On Some Developments of the Erdős-Ginzburg-Ziv Theorem II (personal/extended copy)

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1 On Some Developments of the Erdős-Gnzburg-Zv Theorem II (personal/extended copy) Are Balostock, Paul Derker, Davd Grynkewcz and Mark Lotspech August 19, 2001 Abstract. Let S be a sequence of elements from the cyclc group Z m. We say S s zsf (zero-sum free) f there does not exst an m-term subsequence of S whose sum s zero. Denote by g(m, k) the least nteger such that every sequence S wth at least k dstnct elements and length g(m, k) must contan an m-term subsequence whose sum s zero. Furthermore, denote by E(m, s) the set of all equvalence classes of zsf sequences wth length s, up to order and affne transformaton, that are not a proper subsequence of another zsf sequence. In ths paper, we frst fnd for a sequence S of suffcent length, S 2m m 2, necessary and suffcent condtons n terms of a system of nequaltes over the ntegers for S to be zsf. Among the consequences, we determne g(m, k) for large m, namely g(m, k) = 2m k2 2k+5 provded m k 2 2k 3, whch n turn resolves two conjectures of the frst and fourth authors. Next, usng ndependent methods, we evaluate g(m, 5) for every m 5. We conclude wth the lst of E(m, s) for every m and s satsfyng 2m 2 s max{2m 8, 2m m 2}. 1

2 1 Introducton Let S be a sequence of elements from the cyclc group Z m. We say S s zsf (zero-sum free) f there does not exst an m-term subsequence of S whose sum s zero. Let g(m, k) (g (m, k)) denote the least nteger such that every sequence S wth at least (wth exactly) k dstnct elements and length g(m, k) (g (m, k)) must contan an m-term subsequence whose sum s zero. By an affne transformaton n Z m, we mean a map of the form x ax + b, wth a, b Z m and gcd(a, m) = 1. Furthermore, let E(m, s) denote the set of all equvalence classes of zsf sequences S of length s, up to order and affne transformaton, that are not a proper subsequence of another zsf sequence. Usng the above notaton, the renowned Erdős-Gnzburg-Zv theorem, [1], [11], states that g(m, 1) = g(m, 2) = 2m 1 for m 2. The functon g(m, k) was ntroduced n [], where t was shown that g(m, ) = 2m 3 for m. Furthermore, based on a lower bound constructon the authors conjectured the value of g(m, k) for fxed k and suffcently large m. Concernng the upper bound, they establshed an upper bound for m prme modulo the affrmaton of the Erdős-Helbronn conjecture (EHC). Snce then, the EHC has been affrmed [9], [2], moreover, the bound gven n [] was extended for nonprmes n [17]. As t wll later be seen, t s worthwhle to menton that the affrmaton of the EHC has resulted n several attempted generalzatons and related results [6], [1], [23], [25]. Other relevant developments concernng g(m, k) appear n [3], [5], [7], [13], [15], [16], [23]. For example, the value g(m, 3) = 2m 2 was determned n [3], and the closely related functon g (m, k), ntroduced n [5] and further nvestgated n [], [15], [16], was determned for all m and k satsfyng k > m Ths paper started under the authorshp of the frst, second and fourth authors, and was cted n [] as n preparaton. Actually, a rough draft was ready determnng g(m, 5) usng the methods of []. The motvaton for the current paper s twofold. Frst, the thrd author was able to determne exactly g(m, k) for fxed k and large m by mprovng the known lower bound constructon and adaptng the proof used by Gao and Hamdoune [17] to obtan a better upper bound than the one conjectured n []. Consequently, the followng conjecture of [] has been affrmed. Conjecture 1.1. For every k, k 2, there exsts an nteger m 0 = m 0 (k) such that f m > m 0, then (a) g(m, k) = 2m c, where c = c(k) s ndependent of m (b) g (m, k) = g(m, k). 2

