Extremal product-one free sequences in some non-abelian ... · Muito obrigado tamb em, Paulo Ribenboim, pelas dicas sobre as universidades canadenses. I would like to thank the guys

Universidade Federal de Minas Gerais

Instituto de Ciencias Exatas

Departamento de Matematica

Extremal product-one free sequences insome non-abelian groups and shifted Turan

sieve method on tournaments

Savio Ribas

Orientador:Fabio Enrique Brochero Martınez

Orientadora:Yu-Ru Liu

Belo Horizonte

2017

Universidade Federal de Minas Gerais

Instituto de Ciencias Exatas

Departamento de Matematica

Extremal product-one free sequences in some

non-abelian groups and shifted Turan sieve method

on tournaments

Savio Ribas ∗

Tese apresentada ao Departamento de Matematica da Universidade Federal de MinasGerais como parte dos requisitos para obtencao do grau de Doutor em Matematica.

Orientador: Fabio Enrique Brochero Martınez (UFMG)

Orientadora: Yu-Ru Liu (University of Waterloo)

Belo Horizonte, 06 de abril de 2017

Banca de defesa:

Fabio Enrique Brochero Martınez (UFMG)

John William MacQuarrie (UFMG)

Carlos Gustavo Tamm de Araujo Moreira (IMPA)

Eduardo Tengan (UFSC)

Hemar Teixeira Godinho (UnB)

∗ Pesquisa financiada pela CAPES (atraves do programa PICME) e Ciencia Sem Fronteiras/CNPq.

i

Dedico a minha esposa, Karine,e a meus pais, Rovia e Antonio

Acknowledgement

Finalmente, a ultima etapa da preparacao academica esta chegando ao fim, e isso seria

praticamente impossıvel sem a ajuda de varias das pessoas que me rodeiam. Comecarei

pelas mais importantes: Mae, obrigado pelo exemplo que nos deu e por acreditar que a

educacao poderia nos transformar. Lembro bem tudo o que voce passou no inıcio do meu

curso para me manter estudando e sou eternamente grato. Pai, onde quer que voce esteja,

sei que esta torcendo por mim, como sempre fez (se a defesa fosse numa quadra ou estadio,

voce estaria sentado na primeira fileira atras da banca e gritando para eles me aprovarem

logo). Voces passaram por varios apertos para que um dia eu chegasse aqui. Amo voces e

tudo que fizeram por mim! Karine, tesouro, carinho, rainha, coracao, LOTERIA! Sei que

voce nao gosta quando eu tento colocar humor em coisas profissionais, mas mesmo assim

eu dedico as frases motivadoras dessa tese a voce :-) Obrigado por me acompanhar nessa

caminhada chamada vida! Sabir, eu queria marcar a defesa para ontem, mas pela ordem

natural era melhor eu nao fazer isso, nao e? Obrigado pelos ensinamentos de irmao mais

velho, doto! Tiene, quando todos ouvirem o que voce tem a dizer, o mundo sera mais

justo. Obrigado pela forca e por tentar enfiar alguma didatica na minha cabeca! Lili,

eu nunca me esquecerei das aulas particulares, normais da escola ou de treinamento para

olimpıadas durante o ensino fundamental. Ali aprendi a gostar de matematica e a querer

fazer disso uma profissao! Quero ensinar com o mesmo entusiasmo que voce! Amo todos

voces! Agradeco tambem a toda minha famılia: Vo Vicente, tios, tias, primos, sogros

Conceicao e Alberto, e cunhados Kleriston, Kesley e Karem. Valeu!

Fabio, nao tenho palavras para agradecer toda a sua paciencia e dedicacao, e ainda

repleto de bom humor (e mesmo?), o que deixa a caminhada mais leve! Sem voce tudo

seria muito mais difıcil ou mesmo impossıvel: PECI, mestrado, doutorado, sanduıche,

IFMG... A primeira vez que fui te procurar para me orientar eu era ainda um menino, e

agora saio ja casado, concursado e ate com uns artigos submetidos (que ate pouco tempo

eu nao acreditava que fosse possıvel...). Desculpe-me por nao conseguir chegar tao cedo

(“Savio, eu chego 7h”, “Ok, a gente se encontra entao as 10h”) e muito obrigado por tudo!

Gostaria de agradecer tambem a meus grandes amigos de curso. Alguns me acom-

iii

panharam desde o inıcio da graduacao e pouco a pouco surgiram alguns agregados que

estao sempre presentes (tem tambem aqueles que estudaram na escola): Tulio (chic-chic)

e Luana (claro?), Vlad (po, cara, 60 e muito ruim), Remer Cueio (e bao ne?), Luis Felipe,

Paim e Claudiao, Dani e Bida, Nat e Prıncipe, Vinicao (Girafales), Danilo e Ali...sta,

obrigado pela grande parceria desde sempre! Agradeco tambem a D. Sonia, Luiz e Julia

pela forca! Levarei voces para a vida toda! Aos companheiros da antiga 1001 (atual 3105),

da turma 2013/1 (pior turma) e demais amigos da pos: Rodrego (estudou nada, hein!?),

Cristhiano (ah!), Letıcia (minha amiga tem 2 mestrados), Lucas (irmao), Lilian (irma),

1garetti, Pedro F., Pedro H., Jose (tranca), Allan (fechamento (nao e pela musica!!!)),

Felepe (cabra macho) e Javier (cao). Agradeco tambem a todos os demais colegas do

Departamento de Matematica da UFMG, em especial aos Profs. Renato e Marcelo Terra

que resolveram varias burocracias e a Andrea e a Kelli que estao sempre prontas a ajudar.

Eu me despeco dessa universidade apos 8 anos, mas ja quero voltar. Aos amigos de fora

da matematica que tambem tem seu lugar de grande destaque na galeria de agradecimen-

tos: Saulo e Ingrid, Flavia, Fernanda e toda a famılia, Michele, Lucas Assis, Marcus e

Gabi, valeu por tudo! Muito obrigado tambem, Paulo Ribenboim, pelas dicas sobre as

universidades canadenses.

I would like to thank the guys from University of Waterloo, Canada, who welcomed me

so well. Thank you very much Yu-Ru Liu for accepting me as a student during my visitor

time. This opportunity you provided me helped me to grow a lot. Thanks Wentang Kuo

and Kevin Zhou for working together, our work resulted in Chapter 5. Thanks Stanley for

providing me a sweet home and for the mathematical discussions that resulted in Chapter

4. Gracias Ignacio (Nacho, how are you?) for the discussions watered on beers on snowy

nights! Thanks Samin, Zia and Shubham, partners from sweet home. Thanks Alejandra,

Omar, Robert, Blake, Anton, Chad, Diana, Julian, J. C., Pavlina, Lis and Cam Stewart.

I know I should know more English, but you have all been very patient with me! I am

eternally grateful! I would also like to thank all other members of the Department of Pure

Mathematics for helping to make my exchange one of the best experiences in my life! E

nesse paragrafo canadense nao poderia faltar o paulista quase mineiro Danilo. Valeu pelos

almocos e pelo hockey (Go, Rangers! Make some noise!).

Agradeco a banca: John, Gugu, Tengan e Hemar, pelos valiosos comentarios e su-

gestoes. Gostaria de agradecer tambem a CAPES pela bolsa de doutorado no Brasil,

ao programa Ciencia Sem Fronteiras do CNPq pela bolsa de sanduıche e ao programa

PICME (em particular, a Sylvie pelo excelente trabalho). Sem esse suporte financeiro

essa pesquisa nao teria sido possıvel. Um agradecimento especial vai para duas figuras

importantıssimas na historia desse paıs, que olharam com carinho para o ensino publico,

para as pesquisas e para as famılias que mais necessitam: Obrigado Lula e Dilma!

iv

“Esse e de laranja que parece de limao, mas tem gosto de tamarindo” (Chaves)

“Teria sido melhor ir ver o filme do Pele” (Chaves)

“Ja chegou o disco voador” (Chaves)

Seu Barriga: “Chaves, pegue um balde, rapido!”

Chaves: “Serve um balde vermelho?”

Seu Barriga: “Serve!”

Chaves: “Mas nao tem vermelho!”

“Outra vez flores? (...) Outra vez cafe? E por isso que ele so te traz flores” (Quico)

Chaves: “Eu sei que o Homem Invisıvel esta aqui!”

Quico: “Por que?”

Chaves: “Porque nao estou vendo ele!”

“Nao ha nada mais trabalhoso do que viver sem trabalhar” (Seu Madruga)

“Palma, palma! Nao priemos canico” (Chapolin)

“Meus movimentos sao friamente calculados” (Chapolin)

“Nao contavam com minha astucia” (Chapolin)

Published, submitted and inpreparation works

The following works were submitted or are in advanced stage of preparation:

[A] Brochero Martınez, F. E., Ribas, Savio; Extremal product-one free sequences in Cqos

Cm. Submitted. Available at: https://arxiv.org/pdf/1610.09870.pdf (2016).

See [11].

[B] Brochero Martınez, F. E., Ribas, Savio; Extremal product-one free sequences in Dihe-

dral and Dicyclic Groups. Submitted. Available at: https://arxiv.org/pdf/1701.

08788.pdf (2017). See [12].

[C] Brochero Martınez, F. E., Ribas, Savio; Improvement of Alon-Dubiner constants for

a zero-sum problem in Zdp. In preparation.

[D] Kuo, W., Liu, Y.-R., Ribas, Savio, Zhou, K.; The shifted Turan sieve method on

tournaments. Submitted. See [53].

https://arxiv.org/pdf/1610.09870.pdf



Resumo

Essa tese esta dividida em duas partes:

• Parte I: Problemas de soma-zero inversos.

Nos comecamos apresentando uma visao geral sobre a Teoria de Soma-Zero. Em

particular, apresentamos os principais resultados e conjecturas acerca dos seguintes

invariantes: Constante de Davenport, Constante de Erdos-Ginzburg-Ziv e Constante

η (sem pesos ou com pesos {±1}). Depois, focamos na constante de Davenport: esse

invariante denota o menor inteiro positivo D(G) tal que qualquer sequencia S de

elementos de G, com comprimento |S| ≥ D(G), contem uma subsequencia com

produto 1 em alguma ordem, onde G e um grupo finito escrito multiplicativamente.

J. Bass [6] determinou a constante de Davenport dos grupos metacıclicos (em alguns

casos especiais) e dicıclicos e J. J. Zhuang e W. D. Gao [88] determinaram a constante

de Davenport dos grupos diedrais. Em conjunto com F. E. Brochero Martınez

(ver [11] e [12]), para cada um desses grupos nao abelianos nos exibimos todas as

sequencias de tamanho maximo que sao livres de subsequencias com produto 1.

Nos concluımos apresentando as Propriedades B, C e D, que sao, em geral, conjec-

turas de extrema importancia no estudo dos problemas inversos.

• Parte II: O crivo de Turan deslocado aplicado a torneios.

Nos comecamos apresentando uma visao geral sobre alguns problemas famosos que

podem ser resolvidos parcialmente ou totalmente usando a Teoria do Crivo. Apos

isso, construımos uma versao deslocada do crivo de Turan, que foi desenvolvido por

Y.-R. Liu e M. R. Murty (ver [54] e [55]), e o aplicamos a problemas de contagem de

torneios em grafos, isto e, grafos completos direcionados, de acordo com o numero

vi

vii

de ciclos. Mais precisamente, obtemos cotas superiores para o numero de torneios

que contem um numero pequeno de r-ciclos restritos (no caso de torneios normais

e multipartidos) ou irrestritos (no caso de torneios bipartidos), como foi feito em

conjunto com W. Kuo, Y.-R. Liu e K. Zhou [53]. Depois, mostramos como futura-

mente a teoria do crivo pode ser util em outros contextos da Matematica, como nos

problemas de soma-zero.

Palavras-chave: Problemas de soma-zero, constante de Davenport, problemas de soma-

zero inverso, constante de Davenport inversa, grupo metacıclico, grupo diedral, grupo

dicıclico; teoria do crivo, conjectura dos primos gemeos, conjectura de Goldbach, crivo

combinatorio, crivo de Turan, crivo de Turan deslocado, torneios, ciclos.

Abstract

This thesis is divided into two parts:

• Part I: Inverse zero-sum problems.

We start presenting an overview on Zero-Sum Theory. In particular, we present

the main results and conjectures concerning the following invariants: Davenport

constant, Erdos-Ginzburg-Ziv constant and η constant (either with weights {±1}or unweighted as well). Afterwards, we focus on Davenport constant: this invariant

denotes the smallest positive integer D(G) such that every sequence of elements in

G of length |S| ≥ D(G) contains a product-1 subsequence in some order, where G

is a finite group written multiplicatively.

J. Bass [6] determined the Davenport constant of metacyclic (in some special cases)

and dicyclic groups, and J. J. Zhuang and W. D. Gao [88] determined the Davenport

constant of dihedral groups. In a joint work with F. E. Brochero Martınez (see [11]

and [12]), for each of these non-abelian groups we exhibit all sequences of maximum

length that are free of product-1 subsequences.

We conclude by presenting Properties B, C, and D, which are, in general, conjectures

of extreme importance in the study of inverse problems.

• Part II: Shifted Turan sieve on tournaments.

We start presenting an overview on some famous problems which can be partially

or totally resolved using Sieve Theory. Afterwards, we construct a shifted version of

the Turan sieve method, developed by Y.-R. Liu and M. R. Murty (see [54] and [55]),

and apply it to counting problems on tournaments in graph theory, i.e., complete

directed graph, according to the number of cycles. More precisely, we obtain upper

viii

ix

bounds for the number of tournaments which contain a small number of restricted

r-cycles (in case of normal or multipartite tournaments) or unrestricted r-cycles

(in case of bipartite tournaments), as done in [53]. Then, we show how the sieve

theory may in the future be useful in other mathematical branches, such as zero-sum

problems.

Key-words: Zero-sum problems, Davenport constant, inverse zero-sum problems, in-

verse Davenport constant, metacyclic group, dihedral group, dicyclic group; sieve theory,

twin prime conjecture, Goldbach’s conjecture, combinatorial sieve, Turan sieve, shifted

Turan sieve, tournaments, cycles.

Contents

Part I Inverse zero-sum problems 1

Notations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2

1. Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

1.1 Inverse zero-sum problems . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

2. Inverse zero-sum problems in some non-abelian groups . . . . . . . . . . 10

2.1 Extremal product-one free sequences in Metacyclic Groups . . . . . . . . . 10

2.1.1 Auxiliary results . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

2.1.2 Sequences in C5 os Cm . . . . . . . . . . . . . . . . . . . . . . . . . 16

2.1.3 Proof of Theorem 2.1.1 . . . . . . . . . . . . . . . . . . . . . . . . . 17

2.2 Extremal product-one free sequences in Dihedral Groups . . . . . . . . . . 25

2.2.1 Proof of Theorem 2.2.1 . . . . . . . . . . . . . . . . . . . . . . . . . 26

2.3 Extremal product-one free sequences in Dicyclic Groups . . . . . . . . . . . 30

2.3.1 Proof of Theorem 2.3.1: Case n ≥ 4 . . . . . . . . . . . . . . . . . . 32

2.3.2 Proof of Theorem 2.3.1: Case n = 3 . . . . . . . . . . . . . . . . . . 33

2.3.3 Proof of Theorem 2.3.1: Case n = 2 . . . . . . . . . . . . . . . . . . 35

3. The Properties B, C, D . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38

Part II Shifted Turan sieve on tournaments 40

Notations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

4. Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42

4.1 Twin prime conjecture . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43

4.2 Goldbach conjecture . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45

4.3 Other problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46

5. The shifted Turan sieve method on tournaments . . . . . . . . . . . . . . 48

5.1 The shifted Turan sieve . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48

5.2 r-cycles on tournaments . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51

5.3 Restricted r-cycles on t-partite tournaments . . . . . . . . . . . . . . . . . 59

5.4 Unrestricted 2r-cycles on bipartite tournaments . . . . . . . . . . . . . . . 64

6. Future applications of Sieve Theory . . . . . . . . . . . . . . . . . . . . . . 74

Appendix 76

Notations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77

A. Erdos-Ginzburg-Ziv theorem . . . . . . . . . . . . . . . . . . . . . . . . . . 78

A.1 Chevalley-Warning theorem . . . . . . . . . . . . . . . . . . . . . . . . . . 78

A.2 Proof of Erdos-Ginzburg-Ziv theorem . . . . . . . . . . . . . . . . . . . . . 79

B. Cauchy-Davenport inequality . . . . . . . . . . . . . . . . . . . . . . . . . . 81

B.1 The e-transform . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82

B.2 Proof of Cauchy-Davenport inequality . . . . . . . . . . . . . . . . . . . . . 82

B.3 Another proof of Erdos-Ginzburg-Ziv theorem . . . . . . . . . . . . . . . . 84

C. Vosper’s theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85

D. An elliptic curve modulo p . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90

D.1 Multiplicative characters . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90

D.2 The equation xn ≡ a (mod p) . . . . . . . . . . . . . . . . . . . . . . . . . 92

D.3 The sums of Gauss and Jacobi . . . . . . . . . . . . . . . . . . . . . . . . . 93

D.4 The equation a1xl11 + · · ·+ arx

lrr ≡ b (mod p) . . . . . . . . . . . . . . . . . 97

E. {±1}-weighted Davenport constant . . . . . . . . . . . . . . . . . . . . . . 99

F. Conditions for a zero-sum modulo n: Proof of Theorem 1.1.1 . . . . . . 100

References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 104

Part I

INVERSE ZERO-SUM PROBLEMS

Notations

• N = {1, 2, 3, . . . } is the set of positive integers;

• N0 = {0, 1, 2, 3, . . . } is the set of non-negative integers;

• Z = {. . . ,−2,−1, 0, 1, 2, . . . } is the set of integers;

• Zn = (Z/nZ) is the set of residue classes modulo n;

• Cn ' Zn is the cyclic group of order n written multiplicatively;

• Z∗n = {1 ≤ a ≤ n| gcd(a, n) = 1} is the set of invertibles modulo n;

• If X and Y are non-empty subsets of a group G written additively then the sum-set

is defined by X + Y = {a+ b ∈ G | a ∈ X, b ∈ Y };

• If G is written multiplicatively then the product-set is defined by X · Y = {a · b ∈G | a ∈ X, b ∈ Y };

• Let G be a finite group written multiplicatively and S = (g1, g2, . . . , gl) be a se-

quence of elements of G. T is a subsequence of S if T = (gn1 , gn2 , . . . , gnk), where

{n1, n2, . . . , nk} is a subset of {1, 2, . . . , l}. We say that T is a product-1 subsequence

if gσ(n1)gσ(n2) . . . gσ(nk) = 1 for some permutation σ of {n1, . . . , nk}, and if there are

no such product-1 subsequences then S is free of product-1 subsequences. If G is

written additively then we exchange “product-1” by “zero-sum” in our definitions.

• Suppose that S1 = (gi1 , . . . , giu) and S2 = (gj1 , . . . , gjv) are subsequences of S. Then

– π(S) = g1g2 . . . gl denotes the product of the elements in S in the order that

they appear;

– πn(S) = gn+1 . . . glg1 . . . gn, for 0 ≤ n ≤ l − 1, denotes the product of the

elements in S with a n-shift in the indices;

Notations 3

– |S| = l denotes the length of the sequence S;

– SS−11 denotes the subsequence formed by the elements of S without the ele-

ments of S1;

– S1 ∩ S2 denotes the intersection of the subsequences S1 and S2. In the case

that S1 ∩ S2 = ∅, we say that S1 and S2 are disjoint subsequences;

– S1S2 = (gi1 , . . . , giu , gj1 , . . . , gjv) denotes the concatenation of S1 and S2;

– Sk = SS . . . S denotes the concatenation of k identical copies of S’s.

1Introduction

Given a finite group G written multiplicatively, the Zero-Sum Problems study condi-

tions to ensure that a given sequence in G has a non-empty subsequence with prescribed

properties (such as length, repetitions, weights) such that the product of its elements, in

some order, is equal to the identity of the group.

One of the first problems of this type is the remarkable Theorem of Erdos-Ginzburg-

Ziv (see [27]): Given 2n−1 integers, it is possible to select n of them, such that their sum

is divisible by n, or in group theory language, every sequence with l ≥ 2n − 1 elements

in a finite cyclic group of order n has a subsequence of length n, the product of whose n

elements being the identity. In this theorem, the number 2n − 1 is the smallest integer

with this property, since the sequence

(0, 0, . . . , 0︸︷︷︸n−1 times

, 1, 1, . . . , 1︸︷︷︸n−1 times

)

has no subsequence of length n and sum 0 ∈ Zn. For two different proofs of this theorem,

see Appendices A and B.3.

Traditionally, this class of problems have been extensively studied for abelian groups.

For an overview on Zero-Sum Theory for finite abelian groups, see for example the surveys

of Y. Caro [17] and W. Gao and A. Geroldinger [34].

Related to the Erdos-Ginzburg-Ziv theorem, we define the EGZ constant of a finite

group G (written multiplicatively) as follows: Let s(G) be the smallest positive integer

s such that every sequence with s(G) elements in G (repetition allowed) contains some

subsequence of length exp(G) such that the product of its terms in some order is 1, where

exp(G) is the exponent of G.

5

It is valid that

2d(n− 1) + 1 ≤ s(Cdn) ≤ nd(n− 1) + 1. (1.1)

The first inequality follows from the fact that the sequence formed by n − 1 copies of

every vector with entries 1 or g (where g generates Cn) is free of product-1 subsequences

of length n, and the second inequality follows from Pigeonhole Principle. In particular,

since this problem is multiplicative over n (see Equation (1.3) and Appendix A), we obtain

s(Cd2k) = 2d(2k − 1) + 1. (1.2)

In [50], Kemnitz conjectured that s(C2n) = 4n − 3 and it has been proved independently

by C. Reiher [71] and C. di Fiore [72] (using Chevalley-Warning theorem, see Appendix

A.1). Furthermore, it is also valid that s(Cm × Cn) = 2m+ 2n− 3 provided 1 ≤ m|n.

The problem of determining s(Cdn) precisely for all n ≥ 3 and d ≥ 3 seems extremely

difficult. In [25], C. Elsholtz showed that the lower bound in (1.1) is not tight in general,

i.e. he proved that if n ≥ 3 is odd then s(Cdn) ≥ 1.125bd/3c(n − 1)2d + 1. In particular,

s(C3n) ≥ 9n− 8 if n is odd. In [4], N. Alon and M. Dubiner gave an explicit linear upper

bound by proving that s(Cdp ) ≤ cdp, where c2 = 4, cd = 256cd−1 log2[(32d)d]+d+1 for d ≥ 3

and p is a sufficiently large prime. In particular, s(C3p) ≤ 20234p and cd ≤ (cd log2 d)d for

a certain absolute constant c > 0. In a joint work with F. E. Brochero Martınez, we hope

to lower the bound in such way that s(C3p) ≤ 1000p for sufficiently large primes p. The

above inequalities must also be valid for n with large prime factors instead of p sufficiently

large prime, since

s(Cdmn) ≤ min{s(Cd

n) + n[s(Cdm)− 1], s(Cd

m) +m[s(Cdn)− 1]}. (1.3)

It is also conjectured that

s(C3n) =

8n− 7 if n is even;

9n− 8 if n is odd.

This conjecture is confirmed if n = 2k for some k ≥ 1 (by Equation (1.2)), for n = 2k × 3

for some k ≥ 1 and for n = 3k × 5l for some k, l ∈ N0 with k + l ≥ 1 [37].

In [6], J. Bass calculated the EGZ constant of some non-abelian groups, such as di-

cyclic groups and special cases of metacyclic groups, and in [88], J. Zhuang and W. Gao

calculated the Davenport constant of the dihedral groups (see Sections 2.1, 2.2 and 2.3

for definitions).

We may also consider the weighted problem: Let A ⊂ {1, 2, . . . , exp(G)− 1} where G

6

is a finite group, and S be a sequence in G. We say that S is an A-product-1 sequence if

we can find εj ∈ A for 1 ≤ j ≤ |S| and a permutation σ of {1, 2, . . . , |S|} such that

gε1σ(1) · gε2σ(2) . . . g

ε|S|σ(|S|) = 1.

If S does not have subsequences T which are A-product-1 sequences then S is said to be

free of A-product-1 subsequences. The A-weighted EGZ constant is the smallest integer

sA(G) such that if S is a sequence in G with |S| ≥ sA(G) then S has some non-empty

A-product-1 subsequence of length exp(G).

This is a very difficult problem for a general weight set A, but it has been studied

in some specific cases, for example A = {1, exp(G) − 1} = {±1} (mod exp(G)). In [1],

S. D. Adhikari et al. proved that s{±1}(Cn) = n + blog2 nc and, furthermore, we have

s{±1}(C2n) = 2n− 1 by a simple application of Chevalley-Warning theorem.

Another important type of Zero-Sum Problem is to determine the Davenport constant

of a finite group G (written multiplicatively): This constant, denoted by D(G), is the

smallest positive integer d such that every sequence with d elements in G (repetition

allowed) contains some subsequence such that the product of its terms in some order is 1.

It is easy to show that D(G) ≤ |G| for all finite groups G, and it implies that D(G) is

well defined. In fact, let g1, g2, . . . , g|G| ∈ G. Consider the following |G| products:

g1, g1g2, g1g2g3, . . . , g1 . . . g|G|.

If they are all distinct then one of them is 1 ∈ G. Otherwise, there exist 1 ≤ r < s ≤|G| such that g1 . . . gr = g1 . . . gs, which implies gr+1 . . . gs = 1 and we are done. As a

consequence, D(Cn) = n, since |Cn| = n and the sequence (g)n−1 is free of product-1

subsequences, where g is a generator of Cn.

In a pair of papers ([66] and [67]), J. Olson considered the generalization for larger

ranks. He proved that D(Cm×Cn) = m+n−1 where 1 ≤ m|n, and D(Cpe1×· · ·×Cper ) =

1 +∑r

i=1(pei − 1). In general, if n1|n2| . . . |nr then the following lower bound

D(Cn1 × · · · × Cnr) ≥ 1 +r∑i=1

(ni − 1) (1.4)

holds, since the sequence (g1)n1−1 . . . (gr)

nr−1 is free of product-1 subsequences, where gi

generates Cni . However, for r ≥ 4 there are groups such that this lower bound is not tight

(see, for example, [39] and [58]).

In [81], P. van Emde Boas and D. Kruyswijk showed that if G is a finite abelian

7

group then D(G) < exp(G)[1 + log

(|G|

exp(G)

)]. This result can be sharpened to give the

bound exp(G)[γ + ε+ log

(|G|

exp(G)

)], where γ is the Euler-Mascheroni constant, provided

exp(G) and |G|/ exp(G) are each sufficiently large as a function of ε > 0. In [3], W. R.

Alford, A. Granville and C. Pomerance used the first inequality with G = Z∗L as a tool to

prove that there are infinitely many Carmichael numbers.

