In this paper, we show that the bound given in Theorem A also holds for a class of abelian groups of rank 4.

Transcription

1 SUBSEQUENCE SUMS OF ZERO-SUM FREE SEQUENCES OVER FINITE ABELIAN GROUPS YONGKE QU, XINGWU XIA, LIN XUE, AND QINGHAI ZHONG Abstract. Let G be a finite abelian group of rank r and let X be a zero-sum free sequence over G whose support supp(x) generates G. In 009, Pixton proved that Σ(X) r 1 ( X r + ) 1 for r 3. In this paper, we show that this result also holds for abelian groups G of rank 4 if the smallest prime p dividing G satisfies p Introduction Let G be a finite abelian group. By a sequence over G we mean a finite sequence of terms from G where repetition is allowed and the order is disregarded. In 197, Eggleton and Erdős[1] first tackled the problem of determining the minimal cardinality of the set Σ(X) of subsums of zero-sum free sequences over G. In 1977, Olson and White[6] obtained a lower bound for Σ(X), where X is a zero-sum free sequence over G and supp(x) generates a noncyclic group. Subsequently, several authors [, 7, 8, 9] obtained a huge variety of results. In particular, we refer the reader to Part II of the recent monograph [5] by Grynkiewicz. In 009, Pixton [7, Lemma 1.1, Theorem 1.3 and Theorem 1.7] proved the following theorem Theorem A. Let G be a finite abelian group of rank r 3, and let X be a zerosum free sequence over G whose support generates G. Then Σ(X) r 1 ( X r + ) 1 where X denotes the length of X. In this paper, we show that the bound given in Theorem A also holds for a class of abelian groups of rank 4. Main Theorem 1.1. Let G be a finite abelian group of rank 4 and let X be a zero-sum free sequence over G whose support generates G. If the smallest prime p dividing G satisfies p 13, then Σ(X) 8 X 17. Let G be a nontrivial finite abelian group, say G C n1... C nr with 1 < n 1... n r. Then r = r(g) is the rank of G. Suppose that r = 4. Then the lower bound given in Theorem 1.1 is best possible. Indeed, if (e 1, e, e 3, e 4 ) is a basis of G, then the sequence S = e 1 e e 3 e ord(e 4) Mathematics Subject Classification. Primary 11B50; Secondary 11P99. Key words and phrases. zero-sum free sequence, finite abelian group, Davenport s constant. 1

2 Y. K. QU, X. W. XIA, L. XUE, AND Q. ZHONG is zero-sum free, S = ord(e 4 ) +, and Σ(S ) = 3 ord(e 4 ) 1 = 8 S 17. We do not know whether the result holds true for groups G having a prime divisor q which is smaller than 13. Note that we have no information on the maximal length or on the structure of zero-sum free sequences over groups of rank 4. We only want to recall that there are zero-sum free sequences S over groups of rank 4 whose lengths are strictly larger than 4 i=1(n i 1) (see [4]).. Preliminaries Let N denote the set of positive integers and N 0 = N {0}. Let Z denote the set of integers. For a, b Z with a b, we define [a, b] = {x Z a x b}. Let G be an additively written finite abelian group. Let F (G) be the free abelian monoid, multiplicatively written, with basis G. The elements of F (G) are called sequences over G. We write a sequence S F (G) in the form S = g G g v g(s ), with v g (S ) N 0 for all g G. We call v g (S ) the multiplicity of g in S. We say that S contains g if v g (S ) > 0. The unit element 1 F (G) is called the empty sequence. A sequence S 1 is called a subsequence of S if S 1 S in F (G) (equivalently, v g (S 1 ) v g (S ) for all g G). Let S 1, S F (G), we denote by S 1 S the sequence g v g(s 1 )+v g (S ) F (G), g G and we denote by S 1 S 1 the sequence g v g(s 1 ) min{v g (S 1 ),v g (S )} F (G). g G If a sequence S F (G) is written in the form S = g 1... g l, we tacitly assume that l N 0 and g 1,..., g l G. For a sequence we call S = g 1... g l = g G g v g(s ) F (G), S = l = g G v g (S ) N 0 the length of S, σ(s ) = l g i = v g (S )g G the sum of S, i=1 g G supp(s ) = {g G v g (S ) > 0} G the support of S, Σ(S ) = { i I g i I [1, n]} the set of sub(sequence) sums of S. The sequence S is called a zero-sum sequence if σ(s ) = 0, a zero-sum free sequence if 0 Σ(S ), a minimal zero-sum sequence if S 1, σ(s ) = 0, and S contains no proper and nontrivial zero-sum subsequence.

