Some zero-sum problems and Ramsey-type

results in Additive Combinatorics

Sukumar Das Adhikari

July 12, 2013

Abstract

The early Ramsey-type results preceding the result of Ramsey(these include results of Schur, van der Waerden and Hilbert) belongto the area of combinatorial number theory. In the first two sectionsof this article, we shall take up these results and some generalizationsof them. We shall try to have a glimpse of applications of elementaryalgebraic and topological methods in dealing with these problems. Inthe last two sections, we take up zero-sum problems in additive com-binatorics. Here we shall see proofs of some early results, includingthe EGZ theorem, by various methods; we shall also discuss somegeneralizations.

Some notations and terminologies. Here the symbols Z+,N, Z, Q andR will denote respectively the set of positive integers, the set of non-negativeintegers, the set of integers, the set of rationals and the set of real numbers.For any prime power q, Fq will denote the finite field with q elements andthe symbol F∗

q will denote the multiplicative group of non-zero elements ofFq. In general, for any field K, K∗ will denote the multiplicative group ofits non-zero elements. For a finite Set A, |A| will be the number of elementsof A. If r ∈ Z+, then an r-colouring of a set S is a map χ : S → 1, · · · , r.Since S = χ−1(1)∪χ−1(2)∪ · · · ∪χ−1(r), an r-colouring of a set S is nothingbut a partition of S into r parts. If s is an element of S, then χ(s) is calledthe colour of s. A set T ⊂ S is called monochromatic with respect to acolouring χ if χ is constant on T . For two non-zero integers a, b, we shallconsider gcd (a, b) to be positive.

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1 Early Ramsey-type theorems and

some generalizations: Part I

In most of the early Ramsey-type results, one sees that if a large structure isdivided into finitely many parts, at least one of the parts will retain certainregularity properties of the original structure. Sometimes, if the ‘size’ of asubstructure is big enough, certain regularities are unavoidable. “Completedisorder is impossible”: the philosophy of Ramsey Theory is often describedby this statement of Motzkin. To become acquainted with many facets ofthe theory, one may look into the Graham-Rothschild-Spencer monograph[37] and the collection of articles [53] edited by Nesetril and Rodl. We alsorefer to a recent article of Soifer [68] giving detailed history of the emergenceof the subject.

Theorem 1. (Generalized pigeonhole principle) Let A and B be finitenonempty sets where B = b1, b2, · · · , br. Let f be a function from A to B.Then, for non-negative integers a1, a2, · · · , ar, the equality

|A| = a1 + a2 + · · ·+ ar − r + 1

would imply that ∃ i ∈ 1, · · · , r such that |f−1(bi)| ≥ ai.

Proof: If there is no such i, then

|A| =r∑

i=1

|f−1(bi)| ≤r∑

i=1

(ai − 1) =r∑

i=1

ai − r

– a contradiction to our assumption.

Remark 1. Clearly, it is interesting only when |A| > |B|; when |A| = |B|+1,it follows that there is bi ∈ B with |f−1(bi)| ≥ 2, which is the commonlyknown version of the pigeonhole principle.

Exercise 1. Show that given any five integer lattice points on the plane themidpoint of the line segment determined by some two distinct points amongthem will also be an integer lattice point.

Exercise 2. (Mantel) Let n ≥ 2 be a positive integer. Show that if a simplegraph G of order 2n contains n2 + 1 edges, then G contains a triangle.

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Exercise 3. (Erdos and Szekeres) Given a sequence of mn + 1 distinct realnumbers, show that if it does not contain a monotone increasing subsequenceof length m+ 1, then it must contain a monotone decreasing subsequence oflength n+ 1.

A well-known example. If you have a party of at least 6 people, you canguarantee that there will be a group of 3 people who all know each other, ora group of 3 people who all do not know each other.

Proof by graphs:

(joining by red line when two people know each other)

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The following result [59] generalizes the above example.

Theorem 2. (Ramsey’s Theorem) Let k, r, l(≥ k) be positive integers.Then there exists a positive integer n = n(k, r, l) such that for any r-colouringof the k-subsets of the set [n] = 1, 2, · · · , n, there is an l-subset of [n] allof whose k- subsets are of the same colour.

We shall give a proof of the following general version.

Theorem 3. (Generalized Ramsey’s Theorem) Given positive integersk, r, l1, · · · , lr, there exists a positive integer n = n(k, r; l1, · · · , lr) such thatfor any r-colouring of the k-subsets of the set [n] with colours c1, · · · , cr , forsome i ∈ 1, · · · , r, there is an li-subset of [n] all of whose k- subsets are ofthe colour ci.

Proof: Once a positive integer n = n(k, r; l1, · · · , lr) exists satisfying therequirement for positive integers k, r, l1, · · · , lr, we assume that we are takingthe smallest such n.

For k = 1, it is Theorem 1.Let k > 1 and assume that the numbers n(k − 1, r; l1, · · · , lr) are defined

for all li’s.Again, for any given k, if li = k for all i ∈ 1, · · · , r, then the result

follows trivially.

Assuming that the numbers

ai = n(k, r; l1, · · · , li−1, li − 1, li+1, · · · , lr)

are defined for i = 1, · · · , r, we have to show that n(k, r; l1, l2, · · · , lr) exists.

Let n = 1 + n(k − 1, r; a1, · · · , ar).

Let an r-colouring c of the k-subsets of the set [n] with colours c1, · · · , crbe given. Let m ∈ [n] and Y = [n] \ m. An induced r-colouring c∗ ofthe (k − 1)-subsets of Y is now defined in the following manner. Given a(k − 1)-subset S of Y , the colour of S is c∗i if and only if that of S ∪ m isci in the original colouring.

By the definition of n, for some i there is a set Ai (⊂ [n] \ m) of size aiall of whose (k − 1)-subsets have the same colour c∗i in the c∗- colouring.

By the definition of ai, the set Ai will contain either a set of size lj all ofwhose k-subsets have colour cj for some j 6= i or a set X say of size li − 1 allof whose k-subsets have colour ci.

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In the first case we are through. In the second case, X ∪ m is a setof size li. Given a k-subset K of X ∪ m, if m does not belong to K thenby the assumption in the above paragraph, K is of colour ci. On the otherhand, if m ∈ K, then since all (k − 1) -subsets of Ai are of colour c∗i , so isthe set K \ m and therefore by the definition of the induced colouring c∗,K is of colour ci. This completes the proof.

The following theorem of Erdos and Szekeres [25] played a major rolein popularizing Ramsey’s theorem and the subsequent emergence of a newbranch of Combinatorics; we leave its proof as an exercise.

Theorem 4. (Erdos-Szekeres Theorem) For a given positive integer n(≥3), there is an integer N = N(n) such that any collection of N or more pointsin the plane, no three on a line, has a subset of n points forming a convexn-gon.

Now we take up a result of Schur [67], which appeared in 1916. This isone among the Ramsey-type results which appeared before Ramsey’s 1930paper. After the branch of combinatorics called Ramsey theory grew up,with hindsight, one could see the unifying feature of the seemingly unrelatedresults of early Ramsey theory.

Theorem 5. (Schur’s Theorem) Given a positive integer r, there is N(r) ∈Z+, such that for any r-colouring of [N(r)], ∃ x, y, z ∈ [N(r)] of the samecolour such that x + y = z; the situation is described by saying that theequation x+ y = z has a monochromatic solution.

