Zero-sumfree sequences in cyclic groups and some arithmetical application

J

OURNAL DE

T

HÉORIE DES

N

OMBRES DE

B

ORDEAUX

A

LFRED

G

EROLDINGER

Y

AHYA

O

ULD

H

AMIDOUNE

Zero-sumfree sequences in cyclic groups and

some arithmetical application

Journal de Théorie des Nombres de Bordeaux, tome

14, n

o

_{1 (2002),}

p. 221-239

<http://www.numdam.org/item?id=JTNB_2002__14_1_221_0>

© Université Bordeaux 1, 2002, tous droits réservés.

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Article numérisé dans le cadre du programme Numérisation de documents anciens mathématiques

Zero-sumfree

_sequences

in

_cyclic

_groups

and

some

arithmetical

application

par ALFRED GEROLDINGER et YAHYA OULD HAMIDOUNE

RÉSUMÉ. Nous montrons _que dans un _groupe

_cyclique

d’ordre _n,

toute suite S de

_longueur

_|S|~

sans sous-suite non vide

de somme nulle contient un élément d’ordre n _ayant une

_grande

multiplicité.

ABSTRACT. We show that in a

_cyclic

_group with n elements _every

zero-sumfree sequence S with

length

|S|~

contains some

element of order n with

_high

_{multiplicity.}

1. Introduction

Let G be a finite

_cyclic

_group with

IGI

= _n and let S be _a

sequence in

G. We say that S is

zero-sumfree,

if no

_non-empty

subsequence

of S sums

to zero. An _easy observation shows that r~ -1 is the maximal

_length

of a zero-sumfree _sequence and if S is a zero-sumfree _sequence with

length

= n -1,

then S consists of one

_{element g}

E G with order n which is

repeated n -

1 times.

_{Investigations}

of the structure of

_long

zero-sumhee

sequences were started in the seventies

_by

P. Erdos et al. In

_[BEN75]

it is

proved

that a zero-sumfree _sequence with

_length

contains some

element with

_multiplicity

_{2 ~ S ~ - n}

+ 1. In a recent _paper

by

W. Gao and the first author it is

_proved

that a zero-sumfree _sequence S with

2

contains some element of order n

(see

[GG98]).

Using

a new

approach

we can

sharpen

this result and show that a zero-sumfree _sequence S with

length

contains some element of order n with

_high

_{multiplicity.}

It is easy to

_verify

that this result is best

possible

(see

Theorem 3.12 and Remark

_3.13).

Besides of

_being

of interest for its own

right,

_any information about the

structure of zero-sumfree _sequences in a finite abelian _group G

gives

infor-mation about the

_{non-uniqueness}

of factorizations in a Krull monoid

_having

divisor class _group G. For this connnection and some

_general

information on factorization

_{theory we}

refer to the _survey articles in

_[And97].

Let H

be a Krull monoid with

cyclic

divisor class G such that _every class con-tains a

_prime

divisor and set = n. Then the sets of

_lengths

are almost arithmetical

_{multiprogressions}

and let

_{A* (G)}

denote the set of

_possible

dif-ferences. It is easy to see that

maxA*(G)

= n - 2 and

having

the

sharp

result of Theorem 3.12 at our

_disposal

we can show that the second

_largest

value in

_A*(G)

_equals

_{[!Ij - 1}

_(Theorem

_4.4).

2. Preliminaries

Let N denote the

_{positive integers}

and let

_No

= N U

101.

For some real number x E R let

_{l x J}

E Z denote the

_{largest integer}

with

_{Lx J}

x

and E Z the smallest

_integer

with x For _a, b E Z we set

a,b

Throughout,

all abelian _groups will be written

_additively.

Let G be an abelian

_Go

C G a finite

subset,

:F(Go)

the free abelian monoid with basis

_Go

and

a _sequence in

Go

where _{gl, ... , gl} E

Go

and for

_{afl g}

E

_Go.

We use the same notations as in

[GG99].

In

_particular,

= 1 =

¿9EGo

denotes the

_lengths

of

_S,

_{supp(S) =}

_{g

E

_{Go ~}

_I

&#x3E;

_0}

C

Go

the

_support

of

_S,

_{a(S) _}

G the sum of S and

The sequence S is called

.

zero-sum free

if 0

.

sqvaref ree

=1 for

_{all g}

_E

supp(S),

o a zero-sum _sequences, if =

_0,

9 a minimal zero-sum _sequence, if S is a zero-sum _sequence and _every

proper

subsequence

is zero-sumfree.

