On short zero-sum subsequences of zero-sum sequences

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1 On short zero-sum subsequences of zero-sum sequences Yushuang Fan Center for Combinatorics Nankai University, LPMC-TJKLC Tianjin, P. R. China Guoqing Wang Department of Mathematics Tianjin Polytechnic University Tianjin, P. R. China Weidong Gao Center for Combinatorics Nankai University, LPMC-TJKLC Tianjin, P. R. China Qinghai Zhong Center for Combinatorics Nankai University, LPMC-TJKLC Tianjin, P. R. China Jujuan Zhuang Department of Mathematics Dalian Maritime University Dalian, P. R. China Submitted: Aug 4, ; Accepted: Aug 3, ; Published: Sep 6, Mathematics Subject Classifications: B75, P99 Abstract Let G be a finite abelian group of exponent exp(g). By D(G) we denote the smallest integer d N such that every sequence over G of length at least d contains a nonempty zero-sum subsequence. By η(g) we denote the smallest integer d N such that every sequence over G of length at least d contains a zero-sum subsequence T with length T [, exp(g)], such a sequence T will be called a short zero-sum sequence. Let C (G) denote the set consists of all integer t [D(G) +, η(g) ] such that every zero-sum sequence of length exactly t contains a short zero-sum subsequence. In this paper, we investigate the question whether C (G) for all non-cyclic finite abelian groups G. Previous results showed that C (G) for the groups Cn (n 3) and C3 3. We show that more groups including the groups C m C n with 3 m n, C 3, C 3 3 a 5 b 3 a, C4 3 a and Cr (b ) have this property. We b Corresponding author: Guoqing Wang the electronic journal of combinatorics 9(3) (), #P3

2 also determine C (G) completely for some groups including the groups of rank two, and some special groups with large exponent. Keywords: Zero-sum sequence; short zero-sum sequence; short free sequence; zerosum short free sequence; Davenport constant Introduction Let G be an additive finite abelian group of exponent exp(g). We call a zero-sum sequence S over G a short zero-sum sequence if S exp(g). Let η(g) be the smallest integer d such that every sequence S over G of length S d contains a short zero-sum subsequence. Let D(G) be the Davenport constant of G, i.e., the smallest integer d such that every sequence over G of length at least d contains a nonempty zero-sum subsequence. Both D(G) and η(g) are classical invariants in combinatorial number theory. For detail on terminology and notation we refer to Section. By the definition of η(g) we know that for every integer t [, η(g) ], there is a sequence S over G of length exactly t such that S contains no short zero-sum subsequence. In this paper, we consider the following problem related to D(G) and η(g), which was first investigated by Emde Boas in the late sixties. Given a finite abelian group, what are integers exp(g) + t η(g), if any, such that every zero-sum sequence S over G of length S = t contains a short zero-sum subsequence. Denote by C (G) the set of all those integers t. It will be readily seen that C (G) [D(G) +, η(g) ]. In 969, Emde Boas and D. Kruyswijk [7] proved that 4 C (C3). 3 In 997, the second author of this paper showed that [q, 3q 3] C (Cq ), where q is a prime power. Let us first make some easy observations on C (G). Note that for every t [, D(G)] there exists a minimal zero-sum sequence over G of length t. So, C (G) [D(G) +, η(g) ] follows from the easy fact that D(G) exp(g). If G = C C m then D(G) + = m + and η(g) = m +. Therefore, by the definition we have C (C C m ) =. We suggest the following Conjecture. Let G be a non-cyclic finite abelian group. If G C C m then C (G). In this paper we shall determine C (G) completely for some groups. Theorem. Let G be a non-cyclic finite abelian group, and let r(g) be the rank of G. Then,. C (G) = [D(G) +, η(g) ] if r(g) =.. C (G) = [D(G) +, η(g) ] if G = C mp n H with p a prime, H a p-group and p n D(H). 3. C (C 4 3) = {η(c 4 3), η(c 4 3) } = {37, 38}. the electronic journal of combinatorics 9(3) (), #P3

3 4. C (C r ) = {η(c r ) 3, η(c r ) }, where r 3. We also confirm Conjecture for more groups other than those listed in Theorem. Theorem 3. If G is one of the following groups then C (G).. G = C 3 3 a 5 b where a or b.. G = C3 3 a where a G = C3 4 a where a. 4. G = C r a where 3 r a, or a = and r G = C 3 k where k = 3n 5 n 7 n 3 n 4 3 n 5, n, n 3 + n 4 + n 5 3, and n + n + 34(n 3 + n 4 + n 5 ). The rest of this paper is organized as follows: In Section we introduce some notations and some preliminary results; In Section 3 we prove three lemmas connecting C (G) with property C; In Section 4 we shall derive some lower bounds on min{c (G)}; In Section 5 we study C (G) with focus on the groups C r 3; In Section 6 and 7 we shall prove Theorem and Theorem 3, respectively; and in the final Section 8 we give some concluding remarks and some open problems. Notations and some preliminary results Our notations and terminologies are consistent with [] and [3]. We briefly gather some key notions and fix the notations concerning sequences over finite abelian groups. Let Z denote the set of integers. Let N denote the set of positive integers, and N = N {}. For real numbers a, b R, we set [a, b] = {x Z : a x b}. Throughout this paper, all abelian groups will be written additively, and for n, r N, we denote by C n a cyclic group with n elements, and denote by C r n the direct sum of r copies of C n. Let G be a finite abelian group and exp(g) its exponent. By r(g) we denote the rank of G. A sequence S over G will be written in the form S = g... g l = g G g vg(s), with v g (S) N for all g G, and we call S = l N the length and σ(s) = l g i = v g (S)g G the sum of S. g G i= Let supp(s) = {g G : v g (S) > }. We call S a square free sequence if v g (S) for every g G. So, a square free sequence over G is actually a subset of G. A sequence T the electronic journal of combinatorics 9(3) (), #P3 3

