Inductive Methods and Zero-Sum Free Sequences
Gautami Bhowmik, Immanuel Halupczok and Jan-Christoph Schlage-Puchta
Abstract. A fairly long-standing conjecture is that the Davenport constant of a groupG D Zn1˚ ˚Znkwithn1j jnkis1CPk
iD1.ni 1/. This conjecture is false in general, but it remains to know for which groups it is true. By using inductive methods we prove that for two fixed integerskand`it is possible to decide whether the conjecture is satisfied for all groups of the formZ`k˚Znwithnco-prime tok.
We also prove the conjecture for groups of the formZ3˚Z3n˚Z3n;wherenis co-prime to6, assuming a conjecture about the maximal zero-sum free sets inZ2n.
Keywords. Zero-sum sequences, Davenport constant, inductive method, decidability.
AMS classification. 11B50, 20K01, 68R05.
1 Introduction and Results
LetGbe a finite abelian group written additively, anda1; : : : ; aka sequence of ele- ments inG. We say that this sequence contains a zero-sum if there is some non-empty subsequence1i1< i2< < i`ksatisfyingai1C Cai` D0; otherwise it is called zero-sum free. Denote byD.G/the least integerksuch that every sequence of lengthkcontains a zero-sum. This number is usually called Davenport’s constant, since the question of whether zero-sums exist was studied by Davenport in the con- text of algebraic number theory (where G is the class group of some number field, the elementsai are given ideal classes from which one wants to construct a principal ideal). This line of research was continued in the study of domains with non-unique factorisation, for an overview see [12]. Among applications, Brüdern and Godinho [6] discovered that the existence of zero-sums can be used to simplifyp-adic forms, which led to considerable progress towards Artin’s conjecture onp-adic forms.
To avoid cumbersome notation we shall from now on always talk about multi-sets instead of sequences; in the sequel all sets are multi-sets unless stated otherwise. We shall write the multiplicity of an element as its exponent, e.g.¹an; bmºis a multi-set containing nCmelements, nof which are equal toa, and mare equal tob. We believe that the imprecision implied by the non-standard use of equality is more than outweighed by easier readability.
The second author was supported by the Agence National de la Recherche (contract ANR-06-BLAN-0183-01).
One approach to bound D.G/is the so called inductive method, which runs as follows: IfN < Gis a subgroup andnan integer such that every sequence of lengthn inG=N contains a system ofD.N /disjoint zero-sums, thenD.G/n. Indeed, given a multi-set inG, each zero-sum of its image inG=N defines an element in N, and choosing a zero-sum among these elements defines a zero-sum inG. Unfortunately, in general this method does not give the exact value for D.G/. For example, for G D Z23˚Z3n, Delorme, Ordaz and Quiroz showed thatD.G/ 3nC5, which is 1 more than the exact value. The sub-optimality of this method stems from the fact that in general we have many ways to choose a system of disjoint zero-sums in G=N, and it suffices to show that one of these systems yields a zero-sum inN. If the structure of all zero-sum free subsets inN of size close toD.N /is sufficiently well understood one can use this information to choose an appropriate system of subsets inG=N. In this way one can show that for groups of the formG D Z23˚Z3nwe always haveD.G/D3nC4(confer [4]), the corresponding lower bound being given by the multiset¹.1; 0; 0/2; .0; 1; 0/2; .0; 0; 1/3n 1º. In fact, this example immediately generalises to arbitrary finite groups: IfGDZn1˚ ˚Znkwithn1j jnk, then D.G/M.G/WD 1CPk
iD1.ni 1/. The conjecture thatD.G/DM.G/, which we shall refer to as the main conjecture, is proven for groups of rank 2, and fails for infinitely many groups of rank 4. It is not yet known whether it holds true for all groups of rank 3.
In this article we generalise the improved inductive method to other sequences of groups. We first give a decidability result. Suppose k; ` 2 N are fixed. Then one can check the main conjecture for all groups of the formG WD Z`k ˚Zn at once (in a finite amount of time), wherenruns through all numbers co-prime tok. Note thatG Š Z` 1k ˚Zk n, soM.G/ D.` 1/.k 1/Ck n. Moreover, if the main conjecture does fail for some of the groupsZ`k˚Zn, then we give a description of the set of numbersnwhere it fails.
It turns out that the same proof actually yields a bit more: if the main conjecture happens to be false forGone can ask about the differenceD.G/ M.G/. Our results not only apply to the set of thosenwhere the main conjecture fails, but also to set of suchnwhereD.G/ M.G/ > ıfor any fixedı. Here is the precise statement:
Theorem 1.Supposek 2, ` 1 andı are three integers. Let N be the set of integersnco-prime toksuch thatD.Z`k˚Zn/ > k nCı. Then eitherN is finite, or there exists an integerd > 0and a setT of divisors ofd containing1such thatN differs from the set
N0 WD ¹x2NW.x; d /2Tº only in finitely many elements.
In addition, there is an algorithm which, givenk,`andı, prints outN if the latter is finite. Otherwise its output is d, T and the set of elements in whichN and N0 differ.
ChoosingıD.` 1/.k 1/yields:
Corollary 2.Supposek 2; ` 1are two integers. LetN be the set of integersn co-prime toksuch that the main conjecture fails forZ`k˚Zn. ThenN has the form described in Theorem 1, and there is an algorithm which, givenkand`, describesN as above.
In theory, this means that a computer can be programmed to prove statements of the form “the main conjecture is true forZ`k˚Znfor allnco-prime tok”. However, the reader should be aware that the existence of an algorithm often sounds better than it is: a straight-forward application of our algorithm would require astronomical running time even for very smallk and`(see constants appearing in Proposition 11). Still, we believe that by combining computer search with manual arguments one can prove the main conjecture for certain series of groups. In fact, in [4] the methods of this theorem have been explicitly applied to prove the main conjecture in the casek D3,
`D3.
In the theorem, we mention that the setT of divisors contains1. This is helpful to get a statement of the form “if there is a counter-example to the main conjecture, then there is a small one”; indeed, Proposition 11 is such a statement.
