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

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

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 2 + 1. 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

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 1 3 2 ). 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

(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}. 3.1. 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:

() 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

() 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

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 2 + 1 < 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

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 + 1 + 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+3 8 + 1 or k s even and m k2 +2k 8 + 1, then g (m, k) 2m k2 2k+5. 8

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 2 1 8 (1) m k2 +k+3 8 (2)(3)... ( k + 1 2 ). 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 2 + 3 +... + (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 2 2 + 2w 2 m u 1 w 2. (8) Inequaltes (7) and (8) mply c 2 1 c 1 2 + c2 2 c 2 2 m v 1 w 1 + m u 1 w 2 = l 2, 9

whch, n turn, yelds l (c 1 + c 2 ) 2 + c 1 + c 2 2 + 2 (k 2)2 + k 2 2 + 2 = k2 2k + + 1 > 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

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

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

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

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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 5 2 2 ( 1)0 m 3 1 m 3 2 0 m 1 1 m 3 m 12 2m 5 0 m 1 1 m 7 2 3 ( 1)0 m 3 1 m 5 2 2 0 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 7 2 3 ( 1) 2 0 m 5 1 m 5 2 2 0 m 1 1 m 8 2 2 3 m 16 2m 6 ( 1)0 m 3 1 m 6 23 ( 2)0 m 1 m 5 2 2 0 m 1 1 m 7 2 0 m 1 1 m 7 3 2 ( 1)0 m 3 1 m 5 ( 2)0 m 1 m 3 0 m 1 1 m 6 5 0 m 1 1 m 11 2 5 ( 1)0 m 3 1 m 9 2 ( 1) 2 0 m 5 1 m 7 2 3 0 m 1 1 m 10 2 3 3 ( 1)0 m 3 1 m 8 2 2 3 ( 2)0 m 1 m 7 2 3 ( 1) 2 0 m 5 1 m 6 23 0 m 1 1 m 9 23 2 m 20 2m 7 0 m 1 1 m 9 2 2 ( 1)0 m 3 1 m 7 2 ( 1)0 m 3 1 m 7 3 2 ( 2)0 m 1 m 6 23 ( 1) 2 0 m 5 1 m 5 0 m 1 1 m 8 25 0 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 7 6 0 m 1 1 m 13 2 6 ( 1)0 m 3 1 m 11 2 5 ( 1) 2 0 m 5 1 m 9 2 ( 1) 3 0 m 7 1 m 7 2 3 0 m 1 1 m 12 2 3 ( 1)0 m 3 1 m 10 2 3 3 ( 2)0 m 1 m 9 2 ( 1) 2 0 m 5 1 m 8 2 2 3 ( 2)( 1)0 m 6 1 m 7 2 3 0 m 1 1 m 11 2 3 0 m 1 1 m 11 2 2 3 2 ( 1)0 m 3 1 m 9 2 2 ( 1)0 m 3 1 m 9 23 2 ( 2)0 m 1 m 8 2 2 3 ( 1) 2 0 m 5 1 m 7 2 ( 1) 2 0 m 5 1 m 7 3 2 m 2 2m 8 ( 3)0 m 5 1 m 7 2 3 ( 2)( 1)0 m 6 1 m 6 23 0 m 1 1 m 10 2 2 5 0 m 1 1 m 10 23 0 m 1 1 m 10 3 3 ( 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 7 3 2 ( 1) 2 0 m 5 1 m 6 5 ( 3)0 m 5 1 m 6 23 0 m 1 1 m 9 26 0 m 1 1 m 9 35 0 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 8 7 17

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 253-57 Caltech Pasadena, CA 91125, USA Mark Lotspech Department of Mathematcs Albertson College Caldwell, ID 83605, USA 18