3 SUBSEQUENCE SUMS OF ZERO-SUM FREE SEQUENCES 3 If G 1 is a group and ϕ : G G 1 a map, then ϕ(s ) = ϕ(g 1 )... ϕ(g l ) is a sequence over G 1. We denote by D(G) = max{ S S is a minimal zero-sum sequence over G} the Davenport constant of G. For any integer-valued function f : A Z defined on a finite set A, we denote by min( f ) = min{ f (a) a A} and by max( f ) = max{ f (a) a A}. Next we list some necessary lemmas. Lemma.1. ([7, Lemma 4.4]) Let G be a finite abelian group and X G \ {0} be a generating set for G. Suppose that f : G Z is a function on G. Then max{ f (g + x) f (g), 0} (max( f ) min( f )) X. x X, g G Lemma.. Let G be a finite abelian group, H G a subgroup, S G a subset, and let f : G/H Z be defined by f (a + H) = (a + H) S for all a G. Suppose that X G\{0} is a generating set for G and satisfies (S + x)\s 7 for all x X. Then min( f ) max( f ) 7. In particular, if there exists an element b G such that f (b + H) 8, then f (a + H) 1 for all a G. Proof. Obviously, the assertion holds for H = {0} and for H = G. Suppose that {0} H G. Let A G be such that G = a A(a + H) and A = G/H. Since {x + H x X} is a generating set for G/H, choose X X such that {x + H x X } is a generating set for G/H and {x + H x X } = X. By (S + x) \ S 7 for all x X, 7 X (S + x) \ S x X = ( (S + x) (a + H) ) \ ( S (a + H) ) = x X a A max{ f (a x + H) f (a + H), 0} x X a A x X a A max{ f (a + H + x + H) f (a + H), 0}. Since {x + H x X } is a generating set for G/H, we get {x + H x X } is also a generating set for G/H. Therefore, by Lemma.1, 7 X max{ f (a + H + x + H) f (a + H), 0} (max( f ) min( f )) X. x X a A By X 0, we obtain min( f ) max( f ) 7.

4 4 Y. K. QU, X. W. XIA, L. XUE, AND Q. ZHONG In particular, if f (b + H) 8 for some b G, then for all a G, f (a + H) min( f ) max( f ) 7 f (b + H) 7 1. We also need the following simple and well-known result. Lemma.3. Let G be a finite abelian group and S be a zero-sum free sequence over G. Then (1) Σ(S ) S, () D(G) G. Proof. (1) Suppose S = g 1... g l. Then g 1, g 1 + g,..., g g l are all distinct. It follows that Σ(S ) l = S. () Assume to the contrary that X is a zero-sum free sequence over G with length G. Thus by (1), Σ(X) G which implies that 0 Σ(X), a contradiction. Lemma.4. Let G be a finite abelian group and X = X 1 X be a zero-sum free sequence over G, then (1) Σ(X) Σ(X 1 ) + Σ(X ). () Let H = supp(x 1 ) and let ϕ : G G/H be the canonical epimorphism. If ϕ(x ) is a zero-sum free sequence over G/H, then Σ(X) ( Σ(X 1 ) + 1)( X + 1) 1. Proof. (1) This follows by [3, Theorem 5.3.1]. () Since ϕ(x ) is a zero-sum free sequence over G/H, we get H Σ(X ) = and Σ(ϕ(X )) X by Lemma.3.1. Thus for any a Σ(X ), Σ(X 0 ) (a + H) (Σ(X 1 ) + a) {a} = Σ(X 1 ) + 1. Therefore Σ(X 0 ) Σ(X 0 ) H + a+h Σ(ϕ(X )) Σ(X 1 ) + ( Σ(X 1 ) + 1) Σ(ϕ(X )) ( Σ(X 1 ) + 1)( X + 1) The proof of Theorem 1.1 Σ(X 0 ) (a + H) For the simplicity of formulations, we define C-sequences and C-groups. To begin with, a sequence X over a finite abelian group G is called a C-sequence if the following three conditions hold:

5 (i) supp(x) = G, (ii) X is zero-sum free, (iii) (X) 8 X 18. SUBSEQUENCE SUMS OF ZERO-SUM FREE SEQUENCES 5 Furthermore, a finite abelian group G is called a C-group if the following three conditions hold: (i) r(g) = 4, (ii) the smallest prime p dividing G satisfies p 13, (iii) there exists a C-sequence over G. Proof of Theorem 1.1. If Theorem 1.1 does not hold, then there exists a C- group. Let G 0 be the C-group with minimal order and let X 0 be a C-sequence over G 0 with minimal length. We proceed by the following four claims: Claim A. Let X be a zero-sum free sequence over G 0 and H = supp(x) with r = r(h). If H < G 0 or X < X 0, then Σ(X) r 1 ( X r + ) 1. Proof. By Theorem A and these hypothesis about G 0 and X 0, it follows directly! Claim B. (1) Let H be a subgroup of G 0. Then for any a G 0, Σ(X 0 ) (a + H) max g G 0 Σ(X 0 ) (g + H) 7 Σ(X 0 ) H 7. () Suppose that X 0 has a factorization X 0 = X 1 X such that H = supp(x 1 ) is a proper subgroup of G. If Σ(X 1 ) 7, then Σ(X 0 ) (Σ(X 1 ) + 1) G/H 1. Proof. (1) For any x X 0, denote H x = supp(x 0 x 1 ). Then r(h x ) r(g 0 ) 1 = 3. By Claim A and X 0 x 1 < X 0, we get Σ(X 0 x 1 ) min { 4( X ) 1, 8( X ) 1 } = 4 X 0 9. If H x G 0, then x H x. Thus Σ(X 0 ) Σ(X 0 x 1 ) X 0 17 by Lemma.4., a contradiction to that X 0 is a C-sequence. Therefore H x = G 0 and r(h x ) = 4. By Claim A and X 0 x 1 < X 0, Σ(X 0 x 1 ) 8( X 0 1) 17 = 8 X 0 5. Let S = Σ(X 0 ). Then S 8 X 0 18 and for all x X 0, (S + x) \ S = S \ (S x) S \ Σ(X 0 x 1 ) S Σ(X 0 x 1 ) (8 X 0 18) (8 X 0 5) 7.

6 6 Y. K. QU, X. W. XIA, L. XUE, AND Q. ZHONG By supp(x 0 ) = G 0 and Lemma., for any a G, Σ(X 0 ) (a + H) max g G 0 Σ(X 0 ) (g + H) 7 Σ(X 0 ) H 7. () Since H is a proper subgroup of G 0, there exists x X such that x H. Then Σ(X 0 ) (x + H) (Σ(X 1 ) + x) {x} Σ(X 1 ) For any a G 0 \H, we get that Σ(X 0 ) (a+h) Σ(X 0 ) (x+h) 7 1 by (1) which implies that Σ(X ) (a + H). Choose b Σ(X ) (a+h). Then Σ(X 0 ) (a+h) (Σ(X 1 )+b) {b} = Σ(X 1 ) + 1 for all a G 0 \ H. Therefore Σ(X 0 ) Σ(X 1 ) + ( Σ(X 1 ) + 1)( G/H 1) ( Σ(X 1 ) + 1)( G/H ) 1. Claim C. Let X be a subsequence of X 0. If H = supp(x) is a proper subgroup of G 0, then r(h) 3. Proof. Assume to the contrary that r(h) = 4. Then X 4. Let ϕ : G 0 G 0 /H denote the canonical epimorphism from G 0 to G 0 /H with ker(ϕ) = H. Then ϕ(x 0 ) is a sequence over G 0 /H. We can get a factorization of X 0, X 0 = X X 1... X α X, satisfying that for 1 i α, ϕ(x i ) is a minimal zero-sum sequence over G 0 /H and ϕ(x ) is a zero-sum free sequence over G 0 /H. Thus Σ(ϕ(X )) X and X 0 X + αd(g/h) + X X + α G/H + X by Lemma.3. Let Y = X σ(x 1 )... σ(x α ). Then Y is a zero-sum free sequence over H. By H < G 0 and Claim A, we have Σ(X 0 ) H Σ(Y) H 8 Y 17. For any a Σ(X ), we get a H and Σ(X 0 ) (a + H) Σ(Y a) (a + H) Σ(Y) H Y 16. Let A Σ(X ) satisfy {a + H a Σ(X )} = {a + H a A } and A = ϕ(σ(x )). Let A G 0 be a subset with A A such that G 0 = a A (a + H) and A = G 0 /H. Then for any b A \ (A H), by Claim B.1. Σ(X 0 ) (b + H) Σ(X 0 ) H 7 8 Y 4,