Proof: Let N(r) = n(2, r, 3), where n(k, r, l) is as in Theorem 2.Let χ : [N(r)] → [r] an r-colouring.This induces an r-colouring χ∗ of the collection of 2-element subsets of

[N(r)] as follows:

χ∗(i, j) = χ(|i− j|), i 6= j ∈ [N ].

By definition of N(r), ∃ a 3-element subset a, b, c of [N(r)] with a <b < c such that χ∗(a, b) = χ∗(b, c) = χ∗(c, a), that is, χ(b − a) =χ(c− b) = χ(c− a).

Since (b − a) + (c − b) = (c − a), we get a monochromatic solution ofx+ y = z.

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Remark 2. From the above proof of Schur’s theorem, in the monochromaticsolution x, y, z one may have x = y. For a proof of a strong version givinga monochromatic solution with distinct integers x, y, z, one may see [68].

Remark 3. Schur [67] used Theorem 5 to show that for any m ∈ Z+, ∃M = M(m) ∈ Z+ such that for primes p > M , the congruence

xm + ym ≡ zm (mod p)

always has nontrivial solutions.We now state van der Waerden’s theorem [72], which is yet another

Ramsey-type result which appeared before Ramsey’s paper.

Theorem 6. (van der Waerden’s Theorem) Given k, r ∈ Z+, there existsW (k, r) ∈ Z+ such that for any r-colouring of 1, 2, · · ·W (k, r), there is amonochromatic arithmetic progression of k terms.

Exercise 4. From Theorem 6 deduce that given k, r, s ∈ Z+, there existsN = N(k, r, s) ∈ Z+ such that for any r-colouring of [N ], there are a, d ∈ Z+

such that the set

a, a+ d, · · · , a+ kd ∪ sd ⊂ [N ]

is monochromatic.

Schur’s theorem and van der Waerden’s Theorem give rise to the naturalquestion about the existence of monochromatic solutions of the general linearhomogeneous equation

(1) c1x1 + · · ·+ cnxn = 0, ci( 6= 0) ∈ Z.

The following abridged version of a theorem of Rado (see [56], [57], [58])answers to that question.

Theorem 7. (Rado) For n ≥ 2, equation (1) has monochromatic solutionfor any finite colouring of Z, if and only if there is a non-empty subset I ⊂1, · · · , n, such that

∑

i∈I ci = 0.

Proof: Fix a prime p > 0. For q ∈ Q∗, write q = pja

b, where j ∈ Z, a ∈

Z, b ∈ Z+, gcd(a, b) = 1 and p does not divide ab. The super modulo colourSp on Q∗ is defined by:

Sp(q) =a

b(mod p) ∈ F∗

p.

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Now, Sp is a (p− 1)-colouring of Q∗ with the property that

Sp(x) = Sp(y) ⇒ Sp(αx) = Sp(αy) ∀ α ∈ Q∗.

Assume that for no non-empty subset I of 1, · · · , n,∑

i∈I ci ≡ 0 (mod p),and if possible let x1, · · · , xn ∈ Z+ be a monochromatic solution w.r.t. Sp

of equation (1). We can assume that gcd (x1, · · · , xn) = 1. Rearranging theindices if necessary, there is k, with n ≥ k ≥ 1 such that

p 6 |xi for i = 1, · · · , k

and p|xi for k < i ≤ n.

Reducing equation (1) modulo p, we have

0 =

k∑

i=1

cixi =

(k∑

i=1

ci

)

x1.

Since p 6 |x1, the above implies∑k

i=1 ci = 0, which is a contradiction toour assumption.

Therefore, if there is no non-empty subset I ⊂ 1, · · · , n, such that∑

i∈I ci = 0 then ∑

i∈I ci : ∅ 6= I ⊂ [n] is a set consisting of finitelymany non-zero integers and hence some prime p will not divide

∑

i∈I ci forany (∅ 6=)I ⊂ [n] and by the above observation, there is no monochromaticsolution w.r.t. Sp of the equation (1).

This proves that the condition is necessary.Now, reordering if necessary, we assume that for some k ≥ 2,

(2) c1 + · · ·+ ck = 0.

Suppose that a finite colouring of Z+ is given.If k = n, then x1 = · · · = xn = 1 would be monochromatic solution to

equation (1). So we assume k < n.Let

(3) A = gcd (c1, · · · , ck) and B = ck+1 + · · ·+ cn,

where B can be assumed to be nonzero, for otherwise once again x1 = · · · =xn = 1 would be a monochromatic solution. We have A > 0 and multiplyingby (−1), if necessary, we can also assume that B > 0.

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Writing − t =B

gcd (A,B),

we have

(4) −At =AB

gcd (A,B)= BD, say.

Also, ∃λ1, · · · , λk ∈ Z such that

(5) c1λ1 + · · ·+ ckλk = At.

We choose l ∈ Z+ such that γi = l + λi ≥ 1 ∀ i = 1, · · ·k. Let r ∈ Z+

be such that r ≥ max γi. Now, by Exercise 4, there are a and d such thata+ d, · · · , a+ rd ∪ dD is monochromatic.

Let

(6) xi =

a+ γid 1 ≤ i ≤ kdD k < i.

Now,

n∑

i=1

cixi = (a + ld)(c1 + · · ·+ ck) + d

k∑

i=1

ciλi + dD

n∑

i=k+1

ci

= 0 + dAt+ dDB (by equations (2), (3) and (5))

= 0. (by equation (4))

In other words, equation (6) provides us with a monochromatic solution toequation (1).

Schur’s theorem, as in Remark 2, can be restated by saying that givena positive integer r, there is a positive integer n = n(r) such that for anyr-colouring of [n], ∃ x, y with x < y, such that the elements x, y and x + yare in [n] and are of the same colour.

The above statement is generalized in the following theorem:

Theorem 8. (Folkman-Sanders) Given any two positive integers r andk, there is a positive integer n = n(r, k) such that if [n] is r-coloured, thereare positive integers a1 < a2 < · · · < ak satisfying

∑

1≤i≤k ai ≤ n such theelements

∑

i∈I ai are identically coloured as I varies over different non-emptysubsets of 1, 2, · · · , k.

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As has been observed in [37], the Folkman-Sanders theorem also followsfrom Rado’s theorem (apart from the original papers of Rado, the completestatement of Rado’s theorem can be found in [37]). One comes to know fromSoifer [68] that Arnautov was yet another mathematician to discover thisresult independently and Soifer calls it Arnautov- Folkman-Sanders theorem.

At the end of this section, we take up a theorem of Hilbert [44] which isgenerally accepted to be the earliest Ramsey-type theorem. We give a proofof a theorem of Hilbert [44] by topological methods. We shall not have muchscope to dwell on the application of dynamical systems to Ramsey Theory,an interesting area which is very active today (see [28] and [14] for instance).

Let T be a continuous map of a topological space X into itself. Then apoint x ∈ X is called a recurrent point for T if for any neighbourhood V ofx, ∃ n ≥ 1 with T n(x) ∈ V .

Now, let X be a compact topological space and T : X → X a continuousmap. Let F denote the family of nonempty closed subsets of X invariantunder T . Ordering by inclusion, we observe that because of compactness ofX , by the finite intersection property, the intersection of a totally orderedchain in F belongs to F . Hence, by Zorn’s lemma, F has a minimal elementY0.

Let y ∈ Y0 and Y=T n(y) : n ≥ 1, the forward orbit closure of y. Now,Y0 being T invariant, T n(y) : n ≥ 1 ⊂ Y0 and therefore, Y0 being closed,T n(y) : n ≥ 1 ⊂ Y0. But by definition, Y is nonempty and closed. Further,since T is continuous, Y is invariant under T . Therefore, Y ∈ F and byminimality Y = Y0. Hence y ∈ Y which means that each neighbourhood ofy contains T n(y) for some n ≥ 1.