In [6], J. Bass also calculated the Davenport constant of some non-abelian groups,

such as dicyclic groups and special cases of metacyclic groups, and in [88], J. Zhuang and

W. Gao also calculated the Davenport constant of the dihedral groups (see Sections 2.1,

2.2 and 2.3 for definitions and values).

As before, we may also consider the weighted problem. The A-weighted Davenport

constant is the smallest integer DA(G) such that if S is a sequence in G with |S| ≥ DA(G)

then S has some non-empty A-product-1 subsequence. The weighted Davenport constant

also has been studied in some specific cases. In [1], S. D. Adhikari et al. proved that

D{±1}(Cn) = blog2 nc+ 1. This result, which is going to be used in Section 2.2, is proved

in Appendix E.

There is also another central invariant in Zero-Sum Problems, called η constant of

a group G (written multiplicatively). This constant, denoted by η(G), is the smallest

positive integer η such that every sequence with η elements in G (repetition allowed)

possesses some subsequence of length at most exp(G) such that the product of its terms

in some order is 1.

Since exp(Cn) = |Cn| = n = D(Cn), we obtain η(Cn) = n. In [32], W. Gao and A.

Geroldinger proved that η(C2n) = 3n − 2 and, furthermore, η(Cm × Cn) = 2m + n − 2

where 1 ≤ m|n. W. Gao, et al. [37] proved that η(C3n) = 8n − 7 if n = 3k × 5l for some

k, l ∈ N0 with k + l ≥ 1, η(C3n) = 7n− 6 if n = 2k for some k ≥ 1 or n = 2k × 3 for some

k ≥ 0. Also, it has been proved (see [37]) that η(Cdn) ≥ (2d − 1)(n− 1) + 1, and if n = 2k

for some k ≥ 1 then the equality occurs. In addition, if n is odd then η(C3n) ≥ 8n− 7.

By definition, we obtain D(G) ≤ η(G) ≤ s(G) for every finite group G. In addition,

W. A. Schmid and J. J. Zhuang [75] proved that if G is a p-group for some odd prime p

and D(G) ≤ 2 exp(G)− 1 then

2D(G)− 1 ≤ η(G) + exp(G)− 1 ≤ s(G) ≤ D(G) + 2 exp(G)− 2, (1.5)

connecting all these constants. In particular, if D(G) = 2 exp(G)− 1 then s(G) = η(G) +

exp(G) − 1 = 4n − 3. It is conjectured that the equality holds at the lower bounds for

1.1 Inverse zero-sum problems 8

every finite abelian p-group, i.e., 2D(G)− 1 = η(G) + exp(G)− 1 = s(G). The inequality

s(G) ≥ η(G) + exp(G)− 1

holds for every finite group G, since if T is a sequence with |T | = η(G)−1 which is free of

product-1 subsequences with length at most exp(G) then the sequence S = (1)exp(G)−1T

satisfies |S| = η(G) + exp(G)− 2 and has no product-1 subsequences of length equals to

exp(G). It is also conjectured that the last inequality is an equality for every finite group

G, and this is confirmed for all groups where the values of these constants are known,

even the non-abelian ones.

1.1 Inverse zero-sum problems

Let G be a finite group. By the definition of Davenport constant, there exist sequences

S of G with D(G) − 1 elements that are free of product-1 subsequences, i.e., there exists

S = (x1, . . . , xD(G)−1) sequence of G such that xi1xi2 · · ·xik 6= 1 for every non-empty

subset {i1, i2, . . . , ik} ⊂ {1, 2, . . . , D(G)− 1} . The Inverse Zero-Sum Problems study the

structure of these extremal sequences which are free of product-1 subsequences with some

prescribed property. For an overview on the inverse problems, see articles such as [36],

[73] and [35].

The inverse problems associated to the Davenport constant are solved for a few abelian

groups. The following theorem resolves the issue for the cyclic group Cn (see Appendix

F).

Theorem 1.1.1 ([10], see also [36, Theorem 2.1]). Let S be a sequence in Cn free of

product-1 subsequences, where n ≥ 3. Suppose that |S| ≥ (n + 1)/2. Then there exists

some g ∈ S with multiplicity ≥ 2|S| − n+ 1. In particular, D(Cn) = n and the following

statements hold:

(a) If |S| = n− 1 then S = (g)n−1, where ord(g) = n;

(b) If |S| = n− 2 then either S = (g)n−3(g2) or S = (g)n−2, where ord(g) = n;

(c) If |S| = n−3 then S has one of the following forms: (g)n−5(g2)2, (g)n−4(g2), (g)n−4(g3)

or (g)n−3, where ord(g) = n.

Observe that in the cyclic case, sequences free of product-1 subsequences contain an

element repeated many times. It is natural to ask if this is true in a general case.

For non-abelian groups, nothing about inverse problems was known, but in a joint

work with F. E. Brochero Martınez we solved the inverse problem associated to certain

1.1 Inverse zero-sum problems 9

types of metacyclic groups (see [11]) and all types of dihedral groups and dicyclic groups

(see [12]). Our results are reproduced in the next chapter (Sections 2.1, 2.2 and 2.3,

respectively), where we characterize the maximal sequences which are free of product-1

subsequences, and we show that these sequences have a property similar to the cyclic case.

In Chapter 3, we present the Properties B, C, D, which are directly related to the

inverse zero-sum problems. In the appendix, we present all the auxiliary results already

known and that will be used in the next chapter.

2Inverse zero-sum problems in some

non-abelian groups

2.1 Extremal product-one free sequences in Metacyclic

Groups

Let q be a prime, m ≥ 2 be a divisor of q−1 and s ∈ Z∗q such that ordq(s) = m. Denote

by Cq os Cm the metacyclic group written multiplicatively, i.e., the group generated by x

and y with relations:

xm = 1, yq = 1, yx = xys, where sm ≡ 1 (mod q). (2.1)

For the metacyclic group Cq os Cm, let

• HM be the normal cyclic subgroup of order q generated by y;

• NM = (Cq os Cm) \HM = N1 ∪N2 ∪ · · · ∪Nm−1, where Ni = xi ·HM .

In [6], J. Bass showed that D(Cq os Cm) = m + q − 1 where q ≥ 3 is a prime and

ordq(s) = m ≥ 2. In a joint work with F. E. Brochero Martınez [11], we proved the

following result, reproduced here:

Theorem 2.1.1. Let q be a prime, m ≥ 2 be a divisor of q − 1 and s ∈ Z∗q such that

ordq(s) = m. Let SM be a sequence in the metacyclic group Cq os Cm with m + q − 2

elements.

1. If (m, q) 6= (2, 3) then the following statements are equivalent:

2.1 Extremal product-one free sequences in Metacyclic Groups 11

(i) SM is free of product-1 subsequences;

(ii) For some 1 ≤ t ≤ q − 1, 1 ≤ i ≤ m − 1 such that gcd(i,m) = 1 and

0 ≤ ν1, . . . , νm−1 ≤ q − 1,

SM = (yt, yt, . . . , yt︸︷︷︸q−1 times

, xiyν1 , xiyν2 , . . . , xiyνm−1).

2. If (m, q) = (2, 3) then SM is free of product-1 subsequences if and only if

SM = (yt, yt, xyν) for t ∈ {2, 3} and ν ∈ {0, 1, 2} or SM = (x, xy, xy2).

2.1.1 Auxiliary results

In this subsection we present the auxiliary theorems and lemmas that we use through-

out this chapter.

A very fundamental result on sum-sets is the Cauchy-Davenport theorem, which gives

a lower bound for the number of elements of a sum-set in Zq depending on the cardinality

of each set. For its proof, see Appendix B.

Theorem 2.1.2 (Cauchy-Davenport inequality, [65, p. 44-45]). For q a prime number

and for any r non-empty sets X1, . . . , Xr ⊂ Zq,

|X1 + · · ·+Xr| ≥ min{q, |X1|+ · · ·+ |Xr| − r + 1}.

Looking at the inequality above in the case that r = 2 we get |X+Y | ≥ min{q, |X|+|Y | − 1}. A pair of subsets X, Y of Zq is called a critical pair if the equality |X + Y | =

min{q, |X|+|Y |−1} occurs. The following theorem provides criteria for a pair X, Y ⊂ Zqto be a critical pair. For its proof, see Appendix C.

Theorem 2.1.3 (Vosper, [84]). Let q be a prime number and let X, Y non-empty subsets

of Zq. Then

|X + Y | = min{q, |X|+ |Y | − 1}

if and only if one of the following conditions is satisfied:

(a) |X|+ |Y | > q;

(b) min{|X|, |Y |} = 1;

(c) |X + Y | = q − 1 and Y = Zq \ {c− a| a ∈ X}, where {c} = Zq \ (X + Y );

(d) X and Y are arithmetic progressions with the same common difference.


Notice that assertion (a) above is the only one giving the equality |X + Y | = q, that

is, X + Y = Zq. Assertion (b) means that we just translate the set Y , supposing |X| = 1.

The non-trivial cases yielding a critical pair are (c) and (d).

Now, suppose that A is a sequence in Cq os Cm and that A has no elements in the

normal subgroup HM , but that the product of its elements is in HM . The next lemma

shows that if A is minimal with these properties then it is possible to generate at least

|A| distinct products in HM , just shifting the order of the product. Its proof is contained

in the proof of Lemma 14 in [6] and we reproduce in detail here.

Lemma 2.1.4. Let A = (a1, a2, . . . , al) be a sequence in NM ⊂ CqosCm, where ordq(s) =

m, such that π(A) ∈ HM but no subsequence of A has product in HM . Then:

(a) πn(A) ∈ HM for every 0 ≤ n ≤ l − 1;

(b) πi(A) 6= πj(A) for every 0 ≤ i < j ≤ l − 1.

Proof. (a) Since π(A) ∈ HM , HM C Cm os Cq and

πn(A) = an+1 . . . ala1 . . . an = (a1 . . . an)−1π(A)(a1 . . . an),

we obtain πn(A) ∈ HM for every 0 ≤ n ≤ l − 1.

(b) Suppose that πi(A) = πj(A) = yu for some 0 ≤ i < j ≤ l − 1. If u ≡ 0 (mod q) then

A contains a product-1 subsequence. So we assume u 6≡ 0 (mod q).

We claim that z = ai+1 . . . aj ∈ HM , and this contradicts the minimality of A. In

fact, we have

zyu = ai+1 . . . ajaj+1 . . . ala1 . . . aiai+1 . . . aj = yuz.

Since HM C Cm os Cq, there exist v ∈ Zq such that zyz−1 = yv, which implies

yu = zyuz−1 = yuv =⇒ u ≡ uv (mod q).

Since q is prime and u 6≡ 0 (mod q), we obtain v ≡ 1 (mod q) and so zy = yz. On

the other hand, z = xαyβ for some α ∈ Zm and β ∈ Zq. Hence

xαyβ = z = y−1zy = y−1xαyβ+1 =⇒ yxα = xαy.

Since x−1yx = ys we obtain ysα

= x−αyxα = y, thus sα ≡ 1 (mod q). Since ordq(s) =

m we obtain α ≡ 0 (mod m), which implies ai+1 . . . aj = z = yβ ∈ HM .


Throughout this section, we deal with many expressions of the type a0 + a1s + · · · +am−1s

m−1 (mod q), where a0, a1, . . . , am−1 ∈ Zq. The following lemma explains why we

can assume without loss of generality that this kind of expression has a value different

than 0 under a weak assumption.

Lemma 2.1.5. Let A = (a0, a1, . . . , am−1) be a sequence in Zq with at least 2 distinct

elements, say ai 6≡ aj (mod q). Suppose that ordq(s) = m. Then

m−1∑k=0

aksk 6≡ ais

j + ajsi +

m−1∑k=0k 6=i,j

aksk (mod q).

In particular, if one of them is 0 modulo q then the other is not 0 modulo q.

Proof. Since at least two elements are distinct, there exists 0 ≤ i ≤ m − 1 such that

ai 6≡ ai+1 (mod q) (assuming that the index of the coefficients are taken modulo m, i.e.,

a0 = am). Since ordq(s) = m, we may translate the coefficients by multiplying by s,

therefore we may assume without loss of generality that i = 0, i.e., a0 6≡ a1 (mod q).

Now, ifm−1∑k=0

aksk ≡ a0s+ a1 +

m−1∑k=2

aksk (mod q)

then

a0 + a1s ≡ a1 + a0s (mod q) ⇐⇒ (a0 − a1)(1− s) ≡ 0 (mod q),

a contradiction.

One of the assertions from Vosper’s theorem says that if X and Y are arithmetical

progressions then X and Y form a critical pair. On the other hand, if α = a0 + a1s +

· · · + am−1sm−1 ∈ Zq then the set {αsj}j=0,1,...,m−1 is a geometric progression. Since we

have the freedom to choose the order of the products, it is expected to deal with critical

pairs that are formed by geometric progressions. The following lemma shows under some

conditions that, in Zq, an arithmetic progression cannot be a geometric progression.

Lemma 2.1.6. Let q ≥ 5 be a prime number, s ∈ Z∗q \ {1} and 2 ≤ k ≤ q − 1. Let

A = {1, 2, 3, . . . , k − 1} ⊂ Zq,

B = {k, k + 1, k + 2, . . . , q − 1} ⊂ Zq.

Then A and B are not invariant by multiplication by s.


Proof. Since A and B are complementary in Z∗q, they are both s-invariants or not simul-

taneously. Suppose that they are s-invariants.

As 1 · s ∈ A and (q − 1) · s ∈ B, we obtain 2 ≤ s ≤ min{k − 1, q − k}, therefore we

may assume without loss of generality k − 1 ≤ q − k. So 2 ≤ s ≤ k − 1 ≤ (q − 1)/2.

Let c ≡ sordq(s)−1 (mod q) be the inverse of s modulo q. Since sj ∈ A for all j ∈ N and

s 6≡ 1 (mod q), we obtain c ∈ A and c 6≡ 1 (mod q), thus c − 1 ∈ A, which implies

s · (c− 1) ≡ 1− s ∈ A. But 1− s has a representative in B, which is a contradiction.

The next lemma gives both upper and lower bounds for the number of solutions

(z, w) ∈ Z2q of the equation az2 − bw4 ≡ c (mod q) when q ≡ 1 (mod 4) and a, b ∈ Z∗q

are fixed. We use the upper bound to show that the set {c+ bw4| w ∈ Z∗q} contains both

quadratic and non-quadratic residues modulo q.

Lemma 2.1.7. Let q ≡ 1 (mod 4) be a prime number, a, b, c ∈ Z∗q and N be the number

of solutions (z, w) ∈ Z2q of az2 − bw4 ≡ c (mod q). Then

|N − q| < 3√q.

Proof. Direct consequence of Corollary D.4.2.

Corollary 2.1.8. Let q ≡ 1 (mod 4) be a prime such that q ≥ 13 and let b, c ∈ Z∗q .

Then there exist w1, w2 ∈ Z∗q such that c + bw41 is a quadratic residue and c + bw4

2 is a

non-quadratic residue.

Proof. Fix a ∈ Z∗q and denote by N the number of solutions of

az2 − bw4 ≡ c (mod q). (2.2)

Let w0 ∈ Z∗q such that ordq(w0) = 4. Since each solution of Equation (2.2) generates

seven other solutions (by switching z by −z and multiplying w by powers of w0), the

number of elements in the set

C = {az2| z ∈ Z∗q} ∩ {c+ bw4| w ∈ Z∗q}

is at most N/8. By the previous lemma, it follows that

|C| <q + 3

√q

8<q − 1

4= |{c+ bw4| w ∈ Z∗q}|,

therefore

{c+ bw4| w ∈ Z∗q} 6⊂ {az2| z ∈ Z∗q}.


Thus, selecting a being either a square or not, we obtain that the set {c+bw4| w ∈ Z∗q}contains quadratic residues and non-quadratic residues.

It follows from the corollary above and Lemma 2.1.5 that, in the case m = (q − 1)/2,

it is possible to get more than the m distinct values which were provided by Lemma 2.1.4.

Corollary 2.1.9. Let q ≡ 1 (mod 4) be a prime number such that q ≥ 13, s ∈ Z∗q, such

that ordq(s) = q−12

= m and a0, a1, . . . am−1 ∈ Zq. Suppose that each of the sets

{a0, a2, a4, . . . , am−2} and {a1, a3, a5, . . . , am−1}

has at least two distinct elements modulo q. Then the set

A ={aσ(0) + aσ(1)s+ · · ·+ aσ(m−1)s

m−1 ∈ Zq | σ is a permutation of (0, 1, . . . ,m− 1)}

has at least q − 1 distinct elements.

Proof. By Lemma 2.1.5, we can suppose without loss of generality that

α := a0 + a1s+ a2s2 + · · ·+ am−1s

m−1 6≡ 0 (mod q).

Notice that we can obtain αsj by shifting the coefficients, so αsj ∈ A for all 0 ≤ j ≤m− 1. By Lemma 2.1.4, these m elements are distinct.

Since ordq(s) = (q − 1)/2, s is a square modulo q and, indeed, s generates the

quadratic residues modulo q, therefore the elements αsj are all quadratic residues or

all non-quadratic residues depending on whether α is a square or not. Define

b ≡ a1s+ a3s3 + · · ·+ am−1s

m−1 (mod q),

c ≡ a0 + a2s2 + · · ·+ am−2s

m−2 (mod q).

In the same way, by Lemma 2.1.5 we can assume without loss of generality that b, c 6≡ 0

(mod q). Since c + bs2j ∈ A for all 0 ≤ j ≤ m/2, Corollary 2.1.8 tells us that the set

{c+ bs2j| 0 ≤ j ≤ m/2} contains a quadratic residue and a non-quadratic residue modulo

q. By multiplying by s, we obtain all quadratic residues and all non-quadratic residues

modulo q. Therefore, |A| ≥ q − 1.

Analogously to the previous corollary but now in the case m = q − 1, the next result

states that it is possible to split the coefficients into parts such that each part generates

at least m distinct values, also improving the result of Lemma 2.1.4.


Corollary 2.1.10. Let q ≡ 1 (mod 4) be a prime number such that q ≥ 13, s be a

generator of Z∗q and a0, a1, . . . aq−2 ∈ Zq. Suppose that each of the sets

{a0, a4, a8, . . . , aq−5}, {a1, a5, a9, . . . , aq−4}, {a2, a6, a10, . . . , aq−3} and {a3, a7, a11, . . . , aq−2}

has at least two distinct elements modulo q. Then each of the sets

Ae ={aσ(0) + aσ(2)s

2 + · · ·+ aσ(q−3)sq−3 ∈ Zq | σ is a permutation of (0, 2, . . . , q − 3)

},

Ao ={aτ(1)s+ aτ(3)s

3 + · · ·+ aτ(q−2)sq−2 ∈ Zq | τ is a permutation of (1, 3, . . . , q − 2)

}have at least q − 1 distinct elements.

Proof. This follows directly from the previous corollary.

2.1.2 Sequences in C5 os Cm

In our proof of Theorem 2.1.1, for the case q ≡ 1 (mod 4), we use Corollaries 2.1.9

and 2.1.10, where the hypothesis q ≥ 13 is necessary. In this subsection, we consider the

remaining case, i.e., q = 5. There are two possibilities for m, namely, m = 2 and m = 4.

For m = 2 the only possible value for s is 4 (mod 5), therefore we obtain the dihedral

group of order 10 (see Section 2.2). The next proposition deals with this case and shows

that if a sequence S is free of product-1 subsequences and satisfies some assumptions then

SM must contain an element in HM , the normal subgroup.

Proposition 2.1.11. Let SM be a sequence in C5 o4 C2 free of product-1 subsequences

and suppose that |SM | = 5. Then SM ∩HM 6= ∅.

Proof. Suppose that SM = (xyα1 , . . . , xyα5), where 0 ≤ α1 ≤ · · · ≤ α5 ≤ 4. If there exist

two identical elements in SM then their product is 1 and so SM is not free of product-1

subsequences. Thus, (α1, . . . , α5) = (0, 1, 2, 3, 4), therefore x · xy4 · xy · xy2 = 1. Hence,

SM is not free of product-1 subsequences, a contradiction.

For m = 4, the cases to consider are s ≡ 2 (mod 5) or s ≡ 3 (mod 5). The proposition

below shows that if SM is free of product-1 subsequences then SM cannot belong to a single

coset Ni, where gcd(i, 4) = 1.

Proposition 2.1.12. Let s ∈ {2, 3} and let SM be a sequence in C5 os C4 such that

|SM | = 7. If every element of SM belongs to a coset Ni, where i ∈ {1, 3}, then SM is not

free of product-1 subsequences.


Proof. Let SM = (xiyα1 , . . . , xiyα7), where i ∈ {1, 3} and 0 ≤ α1 ≤ · · · ≤ α7 ≤ 4. Notice

that at most three of the αj’s are equal, otherwise the product of four identical elements

would be 1. This implies that there are at least three distinct elements. Also, there are

at least a pair of elements repeating two or three times. Since ord5(s) = 4, it follows that

s3i0 + s2i0 + si0 + 1 ≡ 0 (mod 5). Therefore, for all β ∈ Zq it holds that:

xi0ya1 · xi0ya2 · xi0ya3 · xi0ya4 = ya1s3i0+a2s2i0+a3si0+a4

= ya1s3i0+a2s2i0+a3si0+a4−β(s3i0+s2i0+si0+1)

= xi0ya1−β · xi0ya2−β · xi0ya3−β · xi0ya4−β,

thus we may assume without loss of generality that the element that repeats most is

α1 = 0. If there is another pair of identical elements, say 1 ≤ αj = αj+1 = λ ≤ 4 for some

3 ≤ j ≤ 6, then we may choose A1 = (x, xyλ, x, xyλ) or A1 = (x3, x3yλ, x3, x3yλ). Since

ord5(s) = 4 and gcd(i, 4) = 1, it follows that λs2i+λ ≡ 0 (mod 5), and so xi·xiyλ·xi·xiyλ =

1. Otherwise, the only possibility is (α1, . . . , α7) = (0, 0, 0, 1, 2, 3, 4) and we may choose,

for example,

A1 = (x, x, xy, xy2) or A1 = (x, x, xy, xy3) when i = 1,

A1 = (x3, x3, x3y, x3y2) or A1 = (x3, x3, x3y, x3y3) when i = 3,

as the following table shows:

s = 2 s = 3

i = 1 si ≡5 2⇒ x · x · xy · xy3 = 1 si ≡5 3⇒ x · x · xy · xy2 = 1

i = 3 si ≡5 3⇒ x3 · x3 · x3y · x3y2 = 1 si ≡5 2⇒ x3 · x3 · x3y · x3y3 = 1

Thus, SM is not free of product-1 subsequences.

2.1.3 Proof of Theorem 2.1.1

In this subsection we investigate the sequences of Cq os Cm (where ordq(s) = m and

q is a prime) with m + q − 2 elements which are free of product-1 subsequences. Notice

that if (m, q) = (2, 3) then s ≡ 2 (mod 3) and the sequence S = (x, xy, xy2) provides the

only counter example.

It is easy to check that (ii) implies (i), therefore we just need to prove that (i) implies

(ii). Let us assume that SM is a sequence free of product-1 subsequences in Cq os Cm.


Let us also define k ∈ Z by the equation

|SM ∩HM | = q − k.

If k ≤ 0, since D(HM) = D(Cq) = q, there exists a product-1 subsequence in HM .

If k = 1 then |SM ∩ HM | = q − 1 and |SM ∩ NM | = m − 1. By Theorem 1.1.1, the

elements of SM ∩HM must all be equal, say, SM ∩HM = {yt}q−1 and the other elements

of SM must be in the same Ni, where gcd(i,m) = 1, since (Cq os Cm)/HM ' Cm.

From now on, assume k ≥ 2. In this case, we are going to prove that SM is not free of

product-1 subsequences. We have

|SM ∩NM | = m+ k − 2 ≥ m = D(Cm) = D((Cq os Cm)/HM).

Let A = (a1, . . . , al) be a subsequence of SM ∩ NM such that π(A) ∈ HM ' Cq but

no subsequence of A has product in HM , as in Lemma 2.1.4. We have that 2 ≤ |A| ≤ m.

Let us denote A by A1. If |(SM ∩ NM)A−11 | ≥ m then we can construct A2 with the

same property and repeat this argument replacing successively (SM ∩ NM)(A1 . . . Aj)−1

by (SM ∩NM)(A1 . . . AjAj+1)−1 and so on, until |(SM ∩NM)(A1 . . . Ar)

−1| ≤ m− 1. This

implies thatr∑i=1

|Ai| ≥ |SM ∩NM | − (m− 1) = k − 1.

We construct the following set of products:

R = ({πj(A1)}j=0,1,...,|A1|−1)·({1} ∪ {πj(A2)}j=0,1,...,|A2|−1) · . . .

· · · · ({1} ∪ {πj(Ar)}j=0,1,...,|Ar|−1) ⊂ HM .

By the Cauchy-Davenport inequality we obtain

|R| ≥ min

{q, |A1|+

r∑i=2

(|Ai|+ 1)− r + 1

}= min

{q,

r∑i=1

|Ai|

}.

If this minimum is q then R = HM 3 1 and SM is not free of product-1 subsequences.

On the other hand, if this minimum is∑r

i=1 |Ai| ≥ k − 1 then we obtain at least k − 1

distinct elements in HM arising from SM ∩ NM . Suppose without loss of generality that

SM ∩HM = (h1, h2, . . . , hq−k). If∑r

i=1 |Ai| ≥ k then, by the Pigeonhole Principle, either

1 ∈ R or R contains the inverse of one of the products h1, (h1h2), . . . , (h1 . . . hq−k) (which

we may assume that are all distinct amongst themselves), hence SM is not free of product-1

subsequences.


Therefore, suppose that∑r

i=1 |Ai| = k − 1, namely, R = {g1, g2, . . . , gk−1}. If there

exist 1 ≤ n1 ≤ k−1 and 1 ≤ n2 ≤ q−k such that gn1 = (h1 . . . hn2)−1 then gn1h1 . . . hn2 = 1

and so SM is not free of product-1 subsequences. Therefore,

{g1, . . . , gk−1, h−11 , (h1h2)−1, . . . , (h1 . . . hq−k)

−1} = {y, y2, . . . , yq−1}.

If hi 6= hj for some 1 ≤ i < j ≤ q − k, say without loss of generality that h1 6= h2,

then either the set

{g1, . . . , gk−1, h−11 , h−12 , (h1h2)−1, . . . , (h1 . . . hq−k)

−1}

has q elements, in particular, it has the element 1, or it contains two identical elements.

In any case, SM has a product-1 subsequence. Hence,

h1 = h2 = · · · = hq−k = yt

for some 1 ≤ t ≤ q − 1 and the only possibility for R not to contain the inverse of any of

the elements yt, y2t, . . . , y(k−1)t is:

R = {g1, g2, . . . , gk−1} = {yt, y2t, . . . , y(q−k)t}.