4 over G is called a subsequence of S if v g (T ) v g (S) for every g G, and denote by T S. For every r [, l], define (S) = {σ(t ) : T S, T r} r and define (S) = {σ(t ) : T S, T }. The sequence S is called a zero-sum sequence if σ(s) =. a short zero-sum sequence over G if it is a zero-sum sequence of length S [, exp(g)]. a short free sequence over G if S contains no short zero-sum subsequence. So, a zero-sum sequence over G which contains no short zero-sum subsequence will be called a zero-sum short free sequence over G. For every element g G, we set g + S = (g + g )... (g + g l ). Every map of abelian groups ϕ : G H extents to a map from the sequences over G to the sequences over H by ϕ(s) = ϕ(g )... ϕ(g l ). If ϕ is a homomorphism, then ϕ(s) is a zero-sum sequence if and only if σ(s) ker(ϕ). In the rest of this section we gather some known results which will be used in the sequel. We shall study C (G) by using the following property which was first introduced and investigated by Emde Boas and Kruyswijk [7] in 969 for the groups C p with p a prime, and was investigated in 7 for the groups C r n by the second author, Geroldinger and Schmid []. Property C: We say the group Cn r has property C if η(cn) r = c(n ) + for some positive integer c, and every short free sequence S over Cn r of length S = c(n ) has the form S = c i= gn i where g,..., g c are pairwise distinct elements of Cn. r It is conjectured that every group of the form C r n has Property C(see [], Section 7). We need the following result which states that Property C is multiple. Lemma 4. ([], Theorem 3.) Let G = C r mn with m, n, r N. If both C r m and C r n have Property C and η(c r m) m for some c N then G has Property C. = η(cr n) n We also need the following old easy result. = η(cr mn) mn = c the electronic journal of combinatorics 9(3) (), #P3 4

5 Lemma 5. ([]) D(C 3 n) 3n. Definition 6. Let G be a finite abelian group. Define g(g) to be the smallest integer t such that every square free sequence over G of length t contains a zero-sum subsequence of length exp(g). Let f(g) be the smallest integer t such that every square free sequence over G of length t contains a short zero-sum subsequence. We now gather some known results on Property C, η(g), g(g) and f(g) which will be used in the sequel. Lemma 7. Let r, t N, and let n 3 be an odd integer. Then,. η(c 3 n) 8n 7. ([6], or [5], Theorem.). η(c 4 n) 9n 8. ([5], Theorem.3) 3. η(c 3 3) = 7. ([9], or [6], page 3) 4. η(c 4 3) = 39 and g(c 4 3) =. ([9], or [6], page 3) 5. η(c 3 5) = 33 = ([], Theorem.7) 6. η(c r t ) = ( r )( t ) +. ([8]) 7. η(c 3 3 α) = 7(3 α ) + where α. ([], Theorem.8) 8. C 3 5 has Property C. ([], Theorem.9) 9. η(c r 3) = f(c r 3). ([8]). C r 3 has Property C. ([8]) Lemma 8. ([5], Lemma 5.4) Let r [3, 5], and let S and S be two square free sequences over C3 r of length S = S = g(c3) r. Suppose that both S and S contain no zero-sum subsequence of length 3. Then S = ϕ(s) + a, where ϕ is an automorphism of C3 r and a C3. r Lemma 9. ([], Lemma ) Let T be a square free sequence over C3 3 of length 8. If T contains no short zero-sum subsequence then there exists an automorphism ϕ of C3 3 such that ϕ(t ) =. the electronic journal of combinatorics 9(3) (), #P3 5

6 Lemma. ([3]; [5], page 8) The following square free sequence over C 4 3 of length contains no zero-sum subsequence of length 3.. Lemma. ([5], Theorem 5.) Every sequence S over C n of length S = 3n contains a zero-sum subsequence of length n or n. Lemma. ([], Lemma 4.5) Let G be a finite abelian group, and let H be a proper subgroup of G with exp(g) = exp(h) exp(g/h). Then η(g) (η(h) ) exp(g/h) + η(g/h). Lemma 3. Let p be a prime and let H be a finite abelian p-group such that p n D(H). Let n, n, m, n N with n n. Then,. D(C n C n ) = n + n. ([]). D(C mp n H) = mp n + D(H). ([7]) 3. Let G = C p e C p er with e i N. Then, D(G) = + r i= (pe i ). ([]) 4. η(c n C n ) = n + n. ([4]) 5. Let G = H C n with exp(h) n. Then, η(g) (D(H) ) + n. ([5]) We also need the following easy lemma Lemma 4. ([6] Lemma 4..) Let G be a finite abelian group. Then, s(g) η(g) + exp(g). We shall show that the following property can also be used to study C (G). Property D : Let c, n N. We say that C r n has property D with respect to c if every sequence of the form g c i= gn i contains a zero-sum subsequence of length exactly n, where g, g i C r n for all i [, c]. Lemma 5. ([8], page 8) Let m = 3 a 5 b with a, b nonnegative integers. Let n 65 be an odd positive integer such that C 3 p has Property D with respect to 9 for all prime divisors p of n. If m 57 n 7 (n 7)n 64 then s(c 3 mn) = 9mn 8. the electronic journal of combinatorics 9(3) (), #P3 6

