On Normalized Integral Table Algebras (Fusion Rings): Generated by a Faithful Non-real Element of Degree 3

On Normalized Integral Table Algebras (Fusion Rings) For further volumes: www.springer.com/series/6253 Algebra and A...

Author: Zvi Arad | Xu Bangteng | Guiyun Chen | Effi Cohen | Arisha Haj Ihia Hussam | Mikhail Muzychuk

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On Normalized Integral Table Algebras (Fusion Rings)

For further volumes: www.springer.com/series/6253

Algebra and Applications Volume 16

Series Editors: Alice Fialowski Eötvös Loránd University, Budapest, Hungary Eric Friedlander Northwestern University, Evanston, USA John Greenlees Sheffield University, Sheffield, UK Gerhard Hiß Aachen University, Aachen, Germany Ieke Moerdijk Utrecht University, Utrecht, The Netherlands Idun Reiten Norwegian University of Science and Technology, Trondheim, Norway Christoph Schweigert Hamburg University, Hamburg, Germany Mina Teicher Bar-llan University, Ramat-Gan, Israel Alain Verschoren University of Antwerp, Antwerp, Belgium Algebra and Applications aims to publish well written and carefully refereed monographs with up-to-date information about progress in all fields of algebra, its classical impact on commutative and noncommutative algebraic and differential geometry, K-theory and algebraic topology, as well as applications in related domains, such as number theory, homotopy and (co)homology theory, physics and discrete mathematics. Particular emphasis will be put on state-of-the-art topics such as rings of differential operators, Lie algebras and super-algebras, group rings and algebras, C ∗ -algebras, Kac-Moody theory, arithmetic algebraic geometry, Hopf algebras and quantum groups, as well as their applications. In addition, Algebra and Applications will also publish monographs dedicated to computational aspects of these topics as well as algebraic and geometric methods in computer science.

Zvi Arad Xu Bangteng Guiyun Chen Effi Cohen Arisha Haj Ihia Hussam Mikhail Muzychuk

On Normalized Integral Table Algebras (Fusion Rings) Generated by a Faithful Non-real Element of Degree 3

Zvi Arad Department of Mathematics Bar Ilan University Ramat Gan, 52900 Israel and Netanya Academic College 1 University Street Netanya Israel [email protected] Xu Bangteng Department of Mathematics and Statistics Eastern Kentucky University Richmond, KY, 40475 USA [email protected] Guiyun Chen Department of Mathematics Southwest University Chongqing, 400715 People’s Republic of China [email protected]

Effi Cohen Department of Mathematics Bar Ilan University Ramat Gan, 52900 Israel [email protected] Arisha Haj Ihia Hussam Department of Mathematics Alqasemi Academic College of Education Baqa El-Gharbieh, 30100 Israel [email protected] Mikhail Muzychuk Netanya Academic College Kibbutz Galuyot 16 Netanya, 42 365 Israel [email protected]

ISSN 1572-5553 e-ISSN 2192-2950 ISBN 978-0-85729-849-2 e-ISBN 978-0-85729-850-8 DOI 10.1007/978-0-85729-850-8 Springer London Dordrecht Heidelberg New York British Library Cataloguing in Publication Data A catalogue record for this book is available from the British Library Library of Congress Control Number: 2011934389 Mathematics Subject Classification: 05E10, 05E30, 13A99, 20C15, 20N20, 81T40 © Springer-Verlag London Limited 2011 Apart from any fair dealing for the purposes of research or private study, or criticism or review, as permitted under the Copyright, Designs and Patents Act 1988, this publication may only be reproduced, stored or transmitted, in any form or by any means, with the prior permission in writing of the publishers, or in the case of reprographic reproduction in accordance with the terms of licenses issued by the Copyright Licensing Agency. Enquiries concerning reproduction outside those terms should be sent to the publishers. The use of registered names, trademarks, etc., in this publication does not imply, even in the absence of a specific statement, that such names are exempt from the relevant laws and regulations and therefore free for general use. The publisher makes no representation, express or implied, with regard to the accuracy of the information contained in this book and cannot accept any legal responsibility or liability for any errors or omissions that may be made. Cover design: VTeX UAB, Lithuania Printed on acid-free paper Springer is part of Springer Science+Business Media (www.springer.com)

Preface

Arad and Chen proved in [AC] and [AC1] that a Normalized Integral Table Algebra (fusion ring) (A, B) generated by a non-real faithful element b3 ∈ B of degree three the non-identity elements of which have minimal degree 3 satisfies the condition b3 b¯3 = 1 + b8 where b8 ∈ B is an element of degree 8. They also showed that the general case naturally splits into four main sub-cases: (1) (2) (3) (4)

(A, B) ∼ =x (CH(PSL(2, 7), Irr(PSL(2, 7))); b32 = b4 + b5 where b4 , b5 ∈ B are elements of degrees 4 and 5; b32 = b¯3 + b6 where b6 ∈ B is a non-real element of degree 6; b32 = c3 + b6 where c3 , b6 ∈ B are elements of degrees 3 and 6, c3 = b3 , b¯3 .

The cases (1), (3) and (4) are considered in Chap. 2. Chapter 3 deals with the case (2). Chapters 4 and 5 analyze the most complicated case—the third one. We developed new original methods for enumerating NITAs in the title. Using the developed technique we settled the above cases almost completely.

v

Acknowledgements

Zvi Arad is sincerely grateful to Eastern Kentucky University (EKU) for inviting him to head the Wilson Endowment Chair in the Department of Mathematics and Statistics in Spring 2009. The warm hospitality and friendly atmosphere there enabled him to conduct the joint research of Chap. 3 in the spring semester of 2009. Effie Cohen submitted Chaps. 4 and 5 of this book in partial fulfillment of the requirements for the Ph.D. degree in the Department of Mathematics, Bar-Ilan University, Ramat Gan, Israel, under the supervision of Prof. Zvi Arad and Prof. Malka Schaps. The authors thank the anonymous referees for their helpful suggestions and comments. The author thanks Mrs. Miriam Beller for her great efforts to print the first version of the book and for her helpful remarks.

vii

Contents

1

Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

1 6

2

Splitting of the Main Problem into Four Sub-cases . . . . . . . . . 2.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2 Two NITA Generated by a Non-real Element of Degree 3 not Derived from a Group and Lemmas . . . . . . . . . . . . . . 2.3 NITA Generated by b3 and Satisfying b32 = b4 + b5 . . . . . . . 2.4 General Information on NITA Generated by b3 and Satisfying b32 = b¯3 + b6 and b32 = c3 + b6 . . . . . . . . . . . . . . . . . . 2.5 NITA Generated by b3 Satisfying b32 = b¯3 + b6 and b6 Nonreal and b10 ∈ B is Real . . . . . . . . . . . . . . . . . . . . . . . . 2.6 NITA Generated by b3 Satisfying b32 = c3 + b6 , c3 = b3 , b¯ 3 , b6 Non-real, (b3 b8 , b3 b8 ) = 4 and c32 = r3 + s6 . . . . . . . . . . . 2.7 Structure of NITA Generated by b3 and Satisfying b32 = c3 + b6 , c3 = b3 , b¯3 , (b3 b8 , b3 b8 ) = 3 and c3 Non-real . . . . . . . . . . 2.8 Structure of NITA Generated by b3 and Satisfying b32 = c3 + b6 , c3 = b3 , b¯3 , (b3 b8 , b3 b8 ) = 3 and c3 Real . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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A Proof of a Non-existence of Sub-case (2) . . . . . 3.1 Introduction . . . . . . . . . . . . . . . . . . . 3.2 Preliminary Results . . . . . . . . . . . . . . . 3.3 Case z = z3 . . . . . . . . . . . . . . . . . . . . 3.4 Cases z = z4 , z = z5 , z = z6 , z = z7 , and z = z8 3.5 Case z = z9 . . . . . . . . . . . . . . . . . . . .

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151 151 152 158 174 180

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Preliminary Classification of Sub-case (3) 4.1 Introduction . . . . . . . . . . . . . . 4.2 Preliminary Results . . . . . . . . . . 4.3 Case R15 = x5 + x10 . . . . . . . . . . 4.4 Case R15 = x6 + x9 . . . . . . . . . .

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187 187 188 190 195

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Contents

4.5 Case R15 = x7 + x8 . . . . . . . . . . . . . . . . . . . . . . . . . 209 4.6 Case (b3 x7 , b3 x7 ) = 3 . . . . . . . . . . . . . . . . . . . . . . . . 233 4.7 Case b3 b10 = b15 + x5 + y5 + z5 . . . . . . . . . . . . . . . . . . 238 5

Finishing the Proofs of the Main Results 5.1 Introduction . . . . . . . . . . . . . 5.2 Proof of Theorem 5.1 . . . . . . . . 5.3 Proof of Theorem 5.2 . . . . . . . . References . . . . . . . . . . . . . .

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243 243 247 268 271

Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 273

Chapter 1

Introduction

The study of the properties of products of conjugacy classes of finite groups is an old branch of finite group theory. This topic was extensively studied in the 1980’s. The book [AH2], Products of Conjugacy Classes in Groups, edited by Z. Arad and M. Herzog, provides a comprehensive picture of the results obtained during this period. It was realized by several researchers that this investigation could be extended to products of irreducible characters. We refer the reader to the papers [AH1, BC]. This led to the notion of Table Algebra, introduced by Z. Arad and H. Blau in [AB], in order to study in a uniform way the decomposition of products of conjugacy classes Cla(G) and irreducible characters Irr(G) of finite groups G. Since then the theory of Table Algebras has been extensively developed. Today the theory of Table Algebras is an important branch of modern algebra. This book gives a classification of Normalized Integral Table Algebras (Fusion Rings) generated by a faithful non-real element of degree 3. It contains an Introduction and 4 chapters. Zvi Arad is a co-author of each chapter of the book and planned the outline of attack to the solution of the classification of the family of Table Algebras. The co-authors of the 5 chapters are as follows: The Introduction (Chap. 1) was written by Zvi Arad and Effi Cohen. Chapter 2 is a joint research of Zvi Arad, Guiyun Chen and Arisha Haj Ihia Hussam. Chapter 3 is a joint research of Zvi Arad and Xu Bangteng. Chapter 4 is a joint research of Zvi Arad and Effi Cohen. Chapter 5 is a joint research of Zvi Arad, Effi Cohen and Mikhail Muzychuk. We state the definition of Table Algebra and give two important examples of Table Algebras. Definition 1.1 Let B := {b1 = 1, b2 , . . . , bk } be a distinguished basis of a finite dimensional, associative and commutative algebra A over the complex field C with identity element 1A . Then (A, B) is a Table Algebra (B is a table basis and |B| is the dimension of the Table Algebra (A, B)) if the following hold: TA1. For all i, j , m, we have bi bj = km=1 λij m bm with λij m a nonnegative real number. Z. Arad et al., On Normalized Integral Table Algebras (Fusion Rings), Algebra and Applications 16, DOI 10.1007/978-0-85729-850-8_1, © Springer-Verlag London Limited 2011

1

2

1

Introduction

TA2. There is an algebra automorphism (denoted by –) of A whose order divides 2, such that bi ∈ B implies that b¯i ∈ B (then i¯ is defined by bi¯ = b¯i and bi ∈ B is called real if bi¯ = bi ). TA3. There is a coefficient function g : B × B −→ R+ such that λij m = g(bi , bm )λj¯mi where λij m , λj¯mi are defined in TA1 and TA2 for all i, j , m. Remark For all i, j , m as in TA1, we define Irr(bi bj ) to be the set {bm |λij m = 0}. The methods of research that were developed and the results of Table Algebras provide, in many cases, surprising and innovative results that were not known before. The results can be used in finite group theory and graph theory. Definition 1.2 Let (A, B) be a Table Algebra. A subset D ⊆ B is called a table subset of B if D = ∅ and Irr(bi bj ) ⊆ D for all bi , bj ∈ D. A subalgebra of A generated by some table subset of B is called a table subalgebra of (A, B). We say that (A, B) is simple if the only table subsets of B are B and {1}. Definition 1.3 Let (A, B) be a Table Algebra and fix c ∈ B. The set generated by i c is defined as Bc := { ∞ i=1 Irr(c )}. An element b ∈ B will be called faithful if Bb = B. Note that Bc is a table subset of B. Definition 1.4 If (A, B) is a Table Algebra, then, by [AB], there exists a unique ¯ > 0, ∀b ∈ B. We call | | the algebra homomorphism | | : A −→ R such that |b| = |b| degree homomorphism. The positive real numbers {|b|}b∈B are called the degrees of (A, B). An element bi ∈ B is called standard if |bi | = λi i1 ¯ . A Table Algebra (A, B) is called standard if all elements in B are standard. If for all bi ∈ B, λi i1 ¯ = 1, then (A, B) is a Normalized Table Algebra. A Table Algebra (A, B) is called homogenous of degree λ if |x| = λ for all x ∈ B{1}. A Table Algebra is called Integral if for all b ∈ B |b| ∈ N and for all i, j , m, λij m ∈ N ∪ {0}. The number 2 o(B) := ki=1 |bλ i¯| is called the order of B. In particular if (A, B) is Normalized, i i1 then o(B) = b∈B |b|2 . Definition 1.5 Two Table Algebras (A, B) and (A , B ) are called isomorphic (denoted B ∼ = B ) when there exists an algebra isomorphism ψ : A −→ A such that ψ(B) is a rescaling of B ; the algebras are called strictly isomorphic (denoted B∼ =x B means that B and B yield the =x B ) when ψ(B) = B . Therefore, B ∼ same structure constants. Table Algebras, as defined, may be considered a special class of C-algebras, introduced by Y. Kawada [K] and G. Hoheisel [H]. More precisely, a Table Algebra is a C-algebra where the structure constants are nonnegative. Each finite group yields two natural Table Algebras: the Table Algebra of conjugacy classes and the Table Algebra of irreducible characters.

1 Introduction

3

Example 1.1 Let G be a finite group. Let Ch(G) denote the set of all complex valued class functions on G, a commutative algebra under pointwise addition and multiplication. Let Irr(G) be the set of irreducible characters of G, a basis for Ch(G). Then (Ch(G), Irr(G)) satisfies all the axioms TA1–TA3 where—extends linearly from χi −→ χ¯ i (complex conjugate characters) and |χi | = χi (1) for all χi ∈ Irr(G). Here each λij m is a nonnegative integer and each g(χi , χj ) = 1. (Clearly, Example 1.1 is a Normalized Integral Table Algebra.) Example 1.2 Let G be a finite group and let Z(CG) be the center of the complex group algebra. If C1 = 1, C2 , . . . , Ck are the conjugacy classes of G. Let bi = g∈Ci g for each Ci , and let Cla(G) = {b1 , . . . , bk }. Then (Z(CG), Cla(G)) satisfies TA1–TA3, where—extends linearly from bi −→ b¯i = g∈Ci g −1 , and |bi | = |Ci |. Here also each λij m is a nonnegative integer, but g(bi , bm ) = |Ci |/|Cm |. (Clearly, Example 1.2 is a Standard Table Algebra.) “Groups occur as the algebraic abstraction of sets of symmetries, closed under composition and inversion, of geometric spaces. Table algebras may be regarded as playing a role analogous to that of groups for a generalization of geometric spaces; namely, for highly symmetric combinatorial configurations. Perhaps the most central instance of this perspective is the view of a Table Algebra as an abstraction of the adjacency algebra of an association scheme.” See [B2]. We can find the axioms of the Table Algebra (A, B) in many different types of algebras which were the subjects of intensive research over the last hundred years. In the excellent survey article [B2] of Harvey Blau, there is a comprehensive picture of Table Algebras and an explanation of the strong connection between Table Algebras to different types of algebras. The following is a partial list of those algebras: Schur rings, C-algebras, association schemes (Banai Itô), hypergroups (Wildberger), posets, Bose-Mesner algebras, Hopf algebras, distance regular graphs, products of conjugacy classes, products of ordinary characters, fusion rings (a concept from physics), fusion rule algebras, pseudo groups (Brauer), coherent algebras (Higman), generalized circulants of finite groups (Kazhdan-Lusztig). Normalized Integral Table Algebras are also known as integral commutative fusion rings, or integral fusion rule algebras. These are an important part of the theory of fusion categories, which is now a very active research area. It is related to conformal field theory and quantum physics, as well as to Hopf algebras, and is pursued independently of Table Algebras. Table Algebras generated by elements of small degree were extensively investigated, as follows: Homogenous Integral Table Algebras of degrees 1, 2, 3, 4; see [AB, A, B, ABFMMX] and [AEM]. Integral Table Algebras (denoted ITA) generated by a faithful element of degree 2 without non-identity basis elements of degree 1; see [B]. ITA generated by a faithful non-real element of degree 3, see [B1].

4

1

Introduction

Standard Integral Table Algebras (denoted SITA) generated by a faithful non-real element of degree 3; see [AAFM]. SITA generated by a faithful non-real element of degree 4; see [AM]. Normalized Integral Table Algebras (denoted NITA) generated by a faithful element of degree two without non-identity elements of degree 1; see [FK]. NITA generated by a faithful element of degree 3 without non-identity basis elements of degree 1 or 2; see [AC] and [AC1]. In [AC] and [AC1], the classification has not been completed. However, Arad and Chen have proved the following Theorem. We use the following notation. If (A, B) is a Table Algebra then bn , cn ∈ B denote elements of B of degree n. Theorem Let (A, B) be NITA generated by a non-real element b3 ∈ B of degree 3 and without non-identity basis elements of degree 1 or 2. Then b3 b¯3 = 1 + b8 , b8 ∈ B and one of the following holds: ∼x (CH(PSL(2, 7)), Irr(PSL(2, 7))). 1. (A, B) = 2. b32 = b4 + b5 where b4 , b5 ∈ B. 3. b32 = b¯3 + b6 where b6 ∈ B is a non-real element. 4. b32 = c3 + b6 , where c3 , b6 ∈ B and c3 = b3 , b¯3 . In this book we investigate Cases 2–4 of the above theorem, and we have made progress in order to complete the classification. To this end, we use the following definitions: Definition 1.6 Let B := {b1 = 1, b2 , . . . , bl , . . . } be a distinguished basis of a countable dimensional, associative and commutative algebra A over the complex field C with identity element 1A . If a ∈ A and bi ∈ B, let k(bi , a) be the coefficient of bi in a. Then (A, B) is a Countable Table Algebra (B is a table basis and |B| is the dimension of the Table Algebra (A, B)) if the following hold: TA1. For all i, j, m ∈ N bi bj = km=1 bij m bm where k ∈ N with bij m a nonnegative real number. TA2. There is an algebra automorphism (denoted by –) of A whose order divides 2, such that bi ∈ B implies that b¯i ∈ B (then i¯ is defined by bi¯ = b¯i and bi ∈ B is called real if bi¯ = bi ). TA3. There is a function g : B × B −→ R+ such that bij m = g(bi , bm )bj¯mi where bij m , bj¯mi are defined in TA1 and TA2 for all i, j , m. TA4. There exists a unique algebra homomorphism | | : A −→ C such that |b| = ¯ > 0, b ∈ B. We call it the degree homomorphism. The positive real num|b| bers {|b|}b∈ B are called the degrees of (A, B). If for all bi ∈ B, bi i1 ¯ = 1, then (A, B) is a Normalized Countable Table Algebra (denoted C-NTA) and if in addition for all b ∈ B |b| ∈ N and for all i, j , m, bij m ∈ N ∪ {0}, then (A, B) is a Normalized Integral Countable Table Algebra (denote by C-NITA).

1 Introduction

5

Note 1.1 If the dimension of a Countable Table Algebra is finite then by [AB, Lemma 2.9], axiom TA4 is a consequence of axioms TA1–TA3. In Chaps. 2–4 we assume that the Table Algebras are NITA of finite dimension. If we assume instead that the Table Algebras are C-NITA, then the same proofs of Chaps. 2–4 give us C-NITA of finite dimension, i.e., Table Algebras. Definition 1.7 Let D = {(k, n)|n ∈ N, k is an odd number with −1 ≤ k ≤ 2n + 1}. Define f : D → N ∪ {0} by f (k, n) =

(n + 1)(k + 1)(2n − k + 1) . 8

By definition, f (−1, n) = 0, f (2n + 1, n) = 0, f (1, 1) = 1 and f (1, 2) = 3. Definition 1.8 Let (A, B) be a C-NITA with a faithful non-real element b3 such that b32 = b¯3 + b6 where b6 ∈ B. For t ∈ N with t ≥ 2, or t = ∞, define an array Ct as Ct = {b(2j +1,i) }t+1;i i=1;j =−1 , where b(−1,i) = b(2i+1,i) = 0 for 1 ≤ i ≤ t + 1, b(k,i) ∈ B for all 1 ≤ i ≤ t + 1 and odd k with 1 ≤ k ≤ 2i − 1, and the following properties hold: 1. b(1,1) = 1, b(1,2) = b3 and b(k,i) = b¯(2i−k,i) for all 1 ≤ i ≤ t + 1 and odd k with 1 ≤ k ≤ 2i − 1. 2. b3 b(k,i) = b(k−2,i−1) + b(k+2,i) + b(k,i+1) for all 2 ≤ i ≤ t and odd k with 1 ≤ k ≤ 2i − 1. Remark If an array Ct exists for t > 2, then properties 1 and 2 imply that Ct−1 exists, with Ct−1 ⊆ Ct in the obvious sense. It is a straightforward consequence of properties 1 and 2 and induction on i that |b(k,i) | = f (k, i),

for all 1 ≤ i ≤ t + 1 and odd k with 1 ≤ k ≤ 2i + 1.

Definition 1.9 Let (A, B) be a C-NITA as in Definition 1.8. If for some n ∈ N there exists an array Cn for (A, B), but there does not exist an array Cn+1 , then n is called the stopping number of (A, B). All such Table Algebras with stopping number n ∈ N are denoted (A, B, Cn ). If there are arrays Cn for all n ∈ N, n ≥ 2, then the array C∞ exists. In this case, we say that the stopping number for (A, B) is ∞, and denote such a Table Algebra as (A, B, C∞ ). In either case, we say that (A, B) is a Cn -Table Algebra.

6

1

Introduction

Remarks a) We will show in Theorem 5.1 that there exists no C-NITA (A, B, Cn ) such that ∞ > n ≥ 43. The classification of C-NITA (A, B, Cn ) such that 4 ≤ n ≤ 42 is still open. If there exists a C-NITA (A, B, Cn ), it could be an infinite dimensional C-NITA. For a given n ∈ N, there may exist several (perhaps infinitely many) nonisomorphic (A, B, Cn ) of either finite or infinite dimension. b) A C-NITA (A, B) is called a C∞ -Table Algebra (denoted by (A, B, C∞ )) if C∞ exists and 1, 2 of Definition 1.8 hold. In this case b3 faithful implies that B = C∞ \ {0}. We will show in Chap. 5, Theorem 5.2 that (A, B, C∞ ) is a uniquely determined C-NITA. Note that (A, B, C∞ ) is an infinite dimensional C-NITA as defined in Definitions 1.8 and 1.9. Now we state the main theorem of this book. Main Theorem Let (A, B) be a C-NITA generated by a non-real element b3 ∈ B of degree 3 and without non-identity basis elements of degree 1 and 2. Then b3 b¯3 = 1 + b8 , b8 ∈ B, and (A, B) is of one of the following types: (1) (A, B) ∼ =x (A, B, C2 ) which is the unique C2 -Table Algebra and is strictly isomorphic to (CH PSL(2, 7), Irr PSL(2, 7)) of dimension 6. (2) (A, B) ∼ =x (A, B, C3 ) which is the unique C3 -Table Algebra and is strictly isomorphic to (CH(3 · A6 ), Irr(3 · A6 )) of dimension 17. (3) There exists a unique C∞ -Table Algebra which is strictly isomorphic to CNITA generated by the non-real polynomial representation of dimension 3 of SL(3, C). (4) There exists no C-NITA (A, B, Cn ) such that 43 ≤ n and the classification of the Table Algebra (A, B, Cn ) such that 4 ≤ n ≤ 42 is still open. (5) There exist c3 , b6 ∈ B, c3 = b3 or b¯3 , such that b32 = c3 + b6 and either (b3 b8 , b3 b8 ) = 3 or 4. If (b3 b8 , b3 b8 ) = 3 and c3 is non-real, then (A, B) ∼ =x (A(3 · A6 · 2), B32 ) of dimension 32. (See Theorem 2.9 of Chap. 2 for the definition of this specific NITA.) If (b3 b8 , b3 b8 ) = 3 and c3 is real, then (A, B) ∼ =x (A(7 · 5 · 10), B22 ) of dimension 22. (See Theorem 2.10 of Chap. 2 for the definition of this specific NITA.) The Case (b3 b8 , b3 b8 ) = 4 is still open. For the complete classification, we will have to solve the open problems in Cases 4 and 5 of the Main Theorem.

References [A]

Arad, Z.: Homogeneous integral table algebras of degrees two, three and four with a faithful element. In: Group St. Andrews 1997 in Bath, I. London Math. Soc. Lecture Notes Series, vol. 260, pp. 20–29 (1999)

References [AAFM]

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Arad, Z., Arisha, H., Fisman, E., Muzychuk, M.: Integral standard table algebras with a faithful nonreal element of degree 3. J. Algebra 231(2), 473–483 (2000) [AB] Arad, Z., Blau, H.I.: On table algebras and applications to finite group theory. J. Algebra 138, 137–185 (1991) [ABFMMX] Arad, Z., Blau, H.I., Fisman, E., Miloslavsky, V., Muzychuk, M., Xu, B.: Homogeneous Integral Table Algebra of Degree Three, A Trilogy, Memoirs of the AMS, vol. 144 (2000), no. 684 [AC] Arad, Z., Chen, G.: On normalized table algebras generated by a faithful non-real element of degree 3. In: Shum, K.P. (ed.) Advances in Algebra, Proc. of the ICN Satellite Conf. in Algebra and Related Topics, pp. 13–37. World Scientific, Singapore (Dec. 2003) [AC1] Arad, Z., Chen, G.: On four normalized table algebras generated by a faithful nonreal element of degree 3. J. Algebra 283, 457–484 (2005) [AEM] Arad, Z., Erez, Y., Muzychuk, M.: On homegeneous standard integral table algebras of degree 4. Comm. Algebra 34, 463–519 (2006) [AH1] Arad, Z., Herzog, M. (eds.): Survey on Products of Conjugacy Classes in Groups. Mathematical Lecture Notes Series, vol. 2, pp. 2.01–2.04 (1984) [AH2] Arad, Z., Herzog, M. (eds.): Products of Conjugacy Classes in Groups. LNM, vol. 1112. Springer Verlag, Berlin (1985) [AM] Arad, Z., Muzychuk, M. (eds.): Standard Integral Table Algebras Generated by a Non-real Element of Small Degree. Lecture Notes in Mathematics (2002) [B] Blau, H.I.: Homogenous integral table algebras of degree two. Algebra Colloq. 4(4), 393–408 (1997) [B1] Blau, H.I.: Integral table algebras and Bose-Mesner algebras with a faithful nonreal element of degree 3. J. Algebra 231(2), 484–545 (2000) [B2] Blau, H.I.: Table algebras. Eur. J. Comb. 30, 1426–1455 (2009) [BC] Blau, H.I., Chillag, D.: On powers of characters and powers of conjugacy classes of a finite group. Proc. Am. Math. Soc. 298, 7–10 (1986) [FK] Fröhlich, J., Kerler, T.: Quantum Groups, Quantom Categories and Quantum Field Theory, Lecture Notes in Mathematics, vol. 1542 (1993) [H] Hoheisel, G.: Uber Charaktere. Monatsch. f. Math. Phys. 48, 448–456 (1939) [K] Kawada, Y.: Uber den Dualitatssatz der Charaktere nichtcommutativer Gruppen. Proc. Phys. Math. Soc. Jpn. 24(3), 97–109 (1942)

Chapter 2

Splitting of the Main Problem into Four Sub-cases

2.1 Introduction Each of Chaps. 2–5 of this book is self-contained. Therefore the definitions in the different parts of the book may sometimes appear with different signs. The concept of Table Algebra was introduced by Z. Arad and H. Blau in [AB] in order to study in a uniform way properties of products of conjugacy classes and of irreducible characters of a finite group. Definition 2.1 A Table Algebra (A, B) is a finite dimensional, commutative algebra A with identity element 1 over the complex numbers C, and a distinguished base B = {b1 = 1, b2 , . . . , bk } such that the following properties hold (where (bi , a) denotes the coefficient of bi in a ∈ A, a written as a linear combination of B; and where R ∗ denotes R + ∪ {0}, the set of non-negative real numbers): (I) For all i, j , m, bi bj = m δij m bm with δij m a nonnegative real number. (II) There is an algebra automorphism (denoted by − ) of A, whose order divides 2, such that bi ∈ B implies that b¯i ∈ B. (Then i¯ is defined by b¯i = bi¯ .) (III) Hypothesis (II) holds and there is a function g : B × B → R + (the positive reals) such that (bm , bi bj ) = g(bi , bm ) · (bi , b¯j bm ), where g(bi , bm ) is independent of j for all i, j , m. B is called the table basis of (A, B). Clearly 1 ∈ B, we always use b1 to denote base element 1, and B to denote B \ {b1 }. The elements of B are called the irreducible components of (A, B), and nonzero nonnegative linear combinations of elements of B with coefficients in R + are called components. If a = km=1 λm bm is a component (λm ∈ R ∗ ) then Supp{a} = {bm |λm = 0} is called the set of irreducible constituents of a. An element a ∈ A is called a real element if a = a. ¯ Two Table Algebras (A, B) and (A , B ) are called isomorphic (denoted B ∼ = B ) when there exists an algebra isomorphism ψ : A → A such that ψ(B) is a rescaling of B ; and the algebras are called exactly isomorphic (denoted B ∼ =x B ) when ψ(B) = B . So B ∼ =x B means that B and B yield the same structure constants. Z. Arad et al., On Normalized Integral Table Algebras (Fusion Rings), Algebra and Applications 16, DOI 10.1007/978-0-85729-850-8_2, © Springer-Verlag London Limited 2011

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2 Splitting of the Main Problem into Four Sub-cases

Proposition 2.2 of [AB] shows that if (A, B) is a Table Algebra, then there exists a basis B , which consists of suitable positive real scalar multiples bi of the elements bi of B, g(bi , bj ) = 1 for any bi , bj ∈ B . Such a basis B is called a normalized basis. Now Supp(bi1 bi2 · · · bit ) consists of the corresponding scalar multiples of Supp(bi1 bi2 · · · bit ), for any sequence i1 , i2 , . . . , it , of indices. So in the proof of any proposition which identifies the irreducible constituents of a product of basis elements, we may assume that B is normalized. Suppose that B is normalized. It follows from (III) and [AB, Sect. 2] that A has a positive definite Hermitian form, with B as an orthonormal basis, and such that ¯ c) (a, bc) = (a b, for all a, b, c ∈ RB. It follows from Sect. 2 in [AB] that there exists an algebra homomorphism f from A to C such that f (B) ⊆ R + . For an element b ∈ B, f (b) is called the degree of b. Each finite group G yields two natural Table Algebras: the Table Algebra of conjugacy classes (denoted by (ZC(G), Cl(G))) and the Table Algebra of generalized characters (denoted by (Ch(G), Irr(G))). Both Table Algebras arising from group theory have an additional property: their structure constants and degrees are non-negative integers. Such algebras were defined in [B1] as Integral Table Algebras (denoted ITA). Each of the elements of a Table Algebra is contained in a unique table subalgebra which may be considered as a table subalgebra generated by this element. So it is natural to start the study of Integral Table Algebras from those which are generated by a single element. Normalized Integral Table Algebras (denoted NITA) generated by an element of degree 2 were completely classified by H. Blau in [B]. The finite linear groups in dimension n ≤ 7 have been completely classified. See Feit [F]. Representation theory and properties of finite groups are heavily used for the proofs. For n = 2, 3, 4 see Blichfeldt [Bl]. For n = 5 see Brauer [Br]. For n = 6 see Lindsey [L]. For n = 7 see Wales [W]. In order to generalize these results to Normalized Integral Table Algebras, the authors began to classify Normalized Integral Table Algebras (A, B) generated by a faithful nonreal element of degree 3 and without nonidentity irreducible elements of degree 1 or 2 in [CA], and arrived at the following theorem: Theorem 2.1 Let (A, B) be NITA generated by a nonreal element b3 ∈ B of degree 3 and without non-identity basis element of degree 1 or 2. Then b3 b¯3 = 1 + b8 and one of the following holds: (1) (A, B) ∼ =x (Ch(PSL(2, 7)), Irr(PSL(2, 7))); (2) b32 = b4 + b5 , where b4 ∈ B and b5 ∈ B; (3) b32 = b¯3 + b6 , where b6 ∈ B is a nonreal element; (4) b32 = c3 + b6 , where c3 , b6 ∈ B and c3 = b3 , b¯3 . The purpose of this chapter is to investigate NITA generated by a nonreal element b3 that satisfies one of the conditions (2), (3) and (4) in Theorem 2.1. The following theorem is proved:

2.2 Two NITA Generated by a Non-real Element of Degree 3

11

Main Theorem 1 Let (A, B) be a NITA generated by a nonreal element b3 ∈ B of degree 3 and without non-identity basis element of degree 1 or 2. Then b3 b¯3 = 1 + b8 , b8 ∈ B and one of the following holds: (1) There exists a real element b6 ∈ B such that b32 = b¯3 + b6 and (A, B) ∼ =x (Ch(PSL(2, 7)), Irr(PSL(2, 7))). (2) There exists b4 , b5 ∈B such that b32 = b4 +b5 , (b3 b8 , b3 b8 ) = 3 and (b42 , b42 ) = 3. (3) There exists b6 , b10 , b15 ∈ B, where b6 is nonreal such that b32 = b¯3 + b6 ,

b¯3 b6 = b3 + b15 ,

b3 b6 = b8 + b10

and (b3 b8 , b3 b8 ) = 3. Moreover if b10 is real, then (A, B) ∼ =x (CH(3 · A6 ), Irr(3 · A6 )). (4) There exists c3 , b6 ∈ B, c3 = b3 , b¯3 , such that b32 = c3 + b6 and either (b3 b8 , b3 b8 ) = 3 or 4. When (b3 b8 , b3 b8 ) = 3. If c3 is nonreal, then (A, B) is exactly isomorphic to (A(3 · A6 · 2), B32 ) (see Theorem 2.9). If c3 is real, then (A, B) is exactly isomorphic to the NITA (A(7 · 5 · 10), B22 ) of dimension 22 (see Theorem 2.10). Proof The theorem follows from Theorem 2.1 and Theorems 2.2, 2.7, 2.9 and 2.10 below.

2.2 Two NITA Generated by a Non-real Element of Degree 3 not Derived from a Group and Lemmas Here we give two examples of NITA not induced from group theory, which has either 32 or 22 basis elements. The NITA of dimension 32 contains a subalgebra which is strictly isomorphic to the algebra of dimension 17 of characters of the group 3 · A6 , we denote it as (A(3 · A6 · 2), B32 ), where its basis is B32 . For B32 there are three table subsets: {1} ⊆ C ⊆ D ⊆ B, where C = {1, b8 , x10 , b5 , c5 , c8 , x9 }, D = C ∪ {c3 , c¯3 , d3 , d¯3 , c9 , c¯9 , b6 , b¯6 , y5 , y¯15 }, B32 = D ∪ {b3 , b¯3 , r3 , x6 , x¯6 , s6 , b9 , b¯9 , x15 , x¯15 , t15 , d9 , y3 , z3 , z¯ 3 }. From the equations below one can check that B32 has the following table subsets: {1} ⊆ C ⊆ E = C ∪ {r3 , s6 , t15 , d9 , y3 } ⊆ B32 . The table subsets E and D are two maximal table subsets of B32 . The structure algebra constants of C: b52 = 1 + x9 + x10 + b5 , b5 c5 = c8 + b8 + x9 , b5 c8 = c5 + c8 + b8 + x9 + x10 , b5 b8 = c5 + c8 + b8 + x9 + x10 , b5 x9 = c8 + c5 + b8 + b5 + x9 + x10 , b5 x10 = c8 + b8 + x9 + b5 + 2x10 ,

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c52 = 1 + x9 + x10 + c5 , c5 c8 = b5 + c8 + b8 + x9 + x10 , c5 b8 = b5 + c8 + b8 + x9 + x10 , c5 x9 = b8 + c8 + b5 + c5 + x9 + x10 , c5 x10 = b8 + c8 + x9 + c5 + 2x10 , b82 = 1 + b5 + c5 + c8 + 2b8 + x9 + 2x10 , b8 x9 = b5 + c5 + 2c8 + b8 + 2x9 + 2x10 , b8 x10 = b5 + c5 + 2c8 + 2b8 + 2x9 + 2x10 , c82 = 1 + b5 + c5 + 2c8 + b8 + x9 + 2x10 , c8 b8 = b5 + c5 + c8 + b8 + 2x9 + 2x10 , c8 x9 = b5 + c5 + c8 + 2b8 + 2x9 + 2x10 , c8 x10 = b5 + c5 + 2c8 + 2b8 + 2x9 + 2x10 , x92 = 1 + b5 + c5 + 2c8 + 2b8 + 2x9 + 2x10 , x9 x10 = b5 + c5 + 2c8 + 2b8 + 2x9 + 3x10 , 2 = 1 + 2b + 2c + 2c + 2b + 3x + 2x . x10 5 5 8 8 9 10

The structure algebra constants of D (since C ⊆ D we will not repeat the equations of C as shown above): c3 c¯3 = 1 + b8 , c32 = c¯3 + b¯6 , c 3 d3 = x 9 , c3 d¯3 = c¯9 , c3 b6 = c¯3 + y¯15 , c3 b¯6 = b8 + x10 , c3 c9 = d3 + c¯9 + y¯15 , c3 c¯9 = c8 + x9 + x10 , c3 y15 = b¯6 + 2y¯15 + c¯9 , c3 y¯15 = b5 + c5 + c8 + b8 + x9 + x10 , c3 b8 = c3 + b6 + y15 , c3 x10 = b6 + c9 + y15 , c3 b5 = y15 , c3 c5 = y15 , c3 c8 = y15 + c9 , c3 x9 = y15 + c9 + d¯3 ,


d3 d¯3 = 1 + c8 , d32 = d¯3 + b6 , d3 b6 = c8 + x10 , d3 b¯6 = d¯3 + y15 , d3 c9 = b8 + x9 + x10 , d3 c¯9 = c3 + y15 + c9 , d3 y15 = b5 + c5 + c8 + b8 + x9 + x10 , d3 y¯15 = 2y15 + b6 + c9 , d3 b8 = c¯9 + y¯15 , d3 x10 = y¯15 + b¯6 + c¯9 , d3 b5 = y¯15 , d3 c5 = y¯15 , d3 c8 = y¯15 + b6 + d3 , d3 x9 = y¯15 + c¯3 + c¯9 , b6 b¯6 = 1 + b5 + c5 + c8 + b8 + x9 , b62 = 2b¯6 + y¯15 + c¯9 , b6 c9 = 2y¯15 + 2c¯9 + b¯6 , b6 c¯9 = b5 + c5 + c8 + b8 + 2x9 + x10 , b6 y¯15 = b5 + c5 + 2c8 + 2b8 + 2x9 + 3x10 , b6 y15 = 4y¯15 + b¯6 + c¯3 + d3 + 2c¯9 , b6 b5 = b6 + y15 + c9 , b6 c5 = y15 + b6 + c9 , b6 c8 = d¯3 + b6 + 2y15 + c9 , b6 b8 = c3 + b6 + c9 + 2y15 , b6 x9 = 2y15 + b6 + 2c9 , b6 x10 = 3y15 + c3 + d¯3 + c9 , y15 y¯15 = 1 + 3b5 + 3c5 + 5c8 + 5b8 + 6x9 + 6x10 , 2 = 9y¯ + 4b¯ + 2c¯ + 2d + 6c¯ , y15 15 6 3 3 9 y15 b5 = 3y15 + b6 + c3 + d¯3 + 2c9 , y15 c5 = 3y15 + b6 + c3 + d¯3 + 2c9 , y15 c8 = 5y15 + c3 + d¯3 + 2b6 + 3c9 , y15 b8 = 5y15 + c3 + d¯3 + 2b6 + 3c9 , y15 x9 = 6y15 + 2b6 + d¯3 + c3 + 3c9 , y15 x10 = 6y15 + 3b6 + c3 + d¯3 + 4c9 ,

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c9 c¯9 = 1 + 2c8 + 2b8 + 2x9 + 2x10 + c5 + b5 , c92 = 3y¯15 + d3 + c¯3 + 2b¯6 + 2c¯9 , c9 y15 = 6y¯15 + 2b¯6 + c¯3 + d3 + 3c¯9 , c9 y¯15 = 2b5 + 2c5 + 3c8 + 3b8 + 3x9 + 4x10 , c9 b5 = 2y15 + b6 + c9 , c9 c5 = 2y15 + b6 + c9 , c9 b8 = 3y15 + d¯3 + b6 + 2c9 , c9 c8 = 3y15 + b6 + c3 + 2c9 , c9 x9 = c3 + 2c9 + 3y15 + d¯3 + 2b6 , c9 x10 = c3 + 2c9 + 4y15 + d¯3 + b6 . From the equations one can see that C and D are table subsets of B32 . Furthermore, the NITA subalgebra (D , D) of dimension 17 is strictly isomorphic to (CH (3 · A6 ), Irr(3 · A6 )) while our NITA of dimension 32 is not induced from a finite group G. The following equations describe other products of basis elements in B32 : b3 b¯3 = 1 + b8 , b32 = c3 + b6 , b3 r3 = c¯3 + b¯6 , b3 x6 = c3 + y15 , b3 x¯6 = b8 + x10 , b3 s6 = c¯3 + y¯15 , b3 b9 = y15 + c9 + d¯3 , b3 b¯9 = x9 + c8 + x10 , b3 x15 = b6 + 2y15 + c9 , b3 x¯15 = b5 + c5 + b8 + c8 + x9 + x10 , b3 t15 = b¯6 + c¯9 + 2y¯15 , b3 d9 = c¯9 + y¯15 + d3 , b3 y3 = c¯9 , b3 z3 = x9 , b3 z¯ 3 = c9 , b3 b8 = b3 + x6 + x15 , b3 x10 = x6 + x15 + b9 , b3 b5 = x15 , b3 c5 = x15 , b3 c8 = x15 + b9 , b3 x9 = z¯ 3 + b9 + x15 , b3 c3 = r3 + s6 , b3 c¯3 = b¯3 + x¯6 , b3 d3 = b¯9 , b3 d¯3 = d9 , b3 c9 = t15 + d9 + y3 , b3 c¯9 = b¯9 + x¯15 + z3 , b3 b6 = r3 + t15 , b3 b¯6 = b¯3 + x¯15 , b3 y15 = s6 + 2t15 + d9 , b3 y¯15 = x¯6 + 2x¯15 + b¯9 ,


r32 = 1 + b8 , r3 x6 = c¯3 + y¯15 , r3 s6 = b8 + x10 , r3 b9 = d3 + c¯9 + y¯15 , r3 x15 = b¯6 + c¯9 + 2y¯15 , r3 t15 = b5 + c5 + b8 + c8 + x9 + x10 , r3 d9 = c8 + x9 + x10 , r3 y3 = x 9 , r3 z3 = c9 , r3 b8 = s6 + r3 + t15 , r3 x10 = s6 + t15 + d9 , r3 b5 = t15 , r3 c5 = t15 , r3 c8 = t15 + d9 , r3 x9 = t15 + d9 + y3 , r3 c3 = b¯3 + x¯6 , r3 d3 = b9 , r3 c9 = b¯9 + x¯15 + z3 , r3 b6 = x¯15 + b¯3 , r3 y15 = x¯6 + 2x¯15 + b¯9 , x62 = 2b6 + y15 + c9 , x6 x¯6 = 1 + b8 + b5 + c5 + c8 + x9 , x6 s6 = 2b¯6 + c¯9 + y¯15 , x6 b9 = b¯6 + 2c¯9 + 2y¯15 , x6 b¯9 = b5 + c5 + x10 + b8 + c8 + 2x9 , x6 x15 = 4y15 + b6 + 2c9 + d3 + c3 , x6 x¯15 = b5 + c5 + 2b8 + 3x10 + 2c8 + 2x9 , x6 t15 = c¯3 + d3 + 2c¯9 + 4y¯15 + b¯6 , x6 d9 = b¯6 + 2c¯9 + 2y¯15 , x6 y3 = d3 + y¯15 , x6 z3 = x10 + c8 , x6 z¯ 3 = d¯3 + y15 , x6 b8 = b3 + x6 + 2x15 + b9 , x6 x10 = b9 + b3 + z¯ 3 + 3x15 , x6 b5 = x6 + b9 + x15 , x6 c5 = x6 + b9 + x15 , x6 c8 = x6 + b9 + 2x15 + z¯ 3 , x6 x9 = 2b9 + x6 + 2x15 , x6 c3 = r3 + t15 , x6 c¯3 = b¯3 + x¯15 , x6 d3 = z3 + x¯15 , x6 d¯3 = y3 + t15 , x6 c9 = 2t15 + 2d9 + s6 , x6 c¯9 = 2x¯15 + x¯6 + 2b¯9 , x6 b6 = 2s6 + t15 + d9 , x6 b¯6 = 2x¯6 + b¯9 + x¯15 , x6 y15 = r3 + 4t15 + s6 + 2d9 + y3 , x6 y¯15 = b¯3 + z3 + x¯6 + 4x¯15 + 2b¯9 ,

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s62 = 1 + b5 + c5 + b8 + c8 + x9 , s6 b9 = b¯6 + 2c¯9 + 2y¯15 , s6 x15 = c¯3 + d3 + b¯6 + 2c¯9 + 4y¯15 , s6 t15 = b5 + c5 + 2b8 + 2c8 + 2x9 + 3x10 , s6 d9 = b5 + c5 + 2x9 + b8 + c8 + x10 , s6 y3 = c8 + x10 , s6 z3 = d¯3 + y15 , s6 b8 = r3 + s6 + 2t15 + d9 , s6 x10 = r3 + y3 + d9 + 3t15 , s6 b5 = s6 + d9 + t15 , s6 c5 = s6 + t15 + d9 , s6 c8 = y3 + s6 + d9 + 2t15 , s6 x9 = s6 + 2d9 + 2t15 , s6 c3 = b¯3 + x¯15 , s6 d3 = z¯ 3 + x15 , s6 c9 = x¯6 + 2b¯9 + 2x¯15 , s6 b6 = 2x¯6 + b¯9 + x¯15 , s6 y15 = b¯3 + z3 + 4x¯15 + x¯6 + 2b¯9 , b9 b¯9 = 1 + b5 + c5 + 2b8 + 2c8 + 2x9 + 2x10 , b92 = d¯3 + c3 + 2b6 + 2c9 + 3y15 , b9 x15 = 6y15 + 3c9 + d¯3 + 2b6 + c3 , b9 x¯15 = 2b5 + 2c5 + 3c8 + 3b8 + 4x10 + 3x9 , b9 t15 = d3 + c¯3 + 6y¯15 + 3c¯9 + 2b¯6 , b9 d9 = c¯3 + d3 + 2b¯6 + 2c¯9 + 3y¯15 , b9 y3 = c¯3 + y¯15 + c¯9 , b9 z3 = b8 + x9 + x10 , b9 z¯ 3 = c9 + c3 + y15 , b9 b8 = z¯ 3 + x6 + 2b9 + 3x15 , b9 x10 = z¯ 3 + b3 + 2b9 + x6 + 4x15 , b9 b5 = x6 + b9 + 2x15 , b9 c5 = b9 + 2x15 + x6 , b9 c8 = r3 + t15 + 2b9 + x6 + 2x15 , b9 x9 = 2b9 + 3x15 + z¯ 3 + b3 + 2x6 , b9 c3 = t15 + d9 + y¯3 , b9 c¯3 = z3 + b¯9 + x¯15 , b9 d3 = b¯3 + b¯9 + x¯15 , b9 d¯3 = r3 + t15 + d9 , b9 c9 = r3 + y3 + 2s6 + 2d9 + 3t15 , b9 c¯9 = b¯3 + z3 + 2x¯6 + 3x¯15 + 2b¯9 , b9 b6 = s6 + 2d9 + 2t15 , b9 b¯6 = 2b¯9 + 2x¯15 + x¯6 , b9 y¯15 = 6x¯15 + b¯3 + z3 + 2x¯6 + 3b¯9 , b9 y15 = r3 + y3 + 2s6 + 6t15 + 3d9 ,

2.2 Two NITA Generated by a Non-real Element of Degree 3 2 = 2c + 2d¯ + 4b + 6c + 9y , x15 3 3 6 9 15 x15 x¯15 = 1 + 6x9 + 3b5 + 3c5 + 5b8 + 5c8 + 6x10 , x15 t15 = 4b¯ 6 + 6c¯9 + 9y¯15 + 2c¯3 + 2d3 , x15 d9 = c¯3 + d3 + 2b¯6 + 3c¯9 + 6y¯15 , x15 y3 = b¯6 + c¯9 + 2y¯15 , x15 z¯ 3 = b6 + c9 + 2y15 , x15 z3 = b5 + c5 + b8 + c8 + x9 + x10 , x15 b8 = b3 + z¯ 3 + 2x6 + 5x15 + 3b9 , x15 x10 = b3 + z¯ 3 + 3x6 + 4b9 + 6x15 , x15 b5 = z¯ 3 + b3 + x6 + 2b9 + 3x15 , x15 c5 = b3 + z¯ 3 + x6 + 2b9 + 3x15 , x15 c8 = b3 + z¯ 3 + 2x6 + 3b9 + 5x15 , x15 x9 = b3 + z¯ 3 + 2x6 + 3b9 + 6x15 , x15 c3 = 2t15 + s6 + d9 , x15 c¯3 = 2x¯ 15 + x¯6 + b¯9 , x15 d3 = x¯6 + 2x¯15 + b¯9 , x15 d¯3 = s6 + 2t15 + d9 , x15 c9 = r3 + y3 + 2s6 + 6t15 + 3d9 , x15 c¯9 = b¯3 + z3 + 3b¯9 + 2x¯6 + 6x¯15 , x15 b6 = r3 + s6 + y3 + 2d9 + 4t15 , x15 b¯6 = b¯3 + z3 + 2b¯9 + 4x¯15 + x¯6 , x15 y15 = 2y3 + 2r3 + 4s6 + 6d9 + 9t15 , x15 y¯15 = 2z3 + 9x¯15 + 2b¯3 + 4x¯6 + 6b¯9 , 2 = 1 + 5b + 5c + 3b + 3c + 6x + 6x , t15 8 8 5 5 10 9 t15 d9 = 2b5 + 2c5 + 3b8 + 3c8 + 3x9 + 4x10 , t15 y3 = b5 + c5 + b8 + c8 + x9 + x10 , t15 z3 = b6 + c9 + 2y15 , t15 b8 = r3 + y3 + 5t15 + 2s6 + 3d9 , t15 x10 = r3 + y3 + 4d9 + 3s6 + 6t15 , t15 b5 = r3 + y3 + s6 + 2d9 + 3t15 , t15 c5 = r3 + y3 + s6 + 2d9 + 3t15 , t15 c8 = r3 + y3 + 2s6 + 3d9 + 5t15 , t15 x9 = r3 + y3 + 3d9 + 2s6 + 6t15 , t15 c3 = x¯6 + b¯9 + 2x¯15 , t15 d3 = x6 + b9 + 2x15 , t15 c9 = b¯3 + z3 + 3b¯9 + 2x¯6 + 6x¯15 , t15 b6 = b¯3 + z3 + x¯6 + 2b¯9 + 4x¯15 , t15 y15 = 2b¯ 3 + 2z3 + 4x¯6 + 6b¯9 + 9x¯15 ,

17

18


d92 = 1 + b5 + c5 + 2b8 + 2c8 + 2x9 + 2x10 , d9 y3 = x9 + b8 + x10 , d9 z3 = c3 + c9 + y15 , d9 b8 = y3 + s6 + 2d9 + 3t15 , d9 x10 = y3 + r3 + s6 + 2d9 + 4t15 , d9 b5 = s6 + d9 + 2t15 , d9 c5 = s6 + d9 + 2t15 , d9 c8 = r3 + s6 + 2d9 + 3t15 , d9 x9 = r3 + y3 + 2s6 + 2d9 + 3t15 , d9 c3 = b¯9 + z3 + x¯15 , d9 d3 = b3 + b9 + x15 , d9 c9 = b¯3 + z3 + 2x¯6 + 2b¯9 + 3x¯15 , d9 b6 = x¯6 + 3b¯9 + 2x¯15 , d9 y15 = b¯3 + z3 + 2x¯6 + 3b¯9 + 6x¯15 , y32 = 1 + c8 , y3 z3 = d¯3 + b6 , y3 b8 = d9 + t15 , y3 x10 = s6 + t15 + d9 , y3 b5 = t15 , y3 c5 = t15 , y3 c8 = s6 + t15 + y3 , y3 x9 = t15 + d9 + r3 , y3 c3 = b¯9 , y3 d3 = x6 + z¯ 3 , y3 c9 = b¯3 + b¯9 + x¯15 , y3 b6 = x¯15 + z3 , y3 y15 = x¯6 + b¯9 + 2x¯15 , z3 z¯ 3 = 1 + c8 , z32 = d3 + b¯6 , z3 b8 = b¯9 + x¯15 , z3 x10 = x¯6 + b¯9 + x¯15 , z3 b5 = x¯15 , z3 c5 = x¯15 , z3 c8 = x¯6 + x¯15 + z3 , z3 x9 = b¯3 + b¯9 + x¯15 , z3 c3 = b9 , z3 c¯3 = d9 , z3 d3 = s6 + y3 , z3 d¯3 = z¯ 3 + x6 , z3 c9 = b3 + b9 + x15 , z3 c¯9 = r3 + d9 + t15 , z3 b6 = z¯ 3 + x15 , z3 b¯6 = y3 + t15 , z3 y15 = x6 + b9 + 2x15 , z3 y¯15 = s6 + 2t15 + d9 .


19

Now let us see the following example of NITA of dimension 22, which satisfies b32 = r3 + s6 , r3 is real, denoted as (A(7 · 5 · 10), B22 ), where B22 = {1, b8 , x10 , b5 , c5 , c8 , x9 , r3 , y3 , s6 , t15 , d9 , b3 , b¯3 , t6 , t¯6 , b15 , b¯15 , y9 , y¯9 , x3 , x¯3 }. The interesting thing is that this example also contains table subsets: {1} ⊆ C ⊆ E ⊆ B22 , where C and E are as defined in the previous example of dimension 32. In particular both examples of dimensions 22 and 32 each contain the same sub-table algebra of dimension 12 generated by the basis E. Also E is a maximal table subset of B22 . The equations of products of the basis elements of C and E are as in the example of A(3 · A6 · 2, 32) described above b3 b¯3 = 1 + b8 , b32 = r3 + s6 , b3 t6 = b8 + x10 , b3 t¯6 = r3 + t15 , b3 b15 = 2t15 + s6 + d9 , b3 b¯15 = b8 + x10 + b5 + c5 + c8 + x9 , b3 y9 = t15 + d9 + y3 , b3 y¯9 = c8 + x9 + x10 , b3 x3 = d9 , b3 x¯3 = x9 , b3 b8 = b3 + t¯6 + b15 , b3 x10 = b15 + t¯6 + y9 , b3 b5 = b15 , b3 c5 = b15 , b3 c8 = b15 + y9 , b3 x9 = b15 + y9 + x3 , b3 r3 = b¯3 + t6 , b3 y3 = y¯9 , b3 s6 = b¯3 + b¯15 , b3 t15 = 2b¯15 + t6 + y¯9 , b3 d9 = b¯15 + y¯9 + x¯3 ,

20


t6 t¯6 = 1 + b5 + c5 + b8 + c8 + x9 , t62 = 2s6 + t15 + d9 , t6 b15 = 2b8 + 2c8 + b5 + c5 + 3x10 + 2x9 , t6 b¯15 = c3 + b6 + 4t15 + y3 + 2d9 , t6 y9 = b8 + b5 + c5 + c8 + 2x9 + x10 , t6 y¯9 = s6 + 2d9 + 2t15 , t6 x3 = x10 + c8 , t6 x¯3 = t15 + y3 , t6 b8 = b¯3 + 2b¯15 + t6 + y¯9 , t6 x10 = 3b¯ 15 + y¯9 + b¯3 + x¯3 , t6 b5 = b¯15 + t6 + y¯9 , t6 c5 = b¯15 + t6 + y¯9 , t6 c8 = 2b¯ 15 + t6 + y¯9 + x¯3 , t6 x9 = 2b¯ 15 + t6 + 2y¯9 , t6 r3 = b15 + b3 , t6 y3 = b15 + x3 , t6 s6 = 2t¯6 + b15 + y9 , t6 t15 = 2y9 + x3 + b3 + t¯6 + 4b15 , t6 d9 = 2y9 + 2b15 + t¯6 , b15 b¯15 = 1 + 3b5 + 3c5 + 5b8 + 5c8 + 6x10 + 6x9 , 2 = 2r + 2y + 4s + 6d + 9t , b15 3 3 6 9 15 b15 y9 = r3 + y3 + 2s6 + 3d9 + 6t15 , b15 y¯9 = 2b5 + 2c5 + 3b8 + 3c8 + 3x9 + 4x10 , b15 x3 = s6 + 2t15 + d9 , b15 x¯3 = c5 + c8 + x9 + b8 + x10 + b5 , b15 b8 = 5b15 + 3y9 + x3 + b3 + 2t¯6 , b15 x10 = 6b15 + 4y9 + b3 + x3 + 3t¯6 , b15 b5 = b3 + x3 + t¯6 + 2y9 + 3b15 , b15 c5 = 3b15 + 2y9 + b3 + x3 + t¯6 , b15 c8 = 5b15 + 3y9 + x3 + b3 + 2t¯6 , b15 x9 = 6b15 + 3y9 + x3 + b3 + 2t¯6 , b15 r3 = 2b¯ 15 + t6 + y¯9 , b15 y3 = 2b¯ 15 + y¯9 + t6 , b15 s6 = 4b¯ 15 + 2y¯9 + b¯3 + x¯3 + t6 , b15 t15 = 9b¯ 15 + 6y¯9 + 2b¯3 + 2x¯3 + 4t6 , b15 d9 = 6b¯ 15 + 3y¯9 + b¯3 + x¯3 + 2t6 ,


21

y9 y¯9 = 1 + b5 + c5 + 2b8 + 2c8 + 2x9 + 2x10 , y92 = r3 + y3 + 2s6 + 2d9 + 3t15 , y9 x3 = r3 + d9 + t15 , y9 x¯3 = b8 + x10 + x9 , y9 b8 = t¯6 + 2y9 + 3b15 + x3 , y9 x10 = t¯6 + 2y9 + b3 + x3 + 4b15 , y9 b5 = t¯6 + y9 + 2b15 , y9 c5 = t¯6 + y9 + 2b15 , y9 c8 = 3b15 + b3 + 2y9 + t¯6 , y9 x9 = 2t¯6 + 2y9 + 3b15 + x3 + b3 , y9 r3 = b¯15 + y¯9 + x¯3 , y9 y3 = b¯15 + b¯3 + y¯9 , y9 s6 = 2b¯ 15 + t6 + 2y¯9 , y9 t15 = 6b¯ 15 + 2t6 + b¯3 + 3y¯9 + x¯3 , y9 d9 = 3b¯ 15 + 2t6 + b¯3 + 2y¯9 + x¯3 , x3 x¯3 = 1 + c8 , x32 = s6 + y3 , x3 b8 = b15 + y9 , x3 x10 = b15 + t¯6 + y9 , x3 b5 = b15 , x3 c5 = b15 , x3 c8 = x3 + b15 + t¯6 , x3 x9 = b3 + y9 + b15 , x3 r3 = y¯9 , x3 y3 = x¯3 + t6 , x3 s6 = b¯15 + x¯3 , x3 t15 = 2b¯ 15 + t6 + y¯9 , x3 d9 = b¯15 + b¯3 + y¯9 . If we define a Table Algebra (A, B) for R, S ⊆ B, R ∗ S = {∪ Supp(R · S)|r ∈ R, s ∈ S}, then: Cb3 ∗ C b¯3 = Cr32 = Cc3 ∗ C c¯3 = C, C is an identity. (Cb3 )2 = C c¯3 , Cb3 ∗ Cc3 = Cr3 , Cb3 ∗ C c¯3 = C b¯3 and Cb3 Cr3 = C c¯3 . For the definitions of abelian table algebra, quotient of table algebra and its basis and the sub-table algebra generated by an element b ∈ B denoted by b = Bb , see [AB] and [B1]. The definitions of a linear and faithful element b ∈ Bcan be also found there. ∞ i i One can derive from Table 2.1 that B32 = 10 Bb = i=1 i=0 Bb3 and b3 is a 3 faithful element of B32 that generates B32 . Also the quotient table algebra with basis B32 /C is strictly isomorphic to the Table Algebra induced by the cyclic group Z6 of order 6. i One can derive from Table 2.2 that B22 = 7i=1 Bb3i = ∞ i=0 Bb3 and b3 is a faithful element of B22 generates B22 . Also the quotient table algebra with basis

22


Table 2.1 NITA (A(3 · A6 · 2), B32 ) The powers of b3

Constituents of the powers

Basis of quotient of B32 /C

b32

c3 , b6

C = {1, b8 , x10 , b5 , c5 , c8 , x9 }

b33

r3 , s6 , t15 c¯3 , b¯6 , y¯15 , c¯9

Cb3 = {b3 , x6 , x15 , b9 , z¯ 3 } C b¯3 = {b¯3 , x¯6 , x¯15 , b¯9 , z3 }

b¯ 3 , x¯6 , x¯15 , b¯9 , z3

Cr3 = {r3 , t15 , d9 , y3 , s6 } Cc3 = {c3 , b6 , y15 , c9 , d¯3 }

b34 b35 b36 b37 b38 b39 b310

1, b8 , x10 , b5 , c5 , c8 , x9 b3 , x6 , x15 , b9 , z¯ 3 c3 , b6 , y15 , c9 , d¯3

C c¯3 = {c¯3 , b¯6 , y¯15 , c¯9 , d3 }

r3 , s6 , t15 , d9 , y3 c¯3 , b¯6 , y¯15 , c¯9 , d3

Table 2.2 NITA (A(7 · 5 · 10), B22 ) The powers of b3

New constituents of the powers

Basis of quotient of B32 /C

b32

r3 , s6 b¯3 , t6 , b¯ 15

C = {1, b8 , x10 , b5 , c5 , c8 , c9 } C b¯3 = {b¯3 , t6 , b¯15 , y¯9 , x¯3 }

1, b8 , x10 , b5 , c5 , c8 , x9 b3 , t¯6 , b15 , y9 , x3

Cr3 = {r3 , s6 , t15 , d9 , y3 } Cb3 = {b3 , t¯6 , b15 , y9 , x3 }

b33 b34 b35 b36 b37

r3 , s6 , t15 , d9 , y15 , y3 b¯3 , t6 , b¯ 15 , y¯9 , x¯3

B22 /C is strictly isomorphic to the Table Algebra induced by the cyclic group Zn of order 4. In order to prove the Main Theorem 2.1, the following fact is frequently used: Lemma 2.1 Let em , fm , un , vn ∈ B such that em e¯m = fm f¯m . Then (em un , em un ) = (e¯m un , e¯m un ) = (fm vn , fm vn ) = (f¯m vn , f¯m vn ). Proof (1) follows from (em un , em un ) = (em e¯m , un u¯ n ) = (fm f¯m , un u¯ n ) = (fm un , fm un ). Note that (t3 t¯3 , b8 ), (s4 s¯4 , b8 ) ≤ 1, and we obtain (2). Lemma 2.2 Let b3 , t3 , s4 ∈ B, b3 b¯3 = 1 + b8 . Then (b3 t3 , b3 t3 ), (b3 s4 , b3 s4 ) ≤ 2. If (b3 t3 , b3 t3 ) = 2, then t3 t¯3 = 1 + b8 . Proof If (b3 x4 , b3 x4 ) ≥ 3, then (x4 x¯4 , b8 ) ≥ 2 for b3 b¯3 = 1 + b8 , a contradiction. (1) follows. If (b3 c3 , b3 c3 ) = 2, then (b3 b¯3 , c3 c¯3 ) = 2. (2) follows from b3 b¯3 = 1 + b8 .

2.3 NITA Generated by b3 and Satisfying b32 = b4 + b5

23

2.3 NITA Generated by b3 and Satisfying b32 = b4 + b5 In this section, we shall investigate NITA generated by b3 and satisfying b32 = b4 + b5 , and obtain the following theorem: Theorem 2.2 Let (A, B) be a NITA generated by a nonreal element b3 ∈ B of degree 3 and without non-identity basis element of degree 1 or 2. Then b3 b¯3 = 1 + b8 , b8 ∈ B. Assume that b32 = b4 + b5 , b4 ∈ B and b5 ∈ B. Then (b¯5 b5 , b8 ) = 1, (b42 , b42 ) = 3 and (b3 b8 , b3 b8 ) = 3. Now we start to investigate NITA satisfying b3 b¯3 = 1 + b8 , b32 = b4 + b5 .

(2.1) (2.2)

Proof Based on the above equations, one may set b¯3 b4 = b3 + b9 ,

some b9 ∈ N ∗ B.

(2.3)

Since (b4 b¯4 , b8 ) ≤ 1, we have that (b¯3 b4 , b¯3 b4 ) = 2, b9 ∈ B and b4 b¯4 = 1 + b8 + c7 ,

where c7 ∈ N ∗ B.

(2.4)

Collecting the above equations, checking the associative laws for basis elements and making the necessary calculations and assumptions, we have the following lemma. Lemma 2.3 The following equations always hold: b¯3 b4 = b3 + b9 , b4 b¯4 = 1 + b8 + c7 , c7 ∈ N ∗ B, b¯3 b5 = b3 + x12 , x12 ∈ N ∗ B b3 b8 = b3 + b9 + x12 b3 c7 = b9 + f12 , f12 ∈ N ∗ B b4 b¯9 = b3 + b9 + x12 + f12 b4 b8 = b5 + b3 b9 b¯4 b5 = b8 + z12 , z12 ∈ N ∗ B b¯3 b9 = c7 + b8 + z¯ 12 b82 = 1 + c7 + b8 + z¯ 12 + b¯3 x12 b8 c7 = b8 + z¯ 12 + b¯3 f12 b5 b¯9 = b9 + b3 z12 b9 b¯9 = 1 + b8 − z12 + z¯ 12 + b¯3 f12 + b¯3 x12

assumption, calculating b32 b¯3 = b3 (b3 b¯3 ), assumption, calculating b3 (b¯4 b4 ) = (b3 b¯4 )b4 , calculating b3 (b¯3 b4 ) = (b3 b¯3 )b4 , assumption, calculating b¯32 b4 = b¯3 (b¯3 b4 ), calculating (b3 b¯3 )b8 = (b3 b8 )b¯3 , calculating (b3 b¯3 )c7 = (b3 b7 )b¯3 , calculating (b¯3 b4 )b¯5 = b¯ 3 (b4 b¯5 ), calculating (b4 b¯4 )b8 = b¯ 4 (b4 b8 ).

24


Proof Based on the known equations, checking associative laws one by one and making necessary assumptions, one can easily obtain the equations above. The proof of the last equation follows from (b4 b¯4 )b8 = b8 + b82 + b8 c7 = b8 + 1 + c7 + b8 + z¯ 12 + b¯3 x12 + b8 + z¯ 12 + b¯3 f12 , b¯4 (b4 b8 ) = b¯4 b5 + (b¯4 b3 )b9 = b¯4 b5 + b¯3 b9 + b9 b¯9 = b8 + z12 + c7 + b8 + z¯ 12 + b9 b¯9 . Obviously (b5 b¯5 , b8 ) ≤ 3, c7 is either c3 + c4 or irreducible. If c7 ∈ B, then there exist c3 , c4 ∈ B such that c7 = c3 + c4 . First we have the following lemma: Lemma 2.4 There exists no NITA generated by b3 and satisfying b3 b¯3 = 1 + b8 , b32 = b4 + b5 and c7 = c3 + c4 . Proof By the assumption, we have that b4 b¯4 = 1 + b8 + c3 + c4 .

(2.5)

Then b32 b¯4 = b4 b¯4 + b5 b¯4 = 1 + c3 + c4 + b8 + b¯4 b5 , b3 (b3 b¯4 ) = b3 b¯3 + b3 b¯9 = 1 + b8 + b3 b¯9 . Hence (c3 , b3 b¯9 ) = (c4 , b3 b¯9 ) = 1. Therefore b3 c3 = b9 ,

(2.6)

b3 c4 = d3 + b9 ,

some d3 ∈ B, d3 = b3 .

(2.7)

Now one has (b3 d¯3 , c4 ) = 1. Then b3 d¯3 = c4 + d5 ,

some d5 ∈ B,

(2.8)

from which (b3 b¯3 , d3 d¯3 ) = 2 follows. Hence d3 d¯3 = 1 + b8 .

(2.9)


25

By (2.5), we have that (b4 c3 , b4 ) = 1. Let y8 ∈ N ∗ B.

b4 c3 = b4 + y8 ,

(2.10)

Since (b3 b¯3 )c3 = c3 + c3 b8 , b¯3 (b3 c3 ) = b¯ 3 b9 , one has that b¯3 b9 = c3 + c3 b8 . But (c4 , b¯3 b9 ) = 1. Then (c4 , c3 b8 ) = 1 and so (c4 c3 , b8 ) = 1. Thus c3 c4 = x4 + b8 ,

some x4 ∈ B.

(2.11)

By (2.10) we have that 2 ≤ (b4 c3 , b4 c3 ) = (b4 b¯4 , c32 ) = (1 + c3 + c4 + b8 , c32 ), which implies that c32 equals one of 1 + b8 , c32

1 + c3 + x5 ,

1 + c 4 + y4 ,

1 + 2c4 .

By (2.11), we know that c32 cannot be 1 + c4 + y4 or 1 + 2c4 . And by (2.6), = 1 + b8 . Hence c32 = 1 + c3 + x5 .

(2.12)

Therefore (b4 c3 , b4 c3 ) = (c32 , b4 b¯4 ) = 2, which implies that y8 ∈ B. By Lemmas 2.3, (2.8) and (2.9), we have b3 (b3 b¯3 ) = b3 + b3 b8 = 2b3 + b9 + x12 , b¯ 3 (b32 ) = b¯3 b4 + b¯3 b5 = b3 + b9 + b¯3 b5 , (b3 d¯3 )d3 = c4 d3 + d5 d3 . Hence b3 b8 = b9 + b¯3 b5 , and b9 ∈ Supp{c4 d3 + d5 d3 }. Assume that b9 ∈ Supp{d3 d5 }. By (2.8), it follows that d3 d5 = b9 + b3 + l3 ,

l3 ∈ B,

b¯ 3 b5 = c4 d3 + l3 , so that b3 b8 = b9 + l3 + c4 d3 , which implies that b8 ∈ Supp{b3 l¯3 }. Hence l3 = b3 . So (d3 d5 , b3 ) = 3, which implies that (b4 d¯3 , d5 ) = 2, a contradiction to (2.8). Thus b9 ∈ Supp{c4 d3 } and c4 d3 = b3 + b9 .

(2.13)

Thus (d3 d¯3 , c42 ) = 2, we may assume that c42 = 1 + b8 + f7 ,

some f7 ∈ N ∗ B.

(2.14)

Then c4 ∈ Supp{c4 x5 }. Otherwise, c4 ∈ Supp{c4 x5 }, which implies that x5 ∈ Supp{c42 }, and so f7 = x5 + x2 , a contradiction.

26


Since c32 c4 = c4 + c3 c4 + x5 c4 = c4 + x4 + b8 + c4 x5 , c3 (c3 c4 ) = c3 b8 + c3 x4 , we have that c3 b8 + c3 x4 = c4 + x4 + b8 + c4 x5 . But the left hand side of the above equation contains two c4 by (2.11). Hence c4 = x4 by c4 ∈ Supp{c4 x5 }. Therefore c3 c4 = b8 + c4 ,

(2.15)

which implies that (c42 , c3 ) = 1 and c42 = 1 + b8 + c3 + e4 ,

some e4 ∈ B.

Now we have the following equations: b3 c42 = b3 + b3 b8 + b3 c3 + b3 e4 = 2b3 + 2b9 + x12 + b3 e4 , by (2.7) and (2.13)

c4 (b3 c4 ) = c4 b9 + c4 d3 = c4 b9 + b9 + b3 .

Then c4 b9 = b3 + b9 + x12 + b3 e4 . Since d3 ∈ Supp{c4 b9 }, one has that d3 ∈ Supp{x12 } or b3 e4 . If d3 ∈ Supp{x12 }, then d3 ∈ Supp{b3 b8 }, which implies that b3 d¯3 = 1 + b8 . Thus b3 = d3 , which implies that b3 b¯3 = c4 + d5 , a contradiction. Hence d3 ∈ Supp{b3 e4 }, e4 ∈ Supp{b3 d¯3 }. Therefore by (2.8) e4 = c4 and by (2.7) c42 = 1 + c3 + c4 + b8 = b4 b¯4 , c4 b9 = b3 + d3 + 2b9 + x12 . Furthermore b3 c42 = b3 + b3 b8 + b3 c4 + b3 c3 = b3 + b3 + b9 + x12 + d3 + b9 + b9 , (b3 b¯4 )b4 = b¯3 b4 + b¯9 b4 = b3 + b9 + b4 b¯9 , (b3 c4 )c4 = c4 b9 + d3 c4 = c4 b 9 + b 3 + b 9 .


27

Thus b4 b¯9 = b3 + d3 + 2b9 + x12 = c4 b9 . Hence by (2.6) (b3 c3 )c4 = b9 c4 = b3 + d3 + 2b9 + x12 . But by (2.7) (b3 c4 )c3 = c3 b 9 + c3 d 3 . Since d3 ∈ Supp{c3 d3 }, we have that d3 ∈ Supp{c3 b9 } c3 d3 = b9 , c3 b9 = b3 + d3 + b9 + x12 .

(2.16)

Therefore (b3 c3 , c3 d3 ) = 1, so (b3 d¯3 , c32 ) = 1. By (2.8) and (2.12), we have that (b3 d¯3 , x5 ) = 1, which implies that x5 = d5 . Consequently d5 is real. Then b3 d¯3 = c4 + d5 = b¯3 d3 , c32 = 1 + c3 + d5 . Moreover (b32 , d32 ) = (b3 d¯3 , b¯3 d3 ) = 2. So d32 = 1 + 2b4 or b4 + b5 . Since (2.9) implies that d3 is nonreal. So d32 = b4 + b5 . By (2.8), we may state that b¯3 d5 = d¯3 + u12 ,

u12 ∈ N ∗ B.

Then b¯3 c32 = b¯3 + b¯3 c3 + b¯3 d5 = b¯3 + b¯9 + d¯3 + u12 , (b¯3 c3 )c3 = c3 b¯9 . Thus c3 b9 = b3 + d3 + b9 + u¯ 12 . Hence x12 = u¯ 12 by (2.16) and so b3 d5 = d3 + x12 . Since c4 c32 = c4 + c3 c4 + d5 c4 and c3 (c3 c4 ) = c3 c4 + c3 b8 by (2.15), we have that c3 b8 = c4 + c4 d5 . Now, by c42 = b4 b¯4 , we have that c3 c42 = c3 + c32 + c3 c4 + c3 b8 = c3 + 1 + c3 + d5 + c4 + b8 + c4 + c4 d5 ,

28


(c3 b4 )b¯4 = b4 b¯4 + y8 b¯4 = 1 + c3 + c4 + b8 + b¯4 y8 . So b¯4 y8 = c3 + c4 + d5 + c4 d5 . We know that y8 is irreducible, then (c4 b4 , y8 ) = 1. But (b4 c4 , b4 c4 ) = 4. Hence (c4 b4 − y8 , c4 b4 − y8 ) = 3, which is impossible for L1 (B) = {1} and L2 (B) = ∅. The lemma follows. Now we begin to investigate NITA such that c7 ∈ B is based on the value of (b5 b¯5 , b8 ). Lemma 2.5 There exists no NITA such that c7 ∈ B and (b5 b¯5 , b8 ) = 3. Proof It is easy to see that (b5 b¯5 , b8 ) = 3 implies that (b¯3 b5 , b¯3 b5 ) = 4. By Lemma 2.3, we may assume that x12 = x + y + z,

x, y, z ∈ B,

b¯3 b5 = b3 + x + y + z, b3 b8 = b3 + b9 + x + y + z. From the last equation above, one has that b8 is contained in b¯3 x, b¯ 3 y and b¯3 z. Hence no one of x, y and z has degree 3. Therefore x12 = x4 + y4 + z4 , b¯3 x4 = r4 + b8 ,

some r4 ∈ B,

b¯3 y4 = s4 + b8 ,

some s4 ∈ B,

b¯ 3 z4 = t4 + b8 ,

some t4 ∈ B.

Furthermore, b3 x4 , b3 y4 and b3 z4 have exactly two constituents. Let b3 x4 = b5 + r7 ,

some r7 ∈ B,

b3 y4 = b5 + s7 ,

some s7 ∈ B,

b3 z4 = b5 + t7 ,

some t7 ∈ B.

Since b¯3 (b3 x4 ) = b¯3 b5 + b¯3 r7 = b3 x4 + y4 + z4 + b¯3 r7 , b3 (b¯3 x4 ) = b3 b8 + b3 r4 = b3 + b9 + x4 + y4 + z4 + b3 r4 ,


29

we have that b¯3 r7 = b9 + b3 r4 . By the same reasoning we have that b¯ 3 s7 = b9 + b3 s4 and b¯3 t7 = b9 + b3 t4 . Hence r7 , s7 , t7 ∈ Supp{b3 b9 }. If r7 , s7 and t7 are distinct, then b3 b9 = b4 + r7 + s7 + t7 + e2 ,

some e2 ∈ B,

a contradiction. Hence at least two of r7 , s7 and t7 are equal, without loss of generality, let r7 = s7 . Then (b3 x4 , b3 y4 ) = (b¯3 x4 , b¯3 y4 ) = 2. Thus r4 = s4 . By Lemma 2.3, we have b82 = 1 + 4b8 + c7 + z12 + 2r4 + t4 . So z¯ 12 = 2r4 + t4 . Lemma 2.3 implies that b¯3 b9 = c7 + b8 + 2r4 + t4 . Then (b¯3 b9 , r4 ) = (b9 , b3 r4 ) ≥ 2, a contradiction, from which the lemma follows. Lemma 2.6 There exists no NITA such that c7 ∈ B and (b5 b¯5 , b8 ) = 2. Suppose that (b5 b¯5 , b8 ) = 2. Let b5 b¯5 = 1 + 2b8 + d8 ,

some d8 ∈ N ∗ B.

Since (b¯3 b5 , b¯3 b5 ) = (b3 b¯3 , b5 b¯5 ) = 3, by Lemma 2.3, we have that x12 = x + y and b¯3 b5 = b3 + x + y, b3 b8 = b3 + b9 + x + y. Thus b8 ∈ Supp{b¯3 x}, Supp{b¯3 y}, which implies that degrees of x and y are ≥ 4. We have three possibilities: (f (x), f (y)) = (4, 8), (5, 7), (6, 6). We shall divide the proof into three propositions, Propositions 2.1, 2.2, 2.3 from which the lemma follows. Proposition 2.1 There exists no NITA such that b5 b¯5 = 1 + 2b8 + d8 and (f (x), f (y)) = (4, 8). Proof Assume (f (x), f (y)) = (4, 8). Then x12 = x4 + y8 and we may set b¯ 3 x4 = b8 + x4 , b¯ 3 y8 = b8 + y16 ,

x4 ∈ B, y16 ∈ N ∗ B,

Then b82 = 1 + c7 + 3b8 + x4 + z¯ 12 + y16 .

(2.17)

30


Hence (b3 b¯3 )2 = 2 + 5b8 + c7 + x4 + z¯ 12 + y16 , b32 b¯32 = b4 b¯4 + b4 b¯5 + b¯4 b5 + b5 b¯5 = 1 + b8 + c7 + b4 b¯5 + b¯4 b5 + 1 + 2b8 + d8 , so z12 + d8 = x4 + y16 . Now we have two cases, x4 ∈ Supp{d8 } or x4 ∈ Supp{d8 }. Case I Suppose that x4 ∈ Supp{z12 }. Then there exists s8 such that z12 = x4 + s8 , y16 = d8 + s8 . Since (b¯4 b5 , b¯4 b5 ) = (b4 b¯4 , b5 b¯5 ) = 3, we have that s8 ∈ B and b¯3 b9 = c7 + b8 + x¯ 4 + s¯8 . Hence b9 ∈ Supp{b3 x4 }, which means that (b3 x¯4 , b3 x¯4 ) = 2. So (b3 x4 , b3 x4 ) = 2. Since b¯ 3 x4 = b8 + x4 , we may set b3 x¯4 = a3 + b9 and b3 x4 = x4 + e8 , for some a3 , e8 ∈ B. Suppose that b3 a¯ 3 = x4 + l5 , for some l5 ∈ B. Then b3 (b¯3 x4 ) = b3 b¯9 + b3 a¯ 3 = c7 + b8 + x4 + s8 + x4 + l5 , b¯3 (b3 x4 ) = b¯3 x4 + b¯3 e8 = b8 + x4 + b¯3 e8 . Hence b¯3 e8 = c7 + s8 + x4 + l5 . Therefore e8 ∈ Supp{b3 c7 }. Let b3 c7 = b9 + e8 + e4 , e4 ∈ B. By (b3 b¯3 )a¯ 3 = b¯3 (b3 a¯ 3 ), we obtain a¯ 3 b8 = b¯ 9 + b¯3 l5 . Since (a¯ 3 b8 , a¯ 3 b8 ) = (a3 a¯ 3 , b82 ) = (b3 b¯3 , b82 ) = 4, we have that (b¯3 l5 , b¯3 l5 ) = 3. So (b3 l5 , b3 l5 ) = 3. Let b3 l5 = e8 + α3 + β4 , where α3 , β4 ∈ B. Obviously α3 = b3 . b3 (b¯3 e8 ) = b3 c7 + b3 b8 + b3 x4 + b3 l5 = 2b9 + 3e8 + e4 + b3 + 2x4 + y8 + α3 + β4 , (b3 b¯3 )e8 = e8 + e8 b8 .


31

Then e8 b8 = 2b9 + 2e8 + e4 + b3 + 2x4 + y8 + α3 + β4 . Hence e8 ∈ Supp{b3 b8 }, which implies that e8 = y8 . Hence b3 c7 = b9 + y8 + e4 , so that c7 ∈ Supp{b¯3 y8 }, which implies that c7 ∈ Supp{y16 } = {d8 , s8 }, a contradiction. Case II NITA satisfying x4 ∈ Supp{d8 }. If x4 ∈ d8 , then d8 = x4 + x¯4 (x4 maybe real). Then b5 b¯5 = 1 + 2b8 + x4 + x¯4 . Hence y16 = x¯4 + z12 and b82 = 1 + c7 + 3b8 + x4 + x¯4 + z12 + z¯ 12 , b¯3 y8 = b8 + x¯4 + z12 , b9 b¯9 =

1 + 3b8 + x4

(2.18) (2.19)

+ z12 + x¯4

+ b¯3 f12 ,

(2.20)

by (2.17). Therefore (y8 , b3 x¯4 ) = 1. Let b3 x¯4 = y8 + t4 , b3 x4 = x4 + j8 ,

where t4 , j8 ∈ B.

(2.21)

Since b3 (b5 b¯5 ) = b3 + 2b3 b8 + b3 x4 + b3 x¯4 = 3b3 + 3b9 + 2x4 + 3y8 + t4 + x4 + j8 , b5 (b3 b¯5 ) = b5 b¯3 + b5 x¯4 + b5 y¯8 = b3 + x4 + y8 + b5 x¯4 + b5 y¯8 , we have that x¯ 4 b5 + y¯8 b5 = 2b3 + 2b9 + 2x4 + 2y8 + t4 + j8 . Since x4 (b¯3 b4 ) = b3 x4 + b9 x4 = x4 + j8 + b9 x4 , (b¯3 x4 )b4 = y¯8 b4 + t¯4 b4 , we have that x4 ∈ Supp{b4 y¯8 } or Supp{t¯4 b4 }. We assert that x4 ∈ Supp{b4 t¯4 }. Otherwise x4 ∈ Supp{b4 y¯8 }, and y8 ∈ Supp{b4 x¯4 }. But b9 ∈ Supp{b4 x¯4 } by Lemma 2.3, a contradiction. Hence x4 ∈ Supp{b4 t¯4 }, which implies that t4 ∈ Supp{b4 x¯4 }. Therefore b4 x¯4 = t4 + b9 + r3 ,

r3 ∈ B.

(2.22)

32


Since b¯3 (b4 b8 ) = b¯3 b5 + b¯3 (b3 b9 ) = b3 + x4 + y8 + b9 + b3 (b¯3 b9 ) = b3 + x4 + y8 + b9 + b3 c7 + b3 b8 + b3 z¯ 12 = b3 + x4 + y8 + b9 + b9 + f12 + b3 + x4 + y8 + b9 + b3 z¯ 12 , b4 (b¯3 b8 ) = b4 b¯3 + b4 x¯4 + b4 y¯8 + b4 b¯9 = b3 + b9 + t4 + b9 + r3 + b4 y¯8 + b3 + b9 + x4 + y8 + f12 , (b¯3 b4 )b8 = b3 b8 + b8 b9 = b3 + x4 + y8 + b9 + b8 b9 , we have that x4 b8 = b8 + z12 + b3 t¯4 , b8 b9 = b3 + x4 + y8 + b9 + y12 + b3 z¯ 12 , b4 y¯8 + b9 + t4 + r3 = x4 + y8 + b3 z¯ 12 . We continue the proof based on t4 = x4 or t4 = x4 . Subcase I There exists no NITA such that t4 = x4 . If t4 = x4 , then (x4 , b3 x¯4 ) = 1 by b3 x¯4 = y8 + t4 , which implies that x4 ∈ Supp{b3 x¯4 } = {b8 , x¯4 }. And b¯ 3 t4 = b¯3 x4 = b8 + x4 . Hence x4 is real and b5 b¯5 = 1 + 2b8 + 2x4 ,

(2.23)

b3 x4

(2.24)

= x4 + y8 .

Thus b¯3 (b3 x4 ) = b¯3 y8 + b¯3 x4 = b8 + x4 + z¯ 12 + b8 + x4 ,

(b3 b¯3 )x4 = x4 + x4 b8 . So

x4 b8 = x4 + 2b8 + z12 . Since b3 (b5 b¯5 ) = b3 + 2b3 b8 + 2b3 x4 = b3 + 2b3 + 2b9 + 2x4 + 2y8 + 2x4 + 2y8 , b5 (b3 b¯5 ) = b¯3 b5 + x¯4 b5 + y¯8 b5 = b3 + x4 + y8 + x¯4 b5 + y¯8 b5 ,


33

we have that x¯4 b5 + y¯8 b5 = 2b3 + 2b9 + 3x4 + 3y8 . Since b3 ∈ Supp{x¯4 b5 } and b3 ∈ Supp{y¯8 b5 }, there are exactly two possibilities: x¯ 4 b5 = b3 + b9 + y8 ,

y¯8 b5 = b3 + b9 + 3x4 + 2y8 ,

x¯4 b5 = b3 + b9 + 2x4 ,

y¯8 b5 = b3 + b9 + x4 + 3y8 .