Let C := (SM ∩NM)(A1 . . . Ar)−1. Notice that C is free of subsequences with product

in HM . Since

|C| = |SM ∩NM | −r∑j=1

|Aj| = m+ k − 2− (k − 1) = m− 1,

we conclude, by Theorem 1.1.1, that C does not have subsequences with product in HM if

these m− 1 elements are in the same class Ni0 , where 1 ≤ i0 ≤ m− 1 and gcd(i0,m) = 1,

because ordm(i0) must be m. Thus, we assume that C is a sequence in SM ∩Ni0 , namely,

C = (xi0yt1 , xi0yt2 , . . . , xi0ytm−1).

If there exist 1 ≤ j ≤ r such that Aj has at least one element out of Ni0 , then we may

select xi1yθ ∈ Aj with i1 6= i0. If 2 ≤ l ≤ m − 1 is such that i0l ≡ i1 (mod m) then we

may replace Aj by

Aj = (Aj \ {xi1yθ}) ∪ (xi0yt1 , . . . , xi0ytl︸︷︷︸∈C

),


which also has product in HM , since

l∏n=1

xi0ytn = xi1yT

for some T ∈ Zq and HM C Cq os Cm. Since |Aj| > |Aj|, the new set R generated by this

change has more than k−1 elements. This replacement may mean that Aj is not minimal

anymore and in this case we break Aj into its minimal components with product in HM ,

say, Aj1 , Aj2 , . . . , Ajδ . We observe that |Aj| = |Aj1| + |Aj1 | + · · · + |Ajδ |, and repeat the

previous arguments with A1, . . . , Aj−1, Aj+1, . . . , Ar and the new Aji ’s. Since

r∑n=1n 6=j

|An|+δ∑

n=1

|Ajn| >r∑

n=1

|Aj| = k − 1,

there exists a product-1 subsequence in SM .

Hence, SM ∩NM must be a sequence in Ni0 , so

Aj (mod HM) = {xi0}m for all 1 ≤ j ≤ r, and C = (xi0yt1 , . . . , xi0ytm−1).

By double counting the number of elements in SM ∩Ni0 we conclude that m+k−2 ≡m − 1 (mod m), that is, k ≡ 1 (mod m). Since k ≥ 2 and k ≡ 1 (mod m), we have

m+ 1 ≤ k ≤ q.

Observe that the case m+ 1 ≤ k < q is not possible. In fact, in this case, if

Aj = (xi0yη1 , . . . , xi0yηm)

then

π(Aj) = yη1si0(m−1)+η2si0(m−2)+···+ηm−1si0+ηm

and, more generally,

πn(Aj) =(yη1s

i0(m−1)+η2si0(m−2)+···+ηm−1si0+ηm)si0n

.

We claim that R is invariant under taking powers of si0 . In fact, since gcd(i0,m) = 1

we obtain ordq(si0) = ordq(s) = m. An element in R is of the form

yjt = πj1(Aν1) . . . πju(Aνu),


where 1 ≤ j ≤ k − 1. Taking powers of si0 in both sides, we obtain that

ysi0jt = πj1+1(Aν1) . . . πju+1(Aνu)

belongs to R, because it can be obtained by the product of some Ai’s in some order.

Therefore, the claim is proved.

Looking at the exponent of y, the above claim implies that the set {t, 2t, . . . , (k− 1)t}is si0-invariant modulo q, and so {1, 2, . . . , k−1} is si0-invariant, which contradicts Lemma

2.1.6.

From now on, we assume k = q and SM is a sequence in Ni0 . As |Aj| = m for 1 ≤ j ≤ r

and |C| = m− 1 we obtain mr + m− 1 = |SM | = m + q − 2, therefore mr = q − 1. We

consider the following cases depending on whether r ≥ 3, r = 2 or r = 1:

(1) Case r ≥ 3: This implies that 3m ≤ q − 1. Since the equality in the Cauchy-

Davenport inequality occurs, Vosper’s theorem with the sets

A = {πn(A1)}n and B = ({1} ∪ {πn(A2)}n) · · · · · ({1} ∪ {πn(Ar)}n)

says that at least one of the following statements hold:

(i) |A|+ |B| > q;

(ii) min{|A|, |B|} = 1;

(iii) |A · B| = q − 1 and A = HM \ ({b−1| b ∈ B} ∪ {1});

(iv) A and B are geometric progressions with the same common ratio, or in other

words, the exponents of the elements in A and B form arithmetic progressions

with the same common difference.

Since |A| = m and |B| = q−k = q−m, we have |A|+ |B| = q, thus (i) is not possible,

and since min{|A|, |B|} = min{m, q −m} ≥ min{m, 2m + 1} = m ≥ 2, (ii) does not

hold. In order to discard item (iii), define

B1 = {1} ∪ {πn(A2)}n,

B2 = ({1} ∪ {πn(A3)}n) · · · · · ({1} ∪ {πn(Ar)}n).

As 3m ≤ mr = q − 1, we have

q −m = |B| = |B1 · B2| = |B1|+ |B2| − 1 < q − 1,

|B1| = m+ 1,

|B2| = q − 2m ≥ m+ 1.


Therefore, the items (i), (ii) and (iii) from Vosper’s theorem are false, hence the

exponents of the elements in each of the sets B1 and B2 form arithmetic progressions,

say

B1 = {y−av, . . . , y−v, 1, yv, . . . , y(m−a−1)v}.

Looking at the Vosper’s equality involving A and B, item (iv) tells us that the expo-

nents of the elements in A also form an arithmetic progression with the same common

difference, say

A = {yw, yw+v, . . . , yw+(m−1)v}.

By switching A1 and A2, the only possibility is that {πn(A1)} = {yv, y2v, . . . , ymv} for

some 1 ≤ v ≤ q − 1.

On the other hand, the exponents of the elements from the set {πn(A1)}n are si0-

invariant, that is, the exponents of y in R are invariant by multiplication by si0 . By

Lemma 2.1.6, the set {πn(A1)}n cannot be of the above form.

(2) Case r = 2: In this case, m = (q−1)/2 and, in particular, si0 generates the quadratic

residues modulo q. The case (m, q) = (2, 5) follows from Proposition 2.1.11, therefore

we may assume q ≥ 7. If there exist m identical elements then their product is 1,

thus there are at most m− 1 identical elements. Since

m+ q − 2

m− 1=

3m− 1

m− 1> 3,

there are at least 4 distinct elements among SM = {xi0yα1 , . . . , xi0yα3m−1}. We split

this case into the two subcases q ≡ 1 (mod 4) and q ≡ 3 (mod 4).

(2.1) Subcase q ≡ 1 (mod 4) and q ≥ 13: We may choose

A1 = (xi0ya0 , xi0ya1 , . . . , xi0yam−1),

A2 = (xi0yb0 , xi0yb1 , . . . , xi0ybm−1)

to be disjoint subsequences of SM such that it is possible to split each one into

two subsequences of the same size, say

A1 = (xi0ya0 , xi0ya2 , . . . , xi0yam−2) ∪ (xi0ya1 , xi0ya3 , . . . , xi0yam−1),

A2 = (xi0yb0 , xi0yb2 , . . . , xi0ybm−2) ∪ (xi0yb1 , xi0yb3 , . . . , xi0ybm−1),

where each partition has at least 2 distinct elements. This is possible since SM

has at most m − 1 identical elements. From Corollary 2.1.9, A1 and A2 each


generate at least q − 1 elements in HM , therefore the product-set A1 · A2 has

to be HM by the Cauchy-Davenport inequality, which implies that 1 ∈ A1 ·A2.

Thus SM is not free of product-1 subsequences.

(2.2) Subcase q ≡ 3 (mod 4) and q ≥ 7: It is known that −1 is not a quadratic

residue modulo q. Let

A1 = {xi0ya0 , xi0ya1 , . . . , xi0yam−1},

A2 = {xi0yb0 , xi0yb1 , . . . , xi0ybm−1},

where A1 and A2 are disjoint subsequences of SM and each one has at least 2

distinct elements. It is enough to consider the exponents of y and construct the

following sets of exponents:

X = {sni0(a0 + a1si0 + · · ·+ am−2s

i0(m−2) + am−1si0(m−1)) ∈ Zq| 0 ≤ n ≤ m− 1},

Y = {sni0(b0 + b1si0 + · · ·+ bm−2s

i0(m−2) + bm−1si0(m−1)) ∈ Zq| 0 ≤ n ≤ m− 1}.

Suppose that 0 6∈ X, 0 6∈ Y and 0 6∈ X + Y (otherwise we are done). Since

gcd(i0,m) = 1 and ordq(s) = m, we have ordq(si0) = m, therefore by Lemma

2.1.4 we obtain |X| = |Y | = m. Let

α = a0 + a1si0 + · · ·+ am−2s

i0(m−2) + am−1si0(m−1) ∈ X.

If −α ∈ Y then 0 ∈ X + Y , a contradiction, therefore −α ∈ X. Hence, there

exist 0 ≤ n ≤ m− 1 such that αs2ni0 ≡ −α (mod q), which implies s2ni0 ≡ −1

(mod q), so −1 is a quadratic residue modulo q, which is also a contradiction.

(3) Case r = 1: In this case, m = q − 1 and, in particular, m is even and si0 generates

Z∗q. The cases where (m, q) = (4, 5) follow from Proposition 2.1.12, therefore we may

assume q ≥ 7. If there exist m identical elements then their product is 1, thus there

are at most m− 1 identical elements. Since

m+ q − 2

m− 1=

2m− 1

m− 1> 2,

there are at least 3 distinct elements among SM = {xi0yα1 , . . . , xi0yα2m−1}. Again, we

split up this case into q ≡ 1 (mod 4) and q ≡ 3 (mod 4), as follows:

(3.1) Subcase q ≡ 1 (mod 4) and q ≥ 13: We may choose


A1 = {xi0ya0 , xi0ya1 , . . . , xi0yam−1}

to be a subsequence of SM that we can further split into two subsequences of

the same size (q − 1)/2, say

A1 = {xi0ya0 , xi0ya2 , . . . , xi0yam−2} ∪ {xi0ya1 , xi0ya3 , . . . , xi0yam−1},

satisfying a0 6≡ a2 (mod q) and a1 6≡ a3 (mod q). Since si0 generates Z∗q, s2i0

generates the quadratic residues of Z∗q. From Corollary 2.1.10, there exist two

permutations σ of (0, 2, . . . ,m− 2) and τ of (1, 3, . . . ,m− 1) such that

1 ≡ aσ(0) + aσ(2)s2i0 + · · ·+ aσ(2)s

(m−2)i0 (mod q),

−1 ≡ aτ(1)si0 + aτ(3)s

3i0 + · · ·+ aτ(m−1)s(m−1)i0 (mod q).

Therefore

1 = y · y−1 = yaσ(0)+aσ(2)s2i0+···+aσ(2)s(m−2)i0 · yaτ(1)si0+aτ(3)s3i0+···+aτ(m−1)s

(m−1)i0

= xi0yσ(0) · xi0yτ(1) · xi0yσ(2) · xi0yτ(3) . . . xi0yσ(m−2) · xi0yτ(m−1),

thus SM is not free of product-1 subsequences.

(3.2) Subcase q ≡ 3 (mod 4) and q ≥ 7: It is known that −1 is not a quadratic

residue modulo q. Let

A1 = {xi0ya0 , xi0ya1 , . . . , xi0yam−1},

where at least 3 of the aj’s are distinct. By considering the set of all products

obtained by changing the order of the elements of A1, we obtain the set

C = {απ = aπ(0) + aπ(1)si0 + · · ·+ aπ(m−1)s

(m−1)i0 ∈ Zq|

π is a permutation of (0, 1, . . . ,m− 1)}.

Hence, it is enough to consider the exponents απ of y. Reindexing the aj’s, we

may construct the set of exponents

X = {s2ni0(a0 + a2s2i0 + · · ·+ am−2s

i0(m−2)) ∈ Zq| 0 ≤ n ≤ m− 1},

Y = {s2ni0(a1si0 + a3s3i0 + · · ·+ am−1s

i0(m−1)) ∈ Zq| 0 ≤ n ≤ m− 1},

where X and Y each have at least 2 distinct elements. Clearly, X + Y ⊂ C.Suppose that 0 6∈ X+Y (otherwise we are done). By Lemma 2.1.4, |X+Y | = m,

2.2 Extremal product-one free sequences in Dihedral Groups 25

thus X + Y = Z∗q. By Lemma 2.1.5, we may assume without loss of generality

that |X| = m/2 and |Y | = m/2. Let

α = a0 + a2s2i0 + · · ·+ am−2s

i0(m−2) ∈ X.

If −α ∈ Y then 0 ∈ X + Y , a contradiction. Therefore −α ∈ X. Hence, there

exists 0 ≤ n ≤ m− 1 such that αs2ni0 ≡ −α (mod q), which implies s2ni0 ≡ −1

(mod q), so −1 is a quadratic residue modulo q, but this is impossible.

2.2 Extremal product-one free sequences in Dihedral

Groups

Let n ≥ 2. Denote by D2n ' (Cn o−1 C2) the Dihedral Group of order 2n, i.e., the

group generated by x and y with relations:

x2 = yn = 1, yx = xy−1.

For the dihedral group D2n, let

• HD be the normal cyclic subgroup of order n generated by y;

• ND = D2n \HD = x ·HD.

The product of any even number of elements in ND is in HD, since

xyα · xyβ = yβ−α. (2.3)

In [88], J. Zhuang and W. Gao showed that D(D2n) = n + 1 where n ≥ 3. Their

argument uses the fact that

D(G) ≤⌈|G|+ 1

2

⌉(2.4)

provided G is a finite non-abelian group (see, for example, [68]), therefore D(D2n) ≤ n+1.

The lower bound follows from the fact that the sequence S = (y)n−1(x) does not have

product-1 subsequences and |S| = n. In a joint work with F. E. Brochero Martınez [12],

we proved the following result, reproduced here:

Theorem 2.2.1. Let SD be a sequence in the dihedral group D2n with n elements, where

n ≥ 3.


1. If n ≥ 4 then the following statements are equivalent:

(i) SD is free of product-1 subsequences;

(ii) For some 1 ≤ t ≤ n− 1 with gcd(t, n) = 1 and 0 ≤ s ≤ n− 1,

SD = (yt, yt, . . . , yt︸︷︷︸n−1 times

, xys).

2. If n = 3 then SD is free of product-1 subsequences if and only if

SD = (yt, yt, xyν) for t ∈ {2, 3} and ν ∈ {0, 1, 2} or SD = (x, xy, xy2).

Notice that this theorem reduces to Theorem 2.1.1 in the case n prime taking m = 2,

which implies s ≡ −1 (mod n). For n ≥ 4, it is easy to check that (ii) =⇒ (i). Observe

that the case n = 3 is exactly the item (2) of Theorem 2.1.1, since D6 ' C3 o−1 C2. For

n = 2, we have D4 ' Z22 and it is easy to check that SD is free of product-1 subsequences

if and only if SD = (x, y), SD = (xy, y) or SD = (x, xy).

The main technical difficulty in our present proof lies in the fact that n may not be

prime, therefore we cannot use Cauchy-Davenport inequality and Vosper’s theorem, as

used in the proof of Theorem 2.1.1. Instead, we now exhibit the product-1 subsequences in

cases not covered by those given forms. In the next section we prove Theorem 2.2.1, solving

the extremal inverse zero-sum problem associated to Davenport constant for dihedral

groups.

The only auxiliary result we are going to use is the following theorem, known as

“Davenport constant of Zn with weights {±1}”. For its proof, see Appendix E.

Theorem 2.2.2 ([1, Lemma 2.1]). Let n ∈ N and (y1, . . . , ys) be a sequence of integers

with s > log2 n. Then there exist a nonempty J ⊂ {1, 2, 3, . . . , s} and εj ∈ {±1} for each

j ∈ J such that ∑j∈J

εjyj ≡ 0 (mod n).

2.2.1 Proof of Theorem 2.2.1

We just need to show that 1.(i) =⇒ 1.(ii). Let SD be a sequence inD2n with n elements

that is free of product-1 subsequences. If SD ∩ ND contains two identical elements then

SD is not free of product-1 subsequences by Equation (2.3). Hence, we assume that the

elements of SD ∩ND are all distinct.

From now on, we consider some cases, depending on the cardinality of SD ∩HD:


(a) Case |SD ∩HD| = n: In this case, SD is contained in the cyclic subgroup of order n.

Since D(Zn) = n, SD contains some non-empty subsequence with product 1.

(b) Case |SD∩HD| = n−1: In this case, by Theorem 1.1.1, the elements of SD∩HD must

all be equal, say, SD ∩HD = (yt)n−1 where gcd(t, n) = 1, and so SD = (yt)n−1(xys).

(c) Case |SD ∩HD| = n− 2: In this case, by Theorem 1.1.1, SD must have one of these

forms:

(c-1) Subcase SD = (yt)n−2(xyu, xyv): Notice that we can obtain the products

xyu · yt · yt . . . yt︸︷︷︸k times

·xyv = yv−u−kt.

Since 1 ≤ k ≤ n − 2, it’s enough to take k ≡ (v − u)t−1 (mod n). The only

problem occurs when k = n − 1, that is, when v − u + t ≡ 0 (mod n), but in

this case we switch u and v and so

xyv · yt · xyu = yu−v−t = 1.

(c-2) Subcase SD = (yt)n−3(y2t, xyu, xyv): Notice that we can obtain the products

xyu · yt · yt . . . yt︸︷︷︸k times

·xyv = yv−u−kt.

Since 1 ≤ k ≤ n− 1, it’s enough to take k ≡ (v − u)t−1 (mod n).

(d) Case |SD ∩ HD| = n − 3: In this case, by Theorem 1.1.1, SD must contain at least

n− 5 copies of some yt, where gcd(t, n) = 1. Also, suppose that

SD ∩ND = (xyα, xyβ, xyγ).

By renaming z = yt, we may assume without loss of generality that t = 1. By

Pigeonhole Principle it follows that there exist two exponents of y with difference in

{1, 2, . . . , bn/3c} (mod n), say,

α− β ∈ {0, 1, 2, . . . , bn/3c} (mod n).

We may ensure that

xyβ · yr · xyα = yα−β−r = 1


for some 1 ≤ r ≤ n/3 provided there are enough y’s. But if n ≥ 8 then r ≤ n/3 ≤n − 5, so there are enough y’s and the theorem follows in these cases. Since the

theorem is already proved for n prime, it only remains to prove for n ∈ {4, 6}.

For n = 4, we have SD = (xyα, xyβ, xyγ, y) and some of the α, β, γ are consecutives

modulo 4. Without loss of generality, suppose α−β ≡ 1 (mod 4), so xyβ ·y ·xyα = 1.

For n = 6, we have the cases

SD = (xyα, xyβ, xyγ, y, y, y);

SD = (xyα, xyβ, xyγ, y, y, y2);

SD = (xyα, xyβ, xyγ, y, y, y3); or

SD = (xyα, xyβ, xyγ, y, y2, y2).

If the set {α, β, γ} (mod 6) contains two consecutive elements, say α ≡ β+1 (mod 6),

then

xyα · xyβ · y = yβ+1−α = 1.

Otherwise, the only possibilities are {α, β, γ} (mod 6) = {0, 2, 4} or {α, β, γ} (mod 6) =

{1, 3, 5}. Suppose that α ≡ β + 2 (mod 6) and notice that in any option for SD it is

possible to take a product y2 coming from SD ∩HD. Therefore

xyα · xyβ · y2 = yβ+2−α = 1.

(e) Case |SD ∩HD| = n− k, 4 ≤ k ≤ n: Suppose that

SD ∩HD = (yt1 , yt2 , . . . , ytn−k),

SD ∩ND = (xyα1 , xyα2 , . . . , xyαk).

It follows from Theorem 2.2.2 that if

bk/2c > blog2 nc (2.5)

then there exist a linear combination of a subset of

{(α1 − α2), (α3 − α4), . . . , (α2bk/2c−1 − α2bk/2c)}

with coefficients ±1 summing 0. Suppose without loss of generality that, in this


combination,

(α1 − α2), (α3 − α4), . . . , (α2u−1 − α2u)

appear with sign −1 and

(α2u+1 − α2u+2), (α2u+3 − α2u+4), . . . , (α2v−1 − α2v),

appear with sign +1. Then

(xyα1 · xyα2) . . . (xyα2u−1 · xyα2u) · (xyα2u+2 · xyα2u+1) · (xyα2v · xyα2v−1) =

yα2−α1 . . . yα2u−α2u−1 · yα2u+1−α2u+2 . . . yα2v−1−α2v = 1

Thus the theorem is true for k > 2blog2 nc + 1.

Hence, we may assume 4 ≤ k ≤ 2blog2 nc+ 1. Theorem 1.1.1 implies without loss of

generality that ti = t for 1 ≤ i ≤ n − 2k + 1. By renaming z = yt, we may assume

without loss of generality that t = 1. Since k ≥ 4, Pigeonhole Principle implies that

there exist αi, αj such that

αi − αj ∈ {1, 2, . . . , bn/4c}.

Notice that if

n− 2k + 1 ≥ n/4 (2.6)

then

xyαj · yr · xyαi = yαi−αj−r = 1

for some 0 ≤ r ≤ bn/4c. But if n ≥ 8 then 3n ≥ 8 log2 n ≥ 4(k − 1), therefore

Equation (2.6) holds in this case. Since the theorem is already proved for n prime, it

only remains to prove for n ∈ {4, 6}.

For n = 4, the only possibility is SD = (x, xy, xy2, xy3). Thus, x · xy · xy3 · xy2 = 1.

For n = 6, there are three subcases to consider:

• Subcase k = 4: Let SD = (yt1 , yt2 , xyα1 , xyα2 , xyα3 , xyα4). Then either there

are two pairs of consecutive αi’s modulo 6 or there are three consecutive αi’s

modulo 6.

– If there are two pairs of consecutive αi’s, say α1 + 1 ≡ α2 and α3 + 1 ≡ α4

(mod 6), then

xyα1 · xyα2 · xyα4 · xyα3 = 1.

2.3 Extremal product-one free sequences in Dicyclic Groups 30

– If there are three consecutive αi’s, say α1 + 2 ≡ α2 + 1 ≡ α3 (mod 6), then

α4 can be any element in the set {α3 + 1, α3 + 2, α3 + 3}. If α4 ≡ α3 + 1

or α4 ≡ α3 + 3 (mod 6) then we return to the previous item. Therefore we

may assume α4 ≡ α3 +2 (mod 6). Taking the products xyαi ·xyαj = yαj−αi ,

we can get any element in {y, y2, y3, y4, y5}. For example,

xyα1 · xyα2 = y,

xyα1 · xyα3 = y2,

xyα2 · xyα4 = y3,

xyα1 · xyα4 = y4, and

xyα2 · xyα1 = y5.

Let i, j such that αi − αj ≡ t1 (mod 6). Then

xyαi · xyαj · yt1 = yαj−αi+t1 = 1.

• Subcase k = 5: Let SD = (yt, xyα1 , xyα2 , . . . , xyα5). In this case, there are four

consecutive αi’s modulo 6, say α1 + 3 ≡ α2 + 2 ≡ α3 + 1 ≡ α4 (mod 6). Then

xyα1 · xyα2 · xyα3 · xyα4 = yα2−α1+α4−α3 = 1.

• Subcase k = 6: The only possibility is SD = (x, xy, xy2, xy3, xy4, xy5). So

x · xy · xy3 · xy2 = 1.

2.3 Extremal product-one free sequences in Dicyclic

Groups

Let n ≥ 2. Denote by Q4n the Dicyclic Group, i.e., the group generated by x and y

with relations:

x2 = yn, y2n = 1, yx = xy−1.

We have Z(Q4n) = {1, yn}, where Z(G) denotes the center of a group G. In addition:

Q4n/{1, yn} ' D2n,


where D2n is the dihedral group of order 2n.

For the dicyclic group Q4n, let

• HQ be the normal cyclic subgroup of order 2n generated by y;

• NQ = Q4n \HQ = x ·HQ.

The product of any even number of elements in NQ is in HQ, since

xyα · xyβ = yβ−α+n. (2.7)

In [6], J. Bass showed that D(Q4n) = 2n+1 where n ≥ 3. His argument uses Inequality

(2.4), which implies D(Q4n) ≤ 2n + 1. The lower bound follows from the fact that the

sequence S = (y)2n−1(x) does not have product-1 subsequences and |S| = 2n. Almost

as a consequence of Theorem 2.2.1, we obtain the following result, proved in [12] and

reproduced here:

Theorem 2.3.1. Let SQ be a sequence in the dicyclic group Q4n with 2n elements, where

n ≥ 2.

1. If n ≥ 3 and |SQ| = 2n then the following statements are equivalent:

(i) SQ is free of product-1 subsequences;

(ii) For some 1 ≤ t ≤ n− 1 with gcd(t, 2n) = 1 and 0 ≤ s ≤ 2n− 1,

SQ = (yt, yt, . . . , yt︸︷︷︸2n−1 times

, xys).

2. If n = 2 then SQ is free of product-1 subsequences if and only if, for some r ∈ Z∗4and s ∈ Z4, SQ has one of the forms

(yr, yr, yr, xys), (yr, xys, xys, xys) or (xys, xys, xys, xyr+s).

Again, if n ≥ 3 then it is easy to check that (ii) =⇒ (i), therefore we just need

to show that (i) =⇒ (ii). If n = 2 then it is easy to check that the sequences of the

form (yr, yr, yr, xys), (yr, xys, xys, xys), (xys, xys, xys, xyr+s) are free of product-1 subse-

quences, therefore we just need to show that the other sequences with 2n elements have

subsequences with product 1.

Also note that if n = 2 then Q8 is isomorphic to the Quaternion Group, i.e. the group

defined by

〈e, i, j, k| i2 = j2 = k2 = ijk = e, e2 = 1〉.


This isomorphism may be described, for example, by x 7→ i, y 7→ j (or its natural permu-

tations, by the symmetry of i, j, k). The above theorem says that, in terms of Quarternion

Group, extremal sequences free of product-1 subsequence are of the forms

±(i, i, i,±j), ±(i, i, i,±k), ±(j, j, j,±i),±(j, j, j,±k), ±(k, k, k,±i), ±(k, k, k,±j).

The main technical difficulty in our present proof lies on the fact that n may not be

prime, therefore we cannot use Cauchy-Davenport inequality and Vosper’s theorem, as

used in the proof of Theorem 2.1.1. Instead, we now exhibit the product-1 subsequences

in cases not covered by those given forms. The proof of Theorem 2.3.1 is split up into

Subsections 2.3.1, 2.3.2 and 2.3.3, where we solve the extremal inverse zero-sum problem

associated to the Davenport constant for dicyclic groups of orders 4n, where n ≥ 4, n = 3

and n = 2, respectively. The case n ≥ 4 is a direct consequence of Theorem 2.2.1 and the

case n = 3 follows from Theorem 2.2.1, but in the special case (2). The proof in the case

n = 2 is done manually, using only Theorem 1.1.1 to reduce the number of cases, without

using Theorem 2.2.1.

2.3.1 Proof of Theorem 2.3.1: Case n ≥ 4

We just need to prove that (i) =⇒ (ii). Let SQ be a sequence in Q4n with 2n ele-

ments that is free of product-1 subsequences. We consider some cases depending on the

cardinality of SQ ∩HQ:

(1.1) Case |SQ ∩HQ| = 2n: In this case, SQ is contained in the cyclic subgroup of order

2n. Since D(HQ) = D(Z2n) = 2n, SQ contains some non-empty subsequence with

product 1.