7 3 Three lemmas connecting C (G) with Property C Lemma 6. Let G = C r n with η(g) = c(n ) + for some c N. If c n and if G has Property C then η(g) C (G). Proof. Let S be a zero-sum sequence over G of length S = η(g) = c(n ). We need to show that S contains a short zero-sum subsequence. If S = c i= gn i for some g i G, then (n )(g + g + + g c ) = σ(s) = = n(g + g + + g c ). It follows that g + g + + g c =. Therefore, g g... g c is a zero-sum subsequence of S of length c n and we are done. Otherwise, S c i= gn i for any g i G. It follows from G having Property C that S contains a short zero-sum subsequence. Lemma 7. Let G be a finite abelian group, and let H be a proper subgroup of G with exp(g) = exp(h) exp(g/h). Suppose that the following conditions hold. (i) η(g) = (η(h) ) exp(g/h) + η(g/h); (ii) G/H = C r n has Property C; (iii) There exist t [, exp(g/h) ] and t {, } such that t t and such that [η(g/h) t, η(g/h) t ] C (G/H). Then, [η(g) t, η(g) t ] C (G). Proof. To prove this lemma, we assume to the contrary that there is a zero-sum short free sequence S over G of length η(g) t for some t [t, t ]. Let ϕ be the natural homomorphism from G onto G/H. Note that S = η(g) t = (η(h) ) exp(g/h) + (η(g/h) t). (3.) This allows us to take an arbitrary decomposition of S η(h) S = S (3.) with and for every i [, η(h) ]. Combining (3.), (3.), (3.3) and (3.4) we infer that i= S i S i [, exp(g/h)] (3.3) σ(s i ) ker(ϕ) = H (3.4) S η(g/h) t η(g/h) t (3.5) and σ ( ϕ(s ) ) =. (3.6) the electronic journal of combinatorics 9(3) (), #P3 7

8 Claim. ϕ(s ) contains no zero-sum subsequence of length in [, exp(g/h)]. Proof of the claim. Assume to the contrary that, there exists a subsequence S η(h) (say) of S of length S η(h) [, exp(g/h)] such that σ(s η(h) ) ker(ϕ) = H. It follows that the η(h) sequence U = σ(s i ) contains a zero-sum subsequence W = i I σ(s i) over H with i= I [, η(h)] and W = I [, exp(h)]. Therefore, the sequence i I S i is a zero-sum subsequence of S over G with i I S i I exp(g/h) exp(h) exp(g/h) = exp(g), a contradiction. This proves the claim. By (3.5), (3.6), the above claim and Condition (iii), we conclude that and t = This together with Condition (ii) implies that S = η(g/h). (3.7) ϕ(s ) = x n... x n c (3.8) where c = η(g/h) and x n,..., x c are pairwise distinct elements of the quotient group G/H. So, we just proved that every decomposition of S satisfying conditions (3.3) and (3.4) has the properties (3.5)-(3.8). Since t t exp(g/h), it follows from (3.), (3.3) and (3.7) that S i [, exp(g/h)] for all i [, η(h) ]. Moreover, since t t =, it follows that there exists j [, η(h) ] such that S j exp(g/h). Without loss of generality we assume that S [, exp(g/h) ]. Suppose that there exists h supp(ϕ(s )) supp(ϕ(s )). By (3.8), we have that the sequence S S contains a subsequence ( S ) with ϕ(s ) = h n. Let S = S S S. We get a decomposition S = S η(h) S i S satisfying (3.3) and (3.4). But S = i= S + S S (n ) + (η(g/h) ) n = η(g/h), a contradiction on (3.7). Therefore, supp(ϕ(s )) supp(ϕ(s )) =. Take a term g S. Since ϕ(g) / supp(ϕ(s )) and S g = η(g/h), it follows from the above claim that S g contains a subsequence S with g S (3.9) and and S exp(g/h) (3.) σ(s ) ker(ϕ). (3.) the electronic journal of combinatorics 9(3) (), #P3 8

9 Let S = S S S. By (3.8), (3.9), (3.) and (3.), we conclude that supp(ϕ(s )) c +, a contradiction with (3.8). This proves the lemma. From Lemma 7, we immediately obtain the following Lemma 8. Let r N, and let G = Cn r, G = Cn r and G = Cn r n. Suppose that the following conditions hold. η(g (i) ) n = η(g ) n = η(g) n n = c for some c N; (ii) G has Property C; (iii) There exist t [, n ], t {, } such that t t and such that [η(g ) t, η(g ) t ] C (G ). Then, [η(g) t, η(g) t ] C (G). 4 Some lower bounds on min{c (G)} In this section we shall prove the following Proposition 9. Let G = Cn r with n 3, r 3, and let α r r (mod n) with α r [, n ]. Then,. C (G) [( r )(n ) α r +, η(g) ] if α r.. C (G) {( r )(n ) n, ( r )(n ) n + } if α r =. Note that α r if and only if n k, or n = k and r < k; and α r = if and only if n = k and k r. For every r N, let G = C r n =< e > < e r > with < e i >= C n for every i [, r], and let S r = ( ) n e i. =I [,r] We can regard Cn r as a subgroup of Cn r+ and therefore S r+ has the following decomposition S r+ = S r (S r + e r+ )e n r+. Since the proof of Proposition 9 is somewhat long, we split the proof into lemmas begin with the following easy one Lemma. S r is a short free sequence over C r n of length S r = ( r )(n ) and of sum σ(s r ) = r (e + + e r ) = α r (e + + e r ). Proof. Obviously. i I the electronic journal of combinatorics 9(3) (), #P3 9