If G =

Z /nZ

for _{some n E} N and S = ₊

nZ)

_E

7(Z /nZ)

where

a1,... ad E

~1, n~,

we set

Clearly,

=

a(S),

S is _a zero-sum _sequence if and

_only

if 0 mod _n, and if

_uz(S)

_n, then S is zero-sumfree.

We denote

_by

_,A(Go)

the set of all minimal zero-sum _sequences in

Go.

This is a finite set and

denotes

_{DavenPort’s}

constant of

_Go.

If G is

_cyclic,

then

_{D(G) =}

_IGI

and

1 is the maximal

_length

of a zero-sumfree _{sequence in} G.

3. Zero-sumfree sequences

We shall use the

following

two well-known addition theorems. The first one is a

special

case of Chowla’s Addition Theorem

(we

give

a

simple

proof)

and the second one follows from a result of L. Moser and P. Scherk

(cf.

[MS55]).

Proposition

3.1. Let G be a

finite cyclic

_group, a E G with

_{ord(a) =}

_IGI

and let B be a subset

_of

G. Then

110,

a}

+

_IBI

+

_11.

Proof. Suppose

that

_{+ B) ~B~.}

Since B C

_{{0, a}}

_{+ B,}

it follows

that B =

10, al

₊ B whence a + B C B. Therefore

_ja

+ B C B for _every

j

E N and thus

whence +

_{~B~ &#x3E;}

0

Proposition

3.2. Let G be a

finite

abedian _group and S E a

zero-sumfree

sequence.

If S =

with E

_.’(G),

then

.

Lemma 3.3. Let G be a

finite cyclic

_group with 4 and S E

a

_zero-sumfree

_sequence such that

vg (S)

&#x3E; 0

_for

some _g E G with

ord(g)

= _n. Then

Proof.

Without restriction we _{may suppose} that

where k =

V1+nZ(S)

_E

N, 1

_E

No

and _{al, ... , al E}

[2, n - 1].

Case 1: For _{every i} E

_{[1, 1]}

we have _{ai E}

[2, k].

Then

Since S is

_{zero-sumfree,}

it follows that k + ai n whence

Case 2: There exists some i E

_{~1, l~}

such

_{that ai V [2, k~.}

Since S is

and

Applying

Proposition

3.2 we infer that

Definition 3.4. For a finite abelian _group

G,

a

_non-empty

subset

_Go

c G and some k E N we set

f (Go, k)

= _: S _E

zero-sumfree, squarefree

and =

k}

The above invariant was introduced

by

Eggleton

and Erd6s in

[EE72].

For information around it and some recent results we refer to

_[GG98],

sec-tion three. Here we

_only

need the

_following

lemma which is well-known. For convenience we

provide

its

simple

proof.

Lemma 3.5. Let G be a

finite

abelian _group and

Go

C G a subset which contains a

square-free zero-samfree

_sequence

of

length

three but no elements

of

order two. Then

_{f (Go,1)}

_=1,

_{f (Go,}

₂₎

= 3 and

f (Go, 3) &#x3E;

6.

Proof.

Clearly,

f (Go,1)

= 1 and if S = a - b E

_7(Go)

with then

E(S) = {a, b, a + b},

( =

3 and thus

_{f (Go, 2)}

= 3.

Let S = _{al - a2 - a3 E}

Y(Go)

be _a

squarefree,

zero-sumfree _sequence.

Clearly,

M =

la,

_{+ a2, ai + a3~ a2 + a3~ ai + a2}

+ a3}

_C

E(S)

and

IMI

= 4.

We assert that at most one of the elements _{al, a2, a3} lies in M. This

_implies

that

_I

&#x3E; 6 and we are done. Assume to the

_contrary,

that there are

i, j E

~1, 3~

= such that E M.. Then

aj + ak and

aj = _ai + _ak whence

2ak

=

0,

_a contradiction. D

Next we introduce a

key

notion of this _paper.

