extremal product-one free sequences in some non-abelian ...€¦ · universidade federal de minas...

Universidade Federal de Minas Gerais

Instituto de Ciências Exatas

Departamento de Matemática

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

sieve method on tournaments

Sávio Ribas

Orientador:Fabio Enrique Brochero Mart́ınez

Orientadora:Yu-Ru Liu

Belo Horizonte

2017

Universidade Federal de Minas Gerais

Instituto de Ciências Exatas

Departamento de Matemática

Extremal product-one free sequences in some

non-abelian groups and shifted Turán sieve method

on tournaments

Sávio Ribas ∗

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

Orientador: Fabio Enrique Brochero Mart́ınez (UFMG)

Orientadora: Yu-Ru Liu (University of Waterloo)

Belo Horizonte, 06 de abril de 2017

Banca de defesa:

Fabio Enrique Brochero Mart́ınez (UFMG)

John William MacQuarrie (UFMG)

Carlos Gustavo Tamm de Araújo Moreira (IMPA)

Eduardo Tengan (UFSC)

Hemar Teixeira Godinho (UnB)

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

i

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

Acknowledgement

Finalmente, a última etapa da preparação acadêmica está chegando ao fim, e isso seria

praticamente imposśıvel sem a ajuda de várias das pessoas que me rodeiam. Começarei

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

educação poderia nos transformar. Lembro bem tudo o que você passou no ińıcio do meu

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

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

você estaria sentado na primeira fileira atrás da banca e gritando para eles me aprovarem

logo). Vocês passaram por vários apertos para que um dia eu chegasse aqui. Amo vocês e

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

você não gosta quando eu tento colocar humor em coisas profissionais, mas mesmo assim

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

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

natural era melhor eu não fazer isso, não é? Obrigado pelos ensinamentos de irmão mais

velho, dotô! Tiene, quando todos ouvirem o que você tem a dizer, o mundo será mais

justo. Obrigado pela força e por tentar enfiar alguma didática na minha cabeça! Lili,

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

olimṕıadas durante o ensino fundamental. Ali aprendi a gostar de matemática e a querer

fazer disso uma profissão! Quero ensinar com o mesmo entusiasmo que você! Amo todos

vocês! Agradeço também a toda minha famı́lia: Vô Vicente, tios, tias, primos, sogros

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

Fabio, não tenho palavras para agradecer toda a sua paciência e dedicação, e ainda

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

seria muito mais dif́ıcil ou mesmo imposśıvel: PECI, mestrado, doutorado, sandúıche,

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

agora saio já casado, concursado e até com uns artigos submetidos (que até pouco tempo

eu não acreditava que fosse posśıvel...). Desculpe-me por não conseguir chegar tão cedo

(“Sávio, eu chego 7h”, “Ok, a gente se encontra então às 10h”) e muito obrigado por tudo!

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

iii

panharam desde o ińıcio da graduação e pouco a pouco surgiram alguns agregados que

estão sempre presentes (tem também aqueles que estudaram na escola): Tulio (chic-chic)

e Luana (claro?), Vlad (pô, cara, 60 é muito ruim), Remer Cueio (é bão né?), Luis Felipe,

Paim e Claudião, Dani e Bida, Nat e Pŕıncipe, Vinição (Girafales), Danilo e Ali...sta,

obrigado pela grande parceria desde sempre! Agradeço também a D. Sônia, Luiz e Julia

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

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

Cristhiano (ah!), Let́ıcia (minha amiga tem 2 mestrados), Lucas (irmão), Lilian (irmã),

1garetti, Pedro F., Pedro H., Jose (trança), Allan (fechamento (não é pela música!!!)),

Felepe (cabra macho) e Javier (cão). Agradeço também a todos os demais colegas do

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

que resolveram várias burocracias e a Andréa e a Kelli que estão sempre prontas a ajudar.

