨆ aA2 = B1 ⨆ bB2 .

.

(4.15)

Let us show that a and b freely generate a free subgroup in G. We have A1 = G \ aA2 = aA1 ⨆ aB1 ⨆ aB2 so that A1 ⊃ aA1 ⨆ aB1 ⨆ aB2 . Thus, by using induction, we obtain A1 ⊃ aA1 ⊃ a 2 A1 ⊃ . . . ⊃ (a n A1 ⨆ a n (B1 ⨆ B2 ))

.

for all integers n ≥ 1. Similarly, from A2 = G \ a −1 A1 = a −1 A2 ⨆ a −1 B1 ⨆ a −1 B2 , we deduce A2 ⊃ a −1 A1 ⊃ a −2 A1 ⊃ . . . ⊃ (a −n A1 ⨆ a −n (B1 ⨆ B2 ))

.

4.2 Exercises

239

for all integers n ≥ 1. Thus, setting A := A1 ⨆ A2 and B := B1 ⨆ B2 , we have A ∩ B = ∅ and anB ⊂ A

.

(4.16)

for all n ∈ Z \ {0}. Exchanging the roles of the Ak s and Bk s, we similarly get bn A ⊂ B for all n ∈ Z \ {0}. We may then apply Klein’s Ping-Pong lemma ([CAG, Theorem D.5.1]), and deduce that a and b generate a free subgroup of rank 2. In the second case, writing {k ∈ K : Ak /= ∅} = {k1 , k2 } and {k ∈ K : Bk /= ∅} = {k1 , k3 }, we have G = k1 Ak1 ⨆ k2 Ak2 ⨆ k1 Bk1 ⨆ k3 Bk3 = Ak1 ⨆ Ak2 = Bk1 ⨆ Bk3 .

.

Setting a := k2−1 k1 , b := k3−1 k1 , A1 := Ak1 , A2 := k1−1 k2 Ak2 , B1 := Bk1 , and B2 := k1−1 k3 Bk3 , we get again (4.15). Therefore a and b generate a free subgroup of rank 2 in G. The last possibility, namely that |{k ∈ K : Ak /= ∅} ∩ {k ∈ K : Ak /= ∅}| = 2 cannot occur. Indeed, otherwise, after taking K ' := {k ∈ K : Ak /= ∅} = {k ∈ K : Bk /= ∅}, we would get that (K ' , (Ak )k∈K ' , (Bk )k∈K ' ) is a paradoxical decomposition of G with |K ' | = 2, contradicting Exercise 4.57. (f) Let P = (K, (Ak )k∈K , (Bk )k∈K ) be a paradoxical decomposition of G. Let us show that m(P) ≥ 3. Suppose by contradiction that m(P) = 2. Then there exist k1 , k2 ∈ K such that G = k1 Ak1 ⨆ k2 Ak2 ⨆ (⨆k∈K kBk ) = Ak1 ⨆ Ak2 = ⨆k∈K Bk .

.

This gives us ⎞ ⎛ G = Ak1 ⨆ k1−1 k2 Ak2 ⨆ ⨆k∈K k1−1 kBk = Ak1 ⨆ Ak2 = ⨆k∈K Bk .

.

Setting h := k1−1 k2 , we deduce that ⎞ ⎛ Ak2 = G \ Ak1 = hAk2 ⨆ ⨆k∈K k1−1 kBk .

.

By induction, we obtain ⎞ ⎛ Ak2 ⊃ hAk2 ⊃ h2 Ak2 ⊃ · · · ⊃ hn Ak2 ⨆ ⨆k∈K hn−1 k1−1 kBk

.

for all n ≥ 1. Since h has finite order, there exists⎛an integer n0 ≥ 1⎞such that hn0 = 1G . But then, we would get Ak2 ⊃ Ak2 ⨆ ⨆k∈K hn0 −1 k1−1 kBk , forcing Bk = ∅ for all k ∈ K, a contradiction.

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4 Amenable Groups

Similarly, one proves that n(P) ≥ 3. As a consequence, c(P) = m(P) + n(P) ≥ 6. ■ Comment The notions of complexity of a paradoxical decomposition and of Tarski number of a group were introduced by Grigorchuk, de la Harpe, and the first named author in [CecGH2] and [CecGH1]. There are other definitions of Tarski numbers, see [ErsGS, Appendix A]. The result in (c), namely that for any group G there exists a finitely generated subgroup H of G satisfying T (H ) = T (G), can be interpreted as a quantitative version of the fact that a group is amenable if and only if it is locally amenable. The result in (e), namely that a group G has Tarski number T (G) = 4 if and only if G contains a free subgroup of rank 2, was proved by Jónsson, a student of Tarski, in an unpublished work from the 1940s (see the notes in Chapter 4 in Wagon’s book [Wag]). The result in (f), namely that the Tarski number of every torsion group G satisfies T (G) ≥ 6, was proved in [CecGH2], [CecGH1]. In [CecGH2] and [CecGH1], it was also shown that for the free Burnside groups B(m, n) := of rank m ≥ 2 and exponent n ≥ 665 odd, one has 6 ≤ T (B(m, n)) ≤ 14. The lower bound follows from (f) since B(m, n) is a torsion group while the proof of the upper bound involves spectral analysis and Cheeger-Buser type isoperimetric inequalities, the Grigorchuk cogrowth formula, and Adyan’s cogrowth estimates [Ady] for B(m, n). Recently, there has been some progress in the understanding of Tarski numbers. For example, it is shown in [ErsGS] (see also [OzaS]) that there are 2-generated non-amenable groups with arbitrarily large Tarski numbers, and groups with Tarski number 5 and 6. However, the problem of determining the set of integers that can occur as Tarski numbers of groups remains open. In particular, it is not known if there exists a group with Tarski number 7. The notions of paradoxical decompositions and Tarski numbers can be extended in an obvious way to the more general setting of group actions on sets. In [ErsGS], it is shown that every integer ≥ 4 is the Tarski number of some faithful transitive action of a finitely generated free group. Exercise 4.61 Let G be a group. A strong (left) paradoxical decomposition of G is a tuple Q = (I, (A'i )i∈I , (gi )i∈I ; J, (Bj' )j ∈J , (hj )j ∈J ), where I (resp. J ) is a finite index set, A'i ⊂ G (resp. Bj' ⊂ G) and gi ∈ G (resp. hj ∈ G) for all i ∈ I (resp. j ∈ J ), such that ⎞ ( ) ⎛ G = ⨆i∈I A'i ⨆ ⨆j ∈J Bj' = ⨆i∈I gi A'i = ⨆j ∈J hj Bj' .

.

The number c' (Q) := |I |+|J | is called the complexity of Q. Let T ' (G) := inf c(Q), where the infimum is taken over all strong paradoxical decompositions Q of G and the convention that T ' (G) = +∞ if G admits no strong paradoxical decompositions is used. (a) Show that G admits a paradoxical decomposition if and only if it admits a strong paradoxical decomposition. (b) Show that G is amenable if and only if it does not admit any strong paradoxical decomposition.

4.2 Exercises

241

(c) Show that T (G) = T ' (G). (d) Suppose that G is non-amenable. Let Q = (I, (A'i )i∈I , (gi )i∈I ; J, ' (Bj )j ∈J , (hj )j ∈J ) be a strong paradoxical decomposition of G such that c' (Q) = T (G) and denote by H (Q) ⊂ G the subgroup generated by the elements gi , hj , i ∈ I, j ∈ J . Show that one has T (H (Q)) = T (G). (e) Let Q = (I, (A'i )i∈I , (gi )i∈I ; J, (Bj' )j ∈J , (hj )j ∈J ) be a strong paradoxical decomposition of G. Show that there exists a strong paradoxical decomposition Q ' = (I, (A'i )i∈I , (gi )i∈I ; J, (Bj' )j ∈J , (hj )j ∈J ) such that gi = hj = 1G for some i ∈ I and j ∈ J . (f) Let m ≥ 1 and suppose that every subgroup H ⊂ G generated by m elements is amenable. Show that T (G) ≥ m + 3. (g) Recover from (f) the fact already established in Exercise 4.60(d) that one always has T (G) ≥ 4. Solution (a) Suppose first that G admits a paradoxical decomposition P = (K, (Ak )k∈K , (Bk )k∈K ). Set I := {k ∈ K : Ak /= ∅} (resp. J := {k ∈ K : Bk /= ∅}), A'i := iAi (resp. Bj' := j Bj ), and gi := i −1 ∈ K −1 ⊂ G for all i ∈ I (resp. hj := j −1 ∈ K −1 ⊂ G for all j ∈ J ). It is straightforward that Q = Q(P) := (I, (A'i )i∈I , (gi )i∈I ; J, (Bj' )j ∈J , (hj )j ∈J ) is a strong paradoxical decomposition. Note that c' (Q) = |I | + |J | = c(P). Conversely, suppose that Q = (I, (A'i )i∈I , (gi )i∈I ; J, (Bj' )j ∈J , (hj )j ∈J ) is a strong paradoxical decomposition of G. Consider the finite subset K ⊂ G defined by K := {gi−1 : i ∈ I } ∪ {h−1 j : j ∈ J }. For k ∈ K, set Ak :=



gi A'i

.

gi−1 =k

and Bk :=



.

h j Bj .

h−1 j =k

Observe that Ak = ∅ (resp. Bk = ∅) if there is no i ∈ I (resp. j ∈ J ) such that gi−1 = k (resp. h−1 j = k). Clearly P = P(Q) := (K, (Ak )k∈K , (Bk )k∈K ) is a paradoxical decomposition of G. Note that c(P) ≤ c' (Q). (b) This immediately follows from (a) since G is amenable if and only if it does not admit any paradoxical decomposition [CAG, Theorem 4.9.2]. (c) If P is a paradoxical decomposition of G, then the strong paradoxical decomposition Q(P) of G constructed in (a) is such that c' (Q(P)) = c(P). We deduce that T ' (G) ≤ T (G). Supose now that Q is a strong paradoxical decomposition of G. Then the paradoxical decomposition P = P(Q) of G constructed in (a) satisfies c(P) ≤ c' (Q). Therefore, we have T (G) ≤ T ' (G).

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4 Amenable Groups

This shows that T (G) = T ' (G). (d) Set H := H (Q). Also, for i ∈ I (resp. j ∈ J ), set A''i := A'i ∩ H (resp. '' Bj := Bj' ∩ H ) and observe that gi A''i = gi (A'i ∩ H ) = (gi A'i ) ∩ H , since gi ∈ H (resp. hj Bj'' = hj (Bj' ∩ H ) = (hj Bj' ) ∩ H , since hj ∈ H ). We deduce that ⎞ ) ⎛ ( H = ⨆i∈I A''i ⨆ ⨆j ∈J Bj'' = ⨆i∈I gi A''i = ⨆j ∈J hj Bj'' .

.

This shows that QH := (I, (A''i )i∈I , (gi )i∈I ; J, (Bj'' )j ∈J , (hj )j ∈J ) is a strong paradoxical decomposition of H . Using (c), we then have that T (H ) = T ' (H ) ≤ c' (QH ) = |I | + |J | = T ' (G) = T (G). Since T (G) ≤ T (H ) by Exercise 4.60(a), we deduce that T (H ) = T (G). gi for all i ∈ I (e) Pick i0 ∈ I (resp. j0 ∈ J ) and set gi' := gi−1 0 −1 ' := (resp. hj hj0 hj for all j ∈ J ). It is straighforward that Q ' := ' ' (I, (Ai )i∈I , (gi )i∈I ; J, (Bj' )j ∈J , (h'j )j ∈J ) is a strong paradoxical decomposition of G. Moreover gi'0 = 1G (resp. h'j0 = 1G ). (f) Since T (G) = +∞ if G is amenable, we can assume that G is non-amenable. Let Q = (I, (A'i )i∈I , (gi )i∈I ; J, (Bj' )j ∈J , (hj )j ∈J ) be a strong paradoxical decomposition of G such that |I | + |J | = T (G). By (e), we can assume that gi0 = 1G (resp. hj0 = 1G ) for some i0 ∈ I (resp. j0 ∈ J ). Then H (Q) is generated by the elements gi' , h'j , i ∈ I \ {i0 } and j ∈ J \ {j0 }. Therefore, H (Q ' ) can be generated by a set of cardinality (|I | − 1) + (|J | − 1) = |I | + |J | − 2 = T (G) − 2. As H (Q) is non-amenable by (d), we have T (G) − 2 ≥ m + 1 and hence T (G) ≥ m + 3. (g) It suffices to observe that any group G satisfies the hypothesis of (f) for m = 1 since every cyclic group is amenable. ■ Comment The result in (e) is an observation of Ozawa in a MATHOVERFLOW post [OzaS]. Ozawa also asked if it could be used to prove the existence of nonamenable groups with arbitrary large Tarski numbers. This was answered in the affirmative by Sapir (see [OzaS] and [ErsGS]).

Chapter 5

The Garden of Eden Theorem

This chapter is devoted to the Garden of Eden theorem and some of its generalizations. The homoclinicity relation for a dynamical system is introduced and pre-injectivity of maps is investigated. Topological entropy of dynamical systems with amenable acting group and entropy of subshifts over amenable groups are studied and explicitly computed for some examples including the golden mean, the even, and the Morse subshifts. The Moore and Myhill properties for subshifts are discussed. A Garden of Eden theorem for strongly irreducible subshifts of finite type over amenable groups is established. Post-surjectivity of cellular automata, a notion dual to pre-injectivity which was introduced by Capobianco, Kari, and Taati, is also investigated.

5.1 Summary 5.1.1 Interiors, Closures, and Boundaries Let G be a group. Let E and .Ω be subsets of G. The E-interior .Ω −E , the E-closure +E , and the E-boundary .∂ (Ω) of .Ω are the subsets of G defined respectively by .Ω E Ω −E := {g ∈ G : gE ⊂ Ω} =



.

Ωe−1 ,

e∈E

Ω

+E

:= {g ∈ G : gE ∩ Ω /= ∅} =



Ωe−1 = ΩE −1 ,

e∈E

∂E (Ω) := Ω +E \ Ω −E .

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 T. Ceccherini-Silberstein, M. Coornaert, Exercises in Cellular Automata and Groups, Springer Monographs in Mathematics, https://doi.org/10.1007/978-3-031-10391-9_5

243

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5 The Garden of Eden Theorem

One has (cf. [CAG, Proposition 5.4.2]) (G \ Ω)+E = G \ Ω −E and (G \ Ω)−E = G \ Ω +E .

.

(5.1)

Let A and B be sets and let .τ : AG → B G be a cellular automaton with memory set S. If two configurations .x1 , x2 ∈ AG coincide on a subset .Ω ⊂ G, then the configurations .τ (x1 ) and .τ (x2 ) coincide on .Ω −S . By using (5.1), one deduces that if .x1 and .x2 coincide on .G \ Ω then .τ (x1 ) and .τ (x2 ) coincide on .G \ Ω +S .

5.1.2 Tilings Let G be a group. Let E and .E ' be two subsets of G. A subset .T ⊂ G is called an ' .(E, E )-tiling of G provided that the sets tE, .t ∈ T , are pairwise disjoint and the sets .tE ' , .t ∈ T , cover G, in formulæ: (T1) .tU 1 E ∩ t2 E = ∅ for all .t1 , t2 ∈ T with .t1 /= t2 ; (T2) . t∈T tE ' = G. If E is non-empty and .E ' := EE −1 = {g1 g2−1 : g1 , g2 ∈ E} then, using the Zorn lemma, one shows the existence of an .(E, E ' )-tiling of G (cf. [CAG, Proposition 5.6.3]). Suppose that G is amenable. Let .F = (Fj )j ∈J be a right Følner net for G. Let E and .E ' be finite subsets of G and suppose that .T ⊂ G is an .(E, E ' )-tiling of G. For each .j ∈ J set Tj := T ∩ Fj−E = {t ∈ T : tE ⊂ Fj }.

.

Then there exists a real number .α > 0 and an index .j0 ∈ J such that |Tj | ≥ α|Fj |

.

(5.2)

for all .j ≥ j0 (cf. [CAG, Proposition 5.6.4]).

5.1.3 Pre-injective Maps Let G be a group and let A be a set. Two configurations .x1 , x2 ∈ AG are said to be almost equal if the set .{g ∈ G : x1 (g) /= x2 (g)} is finite. Clearly, being almost equal is an equivalence relation on .AG . Given a subset .X ⊂ AG and a set Z, a map .f : X → Z is called pre-injective if .f (x1 ) = f (x2 ) implies .x1 = x2 whenever .x1 , x2 ∈ X are almost equal. This amounts to saying that the restriction of f to each almost equality class is injective. It is clear that every injective map .f : X → Z is pre-injective.

5.1 Summary

245

Two distinct patterns .p1 , p2 : Ω → A with the same support .Ω ⊂ G are said to be mutually erasable for .τ if they satisfy the following condition: if .x1 , x2 ∈ AG are such that .x1 |Ω = p1 , .x2 |Ω = p2 , and .x1 |G\Ω = x2 |G\Ω , then .τ (x1 ) = τ (x2 ). It turns out that .τ is pre-injective if and only if it does not admit mutually erasable patterns [CAG, Proposition 5.5.2].

5.1.4 Garden of Eden Configurations Let G be a group and let A be a set. Let .τ : AG → AG be a cellular automaton. A configuration .x ∈ AG which is not in the image of .τ , that is, such that .x ∈ AG \ τ (AG ), is called a Garden of Eden configuration for .τ . Note that the existence of a Garden of Eden configuration for .τ is equivalent to the non-surjectivity of .τ . A pattern .p : Ω → A is called a Garden of Eden pattern for .τ if there is no configuration .x ∈ AG such that .τ (x)|Ω = p. In other words, p is a Garden of Eden pattern for .τ if every configuration extending p is a Garden of Eden configuration for .τ . The existence of a Garden of Eden pattern implies the existence of a Garden of Eden configuration. The converse is also true whenever the alphabet set A is finite [CAG, Proposition 5.1.1].

5.1.5 Entropy Let G be a group and let A be a set. For .E ⊂ G, we denote by .πE : AG → AE the canonical projection (restriction map). We thus have .πE (x) = x|E for all .x ∈ AG . Suppose now that the group G is amenable and that the alphabet A is finite. Let .F = (Fj )j ∈J be a right Følner net for G. Recall from Chap. 4 that this means that J is a directed set and that .(Fj )j ∈J is a family of nonemty finite subsets of G such that .

lim

j ∈J

|Fj g \ Fj | =0 |Fj |

for all .g ∈ G (here .| · | is used to denote cardinality of finite sets). Given a subset .X ⊂ AG , the entropy .entF (X) of X with respect to the right Følner net .F is defined by .

entF (X) := lim sup j

log |πFj (X)| |Fj |

.

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5 The Garden of Eden Theorem

We list below some general properties of entropy (cf. Proposition 5.7.2, Proposition 5.7.3, and Corollary 5.7.5 in [CAG]): entF (AG ) = log |A|; if .X ⊂ Y ⊂ AG , then .entF (X) ≤ entF (Y ); if .X ⊂ AG , then .entF (X) ≤ log |A|; if B is a finite set, .τ : AG → B G is a cellular automaton, and .X ⊂ AG , then .entF (τ (X)) ≤ entF (X); (Ent5) if .X ⊂ AG is a G-invariant subset and there exists a finite subset .E ⊂ G such that .πE (X)  AE , then .entF (X) < log |A|. (Ent1) (Ent2) (Ent3) (Ent4)

.

5.1.6 The Garden of Eden Theorem The Garden of Eden Theorem [CAG, Theorem 5.3.1] states that if G is an amenable group and A is a finite set, then a cellular automaton .τ : AG → AG is surjective if and only if it is pre-injective. In fact, we have the following more complete formulation (cf. Proposition 5.1.1, Proposition 5.5.2, and Theorem 5.8.1 in [CAG]): Theorem (Garden of Eden Theorem) Let G be an amenable group, let .F be a right Følner net for G, and let A be a finite set. Let .τ : AG → AG be a cellular automaton. Then the following conditions are equivalent: (GOE1) (GOE2) (GOE3) (GOE4) (GOE5) (GOE6)

τ is pre-injective; there exist no pairs of mutually erasable patterns for .τ ; there exist no Garden of Eden patterns for .τ ; there exist no Garden of Eden configurations for .τ ; .τ is surjective; G .entF (τ (A )) = log |A|. .

The amenability assumption on the group G in the statement of the Garden of Eden theorem cannot be removed. For instance, if .G = F2 , the free group of rank 2, there exist a cellular automaton .τ : AG → AG , with .A = {0, 1}, which is surjective (resp. pre-injective) but not pre-injective (resp. not surjective) [CAG, Sections 5.10 and 5.11]. More generally, if G is non-amenable, there exists a finite set A and a cellular automaton .τ : AG → AG which is surjective but not pre-injective [CAG, Theorem 5.12.1].

5.2 Exercises Exercise 5.1 (Homoclinic Points) Let X be a uniform space equipped with an action of a group G. One says that two points x, y ∈ X are homoclinic if they satisfy the following condition: for every entourage U of X, there exists a finite subset Ω = Ω(U ) ⊂ G such that (gx, gy) ∈ U for all g ∈ G \ Ω.

5.2 Exercises

247

(a) Show that homoclinicity is an equivalence relation on X. (b) Suppose that the uniform structure on X is induced by a metric d. Show that x, y ∈ X are homoclinic if and only if the following holds: (MH) for every ε > 0 there exists a finite subset Ω = Ω d (ε) ⊂ G such that d(gx, gy) < ε for all g ∈ G \ Ω. (c) Let Y be a uniform space equipped with an action of G and suppose that there exists a uniformly continuous G-equivariant map f : X → Y . Show that if x, y ∈ X are homoclinic then f (x), f (y) ∈ Y are homoclinic. Solution (a) Homoclinicity is a reflexive relation since every entourage of X contains the diagonal (cf. [CAG, (UNI-1)]). It is symmetric since every entourage contains a symmetric one (cf. [CAG, (UNI-3) and (UNI-4)]). Transitivity follows from the fact that if U is an entourage of X, then there is an entourage V of X such that V ◦V ⊂ U (cf. [CAG, (UNI-5)]). (b) Suppose first that x and y are homoclinic. Let ε > 0. Setting Vε := {(x1 , x2 ) ∈ X × X : d(x1 , x2 ) < ε} and Ω = Ω d (ε) := Ω(Vε ) ⊂ G we have d(gx, gy) < ε for all g ∈ G \ Ω, showing that condition (MH) is satisfied. Conversely, suppose that x, y ∈ X satisfy condition (MH). Let U be an entourage of X. Since the entourages Vε := {(x1 , x2 ) ∈ X × X : d(x1 , x2 ) < ε}, ε > 0, constitute a base for the uniform structure on X associated with the metric d (cf. [CAG, Example B.1.5]), we can find εU > 0 such that VεU ⊂ U . Setting Ω = Ω(U ) := Ω d (εU ) ⊂ G we have (gx, gy) ∈ VεU ⊂ U for all g ∈ G \ Ω, showing that x and y are homoclinic. (c) Let x, y ∈ X and suppose that they are homoclinic. Let V be an entourage of Y . Since f is uniformly continuous, the set U := (f ×f )−1 (V ) is an entourage of X. As x and y are homoclinic, there exists a finite subset Ω ⊂ G such that (gx, gy) ∈ U for all g ∈ G\Ω. Using the G-equivariance of f , we have (g(f (x)), g(f (y))) = (f (gx), f (gy)) = (f × f )(gx, gy) ∈ (f × f )(U ) ⊂ V for all g ∈ G \ Ω. As V was arbitrary, this shows that f (x) and f (y) are homoclinic. ■ Exercise 5.2 Let G be a group and let A be a set. Equip AG with its prodiscrete uniform structure and the shift action of G. (a) Show that two configurations in AG are homoclinic if and only if they are almost equal. (b) Let X ⊂ AG be a subset, let Z be a set, and let f : X → Z be a map. Show that f is pre-injective if and only if its restriction to each homoclinicity class in X is injective. Solution (a) Let x, y ∈ AG and suppose first that x and y are almost equal. This means that there exists a finite subset Ω ⊂ G such that x(g) = y(g) for all g ∈ G\Ω. Now let U ⊂ AG × AG be an entourage of AG . By definition of the prodiscrete uniform structure, there is a finite subset Λ ⊂ G such that W := {(x, y) ∈ AG × AG : x|Λ = y|Λ } ⊂ U . Observe now that (gx, gy) ∈ W ⊂ U for all g ∈ G \ ΛΩ −1 . As the set ΛΩ −1 is finite, we deduce that x and y are homoclinic.

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5 The Garden of Eden Theorem

Conversely, suppose that x and y are homoclinic. Then there exists a finite subset Ω ⊂ G such that (gx, gy) ∈ {(x, y) ∈ AG × AG : x(1G ) = y(1G )} for all g ∈ G \ Ω. This implies that x(g) = y(g) for all g ∈ G \ Ω −1 . Therefore x and y are almost equal. (b) This follows immediately from the definition of pre-injectivity and (a). ■ Exercise 5.3 Let G be a group and let A be a finite set. Let τ : AG → AG be a cellular automaton admitting a memory set M ⊂ G such that |M| = 1. Let μ : AM → A denote the associated local defining map. Show that the following conditions are equivalent: (i) (ii) (iii) (iv) (v)

τ is pre-injective; τ is injective; μ is injective μ is surjective; τ is surjective.

Solution Let m ∈ G such that M = {m}. Suppose (i). Let x, y ∈ AG such that τ (x) = τ (y). Let g ∈ G. Consider the configuration y ' ∈ AG defined by setting ⎧ y ' (h) :=

x(gm)

if h = gm

y(h)

otherwise,

.

for all h ∈ G. Observe that y ' is almost equal to y. We have τ (y ' )(g) = μ((g −1 y ' )|M ) = μ(y ' (gm)) = μ(x(gm)) = τ (x)(g) = τ (y)(g)

.

and, for k /= g, τ (y ' )(k) = μ((k −1 y ' )|M ) = μ(y ' (km)) = μ(y(km)) = τ (y)(k).

.

Thus τ (y ' ) = τ (y). Since y ' is almost equal to y and τ is pre-injective, we deduce that y ' = y. This shows that x(gm) = y ' (gm) = y(gm). Since g was arbitrary, this shows that x = y. We deduce that τ is injective. Since injectivity implies preinjectivity, this shows the equivalence (i) ⇐⇒ (ii). Suppose (ii) and let us show that μ is injective. Let a, b ∈ A such that μ(a) = μ(b) and consider the configurations x, y ∈ AG defined by setting x(g) := a for all g ∈ G and y(g) := a for all g ∈ G \ {m} and y(m) := b. Then x and y are almost equal and τ (x)(g) = μ((g −1 x)|M ) = μ(x(gm)) = μ(y(gm)) = μ((g −1 y)|M ) = τ (y)(g)

.

for all g ∈ G. Since τ is pre-injective, we deduce that x = y so that, in particular a = x(m) = y(m) = b. This shows that μ is injective. Conversely, suppose that μ

5.2 Exercises

249

is injective. Let x, y ∈ AG such that τ (x) = τ (y). For g ∈ G we have μ(x(gm)) = μ((g −1 x)|M ) = τ (x)(g) = τ (y)(g) = μ((g −1 y)|M ) = μ(y(gm))

.

so that x(gm) = y(gm), by injectivity of μ. Since g was arbitrary, it follows that x = y. This shows the equivalence (ii) ⇐⇒ (iii). The equivalence (iii) ⇐⇒ (iv) is obvious, since A is finite. Suppose (iv) and let us show that τ is surjective. Given y ∈ AG we define x ∈ AG by setting x(g) := μ−1 (y(gm−1 ))

.

for all g ∈ G. We then have τ (x)(g) = μ((g −1 x)|M ) = μ(x(gm)) = μ(μ−1 (y(gmm−1 ))) = y(g)

.

for all g ∈ G, so that τ (x) = y. This shows that τ is surjective. Conversely, suppose that μ is not surjective and let b ∈ A \ μ(A). For each a ∈ A let xa ∈ AG denote the constant configuration defined by setting xa (g) := a for all g ∈ G. Then, for all x ∈ AG we have τ (x)(1G ) = μ(x(m)) /= b = xb (1G ). This shows that xb ∈ AG \ τ (AG ). We deduce that τ is not surjective. Thus the outstanding equivalence (iv) ⇐⇒ (v) follows as well. ■ Exercise 5.4 Let G be an infinite group and let A be a finite set. Let τ : AG → AG be a non-surjective cellular automaton. Show that there are uncountably many Garden of Eden configurations in AG . Solution Let Γ := AG \ τ (AG ) denote the set of Garden of Eden configurations in AG . Since AG is compact Hausdorff and τ is continuous, the set τ (AG ) is closed in AG . As τ is not surjective, it follows that Γ is a non-empty open subset of AG . Thus if we consider a configuration x ∈ Γ , there exists a finite subset Ω ⊂ G such that Γ contains the set V (x, Ω) consisting of all configurations y ∈ AG such that x|Ω = y|Ω . Observe that A has more than one element since otherwise τ would be surjective. As the set G \ Ω is infinite, this implies that AG\Ω is uncountable. Since the map ρ : V (x, Ω) → AG\Ω , defined by ρ(y) := y|G\Ω is a bijection, we deduce that the set V (x, Ω) is uncountable. As V (x, Ω) ⊂ Γ , this implies that Γ is itself uncountable. ■ Exercise 5.5 (Life on Z) Let A := {0, 1} and consider the cellular automaton τ : AZ → AZ with memory set S := {−1, 0, 1} and local defining map μ : AS → A given by ⎧ μ(y) :=

for all y ∈ AS .

∑

y(s) = 2

1

if

0

otherwise

.

s∈S

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5 The Garden of Eden Theorem

(a) Show that τ is not pre-injective. (b) Deduce from (a) that τ is not surjective either. (c) It follows from [CAG, Proposition 5.1.1] that τ admits a Garden of Eden pattern. Check that the map p : {0, 1, 2, 3, 4, 5, 6, 7, 8} → A defined by p(0) = p(1) = p(3) = p(5) = p(8) := 1 and p(2) = p(4) = p(6) = p(7) := 0 is a Garden of Eden pattern for τ . Solution (a) Consider the configurations c0 , c1 ∈ AZ respectively defined by c0 (n) := 0 for all n ∈ Z and c1 (0) := 1 and c1 (n) := 0 for all n ∈ Z \ {0}. We have that c0 and c1 are distinct and almost equal, moreover τ (c0 ) = c0 = τ (c1 ). This shows that τ is not pre-injective. (b) Since Z is amenable and, by (a), the cellular automaton τ is pre-injective, the Garden of Eden theorem ensures us that τ is not surjective. (c) Suppose by contradiction that there exists a configuration x ∈ AZ such that the restriction of τ (x) to the set {0, 1, 2, 3, 4, 5, 6, 7, 8} is equal to the pattern p. Since p(0) = 1, we have three possibilities for x|{−1,0,1} , namely, x(−1)x(0)x(1) ∈ {011, 110, 101}. Suppose first that x(−1) = 0 and x(0) = x(1) = 1. Since p(2) = 0 and x(1) = 1, we either have x(2) = x(3) = 1 or x(2) = x(3) = 0. In the first case, τ (x)(1) = 0 /= 1 = p(1), a contradiction. In the second case, no matter x(4), we have τ (x)(3) = 0 /= 1 = p(3), a contradiction. Suppose now that x(−1) = x(0) = 1 and x(1) = 0. From the definition of p, we recursively deduce that x(2) = x(4) = x(6) = 1 and x(3) = x(5) = x(7) = x(8). Now, no matter x(9), we have τ (x)(8) = 0 /= 1 = p(8), a contradiction. Finally, suppose that x(−1) = 1, x(0) = 0, and x(1) = 1. From the definition of p, we recursively deduce that x(2) = x(3) = 1 and x(4) = x(5) = 0. Then, no matter the choice of x(6), we have τ (x)(5) = 0 /= 1 = p(5), again a contradiction. This shows that p is a Garden of Eden pattern for τ . ■ Comment Observe that τ is an elementary cellular automaton. In Wolfram’s notation, it is Rule 104. Exercise 5.6 Let G := Z, let A := {0, 1}, let S := {−1, 0, 1}, and let τ : AG → AG denote the associated majority action cellular automaton (cf. [CAG, Example 1.4.3.(c)]). Thus, one has ⎧ τ (x)(n) :=

1

if

0

if

.

∑ ∑

s∈S

x(n + s) ≥ 2

s∈S

x(n + s) ≤ 1,

for all x ∈ AG and n ∈ Z. As observed in [CAG, Example 5.2.1.(c)], the celluar automaton τ is not pre-injective so that, by amenability of Z and the Garden of Eden Thoerem, τ is not surjective either (cf. [CAG, Example 5.3.2.(c)]). It then follows from [CAG, Proposition 5.1.1] that τ admits a Garden of Eden pattern.

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Check that the map p : {1, 2, 3, 4, 5} → A defined by p(1) = p(3) = p(4) := 0 and p(2) = p(5) := 1 is a Garden of Eden pattern for τ . Solution Suppose by contradiction that p is not a Garden of Eden pattern for τ . This means that there exists a configuration x ∈ {0, 1}Z whose image y := τ (x) satisfies y(1)y(2)y(3)y(4)y(5) = 01001.

.

We cannot have x(0) = 1 since y(1) = 0 would give us x(1) = x(2) = 0 and hence y(2) = 0. Therefore x(0) = 0. We cannot have x(1) = 0 since y(2) = 1 would imply x(2) = x(3) = 1 and hence y(3) = 1. Therefore x(1) = 1. We have x(2) = 0 since x(0) = 0, x(1) = 1, and y(1) = 0. We have x(3) = 1 since x(1) = 1, x(2) = 0, and y(2) = 1. We have x(4) = 0 since x(2) = 0, x(3) = 1, and y(3) = 0. We have x(5) = 0 since x(3) = 1, x(4) = 0, and y(4) = 0. From x(4) = x(5) = 0, we deduce y(5) = 0, a contradiction. ■ Comment In Wolfram’s notation, the elementary cellular automaton τ is Rule 232 (cf. Exercise 1.35). Exercise 5.7 Let (A, +) be an abelian group (not necessarily finite). Let τ : AZ → AZ be the cellular automaton defined by τ (x)(n) := x(n − 1) + x(n) + x(n + 1) for all x ∈ AZ and n ∈ Z. Show that τ is surjective and pre-injective. Solution Let y ∈ AZ . We recursively define x ∈ AZ as follows. First we set x(0) = x(1) := 0. Then suppose that for n ≥ 2 we have defined x(n − 2) and x(n − 1), we then set x(n) := y(n−1)−x(n−1)−x(n−2). Similarly, suppose that for m ≤ 0 we have defined x(m) and x(m + 1). We then set x(m − 1) := y(m) − x(m) − x(m + 1). It is clear that τ (x) = y. This shows that τ is surjective. Suppose that x1 , x2 ∈ AZ are almost equal and satisfy τ (x1 ) = τ (x2 ). As x1 and x2 are almost equal, there exists n0 ∈ Z such that x1 (n) = x2 (n) for all n ≤ n0 . Since τ (x1 ) = τ (x2 ), we have x1 (n − 1) + x1 (n) + x1 (n + 1) = x2 (n − 1) + x2 (n) + x2 (n + 1). Using induction on n, we deduce that x1 (n + 1) = x2 (n + 1) for all n ≥ n0 . Therefore x1 = x2 , showing that τ is pre-injective. ■ Exercise 5.8 Take G := Z and A := Z/2Z = {0, 1}. Let x1 ∈ AG be the configuration defined by x1 (n) := 1 for all n ∈ Z. Let x2 ∈ AG be the configuration defined by x2 (0) := 1 and x2 (n) := 0 for all n ∈ Z \ {0}. Let f : AG → AG be the map defined by f (x2 ) := x1 and f (x) := x for all x ∈ AG \ {x2 }. Let g : AG → AG be the map defined by g(x)(n) := x(n) + x(n + 1) for all n ∈ Z and x ∈ AG . Verify that the maps f and g are both pre-injective but that the map g ◦ f is not pre-injective. Solution If x, y ∈ AG satisfy x /= y and f (x) = f (y), then (x, y) = (x1 , x2 ) or (x, y) = (x2 , x1 ). As the configurations x1 and x2 are not almost equal, we deduce that f is pre-injective. Let x0 ∈ AG be the configuration defined by x0 (n) := 0 for all n ∈ Z. Observe that g is an endomorphism of the additive group AG = (Z/2Z)Z with kernel

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5 The Garden of Eden Theorem

{x0 , x1 }. Thus if x, y ∈ AG are two distinct configurations such that g(x) = g(y), then x − y = x1 , so that x and y are not almost equal. This shows that g is preinjective. We have f (x0 ) = x0 and hence g ◦ f (x0 ) = g(x0 ) = x0 . On the other hand, we have f (x2 ) = x1 and hence g ◦ f (x2 ) = g(x1 ) = x0 . Thus g ◦ f (x0 ) = g ◦ f (x2 ). As the configurations x0 and x2 are almost equal and distinct, this shows that g ◦ f is not pre-injective. ■ Comment Observe that g is a cellular automaton but f is not (cf. Exercise 5.10). Exercise 5.9 Let G be a group. Let A and B be sets. (a) Let τ : AG → B G be a cellular automaton. Suppose that the configurations x, x ' ∈ AG are almost equal. Show that the configurations τ (x), τ (x ' ) ∈ B G are almost equal. (b) Let X ⊂ AG , Y ⊂ B G be subshifts and let σ : X → Y be a cellular automaton. Suppose that the configurations x, x ' ∈ X are almost equal. Show that the configurations τ (x), τ (x ' ) ∈ Y are almost equal. Solution (a) Let S ⊂ G be a memory set for τ . Since x and x ' are almost equal, there exists a finite subset Ω ⊂ G such that x and x ' coincide on G \ Ω. This implies that τ (x) and τ (x ' ) coincide on G \ ΩS −1 . As the set ΩS −1 is finite, we deduce that the configurations τ (x) and τ (x ' ) are almost equal. (b) Since σ : X → Y is a cellular automaton, there exists a cellular automaton τ : AG → B G such that τ (X) ⊂ Y and τ |X = σ . By (a), the configurations τ (x) and τ (x ' ) are almost equal. As σ (x) = τ (x) and σ (x ' ) = τ (x ' ), we deduce that the configurations σ (x) and σ (x ' ) are almost equal. ■ Exercise 5.10 Let G be a group and let A, B be sets. Let X ⊂ AG , Y ⊂ B G be subshifts and let Z be a set. Suppose that τ : X → Y is a pre-injective cellular automaton and that f : Y → Z is a pre-injective map. Show that the composite map f ◦ τ : X → Z is pre-injective. Solution Let x, x ' ∈ X be almost equal configurations such that (f ◦ τ )(x) = (f ◦ τ )(x ' ). By Exercise 5.9(b), the configurations τ (x) and τ (x ' ) are almost equal. As f (τ (x)) = (f ◦ τ )(x) = (f ◦ τ )(x ' ) = f (τ (x ' )) and f is pre-injective, we deduce that τ (x) = τ (x ' ). Since τ is itself pre-injective, it follows that x = x ' . This shows that f ◦ τ is pre-injective. ■ Exercise 5.11 Let G be a group and let A, B, C be sets. Let X ⊂ AG , Y ⊂ B G , and Z ⊂ C G be subshifts. Suppose that τ : X → Y and σ : Y → Z are pre-injective cellular automata. Show that the composite cellular automaton σ ◦ τ : X → Z is pre-injective. Solution This follows from Exercise 5.10.

■

Exercise 5.12 Let G be a locally finite group and let A be a set. Show that every pre-injective cellular automaton τ : AG → AG is injective.

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Solution Let τ : AG → AG be a pre-injective cellular automaton and let S ⊂ G be a memory set for τ . Let H denote the subgroup of G generated by S and consider the restriction cellular automaton τH : AH → AH associated with τ . The pre-injectivity of τ implies that of τH by [CAG, Proposition 5.2.2]. Since G is locally finite, the group H is finite. Consequently, the pre-injectivity of τH implies its injectivity. Finally, the injectivity of τH implies that of τ by [CAG, Proposition 1.7.4.(i)]. ■ Exercise 5.13 (The Garden of Eden Theorem for Locally Finite Groups) Let G be a locally finite group and let A be a finite set. Let τ : AG → AG be a cellular automaton. Show that the following conditions are all equivalent: (C1) τ is pre-injective; (C2) τ is injective; (C3) τ is surjective. Solution As every locally finite group is amenable, we already know, by the Garden of Eden theorem, that (C1) is equivalent to (C3) but we shall give an alternative proof of it. Let S ⊂ G be a memory set for τ and let H denote the subgroup of G generated by S. Consider the restriction cellular automaton τH : AH → AH associated with τ . As G is locally finite, the subgroup H is finite. Therefore the set AH is itself finite. Thus pre-injectivity, injectivity, and surjectivity are all equivalent conditions for τH . On the other hand, we know from Proposition 5.22 (resp. Proposition 1.7.4) in [CAG] that τ is pre-injective (resp. injective, resp. surjective) if and only if τ is pre-injective (resp. injective, resp. surjective). This shows that conditions (C1), (C2), and (C3) are all equivalent. ■ Exercise 5.14 (Life in a Tree) Let G := F4 denote the free group based on four generators a, b, c, d. Let A := {0, 1}. Consider the cellular automaton τ : AG → AG with memory set S := {1G , a, b, c, d, a −1 , b−1 , c−1 , d −1 }

.

and local defining map μ : AS → A given by

μ(p) :=

.

⎧ ⎪ ⎪ ⎪ ⎪ ⎨1 ⎪ ⎪ ⎪ ⎪ ⎩

0

⎧∑ ⎪ ⎪ ⎨ s∈S p(s) = 3 if or ⎪ ⎪ ⎩∑ p(s) = 4 and p(1 ) = 1, G s∈S otherwise

for all p ∈ AS (cf. [CAG, Example 1.4.3(a)]). Show that τ is surjective but not pre-injective. Solution Let x0 ∈ AG be the configuration defined by x0 (g) := 0 for all g ∈ G. Let x1 ∈ AG be the configuration defined by x1 (1G ) := 1 and x1 (g) := 0 for

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5 The Garden of Eden Theorem

all g ∈ G \ {1G }. The configurations x0 and x1 are almost equal but distinct. As τ (x0 ) = τ (x1 ), we deduce that τ is not pre-injective. Let us show now that τ is surjective. Let y ∈ AG . We can construct a configuration x ∈ AG such that y = τ (x) as follows. Consider the map ψ : G \ {1G } → G that associates with each g ∈ G \ {1G } the element of G obtained by suppressing the last factor in the reduced form of g. Thus if g ∈ G \ {1G } has length n and reduced form g = s1 s2 . . . sn , with si ∈ S \ {1G } for 1 ≤ i ≤ n and si si+1 /= 1G for 1 ≤ i ≤ n − 1, then ψ(g) = s1 s2 . . . sn−1 . Given h ∈ G, the elements of its pre-image set ψ −1 (h) are called the children of h. We construct x by induction on the length of the elements of G. We first define x(1G ) := y(0). Then we define x to take the value y(0) at 2 of the children of 1G and the value 0 at the 5 other children. Suppose now n ≥ 1 and that x(g) has been already defined for all g ∈ G with length at most n. Consider an element h ∈ G with length n. If y(h) = 0, we extend x by setting x(g) := 0 for each of the 7 children g of h. Suppose now y(h) = 1. If x(h) = 1, or if x(h) = 0 and x(ψ(h)) = 1, we set x(g) := 1 for 2 of the 7 children g of h and x(g) := 0 for the 5 others. Finally, if x(h) = x(ψ(h)) = 0, we define x(g) := 1 for 3 of the 7 children of h and x(g) := 0 for the 4 others. This defines x by induction on the whole of G. Clearly, τ (x) = y. This shows that τ is surjective. ■ Exercise 5.15 Let G be a group. Let E, F , and Ω be subsets of G. (a) Show that (Ω −E )−F = Ω −F E and (Ω +E )+F = Ω +F E . (b) Let G := Z, E := {−2, −1}, F := {1, 2, 3}, and Ω := {−2, −1, 1}. Determine the three sets ∂F (∂E (Ω)), ∂E (∂F (Ω), and ∂E+F (Ω), and check that these sets are all distinct. Solution (a) We have (Ω −E )−F =

.

    ( Ωe−1 )f −1 = Ω(f e)−1 = Ωg −1 = Ω −F E . f ∈F e∈E

f ∈F e∈E

g∈F E

Moreover, (Ω +E )+F = (ΩE −1 )F −1 = ΩE −1 F −1 = Ω(F E)−1 = Ω +F E .

.

(b) We have Ω +E = {−1, 0, 1, 2, 3} and Ω −E = {0}, so that ∂E (Ω) = \ Ω −E = {−1, 1, 2, 3}. Moreover, (∂E (Ω))+F = {−4, −3, −2, −1, 0, 1, 2} and (∂E (Ω))−F = {0}, so that ∂F (∂E (Ω)) = (∂E (Ω))+F \ (∂E (Ω))−F = {−4, −3, −2, −1, 1, 2}. We have Ω +F = {−5, −4, −3, −2, −1, 0} and Ω −F = ∅, so that ∂F (Ω) = Ω +F \ Ω −F = {−5, −4, −3, −2, −1, 0}. Moreover, (∂F (Ω))+E = {−4, −3, −2, −1, 0, 1, 2} and (∂F (Ω))−E = {−3, −2, −1, 0, 1}, so that ∂E (∂F (Ω)) = (∂F (Ω))+E \ (∂F (Ω))−E = {−4, 2}.

Ω +E

5.2 Exercises

255

Finally, we have E + F = {−1, 0, 1, 2}. Thus, Ω +(E+F ) = {−4, −3, −2, −1, 0, 1, 2}, Ω −(E+F ) = ∅, and therefore ∂E+F (Ω) = Ω +(E+F ) \ Ω −(E+F ) = {−4, −3, −2, −1, 0, 1, 2}. ■ Exercise 5.16 Let G be a group and let E ⊂ G. Determine the sets ∂E (∅) and ∂E (G). Solution We have ∅+E = ∅E −1 = ∅, so that ∂E (∅) = ∅+E \ ∅−E = ∅. two cases. Firstly, for We claim that ∂E (G) = ∅. To see this, we distinguish   E /= ∅, we have G+E = GE −1 = G and G−E = e∈E Ge−1 = e∈E G = G, so that ∂E (G) = G+E \ G−E = G \ G = ∅. On the other hand, for E = ∅, we have G+E = GE −1 = ∅, so that ∂E (G) = G+E \ G−E = ∅. This proves our claim. ■ Exercise 5.17 Let G be a group. Let E and Ω be subsets of G. Show that ∂E (G \ Ω) = ∂E (Ω). Solution Using (5.1), we get ∂E (G \ Ω) = (G \ Ω)+E \ (G \ Ω)−E

.

= (G \ Ω −E ) \ (G \ Ω +E ) = Ω +E \ Ω −E = ∂E (Ω). ■ Exercise 5.18 Let G be a group. Let E, Ω1 and Ω2 be subsets of G. (a) Show that ∂E (Ω1 ∪ Ω2 ) ⊂ ∂E (Ω1 ) ∪ ∂E (Ω2 ). (b) Give an example showing that one may have ∂E (Ω1 ∪ Ω2 ) /= ∂E (Ω1 ) ∪ ∂E (Ω2 ). Solution (a) We have (Ω1 ∪ Ω2 )+E = (Ω1 ∪ Ω2 )E −1 = Ω1 E −1 ∪ Ω2 E −1 = Ω1+E ∪   +E Ω2 and (Ω1 ∪ Ω2 )−E = e∈E (Ω1 ∪ Ω2 )e−1 = e∈E (Ω1 e−1 ∪ Ω2 e−1 ) ⊃   ( e∈E Ω1 e−1 ) ∪ ( e∈E Ω2 e−1 ) = Ω1−E ∪ Ω2−E . This gives us ∂E (Ω1 ∪ Ω2 ) = (Ω1 ∪ Ω2 )+E \ (Ω1 ∪ Ω2 )−E

.

= (Ω1+E ∪ Ω2+E ) \ (Ω1 ∪ Ω2 )−E ⊂ (Ω1+E ∪ Ω2+E ) \ (Ω1−E ∪ Ω2−E ) ⊂ (Ω1+E \ Ω1−E ) ∪ (Ω2+E \ Ω2−E ) = ∂E (Ω1 ) ∪ ∂E (Ω2 ), and hence ∂E (Ω1 ∪ Ω2 ) ⊂ ∂E (Ω1 ) ∪ ∂E (Ω2 ). (b) Let us take G := Z. For m, n ∈ Z, write [m, n] := {k ∈ Z : m ≤ k ≤ n}. Set E := [−1, 1], Ω1 := [1, 3], and Ω2 := [3, 5]. We then have Ω1+E = [0, 4], Ω1−E = {2}, Ω2+E = [2, 6], and Ω1−E = {4}, so that ∂E (Ω1 ) = [0, 2] ∪ [3, 4] and ∂E (Ω2 ) = [2, 3] ∪ [5, 6].

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5 The Garden of Eden Theorem

On the other hand, Ω1 ∪Ω2 = [1, 5], (Ω1 ∪Ω2 )+E = [0, 6], and (Ω1 ∪Ω2 )−E = ■ [2, 4], so that ∂E (Ω1 ∪ Ω2 ) = [0, 6] \ [2, 4] /= [0, 6] = ∂E (Ω1 ) ∪ ∂E (Ω2 ). Exercise 5.19 Let G be a group. Let E, Ω1 and Ω2 be subsets of G. (a) Show that ∂E (Ω1 \ Ω2 ) ⊂ ∂E (Ω1 ) ∪ ∂E (Ω2 ). (b) Give an example showing that one may have ∂E (Ω1 \ Ω2 ) /= ∂E (Ω1 ) ∪ ∂E (Ω2 ). Solution (a) We have (Ω1 \ Ω2 )+E = (Ω1 ∩ (G \ Ω2 ))+E

.

= (Ω1 ∩ (G \ Ω2 ))E −1 = Ω1 E −1 ∩ (G \ Ω2 )E −1 = Ω1+E ∩ (G \ Ω2 )+E . Using (5.1), we deduce that (Ω1 \ Ω2 )+E = Ω1+E ∩ (G \ Ω2−E ).

.

On the other hand, we have (Ω1 \ Ω2 )−E = (Ω1 ∩ (G \ Ω2 ))−E  = (Ω1 ∩ (G \ Ω2 ))e−1

.

e∈E

=



(Ω1 e−1 ∩ (G \ Ω2 )e−1 )

e∈E

⎛

=



⎞

⎛

Ω1 e−1 ∩

e∈E

=

Ω1−E



⎞ (G \ Ω2 )e−1

e∈E

∩ (G \ Ω2 )

−E

.

Using (5.1), we deduce that (Ω1 \ Ω2 )−E = Ω1−E ∩ (G \ Ω2+E ).

.

5.2 Exercises

257

As a consequence, ∂E (Ω1 \ Ω2 )

.

= (Ω1 \ Ω2 )+E \ (Ω1 \ Ω2 )−E = (Ω1+E ∩ (G \ Ω2−E )) \ (Ω1−E ∩ (G \ Ω2+E )) ⎞ ⎛ ⎞ ⎛ = (Ω1+E ∩ (G \ Ω2−E )) \ Ω1−E ∪ Ω1+E ∩ (G \ Ω2−E ) \ (G \ Ω2+E ) ⎞ ⎛ ⎞ ⎛ ⊂ Ω1+E \ Ω1−E ∪ (G \ Ω2−E ) \ (G \ Ω2+E ) = (Ω1+E \ Ω1−E ) ∪ ∂E (G \ Ω2 ) = ∂E (Ω1 ) ∪ ∂E (Ω2 ), where in the last step we used Exercise 5.17. (b) Let us take G := Z, E := {−1, 0, 1}, Ω1 := {1, 2, 3, 4}, and Ω2 := {4, 5, 6, 7}. Then, Ω1 \ Ω2 = {1, 2, 3}, (Ω1 \ Ω2 )+E = {0, 1, 2, 3, 4}, and (Ω1 \Ω2 )−E = {2}. Thus, ∂E (Ω1 \Ω2 ) = {0, 1, 3, 4}. On the other hand, (Ω1 )+E = {0, 1, 2, 3, 4, 5}, (Ω1 )−E = {2, 3}, so that ∂E (Ω1 ) = (Ω1 )+E \ (Ω1 )−E = {0, 1, 4, 5}. Similarly, ∂E (Ω2 ) = {3, 4, 7, 8}. We deduce that ∂E (Ω1 \ Ω2 ) = {0, 1, 3, 4} /= {0, 1, 3, 4, 5, 7, 8} = ∂E (Ω1 ) ∪ ∂E (Ω2 ). ■ Exercise 5.20 Let G be a group. Let E and Ω be subsets of G. Show that Ω −gE Ω −E g −1 , Ω +gE = Ω +E g −1 , and ∂gE (Ω) = ∂E (Ω)g −1 for all g ∈ G.   −1 = −1 Solution Let g ∈ G. Then Ω −gE = h∈gE Ωh e∈E Ω(ge)  −1 −1 −E −1 +gE −1 −1 = Ω g . Similarly, Ω = Ω(gE) = ΩE g −1 e∈E Ωe g +E −1 +gE −gE +E −1 Ω g . We deduce that ∂gE (Ω) = Ω \Ω = Ω g \ Ω −E g −1 +E −E −1 −1 \ Ω )g = ∂E (Ω)g . (Ω

= = = = ■

Exercise 5.21 Let G be a group. Show that a subset R ⊂ G is syndetic (cf. Exercise 3.48) if and only if there exists a finite subset S ⊂ G such that the set Ω := R −1 satisfies Ω +S = G. Solution Let R ⊂ G. By definition, the subset R is syndetic if and only if there exists a finite subset K ⊂ G such that the set Kg meets R for every g ∈ G. Taking inverses and setting S := K −1 , we see that R is syndetic if and only if there exists a finite subset S ⊂ G such that g −1 S meets Ω. This is equivalent to the condition that g −1 ∈ Ω +S for all g ∈ G. We deduce that R is syndetic if and only if there exists a finite subset S ⊂ G such that Ω +S = G. ■ Exercise 5.22 Let G be a group. Let E and Ω be subsets of G. Show that Ω ⊂ −1 (Ω +E )−E .

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5 The Garden of Eden Theorem

 −1 −1 Solution We have Ω +E = ΩE and (Ω +E )−E = e∈E (ΩE)e−1 . Let g ∈ Ω. Given e ∈ E, we have g = (ge)e−1 ∈ (ΩE)e−1 . Since e was arbitrary, we have  −1 −1 g ∈ e∈E (ΩE)e−1 = (Ω +E )−E . We deduce that Ω ⊂ (Ω +E )−E . ■ Exercise 5.23 Let G be a group and let A be a set. Let τ : AG → AG be a cellular automaton with memory set M. Let Ω be a subset of G. Show that if two −1 configurations x1 , x2 ∈ AG coincide on Ω +M then the configurations τ (x1 ) and τ (x2 ) coincide on Ω. Solution Let μ : AM → A be the local defining map for τ associated with M. −1 Suppose that x1 , x2 ∈ AG coincide on Ω +M = ΩM and let g ∈ Ω. Then we have (g −1 x1 )(m) = x1 (gm) = x2 (gm) = (g −1 x2 )(m) for all m ∈ M, that is, (g −1 x1 )|M = (g −1 x2 )|M . We deduce that τ (x1 )(g) = μ((g −1 x1 )|M ) = μ((g −1 x2 )|M ) = τ (x2 )(g).

.

As g ∈ Ω was arbitrary, this shows that τ (x1 ) and τ (x2 ) coincide on Ω.

■

Exercise 5.24 Let (A, +) be an abelian group, let k be a positive integer, and let ϕ : Ak → A be a map. Consider the map τ : AZ → AZ defined by τ (x)(n) := x(n) + ϕ(x(n + 1), x(n + 2), . . . , x(n + k))

.

for all n ∈ Z and x ∈ AZ . (a) Show that τ is a cellular automaton over the group Z and the alphabet A. (b) Show that τ is pre-injective. (c) Show that the image of τ is dense in AZ for the prodiscrete topology. (d) Show that if A is finite then τ is surjective. (e) Take A := Z, k := 1, and ϕ : A → A given by ϕ(a) := −3a for all a ∈ A, so that τ (x)(n) = x(n) − 3x(n + 1) for all n ∈ Z and x ∈ AZ . Show that the cellular automaton τ is injective but not surjective and that the image of τ is not closed in AZ for the prodiscrete topology. Solution (a) The map τ is the cellular automaton admitting S := {0, 1, 2, . . . , k} as a memory set and the map μ : AS → A, defined by μ(p) := p(0) + ϕ(p(1), p(2), . . . , p(k)) for all p ∈ AS , as the associated local defining map. (b) Suppose that x1 , x2 ∈ AZ are almost equal configurations that satisfy τ (x1 ) = τ (x2 ). Since x1 and x2 are almost equal, there exists n0 ∈ Z such that x1 (n) = x2 (n) for all n ≥ n0 . Using decreasing induction on n and the fact that τ (x1 )(n) = τ (x2 )(n), we get x1 (n) = x2 (n) for all n < n0 . Thus x1 = x2 . This shows that τ is pre-injective. (c) Let y ∈ AZ and let F be a finite subset of Z. Choose M ∈ Z such that F ⊂ (−∞, M]. Define by decreasing induction the configuration x ∈ AZ by x(n) := 0 for all n ≥ M + 1 and x(n) := y(n) − ϕ(x(n + 1), x(n + 2), . . . , x(n + k)) for all n ≤ M. We then have τ (x)(n) = y(n) for all n ≤ M, so that the configurations τ (x)

5.2 Exercises

259

and y coincide on (−∞, M] and hence on F . Thus y is in the closure of τ (AZ ). This shows that τ (AZ ) is dense in AZ for the prodiscrete topology. (d) Since the cellular automaton τ is pre-injective by (b), it is surjective if A is finite by the Garden of Eden theorem. (e) We first observe that if we equip AZ with its canonical product abelian group structure then τ is a group endomorphism of AZ . Suppose that x ∈ AZ is in the kernel of τ . This means that x(n) − 3x(n + 1) = 0 for all n ∈ Z. By induction, we get x(n) = 3k x(n + k) ∈ 3k Z for all k ∈ Z, so that x(n) = 0 for all n ∈ Z. This shows that the kernel of τ is reduced to the 0-configuration and hence that τ is injective. Consider now the constant configuration y ∈ AZ defined by y(n) := 1 for all n ∈ Z. Let us show that y is not in the image of τ . Suppose on the contrary that there exists x ∈ AZ such that y = τ (x). This means that x(n) − 3x(n + 1) = 1 for all n ∈ Z. This implies x(n) − x(n + 1) − 3(x(n + 1) − x(n + 2)) = 0 for all n ∈ Z, showing that the configuration n |→ x(n) − x(n + 1) is in the kernel of τ and therefore is identically 0. We deduce that the configuration x is constant. This leads to a contradiction. Indeed, if a ∈ Z satisfies x(n) = a for all n ∈ Z, then τ (x)(n) = −2a /= 1 = y(n). This shows that τ is not surjective. The set τ (AZ ) is dense in AZ by (c). As τ is not surjective, we deduce that τ (AZ ) is not closed in AZ for the prodiscrete topology. ■ Exercise 5.25 Let G := Z and A := {0, 1}. Verify that, among the 16 cellular automata τ : AZ → AZ with memory set S := {0, 1} ⊂ Z, there are exactly 6 of them which are surjective, 4 of them which are injective, 4 of them which are bijective, and 10 of them which are neither surjective nor injective. Solution Denote by C the set of cellular automata τ : AZ → AZ admitting S as a memory set. By the Garden of Eden theorem, a cellular automaton τ ∈ C is injective if and only if it is bijective. We denote by Csur (resp. Cbij ) the subset of C consisting of all surjective (resp. bijective) cellular automata. We have Cbij ⊂ Csur . We use a numbering for elements of C similar to that of Wolfram for elementary cellular automata (see the comments to Exercise 1.35). More precisely, for 0 ≤ k ≤ 15, we denote by τk the cellular automaton in C whose local defining map μ : AS → A satisfies k=

∑

.

μ(i, j )22i+j .

i,j ∈A

In other words, k is the integer between 0 and 15 whose 4-digits binary expansion is k = μ(1, 1)μ(1, 0)μ(0, 1)μ(0, 0)2 .

.

Consider now the involution ι = τ3 : AZ → AZ defined by ι(x)(n) := 1 − x(n) for all x ∈ AZ and n ∈ Z. It induces involutions L, R : C → C respectively defined by L(τ ) := ι ◦ τ and R(τ ) := τ ◦ ι for all τ ∈ C . Clearly, each of these

260

5 The Garden of Eden Theorem

two involutions leaves invariant Cbij and Csur . We have L(τk ) = τ15−k for all 0 ≤ k ≤ 15. On the other hand, R(τk ) = τk˜ , where 0 ≤ k˜ ≤ 15 is the integer whose 4-digits binary expansion is obtained from the 4-digits expansion of k by reading the digits in reverse order. As L and R commute, they generate a subgroup Γ of Sym(C ) isomorphic to Z/2Z ⊕ Z/2Z. The group Γ leaves each of the sets Cbij and Csur invariant. The orbit of τ ∈ C under the action of Γ is {τ, ι ◦ τ, τ ◦ ι, ι ◦ τ ◦ ι} and has cardinality 2 or 4. These orbits are O1 := {τ0 , τ15 }, O2 := {τ1 , τ7 , τ8 , τ14 }, O3 := {τ2 , τ4 , τ11 , τ13 }, O4 := {τ3 , τ12 }, O5 := {τ5 , τ10 }, and O6 := {τ6 , τ9 }. The cellular automaton τ0 is constant, so that O1 ⊂ C \ Csur . The cellular automaton τ1 is not surjective since the word 101 cannot appear in a configuration belonging to the image of τ1 . Therefore O2 ⊂ C \ Csur . The cellular automaton τ2 is not surjective since the word 11 cannot appear in a configuration belonging to the image of τ2 . Therefore O3 ⊂ C \ Csur . The cellular automaton τ12 is the identity map, so that O4 ⊂ Cbij . The cellular automaton τ10 is the one-step left shift map, so that O5 ⊂ Cbij . The cellular automaton τ6 , which is given by τ6 (x)(n) = x(n) + x(n + 1) mod 2 for all x ∈ AZ and n ∈ Z, is surjective but not injective (see [CAG, Example 3.3.8]), so that O6 ⊂ Csur \ Cbij . Thus, the set Cbij = O4 ∪ O5 = {τ3 , τ5 , τ10 , τ12 } has cardinality 4, the set Csur = O4 ∪ O5 ∪ O6 = {τ3 , τ5 , τ6 , τ9 , τ10 , τ12 } has cardinality 6, and the set C \ Csur = O1 ∪ O2 ∪ O3 = {τ0 , τ1 , τ2 , τ4 , τ7 , τ8 , τ11 , τ13 , τ14 , τ15 } has cardinality 10. ■ Exercise 5.26 Let G be an amenable group, let A be a finite set, let F be a right Følner net for G, and let X ⊂ AG be a subset. Let Ω be a finite subset of G such that 1G ∈ Ω. Show that .

lim sup

log |πFj Ω (X)| |Fj |

j

= entF (X).

Solution Since 1G ∈ Ω, we have Fj Ω ⊃ Fj and hence |πFj Ω (X)| ≥ |πFj (X)|. We deduce that .

lim sup j

log |πFj Ω (X)| |Fj |

≥ lim sup

log |πFj (X)|

j

|Fj |

= entF (X).

To prove the reverse inequality, observe that πFj Ω (X) ⊂ πFj (X) × AFj Ω\Fj . It follows that .

log |πFj Ω (X)| ≤ log |πFj (X)| + |Fj Ω \ Fj | · log |A|.

Note that, since F is a right Følner net, we have .

lim j

|Fj Ω \ Fj | = 0. |Fj |

5.2 Exercises

261

We deduce that .

lim sup

log |πFj Ω (X)| |Fj |

j

≤ lim sup j

log |πFj (X)| |Fj |

= entF (X). ■

Exercise 5.27 (Topological Entropy) Let G be an amenable group, let F = (Fj )j ∈J be a right Følner net for G, and let X be a compact topological space equipped with a continuous action of G. Suppose that U = (Ui )i∈I is an open cover of X. By compactness, U admits a finite subcover. We denote by N(U ) the minimal cardinality of a finite subcover of U . Thus, N(U ) is the smallest U integer n ≥ 0 such that there is a finite subset I0 ⊂ I of cardinality |I0 | = n with i∈I0 Ui = X. (a) Let U = (Ui )i∈I and V = (Vk )k∈K be two open covers of X. Suppose that V is finer than U , that is, for each k ∈ K, there exists i ∈ I such that Vk ⊂ Ui . Show that N (U ) ≤ N(V ). (b) Suppose that U = (Ui )i∈I is an open cover of X which forms a partition of X (i.e., such that Ui1 ∩ Ui2 = ∅ for all i1 , i2 ∈ I with i1 /= i2 ). Let I0 := {i ∈ I : Ui /= ∅}. Show that N(U ) = |I0 |. (c) Let U = (Ui )i∈I be an open cover of X and let F be a non-empty finite subset of G. Consider the family U (F ) = (Wα )α∈I F indexed by the set I F = {α : F → I } defined by Wα =



gUα(g)

.

g∈F

for all α ∈ I F . Show that U (F ) is an open cover of X and that N(U (F ) ) ≤ N (U )|F | . (d) Let U be an open cover of X. Define hF (U ) := lim sup

.

j

log N(U (Fj ) ) . |Fj |

Show that hF (U ) ≤ log N(U ). (e) The topological entropy of the dynamical system (X, G) is the quantity hF (X, G) defined by hF (X, G) = sup hF (U ),

.