Definition 3.6. Let G be a finite abelian _group, S E a

_non-empty

sequence, k e N and

IS

I

=

kq -

_r with _q E N and r E

_{0, k -1}

An

optimal

k-partition

of S is a

product decompositon

S = where

E are

squarefree

and

1811 I

= k

(then

clearly

and

_ISql

= k - r _E

[1,

The reason

_why

we are interested in

optimal k-partitions

lies in the

_fact,

that an

optimal k-partition

of a zero-sumfree _sequence S

gives

a

_large

lower bound for This will be done in the next lemma.

Lemma 3.7. Let G be a

finite

abelian _group, S E and k E N.

_If

S

is

_zero-sumfree

and has an

optimal k-partition,

then

for

Go

=

supp(S)

we have

Proof.

Let S = be _an

optimal k-partition

of S.

Using Proposition

3.2 we infer that

Proposition

3.8. Let G be a

finite

abelian _groups, S E a

_non-empty

sequence, k e N and

= kq -

r where r E

_{~0, k - 1].}

Then the

_following

conditions are

_equivalent:

1. S has an

optimal

k-partition.

2.

_g

E

_G}

_q and

_~{g

e

_{G ~}

=

q}~ k -

_r.

Proof.

1. ~ 2.

_Suppose

that S has an

_{optimal k-partition,}

_say S =

Tjq 1 Si

where all

_Si

E

_F(G)

_{{ 1 }}

are

squarefree

and

IS11

_ ’ _

=

k.

If S is a

product

of t

_{squarefree subsequences}

for

_{any t}

E

_N,

then

_clearly

g

E

_G}

t whence E

_G}

q.

Let

_{g

E G

_I

=

9} =

{gl, ... , gl}.

If 1 =

0,

then

clearly 1

=

0 k - r.

Suppose 1

&#x3E; 0 and set

_So

= Since

I

S and

Sl, ... ,

Sq

are

squarefree,

it follows that

So ~

Sz

for _{every i} E

_{[1, q]}

whence

T = Thus it follows that

whence = _r.

2. - 1. Let

_{9

E G

_I

=

9} = {9¡,...,

and

So

=

(clearly,

So

= 1 is the

_empty

_sequence if and

only

if l = 0).

Then S =

for some T E with

_{gcd(So, T)}

= 1

(i.e.,

So

and T have no elements

in

_common)

and _{q - 1.} We show that T _may be

where

_{T1,... Tq}

are

_squarefree,

IT11

= ... =

ltq-ll

= k - l and

ITql =

k - l - r &#x3E; 0. Then

obviously

is an

_{optimal k-paxtition}

of S and we are done. In order to show

_(*)

we write T in the form

where _{al, ... , au} are

_{pairwise distinct, q -1 &#x3E;}

_{ml &#x3E; ~ ~ ~ &#x3E;}

1 and

91, ... ,

_9)T)

are numbered in the

following

_{way: a1} = _gl = ... =

9m¡ , a2 =

gml+l = ... =

9rri1+m2 and so on.

We describe how to build

_subsequences

_{Tl, ... ,}

_Tq.

We

_give

the first element _g, to

_Tl,

the second element _g2 to

_T2

and

_{finally gq}

to

_Tq.

Then we

_give

_9q+1 to

_Ti,

...., g2q to

Tq.

We do this k - I - r times. After this we have

_spent

_{(k - I -}

_r)q

elements of T. From now on we

_{only give}

elements

to

_{Tl, ... ,}

_{Tq_1.}

We

_give

_9(k-l-r)q+l

to

_Tl,

and We

repeat

this

_procedure

r times. Hence for _every

z E ~1, q -1~

the _sequence

₁₁

consists of

_{(1~ - t - r)}

+ r elements and

_Tq

consists of

(k - l -

r)

elements. Since

_{max{ Vg}

_{(T) g}

E

_G

_{q -1}

and T is

_{suitably numbered,}

all _sequences

11

are

_squarefree.

Thus T = and

(*)

holds. 0

Corollary

3.9. Let G be a

_finite

abelian _group and S E

_{7(G) ) 111.}

1. 8 has an

optimal2-partition if and only

E

2.

_If

S has no

_optimal

_3-partition,

then one

_of

the

_following

conditions holds:

(a)

g

E

ll

+ 1.

(a)

maxfvg

(S) I

g E

_-

3

(b)

There are two E

_supp(S)

such that

v () =

h

(S) =

₃

-Proof.