(1.2) Case |SQ ∩HQ| = 2n− 1: In this case, by Theorem 1.1.1, the elements of SQ ∩HQ

must all be equal, say, SQ ∩ HQ = (yt)2n−1 where gcd(t, 2n) = 1, and so SQ =

(yt)2n−1(xys).

(1.3) Case |SQ ∩ HQ| = 2n − 2: In this case, let S1 be a subsequence of SQ such that

|S1∩HQ| = n−2 and let S2 = SQS−11 . Then S2 is a sequence in HQ with n elements.

Since

Q4n/{1, yn} ' D2n,

Theorem 2.2.1 tells us that S1 and S2 must contain subsequences T1 and T2, respec-

tively, with products in {1, yn}. If some of these products is 1 then we are done.


Otherwise, both products are yn, therefore∏z∈T1T2

z = y2n = 1,

thus SQ is not free of product-1 subsequences.

(1.4) Case |SQ ∩ HQ| = 2n − 3: In this case, let S1 be a subsequence of SQ such that

|S1 ∩ HQ| = n − 3 and let S2 = SQS−11 . Then S2 is a sequence in HQ with n

elements. The argument is similar to the above case, thus SQ is not free of product-

1 subsequence.

(1.5) Case |SQ ∩HQ| = 2n− k, 4 ≤ k ≤ 2n: In this case, let S1 be a subsequence of SQ

such that |S1 ∩HQ| ≤ n− 2 and S2 = SQS−11 is such that |S2 ∩HQ| ≤ n− 2. The

argument is similar to the above cases, thus SQ is not free of product-1 subsequence.

Therefore, the proof for n ≥ 4 is complete.

2.3.2 Proof of Theorem 2.3.1: Case n = 3

We have Q12 = 〈x, y| x2 = y3, y6 = 1, yx = xy5〉 and we just need to prove that

(i) =⇒ (ii). Let SQ be a sequence in Q12 with 6 elements that is free of product-1

subsequences. We consider some cases depending on the cardinality of SQ ∩HQ:

(2.1) Case |SQ ∩HQ| = 6: In this case, SQ must contain a product-1 subsequence, since

D(HQ) = D(Z6) = 6.

(2.2) Case |SQ ∩HQ| = 5: In this case, Theorem 1.1.1 says that SQ ∩HQ = (yr)5 where

r ∈ {1, 5}, therefore SQ is of the form (yr)5(xys).

(2.3) Case |SQ ∩ HQ| = 4: In this case, we decompose SQ = S1S2 where |Si| = 3 for

i ∈ {1, 2}, |S1 ∩HQ| = 1 and |S2 ∩HQ| = 3, and use the same argument than item

(1.3), therefore SQ is not free of product-1 subsequences.

(2.4) Case |SQ ∩ HQ| = 3: In this case, we decompose SQ = S1S2 where |Si| = 3 for

i ∈ {1, 2}, |S1 ∩HQ| = 1 and |S2 ∩HQ| = 2, therefore

S1 (mod {1, y3}) = (yr, xyu, xyv) and

S2 (mod {1, y3}) = (yt, yt, xys) for r, t ∈ {1, 2} and s, u, v ∈ {0, 1, 2}.


Notice that S1 contains a subsequence with product in {1, y3}. Observe that if r 6= t

then we could also decompose SQ = S ′1S′2 where

S ′1 = S1(yt)(yr)−1 and S ′2 = S2(y

r)(yt)−1.

So, S ′1 and S ′2 have subsequences with products in {1, y3} and we can use the same

argument than item (1.3). Therefore, r = t and we can decompose SQ = S ′′1S′′2 such

that

S ′′1 (mod {1, y3}) = (yt)3 and S ′′2 (mod {1, y3}) = (xys, xyu, xyv).

Notice that S ′′1 contains a subsequence with product in {1, y3} and S ′′2 does not

contains a subsequence with product in {1, y3} if and only if S ′′2 (mod {1, y3}) =

(x, xy, xy2). Hence, the only possibility for SQ (mod {1, y3}) is:

SQ (mod {1, y3}) = (yt)3(x, xy, xy2),

and so the possibilities for SQ ∩HQ are

(y)3, (y2)3, (y4)3, (y5)3, (y)2(y4), (y)(y4)2, (y2)2(y5) or (y2)(y5)2.

Notice that the second, third, fifth and eighth possibilities contain subsequences

with product 1, therefore it only remains

(y)3, (y5)3, (y)(y4)2 or (y2)2(y5).

Observe that in every case, it is possible to find a subsequence with product either y

or y5. We claim that SQ ∩NQ contains subsequences with product y and y5, which

we can join with those y or y5 coming from SQ∩HQ to get a product-1 subsequence.

For this, a sufficient condition is the existence of two elements in SQ∩NQ such that

the exponents of y have difference 2, since xyα · xyα+2 = y and xyα+2 · xyα = y5. In

fact, if xyβ ∈ SQ ∩NQ and xyβ−2, xyβ+2 6∈ SQ ∩NQ then xyβ+1, xyβ−1 ∈ SQ ∩NQ,

and so (β + 1)− (β − 1) = 2.

(2.5) Case |SQ ∩ HQ| = 2: In this case, we decompose SQ = S1S2 where |Si| = 3 and

|Si ∩HQ| = 1, and use the same argument than item (1.3), therefore SQ is not free

of product-1 subsequences.


(2.6) Case |SQ ∩ HQ| = 1: In this case, we decompose SQ = S1S2 where |Si| = 3,

|S1 ∩ HQ| = 1 and |S2 ∩ HQ| = 0. Notice that S2 contains a subsequence with

product in {1, y3}. The same argument than item (1.3) does not apply if and only if

S2 (mod {1, y3}) = (x, xy, xy2). Observe that we could also decompose SQ = S ′1S′2

in such way that |S ′i| = 3 for i ∈ {1, 2}, |S ′1 ∩ HQ| = 1, |S ′2 ∩ HQ| = 0 and S ′2

(mod {1, y3}) contains two elements with the same exponent in y. Therefore, the

same argument than item (1.3) applies for S ′1 and S ′2.

(2.7) Case |SQ ∩ HQ| = 0: In this case, we decompose SQ = S1S2 where |Si| = 3 and

|Si ∩ HQ| = 0 for i ∈ {1, 2}. Notice that Si contains a subsequence with product

in {1, y3} if and only if Si (mod {1, y3}) is not of the form (x, xy, xy2). Hence, the

same argument than item (1.3) does not apply if and only if Si (mod {1, y3}) =

(x, xy, xy2) for i ∈ {1, 2}. Observe that we could also decompose SQ = S ′1S′2 in

such way that S ′1 (mod {1, y3}) = (x, x, xy) and S ′2 (mod {1, y3}) = (xy, xy2, xy2).

Therefore, the same argument than item (1.3) applies for S ′1 and S ′2.

Therefore, the proof for n = 3 is complete.

2.3.3 Proof of Theorem 2.3.1: Case n = 2

We have Q8 = 〈x, y| x2 = y2, y4 = 1, yx = xy3〉. Suppose that SQ is a sequence in

Q8 with 4 elements that is free of product-1 subsequences. We want to show that SQ has

some of those forms given in item (2). For this, we consider some cases depending on the

cardinality of SQ ∩HQ:

(3.1) Case |SQ ∩HQ| = 4: In this case, SQ must contain a product-1 subsequence, since

D(HQ) = D(Z4) = 4.

(3.2) Case |SQ ∩HQ| = 3: In this case, Theorem 1.1.1 says that SQ ∩HQ = (yr)3 where

r ∈ {1, 3}, therefore SQ is of the form (yr)3(xys).

(3.3) Case |SQ ∩HQ| = 2: In this case, the only possibilities for SQ ∩HQ making SQ be

free of product-1 subsequences are

(y)2, (y, y2), (y3)2 and (y2, y3),

and in all these possibilities SQ ∩ HQ possesses a subsequence with product y2,

namely y · y, y3 · y3 or y2 itself. On the other hand, the possibilities for SQ ∩NQ are

(x, xy), (x, xy2), (x, xy3), (xy, xy2), (xy, xy3), (xy2, xy3) and (xys, xys)


for s ∈ Z4.

The cases (xys, xys) can be eliminated, since xys · xys · y2 = 1.

The cases (x, xy2) and (xy, xy3) can also be eliminated, since xys · xys+2 = 1.

The other cases can be eliminated by the following table:

XXXXXXXXXXXXXXXSQ ∩NQ

SQ ∩HQ(y)2 (y, y2) (y3)2 (y2, y3)

(x, xy) x · xy · y = 1 x · xy · y = 1 xy · x · y3 = 1 xy · x · y3 = 1

(x, xy3) x · y · xy3 = 1 x · y · xy3 = 1 x · xy3 · y3 = 1 x · xy3 · y3 = 1

(xy, xy2) xy · xy2 · y = 1 xy · xy2 · y = 1 xy2 · xy · y3 = 1 xy2 · xy · y3 = 1

(xy2, xy3) xy2 · xy3 · y = 1 xy2 · xy3 · y = 1 xy3 · xy2 · y3 = 1 xy3 · xy2 · y3 = 1

(3.4) Case |SQ ∩ HQ| = 1: In this case, the possibilities for SQ ∩ HQ are (y), (y2) and

(y3). On the other hand, SQ ∩ NQ has three elements and, by Equation (2.7), we

may assume that x ∈ SQ. Thus, the possibilities for SQ ∩NQ are

(x, x, x), (x, x, xy), (x, x, xy2), (x, x, xy3), (x, xy, xy), (x, xy, xy2), (x, xy, xy3),

(x, xy2, xy2), (x, xy2, xy3), and (x, xy3, xy3).

If SQ contains (x, xy2) or (xy, xy3) then SQ is not free of product-1 subsequences,

since

x · xy2 = 1 = xy · xy3. (2.8)

Therefore, the remainder possibilities are (x, x, x), (x, x, xy), (x, x, xy3), (x, xy, xy)

and (x, xy3, xy3). Notice that these last four possibilities contains two identical

terms and two terms such that the exponents of y have difference 1 modulo 4. Since

xyα · xyα+1 · y = 1

xyα · xyα · y2 = 1

xyα+1 · xyα · y3 = 1,

we may discard these four cases. Therefore, the only remainder possibility is

(x, x, x), and so SQ = (y)(xys)3 or SQ = (y3)(xys)3.

(3.5) Case |SQ ∩ HQ| = 0: In this case, we also may assume x ∈ SQ. Therefore, the

possibilities for SQ are

(x, x, x, x), (x, x, x, xy), (x, x, x, xy2), (x, x, x, xy3), (x, x, xy, xy), (x, x, xy, xy2),


(x, x, xy, xy3), (x, x, xy2, xy2), (x, x, xy2, xy3), (x, x, xy3, xy3), (x, xy, xy, xy),

(x, xy, xy, xy2), (x, xy, xy, xy3), (x, xy, xy2, xy2), (x, xy, xy2, xy3),

(x, xy2, xy2, xy2), (x, xy2, xy2, xy3), (x, xy2, xy3, xy3), and (x, xy3, xy3, xy3).

If SQ contains two pairs of identical elements, say (x, x, xα, xα), then

x · x · xyα · xyα = 1,

so we remove (x, x, x, x), (x, x, xy, xy), (x, x, xy2, xy2) and (x, x, xy3, xy3).

If SQ contains some of the pairs (x, xy2) or (xy, xy3) then, by equations in 2.8, we

may remove other 11 possibilities, thus it only remains (x, x, x, xy), (x, x, x, xy3),

(x, xy, xy, xy) and (x, xy3, xy3, xy3). Therefore, SQ = (xys)(xyr+s)3 for r ∈ Z∗4 and

s ∈ Z4.

3The Properties B, C, D

First of all, let us give a definition: A minimal zero sequence S in a finite abelian group

G is a sequence such that the product of its elements is 1, but each proper subsequence

is free of product-1 subsequences.

In this chapter, we give a key definition of some properties that have a great importance

in the study of inverse zero-sum problems. So let G be a finite group. We say that G has

• Property B if every minimal zero sequence S in G with length |S| = D(G) − 1

contains some element with multiplicity exp(G)− 1;

• Property C if every sequence S in G with length |S| = η(G) − 1 which contains

no product-1 subsequences of length at most exp(G) has the form S = T exp(G)−1 for

some subsequence T of S;

• Property D if every sequence S in G with length |S| = s(G) − 1 which contains

no product-1 subsequences of length exp(G) has the form S = T exp(G)−1 for some

subsequence T of S.

These properties are well-studied for abelian groups of rank at most 2 and we believe

that they are actually theorems, at least for groups of the form G = Cdn, d ∈ N. Notice

that Theorem 1.1.1 states that Cn has Properties B and C.

It has been shown [36, Theorem 3.2] that Properties C and D are both multiplicative,

thus the verification of these properties are essentially reduced to the case of elementary

p-groups. Also, it has been proved [35] that Property B is multiplicative in the case

G = C2n. Again for the group C2

n, it is known that Property B implies Property C (see

[36]) and Property D implies Property C (see [33]). In [70], C. Reiher proved that if p is

a prime then C2p possesses Property B, which implies that C2

p also possesses Property C.

39

The conclusion is that C2n has Properties B and C. In [74], W. A. Schmid discusses the

case Cn×Cm, where n|m. Not much is known about groups of rank ≥ 3, only few specific

cases (see, for example, [74]). In [32, Theorem 6.4], W. Gao and A. Geroldinger proved

that Property B holds for certain groups such as p-groups, cyclic groups, groups with

rank two and groups that are the direct sum of two elementary p-groups, thus Property

C holds for the same groups as well.

Another type of inverse zero-sum problem, associated to the Erdos-Ginzburg-Ziv the-

orem, was proved by A. Bialostocki and P. Dierker in [9]. They established that if S is

a sequence in Cn with 2n − 2 elements and S is free of product-1 subsequences with n

elements, then

S = (g, h)n−1

for some g, h ∈ Cn with ord(gh−1) = n. Therefore, the group Cn possesses Property D.

However, in general very little is known about Property D. For example, if n ∈ {2, 3, 5, 7}then the group C2

n has Property D (see [76]) and if n = 5k then C3n has Property D (see

[37]).

There is still much to be done in this branch of zero-sum and inverse problems. Very

little is known about groups with rank greater than or equal to 3 and we hope that soon

we can contribute more, for example, by completing the verification of all these properties

for groups with rank 2, improving the current results regarding the groups with rank

greater than or equal to 3, and further, calculating D(G), s(G) and η(G) for more groups,

abelian or not.

Part II

SHIFTED TURAN SIEVE ON

TOURNAMENTS

Notations

In addition to the notations introduced in Part I, we shall also use the following:

• R is the set of real numbers;

• R+ is the set of positive real numbers;

• f(x) = O(g(x)) or f(x)� g(x) ⇐⇒ there exist C > 0 such that |f(x)| ≤ C · g(x)

for all x sufficiently large;

• f(x) = o(g(x)) ⇐⇒ limx→∞

f(x)

g(x)= 0;

• f(x) ∼ g(x) ⇐⇒ limx→∞

f(x)

g(x)= 1;

• Given m1, . . . ,mt ∈ N0 and 1 ≤ k ≤ t, let σk = σk(m1, . . . ,mt) be the coefficients

defined according to the following equation in Z[x]:

t∏i=1

(x+mi) =t∑

k=0

σkxt−k. (4.0)

4Introduction

Sieve theory is a set of general techniques in Number Theory (and currently in Com-

binatorics), designed to count, or more realistically, to estimate the size of some sets. The

primordial example of sieve is the sieve of Eratosthenes (around 3rd century B.C.), used

to generate prime numbers. Later A. M. Legendre generalized it in his study of the prime

counting function π(x) = #{p ≤ x | p is a prime}. Sieve methods bloomed and became

a topic of intense investigation after the pioneering work of V. Brun [16]. Since then, the

sieves have had a long and fruitful history.

There are a huge amount of problems that can be attacked via sieve methods. Four

of them became very famous in 1912 when, in his speech at the International Congress

of Mathematics, E. Landau characterised them as “unattackable at the present state of

mathematics”and now they are known as Landau’s problems. Currently, all four problems

remain unresolved. Landau’s problems are:

(a) Twin prime conjecture;

(b) Goldbach’s conjecture;

(c) n2 + 1 conjecture.

(d) Legendre’s conjecture;

In Sections 4.1 and 4.2 of this Introduction, we describe respectively the Twin prime

conjecture and Goldbach’s conjecture and the sieves used to provide partial solutions. In

Section 4.3, we describe the other Landau’s problems and some open problems on primes

related to sieve methods.

4.1 Twin prime conjecture 43

4.1 Twin prime conjecture

A twin prime is a prime p such that either p− 2 or p+ 2 is also prime. For example,

the smallests twin prime pairs are (3, 5), (5, 7), (11, 13), (17, 19), (29, 31) and (41, 43).

The current largest twin prime pair known is 2, 996, 863, 034, 895 × 21,290,000 ± 1. This

pair was discovered in September, 2016. Each number has 388, 342 decimal digits (see

http://www.primegrid.com/forum_thread.php?id=7021).

The question of whether there exist infinitely many twin prime pairs is known as the

Twin Prime Conjecture and has been one of the most famous open problems in Number

Theory for many years. In this section we discuss the main advances on this conjecture

and its relation to the sieve theory.

In 1919, V. Brun [14] created a new counting method called Brun Sieve to asymptot-

ically upper bound the twin primes up to x. This sieve involves the Inclusion-Exclusion

Principle and the Mobius function, but it is difficult to calculate the exact bound. Instead,

he used an expert truncation to get some inequalities involving Mobius function and it

gives the following bound:

#{p ≤ x | p and p+ 2 are both primes} � x(log log x)2

(log x)2.

In particular, this upper bound implies that:∑p is a prime;

p+2 is also prime

1

p<∞,

while the sum of reciprocals of prime numbers diverges (see [28]). With some refinements

on Brun sieve, it has been proved (see [15]) that:

#{p ≤ x | p and p+ 2 are both primes} � x

(log x)2.

Let p1 = 2, p2 = 3, p3 = 5, . . . be the sequence of primes in ascending order. Notice

that twin prime conjecture is equivalent to lim infn→∞(pn+1 − pn) = 2. The simplest

version of Prime Number Theorem (see [23] and [43]) states that:

limx→∞

π(x)

x/ log x= 1,

where π(x) = #{p ≤ x | p prime} is the prime counting function, and this implies that

lim infn→∞

pn+1 − pnlog pn

≤ 1.

http://www.primegrid.com/forum_thread.php?id=7021

4.1 Twin prime conjecture 44

In 1940, Erdos [26] improved this result for

lim infn→∞

pn+1 − pnlog pn

< 1.

Many improvements were done in order to lower the bound, until 2005 when D. A.

Goldston, J. Pintz, and C. Y. Yildirim [41] showed that

lim infn→∞

pn+1 − pnlog pn

= 0,

creating the GPY method. It was strengthened in 2006 (see [42]) by

lim infn→∞

pn+1 − pn√log pn(log log pn)2

<∞.

The GPY method depends on two main ingredients: Selberg Sieve (see, for example,

[21, Chapter 7]) and Bombieri-Vinogradov Theorem ([21, Chapter 9]). This method relies

on the distribution of primes in arithmetic progressions.

Given θ > 0, we say the primes have “level of distribution θ” if, for any A > 0, we have∑q≤xθ

maxgcd(a,q)=1

∣∣∣∣π(x; q, a)− π(x)

ϕ(q)

∣∣∣∣� x

(log x)A, (4.1)

where π(x; q, a) = #{p ≤ x | p prime and p ≡ a (mod q)}. Bombieri-Vinogradov theo-

rem establishes that the primes have level of distribution θ for any θ < 1/2. This theorem

basically states that the Generalized Riemann Hypothesis holds in average. P. D. T. A.

Elliott and H. Halberstam [24] conjectured that Inequality (4.1) could be extended to any

θ < 1. In their work, D. A. Goldston, J. Pintz and C. Y. Yildirim also established that if

the primes have“level of distribuition θ”for some θ > 1/2 then lim infn→∞(pn+1−pn) <∞.

In May, 2013, Y. Zhang [87] established a suitably weakened form of Inequality (4.1)

for some θ > 1/2, and so it was possible to show that

lim infn→∞

(pn+1 − pn) ≤ 7× 107.

For a few months there was a collective effort to lower the above bound in a project

called Polymath (https://en.wikipedia.org/wiki/Polymath_Project, see also [69]),

created by Tim Gowers and with great collaboration of Terence Tao. In November, 2013,

the bound was already 4680. At that time, J. Maynard [57] and T. Tao (unpublished)

independently announced an adjustment in Selberg sieve and obtained the bound of 600

without using the improvement of Bombieri-Vinogradov theorem given by Y. Zhang. Af-

terwards, many refinements were done until the bound got down to 246. Maynard’s work

also establishes that for any m ≥ 2, the quantity lim infn→∞(pn+m − pn) is finite and,

https://en.wikipedia.org/wiki/Polymath_Project

4.2 Goldbach conjecture 45

under Elliott-Halberstam conjecture, we have lim infn→∞(pn+1 − pn) ≤ 12.

On the other hand, in 1966, J. Chen [19] showed that there are infinitely many primes

p such that p− 2 possesses at most two prime factors. He used a sieve called Linear Sieve

(see, for example, [29]). This sieve is known to provide lower bounds.

There are also stronger conjectures concerning prime gaps. For example, one of the

Hardy-Littlewood conjectures (known as k-Tuple Conjecture, see [44]) claims that the

number of twin primes up to x is asymptotic to

2∏p primep>2

(1− 1

(p− 1)2

)· x

(log x)2.

In 1849, A. Polignac conjectured that, for any positive even number 2n, there are

infinitely many prime gaps of size 2n, or in other words, there are infinitely many cases

of two consecutive primes with difference 2n. The results mentioned above imply that

Polignac’s Conjecture holds for some 2 ≤ 2n ≤ 246.

4.2 Goldbach conjecture

One of the oldest and best-known unsolved problems in Number Theory is Goldbach’s

Conjecture. It possesses two versions that state:

• Goldbach’s Conjecture: Every even integer greater than 2 can be expressed as the

sum of two primes.

• Ternary (or Weak) Goldbach’s Conjecture: Every odd integer greater than 5 can be

expressed as the sum of three primes.

Adding the prime 3, it is obvious that Goldbach conjecture implies ternary Goldbach

conjecture. The first examples of even numbers that are sum of two primes are: 4 = 2+2,

6 = 3+3, 8 = 3+5, 10 = 3+7 = 5+5, 12 = 5+7, 14 = 3+11 = 7+7, 16 = 3+13 = 5+11,

18 = 5 + 13 = 7 + 11 and 20 = 3 + 17 = 7 + 13.

In 1937, I. M. Vinogradov [83] proved that all sufficiently large odd numbers can be

expressed as the sum of three primes, using a method called Hardy-Littlewood Method or

simply Circle Method (see, for example, [82]), which is different than the sieve method.

Vinogradov’s original proof did not give a bound for “sufficiently large”, but in 1956, his

student K. Borozdin proved that 3315 is large enough. This number has 6, 846, 169 decimal

digits, so checking every number under this figure would be completely infeasible. The

computations have only reached numbers up to 1030. In 2013, H. A. Helfgott released a

series of papers (see [46], [47], [48]) proving the ternary Goldbach’s conjecture.

4.3 Other problems 46

Towards the stronger conjecture, in 1966, Chen [19] proved that all sufficiently large

even number can be expressed as the sum of either two primes or a prime and a number

with at most two prime factors. His proof is almost identical to the proof of Chen’s

theorem stated in previous section. In 1975, H. L. Montgomery and R. C. Vaughan [59]

showed the set of even integers which are not the sum of two primes has density zero.

4.3 Other problems

Several other problems in Number Theory and other branches of mathematics can be

partially or totally solved using sieve method. In this section we introduce the Landau’s

problems (c) and (d) and also other problems which sieve method can be used to handle.

The Landau’s problem (c) asks whether there exist infinitely many primes p such that

p = n2 + 1. It is known since P. Fermat that every prime p ≡ 1 (mod 4) can be written

in the form p = a2 + b2 for some a, b ∈ Z, but no primes p ≡ 3 (mod 4) can be written as

p = a2 + b2. In 1998, using sieve method, J. Friedlander and H. Iwaniec [30] showed that

there exist infinitely many primes p of the form p = a2 + b4.

The Landau’s problem (d), called Legendre’s conjecture, asks whether for all n ∈ Nthere exist a prime p with n2 ≤ p ≤ (n + 1)2. In 1845, J. Bertrand [8] conjectured that

for all n ≥ 2 there exists a prime p such that n < p < 2n. This fact was proved by

P. Chebyshev [18] in 1852. The Prime Number theorem implies that for all ε > 0 there

exists a prime p with n < p < (1 + ε)n for n sufficietly large and, actually, the number

of primes between n2 and (n + 1)2 is asymptotic to n/ log n. If Legendre’s conjecture

is true then pn+1 − pn = O(√pn). H. Cramer conjectured that these gaps are always

much smaller, pn+1 − pn = O((log pn)2). If Cramer’s conjecture holds then Legendre’s

conjecture would follow for all n sufficiently large. H. Cramer [22] also proved that the

Riemann Hypothesis implies a weaker bound of O(√pn log pn) on the size of the largest

prime gaps. It is conjectured that there is always a prime in the interval [n, n+ n1/2] for

n sufficiently large. In 2001, R. C. Baker, G. Harman and J. Pintz [5] showed that there

is always a prime in the interval [n, n+ n21/40] for n sufficiently large.

Other related problems that can be thought to be solved through sieve method are

the followings:

1. Are there infinitely many prime numbers p such that 4p+ 1 is also prime?

2. Are there infinitely many prime numbers p such that 2p+ 1 is also prime?

The first one is related to an important conjecture, known as Artin’s primitive root

conjecture. This conjecture states that, for all a ∈ Z with a 6∈ {−1, 0, 1}, there exists

4.3 Other problems 47

infinitely many primes p such that a is a primitive root modulo p. It has been proved (see

[13, Chapter 1]) that if the problem (1) is true then Artin’s conjecture holds for a = 2.

The primes p such that 2p+ 1 is also prime are called Sophie Germain primes. In the

early 19th century, S. Germain proved that if p > 2 is a Sophie Germain prime and p does

not divide the product xyz then the Fermat’s equation

xp + yp = zp

has no solution. A.-M. Legendre proved that the Fermat’s equation has no solution when

4p + 1, 8p + 1, 10p + 1, 14p + 1 or 16p + 1 is also prime. So, they got the Fermat’s Last

Theorem for all prime p < 100. Sophie Germain primes also have a great importance in

Cryptography and Primality Tests (see, for example, [2]). Using Brun sieve, it is possible

to show that the number of pair of primes (p, q) that satisfy the conditions ap − bq = c

and p ≤ x is bounded above by O(x/(log x)2).