3 Thus, the g(m, k) problem for fxed k and large m has been put to rest. Second, as several recent works, e.g. [21], [27], use the known values of g(m, k) and E(m, s) for k and s 2m 3, and as t s lkely that g(m, 5) wll be needed for further zero-sum applcatons, we determne g(m, 5) for every m 5. The paper s organzed as follows. In secton 2, defntons and notaton are ntroduced and known results needed later n the paper are lsted. In Secton 3, frst the upper bound proof of [17] s adapted to fnd for a sequence S wth S 2m m 2, necessary and suffcent condtons n terms of a system of nequaltes over the ntegers for S to be zsf. Ths result and a lower bound constructon mply the value of g(m, k) for fxed k and large m. Followng s the affrmaton of Conjecture 1.1. Secton contans the evaluaton of g(m, 5) for every m 5. The paper concludes wth an appendx lstng the elements of E(m, s) for every m and s satsfyng 2m 2 s max{2m 8, 2m m 2}. 2 Prelmnares Let G denote an abelan group of order m. As we work smultaneously wth the cyclc group Z m of resdue classes modulo m and the addtve group of the ntegers Z, for α Z m we denote by α the least postve nteger that s congruent to α modulo m. A sequence S of elements from G s abbrevated as a strng usng exponental notaton (e.g. the sequence 1, 1, 1, 2, 2, 2, 2 s abbrevated by ). Furthermore, the length of S s denoted by S. Let t be a nonnegatve nteger. Denote by S the set of all sums over nonempty subsequences of S, and by t S ( t S) the set of all sums over subsequences of S of length exactly (at least) t. If A, B G, then ther sumset, A + B, s the set of all possble parwse sums,.e. {a + b a A, b B}. Followng Kemperman [2], a set A G s H-perodc f t s the unon of H-cosets for some nontrval subgroup H G. Furthermore, an n-set partton of S s a sequence of n nonempty subsequences of S, parwse dsjont as sequences, such that every term of S belongs to exactly one subsequence, and the terms n each subsequence are dstnct. be consdered sets. ϕ(1, 1, 1, 2, 2, ) = {1, 2, }). Thus such subsequences can Fnally let ϕ be the map that takes a sequence to ts underlyng set (e.g. Frst, we state three well know theorems, the frst one s a generalzed verson of what s known as the Caveman Theorem [12], followed by a generalzed form of the Erdős-Gnzburg-Zv Theorem 3

4 (EGZ) [1], [11], and the affrmed EHC [2], [9]. The orgnal EGZ Theorem s Theorem 2.2 wth r = 1; and Theorem 2.2 s obtaned by r applcatons of the EGZ Theorem. Theorem 2.1 s smlarly obtaned, where the orgnal Caveman Theorem s the case S = m. Theorem 2.1. Let S be a sequence of elements from an abelan group G of order m. If S m, then there exsts a subsequence of S of length r whose terms sum to zero, where r satsfes S (m 1) r S. Theorem 2.2. Let r be a postve nteger. If S s a sequence of (r + 1)m 1 elements from an abelan group G of order m, then S contans an rm-term subsequence whch sums to zero. Theorem 2.3. Let S be a sequence of k dstnct elements from Z m. If m s prme, then h S mn{m, hk h 2 + 1} The followng two theorems of Gao [18], [19], respectvely, are central to the proof of Theorem Theorem 2.. Let l and m be postve ntegers satsfyng 2 l m +2, and let S be a sequence of elements from Z m satsfyng S = 2m l. If 0 / m S, then up to order and affne transformaton S = 0 u 1 v c 1... c w, where m 2l + 3 v u m 1 and w l 2. Theorem 2.5. Let m, n and h be postve ntegers. Suppose G s a fnte abelan group of order m, and g G. Furthermore, let S = 0 h a 1... a n be a sequence of elements from G such that the multplcty of every element n the subsequence T = a 1... a n s at most h. If g m h T, then g m S. Next, we state the Cauchy-Davenport Theorem [8], [26], and a recently proved composte analog of t [20]. Theorem 2.6. For postve ntegers n and m, let A,..., A n Z p. If m s prme, then n A mn{m, n A n + 1}. Theorem 2.7. Let n be an nteger and let S be a sequence of elements from an abelan group G of order m such that S n and every element of S appears at most n tmes n S. Furthermore, let p be the smallest prme dvsor of m. Then ether:

5 () there exsts an n-set partton, A 1,..., A n, of S, such that n n A mn{m, (n + 1)p, A n + 1} () there exsts α G and a nontrval proper subgroup H of ndex a, such that all but at most a 2 terms of S are from the coset α + H, and there exsts an n-set partton, A 1,..., A n of the n subsequence of S consstng of terms of S from α + H such that A = nα + H. When usng the above theorem, the followng two basc propostons about n-set parttons and sumsets are often used, and for the sake of clarty, we provde ther proofs. Proposton 2.1. A sequence S has an n-set partton A f and only f the multplcty of each element n S s at most n and S n. Furthermore, a sequence S wth an n-set partton has an n-set partton A = A 1,..., A n such that A A j 1 for all and j satsfyng 1 j n. Proof. Suppose S has an n-set partton, then from the defnton the multplcty of each element n S s at most n, and snce empty sets are not allowed n the n-set partton, t follows that S n. Next suppose that the multplcty of each element n S s at most n and S n. Let ϕ(s) = {s 1,..., s u }, and rearrange the terms of S so that all the terms that are equal to s 1 come frst, followed by all the terms that are equal to s 2, and so forth, termnatng wth the terms equal to s u. Let us denote ths new sequence by S = x 1 x 2... x kn+r, where S = S = kn + r and 0 r < n. Consder the followng sequence A of n subsequences of S wrtten vertcally. x 1 x n+1 x r x n+r x r+1 x n+r+1 A =. x..... (k 1)n+1 x (k 1)n+r x.... (k 1)n+r+1 x kn x kn+1 x kn+r.. We wll show that A s an n-set partton of S and hence of S. Indeed, snce S n, t follows that none of the sets n A are empty. Furthermore, n vew of the defnton of S and the fact that the maxmum multplcty of a term n S does not exceed n, t follows that x j1 n+ x j2 n+, for every and every j 1 j 2. Thus A s an n-set partton of S. The furthermore more part s clear from the defnton of A. Proposton 2.2. Let S be a sequence of elements from a fnte abelan group G, and let A = A 1... A n be an n-set partton of S, where n A = r, and s s the cardnalty of the largest set n A. Furthermore, let a 1... a n be a subsequence of S such that a A for = 1,..., n. 5 x n x 2n