U

where U ranges over all open covers of X. Suppose that G acts continuously on another compact space Y and that the dynamical systems (X, G) and (Y, G) are topologically conjugate, i.e., there is a G equivariant homeomorphism ϕ : X → Y . Show that hF (X, G) = hF (Y, G).

262

5 The Garden of Eden Theorem

(f) Let A be a finite set and let X ⊂ AG be a subshift. Show that if X is equipped with the action of G induced by the G-shift on AG , then one has hF (X, G) = entF (X).

.

Solution U (a) Let K0 ⊂ K be a finite subset such that k∈K0 Vk = X and |K0 | = N(V ). Since V is finer than U , for every k ∈ K0 there existsU i(k) ∈ I such that Vk ⊂ Ui(k) . We deduce that I0 := {i(k) : k ∈ K0 } satisfies that i∈I0 Ui = X. Consequently, | = N(V ). we have N(U ) U ≤ |I0 | ≤ |K0U (b) We have i∈I0 Ui = i∈I Ui = X so that U0 := (Ui )i∈I0 is a subcover of U . As U forms a partition of X, it is clear that U0 is the unique minimal subcover of U . This implies N(U ) = |I0 |. (c) For every α ∈ I F and any g ∈ F , the set gUα(g) is open in X since Uα(g) is open in X and the action of G on X is continuous. As any finite intersection of open subsets of a topological space is itself open, we deduce that Wα is open in X for every α ∈ I F . On the other hand, given x ∈ X, we have x ∈ Wα by taking α : F → I such that g −1 x ∈ Uα(g) for all g ∈ F . This shows that U (F ) is an open cover of X. Let U0 = (Ui )i∈I0 be a finite subcover of U such that |I0 | = N(U ). Observe (F ) (F ) that U0 is a finite subcover of U (F ) since I0F ⊂ I F . As U0 has cardinality |I0F | = N(U )|F | , we deduce that N(U (F ) ) ≤ N(U )|F | . (d) For every j ∈ J , we have log N(U (Fj ) ) ≤ |Fj | log N(U ) by (c). It follows that hF (U ) = lim sup

.

j

|Fj | log N(U ) log N(U (Fj ) ) ≤ lim sup = N(U ). |Fj | |Fj | j

(e) The fact that hF (X, G) = hF (Y, G) is clear from the definition of topological entropy. Indeed, since ϕ is a homeomorphism, it induces a one-toone correspondence ϕ∗ between open covers of X and open covers of Y , given by U = (Ui )i∈I |→ ϕ∗ (U ) := (ϕ(Ui ))i∈I . Moreover, the G-equivariance of ϕ implies that ϕ∗ (U (F ) ) = (ϕ∗ (U ))(F ) for every open cover U of X and any non-empty finite subset F ⊂ G. (f) Consider the open cover T = (Ta )a∈A of X defined by Ta := {x ∈ X : x(1G ) = a}. Since Ta ∩ Tb = ∅ for a, b ∈ A distinct, T is a partition of X. Let F ⊂ G be a non-empty finite subset. For p ∈ AF and g ∈ F , we have gTp(g) = g{x ∈ X : x(1G ) = p(g)} = {x ∈ X : x(g) = p(g)}.

.

We deduce that T (F ) = (Cp )p∈AF , where Cp :=



.

g∈F

gTp(g) = {x ∈ X : x|F = p}.

5.2 Exercises

263

Clearly, T (F ) is also a partition of X. Note that Cp /= ∅ if and only if p ∈ πF (X), where πF : AG → AF denotes the projection map. We deduce from (b) that N (T (F ) ) = |πF (X)|. Thus, hF (X, G) ≥ hF (T ) = lim sup

.

j

log |πFj (X)| log N(T (Fj ) ) = lim sup = entF (X). |Fj | |Fj | j

To prove hF (X, G) ≤ entF (X), consider an arbitrary open cover U = (Ui )i∈I of X. Let us show that there exists a finite subset Ω ⊂ G with 1G ∈ Ω such that the open cover T (Ω) is finer than U . First observe that, for every x ∈ X, there exists i(x) ∈ I and a finite subset Ω(x) ⊂ G such that x ∈ Cx|Ω(x) ⊂ Ui(x) ,

.

where, as above, Cx|Ω(x) denotes the set consisting of all configurations in X that coincide with x on Ω(x). Since the family (Cx|Ω(x) )x∈X is an U open cover of X and X is compact, there exists a finite subset X0 ⊂ X such that x∈X0 Cx|Ω(x) = X. ⎛U ⎞ Define Ω := Ω(x) ∪ {1G }. Recall that T (Ω) = (Cp )p∈AΩ . Let p ∈ AΩ x∈X0 such that Cp /= ∅ and pick y ∈ Cp . Then there exists x ∈ X0 such that y ∈ Cx|Ω(x) . As Ω(x) ⊂ Ω, we have Cp ⊂ Cx|Ω(x) ⊂ Ui(x) .

.

This shows that T (Ω) is finer than U . We claim that, for every non-empty finite subset F ⊂ G, the open cover T (F Ω) is finer than U (F ) . Indeed, T (F Ω) = (Cq )q∈AF Ω , where Cq := {x ∈ X : x|F Ω = q}. Let q ∈ AF Ω . For every g ∈ F , define qg ∈ AΩ by qg (ω) := q(gω) for all ω ∈ Ω. Since the cover T (Ω) = (Cp )p∈AΩ is finer than U = (Ui )i∈I , there exists, for every g ∈ F , an index α(g) ∈ I such that Cqg ⊂ Uα(g) . We deduce that Cq = {x ∈ X : x|F Ω = q}  {x ∈ X : (g −1 x)|Ω = qg } =

.

g∈F

=



g{x ∈ X : x|Ω = qg }

g∈F

=



gCqg

g∈F

⊂



g∈F

This proves our claim.

gUα(g) = Wα .

264

5 The Garden of Eden Theorem

By (a), we have that N(U (Fj ) ) ≤ N(T (Fj Ω) ) = |πFj Ω (X)| for every j ∈ J . It follows that hF (U ) = lim sup

.

j

log |πFj Ω (X)| log N(U (Fj ) ) ≤ lim sup = entF (X), |Fj | |Fj | j

where the last equality follows from Exercise 5.26. Since the open cover U was arbitrary, this shows that hF (X, G) ≤ entF (X). We conclude that hF (X, G) = entF (X). ■ Comment Topological entropy was first defined for continuous self-mappings of compact spaces by Adler, Konheim, and McAndrew in [AdlKM]. Given a compact space X equipped with a continuous action of an amenable group G, it can be shown, by using a result due to Ornstein and Weiss [OrnW] (see also [Gro4, Section 1.3.1], [Kri], [CecCK]), that for every open cover U of X, the quantity hF (U ) is in fact a true limit and it is independent of the choice of the right Følner net F . As a consequence, the topological entropy hF (X, G) is also independent of the choice of the right Følner net F . Exercise 5.28 Let G be an amenable group and let F = (Fj )j ∈J be a right Følner net for G. Let A and B be two finite sets and let X ⊂ AG and Y ⊂ B G be two subshifts. Suppose that the subshifts X and Y are topologically conjugate, i.e., there exists a G-equivariant homeomorphism ϕ : X → Y . Show that entF (X) = entF (Y ). Solution This immediately follows from (e) and (f) in Exercise 5.27.

■

Exercise 5.29 Let G be an amenable group, let F = (Fj )j ∈J be a right Følner net for G, and let A be a finite set. Let X be a subset of AG and let X denote the closure of X in AG for the prodiscrete topology. Show that one has entF (X) = entF (X). Solution For every finite subset F ⊂ G, if we denote by πF : AG → AF the projection map, we have πF (X) = πF (X). Indeed, πF (X) ⊂ πF (X) since X ⊂ X. On the other hand, if x ∈ X, there exists xF ∈ X such that πF (x) = πF (xF ), showing that πF (X) ⊂ πF (X). Thus, .

entF (X) = lim sup

log |πFj (X)| |Fj |

j

= lim sup

log |πFj (X)|

j

|Fj |

= entF (X). ■

Exercise 5.30 Let G be a finite group, let F = (Fj )j ∈J be a right Følner net for G, and let A be a finite set. Let X ⊂ AG be a subset. Show that .

entF (X) =

log |X| . |G|

5.2 Exercises

265

Solution Observe that X is finite since A and G are both finite. It follows from Exercise 4.51 that there exists j0 ∈ J such that Fj = G for all j ≥ j0 . As a consequence, we have |πFj (X)| = |πG (X)| = |X| for all j ≥ j0 , so that .

entF (X) = lim sup

log |πFj (X)| |Fj |

j

= lim sup j

log |X| log |X| = . |G| |G| ■

Exercise 5.31 Let G be an infinite amenable group, let F = (Fj )j ∈J be a right Følner net for G, and let A be a finite set. Let X ⊂ AG be a non-empty finite subset. Show that entF (X) = 0. Solution By Exercise 4.52, we have that limj |Fj | = ∞. As |πFj (X)| ≤ |X| for all j ∈ J , we deduce that 0 ≤ entF (X) = lim sup

.

j

showing that entF (X) = 0.

log |πFj (X)| |Fj |

≤ lim sup j

log |X| = 0, |Fj | ■

Exercise 5.32 Let G be an infinite amenable group and let F = (Fj )j ∈J be a right Følner net for G. Let A := {0, 1} and let X ⊂ AG denote the at-most-one-one subshift (cf. Exercise 1.71). Show that entF (X) = 0. Solution Let Ω ⊂ G be a finite subset and denote by πΩ : AG → AΩ the canonical projection map. A pattern p ∈ AΩ is in πΩ (X) if and only if p takes the value 1 at most once on Ω. Therefore |πΩ (X)| = |Ω|+1. As limj |Fj | = ∞ by Exercise 4.52, we deduce that ( ) log |πFj (X)| log |Fj | + 1 = lim sup = 0. . entF (X) = lim sup |Fj | |Fj | j j ■ Exercise 5.33 Let G be an amenable group, let F = (Fj )j ∈J be a right Følner net for G, and let A be a finite set. Let X and Y be two non-empty subshifts of AG . (a) Suppose that G is infinite or that either X or Y has more thant one configuration. Show that entF (X ∪ Y ) ≤ entF (X) + entF (Y ). (b) Suppose that G is finite. Let A := {0, 1} and set X := {x0 } and Y := {x1 }, where xi : G → A is the constant configuration xi (g) = i for all g ∈ G, i = 0, 1. Check that entF (X ∪ Y ) /≤ entF (X) + entF (Y ). Solution (a) For j ∈ J , let πj : AG → AFj denote the canonical projection map. We have πj (X ∪ Y ) ⊂ πj (X) ∪ πj (Y ) and hence |πj (X ∪ Y )| ≤ |πj (X) ∪ πj (Y )| ≤ |πj (X)| + |πj (Y )|.

266

5 The Garden of Eden Theorem

Suppose first that both X and Y have more than one configuration. Observe that a + b ≤ ab for all a, b ∈ [2, ∞). Indeed, to establish this inequality, we can assume by symmetry a ≤ b and we then have a + b ≤ 2b ≤ ab. We deduce that .

log(a + b) ≤ log(ab) = log(a) + log(b)

(5.3)

for all a, b ∈ [2, ∞). For all j ∈ J , we have that |πj (X)| ≥ 2 (resp. |πj (Y )| ≥ 2) since X (resp. Y ) is G-invariant with more than one configuration. By using (5.3), we deduce that .

log |πj (X ∪ Y )| ≤ log(|πj (X)| + |πj (Y )|) ≤ log |πj (X)| + log |πj (Y )|.

This gives us .

entF (X ∪ Y ) = lim sup j

≤ lim sup j

log |πj (X ∪ Y )| |Fj | log |πj (X)| log |πj (Y )| + lim sup |Fj | |Fj | j

= entF (X) + entF (Y ). This establishes the required inequality in that case. Suppose now that G is infinite and that at least one of the two subshifts X and Y , say Y , is reduced to a single configuration. We then have entF (Y ) = 0 by Exerlog(|πj (X)|+1) cise 5.31. As limj |Fj | = ∞ by Exercise 4.52, we have that lim supj = |Fj | lim supj

log |πj (X)| |Fj |

= entF (X). Keeping in mind that |πj (Y )| = 1, we deduce that

entF (X ∪ Y ) = lim sup j

≤ lim sup .

j

≤ lim sup j

log |πj (X ∪ Y )| |Fj | log(|πj (X)| + |πj (Y )|) |Fj | log(|πj (X)| + 1) |Fj |

= entF (X) = entF (X) + entF (Y ). This gives us entF (X ∪ Y ) ≤ entF (X) + entF (Y ). Note that equality holds in this case since X ⊂ X ∪ Y so that entF (X) + entF (Y ) = entF (X) ≤ entF (X ∪ Y ). (b) It follows from Exercise 5.30 that entF (X ∪ Y ) = (log |X ∪ Y |)/|G| = (log 2)/|G| and entF (X) = entF (Y ) = 0. ■ Exercise 5.34 Let G be an amenable group, let A be a finite set, let F be a right Følner net for G, and let X ⊂ AG be a subshift. Let F be a non-empty finite

5.2 Exercises

267

subset of G and consider the subshift X[F ] ⊂ B G , where B := AF , introduced in Exercise 1.95. Show that entF (X[F ] ) = entF (X). Solution It follows from Exercise 1.30 that the map ΦF : AG → B G defined therein yields a topological conjugacy from X onto X[F ] . We deduce that ■ entF (X[F ] ) = entF (X) by using Exercise 5.28. Exercise 5.35 Let G be an amenable group and let H be a finite index subgroup of G. Let T ⊂ G be a complete set of representatives for the right cosets of H in G. Let FH = (Fj )j ∈J be a right Følner net for H . (a) Show that FG := (Fj T )j ∈J is a right Følner net for G. (b) Let A be a finite set and set B := AT . Let X ⊂ AG be a subshift and consider the subshift X(H,T ) ⊂ B H (cf. Exercise 1.72). Show that entFH (X(H,T ) ) = [G : H ] entFG (X). Solution (a) This follows from Exercise 4.53 and the fact that if F = (Fj )j ∈J (resp. T ⊂ G) is a left Følner net for H (resp. a complete set of representatives for the left cosets of H for G) then (Fj−1 )j ∈J (resp. T −1 ⊂ G) is a right Følner net for H (resp. a complete set of representatives for the right cosets of H in G). (b) Recall that Y := X(H,T ) = Ψ (X), where Ψ : AG → B H is the H -equivariant uniform isomorphism defined by setting Ψ (x)(h)(t) := x(ht) for all h ∈ H and t ∈ T (cf. Exercise 1.72). For j ∈ J denote by πjH : B H → B Fj (resp. πjG : AG → AFj T ) the projection map. Given x ∈ X and setting y := Ψ (x) ∈ Y , we have πjH (y) = y|Fj = (y(h))h∈Fj ∈ B Fj . Thus, if x ' ∈ X and y ' := Ψ (x ' ) ∈ Y we have πjH (y) = πjH (y ' ) if and only if y(h) = y ' (h) for all h ∈ Fj , that is, x(ht) = x ' (ht) for all h ∈ Fj and t ∈ T . We deduce that |πjH (Y )| = |πjG (X)|. As a consequence, using (a) and keeping in mind that |T | = [G : H ] and |Fj T | = |Fj | · |T | for all j ∈ J , we have

.

entFH (Y ) = lim sup j

log |πjH (Y )| |Fj |

= lim sup

log |πjG (X)|

j

= lim sup j

|Fj |

|T | log |πjG (X)| |Fj T |

= [G : H ] entFG (X). ■

Exercise 5.36 A sequence (un )n≥1 of real numbers is said to be subadditive (resp. submultiplicative) if it satisfies un+m ≤ un + um (resp. un+m ≤ un um ) for all n, m ≥ 1. (a) Let (un )n≥1 be a subadditive⎛ sequence of real numbers such that un ≥ 0 for un ⎞ all n ≥ 1. Show that the sequence is convergent and that one has n n≥1 .

un un = inf . n→∞ n n≥1 n lim

268

5 The Garden of Eden Theorem

(b) Let (vn )n≥1 be a submultiplicative ⎛ sequence ⎞ of real numbers such that vn ≥ 1 log vn for all n ≥ 1. Show that the sequence is convergent and that one has n n≥1 .

lim

n→∞

log vn log vn = inf . n≥1 n n

(c) Let A be a finite set and let X ⊂ AZ be a non-empty subshift. Consider the Følner sequence F for Z defined by F = (Fn )n≥1 , where Fn := {0, 1, . . . , n − 1} ∗ for all n ≥ 1. For n ≥ 1, let Ln (X) ⊂ ⎞ set of words of length n that ⎛ A denote the log |Ln (X)| is convergent and that one appear in X. Show that the sequence n n≥1 has .

log |Ln (X)| log |Ln (X)| = inf . n→∞ n≥1 n n

entF (X) = lim

(5.4)

Solution (a) Let n and k be integers such that n > k ≥ 1. Then there are unique integers q ≥ 1 and r ∈ {1, . . . , k} such that n = qk + r. The subadditivity of the sequence (un ) gives us .

uqk+r un quk + ur quk ur uk max(u1 , . . . , uk ) = ≤ ≤ + ≤ + . n n n qk n k n

By letting n tend to infinity, we deduce that .

lim sup n→∞

uk un ≤ n k

for all k ≥ 1. Consequently, we have that .

lim sup n→∞

un un ≤ λ := inf . n≥1 n n

Since λ ≤ lim inf

.

n→∞

un un ≤ lim sup , n n→∞ n

it follows from inequality (5.5) that .

lim sup n→∞

This shows that the sequence

un un = lim inf = λ. n→∞ n n

⎛u ⎞ n

n

converges to λ.

(5.5)

5.2 Exercises

269

(b) This follows from (a) by taking un := log vn . (c) Let n, m ≥ 1. As X is invariant under the Z-shift, every w ∈ Ln+m (X) factorizes in the form w = uv with u ∈ Ln (X) and v ∈ Lm (X). Therefore that the⎞ sequence (|Ln (X)|)n≥1 is sub|Ln+m (X)| ≤ |Ln (X)| · |Lm (X)|, showing ⎛ log |Ln (X)| multiplicative. By (b), the sequence is convergent. Formula 5.4 n n≥1 then follows from the fact that Ln (X) is in bijection with πFn (X). ■ Comment The convergence result for subadditive sequences stated in (a) is known as Fekete’s lemma and is named after Fekete [Fek, Satz 2] (see also [PolS, p. 198]). Exercise 5.37 (Entropy of the Golden Mean and Even Subshifts) Let A := {0, 1}. Let X ⊂ AZ and Y ⊂ AZ denote the golden mean subshift and the even subshift, respectively. Consider the Følner sequence F = (Fn )n≥1 for Z given by Fn := {0, 1, . . . , n − 1}. The goal √ of the exercise is to show that entF (X) = 1+ 5 is the golden mean. entF (Y ) = log ϕ, where ϕ := 2 (a) A sequence (fn )n≥1 of real numbers is called a Fibonacci sequence if it satisfies the recurrence relation fn+2 = fn+1 + fn

.

(5.6)

for all n ≥ 1. Let (fn )n≥1 be a Fibonacci sequence and write a :=

.

f2 (ϕ − 1) + f1 (2 − ϕ) f2 ϕ − f1 (ϕ + 1) and b := . √ √ 5 5

(5.7)

Show that fn = aϕ n + b(1 − ϕ)n

.

(5.8)

for all n ∈ N. (b) Let (fn )n≥1 be a Fibonacci sequence such that f1 , f2 > 0. Show that .

lim

n→∞

log fn = log ϕ. n

(5.9)

(c) For n ≥ 1, let Ln (X) = πFn (X) ⊂ A∗ denote the set of words of length n that appear in X. Show that the sequence (|Ln (X)|)n≥1 is a Fibonacci sequence and deduce that entF (X) = log ϕ. (d) For n ≥ 1, let Ln (Y ) = πFn (Y ) ⊂ A∗ denote the set of words of length n that appear in Y . Show that the sequence (1 + |Ln (Y )|)n≥1 is a Fibonacci sequence and deduce that entF (Y ) = log ϕ.

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5 The Garden of Eden Theorem

Solution √ (a) Set ψ := 1 − ϕ = 1−2 5 and observe that ψ = −ϕ −1 . It is straightforward to check that ϕ and ψ are the roots of the equation x 2 = x + 1. As a consequence, ϕ and ψ satisfy the equations x n+2 = x n+1 + x n for all n ≥ 1. Thus, the sequences (ϕ n )n≥1 and (ψ n )n≥1 are both Fibonacci sequences. By linearity of the recurrence relation (5.6), it is clear that any linear combination of Fibonacci sequences is still a Fibonacci sequence. Thus the sequence (fn' )n≥1 , defined by fn' := aϕ n + bψ n , where a and b are as in (5.7), is a Fibonacci sequence as well. We have f1' = aϕ + bψ f2 (ϕ − 1)ϕ + f1 (2 − ϕ)ϕ f2 ϕψ − f1 (ϕ + 1)ψ + √ √ 5 5

=

2ϕ − ϕ 2 + 1 − (1 − ϕ) ϕ2 − ϕ − 1 + f2 √ √ 5 5 2ϕ − 1 = f1 √ 5 = f1

.

= f1 and f2' = aϕ 2 + bψ 2 = a + b + aϕ + bψ .

f2 (ϕ − 1) + f1 (2 − ϕ) f2 ϕ − f1 (ϕ + 1) + + f1 √ √ 5 5 2ϕ − 1 2ϕ − 1 = f2 √ + f1 − f1 √ 5 5 =

= f2 − f1 + f1 = f2 . Since any Fibonacci sequence is uniquely determined by its two first terms, we deduce that fn' = fn for all n ≥ 1, and (5.8) follows. (b) This is an immediate consequence of (5.8) which yields ⎞ ⎛ 1 ψn log fn = log ϕ + log a + b n → log ϕ . n n ϕ as n → ∞ (note that ψ n /ϕ n = (−1)n ϕ −2n → 0).

5.2 Exercises

271

(c) The set Ln (X) consists of all words of length n on the alphabet A with no two consecutive 1s. Consider the map αn : Ln (X) → Ln+2 (X) defined by αn (u) := u00 for all u ∈ Ln (X) and the map βn : Ln+1 (X) → Ln+2 (X) defined by ⎧ βn (v) :=

v1

if the word v ends by 0,

v0

if the word v ends by 1

.

for all v ∈ Ln+1 (X). The maps αn and βn are clearly injective. Moreover, their images form a partition of Ln+2 (X). As a consequence, we have |Ln+2 (X)| = |Ln+1 (X)| + |Ln (X)|. This shows that the sequence (Ln (X))n≥1 is a Fibonacci sequence and we deduce from (b) that .

entF (X) = lim

n→∞

log |Ln (X)| = ϕ. n

(d) The set Ln (Y ) consists of all words of length n on the alphabet A with an even number of 0s between any two 1s. Consider the map αn : Ln (X) → Ln+2 (X) defined by ⎧ αn (u) :=

u10

if u = 0n or u = w10h with w ∈ A∗ and h ∈ N even

u00

if u = w10h with w ∈ A∗ and h ∈ N odd

.

for all u ∈ Ln (X), and the map βn : Ln+1 (X) → Ln+2 (X) defined by ⎧ βn (v) :=

v1

if v = 0n+1 or v = w10h with w ∈ A∗ and h ∈ N even

v0

if v = w10h with w ∈ A∗ and h ∈ N odd

.

for all v ∈ Ln+1 (X). The maps αn and βn are clearly injective. Moreover, the sets {0n+2 }, αn (Ln (Y )), and βn (Ln+1 (Y ) form a partition of the set Ln+2 (Y ). As a consequence, we have |Ln+2 (Y )| = 1 + |Ln+1 (Y )| + |Ln (Y )| for all n ≥ 1. This implies that the sequence (1 + |Ln (Y )|)n≥1 is a Fibonacci sequence and we deduce from (b) that .

entF (Y ) = lim

n→∞

log(1 + |Ln (Y )|) log |Ln (Y )| = lim = ϕ. n→∞ n n ■

Comment Formula (5.8) is known as Binet’s Fibonacci number formula (or Binet’s formula, for short) after Binet. Exercise 5.38 (Entropy of the Morse Subshift) Let A := {0, 1} and let X ⊂ AZ denote the Morse subshift (cf. Exercise 3.52). For n ≥ 1 let Ln (X) ⊂ A∗ denote the set of words of length n that appear in X.

272

5 The Garden of Eden Theorem

(a) Show that |Ln (X)| ≤ 8n for all n ≥ 1. (b) Let F := (Fn )n≥1 denote the Følner sequence for Z defined by Fn := {0, 1, . . . , n − 1}. Show that entF (X) = 0. Solution (a) Let n ≥ 1 and let w ∈ Ln (X). Consider the Thue-Morse sequence T : N → A introduced in Exercise 3.51. By definition of the Morse subshift, the word w appears in T , that is, there exists i ∈ N such that w = T (i)T (i + 1) · · · T (i + n − 1). Let k ≥ 1 denote the unique integer such that 2k−1 ≤ n < 2k . Consider the words u := T (0)T (1) · · · T (2k − 1) and v := ι(u), where ι : A∗ → A∗ denotes the unique monoid automorphism such that ι(0) = 1 and ι(1) = 0. Let h ∈ N such that h2k ≤ i < (h + 1)2k . Then h2k ≤ i ≤ i + n − 1 < (h + 1)2k + 2k − 1 = h2k + 2k + 2k − 1 and it follows from Exercise 3.51(e) (with k and n therein replaced by h and k, respectively) that w is a subword of one of the words uv, vv, vu, or uu. Altogether, this gives us at most 4 · 2k = 8 · 2k−1 ≤ 8n distinct possibilities for w. Therefore |Ln (X)| ≤ 8n. (b) For all n ≥ 1, we have |Ln (X)| ≤ 8n by (a) and hence log |Ln (X)| ≤ log n + 3 log 2. It follows that .

log n log 2 log |Ln (X)| ≤ +3 , n n n

so that .

entF (X) = lim

n→∞

log |Ln (X)| = 0. n ■ AZ

Exercise 5.39 Let A := {0, 1, 2}. Let Y := {y ∈ : y(g) ∈ {1, 2} for all g ∈ Z} and X := Y ∪ {x0 }, where x0 ∈ AZ denotes the constant configuration defined by x0 (g) := 0 for all g ∈ Z. (a) Show that X and Y are subshifts of finite type of AZ and that one has the strict inclusion Y  X. (b) Show that X is not topologically transitive. (c) Show that X is not irreducible. (d) Check that X and Y have the same entropy entF (X) = entF (Y ) = log 2 with respect to the Følner sequence F = (Fn )n≥1 for Z given by Fn := {0, 1, . . . , n−1}. Solution (a) Clearly, the subset Y (resp. X) is the subshift of finite type admitting the set {11, 12, 21, 22} (resp. {00, 11, 12, 21, 22}) as a defining set of admissible words. We have Y  X since x0 ∈ / Y. (b) If x ∈ X, then the orbit closure of x is either reduced to {x0 } or contained in Y . Consequently, there is no x ∈ X whose orbit is dense in X. It follows that X is not topologically transitive (cf. Exercise 1.63(b)).

5.2 Exercises

273

(c) Since X is not topologically transitive, it is not irreducible either. For a more direct proof of the non-irreducibility of X, it suffices to observe that the one-letter words u := 0 and v := 1 both appear in X while there is no word w ∈ A∗ such that uwv appears in X. (d) For each n ≥ 1, let πn : AZ → AFn denote the projection map. We have πn (Y ) = 2n and πn (X) = 2n + 1. Therefore .

log 2n log |πn (Y )| = = log 2 |Fn | n

and .

log(2n + 1) log(1 + 2−n ) log |πn (X)| = = log 2 + |Fn | n n

both converge to log 2 as n → ∞. We deduce that entF (Y ) = entF (X) = log 2.

■

Exercise 5.40 Let G be an amenable group, let A be a finite set, and let F = (Fj )j ∈J be a right Følner net for G. Suppose that X ⊂ AG is a strongly irreducible subshift containing at least two distinct configurations. Show that one has entF (X) > 0. Solution Given a subset Ω ⊂ G, we denote by πΩ : AG → AΩ the projection map. Let x1 , x2 ∈ X such that x1 /= x2 . Then there exists g0 ∈ G such that x1 (g0 ) /= x2 (g0 ). Let Δ ⊂ G be a finite subset such that X is Δ-irreducible. We can assume 1G ∈ Δ. Let E := g0 Δ and E ' := EE −1 = {ab−1 : a, b ∈ E}. Let T ⊂ G be an (E, E ' )-tiling. For each j ∈ J , define Tj := {g ∈ T : gE ⊂ Fj }. Observe that gg0 ∈ gg0 Δ = gE for all g ∈ G, since 1G ∈ Δ. As the sets gE, g ∈ T , / g2 g0 Δ for all distinct g1 , g2 ∈ Tj . are pairwise disjoint, we deduce that g1 g0 ∈ Since X is Δ-irreducible, it follows that, for every map ε : Tj → {1, 2}, we can find a configuration x ∈ X such that x(gg0 ) = (gxε(g) )(gg0 ) for all g ∈ Tj . As (gxε(g) )(gg0 ) = xε(g) )(g0 ), this implies |πFj (X)| ≥ 2|Tj |

.

for all j ∈ J.

By(5.2), there exists a real number α > 0 and an element j0 ∈ J such that |Tj | ≥ α|Fj | for all j ≥ j0 . We deduce that .

entF (X) = lim sup j

log |πFj (X)| |Fj |

) ( log 2α|Fj | ≥ lim sup = α log 2 > 0. |Fj | j ■

Comment This is taken from [CecC3, Proposition 4.5].

274

5 The Garden of Eden Theorem

Exercise 5.41 Let G be an amenable group, let F = (Fj )j ∈J be a right Følner net for G, and let A be a finite set. Let X ⊂ AG be a strongly irreducible subshift. Given a subset Ω ⊂ G, denote by πΩ : AG → AΩ the projection map. Let Δ be a finite subset of G such that 1G ∈ Δ and X is Δ-irreducible. Let D be a finite subset of G. Set E := D +Δ and E ' := EE −1 = {ab−1 : a, b ∈ E}. Let T ⊂ G be an (E, E ' )-tiling. Suppose now that Z is a subset of X such that πgD (Z)  πgD (X)

.

for all g ∈ T .

(5.10)

(a) Write χ := |πE (X)| and Tj := {g ∈ T : gE ⊂ Fj }. Show that |πFj (Z)| ≤ (1 − χ −1 )|Tj | |πFj (X)|.

.

for all j ∈ J . (b) Deduce from (a) that entF (Z) < entF (X). Solution (cf. [CAG, Proposition 5.7.4] for X = AG ) Observe that |πgE (X)| = |πE (X)| = χ

.

for all g ∈ G,

(5.11)

since X is G-invariant. As 1G ∈ Δ, we have gD ⊂ gDΔ−1 = gD +Δ = gE ⊂ Fj for all g ∈ Tj . Let j ∈ J and consider the subset Qj ⊂ πFj (X) consisting of all paterns q ∈ πFj (X) such that q|gD ∈ πgD (Z) for all g ∈ Tj . We claim that |Qj | ≤ (1 − χ −1 )|Tj | |πFj (X)|.

.

(5.12)

To prove our claim, suppose that Tj = {g1 , g2 , . . . , gm }, where m := |Tj |. Consider, (i) for each i ∈ {0, 1, . . . , m}, the subset Qj ⊂ πFj (X) consisting of all paterns (i−1) q ∈ πFj (X) such that q|gk D ∈ πgk D (Z) for all 1 ≤ k ≤ i. Note that Q(i) j ⊂ Qj for all 1 ≤ i ≤ m. Let us show by induction on i that −1 i |Q(i) j | ≤ (1 − χ ) |πFj (X)|

.

(5.13)

(m)

for all i ∈ {0, 1, . . . , m}. This will prove (5.12) since Qj

= Qj . (i) For i = 0, we have Qj = πFj (X) so that (5.13) is trivially satisfied. Suppose (i−1) | ≤ (1 − χ −1 )i−1 |πFj (X)| has been established for now that the inequality |Qj (i−1) ⊂ some i ≤ m − 1. Consider the projection map ρi : AFj → AFj \gi E . As Qj (i−1) (i−1) (i−1) (i−1) ) × πgi E (X), we have |Qj | ≤ |ρi (Qj ) × πgi E (X)| = |ρi (Qj |· ρi (Qj |πgi E (X)|. Since |πgi E (X)| = χ by (5.11), this gives us |ρi (Qj(i−1) )| ≥ χ −1 |Qj(i−1) |.

.

(5.14)

5.2 Exercises

275

On the other hand, it follows from (5.10) that πgi D (Z)  πgi D (X). Thus, we can find a configuration x1 ∈ X such that πgi D (x1 ) ∈ / πgi D (Z). As gi D does not meet (i−1) (Fj \ gi E)Δ and X is Δ-irreducible, we can find, for each pattern p ∈ ρi (Qj ), a configuration x ∈ X such that πFj \gi E (x) = p and πgi D (x) = πgi D (x1 ) ∈ / πgi D (Y ). We deduce that (i−1) (i−1) |Qj(i−1) | − |Q(i) \ Q(i) )|. j | = |Qj j | ≥ |ρi (Qj

.

Combining this inequality with (5.14), we obtain (i−1) |Q(i) | − |ρi (Qj(i−1) )| ≤ (1 − χ −1 )|Qj(i−1) |, j | ≤ |Qj

.

which implies |Qj | ≤ (1 − χ −1 )i |πFj (X)| by our induction hypothesis. This completes the induction argument and therefore establishes (5.12). As πFj (Z) ⊂ Qj , we conclude that (i)

|πFj (Z)| ≤ |Qj | ≤ (1 − χ −1 )|Tj | |πFj (X)|.

.

(b) As entF (∅) = −∞, we can assume Z /= ∅. We then have χ ≥ 2 so that log(1 − χ −1 ) < 0. We deduce from (a) that .

log |πFj (Z)| ≤ |Tj | log(1 − χ −1 ) + log |πFj (X)|

for all j ∈ J . By (5.2), there exists a real number α > 0 and an element j0 ∈ J such that |Tj | ≥ α|Fj | for all j ≥ j0 . It follows that .

log |πFj (Z)| |Fj |

≤

log |πFj (X)| |Tj | log(1 − χ −1 ) + |Fj | |Fj |

≤ α log(1 − χ −1 ) +

log |πFj (X)| |Fj |

for all j ≥ j0 . This implies .

entF (Z) = lim sup j

log |πFj (Z)| |Fj |

≤ α log(1 − χ −1 ) + lim sup

log |πFj (X)|

j

|Fj |

= α log(1 − χ −1 ) + entF (X) and, keeping in mind that α log(1−χ −1 ) < 0, we deduce that entF (Z) < entF (X). ■

276

5 The Garden of Eden Theorem

Comment See [CecC3, Lemma 4.1]. Exercise 5.42 Let G be an amenable group, let F = (Fj )j ∈J be a right Følner net for G, and let A be a finite set. Suppose that X and Y are subshifts of AG with X strongly irreducible and Y  X. Show that entF (Y ) < entF (X). Solution Given a subset Ω ⊂ G, we denote by πΩ : AG → AΩ the projection map. As Y  X, there exists a configuration x0 ∈ X such that x0 ∈ / Y . Since Y is closed in AG , there exists a finite subset D ⊂ G such that πD (x0 ) ∈ / πD (Y ). Thus, πD (Y )  πD (X). Since X and Y are G-invariant, this implies πgD (Y )  πgD (X) for all g ∈ G. By applying the result of Exercise 5.41(b) with Z = Y , we deduce that entF (Y ) < entF (X). ■ Comment See [Fio2, Lemma 4.4] and [CecC3, Proposition 4.2]. Exercise 5.43 Let G be a group and let A be a finite set. Let Δ be a finite subset of G and let X ⊂ AG be a Δ-irreducible subshift. Suppose that (Ωi )i∈I is a family of (possibly infinite) subsets of G, indexed by a (possibly infinite) set I , such that Ωj+Δ ∩ Ωk = ∅

.

for all distinct j, k ∈ I.

(5.15)

Suppose also that (xi )i∈I is a family of configurations in X. Denote by Pf (G) the set of all finite subsets of G. For each Λ ∈ Pf (G) let X(Λ) ⊂ X denote the set consisting of all configurations in X which coincide with xi on Λ ∩ Ωi for all i ∈ I . (a) Show that X(Λ) is a closed non-empty subset of X for each Λ ∈ Pf (G). (b) Let n be a positive integer and let Λ1 , Λ2 , . . . , Λn ∈ Pf (G). Show that X(Λ1 ) ∩ X(Λ2 ) ∩ · · · ∩ X(Λn ) /= ∅. (c) Show that there exists a configuration x ∈ X such that x coincides with xi on Ωi for all i ∈ I . Solution  (a) Let Λ ∈ Pf (G). We have X(Λ) = i∈I Ci , where Ci is the subset of X consisting of all configurations that coincide with xi on Λ∩Ωi . As each Ci is closed in X, we deduce that X(Λ) is closed in X. The sets Ψi := Λ ∩ Ωi , i ∈ I , are all contained in Λ and therefore finite. On the other hand, if j, k ∈ I are distinct, we have Ψj+Δ ∩ Ψk ⊂ Ωj+Δ ∩ Ωk and hence Ψj+Δ ∩Ψk = ∅ by (5.15). We deduce from the Δ-irreducibility of X and an obvious induction argument on the cardinality of the set {Ψi : i ∈ I } that there exists x ∈ X such that x coincides with xi on Ψi for all i ∈ I , that is, x ∈ X(Λ). Therefore X(Λ) /= ∅. (b) Using (a) we have X(Λ1 ) ∩ X(Λ2 ) ∩ · · · ∩ X(Λn ) = X(Λ1 ∪ Λ2 ∪ · · · ∪ Λn ) /= ∅.

.

5.2 Exercises

277

(c) By using (a) and (b), we see that (X(Λ))Λ∈Pf (G) is a family of closed subsets of X with the finite intersection property. Since X is compact, this family has a nonempty intersection. This means that there exists a configuration x ∈ X such that x ∈ X(Λ) for every finite subset Λ ⊂ G. Clearly, such a configuration has the required properties. ■ Comment See [CecC3, Lemma 4.6]. Exercise 5.44 Let G be a group and let A be a finite set. Let X ⊂ AG be a strongly irreducible subshift. Suppose that Δ is a finite subset of G such that X is Δ-irreducible. Show that if Ω1 and Ω2 are (possibly infinite) subsets of G such that Ω1+Δ ∩ Ω2 = ∅, then, given any two configurations x1 and x2 in X, there exists a configuration x ∈ X which coincides with x1 on Ω1 and with x2 on Ω2 . Solution The solution is similar to that given above for Exercise 5.43. Let Pf (G) denote the set consisting of all finite subsets of G. For each Ω ∈ Pf (G), consider the subset X(Ω) ⊂ X consisting of all configurations in X which coincide with x1 on Ω ∩ Ω1 and with x2 on Ω ∩ Ω2 . We observe that (X(Ω))Ω∈Pf (G) is a family of closed subsets of X having the finite intersection property. By the compactness  of X, this family has a non-empty intersection. If x ∈ Ω∈Pf (G) X(Ω), then the configuration x has the required property. ■ Comment In the particular case when Δ is symmetric, that is, Δ = Δ−1 , the result can be deduced from Exercise 5.43 by taking I := {1, 2}. Exercise 5.45 Let G be a group and let A be a finite set. Let X ⊂ AG be a strongly irreducible subshift. Let x ∈ X and let CX (x) denote the set consisting of all y ∈ X such that y is almost equal to x. Show that CX (x) is dense in X. Solution Let y ∈ X and let Ω be a finite subset of G. Let us show that there exists a configuration z ∈ CX (x) which coincides with y on Ω. Let Δ be a finite subset of G such that X is Δ-irreducible. By Exercise 5.44, there exists z ∈ X such that z coincides with y on Ω and coincides with x on G \ Ω +Δ = G \ ΩΔ−1 . Such a configuration z has the required properties. This shows that CX (x) is dense in X. ■ Exercise 5.46 Let G be an amenable group and let F = (Fj )j ∈J be a right Følner net for G. Let A and B be finite sets. Suppose that X ⊂ AG and Y ⊂ B G are subshifts with X strongly irreducible and entF (Y ) < entF (X). Let τ : X → Y be a cellular automaton. For Ω ⊂ G, let πΩ : AG → AΩ and ηΩ : B G → B Ω denote the projection maps. (a) Let S be a finite subset of G that is a memory set for τ . Up to enlarging the subset S if necessary, one can also suppose that 1G ∈ S and X is S-irreducible. Show that there exists j0 ∈ J such that |η +S 2 (Y )| < |πFj0 (X)|. Fj

0

(b) Fix an arbitrary configuration x0 ∈ X and consider the finite subset Z ⊂ X consisting of all configurations z ∈ X which coincide with x0 on G \ Fj+S . Show 0 that |Z| ≥ |πFj0 (X)|.

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5 The Garden of Eden Theorem

(c) Show that τ (z) coincides with τ (x0 ) on G \ Fj+S for all z ∈ Z. 0 (d) Show that |τ (Z)| = |η +S 2 (τ (Z))|. 2

Fj

0

(e) Show that there exist distinct configurations z1 , z2 ∈ Z such that τ (z1 ) = τ (z2 ). (f) Show that τ is not pre-injective. Solution (a) By definition, we have .

entF (X) = lim sup

log |πFj (X)|

j

|Fj |

.

On the other hand, by taking Ω := S −1 S −1 , we get Fj+S = Fj Ω so that we deduce from Exercise 5.26 that 2

.

entF (Y ) = lim sup

log |ηFj Ω (Y )| |Fj |

j

log |η = lim sup j

Fj+S

2

(Y )|

|Fj |

.

As entF (Y ) < entF (X), we deduce that there exists j0 ∈ J such that log |η

Fj+S

2

(Y )|

0

.


0 such that .γ (n) ≤ cγ ' (cn) for all .n ≥ 1. If .γ < γ ' and .γ ' < γ one then writes .γ ∼ γ ' and one says that the two growth functions are equivalent. One has that .∼ is indeed an equivalence relation: we denote by ' ' .[γ ] := {γ : γ ∼ γ } the .∼-equivalence class of the growth function .γ . Moreover, given growth functions .γ1 , γ1' , γ2 , γ2' such that .γ1 ∼ γ1' , .γ2 ∼ γ2' , and .γ1 < γ2 , then ' ' .γ < γ : we then write .[γ1 ] < [γ2 ] [CAG, Proposition 6.4.3]. 1 2 A growth function .γ : N → N is said to be polynomial (resp. exponential) provided that there exists an integer .d ≥ 1 such that .γ (n) ∼ nd (resp. .γ (n) ∼ exp(n)). Let .p(x) ∈ R[x] be a polynomial of degree .d ≥ 1 such that .p(n) ≥ 0 for ' ' all .n ∈ N. Then .[p(n)] = [nd ] < [nd ] < [exp(n)] and .[nd ] /= [nd ] /= [exp(n)] for all integers .d ' > d [CAG, Examples 6.4.4]. Given two finite symmetric generating subsets .S, S ' ⊂ G, the associated growth functions are equivalent: .γS ∼ γS ' [CAG, Corollary 6.4.5]. We thus denote by G .γ (G) := [γ ] the equivalence class of the growth functions of G. It is called the S growth type of G. One says that a finitely generated group G has polynomial growth (resp. sub-exponential growth, resp. exponential growth) if .γ (G) < [nd ] for some integer .d ≥ 1 (resp. .γ (G) /= [exp(n)], resp. .γ (G) = [exp(n)]). One says that a finitely generated group has intermediate growth if it has sub-exponential growth without having polynomial growth. For instance, the free abelian group .Zd of rank d d .d ≥ 1 has polynomial growth: .γ (Z ) = [n ], while a finitely generated non-abelian free group F has exponential growth: .γ (F ) = [exp(n)] [CAG, Examples 6.4.11]. The growth rate of G with√respect to a finite symmetric subset .S ⊂ G is the n limit .λS = λG S := limn→∞ γS (n) ∈ [1, +∞) [CAG, Proposition 6.5.2]. One has .λS = 1 (resp. .λS > 1) if and only if G has sub-exponential growth (resp. exponential growth) [CAG, Proposition 6.5.4]. .

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Let .H ⊂ G be a finitely generated subgroup (resp. K be a quotient of G). Then .γ (H ) < γ (G) (resp. .γ (K) < γ (G)) and we have .γ (H ) = γ (G) (resp. .γ (K) = γ (G)) provided that H (resp. .ker(G → K)) has finite index in G [CAG, Propositions 6.6.3, 6.6.6, and 6.6.8]. Given growth functions .γ1 , γ1' , γ2 and .γ2' , the products .γ1 γ2 and .γ1' γ2' are growth functions and if .γ1 ∼ γ1' and .γ2 ∼ γ2' , then ' ' .γ1 γ2 ∼ γ γ . We thus define .[γ1 ][γ2 ] := [γ1 γ2 ] [CAG, Lemma 6.6.9]. If .G1 and 1 2 .G2 are finitely generated groups so is their direct product .G1 × G2 and one has .γ (G1 × G2 ) = γ (G1 )γ (G2 ) [CAG, Proposition 6.6.10]. All finitely generated nilpotent groups have polynomial growth [CAG, Theorem 6.8.1]. All finitely generated groups of subexponential growth are amenable [CAG, Theorem 6.11.2].

6.1.5 The Grigorchuk Group Let .Σ := {0, 1} and denote by .Σ ∗ the monoid consisting of all words over .Σ. The subgroup .G ⊂ Sym(Σ ∗ ) generated by the elements .a, b, c, d ∈ Sym(Σ ∗ ) recursively defined by setting • • • •

a(ε) := ε and .a(0w) := 1w, .a(1w) := 0w b(ε) := ε and .b(0w) := 0a(w), .b(1w) := 1c(w) .c(ε) := ε and .c(0w) := 0a(w), .c(1w) := 1d(w) .d(ε) := ε and .d(0w) := 0w, .d(1w) := 1b(w) . .

for all .w ∈ Σ ∗ , is called the Grigorchuk group [CAG, Section 2.9]. Given .n ∈ N, the set .Hn := {g ∈ G : g(w) = w for all w ∈ Σ n } is a finite-index normal subgroup of G. One has .G = H0 ⊃ H1 ⊃ H2 ⊃ · · · ⊃ Hn ⊃ Hn+1 ⊃ · · · (cf. [CAG, Proposition 6.9.4]). The subgroup .H1 has index 2 in G and is generated by the elements .b, c, d, aba, aca, ada. It is the normal closure in G of the elements b, c, and d (cf. [CAG, Proposition 6.9.4]). Let .h ∈ H1 . For every .w ∈ Σ ∗ there exist .w0 , w1 ∈ Σ ∗ with .l(w0 ) = l(w1 ) = l(w) such that .h(0w) = 0w0 and .h(1w) = 1w1 . Denote by .h0 , h1 ∈ Sym(Σ ∗ ) the maps defined by .h0 (w) := w0 and .h1 (w) := w1 . We thus have .h(0w) = 0h0 (w) and .h(1w) = 1h1 (w) for all .w ∈ Σ ∗ . Denote by .φ0 : H1 → Sym(Σ ∗ ) (resp. ∗ .φ1 : H → Sym(Σ )) the map defined by .φ0 (h) := h0 (resp. .φ1 (h) := h1 ) and by .φ : H1 → Sym(Σ ∗ ) × Sym(Σ ∗ ) the product map .φ(h) := (h0 , h1 ). From the recursive definition of the elements .a, b, c, and d, one has

.

φ(b) = (a, c), φ(c) = (a, d), φ(d) = (1G , b),

φ(aba) = (c, a), φ(aca) = (d, a), φ(ada) = (b, 1G ).

The maps .φ0 , φ1 : H1 → G are surjective group homomorphisms and the map φ : H1 → G × G is an injective group homomorphism (cf. [CAG, Proposition 6.9.7]).

.

6.1 Summary

301

The Grigorchuk group G is an infinite periodic, residually finite, finitely generated group of intermediate growth [CAG, Theorem 6.9.8, Corollary 6.9.5, and Theorem 6.9.17]. The group G is amenable since every finitely generated group of subexponential growth is amenable.

6.1.6 The Kesten-Day Characterization of Amenability E Let G be a countable group. We denote by .l2 (G) := {x ∈ RG : g∈G x(g)2 < ∞} the Hilbert space consisting of all square-summable real functions on G equipped E with the scalar product .(·, ·) defined by setting .(x, y) := g∈G x(g)y(g) for all 2 2 2 .x, y ∈ l (G). The .l -norm of an element .x ∈ l (G) is the non-negative number 1 ( ) E 1 2 2 2 .||x|| := (x, x) 2 = g∈G x(g)y(g) . A linear map .T : l (G) → l (G) is said to be bounded (or continuous) provided that the quantity ||T x|| , x∈l2 (G) ||x||

||T || := sup ||T x|| = sup

.

x∈l2 (G) ||x||≤1

x/=0

called the norm of T , is finite. Let .T : l2 (G) → l2 (G) be a bounded linear operator. The real spectrum of T is the closed subset σ (T ) := {λ ∈ R : (T − λ Idl2 (G) ) is not bijective}

.

and one has .σ (T ) ⊂ [−||T ||, ||T ||] (cf. [CAG, Proposition I.3.2]). Suppose that G is finitely generated and let .S ⊂ G be a finite (not necessarily symmetric) generating subset. The map .MS : l2 (G) → l2 (G) defined by setting MS (x)(g) :=

.

1 E x(gs) |S| s∈S

for all .x ∈ l2 (G) and .g ∈ G, is called the Markov operator associated with S. It is linear and bounded, in fact .||MS || ≤ 1 [CAG, Theorem 6.12.1]. The map ) ( (2) (2) 2 2 .Δ S : l (G) → l (G), defined by setting .ΔS := |S| Idl2 (G) −MS , so that (2)

ΔS (x)(g) = |S|x(g) −

.

E

x(gs)

s∈S

for all .x ∈ l2 (G) and .g ∈ G, is linear and bounded, and it is called the .l2 -Laplace operator associated with S. The following theorem yields a spectral characterization of amenability for finitely generated groups [CAG, Theorem 6.12.9].

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Theorem (Kesten-Day Characterization of Amenability) Let G be a finitely generated group and let .S ⊂ G be a finite (not necessarily symmetric) generating subset. The following conditions are equivalent: (i) (ii) (iii) (iv)

||MS || = 1; 1 ∈ σ (MS ); (2) .0 ∈ σ (Δ S ); G is amenable. . .

6.1.7 Quasi-Isometries Let G and H be two groups. A uniform embedding of G into H is a map .ϕ : G → H satisfying the following conditions: (UE-1) for every finite subset .K ⊂ G the set .F = F (K, ϕ) ⊂ H defined by F := {ϕ(g1 )−1 ϕ(g2 ) : g1 , g2 ∈ G such that g1−1 g2 ∈ K}

.

is finite; (UE-2) for every finite subset .F ⊂ H the set .K = K(F, ϕ) ⊂ G defined by K := {g1−1 g2 : g1 , g2 ∈ G such that ϕ(g1 )−1 ϕ(g2 ) ∈ F }

.

is finite. A uniform embedding .ϕ : G → H is called a quasi-isometry from G to H if there exists a finite subset .C ⊂ H such that .ϕ(G)C = H . If there exists a quasi-isometry from G to H , one says that the groups G and H are quasi-isometric. Compositions of uniform embeddings (resp. quasi-isometries) are uniform embeddings (resp. quasi-isometries) [CAG2, Proposition 6.13.7]. Moreover, the quasi-isometry relation is an equivalence relation on the class of groups. If two groups G and H are quasi-isometric, then G is finitely generated if and only if H is finitely generated [CAG2, Proposition 6.13.13]. In other words, the property for a group of being finitely generated or not is a quasi-isometry invariant. The property for a group of being amenable or not is also a quasi-isometry invariant [CAG2, Theorem 6.13.19]. Suppose that the groups G and H are finitely generated and let .dG and .dH denote the word metric associated with two finite symmetric generating subsets .SG ⊂ G and .SH ⊂ H . One says that a map .ϕ : G → H is a quasi-isometric embedding if there exist constants .α ≥ 1 and .β ≥ 0 such that .

1 dG (g1 , g2 ) − β ≤ dH (ϕ(g1 ), ϕ(g2 )) ≤ αdG (g1 , g2 ) + β α

(6.1)

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for all .g1 , g2 ∈ G. This definition does not depend on the choices of .SG and SH since the metrics associated with two finite symmetric generating subsets of a finitely generated group are Lipschitz-equivalent. Moreover, the term .β appearing in the right-hand side of the second inequality in (6.1) can be omitted, this leads to an equivalent definition. Every quasi-isometric embedding .ϕ : G → H is a uniform embedding but the converse is false in general. A map .ϕ : G → H is a quasiisometry if and only if it is a quasi-isometric embedding and there exists a constant .δ ≥ 0 such that for every .h ∈ H there exists .g ∈ G such that .dH (ϕ(g), h) ≤ δ. The growth type of a finitely generated group is an invariant of quasi-isometry [CAG2, Corollary 6.14.10]. .

6.2 Exercises Exercise 6.1 Let S = {s1 , s2 , . . . , sn } be a finite subset of Z. Show that S generates Z if and only if gcd(s1 , s2 , . . . , sn ) = 1. Solution Suppose first that S generates Z. This means in particular that 1 ∈ Z can be written in the form 1 = m1 s1 +m2 s2 +· · ·+mn sn for some m1 , m2 , . . . , mn ∈ Z. This equality implies gcd(s1 , s2 , . . . , sn ) = 1. Conversely, suppose that gcd(s1 , s2 , . . . , sn ) = 1. Then, by Bézout’s theorem, there exist u1 , u2 , . . . , un ∈ Z such that u1 s1 + u2 s2 + · · · + un sn = 1. We deduce that every z ∈ Z satisfies z = (zu1 )s1 + (zu2 )s2 + · · · + (zun )sn , showing that S generates Z. O Exercise 6.2 Let G be a finitely generated group and let S be a finite symmetric generating subset of G. Let x ∈ G and denote by Lx , Rx : G → G the maps defined by Lx (g) := xg and Rx (g) := gx for all g ∈ G. (a) Show that the map g |→ dS (g, Rx (g)) is constant on G. (b) Show that if x is in the center of G, then Rx is an isometry of (G, dS ). (c) Show that one has supg∈G dS (g, Lx (g)) < ∞ if and only if the conjugacy class of x in G is finite. Solution (a) For every g ∈ G, we have that dS (g, Rx (g)) = dS (g, gx) = lS (g −1 gx) = lS (x) is independent of g. (b) Suppose now that x is in the center of G. Then for all g1 , g2 ∈ G we have dS (Rx (g1 ), Rx (g2 )) = dS (g1 x, g2 x) = dS (xg1 , xg2 ) = lS ((xg1 )−1 xg2 ) = lS (g1−1 x −1 xg2 ) = lS (g1−1 g2 ) = ds (g1 , g2 ). It follows that Rx is an isometry. (c) Let g ∈ G. Then dS (g, Lx (g)) = dS (g, xg) = lS (g −1 xg). Thus, denoting by C (x) the conjugacy class of x in G, we have that {dS (g, Lx (g)) : g ∈ G} = {lS (c) : c ∈ C (x)}.

.

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6 Finitely Generated Groups

Suppose first that C (x) is finite. Then supg∈G dS (g, Lx (g)) = maxc∈C (x) lS (c) is finite. Conversely, suppose that supg∈G dS (g, Lx (g)) =: n < ∞. Then lS (c) ≤ n for all c ∈ C (x), so that C (x) ⊂ BS (n). As the ball BS (n) is finite, this implies that C (x) is itself finite. O Exercise 6.3 Suppose that S and S ' are two finite symmetric generating subsets of a group G with S ⊂ S ' . (a) Show that one has lS ' (g) ≤ lS (g) for all g ∈ G. (b) Show that one has dS ' (g, h) ≤ dS (g, h) for all g, h ∈ G. (c) Show that BS (n) ⊂ BS ' (n) for all n ∈ N. (d) Show that γS (n) ≤ γS ' (n) for all n ∈ N. (e) Show that λS ≤ λS ' . Solution (a) Let g ∈ G with lS (g) = n. Then there exist s1 , s2 , . . . , sn ∈ S such that g = s1 s2 · · · sn . As S ⊂ S ' , we have s1 , s2 , . . . , sn ∈ S ' , so that lS ' (g) ≤ n = lS (g). (b) Using (a), we get dS ' (g, h) = lS ' (g −1 h) ≤ lS (g −1 h) = dS (g, h) for all g, h ∈ G. (c) Suppose that g ∈ BS (n) for some n ∈ N. This means that lS (g) ≤ n. Using (a), we get lS ' (g) ≤ lS (g) ≤ n, showing that g ∈ BS ' (n). This shows that BS (n) ⊂ BS ' (n). (d) Using (c), we get γ√S (n) = |B√ S (n)| ≤ |BS ' (n)| = γS ' (n). (e) Using (d), we get n γS (n) ≤ n γS ' (n) for all n ∈ N and hence R R n γS (n) ≤ lim n γS ' (n) = λS ' . .λS = lim n→∞

n→∞

O Exercise 6.4 Let G1 and G2 be two finitely generated groups and let G := G1 ×G2 . Let S1 (resp. S2 ) be a finite symmetric generating subset of G1 (resp. G2 ). Show that S := (S1 × {1G2 }) ∪ ({1G1 } × S2 ) is a finite symmetric generating subset of G and G1 G2 that one has lG S (g) = lS1 (g1 ) + lS2 (g2 ) for all g = (g1 , g2 ) ∈ G. Solution We first observe that since S1 ⊂ G1 (resp. S2 ⊂ G2 ) is symmetric, we have S1 = S1−1 (resp. S2 = S2−1 ). As a consequence, S −1 = ( )−1 (S1 × {1G2 }) ∪ ({1G1 } × S2 ) = (S1 × {1G2 })−1 ∪ ({1G1 } × S2 )−1 = (S1−1 × {1G2 }−1 ) ∪ ({1G1 }−1 × S2−1 ) = (S1 × {1G2 }) ∪ ({1G1 } × S2 ) = S. This shows that S is symmetric. Let g = (g1 , g2 ) ∈ G = G1 × G2 . Since S1 (resp. S2 ) is a finite symmetric generating subset of G1 (resp. G2 ) we can find s1,1 , s1,2 , . . . , s1,l1 ∈ S1 (resp. G2 1 s2,1 , s2,2 , . . . , s2,l2 ∈ S2 ), where l1 := lG S1 (g1 ) (resp. l2 := lS2 (g2 )) such that g1 = s1,1 s1,2 · · · s1,l1 (resp. g2 = s2,1 s2,2 · · · s2,l2 ). We deduce that g = (g1 , g2 )

.

= (s1,1 , 1G2 )(s1,2 , 1G2 ) · · · (s1,l1 , 1G2 ) · (1G1 , s2,1 )(1G1 , s2,2 ) · · · (1G1 , s2,l2 ),

6.2 Exercises

305

G1 G2 showing that S generates G and lG S (g) ≤ l1 + l2 = lS1 (g1 ) + lS2 (g2 ). To show the reverse inequality, set l := lG S (g) and let s1 , s2 , . . . , sl ∈ S such that g = s1 s2 · · · sl . Each such si ∈ S is of the form either si = (s1,i , 1G2 ) or si = (1G1 , s2,i ) for some unique s1,i ∈ S1 or s2,i ∈ S2 . Accordingly, we partition the set I := {1, 2, . . . , l} by setting I1 := {i ∈ I : si = (s1,i , 1G2 )} and I2 := I \ I1 = {i ∈ I : si = (1G1 , s2,i )}. || || 1 We deduce that g1 = i∈I1 s1,i and g2 = i∈I2 s2,i . As a consequence, lG S1 (g1 ) ≤ 2 |I1 | and lG S2 (g2 ) ≤ |I2 | so that

G2 G 1 lG S1 (g1 ) + lS2 (g2 ) ≤ |I1 | + |I2 | = |I | = l = lS (g).

.

G1 G2 This proves that lG S (g) = lS1 (g1 ) + lS2 (g2 ).

O

Exercise 6.5 (Direct Product of Labeled Graphs) Let S1 and S2 be two sets. For i = 1, 2, let Gi = (Qi , Ei ) be an Si -labeled graph. We define their direct product G1 × G2 as the S-labeled graph G = (Q, E) with (1) S := S1 U S2 , where U denotes a disjoint union; (2) Q := Q(1 × Q2 ; ) (3) E := { (q1 , q2 ), s, (q1' , q2' ) : either q1 = q1' and (q2 , s, q2' ) ∈ E2 , or q2 = q2' and (q1 , s, q1' ) ∈ E1 }. Suppose that S1 (resp. S2 ) is endowed with an involution ι1 : S1 → S1 (resp. ι2 : S2 → S2 ) and that G1 (resp. G2 ) is edge-symmetric with respect to ι1 (resp. ι2 ). Denote by ι : S → S the map defined by ι(s) := ιi (s) if s ∈ Si , i = 1, 2, and observe that ι is an involution. Show that G is edge-symmetric with respect to ι. Solution Let s ∈ S and let i ∈ {1, 2} such that s ∈ Si . As ιi (s) ∈ Si and ιi is an involution, we have ι2 (s) = ι(ιi (s)) = ι2i (s) = s. This shows that ι is ( an involution. ) Let e = (q1 , q2 ), s, (q1' , q2' ) ∈ E. If q1 = q1' and e2 := (q2 , s, q2' ) ∈ E2 , ( ) so that s ∈ S2 , then e−1 = (q1' , q2' ), ι(s), (q1 , q2 ) satisfies that q1' = q1 and (q2' , ι(s), q2 ) = (q2' , ι2 (s), q2 ) = e2−1 ∈ E2 . If q2 = q2' and e1 := (q1 , s, q1' ) ∈ E1 , ( ) so that s ∈ S1 , then e−1 = (q1' , q2' ), ι(s), (q1 , q2 ) satisfies that q2' = q2 and (q1' , ι(s), q1 ) = (q1' , ι1 (s), q1 ) = e1−1 ∈ E1 . In either case, e−1 ∈ E. This shows that G is edge-symmetric with respect to ι. O Exercise 6.6 Let G1 and G2 be two finitely generated groups and let S1 ⊂ G1 and S2 ⊂ G2 be two finite and symmetric generating subsets such that 1G1 ∈ / S1 and 1G2 ∈ / S2 . Consider the direct product group G := G1 × G2 together with the finite symmetric generating subset S := (S1 × {1G2 }) ∪ ({1G1 } × S2 ). Denote by CS1 (G1 ), CS2 (G2 ) and CS (G) the corresponding Cayley graphs. If we identify S1 with S1 × {1G2 } (resp. S2 with {1G1 } × S2 ), we may regard CS1 (G1 ) (resp. CS2 (G2 )) as an (S1 × {1G2 })-labeled graph (resp. ({1G1 } × S2 )-labeled graph). Show that CS (G) = CS1 (G1 ) × CS2 (G2 ).

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6 Finitely Generated Groups

Solution With the notation in Exercise 6.5 with Gi := CSi (Gi ), i = 1, 2, and G := CS (G), we have (1) S = (S1 × {1G2 }) ∪ ({1G1 } × S2 ) = S1 U S2 ; (2) Q = G( = G1 × G2 = Q1 )× Q2 ; (3) E = { (q1 , q2 ), s, (q1' , q2' ) ∈ Q × S × Q: either q1 = q1' and q2' = q2 s for some s ∈ S2 (equivalently, (q2 , s, q2' ) ∈ E2 ), or q2 = q2' and q1' = q1 s for some s ∈ S1 (equivalently, (q1 , s, q1' ) ∈ E1 )}. Thus, CS (G) equals the direct product CS1 (G1 ) × CS2 (G2 ).

O

S'

Exercise 6.7 Let G := Z, let S := {1, −1}, and let := {2, −2, 3, −3}. Find the best possible positive constants C1 and C2 such that C1 lS (g) ≤ lS ' (g) ≤ C2 lS (g) for all g ∈ G. Solution Observe that S and S ' are both generating sets for Z by Exercise 6.1. Let g ∈ Z. Clearly lS (g) = |g| and lS ' (g) = min{|m| + |n| : (m, n) ∈ Z2 and g = 2m + 3n}. We have g = −2g + 3g, so that lS ' (g) ≤ | − g| + |g| = 2|g| = 2lS (g). Thus, by taking C2 := 2, we have lS ' (g) ≤ C2 lS (g) for all g ∈ Z. As 1 = −2 + 3 and 1 ∈ / S ' ∪ {0}, we see that lS ' (1) = 2. Since lS (1) = 1, this shows that the constant C2 = 2 is optimal. If m, n ∈ Z satisfy g = 2m + 3n, then lS (g) = |g| = |2m + 3n| ≤ 2|m| + 3|n| ≤ 3(|m| + |n|), so that lS (g) ≤ 3lS ' (g). Therefore, C1 := 1/3 satisfies C1 lS (g) ≤ lS ' (g) for all g ∈ Z. This constant C1 is optimal since lS (3) = 3 and lS ' (3) = 1. O Exercise 6.8 (Growth of Zd ) Let G := Zd , where d ≥ 1 is an integer. Consider the finite symmetric generating subset S ⊂ G defined by S := {±(1, 0, 0, . . . , 0), ±(0, 1, 0, . . . , 0), . . . , ±(0, 0, . . . , 0, 1)} ⊂ Zd .