10 Lemma. Let G = C r n with r. Then for every m [, n ] and every i [, r], the sequence S r (e m i ) (me i ) contains no short zero-sum subsequence. Proof. Without loss of generality, we assume that i = r. Assume to the contrary that S r (e m r ) (me r ) contains a short zero-sum subsequence U. Since S r contains no short zero-sum subsequence we infer that me r U. Therefore, U = (me r )U (U + e r )e k r with U S r and U S r and k [, n m]. It follows that U U is zero-sum and U U n. Since every element in supp(s r ) occurs n times in S r, it follows from U U n that U U S r. Therefore, U U is a short zero-sum subsequence of S r, a contradiction with Lemma. Let A be a set of zero-sum sequences over G. Define L(A) = { T : T A}. In this section below we shall frequently use the following easy observation. Lemma. Let G be a finite abelian group, and let a, b N with a b. If there exists a set A of zero-sum short free sequences over G such that [a, b] L(A), then C (G) [a, b] =. Proof. It immediately follows from the definition of C (G). Lemma 3. Let G = C r n with n, r 3. Then,. C (G) [ S r (3n 3) α r, S r α r ] = if α r.. C (G) [ S r (3n 3), S r (n + )] = if α r =. Proof. Recall that S r = ( r )(n ). We split the proof into three steps. Step. In this step we shall prove that no matter α r = or not. Let C (G) [ S r (3n 3) α r, S r (n + ) α r ] = A = { S r ( (e + + e r ) αr W e m 3 ) (me3 ) : W S, σ(w ) =, m [, n ] }. It follows from Lemma that every sequence in A is zero-sum short free. Since L({W : W S, σ(w ) = }) = [n +, n ], we conclude easily that L(A) = [ S r (3n 3) α r, S r (n + ) α r ]. Now the result follows from Lemma and Conclusion follows. Step. We show that C (G) [ S r (n + α r ), S r (r )α r ] = for α r. the electronic journal of combinatorics 9(3) (), #P3

11 Let A = { ( ) S r (e + e ) αr e αr 3... e αr r e m (me ) : m [, n ] } and A = { S r ( (e + e ) αr (e + e 3 )e αr 3 e αr 4... e αr r ) e n }. It is easy to see that every sequence in A A is zero-sum short free by Lemma and Lemma. Note that L(A ) L(A ) = [ S r (r )α r n +, S r (r )α r ] { S r (r )α r n + } = [ S r (r )α r n +, S r (r )α r ]. Since r 3, we have S r (r )α r n + S r (n + α r ). Therefore, L(A A ) = L(A ) L(A ) [ S r (n + α r ), S r (r )α r ]. Again the result follows from Lemma. Step 3. We prove C (G) [ S r (r )α r, S r α r ] = for α r. Let A = { S r ( (e + + e r ) k (e... e r ) k (e + + e k3 )e k e r ) : k [, α r ], k [, α r ], k + k = α r, k 3 [, r] }. Then every sequence in A is zero-sum short free by Lemma and by Lemma, and L(A) = { S r k rk (r k 3 ) : k + k = α r, k [, α r ], k 3 [, r]} = { S r α r ((r )k + (r k 3 )) : k [, α r ], k 3 [, r]} = [ S r rα r, S r α r ]. Now the result follows from Lemma and the proof is completed. Lemma 4. Let n, r N with n 3 and r 3, and let G = Cn. r If α r then C (G) [( r )(n ) α r +, η(g) ]. Proof. It suffices to show that C (G) [n +, S r α r ] =. We proceed by induction on r. Suppose first that r = 3. By Lemma 3 and the definition of C (C 3 n), we only need to prove C (G) [D(C 3 n) +, S 3 (3n 3) α 3 ] =. By Lemma 5 we have D(C 3 n) + 3n. So, it suffices to prove that C (G) [3n, S 3 (3n 3) α 3 ] = C (G) [3n, 4n 4 α 3 ] =. the electronic journal of combinatorics 9(3) (), #P3

12 If n = 3, then [3n, 4n 4 α 3 ] = and the result follows. Now assume n 4. It follows from α 3 that n 5. Thus, α 3 = n 4 and [3n, 4n 4 α 3 ] = {3n }. Let T = (e + e ) (e + e 3 ) n e n e n e 3. Then T is zero-sum short free over Cn 3 of length T = 3n. Now the result follows from Lemma. This completes the proof for r = 3. Now assume that r 4. By the induction hypothesis there exists a set A r of zero-sum short free sequences over Cn r such that Recall that C r n C r n = C r n L(A r ) = [n +, S r α r ]. e r. Let A r = { W (W + e r )e l r : W A r, W A r, l [, n ], W + l (mod n) }. Then, every sequence in A r is zero-sum short free over C r n and L(A r ) = { W + W + l : W A r, W A r, l [, n ], W + l (mod n)} = { W + kn : W A r, k [, S r α r ] n [3n +, S r α r ]. It follows that Note that L(A r ) L(A r ) [n +, S r α r ]. Therefore, Now the result follows from Lemma 3. S r α r = S r (n ) α r S r 3(n ). L(A r ) L(A r ) [n +, S r 3(n )]. Lemma 5. Let n, r, k N with k, r k + and n = k, and let G = C r n. Then, C (G) {( r )(n ) n, ( r )(n ) n + }. Proof. Since r k + we have that α r =. By Lemma 7 we have S r = ( r )(n ) = η(g). So, it suffices to show that C (G) ([n+, η(g) (n+)] [η(g) n+, η(g) ]) =. Since r k + we have σ(s r ) =. the electronic journal of combinatorics 9(3) (), #P3

13 Step. We show C (G) [n +, S r (n + )] =. We proceed by induction on r. Suppose first that r = k +. If r = k + = 3, we only need to prove C (G) [3n, 4n 5] = by Lemma 3 and Lemma 5. Let A ={(e + e + e 3 )(e + e ) n (e + e 3 ) n m (e + e 3 )e m e n e m 3 : m [, n ]} {(e + e ) (e + e 3 ) n e n e n e 3 }. Then every sequence in A is zero-sum short free and L(A) = [3n, 4n 3] and we are done. If r = k + > 3, we have α r and r 3, then by Lemma 4 there exists a set A of zero-sum short free sequences over Cn r such that L(A) [n +, S r α r ]. Let Since B = A { W (W + e r )e l r : W A, W A, l [, n ], W + l (mod n) }. S r α r + S r α r + α r = S r 3n/, we have L(B) [n+, S r 3n/]. It follows from Lemma 3 that C (Cn) [n+, r S r (n + )] =. Now assume that r > k+. By the induction hypothesis, we conclude that there exists a set A of zero-sum short free sequences over Cn r such that L(A) [n+, S r (n+)]. Define a set B of zero-sum short free sequences over Cn r as follows B = { W (W + e r )e l r : W A, W A, l [, n ], W + l (mod n) }. It is easy to see that Let L(B) [ S r n, S r (n + )] = [ S r n, S r (3n + )]. C = {T : T S, σ(t ) = }; C = {(e + e 3 ) n m e m e n (e + e )e m 3 : m [, n ]}; C 3 = {(e + e ) (e + e 3 ) n e n e n e 3 }; C 4 = {(e + e + e 3 )(e + e ) n (e + e 3 ) n m (e + e 3 )e m e n e m 3 : m [, n ]}. Then every sequence in 4 i=c i is zero-sum short free. Clearly, L(C ) = [n +, n ]; L(C ) = [n, 3n ]; L(C 3 ) = {3n }; L(C 4 ) = [3n, 4n 3]. the electronic journal of combinatorics 9(3) (), #P3 3