6 () There exsts a subsequence S of S and an n -set partton A = A 1... A n of S, whch s a subsequence of the n-set partton A = A 1... A n, such that n r s + 1 and n A = r. () There exsts a subsequence S of S of length at most n + r 1, and an n-set partton A = A 1... A n of S, where A A n for = 1,..., n, such that A = n A. Furthermore, a A for = 1,..., n. Proof. We frst prove (). Assume w.l.o.g. that A 1 = s. We wll construct the n -set partton A n n steps as follows; and S wll be mpled mplctly. Denote by A (k) = A 1... A a k the constructed sequence after k steps, and hence A = A (n) and n = a n. Let A (1) = A 1, and for k = 1, 2,..., n 1, let A (k) A (k+1) = A (k) A k+1 f a k A + A k+1 = a k A f a k A + A k+1 > a k A. It s easly seen by the above algorthm that an A = n A = r. Furthermore, snce each kept term ncreases the cardnalty of the sumset of the prevous terms of A by at least one, and snce A 1 = s, t follows that at most r s terms, excludng A 1, were kept, and thus a n = n 1 + r s. The proof of () s smlar to that of (). Frst, for = 1,..., n, let the elements of A be {a () 1... a() A }, where a() 1 = a. We wll construct the n-set partton A n a two loop algorthm. The outer loop has n steps, where at the th step the set A s constructed usng the nner loop. In turn, the nner loop, at the th step of the outer loop, constructs A n A steps. For a gven, where 1 n, let A (k) denote the set constructed after k steps of the nner loop at the th step of the outer loop, and hence A = A ( A 1 ) 1... A ( An ) n wth S mpled mplctly. For a gven j, where 1 j n, let A (1) j A (k+1) j = = {a j }, and for k = 1,..., A j 1, let j 1 j f A (k) A (k) j {a (j) j 1 k+1 } f A ( A ) A ( A ) + A (k) j 1 j = j 1 + A (k) j < A ( A ) A ( A ) + (A (k) j + (A (k) j {a (j) k+1 }) {a (j) k+1 }). It s easly seen by the above algorthm that A ( A ) A and a A ( A ) for = 1,..., n, and that n A ( A ) = n A = r; and snce n A ( A ) n A, t follows that n A ( A ) = n A. Furthermore, snce each kept element a (j) k, where k 1 f j 1, ncreases the cardnalty of the sumset by at least one, t follows that at most r 1 terms, excludng the a s, were kept, and hence S n + r 1. 6

7 We wll also need the followng theorem of [16]. Theorem 2.8. Let m and k be ntegers wth m k 2 and m 5. (a) If m < k m 1, then g(m, k) = m + 2. m m odd (b) If k = m, then g(m, k) = m + 1 m even. We conclude the prelmnares wth a theorem of Eggleton and Erdős [10]. Theorem 2.9. Let S be a sequence of k dstnct elements from a fnte abelan group. If 0 / S and k, then S 2k. 3 A Theorem of Gao and Hamdoune Revsted Theorem 3.1 gves necessary and suffcent condtons for a sequence S of suffcent length to be zsf. More precsely, t reduces the problem of determnng extremal zsf sequences of suffcent length to the problem of fndng nteger parttons wth a fxed number of parts and all parts greater than 1. Its proof s an adapton of a proof of Gao and Hamdoune [17]. Theorem 3.1. For ntegers m and l, let S be a sequence of elements from Z m, satsfyng S = 2m l 2m m 2. The sequence S does not contan an m-term zero-sum subsequence f and only f there exsts a sequence S = 0 u 1 v a 1... a w1 b 1... b w2, where 1 < a m 2 and 1 b < m 2, that s equvalent to S up to order and affne transformaton, and for whch the followng four nequaltes are satsfed, w 1 a m v 1 and w 2 b m u 1 w 2, (1) m 2l + 3 v u m 1 and w 1 + w 2 l 2. (2) Moreover, equalty holds n both nequaltes of (1) f and only f S belongs to an equvalence class of E(m, 2m l). Proof. Frst, suppose S s a sequence of elements from Z m, satsfyng S = 2m l 2m m 2, and 0 / m S. Hence from Theorem 2. t follows that S s equvalent, up to order and affne transformaton, to a sequence S = 0 u 1 v a 1... a w1 b 1... b w2 satsfyng the nequaltes n (2), where 7