Sieve theory also guarantees some strong results such as Bombieri-Vinogradov theorem

that we saw before. The main tool used to prove this theorem is the Large Sieve. Large

Sieve is also used to prove the Brun-Titchmarsh inequality (see, for example, [79] and

[60]), that states:

π(x+ y; q, a)− π(x; q, a) ≤ 2y

ϕ(q) log(y/q),

provided gcd(a, q) = 1 and 1 ≤ q < y ≤ x.

In the next chapter, in a joint work with W. Kuo, Y.-R. Liu and K. Zhou, we present

the Turan sieve in a combinatorial context, as developed by Y.-R. Liu and M. R. Murty

(see [54] and [55]). We also construct its shifted version, and apply it to counting pro-

blems on tournaments in graph theory (i.e., complete directed graph) according to the

number of cycles. More precisely, we obtain upper bounds for the number of tournaments

which contain a small number of restricted r-cycles (in case of normal or multipartite

tournaments) or unrestricted r-cycles (in case of bipartite tournaments), as done in [53].

In Chapter 6, we apply the Turan sieve to a certain zero-sum problem concerning

EGZ constant, defined in Chapter 1. This application does not produce any extraordinary

result, because of the lack of knowledge on the number of solutions of a given system of

equations modulo n, but we hope that in the future this may perhaps be availed as a link

between zero-sum problems and sieve methods.

5The shifted Turan sieve method on

tournaments

5.1 The shifted Turan sieve

In 1934, P. Turan [80] gave a very simple proof of a celebrated result of G. H. Hardy

and S. Ramanujan [45] that the normal order of distinct prime factors of a natural number

n is log log n. If ω(n) denotes the number of distinct prime factors of n, P. Turan proved

that ∑n≤x

(ω(n)− log log x

)2 � x log log x;

from which the normal order of ω(n) is easily deduced. The above expression states that,

for n ≤ x, the “average” of ω(n) is log log x and the “expectation” is bounded above by

x log log x. Turan’s original derivation of the Hardy-Ramanujan theorem was essentially

probabilistic and concealed in it an elementary sieve method. In [54], Y.-R. Liu and M.

R. Murty introduced the Turan sieve method and applied it to probabilistic Galois theory

problems. This method was further generalized to a bipartite graph in [55] to investigate

a variety of combinatorial questions, including graph colourings, Latin squares, etc. In a

joint work with W. Kuo, Y.-R. Liu and K. Zhou [53], we construct a ‘shifted’ version of

the Turan sieve method and apply it to problems on tournaments.

Let X be a bipartite graph with finite partite sets (A,B). For a ∈ A and b ∈ B, we

write a ∼ b if there is an edge that joins a and b. For b ∈ B, we define the degree of b to

be the number of neighbors of b in A, i.e.,

deg b := #{a ∈ A | a ∼ b

}.

5.1 The shifted Turan sieve 49

For b1, b2 ∈ B, we define the number of common neighbors of b1 and b2 to be

n(b1, b2) := #{a ∈ A | a ∼ b1 and a ∼ b2

}.

For each a ∈ A, analogously to deg b for b ∈ B, we define ω(a) to be the number of

neighbors of a in B, i.e.,

ω(a) := #{b ∈ B | a ∼ b

}.

Example 5.1.1. If the partite sets A and B are given as in the drawing:

A B

abb1

b2

Figure 5.1: Example of a ∼ b, ω(a), deg b and n(b1, b2).

then a ∼ b, ω(a) = 4, deg b = 3 and n(b1, b2) = 2.

Notice that ∑a∈A

ω(a) =∑a∈A

∑b∈Ba∼b

1 =∑b∈B

∑a∈Aa∼b

1 =∑b∈B

deg b

and ∑a∈A

ω2(a) =∑a∈A

∑b1,b2∈Ba∼b1a∼b2

1 =∑

b1,b2∈B

∑a∈Aa∼b1a∼b2

1 =∑

b1,b2∈B

n(b1, b2).

Combining the above results, Y.-R. Liu and M. R. Murty proved the following theorem

[55, Theorem 1].

Theorem 5.1.2. Suppose that A and B are partite sets of a bipartite graph X. Then

∑a∈A

(ω(a)− 1

|A|∑b∈B

deg b)2

=∑

b1,b2∈B

n(b1, b2)−1

|A|

(∑b∈B

deg b)2.

Example 5.1.3. If we take

A = {n ∈ N;n ≤ x},

B = {p ≤ x; p is a prime},

n ∼ p ⇐⇒ p|n,

5.1 The shifted Turan sieve 50

then we obtain the Turan’s proof of Hardy-Ramanujan theorem. In fact, for n ∈ A and

p ∈ B we obtain:

ω(n) = #{p ≤ x; p is a prime and p|n},

deg p = #{n ∈ N; n ≤ x and p|n},

n(p1, p2) = #{n ∈ N; n ≤ x, p1|n and p2|n},

|A| = bxc.

From Mertens’ theorem (see, for example, [21, §5]) we obtain∑p∈B

deg p =∑p∈B

⌊x

p

⌋=∑p∈B

(x

p+O(1)

)= x log log x+O(x)

and since ( ∑p≤√x

1

p

)2

≤∑

p1,p2≤x

1

p1p2≤

(∑p≤x

1

p

)2

,

we find that ∑p1,p2∈B

n(p1, p2) =∑

p1,p2∈Bp1 6=p2

n(p1, p2) +∑p1∈B

n(p1, p1)

=∑

p1,p2∈B

x

p1p2+∑p1≤x

x

p1+O(x)

= x(log log x)2 +O(x log log x)

By Theorem 5.1.2 we obtain∑n≤x

(ω(n)− log log x

)2= x(log log x)2 +O(x log log x)− 1

x(x log log x+O(x))2

= O(x log log x),

proving the Hardy-Ramanujan theorem.

From Theorem 5.1.2, Y.-R. Liu and M. R. Murty derive the Turan sieve method [55,

Corollary 1] which states that:

Corollary 5.1.4. (The Turan sieve)

#{a ∈ A |ω(a) = 0

}≤ |A|2 ·

∑b1,b2∈B

n(b1, b2)(∑b∈B

deg b)2 − |A|.

5.2 r-cycles on tournaments 51

The extension of sieve methods to a combinatorial setting has been attempted before.

For example, R. Wilson [85] and T. Chow [20] have formulated the Selberg sieve in a

combinatorial context (see also [56, §2]). However, due to the fact that the Mobius

function of a lattice is difficult to compute in the abstract setting, it is not clear how

one can apply the Selberg sieve to general combinatorial problems. This obstruction is

eliminated by the Turan sieve as the bound in Corollary 5.1.4 does not involve the Mobius

function, and thus it can be applied to many questions in combinatorics.

For combinatorial applications, one could be interested in estimating the number of

a ∈ A with ω(a) > 0. Thus, to allow us more flexibility in some counting questions, we

construct a “shifted” version of the Turan sieve. For a fixed integer k ∈ N0, notice that

#{a ∈ A |ω(a) = k

}·(k − 1

|A|∑b∈B

deg b)2≤∑a∈A

(ω(a)− 1

|A|∑b∈B

deg b)2.

Combining this inequality with Theorem 5.1.2, we obtain the following theorem:

Theorem 5.1.5. (The shifted Turan sieve) For k ∈ N0, we have

#{a ∈ A |ω(a) = k

}≤

|A|2∑

b1,b2∈B

n(b1, b2)− |A|(∑b∈B

deg b)2

(|A| · k −

∑b∈B

deg b)2 .

We note that for k = 0, Theorem 5.1.5 implies Corollary 5.1.4. In this chapter, we

apply Theorem 5.1.5 to some counting problems on tournaments, reproducing the results

of [53].

5.2 r-cycles on tournaments

Given a set W , a digraph on W is a set of ordered pairs (x, y) where x, y ∈ W and

x 6= y. A tournament T on W is a digraph such that for all x, y ∈ W , x 6= y, either

(x, y) ∈ T or (y, x) ∈ T , but not both. Intuitively, W can be thought of as a set of

players (vertices) and (x, y) ∈ T denotes the game (a directed edge) that x is defeated by

y (written as x→ y). Let Tn denote the set of all tournaments on W = {1, 2, · · · , n}.

Example 5.2.1. Some tournaments are graphically given by


x1 x2

x3

x1 x2

x3

x1 x2

x3x4

x1 x2

x3

x4

x5

Figure 5.2: Graphical examples of small tournaments.

For a tournament T ∈ Tn and r ∈ N with 3 ≤ r ≤ n, suppose that V = {x1, x2, . . . , xr}⊆ {1, 2, · · · , n} is a set of vertices such that x1 → x2 → · · · → xr → x1. We call this

subgraph an r-cycle on T and denote it by {V, τ}, where τ represents the collection of

games x1 → x2 → · · · → xr → x1, and we say τ generates T |V , the restriction of T on V .

Example 5.2.2. Some 3, 4 and 5-cycles contained in the tournaments from previous

example are graphically given by

Figure 5.3: Some cycles contained in the tournaments from Example 5.2.1.

One of our main results in this chapter is the following:

Theorem 5.2.3. Let k ∈ N0 and r ∈ N with r ≥ 3. For n ∈ N with n ≥ r, we have

#{T ∈ Tn |T contains exactly k r-cycles

}≤ 2(n2) ·

(n

r

)r! ·

{2nr−3 +O

(1n4

(4nre

)r)[2rk − (r − 1)!

(nr

)]2}.

In this section, we prove the above theorem. In Sections 5.3 and 5.4 we consider

analogous questions for restricted r-cycles on t-partite tournaments and unrestricted 2r-

cycles on bipartite tournaments. We remark here that estimates on the number of 3-cycles

on a tournament can be found in [7], [31], [51], [62], [64] and [77] (see also [61, §5 §6]),

and estimates on the number of 4-cycles on a tournament can be found in [52] and [78].

Although research on 3-cycles and 4-cycles has been active, the previous results are focused

on the number of cycles on “one tournament”. Thus, the theorems in this chapter are the

first which deal with “all tournaments” at once.

The main technical difficulty in applying Theorem 5.1.5 lies in the counting of n(b1, b2).

In the earlier applications of the Turan sieve method on combinatorial problems, such

estimates were often done by considering various cases and their subcases (see the Latin


square counting in [55]). This approach could become computationally infeasible if the

structure of the partite set B is more involved. In this chapter, we develop a new counting

method to estimate n(b1, b2). The central idea is to first “omit some existing cases” and

“include some non-existing cases”to get the expected main contribution. Then we compare

the “under-counting” and “over-counting” of the main contribution with the actual case

to get the correct estimate. Such an approach greatly simplifies many of our calculations.

For example, in Section 5.2, for the case l = 2, since one can argue that the numbers of

under-counting and over-counting are the same, only the estimate of the expected main

term is required.

By applying the shifted Turan sieve we derive an asymptotic upper bound for the

number of tournaments in Tn which contain a fixed number of r-cycles.

Proof. (of Theorem 5.2.3) Let A = Tn and B the set of all r-cycles. An element of B

can be denoted by (V, τ), where V ⊆ {1, 2, . . . , n}, |V | = r, and τ is a permutation of V .

Since there are(nr

)choices for V and (r− 1)! ways to form an r-cycle with vertices on V ,

we have

|A| = 2(n2) and |B| =(n

r

)(r − 1)!. (5.1)

For a = Ta ∈ A and b = (Vb, τb) ∈ B, we say a ∼ b if τb generates Ta|Vb . Thus,

ω(a) = # of r-cycles contained in Ta

and

deg b = #{a ∈ A | τb generates Ta|Vb

}.

Since τb generates an r-cycle on Ta|Vb , it determines r games of Ta. Thus, deg b = 2(n2)−r

and it follows that ∑b∈B

deg b =

(n

r

)2(n2)−r(r − 1)!.

For b1 = (Vb1 , τb1) and b2 = (Vb2 , τb2) ∈ B, consider

n(b1, b2) = #{a ∈ A | τb1 generates Ta|Vb1 and τb2 generates Ta|Vb2

}.

For 0 ≤ l ≤ r, suppose that |Vb1 ∩ Vb2 | = l. We consider the following five possibilities

for l:

(i) l = 0; in this case, there are(nr

)ways to choose Vb1 and

(n−rr

)ways to choose Vb2 .

Also, there are (r − 1)! ways to construct each cycle and there are 2r determined


games. Thus,

∑b1,b2∈B

|Vb1∩Vb2 |=0

n(b1, b2) =

(n

r

)(n− rr

)2(n2)−2r(r − 1)!2.

(ii) l = 1; in this case, there are(nr

)ways to choose Vb1 ,

(r1

)ways to choose the intersec-

tion point and(n−rr−1

)ways to choose the other points of Vb2 . Also, there are (r − 1)!

ways to construct each cycle and there are 2r determined games. Thus,

∑b1,b2∈B

|Vb1∩Vb2 |=1

n(b1, b2) =

(n

r

)(n− rr − 1

)(r

1

)2(n2)−2r(r − 1)!2.

b1 b2

Figure 5.4: Case l = 1.

(iii) l = 2; in this case, there are(nr

)choices for Vb1 and

(n−rr−2

)(r2

)choices for Vb2 . For

fixed Vb1 and Vb2 , observe that the game between their two intersection points, say

x and y, may or may not belong to either cycle. We first ignore the possibility that

τb1 and τb2 share the game x→ y or y → x, and denote by D2 := D2,b1,b2 the number

of tournaments where the games in τb1 and τb2 are chosen independently. Thus, we

have

D2 = 2(n2)−2r(r − 1)!2.

This term gives the expected main contribution and we now estimate its difference

from the actual case. Let Gx→y := Gx→y;b1,b2 be the collection of all tournaments

with fixed Vb1 and Vb2 which share the game x → y. We notice that for each

tournament in Gx→y, since there are (2r−1) games determined by τb1 and τb2 , there

are 2(n2)−(2r−1) = 2 · 2(n2)−2r possible games aside from τb1 and τb2 . However, in the

counting of D2, for such a tournament, we only count 2(n2)−2r possible games aside

from τb1 and τb2 . Thus we “undercount” some games for tournaments in Gx→y. On

the other hand, in the counting of D2, we “overcount” invalid tournaments which

have x→ y in τb1 and y → x in τb2 . However, by reversing the direction of the cycle

τb2 , these cases are in one-to-one correspondence with the cases that x→ y belongs

to both cycles. In other words, the undercounting of tournaments in Gx→y balances


out its overcounting. The same conclusion holds for Gy→x. Since the numbers of

undercounting and overcounting in D2 are the same, we have

∑b1,b2∈B

|Vb1∩Vb2 |=2

n(b1, b2) =

(n

r

)(n− rr − 2

)(r

2

)·D2 =

(n

r

)(n− rr − 2

)(r

2

)2(n2)−2r(r − 1)!2.

b1 b2

Figure 5.5: No game sharing =⇒ no need to undercounting or overcounting.

b1 b2 b1 b2 b1 b2

Figure 5.6: Undercounting and overcounting.

(iv) l = 3; in this case, there are(nr


(n−rr−3

)(r3

)choices for Vb2 . For

fixed Vb1 and Vb2 , observe that the game between their three intersection points, say

x, y and z, may or may not belong to either cycle. Similar to the case l = 2, we first

ignore the sharing games among {x, y, z} and denote by D3 := D3,b1,b2 , the number

of tournaments where the games in τb1 and τb2 are chosen independently. Thus, we

have

D3 = 2(n2)−2r(r − 1)!2.

We now consider the undercounting and overcounting in D3. There are three cases:

(a) τb1 and τb2 have no game among {x, y, z}: in this case, there is no undercounting

or overcounting in D3.

(b) τb1 and τb2 share one game among {x, y, z}, say x → y: we have seen in (iii)

that in this case, the numbers of undercounting and overcounting in D3 are the

same. Thus, there is no adjustment required here.

b1 b2

Figure 5.7: Case (iv), (b). The black game is the only common.


(c) τb1 and τb2 share two or more games among {x, y, z}: we notice that if τb1 and

τb2 share two games, since they are r-cycles and the games among {x, y, z} do

not form a cycle, we need to have r ≥ 4. On the other hand, if τb1 and τb2 share

three games among {x, y, z}, it means that they form a 3-cycle and we need to

have r = 3.

(c-1) Suppose that τb1 and τb2 share two games and r ≥ 4. Notice that there

are 3! = 6 possible choices for two games among {x, y, z}. Fix one of such

games, say x→ y → z. Let Gx→y→z = Gx→y→z;b1,b2 be the collection of all

tournaments with fixed Vb1 and Vb2 which contains the game x→ y → z.

We notice that for each tournament in Gx→y→z, since there are (2r − 2)

games determined by τb1 and τb2 , there are 2(n2)−(2r−2) = 4 ·2(n2)−2r possible

games aside from τb1 and τb2 . However, in the counting of D3, for such a

tournament, we only count 2(n2)−2r possible games aside from τb1 and τb2 .

Thus we “undercount” 3 · 2(n2)−2r possible games for each tournament in

Gx→y→z. On the other hand, in the counting of D3, we “overcount” invalid

tournaments which have x→ y → z in τb1 and z → y → x in τb2 . However,

by reversing the direction of the cycle τb2 , these cases are in one-to-one

correspondence with the cases that x → y → z belongs to both cycles.

Thus for a tournament in Gx→y→z, the difference between undercounting

games and overcounting games in D3 is 2 ·2(n2)−2r. Furthermore, for such a

tournament, there are ((r−3)!)2 choices for τb1 and τb2 . The same argument

can be applied to all other five permutations of {x, y, z}. Combining this

with (a) and (b), we see that if r ≥ 4,

∑b1,b2∈B

|Vb1∩Vb2 |=3

n(b1, b2) =

(n

r

)(n− rr − 3

)(r

3

)·(D3 + 6 · 2 · 2(n2)−2r((r − 3)!)2

)

=

(n

r

)(n− rr − 3

)(r

3

)2(n2)−2r[(r − 1)!2 + 12(r − 3)!2].

(c-2) Suppose that τb1 and τb2 share three games and r = 3. In this case,

the games among {x, y, z} form a 3-cycle and there are 2 possible choices

for such a cycle. Fix one of such cycles, say x → y → z → x. Let

Gx→y→z→x = Gx→y→z→x;b1,b2 be the collection of all tournaments with

fixed Vb1 and Vb2 which contains the cycle x → y → z → x. We no-

tice that for each tournament in Gx→y→z→x, since there are 3 = (2r − 3)

games determined by τb1 and τb2 , there are 2(n2)−(2r−3) = 8 ·2(n2)−2r possible


games aside from τb1 and τb2 . However, in the counting of D3, for such a

tournament, we only count 2(n2)−2r possible games aside from τb1 and τb2 .

Thus, we “undercount” 7 · 2(n2)−2r possible games for each tournament in

Gx→y→z→x. On the other hand, in the counting of D3, we “overcount” in-

valid tournaments which have x→ y → z → x in τb1 and x→ z → y → x

in τb2 . However, by reversing the direction of the cycle τb2 , these cases are

in one-to-one correspondence with the case that x→ y → z → x belong to

both cycles. Thus, for a tournament in Gx→y→z→x, the difference between

undercounting games and overcounting game in D3 is 6 · 2(n2)−2r. Further-

more, for such a tournament, there are 12 = ((r− 3)!)2 choices for τb1 and

τb2 . Combining this with (a) and (b), we see that for r = 3,

∑b1,b2∈B

|Vb1∩Vb2 |=3

n(b1, b2) =

(n

r

)(n− rr − 3

)(r

3

)·(D3 + 2 · 6 · 2(n2)−2r((r − 3)!)2

)

=

(n

r

)(n− rr − 3

)(r

3

)2(n2)−2r[(r − 1)!2 + 12(r − 3)!2].

We note that this formula is exactly the same as the one in (c-1).

(v) 4 ≤ l ≤ r; in these cases, there are(nr

)choices for Vb1 ,

(n−rr−l

)(rl


the number of ways to form each cycle is at most (r − 1)!. Since the number of

determined games is at least (2r − (l − 1)), we have

∑b1,b2∈B

4≤|Vb1∩Vb2 |≤r

n(b1, b2) ≤(n

r

)2(n2)−2r(r − 1)!2

∑4≤l≤r

(n− rr − l

)(r

l

)2l−1.

Combining all the above five possibilities, we obtain

∑b1,b2∈B

n(b1, b2) ≤ 2(n2)−2r(n

r

)(r − 1)!2

{(n− rr

)+

(n− rr − 1

)(r

1

)+

(n− rr − 2

)(r

2

)

+

(n− rr − 3

)(r

3

)[1 +

12

(r − 1)2(r − 2)2

]+∑4≤l≤r

(n− rr − l

)(r

l

)2l−1

}

= 2(n2)−2r(n

r

)(r − 1)!2

[(n

r

)+

12

(r − 1)2(r − 2)2

(n− rr − 3

)(r

3

)+O(nr−44r)

].

By applying Theorem 5.1.5 and Stirling’s approximation, Theorem 5.2.3 follows.

We note that in the proof of Theorem 5.2.3, we have calculated explicitly the cases


r = 3 and r = 4. Thus, we can derive from the above estimates that

Corollary 5.2.4. Let k ∈ N0. We have

#{T ∈ Tn |T contains exactly k 3-cycles

}≤ 2(n2) ·

{316

(n3

)(k − 1

4

(n3

))2}

and

#{T ∈ Tn |T contains exactly k 4-cycles

}≤ 2(n2) ·

{364

(n4

)(4n− 11)(

k − 38

(n4

))2}.

Remark 5.2.5. For fixed k ∈ N0 and r ∈ N with r ≥ 3, we see from Theorem 5.2.3 that

for n sufficiently large, we have

#{T ∈ Tn |T contains exactly k r-cycles

}� 2(n2) ·

{1

n3

}.

Thus, as n→∞, the probability that a tournament contains exactly k r-cycles is 0. One

can obtain various conclusions that are much stronger than the above one from Theorem

5.2.3. For example, let f : N → R+ such that f(n) = o(n3) and 3 ≤ r ≤ (log n)1−ε for

any ε > 0. By Theorem 5.2.3, we can show that as n→∞,

Prob{T ∈ Tn contains at most f(n) r-cycles

}−→ 0.

In fact, this probability is bounded above by

f(n)∑k=1

(n

r

)r!

{2nr−3 +O

(1n4

(4nre

)r)[2rk − (r − 1)!

(nr

)]2}. (5.2)

The contribution of error term has to be negligible compared to the main term, i.e.,(4r

e

)r= o(n) as n→∞,

which is true because r ≤ (log n)1−ε. Hence, by expression (5.2), the desired proportion is

�f(n)∑k=1

1

n3= o(1) as n→∞.

5.3 Restricted r-cycles on t-partite tournaments 59

5.3 Restricted r-cycles on t-partite tournaments

For t ≥ 2, let X1, X2, . . . , Xt be t pairwise disjoint sets of points with Xi = {xi,j; 1 ≤j ≤ mi}, 1 ≤ i ≤ t. An m1 × m2 × · · · × mt t-partite tournament is a simple directed

complete t-partite graph on X1 ×X2 × · · · ×Xt. Thus, each pair of points (xi,di , xj,dj) ∈Xi×Xj, i 6= j, is joined by a directed edge, which is either xi,di → xj,dj or xj,dj → xi,di , but

not both. Let Tm1,··· ,mt denote the set of all t-partite tournaments on X1×X2× · · · ×Xt.

Example 5.3.1. Some partite tournaments are graphically given by

X1 X2

X1

X2

X4

X3

Figure 5.8: A bipartite tournament in T4,3 and a 4-partite tournament in T2,2,3,2.

For a tournament T ∈ Tm1,··· ,mt and r ∈ Z with 3 ≤ r ≤ t, we denote by (x1, · · · , xr, τ)

a restricted r-cycle on T with {x1, · · · , xr} taken from distinct partite sets Xi, 1 ≤ i ≤ t,

and τ representing the games x1 → x2 → · · · → xr → x1. In this case, we say τ generates

T |{x1,··· ,xr}, the restriction of T on {x1, · · · , xr}.

Example 5.3.2. Bipartite tournaments does not contain restricted cycles of any size.

Nevertheless, for t ≥ 3 the t-partite tournaments may contain some cycles or not. Two

of the restricted cycles in the 4-partite tournament from previous example are exhibited

below:

X1

X2

X4

X3

X1

X2

X4

X3

Figure 5.9: Restricted 3 and 4-cycles in the 4-partite tournament of Example 5.3.1.

In [40], W. D. Goddard et al. investigated some properties of the cycle structure of a

t-partite tournament.


In this section, we apply the shifted Turan sieve to derive an upper bound for the

number of t-partite tournaments which contain a fixed number of restricted r-cycles. In

the special case when t = r, we can obtain the following theorem which is sharper than

its generalization.

Theorem 5.3.3. Let k ∈ N0 and r ∈ N with r ≥ 3. For m1, · · · ,mr ∈ N, write

σs = σs(m1, · · · ,mr), 0 ≤ s ≤ r, as in the Notations of Part II. We have

#{T ∈ Tm1,...,mr ; T contains exactly k restricted r-cycles

}≤ 2σ2 · (r − 3)!2σr ·

{12σr−3 +O(2rr4σr−4)

[2rk − (r − 1)!σr]2

}.

Proof. Let A = Tm1,...,mr and B the set of all r-cycles. An element of B can be denoted

by (x1, · · · , xr, τ), where {x1, · · · , xr} are taken from distinct partite sets Xi (1 ≤ i ≤ r),

and τ is a permutation of {x1, · · · , xr}. Since there are σr choices for {x1, · · · , xr} and

(r − 1)! ways to form an r-cycle with vertices x1, · · · , xr, we have

|A| = 2σ2 and |B| = σr(r − 1)!.

For a = Ta ∈ A and b = (x1,b, . . . , xr,b, τb) ∈ B, we say a ∼ b if τb generates

Ta|{x1,b,...,xr,b}. Thus,

ω(a) = # of r-cycles contained in Ta

and

deg b = #{a ∈ A | τb generates Ta|{x1,b,...,xr,b}

}.

Since τb generates an r-cycle on Ta|{x1,b,··· ,xr,b}, it determines r games of Ta. Thus,

deg b = 2σ2−r and it follows that∑b∈B

deg b = σr2σ2−r(r − 1)!.

For b1 = (x1,b1 , . . . , xr,b1 , τb1) and b2 = (x1,b2 , . . . , xr,b2 , τb2) ∈ B, consider

n(b1, b2) = #{a ∈ A | τb1 generates Ta|{x1,b1 ,...,xr,b1} and τb2 generates Ta|{x1,b2 ,...,xr,b2}

}.

