Sumsets and Zero-sum Problems

Zhi-Wei Sun

Nanjing Universityzwsun@nju.edu.cn

http://math.nju.edu.cn/∼zwsun

June 12, 2020

Abstract

In this talk we introduce sumsets and zero-sum problems. Inparticular, we focus on the Cauchy-Davenport theorem,Chevalley-Warning theorem and Erdős-Ginzburg-Ziv theorem.

2 / 48

Part I. Sumsets

3 / 48

A Basic Result

Theorem: Let A and B be subsets of a multiplicative group G . If|A|+ |B| > |G |, then

AB := {ab : a ∈ A and b ∈ B}coincides with G .

Proof. Suppose that AB 6= G . Then g 6∈ AB for some g ∈ G . Forany a ∈ A and b ∈ B we have g 6= ab and hence gb−1 6= a. SoA ∩ gB−1 = ∅ and hence

|G | > |A ∩ gB−1| = |A|+ |gB−1| = |A|+ |B|. �If A1, . . . ,An are subsets of an additive group G , then we call

A1 + . . .+ An := {a1 + . . .+ an : a1 ∈ A1, . . . , an ∈ An}the sumset of A1, . . . ,An. When A1 = . . . = An = A, we write nAfor A1 + . . .+ An.

If A and B are subsets of an additive group G , then

|A|+ |B| > |G | =⇒ A + B = G .4 / 48

Additive bases

Let A ⊆ N = {0, 1, 2, . . .} and h ∈ Z+ = {1, 2, 3, . . .}. If

hA = {a1 + . . .+ ah : a1, . . . , ah ∈ A}

coincides with A, then we say that A is an additive base of order h.If hA contains all sufficiently large natural numbers, then A iscalled an asymptotic additive base of order h.

Lagrange’s Four-square Theorem: Each n ∈ N is the sum offour squares. In other words, the set {02, 12, 22, . . .} is an additivebase of order 4.

5 / 48

Shnirel’man’s density and Mann’s theorem

Goldbach’s Conjecture: Let P be the set of all primes. ThenP + P ⊇ {4, 6, 8, . . .}.Shnirel’man’s Density for A ⊆ N (L. G. Shnirel’man, 1933):

d(A) := infn>1

|{a ∈ A : 1 6 a 6 n}|n

.

Though d(P) = 0, Shnirel’man showed in 1933 that d(P + P) > 0,and that there exists a constant c > 0 such that each integergreater than one can be expressed as a sum of at most c primes;this is the first important progress on Goldbach’s conjecture.

Mann’s Theorem (conjectured by Shnirel’man and proved byMann in 1942): Let A and B be subsets of N containing 0. Then

d(A + B) > min{1, d(A) + d(B)}.

Remark. Any A ⊆ N with 0 ∈ A and d(A) > 0 is an additive baseof order h for some h ∈ Z+. In fact, take h ∈ Z+ withh > 1/d(A), then hA = N since d(hA) > min{1, hd(A)} = 1.

6 / 48

Sumsets over ZObserve that {1, . . . , k}+ {1, . . . , l} = {2, . . . , k + l}.

Theorem. Let A and B be finite nonempty subsets of Z. Then|A + B| > |A|+ |B| − 1.

Proof. Write A = {a1, . . . , ak} with a1 < . . . < ak , andB = {b1, . . . , bl} with b1 < . . . < bl . Without loss of generality,we suppose that k 6 l . Note that

ai + bi < ai + bi+1 < ai+1 + bi+1 (i = 1, . . . , k − 1)

andak + bk < ak + bk+1 < . . . < ak + bl .

So we have found 2(k − 1) + l − (k − 1) = k + l − 1 distinctnumbers in A + B. Thus |A + B| > |A|+ |B| − 1. �

It can be shown further that |A + B| = |A|+ |B| − 1 if and only ifA and B are arithmetic progressions with the same commondifference.

7 / 48

A general theorem

Using the above result, by induction one gets the following generalresult.

Theorem. Let A1, . . . ,An (n > 1) be finite nonempty subsets ofZ. Then

|A1 + . . .+ An| > |A1|+ . . .+ |An| − n + 1,

and equality holds if and only if A1, . . . ,An are arithmeticprogressions with the same difference.

Corollary. Let A be a finite subsets of Z and let n ∈ {2, 3, . . .}.Then

|nA| > n|A| − n + 1,

and equality holds if and only if A is an AP (arithmeticprogression).

8 / 48

Freiman’s Theorem

Freiman’s Theorem (Freiman, 1966). Let A be a finite nonemptysubset of Z with |A + A| 6 c |A|. Then A is contained in ann-dimensional AP

Q = Q(a; q1, . . . , qn; l1, . . . , ln) = {a+x1q1+· · ·+xnqn : 0 6 xi < li}

with |Q| 6 c ′|A|, where c ′ and n only depend on c .

