zero-sum problems in finite abelian groups and affine caps

28 Pages·2007·0.32 MB·English

TheQuarterlyJournalofMathematicsAdvanceAccesspublishedonMonth00,2007
Quart.J.Math.58(2007),159–186;doi:10.1093/qmath/ham003
ZERO-SUM PROBLEMS IN FINITEABELIAN GROUPS
ANDAFFINE CAPS
byYVESEDEL†
(MathematischesInstitutderUniversität,ImNeuenheimerFeld288,69120Heidelberg,Germany)
CHRISTIANELSHOLTZ‡
(DepartmentofMathematics,RoyalHolloway,UniversityofLondon,Egham,
SurreyTW200EX,UK)
ALFREDGEROLDINGER§
(InstitutfürMathematikundWissenschaftlichesRechnen,Karl-Franzens-UniversitätGraz,
Heinrichstrasse36,8010Graz,Austria)
SILKEKUBERTIN¶
(ZurSeebecke6,31311Uetze-Hänigsen,Germany)
andLAURENCERACKHAM(cid:2)
(DepartmentofMathematics,RoyalHolloway,UniversityofLondon,Egham,
SurreyTW200EX,UK)
[Received14November2006]
Abstract
For a finite abelian group G, let s(G) denote the smallest integer l such that every sequence S
overGoflength|S|≥lhasazero-sumsubsequenceoflengthexp(G).Wederivenewupperand
lowerboundsfors(G),andallourboundsaresharpforspecialtypesofgroups.Theresultsare
not restricted to groups G of the form G=Cr, but they respect the structure of the group. In
n
particular,weshows(C4)≥20n−19foralloddn,whichissharpifnisapowerof3.Moreover,
n
weinvestigatetherelationshipbetweenextremalsequencesandmaximalcapsinfinitegeometry.
1. Introductionandmainresults
Let G be a finite abelian group. We denote by s(G) (or η(G) respectively) the smallest integer
l ∈NsuchthateverysequenceS overGoflength|S|≥l hasazero-sumsubsequenceT oflength
|T|=exp(G)(orazero-sumsubsequenceT oflength|T|∈[1,exp(G)]respectively);fordetailson
†E-mail:[email protected]
‡Correspondingauthor.E-mail:[email protected]
§E-mail:[email protected]
¶E-mail:[email protected]
(cid:2)E-mail:[email protected]
159
©2007.PublishedbyOxfordUniversityPress.Allrightsreserved
Forpermissions,pleaseemail:[email protected]
160 Y.EDELetal.
terminologyandnotationwerefertosection2.Theinvestigationoftheseinvariantshasalongtradition
incombinatorialnumbertheoryaswellasinfinitegeometry(foranoverview,see[32,section5.7]
andsection5).AsalreadypointedoutbyHarborth,s(Cr)isthesmallestintegerlsuchthateveryset
n
ofl latticepointsinr-dimensionaleuclideanspacecontainsnelementswhichhaveacentroidwith
integral coordinates.This geometric interpretation was a main reason why emphasis was formerly
placedongroupsoftheformCr.Inthemeantime,newapplications(forexample,inthetheoryof
n
non-uniquefactorizations[32])causedtheneedforinvestigationsoftheinvariantss(G)andη(G)
forgeneralfiniteabeliangroups.
Inthepresentpaper,suchinvestigationsarecarriedoutforthefirsttimeinasystematicway.In
Theorems1.1and1.2,webrieflysummarizethepresentstateofknowledge,andthenwediscussthe
newresults.
For finite abelian groups of rank at most 2, both invariants η(G) and s(G) are completely
determined.
THEOREM1.1 LetG=Cn1 ⊕Cn2 with1≤n1|n2.Then
η(G)=2n +n −2 and s(G)=2n +2n −3.
1 2 1 2
AproofofTheorem1.1wasrecentlygivenin[32,Theorem5.8.3].ItcontainstheresultbyReiher
[56], which states that s(C ⊕C )=4p−3 for all p ∈P [56, 59], and it contains the theorem
p p
of Erdo˝s, Ginzburg and Ziv (set n =1; see [19] for the original paper; for various proofs see [1;
1
51,section2.4]).