8 1 < a m 2 and 1 b < m 2. Snce the fact that S s zsf mples the fact that S s zsf, t follows from Theorem 2.5 that for any gven subsequence T of a 1... a w1 b 1... b w2, ether t T t m v 1 or t T t u T, and (3) ether t T t m u 1 T or t T t v + 1. () Inducton on r, n vew of (3) and the followng three nequaltes, () l m +2, () m v 1 m 2 (follows from (2) and l m + 2), () 3m l + 5 u + 2v (follows from (2)), mples r a = r a m v 1, (5) for every r satsfyng 1 r w 1. Smlarly, nducton on r, n vew of () and the nequaltes (), () and (), and the fact that u v, mples r b = for every r satsfyng 1 r w 2. Hence (5) and (6) mply (1). r b m u 1 r, (6) Next suppose S s an arbtrary sequence of resdues from Z m that satsfes (1) and (2). Actually, we wll use only the fact that (1) s satsfed and v u m 1. It follows from (1) that any m-term zero-sum modulo m subsequence of S must be zero-sum n Z as well. In addton, t follows from (1) that the longest zero-sum n Z subsequence of S that does not contan a zero s of length w 2 + w 2 b m u 1. Hence any m-term zero-sum subsequence must use at least u + 1 zeros, whch exceeds the multplcty of zero n S. Thus S s zsf, and as affne transformatons and reorderng preserve m-term zero-sum subsequences, the proof of the man part of the theorem s complete. Notce that the two nequaltes n (1) are nterchanged by the affne transformaton whch nterchanges 0 and 1. Hence, the moreover part of the theorem s easly deduced from the man part of the theorem. Theorem 3.2. Let m k 2 be postve ntegers. If k s odd and m k2 +k or k s even and m k2 +2k 8 + 1, then g (m, k) 2m k2 2k+5. 8

9 Proof. If k s even, consder the sequence S 0 = ( k 2 2 )... ( 2)( 1)(0)m k 2 +2k 8 (1) m k2 +2k 8 (2)(3)... ( k 2 ), and f k s odd, consder the sequence S 1 = ( k 3 2 )... ( 2)( 1)(0)m k (1) m k2 +k+3 8 (2)(3)... ( k ). It follows from the hypotheses that both strngs are well defned. Snce both S 1 and S 2 satsfy (1), and snce v u m 1, where u and v are the multplctes of 0 and 1 respectvely, t follows from the proof of the second drecton of Theorem 1 that S 1 and S 2 are zsf. We conclude the secton wth Theorem 3.3, whch determnes the value of g(m, k) for fxed k and suffcently large m, dsprovng Conjecture 5.1 of [], and provng Conjectures 1.1(a) and 1.1(b) n parts (a) and (b) respectvely. Agan, ts proof s an adapton of the proof n [17]. Theorem 3.3. Let m k 2 be postve ntegers. If k s even and m k 2 2k or k s odd and m k 2 2k 3, then (a) g(m, k) = 2m k2 2k+5. (b) g (m, k) = g(m, k) Proof. From Theorem 3.2, and from the trval fact that g (m, k) g(m, k), t suffces for both parts (a) and (b) to show g(m, k) 2m k2 2k+5. Assume to the contrary that there s a sequence S of elements from Z m, wth S = 2m k2 2k+5, and 0 / m S. From the hypotheses and the fact that k 2 0 or 1 mod (), t follows that k2 2k+5 m + 2. Hence from Theorem 3.1 t follows that w.l.o.g. S satsfes (1) and (2). Let c 1 = {a 1,..., a w1 } and c 2 = {b 1,..., b w2 }. It follows from the frst nequalty n (1), that (c 1 + 1) + 2(w 1 c 1 ) m v 1, mplyng that c 2 1 c 1 2 Lkewse from the second nequalty n (1), t follows that + 2w 1 m v 1. (7) c 2 2 c w 2 m u 1 w 2. (8) Inequaltes (7) and (8) mply c 2 1 c c2 2 c 2 2 m v 1 w 1 + m u 1 w 2 = l 2, 9