For i = 1, . . . , r, suppose

|{xi,b1} ∩ {xi,b2}| = Ni, (5.3)

where Ni ∈ {0, 1}. Let M(N1, N2, . . . , Nr) denote the collection of all pairs (b1, b2) ∈ B2

such that Equation (5.3) holds. By counting the number of 1 in Ni (1 ≤ i ≤ r), up to


symmetry, there are (r+ 1) distinct possibilities for the r-tuple (N1, . . . , Nr), of which we

group into five cases:

(i) (N1, N2 . . . , Nr) = (0, 0, . . . , 0); in this case, there are σr choices for Vb1 , (m1 −1)(m2− 1) . . . (mr− 1) choices for Vb2 and (r− 1)! choices for each cycle. Also, there

are 2r games determined by τb1 and τb2 . Thus, we have∑(b1,b2)∈M(0,··· ,0)

n(b1, b2) = σr(m1 − 1)(m2 − 1)(m3 − 1) . . . (mr − 1)2σ2−2r(r − 1)!2.

(ii) (N1, N2, N3, . . . , Nr) = (1, 0, 0, . . . , 0); in this case, there are σr choices for Vb1 , (m2−1)(m3− 1) . . . (mr− 1) choices for Vb2 and (r− 1)! choices for each cycle. Also, there

are 2r games determined by τb1 and τb2 . Thus, we have∑(b1,b2)∈M(1,0,··· ,0)

n(b1, b2) = σr(m2 − 1)(m3 − 1) . . . (mr − 1)2σ2−2r(r − 1)!2.

X1

X2 X3

X4

Figure 5.10: Case (ii). Blue game means cycle b1 and red game means cycle b2.

(iii) (N1, N2, N3, N4, . . . , Nr) = (1, 1, 0, 0, . . . , 0); in this case, there are σr choices for Vb1

and (m3 − 1)(m4 − 1) . . . (mr − 1) choices for Vb2 . Also, using the same argument

as (iii) in the proof of Theorem 5.2.3, for fixed Vb1 and Vb2 , there are 2σ2−2r(r− 1)!2

many choices for (b1, b2) ∈M(1, 1, 0, · · · , 0). Thus, we have∑(b1,b2)∈M(1,1,0,··· ,0)

n(b1, b2) = σr(m3 − 1)(m4 − 1) . . . (mr − 1)2σ2−2r(r − 1)!2.


X1

X2 X3

X4

Figure 5.11: Case (iii), (b). Blue game means cycle b1 and red game means cycle b2.

(iv) (N1, N2, N3, N4, N5, . . . , Nr) = (1, 1, 1, 0, 0, . . . , 0); in this case, there are σr choices

for Vb1 and (m4 − 1)(m5 − 1) . . . (mr − 1) choices for Vb2 . Also, using the same

argument as (iv) in the proof of Theorem 5.2.3, for fixed Vb1 and Vb2 , there are

2σ2−2r[(r − 1)!2 + 12(r − 3)!2] many choices for (b1, b2) ∈ M(1, 1, 1, 0, · · · , 0). Thus,

we have∑(b1,b2)∈M(1,1,1,0,··· ,0)

n(b1, b2) = σr(m4−1)(m5−1) . . . (mr−1)2σ2−2r[(r − 1)!2 + 12(r − 3)!2

].

X1

X2 X3

X4

Figure 5.12: Case (iv), (d). Blue game means cycle b1 and red game means cycle b2.

(v) (N1, N2, . . . , Nl, Nl+1, Nl+2, . . . , Nr) = (1, 1, . . . , 1︸︷︷︸l≥4 times

, 0, 0, . . . , 0); in these cases, there

are σr choices for Vb1 , (ml+1 − 1) . . . (mr − 1) choices for Vb2 and at most (r − 1)!2

choices for each cycle. Since there are at least (2r − (l − 1)) determined games, we

have ∑4≤l≤r

(b1,b2)∈M(1,1,...,1︸︷︷︸l≥4 times

,0,0,...,0)

n(b1, b2) ≤ σr2σ2−2r(r − 1)!2

∑4≤l≤r

(ml+1 − 1) . . . (mr − 1)2l−1.


Combining all these possibilities and their symmetrical cases, we get∑b1,b2∈B

n(b1, b2) = 2σ2−2rσr(r − 1)!2

×

{[σr − σr−1 + σr−2 − σr−3 +O(σr−4)︸︷︷︸

symmetrical sum of(0,...,0)

]+[σr−1 − 2σr−2 + 3σr−3 +O(σr−4)︸︷︷︸

symmetrical sum of(1,0,0,...,0)

]

+[σr−2 − 3σr−3 +O(σr−4)︸︷︷︸

symmetrical sum of(1,1,0,0,...,0)

]+

[(1 +

12

(r − 1)2(r − 2)2

)σr−3 +O(σr−4)︸︷︷︸

symmetrical sum of(1,1,1,0,0,...,0)

]

+O

( ∑4≤l≤r

2l−1σr−4

)︸︷︷︸

symmetrical sum ofremainder cases

}

= 2σ2−2rσr(r − 1)!2[σr +

12σr−3(r − 1)2(r − 2)2

+O(2rσr−4)

].

(5.4)

By applying Theorem 5.1.5, Theorem 5.3.3 follows.

We now state the result for general t-partite tournaments with t ≥ r. The following

theorem is more general, but less sharp than Theorem 5.3.3. Since its proof is in the same

spirit of Theorem 5.3.3, we will only provide a sketch of the proof below.

Theorem 5.3.4. Let k ∈ N0 and r, t ∈ N with 3 ≤ r ≤ t. For m1, · · · ,mt ∈ N, write

σs = σs(m1, · · · ,mt) (0 ≤ s ≤ t). We have

#{T ∈ Tm1,...,mt ; T contains exactly k restricted r-cycles

}≤ 2σ2 · (r − 3)!2σr ·

{12(r3

)σr−3 +O(2rr8σr−4)

[2rk − (r − 1)!σr]2

}.

Sketch of the proof. Let A and B be defined as in the proof of Theorem 5.3.3. Thus,

we have

|A| = 2σ2 and∑b∈B

deg b = σr2σ2−r(r − 1)!.

It remains to consider the sum of n(b1, b2). For b1, b2 ∈ B, write bi = {x1,bi · · · , xr,bi , τbi}(1 ≤ i ≤ 2). Assume xj,b1 (1 ≤ j ≤ r) are in X1, · · · , Xr and xj,b2(1 ≤ j ≤ r) are in

X1, · · · , Xl, Xul+1, · · · , Xur for some 0 ≤ l ≤ r and uj ∈ {r + 1, · · · , t} (l + 1 ≤ j ≤ r).

For v ∈ N, define σ[v]s = σs(m1, . . . ,mv, 0, . . . , 0). Thus σ

[t]s = σs. Then using the same

5.4 Unrestricted 2r-cycles on bipartite tournaments 64

argument as the one to prove (5.4), we obtain that∑b1,b2∈B

b1∈X1,...,Xrb2∈X1,...,Xl,Xul+1

,...,Xur

n(b1, b2) ≤ 2σ2−2rm1 . . .mr(r − 1)!2×

×

[σ[l]l +

12σ[l]l−3

(r − 1)2(r − 2)2+O

(2rσ

[l]l−4)]mul+1

. . .mur .

We now sum over the symmetric cases of b2. Notice that the term with the factor

σ[l]l−3mul+1

. . .mur appears at most(r3

)times. Also, the error term appears at most

(r4

)=

O(r4) times. Thus, we can show that

∑b1,b2∈B

n(b1, b2) ≤ 2σ2−2rσr(r − 1)!2

[σr +

12(r3

)σr−3

(r − 1)2(r − 2)2+O

(2rr4σr−4

)].

Then Theorem 5.3.4 follows from Theorem 5.1.5.

Remark 5.3.5. For fixed k ∈ N0 and r, t ∈ N with 3 ≤ r ≤ t, we see from Theorem 5.3.4

that for m1, · · · ,mt sufficiently large, we have

#{T ∈ Tm1,··· ,mt ; T contains exactly k restricted r-cycles

}� 2σ2 ·

{σr−3σr

}.

Thus, as one of m1, · · · ,mt → ∞, the probability that a tournament contains exactly k

restricted r-cycles is 0.

Remark 5.3.6. Similar to the Remark 5.2.5, one can obtain various conclusions that are

stronger than the above one from Theorem 5.3.4. For example, let f : N→ R+ such that

f(n) = o(n3) and 3 ≤ r ≤ log n. Then one can show that

Prob{T ∈ Tm1,··· ,mt contains at most f(n) restricted r-cycles

}−→ 0 as n→∞.

5.4 Unrestricted 2r-cycles on bipartite tournaments

An r-cycle is called an unrestricted r-cycle on a t-partite tournament T if it is

allowed that the partite sets X1, X2, . . . , Xt contains zero, one or more vertices in the r-

cycle. Given a positive integer r and a nonnegative integer k, we would like to study how

many t-partite tournaments have exactly k restricted r-cycles or unrestricted r-cycles. In

previous section (see also [53]), we use the Turan sieve successfully to give an upper of

the number of the tournaments with exactly k restricted r-cycles.


Example 5.4.1. We may find some examples of unrestricted cycles in bipartite tourna-

ments. Below, we highlight some of the cycles of the bipartite tournament from Example

5.3.1:

X1 X2 X1 X2

Figure 5.13: Unrestricted 4 and 6-cycles.

However, for the unrestricted cycle case, the problem becomes much more difficult and

very hard to approach. With the new counting method, we are able to handle the case of

the bipartite tournaments; i.e, t = 2. We remark that estimates on the number of 4-cycles

on a bipartite tournament can be found in [63] (see also [86]). We give an upper bound for

the number of bipartite tournaments which have exactly k unrestricted 2r-cycles, where

k ≥ 0 and r ≥ 2 are fixed integers, as a function of the number of vertices m and n in the

partite sets. More precisely, we prove the following theorem:

Theorem 5.4.2. Let Tm,n be the collection of all bipartite tournaments on sets P and Q,

where |P | = m and |Q| = n. For a fixed k ∈ N0, we have

#{T ∈ Tm,n;T contains exactly k unrestricted 2r-cycles}

≤ 2mn(m

r

)(n

r

)mr−3nr−3r!2

{2(m2n+mn2) +O (24r(m2 + n2))[

22rk − (r − 1)!r!(mr

)(nr

)]2}.

First, we introduce some preliminary definitions and notation.

Definition 5.4.3. An m×n bipartite tournament T is a simple directed complete bipartite

graph with partite sets P = {p1, . . . , pm} and Q = {q1, . . . , qn}. Denote Tm,n to be the set

of all m× n bipartite tournaments.

Definition 5.4.4. Let T ∈ Tm,n be an m × n bipartite tournament with partite sets P

and Q, and let 2 ≤ r ≤ min{m,n} be an integer. Suppose that P ′ = {p1, . . . , pr} ⊂ P and

Q′ = {q1, . . . , qr} ⊂ Q are subsets of vertices and σ, τ ∈ Sr such that

pσ(1) → qτ(1) → pσ(2) → qτ(2) → · · · → pσ(r) → qτ(r) → pσ(1).

Then we say that (P ′, Q′, τ, σ) = pσ(1) → qτ(1) → pσ(2) → qτ(2) → · · · → pσ(r) → qτ(r) →pσ(1) is a 2r-cycle on T . If σ, τ, P ′, Q′ are fixed and (P ′, Q′, τ, σ) is a 2r-cycle on T , then


we say that (σ, τ) generates T |P ′,Q′ (the restriction of T to P ′, Q′).

Intuitively, a bipartite tournament T ∈ Tm,n can be thought as two sets of players P

and Q such that each player in one set plays a game with every player from the other

set, and the direction of the edges points to the winner of the game. Suppose we fix

2 ≤ r ≤ min{m,n} and a non-negative integer k. We consider the tournaments in Tm,n

which contain exactly k 2r-cycles. Using Theorem 5.1.5, we will derive an asymptotic

upper bound for the number of such bipartite tournaments.

Notation: We write (x1, . . . , xn)↔ (y1, . . . , yn) to mean that (x1, . . . , xn) is a permutation

of (y1, . . . , yn).

Lemma 5.4.5. Let T ∈ Tm,n, P ′ and Q′ be vertex subsets of T , and fix 2 ≤ r ≤min{m,n}. Then there are r!(r − 1)! distinct 2r-cycles (P ′, Q′, σ, τ) for σ, τ ∈ Sr.

Proof. There are (r!)2 distinct choices for σ, τ ∈ Sr. It is easily shown that (P ′, Q′, σ, τ) =

(P ′, Q′, σ′, τ ′) if and only if the alternating concatenations of (σ, τ) and (σ′, τ ′) differ by

even cyclic shift. There are r distinct cyclic shifts on Sr, so the number of distinct 2r-cycles

is r!(r − 1)!.

Let A = Tm,n and B be the set of all 2r-cycles on Tm,n. Suppose we pick P ′ =

{pi1 , . . . , pir} ⊆ P and Q′ = {qj1 , . . . , pjr} ⊆ Q. Note that a 2r-cycle on these vertices

means that

pa1 → qb1 → pa2 → qb2 → · · · → par → qbr → pa1

where (a1, . . . , ar−1)↔ (i2, . . . , ir), and (b1, . . . , br)↔ (j1, . . . , jr). Thus given {pi1 , . . . , pir} ⊆P and {qj1 , . . . , pjr} ⊆ Q there are (r − 1)!r! possible 2r-cycles on these vertices. This

means that

|A| = 2mn and |B| = (r − 1)!r!

(m

r

)(n

r

).

For a = Ta ∈ A, b = (P ′b, Q′b, σb, τb), we say that a ∼ b if Ta|P ′b,Q′b is generated by

(σb, τb). Thus,

ω(a) = the number of 2r-cycles contained in Ta.

For a fixed 2r-cycle b = (P ′b, Q′b, σb, τb) ∈ B, we have

deg b = #{a ∈ A;Ta|P ′b,Q′b is generated by (σb, τb)}.

Since (σb, τb) fixes 2r games of Ta, there are 2mn−2r choices for the remaining mn− 2r


games, so deg b = 2mn−2r for all b ∈ B and

∑b∈B

deg b = (r − 1)!r! · 2mn−2r(m

r

)(n

r

).

Let b1 = (P ′b1 , Q′b1, σb1 , τb1) ∈ B, b2 = (P ′b2 , Q

′b2, σb2 , τb2) ∈ B. Consider

n(b1, b2) = #{a ∈ A; (σb1 , τb1) generates Ta|P ′b1 ,Q′b1 and (σb2 , τb2) generates Ta|P ′b2 ,Q′b2}.

Let M(r1, r2) denote the number of pairs (b1, b2) ∈ B2 such that |P ′b1 ∩ P′b2| = r1 and

|Q′b1 ∩Q′b2| = r2. By Theorem 5.1.5, the quantity we are interested in estimating is

#{T ∈ Tm,n;T contains exactly k 2r-cycles} ≤

∑b1,b2∈B

n(b1, b2)− |A|−1(∑b∈B

deg b

)2

(k − |A|−1

∑b∈B

deg b

)2 .

As we are interested in an asymptotic upper bound, we will consider the following

cases for the pair (r1, r2) to obtain the leading terms of the numerator above:

(i) r1 = 0 or r2 = 0. Then b1 and b2 together fix 4r games and there are (r− 1)!r! ways

to choose each cycle since b1 and b2 cannot share any edges. Thus, we have

M(0, r2) = [(r−1)!r!]2(m

r

)(n

r

)(m− rr

)(r

r2

)(n− rr − r2

)and n(b1, b2) = 2mn−4r.

A similar equation holds for M(r1, 0). Summing over the possibilities with r1 = 0

or r2 = 0, we obtain

∑b1,b2∈Br1=0

n(b1, b2) = 2mn−4r[(r − 1)!r!]2(m

r

)(n

r

)(m− rr

)(n

r

),

∑b1,b2∈Br1>0,r2=0

n(b1, b2) = 2mn−4r[(r − 1)!r!]2(m

r

)(n

r

)(n− rr

)[(m

r

)−(m− rr

)].

(ii) (r1, r2) = (1, 1). There are(mr

)(nr

)(r1

)2(m−rr−1

)(n−rr−1

)ways to choose the vertices for b1

and b2 so that |P ′b1 ∩ P′b2| = |Q′b1 ∩ Q

′b2| = 1. Fix a choice of vertices and suppose

that p ∈ P ′ and q ∈ Q′ are the only shared vertices between b1 and b2.

If b1 and b2 share no games then b1 and b2 fix 4r games and n(b1, b2) = 2mn−4r, and


if b1 and b2 share one game then b1 and b2 fix 4r−1 games and n(b1, b2) = 2mn−4r+1.

Since r1 = r2 = 1 there is at most one shared game between b1 and b2. Let M0,M1

be the number of ways to choose the cycles on the vertices of b1 and b2 so that they

have 0 and 1 games in common, respectively. There are (r − 1)!r! ways to create a

2r-cycle from the vertices in b1 and from the vertices in b2, independently of each

other, for a total of [(r − 1)!r!]2 possibilities. Among these possibilities:

(a) There are M0 possibilities where at most one of b1, b2 contains a game in {p→q, q → p}, since in this case b1 and b2 have no games in common and the union

of the two cycles represents a valid subgraph of the bipartite tournament.

(b) There are M1 possibilities where b1 and b2 both contain p → q or q → p,

since the union of these two cycles represents a valid subgraph of the bipartite

tournament. There are an additional M1 possibilities where b1 contains one of

the games in {p → q, q → p} and b2 contains the other, since the union of b1

with the inverse of b2 (Where the orientation of every edge in b2 is switched)

represents a valid subgraph of the bipartite tournament.

Since these cases are exhaustive, we have M0 + 2M1 = [(r − 1)!r!]2 and thus

∑b1,b2∈Br1=r2=1

n(b1, b2) = 2mn−4r[(r − 1)!r!]2(m

r

)(n

r

)(r

1

)2(m− rr − 1

)(n− rr − 1

).

(iii) (r1, r2) = (1, 2). The case for (r1, r2) = (2, 1) is analogous by interchanging P and

Q. There are(mr

)(nr

)(r1

)(r2

)(m−rr−1

)(n−rr−2

)ways to choose the vertices for b1 and b2 so

that |P ′b1 ∩ P′b2| = 1 and |Q′b1 ∩ Q

′b2| = 2. Fix a choice of vertices and suppose that

p ∈ P and q, q′ ∈ Q are the only shared vertices between b1 and b2.

If b1 and b2 share 0,1,2 games then the two cycles together fix 4r, 4r − 1, 4r − 2

games and n(b1, b2) = 2mn−4r, 2mn−4r+1, 2mn−4r+2, respectively. Let M0,M1,M2 be

the number of ways to choose the cycles on the vertices of b1 and b2 so that they have

0, 1, 2 games in common, respectively. There are (r− 1)!r! ways to create a 2r-cycle

from the vertices in b1 and (r − 1)!r! ways to create a 2r-cycle from the vertices in

b2, independently of each other, for a total of [(r− 1)!r!]2 possibilities. Among these

possibilities:

(a) There are M0 possibilities such that at most one of b1, b2 contains a game in

{p → q, q → p} or in {p → q′, q′ → p}, since in this case b1 and b2 have no


games in common and the union of the two cycles represents a valid subgraph

of the bipartite tournament.

(b) There are M1 possibilities where b1 and b2 both contain the same game in {p→q, q → p} but at most one of b1, b2 contains a game in {p → q′, q′ → p} (and

vice versa), since the union of these two cycles represents a valid subgraph of

the bipartite tournament. There are an additional M1 possibilities where b1

contains one of the games in {p → q, q → p}, b2 contains the other game in

{p → q, q → p}, but at most one of b1, b2 contains a game in {p → q′, q′ → p}(and vice versa), since the union of b1 with the inverse of b2 represents a valid

subgraph of the bipartite tournament.

(c) There are M2 possibilities where b1 and b2 both contain q′ → p→ q or q → p→q′, since the union of these two cycles represents a valid subgraph of the bipartite

tournament. There are an additional M2 possibilities where b1 contains one of

q′ → p→ q and q → p→ q′ and b2 contains the other, since the union of b1 with

the inverse of b2 represents a valid subgraph of the bipartite tournament.

Since these cases are exhaustive, we have M0 + 2M1 + 2M2 = [(r − 1)!r!]2 and thus∑b1,b2∈B

(r1,r2)=(1,2)

n(b1, b2) =

= (2mn−4rM0 + 2mn−4r+1M1 + 2mn−4r+2M2)

(m

r

)(n

r

)(r

1

)(r

2

)(m− rr − 1

)(n− rr − 2

)= 2mn−4r

{[(r − 1)!r!]2 + 2M2

}(mr

)(n

r

)(r

1

)(r

2

)(m− rr − 1

)(n− rr − 2

).

We now computeM2 explicitly. Note that if b1 and b2 share two games then they both

must contain either q → p→ q′ or q′ → p→ q. For each of b1 and b2 we can contract

these two common games to a single vertex in Q. Since there are (r − 2)!(r − 1)!

possible 2r-cycles on vertex sets of size r − 1 and r − 1, there are [(r − 2)!(r − 1)!]2

such 2r-cycles on the vertices of b1 and b2. Thus M2 = 2[(r − 2)!(r − 1)!]2 and∑b1,b2∈B

(r1,r2)=(1,2)

n(b1, b2) =

= 2mn−4r[(r − 2)!(r − 1)!]2[(r − 1)2r2 + 4

](mr

)(n

r

)(r

1

)(r

2

)(m− rr − 1

)(n− rr − 2

)and an analogous result holds for (r1, r2) = (2, 1).


Denote the sum of the n(b1, b2)’s in (i), (ii), (iii) and all of their symmetrical cases by

S, and let C := 2mn−4r(mr

)(nr

)[(r − 1)!r!]2. Then

S = C·

{(m− rr

)(n

r

)+

(n− rr

)[(m

r

)−(m− rr

)]+ r2

(m− rr − 1

)(n− rr − 1

)

+r4 − 2r3 + r2 + 4

2(r − 1)

[(m− rr − 1

)(n− rr − 2

)+

(m− rr − 2

)(n− rr − 1

)]}.

If r1 + r2 ≥ 4 with r1, r2 > 0, then b1 and b2 fix at least 2r games, and there are at

most [(r − 1)!r!]2 possibilities for the cycles b1 and b2, so

∑b1,b2∈Br1+r2≥4r1,r2>0

n(b1, b2) ≤ 2mn−2r[(r − 1)!r!]2(m

r

)(n

r

)(r

r1

)(r

r2

)(m− rr − r1

)(n− rr − r2

)

≤ 22rC

(r

r1

)(r

r2

)[r−r1−1∏i=0

(m− r − i)

][r−r2−1∏j=0

(n− r − j)

]

≤ 24rr−2C

[r−r1−1∏i=0

(m− r − i)

][r−r2−1∏j=0

(n− r − j)

].

Note that since r1, r2 > 0, we have mr−r1nr−r2 ≤ max(mr−1nr−3,mr−2nr−2,mr−3nr−1).

Since mr−1nr−3 +mr−3nr−1 ≥ 2mr−2nr−2, for every pair (r1, r2) satisfying r1 + r2 ≥ 4 and

r1, r2 > 0 we have ∑b1,b2∈Br1+r2≥4r1,r2>0

n(b1, b2) = C ·O(24rr−2(mr−1nr−3 +mr−3nr−1))

so that ∑r1+r2≥4r1,r2>0

∑b1,b2∈B(r1,r2)

n(b1, b2) ≤ r2 maxr1+r2≥4r1,r2>0

∑b1,b2∈B(r1,r2)

n(b1, b2)

= C ·O(24r(mr−1nr−3 +mr−3nr−1).

It follows that∑b1,b2∈B

n(b1, b2) = S + C ·O(24r(mr−1nr−3 +mr−3nr−1)).


We now return to computing the numerator obtained from the shifted Turan sieve:

∑b1,b2∈B

n(b1, b2)− |A|−1(∑b∈B

deg b

)2

=∑

b1,b2∈B

n(b1, b2)− C(m

r

)(n

r

)= C

[S

C−(m

r

)(n

r

)+O

(24r(mr−1nr−3 +mr−3nr−1)

)]. (5.5)

We will need the following identities to compare the degree 2r, 2r − 1, and 2r − 2

terms of the polynomial in S/C with the corresponding terms of the polynomial(mr

)(nr

).

Lemma 5.4.6. Let r ∈ N and let p(m) :=(mr

)−(m−rr

). Denote the coefficient of mk in

p by p(k). Then p is a degree r − 1 polynomial with

p(r−1) =1

r!· r2

p(r−2) = − 1

r!· r

2(r − 1)(2r − 1)

2

|p(r−3)| ≤ 1

r!· 8r6 =

1

r!·O(r6).

Proof. The first two equalities are easily shown by Vieta’s formula and induction. For the

inequality involving p(r−3), we note that the coefficient of mr−3 in r!(mr

)and r!

(m−rr

)are

both negative, and the former is less in absolute value than the latter. The coefficient of

mr−3 in r!(m−rr

)is the sum of the products of the numbers {−r,−(r+ 1), . . . ,−(2r− 1)}

taken three at a time, and this is at least −r3(2r − 1)3 > −8r6. The result follows.

Using the lemma, we see that

D1 :=

(m

r

)(n

r

)−(m− rr

)(n

r

)−(n− rr

)[(m

r

)−(m− rr

)]=

[(m

r

)−(m− rr

)]·[(n

r

)−(n− rr

)]=

1

r!2

[r2mr−1 − r2(r − 1)(2r − 1)

2mr−2 +O(r6mr−3)

]×[r2nr−1 − r2(r − 1)(2r − 1)

2nr−2 +O(r6nr−3)

]


Expanding the other terms of S/C, we have(m− rr − 1

)(n− rr − 1

)=

1

r!2

[rmr−1 − r(3r − 2)(r − 1)

2mr−2 +O(r7mr−3)

]×[rnr−1 − r(3r − 2)(r − 1)

2nr−2 +O(r7mr−3)

],(

m− rr − 1

)(n− rr − 2

)=

1

r!2

[rmr−1 − r(3r − 2)(r − 1)

2mr−2 +O(r7mr−3)

]×[(r − 1)rnr−2 +O(r4nr−3)

],(

m− rr − 2

)(n− rr − 1

)=

1

r!2

[rnr−1 − r(3r − 2)(r − 1)

2nr−2 +O(r7nr−3)

]×[(r − 1)rmr−2 +O(r4mr−3)

].

Thus

D2 := D1 − r2(m− rr − 1

)(n− rr − 1

)=

1

r!2

[r4(r − 1)2

2(mr−2nr−1 +mr−1nr−2) +O

(r10(mr−1nr−3 +mr−3nr−1)

)].

Finally, we have

S

C−(m

r

)(n

r

)=r4 − 2r3 + r2 + 4

2(r − 1)

[(m− rr − 1

)(n− rr − 2

)+

(m− rr − 2

)(n− rr − 1

)]−D2

=1

r!2

[1

2(r6 − 2r5 + r4 + 4r2)(mr−1nr−2 +mr−2nr−1)

+O(mr−1nr−3 +mr−3nr−1)

]−D2

=1

r!2[2r2(mr−1nr−2 +mr−2nr−1

)+O

(r10(mr−1nr−3 +mr−3nr−1)

)].

Combining this equation with Equation (5.5) and applying Theorem 5.1.5, we obtain

Theorem 5.4.2.