.

(a) Show that if g = (a1 , a2 , . . . , ad ) ∈ Zd then lS (g) = |a1 | + |a2 | + · · · + |ad |. (b) Let n ∈ N. Set P0 (n) := 1 and, for all integers t ≥ 1 denote by Pt (n) the number of distinct t-tuples (a1 , a2(, .). . , at ) of positive integers such that a1 + a2 + · · · + at ≤ n. Show that Pt (n) = nt for 1 ≤ t ≤ n. (c) For n, t ∈ N and tE≥ 1 denote by Nt (n) the number of all d-tuples (a1 , a2 , . . . , ad ) ∈ Zd with di=1 |ai | ≤ n and exactly t many of the ai ’s nonzero. E d Show that γSZ (n) = dt=0 Nt (n). (d) Let 0 ≤ t ≤ n and let I be a subset of {1, 2, . . . , n} such that |I | = t. Show that there are precisely Pt (n) distinct elements g = (a1 , a2 , . . . , ad ) ∈ Nm with I = {i : ai > 0} and such that lS (g) ≤ n. ( )( ) (e) Deduce from (d) that there are exactly dt nt elements g = (a1 , a2 , . . . , ad ) ∈ Nm with |{i : ai > 0}| = t such that ( )(lS) (g) ≤ n. (f) Deduce from (e) that Nt (n) = 2t dt nt . ( )( ) E d (g) Deduce from (f) and (c) that γSZ (n) = dt=0 2t dt nt .

6.2 Exercises

307

Solution (a) For i = 1, 2, . . . , d, let Gi := Z and Si := {±1}. Observe that lSi (ai ) = |ai | for all ai ∈ Gi and i = 1, 2, . . . , d. Moreover, S = Udi=1 {1G1 } × · · · × {1Gi−1 } × Si × {1Gi+1 } × · · · × {1Gd }.

.

It then follows from an inductive argument applied to Exercise 6.4 that G1 G2 Gm lG S (g) = lS1 (a1 ) + lS2 (a2 ) + · · · + lSm (am ) = |a1 | + |a2 | + · · · + |ad |

.

for all g = (a1 , a2 , . . . , ad ) ∈ G. (b) For 1 ≤ t ≤ n set A (t, n) := {(a1 , a2 , . . . , at ) ∈ Nt : ai ≥ 1 and a1 + a2 + · · · + at ≤ n}

.

and C (t, n) := {B ⊂ {1, 2, . . . , n} : |B| = t}.

.

We claim that the map (a1 , a2 , . . . , at ) |→ {a1 , a1 + a2 , . . . , a1 + a2 + · · · + at } establishes a bijection between A (t, n) and C (t, n). Indeed, we first observe that since ai ≥ 1 for all i = 1, 2, . . . , t, we have a1 < a1 + a2 < · · · < a1 + a2 + · · · + at , so that the t elements a1 , a1 + a2 , . . . , a1 + a2 + · · · + at are all distinct. Moreover, since a1 + a2 + · · · + at ≤ n, they all belong to {1, 2, . . . , n}. Thus, {a1 , a1 + a2 , . . . , a1 + a2 + · · · + at } ∈ C (t, n). The map is injective: indeed, if (a1 , a2 , . . . , at ), (a1' , a2' , . . . , at' ) ∈ A (t, n) have the same image in C (t, n), one recursively finds a1 = a1' , a2 = a2' , . . . , at = at' . The map is surjective: given B ∈ C (t, n) and denoting by b1 , b2 , . . . , bt its elements, written in increasing order (bi < bj if i < j ), and setting a1 := b1 and ai := bi − bi−1 for i = 2, 3, . . . , t, we have ai ≥ 1 for all i = 1, 2, . . . , t, a1 + a2 + · · · + at = b1 + (b2 − b1 ) + · · · + (bt − bt−1 ) = bt ≤ n, and (a1 , a2 , . . . , at ) |→ B. We deduce that Pt (n) = |A (t, n)| = |C (t, n)| =

.

( ) n t

for all 1 ≤ t ≤ n. (c) For g = (a1 , a2 , . . . , ad ) ∈ Zd , we set t (g) := |{i ∈ {1, 2, . . . , d} : ai /= 0}|. Note that 0 ≤ t (g) ≤ d. Keeping in mind (a), we then have γSZ (n) = |BSZ (n)| = |{(a1 , a2 , . . . , ad ) ∈ Zd : d

d

d E

.

|ai | ≤ n}|

i=1

= | Udt=0 {g = (a1 , a2 , . . . , ad ) ∈ Zd :

d E i=1

|ai | ≤ n and t (g) = t}|

308

6 Finitely Generated Groups

=

d E

|{g = (a1 , a2 , . . . , ad ) ∈ Zd :

t=0

=

d E

d E

|ai | ≤ n and t (g) = t}|

i=1

Nt (n).

t=0

(d) Let BSN (n; I ) denote the set of all elements g = (a1 , a2 , . . . , ad ) ∈ Nd such that I = {i : ai > 0} and lS (g) ≤ n. It is clear that the map which associates d with g = (a1 , a2 , . . . , ad ) ∈ BSN (n; I ) the element a(g) := (ai )i∈I ∈ A (t, n) is a bijection. We deduce from (b) that d

|BSN (n; I )| = |A (t, n)| = Pt (n). d

.

(e) Let BSN (n; t) denote the set of all elements g = (a1 , a2 , . . . , ad ) ∈ Nd such that I (g) := {i : ai > 0} satisfies |I (g)| ( ) = t, and lS (g) ≤ n. Keeping in mind that, for each t = 0, 1, . . . , n, there are nt subsets I ⊂ {1, 2, . . . , n} of cardinality |I | = t, we deduce from (d) that d

|BSN (n; t)| = |{g ∈ BSN (n; t)} d

d

= | UI ⊂{1,2,...,n} {g ∈ BSN (n; t) : I (g) = I }| d

|I |=t

=

E

|{g ∈ BSN (n; t) : I (g) = I }| d

I ⊂{1,2,...,n} |I |=t

.

=

E

|BSN (n; I )| d

I ⊂{1,2,...,n} |I |=t

=

( )( ) ( ) n d n . Pt (n) = t t t

(f) The map ϕ : {−1, 1}t × BSN (n; t) → BSZ (n; t) defined by setting ϕ(E, g) := d (b1 , b2 , . . . , bd ) where, for g = (a1 , a2 , . . . , an ) ∈ BSN (n; I ) with |I | = t, one sets bi := E(i)ai if i ∈ I and bi := ai = 0 otherwise, is clearly a bijection. We deduce from (d) that d

d

Nt (n) = |BSZ (n; t)| = |{−1, 1}t × BSN (n; t)| = 2t × |BSN (n; t)| = 2t d

.

d

d

( )( ) n d . t t

6.2 Exercises

309

(g) From (f) and (c) we immediately deduce that γSZ (n) = d

d E

.

Nt (n) =

t=0

( )( ) d E d n . 2t t t t=0

O Exercise 6.9 (Growth Series) Let G be a finitely generated group and let S ⊂ G be a finite symmetric generating subset. For n ∈ N we denote by ΣS (n) := {g ∈ G : lS (g) = n} = {g ∈ G : dS (g, 1G ) = n}

.

the sphere of radius n centered at 1G . The function σS : N → N, defined by σS (n) := |ΣS (n)| for all n ∈ N, is called the spherical growth function of G relative to S. The formal power series BS (z) :=

∞ E

.

γS (n)zn

n=0

and AS (z) :=

∞ E

.

σS (n)zn

n=0

are called the growth series and the spherical growth series of G corresponding to the generating subset S, respectively. E (a) Show that γS (n) = Eni=0 σS (i) for all n ∈ N. (b) Show that AS (z) = g∈G zlS (g) . (c) Show that AS (z) = (1 − z)BS (z). (d) Let G := Z and let S := {−1, 1}. Compute the growth series AS (z) and BS (z). (e) Let G1 (resp. G2 ) be a finitely generated group and let S1 ⊂ G1 (resp. S2 ⊂ G2 ) denote a finite symmetric generating subset. Let G := G1 × G2 and let S := (S1 × {1G2 }) ∪ ({1G1 } × G2 ). Recall (cf. [CAG, Proposition 6.6.10]) that S is a finite symmetric generating subset of G. Show that AS (z) = AS1 (z)AS2 (z)

.

and

BS (z) = (1 − z)BS1 (z)BS2 (z).

(f) Let d ≥ 1 be an integer. Let G := Zd and let S := {e1 , −e1 , e2 , −e2 , . . . , ed , −ed }, where ei := (0, 0, . . . , 0, 1, 0, 0 . . . , 0) ∈ Zd with 1 being placed in the ith coordinate position, for i = 1, 2, . . . , d. Compute the growth series AS (z) and BS (z). (g) Suppose that G is a free group of finite rank r ≥ 1 and let X ⊂ G be a free base for G. Set S := X ∪ X−1 . Compute the growth series AS (z) and BS (z).

310

6 Finitely Generated Groups

Solution (a) It suffices to observe that the ball BS (n) is the disjoint union of the spheres ΣS (i), 0 ≤ i ≤ n. (b) We have E .

zlS (g) =

g∈G

∞ E E n=0

∞ E E

zlS (g) =

g∈G lS (g)=n

n=0

zn =

g∈G lS (g)=n

∞ E

σS (n)zn = AS (z).

n=0

(c) We have γS (0) = σS (0) = 1. Moreover, it follows from (a) that σS (n) = γS (n) − γS (n − 1) for all n ≥ 1. Summing up, we obtain AS (z) =

∞ E

σS (n)zn

n=0

=1+

∞ E

σS (n)zn

n=1

=1+ .

∞ E

(γS (n) − γS (n − 1)) zn

n=1

=1+

∞ E

γS (n)zn − z

∞ E

n=1

=

∞ E

γS (n − 1)zn−1

n=1

γS (n)zn − z

n=0

∞ E

γS (n)zn

n=0

= BS (z) − zBS (z) = (1 − z)BS (z). (d) We have σS (0) = 1 and σS (n) = 2 for all n ≥ 1. As a consequence, AS (z) = 1 + 2

∞ E

.

n=1

zn = 1 +

1+z 2z = . 1−z 1−z

1+z . (1 − z)2 (e) We have ΣS (n) = Uni=0 ΣS1 (i) × ΣS2 (n − i) for all n ∈ N. As a consequence,

It then follows from (c) that BS (z) =

AS (z) =

∞ E

.

n=0

σS (n)zn =

n ∞ E E n=0 i=0

σS1 (i)zi σS2 (n − i)zn−i = AS1 (z)AS2 (z).

6.2 Exercises

311

It then follows from (c) that BS (z) =

.

AS1 (z)AS2 (z) AS (z) = = (1 − z)BS1 (z)BS2 (z). 1−z 1−z

(f) It follows from (d), (e), and (c) that ( AS (z) =

.

1+z 1−z

)d and

BS (z) =

(1 + z)d . (1 − z)d+1

(g) It follows from the uniqueness of reduced forms in free groups that σS (0) = 1 and σS (n) = 2r(2r − 1)n−1 for all n ≥ 1. As a consequence, ∞

AS (z) = 1 +

.

2r E ((2r − 1)z)n 2r − 1 n=1

=1+

(2r − 1)z 2r · 2r − 1 1 − (2r − 1)z

=1+

2rz 1 − (2r − 1)z

=

1+z . 1 − (2r − 1)z

It then follows from (c) that BS (z) =

.

1+z . (1 − z)(1 − (2r − 1)z) O

Comment For more on growth series, see [Har1, Section VI.A] and [Man, Chapter 14]. Exercise 6.10 Suppose that the maps γ , γ ' : N → [0, +∞) are such that γ < γ ' . R √ n n ' Show that if lim supn→∞ γ (n) > 1 then lim supn→∞ γ (n) > 1. Solution As γ < γ ' , there exists an√integer c ≥ 1 such that γ (n) ≤ cγ ' (cn) for all n ≥ 1. Suppose that lim supn→∞ n γ (n) >√1. This implies that there exist a real ≥ 1R such that n γR (n) ≥ λ for all number λ > 1 and an integer n0 √ R √n ≥ n0 . We deduce cn n cn that cγ ' (cn) ≥ λ and hence c γ ' (cn) = cn cγ ' (cn) ≥ cRλ for all n√≥ 1 for √ all n ≥ 1. Since cn c → 1 as n → ∞, this implies lim supn→∞ n γ ' (n) ≥ c λ > 1. O ( ) 21 Exercise 6.11 Let M := ∈ SL2 (Z). Consider the semidirect product 11 G := Z2 Xα Z, where α : Z → Aut(Z2 ) is the group homomorphism defined by

312

6 Finitely Generated Groups

( ) ( ) x x z := M α(z) for all x, y, z ∈ Z. Recall (cf. Exercise 2.30) that y y [( ( ) ) ] x .G = , z : x, y, z ∈ Z y with the multiplication defined by (( ) ) (( ) ) (( ) ( ) ) x1 x2 x1 z1 x2 , z1 · , z2 = +M , z 1 + z2 . y1 y2 y1 y2 for all x1 , x2 , y1 , y2 , z1 , z2 ∈ Z. (a) Show that Mn

.

( ) ( ) 1 f = 2n+1 f2n 0

for all n ∈ N, where (fk )k∈N is the Fibonacci sequence which is inductively defined by f0 := 0, f1 := 1, and fk := fk−2 + fk−1 for all k ≥ 2. (b) Show that n−1 E .

f2i+1 = f2n

(6.2)

i=0

for every integer n ≥ 1. (c) Deduce from (a) and (b) that, for any integer n ≥ 1, the set A(n) ⊂ G defined by A(n) :=

[( E n−1

.

i=0

( ) ) ] 1 ui M , 0 : ui ∈ {0, 1} for 1 ≤ i ≤ n 0 i

has cardinality |A(n)| = 2n . (d) Consider the subset S ⊂ G defined by S := {s, t, z, s −1 , t −1 , z−1 }, where (( ) ) (( ) ) (( ) ) 1 0 0 .s := , 0 , t := , 0 , and z = ,1 . 0 1 0 Show that S is a finite symmetric generating subset of G and that one has A(n) ⊂ BSG (3n − 2) for all n ≥ 1. (e) Deduce from (c) and (d) that G has exponential growth.

6.2 Exercises

313

Solution (a) We prove the statement by induction. For n = 0 we have M0

.

( ) ( )( ) ( ) ( ) 1 10 1 1 f = = = 1 f0 0 01 0 0

( ) 1 and the base of the induction follows. Let n ≥ 0 and suppose that M n = 0 ) ( f2n+1 . Then, after observing that 2f2n+1 + f2n = f2n+1 + (f2n+1 + f2n ) = f2n f2n+1 + f2n+2 = f2n+3 , we get M n+1

.

( ) ( ) ( ) ( ) ( ) ( ) 1 1 2f2n+1 + f2n f 21 f = 2n+3 . = M · Mn = · 2n+1 = f2n f2n + f2n+1 f2n+2 0 0 11

This completes the inductive argument. (b) We proceed by induction on n ≥ 1. For n = 1, Ewe have f1 = 1 = f2 and the base of the induction follows. Suppose now that n−1 i=0 f2i+1 = f2n for some n ≥ 1. We then have n E .

i=0

f2i+1 =

n−1 E

f2i+1 + f2n+1 = f2n + f2n+1 = f2n+2 = f2(n+1) .

i=0

This proves (6.2). (c) Consider the map ϕ : {0, 1}n → Z2 defined by ϕ(u) :=

n−1 E

.

i=0

( ) 1 ui M 0 i

for all u = (ui )0≤i≤n−1 ∈ {0, 1}n . Note that ) (En−1 ui f2i+1 i=0 En−1 .ϕ(u) = i=0 ui f2i

(6.3)

by (a). We claim that ϕ is injective. Indeed, suppose it is not, i.e., there exist u = (ui )0≤i≤n−1 , v = (vi )0≤i≤n−1 ∈ {0, 1}n with u /= v and ϕ(u) = ϕ(v). By (6.3), this implies in particular n−1 E .

i=0

ui f2i+1 =

n−1 E i=0

vi f2i+1 .

(6.4)

314

6 Finitely Generated Groups

Consider the largest integer k ≤ n − 1 such that uk /= vk . By symmetry, we may assume uk = 1 and vk = 0. It then follows from (6.4) that k E .

ui f2i+1 =

i=0

k−1 E

vi f2i+1 .

(6.5)

i=0

On the other hand, we have k E .

ui f2i+1 ≥ uk f2k+1 = f2k+1

i=0

and k−1 E .

vi f2i+1 ≤

i=0

k−1 E

f2i+1 = f2k ,

i=0

where the last equality follows from (b). This clearly contradicts (6.5) since f2k < f2k+1 for all k. Therefore ϕ is injective. As A(n) = {(ϕ(u), 0) : u ∈ {0, 1}n }, we deduce that |A(n)| = |{0, 1}n | = 2n . (d) The subset S is clearly symmetric. Moreover, since {s, t, s −1 , t −1 } generates 2 Z and {z, z−1 } generates Z, we deduce from Exercise 2.32 that S generates G. ( ) En−1 i 1 . Note that Let now a = (ϕ(u), 0) ∈ A(n), where ϕ(u) = i=0 ui M 0 ( ) ) ( 1 ui M i , 0 = zi s ui z−i for all 0 ≤ i ≤ n − 1. As 0 a=

n−1 | | (

ui M i

.

i=0

=

n−1 | |

( ) ) 1 ,0 0

zi s ui z−i

i=0

= s u0 · zs u1 z−1 · z2 s u2 z−2 · . . . · zn−2 s un−2 z−n+2 · zn−1 s un−1 z−n+1 = s u0 zs u1 zs u2 · · · zs un−2 zs un−1 z−n+1 , we deduce that lS (a) ≤ 2(n − 1) + 1 + (n − 1) = 3n − 2. This shows A(n) ⊂ BSG (3n − 2). (e) We have 2n = |A(n)| ≤ |BSG (3n − 2)| = bSG (3n − 2), showing that 2n ≺ G bS (n). It follows that G has exponential growth. O Comment The group G := Z2 Xα Z is metabelian (and hence solvable) since we have a short exact sequance 0 → Z2 → G → Z → 0. The matrix M induces a

6.2 Exercises

315

group automorphism of the torus R2 /Z2 which is known as the Arnold cat map. It is a very popular example of an Anosov diffeomorphism in the theory of hyperbolic dynamical systems. Exercise 6.12 Show that any two of the matrices S, T , U ∈ SL2 (Z) defined by S :=

.

( ) 0 −1 1 0

T :=

( ) 11 01

( and

U :=

10 11

)

generate the group SL2 (Z). Solution Let us first show that the set {S, T , U } generates SL2 (Z). Denote by H the subgroup of SL2 (Z) generated by {S, T , U }. Let M=

.

( ) ab cd

be an arbitrary matrix in SL2 (Z). We shall prove that M ∈ H by showing that there exists a matrix P ∈ H such that MP ∈ H . We have ad − bc = det(M) = 1 so that gcd(a, b) = 1. ( ) If we multiply M on the right by S, the first row of M becomes b −a . Thus, by multiplying M on the right by a suitable power of S, we can obtain a matrix in which the two entries on the first row are non-negative. Consequently, we may assume a ≥ 0 and b ≥ 0. ( ( ) ) 1 −q 1 0 After multiplying M on the right by T −q = (resp. U −q = ), 0 1 −q 1 ( ) ( ) the first row of M becomes a b − aq (resp. a − bq b ). Thus, we can perform the Euclidean algorithm to the first row of M by multiplying M on the right by a finite sequence of powers of T and U . At the end we will arrive at ( of) the(algorithm, ) a matrix N ∈ SL2 (Z) whose first row is either 1 0 or 0 1 . In the first case, ( N=

.

) 10 = Un n0

for some n ∈ Z, while in the second case ( N=

.

) ( )( ) 0 1 1 0 0 1 = = U m S −1 −1 m m1 −1 0

for some m ∈ Z. Thus, we have N ∈ H in either case. This shows that M ∈ H and hence that SL2 (Z) is generated by the set {S, T , U }. As S = U T −1 U , T = S −1 U −1 S, and U = ST −1 S −1 , we conclude that any two O among the matrices S,T , and U suffice to generate SL2 (Z).

316

6 Finitely Generated Groups

Exercise 6.13 (Elements of Finite Order in SL2 (Z)) (a) Show that for every n ∈ {1, 2, 3, 4, 6} there exists an element of order n in the group SL2 (Z). (b) Suppose that M ∈ SL2 (Z) has finite order n. Show that one has n ∈ {1, 2, 3, 4, 6} and that n = 1 (resp. 2, resp. 3, resp. 4, resp. 6) if and only if Tr(M) = 2 (resp. −2, resp. 1, resp. 0, resp. 1). (c) Let m ≥ 3 be an integer and let ρ : SL2 (Z) → SL2 (Z/mZ) denote the canonical group homomorphism given by reduction modulo m. Show that the kernel of ρ is a finitely generated torsion-free group. Solution (a) It suffices to observe that the matrices ( I,

.

−I,

) 0 1 , −1 −1

( ) 0 −1 , 1 0

(

0 −1 1 1

)

are all in SL2 (Z) and have order 1,2,3,4, and 6, respectively. (b) Suppose that M ∈ SL2 (Z) has finite order n. Then the minimal polynomial of M divides Xn −1. We deduce that all the roots in C of the minimal polynomial of M are simple. Therefore, M is diagonalizable, i.e., there exists a matrix P ∈ GL2 (C) such that ( ) λ0 .M = P P −1 , 0λ where λ and λ are the roots in C of the characteristic polynomial of M. We have M n = I and hence λn = 1. Therefore λ = cos θ + i sin θ for some θ ∈ R. We then have 2 cos θ = λ + λ = Tr(M) ∈ Z. As −2 ≤ 2 cos θ ≤ 2, we deduce that Tr(M) ∈ {−2, −1, 0, 1, 2}. If Tr(M) = 2, then cos θ = 1 and hence λ = λ = 1, so that M = I has order 1. If Tr(M) = −2, then cos θ = −1 and hence λ = λ = −1, so that M = −I has order 2. If Tr(M) = −1 (resp. 0, resp. 1) then cos θ = −1/2 (resp. 0, resp. 1/2) so that either λ or λ equals ei2π/3 (resp. eiπ/2 , resp. eiπ/3 ). It follows that M is conjugate in GL2 (Z) to the matrix .

( i2π/3 )( ( iπ/2 ) ( iπ/3 )) 0 0 0 e e e resp. , resp. 0 e−i2π/3 0 e−iπ/2 0 e−iπ/3

and hence that n = 3 (resp. 4, resp. 6). (c) Suppose that M ∈ SL2 (Z) is in the kernel of ρ and has finite order n. Then Tr(M) is congruent to 2 = Tr(I ) modulo m. As Tr(M) ∈ {−2, −1, 0, 1, 2} by (b) and m ≥ 3, we deduce that Tr(M) ∈ {−2, −1, 2}. We cannot have Tr(M) = −2 since otherwise, as observed in the solution of (b), we would have M = −I , contradicting the fact that M is congruent to I modulo m. Suppose now Tr(M) =

6.2 Exercises

317

−1. Then we must have m = 3 and M of the form M=

.

( ) 1 + 3a 3b 3c −2 − 3a

for some a, b, c ∈ Z. But this gives us det(M) = −2 − 9a + 9k 2 − 9bc which is congruent to −2 modulo 9, contradicting det(M) = 1. Therefore Tr(M) = 2 and hence M = I by (b). This shows that the kernel of ρ is torsion-free. O Exercise 6.14 Consider the matrices S, V ∈ SL2 (Z) defined by S :=

.

( ) 0 −1 1 0

and

V :=

( ) 0 −1 . 1 1

(a) Show that S has order 4 and that V has order 6. (b) Show that S and V generate SL2 (Z). Solution (a) Denoting by I the identity matrix in SL2 (Z), it is immediate to check that S 2 = V 3 = −I . This implies that S has order 4 and that V has order 6. (b) Consider the matrix T ∈ SL2 (Z) defined by T :=

.

( ) 11 . 01

The group SL2 (Z) is generated by the matrices S and T by Exercise 6.12. As T = SV −2 , the matrices S and V also generate SL2 (Z). O Exercise 6.15 Show that every group homomorphism SL2 (Z) → Z is trivial. Solution This immediately follows from Exercise 6.14 which tells us that SL2 (Z) can be generated by two elements of finite order. O Comment The homological interpretation of this result is that H 1 (SL2 (Z), Z) = 0 (see [Bro]). Exercise 6.16 (Indecomposable Groups) A group G is called indecomposable if G is non-trivial and there do not exist non-trivial groups H and K such that the group H × K is isomorphic to G. Show that the group SL2 (Z) is indecomposable. Solution Suppose that there exist groups H, K and a group isomorphism ϕ : SL(2 (Z) → ) H × K. Let us show that either H or K is trivial. The matrix 0 −1 S := is an element of order 4 in SL2 (Z). Therefore (x, y) := ϕ(S) has 1 0 order 4 in H × K. We must have x 2 = 1H or y 2 = 1K . Otherwise, (x 2 , 1K ) and (1H , y 2 ) would be two distinct elements of order 2 in H × K, a contradiction since SL2 (Z) contains only one element of order 2, namely S 2 = −I . After possibly exchanging H and K, we may assume y 2 = 1K . Then x 2 /= 1H since otherwise

318

6 Finitely Generated Groups

(x, y) would have order 2. Moreover, y = 1K since otherwise (x 2 , 1K ) and (x 2 , y) would be two distinct elements of order 2. ( Thus,)ϕ(S) = (x, 1K ) and x has order ab 4 in H . Now observe that a matrix M := ∈ SL2 (Z) commutes with S if cd and only if d = a and c = −b. This implies det(M) = a 2 + b2 = 1 and hence (a, b) ∈ {(1, 0), (0, −1), (−1, 0), (0, 1)}. We deduce that the centralizer of S in SL2 (Z) is reduced to {I, S, −I, −S}. As all the elements (x k , z) for 0 ≤ k ≤ 3 and z ∈ K are distinct and commute with (x, 1K ), this implies that K = {1K }. This shows that SL2 (Z) is indecomposable. O Comment The study of indecomposable groups and decomposition of groups into direct factors is a classical subject in group theory and has been investigated by many authors (see [CorH], [Robi, Chapter 3], [Rot, Chapter 6] and the references therein). Exercise 6.17 Show that the matrices R, T ∈ GL2 (Z) defined by .R :=

( ) 01 10

and

T :=

( ) 11 01

generate the group GL2 (Z). ( ) 0 −1 Solution Setting S := , it follows from Exercise 6.12 that the group 1 0 SL2 (Z) is generated by the set {S, T }. As SL2 (Z) is a subgroup of index 2 in GL2 (Z) and R ∈ / SL2 (Z), we deduce that the set {R, S, T } generates GL2 (Z). Since S = T −1 RT RT −1 , the set {R, T } is sufficient to generate GL2 (Z). O Exercise 6.18 (Free Groups Are Torsion-Free) Let F be a free group and let X ⊂ F be a base for F . Recall that every element g ∈ G can be uniquely written in its X-reduced form, i.e., in the form g = x1 x2 · · · xn

.

(6.6)

with n ≥ 0, xi ∈ X ∪ X−1 for 1 ≤ i ≤ n, and xi+1 /= xi−1 for 1 ≤ i ≤ n − 1. One says that g is cyclically reduced (with respect to X) if its reduced form (6.6) satisfies xn /= x1−1 . (a) Show that every element in F is conjugate to some cyclically reduced element. (b) Deduce from (a) that F is torsion-free. Solution (a) Let g ∈ F and write g in its reduced form (6.6). Let k be the largest integer such that −1 −1 x1 x2 · · · xk = xn−1 xn−1 · · · xn−k+1 .

.

(6.7)

6.2 Exercises

319

Observe that 2k ≤ n − 1 since otherwise (6.7) would imply, by uniqueness of −1 −1 reduced forms, xn/2 = xn/2+1 (for n even) or x(n+1)/2 = x(n+1)/2 (for n odd), contradicting reductedness of (6.6). Let g := xk+1 xk+2 · · · xn−k .

.

Observe that g is cyclically reduced by maximality of k. Moreover, we have g = hgh−1 by taking h := x1 x2 · · · xk . Therefore g is conjugate to g. (b) As conjugate elements have the same order, it suffices to show that every non-trivial cyclically reduced element g ∈ F has infinite order. This is clear since if g is cyclically reduced with reduced form (6.6), then, for any integer m ≥ 1, g m = x1 x2 · · · xn x1 x2 · · · xn · · · x1 x2 · · · xn

.

is a reduced form, so that g m /= 1F .

O

Exercise 6.19 (Schreier Lemma) Let G be a group and let H be a subgroup of G. Let S ⊂ G be a complete set of representatives for the right cosets of H in G such that 1G ∈ S. For g ∈ G, let g ∈ S denote the representative of the right coset H g. (a) Show that hg = g for all h ∈ H and g ∈ G. (b) Show that gg −1 and g(g)−1 are both in H for all g ∈ G. (c) Show that g1 g2 = g1 g2 for all g1 , g2 ∈ G. (d) Let s ∈ S and x ∈ G. Show that the element t ∈ S defined by t := sx −1 satisfies sx −1 (sx −1 )−1 = (tx(tx)−1 )−1 .

.

(6.8)

(e) Let X ⊂ G be a generating subset of G. Show that the set Y = Y (X, S) := {sx(sx)−1 : s ∈ S and x ∈ X}

.

(6.9)

is a generating subset of H . A complete set of representatives for the right cosets of a subgroup H of a group G is called a right transversal of H in G. The elements in the set Y (X, S) are called the Schreier generators of H (relative to the right transversal S and the generating subset X of G). Solution (a) This is clear since H hg = H hg = H g = H g for all h ∈ H and g ∈ G. (b) Let g ∈ G and let h ∈ H such that g = hg. Then gg −1 = h−1 ∈ H and g(g)−1 = h ∈ H . (c) This is clear since H g1 g2 = H g1 g2 = H g1 g2 = H g1 g2 for all g1 , g2 ∈ G. (d) By using (c), we get tx = sx −1 x = sx −1 x = s = s, so that tx(tx)−1 = txs −1 . It follows that tx(tx)−1 ·sx −1 (sx −1 )−1 = txs −1 ·sx −1 (sx −1 )−1 = 1G . This gives us (6.8).

320

6 Finitely Generated Groups

(e) The inclusion Y ⊂ H immediately follows from (b). Let now h ∈ H . Since X generates G, we can find and integer m ≥ 0 and a1 , a2 , . . . , am ∈ X ∪ X−1 such that h = a1 a2 · · · am . Keeping in mind (c) we have (

.

)−1 ) ( )−1 ) ( ( )−1 ) ( ( · ··· · a1 a2 a3 a1 a2 a3 · a1 a2 a1 a2 1G a1 1G a1 ( )−1 ) ( · · · · (a1 a2 · · · am−1 am ) · a1 a2 · · · am−1 am ) ( ) ( ) ( = a1 (a1 )−1 · a1 a2 (a1 a2 )−1 · a1 a2 a3 (a1 a2 a3 )−1 · · · · ( ) · · · · (a1 a2 · · · am−1 am ) · (a1 a2 · · · am−1 am )−1 = a1 a2 · · · am (a1 a2 · · · am )−1 = h,

where the last equality follows from the fact that a1 a2 · · · am = h = 1G . This shows that we can write h as a product of elements from the set {sa(sa)−1 : s ∈ S, a ∈ X ∪X−1 }. As, by (d), this generating set equals Y ∪Y −1 , this shows that Y generates H. O Comment The result in (e) is known as Schreier’s lemma after Schreier [Schr]. An immediate consequence of Schreier’s lemma is that every finite index subgroup of a finitely generated group is itself finitely generated [CAG, Proposition 6.6.6]. Exercise 6.20 (The Sanov Subgroup) Let F denote the subgroup of SL2 (Z) generated by the matrices A, B ∈ SL2 (Z) defined by ( A :=

.

) 12 , 01

( B :=

) 10 . 21

(a) Show that F is a free group of rank 2 which is freely generated by A and B. (b) Show that F is a normal subgroup of index 12 of SL2 (Z). (c) Show that F consists of all matrices .

( ) ab ∈ SL2 (Z) cd

such that a ≡ d ≡ 1 mod 4 and b ≡ c ≡ 0 mod 2. Solution (a) This follows from Lemma 2.3.2 in [CAG]. (b) Consider the group homomorphism ρ : SL2 (Z) → SL2 (Z/2Z) given by reduction modulo 2. As [( ) ( ) ( ) ( ) ( ) ( )] 10 01 11 10 11 01 . SL2 (Z/2Z) = , , , , , , 01 10 01 11 10 11

6.2 Exercises

321

the restriction of ρ to the set Σ ⊂ SL2 (Z) given by Σ :=

.

[( ) ( ) ( ) ( ) ( ) ( )] 10 0 −1 11 10 1 −1 0 −1 , , , , , 01 1 0 01 11 1 0 1 1

is bijective. Consequently, ρ is a group epimorphism and the kernel K of ρ is a normal subgroup of SL2 (Z) of index | SL2 (Z/2Z)| = 6. Consider the matrix C ∈ SL2 (Z) defined by ( ) −1 0 .C := −I = . 0 −1 Clearly A, B, C ∈ K. Let us show that the group K is generated by the set {A, B, C}. We shall apply Schreier’s lemma (cf. Exercise 6.19(e)). We know from Exercise 6.12 that the set X := {T , U } ⊂ SL2 (Z), where ( T :=

.

11 01

)

( and

U :=

) 10 , 11

generates SL2 (Z). On the other hand, Σ is a complete set of representatives for the 6 cosets of K in SL2 (Z}. For each M ∈ SL2 (Z), denote by M the unique matrix in Σ such that KM = KM. By Schreier’s lemma, the set } { Y := M(M)−1 : M ∈ ΣX

.

is a generating subset for K. Observe that M = M for all M ∈ Σ, so that the subset Z ⊂ Y defined by } { −1 .Z := M(M) : M ∈ ΣX \ Σ is also a generating subset for K. After computation, we get [( ) ( ) ( ) ( ) ( ) ( )] 11 0 −1 12 11 10 0 −1 .ΣT = , , , , , 01 1 1 01 12 11 1 2 and [( ) ( ) ( ) ( ) ( ) ( )] 10 −1 −1 21 10 0 −1 −1 −1 .ΣU = , , , , , . 11 1 0 11 21 1 0 2 1 Consequently, as ΣX = ΣU ∪ ΣT , we have [( ) ( ) ( ) ( ) ( ) ( ) ( )] 12 11 0 −1 −1 −1 21 10 −1 −1 .ΣX \ Σ = , , , , , , . 01 12 1 2 1 0 11 21 2 1

322

6 Finitely Generated Groups

Since ( ) ( )−1 12 10 01 01 ( )( )−1 11 1 −1 12 1 0 ( )( )−1 0 −1 0 −1 1 2 1 0 ( )( )−1 −1 −1 1 −1 . 1 0 1 0 ( )( )−1 21 0 −1 11 1 1 ( ) ( )−1 10 10 21 01 ( ) ( )−1 −1 −1 11 2 1 01

( )( ) ( ) 12 10 12 = 01 01 01 ( )( ) ( ) 11 0 1 −1 2 = = 12 −1 1 −2 3 ( )( ) ( ) 0 −1 0 1 1 0 = = 1 2 −1 0 −2 1 ( )( ) ( ) −1 −1 0 1 1 −2 = = 1 0 −1 1 0 1 ( )( ) ( ) 21 0 1 12 = = 11 −1 0 01 ( )( ) ( ) 10 10 10 = = 21 01 21 ( )( ) ( ) −1 −1 1 −1 −1 0 = = , 2 1 0 1 2 −1 =

we have [( Z=

.

) ( ) ( ) ( ) ( ) ( ) ( )] 12 −1 2 1 0 1 −2 −1 2 10 −1 0 , , , , , , 01 −2 3 −2 1 0 1 0 1 21 2 −1

= {A, BCA−1 , B −1 , A−1 , A, B, B −1 C} = {A, A−1 , B, B −1 , B −1 C, BCA−1 }. This shows that the set {A, B, C} generates K. We have C ∈ / F since C has order 2 and all free groups are torsion-free by Exercise 6.18. As K is a normal subgroup of SL2 (Z) of index 6 and C is in the center of SL2 (Z), we deduce that K = F ⊕ {I, C} and that F is a normal subgroup of index 12 in SL2 (Z). (c) Let F ' denote the set consisting of all matrices .

( ) ab ∈ SL2 (Z) cd

such that a ≡ d ≡ 1 mod 4 and b ≡ c ≡ 0 mod 2. It is immediate to check that F ' is a subgroup of SL2 (Z). Note that F ⊂ F ' since A, B ∈ F ' . Moreover, we

6.2 Exercises

323

( ) 12 clearly have ⊂ K and the inclusion is strict since the matrix is in K but 03 O not in F ' . As F is a subgroup of index 2 in K, we deduce that F = F ' . F'

Comment The group F is called the Sanov subgroup of SL2 (Z) after Sanov [San]. The fact that SL2 (Z) contains a finite index free subgroup of rank 2 shows that the groups SL2 (Z) and F2 are commensurable and hence quasi-isometric. Note that SL2 (Z/2Z) ∼ = Sym(3), so that SL2 (Z)/K ∼ = Sym(3) and SL2 (Z)/F ∼ = Sym(3) × Z/2Z. An alternative proof of the fact that the set {A, B, C} generates K can be given by using a variant of the Euclidean algorithm and following an argument similar to the one employed in Exercise 6.12 for showing that the set {S, T } generates SL2 (Z). One can also prove it by using the geometric properties of the action of K ⊂ SL2 (Z) by homographies on the Poincaré half-plane H := {z ∈ C : Im(z) > 0} ⊂ C ∪ {∞} = P1 (C), after observing that a fundamental region for the action of K on H is the ideal hyperbolic quadrilateral with vertices −1, 0, 1, and ∞ and that the quotient map H → H /K is the universal cover of a thrice-punctured sphere (see for example [Leh]). Given integers n ≥ 1 and d ≥ 2, the kernel of the group homomorphism SLd (Z) → SLd (Z/nZ) is called the principal congruence subgroup of SLd (Z) of level n. Thus, the subgroup K is the principal congruence subgroup of level 2 of SL2 (Z). Note that K is not torsion-free since K ∼ = F2 × Z/2Z. On the other hand, it follows from Exercise 6.13(c) that all principal subgroups of level n ≥ 3 of SL2 (Z) are torsion-free. Using for example the action of SL2 (Z) on its tree (see [Ser3]) or on Poincaré’s half-plane (see [Leh]), one can actually show that all principal congruence subgroups of level n ≥ 3 of SL2 (Z) are free groups of finite rank. A subgroup Λ ⊂ SLd (Z) is called a congruence subgroup if there exists a principal congruence subgroup Γ such that Γ ⊂ Λ. The least integer n such that there exists a principal congruence subgroup Γ of level n such that Γ ⊂ Λ is then called the level of the congruence subgroup Λ. For instance, it immediately follows from (c) that the subgroup F ⊂ SL2 (Z) is a non-principal congruence subgroup of level 4. As the groups SLd (Z/nZ) are finite, it is clear that all congruence subgroups Λ ⊂ SLd (Z) are of finite index. The congruence subgroup problem asks whether the converse is also true, i.e., all finite index subgroups of SLd (Z) are congruence subgroups. It turns out that the answer to the congruence subgroup problem for SL2 (Z) is negative. The existence of finite index subgroups of SL2 (Z) that are not congruence subgroups was already known to Klein in 1880 [Kle, p. 63]. The first explicit examples of such subgroups appeared in two articles published in Mathematische Annalen by Fricke and Pick in 1887 (a larger class of examples in the same spirit was later given by Reiner [Rei]). In [LubS], there is a proof of the existence of non-congruence finite index subgroups of SL2 (Z) that is based on comparison of subgroup growth estimates. In the case d ≥ 3, the congruence subgroup problem for SLd (Z) was answered affirmatively by Bass et al. [BasLS] and independently by Mennicke [Men]. The congruence subgroup problem has been

324

6 Finitely Generated Groups

investigated for more general arithmetic groups and there is a vast literature devoted to the subject (see for example [Har1, Section III.D], [Jon1], [Sou, Section 5], [Sur] and the references therein). Exercise 6.21 Let G := SL2 (Z) and let n ≥ 1 be an integer. Consider the group homomorphism ρn : G → SL2 (Z/nZ) given by reduction of entries modulo n. Let Kn ⊂ G denote the kernel of ρn . (a) Show that ρn is surjective. (b) Show that the index of Kn in G is given by the formula [G : Kn ] = n

3

.

| | ( p

1 1− 2 p

) ,

(6.10)

where p runs over all prime factors of n. Solution (a) Let ( A=

.

αβ γ δ

) ∈ SL2 (Z/nZ).

Let a, b, c, d ∈ {1, 2, . . . , n} representing α, β, γ , δ, respectively. Then the matrix M :=

.

( ) ab cd

reduces to A modulo n but may fail to be in SL2 (Z). However, we have det(M) = 1 + kn for some k ∈ Z. Suppose that p is a prime dividing c. Since ad −bc = 1+kn, the prime p cannot divide both d and n. Set λp := 0 if p does not divide d and λp := 1 if p divides d. Then d + λp n is not divisible by p. By the Chinese remainder theorem, we can find λ ∈ Z such that λ ≡ λp mod p for all primes p dividing c. It follows that gcd(d + λn, c) = 1. Thus, by Bézout’s theorem, we can find u, v ∈ Z such that u(d + λn) − vc = −k. Then the matrix ( .N :=

) a + un b + vn c d + λn

satisfies .

det(N ) = (a + un)(d + λn) − (b + vn)c = ad − bc + (u(d + λn) − vc)n = 1 + kn − kn = 1.

6.2 Exercises

325

Therefore N ∈ SL2 (Z). As the coefficients of N and M are congruent modulo n, we have ρn (N ) = A. This shows that ρn is surjective. (b) Since ρn is surjective by (a), we have [G : Kn ] = | SL2 (Z/nZ)|. Suppose first that n = p is a prime. The cardinality of GL2 (Z/pZ) is equal to the number of bases in the Z/pZ-vector plane (Z/pZ)2 . Therefore | GL2 (Z/pZ)| = (p2 − 1)(p2 − p) = p(p2 − 1)(p − 1). As SL2 (Z/pZ) is the kernel of the surjective group homomorphism det : GL2 (Z/pZ) → (Z/pZ)∗ := Z/pZ \ {0}, we deduce that | SL2 (Z/pZ)| = p(p2 − 1) = p3 (1 − 1/p2 ). This shows (6.10) in the case when n is a prime. Let us now establish formula (6.10) for n = pk , where p is a prime and k is a positive integer. Consider the ring epimorphism π : Z/pk+1 Z → Z/pk Z given by reduction modulo pk . Observe that the kernel of π is the ideal of Z/pk+1 Z generated by pk and that π induces a group homomorphism φ : SL2 (Z/pk+1 Z) → SL2 (Z/pk Z). We clearly have ρpk = φ ◦ ρpk+1 . As ρpk is surjective by (a), we deduce that φ is surjective. Let us show that the kernel of φ has cardinality p3 . This will imply that | SL2 (Z/pk+1 Z)| = p3 | SL2 (Z/pk Z)| and formula (6.10) will then follow by induction on k since we have already proved it for k = 1. To compute the cardinality of ker(φ), we observe that ker(φ) consists of all matrices of the form ) ( 1 + pk α pk β ∈ SL2 (Z/pk+1 Z), .A = pk γ 1 + pk δ where α, β, γ , δ ∈ {1, 2, . . . , p}. As det(A) = (1 + pk α)(1 + pk δ) − p2k βγ = 1 + pk (α + δ) in Z/pk+1 Z, the condition det(A) = 1 is equivalent to δ = p − α. As α, β, γ ∈ {1, 2, . . . , p} can be chosen arbitrarily, we conclude that | ker(φ)| = p3 . This completes the proof of (6.10) in the case when n is a prime power. We treat now the general case. Recall that it follows from the|| Chinese remainder theorem that the ring Z/nZ is isomorphic to the product ring p Z/pkp Z, where p runs over all primes dividing n and kp ≥ 1 denotes the largest integer such that pkp divides || n. This implies that the group SL2 (Z/nZ) is isomorphic to the product group p SL2 (Z/pkp Z). Using the previous case, we deduce that | | || | | | | kp .| SL2 (Z/nZ)| = | SL2 (Z/p Z)| |p | | | | | | | = |SL2 (Z/pkp Z)| p

=

| | p

) ( 1 p3kp 1 − 2 p

326

6 Finitely Generated Groups

( | |

)3

| | (

1 1− 2 = p p p p ( ) | | 1 1− 2 . = n3 p p kp

)

O

This gives us (6.10).

Exercise 6.22 (The Projective Modular Group) Consider the modular group G := SL2 (Z) and let Z(G) denote its center. (a) Show that Z(G) = {I, −I }, where I denotes the identity matrix in G. (b) Let Γ := G/Z(G) = SL2 (Z)/{I, −I } and denote by π : G → Γ the canonical group epimorphism. Consider the matrices S, V ∈ SL2 (Z) defined by S :=

.

( ) 0 −1 1 0

and

V :=

( ) 0 −1 . 1 1

Show that the elements α := π(S) and β := π(V ) have order 2 and 3, respectively, in Γ . (c) Show that α and β generate Γ . (d) Show that Γ is residually finite. (e) Show that the subgroup Φ ⊂ Γ generated by the elements γ1 := βαβα and γ2 := β −1 αβ −1 α is a normal subgroup of index 6 of Γ and that Φ is free of rank 2. (f) Show that Γ is non-amenable. (g) Consider the finite symmetric generating subset X ⊂ Γ defined by X := {α, β, β −1 }. Given an integer n ≥ 0, one says that a sequence (xi )1≤i≤n of elements of X is admissible if, for all 1 ≤ i ≤ n − 1, one has xi = α if and only if xi+1 ∈ {β, β −1 }. Suppose that (xi )1≤i≤n is an admissible sequence of elements of X such that n ≥ 1. By playing ping-pong on the real projective line, show that one has x1 x2 · · · xn /= 1Γ .

.

(h) Let F (x, y) denote the free group of rank 2 based on the symbols x and y. Show that there is a unique group homomorphism ψ : F (x, y) → Γ such that ψ(x) = α and ψ(y) = β. (i) Show that ψ is surjective and that the kernel of ψ is the normal closure in F (x, y) of the set {x 2 , y 3 }. (j) Show that Γ admits the presentation Γ = (x, y : x 2 = y 3 = 1).

.

(k) Let γ ∈ Γ and let lX (γ ) denote the word length of γ with respect to X. Show that there exist a unique integer n ≥ 0 and a unique admissible sequence

6.2 Exercises

327

(xi )1≤i≤n of elements of X such that γ = x1 x2 · · · xn , and that one has n = lX (γ ). This sequence (xi )1≤i≤n is called the normal form of γ . (l) Draw the Cayley graph of Γ with respect to X. Solution

( ) ab (a) We clearly have I, −I ∈ Z(G). Conversely, suppose that M := ∈ cd ( ) 11 Z(G). The fact that M commutes with gives us a = d and c = 0. Using then 01 ( ) 10 the fact that M commutes with , we get b = 0. As det(M) = a 2 = 1, we 11 deduce that M = I or M = −I . This shows that Z(G) = {I, −I }. (b) We have S ∈ / Z(G) and S 2 = −I ∈ Z(G). Therefore α has order 2. On the other hand, we have V ∈ / Z(G) while V 3 = −I ∈ Z(G). It follows that β has order 3. (c) The matrices S and V generate SL2 (Z) by Exercise 6.14. As π is a surjective group homomorphism, it follows that α and β generate Γ . (d) Let γ ∈ Γ such that γ /= 1Γ . Choose M ∈ G such that π(M) = γ . Then M /= I and M /= −I . Then we can find an even integer n ≥ 2 such that, denoting by ϕ : G → SL2 (Z/nZ) the group homomorphism given by reduction modulo n, we have ϕ(M) /= 1SL2 (Z/nZ) . As Z(G) is contained in the kernel of ϕ, there is a group homomorphism φ : Γ → SL2 (Z/nZ) such that φ ◦ π = ϕ. We then have φ(γ ) = π(M) /= 1SL2 (Z/nZ) . This shows that Γ is residually finite. (e) Consider the group homomorphism ρ : SL2 (Z) → SL2 (Z/2Z) given by reduction modulo 2. We have seen in the solution to Exercise 6.20 that ρ is surjective and that the kernel K of ρ satisfies K = F ∪ (−F ), where F is the free group of rank 2 generated by the matrices A :=

.

( ) 12 01

and

B :=

( ) 10 . 21

Observe that A = V SV S and B = V −1 SV −1 S, so that ρ(A) = γ1 and ρ(B) = γ2 . As −I ∈ / F (since free groups are torsion-free by Exercise 6.18(b)), we deduce that ρ induces a group isomorphism from F onto Φ. Therefore Φ is free of rank 2. On the other hand, we have π(K) = Φ. As K is normal in SL2 (Z), we deduce that Φ is normal in Γ . Moreover, π induces a group isomorphism SL2 (Z)/K → Γ /Φ. Since K has index | SL2 (Z/2Z)| = 6 in SL2 (Z), it follows that Φ also has index 6 in Γ . (f) By (e), the group Γ contains a free subgroup of rank 2. This implies that Γ is non-amenable [CAG, Corollary 4.5.2]. The fact that Γ is non-amenable can also be deduced from Exercise 4.5. Indeed, G is non-amenable by [CAG, Example 4.5.3.(b)] while Z(G) is amenable since it is finite. (g) Let us set γ := x1 x2 · · · xn .

328

6 Finitely Generated Groups

The group G naturally acts on the projective line P1 (R) = R ∪ {∞}. This action is given by the formula Mz =

.

az + b , cz + d

( ) ab ∈ G and z ∈ P1 (R). The action of Z(G) on P1 (R) is trivial so cd that there is an induced action of Γ on P1 (R). For all z ∈ P1 (R), we have

where M =

1 αz = − , z

1 βz = 1 − , z

.

and

β −1 z =

1 . 1−z

Consider the open intervals I1 , I2 ⊂ R defined by I1 := {z ∈ R : z < 0} and I2 := {z ∈ R : 0 < z}. Observe the inclusions αI2 ⊂ I1 ,

.

βI1 ⊂ I2 ,

β −1 I1 ⊂ I2 .

(6.11)

Suppose first that n is odd. Then either x1 = xn = α or x1 , xn ∈ {β, β −1 }. By the above observations, a ping-pong-type argument gives us γ I2 ⊂ I1 in the first case and γ I1 ⊂ I2 in the second. As the sets I1 and I2 are disjoint and non-empty, this implies γ /= 1Γ . Suppose now that n is even. After replacing γ by γ −1 = xn−1 · · · x2−1 x1−1 if necessary, we may assume x1 = α. Then xn = β ε for some ε ∈ {−1, 1}. By applying the first part of the proof to the conjugate γ ' := β −ε γβ ε = β −ε x1 x2 · · · xn−1 β 2ε = β −ε x1 x2 · · · xn−1 β −ε , we get γ ' /= 1Γ and hence γ /= 1Γ . (h) This follows from the universal property of free groups. (i) The homomorphism ψ is surjective since α and β generate Γ by (c). Let N denote the normal closure in F (x, y) of the set {x 2 , y 3 }. We have N ⊂ ker(ψ) since α has order 2 and β has order 3. Conversely, suppose that f ∈ ker(ψ) and let us show that f ∈ N . Write f in reduced form, i.e., in the form f = uk11 uk22 · · · uknn

.

with n ≥ 0, ui ∈ {x, y} and ki ∈ Z \ {0} for 1 ≤ i ≤ n, and ui /= ui+1 for 1 ≤ i ≤ n − 1 (cf. [CAG, Corollary D.3.4]). Observe that f1 Nf2 = Nf1 f2 for all f1 , f2 ∈ F (x, y) since N is normal in F (x, y). Therefore, f1 f2 ∈ N implies f1 gf2 ∈ N for all g ∈ N. Thus, as x 2 , y 3 ∈ N, we can assume that ki = 1 if ui = x and ki ∈ {−1, 1} if ui = y. But then, writing xi := ψ(uki i ) for 1 ≤ i ≤ n, we get 1Γ = ψ(f ) = x1 x2 · · · xn .

.

As (xi )1≤i≤n is an admissible sequence of elements of X, this implies n = 0 by (g), so that f = 1F ∈ N. (j) This follows from (k).

6.2 Exercises

329

(k) Let n := lX (γ ). Then there exists a sequence (xi )1≤i≤n of elements of X such that γ = x1 x2 · · · xn .

.

(6.12)

Let 1 ≤ i ≤ n − 1. If xi = α, we cannot have xi+1 = α. Otherwise, we could suppress the factor xi xi+1 in the right-hand side of (6.12), contradicting the minimality of n = lX (γ ). Similarly, if xi ∈ {β, β −1 }, we cannot have xi+1 ∈ {β, β −1 }. Otherwise, we could either suppress the factor xi xi+1 or replace it by β or β −1 . This shows that the sequence (xi )1≤i≤n is admissible and proves existence. To establish uniqueness, suppose that (yj )1≤j ≤m is another admissible sequence such that γ = y1 y2 · · · ym . Observe that n ≤ m by minimality of n. We claim that xn = ym . Indeed, we have −1 x1 x2 · · · xn ym · · · y2−1 y1−1 = γ γ −1 = 1Γ .

.

(6.13)

−1 ∈ {β, β −1 }. This would imply that the sequence If xn /= ym , we would have xn ym −1 , and z := (zk )1≤k≤n+m−1 defined by zk := xk for 1 ≤ k ≤ n − 1, zn := xn ym k −1 ym+n−k , for n + 1 ≤ j ≤ n + m − 1, is admissible. As z1 z2 · · · zm+n−1 = 1Γ by (6.13), this would contradict (g) since n + m − 1 ≥ 1. Thus, xn = ym and therefore

x1 x2 · · · xn−1 = y1 y2 · · · ym−1 .

.

By an obvious induction, we successively get xn−1 = ym−1 , xn−2 = ym−2 , . . . , x1 = ym−n+1 . This yields y1 y2 · · · ym−n = 1Γ , so that m = n by using again (g). This shows uniqueness. (l) The Cayley graph CX (Γ ) of the group Γ = PSL2 (Z) with respect to the symmetric generating subset X = {α, β, β −1 } ⊂ Γ is represented in Fig. 6.1. The edges corresponding to the sides of the triangles are labeled β or β −1 depending on the orientation while all the other edges are labeled α. O Comment The group Γ := SL2 (Z)/{I, −I } is called the projective modular group and denoted by PSL2 (Z). The proof of (g) is based on a note of Alperin [Alp]. The property established in (g) amounts to saying that Γ is the free product of the cyclic subgroup A generated by α and of the cyclic subgroup B generated by β. This is written Γ = A ∗ B or Γ = Z/2Z ∗ Z/3Z (see [LynS, Chapter 4], [Robi, Chapter 6], [Rot, Chapter 11], or [Ser3, Chapter 1] for the general definition and properties of free products of groups). By looking at the action of Γ on the Poincaré halfplane, one sees that Γ is the hyperbolic triangle group of type (2, 3, ∞) (cf. [Ser1, Chapter 7]). Exercise 6.23 (Growth of the Projective Modular Group) Let G := PSL2 (Z) denote the projective modular group and consider the finite symmetric generating subset S := {α, β, β −1 } ⊂ G, where α and β are as in Exercise 6.22(g).

330

6 Finitely Generated Groups

b–1ab

bab–1 b–1a

b–1ab–1

ba b

b–1

bab

1G a ab–1 ab–1a

ab aba

Fig. 6.1 The Cayley graph of the projective modular group PSL2 (Z)

(a) Compute the spherical growth series AS . (b) Compute the growth series BS . (c) Compute the growth function γS (n). √ (d) Compute the growth rate λS := limn→∞ n γS (n). (e) Show that there exist constants C1 , C2 > 0 such that C1 λnS ≤ γS (n) ≤ C2 λnS for all n ∈ N. (f) Show that the sequence (γS (n)/λnS )n∈N is not convergent. Solution (a) It follows from Exercise 6.22(i) that for every g ∈ G there exists a unique admissible sequence (xi )1≤i≤n of elements in S such that g = x1 x2 · · · xn and n = lX (g). As a consequence, σS (n) is equal to the number a(n) of admissible sequences (xi )1≤i≤n of elements in S. For each n ≥ 1, let aα (n) (resp. aβ (n)) denote the number of admissible sequences (xi )1≤i≤n in S such that xn = α (resp. xn ∈ {β, β −1 }). Note that a(0) = 1 and a(n) = aα (n) + aβ (n) for all n ≥ 1. Moreover, we have { aα (1) = 1 and aβ (1) = 2 .

aα (n + 1) = aβ (n) and aβ (n + 1) = 2aα (n)

6.2 Exercises

331

for all n ≥ 1. Therefore, the formal power series ∞ E

Aα (z) :=

.

aα (n)zn

Aβ (z) :=

and

n=1

∞ E

aβ (n)zn

n=1

satisfy Aα (z) = z + z

∞ E

.

aα (n + 1)zn

n=1

=z+z

∞ E

aβ (n)zn

n=1

= z(1 + Aβ (z)) and Aβ (z) = 2z + z

∞ E

.

aβ (n + 1)zn

n=1

= 2z + z

∞ E

2aα (n)zn

n=1

= 2z(1 + Aα (z)). This gives us Aα (z) = z(1 + 2z(1 + Aα (z)))

.

and therefore Aα (z) =

.

z(1 + 2z) 1 − 2z2

and

Aβ (z) =

2z(1 + z) . 1 − 2z2

Finally, we obtain AS (z) = 1 + Aα (z) + Aβ (z) =

.

(1 + z)(1 + 2z) . 1 − 2z2

(b) From Exercise 6.9(c), we deduce that BS (z) =

.

(1 + z)(1 + 2z) AS (z) = . 1−z (1 − z)(1 − 2z2 )

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6 Finitely Generated Groups

(c) We have BS (z) =

.

(1 + z)(1 + 2z) (1 − z)(1 − 2z2 )

=−

7 + 10z 6 + 1−z 1 − 2z2

= −6

∞ E

zk + (7 + 10z)

k=0

= −6

∞ E

∞ E

(2z2 )k

k=0

zk +

k=0

∞ ∞ E E (7 · 2k )z2k + (10 · 2k )z2k+1 . k=0

k=0

As BS (z) =

∞ E

.

γS (n)zn ,

n=0

we deduce that γS (n) =

.

{ 7 · 2n/2 − 6 10 · 2(n−1)/2

if n is even −6

if n is odd.

√ (d) We have γS (n) = ( 2)n ρn , where { ρn :=

.

7 − 6 · 2−n/2

if n is even

10 · 2−1/2

if n is odd.

− 6 · 2−n/2

As the sequence (ρn )n∈N is bounded, it follows that infinity. This implies λS = lim

.

n→∞

R n

γS (n) =

√ n ρ tends to 1 as n goes to n

√ 2.

(e) We have 1 ≤ ρn ≤ 10 for all n ∈ N. Thus, we may take C1 := 1 and C2 := 10. (f) We have .

lim inf n→∞

γS (n) γS (2n) γS (2n + 1) γS (n) ≤ lim = 7 < 10 = lim ≤ lim sup n . 2n+1 n→∞ λ2n n→∞ λnS λS n→∞ λ S S

This implies that the sequence (γS (n)/λnS )n∈N is not convergent.

O

6.2 Exercises

333

Comment These formulas are taken from an unpublished work of Machì (see [Har1, Section VI.7]). Note that the inequality λS > 1 is in accordance with the fact that G is non-amenable (cf. Exercise 6.22(f)) and therefore has exponential growth. Exercise 6.24 (Presentation of SL2 (Z)) Let G := SL2 (Z). Consider the matrices S, V ∈ G defined by ( ) 0 −1 .S := 1 0

and

( ) 0 −1 V := 1 1

n := {S, V , V −1 }. Given an integer n ≥ 0, one says that a sequence and let X n is admissible if, for all 1 ≤ i ≤ n − 1, one has xi = S if (xi )1≤i≤n of elements of X −1 and only if xi+1 ∈ {V , V }. (a) Let M ∈ G. Show that there exist an unique integer n ≥ 0, an unique integer n such that ε ∈ {1, −1}, and a unique admissible sequence (xi )1≤i≤n of elements of X M = εx1 x2 · · · xn .

.

This expression is called the normal form of M. (b) Let F (x, y) denote the free group of rank 2 based on the symbols x and n : F (x, y) → G such that y. Show that there is a unique group homomorphism ψ n n ψ (x) = S and ψ (y) = V . n is surjective and that the kernel of ψ n is the normal closure in (c) Show that ψ 4 −3 2 F (x, y) of the set {x , y x }. (d) Show that G admits the presentation G = (x, y : x 2 = y 3 , x 4 = 1).

.

Solution (a) Let I denote the identity matrix in SL2 (Z). Consider the projective modular group Γ := PSL2 (Z) = SL2 (Z)/{I, −I } and the canonical group epimorphism π : G → Γ . Let α, β ∈ Γ be the elements defined by α := π(S) and β := π(V ). Let M ∈ SL2 (Z). By Exercise 6.22(k), there exist an integer n ≥ 0 and an admissible sequence (yi )1≤i≤n of elements of X := {α, β, β −1 } such that π(M) = y1 y2 · · · yn .

.

Consider the sequence (xi )1≤i≤n , where xi := S (resp. xi := V , resp. xi := V −1 ) if and only if yi = α (resp. yi = β, resp. yi = β −1 ). Clearly, (xi )1≤i≤n is an n Moreover, we have admissible sequence of elements of X. π(M) = π(x1 )π(x2 ) · · · π(xn ) = π(x1 x2 · · · xn ).

.

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6 Finitely Generated Groups

As ker(π ) = {I, −I }, this implies M = εx1 x2 · · · xn

.

for some ε ∈ {1, −1}. This shows existence of the normal form. To prove uniqueness, suppose that M = εx1 x2 · · · xn = ε' x1' x2' · · · xn' ' ,

.

(6.14)

where n, n' ≥ 0, ε, ε' ∈ {1, −1}, and (xi )1≤i≤n , (xj' )1≤j ≤n' are admissible n By applying π , we obtain sequences of elements of X. π(x1 )π(x2 ) · · · π(xn ) = π(x1' )π(x2' ) · · · π(xn' ' ).

.

Since (π(xi ))1≤i≤n and (π(xj' ))1≤j ≤n' are both admissible sequences of elements of X, it follows from the uniqueness of the normal form for elements of Γ established in Exercise 6.22(k) that n = n' and π(xi ) = π(xi' ) for all 1 ≤ i ≤ n. This implies xi = xi' for all 1 ≤ i ≤ n. Finally, we then deduce from (6.14) that ε = ε' . This shows uniqueness of the normal form in G. (b) This follows from the universal property of free groups. n is surjective since S and V generate G by Exer(c) The homomorphism ψ n denote the normal closure in F (x, y) of the set {x 4 , y −3 x 2 }. cise 6.14(b). Let N n ⊂ ker(ψ n) since S has order 4 and S 2 = V 3 . We have N Consider the unique homomorphism ψ : F (x, y) → Γ such that ψ(x) = α and n by uniqueness of ψ. Therefore ker(ψ n) ⊂ ker(ψ). ψ(y) = β. We have ψ = π ◦ ψ Moreover, it follows from Exercise 6.22(i) that ker(ψ) = N, where N is the n(x 2 ) = S 2 = −I and ψ n(y 3 ) = normal closure in F (x, y) of the set {x 2 , y 3 }. As ψ 3 n is {I, −I }. This implies that ker(ψ n) is a V = −I , the image of ker(ψ) under ψ n ⊂ ker(ψ n) ⊂ ker(ψ) = N. Therefore, subgroup of index 2 in ker(ψ). We have N n = ker(ψ n has index at most 2 in N. Let n), it suffices to show that N to prove that N n n ∈ N/N n ⊂ F (x, y)/N n. Clearly z commutes with x N n. Since y −3 x 2 ∈ N, z := x 2 N 3 n which shows that z also commutes with y N n. Thus, z is in the we have z = y N, n. Now let n ∈ N. By definition of N, we can write n as a center of F (x, y)/N n = zk for some product of conjugates of x 2 , y 3 , x −2 , or y −3 . We deduce that nN 2 4 n = N, n it follows that nN n ∈ {N, n z}. This shows that N n has k ∈ Z. As z = x N index at most 2 in N. (d) This follows from (c). O Comment Let A ⊂ G (resp. B ⊂ G) denote the cyclic subgroup of order 4 (resp. 6) generated by S (resp. V ) and let C := A ∩ B = {I, −I }. The property established in (a) amounts to saying that G is the free product of A and B amalgamated along C. This is written G = A ∗C B or G = (Z/4Z) ∗Z/2Z (Z/6Z) (see [LynS, Chapter 4], [Robi, Chapter 6], [Rot, Chapter 11], or [Ser3, Chapter 1] for the general definition and properties of free products with amalgamation of groups).