Thus each point of the non-empty set Y0 is a recurrent point for T andwe have proved the following:

Proposition 1. For a continuous map T from a compact topological spaceX into itself, the set of recurrent points for T is nonempty.

Let Ω be the collection of sequences made of finitely many symbols. Inother words, if we denote the finite set of symbols by Λ, then Ω consists ofall maps f : Z+ → Λ.

For a, b ∈ Λ the metric d defined below gives the discrete topology on Λ.

d(a, b) =

0 if a = b1 if a 6= b.

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The space Ω = ΛZ+

with the product topology is metrizable. If forω, ω′ ∈ Ω one defines

D(ω, ω′) =∞∑

n=1

d (ω(n), ω′(n))

2n,

then it is easy to verify that the metricD corresponds to the product topologyon Ω. By Tychonoff’s theorem, (Ω, D) is therefore compact.

Also the semigroup Z+ acts on the elements of Λ in the obvious way byshifting. That is by the action of n ∈ Z+, an element ω ∈ Ω goes to ω′ ∈ Ωwhere ω′(m) = ω(m+ n). The map on Ω corresponding to n = 1 (which infact determines the action of the semigroup Z+) will be called the shift mapand we shall denote it by σ. The map σ : Ω → Ω is continuous.

The space Ω endowed with the metric D and the Z+ action is a symbolicflow in the terminology of Dynamical Systems. In a symbolic flow, by sayingthat a point is recurrent one means that it is recurrent for the shift map.

Λ is sometimes called the alphabet and a finite sequence of elements in Λis called a word. It is clear that two points ω, ω′ ∈ Ω are close if they agreeon a large initial block of numbers (1, 2, · · · , N). Hence it follows that:

Proposition 2. In a symbolic flow, a sequence ω ∈ Ω is recurrent if andonly if every word occurring in ω occurs a second time.

Consequently, a most general recurrent point ω will look like:

(7) ω = [(aω(1)a)ω(2)(aω(1)a)]ω(3)[(aω(1)a)ω(2)(aω(1)a)] · · ·

where a = ω(1) ∈ Λ and ω(i) ’s are arbitrary words composed of elementsof Λ.

Theorem 9. (Hilbert) Given a finite colouring on Z+ and a positive in-teger l, one can find l elements m1 ≤ m2 ≤ · · · ≤ ml in Z+ such that ifP (m1, · · · , ml) denotes the set of sums

∑l

i=1 c(i)mi, c(i) = 0 or 1, theninfinitely many translates of P (m1, · · · , ml) are of the same colour.

Proof: Let χ : Z+ → c1, c2, · · · , cq be a q-colouring on Z+. Let Λ =1, 2, · · · , q and Ω be defined as above. We consider the element ξ ∈ Ωwhere

ξ(n) = i ⇐⇒ n ∈ χ−1(ci), i = 1, 2 · · · q.

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Case I. (ξ is a recurrent point)Let ξ has the form given in (7) and

ω0 = a

ω1 = ω0ω(1)ω0

ω2 = ω1ω(2)ω1

· · · · · ·

ωn = ωn−1ω(n)ωn−1

Now, denoting the length of the word ωnω(n+1) by mn+1, we see that if

some symbol occurs at position p in wn, then it occurs at positions p andp + mn+1 in ωn+1 = ωnω

(n+1)ωn. Thus the symbol a occurs at positions1, 1 + m1, 1 + m2, 1 + m1 + m2 and in general at positions belonging to1 + P (m1, · · ·ml) for any l. Since every finite configuration occurs infinitelyoften in ξ, it is now clear that χ−1(a) contains infinitely many translates ofP (m1, · · ·ml).

Case II. (ξ is not a recurrent point)In this case we consider the forward orbit closure X of ξ in Ω. The shift

operator takes points of X to X and therefore by Proposition 1 there is arecurrent point say w for the shift operator σ in X . Let w be of the formgiven in (7). Therefore, there exists a sequence of positive integers nk suchthat σnk(ξ) → w. If a is the leading symbol in w, then arguing as before, aoccurs at positions belonging to 1+P (m1, · · ·ml). Choose k such that σnk(ξ)agrees with w for (1 +m1 + · · ·+ml) terms. Then ξ(nk + p) = a wheneverp ∈ (1 + P (m1, · · ·ml)). One can assume that nk → ∞. For, otherwise, afinite translate of ξ would be recurrent and one could then invoke the firstcase. Therefore we obtain that 1+nk+P (m1, · · ·ml) ∈ χ−1(a) for a sequencenk where nk → ∞ and this proves the theorem.

The following result of Hindman [45] implies Arnautov-Folkman-Sanderstheorem (Theorem 8) as well as Hilbert’s theorem (Theorem 9); this was firstconjectured by Sanders in his Ph.D. thesis [64].

Theorem 10. (Hindman ) Given any finite colouring of Z+, there alwaysexist an infinite subset of Z+ all the finite sums of which are of the samecolour.

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The proof of Hindman was later simplified by Baumgartner [13]. Fursten-berg and Weiss [29] used the notion of proximality to establish Hindman’stheorem. The Graham-Rothschild-Spencer monograph [37] contains twotopological proofs of Hindman’s theorem; one of them is the Furstenberg-Weiss proof, the other being the ultrafilter-proof due to Glazer. The proofof Baumgartner is also available in [37].

2 Early Ramsey-type theorems and

some generalizations: Part II

The theorem of van der Waerden was a prelude to a very important themewhere interplay of several areas of Mathematics would be seen.

The following result (see Anderson [11], see also [48], [1]) generalizes vander Waerden’s Theorem to higher dimensions.

Theorem 11. (Grunwald) Let d, r ∈ Z+, the set of positive integers. Thengiven any finite set S ⊂ (Z+)d, and an r-colouring of (Z+)d, there exists apositive integer ‘a’ and a point ‘v’ in (Z+)d such that the set aS + v ismonochromatic.

Proof: The theorem will follow from the following statement:

A(S): Given any finite set S ⊂ (Z+)d, for each k ∈ Z+, ∃ n = n(k) such

that for every k-colouring of Bndef= (a1, . . . , ad) : ai ∈ Z+, 1 ≤ ai ≤ n, Bn

contains a monochromatic subset of the form aS + v for some a ∈ Z+ andv ∈ Bn.

Since A(S) is obviously true if |S| = 1, it is enough to show that A(S) ⇒A(S ∪ s) for any s ∈ (Z+)d.

For the induction procedure, once A(S) is established for a given S, weprove the following intermediate statement C(p) corresponding to a positiveinteger p. Once C(p) is established for any positive integer p, it will lead tothe statement A(S ∪ s) and we shall be through.

C(p): Let S ⊂ (Z+)d be fixed for which A(S) is true. Then for givenk ∈ Z+ and s ∈ (Z+)d, ∃n = n(p, k, s) ∈ Z+ such that for each k-colouringof Bn, there are positive integers a0, a1, . . . , ap and a point u ∈ (Z+)d such

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that the each of the (p+ 1) sets

Tqdef= u+

∑

0≤i<q

aiS +

(∑

q≤i≤p

ai

)

s, 0 ≤ q ≤ p,

are monochromatic subsets of Bn.

C(0) holds trivially and we have to show that C(p) ⇒ C(p+ 1).