1. Let

₁₈₁

=

2q -

_r with _{r E}

[0, 1]

whence _q = If S has

2

an

_{optimal 2-partition,}

then E

_{G q}

_{by Proposition}

3.8.

Conversely,

suppose that

I 9

E

G} ~

q,

We have to show that 2 - 0. Then the assertion follows from

implies

r = 0 and 2 =

ISol

whence

2 -ISol-

r &#x3E; 0.

2. Let

₁₈₁

=

3q -

r with r E

_{~0, 2~}

_{whence q}

=

Suppose

that S has

no

optimal 3-partition

and that

g

E

_G}

q. We have to show

that condition

_b)

holds. As above we set

Then

_Proposition

3.8

_implies

that

₁₈₀₁

&#x3E; 3 - r whence 2. Since

3q -

r =

181

=

ql801

₊

ITI,

it follows that

isol

= 2 whence _r = 2 and

!S+2 _r-.

q = 3

* 0

Lemma 3.10. Let G be a

cyclic

_groups with even order _{n, g} E G with

ord(g)

= 11,

and S E

_{7(G )}

_{g})

a _sequence such that

vh(S)

2

_for

all h E G with 2h =

g. Then S may be written in the

form

+

where

_I

E

_{(1, 2),}

E

_{(g) B {g}}

_for

all i e

_[1~],

2 and

_IUI

= 2

implies

that

_Q(U)

= _g.

Proof.

We

_proceed

_by

induction on The assertion is clear for 2.

Suppose

3. We construct a

_subsequence

_To

of S with

_[To

_[

E

_{(1, 2}

and

(g) B {g}.

Then the assertion follows

_by

induction

_hypothesis.

If there exists some b E

_{supp(8) n}

_(g),

then we set

To

= b.

Suppose

that

supp( 8)

c

_{G B {g}.}

We decide two cases.

Case 1: 8 is

_squarefree.

Then there are

pairwise

distinct elements _a,

b,

c E

supp(8)

and

_clearly

we have

_{a + b, a + c}

E

_(g).

g, we set

_To

= a-6.

If

Case 2: 8 not

_squarefree.

Then there is some a E

_{supp( 8)}

with

_S.

If

_{2a #}

_g, then we set

To

=

a2.

Suppose

2a =

g. Then

va(S)

= 2 whence

there is some such that

a2 .

_b

_S.

Thus a + b E

_{(g) B {g}}

and we set

0

Proposition

3.11. Let G be a

finite cyclic

_group with

~G~

_{= n &#x3E;}

_4,

8 E

.~’(G)

a

zero-sumfree

_sequence with

nt1

and 9

e G with

Then

_ord(g)

E

_{{n, 2, 3 }}

_and,

_{if ord(g) = 2,}

then there is some h E

_supp(S)

with

_ord(h)

= nand 3.

whence

Let H denote the

_subgroup

_generated

_{by g}

whence =

ord(g).

Suppose

that

_{= ~}

and assume to the

_contrary

that 2 for all h E G with

_ord(h)

= n.

_Applying

Lemma 3.10 to S we see that S

may be written in the form

where

_I

E

_{~1, 2~,}

_Q(Ti)

E

_{(g) B {g}}

for all i E

_{[1, t],}

2 and

_{I Ul}

= 2

implies

that

_cr(!7)

= _g.

First we consider the case 2. Then

is a zero-sumfree _sequence with =

vg(S).

Thus Lemma 3.3

_implies

that

a contradiction.

is a zero-sumfree _sequence with

v9(S’)

=

v9(S)+1.

Thus Lemma 3.3

_implies

that

a contradiction.

Assume now that

IHI

I

E ~ 4 , 5 }

and write S in the form

Since n &#x3E; 4 and

it follows that

We assert that there are elements _a, b E G such that ab

I

Sl

and a +

H,

b + H are

_generators

of

_G/H.

Since

Sl

E

_{7(G ) H)}

and

_{1811 ~}

_3,

there exist two elements _a, b such that ab

_I

_Sl

and a +

_H,

b + H E

_{G/H B}

_{H}.

If =

5,

then _{a +}

H,

b ₊ H _are

generating

elements. Assume to the

contrary,

that = 4 and the assertion does not hold. Then

Sl

= _{a -}

S2

where

_S2

E

_T(H’)

for a

subgroup

H’ with H

C H’

_C

G. Then

S1S2

is a zero-sumfree _{sequence in} H’ with

a contradiction.