Remark. For a proof of Freiman’s theorem, one may visithttp://maths.nju.edu.cn/∼zwsun/Szemeredi.pdf. Thisdeep theorem plays a crucial role in the Fields medalist W. T.Gowers’ quantitative proof [Geom. Func. Analysis Appl., 2001] ofthe famous Szemerédi theorem. Ben Green and I. Z. Ruzsa [J.London Math. Soc. 2007] extended Freiman’s theorem to anyabelian group.

9 / 48

Szemerédi’s TheoremThe following deep result conjectured by P. Erdős and P. Turán in1936, implies the classical van der Waerden theorem.

Szemerédi’s Theorem (Acta Arith. 1975). Let 0 < δ 6 1 andk ∈ {3, 4, . . .}. If n ∈ Z+ is sufficiently large, then A ⊆ {1, . . . , n}with |A| > δn contains an AP of length k.In 1956 K. Roth proved this result for k = 3 by the circle methodin analytic number theory. In 1969 E. Szemerédi handled the casek = 4 by a combinatorial method. The case of general k wassettled by Szemerédi in 1975 in a paper which was regarded as “amasterpiece of combinatorial reasoning” by R. L. Graham. In 1977H. Furstenberg used ergodic theory to give a new proof ofSzemerédi’s theorem. In 2001 W. T. Gowers employed Fourieranalysis and combinatorics (including Frieman’s theorem onsumsets) to reprove the theorem with explicit bounds.

Szemerédi’s theorem plays an important role in the proof of thefollowing result.

Green-Tao Theorem There are arbitrarily long APs of primes. 10 / 48

Dyson’s g -transformation

F. Dyson simplified Mann’s proof by introducing the so-calledDyson g-transformation which plays an important role in inductionproofs of some additive results.

Let A and B be nonempty subsets of an abelian group G . Forg ∈ G we let

A(g) = A ∪ (g + B) ⊇ A and B(g) = (A− g) ∩ B ⊆ B,

and call the pair (A(g),B(g)) the Dyson g-transformation of thepair (A,B).

We have A(g) + B(g) ⊆ A + B since for x ∈ g + B and y ∈ B(g)clearly x + y = (y + g) + (x − g) ∈ A + B.

If A and B are finite, then |A(g)|+ |B(g)| = |A|+ |B| since

|A(g)\A| = |(g+B)\A| = |g+(B\(A−g))| = |B\(A−g)| = |B\B(g)|.

11 / 48

Cauchy-Davenport Theorem

Let p be a prime. Then Fp = Z/pZ = {ā = a + pZ : a ∈ Z} is afield with p elements. If A = {1̄, . . . , k̄} and B = {1̄, . . . , l̄} with|A| = k 6 p and |B| = l 6 p, then A + B = {2̄, . . . , k + l} andhence

|A + B| = min{p, k + l − 1} = min{p, |A|+ |B| − 1}.Cauchy-Davenport Theorem. Let p be any prime. If A and Bare nonempty subsets of Fp = Z/pZ, then

|A + B| > min{p, |A|+ |B| − 1}.Remark. This theorem was first proved by Cauchy in 1813, andthen rediscovered by Davenport in 1935.

By induction, the Cauchy-Davenport theorem can be extended tosumsets of n subsets of Fp.Theorem. Let p be a prime and let A1, . . . ,An be nonemptysubsets of Fp = Z/pZ. Then

|A+ . . .+ An| > min{p, |A1|+ . . .+ |An| − n + 1}.12 / 48

Proof of the Cauchy-Davenport theoremChoose b0 ∈ B and set B ′ = B − b0. Then 0 ∈ B ′, |B ′| = |B| and|A + B| = |A + B ′|. So, without loss of generality, we may simplyassume 0 ∈ B. If |A|+ |B| > p, then A + B = Fp and hence|A + B| = p = min{p, |A|+ |B| − 1}.Below we show |A|+ |B| 6 p ⇒ |A + B| > |A|+ |B| − 1. Supposethat this is not true and choose nonempty A,B ⊆ Fp with |B|minimal such that |A|+ |B| 6 p and |A + B| < |A|+ |B| − 1.Clearly |A|, |B| > 1. As |A + B| < p and 0 ∈ B, we have A 6= Fp.Choose b ∈ B \ {0}. If A + b ⊆ A, then A + 2b ⊆ A + b ⊆ A, . . .,A + (p − 1)b ⊆ A. For any a ∈ A, {a + kb : 0 6 k 6 p − 1} ⊆ Awhich contradicts A 6= Fp. So A + b 6⊆ A.Choose g ∈ A with g + b 6∈ A and consider the Dysong -transformation (A(g),B(g)) of the pair (A,B). Note that

|A(g) + B(g)| 6 |A + B| < |A|+ |B|−1 = |A(g)|+ |B(g)|−1 < p,

Also, 0 ∈ B(g) = B ∩ (A− g) and |B(g)| < |B|. This contradictsthe choice of B.