Let n = n(p, k, s) be the integer specified for C(p). Now, given a k-colouring of (Z+)d, we define the associated colouring of (Z+)d such thattwo points u and v will have the same colour in this new colouring iff thelattice points in the cubes u+Bn and v +Bn are identically coloured in theoriginal k- colouring of (Z+)d.

Clearly, this associated colouring of (Z+)d is a k′- colouring where k′ def=

knd

.

Now, from A(S), it follows that ∃ an integer n′ = n′(k′) such that forevery k′-colouring of Bn′, Bn′ contains a monochromatic subset of the forma′S + v′ for some a′( 6= 0) ∈ Z+ and v′ ∈ Bn′ .

Let N = n + n′ + 1. Let a k-colouring of BN be given. In an arbitraryway we extend this to a k-colouring of (Z+)d. Now, corresponding to theassociated k′- colouring of (Z+)d, Bn′ contains a monochromatic subset ofthe form a′S + v′. This means that the |S|-cubes Bn + a′t + v′, t ∈ S arecoloured identically in the original k- colouring. We observe that all thesecubes lie in BN . By C(p), for any t ∈ S, the cube Bn + a′t + v′ containsmonochromatic sets

Tq(t) = a′t+ v′ + u+∑

0≤i<q

aiS +

(∑

q≤i≤p

ai

)

s, 0 ≤ q ≤ p.

Setting b0 = a′ and bi = ai−1, 1 ≤ i ≤ p+ 1, we claim that the sets

T ′q = (v′ + u) +

∑

0≤i<q

biS +

(∑

q≤i≤p+1

bi

)

s, 0 ≤ q ≤ p + 1

are monochromatic.

For q = 0,

T ′0=(v′ + u) +

(∑

0≤i≤p+1

bi

)

s

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is a singleton and the claim is established.

For q ≥ 1, T ′q = ∪t∈STq−1(t). Since Bn + a′t + v′ are identically coloured

for different t’s belonging to S, it follows that the monochromatic sets Tq−1(t)are of the same colour and hence T ′

q = ∪t∈STq−1(t) is a monochromatic subsetof BN .

Thus C(p+ 1) holds with n(p+ 1, k, s) = N .

Now that C(p) is established for all integers p ≥ 0, the particular casep = k gives us an integer n = n(k, k, s) such that given any k-colouring ofBn, ∃(k + 1) monochromatic sets T0, . . . , Tk in Bn. By pigeonhole principal,two of these sets, say Tr, Tq with r < q are of the same colour.

Writing

Tr = u+∑

0≤i<r

aiS +

(∑

r≤i<q

ai

)

s+

(∑

q≤i<k+1

ai

)

s

and

Tq = u+∑

0≤i<r

aiS +∑

r≤i<q

aiS +

(∑

q≤i<k+1

ai

)

s,

and choosing s0 ∈ S (S being nonempty), it follows that the set

T = u+

(∑

0≤i<r

ai

)

s0 +

(∑

r≤i<q

ai

)

(S ∪ s) +

(∑

q≤i<k+1

ai

)

s

is contained in BN and is monochromatic.

Setting a =∑

r≤i<q ai and v = u +(∑

0≤i<r ai)s0 +

(∑

q≤i<k+1 ai

)

s,

T = a(S ∪ s) + v and this establishes A(S ∪ s).

We discuss the theorem of Hales and Jewett [40] which reveals the com-binatorial nature of van der Waerden’s theorem. As has been said in [37]:

“the Hales-Jewett theorem strips van der Waerden’s theorem of its unessen-tial elements and reveals the heart of Ramsey theory”.

LetCn

t = x1x2 . . . xn : xi ∈ 1, 2, . . . t,

the collection of words of length n over the alphabet of t symbols 1, 2, . . . t.

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A combinatorial line in Cnt is a set of t points in Cn

t ordered asX1, X2, . . . ,Xt where Xi = xi1xi2 . . . xin such that for j belonging to a nonempty subsetI of 1, . . . n we have xsj = s for 1 ≤ s ≤ t and x1j = · · · = xtj = cj forsome cj ∈ 1, . . . t for j belonging to the complement (possibly empty) of Iin 1, . . . n.

For example, for t = 3 and n = 5, the following is a combinatorial line inC5

3 :

111222122231322

Theorem 12. (Hales-Jewett) Given any two positive integers r and t,there exists a positive integer n = HJ(r, t) such that if Cn

t is r-coloured thenthere exists a monochromatic combinatorial line.

Observing the above example of the combinatorial line in C53 , if we identify

the collection of words in C53 with the set of integers in their usual expres-

sion in decimal system, it is easy to see that the above combinatorial linecorresponds to a three term arithmetic progression with common difference10100.

Thus, if we consider an r-colouring of Cnt (with suitably large t), induced

from a given r-colouring of the integers which have base d representation withd ≥ t, Hales-Jewett Theorem would imply the existence of a monochromaticarithmetic progression of t terms.

Erdos and Turan [26] conjectured that any subset of Z+ with positiveupper natural density contains arithmetic progressions of arbitrary length,where, for A ∈ Z+, the upper natural density d(A) of A is defined by

d(A) = lim supN→∞

|A ∩ [N ]|

N,

where [N ] denotes the set 1, 2, . . . , N.

Remark 4. In connection with Schur’s theorem, the situation is quite dif-ferent; though the set of even integers and the set of odd integers have thesame upper natural density 1/2 in Z+, there is no solution of x + y = z inthe subset of odd positive integers.

15

Towards the above mentioned conjecture of Erdos and Turan, in 1953Roth [63] proved that any subset A of the set Z+ of positive integers withpositive upper natural density will always contain a three-term arithmeticprogression. Later, Szemeredi first improved [69] Roth’s result to that of Apossessing a four-term arithmetic progression and finally in 1974, in a famouspaper [70] proved the general Erdos-Turan conjecture by a sophisticated com-binatorial argument. Later, Furstenberg [27] gave an ergodic theoretic proofof Szemeredi’s theorem which opened up the subject of Ergodic RamseyTheory (see [28], [14] and [50]). There have been other important proofs ofSzemeredi’s theorem since then.

Defining rk(n) to be the smallest integer such that whenever A ⊂ [n] satis-fies |A| > rk(n), A contains an arithmetic progression of k terms, Szemeredi’sresult [70] implies that

rk(n) = o(n).

For the case k = 3, Roth’s proof [63] gave r3(n) = O(

nlog logn

)

; suc-

cessive improvements in this direction were obtained by Heath-Brown [43],Szemeredi [71] and Bourgain [17], the result of Bourgain being

r3(n) = O

(

n ·

√

log log n

logn

)

.

Regarding estimates of rk(n) for k > 3, Gowers [35] has made a remark-able breakthrough while establishing

r4(n) <n

(log log n)dfor some absolute constant d > 0,

where the method seems to go through for rk(n) for k ≥ 4. Apart from theoriginal paper [35], we recommend the two beautiful articles [36] and [17] forgetting an idea as well as the background of the proof.

Conjecture (Erdos). If A ⊂ Z+ satisfies

∑

a∈A

1

a= ∞,

then A contains arithmetic progressions of arbitrary length.

The Green-Tao Theorem [38] we have seen a major breakthrough. Greenand Tao showed that the set of primes contains arithmetic progressions ofarbitrary length.

16

3 Zero-sum problems: Part I

Considering the sums s1 = a1, s2 = a1 + a2, · · · , sn = a1 + · · · + an, if nosi is 0 (mod n), then at least two of the si’s are equal modulo n. Thus, ∃nonempty I ⊂ 1, · · · , n such that

(8)∑

i∈I

ai = 0 (mod n).