If x¯4 b5 = b3 + b9 + y8 and y¯8 b5 = b3 + b9 + 3x4 + 2y8 , then (y¯8 b5 , x4 ) = 3, which implies that (b5 x¯4 , y8 ) = 3, a contradiction. If x¯4 b5 = b3 + b9 + 2x4 and y¯8 b5 = b3 + b9 + x4 + 3y8 , then (x¯4 b5 , x4 ) = 3. So (x42 , b5 ) = 2. It follows that (x42 , x42 ) ≥ 5, which is impossible for either x4 x¯4 = 1 + b8 + r7 , r7 ∈ B or x4 x¯4 = 1 + b8 + r4 + r3 , r3 , r4 ∈ B. Hence Subcase I follows. Subcase II There exists no NITA such that t4 = x4 . If t4 = x4 , by b4 y¯8 + b9 + t4 + r3 = x4 + y8 + b3 z¯ 12 , we have that r3 + t4 + b9 is a part of b3 z¯ 12 , x4 + y8 is a part of b4 y¯8 and b3 + x4 + y8 + b9 + y12 + r3 + t4 + b9 is a part of b8 b9 . If b9 ∈ Supp{y12 }, then there exists h3 ∈ B such that y12 = h3 + b9 . So b3 c7 = h3 + 2b9 , b3 h¯ 3 contains c7 , which is impossible for L2 (B) = ∅. Hence b9 ∈ Supp{y12 }. Since (b9 b¯9 , b8 ) = 3, so (b8 b9 , b9 ) = 3, which implies that (b3 z¯ 12 , b9 ) = 2 and there exists x11 ∈ N ∗ B of degrees 11 such that b3 z¯ 12 = 2b9 + t4 + r3 + x11 , b4 y¯8 = x4 + y8 + b9 + x11 . The following leads to a contradiction based on the z12 representation. Since (z12 , z12 ) = 2, we have that z12 = α + β, where (f (α), f (β)) = (3, 9), (4, 8), (5, 7) or (6, 6). Step 1 There exists no NITA such that z12 = α3 + β9 . If z12 = α3 + β9 , then b3 (α¯ 3 + β¯9 ) = r3 + t4 + b9 + x11 , b3 α¯ 3 = b9 and b¯3 (b3 α¯ 3 ) = c7 + b8 + α¯ 3 + β¯9 , (b¯3 b3 )α¯ 3 = α¯ 3 + α¯ 3 b8 . Thus α3 b8 = c7 + b8 + β9 , which implies that (α3 b8 , α3 b8 ) = 3. Hence (α3 α¯ 3 , b82 ) = 3. It follows that α3 α¯ 3 = 1 + x4 + x¯4 by the expression b82 (see (2.18)), which contradicts Theorem 2.1. Step 2 There exists no NITA such that z12 = α4 + β8 . If z12 = α4 + β8 , then b3 (α¯ 4 + β¯8 ) = r3 + t4 + 2b9 + x11 . Since b9 ∈ Supp{b3 α¯ 4 }, we have that b3 α¯ 4 = r3 + b9 ,

b3 β¯8 = t4 + b9 + x11 .

34


Let b¯3 r3 = α¯ 4 + e5 . Then b¯3 (b3 α¯ 4 ) = b¯3 b9 + b¯3 r3 = c7 + b8 + α¯ 4 + β¯8 + α¯ 4 + e5 , α¯ 4 (b3 b¯3 ) = α¯ 4 + α¯ 4 b8 , which means that α4 b8 = c7 + b8 + β8 + e5 . Since b¯3 y8 = α4 + β8 + x¯ 4 + b8 , we have that b3 α4 = y8 + γ4 , γ4 ∈ B. Since b¯3 (b3 α4 ) = b¯3 y8 + b3 γ4 = b8 + α4 + β8 + x¯4 + b3 γ4 , (b¯3 b3 )α4 = α4 + α4 b8 = 2α4 + c7 + b8 + β8 + e5 , we have that x¯4 + b3 γ4 = α4 + c7 + e5 , which implies that x¯ 4 = α4 . But by (2.21) b3 x4 = j8 + x4 , j8 ∈ B and b3 α¯ 4 = b9 + r3 , a contradiction. Step 3 There exists no NITA such that z12 = α5 + β7 . If z12 = α5 + β7 , then b3 (α¯ 5 + β¯7 ) = 2b9 + r3 + t4 + 2b9 + x11 . We need a sum of degree 6 to make up b3 α¯ 5 for b9 ∈ Supp{b3 α¯ 5 }. If r3 ∈ Supp{b3 β¯7 }, then b¯3 r3 contains an element of degree 2, a contradiction. Hence r3 ∈ Supp{b3 α¯ 5 }. We may set b3 α¯ 5 = b9 + r3 + δ3 ,

δ3 ∈ Supp{x11 }.

Furthermore b¯3 r3 = α¯ 5 + d4 , for some d4 ∈ B. It follows that r3 r¯3 = 1 + b8 . Now we have that (α5 α¯ 5 , b8 ) = 2 and 2b3 + x4 + y8 + b9 = b3 (b3 b¯3 ) = r3 (b3 r¯3 ) = r3 α5 + r3 d¯4 , from which the following holds: r3 α5 + r3 d¯4 = 2b3 + x4 + y8 + b9 . Comparing the two sides, the following equations hold: r3 d¯4 = b3 + b9 , r3 α5 = b3 + x4 + y8 . Therefore


35

b¯3 (r3 α5 ) = b3 b¯3 + x4 b¯3 + y8 b¯3 = 1 + b8 + b8 + x4 + b8 + α5 + β7 + x¯4 , α5 (b¯3 r3 ) = α5 α¯ 5 + d4 α5 . Since (α5 α¯ 5 , b8 ) = 2, we can only have that α5 α¯ 5 = 1 + 2b8 + x4 + x¯4 , d4 α5 = b8 + α5 + β7 . Hence d4 = x4 or x¯4 . So b¯3 r3 = α5 + x4 or α5 + x¯4 . If b¯3 r3 = α5 + x¯4 , then r3 ∈ Supp{b3 x¯4 } = {y8 , t4 } by (2.21), a contradiction. Hence b¯3 r3 = α5 + x4 , which implies that r3 ∈ Supp{b3 x4 } = {x4 , j8 }, a contradiction. Step 4 There exists no NITA such that z12 = α6 + β6 . If z12 = α6 + β6 , then b3 (α¯ 6 + β¯6 ) = 2b9 + t4 + r3 + x11 . Hence α6 = β6 . Without loss of generality, let r3 ∈ Supp{b3 α6 }. We may set b¯ 3 r3 = α¯ 6 +d3 , which implies that r3 r¯3 = 1+b8 . Therefore we have the following equation: α6 r3 + d¯3 r3 = (b3 r¯3 )r3 = (b3 b¯3 )b3 = 2b3 + b9 + x4 + y8 . It is easy to see that we cannot sum up d¯3 r3 , a contradiction. Proposition 2.1 follows. Proposition 2.2 There exists no NITA such that (f (x), f (y)) = (5, 7). Proof If (f (x), f (y)) = (5, 7), that is to say, x12 = x5 + y7 , and b¯3 b5 = b3 + x5 + y7 , b3 b8 = b3 + b9 + x5 + y7 . So we may set b¯3 x5 = b8 + e7 ,

e7 ∈ N ∗ B,

b3 x5 = b5 + e10 ,

e10 ∈ N ∗ B,

b¯3 y7 = b8 + y13 ,

y13 ∈ N ∗ B,

b3 y7 = b5 + y16 ,

y16 ∈ N ∗ B.

36


Hence b¯3 (b3 b8 ) = b¯3 b3 + b¯3 b9 + b¯3 x5 + b¯3 y7 = 1 + b8 + c7 + b8 + z¯ 12 + b8 + e7 + b8 + y13 , (b3 b¯3 )b8 = b8 + b82 , (b¯ 3 b5 )b3 = b32 + b3 x5 + b3 y7 = b4 + 3b5 + e10 + y16 , (b3 b¯3 )b5 = b5 + b5 b8 , and one has that b82 = 1 + c7 + e7 + 3b8 + z¯ 12 + y13 , b5 b8 = b4 + 2b5 + e10 + y16 . On the other hand, (b3 b¯3 )2 = 1 + 2b8 + 1 + c7 + e7 + 3b8 + z¯ 12 + y13 , b32 b¯32 = 1 + b8 + c7 + b8 + z12 + b8 + z¯ 12 + 1 + 2b8 + d8 , which implies that z12 + d8 = e7 + y13 . Step 1 There exists no NITA such that e7 ∈ Supp{z12 }. Suppose that e7 ∈ Supp{z12 }. Since e7 ∈ Supp{d8 }, there exist α3 ∈ Supp{z12 }, g4 ∈ Supp{d8 } or α4 ∈ Supp{z12 }, g3 ∈ Supp{d8 } such that e7 = α3 + g4 or α4 + g3 . Substep 1 There exists no NITA such that e7 = α3 + g4 . If e7 = α3 + g4 , then there exist z9 ∈ N ∗ B and h4 ∈ B such that z12 = α3 + z9 and d8 = g4 + h4 . Furthermore, b¯4 b5 = α3 + z9 + b8 , b¯3 b9 = c7 + b8 + α¯ 3 + z¯ 9 , b¯3 x5 = b8 + α3 + g4 , Hence b3 α3 = x5 + i4 ,

i4 ∈ B,

b3 α¯ 3 = b9 , which is impossible for (b3 α¯ 3 , b3 α¯ 3 ) = 1 and (b3 α3 , b3 α3 ) = 2. Substep 2 There exists no NITA such that e7 = α4 + g3 . If e7 = α4 + g3 , then there exist z8 ∈ Supp{N ∗ B} and h5 ∈ B such that z¯ 12 = α4 + z8 , d8 = g3 + h5 . Hence


37

b¯3 x5 = g3 + α4 + b8 , b5 b¯5 = 1 + 2b8 + g3 + h5 , b¯3 y7 = b8 + z8 + h5 . So there exists l4 ∈ B and n8 ∈ N ∗ B such that b3 g3 = x5 + l4 and b3 h5 = y7 + n8 . Hence g3 g¯ 3 = 1 + b8 and b¯3 l4 = g3 + m9 . Here m9 ∈ B, otherwise l¯4 l4 will contains two b8 , which is a contradiction. Since b3 (b3 b¯3 ) = 2b3 + b9 + x5 + y7 , (b3 g3 )g¯3 = g¯ 3 x5 + g¯3 l4 , it follows that g¯ 3 l4 = b3 + b9 and g¯ 3 x5 = b3 + x5 + y7 = b¯3 b5 .

(2.25)

We assert that Supp{n8 } ∩ {b3 , x5 , y7 , l4 } = ∅. Obviously y7 ∈ Supp{n8 }. By the expressions of b3 h5 and b3 b¯3 , we have that b3 ∈ Supp{n8 }. If x5 ∈ Supp{n8 }, then h5 ∈ Supp{b¯3 x5 }, a contradiction. If l4 ∈ Supp{n8 }, then h5 ∈ Supp{b¯3 l4 }, a contradiction. Since b3 (b5 b¯5 ) = b3 + 2b3 b8 + b3 g3 + b3 h5 = 3b3 + 2b9 + 2x5 + 2y7 + x5 + l4 + y7 + n8 , (b3 b¯5 )b5 = b5 (b¯3 + x¯5 + y¯7 ) = b3 + x5 + y7 + x¯5 b5 + y¯7 b5 , it holds that b5 x¯5 + b5 y¯7 = 2b3 + 2b9 + 2x5 + 2y7 + l4 + n8 .

(2.26)

Now we assert that (b5 y¯7 , b9 ) = 1. In fact, it follows from that b5 b¯9 = b9 + b3 α4 + b3 z8 by Lemma 2.3 and (b3 α4 , y7 ) = 0, (b3 z8 , y7 ) = 1. Therefore (b5 x¯5 , b3 ) = (b5 x¯5 , b9 ) = 1 and (b5 y¯7 , b3 ) = (b5 y¯7 , b9 ) = 1. By (2.25), (g¯ 3 x5 , b¯3 b5 ) = 3, so (b3 g¯ 3 , b5 x¯5 ) = 3. By (2.26), b5 x¯5 cannot contain three multiples of a constituent of degree 3 or two multiples of a constituent of degree 6. Then the common constituents of b3 g¯3 and b5 x¯5 are degree 5 or 4, since b3 g¯ 3 can now only have constituents of degree 5 and 4. From (b3 g¯3 , b5 x¯5 ) = 3 and (2.26), we have that b5 x¯5 must contain 2x5 and some element v4 of degree 4 from l4 + n8 . This leads to b5 x¯5 = b3 + b9 + 2x5 + v4 , a contradiction for the left hand side having degree 26.

38


Step 2 There exists no NITA such that e7 ∈ Supp{z12 }. If e7 ∈ Supp{z12 }, then there exists w5 ∈ B such that z12 = e7 + w5 . So y13 = d8 + w5 and b¯3 x5 = b8 + e7 ,

(2.27)

b¯3 y7 = b8 + d8 + w5 ,

(2.28)

b¯4 b5 = w5 + e7 + b8 ,

(2.29)

b¯3 b9 = w¯ 5 + c7 + e¯7 + b8 .

(2.30)

Since (b¯4 b5 , b¯4 b5 ) = 3, we have that e7 , w5 ∈ B. By Lemma 2.3 b5 b¯9 = b9 + b3 e7 + b3 w5 .

(2.31)

Since y7 ∈ Supp{b3 w5 }, b9 ∈ Supp{b3 w¯ 5 }, and x5 ∈ Supp{b3 e7 }, we may set b3 w5 = y7 + v8 ,

v8 ∈ N ∗ B,

(2.32)

b3 w¯ 5 = b9 + d6 ,

d8 ∈ N ∗ B,

(2.33)

∗

b3 c7 = b9 + f12 ,

f12 ∈ N B,

(2.34)

b3 e7 = x5 + h16 ,

h16 ∈ N ∗ B,

(2.35)

b3 e¯7 = b9 + g12 ,

∗

g12 ∈ N B.

(2.36)

Case 1 There exists no NITA such that either v8 or d6 is reducible. It is easy to see that when either v8 or d6 is reducible, this will imply that the other one is also reducible. Let d6 = d3 + g3 . Then b3 w¯ 5 = d3 + g3 + b9 . It is easy to show that d3 = g3 . Hence we may set b3 d¯3 = w5 + δ4 , b3 g¯3 = w5 + ψ4 , which implies that d3 d¯3 = g3 g¯ 3 = 1 + b8 . If δ4 = ψ4 , then b3 d¯3 = b3 g¯ 3 , (b3 d¯3 , b3 g¯ 3 ) = 2, which implies that d3 g¯ 3 = 1 + b8 . Thus g3 = d3 and b3 w¯ 5 = b9 + 2g3 , so that (b3 g¯ 3 , w5 ) = 2, a contradiction. Hence δ4 = ψ4 . We calculate b32 b¯3 = b3 d3 d¯3 = b3 g3 g¯ 3 . We have that (b3 b¯3 )b3 = 2b3 + b9 + x5 + y7 , (b3 d¯3 )d3 = w5 d3 + δ4 d3 , (b3 g¯ 3 )g3 = w5 g3 + ψ4 g3 .


39

It is easy to see that δ4 d3 = ψ4 g3 = b3 + b9 , w5 d3 = w5 g3 = b3 + x5 + y7 . On the other hand, we have that w5 w¯ 5 = 1 + 2b8 + u8 , b¯ 3 (g3 ψ4 ) = b¯3 (d3 δ4 ) = b¯ 3 b3 + b¯3 b9 = 1 + b8 + c7 + e¯7 + w¯ 5 + b8 , (b¯3 g3 )ψ4 = w¯ 5 ψ4 + ψ¯ 4 ψ4 , (b¯3 d3 )δ4 = w¯ 5 δ4 + δ¯4 δ4 , so that w¯ 5 ψ4 = w¯ 5 δ4 = e¯7 + w¯ 5 + b8 , ψ4 ψ¯ 4 = δ4 δ¯4 = 1 + b8 + c7 . Hence ψ4 , δ4 ∈ Supp{w5 w¯ 5 }. Therefore w5 w¯ 5 = 1 + 2b8 + δ4 + ψ4 . But (g3 d¯3 , w5 w¯ 5 ) = (g3 w5 , d3 w5 ) = 3. We have that 1 ∈ Supp{g3 d¯3 }, which implies that g3 = d3 , a contradiction. Case 2 There exists no NITA such that v8 and d6 are irreducible. Suppose v8 and d6 are irreducible. We assert that f12 is irreducible. Otherwise, let f12 = r + s. Suppose that c7 ∈ Supp{b¯ 3 r}, which implies that f (r) ≥ 4. By Lemma 2.3, one has that b8 c7 = e7 + w5 + b8 + b3 f¯12 , b¯9 b9 = 1 + d8 + 3b8 + e7 + w5 + b3 f¯12 . Subcase 1 w5 is not real. If w5 is real, then y7 + v8 = b3 w5 = b3 w¯ 5 = b9 + d6 by (2.32) and (2.33), a contradiction. Subcase 2 f12 ∈ B. By the expression b8 c7 , we have that w¯ 5 ∈ Supp{b3 f¯12 }. If f12 is irreducible, then f12 ∈ Supp{b3 w5 } = {y7 , v8 } by (2.32), a contradiction. So f12 is reducible.

40


Subcase 3 e7 is not real. If e7 is real, then there exists an element u7 ∈ N ∗ B by (2.27) and (2.30), such that b3 e7 = x5 + b9 + u7 . Hence (e72 , b8 ) ≥ 2, so (e7 b8 , e7 ) ≥ 2. By the expressions of b8 c7 and b82 , we have that (e7 b8 , c7 ) = 1, (e7 b8 , b8 ) ≥ 2. Thus there exists an element x19 ∈ N ∗ B of degree 19 such that e7 b8 = 2e7 + 2b8 + c7 + x19 . On the other hand, from (2.27) and (2.30), we have that b¯3 (b3 e7 ) = b¯3 x5 + b¯3 b9 + b¯3 u7 = b8 + e7 + c7 + e7 + w¯ 5 + b8 + b¯3 u7 , (b¯3 b3 )e7 = e7 + b8 e7 = 3e7 + 2b8 + c7 + x19 . Then w¯ 5 + b¯3 u7 = e7 + x19 . Since the right side is real, we have that w5 ∈ Supp{b¯3 u7 } by Substep 1, which implies that there exists a constituent t of u7 such that w5 ∈ Supp{b¯3 t}. Then t ∈ Supp{b3 w5 }, so t = y7 or v8 . It follows that u7 = y7 . By (2.28) implies that w5 + w¯ 5 + b8 + d8 = w¯ 5 + b¯3 y7 = e7 + x19 , a contradiction. Subcase 4 Case 2 follows. By Subcases 1, 2, 3, we have that e7 and w5 are not real, and f12 is reducible. Note that b8 c7 is real, by the expression b8 c7 , we have that b3 f¯12 contains w¯ 5 and e¯7 . Hence there exists a constituent r of f12 such that b3 r¯ contains w¯ 5 , which implies that r ∈ Supp{b3 w5 }. Therefore r = y7 or v8 . If r = y7 , then b3 y¯7 = b8 + d8 + w¯ 5 by (2.28). But (2.34) means that b3 y¯7 contains c7 , a contradiction. If r = v8 , then there exists s4 such that f12 = v8 + s4 . Since c7 ∈ Supp{b3 s¯4 } by (2.34), we have that e¯7 ∈ Supp{b3 s¯4 }. So e¯7 ∈ Supp{b3 v¯8 }. Hence there exists a basis element a5 of degree 5 such that b3 v¯8 = w¯ 5 + e¯7 + c7 + a5 ,

(2.37)

b3 s¯4 = c7 + c5 ,

(2.38)

b3 c7 = v8 + s4 + b9 .

(2.39)

Checking b3 (b5 b¯5 ) = b5 (b3 b¯5 ), this implies that b5 x¯5 + b5 y¯7 = 2b3 + 2b9 + x5 + y7 + b3 d8 .

(2.40)


41

We assert that d8 ∈ B. In fact, if d8 ∈ B, then d8 = 2x4 or x4 + y4 . If d8 = 2x4 , then b¯3 y7 = b8 + 2x4 + w5 by (2.28). Hence (b3 x4 , y7 ) = 2, which is impossible. If d8 = x4 + y4 , then b¯3 y7 = b8 + x4 + y4 + w5 by (2.28). Let b3 x4 = y7 + p5 and b3 y4 = y7 + q5 . By (2.40), it holds that b5 x¯5 + b5 y¯7 = 2b3 + 2b9 + x5 + 3y7 + p5 + q5 . But (b5 x¯5 , b3 ) = (b5 x¯5 , b9 ) = 1. So it is necessary to find elements from the right hand side of the above equation such that their sum having degree 13 yields b5 x¯5 . This is impossible. Therefore d8 ∈ B. Since b¯3 (b3 w5 ) = b¯3 y7 + b¯3 v8 = b8 + d8 + w5 + c7 + w5 + e7 + a¯ 5 , b3 (b¯3 w5 ) = b3 b¯9 + b3 d¯6 = w5 + c7 + e7 + b8 + b3 d¯6 , then b3 d¯6 = d8 + w5 + a¯ 5 .

(2.41)

Now we assert that c5 and e5 are nonreal. Otherwise, d6 , v8 , ∈ Supp{b3 a5 } by (2.37) and (2.41). So 1 ∈ Supp{b3 a5 }, a contradiction. Hence c5 and a5 are nonreal, and c5 = a¯ 5 by the expression b82 . Hence b3 d¯6 = d8 + w5 + c5 . Associated with (2.38), we have that (b3 c¯5 , d6 ) = (b3 c¯5 , s4 ) = 1, which implies that (b3 c¯5 , b3 c¯5 ) = 3. Furthermore (b3 c5 , b3 c5 ) = 3. Let b3 c5 = v8 + δ3 + λ4 . Thus we may set b¯3 δ3 = c5 + Δ4 . Then δ¯3 δ3 = 1 + b8 , and b3 (δ3 δ¯3 ) = b3 + b3 b8 = 2b3 + b9 + x5 + y7 , (b3 δ¯3 )δ3 = c¯5 δ3 + Δ¯ 4 δ3 . We have that Δ¯ 4 δ3 = b3 + b9

and c¯5 δ3 = b3 + x5 + y7 .

Furthermore b¯3 (Δ¯ 4 δ3 ) = b¯3 b3 + b¯3 b9 = 1 + 2b8 + c7 + w¯ 5 + e¯7 , (b¯ 3 δ3 )Δ¯ 4 = c5 Δ¯ 4 + Δ4 Δ¯ 4 .

(2.42)

42


It follows that Δ4 Δ¯ 4 = 1 + b8 + c7

and

c5 Δ¯ 4 = b8 + w¯ 5 + e¯7 = b4 b¯5 .

Hence (b3 c5 )Δ¯ 4 = b3 (c5 Δ¯ 4 ) = b3 (b4 b¯5 ) = b4 (b3 b¯5 ). But b4 (b3 b¯5 ) = b4 b¯3 + b4 x¯5 + b4 y¯7 = b3 + b9 + b4 x¯5 + b4 y¯7 , (b3 c5 )Δ¯ 4 = v8 Δ¯ 4 + δ3 Δ¯ 4 + λ4 Δ¯ 4 . So b3 is a constituent of one of v8 Δ¯ 4 , δ3 Δ¯ 4 and λ4 Δ¯ 4 . Then one of v8 , δ3 and λ4 is a constituent of b3 Δ4 . On the other hand δ3 ∈ Supp{b3 Δ4 } by (2.42). But b3 Δ4 contains at most two constituents, a contradiction. Subcase 4 follows, which concludes Step 2. Consequently the proposition follows. Proposition 2.3 There exists no NITA such that (f (x), f (y)) = (6, 6). Proof If x12 = x6 + y6 , then b¯3 b5 = b3 + x6 + y6 , b3 b8 = b3 + b9 + x6 + y6 . Assume that b¯3 x6 = b8 + h10 ,

for some h10 ∈ N ∗ B,

b¯3 y6 = b8 + i10 ,

for some i10 ∈ N ∗ B,

b3 x6 = b5 + g13 ,

for some g13 ∈ N ∗ B,

b3 y6 = b5 + f13 ,

for some f13 ∈ N ∗ B.

Therefore, b82 = 1 + c7 + 3b8 + h10 + i10 + z¯ 12 . By the above equations and Lemma 2.3, one has that (b¯3 b3 )2 = 2 + c7 + 5b8 + h10 + i10 + z¯ 12 , b¯ 32 b32 = b¯4 b4 + b¯4 b5 + b¯5 b4 + b¯5 b5 = 2 + 5b8 + c7 + z¯ 12 + z12 + d8 . Hence d8 + z12 = h10 + i10 . We assert that d8 cannot be totally contained in one of h10 and i10 . Otherwise, without loss of generality, assume d8 is contained in h10 . Then h10 contains an


43

element of degree 2, a contradiction. Thus d8 must be reducible and cannot be contained in one of h10 and i10 . For d8 , there are four possibilities: d8 = c3 + d5 ,

c 4 + d4 ,

c4 + c¯4 ,

and 2c4 .

Step 1 There exists no NITA such that d8 = c3 + d5 . If d8 = c3 + d5 , then c3 + d5 + z12 = h10 + i10 . Without loss of generality, let h10 = c3 + h7 ,

i10 = d5 + i5 .

Then z12 = h7 + i5 and b¯ 3 x6 = c3 + h7 + b8 ,

(2.43)

b¯3 y6 = d5 + i5 + b8 ,

(2.44)

b¯4 b5 = b8 + i5 + h7 ,

(2.45)

b¯3 b9 = c7 + b8 + i¯5 + h¯ 7 ,

(2.46)

b82

= 1 + c7 + 3b8 + c3 + h7 + d5 + i5 + h¯ 7 + i¯5 .

(2.47)

Hence c3 is real. By (2.45) and (b¯4 b5 , b¯4 b5 ) = 3, we have that h7 ∈ B. Set b3 c3 = x6 + u3 , u3 ∈ B by Lemma 2.2. Thus (b3 c3 , b3 c3 ) = 2 so that c32 = 1 + b8 . Moreover, (b3 c3 )c3 = x6 c3 + u3 c3 , b3 c32 = b32 b¯3 = b3 + b3 b8 = 2b3 + b9 + x6 + y6 . Hence c3 u3 = b3 + x6 or b3 + y6 , c3 x6 = b3 + b9 + y6 or b3 + b9 + x6 . If c3 u3 = b3 + x6 , c3 x6 = b3 + b9 + y6 , then c3 ∈ Supp{u¯ 3 x6 }. But (b3 b¯3 )(c32 ) = 2 + 5b8 + c7 + c3 + d5 + h7 + h¯ 7 + i5 + i¯5 , (b3 c3 )(b¯3 c3 ) = x6 x¯6 + u3 x¯6 + u¯ 3 x6 + 1 + b8 . Note that u3 x¯6 + u¯ 3 x6 contains two c3 , and so we come to a contradiction. Therefore c3 u3 = b3 + y6 , c3 x6 = b3 + b9 + x6 . Let b¯3 u3 = c3 + w6 . Then u3 u¯ 3 = 1 + b8 by Lemma 2.2 and

44


(c3 b3 )b¯3 = b¯3 x6 + b¯3 u3 = b8 + c3 + h7 + c3 + w6 . Since c3 (b3 b¯3 ) is real, it follows that h7 and w6 are real. Since (b¯3 u3 )c3 = c32 + c3 w6 = 1 + b8 + c 3 w 6 , (b¯3 c3 )u3 = x¯6 u3 + u¯ 3 u3 = 1 + b8 + x¯6 u3 , b¯3 (c3 u3 ) = b¯3 b3 + b¯3 y6 = 1 + b8 + b8 + d5 + i5 , we have that c3 w6 = x¯6 u3 = d5 + i5 + b8 . Since c3 , w6 are real, i5 is also real by the above equation. By (2.43), (2.44) and (2.46), one has that b3 i5 = y6 + b9 , b3 h7 = b9 + x6 + l6 ,

for some l6 ∈ N ∗ B.

Considering c32 u3 u¯ 3 , we have that (c3 u3 )(c3 u¯ 3 ) = b3 b¯3 + b¯3 y6 + b3 y¯6 + y6 y¯6 = 1 + b8 + 2b8 + 2d5 + 2i5 + y6 y¯6 , c32 (u3 u¯ 3 ) = 2 + 5b8 + c7 + c3 + 2i5 + 2h7 + d5 . Comparing the number of elements of degree 5, we come to a contradiction. Step 1 follows. Step 2 There exists no NITA such that d8 = 2c4 . If d8 = 2c4 , then 2c4 + z12 = h10 + i10 . There exist h6 and i6 such that z12 = h6 + i6 . Since (b¯4 b5 , b¯4 b5 ) ≤ 4, we obtain that h6 , i6 ∈ B. Furthermore b¯3 x6 = b8 + c4 + h6 ,

b¯3 y6 = b8 + c4 + i6 .

Thus b3 c4 = x6 + y6 . Then we can set c42 = 1 + b8 + g7 ,

g7 ∈ N ∗ B.

Since b5 b¯5 = 1 + 2b8 + 2c4 , (c4 b5 , b5 ) = 2. Furthermore, (c4 b5 , c4 b5 ) ≥ 5. Hence c42 = 1 + b8 + c4 + e3


45

and (c4 b5 , c4 b5 ) = 5. There exists t10 ∈ B such that c4 b5 = 2b5 + t10 . Since b3 (b5 b¯5 ) = b3 + 2b3 b8 + 2b3 c4 = 3b3 + 2b9 + 4x6 + 4y6 , b5 (b3 b¯5 ) = b5 (b¯3 + x¯6 + y¯6 ) = b3 + x6 + y6 + x¯6 b5 + y¯6 b5 , we have that b5 x¯6 + b5 y¯6 = 2b3 + 2b9 + 3x6 + 3y6 . On the other hand, b¯ 3 (c4 b5 ) = 2b¯ 3 b5 + b¯3 t10 = 2b3 + 2x6 + 2y6 + b¯3 t10 , (b¯3 c4 )b5 = x¯6 b5 + y¯6 b5 = 2b3 + 2b9 + 3x6 + 3y6 . Hence b¯3 t10 = 2b9 + x6 + y6 , from which it follows that (b3 b9 , t10 ) = 2. Therefore (b3 b9 , b3 b9 ) ≥ 5. But (b¯3 b9 , b¯3 b9 ) = 4 by Lemma 2.3 and h6 , i6 ∈ B, a contradiction. Step 3 There exists no NITA such that d8 = c4 + d4 . If d8 = c4 + d4 , then there exist h6 , i6 ∈ N ∗ B such that h10 = c4 + h6 , i10 = d4 + i6 , z12 = h6 + i6 and b¯3 x6 = c4 + h6 + b8 ,

(2.48)

b¯3 y6 = d4 + i6 + b8 , b¯4 b5 = b8 + i6 + h6 ,

(2.49) (2.50)

b¯3 b9 = c7 + b8 + i¯6 + h¯ 6 ,

(2.51)

b82 = 1 + c7 + 3b8 + c4 + h6 + d4 + i6 + h¯ 6 + i¯6 .

(2.52)

By (2.50) and (b¯4 b5 , b¯4 b5 ) = 3, we have that i6 , h6 ∈ B. Now we may set b3 c4 = x6 + z6 ,

z6 ∈ B,

(2.53)

b3 d4 = y6 + p6 ,

p6 ∈ B.

(2.54)

Substep 1 h6 and i6 are nonreal. If h6 is real, then there exist e3 and f3 such that b¯ 3 h6 = b¯9 + x¯6 + e3 , b3 e3 = f3 + h6 . Thus e3 e¯3 = 1 + b8 by Lemma 2.2. Since

46


b3 (e3 e¯3 ) = 2b3 + b9 + x6 + y6 , (b3 e3 )e¯3 = h6 e¯3 + f3 e¯3 , hence f3 e¯3 = b3 + x6 or b3 + y6 , h6 e¯3 = b3 + b9 + y6 or b3 + b9 + x6 . Therefore b¯3 (f3 e¯3 ) = b¯3 b3 + b¯3 x6 or b¯3 b3 + b¯3 y6 = 1 + 2b8 + c4 + h6 or 1 + 2b8 + d4 + i6 , f3 (b¯3 e¯3 ) = f3 (h6 + f¯3 ) = 1 + b8 + f3 h6 . Thus f3 h6 = b8 + c4 + h6 or b8 + d4 + i6 . On the other hand, one has the following equations: (b3 b¯3 )h6 = h6 + h6 b8 , (b3 h6 )b¯3 = (e¯3 + x6 + b9 )b¯3 = b¯3 b9 + b¯3 x6 + b¯3 e¯3 = c7 + 2b8 + 3h6 + i¯6 + c4 + f¯3 . Then h6 b8 = c7 + c4 + 2b8 + 2h6 + i¯6 + f¯3 ,

(2.55)

which implies that i6 and f3 are real. Now let us calculate (b3 b¯3 )(e3 e¯3 ): (b3 b¯3 )(e3 e¯3 ) = 2 + 5b8 + c7 + c4 + d4 + 2h6 + 2i6 , (b3 e3 )(b¯3 e¯3 ) = (h6 + f3 )2 = h26 + 2f3 h6 + f32 = 1 + h26 + 3b8 + 2c4 + 2h6 or 1 + h26 + 3b8 + 2d4 + 2i6 . Comparing the number of elements of degree 4, we have that c4 = d4 , a contradiction. Symmetrically, we can prove that i6 is nonreal.


47

Substep 2 There exists no NITA such that d8 = c4 + d4 and h6 , i6 are nonreal. Since (b3 b¯3 )c4 = c4 + c4 b8 and b¯ 3 (b3 c4 ) = b¯3 x6 + b¯3 z6 = c4 + h6 + b8 + b¯3 z6 , we have that c4 b8 = b8 + h6 + b¯3 z6 , Since h6 is nonreal and c4 b8 is real, h¯ 6 ∈ Supp{b¯3 z6 }. By (2.53), there exists r8 ∈ N ∗ B such that b¯3 z6 = c4 + h¯ 6 + r8 . By (2.51), there exists α3 such that b3 h¯ 6 = b9 + z6 + α3 . Thus there exists γ3 such that b3 α¯ 3 = h6 + γ3 . Hence α3 α¯ 3 = 1 + b8 by Lemma 2.2. Since b3 (α3 α¯ 3 ) = 2b3 + b9 + x6 + y6 , (b3 α3 )α¯ 3 = h6 α¯ 3 + γ3 α¯ 3 , we have that γ3 α¯ 3 = b3 + x6 or b3 + y6 , h6 α¯ 3 = b3 + b9 + y6 or b3 + b9 + x6 . Hence b¯3 (γ3 α¯ 3 ) = b¯3 b3 + b¯3 x6 or b¯3 b3 + b¯3 y6 = 1 + 2b8 + c4 + h6 or 1 + 2b8 + d4 + i6 , γ3 (b¯3 α¯ 3 ) = γ3 h¯ 6 + γ3 γ¯3 = 1 + b8 + γ3 h¯ 6 . Therefore γ3 h¯ 6 = b8 + c4 + h6 or b8 + d4 + i6 . Since (b3 b¯3 )(α3 α¯ 3 ) = 2 + 5b8 + c7 + c4 + d4 + 2h6 + 2i6 , (b3 α3 )(b¯3 α¯ 3 ) = (h6 + γ3 )(h¯ 6 + γ¯3 ) = h6 h¯ 6 + γ3 h¯ 6 + γ¯3 h6 + γ3 γ¯3 = 1 + 3b8 + 2c4 + h6 + h¯ 6 + h6 h¯ 6 or 1 + 3b8 + 2d4 + i6 + i¯6 + h6 h¯ 6 ,

48


comparing the number of degree 4, we come to c4 = d4 , a contradiction. So Substep 2 follows. Step 4 There exists no NITA such that d8 = c4 + c¯4 . If d8 = c4 + c¯4 , then b5 b¯5 = 1 + 2b8 + c4 + c¯4 . And (2.48), (2.49), (2.50), (2.51), (2.52), (2.53) and (2.54) still hold with d4 = c¯4 . Thus b¯3 (b3 c4 ) = b¯3 x6 + b¯3 z6 , = c4 + h6 + b8 + b¯3 z6 , b3 (b¯3 c4 ) = b3 y¯6 + b3 p¯ 6 = c4 + i¯6 + b8 + b3 p¯ 6 . Then h6 + b¯3 z6 = i¯6 + b3 p¯ 6 .

(2.56)

Substep 1 There exists no NITA such that h6 = i¯6 . If h6 = i¯6 , then it follows from (2.48) and (2.51) that there is α3 ∈ B such that b3 h6 = b9 + x6 + α3 , b3 h¯ 6 = b9 + y6 + β3 . It is easy to see that α3 = β3 . Since z12 = h6 + h¯ 6 , this means by Lemma 2.3 that b5 b¯9 = x6 + y6 + 3b9 + α3 + β3 . Thus (b5 b¯9 , b5 b¯9 ) = 13. By Lemma 2.3, we have that b9 b¯9 = 1 + 3b8 + c4 + c¯4 + h6 + h¯ 6 + b¯3 f12 . But b5 b¯5 = 1 + 2b8 + c4 + c¯4 . Since (b9 b¯9 , b8 ) = 3, we have that 13 = (b5 b¯9 , b5 b¯9 ) = 9 + (c4 , b¯3 f12 ) + (c¯4 , b¯3 f12 ). We conclude that (c4 , b¯3 f12 ) = (c¯4 , b¯3 f12 ) = 2. Thus (b3 c4 , f12 ) = (b3 c¯4 , f12 ) = 2. From (2.53) and (2.54), it follows that (x6 + z6 , f12 ) = (y6 + p6 , f12 ) = 2. Since x6 = y6 by (b3 b¯3 , b5 b¯5 ) = 3 and z6 = x6 , y6 = p6 by Lemma 2.2, it follows that f12 = 2z6 and z6 = p6 . Hence b3 c7 = b9 + 2z6 by Lemma 2.3. Therefore (b3 c7 , b3 c7 ) = 5 and (b¯3 z6 , c7 ) = 2. So b¯3 z6 = 2c7 + c4 . But b8 c7 = b8 + h6 + h¯ 6 + 2b¯3 z6 = b8 + h6 + h¯ 6 + 4c7 + 2c4 ,


49

This is impossible for b8 c7 is real and c4 is not real. Substep 2 There exists no NITA such that h6 = i¯6 . Suppose that h6 = i¯6 . By (2.56), it follows that (b3 p¯6 , h6 ) = (p6 , b3 h¯ 6 ) = (i¯6 , b¯3 z6 ) = (b3 i¯6 , z6 ) = 1. But b9 ∈ Supp{b3 h¯ 6 } by (2.51) and (c4 , b6 p¯ 6 ) = (c4 , b¯3 z6 ) = 1 by (2.53) and (2.54). By (2.56), (2.51) and (2.54), there exist δ3 , z3 and v8 ∈ N ∗ B such that b3 h¯ 6 = b9 + p6 + δ3 ,

(2.57)

b¯3 z6 = i¯6 + c4 + v8 ,

(2.58)

b3 p¯ 6 = h6 + c4 + v8 .

(2.59)

Hence b¯3 δ3 = h¯ 6 + γ3 , some γ3 ∈ B. Therefore (b3 γ3 , b3 γ3 ) ≥ 2. It follows that δ3 δ¯3 = γ3 γ¯3 = 1 + b8 by Lemma 2.2. Since (b3 b¯3 )c4 = c4 + c4 b8 , b3 (b¯3 c4 ) = b3 y¯6 + b3 p¯ 6 = c4 + i¯6 + b8 + h6 + c4 + v8 , we have that c4 b8 = c4 + h6 + i¯6 + v8 + b8 . Then (c4 b8 , c4 b8 ) ≥ 5. Let c4 c¯4 = 1 + b8 + r7 , r∈ N ∗ B and r7 is real. By (2.52) and c4 is not real, we have that r7 = c7 and then (c4 b8 , c4 b8 ) = 5. Consequently v8 ∈ B. By b3 (δ3 δ¯3 ) = (b3 δ¯3 )δ3 , we have that δ3 γ¯3 = b3 + x6 or b3 + y6 , δ3 h¯ 6 = b3 + b9 + y6 or b3 + b9 + x6 . Without loss of generality, let δ3 γ¯3 = b3 + x6 . Then 1 + 2b8 + b82 = (δ3 δ¯3 )(γ3 γ¯3 ) = (δ3 γ¯3 )(δ¯3 γ3 ) = b3 b¯3 + b3 x¯6 + b¯3 x6 + x6 x¯6 = 1 + 3b8 + c4 + c¯4 + h6 + h¯ 6 + x6 x¯6 . It holds by (2.52) that x6 x¯6 = 1 + 2b8 + c7 + i6 + i¯6 . Furthermore (b5 b¯5 , x6 x¯6 ) = 5. On the other hand, the following equations hold: b3 (b5 b¯5 ) = b3 + 2b3 b8 + b3 c4 + b3 c¯4 = 3b3 + 2b9 + 2x6 + 2y6 + x6 + z6 + y6 + p6 ,

50


(b3 b¯5 )b5 = b¯3 b5 + x¯6 b5 + y¯6 b5 = b3 + x6 + y6 + x¯6 b5 + y¯6 b5 . Hence x¯ 6 b5 + y¯6 b5 = 2b3 + 2b9 + 2x6 + 2y6 + z6 + p6 .

(2.60)

It is proved that (b5 b¯5 , x6 x¯6 ) = 5. By the above equation and (b5 x¯5 , b3 ) = (b5 y¯5 , b3 ) = 1, we can only have that b5 x¯6 = b3 + b9 + x6 + y6 + z6 or b3 + b9 + x6 + y6 + p6 , b5 y¯6 = b3 + b9 + x6 + y6 + p6 or b3 + b9 + x6 + y6 + z6 . Hence b¯3 (b5 x¯6 ) = b¯3 b3 + b¯3 b9 + b¯3 x6 + b¯3 y6 + b¯3 z6 or b¯3 b3 + b¯3 b9 + b¯3 x6 + b¯3 y6 + b¯3 p6 = 1 + b8 + c7 + b8 + h¯ 6 + i¯6 + c4 + h6 + b8 + c¯4 + i6 + b8 + i¯6 + c4 + v8 or 1 + b8 + c7 + b8 + h¯ 6 + i¯6 + c4 + h6 + b8 + c¯4 + i6 + b8 + h¯ 6 + c¯4 + v¯8 , (b¯3 b5 )x¯6 = b3 x¯6 + x6 x¯6 + y6 x¯6 = c¯4 + h¯ 6 + b8 + 1 + 2b8 + c7 + i6 + i¯6 + y6 x¯6 . Therefore y6 x¯6 = 2c4 + h6 + b8 + i¯6 + v8 or c4 + c¯4 + h6 + b8 + h¯ 6 + v¯8 . If y6 x¯6 = 2c4 + h6 + b8 + i¯6 + v8 , then (c4 x6 , y6 ) = 2. Hence (c4 x6 , c4 x6 ) ≥ 5, a contradiction. Therefore y6 x¯6 = c4 + c¯4 + h6 + b8 + h¯ 6 + v¯ 8 . Thus (c¯4 x6 , y6 ) = 1. Consider the following equations: b3 (c4 c¯4 ) = b3 + b3 b8 + b3 c7 = 2b3 + b9 + x6 + y6 + b9 + f12 , (b3 c4 )c¯4 = c¯4 x6 + c¯4 z6 , (b3 c¯4 )c4 = c4 y6 + c4 p6 .

(2.61)


51

By the expression x6 x¯6 , we have that (x6 , c¯4 x6 ) = 0. Consequently, (x6 , c¯4 z6 ) = 1 by the above equations. Since b3 c7 = b9 + f12 and L1 (B) = {1} and L2 (B) = ∅, so the constituents of f12 are of degrees larger than 3. Since (c¯4 x6 , b3 ) = (c¯4 z6 , b3 ) = 1 by (2.53), (c¯4 x6 , c¯4 x6 ) = 4 and it is proved that (c¯4 x6 , y6 ) = 1, we have that c¯4 x6 = b3 + y6 + b9 + g6 ,

g6 ∈ B.

Consequently c¯4 z6 = b3 + x6 + b9 + f6 , where f12 = f6 + g6 and f6 ∈ N ∗ B. Let b¯3 g6 = c7 + x11 and b¯3 f6 = c7 + y11 , x11 , y11 ∈ N ∗ (B). Then b¯3 (c¯4 x6 ) = b¯3 b3 + b¯ 3 y6 + b¯3 b9 + b¯3 g6 = 1 + b8 + c¯4 + i6 + b8 + c7 + b8 + h¯ 6 + i¯6 + c7 + x11 , c¯4 (b¯3 x6 ) = c¯4 c4 + c¯4 h6 + c¯4 b8 = 1 + b8 + c7 + c¯4 h6 + c¯4 + h¯ 6 + i6 + v¯ 8 + b8 , b¯ 3 (c¯4 z6 ) = b¯3 b3 + b¯ 3 x6 + b¯3 b9 + b¯3 f6 = 1 + b8 + c4 + h6 + b8 + c7 + b8 + h¯ 6 + i¯6 + c7 + y11 , c¯4 (b¯3 z6 ) = c¯4 c4 + c¯4 i¯6 + c¯4 v8 = 1 + b8 + c7 + c¯4 i¯6 + c¯4 v8 . Hence c¯4 h6 + v¯8 = b8 + i¯6 + c7 + x11 , c¯4 i¯6 + c¯4 v8 = c4 + h6 + 2b8 + i¯ + h¯ 6 + c7 + y11 .

(2.62) (2.63)

We assert that v8 = b8 . Otherwise, if v8 = b8 , then (c4 b8 , b8 ) = 2. So (b82 , c4 )=2, a contradiction by (2.52). Hence v¯ 8 ∈ Supp{x11 } by (2.62). Let x11 = v¯ 8 + x3 , then it follows by (2.62) that c¯4 h6 = b8 + i¯6 + c7 + x3 . By (2.63), we have that h¯ 6 ∈ Supp{c¯4 i¯6 + c¯4 v8 }. So (c¯4 h6 , i6 )≥1 or (c¯4 h6 , v¯ 8 )≥1. The latter case will lead to v8 = b8 , a contradiction. Hence (c¯4 h6 , i6 ) ≥ 1. Thus i6 = i¯6 and (c¯4 h6 , i6 ) = 1. Furthermore, x6 x¯6 = 1 + 2b8 + c7 + 2i6 . And it follows by (2.49) and (2.51) that b3 i6 = b9 + y6 + z3 ,

some z3 ∈ B.

Thus b3 (x6 x¯6 ) = b3 + 2b3 b8 + b3 c7 + 2b3 i6 = 3b3 + 2b9 + 2x6 + 2y6 + b9 + g6 + f6 + 2b9 + 2y6 + 2z3 ,

52


x6 (b3 x¯6 ) = x6 c¯4 + x6 h¯ 6 + x6 b8 = b3 + y6 + b9 + g6 + x6 h¯ 6 + x6 b8 . Then x6 h¯ 6 + x6 b8 = 2b3 + 2b9 + 2x6 + 3y6 + f6 + 2b9 + 2z3 . By the expression y6 x¯6 and v8 = b8 , we have that (b8 x6 , y6 ) = 1. Hence (x6 h¯ 6 , y6 ) = 2 and so (x¯6 y6 , h¯ 6 ) = 2. Again by the expression y6 x¯6 , we have that h6 = h¯ 6 . Now (2.48) implies that x6 ∈ Supp{b3 h6 }. Then p6 = x6 by (2.57). Consequently b3 c¯4 = x6 + y6 , from which we have (bc4 , b¯3 x6 ) = 1. Thus c4 = c¯4 by (2.48). Step 4 follows. This completes the proof of Lemma 2.6. Therefore (b5 b¯5 , b8 ) = 1 and (b3 b¯3 , b5 b¯5 ) = 2. Consequently by Lemma 2.3 we have x12 ∈ B and (b3 b8 , b3 b8 ) = 3. Thus Theorem 2.2 holds. Furthermore, in Lemma 2.3 we proved that c7 ∈ B.

2.4 General Information on NITA Generated by b3 and Satisfying b32 = b¯3 + b6 and b32 = c3 + b6 Since the structure of NITA generated by b3 and satisfying b32 = b¯3 + b6 and b32 = c3 + b6 are related, we investigate them simultaneously. Some results are written in the same proposition. The purpose of this section is to prove (3) and (4) of the Main Theorem 1. The following lemmas will be useful during our investigation. Lemma 2.7 (1) If b32 = b¯3 + b6 , then b3 b8 = b6 + b¯3 b6 ; (2) If b32 = c3 + b6 , then there exists an element x6 ∈ B such that b¯3 c3 = b3 + x6 and c3 c¯3 = 1 + b8 . Furthermore, b3 b8 = x6 + b¯3 b6 and c3 b8 = b6 + b3 x6 . Proof (1) follows from (b3 b¯3 )b3 = b32 b¯3 . (2) Since b3 ∈ Supp{b¯3 c3 }, (b3 b¯3 , c3 c¯3 ) ≥ 2. But (b3 b¯3 , c3 c¯3 ) ≤ 2, so (b3 b¯3 , c3 c¯3 ) = 2 and c3 c¯3 = 1 + b8 ,

b¯3 c3 = b3 + x6 ,

x6 ∈ B.

Checking (b3 b¯3 )b3 = b32 b¯3 and c3 (c3 c¯3 ) = b3 (b¯3 c3 ), one has (2).

Lemma 2.8 If (b3 b8 , b3 b8 ) ≥ 4, then any constituent of b3 b8 and b¯ 3 b6 is either b3 or of degree > 3 and (b3 b8 , b3 b8 ) = 4, 5. Proof Since (b3 b8 , b3 ) = 1, the degree of the sum of the remaining constituents of b3 b8 must be 21. On the other hand, any constituent yn of b3 b8 must have degree ≥ 4. Otherwise, n = 3 and b3 y¯3 = 1 + b8 , which implies that b3 = y3 , a contradiction.

2.4 General Information on NITA Generated by b3

53

Theorem 2.3 There exists no NITA generated by b3 and satisfying b32 = b¯3 + b6 or b32 = c3 + b6 and (b3 b8 , b3 b8 ) = 2. Proof Since b3 b¯3 = 1 + b8 and (b3 b8 , b3 b8 ) = 2, we may assume that b3 b8 = b3 + b21 , where b21 ∈ B. By Lemma 2.7, one has that b3 + b21 = b6 + b¯3 b6 or x6 + b¯3 b6 ,

which is impossible.

Theorem 2.4 There is no NITA generated by b3 and satisfying b32 = b¯3 + b6 or b32 = c3 + b6 and (b3 b8 , b3 b8 ) = 5. Proof (1) There is no NITA satisfying b32 = b¯3 + b6 and (b3 b8 , b3 b8 ) = 5. By b32 b¯3 = (b3 b¯3 )b3 , we have that b3 b8 = b6 + b¯3 b6 , which implies that b8 ∈ Supp{b¯3 b6 }. But b3 ∈ Supp{b¯ 3 b6 }. We need a sum with degree 7 of two constituents to make up b¯3 b6 , which is impossible by Lemma 2.7. (2) There is no NITA satisfying b32 = c3 + b6 and (b3 b8 , b3 b8 ) = 5. By Lemma 2.7 and (b3 b8 , b3 b8 ) = 5, we may assume that (A)

b3 b8 = b3 + x6 + x4 + y4 + z7 ,

(B)

b3 b8 = b3 + x6 + x4 + y5 + z6 ,

(C)

b3 b8 = b3 + x6 + x5 + y5 + z5 .

Since b¯3 (b¯3 c3 ) = b3 b¯3 + x6 b¯3 , b¯32 c3 = c3 c¯3 + c3 b¯6 = 1 + b8 + c3 b¯6 , we have that b¯3 x6 = c3 b¯6 . Since (c3 b8 , c3 b8 ) = (b3 b8 , b3 b8 ) = 5 by Lemma 2.7 and c3 b8 = b6 + b3 x6 by Lemma 2.7, we have (b3 x6 , b3 x6 ) = 4. Furthermore, (b¯3 x6 , b¯3 x6 ) = 4. But (b¯3 x6 , b8 ) = 1. We may assume c3 b¯6 = b¯3 x6 = b8 + m3 + n3 + v4 , where m3 , n3 , v4 ∈ B and m3 , n3 , v4 are distinct. We may assume c3 m ¯ 3 = b6 + m∗3 ,

m∗3 ∈ B,

c3 n¯ 3 = b6 + n∗3 . ¯ 3 , c3 m ¯ 3 ) = 2. So m3 m ¯ 3 = 1 + b8 . Moreover, Hence (c3 m ¯3 = m ¯3 +m ¯ 3 b8 , (c3 c¯3 )m

54


(c3 m ¯ 3 )c¯3 = c¯3 b6 + c¯3 m∗3 = b8 + m ¯ 3 + n¯ 3 + v¯ 4 + c¯3 m∗3 . Then m ¯ 3 b8 = b8 + n¯ 3 + v¯4 + c¯3 m∗3 , which implies that m3 n¯ 3 = 1 + b8 . So m3 = n3 , a contradiction.

Theorem 2.5 There exists no NITA generated by b3 and satisfying b32 = b¯3 + b6 , (b3 b8 , b3 b8 ) = 4. Proof Since b3 b8 = b6 + b¯3 b6 by Lemma 2.7 and by assumption that (b3 b8 , b3 b8 ) = 4, then b8 ∈ Supp{b¯3 b6 } and (b¯ 3 b6 , b¯3 b6 ) = 3. But b3 ∈ Supp{b¯3 b6 }. Thus there exists p7 ∈ B such that b¯3 b6 = b3 + b8 + p7 ,

b3 b8 = b3 + b6 + b8 + p7 .

Hence (b3 b¯3 )b8 = b8 + b82 , b¯3 (b3 b8 ) = b¯3 b3 + b¯3 b6 + b¯3 b8 + b¯3 p7 = 1 + b8 + b3 + b8 + p7 + b¯3 + b¯6 + b¯8 + p¯ 7 + b¯3 p7 . Consequently, b82 = 1 + b3 + b¯3 + p7 + p¯ 7 + 2b8 + b¯6 + b¯3 p7 . On the other hand, (b3 b¯3 )2 = 1 + 2b8 + b82 , b32 b¯32 = b3 b¯3 + b¯3 b6 + b3 b¯6 + b6 b¯6 = 1 + b8 + b3 + b8 + p7 + b¯3 + b8 + p¯ 3 + b6 b¯6 . We have that b82 = b3 + b¯3 + p7 + p¯ 7 + b8 + b6 b¯6 . Hence b6 b¯6 = 1 + b¯6 + b8 + b¯3 p7 . Since b¯32 b6 = b3 b6 + b¯6 b6 = b3 b6 + 1 + b8 + b¯6 + b¯3 p7 , b¯3 (b¯3 b6 ) = b¯3 b3 + b¯3 b8 + b¯3 p7 = 1 + b8 + b¯3 + b¯6 + b8 + p¯ 7 + b¯3 p7 , it holds that b3 b6 = b¯3 + p¯ 7 + b8 . Now the following equations follow: b3 (b¯3 b6 ) = b32 + b3 b8 + b3 p7 = b¯3 + b6 + b3 + b6 + b8 + p7 + b3 p7 , (b3 b6 )b¯3 = b¯32 + b¯3 p¯7 + b8 b¯3 = b3 + b¯6 + b¯3 + b¯6 + b8 + p¯ 7 + b¯3 p¯ 7 .

2.5 NITA Generated by b3 Satisfying b32 = b¯3 + b6

55

Then 2b6 + p7 + b3 p7 = 2b¯6 + p¯ 7 + b¯3 p¯ 7 . Since b6 is nonreal, we have that (b3 p7 , b¯6 ) = 2, a contradiction to the expression b¯3 b6 . The theorem follows. Theorem 2.6 c32 = r4 + s5 .

There exists no NITA satisfying b32 = c3 + b6 , (b3 b8 , b3 b8 ) = 4 and

Proof Let us investigate Sub-NITA generated by c2 . By assumptions, we have that (b3 b¯3 , b3 b¯3 ) = (c3 c¯3 , c3 c¯3 ), hence b3 b¯3 = c3 c¯3 . Therefore (c3 c¯3 , b82 ) = (b3 b¯3 , b82 ) = 4, thus (c3 b8 , c3 b8 ) = 4, a contradiction to Theorem 2.2.

2.5 NITA Generated by b3 Satisfying b32 = b¯3 + b6 and b6 Nonreal and b10 ∈ B is Real Now we discuss NITA satisfying the following hypothesis. Hypothesis 2.1 Let (A, B) be a NITA generated by a nonreal element b3 ∈ B satisfies b3 b¯3 = 1 + c8 and b32 = b¯ 3 + b6 , where b6 ∈ B is nonreal. Here we still assume that L(B) = 1 and L2 (B) = ∅. In this section, we use the symbols bi , ci , di to denote elements of B with degree i, where i ≥ 2. By the associative law and Hypothesis 2.1, we have b3 (b3 b¯3 ) = (b32 )b¯3 , b3 + c8 b3 = b¯32 + b6 b¯3 , so c8 b3 = b¯6 + b6 b¯3 . Hence (b3 b6 , c8 ) = (b3 c8 , b¯6 ) = 1, so b3 b6 = c8 + b10 , therefore b10 ∈ B since (b3 c8 , b3 c8 ) = 3. Now we shall deal with the case b10 = b¯10 ∈ B. Lemma 2.9 Let (A, B) satisfy Hypothesis 2.1 and b¯10 = b10 ∈ B, then b3 b6 = c8 + b10 ; b¯3 b6 = b3 + b15 , where b15 ∈ B; b3 c8 = b3 + b¯6 + b15 ; b3 b10 = b15 + b¯6 + c¯9 , c9 ∈ B; b3 b15 = 2b¯15 + b6 + c9 ; b62 = 2b¯6 + c¯9 + b15 ; b6 c8 = b¯3 + 2b¯15 + b6 + c9 ; b¯6 b6 = 1 + c8 + x + y + z + w, where x, y, z, w ∈ B, |x + y + z + w| = 27, and the degrees of x, y, z and w ≥ 5. Moreover, x + y + z + w = x + y + z + w; 9) c82 = 1 + 2c8 + 2b10 + x + y + z + w; 1) 2) 3) 4) 5) 6) 7) 8)

56


10) b3 b¯15 = b10 + c8 + x + y + z + w; 11) (c¯9 b3 , c9 ) = 1 or (b6 c9 , c¯9 ) ≥ 1. Proof We have seen that c8 b3 = b¯6 + b6 b¯3

(1)

b3 b6 = c8 + b10 ,

(2)

and where b10 is real. So we obtain (b¯3 b6 , b¯3 b6 ) = 2 and consequently we obtain b¯3 b6 = b3 + b15 ,

(3)

c8 b3 = b3 + b¯6 + b15 .

(4)

where b15 ∈ B. Hence by (1)

By the associative law and (2), (3), we have b3 (b¯3 b6 ) = (b3 b6 )b¯3 and b32 + b15 b3 = c8 b¯3 + b10 b¯3 , so by (4) and Hypothesis 2.1 we have b15 b3 = b¯15 + b¯3 b10 .

(5)

If x ∈ B and (b15 b3 , x) = (b15 , b¯3 x), so |x| ≥ 5, ∀x ∈ B. So every constituent of ¯b3 b10 has degrees ≥ 5. Since b10 is real, the constituents of b3 b10 also have degrees ≥ 5, so by the associative law and Hypothesis 2.1 and (4), (2): (b¯3 b3 )c8 = (b3 c8 )b¯3 , c8 + c82 = b3 b¯3 + b¯6 b¯3 + b15 b¯3 , c82 = 1 + c8 + b10 + b¯3 b15 .

(6)

By the associative law and Hypothesis 2.1 and (2): (b3 b¯3 )(b3 b¯3 ) = b32 b¯32 , 1 + 2c8 + c82 = b¯3 b3 + b¯3 b¯6 + b6 b3 + b6 b¯6 , c82 = 2b10 + c8 + b6 b¯6 ;

(7)

so we get 1 = (b¯ 3 b15 , b10 ) = (b15 , b3 b10 ). By (2), (b3 b10 , b¯6 ) = (b3 b6 , b10 ) = 1. b3 b10 have constituents of degree ≥ 5 as we have proved. Therefore b3 b10 = b15 + b¯6 + c¯9

(8)


57

where c9 ∈ B. So by (5), b3 b15 = 2b¯15 + b6 + c9 . By the associative law and Hypothesis 2.1 and (2), (3), (4): (b32 )b6 = (b3 b6 )b3 , b¯3 b6 + b62 = c8 b3 + b10 b3 , consequently b62 = 2b¯ 6 + b15 + c¯9 . Now b3 (b¯3 b6 ) = (b3 b¯3 )b6 , b32 + b15 b3 = b6 + b6 b8 . Thus c8 b6 = b¯3 + 2b¯15 + b6 + c9 . So (b¯6 b6 , c8 ) = (b6 , b6 c8 ) = 1. Since (b62 , b62 ) = 6 = (b¯6 b6 , b¯6 b6 ), hence b6 b¯6 = 1 + c8 + x + y + z + w where x, y, z, w ∈ B. Therefore by (7), (6) c82 = 1 + 2c8 + 2b10 + x + y + z + w, and b3 b¯5 = b10 + c8 + x + y + z + w. ¯ b15 ), so |m| ≥ 5, hence So (b3 b¯15 , m) = (b3 m, |x|, |y|, |z|, |w| ≥ 5. By (8), (b3 c9 , b10 ) = (b3 b10 , c¯9 ) = 1, hence b3 c9 = b10 + α,

(9)

58


where α ∈ N ∗ B. By the associative law and Hypothesis 2.1: (b32 )c9 = (b3 c9 )b3 , c9 b¯3 + c9 b6 = b10 b3 + αb3 = b15 + b¯6 + c¯9 + αb3 ; hence (c9 b¯3 , c¯9 ) ≥ 1, or (b6 c9 , c¯9 ) ≥ 1. By (9) we conclude that either (c9 b¯3 , c¯9 ) = 1 or (b6 c9 , c¯9 ) ≥ 1. We start to investigate the case (c¯9 b3 , c9 ) = 1 of Lemma 2.9. Lemma 2.10 Let (A, B) satisfy Hypothesis 2.1. Assume that b¯10 = b10 ∈ B and (c¯9 b3 , c9 ) = 1. Then we have the following: 1) 2) 3) 4) 5) 6) 7) 8) 9) 10) 11) 12) 13) 14) 15) 16)

b3 c¯9 = b¯15 + c9 + c3 , c3 , c9 ∈ B; b6 b10 = 3b¯15 + b¯3 + c3 + c9 , b15 ∈ B; c8 b15 = 5b15 + b3 + c¯3 + 2b¯6 + 3c¯9 ; b3 c¯3 = c9 ; b3 c3 = b9 , b¯9 = b9 ∈ B; b3 c9 = b9 + b10 + b8 , b9 = b¯9 , b¯8 = b8 ∈ B; c¯3 b6 = b10 + b8 ; b6 b¯15 = 4b15 + b¯6 + b3 + c¯3 + 2c¯9 ; c3 c¯3 = 1 + b8 ; b¯6 b6 = 1 + c8 + b9 + b8 + c5 + b5 , c5 + b5 = c5 + b5 ; c82 = 1 + 2c8 + 2b10 + b9 + b8 + b5 + c5 ; b3 b¯15 = b10 + c8 + b9 + b8 + b5 + c5 ; b6 b15 = 2c8 + 3b10 + 2b9 + 2b8 + b5 + c5 ; c8 b10 = 2c8 + 2b10 + 2b9 + 2b8 + b5 + c5 ; b6 c¯9 = b10 + c8 + 2b9 + b8 + b5 + c5 ; 2 = 1 + 2c + 2b + 3b + 2b + 2b + 2c . b10 8 10 9 8 5 5

Proof By 5) in Lemma 2.9, (b¯3 c9 , b15 ) = (c9 , b3 b15 ) = 1. So by the assumption in the lemma, (b3 c¯9 , c9 ) = 1, hence b3 c¯9 = b¯15 + c9 + c3 ,

(1)

where c3 ∈ B. By the associative law and Hypothesis 2.1 and 4) in Lemma 2.9 (b32 )b10 = (b3 b10 )b3 , b¯3 b10 + b6 b10 = b15 b3 + b¯6 b3 + c¯9 b3 ; by 2), 4), 5) in Lemma 2.9 and (1), b6 b10 = 3b¯15 + b¯3 + c3 + c9 . By the associative law and Hypothesis 2.1 and 5) in Lemma 2.9:

(2)


59

(b¯3 b3 )b15 = (b3 b15 )b¯3 , c8 b15 + b15 = 2b¯ 15 b¯3 + b¯3 b6 + c9 b¯3 ; by (1) and 2), 5) in Lemma 2.9, c8 b15 = 5b15 + b3 + c¯3 + 2b¯6 + 3c¯9 .

(3)

By (1), (c¯3 b3 , c9 ) = (b3 c¯9 , c3 ) = 1, hence c¯3 b3 = c9 .

(3 )

So (c3 b3 , c3 b3 ) = (b3 c¯9 , c3 ) = 1, hence c3 b3 = b9 .

(4)

By the associative law and Hypothesis 2.1 and (3) c¯3 (b32 ) = (c¯3 b3 )b3 , c¯3 b¯3 + c¯3 b6 = c9 b3 , hence c9 b3 = b¯ 9 + c¯3 b6 .

(5)

By (1) and (4) in Lemma 2.9, (c9 b3 , b10 ) = (b3 b10 , c¯9 ) = 1, (b3 c9 , b3 c9 ) = (b3 c¯9 , b3 c¯9 ) = 3, hence c9 b3 = b¯ 9 + b10 + b8 ,

(6)

c¯3 b6 = b10 + b8 .

(7)

where b8 ∈ B. So

By the associative law and Hypothesis 2.1 and 4) in Lemma 2.9: (b¯3 b3 )b10 = (b3 b10 )b¯3 , b10 + c8 b10 = b15 b¯3 + b¯6 b¯3 + c¯9 b¯3 , by 1), 10) in Lemma 2.9 and (6) c8 b10 = c8 + x¯ + y¯ + z¯ + w¯ + b¯8 .

(8)

b10 , c8 are reals, so if b¯8 = b8 , then (c8 b10 , b8 ) = 0, and without loss of generality x¯ = b8 , so x = b8 and by 8) in Lemma 2.9 and without loss of generality y = b8 ,

60


so by (8) (c8 b10 , b¯8 ) = 2, hence (c8 b10 , b8 ) = 2, so without loss of generality z¯ = b8 , hence z = b¯8 and by 8) in Lemma 2.9 without loss of generality w = b8 then (c8 b10 , b8 ) = 3, a contradiction. Hence b8 = b¯8 . In the same way b9 = b¯9 . By the associative law and 2), 6) in Lemma 2.9: (b62 )b¯3 = (b6 b¯3 )b6 , 2b¯6 b¯3 + c¯9 b¯3 + b15 b¯3 = b3 b6 + b15 b6 , and by Lemma 2.9 and (6) b6 b15 = 2c8 + 3b10 + b8 + b9 + x + y + w + z.

(9)

By the associative law and the Hypothesis 2.1 and 4) in Lemma 2.9: b10 (b3 b¯3 ) = (b¯3 b10 )b3 , b10 + c8 b10 = b3 b¯15 + b3 b6 + b3 c9 ; by (6) and 1), 10) in Lemma 2.9 c8 b10 = 2c8 + 2b10 + x + y + z + w + b8 + b9 .

(10)

By the associative law and 2), 3), 6) in Lemma 2.9 and (3): c8 (b¯3 b6 ) = (b¯3 c8 )b6 , b3 c8 + b15 c8 = b¯3 b6 + b62 + b¯15 b6 , b6 b¯15 = 4b15 + b¯6 + b3 + c¯3 + 2c¯9 . By the associative law and (1), (4), (6) and Hypothesis 2.1: (b32 )c¯9 = (b3 c¯9 )b3 , b¯3 c¯9 + b6 c¯9 = b¯15 b3 + c9 b3 + c3 b3 , b6 c¯9 = b10 + c8 + x + y + z + w + b9 .

(11)

By the associative law and 1) in Lemma 2.9 and (2): (b3 b6 )b10 = (b6 b10 )b3 , 2 = 3b¯ b + b¯ b + c b + c b , c8 b10 + b10 15 3 3 3 3 3 9 3

by (4), (6), (10) and Hypothesis 2.1 2 b10 = 1 + 2c8 + b9 + 2b10 + 2x + 2y + 2z + 2w.

(12)

2 ) = 10, so By (2), (b6 b10 , b6 b10 ) = 12. But by 8) in Lemma 2.9 and (12) (b6 b¯6 , b10 we obtain, without loss of generality,

x = b9 .

(13)


61

By (4) and that b9 = b¯9 , (c3 b3 , c¯3 b¯3 ) = (b32 , c¯32 ) = 1, but by (3 ) (c32 , b3 ) = (c3 b¯3 , c¯3 ) = 0, so by Hypothesis 2.1 (c¯32 , b6 ) = 1.

(14)

Hence (c3 c¯3 , c3 c¯3 ) = 2, so c3 c¯3 = 1 + t8 .

(15)

By (4) (c3 b3 , c3 b3 ) = (c3 c¯3 , b3 b¯3 ) = 1, therefore by hypothesis t8 = c8 . By (12), (13) and 9) in Lemma 2.9 2 (b10 , c82 ) = 18 = (c8 b10 , c8 b10 ).

So by (8) we conclude that, without loss of generality, y = b8 .

(16)

So by (13), (16) and 8) in Lemma 2.9, we obtain |z| = |w| = 5, set z = b5 and w = c5 . By (7), (c¯3 b6 , c¯3 b6 ) = (c¯3 c3 , b¯6 b6 ) = 2, so by (15) and 8) in Lemma 2.9 and by (16) t8 = b8 . So c3 c¯3 = 1 + b8 . By 8), 9), 10) in Lemma 2.9 and (10), (11), (12), (13), (16), (9) b¯6 b6 = 1 + c8 + b9 + b8 + c5 + b5 , c82 = 1 + 2c8 + 2b10 + b9 + b8 + b5 + c5 , b3 b¯15 = b10 + c8 + b9 + b8 + b5 + c5 , b6 b15 = 2c8 + 3b10 + 2b9 + 2b8 + b5 + c5 , c8 b10 = 2c8 + 2b10 + 2b9 + 2b8 + b5 + c5 , b6 c¯9 = b10 + c8 + 2b9 + b8 + b5 + c5 , 2 = 1 + 2c + 2b + 3b + 2b + 2b + 2c . b10 8 10 9 8 5 5

Lemma 2.11 Let (A, B) satisfy Hypothesis 2.1. Assume that b¯ 10 = b10 ∈ B and (c¯9 b3 , c9 ) = 1. Then we obtain the following equations: 1) b3 b9 = b15 + c¯9 + c¯3 ; 2) c3 c8 = c9 + b¯15 ;

62

3) 4) 5) 6) 7) 8) 9) 10) 11) 12) 13) 14) 15) 16) 17) 18)


b3 b8 = b15 + c¯9 ; b3 c5 = b15 ; b3 b5 = b15 ; b6 b5 = b¯15 + b6 + c9 ; b6 c9 = 2b15 + 2c¯9 + b¯6 ; c3 b6 = c¯3 + b15 ; b6 b9 = 2b¯15 + b6 + 2c9 ; b6 c5 = b¯15 + b6 + c9 ; b6 b8 = 2b¯15 + b6 + c3 + c9 ; c3 b10 = b¯15 + b6 + c9 ; c3 c9 = b15 + c¯9 + b3 ; c3 b8 = c3 + b6 + b¯15 ; c3 b9 = b¯15 + b¯3 + c9 ; c3 b¯15 = 2b15 + b¯6 + c¯9 ; b8 b¯15 = 5b¯15 + b¯3 + c3 + 2b6 + 3c9 ; b8 c9 = 3b¯15 + 2c9 + b6 + b¯3 .

Proof By 5), 6), 12), in Lemma 2.10 (b3 b9 , c¯3 ) = (b3 c3 , b9 ) = 1, (b3 b9 , c¯9 ) = (b3 c9 , b9 ) = 1, (b3 b9 , b15 ) = (b3 b¯15 , b9 ) = 1, then b3 b9 = b15 + c¯9 + c¯3 .

(1)

By the associative law and 1), 4) in Lemma 2.10 and Hypothesis 2.1, (b¯3 c3 )b3 = (b¯3 b3 )c3 , c¯9 b3 = c3 + c3 c8 , so c3 c8 = c9 + b¯15 .

(2)

By 6), 12) in Lemma 2.10, (b3 b8 , b15 ) = (b3 b¯15 , b8 ) = 1, (b3 b8 , c¯9 ) = (b3 c9 , b8 ) = 1, then b3 b8 = b15 + c¯9 . By the associative law and 2) in Lemma 2.9 and 10) in Lemma 2.10: (b6 b¯6 )b3 = (b3 b¯6 )b6 , b3 + c8 b3 + b9 b3 + b8 b3 + c5 b3 + b5 b3 = b¯3 b6 + b¯15 b6 ,

(3)


63

by 2), 3) in Lemma 2.9 and (3), (1) and 8) in Lemma 2.10: c5 b3 + b5 b3 = 2b15 , so c5 b3 = b15 ,

(4)

b5 b3 = b15 .

(5)

By the associative law and Hypothesis 2.1 and (5) and 5) in Lemma 2.9 (b32 )b5 = (b3 b5 )b3 = b3 b15 , b¯3 b5 + b6 b5 = 2b¯15 + b6 + c9 , by (5) (b¯ 3 b5 , b¯3 b5 ) = (b5 b3 , b5 b3 ) = 1. So b¯3 b5 = b¯15 , therefore b6 b5 = b¯15 + b6 + c9 .

(6)

By the associative law and 4), 7) in Lemma 2.10: (b3 c¯3 )b6 = (c¯3 b6 )b3 , c9 b6 = b10 b3 + b8 b3 , by 4) in Lemma 2.9 and by (3) b6 c9 = 2b15 + 2c¯9 + b¯6 .

(7)

By the associative law and the Hypothesis 2.1 and 5) in Lemma 2.10: (b32 )c3 = (b3 c3 )b3 ,

b¯3 c3 + b6 c3 = b9 b3 ,

so by (1) and 4) in Lemma 2.10 b6 c3 = c¯3 + b15 .

(8)

10), 13), 15) in Lemma 2.10, (b6 b9 , b¯15 ) = (b6 b15 , b9 ) = 2, (b6 b9 , b6 ) = (b9 , b¯6 b6 ) = 1, (b6 b9 , c9 ) = (b6 c¯9 , b9 ) = 2, so b6 b9 = 2b¯15 + b6 + 2c9 .

(9)

64


By the associative law and Hypothesis 2.1 and (4): b¯3 c5 + b6 c5 = b15 b3 ,

(b32 )c5 = (b3 c5 )b3 , by 5) in Lemma 2.9

b¯3 c5 + b6 c5 = 2b¯ 15 + b6 + c9 , (c5 b¯3 , c5 b¯3 ) = (c5 b3 , c5 b3 ) = 1, so b¯3 c5 = b¯15 , hence by 7), 10), 13), and 15) in Lemma 2.10 b6 c5 = b¯ 15 + b6 + c9 ,

(10)

(b6 b8 , b6 ) = (b8 , b¯6 b6 ) = 1,

(b6 b8 , c3 ) = (b6 c¯3 , b8 ) = 1, (b6 b8 , c9 ) = (b8 , b¯6 c9 ) = 1,

(b6 b8 , b¯15 ) = (b6 b15 , b8 ) = 2,

so b6 b8 = 2b¯15 + b6 + c3 + c9 .

(11)

By the associative law and 1) in Lemma 2.9 and (8) c3 (b3 b6 ) = (c3 b6 )b3 ,

c3 c8 + c3 b10 = c¯3 b3 + b15 b3 ,

by (2), 4) in Lemma 2.10 and 5) in Lemma 2.9 c3 b10 = b¯15 + b6 + c9 .

(12)

By the associative law and 4), 9) in Lemma 2.10: (c3 c¯3 )b3 = (c¯3 b3 )c3 ,

b3 + b8 b3 = c9 c3 ,

by (3) c3 c9 = b15 + c¯9 + b3 . By the associative law and 6) in Lemma 2.10 and (13) (b3 c9 )c3 = (c3 c9 )b3 , b9 c3 + b10 c3 + b8 c3 = b15 b3 + c¯9 b3 + b32 ; by (12) and 5) in Lemma 2.9 and 1) in Lemma 2.10 and Hypothesis 2.1 b9 c3 + b8 c3 + b15 + b6 + c9 = 2b¯15 b6 + c9 + b¯15 + c9 + c3 + b¯3 + b6 , b9 c3 + b8 c3 = 2b¯15 + c9 + c3 + b¯3 + b6 ; by 5), 7), 9) in Lemma 2.10

(13)


65

(b8 c3 , b¯3 ) = (c3 b3 , b8 ) = 0, (b8 c3 , c3 ) = (b8 , c¯3 c3 ) = 1, (b8 c3 , b6 ) = (b8 , c¯3 b6 ) = 1. So b8 c3 = c3 + b6 + b¯15 ,

(14)

b9 c3 = b¯15 + b¯3 + c9 .

(15)

hence

By the associative law and 2) in Lemma 2.9 and 7) in Lemma 2.10: c3 (b3 b¯6 ) = (c3 b¯6 )b3 ,

c3 b¯3 + c3 b¯15 = b10 b3 + b8 b3 ,

by 4) in Lemma 2.10 and 4) in Lemma 2.9 and (3): c3 b¯15 = 2b15 + b¯6 + c¯9 .

(16)

By the associative law and 9) in Lemma 2.10: (c3 b¯15 )c¯3 = (c3 c¯3 )b¯15 ,

2b15 c¯3 + b¯6 c¯3 + c¯9 c¯3 = b¯15 + b8 b¯15 ,

by (16) and (8), (13) b8 b¯15 = 5b¯15 + b¯3 + c3 + 2b6 + 3c9 .

(17)

By the associative law and (13) and 9) in Lemma 2.10: (c3 c¯3 )c9 = (c3 c9 )c¯3 ,

c9 + b8 c9 = b15 c¯3 + c¯9 c¯3 + b3 c¯3 ,

by (13), (16) and 4) in Lemma 2.10: c9 + b8 c9 = 2b¯15 + b6 + c9 + b¯15 + c9 + b¯3 + c9 , then b8 c9 = 3b¯15 + 2c9 + b6 + b¯3 .

(18)

Lemma 2.12 Let (A, B) satisfy Hypothesis 2.1. Assume that b¯10 = b10 ∈ B and (c¯9 b3 , c9 ) = 1. Then we obtain the following equations: 1) 2) 3) 4) 5)

c3 c¯9 = b10 + b9 + c8 ; c3 b15 = b5 + c5 + c8 + b8 + b9 + b10 ; c32 = c¯3 + b¯6 ; b¯15 b15 = 1 + 3b5 + 3c5 + 5c8 + 5b8 + 6b6 + 6b10 ; 2 = 9b¯ + 4b + 2b¯ + 2c + 6c ; b15 15 6 3 3 9

66

6) 7) 8) 9) 10) 11) 12) 13) 14) 15) 16) 17) 18)


b10 b15 = 6b15 + 3b¯6 + b3 + c¯3 + 4c¯9 ; c9 b15 = 2b5 + 2c5 + 3c8 + 3b8 + 3b9 + 4b10 ; c9 b¯15 = 6b15 + 2b¯6 + b3 + c¯3 + 3c¯9 ; c92 = 3b15 + b3 + c¯3 + 2b¯6 + 2c¯9 ; c9 c¯9 = 1 + 2c8 + 2b9 + 2b10 + c5 + b5 + 2b8 ; c8 c9 = 3b¯15 + b6 + c3 + 2c9 ; b9 c9 = b¯3 + 2c9 + 3b¯15 + c3 + 2b6 ; c9 b10 = b¯3 + 2c9 + 3b¯15 + c3 + 2b6 ; b5 c9 = 2b¯15 + b6 + c9 ; c5 c9 = 2b¯15 + b6 + c9 ; b5 = b¯5 , c5 = c¯5 ; c3 b5 = b¯15 ; c3 c5 = b¯15 .

Proof By 2), 12), 15) in Lemma 2.11, (c3 c¯9 , b10 ) = (c3 b10 , c9 ) = 1, (c3 c¯9 , b9 ) = (c3 b9 , c9 ) = 1, (c3 c¯9 , c8 ) = (c3 c8 , c9 ) = 1, then c3 c¯9 = b10 + b9 + c8 .

(1)

By the associative law and 3) and 14) in Lemma 2.11: (b3 b8 )c3 = (c3 b8 )b3 ,

b15 c3 + c¯9 c3 = c3 b3 + b6 b3 + b¯15 b3 ,

by 5), 12) in Lemma 2.10 and (1) and 1) in Lemma 2.9 c3 b15 = b5 + c5 + c8 + b8 + b9 + b10 .

(2)

By the associative law and (1) and 1) Lemma 2.10: (b3 c¯9 )c3 = (c3 c¯9 )b3 ,

b¯15 c3 + c9 c3 + c32 = b10 b3 + b9 b3 + c8 b3 ;

by 1), 13), 16) in Lemma 2.11 and 3), 4) in Lemma 2.9 c32 = c¯3 + b¯6 .

(3)

By the associative law and 2) in Lemma 2.9 and 8) in Lemma 2.10: (b3 b¯6 )b15 = (b¯6 b15 )b3 , b¯ 3 b15 + b¯15 b15 = 4b¯15 b3 + b6 b3 + b¯3 b3 + c3 b3 + 2c9 b3 . By 6), 5), 12) in Lemma 2.10 and 1) in Lemma 2.9 and Hypothesis 2.1, we obtain that b¯15 b15 = 1 + 3b5 + 3c5 + 5c8 + 5b8 + 6b6 + 6b10 . By the associative law and 2) in Lemma 2.9 and 13) in Lemma 2.10:

(4)


67

(b3 b¯6 )b¯15 = b3 (b¯6 b¯15 ), 2 = 2c b + 3b b + 2b b + 2b b + b b + c b ; b¯3 b¯15 + b¯15 8 3 10 3 9 3 8 3 5 3 5 3

by 3), 4) in Lemma 2.9 and 1, 3), 4), 5) in Lemma 2.11 2 = 9b¯15 + 4b6 + 2b¯3 + 3c3 + 6c9 . b15

(5)

By 4), 12), 13) in Lemma 2.10 and 12) in Lemma 2.11 and 5) in Lemma 2.9: (b10 b15 , b3 ) = (b10 , b¯15 b15 ) = 6,

(b10 b15 , b¯6 ) = (b15 b6 , b10 ) = 3,

(b10 b15 , b3 ) = (b¯15 b3 , b10 ) = 1,

(b10 b15 , c¯3 ) = (b10 c3 , b¯15 ) = 1,

(b10 b15 , b¯3 c3 ) = (b15 b3 , b10 c3 ) = 4, so (b10 b15 , c¯9 ) = 4. Hence b10 b15 = 6b15 + 3b¯6 + b3 + c¯3 + 4c¯9 .

(6)

By the associative law and 4) in Lemma 2.10 and 16) in Lemma 2.11: (b3 c¯3 )b15 = b3 (c¯3 b15 ),

c9 b15 = 2b¯15 b3 + b3 b6 + b3 c9 ;

by 6), 12) in Lemma 2.10 and 1) in Lemma 2.9: c9 b15 = 2b5 + 2c5 + 3c8 + 3b8 + 3b9 + 4b10 .

(7)

By 4) and Lemma 2.10 and (2) we get: (b3 c¯3 )b¯15 = (c¯3 b¯15 )b3 , c9 b¯15 = b5 b3 + c5 b3 + c8 b3 + b8 b3 + b9 b3 + b10 b3 ; by 1), 3), 4), 5) in Lemma 2.11 and 3), 4) in Lemma 2.9, c9 b¯15 = 6b15 + 2b¯6 + b3 + c¯3 + 3c¯9 .

(8)

By the associative law and 4) in Lemma 2.10 and (1): (b3 c¯3 )c9 = b3 (c¯3 c9 ),

c92 = b10 b3 + b9 b3 + c8 b3 ;

by 3), 4) in Lemma 2.9 and 1) in Lemma 2.11 c92 = 3b15 + b3 + c¯3 + 2b¯6 + 2c¯9 . By the associative law and 4) in Lemma 2.10 and 13) in Lemma 2.11: (b3 c¯3 )c¯9 = b3 (c¯3 c¯9 ),

c9 c¯9 = b3 b¯15 + b3 c9 + b3 b¯3 ;

(9)

68


by 6), 12) in Lemma 2.10 and Hypothesis 2.1: c9 c¯9 = 1 + 2c8 + 2b9 + 2b10 + c5 + b5 + 2b8 .

(10)

By the associative law and 4) in Lemma 2.10 and 2) in Lemma 2.11: (b3 c¯3 )c8 = (c¯3 c8 )b3 ,

c9 c8 = c¯9 b3 + b15 b3 ;

by 1) in Lemma 2.10 and 5) in Lemma 2.9: c9 c8 = 3b¯15 + b6 + c3 + 2c9 .

(11)

By the associative law and 5) in Lemma 2.10 and 13) in Lemma 2.11: (b3 c3 )c9 = (c3 c9 )b3 ,

b9 c9 = b15 b3 + c¯9 b3 + b32 ;

by 5) in Lemma 2.9 and Hypothesis 2.1 and 1) in Lemma 2.10 b9 c9 = b¯3 + 2c9 + 3b¯15 + c3 + 2b6 .

(12)

By the associative law and 4) in Lemma 2.10 and 12) in Lemma 2.11: c9 b10 = b15 b3 + b¯6 b3 + c¯9 b3 ;

(b3 c¯3 )b10 = (c¯3 b10 )b3 ,

by 2) and 5) in Lemma 2.9 and 1) in Lemma 2.10 c9 b10 = b¯3 + 2c9 + 4b¯15 + c3 + b6 .

(13)

By (2), (c¯3 b5 , b15 ) = (b5 , c3 b15 ) = 1,

(c¯3 c5 , b15 ) = (c5 , c3 b15 ) = 1.

So c¯3 b5 = b15 ,

(14)

c¯3 c5 = b15 .

(15)

Hence by 4), 5) in Lemma 2.10 and the associative law: (c¯3 b5 )b3 = (b3 c¯5 )b5 , b15 b3 = c9 b5 ,

(c¯3 c5 )b3 = (b3 c¯3 )c5 , b15 b3 = c9 c5 .

By 5) in Lemma 2.9 we get: b5 c9 = 2b¯ 15 + b6 + c9 ,

(16)

c5 c9 = 2b¯15 + b6 + c9 .

(17)


69

By the associative law and 12) in Lemma 2.10 and 5) in Lemma 2.11: (b3 b¯15 )b5 = (b3 b5 )b¯15 = b15 b¯15 , b10 b5 + c8 b5 + b9 b5 + b8 b5 + b52 + c5 b5 = b15 b¯15 , so b5 is real and so c5 is real. So by (14), (15) c3 b5 = b¯15 , c3 b5 = b15 .