The proof of Theorem 1 makes much use of the simple structure ofZnwhere there is essentially one single example of a large zero-sum free set. In our next theorem, we would like to replaceZn by a larger group. However, for non-cyclic groups the structure of maximal zero-sum free sets is less clear and there are essentially different possibilities for such sets. Due to this complication, we can only deal with groups of rank 2. Though the structure of maximal zero-sum free sets is not known, there is a plausible conjecture concerning these sets. We say that an integernsatisfies property B if every zero-sum free subsetAZ2nof cardinality2n 2contains an elementa with multiplicityn 2.
Conjecture 3.Every integernsatisfies propertyB.
This conjecture is known to hold in several cases.
Proposition 4. (1) Ifnandmsatisfy propertyB, then so doesnm.
(2) All prime numbers up to23satisfy propertyB.
The first statement is essentially due to Gao, Geroldinger and Grynkiewicz [11], the second is proven in [3].
Theorem 5.Letnbe an integer co-prime to6such thatB.n/holds true. ThenD.Z3˚ Z23n/D6nC1.
We remark that even the simplest case dealt by this theorem, that isZ3˚Z215, was till now undecided.
Although we tried to prove as much as possible by hand, the proof of this theo- rem needs a lemma on subsets of Z33 which we could only prove by massive case distinction, which has been done by our computer.
2 Auxiliary Results
For an abelian groupG, we denote byDm.G/the minimalnsuch that any subset of Gof cardinalityncontainsmdisjoint zero-sums.
Lemma 6.The following both statements hold:
(1) For integerskand`, there exists a constantc.k; `/such thatDm.Z`k/kmC c.k; `/.
(2) We haveDm.Z23/D3mC2:
Proof. (1) Given a multi-setAZ`k, form as many zero-sums as possible which are of the form¹akºfor somea 2Z`k. For eacha 2Z`k, there are at mostk 1copies ofa inA which we can not use in this way, soc.k; `/ WD .k 1/k` is certainly sufficient.
(2) It is easy to check that every subset of 5 elements contains a zero-sum, and that every subset of 7 elements contains a zero-sum of length3. Our claim now follows by induction onm.
Lemma 7.Letk; `be integers,A2Zk`a matrix,b 2Zk a vector. Then either(a) there exists an integerdand a setT of divisors ofdincluding1, such that the system Ax Dbis solvable inZn if and only if.d; n/ 2T or(b)there exists a finite set of integersN, such that the above system is solvable if and only ifn2N.
If all entries in A are of modulus M, and all entries of b are of modulus N, then in case(a) d min.k; `/ŠMmin.k;`/, and there is a polynomialp, in- dependent of k,`,M andN, such that in case(b), every elementx 2 N satisfies xN 2p.k`logM /.
Proof. Computing the Smith normal form of the matrix A, we see that there exist invertible matricesP; QoverZ, such thatDWDPAQ 1has non-zero entries at most on the diagonaldi i,i k, and these entries satisfydi i jdiC1;iC1. Since every matrix invertible overZis also invertible overZn, the equationAxDbis solvable inZnif and only if the equationDxDb0is solvable, whereb0 DP b. A necessary condition for solvability is that in every row containing only zeros inD, the corresponding entry ofb0 vanishes, that is,n j bj0 for everyj such thatj > m, wheremis the greatest integer such thatdmm ¤0. If one of thesebj0 does not vanish, then there are at most finitely manynfor which the equation is solvable, and our claim is true. If all these bj0 equal zero, the system is equivalent to the systemdi ixi Dbi0, which is solvable if
and only if.n; di i/j bi0. We taked to bedmm. Sincedi i jd for eachi m, the set ofnfor which the system is solvable is of the form¹nW.n; d /2Tºfor some setT. Moreover.n; d /D1implies.n; di i/jbi0, so12T.
For the numerical bounds note that d equals the greatest common divisor of all m m sub-determinants of A. Since the Q-rank of A equals m, there exists a non-vanishing sub-determinant, containing only entries M, which is therefore mŠMmmin.k; `/ŠMmin.k;`/.
The entries in the set N are bounded by the entries in P b, which in turn are bounded bykN times the entries ofP. A general estimate for the entries of such transformation matrices was obtained by Kannan and Bachem [13, Theorem 5]. They found a polynomial algorithm which takes an`0 `0-matrixAwith integral entries, transforms it into Smith normal formPAQ 1, and returns the transformation matri- cesPandQ. To apply this to our case, we enlarge ourAto a square matrix by adding zeros (i.e.,`0Dmax.k; `/). Then the size of the input data is.`0/2logM, so the size of the output data – and in particular the number of digits of the entries ofP andQ – is bounded byp.`0logM /for some polynomialp. After possibly changingp, this yields the claim.
Corollary 8.Consider the system Ax D b as in the previous lemma, set m WD min.k; `/, and suppose that there are infinitely manynsuch that this system is solv- able inZn. Then for eachz z0 D max 21;mlog.mM /log2
the system is solvable for somen2Œz; 2z.
Proof. If the system has infinitely many solutions, then there exists an integerd mŠMm such that the system is solvable inZnwhenever.n; d / D1. If the system is unsolvable for alln2Œz; 2z, then in particulard is divisible by all prime numbers in this interval. Since forz 21, the product of all prime numbers inŒz; 2zis2z, our claim follows.
The following result is essentially due to Bovey, Erd˝os and Niven [5].
Lemma 9.LetA Zn be a zero-sum free multi-set containingN elements, where N 2n=3. Then there exists an elementaofZn, which occurs inAwith multiplicity greater than2N n. Moreover,ais a generator ofZn.
Proof. The statement on the multiplicity is [5]. Now suppose thatais not a generator ofZn, and letH be the subgroup generated bya. Denote bymthe multiplicity of a. Among.Zn W H /elements ofZn=H we can choose a zero-sum, that is, among theN melements ofAn ¹amºwe can choose a system ofb.ZN m
nWH /cdisjoint sets, each one adding up to an element inH. SinceAis zero-sum free, we cannot obtain jHj elements in this way, that is,mC b.ZN m
nWH /c jHj 1, which implies.Zn W H /mCN m < n. Since m 2N nC1, and .Zn W H / 2, we obtain 3N C1 < 2n, contradictingN 2n=3.