Hence there are

a, b E G having

the above

properties,

and we choose some c E G such that abc ~

_Sl.

_{By Proposition}

3.1 we infer that

This

_implies

that Thus we _may infer that

a contradiction. 0

Theorem 3.12. Let G be a

cyclic

_group with

IGI

= n &#x3E; 3 and S _E

a

zero-sumjbee

_sequence.

If

nt1,

then there exists some _{g e}

supp(S)

with

Proof.

The assertion is

_obvious,

if n = 3. Hence _suppose that n &#x3E; 4 and

let S be a zero-sumfree _sequence with

-First we show that

supp(S)

does not contain an element of order two.

Assume to the

_contrary

that S

_{= gT}

where

_2g

= 0. Then n is

even, say

n =

2m,

and _{m +} 1. Let

p : G -

G denote the

multiplication by

2.

Since

_D(p(G))

=

m, the sequence

cp(T)

contains a

_subsequence

with sum

zero, say = 0 for some

_subsequence

T’ of T. Then either T’ or

gT’

has sum zero in G.

For k E

_[1,3]

we set

_a(k)

_{= f (supp(S),}

_k).

Then Lemma 3.5

_implies

that

a(1)

=

1, a(2)

= 3 and

a(3) &#x3E;

6.

We set

whence

We show that either there exists _{some g} E G such that

Assume to the

_contrary

that this does not hold. Then

_{by Corollary}

3.9 S has an

optimal

_3-partition

whence Lemma 3.7

_implies

that

a contradiction.

Now we first _suppose that there exist _g E G such that

and we

_{apply Proposition}

3.11.

If one of the two elements has order

_~,

then n is even and

_{by Proposition}

3.11 there exists some a E

_supp(S)

with

_ord(a)

= n and

va (S) &#x3E;

3.

Suppose

that

_{{ord(g), ord(h)}}

C

_{1!1, nl.}

_{If ord(g) = ord(h)}

_{= 3 ,}

then S would not be zero-sumfree whence either

_ord(g)

= n or

ord(h)

= _n, and we are done.

Now we _suppose that there exists _{some g} E G such that

Then

and

_Proposition

3.11

_implies

that

_ord(g)

E

_In,

_{2 ,}

_~}.

If

_ord(g)

E

then the assertion follows.

Assume to the

_contrary

that

_ord(g)

=

n.

We set H =

(g)

and write S

in the form

Since 11

=

ord(g)

_&#x3E; 2 and

it follows that 3.

First show that n &#x3E; 24. There exist t &#x3E;

_{disjoint subsequences}

zero-sumfree,

the _sequence

_So

_{o,(Qi) E}

is zero-sumfree whence +

which

_implies

that n &#x3E; 23. Since n is a

_multiple

of

_3,

we obtain that n &#x3E; 24.

Case 1: There exists some b E G such that + 2.

( ej

If there is some c E

_supp(Sl)

with

_c

_(b),

then S =

(b28o)(b

2 C)S3

for some

S3

E

_7(G)

and we infer that

a contradiction. Thus it follows that

Sl

E

_7"((6)).

If 9 f/. (b),

then we write S in the form S =

(b2Sog-1)(Slgb-2),

_apply

Proposition

3.2 and infer that

a contradiction.

Suppose that g

E

_(b).

_{Then y}

=

ord(g)

divides

ord(b)

and since b _E

G B

H,

it follows that

_{(g) # (b)}

whence

_(b)

= G and

ord(b)

= _n.

Then

and Lemma 3.3

_implies

that

Clearly,

we have b ~

b’)1 ~ 3 [So I

and so we obtain that

whence

Case 2. For every c E G we have

r1¥l

+ 1.

We assert that there are elements

a, b E G

such that

ab ~

I

Si,

such that for _every c E G we have

Such a choice _may be done in the

following

way.

Suppose

that S =

with

_pairwise

_p distinct

_{hl,..., hi}

and

kl &#x3E; &#x3E; &#x3E;

- - 1. If

_kl

= Sl

2 +

_1,

then we set a = b =

hl,

and

obviously

the assertion holds. If

k1

r1¥l,

2

then 1

_k2

r1¥l

We set a =

hi, b

=

h2,

and

obviously,

the assertion

2 holds

_again.

