aa r X i v : . [ m a t h . N T ] S e p Subsequence Sums of Zero-sum free Sequences II
Pingzhi Yuan
School of Mathematics, South China Normal University , Guangzhou 510631, P.R.CHINAe-mail [email protected]
Abstract
Let G be a finite abelian group, and let S be a sequence of elements in G .Let f ( S ) denote the number of elements in G which can be expressed as thesum over a nonempty subsequence of S . In this paper, we determine all thesequences S that contains no zero-sum subsequences and f ( S ) ≤ | S | − MSC:
Primary 11B75; Secondary 11B50.
Key words:
Zero-sum problems, Davenport’s constant, zero-sum freesequences.
Let G be a finite abelian group (written additively)throughout the present paper. F ( G ) denotes the free abelian monoid with basis G , the elements of which arecalled sequences (over G ). A sequence of not necessarily distinct elements from G will be written in the form S = g · · · · · g k = Q ki =1 g i = Q g ∈ G g v g ( S ) ∈ F ( G ), where v g ( S ) ≥ multiplicity of g in S . Denote by | S | = k the number ofelements in S (or the length of S ) and let supp( S ) = { g ∈ G : v g ( S ) > } be the support of S .We say that S contains some g ∈ G if v g ( S ) ≥ T ∈ F ( G )is a subsequence of S if v g ( T ) ≤ v g ( S ) for every g ∈ G , denoted by T | S . If T | S , then let ST − denote the sequence obtained by deleting the terms of T from S . Furthermore, by σ ( S ) we denote the sum of S , (i.e. σ ( S ) = P ki =1 g i = P g ∈ G v g ( S ) g ∈ G ). By P ( S ) we denote the set consisting of all elements whichcan be expressed as a sum over a nonempty subsequence of S , i.e. X ( S ) = { σ ( T ) : T is a nonempty subsequence of S } . We write f ( S ) = | P ( S ) | , < S > for the subgroup of G generated by all theelements of S . Supported by NSF of China (No. 10571180). S be a sequence over G . We call S a zero − sum sequence if σ ( S ) = 0,a zero − sum f ree sequence if σ ( W ) = 0 for any subsequence W of S , and squaref ree if v g ( S ) ≤ g ∈ G . We denote by A ⋆ ( G ) the set of allzero-sum free sequences in F ( G ).Let D ( G ) be the Davenport’s constant of G , i.e., the smallest integer d suchthat every sequence S over G with | S | ≥ d satisfies 0 ∈ P ( S ). For every positiveinteger r in the interval { , . . . , D ( G ) − } , let f G ( r ) = min S, | S | = r f ( S ) , (1)where S runs over all zero-sum free sequences of r elements in G . How does thefunction f G behave?In 2006, Gao and Leader proved the following result. Theorem A [5]
Let G be a finite abelian group of exponent m . Then(i) If ≤ r ≤ m − then f G ( r ) = r .(ii) If gcd(6 , m ) = 1 and G is not cyclic then f G ( m ) = 2 m − . Recently, Sun[10] showed that f G ( m ) = 2 m − , m ) = 1.Using some techniques from the author [11], the author [12] proved the fol-lowing two theorems. Theorem B [12, 8]
Let S be a zero-sum free sequence over G such that < S > is not a cyclic group, then f ( S ) ≥ | S | − . Theorem C [12]