13 / 48

Pollard’s Theorem

Theorem (Pollard, 1974). Let p be a prime and let A and B benonempty subsets of Fp. For t = 1, . . . ,min{|A|, |B|} define

Nt = |{g ∈ Fp : |{(a, b) : a ∈ A, b ∈ B, a + b = g}| > t}|.

ThenN1 + . . .+ Nt

t> min{p, |A|+ |B| − t}

for all t = 1, . . . ,min{|A|, |B|}.

This theorem in the case t = 1 gives the Cauchy-Davenporttheorem.

14 / 48

Kneser’s Theorem

In 1953 Kneser extended the Cauchy-Davenport theorem togeneral abelian groups.

Kneser’s Theorem. Let G be an additive abelian group. Let Aand B be finite nonempty subsets of G , and let H = H(A + B) bethe stablizer {g ∈ G : g + A + B = A + B}. If|A + B| 6 |A|+ |B| − 1, then

|A + B| = |A + H|+ |B + H| − |H|.Corollary. Let G be an additive abelian group. Let p(G ) be theleast order of a nonzero element of G , or p(G ) = +∞ if G istorsion-free. Then, for any finite nonempty subsets A and B of G ,we have

|A + B| > min{p(G ), |A|+ |B| − 1}.Proof. Suppose that |A + B| < |A|+ |B| − 1. ThenH = H(A + B) 6= {0} by Kneser’s theorem. Therefore |H| > p(G )and hence

|A + B| = |A + H|+ |B + H| − |H| > |A + H| > |H| > p(G ). 15 / 48

Part II. Finite Fields and Chevalley-Warning Theorem

16 / 48

Finite Fields

For a field F with identity e, if e, 2e = e + e, 3e = e + e + e, . . .are all nonzero then we say that the characteristic of F is zero, ifne = 0 for some n ∈ Z+ then the least positive integer p withpe = 0 is called characteristic of F . The characteristic ch(F ) of afield F is either zero or a prime. (Note that (me)(ne) = (mn)e.)

For any prime p, Zp = Z/pZ = {ā = a + pZ} is a field ofcharacteristic p.Let p be a prime and let n ∈ Z+. It is known that there is a monicirreducible polynomial f (x) ∈ Zp[x ] of degree n. Then

Zp[x ]/(f (x)) ={P(x) mod f (x) : P(x) ∈ Zp[x ]}is a field with pn elements.

A field F with |F | = q ∈ Z+ exists if and only if q = pn for someprime p and positive integer n. If q > 1 is a prime power, then anytwo fields of order q are isomorphic, and we write Fq to denote the(unique) field F of order q. F ∗ = F \ {0} is a cyclic group of orderq − 1.

17 / 48

Chevalley-Warning Theorem

Chevalley-Warning Theorem. Let F be any finite field ofcharacteristic p. Let fi (x1, . . . , xn) ∈ F [x1, . . . , xn] for alli = 1, . . . ,m. Let V = Z (f1, . . . , fm) be the set of solutions to thesystem of equations

f1(x1, . . . , xn) = 0,

f2(x1, . . . , xn) = 0,

. . . . . .

fm(x1, . . . , xn) = 0

(∗)

over F n. If∑m

i=1 deg(fi ) < n, then p | |V |, and in particular (∗)cannot have a unique solution over F n.

18 / 48

Proof

Write |F | = q = pa with a ∈ Z+. Then F ∗ = F \ {0} is a cyclicgroup of order q − 1. Let g be a primitive element of F (i.e., agenerator of the cyclic group F ∗). For any k ∈ N we have

gk∑x∈F

xk =∑x∈F

(gx)k =∑y∈F

yk

and hence (gk − 1)∑

x∈F xk = 0. If q − 1 - k , then gk 6= 1 and

hence∑

x∈F xk = 0. Note also that

∑x∈F x

0 = q1 = 0. So∑x∈F

xk = 0 for all k = 0, . . . , q − 2.

19 / 48

Continue the proofFor x1, . . . , xn ∈ F , clearly

m∏i=1

(1− fi (x1, . . . , xn)q−1

)=

{1 if (x1, . . . , xn) ∈ V ,0 otherwise.

Thus

|V |1 =∑

x1,...,xn∈F

m∏i=1

(1− fi (x1, . . . , xn)q−1

).