Lemma 2.13 Let (A, B) satisfy Hypothesis 2.1. Assume that b¯10 = b10 ∈ B and (c¯9 b3 , c9 ) = 1. Then we obtain the following equations: 1) 2) 3) 4) 5) 6) 7) 8) 9) 10) 11) 12) 13) 14) 15) 16) 17) 18) 19) 20) 21)

b5 c8 = c5 + c8 + b8 + b9 + b10 ; b5 b8 = c5 + c8 + b8 + b9 + b10 ; b5 b9 = b10 + c8 + b9 + b8 + b5 + c5 ; b5 b10 = c8 + b8 + b9 + b5 + 2b10 ; b52 = 1 + b10 + b9 + b5 ; c5 b5 = b8 + c8 + b9 ; c8 b8 = b5 + c5 + c8 + b8 + 2b9 + 2b10 ; c8 b9 = 2b8 + 2b9 + 2b10 + b5 + c5 + c8 ; b92 = 1 + b5 + c5 + 2c8 + 2b8 + 2b9 + 2b10 ; b82 = 1 + b5 + c5 + c8 + 2b8 + b9 + 2b10 ; b8 b9 = b5 + c5 + 2c8 + b8 + 2b9 + 2b10 ; b8 b10 = b5 + c5 + 2c8 + 2b8 + 2b9 + 2b10 ; c5 c8 = b5 + c8 + b8 + b9 + b10 ; c5 b8 = b5 + c8 + b8 + b9 + b10 ; c5 b9 = b10 + c8 + b9 + b8 + b5 + c5 ; c5 b10 = b8 + c8 + b9 + c5 + 2b10 ; c52 = 1 + b9 + b10 + c5 ; b5 b15 = 3b15 + b¯6 + b3 + c¯3 + 2c¯9 ; c5 b15 = 3b15 + b¯6 + b3 + c¯3 + 2c¯9 ; b9 b15 = 6b15 + 3c¯9 + b3 + 2b¯6 + c¯3 ; b9 b10 = b5 + c5 + 2c8 + 2b8 + 2b9 + 3b10 .

Proof By the associative law and Hypothesis 2.1 and 12) in Lemma 2.10: b5 (b3 b¯3 ) = (b¯3 b5 )b3 = b¯15 b3 , b5 + b5 c8 = b¯15 b3 = b10 + c8 + b9 + b8 + b5 + c5 . Thus c8 b5 = c5 + c8 + b8 + b9 + b10 .

(1)

By the associative law and 9) in Lemma 2.10 and 18) in Lemma 2.12: (c3 c¯3 )b5 = (c¯3 b5 )c3 ,

b5 + b5 b8 = b15 c3 ;

by 2) in Lemma 2.12 b5 b8 = c5 + c8 + b8 + b9 + b10 .

(2)

70


By the associative law and 5), 12) in Lemma 2.10 and 18) in Lemma 2.12: (b3 c3 )b5 = (c3 b5 )b3 , b9 b5 = b¯15 b3 = b10 + c8 + b9 + b8 + b5 + c5 .

(3)

By the associative law and 1) in Lemma 2.9 and 5) in Lemma 2.11: (b3 b6 )b5 = (b3 b5 )b6 ,

c8 b5 + b10 b5 = b15 b6 ;

by (1) and 13) in Lemma 2.10: b5 b10 = c8 + b8 + b9 + b5 + 2b10 .

(4)

By the associative law and 12) in Lemma 2.10 and 5) in Lemma 2.11: (b3 b¯15 )b5 = (b3 b5 )b¯15 , b10 b5 + c8 b5 + b9 b5 + b8 b5 + b52 + c5 b5 = b15 b¯15 ; so by 4) in Lemma 2.12 and (1), (2), (3), (4), b52 + c5 b5 = 1 + b5 + 2b9 + b10 + b8 + c8 , and by (2), (4), (3), (1), (b52 , b10 ) = (b5 , b5 b10 ) = 1,

(b52 , b9 ) = (b5 , b5 b9 ) = 1.

Therefore b52 = 1 + b10 + b9 + b5 ,

(5)

c5 b5 = b8 + c8 + b9 .

(6)

By the associative law and Hypothesis 2.1 and 3) in Lemma 2.11: (b3 b¯3 )b8 = (b¯3 b8 )b3 ,

b8 + c8 b8 = b¯ 15 b3 + c9 b3 ;

by 6), 12) in Lemma 2.10, we obtain c8 b8 = b5 + c5 + c8 + b8 + 2b9 + 2b10 . By the associative law and Hypothesis 2.1 and 1) in Lemma 2.11: (b3 b¯3 )b9 = (b¯3 b9 )b3 ,

b9 + c8 b9 = b¯15 b3 + c9 b3 + c3 b3 ;

by 5), 6), 12) in Lemma 2.10 we obtain b9 + c8 b9 = b10 + c8 + b9 + b8 + b5 + c5 + b9 + b10 + b8 + b9 .

(7)


71

By 5) in Lemma 2.10 and 15) in Lemma 2.11 we obtain c8 b9 = 2b8 + 2b9 + 2b10 + b5 + c5 + c8 , (b3 c3 )b9 = (c3 b9 )b3 ,

(8)

b92 = b¯15 b3 + b¯3 b3 + c9 b3 ;

by 12) in Lemma 2.10 and Hypothesis 2.1 and 6) in Lemma 2.10: b92 = 1 + b5 + c5 + 2c8 + 2b8 + 2b9 + 2b10 .

(9)

By the associative law and 9) in Lemma 2.10 and 14) in Lemma 2.11: (c3 c¯3 )b8 = c3 (c¯3 b8 ),

b8 + b82 = c3 c¯3 + c3 b¯6 + c3 b15 ;

by 7), 9) in Lemma 2.10 and 2) in Lemma 2.12: b82 = 1 + b5 + c5 + c8 + 2b8 + b9 + 2b10 .

(10)

By the associative law and 9) in Lemma 2.10 and 15) in Lemma 2.11: (c3 c¯3 )b9 = (c¯3 b9 )c3 ,

b9 + b8 b9 = b15 c3 + b3 c3 + c¯9 c3 ;

by 1), 2) in Lemma 2.12 and 5) in Lemma 2.10: b9 + b8 b9 = b5 + c5 + c8 + b8 + b9 + b10 + b9 + b10 + b9 + c8 , b8 b9 = b5 + c5 + 2c8 + b8 + 2b9 + 2b10 .

(11)

By the associative law and 9) in Lemma 2.10, 12) in Lemma 2.11: (c3 c¯3 )b10 = (c¯3 b10 )c3 ,

b10 + b8 b10 = b15 c3 + b¯6 c3 + c¯9 c3 ,

so by 1), 2) in Lemma 2.12 and 7) in Lemma 2.10: b8 b10 = b5 + c5 + 2c8 + 2b8 + 2b9 + 2b10 .

(12)

By the associative law and Hypothesis 2.1 and 4) in Lemma 2.11, 12) in Lemma 2.10: (b3 b¯3 )c5 = (b¯3 c5 )b3 , c5 + c5 c8 = b¯15 b3 = b10 + c8 + b9 + b8 + b5 + c5 , c5 c8 = b10 + c8 + b9 + b8 + b5 .

(13)

By the associative law and 9) in Lemma 2.10 and 2), 18) in Lemma 2.12: (c3 c¯3 )c5 = (c¯3 c5 )c3 , c5 + c5 b8 = b15 c3 = b5 + c5 + c8 + b8 + b9 + b10 , c5 b8 = b5 + c8 + b8 + b9 + b10 .

(14)

72


By the associative law and 5), 12) in Lemma 2.10, 18) in Lemma 2.12: (b3 c3 )c5 = (c3 c5 )b3 , b9 c5 = b¯15 b3 = b10 + c8 + b9 + b8 + b5 + c5 .

(15)

By the associative law and 1) in Lemma 2.9 and 10) in Lemma 2.11: (b3 b6 )c5 = (b6 c5 )b3 , c8 c5 + b10 c5 = b¯15 b3 + b3 b6 + b3 c9 . By 6), 12) in Lemma 2.10 and (13) and 1) in Lemma 2.9 we get: b10 + c8 + b9 + b8 + b5 + b10 c5 = b10 + c8 + b9 + b8 + b5 + c5 + c8 + b10 + b9 + b10 + b, b10 c5 = c5 + c8 + 2b10 + b9 + b8 .

(16)

By the associative law and 10) in Lemma 2.10, 10) in Lemma 2.11: (b6 b¯6 )c5 = b¯6 (b6 c5 ), c5 + c5 c8 + b9 c5 + b8 c5 + c52 + b5 c5 = b¯6 b¯15 + b¯6 b6 ; by (13), (15), (14), (6) and 10), 13) in Lemma 2.10: c52 = 1 + b9 + b10 + c5 .

(17)

By the associative law and 18) in Lemma 2.12 and (5): c3 b52 = (c3 b5 )b5 , c3 + b10 c3 + b9 c3 + b5 c3 = b¯15 b5 ; so by 12), 15) in Lemma 2.11 and 17) in Lemma 2.12: b5 b¯15 = 3b¯15 + b6 + b¯3 + c3 + 2c9 .

(18)

By the associative law and 18) in Lemma 2.12 and (17): (c52 )c3 = (c5 c3 )c5 ,

c3 + b9 c3 + b10 c3 + c5 c3 = b¯15 c5 ;

so by 12), 15) in Lemma 2.11 and 18) in Lemma 2.12: b¯15 c5 = 3b¯15 + b6 + b¯3 + c3 + 2c9 . By the associative law and 4) in Lemma 2.11 and (15): (b3 c5 )b9 = b3 (c5 b9 ), b15 b9 = b3 b8 + c8 b3 + b5 b3 + c5 b3 + b9 b3 + b10 b3 ; by 3), 4) in Lemma 2.9 and 1), 3), 4), 5) in Lemma 2.11:

(19)


73

b15 b9 = b15 + c¯9 + b3 + b¯6 + b15 + 2b15 + 2b15 + 2c¯9 + c¯3 + b¯6 , b15 b9 = 6b15 + 3c¯9 + b3 + 2b¯6 + c¯3 . By the associative law and 5) in Lemma 2.10 and 4) in Lemma 2.9: b10 (b3 c3 ) = (b10 b3 )c3 , b10 b9 = b15 c3 + b¯6 c3 + c¯9 c3 , b10 b9 = b5 + c5 + c8 + b8 + b9 + b10 + b8 + c8 + b9 + b10 , b10 b9 = b5 + c5 + 2c8 + 2b8 + 2b9 + 3b10 . This completes the investigation of 11) in Lemma 2.9, subcase (c¯9 b3 , c9 ) = 1. Thus we have the remaining case that (b3 c¯9 , c9 ) = 0. In the remainder of this section, we deal with the investigation of 11) in Lemma 2.9, subcase (b6 c9 , c¯9 ) ≥ 1. Lemma 2.14 Let (A, B) satisfy Hypothesis 2.1. Assume that b¯ 10 = b10 ∈ B and (b6 c9 , c¯9 ) ≥ 1. Then we obtain: 1) 2) 3) 4) 5) 6) 7) 8)

b3 c9 = b10 + x¯ + s, s ∈ B, |x¯ + s| = 17, |¯s | = |s|; (c8 b15 , c¯9 ) = (c8 c9 , b¯15 ) = 2; b¯6 c¯9 = 3c9 + b¯15 + b6 + t6 , t6 ∈ N ∗ B; c8 b10 = 2c8 + 2b10 + 2x + y + z + w + s; b6 b15 = 3b10 + 2c8 + x + y + z + w + x¯ + s; b¯3 c9 = b15 + ε + θ , ε, θ ∈ B, |ε + θ | = 12; b6 b10 = b¯3 + 3b¯15 + ε¯ + θ¯ ; c8 b15 = 5b15 + 2b¯6 + 2c¯9 + b3 + ε + θ .

Proof By the associative law and Hypothesis 2.1 and 10) in Lemma 2.9: (b32 )b¯15 = (b3 b¯15 )b3 , b¯3 b¯15 + b6 b¯15 = b10 b3 + c8 b3 + b3 (x + y + z + w). So by 5), 4), 3) in Lemma 2.9, b6 b¯15 = b3 + b¯6 + b3 (x + y + z + w).

(1)

By the associative law and Hypothesis 2.1 and 4) in Lemma 2.9: (b32 )b10 = (b3 b10 )b3 , b¯3 b10 + b6 b10 = b15 b3 + b¯6 b3 + c¯9 b3 ; by 4), 5), 2) in Lemma 2.9: b6 b10 = b¯ 3 + 2b¯15 + b3 c¯9 .

(2)

74


By the associative law and (2) and 4) in Lemma 2.9: (b3 b10 )b¯6 = (b¯6 b10 )b3 , b¯62 + b15 b¯6 + b¯6 c¯9 = b32 + 2b15 b3 + (b¯3 b3 )c9 . So by 5), 6) in Lemma 2.9 and Hypothesis 2.1: 2b6 + c9 + b¯15 + b¯3 + b6 + b¯3 (x + y + z + w) + b¯6 c¯9 = b¯3 + b6 + 4b¯15 + 2b6 + 2c9 + c9 + c8 c9 , b¯6 c¯9 + b¯3 (x + y + z + w) = 3b¯15 + 2c9 + c8 c9 .

(3)

By the associative law and 6), 7) in Lemma 2.9: (b62 )c8 = (b6 c8 )b6 , 2b¯6 c8 + c¯9 c8 + b15 c8 = b¯3 b6 + 2b¯15 b6 + b62 + c9 b6 ; by 7), 2), 6) in Lemma 2.9 and (3), (1) c8 b15 = b3 + 2b¯6 + c¯9 + b15 + b3 (x + y + z + w).

(4)

By the associative law and Hypothesis 2.1: (b3 b¯3 )c9 = (b¯3 c9 )b3 ,

c9 + c9 c8 = (b¯3 c9 )b3 ,

but by 5) in Lemma 2.9, (b¯3 c9 , b15 ) = (c9 , b3 b15 ) = 1. So c9 + c9 c8 = b15 b3 + α1 b3 ; hence by 5) in Lemma 2.9 we obtain that: c9 c8 = 2b¯15 + b6 + α1 b3 .

(5)

(c8 c9 , b¯15 ) ≥ 2.

(6)

Hence

By 10) in Lemma 2.9, (b¯3 x, b¯15 ) = 1,

(b¯3 y, b¯15 ) = 1,

(b¯ 3 z, b¯15 ) = 1,

(b¯3 w, b¯15 ) = 1.

So by (3), (b¯6 c¯9 , b¯15 ) ≥ 1.

(7)


75

By (6), (c8 b15 , c¯9 ) ≥ 2, so by (4), (b3 (x + y + z + w), c¯9 ) = 0. Hence one of b3 x, b3 y, b3 z, and b3 w contains c¯9 ; thus, without loss of generality, we may assume that (b3 x, c¯9 ) = 0.

(8)

By the associative law and Hypothesis 2.1 and 4) in Lemma 2.9: (b3 b¯3 )b10 = (b3 b10 )b¯3 , b10 + c8 b10 = b15 b¯3 + b¯6 b¯3 + c¯9 b¯3 ; by 10), 1) in Lemma 2.9 c8 b10 = 2c8 + b10 + x + y + z + w + b¯3 c¯9 .

(9)

By (8) and 4) in Lemma 2.9, (b3 c9 , b10 ) = (b3 b10 , c¯9 ) = 1, (b3 c9 , x) ¯ = 0; so b3 c9 = b10 + x¯ + α,

α ∈ N ∗ B.

(10)

¯ = 0. By (9) since c8 , b10 , x + y + z + w are reals then Then (b3 x, c¯9 ) = (b3 c9 , x) b3 c9 is real. By 10) in Lemma 2.9, (b3 x, b15 ) = 1; therefore b3 x = c¯9 + b15 + β,

(11)

where β ∈ N ∗ B. So |x| ≥ 8. Hence by (10) 3 ≤ (b3 c9 , b3 c9 ) ≤ 5. Thus 3 ≤ (b3 b¯3 , c9 c¯9 ) ≤ 5. So by Hypothesis 2 ≤ (c9 c¯9 , c8 ) ≤ 4. Thus 2 ≤ (c9 , c9 c8 ) ≤ 4.

(12)

If (c8 b15 , c¯9 ) ≥ 4 then by (4), we have that (b3 (x + y + z + w), c¯9 ) ≥ 3. Since 1 + (c8 , b15 b¯15 ) = (1 + c8 , b15 b¯15 ) = (b3 b¯3 , b15 b¯15 ) = (b3 b15 , b3 b15 ) = 6 by 10) in Lemma 2.9, we have that (c8 b15 , b15 ) = 5. Hence we obtain, without loss of generality, by (11) that b3 y = c¯9 + b15 + β1 ,

b3 z = c¯9 + b15 + β2 .

So the degrees of x, y and z are all larger or equal to 8, but |x + y + z + w| = 27; hence |x| = |y| = |z| = 8 by (11) and (b3 c9 , x) = (b3 c9 , y) = (b3 c9 , z) = 1, a contradiction to (10). If (c8 b15 , c¯9 ) = 3. Then by (5) and (12) c8 c9 = 3b¯15 + b6 + 2c9 + c3 ,

(13)

76


where c3 ∈ B. By (4), (b3 (x + y + z + w), c¯9 ) = 2, so by (3) (b¯ 6 c¯9 , c9 ) = 2 and by 10), 6) in Lemma 2.9, (b¯6 c¯9 , b6 ) = 1, (b¯3 (x + y + z + w), b¯15 ) = 4, so by (3), (13) (b¯ 6 c¯9 , b¯15 ) = 2. Hence b6 c9 = b¯6 + 2c¯9 + 2b15 .

(14)

So by (3), b¯3 (x + y + z + w) = 4b¯15 + 2c9 + c3 . Then we obtain, without loss of generality, that the only possibility is x = b8 ,

y = b9 ,

z = b5 ,

w = c5 ,

when b8 , b9 , b5 , c5 are reals. So b¯3 b8 = b¯15 + c9 ,

(15)

b¯ 3 b9 = b¯15 + c9 + c3 ,

(16)

b¯3 b5 = b¯15 ,

(17)

b¯ 3 c5 = b¯15 .

(18)

So (b3 c9 , b8 ) = 1, and by 4) in Lemma 2.9, (b3 c9 , b10 ) = 1. Thus b3 c9 = b10 + b8 + b9 .

(19)

So by the associative law and Hypothesis 2.1 (b32 )c9 = (b3 c9 )b3 , c9 b¯3 + c9 b6 = b10 b3 + b8 b3 + b9 b3 . Therefore by (15), (16), (14) c9 b¯3 = b15 + c¯9 + c¯3 which contradicts the assumption in the lemma, we conclude that (c8 b15 , c¯9 ) ≤ 2 and by (8) and (4), (c8 b15 , c¯9 ) = 2.

(20)

(b3 (x + y + z + w), c9 ) = 1,

(21)

By (4) and by (3), (12) we obtain 3 ≤ (b¯6 c¯9 , c9 ) ≤ 5, by (20) (c8 c9 , b¯15 ) = 2. So by (3) and that (b¯ 3 (x + y + z + w), b¯15 ) = 4 we get (b¯6 c¯9 , b¯15 ) = 1. And by 6) in Lemma 2.9, (b¯ 6 c¯9 , b6 ) = 1. So (b¯6 c¯9 , c9 ) = 3. Then by (3), (21), (c8 c9 , c9 ) = 2. So (c8 , c¯9 c9 ) = 2, and by Hypothesis 2.1 (c¯9 c9 , b¯3 b3 ) = 3. Thus (b3 c9 , b3 c9 ) = 3. So by 4) in Lemma 2.9 and (10), we obtain that b3 c9 = b10 + x¯ + s,

(22)

b¯6 c¯9 = 3c9 + b¯15 + b6 + t6 ,

(23)

s ∈ B. Also


77

where t6 ∈ N ∗ B, t6 = b6 , b¯6 . By (9) and (22), ¯ c8 b10 = 2c8 + 2b10 + 2x + y + z + w + S.

(24)

By the associative law and 5) in Lemma 2.9: (b32 )b15 = (b3 b15 )b3 , b¯3 b15 + b6 b15 = 2b¯15 b3 + b6 b3 + c9 b3 ; by (22) and 1), 10) in Lemma 2.9: b6 b15 = 3b10 + 2c8 + x + y + z + w + x¯ + S.

(25)

By the associative law and Hypothesis 2.1 and 5) in Lemma 2.9: Since b15 (b3 b¯3 ) = b¯3 (b15 b3 ), then c8 b15 = 4b15 + 2b¯6 + 2c¯9 + b3 + b¯3 c9 .

(26)

By (22) and 5) in Lemma 2.9: b¯3 c9 = b15 + ε + θ,

(27)

where ε, θ ∈ B, |ε + θ | = 12, so by (26) c8 b15 = 5b15 + 2b¯6 + 2c¯9 + b3 + ε + θ. By the associative law and Hypothesis 2.1 and 4) in Lemma 2.9: (b32 )b10 = b3 (b3 b10 ), b10 b¯3 + b10 b6 = b15 b3 + b¯6 b3 + c¯9 b3 ; so by (27) and 5), 2), 4) in Lemma 2.9: b6 b10 = b¯ 3 + 3b¯15 + ε¯ + θ¯ . Lemma 2.15 Let (A, B) satisfy Hypothesis 2.1. Assume that b¯ 10 = b10 ∈ B and (b3 c¯9 , c9 ) = 0. Then (b6 c9 , c¯9 ) ≥ 1 is impossible. Proof We assume that our assumption is possible. By Hypothesis 2.1 and 10) in Lemma 2.9 and the associative law: (b3 b¯3 )b15 = b3 (b¯3 b15 ), b15 + c8 b15 = b3 b10 + b3 c8 + b3 x¯ + b3 y¯ + b3 z¯ + b3 w. ¯

78


By 3), 4) in Lemma 2.9, 8) in Lemma 2.14: b15 + 5b15 + 2b¯6 + 2c¯9 + b3 + ε + θ = b15 + b¯6 + c¯9 + b3 + b¯6 + b15 + b3 x¯ + b3 y¯ + b3 z¯ + b3 w, ¯ ¯ 4b15 + c¯9 + ε + θ = b3 x¯ + b3 y¯ + b3 z¯ + b3 w.

(1)

By 10) in Lemma 2.9, and 1) in Lemma 2.14: (b3 x, ¯ b15 ) = (b¯15 b3 , x) = 1,

(b3 y, ¯ b15 ) = (b3 b¯15 , y) = 1,

(b3 z¯ , b15 ) = (b¯15 b3 , z) = 1,

(b3 w, ¯ b15 ) = (b3 b¯15 , w) = 1,

(b3 x, ¯ c¯9 ) = (b3 c9 , x) = 1, so x = x. ¯ If otherwise two of x, y, z, w have degrees ≥ 8, this is impossible. Then b3 x = c¯9 + b15 ,

(2)

and then x = x8 . So |y¯ + z¯ + w| ¯ = 19, and also we get 5 ≤ |y|, ¯ |¯z|, |w| ¯ ≤ 9 and also, without loss of generality, |y| ¯ = 5, hence y¯ = y¯5 , so b3 y¯5 = b15 .

(3)

|¯z + w| ¯ = 14.

(4)

So

By 1) in Lemma 2.14 and the fact that x = x8 , we obtain that b3 c9 = b10 + x8 + s9 .

(5)

So by the associative law and Hypothesis 2.1 and (5): (b32 )c9 = b3 (b3 c9 ), c9 b¯3 + b6 c9 = b3 b10 + b3 x8 + s9 b3 . Then by (2) and (6), 3) in Lemma 2.14, 4) in Lemma 2.9: b15 + ε + θ + 3c¯9 + b15 + b¯6 + t¯6 = b15 + b¯6 + c¯9 + b15 + c¯9 + s9 b3 , s9 b3 = ε + θ + c¯9 + t¯6 .

(6)

If |z| = 5, then by (4) |w| = 9, hence z = z¯ 5 , w = w¯ 9 . So b3 z5 = b15 , b3 w9 = b15 + ε + θ. So by (6)

(7)


(b¯3 ε, s9 ) = (ε, b3 s9 ) = 1, (b¯3 ε, w9 ) = (ε, b3 w9 ) = 1,

79

(b¯3 θ, s9 ) = (θ, b3 s9 ) = 1, (b¯3 θ, w9 ) = (θ, b3 w9 ) = 1.

Hence b¯ 3 ε = s9 + w9 , b¯3 θ = s9 + w9 ,

(8) (9)

and hence ε = ε6 , θ = θ6 , by 4) in the Lemma 2.14 s9 = s¯9 . So by the associative law and Hypothesis 2.1: ε¯ 6 (b32 ) = b3 (¯ε6 b3 ),

ε¯ 6 b¯3 + ε¯ 6 b6 = s9 b3 + w9 b3 ; so by (6), (7) ε¯ 6 b¯3 + ε¯ 6 b6 = ε6 + θ6 + c¯9 + t¯6 + b15 + ε6 + θ6 , ε¯ 6 b¯3 + ε¯ 6 b6 = 2ε6 + 2θ6 + c¯9 + t¯6 + b15 . By 6) in Lemma 2.14 (¯ε6 b¯3 , c¯9 ) = (ε6 , b¯3 c9 ) = 1 and (¯ε6 b¯3 , δ) = 1, where δ = ε6 or θ6 , so ε¯ 6 b¯3 = δ + c¯9 + m3 , so t¯6 = m3 + k3 , and (b3 m3 , ε¯ 6 ) = (b¯3 ε¯ 6 , m3 ) = 1. Therefore (b3 m3 , b3 m3 ) = (b3 b¯3 , m3 m ¯ 3 ) > 1. In (6) we get (b¯3 m3 , s9 ) = (m3 , b3 s9 ) = 1, then ¯ 3 m3 ) = 1, (b¯3 m3 , b¯3 m3 ) = (b¯3 b3 , m a contradiction. If |¯z| = 6 then by (4) |w| ¯ = 8, hence z = z¯ , w = w. ¯ So b3 z6 = b15 + ε3 , b3 w8 = b15 + θ9 . So (ε3 b¯3 , z6 ) = (ε3 , b3 z6 ) = 1, hence (ε3 b¯3 , ε3 b¯3 ) > 1. By (6), (s9 b3 , ε3 ) = (s9 , b¯3 z6 ) = 1, hence b¯3 ε3 = s9 , so (ε3 b¯3 , ε3 b¯3 ) = 1, a contradiction. If |¯z| = 7 then by (4) |w| ¯ = 7, so z = z¯ ,

w = w¯ or z¯ = w, b3 z¯ 7 = b15 + ε6 , b3 w¯ 7 = b15 + θ6 .

(10) (11)

So (b¯3 ε6 , z¯ 7 ) = (ε6 , b3 z¯ 7 ) = 1. By (6) (b¯3 ε6 , s9 ) = (ε6 , b3 s9 ) = 1, hence b¯3 ε6 = z¯ 7 + s9 + m2 , a contradiction.

80


Consequently, 11) in Lemma 2.9 subcase (b6 c9 , c¯9 ) ≥ 1 is impossible and we have only the case (b3 c¯9 , c9 ) = 1. We can now state Theorem 2.7. Theorem 2.7 Let (A, B) be a NITA generated by a nonreal element b3 ∈ B satisfying b3 b¯3 = 1 + c8 and b32 = b¯3 + b6 , where b6 ∈ B is nonreal and satisfying Hypothesis 2.1. Then b3 b6 = c8 + b10 and if b10 = b¯10 then (A, B) ∼ = (CH(3A6 ), Irr(3A6 )). (A, B) is a Table Algebra of dimension 17: B = {1, b3 , b¯3 , c3 , c¯3 , b6 , b¯6 , c9 , c¯9 , b15 , b¯15 , b5 , c5 , b8 , c8 , b9 , b10 } and B has an increasing series of table subsets {b1 } ⊆ {b1 , b5 , c5 , b8 , c8 , b9 , b10 } ⊆ B defined by 1) 2) 3) 4) 5) 6) 7) 8) 9) 10) 11) 12) 13) 14) 15) 16) 18) 19) 20) 21) 22) 23) 24) 25) 26) 27) 28) 29) 30) 31) 32) 33) 34) 35) 36) 37)

b3 b¯3 = 1 + c8 ; b32 = b¯3 + b6 ; b3 c3 = b9 ; b3 c¯3 = c9 ; b3 b6 = c8 + b10 ; b3 b¯6 = b¯3 + b¯15 ; b3 c9 = b9 + b10 + b8 ; b3 c¯9 = b¯15 + c9 + c3 ; b3 b15 = 2b¯15 + b6 + c9 ; b3 b¯15 = b10 + c8 + b9 + b8 + b5 + c5 ; b3 b5 = b15 ; b3 c5 = b15 ; b3 b8 = b15 + c¯9 ; b3 c8 = b3 + b¯6 + b15 ; b3 b9 = b15 + c¯9 + c¯3 ; b3 b10 = b15 + b¯6 + c¯9 ; c32 = c¯3 + b¯6 ; c3 b6 = c¯3 + b15 ; c3 b¯6 = b10 + b8 ; c3 c9 = b3 + b15 + c¯9 ; c3 c¯9 = c8 + b9 + b10 ; c3 b15 = b5 + c5 + c8 + b8 + b9 + b10 ; c3 b¯15 = 2b15 + b¯6 + c¯9 ; c3 b5 = b¯15 ; c3 c5 = b¯15 ; c3 b8 = c3 + b6 + b¯15 ; c3 c8 = c9 + b¯15 ; c3 b9 = b15 + b¯3 + c9 ; c3 b10 = b¯15 + b6 + c9 ; b6 b¯6 = 1 + c8 + b9 + b8 + c5 + b5 ; b6 c9 = 2b15 + 2c¯9 + b¯6 ; b6 c¯9 = b10 + c8 + 2b9 + b8 + b5 + c5 ; b6 b15 = 2c8 + 3b10 + 2b9 + 2b8 + b5 + c5 ; b6 b¯15 = 4b15 + b¯6 + b3 + c¯3 + 2c¯9 ; b6 b5 = b¯15 + b6 + c9 ; b6 c5 = b¯15 + b6 + c9 ;


38) 39) 40) 41) 42) 43) 44) 45) 46) 47) 48) 49) 50) 51) 52) 53) 54) 55) 56) 57) 58) 59) 60) 61) 62) 63) 64) 65) 66) 67) 68) 69) 70) 71) 72) 73) 74) 75) 76) 77) 78) 79) 80) 81)

b6 b8 = 2b¯15 + b6 + c3 + c9 ; b6 c8 = b¯3 + 2b¯15 + b6 + c9 ; b6 b9 = b6 + 2c9 + 2b¯15 ; b6 b10 = 3b¯15 + b¯3 + c3 + c9 ; b62 = 2b¯6 + c¯9 + b15 ; c9 c¯9 = 1 + 2c8 + 2b9 + 2b10 + c5 + b5 + 2b8 ; c9 b15 = 2b5 + 2c5 + 3c8 + 3b8 + 3b9 + 4b10 ; c9 b¯15 = 6b15 + 2b¯6 + b3 + c¯3 + 3c¯9 ; c9 b5 = 2b¯15 + b6 + c9 ; c9 c5 = 2b¯15 + b6 + c9 ; c9 b8 = 3b¯15 + b¯3 + b6 + 2c9 ; c9 c8 = 3b¯15 + c3 + b6 + 2c9 ; c9 b9 = b¯3 + 2c9 + 3b¯15 + c3 + 2b6 ; c9 b10 = b¯3 + 2c9 + 4b¯15 + c3 + b6 ; c92 = 3b15 + b3 + c¯3 + 2b¯6 + 2c¯9 ; b15 b¯15 = 1 + 3b5 + 3c5 + 5c8 + 5b8 + 6b9 + 6b10 ; 2 = 9b¯ + 4b + 2b¯ + 2c + 6c ; b15 15 6 3 3 9 b15 b5 = 3b15 + b¯6 + b3 + c¯3 + 2c¯9 ; b15 c5 = 3b15 + b¯6 + b3 + c¯3 + 2c¯9 ; b15 b8 = 5b15 + b3 + c¯3 + 2b¯6 + 3c¯9 ; b15 c8 = 5b15 + b3 + c¯3 + 2b¯6 + 3c¯9 ; b15 b9 = 6b15 + 3c¯9 + b3 + 2b¯6 + c¯3 ; b15 b10 = 6b15 + 3b¯6 + +b3 + c¯3 + 4c¯9 ; b52 = 1 + b10 + b9 + b5 ; b5 c5 = b8 + c8 + b9 ; b5 b8 = c5 + c8 + b8 + b9 + b10 ; b5 c8 = c5 + b8 + b9 + b10 + c8 ; b5 b9 = c8 + c5 + b8 + b5 + b9 + b10 ; b5 b10 = c8 + b8 + b9 + b5 + 2b10 ; c52 = 1 + b9 + b10 + c5 ; c5 b8 = b5 + c8 + b8 + b9 + b10 ; c5 c8 = b5 + c8 + b8 + b9 + b10 ; c5 b9 = b8 + c8 + b5 + c5 + b9 + b10 ; c5 b10 = b8 + c8 + b9 + c5 + 2b10 ; b82 = 1 + b5 + c5 + c8 + 2b8 + b9 + 2b10 ; b8 c8 = b5 + c5 + c8 + b8 + 2b9 + 2b10 ; b8 b9 = b5 + c5 + 2c8 + b8 + 2b9 + 2b10 ; b8 b10 = b5 + c5 + 2c8 + 2b8 + 2b9 + 2b10 ; c82 = 1 + 2c8 + 2b10 + b9 + b8 + b5 + c5 ; c8 b9 = b5 + c5 + c8 + 2b8 + 2b9 + 2b10 ; c8 b10 = 2c8 + 2b10 + 2b9 + 2b8 + b5 + c5 ; b9 b10 = b5 + c5 + 2c8 + 2b8 + 2b9 + 3b10 ; b92 = 1 + b5 + c5 + 2c8 + 2b8 + 2b9 + 2b10 ; 2 = 1 + 2c + 2b + 3b + 2b + 2b + 2c . b10 8 10 9 8 5 5

81

82


Proof The theorem follows by Lemmas 2.9–2.15.

In this section, we do not have information about what NITA satisfying b10 nonreal. But we conjecture: Conjecture 2.1 There exists no NITA satisfying Hypothesis 2.1 and b3 b6 = c8 + b10 , where b10 is nonreal.

2.6 NITA Generated by b3 Satisfying b32 = c3 + b6 , c3 = b3 , b¯ 3 , b6 Non-real, (b3 b8 , b3 b8 ) = 4 and c32 = r3 + s6 Now we discuss NITA satisfying the following hypothesis. Hypothesis 2.2 Let (A, B) be a NITA generated by a nonreal element b3 ∈ B satisfying b32 = c3 + b6 , c3 = b3 , b¯3 , b6 nonreal, (b3 b8 , b3 b8 ) = 4 and c32 = r3 + s6 . Here we still assume that L(B) = 1 and L2 (B) = ∅. In this section, we are not able to classify NITAs satisfying Hypothesis 2.2, but we have the following theorem. Theorem 2.8 There exists no NITA satisfying Hypothesis 2.2 and c3 generates a sub-NITA isomorphic to one of the four known NITAs in the Main Theorem 1. Proof By assumptions we have that (1) b3 b¯3 = 1 + b8 = c3 c¯3 ; (2) b32 = c3 + b6 ; (3) (b3 b8 , b3 b8 ) = 4 and (4) c32 = r3 + s6 . Step 1 There exists no NITA satisfying hypotheses (1) to (4) and containing a NITA strictly isomorphic to (Ch(PSL(2, 7)), Irr(PSL(2, 7))). By [CA], (Ch(PSL(2, 7)), Irr(PSL(2, 7))) the sub-NITA generated by c3 which is of dimension 6 and has base elements: 1, c3 , c¯3 , s6 , b7 , b8 . Moreover, we have the following equations: (a1) c3 b8 = c3 + s6 + b7 + b8 , (a2) s62 = 1 + 2s6 + b7 + b8 , (a3) c3 s6 = c¯3 + b7 + b8 . It follows by (1) that (b3 c¯3 , b3 c¯3 ) = (b3 b¯3 , c3 c¯3 ) = 2, hence we obtain equation (a4): b3 c¯3 = b¯3 + t6 , where t6 ∈ B. By the associative law and equations (a1) and (a4), we have (b3 b¯3 )c¯3 = c¯3 + c¯3 b8 = c¯3 + c¯3 + s6 + b7 + b8 , (b3 c¯3 )b¯3 = b¯32 + b¯3 t6 = c¯3 + b¯6 + b¯3 t6 . Then we get b¯6 = s6 , so b6 is real. Moreover we have equation (a5): b¯3 t6 = c¯3 + s6 + b7 + b8 . By (b3 b¯3 )b3 = b32 b¯3 , we arrive at b3 b8 = t¯6 + b¯3 s6 . Now by

2.7 Structure of NITA Generated by b3

83

(3) we obtain equation (a6): b¯ 3 s6 = b3 + x + y, where distinct b3 , x and y such that x + y is of degree 15. Hence we have equation (a7): b3 b8 = b3 + t¯6 + x + y. Therefore by (a2), (a3), (a5) and (a6) b32 s6 = (c3 + s6 )s6 = c3 s6 + s6 s6 = c¯3 + b7 + b8 + 1 + 2s6 + b7 + b8 , ¯ = 1 + b8 + b3 x¯ + b3 y. ¯ b3 (b3 s6 ) = b3 (b¯3 + x¯ + y) So b3 x¯ + b3 y¯ = c¯3 + 2s6 + 2b7 + b8 , but by (a7), the left hand side of this equation contains two b8 , and the right side only one, a contradiction. Step 2 There exists no NITA satisfying hypotheses (1) to (4) and containing a subNITA generated by c3 which is a NITA of dimension 17, 32 or 22. By [CA], we see that for the NITAs of dimensions 17, 32 and 22, one can find new base elements: b5 , c5 , s6 , x8 , b9 and b10 such that the following equations hold: (a1) c3 b8 = c3 + s¯6 + b15 , (a2) s6 s¯6 = 1 + b5 + c5 + b8 + x8 + b9 , (a3) c3 s6 = b8 + b10 . Suppose b3 c¯3 = b¯3 + t6 , where t6 ∈ B. Then by hypothesis and equation (a1) it follows that (b3 b¯3 )c¯3 = c¯3 + c¯3 b8 = c¯3 + c¯3 + s6 + b¯15 , (b3 c¯3 )b¯3 = b¯32 + b¯3 t6 = c¯3 + b¯6 + b¯3 t6 . We get s6 = b¯6 and equation (a4): b¯3 t6 = c¯3 + b¯15 . Hence b3 +b3 b8 = (b3 b¯3 )b3 = b32 b¯3 = b¯3 c3 + b¯3 s¯6 , and we obtain the equation (a5): b3 b8 = t¯6 + b¯3 s¯6 . Now by (3) we obtain equation (a6): b¯3 s¯6 = b3 + x + y, where distinct b3 , x and y such that x + y is of degree 15. Hence we have equation (a7): b3 b8 = b3 + t¯6 + x + y. Therefore b32 s6 = (c3 + s¯6 )s6 = c3 s6 + s6 s¯6 = b8 + b10 + 1 + b5 + c5 + b8 + x8 + b9 , ¯ = 1 + b8 + b3 x¯ + b3 y. ¯ b3 (b3 s6 ) = b3 (b¯3 + x¯ + y) So b3 x¯ + b3 y¯ = 1 + b5 + c5 + b8 + x8 + b9 , which implies x8 = b8 by (a7), a contradiction. This concludes the proof.

2.7 Structure of NITA Generated by b3 and Satisfying b32 = c3 + b6 , c3 = b3 , b¯ 3 , (b3 b8 , b3 b8 ) = 3 and c3 Non-real In this section we classify NITA generated by b3 and satisfying b32 = c3 + b6 , c3 = b3 , b¯3 and c3 nonreal and (b3 b8 , b3 b8 ) = 3. First, we state Hypothesis 2.3. Hypothesis 2.3 Let (A, B) be a NITA generated by a nonreal element b3 ∈ B such that b3 b¯3 = 1 + b8 and b32 = c3 + b6 , c3 , b6 ∈ B, c3 = b¯3 , c3 = c¯3 and (b3 b8 , b3 b8 ) = 3.

84


Lemma 2.16 Let (A, B) be a NITA satisfying Hypothesis 2.3. Then c3 c¯3 = b3 b¯3 = 1 + b8 and b3 b8 = b3 + x6 + y15 and c3 b8 = c3 + y6 + z15 , where b8 , x6 , y15 ∈ B, and there exist r3 , s6 , u3 , v6 , w3 , z6 ∈ B such that b32 = r3 + s6 ,

b¯3 r3 = b3 + x6 ,

b¯3 s6 = b3 + y15 ,

b3 c3 = u3 + v6 ,

b¯3 u3 = c3 + y6 ,

b¯3 v6 = c3 + z15 ,

b¯ 3 c3 = w3 + z6 ,

b3 w3 = c3 + y6 ,

b3 z6 = c3 + z15 .

Proof It is obvious that (b32 , b32 ) = 2. So b32 = r3 + s6 or r4 + s5 . Hence b32 b¯3 = b¯3 r3 + b¯3 s6 or b¯3 r4 + b¯3 s6 , (b3 b¯3 )b3 = b3 + b3 b8 = 2b3 + x6 + y15 . Since b3 ∈ Supp{b¯3 r3 } and Supp{b¯3 s6 } or b3 ∈ Supp{b¯3 r4 } and Supp{b¯3 s5 }, the first part of the lemma follows. Since b3 b¯3 = c3 c¯3 = 1 + b8 , it follows that b3 c3 = u3 + v6 or u4 + v5 and b¯3 c3 = w3 + z6 or w4 + z5 . The second and third parts of the lemma follow from (b3 b¯3 )c3 = (b3 c3 )b¯3 = (b¯3 c3 )b3 . Lemma 2.17 Let (A, B) satisfy Hypothesis 2.3. Then there are nonreal base elements x6 , y15 , x15 and real basis elements r3 , s6 , d9 , x10 , t15 such that the following equations hold: b3 b¯3 = 1 + b8 , b3 c¯3 = b¯3 + x¯6 , b3 b8 = b3 + x6 + x15 , b3 b¯6 = b¯3 + x¯15 , b3 c3 = r3 + s6 , b3 s6 = c¯3 + y¯15 , b3 y¯15 = x¯6 + 2x¯15 + b¯9 , b3 b6 = r3 + t15 , b3 t15 = b¯6 + c¯9 + 2y¯15 ,

b32 = c3 + b6 , b3 x6 = c3 + y15 , b3 x¯6 = b8 + x10 , b3 c6 = c¯3 + y¯15 , b3 r3 = c¯3 + b¯6 , b3 x10 = x6 + x15 + b9 , b3 y15 = s6 + 2t15 + d9 , b3 x15 = b6 + 2y15 + c9 ,

c3 c¯3 = 1 + b8 , c3 s6 = b¯3 + x¯15 , c3 b8 = c3 + b6 + y15 , c3 x6 = r3 + t15 , c3 y15 = b¯6 + c¯9 + 2y¯15 , c3 t15 = x¯6 + b¯9 + 2x¯15 , c3 x¯15 = 2x15 + x6 + b9 ,

c3 r3 = b¯3 + x¯6 , c32 = c¯3 + b¯6 , c3 x¯6 = b3 + x15 , c3 x10 = b6 + y15 + c9 , c3 b6 = c¯3 + y¯15 , c3 x15 = 2t15 + s6 + d9 , c3 b¯6 = b8 + x10 ,


85

r3 c¯3 = c3 + x6 , r3 x¯6 = c3 + y15 , r3 b6 = x¯15 + b¯3 , r3 x¯15 = b6 + c9 + 2y15 , r3 x10 = s6 + t15 + d9 ,

r3 b¯3 = c3 + b6 , r3 y15 = x¯6 + 2x¯15 + b¯9 , r3 b8 = s6 + r3 + t15 ,

x62 = 2b6 + y15 + c9 ,

x6 b8 = b3 + x6 + 2x15 + b9 ,

b62 = 2b¯6 + y¯15 + c¯9 , b6 x6 = 2s6 + t15 + d9 ,

b6 x¯6 = 2x6 + b9 + x15 , b6 b8 = c3 + b6 + c9 + 2y15 ,

s6 b¯6 = 2x6 + x15 + b9 , s6 c¯3 = b3 + x15 ,

s6 b8 = r3 + s6 + 2t15 + d9 ,

b8 y15 = c3 + 2b6 + 2c9 + 4y15 + b¯3 d9 . Proof By Hypothesis 2.3 (b3 b8 , b3 b8 ) = 3. Then the following equations hold by Lemma 2.7: b3 b¯3 = 1 + b8 , b32

= c 3 + b6 ,

c3 c¯3 = 1 + b8 , b¯3 c3 = b3 + x6 .

Let b3 x6 = c3 + y15 , then (b3 b¯3 )c3 = c3 + c3 b8 , (b¯3 c3 )b3 = b32 + b3 x6 = c3 + b6 + c3 + y15 . Hence c3 b8 = c3 + b6 + y15 . Therefore y15 ∈ B by Lemma 2.7. Let c3 x¯6 = b3 + x15 , x15 ∈ N ∗ B. Since (c3 c¯3 )b3 = (b3 c¯3 )c3 , it follows that x15 ∈ B and b3 b8 = b3 + x6 + x15 by b32 b¯3 = (b3 b¯3 )b3 . Since y15 ∈ B, so (b¯3 x6 , b¯3 x6 ) = (b3 x6 , b3 x6 ) = 2. Thus b¯3 x6 = b8 + x10 ,

x10 ∈ B.

Since b32 c¯3 = (b3 c¯3 )b3 and b32 b¯3 = (b3 b¯3 )b3 , one has that c3 b¯6 = b8 + x10 ,

b¯3 b6 = b3 + x15 .

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By Lemma 2.16, there exist w3 , z6 , r3 , s6 ∈ B such that c32 = w3 + z6 ,

w3 c¯3 = c3 + b6 ,

z6 c¯3 = c3 + y15 ,

b3 c3 = r3 + s6 ,

b3 r3 = w3 + z6 ,

b3 s6 = w3 + z15 .

By Lemma 2.16 again, it follows that r3 c¯3 = b3 + x6 ,

c¯3 s6 = b3 + x15 ,

r3 b¯3 = c3 + b6 ,

b¯3 s6 = c3 + y15 .

Thus (c¯3 r3 , c¯3 r3 ) = 2, which implies that r3 r¯3 = 1 + b8 . Since (b3 r3 )c¯3 = w3 c¯3 + z6 c¯3 = 2c3 + b6 + y15 , (b3 c¯3 )r3 = r3 b¯3 + r3 x¯6 = c3 + b6 + r3 x¯6 , which implies that r3 x¯6 = c3 + y15 . We assert that x10 is real. In fact, b¯3 (r3 c¯3 ) = b3 b¯3 + x6 b¯3 = 1 + b8 + b8 + x10 , (b¯ 3 r3 )c¯3 = c3 c¯3 + b6 c¯3 = 1 + b8 + b8 + x¯10 , so it follows that x10 is real. Since b¯3 (w3 c¯3 ) = c3 b¯3 + b6 b¯3 = b3 + x6 + b3 + x15 , (b¯3 c¯3 )w3 = r¯3 w3 + s¯6 w3 , we have that r¯3 w3 = b3 + x6 ,

w3 s¯6 = b3 + x15 .

Consequently b¯3 w3 = r3 + s6 . Now we have that (b3 b¯3 )w3 = (b¯3 w3 )b3 and w3 b8 = w3 + z6 + z15 . Then w3 w¯ 3 = 1 + b8 .


87

Since (b3 b¯3 )2 = 1 + 2b8 + b82 , b32 b¯32 = (c3 + b6 )(c¯3 + b¯6 ) = 1 + b8 + c3 b¯6 + c¯3 b6 + b6 b¯6 = 1 + 3b8 + 2x10 + b6 b¯6 , so b82 = 2x10 + b8 + b6 b¯6 . By b32 b¯6 = (b3 b¯6 )b3 , we have that x10 + b6 b¯6 = 1 + b3 x¯15 . Then (b3 x10 , x15 ) = 1. But (b3 x10 , x6 ) = 1. There exists b9 ∈ B such that b3 x10 = x6 + x15 + b9 ,

b9 ∈ N ∗ B.

It is easy to see that b3 x10 cannot have constituents of degree 3 and 4 by L1 (B) = {1} and L2 (B) = ∅. Thus b9 ∈ B. Since (b¯3 x6 )b3 = b3 b8 + b3 x10 = b3 + x6 + x15 + x6 + x15 + b9 , (b3 b¯3 )x6 = x6 + x6 b8 , we have x6 b8 = b3 + x6 + 2x15 + b9 . Multiplying both sides of the equation x10 + b6 b¯6 = 1 + b3 x¯15 by b3 , we have that b3 x10 + (b3 b¯6 )b6 = b3 + b32 x¯15 , so x6 + x15 + b9 + b6 b¯3 + b6 x¯15 = c3 x¯15 + b6 x¯15 . Hence c3 x¯15 = 2x15 + x6 + b9 . Now checking the associative law of (b3 c¯3 )b6 = (c¯3 b6 )b3 , (r3 r¯3 )x6 = (¯r3 x6 )r3 and (r3 b¯3 )x¯6 = (r3 x¯6 )b¯3 , we obtain b6 x¯6 = 2x6 + b9 + x15 , r3 y¯15 = x6 + 2x15 + b9 , b¯3 y15 = x6 + 2x15 + b9 .

88


Hence c3 y¯15 + b6 y¯15 = b32 y¯15 = b3 (b3 y¯15 ) = b3 x¯6 + 2b3 x¯15 + b3 b¯9 . Since (b3 x¯6 , b8 ) = (b3 x¯6 , x10 ) = (b3 x¯15 , b8 ) = (b3 x¯15 , x10 ) = (b3 b¯9 , x10 ) = 1 and (b3 b¯9 , b8 ) = 0, we have that (c3 y¯15 + b6 y¯15 , b8 ) = 3

and (c3 y¯15 + b6 y¯15 , x10 ) = 4.

From the degrees, we know that (c3 y¯15 , x10 ) ≤ 1. If (c3 y¯15 , x10 ) = 0, then (b6 y¯15 , x10 ) = 4, so b6 x10 = 4y15 . But (b6 x10 , c3 ) = 1, a contradiction. Hence (c3 y¯15 , x10 ) = 1 and (b6 x10 , y15 ) = (b6 y¯15 , x10 ) = 3. Since (c3 x10 , c3 x10 ) = (b3 x10 , b3 x10 ) = 3 and (c3 x10 , b6 ) = 1, there exists c9 ∈ B such that c3 x10 = b6 + y15 + c9 . Then (b¯3 c3 )x6 = b3 x6 + x62 = c3 + y15 + x62 , (b¯3 x6 )c3 = c3 b8 + c3 x10 = c3 + b6 + y15 + b6 + y15 + c9 , (c3 c¯3 )b6 = b6 + b6 b8 , (c¯3 b6 )c3 = b8 c3 + x10 c3 = c3 + b6 + y15 + y15 + b6 + c9 , (c3 b6 )c¯3 = c3 + b6 + c¯3 z15 , which implies that x62 = 2b6 + y15 + c9 , b6 b8 = c3 + b6 + c9 + 2y15 , c3 z¯ 15 = b¯6 + c¯9 + 2y¯15 . Since (b3 c¯3 )2 = (b¯ 3 + x¯6 )2 = b¯32 + 2b¯3 x¯6 + x¯62 = 2c¯3 + 3b¯6 + 3y¯15 + c¯9 , b32 c¯32 = c3 w¯ 3 + c3 z¯ 6 + w¯ 3 b6 + b6 z¯ 6 = c¯3 + b¯6 + c¯3 + y¯15 + b6 w¯ 3 + b6 z¯ 6 ,


89

but (w¯ 3 b6 , c¯3 ) = 1, we have that w3 b¯6 = c3 + y15 , b6 z¯ 6 = 2b¯6 + y¯15 + c¯9 . Since (b3 r¯3 )b¯6 = c¯3 b¯6 + b¯62 = w¯ 3 + z¯ 15 + b¯62 , (b3 b¯6 )¯r3 = b¯3 r¯3 + x¯15 r¯3 = w¯ 3 + z¯ 6 + r¯3 x¯15 , and (b¯ 62 , z¯ 6 ) = 2, we have that (¯r3 x¯15 , z¯ 6 ) = 1. Hence (r3 z¯ 6 , x¯15 ) = 1. But (r3 z¯ 6 , b¯3 ) = 1 by b3 r3 = w3 + z6 . Thus r¯3 z6 = b3 + x15 . From the following equations b¯ 3 (w3 b¯6 ) = b¯3 (c3 + y15 ) = b3 + x6 + x6 + 2x15 + b9 , (b¯3 w3 )b¯6 = r3 b¯6 + s6 b¯6 , and (r3 b¯6 , r3 b¯6 ) = (b3 b¯6 , b3 b¯6 ) = 2, (r3 b¯6 , b3 ) = 1, we obtain r3 b¯6 = b3 + x15 ,

s6 b¯6 = 2x6 + x15 + b9 .

By (b3 c3 )¯r3 = (c3 r¯3 )b3 and c¯3 (w3 b8 ) = (c¯3 b8 )w3 , we have that r¯3 s6 = b8 + x10 ,

w3 y¯15 = b6 + c9 + 2y15 .

Since (¯r3 b8 , r¯3 b8 ) = (r3 r¯3 , b82 ) = (b3 b¯3 , b82 ) = (b3 b8 , b3 b8 ) = 3 by Lemma 2.16 and (¯r3 b8 , s¯6 ) = (¯r3 b8 , r¯3 ) = 1, there exists t15 ∈ B such that r¯3 b8 = s¯6 + r¯3 + t15 . Since (w3 b8 )¯r3 = w3 r¯3 + z6 r¯3 + z15 r¯3 = b3 + x6 + b3 + x15 + r¯3 z15 , (¯r3 w3 )b8 = b3 b8 + x6 b8 = b3 + x6 + x15 + b3 + x6 + 2x15 + b9 , (¯r3 b8 )w3 = s¯6 w3 + r¯3 w3 + t15 w3 = b3 + x15 + b3 + x6 + w3 t15 ,

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we have that r3 z¯ 15 = x¯6 + 2x¯15 + b¯9 , w3 t15 = x6 + b9 + 2x15 . Since (b¯3 r3 )b8 = c3 b8 + b8 b6 = c3 + b6 + y15 + 2y15 + b6 + c3 + c9 , (b¯3 b8 )r3 = r3 b¯3 + r3 x¯6 + r3 x¯15 = c3 + b6 + c3 + y15 + r3 x¯15 , (r3 b8 )b¯3 = r3 b¯3 + s6 b¯3 + t¯15 b¯3 = c3 + b6 + c3 + y15 + b¯3 t¯15 , we have that w3 = c¯3 , z6 = b¯6 , z15 = y¯15 and r3 x¯15 = b6 + c9 + 2y15 , b3 t15 = b¯6 + c¯9 + 2y¯15 . Furthermore, b3 r3 = c¯3 + b¯6 , and we have that (b3 c3 , r¯3 ) = 1. But (b3 c3 , r3 ) = 1; hence r3 = r¯3 . Moreover r¯3 + s¯6 = b3 w¯ 3 = b3 c3 = r3 + s6 , which implies that s6 = s¯6 . Therefore t15 is real by the expression r3 b8 , and c3 t15 = x¯6 + c¯9 + 2x¯15 . Now we have that t15 ∈ Supp{c3 x6 }. Thus c3 x6 = r3 + t15 . Since (c3 r3 )b¯6 = b¯3 b¯6 + x¯6 b¯6 , (r3 b¯6 )c3 = b3 c3 + c3 x15 = r3 + s6 + c3 x15 , (c3 b¯6 )r3 = r3 b8 + r3 x10 = r3 + s6 + t15 + r3 x10 and (r3 x10 , s6 ) = 1, we have s6 ∈ Supp{c3 x15 }. But (c3 x15 , t15 ) = (c3 t15 , x¯15 ) = 2. Let c3 x15 = 2t15 + s6 + d9 ,

d9 ∈ N ∗ B.

Then r3 x10 = s6 + t15 + d9 and by (c3 r3 )b¯6 = (c3 b¯6 )r3 we obtain that b¯3 b¯6 + x¯6 b¯6 = r3 + s6 + t15 + s6 + t15 + d9 .


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Furthermore d9 is real for r3 and x10 are real, and b3 b6 = r3 + t15 , x6 b6 = 2s6 + t15 + d9 . Since (b3 b¯3 )b6 = b6 + b6 b8 = b6 + c3 + b6 + c9 + 2y15 , b3 (b¯3 b6 ) = b32 + b3 x15 = c3 + b6 + b3 x15 , (b3 b6 )b¯3 = r3 b¯3 + t15 b¯3 = c3 + b6 + b¯ 3 t15 , we have that b3 x15 = b6 + 2y15 + c9 , b3 t15 = b¯6 + c¯9 + 2y¯15 . Since b32 x6 = c3 x6 + b6 x6 = r3 + t15 + 2s6 + t15 + d9 , (b3 x6 )b3 = b3 c3 + b3 y15 = r3 + s6 + b3 y15 , we have that b3 y15 = 2t15 + s6 + d9 . Since (b3 y15 , b3 y15 ) = (b3 y¯15 , b3 y¯15 ) = 6, we have that d9 ∈ B. Checking the associative law of (b3 b¯3 )s6 = (b3 s6 )b¯3 , we have s6 b8 = r3 + s6 + 2t15 + d9 . Since (b3 b¯3 )y15 = y15 + b8 y15 , (b3 y15 )b¯3 = b¯3 s6 + 2b¯3 t15 + b¯3 d9 = c3 + y15 + 2b6 + 2c9 + 4y15 + b¯3 d9 , it follows that b8 y15 = c3 + 2b6 + 2c9 + 4y15 + b¯3 d9 . This completes the proof of the lemma.

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Remark It always follows that (c3 c9 , y¯15 ) = 1 by Lemma 2.17. So (c3 c9 , c3 c9 ) ≥ 2. In the following, we shall investigate the expression c3 c9 . Lemma 2.18 There exist no x4 , y8 ∈ B such that c3 c9 = x4 + y8 + y¯15 . Proof If there exist no x4 , y8 ∈ B such that c3 c9 = x4 + y8 + y¯15 , then there exists z3 ∈ B such that c3 x¯4 = c¯9 + z3 . Furthermore, there is y5 ∈ B such that c3 z¯ 3 = x4 + y5 . Thus z3 z¯ 3 = 1 + b8 . By Lemma 2.16 we have that c3 (z3 z¯ 3 ) = c3 + c3 b8 = 2c3 + b6 + y15 , z3 (c3 z¯ 3 ) = z3 x4 + z3 y5 . It must follow that z3 x4 = 2c3 + b6 , which implies that (z3 c¯3 , x4 ) = 2, a contradiction. Lemma 2.19 There exist no x5 , y7 ∈ B such that c3 c9 = x5 + y7 + y¯15 . Proof If some x5 , y7 ∈ B such that c3 c9 = x5 + y7 + y¯15 . Then (c3 x¯5 , c¯9 ) = 1. There are three possibilities: (I) c3 x¯5 = c¯9 + 2x3 , (II) c3 x¯5 = c¯9 + x3 + y3 , (III) c3 x¯5 = c¯9 + y6 . It is easy to see that (I) will lead to a contradiction. If (II) follows, then c3 x¯5 = c¯9 + x3 + y3 . Set c3 x¯3 = x5 + z4 , c3 y¯3 = x5 + t4 , where z4 , t4 ∈ B. Hence c¯3 (c3 x¯3 ) = c¯3 x5 + c¯3 z4 = c9 + x¯ 3 + y¯3 + c¯3 z4 , (c3 c¯3 )x¯3 = x¯3 + x¯3 b8 . Therefore (x3 y¯3 , b8 ) = 1, which implies that y¯3 x3 = 1 + b8 . So y3 = x3 , a contradiction. If (III) follows, then c¯3 (c3 x¯5 ) = c¯3 c¯9 + c¯3 y6 = y15 + x¯5 + y¯7 + c¯3 y6 , (c3 c¯3 )x¯5 = x¯5 + x¯5 b8 .


93

Hence x¯5 b8 = y15 + y¯7 + c¯3 y6 . Therefore (b8 y15 , x¯5 ) = 1. By Lemma 2.17, we have that (x¯5 , b¯3 d9 ) = 1. Since (b¯3 d9 , y15 ) = 1, we may set b¯3 d9 = x¯5 + y15 + e7 ,

e7 ∈ N ∗ B.

If e7 = m3 + n4 , then b3 n4 = d9 + p3 , some p3 ∈ B. Let b¯3 p3 = n4 + z5 , some z5 ∈ B. Then p3 p¯ 3 = 1 + b8 . Furthermore, b3 (p3 p¯3 ) = b3 + b3 b8 = 2b3 + x6 + x15 , (b3 p¯3 )p3 = n¯ 4 p3 + z¯ 5 p3 . Note that b3 ∈ Supp{n¯ 4 p3 } and b3 ∈ Supp{¯z5 p3 }, we come to a contradiction. If e7 ∈ B, then b¯3 d9 = x¯5 + y15 + e7 . Let b3 x¯5 = d9 + u6 . We assert that u6 ∈ B. Otherwise, let u6 = r3 + s3 , where r3 , s3 ∈ B, then b3 r¯3 = x5 + u4 . Hence r3 r¯3 = 1 + b8 . Moreover, b3 (r3 r¯3 ) = b3 + b3 b8 = 2b3 + x6 + x15 , r3 (b3 r¯3 ) = r3 x5 + r3 u4 , which is impossible for b3 ∈ Supp{r3 x5 } and b3 ∈ Supp{r3 u4 }. Now we have that b¯3 (c3 x¯5 ) = b¯3 c¯9 + b¯3 y6 , (b¯3 c3 )x¯5 = b3 x¯5 + x6 x¯5 = d9 + u6 + x6 x¯5 . If d9 , u6 ∈ Supp{b¯3 y6 }, then b¯ 3 y6 = d9 + u6 + t3 . Let b3 t3 = y6 + p3 . Hence ¯ t3 t3 = 1 + b8 . Moreover, b3 (t3 t¯3 ) = 2b3 + x6 + x15 , t¯3 (b3 t3 ) = t¯3 y6 + t¯3 p3 . So t¯3 y6 = b3 + x15 . But 3 = (b¯3 y6 , b¯3 y6 ) = (t¯3 y6 , t¯3 y6 ) = 2 by Lemma 2.1, a contradiction. If u6 ∈ Supp{b¯3 y6 }, d9 ∈ Supp{b¯3 c¯9 } or u6 ∈ Supp{b¯3 c¯9 }, d9 ∈ Supp{b¯3 y6 }. If the former one follows, then there exists a q3 such that b3 c9 = t15 + d9 + q3 . Hence c9 ∈ Supp{b¯3 d9 }, a contradiction.

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If u6 ∈ Supp{b¯3 c¯9 }, d9 ∈ Supp{b¯3 y6 }, then b3 d9 = y¯15 + y6 + r6 , where r6 ∈ N ∗ B. But d9 is real, b¯3 d9 = x¯5 + y15 + e7 , a contradiction. This completes the proof. Lemma 2.20 There exist no m6 and n6 such that c3 c9 = y¯15 + m6 + n6 . Proof If there exist m6 and n6 such that c3 c9 = y¯15 + m6 + n6 . Then m6 = b6 , b¯6 , n6 = b6 , b¯6 by Lemma 2.17 and c3 m ¯ 6 has three possibilities: (I) c3 m ¯ 6 = c¯9 + x3 + y3 + z3 , x3 , y3 , z3 distinct ¯ 6 = c¯9 + x3 + y6 , x3 and y6 ∈ B, (II) c3 m (III) c3 m ¯ 6 = c¯9 + x4 + y5 , x4 and y5 ∈ B. Suppose (I) follows. We may assume that c3 x¯3 = m6 + r3 , where r3 ∈ B. Then x¯3 + x¯3 b8 = x¯ 3 (c3 c¯3 ) = (x¯3 c3 )c¯3 = m6 c¯3 + r3 c¯3 = c9 + x¯3 + y¯3 + z¯ 3 + r3 c¯3 . Hence y¯3 ∈ Supp{x¯3 b8 }, which implies that x3 = y3 , a contradiction. If (II) follows, then (c3 x¯3 , c3 x¯3 ) = 2, from which x3 x¯3 = 1 + b8 follows. Also we may set c3 m ¯ 3 = x3 + t6 , where t6 ∈ B. Hence (c3 c¯3 )x¯3 = x¯3 + x¯3 b8 , c¯3 (c3 x¯3 ) = c¯3 m6 + c¯3 m3 = x¯4 + y¯5 + c9 + x¯3 + t¯6 . Therefore x3 b8 = x4 + y5 + c¯9 + t6 . We have that (x3 b8 , x3 b8 ) = 4. But by Lemma 2.8, (x3 b8 , x3 b8 ) = (b3 b8 , b3 b8 ) = 3, a contradiction. If (III) follows, set c3 x¯4 = m6 + p6 , where p6 ∈ B. By c¯3 (c3 m ¯ 6 ) = (c3 c¯3 )m ¯ 6, one has that ¯ 6 + n¯ 6 + p¯ 6 + c¯3 y5 . m ¯ 6 b8 = y15 + m ¯ 6 ) = 1. By the expression b8 y15 in Lemma 2.17, (m ¯ 6 , b¯3 d9 ) = 1. Hence (b8 y15 , m ∗ ¯ But (y15 , b3 d9 ) = 1. There exists q6 ∈ N B such that ¯ 6 + q6 . b¯3 d9 = y15 + m

(2.64)

If q6 ∈ B, then q = 2z3 or q6 = w3 + u3 . If the first one holds, then (b¯3 d9 , z3 ) = 2 and (b3 z3 , d9 ) = 2, which is impossible. So the second one follows. Consequently b3 w3 = d9 . Hence w3 + w3 b8 = w3 (b3 b¯3 ) = b¯3 (b3 w3 ) = b¯3 d9 = y15 + m ¯ 6 + w3 + u3 .


95

We can see that w3 u¯ 3 = 1 + b8 . Then w3 = u3 , a contradiction. Hence p6 ∈ B. It follows by Lemma 2.17 and (2.64) that ¯ 6 + q6 . b8 y15 = 5y15 + 2c9 + 2b6 + c3 + m Since c3 (b¯3 d9 ) = c3 y15 + c3 m ¯ 6 + c3 q6 = b¯6 + c¯9 + 2y¯15 + x4 + y5 + c¯9 + c3 q6 , (c3 b¯3 )d9 = b3 d9 + x6 d9 = y¯15 + m6 + q¯6 + x6 d9 , ¯ 6 , q¯6 ) ≥ 1, a contradiction to our assumpwe have that (c3 q6 , m6 ) ≥ 1. Hence (c3 m tion. This completes the proof. Lemma 2.21 There exists no NITA such that (c3 c9 , c3 c9 ) ≥ 4. Proof The proof is given in three steps. Step 1 The following equations hold: (i) c3 c9 = y¯15 + y3 + z3 + w3 + u3 , (ii) c3 c9 = y¯15 + y3 + y3 + y6 , (iii) c3 c9 = y¯15 + y3 + y4 + y5 , (iv) c3 c9 = y¯15 + y4 + z4 + w4 . For any z ∈ Supp{c3 c9 } \ {y¯15 }, it is sufficient to prove that (c3 c9 , z) = 1. Otherwise (c3 c9 , z) ≥ 2. Since c3 c9 − y¯15 is of degree 12, one has that (c3 c9 , z) = 2 and z is of degree 6. So c3 c9 = y¯15 + 2z6 , c3 z¯ 6 = 2c¯9 . Hence z¯ 6 + z¯ 6 b8 = (c3 c¯3 )¯z6 = c¯3 (c3 z¯ 6 ) = 2c¯3 c¯9 = 2y15 + 4¯z6 . Thus z¯ 6 b8 = 2y15 + 3¯z6 . Consequently (b8 y15 , z¯ 6 ) = 2. By the expression b6 b8 in Lemma 2.17, z6 = b6 , b¯6 and the expression b8 y15 in Lemma 2.17, we have that b¯3 d9 = y15 + 2¯z6 , b8 y15 = 5y15 + 2c9 + 2b6 + c3 + 2¯z6 . Therefore b3 z¯ 6 = 2d9 . Since (b3 b¯3 )d9 = d9 + b8 d9 , b3 (b¯3 d9 ) = b3 y15 + 2b3 z¯ 6 = s6 + 2t15 + 5d9 ,

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it follows that b8 d9 = s6 + 2t15 + 4d9 . Consequently, b8 (b3 z¯ 6 ) = 2b8 b9 = 2s6 + 4t15 + 8d9 , (b3 b8 )¯z6 = b3 z¯ 6 + x6 z¯ 6 + x15 z¯ 6 = 2d9 + x6 z¯ 6 + x15 z¯ 6 . Then x6 z¯ 6 + x15 z¯ 6 = 2s6 + 4t15 + 6d9 . We have exactly three possibilities: (I) x6 z¯ 6 = 4d9 , x15 z¯ 6 = 2s6 + 4t15 + 2d9 , x15 z¯ 6 = s6 + 2t15 + 6d9 , (II) x6 z¯ 6 = s6 + 2t15 , (III) x6 z¯ 6 = 2s6 + t15 + d9 , x15 z¯ 6 = 3t15 + 5d9 . The first case implies that b¯3 (x6 z¯ 6 ) = 4b¯3 d9 = 4y15 + 8¯z6 , (b¯3 x6 )¯z6 = b8 z¯ 6 + x10 z¯ 6 = 2y15 + 3¯z6 + x10 z¯ 6 . So x10 z¯ 6 = 2y15 + 5¯z6 . Thus (z6 z¯ 6 , x10 ) = 5, a contradiction. The second and the third cases mean (z6 d9 , x15 ) = 6 and (z6 d9 , x15 ) = 5, respectively. This is impossible. Step 2 No NITA satisfies either (i), (ii) or (iii). It is sufficient to show that y3 ∈ Supp{c3 c9 }. If y3 ∈ Supp{c3 c9 }, then c¯3 y3 = c9 . Hence (b¯3 y3 , b¯3 y3 ) = (c¯3 y3 , c¯3 y3 ) = 1 by Lemma 2.1. Let b¯3 y3 = g9 , g9 ∈ B. Then b3 c9 = b3 (c¯3 y3 ) = (b3 c¯3 )y3 = b¯3 y3 + x¯6 y3 = g9 + x¯6 y3 . Since (t15 , b3 c9 ) = 1, we have that b3 c9 = g9 + t15 + α3 , some α3 ∈ B. Therefore (b3 c9 , b3 c9 ) = 3. But (b3 c9 , b3 c9 ) = (c3 c9 , c3 c9 ) by Lemma 2.1, which is impossible for (c3 c9 , c3 c9 ) ≥ 4. Step 3 No NITA satisfies (iv). If (iv) holds, then there exists m3 such that c¯3 y4 = c9 + m3 . Thus c3 m3 = y4 + k5 , some k5 ∈ B. So (c3 m3 , c3 m3 ) = 2, which implies that m3 m ¯ 3 = 1 + b8 . Hence ¯ 3 k5 = c3 (m3 m ¯ 3 ) = c3 + c3 b8 = (c3 c¯3 )m ¯ 3 = 2c3 + b6 + y15 , m ¯ 3 y4 + m ¯ 3 y4 } and Supp{m ¯ 3 k5 }. The lemma follows. which is impossible for c3 ∈ Supp{m Lemma 2.22 There exists no NITA such that (c3 c9 , c3 c9 ) = 2.


97

Proof If (c3 c9 , c3 c9 ) = 2, then c3 c9 = y¯15 + b12 , b12 ∈ B and c3 c¯9 = x10 + x17 , x17 ∈ B. Since c¯3 (c3 c9 ) = c¯3 y¯15 + c¯3 b12 = b6 + c9 + 2y15 + c¯3 b12 , we have that b8 c9 = b6 + 2y15 + c¯3 b12 . On the other hand, since c9 + b8 c9 = (c3 c¯3 )c9 = c3 (c¯3 c9 ) = c3 (x¯10 + x¯17 ) = b6 + y15 + c9 + c3 x¯17 , we have that b8 c9 = b6 + y15 + c3 x¯17 . By the expression b8 y15 in Lemma 2.17, (b8 c9 , y15 ) = 2. Hence y15 ∈ Supp{c3 x¯17 }, which implies that x17 ∈ Supp{c3 y¯15 }. By the expressions of c3 b8 and c3 x10 in Lemma 2.17, (c3 y¯15 , b8 ) = (c3 y¯15 , x10 ) = 1. There exists t10 ∈ N ∗ B such that c3 y¯15 = x17 + b8 + x10 + t10 , some t10 ∈ N ∗ B. The expression c3 y15 in Lemma 2.17 means that (c3 y15 , c3 y15 ) = 6. Hence (c3 y¯15 , c3 y¯15 ) = 6. So t10 = x3 + y3 + z4 . This leads to y15 ∈ Supp{c3 y¯3 }, a contradiction. The lemma follows. Lemma 2.23 c3 c9 = d3 + c¯9 + y¯15 . Proof First, by the previous lemma in this section, we know that (c3 c9 , c3 c9 ) = 3 and c3 c9 is a sum of irreducible base elements of degree 3, 9 and 15. So we may assume that c3 c9 = d3 + y9 + y¯15 , where d3 , y9 ∈ B. Then c3 d¯3 = c¯9 . Checking the associative law for c¯3 (c3 d¯3 ) = (c3 c¯3 )d¯3 , it follows that d¯3 b8 = y¯9 + y15 . Since b6 (b3 x¯6 ) = b6 b8 + b6 x10 = c3 + b6 + c9 + 2y15 + b6 x10 , (b6 x¯6 )b3 = 2b3 x6 + b3 b9 + b3 x15 = 2c3 + 2y15 + b3 b9 + b6 + 2y15 + c9 , we have that b6 x10 = b3 b9 + c3 + 2y15 . Since (c32 )d¯3 = c¯3 d¯3 + b¯6 d¯3 , c3 (c3 d¯3 ) = c3 c¯9 and x10 ∈ Supp{c3 c¯9 } by Lemma 2.17, we obtain x10 ∈ Supp{b¯ 6 d¯3 }. So (b6 x10 , d¯3 ) = 1. Furthermore (b3 b9 , d¯3 ) = 1. Thus b3 d3 = b¯9 . Therefore b¯3 b¯9 = b¯3 (b3 d3 ) = (b¯ 3 b3 )d3 = d3 + d3 b8 = d3 + y9 + y¯15 , which implies that b3 b9 = d¯3 + y¯9 + y15 , b6 x10 = d¯3 + c3 + y¯9 + 3y15 .

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Since (r3 b3 )d3 = c9 + b¯6 d3 and (b3 d3 )r3 = r3 b¯9 , we have that c9 ∈ Supp{r3 b¯9 }. But y15 ∈ Supp{r3 b¯9 } by Lemma 2.17. Thus there exists m3 ∈ B such that r3 b¯9 = m3 + c9 + y15 . Hence b9 ∈ Supp{r3 c¯9 }. But (r3 c¯9 , x15 ) = 1 by Lemma 2.17. Therefore r3 c¯9 = n3 + b9 + x15 ,

n3 ∈ B.

By r3 real, it follows that r3 n3 = c¯9 . Furthermore, r3 (r3 c¯9 ) = r3 x15 + r3 b9 + r3 n3 = b¯6 + c¯9 + 2y¯15 + c¯9 + y¯15 + m ¯ 3 + c¯9 , r32 c¯9 = c¯9 + b8 c¯9 , so that b8 c¯9 = b¯6 + 2c¯9 + 3y¯15 + m ¯ 3. Hence (b8 y15 , c9 ) = 3. By the expression b8 y15 in Lemma 2.17, we have (b¯3 d9 , c9 ) = 1. From the expression b3 t15 in Lemma 2.17, one has that (t15 , b3 c9 ) = 1. Hence there exists u3 ∈ B such that b3 c9 = u3 + d9 + t15 . Thus b¯3 u3 = c9 . Moreover, c3 b9 = (b¯ 3 d¯3 )c3 = b¯ 3 (c3 d¯3 ) = b¯3 c¯9 = u¯ 3 + d9 + t15 . Thus u3 c3 = b¯9 . Since c¯3 (b3 u¯ 3 ) = c¯3 c¯9 = d¯3 + y¯9 + y15 , (b3 c¯3 )u¯ 3 = b¯3 u¯ 3 + x¯6 u¯ 3 , we have that b3 u3 = y9

and u3 x6 = y¯15 + d¯3 .

Hence c3 y9 = c3 (b3 u3 ) = b3 (u3 c3 ) = b3 b¯9 . Since (x10 , b3 b¯9 ) = 1 by Lemma 2.17, one has that (x10 , c3 y9 ) = 1, so (y¯9 , c3 x10 ) = 1. Therefore y¯9 = c9 by the expression c3 x10 (see Lemma 2.17). The lemma follows.


99

Lemma 2.24 A NITA generated by c3 is isomorphic to the algebra of characters of 3 · A6 : (Ch(3 · A6 ), Irr(3 · A6 )). Furthermore, we have all the equations of products of base elements in the table subset D listed in Sect. 2.2. Proof By Lemmas 2.17 and 2.23 the sub-algebra generated by c3 satisfies the following equations: c3 c¯3 = 1 + b8 , c3 b¯6 = b8 + x10 , x10 real,

c32 = c¯3 + b¯6 ,

c3 b6 = c¯3 + y¯15 ,

c3 x10 = b6 + c9 + y15 ,

c3 y15 = b¯6 + c¯9 + 2y¯15 .

Changing c3 into b3 , b6 into b¯6 , y15 into b15 , x10 into b10 , we can see that the above equations are the same as those in Hypothesis 2.1 and Lemma 2.9. Hence the NITA generated by c3 is exactly the Table Algebra of characters of 3 · A6 by Theorem 2.7. Of course, we have equations of products of base elements in the table subset D listed in Sect. 2.2. Now we continue our calculations, and we shall construct a new NITA which is not derived from groups. Theorem 2.9 If c3 is nonreal, (b3 b8 , b3 b8 ) = 3. Then the NITA generated by b3 with b32 = c3 + b6 is isomorphic to the NITA in (A(3 · A6 · 2), B32 ), the NITA of dimension 32 in Sect. 2.2. Proof By Lemma 2.24, we may say that we have found a table subset: D = {1, b8 , x10 , b5 , c5 , c8 , x9 , c3 , c¯3 , d3 , d¯3 , c9 , c¯9 , b6 , b¯6 , y5 , y¯15 }. To prove that the NITA is isomorphic to NITA of dimension 32 in Sect. 2.2, it is sufficient to find all the basis elements outside D and the remaining expressions of all products of basis elements. Until now we have found many products of basis elements in Lemmas 2.16, 2.17 and 2.23. It remains to find the rest of them. In order to make the proof read easily, we shall divide it into several steps: Step 1 b3 b9 = y15 + c9 + d¯3 , b3 d3 = b¯ 9 , b3 b¯9 = x9 + c8 + x10 , b3 d9 = c¯9 + d3 + y¯15 , b3 d¯3 = d9 , b3 c¯9 = b¯9 + x¯15 + z3 , b3 z¯ 3 = c9 , b3 x¯15 = b5 + c5 + b8 + c8 + x9 + x10 , b3 z3 = x9 , b3 y3 = c¯9 . Since by Lemma 2.17 b32 x10 = c3 x10 + b6 x10 = b6 + c9 + y15 + 3y15 + c3 + d¯3 + c9 , (b3 x10 )b3 = x6 b3 + x15 b3 + b9 b3 = c3 + y15 + b6 + 2y15 + c9 + b3 b9 , we have that b3 b9 = y15 + c9 + d¯3 . Furthermore, b3 d3 = b¯ 9 . So b3 b¯9 = (b3 d3 )b3 = b32 d3 = c3 d3 + b6 d3 = x9 + c8 + x10 .

100


Set b3 d¯3 = y9 . Since b3 d3 = b¯9 , then y9 ∈ B. And b3 y9 = (b3 d¯3 )b3 = b32 d¯3 = c3 d¯3 + b6 d¯3 = c¯9 + d3 + y¯15 . We have that (b3 y15 , y¯9 ) = 1. But b3 y15 = s6 + 2t15 + d9 , thus d9 = y¯9 and b¯3 d3 = d9 . Since (b3 x15 , c9 ) = (b3 b9 , c9 ) = 1, there exists z3 ∈ B such that b3 c¯9 = b¯9 + x¯15 + z3 . Hence b3 z¯ 3 = c9 . Since b32 b¯6 = c3 b¯6 + b6 b¯6 = b8 + x10 + 1 + b8 + b5 + c5 + c8 + x9 , (b3 b¯6 )b3 = b3 b¯3 + b3 x¯15 = 1 + b8 + b3 x¯15 , we have that b3 x¯15 = b5 + c5 + b8 + c8 + x9 + x10 . Now we have b32 c¯9 = c3 c¯9 + b6 c¯9 = c8 + x9 + x10 + 2x9 + b8 + x10 + c8 + b5 + c5 , (b3 c¯9 )b3 = b3 b¯9 + b3 x¯15 + b3 z3 = c8 + x9 + x10 + b5 + c5 + b8 + c8 + x9 + x10 + b3 z3 , which imply b3 z3 = x9 . Step 2 b3 c9 = t15 + d9 + y3 , y3 real, d3 x¯6 = t15 + y¯3 , c3 b9 = t15 + d9 + y3 , c3 y3 = b¯9 , b3 y3 = c¯9 and b8 t15 = r3 + y¯3 + 5t15 + 2s6 + 3d9 . We have that b3 c9 = b3 (d3 c¯3 ) = (b3 c¯3 )d3 = b¯3 d3 + x¯3 d3 = d9 + x¯6 d3 . Hence (b3 c9 , d9 ) = 1. But by Lemma 2.17 (b3 t15 , c¯9 ) = 1. Then there exists y3 ∈ B such that b3 c9 = t15 + d9 + y3 ,

d3 x¯6 = t15 + y¯3 .

Consequently b3 y¯3 = c¯9 . Since b3 c9 = (d3 c¯3 )b3 = (b3 d3 )c¯3 = c¯3 b¯9 , we have that c3 b9 = t15 + d9 + y3 .


101

Consequently c3 y¯3 = b¯ 9 . Since b32 c9 = c3 c9 + b6 c9 = d3 + c¯9 + y¯15 + b¯6 + 2c¯9 + 2y¯15 , (b3 c9 )b3 = b3 t15 + b3 d9 + b3 y3 = b¯6 + c¯9 + 2y¯15 + d3 + c¯9 + y¯15 + b3 y3 , then b3 y3 = c¯9 . Since we have that (b3 b¯3 )t15 = t15 + b8 t15 (b3 t15 )b¯3 = b¯3 b¯6 + b¯3 c¯9 + 2b¯3 y¯15 = r3 + t15 + d9 + y¯3 + t15 + 2s6 + 4t15 + 2d9 , so b8 t15 = r3 + y¯3 + 5t15 + 2s6 + 3d9 , which implies that y3 is real. Step 3 b3 d9 = c¯9 + y¯15 + d3 , r3 t15 = b5 + c5 + b8 + c8 + x9 + x10 , r3 d9 = c8 + x9 + x10 . Since b32 y15 = c3 y15 + b6 y15 = b¯6 + c¯9 + 2y¯15 + c¯3 + d3 + b¯6 + 2c¯9 + 4y¯15 , (b3 y15 )b3 = b3 s6 + 2b3 t15 + b3 d9 = c¯3 + y¯15 + b3 d9 + 2b¯6 + 2c¯9 + 4y¯15 , then b3 d9 = c¯9 + y¯15 + d3 . Since r32 b8 = (r3 b8 )r3 and r32 x10 = (r3 x10 )r3 , we have that r3 t15 = b5 + c5 + b8 + c8 + x9 + x10 , r3 d9 = c8 + x9 + x10 . Step 4 r3 d3 = b9 and r3 b9 = d3 + c¯9 + y¯15 . There are three possibilities: r3 d3 = t9 , m3 + n6 , m4 + n5 . Thus r32 d3 = d3 + d3 b8 = d3 + c¯9 + y¯15 , (r3 d3 )r3 = r3 t9 or m3 r3 + n6 r3 or m4 r3 + n5 r3 , If r3 d3 = m3 + n6 , then m3 r3 = c¯9 and r3 n6 = d3 + y¯15 . But r3 y¯15 = x6 + 2x15 + b9 . Hence n6 = x6 , so that (r3 x6 , d3 ) = 1, a contradiction.

102


If r3 d3 = m4 + n5 , then m4 r3 = d3 + c¯9 and r3 n5 = y¯15 . But (r3 y¯15 , n5 ) = 0, a contradiction. Hence r3 d3 = t9 and r3 t9 = d3 + c¯9 + y¯15 , which implies that (r3 y¯15 , t9 ) = 1. Then t9 = b9 . Step 5 x6 y15 = r3 + 4t15 + s6 + 2d9 + y3 , b8 x15 = b3 + z¯ 3 + 2x6 + 5x15 + 3b9 , b8 b9 = z¯ 3 + x6 + 2b9 + 3x15 , b8 d9 = y3 + s6 + 2d9 + 3t15 , y3 b8 = d9 + t15 and b6 b9 = s6 + 2d9 + 2t15 and b6 x15 = r3 + s6 + y3 + 2d9 + 4t15 . The above equations follow from b3 x62 = (b3 x6 )x6 , (b3 b¯3 )x15 = (b3 x15 )b¯3 , (b3 b¯3 )b9 = b3 b9 )b¯3 , (b3 b¯3 )d9 = (b3 d9 )b¯3 , (b3 b¯3 )y3 = (b3 y3 )b¯3 , b32 b9 = (b3 b9 )b3 and b32 x15 = (b3 x15 )b3 . Step 6 c3 d9 = b¯9 + z3 + x¯15 , b6 d9 = x¯6 + 3b¯9 + 2x¯15 , z3 c¯3 = d9 . Since b32 d9 = c3 d9 + b6 d9 , (b3 d9 )b3 = b3 d3 + b3 c¯9 + b3 y¯15 = 2b¯9 + 3x¯15 + z3 + x¯6 , and (c3 x15 , d9 ) = 1, from which Step 6 follows. Step 7 b8 x15 = b3 + z¯ 3 + 2x6 + 5x15 + 3b9 , b3 x9 = z¯ 3 + b9 + x15 , b3 c8 = x15 + b9 , b3 b5 = x15 and b3 c5 = x15 . By Lemma 2.17 and Step 1, we have (b3 x15 )b¯3 = b6 b¯3 + 2y15 b¯3 + c9 b¯3 = b3 + x15 + x6 + 2x15 + b9 + b9 + x15 + z¯ 3 , (b3 b¯3 )x15 = x15 + b8 x15 , hence b8 x15 = b3 + z¯ 3 + 2x6 + 5x15 + 3b9 . Consequently, we have the following equations: b82 b3 = b3 + 2b3 b8 + 2b3 x10 + b3 b5 + b3 c5 + b3 c8 + b3 x9 = 3b3 + 2x6 + 2x15 + 2x6 + 2x15 + 2b9 + b3 b5 + b3 c5 + b3 c8 + b3 x9 , (b3 b8 )b8 = b3 b8 + x6 b8 + x15 b8 = b3 + x6 + x15 + b3 + x6 + b9 + 2x15 + b3 + z¯ 3 + 2x6 + 5x15 + 3b9 , which imply that b3 b5 + b3 c5 + b3 c8 + b3 x9 = 4x15 + 2b9 + z¯ 3 . By the expression b3 x¯15 in Step 3, it can be easily shown that Step 7 follows. Step 8 r3 b5 = t15 , r3 c5 = t15 , r3 c8 = t15 + d9 , r3 x9 = t15 + d9 + y3 and r3 y3 = x9 .