A generalization of the above problem to an arbitrary finite abelian groupG, is to determine the Davenport’s constant D(G) which is the smallestpositive integer s such that for any sequence g1, g2, · · · , gs of (not necessar-ily distinct) elements of G, there is a nonempty I ⊂ 1, · · · , s such that∑

∈I gi = 0.

This constant, though attributed to Davenport, seems to have been firststudied by K. Rogers [61] in 1962. This particular reference was somehowmissed-out by most of the authors in this area.

Apart from its interest in zero-sum problems of additive number theoryand non-unique factorization in algebraic number theory (see [34]), this con-stant has other applications. One of the most important applications of thiscan be seen in the proof of the infinitude of Carmichael numbers by Alford,Granville and Pomerance [3], where some knowledge of zero-sum sequencesin the group of units of Zn is required. Zero-sum results have been also seento be useful in an interesting paper of Brudern and Godinho [18].

Given a finite abelian group G = Zn1×Zn2

×· · ·×Znr, with n1|n2| · · · |nr,

writingM(G) = 1+∑r

i=1(ni−1), it is trivial to see thatM(G) ≤ D(G) ≤ |G|.

When G = Zn, observing that the sequence

(1, 1, . . . , 1︸ ︷︷ ︸

(n−1) times

)

does not have any non-empty subsequence whose sum is zero, one has theequality

(9) D(Zn) = n.

It is known that D(G) = |G| holds if and only if G = Zn.

17

Olson [54] [55] proved that D(G) = M(G) for all finite abelian groups ofrank 2 and for all p-groups.

In the following theorem, we write the result of Olson [54] in the case forabelian p-groups; we shall see a proof of this in the next section.

Theorem 13. (Olson) For a finite abelian p-group G = Zpr1 × Zpr2 ×· · · × Zprm , given a sequence S = (s1, s2, · · · , sk) of elements in G, such thatk ≥ 1 +

∑m

i=1(pri − 1), writing fe(S) for the number of subsequences of even

length of S which sum up to zero and fo(S) for the number of subsequencesof odd length which sum up to zero, we have

fe(S)− fo(S) ≡ −1 (mod p).

In particular, we have

D(G) = M(G) = 1 +m∑

i=1

(pri − 1).

We take up generalization of the problem in another direction, where oneasks about zero-sum subsequence of a prescribing size.

In this particular direction, a theorem of Erdos, Ginzburg and Ziv [24](henceforth, referred to as the EGZ theorem) says that

Theorem 14. (EGZ theorem) For any positive integer n, any sequencea1, a2, · · · , a2n−1 of 2n − 1 integers has a subsequence of n elements whosesum is 0 modulo n.

A prototype of zero-sum theorems, the EGZ theorem continues to play acentral role in the development of this area of combinatorics.

The article [7] of Alon and Dubiner contains five proofs of Theorem 14,among other things. One may also find several proofs in the books [1], [52],[34]. Here we shall see some of these proofs.

We observe that it is enough to prove Theorem 14 when n = p, a prime.The case n = 1 is trivial.Assume that the result is true when n is a prime and proceed by induction

on the number of prime factors (counted with multiplicity) of n.

If n > 1 is not a prime, we write n = mp where p is prime and assumethat the result is true for all integers with number of prime factors less thanthat of n.

18

Since each subsequence of 2p−1 members of the sequence a1, a2, · · · , a2n−1

has a subsequence of p elements whose sum is 0 modulo p, we go on repeatedlyomitting such subsequences of p elements having sum equal to 0 modulo p.

After removing 2m− 2 such sequences one after another, we are left with2pm− 1 − (2m − 2)p = 2p− 1 elements and we can have at least one moresubsequence of p elements with the property that sum of its elements is equalto 0 modulo p.

Thus, one can find 2m − 1 pairwise disjoint subsets I1, I2, · · · , I2m−1 of1, 2, · · · , 2mp− 1 with |Ii| = p and

∑

j∈Iiaj ≡ 0(mod p) for each i.

Writing bi =1p

∑

j∈Iiaj , by the induction hypothesis, this new sequence

has a subsequence of m elements whose sum is divisible by m. The unionof the corresponding sets Ii will supply the desired subsequence of mp = nelements of the original sequence such that the sum of the elements of thissubsequence is divisible by n.

We state a couple of theorems which will be used by us.We shall need the following generalized version of Cauchy-Davenport in-

equality [20], [21] (one can also look into [49] or [52] for instance):

Theorem 15. (Cauchy-Davenport) Let A1, A2, · · · , Ah be non-empty sub-sets of Fp, the finite field with p elements. Then

|h∑

j=1

Aj | ≥ min

(

p,

h∑

j=1

|Aj| − h + 1

)

.

Another result which will be required is the following, for a proof of which,one may look into [1], [46] or [52] for instance.

Theorem 16. (Chevalley-Warning) Let fi(x1, x2, · · · , xn), i = 1, · · · , r,be r polynomials in Fq[x1, x2, · · · , xn] such that the sum of the degrees ofthese polynomials is less than n and fi(0, 0, · · · , 0) = 0, i = 1, · · · , r. Thenthere exists (α1, α2, · · · , αn) ∈ Fn

q with not all αi’s zero, which is a commonsolution to the system fi(x1, x2, · · · , xn) = 0, i = 1, · · · , r.

Proofs of Theorem 14 in the case n = p, a prime:

Proof I. This proof involves an application of Theorem 15.

19

Consider representatives modulo p in the interval 0 ≤ aj ≤ p− 1 for thegiven elements and rearranging, if necessary, we assume that

0 ≤ a1 ≤ a2 ≤ · · · ≤ a2p−1 ≤ p− 1.

We can assume that

aj 6= aj+p−1, for j = 1, · · · , p− 1,

for otherwise, the p elements aj, aj+1, · · · , aj+p−1 being equal, the result holdstrivially.

Writing Aj := aj, aj+p−1, for j = 1, · · · , p− 1, by Theorem 15,

|

p−1∑

j=1

Aj| ≥ min

(

p,

p−1∑

j=1

|Aj | − (p− 1) + 1

)

= p.

This implies that −a2p−1 ∈∑p−1

j=1 Aj, and we are through.

Proof II. Following Alon [4] (also see [7]), we now give a proof by applyingTheorem 16.

Given a sequence a1, a2, · · · , a2p−1 with ai ∈ Fp, consider the system ofpolynomial equations:

2p−1∑

i=1

aixp−1i = 0,

2p−1∑

i=1

xp−1i = 0.

Since the sum of degrees of the two equations is 2(p−1) which is less than2p − 1 and x1 = x2 = · · · = x2p−1 = 0 is a solution, we can apply Theorem16. Thus there is a nontrivial solution (α1, · · · , α2p−1) of the above system.

By Fermat’s little theorem, writing I = i : αi 6= 0, from the firstequation it follows that

∑

i∈I ai = 0 and from the second equation we have|I| = p.

Proof III. By Theorem 13 of Olson, the sequence

(a1, 1), (a2, 1), · · · , (a2p−1, 1)

20

of 2p−1 elements in Zp⊕Zp has a non-empty zero-sum subsequence and thedesired result follows once again by observing that there is only one multipleof p among the integers 1, 2, . . . , 2p− 1.

Proof IV. We start with the following definition.

Given an n by n matrix A = (aij) over a field F , its permanent, denotedby per A, is defined by

per A =∑

a1 α(1)a2 α(2) · · · an α(n),

where the summation is taken over all permutations α of [n].