Corollary 10.LetA Znbe a subset withjAj 3n=4. ThenAis zero-sum free if and only if0…Aand there exists some invertible˛2Zn, such thatP
a2A.˛a/
n 1, where W Zn ! N is the map sendingx to the least non-negative residue contained in the classx.
Proof. Obviously, if0…AandP
a2A.˛a/n 1, thenAis zero-sum free. Hence, we now assume thatAis zero-sum free and bound the sum. In view of Lemma 9 we may assume without loss thatAcontains the element 1 with multiplicitym > n=2. If Acontains an element in the intervalŒn=2; n, this element can be combined with a certain multiple of 1 to get a zero-sum. Letx1; : : : ; xkbe the list of all elements inA different from 1. EitherP
.xi/n m 1, which is consistent with our claim, or there is a least`such thats DP`
iD1.xi/ > n m 1. Since no singlexi satisfies .xi/ > n=2, we haves 2 Œn m; n 1, hence,s can be combined with a certain multiple of 1 to get a zero-sum, which is a contradiction.
3 Proof of Theorem 1
Proof of Theorem 1. Letkand`be fixed once and for all. We want to describe the set ofnco-prime toksuch thatD.Z`k˚Zn/ > k nCıholds. More precisely, it suffices to describe this set fornsufficiently big, as long as the bound onnis computable.
By definition,D.Z`k˚Zn/ > k nCıholds if and only if there exists a zero-sum free setA Z`k ˚Zn of cardinalityk nCı. Such a setAcan be described by its projectionAontoZ`k and the multi-functionf W A ! Zn such that.a; f .a// 2A is the preimage ofa2A. Using this description, the existence of a setAas above is equivalent to the existence of a setAZ`kof cardinalityk nCıand a multi-function f WA!Zn(call.A; f /a “candidate”) such that the following condition holds:
For any zero-sumZA, the sumP
a2Zf .a/is not equal to zero. () The sumP
a2Zf .a/will often simply be called the “Zn-sum ofZ”. Moreover, we will use the following terminology: A “constant” is a value which only depends on k,`andı(but not onn); “bounded” means bounded by a constant (in the sense just described), and “almost all” means that the number of exceptions is bounded.
Here is the main part of the proof. We initially skip the proofs of the two following steps:
(1) Suppose .A; f / is a candidate and .Zi/im is a system of m disjoint zero- sum subsets of A (for somem 2 N). From this we can form the multi-set B WD B..Zi/i/WD ¹P
a2Zif .a/W1imº Zn. If.A; f /satisfies (), thenBhas to be zero-sum free.
We will find a constant cdefect such that for m WD n cdefect, we also have the converse:.A; f /satisfies () if and only if for all systems.Zi/imofmDn cdefect
disjoint zero-sum subsets ofA, the corresponding setB..Zi/i/is zero-sum free. From now on, we fixmlike this.
(2) We will show that if a candidate .A0; f0/ satisfying () exists, then there does already exist a candidate .A; f /of a particular form. Candidates of this form will be called “main candidates”, and they are defined as follows. We will fix a suitable constantcvar..A; f /is a main candidate if there exists an elementa02Zlk such that there are at least jAj cvar occurrences ofa0 in Awithf .a0/ D 1k. Note that k1 does make sense askandnare co-prime. (Right now, we could as well have written f .a0/D1instead off .a0/D k1, but later, 1k will be more handy.)
The remainder of the proof goes as follows:
(3) A “datum for a main candidate” is a tuple .a0; .aj/j; .fj/j/, wherea0 2 Z`k, .aj/j 2.Z`k/cvar, and.fj/j 2.Z`n/cvar. Such a datum yields a main candidate.A; f / in the following way:ADA0[A?, whereA0WD ¹a0k nCı cvarºandA? WD ¹aj W1 j cvarº,f .a0/ D 1k for eacha02 A0, andf .aj/ Dfj foraj 2A?. Each main candidate can be described by such a datum.
Only the.fj/j part of such a datum depends onn. Our goal now is to verify that after fixing a0 and .aj/j, whether () holds for the corresponding main candidate depends on.fj/j in a simple way: we will construct systems of linear equations over Zsuch that () holds if and only if the tuple.fj/j is a solution of one of these systems modulon. Then the theorem will follow using Lemma 7.
(4) Fix a datum.a0; .aj/j; .fj/j/and the corresponding main candidate.A; f / as in step (3). We claim that to check whether.A; f /satisfies (), it suffices to consider systems .Zi/im where for any i > cvar, we have Zi D ¹ak0º. Indeed, suppose that.Zi/im is an arbitrary system ofmdisjoint zero-sums and thatB..Zi/i/does contain a zero-sum; denote byJ the set of indices such that this zero-sum consists of theZn-sums of the setsZj,j 2J. We will modify.Zi/imuntil it satisfies the condition of the claim, keeping the zero-sum intact.
By renumbering the setsZi, we may supposeZi A0fori > cvar; in particular, Zi D ¹ar0ikº for some integers ri. Now we replace each of these sets Zi by its subset¹a0kº. To compensate for this in the zero-sum, we have to find ani0 2J with i0 cvar; then we can repair the zero-sum by adding toZi0 all the elements which we removed fromZi,i 2J,i > cvar.
Suppose that such ani0does not exist. Then our zero-sum isP
i2J
P
a2Zif .a/D P
i2J jZij1k D P
i2J ri. However, this can not be zero in Zn, as P
i >cvarri
k nCı cvar
k < n; for the last inequality, we suppose without losscvar> ı.