103

Since b82 r3 = r3 + 2r3 b8 + 2r3 x10 + r3 b5 + r3 c5 + r3 c8 + r3 x9 = r3 + 2r3 + 2s6 + 2t15 + 2s6 + 2t15 + 2d9 + r3 b5 + r3 c5 + r3 c8 + r3 x9 , (r3 b8 )b8 = r3 b8 + b8 s6 + b8 t15 = r3 + s6 + t15 + r3 + s6 + 2t15 + d9 + r3 + y3 + 5t15 + 2s6 + 3d9 , it holds that r3 b5 + r3 c5 = r3 c8 + r3 x9 = 4t15 + 2d9 + y3 . But (r3 t15 , x9 ) = 1. The first four equations follow from the above equation. Consequently, r3 y3 = x9 . Step 9 r3 c9 = b¯9 + x¯15 + z3 , r3 z3 = c9 , x6 y¯15 = b¯ 3 + z3 + x¯6 + 4x¯15 + 2b¯9 , x6 c9 = 2t15 + 2d9 + s6 , x6 x10 = b9 + b3 + z¯ 3 + 3x15 , x6 x¯6 = 1 + b8 + b5 + c5 + c8 + x9 , x6 s6 = 2b¯6 + c¯9 + y¯15 and x6 c¯9 = 2x¯15 + x¯6 + 2b¯9 . Since r32 b9 = (r3 b9 )r3 , we have the first equation, from which the second equation follows. The rest of the equations follow from associativity of (c3 b8 )x¯6 = b8 (c3 x¯6 ), (d3 b8 )x¯6 = b8 (d3 x¯6 ), (b3 x¯6 )x6 = (b3 x6 )x¯6 , (b3 c¯3 )x6 = (b3 x6 )c¯3 , (b3 c3 )x6 = b3 (c3 x6 ) and (r3 c3 )c9 = r3 (c3 c9 ), respectively. Step 10 x6 y3 = d3 + y¯15 . By known equations, we have that (x6 y3 , y¯15 ) = (x6 y15 , y3 ) = 1 and (x6 y3 , d3 ) = (x6 d¯3 , y3 ) = 1, from which Step 10 follows. Step 11 x6 b5 = x6 + b9 + x15 , x6 c5 = x6 + b9 + x15 , x6 c8 = x6 + b9 + 2x15 + z3 and z3 x6 = x10 + c8 . Since b82 x6 = x6 + 2x6 b8 + 2x6 x10 + x6 b5 + x6 c5 + x6 c8 + x6 x9 = x6 + 2b3 + 2x6 + 2b9 + 4x15 + 2b9 + 2b3 + 2¯z3 + 6x15 + + x6 b5 + x6 c5 + x6 c8 + x6 + 2b9 + 2x15 , (b8 x6 )b8 = b3 b8 + x6 b8 + b9 b8 + 2b8 x15 = b3 + x6 + x15 + b3 + x6 + b9 + 2x15 + z¯ 3 + x6 + 2b9 + + 3x15 + 2b3 + 2¯z3 + 4x6 + 10x15 + 6b9 , we have that x6 b5 + x6 c5 + x6 c8 = 3x6 + 3b9 + z¯ 3 + 4x15 .

104


Table 2.3 Further calculations Values of inner products

New equations

(z3 b9 , b8 ) = (b8 b9 , z¯ 3 ) = 1 (z3 x15 , b8 ) = (b8 x15 , z¯ 3 ) = 1

z3 b8 = b¯9 + x¯15

(x6 x10 , b9 ) = (x6 c5 , b9 ) = (x6 b5 , b9 ) = 1 (x6 b8 , b9 ) = (x6 c8 , b9 ) = 1, (x6 x9 , b9 ) = 2

x6 b¯9 = b5 + c5 + x10 + b8 + c8 + 2x9

(x6 b8 , x15 ) = 2, (x6 x10 , x15 ) = 3, (x6 b5 , x15 ) = (x6 c5 , x15 ) = 1

x6 x¯15 = b5 + c5 + 2b8 + 3x10 + 2c8 + 2x9

(x6 c8 , x15 ) = (x6 x9 , x15 ) = 2 (x6 c3 , t15 ) = (x6 d¯3 , t15 ) = (x6 b6 , t15 ) = 1 (x6 c9 , t15 ) = 2, (x6 y15 , t15 ) = 4 (x6 z¯ 3 , d¯3 ) = (x6 y¯15 , z3 ) = 1 (x6 d3 , x¯15 ) = (x6 b¯6 , x¯15 ) = (x6 c¯3 , x¯15 ) = 1 (x6 c¯9 , x¯15 ) = 2, (x6 y¯15 , x¯15 ) = 4

x6 t15 = c¯3 + d3 + 2c¯9 + 4y¯15 + b¯6 x6 z¯ 3 = d¯3 + y15 x6 x15 = 4y15 + b6 + 2c9 + d3 + c3

By the expression x6 x¯6 in Step 9, we have that (x6 b5 , x6 ) = (x6 c5 , x6 ) = (x6 c8 , x6 ) = 1. The first three equations of Step 11 follow by comparing the degrees. Hence (x6 z3 , c8 ) = 1. In addition to (z3 x6 , x10 ) = (x6 x10 , z¯ 3 ) = 1, from which the last equation follows. Step 12 x6 x9 = x6 + 2b9 + 2x15 , x6 d9 = b¯6 + 2c¯9 + 2y¯15 , x6 d3 = z3 + x¯15 and x6 b9 = 2c¯9 + b¯6 + 2y¯15 . It follows from (b3 z3 )x6 = b3 (z3 x6 ), (z3 c¯3 )x6 = c¯3 (z3 x6 ), (r3 c3 )d¯3 = r3 (c3 d¯3 ) and (r3 c3 )b¯9 = (r3 b¯9 )c3 , respectively. Step 13 Based on the equations known by us, we can check the inner products of some products of basis elements and obtain Table 2.3. Step 14 Repeatedly considering the associativity of basis elements, we can obtain new equations. Since the process is trivial and repeated, the concrete process is omitted and the correspondence between the associativity of basis elements and new equations are listed in the following table. Occasionally, we need to check the inner products based on the old and new equations to obtain a newer equation. Table 2.4 is an explanation.


105

Table 2.4 Further calculations Associativity checked or inner products

New equations

(b3 c3 )d¯3 = b3 (c3 d¯3 )

d3 s6 = z¯ 3 + x15

(b3 c3 )y3 = b3 (c3 y3 )

y3 s6 = c8 + x10

(b3 c3 )z3 = (b3 z3 )c3

z3 s6 = d¯3 + y15

(b3 c3 )¯z3 = b3 (c3 z¯ 3 )

z¯ 3 s6 = d3 + y¯15

(b3 c3 )s6 = b3 (c3 s6 )

s62 = 1 + b5 + c5 + b8 + c8 + x9 b9 s6 = b¯6 + 2c¯9 + 2y¯15

(b3 c3 )b9 = b3 (c3 b9 ) (b3 c3 )x15 = b3 (c3 x15 )

s6 x15 = c¯3 + d3 + b¯6 + 2c¯9 + 4y¯15

(b3 c3 )t15 = b3 (c3 t15 )

s6 t15 = b5 + c5 + 2b8 + 2c8 + 2x9 + 3x10

(b3 c3 )d9 = b3 (c3 d9 )

s6 d9 = b5 + c5 + 2x9 + b8 + c8 + x10

(b3 c3 )x10 = b3 (c3 x10 )

s6 x10 = r3 + y3 + d9 + 3t15

(b3 c3 )b5 = b3 (c3 b5 )

s6 b5 = s6 + d9 + t15

(b3 c3 )c5 = b3 (c3 c5 )

s6 c5 = s6 + t15 + d9

(b3 c3 )c8 = b3 (c3 c8 )

s6 c8 = y3 + s6 + d9 + 2t15

(b3 c3 )x9 = b3 (c3 x9 )

s6 x9 = s6 + 2d9 + 2t15

(b3 c3 )c9 = b3 (c3 c9 )

s6 c9 = x¯6 + 2b¯9 + 2x¯15

(s6 b3 , y¯15 ) = 1 (s6 b9 , y¯15 ) = 2, (s6 x15 , y¯15 ) = 4 (s6 z¯ 3 , y¯15 ) = (s6 x6 , y¯15 ) = 1

s6 y15 = 2b¯ 9 + 4x¯15 + z3 + x¯6 + b¯3

(b3 d3 )b9 = d3 (b3 b9 ) (b3 d3 )b¯9 = d3 (b3 b¯9 )

b9 b¯9 = 1 + b5 + c5 + 2b8 + 2c8 + 2x9 + 2x10 b2 = d¯3 + c3 + 2b6 + 2c9 + 3y15

(b3 d3 )x¯15 = d3 (b3 x¯15 )

b9 x15 = 6y15 + 3c9 + d¯3 + 2b6 + c3

(b3 d3 )x15 = d3 (b3 x15 ) (b3 d3 )t15 = d3 (b3 t15 )

b9 x¯15 = 2b5 + 2c5 + 3c8 + 3b8 + 4x10 + 3x9 b9 t15 = d3 + c¯3 + 6y¯15 + 3c¯9 + 2b¯6

(b3 d3 )d9 = d3 (b3 d9 )

b9 d9 = c¯3 + d3 + 2b¯6 + 2c¯9 + 3y¯15

(b3 d3 )y3 = d3 (b3 y3 )

b9 y3 = c¯3 + y¯15 + c¯9

(b3 d3 )z3 = d3 (b3 z3 )

b9 z¯ 3 = c9 + c3 + y15

(b3 d3 )¯z3 = d3 (b3 z¯ 3 )

b9 z3 = b8 + x9 + x10

(b3 d3 )x10 = b3 (d3 x10 )

b9 x10 = z¯ 3 + b3 + 2b9 + x6 + 4x15 b¯9 c9 = b3 + z¯ 3 + 2x6 + 3x15 + 2b9

(b3 d3 )b5 = b3 (d3 b5 ) (b3 d3 )c5 = b3 (d3 c5 ) (b3 d3 )b6 = b3 (d3 b6 ) (b3 d3 )c8 = b3 (d3 c8 ) (b3 d3 )x9 = b3 (d3 x9 ) (b3 d3 )y¯15 = b3 (d3 y¯15 ) (b3 d3 )d3 = b3 (d32 ) (b3 d3 )c3 = b3 (d3 c3 )

9

b9 c9 = r3 + y3 + 2s6 + 2d9 + 3t15 b¯9 b6 = 2b9 + 2x15 + x6 b¯9 y15 = 6x15 + b3 + z¯ 3 + 2x6 + 3b9 b¯9 x9 = 2b¯9 + 3x¯15 + z3 + b¯3 + 2x¯6 b¯9 y¯15 = r3 + y3 + 2s6 + 6t15 + 3d9 b¯9 d3 = r3 + t15 + d9 b¯9 c3 = z¯ 3 + b9 + x15 (continued on the next page)

106


Table 2.4 (Continued) Associativity checked or inner products

New equations

(b3 d3 )d¯3 = b3 (d3 d¯3 ) (b3 d3 )c3 = b3 (d3 c3 )

b9 d3 = b¯3 + b¯9 + x¯15 b9 c¯3 = z3 + b¯9 + x¯15

(b3 b5 )x15 = (b3 x15 )b5

2 = 2c + 2d¯ + 4b + 6c + 9y x15 3 3 6 9 15

(b3 b5 )x¯15 = (b3 x¯ 15 )b5 (b3 b5 )t15 = (b3 t15 )b5

x15 x¯15 = 1 + 6x9 + 3b5 + 3c5 + 5b8 + +5c8 + 6x10 x15 t15 = 4b¯ 6 + 6c¯9 + 9y¯15 + +2c¯3 + 2d3

(b3 b5 )d9 = (b3 d9 )b5

x15 d9 = c¯3 + d3 + 2b¯6 + 3c¯9 + 6y¯15

(b3 b5 )z3 = (b3 z3 )b5

x15 z3 = b5 + c5 + b8 + c8 + x9 + x10

(b3 b5 )b5 = b3 b52

x15 b5 = z¯ 3 + b3 + x6 + 2b9 + 3x15

(b3 b5 )x10 = (b3 x10 )b5

x15 x10 = b3 + z¯ 3 + 3x6 + 4b9 + 6x15

(b3 b5 )c5 = (b5 c5 )b3

x15 c5 = b3 + z¯ 3 + x6 + 2b9 + 3x15

(b3 b5 )c8 = (b5 c8 )b3

x15 c8 = b3 + z¯ 3 + 2x6 + 3b9 + 5x15

(b3 b5 )x9 = (b5 x9 )b3

x15 x9 = b3 + z¯ 3 + 2x6 + 3b9 + 6x15 x15 d3 = x¯6 + 2x¯15 + b¯9

(b3 b5 )d3 = (b5 d3 )b3 (b3 b5 )d¯3 = (b5 d¯3 )b3 (b3 b5 )c9 = (b5 c9 )b3 (b3 b5 )c¯9 = (b5 c¯9 )b3 (b3 b5 )b¯6 = (b5 b¯6 )b3

x15 d3 = s6 + d9 + 2t15 x15 c9 = r3 + y3 + 2s6 + 6t15 + 3d9 x15 b¯9 = b¯3 + z3 + 2x¯ 6 + 3b¯ 9 + 6x¯15 x15 b¯6 = b¯3 + z3 + 2b¯ 9 + 4x¯ 15 + x¯6

(b3 b5 )y15 = (b5 y15 )b3

x15 y15 = 2y3 + 2r3 + 4s6 + 6d9 + 9t15

(b3 b6 )t15 = (b3 t15 )b6

2 = 1 + 5b + 5c + 3b + 3c + 6x + 6x t15 8 8 5 5 10 9

(b3 b6 )d9 = (b3 d9 )b6

d9 t15 = 2b5 + 2c5 + 3b8 + 3c8 + 3x9 + 4x10

(b3 b6 )y3 = (b3 y3 )b6

y3 t15 = b5 + c5 + b8 + c8 + x9 + x10

(b3 b6 )z3 = (b3 z3 )b6

z3 t15 = b6 + c9 + 2y15

(b3 b6 )x10 = (b6 x10 )b3

t15 x10 = r3 + y3 + 4d9 + 3s6 + 6t15

(b3 b6 )b5 = (b6 b5 )b3

t15 b5 = r3 + y3 + s6 + 2d9 + 3t15

(b3 b6 )c5 = (b6 c5 )b3

t15 c5 = r3 + y3 + s6 + 2d9 + 3t15

(b3 b6 )c8 = (b6 c8 )b3

t15 c8 = r3 + y3 + 2s6 + 3d9 + 5t15

(b3 b6 )x9 = (b6 x9 )b3

t15 x9 = r3 + y3 + 3d9 + 2s6 + 6t15

(b3 b6 )d3 = (b6 d3 )b3

t15 d3 = x6 + b9 + 2x15 t15 c9 = b¯3 + z3 + 3b¯9 + 2x¯6 + 6x¯15

(b3 b6 )c9 = (b6 c9 )b3 (b3 b6 )b6 = b3 b62 (b3 b6 )y15 = (b6 y15 )b3

t15 b6 = b¯3 + z3 + x¯6 + 2b¯9 + 4x¯15 t15 y15 = 2b¯ 3 + 2z3 + 4x¯6 + 6b¯9 + 9x¯15

(b3 d¯3 )d9 = (b3 d9 )d¯3 (b3 d¯3 )x10 = (b3 x10 )d¯3

d92 = 1 + b5 + c5 + 2b8 + 2c8 + 2x9 + 2x10

(b3 d¯3 )b5 = (b3 b5 )d¯3 (b3 d¯3 )c5 = (b3 c5 )d¯3

d9 b5 = s6 + d9 + 2t15

d9 x10 = y3 + r3 + s6 + 2d9 + 4t15 d9 c5 = s6 + d9 + 2t15 (continued on the next page)


107


New equations

(b3 d¯3 )c8 = (b3 c8 )d¯3 (b3 d¯3 )d3 = (b3 d3 )d¯3

d9 c8 = r3 + s6 + 2d9 + 3t15

(b3 d¯3 )c9 = (b3 c9 )d¯3 (b3 d¯3 )y15 = (b3 y15 )d¯3

d9 c9 = b3 + z3 + 2x¯6 + 2b¯9 + 3x¯15 d9 y15 = b¯3 + z3 + 2x¯6 + 3b¯ 9 + 6x¯ 15

(b3 d¯3 )x9 = (d¯3 x9 )b3

d9 x9 = r3 + y3 + 2s6 + 2d9 + 3t15

(x6 d3 )z3 = (z3 x6 )d3

z32 = d3 + b¯6

(x6 d3 )¯z3 = (¯z3 x6 )d3 (x6 d¯3 )z3 = (y3 x6 )d¯3

z3 z¯ 3 = 1 + b8

(x6 d3 )y3 = (y3 x6 )d3 (x6 d3 )d¯3 = (d3 d¯3 )x6

d9 d3 = b3 + b9 + x15

y32 = 1 + c8 y3 z3 = d¯3 + b6 z3 d¯3 = z¯ 3 + x6

(x6 d3 )c¯9 = (d3 c¯9 )x6 (x6 d3 )b¯6 = (d3 b¯6 )x6

z3 c¯9 = r3 + d9 + t15 z3 b¯6 = y3 + t15

(x6 d3 )b6 = (d3 b6 )x6

z3 b6 = z¯ 3 + x15

(r3 x10 , s6 ) = (s6 b8 , r3 ) = 1

r3 s6 = x10 + b8

(d9 y15 , x¯6 ) = 2, (d9 b6 , x¯6 ) = 1, (d9 c9 , x¯6 ) = 2

x6 d9 = b¯6 + 2y¯15 + 2c¯9 x6 d¯3 = y3 + t15

(x6 y3 , d3 ) = (x6 t15 , d3 ) = 1 (r3 c3 )d¯3 = r3 (c3 d¯3 )

x6 d3 = z3 + x¯15

(x¯6 b6 , b9 ) = 1, (b¯9 y15 , x6 ) = (b¯9 c9 , x6 ) = 2 (x6 b¯9 , x9 ) = (x6 x¯15 , x9 ) = 2, (x6 x¯6 , x9 ) = 1

x6 b9 = b6 + 2y15 + 2c9

(b9 b¯9 , c8 ) = 2, (b9 x¯15 , c8 ) = 1, (x6 c8 , b9 ) = 1 (b9 b¯9 , b8 ) = 2, (b9 x¯15 , b8 ) = 3,

b9 c8 = 2b9 + 3x15 + x6 + b3

(b9 z3 , b8 ) = (x6 b8 , b9 ) = 1 (b9 b¯9 , c5 ) = (x6 c5 , b9 ) = 1, (b9 x¯15 , c5 ) = 2

b9 b8 = 2b9 + 3x15 + z¯ 3 + x6

x6 x9 = 2b9 + x6 + 2x15

(x15 c9 , y3 ) = (x15 b6 , y3 ) = 1, (x15 y15 , y3 ) = 2

c5 b9 = b9 + 2x15 + x6 x15 y3 = b¯6 + c¯9 + 2y¯15

(b3 b5 )¯z3 = (b3 z¯ 3 )b5 (b3 b5 )d¯3 = (b5 d¯3 )b3

x15 z¯ 3 = b6 + c9 + 2y15 x15 d¯3 = s6 + 2t15 + d9

(b3 b5 )c¯9 = (b5 c¯9 )b3 (b3 b5 )y15 = (b5 y15 )b3

x15 c¯9 = b¯3 + z3 + 3b¯9 + 2x¯6 + 6x¯15 x15 y¯15 = 9x¯15 + 2z3 + 2b¯3 + 4x¯6 + 6b¯ 9

(x6 t15 , c3 ) = (b9 t15 , c¯3 ) = 1, (x15 t15 , c¯3 ) = 2

t15 c3 = x¯6 + b¯9 + 2x¯15

(d9 b8 , y3 ) = (d9 x10 , y3 ) = (d9 x9 , y3 ) = 1

d9 y3 = x9 + b8 + x10

(d9 y15 , z3 ) = (d9 c3 , z3 ) = (d9 c9 , z3 ) = 1

d9 z3 = c3 + c9 + y15

(b3 b¯3 )d9 = (b3 d9 )b¯3 (b9 d9 , c¯3 ) = (d9 z3 , c3 ) = (x15 d9 , c¯3 ) = 1

d9 b8 = y3 + s6 + 2d9 + 3t15 d9 c3 = b¯9 + z3 + x¯15

(d9 b8 , y3 ) = (t15 b8 , y3 ) = 1

y3 b8 = d9 + t15

(s6 x10 , y3 ) = (t15 x10 , y3 ) = (d9 x10 , y3 ) = 1

y3 x10 = s6 + t15 + d9 (continued on the next page)

108



New equations

(t15 y3 , b5 ) = 1

y3 b5 = t15

(t15 y3 , c5 ) = 1

y3 c5 = t15

(t15 y3 , c8 ) = (s6 y3 , c8 ) = (y3 c8 , y3 ) = 1

y3 c8 = s6 + t15 + y3

(t15 y3 , x9 ) = (d9 y3 , x9 ) = (r3 y3 , x9 ) = 1

y3 x9 = t15 + d9 + r3 y3 c3 = b¯9

(b9 y3 , c¯3 ) = 1 (b9 y3 , d¯3 ) = 1

y3 d3 = b¯9 y3 c9 = b¯3 + b¯9 + x¯15

(b3 y3 , c9 ) = (b9 c9 , y3 ) = (x15 c9 , y3 ) = 1 (x15 y3 , b6 ) = (y3 z3 , b6 ) = 1 (x6 y3 , y¯15 ) = (b9 y3 , y¯15 ) = 1, (x15 y3 , y¯15 ) = 2

y3 b6 = x¯15 + z3 y3 y15 = x¯6 + b¯9 + 2x¯15

(x6 x10 , z3 ) = (b9 x10 , z3 ) = (x15 x10 , z3 ) = 1

z3 x10 = x¯6 + b¯9 + x¯15

(x15 b5 , z3 ) = 1

z3 b5 = x¯15

(x15 c5 , z3 ) = 1

z3 c5 = x¯15

(x6 c8 , z3 ) = (x15 c8 , z3 ) = (z3 z¯ 3 , c8 ) = 1

z3 c8 = x¯6 + x¯15 + z3 z3 x9 = b¯3 + b¯9 + x¯15

(b3 x9 , z3 ) = (b9 x9 , z3 ) = (x15 x9 , z3 ) = 1 (b9 c3 , z3 ) = 1 (s6 d¯3 , z3 ) = (d3 y3 , z¯ 3 ) = 1

z3 c3 = b9 z3 d3 = s6 + y3

(b¯3 c9 , z¯ 3 ) = (b¯9 c9 , z¯ 3 ) = (x15 c¯9 , z3 ) = 1

z3 y15 = b3 + b9 + x15

(x6 y¯15 , z3 ) = (b9 y¯15 , z3 ) = 1, (x15 y¯15 , z3 ) = 2

z3 y15 = x6 + b9 + 2x15

(s6 y15 , z3 ) = (d9 y15 , z3 ) = 1, (t15 y15 , z3 ) = 2

z3 y¯15 = s6 + 2t15 + d9

2.8 Structure of NITA Generated by b3 and Satisfying b32 = c3 + b6 , c3 = b3 , b¯ 3 , (b3 b8 , b3 b8 ) = 3 and c3 Real In this section we discuss NITA satisfying the following hypothesis. Hypothesis 2.4 Let (A, B) be a NITA generated by a nonreal element b3 ∈ B of degree 3 and without non-identity basis element of degree 1 or 2, such that: b¯3 b3 = 1 + b8 and b32 = c3 + b6 ,

c3 = c¯3

where bi , ci , di ∈ B are of degree i. Lemma 2.25 Let (A, B) satisfy Hypothesis 2.4, then 1) 2) 3) 4)

c32 = 1 + b8 , b¯3 c3 = b3 + x6 , x6 = x¯6 ∈ B, b8 b3 = x6 + b6 b¯3 , b8 + b82 = c3 b¯6 + b6 c3 + b6 b¯6 ,


5) 6) 7)

109

b¯6 c3 = x6 b¯3 , c3 b8 = b6 + x6 b3 , (b8 b3 , b8 b3 ) = 3, 4.

Proof By Hypothesis 2.4, we have (b¯3 c¯3 , b3 ) = (c3 , b32 ) = 1. So (b¯3 b¯3 , c32 ) = (b¯3 c3 , b¯3 c3 ) ≥ 2, hence c32 = 1 + b8 .

(1)

b¯3 c3 = b3 + x6 .

(2)

Furthermore, we have If x6 = x¯6 , then 1 = (b¯3 c3 , b3 c3 ) = (c32 , b32 ), a contradiction By the associative law (b¯3 b3 )b3 = b¯3 b32 and Hypothesis 2.4,

to Hypothesis 2.4. one concludes that b3 + b8 b3 = c3 b¯3 + b6 b¯3 . So b8 b3 = x6 + b6 b¯3 holds by (2). By the associative law (b¯3 b3 )2 = b¯32 b32 and Hypothesis 2.4 and (1), we see that 1 + 2b8 + b82 = (c3 + b¯6 )(c3 + b6 ), 1 + 2b8 + b82 = c32 + c3 b6 + b¯6 c3 + b¯6 b6 , b82 + b8 = c3 b6 + b¯6 c3 + b¯6 b6 . By the associative law (b¯6 b3 )b3 = b¯3 b32 and Hypothesis 2.4 and (2), it holds that b3 + b8 b3 = c3 b¯3 + b6 b¯3 ; hence b8 b3 = x6 + b6 b¯3 .

(3)

By Hypothesis 2.4, (b6 b¯3 , b3 ) = (b6 , b32 ) = 1, so by (3) (b8 b3 , b8 b3 ) ≥ 3, we shall show that (b8 b3 , b8 b3 ) = 3, 4. By the associative law and Hypothesis 2.4 and (1) and (2), we have that (b¯32 )c3 = b¯3 (b¯3 c3 ), c32 + b¯6 c3 = b3 b¯3 + x6 b¯3 ; thus b¯6 c3 = x6 b¯3 .

(4)

By (3), (x6 b¯3 , b8 ) = (x6 , b3 b8 ) = 1 holds. So (x6 b¯3 , x6 b¯3 ) ≥ 2. We shall show that (x6 b¯3 , x6 b¯3 ) = 2, 3. If (x6 b¯3 , x6 b¯3 ) = 4, then x6 b¯3 = b8 + g3 + w3 + v4 ; hence b3 g3 = x6 + w3 ,

b3 w3 = x6 + k3 ,

b3 v4 = x6 + z6 .

Therefore 1 ≤ (b3 g3 , b3 w3 ) = (b3 b¯3 , g¯ 3 w3 ). We obtain w3 = g3 by Hypothesis 2.4, hence 2 = (x6 b¯3 , w3 ) = (x6 , b3 w3 ), so that b3 w3 = 2x6 , a contradiction by (4). Now we can state that (b¯ 6 c3 , b¯6 c3 ) = (b¯6 b6 , c32 ) = 2, 3. By (1), it follows that (b¯6 b6 , b3 b¯3 ) = (b6 b¯3 , b6 b¯3 ) = 2 or 3. Therefore (b8 b3 , b8 b3 ) = 3, 4 by (3). By the associative law (b¯3 c3 )b3 = c3 (b¯3 b3 ) and Hypothesis 2.4, we get b32 + x6 b3 = c3 + c3 b8 and c3 b8 = b6 + x6 b3 .

110


Lemma 2.26 Let (A, B) satisfy Hypothesis 2.4 and (b8 b3 , b8 b3 ) = 3, then we get: 1) b6 b¯3 = b3 + x15 , 2) b8 b3 = x6 + b3 + x15 , x6 = x¯6 , 3) x¯6 b3 = b8 + x10 = b6 c3 , x6 = b6 , x10 = x¯10 , 4) x6 c3 = b¯3 + x¯15 , 5) b3 x6 = c3 + y15 , 6) b8 c3 = b6 + c3 + y15 , b6 = b¯6 , y15 = y¯15 . Proof By our assumption in this lemma and part 3) in Lemma 2.25, and Hypothesis 2.4: b6 b¯3 = b3 + x15 ,

(1)

then b8 b3 = x6 + b3 + x15 by (1), (b6 b¯6 , b3 b¯3 ) = (b6 b¯3 , b6 b¯3 ) = 2. Then by 1) in Lemma 2.25, 2 = (b6 b¯6 , c32 ) = (b¯6 c3 , b¯6 c3 ).

(2)

By 6) in Lemma 2.25, (b¯6 c3 , b8 ) = (c3 b8 , b6 ) = 1, so by (2) b¯ 6 c3 = b8 + x10 .

(3)

So by 5) in Lemma 2.25 x6 b¯3 = b8 + x10 ; so by (1) we get that x6 = b6 . By the associative law Lemma 2.25 and (2), it follows that

(4) (c32 )b¯3

= (c3 b¯3 )c3 and 1), 2) in

b¯3 + b¯ 3 b8 = b3 c3 + x6 c3 , b¯3 + x¯6 + b¯3 + x¯15 = b¯3 + x¯6 + x6 c3 , so x6 c3 = b¯3 + x¯15 .

(5)

By (4) and 2) in Lemma 2.25: 2 = (x¯6 b3 , x6 b3 ) = (x6 b3 , x6 b3 ),

(b3 x6 , c3 ) = (x6 , b¯3 c3 ) = 1.

Therefore b3 x6 = c3 + y15 .

(6)


111

By the associative law (b3 b¯3 )c3 = (b¯3 c3 )b3 and 2) in Lemma 2.25 and Hypothesis 2.4, one has c3 + b8 c3 = b32 + x6 b3 , c3 + b8 c3 = c3 + b6 + c3 + y15 , b8 c3 = b6 + c3 + y15 . b8 , c3 are reals, then b6 = b¯6 ,

y15 = y¯15 .

By (3), we have x10 = x¯10 .

Lemma 2.27 Let (A, B) satisfy Hypothesis 2.4 and (b8 b3 , b8 b3 ) = 3, then Lemma 2.26 holds and we get: 1) 2) 3) 4) 5) 6) 7) 8) 9) 10)

x10 b3 = x15 + x6 + x9 , x10 = x¯10 , x6 b8 = 2x15 + b3 + x6 + x9 , x¯15 c3 = 2x15 + x6 + x9 , y15 b¯3 = 2x15 + x6 + x9 , x6 b6 = 2x¯6 + x¯15 + x¯9 , c3 x10 = b6 + y15 + y9 , x62 = 2b6 + y15 + y9 , x15 b3 = 2y15 + b6 + y9 , b8 b6 = 2y15 + b6 + c3 + y9 , y9 = y¯9 , b8 x15 = 4x15 + 2x6 + 2x9 + b3 + y9 b¯3 .

Proof By the associative law (b32 )b6 = (b3 b6 )b3 and Hypothesis 2.4 and 1), 3) in Lemma 2.26, we get c3 b6 + b62 = b¯3 b3 + x¯15 b3 , b8 + x10 + b62 = 1 + b8 + x¯15 b3 . Hence x10 + b62 = 1 + x¯15 b3 . By 3) in Lemma 2.26, (b3 x10 , x6 ) = (x10 , b¯3 x6 ) = 1, hence by (1), (x10 b3 , x15 ) = (b3 x¯15 , x10 ) = 1. So x10 b3 = x15 + x6 + x9 ,

(1)

112


where x9 is a linear combination of B. We shall show that x9 ∈ B. If x10 b3 has an element of degree 3, then 1 = (x10 b3 , y3 ) = (x10 ; b¯ 3 y3 ), hence b¯3 y3 = x10 , a contradiction. So x10 b3 = x15 + x6 + x9 ,

(2)

where x9 ∈ B. By the associative law and Hypothesis 2.4 and 2), 3) in Lemma 2.26, we get (b3 x¯6 )b¯3 = x¯6 (b3 b¯3 ), b8 b¯3 + x10 b¯3 = x¯6 + x¯6 b8 , x¯6 + x¯6 b8 = x¯6 + b¯3 + x¯15 + x¯15 + x¯6 + x¯9 . Hence x¯6 b8 = 2x¯15 + b¯3 + x¯6 + x¯9 .

(3)

By the associative law (b¯3 c3 )b8 = b¯3 (b8 c3 ) and 1), 2), 6) in Lemma 2.26 and 2) in Lemma 2.25, we obtain b3 b8 + x6 b8 = b6 b¯3 + c3 b¯3 + y15 b¯3 , x6 + b3 + x15 + 2x15 + b3 + x6 + x9 = b3 + x15 + b3 + x6 + y15 b¯3 . So y15 b¯3 = 2x15 + x6 + x9 .

(4)

By the associative law (b¯3 c3 )b8 = (b¯ 3 b8 )c3 and 1), 2), 6) in Lemma 2.26 and in Lemma 2.25 and (3), the following equations hold: b3 b8 + x6 b8 = x¯6 c3 + b¯3 c3 + x¯15 c3 , x6 + b3 + x15 + 2x15 + b3 + x6 + x9 = b3 + x15 + b3 + x6 + x¯15 c3 . So x¯15 c3 = 2x15 + x6 + x9 .

(5)

By the associative law (b¯3 c3 )b6 = c3 (b¯3 b6 ) and 2) in Lemma 2.25 and 1) in Lemma 2.26 and (5), one should have b3 b6 + x6 b6 = b3 c3 + x15 c3 , b¯ 3 + x¯15 + x6 b6 = b¯3 + x¯6 + 2x¯15 + x¯6 + x¯9 , x6 b6 = 2x¯6 + x¯ 15 + x¯9 .

(6)


113

By the associative law and 5), 4) in Lemma 2.26 and 1) in Lemma 2.25, we get (b3 x6 )c3 = b3 (x6 c3 ), c32 + y15 c3 = b3 b¯3 + b3 x¯15 and y15 c3 = b3 x¯15 . Now by (2) it follows that (x¯ 15 b5 , x10 ) = (b3 x10 , x15 ) = 1. So 1 = (y15 c3 , x10 ) = (y15 , c3 x10 ).

(7)

(c3 x10 , b6 ) = (c3 b6 , x10 ) = 1.

(8)

By 3) in Lemma 2.26,

2 ) = (b b¯ , x 2 ) = By (2) and 1) in Lemma 2.25, we have (c3 x10 , c3 x10 ) = (c32 , x10 3 3 10 (b3 x10 , b3 x10 ) = 3. Therefore by (7) and (8), the following equation holds:

c3 x10 = b6 + y15 + y9 ,

(9)

where y9 ∈ B. By the associative law and 3), 5), 6) in Lemma 2.26 and 2) in Lemma 2.25 and (9), we have that (x¯6 b3 )c3 = x¯6 (b3 c3 ) and b8 c3 + x10 c3 = x¯6 b¯3 + x¯62 , b6 + c3 + y15 + b6 + y15 + y9 = c3 + y15 + x¯62 , so x62 = 2b6 + y15 + y9 .

(10)

By the associative law and 3), 4), 6) in Lemma 2.26 and Hypothesis 2.4 and (9), we have (x¯6 b3 )c3 = (x¯6 c3 )b3 . Hence b8 c3 + x10 c3 = b32 + x15 b3 , b6 + c3 + y15 + b6 + y15 + y9 = c3 + b6 + x15 b3 , so x15 b3 = 2y15 + b6 + y9 .

(11)

By the associative law and 2), 5), 6) in Lemma 2.26 and (11) and the Hypothesis 2.4, b3 (b8 b3 ) = b8 (b32 ) holds, from which we conclude x6 b3 + b32 + x15 b3 = b8 c3 + b8 b6 , c3 + y15 + c3 + b6 + 2y15 + b6 + y9 = b6 + c3 + y15 + b8 b6 , so b8 b6 = 2y12 + b6 + c3 + y9 .

(12)

By the associative law and 1), 2) in Lemma 2.26 and (4) and 2) in Lemma 2.25: (b6 b¯3 )b8 = b¯3 (b6 b8 ),

b3 b8 + x15 b8 = 2y12 b¯3 + b6 b¯3 + c3 b¯3 + y9 b¯3 .

114


Lemma 2.28 Let (A, B) satisfy Hypothesis 2.4 and (b8 b3 , b3 b8 ) = 3, then Lemma 2.27 holds and x9 c3 = x¯15 + d3 + d9 , d3 = c3 , b3 , b¯3 . Proof By 3) in Lemma 2.27, (c3 x9 , x¯15 ) = (x9 , c3 x¯15 ) = 1, so c3 x9 = x¯15 + α,

(1)

where α is a linear combination at B. We shall show that α has neither degree 4 nor two elements of degree 3. If x4 ∈ α, then 1 ≤ (c3 x9 , x4 ) = (x9 , c3 x4 ), hence 1 = (x9 , c3 x4 ). So c3 x4 = x9 + m3 .

(2)

So (c3 m3 , x4 ) = (m3 , c3 x4 ) = 1, hence c3 m3 = y4 + k5 .

(3)

Hence 2 = (c3 m3 , c3 m3 ) = (c32 , m ¯ 3 m3 ), therefore m ¯ 3 m3 = 1 + b8 . So by the associative law and (3) and 6) in Lemma 2.26, ¯ 3 (m3 c3 ), c3 + c3 b8 = y4 m ¯ 3 + k5 m ¯ 3 . Hence we get (m ¯ 3 m3 )c3 = m ¯ 3 + k5 m ¯ 3. 2c3 + b6 + y15 = y4 m

(4)

By (3) (y4 m ¯ 3 , c3 ) = (y4 , m3 c3 ) = 1,

(k5 m ¯ 3 , c3 ) = (k5 , m3 c3 ) = 1;

hence by (4), we get a contradiction. If x3 , y3 ∈ α, then by (1), 1 = (c3 x9 , x3 ) = (x9 , c3 x3 ), 1 = (c3 x9 , y3 ) = (x9 , c3 y3 ). So c3 x3 = x9 ,

c3 y3 = x9 ,

hence 1 = (c3 x3 , c3 y3 ) = (c32 , x¯3 y3 ). Therefore by 1) in Lemma 2.25, x¯3 y3 = b8 + 1, so x3 = y3 ; then by (1) 2 = (c3 x9 , x3 ) = (x9 , c3 x3 ), a contradiction. Therefore by (1) one of the following holds: a) α = d3 + d9 or b) α = d5 + d7 or c) α = d6 + k6 or d) α = 2d6 or e) α = d12 .


115

We shall show that a) is the only possibility. Assume b) holds, then by (1) we get c3 x9 = x¯15 + d5 + d7 . So (c3 d5 , x9 ) = (d5 , c3 x9 ) = 1. Therefore c3 d5 = x9 + β, where β = 2x3 or β = x3 + y3 or β = m6 . We shall show that these possibilities are impossible if β = 2x3 , then c3 d5 = x9 + 2x3 . Therefore (c3 x3 , d5 ) = (x3 , c3 d5 ) = 2, so c3 x3 = 2d5 , a contradiction. If β = x3 + y3 , then c3 d5 = x9 + x3 + y3 . Hence (c3 x3 , d5 ) = (x3 , c3 d5 ) = 1, (c3 y3 , d5 ) = (y3 , c3 d5 ) = 1, so 1 ≤ (c3 x3 , c3 y3 ) = (c32 , x¯3 y3 ). Then by 1) in Lemma 2.25 x3 = y3 a contradiction. If β = m6 , then c3 d5 = x9 + m6 , so by (1) and (6), we have (c32 )d5 = c3 (c3 d5 ), d5 + b8 d5 = x9 c3 + m6 c3 and b8 d5 = x¯15 + d7 + m6 c3 . So (b8 x15 , d¯5 ) = (b8 d5 , x¯15 ) = 1, hence by 10) in Lemma 2.27: (y9 b¯3 , d¯5 ) = 1 and by 8) in Lemma 2.27: (y9 b¯3 , x15 ) = (y9 , b3 x15 ) = 1, so y9 b¯3 = x15 + d¯5 + k7 , where k7 is a linear combination. When k7 = e3 + m4 , then y9 b¯3 = x15 + d¯5 + e3 + t3 , hence (b3 m4 , y9 ) = (m4 , b¯3 y9 ) = 1, so b3 m4 = y9 + t3 , hence (t3 b¯3 , m4 ) = (t3 , b3 m4 ) = 1, therefore t3 b¯3 = m4 + t5 ,

(5)

so 2 = (t3 b¯3 , t3 b¯3 ) = (t3 t¯3 , b3 b¯3 ), so by Hypothesis 2.4 t¯3 t3 = 1 + b8 . Hence by the associative law and (5) we obtain: (t¯3 t3 )b3 = (t¯3 b3 )t3 ,

b3 + b 8 b 3 = m ¯ 4 t3 + t¯5 t3 .

By 2) in Lemma 2.26, it follows that 2b3 + x6 + x15 = m ¯ 4 t3 + t¯5 t3 , so m ¯ 4 t3 = ¯ 4 t3 , m ¯ 4 t3 ) = (m ¯ 4 m4 , t¯5 t3 ), and then m ¯ 4 m4 = 1 + 4b8 , a 2b3 + x6 . Hence 5 = (m contradiction. If y9 b¯3 = x15 + d¯5 + k7 ,

(6)

where k7 ∈ B, then (b3 d¯5 , y9 ) = (d¯5 , b¯3 y9 ) = 1; hence b3 d¯5 = y9 + γ , where γ = v6 or γ = v3 + w3 . When b3 d¯5 = y9 + v3 + w3 , then (b3 v¯3 , d5 ) = (b3 d¯5 , v3 ) = 1. Also ¯ 3 , d5 ) = 1, so 1 ≤ (b3 v¯3 , b3 m ¯ 3 ) = (b3 b¯3 , v3 m ¯ 3 ), by Hypothesis 2.4, v3 = w3 , (b3 m then 2 = (b3 d¯5 , v3 ) = (d¯5 , b¯3 v3 ), so b¯3 v3 = 2d¯5 , a contradiction. When b3 d¯5 = y9 + v6 , then by the associative law and (6) and our assumption: (b32 )d¯5 = b3 (b3 d¯5 ), d¯5 c3 + d¯5 b6 = y9 b3 + v6 b3 , ¯ 6 + d¯5 b6 = x¯15 + d5 + k¯7 + v6 b3 . x¯9 + m Hence ¯ 6 + t3 , v6 b3 = x¯9 + m

(7)

116


so (t3 b¯3 , v6 ) = (t3 , b3 v6 ) = 1, hence t3 b¯3 = v6 + p3 . Therefore 2 = (b¯ 3 t3 , b¯3 t3 ) = (t3 t¯3 , b3 b¯3 ), Now by the Hypothesis 2.4 we get t3 t¯3 = 1 + b8 .

(8)

Again by the associative law: b3 (t3 t¯3 ) = t3 (b3 t¯3 ),

b3 + b3 b8 = t3 v¯6 + t3 p¯ 3 ;

and by 2) in Lemma 2.26, we have 2b3 + x6 + x15 = t3 v¯6 + t3 p¯ 3 , but (t3 v¯6 , b3 ) = (t3 b¯3 , v6 ) = 1, (t3 p¯3 , b3 ) = (t3 b¯3 , p3 ) = 1. This means v¯ 6 t3 = b3 + x15 , t3 p¯ 3 = b3 + x6 . Then by (7) and (8), it follows that 2 = (v¯6 t3 , v¯6 t3 ) = (v¯6 v6 , t¯3 t3 ) = (v¯6 v6 , b¯3 b3 ) = (v6 b3 , v6 b5 ) = 3, a contradiction. Assume c) holds. Then by (1) one has that c3 x9 = x¯15 + d6 + k6 , so (c3 d6 , x9 ) = (d6 , c3 x9 ) = 1; hence c3 d6 = x9 + γ , where γ = m9 or γ = m3 + v3 + k3 or γ = v3 + y6 or γ = v4 + v5 or γ = 2v3 + m3 or γ = 3v3 . If γ = m9 , then c3 d6 = x9 + m9 , so by the associative law and 1) in Lemma 2.26: (c32 )d6 = (c3 d6 )c3 , d6 + d6 b8 = x9 c3 + m9 c3 , d6 + d6 b8 = x¯ 15 + d6 + k6 + m9 c3 , d6 b8 = x¯ 15 + k6 + m9 c3 .

(9)

We shall prove that d6 = x6 , x¯6 , k6 = x6 , x¯6 . By 4) in Lemma 2.26 and our assumption if d6 = x6 , 1 = (c3 x9 , x6 ) = (x9 , c3 x6 ) = 0,

1 = (c3 x9 , x¯6 ) = (c3 x6 , x¯9 ) = 0,

a contradiction, so d6 = x6 , x¯6 . In the same way k6 = x6 , x¯6 . By (9), (b8 x15 , d¯6 ) = (b8 d6 , x¯15 ) = 1, in the same way (b8 x15 , k¯6 ) = (b8 k6 , x¯15 ) = 1, then by 8), 10) in Lemma 2.27, (y9 b¯3 , x15 ) = (y9 , b3 x15 ) = 1, (y9 b¯3 , d¯6 ) = 1, (y9 b¯3 , k¯6 ) = 1. So y9 b¯3 = x15 + d¯6 + k¯6 .

(10)

So by the associative law and 2) in Lemma 2.25: (b¯3 c3 )y9 = c3 (b¯3 y9 ), b3 y9 + y9 x6 = x15 c3 + d¯6 c3 + k¯6 c3 , x¯15 + d6 + k6 + y9 x6 = 2x¯15 + x¯6 + x¯9 + x¯9 + m ¯ 9 + k¯6 c3 , d6 + k6 + y9 x6 = x¯15 + x¯6 + 2x¯9 + m ¯ 9 + k¯6 c3 . And by 3) in Lemma 2.27, (10) and the fact that c3 d6 = x9 + m9 , since k6 , d6 = x6 , then 1 ≤ (k¯6 c3 , d6 ) = (k¯6 , c3 d6 ), a contradiction to the assumption that c3 d6 = x9 + m9 . If γ = m3 + v3 + k3 , then c3 d3 = x9 + m3 + v3 + k3 , so (c3 m3 , d3 ) =


117

(m3 , c3 d3 ) = 1, (c3 v3 , d3 ) = (v3 , c3 d3 ) = 1. Hence 1 ≤ (c3 m3 , c3 v3 ) = (m3 v¯3 , c32 ), then by 1) in Lemma 2.25 m3 = v3 , a contradiction. If γ = v3 + y6 , then c3 d6 = x9 + v3 + y6 , so (c3 v3 , d6 ) = (v3 , c3 d6 ) = 1. Hence c3 v3 = d6 + t3 , so by the associative law and 1) in Lemma 2.25, we obtain equations c32 v3 = c3 (c3 v3 ) and v3 + v3 b8 = d6 c3 + t3 c3 , v3 + v3 b8 = x9 + v3 + y9 + t3 c3 . Hence v3 b8 = x9 + y6 + t3 c3 . Since (t3 c3 , v3 ) = (t3 , c3 v3 ) = 1, then (v3 b8 , v3 b8 ) ≥ 4. On the other hand, 2 = (c3 v3 , c3 v3 ) = (c32 , v¯ 3 v3 ), so by 1) in Lemma 2.25, v¯3 v3 = 1 + b8 = b¯3 b3 . Therefore (v3 b8 , v3 b8 ) = (b82 , v3 v¯3 ) = (b82 , b¯3 b3 ) = (b8 b3 , b8 b3 ), by 2) in Lemma 2.26, (b8 b3 , b8 b3 ) = 3, a contradiction. If γ = v4 +v5 , then c3 d6 = x4 + v4 + v5 , so by the associative law and 1) in Lemma 2.25: c32 d6 = c3 (c3 d6 ), d6 + d6 b8 = x9 c3 + v4 c3 + v5 c3 . Hence by our assumption we have d6 + d6 b8 = x¯15 + d6 + k6 + v4 c3 + v5 c3 , thus d6 b8 = x¯15 + k6 + v4 c3 + v5 c3 .

(11)

We shall show that d6 = x6 , x¯6 , k6 = x6 , x¯6 . By 4) in Lemma 2.26, if d6 = x6 , or d6 = x¯6 , then 1 = (c3 x9 , x6 ) = (x9 , c3 x6 ) = 0, 1 = (c3 x9 , x¯6 ) = (c3 x6 , x¯9 ) = 0, a contradiction. In the same way k6 = x6 , x¯6 . By (12) and 10) in Lemma 2.27, (b8 x15 , d¯6 ) = (b8 d6 , x¯15 ) = 1. In the same way (b8 x15 , k¯6 ) = 1 and by 10), 8) in Lemma 2.27 (y9 b¯3 , k¯6 ) = 1, (y9 b¯3 , d¯6 ) = 1, (y9 b¯3 , x15 ) = 1, so y9 b¯3 = x15 + k¯6 + d¯6 . Hence by the associative law an 2) in Lemma 2.25: (b¯3 c3 )y9 = (b¯3 y9 )c3 , b3 y9 + x6 y9 = x15 c3 + k¯6 c3 + d¯6 c3 , so by 3) in Lemma 2.27 and our assumption: x¯15 + k6 + d6 + x6 y9 = 2x¯15 + x¯6 + x¯ 9 + k¯6 c3 + d¯6 c3 , k6 + d6 + x6 y9 = x¯15 + x¯6 + x¯9 + v¯ 4 + v¯5 + x¯9 + k¯6 c3 . Hence 1 = (k¯6 c3 , d6 ) = (k¯6 , c3 d6 ), a contradiction to our assumption. If γ = 2v3 + m3 , or γ = 3v3 , then c3 d6 = x9 + 2v3 + m3 or c3 d6 = x9 + 3v3 . In two cases, it always follows that (c3 v3 , d6 ) = (v3 , c3 d6 ) ≥ 2, a contradiction. Assume then by (1) c3 x9 = x¯15 + 2d6 . Hence (c3 d6 , x9 ) = (d6 , c3 x9 ) = 2, therefore c3 d6 = 2x9 . So by the associative law and 1) in Lemma 2.25, we have that (c32 )d6 = c3 (c3 d6 )

118


d6 + d6 b8 = 2x9 c3 = 2x¯15 + 4d6 , d6 b8 = 2x¯15 + 3d6 . So (b8 x15 , d¯6 ) = (b8 d6 , x¯15 ) = 2.

(12)

By 4) in Lemma 2.26, if x6 = d6 , then 2 = (c3 x9 , x6 ) = (x9 , c3 x6 ) = 0, then d6 = x6 , also d6 = x¯6 . Then by (12) and 8), 10) in Lemma 2.27: (y9 b¯3 , x15 ) = (y9 , b3 x15 ) = 1, (y9 b¯3 , d¯6 ) = 2. So y9 b¯3 = x15 + 2d¯6 . Hence by the associative law and 2) in Lemma 2.25: y9 (b¯3 c3 ) = (y9 b¯3 )c3 , y9 b3 + y9 x6 = x15 c3 + 2d¯6 c3 , then by 3) in Lemma 2.27, x¯15 + 2d6 + y9 x6 = 2x¯15 + x¯6 + x¯9 + 2x¯9 , a contradiction. Assume e) then by 1), c3 x9 = x¯15 + d12 . Then by 1) in Lemma 2.25, 2 = (x9 c3 , x9 c3 ) = (x9 x¯9 , c32 ) = (x9 x¯9 , b3 b¯3 ) = (x9 b¯3 , x9 b¯3 ). So by 1) in Lemma 2.27, (x9 b¯3 , x10 ) = (x9 , b3 x10 ) = 1, hence x9 b¯3 = x10 + x17 (13). By the associative law and 1) in Lemma 2.25: (c32 )x9 = (b3 b¯3 )x9 = b3 (b¯3 x9 ) = (c3 x9 )c3 , x10 b3 + b3 x17 = x¯15 c3 + d12 c3 . By 1), 3) and Lemma 2.27, x15 + x6 + x9 + b3 x17 = 2x15 + x6 + x9 + d12 c3 , b3 x17 = x15 + d12 c3 . Then (x15 b¯3 , x17 ) = (x15 , b3 x17 ) = 1 and by 2) in Lemma 2.26 and 1), 8) in Lemma 2.27: (b¯3 x15 , b8 ) = (x15 , b3 b8 ) = 1, (b¯3 x15 , x10 ) = (x15 , b3 x10 ) = 1, (b¯3 x15 , b¯3 x15 ) = (x15 b3 , x15 b3 ) = 6. Therefore the only possibility is b¯3 x15 = b8 + x10 + x17 + m3 + z3 + z4 . So (b3 m3 , x15 ) = (m3 , b¯3 x15 ) = 1, a contradiction. Hence the only possibility is that α = d3 + d9 , hence by 1) x9 c3 = x¯ 15 + d3 + d9 . If d3 = c3 , then (c32 , x9 ) = (c3 , c3 x9 ) = 1, a contradiction to 1) in Lemma 2.25. If d3 = b3 , then (c3 b3 , x9 ) = (b3 , c3 x9 ) = 1 a contradiction to 2) in Lemma 2.25 in the same way. If d3 = b¯3 , we get a contradiction so d3 = c3 , b3 , b¯3 . Lemma 2.29 Let (A, B) satisfy Hypothesis 2.4 and (b8 b3 , b8 b3 ) = 3, then Lemma 2.28 holds and we obtain the following: 1) 2) 3) 4) 5) 6)

d3 c3 = x9 , x9 = x¯9 , d3 b8 = x¯15 + x¯9 , b3 d¯3 = y9 , y9 = y¯9 , y9 = x9 , x9 c3 = x¯15 + d3 + x¯9 , y9 b¯3 = d¯3 + x9 + x15 , d¯3 b6 = d3 + x¯15 ,


7) 8) 9) 10) 11) 12) 13) 14) 15) 16) 17)

119

b8 x15 = 5x15 + 3x9 + 2x6 + b3 + d¯3 , x9 b8 = 3x15 + 2x9 + x6 + d¯3 , y15 x6 = 4x¯15 + 2x¯9 + x¯6 + b¯3 + d3 , b6 x15 = 4x¯15 + x¯6 + 2x¯9 + b¯3 + d3 , d3 y15 = 2x15 + x6 + x9 , x9 x15 = 6x¯15 + 2x¯6 + 3x¯9 + b¯ 3 + d3 , d3 b8 = x¯15 + x¯9 , d¯3 d3 = 1 + c8 , c8 = b8 , d32 = b6 + y3 , d3 = c3 , b3 , b¯3 , y3 = b3 , c3 , b¯3 , b3 d3 = z9 , z9 = y9 , x9 , x¯9 , z9 = z¯ 9 , y32 = 1 + c8 .

Proof By Lemma 2.28, (c3 d3 , x9 ) = (d3 , c3 x9 ) = 1, hence d3 c 3 = x 9 .

(1)

By the associative law and 1) in Lemma 2.28, we have d3 (c32 ) = x9 c3 and d3 + d3 b8 = x¯15 + d3 + d9 , which leads to d3 b8 = x¯15 + d9 .

(2)

So (b8 x15 , d¯3 ) = (b8 d3 , x¯15 ) = 1 and by 10) in Lemma 2.27, (y9 b¯3 , d¯3 ) = 1 = (y9 , b3 d¯3 ). So b3 d¯3 = y9 .

(3)

By the associative law (b¯3 b3 )d¯3 = b¯3 (b3 d¯3 ) and Hypothesis 2.4, one has d¯3 + d¯3 b8 = y9 b¯3 . By (2), we get y9 b¯3 = d¯3 + d¯9 + x15 .

(4)

By the associative law and Hypothesis 2.4 and (3) we have (b32 )d¯3 = (b3 d¯3 )b3 = y9 b3 , d¯3 c3 + d¯3 b6 = y9 b3 = d3 + d9 + x¯15 . So d¯3 c3 = d9 ,

(5)

d¯3 b6 = d3 + x¯15 .

(6)

Hence by (1) we obtain d¯9 = x9 . By Lemma 2.28, we may assume x9 c3 = x¯15 + d3 + x¯9 . By 10) in Lemma 2.27 and (4) we get b8 x15 = 5x15 + 3x9 + 2x6 + b3 + d¯3 . By the associative law and 1) in Lemma 2.25 and Lemma 2.28: (c32 )x9 = (c3 x9 )c3 ,

(7)

120


x9 + x9 b8 = x¯15 c3 + d3 c3 + x¯9 c3 . By (1) and 3) in Lemma 2.27 and Lemma 2.28, we obtain x9 + x9 b8 = 2x15 + x6 + x9 + x15 + d¯3 + x9 , so x9 b8 = 3x15 + 2x9 + x6 + d¯3 .

(8)

By the associative law and 4) in Lemma 2.27 and 2) in Lemma 2.25, we have that y15 (b¯3 c3 ) = (y15 b¯3 )c3 , y15 b3 + y15 x6 = 2x15 c3 + x6 c3 + x9 c3 . Moreover by 4) in Lemma 2.26 and 3), 4) in Lemma 2.27 and Lemma 2.28, we get y15 b3 + y15 x6 = 4x¯15 + 2x¯6 + 2x¯9 + b¯3 + x¯15 + x¯15 + d3 + x¯9 , then y15 x6 = 4x¯15 + 2x¯9 + x¯6 + b¯3 + d3 .

(9)

By the associative law and 8) in Lemma 2.27 and Hypothesis 2.4, we obtain (b32 )x15 = (b3 x15 )b3 , x15 c3 + b6 x15 = 2y15 b3 + b6 b3 + y9 b3 , By 3), 4) in Lemma 2.27 and 10) in Lemma 2.26 and (4), we have 2x¯15 + x¯6 + x¯9 + b6 x15 = 4x¯15 + 2x¯6 + 2x¯9 + b¯3 + x¯15 + d3 + x¯9 + x¯15 , so b6 x15 = 4x¯15 + x¯6 + 2x¯ 9 + b¯3 + d3 . By the associative law and 6) in Lemma 2.26 and (1), (6) and (8), we have (c3 b8 )d3 = (c3 d3 )b8 = x9 b8 . Hence b6 d3 + c3 d3 + y15 d3 = 3x15 + 2x9 + x6 + d¯3 , d¯3 + x15 + x9 + y15 d3 = 3x15 + 2x9 + x6 + d¯3 , y15 d3 = 2x15 + x6 + x9 . So by the associative law and (1), it follows that (d3 c3 )y15 = c3 (d3 y15 ) and x9 y15 = 2x15 c3 + x9 c3 + x6 c3 . Hence by 3) in Lemma 2.27 and 4) in Lemma 2.26, we obtain that x9 c3 = x¯15 + d3 + x¯9 ,

x9 x15 = 6x¯15 + 2x¯6 + 3x¯9 + b¯3 + d3 .

By the associative law 1) in Lemma 2.25 and Lemma 2.28 and (1), one has that d3 (c32 ) = (d3 c3 )c3 , d3 + d3 b8 = c3 x9 = x¯15 + d3 + x¯9 , and d3 b8 = x¯15 + x¯9 . Then we have by (6) that (d32 , b6 ) = (d3 , d¯3 b6 ) = 1, so d32 = b6 + y3 ,

(10)

hence (d3 d¯3 , d3 d¯3 ) = (d32 , d32 ) = 2. Now by (1) and 1) in Lemma 2.25, we get (d3 c3 , d3 c3 ) = (d¯3 d3 , c32 ) = 1. Hence d3 d¯3 = 1 + c8 ,

(11)


121

where c8 = b8 by the main theorem in [CA], so d3 = c3 , b3 , b¯3 . Also y3 = c3 , b3 , b¯3 , since if y3 = c3 , then 1 = (d32 , c3 ) = (d¯3 c3 , d3 ); therefore (d¯3 d3 , c3 d¯3 ) = (d¯3 c3 , d¯3 c3 ) ≥ 2, a contradiction. In the same way if y3 = b3 , b¯3 . If x9 = x¯9 , then by (1) we have 1 = (c3 d3 , c3 d¯3 ) = (d¯32 , c32 ), a contradiction to 1) in Lemma 2.25 and that d32 = b6 + y3 , so x9 = x¯9 . By (11) and Hypothesis 2.4, it holds that 1 = (d¯3 d3 , b¯3 b3 ) = (d3 b3 , d3 b3 ), hence d3 b3 = z9 . By (10) and Hypothesis 2.4, we obtain 1 = (d32 , b¯32 ) = (d3 b3 , d¯3 b¯3 ), so z9 = z¯ 9 , hence z9 = x9 , x¯9 . If z9 = y9 , then by (3) we have 1 = (d3 b3 , b3 d¯3 ) = (d32 , b¯3 b3 ), a contradiction to (10) and Hypothesis 2.4; hence z9 = y9 . By the associative law and (11), (6) and (10): (d3 b6 )d¯3 = b6 (d3 d¯3 ), d¯32 + x15 d¯3 = b6 + c8 b6 , then c8 b6 = y¯3 + x15 d¯3 since c8 , b6 are reals and (x15 d¯3 , y3 ) = (x15 , d3 y3 ) = 0. So y3 = y¯3 , by (10), (d¯3 y3 , d3 ) = 1, hence 2 ≤ (d¯3 y3 , d¯3 y3 ) = (d3 d¯3 , y32 ), so y32 = 1 + c8 .

Lemma 2.30 Let (A, B) satisfy Hypothesis 2.4 and (b8 b3 , b8 b3 ) = 3, then Lemma 2.29 holds and we obtain the following: 1) 2) 3) 4) 5) 6) 7) 8) 9) 10)

b82 = 1 + 2b8 + 2x10 + z4 + c8 + b5 + c5 , x¯15 b3 = x10 + b8 + c8 + z9 + b5 + c5 , either b5 , c5 are reals or b5 = c¯5 , b62 = 1 + b8 + z9 + c8 + b5 + c5 , x¯ 6 x6 = 1 + b8 + c8 + z9 + b5 + c5 , 2 = 1 + 2b + 2x + 2c + 2b + 2c + 3z , x10 8 10 8 5 5 9 x10 x¯6 = b¯3 + 3x¯15 + d3 + x¯9 , b8 x10 = 2b8 + x10 + x¯9 b3 + c8 + z9 + b5 + c5 , c3 y15 = x10 + b8 + c8 + z9 + b5 + c5 , c3 b5 = y15 , c3 c5 = y15 .

Proof First we note that either b5 and c5 are reals or b5 = c¯5 holds. But in these two cases, it always follows that b5 + c5 = b¯5 + c¯5 = b5 + c5 . By 2) in Lemma 2.26 and 1), 8) in Lemma 2.27, we have (x¯15 b3 , x¯15 b3 ) = 6, (x¯15 b3 , x10 ) = (b3 x10 , x15 ) = 1, (x¯15 b3 , b8 ) = (b3 b8 , x15 ), then x¯15 b3 = x10 + b8 + x + y + z + w,

(1)

122


where |x + y + z + w| = 27. Here the case x¯15 b3 = x10 + b8 + 2x is impossible since |2x| = 27 implies that |x| = 13.5. But x is an integer, a contradiction. By the associative law (b¯3 b8 )b3 = b8 (b¯3 b3 ) and Hypothesis 2.4 and 2), 3) in Lemma 2.26, we get x¯6 b3 + b¯3 b3 + x¯15 b3 = b8 + b82 ,

b8 + x10 + 1 + b8 + x¯15 b3 = b8 + b82 ,

1 + x10 + b8 + x¯15 b3 = b82 . Thus b82 = 1 + 2b8 + 2x10 + x + y + z + w.

(2)

So by 4) in Lemma 2.25 and 3) in Lemma 2.26, b62 = 1 + b8 + x + y + z + w.

(3)

By the associative law (b¯3 c3 )(b3 c3 ) = (b¯ 3 b3 )c32 and 1), 2) in Lemma 2.25 and Hypothesis 2.4 and 3) in Lemma 2.26: (b3 + x6 )(b¯3 + x¯6 ) = (1 + b8 )2 = 1 + 2b8 + b82 , b3 b¯3 + b3 x¯6 + b¯3 x6 + x6 x¯6 = 1 + 2b8 + b82 . So by (2) and (1) and 3) in Lemma 2.26: x6 x¯6 = 1 + b8 + x + y + z + w.

(4)

By the associative law (x¯6 b8 )b3 = (x¯6 b3 )b8 and 3) in Lemma 2.26 and 2) in Lemma 2.27: 2x¯ 15 b3 + b¯3 b3 + x¯6 b3 + x¯9 b3 = b82 + x10 b8 , then by 3) in Lemma 2.26 and (1) and (2) we have 2x10 + 2b8 + 2x + 2y + 2z + 2w + 1 + b8 + b8 + x10 + x¯9 b3 = 1 + 2b8 + 2x10 + x + y + z + w + x10 b8 , so 2b8 + x10 + x¯9 b3 + x + y + z + w = x10 b8 .

(5)

By the associative law (x¯62 )b3 = x¯6 (x¯6 b3 ) and 7) in Lemma 2.27 and 3) in Lemma 2.26, we get 2b6 b3 + y15 b3 + y9 b3 = x¯6 b8 + x¯6 x10 . Then by 1) in Lemma 2.26 and 2), 4) in Lemma 2.27 and 5) in Lemma 2.29, we have the equation: 2b¯ 3 + 2x¯15 + 2x¯15 + x¯6 + x¯9 + d3 + x¯15 + x¯9 = 2x¯15 + b¯3 + x¯6 + x¯9 + x10 x¯6 , so b¯3 + 3x¯15 + d3 + x¯9 = x10 x¯6 .

(6)


123

By the associative (x¯6 b3 )x10 = b3 (x¯6 x10 ) and 3) in Lemma 2.26, we get b8 x10 + 2 = b¯ b + 3x¯ b + d b + x¯ b . Then by (3) and Hypothesis 2.4 and (1) and 5) x10 3 3 15 3 3 3 9 3 in Lemma 2.29, it follows that 2 2b8 + x10 + x¯9 b3 + x + y + z + w + x10

= 1 + b8 + 3x10 + 3b8 + 3x + 3y + 3z + 3w + z9 + x¯9 b3 . Hence 2 x10 = 1 + 2b8 + 2x10 + z9 + 2x + 2y + 2z + 2w.

(7)

By 13) and 14) in Lemma 2.29, one has that (d3 d¯3 , b82 ) = (d3 b8 , d3 b8 ) = 2, hence we obtain by (2) that (b82 , c8 ) = 1. Without loss of generality, say x = c8 . By (4) and 2 , x x¯ ) = (x x¯ , x x¯ ). But by (6), (x x¯ , x x¯ ) = 12 (7), it holds that 11 ≤ (x10 6 6 10 6 10 6 10 6 10 6 holds, thus we may assume y = z9 . By (1), (b¯3 α, x¯15 ) = (α, b3 x¯15 ) = 1 holds, where α ∈ {x, y, z, w}. So |α| ≥ 5. Hence z = b5 and w = c5 . Hence by (1), (3), (4), (5), (6) and (7), we obtain x¯ 15 b¯3 = x10 + b8 + c8 + z9 + b5 + c5 , b82 = 1 + 2b8 + 2x10 + c8 + z9 + b5 + c5 , b62 = 1 + b8 + c8 + z9 + b5 + c5 , x6 x¯6 = 1 + b8 + c8 + z9 + b5 + c5 , 2 x10 = 1 + 2b8 + 2x10 + 3z9 + 2c8 + 2b5 + 2c5 ,

b8 x10 = 2b8 + x10 + x¯9 b3 + c8 + z9 + b5 + c5 . So by the associative law (b¯3 x¯6 )c3 = x¯ 6 (b¯3 c3 ) and 1), 2) in Lemma 2.25 and 3), 5) in Lemma 2.26: c32 + c3 y15 = b3 x¯6 + x6 x¯6 , 1 + b8 + c3 y15 = b8 + x10 + 1 + b8 + c8 + z9 + b5 + c5 , c3 y15 = x10 + b8 + c8 + z9 + b5 + c5 . So (c3 b5 , y15 ) = (b5 , c3 y15 ) = 1, (c3 c5 , y15 ) = (y15 c3 , c5 ) = 1, hence c3 b5 = y15 , c3 c5 = y15 . Lemma 2.31 Let (A, B) satisfy Hypothesis 2.4 and (b8 b3 , b8 b3 ) = 3, then Lemma 2.30 holds and we obtain the following: 1) 2) 3) 4) 5)

b3 z9 = x9 + d¯3 + x15 , x¯9 b3 = z9 + x10 + c8 , b8 x10 = 2b8 + 2x10 + 2z9 + 2c8 + b5 + c5 , b3 c8 = x9 + x15 , d3 x6 = x10 + c8 ,

124

6) 7) 8) 9)


d3 b6 = d¯3 + x15 , d3 c8 = x¯15 + x¯6 + d3 , y3 d¯3 = x¯6 + d3 , d3 x¯6 = y15 + y3 .

Proof By the associative law (b32 )d3 = b3 (b3 d3 ) and 6), 4) in Lemma 2.29 and Hypothesis 2.4, we get c3 d3 + b6 d3 = b3 z9 . Then by 1), 6) in Lemma 2.29 it follows that x9 + d¯3 + x15 = b3 z9 .

(1)

Then (b¯3 x9 , z9 ) = (b3 z9 , x9 ) = 1. And by 1) in Lemma 2.27, (x¯9 b3 , x10 ) = (b3 x10 , x9 ) = 1 holds. It follows by 4) in Lemma 2.29 and 1) in Lemma 2.25 and Hypothesis 2.4 that (x¯9 b3 , x¯9 b3 ) = (x¯ 9 x9 , b¯3 b3 ) = (x¯9 x9 , c32 ) = (x9 c3 , x9 c3 ) = 3. Hence x¯ 9 b3 = z9 + x10 + d8 .

(2)

By 1) and 5) in Lemma 2.30, we get 2 (b82 , x10 ) = 18 = (b8 x10 , b8 x10 ).

(3)

By 7) in Lemma 2.30 and (2), we have b8 x10 = 2b8 + 2x10 + 2z9 + c8 + b5 + c5 + d8 , then by (3) we obtain that c8 = d8 , so b8 x10 = 2b8 + 2x10 + 2z9 + 2c8 + b5 + c5 .

(4)

Now we have (b3 c8 , x9 ) = (b3 x¯9 , c8 ) = 1 by (2) and (b3 c8 , x15 ) = (b3 x¯15 , c8 ) = 1 by 2) in Lemma 2.30, so b3 c8 = x9 + x15 .

(5)

By the associative law d3 (c3 b¯3 ) = (d3 c3 )b¯3 and 1) in Lemma 2.29 and 2) in Lemma 2.25 and (2), we see that d3 (c3 b¯3 ) = (d3 c3 )b¯3 , d3 b3 + d3 x6 = x9 b¯3 = z9 + x10 + c8 . Hence by 14) in Lemma 2.29 we have d3 x6 = x10 + c8 .

(6)

By the associative law (b32 )d3 = (b3 d3 )b3 and 14) in Lemma 2.29 and (1) and Hypothesis 2.4, it follows d3 c3 + d3 b6 = z9 b3 = x9 + d¯3 + x15 . Hence by 1) in Lemma 2.29 it holds that: d3 b6 = d¯3 + x15 .

(7)

By the associative law (d¯3 d3 )d3 = d¯3 (d32 ) and 12) and 13) in Lemma 2.29, we have d3 + d3 c8 = b6 d¯3 + y3 d¯3 , and get d3 c8 = x¯15 + y3 d¯3 by (7) and (d3 c8 , x¯6 ) = (d3 x6 , c8 ) = 1 by (6). By 12) in Lemma 2.29, (d3 c8 , d3 ) = (c8 , d¯3 d3 ) = 1 holds, hence d3 c8 = x¯15 + x¯6 + d3 .

(8)


125

Therefore y3 d¯3 = x¯ 6 + d3 .

(9)

Moreover (d3 x¯6 , y3 ) = (x¯6 , d¯3 y3 ) = 1, and by 11) in Lemma 2.29, it follows that (d3 x¯6 , y15 ) = 1, so d3 x¯6 = y15 + y3 . Lemma 2.32 Let (A, B) satisfy Hypothesis 2.4 and (b8 b3 , b8 b3 ) = 3, then Lemma 2.31 holds and we obtain the following: 1) 2) 3) 4) 5) 6) 7) 8) 9) 10) 11) 12) 13)

x9 b3 = y9 + y15 + y3 , x9 d¯3 = y15 + c3 + y9 , x15 d¯3 = y15 + y6 + y9 , c8 b6 = 2y15 + y3 + y9 + b6 , z9 b6 = 2y15 + 2y9 + b6 , x10 b6 = 3y15 + y3 + c3 + y9 , y3 b6 = c8 + x10 , d3 x15 = b8 + z9 + b5 + c5 + x10 + c8 , y15 b6 = 2b8 + 3x10 + 2z9 + 2c8 + b5 + c5 , y9 b6 = b8 + 2z9 + b5 + c5 + x10 + c8 , x¯15 x15 = 5b8 + 6x10 + 6z9 + 5c8 + 3b5 + 3c5 + 1, b6 b5 = y15 + y9 + b6 , b3 y3 = x¯9 .

Proof By the associative law d3 (c3 b3 ) = (d3 c3 )b3 = x9 b3 and 1) in Lemma 2.29 and 2) in Lemma 2.25, we have that d3 b¯3 + d3 x¯6 = x9 b3 and by 3) in Lemma 2.29 that y9 + d3 x¯6 = x9 b3 . Finally, we obtain x9 b3 = y9 + y15 + y3 ,

(1)

by 9) in Lemma 2.31. By the associative law b3 (c8 d¯3 ) = d¯3 (b3 c8 ) and 4), 7) in Lemma 2.31, we get x15 b3 + b3 x6 + b3 d¯3 = x9 d¯3 + x15 d¯3 , and by 5) in Lemma 2.26 and 3) in Lemma 2.29 and 8) in Lemma 2.27, we get 2y15 +b6 +y9 +c3 +y15 = x9 d¯3 +x15 d¯3 , x9 d¯3 + x15 d¯3 = 3y15 + 2y9 + b6 + c3 . Now from 10), 11), 1) in Lemma 2.29, we see that (x9 d¯9 , c3 ) = (x9 , d3 c3 ) = 1 and (x9 d¯3 , y15 ) = (x9 , d3 y15 ) = 1 and (x9 d¯3 , b6 ) = (x9 , d3 b6 ) = 0, hence x9 d¯3 = y15 + c3 + y9 ,

(2)

x15 d¯3 = 2y15 + y9 + b6 .

(3)

which gives By the associative law (d¯3 d3 )b6 = d¯3 (d3 b6 ) and 6), 12) in Lemma 2.29, one has that b6 + c8 b6 = d¯32 + d¯3 x15 . And by (3) and 13) in Lemma 2.29, we obtain b6 + c8 b6 = b6 + y3 + 2y15 + y9 + b6 , hence c8 b6 = 2y15 + y3 + y9 + b6 .

(4)

126


By the associative law (d3 b3 )b6 = d3 (b3 b6 ) and 1) in Lemma 2.26 and 3) and 16) in Lemma 2.29, we obtain z9 b6 = d3 b¯3 + d3 x¯15 , and hence by (3) and 3) in Lemma 2.29, we have z9 b6 = 2y15 + 2y9 + b6 .

(5)

By the associative law (x6 b6 )d3 = b6 (x6 d3 ) and 5) in Lemma 2.27 and 5) in Lemma 2.31, we have 2x¯6 d3 + x¯15 d3 + x¯9 d3 = x10 b6 + c8 b6 . Then by 9) in Lemma 2.31 and (2), (3) and (4), it follows that 2y15 + 2y3 + 2y15 + y9 + b6 + y15 + c3 + y9 = x10 b6 + 2y15 + y3 + y9 + b6 . Hence x10 b6 = 3y15 + y3 + c3 + y8 .

(6)

Now we have by (4) (y3 b6 , c8 ) = (y3 , b6 c8 ) = 1, (y3 b6 , x10 ) = (y3 , b6 x10 ) = 1 by (4) and (6), from which we get y3 b6 = c8 + x10 .

(7)

By the associative law (d32 )b6 = d3 (d3 b6 ) and 15) in Lemma 2.29 and 6) in Lemma 2.31, we obtain b62 + y3 b6 = d3 d¯3 + x15 d3 . Moreover by 14) in Lemma 2.29 and 3) in Lemma 2.30 and (7), we have 1 + b8 + z9 + c8 + b5 + c5 + x10 + c8 = 1 + c8 + x15 d3 , b8 + z9 + b5 + c5 + x10 + c8 = x15 d3 .

(8)

By the associative law and 5) in Lemma 2.26 and 5) in Lemma 2.27: b3 (x6 b6 ) = b6 (b3 x6 ), 2x¯6 b3 + x¯15 b3 + x¯9 b3 = b6 c3 + y15 b6 . By 2) in Lemma 2.31 and 2) in Lemma 2.30 and 3) in Lemma 2.26: 2b8 + 2x10 + x10 + b8 + z9 + c8 + b5 + c5 + x10 + z9 + c8 = b8 + x10 + y15 b6 , y15 b6 = 2b8 + 3x10 + 2z9 + 2c8 + b5 + c5 .

(9)

By 3) in Lemma 2.29 and 1) in Lemma 2.26: (b3 b6 )d¯3 = (b3 d¯3 )b6 = y9 b6 , b¯3 d¯3 + x¯15 d¯3 = y9 b6 , by 16) in Lemma 2.29 and (8) b8 + 2z9 + b5 + c5 + x10 + c8 = y9 b6 .

(10)


127

By the associative law and 8) in Lemma 2.27 and 1) in Lemma 2.26: (b3 b6 )x15 = b6 (b3 x15 ), b¯3 x15 + x¯15 x15 = 2y15 b6 + b62 + y9 b6 , so by (10), (9) and 2), 3) in Lemma 2.30 x10 +b8 +z9 +c8 +b5 +c5 + x¯15 x15 = 4b8 +6x10 +4z9 +4c8 +2c5 +1+b8 +z9 +c8 +b5 +c5 +b8 +2z9 +c8 +x10 +b5 +c5 , x¯15 x15 = 1+5b8 +6x10 +6z9 +5c8 +3b5 +4c5 . By (9) and (10), we get (b6 b5 , y9 ) = (b6 y9 , b5 ) = 1,

(b6 b5 , y15 ) = (b5 , y15 b6 ) = 1

and by (3) (b5 b6 , b6 ) = (b5 , b62 ) = 1, so b5 b6 = y9 + y15 + b6 . By (1) (b3 y3 , x¯9 ) = (b3 x9 , y3 ) = 1, then b3 y3 = x¯9 .

Lemma 2.33 Let (A, B) satisfy Hypothesis 2.4 and (b8 b3 , b8 b3 ) = 3, then Lemma 2.32 holds and we obtain the following: 1) 2) 3) 4) 5) 6) 7) 8) 9) 10) 11) 12)

y9 d3 = b3 + x15 + x9 , z9 d3 = b¯3 + x¯15 + x¯9 , z9 y9 = 3y15 + 2b6 + 2y9 + c3 + y3 , c8 b8 = 2x10 + 2z9 + b8 + c8 + b5 + c5 , y9 c3 = z9 + x10 + c8 , x¯15 x6 = 2b8 + 3x10 + 2z9 + 2c8 + b5 + c5 , x10 d3 = x¯15 + x¯6 + x¯9 , d3 x9 = x10 + b8 + z9 , y3 y9 = z9 + b8 + x10 , b6 y9 = 2z9 + b8 + x10 + b5 + c5 + c8 , 2 = 9y + 4b + 6y + 2c + 2y , x15 15 6 9 3 3 x15 x6 = c3 + 4y15 + b6 + 2y9 + y3 .

Proof By the associative law and 13), 3) in Lemma 2.29: (b3 d¯3 )d3 = (b¯3 d3 )d3 = b¯ 3 (d32 ),

y9 d3 = b6 b¯3 + y3 b¯3 ,

128


then by 1) in Lemma 2.26 and 13) in Lemma 2.32, y9 d3 = b3 + x15 + x9 .

(1)

By the associative law and 16), 15) in Lemma 2.29: (d32 )b3 = (d3 b3 )d3 , b 6 b 3 + y 3 b 3 = z9 d 3 , then by 13) in Lemma 2.32 and 1) in Lemma 2.26, z9 d3 = b¯3 + x¯15 + x¯9 .

(2)

By 16) in Lemma 2.29 and (1): (b3 d3 )y9 = z9 y9 = (d3 y9 )b3 = b32 + x15 b3 + x9 b3 , by 8) in Lemma 2.27 and Hypothesis 2.4 and 1) in Lemma 2.32: c3 + b6 + 2y15 + b6 + y9 + y9 + y15 + y3 = z9 y9 , 3y15 + 2b6 + 2y9 + c3 + y3 = z9 y9 .

(3)

By the associative law and Hypothesis 2.4 and 4) in Lemma 2.31: (b3 b¯3 )c8 = b¯3 (b3 c8 ), c8 + c8 b8 = x9 b¯3 + x15 b¯3 , then by 2) in Lemma 2.31 and 2) in Lemma 2.30: c8 b8 = 2x10 + 2z9 + b8 + c8 + b5 + c5 .

(4)

By the associative law and 2) in Lemma 2.25 and 3) in Lemma 2.29: (b3 c3 )d¯3 = (b3 d¯3 )c3 , b¯3 d¯3 + x¯6 d¯3 = y9 c3 , then by 16) in Lemma 2.29 and 5) in Lemma 2.31: y9 c3 = z9 + x10 + c8 .

(5)

By the associative law and 7) in Lemma 2.27 and 4) in Lemma 2.26: (x62 )c3 = x6 (x6 c3 ), 2b6 c3 + y15 c3 + y9 c3 = b¯ 3 x6 + x¯15 x6 , then by (5) and 3) in Lemma 2.26 and 8) in Lemma 2.30: 2b8 + 2x10 + x10 + b8 + z9 + c8 + b5 + c5 + z9 + x10 + c8 = b8 + x10 + x¯15 x6 , x¯15 x6 = 2b8 + 3x10 + 2z9 + 2c8 + b5 + c5 .

(6)


129

By the associative law, 9) in Lemma 2.31 and 3) in Lemma 2.26: (x¯6 b3 )d3 = (x¯6 d3 )b3 , b8 d3 + x10 d3 = y15 b3 + y3 b3 , then by 4) in Lemma 2.27 and 13) in Lemma 2.32 b8 d3 + x10 d3 = 2x¯15 + x¯6 + x¯9 + x¯9 , by 13) in Lemma 2.29 x10 d3 = x¯15 + x¯6 + x¯9 .

(7)

So (d3 x9 , x10 ) = (d3 x10 , x¯9 ) = 1, and by 13) in Lemma 2.29, (d3 x9 , b8 ) = (d3 b8 , x¯9 ) = 1; and by (2) (d3 x9 , z9 ) = (d3 z9 , x¯9 ) = 1, so d3 x9 = x10 + b8 + z9 .

(8)

By the associative law and 15) in Lemma 2.29 and (1): (d32 )y9 = d3 (d3 y9 ), b6 y9 + y3 y9 = b3 d3 + x15 d3 + x9 d3 , then by 16) in Lemma 2.29 and (8) and 8) in Lemma 2.32: b6 y9 + y3 y9 = z9 + b8 + z9 + b5 + c5 + x10 + c8 + x10 + b8 + z9 , b6 y9 + y3 y9 = 3z9 + 2b8 + b5 + c5 + c8 + 2x10 . By 5), 6) in Lemma 2.32 and 9) in Lemma 2.27: (b6 y9 , b8 ) = (y9 , b6 b8 ) = 1, (b6 y9 , x10 ) = (y9 , b6 x10 ) = 1, (b6 y9 , z9 ) = (y9 , z9 b6 ) = 2; hence y3 y9 = z9 + b8 + x10 .

(9)

b6 y9 = 2z9 + b8 + x10 + b5 + c5 + c8 .

(10)

Therefore

130


By the associative law and 10) in Lemma 2.29 and 1) in Lemma 2.26: (b6 b¯3 )x15 = b¯3 (b6 x15 ),

2 b3 x15 + x15 = 4x¯15 b¯3 + x¯6 b¯3 + 2x¯9 b¯3 + b¯32 + d3 b¯3 ,

by 8) in Lemma 2.27 and 5) in Lemma 2.26 and 1) in Lemma 2.32 and Hypothesis 2.4 and 3) in Lemma 2.29: 2 x15 = 6y15 + 3b6 + 3y9 + c3 + y15 + 2y9 + 2y15 + 2y3 + c3 + b6 + y9 , 2 x15 = 9y15 + 4b6 + 6y9 + 2c3 + 2y3 .

By the associative law and 5) in Lemma 2.27 and 1) in Lemma 2.26: (x¯6 b6 )b3 = x¯6 (b6 b3 ), 2x6 b3 + x15 b3 + x9 b3 = b¯3 x¯6 + x¯15 x¯6 , then by 5) in Lemma 2.26 and 8) in Lemma 2.27 and 1) in Lemma 2.32: 2c3 + 2y15 + 2y15 + b6 + y9 + y15 + y3 = c3 + y15 + x¯15 x¯6 , x15 x6 = c3 + 4y15 + b6 + 2y9 + y3 . Lemma 2.34 Let (A, B) satisfy Hypothesis 2.4 and (b8 b3 , b8 b3 ) = 3, then Lemma 2.33 holds, and we obtain the following: 1) 2) 3) 4) 5) 6) 7) 8) 9) 10) 11) 12) 13)

y3 c3 = z9 , z9 c3 = y9 + y15 + y3 , y3 b8 = y9 + y15 , z9 b8 = 2z9 + 2x10 + 2c8 + b8 + b5 + c5 , z92 = 1 + 2x10 + 2z9 + 2b8 + 2c8 + b5 + c5 , c82 = 1 + 2c8 + 2x10 + b8 + z9 + b5 = c5 , c8 c3 = y15 + y9 , x92 = 2b6 + 3y15 + 2y9 + y3 + c3 , c8 y3 = b6 + y3 + y15 , x10 y3 = b6 + y3 + y15 , d3 y3 = d¯3 + x6 , x¯6 y3 = d¯3 + x15 , x15 y3 = 2x¯15 + x¯6 + x¯9 .

Proof By the associative law and Hypothesis 2.4 and 13) in Lemma 2.32: y3 (b32 ) = x¯ 9 b3 , y3 c3 + y3 b6 = x¯9 b3 , then by 2) in Lemma 2.31 y3 c3 + y3 b6 = z9 + x10 + c8 ,


131

then by 7) in Lemma 2.32 y3 c3 = z9 .

(1)

By 1), 16) in Lemma 2.29 and the associative law and 1) in Lemma 2.32: b3 (c3 d3 ) = (b3 c3 )d3 = (b3 d3 )c3 = z9 c3 , b3 x9 = z9 c3 , then z9 c3 = y9 + y15 + y3 .

(2)

So by (1) and 1) in Lemma 2.25 and the associative law: y3 (c32 ) = z9 c3 , y3 + y3 b8 = y9 + y15 + y3 . Thus y3 b8 = y9 + y15 .