7 SUBSEQUENCE SUMS OF ZERO-SUM FREE SEQUENCES 7 Therefore, Σ(X 0 ) = Σ(X 0 ) (a + H) a A 8 Y 17 + (8 Y 16) Σ(ϕ(X )) + (8 Y 4)( G/H 1 Σ(ϕ(X )) ) (8 Y 4) G/H + 8 Σ(ϕ(X )) + 7 8( X 3)( G / H 1) + 8( X + α G / H + X ) 17 8 X A contradiction. Claim D. Let Y be a subsequence of X 0 with length 4. Then supp(y) = G 0. Proof. Let X be the longest subsequence of X 0 such that supp(x) G 0. Denote H = supp(x). Then r(h) = 3 by Claim C and G 0 /H 13 since G 0 is a C-group. Let ϕ : G 0 G 0 /H denote the canonical epimorphism. We only need to prove that X 3. Assume to the contrary that X 4. We distinguish three cases to finish the proof. CASE 1: X 0 ( X 1) G 0/H +4. By H < G 0 and Claim A, Σ(X) 4( X 1) Then by Claim B., Σ(X 0 ) ( Σ(X) + 1) G 0 /H 1 4( X 1) G 0 /H 1, which implies that Σ(X 0 ) 8 X 0 17 by X 0 ( X 1) G 0/H +4, a contradiction. CASE : There exists no zero-sum free subsequence of ϕ(x 0 X 1 ) with length 6. Since ϕ(x 0 ) is a sequence over G 0 /H, we can get a factorization of X 0, X 0 = X X 1... X α X, satisfying that for 1 i α, ϕ(x i ) is a minimal zero-sum sequence over G 0 /H and ϕ(x ) is a zero-sum free sequence over G 0 /H. Thus X 0 = X + X X α + X X + X + 6α and Σ(ϕ(X )) X by Lemma.3. Let Y = X σ(x 1 )... σ(x α ). Then Y is a zero-sum free sequence over H. By Claim A and H < G 0, we have Σ(X 0 ) H Σ(Y) H 4 Y 5. For any a Σ(X ), we obtain a H and Σ(X 0 ) (a + H) Σ(Y a) (a + H) Σ(Y) H Y 4. Let A Σ(X ) satisfy {a + H a Σ(X )} = {a + H a A } and A = ϕ(σ(x )). Let A G 0 be a subset with A A such that G 0 = a A (a + H) and A = G 0 /H. Then for any b A \ (Σ(X ) H), Σ(X 0 ) (b + H) Σ(X 0 ) H 7 4 Y 1,