(5) From now on, we only consider systems.Zi/imas in step (4), i.e., withZi D
¹ak0º for i > cvar. These are in bijection to the systems .Zi/icvar ofcvar disjoint
zero-sums ofAn ¹ak.m c0 var/º D A? [ ¹akc0 defectCıC.k 1/cvarº DW A??. We see that the set B WD B..Zi/im/ Zn corresponding to such a system is of the form
¹b1; : : : ; bcvar; 1m cvarº, wherebi DP
a2Zif .a/. This sum equalsP
¹jWaj2ZiºfjC
1
kzi wherezi D jZi\.A??nA?/j.
(6) Supposem 34n, i.e.,n 4cdefect. Then we can apply Corollary 10 to the set B and get that it is zero-sum free if and only if bi ¤ 0for all i cvar and there exists some˛ 2 Zn such thatP
b2B.˛b/ < n(withW Zn ! N defined as in Corollary 10). Supposingm cvar n=2, we get that only˛D1is possible, and the condition becomesPcvar
iD1.bi/ < n .m cvar/DcdefectCcvar.
(7) This can be reformulated as follows: Set C0 WD ¹.ci/icvar 2 Zcvar W ci 1and Pcvar
iD1ci < cdefectCcvarº (note that C0 does not depend on n), and denote by W Zcvar Zcnvarthe projection. ThenB is zero-sum free if and only if.bi/i D ..ci/i/for some .ci/i 2 C0. Moreover, we rewrite the equationbi D .ci/ as P
¹jWaj2Ziºkfj D.kci zi/.
(8) Putting all this together, we have: For sufficiently large n, there exists a pair .A; f /satisfying () if and only if:
Ex. main cand. s. th.
‚ _ …„ ƒ
a02Z`k .aj/j2.Z`k/cvar
9.fj/j 2Zcnvar
for all relevant zero-sum systems
‚ …„^ ƒ
.Zi/isystem ofcvardisjoint zero-sums inA??
Bis zero-sum free
‚_ …„ ƒ
.ci/i2C0
^
1icvar
X
¹jWaj2Ziº
kfj D.kci zi/ :
We used big conjunctionsV
and disjunctionsW
as notation for some of the universal and existential quantifiers to emphasise that their range is finite and independent ofn.
Putting this formula into disjunctive normal form and moving the existential quan- tifier inside theW
, we get that there exists a pair.A; f /satisfying () if and only if at least one of a finite number of systems of linear equations (with coefficients inZnot depending onn) has a solution inZn.
By Lemma 7, each system either contributes only finitely many integers nsuch that.A; f /satisfies (), or the contributed set has the form¹nW.n; d /2Tºfor some integerdand some setT of divisors ofd containing1. The union of sets of this form again has this form, so the first part of the theorem is proven.
Concerning the algorithm it is enough to find computable bounds for the following:
a boundn0such that the above formula holds for allnn0; a boundn1such that if the system of equations is solvable modulononly for finitely manyn, then thesenare at mostn1; a boundd0such that if the system of equations is solvable for infinitely manyn, thend d0.
Clearly, all bounds which appear in this proof are computable, so we do get this result. In Section 3.1, we will even determine such bounds explicitly.
Now let us fill in the two remaining steps.
(1) LetAZ`kbe of cardinalityk nCı, and supposeZAis any zero-sum subset.
We will construct a large system.Zi/i of disjoint zero-sums inAsuch thatZcan be written as union of some of these zero-sums Zi. This then implies the first step: if B..Zi/i/is zero-sum free, then in particular the sumP
a2Zf .a/is not zero.
By Lemma 6 we can find at leastbjZj kc.k;`/cdisjoint zero-sums inZand at least bjAnZjkc.k;`/cdisjoint zero-sums inAnZ. We may suppose thatZis the union of the zero-sums we found inside. Together, we get bjZj kc.k;`/c C bjAnZjkc.k;`/c bjAj 2c.k;`/k c 1DWmDWn cdefectdisjoint zero-sums inA. Note thatcdefectdoes not depend onn.
The second step requires some more work, so we decompose it into several sub- steps. We suppose that.A; f /is a candidate satisfying (). In the first four substeps, we prove some properties of.A; f /; in the last substep, we use this to construct an- other candidate.A0; f0/which will be a main candidate satisfying ().
(2.1) Claim: There is a constant cmore such that in any system.Zi/i ofmdisjoint zero-sums ofA, at mostcmoresetsZi have more thankelements.
Let .Zi/i be given and let r be the number of sets with more than k elements.
Together, these sets have at leastr.kC1/elements. Remove these big sets from our system and instead use Lemma 6 to repartition them into disjoint zero-sums. After that, we have a new system.Zi0/iconsisting ofm r old sets andbr.kC1/ c.k;`/
k c D
r C br c.k;`/k cnew ones. By (),B..Zi0/i/does not contain a zero-sum, so this new system consists of at most n 1sets; this implies mC br c.k;`/k c n 1, i.e., r < cdefectkCc.k; `/DWcmore.
(2.2) Claim: Suppose that nis sufficiently large. Then for any system .Zi/i ofm disjoint zero-sums inA, almost all elements of the sum-setB WDB..Zi/i/are equal to one single elementb2Znwhich generatesZn.
This follows from Lemma 9. We needjBj D n cdefect 23n, i.e.,n3cdefect. And we get an elementbwith multiplicity at least2jBj nC1Dm cdefectC1DW m cws(ws = wrong sum).
(2.3) Claim: Ifn 0, then the prevalent valuebin B..Zi/i/is the same for any system.Zi/iofmdisjoint zero-sums ofA.
Suppose.Zi/i and.Zi0/i are two different systems of disjoint zero-sums, and de- note the prevalent values of B..Zi/i/ andB..Zi0/i/ byb andb0 respectively. We choosecwsC1of the setsZi which all have cardinality at mostk and all haveZn- sumb. This is possible ifm cmoreC2cwsC1. Without loss, our chosen sets are Z1; : : : ; ZcwsC1.