(3)

By (2) and the associative law and 1) in Lemma 2.25: z9 (c32 ) = y9 c3 + y15 c3 + y3 c3 , z9 + z9 b8 = y9 c3 + y15 c3 + y3 c3 , so by (1) and 8) in Lemma 2.30 and 5) in Lemma 2.33: z9 + z9 b8 = z9 + x10 + c8 + x10 + b8 + c8 + z9 + b5 + c5 + z9 , z9 b8 = 2z9 + 2x10 + 2c8 + b8 + b5 + c5 .

(4)

By 17) in Lemma 2.29 and 1) in Lemma 2.25 and (1): y32 c32 = z92 , (1 + c8 )(1 + b8 ) = z92 , 1 + b8 + c8 + c8 b8 = z92 , by 4) in Lemma 2.33 z92 = 1 + 2x10 + 2z9 + 2b8 + 2c8 + b5 + c5 . By the associative law and 14) in Lemma 2.29 and 7) in Lemma 2.31: (d¯3 d3 )c8 = d¯3 (d3 c8 ), c8 + c82 = d¯3 x¯15 + x¯6 d¯3 + d3 d¯3 ,

(5)

132


then by 5) in Lemma 2.31 and 8) in Lemma 2.32: c8 + c82 = b8 + z9 + b5 + c5 + x10 + c8 + x10 + c8 + 1 + c8 , c82 = 1 + 2c8 + 2x10 + b8 + z9 + b5 + c5 .

(6)

By 1), 12) in Lemma 2.29 and the associative law: d¯3 (d3 c3 ) = c3 + c8 c3 , d¯3 x9 = c3 + c8 c3 , then by 2) in Lemma 2.32 c8 c3 = y15 + y9 .

(7)

By the associative law and 13) in Lemma 2.32 and 17) in Lemma 2.29 and Hypothesis 2.4: b32 y32 = x¯ 92 , (c3 + b6 )(1 + c8 ) = x¯92 , c3 + b6 + c3 c8 + b6 c8 = x¯92 , then by (7) and 4) in Lemma 2.32: c3 + b6 + y15 + y9 + 2y15 + y3 + y9 + b6 = x¯92 , x92 = 2b6 + 3y15 + 2y9 + y3 + c3 . By the associative law and 4), 7) in Lemma 2.32 and 17) in Lemma 2.29: (y32 )b6 = y3 (y3 b6 ), b6 + c8 b6 = c8 y3 + x10 y3 , 2y15 + y3 + y9 + 2b6 = c8 y3 + x10 y3 , by 9) in Lemma 2.33, 4) in Lemma 2.32 and 15) in Lemma 2.29: (c8 y3 , b6 ) = (y3 , c8 b6 ) = 1, (c8 y3 , y3 ) = (c8 , y32 ) = 1, (c8 y3 , y9 ) = (c8 , y3 y9 ) = 0. So c8 y3 = b6 + y3 + y15 ,

(8)

x10 y3 = b6 + y9 + y15 .

(9)

then

By the associative law and 8) in Lemma 2.31 and 17) in Lemma 2.29: (y32 )d¯3 = y3 (y3 d¯3 ), d¯3 + d¯3 c8 = x¯6 y3 + d3 y3 ,


133

by 7) in Lemma 2.31, 2d¯3 + x15 + x6 = x¯6 y3 + d3 y3 , by 15) in Lemma 2.29, (d3 y3 , d¯3 ) = (d32 , y3 ) = 1. So d3 y3 = d¯3 + x6 ,

(10)

x¯6 y3 = d¯3 + x15 .

(11)

then

By the associative law and 6) in Lemma 2.31 and 7) in Lemma 2.32: (d3 b6 )y3 = d3 (y3 b6 ), ¯ d3 y3 + x15 y3 = c8 d3 + x10 d3 , then by 7) in Lemma 2.33 and 7) in Lemma 2.31 and (10): d3 + x¯6 + x15 y3 = x¯15 + x¯6 + d3 + x¯15 + x¯6 + x¯9 , y15 y3 = 2x¯15 + x¯6 + x¯9 . Lemma 2.35 Let (A, B) satisfy Hypothesis 2.4 and (b8 b3 , b8 b3 ) = 3, then Lemma 2.34 holds and we obtain the following: (1) (2) (3) (4) (5) (6) (7) (8) (9) (10)

z9 x10 = 3x10 + 2b8 + 2z9 + 2c8 + b5 + c5 , z9 c8 = 2x10 + 2b8 + 2z9 + b5 + c5 + c8 , c8 x10 = 2x10 + 2c8 + 2b8 + 2z9 + b5 + c5 , y92 = 1 + 2z9 + 2b8 + 2c8 + 2x10 + b5 + c5 , x9 c8 = 3x15 + x6 + 2x9 + b3 , x¯9 y3 = b3 + x9 + x15 , x¯9 x9 = 1 + 2b8 + 2x10 + 2c8 + 2z9 + b5 + c5 , x9 y9 = d3 + 3x¯15 + 2x¯9 + b¯3 + 2x¯6 , z9 y3 = y9 + c3 + y15 , x9 z9 = 3x15 + 2x9 + 2x6 + b3 + d¯3 .

Proof By the associative law and 7) in Lemma 2.33 and 16) in Lemma 2.29: (b3 d3 )x10 = b3 (d3 x10 ), z9 x10 = x¯15 b3 + x¯6 b3 + x¯9 b3 . Then by 2) in Lemma 2.30 and 3) in Lemma 2.26 and 2) in Lemma 2.31: z9 x10 = x10 + b8 + c8 + z9 + b5 + c5 + b8 + x10 + z9 + x10 + c8 , z9 x10 = 3x10 + 2b8 + 2z9 + 2c8 + b5 + c5 . By the associative law and 16) in Lemma 2.29 and 7) in Lemma 2.31: (b3 d3 )c8 = b3 (d3 c8 ), z9 c8 = x¯15 b3 + x¯6 b3 + d3 b3 .

(1)

134


Then by 2) in Lemma 2.30 and 3) in Lemma 2.26 and 16) in Lemma 2.29, we get z9 c8 = x10 + b8 + c8 + z9 + b5 + c5 + b8 + x10 + z9 and z9 c8 = 2x10 + 2b8 + 2z9 + b5 + c5 + c8 .

(2)

By the associative law and 7) in Lemma 2.33 and 14) in Lemma 2.29: (d¯3 d3 )x10 = d¯3 (d3 x10 ), x10 + c8 x10 = d¯3 x¯15 + d¯3 x¯6 + d¯3 x¯9 , then by 5) in Lemma 2.31 and 8) in Lemma 2.32 and 8) in Lemma 2.33, it follows that x10 + c8 x10 = b8 + z9 + b5 + c5 + x10 + c8 + x10 + c8 + x10 + b8 + z9 and c8 x10 = 2x10 + 2c8 + 2b8 + 2z9 + b5 + c5 .

(3)

By the associative law and 3), 15) in Lemma 2.29 and Hypothesis 2.4, one has b32 d¯32 = y92 , (c3 + b6 )(b6 + y3 ) = y92 , c3 b6 + c3 y3 + b62 + b6 y3 = y92 , then by 3) in Lemma 2.26 and 1) in Lemma 2.34 and 3) in Lemma 2.30 and 7) in Lemma 2.32, we have y92 = b8 + x10 + z9 + 1 + b8 + z9 + c8 + b5 + c5 + c8 + x10 . So y92 = 1 + 2z9 + 2b8 + 2c8 + 2x10 + b5 + c5 .

(4)

By the associative law and 13) in Lemma 2.32 and 9) in Lemma 2.34: (b3 y3 )c8 = b3 (y3 c8 ), x¯9 c8 = b6 b3 + y3 b3 + y15 b3 , then by 13) in Lemma 2.32 and 1) in Lemma 2.26 and 4) in Lemma 2.27, x¯9 c8 = b¯3 + x¯15 + x¯9 + 2x¯15 + x¯6 + x¯9 and x¯9 c8 = b¯3 + 3x¯15 + 2x¯9 + x¯6 .

(5)

By the associative law and 13) in Lemma 2.32 and 17) in Lemma 2.29: (y32 )b3 = y3 (y3 b3 ), b3 + c8 b3 = x¯9 y3 , then 4) in Lemma 2.31 x¯ 9 y13 = b3 + x9 + x15 .

(6)


135

By the associative law and 13) in Lemma 2.32 and (6): (b3 y3 )x9 = b3 (y3 x9 ), x¯9 x9 = b3 b¯3 + b3 x¯9 + b3 x¯15 , then by Hypothesis 2.4 and 2) in Lemma 2.31 and 2) in Lemma 2.30: x¯9 x9 = 1 + b8 + z9 + x10 + c8 + x10 + b8 + c8 + z9 + b5 + c5 , x¯9 x9 = 1 + 2b8 + 2x10 + 2x10 + 2c8 + 2z9 + b5 + c5 .

(7)

By the associative law and 13) in Lemma 2.32 and 9) in Lemma 2.33: (b3 y3 )y9 = b3 (y3 y9 ), x¯9 y9 = z9 b3 + b8 b3 + x10 b3 , then by 1) in Lemma 2.31 and 2) in Lemma 2.26 and 1) in Lemma 2.27: x¯9 y9 = x9 + d¯3 + x15 + x6 + b3 + x15 + x15 + x6 + x9 = 2x9 + 3x15 + 2x6 + d¯3 + b3 . (8) By the associative law and 16) in Lemma 2.29 and 11) in Lemma 2.34: (b3 d3 )y3 = (d3 y3 )b3 , z9 y3 = d¯3 b3 + x6 b3 , then by 3) in Lemma 2.29 and 5) in Lemma 2.26 z9 y3 = y9 + c3 + y15 .

(9)

By the associative law and 13) in Lemma 2.32 and (9): (z9 y3 )b3 = z9 (y3 b3 ), y9 b3 + c3 b3 + y15 b3 = z9 x¯9 , then by 5) in Lemma 2.29 and by 2) in Lemma 2.35 and 4) in Lemma 2.27: z9 x¯9 = d3 + x¯9 + x¯15 + b¯3 + x¯6 + 2x¯15 + x¯6 + x¯9 = 3x¯15 + 2x¯9 + 2x¯6 + b¯3 + d3 . (10) Lemma 2.36 Let (A, B) satisfy Hypothesis 2.4 and (b8 b3 , b8 b3 ) = 3, then Lemma 2.25 holds and we obtain the following: (1) (2) (3) (4)

x15 c8 = 5x15 + 2x6 + 3x9 + b3 + d¯3 , x15 x9 = 6y15 + 3y9 + 2b6 + c3 + y − 3, x¯9 x15 = 3b8 + 4x10 + 3z9 + 3c8 + 2b5 + 2c5 , x15 y15 = 4x¯6 + 9x¯15 + 6x¯9 + 2b¯3 + 2d3 ,

136

(5) (6) (7) (8) (9) (10) (11)


x15 y9 = 6x¯15 + 2x¯6 + 3x¯9 + b¯3 + d3 , x¯9 x6 = 2z9 + x10 + b8 + c8 + b5 + c5 , x6 x9 = 2y9 + 2z15 + b6 , x6 y9 = 2x¯15 + 2x¯9 + x¯6 , x6 z9 = 2x15 + 2x9 + x6 , x6 c8 = 2x15 + x6 + x9 + d¯3 , y15 y9 = 4x10 + 3b8 + 3z9 + 3c8 + 2c5 + 2b5 .

Proof By the associative law and 6), 7) in Lemma 2.31: (d3 b6 )c8 = b6 (d3 c8 ), d¯3 c8 + x15 c8 = x¯15 b6 + x¯6 b6 + d3 b6 , then by 7) in Lemma 2.31 and 6), 10) in Lemma 2.29, 5) in Lemma 2.27: x15 + x6 d¯3 + x15 c8 = 4x15 + x6 + 2x9 + b3 + d¯3 + 2x6 + x15 + x9 + d¯3 + x15 , x15 c8 = 5x15 + 2x6 + 3x9 + b3 + d¯3 .

(1)

By the associative law and 8) in Lemma 2.33 and 6) in Lemma 2.31: (d3 b6 )x9 = b6 (d3 x9 ), d¯3 x9 + x15 x9 = x10 b6 + b8 b6 + z9 b6 , then by and 9) in Lemma 2.27 and 2), 5), 6) in Lemma 2.32: y15 + c3 + y9 + x15 x9 = 3y15 + y3 + c3 + y9 + 2y15 + b6 + c3 + y9 + 2y15 + 2y9 + b6 , x15 x9 = 6y15 + 3y9 + 2b6 + c3 + y3 .

(2)

By 13) in Lemma 2.32 and 13) in Lemma 2.34: b3 (y3 x15 ) = (b3 y3 )x15 = x¯9 x15 , 2x¯15 b3 + x¯6 b3 + x¯9 b3 = x¯9 x15 . and 2) in Lemma 2.31 and 3) in Lemma 2.26 and 2) in Lemma 2.30: 2x10 + 2b8 + 2c8 + 2z9 + 2b5 + 2c5 + b8 + x10 + z9 + x10 + c8 = x¯9 x15 , x¯ 9 x15 = 3b8 + 4x10 + 3z9 + 3c8 + 2b5 + 2c5 .

(3)

By 11), 12) in Lemma 2.33 and by (2) and 8) in Lemma 2.27 and 3) in Lemma 2.32: (x15 y15 , x¯6 ) = (x15 x6 , y15 ) = 4, (x15 y15 , x¯9 ) = (x15 x9 , y15 ) = 6,

2 , y ) = 9, (x15 y15 , x¯15 ) = (x15 15

(x15 y15 , b¯3 ) = (x15 b3 , y15 ) = 2,

(x15 y15 , d3 ) = (x15 d¯3 , y15 ) = 2.


137

Then x15 y15 = 4x¯6 + 9x¯15 + 6x¯9 + 2b¯3 + 2d3 .

(4)

By (2) and 11), 12) in Lemma 2.33 and 8) in Lemma 2.27 and 3) in Lemma 2.32: 2 ) = 6, (x15 y9 , x¯15 ) = (y9 , x15

(x15 y9 , x¯9 ) = (x15 x9 , y9 ) = 3,

(x15 y9 , x¯6 ) = (x15 x6 , y9 ) = 2, (x15 y9 , b¯3 ) = (x15 b3 , y9 ) = 1,

(x15 y9 , d3 ) = (x15 d¯3 , y9 ) = 1. Then x15 y9 = 6x¯15 + 2x¯6 + 3x¯9 + b¯3 + d3 .

(5)

By the associative law and 13) in Lemma 2.32 and 12) in Lemma 2.34: b3 (y3 x6 ) = (b3 y3 )x6 = x¯9 x6 , b3 d3 + b3 x¯15 = x¯9 x6 , then by 16) in Lemma 2.29 and 2) in Lemma 2.30 2z9 + x10 + b8 + c8 + b5 + c5 = x¯9 x6 .

(6)

By the associative law and 13) in Lemma 2.32 and 12) in Lemma 2.34: b3 (y3 x¯6 ) = x¯9 x¯6 , b3 d¯3 + x15 b3 = x¯9 x¯6 , so by 3) in Lemma 2.29 and 8) in Lemma 2.27 x¯6 x¯9 = 2y9 + 2y15 + b6 .

(7)

By the associative law and 3) in Lemma 2.29 and 9) in Lemma 2.31: b3 (d¯3 x6 ) = (b3 d¯3 )x6 = x6 y9 , b3 y15 + b3 y3 = x6 y9 .

(8)

By 16) in Lemma 2.29 and 5) in Lemma 2.31 and the associative law: x6 (b3 d3 ) = b3 (x6 d3 ), x6 z9 = x10 b3 + c8 b3 , then by 1) in Lemma 2.27 and 4) in Lemma 2.31: x6 z9 = x15 + x6 + x9 + x9 + x15 = 2x15 + 2x9 + x6 .

(9)

138


By (1) and 4) in Lemma 2.30 and 5) in Lemma 2.35 and 7) in Lemma 2.31: (x6 c8 , x15 ) = (x6 , c8 x15 ) = 2, (x6 c8 , x9 ) = (x6 , c8 x9 ) = 1,

(x6 c8 , x6 ) = (x6 x¯6 , c8 ) = 1, (x6 c8 , d¯3 ) = (c8 d3 , x¯6 ) = 1,

then x6 c8 = 2x15 + x6 + x9 + d¯3 .

(10)

By the associative law and 9) in Lemma 2.31 and 1) in Lemma 2.33: (d3 x¯6 )y9 = (d3 y9 )x¯6 , y15 y9 + y9 y3 = b3 x¯6 + x15 x¯6 + x9 x¯6 , then by (6) and 6), 9) in Lemma 2.33 and 3) in Lemma 2.26: y15 y9 + z9 + b8 + x10 = b8 + x10 + 2b8 + 3x10 + 2z9 + 2c8 +b5 + c5 + 2z9 + x10 + b8 + c8 + b5 + c5 , y15 y9 = 4x10 + 3b8 + 3z9 + 3c8 + 2c5 + 2b5 . Lemma 2.37 Let (A, B) satisfy Hypothesis 2.4 and (b8 b3 , b8 b3 ) = 3, then Lemma 2.36 holds and we obtain the following: 1) 2) 3) 4) 5) 6) 7) 8) 9) 10) 11) 12) 13) 14) 15) 16)

y3 y15 = x10 + b8 + b5 + c5 + c8 + z9 , 2 = 1 + 3b + 6x + 6z + 5b + 3c + 5c , y15 5 10 9 8 5 8 x10 y15 = 6y15 + 3b6 + 4y9 + c3 + y3 , z9 x10 = 3x10 + b5 + c5 + 2b8 + 2z9 + 2c8 , y9 x10 = 4y15 + 2y9 + y3 + b6 + c3 , y9 c8 = 3y15 + 2y9 + b6 + c3 , z9 y15 = 6y15 + 2b6 + 3y9 + c3 + y3 , x10 x15 = 3x6 + 6x15 + b3 + d¯3 + 4x9 , x10 x9 = 4x15 + x6 + 2x9 + b3 + d¯3 , b6 x9 = 2x¯15 + 2x¯9 + x¯6 , b8 y9 = 3y15 + y3 , x¯6 d3 = y15 + y3 , b8 y15 = 5y15 + 2b6 + 3y9 + c3 + y3 , x15 z9 = 2x6 + 6x15 + b3 + d¯3 + 3x9 , x9 y15 = 6x¯15 + 2x¯6 + 3x¯9 + b¯ 3 + d3 , y15 c8 = 2b6 + 5y15 + c3 + y3 + 3y9 .

Proof By the associative law and 9) in Lemma 2.31 and 15) in Lemma 2.29 and 7) in Lemma 2.27:


139 2 d32 x¯62 = y15 + 2y15 y3 + y32 ,

2 (b6 + y3 )(2b6 + y15 + y9 ) = y15 + 2y15 y3 + y32 , 2 + 2y15 y3 + y32 , 2b62 + y15 b6 + b6 y9 + 2b6 y3 + y3 y15 + y3 y9 = y15

by 3) in Lemma 2.30 and 7), 9), 10) in Lemma 2.32 and 9) in Lemma 2.33 and 17) in Lemma 2.29: 2 y15 + 1 + c8 + y15 y3 = 2 + 2b8 + 2z9 + 2c8 + 2b5 + 2c5 + 2b8 + 3x10 + 2z9

+ 2c8 + b5 + c5 + b8 + 2z9 + b5 + c5 + x10 + c8 .

(1)

By the associative law and 11), 15) in Lemma 2.29: (d32 )y15 = (d3 y15 )d3 ,

b6 y15 + y3 y15 = 2x15 d3 + x6 d3 + x9 d3 ,

so by 8), 9) in Lemma 2.32 and 5) in Lemma 2.31 and 8) in Lemma 2.33: 2b8 + 3x10 + 2z9 + 2c8 + b5 + c5 + y3 y15 = 2b8 + 2z9 + 2b5 + 2c5 + 2x10 + 2c8 + x10 + c8 + x10 + b8 + z9 , y3 y15 = x10 + b8 + b5 + c5 + c8 + z9 .

(2)

2 = 1 + 3b5 + 6x10 + 6z9 + 5b8 + 3c5 + 5c8 . y15

(3)

Then by (1),

By 9) in Lemma 2.32 and 11) in Lemma 2.36 and 8) in Lemma 2.30 and (2): 2 ) = 6, (x10 y15 , y15 ) = (x10 , y15

(x10 y15 , y9 ) = (x10 , y15 y9 ) = 4,

(x10 y15 , b6 ) = (x10 , y15 b6 ) = 3, (x10 y15 , c3 ) = (x10 , y15 c3 ) = 1,

(y15 x10 , y3 ) = (x10 , y15 y3 ) = 1. So x10 y15 = 6y15 + 3b6 + 4y9 + c3 + y3 .

(4)

By 16) in Lemma 2.29 and 7) in Lemma 2.33: b3 (d3 x10 ) = (b3 d3 )x10 = z9 x10 , x¯15 b3 + x¯6 b3 + x¯9 b3 = z9 x10 , then by 2) in Lemma 2.30 and 3) in Lemma 2.26 and 2) in Lemma 2.31: x10 + b8 + c8 + z9 + b5 + c5 + b8 + x10 + z9 + x10 + c8 = z9 x10 , z9 x10 = 3x10 + b5 + c5 + 2b8 + 2z9 + 2c8 .

(5)

140


Lemma 2.38 Let (A, B) satisfy Hypothesis 2.4 and (b8 b3 , b8 b3 ) = 3, then b5 = c¯5 . Proof Assume b5 = c¯5 . By 2) in Lemma 2.30, (b¯3 b5 , x¯15 ) = (b5 , b3 x¯15 ) = 1, (b3 b5 , x15 ) = (b3 x¯15 , b¯5 ) = 1, then b¯3 b5 = x¯15 ,

(1)

b3 b5 = x15 .

(2)

By 9), 10) in Lemma 2.32 and 3) in Lemma 2.30: (b6 b5 , y15 ) = (b5 , b6 y15 ) = 1, (b6 b5 , y9 ) = (b5 , b6 y9 ) = 1, (b6 b5 , b6 ) = (b5 , b62 ) = 1, then b6 b5 = y15 + y9 + b6 .

(3)

By 4) in Lemma 2.30 and 6) in Lemma 2.33 and 6) in Lemma 2.36: (x6 b5 , x6 ) = (b5 , x¯6 x6 ) = 1, (x6 b5 , x15 ) = (b5 , x¯6 x15 ) = 1, (x6 b5 , x9 ) = (b5 , x¯6 x9 ) = 1, then x6 b5 = x6 + x15 + x9 .

(4)

By 6) in Lemma 2.33 and 6) in Lemma 2.36 and 4) in Lemma 2.30: (x¯6 b5 , x¯15 ) = (b5 , x6 x¯15 ) = 1, (x¯6 b5 , x¯9 ) = (x¯6 x9 , b¯5 ) = 1, (x¯6 b5 , x¯6 ) = (x¯6 x6 , b¯5 ) = 1 then x¯6 b5 = x¯15 + x¯9 + x¯6 .

(5)

By 6) in Lemma 2.33 and 8), 11) in Lemma 2.32 and 2) in Lemma 2.30 and 3) in Lemma 2.36: (x15 b5 , x6 ) = (b5 , x¯15 x6 ) = 1, (x15 b5 , d¯3 ) = (x15 d3 , b¯5 ) = 1, (x15 b5 , x15 ) = (b5 , x¯15 x15 ) = 3, (x15 b5 , x9 ) = (b5 , x¯15 x9 ) = 2, (x15 b5 , b3 ) = (x15 b¯3 , b¯5 ) = 1,


141

then x15 b5 = x6 + 3x15 + 2x9 + b3 + d¯3 .

(6)

By 6) in Lemma 2.33 and 8), 11) in Lemma 2.32 and 3) in Lemma 2.36 and 2) in Lemma 2.30: (x¯15 b5 , x¯6 ) = (b5 , x15 x¯6 ) = 1, (x¯15 b5 , d3 ) = (b5 , x15 d3 ) = 1, (x¯15 b5 , x¯15 ) = (x¯15 x15 , b¯5 ) = 3, (x¯15 b5 , x¯9 ) = (x¯15 x9 , b¯5 ) = 1, (x¯15 b5 , b¯3 ) = (x¯15 b3 , b¯5 ) = 1, then x¯15 b5 = x¯6 + 3x¯15 + 2x¯9 + b¯3 + d3 .

(7)

By 3), 6) in Lemma 2.36 and 7) in Lemma 2.35: (x9 b5 , x15 ) = (x9 x¯15 , b¯5 ) = 2, (x9 b5 , x6 ) = (b5 , x¯9 x6 ) = 1, (x9 b5 , x9 ) = (x9 x¯9 , b¯5 ) = 1, then x9 b5 = 2x15 + x6 + x9 .

(8)

By 3), 6) in Lemma 2.36 and 7) in Lemma 2.35: (x¯9 b5 , x¯15 ) = (x¯9 x15 , b¯5 ) = 2, (x¯9 b5 , x¯6 ) = (b5 , x¯9 x6 ) = 1, (x9 b5 , x9 ) = (x9 x¯9 x9 , b¯5 ) = 1, then x¯9 b5 = 2x¯15 + x¯6 + x¯9 .

(9)

By 9) in Lemma 2.32 and 1), 2) in Lemma 2.37 and 11) in Lemma 2.36 and 8) in Lemma 2.30 (y15 b5 , b6 ) = (b5 , y15 b6 ) = 1, 2 ) = 3, (y15 b5 , y15 ) = (b5 , y15

(y15 b5 , y3 ) = (y15 y3 , b¯5 ) = 1, (y15 b5 , c3 ) = (b5 , y15 c3 ) = 1, (y15 b5 , y9 ) = (b5 , y15 y9 ) = 2,

142


then y15 b5 = b6 + 3y15 + 2y9 + c3 + y3 .

(10)

By 11) in Lemma 2.36 and 10) in Lemma 2.32 and 4) in Lemma 2.35: (y9 b5 , y15 ) = (b5 , y9 y15 ) = 2, (y9 b5 , b6 ) = (b5 , y9 b6 ) = 1, (y9 b5 , y9 ) = (b5 , y92 ) = 1, then y9 b5 = 2y15 + b6 + y9 .

(11)

By 8) in Lemma 2.32, (d3 b5 , x¯15 ) = (d3 x15 , b¯5 ) = 1, (d¯3 b5 , x15 ) = (b5 , d3 x15 ) = 1, then d3 b5 = x¯15 ,

(12)

d¯3 b5 = x15 .

(13)

Then by 16) in Lemma 2.29 and associative law: b3 (b5 d3 ) = (b3 d3 )b5 ,

x¯15 b3 = z9 b5 ,

then by 2) in Lemma 2.30, z9 b5 = x10 + b8 + c8 + z9 + b5 + c5 .

(14)

By 1) in Lemma 2.37, (y3 b5 , y15 ) = (b5 , y3 y15 ) = 1, then y3 b5 = y15 .

(15)

Then 1 = (y3 b5 , y3 b¯5 ) = (b52 , y32 ), then by 17) in Lemma 2.29, (b52 , c8 ) = 1,

(16)

By (1), (2) we get 1 = (b3 b¯5 , b3 b5 ) = (b3 b¯3 , b52 ), then by Hypothesis 2.4, (b52 , b8 ) = 1.

(17)

By (2), (12) 1 = (d¯3 b¯5 , b3 b5 ) = (d¯3 b¯3 , b52 ), then by 16) in Lemma 2.29 (b52 , z9 ) = 1, so by (16), (17), we obtain b52 = c8 + b8 + z9 .

(18)


143

By (1) 1 = (b¯3 b5 , b¯3 b5 ) = (b¯3 b3 , b¯5 b5 ), then by Hypothesis 2.4, (b5 b¯5 , b8 ) = 0.

(19)

By (2), (4), 1 = (x6 b5 , b3 b5 ) = (x6 b¯3 , b¯5 b5 ), then by 3) in Lemma 2.26 and (19): (b¯5 b5 , x10 ) = 1. By (13), 3 = (b6 b5 , b6 b5 ) = (b62 , b¯5 b5 ), so by 3) in Lemma 2.30, m5 = b5 , or m5 = b¯5 , in the two cases (b¯5 b5 , b5 ) = 0 or (b¯5 b5 , b¯5 ) = 0, a contradiction to (18). Lemma 2.39 Let (A, B) satisfy Hypothesis 2.4 and (b8 b3 , b8 b3 ) = 3, then Lemma 2.37 holds and we obtain the following: 1) 2) 3) 4) 5) 6) 7) 8) 9) 10) 11) 12) 13) 14) 15) 16) 17) 18) 19) 20) 21) 22) 23) 24) 25) 26) 27) 28) 29)

b3 b5 = x15 ; b3 b5 = x15 ; b6 b5 = b6 + y15 + y9 ; b6 c5 = b6 + y15 + y9 ; b5 x6 = x15 + x9 + x6 ; c5 x6 = x15 + x9 + x6 ; x15 b5 = x6 + d¯3 + 3x15 + b3 + 2x9 ; x15 c5 = x6 + d¯3 + 3x15 + b3 + 2x9 ; x9 b5 = 2x15 + x6 + x9 ; x9 c5 = 2x15 + x6 + x9 ; y15 b5 = b6 + y3 + 3y15 + 2y9 + c3 ; y15 c5 = b6 + y3 + 3y15 + 2y9 + c3 ; y9 b5 = 2y15 + b6 + y9 ; y9 c5 = 2y15 + b6 + y9 ; d3 b5 = x¯15 ; d3 c5 = x¯15 ; z9 b5 = x10 + b8 + c8 + z9 + b5 + c5 ; z9 c5 = x10 + b8 + c8 + z9 + b5 + c5 ; y3 b5 = y15 ; y3 c5 = y15 ; c5 b5 = c 8 + b 8 + z 9 ; b8 b5 = x10 + b8 + c8 + c5 + z9 ; b8 c5 = x10 + b8 + c8 + c5 + z9 ; b52 = 1 + b5 + z9 + x10 ; c52 = 1 + c5 + z9 + x10 ; b5 x10 = 2x10 + b8 + z9 + b5 + c8 ; c5 x10 = 2x10 + b8 + z9 + c5 + c8 ; c8 b5 = b8 + z9 + c5 + x10 + c8 ; c8 c5 = b8 + z9 + x10 + c8 + b5 .

Proof By 2) in Lemma 2.30, (b3 b5 , x15 ) = (b3 x¯15 , b5 ) = 1, (b3 c5 , x15 ) = (b3 x¯15 , c5 ) = 1, then b3 b5 = x15 ,

(1)

144


b3 c5 = x15 .

(2)

By 9), 10) in Lemma 2.32 and 3) in Lemma 2.30: (b6 b5 , b6 ) = (b62 , b5 ) = 1, (b6 b5 , y15 ) = (b5 , b6 y15 ) = 1, (b6 b5 , y9 ) = (b5 , b6 y9 ) = 1,

(b6 c5 , b6 ) = (b62 , c5 ) = 1, (b6 c5 , y15 ) = (c5 , b6 y15 ) = 1, (b6 c5 , y9 ) = (c5 , b6 y9 ) = 1,

then b5 b6 = b6 + y15 + y9 ,

(3)

b6 c5 = y15 + y9 + b6 .

(4)

By 4) in Lemma 2.30 and 6) in Lemma 2.33 and 6) in Lemma 2.36: (b5 x6 , x6 ) = (b5 , x¯6 x6 ) = 1, (b5 x6 , x15 ) = (x6 x¯15 , b5 ) = 1, (b5 x6 , x9 ) = (x6 x¯9 , b5 ) = 1,

(c5 x6 , x6 ) = (c5 , x¯6 x6 ) = 1, (c5 x6 , x15 ) = (x6 x¯15 , c5 ) = 1, (c5 x6 , x9 ) = (x6 x¯9 , c5 ) = 1,

then b5 x6 = x6 + x15 + x9 ,

(5)

c5 x6 = x15 + x9 + x6 .

(6)

By 6) in Lemma 2.33 and 8), 11) in Lemma 2.32 and 2) in Lemma 2.30 and 3) in Lemma 2.36: (x15 b5 , x6 ) = (b5 , x¯15 x6 ) = 1,

(x15 c5 , x6 ) = (c5 , x¯15 x6 ) = 1,

(x15 b5 , d¯3 ) = (x15 d3 , b5 ) = 1,

(x15 c5 , d¯3 ) = (x15 d3 , c5 ) = 1,

(x15 b5 , x15 ) = (x15 x¯15 , b5 ) = 3,

(x15 c5 , x15 ) = (x15 x¯15 , c5 ) = 3,

(x15 b5 , b3 ) = (b5 , x¯15 b3 ) = 1,

(x15 c5 , b3 ) = (c5 , x¯15 b3 ) = 1,

(x15 b5 , x9 ) = (b5 , x¯15 x9 ) = 2,

(x15 c5 , x9 ) = (x15 x¯9 , c5 ) = 2,

then x15 b5 = x6 + d¯3 + 3x15 + b3 + 2x9 ,

(7)

x15 c5 = x6 + d¯3 + 3x15 + b3 + 2x9 .

(8)

By 3), 6) in Lemma 2.36 and 7) in Lemma 2.35: (x9 b5 , x15 ) = (b5 , x¯9 x15 ) = 2,

(x9 c5 , x15 ) = (c5 , x¯9 x15 ) = 2,

(x9 b5 , x6 ) = (b5 , x¯9 x6 ) = 1,

(x9 c5 , x6 ) = (c5 , x¯9 x6 ) = 1,

(x9 b5 , x9 ) = (b5 , x¯9 x9 ) = 1,

(x9 c5 , x9 ) = (c5 , x¯9 x9 ) = 1,


145

then x9 b5 = 2x15 + x6 + x9 ,

(9)

x9 c5 = 2x15 + x6 + x9 .

(10)

By 9) in Lemma 2.32 and 1), 2) in Lemma 2.37 and 11) in Lemma 2.36 and 8) in Lemma 2.30: (y15 b5 , b6 ) = (y15 b6 , b5 ) = 1,

(y15 c5 , b6 ) = (y15 b6 , c5 ) = 1,

(y15 b5 , y3 ) = (y15 y3 , b5 ) = 1,

(y15 c5 , y3 ) = (y15 y3 , c5 ) = 1,

2 ) = 3, (y15 b5 , y15 ) = (b5 , y15

2 , c ) = 3, (y15 c5 , y15 ) = (y15 5

(y15 b6 , c3 ) = (y15 c3 , b6 ) = 1,

(y15 c5 , c3 ) = (y15 c3 , c5 ) = 1,

(y15 b5 , y9 ) = (y15 y9 , b5 ) = 2,

(y15 c5 , y9 ) = (c5 , y15 y9 ) = 2,

then y15 b5 = b6 + y3 + 3y15 + 2y9 + c3 ,

(11)

y15 c5 = b6 + y3 + 3y15 + 2y9 + c3 .

(12)

By 11) in Lemma 2.36 and 10) in Lemma 2.32 and 4) in Lemma 2.35: (y9 b5 , y15 ) = (b5 , y9 y15 ) = 2, (y9 b5 , b6 ) = (y9 b6 , b5 ) = 1, (y9 b5 , y9 ) = (b5 , y92 ) = 1,

(y9 c5 , y15 ) = (c5 , y9 y15 ) = 2, (y9 c5 , b6 ) = (y9 b6 , c5 ) = 1, (y9 c5 , y9 ) = (c5 , y92 ) = 1,

then y9 b5 = 2y15 + b6 + y9 ,

(13)

y9 c5 = 2y15 + b6 + y9 .

(14)

By 8) in Lemma 2.32, (d3 b5 , x¯15 ) = (d3 x15 , b5 ) = 1, (d3 c5 , x¯15 ) = (d3 x15 , c5 ) = 1. Then d3 b5 = x¯ 15 ,

(15)

d3 c5 = x¯15 .

(16)

So by the associative law and 16) in Lemma 2.29, we get (b3 d3 )b5 = b3 (d3 b5 ), (b3 d3 )c5 = b3 (d3 c5 ), z9 b5 = b3 x¯15 , z9 c5 = b3 x¯15 , then by 2) in Lemma 2.30, z9 b5 = x10 + b8 + c8 + z9 + b5 + c5 ,

(17)

z9 c5 = x10 + b8 + c8 + z9 + b5 + c5 .

(18)

146


By 1) in Lemma 2.37, (y3 b5 , y15 ) = (b5 , y3 y15 ) = 1, (y3 c5 , y15 ) = (c5 , y3 y15 ) = 1, then y3 b5 = y¯15 ,

(19)

y3 c5 = y¯15 .

(20)

So (y32 , b52 ) = (y3 b5 , y3 b5 ) = 1, (y32 , c5 b5 ) = (y3 b5 , y3 c5 ) = 1, then by 17) in Lemma 2.29, (c5 b5 , c8 ) = 1. By (1), (2), (b¯3 b3 , b5 c5 ) = (b3 b5 , b3 c5 ) = 1. then by Hypothesis 2.4, (c5 b5 , b8 ) = 1. By (17), c5 b5 = c8 + b8 + z9 .

(21)

By the associative and Hypothesis 2.4 and (1), (2), it follows that (b3 b¯3 )b5 = (b3 b5 )b¯3 , (b3 b¯3 )c5 = (b3 c5 )b¯3 , b5 + b8 b5 = x15 b¯3 , c5 + b8 c5 = x15 b¯3 , then by 2) in Lemma 2.30, b8 b5 = x10 + b8 + c8 + c5 + z9 ,

(22)

b8 c5 = x10 + b8 + c8 + b5 + z9 .

(23)

2 ) = 2, (x c , x ) = (c , x 2 ) = 2. By 5) in Lemma 2.30, (x10 b5 , x10 ) = (b5 , x10 10 5 10 5 10 By (22) and (23), (x10 b5 , b8 ) = (x10 , b5 b8 ) = 1, (x10 c5 , b8 ) = (x10 , c5 b8 ) = 1. By (17), (18), (x10 b5 , z9 ) = (x10 , z9 b5 ) = 1, (x10 c5 , b8 ) = (x10 , c5 b8 ) = 1, (b52 , z9 ) = (b5 , b5 z9 ) = 1 and (c52 , z9 ) = (c5 , c5 z9 ) = 1. By (19), (20), we obtain (y32 , b52 ) = (y3 b5 , y3 b5 ) = 1, (y32 , c52 ) = (y3 c5 , y3 c5 ) = 1, then by 17) in Lemma 2.29 (b52 , c8 ) = 0, (c52 , c8 ) = 0. By (1) and (2), we see that (b¯3 b3 , b52 ) = (b3 b5 , b3 b5 ) = 1, (b¯3 b3 , c52 ) = (b3 c5 , b3 c5 ) = 1, then by Hypothesis 2.4, (b52 , b8 ) = 0, (c52 , b8 ) = 0 holds. By (5), (6), (15) and (16), (b52 , d3 x6 ) = (b5 d¯3 , b5 x6 ) = 1, (c52 , d3 x6 ) = (c5 d¯3 , c5 x6 ) = 1, so by 5) in Lemma 2.31, (b52 , x10 ) = 1, (c52 , x10 ) = 1. By (5), (21) and (6) we have the following inner-products: (b52 , x¯6 x6 ) = (b5 x6 , b5 x6 ) = 3, (c52 , x¯6 x6 ) = (c5 x6 , c5 x6 ) = 3. (c52 , b5 ) = (c5 , c5 b5 ) = 0 and (b52 , c5 ) = (b5 , b5 c5 ) = 0, then by 4) in Lemma 2.30 (b52 , b5 ) = 1, (c52 , c5 ) = 1, hence

b52 = 1 + b5 + z9 + x10 ,

c52 = 1 + b5 + z9 + x10 .

So (b5 x10 , b5 ) = (x10 , b52 ) = 1, (c5 x10 , c5 ) = (x10 , c52 ) = 1. By 14) in Lemma 2.29 and (15), (16), one has b5 (d¯3 d3 ) = d¯3 (b5 d3 ), c5 (d¯3 d3 ) = d¯3 (c5 d3 ). Hence b5 + c8 b5 = x¯15 d¯3 , c5 + c8 c5 = d¯3 x¯15 . Therefore by 8) in Lemma 2.32, we have the equations c8 b5 = b8 + z9 + c5 + x10 + c8 and c8 c5 = b8 + z9 + x10 + c8 + b5 . Consequently (b5 x10 , c8 ) = (x10 , b5 c8 ) = (x10 , c5 c8 ) = 1. Finally, we get b5 x10 = c8 + 2x10 + b8 + z9 + b8 + z9 + b5 and c5 x10 = 2x10 + b8 + z9 + c5 + c8 .


147

Theorem 2.10 Let (A, B) be a NITA generated by a nonreal element b3 ∈ B of degree 3 and without non-identity basis element of degree 1 or 2, such that: b¯3 b3 = 1 + b8 and b32 = c3 + b6 , c3 = c¯3 , (b8 b3 , b8 b3 ) = 3. Then (A, B) is a Table Algebra of dimension 22: B = {b1 , b¯3 , b3 , c3 , b6 , x6 , x¯6 , b8 , x15 , x¯15 , x10 , x9 , x¯9 , y15 , y9 , d3 , d¯3 , z9 , c8 , b5 , c5 , y3 } and B has increasing series of table subsets {b1 } ⊆ {b1 , b8 , x10 , z9 , c8 , b5 , c5 } ⊆ B defined by: 1) 2) 3) 4) 5) 6) 7) 8) 9) 10) 11) 12) 13) 14) 15) 16) 17) 18) 19) 20) 21) 22) 23) 24) 25) 26) 27) 28) 29) 30) 31) 32) 33) 34) 35)

b3 b¯3 = 1 + b8 ; b32 = c3 + b6 ; b3 c3 = b¯3 + x¯6 ; b3 b6 = b¯3 + x¯15 ; b3 x6 = c3 + y15 ; b3 x¯6 = b8 + x10 ; b3 b8 = x6 + b3 + x15 ; b3 x15 = 2y15 + b6 + y9 ; b3 x10 = x15 + x6 + x9 ; b3 x¯15 = x10 + b8 + z9 + c8 + b5 + c5 ; b3 x9 = y9 + y15 + y3 ; b3 x¯9 = x10 + z9 + c8 ; b3 y15 = 2x¯ 15 + x¯6 + x¯9 ; b3 y9 = d3 + x¯15 + x¯9 ; b3 d3 = z9 ; b3 d¯3 = y9 ; b3 z9 = x9 + d¯3 + x15 ; b3 c8 = x15 + x9 ; b3 b5 = x15 ; b3 c5 = x15 ; b3 y3 = x¯9 ; c32 = 1 + b8 ; c3 b6 = b8 + x10 ; c3 x6 = b¯3 + x¯15 ; c3 b8 = b6 + c3 + y15 ; c3 x15 = 2x¯15 + x¯6 + x¯9 ; c3 x10 = b6 + y15 + y9 ; c3 x9 = d3 + x¯15 + x¯9 ; c3 y15 = x10 + b8 + z9 + c8 + b5 + c5 ; c3 y9 = z9 + x10 + c8 ; c3 d3 = x9 ; c3 z9 = y3 + y9 + y15 ; c3 c8 = y9 + y15 ; c3 b5 = y15 ; c3 c5 = y15 ;

148

36) 37) 38) 39) 40) 41) 42) 43) 44) 45) 46) 47) 48) 49) 50) 51) 52) 53) 54) 55) 56) 57) 58) 59) 60) 61) 62) 63) 64) 65) 66) 67) 68) 69) 70) 71) 72) 73) 74) 75) 76) 77) 78) 79) 80)


c3 y 3 = z 9 ; b62 = 1 + b8 + z9 + c8 + b5 + c5 ; b6 x6 = 2x¯6 + x¯15 + x¯9 ; b6 b8 = 2y15 + b6 + y9 + c3 ; b6 x15 = 4x¯ 15 + x¯6 + 2x¯9 + b¯ 3 + d3 ; b6 x10 = c3 + 3y15 + y9 + y3 ; b6 x9 = 2x¯15 + 2x¯9 + x¯ 6 ; b6 y15 = 2b8 + 3x10 + 2z9 + 2c8 + b5 + c5 ; b6 y9 = b8 + 2z9 + c8 + x10 + b5 + c5 ; b6 d3 = d¯3 + x15 ; b6 z9 = 2y15 + 2y9 + b6 ; b6 c8 = b6 + y3 + y9 + 2y15 ; b6 b5 = y15 + y9 + b6 ; b6 c5 = y15 + b6 + y9 ; b6 y3 = x10 + c8 ; x62 = 2b6 + y15 + y9 ; x6 x¯6 = 1 + b8 + z9 + c8 + b5 + c5 ; x6 b8 = b3 + 2x15 + x6 + x9 ; x6 x15 = 4y15 + b6 + 2y9 + c3 + y3 ; x6 x10 = b3 + 3x15 + d¯3 + x9 ; x6 x¯15 = 2b8 + 3x10 + 2z9 + 2c8 + b5 + c5 ; x6 x9 = 2y15 + 2y9 + b6 ; x6 x¯9 = x10 + b8 + 2z9 + c8 + b5 + c5 ; x6 y15 = 4x¯ 15 + 2x¯9 + x¯6 + b¯ 3 + d3 ; x6 y9 = 2x¯15 + 2x¯9 + x¯ 6 ; x6 d3 = x10 + c8 ; x6 d¯3 = y15 + y3 ; x6 z9 = 2x9 + 2x15 + x6 ; x6 c8 = 2x15 + x6 + x9 + d¯3 ; x6 b5 = x6 + x15 + x9 ; x6 c5 = x15 + x9 + x6 ; x6 y3 = d3 + x¯15 ; b82 = 1 + 2b8 + 2x10 + z9 + c8 + b5 + c5 ; b8 x10 = 2x10 + 2z9 + 2c8 + 2b8 + b5 + c5 ; b8 x9 = 3x15 + 2x9 + x6 + d¯3 ; b8 y15 = 5y15 + 2b6 + 3y9 + c3 + y3 ; b8 y9 = 3y15 + 2y9 + b6 + y3 ; b8 d3 = x¯15 + x¯9 ; b8 z9 = 2x10 + b8 + 2c8 + 2z9 + b5 + c5 ; b8 c8 = 2x10 + 2z9 + b8 + c8 + b5 + c5 ; b8 b5 = b8 + z9 + c5 + c8 + x10 ; b8 c5 = b8 + x10 + c8 + z9 + b5 ; b8 y3 = y9 + y15 ; b8 x15 = 5x15 + 2x6 + 3x9 + b3 + d¯3 ; 2 = 9y + 4b + 6y + 2c + 2y ; x15 15 6 9 3 3


81) 82) 83) 84) 85) 86) 87) 88) 89) 90) 91) 92) 93) 94) 95) 96) 97) 98) 99) 100) 101) 102) 103) 104) 105) 106) 107) 108) 109) 110) 111) 112) 113) 114) 115) 116) 117) 118) 119) 120) 121) 122) 123) 124) 125)

x15 x10 = 3x6 + 6x15 + b3 + d¯3 + 4x9 ; x15 x¯15 = 5b8 + 6x10 + 6z9 + 5c8 + 3b5 + 3c5 + 1; x15 x9 = 6y15 + 3y9 + 2b6 + c3 + y3 ; x15 x¯9 = 3b8 + 4x10 + 3z9 + 3c8 + 2b5 + 2c5 ; x15 y15 = 4x¯6 + 9x¯15 + 6x¯9 + 2b¯3 + 2d3 ; x15 y9 = 6x¯ 15 + 2x¯6 + 3x¯9 + b¯3 + d3 ; x15 d3 = b8 + z9 + b5 + c5 + x10 + c8 ; x15 d¯3 = 2y15 + y9 + b6 ; x15 z9 = 2x6 + 6x15 + b3 + d¯3 + 3x9 ; x15 c8 = 5x15 + 2x6 + 3x9 + b3 + d¯3 ; x15 b5 = x6 + 3x15 + 2x9 + b3 + d¯3 ; x15 c5 = x6 + 3x15 + 2x9 + b3 + d¯3 ; x15 y3 = x¯ 6 + 2x¯15 + x¯9 ; 2 = 1 + 2b + 2x + 2c + 2b + 2c + 3z ; x10 8 10 8 5 5 9 x10 x9 = b3 + 4x15 + d¯3 + 2x9 + x6 ; x10 y15 = 6y15 + 3b6 + 4y9 + c3 + y3 ; x10 y9 = 4y15 + 2y9 + y3 + b6 + c3 ; x10 d3 = x¯ 15 + x¯6 + x¯9 ; x10 z9 = 3x10 + b5 + c5 + 2b8 + 2z9 + 2c8 ; x10 c8 = 2x10 + 2c8 + 2b8 + 2z9 + b5 + c5 ; x10 b5 = b8 + 2x10 + z9 + c8 + b5 ; x10 c5 = b8 + z9 + c5 + 2x10 + c8 ; x10 y3 = y15 + y9 + b6 ; x92 = y3 + c3 + 2b6 + 3y15 + 2y9 ; x9 x¯9 = 1 + 2b8 + 2x10 + 2c8 + 2z9 + b5 + c5 ; x9 y15 = 6x¯ 15 + 2x¯6 + 3x¯9 + b¯3 + d3 ; x9 y9 = d3 + 3x¯15 + 2x¯9 + b¯3 + 2x¯6 ; x9 d3 = x10 + b8 + z9 ; x9 d¯3 = y9 + c3 + y15 ; x9 z9 = 3x15 + 2x9 + 2x6 + b3 + d¯3 ; x9 b5 = 2x15 + x6 + x9 ; x9 c5 = 2x15 + x6 + x9 ; x9 y3 = b¯3 + x¯9 + x¯15 ; 2 = 1 + 3b + 6x + 6z + 5b + 3c + 5c ; y15 5 10 9 8 5 8 y15 y9 = 4x10 + 3b8 + 3z9 + 3c8 + 2c5 + 2b5 ; y15 d3 = 2x15 + x6 + x9 ; y15 z9 = 6y15 + 2b6 + 3y9 + c3 + y3 ; y15 c8 = 2b6 + 5y15 + c3 + y3 + 3y9 ; y15 b5 = b6 + 3y15 + 2y9 + c3 + y3 ; y15 c5 = b6 + c3 + 3y15 + y3 + 2y9 ; y15 y3 = b5 + b8 + z9 + c5 + x10 + c8 ; y92 = 1 + 2z9 + 2b8 + 2c8 + 2x10 + b5 + c5 ; y9 d3 = b3 + x15 + x9 ; y9 z9 = 3y15 + 2b6 + 2y9 + c3 + y3 ; y9 c8 = 3y15 + 2y9 + b6 + c3 ;

149

150

126) 127) 128) 129) 130) 131) 132) 133) 134) 135) 136) 137) 138) 139) 140) 141) 142) 143) 144) 145) 146) 147) 148) 149) 150)


y9 b5 = 2y15 + b6 + y9 ; y9 c5 = 2y15 + y9 + b6 ; y9 y3 = z9 + b8 + x10 ; d3 d¯3 = 1 + c8 ; d32 = b6 + y3 ; d3 z9 = b¯3 + x¯15 + x¯9 ; d3 c8 = x¯15 + d3 + x¯6 ; d3 b5 = x¯15 ; d3 c5 = x¯15 ; d3 y3 = d¯3 + x6 ; z92 = 1 + 2x10 + 2z9 + 2b8 + 2c8 + b5 + c5 ; z9 c8 = 2x10 + 2b8 + 2z9 + b5 + c5 + c8 ; z9 b5 = x10 + b8 + z9 + c8 + b5 + c5 ; z9 c5 = x10 + b8 + z9 + c8 + b5 + c5 ; z9 y3 = y9 + c3 + y15 ; c82 = 1 + 2c8 + 2x10 + b8 + z9 + b5 + c5 ; c8 b5 = b8 + z9 + c5 + x10 + c8 ; c8 c5 = b8 + z9 + x10 + c8 + b5 ; c8 y3 = b6 + y3 + y15 ; b52 = 1 + b5 + x10 + z9 ; b5 c5 = b8 + z9 + c8 ; b5 y3 = y15 ; c52 = 1 + z9 + x10 + c5 ; c5 y3 = y15 ; y32 = 1 + c8 .

Proof The equations hold by Lemmas 2.25–2.37 and 2.39.

References [AB] [B] [B1] [Bl] [Br] [CA] [F]

[L] [W]

Arad, Z., Blau, H.: On table algebras and application to finite group theory. J. Algebra 138, 137–185 (1991) Blau, H.I.: Integral table algebra, affine diagrams, and the analysis of degree two. J. Algebra 178, 872–918 (1995) Blau, H.I.: Quotient structures in C-algebra. J. Algebra 177, 297–337 (1995) Blichfeldt, H.F.: Finite Collineation Groups. University of Chicago Press, Chicago (1917) Brauer, R.: Über endliche lineare Gruppen von Primzahlgrad. Math. Ann. 169, 73– 96 (1967) Chen, G.Y., Arad, Z.: On four normalized table algebras generated by a faithful nonreal element of degree 3. J. Algebra 283, 457–484 (2005) Feit, W.: The current situation in the theory of finite simple groups, Actes du Congrès International des Mathématiciens (Nice, 1970), Tome 1, pp. 55–93. Gauthier-Villars, Paris (1971) Lindsey, J.H.: On a projective representation of Hall-Janko group. Bull. Am. Math. Soc. 74, 1094 (1968) Wales, D.B.: Finite linear groups of degree seven I. Can. J. Math. 21, 1025–1041 (1969)

Chapter 3

A Proof of a Non-existence of Sub-case (2)

3.1 Introduction In Chap. 3 we shall freely use the definitions and notation used in Chap. 2. The Main Theorem 1 of Chap. 2 left three unsolved problems for the complete classification of the Normalized Integral Table Algebra (NITA) (A, B) generated by a faithful non-real element of degree 3 with L1 (B) = 1 and L2 (B) = ∅. In this chapter we solve the first problem, and prove that the NITA which satisfies case (2) of the Main Theorem 1 of Chap. 2 does not exist. Consequently, we can state the Main Theorem 2 of this chapter as follows. Main Theorem 2 Let (A, B) be a NITA that is generated by a non-real element b3 ∈ B of degree 3 and has no non-identity basis elements of degrees 1 or 2. Then b3 b¯3 = 1 + b8 , b8 ∈ B, and one of the following holds: (1) There exists a real element b6 ∈ B such that b32 = b¯3 + b6 and (A, B) ∼ =x CH PSL(2, 7), Irr PSL(2, 7) . (2) There exist b6 , b10 , b15 ∈ B, where b6 is non-real, such that b32 = b¯3 + b6 , b¯3 b6 = b3 + b15 , b3 b6 = b8 + b10 , and (b3 b8 , b3 b8 ) = 3. Moreover, if b10 is real then (A, B) ∼ =x (CH(3 · A6 ), Irr(3 · A6 )) of dimension 17. (3) There exist c3 , b6 ∈ B, c3 = b3 or b¯3 , such that b32 = c3 + b6 and (b3 b8 , b3 b8 ) = 3 or 4. If (b3 b8 , b3 b8 ) = 3 and c3 is non-real, then (A, B) ∼ =x (A(3 · A6 · 2), B32 ) of dimension 32. (See Theorem 2.9 of Chap. 2 for the definition of this specific NITA.) If (b3 b8 , b3 b8 ) = 3 and c3 is real, then (A, B) ∼ =x (A(7 · 5 · 10), B22 ) of dimension 22. (See Theorem 2.10 of Chap. 2 for the definition of this specific NITA.) In the above Main Theorem 2, we still have 2 open problems in the cases (2) and (3) that have not been solved. In case (2) we must classify the NITA such that b10 is non-real, and in case (3) we must classify the NITA such that (b3 b8 , b3 b8 ) = 4. In Chaps. 4 and 5, the first problem will be almost completely solved. Z. Arad et al., On Normalized Integral Table Algebras (Fusion Rings), Algebra and Applications 16, DOI 10.1007/978-0-85729-850-8_3, © Springer-Verlag London Limited 2011

151

152

3


Let us emphasize that the NITA’s of dimensions 7 and 17 in the cases (1) and (2) of the Main Theorem 2 are strictly isomorphic to the NITA’s induced from finite groups G via the basis of the irreducible characters of G. However, the NITA’s of dimensions 22 and 32 are not induced from finite groups as described in Chap. 2. The Main Theorem 2 follows from the Main Theorem 1 in Chap. 2 and the following theorem: Theorem 3.1 There exists no NITA (A, B) that is generated by a non-real element b3 ∈ B of degree 3 and has no non-identity basis elements of degrees 1 or 2 such that b3 b¯3 = 1 + b8 , b8 ∈ B, and b32 = b4 + b5 ,

b4 , b5 ∈ B,

(b3 b8 , b3 b8 ) = 3,

and

(b42 , b42 ) = 3.

The rest of this chapter is devoted to proving the above Theorem 3.1.

3.2 Preliminary Results We are going to prove Theorem 3.1, by contradiction. That is, for the rest of this chapter we shall always assume that (A, B) is a NITA that is generated by a nonreal element b3 ∈ B of degree 3 and has no non-identity basis elements of degrees 1 or 2 such that b3 b¯3 = 1 + b8 , b32 = b4 + b5 ,

b8 ∈ B,

(3.1)

b4 , b5 ∈ B,

(3.2)

and (b3 b8 , b3 b8 ) = 3, (b42 , b42 ) = 3. Then we are going to derive a contradiction. From (b3 b¯3 )b3 = b32 b¯3 and (3.1), (3.2), we obtain that b¯3 b4 = b3 + x9 ,

x9 ∈ B,

(3.3)

b¯3 b5 = b3 + y12 ,

y12 ∈ B,

(3.4)

and b3 b8 = b3 + x9 + y12 .

(3.5)

Now 2 = (b¯3 b4 , b¯3 b4 ) = (b¯3 b3 , b¯4 b4 ) implies that (b¯4 b4 , b8 ) = 1. Hence from (b¯ 4 b4 , b¯4 b4 ) = (b42 , b42 ) = 3, we see that b4 b¯4 = 1 + b7 + b8 ,

b7 ∈ B is real.

(3.6)

Furthermore, b32 b¯4 = (b3 b¯4 )b3 and (3.6), (3.3), (3.1) imply that b7 + b¯4 b5 = b3 x¯9 .

(3.7)

3.2 Preliminary Results

153

Since (b3 x¯9 , b8 ) = 1 by (3.5), and b7 , b8 are real by (3.6), from the above equality we may assume that b4 b¯5 = b8 + Σ12 ,

Σ12 ∈ N(B),

(3.8)

and b¯3 x9 = b7 + b8 + Σ12 .

(3.9)

Thus, from (3.4), (3.2), (3.1), (3.8), and b¯3 (b¯3 b5 ) = b¯32 b5 , we see that 1 + b¯3 y12 = Σ¯ 12 + b5 b¯5 .

(3.10)

Note that (b3 b4 , b3 b4 ) = (b3 b¯3 , b4 b¯4 ) = 2 by (3.1) and (3.6). So we may set b3 b4 = z + w,

z, w ∈ B.

(3.11)

Thus, b¯3 z + b¯3 w = b¯3 (b3 b4 ) = (b3 b¯3 )b4 = b4 + b4 b8 . But (b4 b8 , b5 ) = 1 by (3.8). Thus by (3.8), without loss of generality, we may assume that (b¯3 z, b5 ) = 1. Since (b3 b5 , b3 b5 ) = (b¯3 b5 , b¯3 b5 ) = 2 by (3.4), we can set b3 b5 = z + u,

u ∈ B.

(3.12)

Note that in (3.11) and (3.12), the degrees of z, w, and u are not yet determined. According to the possibilities of their degrees, we will have difference cases to consider. But let us first prove the following proposition. Proposition 3.1 Let Σ12 be the same as in (3.8). Then Σ12 ∈ B. Proof Toward a contradiction, assume that Σ12 ∈ / B. Then (b4 b¯5 , b4 b¯5 ) ≥ 3 by (3.8), and (b3 Σ12 , x9 ) ≥ 2

(3.13)

by (3.9). Note that b3 (b¯3 b5 ) = (b3 b¯3 )b5 and (3.4), (3.1), (3.2) imply that b5 b8 = b4 + b3 y12 , and (b3 y12 , b5 ) = 1 by (3.4). So (b5 b¯5 , b8 ) = (b5 b8 , b5 ) = 1. Thus, 3 ≤ (b4 b¯5 , b4 b¯5 ) = (b4 b¯4 , b5 b¯5 ) = 2 + (b7 , b5 b¯5 ). Hence (b7 , b5 b¯5 ) ≥ 1. But L2 (B) = ∅. So (b7 , b5 b¯5 ) = 1. Thus, we must have that (b4 b¯5 , b4 b¯5 ) = 3. Therefore, we may assume that Σ12 = r + s,

r, s ∈ B,

(3.14)

and b5 b¯5 = 1 + b7 + b8 + Σ9 ,

(3.15)

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3


where Σ9 ∈ N(B) is real. Furthermore, (b¯3 y12 , b7 ) ≥ 1 by (3.10), and hence (b3 b7 , y12 ) ≥ 1. Thus, (b3 b7 , y12 ) = 1, (b¯3 y12 , b7 ) = 1, and b7 = r¯ or s¯ . But we also have (b3 b7 , x9 ) = 1 by (3.7). Therefore, b3 b7 = x9 + y12 .

(3.16)

Thus, from b3 (b4 b¯4 ) = b4 (b3 b¯4 ), (3.6), (3.3), and (3.5), we obtain that b4 x¯9 = b3 + x9 + 2y12 .

(3.17)

From (3.10), (3.14), and (3.15), we see that b¯3 y12 = r¯ + s¯ + b7 + b8 + Σ9 .

(3.18)

Thus, (b¯3 b4 )x¯9 = (b4 x¯9 )b¯3 and (3.3), (3.9), (3.14), (3.17), (3.1), (3.18) yield that x9 x¯9 = 1 + 2b7 + 3b8 + 2Σ9 + r + s + r¯ + s¯ .

(3.19)

From (3.8), (3.5), (3.3), (3.4), and b¯3 (b4 b¯5 ) = (b¯ 3 b4 )b¯5 , we obtain that b¯5 x9 = x¯ 9 + b¯3 r + b¯ 3 s.

(3.20)

(b¯5 x9 , y¯12 ) = 2.

(3.21)

Thus by (3.18)

Note that the degrees of both r and s are at least 4 by (3.18). If one of r and s has degree 4, without loss of generality, we may assume that r = r4 and s = s8 . Then by (3.18), again, we see that b3 r¯4 = y12 . So (r4 x¯9 , b3 b¯4 ) = (b4 x¯9 , b3 r¯4 ) = 2 by (3.17). That is, (r4 x¯9 , b¯3 + x¯9 ) = 2. Note that (b3 r4 , b3 r4 ) = (b3 r¯4 , b3 r¯4 ) = 1. So (r4 x¯9 , b¯3 ) = (b3 r4 , x9 ) = 0. Thus, (r4 x¯9 , x¯9 ) = 2. On the other hand, (b¯5 x9 , b¯3 r4 ) = (b¯5 x9 , y¯12 ) = 2 by (3.21). So (r4 x¯9 , b3 b¯5 ) = 2, and hence (r4 x¯9 , b¯3 ) = 0 forces that (r4 x¯9 , y¯12 ) = 2 by (3.4). Therefore, r4 x¯9 = 2x¯9 + 2y¯12 + · · · , a contradiction. This proves that the degrees of both r and s are at least 5. Now we claim that Σ9 ∈ B. If Σ9 ∈ / B, then (3.18) forces that Σ9 = c4 + d5 , where c4 , d5 ∈ B are real, and b3 c4 = y12 ,

b3 d5 = y12 + e3 ,

e3 ∈ B.

Thus, c4 real implies that (c4 x¯9 , b3 b¯5 ) = (b5 x¯9 , b3 c4 ) = 2 by (3.21). But (c4 x¯9 , b¯3 ) = (b3 c4 , x9 ) = 0. So by (3.4), (c4 x¯9 , y¯12 ) = 2. From (3.19) we also have that (c4 x¯9 , x¯9 ) = (x9 x¯9 , c4 ) ≥ 2.


155

Thus, c4 x¯9 = 2y¯12 + 2x¯9 + · · · , a contradiction. This proves that Σ9 ∈ B. Therefore, (b5 b¯5 , x9 x¯9 ) = 8 by (3.15) and (3.19). So (b5 x¯9 , b5 x¯9 ) = 8.

(3.22)

Since (b¯3 r + b¯3 s, y¯12 ) = 2 by (3.18), (3.20) and (3.22) force that (b¯3 r + b¯ 3 s, x¯9 ) = 0, and (b¯ 3 r + b¯3 s, b¯3 r + b¯3 s) = 7. Hence by b¯3 r + b¯3 s = 2y¯12 + sum of three distinct elements of B. Recall that the degrees of both r and s are at least 5. So without loss of generality, we may assume that (b3 r¯ , b3 r¯ ) = 2 and (b3 s¯ , b3 s¯ ) = 3. In the following, we show that c3 , d4 , e5 ∈ B. (3.23) If (3.23) does not hold, then (b3 r¯ , b3 r¯ ) = 2 and (b3 s¯ , b3 s¯ ) = 3 force that

r = r5 ,

s = s7 ,

b3 r¯5 = y12 + c3 ,

r = r6 ,

b3 s¯7 = y12 + d4 + e5 ,

s = s6 ,

b3 s¯6 = y12 + d3 + e3 ,

b3 r¯5 = y12 + c6 , c6 , d3 , e3 ∈ B,

d3 = e3 .

Hence (b¯3 d3 , s¯6 ) = 1 and (b¯3 e3 , s¯6 ) = 1. So (b¯3 d3 , b¯3 e3 ) ≥ 1. Thus, (b3 b¯3 , d3 e¯3 ) ≥ 1. But d3 = e3 . So (1, d3 e¯3 ) = 0 and (b8 , d3 e¯3 ) = 1, a contradiction to the assumption that L1 (B) = {1}. This proves (3.23). Since (b3 c¯3 , r5 ) = 1 by (3.23), we have that (c3 c¯3 , b3 b¯3 ) = (b3 c¯3 , b3 c¯3 ) = 2. Thus, c3 c¯3 = 1 + b8 .

(3.24)

Hence (3.1) and (3.16) imply that (c3 b7 , c3 b7 ) = (c3 c¯3 , b72 ) = (b3 b¯3 , b72 ) = (b3 b7 , b3 b7 ) = 2. Similarly, (c3 b8 , c3 b8 ) = 3 by (3.1) and (3.5). That is, (c3 b7 , c3 b7 ) = 2

and (c3 b8 , c3 b8 ) = 3.

(3.25)

Furthermore, (b3 b4 )b¯4 = b3 (b4 b¯4 ) and (3.11), (3.6), (3.5) yield that zb¯4 + w b¯4 = 2b3 + 2x9 + 2y12 . Hence the degree of z must be one of 3, 6, and 9. According to the degree of z, we have three cases to consider. In the following, we will derive a contradiction for each case. Note that (zb¯4 , b3 ) = (w b¯4 , b3 ) = 1 by (3.11). Case A Assume that z = z3 ,

w = w9 ,

z3 b¯4 = b3 + x9 ,

w9 b¯4 = b3 + x9 + 2y12 .

(3.26)

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3


By (3.11) and (3.12) we obtain that b¯3 z3 = b4 + b5 .

(3.27)

Thus, (z3 z¯ 3 , b3 b¯3 ) = (b¯3 z3 , b¯3 z3 ) = 2. Hence z3 z¯ 3 = 1 + b8 . Therefore, (3.27) and (3.5) yield that b¯4 z3 + b¯5 z3 = (b3 z¯ 3 )z3 = b3 (¯z3 z3 ) = 2b3 + x9 + y12 . So b¯4 z3 = b3 + x9 , and by (3.9), b3 (b4 z¯ 3 ) = b3 (b¯3 + x¯9 ) = 1 + b7 + 2b8 + Σ¯ 12 . But, on the other hand, from (3.27), (3.6), and (3.8), we see that b3 (b4 z¯ 3 ) = b4 (b3 z¯ 3 ) = b4 (b¯4 + b¯5 ) = 1 + b7 + 2b8 + Σ12 . Thus, Σ¯ 12 = Σ12 , and (3.10) implies that (b3 Σ12 , y12 ) = (b¯3 y12 , Σ12 ) ≥ 2 (because Σ12 ∈ / B). But (b3 Σ12 , x9 ) ≥ 2 by (3.13). Hence b3 Σ12 = 2x9 + 2y12 + · · · , a contradiction. Case B Assume that z = z6 ,

w = w6 ,

u = u9 ,

z6 b¯4 = b3 + x9 + y12 ,

w6 b¯4 = b3 + x9 + y12 .

(3.28)

Note that (3.12), (3.8), (3.5), (3.28), (3.14), (3.23), and (b3 b5 )b¯4 = (b¯ 4 b5 )b3 yield that b¯4 u9 = 2y12 + c3 + d4 + e5 .

(3.29)

Thus, c3 b4 = u9 + p3 ,

p3 ∈ B.

(3.30)

Also (b¯4 p3 , c3 ) = 1 by (3.30). So 1 < (b¯4 p3 , b¯4 p3 ) = (p3 p¯ 3 , b4 b¯4 ) ≤ 2. Hence (b¯ 4 p3 , b¯4 p3 ) = 2, and b¯4 p3 = c3 + q9 , q9 ∈ B. Therefore from (c3 b4 )b¯4 = c3 (b4 b¯4 ) and (3.30), (3.29), (3.6), we obtain that c3 b7 + c3 b8 = 2y12 + c3 + d4 + e5 + q9 .


157

Since (c3 b7 , c3 b7 ) = 2 and (c3 b8 , c3 b8 ) = 3 by (3.25), the above equality cannot be true, a contradiction. Case C Assume that w3 b¯4 = b3 +x9 . (3.31) Note that (3.12), (3.8), (3.5), (3.31), (3.14), (3.23), and (b3 b5 )b¯4 = (b¯ 4 b5 )b3 yield that z = z9 ,

w = w3 ,

u = u6 ,

z9 b¯4 = b3 +x9 +2y12 ,

b¯4 u6 = y12 + c3 + d4 + e5 .

(3.32)

Since (c3 b4 , c3 b4 ) = (c3 c¯3 , b4 b¯4 ) = 2 by (3.24) and (3.6), from (3.32) we may set c3 b4 = u6 + p6 ,

p6 ∈ B.

(3.33)

Thus, (c3 b4 )b¯4 = c3 (b4 b¯4 ) and (3.33), (3.32) yield that c3 b7 + c3 b8 = y12 + d4 + e5 + b¯4 p6 .

(3.34)

Since (c3 b7 , c3 b7 ) = 2 and (c3 b8 , c3 b8 ) = 3 by (3.25), the above equality forces that (b¯4 p6 , b¯4 p6 ) = 2. But (b¯4 p6 , c3 ) = 1 by (3.33). So we may assume that b¯4 p6 = c3 + q21 , q21 ∈ B. Hence (3.34) yields that c3 b7 + c3 b8 = y12 + d4 + e5 + c3 + q21 , a contradiction to (3.25). Therefore, we must have that Σ12 ∈ B.

From Proposition 3.1, we may assume that Σ12 = b12 ∈ B. Hence (3.9) implies that b¯3 x9 = b7 + b8 + b12 ,

(3.35)

b4 b¯5 = b8 + b12 .

(3.36)

and (3.8) yields that

Furthermore, by (3.10) we see that 1 + b¯3 y12 = b¯12 + b5 b¯5 .

(3.37)

Since (b3 b7 , x9 ) = 1 by (3.35), we may assume that ∗ , b3 b7 = x9 + Σ12

∗ Σ12 ∈ N(B).

(3.38)

Thus, from (3.6), (3.3), (3.5), (3.38), and b3 (b4 b¯4 ) = b4 (b3 b¯4 ), we see that ∗ . b4 x¯9 = b3 + y12 + x9 + Σ12

(3.39)

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Table 3.1 Splitting to seven cases Case 1

z = z3

w = w9

Case 2

z = z4

w = w8

Case 3

z = z5

w = w7

Case 4

z = z6

w = w6

Case 5

z = z7

w = w5

Case 6

z = z8

Case 7

z = z9

b¯3 z3 = b4 + b5 , b¯3 w9 = b4 + σ + θ b¯3 z4 = b4 + b5 + σ3 , b¯4 w8 = b4 + θ20 b¯3 z5 = b4 + b5 + σ6 , b¯3 z6 = b4 + b5 + σ9 ,

b¯3 w7 = b4 + θ17 b¯3 w6 = b4 + θ14

w = w4

b¯3 z7 = b4 + b5 + σ12 , b¯3 z8 = b4 + b5 + σ15 ,

b¯ 3 w5 = b4 + θ11 b¯ 3 w4 = b4 + θ8

w = w3

b¯3 z9 = b4 + b5 + σ18 ,

b¯ 3 w3 = b4 + θ5

Therefore, (3.11), (3.6), (3.38), (3.5), and (b3 b4 )b¯4 = (b4 b¯4 )b3 force that ∗ , zb¯4 + w b¯ 4 = 2b3 + 2x9 + y12 + Σ12

(3.40)

and (3.12), (3.36), (3.5), and (b3 b5 )b¯4 = b3 (b¯4 b5 ) yield that zb¯4 + ub¯4 = b3 + x9 + y12 + b3 b¯12 .

(3.41)

Note that (b3 x9 , b3 x9 ) = (b¯3 x9 , b¯3 x9 ) = 3 by (3.35), and (b3 x9 , b4 ) = 1 by (3.3). So we may assume that b3 x9 = b4 + σ + θ,

σ, θ ∈ B,

and

|σ | + |θ | = 23.

(3.42)

Thus, from (3.11), (3.3), (3.2), and b¯3 (b3 b4 ) = (b¯ 3 b4 )b3 , we get that b¯3 z + b¯3 w = 2b4 + b5 + σ + θ.

(3.43)

Recall that (b¯3 z, b4 ) = (b¯3 z, b5 ) = 1 by (3.11) and (3.12), respectively. So from (3.43), we have the cases in Table 3.1. In the rest of this chapter, we derive a contradiction for each case in Table 3.1. Here let us summarize the identities obtained in this section. b3 b¯3 = 1 + b8 , b¯3 b5 = b3 + y12 ,

b32 = b4 + b5 , b3 b8 = b3 + x9 + y12 , b4 b¯5 = b8 + b12 ,

b¯3 x9 = b7 + b8 + b12 , b3 b4 = z + w,

b3 b5 = z + u,

∗ . b4 x¯9 = b3 + y12 + x9 + Σ12

b¯3 b4 = b3 + x9 , b4 b¯4 = 1 + b7 + b8 , b3 x9 = b4 + σ + θ, ∗ , b3 b7 = x9 + Σ12

3.3 Case z = z3 In this section we assume that z = z3 . Then we have that z = z3 ,

w = w9 ,

u = u12 ,

b¯3 z3 = b4 + b5 ,

b¯3 w9 = b4 + σ + θ. (3.44)

3.3 Case z = z3

159

Since (b3 b¯3 , z3 z¯ 3 ) = (b¯ 3 z3 , b¯3 z3 ) = 2 by (3.44), we see that z3 z¯ 3 = 1 + b8 .

(3.45)

Also note that (b3 z3 , b3 z3 ) = (b¯3 z3 , b¯3 z3 ) = 2 by (3.44). So we may set b3 z3 = μ + ν, μ, ν ∈ B and |μ| < |ν|. Then from (3.44), (3.11), (3.12) and (b3 z3 )b¯3 = (b¯3 z3 )b3 , we obtain that b¯3 μ + b¯3 ν = 2z3 + w9 + u12 .

(3.46)

But (b¯3 μ, z3 ) = (b¯3 ν, z3 ) = 1. So we must have either μ = μ4 and ν = ν5 , or μ = 1 and ν = ν8 . In the following, we derive a contradiction for each of these two cases. Subcase 1A μ = μ4 and ν = ν5 . Thus, b3 z3 = μ4 + ν5 ,

(3.47)

b¯3 μ4 = z3 + w9 ,

(3.48)

b¯3 ν5 = z3 + u12 .

(3.49)

and (3.46) implies that

and On the other hand, from (b¯ 3 z3 )b3 = (b3 b¯3 )z3 , (3.44), (3.1), (3.11), and (3.12), we see that z3 b8 = z3 + w9 + u12 .

(3.50)

Furthermore, b¯32 μ4 = b¯ 3 (b¯3 μ4 ) and (3.44), (3.2), (3.48) imply that b¯4 μ4 + b¯5 μ4 = 2b4 + b5 + σ + θ. If (b¯4 μ4 , b4 ) ≥ 2, then (b42 , μ4 ) ≥ 2, and hence (b42 , b42 ) ≥ 5, a contradiction to the assumption that (b42 , b42 ) = 3. Thus, (b¯ 4 μ4 , b4 ) ≤ 1. So (b¯5 μ4 , b4 ) ≥ 1, and hence (b4 b5 , μ4 ) ≥ 1. But (b4 b5 , b4 b5 ) = (b4 b¯5 , b4 b¯5 ) = 2 by (3.36). So we must have that (b4 b5 , μ4 ) = 1. Thus, we may assume that b4 b5 = μ4 + n16 ,

n16 ∈ B.

Hence from (3.12), (3.47), (3.2), (3.51), and b3 (b3 b5 ) = b32 b5 , we get that μ4 + ν5 + b3 u12 = μ4 + n16 + b52 . Since (b3 u12 , ν5 ) = 1 by (3.49), there exists Σ15 ∈ N(B) such that b3 u12 = ν5 + n16 + Σ15

and

b52 = 2ν5 + Σ15 .

(3.51)

160

3


Therefore (b3 u12 , b3 u12 ) < (b52 , b52 ).

(3.52)

Note that (3.11), (3.2) and b3 (b3 b4 ) = b32 b4 yield that b3 z3 + b3 w9 = b42 + b4 b5 , and z3 (b¯3 b4 ) = (b¯ 3 z3 )b4 implies that b3 z3 + z3 x9 = b42 + b4 b5 by (3.3) and (3.44). Thus, b3 w9 = z3 x9 . However, from z3 (b3 b8 ) = (z3 b8 )b3 , (3.5), and (3.50), we get that b3 z3 + z3 x9 + z3 y12 = b3 z3 + b3 w9 + b3 u12 . So z3 y12 = b3 u12 , and hence by (3.52), (z3 y12 , z3 y12 ) < (b52 , b52 ).

(3.53)

Since b3 b¯3 = z3 z¯ 3 by (3.1) and (3.45), we must have that (z3 y12 , z3 y12 ) = (z3 z¯ 3 , y12 y¯12 ) = (b3 b¯3 , y12 y¯12 ) = (b¯3 y12 , b¯3 y12 ). So (b¯3 y12 , b¯3 y12 ) < (b52 , b52 ) by (3.53). But, on the other hand, from (3.37) we also have that (b¯3 y12 , b¯3 y12 ) ≥ (b5 b¯5 , b5 b¯5 ) = (b52 , b52 ), a contradiction. Subcase 1B μ = 1 and ν = ν8 . Thus, we must have that z3 = b¯3 , and b¯32 = b4 + b5 by (3.44). Hence both b4 and b5 are real by (3.2), and b12 is real by (3.36). Thus, we see that w9 = x¯9 by (3.3) and (3.11), u12 = y¯12 by (3.4) and (3.12), and b¯ 3 x¯9 = b4 + σ + θ by (3.44). So b3 x9 = b4 + σ + θ . Since |σ | + |θ | = 23, (3.42) implies that both σ and θ are real. Hence, b3 x9 = b4 + σ + θ . Since (b¯3 y12 , b8 ) = 1 by (3.5), we see that (b52 , b8 ) = 1 by (3.37). So we may assume that b52 = 1 + b8 + Σ16 ,

Σ16 ∈ N(B) is real.

(3.54)

Then by (3.37), b¯3 y12 = b8 + b12 + Σ16 .

(3.55)

Hence the degree of any component of Σ16 is at least 4 by (3.55). Our goal is to prove that (Σ16 , Σ16 ) = 2. But we need first to prove the next lemma: Lemma 3.1 Let Σ16 be the same as in (3.54). Then (Σ16 , Σ16 ) ≤ 3.

3.3 Case z = z3

161

Proof Toward a contradiction, assume that (Σ16 , Σ16 ) ≥ 4. Then (b¯3 y12 , Σ16 ) ≥ (Σ16 , Σ16 ) ≥ 4 by (3.55). So (b3 Σ16 , y12 ) ≥ 4. Hence (b3 Σ16 , y12 ) = 4, b3 Σ16 = 4y12 , and (Σ16 , Σ16 ) = 4. Thus, either Σ16 has four distinct components of degree 4, or Σ16 = 2x8 , x8 ∈ B. Furthermore, (b5 y¯12 , b¯3 b5 ) = (b¯3 y12 , b52 ) ≥ 5 by (3.54) and (3.55). But (b5 y¯12 , b3 ) = 1 by (3.4). So we must have that b5 y¯12 = b3 + 4y12 + · · · .

(3.56)

If Σ16 has four distinct components of degree 4, then for a component x4 of Σ16 , b3 Σ16 = 4y12 implies that b3 x4 = y12 . Thus, (3.4) and (3.56) imply that (x4 y12 , b3 + y12 ) = (x4 y12 , b¯3 b5 ) = (b3 x4 , b5 y¯12 ) = 4.

(3.57)

Note that by (3.3), (3.4), and (3.55), (b4 y¯12 , y12 ) = (b4 y¯12 , b3 + y12 ) = (b4 y¯12 , b¯3 b5 ) = (b4 b5 , b¯3 y12 ) = 2. So (x4 y12 , b3 + x9 ) = (x4 y12 , b¯3 b4 ) = (b3 x4 , b4 y¯12 ) = (y12 , b4 y¯12 ) = 2.

(3.58)

Hence (3.57) and (3.58) force that (x4 y12 , b3 ) = 1, and b¯3 x4 = y¯12 .

x4 y12 = b3 + x9 + 3y12 ,

(3.59)

Therefore, (y¯12 y12 , x¯4 x4 ) = (x4 y12 , x4 y12 ) = 11. But from (b¯3 x4 )y12 = b¯ 3 (x4 y12 ), (3.59), (3.1), (3.35), and (3.55), we get that y¯12 y12 = 1 + b7 + 5b8 + 4b12 + 3Σ16 . Since we assume that Σ16 has four distinct components of degree 4, and (x¯4 x4 , b8 ) = 0 by (3.59), the above equality implies that it is impossible that (y¯12 y12 , x¯4 x4 ) = 11, a contradiction. If Σ16 = 2x8 , x8 ∈ B, then b3 Σ16 = 4y12 implies that b3 x8 = 2y12 , and similar to the proof in the above paragraph, we have that x8 y12 = 2b3 + 2x9 + 6y12 ,

b¯3 x8 = 2y¯12 ,

(y¯12 y12 , x¯8 x8 ) = 44.

(3.60)

Hence from (b¯3 x8 )y12 = b¯3 (x8 y12 ), (3.60), (3.1), (3.35), and (3.55), we see that y¯12 y12 = 1 + b7 + 5b8 + 4b12 + 6x8 . Since (x¯8 x8 , b¯3 b3 ) = (b3 x8 , b3 x8 ) = 4, we may set x¯8 x8 = 1 + 3b8 + Σ39 ,

Σ39 ∈ N(B).

162

3


But b3 x¯8 = 2y12 by (3.60), and b3 b8 = b3 + x9 + y12 by (3.5). Multiplying both sides of the above equation by b3 , we obtain that 1 x8 y12 = (4b3 + 3x9 + 3y12 + b3 Σ39 ). 2 Recall that (b3 b7 , x9 ) = 1 by (3.38), and (b3 b12 , x9 ) = 1 by (3.35). So the above equality and (3.60) force that (Σ39 , b7 + b12 ) ≤ 1. Therefore, (y¯12 y12 , x¯8 x8 ) = 44 yields that x¯8 x8 = 1 + 3b8 + b12 + 4x8 . However, the above equality is impossible, a contradiction. So Lemma 3.1 holds.

Since (b3 b8 , y12 ) = 1 by (3.5), we see that (Σ16 , b8 ) = 0 by (3.55). If (Σ16 , b12 ) = 1, then (b3 b12 , y12 ) = 2 by (3.55). But we also have (b3 b12 , x9 ) = 1 by (3.35). Hence b3 b12 = x9 + 2y12 + x3 for some x3 ∈ B, a contradiction. Therefore, we must have that (Σ16 , b12 ) = 0. Thus, Lemma 3.1 and (3.55) yield that (b¯3 y12 , b¯3 y12 ) ≤ 5. Since (b3 b12 , x9 ) = 1 by (3.35) and (b3 b12 , y12 ) = 1 by (3.55), we may set b3 b12 = x9 + y12 + Σ15 ,

Σ15 ∈ N(B).

(3.61)

Thus, from (3.35), (3.39), (3.38), (3.5), (3.61), and b¯32 x9 = b¯ 3 (b¯3 x9 ), we get that b5 x¯9 = 2x9 + y12 + Σ15 .

(3.62)

Hence b32 x9 = b3 (b3 x9 ) and (3.42), (3.2), (3.3), (3.39), (3.62) imply that ∗ b3 σ + b3 θ = 2x¯9 + 2y¯12 + Σ¯ 12 + Σ¯ 15 .

(3.63)

Furthermore, from b3 (b4 b5 ) = b4 (b3 b5 ), (3.4), (3.3), (3.36), (3.5), and (3.61), we obtain that b4 y¯12 = x9 + 2y12 + Σ15 .

(3.64)

Lemma 3.2 Let Σ16 be the same as in (3.54). Then (Σ16 , Σ16 ) = 2. Assume that Σ16 = p + q, p, q ∈ B. Then |p| ≥ 5, |q| ≥ 5, and b¯3 y12 = b8 + b12 + p + q,

b3 y12 = σ + θ + b5 + x8 ,

x8 ∈ B.

Proof Let us first prove that (b3 σ, y¯12 ) = 1 and (b3 θ, y¯12 ) = 1. If (b3 σ, y¯12 ) ≥ 2, then (b3 y12 , σ ) ≥ 2. Since (b3 y12 , b5 ) = 1 by (3.4), we must have that b3 y12 = 2σ + b5 + · · · .

3.3 Case z = z3

163

Thus, (b3 y12 , b3 y12 ) ≥ 6. So (b¯3 y12 , b¯3 y12 ) = (b3 y12 , b3 y12 ) ≥ 6. But we have already shown that (b3 y12 b3 y12 ) ≤ 5, a contradiction. This proves that (b3 σ, y¯12 ) < 2. Similarly, we can prove that (b3 θ, y¯12 ) < 2. Therefore, (3.63) forces that (b3 σ, y¯12 ) = 1 and (b3 θ, y¯12 ) = 1. Thus, we have that (b3 y12 , σ ) = 1, and (b3 y12 , θ ) = 1. So we can set b3 y12 = σ + θ + b5 + x8 ,

x8 ∈ N(B).

(3.65)

Now we prove that x8 ∈ B. If x8 ∈ / B, then we must have that x8 = s4 + t4 , s4 , t4 ∈ B, and s4 = t4 . Hence b¯3 s4 = b¯3 t4 = y12

and

(Σ16 , Σ16 ) = 3.