Remark 5.4.7. If we fix k, r ∈ N0 where r ≥ 2 and let m,n→∞ then we obtain

#{T ∈ Tm,n;T contains exactly k 2r-cycles} � 2mn ·{m+ n

m2n2

}.

Remark 5.4.8. In the special case that m = n, if f : N≥2 → R+ satisfies f(n) = o(n3)

and 2 ≤ r ≤ (log n)/3 then the probability of a random bipartite tournament T ∈ Tn,n


containing less than f(n) many unrestricted 2r-cycles approaches 0 when n→∞.

We note that in the proof of last theorem, we have calculated explicitly the case r = 2.

Thus, we can derive from the above estimates that

Corollary 5.4.9. Let Tm,n be the collection of all bipartite tournaments on sets P and

Q, where |P | = m and |Q| = n. For a fixed k ∈ N0, we have

#{T ∈ Tm,n; T contains exactly k 4-cycles} ≤ 2mn(m

2

)(n

2

){2m+ 2n+ 1[8k −

(m2

)(n2

)]2}.

6Future applications of Sieve Theory

We saw in the last chapter that the setting of the shifted Turan sieve is rather flexible,

since we are free to choose the partite sets A and B. But when the structure of the sets

A and B is complicated, the calculation of |A|,∑

b∈B deg b or∑

b1,b2∈B n(b1, b2) can be

difficult.

In other types of sieve also there is a limitation. For example, the difficulty to lower

the bound in Maynard’s work or to obtain the Twin Prime Conjecture or Goldbach’s

conjecture through Chen’s theorems ilustrates this limitation. See, for example, the parity

problem in an excelent text from T. Tao in https://terrytao.wordpress.com/2007/06/

05/open-question-the-parity-problem-in-sieve-theory/, or for some other texts,

see https://terrytao.wordpress.com/tag/parity-problem/.

We believe that the combinatorial Turan sieve will have more applications in the

future. The purpose of this chapter is mainly to introduce it as a viable tool to deal with

combinatorial questions. We now apply the Turan sieve to a zero-sum problem concerning

EGZ constant. Nowadays, this application does not produce any good results, however

we expect that this connection between sieve theory and zero-sum problems will soon

generate better theorems than those that currently exist.

Let σ : {0, 1, 2, . . . , nd−1} → Zdn denote the natural bijection that writes the numbers

in the base n and separates them by comma, resulting in a vector. For example, σ(0) =

(0, . . . , 0︸︷︷︸d times

), σ(1) = (0, . . . , 0︸︷︷︸d−1 times

, 1), σ(n) = (0, . . . , 0︸︷︷︸d−2 times

, 1, 0) and σ(nd − 1) = (n− 1, . . . , n− 1︸︷︷︸d times

).

https://terrytao.wordpress.com/2007/06/05/open-question-the-parity- problem-in-sieve-theory/

https://terrytao.wordpress.com/2007/06/05/open-question-the-parity- problem-in-sieve-theory/

https://terrytao.wordpress.com/tag/parity-problem/

75

Let N ≥ n be a positive integer and define the partite sets as follows:

A =

(a0, a1, . . . , and−1) ∈ Nnd

0

∣∣∣ nd−1∑i=0

ai = N

,

B =

(b0, b1, . . . , bnd−1) ∈ Nnd

0

∣∣∣ nd−1∑i=0

bi = n andnd−1∑i=0

biσ(j) ≡ (0, . . . , 0︸︷︷︸d times

) (mod n)

.

The idea behind the elements a = (a0, a1, . . . , and−1) ∈ A and b = (b0, b1, . . . , bnd−1) ∈B is the following: For 0 ≤ i ≤ nd− 1, the number ai ∈ N0 will denote the multiplicity of

σ(ai) in a sequence with N elements. Therefore, the partite set A can be seen as the set

of all sequences with N elements. On the other hand, for 0 ≤ j ≤ nd − 1, the number bj

will denote the multiplicity of σ(bj) in a zero-sum sequence with n elements. Thus, the

partite set B can be seen as the set of all sequences with n elements whose sum is 0 ∈ Zdn.

The only condition that an element a = (a0, a1, . . . , and−1) ∈ Nnd

0 must satisfy in order

to be in A is∑nd−1

i=0 ai = N , therefore we obtain

|A| =(N + nd − 1

nd − 1

).

We say that a ∼ b if aj ≥ bj for all 0 ≤ j ≤ nd − 1, or in other words, if a sequence a

with N elements contains a subsequence b with n elements whose sum is 0. Thus

ω(a) = #{subsequences of a with n elements whose sum is 0 ∈ Zdn},

deg b = #{sequences with N elements containing b}.

Hence, the equality ω(a) = 0 means that a is free of zero-sum subsequences.

Having fixed a sequence b = (b0, b1, . . . , bnd−1) ∈ B (with n elements), we add some

elements in order to obtain a sequence a = (b0 +c0, b1 +c1, . . . , bnd−1 +cnd−1) ∈ A (with N

elements) such that a ∼ b. Thus, we have c0 + c1 + · · ·+ cnd−1 = N − n, and the number

of elements that we may add is exactly the number of solutions of this equation, i.e.:

deg b =

(N − n+ nd − 1

nd − 1

)∀ b ∈ B.

However, reasonable estimates for |B| and n(b1, b2), where b1, b2 ∈ B, seem difficult

to achieve. It is enough to have a good lower bound for |B| and a good upper bound for

n(b1, b2). The Turan sieve (Corollary 5.1.4) would provide an upper bound for the number

of sequences in Zdn with N elements that are free of zero-sum subsequences of length n. We

hope that soon we can calculate precisely these values and take, for example, N = 8n− 8

if n is even and N = 9n− 9 if n is odd in order to study s(Z3n) in a non-trivial way.

APPENDIX

Notations

In addition to the notations introduced in Parts I and II, we shall also use the following:

• C is the set of complex numbers;

• If z ∈ C then z denotes the conjugate of z;

• Zn = (Z/nZ) is the set of residue classes modulo n;

• If G is a group and A is a ring then G∗ = G \ {e}, where e is the identity of G, and

A∗ = A \ {0};

• If G is a group and X ⊂ G then X = G \X is the complementar of X in G;

AErdos-Ginzburg-Ziv theorem

In this chapter, we give a proof of the remarkable theorem due to Erdos, Ginzburg

and Ziv [27] using Chevalley-Warning theorem. In Appendix B.3 we give a proof that

uses another tool: Cauchy-Davenport inequality.

A.1 Chevalley-Warning theorem

Lemma A.1.1. Let p a prime. Then

1t + 2t + · · ·+ (p− 1)t (mod p) =

0 if (p− 1) 6 | t,

p− 1 if (p− 1) | t.

Proof. The case (p − 1)|t is obvious. So let us assume (p − 1) 6 | t. Let g be a primitive

root modulo p. We have

1t + 2t + · · ·+ (p− 1)t ≡ 1 + gt + g2t + · · ·+ g(p−2)t (mod p).

Let S = 1 + gt + g2t + · · ·+ g(p−2)t. Then

gtS ≡ gt + g2t + · · ·+ g(p−2)t + 1 ≡ S (mod p) ⇐⇒ (gt − 1)S ≡ 0 (mod p).

Since g is a primitive root modulo p, the only possibility is S ≡ 0 (mod p).

Theorem A.1.2 (Chevalley-Warning). Let p a prime and let

f1(x1, . . . , xn), . . . , fk(x1, . . . , xn) ∈ Z[x1, . . . , xn]

A.2 Proof of Erdos-Ginzburg-Ziv theorem 79

such thatk∑i=1

deg(fi) < n.

Then the number of solutions of the system

fi(x1, . . . , xn) ≡ 0 (mod p) para todo 1 ≤ i ≤ k

is a multiple of p. In particular, if (0, . . . , 0) ∈ (Zp)n is a solution then there exist a

solution (x1, . . . , xn) 6= (0, . . . , 0) in (Zp)n.

Proof. Let N denote the number of solutions of the above system and let

g(x1, . . . , xn) =k∏j=1

(1− fj(x1, . . . , xn)p−1

).

By Fermat’s Little Theorem, we obtain

N ≡∑

(x1,...,xn)∈Znp

g(x1, . . . , xn) (mod p). (A.1)

Also, the hypothesis implies that deg(g) ≤∑

1≤j≤k(p− 1) deg(fj) < (p− 1)n, hence each

monomial cxi11 xi22 . . . x

inn of g is such that

∑1≤j≤n ij < (p − 1)n, therefore by Pigeonhole

Principle there exist 1 ≤ r ≤ n with 0 ≤ ir < p− 1. By previous lemma,∑

xr∈Zp xirr ≡ 0

(mod p) and so∑(x1,...,xn)∈Znp

cxi11 xi22 . . . x

inn ≡ c

∑x1∈Zp

xi11∑x2∈Zp

xi22 · · ·∑xn∈Zp

xinn ≡ 0 (mod p).

Thus, by Equation (A.1) we obtain N ≡ 0 (mod p).

A.2 Proof of Erdos-Ginzburg-Ziv theorem

Theorem A.2.1 (Erdos-Ginzburg-Ziv). Let n ∈ N. Given a1, a2, . . . , a2n−1 ∈ Z, there

exist 1 ≤ i1 < i2 < · · · < in ≤ 2n− 1 such that ai1 + ai2 + · · ·+ ain is divisible by n.

Proof. We first show that the theorem is totally multiplicative, i.e. if it holds for m

and n then it holds for mn. Let a1, a2, . . . , a2mn−1 ∈ Z. By hypothesis, for each A ⊂{1, 2, . . . , 2mn − 1} with |A| = 2n − 1, there exist B ⊂ A with |B| = n and

∑i∈B xi

divisible by n. Inductively, we construct Bj, 1 ≤ j ≤ 2m− 1 as follows:

A.2 Proof of Erdos-Ginzburg-Ziv theorem 80

1. We select any Aj ⊂ {1, 2, . . . , 2mn− 1} \⋃k<j Bk with |A| = 2n− 1;

2. We select Bj ⊂ Aj with |Bj| = n and∑

i∈Bj xi ≡ 0 (mod n).

Notice that if j ≤ 2m− 1 then∣∣∣∣∣{1, 2, . . . , 2mn− 1} \⋃k<j

Bj

∣∣∣∣∣ = 2mn− 1− (j − 1)n ≥ 2mn− 1− (2m− 2)n = 2n− 1,

and it guarantee this construction until j = 2m − 1. Let bj = 1n

∑i∈Bj ai ∈ Z. By

hypothesis, there exist C ⊂ {1, 2, . . . , 2m− 1} with |C| = m and∑

j∈C bj ≡ 0 (mod m).

Thus, ∑j∈C

∑i∈Bj

ai = n∑j∈C

bj ≡ 0 (mod mn)

and it has mn summands.

Now, it is enough to prove the theorem for n = p prime. Let

f1(x1, . . . , x2p−1) = xp−11 + xp−12 + · · ·+ xp−12p−1,

f2(x1, . . . , x2p−1) = a1xp−11 + a2x

p−12 + · · ·+ a2p−1x

p−12p−1.

Notice that (0, . . . , 0) is a solution of the system f1 ≡ f2 ≡ 0 (mod p). By Chevalley-

Warning theorem, there exist a non-trivial solution. Observe that f1 ≡ 0 (mod p) and

Fermat’s little theorem ensure that the number of non-zero coordinates xi is exactly p and

f2 ≡ 0 (mod p), Fermat’s little theorem and previous observation ensure that the sum of

the respectives ai’s is 0 ∈ Zp.

BCauchy-Davenport inequality

Let G be a finite abelian group and let A1, . . . , An, A,B be finite subsets of G. We

define the sum-set

A1 + · · ·+ An = {a1 + · · ·+ an; ai ∈ Ai, 1 ≤ i ≤ n}.

We also define −A = {−a; a ∈ A}, and so A−B = {a− b; a ∈ A, b ∈ B}. If c ∈ G we

define c+ A = {c}+ A, the translation by c.

For each g ∈ G we denote by rA,B(g) = #{(a, b) ∈ A × B; g = a + b} as the number

of representations of g as a sum of an element in A with an element in B.

Lemma B.0.1. Let G be a finite abelian group and let A,B be subsets of G such that

|A|+ |B| ≥ |G|+ t. Then rA,B(g) ≥ t for all g ∈ G. In particular, if |A|+ |B| > |G| then

A+B = G.

Proof. We have

|G| ≥ |A ∪ (g −B)| = |A|+ |g −B| − |A ∩ (g −B)| = |A|+ |B| − |A ∩ (g −B)|,

therefore rA,B(g) = |A ∩ (g −B)| ≥ |A|+ |B| − |G| ≥ t.

From now on, we may assume |A|+ |B| ≤ |G| in order to study A+B in a non-trivial

way.

B.1 The e-transform 82

B.1 The e-transform

Let (A,B) be a ordered pair of non-empty subsets in an abelian group G. For each

e ∈ G, we define the e-transform of (A,B) as being the pair (A(e), B(e)) of subsets of G

given by

A(e) = A ∪ (e+B) and B(e) = B ∩ (−e+ A).

Proposition B.1.1. Let A,B be non-empty subsets of an abelian group G. For each

e ∈ G, the e-transform of (A,B) satisfies:

1. A ⊂ A(e) and B(e) ⊂ B;

2. A(e) +B(e) ⊂ A+B

3. A(e) \ A = e+ (B \B(e));

4. |A(e)|+ |B(e)| = |A|+ |B| provided A,B are finite;

5. If e ∈ A and 0 ∈ B then e ∈ A(e) and 0 ∈ B(e).

Proof. 1. It follows directly from definition;

2. Let a′ ∈ A(e) and b′ ∈ B(e). We have a′ ∈ A or a′ ∈ e + B. If a′ ∈ A then

b′ ∈ B(e) ⊂ B and so a′ + b′ ∈ A + B. On the other hand, if a′ ∈ e + B then

b′ ∈ B(e) ⊂ −e + A, so there exist a ∈ A and b ∈ B such that a′ = b + e and

b′ = a− e. Therefore, a′ + b′ = (b+ e) + (a− e) = a+ b ∈ A+B;

3.

A(e) \ A = (B + e) \ A = {b+ e; b ∈ B and b+ e 6∈ A}

= e+ {b ∈ B; b 6∈ A− e} = e+ {b ∈ B; b 6∈ B(e)} = e+ (B \B(e));

4. By (ii), we have |A(e)|−|A| = |A(e)\A| = |e+(B\B(e))| = |B\B(e)| = |B|−|B(e)|;

5. It follows directly from definition.

B.2 Proof of Cauchy-Davenport inequality

In this section, we present the proof of Cauchy-Davenport inequality (reproducing the

result of [65, p. 44-45]). This theorem gives the tight bound for the cardinality of a sum-

B.2 Proof of Cauchy-Davenport inequality 83

set in Zp, where p is a prime. But before we need the I. Chowla’s theorem. In the end,

we exhibit another proof of Erdos-Ginzburg-Ziv theorem.

Theorem B.2.1 (I. Chowla). Let m ≥ 2 and A,B be non-empty subsets of Zm. If 0 ∈ Band (b,m) = 1 for all b ∈ B∗ then

|A+B| ≥ min{m, |A|+ |B| − 1}.

Proof. From Lemma B.0.1, we just need to show the theorem for |A| + |B| ≤ m. Notice

that the theorem trivially holds for min{|A|, |B|} = 1. Thus, assume that there exist

A,B ⊂ Zm with min{|A|, |B|} ≥ 2 such that |A+B| < |A|+ |B| − 1. The last inequality

together with |A| + |B| ≤ m says that A 6= Zm, since othewise we would have |A| = m

and |B| ≥ 2.

Now, we may choose a pair (A,B) such that |B| ≥ 2 is the smallest possible. Let

b∗ ∈ B∗. If a + b∗ ∈ A for all a ∈ A then a + jb∗ ∈ A for all j ≥ 0. Since (b∗,m) = 1,

we have {a + jb∗; 0 ≤ j ≤ m − 1} = Zm, therefore A = Zm, contradiction. Thus, there

exist e ∈ A such that e+ b∗ 6∈ A. By applying the e-transform in (A,B) and by previous

lemma, we obtain:

|A(e) +B(e)| ≤ |A+B| < |A|+ |B| − 1 = |A(e)|+ |B(e)| − 1.

Since e ∈ A and 0 ∈ B, the item (v) from previous lemma implies that 0 ∈ B(e).

Furthermore, (b,m) = 1 for all b ∈ B(e)∗, since B(e) ⊂ B. But b∗ 6∈ A − e and so

b∗ 6∈ B(e). Therefore, |B(e)| < |B| and |B| is not the smallest possible, contradiction.

Theorem B.2.2 (Cauchy-Davenport). Let p be a prime and A,B be non-empty subsets

of Zp. Then

|A+B| ≥ min{p, |A|+ |B| − 1}.

In particular, if A1, . . . , An are non-empty subsets of Zp then

|A1 + · · ·+ An| ≥ min

{p,

n∑i=1

|Ai| − n+ 1

}.

Proof. Let b0 ∈ B. We have 0 ∈ −b0 + B and (b, p) = 1 for all 0 6= b ∈ B, thus A and

−b0 +B satisfy I. Chowla’s theorem. Hence

|A+B| = |A+ (−b0 +B)| ≥ min{p, |A|+ | − b0 +B| − 1} = min{p, |A|+ |B| − 1}.

B.3 Another proof of Erdos-Ginzburg-Ziv theorem 84

The other part follows immediately by induction.

B.3 Another proof of Erdos-Ginzburg-Ziv theorem

This is the original proof of Erdos-Ginzburg-Ziv theorem. It makes use of Cauchy-

Davenport inequality and we reproduce here:

Proof. As in Section A.2, we assume that n = p is a prime. Without loss of generality,

we also may assume 0 ≤ a1 ≤ a2 ≤ · · · ≤ a2p−1 ≤ p − 1. If ai = ai+p−1 for some

1 ≤ i ≤ p− 1 then ai + ai+1 + · · ·+ ai+p−1 = pai ≡ 0 (mod p), so we are done. Otherwise,

let Ai = {ai, ai+p−1} for 1 ≤ i ≤ p− 1. By Cauchy-Davenport inequality, we obtain

|A1 + A2 + · · ·+ Ap−1| ≥ min

{p,

p−1∑i=1

(|Ai| − 1) + 1

}= p,

therefore A1 + A2 + · · · + Ap−1 = Zp. In particular, −a2p−1 can be written as the sum

of exactly p − 1 elements of the elements of the sequence (a1, a2, . . . , a2p−1) and we are

done.

CVosper’s theorem

In this chapter we present the tightness of Cauchy-Davenport inequality, expliciting

via Vosper’s theorem [84] the cases where the equality |A + B| = |A| + |B| − 1 occurs.

Observe that Lemma B.0.1 provides an example of sufficient conditions for the equality

|A+B| = p.

Theorem C.0.1. Let p be a prime and let A,B be non-empty subsets of the group Zpsuch that A+B 6= Zp. Then

|A+B| = |A|+ |B| − 1

if and only if at least one of the following conditions holds:

(i) min{|A|, |B|} = 1;

(ii) |A+B| = p− 1 and B = c− A, where {c} = Zp \ (A+B);

(iii) A and B are arithmetic progressions with the same common difference.

Proof. (⇐):

(i) If min{|A|, |B|} = 1 then |A+B| = |A|+ |B| − 1.

(ii) Let c ∈ Zp, A be a subset with 2 ≤ |A| ≤ p−1 and B = c− A. Then c 6∈ A+B and

so |A+B| ≤ p−1. Since |B| = |c− A| = p−|c−A| = p−|A|, the Cauchy-Davenport

inequality implies that

p− 1 = |A|+ |B| − 1 ≤ |A+B| ≤ p− 1,

therefore |A+B| = |A|+ |B| − 1.

86

(iii) If A,B are arithmetic progressions in Zp with the same common difference d then

there exist a, b ∈ Zp and positive integers k, l with k + l ≤ p such that

A = {a+ id; 0 ≤ i ≤ k − 1};

B = {b+ jd; 0 ≤ j ≤ l − 1}.

If d ≡ 0 (mod p) then |A| = |B| = 1 and we are done. Therefore d ∈ Z∗p, so

A+B = {a+ b+ id; 0 ≤ i ≤ k + l − 2}.

Hence, |A+B| = k + l − 1 = |A|+ |B| − 1.

Thus, the equality in Cauchy-Davenport occurs provided (i), (ii) or (iii).

(⇒): Conversely, suppose that |A+B| = |A|+ |B| − 1. If min{|A|, |B|} = 1 then we

are done because (A,B) is of the form (i). If |A+B| = p− 1 then A+B = {c} for some

c ∈ Zp. Since c 6∈ A+B, it follows that B ∩ (c− A) = ∅, hence B ⊂ c− A. Therefore

|B| ≤ |c− A| = p− |c− A| = p− |A|.

Since

p− 1 = |A+B| = |A|+ |B| − 1 ≤ p− 1,

it follows that |B| = p − |A|, hence B = c− A and the pair (A,B) is of the form (ii).

From now on, we may assume

min{|A|, |B|} ≥ 2 and |A+B| < p− 1.

In order to show that A and B are arithmetic progressions with the same common differ-

ence, we need some lemmas:

Lemma C.0.2. Let A and B be subsets of Zp such that min{|A|, |B|} ≥ 2 and |A+B| =|A|+ |B|−1 < p−1. If A is an arithmetic progression then B is an arithmetic progression

with the same common difference. In particular, if min{|A|, |B|} = 2 then A and B

are arithmetic progressions with the same common difference, since every set with two

elements is an arithmetic progression.

Proof. Let |A| = k and |B| = l. There exist a0 ∈ Zp and d ∈ Z∗p such that A =

{a0 + id; 0 ≤ i ≤ k − 1}. Define

A′ = {(a− a0)d−1; a ∈ A} = {i+ pZ; 0 ≤ i ≤ k − 1} ⊂ Zp.

87

Select b0 ∈ B and define

B′ = {(b− b0)d−1; b ∈ B}.

We have 0 ∈ B′, |A′| = |A| = k ≥ 2, |B′| = |B| = l ≥ 2, A′ + B′ = {(c− a0 − b0)d−1; c ∈A+B} and so |A′+B′| = |A+B| = |A|+ |B|− 1 = |A′|+ |B′|− 1 < p− 1. Therefore, we

may assume that A = A′ and B = B′ without loss of generality. We would like to show

that B = {b, b+ 1, b+ 2, . . . , b+ l − 1} for some b ∈ B.

Denote B = {b0, b1, . . . , bl−1}, where 0 ≤ b1 < b2 < · · · < bl−1 ≤ p − 1 and suppose

that bl = p. Since every element of A+B is of the form bj + i+pZ for some 0 ≤ i ≤ k− 1

and 0 ≤ j ≤ l − 1, we obtain that

A+B ⊂l−1⋃j=0

[bj, bj + min{k − 1, bj+1 − bj − 1}] + pZ.

The above union is disjoint, therefore

k+ l−1 = |A+B| =l−1∑j=0

(1 + min{k−1, bj+1− bj−1}) = l+l−1∑j=0

min{k−1, bj+1− bj−1}.

If bj+1 − bj − 1 ≤ k − 1 for all 0 ≤ j ≤ l − 1 then k + l − 1 = bl − b0 = p, which is a

contradiction. Therefore bj0+1 − bj0 − 1 > k − 1 for some 0 ≤ j0 ≤ l − 1, hence

k + l − 1 = |A+B| = k + l − 1 +l−1∑j=0j 6=j0

min{k − 1, bj+1 − bj − 1},

which implies bj+1 = bj + 1 for all 0 ≤ j ≤ l − 1 with j 6= j0. Thus B = {bj0+1, bj0+1 +

1, . . . , bj0+1 + l − 1} is an arithmetic progression with common difference 1.

Lemma C.0.3. Let A and B be subsets of Zp such that min{|A|, |B|} ≥ 2 and |A+B| =|A|+ |B| − 1 < p− 1. If D = A+B then |D − A| = |D|+ |A| − 1.

Proof. Let |A| = k and |B| = l. Since k + l − 1 ≤ p− 2, we obtain that |D| = |A+B| =p− (k + l − 1) ≥ 2. By Cauchy-Davenport inequality, it follows that

|D − A| ≥ min{p, |D|+ |A| − 1} = min{p, (p− k − l + 1) + k − 1} = p− l.

On the other hand, since D∩(A+B) = ∅ we obtain B∩(D−A) = ∅ and so D−A ⊂ B.

Thus, |D − A| ≤ |B| = p− |B| = p− l. Therefore, |D − A| = p− l = |D|+ |A| − 1.

88

Lemma C.0.4. Let A and B be subsets of Zp such that min{|A|, |B|} ≥ 2 and |A+B| =|A|+ |B| − 1 < p− 1. If A+B is an arithmetic progression then A and B are arithmetic

progressions with the same common difference.

Proof. Since A + B is an arithmetic progression then D = A+B is also an arithmetic

progression. By previous lemma, |D − A| = |D| + |A| − 1 and by Lemma C.0.2, the set

−A is an arithmetic progression. Therefore A is an arithmetic progression and, again by

Lemma C.0.2, B is an arithmetic progression with the same common difference.

Lemma C.0.5. Let A and B be subsets of Zp such that |A| = k ≥ 2, |B| = l ≥ 3, 0 ∈ Band |A + B| = |A| + |B| − 1 < p − 1. Then there exists a congruence class e ∈ A with

the property that the e-transform (A(e), B(e)) satisfies |A(e) +B(e)| = |A(e)|+ |B(e)|− 1

and 2 ≤ |B(e)| < |B|.

Proof. If (A(e), B(e)) is any e-transform of the critical pair (A,B) then, by Proposition

B.1.1 and Cauchy-Davenport inequality, it follows that

|A|+ |B| − 1 = |A(e)|+ |B(e)| − 1 ≤ |A(e) +B(e)| ≤ |A+B| = |A|+ |B| − 1,

therefore |A(e)| + |B(e)| − 1 = |A(e) + B(e)| = |A + B|. By Proposition B.1.1, we have

A(e) + B(e) = A+ B. Let X = {e ∈ A;B(e) 6= B}. We have |B(e)| < |B| for all e ∈ X.

We would like to show that |X| ≥ 2. Let Y = A \X = {e ∈ A;B(e) = B}.If Y = ∅ then X = A and |X| ≥ 2. Assume that Y 6= ∅ and choose e ∈ Y . Then

B = B(e) = B∩ (−e+A), which implies B ⊂ −e+A. Therefore e+B ⊂ A for all e ∈ Y ,

and so Y +B ⊂ A. By Cauchy-Davenport inequality,

k = |A| ≥ |B + Y | ≥ min{p, |Y |+ l − 1} = |Y |+ l − 1 = k − |X|+ l − 1,

therefore |X| ≥ l − 1 ≥ 2.