14 Let Then, Let C = 4 i=c i. L(C) [n +, 4n 3]. D = {S r T : T C}. Then every sequence in D is zero-sum short free, and This completes the proof of Step. L(D) [ S r (4n 3), S r (n + )] [ S r 3n, S r (n + )]. Step. We prove C (G) [η(g) n +, η(g) ] =. Let A = {S r (e m r ) (me r ) : m [, n ]}. Then every sequence in A is zero-sum short free by Lemma, and This completes the proof. L(A) = [ S r n +, S r ] = [η(g) n +, η(g) ]. Proof of Proposition 9.. It is just Lemma 4.. Since α r =, we have n = k for some k [, r ], now the result follows from Lemma 5. 5 On the groups C r 3 In this section we shall study C (G) with focus on G = C r 3. Proposition 6. Let r, t N. Then,. C (C 3 3) [η(c 3 3) 4, η(c 3 3) ].. C (C5) 3 [η(c5) 3 5, η(c5) 3 ]. { [η(c 3. C (C r r ) t) ( t r ), η(c r ) ], if r t, t t [η(c r ) ( t + ), η(c r t ) t ], if r > t. t 4. C (C 3 6) {η(c 3 6), η(c 3 6) }. Proof. Conclusions, and 4 follow from Lemma 7 and Proposition 9. So, it remains to prove Conclusion 3. If r t then applying Proposition 9 with α r = t r, it follows from Conclusion 6 of Lemma 7 that C (C r t ) [( r )( t ) ( t r )+, η(c r t ) ] = [η(c r t ) ( t r ), η(c r t ) ]. If r > t then applying Proposition 9 with α r = we get, C (C r t ) [η(c r t ) ( t + ), η(c r t ) t ]. the electronic journal of combinatorics 9(3) (), #P3 4

15 Lemma 7. Let G = C r 3 with r 3, and let S be a sequence over G. Then,. If S is a short free sequence over G of length S = η(g), then (S) = C r 3 \ {}.. Let T be a square free and short free sequence over G, and let S = T. Then, for every g supp(s) we have, (S g ) = \ {g}. (S) 3. If every short free sequence of length η(g) has sum zero, then η(g) C (G). Proof. Conclusions and are obvious. To prove Conclusion 3, we assume to the contrary that η(g) / C (G), i.e., there exists a zero-sum short free sequence S over G of length S = η(g). By Lemma 7, we have η(g) = (f(g) ) +. This forces that S = g... g f(g) g f(g) for some distinct elements g,..., g f(g) with g... g f(g) contains no short zero-sum subsequence. Put T = S g f(g). Then T = η(g). But T contains no short zero-sum subsequence and σ(t ) = g f(g), a contradiction. Lemma 8. Every short free sequence over C 3 3 of length 6 has sum zero. Proof. Let S be an arbitrary short free sequence over C 3 3 of length S = 6. From Lemma 7 we obtain that S = T, where T is a square free and short free sequence over C 3 3 of length 8. It follows from Lemma 9 that σ(t ) =. Therefore, σ(s) = σ(t ) =. Lemma 9. The following two conclusions hold.. {4, 5} = {η(c 3 3) 3, η(c 3 3) } C (C 3 3).. {37, 38} = {η(c 4 3), η(c 4 3) } C (C 4 3). Proof.. The conclusion 4 C (C 3 3) is due to Emde Boas and D. Kruyswijk [7]. Now 5 C (C 3 3) follows from Conclusion 3 of Lemma 7, Lemma 7 and Lemma 8.. Denote by U the square free sequence over C3 4 given in Lemma. It follows from Conclusion 4 of Lemma 7 that U is a square free sequence of maximum length which contains no zero-sum subsequence of length 3. Choose an arbitrary square free sequence T over C3 4 of length f(c3) 4 such that T contains no short zero-sum subsequence. By Lemma 7, we have T = 9. Claim. σ(t ) / supp(t ) {}. Proof of the claim. Put S = T. It follows from Conclusion 4 of Lemma 7 that S is a square free sequence over C3 4 of maximum length which contains no zero-sum subsequence of length 3. By Lemma 8, there exists an automorphism ϕ of C3 4 and some g C3 4 such that S = ϕ(u g). Since S, it follows that g U. Thus, σ(t ) = σ(s) = σ(ϕ(u g)) = ( ) ϕ(σ(u g)) = ϕ(σ(u) g) = ϕ(σ(u) + g). It is easy to check that σ(u) =. Since ) σ(u) = / supp(u), it follows that σ(t ) = ϕ(σ(u) + g) = ϕ( σ(u) g) / ( ϕ(supp(u) g) = supp(s) = supp(t ) {}. This proves the claim. the electronic journal of combinatorics 9(3) (), #P3 5