Since a + H and b + H are

_generating

elements of

G/H,

Proposition

3.1

The _sequence satisfies the

_assumptions

of

_Corollary

3.9 with k = 2. We have

and

and let r’ defined

_by

Thus we obtain that

Summing

up we infer that

whence

and thus

The final

_{inequality implies}

that

u whence

_equality

holds.

Since

_{IS11 ~}

_3,

we _may set

_Sl

= abcT with c E

_G,

T E

_7(G)

and infer that

a contradiction. 0

Remark 3.13. We discuss some

examples

which show that the above re-sult is

_sharp

in

_general.

1. If n is even but not a _power of

2,

then the

assumption

"~S~ I

&#x3E;

nt1"

cannot be weakened.

_Suppose

n =

2~m

where 1 m is odd. Then

is a _sequence with

_length

181

_{= 2}

which does not contain an element of order n. We assert that S is zero-sumfree. Assume to the

_contrary,

that S contains a zero-sum

subsequence

T. Then T =

(2

₊

(m

₊

nZ)

with i E

_{[1, ~ -}

_1]

_{and n}

_O’z(8)

2n whence

_az(T)

= n, a contradiction.

2. If n is _even, then the assertion

_{"v9 (S) &#x3E;}

3" cannot be enbettered.

Suppose n

= 2m for some m &#x3E; 2 and consider

Then = 2~n - 1 _n whence S is

zero-sumfree,

= ~

+ 1 and 3.

3. If n is

_odd,

then the assertion

_{"v9(S) &#x3E;}

cannot be enbettered. Let n = 6k - r be odd with r e

_{(0, 5~}

_(whence

k = and

Then = k ₊ 2k ₊

3(nt1 -

2k)

_n whence S is zero-sumfree and

9 C

_G}

=

n + 1

2k}

= k =

n+5

.

max{Vg(S)

I

9 E

_G}

=

max{k,

z

2k}

= k = ! 6 4. On

Let G be a finite abelian _group

Go

c G a

_non-empty

subset. We

briefly

recall some basic

_terminology

from factorization

_theory.

To do _so, we restrict to block monoids.

_However,

matters are similar for Krull monoids

having

finite divisor class _groups. For details the reader is referrred to the

Let denote the set of all zero-sum _sequences in

Go.

Then C

F(Go)

is a submonoid - called the block monoid over

Go -

and is

precisely

the set of irreducible elements of

_B(Go).

Let B Then

there are minimal zero-sum _sequences

Ul, ... ,

Uk

E

_A(Go)

such that

B = Ul ... Uk.

Such a

product decompostion

is called a

f actorization

of A and k is called the

_length

of this factorization. Then

L(B) =

N I B

has a factorization of

length

11

C N

is a finite subset of the

_positive

_integers

and is called the set

_{of lengths}

of B. For _any finite subset L =

{a?0)"’ ?

_C Z with _xl ... let

denote the set

_of

distances of L.

_Clearly,

1 if and

_only

if L is an arithmetical

_progression.

We define

and say that

Go

is

_{half-factorial,}

if

_0(Go)

= 0

_{(equivalently, IL(B)I}

= 1 for

all B E

_B(Go)).

We shall need the

_following

_properties

of

_0(Go).

Lemma 4.1. Let G be a

finite

abedian _group and

Go

C G a subset with

0.

1.

_{max A (Go) D(Go) -}

2.

2.

_{min A (GO)}

Proof.

See

_Proposition

3 and

_Proposition

4 in

_[Ger88].

0 Lemma 4.2. Let G =

7G/n7G

with n &#x3E; 4 and

Go

_C G a subset with

0

and 1 + nZ E

_Go.

Then

Proof.

See

_Proposition

7 in

_[Ger87].

D

Let G be a finite abelian group and

_Go

C G a subset which is not

half-factorial. Then for _every N E N there exists some B E

_B(Go)

such that N.

_Moreover,

there exists some M E N‘ such that for _every B E

B(Go) the

set of

_lengths

_L(B)

is an almost arithmetical

multiprogression

with bound M. The set

_{0* (G)}

defined below describes the set of

_possible

differences which appear in such

multiprogressions

(cf.