The following is a variant of the permanent lemma; the statement here isnot as in [5]; the statement in [5] is a slightly extended version of a lemmafirst proved in [9] (see also [8]).

The permanent lemma. (Alon) Let A = [ai,j ] be a m × m matrix overFp such that per A 6= 0.

Then for any c1, · · · , cm ∈ Fp, there are ǫ1, ǫ2, · · · , ǫm ∈ 0, 1 such that∑m

j=1 ǫjai,j 6= ci for all 1 ≤ i ≤ m.

Proof: By induction on m it is rather easy to observe that if

P (x1, · · · , xm) =∑

U⊂[m]

bU∏

i∈U

xi

is a multilinear polynomial over a commutative ring with identity such thatP (x1, · · · , xm) = 0 for each (x1, · · · , xm) ∈ 0, 1m, then P is identicallyzero, that is, bU = 0 for all U ⊂ [m].

If possible let the lemma be false. Then there are no ǫ1, ǫ2, · · · , ǫm asabove.

We consider the polynomial

P (x1, · · · , xm) =

m∏

i=1

(m∑

j=1

ai,jxj − ci

)

.

By our assumption, P (x1, · · · , xm) = 0 for each (x1, · · · , xm) ∈ 0, 1m.

From P (x1, · · · , xm), we obtain the multilinear polynomial P ′(x1, · · · , xm)by first writing P in the standard form as a sum of monomials and then re-placing each monomial of the form bU

∏

i∈U xδii , with δi ≥ 1, by bU

∏

i∈U xi.

21

Clearly, P ′(x1, · · · , xm) = P (x1, · · · , xm) = 0 for each (x1, · · · , xm) ∈0, 1m. Therefore, by the observation made in the beginning of the proof,P ′ is identically zero.

But, the coefficient of∏m

i=1 xi in P ′(x1, · · · , xm) (which is the same asthe coefficient of

∏m

i=1 xi in P (x1, · · · , xm)) is per A 6= 0 and we arrive at acontradiction.

Hence the lemma.

Following the treatment in [7], we shall now see how to derive the EGZtheorem from the above lemma.

For a prime p, we are given a sequence a1, a2, · · · , a2p−1, of elements inFp. We identify them with the integers in the set 0, 1, · · · , p− 1 representingthem (mod p) and by rearranging the elements, if necessary, we assume thata1 ≤ a2 ≤ · · · ≤ a2p−1.

If there is an i ≤ p − 1, such that ai = ai+p−1 (which means that ai =ai+1 = · · · = ai+p−1), then taking I = i, i+1, · · · , i+ p− 1, the theorem isestablished.

If there is no such i, we define

bi = ai − ai+p−1( 6= 0) ∀ i, 1 ≤ i ≤ p− 1.

Let c1, c2, · · · , cp−1 be the set of all elements in Fp besides the element(

−∑2p−1

j=p aj

)

.

Let A = [ai,j ] be the (p− 1)× (p− 1) matrix over Fp defined by ai,j = bjfor all 1 ≤ i, j ≤ p− 1, so that

per A = (p− 1)! ·

p−1∏

j=1

bj 6= 0.

Therefore, by the permanent lemma, there are ǫ1, ǫ2, · · · , ǫp−1 ∈ 0, 1such that

∑p−1j=1 ǫjai,j 6= ci for all 1 ≤ i ≤ p− 1 and hence

p−1∑

j=1

ǫjai,j = −

2p−1∑

j=p

aj.

Rearranging, a2p−1 +

p−1∑

j=1

(aj+p−1 + ǫi(aj − aj+p−1)) = 0.

22

Since each term ai+p−1 + ǫi(ai − ai+p−1) is either ai+p−1 or ai, this gives ap-subset of the sequence ai the sum of whose elements is 0, as required.

If G is a finite abelian group with exp(G) = n, then the Erdos-Ginzburg-Ziv constant s(G) is defined to be the least integer k such that any sequenceS with length |S| ≥ k of elements in G has a zero-sum subsequence of lengthexp(G) = n.

In general, for a finite abelian group G with exp(G) = n, smn(G) is defined(see [31], for instance) to be the least integer k such that any sequence S withlength k of elements in G has a zero-sum subsequence of length mn. Puttingm = 1, one observes that the constant s(G) is the same as sn(G). We shalluse the notation E(G) for the constant s|G|(G).

Harborth [42] and Kemnitz [47] had considered the problem of determin-ing s(G) where G = Zd

n; that is, the problem of finding the smallest naturalnumber k such that any sequence of k elements in Zd

n has a subsequence oflength n summing up to 0, the identity element of the group Zd

n.

In geometric terms, s(Zdn

)is the smallest positive integer k such that

given a sequence x1, x2, . . . , xk of elements of Zd, not necessarily distinct,there exists a subsequence xi1 , xi2 , . . . , xin of length n such that its centroid(xi1 + xi2 + · · ·+ xin)/n also belongs to Zd.

Observing that the number of elements of Zdn having coordinates 0 or 1

is 2d and considering a sequence where each of these elements are repeated(n− 1) times, we observe that

(10) 1 + 2d(n− 1) ≤ s(Zdn).

Again, observing that in any sequence of 1 + nd(n − 1) elements of Zdn,

there will be at least one element appearing at least n times, we have

(11) s(Zdn) ≤ 1 + nd(n− 1).

EGZ theorem says that in the case d = 1, the lower bound 2n− 1 givenby (10) is the exact value of s(Zn).

There is an early survey on the zero-sum problems by Caro [19]. Forrecent developments, one may look into the expository article of Gao andGeroldinger [31], the book of Geroldinger and Halter-Koch [34] and the sur-vey of Geroldinger [33].

That the trivial lower bound in (10) is the exact value in the case d = 2as well, that is

s(Z2n) = 4n− 3,

23

had been known as the Kemnitz Conjecture in the literature.Kemnitz [47] had verified the conjecture when n is of the form 2e3f5g7h.For general n, Alon and Dubiner [7] obtained the upper bound

s(Z2n) ≤ 6n− 5.

Later Ronyai [62] proved that

s(Z2p) ≤ 4p− 2,

for a prime p, and using this one easily obtains

s(Z2n) ≤

41

10n,

by induction on the number of primes dividing n and using the followinginequality of Harborth [42]:

(12) s(Zdmn) ≤ min(s(Zd

n) + n(s(Zdm)− 1), s(Zd

m) +m(s(Zdn)− 1)).

Finally, Reiher [60] established the Kemnitz Conjecture (see Savchev andChen [65] for the related structural analysis).

Geroldinger and Halter-Koch [34] have a more general result which saysthat for positive integers m,n with m|n,

s(Zm ⊕ Zn) = 2m+ 2n− 3.

Theorem 17. (Reiher) For a positive integer n, we have

s(Z2n) = 4n− 3.

Proof. It suffices to show that s(Z2p) ≤ 4p− 3.

Let X be a sequence of 4p− 3 elements of Z2p.

For any subsequence J of X , let (n, J) denote the number of zero-sumsubsequences of J of length n.

In particular, we have (p,X) = 0 and hence by a result of Alon andDubiner [7], (3p,X) = 0.

Now, we deduce the following consequence of Theorem 13.

For any subsequence J = (a1, b1), (a2, b2), · · · , (a(3p−3), b(3p−3)) of X , wehave

(13) 1− (p− 1, J) + (2p− 1, J) + (2p, J) ≡ 0 (mod p).