6.2 Exercises

335

Finite presentations for SLn (Z) have been given by Nielsen [Nie2] for n = 3 and by Magnus [Mag2] for all n ≥ 3 (cf. [CoxM, Chapter 7], [Mil1, Corollary 10.3], [Sou]). Exercise 6.25 (Commutator Subgroups of PSL2 (Z) and SL2 (Z)) Recall the notation introduced in Exercise 6.22. The modular group is G := SL2 (Z) and the projective modular group is Γ := PSL2 (Z). The matrices S, V , T ∈ G are defined by ( ) 0 −1 .S := , 1 0

(

) 0 −1 V := , 1 1

( ) 11 T := , 01

and the elements α, β ∈ Γ are given by α := π(S) and β := π(V ), where π : G → Γ is the canonical group epimorphism. (a) Show that there exists a surjective group homomorphism η : Γ → Z/6Z. (b) Show that ker(η) is generated by the elements c1 , c2 ∈ Γ defined by c1 := αβαβ −1 and c2 := αβ −1 αβ. (c) Show that ker(η) = [Γ, Γ ]. (d) Show that [Γ, Γ ] is a free group of rank 2 based on {c1 , c2 }. (e) Show that Γ /[Γ, Γ ] is a cyclic group of order 6 generated by the class of αβ. (f) Show that there is a surjective group homomorphism n η : G → Z/12Z. (g) Consider the matrices C1 , C2 ∈ G defined by ( C1 := −

.

) 21 11

( and

C2 := −

) 11 . 12

Show that the commutator subgroup [G, G] is a free group of rank 2 based on η). {C1 , C2 } and that one has [G, G] = ker(n (h) Show that G/[G, G] is a cyclic group of order 12 generated by the class of T. Solution (a) By Exercise 6.22(i), there is a group isomorphism ψ : F (x, y)/N → Γ , where F (x, y) denotes the free group based on two generators x, y and N is the normal closure of the set {x 2 , y 3 } in F (x, y), such that ψ(xN) = α and ψ(yN) = β. By the universal property of free groups, there is a group homomorphism φ : F (x, y) → Z/6Z such that φ(x) = 3 + 6Z and φ(y) = 2 + 6Z. Observe that φ(y −1 x) = 1 + 6Z generates Z/6Z. Therefore φ is surjective. As φ(x 2 ) = φ(y 3 ) = 6Z is the identity element of Z/6Z, the homomorphism φ induces a surjective group homomorphism ξ : F (x, y)/N → Z/6Z. The composite η := ξ ◦ψ −1 : Γ → Z/6Z is a group homomorphism with the required properties. (b) We first observe that ker(η) has index |Z/6Z| = 6 in Γ and that a complete set of representatives of the cosets of ker(η) in Γ is the set Σ ⊂ Γ given by Σ := {1Γ , β −1 α, β, α, β −1 , βα}.

.

336

6 Finitely Generated Groups

For each γ ∈ Γ , denote by γ ∈ Σ the representative of the coset γ ker(ρ). As the set X := {α, β} generates Γ , it follows from Exercise 6.19(e) that the subset Z ⊂ ker(ρ) defined by Z := {γ γ −1 : γ ∈ ΣX} \ {1Γ }

.

is a generating subset of ker(η). We have Σα = {α, β −1 , βα, 1Γ , β −1 α, β} and Σβ = {β, β −1 αβ, β −1 , αβ, 1Γ , βαβ}.

.

As all the elements of ΣX = Σα ∪ Σβ belong to Σ with the exception of β −1 αβ, αβ, and βαβ, we deduce that Z = {β −1 αβ β −1 αβ

−1

.

, αβ αβ

−1

, βαβ βαβ

−1

}

= {β −1 αβα −1 , αβ(βα)−1 , βαβ(β −1 α)−1 } = {β −1 αβα, αβαβ −1 , βαβαβ} = {c2−1 , c1 , c1−1 c2 }. It follows that c1 and c2 generate ker(η). (c) As α has order 2, we have c1 = [α, β] and c2 = [α, β −1 ]. Therefore c1 and c2 are in [Γ, Γ ]. Since c1 and c2 generate ker(η) by (b), this implies ker(η) ⊂ [Γ, Γ ]. On the other hand, we have [Γ, Γ ] ⊂ ker(η) since Im(η) = Z/6Z is abelian. This shows that [Γ, Γ ] = ker(η). (d) We have c1 c2 = αβαβ −1 αβ −1 αβ,

.

c1 c2−1 = αβαβ −1 (αβ −1 αβ)−1 = αβαβαβα, c2 c1 = αβ −1 αβαβαβ −1 , c2 c1−1 = (c1 c2−1 )−1 = αβ −1 αβ −1 αβ −1 α, c1−1 c2 = (αβαβ −1 )−1 αβ −1 αβ = βαβαβ, c1−1 c2−1 = (c2 c1 )−1 = βαβ −1 αβαβ −1 α, c2−1 c1 = (c1−1 c2 )−1 = β −1 αβ −1 αβ −1 , c2−1 c1−1 = (c1 c2 )−1 = β −1 αβαβαβ −1 α. We deduce that if γ ∈ Γ satisfies γ = uk11 uk22 · · · uknn ,

.

6.2 Exercises

337

where n ≥ 1, ui ∈ {c1 , c2 } and ki ∈ Z \ {0} for all 1 ≤ i ≤ n, and ui /= ui+1 for 1 ≤ i ≤ n − 1, then the normal form of γ has length at least 4(|k1 | + |k2 | + · · · + |kn |) − 3(n − 1) ≥ n + 3 ≥ 1. Therefore γ /= 1Γ . This shows that the subgroup of Γ generated by c1 and c2 is free with base {c1 , c2 } (cf. [CAG, Corollary D.3.4]). (e) We have Γ /[Γ, Γ ] = Γ / ker(η) ∼ = Im(η) = Z/6Z. Therefore Γ /[Γ, Γ ] is cyclic of order 6. The class of αβ generates Γ /[Γ, Γ ] since η(αβ) = η(α)+η(β) = 5 + 6Z generates Z/6Z. n : F (s, v)/N → G, (f) By Exercise 6.24(d), there is a group isomorphism ψ where F (s, v) denotes the free group based on two generators s, v and N is the n(sN) = S and ψ n(vN) = normal closure of the set {s 2 v −3 , s 4 } in F (s, v), such that ψ V . By the universal property of free groups, there is a group homomorphism n: F (s, v) → Z/12Z such that φ n(s) = 3 + 12Z and φ n(v) = 2 + 12Z. Observe φ 2 −3 4 n n that φ (s v ) = φ (s ) = 12Z is the identity element of Z/12Z. Therefore, the n induces a group homomorphism n homomorphism φ ξ : F (s, v)/N → Z/12Z. The n−1 : G → Z/12Z is a group homomorphism satisfying composite n η := n ξ ◦ψ η(S) − n η(V ) = 1 + 12Z n η(S) = 3 + 12Z and n η(V ) = 2 + 12Z. Since n η(SV −1 ) = n generates Z/12Z, the homomorphism n η is surjective. (g) We have [G, G] ⊂ ker(n η) since Z/16Z is abelian. As n η(−I ) = n η(S 2 ) = 2n η(S) = 6 + 12Z /= 12Z, we have −I ∈ / ker(n η) and hence −I ∈ / [Γ, Γ ]. It follows that π induces by restriction a group isomorphism [G, G] → [Γ, Γ ]. An easy computation shows that C1 = [S, V ] and C2 = [S, V −1 ]. Consequently, we have −C1 , −C2 ∈ [G, G]. Moreover, π(C1 ) = π([S, V ]) = [π(S), π(V )] = [α, β] = c1 and π(C2 ) = π([S, V −1 ]) = [π(S), π(V )−1 ] = [α, β −1 ] = c2 . As [Γ, Γ ] is a free group with base {c1 , c2 }, this shows that [G, G] is a free group based on {C1 , C2 }. Observe that ker(n η) has index |Z/12Z| = 12 in G. On the other hand, π induces a surjective group homomorphism G/[G, G] → Γ /[Γ, Γ ] whose kernel {[G, G], −[G, G]} has two elements and [Γ, Γ ] is of index 6 in Γ by (e). Thus, the commutator subgroup [G, G] is also of index 12 in G. Since [G, G] ⊂ ker(n η), we conclude that [G, G] = ker(n η). (h) We have G/[G, G] = G/ ker(n η) ∼ η) = Z/12Z. Therefore G/[G, G] = Im(n is cyclic of order 12. The class of T generates G/[G, G] since T = SV −2 so that n η(T ) = n η(SV −2 ) = n η(S) − 2n η(V ) = −1 + 12Z generates Z/12Z. O Comment One can also prove (d) by playing ping-pong with c1 and c2 on the real projective line. The homological interpretation of (e) and (h) is that H1 (PSL2 (Z), Z) = Z/6Z and H1 (SL2 (Z), Z) = Z/12Z (see [Bro]). Exercise 6.26 (Short-Lex Ordering) Let A be a set. Recall that it follows from the Axiom of Choice that A admits a well-ordering, that is, a total ordering ≤ with the property that every non-empty subset of A has a minimal element. Consider the monoid A∗ consisting of all words on the alphabet A and denote by |u| the length of a word u ∈ A∗ . Define a relation < on A∗ by setting, for all u, v ∈ A∗ , u ≺ v if either |u| < |v| or there exist w, u' , v ' ∈ A∗ with |u' | = |v ' | and x, y ∈ A with x < y, such that u = u' xw and v = v ' yw.

338

6 Finitely Generated Groups

Show that < is a well-ordering on A∗ . Solution To show that < is an ordering, we only need to check for the transitivity property since the reflexive and anti-symmetric properties are trivially satisfied. Thus, let u, v, z ∈ A∗ and suppose that u ≺ v and v ≺ z. If |u| < |v| or |v| < |z|, then |u| < |z| and hence u ≺ z. Thus, we may suppose that |u| = |v| = |z|. Then there exist w, u' , v ' ∈ A∗ (resp. w ' , v '' , z'' ∈ A∗ ) and x, y ∈ A with x < y (resp. s, t ∈ A with s < t) such that u = u' xw and v = v ' yw (resp. v = v '' sw ' and z = z'' tw ' ). Note that |u' | = |v ' | and |v '' | = |z'' |. If |w| = |w' |, then indeed w = w ' and y = s so that x < y = s < t, and therefore u ≺ z. Suppose now that |w| < |w ' |. Since v = v ' yw = v '' sw ' , there exists w '' ∈ A∗ such that w ' = w '' yw. We thus have v ' yw = v = (v '' sw '' )yw and z = z'' tw ' = (z'' tw '' )yw, so that |u' | = |v ' | = |v '' sw '' | = |z'' tw '' |. We deduce that u = u' xw ≺ (z'' tw '' )yw = z'' t (w '' yw) = z'' tw ' = z. Finally, suppose that |w' | < |w|. Since v = v '' sw ' = v ' yw, there exists w ''' ∈ A∗ such that w = w ''' sw ' . We thus have u = (u' xw '' )sw ' and v = v ' yw '' sw ' = v '' sw ' so that, in particular, |u' xw ''' | = |v '' | = |z'' |. We deduce that u = u' xw = u' x(w ''' sw ' ) = (u' xw ''' )sw ' ≺ z'' tw ' = z. This proves the transitivity property of < and therefore that < is an ordering. Let us show that < is total. Let u, v ∈ A∗ . If |u| /= |v|, say, |u| < |v|, we have u ≺ v. Suppose now that |u| = |v| =: n and u /= v. Then u = x1 x2 · · · xn and v = y1 y2 · · · yn with x1 , x2 , . . . , xn , y1 , y2 , · · · , yn ∈ A. Let 1 ≤ i ≤ n be the maximal index such that xi /= yi . Then, setting u' := x1 x2 · · · xi−1 , v ' := y1 y2 · · · yi−1 , x := xi , y := yi , and w := xi+1 xi+2 · · · xn = yi+1 yi+2 · · · yn , we have u = u' xw and v = v ' yw. Since ≤ is a total ordering on A and x /= y we necessarily have either x < y, so that u ≺ v, or y < x, so that v ≺ u. This shows that < is a total ordering on A∗ . We are only left to show that < is a well ordering. Let then U ⊂ A∗ be a nonempty set and let us show that U has a minimal element. The set {|u| : u ∈ U } ⊂ N is non-empty and therefore (the usual ordering on N is a well-ordering) it has a minimal element l. Let U (l) := {u ∈ U : |u| = l}. If l = 0, then ε ∈ U is the seeked minimal element. Suppose that l ≥ 1. Then, for all u ∈ U (l), there exist xiu ∈ A, 1 ≤ i ≤ l, such that u = x1u x2u · · · xlu . Consider the set Al := {xlu : u ∈ U (l)} ⊂ A. Clearly, Al is non-empty. Since ≤ is a well-ordering on A, there exists a minimal element xl ∈ Al . If l = 1, we are done, since x1 = xl is the minimal element of U (l) and therefore of U . If l ≥ 2, let Ul := {u ∈ U (l) : xlu = xl } ⊂ u U (l). Note that Ul /= ∅, and set Al−1 := {xl−1 : u ∈ Ul } ⊂ A. As Al−1 is non-empty, it admits a minimal element xl−1 ∈ Al−1 . If l = 2, we are done, since x1 x2 = xl−1 xl is the least element of U (l) and therefore of U . Continuing this way, we eventually find a minimal element u = x1 x2 · · · xl−1 xl ∈ U (l). Such an element is also minimal in U . This shows that < is a well-ordering. O

6.2 Exercises

339

Comment The ordering < is called the right short-lex ordering on A∗ associated with the well-ordering ≤ on A. Alternatively, setting, for all u, v ∈ A∗ , u ≺ v if either |u| < |v| or there exist w, u' , v ' ∈ A∗ with |u' | = |v ' | and x, y ∈ A with x < y, such that u = wxu' and v = wyv '

defines another well-ordering on A∗ , which is called the left short-lex ordering associated with ≤. Exercise 6.27 (Schreier Systems) Let A be a set. One says that a word u ∈ A∗ is a prefix of a word w ∈ A∗ if there exists a word v ∈ A∗ such that w = uv. A subset L ⊂ A∗ is said to be prefix-closed provided that, given any word w ∈ L, every prefix of w is itself in L. Let F be a free group based on X ⊂ F . Setting A := X ∪ X−1 , one can regard F as the set of all reduced words in A∗ . Let H be a subgroup of F . Show that there exists a complete set of representatives for the right cosets of H in F which is prefix-closed. Such a prefix-closed complete set of representatives for the right cosets of H in the free group F equipped with the base X is called a Schreier system. Solution Choose a well-ordering ≤ on A and let < denote the right short-lex ordering on A∗ associated with ≤ (cf. Exercise 6.26). The restriction of < to F yields a well-ordering on F . For each g ∈ F , denote by g the minimal element of the right coset H g. Then S := {g : g ∈ F } is a complete set of representatives of the right cosets of H in F . We claim that S is prefix-closed. Let s ∈ S, say s = a1 a2 · · · an with ai ∈ A for i = 1, 2, . . . , n and l(s) = n. Consider the element t := a1 a2 · · · an−1 ∈ S. Since H t = H a1 a2 · · · an−1 , we have t < a1 a2 · · · an−1

.

(6.15)

so that, in particular, l(t) ≤ l(a1 a2 · · · an−1 ) = n − 1. Now, H s = (H a1 a2 · · · an−1 )an = H tan so that n = l(s) ≤ l(tan ) ≤ l(t) + l(an ) ≤ (n − 1) + 1 = n. We deduce that l(t) = n − 1 and l(tan ) = n, that is, tan is reduced. Thus, since (a1 a2 · · · an−1 )an = s < tan , we get a1 a2 · · · an−1 < t,

.

(6.16)

by definition of the short-lex ordering. From (6.15) and (6.16) we deduce that O a1 a2 · · · an−1 = t ∈ S. By induction, this shows that S is prefix-closed. Exercise 6.28 (The Nielsen-Schreier Theorem) Let F be a free group and let H ⊂ F be a subgroup. Let X ⊂ F be a base of F . Let also S ⊂ F be a Schreier system of representatives of the right cosets of H in F with respect to X (cf. Exercise 6.27). Recall that one can identify each element g of F with its reduced form, which is a word w on the alphabet X ∪ X−1 . We denote by l(g) ∈ N the length of w, and, provided that g /= 1F , we denote by α(g) ∈ X ∪ X−1 (resp. ω(g) ∈ X ∪ X−1 ) the first letter (resp. the last letter) of w. For g ∈ F , let g denote

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6 Finitely Generated Groups

the representative in S of the right coset H g. Recall (cf. Exercise 6.19) that the set U := Y (X, S) \ {1F } ⊂ H (cf. (6.9)) is a generating subset of H . (a) Let g1 , g2 ∈ F \ {1F } such that ω(g1 ) /= α(g2 )−1 . Show that l(g1 g2 ) = l(g1 ) + l(g2 ). (b) Let s ∈ S and a ∈ X ∪ X−1 . Suppose that u := sa(sa)−1 /= 1F . Show that l(u) = l(s) + 1 + l(sa). (c) Let s1 , s2 ∈ S and a1 , a2 ∈ X ∪ X−1 . Set t1 := s1 a1 , t2 := s2 a2 , u1 := s1 a1 (s1 a1 )−1 = s1 a1 t1−1 , and u2 := s2 a2 (s2 a2 )−1 = s2 a2 t2−1 . Suppose that u1 , u2 /= 1F and u2 /= u−1 1 . Show that: (i) if l(t1 ) < l(s2 ), then t1 a1−1 is not a prefix of s2 ; (ii) if l(t1 ) > l(s2 ), then s2 a2 is not a prefix of t1 ; (iii) if l(t1 ) = l(s2 ), then a1 t1−1 s2 a2 /= 1F . Deduce that setting w := t1−1 s2 ∈ F one has u1 u2 = s1 a1 wa2 t2−1 and l(u1 u2 ) = l(s1 ) + 1 + l(w) + 1 + l(t1 ). (d) Let u1 , u2 , . . . un ∈ U ∪U −1 . Suppose that ui /= u−1 i+1 for i = 1, 2, . . . , n−1. Show that l(u1 u2 · · · un ) ≥ n − 1. (e) Show that H is a free group based on U . Solution (a) Let w1 = a1 a2 · · · am and w2 = b1 b2 · · · bn , with a1 , a2 , . . . , am , b1 , b2 , . . . , bn ∈ X ∪ X−1 , denote the reduced words representing g1 and g2 , respectively. By hypothesis, am = ω(g1 ) /= α(g2 )−1 = b1−1 so that the concatenation w1 w2 is reduced. As a consequence, l(g1 g2 ) = l(w1 w2 ) = l(a1 a2 · · · am b1 b2 · · · bn ) = m + n = l(g1 ) + l(g2 ). (b) We first show that l(sa) = l(s) + 1. If not, by (a) there exists t ∈ F such that, in reduced form, s = ta −1 . Note that t ∈ S because S is a Schreier system. But then sa = ta −1 a = t = t, which gives sa(sa)−1 = ta −1 at −1 = 1F , contradicting our assumptions. As a consequence, the word sa ∈ (X ∪ X−1 )∗ is reduced. We now show that l(a(sa)−1 ) = 1 + l(sa). If not, by (a) there exists t ∈ F such that, in reduced form, (sa)−1 = a −1 t −1 , so that sa = ta. As before, t ∈ S. Moreover, from sa = ta we deduce that H sa = H ta, hence t = s. This gives sa(sa)−1 = saa −1 s −1 = 1F , contradicting our assumptions. As a consequence, the word a(sa)−1 ∈ (X ∪ X−1 )∗ is reduced, in particular α((sa)−1 ) /= a −1 . Setting g1 := sa and g2 := (sa)−1 , we have l(g1 ) = l(sa) = l(s) + 1 so that g1 /= 1F . If g2 = 1F , then l(sa) = 0 so that l(u) = l(g1 g2 ) = l(g1 ) = l(s) + 1 = l(s) + 1 + l(sa). If g2 /= 1F then ω(g1 ) = a /= α((sa)−1 )−1 = α(g2 )−1 so that, by (a), l(u) = l(g1 g2 ) = l(g1 ) + l(g2 ) = l(s) + 1 + l(sa). (c) We argue by contradiction. (i) Since S is a Schreier system, the prefix t1 a1−1 = s1 a1 a1−1 would belong to S. Hence, using Exercise 6.19(c), s1 a1 a1−1 = s1 a1 a1−1 = s1 a1 a1−1 = s1 = s1 . We deduce that s1 a1 = s1 a1 , so that u1 = s1 a1 (s1 a1 )−1 = 1F , a contradiction. (ii) Again, since S is a Schreier system, the prefix s2 a2 would belong to S. Hence, t2 = s2 a2 and u2 = s2 a2 t2−1 = 1F , a contradiction.

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(iii) Suppose that a1 t1−1 s2 a2 = 1F . This means that s1 a1 = t1 = s2 and a1 =

a2−1 . Using Exercise 6.19(c) once again, s2 a2 = s1 a1 a2 = s1 a1 a1−1 = s1 a1 a1−1 = s1 = s1 . This gives us u1 = u−1 2 , a contradiction. Finally, when performing the multiplication of u1 and u2 , some cancellation may occur. Roughly speaking, we show that such cancellation stops before reaching a1 (from the right) and a2 (from the left). Let z denote the maximal suffix of u1 which equals the inverse of a prefix of u2 . We distinguish the following three cases.

• If l(t1 ) < l(s2 ), then it follows from (i) that z is a suffix of t1−1 so that l(z) ≤ l(t1 ) < l(s2 ). As a consequence, z−1 is a prefix of s2 . • If l(t1 ) > l(s2 ), then it follows from (ii) that z−1 is a prefix of s2 so that l(z) ≤ l(s2 ) < l(t1 ). As a consequence, z is a suffix of t1−1 . • If l(t1 ) = l(s2 ), then it follows from (iii) that z is a suffix of t1−1 and z−1 is a prefix of s2 . In either case, there exist w1 , w2 ∈ F2 such that t1−1 = w1 z and s2 = z−1 w2 . We then have w = t1−1 s2 = (w1 z)(z−1 w2 ) = w1 w2 and l(w) = l(w1 ) + l(w2 ), by maximality of z. As a consequence, u1 u2 = s1 a1 wa2 t2−1 and l(u1 u2 ) = l(s1 ) + 1 + l(w) + 1 + l(t1 ). This proves (c). (d) We first observe that, by virtue of Exercise 6.19(d), we may assume that each ui is of the form ui = si ai (si ai )−1 , where si ∈ S and ai ∈ X ∪ X−1 . It follows from (c) that there exist wi ∈ F such that ui ui+1 = si ai wi ai+1 (si+1 ai+1 )−1 and l(ui ui+1 ) = l(s1 ) + 1 + l(wi ) + 1 + l(si+1 ai+1 ), for all i = 1, 2, . . . , n − 1. Thus u1 u2 · · · un = s1 a1 w1 a2 w2 · · · an−1 wn−1 (sn an )−1 and l(u1 u2 · · · un ) = l(s1 )+1+ l(w1 ) + 1 + l(w2 ) + · · · + 1 + l(wn−1 ) + l(sn an ) ≥ n − 1. (e) We already know (cf. Exercise 6.19(d)) that U generates H . Thus, by virtue of [CAG, Corollary D.3.4], in order to show that U is a free base for H it suffices to show that every nonempty reduced product u1 u2 · · · un of elements in U ∪ U −1 is different from 1F . As l(u1 u2 · · · un ) ≥ n − 1 by (d), we deduce that u1 u2 · · · un /= 1F O Comment The result in (e), namely that subgroups of free groups are free, is called the Nielsen-Schreier theorem. It was first proved for finitely generated free groups by Nielsen [Nie1] and subsequently established in its full generality by Schreier [Schr]. In [Ser3, Theorem 5], Serre presented a geometrical proof of the Nielsen-Schreier theorem based on the Bass-Serre theory of group actions on trees. Exercise 6.29 (Schreier Index-Rank Formula) Let F be a free group of finite rank and let H be a finite index subgroup of F . (a) Show that H is a free group of finite rank. (b) Let X ⊂ F be a free base for F and let S ⊂ F be a Schreier system of representatives of the right cosets of H in F with respect to X (cf. Exercise 6.27). Recall that one can identify each element of F with its reduced form, which is a word on the alphabet X ∪ X−1 . For g ∈ F , let g denote the representative in S of

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the right coset H g. Set Z := {(s, x) ∈ S × X : sx ∈ / S}. Show that the map from Z into H , given by (s, x) |→ sx(sx)−1 , is injective. (c) Let n (resp. r) denote the rank of F (resp. H ). Show that one has r = 1 + [F : H ](n − 1).

.

(6.17)

Solution (a) The subgroup H is free by the Nielsen-Schreier theorem (see Exercise 6.28). On the other hand, H is finitely generated since every finite index subgroup of a finitely generated group is itself finitely generated [CAG, Proposition 6.6.6]. Therefore H is a free group of finite rank. (b) For g ∈ F , let l(g) denote the word length of g. It follows from Exercise 6.28(b) that if (s, x) ∈ Z then u := sx(sx)−1 satisfies l(u) = l(s) + 1 + l(sx). Identifying s, sx, and u with their reduced forms, this implies in particular that s is a prefix of sx and that sx is a prefix of u. We deduce that the maps (s, x) |→ sx and (s, x) |→ sx(sx)−1 are injective on Z. (c) It follows from Exercise 6.28(e) that U := {sx(sx)−1 : (s, x) ∈ Z} is a base for H . Therefore r = |Z| by (b). Let T := S \ {1F }. Consider the map ϕ : T → (S × X) \ Z defined as follows. Let t ∈ T and denote by y ∈ X ∪ X−1 the last letter of t. We set ϕ(t) := (ty −1 , y) if y ∈ X and ϕ(t) := (t, y −1 ) if y ∈ X−1 . Observe that the map ϕ is well defined since S is prefix-closed. Note also that ϕ is bijective with inverse map ϕ −1 : (S × X) \ Z → T given by (s, x) |→ t, where t := s if s /= 1F ends with the letter x −1 , and t := sx in all other cases. We deduce that |(S × X) \ Z| = |T | = |S| − 1, so that r = |Z| = |(S × X) \ ((S × X) \ Z)|

.

= |S × X| − |(S × X) \ Z| = |S| · |X| − (|S| − 1) = [F : H ]n − ([F : H ] − 1) = 1 + [F : H ](n − 1). This shows (6.17).

O

Exercise 6.30 (Cofinitely Hopfian Groups) A group G is said to be cofinitely Hopfian if every endomorphism of G whose image has finite index in G is an automorphism of G. (a) Show that every cofinitely Hopfian group is Hopfian. (b) Show that every non-trivial finite group is Hopfian but not cofinitely Hopfian. (c) Show that the group Z is Hopfian but not cofinitely Hopfian. (d) Show that every free group of finite rank n ≥ 2 is cofinitely Hopfian.

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Solution (a) Let G be a cofinitely Hopfian group and let φ : G → G be a surjective endomorphism. Then [G : φ(G)] = [G : G] = 1 < ∞. As G is cofinitely Hopfian, we deduce that φ is an automorphism of G. This shows that G is Hopfian. (b) Let G be a non-trivial finite group. The group G is Hopfian since every finite group is Hopfian [CAG, Example 2.4–2.(a)]. Let ψ : G → G be the endomorphism defined by setting ψ(g) := 1G for all g ∈ G. Then [G : ψ(G)] = [G : {1G }] = |G| < ∞. However, since G is assumed to be non-trivial, ψ is not an automorphism of G. Therefore, G is not cofinitely Hopfian. (c) The group Z is Hopfian by using [CAG, Corollary 2.4.5] (or, alternatively, by observing that there are only two surjective endomorphisms of Z, namely IdZ and − IdZ , and that they are both injective). The endomorphism φ : Z → Z, defined by φ(n) := 2n for all n ∈ Z, has an image subgroup of index 2 in Z. However, φ is not an automorphism of Z since it is not surjective. Therefore, Z is not cofinitely Hopfian. (d) Let F be a free group of finite rank n ≥ 2. Let φ : F → F be an endomorphism such that [F : φ(F )] < ∞. By Exercise 6.29(c), the group φ(F ) is free of finite rank r with r = 1+[F : φ(F )](n−1). If {x1 , x2 , . . . , xn } is a free base for F , then the set {φ(x1 ), φ(x2 ), . . . , φ(xn )} generates φ(F ). Using Exercise 2.27, we deduce that r ≤ n. As [F : φ(F )](n − 1) = r − 1 and n ≥ 2, this implies n = r and [F : φ(F )] = 1. Therefore φ(F ) = F . This shows that φ is surjective. As all free groups of finite rank are Hopfian [CAG, Corollary 2.4.5], we deduce that φ is an automorphism of F . Therefore, F is cofinitely Hopfian. O Comment We have seen in [CAG, Remark 2.4–6.(a)] that free groups of infinite rank are not Hopfian. Combining this observation with (a), (c), and (d), we deduce that a free group is cofinitely Hopfian if and only if it has a finite rank n ≥ 2. In [Sel1], Zlil Sela established the Hopficity of all torsion-free groups that are word hyperbolic in the sense of Gromov [Gro2]. Sela’s result was extended in [BriHM, Theorem 4.3], where it is shown that every non-elementary torsion-free word hyperbolic group is cofinitely Hopfian (see also [BriGHM]). This covers (d) since all free groups of finite rank are word hyperbolic. Note that every non-trivial torsion-free elementary word hyperbolic group is infinite cyclic and consequently Hopfian but not cofinitely Hopfian (cf. (c)). The question whether or not all word hyperbolic groups are residually finite remains open. Exercise 6.31 (Rank of a Finitely Generated Group) Let G be a finitely generated group. The minimal cardinality |X| of a (not necessarily symmetric) generating subset X ⊂ G is called the rank of G and it is denoted by rk(G). (a) Show that rk(G) = 0 (resp. rk(G) = 1) if and only if G is a trivial group (resp. is a nontrivial cyclic group). (b) Let n ≥ 0 be an integer. Show that if F is a free group of rank n, then rk(F ) = n. (c) Let π : G → K be a group epimorphism of G onto a group K. Show that rk(K) ≤ rk(G).

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(d) Show that rk(G) is equal to the minimal n ∈ N such that there exist a free group F of rank n and an group epimorphism π : F → G. (e) Let d ≥ 1 be an integer. Show that rk(Zd ) = d. (f) Let G1 and G2 be two finitely generated groups. Show that rk(G1 × G2 ) ≤ rk(G1 ) + rk(G2 ) and provide an example showing that one may have a strict inequality. (g) Let H be a finite index subgroup of G. Show that H is finitely generated and that one has rk(H ) ≤ 1 + [G : H ](rk(G) − 1). Solution (a) Recall that given a subset X ⊂ G, the subgroup (X) ⊂ G generated by X is the intersection of all the subgroups H ⊂ G such that X ⊂ H . In particular, (∅) = {1G }. Suppose that G is a trivial group. Then G = (∅) so that rk(G) = |∅| = 0. Conversely, if rk(G) = 0, then G is generated by ∅, so that G = {1G } is a trivial group. Suppose now that G is a nontrivial cyclic group. As G is cyclic, there exists an element g ∈ G which generates G. It follows that rk(G) ≤ |{g}| = 1. Since G is non-trivial, we have rk(G) = 1. The converse is obvious. (b) Let F be a free group of rank n. This means that there exists a base X ⊂ F such that |X| = n. As X generates F (cf. [CAG, Proposition D.2.3]), we have rk(F ) ≤ |X| = n. Consequently, if Y ⊂ F is a finite generating subset such that |Y | = rk(F ), it follows from Exercise 2.26 that |Y | = n. This shows that rk(F ) = n. (c) Let X ⊂ G be a generating subset of G such that |X| = rk(G). Then Y := π(X) ⊂ K is a generating subset of K. Therefore, one has rk(K) ≤ |Y | ≤ |X| = rk(G). (d) Denote by n(G) the minimal n ∈ N such that there exist a free group F of rank n and an epimorphism π : F → G. Let X ⊂ G be a finite generating subset such that |X| = rk(G). Let F be a free group of rank rk(G) and let Y ⊂ F be a free base. Then |Y | = rk(G) and there exists a bijective map f : Y → X. By the universal property of free groups, f extends to a group homomorphism φ : F → G. Since f (Y ) = X ⊂ G is a generating subset, φ is an epimorphism. This shows that n(G) ≤ rk(G). On the other hand, by definition of n(G), we can find a group epimorphism φ ' : F ' → G where F ' is a free group of rank n(G). It follows from (c) and (b) that rk(G) ≤ rk(F ' ) = n(G). This shows that n(G) = rk(G). (e) The set X := {ei : 1 ≤ i ≤ d} ⊂ Zd , where ei ∈ Zd is the element whose i-th coordinate is 1 and all other coordinates are 0, generates Zd . As X has cardinality d, we deduce that rk(Zd ) ≤ d. Suppose now that Y ⊂ Zd is a finite generating subset. Consider the Q-vector space V := Qd . Observe that Y ⊂ Zd ⊂ V . We claim that the Q-linear span of Y is the whole of V . To see this, let v = (v1 , v2 , . . . , vd ) ∈ V and denote by n ∈ N \ {0} the lowest common denominator of the numbers d v1 , v2 , . . . , vd ∈ Q. Then nv = (nv1 , nv2 , . . . , nvdE ) ∈ Zd . As Y generates EZ , there exists a family of integers (ay )y∈Y such that nv = y∈Y ay y. Then v = y∈Y by y, where by := ay /n ∈ Q for all y ∈ Y . This proves the claim. From basic linear algebra, we deduce that d = dimQ (V ) ≤ |Y |. This shows that rk(Zd ) = d.

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(f) Let Xi ⊂ Gi be a finite generating subset of Gi such that |Xi | = rk(Gi ), for i = 1, 2. As X := (X1 × {1G2 }) ∪ ({1G1 } × X2 ) generates G1 × G2 , we have rk(G1 × G2 ) ≤ |X| = |X1 | + |X2 | = rk(G1 ) + rk(G2 ). Let G1 := Z/2Z and G2 := Z/3Z. As G1 × G2 is a cyclic group of order 6, we deduce from (a) that rk(G1 × G2 ) = 1 < 2 = rk(G1 ) + rk(G2 ). (g) The group H is finitely generated since every finite index subgroup of a finitely generated group is itself finitely generated [CAG, Proposition 6.6.6]. It follows from (d) that there exist a free group F of rank rk(G) and a group epimorphism φ : F → G. Consider the subgroup K := φ −1 (H ) ⊂ F . The map Kf |→ H φ(f ), f ∈ F , yields a bijection from the set of right cosets of K in F onto the set of right cosets of H in G. Therefore [F : K] = [G : H ] (third isomorphism theorem). Observe now that the group K is free of rank 1 + [F : K](rk(F ) − 1) by Schreier’s index formula (cf. Exercise 6.29(b)). As φ induces by restriction a group epimorphism of K onto H , we deduce from (d) that .

rk(H ) ≤ rk(K) = 1 + [F : K](rk(F ) − 1) = 1 + [G : H ](rk(G) − 1). O

Comment There are only countably many isomorphism classes of groups of rank less than 2 since any such group is cyclic and hence isomorphic to Z/nZ for some n ∈ N. On the other hand, Neumann [Neu1] (see also [Har1, Section III.B]) proved that there are uncountably many pairwise non-isomorphic groups of rank 2. Exercise 6.32 (A Theorem of Ralph Strebel) Let G be a finitely generated residually finite group. Suppose that for every finite index subgroup H of G, the inequality established in Exercise 6.31(g) is actually an equality, i.e., one has .

rk(H ) = 1 + [G : H ](rk(G) − 1).

(6.18)

The goal of this exercise is to show that G is free. (a) Let A be a set. Recall that it follows from the Axiom of Choice that A admits a well-ordering, that is, a total ordering ≤ with the property that every non-empty subset of A has a minimal element. Let (≤i )i≥1 be a sequence of well-orderings on A. Consider the monoid A∗ consisting of all words on the alphabet A and denote by |u| the length of a word u ∈ A∗ . Define a relation < on A∗ by setting, for all u, v ∈ A∗ , u ≺ v if either |u| < |v| or there exist w, u' , v ' ∈ A∗ and x, y ∈ A such that u = u' xw and v = v ' yw with |u' | = |v ' | and x 1. n→∞

This shows that the group G has exponential growth.

O

Comment The Baumslag-Solitar group BS(1, m) is metabelian by Exercise 4.45(a) and therefore amenable. Consequently, it contains no free subgroups of rank 2. On the other hand, if n and m are integers such that |n|, |m| ≥ 2, it can be shown that the Baumslag-Solitar group BS(n, m) := (a, b : abn a −1 = bm ) contains a free subgroup of rank 2 (this can be deduced from Britton’s lemma for HNN-extensions,

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see [LynS, Chapter 4]). Thus, for |n|, |m| ≥ 2, the group BS(n, m) is non-amenable (and hence has exponential growth). Exercise 6.37 Let G be a finitely generated group. Suppose that there exist a set A with more than one element and an injective monoid homomorphism ϕ : A∗ → G. Show that G has exponential growth. Solution Let a, b ∈ A be two distinct elements and set B := {a, b}. Let H ⊂ G denote the subgroup generated by α := ϕ(a) and β := ϕ(b). Restricting ϕ to B ∗ ⊂ A∗ yields an injective monoid homomorphism B ∗ → H . Setting S := {α, α −1 , β, β −1 } ⊂ H , we have that γSH (n) ≥ |ϕ(B n )| = |B n | = 2n , so that H has exponential growth. We deduce (cf. [CAG, Corollary 6.6.4]) that G has itself exponential growth. O Exercise 6.38 Let G be a group acting on a set X. Suppose there exist g1 , g2 ∈ G and non-empty disjoint subsets X1 , X2 ⊂ X such that g1 (X1 ∪ X2 ) ⊂ X1

.

and

g2 (X1 ∪ X2 ) ⊂ X2 .

(6.20)

(a) Let A := {a1 , a2 } be a set with two distinct elements. Show that the unique monoid homomorphism ϕ : A∗ → G such that ϕ(a1 ) = g1 and ϕ(a2 ) = g2 is injective. (b) Deduce that if G is finitely generated, then G has exponential growth. Solution (a) Denote by |w| the length of a word w ∈ A∗ . In order to prove that ϕ is injective, let us show, by induction on n := min(|w|, |w' |), that if w, w ' ∈ A∗ satisfy w /= w ' then ϕ(w) /= ϕ(w ' ). By symmetry, we can assume |w| ≤ |w' |. If n = 0, then w is the empty word E and ϕ(w) = 1G . On the other hand, |w ' | ≥ 1 and, denoting by aj the first letter of w ' , we deduce from (6.20) and a ping-pong-type argument that ϕ(w' )(X1 ∪ X2 ) ⊂ Xj . As X1 and X2 are disjoint and non-empty, this implies ϕ(w' ) /= 1G and hence ϕ(w ' ) /= ϕ(w). Suppose now that n ≥ 1. Let ai (resp. aj ) denote the first letter of w (resp. w ' ). Then ϕ(w)(X1 ∪ X2 ) ⊂ Xi and ϕ(w ' )(X1 ∪ X2 ) ⊂ Xj . We deduce that ϕ(w) /= ϕ(w ' ) if ai /= aj . In the case when ai = aj , we have w = ai v and w ' = ai v ' , where v, v ' ∈ A∗ have length |v| = |w| − 1 and |v ' | = |w ' | − 1. Moreover, ϕ(w) = ϕ(ai )ϕ(v) and ϕ(w ' ) = ϕ(ai )ϕ(v ' ), so that ϕ(v) /= ϕ(v ' ). By induction, we deduce that v /= v ' . It follows that w /= w ' . This shows that ϕ is injective. (b) This follows from (a) and Exercise 6.37. O Exercise 6.39 (Affine Representation of BS(1, m)) Let m be an integer such that |m| ≥ 2. Consider the group G given by the presentation G = (a, b : aba −1 = bm ) (cf. Exercise 2.13, Exercise 4.45, and Exercise 6.36). (a) Let α, β : R → R be the maps respectively defined by α(t) = mt and β(t) = t + 1 for all t ∈ R. Show that there is a unique group homomorphism ϕ : G → Sym(R) satisfying ϕ(a) = α and ϕ(b) = β. (b) Show that ϕ is injective.

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(c) Let n0 be an integer such that |m|n0 ≥ 3. Consider the elements λ, μ ∈ Sym(R) respectively defined by λ = α −n0 and μ = βλβ −1 . Check that the open intervals I, J ⊂ R defined by I := (−1/2, 1/2) and J := (1/2, 3/2) satisfy λ(I ∪ J ) ⊂ I and μ(I ∪ J ) ⊂ J . (d) Use (c) to get another proof of the fact that G has exponential growth. Solution (a) Uniqueness follows from the fact that a and b generate G. To prove existence, it suffices to observe that (αβα −1 )(t) = (αβ)(t/m) = α(t/m + 1) = t + m = β m (t)

.

for all t ∈ R, which shows that α and β satisfy the relation αβα −1 = β m . (b) Let g ∈ ker(ϕ). By Exercise 2.13(a), we can find i, j, k ∈ Z, with i ≤ 0 and k ≥ 0, such that g = a i bj a k . It follows that for any t ∈ R we have t = ϕ(g)(t) = ϕ(a i bj a k )(t) = (α i β j α k )(t)

.

= (α i β j )(mk t) = α i (mk t + j ) = mi+k t + mi j. Taking t = 0 forces j = 0 and then taking t = 1 forces k = −i, so that g = 1G . This shows that ϕ is injective. (c) Let t ∈ I ∪ J = (−1/2, 3/2) \ {1/2}. We have λ(t) = α −n0 (t) = m−n0 t, so that |λ(t)| = |t|/(|m|n0 ) < 1/2. Therefore λ(t) ∈ (−1/2, 1/2) = I . This shows that λ(I ∪ J ) ⊂ I . On the other hand, we have μ(t) = (βλβ −1 )(t) = (βλ)(t − 1) = β(m−n0 (t − 1)) = m−n0 (t − 1) + 1.

.

As t − 1 ∈ (−3/2, 1/2) and |m|−n0 ≤ 1/3, we deduce that m−n0 (t − 1) ∈ (−1/2, 1/2), so that μ(t) = m−n0 (t − 1) + 1 ∈ (1/2, 3/2) = J . This shows that μ(I ∪ J ) ⊂ J . (d) As the groups G and ϕ(G) are isomorphic by (b), we deduce from (c) and Exercise 6.20(b) that G has exponential growth (cf. Exercise 6.36). O Exercise 6.40 (Growth of the Lamplighter Group) Let L := (Z/2Z) s Z denote the lamplighter group. Recall that L is the semidirect product of a normal subgroup H := ⊕n∈Z An , where each An is a subgroup of order 2, with an infinite cyclic subgroup N generated by an element t which satisfies tat −1 = (an−1 )n∈Z for all a = (an )n∈Z ∈ H . Let s denote the non-trivial element of A0 . (a) Show that S := {s, t, t −1 } is a symmetric generating subset of L. (b) For each integer n ≥ 1, let Hn := ⊕n−1 k=0 Ak . Prove that Hn is the subgroup of H generated by the elements s, tst −1 , t 2 st −2 , . . . , t n−1 st −n+1 . (c) Deduce from (b) that 2n ≤ γSL (3n − 2) for all n ≥ 0. (d) Deduce from (c) that L has exponential growth.

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Solution (a) First observe that s has order 2 so that s = s −1 . Therefore S is a finite symmetric subset of L. On the other hand, as L is the semidirect product of H and N, every element g ∈ L can be written in the form g = t k h for some k ∈ Z and h ∈ H . Thus, L is generated by t and H . Since H = ⊕n∈Z An and t n st −n is the non-trivial element of An , every element h ∈ H can be written as a finite product of elements of the form t n st −n . This shows that S generates L. (b) This is clear since, for each 0 ≤ k ≤ n − 1, the subgroup Ak is the cyclic group of order 2 generated by t k st −k . || n (c) The subgroup Hn has cardinality |Hn | = n−1 k=0 |Ak | = 2 . On the other hand, for every element h ∈ Hn , there exist εk ∈ {0, 1}, 0 ≤ k ≤ n − 1, such that h = s ε0 ts ε1 t −1 t 2 s ε2 t −2 t 3 s ε3 t −3 · · · t n−1 s εn−1 t −n+1

.

= s ε0 ts ε1 ts ε2 ts ε3 · · · ts εn−1 t −n+1 , showing that lS (h) ≤ 3n − 2. Therefore Hn ⊂ BS (3n − 2), so that γS (3n − 2) = |BS (3n − 2)| ≥ |Hn | = 2n .

.

(d) From (c), we deduce that R γS (3n − 2)

λS = lim

3n−2

≥ lim

3n−2

.

n→∞

n→∞

√ 2n n

1

= lim 2 3n−2 = 2 3 > 1. n→∞

O

This shows that L has exponential growth.

Exercise 6.41 (Growth of the Heisenberg Group) Let G := HZ denote the Heisenberg group over the ring of integers. Recall that G is the subgroup of SL3 (Z) consisting of all matrices of the form (

1y ( := .M(x, y, z) 01 00

) z x) 1

(x, y, z ∈ Z).

Let us set A := M(1, 0, 0), B := M(0, 1, 0), C := M(0, 0, 1), and S := {A, A−1 , B, B −1 , C, C −1 }. (a) Verify that M(x, y, z) = Ax B y C z for all x, y, z ∈ Z. (b) Show that S is a finite symmetric generating subset of G. (c) Show that C z Ax = Ax C z , C z B y = B y C z , and B y Ax = Ax B y C xy for all x, y, z ∈ Z.

6.2 Exercises

355

(d) Deduce from (c) that if P ∈ G satisfies lS (P ) ≤ n, then there exist x, y, z ∈ Z with |x| ≤ n, |y| ≤ n, and |z| ≤ n2 + n, such that P = Ax B y C z . (e) Deduce from (d) that there exists a constant C1 > 0 such that γS (n) ≤ C1 n4 for all n ≥ 1. (f) Let n, x, y, z be integers such that n ≥ 1, 0 ≤ x ≤ n, 0 ≤ y ≤ n, and 0 ≤ z ≤ n2 . Show that there exist integers q, r with 0 ≤ q ≤ n and 0 ≤ r ≤ n − 1 such that Ax B y C z = Ax−q B n Aq B y−n C r .

.

(g) Deduce from (f) that γS (5n) ≥ (n + 1)2 (n2 + 1) for all n ≥ 0. (h) Deduce from (g) that there exists a constant C2 > 0 such that γS (n) ≥ C2 n4 for all n ≥ 0. (i) Show that γ (G) ∼ n4 . (j) Show that the map ϕ : Z → G, defined by ϕ(k) := C k for all k ∈ Z, is a uniform embedding but not a quasi-isometric embedding. Solution (a) It is immediate to check that M(x, 0, 0)M(x ' , 0, 0) = M(x + x ' , 0, 0) for all x, x ' ∈ Z. Thus the map Z → G given by x |→ M(x, 0, 0) is a group homomorphism. It follows that M(x, 0, 0) = M(1, 0, 0)x = Ax for all x ∈ Z. Similarly, one verifies that the map Z → G given by y |→ M(0, y, 0) and the map Z → G given by z |→ M(0, 0, 1) are group homomorphisms, so that M(0, y, 0) = M(0, 1, 0)y = B y and M(0, 0, z) = M(0, 0, 1)z = C z for all y, z ∈ Z. Finally, we have that Ax B y C z = M(x, 0, 0) M(0, y, 0) M(0, 0, z) ( )( )( ) 100 1y0 10z = (0 1 x ) (0 1 0) (0 1 0) 001 001 001 ( )( ) 1y0 10z = (0 1 x ) (0 1 0) 001 001 ( ) 1y z = (0 1 x ) = M(x, y, z). 001

.

(b) The set S is clearly finite and symmetric. The fact that S generates G follows from (a).

356

6 Finitely Generated Groups

(c) For all x, y, z ∈ Z, we have (

C z Ax

.

Ax C z

CzBy

By Cz

B y Ax

)( ) ( ) 10z 100 10z = M(0, 0, z)M(x, 0, 0) = (0 1 0) (0 1 x ) = (0 1 x ) , 001 001 001 ( ) 10z = Ax B 0 C z = M(x, 0, z) = (0 1 x ) , 001 ( )( ) ( ) 10z 1y0 1yz = M(0, 0, z)M(0, y, 0) = (0 1 0) (0 1 0) = (0 1 0) , 001 001 001 ( ) 1yz 0 y z = A B C = M(0, y, z) = (0 1 0) , 001 ( )( ) ( ) 1y0 100 1 y xy = M(0, y, 0)M(x, 0, 0) = (0 1 0) (0 1 x ) = (0 1 x ) , 001

001

00 1

(

x

y

A B C

xy

) 1 y xy = M(x, y, xy) = (0 1 x ) , 00 1

where we used (a) for computing Ax C z , B y C z , and Ax B y C xy . This shows that C z Ax = Ax C z , C z B y = B y C z , and B y Ax = Ax B y C xy . (d) Let P ∈ G such that lS (P ) ≤ n. This means that there exist k ≤ n and M1 , M2 , . . . , Mk ∈ S such that P = M1 M 2 · · · M k .

.

(6.21)

By (c), the matrix C commutes with both A and B. Thus, we can move all occurrences of C and C −1 to the right in the right-hand side of (6.21). We also have B ε Aη = Aη B ε C εη for all ε, η ∈ {−1, 1}. This allows us to move all occurrences of A and A−1 on the left up to producing a new occurrence of C or C −1 for each replacement of B ε Aη by Aη B ε . At the end of this process, we obtain an expression of the form P = Ax B y C z , where |x| ≤ k ≤ n, |y| ≤ k ≤ n, and |z| ≤ k 2 + k ≤ n2 + n. (e) Let n ≥ 1. If M(x, y, z) ∈ BS (n), it follows from (a) and (d) that |x| ≤ n, |y| ≤ n, and |z| ≤ n2 + n. Consequently, we have γS (n) = |BS (n)| ≤ (2n + 1)2 (2n2 + 2n + 1). As 2n + 1 ≤ 3n ≤ 3n2 , this implies γS (n) ≤ 45n4 . Thus, we can take C1 = 45.

6.2 Exercises

357

(f) As n ≥ 1 and 0 ≤ z ≤ n2 , Euclidean division of z by n yields integers q and r such that z = qn + r, 0 ≤ q ≤ n, and 0 ≤ r ≤ n − 1. By (c), we have B n Aq = Aq B n C nq . Using the fact that C commutes with A and with B, this gives us Ax−q B n Aq B y−n C r = Ax−q Aq B n C nq B y−n C r = Ax B y C nq+r = Ax B y C z .

.

(g) Let n, x, y, z as in (f). As −n ≤ x − q ≤ n, 0 ≤ q ≤ n, −n ≤ y − n ≤ 0, and 0 ≤ r ≤ n, the equality Ax B y C z = Ax−q B n Aq B y−n C r implies that lS (Ax B y C z ) ≤ 5n. Therefore, Ax B y C z ∈ BS (5n) for all 0 ≤ x ≤ n, 0 ≤ y ≤ n, and 0 ≤ z ≤ n2 . It follows that γS (5n) = |BS (n)| ≥ (n + 1)2 (n2 + 1)

.

for all n ≥ 1. Note that the inequality remains valid for n = 0. (h) Let n ≥ 0 and consider the largest integer m ≥ 0 such that 5m ≤ n. As γS is increasing, we have γS (n) ≥ γS (5m). Using (g), this implies γS (n) ≥ (m + 1)2 (m2 + 1). As n ≤ 5m + 4 by the choice of m, we have m ≥ (n − 4)/5, so that m + 1 ≥ (n + 4)/5 ≥ n/5 and m2 + 1 ≥ (m + 1)2 /2 ≥ n2 /50. This gives us γS (n) ≥ (m + 1)2 (m2 + 1) ≥ n2 /25 · n2 /50 = n4 /1250 and we may take C2 := 1/1250. (i) We have γ (G) < n4 by (e) and n4 < γ (G) by (h). Thus, γ (G) ∼ n4 . (j) We have ϕ(k1 + k2 ) = C k1 +k2 = C k1 C k2 = ϕ(k1 )ϕ(k2 ) for all k1 , k2 ∈ Z. Therefore ϕ is a group homomorphism. On the other hand, the map ϕ is injective since C has infinite order in G. As ϕ is an injective group homomorphism, we deduce from [CAG2, Example 6.13.5] that it is a uniform embedding. Suppose by contradiction that ϕ is a quasi-isometric embedding. Then, there exist constants α ≥ 1 and β ≥ 0 such that α|k| − β ≤ lS (C k ) ≤ α|k| + β.

(6.22)

.

for all k ∈ Z. By (c), we have B k Ak = Ak B k C k and hence C k = B −k A−k B k Ak 2 for all k ∈ Z. It follows that lS (C k ) ≤ 4|k|. Thus, the inequality on the left of (6.22) implies αk 2 − β ≤ 4|k| for all k ∈ Z, which is clearly impossible. O 2

2

Exercise 6.42 Show that the Grigorchuk group is Hopfian. Solution Since the Grigorchuk group is finitely generated [CAG, Definition 6.9.2] and residually finite [CAG, Corollary 6.9.5], it is Hopfian by Mal’cev’s theorem [CAG, Theorem 2.4.3]. O Exercise 6.43 Let G be the Grigorchuk group and let Σ := {0, 1}. Recall that G naturally acts on Σ ∗ . (a) Show that G acts transitively on Σ n for all n ∈ N. (b) Use (a) to recover the fact that G is infinite.

358

6 Finitely Generated Groups

Solution (a) We use induction on n. For n = 0 we have Σ 0 = {E} and there is nothing to prove. Let n ≥ 1 and suppose that G acts transitively on Σ n−1 . Let w1 , w2 ∈ Σ n and let us show that there exists g ∈ G such that g(w1 ) = w2 . Suppose first that w1 and w2 start with the same letter, that is, there exist σ ∈ Σ and u1 , u2 ∈ Σ n−1 such that w1 = σ u1 and w2 = σ u2 . By induction, there exists h ∈ G such that h(u1 ) = u2 . As the map φσ : H1 → G is surjective (cf. [CAG, Proposition 6.9.7]), there exists g ∈ H1 ⊂ G such that φσ (g) = h. We then have g(w1 ) = g(σ u1 ) = σ h(u1 ) = σ u2 = w2 . Suppose now that w1 and w2 start with different letters. Then w1' := a(w1 ) ∈ Σ n and w2 do start with the same letter and, by the first part of the proof, we can find g ' ∈ G such that g ' (w1' ) = w2 . But then the element g := g ' a ∈ G satisfies g(w1 ) = (g ' a)(w1 ) = g ' (w1' ) = w2 . This shows that the action of G on Σ n is transitive for all n ∈ N. (b) A finite group H cannot act transitively on a finite set X of cardinality |X| > |H |. Indeed, if H acts transitively on X and x ∈ X, then the orbit map H → X, given by h |→ hx, is surjective. As Σ n has cardinality 2n , it then follows from (a) that G is infinite. O Exercise 6.44 (The Automorphism Group of a Regular Rooted Tree) Let Σ be a set. Let Σ ∗ denote the free monoid based on Σ, i.e., the set consisting of all words on the alphabet Σ with the concatenation of words as the monoid operation. Recall that one says that a word u ∈ Σ ∗ is a prefix of a word w ∈ Σ ∗ , and one writes u < w, if there exists v ∈ Σ ∗ such that uv = w. The relation < is a partial order on Σ ∗ and the empty word E ∈ Σ ∗ is the least element of (Σ ∗ , 0 for all n ≥ n0 (p, q). It follows from Exercise 6.88(c) that there exists a synchronizing word u := wp ∈ LG (resp. v := wq ∈ LG ) which focuses on p (resp. q). Since X is topologically mixing, we can find and integer n0 (u, v) ≥ 0 such that for all m ≥ n0 (u, v) there exists w ∈ A∗ such that |w| = m and uwv ∈ L(X). Let then π be a finite path in G such that λ(π ) = uwv. We have a unique decomposition π = π1 π2 π3 where λ(π1 ) = u, λ(π2 ) = w, and λ(π3 ) = v. Since u is synchronizing and focuses on p, we have π2− = π1+ = p. On the other hand, since v is synchronizing and focuses on q, so is wv (by Exercise 6.88(a)). We thus have (π2 π3 )+ = π3+ = q. In conclusion, the finite path π2 π3 connects p and q and has length l(π2 π3 ) = l(π2 ) + l(π3 ) = |w| + |v| = m + |v|. We deduce that, setting n0 (p, q) := n0 (u, v) + |v| one has (n) > 0 for all n ≥ n0 (p, q). Taking n0 := maxp,q∈Q n0 (p, q) we then have that bpq for all n ≥ n0 , all the entries of B n are positive. This shows that B is a primitive matrix (cf. Exercise 6.92(a)). O

438

6 Finitely Generated Groups

Comment The equivalence (i) ⇐⇒ (iii) is [LinM, Exercise 4.5.16(a)]. The equivalence (ii) ⇐⇒ (iii) is [CecC3, Corollary 1.3]. Exercise 6.97 Let A be a finite set and let X ⊂ AZ be a sofic subshift. Show that X is topologically mixing if and only if it is strongly irreducible. Solution Every strongly irreducible (resp. topologically mixing) subshift is irreducible (cf. Exercises 1.64(b) and 1.56(e)). Thus, since X ⊂ AZ is irreducible and sofic, there exists a connected, deterministic, follower separated finite A-labeled graph G such that X = X G (cf. Exercise 6.89). The equivalence in the statement then follows from Exercise 6.96. O Comment As every subshift of finite type is sofic, the above result extends Exercise 1.80. Recall (cf. Exercise 1.86) that there exist topologically mixing subshifts over Z which are not strongly irreducible. Exercise 6.98 (Period of a Subshift) Let A be a finite set and let X ⊂ AZ be a nonempty subshift. For each integer n ≥ 1, let pern (X) := | Fix(nZ) ∩ X| denote the number of n-periodic configurations in X. Set Per(X) := {n ≥ 1 : pern (X) /= 0}. The period per(X) of X is defined by per(X) := d, where d is the greatest common divisor of all the elements in Per(X) if Per(X) /= ∅ and d := ∞ if Per(X) = ∅. (a) Suppose that X is sofic (e.g., of finite type). Show that per(X) < ∞. (b) Suppose that X is the labeled edge subshift associated with a finite A-labeled graph G . Show that Per(G ) ⊂ Per(X). (c) Suppose that X is of finite type and that m ≥ 2 is an integer such that the set {1, 2, . . . , m} is a memory set for X. Let G = (Q, E) denote the De Bruijn graph associated with the pair (X, m). Show that Per(X) = Per(G ) and per(X) = per(G ). (d) Let B be another finite set and let τ : X → B Z be a cellular automaton. Let Y := τ (X) ⊂ B Z denote the image subshift. Show that Per(X) ⊂ Per(Y ). Solution (a) This follows from Exercise 1.96. (b) Given a closed path π = (e1 , e2 , . . . , en ) in G of length n ≥ 1, the sequence π ∞ = (fk )k∈Z defined by fk := ei if k ≡ i mod n, 0 ≤ i ≤ n, yields a bi-infinite path in G whose label x := λ(π ∞ ) ∈ X is n-periodic. This shows the inclusion Per(G ) ⊂ Per(X). (c) Let n ∈ Per(X). This means that X contains an n-periodic configuration x. For i = 1, 2, . . . , n + 1, set qi := x(i)x(i + 1) · · · x(i + m − 1) ∈ Lm−1 (X) = Q

.

and ei := (qi , x(i + m), qi+1 ) ∈ E.

.

Then ω(ei ) = qi+1 = α(ei+1 ) for all 1 ≤ i ≤ n − 1, and ω(en ) = qn+1 = x(n + 1)x(n + 2) · · · x(n + m − 1) = x(1)x(2) · · · x(m − 1) = q1 = α(e1 ). Thus π := (e1 , e2 . . . , en ) is a closed path in G of length n. This shows the inclusion

6.2 Exercises

439

Per(X) ⊂ Per(G ). On the other hand, it follows from Exercise 6.72(b) that X is the labeled edge subshift associated with G . Thus, using (b), we deduce that Per(X) = Per(G ) and therefore per(X) = per(G ). (d) Let n ∈ Per(X). This means that X contains an n-periodic configuration x. As nx = x, we have nτ (x) = τ (nx) = τ (x) by using the Z-equivariance of τ . This shows that τ (x) ∈ Y is n-periodic. We deduce that Per(X) ⊂ Per(Y ). O Exercise 6.99 (The Nth Higher Block Subshift) Let A be a finite set and let N be a positive integer. Consider the map ΦN : AZ → (AN )Z defined by ΦN (x)(n) := (x(n), x(n + 1), x(n + 2), . . . , x(n + N − 1)) for all x ∈ AZ and n ∈ Z. Consider also the map ϕN : A∗ → (AN )∗ defined by ϕN (w) := E (the empty word) if w has length l(w) < N and ϕN (w) := (a1 , a2 , . . . , aN )(a2 , a3 , . . . , aN+1 ) · · · (am+1 , am+2 , . . . , am+N ) (6.67)

.

if l(w) ≥ N and w = a1 a2 · · · am+N , with ai ∈ A for 1 ≤ i ≤ m + N and m ≥ 0. Let X ⊂ AZ be a subshift and set X [N] := ΦN (X) ⊂ (AN )Z . For L ⊂ A∗ , set L[N] := ϕN (L) ⊂ (AN )∗ . (a) Show that X[N] is a subshift of (AN )Z (it is called the Nth higher block subshift of X). (b) Show that the restriction of ϕN to {w ∈ A∗ : l(w) ≥ N} is injective and deduce that L(X[N] ) = (L(X))[N] . (c) Show that X is topologically transitive (resp. topologically mixing, resp. strongly irreducible, resp. of finite type, resp. sofic, resp. irreducible) if and only if X[N] is topologically transitive (resp. topologically mixing, resp. strongly irreducible, resp. of finite type, resp. sofic, resp. irreducible). (d) Consider the Følner sequence F = (Fn )n≥1 for Z defined by Fn := {0, 1, . . . , n − 1} for every n ≥ 1. Show that entF (X) = entF (X[N] ). Solution (a) Setting F := {0, 1, . . . , N − 1} ⊂ Z and after identifying AF and AN , the map ΦN : AZ → (AN )Z is nothing but the Z-equivariant uniform embedding Φ defined in Exercise 1.30. Moreover, with the notation in Exercise 1.95, we have X [N] = X[F ] . It then follows from Exercise 1.95(a) that X[N] is a subshift of (AN )Z . (b) For i = 1, 2, . . . , N , let πi :− AN → A denote the projection on the i-th coordinate, that is, πi (a1 , a2 , . . . , aN ) := ai for all (a1 , a2 , . . . , aN ) ∈ AN . Given u ∈ (AN )∗ \ {E}, there exists unique m ≥ 0 and b1 , b2 , . . . , bm+1 ∈ AN such that u = b1 b2 · · · bm+1 . Define ψN (u) := π1 (b1 )π1 (b2 ) · · · π1 (bm )π1 (bm+1 )π2 (bm+1 ) · · · πN (bm+1 ) ∈ A∗ . If w = a1 a2 · · · am+N , with m ≥ 0, and u = b1 b2 · · · bm+1 := ϕN (w), we thus have bi = (ai , ai+1 , . . . , ai+N−1 ) so that ψN (u) = π1 (b1 )π1 (b2 ) · · · π1 (bm )π1 (bm+1 )π2 (bm+1 ) · · · πN (bm+1 ) .

= a1 a2 · · · am am+1 · · · am+N = w.

440

6 Finitely Generated Groups

This shows that the restriction of ϕN to {w ∈ A∗ : l(w) ≥ N} is injective with inverse ψN : (AN )∗ \ {E} → {w ∈ A∗ : l(w) ≥ N}. Let now w ∈ L(X[N] ). This means that there is a configuration y ∈ X[N] , integers m ∈ N and n ∈ Z such that w = y(n)y(n + 1) · · · y(n + m) ∈ (AN )∗ . −1 If x := ΦN (y) ∈ X we have y(i) = (x(i), x(i + 1), . . . , x(i + N − 1)) ∈ AN for all i ∈ Z. Setting u := x(n)x(n + 1) · · · x(n + m − 1) ∈ L(X) we then have w = ϕN (u) ∈ (L(X))[N] . This shows the inclusion L(X[N] ) ⊂ (L(X))[N] . Conversely, let u ∈ (L(X))[N] and let us show that u ∈ L(X [N] ). If u = ε, there is nothing to prove. Otherwise, there exists w ∈ L(X) with m := l(w) ≥ N such that u = ϕN (w). We can then find a configuration x ∈ L(X) and n ∈ Z such that w = x(n)x(n + 1) · · · x(n + m − 1). Set y := ΦN (x) ∈ X[N] . It is clear that u = ϕN (w) = (x(n), x(n + 1), . . . , x(n + N − 1)) · (x(n + 1), x(n + 2), . . . , x(n + N)) · · · .

· · · (x(n + m), x(n + m + 1), . . . , x(n + m + N − 1)) = y(n)y(n + 1) · · · y(n + N − 1), so that u ∈ L(X[N] ). This shows the inclusion (L(X))[N] ⊂ L(X[N] ). We deduce that L(X[N] ) = (L(X))[N] . (c) This follows from Exercise 1.95(c), (d). (d) This follows from Exercise 5.34. O

Exercise 6.100 (The Nth Higher Edge Graph) Let A be a finite set and let G = (Q, E) be an A-labeled graph. For N a positive integer, the Nth higher edge graph associated with G is the AN -labeled graph G [N] = (Q[N] , E [N] ) defined as follows. For N = 1, one has G [1] := G and, for N ≥ 2, the vertex set Q[N] is the set of all paths of length N − 1 in G and E [N] := {(π,(λ(e1 ), λ(e2 ), . . . , λ(eN )), π ' ) ∈ Q[N] × AN × Q[N] : .