Fromnowonweconsiderfiniteabeliangroupsofranklargerthan2,andwestartwiththediscussion
oflowerbounds.
THEOREM1.2 Letn,r ∈N.
(1) η(Cr)≥(2r −1)(n−1)+1ands(Cr)≥2r(n−1)+1.
n n
(2) If n is odd, then there exists a sequence T ∈F(C3) of length |T|=9 such that Tn−1 has
n
nozero-sumsubsequenceoflengthn.Inparticular,wehaveη(C3)≥8n−7ands(C3)≥
n n
9n−8.
The first result is due to Harborth (see [34, Hilfssatz 1] or Proposition 3.1 for a generalization)
andthesecondduetoElsholtz([17],seealsoLemma3.4forasimpleralternativeproof).Notethat
inthesepapersonlytheresultfors(Cr)isformulated,buttheproofsandLemma2.3(2)showthe
n
lowerboundforη(Cr).
n
GaoandThangadurai[27]conjecturethatthelowerboundsgiveninTheorem1.2(2)aretheprecise
values,thatis,
η(Cn3)=8n−7 and s(Cn3)=9n−8 foralloddn∈N≥3
(seealsoCorollary4.5).
Beforediscussingournewresults,weconsidertheinverseproblemsassociatedtoinvariantss(G)
andη(G).Inotherwords,westudythestructureofsequencesS ∈F(G)oflength|S|=s(G)−1
(or|S|=η(G)−1respectively),whichhavenozero-sumsubsequenceT oflength|T|=exp(G)
(ornozero-sumsubsequenceT oflength|T|∈[1,exp(G)]respectively).Theseproblemswerefirst
studied for groups of the form G=C ⊕C by van Emde Boas [18]. Suppose that G=Cr with
n n n
n≥2andr ∈N.ItisgenerallybelievedthatGhasthefollowingtwoproperties.
ZERO-SUMPROBLEMS 161
PROPERTY C Every sequence S ∈F(G) of length |S|=η(G)−1 which has no short zero-sum
subsequencehastheformS =Tn−1forsomesequenceT ∈F(G).
PROPERTYD Every sequence S ∈F(G) of length |S|=s(G)−1 which has no zero-sum subse-
quenceoflengthnhastheformS =Tn−1forsomesequenceT ∈F(G).
If n≥2 and r =1, then Property C holds (trivially) and so does Property D, as was proved
independently by several authors [4;24, Theorem 1; 54]. For a detailed discussion of these two
propertiesinthecaser =2,see[26];inthecaser ≥3,see[28].Clearly,ifPropertyDholdsthen
n−1dividess(Cr)−1and,byLemma2.3(2),wehaveη(Cr)=s(Cr)−n+1.
n n n
InTheorem1.3wepresentnewlowerboundsforη(C4)ands(C4),andweconjecturethatthese
n n
lower bounds give the precise value for all odd n∈N≥3 (see Corollary 4.5). The sequence S we
construct for the proof of Theorem 1.3 has the form S =Tn−1 for some T ∈F(Cr), supporting
n
the conjecture that Property D holds for C4. The proof of Theorem 1.3 is given in section 3, and
n
subsequentlyweshowhowtolifttheboundsofTheorem1.3andTheorem1.2togroupsofhigher
rank(seeProposition3.5andCorollary3.6).
THEOREM1.3 Letnbeanoddintegerwithn≥3.ThenthereexistsasequenceT ∈F(Cn4)oflength
|T|=20suchthatTn−1 hasnozero-sumsubsequenceoflengthn.Inparticular,wehaveη(C4)≥
n
19n−18ands(C4)≥20n−19.
n
Nowwediscusstheupperbounds.GaoandYang(see[30]fortheoriginalpaper(inChinese)or
[32, Theorem 5.7.4]) proved that s(G)≤|G|+exp(G)−1 for every finite abelian group G. The
upperboundsforgroupsGoftheformG=CrweregivenbyAlon,DubinerandrecentlybyKubertin
n
(seeRemarks3.7).