As∑m

i=1 deg(fi ) < n, we may write

m∏i=1

(1− fi (x1, . . . , xn)q−1

)=

∑j1+...+jn

Part III. Zero-sum Problems

21 / 48

Zero-sum sequences on an abelian group

Let G be an additive abelian group. For a sequence (ak)16k6n ofelements of G if a1 + . . .+ an = 0 then the sequence is said to bea zero-sum sequence.

Let G be an additive abelian group of order n. Let a1, . . . , an ∈ G .Then

s0 = 0, s1 = a1, s2 = a1 + a2, . . . , sn = a1 + . . .+ an

cannot be pairwise distinct as they all belong to the group G . So,for some 0 6 i < j 6 n we have ai+1 + . . .+ aj = sj − si = 0.

22 / 48

EGZ Theorem

Erdős-Ginzburg-Ziv Theorem [Bull. Research Council Israel,1961]. Let a1, . . . , a2n−1 ∈ Z. Then

∑i∈I ai ≡ 0 (mod n) for some

I ⊆ {1, . . . , 2n − 1} with |I | = n.

In the group-theoretic language, this can be restated as follows.

Another Version of the EGZ Theorem. Let (ak)16k62n−1 be asequence of elements of the cyclic group G = Z/nZ. Then it has azero-sum subsequence of length n

Note that we cannot replace 2n − 1 in the EGZ Theorem by2n− 2. For example, if a1 = . . . , an−1 = 1 and an = . . . , a2n−2 = 0then

∑i∈I ai 6≡ 0 (mod n) for all n-subsets of {1, . . . , 2n − 2}.

23 / 48

Reduction of the EGZ theorem to the case with n primeSuppose that the EGZ theorem holds whenever n is prime. Now weuse induction to show the EGZ theorem for any positive integer n.

The case n = 1 is trivial since a1 ∈ Z is congruent to 0 mod 1.Now let n ∈ {2, 3, . . .} and assume that the EGZ theorem holds forall smaller values of n. If n is prime, it is okay. Now let n = dqwith 1 < d 6 q < n.

As 2n − 1 = 2dq − 1 > 2q − 1, for some I1 ⊆ {1, . . . , 2n − 1} with|I1| = q such that

∑i∈I1 ai ≡ 0 (mod 2q − 1). Since

2n − 1− |I1| = (2d − 1)q − 1 > 2q − 1, we can selectI2 ⊆ {1, . . . , 2n − 1} \ I1 with |I2| = q such that

∑i∈I2 ai ≡ 0

(mod q). Continue this process until we find pairwise disjointq-subsets I1, . . . , I2d−1 of {1, . . . , 2n − 1} such that∑

i∈Ij

ai ≡ 0 (mod q) for all j = 1, . . . , 2d − 1.

Note that 2n − 1− (2d − 1)q = q − 1 < 2q − 1.24 / 48

Reduction of the EGZ theorem to the prime case

For each j = 1, . . . , 2d − 1 write∑i∈Ij

ai = qbj with bj ∈ Z.

By the induction hypothesis, for some d-subset J of{1, . . . , 2d − 1} we have

∑j∈J bj ≡ 0 (mod d). Let I =

⋃j∈J Ij .

Then |I | =∑

j∈J |Ij | = dq = n and∑i∈I

ai =∑j∈J

∑i∈Ij

ai =∑j∈J

qbj = q∑j∈J

bj ≡ 0 (mod n).

This concludes the induction step.

So, to finish the proof of the EGZ theorem, we only need to handlethe prime case.

25 / 48

A LemmaLemma. Let p be a prime. If none of a1, . . . , ak ∈ Z (k < p) isdivisible by p, then |Sk | > k + 1, where

Sk =

{∑i∈I

ai mod p : I ⊆ {1, . . . , k}

}.

First Proof (by EGZ). As p - a1, we have |S1| = 2 = 1 + 1.Now let 1 < k < p and assume that |Sk−1| > k. ClearlySk−1 ⊆ Sk . If Sk−1 6= Sk , then |Sk | > |Sk−1|+ 1 > k + 1.Now suppose that Sk−1 = Sk . Then ak mod p ∈ Sk = Sk−1,2ak mod p ∈ Sk = Sk−1, . . ., pak mod p ∈ Sk . Hence|Sk | > p > k + 1.Second Proof. Let ā = a + pZ for a ∈ Z. ThenAi = {0̄, ai} ⊆ Z/pZ and |Ai | = 2 for all i = 1, . . . , k. By theCauchy-Davenport theorem,

|A1+. . .+Ak | > min{p, |A1|+. . .+|Ak |−k+1} = min{p, k+1} = k+1.26 / 48

Proof of the EGZ theorem in the prime caseProof of the EGZ Theorem in the prime case. Let p be aprime, and let a1, . . . , a2p−1Z. We want to show that

∑i∈I ai ≡ 0

(mod p) for some p-subset of {1, . . . , 2p − 1}. Without loss ofgenerality, we suppose that a′1 6 . . . 6 a

′2p−1, where a

′i denotes the

least nonnegative residue of ai mod p.