π = (e1 , e2 , . . . , eN−1 ), π ' = (e2 , e3 , . . . , eN−1 , eN )}. [N]

(a) With the notation from Exercise 6.99, show that XG = (XG )[N] . (b) We know from Exercise 6.99(c) that if X ⊂ AZ is a sofic subshift, then the subshift X[N] ⊂ (AN )Z is also sofic. Using (a), give another proof of this fact. (c) Suppose that G is connected . Show that G [N] is connected. (d) Suppose that G is deterministic. Show that G [N] is deterministic. Solution (a) For N = 1, there is nothing to prove. So we suppose that N ≥ 2. Let y ∈ [N] XG . This means that there exists a bi-infinite path π [N] = (ei[N] )i∈Z in G [N] such i ∈ E such that that y = λ[N] (π [N] ). For i ∈ Z, let e1i , e2i , . . . , eN i i i ei[N] = ((e1i , e2i , . . . , eN−1 ), (λ(e1i ), λ(e2i ), . . . , λ(eN )), (e2i , e3i , . . . , eN )),

.

6.2 Exercises

441

and observe that e1i+1 = e2i . It is then clear that π := (e1i )i∈Z is a bi-infinite path in G . Moreover, setting x := λ(π ) ∈ XG we have y = ΦN (x) ∈ (XG )[N] . This shows [N] that XG ⊂ (XG )[N] . Conversely, let y ∈ (X G )[N] . This means that there exists x ∈ XG such that y = ΦN (x). Let π = (ei )i∈Z be a bi-infinite path in G such that λ(π ) = x. For i ∈ Z, consider the finite path πi := (ei , ei+1 , . . . , ei+N−2 ) in G and set ei[N] := (πi , (λ(ei ), λ(ei+1 ), . . . , λ(ei+N−2 )), πi+1 ) ∈ E [N] . Then π [N] = (ei[N] )i∈Z is a [N] bi-infinite path in G [N] and y = ΦN (x) = λ[N] (π [N] ) ∈ XG . This shows that [N] [N] (XG )[N] ⊂ XG . We deduce that (XG )[N] = XG . (b) Suppose that X ⊂ AZ is a sofic subshift. It follows from Exercise 6.78 that there exists a finite A-labeled graph G such that X = XG . By (a), the subshift [N] X[N] ⊂ (AN )Z satisfies X[N] = XG . Since G [N] is a finite AN -labeled graph, it follows from Exercise 6.78 that X[N] is itself sofic. (c) If N = 1, we have G [N] = G and there is nothing to prove. ' Suppose that N ≥ 2. Let π = (e1 , e2 , . . . , eN−1 ), π ' = (e1' , e2' , . . . , eN −1 ) ∈ ' [N] + ' − Q and set p := π = ω(eN−1 ), q := (π ) = α(e1 ) ∈ Q. Since G is connected, we can find a finite path ϕ = (f1 , f2 , . . . , fn ) in G such that ϕ − = α(f1 ) = p and ϕ + = ω(fn ) = q. Consider the finite path ' (h1 , h2 , . . . , hn+2N−2 ) := (e1 , e2 , . . . , eN−1 , f1 , f2 , . . . , fn , e1' , e2' , . . . , eN −1 )

.

in G . We then define the vertices π 1 := π = (h1 , h2 , . . . , hN−1 ), π 2 := (h2 , h3 , . . . , hN ), . . ., π n+N := π ' = (hn+N , hn+N+1 , . . . , hn+2N−2 ) ∈ Q[N] , as well as the edges e1 := (π 1 , (λ(h1 ), λ(h2 ), . . . , λ(hN )), π 2 ), e2 := (π 2 , (λ(h2 ), λ(h3 ), . . . , λ(hN+1 )), π 3 ), . . ., en+N−1 := (π n+N−1 , (λ(hn+N−1 ), λ(hn+N ), . . . , λ(hn+2N−2 )), π n+N ) ∈ E [N] . It is then clear that (e1 , e2 , . . . , em−N+1 ) is a finite path in G [N] connecting π = 1 π and π ' = π m−N+2 . This shows that G is connected. (d) If N = 1, we have G [N] = G and there is nothing to prove. Suppose that N ≥ 2. Let e = (π, (a1 , a2 , . . . , aN ), π ' ) and f = (π, (a1 , a2 , . . . , aN ), π '' ) be in E [N] . Let us show that π ' = π '' so that e = f . Writing π = (e1 , e2 , . . . , eN−1 ), π ' = (e2 , e3 , . . . , eN−1 , eN ), and ' ) we have a = λ(e ) for i = 1, 2, . . . , N − 1 and π '' = (e2 , e3 , . . . , eN−1 , eN i i ' ' ) and G is deterministic, aN = λ(eN ) = λ(eN ). Since α(eN ) = ω(eN−1 ) = α(eN ' ' we deduce that ω(eN ) = ω(eN ), that is, eN = eN . Consequently, π ' = π '' and e = f , showing that G [N] is deterministic. O Exercise 6.101 Let G = (Q, E) be a finite labeled graph. For each integer n ≥ 1, let Pn = Pn (G ) denote the set consisting of all paths of length n in G . Show that the limit R n .ρ(G ) := lim |Pn | (6.68) n→∞

exists and is finite.

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6 Finitely Generated Groups

Solution Let m and n be positive integers. A path of length m + n is uniquely determined by its initial path of length m and the subsequent path of length n. This yields an injective map Pm+n → Pm × Pn . We deduce that |Pm+n | ≤ |Pm × Pn | = |Pm | · |Pn |. Thus, the sequence (|Pn |)n≥1 is submultiplicative. This implies that the limit in (6.68) exists and is finite (cf. [CAG, Lemma 6.5.1]). O Exercise 6.102 Let A be a finite set and let G = (Q, E) be a finite A-labeled graph. Let X ⊂ AZ and Y ⊂ E Z denote respectively the labeled edge subshift and the edge subshift associated with G (cf. Exercise 6.69). Let ρ(G ) be the quantity defined by (6.68). Consider the Følner sequence F = (Fn )n≥1 for Z defined by Fn := {0, 1, . . . , n − 1} for every n ≥ 1. (a) Show that one has entF (X) ≤ log (ρ(G )). (b) Show that if G is deterministic then one has entF (X) = log (ρ(G )). (c) Show that one has entF (Y ) = log (ρ(G )). Solution (a) For n ≥ 1, consider the set Ln (X) ⊂ A∗ consisting of all words w ∈ L(X) of length |w| = n that appear in X, and the set Pn (G ) consisting of all paths π = (e1 , e2 , . . . , en ) in G of length n. As every w ∈ Ln (X) is the label of some path π ∈ Pn (G ), we have |Ln (X)| ≤ |Pn (G )|. We deduce that log |Ln (X)| n log |Pn (G )| ≤ lim n→∞ n R n = lim log |Pn (G )| n→∞ ) ( R = log lim n |Pn (G )|

entF (X) = lim

n→∞

.

n→∞

= log (ρ(G )) . (b) Suppose that G = (Q, E) is deterministic. Then, given q ∈ Q and w ∈ Ln (X), there is at most one path π ∈ Pn (G ) starting at q with label w. Thus, there are at most |Q| paths π ∈ Pn (G ) with label w. We deduce that |Pn (G )| ≤ |Q| · |Ln (X)|. This implies .

log (ρ(G )) = lim log n→∞

R n |Pn (G )|

R ≤ lim log n |Q| · |Ln (X)| n→∞ ) ( R R ≤ lim log n |Q| + log n |Ln (X)| n→∞

= lim log n→∞

R n |Ln (X)|

6.2 Exercises

443

= lim

n→∞

log |Ln (X)| n

= entF (X). Using (a), we conclude that entF (X) = log (ρ(G )). (c) By definition, Y is the edge labeled graph of the finite E-labeled graph G ' = (Q' , E ' ), where Q' := Q and E ' ⊂ Q × E × Q consists of all the triples (q, e, p) such that e ∈ E satisfies e = (q, a, p) for some a ∈ A. We clearly have |Pn (G ' )| = |Pn (G )| for all n ≥ 1. Therefore ρ(G ' ) = ρ(G ). As G ' is deterministic, we deduce from (b) that entF (Y ) = ρ(G ' ) = ρ(G ). O Exercise 6.103 (Scarabotti’s Lemma) Let G = (Q, E) be a finite labeled graph. Suppose that G is connected and let e ∈ E. Show that there exists a positive integer n0 = n0 (G ) such that if π is any path in G of length n ≥ n0 , then there exists a path π ' = (e1' , e2' , . . . , en' ) with the same length n, the same initial and terminal vertices as π, and which contains e (i.e., such that ei' = e for some 1 ≤ i ≤ n). Solution Given a path π = (e1 , e2 , . . . , en ) in G , denote by πQ = (q0 , q1 , . . . , qn ) the associated sequence of visited vertices. We define a decomposition of π as follows. Let i1 be the largest index such that the vertices q0 , q1 , . . . , qi1 −1 are all distinct. Then qi1 = qj1 for a suitable 1 ≤ j1 < i1 and we set r1 := (e1 , e2 , . . . , ej1 ) and c1 := (ej1 +1 , ej1 +2 , . . . , ei1 ). Continuing this way, we obtain a decomposition of the path π = r1 c1 r2 c2 · · · rk ck rk+1 , where the c1 , c2 , . . . , ck are simple closed paths and r1 , r2 , . . . , rk+1 are simple (possibly empty) paths. With this notation, given 1 ≤ s ≤ k, we say that the path πs := r1 c1 r2 c2 rs rs+1 cs+1 · · · ck rk is obtained from π by collapsing the sth simple closed path cs . Similarly, if π '' is a closed path such that (π '' )− = π − and d is a positive integer, we say that the path (π '' )d π is obtained from π by adding d copies of π '' in front of π. Now, since G is connected, we can find a closed path π '' with initial (= terminal) vertex π − containing the edge e. Let m denote the length of π '' . By the pigeon-hole principle, if n is large enough, then in the decomposition π = r1 c1 r2 c2 · · · rk ck rk+1 there exists a simple closed path c ∈ {c1 , c2 , . . . , ck } that occurs at least m times. Let d denote the length of c. It is then clear that adding d copies of π '' in front of π and collapsing the first m copies of c in π, we obtain the desired path π ' . O Comment This result is due to Scarabotti [Sca, Lemma]. Exercise 6.104 Let G = (Q, E) be a finite connected labeled graph. Let e ∈ E and denote by H = (Q' , E ' ) the labeled subgraph of G obtained by removing the edge e, so that Q' := Q and E ' := E \ {e}. Let n0 = n0 (G ) denote the positive integer given by Exercise 6.103 and set α := |Pn0 (G )|−1 . Let k be a positive integer. (a) Express every path π ∈ Pkn0 (G ) as the composition π = π1 π2 · · · πk , where πi ∈ Pn0 (G ) for 1 ≤ i ≤ k, and denote by ϕi the set of all paths π ∈ Pkn0 (G ) such

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6 Finitely Generated Groups

i (G ) := P that πi contains e. For 0 ≤ i ≤ k, set Pkn kn0 (G ) \ 0 one has

Ui

j =1 ϕj .

Show that

i i |Pkn (G ) ∩ ϕi+1 | ≥ α|Pkn (G )| 0 0

.

for all 0 ≤ i ≤ k − 1. (b) Show that k |Pkn (G )| ≤ (1 − α)k |Pkn0 (G )|. 0

.

(c) Show that ρ(H ) ≤ (1 − α)1/n0 ρ(G ). (d) Show that ρ(H ) < ρ(G ). Solution i (G ), it follows from (a) Let 0 ≤ i ≤ k − 1. Given π = π1 π2 · · · πk ∈ Pkn 0 the definition of n0 that there exists a path σ ∈ Pn0 (G ) containing e such that − + σ − = πi+1 and σ + = πi+1 . Then the path π ' := π1 · · · πi σ πi+2 · · · πk , obtained i ∩ϕ from π by replacing the subpath πi+1 by σ , belongs to Pkn i+1 . We thus get an 0 i i injective map Pkn0 (G ) → (Pkn0 ∩ ϕi+1 ) × Pn0 (G ) given byπ |→ (π ' , πi+1 ). This i (G )| ≤ |P i ∩ ϕ i i implies |Pkn i+1 | · |Pn0 (G )| and hence |Pkn0 ∩ ϕi+1 | ≥ α|Pkn0 (G )|. kn0 0 (b) Let us show, by induction on i, that i |Pkn (G )| ≤ (1 − α)i |Pkn0 (G )| 0

(6.69)

.

i (G ) = P for all i = 0, 1, 2, . . . , k. For i = 0, we have Pkn kn0 (G ) and there is 0 nothing to prove. Suppose now that (6.69) holds for some i ≤ k − 1. Then

i+1 |Pkn (G )| = |Pkn0 (G ) \ 0

i+1 | |

.

=

( ϕj | = | (Pkn0 (G ) \

j =1

i | |

) ϕj ) \ ϕi+1 |

j =1

i (G ) \ ϕi+1 | |Pkn 0

i i = |Pkn (G )| − |Pkn (G ) ∩ ϕi+1 | 0 0 i i ≤ |Pkn (G )| − α|Pkn (G )| 0 0

(by (a))

i = (1 − α)|Pkn (G )| 0

≤ (1 − α)(1 − α)i |Pkn0 (G )|

(by induction)

= (1 − α)i+1 |Pkn0 (G )|. k (G )| ≤ (1 − This ends the induction argument. Taking i := k, (6.69) yields |Pkn 0 α)k |Pkn0 (G )|.

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445

k (G ). Using (b), we deduce that (c) Observe that Pkn0 (H ) = Pkn 0 k |Pkn0 (H )| = |Pkn (G )| ≤ (1 − α)k |Pkn0 (G )|. 0

.

Taking (kn0 )th roots, we get |Pkn0 (H )|1/(kn0 ) ≤ (1 − α)1/n0 |Pkn0 (G )|1/(kn0 )

.

and, taking the limit for k → ∞, this gives us ρ(H ) = lim |Pkn0 (H )|1/(kn0 ) ≤ (1 − α)1/n0 lim |Pkn0 (G )|1/(kn0 )

.

k→∞

k→∞

= (1 − α)

1/n0

ρ(G ).

(d) Keeping in mind that α = |Pn0 (G )|−1 , so that 0 < α < 1, we have (1 − < 1. From (c), we thus deduce that

α)1/n0

ρ(H ) ≤ (1 − α)1/n0 ρ(G ) < ρ(G ).

.

O Comment This result is usually proved by means of the Perron-Frobenius Theory (see [LinM, Theorem 4.4.7]). Our proof is based on Fabio Scarabotti’s argument in [Sca, Theorem] which avoids the Perron-Frobenius theorem and makes use instead of a technique developed by Gromov in [Gro3]. Exercise 6.105 Let A be a finite set. Let X ⊂ AZ be a subshift and let w ∈ L(X) \ {E} be a non-empty word appearing in X. Let Xw ⊂ AZ be the set consisting of all configurations x ∈ X such that w does not appear in x. (a) Show that Xw is a subshift of AZ and that one has Xw G X. (b) Let N ≥ 1 denote the length of w. Using the notations introduced in Exercise 6.99, show that one has (Xw )[N] = (X[N] )ϕN (w) . (c) Suppose that X is an irreducible sofic subshift of AZ . It follows from Exercise 6.89 that there exists a deterministic connected A-labeled graph G such that X = XG . Let H denote the labeled subgraph of the Nth higher edges graph G [N] (cf. Exercise 6.100) obtained by removing all edges in E [N] labeled by ϕN (w). Show that (X[N] )ϕN (w) = XH . (d) Suppose that X is an irreducible sofic subshift of AZ . Consider the Følner sequence F = (Fn )n≥1 for Z, where Fn := {0, 1, . . . , n − 1} for every n ≥ 1. Show that entF (Xw ) < entF (X). Solution (a) If F ⊂ A∗ is a defining set of forbidden words for X, it is clear that Xw is the subshift admitting F ∪{w} as a defining set of forbidden words (cf. Exercise 1.77(a)). As w ∈ L(X), there exists x ∈ X such that w appears in X. We then have x ∈ / Xw by definition of Xw . This shows that Xw G X.

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6 Finitely Generated Groups

(b) Write w = a1 a2 · · · aN with ai ∈ A for 1 ≤ i ≤ N. Let y ∈ (Xw )[N] . This means that there exists x ∈ Xw such that y = ΦN (x) ∈ ΦN (Xw ) ⊂ ΦN (X) = X [N] . Thus, y(n) = ΦN (x)(n) = (x(n), x(n + 1), . . . , x(n + N)) /= (a1 , a2 , . . . , aN ) = ϕN (w) for all n ∈ Z. This shows that y ∈ (X[N] )ϕN (w) , and the inclusion (Xw )[N] ⊂ (X [N] )ϕN (w) follows. Conversely, suppose that y ∈ (X [N] )ϕN (w) . This means, on the one hand, that there exists x ∈ X such that y = ΦN (x) ∈ ΦN (X) = X[N] and, on the other hand, that (x(n), x(n + 1), . . . , x(n + N)) = ΦN (x)(n) = y(n) /= ϕN (w) = (a1 , a2 , . . . , aN )

.

for all n ∈ Z, equivalently, w does not appear in x. Thus x ∈ Xw and therefore y = ΦN (x) ∈ ΦN (Xw ) = (Xw )[N] , and the inclusion (X[N] )ϕN (w) ⊂ (Xw )[N] follows as well. We deduce that (Xw )[N] = (X[N] )ϕN (w) . [N] (c) By virtue of Exercise 6.100, the higher edge graph G [N] satisfies XG = X[N] . Since H is the subgraph of G [N] obtained by removing all edges labeled . To show the converse inclusion, let by ϕN (w), we clearly have XH ⊂ Xϕ[N] N (w) [N]

x ∈ Xϕ[N] ⊂ X[N] = XG . Thus, there exists a bi-infinite path π = (en )n∈Z in N (w) G [N] such that x = λ(π ) and λ(en ) = x(n) /= ϕN (w) for all n ∈ Z. This means that in fact π is in H so that x ∈ XH . This shows that (X[N] )ϕN (w) ⊂ XH . We deduce that (X[N] )ϕN (w) = XH . (d) Since G is connected and deterministic, it follows from Exercise 6.100(c), (d) that the Nth higher edge graph G [N] is connected and deterministic so that its subgraph H is deterministic as well. We deduce that .

entF (Xw ) = entF ((Xw )[N] )

(by Exercise 6.99(d))

= entF ((X[N] )ϕN (w) )

(by (b))

= entF (XH )

(by (c))

= log ρ(H )

(by (6.102)(b))

< log ρ(G

[N]

= entF (XG

(by Exercise 6.104(f))

)

[N]

)

(by (6.102)(b))

= entF ((XG )[N] )

(by Exercise 6.100(a))

= entF (XG )

(by Exercise 6.99(d))

= entF (X). O Exercise 6.106 Let A be a finite set. Let X ⊂ AZ be an irreducible sofic subshift. Consider the Følner sequence F = (Fn )n≥1 for Z, where Fn := {0, 1, . . . , n−1} for

6.2 Exercises

447

every n ≥ 1. Show that if Y ⊂ AZ is a subshift such that Y G X, then entF (Y ) < entF (X). Solution Since Y G X, it follows from Exercise 1.74 that L(Y ) G L(X). If w ∈ L(X) \ L(Y ), we have Y ⊂ Xw G X (cf. Exercise 6.105). Using the monotonicity property of entropy (cf. [CAG, Proposition 5.7.2.(ii)]), we deduce that entF (Y ) ≤ entF (Xw ) < entF (X), where the strict inequality follows from Exercise 6.105(d). O Exercise 6.107 Let A be a finite set and let X ⊂ AZ be an irreducible sofic subshift. Use the result of Exercise 6.106 to give an alternative proof of the fact, already established in Exercise 6.90(b), that the subshift X is surjunctive. Solution Let τ : X → X be an injective cellular automaton. Consider the Følner sequence F = (Fn )n≥1 for Z, where Fn := {0, 1, . . . , n − 1} for every n ≥ 1. Since the subshifts X and τ (X) are topologically conjugate by τ , we have entF (X) = entF (τ (X)) by Exercise 5.28. Using Exercise 6.106, we deduce that τ (X) = X. This shows that τ is surjective and hence that X is surjunctive. O Exercise 6.108 Let A and B be finite sets. Suppose that X ⊂ AZ is a splicable subshift and that Y ⊂ B Z is an irreducible sofic subshift such that entF (X) = entF (Y ), where F = (Fn )n≥1 is the Følner sequence for Z defined by Fn := {0, 1, . . . , n − 1} for every n ≥ 1. Show that every pre-injective cellular automaton τ : X → Y is surjective. Solution If τ : X → Y is a pre-injective cellular automaton then one has entF (τ (X)) = entF (X) by Exercise 5.60. As entF (X) = entF (Y ), this implies entF (τ (X)) = entF (Y ) and hence τ (X) = Y by Exercise 6.106. O Exercise 6.109 Let A and B be finite sets. Let F be a Følner net for Z. Suppose that X ⊂ AZ is a sofic subshift. Let τ : X → B Z be a pre-injective cellular automaton. Show that one has entF (τ (X)) = entF (X). Solution Since X is sofic, it follows from Exercise 6.86 that there exist a finite set C, a subshift of finite type Z ⊂ C Z with entF (X) = entF (Z), and a preinjective and surjective cellular automaton σ : Z → X. Then the composite cellular automaton τ ◦ σ : Z → B Z is pre-injective by Exercise 5.10. As the subshift Z is of finite type, it is splicable by Exercise 1.98(a). Using Exercise 5.60 we then deduce that .

entF (τ (X)) = entF (τ (σ (Z))) = entF ((τ ◦ σ )(Z)) = entF (Z) = entF (X). O

Comment Cf. [LinM, Theorem 8.1.16]. Exercise 6.110 Let A and B be finite sets. Suppose that X ⊂ AZ is a sofic subshift and that Y ⊂ B Z is an irreducible sofic subshift such that entF (X) = entF (Y ), where F = (Fn )n≥1 is the Følner sequence for Z defined by Fn := {0, 1, . . . , n−1}

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6 Finitely Generated Groups

for every n ≥ 1. Show that every pre-injective cellular automaton τ : X → Y is surjective. Solution Let τ : X → Y be a pre-injective cellular automaton. It follows from Exercise 6.109 that entF (τ (X)) = entF (X) = entF (Y ). Since Y is an irreducible sofic subshift, we deduce from Exercise 6.106 that τ (X) = Y . This shows that τ : X → Y is surjective. O Exercise 6.111 Let A be a finite set. Show that every irreducible sofic subshift X ⊂ AZ has the Myhill property. Solution This immediately follows from Exercise 6.110 after taking X = Y .

O

Comment This result is due to Fiorenzi [Fio1, Corollary 2.21]. Note that the hypothesis that X is irreducible cannot be removed. Indeed, the subshift X ⊂ {0, 1}Z consisting of the two constant configurations is of finite type (cf. Exercise 1.57(e)) and therefore sofic but it does not have the Myhill property (cf. Exercise 5.48(b)). Exercise 6.112 Let A := {0, 1} and let X ⊂ AZ denote the even subshift. Show that X has the Myhill property but not the Moore property. Solution Since X is an irreducible sofic subshift of AZ (cf. Exercises 1.85(d) and 1.97(d)), it has the Myhill property by Exercise 6.111. On the other hand, the subshift X does not have the Moore property by Exercise 6.74(f). O Comment The fact that X has the Myhill property can also be deduced from Exercise 5.50 since X is strongly irreducible by Exercise 1.85(c). Exercise 6.113 (Diamonds in Labeled Graphs) Let A be a finite set and let G = (Q, E) be a finite A-labeled graph. A diamond in G is an ordered pair (π1 , π2 ) of distinct finite paths in G with the same starting and terminal vertices and the same label (see Fig. 6.14), that is, such that π1 /= π2 , π1− = π2− , π1+ = π2+ , and λ(π1 ) = λ(π2 ).

.

(a) Let (π1 , π2 ) be a diamond in G . Show that π1 and π2 have the same length. (b) Suppose that G has no diamonds and that π1∞ , π2∞ ∈ E Z are bi-infinite paths in G that are almost equal and have the same label λ(π1∞ ) = λ(π2∞ ) ∈ AZ . Show that one has π1∞ = π2∞ . Fig. 6.14 A diamond in a labeled graph

6.2 Exercises

449

(c) Suppose that G is essential. Show that the following conditions are equivalent: (i) G has no diamonds; (ii) if π1∞ , π2∞ ∈ E Z are bi-infinite paths in G that are almost equal and have the same label λ(π1∞ ) = λ(π2∞ ) ∈ AZ then π1∞ = π2∞ . (d) Suppose that G is deterministic. Show that G has no diamonds. (e) Let X ⊂ AZ be a subshift of finite type. Let m ≥ 2 be an integer such that {1, 2, . . . , m} ⊂ Z is a memory set for X. Show that the De Bruijn graph G associated with the pair (X, m) has no diamonds. (f) Let X ⊂ AZ be a subshift of finite type. Let B be another finite set and let τ : AZ → B Z be a cellular automaton. Let n, m ∈ Z with m ≥ 2 such that the set {n, n + 1, . . . , n + m − 1} ⊂ Z is a memory set for both X and τ and let G = (Q, E) denote the De Bruijn graph associated with the quadruple (X, τ, n, m). Show that the following conditions are equivalent: (1) the restriction of τ to X is pre-injective; (2) G has no diamonds. Solution (a) We have l(π1 ) = |λ(π1 )| = |λ(π2 )| = l(π2 ). (b) Suppose by contradiction that π1∞ /= π2∞ . Then F := {k ∈ Z : ∞ π1 (k) /= π2∞ (k)} is a non-empty finite subset of Z. Setting m0 := min(F ) − 1 and n0 := max(F ) + 1, it is clear that the finite paths π1 , π2 in G defined by πi := (πi∞ (k + m0 − 1))1≤k≤n0 −m0 +1 for i ∈ {1, 2}, form a diamond (π1 , π2 ) in G . This is a contradiction. (c) The implication (i) =⇒ (ii) follows from (b). Conversely, suppose that G admits a diamond (π1 , π2 ). Write π1 = (e1 , . . . en ) and π2 = (f1 , . . . , fn ). Since G is essential, it follows from Exercise 6.83(a) that the finite path π1 can be extended to a bi-infinite path π1∞ ∈ E Z . Then π1∞ (k) = ek for all 1 ≤ k ≤ n. Define π2∞ ∈ E Z by setting π2∞ (k) := fk for all 1 ≤ k ≤ n and π2∞ (k) := π1∞ (k) for all k ∈ Z \ {1, . . . , n}. It is clear that π2∞ is a bi-infinite path in G . Moreover, π1∞ and π2∞ are almost equal and have the same label. As π1∞ /= π2∞ (since π1 /= π2 ), this shows that (ii) implies (i). (d) This follows immediately from Exercise 6.82(a). (e) This follows immediately from (d) since the De Bruijn graph G associated with the pair (X, m) is deterministic by Exercise 6.82(b). (f) Let G ' = (Q' , E ' ) denote the De Bruijn graph associated with the quadruple (X, τ, n, m). Recall from Exercise 6.73 that Q' = Q and that there is a bijective map ϕ : E ' → E such that ϕ(e' ) has the same initial vertex and the same terminal vertex as e' for all e' ∈ E ' . This map induces a bijective map ϕ∗ from the set of bi-infinite paths in G ' onto the set of bi-infinite paths in G . Moreover, it was observed in the π ∞ in G ' has label x ∈ X ⊂ AZ proof of Exercise 6.73(c) that if a bi-infinite path n ∞ then the label of the bi-infinite path π in G defined by π ∞ := ϕ∗ (n π ∞ ) is the

450

6 Finitely Generated Groups

configuration τ (x) ∈ B Z shifted on the left by m + n − 1. Note that the labeled graph G ' has no diamonds by (d) since it is deterministic by Exercise 6.82(b). Suppose now that there exists a diamond (π1 , π2 ) in G . Since G is essential by Exercise 6.83(e), it follows from the implication (ii) =⇒ (i) established in (c) that there exist two distinct bi-infinite paths π1∞ and π2∞ in G that are almost equal and have the same label. Then the bi-infinite paths n π1∞ and n π2∞ in G ' , defined by n π1∞ := ϕ∗−1 (π1∞ ) and n π2∞ := ϕ∗−1 (π2∞ ), are also distinct and almost equal. As n π1∞ ∞ and n π2 are almost equal, their labels x1 , x2 ∈ X are almost equal. By applying (b), we have x1 /= x2 since G has no diamonds. On the other hand, τ (x1 ) = τ (x2 ) since π1∞ and π2∞ have the same label in G . We deduce that the restriction of τ to X is not pre-injective. This shows that (2) implies (1). Conversely, suppose that G has no diamonds. Let x1 , x2 ∈ X such that x1 and x2 are almost equal and have the same image under τ . For k ∈ Z and i ∈ {1, 2}, consider the word qki ∈ Lm−1 (X) defined by qki := xi (k + n)xi (k + n + 1) · · · xi (k + n + m − 2).

.

Observe that i eki := (qk , (τ (xi ))(k), qk+1 )∈E

.

for all k ∈ Z and i ∈ {1, 2}. Thus, for each i ∈ {1, 2}, the sequence πi∞ ∈ E Z , defined by πi∞ (k) := eki for all k ∈ Z, is a bi-infinite path in G . Observe that π1∞ and π2∞ are almost equal since x1 and x2 are almost equal. On the other hand, π1∞ and π2∞ have the same label since τ (x1 ) = τ (x2 ). Since G has no diamonds by our assumption, it follows that π1∞ = π2∞ by (b). This implies qk1 = qk2 and hence x1 (k) = x2 (k) for all k ∈ Z. Therefore x1 = x2 . This shows that the restriction of τ to X is pre-injective. Thus, (1) implies (2). This completes the proof that conditions (1) and (2) are equivalent. O Exercise 6.114 (The Nth Higher Power Subshift) Let A be a finite set and let N be a positive integer. Consider the map ΨN : AZ → (AN )Z defined by ΨN (x)(n) := (x(nN), x(nN + 1), . . . , x((n + 1)N − 1))

.

for all x ∈ AZ and n ∈ Z. Consider also the map ψN : A∗ → (AN )∗ defined as follows. For w ∈ A∗ , set ψN (w) := (a1 , a2 , . . . , aN )(aN+1 , aN+2 , . . . , a2N )

.

· · · (a(n−1)N +1 , a(n−1)N +2 , . . . , anN ) if w = a1 a2 · · · anN for some n ≥ 1 and ψN (w) := E otherwise. Let X ⊂ AZ be a subshift and set X(N ) := ΨN (X) ⊂ (AN )Z . For L ⊂ A∗ , set L(N ) := ψN (L) ⊂ (AN )∗ .

6.2 Exercises

451

(a) Show that X(N ) is a subshift of (AN )Z (it is called the Nth higher power subshift of X). (b) Show that the restriction of the map ψN to {w ∈ A∗ : l(w) is a multiple of N } is injective and that L(X(N ) ) = (L(X))(N ) . (c) Show that X is of finite type if and only if X(N ) is of finite type. (d) Consider the Følner sequence F := (Fn )n≥1 for Z, where Fn := {0, 1, 2, . . . , n − 1}. Show that entF (X(N ) ) = N entF (X). Solution (a) Let G := Z. Then H := NZ is a finite index subgroup and T := {0, 1, . . . , N − 1} ⊂ Z is a complete set of representatives for the cosets of H in G. Thus, after identifying AT and AN and modulo the group isomorphism G ∃ n |→ N n ∈ H , the map ΨN : AZ → (AN )Z is nothing but the Zequivariant uniform embedding Ψ defined in Exercise 1.32. Moreover, with the notation in Exercise 1.72 we have that X(N ) equals X(H,T ) . It then follows from Exercise 1.72(a) that X (N ) is a subshift of (AN )Z . (b) If the length of a word w ∈ A∗ is a multiple of N, then each letter of w appears in one of the N components of one of the letters of ψN (w). This implies that the restriction of ψN to {w ∈ A∗ : l(w) is a multiple of N} is injective. Let u ∈ L(X(N ) ). Thus, there exist y ∈ X(N ) and m ∈ Z such that u = y(m)y(m + 1) · · · y(m + k − 1), where k := l(u). If x := ΨN−1 (y) ∈ X, then we have y(n) = ΨN (x)(n) = (x(nN), x(nN + 1), . . . , x((n + 1)N − 1)) for all n = m, m + 1, . . . , m + k − 1. It is then clear that setting w := x(mN)x(mN + 1) · · · x((m + k)N − 1) ∈ L(X), we have u = ψN (w) ∈ ψN (L(X)) = (L(X))(N ) . This shows the inclusion L(X(N ) ) ⊂ (L(X))(N ) . Conversely, suppose that u ∈ (L(X))(N ) and let w := ψN−1 (u) ∈ L(X). Then there exist x ' ∈ X and n' ∈ Z such that w = x ' (n' )x ' (n' + 1) · · · x ' (n' + kN − 1), where k := l(w)/N . Let 0 ≤ r < N and n ∈ Z such that n' = nN + r, and set x := rx ' ∈ X. Then, w = x(n' )x(n' + 1) · · · x(n' + kN − 1) = x ' (nN + r)x ' (nN + 1+r) · · · x ' ((n+k)N −1+r) = x(nN)x(nN +1) · · · x((n+k)N −1). Then setting y := ψN (x) ∈ X(N ) we have u = y(n)y(n + 1) · · · y(n + k − 1) ∈ L(X(N ) ). This shows the inclusion (L(X))(N ) ⊂ L(X(N ) ). We deduce that L(X(N ) ) = (L(X))(N ) . (c) This follows from Exercise 1.72(b). (d) This follows from Exercise 5.35(b) after observing that N = [Z : NZ] = [G : H ]. O Exercise 6.115 (The Nth Higher Power Graph) Let A be a set and let G = (Q, E) be an A-labeled graph. Let N be a positive integer. The Nth higher power graph associated with G is the AN -labeled graph G (N ) = (Q(N ) , E (N ) ) defined as follows. The vertex set of G (N ) is Q(N ) := Q and its edge set is E (N ) := {(π − , λ(π ), π + ) ∈ Q × AN × Q : π a path of length N in G },

.

where π − ∈ Q (resp. π + ∈ Q) is the initial (resp. terminal) vertex of the path π and λ(π ) ∈ An is its label.

452

6 Finitely Generated Groups

Recall that, given a labeled graph H , we denote by BH its adjacency matrix (cf. Exercise 6.91) and by XH the labeled edge subshift associated with H . Recall also that, given a subshift X ⊂ AZ , we denote by X(N ) ⊂ (AN )Z the Nth higher power subshift of X (cf. Exercise 6.114). (a) Show that BG (N) = (BG )N . (b) Suppose that G is connected. Show that one has per(G ) = per(BG ) and that G (N ) is connected if and only if the integers N and per(G ) are relatively prime. (N) (c) Show that X G = (XG )(N ) . (d) Show that ρ(G (N ) ) = ρ(G )N . (e) Show that if X ⊂ AZ is a sofic subshift, then the subshift X(N ) ⊂ (AN )Z is also sofic. (f) Suppose that X ⊂ AZ is an irreducible subshift of finite type such that the integers per(X) and N are relatively prime. Show that the subshift X (N ) is irreducible. Solution ' ) (a) Let us set B = (bq,p )q,p∈Q := BG and B ' = (bq,p q,p∈Q := BG (N) . Given ' q, p ∈ Q, the entry bq,p is equal to the number of edges e(N ) ∈ E (N ) with initial vertex q and terminal vertex p. By definition of the graph G (N ) , this is equal to the number of paths π of length N in G connecting q to p. Now, each such path π is of the form π = (e1 , e2 , . . . , eN ) with ei = (qi , aE i , qi+1 ) for i = 1, 2, . . . , N such ' that q1 = q and qN+1 = p. It follows that bq,p = q2 ,q3 ,...,qN bqq2 bq2 q3 · · · bqN p = (N )

bq,p , showing that ' (N ) BG (N) = B ' = (bq,p )q,p∈Q = (bq,p )q,p∈Q = B N = (BG )N .

.

(b) The equality per(G ) = per(BG ) follows from Exercise 6.95(b). By Exercise 6.95(a), the graph G (N ) is connected if and only if its adjacency matrix is irreducible. As the adjacency matrix of G (N ) is BGN by (a), we deduce from Exercise 6.94 that G (N ) is connected if and only if N and per(BG ) are relatively prime. (N) (N ) (c) Let y ∈ XG . This means that there is a bi-infinite path π (N ) = (en )n∈Z in (N ) (N ) (N ) such that y = λ(π ), that is, y(n) = λ(en ) for all n ∈ Z. For each n ∈ Z, G (N ) n there exists a path πn = (e0n , e1n , . . . , eN = −1 ) of length N in G such that en n n n − + N (πn , λ(πn ), πn ), where, in particular, λ(πn ) = λ(e0 )λ(e1 ) · · · λ(eN −1 ) ∈ A . As (N )

n a consequence, y(n) = λ(en ) = λ(πn ) = λ(e0n )λ(e1n ) · · · λ(eN −1 ). Keeping in − n+1 n + mind that ω(eN −1 ) = πn = πn+1 = e0 for all n ∈ Z, setting

ekN+r := erk

.

for all k ∈ Z and 0 ≤ r ≤ N − 1 defines a bi-infinite path π = (ei )i∈Z in G . Moreover, setting x := λ(π ) ∈ XG , one has y = ΨN (x) ∈ X (N ) = (XG )(N ) . This (N) shows the inclusion XG ⊂ (XG )(N ) .

6.2 Exercises

453

Conversely, suppose that y ∈ (XG )(N ) and let x := ΨN−1 (y) ∈ XG . Then there exists a bi-infinite path π = (ek )k∈Z in G such that x = λ(π ), that is, x(k) = λ(ek ) for all k ∈ N. For k ∈ Z we have that πk := (ekN , ekN+1 , . . . , e(k+1)N−1 ) is a path of length N in G . After observing that πk+ = ω(e(k+1)N−1 ) = α(e(k+1)N ) = − , we have that setting ek := (πk− , λ(πk ), πk+ ) ∈ E (N ) defines a bi-infinite path πk+1

π (N ) = (ek )k∈Z in G (N ) . Let z := λ(π (N ) ) ∈ XG . We have z(k) = λ(π (N ) )(k) = λ(ek ) = λ(πk ) = λ(ekN )λ(ekN+1 ) · · · λ(e(k+1)N−1 ) = x(kN )x(kN + 1) · · · x((k + (N) 1)N − 1) = ΨN (x)(k) = y(k). As a consequence y = z ∈ XG . This shows that (N) (XG )(N ) ⊂ XG . (N) We deduce that XG = (XG )(N ) . (d) Let π ∈ PnN (G ), say π = (e1 , e2 , . . . , enN ). Set πi := (e(i−1)N+1 , e(i−1)N+2 , . . . , eiN ) ∈ PN (G ) and pi := πi− = α(e(i−1)N+1 ) and qi := πi+ = ω(eiN ) for all (N ) i = 1, 2, . . . , N. Then, setting ei := (pi , λ(πi ), qi ) ∈ E (N ) for i = 1, 2, . . . , n, (N ) (N ) (N ) defines a path π (N ) = (e1 , e2 , . . . , en ) ∈ Pn (G (N ) ). Conversely, let π (N ) ∈ (N ) (N ) Pn (G (N ) ), say π (N ) = (e1 , e2 , . . . , en(N ) ), where ei(N ) = (pi , λ(πi ), qi ) with πi ∈ PN (G ) such that πi− = pi (resp. πi+ = qi ) for all i = 1, 2, . . . , n and πi+ = (N ) − ) = πi+1 for all i = 1, 2, . . . , n − 1. Then π := π1 π2 · · · πn ∈ ω(ei(N ) ) = α(ei+1 PnN (G ). This shows that the map π |→ π (N ) yields a bijection from PnN (G ) onto Pn (G (N ) ). As a consequence (N)

/

ρ(G (N ) ) = lim

n

n→∞

R n

= lim

n→∞

.

= lim

n→∞

=

|Pn (G (N ) )| |PnN (G ))|

)N (R n |Pn (G ))|

(

lim

n→∞

)N R n |Pn (G ))|

= ρ(G )N . (e) Suppose that X ⊂ AZ is a sofic subshift. By Exercise 6.78, there exists a finite A-labeled graph G such that X = XG . Consider the AN -labeled graph G (N ) . (N) It follows from (c) and Exercise 6.69(b) that X (N ) = (XG )(N ) = XG is also sofic. (f) Let m ≥ 2 be an integer such that the set {1, 2, . . . , m} is a memory set for X and let G = (Q, E) denote the De Bruijn graph associated with the pair (X, m). Also let B = BG denote the adjacency matrix of G . Since X is irreducible, the graph G is connected by Exercise 6.72(b) and the matrix BG is irreducible by Exercise 6.95(a). Moreover, per(X) = per(G ) = per(B) by Exercises 6.98(b) and 6.95(b). Since per(X) and N are assumed to be relatively prime, it follows from (b) that G (N ) is connected. By using (c) and Exercise 6.69(f), we conclude that (N) O X(N ) = (XG )(N ) = XG is irreducible.

454

6 Finitely Generated Groups

Exercise 6.116 Let A, B be finite sets and let X ⊂ AZ be an irreducible subshift of finite type. Let τ : AZ → B Z be a cellular automaton and suppose that the restriction of τ to X is not pre-injective. Consider the subshift Y ⊂ B Z defined by Y := τ (X). Let m, n ∈ Z with m ≥ 2 such that {n, n + 1, . . . , n + m − 1} ⊂ Z is a memory set for both X and τ and let G = (Q, E) (resp. H = (QH , EH )) denote the De Bruijn graph associated with the pair (X, m) (resp. with the quadruple (X, τ, n, m)). (a) Show that G and H are connected. (b) Show that one has per(X) = per(G ) = per(H ) < ∞. (c) Show that there exists a diamond (π1 , π2 ) in H such that N := l(π1 ) = l(π2 ) is relatively prime with per(G ). (d) Consider the Nth higher power graph G (N ) = (Q(N ) , E (N ) ) of G . Regard π1 as a path in G and consider the edge e1 := (q, λ(π1 ), p) ∈ E (N ) , where q ∈ Q(N ) = Q is the initial vertex of π1 , p ∈ Q(N ) = Q is its terminal vertex, and λ(π1 ) ∈ AN is the label of π1 in G . Let K = (QK , EK ), where QK := Q(N ) = Q and EK := E (N ) \ {e1 }, be the AN -labeled graph obtained from G (N ) by removing the edge e1 . Show that the labeled edge subshift Z ⊂ (AN )Z associated with K satisfies Z ⊂ X(N ) . (e) Let τ (N ) : (AN )Z → (B N )Z denote the cellular automaton induced by τ , i.e., using the notations of Exercise 1.73, τ (N ) := τ (T ) with G := Z, H := NZ, and T := {0, 1, . . . , N − 1} ⊂ G. Show that τ (N ) (Z) = τ (N ) (X(N ) ) = Y (N ) . (f) Let F = (Fj )j ≥1 be the Følner sequence for Z defined by Fj := {0, 1, . . . , j − 1} for all j ≥ 1. Show that entF (Y ) < entF (X). Solution (a) As X is irreducible, the graph G (resp. H ) is connected by Exercise 6.72(c) (resp. Exercise 6.73(d)). (b) We have per(X) < ∞ since X is of finite type by Exercise 6.98(a). The subshift X is the labeled edge subshift associated with G by Exercise 6.72(b) and we have per(X) = per(G ) by Exercise 6.98(c). On the other hand, we have per(G ) = per(H ) since G and H have the same sets of vertices and the same set of edges, and therefore of finite paths, up to labeling by Exercise 6.73(b). This shows that per(X) = per(G ) = per(H ) < ∞. (c) Since the restriction of τ to X is not pre-injective, there exists a diamond (π1' , π2' ) in H by Exercise 6.113(f). On the other hand, as H is connected by (a), given any k ∈ N, we can find a finite path π of length k in H whose initial vertex is the common terminal vertex of π1' and π2' . Setting N ' := l(π1' ) = l(π2' ), we obtain by concatenation two finite paths π1 := π1' π and π2 := π2' π with length N := N ' +k = l(π1 ) = l(π2 ) such that (π1 , π2 ) is also a diamond in H . Choosing k ∈ N such that N ' + k is a prime that does not divide per(G ), ensures that the integers N and per(G ) are relatively prime. (d) We know from Exercise 6.72(b) that X is the labeled edge subshift associated with G . By Exercise 6.115(c), it follows that the labeled edge subshift associated with G (N ) is X(N ) . Since K is a subgraph of G (N ) , we deduce that Z ⊂ X(N ) . (e) In the following, we denote by λ (resp. λ' ) the labeling in the graph G (resp. H ). Recall that Y is the labeled edge subshift associated with H by

6.2 Exercises

455

Exercise 6.73(c). We have that τ (N ) (X(N ) ) = Y (N ) by Exercise 1.73(b). Since Z ⊂ X(N ) by (d), we have τ (N ) (Z) ⊂ τ (N ) (X(N ) ) = Y (N ) . Let now y ∈ Y (N ) . By Exercise 6.115(c), Y (N ) is the labeled edge subshift associated with H (N ) , the N th power graph of H . Thus, there exists a bi-infinite path π in H (N ) with label y ∈ (B N )Z . Consider the corresponding bi-infinite path π in G (N ) . Observe that the edges e1 ∈ E (N ) and e2 := (q, λ(π2 ), p) ∈ EK ⊂ E (N ) , when viewed in the graph H (N ) , have the same label λ' (π1 ) = λ' (π2 ) ∈ B N since (π1 , π2 ) is a diamond in H . As a consequence, by replacing all occurrences of the edge e1 in π by the edge e2 , we obtain a bi-infinite path π ' in K such that λ' (π ' ) = λ' (π ). The configuration z := λ(π ) ∈ Z then satisfies τ (N ) (z) = λ' (π ' ) = λ' (π ) = y. This shows that Y (N ) ⊂ τ (N ) (Z). We deduce that τ (N ) (Z) = Y (N ) . (f) Since G is connected by (a) and gcd(N, per(G )) = 1 by our choice of N, we deduce from Exercise 6.115(b) that G (N ) is also connected. Thus, by applying Exercise 6.104(d), we get ρ(H ) < ρ(G (N ) ). As log(ρ(G (N ) )) = N log(ρ(G )) by Exercise 6.115(d), this gives us log(ρ(H )) < N log(ρ(G )).

.

(6.70)

The graph G is deterministic by Exercise 6.82(b). As X is the labeled edge subshift associated with G by Exercise 6.72(b), we deduce from Exercise 6.102(b) that .

entF (X) = log(ρ(G )).

(6.71)

Finally, we get 1 entF (Y (N ) ) (by Exercise 6.114(d)) N 1 = entF (τ (N ) (Z)) (by (e)) N 1 entF (Z) ≤ N

entF (Y ) =

(by [CAG, Proposition 5.7.3]) ≤

1 log(ρ(H )) (by Exercise 6.102(a)) N

< log(ρ(G )) (by (6.70)) = entF (X) (by (6.71)). This implies entF (Y ) < entF (X).

O

Exercise 6.117 Let A and B be finite sets. Suppose that X ⊂ AZ and Y ⊂ B Z are two subshifts such that X is irreducible of finite type and entF (X) = entF (Y ), where F = (Fn )n≥1 is the Følner sequence for Z defined by Fn := {0, 1, . . . , n−1}

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6 Finitely Generated Groups

for all n ≥ 1. Show that every surjective cellular automaton τ : X → Y is preinjective. Solution If τ : X → Y is a cellular automaton which is not pre-injective, it follows from Exercise 6.116(f) and our assumptions that entF (τ (X)) < entF (X) = entF (Y ). This shows that τ (X) G Y , so that τ is not surjective. O Comment This is the implication (5) =⇒ (4) in [LinM, Theorem 8.1.16]. Note that the irreducibility of X cannot be removed from the hypotheses. Indeed, take A = B := {0, 1} and consider the subshifts X, Y ⊂ AZ where X is the subshift of finite type admitting {10} as a defining set of forbidden words and Y is reduced to the identically-0 configuration. Then entF (X) = entF (Y ) = 0 and the unique map τ : X → Y is a surjective cellular automaton. However, τ is not pre-injective since the configurations xn ∈ X, n ∈ Z, defined by setting xn (k) := 0 for k ≤ n and xn (k) := 1 for k > n, are all almost equal and have the same image under τ . Exercise 6.118 Let A, B be finite sets and let X ⊂ AZ be an irreducible subshift of finite type. Let τ : X → B Z be a cellular automaton and consider the subshift Y ⊂ B Z defined by Y := τ (X). Suppose that every periodic configuration y ∈ Y has a countable (e.g. finite) pre-image set τ −1 (y) ⊂ AZ . Show that τ is pre-injective. Solution Suppose by contradiction that τ is not pre-injective. Let m, n ∈ Z with m ≥ 2 such that {n, n + 1, . . . , n + m − 1} ⊂ Z is as a memory set for both X and τ . Consider the De Bruijn graph G = (Q, E) associated with the quadruple (X, τ, n, m) and the De Bruijn graph G ' associated with the pair (X, m). Recall that G and G ' have the same set of vertices and the same set of edges up to labeling (cf. Exercise 6.73(b)) so that there is a natural identification between bi-infinite paths in G and bi-infinite paths in G ' . Recall also that if the label of a bi-infinite path in G ' is the configuration x ∈ AZ then its label in G is the configuration τ (x) ∈ B Z shifted on the left by n + m − 1 (see the solution to Exercise 6.73(b)). Since τ is not pre-injective, there exists a diamond (π1 , π2 ) in G by Exercise 6.113(f). Let q ∈ Q (resp. p ∈ Q) denote the common initial (resp. terminal) vertex of π1 and π2 . As X is irreducible, the graph G is connected by Exercise 6.73(d). Therefore, there exists a finite path π0 in G with initial vertex p and terminal vertex q. Write N := l(π1 ) = l(π2 ) and P := N + l(π0 ). By concatenation, we get finite closed paths γ1 := π1 π0 , γ2 := π2 π0 in G of length P . We have γi : {1, 2, . . . , P } → E for i ∈ {1, 2}. For every f : Z → {1, 2}, consider the bi-infinite path ψf : Z → E defined by ψf (s + kP ) := γf (k) (s) for all k ∈ Z and s ∈ {1, 2, . . . , P }. Observe that ψf ∈ E Z is P -periodic. Moreover, since π1 /= π2 , the map f |→ ψf is an injection from {1, 2}Z into the set of bi-infinite paths in G . Thus, taking the labels of the paths ψf in G ' , we get an uncountable subset of P -periodic configurations in X having all the same image y ∈ Y under τ . As y is P -periodic, this yields a contradiction. O Exercise 6.119 Let A and B be finite sets. Suppose that X ⊂ AZ and Y ⊂ B Z are two irreducible subshifts of finite type such that entF (X) = entF (Y ), where F = (Fn )n≥1 is the Følner sequence for Z defined by Fn := {0, 1, . . . , n − 1} for

6.2 Exercises

457

all n ≥ 1. Let τ : X → Y be a cellular automaton. Show that τ is surjective if and only if it is pre-injective. Solution The fact that the pre-injectivity of τ implies its surjectivity follows from Exercise 6.110 since every subshift of finite type is sofic. The converse implication follows from Exercise 6.117. O Exercise 6.120 (The Garden of Eden Theorem for Irreducible Subshifts of Finite Type Over Z) Let A be a finite set. Let X ⊂ AZ be an irreducible subshift of finite type and let τ : X → X be a cellular automaton. Show that τ is surjective if and only if it is pre-injective. Solution This follows from Exercise 6.119 after taking A = B and X = Y .

O

Comment This result is Corollary 2.19 in [Fio1]. It may be rephrased by saying that every irreducible subshift of finite type over the group Z has both the Moore and the Myhill properties. This becomes false if Z is replaced by Z2 (see Exercise 5.58 for 2 an example of a irreducible subshift of finite type X ⊂ {0, 1}Z which does not have the Myhill property). Exercise 6.121 Let A := {0, 1}. Consider the subset X ⊂ AZ consisting of all configurations x ∈ AZ such that the set {n ∈ Z : x(n) = 1} is an interval of Z. (a) Show that X is a sofic subshift. (b) Show that X is neither of finite type nor topologically transitive. (c) Give a labeled graph G such that X is the labeled edge subshift associated with G . (d) Consider the Følner sequence F for Z defined by F = (Fn )n≥1 , where Fn := {0, 1, . . . , n − 1} for all n ≥ 1. Compute entF (X), (e) Let σ : AZ → AZ be the cellular automaton with memory set S := {0, 1} and local defining map μ : AS → A given by { μ(y) :=

1

if (y(0), y(1)) = (0, 1),

y(0)

otherwise.

.

Check that σ (X) ⊂ X. (f) Show that the cellular automaton τ = σ |X : X → X is injective (and therefore pre-injective) but not surjective. (g) Show that the subshift X is not surjunctive. (h) Show that the subshift X has neither the Myhill property nor the Moore property. Solution (a) Let B := {0, 1, 2}. Consider the subshift of finite type Z ⊂ B Z consisting of all z ∈ B Z such that (z(n), z(n + 1)) ∈ {(0, 0), (0, 1), (1, 1), (1, 2), (2, 2)} for all n ∈ Z. Clearly X = ρ(Z) where ρ : B Z → AZ is the cellular automaton with memory set M := {0} and local defining map f : B M = B → A given by f (0) = f (2) := 0 and f (1) := 1. This shows that X is a sofic subshift.

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6 Finitely Generated Groups

Fig. 6.15 The A-labeled graph G satisfies XG = X, where X ⊂ AZ is the subshift consisting of all configurations x ∈ AZ such that {n ∈ Z : x(n) = 1} is an interval of Z

(b) Suppose by contradiction that X is of finite type. Then there is an integer n ≥ 0 and a set L ⊂ A∗ of words of length n such that L is a defining set of forbidden words for X. As the word w := 10n−1 1 is a forbidden word for X although no subword of w of length n is in L, this yields a contradiction. This shows that the subshift X is not of finite type. Consider the non-empty open subsets U, V ⊂ X defined by U := {x ∈ X : x(0) = x(1) = 1} and

.

V := {x ∈ X : x(−1) = x(1) = 0 and x(0) = 1}. Clearly V is reduced to a single configuration x1 and the Z-orbit of x1 does not meet U . Consequently, the subshift X is not topologically transitive. (c) Direct inspection shows that the labeled graph G in Fig. 6.15 has the desired property. This yields an alternative proof of (a). (d) Let Ln (X) ⊂ A∗ denote the set consisting of all words of length n appearing in X. We have Ln (X) = {0i 1j 0k : i, j, k ∈ N and i + j + k = n}, so that |Ln (X)| ≤ n2 . This gives us .

log |Ln (X)| = 0. n→∞ n

entF (X) = lim

(e) The effect of applying σ to a configuration in X consists in adding one 1 on the left to the string of 1s appearing in x if this operation is possible. More precisely, let x ∈ X and let I := {n ∈ Z : x(n) = 1}. If the interval I has a minimal element n0 then σ (x)(n0 − 1) = 1 and σ (x)(n) = x(n) for all n ∈ Z \ {n0 − 1}. Otherwise, we have σ (x) = x. It follows that σ (x) ∈ X for all x ∈ X. (f) The description of τ given in (e) shows that it is injective. On the other hand, the configuration x0 ∈ X defined by x0 (0) = 1 and x0 (n) = 0 for all n ∈ Z \ {0} is clearly not in the image of τ . Therefore τ is not surjective. (g) The subshift X is not surjunctive since the cellular automaton τ : X → X is injective but not surjective by (f). (h) The subshift X does not have the Myhill property since it is not surjunctive by (g). Consider the cellular automaton σ ' : AZ → AZ with memory set S := {0, 1} and local defining map μ' : AS → A given by { '

μ (y) :=

.

0

if (y(0), y(1)) = (1, 0),

y(0)

otherwise.

6.2 Exercises

459

The effect of applying σ ' to a configuration x ∈ X is to replace by 0 the 1 located at the far right (if any) of the string of 1s in x. Therefore σ ' (X) ⊂ X. Consider the cellular automaton τ ' = σ ' |X : X → X. Clearly τ ' is surjective. On the other hand, the configurations x0 , x1 ∈ X defined by x0 (n) := 0 for all n ∈ Z, x1 (0) := 1, and x1 (n) := 0 for all n ∈ Z \ {0}, are almost equal and satisfy τ ' (x0 ) = τ ' (x1 ). As x0 /= x1 , we deduce that τ ' is not pre-injective. This shows that X does not have the Moore property. O Exercise 6.122 (Follower Set of a Subshift) Let A be a finite set and let X ⊂ AZ be a subshift. Given a word u ∈ L(X) the set FX (u) := {v ∈ A∗ : uv ∈ L(X)}

.

is called the follower set of the word u in X. The set FX := {FX (u) : u ∈ L(X)} is called the follower set of X. (a) Show that FX (u) ⊂ L(X) for all u ∈ L(X). (b) Show that L(X) ∈ FX if X /= ∅. (c) Let Y ⊂ AZ be another subshift. Show that FX = FY if and only if X = Y . (d) Determine the follower set of X in each of the following cases: (1) (2) (3) (4)

X X X X

:= AZ is the full subshift; ⊂ {0, 1}Z is the golden mean subshift; ⊂ {0, 1}Z is the at-most-one-one subshift; ⊂ {0, 1}Z is the even subshift.

(e) Show that if X is of finite type then FX is finite. Solution (a) This follows immediately from the fact that if u, v ∈ A∗ satisfy uv ∈ L(X), then v ∈ L(X) (cf. Exercise 1.74(e)). (b) Suppose X /= ∅. Then L(X) contains the empty word E. For every u ∈ L(X), we have Eu = u ∈ L(X), so that FX (E) = L(X). This shows L(X) ∈ FX . (c) Since F∅ = ∅, we may suppose that X, Y /= ∅. It is clear that if X = Y then FX = FY . To show the converse implication, suppose that FX = FY . It follows from (a) and (b) that L(X) (resp. L(Y )) is the largest element in FX (resp. FY ) with respect to set inclusion. We deduce that L(X) = L(Y ). This implies X = Y by Exercise 1.74(d). (d)We have the following. (1) It is clear that FX (u) = A∗ = L(X) for all u ∈ L(X). We deduce that FX = {A∗ }. (2) In this case, L(X) consists of all the words in {0, 1}∗ with no two consecutive 1s. Let u ∈ L(X). If the last letter of the word u is 1, then FX (u) = 0A∗ ∩ L(X) = 0L(X). Otherwise, FX (u) = L(X). We deduce that FX = {L(X), 0L(X)}. (3) In this case, L(X) consists of all the words in {0, 1}∗ with at most one 1. If u ∈ L(X) contains only 0s then FX (u) = L(X). Otherwise, FX (u) = {0n : n ∈ N}. It follows that FX = {L(X), {0n : n ∈ N}}.

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6 Finitely Generated Groups

(4) Let u = a1 a2 . . . am ∈ L(X), where m = l(u) and ai ∈ A := {0, 1} for all 1 ≤ i ≤ m. If am = 1, then FX (u) = {0n : n ∈ N} ∪ ({02n 1 : n ∈ N}A∗ ∩ L(X)). If a1 = a2 = · · · = am = 0, then FX (u) = L(X). Finally, if there exists 1 ≤ i < m such that ai = 1 and ai+1 = ai+2 = · · · = am = 0, then FX (u) = {0n : n ∈ N} ∪ ({02n 1 : n ∈ N}A∗ ∩ L(X)) if m − i is even and FX (u) = {0n : n ∈ N} ∪ ({02n+1 1 : n ∈ N}A∗ ∩ L(X)) if m − i is odd. We deduce that FX = {L(X), {0n : n ∈ N}∪({02n 1 : n ∈ N}A∗ ∩L(X)), {0n : n ∈ N}∪({02n+1 1 : n ∈ N}A∗ ∩L(X))}. (e) Suppose that X is of finite type. It follows from Exercise 1.79 that there exists an integer n0 ≥ 0 such that if the words u, v, w ∈ A∗ satisfy uw, wv ∈ L(X) and w has length l(w) ≥ n0 , then one has uwv ∈ L(X). For i ∈ N we set Li (X) := {w ∈ L(X) : l(w) = i}. Let us show that FX = {FX (u) : u ∈ Li (X), 0 ≤ i ≤ n0 }.

.

(6.72)

Denote by FX' the right hand side of (6.72). The inclusion FX ⊃ FX' is obvious. Conversely, let u ∈ L(X). If l(u) ≤ n0 then FX (u) ∈ FX' . If l(u) > n0 , say u = vw with v, w ∈ A∗ and l(w) = n0 , then w ∈ Ln0 (X) by Exercise 1.74(e) and, by the definition of n0 , we have FX (u) = FX (w) ∈ FX' . This shows that FX ⊂ FX' , and (6.72) follows. We deduce that |FX | = |{FX (u) : u ∈ Li (X), 0 ≤ i ≤ n0 }| ≤ |{u ∈ L(X) : l(u) ≤ n0 }| ≤ |{u ∈ A∗ : l(u) ≤ n0 }| = 1 + |A| + |A|2 + · · · + |A|n0 . O Exercise 6.123 Let A be a finite set and let X ⊂ AZ be a subshift. Let L(X) ⊂ A∗ denote the language of X. For u ∈ L(X) set Au := {a ∈ A : ua ∈ L(X)}. (a) Show that Au /= ∅ for all u ∈ L(X). (b) Show that Au = FX (u) ∩ A for all u ∈ L(X). (c) Let u, v ∈ L(X) such that FX (u) = FX (v). Show that Au = Av and that FX (ua) = FX (va) for all a ∈ Au = Av . (d) Show that X is sofic if and only if the set FX is finite. (e) Let X ⊂ {0, 1, 2}Z denote the context-free subshift. Use (d) to recover the fact that X is not sofic (cf. Exercise 6.80(b)). Solution (a) Let u ∈ L(X). It follows from Exercise 1.74(e) that there exists a ∈ A such that ua ∈ L(X). This shows that Au /= ∅. (b) This is obvious from the definitions. (c) Using (b), we get Av = FX (v)∩A = FX (u)∩A = Au . Moreover, FX (ua) = {w ∈ A∗ : aw ∈ FX (u)} = {w ∈ A∗ : aw ∈ FX (v)} = FX (va) for all a ∈ Au = Av . (d) Suppose first that X is sofic. Then there exists an essential finite A-labeled graph G = (Q, E) such that X = X G (cf. Exercise 6.83). Let u ∈ L(X). Let Qu ⊂ Q denote theUset of terminal vertices of all finite paths π in G whose label is u. Then FX (u) = q∈Qu FG (q) (where FG (q) denotes the follower set of q ∈ Q (cf. Exercise 6.87)). This shows that |FX | ≤ 2|Q| .

6.2 Exercises

461

Conversely, suppose that the set FX is finite. We construct an essential finite A-labeled graph G = (Q, E) such that X = XG in the following way. We first set Q := FX . Let q ∈ Q. Then q = FX (u) for some u ∈ L(X). By (c), the set Aq := Au ⊂ A and, for a ∈ Aq , the set qa := FX (ua) ∈ Q are all well defined, i.e., do not depend on the choice of u. We then set E := {e = (q, a, qa ) : q ∈ Q, a ∈ Aq } ⊂ Q × A × Q. This completes the construction of G . Let us show that G is essential. By virtue of Exercise 6.83(a), it suffices to show that every vertex q ∈ Q is not stranded. Let q ∈ Q. Then q = FX (u) for some u = a1 a2 · · · an ∈ L(X), where ai ∈ A for all 1 ≤ i ≤ n and n is the length of u. Setting v := a1 a2 · · · an−1 , we have v ∈ L(X) so that if p := FX (v) ∈ Q then an ∈ Ap and e := (p, an , q) ∈ E, and thus q = ω(e). On the other hand, Aq = Au /= ∅ by (a). If a ' ∈ Aq , then e' := (q, a ' , qa ' ) ∈ E and thus q = α(e' ). This shows that q is not stranded. We deduce that G is essential. By virtue of Exercise 6.83(d), in order to show that X = XG it suffices to show that L(X) = LG . Let u = a1 a2 · · · an ∈ L(X). Let an+1 ∈ Au and, for i = 1, 2, . . . , n + 1 (resp. i = 1, 2, . . . , n), set qi := FX (a1 a2 · · · ai ) (resp. ei := (qi , ai , qi+1 )). It is then clear that the finite path π := (e1 , e2 , . . . , en ) in G satisfies u = a1 a2 · · · an = λ(π ) ∈ LG . This shows the inclusion L(X) ⊂ LG . Conversely, let u = a1 a2 · · · an ∈ LG . This means that there exists a finite path π = (e1 , e2 , . . . , en ) in G such that u = λ(π ). Set ei := (qi , ai , qi+1 ) for all i = 1, 2, . . . , n and let v ∈ L(X) such that q1 = FX (v). Then qi+1 = FX (va1 a2 · · · ai ) for all i = 1, 2, . . . , n − 1 and, eventually, qn = FX (va1 a2 · · · an ) = FX (vu). In particular, vu ∈ L(X) so that, by Exercise 1.74(e), u ∈ L(X). This shows that LG ⊂ L(X). We deduce that L(X) = LG so that, G being essential, X = XG . Thus, X is sofic by Exercise 6.78. (e) Recall (cf. Exercise 6.80) that X consists of all configurations x ∈ {0, 1, 2}Z satisfying the following condition: if x(n) = 0, x(n + 1) = x(n + 2) = · · · = x(n + h) = 1, x(n + h + 1) = x(n + h + 2) = · · · = x(n + h + k) = 2, and x(n + h + k + 1) = 0 for some n ∈ Z and h, k ∈ N, then h = k. Observe that 01n ∈ L(X) for all n ∈ N. Suppose that 0 ≤ n < m are integers. Then 2n 1 ∈ FX (01n ) \ FX (01m ) and therefore FX (01n ) /= FX (01m ). We deduce that the set {FX (01n ) : n ∈ N} is infinite. This implies that FX is infinite and hence that X is not sofic by (d). O Exercise 6.124 (Left-Infinite Configurations and Their Follower Sets) Let A be a finite set. A map p : − N → A, i.e., an element p ∈ A−N , is called a left-infinite configuration over the alphabet A. Given a configuration x ∈ AZ , the left-infinite configuration x − := x|−N ∈ A−N is called the left-infinite configuration associated with x. Let X ⊂ AZ be a subshift. One says that a left-infinite configuration p ∈ A−N appears in X provided there exists x ∈ X such that p = x − . We denote by X− := {x − : x ∈ X} ⊂ A−N the set of all left-infinite configurations that appear in X.