We derive new upper bounds for both η(G) and s(G). The first (Theorem 1.4) rests on upper
bounds for s(Cr) for primes p ∈P dividing exp(G), and the second (Theorem 1.5) is valid for
p
groupswithlargeexponent(asusual,D(G)denotestheDavenportconstantofG;seeDefinition2.1
andthesubsequentremarks).
THEOREM1.4 Let G=Cn1 ⊕···⊕Cnr withr =r(G)and1<n1|···|nr.Letc1,...,cr ∈Nsuch
thatforallprimesp ∈Pwithp|n andalli ∈[1,r],wehaves(Ci)≤c (p−1)+1.Then
r p i
(cid:2)r
s(G)≤ (cr+1−i −cr−i)ni −cr +1, wherec0 =0.
i=1
Inparticular,ifn =···=n =n,thens(G)≤c (n−1)+1.
1 r r
THEOREM1.5 Let G=H ⊕Cn be a finite abelian group where H ⊂G is a subgroup, exp(G)=
n≥2andD(G)≤2n−1.
(1) IfD(G⊕C )≤3n−1,then2(D(H)−1)+n≤η(G)≤D(G⊕C ).
n n
(2) IfGisap-groupforsomeoddprimep,then
D(G⊕C )+D(H)−1≤s(G)≤D(G⊕C )+n.
n n
162 Y.EDELetal.
The proofs of Theorems 1.4 and 1.5 will be given in section 4. Theorem 1.4 is proved by the
induction method and provides the best known upper bound for groups G which are not of the
form G=Cr.The main part ofTheorem 1.5 is the upper bound in 1.5(2), which generalizes (for
n
odd primes) the recent results of Gao and Zhou [31, Theorem 1.5;23, Proposition 3.1]. Its proof
usesthepolynomialmethod,firstdevelopedbyRónyai[57]andlatergeneralizedbySun[61],and
Kubertin[45].
AftertheproofofTheorem1.5wepresentspecialtypesofgroupsforwhichthelowerandupper
boundsderivedinthispapercoincide(seeCorollaries4.4,4.5and4.6).Thequalityofthebounds
in Theorem 1.5 can immediately be seen by considering the following most simple case. If (in
Theorem1.5)H =C forsomemdividingn,then2(D(H)−1)+n=2m+n−2=η(G)(see
m
Theorem1.1).Moreover,ifH =C andnisaprimepower,then(againbyTheorem1.1)η(G)=
n
3n−2=D(G⊕C )ands(G)=4n−3=D(G⊕C )+n−1=D(G⊕C )+D(H)−1.
n n n
Insection5wediscussfurtherapplicationsofinvariantss(G)andη(G).Specialemphasisisgiven
totheroleofinvariants(Cr)infinitegeometryandproblemsonarithmeticprogressions.Wegivea
n
detaileddiscussionofthehistoryoftheassociatedgeometricproblems.
2. Notationandsomepreparatoryresults
LetNdenotethesetofpositiveintegers,P⊂NthesetofallprimenumbersandletN =N∪{0}.
0
For integers a,b∈Z we set [a,b]={x ∈Z|a ≤x ≤b}, and for c∈N let N≥c =N\[1,c−1].
Throughout,allabeliangroupswillbewrittenadditively,andforn∈N,letC denoteacyclicgroup
n
with n elements. For p ∈P, let F =Z/pZ, and for a power q of p, let F denote a field with q
p q
elementssuchthatF ⊃F .
q p
LetGbeanadditivefiniteabeliangroup.If|G|>1,thenthereareuniquelydeterminedintegers
r,n ,...,n with 1<n |···|n such that G∼=C ⊕···⊕C .Then r =r(G) is the rank of G,
1 r 1 r n1 nr
and n =exp(G) is the exponent of G.An r-tuple (e ,...,e ) in G\{0} is called a basis of G if
r 1 r
G=(cid:11)e (cid:12)⊕...⊕(cid:11)e (cid:12).Forn∈N,wesetnG={ng|g ∈G}.