16 From Conclusions 4, 9, of Lemma 7 and the above claim, we derive that every short free sequence over C3 4 of length η(c3) 4 = 38 has a nonzero sum. This is equivalent to that every zero-sum sequence over C3 4 of length η(c3) 4 contains a short zero-sum subsequence. Hence, 38 = η(c3) 4 C (C3). 4 Suppose that 37 = η(c3) 4 / C (G), that is, there exists a zero-sum short free sequence V over C3 4 of length V = η(c3) 4 = 37. Since v g (V ) for every g supp(v ), we have supp(v ) 9. On the other hand, by Conclusion 4 and 9 of Lemma 7, we can derive that supp(v ) f(c3) 4 = η(c4 3 ) = 9. Thus, V = W h, where h W and W is a square free and short free sequence over G of length f(c3) 4 = 9. It follows from σ(v ) = that σ(w ) = h supp(w ), a contradiction with the claim above. Proposition 3. Let G = C r 3 with r 3. If there is a short free sequence S over G of length S = η(g) such that σ(s), then. {η(g), η(g) 3} C (G).. {η(g) 3, η(g) 4} C (G). Proof.. Since σ(s), it follows from Lemma 7 that there exists a subsequence W of S of length W {, } such that σ(s) = σ(w ). Therefore, σ(s W ) =, S W {η(g) 3, η(g) } and S W contains no short zero-sum subsequence. Hence, η(g) / C (G) or η(g) 3 / C (G).. By Conclusion of Lemma 7, we have that S = T, where T is a square free sequence over G. Choose g supp(s) such that σ(s g ). Since σ(s g ) = σ(s) g g, it follows from Conclusion of Lemma 7 that σ(s g ) (S g ) = C r 3 \ {, g}. Similarly to Conclusion, we infer that η(g) 3 / C (G) or η(g) 4 / C (G). Proposition 3. C (C 4 3) = {37, 38}. Proof. By Proposition 9, we have C (C 4 3) [3, η(c 4 3) ] = [3, 38]. (5.) We show next that Put [3, 36] C (C 4 3) =. (5.) T = ; T 3 = ; the electronic journal of combinatorics 9(3) (), #P3 6

17 T 4 = ; T 5 = ; T 6 = ; T 7 = ; T 8 =. Let U be the square free sequence given in Lemma. Then σ(u) =. Let S = U. We see that S is a short free sequence of length 38 = η(c 4 3). By removing T i from S, we obtain that the resulting sequence S i is a zero-sum short free sequence of length η(g) i = 38 i. This proves (5.). Combining (5.), (5.) and Lemma 9, we conclude that C (C 4 3) = {η(g), η(g) } = {37, 38}. 6 Proof of Theorem In this section we shall prove Theorem and we need the following lemma. Lemma 3. Let p be a prime and let H be a finite abelian p-group such that p n D(H). Then,. Every sequence S over C p n H of length S = p n + D(H) contains a zero-sum subsequence T of length T {p n, p n }.. η(c mp n H) mp n + p n + D(H). the electronic journal of combinatorics 9(3) (), #P3 7

18 Proof.. Let S = g... g l be a sequence over G = C p n ( ) H of length l = S = p n +D(H). Let α i = C g p n C p n H with C p n. By Conclusion of Lemma i 3, α... α l is a sequence over C p n G of length l = p n + p n + D(H) = D(C p n G). Therefore, α... α l contains a nonempty zero-sum subsequence W (say). By the making of α i we infer that W = p n or W = p n. Let T be the subsequence of S which corresponds to W. Then T is a zero-sum subsequence of S of length T {p n, p n }.. We first consider the case that m =. Let G = C p n H. We want to prove that η(g) p n + D(H). Let S = g... g l be a sequence over G = C p n H of length l = S = p n +D(H). We need to show that S contains a short zero-sum subsequence. It follows from Conclusion that S contains a zero-sum subsequence T of length T {p n, p n }. If T = p n then T itself is a short zero-sum sequence over G and we are done. Otherwise, since p n D(H), it follows from Conclusion 3 of Lemma 3 that T = p n > p n + D(H) = D(G). Therefore, T contains a nonempty proper zero-sum subsequence T. Now either T or T T is a short zero-sum subsequence of S. This proves that η(c p n H) p n +D(H). By Lemma, we have η(c mp n H) (η(c m ) ) exp(c p n H) + η(c p n H) (m )p n + p n + D(H) = mp n + p n + D(H). Lemma 33. Let G be a finite abelian group. Then [D(G) +, min{ exp(g) +, η(g) }] C (G). Proof. If [D(G)+, min{ exp(g)+, η(g) }] = then the conclusion of this lemma hold true trivially. Now assume that [D(G) +, min{ exp(g) +, η(g) }]. Let S be an arbitrary zero-sum sequence over G of length S [D(G)+, min{ exp(g)+, η(g) }]. It suffices to show that S contains a short zero-sum subsequence. Since S D(G) +, it follows that S contains a zero-sum subsequence T of length T [, S ]. Then σ(st ) =. Since S exp(g) +, we infer that T [, exp(g)] or ST [, exp(g)]. This proves the lemma. Proof of Theorem,. By the definition of C (G) we have, C (G) [D(G)+, η(g) ]. So, we need to show [D(G) +, η(g) ] C (G). Suppose first that G = C n C n. By Conclusions and 4 of Lemma 3, we have D(G) = n and η(g) = 3n. Let S be a zero-sum sequence over G of length S [n, 3n 3]. We need to show S contains a short zero-sum subsequence. We may assume that v (S) =. the electronic journal of combinatorics 9(3) (), #P3 8