Let S be a zero-sum free sequence over G such that < S > is not a cyclic group and f ( S ) = 2 | S | − . Then S is one of the following forms(i) S = a x ( a + g ) y , x ≥ y ≥ , where g is an element of order 2.(ii) S = a x ( a + g ) y g, x ≥ y ≥ , where g is an element of order 2.(iii) S = a x b, x ≥ . However, Theorem B is an old theorem of Olson and White [8] which has beenoverlooked by the author. For more recent progress on this topic, see [4, 9, 13].The main purpose of the present paper is to determine all the sequences S over a finite abelian group such that S contains no zero-sum subsequences and f ( S ) ≤ | S | −
1. To begin with, we need the notation of g -smooth. Definition 1.1 [7, Definition 5.1.3]
A sequence S ∈ F ( G ) is called smooth if S = ( n g )( n g ) · · · · · ( n l g ) , where | S | ∈ N , g ∈ G, n ≤ · · · ≤ n l , n = n + · · · + n l < ord ( g ) and P ( S ) = { g, . . . , ng } ( in this case we say moreprecisely that S is g -smooth). We have
Theorem 1.1
Let G be a finite abelian group and let S be a zero-sum free se-quence over G with f ( S ) ≤ | S | − . Then S has one of the following forms:(i) S is a -smooth for some a ∈ G . ii) S = a k b , where k ∈ N and a, b ∈ G are distinct.(iii) S = a k b l , where k ≥ l ≥ and a, b ∈ G are distinct with a = 2 b .(iv) S = a k b l ( a − b ) , where k ≥ l ≥ and a, b ∈ G are distinct with a = 2 b . For a sequence S over G we call h ( S ) = max { v g ( S ) | g ∈ G } ∈ [0 , | S | ] the maximum of the multiplicities of S. Let S = a x b y T with x ≥ y ≥ h ( T ), then Theorem 1.1(i) can be stated moreprecisely as that S is a -smooth or b -smooth. Let ∅ 6 = G ⊆ G be a subset of G and k ∈ N . Define f ( G , k ) = min { f ( S ) : S ∈ F ( G ) zero − sumfree , squarefree and | S | = k } and set f ( G , k ) = ∞ , if there are no sequences over G of the above form. Lemma 2.1
Let G be a finite abelian group.1. If k ∈ N and S = S · · · · · S k ∈ A ⋆ ( G ) , then f ( S ) ≥ f ( S ) + · · · + f ( S k ) .
2. If G ⊂ G , k ∈ N and f ( G , k ) > , then f ( G , k ) = 1 , if k = 1 , = 3 , if k = 2 , ≥ , if k = 3 , ≥ , if k = 3 and g = 0 for all g ∈ G , ≥ k , if k ≥ . Proof.
1. See [6, Theorem 5.3.1].2. See [6, Corollary 5.3.4]. ✷ Lemma 2.2
Let a, b be two distinct elements in an abelian group G such that a b ∈ A ⋆ ( G ) , a = 2 b, a = 2 b , and b = 2 a . Then f ( a b ) = 8 . roof. It is easy to see that a, a, b, b, a + b, a + 2 b, a + b, a + 2 b are all thedistinct elements in P ( a b ). We are done. ✷ Lemma 2.3
Let S = a k b be a zero-sum free sequence over G . If S = a k b is not a -smooth, then f ( S ) = 2 k + 1 .Proof. The assertion follows from the fact that a, . . . , ka, b, a + b, . . . , ka + b areall the distinct elements in P ( a k b ). ✷ Lemma 2.4 [10, Lemma 4]
Let S be a zero-sum free sequence over G . If thereis some element g in S with order , then f ( S ) ≥ | S | − . Lemma 2.5