1 r
WedenotebyF(G)thefree(abelian,multiplicative)monoidwithbasisG.AnelementS ∈F(G)
iscalledasequenceoverGandwillbewrittenintheform
(cid:3) (cid:3)l
S = gvg(S) = gi ∈F(G),
g∈G i=1
wherev (S)iscalledthemultiplicityofginS.AsequenceS(cid:13) ∈F(G)iscalledasubsequenceofS
g
ifthereexistssomeS(cid:13)(cid:13) ∈F(G)suchthatS =S(cid:13)S(cid:13)(cid:13) (equivalently,S(cid:13)|S orv (S(cid:13))≤v (S)forevery
g g
g ∈G).Ifthisholds,thenS(cid:13)(cid:13) =S(cid:13)−1S.Asusual
(cid:2) (cid:2)l
σ(S)= v (S)g = g ∈ G
g i
g∈G i=1
denotesthesumofS,
supp(S)={g ∈G|v (S)>0}⊂G
g
isthesupportofS and (cid:2)
|S|= v (S)=l ∈N
g 0
g∈G
ZERO-SUMPROBLEMS 163
denotes the length of S. Clearly, |S|=0 if and only if S =1 is the empty sequence.We say that
sequenceS is
• azero-sumsequence(resp.hassumzero)ifσ(S)=0;
• short(inG)if|S|∈[1,exp(G)];
• square-freeifv (S)≤1forallg ∈G.
g
DEFINITION2.1 Wedenoteby
• D(G) the smallest integer l ∈N such that every sequence S ∈F(G) of length |S|≥l has a
zero-sumsubsequence.WecallD(G)theDavenportconstantofG;
• η(G) the smallest integer l ∈N such that every sequence S ∈F(G) of length |S|≥l has a
shortzero-sumsubsequence;
• s(G) the smallest integer l ∈N such that every sequence S ∈F(G) of length |S|≥l has a
zero-sumsubsequenceT oflength|T|=exp(G);
• g(G) the smallest integer l ∈N such that every square-free sequence S ∈F(G) of length
|S|≥lhasazero-sumsubsequenceT oflength |T|=exp(G).
AthoroughtreatmentoftheDavenportconstant,acentralinvariantinzero-sumtheory,maybe
foundin[32,Chapter5],andforsomerecentresultswereferto[11].Apartfrombasicproperties,
we use the following classical results on D(G) (originally due to Kruyswijk and Olson): if G=
C ⊕···⊕C ,wherer =r(G)and1<n |···|n ,then
n1 nr 1 r
(cid:2)r
1+ (n −1)≤D(G),
i
i=1
andequalityholdsifeitherr ≤2orGisap-group[32,Theorems5.5.9and5.8.3].
Letϕ: G→G(cid:13)beamapofabeliangroups.Thenthereisauniquehomomorphismϕ: F(G)→
F(G(cid:13)) with ϕ|G=ϕ. We simply write ϕ instead of ϕ, whence if S =g ·...·g ∈F(G), then
1 l
ϕ(S)=ϕ(g )·...·ϕ(g)∈F(G(cid:13)).
1 l
We start with a simple observation which will be used tacitly throughout the paper. Then we
continuewithalemmarelatinginvariantsη(G),s(G)andg(G).Insections3and4weconcentrate
onη(G)ands(G),andinsection5wemainlydealwithg(G).
LEMMA 2.2 Let G be a finite abelian group with exp(G)=n≥2, g ∈G, ϕ: G→G an
automorphism,andf: G→Gamapdefinedbyf(x)=ϕ(x)−gforeveryx ∈G.
(1) A sequence S ∈F(G) has a zero-sum subsequence of length n if and only if f(S) has a
zero-sumsubsequenceoflengthn.
(2) Let S =hn−1T ∈F(G) with h∈G, f(h)=0 and T ∈F(G). Then S has a zero-sum
subsequenceoflengthnifandonlyiff(T)hasashortzero-sumsubsequence.