[CG97]

and

[GG00] ) .

Now we can formulate the main result in this section which

heavily

rests on Theorem 3.12.

Theorem 4.4. Let G be a

cyclic

_groups with

IGI

= r~ &#x3E; 4. Then

Proo f .

Lemma 4.1

_implies

that

whence

_by

Lemma 4.2 we have n - 2 =

minA(Go)

_E

A**(G).

If n = 2m _{+ 1} with m &#x3E;

2,

we set

_Go

=

{1

₊

nZ, m

₊

n7G}

and assert

that

_{min A(Go)}

= m - 1

= ~ 2 J -

1.

_Clearly,

we have

and

whence Lemma 4.2

_implies

that = m - 1.

If n = 2m for some m &#x3E;

_2,

we set

_Go

=

{1

₊ m +

_nZ,

_{(n - 1)}

+

_n7G}

and assert that

_minA(Go)

= _{m -} 1. Since

the assertion follows from Lemma 4.2. We now come to the

_proof

of

Clearly,

the assertion holds for n E

_{~4, 6J.}

Let n &#x3E; 7 and

_G1

C G a non

half-factorial subset with

_{min A(Gi)}

n - 2. Let

Go

C

_{G1 B}

_{0}

be a minimal non half-factorial subset with d =

minA(Go).

If d _{= n -}

2,

then

n -

2 ~

min A(Gi )

=

gcdA(Go)

_{= n -} 2

whence

_n22.

_Suppose

that d n-2. It suffices to show that

d

implies

that

_{D(Go) &#x3E;}

Therefore there exists a zero-sumfree _sequence

S with

By

Theorem 3.12 there exists some 9 E

_supp(S)

with

_ord(g)

= _n. Since = n -

_2,

it follows that

-g 0

Go.

There exists some _group

automorphism

cp : G -

Z /nZ

with

cp(g)

= 1 + nZ. Since d =

minA(Go) =

we _{may suppose without restriction}

that g

= 1 + nZ and

with _{a2 ... as n - I. We} set

..

with 0

dl

...

_dk.

_By

Lemma 4.2 it follows that d =

gcd{dl, ... , dkl.

Since

it follows that k =

_{l, di}

= d and

Case 1: There exists some a E

_{~2,...

_as~

with

_{gcd(a, n}}

= 1. Then

there exists some

₁₁

E

_{[2, n - 1]}

with

_all

+ 1 = 0 mod n.

Considering

the

following

two irreducible blocks

we infer that

uz(U) =

na &#x3E;

lia

+ 1 =

uz(V)

and thus

lia

₊ 1 = _n. This

implies

that

a contradiction.

’

Case 2: There exists some a E

la2, - - .,a, I

with 1

_{gcdfa, nj}

a.

Then U = _E and

a contradiction.

Case 3: For

_{every i}

E

_{[2, s]}

we have _{ai ~}

n.

Thus for _every U =

(ai

₊ E with

_az(U)

&#x3E; n we have

ki n -1

- and hence

ai

References

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[BEN75] J. D. BOVEY, P. _ERDÖS, I. NIVEN, Conditions for zero sum modulo n. Canad. Math. Bull. 18 _(1975), 27-29.

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[Ger88] A. GEROLDINGER, Über nicht-eindeutige Zerlegungen in irreduzible Elemente. Math. Z. 197 _(1988), 505-529.

[GG98] W. GAO, A. GEROLDINGER, On the structure _{of zerofree} sequences. Combinatorica 18

(1998), 519-527.

[GG99] W. GAO, A. GEROLDINGER, On long minimal zero _sequences in finite abelian _groups. Period. Math. _Hungar. 38 _(1999), 179-211.

[GG00] W. GAO, A. GEROLDINGER, Systems of sets _{of lengths} II. Abh. Math. Sem. Univ.

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[MS55] L. MosER, P. SCHERK, Distinct elements in a set of sums. Amer. Math. _Monthly 62 (1955), 46-47.

Alfred GEROLDINGER Institut fur Mathematik Karl-Franzens Universitat Heinrichstrasse 36 8010 Graz, Austria

E-mail : _{alfred.geroldingerGuni-graz.at}

Yahya Ould HAMIDOUNE E. Combinatoire, Case 189

Universite P. et M. Curie

4, Place Jussieu 75005 Paris, France