So from b¯32 s4 = b¯3 (b¯3 s4 ), (3.2), and (3.55), we see that b4 s4 + b5 s4 = b8 + b12 + Σ16 . If b4 s4 = Σ16 , then (b4 s4 , b52 ) = 3 by (3.54), and hence (b4 b5 , b5 s4 ) = 3. Thus, |b5 s4 | ≥ 24 by (3.36), a contradiction. So b4 s4 = Σ16 . But from (b3 b¯3 )s4 = b3 (b¯3 s4 ), we see that s4 b8 = σ + θ + b5 + t4 .

(3.66)

So (b5 s4 , b8 ) = 1. Thus, b4 s4 = Σ16 forces that (b5 s4 , b12 ) = 0. So we must have that (b4 s4 , b12 ) = 1. Therefore, we may assume that Σ16 = p4 + q + r,

b4 s4 = b12 + p4 ,

b5 s4 = b8 + q + r,

p4 , q, r ∈ B.

Since (s4 t¯4 , b8 ) = 1 by (3.66), s4 = t4 , and L1 (B) = {1}, we see that (b4 s4 , b4 t4 ) = (s4 t¯4 , b42 ) = 1 by (3.6). So in a way that is similar to the above discussions, we get that Σ16 = p4 + q4 + r8 ,

b4 t4 = b12 + q4 ,

p4 , q4 , r8 ∈ B,

b5 t4 = b8 + p4 + r8 ,

p4 = q4 .

Note that b3 p4 = y12 and b3 q4 = y12 by (3.55). So b32 p4 = b3 (b3 p4 ) yields that b4 p4 + b5 p4 = σ + θ + b5 + s4 + t4 . But (b4 p4 , s4 ) = 1, (b5 p4 , t4 ) = 1, and (b5 p4 , b5 ) = 1. Thus, without loss of generality, we may assume that σ = σ12 , θ = θ11 , and b4 p4 = s4 + σ12 ,

b5 p4 = t4 + b5 + θ11 .

Similarly, we also have that b4 q4 = s4 + σ12 ,

b5 q4 = t4 + b5 + θ11 .

164

3


Hence (p4 q¯4 , b42 ) = (b4 p4 , b4 q4 ) = 2, and (p4 q¯4 , b52 ) = (b5 p4 , b5 q4 ) = 3. However, it follows from (p4 q¯4 , b42 ) = 2, p4 = q4 , L1 (B) = {1}, and L2 (B) = ∅ that p4 q¯4 = 2b8 . Hence (p4 q¯4 , b52 ) = 2 by (3.54), a contradiction. This proves that x8 ∈ B. Thus, (b¯3 y12 , b¯3 y12 ) = (b3 y12 , b3 y12 ) = 4, and hence (Σ16 , Σ16 ) = 2. Assume that Σ16 = p + q, p, q ∈ B. Then by (3.55), b¯3 y12 = b8 + b12 + p + q. Hence |p| ≥ 4 and |q| ≥ 4. If |p| = 4, then p = p4 , b3 p4 = y12 , and both p4 and q are real (because Σ16 is real). Thus, (3.2) and b32 p4 = b3 (b3 p4 ) yield that b4 p4 + b5 p4 = σ + θ + b5 + x8 . Since (b5 p4 , b5 ) = 1, we see that (b5 p4 , b5 p4 ) = 2 or 3. If (b5 p4 , b5 p4 ) = 2, then (b4 p4 , b4 p4 ) = 2. Since |σ | + |θ | = 23, without loss of generality, we may assume that σ = σ8 , and b4 p4 = x8 + σ8 . Recall that σ8 is real, (b3 σ8 , x¯9 ) = 1 by (3.42), and (b3 σ8 , y¯12 ) = 1 by (3.65). So we have that b3 σ8 = x¯9 + y¯12 + x3 , where x3 is a ∗ or Σ ¯ 15 by (3.63). But any component of Σ¯ ∗ and Σ¯ 15 has degree component of Σ¯ 12 12 at least 4, a contradiction. If (b5 p4 , b5 p4 ) = 3, then (b4 p4 , b4 p4 ) = 1, and hence we have either b4 p4 = σ or b4 p4 = θ . Thus, (b4 y¯12 , y12 ) = (b4 y¯12 , b3 p4 ) = (b4 p4 , b3 y12 ) = 1, a contradiction to (3.64). This proves that |p| > 4. Similarly we can prove that |q| > 4. So the lemma holds. For the rest of this section, we will set Σ16 = p +q, p, q ∈ B. Then b52 = 1+b8 + p + q by (3.54). From (b¯3 b3 )b8 = b¯3 (b3 b8 ), (3.1), (3.5), (3.35), and Lemma 3.2, we see that b82 = 1 + b7 + 2b8 + 2b12 + p + q. Thus, we have that b52 = 1 + b8 + p + q

and

b82 = 1 + b7 + 2b8 + 2b12 + p + q.

(3.67)

∗ by (3.38). In the following, we prove that Σ ∗ ∈ B. Recall that b3 b7 = x9 + Σ12 12 But first we need to prove the next lemma. ∗ be the same as in (3.38). Then (Σ ∗ , Σ ∗ ) ≤ 2, and either Lemma 3.3 Let Σ12 12 12 ∗ (b3 σ, Σ¯ ) = 0 or (b3 θ, Σ¯ ∗ ) = 0. 12

12

Proof Let us first show that ∗ ∗ (Σ12 , Σ12 ) ≤ 2.

(3.68)

∗ , Σ∗ ) (Σ12 12

≥ 3, and hence (b3 b7 , b4 x¯9 ) ≥ 4 by If (3.68) does not hold, then (3.38) and (3.39). So b7 real implies that (b4 b7 , b3 x9 ) ≥ 4. Also b4 real implies (b4 b7 , b4 ) = 1 by (3.6). But b3 x9 = b4 + σ + θ by (3.42). Hence (b4 b7 , σ + θ ) ≥ 3.

(3.69)

Since |σ | + |θ | = 23 by (3.42), and (b4 b7 , b4 ) = 1, the above inequality forces that either (b4 b7 , σ ) = 0 or (b4 b7 , θ ) = 0. Without loss of generality, we may assume

3.3 Case z = z3

165

that (b4 b7 , θ ) = 0. So (3.69) yields that (b4 b7 , σ ) ≥ 3, and b4 b7 = b4 + 3σ + · · · . Therefore, (b4 b7 , b4 b7 ) ≥ 10. So (b42 , b72 ) ≥ 10, and hence (3.6) forces that |b72 | ≥ 1 + 7 · 9 = 64, a contradiction. Thus, (3.68) must hold. ∗ ) = 0 and (b θ, Σ ¯ ∗ ) = 0, since any component of Σ15 has deIf both (b3 σ, Σ¯ 12 3 12 ∗ , Σ ∗ ) = 2 by (3.68) and (3.63). gree at least 4 by (3.61), we must have that (Σ12 12 ∗ So we may set Σ12 = r + s, r, s ∈ B. Thus, b3 b7 = x9 + r + s by (3.38). So (b¯3 r, b¯3 s) ≥ 1, and hence (b3 r, b3 s) = (b¯3 r, b¯3 s) ≥ 1. Let g be a common component of b3 r and b3 s. Then (b¯3 g, r) ≥ 1, and (b¯3 g, s) ≥ 1. Hence |r| + |s| = 12 ∗ ) = 0 and (b θ, Σ ¯ ∗ ) = 0, from (3.63) we see forces that |g| ≥ 4. Since (b3 σ, Σ¯ 12 3 12 ∗ ∗ that (b3 Σ12 , σ ) = 1 and (b3 Σ12 , θ ) = 1. Therefore, there is h ∈ B such that |h| ≤ 5 and ∗ b3 Σ12 = σ + θ + 2g + h.

Hence it follows from (3.38), (3.2), (3.42), and b32 b7 = b3 (b3 b7 ) that b4 b7 + b5 b7 = b4 + 2σ + 2θ + 2g + h.

(3.70)

Similar to the proof in the above paragraph, we may assume that b4 b7 = b4 + 2σ + · · · and (b4 b7 , θ ) = 0. Note that (b3 b¯3 , b72 ) = (b3 b7 , b3 b7 ) = 3 by (3.38). Hence (b72 , b8 ) = 2 by (3.1). Thus, (b4 b7 , b5 b7 ) = (b4 b5 , b72 ) ≥ 2 by (3.36), a contradiction ∗ ) = 0 or (b θ, Σ ¯ ∗ ) = 0. to (3.70). This proves that either (b3 σ, Σ¯ 12 3 12 From Lemma 3.3 we can prove the following ∗ be the same as in (3.38). Then Σ ∗ ∈ B. Lemma 3.4 Let Σ12 12 ∗ ∈ ∗ , Σ ∗ ) = 2 by / B. Then (Σ12 Proof Toward a contradiction, assume that Σ12 12 (3.68). As in the proof of Lemma 3.3, we may assume that b4 b7 = b4 + 2σ + · · · . ∗ ) = 0, then Then L2 (B) = ∅ forces that either |σ | = 12 or |σ | ≤ 10. If (b3 θ, Σ¯ 12 ∗ + ··· b3 σ = x¯9 + y¯12 + Σ¯ 12

by (3.63) and Lemma 3.2. Since any component of Σ15 has degree at least 4 by (3.61), the above equality implies that either |σ | = 11 or |σ | ≥ 13, a contra∗ ) = 0 by Lemma 3.3. But, on the other diction. Thus, we must have that (b3 σ, Σ¯ 12 2 hand, from (3.38), (3.2), (3.42), and b3 b7 = b3 (b3 b7 ), we obtain that ∗ . b4 b7 + b5 b7 = b4 + σ + θ + b3 Σ12 ∗ ) = 1, a contradiction. So (b4 b7 , σ ) = 2 and the above equality force that (b3 σ, Σ¯ 12 ∗ This proves that Σ12 ∈ B.

166

3


∗ = c ∈ B. Then (3.38) and (3.17) yield that In the following, we denote Σ12 12

b3 b7 = x9 + c12

and

b4 x¯9 = b3 + y12 + x9 + c12 .

(3.71)

Lemma 3.5 Let Σ15 be the same as in (3.61). Then Σ15 ∈ B. Proof It follows from (b¯3 b4 )x¯9 = (b4 x¯9 )b¯3 , (3.3), (3.35), (3.71), (3.1), (3.42), and Lemma 3.2 that x9 x¯9 = 1 + 2b8 + b12 + p + q + b¯3 c12 . Since (b3 p + b3 q, y12 ) = 2 by Lemma 3.2, and |p| + |q| = 16, we must have that (b3 p + b3 q, c12 ) ≤ 2. Thus, (b¯ 3 c12 , p + q) ≤ 2, and hence (b52 , x9 x¯9 ) ≤ 7 by Lemma 3.2 and (3.54). So (b5 x¯9 , b5 x¯9 ) ≤ 7, and (3.62) forces that (Σ15 , Σ15 ) ≤ 2 and (3.54). On the other hand, since (b5 x¯9 , b5 x¯9 ) ≥ 6 by (3.62), we see that (b52 , x9 x¯9 ) ≥ 6, and hence (b¯3 c12 , p + q) ≥ 1 by Lemma 3.2. Therefore, 1, if Σ15 ∈ B; (b¯3 c12 , p + q) = 2, otherwise. Furthermore, since |p| ≥ 5 and |q| ≥ 5 by Lemma 3.2, we see that / B, if Σ15 ∈

then p = p8 ,

q = q8 ,

and

b3 p8 = b3 q8 = y12 + c12 . (3.72)

Toward a contradiction, in the following we assume that Σ15 ∈ / B. Then from the proof above, (Σ15 , Σ15 ) = 2. Since p + q is real, it follows from Lemma 3.2, (3.72), and b¯32 y12 = b¯3 (b¯3 y12 ) that b4 y12 + b5 y12 = b¯3 + 2x¯9 + 4y¯12 + 2c¯12 + Σ¯ 15 . But, on the other hand, from b32 y12 = b3 (b3 y12 ), (3.63), and Lemma 3.2, we see that b4 y12 + b5 y12 = b¯3 + 2x¯9 + 3y¯12 + c¯12 + Σ¯ 15 + b3 x8 . Thus, b3 x8 = y¯12 + c¯12 . Hence b32 x8 = b3 (b3 x8 ) and Lemma 3.2 yield that b4 x8 + b5 x8 = b8 + b12 + p8 + q8 + b3 c¯12 . Since (b3 c¯12 , b12 ) = 0 by (3.61), the above equation implies that (b4 x8 , b12 ) ≤ 1.

(3.73)

Now from (3.42), (3.71), (3.6), (3.1), (3.35), Lemma 3.2, and b4 (b3 x9 ) = b3 (b4 x9 ), we get that b4 σ + b4 θ = 2b8 + 2b12 + p + q + b3 c¯12 .

3.3 Case z = z3

167

Since b4 (b3 y12 ) = b4 σ + b4 θ + b4 b5 + b4 x8 by Lemma 3.2, and b3 (b4 y12 ) = b3 x¯9 + 2b3 y¯12 + b3 Σ¯ 15 by (3.64), it follows that b3 c¯12 + b4 x8 = b7 + p + q + b3 Σ¯ 15 . Since (b3 c¯12 , b12 ) = 0 and (b3 Σ¯ 15 , b12 ) = 2 by (3.61), the above equation forces that (b4 x8 , b12 ) = 2, a contradiction to (3.73). This proves that Σ15 must be an element in B. In the following, we denote Σ15 = c15 ∈ B. Then from (3.61), (3.62), and (3.64), we obtain that b3 b12 = x9 + y12 + c15 ,

b5 x¯9 = 2x9 + y12 + c15 ,

b4 y¯12 = x9 + 2y12 + c15 . (3.74)

Lemma 3.6 Assume that |σ | < |θ |. Then σ = σ11 , θ = θ12 , and b3 σ11 = x¯9 + y¯12 + c¯12 ,

b3 θ12 = x¯ 9 + y¯12 + c¯15 .

Proof Since |σ | < |θ |, we must have that (b3 σ, c¯15 ) = 0. In the following, we show that (b3 θ, c¯12 ) = 0 by (3.42), (3.63), and Lemma 3.2. Toward a contradiction, assume that (b3 θ, c¯12 ) = 0. Then from (3.42), (3.63), and Lemma 3.2, we must have that σ = σ7 , θ = θ16 , and b3 σ7 = x¯9 + y¯12 ,

b3 θ16 = x¯9 + y¯12 + c¯12 + c¯15 .

(3.75)

Thus, b32 σ7 = b3 (b3 σ7 ) yields that b4 σ7 + b5 σ7 = b7 + 2b8 + 2b12 + p + q. Note that (b3 b¯3 )b4 = b3 (b¯3 b4 ) and (3.1), (3.2), (3.3, (3.42) yield that (b4 b8 , σ7 ) = 1. Hence (b4 σ7 , b8 ) = 1 and similarly, (b5 σ7 , b8 ) = 1. But the degrees of p and q are at least 5, and the above equality forces that p = p8 , q = q8 . Without loss of generality, we may assume that b4 σ7 = b8 + b12 + p8 ,

b5 σ7 = b7 + b8 + b12 + q8 .

Note that from b4 (b3 σ7 ) = (b3 b4 )σ7 , we get that σ7 x¯9 = b3 + x9 + 2y12 + c12 + c15 . Hence b3 (b4 σ7 ) = (b3 b4 )σ7 implies that b3 p8 = y12 + c12 . So we may set b3 q8 = y12 + ε12 , ε12 ∈ N(B). Thus, from b3 (b5 σ7 ) = (b3 b5 )σ7 , we have that σ7 y¯12 = b3 + 2x9 + 2y12 + c12 + c15 + ε12 . Note that (b3 b¯3 , σ72 ) = 2, (b42 , σ72 ) = 3, (b52 , σ72 ) = 4, and (b4 b5 , σ72 ) = 2. So there is x5 ∈ B such that σ72 is one of the following: 1 + b7 + b8 + p8 + q8 + b12 + x5 ,

168

3


1 + b7 + b8 + 2p8 + b12 + x5 , 1 + b7 + b8 + 2q8 + b12 + x5 . Hence from (b3 σ7 )σ7 = b3 σ72 , we must have that σ72 = 1 + b7 + b8 + p8 + q8 + b12 + x5

and b3 x5 = c15 .

Thus, (b5 x5 , b4 + σ7 + θ16 ) = (b5 x5 , b¯3 x¯9 ) = (b5 x¯9 , b3 x5 ) = 1. So (b5 x5 , θ16 ) = 1. On the other hand, (b4 x5 , b5 + x8 + σ7 + θ16 ) = (b4 x5 , b¯3 y¯12 ) = (b3 x5 , b4 y¯12 ) = 1. Thus, (b4 x5 , x8 + θ16 ) = 1. But b4 x5 + b5 x5 = b32 x5 = b3 c15 , and (b3 c15 , θ16 ) = 1 by (3.75). So we must have that (b4 x5 , θ16 ) = 0, and hence (b4 x5 , x8 ) = 1. Thus, (b4 x8 , x5 ) = 1. Since b32 x8 = b3 (b3 x8 ) and b3 x8 = y12 + ε12 , we see that b4 x8 + b5 x8 = b5 + x8 + σ7 + θ16 + b3 ε12 . Hence (b3 ε12 , x5 ) ≥ 1. So (b¯3 x5 , ε12 ) ≥ 1. However, (b¯3 x5 , b¯3 x5 ) = (b3 x5 , b3 x5 ) = 1, a contradiction. This proves that (3.75) is not true. So we must have that (b3 θ, c¯12 ) = 0, and the lemma holds by (3.42), (3.63), and Lemma 3.2. For the rest of this section, we will assume that σ = σ11 and θ = θ12 . Note that b32 b8 = b3 (b3 b8 ) yields that b4 b8 + b5 b8 = 2b4 + 2b5 + x8 + 2σ11 + 2θ12 . Since (b4 b8 , b4 ) = 1 and (b4 b8 , b5 ) = 1, the above equality forces that b4 b8 = b4 + b5 + σ11 + θ12

and

b5 b8 = b4 + b5 + σ11 + θ12 + x8 .

(3.76)

Lemma 3.7 Let p and q be the same as in Lemma 3.2. Then p = p8 and q = q8 . Proof Toward a contradiction, we may assume that |p| > |q|. From b¯32 y12 = b¯3 (b¯3 y12 ), Lemma 3.2, and (3.74), we obtain that b5 y12 = b¯3 + x¯9 + b¯3 p + b¯3 q. On the other hand, from b32 y12 = b3 (b3 y12 ), Lemma 3.2, and (3.74), we see that b5 y12 = b¯3 + x¯9 + c¯12 + y¯12 + b3 x8 .

3.3 Case z = z3

169

Thus, the components of b¯3 p and b¯3 q have degrees at least 4. Since |p| > |q|, p and q are real, (b¯3 c12 , p + q) = 1 from the proof of Lemma 3.5, and (b¯3 y12 , p) = (b¯3 y12 , p) = 1 by Lemma 3.2, we must have that p = p10 , q = q6 , and b¯3 p10 = y¯12 + c¯12 + x6 ,

b¯3 q6 = y¯12 + y6 ,

b3 x8 = y¯12 + x6 + y6 ,

x6 , y6 ∈ B.

Thus, b¯ 32 q6 = b¯3 (b¯3 q6 ) and Lemma 3.2 yield that b4 q6 + b5 q6 = b5 + x8 + σ11 + θ12 + b¯3 y6 . Since (b5 q6 , b5 ) = 1 by (3.67), the above equality forces that either (b5 q6 , σ11 ) = 0 or (b5 q6 , θ12 ) = 0. In the following, we derive a contradiction for each of these two cases. First assume that (b5 q6 , σ11 ) = 0. Since (b¯3 y6 , x8 ) = 1, we must have that b4 q6 = σ11 + x8 + d5 ,

b5 q6 = b5 + θ12 + x8 + e5 ,

b¯3 y6 = x8 + d5 + e5 ,

d5 , e5 ∈ B.

Hence it follows from b5 (b3 q6 ) = b3 (b5 q6 ) that c¯12 + b5 y¯6 = c¯15 + y¯12 + b3 e5 . Thus, b3 e5 = c¯12 + x3 ,

b5 y¯6 = c¯15 + y¯12 + x3 ,

x3 ∈ B.

On the other hand, b4 (b3 q6 ) = b3 (b4 q6 ) implies that c¯15 + b4 y¯6 = c¯12 + x6 + y6 + b3 d5 . Thus, b3 d5 = c¯15

and

b4 y¯6 = c¯12 + x6 + y6 .

Therefore, (b5 y¯6 , b4 y¯6 ) = 0, and hence (y6 y¯6 , b4 b5 ) = 0. So (y6 y¯6 , b8 ) = 0 by (3.36). But b¯3 y6 = x8 + d5 + e5 forces that (y6 y¯6 , b8 ) = 2, a contradiction. Now assume that (b5 q6 , θ12 ) = 0. Then (b¯3 y6 , x8 ) = 1 forces that b4 q6 = θ12 + x8 + d4 ,

b5 q6 = b5 + σ11 + x8 + Σ6 ,

b¯3 y6 = x8 + d4 + Σ6 ,

d4 ∈ B, Σ6 ∈ N(B).

Hence from b4 (b3 q6 ) = b3 (b4 q6 ) we see that b4 y¯6 = x6 + y6 + b3 d4 . But (b3 d4 , y6 ) = 1. So (b4 y¯6 , b4 y¯6 ) ≥ 6, and hence (y6 y¯6 , b42 ) ≥ 6. Note that b¯3 y6 = x8 + d4 + Σ6 implies that (y6 y¯6 , b8 ) ≥ 2. Thus, by (3.6) we get that |y6 y¯6 | ≥ 1 + 2 · 8 + 3 · 7 = 38, a contradiction. This proves that |p| = |q| = 8, and the lemma holds.

170

3


Since (b¯3 c12 , p8 + q8 ) = 1 from the proof of Lemma 3.5, and (b¯3 y12 , p8 ) = ¯ (b3 y12 , p8 ) = 1 by Lemma 3.2, without loss of generality, we may set b¯3 p8 = y¯12 + c¯12 ,

b¯3 q8 = y¯12 + ε12 ,

ε12 ∈ N(B).

(3.77)

b5 y12 = b¯3 + x¯9 + c¯12 + 2y¯12 + ε12 .

(3.78)

Thus, from the proof of Lemma 3.7 we see that b3 x8 = y¯12 + ε12 ,

and

Lemma 3.8 There exists real elements t9 , t12 ∈ B such that b3 c¯12 = b7 + p8 + t9 + t12 ,

b4 σ11 = b7 + b8 + b12 + t9 + p8

and b5 σ11 = b7 + b8 + b12 + p8 + q8 + t12 ,

b4 θ12 = b8 + b12 + t12 + p8 + q8 .

Proof From b4 (b3 x9 ) = b3 (b4 x9 ), (3.42), and (3.71), we obtain that b4 σ11 + b4 θ12 = 2b8 + 2b12 + p8 + q8 + b3 c¯12 .

(3.79)

From b4 (b3 y12 ) = b3 (b4 y12 ), Lemma 3.2, and (3.64), we see that b4 σ11 + b4 θ12 + b4 x8 = b7 + 2b8 + 2b12 + 2p8 + 2q8 + b3 c¯15 .

(3.80)

Thus, if τ is a component of b3 c¯12 and τ = b7 or p8 , then τ is also a component of b3 c¯15 . Hence both c¯12 and c¯15 are components of b¯3 τ . So the degree of τ is at least 9. But from (3.71), (3.77), (3.79), and (3.80), b3 c¯12 = b7 + p8 + the linear combination of some common components of b3 c¯12 and b3 c¯15 . Therefore (b3 c¯12 , b3 c¯12 ) = 3 or 4. It follows from b42 b8 = b4 (b4 b8 ), (3.6), (3.67), and (3.36) that b8 + b7 b8 + b12 + p8 + q8 = b4 σ11 + b4 θ12 . Thus, b7 b8 = b8 + b12 + b3 c¯12 . Hence (b7 b8 , b7 b8 ) = 5 or 6. So (b72 , b82 ) = 5 or 6.

(3.81)

In the following, we show that (b4 b7 , σ11 ) = 1. If (b4 b7 , σ11 ) = 1, since (b4 b7 , b4 ) = 1 by (3.6) and L2 (B) = ∅, we must have that (b4 b7 , σ11 ) = 0. Note that b32 b7 = b3 (b3 b7 ) implies that b4 b7 + b5 b7 = b4 + σ11 + θ12 + b3 c12 . So (b3 c12 , σ11 ) = 1 forces that b4 b7 = b4 + θ12 + · · ·

and b5 b7 = 2σ11 + · · · .

3.3 Case z = z3

171

Thus, (b4 b7 , b4 b7 ) ≥ 3, and (b5 b7 , b5 b7 ) ≥ 5. So (b3 b7 , b3 b7 ) = 2 implies that (b72 , b8 ) = 1,

(b72 , b7 ) ≥ 1,

(b72 , p8 + q8 ) ≥ 3.

Therefore, (b72 , b82 ) ≥ 7 by (3.67), a contradiction to (3.81). This proves that (b4 b7 , σ11 ) = 1. Thus, b4 b7 = b4 + σ11 + · · ·

and b5 b7 = σ11 + θ12 + · · · .

Note that b32 θ12 = b3 (b3 θ12 ), (3.2), and Lemma 3.6 imply that b4 θ12 + b5 θ12 = b7 + 2b8 + 2b12 + p8 + q8 + b3 c¯15 . Hence (b4 θ12 , p8 ) ≤ 1. So (b4 σ11 , p8 ) ≥ 1 by (3.79). It follows from b42 b5 = b4 (b4 b5 ), (3.6), (3.36), and (3.76) that b4 b12 = b5 + x8 + b5 b7 = b5 + x8 + σ11 + θ12 + · · · . Thus, b4 σ11 = b7 + b8 + b12 + p8 + · · · . So (3.79) forces that b3 c¯12 = b7 + p8 + t9 + t12 ,

t9 , t12 ∈ B,

and b4 σ11 = b7 + b8 + b12 + t9 + p8 ,

b4 θ12 = b8 + b12 + t12 + p8 + q8 .

Clearly t9 and t12 are real by the above equalities. Furthermore, b32 σ11 = b3 (b3 σ11 ) implies that b5 σ11 = b7 + b8 + b12 + p8 + q8 + t12 . So the lemma holds. Lemma 3.9 There exists d12 ∈ B such that b3 c¯15 = t9 + t12 + b12 + d12 . Furthermore, (b3 c¯15 , b3 c¯15 ) = 4. Proof It follows from Lemma 3.8 and (3.80) that t9 + t12 + b4 x8 = q8 + b3 c¯15 . Thus, q8 is a component of b4 x8 . Furthermore, since (b3 c¯15 , p8 ) = 0 by (3.77), the above equation forces that (b4 x8 , p8 ) = 0. Note that b32 x8 = b3 (b3 x8 ) yields that b4 x8 + b5 x8 = b8 + b12 + p8 + q8 + b3 ε12 by (3.78) and Lemma 3.2. Since (b3 c¯15 , b12 ) = 1 by (3.74), we see that (b4 x8 , b12 ) = 1. Also (b4 x8 , b8 ) = (b4 b8 , x8 ) = 0 by (3.76). Thus, if τ is a component of b4 x8 and τ = q8 or b12 , then τ is a component of both b3 c¯15 and b3 ε12 . But we have either ε12 ∈ B or any component of ε12 has degree at least 4 by (3.78). Hence,

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3


as a common component of b3 c¯15 and b3 ε12 , the degree of tau is at least 7. Furthermore, b4 x8 = b12 + q8 + the linear combination of some common components of b3 c¯15 and b3 ε12 . Thus, we must have that b3 c¯15 = t9 + t12 + b12 + d12 ,

b4 x8 = b12 + d12 + q8 ,

d12 ∈ B.

Hence from Lemmas 3.6 and 3.8, and b32 θ12 = b3 (b3 θ12 ), we get that b5 θ12 = b7 + b8 + 2b12 + t9 + d12 . Note that b12 = d12 or t12 by (3.74). Also d12 = t12 ; otherwise (b¯3 t12 , c¯15 ) = 2 and (b¯3 t12 , c¯12 ) = 1, a contradiction. Thus, (b3 c¯15 , b3 c¯15 ) = 4. From Lemmas 3.8 and 3.9, we have the following equalities: b¯3 t9 = c¯12 + c¯15

and b¯3 t12 = c¯12 + c¯15 + Σ9 ,

Σ9 ∈ N(B).

(3.82)

Lemma 3.10 There exist real elements s4 , s9 , s12 ∈ B such that b3 c12 = σ11 + s12 + s9 + s4 ,

b¯3 s4 = c12 ,

b¯3 s9 = c12 + c15

and b4 s4 = b7 + t9 ,

b5 s4 = p8 + t12 .

Proof Recall that (b3 c12 , σ11 ) = 1 and (b3 c15 , θ12 ) = 1 by Lemma 3.6. Since (b3 c12 , b3 c12 ) = (b3 c¯12 , b3 c¯12 ) = 4, (b3 c15 , b3 c15 ) = (b3 c¯15 , b3 c¯15 ) = 4, and (b3 c12 , b3 c15 ) = (b3 c¯12 , b3 c¯15 ) = 2, we see that b3 c12 and b3 c15 have two common components. Clearly these two components have degrees at least 9. Thus, besides σ11 , b3 c12 has two components of degrees at least 9, and its fourth component has degree at least 4. From the proof of Lemma 3.8, we have the following equations: b4 b7 + b5 b7 = b4 + σ11 + θ12 + b3 c12 , b4 b7 = b4 + σ11 + · · · , b5 b7 = σ11 + θ12 + · · · . Thus, we must have s4 , s9 , s12 ∈ B such that b3 c12 = σ11 + s12 + s9 + s4 , b4 b7 = b4 + σ11 + s9 + s4 , b5 b7 = σ11 + θ12 + s12 . Clearly s4 , s9 , and s12 are all real, and b¯3 s4 = c12 .

(3.83)

3.3 Case z = z3

173

Furthermore, since we have already shown that (b3 c15 , b3 c15 ) = 4, and that (b3 c15 and b3 c¯12 have two common components, we must have that b3 c15 = θ12 + s12 + s9 + e12 ,

e12 ∈ B, and θ12 , s12 , e12 are distinct.

(3.84)

Thus, b¯3 s9 = c12 + c15 . Now from b¯32 s4 = b¯ 3 (b¯3 s4 ), b¯3 s4 = c12 , and Lemma 3.8, we obtain that b4 s4 + b5 s4 = b7 + p8 + t9 + t12 . Hence we must have that b4 s4 = b7 + t9

and

b5 s4 = p8 + t12 .

So the lemma holds. Lemma 3.11 There exist r4 , r5 ∈ B such that b¯3 t12 = c¯12 + c¯15 + r4 + r5 . Proof It follows from (b3 b¯3 )s4 = b3 (b¯3 s4 ), (3.1), and Lemma 3.10 that s4 b8 = σ11 + s12 + s9 . Thus, from (3.6), Lemma 3.10, (3.83), and b4 (b4 s4 ) = b42 s4 , we see that s4 b7 + s12 = b4 + b4 t9 . But b¯32 t9 = b¯3 (b¯3 t9 ), (3.82), and (3.84) yield that b4 t9 + b5 t9 = σ11 + θ12 + 2s9 + 2s12 + s4 + e¯12 . So e12 is real, and we must have that b4 t9 = σ11 + s4 + s12 + s9 , b5 t9 = θ12 + s12 + e12 + s9 , s4 b7 = b4 + s4 + s9 + σ11 . Hence from b¯3 (s4 b7 ) = b7 (b¯3 s4 ) we see that b7 c12 = b3 + 2x9 + 3c12 + c15 + y12 . Note that it follows from (b¯3 b4 )s4 = b¯3 (b4 s4 ) and (3.82) that s4 x9 = x¯9 + c¯12 + c¯15 . Thus, (b7 b¯3 )s4 = b7 (b¯3 s4 ) yields that s4 c¯12 = b3 + x9 + 2c12 + y12 .

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Therefore, (b¯3 s4 )c¯12 = b¯3 (s4 c¯12 ) implies that c12 c¯12 = 1 + 3b7 + 3b8 + 2b12 + 3p8 + q8 + 2t9 + 2t12 . On the other hand, from b¯3 (b5 s4 ) = b5 (b¯3 s4 ) and (3.82), we get that b5 c¯12 = y¯12 + 2c¯12 + c¯15 + Σ9 . Since (b5 c¯12 , b5 c¯12 ) = (b52 , c12 c¯12 ) = 8, the above equality forces that (Σ9 , Σ9 ) = 2. But the degree of any component in Σ9 has to be at least 4 by (3.82). Thus, Σ9 = r4 + r5 , r4 , r5 ∈ B, and hence by (3.82), b¯3 t12 = c¯12 + c¯15 + r4 + r5 .

So the lemma holds.

Now we are ready to reach a contradiction for Subcase 1B. It follows from Lemma 3.11 that b3 r4 = t12 . Thus, b32 r4 = b3 (b3 r4 ) implies that b4 r4 + b5 r4 = c12 + c15 + r¯4 + r¯5 . Therefore, b4 r4 = c12 + r¯4

and b5 r4 = c15 + r¯5 .

So (b4 c12 , r4 ) = 1.

(3.85)

But, on the other hand, b4 (b¯3 s4 ) = b¯3 (b4 s4 ) forces that b4 c12 = x¯9 + 2c¯12 + c¯15 , a contradiction to (3.85).

3.4 Cases z = z4 , z = z5 , z = z6 , z = z7 , and z = z8 In this section we derive contradictions for the cases z = z4 , z = z6 , z = z6 , z = z7 , and z = z8 . Case 2 z = z4 . So we have that z = z4 ,

w = w8 ,

b¯3 z4 = b4 + b5 + σ3 ,

b¯4 w8 = b4 + θ20 .

Hence (b3 b¯3 , z4 z¯ 4 ) = (b¯ 3 z4 , b¯3 z4 ) = 3. So (b8 , z4 z¯ 4 ) = 2, a contradiction.

3.4 Cases z = z4 , z = z5 , z = z6 , z = z7 , and z = z8

175

Case 3 z = z5 . In this case we have the following: w = w7 ,

z = z5 ,

b¯3 z5 = b4 + b5 + σ6 ,

b¯3 w7 = b4 + θ17 .

(3.86)

Note that ∗ z5 b¯4 + w7 b¯4 = 2b3 + 2x9 + y12 + Σ12

by (3.40). Since (z5 b¯4 , b3 ) = 1 and (w7 b¯4 , b3 ) = 1 by (3.11), and any constituent of ∗ has degree at least 4 by (3.38), we have either Σ ∗ = α + α , α , α ∈ B, and Σ12 5 7 5 7 12 z5 b¯4 = b3 + y12 + α5 ,

w7 b¯4 = b3 + 2x9 + α7 ,

∗ = α + Σ , α ∈ B, Σ ∈ N(B), and or Σ12 4 8 4 8

z5 b¯4 = b3 + x9 + Σ8 ,

w7 b¯4 = b3 + x9 + y12 + α4 .

If w7 b¯4 = b3 + 2x9 + α7 , then by (3.1) and (3.35), (w7 b¯4 )b¯3 = 1 + 2b7 + 3b8 + 2b12 + b¯3 α7 . Hence ((w7 b¯4 )b¯3 , b8 ) ≥ 3. But if w7 b¯4 = b3 + x9 + y12 + α4 , then by (3.1) and (3.35), (w7 b¯4 )b¯3 = 1 + b7 + 2b8 + b12 + b¯3 y12 + b¯3 α4 . Since (b¯3 y12 , b8 ) = 1 by (3.5), we also have that ((w7 b¯4 )b¯3 , b8 ) ≥ 3. That is, we always have that ((w7 b¯4 )b¯3 , b8 ) ≥ 3. On the other hand, (3.86) and (3.6) imply that (b¯3 w7 )b¯4 = b4 b¯4 + θ17 b¯4 = 1 + b7 + b8 + θ17 b¯4 . Thus, (θ17 b¯4 , b8 ) ≥ 2. Hence (b4 b8 , θ17 ) ≥ 2, a contradiction. Case 4 z = z6 . So we have the following: z = z6 ,

w = w6 ,

b¯3 z6 = b4 + b5 + σ9 ,

b¯3 w6 = b4 + θ14 .

(3.87)

Note that ∗ z6 b¯4 + w6 b¯4 = 2b3 + 2x9 + y12 + Σ12

(3.88)

z6 b¯4 + u9 b¯4 = b3 + x9 + y12 + b3 b¯12

(3.89)

and

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3


by (3.40) and (3.41), respectively. Since (z6 b¯4 , b3 ) = 1 and (w6 b¯4 , b3 ) = 1 by ∗ is at least 4 by (3.38), from (3.88) (3.11), and the degree of any constituent of Σ12 we have either z6 b¯4 = b3 + x9 + y12 ,

∗ w6 b¯4 = b3 + x9 + Σ12

∗ , z6 b¯4 = b3 + x9 + Σ12

w6 b¯4 = b3 + x9 + y12 .

or

Subcase 4A z6 b¯4 = b3 + x9 + y12

∗ and w6 b¯4 = b3 + x9 + Σ12 .

(3.90)

∗ ∈ B. Toward a contradiction, assume that Σ ∗ ∈ First we prove that Σ12 12 / B. Then by (3.38), ∗ , b7 ) ≥ 2. (b¯3 Σ12

(3.91)

Since (b¯3 w6 )b¯4 = 1 + b7 + b8 + b¯4 θ14 by (3.87) and (3.6), and (b¯4 w6 )b¯3 = 1 + b7 + ∗ by (3.90), (3.1), and (3.35), we see that 2b8 + b12 + b¯3 Σ12 ∗ b¯4 θ14 = b8 + b12 + b¯3 Σ12 .

Thus, (b¯4 θ14 , b7 ) ≥ 2 by (3.91), and hence (b4 b7 , θ14 ) ≥ 2. Therefore, we must have that (b4 b7 , θ14 ) = 2, and b4 b7 = 2θ14 . But, on the other hand, we also have ∗ ∈ B. (b4 b7 , b4 ) = 1 by (3.6), a contradiction. This proves that Σ12 ∗ So we may assume that Σ12 = c12 ∈ B. Thus by (3.38), (3.39), and (3.90), w6 b¯4 = b3 + x9 + c12 . (3.92) From (3.4), (3.2), (3.12), (3.87), and (b3 b5 )b¯3 = (b¯3 b5 )b3 , we get that b3 b7 = x9 + c12 ,

b4 x¯9 = b3 + y12 + x9 + c12 ,

b3 y12 = σ9 + b¯3 u9 .

(3.93)

Hence (b¯3 σ9 , y12 ) ≥ 1. But we also have (b¯3 σ9 , x9 ) = 1 by (3.42). So we may assume that b¯3 σ9 = x9 + y12 + v6 , v6 ∈ N(B). Hence (b3 x9 )b¯3 = b3 + 2x9 + y12 + v6 + b¯3 θ14 by (3.3) and (3.95). But (b¯ 3 x9 )b3 = b3 + 2x9 + y12 + c12 + b3 b12 by (3.35), (3.92), and (3.5). So we have that v6 + b¯3 θ14 = c12 + b3 b12 .

(3.94)

Note that any constituent of b3 b12 has degree at least 4. So the above equality forces that v6 ∈ B. Thus, b¯3 σ9 = x9 + y12 + v6 ,

v6 ∈ B.

(3.95)

Now (3.87), (3.3), (3.4), and (3.95) yield that b¯3 (b¯3 z6 ) = 2b3 + 2x9 + 2y12 + v6 , and (3.2) and (3.90) imply that (b¯3 b¯3 )z6 = b3 + x9 + y12 + b¯5 z6 . Thus, b¯ 5 z6 = b3 + x9 + y12 + v6 .

(3.96)


177

Since (b3 b¯3 , z6 z¯ 6 ) = (b¯3 z6 , b¯3 z6 ) = 3 by (3.87), and (b4 b¯5 , z6 z¯ 6 ) = (b¯4 z6 , b¯5 z6 ) = 3 by (3.90) and (3.96), we see that (z6 z¯ 6 , b8 ) = 2

and (z6 z¯ 6 , b12 ) = 1.

Hence we also have that (z6 z¯ 6 , b¯12 ) = 1. Therefore, b12

is a real element.

Recall that (b3 b12 , x9 ) = 1 by (3.35), and b12 real implies that (b3 b12 , y12 ) ≥ 1 by (3.37). Thus, since any constituent of b¯3 θ14 has degree at least 6, from (3.94) we see that there is v9 ∈ B such that b3 b12 = x9 + y12 + v6 + v9

(3.97)

b¯3 θ14 = x9 + y12 + c12 + v9 .

(3.98)

and

Since (b3 y12 , b5 ) = 1 by (3.4), (b3 y12 , σ9 ) = 1 by (3.95), and (b3 y12 σ14 ) = 1 by (3.98), we must have that (b3 y12 , θ14 ) = 1 by (3.98). Thus, b3 y12 = b5 + σ9 + θ14 + t8 ,

t8 ∈ N(B).

(3.99)

So by (3.93), b¯3 u9 = b5 + θ14 + t8 .

(3.100)

If t8 ∈ / B, since any constituent of b3 y12 has degree at least 4, we see that t8 = c4 + d4 , c4 , d4 ∈ B. Hence (3.99) forces that b¯3 c4 = y12 . Thus, (b3 c4 , b3 c4 ) = (b¯3 c4 , b¯3 c4 ) = 1. But, on the other hand, we also have (b3 c4 , u9 ) = 1 by (3.100), a contradiction. Therefore, we have proved that t8 ∈ B. Since b12 is real, (3.89), (3.90), and (3.97) yield that u9 b¯4 = b3 b12 = x9 + y12 + v6 + v9 . Hence by (3.2), b¯32 u9 = x9 + y12 + v6 + v9 + b¯5 u9 . On the other hand, from (3.100), (3.4), and (3.98), we get that b¯3 (b¯3 u9 ) = b3 + x9 + 2y12 + c12 + v9 + b¯3 t8 . Thus, (b¯3 t8 , v6 ) = 1. So (b3 v6 , t8 ) = 1. But we also have (b3 v6 , σ9 ) = 1 by (3.95). Hence b3 v6 = t8 + σ9 + · · · . Since L1 (B) = {1}, the above equality cannot be true, a contradiction.

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3


Subcase 4B ∗ z6 b¯4 = b3 + x9 + Σ12

and

w6 b¯4 = b3 + x9 + y12 .

(3.101)

So by (3.1), (3.35), and (3.101), (w6 b¯4 )b¯3 = 1 + b7 + 2b8 + b12 + b¯3 y12 . But ¯ (b3 w6 )b¯4 = 1 + b7 + b8 + b¯4 θ14 by (3.87) and (3.6). Hence b¯ 4 θ14 = b8 + b12 + b¯3 y12 . Since (b¯3 y12 , b8 ) = 1 by (3.5), we see that (b¯4 θ14 , b8 ) = 2. Thus, (b4 b8 , θ14 ) = 2. But, on the other hand, from (b3 b¯3 )b4 = (b¯3 b4 )b3 , (3.1), (3.3), (3.2), and (3.42), we see that b4 b8 = b5 + b3 x9 = b4 + b5 + σ9 + θ14 , a contradiction. Case 5 z = z7 . In this case we assume that z = z7 ,

w = w5 ,

b¯3 z7 = b4 +b5 +σ12 ,

and b¯3 w5 = b4 +θ11 . (3.102)

∗ has degree at least 4 by (3.38). So from (3.40), Recall that any constituent of Σ12 ∗ we have either Σ12 = α4 + Σ8 , α4 ∈ B, Σ8 ∈ N(B), and

z7 b¯4 = b3 + x9 + y12 + α4 ,

w5 b¯4 = b3 + x9 + Σ8

(3.103)

w5 b¯4 = b3 + y12 + α5 .

(3.104)

∗ = α + α , α , α ∈ B, and or Σ12 5 7 5 7

z7 b¯4 = b3 + 2x9 + α7 ,

If (3.103) holds, then b¯3 (z7 b¯4 ) = 1 + b7 + 2b8 + b12 + b¯3 y12 + b¯3 α4 by (3.1) and (3.35), and (b¯3 z7 )b¯4 = 1 + b7 + 2b8 + b¯12 + b¯4 σ12 by (3.102), (3.6), and (3.36). Thus, b¯ 12 + b¯4 σ12 = b12 + b¯3 y12 + b¯3 α4 . Since (b¯3 α4 , b7 ) = 1 by (3.38), the above equality implies that (b¯4 σ12 , b7 ) = 1. Hence (b4 b7 , σ12 ) = 1. On the other hand, we also have that (b¯3 w5 )b¯4 = 1 + b7 + b8 + b¯4 θ11 by (3.102) and (3.6), and (w5 b¯4 )b¯3 = 1 + b7 + 2b8 + b12 + b¯ 3 Σ8 by (3.103), (3.1), and (3.35). So b¯ 4 θ11 = b8 + b12 + b¯3 Σ8 .


179

Since (b¯3 Σ8 , b7 ) ≥ 1 by (3.38), the above equality implies that (b¯4 θ11 , b7 ) ≥ 1. Hence (b4 b7 , θ11 ) ≥ 1. But we already have (b4 b7 , σ12 ) = 1. So (b4 b7 , θ11 ) = 1. Note that (b4 b7 , b4 ) = 1 by (3.6). Thus, b4 b7 = b4 + θ11 + σ12 + · · · . Since L1 (B) = {1}, the above equality cannot be true, a contradiction. If (3.104) holds, then (b4 x9 , z7 ) = 2. Recall that (b4 x9 , x9 ) = 1 by (3.17). So b4 x9 = 2z7 + x9 + · · · , and hence (b4 x9 , b4 x9 ) ≥ 6. But, on the other hand, since b¯4 x9 = b3 + x9 + y12 + α5 + α7 by (3.17), we see that (b4 x9 , b4 x9 ) = (b¯4 x9 , b¯4 x9 ) = 5, a contradiction. Case 6 z = z8 . In this case we assume that z = z8 ,

b¯3 z8 = b4 + b5 + σ15 ,

w = w4 ,

and b¯3 w4 = b4 + θ8 . (3.105)

Recall that by (3.40), ∗ . z8 b¯4 + w4 b¯4 = 2b3 + 2x9 + y12 + Σ12

Since (w4 b¯4 , b3 ) = 1 by (3.11), and L1 (B) = {1}, the above equality forces that ∗ Σ12 = α4 + Σ8 ,

α4 ∈ B, Σ8 ∈ N(B),

(3.106)

w4 b¯4 = b3 + x9 + α4 ,

(3.107)

z8 b¯4 = b3 + x9 + y12 + Σ8 .

(3.108)

and Thus, from (3.105), (3.6), (3.8), (3.108), (3.1), (3.9), and (b¯3 z8 )b¯4 = (z8 b¯4 )b¯3 , we get that b¯12 + b¯4 σ15 = b12 + b¯3 y12 + b¯3 Σ8 . But (b¯3 Σ8 , b7 ) ≥ 1 by (3.38) and (3.106). So (b¯4 σ15 , b7 ) ≥ 1, and hence (b4 b7 , σ15 ) ≥ 1. Therefore, we must have that (b4 b7 , σ15 ) = 1.

(3.109)

On the other hand, (b¯3 w4 )b¯4 = (b¯4 w4 )b¯3 yields that b¯4 θ8 = b8 + b12 + b¯3 α4 , by (3.105), (3.107), (3.6), (3.1), and (3.35). Since (b¯3 α4 , b7 ) = 1 by (3.38) and (3.106), the above equality implies that (b¯4 θ8 , b7 ) = 1. Hence (b4 b7 , θ8 ) = 1.

(3.110)

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3


But we also have that (b4 b7 , b4 ) = 1 by (3.6). So by (3.109) and (3.110), b4 b7 = b4 + θ8 + σ15 + · · · . However, L1 (B) = {1} implies that the above equality cannot be true, a contradiction.

3.5 Case z = z9 In this section we assume the following z = z9 ,

w = w3 ,

u = u6 ,

b¯3 z9 = b4 + b5 + σ18 ,

Since (b3 b¯3 , w3 w¯ 3 ) = (b¯ 3 w3 , b¯3 w3 ) = 2 by (3.111), we see that

b¯3 w3 = b4 + θ5 . (3.111)

w3 w¯ 3 = 1 + b8 .

(3.112)

Thus, from (b3 w¯ 3 )w3 = b3 (w3 w¯ 3 ), (3.111), and (3.5), we see that b¯4 w3 + θ¯5 w3 = 2b3 + x9 + y12 . Note that (b¯4 w3 , b3 ) = 1 and (θ¯5 w3 , b3 ) = 1 by (3.111). So we must have that b¯ 4 w3 = b3 + x9

and θ¯5 w3 = b3 + y12 .

Thus, (3.112) yields that (b3 θ5 , b3 θ5 ) = (b3 b¯3 , θ5 θ¯5 ) = (w3 w¯ 3 , θ5 θ¯5 ) = (θ¯5 w3 , θ¯5 w3 ) = 2. So we may set b3 θ5 = w3 + γ12 ,

γ12 ∈ B.

(3.113)

Furthermore, since (b3 w3 , b3 w3 ) = (b¯3 w3 , b¯3 w3 ) = 2 by (3.111), we may assume that b3 w3 = α + β, α, β ∈ B and |α| < |β|. Then from (3.11), (3.111), (3.113), and b¯3 (b3 w3 ) = b3 (b¯3 w3 ), we get that b¯ 3 α + b¯ 3 β = 2w3 + z9 + γ12 .

(3.114)

The above equality forces that either α = α4 and β = β5 , or α = 1 and β = β8 . Subcase 7A α = α4 and β = β5 . Thus, b3 w3 = α4 + β5 ,

(3.115)

b¯3 α4 = w3 + z9 ,

(3.116)

and by (3.114),

3.5 Case z = z9

181

and b¯3 β5 = w3 + γ12 .

(3.117)

Therefore, from (3.115), (3.112), (3.5), and (b3 w3 )w¯ 3 = b3 (w3 w¯ 3 ), we get that α4 w¯ 3 + β5 w¯ 3 = 2b3 + x9 + y12 . Since (α4 w¯ 3 , b3 ) = 1 by (3.116), and (β5 w¯ 3 , b3 ) = 1 by (3.117), the above equality yields that α4 w¯ 3 = b3 + x9

β5 w¯ 3 = b3 + y12 .

and

Thus, (3.111), (3.2), (3.42), and α4 (w¯ 3 b3 ) = (α4 w¯ 3 )b3 force that α4 b¯4 + α4 θ¯5 = 2b4 + b5 + σ18 + θ5 . However, the above equality cannot be true, a contradiction. Subcase 7B α = 1 and β = β8 . Thus, we must have that w3 = b¯3 and β8 = b8 . So (3.114) yields that b¯3 b8 = ¯b3 + z9 + γ12 . Hence (3.5) forces that z9 = x¯ 9

and

γ12 = y¯12 .

Therefore, b3 b4 = x¯9 + b¯3

and

b3 b5 = x¯ 9 + u6

(3.118)

by (3.11) and (3.12), respectively. Furthermore, from (3.111) we see that b¯3 x¯9 = b4 + b5 + σ18 . But, on the other hand, b3 x9 = b4 + σ18 + θ5 by (3.42). So we must have that b4 = b¯4 ,

θ5 = b¯5 ,

σ18 = σ¯ 18 .

(3.119)

Thus, (3.42) implies that b3 x9 = b4 + b¯5 + σ18 .

(3.120)

Note that b¯3 (b3 x9 ) = b3 + 2x9 + u¯ 6 + b¯3 σ18 by (3.120), (3.118), and (3.3). On the ∗ + b b by (3.35), (3.38), and (3.5). other hand, b3 (b¯3 x9 ) = b3 + 2x9 + y12 + Σ12 3 12 Thus, ∗ u¯ 6 + b¯3 σ18 = y12 + Σ12 + b3 b12 .

(3.121)

The above equality forces that (b3 b12 , u¯ 6 ) ≥ 1. Thus, (b3 u6 , b¯12 ) = 1, and hence / B, then t6 = c3 + d3 , c3 , d3 ∈ B, and b3 u6 = b3 u6 = b¯12 + t6 , t6 ∈ N(B). If t6 ∈ b¯12 + c3 + d3 . Therefore, (b¯3 c3 , u6 ) = 1 and (b¯3 d3 , u6 ) = 1. So (b¯3 c3 , b¯3 d3 ) ≥ 1. Hence (b3 b¯3 , c3 d¯3 ) = (b¯ 3 c3 , b¯3 d3 ) implies that (b3 b¯3 , c3 d¯3 ) ≥ 1.

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3


Recall that L1 (B) = {1}. So the above equality forces that c3 = d3 . Thus, b3 u6 = b¯12 + 2c3 , and hence (c3 b¯3 , u6 ) = (b3 u6 , c3 ) = 2, a contradiction. This proves that b3 u6 = b¯12 + t6 ,

t6 ∈ B.

(3.122)

Note that (b¯3 u6 , b5 ) = 1 by (3.118), and (b¯3 u6 , b¯3 u6 ) = (b3 u6 , b3 u6 ) = 2 by (3.122). So we may set b¯3 u6 = b5 + s13 ,

s13 ∈ B.

(3.123)

From (3.118), (3.119), (3.120), and (3.123), we get that b¯3 (b3 b5 ) = b4 + 2b5 + σ18 + s13 . But we also have b3 (b¯3 b5 ) = b4 + b5 + b3 y12 by (3.4) and (3.2). Thus, b3 y12 = σ18 + b5 + s13 .

(3.124)

Therefore, (b¯3 y12 , b¯3 y12 ) = (b3 y12 , b3 y12 ) = 3. Recall that 1 + b¯3 y12 = b¯12 + b5 b¯5 by (3.37), and (b¯3 y12 , b8 ) = 1 by (3.5). So we may assume that b¯3 y12 = b¯12 + b8 + r16 ,

r16 ∈ B.

(3.125)

Hence by (3.37), b5 b¯5 = 1 + b8 + r16 .

(3.126)

Therefore, from (b3 b¯3 )b8 = b¯ 3 (b3 b8 ), (3.1), (3.5), (3.35), and (3.125), we get that b82 = 1 + b7 + 2b8 + b12 + b¯12 + r16 .

(3.127)

Furthermore, (b3 b¯3 )b4 = (b¯3 b4 )b3 yields that b4 b8 = b4 + b5 + b¯5 + σ18

(3.128)

by (3.1), (3.3), (3.2), and (3.120). Also from (3.1), (3.4), (3.2), (3.124), and (b3 b¯3 )b5 = (b¯3 b5 )b3 , we see that b5 b8 = b4 + b5 + σ18 + s13 .

(3.129)

Note that (b4 b¯5 , b82 ) = (b4 b8 , b5 b8 ) = 3 by (3.128) and (3.129). So (3.36) and (3.127) force that b12 = b¯12 .

(3.130)

Moreover, it follows from b32 b5 = b3 (b3 b5 ), (3.2), (3.119), (3.36), (3.118), (3.35), and (3.122) that b52 = b7 + b¯12 + t6 .

(3.131)

∗ ∈ B. If Σ ∗ ∈ ¯ ∗ Now we claim that Σ12 12 / B, then (b3 Σ12 , b7 ) ≥ 2 by (3.38). Since ¯ z9 = x¯9 and b4 = b4 , we have that ∗ + u 6 b4 b3 b¯12 = Σ12

(3.132)

3.5 Case z = z9

183

∗ ∈ ∗ , b¯ ) ≥ 2. Hence b¯ Σ ∗ = by (3.41) and (3.39). Thus, Σ12 / B forces that (b¯3 Σ12 12 3 12 ∗ ∈ B. So we 2b7 + 2b¯12 + · · · , a contradiction. Therefore, we must have that Σ12 ∗ = c ∈ B. Thus, may assume that Σ12 12

b3 b7 = x9 + c12

(3.133)

b4 x¯9 = b3 + x9 + y12 + c12

(3.134)

by (3.38), and by (3.39). From b¯3 (b4 x¯9 ) = b4 (b¯3 x¯9 ), (3.134), (3.1), (3.35), (3.125), (3.120), (3.119), (3.6), and (3.36), we get that b4 σ18 = b8 + b12 + r16 + b¯3 c12 .

(3.135)

Since (b¯3 c12 , b7 ) = 1 by (3.133), the above equality implies that (b4 σ18 , b7 ) = 1. Thus, (b4 b7 , σ18 ) = 1 by (3.119). But we also have (b4 b7 , b4 ) = 1 by (3.6). Hence b4 b7 = σ18 + b4 + v6 , v6 ∈ N(B). By b32 b7 = b3 (b3 b7 ), (3.2), (3.133), and (3.120), we see that b4 b7 + b5 b7 = b4 + b¯5 + σ18 + b3 c12 .

(3.136)

Since any constituent of b3 c12 has degree at least 4, the above equality forces that v6 ∈ B. Thus b4 b7 = σ18 + b4 + v6 ,

v6 ∈ B.

(3.137)

Note that (b3 b¯3 , u6 u¯ 6 ) = (b3 u6 , b3 u6 ) = 2 by (3.122). So (u6 u¯ 6 , b8 ) = 1. Hence we must have that (b5 b¯5 , u6 u¯ 6 ) ≤ 3 by (3.126). Thus, (b5 u6 , b5 u6 ) = (b5 b¯5 , u6 u¯ 6 ) ≤ 3. But from (3.122), (3.2), (3.132), and b32 u6 = b3 (b3 u6 ), we get that b5 u6 = c12 + b3 t6 .

(3.138)

Since (b¯3 t6 , u6 ) = 1 by (3.122), (b¯3 t6 , b¯3 t6 ) ≥ 2. Hence (b3 t6 , b3 t6 ) = (b¯3 t6 , b¯3 t6 ) ≥ 2. So (b5 u6 , b5 u6 ) ≥ 3 by (3.138). Thus, we must have that (b5 u6 , b5 u6 ) = 3, and (b3 t6 , b3 t6 ) = 2. Hence (b¯3 t6 , b¯3 t6 ) = 2, and by (3.122) we may set b¯3 t6 = u6 + d12 ,

d12 ∈ B.

(3.139)

From (3.2), (3.133), (3.120), (3.137), and b32 b7 = b3 (b3 b7 ), we see that v6 + b5 b7 = b¯5 + b3 c12 . Hence (b¯3 v6 , c12 ) = (v6 , b3 c12 ) ≥ 1. Thus, we may assume that b¯3 v6 = c12 + p6 , p6 ∈ N(B). As in the proof of (3.122), we can prove that p6 ∈ B. Therefore, b¯3 v6 = c12 + p6 ,

p6 ∈ B.

(3.140)

Note that b¯3 (b4 b7 ) = (b¯3 b4 )b7 implies that b7 x9 = b¯3 σ18 + b3 + p6

(3.141)

184

3


by (3.137), (3.3), (3.140), and (3.133). Since (b¯3 σ18 , x9 ) = 1 by (3.120), (b7 x9 , x9 ) = 1. So (x9 x¯9 , b7 ) = 1. Similarly, b¯3 (b4 b8 ) = (b¯3 b4 )b8 yields that b8 x9 = b¯3 σ18 + b3 + x9 + u¯ 6

(3.142)

by (3.128), (3.3), (3.4), (3.118), and (3.5). Thus, (b¯3 σ18 , x9 ) = 1 implies that (b8 x9 , x9 ) = 2, and hence (x9 x¯9 , b8 ) = 2. Recall that (b3 b¯12 , y12 ) = 1 by (3.125). Note that c12 = y12 by (3.132), (3.125), and (3.130). So (b4 u6 , y12 ) = 1 by (3.132). Hence we may assume that b4 u6 = y12 + q12 ,

(3.143)

where q12 ∈ N(B). So by (3.132), b3 b¯12 = c12 + y12 + q12 .

(3.144)

It follows from b3 b52 = (b3 b5 )b5 , (3.131), (3.133), (3.144), (3.118), and (3.138) that b5 x¯9 = x9 + c12 + y12 + q12 .

(3.145)

Thus, (b4 b5 , x9 x¯9 ) = (b4 x¯9 , b5 x¯9 ) = 3 by (3.134) and (3.145). So (x9 x¯9 , b8 ) = 2 / B, then (3.145) implies that and (3.36) force that (x9 x¯9 , b¯12 ) = 1. If q12 ∈ (b5 x¯9 , b5 x¯9 ) ≥ 5. Hence (b5 b¯5 , x9 x¯9 ) ≥ 5. But we have already proved that (x9 x¯9 , b8 ) = 2. So (x9 x¯9 , r16 ) ≥ 2. Recall that b12 is not real by (3.130). Then the above discussions yield that x9 x¯9 = 1 + b7 + 2b8 + b12 + b¯12 + 2r16 + · · · . Since L1 (B) = {1}, the above equality cannot be true. This proves that q12 ∈ B and (b5 b¯5 , x9 x¯9 ) = 4. So c12 , y12 , and q12 are distinct, and (b3 b¯12 , b3 b¯12 ) = 3 by (3.144). Hence (b3 b12 , b3 b12 ) = (b3 b¯12 , b3 b¯12 ) = 3. Note that (b3 b12 , u¯ 6 ) = 1 by (3.122), and (b3 b12 , x9 ) = 1 by (3.35). So we may set b3 b12 = u¯ 6 + x9 + n21 ,

n21 ∈ B.

(3.146)

Hence by (3.121), b¯3 σ18 = c12 + y12 + x9 + n21 .

(3.147)

Now (3.141) and (3.147) imply that b7 x9 = b3 + p6 + c12 + y12 + x9 + n21 ,

(3.148)

and (3.142) and (3.147) yield that b8 x9 = b3 + u¯ 6 + c12 + y12 + 2x9 + n21 .

(3.149)

3.5 Case z = z9

185

Recall that (b3 b¯3 , b72 ) = (b3 b7 , b3 b7 ) = 2 by (3.133), and (b4 b¯4 , b72 ) = (b4 b7 , b4 b7 ) = 3 by (3.137). So (b72 , b7 ) = 1, and (b72 , b8 ) = 1. Since (b¯3 c12 , b7 ) = 1 by (3.133), and (b¯3 c12 , b¯12 ) = 1 by (3.144), we see that (b¯3 c12 , b¯3 c12 ) ≥ 3. Hence (b3 c12 , b3 c12 ) = (b¯3 c12 , b¯3 c12 ) ≥ 3. Therefore (3.136), (3.137), and (3.140) force that (b5 b7 , b5 b7 ) = (b3 c12 , b3 c12 ) ≥ 3. So (b5 b¯5 , b72 ) = (b5 b7 , b5 b7 ) ≥ 3. But we have just proved that (b72 , b8 ) = 1. Thus, (b72 , r16 ) ≥ 1. Therefore, L1 (B) = {1} forces that (b72 , r16 ) = 1, and hence (b3 c12 , b3 c12 ) = 3. So we may set b¯3 c12 = b7 + b¯12 + m17 ,

m17 ∈ B.

(3.150)

Hence by (3.135), b4 σ18 = b7 + b8 + b12 + b¯12 + r16 + m17 .

(3.151)

Now it follows from (b3 b¯3 )b7 = b¯ 3 (b3 b7 ), (3.1), (3.133), (3.35), and (3.150) that b7 b8 = b7 + b8 + b12 + b¯12 + m17 .

(3.152)

Furthermore, from (3.134), (3.1), (3.35), (3.125), (3.150), (3.3), and b¯3 (b4 x¯9 ) = (b¯3 b4 )x¯9 , we get that x9 x¯9 = 1 + b7 + 2b8 + b12 + b¯12 + r16 + m17 .

(3.153)

Since we have already known that (b72 , b7 ) = (b72 , b8 ) = (b72 , r16 ) = 1, and (b72 , x9 x¯9 ) = (b7 x9 , b7 x9 ) = 6, by (3.153) we must have that b72 = 1 + b7 + b8 + r16 + m17 .

(3.154)

Therefore, since b4 is real, from (3.137), (3.6), (3.151), (3.154), (3.152), and b4 (b4 b7 ) = b42 b7 , we get that b4 v6 = b7 + m17 .

(3.155)

On the other hand, since (b3 , v6 p¯6 ) = 1 by (3.140), (3.3) yields that (b¯3 b4 , v6 p¯ 6 ) = (b3 + x9 , v6 p¯6 ) ≥ 1. But b4 real implies that (b¯3 p6 , b4 v6 ) = (b¯3 b4 , v6 p¯ 6 ). Thus, (b¯3 p6 , b4 v6 ) ≥ 1.

(3.156)

Note that (b¯3 p6 , b7 ) = 0 by (3.133). So (3.155) and (3.156) force that (b¯ 3 p6 , m17 ) = 1. Since L1 (B) = {1}, the above equality cannot be true, a contradiction. Now we have derived a contradiction for each case in Table 3.1 before the end of Sect. 3.2. Hence, Theorem 3.1 holds.

Chapter 4

Preliminary Classification of Sub-case (3)

4.1 Introduction In Chap. 4 we shall freely use the definitions and notation used in Chap. 2. The Main Theorem 2 of Chap. 3 left two unsolved problems for the complete classification of the Normalized Integral Table Algebra (NITA) (A, B) generated by a faithful nonreal element of degree 3 with L1 (B) = 1 and L2 (B) = ∅. In this chapter we solve the second problem where (b3 b10 , b3 b10 ) = 2, and prove that the NITA that satisfies case (2) of the Main Theorem 1 of Chap. 2 where (b3 b10 , b3 b10 ) = 2 does not exist. Consequently, we can state the Main Theorem 3 of this chapter as follows. Main Theorem 3 Let (A, B) be a NITA generated by a non-real element b3 ∈ B of degree 3 and without non-identity basis element of degree 1 and 2. Then b3 b¯3 = 1 + b8 , b8 ∈ B, and one of the following holds: (1) There exists a real element b6 ∈ B such that b32 = b¯3 + b6 and (A, B) ∼ =x (CH(PSL(2, 7)), Irr(PSL(2, 7))). (2) There exist b6 , b10 , b15 ∈ B, where b6 is non-real, such that b32 = b¯3 + b6 , b¯3 b6 = b3 + b15 , b3 b6 = b8 + b10 , and (b3 b8 , b3 b8 ) = 3. Moreover, if b10 is real then (A, B) ∼ =x (CH(3 · A6 ), Irr(3 · A6 )) of dimension 17, and if b10 is non-real then (b3 b10 , b3 b10 ) = 2 and b15 is a non-real element. (3) There exist c3 , b6 ∈ B, c3 = b3 or b¯3 , such that b32 = c3 + b6 and either (b3 b8 , b3 b8 ) = 3 or 4. If (b3 b8 , b3 b8 ) = 3 and c3 is non-real, then (A, B) ∼ =x (A(3 · A6 · 2), B32 ) of dimension 32. (See Theorem 2.9 of Chap. 2 for the definition of this specific NITA.) If (b3 b8 , b3 b8 ) = 3 and c3 is real, then (A, B) ∼ =x (A(7 · 5 · 10), B22 ) of dimension 22. (See Theorem 2.10 of Chap. 2 for the definition of this specific NITA.) In the above Main Theorem 3 we still have 2 open problems in the cases (2) and (3). In case (2) we must classify the NITA such that b10 is non-real and (b3 b10 , b3 b10 ) = 2, and in case (3) we must classify the NITA such that (b3 b8 , b3 b8 ) = 4. In Chap. 5, the open case (2) will be solved. Z. Arad et al., On Normalized Integral Table Algebras (Fusion Rings), Algebra and Applications 16, DOI 10.1007/978-0-85729-850-8_4, © Springer-Verlag London Limited 2011

187

188

4


Let us emphasize that the NITA’s of dimension 7 and 17 in the cases (1) and (2) of the Main Theorem 2 are strictly isomorphic to the NITA’s induced from finite groups G via the basis of the irreducible characters of G. However, the NITA’s of dimensions 22 and 32 are not induced from finite groups as described in Chap. 2. The Main Theorem 3 follows from the Main Theorem 2 in Chap. 3 and the next theorem. Theorem 4.1 Let (A, B) be a NITA generated by a non-real element b3 ∈ B of degree 3 and without non-identity basis element of degree 1 or 2. Then b3 b¯3 = 1 + b8 , b8 ∈ B. Assume that b32 = b¯3 + b6 ,

b¯3 b6 = b3 + b15 ,

b3 b6 = b8 + b10 ,

and b3 b8 = b3 + b¯ 6 + b15

where b6 , b10 are non-real elements in B and b15 ∈ B. Then b15 is a non-real element and (b3 b10 , b3 b10 ) = 2. The rest of this chapter is devoted to proving the above theorem.

4.2 Preliminary Results For the rest of this chapter, we shall always assume that (A, B) is a NITA generated by a non-real element b3 ∈ B of degree 3 and without non-identity basis element of degree 1 and 2 such that b32 = b¯ 3 + b6 , b3 b¯3 = 1 + b8 ,

(4.1)

b8 ∈ B,

(4.2)

b¯ 3 b6 = b3 + b15 ,

b15 ∈ B,

(4.3)

b3 b6 = b8 + b10 ,

b10 ∈ B,

(4.4)

and b3 b8 = b3 + b¯6 + b15 .

(4.5)

By (4.4), (4.5) and b¯3 (b3 b6 ) = (b3 b¯3 )b6 , we obtain that b6 b8 = b¯3 + b¯15 + b¯3 b10 .

(4.6)

By (4.1), (4.3), (4.4), (4.5) and b3 (b3 b6 ) = b32 b6 we obtain that b62 = b¯6 + b3 b10 . By (4.2) and (4.3), we obtain that (b¯ 3 b6 , b3 b10 ) = (b3 , b3 b10 ) + (b15 , b3 b10 ) = (b15 , b3 b10 );

(4.7)


189

on the other hand, (4.1) and (4.4) imply that (b¯3 b6 , b3 b10 ) = (b32 , b6 b¯10 ) = (b¯ 3 , b6 b¯10 ) + (b6 , b6 b¯10 ) = 1 + (b6 , b6 b¯10 ). Hence 1 + (b6 , b6 b¯10 ) = (b15 , b3 b10 ) which implies that b3 b10 = b15 + R15

where R15 ∈ NB.

(4.8)

The degrees appearing in R15 must all be ≥ 5. By (4.1), (4.3), (4.5) and b3 (b3 b8 ) = b32 b8 , we obtain that b6 b8 = b¯3 + b3 b15 .

(4.9)

b3 b15 = b¯15 + b¯3 b10 .

(4.10)

Now (4.6) implies that (b3 b¯3 )(b3 b¯3 ) = b32 b¯32 ,

By (4.1), (4.2), (4.3), (4.4) and

we obtain that

b82 = b8 + b10 + b¯10 + b6 b¯6 .

(4.11)

By (4.1), (4.2), (4.3), (4.4) and b¯6 b32 = (b3 b¯6 )b3 , we obtain that b¯10 + b6 b¯6 = 1 + b3 b¯15 .

(4.12)

By (4.1), (4.8), (4.10) and b3 (b3 b10 ) = b32 b10 , we obtain that b6 b10 = b¯ 15 + b3 R15 .

(4.13)

The proof that b15 is non-real is simple. Assume henceforth that b15 is a real element and we shall derive a contradiction. By (4.3) we have that (b6 , b3 b15 ) = 1 and together with (4.12), we obtain that (b¯6 , b3 b15 ) = 1 which implies that (b15 , b3 b6 ) = 1, and we have a contradiction to (4.4). Therefore b15

is a non-real element.

(4.14)

The remainder of this chapter will consist of the proof that (b3 b10 , b3 b10 ) = 2. By (4.4) we obtain that (b6 , b¯3 b10 ) = 1 which implies that (b3 b10 , b3 b10 ) ≥ 2. By (4.4) we obtain that (b¯ 6 , b3 b10 ) = 0. Now by (4.3), (4.4) and (4.7) we obtain that (b3 b10 , b3 b10 ) = (b62 , b3 b10 ) = (b¯3 b6 , b¯6 b10 ) = (b3 , b¯6 b10 ) + (b15 , b¯6 b10 ) = 1 + (b15 , b¯6 b10 ). Since (b3 , b¯6 b10 ) = 1, we have that (b15 , b¯6 b10 ) ≤ 3 which implies that (b3 b10 , b3 b10 ) ≤ 4. Therefore 2 ≤ (b3 b10 , b3 b10 ) ≤ 4.

(4.15)

Assume henceforth that b15 = R15 and we shall derive a contradiction. Then by (4.8) we obtain that b3 b10 = 2b15 which implies that (b¯10 , b3 b¯15 ) = 2, and by (4.12) we obtain that (b¯10 , b6 b¯6 ) = 1 which implies that (b10 , b6 b¯6 ) = 1. Now (4.12)

190

4

Table 4.1 Splitting to five cases


Case 1

R15 = x5 + x10 , x5 , x10 ∈ B

Case 2

R15 = x6 + x9 , x6 , x9 ∈ B

Case 3

R15 = x7 + x8 , x7 , x8 ∈ B

Case 4

R15 = x5 + y5 + z5 , x5 , y5 , z5 ∈ B

Case 5

R15 ∈ B, b15 = R15

implies that (b¯15 , b¯3 b10 ) = (b10 , b3 b¯15 ) = 1. By (4.4) we obtain that (b6 , b¯3 b10 ) = 1 and since (b3 b10 , b3 b10 ) = 4, we obtain that b¯3 b10 = b6 + b¯15 + α + β where α, β ∈ B, |α| + |β| = 9 and |α|, |β| ≥ 5, which is a contradiction to our assumption that b15 = R15 . Now by (4.8) and (4.15), we have five cases (see Table 4.1). In the rest of this chapter, we derive a contradiction for Cases 1–4 in the above table. In Case 5, by (4.8) we then have that (b3 b10 , b3 b10 ) = (b15 , b15 ) + (R15 , R15 ) = 2. Then we will have proven Theorem 1.1.

4.3 Case R15 = x5 + x10 In this section we assume that R15 = x5 + x10 . Then by (4.7), (4.8) and (4.13), we have that b3 b10 = b15 + x5 + x10 ,

(4.16)

b62 = b¯6 + b15 + x5 + x10

(4.17)

b6 b10 = b¯15 + b3 x5 + b3 x10 .

(4.18)

and Now by (4.16) we obtain that (b10 , b¯3 x5 ) = 1 which implies that b¯3 x5 = b10 + y5 ,

where y5 ∈ B.

(4.19)

Now (4.1) and b¯32 x5 = b¯3 (b¯3 x5 ) imply that b3 x5 + b¯6 x5 = b¯3 b10 + b¯3 y5 .

(4.20)

By (4.2), (4.16), (4.19) and b3 (b¯3 x5 ) = (b3 b¯3 )x5 , we obtain that b8 x5 = x10 + b15 + b3 y5 .

(4.21)

Now by (4.4), (4.19) and (b¯ 3 b¯6 )x5 = b¯6 (b¯3 x5 ), we obtain that b¯6 b10 + b¯6 y5 = x10 + b15 + b3 y5 + b¯10 x5 .

(4.22)

4.3 Case R15 = x5 + x10

191

By (4.1) and (4.19) we obtain that (b¯3 y5 , b3 x5 ) = (b32 , y5 x¯5 ) = (b¯3 , y5 x¯5 ) + (b6 , y5 x¯5 ) = 1 + (b6 , x¯5 y5 ). Hence (b¯3 y5 , b3 x5 ) = 1 + (b6 , x¯5 y5 ).

(4.23)

By (4.2) and (4.19) we obtain that 2 = (b3 x5 , b3 x5 ) = (b3 b¯3 , x5 x¯5 ) = 1 + (b8 , x5 x¯5 ). Hence (b8 , x5 x¯5 ) = 1

(4.24)

and together with b10 = b¯ 10 , we obtain that (b10 , x5 x¯5 ) = 0.

(4.25)

Now (4.4) implies that (b¯ 3 x¯5 , b6 x¯5 ) = (b3 b6 , x5 x¯5 ) = (b8 , x5 x¯5 ) + (b10 , x5 x¯5 ) = 1. Hence (b¯3 x¯5 , b6 x¯5 ) = 1.

(4.26)

By (4.3) and (4.19) we obtain that (b¯3 y5 , b¯6 x5 ) = (b3 b¯6 , y5 x¯5 ) = (b¯3 , y5 x¯5 ) + (b¯15 , y5 x¯5 ) = 1 + (b¯15 , y5 x¯5 ). Therefore (b¯3 y5 , b¯6 x5 ) = 1 + (b¯15 , y5 x¯5 )

(4.27)

(b¯3 y5 , b¯6 x5 ) ≤ 2.

(4.28)

which implies that By (4.2) we obtain that (b3 y5 , b3 y5 ) = (b3 b¯3 , y5 y¯5 ) = 1 + (b8 , y5 y¯5 ) which implies that (b3 y5 , b3 y5 ) ≤ 4.

(4.29)

Assume henceforth that (b¯3 y5 , b¯3 y5 ) = 4 and we shall derive a contradiction. Then b¯3 y5 = x + y + z + v

where x, y, z, v ∈ B.

Now, since (b3 x5 , b3 x5 ) = 2, by (4.23) we have that (b3 x5 , b¯3 y5 ) = 1. Hence b 3 x5 = x + w

where w ∈ B,

and together with (4.20), we get that x + w + b¯6 x5 = b¯3 b10 + x + y + z + v. Thus (b¯6 x5 , b¯3 y5 ) ≥ 3, and we have a contradiction to (4.28). Assume henceforth that (b¯3 y5 , b¯3 y5 ) = 3 and we shall derive a contradiction. Then b¯3 y5 = x + y + z

where x, y, z ∈ B

192

4


and b3 x 5 = x + w

where w ∈ B.

By (4.20) we have that x + w + b¯6 x5 = b¯3 b10 + x + y + z. Now (4.28) implies that (x, b¯6 x5 ) = 0 and by (4.26), we obtain that (b3 x5 , b¯6 x5 ) = 1 which implies that (w, b¯6 x5 ) = 1 and (w, b¯ 3 b10 ) = 2. Since (b3 b10 , b3 b10 ) = 3 we have a contradiction. Assume henceforth that (b¯3 y5 , b¯3 y5 ) = 1 and we shall derive a contradiction. Then b¯3 y5 = c15 and by (4.23) we obtain that b3 x5 = c15 , which is a contradiction to (4.19). Now (4.29) implies that (b3 y5 , b3 y5 ) = 2.

(4.30)

Assume henceforth that (b¯15 , y5 x¯5 ) = 1 and we shall derive a contradiction. Then (4.27) implies that (b¯3 y5 , b¯6 x5 ) = 2. Now (4.17) and (4.20) imply that b¯6 x5 = b6 + 2x12

where x12 ∈ B

and b¯3 y5 = x12 + c3 where c3 ∈ B. Now by (4.4) we obtain that (b6 , b¯3 x¯5 ) = 0, which is a contradiction to (4.26). Hence (b¯15 , y5 x¯5 ) = 0.

(4.31)

By (4.21) and (4.30) we obtain that (b8 x5 , b3 y5 ) ≥ 2. In addition, (4.5), (4.19) and (4.31) imply that (b8 x5 , b3 y5 ) = (b¯3 b8 , y5 x¯5 ) = (b¯3 , y5 x¯5 ) + (b6 , y5 x¯5 ) + (b¯15 , y5 x¯5 ) = 1 + (b6 , y5 x¯5 ). Hence (b6 , y5 x¯5 ) ≥ 1. By (4.19), (4.23) and (4.30) we have that (b6 , y5 x¯5 ) ≤ 1. Therefore (b6 , y5 x¯5 ) = 1 and

(b¯3 y5 , b3 x5 ) = 2

(4.32)

which implies that b¯3 y5 = b3 x5 .

(4.33)

b¯6 x5 = b¯3 b10 .

(4.34)

In addition, (4.20) implies that

4.3 Case R15 = x5 + x10

193

By (4.19) we obtain that (x5 , b3 y5 ) = 1. Now (4.30) implies that b3 y5 = x5 + y10 ,

where y10 ∈ B.

(4.35)

Now (4.33) implies that b¯3 y5 = b3 x5 = x + y

where x, y ∈ B

(4.36)

and (4.21) implies that b8 x5 = x5 + x10 + y10 + b15 .

(4.37)

Now by (4.2) and b¯3 (b3 x5 ) = (b¯ 3 b3 )x5 , we obtain that b¯3 x + b¯3 y = 2x5 + x10 + y10 + b15 .

(4.38)

Thus we can assume that x = z5 and y = z10 , which imply that b¯ 3 y5 = b3 x5 = z5 + z10 ,

(4.39)

and one of the two following cases hold: Case 1 b¯3 z5 = x5 + x10

and b¯ 3 z10 = x5 + y10 + b15 .

(4.40)

b¯ 3 z5 = x5 + y10

and b¯ 3 z10 = x5 + x10 + b15 .

(4.41)

Case 2

Thus (z10 , b3 b15 ) = 1, by (4.3) we have that (b6 , b3 b15 ) = 1 and together with (4.10), we obtain that b3 b15 = b6 + z10 + x14 + b¯15

where x14 ∈ B,

(4.42)

b¯ 3 b10 = b6 + z10 + x14 ,

(4.43)

b¯6 x5 = b6 + z10 + x14 .

(4.44)

and by (4.34) we obtain that

By (4.43) we obtain that (b10 , b3 z10 ) = 1, by (4.39) we obtain that (y5 , b3 z10 ) = 1, and by (4.40) and (4.41) we obtain that (b3 z10 , b3 z10 ) = 3. Thus b3 z10 = y5 + b10 + x15

where x15 ∈ B.

(4.45)

By (4.9) and (4.42), we obtain that b6 b8 = b¯3 + b¯15 + b6 + z10 + x14 .

(4.46)

194

4


Now by (4.4) we obtain that (b6 b¯6 , b3 z10 ) = (b¯3 b¯6 , b¯6 z10 ) = (b8 , b¯6 z10 )+(b¯10 , b¯6 z10 ) = 1 + (b¯10 , b¯6 z10 ) which implies that (b6 b¯6 , b3 z10 ) ≥ 1. By (4.17) we have that (b6 b¯6 , b6 b¯6 ) = 4. Thus either b6 b¯6 = 1 + b8 + x12 + x15

where x12 , x15 ∈ B

b6 b¯6 = 1 + b8 + y5 + x22

where y5 , x22 ∈ B.

or Assume henceforth that b6 b¯6 = 1 + b8 + y5 + x22 and we shall derive a contradiction. Then by (4.12) we obtain that b¯3 y5 = b¯ 15 , and we have a contradiction to (4.30). Hence b6 b¯6 = 1 + b8 + x12 + x15

(4.47)

b3 b¯15 = b¯10 + b8 + x12 + x15 .

(4.48)

and

By (4.1), (4.35), (4.39) and b3 (b3 y5 ) = b32 y5 , we obtain that b3 y10 = b6 y5 .

(4.49)

By (4.39) we obtain that (y5 , b3 z5 ) = 1, and by (4.40) and (4.41) we obtain that (b3 z5 , b3 z5 ) = 2 which implies that b3 z5 = y5 + t10

where t10 ∈ B.

(4.50)

By (4.19), (4.35), (4.39), (4.45), (4.50) and b3 (b¯3 y5 ) = b¯3 (b3 y5 ), we obtain that b¯3 y10 = y5 + t10 + x15 .

(4.51)

By (4.4), (4.8), (4.12), (4.16), (4.19), (4.43), (4.45), (4.48) and b¯3 (b3 b10 ) = b3 (b¯3 b10 ), we obtain that x12 + b¯3 x10 = b3 x14

(4.52)

which implies that (x14 , b¯3 x12 ) = 1. Now by (4.48) we obtain that b¯3 x12 = t7 + x14 + b¯15

where t7 ∈ B.

(4.53)

By (4.34) we obtain that 3 = (b6 x5 , b6 x5 ) = (b6 b¯6 , x5 x¯5 ). Now by (4.47) we obtain that x5 x¯5 = 1 + b8 + x12 + t4

where t4 ∈ B.

(4.54)

By (4.5), (4.19) ,(4.53), (4.54) and b3 (x5 x¯5 ) = (b3 x¯5 )x5 , we obtain that b¯10 x5 + y¯5 x5 = 2b3 + b¯6 + 2b15 + t¯7 + x¯14 + b3 t4 .

(4.55)

4.4 Case R15 = x6 + x9

195

By (4.35) we obtain that (b3 , y¯5 x5 ) = 1, and by (4.32) we obtain that (b15 , y¯5 x5 ) = (x¯ 14 , y¯5 x5 ) = 0 which implies that b¯10 x5 = b3 + c3 + 2b15 + x¯14

where c3 ∈ B

and (c3 , b3 t4 ) = 1 which implies that (b¯ 3 c3 , b¯3 c3 ) = 2. By (4.2) we obtain that (b¯3 c3 , b¯3 c3 ) = (b3 b¯3 , c3 c¯3 ) = 1 + (b8 , c3 c¯3 ) which implies that c3 c¯3 = 1 + b8 and b¯3 c3 = t4 + u5

where u5 ∈ B.

(4.56)

Now by (4.5) and b3 (c3 c¯3 ) = (b3 c¯3 )c3 we obtain that c3 t4 + c3 u¯ 5 = 2b3 + b¯6 + b15 which implies that c3 u¯ 5 = b15 . By (4.56) we obtain that (b3 , c3 u¯ 5 ) = 1, which is a contradiction.

4.4 Case R15 = x6 + x9 Throughout this section we assume that R15 = x6 + x9 . Then by (4.7), (4.8) and (4.13), we have that b3 b10 = b15 + x6 + x9 , b62

= b¯ 6 + b15 + x6 + x9 ,

(4.57) (4.58)

b6 b10 = b¯15 + b3 x6 + b3 x9 .

(4.59)

(b10 , b¯3 x6 ) = 1

(4.60)

By (4.57) we obtain that which implies that either 2 ≤ (b¯ 3 x6 , b¯3 x6 ) ≤ 3 or b¯3 x6 = b10 + 2b4 . Assume henceforth that b¯3 x6 = b10 + 2b4 and we shall derive a contradiction. Then b3 b4 = 2x6 which implies by (4.2) that 4 = (b3 b4 , b3 b4 ) = (b3 b¯3 , b4 b¯4 ) = 1 + (b8 , b4 b¯4 ). Thus (b8 , b4 b¯4 ) = 3, and we have a contradiction. Now we have that 2 ≤ (b¯3 x6 , b¯3 x6 ) ≤ 3. Assume that (b3 x6 , b3 x6 ) = 2, so we can write that b3 x 6 = c + d

where c, d ∈ B.

(4.61)

196

4


By (4.60) we obtain that b¯3 x6 = b10 + y8

where y8 ∈ B.

(4.62)

Now by (4.2), (4.57) and b3 (b¯3 x6 ) = (b3 b¯3 )x6 we obtain that b8 x6 = x9 + b15 + b3 y8 .

(4.63)

By (4.1), (4.61), (4.62) and b¯3 (b¯3 x6 ) = b¯32 x6 we obtain that b¯ 3 b10 + b¯3 y8 = c + d + b¯6 x6 .

(4.64)

By (4.58) we obtain that (b6 b¯6 , b6 b¯6 ) = 4, by (4.4) we obtain that 2 = (b3 b6 , b3 b6 ) = (b3 b¯3 , b6 b¯6 ) which implies that b6 b¯6 = 1 + b8 + s + t

where s, t are real elements in B.

(4.65)

By (4.61) we obtain that 2 = (b3 x6 , b3 x6 ) = (b3 b¯3 , x6 x¯6 ). Now (4.2) implies that (b8 , x6 x¯6 ) = 1

(4.66)

and together with (4.4), we obtain that (b3 x6 , b¯6 x6 ) = (b3 b6 , x6 x¯6 ) = (b8 , x6 x¯6 ) + (b10 , x6 x¯6 ) = 1 + (b10 , x6 x¯6 ). Thus (b3 x6 , b¯6 x6 ) = 1 + (b10 , x6 x¯6 ).