(cid:4)
Proof. (1) LetS =g ·...·g ∈F(G)andT = g beasubsequenceofS,whereI ⊂[1,l]
1 l i∈I i
with|I|=n.Then,
(cid:2)(cid:5) (cid:6)
σ(f(T))= ϕ(g )−g =ϕ(σ(T)),
i
i∈I
whenceσ(T)=0ifandonlyifσ(f(T))=0.
164 Y.EDELetal.
(2) By (1), S has a zero-sum subsequence of length n if and only if f(S) has a zero-sum
subsequenceoflengthn.Sincef(S)=0n−1f(T),theassertionfollows.
LEMMA2.3 LetGbeafiniteabeliangroupwithexp(G)=n≥2,S ∈F(G),asequencewhichhas
nozero-sumsubsequenceoflengthn,andh=max{v (S)|g ∈G}.
g
(1) D(G)≤η(G)≤s(G)−n+1.
(2) If h=n−1, then η(G)≥|S|−n+2. In particular, if |S|=s(G)−1, then η(G)=
s(G)−n+1.
(3) g(G)≤s(G)≤(g(G)−1)(n−1)+1. If G=Cr, with n≥2 and r ∈N, and s(G)=
n
(g(G)−1)(n−1)+1,thenGhasPropertyD.
(4) IfH isafiniteabeliangroupwith|H|≥handf: [1,h]→H aninjectivemap,then
(cid:3)v(cid:3)g(S)
(g+f(i))∈F(G⊕H)
g∈G i=1
isasquare-freesequencewhichhasnozero-sumsubsequenceoflengthn.Inparticular,if
exp(H)|n,theng(G⊕H)≥s(G),andg(Cr+1)≥s(Cr).
n n
Proof. (1) Straightforward(foradetailedproof,see[32,Lemma5.7.2]).
(2) Let S =gn−1T, with g ∈G and T ∈F(G), and consider the map f: G→G defined by
f(x)=x−g for all x ∈G. By Lemma 2.2(2), f(T) has no short zero-sum subsequence
whenceη(G)≥|f(T)|+1=|S|−n+2.If|S|=s(G)−1,thenη(G)≥s(G)−n+1,
whence(1)impliesthatη(G)=s(G)−exp(G)+1.
(3) Thefirstinequalityfollowsbydefinition.Let
U =gk1 ·...·gkl ∈F(G), wherel,k ,...,k ∈N and g ,...,g ∈G aredistinct,
1 l 1 l 1 l
beasequenceoflength|U|=s(G)−1,whichhasnozero-sumsubsequenceoflengthn.
Clearly, T =g ·...·g is a square-free sequence which has no zero-sum subsequence of
1 l
lengthnwhencel ≤g(G)−1.Therefore,weobtainthat
(cid:2)l
s(G)−1=|U|= k ≤l(n−1)≤(g(G)−1)(n−1).
i
i=1
Furthermore, if G=Cr and equality holds, then k =···=k =n−1, whence G has
n 1 l
PropertyD.
(4) Byconstruction,thegivensequencehasalltheassertedproperties.
AllsequencesS constructedinthispaper,thathavenozero-sumsubsequenceoflengthexp(G),
have the additional property of Lemma 2.3(2). (That is, they have some element with multiplicity
exp(G)−1)whencewealwaysgetη(G)≥|S|−exp(G)+2.Gao[25]conjecturedthatforallfinite
abeliangroupsGwehaveη(G)=s(G)−exp(G)+1.Amongothers,thisholdstrueforallgroups
Gwithr(G)≤2(seeT(cid:5)heorem1.1(cid:6))andexp(G)≤4.LetG=Crwithn≥2andr ∈Nandconsider
n
theinequalitys(G)≤ g(G)−1 (n−1)+1.Thentheequalityholdsforn=2(trivial)andn=3
[34,Hilfssatz3].Ifpisaprimewithp ≥67,theng(C ⊕C )=s(C )=2p−1;seeTheorem1.1
p p p
and[29].