If a′i = a′i+p−1 for some i = 1 . . . , p, then a

′i = . . . = a

′i+p−1 and

hence∑p−1

j=0 ai+j ≡ pai ≡ 0 (mod p).Now suppose that a′i 6= a′i+p−1 for all i = 1, . . . , p. By the Lemma,

S =

{∑i∈I

(ai+p−1 − ai ) mod p : I ⊆ {1, . . . , p − 1}

}has cardinality at least (p − 1) + 1 = p. Thus S contains∑pi=1 ai+p−1 mod p and hence for some I ⊆ {1, . . . , p − 1} we

have ∑i∈I

ai +∑

j∈{1,...,p}\I

aj+p−1 ≡ 0 (mod p).

27 / 48

Another proof

Another Proof via the Chevalley-Warning Theorem.Fp = Z/pZ is a field of characteristic p. Define

f1(x1, . . . , x2p−1) =

2p−1∑k=1

xp−1k , f2(x1, . . . , x2p−1) =

2p−1∑k=1

akxp−1k .

They are polynomials over Fp with deg(f1) + deg(f2) < 2p − 1.Clearly fi (0, . . . , 0) = 0 for i = 1, 2.

In view of the Chevalley-Warning Theorem, for somex1, . . . , x2p−1 ∈ F not all zero we have

f1(x1, . . . , x2p−1) = 0 and f2(x1, . . . , x2p−1) = 0.

Thus I = {1 6 i 6 2p − 1 : xi 6= 0} 6= ∅, |I |1 = 0 and∑i∈I ai = 0. Since p | |I | and 0 < |I | < 2p, we must have |I | = p.

28 / 48

Davenport’s constant

Let G be a finite abelian group. The Davenport constant D(G ) isthe least k ∈ Z+ such that any sequence (a1, . . . , ak) of elementsof G has a nonempty zero-sum subsequence (i.e.,

∑i∈I ai = 0 for

some ∅ 6= I ⊆ {1, . . . , k}). It is clear that D(G ) 6 |G |.In 1966 Davenport showed that if K is an algebraic number fieldwith ideal class group G , then D(G ) is the maximal number ofprime ideals (counting multiplicity) in the decomposition of anirreducible algebraic integer in K .

Clearly, D(Zn) = n for any positive integer n.For each finite abelian group G with |G | > 1, there is a uniquesequence d1, . . . , dr of positive integers with d1 > 1 and di | di+1for all i = 1, . . . , r − 1 such that G ∼= Zd1 ⊕ . . .⊕Zdr ; we call r therank of G , dr the exponent of G , and define

d∗(G ) :=r∑

i=1

(di − 1).

29 / 48

Olson’s theorem

For G = Zpa1 ⊕ . . .+⊕Zpar (where p is a prime anda1, . . . , ar ∈ Z+), For the sequence consisting of pa1 − 1 copies of(1, 0, . . . , 0) ∈ G , pa2−1 copies of (0, 1, 0, . . . , 0), . . ., par−1 copiesof (0, . . . , 0, 1), it has no nonempty zero-sum subsequence. SoD(G ) > d∗(G ).

Olson’s Theorem [J. Number Theory 1(1969)]. Let G be anabelian p-group with |G | > 1, where p is a prime. ThenD(G ) = 1 + d∗(G ). Moreover, for any c , c1, . . . , cd∗(G)+1 ∈ G wehave ∑

I⊆{1,...,d∗(G)+1}∑s∈I cs=c

(−1)|I | ≡ 0 (mod p).

Conjecture (Olson, 1969). For any n, r ∈ Z+ we have

D(Zrn) = 1 + r(n − 1).

30 / 48

Olson’s theorem implies the EGZ theorem

Let q be a power of a prime p, and let c , c1, . . . , c2q−1 ∈ Zq. Then(0, c), (1, c1), . . . , (1, c2q−1) ∈ Zq ⊕ Zq. By Olson’s theorem in thecase G = Z2q = Zq ⊕ Zq, we have∑

I⊆[1,2q−1]q||I |,

∑s∈I cs=c

(−1)|I | ≡ 0 (mod p).

In other words,∣∣∣∣{I ⊆ [1, 2q − 1] : |I | = q and ∑s∈I

cs = c

}∣∣∣∣ ≡ [[c = 0]] (mod p),where for a predicate P we let [[P]] be 1 or 0 according as P holdsor not. Thus, Olson’s theorem implies the EGZ theorem.