10 whch, n turn, yelds l (c 1 + c 2 ) 2 + c 1 + c (k 2)2 + k = k2 2k > l, whch s a contradcton; and the proof s complete. The Erdős-Helbronn Conjecture and g(m, 5) In vew of Theorem 3.3, g(m, 5) has been determned for m 12. In ths secton, we present an abbrevated proof determnng g(m, 5) for all m 5. We wll make use of the followng conjecture, whch can be verfed for k 5 wth some effort by consderng the equatons generated by the 2-sums of a 5-set S wth 2 S < 7. Conjecture.1. Let S be a sequence of k 2 dstnct elements from Z m. If 2 S < 2k 3, then ether 2 S s H-perodc, where H > 2, or there exsts a K-perodc subset T such that S T and 2( T S ) K 2. Proof. For k 3, the conjecture s easy. Suppose k = 3. Let S = x 1 x 2 x 3 x be a sequence of four dstnct nonzero resdue classes from Z m such that 0 / S. Defne four sets of equatons A for = 1, 2, 3, : A = {x 1 = x 2 + x 3, x 2 = x 1 + x 3, x 3 = x 1 + x 2 }, A 3 = {x 1 = x 2 + x, x 2 = x 1 + x, x = x 1 + x 2 }, A 2 = {x 1 = x 3 + x, x 3 = x 1 + x, x = x 1 + x 3 }, and A 1 = {x 2 = x 3 + x, x 3 = x 2 + x, x = x 2 + x 3 }. Observe that f two equatons hold n some A, then two of the three x j s from A must be from the same coset of m 2 Z m, and the thrd x j must be equal to m 2 (due to the symmetry of the four sets and the equatons n each set t suffces to check any two equatons from any set). Hence, t s not dffcult to deduce that more than one equaton can hold n at most one of the A s. Thus we may w.l.o.g. assume that ether n A only one equaton holds, say x 1 = x 2 + x 3, or else w.l.o.g only equatons n A 1 hold. In the former case, t s then easly checked that the only equatons that can hold are the followng: x = x 1 + x 3, x = x 1 + x 2, x 2 = x 3 + x and x 3 = x 2 + x. Hence the followng elements must all be dstnct: x 1, x 2, x 3, x 1 + x 2, x 1 + x 3, x 1 + x, x 1 + x 2 + x 3 and x 1 + x 2 + x 3 + x. In the later case, the followng elements must all be dstnct: x 1, x 2, x 3, x, x 1 + x 2, x 1 + x 3, x 1 + x and x 1 + x 2 + x 3 + x. The case k = 5 follows from the next lemma. 10

11 Lemma.1. Let S = a 1 a 2 a 3 a a 5 be a sequence of fve dstnct resdues from Z m. Then 2 S 7 or there exsts a subgroup H of Z m of cardnalty h = 5 or h = 6, and n addton there exsts an H-perodc set S, and a subset T of S wth T = h 5, such that S = S \ T. Proof. Let A = {a 1 +a 2, a 1 +a 3, a 1 +a, a 1 +a 5, a 2 +a 3, a 2 +a, a 2 +a 5, a 3 +a, a 3 +a 5, a +a 5 } be the set of all 2-sums of S. If any three of the 2-sums n A are all equal to one another, then ths mples that not all the a are dstnct, a contradcton. Hence f 2 S 6, then there must be at least parwse dsjont equaltes among the 2-sums. Snce there are four dstnct a s n each of the four equaltes, t follows by the pgeonhole prncple that one a must occur n all equaltes, say a 1. Thus, snce a 1 + a 2 a 1 + a 5, t follows w.l.o.g. that the equaltes n (9) and (10) hold. Furthermore, one of the equaltes n (11) and one of the equaltes n (12) hold as well. a 1 + a 2 = a 3 + a (9) a 1 + a 5 = a 2 + a 3 (10) a 1 + a 3 = a 2 + a, a 1 + a 3 = a 2 + a 5, a 1 + a 3 = a + a 5 (11) a 1 + a = a 2 + a 3, a 1 + a = a 2 + a 5, a 1 + a = a 3 + a 5. (12) Subsequently, we wll refer to an equaton n a numbered lne by the number of the lne followed by a letter from abc... n lexcographc order, e.g. (11)a, (11)b and (11)c correspond to the equatons a 1 + a 3 = a 2 + a, a 1 + a 3 = a 2 + a 5 and a 1 + a 3 = a + a 5, respectvely. From (9) and (10) t follows that 2a 2 = a + a 5. (13) Observe that (12)a cannot hold, snce f t does, then together wth (9) and (13), t wll mply that a = a 5, a contradcton. Thus ether (12)b or (12)c holds. If (12)b holds, then together wth (9) and (10), equaltes (1)(a), (1)(b) and (1)(c) are mpled; and n turn (1)a and (1)(b) mply (1)(d). 2a 5 = a 3 + a, 2a 1 = a 3 + a 5, 2a 2 + a 3 = 2a 1 + a, a + 2a 1 = 3a 5 (1) If (12)c holds, then together wth (9) and (10), equaltes (15)(a), (15)(b), (15)(c) and (15)(d) are mpled; and n turn (15)a and (15)(b) mply (16). 2a = a 2 + a 5, 2a 5 = a 2 + a, 2a 3 + a 5 = 2a 1 + a 2, 2a 1 + a = 2a 3 + a 2, (15) 11