Let k ≥ l ≥ be two integers, and let a and b be two distinctelements of G such that a k b l ∈ A ⋆ ( G ) and a k b l is not smooth. Then we have(i) If a = 2 b , then f ( a k b l ) ≥ k + l ) .(ii) If a = 2 b , then f ( a k b l ) = 2( k + l ) − .Proof. If nb = sa for any n and s with 1 ≤ n ≤ l and 1 ≤ s ≤ k , then ra + sb, r + s = 0 , ≤ r ≤ k, ≤ s ≤ b are all the distinct elements in P ( a k b l ),and so f ( a k b l ) = kl + k + l ≥ k + l ) . Now we assume that nb = sa for some n and s with 1 ≤ n ≤ l and 1 ≤ s ≤ k .Let n be the least positive integer with nb = sa, ≤ n ≤ l, ≤ s ≤ k . Then n ≥ s ≥ a, . . . , ka, . . . , ( k + [ ln ] s ) a,b, a + b, . . . , b + ka, . . . , b + ( k + [ l − n ] s ) a,. . . . . . ( n − b, . . . , ( n − b + ka, . . . , ( n − b + ( k + [ l − n + 1 n ] s ) a are all the distinct elements in P ( a k b l ), and so f ( a k b l ) = k + [ ln ] s + 1 + k + [ l − n ] s + · · · + 1 + k + [ l − n + 1 n ] s = n ( k − s + 1) + ls + s − . Since n ( k − s + 1) + ls + s − − k + l ) = ( n − k − s ) + ( l − s −
2) + n − f ( a k b l ) ≥ k + l ) − n = s = 2,that is 2 a = 2 b . This completes the proof. ✷ emark: Note that if a k b l ∈ A ⋆ ( G ) , k ≥ l ≥
2, then the conditions that a k b l is smooth and 2 a = 2 b cannot hold simultaneously. Otherwise, we may supposethat 2 a = 2 b and a k b l is a -smooth (the case that a k b l is b -smooth is similar), then b = ta, ≤ t ≤ ( k + 1). It follows that b + ( t − a = 2( t − a = 2 b − a =0 , < t − ≤ k −
1, which contradicts the fact that a k b l ∈ A ⋆ ( G ). Lemma 2.6 [12, Lemma 2.9]
Let S = a k b l g, k ≥ l ≥ be a zero-sum free se-quence over G with b − a = g and ord( g ) = 2 , then f ( S ) = 2( k + l ) + 1 . Lemma 2.7
Let S ∈ F ( G ) and a, g ∈ G such that S = S a ∈ A ⋆ ( G ) , S is g -smooth and S is not g -smooth. Then f ( S ) = 2 f ( S ) + 1 .Proof. If a < g > , then P ( S ) = P ( S ) ∪ { a } ∪ ( P ( S ) + a ), and so f ( S ) =2 f ( S ) + 1.If a ∈ < g > , we let P ( S ) = { g, . . . , ng } , a = tg, t ∈ N , then t ≥ n + 2 byour assumptions. It follows that P ( S ) = { g, . . . , ng, tg, ( t + 1) g, . . . , ( t + n ) g } ,and so f ( S ) = 2 f ( S ) + 1. ✷ Lemma 2.8
Let k ≥ be a positive integer and a, b, c three distinct elements in G such that a k bc ∈ A ⋆ ( G ) and a k bc is not a -smooth. Then f ( a k bc ) ≥ k + 4 .Proof. Observe that f ( a k bc ) ≥ k + 4 when a k bc is b or c -smooth. We considerfirst the case that a k b is a -smooth (the case that a k c is a -smooth is similar). Itis easy to see f ( a k b ) ≥ k + 2, and so f ( a k bc ) = 2 f ( a k b ) + 1 ≥ k + 5 by Lemma2.7. Therefore we may assume that both a k b and a k c are not a -smooth in theremaining arguments. We divide the proof into three cases.(i) If a k ( b + c ) is not a -smooth, then a, . . . , ka, b, b + a, . . . , b + c, b + c + a, . . . , b + c + ka are distinct elements in P ( a k bc ), and so f ( a k bc ) ≥ k + k + 1 + k + 1 ≥ k + 4 . (ii) If neither a k ( b − c ) nor a k ( c − b ) is a -smooth, then a, . . . , ka, b, b + a, . . . , c, c + a, . . . , c + ka, b + c + ka are distinct elements in P ( a k bc ), andso f ( a k bc ) ≥ k + k + 1 + k + 1 + 1 ≥ k + 5 . (iii) If a k ( b + c ) is a -smooth and a k ( b − c ) (or a k ( c − b ) ) is a -smooth, then wehave b + c = sa, b − c = ta, ≤ s, t ≤ k + 1 , s = t. It is easy to see that a, . . . , ka, ( k + 1) a, . . . , ( k + s ) a, c, c + a, . . . , c + ( k + t ) a are all distinct elements in P ( a k bc ), and so f ( a k bc ) = k + s + k + t + 1 ≥ k + 4 . The second equality holds if and only if ( s, t ) = (1 ,
2) or (2 , ✷ Corollary 2.1