24

The proof of the statement (13) is as follows. We consider the sequenceJ1 consisting of the 3p − 3 elements in (Z/pZ)3 given by (a1, b1, 1), · · · ,(a3p−3, b3p−3, 1) along with the element (0, 0, 1) . Now the length of J1 is 3p−2 ≥ 1+ 3(p− 1), hence we can apply Theorem 13 of Olson. Since the thirdco-ordinate of all the above elements is 1, any subsequence of even length ofJ1 which sums up to zero has to be of length 2p while any subsequence of oddlength of J1 which sums up to zero has to be of length p. Any subsequenceof length 2p of J1 which sums up to zero and which does not involve the lastterm (0, 0, 1) gives a subsequence of length 2p of J which sums up to zeroand vice versa. However any subsequence of length 2p of J1 which sums upto zero and which involves the last term (0, 0, 1) actually gives a subsequenceof length 2p − 1 of J which sums up to zero and vice versa. Similar is thecase for the zero-sum subsequences of length p. Now using the above resultof Olson and keeping in mind the fact that (p, J) = 0, we get the desiredresult.

Now,

∑

[1− (p− 1, I) + (2p− 1, I) + (2p, I)] ≡ 0 (mod p).

where I runs over all possible subsequences I of length 3p− 3 of X .

Therefore,

(4p− 3

3p− 3

)

−

(3p− 2

2p− 2

)

(p− 1, X) +

(2p− 2

p− 2

)

(2p− 1, X)

+

(2p− 3

p− 3

)

(2p,X) ≡ 0 (mod p),

which gives

(14) 3− 2(p− 1, X) + (2p− 1, X) + (2p,X) ≡ 0 (mod p).

Further,

(15) (2p, J) ≡ −1 (mod p),

for any subsequence J of X of length 3p− 2 or 3p− 1, and

(16) (2p,X) ≡ −1 (mod p),

25

by considering the relevant sequences in (Z/pZ)3 obtained by putting 1 inthe last coordinate and using Theorem 13.

Once again following the argument employed in deducing (13) and using(16), we have

(17) (p− 1, X)− (2p− 1, X) + (3p− 1, X) ≡ 0 (mod p).

Finally, we prove that

(18) (p− 1, X) ≡ (3p− 1, X) (mod p).

We proceed as follows. Let θ denote the number of partitions of X =A ∪ B ∪ C into disjoint subsequences A, B and C where A is a zero-sumsubsequence of length p − 1 and C is a zero-sum subsequence of length 2p.We count θ in two different ways.

First,

θ =∑

A

(2p,X \ A),

where the summation runs over all zero-sum subsequences of length p− 1.

Using (15), this gives

(19) θ ≡ −(p− 1, X) (mod p).

Again,

θ =∑

A′

(2p, A′),

where the summation runs over all zero-sum subsequences of length 3p − 1and by using (15), we have

(20) θ ≡ −(3p− 1, X) (mod p).

From (19) and (20), we get (18).

Now, adding (14) and (17), by (18), we have

(2p,X) ≡ −3 (mod p),

– contradicting (16).

26

4 Zero-sum problems: Part II

In this section we deal with some results for general finite abelian groups andbounds for some of the combinatorial constants in the case of higher ranks.

We start with the following result of Bollobas and Leader [16] whichclearly implies the EGZ theorem by taking r = n− 1.

Theorem 18. (Bollobas and Leader) Let G be an abelian group of ordern and r be a positive integer. Let A denote the sequence (a1, a2, · · · , an+r) ofn+ r not necessarily distinct elements of G. Then, if 0 is not an n-sum, thenumber of distinct n-sums of A is at least r + 1.

A simple proof of the above theorem has been given by Yu [73] using thefollowing generalization of Cauchy-Davenport Theorem (Theorem 15) due toScherk [66].

Scherk’s Lemma. Let A and B be two subsets of a finite abelian group G.Suppose 0 ∈ A∩B and suppose that the only solution of a+b = 0, a ∈ A, b ∈B is a = b = 0. Then,

|A+B| ≥ |A|+ |B| − 1.

Proof: We closely follow the proof of Scherk [66].

The result is obvious for |B| = 1. Let |B| = n > 1 and assume that theresult holds for |B| ≤ n− 1.

Choose b ( 6= 0) ∈ B.

Since by our assumption, 0 6∈ A+b and 0 ∈ A, from the equality |A+b| =|A|, it follows that ∃t ∈ A such that t + b 6∈ A.

LetB1

def= g ∈ B | t+ g 6∈ A.

Since b ∈ B1, the set B1 is non-empty. Also, since t ∈ A, we have 0 6∈ B1,and hence B1 ( B.

Let

A1def= t+B1 ⊂ A +B,

A2def= A ∪A1,

B2def= B \B1.

27

Now, A1 ∩ A = ∅ and |A1| = |B1| and therefore,

(21) |A2|+ |B2| = |A|+ |A1|+ |B2| = |A|+ |B1|+ |B2| = |A|+ |B|.

We claim that

(22) A2 +B2 ⊂ A+B.

Let a2 ∈ A2, b2 ∈ B2. If a2 ∈ A, then a2 + b2 ∈ A + B2 ⊂ A + B. Ifa2 ∈ A1, then a2 + b2 = (t + b′) + b2, for some b′ ∈ B1. Now, (t + b′) + b2 =(t+ b2) + b′ ∈ A+B1 ⊂ A +B. This establishes the claim (22).

We observe that 0 ∈ A ⊂ A2 and since 0 6∈ B1, 0 ∈ B2. Now, if possiblelet a2 + b2 = 0 with a2 ∈ A2, b2 ∈ B2. If a2 ∈ A1, then arguing as before,a2 + b2 = (t + b2) + b′, where (t + b2) ∈ A and b′ ∈ B1 ⊂ B and hence0 = a2 + b2 = (t + b2) + b′ would imply b′ = 0, contradicting the fact that0 6∈ B1. Therefore, a2 ∈ A. Since B2 ⊂ B, we have a2 = b2 = 0. SinceB1 ( B, by the induction hypothesis,

|A2 +B2| ≥ |A2|+ |B2| − 1,

and therefore from (21) and (22),

|A+B| ≥ |A2 +B2| ≥ |A2|+ |B2| − 1 = |A|+ |B| − 1.

At this point, we introduce some more notations. Given a sequence S =(x1, . . . , xk) of elements of an abelian group G, for any 1 ≤ l ≤ k, we shalluse the notation

∑

l(S) to denote the set of all l-sums of S (by an l-sum onemeans a sum xi1 + · · · + xil of a subsequence of length l of S) and

∑

≤l(S)to denote ∪1≤t≤l

∑

t(S). One simply writes∑

(S) in place of∑

k(S), theset of sums of all the subsequences of S. We shall write

∑S to denote the

sum of all elements of S and |S| to denote the length k of S (since it will beunderstood from the context, the notation will not create confusion with thecardinality of a finite set).

The deduction of the following from Scherk’s Lemma is not difficult; weleave it as an exercise.

Lemma 1. Let S be a sequence of n elements in an abelian group G oforder n and let h be the maximum number that any element of G occurs inS. Then

0 ∈∑

≤h

(S).

28

Now, we state the following result of Gao [30] (see also [34], Proposition5.7.9).

Theorem 19. For a finite abelian group G of order n, we have

(23) E(G) = D(G) + n− 1,

where as has been stated before, E(G) = s|G|(G).

Since it is trivial to observe that for a finite abelian group G of order nD(G) ≤ n, from the above theorem it follows that E(G) ≤ 2n− 1. It shouldbe noted that from the EGZ theorem, by induction on the rank it followsthat for a finite abelian group G of order n, E(G) ≤ 2n− 1.