We would like to show that |B(e)| ≥ 2 for some e ∈ X. Since e ∈ X ⊂ A and 0 ∈ B,

it follows that 0 ∈ B(e). Suppose that B(e) = B ∩ (−e + A) = {0} for all e ∈ X. Then

B∗∩ (−e+A) = ∅ and so (e+B∗)∩A = ∅ for all e ∈ X. Thus (X+B∗)∩A = ∅. Since

X +B∗ ⊂ A+B, we obtain X +B∗ ⊂ (A+B) \A, and by Cauchy-Davenport inequality,

|X|+ l − 2 = |X|+ (l − 1)− 1 ≤ |X +B∗| ≤ |A+B| − |A| = l − 1,

which is a contradiction since |X| ≥ 2.

From now on, we use these lemmas to continue the proof of Vosper’s theorem.

89

Let A and B such that |A + B| = |A| + |B| − 1 and |B| = l ≥ 2. We proceed by

induction on l. If l = 2 then the theorem follows from Lemma C.0.2. Suppose that l ≥ 3

and assume that the theorem holds for all pairs (A,B) such that |B| < l. By Lemma C.0.5,

there exist e ∈ A such that (A(e), B(e)) satisfies |A(e) +B(e)| = |A(e)|+ |B(e)| − 1 with

A(e) + B(e) = A + B and 2 ≤ |B(e)| < l. By hypothesis, A(e) and B(e) are arithmetic

progressions with the same common difference. Therefore, A(e) + B(e) = A + B is an

arithmetic progression and, by Lemma C.0.4, A and B are arithmetic progressions with

the same common difference. This completes the proof of Vosper’s theorem.

DAn elliptic curve modulo p

In this chapter we present a plane algebraic curve modulo prime, namely,

a1xl11 + · · ·+ arx

lrr ≡ b (mod p),

over the conditions li|p − 1 for 1 ≤ i ≤ r. We estimate the number of solutions of this

equation and use a particular case where r = 2, l1 = 2, l2 = 4 (that is, an elliptic curve)

and p ≡ 1 (mod 4) in the proof of Theorem 2.1.1 (see Lemma 2.1.7 and the subsequent

Corollary 2.1.8). This chapter is an adaptation of Chapter 8 from [49], called “Gauss and

Jacobi Sums”.

D.1 Multiplicative characters

A multiplicative character (or just “character”) on Zp is a completely multiplicative

map χ : Z∗p → C∗.For a ∈ Zp, define the Legendre symbol given by (a/p) = a(p−1)/2, i.e.:

(a/p) =

0, if a = 0;

1, if a is a quadratic residue modulo p;

−1, if a is not a quadratic residue modulo p.

The Legendre symbol is an example of such a character.

Another example is the trivial multiplicative character defined by the relation ε(a) = 1

for all a ∈ Z∗p.It is often useful to extend the domain of definition of a character to all of Zp. If

D.1 Multiplicative characters 91

χ 6= ε, we do this by defining χ(0) = 0. For ε we define ε(0) = 1. The usefulness of these

definitions will soon become apparent.

Proposition D.1.1. Let χ be a character and a ∈ Z∗p. Then

1. χ(1) = 1;

2. χ(a) is a (p− 1)st root of unity;

3. χ(a−1) = χ(a)−1 = χ(a).

Proof. 1. χ(1) = χ(1 · 1) = χ(1)χ(1). Since χ(1) 6= 0, we have χ(1) = 1. Here, the 1 on

left-hand side belongs to Zp, whereas the 1 on the right-hand side belongs to C;

2. By Fermat’s little theorem we have ap−1 ≡ 1 (mod p), which implies that 1 = χ(1) =

χ(ap−1) = χ(a)p−1;

3. Since 1 = χ(1) = χ(a−1a) = χ(a−1)χ(a), we have χ(a−1) = χ(a)−1. The other equality

follows from the fact that χ(a) is a complex number of norm 1, from (b).

Proposition D.1.2. Let χ be a character. Then

∑t∈Zp

χ(t) =

0, if χ 6= ε;

p, if χ = ε.

Proof. The χ = ε part is easy to check. So we assume χ 6= ε and in this case there is an

a ∈ Z∗p such that χ(a) 6= 1. Let T =∑

t∈Zp χ(t). Then

χ(a)T =∑t∈Zp

χ(a)χ(t) =∑t∈Zp

χ(at) = T.

Therefore, (χ(a)− 1)T = 0 and so T = 0.

The characters form a group by means of the following definitions:

(a) If χ and λ are characters then χλ : a ∈ Z∗p → χλ(a);

(b) If χ is a character, then χ−1 : a ∈ Z∗p → χ(a)−1.

It is not difficult to check that χλ and χ−1 are characters and that these definitions

make the set of characters into a group with identity ε.

D.2 The equation xn ≡ a (mod p) 92

Proposition D.1.3. The group of characters is a cyclic group of order p− 1. If a ∈ Z∗pand a 6≡ 1 (mod p) then there is a character χ such that χ(a) 6= 1.

Proof. We know that Z∗p is cyclic. Let g ∈ Z∗p be a generator. Suppose that a ≡ gl (mod p)

and χ is a character, then χ(a) = χ(g)l. This shows that χ is completely determined by

the value χ(g). Since χ(g) is a (p− 1)st root of unity and since there are exactly p− 1 of

these, it follows that the character group has order at most p− 1. Now, define a function

λ by the equation λ(gk) = e2πik/(p−1). It is easy to check that λ is well defined and is a

character. We claim that p− 1 is the smallest integer n such that λn = ε. If λn = ε then

λn(g) = ε(g) = 1. However, λn(g) = λ(gn) = e2πin/(p−1). It follows that p − 1 divides

n. Since λp−1(a) = λ(ap−1) = λ(1) = 1 we have λp−1 = ε. We have established that the

characters ε, λ, λ2, . . . , λp−2 are all distinct. Since by the first part of the proof there are

at most p−1 characters, we now have that there are exactly p−1 characters and that the

group is cyclic with λ as a generator. If a ∈ Z∗p and a 6≡ 1 (mod p) then a ≡ gl (mod p)

with p− 1 6 | l. Therefore λ(a) = λ(g)l = e2πil/(p−1) 6= 1 and this concludes the proof.

Corollary D.1.4. If a ∈ Z∗p and a 6≡ 1 (mod p) then∑

χ χ(a) = 0, where the summation

is over all characters.

Proof. Let S =∑

χ χ(a). Since a 6≡ 1 (mod p) there is, by previous proposition, a

character λ such that λ(a) 6= 1. Then

λ(a)S =∑χ

λ(a)χ(a) =∑χ

λχ(a) = S,

since λχ runs over all characters as χ does. It follows that (λ(a) − 1)S = 0 and thus

S = 0.

D.2 The equation xn ≡ a (mod p)

Proposition D.2.1. If a ∈ Z∗p, n|p− 1 and xn ≡ a (mod p) is not solvable then there is

a character χ such that

(a) χn = ε;

(b) χ(a) 6= 1.

Proof. Let g and λ be as in Proposition D.1.3 and set χ = λ(p−1)/n. Then χ(g) =

λ(p−1)/n(g) = λ(g)(p−1)/n = e2πi/n. Now, since a ≡ gl (mod p) for some l and xn = a is not

solvable, we must have n 6 | l. Then χ(a) = χ(g)l = e2πil/n 6= 1. Finally, χn = λp−1 = ε.

D.3 The sums of Gauss and Jacobi 93

Let a ∈ Zp and N(xn = a) denote the number of solutions of the equation xn ≡ a

(mod p). This equation and the characters are related by the following proposition:

Proposition D.2.2. If n|p− 1 then

N(xn = a) =∑χn=1

χ(a),

where the sum is over all characters of order dividing n.

Proof. We claim that there are exactly n characters of order dividing n. Since the value

of χ(g) for such a character must be an nth root of unity, there are at most n such

characters. In Proposition D.2.1, we found a character χ such that χ(g) = e2πi/n. It

follows that ε, χ, χ2, . . . , χn−1 are n distinct characters of order dividing n.

To prove the formula, notice that xn ≡ 0 (mod p) has one solution, namely, x ≡ 0

(mod p), and∑

χn=ε χ(0) = 1, since ε(0) = 1 and χ(0) = 0 for χ 6= ε.

Now, suppose that a 6≡ 0 (mod p) and that bn ≡ a (mod p). If χn = ε then χ(a) =

χn(b) = ε(b) = 1. Thus∑

χn=ε χ(a) = n, which is N(xn = a) in this case.

Finally, suppose that a 6≡ 0 (mod p) and that xn ≡ 0 (mod p) is not solvable. Let

T =∑

χn=ε(a). By Proposition D.2.1, there is a character ρ such that ρ(a) 6= 1 and

ρn = ε. A simple calculation shows that ρ(a)T = T , thus T = 0, as required.

D.3 The sums of Gauss and Jacobi

Definition D.3.1 (Gauss sum). Let χ be a character on Zp and a ∈ Zp. A Gauss sum

is defined by the formula

ga(χ) =∑t∈Zp

χ(t)ζat,

where ζ = e2πi/p.

Proposition D.3.2. The value of ga(χ) is summarized in the following table:

(χ, a) a = 0 a 6= 0

χ = ε g0(ε) = p ga(ε) = 0

χ 6= ε g0(χ) = 0 ga(χ) = χ(a−1)g1(χ)

Proof. • a = 0, χ = ε: g0(ε) =∑

t∈Zp ε(t)ζ0t =

∑t∈Zp 1 = p;

• a 6= 0, χ = ε: ga(ε) =∑

t∈Zp ε(t)ζat =

∑t∈Zp ζ

at = 0;


• a = 0, χ 6= ε: g0(χ) =∑

t∈Zp χ(t)ζ0t =∑

t∈Zp χ(t) = 0;

• a 6= 0, χ 6= ε: χ(a)ga(χ) =∑

t∈Zp χ(at)ζat = g1(χ).

From now on we denote g1(χ) = g(χ).

Proposition D.3.3. If χ 6= ε then |g(χ)| = √p.

Proof. If a 6= 0 then, by Proposition D.3.2, ga(χ) = χ(a−1g(χ) = χ(a)g(χ) and ga(χ) =

χ(a−1)g(χ). Thus ga(χ)ga(χ) = χ(a−1)χ(a)g(χ)g(χ) = |g(χ)|2. Since g0(χ) = 0, we have∑a∈Zp

ga(χ)ga(χ) = (p− 1)|g(χ)|2.

On the other hand, ga(χ)ga(χ) =∑

x∈Zp∑

y∈Zp χ(x)χ(y)ζax−ay. Summing over a we

obtain ∑a∈Zp

ga(χ)ga(χ) =∑x∈Zp

∑y∈Zp

χ(x)χ(y)δ(x− y)p = (p− 1)p,

where δ(z) = 0 if z 6= 0 and δ(z) = 1 if z = 0.

Therefore, (p− 1)|g(χ)|2 = (p− 1)p and the result follows.

Definition D.3.4 (Jacobi sum). Let χ1, . . . , χl be characters on Zp. A Jacobi sum is

defined by the formula

J(χ1, . . . , χl) =∑

t1+···+tl=1

χ1(t1) . . . χl(tl).

It is useful to define another sum, left unnamed:

J0(χ1, . . . , χl) =∑

t1+···+tl=0

χ1(t1) . . . χl(tl).

Proposition D.3.5. (a) J0(ε, . . . , ε) = J(ε, . . . , ε) = pl−1;

(b) If some but not all of the χi are trivial then J0(χ1, . . . , χl) = J(χ1, . . . , χl) = 0;

(c) Assume that χl 6= ε. Then

J0(χ1, . . . , χl) =

0, if χ1 . . . χl 6= ε;

χl(−1)(p− 1)J(χ1, . . . , χl−1), otherwise.


Proof. If t1, . . . , tl−1 are chosen (arbitrarily) in Zp then tl is uniquely determined by the

condition t1 + · · ·+ tl−1 + tl = 0. Thus J0(ε, . . . , ε) = pl−1. Similarly for J(ε, . . . , ε).

For the part (b), assume that χ1, . . . , χs are nontrivial and that χs+1 = · · · = χl = ε.

Then ∑t1+···+tl=0

χ1(t1) . . . χl(tl) =∑

t1,...,tl−1

χ1(t1) . . . χs(ts)

= pl−s−1

(∑t1

χ1(t1)

). . .

(∑ts

χs(ts)

)= 0.

Thus J0(χ1, . . . , χl) = 0. Similarly for J(χ1, . . . , χl).

To prove part (c), notice that

J0(χ1, . . . , χl) =∑s

∑t1+···+tl−1=−s

χ1(t1) . . . χl−1(tl−1)

χl(s).

Since χl 6= ε, χl(0) = 0, so we may assume that s 6= 0 in the above sum. If s 6= 0,

define t′i by ti = −st′i. Then∑t1+···+tl−1

χ1(t1) . . . χl−1(tl−1) = χ1 . . . χl−1(−s)∑

t′1+···+t′l−1=1

χ1(t′1) . . . χl−1(t

′l−1)

= χ1 . . . χl−1(−s)J(χ1, . . . , χl−1).

Combining these results yields

J0(χ1, . . . , χl) = χ1 . . . χl−1(−1)J(χ1, . . . , χl−1)∑s 6=0

χ1 . . . χl(s).

The main result follows since the sum is 0 if χ1 . . . χl 6= ε and p−1 if χ1 . . . χl = ε.

Theorem D.3.6. Assume that χ1, . . . , χr are non-trivial and also that χ1 . . . χr is non-

trivial. Then

g(χ1) . . . g(χr) = J(χ1, . . . , χr)g(χ1 . . . χr).

Proof. Let ψ : Zp → C be defined by ψ(t) = ζt. Then ψ(t1 + t2) = ψ(t1)ψ(t2) and


g(χ) =∑χ(t)ψ(t). We have:

g(χ1) . . . g(χr) =

(∑t1

χ1(t1)ψ(t1)

). . .

(∑t1

χ1(t1)ψ(t1)

)

=∑s

( ∑t1+···+tr=s

χ1(t1) . . . χr(tr)

)ψ(s).

If s = 0 then by part (c) of Proposition D.3.5 and the assumption that χ1 . . . χr 6= ε,∑t1+···+tr=0

χ1(t1) . . . χr(tr) = 0.

If s 6= 0, the substitution ti = st′i shows that∑t1+···+tr=s

χ1(t1) . . . χr(tr) = χ1 . . . χr(s)J(χ1, . . . , χr).

Putting these remarks together, we have

g(χ1) . . . g(χr) = J(χ1, . . . , χr)∑s 6=0

χ1 . . . χr(s)ψ(s)

= J(χ1, . . . , χr)g(χ1 . . . χr).

Corollary D.3.7. Suppose that χ1, . . . , χr are non-trivial and that χ1 . . . χr = ε. Then

g(χ1) . . . g(χr) = χr(−1)pJ(χ1, . . . , χr−1).

Proof. By previous theorem, we have g(χ1) . . . g(χr−1) = J(χ1, . . . , χr−1)g(χ1 . . . χr−1).

Since χ1 . . . χr−1 = (χr)−1, we obtain

g(χ1 . . . χr−1)g(χr) = g(χ−1r )g(χr) = χr(−1)p.

Corollary D.3.8. Let the hypothesis be as in previous corollary. Then

J(χ1, . . . , χr) = −χr(−1)J(χ1, . . . , χr−1).


lrr ≡ b (mod p) 97

For r = 2 we set J(χ1) = 1.

Proof. If r = 2, notice that

J(χ, χ−1) =∑a+b=1

χ(a)χ−1(b) =∑a+b=1b 6=0

χ(ab

)=∑a6=1

χ

(a

1− a

).

Set c = a/(1 − a). If c 6= −1 then a = c/(1 + c). It follows that as a varies over Zp, less

the element 1, that c varies over Zp, less the element −1. Thus

J(χ, χ−1) =∑c6=−1

χ(c) = −χ(−1).

Suppose now that r > 2. In the proof of Theorem D.3.6 we use the hypothesis that

χ1 . . . χr = ε. This yields

g(χ1) . . . g(χr) = J0(χ1, . . . , χr) + J(χ1, . . . , χr)∑s 6=0

ψ(s).

Since∑

s ψ(s) = 0, the above sum is −1. By part (c) of Proposition D.3.5, we ob-

tain J0(χ1, . . . , χr) = χr(−1)(p − 1)J(χ1, . . . , χr), and by previous corollary, we have

g(χ1) . . . g(χr) = χr(−1)pJ(χ1, . . . , χr). Putting these results together proves the corol-

lary.

Theorem D.3.9. Assume that χ1, . . . , χr are non-trivial.

(a) If χ1 . . . χr 6= ε then |J(χ1, . . . , χr)| = p(r−1)/2;

(b) If χ1 . . . χr = ε then |J0(χ1, . . . , χr)| = (p− 1)p(r/2)−1 and |J(χ1, . . . , χr)| = p(r/2)−1.

Proof. If χ is non-trivial then |g(χ)| =√p. Part (a) follows directly from Theorem

D.3.6. Part (b) follows similarly from part (c) of Proposition D.3.5 and from previous

corollary.

D.4 The equation a1xl11 + · · · + arx

lrr ≡ b (mod p)

Consider the equation

a1xl11 + · · ·+ arx

lrr ≡ b (mod p),


lrr ≡ b (mod p) 98

where a1, . . . , ar, b ∈ Z∗p and l1, . . . , lr are divisors of p − 1, and let N be its number of

solutions. We estimate N depending on characters and on Jacobi sums as follows:

Theorem D.4.1. The number N satisfies

N = pr−1 +∑

χ1 . . . χr(b)χ1(a−11 ) . . . χr(a

−1r )J(χ1, . . . , χr),

where the summation is over all r-tuples of characters χ1, . . . , χr, where χlii = ε and χi 6= ε

for 1 ≤ i ≤ r.

In particular, if M0 is the number of such r-tuples with χ1 . . . χr = ε and M1 is the

number of such r-tuples with χ1 . . . χr 6= ε then

|N − pr−1| ≤M0p(r/2)−1 +M1p

(r−1)/2.

Proof. We have

N =∑

N(xl11 = u1) . . . N(xlrr = ur),

where the sum runs over all r-tuples (u1, . . . , ur) such that∑r

i=1 aiui = b. Let χi vary

over the characters of order dividing li. Then

N(xlii = ui) =∑χi

χi(ui),

and thus

N =∑

χ1,...,χr

∑∑aiui=b

χ1(u1) . . . χr(ur).

Let ti = b−1aiui. The inner sum becomes

χ1 . . . χr(b)χ1(a−11 ) . . . χr(a

−1r )J(χ1, . . . , χr).

By Proposition D.3.5, the term with J(ε, . . . , ε) has the value pr−1. If some but not

all the χi are equal to ε then the term has the value zero. In the first case the value is

zero unless χ1 . . . χr = ε. Now, this theorem follows from Theorem D.3.9.

Corollary D.4.2. Let the hypothesis be as in previous theorem. Then

|N − pr−1| ≤ (M0 +M1)p(r−1)/2 ≤ (l1 − 1) . . . (lr − 1)p(r−1)/2.

Proof. The first inequality is obvious from theorem above. The second one follows from the

fact that χlii = ε and χi 6= ε, thus there are li− 1 distinct values for χi, for 1 ≤ i ≤ r.

E{±1}-weighted Davenport constant

The following theorem, which was used in the proof of Theorem 2.2.1, is proved in [1].

We reproduce below. In the same paper, they also proved that s{±1}(Cn) = n+ blog2 nc.

Theorem E.0.1. Let n ∈ N and (y1, . . . , ys) be a sequence of integers with s > log2 n.

Then there exists a non-empty J ⊂ {1, 2, . . . , s} and εj ∈ {±1} for each j ∈ J such that∑j∈J

εjyj ≡ 0 (mod n).

Proof. Consider the sequence of 2s > n integers(∑j∈I

yj

)I⊂{1,2,...,s}

that cannot contain distinct integers modulo n. Therefore, there are J1, J2 ⊂ {1, 2, . . . , s}with J1 6= J2 such that ∑

j∈J1

yj ≡∑j∈J2

yj (mod n).

Set J = (J1 ∪ J2) \ (J1 ∩ J2) andεj = +1 if j ∈ J1εj = −1 if j ∈ J2

It is clear that J is non-empty and it has the required property.

F

Conditions for a zero-sum modulo n:

Proof of Theorem 1.1.1

In this chapter, we prove Theorem 1.1.1. This result is shown in [10] and we reproduce

here. Notice that Theorem 1.1.1 can be rewritten as:

Theorem F.0.1. Let n > 0 and k ≥ 0 be integers with n − 2k ≥ 1. Given any n − kintegers

a1, a2, a3, . . . , an−k (F.1)

there is a non-empty subset of indices I ⊂ {1, 2, . . . , n−k} such that the sum∑

i∈I ai ≡ 0

(mod n) if at most n− 2k of the integers F.1 lie in the same residue class modulo n.

The result is best possible if n ≥ 3k−2 in the sense that if “at most n−2k” is replaced

by “at most n − 2k + 1” the result becomes false. This can be seen by taking aj = 1 for

1 ≤ j ≤ n− 2k + 1 and aj = 2 for n− 2k + 2 ≤ j ≤ n− k, noting that the number of 2’s

here is n− k − (n− 2k + 1) = k − 1 ≤ n− 2k + 1.

Lemma F.0.2. Let n be a positive integer. For i = 1, 2, . . . , r let Ai be a set of νi positive

integers, incongruent modulo n, and none ≡ 0 (mod n). If∑r

i=1 νi ≥ n then the set∑ri=1({0} ∪ Ai) contains some non-zero multiple of n.

Proof. Suppose the result is false. We may presume that Ar is not empty. Let A be

the union of r incongruent residue classes 0, a1, a2, . . . , ar−1 (mod n) and B the union

of s incongruent classes 0, b1, b2, . . . , bs−1 (mod n). Suppose that if a ∈ A and b ∈ B

and a + b ≡ 0 (mod n) then a ≡ b ≡ 0 (mod n). Then A + B is the union of at least

min(n, r+ s− 1) distinct residue classes modulo n. If we apply this result to the two sets

101

{0} ∪ A1 and {0} ∪ A2 we conclude that the set

[{0} ∪ A1

]+[{0} ∪ A2

]contains representatives from at least min(n, ν1 + ν2 + 1) distinct classes modulo n. Con-

tinuing by induction we conclude that

r−1∑i=1

({0} ∪ Ai

)contains representatives from at least

min

(n, 1 +

r−1∑i=1

νi

)≥ n− νr + 1

distinct residue classes modulo n, and so it contains representatives from at least n − νrdistinct non-zero residue classes modulo n. But −Ar has representatives from νr non-zero

residue classes modulo n, and so there must bee a representative from a non-zero residue

class in common because there are only n − 1 such classes. From this observation the

lemma follows.

If S is a set of integers (not necessarily distinct) we let∑S denote the set of distinct

non-zero residue classes (mod n) represented by sums of integers in S. We call a set

of three integers which are incongruent modulo n a triple, and any incongruent pair of

integers a double.

Lemma F.0.3. If S is a triple with no subset having a zero sum modulo n then |∑S| ≥ 5,

and if S does not contain n/2 (mod n) then |∑S| ≥ 6. (The notation |

∑S| has the

usual meaning, the number of elements in∑S.)

Proof. Let S = {a, b, c}. If two of the congruences

a+ b ≡ c, a+ c ≡ b, b+ c ≡ a (mod n) (F.2)

are false, say a + b 6≡ c, a + c 6≡ b, then∑S contains the six distinct non-zero residue

classes a, b, c, a+ b, a+ c, a+ b+ c (mod n), because for example if a+ b+ c ≡ a then we

have the contradiction b+ c ≡ 0 (mod n).

Therefore we may assume that at least two of the congruences in (F.2) hold, say

a + b ≡ c and a + c ≡ b. In this case, by addition we see that 2a ≡ 0, i.e. a ≡ n/2

(mod n), and∑S contains the five distinct non-zero residue classes a, b, c, b+ c, a+ b+ c

102

(mod n), because for example if b+ c ≡ 0 we obtain the contradiction a+ b+ c ≡ 2a ≡ 0

(mod n).

Lemma F.0.4. If S is a double having no zero sum then |∑S| = 3.

Proof. The proof is obvious.

Now we turn to the theorem itself, giving a proof by contradiction. Let a1, a2, . . . , an−k

be a sequence of n − k positive integers with no zero sums modulo n, and such that at

most n− 2k terms belong to the same residue class modulo n. We will show that there is

a partition of the index set {1, 2, . . . , n− k} into disjoint sets P1 ∪ · · · ∪ Pr in such a way

that if i, j ∈ Pr and i 6= j then ai 6≡ aj (mod n), and so that

r∑t=1

|∑

St| ≥ n

where St = {ai; i ∈ Pt} for 1 ≤ t ≤ r. Then Lemma F.0.2 gives us the required contra-

diction.

First we suppose that the sequence (F.1) contains no integer ≡ n/2 (mod n). Select

from the sequence (F.1) in any manner whatsoever, triples of elements if possible, until

all that is left in the sequence (F.1) is a single repeated element a modulo n, or two

repeated elements a and b modulo n, with a 6≡ b. Suppose by this process we get j triples,

with j ≥ 0, and the remaining elements a (with say λ occurrences) and b (with say µ

occurrences), and we may presume λ ≥ µ ≥ 0. Since there are n − k elements in the

sequence (F.1) we see that

n− k = 3j + λ+ µ, j =1

3(n− k − λ− µ).

In addition to the j triples we form µ doubles of the form {a, b} and λ− µ singles of the

form {a}. Call the triplesSi with 1 ≤ i ≤ j, the doubles Si with j + 1 ≤ i ≤ j + µ, and

the singles Si with j + µ+ 1 ≤ i ≤ j + λ. By Lemmas F.0.3 and F.0.4 we conclude that

j+λ∑i=1

|∑

Si| = 6j + 3µ+ (λ− µ) = n+ (n− 2k)− λ ≥ n,

the last inequality holding because n − 2k ≥ λ, there being at most n − 2k identical

elements modulo n in the sequence (F.1).

Finally, suppose that n/2 is in the sequence (F.1). It can occur only once since

n/2 + n/2 ≡ 0 (mod n). By the same process as in the first part we choose j triples

103

S1, . . . , Sj without using the element n/2, so that what remains in the sequence (F.1) are

the element n/2 once, the element a occurring λ times, and the element b occurring µ

times, again with λ ≥ µ ≥ 0. We deal with three special cases: λ > µ; λ = µ > 0;

λ = µ = 0. In all cases we have n− k = 3j + λ+ µ+ 1.

In the case λ > µ we choose µ doubles of the form {a, b} and an additional double

{a, n/2}, and also λ− µ− 1 singles of the form {a}. Thus we get∑i

|∑

Si| ≥ 6j + 3(µ+ 1) + (λ− µ− 1) = n+ (n− 2k)− λ ≥ n

as before.

In the case λ = µ > 0 we have, in addition to the j triples without the element n/2

also the triple {n/2, a, b} and the λ− 1 doubles of the form {a, b}. This gives us∑i

|∑

Si| ≥ 6j + 5 + 3(λ− 1) = n+ (n− 2k)− λ ≥ n.

In the case λ = µ = 0 we have one single {n/2} and we get∑i

|∑

Si| ≥ 6j + 1 = n+ (n− 2k)− 1 ≥ n.

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