462

6 Finitely Generated Groups

Let p ∈ A−N and let w = a1 a2 · · · ak ∈ A∗ , where k := l(w) and ai ∈ A for all 1 ≤ i ≤ k. We denote by pw ∈ A−N the left-infinite configuration defined by setting { (pw)(n) :=

.

p(n + k)

if n ≤ −k

ak+n

if − k + 1 ≤ n ≤ 0

for all n ∈ −N. Given p ∈ X − , the set FX (p) := {w ∈ A∗ : pw ∈ X− } is called the follower set of p in X. Write FX− := {FX (p) : p ∈ X− }. (a) Let x ∈ X. For h, k ∈ Z such that h ≤ k, we write x[h,k] := x(h)x(h + 1) · · · x(k) ∈ A∗ . Show that FX (x[−n−1,0] ) ⊂ FX (x[−n,0] ) for all n ∈ N and that one has U − .FX (x ) = FX (x[−n,0] ). (6.73) n∈N

(b) Show that if FX is finite then FX− ⊂ FX . (c) Let u ∈ L(X) and let n := l(u) denote the length of u. Show that FX (u) =

| |

.

FX (x − ).

(6.74)

x∈X:

x[−n+1,0] =u

(d) Show that the following conditions are equivalent: (i) X is sofic; (ii) the set FX is finite; (iii) the set FX− is finite. Solution (a) We have FX (x[−n−1,0] ) ⊂ FX (x[−n,0] ) for all n ∈ N since x[−n,0] is a suffix of x[−n−1,0] . Let w ∈ FX (x − ). Then, for every n ∈ N, weuhave x[−n,0] w ∈ L(X) and − hence w ∈ FX (x[−n,0] u ). This shows that FX (x ) ⊂ n∈N FX (x[−n,0] ). Conversely, suppose that w ∈ n∈N FX (x[−n,0] ). This means that for every n ∈ N there exists n a configuration x n ∈ X such that x[−n,l] = x[−n,0] w, where l := l(w). Since X is compact, there exists a cluster point x ∈ X of the sequence (x n )n∈N . Clearly, (x)− = x −uand x [1,l] = w. Thus w ∈ FX ((x)− ) = FX (x − ) and the inclusion FX (x − ) ⊃ n∈N FX (x[−n,0] ) follows as well. This shows (6.73). (b) Suppose that FX is finite and let x ∈ X. Then the decreasing sequence (FX (x[−n,0] ))n∈N stabilizes, that is, there exists n0 = n0 (x) ∈ N such that FX (x[−n,0] ) = FX (x[−n0 ,0] ) for all n ≥ n0 . By using (a) we deduce that FX (x − ) = u − n∈N FX (x[−n,0] ) = FX (x[−n0 ,0] ) ∈ FX . This shows that FX ⊂ FX .

6.2 Exercises

463

(c) Let us denote by FX (u)' the right hand side in (6.74). Suppose first that w ∈ FX (u). We then have uw ∈ L(X) so that we can find a configuration x ∈ X such that x(−n + 1)x(−n + 2) · · · x(l) = uw, where n := l(u) and l := l(w). It follows that w ∈ FX (x − ). This proves the inclusion FX (u) ⊂ FX (u)' . Conversely, let w ∈ FX (u)' . This means that there exists x ∈ X such that x(−n+1)x(−n+2) · · · x(0) = u and w ∈ FX (x − ). Thus we can find y ∈ X such that uw = y(−n − l + 1)y(−n − l + 2) · · · y(0) ∈ L(X), so that w ∈ FX (u). The inclusion FX (u) ⊃ FX (u)' follows as well. This proves (6.74). (d) The equivalence (i) ⇐⇒ (ii) follows from Exercise 6.123(d). If FX is finite, then FX− ⊂ FX by (b) so that FX− is also finite. This shows that (ii) implies (iii). Conversely, suppose that the set FX− is finite. Since every element of FX is a union of elements of FX− by (6.74), it then follows that FX is also finite (with − cardinality |FX | ≤ 2|FX | ). This shows that (iii) implies (ii) and completes the proof that all the three conditions are equivalent. O Exercise 6.125 Let A be a finite set. Show that every splicable subshift X ⊂ AZ is sofic. Solution Let X ⊂ AZ be a subshift and suppose that X is not sofic. It then follows from Exercise 6.124(d) that the set FX− is infinite. Thus, there exists a sequence (xi )i∈N in X such that FX (xi− ) /= FX (xj− ) for all distinct i, j ∈ N. Let n ∈ N and set n0 := |A|2n . We claim that there exists m0 ∈ N with m0 ≥ 2n such that, setting wi := (xi )[−m0 ,0] = xi (−m0 )xi (−m0 + 1) · · · x(0) ∈ L(X), we have FX (wi ) /= FX (wj ) for all 0 ≤ i < j ≤ n0 . Let us first show that for all distinct i, j ∈ N there exists m0 (i, j ) ∈ N such that FX ((xi )[−m,0] ) /= FX ((xj )[−m,0] ) for all m ≥ m0 (i, j ). Let i /= j and suppose on the contrary that for every k ∈ N, there exists mk ∈ N such that mk ≥ k and FX ((xi )[−mk ,0] ) = FX ((xj )[−mk ,0] ). Then, using Exercise 6.124(a), we would get FX (xi− ) =

U

.

k∈N

FX ((xi )[−mk ,0] ) =

U

FX ((xj )[−mk ,0] ) = FX (xj− ),

k∈N

a contradiction. In order to complete the proof of the claim, we take m0 := max0≤i 0, a map .ϕ : G → Sym(F ) is called a .(K, ε)-almosthomomorphism if it satisfies the following conditions: (.(K, ε)-AH-1) for all .k1 , k2 ∈ K, one has .dF (ϕ(k1 k2 ), ϕ(k1 )ϕ(k2 )) ≤ ε;

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7 Local Embeddability and Sofic Groups

(.(K, ε)-AH-2) for all .k1 , k2 ∈ K such that .k1 /= k2 , one has .dF (ϕ(k1 ), ϕ(k2 )) ≥ 1 − ε, where .dF denotes the Hamming metric on .Sym(F ). A group G is called sofic if it satisfies the following condition: for every finite subset .K ⊂ G and every .ε > 0, there exist a non-empty finite set F and a .(K, ε)almost-homomorphism .ϕ : G → Sym(F ). Every subgroup of a sofic group is sofic [CAG, Proposition 7.5.4]. A group is sofic if and only if all of its finitely generated subgroups are sofic [CAG, Proposition 7.5.5]. All amenable groups are sofic [CAG, Proposition 7.5.6]. More generally, all LEA-groups are sofic [CAG, Corollary 7.5.11]. In particular, all LEFgroups, all locally residually amenable groups, all locally residually finite groups, all residually amenable groups, and all residually finite groups are sofic [CAG, Corollary 7.5.11]. Every group which is locally embeddable into the class of sofic groups is itself sofic [CAG, Proposition 7.5.10]. The direct product as well as the direct sum of a family of sofic groups are sofic [CAG, Proposition 7.5.7 and Corollary 7.5.8]. The limit of a projective system of sofic groups is also sofic [CAG, Corollary 7.5.9]. Extensions of sofic groups by amenable groups are sofic [CAG, Proposition 7.5.14].

7.1.5 Sofic Groups and Ultraproducts Suppose that we are given a triple .T = (I, ω, F ) consisting of the following data: a set I , an ultrafilter .ω on I , and a family .F = (Fi )i∈I of non-empty finite sets indexed by I . Consider the direct product group PT :=

Π

.

Sym(Fi ).

i∈I

For each .i ∈ I , denote by .dFi the Hamming metric on .Sym(Fi ). Given .α = (αi )i∈I , β = (βi )i∈I ∈ PT , we have .0 ≤ dFi (αi , βi ) ≤ 1 for all .i ∈ I . Thus, it follows from [CAG, Corollary J.2.6] that the Hamming distances .dFi (αi , βi ) have a limit δω (α, β) := lim dFi (αi , βi ) ∈ [0, 1]

.

i→ω

along the ultrafilter .ω. The subset .NT ⊂ PT defined by NT := {α ∈ PT : δω (1PT , α) = 0}.

.

is a normal subgroup of .PT [CAG, Proposition 7.6.2] and the quotient group .PT /NT is sofic [CAG, Theorem 7.6.3]. We have the following characterization of sofic groups [CAG, Theorem 7.6.6].

7.2 Exercises

469

Theorem Let G be a group. The following conditions are equivalent: (a) G is sofic; (b) there exists a triple .T = (I, ω, F ) as above such that G is isomorphic to a subgroup of the quotient group .PT /NT .

7.1.6 Geometric Characterization of Finitely Generated Sofic Groups Given an S-labeled graph .Q = (Q, E), denote by .BQ (q, n) the ball of radius .n ∈ N centered at a vertex .q ∈ Q. If G is a finitely generated group with a finite symmetric generating set S, we write .BS (n) := BC (1G , n), where .C is the Cayley graph of G with respect to S. One has the following characterization of finitely generated sofic groups in terms of the asymptotic geometry of their Cayley graphs [CAG, Theorem 7.7.1]. Theorem Let G be a finitely generated group and let S be a finite symmetric generating subset of G. The following conditions are equivalent: (a) the group G is sofic; (b) for all .ε > 0 and .n ∈ N, there exists a finite S-labeled graph .Q = (Q, E) such that there is a proportion at least .1 − ε of vertices .q ∈ Q for which there exists an S-labeled graph isomorphism .BS (n) → BQ (q, n) sending .1G to q.

7.1.7 Surjunctivity of Sofic Groups A proof of the following statement may be found in [CAG, Theorem 7.8.1]. Theorem (Gromov-Weiss Theorem) Every sofic group is surjunctive.

7.2 Exercises Exercise 7.1 Let G be a group and let C be a class of groups. Show that the following four conditions are equivalent: (i) G is locally embeddable into C ; (ii) for every finite subset K ⊂ G, there exist a group C ∈ C and an injective map ϕ : K → C that satisfies ϕ(k1 k2 ) = ϕ(k1 )ϕ(k2 ) for all k1 , k2 ∈ K such that k1 k2 ∈ K; (iii) for every finite subset K ⊂ G, there exist a group C ∈ C and an injective map ϕ : K ∪ K 2 → C such that ϕ(k1 k2 ) = ϕ(k1 )ϕ(k2 ) for all k1 , k2 ∈ K;

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7 Local Embeddability and Sofic Groups

(iv) for every finite subset K ⊂ G, there exist a group C ∈ C and a map ϕ : K ∪ K 2 → C whose restriction to K is injective and that satisfies ϕ(k1 k2 ) = ϕ(k1 )ϕ(k2 ) for all k1 , k2 ∈ K. Solution Let K ⊂ G be a finite subset. If G is locally embeddable into C then there exists C ∈ C and a K-almosthomomorphism ψ : G → C. Then the restriction map ϕ := ψ|K : K → C is injective and satisfies ϕ(k1 k2 ) = ϕ(k1 )ϕ(k2 ) for all k1 , k2 ∈ K such that k1 k2 ∈ K. This shows that (i) implies (ii). If (ii) is satisfied then there exist C ∈ C and an injective map ϕ : K ∪ K 2 → C such that ϕ(k1 k2 ) = ϕ(k1 )ϕ(k2 ) for all k1 , k2 ∈ K ∪ K 2 such that k1 k2 ∈ K ∪ K 2 . As K 2 ⊂ K ∪ K 2 , it follows that ϕ(k1 k2 ) = ϕ(k1 )ϕ(k2 ) for all k1 , k2 ∈ K. This shows that (ii) implies (iii). The implication (iii) =⇒ (iv) is obvious. If we arbitrarily extend to G any ϕ : K ∪ K 2 → C as in (iv), we obtain a Kalmost-homomorphism ϕ : G → C. This shows that (iv) implies (i) and completes the proof of the equivalence of the four conditions. ■ Exercise 7.2 Let G and C be two groups. Let K be a finite subset of G such that 1G ∈ K and let ϕ : G → C be a K-almost-homomorphism. (a) Show that ϕ(1G ) = 1C . (b) Suppose in addition that K is symmetric. Show that ϕ(k −1 ) = ϕ(k)−1 for all k ∈ K. Solution (a) Since 1G ∈ K, we have ϕ(1G ) = ϕ(1G · 1G ) = ϕ(1G )ϕ(1G ) by (K-AH-1). This implies ϕ(1G ) = 1C . (b) Let k ∈ K. Since K is symmetric, we also have k −1 ∈ K. Using (a) and (KAH-1), we get 1C = ϕ(1G ) = ϕ(k · k −1 ) = ϕ(k)ϕ(k −1 ). This gives us ϕ(k −1 ) = ϕ(k)−1 . ■ Exercise 7.3 Let G and C be two groups. Suppose that K is a finite symmetric subset of G such that 1G ∈ K. Show that a map ϕ : G → C is a K-almosthomomorphism if and only if it satisfies (i) ϕ(k1 k2 ) = ϕ(k1 )ϕ(k2 ) for all k1 , k2 ∈ K and (ii) ϕ(k) /= 1C for all k ∈ K 2 \ {1G }. Solution We start by observing that condition (i) is nothing but condition (K-AH-1) in the definition of a K-almost-homomorphism. Thus, we only need to check that for a map ϕ satisfying (K-AH-1), conditions (K-AH-2) and (ii) are equivalent. Let then ϕ : G → C be a map satisfying condition (K-AH-1). Suppose first that ϕ satisfies (K-AH-2), i.e., it is a K-almost-homomorphism. Let k ∈ K 2 such that ϕ(k) = 1C . Choose k1 , k2 ∈ K such that k = k1 k2 . We then have ϕ(k) = ϕ(k1 k2 ) = ϕ(k1 )ϕ(k2 ) by (K-AH-1). Using Exercise 7.2(b), we deduce that ϕ(k1 ) = ϕ(k2 )−1 = ϕ(k2−1 ). As k1 , k2−1 ∈ K and the restriction of ϕ to K is injective by (K-AH-2), this implies that k1 = k2−1 so that k = k1 k2 = 1G . This shows that ϕ satisfies (ii).

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Conversely, suppose that ϕ satisfies conditions (K-AH-1) and (ii). Let k1 , k2 ∈ K such that ϕ(k1 ) = ϕ(k2 ). Since K is symmetric, we have k2−1 ∈ K. By using (K-AH1) and Exercise 7.2(b), we then get ϕ(k1 k2−1 ) = ϕ(k1 )ϕ(k2−1 ) = ϕ(k2 )ϕ(k2 )−1 = 1C . Since k1 k2−1 ∈ K 2 , this implies k1 k2−1 = 1G by (ii). Therefore k1 = k2 . This shows that the restriction of ϕ to K is injective, i.e., ϕ satisfies (K-AH-2). ■ Exercise 7.4 Let G be a group. Show that the following conditions are equivalent: (i) (ii) (iii) (iv) (v)

G is abelian; G is residually abelian; G is locally residually finite abelian; G is locally embeddable into the class of finite abelian groups; G is locally embeddable into the class of abelian groups.

Solution The implication (i) =⇒ (ii) is obvious (cf. [CAG, Example 7.1.4.(a)]). Suppose (ii) and let H be a finitely generated subgroup of G. Let h ∈ H with h /= 1G . As G is residually abelian, there exist an abelian group A and a group homomorphism φ : G → A such that φ(h) /= 1A . The group φ(H ) is finitely generated abelian and therefore isomorphic to a finite direct product of cyclic groups. As every cyclic group is residually finite abelian and the class of residually finite abelian groups is closed under finite direct products, we deduce that φ(H ) is residually finite abelian. Consequently, there exist a finite abelian group F and a group homomorphism ψ : φ(H ) → F such that ψ(1A ) /= 1F . The composite group homomorphism ϕ := ψ ◦ φ : H → F satisfies ϕ(h) /= 1F . It follows that H is residually finite abelian. Therefore G is locally residually finite abelian. This shows that (ii) =⇒ (iii). As the class of finite abelian groups is closed under finite direct products, the implication (iii) =⇒ (iv) follows from [CAG, Corollary 7.1.14]. The implication (iv) =⇒ (v) is obvious. Thus, to conclude it is enough to prove that (v) =⇒ (i). Let G be a group which is locally embeddable into the class of abelian groups. Let g, h ∈ G and consider the finite set K := {g, h, gh, hg} ⊂ G. By our assumptions on G, there exists an abelian group C and a K-almost-homomorphism ϕ : G → C. We have ϕ(gh) = ϕ(g)ϕ(h) = ϕ(h)ϕ(g) = ϕ(hg), by (K-AH-1) and commutativity of C. From (K-AH-2), we deduce that gh = hg. Since g and h were arbitrary, this shows that G is abelian. ■ Exercise 7.5 Show that every group which is locally embeddable into the class of metabelian groups is itself metabelian. Solution Let G be a group which is locally embeddable into the class of metabelian groups. In order to show that G is metabelian, it suffices to show that the commutator of any two elements of G is central in G. Let g, h, k ∈ G and consider the finite set K := {1G , g, g −1 , h, h−1 , k, k −1 }10 ⊂ G. By our assumptions on G, there exists a metabelian group C and a K-almost-homomorphism ϕ : G → C. After observing that [g, h]k, [g, h]−1 k −1 , [g, h], g, h, and k are all in K, and using repeatedly (K-

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AH-1) and Exercise 7.2(b), we obtain ϕ([[g, h], k]) = ϕ([g, h]k[g, h]−1 k −1 ) = ϕ([g, h]k)ϕ([g, h]−1 k −1 )

.

= ϕ([g, h])ϕ(k)ϕ([g, h]−1 )ϕ(k −1 ) = ϕ([g, h])ϕ(k)ϕ([g, h])−1 ϕ(k)−1 = [ϕ(g), ϕ(h)]ϕ(k)[ϕ(g), ϕ(h)]−1 ϕ(k)−1 = [[ϕ(g), ϕ(h)], ϕ(k)] = 1C , where the last equality follows from the fact that C is metabelian. Since ϕ([[g, h], k]) = 1C = ϕ(1G ) (cf. Exercise 7.2(a)) and 1G , [[g, h], k] ∈ K, we deduce from (K-AH-2) that [[g, h], k] = 1G . As g, h, and k were arbitrary, this shows that G is metabelian. ■ Comment If we fix an integer d ≥ 0, a similar argument shows that every group which is locally embeddable into the class of solvable (resp. nilpotent) groups of solvability (resp. nilpotency) degree at most d is itself solvable (resp. nilpotent) of solvability (resp. nilpotency) degree at most d. Exercise 7.6 Show that every group which is locally embeddable into the class of torsion-free groups is itself torsion-free. Solution Let G be a group which is locally embeddable into the class of torsionfree groups. Suppose that g ∈ G has finite order. Denote by K the subgroup of G generated by g. Since K is finite and G is locally embeddable into the class of torsion-free groups, there exist a torsion-free group C and a K-almosthomomorphism ϕ : G → C. The restriction of ϕ to K yields a group embedding of K into C. As C is torsion-free, we deduce that K = {1G }. This shows that G is torsion-free. ■ Exercise 7.7 Show that a group which is locally embeddable into the class of solvable (resp. nilpotent) groups may fail to be solvable (resp. nilpotent). Solution Let us denote by C the class of solvable (resp. nilpotent) groups. Choose, for each d ∈ N, a solvable (resp. nilpotent)⊕group Hd of solvability (resp. nilpotency) degree d. Consider the group G := d∈N Hd . The class C is closed under finite direct products by Exercise 4.27 (resp. Exercise 4.28). As each Hd is in C and therefore locally embeddable into C , we deduce that G is itself locally embeddable into C by applying [CAG, Corollary 7.1.11]. However, the group G is not in C . Otherwise, denoting by d0 the solvability (resp. nilpotency) degree of G, we would have d ≤ d0 for all d ∈ N by Exercise 4.14 since Hd is a subgroup of G. ■ Exercise 7.8 Let C denote the class of finite cyclic groups. Show that the group G := Z/2Z × Z/2Z is residually C but not locally embeddable into C . Solution Let g ∈ G such that g /= 1G . Choose an element h ∈ G \ {1G , g}. Then H := {1G , h} is a normal subgroup of G with G/H cyclic of order 2. The canonical

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group epimorphism G → G/H sends g to the non-identity element of G/H . This shows that G is residually C . Suppose, by contradiction, that G is locally embeddable into C . Taking K := G, this implies that there exists a cyclic group C and a K-almost-homomorphism ϕ : G → C. As K = G, the map ϕ is an injective group homomorphism. This is impossible since every subgroup of a cyclic group is itself cyclic. ■ Comment This shows in particular that Corollary 7.1.14 in [CAG] becomes false if we remove the hypothesis that C is closed under taking finite direct products. Exercise 7.9 Let G be a group. Denote by (D i (G))i∈N (resp. (C i (G))i∈N ) the derived series (resp. the lower central series) of G. Show ∩ thati G is residually solvable (resp. residually nilpotent) if and only if i∈N D (G) = {1G } ∩ (resp. i∈N C i (G) = {1G }). ∩ ∩ Solution Suppose first that i∈N D i (G) = {1G } (resp. i∈N C i (G) = {1G }). Let g ∈ G such that g /= 1G . By our hypothesis, there exists i0 ∈ N such that g ∈ / / C i0 (G)). As the group G/D i0 (G) (resp. G/C i0 (G)) is solvable D i0 (G) (resp. g ∈ (resp. nilpotent) by Exercise 4.17(a) and g is sent to a non-identity element under the canonical group epimorphism G → G/D i0 (G) (resp. G → G/C i0 (G)), we deduce that G is residually solvable (resp. residually nilpotent). Conversely, suppose that G is residually solvable (resp. residually nilpotent). Let g ∈ G such that g /= 1G . Then there exists a solvable (resp. nilpotent) group H and a group homomorphism φ : G → H such that φ(g) /= 1H . Let d denote the solvability (resp. nilpotency) degree of H . By Exercise 4.13(c), we (G)) = {1H }). We deduce that / D d (G) have φ(D d (G)) = {1H } (resp. φ(C d∩ ∩ g ∈ d i (resp. g ∈ / C (G)). It follows that i∈N D (G) = {1G } (resp. i∈N C i (G) = {1G }). ■ Exercise 7.10 (Free Groups Are Residually p) Let p be a prime. One says that a group G is a residually p-group if G is residually-C for C the class of p-groups (finite groups whose cardinality is a power of p). (a) Show that every free group is a residually p-group. (b) Show that every free group is residually finite-nilpotent. Solution (a) Let F be a free group and let X ⊂ F be a base for F . Let f ∈ F such that f /= 1F . By [CAG, Corollary D.3.2], we can write f in the form f = x1m1 x2m2 · · · xnmn ,

.

where n ≥ 1, xi ∈ X and mi ∈ Z \ {0} for 1 ≤ i ≤ n, and xi /= xi+1 for 1 ≤ i ≤ n − 1. Choose an integer k ≥ 1 such that pk does not divide the product m1 m2 · · · mn and write S := {x1 , x2 , . . . , xn }. Consider now the ring R := Z/pk Z and the subgroup T ⊂ GLn+1 (R) consisting of all upper triangular matrices with only 1

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on the diagonal. The group T is finite with cardinality pn(n+1)k/2 . Therefore T is a p-group. For 1 ≤ u < v ≤ n + 1, we denote by Eu,v the elementary matrix in Matn+1 (R) with 1 at (u, v) and all other entries 0. Note that Eu,v Eu' ,v ' = Eu,w' if v = u' and 0 otherwise. Denoting by In+1 the identity matrix in GLn+1 (R), the matrix In+1 + Euv is in T for all 1 ≤ u < v ≤ n + 1. Define, for each s ∈ S, the matrix As ∈ T by Ps :=

Π

.

(In+1 + Ei,i+1 ).

(7.2)

i:xi =s

Observe that the matrices In+1 + Ei,i+1 appearing in the right-hand side of (7.2) commute since xi /= xi+1 for all i. Moreover, we have Ps = In+1 +

∑

Ei,i+1 .

.

i:xi =s

for all s ∈ S. By the universal property of free groups, there exists a homomorphism φ : F → T such that φ(s) = Ps for all s ∈ S. We then have φ(f ) = Pxm1 1 Pxm2 2 · · · Pxmn n ⎛ ⎞m1 ⎛ ⎞m2 ∑ ∑ = ⎝In+1 + Ei,i+1 ⎠ ⎝In+1 + Ei,i+1 ⎠

.

i:xi =x1

⎛

· · · ⎝In+1 +

∑

⎞mn Ei,i+1 ⎠

i:xi =xn

⎛ = ⎝In+1 + m1

∑

⎞⎛

· · · ⎝In+1 + mn

∑

Ei,i+1 ⎠ ⎝In+1 + m2

i:xi =x1

⎛

i:xi =x2

⎞

∑

⎞ Ei,i+1 ⎠

i:xi =x2

Ei,i+1 ⎠ .

i:xi =xn

After developing, we get φ(f ) = In+1 +

∑

.

αu,v Eu,v ,

1≤u 0 and a '' /= 1. Since f (Fix(g)) = Fix(g), either Fix(f ) = {0, ∞} or f exchanges 0 and ∞. If Fix(f ) = {0, ∞}, then f must be of the form z |→ az for some a ∈ R with a > 0 and a /= 1. If f would exchange 0 and ∞, then f would be of the form z |→ −b/z for some b ∈ R with b > 0. This is impossible since the condition f g = gf would imply −b/a ' = −a ' b, which is impossible since a '' > 0 and a '' /= 1. Thus, f and h also commute in this case. It remains to treat the case when Fix(g) = {α, α}, where α ∈ C \ R. Then Fix(f ) = Fix(g) = Fix(h). As PSL2 (R) acts transitively on the upper half-plane {z ∈ C : Im(z) > 0} ⊂ P 1 (C), we can assume α = i. The elements of PSL2 (R)

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fixing i are represented by matrices in .

SO2 (R) =

⎧(  ) a −b : a, b ∈ R and a 2 + b2 = 1 ⊂ SL2 (R). b a

Since the rotation group SO2 (R) is abelian, we deduce that f h = hf . This shows that PSL2 (R) is commutative-transitive. (f) By [CAG, Lemma 2.3.2], the matrices (

12 .a := 01

)

(

and

10 b := 21

)

generate a free subgroup H of rank 2 in SL2 (R). The matrix −I2 is not in H since the center of a free group of rank 2 is trivial by Exercise 7.32 (we could also use the fact that free groups are torsion-free by Exercise 6.18(b)). Therefore, the image of H in PSL2 (R) is isomorphic to H . As PSL2 (R) is commutative-transitive by (e), we deduce that H is commutative-transitive. This implies that every free group of finite rank is itself commutative-transitive. Indeed, every free group of finite rank can be embedded into a free group of rank 2 by [CAG, Corollary D.5.3]. Now, let F be an arbitrary free group and let X ⊂ F be a base for F . Let f, g, h ∈ F \ {1F } such that fg = gf and gh = hg. Let Y be the set of elements of X appearing in the X-reduced form of f , g, or h. As Y is finite, it generates a free subgroup of finite rank. As every free group of finite rank is commutative-transitive, we deduce that f h = hf . This shows that F is commutative-transitive. (g) If K has characteristic 3, then K contains a subfield isomorphic to Z/3Z so that PSL2 (K) contains a subgroup isomorphic to PSL2 (Z/3Z). As + PSL2 (Z/3Z) is isomorphic to the alternating group Sym+ 4 , and Sym4 is not CSA by Exercise 7.30(c), we deduce that PSL2 (K) is not CSA. Suppose now that the characteristic of K is not equal to 3. Let ρ : SL2 (K) → PSL2 (K) denote the canonical group homomorphism. Consider the matrices A, B ∈ SL2 (K) given by (

1 −1 .A := −1 2

)

(

and

) 0 −1 B := , 1 0

and their images g := ρ(A) and h := ρ(B) in G. We have BAB −1 =

.

( ) 21 = A−1 . 11

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7 Local Embeddability and Sofic Groups

Therefore hgh−1 = ρ(BAB −1 ) = ρ(A−1 ) = g −1 commutes with g. However, we have ( ) ( ) −1 −1 1 −2 .AB = and BA = , 2 1 1 −1 so that AB /= BA and AB /= −BA by our hypothesis on the characteristic of K. Therefore gh = ρ(AB) /= ρ(BA) = hg. It follows that condition (CSA-2) in Exercise 7.30(b) is not satisfied. This shows that PSL2 (K) is not CSA whatever the field K. ■ Comment The fields K for which the group PSL2 (K) is commutative-transitive are characterized in [FinGRS, Theorem 3.4]. Exercise 7.34 Show that every non-abelian commutative-transitive group is indecomposable in the sense of Exercise 6.16. Solution Let G be a commutative-transitive group. Suppose that there exist nontrivial groups H, K and a group isomorphism ϕ : G → H × K. For all h ∈ H and k ∈ K, the elements (h, 1K ) and (1H , k) commute. As H × K is commutativetransitive, we deduce that the groups H and K are both abelian. This shows that G is abelian. ■ Exercise 7.35 Show that the projective modular group PSL2 (Z) is indecomposable. Solution The group PSL2 (Z) is naturally isomorphic to a subgroup of PSL2 (R). As PSL2 (R) is commutative-transitive by Exercise 7.33(e), it follows from Exercise 7.25(c) and Exercise 7.34 that PSL2 (Z) is indecomposable. ■ Exercise 7.36 Show that a group which is locally embeddable into the class of commutative-transitive (resp. CSA) groups is itself commutative-transitive (resp. CSA). Solution Let G be a group which is locally embeddable into the class of commutative-transitive groups. Suppose that g, h, k ∈ G \ {1G } satisfy gh = hg and hk = kh. Setting K := {1G , g, h, k, gh, hg, gk, kg, hk, kh} ⊂ G, we can find a commutative-transitive group C and a K-almost-homomorphism ϕ : G → C. By (K-AH-1), we have ϕ(g)ϕ(h) = ϕ(gh) = ϕ(hg) = ϕ(h)ϕ(g) and ϕ(h)ϕ(k) = ϕ(hk) = ϕ(kh) = ϕ(k)ϕ(h). Therefore ϕ(g)ϕ(h) = ϕ(h)ϕ(g) and ϕ(h)ϕ(k) = ϕ(k)ϕ(h). On the other hand, it follows from Exercise 7.2(a) that ϕ(1G ) = 1C . Since the restriction of ϕ to K is injective by (K-AH-2), we deduce that ϕ(g), ϕ(h), ϕ(k) ∈ C \ {1C }. As C is commutative-transitive, it follows that ϕ(g)ϕ(k) = ϕ(k)ϕ(g). Using again (K-AH-1), we obtain ϕ(gk) = ϕ(g)ϕ(k) = ϕ(k)ϕ(g) = ϕ(kg). As the restriction of ϕ to K is injective, this implies gk = kg. This shows that G is commutative-transitive. The proof that every group which is locally embeddable into the class of CSAgroups is itself CSA is similar after taking K := {1G , g, h, h−1 , k, hk, kh, gk,

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kg, gh, gh−1 , hg, hgh−1 , ghg, gh−1 g, ghgh−1 , hgh−1 g} with g, h, k ∈ G \ {1G } and using the characterisation of CSA-groups given in Exercise 7.30(d). ■ Exercise 7.37 Let C be a class of groups. (a) Show that every group in C is residually C . (b) Let C ' denote the class of residually C groups and let C '' denote the class of residually C ' groups. Show that C '' = C ' . (c) Show that the class of residually C groups is closed under taking direct products. Solution (a) If G ∈ C and g ∈ G\{1G }, then the identity map IdG : G → G is a surjective group homomorphism such that IdG (g) = g /= 1G . Therefore G is residually C . (b) The fact that C ' ⊂ C '' follows from (a). For the reverse inclusion, suppose that G ∈ C '' and let g ∈ G \ {1G }. Then we can find a group C ' ∈ C ' and a surjective group homomorphism φ ' : G → C ' such that c' := φ ' (g) /= 1C ' . As C ' is residually C , we can find a group C ∈ C and a surjective group homomorphism ψ : C ' → C such that ψ(c' ) /= 1C . Then the composite map φ := ψ ◦ φ ' : G → C is a surjective group homomorphism and satisfies φ(g) = ψ(φ ' (g)) = ψ(c' ) /= 1C . This shows that G ∈ C ' . Π (c) Let (Gi )i∈I be a family of residually C groups and set G := i∈I Gi . Let g = (gi )i∈I ∈ G \ {1G }. Then there exists i0 ∈ I such that gi0 /= 1Gi0 . As Gi0 is residually C , there exist a group C ∈ C and a surjective group homomorphism φ : Gi0 → C such that φ(gi0 ) /= 1C . Denoting by π : G → Gi0 the projection homomorphism, we have that the composite map φ ◦ π : G → C is a surjective group homomorphism satisfying (φ ◦ π )(g) = φ(gi0 ) /= 1C . This shows that G is residually C . ■ Exercise 7.38 Let C be a class of groups which is closed under taking subgroups. (a) Show that a group G is residually C if and only if it satisfies the following condition: for every g ∈ G \ {1G }, there exist a group C ∈ C and a group homomorphism φ : G → C such that φ(g) /= 1C . (b) Show that the class of residually C groups is itself closed under taking subgroups. (c) Show that the direct sum of any family of residually C groups is residually C. (d) Show that the projective limit of any projective system of residually C groups is residually C . (e) Show that a group G is residually C if and only if there exists a family (C i Π )i∈I of groups in C such that G is isomorphic to a subgroup of the direct product i∈I Ci . Solution (a) The condition is clearly necessary since it is just the definition of a residually C group with the surjectivity requirement on φ omitted.

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To prove sufficiency, suppose that for every g ∈ G \ {1G }, there exist a group C ∈ C and a group homomorphism φ : G → C such that φ(g) /= 1C . Observe that the group C ' := φ(C) is in C since C is closed under taking subgroups, and that the map φ ' : G → C ' induced by φ is a surjective group homomorphism such that φ ' (g) = φ(g) /= 1C = 1C ' . This shows that G is residually C . (b) Suppose that H is a subgroup of a residually C group G. Let h ∈ H \ {1H } ⊂ G \ {1G }. Since G is residually C , there exist a group C ∈ C and a surjective group homomorphism φ : G → C such that φ(h) /= 1C . The restriction map φ|H : H → C is a group homomorphism satisfying φ|H (h) = φ(h) /= 1C . Therefore H is residually C by (a). ⊕ (c) Let (Gi )i∈I be a family of residually Then the direct sum i∈I Gi Π C groups. Π is a subgroup of the direct product i∈I Gi . As i∈I Gi is residually C by ⊕ Exercise 7.37(c), we deduce that i∈I Gi is itself residually C by applying (b). (d) The proof is as in (c)Π since the projective limit of a projective system (Gi )i∈I of groups is a subgroup of i∈I Gi . (e) Let G be a group. Suppose first that G is residually C . Then, for each g ∈ G \ {1G }, there exist a group Cg ∈ C and a group homomorphism φg : G → Π := Cg such that φ (g) = / 1 . Consider the group H C . As the map g C g g g∈G\{1G } Π φ := g∈G\{1G } φg : G → H is an injective group homomorphism, the group G is isomorphic to a subgroup of H . This shows necessity. The converse implication follows from (b) and Exercise 7.37(c) since every group in C is residually C . ■ Exercise 7.39 (Fully Residually C Groups) Let C be a class of groups. A group G is called fully residually C if for any finite subset K ⊂ G, there exist a group C ∈ C and a surjective group homomorphism φ : G → C whose restriction to K is injective. (a) Show that every group in C is fully residually C . (b) Show that if C is closed under taking subgroups then every subgroup of a fully residually C group is itself fully residually C . (c) Show that every fully residually C group is locally embeddable into C . (d) Show that every fully residually C group is residually C . (e) Show that if C is closed under taking finite direct products then every residually C group is fully residually C . (f) Show that the group Z/2Z × Z/2Z is residually cyclic but not fully residually cyclic. Solution (a) If G ∈ C and K ⊂ G is a finite set then the identity map IdG : G → G is a surjective group homomorphism whose restriction to K is injective. Therefore G is fully residually C . (b) Suppose that C is closed under taking subgroups and let H be a subgroup of a fully residually C group G. Let K ⊂ H be a finite subset. Since G is fully residually C , there exist a group C ∈ C and a surjective group homomorphism φ : G → C whose restriction to K is injective. Then C ' := φ(H ) ∈ C and φ induces

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by restriction a surjective group homomorphism φ ' : H → C ' whose restriction to K is injective. This shows that H is fully residually C . (c) This is obvious from the definitions. (d) Let G be a fully residually C group and let g ∈ G such that g /= 1G . By taking K := {1G , g}, there exist a group C ∈ C and a surjective group homomorphism φ : G → C whose restriction to K is injective. We then have φ(g) /= φ(1G ) = 1C . This shows that G is residually C . (e) Suppose that C is closed under taking finite direct products and let G be a residually C group. Let K ⊂ G be a finite subset. Consider the finite subset Ω ⊂ G defined by Ω := {gh−1 : g, h ∈ K and g /= h}. Since G is residually C , for group homomorphism every ω ∈ Ω, there exist a group Cω ∈ C and a surjective Π φω : G → Cω such that φ(ω) /= 1Cω . The group C := ω∈Ω Cω is in C since C is closed under finite Π direct products. Moreover, the group homomorphism φ : G → C given by φ := ω∈Ω φω is injective on K. This shows that G is fully residually C . (f) This immediately follows from (c) and Exercise 7.8. ■ Exercise 7.40 Let C be a class of groups. Show that a group G is fully residually C if and only if it satisfies the following condition: for every finite subset K ⊂ G \ {1G }, there exist a group C ∈ C and a surjective group homomorphism φ : G → C such that φ(k) /= 1C for all k ∈ K. Solution Suppose first that the group G is fully residually C and let K ⊂ G \ {1G } be a finite subset. Set K ' := K ∪ {1G }. Then there exist a group C ∈ C and a surjective group homomorphism φ : G → C which is injective on K ' . As 1G ∈ K ' and φ(1G ) = 1C , we have φ(k) /= 1C for all k ∈ K. This shows necessity. Conversely, suppose that the condition is satisfied by some group G and let K ⊂ G be a finite subset. Set K ' := KK −1 \ {1G }. Then there exist C ∈ C and a surjective group homomorphism φ : G → C such that φ(k ' ) /= 1C for all k ' ∈ K ' . If k1 and k2 are distinct elements in K, then k1 k2−1 ∈ K ' so that φ(k1 k2−1 ) /= 1C . As φ is a group homomorphism, this implies φ(k1 ) /= φ(k2 ). Thus, the restriction of φ to K is injective. This shows that G is fully residually C . ■ Exercise 7.41 Show that every fully residually surjunctive group is surjunctive. Solution This is exactly Lemma 3.3.4 in [CAG].

■

Comment Recall that no example of a non-surjunctive group has been found up to now. It is even unknown whether or not the class of surjunctive groups is closed under taking finite direct products (see [ArzG]). An affirmative answer to this last question would imply that every residually surjunctive group is fully residually surjunctive and hence surjunctive. Exercise 7.42 (Free Groups Are CSA) Given a group G and an element g ∈ G, we denote by the subgroup of G generated by g and by CG (g) := {h ∈ G : gh = hg} the centralizer of g in G. We say that g is a proper power provided there exist g0 ∈ G and an integer n ≥ 2 such that g = g0n . Let F be a non-trivial free group. (a) Show that every non-trivial abelian subgroup of F is infinite cyclic.

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(b) Let g ∈ F \ {1F }. Show that the subgroups and CF (g) are infinite cyclic and that one has ⊂ CF (g) with equality if and only if g is not a proper power. (c) Let g ∈ F \ {1F }. Show that there exists a unique element g0 ∈ G which is not a proper power and a unique integer n ≥ 1 such that g = g0n . (d) Let A be a subgroup of F . Show that the following conditions are equivalent: (1) A is a maximal abelian subgroup of F ; (2) there exists g ∈ F which is not a proper power such that A = ; (3) there exists g ∈ F \ {1F } such that A = CF (g). (e) Show that if A ⊂ F is a maximal abelian subgroup then A = CF (g) for all g ∈ A \ {1F }. (f) Show that every maximal abelian subgroup of F is malnormal. (g) Deduce that all free groups are CSA. Solution (a) Every subgroup of F is free by Exercise 6.28. As every free group of rank greater than 1 is non-abelian, we deduce that any non-trivial abelian subgroup of F is infinite cyclic (cf. Exercise 6.18(b)). (b) The subgroup is non-trivial abelian and hence infinite cyclic by (a). We have ⊂ CF (g) since g ∈ CF (g). On the other hand, CF (g) is abelian since F is commutative transitive by Exercise 7.33(e). It follows that the subgroup CF (g) is also infinite cyclic. Suppose now that g is a proper power. This means that g = g0n for some g0 ∈ G / , this implies that  CF (g). and n ≥ 2. As g0 ∈ CF (g) but g0 ∈ Conversely, suppose  CF (g). We can write g = g0n , where n ≥ 1 and g0 is one of the two generators of CF (g). We must have n ≥ 2 since otherwise we would have g = g0 and hence = = CF (g), a contradiction. Therefore g is a proper power. (c) We can write g = g0n , where n ≥ 1 and g0 is one of the two generators of CF (g). As CF (g) = CF (g0 ) by the commutative-transitivity of F , we have = CF (g0 ) so that g0 is not a proper power by (a). This shows existence. To prove uniqueness, suppose that g = g1n1 = g2n2 , where g1 , g2 ∈ G are not proper powers and n1 , n2 ≥ 1. Then = CF (g1 ) and = CF (g2 ) by (b). Since g1 , g2 ∈ CF (g), we have CF (g1 ) = CF (g2 ) = CF (g) by commutative-transitivity of F . Thus, = , so that n1 = n2 and g1 = g2 . This proves uniqueness. (d) Let A be a maximal abelian subgroup of F . By (a), there exists g ∈ F \ {1F } such that A = . As ⊂ CF (g) and CF (g) is cyclic and therefore abelian, we have = CF (g) by maximality of A. This implies that g is not a proper power by (b). Thus, (1) implies (2). If g is not a proper power, then = CF (g) by (b). Therefore, (2) implies (3). Finally, the implication (3) =⇒ (1) follows from Exercise 7.24(b) since CF (g) is abelian for every g ∈ F \ {1F }. (d) Let A be a maximal abelian subgroup of F . By (c), there exists g0 ∈ F which is not a proper power such that A = . Let now g ∈ A \ {1F }. Then there exists

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n ∈ Z \ {0} such that g = g0n . We then have A = CF (g0 ) ⊂ CF (g0n ) = CF (g). As CF (g) is abelian and A is maximal abelian, we have A = CF (g). (e) Let A ⊂ F be a maximal abelian subgroup. By (c), there exists g0 ∈ F which is not a proper power such that A = . Let g ∈ F \ A. Note that the conjugate subgroup gAg −1 = ⊂ F is also maximal abelian. Suppose that there exists h ∈ gAg −1 ∩ A such that h /= 1F . It then follows from (d) that A = CF (h) = gAg −1 . As a consequence, gg0 g −1 also generates A, so that either gg0 g −1 = g0 or gg0 g −1 = g0−1 . In the first case, we have gg0 = g0 g and hence g ∈ CF (g0 ) = A, a contradiction. In the second case, g 2 g0 g −2 = g(gg0 g −1 )g −1 = gg0−1 g −1 = (gg0 g −1 )−1 = (g0−1 )−1 = g0 , that is, g 2 g0 = g0 g 2 . By commutative transitivity, gg0 = g0 g, so that, as before, g ∈ CF (g0 ) = A, a contradiction. We deduce that gAg −1 ∩ A = {1F }. This shows that A is malnormal in F . (f) This immediately follows from (e) and the definition of a CSA-group. ■ Exercise 7.43 (Residually Free Groups) (a) Show that the class of residually free groups is closed under taking subgroups, direct products, direct sums, and projective limits. (b) Show that a group G is residually free if and only if there exists a family (F i Π )i∈I of free groups such that G is isomorphic to a subgroup of the direct product i∈I Fi . (c) Show that an abelian group G is residually free if and only if there exists a cyclic groups such that G is isomorphic to a subgroup of family (Ci )i∈I of infinite Π the direct product i∈I Ci . (d) Show that every residually free group is torsion-free. (e) Show that every residually free group is residually finite. (f) Let G be a residually free group and let g, h ∈ G \ {1G }. Show that the subgroup H := ⊂ G generated by g and h is either free of rank 2 or isomorphic to Z or Z2 . (g) Let G be a residually free group and let A be a normal abelian subgroup of G. Show that A is contained in the center Z(G) of G. Solution (a) The fact that the class of residually free groups is closed under taking direct products follows from Exercise 7.37(c). The remaining facts follow from Exercise 7.38 since the class of free groups is closed under taking subgroups by Exercise 6.28. (b) This follows from Exercise 7.38(d). (c) The condition is sufficient by (b) since every infinite cyclic group is free. Conversely, suppose that G is an abelian residually free group. For each g ∈ G \ {1G }, there exist a free group Fg and a group homomorphism φg : G → Fg such that φg (g) /= 1Fg . Since G is abelian, so is the subgroup Cg := φg (G) ⊂ Fg . Note that Cg is non-trivial since 1Cg = 1Fg /= φg (g) ∈ Cg for all g ∈ G\{1G }. As all nontrivial abelian subgroups of a free group are infinite cyclic (cf. Exercise 7.42(a)), we deduce that Cg is infinite cyclic for all g ∈ G \ {1G }. Consider the group H :=

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Π As the map φ := g∈G\{1G } φg : G → H is an injective group homomorphism, the group G is isomorphic to a subgroup of H . (d) This immediately follows from (b) since all free groups are torsion-free (cf. Exercise 6.18(b)) and the class of torsion-free groups is closed under taking subgroups and direct products. (e) This follows from Exercise 7.37(b) since every free group is residually finite [CAG, Theorem 2.3.1]. We could also use (b) together with the fact that the class of residually finite groups is closed under taking subgroups and direct products. (f) The subgroup H is non-trivial and generated by two elements. Moreover, it is torsion-free by (d). Consequently, if H is abelian, it must be isomorphic either to Z or Z2 . Suppose now that H is non-abelian, that is, [g, h] /= 1G . Let us show that H is free with base X := {g, h}. Let f : X → K be a map from X into a group K. Since H is residually free by (a), we can find a free group F and a group homomorphism φ : H → F such that φ([g, h]) /= 1F . As [φ(g), φ(h)] = φ([g, h]) /= 1F , we deduce from Exercise 6.28 that the elements φ(g) and φ(h) generate a non-abelian free subgroup F ' ⊂ F based on the set X' := {φ(g), φ(h)} ⊂ F ' . Consequently, the map f ' : X' → K, defined by setting f ' (φ(g)) := f (g) and f ' (φ(h)) := f (h), extends to a group homomorphism ϕ ' : F ' → K. The composite map ϕ := ϕ ' ◦ φ : H → K satisfies ϕ(g) = ϕ ' (φ(g)) = f ' (φ(g)) = f (g) and, similarly, ϕ(h) = f (h). Thus, ϕ is a group homomorphism extending f . This shows that H is a free group of rank 2 based on X. (g) Suppose that a ∈ A \ Z(G). Then there exists g ∈ G such that [a, g] /= 1G . Since G is residually free, there exist a free group F and a surjective group homomorphism φ : G → F such that φ([a, g]) /= 1F . As [φ(a), φ(g)] = φ([a, g]), we deduce that φ(a) and φ(g) do not commute. Moreover, as φ(A) ⊂ F is a normal abelian subgroup, we have [φ(a), φ(g)φ(a)φ(g)−1 ] = 1F . Since F is a CSA-group by Exercise 7.42(g), this contradicts condition (CSA-2) in Exercise 7.30(d). This shows that A ⊂ Z(G). ■ Π

g∈G\{1G } Cg .

Exercise 7.44 (Fully Residually Free Groups) (a) Show that every free group is fully residually free. (b) Show that every fully residually free group is residually free. (c) Show that the class of fully residually free groups is closed under taking subgroups. (d) Show that every fully residually free group is locally embeddable into the class of free groups. (e) Show that every fully residually free group is a CSA-group. (f) Show that if F is a non-abelian free group, then the group F × Z is residually free but not fully residually free. (g) Show that every abelian residually free group is fully residually free. (h) Let G be a residually free group. Show that the following conditions are equivalent: (1) G is fully residually free; (2) G is locally embeddable into the class of free groups; (3) G is a CSA-group;

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(4) G is commutative-transitive; (5) G contains no subgroup isomorphic to F2 × Z, where F2 denotes the free group of rank two. Solution (a) This follows from Exercise 7.39(a). (b) This follows from Exercise 7.39(d). (c) This follows from Exercise 7.39(b), since the class of free groups is closed under taking subgroups (cf. Exercise 6.28). (d) This follows from Exercise 7.39(c). (e) This follows from (d), Exercise 7.42, and Exercise 7.36. (f) As F and Z are both free and therefore residually free, it follows from Exercise 7.43(a) that their direct product F × Z is residually free. After taking G1 := F and G2 := Z in Exercise 7.26, we deduce that F × Z is not commutative transitive and therefore not a CSA-group either (Exercise 7.30(b)). It follows from (e) that F × Z is not fully residually free. (g) Let G be an abelian residually free group. It follows from Exercise 7.43(c) that there exists a family (Ci )i∈I of infinite cyclic groups such that G is isomorphic Π to a subgroup of the direct product i∈I Ci . As the class of fully residually Π free groups is closed under taking subgroups by (c), it suffices to show that i∈I Ci Π is fully residually free. Let K ⊂ i∈I Ci be a finite subset. Then we can find a finite subset IK ⊂ I such that if g = (gi )i∈I , g ' = (gi' )i∈I ∈ K then there exists j = j (g, g ' ) ∈ IK such that gj /= gj' . Up to renaming the elements in IK , we may suppose that IK = {0, 1, . . . , n}. Denote by ti ∈ Ci a generator of Ci for all i ∈ I . Then, given g ∈ K and i ∈ IK , we can write gi = timi for a unique mi = mi (g) ∈ Z. Let p be an integer such that p > 2|mi (g)| for all g ∈ K and i ∈ IK . For each i ∈ IK , consider the group Π homomorphism φi : Ci → Z defined by setting φi (ti ) := pi . Then the map φ : i∈I Ci → Z, defined by φ(g) :=

n ∑

.

i=0

Π

φi (gi ) =

n ∑

mi (g)pi

i=0

for all g ∈ i∈I Ci , is a group homomorphism which is injective on K. To see this, suppose that g = (gi )i∈I and g ' = (gi' )i∈I are both in K. We have φ(g) = φ(g ' ) if ∑ and only if ni=0 (mi (g) − mi (g ' ))pi = 0 which in turn is equivalent to mi (g) = mi (g ' ) for all i = 0, 1, . . . , n, by our choice of p. It follows from our choice of IK ⊂ I that this is the case exactly if g = g ' . As the group Z is free, this shows that Π i∈I Ci is fully residually free. (h) The implications (1) =⇒ (2), (2) =⇒ (3), and (3) =⇒ (4) hold true for any group and follow from (d), from (e), and from Exercise 7.30(b), respectively. The implication (4) =⇒ (5) follows from F2 × Z being not commutative transitive (cf. the proof of (f)) and the fact that the class of commutative transitive groups is closed under taking subgroups (Exercise 7.25(c)). Suppose that G is not commutative transitive and let a, b, c ∈ G\{1G } such that [a, b] = 1G , [b, c] = 1G ,

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but [a, c] /= 1G . As G is residually free and a and c do not commute, it follows from Exercise 7.43(f) that the subgroup of G they generate is free of rank 2. It also follows from Exercise 7.43(d) that the subgroup generated by b is infinite cyclic. We deduce that the subgroup of G generated by a,b, and c is isomorphic to F2 × Z. This shows the implication (5) =⇒ (4). We are only left to show that (4) =⇒ (1). Suppose that G is commutative-transitive and let us shows that G is fully residually free. If Z(G) /= {1G }, then G is abelian by Exercise 7.25(d) and we are done since all abelian residually free groups are fully residually free by (g). Thus, we may assume that Z(G) = {1G }. Let us show, by induction on n = |K|, that the following holds: (∗) if K ⊂ G \ {1G } is a non-empty finite subset, then there exists g ∈ G \ {1G } with the following property: if N is a normal subgroup of G such that g ∈ / N, then k ∈ / N for all k ∈ K. For |k| = 1, (∗) is trivially satisfied by taking as g the unique element in K. Suppose now that (∗) holds for all subsets of G \ {1G } with cardinality n and let K = {k1 , . . . , kn , kn+1 } ⊂ G \ {1G } with |K| = n + 1. By our induction hypothesis, we can find g ∈ G such that, for any normal subgroup N of G, if g ∈ / N then ki ∈ /N for all 1 ≤ i ≤ n. Consider, for each x ∈ G, the commutator c(x) := [g, xkn+1 x −1 ]. If c(x) = 1G for all x ∈ G, that is, every conjugate of kn+1 commutes with g, then by commutative transitivity of G, the normal closure M ⊂ G of the element kn+1 is abelian and therefore trivial by Exercise 7.43(g) and our assumption Z(G) = {1G }, in contradiction with the fact that 1G /= kn+1 ∈ M. We deduce that there exists y ∈ G \ {1G } such that g ' := c(y) /= 1G . Let now N ⊂ G be a normal subgroup −1 −1 such that g ' ∈ / N. As g ' = [g, ykn+1 y −1 ] = g(ykn+1 y −1 g −1 ykn+1 y ), we have g∈ / N and kn+1 ∈ / N. Consequently, ki ∈ / N for all 1 ≤ i ≤ n + 1. This shows that the element g ' ∈ G \ {1G } has the desired properties and completes our inductive proof of (∗). Now, if K ⊂ G \ {1G } is a non-empty finite subset and g ∈ G \ {1G } is as in (∗), there exists, by residual freeness of G, a free group F and a surjective group homomorphism φ : G → N such that φ(g) /= 1F . By taking N := ker(φ) in (∗), we deduce that φ(k) /= 1F for all k ∈ K. This shows that G is fully residually free by Exercise 7.40. ■ Comment The class of fully residually free groups was introduced by Benjamin Baumslag in [Bau1]. Finitely generated fully residually free groups are also called limit groups (see [ChamG] and the references therein). They are the groups that appear as limits of marked free groups. More precisely, a group G is a limit group if and only if there exist a finitely generated free group F and a sequence (Nn ) of normal subgroups of F such that the sequence (F /Nn ) tends to a group isomorphic to G in the space of F -marked groups. Basic examples of limit groups are provided by finitely generated free groups, finitely generated free abelian groups, and fundamental groups of closed 2-dimensional manifolds of Euler characteristic at most −2. The class of limit groups plays an important role in the solutions of the Tarski problem about the equivalence of the elementary theory of finitely

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generated non-abelian free groups given independently by Kharlampovich and Myasnikov [KhaM] and by Sela [Sel2]. A structure theorem for limit groups established by Kharlampovich and Myasnikov and by Sela shows that they can be obtained recursively from the basic examples of limit groups listed above by applying a finite sequence of free products or amalgamations over Z (see also [Guir]). This structure theorem implies in particular that limit groups are finitely presented. Exercise 7.45 Give an example of a finitely generated LEF-group which is neither residually finite nor amenable. Solution Consider the group G := G1 × G2 , where G1 is the finitely generated group described in [CAG, Proposition 2.6.1] and G2 is a free group of rank 2. The group G is finitely generated since it is the direct product of two finitely generated groups. It is not residually finite since it contains a subgroup isomorphic to G1 , and G1 is not residually finite. It is not amenable since it contains a subgroup isomorphic to G2 , and G2 is not amenable. On the other hand, the group G1 is LEF by [CAG, Proposition 7.3.9] and the group G2 is LEF since all free groups are residually finite. As the class of LEF-groups is closed under direct products, we deduce that G is LEF. ■ Exercise 7.46 Let G be a group, let K ⊂ G be a finite subset, let C be a finite group, and let ϕ : G → C be a K-almost-homomorphism. Denote by L : C → Sym(C) the Cayley group homomorphism, that is, the map defined by L(g)(h) := gh for all g, h ∈ C, and set φ := L ◦ ϕ : G → Sym(C). Show that φ is a (K, ε)almost-homomorphism for all ε > 0. Solution Let ε > 0. Denote by dC the Hamming metric on Sym(C). For all k1 , k2 ∈ K, we have ϕ(k1 k2 ) = ϕ(k1 )ϕ(k2 ) by (K-AH-1). This implies φ(k1 k2 ) = L(ϕ(k1 k2 )) = L(ϕ(k1 )ϕ(k2 )) = L(ϕ(k1 ))L(ϕ(k2 )) = φ(k1 )φ(k2 ) and hence dC (φ(k1 k2 ), φ(k1 )φ(k2 )) = 0 ≤ ε. Therefore φ satisfies ((K, ε)-AH-1). On the other hand, if k1 , k2 ∈ K are distinct, we have ϕ(k1 ) /= ϕ(k2 ) by (K-AH-2). This implies φ(k1 )(c) = ϕ(k1 )c /= ϕ(k2 )c = φ(k2 )(c) for all c ∈ C and hence dC (φ(k1 ), φ(k2 )) = 1 ≥ 1 − ε. Therefore φ satisfies ((K, ε)-AH-2). This shows that φ is a (K, ε)-almost-homomorphism. ■ Exercise 7.47 Let G be a group and let K be a finite subset of G. Let F be a non-empty finite set. Show that if 0 < ε < 2/|F | then every (K, ε)-almosthomomorphism ϕ : G → Sym(F ) is a K-almost-homomorphism of G into the group Sym(F ). Solution Let ϕ : G → Sym(F ) be a (K, ε)-almost-homomorphism with 0 < ε < 2/|F | and let us show that ϕ is a K-almost-homomorphism. We can clearly assume that F has more than one element. Denote by dF the Hamming distance on Sym(F ). By ((K, ε)-AH-1), for all k1 , k2 ∈ K, we have dF (ϕ(k1 k2 ), ϕ(k1 )ϕ(k2 )) ≤ ε < 2/|F |. As dF (α1 , α2 ) = |{x ∈ F : α1 (x) /= α2 (x)}|/|F | ≥ 2/|F | for all distinct α1 , α2 ∈ Sym(F ), this implies ϕ(k1 k2 ) = ϕ(k1 )ϕ(k2 ). On the other hand, by ((K, ε)-AH-2), for all k1 , k2 ∈ K such that k1 /= k2 , we have

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dF (ϕ(k1 ), ϕ(k2 )) ≥ 1 − ε > 1 − 2/|F | ≥ 0. This implies dF (ϕ(k1 ), ϕ(k2 )) > 0 and hence ϕ(k1 ) /= ϕ(k2 ). This shows that ϕ is a K-almost-homomorphism. ■ Exercise 7.48 Let G be a group. Show that G is residually finite if and only if the following condition is satisfied: there exist a directed set I and two families (Fi )i∈I and (ϕi )i∈I indexed by I , where, for each i ∈ I , Fi is a non-empty finite subset of G and ϕi : G → Sym(Fi ) is a group homomorphism, such that, for every g ∈ G\{1G }, the Hamming distance between ϕi (g) and IdFi is eventually equal to 1. Solution Suppose first that the condition is satisfied and let g ∈ G \ {1G }. Then there exists i ∈ I such that the Hamming distance between ϕi (g) and IdFi is equal to 1. This implies in particular that ϕi (g) /= IdFi = 1Sym(Fi ) . As the group Sym(Fi ) is finite, this shows that G is residually finite. Conversely, suppose that G is residually finite. Consider the directed set F consisting of all finite subsets Ω ⊂ G ordered by reverse inclusion. Since G is residually finite, for every Ω ∈ F , there exist a finite group FΩ and a group homomorphism φΩ : G → FΩ such that the restriction of φΩ to Ω is injective [CAG, Lemma 2.7.3]. Let γΩ : FΩ → Sym(FΩ ) denote the Cayley group homomorphism and consider the composite group homomorphism ϕΩ := γΩ ◦ φΩ : G → Sym(FΩ ). We claim that the families (FΩ )Ω∈F and (ϕΩ )Ω∈F have the required properties. To see this, let g ∈ G \ {1G }. If Ω ∈ F satisfies {1G , g} ⊂ Ω, then φΩ (g) /= φΩ (1G ) = 1FΩ . As ϕΩ (g) is the permutation of FΩ given by left multiplication by φΩ (g), we deduce that the Hamming distance between ϕΩ (g) and IdFΩ is equal to 1. ■ Exercise 7.49 (The Hilbert-Schmidt Metric) Let H be a complex Hilbert space of finite dimension n ≥ 1. Let L(H ) denote the C-algebra consisting of all linear maps u : H → H . If u ∈ L(H ), we denote by Tr(u) ∈ C the trace of u and by u∗ ∈ L(H ) its adjoint. For u, v ∈ L(H ), we set H S :=

.

1 Tr(uv ∗ ). n

(a) Show that H S is a scalar product on L(H ). Denote by ║ · ║H S the associated norm and by dH S the associated metric. Thus, one has ║u║H S =

.

√ H S

and

dH S (u, v) = ║u − v║H S

for all u, v ∈ L(H ). (b) Let U(H ) := {u ∈ L(H ) : uu∗ = IdH }. Show that U(H ) is a group for the composition of maps. (c) Show that one has / dH S (u, v) =

.

for all u, v ∈ U(H ).

2(1 −

1 Re(Tr(v −1 u))). n

(7.6)

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(d) Show that dH S induces by restriction a bi-invariant metric on U(H ). Solution (a) For all λ ∈ C and u, u1 , u2 , v ∈ L(H ), we have 1 Tr(vu∗ ) n 1 = Tr((vu∗ )∗ ) n 1 = Tr(uv ∗ ) n

H S =

.

= H S , 1 Tr((u1 + u2 )v ∗ ) n 1 = Tr(u1 v ∗ + u2 v ∗ ) n 1 1 = Tr(u1 v ∗ ) + Tr(u2 v ∗ ) n n

H S =

.

= H S + H S , and 1 Tr((λu)v ∗ ) n 1 = Tr(λuv ∗ ) n λ = Tr(uv ∗ ) n

H S =

.

= λH S , showing that H S is a Hermitian form. On the other hand, if the matrix of u ∈ L(H ) with respect to some orthonormal basis of H is A = (aij )1≤i,j ≤n , then H S =

.

∑ 1 1 ¯ = 1 Tr(AA∗ ) = Tr(A tA) |aij |2 . n n n 1≤i,j ≤n

We deduce that H S ≥ 0 with equality if and only if aij = 0 for all 1 ≤ i, j ≤ n, i.e., if and only if u = 0. This shows that the Hermitian form H S is positive definite, i.e., a scalar product on L(H ).

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7 Local Embeddability and Sofic Groups

(b) If u ∈ U(H ), then uu∗ = IdH so that u is surjective. As H is finitedimensional, this implies that u is bijective, so that U(H ) ⊂ GL(H ) and U(H ) = {u ∈ GL(H ) : u−1 = u∗ }. Clearly IdH ∈ U(H ). On the other hand, if u, v ∈ U(H ), then (uv)−1 = v −1 u−1 = v ∗ u∗ = (uv)∗ , so that uv ∈ U(H ). This shows that U(H ) is a subgroup of GL(H ). Consequently, U(H ) is a group for the composition of maps. (c) For all u, v ∈ U(H ), we have (u − v)(u − v)∗ = (u − v)(u∗ − v ∗ ) = (u − v)(u−1 − v −1 ) = 2 IdH −uv −1 − vu−1 .

.

Therefore √ ║u − v║H S / 1 Tr((u − v)(u − v)∗ )) = n / 1 Tr(2 IdH −uv −1 − vu−1 )) = n / 1 (2n − Tr(uv −1 ) − Tr(vu−1 )) = n / 1 = 2 − (Tr(uv −1 ) + Tr(vu−1 )). n

dH S (u, v) =

.

As Tr(uv −1 ) = Tr(v −1 u) and Tr(vu−1 ) = Tr(u−1 v) = Tr((v −1 u)−1 ) = Tr((v −1 u)∗ ) = Tr(v −1 u), this yields Formula (7.6). (d) Using (7.6), we get, for all u, v, w1 , w2 ∈ U(H ), / dH S (w1 uw2 , w1 vw2 ) =

.