(4.67)

By (4.12) and (4.65) we obtain that b3 b¯15 = b8 + b¯10 + s + t,

(4.68)

which implies that |s|, |t| ≥ 5. Assume henceforth that s = s5 and we shall derive a contradiction. Then by (4.68) we obtain that b3 b¯15 = b8 + b¯10 + s5 + t22 and b3 s5 = b15 . Hence (4.2) and b¯ 3 (b3 s5 ) = (b3 b¯3 )s5 imply that b8 s5 = b8 + b10 + t22 . Since b8 and s5 are real elements and b10 is non-real, we have a contradiction. Assume henceforth that s = s6 and we shall derive a contradiction. Then by (4.68) we obtain that b3 b¯15 = b8 + b¯10 + s6 + t21 and b3 s6 = b15 + c3 where c3 ∈ B. Now b¯3 (b3 s6 ) = (b3 b¯3 )s6 implies that b8 s6 = b8 + b10 + t21 + b¯3 c3 . Since b8 and s6 are reals and b10 is non-real, we obtain that (b¯10 , b¯3 c3 ) = 1 and we have a contradiction, which implies that |s| ≥ 7.

(4.69)

|t| ≥ 7.

(4.70)

In the same way we can show that

Now (4.65) implies that (b6 b¯6 , x6 x¯6 ) ≤ 5. By (4.58) we obtain that (b6 , b¯6 x6 ) = 1 and by (4.64) we obtain that (b6 b¯6 , x6 x¯6 ) = (b¯6 x6 , b¯6 x6 ) ≥ 3. Therefore 3 ≤ (b6 b¯6 , x6 x¯6 ) ≤ 5.

(4.71)

4.4 Case R15 = x6 + x9

197

Assume henceforth that (b10 , x6 x¯6 ) = 1 and we shall derive a contradiction. Then x6 x¯6 = 1 + b8 + b10 + b¯10 + z7 , where z7 ∈ B. Now (4.65) implies that (b¯6 x6 , b¯6 x6 ) ≤ 3 and together with (4.71) we obtain that (b¯6 x6 , b¯6 x6 ) = 3, and by (4.61), (4.67) and (b10 , x6 x¯6 ) = 1 we obtain that b¯6 x6 = b6 + c + d, which is a contradiction. Thus (4.67) implies that (b10 , x6 x¯6 ) = 0

(4.72)

(b3 x6 , b¯6 x6 ) = 1.

(4.73)

(c, b¯6 x6 ) = 1.

(4.74)

and

Now we can assume that In addition, (4.1) and (4.62) imply that (b¯3 y8 , b3 x6 ) = (b32 , y8 x¯6 ) = (b¯3 , y8 x¯6 ) + (b6 , y8 x¯6 ) = (y8 , b¯3 x6 ) + (y8 , b6 x6 ) = 1 + (y8 , b6 x6 ). By (4.62) and (4.73), we obtain that 1 = (b3 x6 , b¯6 x6 ) = (b¯3 x6 , b6 x6 ) = (b10 , b6 x6 ) + (y8 , b6 x6 ) which implies that either (b¯ 3 y8 , b3 x6 ) = 1 or (b¯ 3 y8 , b3 x6 ) = 2. Assume that (b¯3 y8 , b3 x6 ) = 1. Then (4.4), (4.64), (4.74) and (b3 b10 , b3 b10 ) = 3 imply that b¯3 b10 = b6 + c + d which is a contradiction, and we obtain that (b¯3 y8 , b3 x6 ) = 2. Now we have that either (c, b¯3 y8 ) = 2 or (c, b¯3 y8 ) = (d, b¯3 y8 ) = 1, which implies that (b3 y8 , b3 y8 ) ≥ 3. By (4.11), (4.66) and (4.72), we obtain that (b8 x6 , b8 x6 ) = (b82 , x6 x¯6 ) = 1 + (b6 b¯6 , x6 x¯6 ). Now (4.71) implies that (b8 x6 , b8 x6 ) ≤ 6 together with (4.63) and since (b3 y8 , b3 y8 ) ≥ 3, we have that (x9 , b3 y8 ) = (b15 , b3 y8 ) = 0,

(4.75)

(b8 x6 , b8 x6 ) = 2 + (b3 y8 , b3 y8 ).

(4.76)

which implies that Therefore (b3 y8 , b3 y8 ) ≤ 4. If (c, b¯3 y8 ) = 2 then b¯3 y8 = 2c12 and (b¯6 x6 , b¯6 x6 ) = 5. Now (4.64) implies that b¯3 b10 = b6 + d6 + f18 where f18 ∈ B and b¯6 x6 = b6 + c12 + f18 , which is a contradiction. Now we have that (c, b¯3 y8 ) = (d, b¯3 y8 ) = 1 which implies that b¯3 y8 = c + d + e6

where e6 ∈ B.

(4.77)

where f ∈ B,

(4.78)

Now (4.64) and (4.74) imply that b¯3 b10 = b6 + c + f

b¯6 x6 = b6 + e6 + c + f,

(4.79)

b3 b15 = b¯ 15 + b6 + c + f.

(4.80)


198

4


By (4.2), (4.61), (4.63) and b¯3 (b3 x6 ) = (b3 b¯3 )x6 , we obtain that b¯3 c + b¯3 d = x6 + x9 + b15 + b3 y8 .

(4.81)

Now we can assume that (b15 , b¯3 c) = 1. By (4.61) we obtain that (x6 , b¯3 c) = (x6 , b¯3 d) = 1 and by (4.77) we obtain that (b3 y8 , b3 y8 ) = 3 which implies by (4.75) and (4.81) that 2 ≤ (b3 c, b3 c) ≤ 4. By (4.77) and (4.78), we obtain that (b10 , b3 c) = (y8 , b3 c) = 1 which implies that (b3 c, b3 c) = 2. Therefore 3 ≤ (b3 c, b3 c) ≤ 4.

(4.82)

By (4.77) we obtain that (b3 y8 , b3 y8 ) = 3 and by (4.62), we obtain that (y8 , b¯3 x6 ) = 1 which implies that b3 y8 = x6 + α + β

where α, β ∈ B.

(4.83)

b¯3 c + b¯3 d = 2x6 + x9 + b15 + α + β.

(4.84)

Now by (4.81) we obtain that

Assume that (b3 c, b3 c) = 3. Then by (4.61) and (4.80) we obtain that b¯3 c = x6 + b15 + g,

where g ∈ B

(4.85)

which implies that |c| ≥ 9, and by (4.84) we obtain that g + b¯3 d = x6 + x9 + α + β. If g = x9 then by (4.85) we obtain that |c| = 10, and if g = x9 then we obtain that either g = α or g = β. If α = 3 = α3 , then (4.5) and (4.83) imply that α = b3 , y8 = b8 and x6 = b¯6 which implies by (4.57) that b3 b10 = b15 + b¯ 6 + x9 . Thus (b¯ 10 , b3 b6 ) = 1, and we have a contradiction to (4.4). In the same way, we can show that |β| = 3 which implies by (4.83) that 4 ≤ |α|, |β| ≤ 14.

(4.86)

Now (4.85) implies that |g| ≤ 12 and |c| ≤ 11. Therefore 9 ≤ |c| ≤ 11.

(4.87)

Assume that (b3 c, b3 c) = 4. Then by (4.62) and by (4.68), we obtain that b¯3 c = x6 + b15 + g + h

where g, h ∈ B.

(4.88)

Now by (4.81) and (4.83) we obtain that b¯3 c + b¯3 d = 2x6 + x9 + b15 + α + β, which implies that g + h + b¯3 d = x6 + x9 + α + β.

(4.89)

If (x9 , b¯3 d) = 1, then g + h = α + β which implies by (4.83) and (4.88) that |c| = 13. If (x9 , b¯3 d) = 0 then we can assume that g + h = x9 + α. Now from 4 ≤ |α| ≤ 14 and by (4.88) we obtain that 12 ≤ |c| ≤ 14.

(4.90)

4.4 Case R15 = x6 + x9

199

Table 4.2 Splitting to seven cases Case 1

c = c9

d = d9

f = f15

(b3 c9 , b3 c9 ) = 3

Case 2

c = c10

d = d8

f = f14

(b3 c10 , b3 c10 ) = 3

Case 3

c = c11

d = d7

f = f13

(b3 c11 , b3 c11 ) = 3

Case 4

c = c12

d = d6

f = f12

(b3 c12 , b3 c12 ) = 4

Case 5

c = c13

d = d5

f = f11

(b3 c13 , b3 c13 ) = 4

Case 6

c = c14

d = d4

f = f10

(b3 c14 , b3 c14 ) = 4

Case 7

(b3 x6 , b3 x6 ) = 3

From (4.61), (4.78), (4.87) and (4.90) we have seven cases (see Table 4.2). In the following seven sections we derive a contradiction for each case in the above table, and in the following three sections we assume that (b3 c, b3 c) = 3,

(4.91)

so (4.87) implies that 9 ≤ |c| ≤ 11.

4.4.1 Case c = c9 In this section we assume that c = c9 . Then (4.77), (4.78) and (4.91) imply that b3 c9 = y8 + b10 + q9 where q9 ∈ B which implies that (c9 , b¯3 q9 ) = 1.

(4.92)

By (4.85) we obtain that b¯3 c9 = x6 + b15 + g6 . Now by (4.61), (4.77), (4.78), (4.80) and b¯3 (b3 c9 ) = b3 (b¯3 c9 ), we obtain that b¯15 + b3 g6 = e6 + b¯3 q9 which implies that (b¯15 , b¯3 h9 ) = 1, and together with (4.92) we obtain that b¯3 h9 = c9 + b¯15 + c3 where c3 ∈ B, b3 c3 = h9 and (c3 , b3 g6 ) = 1. Since b3 c3 = h9 we obtain that (b3 c3 , b3 c3 ) = 1, which is a contradiction to (c3 , b3 g6 ) = 1.

4.4.2 Case c = c10 In this section we assume that c = c10 . Then (4.77), (4.78) and (4.91) imply that b3 c10 = y8 + b10 + q12

where q12 ∈ B.

(4.93)

By (4.85) we obtain that b¯3 c10 = x6 + b15 + g9 .

(4.94)

Now by (4.61), (4.77), (4.78), (4.80) and b¯3 (b3 c10 ) = b3 (b¯3 c10 ), we obtain that b¯15 + b3 g9 = e6 + b¯3 q12 which implies that (q12 , b3 b¯15 ) = 1,

(4.95)

200

4


and (4.12) implies that q12 is a real element. By (4.62), (4.77), (4.83), (4.93) and b3 (b¯3 y8 ) = b¯3 (b3 y8 ) we obtain that q12 + b3 d8 + b3 e6 = b¯3 α + b¯ 3 β. Now we can assume that (q12 , b¯3 α) = 1 and by (4.83) we obtain that (y8 , b¯3 α) = 1 which implies that |α| ≥ 8, and since q12 is a real element we obtain that (α, ¯ b¯3 q12 ) = 1.

(4.96)

Assume henceforth that α = c¯10 and we shall derive a contradiction. Then by (4.81), (4.83) and (4.94) we obtain that (c¯10 , b¯3 d8 ) = 1, and we have a contradiction to (4.94). By (4.86) we obtain that α = b15 . Now by (4.93), (4.95) and (4.96) we ¯ b¯ 3 q12 ) = 1 and (b¯15 , b¯3 q12 ) = 1 which imply that obtain that (c10 , b¯3 q12 ) = 1, (α, either |α| ≤ 7 or |α| = 11. Now |α| ≥ 8 implies that α = α11 and by (4.83) we obtain that β = β7 . Now we have that b3 q12 = c¯10 + b15 + α11 .

(4.97)

By (4.5), (4.8) and (4.95) we obtain that b¯3 b15 = b8 +b10 +q12 +q15 where q15 ∈ B. Hence (4.2), (4.93) and b¯3 (b3 q12 ) = (b3 b¯3 )q12 imply that b8 q12 = y¯8 + b¯10 + q12 + b8 + b10 + q15 + b¯3 α11 .

(4.98)

By (4.62), (4.77), (4.83), (4.93) and b3 (b¯3 y8 ) = b¯3 (b3 y8 ) we obtain that b¯3 α11 + b¯3 β7 = q12 + b3 d8 + b3 e6

(4.99)

(y8 , b¯3 α11 ) = (y8 , b3 e6 ) = 1.

(4.100)

and If (b¯3 α11 , b3 e6 ) = 1, then by (4.97) we obtain that b¯3 α11 = q12 + y8 + Σ13 where Σ13 ∈ NB and b3 d8 = y8 + Σ13 + ν3 where ν3 ∈ B which implies that (ν3 , b¯3 β7 ) = 1 and we have a contradiction. Now we have that (b¯3 α11 , b3 e6 ) ≥ 2. If b3 e6 = y8 + r10 where r10 ∈ B, then b¯3 α11 = c3 + y8 + r10 + q12 and we have a contradiction. If b3 e6 = y8 + r4 + r6 where r4 , r6 ∈ B, then (r6 , b¯3 α11 ) = 1 and (4.98) implies that r6 is a real element, and we have that (e¯6 , b3 r6 ) = (α11 , b3 r6 ) = 1, which is a contradiction. If b3 e6 = y8 + r5 + γ5 where r5 , γ5 ∈ B then we have that (e¯6 , b3 r5 ) = (α11 , b3 r5 ) = 1 and we have a contradiction.

4.4.3 Case c = c11 In this section we assume that c = c11 . Then (4.77), (4.78) and (4.91) imply that b3 c11 = y8 + b10 + q15

where q15 ∈ B.

(4.101)

4.4 Case R15 = x6 + x9

201

By (4.85) we obtain that b¯3 c11 = x6 + b15 + g12 .

(4.102)

Now by (4.61), (4.77), (4.78), (4.80) and b¯3 (b3 c11 ) = b3 (b¯3 c11 ), we obtain that b¯15 + b3 g12 = e6 + b¯3 q15

(4.103)

(q15 , b3 b¯15 ) = 1.

(4.104)

which implies that

Now (4.12) implies that q15 is a real element, and we obtain that (b15 , b3 q15 ) = 1.

(4.105)

By (4.81), (4.83) and (4.102) we obtain that b3 y8 = x6 + α6 + g12 .

(4.106)

Thus β = g12 . By (4.62), (4.77), (4.101) and b3 (b¯3 y8 ) = b¯3 (b3 y8 ), we obtain that b¯3 α6 + b¯3 g12 = q15 + b3 d7 + b3 e6 .

(4.107)

Now by (4.106) we obtain that (y8 , b¯3 α6 ) = 1. Hence (q15 , b¯3 α6 ) = 0, which implies that 1 = (q15 , b¯3 g12 ) = (b3 q15 , g12 ).

(4.108)

Since q15 is a real element, (4.101) implies that (c¯11 , b3 q15 ) = 1. Now by (4.105) and (4.108) we obtain that b3 q15 = c¯11 + g12 + b15 + z7

where z7 ∈ B

(4.109)

which implies that b¯3 z7 = q15 + z6

where z6 ∈ B.

By (4.103), (4.109) and since q15 is a real element, we have that b3 g12 = e6 + z¯ 7 + c11 + g¯ 12 , which implies that b3 z7 = g¯ 12 + r9 ,

where r9 ∈ B.

Now b3 (b¯3 z7 ) = b¯3 (b3 z7 ) implies that b15 + b3 z6 = e¯6 + b¯3 r9 . Hence 1 = (b15 , b¯3 r9 ) = (r9 , b3 b15 ) and we have a contradiction to (4.80).

(4.110)

202

4


In the following three sections, we assume that (b3 c, b3 c) = 4,

(4.111)

so (4.90) implies that 12 ≤ |c| ≤ 14.

4.4.4 Case c = c12 In this section we assume that c = c12 . Then by (4.61), (4.80), (4.81) and (4.111), we obtain that b¯3 c12 = x6 + x9 + b15 + g6

where g6 ∈ B

(4.112)

and b3 y8 = x6 + g6 + β12

where β12 ∈ B.

(4.113)

Assume henceforth that (b3 g6 , b3 g6 ) = 3 and we shall derive a contradiction. Then (4.112) implies that b3 g6 = c12 + v3 + w3 and b¯3 g6 = y8 + ν + μ where v3 , w3 , ν, μ ∈ B. Now by (4.112), (4.113) and b¯3 (b3 g6 ) = b3 (b¯3 g6 ), we obtain that x9 + b15 + b¯3 v3 + b¯3 w3 = β12 + b3 ν + b3 μ, which is a contradiction. Now we have that (b3 g6 , b3 g6 ) = 2, by (4.112) and (4.113) we obtain that b3 g6 = c12 + v6

where v6 ∈ B

(4.114)

where v10 ∈ B.

(4.115)

and b¯3 g6 = y8 + v10

Now by (4.112), (4.113) and b¯3 (b3 g6 ) = b3 (b¯3 g6 ) we obtain that x9 + b15 + b¯3 v6 = β12 + b3 v10 , which implies that (v10 , b¯3 b15 ) = 1.

(4.116)

In addition, we have that (b10 , b¯3 b15 ) = (b8 , b¯3 b15 ) = 1 which implies that b¯3 b15 = b8 + b10 + v10 + v17

where v17 ∈ B.

(4.117)

Now by (4.12) we obtain that v10 and v17 are real elements. By (4.77), (4.78) and (4.111) we obtain that b3 c12 = y8 + b10 + q + u where q, u ∈ B.

(4.118)

Now by (4.61), (4.77), (4.78), (4.80), (4.112), (4.114) and b¯3 (b3 c12 ) = b3 (b¯3 c12 ), we obtain that e6 + b¯3 q + b¯3 u = b¯15 + c12 + v6 + b3 x9 .

4.4 Case R15 = x6 + x9

203

Now we can assume that (q, b3 b¯15 ) = 1; in addition q = y8 , b10 , and by (4.118) we obtain that |q| = 17. Thus q = v17 which implies that q = v10 . Now since v10 is a real element, (4.118) implies (c¯12 , b3 v10 ) = 1 and by (4.115) we obtain that (g6 , b3 v10 ) = 1, which is a contradiction to (4.116).

4.4.5 Case c = c13 In this section we assume that c = c13 . Then by (4.77) we obtain that b¯3 y8 = d5 + e6 + c13 .

(4.119)

By (4.61) and (4.81) we obtain that either b¯3 d5 = x6 + x9

(4.120)

or b¯3 d5 = x6 + z9

where z9 ∈ B.

(4.121)

Now we have that (b3 d5 , b3 d5 ) = 2 and together with (4.119), we obtain that b3 d5 = y8 + z7

where z7 ∈ B.

(4.122)

Assume that b¯3 d5 = x6 + z9 . Then by (4.81) we obtain that b¯3 c13 = x6 + x9 + b15 + r9

where r9 ∈ B

(4.123)

and b3 y8 = x6 + z9 + r9 .

(4.124)

Now by (4.1), (4.121), (4.122) and b3 (b3 d5 ) = b32 d5 , we obtain that r9 + b3 z7 = b6 d5 . By (4.65), (4.69) and (4.70), we obtain that (b6 d5 , b6 d5 ) ≤ 3 which implies that (r9 , b3 z7 ) = 0.

(4.125)

By (4.77), (4.78) and (4.123), we obtain that b3 c13 = y8 + b10 + q + u where q, u ∈ B.

(4.126)

Now by (4.62), (4.119), (4.122), (4.124), and b3 (b¯3 y8 ) = b¯3 (b3 y8 ) we obtain that q + u + y8 + z7 + b3 e6 = b¯3 z9 + b¯3 r9 ,

(4.127)

and together with (4.125), we obtain that (z9 , b3 z7 ) = 1.

(4.128)

204

4


By (4.126) we obtain that |q| + |u| = 21,

(4.129)

and by (4.124) we obtain that (y8 , b¯3 z9 ) = (y8 , b¯3 r9 ) = 1. Now we can assume by (4.127) that (q, b¯3 r9 ) = (u, b¯ 3 z9 ) = 1,

(4.130)

and together with (4.128) we obtain that |u| ≤ 12. By (4.80) we obtain that b3 b15 = b¯15 + b6 + c13 + f11 , and by (4.78) we obtain that b¯3 b10 = b6 + c13 + f11 . Now by (4.61), (4.119), (4.123), (4.126) and b3 (b¯3 c13 ) = b¯3 (b3 c13 ), we obtain that b¯ 15 + b3 x9 + b3 r9 = e6 + b¯3 q + b¯3 u which implies that either (b¯15 , b¯3 q) = 1 or (b¯15 , b¯3 u) = 1. Assume henceforth that (u, b3 b¯15 ) = 1 and we shall derive a contradiction. Then by (4.12) we obtain that u is a real element. By (4.130) we obtain that (z9 , b3 u) = 1, by (4.126) we obtain that (c¯13 , b3 u) = 1, and by the assumption we obtain that (b15 , b3 u) = 1 which implies that |u| ≥ 14, and we have a contradiction to |u| ≤ 12. Now we obtain that (q, b3 b¯15 ) = 1

(4.131)

which implies by (4.12) that q is a real element. By (4.130) we obtain that (r9 , b3 q) = 1, by (4.126) we obtain that (c¯13 , b3 q) = 1, and together with (b15 , b3 q) = 1 we obtain that |q| ≥ 14. Since (c13 , b¯3 u) = 1 we obtain that |u| ≥ 6. Now (4.129) implies that |q| ≤ 15, and we obtain that 14 ≤ |q| ≤ 15

(4.132)

6 ≤ |u| ≤ 7.

(4.133)

and

By (4.123), (4.124) and (4.130) we obtain that b¯ 3 r9 = y8 + q + λ

where λ ∈ B and 4 ≤ λ ≤ 5

and b3 r9 = c13 + m + n

where m, n ∈ B.

Now by (4.123), (4.124) and b3 (b¯3 r9 ) = b¯3 (b3 r9 ), we obtain that z9 + b3 q + b3 λ = x9 + b15 + b¯3 m + b¯ 3 n. Since (r9 , b3 λ) = 1 and 4 ≤ λ ≤ 5, we obtain that (x9 , b3 λ) = 0 which implies that either x9 = z9 or (x9 , b3 q) = 1. If x9 = z9 then by (4.124), (4.128) and (4.57), we obtain that (y8 , b¯3 x9 ) = (z7 , b¯3 x9 ) = (b10 , b¯3 x9 ) = 1 and we have a contradiction.

4.4 Case R15 = x6 + x9

205

Now we have that (x9 , b3 q) = 1. By (4.126), (4.130) and (4.131) we obtain that (b15 , b3 q) = (c¯13 , b3 q) = (r9 , b3 q) = 1, and we have a contradiction to (x9 , b3 q) = 1. Now we have that b¯3 d5 = x6 + x9 .

(4.134)

By (4.81) and (4.83) we obtain that b¯3 c13 = x6 + b15 + α + β

(4.135)

b3 y8 = x6 + α + β.

(4.136)

and Now by (4.62), (4.119), (4.122), (4.126) and b¯3 (b3 y8 ) = b3 (b¯3 y8 ) we obtain that y8 + z7 + q + u + b3 e6 = b¯3 α + b¯3 β. Now we can assume that (α, b3 z7 ) = 1. By (4.1), (4.122), (4.134), (4.136) and

b3 (b3 d5 ) = b32 d5 ,

(4.137) we obtain that

x9 + b6 d5 = α + β + b3 z7 . By (4.65), (4.69) and (4.70) we obtain that (b6 d5 , b6 d5 ) ≤ 3 which implies that x9 = α. By (4.57), (4.136) and (4.137) we obtain that (z7 , b¯3 x9 ) = (b10 , b¯3 x9 ) = (y8 , b¯3 x9 ) = 1, which is a contradiction.

4.4.6 Case c = c14 In this section we assume that c = c14 . Then by (4.77) we obtain that b¯3 y8 = d4 + e6 + c14

(4.138)

which implies b3 d4 = y8 + y4

where y4 ∈ B,

(4.139)

and by (4.78) we obtain that b3 c14 = b10 + y8 + q + u where q, u ∈ B.

(4.140)

Now by (4.61) we obtain that b¯3 d4 = x6 + y6

where y6 ∈ B

(4.141)

206

4


and by (4.81) we obtain that b3 y8 = x6 + y6 + y12

where y12 ∈ B.

(4.142)

Now by (4.62) and b¯3 (b3 y8 ) = b3 (b¯3 y8 ) we obtain that b¯3 y6 + b¯3 y12 = q + u + y8 + y4 + b3 e6 .

(4.143)

By (4.139) we obtain that (b3 y4 , b3 y4 ) = 1

(4.144)

(y4 , b¯3 y12 ) = 0,

(4.145)

(y4 , b¯3 y6 ) = 1.

(4.146)

which implies that


By (4.140) we obtain that |q| + |u| = 24, and by (4.142) we obtain that (y8 , b¯3 y12 ) = 1. Assume henceforth that (u, b¯3 y12 ) = (q, b¯3 y12 ) = 1 and we shall derive a contradiction. Then b¯3 y12 = q + u + y8 + z4 where z4 ∈ B which implies that b3 z4 = y12 and by (4.143) and (4.145) we obtain that (z4 , b3 e6 ) = 1, which is a contradiction to b3 z4 = y12 . Now we can assume that (u, b¯ 3 y12 ) = 1 and (q, b¯3 y6 ) = 1. By (4.142) and (4.146) we obtain that b¯3 y6 = y4 + y8 + q6 , and by (4.140) we obtain that b3 c14 = b10 + y8 + q6 + u18 which implies that b¯3 q6 = c14 + z4

where z4 ∈ B

and b3 q6 = y6 + z12

where z12 ∈ B.

Now by b3 (b¯3 q6 ) = b¯3 (b3 q6 ) we obtain that b10 + u18 + b3 z4 = e4 + b¯3 z12 , and we have a contradiction to (4.57).

4.4 Case R15 = x6 + x9

207

4.4.7 Case (b3 x6 , b3 x6 ) = 3 In this section we assume that (b3 x6 , b3 x6 ) = 3 Then b 3 x6 = c + d + e

where c, d, e ∈ B

(4.147)

and by (4.57) we obtain that b¯3 x6 = b10 + v + w

where v, w ∈ B.

(4.148)

Now by (4.57) and b3 (b¯3 x6 ) = b¯3 (b3 x6 ) we obtain that b¯3 c + b¯3 d + b¯3 e = b15 + x6 + x9 + b3 v + b3 w.

(4.149)

Now we can assume (b15 , b¯3 c) = 1

(4.150)

and by (4.147) we obtain that (x6 , b¯3 c) = 1 which implies that |c| ≥ 7, and by (4.147) we obtain that |c| ≤ 12. Now we have that 7 ≤ |c| ≤ 12.

(4.151)

By (4.4), (4.10) and (4.150) we obtain that b¯3 b10 = b6 + c + f

where f ∈ B

(4.152)

and b3 b15 = b¯ 15 + b6 + c + f.

(4.153)

Assume henceforth that v = v3 and we shall derive a contradiction. Then by (4.148) we obtain that b¯3 x6 = b10 + v3 + w5 , b3 v3 = x6 + z3

where z3 ∈ B

and b¯3 v3 = α + β

where α, β ∈ B.

(4.154)

Now by b3 (b¯3 v3 ) = b¯ 3 (b3 v3 ) we obtain that b3 α + b3 β = b10 + v3 + w5 + b¯3 z3 and we can assume that (α, b¯3 b10 ) = 1.

(4.155)

By (4.4) and (4.154) we obtain that α = b6 and |α| ≤ 6, so (4.151) and (4.152) imply that α = c and α = f . Now we have a contradiction to (4.152) and (4.155).

208

4


Hence b¯3 x6 = b10 + v4 + w4

(4.156)

(x6 , b3 b10 ) = (x6 , b3 v4 ) = 1,

(4.157)

(b3 v4 , b3 v4 ) = 2

(4.158)

which implies that

and we obtain and 1 = (b3 b10 , b3 v4 ) = (b¯3 b10 , b¯3 v4 ). By (4.4) we obtain that (b6 , b¯3 v4 ) = 0. By (4.151) and (4.152) we obtain that 12 ≤ f ≤ 17, and together with (b3 v4 , b3 v4 ) = 2 we obtain that (f, b¯ 3 v4 ) = 0. Now we have that (c, b¯3 v4 ) = 1

(4.159)

which implies that |c| ≤ 9, and together with (4.151) we obtain that 7 ≤ c ≤ 9. If c = c7 then by (4.147) and (4.153) we obtain that b¯3 c7 = b15 + x6 , and by (4.152) and (4.159) we obtain that b3 c7 = v4 + b10 , which is a contradiction. If c = c8 then by (4.147) and (4.153), we obtain that b¯3 c8 = b15 + x6 + r3 where r3 ∈ B. Now we have that r3 = b¯3 and b8 = c8 , since b15 is a non-real element we have a contradiction to (4.5). Now we have that c = c9 then by (4.147) and (4.153), we obtain that b¯3 c9 = b15 + x6 + Σ6

where Σ6 ∈ NB,

(4.160)

where Σ13 ∈ NB.

(4.161)

and by (4.152) and (4.159) we obtain that b3 c9 = v4 + b10 + Σ13

By (4.157), (4.158) and (4.159) we obtain that b3 v4 = x6 + m6

where m6 ∈ B

b¯3 v4 = c9 + m3

where m3 ∈ B.

and Now by (4.156) and b¯3 (b3 v4 ) = b3 (b¯3 v4 ) we obtain that w4 + b¯ 3 m6 = Σ13 + b3 m3 . Now (v4 , b3 m3 ) = 1 implies that (w4 , b3 m3 ) = 0, so (w4 , Σ13 ) = 1. By (4.161) we obtain that b3 c9 = v4 + b10 + w4 + w9

where w9 ∈ B,

4.5 Case R15 = x7 + x8

209

and by (4.160) we obtain that b¯3 c9 = b15 + x6 + l3 + n3

where l3 , n3 ∈ B.

Now by (4.149) we obtain that b3 w4 = x6 + l3 + n3 , which is a contradiction to (b3 w4 , b3 w4 ) ≤ 2.

4.5 Case R15 = x7 + x8 Throughout this section we assume that R15 = x7 + x8 . Then by (4.7) and (4.8) we have that b3 b10 = b15 + x7 + x8 ,

(4.162)

b62 = b¯6 + b15 + x7 + x8 .

(4.163)

Now (b6 b¯6 , b6 b¯6 ) = 4 and by (4.4) and (4.2), we obtain that b6 b¯6 = 1 + b8 + s + t

where s, t are real elements in B

(4.164)

and by (4.12), we obtain that b¯3 b15 = b8 + b10 + s + t.

(4.165)

Assume henceforth that s = s21 and we shall derive a contradiction. Then by (4.165) t = t6 and b3 t6 = b15 + c3 where c3 ∈ B. Now b¯3 (b3 t6 ) = (b3 b¯3 )t6 implies that b8 t6 = b8 + b10 + s21 + b¯3 c3 . Since b8 and t6 are real elements and b10 is a non-real element, we obtain that (b¯10 , b¯3 c3 ) = 1 and we have a contradiction. Now we have that |s| = 21.

(4.166)

b6 b10 = b¯15 + b3 x7 + b3 x8 .

(4.167)

(b10 , b¯3 x7 ) = 1

(4.168)

By (4.13) we obtain that

By (4.162) we obtain that which implies that 2 ≤ (b¯ 3 x7 , b¯3 x7 ) ≤ 3. Assume that (b3 x7 , b3 x7 ) = 2. Then we obtain that b 3 x7 = c + d

where c, d ∈ B,

(4.169)

and by (4.162) we obtain that b¯3 x7 = b10 + y11

where y11 ∈ B.

(4.170)

210

4


By (4.1) and (4.162) we obtain that (b3 x7 , b¯3 b10 ) = (b32 , x¯7 b10 ) = (b¯3 , x¯7 b10 ) + (b6 , x¯7 b10 ) = (x7 , b3 b10 ) + (b6 , x¯7 b10 ) = 1 + (b6 , x¯7 b10 ) which implies that (b3 x7 , b¯3 b10 ) ≥ 1. By (4.4) we obtain that (b6 , b¯3 b10 ) = 1 which implies by (4.169) that (b3 x7 , b¯3 b10 ) = 2. Therefore (b3 x7 , b¯3 b10 ) = 1. Now we can assume that b¯3 b10 = b6 + c + e

where e ∈ B, e = c and e = d,

(4.171)

and by (4.10) we obtain that b3 b15 = b¯15 + b6 + c + e.

(4.172)

Now we obtain by (4.169) that (b15 , b¯3 c) = (x7 , b¯3 c) = 1 which implies that 9 ≤ |c| ≤ 17.

(4.173)

By (4.162), (4.169, (4.170) and b3 (b¯3 x7 ) = b¯3 (b3 x7 ) we obtain that b¯3 c + b¯3 d = b15 + x7 + x8 + b3 y11 .

(4.174)

Assume henceforth that (b15 , b3 y11 ) = 1 and we shall derive a contradiction. Then we have that (b15 , b¯3 d) = 1. By (4.171) we obtain that e = d, by (4.169) and (4.3) we obtain that b6 = d and by (4.173) and (4.169) we obtain that d = b¯15 , which is a contradiction to (4.172). Now we have that (b15 , b3 y11 ) = 0.

(4.175)

Assume henceforth that (b¯ 15 , b¯3 y11 ) = 1 and we shall derive a contradiction. Then by (4.12) we obtain that y11 is a real element which implies that (b15 , b3 y11 ) = 1, and we have a contradiction to (4.175). Now we have that (b¯15 , b¯3 y11 ) = 0.

(4.176)

By (4.1), (4.169), (4.170), (4.171) and b¯ 3 (b¯3 x7 ) = b¯32 x7 we obtain that b6 + e + b¯3 y11 = d + b¯6 x7 .

(4.177)

Assume that (u, b¯3 c) = 1 where u ∈ B. Then by (4.174) we obtain that |u| ≥ 5. Assume henceforth that u = u5 and we shall derive a contradiction. Then by (4.174) we obtain that b¯ 3 u5 = y11 + y4 where y4 ∈ B and b3 u5 = c + q where q ∈ B. Now by b3 (b¯3 u5 ) = b¯3 (b3 u5 ) we obtain that b¯3 c + b¯3 q = b3 y11 + b3 y4 . Now by (4.172) we obtain that (b15 , b¯3 c) = 1 and by (4.175) we obtain that (b15 , b3 y11 ) = 0 which implies that (b15 , b3 y4 ) = 1, and we have a contradiction. Hence If (u, b¯3 c) = 1 where u ∈ B

then |u| ≥ 6.

(4.178)

4.5 Case R15 = x7 + x8

211

Assume henceforth that c = c9 and we shall derive a contradiction. Then by (4.169) and (4.172) we obtain that b¯3 c9 = x7 + b15 + u5 where u5 ∈ B, which is a contradiction to (4.178). Now (4.173) implies that 10 ≤ |c| ≤ 17.

(4.179)

(y11 , b3 e) = 1

(4.180)

Assume henceforth that

and we shall derive a contradiction. Then by (4.171) and (4.177) we obtain that (b6 b¯6 , x7 x¯7 ) = (b¯6 x7 , b¯6 x7 ) ≥ 6. By assumption (b3 b¯3 , x7 x¯7 ) = 2 which implies by (4.2) that (b8 , x7 x¯7 ) = 1. Now by (4.164) we obtain that 6 ≤ (b6 b¯6 , x7 x¯7 ) = 2 + (s, x7 x¯7 ) + (t, x7 x¯7 ). Hence (s, x7 x¯7 ) + (t, x7 x¯7 ) ≥ 4 which implies that either |s| ≤ 10 or

|t| ≤ 10.

(4.181)

By (4.3) and (4.169), we obtain that d = b6 and by (4.171) we obtain that d = e. Now (4.177) implies that (d, b¯3 y11 ) = 1 and by (4.180) we obtain that b¯3 y11 = d + e + Σ

where Σ ∈ NB.

(4.182)

where C ∈ NB

(4.183)

Now (4.171) implies that b3 e = b10 + y11 + C and by (4.172) we obtain that b¯3 e = b15 + D

where D ∈ NB.

(4.184)

Now by (4.171), (4.172) and b3 (b¯3 e) = b¯3 (b3 e) we obtain that b¯15 + b3 D = b¯3 y11 + b¯3 C.

(4.185)

(b¯15 , b¯3 C) = (b3 b¯15 , C) = 1.

(4.186)

Now (4.176) implies that

By (4.165), (4.170), (4.183) and since b10 is a non-real element we obtain that e = x¯7 , e = x¯8 and e = b¯15 . Thus (b¯10 , b3 e) = (b¯3 b¯10 , e) = 0 and (b8 , b3 e) = (b¯ 3 b8 , e) = 0. Now by (4.165), (4.183) and (4.186) we can assume that (s, C) = 1 which implies that (s, b3 e) = 1 and |s| ≤ |C|.

(4.187)

By (4.183) we obtain that |C| = 3|e| − 21 which implies that |s| ≤ 3|e| − 21.

(4.188)

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4


By (4.165) and (4.187) we obtain that (b¯3 s, e) = (b¯3 s, b¯ 15 ) = 1,

(4.189)

and together with (4.188) we obtain that 11 ≤ |e|. By (4.171) and (4.179) we obtain that |e| ≤ 14. Therefore 11 ≤ |e| ≤ 14.

(4.190)

Assume henceforth that e = e11 and we shall derive a contradiction. Then by (4.189) we obtain that 11 ≤ |s| and by (4.183) we obtain that |C| = 12 which implies by (4.188) that |s| ≤ 12. Hence 11 ≤ |s| ≤ 12 which implies by (4.164) that 15 ≤ |t| ≤ 16, and we have a contradiction to (4.181). Assume henceforth that e = e14 and we shall derive a contradiction. Then by (4.189) we obtain that 12 ≤ |s| and by (4.183) we obtain that |C| = 21 which implies by (4.188) that |s| ≤ 21. Therefore 12 ≤ |s| ≤ 21. By (4.166) and (4.183) we obtain that 12 ≤ |s| ≤ 15, and by (4.164) we obtain that 12 ≤ |t| ≤ 15, which is a contradiction to (4.181). Assume henceforth that e = e12 and we shall derive a contradiction. Then by (4.189) we obtain that 9 ≤ |s| and by (4.188) we obtain that |s| ≤ 15. Hence 9 ≤ |s| ≤ 15. Now (4.164) and (4.181) imply that 9 ≤ |s| ≤ 10. By (4.189) it is impossible that s = s10 . Therefore s = s9 and b¯3 s9 = e12 + b¯15 and by (4.183) we obtain that b3 e12 = s9 + y11 + b10 + w6

where w6 ∈ B.

Now by (4.2), (4.165) and b3 (b¯3 s9 ) = (b3 b¯3 )s9 we obtain that b8 s9 = b8 + s9 + b10 + w6 + b¯10 + t18 + y11 . Since b8 and s9 are real elements we obtain that y11 is a real element. By (4.172) we obtain that c = c12 and by (4.169) d = d9 . Now (4.170) and (4.182) imply that b¯3 y11 = e12 + d9 + x¯ 7 + w5

where w5 ∈ B.

(4.191)

Now we have that b3 w5 = y11 + v4

where v4 ∈ B

(4.192)

and by (4.174) we obtain that either (w¯ 5 , b¯3 c12 ) = 1 or (w¯ 5 , b¯3 d9 ) = 1. Now by (4.178) we obtain that (w¯ 5 , b¯3 d9 ) = 1 which implies that b¯3 w5 = d¯9 + w6

where w6 ∈ B.

Now by (4.191), (4.192) and b¯3 (b3 w5 ) = b3 (b¯3 w5 ) we obtain that e12 + d9 + x¯7 + w5 + b¯3 v4 = b3 d¯9 + b3 w6 . Since (w5 , b3 w6 ) = 1 we obtain that (e12 , b3 w6 ) = 0

4.5 Case R15 = x7 + x8

213

which implies that (e12 , b3 d¯9 ) = 1. By (4.169) we obtain that (x¯7 , b3 d¯9 ) = 1 and together with (w5 , b3 d¯9 ) = 1, we obtain that b3 d¯9 = w5 + x¯7 + e12 + r3

where r3 ∈ B

which implies that (r3 , b¯3 v4 ) = 1 and b3 r¯3 = d9 , which is a contradiction. Now we have that e = e13 which implies by (4.188) that |s| ≤ 18, and by (4.189) we obtain that 11 ≤ |s|. Therefore 11 ≤ |s| ≤ 18. Now (4.181) and (4.164) imply that 17 ≤ |s| ≤ 18. By (4.183) and (4.187) we obtain that |s| = 17. Thus s = s18 and by (4.165) we obtain that t = t9 and b¯3 t9 = b¯15 + E12

where E12 ∈ NB.

Now by (4.2), (4.165) and b3 (b¯3 t9 ) = (b3 b¯3 )t9 we obtain that b8 t9 = b8 + b¯10 + s18 + b3 E12 . Now since b8 and t9 are real elements and b10 is a non-real element, we obtain that (b¯10 , b3 E12 ) = 1, which is a contradiction to (4.171). Now we have that (y11 , b3 e) = 0.

(4.193)

By (4.4), (4.162), (4.170), (4.171) and b3 (b¯3 b10 ) = b¯3 (b3 b10 ) we obtain that b8 + b3 c + b3 e = b¯3 b15 + b¯3 x8 + y11 .

(4.194)

Now (4.193) implies that (y11 , b3 c) = 1 and together with (4.171) we obtain that b3 c = b10 + y11 + F

where F ∈ NB.

(4.195)

where G12 ∈ NB.

(4.196)

where H ∈ NB.

(4.197)

Now by (4.177) we obtain that b¯3 y11 = c + d + G12 By (4.169) and (4.172) we obtain that b¯3 c = x7 + b15 + H

Now by (4.169), (4.171), (4.172) and b3 (b¯3 c) = b¯3 (b3 c) we obtain that b¯15 + b3 H = G12 + b¯3 F.

(4.198)

If c = c10 then by (4.197) we obtain that H = H8 ∈ B and (c10 , b3 H8 ) = 1, which is a contradiction to (4.198), and if c = c17 then (4.169) implies that d = d4 , which is a contradiction to (4.196). Now (4.179) implies that 11 ≤ |c| ≤ 16. From (4.169) and (4.171) we have seven cases (see Table 4.3). In the following seven sections, we derive a contradiction for each case in Table 4.3.

214

4

Table 4.3 Splitting to seven cases


Case 1

c = c11

d = d10

e = e13

Case 2

c = c12

d = d9

e = e12

Case 3

c = c13

d = d8

e = e11

Case 4

c = c14

d = d7

e = e10

Case 5

c = c15

d = d6

e = e9

Case 6

c = c16

d = d5

e = e8

Case 7

(b3 x7 , b3 x7 ) = 3

4.5.1 Case c = c11 In this section we assume that c = c11 . Then (4.178) and (4.197) imply that b¯3 c11 = x7 + b15 + f11

where f11 ∈ B.

(4.199)

where g12 ∈ B.

(4.200)

Now (4.195) implies that b3 c11 = b10 + y11 + g12 By (4.198) we obtain that b¯15 + b3 f11 = G12 + b¯3 g12

(4.201)

which implies that (g12 , b3 b¯15 ) = 1, and by (4.165) we obtain that b3 b¯15 = b8 + b¯10 + g12 + g15

where g15 ∈ B.

(4.202)

By (4.170), (4.174) and (4.199) we obtain that b3 y11 = x7 + f11 + Σ15

where Σ15 ∈ NB.

(4.203)

By (4.196) we obtain that b¯3 y11 = c11 + d10 + G12 .

(4.204)

Assume henceforth that y11 is a real element and we shall derive a contradiction. Then by (4.203) and (4.204) we obtain that c11 = f¯11 and b3 y11 = x7 + d¯10 + c¯11 + h5

where h5 ∈ B

(4.205)

which implies that b¯3 h5 = y11 + h4

where h4 ∈ B.

(4.206)

By (4.174), (4.199) and (4.205) we obtain that b¯3 d10 = x7 + x8 + d¯10 + h5

(4.207)

4.5 Case R15 = x7 + x8

215

which implies that b3 h5 = d10 + v5

where v5 ∈ B.

(4.208)

Now b3 (b¯3 h5 ) = b¯3 (b3 h5 ) implies that c¯11 + b3 h4 = x8 + b¯ 3 v5 . Hence (v5 , b3 c¯11 ) = 1, which is a contradiction to (4.199). Now we have that y11 is a non-real element. By (4.200) and (4.201) we obtain that b¯ 3 g12 = b¯15 + c11 + Σ10

where Σ10 ∈ NB.

(4.209)

Now by (4.2), (4.200), (4.202) and b3 (b¯3 g12 ) = (b3 b¯3 )g12 , we obtain that b8 g12 = b8 + b10 + b¯10 + y11 + g12 + g15 + b3 Σ10 .

(4.210)

Since b8 and g12 are real elements and y11 is a non-real element, we obtain that (y¯11 , b3 Σ10 ) = 1. Now (4.209) implies that Σ10 ∈ B and by (4.203) we obtain that b3 y11 = Σ¯ 10 + x7 + f11 + r5

where r5 ∈ B.

(4.211)

Now by (4.174) and (4.199) we obtain that b¯3 d10 = r5 + x7 + x8 + Σ¯ 10 .

(4.212)

Now b¯3 r5 = y11 + h4

where h4 ∈ B

b3 r5 = d10 + v5

where v5 ∈ B.

and (4.213)

By b3 (b¯3 r5 ) = b¯3 (b3 r5 ) we obtain that f11 + b3 h4 = x8 + b¯3 v5 which implies that (f11 , b¯3 v5 ) = 1, and we have a contradiction to (4.213).

4.5.2 Case c = c12 In this section we assume that c = c12 . Then (4.195) and (4.197) imply that b¯3 c12 = x7 + b15 + H14

(4.214)

b3 c12 = b10 + y11 + F15 .

(4.215)

and

216

4


Assume henceforth that y11 is a real element and we shall derive a contradiction. Then by (4.170) and (4.196) we obtain that b3 y11 = x7 + d¯9 + c¯12 + h5

where h5 ∈ B.

(4.216)

Now by (4.174) and (4.214) we obtain that H14 + b¯3 d9 = x8 + c¯12 + d¯9 + x7 + h5 which implies that b¯3 c12 = x7 + b15 + d¯9 + h5 , and we have a contradiction to (4.178). Now we have that y11 is a non-real element. By (4.198) we obtain that b¯15 + b3 H14 = G12 + b¯3 F15 ,

(4.217)

which implies that b¯15 ∈ Irr(b¯ 3 F15 ). By (4.5) we obtain that (b8 , F15 ) = (b8 , b3 c12 ) = 0, and by (4.162) we obtain that (b¯10 , F15 ) = (b¯10 , b3 c12 ) = 0. Hence by (4.165) we can assume that s ∈ Irr(F15 )

(4.218)

which implies that |s| ≤ 15, and by (4.164) s is a real element. Now we have that (s, b3 c12 ) = 1

(4.219)

(s, b3 b¯15 ) = 1

(4.220)

9 ≤ |s| ≤ 15.

(4.221)

and by (4.165) we obtain that which implies that 9 ≤ |s|. Hence

Assume henceforth that s = s15 and we shall derive a contradiction. Then by (4.215) and (4.218) we obtain that b3 c12 = b10 + y11 + s15 .

(4.222)

Now (4.165) implies that b¯3 s15 = c12 + b¯15 + Σ18

where Σ18 ∈ NB

(4.223)

where f14 ∈ B.

(4.224)

and (4.214) implies that b¯3 c12 = x7 + b15 + f14

Now by (4.2), (4.165) and b3 (b¯3 s15 ) = (b3 b¯3 )s15 we obtain that b8 s15 = b8 + b10 + b¯10 + y11 + t12 + s15 + b3 Σ18 .

4.5 Case R15 = x7 + x8

217

Since b8 and s15 are real elements and y11 is a non-real element, we obtain that y¯11 ∈ Irr(b3 Σ18 ). By (4.174) we obtain that (f14 , b3 y11 ) = 1, and together with (4.170) we obtain that b3 y11 = x7 + f14 + S12

where S12 ∈ NB.

(4.225)

Since y¯11 ∈ Irr(b3 Σ18 ) we obtain that there exists an element α in B such that α ∈ Irr(Σ¯ 18 ) ∩ Irr(S12 ) which implies that 10 ≤ |α|. Hence S12 = α12 . Now we have that b¯3 s15 = c12 + b¯15 + α¯ 12 + α6

where α6 ∈ B

(4.226)

which implies that b3 α6 = s15 + α3

where α3 ∈ B,

(4.227)

and by (4.225) we obtain that b3 y11 = x7 + f14 + α12 . By (4.196) we obtain that b¯3 y11 = c12 + d9 + β12

where β12 ∈ B.

By (4.169), (4.171) (4.172), (4.224), (4.226) and b3 (b¯3 c12 ) = b¯3 (b3 c12 ), we obtain that b3 f14 = β12 + c12 + α¯ 12 + α6 . Now we have that b¯3 α6 = f14 + α4

where α4 ∈ B.

Now by (4.226), (4.227) and b3 (b¯3 α6 ) = b¯3 (b3 α6 ) we obtain that b¯15 + b¯3 α3 = β12 + b3 α4 , and we have a contradiction. Now we have that |s| = 15, so (4.215), (4.219) and (4.221) imply that b3 c12 = b10 + y11 + s + k

where k ∈ B and 9 ≤ |s| ≤ 11.

Now (4.219) and (4.220) imply that b¯3 s = c12 + b¯15 + T

where T ∈ NB and |T | ≤ 6,

(4.228)

and together with (4.165) and b3 (b¯3 s) = (b3 b¯3 )s we obtain that b8 s = b8 + b10 + b¯10 + y11 + s + t + k + b3 T . Since b8 and s are real elements and y11 is a non-real element, we obtain that y¯11 ∈ Irr(b3 T ), which is a contradiction to (4.228).

218

4



(4.229)

b3 c13 = b10 + y11 + F18 .

(4.230)

and


where h5 ∈ B.

(4.231)

Now by (4.174) and (4.229) we obtain that H17 + b¯3 d8 = x8 + c¯13 + d¯8 + x7 + h5 , and we have a contradiction. Now we have that y11 is a non-real element. By (4.198) we obtain that b¯15 + b3 H17 = G12 + b¯3 F18 . Now we have that b¯15 ∈ Irr(b¯3 F18 ). By (4.5), (4.162) and (4.230) we can assume that s ∈ Irr(F18 )

(4.232)

and together with (4.165) we obtain that 11 ≤ |s| ≤ 18.

(4.233)

Assume henceforth that s = s18 and we shall derive a contradiction. Then (4.230) and (4.232) imply that b3 c13 = b10 + y11 + s18 .

(4.234)

b¯3 c13 = x7 + b15 + f17

(4.235)

b3 b¯15 = b8 + b¯10 + s18 + t9 .

(4.236)

By (4.229) we obtain that


Now we have that b¯3 s18 = c13 + b¯15 + Σ26

where Σ26 ∈ NB.

(4.237)

4.5 Case R15 = x7 + x8

219

By (4.2) and b3 (b¯3 s18 ) = (b3 b¯3 )s18 we obtain that b8 s18 = b8 + t9 + b10 + b¯10 + y11 + s18 + b3 Σ26 . Since b8 and s18 are real elements and y11 is a non-real element, we obtain that y¯11 ∈ Irr(b3 Σ26 ).

(4.238)

By (4.174) and (4.235) we obtain that b¯3 d8 = x7 + x8 + g9

where g9 ∈ B

(4.239)

and b3 y11 = x7 + f17 + g9 .

(4.240)

Now (4.196) implies that b¯3 y11 = c13 + d8 + h12

where h12 ∈ B.

(4.241)

By (4.237) and (4.240) we obtain that g¯ 9 ∈ / Σ26 , and by (4.170) and (4.237) we obtain that x¯7 ∈ / Σ26 . Now (4.238) and (4.240) imply that f¯17 ∈ Σ26 . By (4.237) we obtain that b¯3 s18 = c13 + b¯15 + f¯17 + f9

where f9 ∈ B.

(4.242)

By (4.198) we obtain that b3 f17 = h12 + c13 + f¯17 + f9

(4.243)

which implies that b¯3 f9 = f17 + Σ10

where Σ10 ∈ NB

b3 f9 = s18 + Σ9

where Σ9 ∈ NB.

(4.244)

and Now by (4.242), (4.243) and b3 (b¯3 f9 ) = b¯3 (b3 f9 ) we obtain that h12 + b3 Σ10 = b¯15 + b¯3 Σ9 which implies that b¯15 ∈ Irr(b3 Σ10 ), and by (4.172) we obtain that b¯6 ∈ Σ10 . Now (4.244) implies that (f9 , b3 b¯6 ) = 1, which is a contradiction to (4.3). Now we have that |s| = 18 which implies by (4.230) and (4.233) that 11 ≤ |s| ≤ 12. By (4.165), (4.230) and (4.232) we obtain that b¯3 s = c13 + b¯15 + T

where T ∈ NB and |T | ≤ 8.

Now by (4.2), (4.230), (4.165) and b3 (b¯3 s) = (b3 b¯3 )s we obtain that b8 s = b8 + b10 + b¯10 + y11 + t + F18 + b3 T .

(4.245)

220

4


Since b8 and s are real elements and y11 is a non-real element, we obtain that s = / F18 . Hence y¯11 ∈ b3 T , y¯11 ; and by (4.232) and 11 ≤ |s| ≤ 12, we have that y¯11 ∈ which is a contradiction to (4.245).


(4.246)

b3 c14 = b10 + y11 + F21 .

(4.247)

and

By (4.169) we obtain that b¯3 d7 = x7 + Σ14

where Σ14 ∈ NB.

(4.248)

By (4.198) we obtain that b¯15 + b3 H20 = G12 + b¯3 F21 which implies that b¯15 ∈ Irr(b¯3 F21 ). By (4.5) and (4.162) we obtain that b8 ∈ / / Irr(b3 c14 ), and by (4.165) and (4.247) we can assume that Irr(b3 c14 ) and b¯10 ∈ s ∈ Irr(F21 ).

(4.249)

By (4.165) we obtain that (s, b3 b¯15 ) = 1 and together with (4.166) we obtain that 12 ≤ |s| ≤ 15

(4.250)

and b¯ 3 s = b¯15 + c14 + T

where T ∈ NB.

(4.251)

Assume henceforth that y11 is a real element and we shall derive a contradiction. Then by (4.170) and (4.196) we obtain that either b3 y11 = x7 + d¯7 + c¯14 + h5

where h5 ∈ B

(4.252)

where Σ12 ∈ NB.

(4.253)

or x7 = d¯7 and then b3 y11 = x7 + c¯14 + Σ12

If (4.252) holds then by (4.174) and (4.246), we obtain that H20 + b¯3 d7 = x8 + c¯14 + d¯7 + x7 + h5 .

4.5 Case R15 = x7 + x8

221

By (4.169) we obtain that (x7 , b¯3 d7 ) = 1 which implies that H20 = x8 + d¯7 + h5 , and we have a contradiction to (4.246). Now we have that x7 = d¯7 . By (4.247), (4.249) and (4.250) we obtain that b3 c14 = b10 + y11 + s + g

where g ∈ B and 6 ≤ |g| ≤ 9.

(4.254)

By (4.162), (4.169), (4.170), (4.246), (4.252) and b3 (b¯3 x7 ) = b¯ 3 (b3 x7 ), we obtain that x8 + Σ12 = H20 . By (4.254) we obtain that (b3 c14 , b3 c14 ) = 4. Therefore Σ12 = h12 where h12 ∈ B and H20 = x8 + h12 . Therefore b¯3 c14 = x7 + b15 + x8 + h12

(4.255)

b3 y11 = x7 + c¯14 + h12 .

(4.256)

and

By (4.194), (4.165) and (4.254) we obtain that g + b3 e10 = t + b¯3 x8 . It is obvious that g = t. By (4.165) and (4.250) we obtain that 12 ≤ |t| ≤ 15. Now (4.171) implies that b3 e10 = b10 + t + k

where k ∈ B

(4.257)

and b¯3 x8 = b10 + g + k.

(4.258)

By (4.255) we obtain that b3 x8 = c14 + α + β

where α, β ∈ B.

(4.259)

Assume henceforth that g is a real element and we shall derive a contradiction. By (4.254) and (4.258) we obtain that (c¯14 , b3 g) = (x8 , b3 g) which implies that 9 ≤ |g|. By (4.254) we obtain that g = g9 . Now we have that b3 g9 = x8 + c¯14 + w5

where w5 ∈ B.

By (4.162), (4.255), (4.258), (4.258), (4.259) and b3 (b¯3 x8 ) = b¯ 3 (b3 x8 ), we obtain that x8 + c¯14 + w5 + b3 k5 = h12 + b¯3 α + b¯3 β. Hence (h12 , b3 k5 ) = 1, which is a contradiction to (4.258). Now we have that g is a non-real element. By (4.2), (4.256), (4.170), (4.254) and (b3 b¯3 )y11 = b3 (b¯3 y11 ), we obtain that b8 y11 = b10 + b¯10 + y11 + s + g + b3 h¯ 12 .

222

4


Since b8 and y11 are real elements and g is a non-real element, we obtain that ¯ b3 h¯ 12 ) = 1 and by (4.258) we obtain that (x8 , b3 g) = 1 which im(h12 , b3 g) = (g, plies that 8 ≤ |g|. By (4.254) we obtain that either g = g8 or g = g9 . Assume henceforth that g = g9 and we shall derive a contradiction. Then by (4.258) and (4.257) we obtain that b3 k5 = x8 + h7

where h7 ∈ B

b¯ 3 k5 = e10 + h5

where h5 ∈ B.

and (4.260)

Now by (4.258), (4.257) and b3 (b¯3 k5 ) = b¯3 (b3 k5 ), we obtain that t15 + b3 h5 = g9 + b¯3 h7 . Hence (g9 , b3 h5 ) = 1, which is a contradiction to (4.260). Now we have that g = g8 . By (4.258) we obtain that (x8 , b3 g8 ) = 1, and since (h12 , b3 g8 ) = 1 we obtain that b3 g8 = x8 + h12 + h4

where h4 ∈ B.

By (4.254) we obtain that b¯3 g8 = c14 + m + n

where m, n ∈ B.

Now by (4.254), (4.258) and b3 (b¯3 g8 ) = b¯ 3 (b3 g8 ), we obtain that y11 + s13 + b3 m + b3 n = k6 + b¯3 h4 + b¯3 h12 which implies that b¯3 h12 = g8 + y11 + s13 + γ4 where γ4 ∈ B. Hence b3 γ4 = h12 which implies that (b3 γ4 , b3 γ4 ) = 1. On the other hand, we have that either (m, b¯3 γ4 ) = 1 or (n, b¯ 3 γ4 ) = 1 which implies that (b3 γ4 , b4 γ4 ) = 1, and we have a contradiction. Now we have that y11 is a non-real element. By (4.2), (4.165), (4.247), (4.251) and b3 (b¯3 s) = (b3 b¯3 )s we obtain that b8 s = b8 + b10 + b¯10 + s + t + F21 + b3 T . By (4.249) and (4.250) we obtain that y¯11 ∈ / Irr(F21 ). Now since b8 and s are real elements and y11 is a non-real element we obtain that y¯11 ∈ Irr(b3 T ).

(4.261)

If s = s12 then by (4.251) we obtain that b¯3 s12 = b¯15 + c14 + f7

where f7 ∈ B.

Now we have that (y¯11 , b3 f7 ) = (s12 , b3 f7 ) = 1, which is a contradiction. If s = s13 then by (4.251), we obtain that b¯3 s13 = b¯15 + c14 + f10

where f10 ∈ B.

(4.262)

where h6 ∈ B.

(4.263)

Now (4.261) implies that b3 f10 = s13 + y¯11 + h6

4.5 Case R15 = x7 + x8

223

By (4.170) we obtain that b3 y11 = x7 + f¯10 + Σ16

where Σ16 ∈ NB.

Now by (4.174), (4.248) and (4.246), we obtain that H20 + Σ14 = x8 + f¯10 + Σ16 which implies that b¯3 c14 = x7 + f¯10 + g10 + b15 b3 y11 = x7 + f¯10 + g10 + g6

where g10 ∈ B,

(4.264)

where g6 ∈ B

(4.265)

and b¯3 d7 = x7 + x8 + g6 .

(4.266)

By (4.247) and (4.249) we obtain that b3 c14 = b10 + y11 + s13 + g8

where g8 ∈ B.

By (4.198) and (4.262) we obtain that b3 f¯10 + b3 g10 = G12 + c14 + f10 + b¯3 g8 . By (4.264) we obtain that b¯3 f10 = c¯14 + Q16

where Q16 ∈ NB.

(4.267)

Now by (4.264), (4.263), (4.262), (4.265) and b3 (b¯3 f10 ) = b¯3 (b3 f10 ) we obtain that c14 + f10 + g¯ 6 + b¯3 h6 = b3 Q16

(4.268)

which implies that c14 ∈ Irr(b3 Q16 ). By (4.264) one of the following cases hold: Case 1: x7 ∈ Irr(Q16 ). Case 2: g10 ∈ Irr(Q16 ). Case 3: f¯10 ∈ Irr(Q16 ). Case 1 By (4.169) and (4.268) we obtain that (d7 , b¯3 h6 ) = 1 and by (4.263) we obtain that (f10 , b¯3 h6 ) = 1, which is a contradiction. Case 2 By (4.267) we obtain that b¯3 f10 = c¯14 + g10 + α6

where α6 ∈ B.

Now (4.264) implies that b3 g10 = β6 + f10 + c14

where β6 ∈ B.

(4.269)

By (4.263) we obtain that (f10 , b¯3 h6 ) = 1 which implies that (β6 , b¯3 h6 ) = 0. Since g10 ∈ Irr(Q16 ) we obtain by (4.268) and (4.269) that β6 = g¯ 6 . Now by (4.269) we

224

4


obtain that (g¯ 10 , b3 g6 ) = 1, and by (4.266) we obtain that (d7 , b3 g6 ) = 1, which is a contradiction. Case 3 By (4.267) we obtain that b¯3 f10 = c¯14 + f¯10 + γ6

where γ6 ∈ B.

By (4.263) we obtain that (f10 , b¯3 h6 ) = 1 which implies that (γ6 , b¯3 h6 ) = 1. Since f¯10 ∈ Q16 we obtain by (4.268) that γ6 = g¯ 6 which implies that (f10 , b3 g¯6 ) = 1, and by (4.265) we obtain that (y¯11 , b3 g¯6 ) = 1, which is a contradiction. If s = s14 then by (4.251) we obtain that b¯3 s14 = c14 + b¯15 + T13 . Now by (4.261) we obtain that there exists an element f ∈ B in Irr(T13 ) such that 10 ≤ |f | which implies that f = T13 . Hence b¯3 s14 = c14 + b¯15 + f13

where f13 ∈ B.

(4.270)

where Σ13 ∈ NB.

(4.271)

By (4.170) and (4.261) we obtain that b3 y11 = x7 + f¯13 + Σ13

Now by (4.174), (4.246) and (4.248) we obtain that H20 + Σ14 = x8 + f¯13 + Σ13 which implies that b¯3 c14 = x7 + g7 + f¯13 + b15 b¯3 d7 = x7 + x8 + g6

where g7 ∈ B,

where g6 ∈ B

(4.272)

and b3 y11 = f¯13 + x7 + g6 + g7 .

(4.273)

By (4.247) and (4.249) we obtain that b3 c14 = b10 + y11 + s14 + h7

where h7 ∈ B.

Now we have that b3 g7 = c14 + α7

where α7 ∈ B

b¯3 h7 = c14 + β7

where β7 ∈ B.

and

By (4.198) we obtain that b3 f¯13 + α7 = G12 + c14 + f13 + β7 .

(4.274)

4.5 Case R15 = x7 + x8

225

If α7 ∈ Irr(G12 ) then there exists an element α5 ∈ B such that α5 ∈ Irr(G12 ) which implies that (α5 , b3 f¯13 ) = 1, and we have a contradiction. Hence α7 = β7 . By (4.273) we obtain that 3 ≤ (b3 y11 , b3 y11 ). Now (4.196) and (4.274) imply that b¯3 y11 = c14 + d7 + α6 + β6

where α6 , β6 ∈ B

(4.275)

and b3 f¯13 = c14 + f13 + α6 + β6 .

(4.276)

By (4.270) and (4.273) we obtain that b3 f13 = y¯11 + s14 + Σ14

where Σ14 ∈ NB.

Now by (4.270), (4.272), (4.273) and b3 (b¯3 f13 ) = b¯3 (b3 f13 ) we obtain that g¯ 6 + b¯3 Σ14 = α6 + β6 + b3 α¯ 6 + b3 β¯6 . By (4.276) we obtain that (f13 , b3 α¯ 6 ) = (f13 , b3 β¯6 ) = 1 which implies that (g¯6 , b3 α¯ 6 ) = (g¯ 6 , b3 β¯6 ) = 0. Now we have that either g6 = α¯ 6 or g6 = β¯6 . By (4.273) we obtain that (y¯11 , b3 g¯ 6 ) = 1, and by (4.275) we obtain that (y11 , b3 g¯ 6 ) = 1, which is a contradiction. If s = s15 then by (4.251) we obtain that b¯3 s15 = b¯15 + c14 + T16 . Now by (4.261) we obtain that there exists an element f ∈ B in Irr(T16 ) such that 10 ≤ |f | which implies that f = T16 . Hence b¯3 s15 = c14 + b¯15 + f16

where f16 ∈ B.

(4.277)

where Σ10 ∈ NB.

(4.278)

By (4.261) and (4.170) we obtain that b3 y11 = x7 + f¯16 + Σ10

Now by (4.174), (4.246) and (4.248) we obtain that H20 + Σ14 = x8 + f¯16 + Σ10 which implies that f¯16 ∈ Irr(H20 ) which implies by (4.246) that b¯ 3 c14 = x7 + b15 + f¯16 + f4 where f4 ∈ B, and we have a contradiction since (f4 , b¯3 c14 ) = 1.


(4.279)

b3 c15 = b10 + y11 + F24 .

(4.280)

and

226

4


By (4.169) we obtain that b¯3 d6 = x7 + Σ11

where Σ11 ∈ NB.

(4.281)


where h5 ∈ B.

(4.282)

Now by (4.174) and (4.279) we obtain that H23 + Σ11 = x8 + c¯15 + d¯6 + h5 . Now we have that H23 = x8 + c¯15 and Σ11 = d¯6 + h5 which imply by (4.279) and (4.281) that b¯3 c15 = x7 + x8 + b15 + c¯15 and b¯3 d6 = x7 + d¯6 + h5 .

(4.283)

Now we have by (4.198) that b¯15 + b3 x8 + x¯7 + x¯8 + b¯15 + c15 = x¯7 + h¯ 5 + b¯3 F24 which implies that (x¯8 , b3 h5 ) = (h¯ 5 , b3 x8 ) = 1, and by (4.283) we obtain that (d6 , b3 h5 ) = 1, which is a contradiction. Now we have that y11 is a non-real element. By (4.198) we obtain that b¯15 + b3 H23 = G12 + b¯3 F24 which implies that b¯15 ∈ Irr(b¯3 F24 ). By (4.280) we obtain that (y11 , b3 c15 ) = 1, and by (4.165) and since y11 is a non-real element we obtain that (y11 , b3 b¯15 ) = 0 which / Irr(b3 c15 ) and implies that c15 = b¯ 15 . Now by (4.5) and (4.162) we obtain that b8 ∈ / Irr(b3 c15 ) and by (4.165) (4.280) we can assume that b¯10 ∈ s ∈ Irr(F24 ).

(4.284)

By (4.165) we obtain that (s, b3 b¯15 ) = 1 which implies that b¯3 s = b¯15 + c15 + T

where T ∈ NB

(4.285)

which implies that 10 ≤ |s|;

(4.286)

and by (4.280) we obtain that b3 c15 = b10 + y11 + s + S

where S ∈ NB.

Now by (4.2), (4.165) and b3 (b¯3 s) = (b3 b¯3 )s we obtain that b8 s = b8 + b10 + b¯10 + y11 + s + t + S + b3 T .

(4.287)

4.5 Case R15 = x7 + x8

227

Since b8 and s are real elements and y11 is a non-real element we obtain that either y¯11 ∈ Irr(S) or y¯11 ∈ Irr(b3 T ). By (4.174) we obtain that b¯3 c15 + b¯3 d6 = b15 + x7 + x8 + b3 y11 . If (x8 , b¯3 d6 ) = 1 then by (4.169), we obtain that b¯3 d6 = x7 + x8 + α3

where α3 ∈ B

which implies that (α3 , b3 y11 ) = 1, and we have a contradiction. Hence (x8 , b¯3 c15 ) = 1 which implies that b3 x8 = c15 + Σ9

where Σ9 ∈ NB.

(4.288)

In addition, we have that 2 ≤ (b3 x8 , b3 x8 ) ≤ 3. If y¯11 ∈ S then we obtain by (4.286) and by (4.287) that b3 c15 = b10 + y11 + y¯11 + s13 . Now by (4.285) we obtain that b¯ 3 s13 = b¯15 + c15 + α9

where α9 ∈ B.

By (4.170) and since y¯11 ∈ S we obtain that ∗ b3 y11 = x7 + c¯15 + Σ11 .

By (4.174), (4.279), (4.281) and (4.288) we obtain that b¯3 c15 = x7 + x8 + b15 + c¯15 . Now by (4.169), (4.171), (4.172), (4.196), (4.288) and b3 (b¯3 c15 ) = b¯3 (b3 c15 ) im∗ , which is a contradiction. plies that Σ9 + x¯8 + b¯15 = Σ12 + α9 + Σ¯ 11 Now we have that y¯11 ∈ b3 T which implies that 5 ≤ |T | and by (4.285) we obtain that 12 ≤ |s|. Assume henceforth that (b3 x8 , b3 x8 ) = 2 and we shall derive a contradiction. Then by (4.162) we obtain that b¯3 x8 = b10 + α14

where α14 ∈ B.

Now by (4.165), (4.194) and (4.287) we obtain that S + b3 e9 = t + b10 + α14 . Since 12 ≤ |s| we obtain by (4.287) that |S| ≤ 12 which implies that α14 ∈ Irr(b3 e9 ) and by (4.171) we obtain that (b10 , b3 e9 ) = 1, which is a contradiction.

228

4


Now we have that (b3 x8 , b3 x8 ) = 3. Then by (4.162) we obtain that b¯3 x8 = b10 + α + β

where α, β ∈ B.

(4.289)

Now by (4.5) we obtain that b8 = x8 , and by (4.288) we obtain that b3 x8 = c15 + v4 + v5

where v4 , v5 ∈ B.

(4.290)

Now by (4.165), (4.194) and (4.287) we obtain that S + b3 e9 = b10 + α + β + t. Now we can assume that b3 e9 = b10 + t + β

(4.291)

b3 c15 = b10 + y11 + s + α

(4.292)

and which implies that 5 ≤ |α|, and by (4.289) we obtain that |α| ≤ 10. Therefore 5 ≤ |α| ≤ 10. Now (4.289) implies that 4 ≤ |β| ≤ 9. By (4.165), (4.291) and (4.292) we obtain that 14 ≤ |s| ≤ 19 which implies that α = s, α = y11 and s = y11 . Since (b3 x8 , b3 x8 ) = 3 we obtain by (4.289) that α = b10 . Now by (4.292) we obtain that (b3 c15 , b3 c15 ) = 4. By (4.279) and (4.288) we obtain that b¯3 c15 = b15 + x7 + x8 + f15

where f15 ∈ B.

(4.293)

By (4.198), (4.290) and (4.292) we obtain that b¯15 + c15 + v4 + v5 + b3 f15 = G12 + b¯3 s + b¯3 α.

(4.294)

/ Irr(G12 ) and since 14 ≤ |s| we have that v4 ∈ / By (4.196) we obtain that v4 ∈ Irr(b¯3 s). Therefore (v4 , b¯3 α) = 1 which implies that |α| ≤ 9 and by (4.292) we obtain that (c15 , b¯3 α) = 1 which implies that 8 ≤ |α|. Thus 8 ≤ |α| ≤ 9

(4.295)

5 ≤ |β| ≤ 6,

(4.296)

which implies by (4.289) that

4.5 Case R15 = x7 + x8

229

and by (4.292) we obtain that 15 ≤ |s| ≤ 16. By (4.162), (4.289), (4.290), (4.293) and b3 (b¯3 x8 ) = b¯3 (b3 x8 ) we obtain that f15 + b¯3 v4 + b¯3 v5 = b3 α + b3 β. By (4.289) we obtain that (x8 , b3 α) = (x8 , b3 β) = 1 which implies by (4.296) that (f15 , b3 β) = 0 and (f15 , b3 α) = 1. So 9 ≤ |α| and by (4.295), we obtain that α = α9 which implies by (4.292) that s = s15 . By (4.285) we obtain that b¯3 s15 = c15 + b¯15 + T15 , and by (4.294) we obtain that v4 + v5 + b3 f15 = G12 + T15 + b¯3 α9 . By (4.292) we obtain that (c15 , b¯3 α9 ) = 1 which implies that either b3 f15 = G12 + T15 + c15 + h3 where h3 ∈ B, which is impossible, or v5 ∈ Irr(G12 ). Now we have that v5 ∈ Irr(G12 ). By (4.196) we obtain that b¯3 y11 = c15 + d6 + v5 + v7

where v7 ∈ B

which implies that b3 v5 = y11 + m4

where m4 ∈ B,

and by (4.290) we obtain that b¯3 v5 = x8 + m7

where m7 ∈ B.

Now by (4.290) and b3 (b¯3 v5 ) = b¯3 (b3 v5 ) we obtain that v4 + b3 m7 = d6 + v7 + b¯3 m4 . Thus b¯3 m4 = v4 + v5 + m3

where m3 ∈ B

which implies that (m3 , b3 m7 ) = 1, and we have a contradiction.


(4.297)

b3 c16 = b10 + y11 + F27 .

(4.298)

and

By (4.169) and (4.174) we obtain that b¯3 d5 = x7 + g8

where g8 ∈ B.

(4.299)

230

4



where h5 ∈ B.

(4.300)

Now by (4.174) and (4.297) we obtain that H26 + g8 = x8 + c¯16 + d¯5 + h5 . Now we have that H26 = c¯16 + d¯5 + h5 which implies by (4.297) that h5 ∈ Irr(b¯3 c16 ), which is a contradiction. Now we have that y11 is a non-real element. By (4.198) we obtain that b¯15 + b3 H26 = G12 + b¯3 F27 which implies that b¯15 ∈ Irr(b¯3 F27 ). By (4.5) and (4.162) we obtain that b8 ∈ / / Irr(b3 c16 ) and by (4.298) we can assume that Irr(b3 c16 ) and b¯10 ∈ s ∈ Irr(F27 ).

(4.301)

By (4.165) we obtain that (s, b3 b¯15 ) = 1 which implies that b¯3 s = b¯15 + c16 + T

where T ∈ NB

(4.302)

which implies that 12 ≤ |s|,

(4.303)

and by (4.298) we obtain that b3 c16 = b10 + y11 + s + S

where S ∈ NB.

(4.304)

Now by (4.2), (4.165) and b3 (b¯3 s) = (b3 b¯3 )s we obtain that b8 s = b8 + b10 + b¯10 + y11 + s + t + S + b3 T .

(4.305)

By (4.171) we obtain that b3 e8 = b10 + Σ14

where Σ14 ∈ NB

(4.306)

and by (4.162) we obtain that ∗ b¯3 x8 = b10 + Σ14 .

Now by (4.165) and (4.304) we obtain that ∗ . S + Σ14 = t + Σ14

Now we have the following three cases:

(4.307)

4.5 Case R15 = x7 + x8

231

∗ . Case 1: S = t and Σ14 = Σ14 ∗ Case 2: S = Σ14 and Σ14 = t14 . ∗ = α + β where α, β ∈ B, S = α and Σ = β + t. Case 3: Σ14 14

Case 1 By (4.304) we obtain that b3 c16 = b10 + y11 + s + t,

(4.308)

and by (4.165) we obtain that b¯3 t = b¯15 + c16 + Γ

where Γ ∈ NB

(4.309)

which implies that 12 ≤ |t|. By (4.305) we obtain that y¯11 ∈ Irr(b3 T ),

(4.310)

and by (4.302) we obtain that s ∈ Irr(b3 T ) which implies that 8 ≤ |T | and 13 ≤ |s|. In the same way we can show that y¯11 ∈ Irr(b3 Γ )

(4.311)

and 13 ≤ |t|. Now by (4.165) we can assume that s = s13 and t = t14 . Now (4.302) and (4.309) imply that b¯ 3 s13 = b¯15 + c16 + v8

where v8 ∈ B

and b¯3 t14 = b¯15 + c16 + v11

where v11 ∈ B.

By (4.170), (4.310) and (4.311) we obtain that b3 y11 = x7 + v¯ 8 + v¯ 11 + f7

where f7 ∈ B.

By (4.174), (4.297) and (4.299) we obtain that H26 + g8 = x8 + v¯ 8 + v¯11 + f7 , and together with (4.297) we obtain that (b¯3 c16 , b¯3 c16 ) = 5, which is a contradiction to (4.308). Case 2 By (4.304) we obtain that ∗ , b3 c16 = b10 + y11 + s13 + Σ14

by (4.306) we obtain that b3 e8 = b10 + t14 ,

(4.312)

by (4.172) we obtain that b¯3 e8 = b15 + h9

where h9 ∈ B

(4.313)

232

4


and by (4.302) we obtain that b¯3 s13 = b¯15 + c16 + v8

where v8 ∈ B.

(4.314)

Now (4.305) implies that y¯11 ∈ Irr(b3 v8 ). Therefore b3 v8 = y¯11 + s13 .

(4.315)

Now (4.174) implies that either (v¯8 , b¯3 c16 = 1) or (v¯8 , b¯3 d5 ) = 1. Assume henceforth that (v¯8 , b¯3 d5 ) = 1 and we shall derive a contradiction. Then by (4.169) we obtain that b¯ 3 d5 = x7 + v¯ 8

(4.316)

and by (4.315) we obtain that b¯3 v8 = d¯5 + f19

where f19 ∈ B.

Now by (4.314), (4.316) and b3 (b¯3 v8 ) = b¯3 (b3 v8 ), we obtain that x¯7 + b3 f19 = b¯15 + c16 + b¯3 y¯11 . Hence (f¯19 , b3 b15 ) = 1, which is a contradiction to (4.172). Now we have that (v¯ 8 , b¯3 c16 ) = 1 which implies by (4.315) that b¯3 v8 = c¯16 + γ8

where γ8 ∈ B

(4.317)

and by (4.297) we obtain that b¯3 c16 = x7 + b15 + v¯ 8 + Σ18

where Σ18 ∈ NB.

Now by (4.3), (4.5), (4.162), (4.165), (4.172), (4.313), (4.314) and b3 (b¯3 b15 ) = b¯3 (b3 b15 ) we obtain that x8 + b3 t14 = e¯8 + Σ18 + b15 + h9 , and by (4.312) we obtain that (e¯8 , b3 t14 ) = 1. Therefore (x8 , Σ18 ) = 1 which implies that b¯3 c16 = x7 + b15 + v¯8 + x8 + h10

where h10 ∈ B,

and by (4.174) and (4.299) we obtain that b3 y11 = x7 + g8 + v¯8 + h10 .

(4.318)

Now by (4.314), (4.315), (4.317) and b3 (b¯3 v8 ) = b¯3 (b3 v8 ) we obtain that c16 + v8 + g¯ 8 = x¯8 +b3 γ8 . By (4.317) we obtain that (v8 , b3 γ8 ) = 1 which implies that g8 = x8 . By (4.318) we obtain that (y11 , b¯3 x8 ) = (x8 , b3 y11 ) = 1, and by (4.5) we obtain that b8 = x8 . Now by (4.162) we obtain that (b10 , b¯3 x8 ) = 1, which is a contradiction. Case 3 By (4.306) we obtain that b3 e8 = b10 + β + t

(4.319)

which implies that (e8 , b¯3 t) = 1, and by (4.165) we obtain that (b¯15 , b¯3 t) = 1. Therefore 9 ≤ |t| and |β| ≤ 5. By (4.5) and (4.319) we obtain that b8 = e8 . Therefore |β| = 3 which implies that 4 ≤ |β| ≤ 5.

4.6 Case (b3 x7 , b3 x7 ) = 3

233

Assume henceforth that β = β4 and we shall derive a contradiction. Then (4.319) implies that b¯ 3 β4 = e8 + h4

where h4 ∈ B.

By (4.307) we obtain that b¯3 x8 = b10 + α10 + β4 which implies that b3 β4 = x8 + g4

where g4 ∈ B.

Now (4.319) and b3 (b¯3 β4 ) = b¯ 3 (b3 β4 ) imply that t10 + b3 h4 = α10 + b¯3 g4 which implies that t10 = α10 . By (4.165) we obtain that (b15 , b3 t10 ) = 1 and by (4.304) we obtain that (c¯16 , b3 t10 ) = 1, which is a contradiction. Now we have that β = β5 , and by (4.165) and (4.319) we obtain that t = t9 , s = s18 and α = α9 . By (4.165), (4.172) and (4.319) we obtain that b¯3 t9 = b¯15 + e8 + α4

where α4 ∈ B

and b¯3 e8 = b15 + f4 + f5

where f4 , f5 ∈ B.

Now by (4.5), (4.3) (4.162), (4.165), (4.297), (4.302), (4.172) and b3 (b¯3 b15 ) = b¯ 3 (b3 b15 ), we obtain that x8 + α¯ 4 + T¯23 = f5 + f4 + H26 which implies that f¯5 ∈ Irr(T23 ), and we have a contradiction to (4.302).

4.6 Case (b3 x7 , b3 x7 ) = 3 Since (b3 x7 , b3 x7 ) = 3 we obtain that b3 x7 = c + d + e

where c, d, e ∈ B,

(4.320)

where w, z ∈ B.

(4.321)

and by (4.162) we obtain that b¯3 x7 = b10 + w + z

Now by (4.162) and b3 (b¯3 x7 ) = b¯3 (b3 x7 ) we obtain that b¯3 c + b¯3 d + b¯3 e = b15 + x7 + x8 + b3 z + b3 w.

(4.322)

Now we can assume that (c, b3 b15 ) = 1. By (4.3) we obtain that (b6 , b3 b15 ) = 1, by (4.10) we obtain that (b¯15 , b3 b15 ) = 1 and (b3 b15 , b3 b15 ) = 4. Now since c = b6 , b¯15 we obtain that b3 b15 = b6 + b¯ 15 + c + f

where f ∈ B,

(4.323)

234

4


and by (4.10) we obtain that b¯3 b10 = b6 + c + f.

(4.324)

By (4.320) we obtain that |c| ≤ 13 and (b¯3 c, x7 ) = 1 and by (4.323) we obtain that (b¯3 c, b15 ) = 1 which implies that 9 ≤ |c| ≤ 13.

(4.325)

11 ≤ |f | ≤ 15.

(4.326)

Now (4.324) implies that By (4.321) we obtain that either b¯3 x7 = z4 + w7 or b¯3 x7 = z5 + w6 . Assume henceforth that b¯3 x7 = b10 + z4 + w7 and we shall derive a contradiction. Then b3 z4 = x7 + g5

where g5 ∈ B

and b¯3 z4 = α + β

where α, β ∈ B.

(4.327)

Now b3 (b¯3 z4 ) = b¯3 (b3 z4 ) implies that b3 α + b3 β = b10 + z4 + w7 + b¯3 g5 .

(4.328)

Now we can assume that (b10 , b3 α) = 1.

(4.329)

By (4.327) we obtain that |α| ≤ 9, by (4.4) we obtain that α = b6 and by (4.326) we obtain that f = α. Now we have by (4.324) that c = α, by (4.325) we obtain that c = c9 and by (4.327) we obtain that β = β3 . By (4.328) we obtain that b3 c9 + b3 β3 = b10 + z4 + w7 + b¯3 g5 which implies that (w7 , b3 c9 ) = 1. By (4.327) we obtain that (z4 , b3 c9 ) = 1 and together with (4.329) we obtain that 4 ≤ (b3 c9 , b3 c9 ). By (4.320) and (4.323) we obtain that (b3 c9 , b3 c9 ) ≤ 3, and we have a contradiction. Now we have that b¯3 x7 = b10 + z5 + w6

(4.330)

and b3 z5 = x7 + Σ8

where Σ8 ∈ NB.

(4.331)

Now b3 (b¯3 z5 ) = b¯3 (b3 z5 ) implies that b3 (b¯3 z5 ) = b10 + z5 + w6 + b¯3 Σ8 , by (4.4) we obtain that (b6 , b¯3 z5 ) = 0 and by (4.324) we obtain that either

(c, b¯3 z5 ) = 1

or

By (4.331) we obtain that 2 ≤ (b3 z5 , b3 z5 ) ≤ 3.

(f, b¯3 z5 ) = 1.

(4.332)

4.6 Case (b3 x7 , b3 x7 ) = 3

235

Assume henceforth that (b3 z5 , b3 z5 ) = 3 and we shall derive a contradiction. Then by (4.325), (4.326) and (4.332) we obtain that b¯3 z5 = c9 + m3 + n3

where m3 , n3 ∈ B.

Now (4.330), (4.331) and b3 (b¯3 z5 ) = b¯3 (b3 z5 ) imply that b3 c9 + b3 m3 + b3 n3 = b10 + z5 + w6 + b¯3 Σ8 . Hence b3 c9 = b10 + z5 + w6 + v6 where v6 ∈ B. By (4.320) and (4.323) we obtain that (b3 c9 , b3 c9 ) ≤, 3 and we have a contradiction. Now (4.331) implies that b3 z5 = x7 + g8

where g8 ∈ B.

(4.333)

Assume henceforth that c = c13 and we shall derive a contradiction. Then by (4.324) we obtain that f = f11 and (4.332) implies that b¯3 z5 = f11 + γ4

where γ4 ∈ B.

By (4.320) we obtain that b3 x7 = c13 + d4 + e4 , b¯3 d4 = x7 + q5

where q5 ∈ B

b¯3 e4 = x7 + u5

where u5 ∈ B.

and Now by (4.322) and (4.333) we obtain that b¯3 c13 + q5 + u5 = b15 + x8 + g8 + b3 w6 which implies that (q5 , b3 w6 ) = (u5 , b3 w6 ) = 1, and by (4.330) we obtain that (x7 , b3 w6 ) = 1, which is a contradiction. Now we have by (4.325) that 9 ≤ |c| ≤ 12 and by (4.324) we obtain that 12 ≤ |f | ≤ 15. Assume henceforth that (f, b¯ 3 z5 ) = 1 and we shall derive a contradiction. Then f = f12 and b¯3 z5 = f12 + γ3

where γ3 ∈ B.

(4.334)

By (4.324) we obtain that c = c12 . By (4.1), (4.320), (4.330) and b3 (b3 x7 ) = b32 x7 we obtain that b3 c12 + b3 d + b3 e = b10 + z5 + w6 + b6 x7 . By (4.320) and since c = c12 we obtain that |d|, |e| = 3, 12 which implies by (4.334) that (z5 , b3 d) = (z5 , b3 e) = 0. Hence (z5 , b3 c12 ) = 1 which implies that c12 = f12 and since (b3 b10 , b3 b10 ) = 3 we have a contradiction to (4.324). Now by (4.332) and (4.333) we obtain that b¯3 z5 = c + δ

where δ ∈ B.