Let k ≥ be a positive integer and a, b, c, d four distinct ele-ments in G such that a k bcd ∈ A ⋆ ( G ) and a k bcd is not a -smooth. Then f ( a k bcd ) ≥ k + 6 . Lemma 2.9
Let a, b, x be three distinct elements in G such that a k b l x ∈ A ⋆ ( G ) , k ≥ l ≥ , a = 2 b , and x = a − b , then f ( a k b l x ) ≥ k + l + 1) + 1 .Proof. If there are no distinct pairs ( m, n ) = (0 , , ( m , n ) = (0 , , ≤ m, m ≤ k, ≤ n, n ≤ l such that ma + nb = m a + n b + x , then P ( a k b l x ) = P ( a k b l ) ∪ { x } ∪ ( P ( a k b l ) + x ), and so f ( a k b l x ) = 2 f ( a k b l ) + 1 = 4( k + l ) − ≥ k + l + 1) + 1.If there are two distinct pairs ( m, n ) = (0 , , ( m , n ) = (0 , , ≤ m, m ≤ k, ≤ n, n ≤ l such that ma + nb = m a + n b + x , then x = a − b or x = ua + b, ≤ u ≤ ( k + l −
1) or x = vb, v ≥ x = ta, t ≥ x = ua + b, ≤ u ≤ ( k + l − a, . . . , ( k + l + u ) a, b, · · · , b +( k + l + u ) a are all distinct elements in P ( a k b l x ), and so f ( a k b l x ) = 2( k + l + u ) + 1 ≥ k + l + 1) + 1.Let x = vb, ≤ v ≤ ( k + l ) (the case that x = ta, t ≥ k iseven, then b, . . . , ( k + l + v ) b, a, a + b, . . . , a + ( k + l − v ) b are all distinctelements in P ( a k b l x ), and so f ( a k b l x ) = 2( k + l + v −
1) + 1 ≥ k + l + 1) + 1. If k is odd, then b, . . . , ( k + l + v − b, a, a + b, . . . , a + ( k + l − v ) b are all distinctelements in P ( a k b l x ), and so f ( a k b l x ) = 2( k + l + v −
1) + 1 ≥ k + l + 1) + 1.We are done. ✷ Lemma 2.10
Let a, b, x be three distinct elements in G such that a k b x ∈ A ⋆ ( G ) , k ≥ and a k b x is not a -smooth or b -smooth, then f ( a k b x ) = 2 k + 5 if and only if a = 2 b and x = b − a .Proof. We divide the proof into four cases.
Case 1 a k b is not smooth and 2 b = sa, ≤ s ≤ k . If x = b − a , then a, . . . , ( k + s ) a, b − a, b, . . . , b + ( k + s − a are all the distinct elements in P ( a k b x ), and so f ( a k b x ) = 2( k + s )+1. If x = ta, ≤ t ≤ k , then a, . . . , ( k + s + t ) a, b, . . . , b +( k + t ) a are all the distinct elements in P ( a k b x ), and so f ( a k b x ) =2( k + t )+ s +1. If x = ta + b, ≤ t ≤ k , then a, . . . , ( k + s + t ) a, b, . . . , b +( k + t + s ) a are all the distinct elements in P ( a k b x ), and so f ( a k b x ) = 2( k + t + s ) + 1.Therefore f ( a k b x ) = 2 k + 5 if and only if 2 a = 2 b and x = b − a in this case. Case 2 a k b is not smooth and 2 b = sa, s > k or 2 b < a > , then f ( a k b ) =3 k + 2. If k ≥
3, then f ( a k b x ) ≥ f ( a k b ) + 1 = 3 k + 3 > k + 5. If k = 2 and f ( abx ) = 7, then f ( a b x ) ≥ f ( abx ) + f ( ab ) = 7 + 3 > k + 5. If k = 2 and f ( abx ) = 6 (i.e., x = a + b or x = a − b or x = b − a ), then it is easy to checkthat f ( a b x ) > k + 5. 6 ase 3 a k b is smooth and a k b x is not smooth. If a k b is a -smooth, then f ( a k b x ) = 2 f ( a k b ) + 1 ≥ k + 2 ×
2) + 1 > k + 5. If a k b is b -smooth, then f ( a k b x ) = 2 f ( a k b ) + 1 ≥ k ) + 1 > k + 5. Case 4 a k b x is x -smooth. We have f ( a k b x ) ≥ k + 2 × > k + 5.This completes the proof of the lemma. ✷ To prove the main theorem of the present paper, we still need the following twoobviously facts on smooth sequences.