12 3a 5 = 3a. (16) We proceed by consderng three cases, correspondng to each of the three equaltes n (11). Case 1: (11)a holds. Then (11)a and (9) mply (17)(a) and (17)(b). 2a 1 = 2a, 2a 3 = 2a 2 (17) Suppose (12)c holds. Then (17)(a), (17)(b) and (15)d mply 3a = 3a 2, whch, along wth (16), (17)a and (17)b, mples the theorem wth h = 6. Hence we may assume (12)b holds. Furthermore, (17)(a), (17)(b) and (1)c mply 3a 3 = 3a ; and (17)a and (1)d mply 3a = 3a 5. Thus from (17)(a), (17)(b) and 3a 3 = 3a = 3a 5, t follows that the theorem holds wth h = 6. Case 2: (11)b holds. Then (11)b and (10) mply 2a 3 = 2a 5 and 2a 1 = 2a 2, whch, along wth (1)a, mples a 3 = a, a contradcton, and whch, along wth (15)c, mples 3a 5 = 3a 2. In the later case, we obtan the three equaltes, 2a 3 = 2a 5, 2a 1 = 2a 2, 3a 5 = 3a 2, whch, along wth (16), mply the theorem wth h = 6. Case 3: (11)c holds. Then (11)c, (10) and (9) mply (18)a and (18)b. 2a 3 = a 2 + a 5, 2a 1 = a 2 + a (18) Suppose that (12)c holds. Then (18)a and (15)a mply 2a 3 = 2a ; and (18)b and (15)b mply 2a 5 = 2a 1. Furthermore, (15)b and (15)c mply 53a 5 + 2a 3 = 2a 1 + 2a 2 + a, whch, along wth (18)b and 2a 3 = 2a, mples 3a 2 = 3a 5. Thus 2a 3 = 2a, 2a 5 = 2a 1, 3a 2 = 3a 5 and (16) mply the theorem wth h = 6. So we may assume (12)b holds. Then (18)b and (1)b mply a 2 + a = a 3 + a 5 ; and we conclude n ths case that there are at most 5 dstnct 2-sums. Furthermore, a 2 +a = a 3 +a 5 and (9) mply 2a = a 5 + a 1. Thus (18)a, (18)b, (1)a, (13), and 2a = a 5 + a 1 mply (19)(a), (19)(b) and (19)(c). 3a + a 3 = 3a 5 + a 1, 3a 3 + a = 3a 5 + a 2, 3a 2 = 2a 1 + a 5 (19) We proceed by combnng (19)(a), (19)(b) and (19)(c) wth (18), (13), and 2a = a 5 + a 1, yeldng 20(a), 20(b) and 20(c). 5a = a 5 + 2a 1 a 3, 5a 3 = a 5 + 2a 2 a, 5a 2 = 2a 1 + 2a 5 + a (20) Next (20)a, (20)b and (1)c mply 5a = 5a 3 ; and (20)c and (1)d mply 5a 2 = 5a 5. Furthermore, (20)a, (1)a, and (1)d mply 5a = 5a 5. Therefore t follows that {a 2, a 3, a, a 5 } are four elements 12