8 8 Y. K. QU, X. W. XIA, L. XUE, AND Q. ZHONG by Claim B.1. Therefore, Σ(X 0 ) = Σ(X 0 ) (a + H) a A 4 Y 5 + (4 Y 4) Σ(ϕ(X )) + (4 Y 1)( G/H 1 Σ(ϕ(X )) ) (4 Y 1) G/H + 8 Σ(ϕ(X )) + 7 (4 X + 4α 1) G/H + 8 X + 7. Since X 4, G/H 13, and X 0 X + X + 6α, we have that Σ(X 0 ) 8 X 0 17, a contradiction. CASE 3: X 0 > ( X 1) G/H +4 and there exists a subsequence X 1 of X 0 X 1 such that ϕ(x 1 ) is a zero-sum free subsequence over G 0 /H with length 6. Since X 0 > ( X 1) G/H +4, we obtain X 0 X > X ( G/H 4) G/H Denote X = X 0 (XX 1 ) 1. Then X = X 0 XX 1 > X + 1. Thus X = G 0 since X is the longest subsequence of X 0 such that supp(x) G 0. Then Σ(X ) 8 X 17 by X < X 0 and Claim A. By H < G 0 and Claim A, Σ(X) 4( X 1) 1. It follows that Σ(XX 1 ) 4( X 1)( Σ(ϕ(X 1 )) + 1) 1 8 XX 1 by Lemma.4., X 4 and X 1 = 6. Therefore by Lemma.4.1, Σ(X 0 ) Σ(XX 1 ) + Σ(X ) 8 XX 1 +8 X 17 = 8 X 0 17, a contradiction. Now we finish the proof of Theorem 1.1 by distinguishing the following two cases. Suppose that X Denote X 0 = x 1... x n. Then by Claim D, any four elements of X 0 are independent which implies that x i, x j + x k, 1 i n, 1 j < k n are all different elements in G 0. Therefore, n(n 1) Σ(X 0 ) n + 8 X 0 17, a contradiction. Suppose that X 0 1. Let X be a subsequence of X 0 with length 3 and H = supp(x) is a proper subgroup of G 0. Then by Claim D, the three elements of X must be independent which implies that Σ(X) = 7. It follows by Claim B. that a contradiction. Σ(X 0 ) ( Σ(X) + 1) G 0 /H X 0 17, Acknowledgements. The authors are grateful to the referee for helpful suggestions and comments. This research was supported by NSFC (grant no , ), NSF of Henan Province (grant no ), the Education

9 SUBSEQUENCE SUMS OF ZERO-SUM FREE SEQUENCES 9 Department of Henan Province (grant no. 009A11001), NSF of Luoyang Normal University (grant no ), and the Austrian Science Fund FWF(project no. M1641-N6). References [1] R. B. Eggleton and P. Erdős, Two combinatorial problems in group theory, Acta Arith. 1 (197), [] W. D. Gao and I. Leader, Sums and k-sums in an abelian groups of order k, J. Number Theory 10 (006), 6 3. [3] A. Geroldinger and F. Halter-Koch, Non-Unique Factorizations, Algebraic, Combinatorial and Analytic Theory, Pure and Applied Mathematics, vol. 78, Chapman & Hall/CRC, 005. [4] A. Geroldinger and R. Schneider, On Davenport s constant, J. Comb. Theory, Ser. A 61 (199), [5] D. J. Grynkiewicz. Structural Additive Theory, Developments in Mathematics, Springer, 013. [6] J. E. Olson and E. T. White, Sums from a sequence of group elements, in: Number Theory and Algebra, Academic Press, New York, 1977, pp. 15. [7] A. Pixton, Sequences with small subsum sets, J. Number Theory 19 (009), [8] F. Sun, On subsequence sums of a zero-sum free sequence, Electron. J. Combin. 14 (007), #R5. [9] P. Z. Yuan, Subsequence sums of zero-sum-free sequences, Electron. J. Combin. 16 (009), #R97. Department of Mathematics, LuoYang Normal University, LuoYang 4710, P.R. China address: yongke139@163.com Department of Mathematics, LuoYang Normal University, LuoYang 4710, P.R. China address: @163.com Department of Mathematics, LuoYang Normal University, LuoYang 4710, P.R. China address: lyxuelin@16.com University of Graz, Institute for Mathematics and Scientific Computing, Heinrichstraße 36, 8010 Graz, Austria address: qinghai.zhong@uni-graz.at