Now we do the same for.Zi0/i, i.e., we choose Z10; : : : ; Zc0
wsC1 to have at most k elements each and to haveZn-sum-valuesb0. But in addition, we want that these sets Zj0 (forj cwsC1) are disjoint from the setsZi (for i cwsC1). Each set Zi can intersect at most k of the sets Zj0, so the additional condition forbids at most k .cws C1/ of themsets Zj. Therefore we can find our desired sets if mcmoreC2cwsC1Ck.cwsC1/.
Now we use Lemma 6 to complete our chosen sets.Zi/icwsC1and.Zi0/icwsC1
to a system ofmdisjoint zero-sum sets. By (2.2), there is a prevalent valueb00for this system, which leaves out at mostcwssets. This implies that bothb andb0 are equal tob00.
Without loss, we will now suppose that the prevalentZn-value of anymdisjoint zero-sums is1.
(2.4) Claim: There exists a constantcvar such that for at mostcvar of the elements a 2 A, we havef .a/ ¤ k1. In fact we will choose cvar such that even a slightly stronger statement holds: for eacha2Z`k, letrabe number of copies ofainAwith f .a/D k1. ThenP
a2Z`kk brkac jAj cvar.
Let us call a subsetZA“neat” if it is of the form¹akºfor somea2Z`k. We construct a system.Zi/i ofmdisjoint zero-sums with lots of neat sets in the following way: for each elementa2Z`kwhich appears with multiplicityinA, we formbkcdisjoint sets of the form¹akº. If we get more thanmsets in this way, we choosemof them. If we get less thanmsets, then we use Lemma 6 on the remainder ofAto complete our system.Zi/i. Denote bythe number of neat sets in.Zi/i.
The minimal value of is attained if the multiplicity in A of each a 2 Z`k is congruentk 1modulok. So we getmin¹m;k1.jAj .k 1/k`/º DWm cnn
(nn = not neat; note thatcnnis constant).
Among all systems of m disjoint zero-sums in A which have neat sets, now choose a system.Zi/i where the number of neat setsZi withZn-sum equal to1is minimal. At mostcwssets have not sum1and at mostcnnare not neat, so even in this minimal choice we get at leastm cnn cwsneat sets with sum 1. We fix this system .Zi/ifor the remainder of step (2.4).
Choosea 2Z`k, and letNabe the union of all neat setsZi of the form¹akºwith Zn-sum1. We claim that if there are at least two such neat sets, thenf is constant onNa; in particular this implies that the value off onNa is k1. Supposef is not constant onNa. Then there are two elementsa1; a22Nawithf .a1/¤f .a2/which belong to two different neat setsZi1,Zi2. Modify the system.Zi/i by exchanginga1
anda2. ThenZi1andZi2do not have sum 1 anymore, so the new system contradicts the assumption that the old one had a minimal number of neat sets with sum 1.
Doing the above construction for all a 2 Z`k yields the claim: The unionN WD S
a2Z`kNa contains all neat sets Zi with Zn-sum 1, so it has cardinality at least
k.m cnn cws/. On the other hand, iff is not constant equal to 1k on a setNa, then jNaj Dk, and this can happen for at mostk` 1of these sets. Thusf is equal to k1 on at leastk.m cnn cws/ k.k` 1/DW jAj cvarelements. As these elements are contributed in groups ofk, we also get the slightly stronger statement mentioned at the beginning of this step.
(2.5) Claim: There is a main candidate .A0; f0/ satisfying () (still assuming that .A; f /is an arbitrary candidate satisfying ()).
Recall that.A0; f0/is a main candidate if there is an elementa02Z`ksuch thatA0 contains at leastjA0j cvarcopiesaofa0which moreover satisfyf0.a/D 1k.
We construct.A0; f0/out of.A; f /in the following way. As before, fora 2Z`k letra be number of copies ofa in Awithf .a/ D 1k. Choosea0 2 Z`k such that ra0is maximal; in particularra0 jAjk`cvar. Let.A0; f0/be equal to.A; f /with the following modification: For eacha2Z`k, replacekbrkaccopiesa0 2Aofasatisfying f .a0/ D 1k by the same number of copies a00ofa0, and set f0.a00/ D k1 on these copies. Denote bythe bijection fromAtoA0which describes these replacements.
Step (2.4) ensures that .A0; f0/ is a main candidate; it remains to show that it satisfies (). To this end, for any zero-sumZ0 A0, we construct a zero-sumZA which has the sameZn-sum asZ0. As.A; f /satisfies (), thisZn-sum is not equal to zero, so.A0; f0/satisfies (), too.
So suppose a zero-sum Z0 A0 is given. Consider the setM A0 of copies a0 ofa0with f0.a0/ D k1, and for a 2 Z`k define the subset Ma WD ¹a0 2 M W 1.a0/is a copy ofaº. As jMajis a multiple of k for anya ¤ a0, and assuming jMa
0j D ra0 k 1, inZ0 we may replace elements ofMby other elements of Msuch thatjMa\Z0jis a multiple ofk for anya ¤a0. (This changes neither the sum nor theZn-sum ofZ0.) Now takeZ WD 1.Z0/. As elements are moved by groups ofk,Zhas the same sum asZ0(i.e., zero), and asf0ıDf, it has the same Zn-sum.
3.1 Computation of the Bounds
The proof of Theorem 1 actually gives a little more than just decidability. In fact, for eachk,`andı, there is a computable constantn0, such thatD.Z`k˚Zn/ıCk n holds true for all integersnco-prime tok if and only if it holds true for all integers nn0which are co-prime tok. In this subsection we compute an upper bound for n0(Proposition 11). Unfortunately,D.G/is computable only for very small groups G, while the value forn0 obtained in this subsection is rather large. However, we still believe that the algorithm given above can be performed for several small values of k and`, in particular if one does some manual improvements using the explicit knowledge ofkand`.
We now compute all bounds appearing in the proof of Theorem 1.