(4.335)

By (4.324) we obtain that b3 c = b10 + z5 + D

where D ∈ NB,

(4.336)

where C ∈ NB.

(4.337)

and by (4.320) and (4.323) we obtain that b¯3 c = x7 + b15 + C

236

4


Now by (4.320), (4.323), (4.324) and b3 (b¯3 c) = b¯3 (b3 c), we obtain that d + e + b¯ 15 + b3 C = δ + b¯3 D. By (4.335) we obtain that |δ| = 15 which implies that δ = b¯ 15 . Therefore b¯15 ∈ Irr(b¯ 3 D). Since 9 ≤ |c| ≤ 12 we obtain by (4.5), (4.162) and / Irr(D) and b¯10 ∈ / Irr(D). Therefore by (4.165) we can assume that (4.336) that b8 ∈ s ∈ Irr(D). Now (4.336) implies that b3 c = b10 + z5 + s + E

where E ∈ NB.

(4.338)

Now (4.165) implies that b¯3 s = b¯15 + c + F

where F ∈ NB.

By (4.2), (4.165) and (b3 b¯3 )s = b3 (b¯3 s) we obtain that b8 s = b10 + z5 + E + b8 + b¯10 + s + t + b3 F. Since b8 and s are real elements and z5 is a non-real element, we obtain that either z¯ 5 ∈ Irr(E) or z¯ 5 ∈ Irr(b3 F ). If z¯ 5 ∈ Irr(E) then by (4.338) we obtain that (c, ¯ b3 z5 ) = 1, and since 9 ≤ |c| ≤ 12 we have a contradiction to (4.333). Therefore z¯ 5 ∈ Irr(b3 F ). Now by (4.333) we obtain that either x¯7 ∈ Irr(F ) or g¯ 8 ∈ Irr(F ). If x¯7 ∈ Irr(F ) then (s, b¯3 x7 ) = (x¯7 , b¯3 s) = 1. Since s is a real element and b10 , z5 are non-real elements we obtain that s = b10 , z5 , so (4.330) implies that s = w6 , which is a contradiction to (4.165) and (4.330). Now we have that b¯3 s = b¯15 + c + g¯ 8 + G

where G ∈ NB.

(4.339)

By (4.333) we obtain that b¯3 g8 = z5 + s + H

where H ∈ NB

(4.340)

Assume henceforth that c = c9 and we shall derive a contradiction. Then by (4.337) we obtain that b¯3 c9 = x7 + b15 + u5

where u5 ∈ B,

and by (4.338) we obtain that b3 c9 = b10 + z5 + s12 . By (4.339) we obtain that b¯3 s12 = b¯ 15 + c9 + g¯ 8 + u4

where u4 ∈ B

which implies that b3 u4 = s12 . Now by (4.320), (4.323), (4.324), (4.335) and b3 (b¯3 c9 ) = b¯3 (b3 c9 ) we obtain that d + e + b3 u5 = δ6 + c9 + g¯8 + u4 . Hence (b3 u4 , b3 u4 ) = 1, and we have a contradiction. Assume henceforth that c = c10 and we shall derive a contradiction. Then by (4.337) we obtain that b¯3 c10 = x7 + b15 + u8

where u8 ∈ B

4.6 Case (b3 x7 , b3 x7 ) = 3

237

which implies by (4.338) that b3 c10 = b10 + z5 + s15 . By (4.339) we obtain that b¯ 3 s15 = b¯15 + c10 + g¯ 8 + G12 . Now by (4.320), (4.323), (4.324), (4.335) and b3 (b¯3 c10 ) = b¯ 3 (b3 c10 ), we obtain that d + e + b3 u8 = δ5 + c10 + g¯ 8 + G12 . By (4.320) we obtain that |d| + |e| = 11, and now we can assume that d = δ5 , e = e6 and e6 ∈ Irr(G12 ) which implies that b¯3 s15 = b¯15 + c10 + g¯ 8 + e6 + α6

where α6 ∈ B

and b3 e6 = s15 + α3

where α3 ∈ B.

By (4.320) we obtain that b¯3 e6 = x7 + α11 . Now by (4.320) and b3 (b¯3 e6 ) = b¯3 (b3 e6 ) we obtain that d5 + b3 α11 = b¯15 + g¯ 8 + α6 + b¯3 α3 which implies that (α¯ 11 , b3 b15 ) = 1, which is a contradiction to (4.323). Assume henceforth that c = c11 and we shall derive a contradiction. Then by (4.335) we obtain that δ = δ4 and (z5 , b3 δ4 ) = 1 which implies that (w6 , b3 δ4 ) = 0. By (4.330), (4.333), (4.335), (4.340), (4.338) and b3 (b¯3 z5 ) = b¯3 (b3 z5 ), we obtain that w6 + z5 + H = E + b3 δ4 . Now we have that w6 ∈ Irr(E) which implies that (w6 , b3 c11 ) = 1, and by (4.338) we obtain that b3 c11 = b10 + z5 + s + w6 + Q

where Q ∈ NB.

Hence 4 ≤ (b3 c11 , b3 c11 ). Now (4.337) implies that b¯3 c11 = x7 + b15 + u5 + u6

where u5 , u6 ∈ B.

By (4.320), (4.323), (4.324), (4.335) and b3 (b¯3 c11 ) = b¯3 (b3 c11 ), we obtain that d + e + b¯15 + b3 u5 + b3 u6 = δ4 + b¯3 s + b¯3 w6 . By (4.339) we obtain that g¯ 8 ∈ Irr(b¯3 s), and since (c11 , b3 u5 ) = (c11 , b3 u6 ) = 1 we obtain that (g¯ 8 , b3 u5 ) = (g¯ 8 , b3 u6 ) = 0. Therefore we have that either d = g¯ 8 or e = g¯8 , but by (4.169) we obtain that |d| + |e| = 10 and we have a contradiction. Now we have that c = c12 which implies by (4.335) that δ = δ3 and (z5 , b3 δ3 ) = 1 which implies that (w6 , b3 δ3 ) = 0. By (4.330), (4.333), (4.335), (4.340), (4.338) and b3 (b¯3 z5 ) = b¯3 (b3 z5 ), we obtain that w6 + z5 + H = E + b3 δ3 . Now we have that w6 ∈ Irr(E) which implies that (w6 , b3 c12 ) = 1, and by (4.338) we obtain that b3 c12 = b10 + z5 + s + w6 + Q where Q ∈ NB which implies that |s| ≥ 15. If |s| = 15 then By (4.320), (4.323), (4.324), (4.335) and b3 (b¯3 c12 ) = b¯ 3 (b3 c12 ) we obtain that d + e + b¯15 + b3 C14 = δ3 + b¯3 s15 + b¯3 w6 .

238

4


By (4.320) we obtain that d, e = δ3 and since c = c12 we obtain that (d, b¯3 s15 ) = (e, b¯3 s15 ) = 0. Now we have that (d, b¯3 w6 ) = (e, b¯3 w6 ) = 1, which is a contradiction to (c12 , b¯3 w6 ) = 1. If |s| = 15 then |s| ≤ 11, which is a contradiction to (4.339).

4.7 Case b3 b10 = b15 + x5 + y5 + z5 Since b3 b10 = b15 + x5 + y5 + z5 we obtain that b¯3 x5 = b10 + u5

where u5 ∈ B

(4.341)

b¯3 y5 = b10 + v5

where v5 ∈ B.

(4.342)

and Now we have that 1 ≤ (b3 x5 , b3 y5 ) which implies that b3 x5 = c + d

where c, d ∈ B

(4.343)

where e ∈ B.

(4.344)

and b3 y5 = c + e

Assume henceforth that d = e and we shall derive a contradiction. Then (b3 x5 , b3 y5 ) = 2 and by (4.341) and (4.342), we obtain that b¯3 x5 = b¯3 y5 = b10 + u5

(4.345)

which implies that (x5 , b3 u5 ) = (y5 , b3 u5 ) = 1. Now we have that b3 u5 = x5 + y5 + g5

where g5 ∈ B,

(4.346)

and by b3 (b¯3 x5 ) = b¯3 (b3 x5 ) we obtain that b¯3 c + b¯3 d = b15 + 2x5 + 2y5 + z5 + g5 which implies that |c|, |d| ≥ 5

(4.347)

and we can assume that (b15 , b¯3 c) = 1. By (4.343) and (4.344) we obtain that (x5 , b¯3 c) = (y5 , b¯3 c) = 1 which implies that |c| ≥ 10 and by (4.343) and (4.347) we obtain that c = c10 and d = d5 . Since (c10 , b3 b15 ) = 1 we obtain by (4.10) that (c10 , b¯3 b10 ) = 1 which implies by (4.4) that b¯3 b10 = b6 + c10 + α + β

where α, β ∈ B.

(4.348)

Now by (4.1), (4.341), (4.343) and b¯3 (b¯3 x5 ) = b¯32 x5 we obtain that b¯3 b10 + b¯3 u5 = d5 + c10 + b¯6 x5 . If (d5 , b¯3 b10 ) = 1 then by (4.10) we obtain that (d5 , b3 b15 ) = 1, and we have a contradiction.

4.7 Case b3 b10 = b15 + x5 + y5 + z5

239

Now we have that (d5 , b¯3 b10 ) = 0

(4.349)

which implies that (d5 , b¯3 u5 ) = 1 and by (4.346) we obtain that (b3 u5 , b3 u5 ) = 3 which implies that b¯3 u5 = d5 + γ + δ

where γ , δ ∈ B.

(4.350)

By (4.345), (4.346) and b3 (b¯3 u5 ) = b¯ 3 (b3 u5 ) we obtain that b3 d5 + b3 γ + b3 δ = 2b10 + 2u5 + b¯3 g5 . By (4.349) we obtain that (b10 , b3 γ ) = (b10 , b3 δ) = 1. By (4.350) and (4.4) we obtain that γ , δ = b6 and |γ | + |δ| = 10 which implies that γ , δ = c10 . Now (4.348) implies that γ + δ = α + β and |α| + |β| = 14, which is a contradiction. Now we have that (b¯3 x5 , b¯3 y5 ) = 1. In the same way, we can show that ¯ (b3 x5 , b¯3 z5 ) = (b¯3 y5 , b¯3 z5 ) = 1 and we obtain that b¯3 z5 = b10 + w5 ,

(4.351)

and by (4.343) and (4.344) we obtain that either b3 z5 = d + e or b3 z5 = c + f where f ∈ B. If b3 z5 = d + e then 2|c| = 15, so b3 z5 = c + f

where f ∈ B.

(4.352)

By b3 (b¯3 x5 ) = b¯3 (b3 x5 ), b3 (b¯3 y5 ) = b¯ 3 (b3 y5 ) and b3 (b¯3 z5 ) = b¯3 (b3 z5 ), we obtain that b¯ 3 c + b¯3 d = b15 + x5 + y5 + z5 + b3 u5 , (4.353) b¯3 c + b¯3 e = b15 + x5 + y5 + z5 + b3 v5 and b¯3 c + b¯3 f = b15 + x5 + y5 + z5 + b3 w5 . By b¯3 (b¯3 x5 ) = b¯32 x5 we obtain that b¯3 b10 + b¯3 u5 = c + d + b¯6 x5 .

(4.354)

Assume henceforth that (c, b3 b15 ) = 0 and we shall derive a contradiction. Then (d, b3 b15 ) = (e, b3 b15 ) = (f, b3 b15 ) = 1. In addition, we have that |d| = |e| = |f | which implies by (4.4) and (4.10) that b¯3 b10 = b6 + d8 + e8 + f8 . Now by (4.352) we obtain that c = c7 and by (4.354) we obtain that (c7 , b¯3 u5 ) = 1. By (4.4), (4.341), (4.342), (4.351) and b3 (b¯3 b10 = b¯3 (b3 b10 ), we obtain that b8 + b3 d8 + b3 e8 + b3 f8 = b¯3 b15 + 2b10 + u5 + v5 + w5 which implies that (b¯3 u5 , b¯3 u5 ) = 2. By (4.353) and since (c, b3 b15 ) = 0 we obtain that (b15 , b¯3 d8 ) = 1 and together with (4.343) we obtain that b¯3 d8 = b15 + x5 + g4

where g4 ∈ B,

240

4


and by (4.353) we obtain that (g4 , b3 u5 ) = 1. Now by (4.341) and (b¯3 u5 , b¯3 u5 ) = 2 we obtain that b3 u5 = g4 + x5 , which is a contradiction. Now we have that (b15 , b¯3 c) = 1 and by (4.10) we obtain that (c, b¯3 b10 ) = 1 which implies that b¯3 b10 = b6 + c + α + β

where α, β ∈ B

(4.355)

and b3 b15 = b¯15 + b6 + c + α + β.

(4.356)

|α|, |β| ≥ 6.

(4.357)

Now we have that By (4.343), (4.344) and (4.352) we obtain that (x5 , b¯3 c) = (y5 , b¯3 c) = (z5 , b¯3 c) = 1, and together with (b15 , b¯3 c) = 1 and b3 x5 = c + d we obtain that either c = c10 or c = c12 . By (4.341) we obtain that (x5 , b3 b10 ) = (x5 , b3 u5 ) = 1 which implies that (b¯3 u5 , b¯3 b10 ) ≥ 1.

(4.358)

Assume henceforth that c = c12 and we shall derive a contradiction. Then by (4.356) we obtain that α = α6 , β = β6 and b¯3 α6 = b15 + h3

where h3 ∈ B.

(4.359)

By (4.354) and (4.356) we obtain that b6 + c12 + α6 + β6 + b¯3 u5 = c12 + d3 + b¯6 x5 which implies that (d3 , b¯3 u5 ) = 1.

(4.360)

By (4.4) we obtain that (b6 , b¯3 u5 ) = 0. By (4.355) we obtain that (b10 , b3 α6 ) = 1 and by (4.359) we obtain that (b3 α6 , b3 α6 ) = 2 which implies that 0 = (u5 , b3 α6 ) = (α6 , b¯3 u5 ). In the same way, we can show that (β6 , b¯3 u5 ) = 0 which implies by (4.358) and (4.360) that b¯3 u5 = d3 + c12 . Now we have that (b3 u5 , b3 u5 ) = 2 and by (4.341) we obtain that b3 u5 = x5 + g10

where g10 ∈ B.

By (4.353) we obtain that b¯ 3 c12 + b¯3 d3 = b15 + 2x5 + y5 + z5 + g10 , and we have a contradiction. Now we have that c = c10 which implies that b¯3 c10 = b15 + x5 + y5 + z5

(4.361)

and by (4.354) and (4.355) we obtain that b6 + α + β + b¯3 u5 = d5 + b¯6 x5 . By (4.357) we obtain that α, β = d5 which implies that (d5 , b¯3 u5 ) = 1.

(4.362)

4.7 Case b3 b10 = b15 + x5 + y5 + z5

241

By (4.4) we obtain that (b6 , b¯3 u5 ) = 0. If (α, b¯3 u5 ) = 1 then (b¯6 x5 , b¯6 x5 ) ≥ 7. By (4.2) and (4.343) we obtain that (b8 , x5 x¯5 ) = 1 in addition (b8 , b6 b¯6 ) = 1. Now we have that x5 x¯5 = 1 + b8 + Σ16

where Σ16 ∈ NB

b6 b¯6 = 1 + b8 + Σ27

where Σ27 ∈ NB.

and By (4.12) we obtain that b3 b¯15 = b8 + b¯10 + Σ27 and since (b3 b15 , b3 b15 ) = 5 we have a contradiction. Now (4.358) implies that (u5 , b3 c10 ) = 1. In the same way, we can show that (v5 , b3 c10 ) = (w5 , b3 c10 ) = 1 and by (4.355) we obtain that (b10 , b3 c10 ) = 1 which implies that (b3 c10 , b3 c10 ) = 4, and we have a contradiction to (4.361).

Chapter 5

Finishing the Proofs of the Main Results

5.1 Introduction In Chap. 5 we will freely use the definitions and notation of Chap. 2. The Main Theorem 3 of Chap. 4 left two unsolved problems for the complete classification of the NITA (A, B) generated by a faithful non-real element of degree 3 with L1 (B) = 1 and L2 (B) = ∅. In Chap. 4, the Main Theorem 3 Case (2) states that there exist b6 , b10 , b15 ∈ B, where b6 is non-real, such that b32 = b¯3 + b6 , b¯ 3 b6 = b3 + b15 , b3 b6 = b8 + b10 and (b3 b8 , b3 b8 ) = 3. Moreover, if b10 is real, then (A, B) ∼ =x (CH(3 · A6 ), Irr(3 · A6 )) of dimension 17. If b10 is non-real, then (b3 b10 , b3 b10 ) = 2 and b15 is a non-real element. In this chapter, we will almost completely solve and classify the open Case (2) of Chap. 4 when b10 is a non-real element, (b3 b10 , b3 b10 ) = 2 and b15 is a non-real element. In Chap. 4 we proved under the assumption of the open Case (2) of the Main Theorem 3 that b3 b8 = b3 + b¯6 + b15 , b3 b10 = b15 + d15

where d15 ∈ B where b15 = d15 ,

and b¯3 b10 = b6 + b24

where b24 ∈ B.

A Table Algebra (A, B) is an algebra of a finite dimension. In the following definition, we define a Countable Table Algebra and the dimension of that algebra is not necessarily finite. Definition 5.1 Let B := {b1 = 1, b2 , . . . , bl , . . . } be a distinguished basis of a countable dimensional, associative and commutative algebra A over the complex field C with identity element 1A . If a ∈ A and bi ∈ B, let k(bi , a) be the coefficient of bi in a. Then (A, B) is a Countable Table Algebra (B is a table basis and |B| is the dimension of the Table Algebra (A, B)) if the following hold: Z. Arad et al., On Normalized Integral Table Algebras (Fusion Rings), Algebra and Applications 16, DOI 10.1007/978-0-85729-850-8_5, © Springer-Verlag London Limited 2011

243

244

5


TA1. For all i, j, m ∈ N bi bj = km=1 bij m bm where k ∈ N with bij m a nonnegative real number. TA2. There is an algebra automorphism (denoted by −) of A whose order divides 2, such that bi ∈ B implies that b¯i ∈ B (then i¯ is defined by bi¯ = b¯i and bi ∈ B is called real if bi¯ = bi ). TA3. There is a function g : B × B −→ R+ such that bij m = g(bi , bm )bj¯mi where bij m , bj¯mi are defined in TA1 and TA2 for all i, j , m. TA4. There exists a unique algebra homomorphism | | : A −→ C such that |b| = ¯ > 0, b ∈ B. We call it the degree homomorphism. The positive real num|b| bers {|b|}b∈ B are called the degrees of (A, B). If for all bi ∈ B bi i1 ¯ = 1, then (A, B) is a Countable Normalized Table Algebra (denoted by C-NTA) and if in addition for all b ∈ B |b| ∈ N and for all i, j , m bij m ∈ N ∪ {0} then (A, B) is a Countable Normalized Integral Table Algebra (denoted by C-NITA). Note 5.1 If the dimension of a Countable Table Algebra is finite then by [AB, Lemma 2.9], axiom TA4 is a consequence of axioms TA1–TA3. In Chaps. 2–4, we assumed that Table Algebras are NITA of finite dimension. If we assume instead that the Table Algebras are C-NITA, the same proofs of Chaps. 2–4 give us C-NITA of finite dimension, i.e., Table Algebras. Definition 5.2 Let D = {(k, n) | n ∈ N, k is an odd number with − 1 ≤ k ≤ 2n + 1}. Define f : D → N ∪ {0} by f (k, n) =

(n + 1)(k + 1)(2n − k + 1) . 8

By definition f (−1, n) = 0, f (2n + 1, n) = 0, f (1, 1) = 1 and f (1, 2) = 3. Definition 5.3 Let (A, B) be a C-NITA with a faithful non-real element of degree b3 such that b32 = b¯3 + b6 where b6 ∈ B. For t ∈ N with t ≥ 2, or t = ∞, define an array Ct as Ct = {b(2j +1,i) }t+1;i i=1;j =−1 , where b(−1,i) = b(2i+1,i) = 0 for 1 ≤ i ≤ t + 1, b(k,i) ∈ B for all 1 ≤ i ≤ t + 1 and odd k with 1 ≤ k ≤ 2i − 1, and the following properties hold: 1. b(1,1) = 1, b(1,2) = b3 and b(k,i) = b¯(2i−k,i) for all 1 ≤ i ≤ t + 1 and odd k with 1 ≤ k ≤ 2i − 1. 2. b3 b(k,i) = b(k−2,i−1) + b(k+2,i) + b(k,i+1) for all 2 ≤ i ≤ t and odd k with 1 ≤ k ≤ 2i − 1. Remark If an array Ct exists for t > 2, then properties 1 and 2 imply that Ct−1 exists, with Ct−1 ⊆ Ct in the obvious sense. It is a straightforward consequence of

5.1 Introduction

245

properties 1 and 2 and induction on i that |b(k,i) | = f (k, i),

for all 1 ≤ i ≤ t + 1 and odd k with 1 ≤ k ≤ 2i + 1.

Definition 5.4 Let (A, B) be a C-NITA as in Definition 5.3. If for some n ∈ N there exists an array Cn for (A, B), but there does not exist an array Cn+1 , then n is called the stopping number of (A, B). All such Table Algebras with stopping number n ∈ N are denoted (A, B, Cn ). If there are arrays Cn for all n ∈ N, n ≥ 2, then the array C∞ exists. In this case, we say that the stopping number for (A, B) is ∞, and denote such a Table Algebra as (A, B, C∞ ). In either case, we say that (A, B) is a Cn -Table Algebra. We shall show in Theorem 5.1 that there exists no C-NITA (A, B, Cn ) such that ∞ > n ≥ 43. The classification of C-NITA (A, B, Cn ) such that 4 ≤ n ≤ 42 is still open. If there exists a C-NITA (A, B, Cn ), it could be an infinite dimension CNITA. We shall show in Theorem 5.2 that (A, B, C∞ ) is a uniquely determined CNITA. Note that (A, B, C∞ ) is an infinite dimensional C-NITA as defined in Definition 5.1. In Theorem 2.1 of Chap. 2 we showed that the Table Algebra (CH PSL(2, 7), Irr PSL(2, 7)) is strictly isomorphic to (A, B, C2 ) and (CH(3 · A6 ), Irr(3 · A6 )) is strictly isomorphic to (A, B, C3 ). Moreover, the Main Theorem 3 shows that the Table Algebras above are unique. In the following two examples, we will give an example of a C-NITA (A, B, C2 ) and an example of a C-NITA (A, B, C3 ). Example 5.1 We list part of the structure of the C-NITA (A, B, C2 ). The full list appears in [AC]. It has a basis {1, b3 , b¯3 , b6 , b8 , b7 } of dimension 6 and by Definition 5.3 we obtain that C2 = {0, 1, b3 , b¯3 , b6 , b8 } and the following hold: b32 = b¯ 3 + b6 , b3 b¯3 = 1 + b8 , b3 b6 = b¯3 + b7 + b8 . / Irr(b3 b(1,3) ), we obNote that in the above example, b(1,3) = b(5,3) . Since b(1,4) ∈ tain that all the terms of Definitions 5.3 and 5.4 hold. Example 5.2 We list part of the structure of the C-NITA (A, B, C3 ). The full list appears in Chap. 2. It has a basis {1, b3 , b¯3 , b6 , b¯6 , b8 , b10 , b15 , b¯15 , c3 , c¯3 , b5 , c5 , c8 , b9 , c9 , c¯9 } of dimension 17 and by Definition 5.3 we obtain that C3 = {0, 1, b3 , b¯3 , b6 , b¯6 , c8 , b10 , b15 , b¯15 } and the following hold:

246

5


b32 = b¯ 3 + b6 , b3 b¯3 = 1 + c8 , b3 b6 = c8 + b10 , b3 b¯6 = b¯3 + b¯15 , b3 c8 = b3 + b¯6 + b15 , b3 b15 = b6 + 2b¯15 + c9 . / Irr(b3 b(3,4) ), we obtain Note that in the above example b(1,4) = b(7,4) . Since b(3,5) ∈ that all the terms of Definitions 5.3 and 5.4 hold. Example 5.3 The classification of a Countable Table Algebra (A, B, C4 ) is still open. If there exists a C-NITA (A, B, C4 ) then C4 = {0, 1, b3 , b¯3 , b6 , b¯6 , b8 , b10 , b¯10 , b15 , b¯15 , c15 , c¯15 , b24 , b¯24 , b27 } and the following hold: b32 = b¯ 3 + b6 , b3 b¯3 = 1 + b8 , b3 b6 = b8 + b10 , b3 b¯6 = b¯3 + b¯15 , b3 b8 = b3 + b¯6 + b15 , b3 b10 = b15 + c15 , b3 b¯10 = b¯6 + b¯24 , b3 b15 = b6 + b¯15 + b24 , b3 b¯15 = b8 + b¯10 + b27 . We can show that if b3 c15 = b24 + b21 then the stopping number of (A, B) is not 4 which implies that if there exists such a C-NITA then it is either a Cn -Table Algebra / B and b3 c15 = b24 + where 5 ≤ n or a C∞ -Table Algebra. Now, we have that b21 ∈ Σ21 where Σ21 ∈ NB \ B. By the Main Theorem 3 of Chap. 4 we show that c15 is a non-real element but we still do not know if b24 is a real element. Note 5.2 From the above example we can see that if there exists a C-NITA (A, B) that satisfies the assumptions of the open case (2) of the Main Theorem 3, then either (A, B) ∼ =x (A, B, Cn ) where 4 ≤ n or (A, B) ∼ =x (A, B, C∞ ). Theorem 5.1 There exists no C-NITA (A, B, Cn ) such that 43 ≤ n. Theorem 5.2 There exists a unique C∞ -Table Algebra which is strictly isomorphic to a C-NITA generated by the non-real polynomial representation of dimension 3 of SL(3, C).

5.2 Proof of Theorem 5.1

247

Now, by the Main Theorem 3 of Chap. 4, Theorems 5.2 and 5.1 of this chapter, we state the Main Theorem 4 of this chapter as follows. Main Theorem 4 Let (A, B) be a C-NITA generated by a non-real element b3 ∈ B of degree 3 and without non-identity basis element of degree 1 and 2. Then b3 b¯3 = 1 + b8 , b8 ∈ B, and (A, B) is of one of the following types: (1) (A, B) ∼ =x (A, B, C2 ) which is the unique C2 -Table Algebra and is strictly isomorphic to (CH PSL(2, 7), Irr PSL(2, 7)) of dimension 6. (2) (A, B) ∼ =x (A, B, C3 ) which is the unique C3 -Table Algebra and is strictly isomorphic to (CH(3 · A6 ), Irr(3 · A6 )) of dimension 17. (3) There exists a unique C∞ -Table Algebra which is strictly isomorphic to a CNITA generated by the non-real polynomial representation of dimension 3 of SL(3, C). (4) There exists no C-NITA (A, B, Cn ) such that 43 ≤ n and the classification of the Table Algebra (A, B, Cn ) such that 4 ≤ n ≤ 42 is still open. (5) There exist c3 , b6 ∈ B, c3 = b3 or b¯3 , such that b32 = c3 + b6 and either (b3 b8 , b3 b8 ) = 3 or 4. If (b3 b8 , b3 b8 ) = 3 and c3 is non-real, then (A, B) ∼ =x (A(3 · A6 · 2), B32 ) of dimension 32. (See Theorem 2.9 of Chap. 2 for the definition of this specific NITA.) If (b3 b8 , b3 b8 ) = 3 and c3 is real, then (A, B) ∼ =x (A(7 · 5 · 10), B22 ) of dimension 22. (See Theorem 2.10 of Chap. 2 for the definition of this specific NITA.) The Case (b3 b8 , b3 b8 ) = 4 is still open. By the Main Theorem 4 we obtain that if (A, B) is a C-NITA generated by a nonreal element b3 ∈ B of degree 3 and without non-identity basis element of degree 1 and 2, then b32 = c¯3 + b6 where c3 , b6 ∈ B. If (A, B) is of one of (1)–(4) types of Main Theorem 4, then c3 = b¯3 and if (A, B) is of type (5) of the Main Theorem 4, then c3 = b3 or b¯3 . The rest of this chapter is devoted to prove Theorems 5.1 and 5.2.

5.2 Proof of Theorem 5.1 Throughout this section, (A, B) denotes a C-NITA. We will use some lemmas in order to prove Theorem 5.1. Lemma 5.1 If a, b ∈ B where |a|, |b| = 1 and c, d ∈ Irr(ab), then 1, |b|2 − 1}.

|d| |c|

≤ min{|a|2 −

Proof Assume, without loss of generality, that |a| ≤ |b|. If c = d then |d| |c| = 1, and since |a|, |b| = 1 the Lemma holds. If c = d then |c| + |d| ≤ |a| · |b|. Since c ∈ Irr(ab) we obtain that b ∈ Irr(ac) ¯ which implies that |b| ≤ |a| · |c|. Now we have 2 that |c| + |d| ≤ |a|2 · |c| which implies that |d| |c| ≤ |a| − 1, as required. Lemma 5.2 If (A, B) is a Cn -Table Algebra then b¯3 b(k,i) = b(k,i−1) + b(k−2,i) + b(k+2,i+1) for all 2 ≤ i ≤ n and odd k with 1 ≤ k ≤ 2 − 1.

248

5


Proof Let i ∈ N where 2 ≤ i ≤ n and let k be an odd number where 1 ≤ k ≤ 2i − 1. By Definition 5.3 we obtain that b3 b¯(k,i) = b3 b(2i−k,i) = b(2i−k−2,i−1) + b(2i−k+2,i) + b(2i−k,i+1) = b¯ (k,i−1) + b¯(k−2,i) + b¯(k+2,i+1) ,

as required.

Lemma 5.3 If (A, B) is a Cn -Table Algebra and i, j, k, l ∈ N then b(k,i) = b(l,j ) where 1 ≤ i ≤ n, 1 ≤ j ≤ n + 1, 1 ≤ k ≤ 2i − 1 is an odd number, 1 ≤ l ≤ 2j − 1 is an odd number and either k = l or i = j . Proof We will prove the lemma by induction on n. For n = 1 we obtain that C1 = {b((1,1) , b(1,2) , b(3,2) }. Now by Definition 5.2 f (1, 1) = f (1, 2) and f (1, 1) = f (3, 2) which imply that b(1,1) = b(1,2) and b(1,1) = b(3,2) as required. Example 5.1 implies that the lemma holds for n = 2. Now we assume that the lemma holds for n = t where 2 ≤ t. We will prove the lemma for n = t + 1. By Definition 5.3 we obtain that Ct ⊂ Ct+1 , so by the induction hypothesis we obtain that b(k,i) = b(l,j ) where 1 ≤ i ≤ t, 1 ≤ j ≤ t + 1, 1 ≤ k ≤ 2i − 1 is an odd number, 1 ≤ l ≤ 2j − 1 is an odd number and either k = l or i = j . We will prove that b(k,t+1) = b(l,t+1)

where 1 ≤ k = l ≤ 2t + 1

(5.1)

and b(k,i) = b(l,t+2) where 1 ≤ i ≤ t + 1, 1 ≤ k ≤ 2i − 1, 1 ≤ l ≤ 2t + 3 and k, l are odd numbers. Assume henceforth that b(k,t+1) = b(l,t+1) where 1 ≤ k = l ≤ 2t + 1 are odd numbers, and we will derive a contradiction. So we can assume, without loss of generality, that k < l. By Definition 5.3 we obtain that b3 b(k,t+1) = b(k−2,t) + b(k+2,t+1) + b(k,t+2) and b3 b(l,t+1) = b(l−2,t) + b(l+2,t+1) + b(l,t+2) . Now we have that b(k−2,t) + b(k+2,t+1) + b(k,t+2) = b(l−2,t) + b(l+2,t+1) + b(l,t+2) , and by the induction hypothesis we obtain that b(k−2,t) = b(l−2,t) , b(l+2,t+1) and b(l−2,t) = b(k+2,t+1) which implies that b(k−2,t) = b(l,t+2) and b(l−2,t) = b(k,t+2) . If k = 1 we obtain that b(−1,t) = b(l,t+2) where 1 ≤ l ≤ 2t + 1. By Definition 5.2 we obtain that f (−1, t) = 0 and f (l, t + 2) = 0, which is a contradiction. If k = 1 we obtain that b(k+2,t+1) = b(l+2,t+1) . We repeat that protimes and we obtain that b(k−l+2t+3,t+1) = b(2t+3,t+1) . Since f (2t + 3, cess 2t+3−l 2 t + 1) = 0 and f (k − l + 2t + 3, t + 1) = 0, we have a contradiction. Assume henceforth that b(k,i) = b(l,t+2) where 1 ≤ i ≤ t + 1, 1 ≤ k ≤ 2i − 1, 1 ≤ l ≤ 2t + 3 and k, l are odd numbers, and we will derive a contradiction. Then we have the following four cases:


Case 1: Case 2: Case 3: Case 4:

249

l = 1, 3, t + 3. l = 1. l = 3. l = t + 3.

In the rest of this section, we derive a contradiction for the above cases. Case 1 By Definition 5.3, Lemma 5.2 and b3 (b¯3 b(l−2,t+1) ) = b¯3 (b3 b(l−2,t+1) ), we obtain that b(l−4,t+2) + b3 b(l,t+2) = b(l+2,t+2) + b¯3 b(l−2,t+2) .

(5.2)

If b(l−4,t+2) = b(l+2,t+2) then f (l, t + 2) = f (l − 2, t + 2), and by Definition 5.2 we obtain that (l + 1)(2t − l + 5) = (l − 1)(2t − l + 7) which implies that l = t + 3. Since l = t + 3 we obtain that f (l − 4, t + 2) = f (l + 2, t + 2)

(5.3)

which implies that b(l−4,t+2) = b(l+2,t+2) and (5.2) implies that b(l+2,t+2) ∈ Irr(b3 b(l,t+2) ). By Lemma 5.2 we obtain that b¯3 b(l−2,t+1) = b(l−2,t) + b(l−4,t+1) + b(l,t+2) which implies that b(l−2,t+1) ∈ Irr(b3 b(l,t+2) ).

(5.4)

Now we have the following two cases: Case A. b(l+2,t+2) = b(l−2,t+1) . Case B. b(l+2,t+2) = b(l−2,t+1) . By deriving a contradiction for Cases A and B, we shall show that Case 1 is impossible. Case A Since b(l+2,t+2) = b(l−2,t+1) we obtain that f (l + 2, t + 2) = f (l − 2, t + 1), and by (5.3) we obtain that f (l − 4, t + 2) = f (l + 2, t + 2) which implies that f (l − 4, t + 2) = f (l − 2, t + 1). Now we have that b(l−4,t+2) = b(l−2,t+1) . By Definition 5.3 we obtain that b3 b(l−2,t+1) = b(l−4,t) + b(l,t+1) + b(l−2,t+2) which implies that b(l−2,t+1) ∈ Irr(b¯ 3 b(l−2,t+2) ). Now by (5.2) and since 3f (l, t + 2) = f (l − 2, t + 1) + f (l + 2, t + 2) + f (l, t + 3), we obtain that b3 b(l,t+2) = 2b(l+2,t+2) + Σf (l,t+3)

where Σf (l,t+3) ∈ NB.

250

5


By Definition 5.3 we obtain that b3 b(k,i) = b(k−2,i−1) + b(k+2,i) + b(k,i+1) which implies that 2b(l+2,t+2) + Σf (l,t+3) = b(k−2,i−1) + b(k+2,i) + b(k,i+1) , so Σf (l,t+3) ∈ B which implies that the stopping number is not t + 1, and we have a contradiction. Case B By (5.2) and (5.4) and since b(l+2,t+2) = b(l−2,t+1) we obtain that b3 b(l,t+2) = b(l−2,t+1) + b(l+2,t+2) + Σf (l,t+3)

where Σf (l,t+3) ∈ NB.

By Definition 5.3 we obtain that b3 b(k,i) = b(k−2,i−1) + b(k+2,i) + b(k,i+1) which implies that b(l−2,t+1) + b(l+2,t+2) + Σf (l,t+3) = b(k−2,i−1) + b(k+2,i) + b(k,i+1) . So Σf (l,t+3) ∈ B which implies that the stopping number is not t + 1, and we have a contradiction. Case 2 Since l = 1 we obtain that b(k,i) = b(1,t+2) where 1 ≤ i ≤ t + 1. By Definition 5.3 we obtain that b3 b(1,t+1) = b(3,t) + b(1,t+2) which implies that b(1,t+1) ∈ Irr(b¯ 3 b(1,t+2) ). By Lemma 5.2 we obtain that b¯3 b(k,i) = b(k,i−1) + b(k−2,i) + b(k+2,i+1) . Since i ≤ t + 1 we obtain by the induction hypothesis that b(k,i−1) = b(1,t+1) . Now we have that either b(k−2,i) = b(1,t+1) or b(k+2,i+1) = b(1,t+1) . If b(k−2,i) = b(1,t+1) then (5.1) and the induction hypothesis imply that k = 3 and i = t + 1. Now we have that b(3,t+1) = b(1,t+2) . Thus f (3, t + 1) = f (1, t + 2) and by Definition 5.2 we obtain that (t − 3)(t + 2) = 0. Since t = −2 we have that t = 3. Now we have that b(3,4) = b(1,5) , which is a contradiction to Main Theorem 3 of Chap. 4. If b(k+2,i+1) = b(1,t+1) then (5.1) and the induction hypothesis imply that i = t + 1, so b(k+2,t+2) = b(1,t+1) and by Case 1 we obtain that either k = 1 or k = t + 1. If k = 1 then b(1,t+1) = b(1,t+2) and f (1, t + 1) = f (1, t + 2). Now by Definition 5.2 we obtain that (t + 1)(t + 2) = (t + 2)(t + 3), and we have a contradiction.


251

If k = t + 1 then b(t+3,t+2) = b(1,t+1) , and by Definition 5.3 we obtain that b(t+1,t+2) = b(2t+1,t+1) . Since f (3, 4) = f (5, 3) we have that t = 2. Now we have a contradiction to Case 1. Case 3 Since l = 3 we obtain that b(k,i) = b(3,t+2) . By Definition 5.3, Lemma 5.2 and b3 (b¯3 b(1,t+1) ) = b¯ 3 (b3 b(1,t+1) ), we obtain that b3 b(3,t+2) = b(5,t+2) + b¯ 3 b(1,t+2) which implies that b(5,t+2) ∈ Irr(b3 b(3,t+2) ). By Lemma 5.2 we obtain that b¯3 b(1,t+1) = b(1,t) + b(3,t+2) which implies that b(1,t+1) ∈ Irr(b3 b(3,t+2) ). By Case 1 we obtain that b(1,t+1) = b(5,t+2) . Now we have that b3 b(3,t+2) = b(1,t+1) + b(5,t+2) + Σf (3,t+3)

where Σf (3,t+3) ∈ NB,

and by Definition 5.3 we obtain that b3 bf (k,i) = bf (k−2,i−1) + bf (k+2,i) + bf (k,i+1) which implies that b(1,t+1) + b(5,t+2) + Σf (3,t+3) = b(k−2,i−1) + b(k+2,i) + b(k,i+1) . So Σf (3,t+3) ∈ B which implies that the stopping number is not t + 1, and we have a contradiction. Case 4 Since l = t + 3 we obtain that b(k,i) = b(t+3,t+2) which implies by property no. 2 of Definition 5.3 that b(2i−k,i) = b(t+1,t+2) . If t = 2 we have a contradiction to Case 3, and if t = 2 we have a contradiction to Case 1. Lemma 5.4 If (A, B) is a Cn -Table Algebra where 2 ≤ n, then for 1 ≤ k ≤ 2n + 1 where k is an odd number b3 b(k,n+1) = b(k−2,n) + b(k+2,n+1) + Σf (k,n+2)

where Σf (k,n+2) ∈ NB.

Proof Let (A, B) be a Cn -Table Algebra. If n = 2 then by Example 5.1 the lemma holds, and if n = 3 then by Example 5.2 the lemma holds. Now we can assume that 4 ≤ n. By Lemma 5.2 we obtain that b¯3 b(k−2,n) = b(k−2,n−1) + b(k−4,n) + b(k,n+1) where k is an odd number and 3 ≤ k ≤ 2n + 1. Now we have that b(k−2,n) ∈ Irr(b3 b(k,n+1) )

(5.5)

252

5


where k is an odd number and 3 ≤ k ≤ 2n + 1. By Definition 5.3, Lemma 5.2 and b3 (b¯3 b(k−2,n) ) = b¯3 (b3 b(k−2,n) ), we obtain that b(k−4,n+1) + b3 b(k,n+1) = b(k+2,n+1) + b¯3 b(k−2,n+1) . If b(k−4,n+1) = b(k+2,n+1) then f (k, n + 1) = f (k − 2, n + 1), and by Definition 5.2 we obtain that k = n + 2. Now we have by Lemma 5.3 that if k = n + 2, then b3 b(k,n+1) = b(k−2,n) + b(k+2,n+1) + Σf (k,n+2)

where Σf (k,n+2) ∈ NB. (5.6)

Now we show that the lemma holds for k = n + 2. By (5.6) and since n = 3 we obtain that b3 b(n−2,n+1) = b(n−4,n) + b(n,n+1) + Σf (n−2,n+2) which implies by property no. 2 of Definition 5.3 that b¯3 b(n+4,n+1) = b(n+4,n) + b(n+2,n+1) + Σ¯ f (n−2,n+2) . Now we have that b(n+4,n+1) ∈ Irr(b3 b(n+2,n+1) ), and by Lemma 5.3 and (5.5) we obtain that b3 b(n+2,n+1) = b(n,n) + b(n+4,n+1) + Σf (n+2,n+2)

where Σf (n+2,n+2) ∈ NB.

Now we have that b3 b(k,n+1) = b(k−2,n) + b(k+2,n+1) + Σf (k,n+2) where k is an odd number and 3 ≤ k ≤ 2n + 1. In particular, b3 b(2n−1,n+1) = b(2n−3,n) + b(2n+1,n+1) + Σf (2n−1,n+2)

where Σf (k,n+2) ∈ NB

which implies by property no. 2 of Definition 5.3 that b¯3 b(3,n+1) = b(3,n) + b(1,n+1) + Σ¯ f (2n−1,n+2) . Now we have that b3 b(1,n+1) = b(−1,n) + b(3,n+1) + Σf (1,n+2)

where Σf (1,n+2) ∈ NB

and the lemma holds for k = 1.

Lemma 5.5 If (A, B) is a Cn -Table Algebra where 2 ≤ n, then for 1 ≤ k ≤ 2n + 1 where k is an odd number b3 b(k,n+1) = b(k−2,n) + b(k+2,n+1) + Σf (k,n+2)

where Σf (k,n+2) ∈ NB \ B.


253

Proof Since the stopping number is n, Lemma 5.4 implies that there exists an odd number 1 ≤ l ≤ 2n + 1 such that b3 b(l,n+1) = b(l−2,n) + b(l+2,n+1) + Σf (l,n+2)

where Σf (l,n+2) ∈ NB \ B. (5.7)

Now by property no. 2 of Definition 5.3 and Lemma 5.4, we obtain that b3 b¯(l,n+1) = b3 b(2n+2−l,n+1) = b(2n−l,n) +b(2n+4−l,n+1) +Σ where Σ ∈ NB. Hence b3 b¯(l,n+1) = ¯ n) + b¯(l−2,n+1) + Σ. Since 3f (l, n + 1) = f (l, n) + f (l − 2, n + 1) + f (l + b(l, 2, n + 2), we obtain that b¯3 b(l,n+1) = b(l,n) + b(l−2,n+1) + Σf (l+2,n+2) .

(5.8)

If l = 1 then (b3 b(l,n+1) , b3 b(l,n+1) ) = (b3 b¯(l,n+1) , b3 b¯(l,n+1) ), (5.7) and Lemma 5.3 imply that either Σf (l+2,n+2) ∈ / B or Σf (l+2,n+2) = b(l−2,n+1) . By property no. 3 of Definition 5.3, Lemma 5.2 and b3 (b¯3 b(l,n) ) = b¯3 (b3 b(l,n) ), we obtain that b3 b(l+2,n+1) = b(l,n) + b(l+4,n+1) + Σf (l+2,n+2) . Now if Σf (l,n+2) = b(l−2,n+1) we obtain that b(l+4,n+1) = Σf (l+2,n+2) which implies that f (l + 4, n + 1) = f (l + 2, n + 2) = f (l − 2, n + 1), and this is impossible. So we have that b3 b(l+2,n+1) = b(l,n) + b(l+4,n+1) + Σf (l+2,n+2)

where Σf (l+2,n+2) ∈ NB \ B. (5.9) If l = 1 then Lemma 5.3 and (5.8) imply that b¯3 b(1,n+1) = b(1,n) + Σf (3,n+2) where Σf (3,n+2) ∈ NB \ B. Now by property no. 3 of Definition 5.3, Lemma 5.2 and b¯3 (b3 b(1,n+1) ) = b3 (b¯3 b(1,n+1) ), we obtain that b3 b(3,n+1) = b(1,n) + b(5,n+1) + Σf (3,n+2)

where Σf (3,n+2) ∈ NB \ B. (5.10)

By (5.7), (5.9) and (5.10) we obtain that if b3 b(l,n+1) = b(l−2,n) + b(l+2,n+1) + Σf (l,n+2)

where Σf (l,n+2) ∈ NB \ B,

then b3 b(l+2,n+1) = b(l,n) + b(l+4,n+1) + Σf (l+2,n+2)

where Σf (l+2,n+2) ∈ NB \ B

where 1 ≤ l ≤ 2n − 1. The stopping number is n, which implies that b3 b(2n+1,n+1) = b(2n−1,n) + Σf (2n+1,n+2) where Σf (l+2,n+2) ∈ NB \ B which implies that b3 b¯(1,n+1) = b¯(1,n) + Σ¯ f (2n+1,n+2) . Now (b3 b(1,n+1) , b3 b(1,n+1) ) = (b3 b¯(1,n+1) , b3 b¯(1,n+1) ) and Lemma 5.4 imply that b3 b(1,n+1) = b(3,n+1) + Σf (1,n+2) and either Σf (1,n+2) ∈ /B or b(3,n+1) = Σf (1,n+2) . If b(3,n+1) = Σf (1,n+2) then f (3, n + 1) = f (1, n + 2) which implies that n = 3 which is impossible by Example 5.2 and the lemma holds. Lemma 5.6 If (A, B) is a Cn -Table Algebra, then b(1,l) b(1,m) =

l i=1

b(2i−1,m+l−i)

254

5


and b(3,l) b(1,m) =

l

b(2i−3,m+l−i−1) +

i=2

l−1

b(2i+1,m+l−i)

i=1

where l, m ∈ N, 2 ≤ m, l ≤ m and l + m ≤ n + 2. Proof In the following proof we assume that 2 ≤ m. Let (A, B) be a Cn -Table Algebra. We will prove that b(1,l) b(1,m) =

l

b(2i−1,m+l−i)

(5.11)

i=1

where l, m ∈ N, l ≤ m and l + m ≤ n + 2 by induction on l. Since b(1,1) b(1,m) = b(1,m) we obtain that (5.11) holds for l = 1. Since b(1,2) b(1,m) = b3 b(1,m) = b(3,m) + b(1,m+1) , we obtain that (5.11) holds for l = 2. By Definition 5.3 and b3 (b3 b(1,m) ) = b32 b(1,m) we obtain that b(1,3) b(1,m) = b(1,m+2) + b(3,m+1) + b(5,m) and (5.11) holds for l = 3. Assume that (5.11) holds for 3 ≤ l ≤ t ≤ m − 1 and we will prove (5.11) for l = t + 1, i.e., we will prove that b(1,t+1) b(1,m) =

t+1

b(2i−1,m+t+1−i) .

i=1

By the induction hypothesis and by Definition 5.3, we obtain that b3 (b(1,t) b(1,m) ) =

t

b3 b(2i−1,m+t−i)

i=1

=

t

(b(2i−3,m+t−i−1) + b(2i+1,m+t−i) + b(2i−1,m+t−i+1) )

i=1

and (b3 b(1,t) )b(1,m) = b(3,t) b(1,m) + b(1,t+1) b(1,m) . Thus b(3,t) b(1,m) + b(1,t+1) b(1,m) =

t i=1

(b(2i−3,m+t−i−1) + b(2i+1,m+t−i) + b(2i−1,m+t−i+1) ).

(5.12)


255

By the induction hypothesis, Definition 5.3, Lemma 5.2 and since 3 ≤ t, we obtain that b¯3 (b(1,t−1) b(1,m) ) =

t−1

b¯3 b(2i−1,m+t−i−1)

i=1

=

t−1

(b(2i−1,m+t−i−2) + b(2i−3,m+t−i−1) + b(2i+1,m+t−i) )

i=1

and (b¯3 b(1,t−1) )b(1,m) = b(1,t−2) b(1,m) + b(3,t) b(1,m) =

t−2

b(2i−1,m+t−i−2) + b(3,t) b(1,m) .

i=1

Thus b(3,t) b(1,m) =

t

b(2i−3,m+t−i−1) +

t−1

i=2

b(2i+1,m+t−i) .

i=1

Now by (5.12) we obtain that b(1,t+1) b(1,m) =

t+1

b(2i−1,m+t+1−i) ,

i=1

as required. Lemma 5.7 If (A, B) is a Cn -Table Algebra, then b¯(1,l) b(1,m) =

l

b(2i−1,m−l−1+2i)

i=1

and b¯(3,l) b(1,m) =

l

b(2i−3,m−l+2i−1) +

i=2

l−1

b(2i+1,m−l+2i)

i=1

where l, m ∈ N, 2 ≤ m, l ≤ m and l + m ≤ n + 2. Proof Let (A, B) be a Cn -Table Algebra. We will prove that b¯(1,l) b(1,m) =

l

b(2i−1,m−l−1+2i)

i=1

where l, m ∈ N, l ≤ m and l + m ≤ n + 2 by induction on l.

(5.13)

256

5


Since b¯(1,1) b(1,m) = b(1,m) , we obtain that (5.13) holds for l = 1. Since b¯(1,2) b(1,m) = b¯3 b(1,m) = b(1,m−1) + b(3,m+1) , we obtain that (5.13) holds for l = 2. By Definition 5.3, Lemma 5.2 and b¯3 (b¯3 b(1,m) ) = b¯32 b(1,m) , we obtain that b(1,3) b(1,m) = b(1,m−2) + b(3,m) + b(5,m+2) and (5.13) holds for l = 3. Assume that (5.13) holds for 3 ≤ l ≤ t ≤ m − 1 and we will prove (5.13) for l = t + 1, i.e., we will prove that b¯(1,t+1) b(1,m) =

t+1

b(2i−1,m−t−2+2i) .

i=1

By the induction hypothesis, Definition 5.3 and Lemma 5.2, we obtain that b¯3 (b¯(1,t) b(1,m) ) =

t

b¯3 b(2i−1,m−t−1+2i)

i=1

=

t

(b(2i−1,m−t−2+2i) + b(2i−3,m−t−1+2i) + b(2i+1,m−t+2i) )

i=1

and (b¯3 b¯(1,t) )b(1,m) = b¯(3,t) b(1,m) + b¯(1,t+1) b(1,m) . Thus b¯ (3,t) b(1,m) + b¯(1,t+1) b(1,m) =

t

(b(2i−1,m−t−2+2i) + b(2i−3,m−t−1+2i) + b(2i+1,m−t+2i) ).

(5.14)

i=1

By the induction hypothesis, Definition 5.3, Lemma 5.2 and since 3 ≤ t, we obtain that b3 (b¯(1,t−1) b(1,m) ) =

t−1

b3 b(2i−1,m−t+2i)

i=1

=

t−1 i=1

and

(b(2i−3,m−t+2i−1) + b(2i+1,m−t+2i) + b(2i−1,m−t+2i+1) )


257

(b3 b¯(1,t−1) )b(1,m) = b¯(1,t−2) b(1,m) + b¯(3,t) b(1,m) =

t−2

b(2i−1,m−t+1+2i) + b¯(3,t) b(1,m) .

i=1

Thus b¯(3,t) b(1,m) =

t

b(2i−3,m−t+2i−1) +

i = 1t−1 b(2i+1,m−t+2i) .

i=2

Now by (5.14) we obtain that b¯(1,t+1) b(1,m) =

t+1

b(2i−1,m−t+2i−2)

i=1

as required. Lemma 5.8 If (A, B) is a C2n−1 -Table Algebra then n + 1 ≤ (b¯(1,n+1) b(1,n+2) , b¯(1,n) b(1,n+1) ). Proof By Lemma 5.6 we obtain that b(1,n) b(1,n+1) =

n

b(2i−1,2n+1−i) .

i=1

Now by Definition 5.3 and Lemma 5.5 we obtain that b3 (b(1,n) b(1,n+1) ) =

n

b3 b(2i−1,2n+1−i)

i=1

=

n

(b(2i−3,2n−i) + b(2i−1,2n+2−i) ) +

i=2

n

b(2i+1,2n+1−i)

i=1

+ Σf (1,2n+1) where Σf (1,2n+1) ∈ NB \ B. By Definition 5.3 and Lemma 5.6, we obtain that 2 (b3 b(1,n) )b(1,n+1) = b(3,n) b(1,n+1) + b(1,n+1)

=

n

b(2i−3,2n−i) +

i=2

n−1

2 b(2i+1,2n+1−i) + b(1,n+1) .

i=1

Thus 2 = b(1,n+1)

n+1 i=2

b(2i−1,2n+2−i) + Σf (1,2n+1) .

(5.15)

258

5


By Lemma 5.6 we obtain that b(1,n−1) b(1,n+2) =

n

b(2i−1,2n+1−i) .

i=1

Now by Definition 5.3 and Lemma 5.5 we obtain that b3 (b(1,n−1) b(1,n+2) ) =

n−1

b3 b(2i−1,2n+1−i)

i=1

=

n−1

(b(2i−3,2n−i) + b(2i−1,2n+2−i) ) +

i=2

n−1

b(2i+1,2n+1−i)

i=1

+ Σf (1,2n+1) . By Definition 5.3 and Lemma 5.6, we obtain that (b3 b(1,n−1) )b(1,n+2) = b(3,n−1) b(1,n+2) + b(1,n) b(1,n+2) =

n−1

b(2i−3,2n−i) +

i=2

n−2

b(2i+1,2n+1−i) + b(1,n) b(1,n+2) .

i=1

Thus b(1,n) b(1,n+2) =

n

b(2i−1,2n+2−i) + Σf (1,2n+1) .

i=2

Now by (5.15) we obtain that 2 n + 1 ≤ (b(1,n) b(1,n+2) , b(1,n+1) )

= (b¯ (1,n+1) b(1,n+2) , b¯(1,n) b(1,n+1) ),

as required. Lemma 5.9 If (A, B) is a C2n−1 -Table Algebra, then b¯(1,n+1) b(1,n+2) =

n−1

b(2i−1,2i) + 2b(2n−1,2n) + Σf (2n+1,2n+2)−f (2n−1,2n)

i=1

where Σf (2n+1,2n+2)−f (2n−1,2n) ∈ NB. Proof If (A, B) is a C2n−1 -Table Algebra then by Lemma 5.6 we obtain that b(1,j ) b(1,n+2) =

j i=1

b(2i−1,j +n+2−i)


259

and bf (1,j +1) bf (1,n+1) =

j +1

b(2i−1,j +n+2−i)

i=1

where 1 ≤ j ≤ n − 1. Now by Lemma 5.3 we obtain that for all 1 ≤ j ≤ n − 1 j = (b(1,j ) b(1,n+2) , b(1,j +1) b(1,n+1) ) = (b¯ (1,j ) b(1,j +1) , b¯(1,n+1) b(1,n+2) ). By Lemma 5.7 we obtain that b¯(1,j ) b(1,j +1) =

j

b(2i−1,2i)

where 1 ≤ j ≤ n.

i=1

Now Lemma 5.8 implies that b¯(1,n+1) b(1,n+2) =

n−1

b(2i−1,2i) + 2b(2n−1,2n) + Σf (2n+1,2n+2)−f (2n−1,2n)

i=1

where Σf (2n+1,2n+2)−f (2n−1,2n) ∈ NB, as required.

Lemma 5.10 If (A, B) is a C2n−1 -Table Algebra, then b(1,n+2) b¯(1,n+1) =

n

b(2i−3,2i−2) + b3 Σf (2n+1,2n+1) − Σ¯ f (2n−1,2n+1)

i=2

where Σf (2n+1,2n+1) ∈ NB \ B and Σf (2n−1,2n+1) ∈ NB. Proof If (A, B) is a C2n−1 -Table Algebra then by Definition 5.3, Lemmas 5.5 and 5.7, we obtain that b¯ 3 (b¯(1,n) b(1,n+1) ) =

n

b¯3 b(2i−1,2i)

i=1

=

n

b(2i−1,2i−1) +

i=1

n

b(2i−3,2i) +

i=2

n−1

b(2i+1,2i+1) + Σf (2n+1)

i=1

where Σf (2n+1,2n+1) ∈ NB \ B. By Definition 5.3 and Lemma 5.7 we obtain that (b¯3 b¯(1,n) )b(1,n+1) = b¯(3,n) b(1,n+1) + b¯(1,n+1) b(1,n+1) =

n i=2

bf (2i−3,2i) +

n−1 i=1

b(2i+1,2i+1) + b¯(1,n+1) b(1,n+1) .

260

5


Thus b(1,n+1) b¯(1,n+1) =

n

b(2i−1,2i−1) + Σf (2n+1,2n+1) .

i=1

By Lemmas 5.2 and 5.7, we obtain that (b¯3 b(1,n) )b¯(1,n+1) = b(1,n−1) b¯(1,n+1) + bf (,n+1) b¯(1,n+1) =

n−1

b¯(2i−1,2i+1) + b(3,n+1) b¯(1,n+1) .

i=1

By Definition 5.3, Lemmas 5.5 and 5.7, we obtain that b¯ 3 (b(1,n) b¯(1,n+1) ) =

n

b¯3 b¯(2i−1,2i)

i=1

=

n

b¯(2i−3,2i−1) +

i=2

n

b¯(2i+1,2i) +

i=1

n−1

b¯(2i−1,2i+1)

i=1

+ Σ¯ f (2n−1,2n+1) where Σf (2n−1,2n+1) ∈ NB \ B. Thus b(3,n+1) b¯(1,n+1) =

n

b¯(2i−3,2i−1) +

i=2

n

b¯(2i+1,2i) + Σ¯ f (2n−1,2n+1) .

i=1

Now by Definition 5.3 we obtain that b3 (b(1,n+1) b¯(1,n+1) ) =

n

b3 b(2i−1,2i−1) + b3 Σf (2n+1,2n+1)

i=1

=

n

n

b(2i−3,2i−2) +

i=2

(b(2i+1,2i−1) + b(2i−1,2i) )

i=1

+ b3 Σf (2n+1,2n+1) . By Definition 5.3 we obtain that (b3 b(1,n+1) )b¯(1,n+1) = b(3,n+1) b¯(1,n+1) + b(1,n+2) b¯(1,n+1) =

n i=2

b¯(2i−3,2i−1) +

n i=1

+ b(1,n+2) b¯(1,n+1) .

b¯(2i+1,2i) + Σ¯ f (2n−1,2n+1)


261

Thus b(1,n+2) b¯(1,n+1) =

n

b(2i−3,2i−2) + b3 Σf (2n+1,2n+1) − Σ¯ f (2n−1,2n+1) ,

i=2

as required.

Corollary 5.1 There exists no Ck -Table Algebra where k is an odd number and 43 ≤ k. Proof By Lemmas 5.9 and 5.10, we obtain that 2b(2n−1,2n) + Σf (2n+1,2n+2)−f (2n−1,2n) + Σ¯ f (2n−1,2n+1) = b3 Σf (2n+1,2n+1) . Now if β ∈ Irr(Σf (2n+1,2n+2)−f (2n−1,2n) ), then |β| ≤ f (2n + 1, 2n + 2) − f (2n − 1, 2n) and there exists α ∈ Irr(Σf (2n+1,2n+1) ) such that β ∈ Irr(b3 α). By Lemma 5.5 we obtain that b¯3 b(2n−1,2n) = b(2n−1,2n−1) + b(2n−3,2n) + Σf (2n+1,2n+1) which implies that b(2n−1,2n) ∈ Irr(b3 α). Now by Lemma 5.1 we obtain that f (2n − 1, 2n) ≤8 |β| which implies by Definition 5.2 that n(2n + 1) ≤ 8, 6n + 6 so there exists no Ck -Table Algebra where k is an odd number and 49 ≤ k. Assume henceforth that (A, B) is a C2n−1 -Table Algebra where 43 ≤ 2n − 1 ≤ 47, and we will derive a contradiction. If Σf (2n+1,2n+2)−f (2n−1,2n) ∈ / B, then there exists β1 , β2 ∈ Irr(Σf (2n+1,2n+2)−f (2n−1,2n) ) and α1 , α2 ∈ Σf (2n+1,2n+1) such that β1 , b(2n−1,2n) ∈ Irr(b3 α1 ) and β2 , b(2n−1,2n) ∈ Irr(b3 α2 ).

262

5


Since |β1 | + |β2 | ≤ f (2n + 1, 2n + 2) − f (2n − 1, 2n) and n ≥ 22, we obtain that either f (2n − 1, 2n) ≥8 |β1 | or f (2n − 1, 2n) ≥ 8, |β2 | and we have a contradiction to Lemma 5.1. Note that if there is only one element β in Irr(Σf (2n+1,2n+2)−f (2n−1,2n) ), then Irr(Σf (2n+1,2n+2)−f (2n−1,2n) ) = rβ

where 2 ≤ r

which implies that 2|β| ≤ f (2n + 1, 2n + 2) − f (2n − 1, 2n), and since n ≥ 22 we obtain that f (2n − 1, 2n) ≥ 8. |β| Now we have that Σf (2n+1,2n+2)−f (2n−1,2n) ∈ B. If n = 24 then b14,700 , Σ1,875 ∈ (Irrb3 α) which implies that 5, 525 ≤ |α| ≤ 5, 625. If |α| = 5, 525 then b3 α5,525 = Σ1,875 + b14,700 and α5,525 ∈ Irr(b¯3 Σ1,875 ) which implies that there exists γ ∈ B such that |γ | ≤ 100 and γ ∈ Irr(b¯ 3 Σ1,875 ), which is a contradiction. If |α| = 5, 525 then there exists γ ∈ B such that γ ∈ Irr(b3 α) and |γ | ≤ 300. Since b14,700 ∈ Irr(b3 α), we have a contradiction to Lemma 5.1. If n = 23 then b12,972 , Σ1,728 ∈ Irr(b3 α) which implies that 4, 900 ≤ |α| ≤ 5, 184. If |α| = 4, 900 then b3 α4,900 = Σ1,728 + b12,972


263

and α4,900 ∈ Irr(b¯3 Σ1,728 ) which implies that there exists γ ∈ B such that |γ | ≤ 284 and γ ∈ Irr(b¯ 3 Σ1,728 ), which is a contradiction. If |α| = 4, 900 then there exists γ ∈ B such that γ ∈ Irr(b3 α) and |γ | ≤ 852, since b12,972 ∈ Irr(b3 α), and we have a contradiction to Lemma 5.1. If n = 22 then b11,385 , Σ1,587 ∈ Irr(b3 α) which implies that 4, 324 ≤ |α| ≤ 4, 761. If |α| = 4, 324 then b3 α4,324 = Σ1,587 + b11,385 and α4,324 ∈ Irr(b¯3 Σ1,587 ) which implies that there exists γ ∈ B such that |γ | ≤ 437 and γ ∈ Irr(b¯ 3 Σ1,587 ), which is a contradiction. If |α| = 4, 324 then there exists γ ∈ B such that γ ∈ Irr(b3 α) and |γ | ≤ 1, 311, since b11,385 ∈ Irr(b3 α), and we have a contradiction to Lemma 5.1. So there exists no Ck -Table Algebra where k is an odd number and 43 ≤ k. Lemma 5.11 If (A, B) is a C2n -Table Algebra then n + 2 ≤ (b¯(1,n+2) b(1,n+2) , b¯(1,n+1) b(1,n+1) ). Proof If (A, B) is a C2n -Table Algebra then by Definition 5.3, Lemmas 5.5 and 5.7, we obtain that b3 (b(1,n+1) b¯(1,n+1) ) =

n+1

b3 b(2i−1,2i−1)

i=1

=

n+1

(b(2i−3,2i−2) + b(2i+1,2i−1) ) +

i=2

n

b(2i−1,2i)

i=1

+ Σf (2n+1,2n+2) where Σf (2n+1,2n+2) ∈ NB \ B. By Definition 5.3 and Lemma 5.7, we obtain that (b3 b(1,n+1) )b¯(1,n+1) = b(3,n+1) b¯(1,n+1) + b(1,n+2) b¯(1,n+1) =

n+1 i=2

b¯(2i−3,2i−1) +

n i=1

b¯(2i+1,2i) + b(1,n+2) b¯(1,n+1) .

264

5


Thus b¯(1,n+1) b(1,n+2) =

n

b(2i−1,2i) + Σ(2n+1,2n+2) .

(5.16)

i=1

Now we have that n + 2 ≤ (b¯(1,n+1) b(1,n+2) , b¯(1,n+1) b(1,n+2) ) = (b¯(1,n+2) b(1,n+2) , b¯(1,n+1) b(1,n+1) ),

as required. Lemma 5.12 If (A, B) is a C2n -Table Algebra, then b¯ (1,n+2) b(1,n+2) =

n

b(2i−1,2i−1) + 2b(2n+1,2n+1) + Σf (2n+3,2n+3)−f (2n+1,2n+1)

i=1

where Σf (2n+3,2n+3)−f (2n+1,2n+1) ∈ NB. Proof If (A, B) is a C2n -Table Algebra, then by Lemma 5.6 we obtain that b(1,j ) b(1,n+2) =

j

b(2i−1,j +n+2−i)

where 1 ≤ j ≤ n.

i=1

Now by Lemma 5.3 we obtain that for all 1 ≤ j ≤ n j = (b(1,j ) b(1,n+2) , b(1,j ) b(1,n+2) ) = (b¯ (1,j ) b(1,j ) , b¯(1,n+2) b(1,n+2) ). By Lemma 5.7 we obtain that b¯(1,j ) b(1,j ) =

j

b(2i−1,2i−1)

where 1 ≤ j ≤ n + 1.

i=1

Now Lemma 5.11 implies that b¯(1,n+2) b(1,n+2) =

n

b(2i−1,2i−1) + 2b(2n+1,2n+1) + Σf (2n+3,2n+3)−f (2n+1,2n+1)

i=1

where Σf (2n+3,2n+3)−f (2n+1,2n+1) ∈ NB, as required. Lemma 5.13 If (A, B) is a C2n -Table Algebra, then b(1,n+2) b¯(1,n+2) =

n

b(2i−1,2i−1) + b3 Σ¯ f (2n+1,2n+2) − Σ¯ f (2n−1,2n+1)

i=1

where Σf (2n+1,2n+2) ∈ NB \ B and Σf (2n−1,2n+1) ∈ NB.


265

Proof If (A, B) is a C2n -Table Algebra then by Definition 5.3, Lemmas 5.5 and 5.7, we obtain that b¯3 (b(1,n) b¯(1,n+2) ) =

n

b¯3 b¯(2i−1,2i+1)

i=1

=

n

b¯(2i−3,2i) +

i=2 n−1

+

n

b(2i+1,2i+1)

i=1

b¯(2i−1,2i+2) + Σ¯ f (2n−1,2n+2)

i=1

where Σf (2n−1,2n+2) ∈ NB \ B. By Lemmas 5.2 and 5.7 we obtain that (b¯3 b(1,n) )b¯(1,n+2) = b(1,n−1) b¯(1,n+2) + b(3,n+1) b¯(1,n+2) =

n−1

b¯(2i−1,2i+2) + b(3,n+1) b¯(1,n+2) .

i=1

Thus b(3,n+1) b¯(1,n+2) =

n

b¯(2i−3,2i) +

i=2

n

b(2i+1,2i+1) + Σ¯ f (2n−1,2n+2) .

i=1

Now by Definition 5.3 we obtain that (b3 b(1,n+1) )b¯(1,n+2) = b(3,n+1) b¯(1,n+2) + b(1,n+2) b¯(1,n+2) n n = b(2i+1,2i+1) + Σ¯ (2n−1,2n+2) b¯(2i−3,2i) + i=2

i=1

+ b(1,n+2) b¯(1,n+2) . By Lemma 5.2 and (5.16) of Lemma 5.11 we obtain that b3 (b(1,n+1) b¯(1,n+2) ) =

n

b3 b¯(2i−1,2i) + b3 Σ¯ f (2n+1,2n+2)

i=1

=

n i=1

b(2i−1,2i−1) +

n i=2

+ b3 Σ¯ f (2n+1,2n+2) .

b¯(2i−3,2i) +

n

b(2i+1,2i+1)

i=1

Thus b(1,n+2) b¯(1,n+2) =

n

b(2i−1,2i−1) + b3 Σ¯ f (2n+1,2n+2) − Σ¯ f (2n−1,2n+2) ,

i=1

as required.

266

5


Corollary 5.2 There exists no Ck -Table Algebra where k is even and 44 ≤ k. Proof By Lemmas 5.12 and 5.13 we obtain that n

b(2i−1,2i−1) + b3 Σ¯ f (2n+1,2n+2) − Σ¯ f (2n−1,2n+2)

i=1

=

n

b(2i−1,2i−1) + 2b(2n+1,2n+1) + Σf (2n+3,2n+3)−f (2n+1,2n+1) .

i=1

Hence b¯3 Σ(2n+1,2n+2) = 2b(2n+1,2n+1) + Σf (2n−1,2n+1) + Σf (2n+3,2n+3)−f (2n+1,2n+1) . Now if β ∈ Irr(Σf (2n+3,2n+3)−f (2n+1,2n+1) ) then |β| ≤ f (2n + 3, 2n + 3) − f (2n + 1, 2n + 1) and there exists α ∈ Irr(Σf (2n+1,2n+2) ) such that β ∈ Irr(b¯3 α). By Lemma 5.5 we obtain that b3 b(2n+1,2n+1) = b(2n−1,2n) + b(2n+3,2n+1) + Σf (2n+1,2n+2) where Σf (2n+1,2n+2) ∈ NB \ B which implies that b(2n+1,2n+1) ∈ Irr(b¯3 α). Now by Lemma 5.1 we obtain that f (2n + 1, 2n + 1) ≤8 |β| which implies by Definition 5.2 that (n + 1)3 ≤ 8, (n + 2)3 − (n + 1)3 so there exists no Ck -Table Algebra where k is even and 48 ≤ k. Assume that toward a contradiction that (A, B) C2n -Table Algebra where 44 ≤ / B. Then there exists 2n ≤ 46. If Σf (2n+3,2n+3)−f (2n+1,2n+1) ∈ β1 , β2 ∈ Σf (2n+3,2n+3)−f (2n+1,2n+1) and α1 , α2 ∈ Σf (2n+1,2n+2) such that β1 , b(2n+1,2n+1) ∈ Irr(b¯ 3 α1 ) and β2 , b(2n+1,2n+1) ∈ Irr(b¯ 3 α2 ).


267

Since |β1 | + |β2 | ≤ f (2n + 3, 2n + 3) − f (2n + 1, 2n + 1) and n ≥ 22 we obtain that either f (2n + 1, 2n + 1) ≥8 |β1 | or f (2n + 1, 2n + 1) ≥ 8, |β2 | and we have a contradiction to Lemma 5.1. Note that if there is only one element β in Irr(Σf (2n+3,2n+3)−f (2n+1,2n+1) ) then Irr(Σf (2n+3,2n+3)−f (2n+1,2n+1) ) = rβ

where 2 ≤ r

which implies that 2|β| ≤ f (2n + 3, 2n + 3) − f (2n + 1, 2n + 1) and since n ≥ 22 we obtain that f (2n − 1, 2n) ≥ 8. |β| Now we have that Σf (2n+2,2n+2)−f (2n+1,2n+1) ∈ B. If n = 23 then b13,824 , Σ1,801 ∈ Irr(b¯3 α) which implies that 5, 209 ≤ |α| ≤ 5, 403, so there exists γ ∈ B such that γ ∈ Irr(b¯3 α) and |γ | ≤ 584, since b13,824 ∈ Irr(b¯3 α), and we have a contradiction to Lemma 5.1. If n = 22 then b12,167 , Σ1,657 ∈ Irr(b¯3 α) which implies that 4, 608 ≤ |α| ≤ 4, 971. If |α| = 4, 608 then b¯ 3 α4,608 = Σ1,657 + b12,167 and α4,608 ∈ Irr(b3 Σ1,657 ) which implies that there exists γ ∈ B such that |γ | ≤ 363 and γ ∈ Irr(b3 Σ1,657 ), which is a contradiction. If |α| = 4, 608 then there exists γ ∈ B such that γ ∈ Irr(b¯3 α) and |γ | ≤ 1, 089. Since b12,167 ∈ Irr(b¯3 α), we have a contradiction to Lemma 5.1. So there exists no Ck -Table Algebra where k is even and 44 ≤ k. By Corollaries 5.1 and 5.2, the proof of Theorem 5.1 is completed.

268

5


Fig. 5.1 The representation graph of b1,0

5.3 Proof of Theorem 5.2 Let (A, B) be an Integral Normalized Table Algebra with countable basis B, the elements of which are parametrized by pairs (m, ), 0 ≤ ≤ m, m = 0, 1, . . . in such a way that b1,0 bm, = bm+1, + bm,+1 + bm−1,−1

(5.17)

and bm, = 0 unless 0 ≤ ≤ m. Note that b0,0 = 1. Denote b := b1,0 . It follows ¯ The indexing from (5.17) that bb1,1 = b0,0 + b2,1 = 1 + b2,1 ; therefore b1,1 = b. here is different from Sect. 5.1 of this chapter, i.e., the index pair (k, n) of Sect. 5.1 of this chapter corresponds to a pair (n − 1, (k − 1)/2) used in this section. The representation graph of b1,0 is depicted in Fig. 5.1. Note that it is isomorphic to a representation graph of a SITA generated by a non-real element of degree 3 [B]. Let I denote the set of pairs {(m, ) | 0 ≤ ≤ m, m = 0, 1, . . . }. On the set I we impose a lexicographic linear ordering, that is (a, b) ≤ (c, d) ⇐⇒ (a < c) ∨ (a = c ∧ b ≤ d). Thus (0, 0) < (1, 0) < (1, 1) < (2, 0) < (2, 1) < (2, 2) . . . In what follows we set Im, := {(i, j ) ∈ I | (i, j ) < (m, )}.


269

The linear ordering of I induces a linear order on the set of monomials bi b¯ j , i, j ≥ 0 as follows: bi b¯ j ≤ bm b¯ ⇐⇒ (i + j, j ) ≤ (m + , ). Thus we have the following linear ordering of monomials: ¯ b2 , bb, ¯ bb¯ 2 , b¯ 3 , . . . ¯ b¯ 2 , b3 , b2 b, 1, b, b, A pair (i, j ) will be called a degree of the monomial bi b¯ j . A monomial of highest degree (together with its coefficient) which appears in a polynomial expression p = i ¯j α ij i,j b b with non-zero coefficient will be called a leading term of p. Proposition 5.1 For each pair (s, t), 0 ≤ s, 0 ≤ t it holds that αij bi,j . bs b¯ t = bs+t,t +

(5.18)

(i,j )∈Is+t,t

Proof Induction on s + t. For s + t = 0 and s + t = 1 our statement follows from ¯ Assume that (5.18) holds for all pairs (s , t ) with s + b0,0 = 1, b1,0 = b, b1,1 = b. t = m ≥ 2. Pick an arbitrary pair (s, t) with s + t = m + 1. Consider first the case of s = 0. Then t = m + 1 ≥ 3 which implies that ¯bm+1 = b¯ · b¯ m = b¯ bm,m + αij bi,j (i,j )∈Im,m

= bm+1,m+1 + bm,m−1 +

αij (bi+1,j +1 + bi−1,j + bi,j −1 ).

(i,j )∈Im,m

It follows from (i, j ) ∈ Im,m that (i + 1, j + 1) < (m + 1, m + 1),

(i − 1, j ) < (m + 1, m + 1),

(i, j − 1) < (m + 1, m + 1). Therefore b¯ m+1 = bm+1,m+1 +

βij bi,j .

(i,j )∈Im+1,m+1

Assume now that s ≥ 1. Then bs b¯ t = b · bs−1 b¯ t and by (5.17), we obtain αij bbi,j b · bs−1 b¯ t = bbs+t−1,t + (i,j )∈Is+t−1,t

= bs+t,t + bs+t−1,t+1 + bs+t−2,t−1 + αij (bi+1,j + bi,j +1 + bi−1,j −1 ). (i,j )∈Is+t−1,t

270

5


It follows from (i, j ) ∈ Is+t−1,t that (i + 1, j ), (i, j + 1), (i − 1, j − 1) ∈ Is+t,t . Therefore bs b¯ t = bs+t,t +

βij bi,j .

(i,j )∈Is+t,t

Since the summation in (5.18) is done within a linear ordered sets Is+t,t , it is invertible which gives us bs+t,t = bs b¯ t + βuv bu−v b¯ v . (5.19) (u,v)∈Is+t,t

Theorem 5.3 The Table Algebra (A, B) is isomorphic to Z[x, y]. Its standard basis pu,v (x, y) is defined as follows: p0,0 = 1, p1,0 = x, p1,1 = y and for m ≥ 1 xpm, − pm,+1 − pm−1,−1 , 0 ≤ ≤ m pm+1, = (5.20) ypm,m + pm,m−1 , =m provided that pi,j = 0 for all (i, j ) ∈ I. In particular, the Table Algebra (A, B) is unique if it exists. ¯ Proof It follows from (5.19) that each bi,j , (i, j ) ∈ I is a polynomial in b and b. Therefore the mapping x → b, y → b¯ has a unique extension to an algebra epimorphism from Z[x, y] onto A. Let us denote this epimorphism as ϕ. Then ϕ(x i y j ) = bi b¯ j . It follows from (5.18) and (5.19) that the monomials bi b¯ j , 0 ≤ i, j form a Z-basis of A. Therefore the kernel of ϕ is trivial, which implies that ϕ is an algebra isomorphism. Denote pm, (x, y) := ϕ −1 (bm, ). Then p0,0 = 1, p1,0 = x, p1,1 = y. Now the formulae (5.20) follow directly from (5.17). Theorem 5.3 implies that if a Table Algebra (A, B) exists, then it is exactly isomorphic to (Z[x, y], {pu,v }(u,v)∈I ). In particular, this means that an existence implies uniqueness. Unfortunately, we cannot check directly that the basis pu,v defined by recurrence relations (5.20) is a table basis of Z[x, y], (that is, whether a product of two polynomials from a table basis is a linear combination of pu,v with nonnegative integer coefficients). To show existence it is sufficient to find an example of a group whose irreducible representations satisfy (5.17). It is well-known, [M], that irreducible polynomial representations of a linear group SL3 (C) are parameterized by pairs (m, ) ∈ I. An irreducible character of SL3 (C) corresponding to a pair (m, ), m ≥ ≥ 0 will be denoted as χm, . The value of χm, on a matrix A ∈ SL3 (C) is equal to s(m,,0) (α1 , α2 , α3 ) where s(m,,0) ∈ Δ[x1 , x2 , x3 ] is a Schur function and α1 , α2 , α3 are the eigenvalues of A.

References

271

Note that the irreducible characters χm, are in one-to-one correspondence with Schur functions of the form s(m,,0) . In fact, any Schur function sλ ∈ Λ[x1 , x2 , x3 ] yields an irreducible polynomial character of SL3 (C). Two Schur functions define the same character iff they are congruent modulo the ideal generated by the polynomial x1 x2 x3 − 1. In particular, the characters produced by s(λ1 ,λ2 ,λ3 ) and s(λ1 −λ3 ,λ2 −λ3 ,0) are the same. The product of two Schur functions sλ , sμ ∈ Λ[x1 , x2 , x3 ] is a linear combination of Schur functions with non-negative integer coefficients: ν sλ sμ = cλμ sν . ν ν cλμ

are defined by the Littlewood-Richardson The coefficients the following formula ⎧ ⎨ s(μ1 +1,μ2 ,0) + s(μ1 ,μ2 +1,0) + s(μ1 ,μ2 ,1) s(1,0,0) s(μ1 ,μ2 ,0) = s(μ1 +1,μ2 ,0) + s(μ1 ,μ2 ,1) ⎩ s(μ1 +1,0,0) + s(μ1 ,1,0)

rule [M]. It yields if μ1 > μ2 > 0 if μ1 = μ2 > 0 if μ2 = 0

Taking into account that s(μ1 ,μ2 ,1) and s(μ1 −1,μ2 −1,0) yield the same character, namely, χμ1 −1,μ2 −1 , we obtain that ⎧ ⎨ χμ1 +1,μ2 + χμ1 ,μ2 +1 + χμ1 −1,μ2 −1 if μ1 > μ2 > 0 if μ1 = μ2 > 0 χ1,0 χμ1 ,μ2 = χμ1 +1,μ2 + χμ1 −1,μ2 −1 ⎩ if μ2 = 0 χμ1 +1,0 + χμ1 ,1 This gives us a NITA which satisfies (5.17). This completes the proof of Theorem 5.2 of this chapter. It is worthwhile to note that polynomial irreducible representations of the group SLn (C) form an Infinite Dimensional Normalized Integral Table Algebra which con-

tains a non-real faithful element of degree n.

References [AB] [AC] [B] [M]

Arad, Z., Blau, H.: On table algebras and applications to finite group theory. J. Algebra 138, 137–185 (1991) Arad, Z., Chen, G.: On four normalized table algebras generated by a faithful nonreal element of degree 3. J. Algebra 283, 457–484 (2005) Blau, H.: Table algebras, Eur. J. of Comb. 30, 1426–1455 (2009) Macdonald, I.G.: Symmetric Functions and Hall Polynomials, Oxford Mathematical Monographs, 2nd edn. The Clarendon Press, Oxford University Press (1995)

Index

0–9 3 · A6 , 6, 11, 14, 99, 151, 187, 243, 245, 247 3 · A6 · 2, 6, 11, 19, 22, 99, 187, 247 A (A, B), 245–248, 251–255, 257–259, 261, 263–266, 268, 270 (A, B, C∞ ), 5, 6, 245, 246 (A, B, Cn ), 5, 6, 245–247 Adjacency algebra of an association scheme, 3 Algebra automorphism, 2, 4, 9, 244 Algebra homomorphism, 2, 4, 10, 244 Algebra isomorphism, 2, 9, 270 Array Ct , 5, 244 Association schemes, 3 Associative algebra, 1, 4, 243 B Bose–Mesner algebra, 3 C C-Algebras, 2, 3 C∞ -Table Algebra, 6, 246, 247 Cn -Table Algebra, 5, 245–248, 251–255 C-NITA, 4–6, 244–247 C-NTA, 4, 244 Coefficient, 2, 4, 9, 243, 269–271 Coherent algebra, 3 Commutative algebra, 1, 3, 4, 9, 243 Component, 9, 160, 161, 164, 165, 170–172, 174 Conformal field theory, 3 Countable Normalized Table Algebra, 244 Countable Table Algebra, 4, 5, 243, 244, 246 Cyclic group, 21, 22 D Definition of Table Algebra, 1

Degree, v, 1–4, 10, 11, 33, 34, 37, 51–53, 55, 56, 108, 151–155, 160, 161, 169–172, 174–178, 187–189, 243, 244, 247, 268, 269 Degree homomorphism, 2, 4, 244 Degree of the monomials, 269 Distance regular graphs, 3 E Exactly isomorphic, 9, 11, 270 F Faithful element, 3, 4, 21 Faithful non-real element, v, 271 Finite group, 1–3, 9, 10, 14, 152, 188 Finite group theory, 1, 2 Fusion rings, v, 1, 3 Fusion rule algebras, 3 G Generalized circulants of finite groups circulants, 3 Generated by an element, 10, 21 Graph theory, 2 H Hermitian form, 10 Homogenous Integral Table Algebra, 3 Hopf algebra, 3 Hypergroups, 3 I Integral Table Algebra, 3, 4, 10 Irr, v, 1–4, 6, 10, 11, 14, 80, 82, 151, 187, 216–220, 222–233, 236, 237, 245–247, 249–252, 261–263, 266, 267

Z. Arad et al., On Normalized Integral Table Algebras (Fusion Rings), Algebra and Applications 16, DOI 10.1007/978-0-85729-850-8, © Springer-Verlag London Limited 2011

273

274 Irreducible characters, 1, 3, 9, 152, 188, 270, 271 Irreducible components, 9 Isomorphic Table Algebra, 2, 9, 270 ITA, 3, 10 L L(B), 55, 82 L1 (B), 28, 51, 87, 151, 155, 163, 164, 177, 179, 180, 182, 184–187, 243 L2 (B), 28, 33, 51, 55, 82, 87, 151, 153, 164, 165, 170, 187, 243 Leading term, 269 Linear element, 21 Littlewood–Richardson rule, 271 M Monomials, 269, 270 N NITA, 4–6, 10, 11, 14, 22–24, 28, 29, 31–36, 38, 39, 42–45, 47–49, 52–55, 82–84, 95, 96, 99, 108, 151, 152, 187, 188 Normalized basis, 10 Normalized Countable Table Algebra, 4 Normalized Integral Table Algebras, 1, 3, 4, 10, 151, 187 Normalized Table Algebra, 2 O Order of B, 2 Orthonormal basis, 10 P Polynomial irreducible representation, 271 Polynomial representation, 6, 246, 247, 270 Posets, 3 Products of conjugacy classes of finite groups, 1, 9 Products of irreducible characters, 1, 9

Index Pseudo groups, 3 PSL(2, 7), v, 4, 6, 10, 11, 82, 151, 187, 245, 247 Q Quantum physics, 3 Quotient Table Algebra, 21 R Real element, 9, 11, 151, 170, 172, 177, 187, 189, 196, 200–204, 209, 210, 212–222, 226, 227, 230, 236, 246 Representation theory, 10 Rescaling of B , 2, 9 S Schur function, 270, 271 Schur rings, 3 Simple Table Algebra, 2 SITA, 4, 268 SL(3, C), 6, 246, 247 Standard element, 2 Standard Table Algebra, 2, 3 Stopping number, 5, 245, 246, 250, 251, 253 Strictly isomorphic, 2, 6, 11, 14, 21, 22, 82, 152, 188, 245–247 Structure constants, 2, 9, 10 Supp, 9, 10, 21, 25–27, 29–43, 47, 49, 51–54, 84, 90, 93–98 T Table Algebra, 1–6, 9, 10, 21, 22, 80, 99, 147, 243, 245, 247, 270 Table Algebra of conjugacy classes, 2, 10 Table Algebra of irreducible characters, 2 Table Algebras, 1–3, 5, 9, 10, 244, 245 Table basis, 1, 4, 9, 243, 270 Table subalgebra, 2, 10 Table subset, 2, 11, 14, 19, 80, 99, 147 Theory of fusion categories, 3

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