Fact 1
Let r be a positive integer and a ∈ G . If W T i ∈ A ⋆ ( G ) is a -smoothfor all i = 1 , . . . , r , then S = T · · · · · T r W is a -smooth. Fact 2
Let r, k, l be three positive integers and a, b two distinct elements in G . If S ∈ A ⋆ ( G ) is a -smooth and a k b l T i ∈ A ⋆ ( G ) is a -smooth or b -smooth for all i = 1 , . . . , r , then the sequence Sa k b l T · · · · · T r is a -smooth or b -smooth. Proof of Theorem 1.1:
We start with the trivial case that S = a k with k ∈ N and a ∈ G . Then P ( S ) = { a, . . . , ka } , and since S is zero-sum free, it follows that k < ord ( a ).Thus S is a -smooth.If S = S g ∈ A ⋆ ( G ), where g is an element of order 2, then f ( S ) ≥ | S | − f ( S ) ≥ f ( S ) + 2 since P ( S ) ⊇ P ( S ) ∪ { g, g + σ ( S ) } .If S = S g g ∈ A ⋆ ( G ), where g and g are two elements of order 2, then f ( S ) ≥ | S | since P ( S ) ⊇ P ( S g ) ∪ { g , g + g , g + g + σ ( S ) } . Therefore itsuffices to determine all S ∈ A ⋆ ( G ) such that S does not contain any element oforder 2 and f ( S ) ≤ | S | −
1, and when f ( S ) ≤ | S | −
1, determine all Sg ∈ A ⋆ ( G )such that g is an element of order 2 and f ( Sg ) = 2 | S | + 1.To begin with, we determine all S ∈ A ⋆ ( G ) such that S does not contain anyelement of order 2 and f ( S ) ≤ | S | −
1. Let S = a x b y c z T with x ≥ y ≥ z ≥ h ( T )and a, b, c supp( T ). The case that | supp( S ) | = 2 follows from Lemmas 2.3 and2.5 and the remark after Lemma 2.5. Therefore we may assume that | supp( S ) | ≥ S does not contain any element of order 2 in the following arguments.If x = y = z , then S allows the product decomposition S = S · · · · · S x , where S i = abc · · · · , i = 1 , . . . , x are squarefree of length | S i | ≥
3. By Lemma2.1, we obtain f ( S ) ≥ x X i =1 f ( S i ) ≥ x X i =1 | S i | = 2 | S | . If x ≥ y > z ≥ h ( T ), or x > y ≥ z ≥ h ( T ), then S allows a productdecomposition 7 = T · · · · · T r W having the following properties: • r ≥ i ∈ [2 , r ], S i ∈ F ( G ) is squarefree of length | S i | = 3. • W ∈ F ( G ) has the form W = a k , k ≥ W = a k b, k ≥ W = a k b l , k ≥ l ≥ k is the largest integer in W = a k (or a k b or a k b l , k ≥ l ≥
2) among all such product decompositions. We dividethe remaining proof into three cases.