A bound for Lemma 6: Denote by Dk.Z`k/the least integer n such that every multi-set consisting of nelements inZ`k contains a zero-sum of length k. Then c.k; `/Dk.Z`k/ k, since every multi-set containingk.m 1/CDk.Z`k/elements contains a system ofmdisjoint zero-sums each of lengthk. ForDk.Z`k/we have the trivial bound k`C1, but also the estimate Dk.Z`k/ .256`log`/` k due to Alon and Dubiner [1]. For specific values of k and`, great improvements on both bounds are possible; it is probably at this point that our estimates can be improved most easily. To avoid some awkward expressions in the sequel, we shall express all constants occurring in the proof of Theorem 1 explicitly in terms ofk,`,ıandc.k; `/, and give an explicit estimate using only the boundc.k; `/k`C1. (For the explicit estimates, we use that we may supposek 2,`3,ı2.)
Step (1):cdefectD1C d2c.k;`/ ık e 3k`. Step (2.1):cmoreDkcdefectCc.k; `/4k`C1. Step (2.2):cwsDcdefect 13k`.
Step (2.2) needsn3cdefect. Son9k`suffices.
Step (2.3) needsncdefectCcmoreC2cwsC1Ck.cwsC1/. Son12k`C1 suffices.
Step (2.4):cnnDmax¹0; .k 1/k` 1 1kı cdefectº. The proof of Theorem 1 allows us to assumecdefect D 3k`, which yieldscnn D 0. (However, using more careful estimates forc.k; `/could yield non-zero values forcnn.) Step (2.4):cvarDıCk.cdefectCcnnCcwsCk` 1/7k`C1Cı.
Step (2.5) needsk nCı ck` var k 1. Son8k`suffices.
Step (4) needscvar> ı, which is certainly the case.
Step (6) needsn4cdefect. Son12k`suffices.
Step (6) also needsm cvar n=2, i.e.,n2.cdefectCcvar/. Heren17k`C1C 2ısuffices. This is the largest bound onnof the proof.
Concerning the systems of equations, we get:
Step (7): The coefficients of the equations are all equal tok.
Step (7): The absolute values of the right-hand sides of the equations are bounded by max.k.cdefectCcvar/;jA??nA?j/Dk.cdefectCcvar/9k`C2Ckı.
Step (8): The number of variables in each system of equations iscvar 7k`C1Cı.
Step (8): The left-hand side of any equation is of the formP
jkfj, where the sum runs over a subset of¹1; : : : ; cvarº; thus we may suppose that each system of equation consists of at most2cvar 27k`C1Cıequations.
Hence, we can apply Lemma 7 and Corollary 8 to obtain the following.
Proposition 11.There exists a constantcsuch that the following holds true. Suppose thatk; `; ıare integers such that there exists somen, co-prime tok, satisfyingD.Z`k˚ Zn/ > ıCk n. Denote by N the set of these n, and let n1 be minimum of N. Then we have n1 22c.k`
C1Cı/
. Moreover, if N is infinite, then we have n1 6`.7k`C1Cı/logkı.
Proof. Using the estimates above and Lemma 7, in the case thatN is finite, we obtain the bound
n1.9k`C2Ckı/2p 27k`
C1Cı.7k`C1Cı/logk
22c.k`
C1Cı/
;
and our claim follows in this case. IfN is infinite, we additionally use Corollary 8 to find that the systems of linear equations are solvable for ann2Œz; 2z, provided that z max.z0; 21/, where
z0 log21 cvarlog.cvark/
log21 .7k`C1Cı/log.7k`C2Cık/
3`.7k`C1Cı/logkı;
where we used the fact that we may suppose` 3,ı2. Hence,n12z0. To be sure to get an element ofN inŒz; 2z, we moreover needz 17k`C1C2ı, which is less than the bound just computed. Thus there exists somen2N which is at most two times our bound; this was our claim.
Note that the smallest case of interest would be k D 4; ` D 3; ı D 6, that is, checking D.Z24˚Z4n/ D 4nC6for all oddnup to 3375 would imply that this equation has only finitely many counter-examples. Unfortunately, even the casenD3 has not yet been decided, although it is within reach of modern computers.
4 Proof of Theorem 5
In this section we prove thatB.n/impliesD.Z3˚Z23n/D6nC1ifnis co-prime to6. We suggest that before reading the following lemmas, the reader goes directly to the main proof and starts reading it to get the main idea.
4.1 Lemmas Needed in the Proof
Lemma 12.Among17arbitrary elements inZ33there is a zero-sum of length at most 3, and among ninedistinctelements there is a zero-sum of length at most3. Moreover, up to linear equivalence, there is precisely one set of eight distinct elements without zero-sums of length at most3, which is given as¹x; y; z; xCy; xCyCz; xC2yC z; 2xCz; yC2zº.
Proof. The second part is [4, Lemma 1 (ii)], the first part is folklore (and follows immediately from the second part).
Lemma 13.Suppose thatn5is an integer having propertyB, andBis a subset of Z2nwith either2n 3or2n 4points. Then, with one exception, there always exists a group homomorphismF WZ2n!Znsuch that:
(1) In the case jBj D 2n 3: For anyc withB [ ¹cº zero-sum free, we have F .c/D1.
(2) In the casejBj D 2n 4: For anyc1; c2withB [ ¹c1; c2ºzero-sum free, we haveF .ci/2 ¹0; 1º, and at least one ofF .c1/andF .c2/is equal to1.
The exception isB D ¹bn 21 ; b2n 2º, whereb1andb2generateZ2n.
Proof. Every completion ofBto a zero-sum free set contains an elementbwith mul- tiplicityn 2orn 1such that all other elements of the completion are contained in a co-set ofhbiwhich is a generator ofZ2n=hbi. We will call an element ofBimportant if it could get such an element after completion; i.e., an elementb2Bis important if its multiplicity is at leastn 3in the first case orn 4in the second case, if its order isnand if all other elements ofBare contained in a co-set ofhbiwhich is a generator ofZ2n=hbi. We may suppose thatBcontains at least one important element. We will do case distinctions between the different possibilities for the important elements of B. But before we start, let us have a closer look at what can happen ifBcontains two different important elements, sayb1andb2.