13 from a coset α + H, where H s a subgroup of Z m of cardnalty 5. Then t can be easly verfed that a 1 s the ffth element of α + H, as otherwse 2 S > 5, contradctng the fact that there are at most fve dstnct 2-sums. Thus the theorem holds wth h = 5, completng the proof. Theorem.1. Let m 5. Then g(6, 5) = 8, and f m 6, then g(m, 5) = 2m 5. Proof. For m 6 the result follows from Theorem 2.8. Suppose S s zsf and S = 2m 5. We may w.l.o.g. assume that 0 has the greatest multplcty n S. Case 1: The multplcty of 0 n S s at most m 2. Applyng Conjecture.1 wth k = 5 to all possble 5-sets of ϕ(s) that nclude 0, we can ether fnd a 5-set A ϕ(s) such that 2 A = 3 A 7 and 0 A, or else there exsts a subgroup H of cardnalty h = 5 or h = 6 such that ϕ(s) H. In the latter case, m 10, and so from Theorem 2.2 t follows that any subsequence wth length m+h 1 2m 5 must contan an m-term zero-sum subsequence, a contradcton. So 3 A 7. In vew of the assumpton of the case and Proposton 2.1, there exsts an (m 3)-set partton P of S \ A wth m 7 sets of cardnalty two. Applyng Theorem 2.7 to S \ A, and usng 3 A 7, we obtan an m-term zero-sum subsequence of S, provded concluson () of Theorem 2.7 holds. Hence we are done for m 8. So assume that concluson () of Theorem 2.7 holds wth coset α + H of ndex a, and w.l.o.g assume α = 0. Let P be the (m 3)-set partton mpled by concluson () of Theorem 2.7. Applyng Proposton 2.2() followed by Proposton 2.2() to P we obtan an ( m a 1)-set partton P of a subsequence Q of S \ A of length at most 2 m a 2, whose sumset s also H. Then there exsts a subsequence R of S \ A of length a 1 whose terms are from H and are not used n P. We can apply Theorem 2.1 to a subsequence of S \ {Q R} of length m m a + 1 wth ts terms consdered as elements from Z m /H to obtan a subsequence T of S \ {Q R} whose sum s an element of H and of length r, where r satsfes m m a a + 2 r m m a + 1. Snce the sumset of P s H, we can fnd m a 1 terms from P whch along wth T and an approprate number of terms from R gves an m-term subsequence wth sum zero. Case 2: The multplcty of 0 n S s m 1. Let T be a subsequence of S that conssts of dstnct nonzero resdue classes and 3 zeros. In vew of Proposton 2.1, t follows that there exsts an (m )-set partton P of S \ T wth m 8 cardnalty two sets. Applyng Theorem 2.7 to P and Theorem 2.9 to ϕ(t ) \ {0}, we fnd an m-term zero-sum subsequence provded concluson () of Theorem 2.7 holds. If concluson () of Theorem 2.7 holds nstead, then snce m > a 2 13

14 mples 0 α + H, the arguments from the end of Case 1 complete the proof. References [1] N. Alon and M. Dubner, Zero-sum sets of prescrbed sze, Combnatorcs, Paul Erdos s eghty, Vol. 1, Bolya Soc. Math. Stud., János Bolya Math. Soc., Budapest, 1993, [2] N. Alon, M. B. Nathanson and I. Ruzsa, The polynomal method and restrcted sums of congruence classes, J. Number Theory 56 (1996), no. 2, [3] A. Balostock and P. Derker, On the Erdős-Gnzburg-Zv theorem and the Ramsey numbers for stars and matchngs, Dscrete Math. 110 (1992), no. 1 3, 1 8. [] A. Balostock and M. Lotspech, Developments of the Erdős-Gnzburg-Zv Theorem I, Sets, graphs and numbers, Budapest, 1991, [5] W. Brakemeer, Ene Anzahlformel von zahlen modulo n, Monatsh. Math. 85 (1978), no., [6] H. Cao and Z. Sun, On sums of dstnct representatves, Acta Arth. 87 (1998), no. 2, [7] Y. Caro, Remarks on a zero-sum theorem, J. Combn. Theory Ser. A 76 (1996), no. 2, [8] H. Davenport, On the addton of resdue classes, J. London Math. Socety 10 (1935), [9] J. A. Das da Slva and Y. O. Hamdoune, A note on the mnmal polynomal of the Kronecker sum of two lnear operators, Lnear Algebra Appl. 11 (1990), [10] R. B. Eggleton and P. Erdős, Two combnatoral problems n group theory, Acta Arthmetca 21 (1972), [11] P. Erdős, A. Gnzburg and A. Zv, Theorem n addtve number theory, Bull. Research Councl Israel 10 F (1961), 1 3. [12] P. Erdős and R. L. Graham, Old and new results n combnatoral number theory, Monographe 28 de L Ensegnement Mathematque, Geneve,