ZERO-SUMPROBLEMS 165
3. Lowerbounds
Alllowerboundsforinvariantsη(G)ands(G)areestablishedbyexplicitconstructionsofsequences
S ∈F(G)havingnozero-sumsubsequenceswiththerequiredproperties.Thefirsttworesultswill
be used several times, whereas the specific preparations for the proof of Theorem 1.3 start from
Lemma 3.3 on.A geometric interpretation of the sequences given in Lemma 3.4 andTheorem 1.3
will be offered after Lemma 5.4. Note that in the special setting of p-groups, the bounds given in
Lemma3.2werefirstprovedin[31,Theorem1.5].
PROPOSITION3.1 Let G=Cn1 ⊕···⊕Cnr,wherer =r(G)and1<n1|···|nr,andlet(e1,(cid:7)...,er)
beabasisofGwithord(e )=n foreveryi ∈[1,r].ForasubsetI ⊂[1,r],wesete = e
i i I i∈I i
(inparticular,e∅ =0).
(1) Let
(cid:3)r (cid:3)
U = (ek+eI)nk−1 ∈ F(G).
k=1I⊂[k+1,r]
Thenthefollowingstatementsareequivalent:
(a) U hasnoshortzero-sumsubsequence;
(b) r =1or(r ≥2andn =n ).
2 r
(2) LetH beafiniteabeliangroupwithexp(H)=nbeingamultipleofn andT ∈F(H)such
r
thatTn−1hasnozero-sumsubsequenceoflengthn.Thenthesequence
(cid:8) (cid:9)
(cid:3) (cid:3)r
S = gn−1 (g+ei)ni−1 ∈ F(G⊕H)
g∈T i=1
hasnozero-sumsubsequenceoflengthn,andhence
(cid:8) (cid:9)
(cid:2)r
s(G⊕H)≥η(G⊕H)+n−1≥1+|T| n−1+ (n −1) .
i
i=1
F(cid:7)urthermore, if m∈N and I1,...,Im ⊂[1,r] are pairwise disjoint sets with
(n −1)≥nforallμ∈[1,m],then
i∈Iμ i
(cid:3)(cid:3)m
S =S (g+e ) ∈ F(G⊕H)
Iμ
g∈Tμ=1
hasnozero-sumsubsequenceoflengthn.
(3) IfG=G⊕Ck withk,n∈Nandn |n,then
n r
(cid:8) (cid:9)
(cid:2)r
s(G)≥η(G)+n−1≥1+2k n−1+ (n −1) .
i
i=1
Proof. (1) (a) ⇒ (b) Ifr ≤2,thenthereisnothingtoshow.Ifr ≥3andn <n ,then
2 r
U(cid:13) = ernr−n2−1(e2+er)n2−n1+1(e1+er)(e1+e2+er)n1−1
isashortzero-sumsubsequenceofU,acontradiction.
166 Y.EDELetal.
(b)⇒(a)Ifr =1,thentheassertionisclear.Supposethatr ≥2andn =n =n.Then
2 r
(cid:3) (cid:3)
U = (e1+eI)n1−1 eIn−1 ∈ F(G)
I⊂[2,r] ∅(cid:17)=I⊂[2,r]
andweconsiderazero-sumsubsequence
(cid:3)k (cid:3)l
U(cid:13) = (e +e ) e
1 Iν Iν
ν=1 ν=k+1
ofU with0≤k ≤l ≤n,subsetsI1,...,Ik ⊂[2,r]andnon-emptysubsetsIk+1,...,Il ⊂
[2,r].Wehavetoshowthat|U(cid:13)|=l =0.Assume,onthecontrary,thatl ≥1.Sincev (U(cid:13))≤
e1
v (U)=n −1,thereexistssomeν ∈[1,l]suchthatI isnon-empty.Leti ∈I and
e1 1 ν ν
α =|{j ∈[1,l]|i ∈I }|.
j
Since α ≡0 mod n and 1≤α ≤l ≤n, it follows that α =l =n. Since U(cid:13) has sum zero
weinferthatI =···=I ,whence
1 n
U(cid:13) =(e +e )ken−k.