31 / 48

Representative work of Wei-Dong GaoIn China, Prof. Wei-Dong Gao and his students study zero-sumproblems.

Gao’s Theorem [conjectured by Y. Caro] (Gao, J. Number Theory1996). Let G be a finite additive abelian group. Let E (G ) be thesmallest positive integer k such that for any a1, . . . , ak ∈ G thesequence (a1, . . . , ak) has a zero-sum subsequence of length |G |.Then E (G ) = D(G ) + |G | − 1.

Note that D(Zn) = n and E (Zn) = 2n − 1 = D(Zn) + |Zn| − 1.

References:

1. W. D. Gao, A combinatorial problem on finite abelian groups, J.Number Theory 58(1996), 100–103.

2. M. Devos, Gao’s theorem for nonabelian groups, Open ProblemGarden, http://garden.irmacs.sfu.ca/?q=op/gaos theorem for nonabelian groups

Conjecture: Gao’s theorem holds for any finite group G .32 / 48

A Polynomial Formula

Lemma (Z. W. Sun [Electron. Res. Announc. Amer. Math. Soc.,2003]). Let R be a ring with identity, and let f (x1, . . . , xk) be apolynomial over R. If J ⊆ [1, k] = {1, . . . , k} and |J| > deg f , then∑

I⊆J(−1)|J|−|I |f ([[1 ∈ I ]], . . . , [[k ∈ I ]])

coincides with [∏

j∈J xj ]f (x1, . . . , xk), the coefficient of∏

j∈J xj inf (x1, . . . , xk).

Proof. Write

f (x1, . . . , xk) =∑

j1,...,jk>0

cj1,...,jk

k∏s=1

x jss ,

and observe that if ∅ 6= J ⊆ [1, k] then

0 =∏j∈J

(1− 1) =∑I⊆J

(−1)|I |.

33 / 48

Continue the proof

Therefore∑I⊆J

(−1)|I |f ([[1 ∈ I ], . . . , [k ∈ I ]])

=∑I⊆J

(−1)|I |∑

j1,...,jk>0{s: js 6=0}⊆I

cj1,...,jk

=∑

j1,...,jk>0{s: js 6=0}⊆J

∑{s: js 6=0}⊆I⊆J

(−1)|I |cj1,...,jk

=∑

j1,...,jk>0{s: js 6=0}⊆J

∑I ′⊆J\{s: js 6=0}

(−1)|I ′|(−1)|{s: js 6=0}|cj1,...,jk

=∑

j1,...,jk>0{s: js 6=0}=J

(−1)|J|cj1,...,jk = (−1)|J|[∏j∈J

xj

]f (x1, . . . , xk),

where in the last step we note that if {s: js 6= 0} = J and js > 1for some s then j1 + . . .+ jk > |J| > deg f and hence cj1,...,jk = 0.

34 / 48

A Useful Technique

Let m be an integer and let p be a prime. Fermat’s little theoremtells that we can characterize whether p divides m as follows:

[]p | m]] ≡ 1−mp−1 (mod p).

To handle general abelian p-groups in a similar way, we need tocharacterize whether a given power of p divides a. Thus, thefollowing lemma is of technical importance. It first appeared inSun’s preprint arXiv:math/NT/0305369 dated May 24, 2003.

Lemma [Z. W. Sun, Israel J. Math., 2009] Let p be a prime, andlet a ∈ N and m ∈ Z. Then we have the following congruence(

m − 1pa − 1

)≡ [[pa | m]] (mod p).

35 / 48

Sun’s Proof of the Olson Theorem without Group Rings

Olson’s Theorem [J. Number Theory 1969]. Let G be an abelianp-group isomorphic to Zpa1 ⊕ · · · ⊕ Zpar where Zn = Z/nZ. Givenc , c1, . . . , ck ∈ G with k > 1 +

∑rt=1(p

at − 1), we have∑I⊆[1,k]∑s∈I cs=c

(−1)|I | ≡ 0 (mod p).

In particular, when c = 0 it follows that there is a nonemptyI ⊆ [1, k] with

∑s∈I cs = 0.

Remark. Olson’s theorem determined the Davenport constant forany abelian p-group. Olson used the group ring method in hisproof of this classical theorem.

Now we prove Olson’s theorem without any use of the group ringmethod. Identify c with a vector 〈c(1) mod pa1 , . . . , c(r) mod par 〉,and write cs in the form 〈c(1)s mod pa1 , . . . , c(r)s mod par 〉.