Eu me despeço dessa universidade após 8 anos, mas já quero voltar. Aos amigos de fora

da matemática que também têm seu lugar de grande destaque na galeria de agradecimen-

tos: Saulo e Ingrid, Flávia, Fernanda e toda a famı́lia, Michele, Lucas Assis, Marcús e

Gabi, valeu por tudo! Muito obrigado também, 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 parágrafo canadense não poderia faltar o paulista quase mineiro Danilo. Valeu pelos

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

Agradeço à banca: John, Gugu, Tengan e Hemar, pelos valiosos comentários e su-

gestões. Gostaria de agradecer também à CAPES pela bolsa de doutorado no Brasil,

ao programa Ciência Sem Fronteiras do CNPq pela bolsa de sandúıche e ao programa

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

essa pesquisa não teria sido posśıvel. Um agradecimento especial vai para duas figuras

important́ıssimas na história desse páıs, que olharam com carinho para o ensino público,

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

iv

“Esse é de laranja que parece de limão, mas tem gosto de tamarindo” (Chaves)

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

“Já chegou o disco voador” (Chaves)

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

Chaves: “Serve um balde vermelho?”

Seu Barriga: “Serve!”

Chaves: “Mas não tem vermelho!”

“Outra vez flores? (...) Outra vez café? É por isso que ele só te traz flores” (Quico)

Chaves: “Eu sei que o Homem Inviśıvel está aqui!”

Quico: “Por quê?”

Chaves: “Porque não estou vendo ele!”

“Não há nada mais trabalhoso do que viver sem trabalhar” (Seu Madruga)

“Palma, palma! Não priemos cânico” (Chapolin)

“Meus movimentos são friamente calculados” (Chapolin)

“Não contavam com minha astúcia” (Chapolin)

Published, submitted and inpreparation works

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

[A] Brochero Mart́ınez, F. E., Ribas, Sávio; Extremal product-one free sequences in CqosCm. Submitted. Available at: https://arxiv.org/pdf/1610.09870.pdf (2016).

See [11].

[B] Brochero Mart́ınez, F. E., Ribas, Sávio; 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 Mart́ınez, F. E., Ribas, Sávio; Improvement of Alon-Dubiner constants for

a zero-sum problem in Zdp. In preparation.

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

tournaments. Submitted. See [53].

https://arxiv.org/pdf/1610.09870.pdfhttps://arxiv.org/pdf/1701.08788.pdfhttps://arxiv.org/pdf/1701.08788.pdf

Resumo

Essa tese está dividida em duas partes:

• Parte I: Problemas de soma-zero inversos.

Nós começamos apresentando uma visão geral sobre a Teoria de Soma-Zero. Em

particular, apresentamos os principais resultados e conjecturas acerca dos seguintes

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

η (sem pesos ou com pesos {±1}). Depois, focamos na constante de Davenport: esseinvariante denota o menor inteiro positivo D(G) tal que qualquer sequência S de

elementos de G, com comprimento |S| ≥ D(G), contém uma subsequência comproduto 1 em alguma ordem, onde G é um grupo finito escrito multiplicativamente.

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

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

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

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

sequências de tamanho máximo que são livres de subsequências com produto 1.

Nós conclúımos apresentando as Propriedades B, C e D, que são, em geral, conjec-

turas de extrema importância no estudo dos problemas inversos.

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

Nós começamos apresentando uma visão geral sobre alguns problemas famosos que

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

isso, constrúımos uma versão deslocada do crivo de Turán, 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 é, grafos completos direcionados, de acordo com o número

vi

vii

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

que contêm um número 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 útil em outros contextos da Matemática, 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 metaćıclico, grupo diedral, grupo

dićıclico; teoria do crivo, conjectura dos primos gêmeos, conjectura de Goldbach, crivo

combinatório, crivo de Turán, crivo de Turán 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, Erdős-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 Gis 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 Mart́ınez (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 Turán 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 Turán 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, Turán sieve, shifted

Turán 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 . . . . . . . . . . . . . . . . . . . . . . . . . 162.1.3 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 . . . . . . . . . . . . . . . . . . 322.3.2 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 Turán 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 Turán sieve method on tournaments . . . . . . . . . . . . . . 48

5.1 The shifted Turán 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. Erdős-Ginzburg-Ziv theorem . . . . . . . . . . . . . . . . . . . . . . . . . . 78

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

A.2 Proof of Erdős-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 Erdős-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) . . . . . . . . . . . . . . . . . . . . . . . . . 92D.3 The sums of Gauss and Jacobi . . . . . . . . . . . . . . . . . . . . . . . . . 93