Case 1 W = a k , k ≥
1. If T i = xyz with a
6∈ { x, y, z } for some i, ≤ i ≤ r such that a k xyz is not a -smooth whenever k >
1, then S admits the productdecomposition S = T · · · · · T i − T ′ i T i +1 · · · · · T r , where T i , i = 1 , . . . , r have the properties described above and T ′ i = a k xyz . ByLemma 2.1, and Corollary 2.1, we get f ( S ) ≥ r X j = i f ( T j ) + f ( T ′ i ) ≥ r X j = i | T j | + 2 | T ′ i | = 2 | S | . If T i = axy for some i, ≤ i ≤ r such that a k +1 xy is not a -smooth, then S admits the product decomposition S = T · · · · · T i − T ′ i T i +1 · · · · · T r , where T i , i = 1 , . . . , r have the properties described above and T ′ i = a k +1 xy . ByLemmas 2.1 and 2.8, we get f ( S ) ≥ r X j = i f ( T j ) + f ( T ′ i ) ≥ r X j = i | T j | + 2 | T ′ i | = 2 | S | . Therefore we have proved that if S is not a -smooth and W = a k , then f ( S ) ≥ | S | . Case 2 W = a k b, k ≥ T i = xyz with a
6∈ { x, y, z } for some i, ≤ i ≤ r . If k = 1, then T i W = abxyz . If k = 2, then T i W = abx · ayz . If k ≥ a k − yz, a k − xz , and a k − xy , say, a k − yz is not a -smooth,then T i W = abx · a k − yz . It follows from Lemmas 2.1 and 2.8 that f ( T i W ) ≥ | T i | + 2 | W | , and so f ( S ) ≥ | S | .Let T i = bxy for some i, ≤ i ≤ r , then k ≥
2. If k = 2, then T i W = abx · aby .If k > a k − by (or a k − bx ) is not a -smooth, then T i W = abx · a k − by T i W = aby · a k − bx ). It follows from Lemmas 2.1 and 2.8 that f ( T i W ) ≥ | T i | + 2 | W | , and so f ( S ) ≥ | S | .Let T i = abx for some i, ≤ i ≤ r , then T i W = a k +1 b x . If a k +1 b x is not a -smooth or b -smooth, then by Lemma 2.10 we have f ( T i W ) ≥ | T i | + 2 | W | , andso f ( S ) ≥ | S | .Therefore we have proved that if S is not a -smooth or b -smooth, then f ( S ) ≥ | S | in this case. Case 3 W = a k b l , k ≥ l ≥
2. If 2 a = 2 b and a k b l is not smooth, then byLemma 2.5 we have f ( W ) ≥ | W | and we are done. Note that the conditionsthat 2 a = 2 b and a k b l is smooth cannot hold simultaneous. Here we omit thesimilar arguments as we have done in Case 1. Subcase 1 a = 2 b .Let T i = xyz with a
6∈ { x, y, z } for some i, ≤ i ≤ r , then T i W = abxy · a k − b l − z . It follows from Lemmas 2.1 and 2.9 that f ( T i W ) ≥ | T i | + 2 | W | , andso f ( S ) ≥ | S | .Let T i = byz for some i, ≤ i ≤ r , then k ≥ l + 1, T i W = aby · a k − b l z . Itfollows from Lemmas 2.1 and 2.9 that f ( T i W ) ≥ | T i | +2 | W | , and so f ( S ) ≥ | S | .Let T i = abx for some i, ≤ i ≤ r , then T i W = a k +1 b l +1 x . If a k +1 b x is not a -smooth or b -smooth, then by Lemma 2.10 we have f ( T i W ) ≥ | T i | + 2 | W | , andso f ( S ) ≥ | S | . Subcase 2 a k b l is smooth, a = 2 b , and b = 2 a . Then W = ( a b ) s W , W = a k or W = a k b . If S = SW − W is not a -smooth or b -smooth, then f ( S ) ≥ | S | , and so by Lemmas 2.1 and 2.2 f ( S ) ≥ sf ( a b ) + f ( S ) ≥ s +2 | S | = 2 | S | . If S = SW − W is a -smooth or b -smooth, then S is a -smooth or b -smooth. Subcase 3 a = 2 b .Let T i = xyz with a, b