19 Let T = S 3n S. Then T = 3n and T contains a zero-sum subsequence T of length T {n, n} by Lemma. If T = n then T v (T ) is a short zero-sum subsequence of S and we are done. So, we may assume that T = n. Let T = T T. Now T is a zero-sum subsequence of T of length T = n. If T contains at least one nonzero element then T v (T ) is a short zero-sum subsequence of S and we are done. So, we may assume that T = n. This forces that T = S. It follows from D(G) = n that S contains a zero-sum subsequence S of length S [, n ]. Therefore, either S or SS is a short zero-sum subsequence of S. Now suppose that G = C n C m with n m and n < m. By Conclusions and 4 of Lemma 3, we have that D(G) = n + m < m and m+ > n+m = η(g). It follows from Lemma 33 that [D(G)+, η(g) ] C (G).. By Conclusion of Lemma 3 and Conclusion of Lemma 3, we have that D(C mp n H) = mp n + D(H) and η(c mp n H) mp n + p n + D(H). Suppose m. Then η(c mp n H) mp n. Similarly to the proof of Conclusion, we can prove that [D(C mp n H) +, η(c mp n H) ] C (G), and we are done. So, we may assume m =. Then η(c p n H) p n +D(H) and the proof is similar to that of by using Conclusion of Lemma It is just Proposition Observe that g =. Then, any square free sequence S over C r with v (S) = g C r\{} and S { r 3, r } must have a nonzero sum. It follows from Conclusion 6 of Lemma 7 that {η(c) r 3, η(c) r } = { r 3, r } C (C). r So, C (C) r = { r 3, r } = {η(c) r 3, η(c) r } follows from Proposition 9. 7 Proof of Theorem 3 Lemma 34. If η(cr m) m η(cmn) r = c(mn ) +. = η(cr n) n = c for some c N and if η(c r mn) c(mn ) + then Proof. The lemma follows from Lemma. Lemma 35. C r has Property C. t Proof. It follows from Lemma 4 and Conclusion 6 of Lemma 7 by induction on t. Proposition 36. Let n = 3m, where m is an odd positive integer. Then,. If η(cm) 3 = 8m 7 then η(cn) 3 C (Cn). 3 the electronic journal of combinatorics 9(3) (), #P3 9

20 . If η(c 4 m) = 9m 8 then {η(c 4 n), η(c 4 n) } C (C 4 n). Proof.. By Conclusion 3 of Lemma 7 and Lemma, we have η(c 3 n) (η(c 3 3) ) m + η(c 3 m) = 6m + 8m 7 = 8n 7. Combined with Conclusion of Lemma 7, we have η(c 3 n) n = η(c3 m) m = η(c3 3) 3 = 8. (7.) Now we show η(c 3 n) C (C 3 n) by applying Lemma 8 with G = C 3 3 and t = t =. Conditions (i)-(iii) of Lemma 8 are verified by (7.), Conclusion of Lemma 7, and Conclusion of Lemma 9 respectively. We are done.. The proof is similar to that of Conclusion. Proposition 37. Let α, β N with α. Then,. If α + β then {η(c 3 3 α 5 β ), η(c 3 3 α 5 β ) } C (C 3 3 α 5 β ).. {η(c 4 3 α), η(c4 3 α) } C (C 4 3 α). Proof.. By Conclusions, 3 and 5 of Lemma 7 and Lemma 34, we conclude that η(c 3 3 s 5 t ) 3 s 5 t = 8 (7.) for every s, t N with s + t. Combined with Proposition 36, we have η(c 3 ) 3 α 5 β C (C 3 ). 3 α 5 β By Lemma 4, Conclusions 8, of Lemma 7 and (7.), we have C 3 has Property 3 α 5 β C. Since α + β, we have 8 < 3 α 5 β. Therefore, it follows from (7.) and Lemma 6 that η(c 3 ) C 3 α 5 β (C 3 ). We are done. 3 α 5 β. By Conclusion of Lemma 9, we need only to consider the case that α >. By Conclusions and 4 of Lemma 7 and Lemma 34, we can derive η(c 4 3 α ) 3 α = 9. Combined with Conclusion of Proposition 36, we have {η(c3 4 α), η(c4 3α) } C (C3 4 α), done. Proposition 38. Let m = 3 α 5 β with α N and β N. Let n 65 be an odd positive integer such that C 3 p has Property D with respect to 9 for all prime divisors p of n. If then η(c 3 mn) C (C 3 mn). m 6 57 n 7 (n 7)n the electronic journal of combinatorics 9(3) (), #P3

21 Proof. Let m = m. Then 3 m = 3 α 5 β 57 n 7 and α. (n 7)n 64 By Lemma 5 and Lemma 4 we have s(cm 3 n ) = 9m n 8 and η(cm 3 n ) 8m n 7. It follows from Lemma 7 that η(cm 3 n ) = 8m n 7. Since η(c3) 3 = and η(cm 3 n ) = 8m n 7, it follows from Lemma 34 that η(cmn) 3 = 8mn 7. What s more, C3 3 has Property C and η(c3) 3 C (C3) 3 by Lemma 9. Therefore, η(cmn) 3 C (Cmn) 3 by Lemma 8. Proof of Theorem 3.. If a then it follows from Proposition 37 and Lemma 9. Now assume b. Since η(c 3 3 a 5 b ) = 8(3 a 5 b ) +, it follows from Lemma 6 that η(c 3 3 a 5 b ) C (C 3 3 a 5 b ).. Let G = C 3 3 and G a 3 = C8. 3 By Lemma 35, Conclusions 6, 7 and 8 of Lemma 7, we have that η(g ) = 7(3 a 3 ) +, η(g ) = 7 (8 ) + and G has Property C. Therefore, η(c8) 3 C (C8) 3 by Lemma 6. So, η(c3 3 a) C (C3 3 a) by Lemma The result follows from Proposition Let G = C r a with 3 r a. By Lemma 35 and Conclusions 6 of Lemma 7, we have η(c r a) = (r )( a ) + and C r has Property C. Since a r < a, it follows from Lemma 6 that η(c r a) C (C r a). If G = C r and r 3, then it follows from Conclusion 4 of Theorem. 5. Let m = 3 n 5 n and n = 7 n 3 n 4 3 n 5. It follows from n 3 + n 4 + n 5 3 that n > 65. By the hypothesis of n + n + 34(n 3 + n 4 + n 5 ) we infer that, m = 3 n 5 n 3 n +n (n 3+n 4 +n 5 ) > (n 3+n 4 +n 5 ) n 4 > 6 57 n Since it has been (n 7)n 64 proved that every prime p {3, 5, 7,, 3} has Property D with respect to 9 in [8], it follows from Proposition 38 that η(ck 3) C (Ck 3). 8 Concluding Remarks and Open Problems Proposition 39. Let G be a non-cyclic finite abelian group with exp(g) = n. C (G) {η(g)} doesn t contain n + consecutive integers. Then Proof. Assume to contrary that [t, t + n] C (G) {η(g)} for some t N. By the definition of C (G) we have that t+n < η(g). So, we can choose a short free sequence T over G of length T = t + n. It follows from t + n C (G) {η(g)} that σ(t ). Let g = σ(t ) and let S = T ( g). Since S = t + n C (G) {η(g)}, S contains a short zero-sum subsequence U with ( g) U. Note that t S U t+n and σ(s U ) =. It follows from [t, t + n] C (G) {η(g)} that S U contains a short zero-sum subsequence, which is a contradiction with S U T. Proposition 39 just asserts that C (G) can t contain any interval of length more than exp(g). Proposition 9 shows that C (C r n) could not contain integers much smaller than η(c r n). So, it seems plausible to suggest the electronic journal of combinatorics 9(3) (), #P3