First note that these two elements generate Z2n, as (by the importance ofb1)b2
lies in a co-set ofhb1i generatingZ2n=hb1i. Now b2 determines the co-set ofhb1i and vice versa, so all elements ofB other thanb1andb2 lie in bothb2C hb1iand b1C hb2i; we getB D ¹b1m1; b2m2; .b1Cb2/jBj m1 m2º. In particular, B contains no third important element.
First consider the casejBj D2n 3. We distinguish the following cases:
Bcontains only one important elementb. Then the other elements ofBdefine a co-setLofhbi, and all elementsccompletingBeither are equal tobor lie in L. Ifbhas multiplicityn 1, thenc Dbis impossible, so chooseF such that F .L/D1. Ifbhas multiplicityn 2, then there are only two possibilities forc:
c Dband one other possibility onL(such that the sum ofcand the elements ofB \Lis equal tob). ChooseF to be 1 on these two possibilities. Ifbhas multiplicityn 3, then onlycDbis possible.
In the remaining cases,Bcontains two important elements, soB D ¹bm11; bm22; .b1C b2/m3ºfor some m1; m2; m3 satisfying and m1Cm2 Cm3 D 2n 3. We may supposem1m2.
m1Dn 1: All completions ofB lie inb2C hb1i.
m1 Dm2Dn 2,m3 D1: There are two possible completions: c Db1and cDb2.
m1Dn 2,m2Dn 3,m3D2: There are two possible completions:cDb1
andcDb2 b1.
m1Dm2Dn 3,m3D3: There is no possible completion.
Now consider the casejBj D2n 4. We distinguish the following cases:
Bcontains only one important elementb. Then the other elements ofBdefine a co-setLofhbi, and for all completions¹c1; c2º, bothci lie inL[ ¹bº. If the multiplicity ofbinBisn 1orn 2, we can takeF to be the function which is 1 onL(and0onb). Otherwise at least one of theci is equal tob and the other one either es equal tob, too, or it lies onLand is determined byB. So a functionF exists.
Again, in the remaining cases B D ¹b1m1; b2m2; .b1Cb2/m3º withm1 m2 and m1Cm2Cm3D2n 4.
m1 Dm2 Dn 2,m3D0: This is the exception mentioned in the statement of the lemma.
m1 Dn 2,m2 n 3: There are three types of completions: c1D b1and c2 2b2C hb1i;c1 D c2 D b2; bothci lie in b2C hb1iwith some condition onc1Cc2. (Note that in the casem2Dn 3, we havem3 D1andc1Db2
impliesc2Db1.) So the functionF which mapsb2C hb1ito1does the job.
m1 Dm2 Dn 3,m3D2: There are four possible completions:¹b21º,¹b22º,
¹b1; b2 b1ºand¹b2; b1 b2º. TakeF to mapb1andb2to1.
m1 Dn 3,n2 Dn 4,m3 D3: There are two possible completions: ¹b12º and¹b1; b2 2b1º. (Note that ¹b22º does not work.) TakeF to mapb1 and b2 2b1to 1.
m1Dm2Dn 4,m3D4: No completion is possible.
We will need the following refined version of part 2 of Lemma 13:
Lemma 14.Suppose thatn5is an odd integer having propertyB. Suppose further that B is a subset ofZ2n with2n 4 points. LetC be the set of two-element-sets
¹c1; c2º Z2nsuch thatB[ ¹c1; c2ºis zero-sum free. Then, up to an automorphism ofZ2n,C is a subset of one of the following sets:
(1) C1D®
¹.x1; 1/; .x2; 1/º Wx1; x22Zn¯ . (2) C2 DC20 [C200withC20 D®
¹.1; 0/; .x; 1/º;¹.x; 1/; .1 x; 1/º Wx2Zn¯ and C200D®
¹.0; 1/; .1; y/º;¹.1; y/; .1; 1 y/º Wy2Zn¯ . (3) C3DC30 [C300withC30 D®
¹.1; 0/2º;¹.1; 0/; . 1; 1/º¯ and C300D®
¹.0; 1/2º;¹.0; 1/; .1; 1/º¯ .
Proof. As in the proof of Lemma 13, we consider the different possibilities for the important elements. IfB contains only one important element, we can suppose that it is .1; 0/and that the other elements ofB have y-coordinate one; we denote the multiplicity of .1; 0/ bym1. If there are two important elements, we suppose that B D ¹.1; 0/m1; .0; 1/m2; .1; 1/m3ºwithm1m2.
One important element,m1Dn 1:C DC1.
One important element, m1 D n 2: apply an automorphism of Z2n fixing .1; 0/and mapping the sum of thosen 2elements ofBwithy-coordinate one to.0; 2/. ThenC DC20 C2.
One important element,m1Dn 3: apply an automorphism fixing.1; 0/and mapping the sum of thosen 1elements ofBwithy-coordinate one to.2; 1/.
ThenC DC30 C3.
One important element,m1Dn 4:C D®
¹.1; 0/2º¯ C3.
Two important elements,m1Dm2Dn 2; m3D0:C DC2.
Two important elements,m1Dn 2; m2Dn 3; m3D1: apply an automor- phism fixing.1; 0/and mapping.0; 1/to.12; 1/. ThenC DC20 C2.
Two important elements,m1Dn 2; m2Dn 4; m3D2: apply an automor- phism fixing.1; 0/and mapping.0; 1/to.1; 1/. ThenC DC20 C2.
Two important elements,m1Dm2Dn 3; m3D2:C DC3.
Two important elements,m1Dn 3; m2Dn 4; m3D3: apply an automor- phism fixing.1; 0/and mapping.0; 1/to.1; 1/. ThenC DC30 C3.
Two important elements,m1Dm2Dn 4; m3D4:C D ;.
In addition, we will need the following two lemmas:
Lemma 15.Suppose nis an integer co-prime to6and A Z33has ten elements.
Suppose further thatAhas no zero-sum of length3andAhas no two disjoint zero- sums. Then there is no multi-functiong W A ! Zn (i.e., function which may take different values on different copies of an elementa2A) such that for every zero-sum ZAwe haveP
z2Zg.z/D1.