2(1 − /

=

2(1 − /

=

2(1 − /

=

2(1 − /

=

2(1 −

1 Re(Tr((w1 vw2 )−1 w1 uw2 ))) n 1 Re(Tr(w2−1 v −1 w1−1 w1 uw2 ))) n 1 Re(Tr(w2−1 v −1 uw2 ))) n 1 Re(Tr(v −1 uw2 w2−1 ))) n 1 Re(Tr(v −1 u))) n

= dH S (u, v). This shows that dH S is a bi-invariant metric on U(H ).

■

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Comment The group U(H ) is the unitary group of H and dH S is called the HilbertSchmidt metric on U(H ). Exercise 7.50 Let F be a non-empty finite set and let dF denote the Hamming metric on Sym(F ). We consider the |F |-dimensional Hilbert space H := CF = {x : F → C}. We denote by U(H ) the unitary group of H and by dH S the HilbertSchmidt metric on U(H ) (cf. Exercise 7.49). For α ∈ Sym(F ), define λ(α) : H → H by λ(α)(x) := x ◦ α −1 for all x ∈ H . (a) Show that λ(α) ∈ U(H ) for all α ∈ Sym(F ). (b) Show that the map λ : Sym(F ) → U(H ) is an injective group homomorphism. (c) Show that one has dF (α, β) = 1 −

.

1 Tr(λ(β −1 α)) |F |

(7.7)

for all α, β ∈ Sym(F ). (d) Show that one has dH S (λ(α), λ(β)) =

.

√

2dF (α, β)

(7.8)

for all α, β ∈ Sym(F ). Solution (a) Let α ∈ Sym(F ). For all v, w ∈ C and x, y ∈ H , we have λ(α)(vx +wy) = (vx +wy)◦α −1 = vx ◦α −1 +wy ◦α −1 = vλ(α)(x)+wλ(α)(y).

.

This shows that λ(α) is C-linear. On the other hand, as x ◦ α −1 ◦ α = x for all x ∈ H , we see that λ(α) is bijective with inverse λ(α −1 ). The scalar product on H is given by :=

∑

.

x(f )y(f ) for all x, y ∈ H.

f ∈F

It follows that =

∑

.

x(α −1 (f ))y(f )

f ∈F

=

∑

x(f )y(α(f ))

f ∈F

= .

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7 Local Embeddability and Sofic Groups

Thus (λ(α))∗ = λ(α −1 ) = (λ(α))−1 . As the adjoint of λ(α) is equal to its inverse, we have λ(α) ∈ U(H ). (b) For all α, β ∈ Sym(F ) and x ∈ H , we have λ(α ◦ β)(x) = x ◦ (α ◦ β)−1 = x ◦ β −1 ◦ α −1 = λ(α)(λ(β)(x)) = (λ(α) ◦ λ(β))(x)

.

and hence λ(α ◦ β) = λ(α) ◦ λ(β). This shows that λ : Sym(F ) → U(H ) is a group homomorphism. Let (ef )f ∈F denote the canonical basis of H . Thus, for all f, f ' ∈ F , we have ef (f ' ) = 0 if f /= f ' and ef (f ) = 1. Clearly λ(α)(ef ) = eα(f ) . We deduce that if α is in the kernel of λ then α(f ) = f for all f ∈ F , that is, α = IdF = 1Sym(F ) . This shows that λ is injective. (c) Let M denote the matrix of λ(β −1 α) with respect to the basis (ef )f ∈F . For every f ∈ F , the image of ef under λ(β −1 α) is eβ −1 (α(f )) . Thus, the trace of M is equal to the number of f ∈ F such that β −1 (α(f )) = f , or, equivalently, such that α(f ) = β(f ). We deduce that 1−

.

1 1 Tr(λ(β −1 α)) = 1 − Tr(M) |F | |F | =1− =

1 |{f ∈ F : α(f ) = β(f )}| |F |

1 |{f ∈ F : α(f ) /= β(f )}| |F |

= dF (α, β). This establishes (7.7). (d) Let α, β ∈ Sym(F ). Using Exercise 7.49(c), we get / dH S (λ(α), λ(β)) =

.

2(1 −

1 Re(Tr(λ(β)−1 λ(α)))) |F |

2(1 −

1 Re(Tr(λ(β −1 α)))) |F |

2(1 −

1 Tr(λ(β −1 α))) |F |

/ = / = =

√

2dF (α, β),

where the last equality follows from (c). This establishes Formula (7.8).

■

Exercise 7.51 (Hyperlinear Groups) Let G be a group. Given a finite subset K ⊂ G, a real number ε > 0, and a complex Hilbert space H with finite positive

7.2 Exercises

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dimension, a map ψ : G → U(H ) is called a (K, ε)-almost-homomorphism if it satisfies the following conditions: ((K, ε)-AHU-1) for all k1 , k2 ∈ K, one has dH S (ψ(k1 k2 ), ψ(k1 )ψ(k2 )) ≤ ε; ((K, ε)-AHU-2) for all distinct k1 , k2 ∈ K, one has dH S (ψ(k1 ), ψ(k2 )) ≥ 1 − ε, where dH S denotes the Hilbert-Schmidt metric on U(H ) (cf. Exercise 7.49). A group G is called hyperlinear if it satisfies the following condition: for every finite subset K ⊂ G and every real number ε > 0, there exist a complex Hilbert space H with finite positive dimension and a (K, ε)-almost-homomorphism ψ : G → U(H ). Show that every sofic group is hyperlinear. Solution Let G be a sofic group. Let K ⊂ G be a finite subset and let ε be a real number such that 0 < ε < 1. Choose some real number η such that 0 < η ≤ min(ε2 /2, 1 − (1 − ε)2 /2). Since G is sofic, there exist a non-empty finite set F and a (K, η)-almost-homomorphism ϕ : G → Sym(F ). Consider the |F |-dimensional complex vector space H := CF and the group homomorphism λ : Sym(F ) → U(H ) defined in Exercise 7.50. Denoting by dF the Hamming metric on Sym(F ), it follows from Exercise 7.50(d) that dH S (λ(α), λ(β)) =

.

√

2dF (α, β)

(7.9)

for all α, β ∈ Sym(F ). Consider the map ψ : G → U(H ) defined by ψ := λ ◦ ϕ. For all k1 , k2 ∈ K, we have dH S (ψ(k1 k2 ), ψ(k1 )ψ(k2 )) = dH S (λ(ϕ(k1 k2 )), λ(ϕ(k1 ))λ(ϕ(k2 )))

.

= dH S (λ(ϕ(k1 k2 )), λ(ϕ(k1 )ϕ(k2 )). Thus, using (7.9) and the fact that ϕ is a (K, η)-almost-homomorphism, we get √ √ 2dF (ϕ(k1 k2 ), ϕ(k1 )ϕ(k2 )) ≤ 2η ≤ ε. .dH S (ψ(k1 k2 ), ψ(k1 )ψ(k2 )) = (7.10) On the other hand, for all k1 , k2 ∈ K such that k1 /= k2 , we have dH S (ψ(k1 ), ψ(k2 )) = dH S (λ(ϕ(k1 )), λ(ϕ(k2 )))

.

and therefore, by using again (7.9) and the fact that ϕ is a (K, η)-almosthomomorphism, √ √ (7.11) .dH S (ψ(k1 ), ψ(k2 )) = 2dF (ϕ(k1 ), ϕ(k2 )) ≥ 2(1 − η) ≥ 1 − ε. We deduce from (7.10) and (7.11) that the map ψ : G → U(H ) is a (K, ε)-almosthomomorphism. This shows that the group G is hyperlinear. ■

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Comment The class of hyperlinear groups was introduced by R˘adulescu [Rad] (cf. [Pes], [PesK], and [CaprL]). By adapting the proof of the analogous characterization of sofic groups, it can be shown that a group is hyperlinear if and only if it embeds into a metric ultraproduct of unitary groups of finite rank. It is unknown whether every hyperlinear group is sofic. The fact that every group is hyperlinear is also unknown and turns out to be equivalent to the Connes embedding conjecture for groups, which states that the von Neumann algebra of every group embeds into some ultrapower of the hyperfinite II1 factor. Note that the general Connes embedding conjecture, which states that every II1 factor can be embedded into some ultrapower of the hyperfinite II1 factor has been recently refuted [JiNVWY] (see [Gol] for a nice historical survey). Exercise 7.52 Suppose that a group G contains U a family (Hi )i∈I of subgroups satisfying the following properties: (1) G = i∈I Hi ; (2) For all i, j ∈ I , there exists k ∈ I such that Hi ∪ Hj ⊂ Hk ; (3) Hi is sofic for all i ∈ I . Show that G is sofic. Solution As each Hi is sofic and therefore locally embeddable into the class of sofic groups, it follows from Exercise 7.16 that G is locally embeddable into the class of sofic groups. This implies that G is sofic by [CAG, Proposition 7.5.10]. ■ Exercise 7.53 Show that every virtually sofic group is sofic. Solution Let G be a virtually sofic group. This means that there exists a subgroup H of finite index in G such that H is sofic. By [CAG, Lemma 2.1.10], we can find a normal subgroup K of finite index in G such that K ⊂ H . As every subgroup of a sofic group is itself sofic [CAG, Proposition 7.5.4], the group K is sofic. Since G/K is finite and therefore amenable, we deduce that G is sofic by applying [CAG, Proposition 7.5.14]. ■ Exercise 7.54 (The Weiss Condition) Let G be a finitely generated group and let S be a finite symmetric generating subset of G. Consider the following condition (W(G, S)) for all ε > 0 and n ∈ N, there exists a finite S-labeled graph Q = (Q, E) such that there is a proportion at least 1 − ε of vertices q ∈ Q for which there exists an S-labeled graph isomorphism BS (n) → BQ (q, n) sending 1G to q. It follows from [CAG, Theorem 7.7.1] that condition (W(G, S)) is equivalent to the soficity of the group G. This implies in particular that the fact that condition (W(G, S)) is satisfied or not is actually independent of the choice of S. The purpose of this exercise is to give a direct proof of this independence (without using the notion of soficity). (a) Let S ' ⊂ G be another finite symmetric generating subset of G such that S ' ⊂ S. Suppose that condition (W(G, S)) is satisfied. Show that condition (W(G, S ' )) is also satisfied.

7.2 Exercises

509

(b) Let m ≥ 1 be an integer and let S '' := BS (m) denote the ball of radius m centered at 1G in the Cayley graph of (G, S). Suppose that condition (W(G, S)) is satisfied. Show that S '' is a finite symmetric generating subset of G and that condition (W(G, S '' )) is also satisfied. (c) Deduce that if S ''' ⊂ G is another finite symmetric generating subset of G, then condition (W(G, S)) is satisfied if and only if condition (W(G, S ''' )) is satisfied. Solution (a) Given an S-labeled graph Γ = (V , E) we denote by Γ ' = (V ' , E ' ) the S ' -labeled graph defined by setting V ' := V and E ' := {e ∈ E : λ(e) ∈ S ' }, where λ : E → S is the labelling map for Γ . Given v ∈ V and n ∈ N, we have that BΓ ' (v, n) is the connected component of (BΓ (v, n))' containing v (cf. Exercise 6.81(b)). This way, CS ' (G) = (CS (G))' and BS ' (n) is the connected component of (BS (n))' containing 1G , for all n ∈ N. Let now ε > 0 and let n ∈ N. By virtue of condition (W(G, S)), we can find a finite S-labeled graph Q = (Q, E) and a subset Q(n) ⊂ Q with |Q(n)| ≥ (1 − ε)|Q| such that for each q ∈ Q(n) there exists an S-labeled isomorphism ϕ q : BS (n) → BQ (q, n) mapping 1G to q. For every q ∈ Q(n), the map ϕ q yields, by restriction, an S ' labeled isomorphism of BS (n)' onto (BQ (q, n))' mapping 1G (resp. the connected component containing 1G ) to q (resp. onto the connected component containing q). Consider the S ' -labeled graph Q ' = (Q' , E ' ) and set Q' (n) := Q(n) ⊂ Q = Q' . We then have |Q' (n)| ≥ (1−ε)|Q' | and for every q ∈ Q' (n) the restriction ϕ q |BS ' (n) yields an S ' -labeled isomorphism of BS ' (n) onto (BQ' (q, n)) mapping 1G to q. This shows that condition (W(G, S ' )) is satisfied. (b) We first observe that the subset S '' ⊂ G is finite and generates G since it contains the generating subset S. It is also symmetric since ℓS (g −1 ) = ℓS (g) for all g ∈ G. Given an S-labeled graph Γ = (V , E) we define an S '' -labeled graph Γ '' = (V '' , E '' ) (possibly with multiple edges) by setting V '' := V and declaring (u, s '' , v) ∈ E '' provided that dΓ (u, v) ≤ m and, if ek = (vk−1 , sk , vk ) ∈ E, k = 1, 2, . . . , ℓ (with 1 ≤ ℓ ≤ m), is a sequence of edges such that v0 = u and vℓ = v, then s '' = s1 s2 · · · sℓ ∈ S '' . Note that if v ∈ V = V '' one has BΓ (v, mn) = BΓ '' (v, n) for all n ∈ N. Moreover, (CS (G))'' (resp. (BS (mn))'' ) has no multiple edges: if (u, s1 s2 · · · sℓ , v) and (u, t1 t2 · · · tℓ' , v) are in E(CS (G)'' ) (resp. E(BS (mn)'' )) then s1 s2 · · · sℓ = u−1 v = t1 t2 · · · tℓ' in S '' . In fact, as S '' labeled graphs, we clearly have (CS (G))'' = CS '' (G) and (BS (mn))'' = BS '' (n) for all n ∈ N. Let now ε > 0 and let n ∈ N. By virtue of condition (W(G, S)), we can find a finite S-labeled graph Q = (Q, E) and a subset Q(mn) ⊂ Q such that |Q(mn)| ≥ (1 − ∈)|Q| such that for each q ∈ Q(mn) there exists an S-labeled isomorphism ϕ q : BS (mn) → BQ (q, mn) mapping 1G to q. Consider the S '' -labeled graph Q '' and set Q'' (n) := Q(mn) ⊂ Q = Q'' . Then |Q'' (n)| ≥ (1 − ∈)|Q'' |. Moreover, if q ∈ Q'' (n) the map ϕ q : BS '' (n) = BS (mn) → BQ (q, mn) = BQ'' (q, n) maps 1G to q and, clearly, yields and S '' -labeled isomorphism. This shows that condition (W(G, S '' )) is satisfied.

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7 Local Embeddability and Sofic Groups

(c) Suppose that condition (W(G, S)) is satisfied. Since S generates G, we can find and integer m ≥ 1 such that S ''' ⊂ BS (m). Setting S '' := BS (m), it follows from (b) that condition (W(G, S '' )) is satisfied. Since S ''' ⊂ S '' , it follows from (a) that (W(G, S ''' )) is also satisfied. By symmetry, after exchanging the roles of S and S ''' , the statement follows. ■ Comment Condition (W(G, S)), which is exactly condition (b) in [CAG, Theorem 7.7.1], is named after Weiss who used it in [Weis2] as a definition of soficity and for his proof of the surjunctivity of sofic groups (cf. [CAG, Theorem 7.8.1]). Exercise 7.55 Let G1 and G2 be two finitely generated groups and let S1 and S2 be finite symmetric generated subsets of G1 and G2 , respectively. Set G := G1 × G2 . Then (cf. [CAG, Proposition 6.6.10]) S := (S1 × {1G2 }) ∪ ({1G1 } × S2 ) is a finite symmetric generating subset of G. Suppose that conditions (W(G1 , S1 )) and (W(G2 , S2 )) hold. Give a direct proof of the fact that condition (W(G, S)) holds as well. Solution We start by identifying S1 with S1 × {1G2 } (resp. S2 with {1G1 } × S2 ). It follows from Exercise 6.6 that the Cayley graph CS (G) is isomorphic, as an S-labeled graph, to the S-labeled product CS1 (G1 ) × CS2 (G2 ) (cf. Exercise 6.5) of the Cayley graphs of G1 and G2 with respect to the generating subsets S1 and S2 . In the following, we shall then identify these two S-labeled graphs. Let ε > 0 and let n ∈ N. For i = 1, 2, by condition (W(Gi , Si )), we can find a finite Si -labeled graph Qi = (Qi , Ei ) together with a subset Qi (n) ⊂ Qi of size |Qi (n)| ≥ (1 − ε/2)|Qi |

.

(7.12)

such that for each vertex qi ∈ Qi there exists an Si -labeled graph isomorphism q ϕi i : BSi (n) → BQi (qi , n) sending 1G to qi . Consider the S-labeled graph Q = Q1 × Q2 constructed in Exercise 6.5. Set Q(n) := Q1 (n) × Q2 (n) ⊂ Q1 × Q2 = Q and observe that by virtue of (7.12) we have |Q(n)| = |Q1 (n)|·|Q2 (n)| ≥ (1−ε/2)2 |Q1 |·|Q2 | = (1−ε+ε2 /4)|Q| ≥ (1−ε)|Q|.

.

Let now q = (q1 , q2 ) ∈ Q(n). Modulo the identifications mentioned above, we have that in the Cayley graph CS (G), the ball BS (n) is contained in BS1 (n)×BS2 (n) and, analogously, in Q, the ball BQ (q, n) is contained in BQ1 (q1 , n) × BQ2 (q2 , n). q q It is clear that the product morphism ϕ1 1 × ϕ2 2 establishes an S-labeled graph isomorphism from BS1 (n) × BS2 (n) onto BQ1 (q1 , n) × BQ2 (q2 , n) sending 1G to q q (q1 , q2 ) = q. As a consequence, its restriction ϕq := (ϕ1 1 × ϕ2 2 )|BS (n) establishes an S-labeled graph isomorphism from BS (n) onto BQ (q, n), sending 1G to q. This shows that condition (W(G, S)) holds. ■ Comment This can be also deduced from the fact that a finitely generated group G, with a given finite symmetric generating subset S, is sofic if and only if condition

7.2 Exercises

511

(W(G, S)) is satisfied [CAG, Theorem 7.7.1] since the direct product of two sofic groups is a sofic group [CAG, Proposition 7.5.7]. Exercise 7.56 Let G be a finitely generated residually finite group and let S be a finite symmetric generating subset of G. Give a direct proof of the fact that condition (W(G, S)) is satisfied. Solution Let ε > 0 and let n ≥ 1 be an integer. Set K := BS (n) ⊂ G. Since G is residually finite, by condition (RF3) we can find a finite group F and a group homomorphism φ : G → F such that the restriction φ|K of φ to K is injective. Up to replacing F by φ(G), we may suppose that φ is surjective. Then S ' := ϕ(S) is a finite symmetric generating subset of F . Let Q := CS ' (F ) denote the associated Cayley graph. Note that since φ is injective on K and K ⊃ S, the restriction φ|S yields a bijection between S and S ' . This way, we may identify S ' and S and regard Q as a finite S-labeled graph. Set Q(n) := Q. We have |Q(n)| = |Q| ≥ (1 − ε)|Q|. Let q ∈ Q(n) = Q = F . Then we can find g ∈ G such that φ(g) = q. Let Lg : G → G denote left multiplication by g and observe that Lg |BS (n) , its restriction to BS (n), yields an S-labeled isomorphism from BS (n) = BS (1G , n) onto BS (g, n) mapping 1G to g. Moreover, keeping in mind that φ is a group homomorphism and its restriction to K = BS (n) is injective, we deduce that its restriction φ|BS (g,n) to BS (g, n) yields an S-labeled isomorphism from BS (g, n) onto BQ (q, n) mapping g to q. It follows that the composite map ϕ q := φ|BS (g,n) ◦ Lg |BS (n) yields an Slabeled isomorphism from BS (n) onto BQ (q, n) mapping 1G to q. This shows that condition (W(G, S)) holds. ■ Comment This can be also deduced from the fact that a finitely generated group G, with a given finite symmetric generating subset S, is sofic if and only if condition (W(G, S)) is satisfied [CAG, Theorem 7.7.1] since every residually finite group is sofic [CAG, Corollary 7.5.11]. Exercise 7.57 Let G be a finitely generated amenable group and let S be a finite symmetric generating subset of G. Give a direct proof of the fact that condition (W(G, S)) is satisfied. Solution Let ε > 0 and let n ≥ 1 be an integer. Set E := BS (n) ⊂ G. It follows from [CAG, Corollary 5.4.5] that we can find a non-empty finite subset F ⊂ G such that |∂E (F )| < ε|F |, where ∂E (F ) = F +E \ F −E is the E-boundary of F . Recall that F +E = F E and F −E = {g ∈ G : gE ⊂ F }. Note that, since 1G ∈ BS (n) = E, we have F −E ⊂ F and F ⊂ F +E . We deduce that |F −E | ≥ (1 − ε)|F |.

.

(7.13)

Let then Q = (Q, E) denote the subgraph of CS (G) induced by F , that is, Q := F and E := (F × S × F ) ∩ E(CS (G)). Let us show that Q(n) := F −E ⊂ F = Q meets our requirements. It follows from (7.13) that |Q(n)| ≥ (1 − ε)|Q|. Moreover, for every q ∈ Q(n), left multiplication Lq : G → G by q yields, by restriction, an

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7 Local Embeddability and Sofic Groups

S-labeled isomorphism from BS (n) onto qBS (n) = BS (q, n) = BQ (q, n) mapping 1G to q. This shows that condition (W(G, S)) holds. ■ Comment This can be also deduced from the fact that a finitely generated group G, with a given finite symmetric generating subset S, is sofic if and only if condition (W(G, S)) is satisfied [CAG, Theorem 7.7.1] since every amenable group is sofic [CAG, Proposition 7.5.6].

Chapter 8

Linear Cellular Automata

This chapter is devoted to linear cellular automata and their group ring matricial representations. This includes the study of linear surjunctivity and the linear version of the Garden of Eden theorem. The leitmotif is the interplay between properties of a group and those of its associated group ring. This leads in particular to connections with some of the celebrated Kaplansky conjectures. Noetherian rings, UPR-rings, von Neumann regular rings, unit-regular rings, Boolean rings, strongly regular rings, reversible rings, directly finite and stably finite rings, right-Ore rings, prime rings are investigated. On the other hand, indicable groups, unique-product groups, orderable groups, diffuse groups, the Klein bottle group, the Passman-PromislowHantzsche-Wendt group and, more generally, the Fibonacci groups are studied. A detailed description of the ring of non-commutative formal power series over a finite alphabet set is given. This yields Magnus’ proof of residual nilpotency of free groups and bi-orderability of free groups. Duality for linear cellular automata, a notion introduced by Bartholdi, is also investigated.

8.1 Summary 8.1.1 Rings All rings and algebras are always assumed to be associative and unital (that is, with a multiplicative neutral element) but are possibly non-commutative. Subrings and subalgebras are required to contain the multiplicative neutral element. Morphisms of rings and algebras are required to send the multiplicative neutral element of the source to the multiplicative neutral element of the target. Let R be a ring. One says that R has zero-divisors if there exist nonzero elements .a, b ∈ R such that .ab = 0R . Thus, R has no zero-divisors if and only if .ab = 0R © The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 T. Ceccherini-Silberstein, M. Coornaert, Exercises in Cellular Automata and Groups, Springer Monographs in Mathematics, https://doi.org/10.1007/978-3-031-10391-9_8

513

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8 Linear Cellular Automata

implies .a = 0R or .b = 0R for all .a, b ∈ R. One says that R is directly finite if ab = 1R implies .ba = 1R for all .a, b ∈ R. The ring R is said to be stably finite if the matrix ring .Matd (R) is directly finite for every integer .d ≥ 1. An element .a ∈ R is called nilpotent if there exists an integer .n ≥ 1 such that .a n = 1R . One says that .a ∈ R is an idempotent if .a 2 = a. The elements .0R and .1R are idempotents. They are called the trivial idempotents of R. An idempotent .a ∈ R is called a non-trivial idempotent if .a /= 0R and .a /= 1R . Observing that .a 2 = a implies .a(a − 1R ) = 0R , one deduces that if the ring R has no zero-divisors then it has no non-trivial idempotent. An element .a ∈ R is called left-invertible (resp. right-invertible) if there exists ' ' ' .a ∈ R such that .a a = 1R (resp. .aa = 1R ). An element .a ∈ R is called invertible if it is both left and right-invertible. This amounts to saying that there exists an element .b ∈ R such that .ab = ba = 1R . Such an element b is then unique. It is called the inverse of the invertible element a and one writes .b = a −1 . The invertible elements of R are also called the units of R. They form a group under multiplication and this group is denoted by .U (R). A nonzero ring R in which .U (R) = R \ {0R } is called a division ring. A commutative division ring is called a field. .

8.1.2 Group Rings Let A be an abelian group and let G be a group. The support of a configuration .x ∈ AG is the subset .supp(x) := {g ∈ G : x(g) /= 0A } ⊂ G. The set of configurations G G .x ∈ A with finite support is a subgroup of .A denoted by .A[G]. We have .A[G] =  G ⊕g∈G A ⊂ g∈G A = A . Suppose now that R is a ring. The convolution product of two elements .α, β ∈ R[G] is the element .αβ ∈ R G defined by (αβ)(g) :=



.

α(h1 )β(h2 ).

h1 ,h2 ∈G h1 h2 =g

for all .g ∈ G. Observe that .supp(αβ) ⊂ supp(α) supp(β). Moreover, for all .g ∈ G, one has   .(αβ)(g) = α(h)β(h−1 g) = α(gh)β(h−1 ). h∈G

h∈G

The natural addition on .R[G] together with the convolution product yield a ring structure on .R[G]. The ring .R[G] is called the group ring of G with coefficients in R. There is also a left R-module structure on .R[G] induced by the product left Rmodule structure on .R G . The left R-module .R[G] is free with base .(δg )g∈G , where .δg ∈ R[G] is defined by .δg (h) := 1R if .h = g and .δg (h) := 0R if .h ∈ G \ {g}. One has .δ1G = 1R[G] and .δg δh = δgh for all .g, h ∈ G. If R is a non-trivial ring, then

8.1 Summary

515

the map .g |→ δg defines a group embedding of G into the group of units of .R[G]. This group embedding can be used to regard G as a subgroup of the group of units of .R[G] via the identification .g = δg for all .g ∈ G. In the case when K is a field, the ring .K[G] with its K-vector space structure is a K-algebra, called the group algebra of G with coefficients in K.

8.1.3 NZD-Groups, Unique-Product Groups, and Orderable Groups A group G is called an NZD-group if the group ring .R[G] has no zero-divisors whenever R is a ring without zero-divisors. Every NZD-group is torsion-free [CAG, Example 8.16.1.(b)]. A group G is called a unique-product group if, given any two non-empty finite subsets .A, B ⊂ G, there exists an element .g ∈ G which can be uniquely expressed as a product .g = ab with .a ∈ A and .b ∈ B. Every unique-product group is a NZD-group [CAG, Proposition 8.16.9]. A group G is called orderable if it admits a left-invariant total ordering, that is, a total ordering .≤ such that .g1 ≤ g2 implies .gg1 ≤ gg2 for all .g, g1 , g2 ∈ G. This is equivalent to the existence of a right-invariant total ordering on G, that is, a total ordering .≤ such that .g1 ≤ g2 implies .g1 g ≤ g2 g for all .g, g1 , g2 ∈ G. Every orderable group is a unique-product group [CAG, Proposition 8.16.8]. A group G is called bi-orderable if it admits a total ordering .≤ which is both left and right invariant, i.e., such that .g1 ≤ g2 implies .gg1 ≤ gg2 and .g1 g ≤ g2 g for all .g, g1 , g2 ∈ G. Every torsion-free abelian group is bi-orderable [CAG, Example 8.16.5.(e)]. Of course, every bi-ordorable group is orderable. To summarize, we have the following implications bi-orderable ⇒ orderable ⇒ unique-product ⇒ NZD ⇒ torsion-free.

.

8.1.4 Linear Shift Spaces G Let  G be a group and let V be a vector space over a field K. The set .V = g∈G V = {x : G → V } is equipped with its prodiscrete uniform structure and the G-shift action. The space .V G has an additional structure of a vector space over K inherited from the vector space structure on V . For every .g ∈ G, the map .x |→ gx is a uniformly continuous vector space automorphism of .V G . The subset .V [G] ⊂ V G , consisting of all configurations with finite support, is a vector subspace of .V G and .V [G] is dense in .V G for the prodiscrete topology [CAG, Proposition 8.2.1]. Two configurations .x, x ' ∈ V G are almost equal if and only if ' G is K.x − x ∈ V [G] [CAG, Proposition 8.2.2]. Since the G-shift action on .V

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8 Linear Cellular Automata

linear, .V G has a natural structure of a left .K[G]-module. It turns out that .V [G] is a free submodule of .V G . In fact, if .(ei )i∈I is a vector basis for V , then the family of configurations .(ci )i∈I , where .ci ∈ V [G] is defined by .ci (1G ) := ei and .ci (g) := 0V for all .g ∈ G \ {1G }, is a free basis for the left .K[G]-module .V [G] (cf. [CAG, Proposition 8.7.3]). In particular, if V has finite dimension .dimK (V ) = d < ∞, then .V [G] is isomorphic, as a left .K[G]-module, to .K[G]d .

8.1.5 Linear Cellular Automata Let G be a group and let V be a vector space over a field K. A cellular automaton τ : V G → V G which is K-linear is called a linear cellular automaton. Given a cellular automaton .τ : V G → V G , with memory set .S ⊂ G and local defining map .μ : V S → V , one has that .τ is linear if and only if .μ is K-linear [CAG, Proposition 8.1.1]. The set .LCA(G; V ) of all linear cellular automata .τ : V G → V G has a natural structure of a K-algebra. In fact, .LCA(G; V ) is a subalgebra of .EndK (V G ), the K-algebra of all vector space endomorphisms of .V G [CAG, Proposition 8.1.4]. Given a linear cellular automaton .τ : V G → V G , one has .τ (V [G]) ⊂ V [G] [CAG, Proposition 8.2.3] so that its restriction .τ |V [G] : V [G] → V [G] is a vector space endomorphism of .V [G], that is, .τ |V [G] ∈ EndK (V [G]). The map .τ |→ τ |V [G] is an injective K-algebra homomorphism from .LCA(G; V ) into .EndK (V [G]) [CAG, Proposition 8.2.4]. Moreover, .τ is pre-injective if and only if .τ |V [G] is injective [CAG, Proposition 8.2.5].

.

8.1.6 Restriction and Induction of Linear Cellular Automata Let G be a group and let V be a vector space over a field K. Let .H ⊂ G be a subgroup. Then .LCA(G, H ; V ) := CA(G, H ; V ) ∩ LCA(G; V ), the set of all linear cellular automata admitting a memory set contained in H , is a Ksubalgebra of .LCA(G; V ) and the map .τ |→ τH , where .τH : V H → V H denotes the restriction cellular automaton of .τ ∈ LCA(G, H ; V ), yields a Kalgebra isomorphism of .LCA(G, H ; V ) onto .LCA(H ; V ) with inverse the map G G : V G → V G denotes the induction cellular automaton of .σ |→ σ , where .σ .σ ∈ LCA(H ; V ) [CAG, Proposition 8.3.2].

8.1 Summary

517

8.1.7 Group Ring Representation of Linear Cellular Automata Let G be a group and let V be a vector space over a field K. Given an element .α ∈ EndK (V )[G] (the group algebra of G with coefficients in the K-algebra .EndK (V )), one defines a linear cellular automaton .τα ∈ LCA(G; V ) by setting τα (x)(g) :=



.

α(h)(x(gh))

h∈G

for all .x ∈ V G and .g ∈ G (note that the sum in the right-hand side is actually finite since .α has finite support). The support of .α coincides with the minimal memory set of .τα (cf. [CAG, Proposition 8.5.1]). Moreover, the map .α |→ τα is a Kalgebra isomorphism of .EndK (V )[G] onto .LCA(G; V ) [CAG, Theorem 8.5.2]. In particular, the K-algebra .LCA(G; K) of one-dimensional linear cellular automata is isomorphic to the group algebra .K[G] [CAG, Corollary 8.5.3]. Every linear cellular automaton .τ ∈ LCA(G; V ) is an endomorphism of the G and .LCA(G; V ) is a subalgebra of the K-algebra .End G .K[G]-module .V K[G] (V ) [CAG, Proposition 8.7.1]. If the vector space V is finite dimensional, the restriction map .τ |→ τ |V [G] yields a K-algebra isomorphism of .LCA(G; V ) onto .EndK[G] (V [G]) [CAG, Theorem 8.7.6)]. Moreover, setting .d := dimK (V ) the K-algebras .LCA(G; V ) and .Matd (K[G]) (the K-algebra of .d × d matrices with coefficients in the group algebra .K[G]) are isomorphic [CAG, Corollary 8.7.8].

8.1.8 Matrix Representation of Linear Cellular Automata Let G be a group and let V be a vector space over a field K with finite dimension d := dimK (V ) < ∞. Let .B = (ei )1≤i≤d be a basis of V . Recall that the family .(ci )1≤i≤d , where .ci ∈ V [G] is defined by .ci (1G ) := ei and .ci (g) := 0V for all .g ∈ G \ {1G }, is a free basis of the free left .K[G]-module .V [G]. Let now .τ : V G → V G be a linear cellular automaton. Recall that .τ (V [G]) ⊂ V [G] and that the restriction map .τ |V [G] : V [G] → V [G] is an endomorphism of the left .K[G]-module .V [G]. Let .M ∈ Matd (K[G]) denote the matrix whose entries are given by .

τ (ci ) =



.

Mij cj ,

(8.1)

1≤j ≤d

where .1 ≤ i ≤ d. The matrix .M ∈ Matd (K[G]) is called the matrix of .τ with  ∈ Matd (K[G]) is defined by .M(g)  := M(g −1 ) for all .g ∈ G, respect to .B. If .M  is an isomorphism of K-algebras from .LCA(G; V ) then the map given by .τ |→ M onto .Matd (K[G]).

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8 Linear Cellular Automata

8.1.9 The Closed Image Property for Linear Cellular Automata Let G be a group and let V be a vector space over a field K. When V is finite dimensional, every linear cellular automaton .τ : V G → V G satisfies the closed image property, that is, the vector subspace .τ (V G ) is closed in .V G with respect to the prodiscrete topology [CAG, Theorem 8.8.1]. The finite dimensionality assumption on the alphabet V cannot be removed (cf. [CAG, Example 8.8.3]).

8.1.10 Invertible Linear Cellular Automata Let G be a group and let V be a vector space over a field K. If a linear cellular automaton .τ : V G → V G is invertible, i.e., .τ is bijective and the inverse map .τ −1 : V G → V G is a cellular automaton, then .τ −1 is also a linear cellular automaton. If V is finite-dimensional, then every bijective linear cellular automaton .τ : V G → V G is invertible [CAG, Theorem 8.12.2]. Moreover, the group .ILCA(G; V ) := LCA(G; V ) ∩ ICA(G; V ) of all invertible linear cellular automata is isomorphic to .GLd (K[G]), where .d = dimK (V ). If V is infinite dimensional, there exist bijective linear cellular automata .τ : V G → V G that are not invertible [CAG, Example 8.12.3].

8.1.11 Mean Dimension Let K be a field and let V be a finite-dimensional vector space over K. Suppose that G is an amenable group and let .F = (Fj )j ∈J be a right Følner net for G. The mean dimension of a vector subspace X of .V G (with respect to the right Følner net .F ) is the number .

mdimF (X) := lim sup j

dim(πFj (X)) |Fj |

where, as usual, for a subset .E ⊂ G, we denote by .πE : V G → V E the projection map. We list below some general properties of mean dimension (cf. Proposition 8.9.3, Proposition 8.9.4, and Proposition 8.9.5 in [CAG]): (i) .mdimF (V G ) = dim(V ); (ii) if .X ⊂ Y ⊂ V G are vector subspaces, then .mdimF (X) ≤ mdimF (Y ) ≤ dim(V ); (iii) if .τ : V G → V G is a linear cellular automaton and .X ⊂ V G is a vector subspace, then .mdimF (τ (X)) ≤ mdimF (X); (iv) if .X ⊂ V G is a G-invariant vector subspace and there exists a finite subset E .E ⊂ G such that .πE (X)  V , then .mdimF (X) < dim(V ).

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519

8.1.12 The Garden of Eden Theorem for Linear Cellular Automata The Garden of Eden Theorem for linear cellular automata (or linear Garden of Eden theorem) states that if G is an amenable group and V is a finite dimensional vector space over a field K, then a linear cellular automaton .τ : V G → V G is surjective if and only if it is pre-injective. In fact, we have the following more complete formulation ([CAG, Theorem 8.9.6]): Theorem (Garden of Eden Theorem for Linear Cellular Automata) Let G be an amenable group, let .F be a right Følner net for G, and let V be a finite dimensional vector space over a field K. Let .τ : V G → V G be a linear cellular automaton. Then the following conditions are equivalent: (LGOE1) .τ is surjective; (LGOE2) .mdimF (τ (V G )) = dim(V ); (LGOE3) .τ is pre-injective. The amenability assumption on the group G in the statement of the linear Garden of Eden theorem cannot be removed. For instance, if .G = F2 , the free group of rank 2, there exists a linear cellular automaton .τ : V G → V G , with .dim(V ) = 2, which is surjective but not pre-injective [CAG, Section 8.11]. Moreover, one also has (see [CAG2, Theorem 8.17.1]): Theorem Let G be a non-amenable group and let K be a field. Then there exist a finite-dimensional vector space V over K and a linear cellular automaton .τ : V G → V G which is pre-injective but not surjective. Combining this result with the Garden of Eden theorem for linear cellular automata one obtains the following characterization of amenability in terms of linear cellular automata (see [CAG2, Corollary 8.17.7]): Corollary Let G be a group. Then the following conditions are equivalent: (a) G is amenable; (b) for any field K and any finite-dimensional vector space V over K, every preinjective linear cellular automaton .τ : V G → V G is surjective; (c) there exists a field K such that, for any finite-dimensional vector space V over K, every pre-injective linear cellular automaton .τ : V G → V G is surjective; (d) for any finite field K and any finite-dimensional vector space V over K, every pre-injective linear cellular automaton .τ : V G → V G is surjective; (e) there exists a finite field K such that, for any finite-dimensional vector space V over K, every pre-injective linear cellular automaton .τ : V G → V G is surjective; (f) for any finite set A, every pre-injective cellular automaton .τ : AG → AG is surjective.

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8 Linear Cellular Automata

8.1.13 The Discrete Laplacian Let G be a group and let K be a field. Given a non-empty finite subset .S ⊂ G, the discrete Laplacian over K associated with G and S is the linear cellular automaton K G → K G (with memory set .S ∪ {1 }) defined by .ΔS = Δ : K G S ΔS (x)(g) := |S|x(g) −



.

x(gs)

s∈S

for all .x ∈ K G and .g ∈ G. Note that .ΔS is never injective since all constant maps .x : G → K are in the kernel of .ΔS . Moreover, if the subgroup of G generated by S is finite then .ΔS is neither pre-injective nor surjective [CAG, Proposition 8.13.1]. On the other hand, if the subgroup of G generated by S is infinite then the real discrete Laplacian .ΔR S is both pre-injective and surjective. Collecting these results together we have the following (cf. [CAG, Theorem 8.13.2]): Theorem (Garden of Eden Theorem for Real Discrete Laplacians) Let G be a G G group and let S be a non-empty finite subset of G. Let .ΔR S : R → R denote the associated real discrete Laplacian. Then the following conditions are equivalent: (i) .ΔR S is surjective; (ii) the subgroup of G generated by S is infinite; (iii) .ΔR S is pre-injective.

8.1.14 Linear Surjunctivity A group G is said to be L-surjunctive if, for every field K and any finite-dimensional vector space V over K, every injective linear cellular automaton .τ : V G → V G is surjective (and hence invertible by [CAG, Theorem 8.12.2]). This amounts to saying that the group algebra .K[G] is stably finite for every field K [CAG, Corollary 8.15.6]. Every sofic group is L-surjunctive [CAG, Theorem 8.14.4]. Thus, if a group G is sofic, then the group algebra .K[G] is stably finite for every field K.

8.2 Exercises Exercise 8.1 Let G be a group and let R be a ring. Suppose that G and R have both cardinality 2. Is the ring R[G] isomorphic to the product ring Z/2Z × Z/2Z? Solution The answer is no. Indeed, the ring R = {0, 1} is a field with two elements 0 and 1, while the group G is cyclic of order 2. Let a be the generator of G. Then 1 + a is a nonzero element in R[G] whose square is 0. On the other hand, there are no nonzero nilpotent elements in the ring Z/2Z × Z/2Z. Therefore R[G] and

8.2 Exercises

521

Z/2Z × Z/2Z are not isomorphic as rings. Observe however that R[G] and Z/2Z × Z/2Z are isomorphic as additive groups. ■ Comment All rings with cardinality 4 are commutative. In fact, it is easy to check that, up to ring isomorphism, there are only four rings with cardinality 4, namely Z/4Z, Z/2Z × Z/2Z, Z/2Z[X]/(X2 ), and the field Z/2Z[X]/(1 + X + X2 ). The group ring considered in Exercise 8.1 is isomorphic to the ring Z/2Z[X]/(x 2 ). It is also to the subring of Mat2 (Z/2Z) consisting of all matrices of the form  isomorphic  ab with a, b ∈ Z/2Z. 0a Exercise 8.2 Let G be a group and let R be a nonzero ring. (a) Show that the ring R[G] is finite if and only if both G and R are finite. (b) Show that the ring R[G] is commutative if and only if both G and R are commutative. Solution (a) If G and R are both finite, then R[G], which can be identified with the set of maps from G to R, is finite with cardinality |R[G]| = |G||R| . Conversely, suppose that R[G] is finite. Then G and R are both finite since G embeds into the group of invertible elements of R[G] via the map g |→ 1R g and R embeds as a subring of R[G] via the map r |→ r1G . (b) Suppose first that both R and G are commutative. Let α, β ∈ R[G]. Write α=



.

g∈G

αg g

and

β=



βg g,

g∈G

where αg , βg ∈ R for all g ∈ G and αg = βg = 0 for all but finitely many g ∈ G. We then have  .αβ = αg βh gh g,h∈G

=



βh αg gh

(since R is commutative)

βh αg hg

(since G is commutative)

g,h∈G

=



g,h∈G

= βα. This shows that R[G] is commutative. Conversely, suppose now that R[G] is commutative. Let a, b ∈ R and g, h ∈ G. Consider the elements α, β ∈ R[G] defined by α := a1G and β := b1G . We then have αβ = (ab)1G and βα = (ba)1G . As αβ = βα by commutativity of R[G], we deduce that ab = ba. This shows that R is commutative.

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8 Linear Cellular Automata

Consider now the elements γ , η ∈ R[G] defined by γ := 1R g and η := 1R h. We then have γ η = 1R (gh) and ηγ = 1R (hg). As γ η = ηγ by commutativity of R[G], we conclude that gh = hg (here we use the fact that 1R /= 0R ). This shows that the group G is commutative. ■ Exercise 8.3 (The Augmentation Morphism)  Let G be a group and let R be a ring. Define the map ε : R[G] → R by ε(α) := g∈G α(g) for all α ∈ R[G]. (a) Show that ε is surjective. (b) Show that ε is a ring morphism. (c) Show that ε is a left R-module morphism. (d) Show that the kernel of ε is a free left R-module admitting as a basis the set of elements of the form g − 1G with g ∈ G and g /= 1G . Solution (a) The map ε is surjective since r = ε(r1G )

(8.2)

.

for all r ∈ R. (b) Let α, β ∈ R[G]. Write α=



.

αg g and β =

g∈G



βg g,

g∈G

where αg , βg ∈ R and αg = βg = 0 for all but finitely many g ∈ G. Then α+β =



.

(αg + βg )g,

g∈G

so that ε(α + β) =



.

(αg + βg ) =

g∈G



αg +

g∈G



βg = ε(α) + ε(β).

(8.3)

g∈G

On the other hand, we have 

αβ =

αg βh gh,

.

g,h∈G

and hence ε(αβ) =



.

g,h∈G

αg βh = (

 g∈G

αg )(



βh ) = ε(α) · ε(β).

h∈G

As ε(1R[G] ) = ε(1R · 1G ) = 1R by (8.2), this shows that ε is a ring morphism.

8.2 Exercises

523

(c) For all r ∈ R, we have that rα =



.

rαg g,

g∈G

so that ε(rα) =



.

rαg = r

g∈G



αg = rε(α).

g∈G

As we already know that ε(α + β) = ε(α) + ε(β) for all α, β ∈ R[G] by (8.3), this shows that ε is a left R-module morphism.  (d) Clearly g − 1G ∈ Ker(ε) forall g ∈ G. Suppose now that α = g∈G αg g ∈ Ker(ε). This means that α1G = − g/=1G αg . Therefore α=



.

g/=1G

=

 g/=1G

=



αg g + α1G 1G ⎛ αg g + ⎝−



⎞ αg ⎠ 1G

g/=1G

αg (g − 1G ).

g/=1G

This shows that the elements g − 1G with g ∈ G and g /= 1G form a free basis of the left R-module Ker(ε). ■ Comment The ring morphism ε is called the augmentation morphism on R[G]. Its kernel Ker(ε) ⊂ R[G] is called the augmentation ideal of R[G]. Exercise 8.4 (The Center of a Group Ring) The center of a ring A is the commutative subring B of A consisting of all the elements b ∈ A that satisfy ab = ba for all a ∈ A. Let G be a group and let R be a ring. Let β ∈ R[G]. Show that β is in the center of R[G] if and only if β is constant on each conjugacy class of G and takes all of its values in the center of R. Solution Suppose first that β ∈ R[G] is constant on each conjugacy class of G and takes all of its values in the center of R. Let α ∈ R[G]. For every g ∈ G, we have (βα)(g) =



.

β(gh−1 )α(h)

h∈g

=



h∈G

α(h)β(gh−1 )

(since β takes its values in the center of R).

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8 Linear Cellular Automata

Now, since β is constant on each conjugacy class of G, we have β(gh−1 ) = β(h−1 (gh−1 )h) = β(h−1 g) for all h ∈ G. Thus we obtain 

(βα)(g) =

.

α(h)β(h−1 g) = (αβ)(g)

h∈G

for all g ∈ G. This shows that βα = αβ for all α ∈ R[G]. Thus β is in the center of R[G]. Conversely, let β be an element in the center of R[G]. Suppose that g1 , g2 ∈ G are in the same conjugacy class of G. This means that there exists h ∈ G such that hg1 h−1 = g2 . Let r ∈ R and consider the element α ∈ R[G] taking the value r at h and the value 0R everywhere else. We then have (αβ)(hg1 ) = rβ(g1 )

.

and

(βα)(g2 h) = β(g2 )r.

As αβ = βα and hg1 = g2 h, we deduce that rβ(g1 ) = β(g2 )r.

.

(8.4)

Taking r = 1r in (8.4), we get β(g1 ) = β(g2 ). Therefore β is constant on each conjugacy class of G. On the other hand, equality (8.4) implies that rβ(g) = β(g)r for all r ∈ R and g ∈ G. This shows that β takes all its values in the center of R. ■ Comment If we adopt the notation Z(A) for the center of a ring A, Exercise 8.4 shows that Z(R[G]) is isomorphic, as a Z[R]-module, to the free Z[R]-module based on the set of conjugacy classes of G. Exercise 8.5 (Group Rings of ICC-Groups) Let G be a group and let R be a nonzero ring. Show that G is an ICC-group if and only if the center of R[G] is equal to the center of R. Solution Suppose first that G has the ICC-property. Let β be an element in the center of R[G]. By the result of Exercise 8.4, we know that β is constant on each conjugacy class of G and takes its values in the center of R. Let g ∈ G \ {1G }. As the conjugacy class of g is infinite while the support of β must be finite, we deduce that β(g) = 0R . Thus β = r1G with r in the center of R. Conversely, it is clear that every element in the center of R is also in the center of R[G]. This shows that the center of R[G] coincides with the center of R. To prove the converse implication, suppose now that G does not have the ICCproperty. This means that there exists g ∈ G \ {1G } whose conjugacy class is finite. Consider the element β ∈ R[G] taking the value 1R at every conjugate of G and the value 0R everywhere else. Then, by the result of Exercise 8.4, β is in the center of R[G]. However, β is not in R (here we use the fact that 0R /= 1R ). This shows that if the center of R[G] coincides with the center of R, then G must have the ICC-property. ■

8.2 Exercises

525

Comment Let H be a complex Hilbert space. A bounded operator on H is a continuous linear map u : H → H . Let B(H ) denote the complex vector space consisting of all bounded operators on H . With the adjoint operation and the composition of maps, B(H ) has the structure of a ∗-algebra. The weak topology on B(H ) is the coarsest topology on B(H ) for which all evaluation maps eh : B(H ) → C, h ∈ H , are continuous, where eh (u) := u(h) for all u ∈ B(H ). A von Neumann algebra over H is a ∗-subalgebra of B(H ) which is closed for the weak topology on B(H ) (see e.g. [Dix], [Sak]). A von Neumann algebra is called a factor if its center consists only of scalar operators. Factors are building blocks for von Neumann algebras and have been classified into three types by Murray and von Neumann [MurN1]. Let G be a group and let ℓ2 (G) := {f : G →  C such that g∈G |f (g)|2 < ∞} denote the Hilbert space of square-summable complex-valued functions on G. Viewing C[G] as a ∗-subalgebra of B(ℓ2 (G)) via left-convolution, the weak closure of C[G] in B(ℓ2 (G)) is a von Neumann algebra over ℓ2 (G) which is called the von Neumann algebra of the group G and it is denoted by N(G). It can be shown that N(G) is a factor if and only if the group G has the ICC-property [MurN2, Lemma 5.3] (see also [BekH, Proposition 7.A.1]). All von Neumann algebras of non-trivial ICC-groups are of type II1 in the Murrayvon Neumann classification of factors. Exercise 8.6 Let G be a group, let K be a field, and let d be a positive integer. Show that the K-algebras Matd (K[G]) and Matd (K)[G] are isomorphic. Solution For M ∈ Matd (K[G]) and 1 ≤ i, j ≤ d, denote by Mij ∈ K[G] the entry of M located in the i-th row and j -th column.  Consider the map ϕ : Matd (K[G]) → Matd (K)[G] defined by ϕ(M) := α = g∈G α(g)g, where the matrix α(g) ∈ Matd (K) is given by α(g)ij := Mij (g) for all 1 ≤ i, j ≤ d. Clearly ϕ is well defined, bijective, and satisfies ϕ(1Matd (K[G]) ) = 1Matd (K)[G] , ϕ(λM) = λϕ(M), and ϕ(M + N ) = ϕ(M) + ϕ(N ) for all λ ∈ K and M, N ∈ Matd (K[G]). To show that ϕ is a isomorphism of K-algebras, it remains only to check that ϕ(MN ) = ϕ(M)ϕ(N ) for all M, N ∈ Matd (K[G]). Write MN = P , ϕ(M) = α, ϕ(N ) = β, and ϕ(P ) = γ . We then have Pij =



.

Mik Nkj

1≤k≤d

for all 1 ≤ i, j ≤ d. Thus, for all g ∈ G, γ (g)ij = Pij (g) ⎛ ⎞  =⎝ Mik Nkj ⎠ (g)

.

1≤k≤d

=

 1≤k≤d

Mik Nkj (g)

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8 Linear Cellular Automata

⎛

⎞

 ⎜  ⎟ ⎜ Mik (h1 )Nkj (h2 )⎟ ⎝ ⎠

=

1≤k≤d

⎛ =

h1 ,h2 ∈G h1 h2 =g

⎞

 ⎜  ⎟ ⎜ ⎟. α(h ) β(h ) 1 ik 2 kj ⎝ ⎠ h1 ,h2 ∈G h1 h2 =g

1≤k≤d

We deduce that γ (g) =



α(h1 )β(h2 ) = (αβ)(g)

.

h1 ,h2 ∈G h1 h2 =g

for all g ∈ G, so that γ = αβ. This shows that ϕ(MN ) = ϕ(M)ϕ(N ) and completes the proof that ϕ is an isomorphism of K-algebras. ■ Exercise 8.7 Let G be a group and let S ⊂ G be a non-empty finite subset. Denote by the standard scalar product on R[G] ⊂ ℓ2 (G) and let ║ · ║ denote the associated norm. (a) Show that for all x ∈ R[G] and λ ∈ R, one has =

.

1  |x(g) − x(gs)|2 + λ║x║2 . 2 g∈G s∈S

(b) Show that if λ > 0 then the linear cellular automaton τ : R G → R G defined by τ := ΔS + λ IdRG is both pre-injective and surjective. Solution  (a) Recall that given x, y ∈ R[G] one has := g∈G x(g)y(g). For all x ∈ R[G], we then have 

|x(g) − x(gs)|2 =

g∈G s∈S



 x(g)2 + x(gs)2 − 2x(g)x(gs)

g∈G s∈S

=



x(g)2 +

g∈G s∈S .

=





x(gs)2 − 2

g∈G s∈S

x(g)2 +

s∈S g∈G

= 2|S| · ║x║2 − 2



g∈G s∈S

x(g)x(gs)

g∈G s∈S

x(gs)2 − 2

s∈S g∈G





x(g)x(gs)



g∈G s∈S

x(g)x(gs)

8.2 Exercises

527

so that .

 1  |x(g) − x(gs)|2 = |S| · ║x║2 − x(g)x(gs). 2 g∈G s∈S

g∈G s∈S

Moreover, =



x(g) (ΔS (x)(g))

g∈G

.

=



 x(g) |S|x(g) −

g∈G

= |S| · ║x║2 −





 x(gs)

s∈S

x(g)x(gs).

g∈G s∈S

We deduce that = + λ║x║2 =

.

1  |x(g) − x(gs)|2 + λ║x║2 . 2 g∈G s∈S

(b) Let λ > 0. Let x ∈ R[G] such that τ (x) = 0. Then by using (a), we get 0 = =

.

1  |x(g) − x(gs)|2 + λ║x║2 . 2 g∈G s∈S

This implies ║x║ = 0 and hence x = 0. Thus, τ is pre-injective. In order to prove that τ is surjective, we distinguish two cases depending on the amenability or non-amenability of the group G. If G is amenable, the Linear Garden of Eden theorem ensures that τ , being a pre-injective linear cellular automaton over a finite-dimensional alphabet space, is surjective. Suppose now that G is not amenable. Then it is a consequence of the Kesten-Day Theorem (cf. [CAG, (2) (2) (Theorem 6.12.9]) that σ (ΔS ) ⊂ (0, |S|], where ΔS : ℓ2 (G) → ℓ2 (G) is the ℓ2 -Laplace operator associated with S and σ (·) denotes the ℓ2 -spectrum. It (2) follows that σ (ΔS + λ Idℓ2 (G) ) ⊂ (λ, |S| + λ]. Since λ > 0, we deduce that (2)

(2)

0 ∈ / σ (ΔS + λ Idℓ2 (G) ), that is, ΔS + λ Idℓ2 (G) is bijective. As a consequence,  

(2) R[G] ⊂ ℓ2 (G) = ΔS + λ Idℓ2 (G) (ℓ2 (G)) ⊂ ΔS + λ IdRG (RG ) = τ (RG ). Using the fact taht R[G], the subspace of configurations with finite support, is dense in RG for the prodiscrete topology, and the fact that τ (RG ) is closed in RG for the prodiscrete topology by the closed image property for linear cellular automata, we conclude that RG = τ (RG ). This proves surjectivity of τ also in this case. ■

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Comment The pre-injectivity of τ : RG → RG can be also established via the maximum principle as follows. Suppose that x ∈ R[G] satisfies τ (x) = 0. Let g 0 ∈ G such that |x(g0 )| = maxg∈G |x(g)|. As 0 = τ (x)(g0 ) = (|S| + λ)x(g0 ) −  s∈S x(g0 s), by using the triangle inequality we get (|S| + λ) · |x(g0 )| = |(|S| + λ)x(g0 )| ≤



.

|x(g0 s)| ≤ |S| · |x(g0 )|.

s∈S

We deduce that λ|x(g0 )| ≤ 0. As λ > 0, we have x(g0 ) = 0. It follows that x = 0. The surjectivity result can be generalized in the following setting. Consider a connected locally finite simplicial graph Γ with vertex set V and the associated combinatorial Laplacian ΔΓ : RV → RV defined by ΔΓ (f )(v) := f (v) − 1  V u∼v f (u) for all f ∈ R and v ∈ V , where the notation u ∼ v indicates deg(v) that the vertices u, v ∈ V are adjacent and deg(v) := |{u ∈ V : u ∼ v}| denotes the degree of the vertex v. In [CecCD], it is shown that if the graph Γ is infinite and λ : V → [0, +∞) is a (not necessarily constant) function on its vertex set, then the drifted Laplacian L := ΔΓ + λ IdRV : RV → RV is surjective. Note that if Γ = CS (G) is the Cayley graph of a finitely generated group G with respect to a finite symmetric generating subset S ⊂ G not containing 1G , then Γ is regular with deg(v) = |S| for all v ∈ V = G and |S|ΔΓ = ΔS . Exercise 8.8 Recall that a monoid M is said to be left-cancellative (resp. rightcancellative) if given r, s, t ∈ M such that tr = ts (resp. rt = st) then r = s. A monoid which is both left and right-cancellative is said to be cancellative. (a) Show that every finite left-cancellative (resp. finite right-cancellative, resp. finite cancellative) monoid is a group. (b) Let G be a group and let S ⊂ G be a subset. Show that the following conditions are equivalent: (i) the subgroup H ⊂ G generated by S is infinite; (ii) the submonoid M ⊂ G generated by S is infinite. Solution (a) Let M be a finite left-cancellative monoid. Let x ∈ M and consider the set {x n : n ∈ N} ⊂ M. Since M is finite, there exist positive integers i < j such that x i = x j . This implies x i · x j −i = x j = x i = x i · 1M and hence x j −i = 1M since M is left-cancellative. Note that j − i − 1 ≥ 0. Setting y := x j −i−1 , we then have xy = yx = x j −i = 1M . This shows that every element x ∈ M is invertible and hence that M is a group. The proof for right-cancellative monoids is, mutatis mutandis, identical. Alternatively, if M is a finite right-cancellative monoid, then its opposite monoid is a finite left-cancellative monoid and, by the previous argument, it is a group. Since the opposite monoid of a group is itself a group, we deduce that M is a group. The remaining statement, namely that any finite cancellative monoid is a group, is a weakening of the previous ones. (b) Suppose that M is finite. Since groups are cancellative monoids and the property of being cancellative is inherited by submonoids, we deduce from (a) that

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M is a subgroup of G. It follows that H = M is finite. This shows the implication (i) =⇒ (ii). The converse implication is trivial since M ⊂ H . ■ Exercise 8.9 Let G be a group and let S be a non-empty finite subset of G. Let G G ΔC S : C → C denote the associated complex discrete Laplacian. Show that the following conditions are equivalent: (i) (ii) (iii) (iv)

ΔC S is surjective; the subgroup of G generated by S is infinite; the submonoid of G generated by S is infinite; ΔC S is pre-injective.

Solution The map CG → RG × RG , given by x |→ (x0 , x1 ), where x0 , x1 ∈ RG are defined by setting x0 (g) := (x(g)) = (x(g) + x(g))/2 and x1 (g) := (x(g)) = (x(g) − x(g))/(2i) for all x ∈ CG and g ∈ G, is an R-vector space R R isomorphism. Moreover, it is straightforward that ΔC S = ΔS × ΔS , in the sense that C R G (ΔS (x))k = ΔS (xk ) for all x ∈ C and k = 0, 1. As a consequence, we have that R ΔC S is surjective if and only if ΔS is surjective. Moreover, since two configurations x, y ∈ CG are almost equal if and only if xk , yk ∈ RG are almost equal for k = R 0, 1, we also deduce that ΔC S is pre-injective if and only if ΔS is pre-injective. The equivalence of (i), (ii), and (iv) then follows from the Garden of Eden Theorem for real discrete Laplacians. Finally, the equivalence of (ii) and (iii) directly follows from Exercise 8.8. ■ Exercise 8.10  Let (Ri )i∈I be a family of directly finite rings. Show that the product ring P := i∈I Ri is directly finite. Solution Let a = (ai )i∈I , b = (bi )i∈I ∈ P such that ab = 1P . As ab = (ai bi )i∈I and 1P = (1Ri )i∈I , we have ai bi = 1Ri for all i ∈ I . Since Ri is directly finite, this implies bi ai = 1Ri for all i ∈ I . Therefore ba = (bi ai )i∈I = (1Ri )i∈I = 1P . This shows that P is directly finite. ■ Exercise 8.11 Show that every subring of a directly finite (resp. stably finite) ring is itself directly finite (resp. stably finite). Solution Let R be a ring and let S be a subring of R. Suppose first that R is directly finite. Let a, b ∈ S such that ab = 1S . As S ⊂ R, 1S = 1R , and R is directly finite, this implies ba = 1R = 1S . Thus, S is directly finite. Suppose now that R is stably finite. This means that the ring Matd (R) is directly finite for every d ≥ 1. As Matd (S) is a subring of Matd (R), we deduce from the first part of the proof that Matd (S) is directly finite for every d ≥ 1. This shows that S is stably finite. ■ Exercise 8.12 (Hopfian Modules) Let R be a ring and let M be a left R-module. One says that the module M is Hopfian if every surjective endomorphism of M is injective. One says that the module M is co-Hopfian if every injective

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endomorphism of M is surjective. Show that if the module M is either Hopfian or co-Hopfian, then the ring EndR (M) is directly finite. Solution Let a, b ∈ EndR (M) such that ab = a ◦ b = 1EndR (M) = IdM . Observe that this implies that a is surjective and b is injective. Suppose first that M is Hopfian. It then follows that a is bijective and hence an invertible element in EndK (V ). This gives us ba = a −1 (ab)a = a −1 a = 1EndR (M) . This shows that the ring EndR (M) is directly finite. In the case when M is co-Hopfian, we get that b is bijective and hence invertible in EndR (M). Writing ba = b(ab)b−1 = bb−1 = 1EndR (M) , we see that EndR (M) is directly finite also in this case. ■ Comment In this exercise as well as in other exercises below, we have restricted ourselves to left modules. All definitions, results, and proofs have their obvious analogues for right modules. Actually, a right module over a ring R is the same thing as a left module over the opposite ring R opp . Exercise 8.13 (Noetherian and Artinian Modules) Let R be a ring and let M be a left R-module. One says that the module M is Noetherian if its submodules satisfy the ascending chain condition, i.e., every increasing sequence N1 ⊂ N2 ⊂ . . .

.

of submodules of M stabilizes (i.e. there is an integer i0 ≥ 1 such that Ni = Ni0 for all i ≥ i0 ). One says that the module M is Artinian if its submodules satisfy the descending chain condition, i.e., every decreasing sequence N1 ⊃ N2 ⊃ . . .