1 I1 I1
Then k ≡0modn1 and k ≤ve1+eI1(U)=n1−1 imply that k =0, whence U(cid:13) =eIn1 is a
subsequenceofU,acontradiction.
(2) Assume,onthecontrary,thatShasazero-sumsubsequenceS(cid:13)oflengthn.Sinceeveryzero-
sumsubsequenceofTm,foranym≥n,oflengthnhastheformgn forsomeg ∈supp(T)
itfollowsthatthesequenceS(cid:13)hastheform
(cid:3)r
S(cid:13) =glg (g+ei)li,
i=1
with g ∈supp(T), l ∈[0,n−1] and l ∈[0,n −1] for all i ∈[1,r]. But l <n implies
g i i g
thatthereissomei ∈[1,r]withl ∈[1,n −1],andhenceσ(S(cid:13))(cid:17)=0,acontradiction.Now
i i
thelowerboundsfors(G⊕H)andη(G⊕H)followfromLemma2.3.
(cid:13)
IfS hasazero-sumsubsequenceS oflengthn,then
(cid:3)r (cid:3)m
(cid:13)
S =glg (g+ei)li (g+eIμ)δμ,
i=1 μ=1
(cid:13)
withδ ∈{0,1}andallotherparametersasbefore.SinceS isnotasubsequenceofS,there
μ
(cid:13)
issomeμ∈[1,m]withδ =1.ThisimpliesthatS mustcontainthesequence
μ
(cid:3)
S(cid:13)(cid:13) =(g+eIμ) (g+ei)ni−1,
i∈Iμ
(cid:7)
(cid:13) (cid:13)(cid:13)
andhencen=|S|≥|S |≥1+ (n −1)≥n+1,acontradiction.
i∈Iμ i
ZERO-SUMPROBLEMS 167
(3) Applying(1)tothegroupCk,weobtainasequenceT =0T(cid:13) oflength|T|=2k suchthat
n
U =T(cid:13)n−1 has no short zero-sum subsequence and hence Tn−1 has no zero-sum subse-
q(cid:7)uence of length n. Then (2) gives us a sequence S ∈F(G) of length |S|=|T|(n−1+
r (n −1)),whichhasnozero-sumsubsequenceoflengthn.Nowtheassertionfollows
i=1 i
fromLemma2.3.
LEMMA3.2 LetGbeafiniteabeliangroupwithexp(G)=n≥2andG=H ⊕(cid:11)e(cid:12),whereH ⊂G
isasubgroupande∈Gwithord(e)=n.Then
η(G)≥2(D(H)−1)+n and s(G)≥2(D(H)−1)+2n−1.
Proof. IfT =g ·...·g ∈F(H)isasequenceoflength|T|=l =D(H)−1,whichhasnozero-
1 l
sumsubsequence,thenobviouslythesequence
(cid:3)l (cid:3)l
S =en−1 g (g +e) ∈ F(G)
i i
i=1 i=1
hasnoshortzero-sumsubsequence.Thisimpliesthat
(cid:5) (cid:6)
η(G)≥|S|+1=2 D(H)−1 +n,
(cid:5) (cid:6)
andbyLemma2.3(1)wehaves(G)≥2 D(H)−1 +2n−1.
Fortherestofthissectionweintroducethefollowingnotation.LetG=Cr,withn≥2andr ∈N,
n
andlet(e ,...,e )beabasisofG.Inordertostressthegeometricaspectofthetheory,wewritethe
1 r
elementsg ∈Gascoordinatevectors;thismeansforanelementg ∈Gwithg =a e +···+a e ,
1 1 r r
weset ⎛ ⎞
a
1
⎜ ⎟
⎜⎝ ... ⎟⎠=a1e1+···+arer =g,
a
r
wherea ,...,a ∈[0,n−1].Fori ∈[1,r],wecalla theithcoordinateofganda +···+a ∈N
1 r i 1 r 0
theweightofg.