22 Conjecture 4. Let G C C m, m N be a non-cyclic finite abelian group. Then C (G) [η(g) (exp(g) + ), η(g) ]. Conjecture 4 and Conjecture suggest the following Conjecture 4. Let G C C m, m N be a non-cyclic finite abelian group. Then C (G) exp(g). Conjecture 4. C (G) = [min{c (G)}, max{c (G)}]. The following notation concerning the inverse problem on s(g) was introduced in []. Property D: We say the group Cn r has property D if s(cn) r = c(n ) + for some positive integer c, and every sequence S over Cn r of length S = c(n ) which contains no zero-sum subsequence of length n has the form S = c i= gn i where g,..., g c are pairwise distinct elements of Cn. r Conjecture 43. ([], Conjecture 7.) Every group C r n has Property D. It has been proved in ([], Section 7) that Conjecture 43, if true would imply Conjecture 44. Every group C r n has Property C. Suppose that Conjecture 44 holds true for all groups of the form Cn. r For fixed n, r N and any a N we have that η(cn r a) = c(na, r)(n ) +, where c(n a, r) N depends on n a and r. By Lemma we obtain that the sequence {c(n a, r)} a= is decreasing. Therefore, c(n a, r) n a for all sufficiently large a. Hence, by Lemma 6 we infer that η(cn r a) C (Cn r a) for all sufficiently large a N. Acknowledgments The authors would like to thank the referee for his/her very useful suggestions. This work has been supported by the PCSIRT Project of the Ministry of Science and Technology, the National Science Foundation of China with grant no. 978 and 35, and the Fundamental Research Funds for the Central Universities. References [] G. Bhowmik and J.C. Schlage-Puchta, Davenport s constant for groups of the form Z 3 Z 3 Z 3d, Additive Combinatorics 43 (7), [] R. Chi, S.Y. Ding, W.D. Gao, A. Geroldinger and W.A. Schmid, On zero-sum subsequences of restricted size IV, Acta Math. Hungar. 7 (5), [3] B.L. Davis, D. Maclagan and R. Vakil, The card game set, Math. Intelligencer 5 (3), [4] Y. Edel, Sequences in abelian groups G of odd order without zero-sum subsequences of length exp(g), Des. Codes Cryptogr. 47 (8), the electronic journal of combinatorics 9(3) (), #P3

23 [5] Y. Edel, C. Elsholtz, A. Geroldinger, S. Kubertin and L. Rackham, Zero-sum problems in finite abelian groups and affine caps, Q. J. Math. 58 (7), [6] C. Elsholtz, Lower bounds for multidimentional zero sums, Combinatorica 4 (4), [7] P. van Emde Boas and D. Kruyswijk, A combinatorial problem on finite abelian groups, II. Math. Centre Report ZW [8] Y.S. Fan, W.D. Gao and Q.H. Zhong, On the Erdos-Ginzburg-Ziv constant of finite abelian groups of high rank, J. Number Theory 3 (), [9] W.D. Gao, On a combinatorial problem connected with factorizations, Colloq. Math. 7 (997), [] W.D. Gao and A. Geroldinger, Zero-sum problems in finite abelian groups : a survey, Expo. Math. 4 (6), [] W.D. Gao, Q.H. Hou, W.A. Schmid and R. Thangadurai, On short zero-sum subsequences II, Integers 7 (7), A. [] W.D. Gao, A. Geroldinger and W.A. Schmid, Inverse zero-sum problems, Acta Arith. 8 (7), [3] A. Geroldinger and F. Halter-Koch, Non-Unique Factorizations. Algebraic, Combinatorial and Analytic Theory, Pure and Applied Mathematics, vol. 78, Chapman & Hall/CRC, (6). [4] A. Geroldinger, Additive group theory and non-unique factorizations, Combinatorial Number Theory and Additive Group Theory (9), 86. [5] A. Geroldinger, D.J. Grynkiewicz and W.A. Schmid, Zero-sum problems with congruence conditions, Acta Math. Hungar. 3 (), [6] A. Geroldinger and I.Z. Ruzsa, Combinatorial number theory and additive group theory, Birkhauser (9). [7] H. Harborth, Ein Extremalproblem für Gitterpunkte, J. Reine Angew. Math. 6 (973), [8] H. Harborth, Ein Extremalproblem für Gitterpunkte, Ph.D. Thesis, Technische Univesität Braunschweig (98). [9] A. Kemnitz, On a lattice point problem, Ars Combin. 6b (983), 5 6. [] J.E. Olson, A combinatorial problem on finite abelian groups, I; II, J. Number Theory (969), 8 ; the electronic journal of combinatorics 9(3) (), #P3 3