15 [13] C. Flores and O. Ordaz, On the Erdős-Gnzburg-Zv theorem, Dscrete Math. 152 (1996), no. 1-3, [1] L. Gallardo, G. Grekos, L. Habseger, F. Hennecart, B. Landreau and A. Plagne, Restrcted addton n Z/n and an applcaton to the Erdős-Gnzburg-Zv problem, J. London Math. Soc. (2) 65 (2002), no. 3, [15] L. Gallardo and G. Grekos, On Brakemeer s varant of the Erdős-Gnzburg-Zv problem, Number theory (Lptovský Ján, 1999), Tatra Mt. Math. Publ. 20 (2000), [16] L. Gallardo, G. Grekos and J. Phko, On a varant of the Erdős-Gnzburg-Zv problem, Acta Arth. 89 (1999), no., [17] W. D. Gao and Y. O. Hamdoune, Zero sums n abelan groups, Combn. Probab. Comput. 7 (1998), no. 3, [18] W. D. Gao, An addton theorem for fnte cyclc groups, Dscrete Math. 163 (1997), no. 1-3, [19] W. D. Gao, Addton theorems for fnte abelan groups, J. Number Theory 53 (1995), no. 2, [20] D. Grynkewcz, On a partton analog of the Cauchy-Davenport Theorem. Preprnt. [21] D. Grynkewcz and R. Sabar, Monochromatc and zero-sum sets of nondecreasng modfeddameter. Preprnt. [22] Y. O. Hamdoune, A. S. Lladó and O. Serra, On restrcted sums, Combn. Probab. Comput. 9 (2000), no. 6, [23] Y. O. Hamdoune, O. Ordaz and A. Ortuño, On a combnatoral theorem of Erdős, Gnzburg and Zv, Combn. Probab. Comput. 7 (1998), no., [2] J. H. B. Kemperman, On small sumsets n an abelan group, Acta Math. 103 (1960), [25] V. F. Lev, Restrcted set addton n groups. I. The classcal settng, J. London Math. Soc. (2) 62 (2000), no. 1,

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17 Appendx In the followng table, Theorem 3.1 s used to lst the values of E(m, s) for all m and s satsfyng 2m 2 s max{2m 8, 2m m 2}. Table 1: m s E(m, s) m 2 2m 2 0 m 1 1 m 1 m 2m 3 0 m 1 1 m 3 2 m 8 2m 0 m 1 1 m ( 1)0 m 3 1 m m 1 1 m 3 m 12 2m 5 0 m 1 1 m ( 1)0 m 3 1 m m 1 1 m 6 23 ( 1)0 m 3 1 m 3 0 m 1 1 m 5 0 m 1 1 m 9 2 ( 1)0 m 3 1 m ( 1) 2 0 m 5 1 m m 1 1 m m 16 2m 6 ( 1)0 m 3 1 m 6 23 ( 2)0 m 1 m m 1 1 m m 1 1 m ( 1)0 m 3 1 m 5 ( 2)0 m 1 m 3 0 m 1 1 m m 1 1 m ( 1)0 m 3 1 m 9 2 ( 1) 2 0 m 5 1 m m 1 1 m ( 1)0 m 3 1 m ( 2)0 m 1 m ( 1) 2 0 m 5 1 m m 1 1 m m 20 2m 7 0 m 1 1 m ( 1)0 m 3 1 m 7 2 ( 1)0 m 3 1 m ( 2)0 m 1 m 6 23 ( 1) 2 0 m 5 1 m 5 0 m 1 1 m m 1 1 m 8 3 ( 1)0 m 3 1 m 6 5 ( 2)0 m 1 m 5 0 m 1 1 m m 1 1 m ( 1)0 m 3 1 m ( 1) 2 0 m 5 1 m 9 2 ( 1) 3 0 m 7 1 m m 1 1 m ( 1)0 m 3 1 m ( 2)0 m 1 m 9 2 ( 1) 2 0 m 5 1 m ( 2)( 1)0 m 6 1 m m 1 1 m m 1 1 m ( 1)0 m 3 1 m ( 1)0 m 3 1 m ( 2)0 m 1 m ( 1) 2 0 m 5 1 m 7 2 ( 1) 2 0 m 5 1 m m 2 2m 8 ( 3)0 m 5 1 m ( 2)( 1)0 m 6 1 m m 1 1 m m 1 1 m m 1 1 m ( 1)0 m 3 1 m 8 25 ( 1)0 m 3 1 m 8 3 ( 2)0 m 1 m 7 2 ( 2)0 m 1 m ( 1) 2 0 m 5 1 m 6 5 ( 3)0 m 5 1 m m 1 1 m m 1 1 m m 1 1 m 9 2 ( 1)0 m 3 1 m 7 6 ( 2)0 m 1 m 6 5 ( 3)0 m 5 16 m 5 0 m 1 1 m

18 Are Balostock Department of Mathematcs Unversty of Idaho Moscow, ID 838, USA Paul Derker Department of Mathematcs Unversty of Idaho Moscow, ID 838, USA Davd Grynkewcz Mathematcs Caltech Pasadena, CA 91125, USA Mark Lotspech Department of Mathematcs Albertson College Caldwell, ID 83605, USA 18