D.4 The equation a1xl11 + · · ·+ arxlrr ≡ 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-setis 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 subsequenceif gσ(n1)gσ(n2) . . . gσ(nk) = 1 for some permutation σ of {n1, . . . , nk}, and if there areno 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 theelements 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 casethat 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 Erdős-Ginzburg-

Ziv (see [27]): Given 2n−1 integers, it is possible to select n of them, such that their sumis divisible by n, or in group theory language, every sequence with l ≥ 2n − 1 elementsin 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 integerwith 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 Erdős-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 ofevery 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 independentlyby 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 upperbound by proving that s(Cdp ) ≤ cdp, where c2 = 4, cd = 256cd−1 log2[(32d)d]+d+1 for d ≥ 3and p is a sufficiently large prime. In particular, s(C3p) ≤ 20234p and cd ≤ (cd log2 d)d fora certain absolute constant c > 0. In a joint work with F. E. Brochero Mart́ınez, we hope

to lower the bound in such way that s(C3p) ≤ 1000p for sufficiently large primes p. Theabove inequalities must also be valid for n with large prime factors instead of p sufficiently

large prime, since

s(Cdmn) ≤ min{s(Cdn) + n[s(Cdm)− 1], s(Cdm) +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 × 3for 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-emptyA-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 haves{±1}(C

2n) = 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) iswell 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 aconsequence, D(Cn) = n, since |Cn| = n and the sequence (g)n−1 is free of product-1subsequences, 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 +

∑ri=1(p

ei − 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 toprove 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 provedin 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 − 2where 1 ≤ m|n. W. Gao, et al. [37] proved that η(C3n) = 8n − 7 if n = 3k × 5l for somek, l ∈ N0 with k + l ≥ 1, η(C3n) = 7n− 6 if n = 2k for some k ≥ 1 or n = 2k × 3 for somek ≥ 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 ofproduct-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 toexp(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 existsS = (x1, . . . , xD(G)−1) sequence of G such that xi1xi2 · · ·xik 6= 1 for every non-emptysubset {i1, i2, . . . , ik} ⊂ {1, 2, . . . , D(G)− 1} . The Inverse Zero-Sum Problems study thestructure 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 existssome g ∈ S with multiplicity ≥ 2|S| − n+ 1. In particular, D(Cn) = n and the followingstatements 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 Mart́ınez 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. Denoteby Cq os Cm the metacyclic group written multiplicatively, i.e., the group generated by xand 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 andordq(s) = m ≥ 2. In a joint work with F. E. Brochero Mart́ınez [11], we proved thefollowing 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 thatordq(s) = m. Let SM be a sequence in the metacyclic group Cq os Cm with m + q − 2elements.

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 and0 ≤ 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 cardinalityof 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, thatis, 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 thenormal 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 containedin 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. Infact, 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. Onthe 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 wecan 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 distinctelements, say ai 6≡ aj (mod q). Suppose that ordq(s) = m. Then

m−1∑k=0

aksk 6≡ aisj + 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 thatai 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, if

m−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 wehave 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 wemay 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 ands 6≡ 1 (mod q), we obtain c ∈ A and c 6≡ 1 (mod q), thus c − 1 ∈ A, which impliess · (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∗qare fixed. We use the upper bound to show that the set {c+ bw4| w ∈ Z∗q} contains bothquadratic 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 numberof 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 + bw42 is anon-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) generatesseven other solutions (by switching z by −z and multiplying w by powers of w0), thenumber 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, suchthat 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)sm−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−1sm−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 thequadratic 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−1sm−1 (mod q),

c ≡ a0 + a2s2 + · · ·+ am−2sm−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 moduloq. 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 resultstates 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 agenerator 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 CmIn 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 theremaining 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 subsequencesand 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 propositionbelow 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 notfree 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 some3 ≤ j ≤ 6, then we may choose A1 = (x, xyλ, x, xyλ) or A1 = (x3, x3yλ, x3, x3yλ). Sinceord5(s) = 4 and gcd(i, 4) = 1, it follows that λs

2i+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, xy

3) when i = 1,

A1 = (x3, x3, x3y, x3y2) or A1 = (x

3, 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 = 1i = 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 andq is a prime) with m + q − 2 elements which are free of product-1 subsequences. Noticethat if (m, q) = (2, 3) then s ≡ 2 (mod 3) and the sequence S = (x, xy, xy2) provides theonly 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 elementsof 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 ofproduct-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 butno 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 thesame 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. Thisimplies that

r∑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

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

distinct elements in HM arising from SM ∩ NM . Suppose without loss of generality thatSM ∩HM = (h1, h2, . . . , hq−k). If