Proof of Theorem 19. It is easy to observe that by appending a sequence ofn− 1 zeroes to a sequence of length D(G)− 1 with no zero-sum subsequence(which exists by the definition of D(G)), the resulting sequence of lengthD(G) + n − 2 will have no zero-sum subsequence of length n and henceE(G) ≥ D(G) + n− 1.

We proceed to establish that

(24) E(G) ≤ D(G) + n− 1.

Let A be any sequence of m = D(G)+n−1 elements of G and we have toshow that 0 ∈

∑

n(A). We can assume that 0 is the most repeated elementin the sequence and arrange A to be of the form

A = (a1, a2, . . . , am−h, 0, 0, . . . , 0︸ ︷︷ ︸

h times

),

with all ai 6= 0.

Clearly, we can assume that h ≤ n.

Let W be a subsequence of (a1, a2, . . . , am−h), maximal subject to thecondition that

∑W = 0.

From the maximality ofW and the definition ofD(G), it is easy to observethat m− h− |W | < D(G), so that |W | > m− h−D(G) = n− h− 1.

If |W | ≤ n, then appending n− |W | ≤ h zeros, we see that 0 ∈∑

n(A).

29

If |W | ≥ n + 1, repeatedly applying Lemma 1 to W , we find a systemof disjoint subsequences W1,W2, . . . ,Wv satisfying the conditions

∑Wi = 0

and |Wi| ≤ h, for each i such that

|W −W1 − · · ·Wv| ≤ n− 1

and|W −W1 − · · ·Wv−1| ≥ n.

Therefore,

n− 1 ≥ |W −W1 − · · ·Wv| ≥ n− |Wv| ≥ n− h.

Writing W0 = W − W1 − · · ·Wv, we have n − 1 ≥ |W0| ≥ n − h andas before, appending n − |W0| ≤ h zeros, 0 ∈

∑

n(A). This completes theproof.

The following result of Hamidoune [41] confirmed a conjecture of Caro[19] with some conditions. When n is prime it had already been observedby Alon, Bialostocki and Caro [6]; later Grynkiewicz [39] established theconjecture of Caro in full generality.

Theorem 20. Let G be an abelian group of order n and k a positive integer.Let (w1, w2, ..., wk) be a sequence of integers where each wi is co-prime to n.Then, given a sequence S : (x1, x2, ..., xk+n−1) of elements of G, if x1 is themost repeated element in the sequence, we have

k∑

1

wixσ(i) =

(k∑

1

wi

)

x1,

for some permutation σ of 1, 2, . . . , k + n− 1.

Remark 5. Taking G = Z/nZ, k = n and wi = 1, for all i, in the abovetheorem, we get the EGZ theorem.

The following result of Adhikari, Chintamani, Moriya and Paul [2], in thespirit of Theorem 18 of Bollobas and Leader, implies the above theorem.

Theorem 21. Let (w1, w2, ..., wk) be a sequence of integers where each wi isco-prime to n. Then, given a sequence A : (x1, x2, ..., xk+r) of elements of G,

30

where 1 ≤ r ≤ n − 1, if 0 is the most repeated element in the sequence, and∑k

1 wixσ(i) 6= 0, for all permutations σ of [k + r], we have

∣∣∣∣∣

k∑

1

wixσ(i) : σ is a permutation of [k + r]

∣∣∣∣∣≥ r + 1.

Now, we turn to the constant s(Zdn) general d. The following result of

Alon and Dubiner [10] says that for any fixed d, the growth of s(Zdn) is linear

in n.

Theorem 22. (Alon and Dubiner) There is an absolute constant c > 0so that

(25) s(Zdn) ≤ (cd log2 d)

dn, for all n.

Alon and Dubiner conjecture the existence of an absolute constant c suchthat

s(Zdn) ≤ cdn, for all n and d.

Harborth [42] proved that s(Z33) = 19; this is strictly greater than the

corresponding lower bound given in (10), namely, 17.

The following result of Elsholtz [23] shows that for d ≥ 3 and an oddinteger n ≥ 3, the lower bound given in (10) is strictly less than the exactvalue of s(Zd

n).

Theorem 23. (Elsholtz) For an odd integer n ≥ 3, one has the followinginequality:

s(Zdn) ≥ (1.125)[

d3] (n− 1)2d + 1.

The above result implies that s(Z3n) ≥ 9n − 8. It has been conjectured

(see Gao and Thangadurai [32]) that s(Z3n) = 9n− 8.

An improvement of Theorem 23 has been obtained in [22]. The newingredient in the proof in [22], is a lower bound for s(Z4

n), namely, s(Z4n) ≥

20n − 19 and the authors conjecture that this lower bound give the precisevalue for all odd n ≥ 4.

We now give a proof of the following; we shall closely follow the originalproof of Olson [54].

31

Proof of Theorem 13. We use multiplicative notation for G in this proof.Since an obvious example shows that the right hand side is a lower bound forD(G), a combinatorial interpretation of the following observation establishesthe theorem.

Observation: If k ≥ 1 +∑r

i=1(pei − 1), then given a sequence g1, . . . , gk of

elements of G, in the group-ring R of G over Zp one has

(26) (1− g1)(1− g2) · · · (1− gk) = 0.

If x1, . . . , xr is the standard basis for G, where xi is of order pei, thensince each gj can be can be written as a product of these basis elements,applying the identity 1 − uv = (1 − u) + u(1 − v) repeatedly, the left handside in equation (26) can be expressed as a sum of elements of the form

g

r∏

i=1

(1− xtii ), where g ∈ G and

r∑

i=1

ti = k > (pei − 1).

Since k > (pei − 1), ti ≥ pei for some i, which would imply that

(1− xtii ) = 0.

Hence equation (26) holds.

In the remaining part, we talk about some general results about Daven-port constant. Emde Boas and Kruyswijk [15], Baker and Schmidt [12] andMeshulam [51] gave upper bounds for Davenport constant which involves theexponent of the group and the cardinality of the group G. The best knownbound is due to Emde Boas and Kruyswijk [15]:

(27) D(G) ≤ m

(

1 + log|G|

m

)

,

where m is the exponent of G. We here sketch a proof by Alford, Granvilleand Pomerance [3] which involves modifying the above Proof of Theorem 13.

Sketch of the proof of (27): Let g1, g2, . . . , gk be a sequence of elementsof G where

k ≥ m

(

1 + log|G|

m

)

.

32

We choose a prime q ≡ 1 (mod m).

Our aim is to find elements a1, a2, . . . , ak ∈ F∗q, such that in the group

ring Fq[G] we have

(28) (a1 − g1)(a2 − g2) · · · (ak − gk) = 0.

If no subsequence of g1, g2, . . . , gk has product equal to 1, writing (a1 −g1)(a2 − g2) · · · (ak − gk) =

∑

g∈G kgg, k1 = a1a2 . . . ak 6= 0 would then con-tradict equation (28).

Extending any character χ : G → F∗q in the character group G, to a ring

homomorphism χ : Fq[G] → Fq by defining χ(∑

g∈G kgg) =∑

g∈G kgχ(g),from the orthogonality relations of group characters, it follows that for b ∈Fq[G], b = 0 if and only if χ(b) = 0 for all χ ∈ G.

Applying the greedy algorithm one finds a1, a2, . . . , ak ∈ F∗q so that for

each χ ∈ G, there exists j, 1 ≤ j ≤ k, such that χ(gj) = aj implying that

χ

(k∏

i=1

(ai − gi)

)

= 0,

which in turn imply the equality in (28).

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