36 / 48

Sun’s Proof of the Olson Theorem without Group RingsDefine

f (x1, . . . , xk) =r∏

t=1

(∑ks=1 c

(t)s xs − c(t) − 1pat − 1

).

Clearly deg f 6∑r

t=1(pat − 1) < k = |[1, k]|. Applying the

polynomial formula with J = [1, k] we get∑I⊆[1,k]

(−1)k−|I |f ([[1 ∈ I ]], . . . , [[k ∈ I ]]) = [x1 · · · xk ]f (x1, . . . , xk) = 0,

∑I⊆[1,k]

(−1)k−|I |r∏

t=1

(∑s∈I c

(t)s − c(t) − 1pat − 1

)= 0.

∑I⊆[1,k]

pat |∑

s∈I c(t)s −c(t)

(−1)|I | ≡ 0 (mod p), i.e.,∑I⊆[1,k]∑s∈I cs=c

(−1)|I | ≡ 0 (mod p).

37 / 48

Kemnitz’s ConjectureFor a finite abelian group G , define s(G ) to be the least positiveinteger k such that any sequence (a1, . . . , ak) of elements of G hasa zero-sum subsequence of length exp(G ), where the exponentexp(G ) of G is the least n ∈ Z+ with nx = 0 for all x ∈ G .By the EGZ theorem, s(Zn) = 2n − 1 for any positive integer n.What is the smallest integer l = s(Z2n) such that every sequence ofl elements in Z2n = Zn ⊕ Zn contains a zero-sum subsequence oflength n?

In 1983 Kemnitz [Ars Combin.] conjectured that s(Z2n) = 4n − 3,and the conjecture can be reduced to the case with n prime.

In 1993 Alon and Dubiner showed that s(Z2n) 6 6n − 5. In 2000Rónyai [Combinatorica] was able to prove that s(Z2p) 6 4p − 2 forevery prime p; in 2001 W. D. Gao [J. Combin. Theory Ser. A]used Olson’s group ring approach to deduce that s(Z2q) 6 4q − 2for any prime power q.

These results were obtained by various algebraic methods.38 / 48

Alon-Dubier Lemma

The following lemma plays an indispensable role in the study of theKemnitz conjecture.

Alon-Dubiner Lemma. Let q be a prime power, and letc1, . . . , c3q be elements of Z2q with c1 + . . .+ c3q = 0. Then thereis an I ⊆ [1, 3q] with |I | = q such that

∑i∈I ci = 0.

Proof. As 3q − 1 > 1 + (q − 1) + (q − 1) + (q − 1), by Olson’stheorem there is a nonempty I ⊆ {1, . . . , 3q − 1} such that∑

s∈I cs = 0 in Z2q and also∑

s∈I 1 = 0 in Zq. So q | |I | and hence|I | ∈ {q, 2q}. If |I | = 2q, then Ī = {1, . . . , 3q} \ I has cardinality qand ∑

t∈Ī

ct =

3q∑i=1

ci −∑s∈I

cs = 0.

39 / 48

A Consequence of the Polynomial FormulaIn March 2003 Sun deduced the following result from hispolynomial formula.

Theorem [Z. W. Sun, Electron. Res. Announc. Amer. Math.Soc. 9(2003)]. Let p be a prime and let h > 0 be an integer. Letai , bi ∈ Z for i = 1, . . . , 4ph − 2.

(i) Set I = {I ⊆ [1, 4ph − 2]:∑

i∈I ai ≡∑

i∈I bi ≡ 0 (mod ph)}.Then

|{I ∈ I: |I | = ph}| ≡ |{I ∈ I: |I | = 3ph}|+ 2 (mod p).

(ii) Suppose that∑I ,J⊆[1,4ph−3]

|I |=|J|=ph−1, I∩J=∅

(∏i∈I

ai

)(∏j∈J

bj

)6≡ 2 (mod p).

Then there exists an I ⊆ [1, 4ph − 3] with |I | = ph such that∑i∈I ai ≡

∑i∈I bi ≡ 0 (mod ph).

40 / 48

Reiher’s workIn his paper “On Kemnitz’s conjecture concerning lattice points inthe plane” written in Nov. 2003, C. Reiher, completely proved theKemnitz conjecture which had been open for 20 years! This workrepresents one of the most important achievements in the theory ofzero-sums.

Reiher’s paper has 4 pages. Pages 1–3 are devoted to 5sophisticated corollaries to the Chevalley-Warning theorem whichare needed later. Actually this can be significantly simplified byusing part (i) of our theorem with a4p−2 = b4p−2 = 0.