.

of submodules of M stabilizes (i.e. there is an integer i0 ≥ 1 such that Ni = Ni0 for all i ≥ i0 ). Show that if the module M is Noetherian (resp. Artinian), then it is Hopfian (resp. co-Hopfian). Solution Suppose first that M is Noetherian and let f : M → M be a surjective endomorphism of M. As (ker(f i ))i≥1 is an increasing sequence of submodules of M, there is an integer i0 ≥ 1 such that ker(f i0 ) = ker(f i0 +1 ). Let x ∈ ker(f ). Since f i0 = f ◦ f ◦ · · · ◦ f is surjective, there exists y ∈ M such that x = f i0 (y). We have 0 = f (x) = f i0 +1 (y) and hence y ∈ ker(f i0 +1 ) = ker(f i0 ), so that x = f i0 (y) = 0. It follows that f is injective. This shows that M is Hopfian. Suppose now that M is Artinian and let f : M → M be an injective endomorphism of M. As (Im(f i ))i≥1 is a decreasing sequence of submodules of M, there is an integer i0 ≥ 1 such that Im(f i0 ) = Im(f i0 +1 ). Since f is injective, it follows ■ from Exercise 3.1 that f is surjective. This shows that M is co-Hopfian. Exercise 8.14 (Projective Modules) Let R be a ring and let P be a left R-module. One says that the module P is projective if for every homomorphism of left R-

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→ modules f : P → M and any surjective homomorphism of left R-modules g : M  such that f = g ◦h. M, there exists a homomorphism of left R-modules h : P → M (a) Show that every free left R-module is projective. (b) Suppose that the module P is projective. Show that P is Hopfian if and only if the ring EndR (P ) is directly finite. Solution (a) Let F be a free left R-module with basis B ⊂ F . Let f : F → M and  → M be homomorphisms of left R-modules with g surjective. Since g is g: M  such that g( surjective, for every b ∈ B, we can find m b ∈ M mb ) = f (b). As B is  a basis for F , there exists a unique homomorphism of left R-modules h : F → M such that h(b) = m b for all b ∈ B. We have f = g ◦ h since f and g ◦ h coincide on B, and B generates F . This shows that F is projective. (b) Necessity follows from Exercise (8.12). Conversely, suppose that the ring EndR (P ) is directly finite and let g : P → P be a surjective module endomorphism. Since the module P is projective, there exists a module endomorphism h : P → P such that g ◦ h = IdP . As the ring EndR (P ) is directly finite by our hypothesis, this implies that h ◦ g = IdP and hence that g is injective. This shows that P is Hopfian. ■ Exercise 8.15 Let R be a ring and let d ≥ 1 be an integer. Equip R d with its natural structure of left R-module. Show that R d is Hopfian if and only if the ring Matd (R) is directly finite. Solution Since R d is a free left R-module, it is projective by Exercise 8.14.(a). Thus, we deduce from Exercise 8.14.(b) that R d is Hopfian if and only if its endomorphism ring EndR (R d ) is directly finite. For every matrix M ∈ Matd (R), the map fM : R d → R d , defined by fM (v) := vM for all v = (v1 , . . . , vd ) ∈ R d , is in EndR (R d ). Moreover, the map M |→ fM is a ring isomorphism from Matd (R) onto the opposite ring of EndR (R d ). As a ring is directly finite if and only if its opposite ring is, we conclude that R d is Hopfian if and only if the ring Matd (R) is directly finite. ■ Exercise 8.16 (Injective Modules) Let R be a ring and let Q be a left R-module. One says that the module Q is injective if for every homomorphism f : M → Q and any injective homomorphism g : M → N of left R-modules, there exists a homomorphism h : N → Q such that f = h ◦ g. Suppose that the module Q is injective. Show that Q is co-Hopfian if and only if the ring EndR (Q) is directly finite. Solution Necessity follows from Exercise (8.12). Conversely, suppose that the ring EndR (Q) is directly finite and let g : Q → Q be an injective module endomorphism. Since the module Q is injective, there exists a module endomorphism h : Q → Q such that h ◦ g = IdQ . As the ring EndR (Q) is directly finite by our hypothesis, this implies that g ◦ h = IdQ and hence that g is surjective. This shows that Q is co-Hopfian. ■

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Exercise 8.17 Let M be a left module over a ring R and let N be a submodule of M. Show that M is Noetherian (resp. Artinian) if and only if the modules N and M/N are both Noetherian (resp. Artinian). Solution Set Q := M/N and denote by ρ : M → Q the quotient map. Suppose first that M is Noetherian (resp. Artinian). As every submodule of N is also a submodule of M, it is clear that N is Noetherian (resp. Artinian). Let (Qn )n∈N be an increasing (resp. a decreasing) sequence of submodules of Q. Then (ρ −1 (Qn ))n∈N is an increasing (resp. a decreasing) sequence of submodules of M. Since M is Noetherian (resp. Artinian), there exists n0 ∈ N such that ρ −1 (Qn ) = ρ −1 (Qn0 ) for all n ≥ n0 . As Qn = ρ(ρ −1 (Qn )), it follows that Qn = Qn0 for all n ≥ n0 . This shows that Q is Noetherian (resp. Artinian). Suppose now that N and Q are Noetherian (resp. Artinian). Let (Mn )n∈N be an increasing (resp. a decreasing) sequence of submodules of M. Then (ρ(Mn ))n∈N is an increasing (resp. a decreasing) sequence of submodules of Q. On the other hand, (Mn ∩ N )n∈N is an increasing (resp. a decreasing) sequence of submodules of N. Since Q and N are both Noetherian (resp. Artinian), there exists n0 ∈ N such that ρ(Mn ) = ρ(Mn0 ) and Mn ∩ N = Mn0 ∩ N for all n ≥ n0 . Let n ≥ n0 . In the Noetherian case, take x ∈ Mn . As ρ(Mn ) = ρ(Mn0 ) and N = ker(ρ), there exist y ∈ Mn0 and z ∈ N such that x = y+z. This implies z = x−y ∈ Mn ∩N = Mn0 ∩N so that x ∈ Mn0 , showing Mn = Mn0 . In the Artinian case, the proof is similar but we now take x ∈ Mn0 . As ρ(Mn0 ) = ρ(Mn ) and N = ker(ρ), there exist y ∈ Mn and z ∈ N such that x = y + z. This implies z = x − y ∈ Mn0 ∩ N = Mn ∩ N so that x ∈ Mn , showing Mn0 = Mn . This shows that M is Noetherian (resp. Artinian). ■ Exercise 8.18 (Noetherian Rings) One says that a ring R is left Noetherian if R is Noetherian as a left module over itself. (a) Let R be a left Noetherian ring. Show that R d is Noetherian as a left R-module for each integer d ≥ 1. (b) Show that every left Noetherian ring is stably finite. Solution (a) This immediately follows from Exercise 8.17 by induction on d. Indeed, we have a short exact sequence of left R modules 0 → R d → R d+1 → R → 0 for every d ≥ 1, where the map R d → R d+1 is given by (r1 , . . . , rd ) |→ (r1 , . . . , rd , 0R ) and the map R d+1 → R by (r1 , . . . , rd+1 ) |→ rd+1 . (b) Suppose that the ring R is left Noetherian. Let d ≥ 1 be an integer. Then, the left R-module R d is Noetherian by (a). This implies that R d is Hopfian by Exercise 8.13 and hence that the ring Matd (R) is directly finite by Exercise 8.15. As d was arbitrary, this shows that R is stably finite. ■ Comment Similarly, a ring R is said to be left Artinian if R is Artinian as a left module over itself.

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When a ring R is viewed as a left module over itself, the submodules of R are precisely the left ideals of R. Thus, a ring is left Noetherian (resp. left Artinian) if and only if its left ideals satisfy the ascending (resp. descending) chain condition. A highly non-trivial result in non-commutative ring theory is the HopkinsLevitzki theorem, which says that every left Artinian ring is also left Noetherian (see [Lam2] for a proof). It implies that every left Artinian ring is stably finite. Exercise 8.19 Show that every division ring is stably finite. Solution Let R be a division ring and let d ≥ 1 be an integer. Equip R d with its natural left R-module structure and let f ∈ EndR (R d ) be a surjective endomorphism. By linear algebra, we have dimR (ker(f )) = dimR (R d ) − dimR (Im(f )) = d − d = 0, so that f is injective. Thus, R d is Hopfian. By applying Exercise 8.15, we deduce that the ring Matd (R) is directly finite. This shows that R is stably finite. ■ Comment This can be also directly deduced from Exercise 8.18.(b). Indeed, any division ring R is left Noetherian since the only left ideals of R are {0R } and R itself. Exercise 8.20 (URP-rings) One says that a ring R has the unique rank property (URP for short), or that R is a URP-ring, if it satisfies the following condition: if m and n are positive integers such that R m and R n are isomorphic as left R-modules, then one has m = n. (a) Show that every stably finite nonzero ring is a URP-ring. (b) Show that all finite nonzero rings, all commutative nonzero rings, and all division rings are URP-rings. (c) Show that a ring R is a URP-ring if and only if it satisfies the following condition: if m and n are positive integers such that there exist matrices A ∈ Matm,n (R) and B ∈ Matn,m (R) such that AB = Im and BA = In , then one has m = n. (d) Show that a ring R is a URP-ring if and only if it satisfies the following condition: if m and n are positive integers such that R m and R n are isomorphic as right R-modules, then one has m = n. (e) Suppose that R and S are rings such that there exists a ring morphism ϕ : R → S. Show that if S is a URP-ring then R is also a URP-ring. Solution (a) Let R be a stably finite nonzero ring. Recall that this means that the matrix ring Matd (R) is directly finite for every integer d ≥ 1. Let m and n be positive integers such that R m and R n are isomorphic as left R-modules. By symmetry, we may assume m ≤ n. Let ϕ : R m → R n be an isomorphism of left R-modules. Consider the projection map π : R n = R m × R n−m → R m . Then π ◦ ϕ : R m → R m is a surjective endomorphism of the left R-module R m . As R m is Hopfian by Exercise 8.15, we deduce that π ◦ ϕ is injective. As ϕ is surjective, this implies that π is injective, so that we must have m = n since R is a nonzero ring. This shows that R has the unique rank property.

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(b) This immediately follows from (a) since all finite rings, all commutative rings, and all division rings are stably finite (see [CAG, Proposition 8.15.1], [CAG, Proposition 8.15.2], and Exercise 8.19). Of course, it is also possible to give a direct proof of URP in each case without mentioning stable finiteness at all. (c) Let f : R m → R n be a left R-module morphism. Let B ∈ Matn,m (R) denote the matrix of f with respect to the canonical bases of R m and R n . By definition, the morphism f is an isomorphism if and only if there exists a left R-module morphism g : R n → R m such that g ◦ f = IdR m and f ◦ g = IdR n . By basic linear algebra, this is equivalent to the existence of a matrix A ∈ Matm,n (R) such that AB = Im and BA = In . As, conversely, every matrix B ∈ Matn,m (R) is the matrix of some left R-module morphism f : R m → R n with respect to the canonical bases of R m and R n , this shows the equivalence between URP and the condition stated in terms of matrices. (d) It suffices to observe that, by taking matrices with respect to the canonical bases of R m and R n (viewed as right R-modules), the condition stated in (d) is equivalent to the one stated in (c) in terms of matrices. (e) Suppose that S has URP. Let A ∈ Matm,n (R) and B ∈ Matn,m (R) such that AB = Im and BA = In . Consider the matrix AS ∈ Matm,n (S) (resp. BS ∈ Matn,m (S)) obtained by replacing each entry of A (resp. B) by its image under ϕ. We clearly have AS BS = Im and BS AS = In . As S has URP, we deduce that m = n. This shows that R has URP. ■ Comment URP-rings are called IBN-rings in [Lam1]. Exercise 8.21 Let K be a field and let V be an infinite-dimensional vector space over K. Show that the ring R := EndK (V ) does not have the unique rank property. Solution We first observe that V and V × V are isomorphic as vector spaces over K. Indeed, if B ⊂ V is a base for V , then B × B ⊂ V × V is a base for V × V . As B and B × B have the same cardinality (since B is infinite), we deduce that the vector spaces V and V × V are isomorphic. Using this observation, we deduce that the set consisting of all K-linear maps f : V × V → V , with its natural structure of a left R-module, is isomorphic to both R × R = R 2 and R. Thus the left R-modules R 2 and R are isomorphic, showing that the ring R does not have the unique-rank property. ■ Exercise 8.22 (Monoid Rings) The notion of a group ring extends in an obvious way to monoids. More precisely, given a ring R and a monoid M, one defines the monoid ring R[M] as being the set of all maps with finite support α : M → R equipped with pointwise addition and the convolution product defined by (αβ)(m) :=



.

α(m1 )β(m2 )

m1 ,m2 ∈M m1 m2 =m

for all α, β ∈ R[M] and m ∈ M. One verifies that this defines a ring structure on R[M] as in the case of group rings.

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(a) Let R be a ring and let M be a monoid. Show that if R has URP then R[M] has URP. (b) Consider the symmetric monoid M := Map(N) consisting of all maps f : N → N with the composition of maps as the monoid operation. Show that if R is any URP-ring , then the ring R[M] has URP but is not directly finite. Solution (a) This immediately follows from Exercise 8.20.(e) using the fact that the map ε : R[M] → R, defined by ε(α) := m∈M α(m), is a ring morphism (the proof of this fact is as in Exercise 8.3.(b)). (b) Let R be a ring having URP. The fact that R[M] has URP follows from (a). Consider the elements p, q ∈ Map(N) respectively defined by  p(n) :=

.

n−1

if n ≥ 1

0

if n = 0

and

q(n) := n + 1 for all n ∈ N.

We have qp /= pq = IdN = 1M since qp(0) = 1. Denoting, for each m ∈ M, by δm ∈ R[M] the element defined by δm (m) := 1R and δm (m' ) := 0R if m' ∈ M \{m}, we deduce that δp δq = δpq = δ1M = 1R[M] while δq δp = δqp /= 1R[M] . This shows that R[M] is not directly finite. ■ Comment Another example of a URP-ring that is not directly finite is described in [Lam1, Proposition 1.8]. The submonoid B ⊂ Map(N) generated by the maps p and q defined in (b) is called the bicyclic monoid (see [CecC5] and the references therein). Exercise 8.23 (von Neumann Regular Rings) Let R be a ring. An element a ∈ R is called von Neumann regular if there exists an element x ∈ R such that a = axa. The ring R is said to be von Neumann regular if every element of R is regular. (a) Show that every idempotent element in R is von Neumann regular. (b) Show that 0R and 1R are von Neumann regular. (c) Show that every invertible element in R is von Neumann regular. (d) Show that every division ring is von Neumann regular. (e) Prove that the ring Z is not von Neumann regular. (f) Is every finite commutative ring von Neumann regular? (g) Let R be a ring and let a ∈ R. Show that the following conditions are equivalent: (1) a is von Neumann regular; (2) there exists an idempotent element e ∈ R such that Ra = Re; (3) there exists an idempotent element e ∈ R such that aR = eR. (h) Let V be a vector space over a field K and let R := EndK (V ) denote the endomorphism ring of V . Show that the ring R is von Neumann regular. (i) Is every von Neumann regular ring directly finite? Solution (a) If a ∈ R is idempotent, i.e., a 2 = a, then a is von Neumann regular since a = axa for x = a.

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(b) This follows from (a) since 0R and 1R are idempotent elements in R. (c) If a ∈ R is invertible, then a is von Neumann regular since a = axa for x = a −1 . (d) This immediately follows from (b) and (c) since, in a division ring, every element is either zero or invertible. (e) The element 2 is not von Neumann regular in Z. Indeed, for a = 2, there is no x ∈ Z such that a = axa, since this would give 2 = 4x. Therefore the ring Z is not von Neumann regular. (f) The answer is no. Consider for instance the ring R := Z/4Z = {0, 1, 2, 3}. Then a := 2 satisfies a 2 = 0R . As a /= 0R , there is no x ∈ R such that a = xa 2 . Thus a is not von Neumann regular. This shows that R is not von Neumann regular. (g) Suppose that a ∈ R is von Neumann regular. Then we can write a = axa for some x ∈ R. Let e := xa. Then e2 = xaxa = xa = e. Therefore e is an idempotent. We have Ra = Raxa = Rae ⊂ Re and Re = Rxa ⊂ Ra, so that Ra = Re. This shows that (1) implies (2). Conversely, suppose that there is an idempotent element e ∈ R such that Ra = Re. Then, as e = 1R e ∈ Re = Ra, there exists x ∈ R such that e = xa. On the other hand, as a = 1R a ∈ Ra = Re, there exists y ∈ R such that a = ye. Since e is idempotent, this gives us a = ye = ye2 = ae = axa. Thus a is von Neumann regular. This completes the proof of the equivalence of conditions (1) and (2). From the equivalence of (1) and (2), we also deduce the equivalence of (1) and (3) since it is clear that a is von Neumann regular in R if and only if it is von Neumann regular in the opposite ring R op . (h) Let f ∈ R. Thus f : V → V is K-linear. Let W be a vector subspace of V such that V = W ⊕ Ker(f ). Then f induces by restriction an isomorphism of Kvector spaces g : W → Im(f ). Consider the inverse isomorphism g −1 : Im(f ) → W and extend g −1 to a K-linear map h : V → V . We then have f = f ◦ h ◦ f . This shows that R is von Neumann regular. (i) The answer is no. Let V be an infinite-dimensional vector space over a field K and consider its endomorphism ring R := EndK (V ). Then R is von Neumann regular by (h). However R is not directly finite (cf. [CAG, Example 8.15.3]). Indeed, since V is infinite-dimensional, there exists a vector subspace W  V that is isomorphic to V . Let ϕ : W → V be an isomorphism of K-vector spaces and let f : V → V be a K-linear map extending ϕ. Let g := ι ◦ ϕ −1 , where ϕ −1 : V → W is the inverse of ϕ and ι : W → V is the inclusion map. Then f, g ∈ R satisfy f ◦ g = IdV = 1R . However, g ◦ f /= 1R since Im(g) = W so that g is not surjective. This shows that the ring R is not directly finite. ■ Comment The class of von Neumann regular rings was introduced by von Neumann [vNeu] who simply called them regular rings. A comprehensive source on von Neumann regular rings is Goodearl’s monograph [Goo]. The result of Exercise 8.23.(g) shows that a ring R is von Neumann regular if and only if every principal left (resp. right) ideal of R can be generated by an idempotent. Actually, it can be proved that a ring R is von Neumann regular if and only if every finitely generated left (resp. right) ideal can be generated by an idempotent (see e.g. [Goo,

8.2 Exercises

537

Theorem 1.1]). There are many other characterizations of von Neumann regular rings. One of them says that a ring R is von Neumann regular if and only if it is absolutely flat, i.e., every left (resp. right) R-module is flat. Exercise 8.24 (Unit-Regular Rings) Let R be a ring. An element a ∈ R is called unit-regular if there exists an invertible element u ∈ R such that a = aua. The ring R is said to be unit-regular if every element of R is unit-regular. (a) Let R be a ring. Show that every unit-regular element of R is von Neumann regular (cf. Exercise 8.23). (b) Show that every unit-regular ring is von Neumann regular. (c) Let R be a ring. Show that the following elements are all unit-regular: 0R , any invertible element in R, any idempotent element in R. (d) Show that every division ring is unit-regular. (e) Let R be a ring and let a be a nonzero unit-regular element in R. Suppose that a is not invertible in R. Show that a is a zero-divisor. (f) Let R be a nonzero unit-regular ring without zero-divisors. Show that R is a division ring. (g) Let R be a ring and a ∈ R. Show that the following conditions are equivalent: (1) a is unit-regular; (2) there exists an invertible element u ∈ R and an idempotent element e ∈ R such that a = ue; (3) there exists an invertible element u ∈ R and an idempotent element e ∈ R such that a = eu. (h) Let V be a finite-dimensional vector space over a field K. Let R := EndK (V ) denote the ring of endomorphisms of V . Show that the ring R is unit-regular. (i) Let K be a field and let d ≥ 1 be an integer. Show that the ring Matd (K) of d × d matrices with entries in K is unit-regular. (j) Show that every unit-regular ring is directly finite. (k) Is every directly finite ring unit-regular? (l) Is every von Neumann regular ring unit-regular? Solution (a) This immediately follows from the definitions. (b) This immediately follows from (a). (c) If a = 0R , then a = aua for u = 1R . As 1R is invertible, this shows that 0R is unit-regular. If a ∈ R is invertible, then we can write a = aua with u = a −1 invertible. This shows that a is unit-regular. If a ∈ R is idempotent, we can write a = aua with u = 1R . As 1R is invertible, this implies that a is unit-regular. (d) This immediately follows from (c) since, in a division ring, every element is either zero or invertible. (e) Since a is unit-regular, there exists an invertible element u ∈ R such that a = aua. This gives us a(ua − 1R ) = 0R . We have ua − 1R /= 0R since otherwise we would have a = u−1 and hence a would be invertible. Therefore a is a zerodivisor. (f) It follows from (e) that every element of R is either zero or invertible. Thus R is a division ring.

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8 Linear Cellular Automata

(g) Suppose that a ∈ R is unit-regular. Then we can vrite a = aua for some u ∈ R invertible. We then have (ua)2 = uaua = ua. Therefore e := ua is an idempotent element. As a = u−1 (ua) = u−1 e and u−1 is invertible, this shows that (1) implies (2). Suppose now that a = ue with u ∈ R invertible and e ∈ R idempotent. Then we can write a = (ueu−1 )u. As ueu−1 is an idempotent, this shows that (2) implies (3). Finally, to show that (3) implies (1), suppose that a ∈ R satisfies a = eu with u invertible and e idempotent. Then we can write a = eu = e2 u = (eu)u−1 (eu) = au−1 a. This shows that a is unit-regular and completes the proof. (h) (compare with the solution of Exercise 8.23.(h)) Let f ∈ R. Thus f : V → V is K-linear. Let W be a vector subspace of V such that V = W ⊕ Ker(f ). Then f induces by restriction an isomorphism of K-vector spaces g : W → Im(f ). Consider the inverse isomorphism g −1 : Im(f ) → W . Since the vector space V is assumed to be finite-dimensional, we can extend g −1 to a K-linear isomorphism h : V → V . We then have f = f ◦ h ◦ f . As h is invertible in R, this shows that R is unit-regular. (i) This immediately follows from (g) since the ring Matd (K) is isomorphic to the ring EndK (K d ). (j) Let R be a unit-regular ring and let a, b ∈ R such that ab = 1R . Since a is unit-regular, there exists an invertible element u ∈ R such that a = aua. This yields 1R = ab = auab = au. Therefore a = u−1 is invertible. As ab = 1R ,we deduce that b = a −1 so that ba = 1R . This shows that R is directly finite. (k) The answer is no. The ring Z is directly finite since it is commutative. However, it is not unit-regular, not even von Neumann regular (see Exercise 8.23). (l) The answer is no. Consider an infinite-dimensional vector space V over a field K and its endomorphism ring R := EndK (V ). We know from Exercise 8.23.(h) that R is von Neumann regular. However, the ring R is not unit-regular since it is not directly finite (see the solution of Exercise 8.23.(i)) and every unit-regular ring is directly finite by (j). ■ Comment The class of unit-regular rings was introduced by Ehrlich in [Ehr1]. From Exercise 8.24.(d), it follows that a ring R is unit-regular if and only if every left (resp. every right) principal ideal of R is generated by an idempotent element. It is mentioned in [Ehr2, p. 89] that G. Bergman has constructed a directly finite von Neumann regular ring which is not unit-regular. On the other hand, it is known that every von Neumann regular commutative ring is unit-regular (see [GilH, Lemma 10]). Exercise 8.25 Let K be a field. Show that every finite-dimensional K-algebra is stably finite. Solution Let R be a finite-dimensional K-algebra. Suppose that a, b ∈ R satisfy ab = 1R . Consider the K-linear map ϕ : R → R defined by ϕ(r) := br for all r ∈ R. Observe that ϕ is injective since ϕ(r) = 0R implies br = 0R and hence r = 1R r = abr = 0R . As R is a finite-dimensional vector space over K, we

8.2 Exercises

539

deduce that ϕ is surjective. Thus, there exists r0 ∈ R such that br0 = 1R . Since r0 = 1R r0 = abr0 = a1R = a, we have ba = 1R . This shows that R is directly finite. For every integer d ≥ 1, the K-algebra Matd (R) has finite dimension 2

.

dimK (Matd (R)) = (dimK (R))d < ∞.

By applying the first part of the proof, we deduce that Matd (R) is directly finite. This shows that R is stably finite. ■ Exercise 8.26 Let G be an amenable group and let F be a right Følner net for G. Let V be a finite-dimensional vector space over some field K. Suppose that X and Y are vector subspaces of V G . Show that one has .

mdimF (X + Y ) ≤ mdimF (X) + mdimF (Y ).

(8.5)

Solution Suppose that F = (Fj )j ∈J and denote by πj : V G → V Fj the projection map. As πj is linear, we have πj (X + Y ) ⊂ πj (X) + πj (Y ) and hence .

dim(πj (X + Y )) ≤ dim(πj (X) + πj (Y )) ≤ dim(πj (X)) + dim(πj (Y ))

for all j ∈ J . It follows that .

mdimF (X + Y ) = lim sup j

≤ lim sup j

≤ lim sup j

dim(πj (X + Y )) |Fj | dim(πj (X)) + dim(πj (Y )) |Fj | dim(πj (X)) dim(πj (Y )) + lim sup |Fj | |Fj | j

= mdimF (X) + mdimF (Y ). This shows (8.5).

■

Exercise 8.27 Prove that every finite group is L-surjunctive by using the characterization of L-surjunctivity in terms of stable finiteness of group rings. Solution Let G be a finite group and let K be a field. The group algebra K[G] has finite dimension dimK (K[G]) = |G|. It then follows from Exercise 8.25 that K[G] is stably finite. Since the L-surjunctivity of a group is equivalent to the stable finiteness of its group algebras (cf. [CAG, Corollary 8.15.6]), we conclude that G is L-surjunctive. ■ Comment More directly, the L-surjunctivity of finite groups follows from the fact that every injective endomorphism of a finite-dimensional vector space is surjective

540

8 Linear Cellular Automata

(see [CAG, Section 8.14]). It can also be deduced from the linear version of the Garden of Eden theorem [CAG, Theorem 8.9.6] (since all finite groups are amenable and injective cellular automata are pre-injective) or from the linear version of the Gromov-Weiss theorem [CAG, Theorem 8.14.4] (since all finite groups are sofic). Exercise 8.28 Prove that every abelian group is L-surjunctive by using the characterization of L-surjunctivity in terms of stable finiteness of group rings. Solution Let G be an abelian group and let K be a field. Then the group ring K[G] is commutative. Since every commutative ring is stably finite (cf. [CAG, Proposition 8.15.2]) and the L-surjunctivity of a group is equivalent to the stable finiteness of its group algebras (cf. [CAG, Corollary 8.15.6]), we conclude that G is L-surjunctive. ■ Comment The fact that every abelian group is L-surjunctive can also be deduced from the linear version of the Garden of Eden theorem [CAG, Theorem 8.9.6] (since all abelian groups are amenable and all injective cellular automata are pre-injective), or from the linear version of the Gromov-Weiss theorem [CAG, Theorem 8.14.4] (since all abelian groups are sofic). Exercise 8.29 Let G be a group and let V be a vector space over a field K. Let τ : V G → V G be an linear cellular automaton. (a) Let H be a subgroup of G. Show that the set of configurations that are fixed by H , i.e., the set .

Fix(H ) := {x ∈ V G : gx = x for all g ∈ H }

is a vector subspace of V G . (b) Suppose that H is a finite index subgroup of G and that V is finitedimensional. Show that the vector space Fix(H ) is finite-dimensional. (c) Deduce from (b) that all residually finite groups are L-surjunctive. Solution (a) Since the shift action of G on V G is linear, the map λg : V G → V G , given by λg (x) := gx − x for all x ∈ V G , is linear for each g ∈ G. As .

Fix(H ) =

 g∈H

{x ∈ V G : gx = x} =



ker(λg ),

g∈H

we deduce that Fix(H ) is a vector subspace of V G . (b) Consider the set H \G = {H g : g ∈ G} consisting of all right cosets of H in G and the canonical map ρ : G → H \G given by ρ(g) = H g for all g ∈ G. Given an element y ∈ V H \G , the composite map y ◦ ρ : G → V is an element of Fix(H ). Moreover, the map ρ ∗ : V H \G → Fix(H ), defined by ρ ∗ (y) = y ◦ ρ for all y ∈ V H \G , is bijective [CAG, Proposition 1.3.3]. As ρ ∗ is clearly linear, we deduce

8.2 Exercises

541

that the vector spaces V H \G and Fix(H ) are isomorphic, so that .

dim(Fix(H )) = dim(V H \G ) = |H \G| · dim(V ) = [G : H ] · dim(V ) < ∞.

(c) Suppose now that G is residually finite. Suppose also that V is finitedimensional and that τ is injective. Consider a configuration x ∈ V G with finite G-orbit. This means that the stabilizer H of x is a finite index subgroup of G. We have τ (Fix(H )) ⊂ Fix(H ) since τ is G-equivariant. As τ is injective and Fix(H ) is finite-dimensional by (b), we deduce that τ (Fix(H )) = Fix(H ). Since x ∈ Fix(H ), it follows that x ∈ τ (Fix(H )) ⊂ τ (V G ). Thus, τ (V G ) contains all configurations with finite orbit. As τ (V G ) is closed in V G by the closed image property of linear cellular automata (cf. [CAG, Theorem 8.8.1]) and the set of configurations with finite orbit is dense in V G by residual finiteness of G (cf. [CAG, Theorem 2.7.1]), we deduce that τ (V G ) = V G , that is, τ is surjective. This shows that the group G is L-surjunctive. ■ Comment The fact that every residually finite group is L-surjunctive can also be deduced from the linear version of the Gromov-Weiss theorem [CAG, Theorem 8.14.4] (since all residually finite groups are sofic ([CAG, Corollary 7.5.11])). Exercise 8.30 Show that every virtually L-surjunctive group is L-surjunctive. Solution Let G be a virtually L-surjunctive group. This means that there exists a finite index subgroup H ⊂ G which is L-surjunctive. Let K be a field and let V be a finite-dimensional vector space over K. Suppose that τ : V G → V G is a linear injective cellular automaton. Let us show that τ is surjective. Let T ⊂ G be a complete set of representatives for the right cosets of H in G and set W := V T . By Exercise 1.32, the map Ψ : V G → W H , defined by Ψ (x)(h)(t) := x(ht) for all x ∈ V G , h ∈ H , and t ∈ T , is an H -equivariant uniform isomorphism. Clearly, Ψ is also K-linear. As τ is G-equivariant, uniformly continuous, and K-linear, we deduce that the map σ := Ψ ◦ τ ◦ Ψ −1 : W H → W H , obtained by conjugating τ by Ψ , is injective, H -equivariant, uniformly continuous, and K-linear. Therefore, σ : W H → W H is an injective linear cellular automaton over the group H and the finite-dimensional alphabet space W . As H is surjunctive, we deduce that σ is surjective. It follows that τ = Ψ −1 ◦ σ ◦ Ψ is surjective as well. This shows that G is L-surjunctive. ■ Exercise 8.31 Let G be a group and let K be a field. Let H be a subgroup of G. Let T ⊂ G be a complete set of representatives for the left cosets of H in G. (a) Show that K[G] is a free right K[H ]-module with base T . (b) Suppose that H has finite index n := [G : H ] in G. Show that there is an injective ring homomorphism K[G] → Matn (K[H ]). (c) Use (b) to recover the fact that every virtually L-surjunctive group is Lsurjunctive (cf. Exercise 8.30).

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8 Linear Cellular Automata

Solution (a) First observe that there is a natural structure of a right K[H ]-module on K[G] since K[G] is a right module on itself and K[H ] is a subalgebra of K[G]. As T is a complete set of representatives for the left cosets of H in G, every g ∈ G can be uniquely  written in the form g = th with t ∈ T and h ∈ H . If α ∈ K[G], then α = g∈G αg g, where αg ∈ K for all g ∈ G and αg = 0 for all but finitely many g ∈ G. We deduce that α=



.

αth th =

t∈T h∈H



t(

t∈T



αth h).

h∈H

 As h∈H αth h ∈ K[H ] for all t ∈ T , this formula shows the  that(t)T generates (t) = right K[H ]-module K[G]. On the other hand, if α = t∈T β , where β  (t) (t) (t) = 0 for all but h∈H βh h ∈ K[H ], βh ∈ K for all t ∈ T and h ∈ H , and β finitely many t ∈ T , then α=



.

(t)

βh th,

t∈T h∈H

so that βh(t) = αth for all t ∈ T and h ∈ H . This shows that T is a base for the right K[H ]-module K[G]. (b) For every α ∈ K[G], the map Lα : K[G] → K[G], given by left multiplication by α, is clearly an endomorphism of the right K[H ]-module K[G]. We have L1G = IdK[G] , Lα1 +α2 = Lα1 + Lα2 , and Lα1 α2 = Lα1 ◦ Lα2 for all α1 , α2 ∈ K[G]. Moreover, if α ∈ K[G] is such that Lα = 0, then 0 = Lα (1G ) = α1G = α. Therefore, the map α |→ Lα is an injective ring homomorphism from K[G] into the ring EndK[H ] (K[G]) of endomorphisms of the right K[H ]-module K[G]. By (a), the right K[H ]-module K[G] is free of rank n = [G : H ] = |T |. Thus, the ring EndK[H ] (K[G]) is isomorphic to the matrix ring Matn (K[H ]) (see the discussion preceding Corollary 8.7.8 in [CAG]) and the result follows. (c) Suppose that H is of finite index n in G and that the group ring K[H ] is stably finite. This means that for every d ≥ 1 the matrix ring Matd (K[H ]) is directly finite. As the matrix ring Matd (K[G]) embeds in the matrix ring Matdn (K[H ]) by (b), this implies that Matd (K[G]) is directly finite for every d ≥ 1. Therefore K[G] is stably finite. As a group is L-surjunctive if and only if its group ring with coefficients in any field K is stably finite, we deduce that every virtually L-surjunctive group is L-surjunctive. ■ Exercise 8.32 Let R be a ring and let x ∈ R. Show that x is an idempotent if and only if 1R − x is an idempotent.

8.2 Exercises

543

Solution We have that x 2 = x ⇐⇒ 1R − 2x + x 2 = 1R − x ⇐⇒ (1R − x)2 = 1R − x.

.

Therefore x is an idempotent if and only if 1R − x is an idempotent.

■

Exercise 8.33 (Boolean Rings) A ring is called Boolean if all of its elements are idempotents. Let R be a Boolean ring. (a) Let x ∈ R. Show that 2x = 0R and x = −x. (b) Show that R is commutative and unit-regular. (c) Show that every subring of R is Boolean. (d) Show that every quotient ring of R is Boolean. (e) Show that the following conditions are all equivalent: (1) R has no zerodivisors; (2) R has no non-trivial idempotents; (3) R has at most two elements. (f) Show that if R is finite then |R| = 2n for some integer n ≥ 0. (g) Let I be a prime ideal of R. Show that I is a maximal ideal of R and that the quotient ring R/I is a field with two elements. Solution (a) We have (2x)2 = 2x since 2x is an idempotent. This yields 4x = 2x and hence 2x = 0R . It follows that x = 2x − x = −x. (b) For all x, y ∈ R, we have (x +y)2 = x +y. This gives us x 2 +y 2 +xy +yx = x + y and hence xy + yx = 0R since x and y are idempotent. As −yx = yx by (a), we deduce that xy = −yx = yx. This shows that R is commutative. Every x ∈ R satisfies x = x 2 = x1R x. As 1R is invertible, this shows that R is unit-regular. (c) Let S be a subring of R. If x ∈ S, then x ∈ R and hence x 2 = x. This shows that S is Boolean. (d) Let Q be a quotien ring of R and let ρ : R → Q denote the quotient ring morphism. If x ∈ Q, then x = ρ(y) for some y ∈ R. As y 2 = y since R is Boolean, this implies x 2 = (ρ(y))2 = ρ(y 2 ) = ρ(y) = x. This shows that the ring Q is Boolean. (e) If R has more than 2 elements, then there exists x ∈ R such that x /= 0R and x /= 1R . We then have 1R + x /= 0R , 1R + x /= 1R , and (1R + x)2 = 1R + 2x + x 2 = 1R + x by (a), so that 1R + x is a non-trivial idempotent. This shows that (2) implies (3). As the implications (1) =⇒ (2) and (3) =⇒ (1) are trivial, we conclude that all three conditions are equivalent. (f) From (a) and (b), we deduce that R has a canonical vector space structure over the field with two elements K := Z/2Z. If R is finite then n := dimK (R) < ∞ and R is isomorphic to the vector space K n so that |R| = |K n | = 2n . (g) Since R/I is an integral domain, we deduce from (d) and (e) that |R/I | = 2. It follows that R/I is a field with two elements. In particular, I is a maximal ideal. ■

544

8 Linear Cellular Automata

Exercise 8.34 Let R be a nonzero ring and let G be a group. Show that the ring R[G] is Boolean if and only if the ring R is Boolean and the group G is trivial. Solution The condition is clearly sufficient since the ring R[G] is isomorphic to R when G is a trivial group. Conversely suppose that the ring R[G] is Boolean. As the group G embeds in the group of units of R[G], we deduce that g 2 = g and hence g = 1G for all g ∈ G. Therefore the group G is trivial. Morover, the ring R is isomorphic to R[G] and hence it is Boolean. ■ Exercise 8.35 Let R be a ring and let G be a group. Let H be a finite subgroup of G. Consider the element α ∈ R[G] defined by α := h∈H h. (a) Show that α 2 = |H |2 α. (b) Suppose that H is normal in G. Show that α is in the center of R[G]. (c) Suppose that R is a nonzero ring and α is in the center of R[G]. Show that H is normal in G. (d) Suppose that R is a nonzero ring, |H | is invertible in R[G], and H /= {1G }. Show that the element μ ∈ R[G] defined by μ := |H |−1 α is a non-trivial idempotent of R[G]. Solution (a) We have α2 = (



.

g∈H

=



g)(

h)

h∈H

  ( gh) g∈H h∈H

=

  ( h' )

(since h |→ gh is a permutation of H )

g∈H h' ∈H

=



α

g∈H

= |H |α. (b) For all β = αβ = (



g∈G βg g



.

h)(

h∈H

=



βg (

g∈G

=



g∈G



∈ R[G], we have βg g)

g∈G



hg)

h∈H

βg (



h' ∈H

gh' )

(since H is normal in G)

8.2 Exercises

545

=(

 g∈G

βg g)(



h' )

h' ∈H

= βα. Therefore α is in the center of R[G]. (c) Let g ∈ G. Since α is in the center of R[G], we have αg = gα and hence  .

h∈H

hg =



gh.

h∈H

As R is a nonzero ring, we deduce that for every h ∈ H , there exists h' ∈ H such that hg = gh' . Therefore H is normal in G. (d) Using (a), we get μ2 = (|H |−1 α)2 = |H |−2 α 2 = |H |−1 α = μ,

.

showing that μ is an idempotent element. On the other hand, μ /= 0 and μ /= 1 since the support of μ is H and |H | ≥ 2. Therefore μ is a non-trivial idempotent of R[G]. ■ Exercise 8.36 (Reduced Rings) A ring R is called reduced if a 2 = 0 implies a = 0 for all a ∈ R. (a) Show that if a ring R has no zero-divisors then R is reduced. (b) Give an example of a reduced ring that has zero-divisors. (c) Show that a ring is reduced if and only if it has no nonzero nilpotent elements. (d) Let G be a finite group and let K be a field such that the characteristic of K divides |G|. Show that the ring K[G] is not reduced. Solution (a) Suppose that the ring R has no zero-divisors. This means that ab = 0 implies a = 0 or b = 0 for all a, b ∈ R. In particular, a 2 = 0 implies a = 0 for all a ∈ R. Therefore R is reduced. (b) The ring Z/6Z provides such a example. Indeed, in Z/6Z, all the squares 12 = 1, 22 = 4, 32 = 3, 42 = 4, 52 = 1 are nonzero so that Z/6Z is reduced. On the otherhand, 2 · 3 = 0 in Z/6Z, showing that Z/6Z has zero-divisors. (c) It is obvious that if a ring R has no nonzero nilpotent elements then R is reduced. Conversely, suppose that a ring R is reduced and a ∈ R is nilpotent. This means that there is an integer n ≥ 1 such that a n = 0. Take an integer k such that p := 2k ≥ n. Then a p = a n · a p−n = 0. On the other hand, as a p = (. . . ((a)2 )2 . . . )2 ,   

.

k times

by induction we deduce that a = 0 since R is reduced.

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8 Linear Cellular Automata

 (d) Consider the element α ∈ K[G] defined by α := g∈G g (cf. Exercise 8.35). Then α 2 = |G|α = 0. Thus α is nilpotent. On the other hand, α /= 0 since the support of α is G and hence non-empty. Therefore K[G] is not reduced. ■ Exercise 8.37 (Strongly Regular Rings) A ring R is called strongly regular if for every a ∈ R, there exists x ∈ R such that a = a 2 x. (a) Show that a commutative ring is strongly regular if and only if it is von Neumann regular. (b) Show that a ring is strongly regular if and only if it is von Neumann regular and reduced. Solution (a) This is clear since in a commutative ring R we have axa = a 2 x for all a, x ∈ R. (b) Suppose first that R is a strongly regular ring. Let a ∈ R. Then there exists x ∈ R such that a = a 2 x. This implies that a = 0 if a 2 = 0. Therefore R is reduced. On the other hand, we have (a − axa)2 = a 2 + axa 2 xa − axa 2 − a 2 xa = a 2 + axa 2 − axa 2 − a 2 = 0.

.

Since R is reduced, this implies a − axa = 0 and hence a = axa. We deduce that R is von Neumann regular. Conversely, suppose that the ring R is von Neumann regular and reduced. Let a ∈ R. Since R is von Neumann regular, there exists x ∈ R such that a = axa. We then have (a − a 2 x)2 = a 2 + a 2 xa 2 x − a 3 x − a 2 xa = a 2 + a 3 x − a 3 x − a 2 = 0.

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As R is reduced, this implies a − a 2 x = 0 and hence a = a 2 x. This shows that R is strongly regular. ■ Comment Strongly regular rings are also called strongly von Neumann regular or abelian regular rings. For further results on strongly regular rings, see [Goo], [Lam2], and [Lam1]. Exercise 8.38 (Reversible Rings) A ring R is called reversible if ab = 0R implies ba = 0R for all a, b ∈ R. (a) Show that every reversible ring is directly finite. (b) Let V be a vector space over a field K such that dimK (V ) ≥ 2. Show that the ring EndK (V ) is not reversible. (c) Let K be a field and let n ≥ 2 be an integer. Show that the ring of n × n matrices with entries in K is directly finite but not reversible. (d) Let K be a field. Show that the group ring K[Sym3 ] is directly finite but not reversible.

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(e) Show that every reduced ring is reversible. (f) Show that if a ring R has no zero-divisors then R is reversible. (g) Give examples of rings that are reversible but not reduced. Solution (a) Let R be a reversible ring and let a, b ∈ R such that ab = 1R . We have that (ba − 1R )b = bab − b = b − b = 0R . Since R is reversible, it follows that b(ba − 1R ) = 0R and hence b2 a = b. We deduce that ba = (ab)ba = a(b2 a) = ab = 1R . This shows that R is directly finite. (b) Let u : V → K be a nonzero linear form on V . Since dimK (V ) ≥ 2, there exist two nonzero vectors e1 , e2 ∈ V such that u(e1 ) /= 0 and u(e2 ) = 0. For each i ∈ {1, 2}, consider the endomorphismfi ∈ EndK (V ) defined by fi (x) = u(x)ei for all x ∈ V . We then have f1 ◦ f2 = 0 since (f1 ◦ f2 )(x) = f1 (f2 (x)) = f1 (u(x)e2 ) = u(x)f1 (e2 ) = u(x)u(e2 )e1 = 0

.

for all x ∈ V . On the other hand, f2 ◦ f1 /= 0 since (f2 ◦ f1 )(e1 ) = f2 (f1 (e1 )) = f2 (u(e1 )e1 ) = u(e1 )f2 (e1 ) = u(e1 )2 e2 /= 0.

.

This shows that the ring EndK (V ) is not reversible. (c) The ring Matn (K) is directly finite and even stably finite since it is a finitedimensional K-algebra (see Exercise 8.25). The fact that Matn (K) is not reversible follows from (b) since the rings Matn (K) and EndK (K n ) are isomorphic. (d) Consider the elements a, b ∈ K[Sym3 ] given by a := (1 2) − (1 2 3) and b := 1 + (2 3). We then have ab = ((1 2) − (1 2 3))(1 + (2 3))

.

= (1 2) + (1 2)(2 3) − (1 2 3) − (1 2 3)(2 3) = (1 2) − (1 2 3) + (1 2 3) − (1 2) = 0, and ba = (1 + (2 3))((1 2) − (1 2 3))

.

= (1 2) + (2 3)(1 2) − (1 2 3) − (2 3)(1 2 3) = (1 2) + (1 3 2) − (1 2 3) − (1 3) /= 0. This shows that the ring K[Sym3 ] is not reversible. The K-algebra K[Sym3 ] is finite-dimensional since the group Sym3 is finite. Therefore, K[Sym3 ] is directly finite (and even stably finite) by Exercise 8.25. (e) Let R be a reduced ring and a, b ∈ R such that ab = 0. Then (ba)2 = b(ab)a = 0. Since R is reduced, it follows that ba = 0. This shows that R is reversible.

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(f) This immediately follows from (e) since any ring without zero-divisors is reduced (cf. Exercise 8.36.(a)). (g) The ring Z/4Z is reversible since it is commutative but it is not reduced since 2 ∈ Z/2Z is a nonzero nilpotent element. Another example is provided by the ring K[X]/(X2 ), obtained by quotienting the polynomial ring K[X] over a field K by the ideal generated by X2 . ■ Comment From the results in Exercise 8.36 and Exercise 8.38, we deduce that the following implications hold true for rings: no zero-divisors =⇒ reduced =⇒ reversible =⇒ directly finite

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while all the converse implications are false. The notion of a reversible ring was introduced by Cohn in [Cohn]. In [GutK, Theorem 2.1], Gutan and Kisielewicz proved that if G is a torsion group containing a subgroup that is not normal in G (e.g., G := Sym3 ) and K is any field, then the group ring K[G] is not reversible. Moreover, they completely characterized the pairs (K, G), where K is a field and G is a torsion group, such that K[G] is reversible. They also showed that, up to ring isomorphism, the group ring F2 [Q8 ], where Q8 := {±1, ±i, ±j, ±k} is the quaternion group and F2 = Z/2Z is the field with two elements, is the smallest non-commutative finite group ring, with coefficients in a field, that is reversible. Note that F2 [Q8 ] is an 8-dimensional vector space over F2 and hence has cardinality 256. Every element g ∈ Q8 satisfies g 2 = 1, so that (1 + g)2 = 1 + g 2 = 0 ∈ F2 [Q8 ]. Thus 1 + g is a nonzero nilpotent element in F2 [Q8 ] for every g ∈ Q8 with g /= 1. This shows in particular that the ring F2 [Q8 ] is not reduced. Exercise 8.39 Let K be a field and let V be a vector space over K. (a) Show that the ring EndK (V ) has no zero-divisors (resp. is reduced, resp. is reversible) if and only if dimK (V ) ≤ 1. (b) Show that the ring EndK (V ) is directly finite if and only if dimK (V ) < ∞. Solution (a) If dimK (V ) ≤ 1, then the ring EndK (V ) is either trivial or isomorphic to the field K. In either case, it has no zero-divisors and is reduced and reversible. Suppose now dimK (V ) ≥ 2. Then there exist vectors v1 , v2 ∈ V that are not collinear. Choose a vector subspace W of V such that V = Kv1 ⊕ Kv2 ⊕ W . Consider the endomorphism f ∈ EndK (V ) such that f (v1 ) = v1 and f |Kv2 ⊕W = 0, and the endomorphism g ∈ EndK (V ) such that g(v2 ) = v1 and g|Kv1 ⊕W = 0. Then f ◦ g = g /= 0 but g ◦ f = 0. Therefore the ring EndK (V ) is not reversible. Consequently, it is not reduced either and has zero-divisors. (b) If dimK (V ) = d < ∞, then EndK (V ) is isomorphic to Matd (K) and is therefore directly finite since every field is stably finite by Exercise 8.19. The fact that if V is infinite-dimensional then EndK (V ) is not directly finite was ■ established in the solution to Exercise 8.23(i).

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Exercise 8.40 Let G be a group. Show that G is a unique-product group if and only if it satisfies the following condition: for every non-empty finite subset A of G, there exists g ∈ G such that g can be uniquely written in the form g = a1 a2 with a1 , a2 ∈ A. Solution Necessity is obvious (just take A = B in the definition of a uniqueproduct group). Conversely, suppose that the condition is satisfied and let A, B be two non-empty finite subsets of G. As BA is a non-empty finite subset of G, it follows from our hypothesis that there exists g ∈ G such that g can be uniquely written in the form g = g1 g2 with g1 , g2 ∈ BA. Write g1 = b1 a1 and g2 = b2 a2 with b1 , b2 ∈ B and a1 , a2 ∈ A. We claim that the element g ' := a1 b2 ∈ AB can be uniquely expressed as a product of an element of A followed by an element of B. Indeed, if a1 b2 = ab with a ∈ A and b ∈ B, then g = b1 a1 b2 a2 = b1 aba2 , so that b1 a1 = b1 a and b2 a2 = ba2 by the uniqueness property of g. It follows that a1 = a and b2 = b. This shows that G is a unique-product group. ■ Exercise 8.41 Let G be a group. (a) Let A be a finite subgroup of G. Show that for every g ∈ A, there are exactly |A| distinct ordered pairs (a, b) with a, b ∈ A such that g = ab. (b) Use (a) to show that every unique-product group is torsion-free. Solution (a) This immediately follows from the fact that if g, a ∈ A are fixed, then the equation g = ab has a unique solution b ∈ A given by b = a −1 g. (b) Suppose that G is not torsion-free. This means that there exists g0 ∈ G with finite order n ≥ 2. Then the subgroup A of G generated by g0 is finite with cardinality |A| = n. It then follows from (a) that there is no g ∈ G such that g can be uniquely written in the form g = ab with a, b ∈ A. This shows that G is not unique-product. ■ Comment The fact that every unique-product group is torsion-free can also be deduced from the implications unique-product =⇒ NZD =⇒ torsion-free. Exercise 8.42 Show that every subgroup of a unique-product group is a uniqueproduct group. Solution Let G be a unique-product group and let H be a subgroup of G. Let A and B be two non-empty finite subsets of H . Since G is a unique-product group, there is an element g ∈ G such that g can be uniquely written as a product g = ab with a ∈ A and b ∈ B. We have g ∈ H since H is a subgroup of G and A, B ⊂ H . Therefore H is a unique-product group. ■ Exercise 8.43 Let G be a group. Suppose that G is locally unique-product, i.e., every finitely generated subgroup of G is a unique-product group. Show that G is a unique-product group. Solution Let A and B be non-empty finite subsets of G. Let H denote the subgroup of G generated by A ∪B. Since H is finitely generated, it is a unique-product group.

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As A, B ⊂ H , there is an element h ∈ H that can be uniquely written in the form h = ab with a ∈ A and b ∈ B. This shows that G is a unique-product group. ■ Exercise 8.44 Let G be a group. Suppose that G contains a normal subgroup N such that N and G/N are both unique-product groups. Show that G is a uniqueproduct group. Solution Let A and B be non-empty finite subsets of G. Let ρ : G → G/N denote the canonical group epimorphism. Since G/N is a unique-product group, there exists c ∈ G/N such that c can be uniquely written in the form c = uv with u ∈ ρ(A) and v ∈ ρ(B). Choose a0 ∈ A and b0 ∈ B such that ρ(a0 ) = u and ρ(b0 ) = v. We then have A ∩ ρ −1 (u) = a0 N1 and B ∩ ρ −1 (v) = N2 b0 , where N1 and N2 are finite subsets of N. Since N is a unique-product group, there exists n ∈ N such that n can be uniquely written in the form n = n1 n2 with n1 ∈ N1 and n2 ∈ N2 . Consider the element x := a0 nb0 = a0 n1 n2 b0 . Setting a := a0 n1 ∈ A ∩ ρ −1 (u) and b := n2 b0 ∈ B ∩ ρ −1 (v), we have x = ab ∈ AB. Suppose that x = a ' b' with a ' ∈ A and b' ∈ B. We then have ρ(a ' )ρ(b' ) = ρ(a ' b' ) = ρ(x) = ρ(ab) = ρ(a)ρ(b) = uv = c,

.

so that ρ(a ' ) = u and ρ(b' ) = v. It follows that a ' = a0 n'1 and b = n'2 b0 for some n'1 ∈ N1 and n'2 ∈ N2 . This gives us a0 n1 n2 b0 = x = a ' b' = a0 n'1 n'2 b0 and hence n1 n2 = n'1 n'2 . This last equality implies n1 = n'1 and n2 = n'2 , so that a ' = a0 n1 = a and b' = n2 b0 = b. Consequently, G is a unique-product group. ■ Comment This is Theorem 6.1 in [RudS]. Note that unique-product groups are called Ω-groups in [RudS]. Exercise 8.45 Let G be a group. Suppose that G is residually unique-product, i.e., for every g ∈ G \ {1G }, there exist a unique-product group H and a group homomorphism φ : G → H such that φ(g) /= 1H . (a) Let Ω be a finite subset of G. Show that there exist a unique-product group H and a group homomorphism φ : G → H such that the restriction of φ to Ω is injective. (b) Deduce from (a) that G is a unique-product group. Solution (a) Let Δ ⊂ G denote the finite subset of G consisting of all elements of the form ω−1 ω' , where ω, ω' ∈ Ω and ω /= ω' . As G is residually unique-product, for each δ ∈ Δ, we can find a unique-product group Hδ and a group homomorphism  φδ : G → Hδ such that φδ (δ) /= 1Hδ . Consider the group  H := δ∈Δ Hδ and the group homomorphism φ : G → H given by φ := δ∈Δ φδ . The group H is a unique-product group since any finite direct product of unique-product groups is itself a unique-product group by Exercise 8.44. Moreover, the restriction of φ to Ω is clearly injective. (b) Let A and B be non-empty finite subsets of G and let Ω := A ∪ B ∪ AB. By (a), there exist a unique-product group H and a group homomorphism φ : G → H

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whose restriction to Ω is injective. Let A' := φ(A) and B ' := φ(B). As H is a unique-product group, there exists h ∈ A' B ' which can be uniquely written in the form h = a ' b' with a ' ∈ A' and b' ∈ B ' . If g ∈ AB, a ∈ A, and b ∈ B are the unique elements such that φ(g) = g ' , φ(a) = a ' , and φ(b) = b' , then g = ab and there is no other writing of g as a product of an element of A followed by an element of B. This shows that G is a unique-product group. ■ Exercise 8.46 Show that every group which is locally embeddable into the class of unique-product groups is a unique-product group. Solution Let G be a group which is locally embeddable into the class of uniqueproduct groups. Let A and B be two non-empty finite subsets of G. Set K := A ∪ B ⊂ G. Then, by our assumptions on G, we can find a unique-product group C and a K-almost-homomorphism ϕ : G → C. The set A' := ϕ(A) ⊂ C (resp. B ' := ϕ(B) ⊂ C) is finite and non-empty. Since C is unique-product, there exists c ∈ C which can be uniquely expressed as a product c = a ' b' with a ' ∈ A' and b' ∈ B ' . Since ϕ|A : A → A' (resp. ϕ|B : B → B ' ) is a bijection, there exist unique elements a ∈ A and b ∈ B such that ϕ(a) = a ' and ϕ(b) = b' . Let us show that the expression of the element g := ab ∈ G as a product of an element of A and an element of B is unique. Let a ∈ A and b ∈ B such that g = a b. Setting ' a ' := ϕ(a) ∈ C (resp. b := ϕ(b) ∈ C) and keeping in mind that ϕ is a K-almosthomomorphism, we have '

c = a ' b' = ϕ(a)ϕ(b) = ϕ(ab) = ϕ(g) = ϕ(a b) = ϕ(a)ϕ(b) = a ' b .

.

'

The choice of the element c ∈ C implies that a ' = a ' and b = b' . Therefore, by injectivity of ϕ|K , we necessarily have a = a and b = b. We deduce that G is a unique-product group. ■ Comment This extends Exercise 8.43 since every group which is locally uniqueproduct is locally embeddable into the class of unique-product groups (cf. [CAG, Proposition 7.1.9]). It gives us also an alternative solution for Exercise 8.45. Indeed, the class of unique-product groups is closed under finite direct products by Exercise 8.44, so that every group which is residually unique-product is locally embeddable into the class of unique-product groups (cf. [CAG, Proposition 7.1.14]). Exercise 8.47 Let G be a group. Suppose that every non-trivial finitely generated subgroup of G admits a non-trivial unique-product quotient group. Show that G is a unique-product group. Solution We proceed by contradiction. So, let us assume that G is not a uniqueproduct group. This means that there exist non-empty finite subsets A and B of G such that each fiber of the map A × B → AB, defined by (a, b) |→ ab, has cardinality ≥ 2. Choose subsets A and B satisfying these properties with |A| + |B| minimal. Observe that, for all g1 , g2 ∈ G, we can replace the pair (A, B) by (g1 A, Bg2 ) without changing the above properties. Therefore, there is no loss of generality to assume in addition 1G ∈ A and 1G ∈ B. Let H denote the subgroup

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of G generated by A ∪ B. The group H is not trivial since A and B cannot be both reduced to 1G . Consequently, there exist a non-trivial unique-product group P and a surjective group homomorphism ρ : H → P . Consider the sets A' := ρ(A) and B ' := ρ(B). Since P is a unique-product group, there exists (a0' , b0' ) ∈ A' × B ' such that (a0' , b0' ) is the unique preimage of a0' b0' with respect to the map A' × B ' → A' B ' defined by (a ' , b' ) |→ a ' b' . Consider the sets A1 := A ∩ ρ −1 (a0' ) and B1 := B ∩ ρ −1 (b0' ). Clearly A1 (resp. B1 ) is a non-empty subset of A (resp. B). If (a1 , b1 ) ∈ A1 × B1 , we know there exists (a, b) ∈ A × B such that a1 b1 = ab and (a1 , b1 ) /= (a, b). As a0' b0' = ρ(a1 )ρ(b1 ) = ρ(a1 b1 ) = ρ(ab) = ρ(a)ρ(b), this implies a0' = ρ(a) and b0' = ρ(b), so that (a, b) ∈ A1 × B1 . We deduce that each fiber of the map A1 × B1 → A1 B1 , given by (a1 , b1 ) |→ a1 b1 , has cardinality ≥ 2. Finally, note that we cannot have both A1 = A and B1 = B. Otherwise, the group P would be trivial since 1G ∈ A ∩ B and ρ(a0' ) = 1P = ρ(b0' ). Therefore, we have |A1 | + |B1 | < |A| + |B|, in contradiction with the minimality of |A| + |B|. ■ Comment This is Theorem 1 in [BurnH]. Note that every locally unique-product group as well as every extension of unique-product groups clearly satisfy the hypothesis of the present exercise. Thus, we recover the results of Exercise 8.43 and Exercise 8.44 (cf. [BurnH, Corollary 1]). Exercise 8.48 (Indicable Groups) A group is said to be indicable if it is either trivial or admits an infinite cyclic quotient. A group is said to be locally indicable if all of its finitely generated subgroups are indicable. (a) Show that every free group is indicable and locally indicable. (b) Show that every torsion-free abelian group is locally indicable. (c) Give an example of a torsion-free abelian group that is not indicable. (d) Give an example of an indicable abelian group that is not locally indicable. (e) Show that every locally indicable group is a unique-product group. Solution (a) Suppose that G is a free group and let X ⊂ G be a free base. Any map f : X → Z extends to a group homomorphism ϕ : G → Z by the universal property of free groups. If G is not trivial, that is, X /= ∅, we can take f such that 1 is in its image and then ϕ : G → Z is a surjective group homomorphism. Therefore G is indicable. As every subgroup of a free group is a free group by the Nielsen-Schreier theorem (cf. Exercise 6.28), we deduce that G is also locally indicable. (b) Let G be a torsion-free abelian group and let H be a non-trivial finitely generated subgroup of G. By the structure theorem for finitely generated abelian groups, there exists an integer n ≥ 1 such that H is isomorphic to the group Zn . As the projection on the first factor yields a surjective group homomorphism Zn → Z, we deduce that H is indicable. Therefore G is locally indicable. (c) The additive group Q of rational numbers is torsion-free abelian. It is not indicable since every group homomorphism Q → Z is trivial. More generally, if G is a non-trivial divisible group then G is not indicable. Indeed, suppose that ϕ : G → Z is a surjective homomorphism. Let g ∈ G such that ϕ(g) = 1. As G is divisible,

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we can find h ∈ G such that h2 = g. We deduce that 1 = ϕ(g) = ϕ(h2 ) = 2ϕ(h) ∈ 2Z, a contradiction. (d) Consider the abelian group Z × H , where H := Z/2Z. It is indicable since the projection on the first factor yields a surjective group homomorphism G → Z. It is not locally indicable since {0} × H is a finitely generated subgroup of G that is not indicable. (e) This immediately follows from Exercise 8.47 using the fact that Z is orderable and hence a unique-product group (cf. [CAG, Examples 8.16.5.(b) and Proposition 8.16.8]). ■ Comment The class of locally indicable groups was introduced, with a slightly different terminology, by Higman in [Hig1]. In [Lev], Levi observed that every biorderable group is locally indicable. On the other hand, it can be shown that every locally indicable group is orderable [BurnH, Corollary 2]. Exercise 8.49 Let G be an orderable group and let ≤ be a left-invariant total ordering on G. The subset P ⊂ G defined by P := {g ∈ G : 1G < g} is called the positive cone of (G, ≤). (a) Let g, h ∈ G. Show that one has g < h if and only if g −1 h ∈ P . (b) Show that P is a subsemigroup of G, i.e., P P ⊂ P . (c) Show that P −1 = {g ∈ G : g < 1G }. (d) Show that G is the disjoint union of the sets P , P −1 , and {1G }. Solution (a) If g < h, we get 1G < g −1 h by multiplying both sides of the inequality on the left by g −1 . Conversely, if 1G < g −1 h, we get g < h by multiplying on the left by g. (b) Let g1 , g2 ∈ P . Then 1G < g2 , so that g1 < g1 g2 by left-invariance. As 1G < g1 , this implies 1G < g1 g2 by transitivity. This shows that P is a subsemigroup of G. (c) We have P −1 = {g ∈ G : g −1 ∈ P } = {g ∈ G : 1G < g −1 } = {g ∈ G : g < 1G }, where the last equality follows from (a). (d) This immediately follows from (c), the fact that the ordering ≤ is total, and the definition of P . ■ Exercise 8.50 Let G be a group. Show that G is orderable if and only if there exists a subsemigroup P of G such that G is the disjoint union of the sets P , P −1 , and {1G }. Solution Necessity follows from Exercise 8.49. Conversely, suppose that G contains a subsemigroup P such that G is the disjoint union of P , P −1 , and {1G }. Define a relation ≤ on G by writing g1 ≤ g2 if and only if g2−1 g1 ∈ P ∪ {1G }. Let g1 , g2 , g3 ∈ G. We have g1−1 g1 = 1G so that g1 ≤ g1 . This shows reflexivity. If g1 ≤ g2 and g2 ≤ g1 , then g2−1 g1 ∈ (P ∪ {1G }) ∩ (P ∪ {1G })−1 = (P ∪ {1G }) ∩ (P −1 ∪ {1G }) = {1G }, so that g2−1 g1 = 1G and hence g1 = g2 . This shows antisymmetry. If g1 ≤ g2 and g2 ≤ g3 , then g1 ≤ g3 since g3−1 g1 = (g3−1 g2 )(g2−1 g1 ) and P is a subsemigroup. This shows transitivity.

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If g2−1 g1 ∈ P ∪ {1G } then g1 ≤ g2 . Otherwise, we have g2−1 g1 ∈ P −1 and hence g1−1 g2 = (g2−1 g1 )−1 ∈ P , so that g2 < g1 . We deduce that ≤ is a total ordering on G. Finally, if g1 ≤ g2 then g3 g1 ≤ g3 g2 since (g3 g2 )−1 (g3 g1 ) = g2−1 g1 ∈ P ∪{1G }. This shows left-invariance of ≤. Therefore the group G is orderable. ■ Exercise 8.51 Let G be a bi-orderable group and let ≤ be a total ordering on G that is both left and right invariant. Consider the subset P ⊂ G defined by P := {g ∈ G : 1G < g}. Show that hP h−1 = P for all h ∈ G. Solution Let g ∈ P and h ∈ G. As 1G < g, we have h < hg by left-invariance and hence 1G = hh−1 < hgh−1 by right-invariance. Therefore hgh−1 ∈ P . This shows hP h−1 ⊂ P for all h ∈ G. From h−1 P h ⊂ P , we deduce that P = h(h−1 P h)h−1 ⊂ hP h−1 . This shows that hP h−1 = P for all h ∈ G. ■ Exercise 8.52 Let G be a group. Show that G is bi-orderable if and only if there exists a subsemigroup P of G such that hP h−1 = P for all h ∈ G and G is the disjoint union of the sets P , P −1 , and {1G }. Solution Necessity follows from Exercises 8.49 and 8.51. Conversely, suppose that G contains a subsemigroup P such that hP h−1 = P for all h ∈ G and G is the disjoint union of P , P −1 , and {1G }. Define a relation ≤ on G by writing g1 ≤ g2 if and only if g2−1 g1 ∈ P ∪ {1G }. We have seen in the solution to Exercise 8.50 that ≤ is a left-invariant total ordering on G. Therefore it reamins only to show right-invariance. Let g1 , g2 , g3 ∈ G such that g1 ≤ g2 . Then g2−1 g1 ∈ P ∪ {1G }. We have (g2 g3 )−1 (g1 g3 ) = g3−1 (g2−1 g1 )g3 ∈ (g3−1 P g3 ) ∪ {1G } = P ∪ {1G }, ■ showing that g1 g3 ≤ g2 g3 . This shows that ≤ is also right-invariant. Exercise 8.53 Let G be a bi-orderable group. Suppose that an element g ∈ G satisfies the following property: there exist an integer n ≥ 1 and elements h1 , h2 , · · · , hn ∈ G such that −1 −1 h1 gh−1 1 h2 gh2 · · · hn ghn = 1G .

.

(8.6)

Show that g = 1G . Solution Let ≤ be a total ordering on G that is both left and right invariant and let P := {x ∈ G : 1G < x}. Suppose by contradiction that g /= 1G . Then g ∈ P or g ∈ P −1 by Exercise 8.49.(c). If g ∈ P then hi gh−1 ∈ P for all 1 ≤ i ≤ n i by Exercise 8.51. As P is a subsemigroup of G by Exercise 8.49.(a), we deduce from (8.6) that 1G ∈ P , a contradiction. After inverting both sides of (8.6), we get −1 −1 −1 −1 hn g −1 h−1 n hn−1 g hn−1 · · · h1 g h1 = 1G ,

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so that a similar argument shows that we cannot have g ∈ P −1 . Therefore ■ g = 1G .

8.2 Exercises

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Exercise 8.54 Let G be a bi-orderable group. Show that if g, h ∈ G are such that g n = hn for some integer n ≥ 1, then g = h. Solution Let ≤ be a total ordering on G that is both left and right invariant. We first observe that if a, b, c, d ∈ G satisfy a < b and c < d then ac < bc < bd so that ac < bd. By induction, we deduce that if a < b then a n < bn for every n ≥ 1. ■ Consequently, if g, h ∈ G satisfy g n = hn for some n ≥ 1, then g = h. Exercise 8.55 Let G be a group and let Z(G) denote the center of G. Suppose that G contains a subgroup N ⊂ Z(G) such that both N and G/N are bi-orderable. Show that G is also bi-orderable. Solution Let ≤1 (resp. ≤2 ) be a total ordering on N (resp. G/N) that is both left and right-invariant. Let ρ : G → G/N denote the canonical group epimorphism and consider the binary relation ≤ on G defined by setting

g ≤ h ⇐⇒

.

⎧ ⎪ ⎪ ⎨ρ(g)