∑ri=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) (whichwe 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 thereexist 1 ≤ n1 ≤ k−1 and 1 ≤ n2 ≤ q−k such that gn1 = (h1 . . . hn2)−1 then gn1h1 . . . hn2 = 1and 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 ofthe 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 productin 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 mayselect xi1yθ ∈ Aj with i1 6= i0. If 2 ≤ l ≤ m − 1 is such that i0l ≡ i1 (mod m) then wemay replace Aj by

Ãj = (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 |Ãj| > |Aj|, the new set R generated by thischange has more than k−1 elements. This replacement may mean that Ãj is not minimalanymore and in this case we break Ãj into its minimal components with product in HM ,

say, Ãj1 , Ãj2 , . . . , Ãjδ . We observe that |Ãj| = |Ãj1| + |Ãj1 | + · · · + |Ãjδ |, and repeat theprevious arguments with A1, . . . , Aj−1, Aj+1, . . . , Ar and the new Ãji ’s. Since

r∑n=1n 6=j

|An|+δ∑

n=1

|Ãjn| >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 havem+ 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 Lemma2.1.6.

From now on, we assume k = q and SM is a sequence in Ni0 . As |Aj| = m for 1 ≤ j ≤ rand |C| = m− 1 we obtain mr + m− 1 = |SM | = m + q − 2, therefore mr = q − 1. Weconsider 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

Ã = {πn(A1)}n and B̃ = ({1} ∪ {πn(A2)}n) · · · · · ({1} ∪ {πn(Ar)}n)

says that at least one of the following statements hold:

(i) |Ã|+ |B̃| > q;

(ii) min{|Ã|, |B̃|} = 1;

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

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

words, the exponents of the elements in Ã and B̃ form arithmetic progressions

with the same common difference.

Since |Ã| = m and |B̃| = q−k = q−m, we have |Ã|+ |B̃| = q, thus (i) is not possible,and since min{|Ã|, |B̃|} = min{m, q −m} ≥ min{m, 2m + 1} = m ≥ 2, (ii) does nothold. In order to discard item (iii), define

B̃1 = {1} ∪ {πn(A2)}n,

B̃2 = ({1} ∪ {πn(A3)}n) · · · · · ({1} ∪ {πn(Ar)}n).

As 3m ≤ mr = q − 1, we have

q −m = |B̃| = |B̃1 · B̃2| = |B̃1|+ |B̃2| − 1 < q − 1,

|B̃1| = m+ 1,

|B̃2| = 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 B̃1 and B̃2 form arithmetic progressions,

say

B̃1 = {y−av, . . . , y−v, 1, yv, . . . , y(m−a−1)v}.

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

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

difference, say

Ã = {yw, yw+v, . . . , yw+(m−1)v}.

By switching A1 and A2, the only possibility is that {πn(A1)} = {yv, y2v, . . . , ymv} forsome 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 quadraticresidues 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 − 2m− 1

=3m− 1m− 1

> 3,

there are at least 4 distinct elements among SM = {xi0yα1 , . . . , xi0yα3m−1}. We splitthis 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 hasto 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 quadraticresidue 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−2si0(m−2) + am−1si0(m−1)) ∈ Zq| 0 ≤ n ≤ m− 1},

Y = {sni0(b0 + b1si0 + · · ·+ bm−2si0(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). Sincegcd(i0,m) = 1 and ordq(s) = m, we have ordq(s

i0) = m, therefore by Lemma

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

α = a0 + a1si0 + · · ·+ am−2si0(m−2) + am−1si0(m−1) ∈ X.

If −α ∈ Y then 0 ∈ X + Y , a contradiction, therefore −α ∈ X. Hence, thereexist 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 generatesZ∗q. The cases where (m, q) = (4, 5) follow from Proposition 2.1.12, therefore we mayassume q ≥ 7. If there exist m identical elements then their product is 1, thus thereare at most m− 1 identical elements. Since

m+ q − 2m− 1

=2m− 1m− 1

> 2,

there are at least 3 distinct elements among SM = {xi0yα1 , . . . , xi0yα2m−1}. Again, wesplit 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 twopermutations 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)s3i0 + · · ·+ 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 quadraticresidue 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−2si0(m−2)) ∈ Zq| 0 ≤ n ≤ m− 1},

Y = {s2ni0(a1si0 + a3s3i0 + · · ·+ am−1si0(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 generalitythat |X| = m/2 and |Y | = m/2. Let

α = a0 + a2s2i0 + · · ·+ am−2si0(m−2) ∈ X.