LEMMA3.3 LetG=Cn2 withn≥3odd,
(cid:8) (cid:9) (cid:8) (cid:9) (cid:8) (cid:9) (cid:8) (cid:9)
n−1 n−1 n−1 n−1
0 0 2 2
V = S with S =
2
0 2 0 2
and (cid:8) (cid:9) (cid:8) (cid:9) (cid:8) (cid:9) (cid:8) (cid:9)
n−1 n−1 n−1 n−1
0 1 1 2
W = .
2
1 0 2 1
ThenneitherV norW hasazero-sumsubsequenceoflengthn.
2 2
168 Y.EDELetal.
Proof. Letf: G→Gbedefinedby
⎛ ⎞
(cid:8) (cid:9) n+1 n+1 (cid:8) (cid:9) (cid:8) (cid:9)
a (cid:19)→⎜⎜⎝ 2 2 ⎟⎟⎠ a + 0 .
b n−1 n+1 b 1
2 2
Thenitiseasytocheckthatf(V )=W .ByProposition3.1(1),Shasnoshortzero-sumsubsequence.
2 2
ThusLemma2.2impliesthatV andf(V )havenozero-sumsubsequenceoflengthn.
2 2
LEMMA3.4 LetG=Cn3withn≥3odd,
⎛ ⎞⎛ ⎞⎛ ⎞⎛ ⎞⎛ ⎞⎛ ⎞⎛ ⎞⎛ ⎞⎛ ⎞
1 1 1 1 2 2 2 2 3
⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟
V =V˜n−1 with V˜ =⎝0⎠⎝0⎠⎝2⎠⎝2⎠⎝0⎠⎝1⎠⎝1⎠⎝2⎠⎝1⎠
3 3 3
0 2 0 2 1 0 2 1 1
and ⎛ ⎞⎛ ⎞⎛ ⎞⎛ ⎞⎛ ⎞⎛ ⎞⎛ ⎞⎛ ⎞⎛ ⎞
2 2 2 2 1 1 1 1 0
⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟
W =W˜n−1 with W˜ =⎝0⎠⎝0⎠⎝2⎠⎝2⎠⎝0⎠⎝1⎠⎝1⎠⎝2⎠⎝1⎠.
3 3 3
0 2 0 2 1 0 2 1 1
Then,neitherV norW hasazero-sumsubsequenceoflengthn.Inparticular,wehaveη(C3)≥
3 3 n
8n−7ands(C3)≥9n−8.
n
Proof. Letf: G→Gbedefinedby
⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞
a −1 0 0 a 3
⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟
⎝b⎠(cid:19)→⎝ 0 1 0⎠⎝b⎠+⎝0⎠.
c 0 0 1 c 0
Then it is easy to verify that f(V )=W . Thus by Lemma 2.2(1) and Lemma 2.3(2), it suffices
3 3
toprovethatV hasnozero-sumsubsequenceoflengthn.Assume,onthecontrary,thatthereisa
3
zero-sumsubsequenceV(cid:13)ofV oflengthn.
3 3
WefirstsupposeV(cid:13)containsonlythoseelementsofV thathavefirstcoordinate1or2.Thenthe
3 3
sumofthefirstcoordinatesoftheelementsofV(cid:13) isnor2n.Thisisonlypossibleifeveryelement
3
of V(cid:13) has the same entry in the first coordinate. If all elements of V(cid:13) have first coordinate 1, then
3 3
the sequence formed by the remaining two coordinates is a zero-sum subsequence of V of length
2
n, a contradiction to Lemma 3.3. Similarly, if all elements of V(cid:13) have first coordinate 2, then the
3
sequenceformedbytheremainingtwocoordinatesisazero-sumsubsequenceofW oflengthn,a
2
contradictiontoLemma3.3. ⎛ ⎞
3
ThusV(cid:13)mustcontaintheelementg =⎝1⎠.Sincethesecondcoordinateofgequals1,thesum
3
1
ofthesecondcoordinatesoftheelementsofV(cid:13)mustben.Thesameistrueforthethirdcoordinate.
3
Sincethefirstcoordinateofg equals3,thesumofthefirstcoordinatesoftheelementsofV(cid:13) must
3