Proof. If we would requiregto be a real (i.e., single-valued) function, then this would be [4, Theorem 1]. So the only thing we have to check is that the existence of a multi- functiongimplies the existence of a real functiong0with the same properties.
Defineg0 by taking forg0.a/the mean value of the values ofg.a/. Note first that the maximal multiplicity of points in A is 2 (asA does not contain a zero-sum of length 3), sogcan have at most two values at any point. In particular the mean value makes sense (because2−n).
Now consider any pointawhereghas two different values. The modification does not changeP
z2Zg.z/ifZ does not contain a or ifZ contains both copies of a.
However, no zero-sumZcan contain only one copy ofa, for otherwise, we would get two different values forP
z2Zg.z/, which contradictsP
z2Zg.z/D1.
Lemma 16.Supposenis an integer co-prime to6,A Z33 has thirteen elements, and f W A ! Z2n is a multi-function. Suppose further that A has no zero-sum of length3andAhas no three disjoint zero-sums. LetC be the set of two-element- sets¹P
z2Z1f .z/;P
z2Z2f .z/º, whereZ1andZ2are two disjoint zero-sums inA.
ThenC is not a subset of any of the three setsC1,C2orC3of Lemma 14.
Proof. This has been verified by our computer. For details on how this has been done see Section 5.
Note that concerningC1, this is just an unnecessarily complicated way of saying that there is no functiong W A ! Zn which sends to1any zero-sum ofAwhich is disjoint to another zero-sum.
4.2 The Proof Itself
We are now in a position to prove Theorem 5.
Proof of Theorem 5. Supposenis co-prime to6,B.n/holds true,G D Z3˚Z23n, andA G is a multi-set of M.G/ D 6nC1elements. Suppose Acontains no zero-sum. We have to get to a contradiction.
LetAbe the projection ofAontoZ33, and letf WA ! Z2n be the multi-function such that.a; f .a//is the preimage ofa2Z33inAunder the projection.
We remove zero-sums of length 3fromAas long as possible, ending in a set A with less than 17 points (by Lemma 12). Denote byB the multi-set inZ2n cor- responding to the removed zero-sums: for each removed zero-sumZ A, put the element P
z2Zf .z/ into B. AsA is zero-sum free, so is B. The strategy in the remainder of the proof is to consider zero-sumsZ 2Aand their corresponding el- ementsc D P
z2Zf .z/inZ2n. If we find such ac such thatB [ ¹cºdoes contain a zero-sum, we have our desired contradiction. When using this strategy, we may assume that while passing fromAtoAwe never removed zero-sums of length< 3;
otherwiseAonly gets bigger and the proof gets easier.
HencejAjhas the form3iC1andjBj D2n i. AsBhas no zero-sum, we have jBj 2n 2, soi 2andjAj 7. If jAj D 7, thenA itself still contains a zero-sum, so this is not possible either. ThereforeAconsists of 10, 13 or 16 points.
Suppose first that we end with jAj D 16. Then we have 16 points without a zero-sum of length3. As nine distinct points would contain such a zero-sum (by Lemma 12) there are precisely eight points taken twice. Since the only configuration of eight distinct points without a zero-sum of length3is the one given in Lemma 12, we find thatAequals this set with each point taken twice. But this set contains four
disjoint zero-sums:¹x; y; .xCy/2º,¹x; z2; 2xCzº,¹y; xCyCz; .xC2yCz/2º and¹xCyCz; 2xCz; .yC2z/2º. So we can enlargeBto a set with2n 1elements, which is a contradiction.
Next, suppose thatjAj D10. ThenBconsists of2n 3points inZ2n, and each zero-sum Z inA yields an elementc D P
z2Zf .z/ofZ2n such that B [ ¹cºis zero-sum free. Sincensatisfies propertyB(and is5), we can apply Lemma 13 and obtain a linear functionF WZ2n! Znsuch that for everycas aboveF .c/D1. But nowgWDF ıf is a contradiction to Lemma 15.
Finally, consider the case jAj D 13. ThenB consists of2n 4points inZ2n. We check thatA andf contradict Lemma 16. It is clear thatA does not contain a zero-sum of length3and thatAdoes not contain three disjoint zero-sums.
Denote byC the set of two-element-sets¹P
z2Z1f .z/;P
z2Z2f .z/º, whereZ1
andZ2are two disjoint zero-sums inA. Each¹c1; c2º 2C completesB to a zero- sum free subset of Z2n, so by Lemma 14, C is a subset of one of the three setsCi
mentioned in that lemma. This is exactly what we need to get a contradiction to Lemma 16.
5 Computer Proof of Lemma 16
Recall the statement of the lemma: we are given an integern co-prime to 6, a set A Z33consisting of13elements, and a multi-functionf W A ! Z2n. We suppose thatAhas no zero-sum of length 3and no three disjoint zero-sums. We letC be the set of two-element-sets¹P
z2Z1f .z/;P
z2Z2f .z/º, whereZ1andZ2 are two disjoint zero-sums inA. The statement is thatCis not a subset of any of the three sets C1,C2orC3of Lemma 14:
C1D®
¹.x1; 1/; .x2; 1/º Wx1; x22Zn¯
; C2D®
¹.1; 0/; .x; 1/º;¹.x; 1/; .1 x; 1/º Wx2Zn¯ [®
¹.0; 1/; .1; y/º;¹.1; y/; .1; 1 y/º Wy2Zn¯
; C3D®
¹.1; 0/2º;¹.1; 0/; . 1; 1/º; ¹.0; 1/2º;¹.0; 1/; .1; 1/º¯ :
The program is divided into two parts. First find all possible multi-setsA (up to automorphism ofZ23), regardless of the functionf, and then, for each fixed setAand eachi 2 ¹1; 2; 3º, find all possible functionsf W A! Z2nsuch thatC Ci. If no suchf is found, then the lemma is proven.
5.1 Finding All Multi-SetsA
The program recursively tries every possibility forA by starting with an empty set and successively adding elements. After adding an element, it checks right away ifA still fulfils the above conditions before adding more elements.