If −α ∈ Y then 0 ∈ X + Y , a contradiction. Therefore −α ∈ X. Hence, thereexists 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., thegroup 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. Theirargument 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 Mart́ınez [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). Observethat the case n = 3 is exactly the item (2) of Theorem 2.1.1, since D6 ' C3 o−1 C2. Forn = 2, we have D4 ' Z22 and it is easy to check that SD is free of product-1 subsequencesif 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 integerswith s > log2 n. Then there exist a nonempty J ⊂ {1, 2, 3, . . . , s} and εj ∈ {±1} for eachj ∈ 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 elementsthat is free of product-1 subsequences. If SD ∩ ND contains two identical elements thenSD 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 mustall 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 theseforms:

(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 onlyproblem occurs when k = n − 1, that is, when v − u + t ≡ 0 (mod n), but inthis 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 leastn− 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 thetheorem 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 ispossible 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 ofgenerality that ti = t for 1 ≤ i ≤ n − 2k + 1. By renaming z = yt, we may assumewithout loss of generality that t = 1. Since k ≥ 4, Pigeonhole Principle implies thatthere 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), thereforeEquation (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 thereare 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), thenα4 can be any element in the set {α3 + 1, α3 + 2, α3 + 3}. If α4 ≡ α3 + 1or α4 ≡ α3 + 3 (mod 6) then we return to the previous item. Therefore wemay 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 fourconsecutive 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 ywith 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 thesequence S = (y)2n−1(x) does not have product-1 subsequences and |S| = 2n. Almostas 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 needto show that (i) =⇒ (ii). If n = 2 then it is easy to check that the sequences of theform (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 = 3and n = 2, respectively. The case n ≥ 4 is a direct consequence of Theorem 2.2.1 and thecase 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 order2n. Since D(HQ) = D(Z2n) = 2n, SQ contains some non-empty subsequence withproduct 1.

(1.2) Case |SQ ∩HQ| = 2n− 1: In this case, by Theorem 1.1.1, the elements of SQ ∩HQmust 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 nelements. 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 SQsuch that |S1 ∩HQ| ≤ n− 2 and S2 = SQS−11 is such that |S2 ∩HQ| ≤ n− 2. Theargument 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-1subsequences. 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, sinceD(HQ) = D(Z6) = 6.

(2.2) Case |SQ ∩HQ| = 5: In this case, Theorem 1.1.1 says that SQ ∩HQ = (yr)5 wherer ∈ {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 fori ∈ {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 fori ∈ {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= tthen 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 notcontains 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, whichwe 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 thatthe exponents of y have difference 2, since xyα · xyα+2 = y and xyα+2 · xyα = y5. Infact, 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 freeof 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 withproduct in {1, y3}. The same argument than item (1.3) does not apply if and only ifS2 (mod {1, y3}) = (x, xy, xy2). Observe that we could also decompose SQ = S ′1S ′2in 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, thesame 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 productin {1, y3} if and only if Si (mod {1, y3}) is not of the form (x, xy, xy2). Hence, thesame 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 insuch 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 inQ8 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, sinceD(HQ) = D(Z4) = 4.

(3.2) Case |SQ ∩HQ| = 3: In this case, Theorem 1.1.1 says that SQ ∩HQ = (yr)3 wherer ∈ {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 befree 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), wemay 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 = (y

3)(xys)3.

(3.5) Case |SQ ∩ HQ| = 0: In this case, we also may assume x ∈ SQ. Therefore, thepossibilities 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) − 1contains some element with multiplicity exp(G)− 1;

• Property C if every sequence S in G with length |S| = η(G) − 1 which containsno 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 containsno 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. Noticethat 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 C2n, 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 C2p 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 specificcases (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 typ

extremal product-one free sequences in some non-abelian ...€¦ · universidade federal de minas...

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