A Consequence of the Theorem. Let p be a prime and let h > 0be an integer. Let ai , bi ∈ Z for i = 1, . . . , 4ph − 3. SetI = {I ⊆ [1, 4ph − 3]:

∑i∈I ai ≡

∑i∈I bi ≡ 0 (mod ph)}. Then

|{I ∈ I: |I | = ph}|+ |{I ∈ I: |I | = ph − 1}|≡|{I ∈ I: |I | = 3ph}|+ |{I ∈ I: |I | = 3ph − 1}|+ 2 (mod p).

41 / 48

Reiher’s Lemma

On the last page of his paper, C. Reiher provided a key lemmawhich is obtained by a combinatorial method rather than analgebraic method.

Reiher’s Lemma. Let p be a prime and let ai , bi ∈ Z fori = 1, . . . , 4p − 3. Set

I ={

I ⊆ [1, 4p − 3]:∑i∈I

ai ≡∑i∈I

bi ≡ 0 (mod p)}.

Then, either {I ∈ I: |I | = p} 6= ∅ or

|{I ∈ I: |I | = p − 1}| ≡ |{I ∈ I: |I | = 3p − 1}| (mod p).

Notation. For J ⊆ [1, 4p − 3] and n = 1, 2, . . . let

(n, J) :=

∣∣∣∣{I ⊆ J: |I | = n & ∑i∈I

ai ≡∑i∈I

bi ≡ 0 (mod p)}∣∣∣∣.

42 / 48

An ObservationObservation. We have

|J| ∈ {3p − 1, 3p − 2} ⇒ (2p, J) ≡ (p, J)− 1 (mod p).

In fact, by the polynomial formula we mentioned before,∑I⊆J

(−1)|J|−|I |(1− |I |p−1)(

1−(∑

i∈Iai

)p−1)(1−

(∑i∈I

bi

)p−1)coincides with the coefficient of

∏j∈J xj in the polynomial(

1−(∑

j∈Jxj

)p−1)(1−

(∑j∈J

ajxj

)p−1)(1−

(∑j∈J

bjxj

)p−1),

which is zero since |J| > 3p − 3. Thus∑I⊆J, |I |∈{0,p,2p}∑

i∈I ai≡∑

i∈I bi≡0 (mod p)

(−1)|I | ≡ 0 (mod p),

that is, (0, J)− (p, J) + (2p, J) ≡ 0 (mod p).43 / 48

Sketch of the proof

Sketch of the Proof. Assume that {I ∈ I: |I | = p} = ∅, i.e.,(p, J) = 0 for any J ⊆ [1, 4p − 3]. Let N denote the number ofpartitions [1, 4p − 3] = I1 ∪ I2 ∪ I3 satisfying

|I1| = p − 1, |I2| = p − 2, |I3| = 2p

and furthermore∑i∈I1

ai ≡∑i∈I1

bi ≡ 0 (mod p),∑i∈I3

ai ≡∑i∈I3

bi ≡ 0 (mod p)

(and hence∑

i∈[1,4p−3]\I2 ai ≡∑

i∈[1,4p−3]\I2 bi ≡ 0 (mod p)).

We count N in two ways.

44 / 48

Continue the proof

Observe that

N =∑I1

(2p, [1, 4p−3]\I1) ≡∑I1

(−1) = −(p−1, [1, 4p−3]) (mod p).

On the other hand,

N =∑I2

(2p, [1, 4p − 3] \ I2)

≡∑

[1,4p−3]\I2

(−1) = −(3p − 1, [1, 4p − 3]) (mod p).

So we have the congruence(p − 1, [1, 4p − 3]) ≡ (3p − 1, [1, 4p − 3]) (mod p).

Remark. The prime power version of Reiher’s Lemma also holds.

45 / 48

Conclusion

Combining Reiher’s Lemma, the Alon-Dubiner lemma and theabove-mentioned consequence of the Theorem, we immediatelyobtain the following result of Reiher [Ramanujan J. 2007].

Kemnitz-Reiher Theorem. The Kemnitz conjecture is true, thatis, any sequence of elements in Zn ⊕Zn with length at least 4n− 3contains a zero-sum sequence of length n.

What does Reiher’s solution teach us? When we apply a powerfulalgebraic method in combinatorics, we should also realize itsdisadvantage and should not forget combinatorial methods. Acombination of algebraic methods and combinatorialmethods might be more powerful!

By the way, Sun [Israel J. Math. 2009] established connections ofthe EGZ theorem, Olson’s theorem and the Alon-Dubiner lemmato covering systems of Z by residue classes. But it seems that theKemnitz-Reiher theorem cannot be connected with covers of Z.

46 / 48

Open Problem

How to determine s(Zdn) for anyd , n ∈ Z+?In particular,

how to prove the conjecture thats(Z3p) = 9p − 8 for any prime p > 3?

47 / 48

Thank you!

48 / 48