6∈ { x, y, z } for some i, ≤ i ≤ r , then it is easy tosee that f ( T i W ) = f ( a k b l xyz ) = f ( b k + l xyz ). It follows from Corollary 2.1 that b k + l xyz is b -smooth or f ( T i W ) ≥ | T i | + | W | ).Let T i = bxy with a, b
6∈ { x, y } for some i, ≤ i ≤ r , then f ( T i W ) = f ( a k b l +1 xy ) = f ( b k + l +1 xyz ). It follows from Lemma 2.8 that b k + l +1 xy is b -smooth or f ( T i W ) ≥ | T i | + | W | ).Let T i = abx with a = x, b = x for some i, ≤ i ≤ r , then f ( T i W ) = f ( a k +1 b l +1 x ) = f ( b k + l +3 xyz ). It follows from Lemma 2.3 that b k + l +3 x is b -smooth or f ( T i W ) ≥ | T i | + | W | ). Subcase 4 b = 2 a . Similar to Subcase 3.Therefore we have proved that if S is not a -smooth or b -smooth, then f ( S ) ≥ | S |− f ( S ) = 2 | S |− S = a k b or S = a k b l , a = 2 b, k ≥ l ≥ f ( S ) ≤ | S | −
1, we will determine all Sg ∈ A ⋆ ( G ) such that g is an element of order 2 and f ( Sg ) = 2 | S | + 1.(i) If S is a -smooth (the case that S is b -smooth is similar), we set P ( S ) = { a, . . . , na } , n ≤ | S | −
1, then g P ( S ) since g is an element of order 29nd Sg ∈ A ⋆ ( G ). It follows that P ( Sg ) = P ( S ) ∪ { g } ∪ { g + P ( S ) } , and so f ( Sg ) = 2 n + 1. Therefore f ( Sg ) ≤ | S | + 1 if and only if S = a k .(ii) S = a k b is not smooth, by Lemma 2.8, f ( a k bg ) ≤ k + 1 only if a k bg is a -smooth, which is impossible since g is an element of order 2 and a k bg ∈ A ⋆ ( G ).(iii) S = a k b l , a = 2 b, k ≥ l ≥
2. The result follows from Lemmas 2.5 and2.9.Therefore we have proved that if S = a x b y · · · · ∈ A ⋆ ( G ) , x ≥ y ≥ · · · ,where a, b, . . . are distinct elements of G and f ( S ) ≤ | S | −
1, then S is a -smooth or b -smooth or S = a k b, b P ( a k ) or S = a k b l , k ≥ l ≥ , a = 2 b or S = a k b l , k ≥ l ≥ , a = 2 b, g = a − b . Theorem 1.1 is proved. ✷ Acknowledgement:
The author wishes to thank Alfred Geroldinger forsending the preprint [7] to him. He also thanks the referee for his/her valuablesuggestions.
References [1] J.D. Bovey, P. Erd˝os, and I. Niven,
Conditions for zero sum modulo n ,Canad. Math. Bull. (1975), 27 – 29.[2] B. Bollob´ a s and I. Leader, The number of k -sums modulo k , J. NumberTheory (1999), 27-35.[3] S.T. Chapman and W.W. Smith, A characterization of minimal zero-sequences of index one in finite cyclic groups , Integers (2005), PaperA27, 5pp.[4] W. Gao, Y. Li, J. Peng, and F. Sun,
On subsequence sums of a zero-sumfree sequence II , the Electronic Journal of Combinatorics (2008), ♯ R117.[5] W.D. Gao and I. Leader, sums and k -sums in an abelian groups of order k ,J. Number Theory (2006), 26-32.[6] A. Geroldinger and F. Halter-Koch, Non-Unique Factorizations. Algebraic,Combinatorial and Analytic Theory , Pure and Applied Mathematics, Vol.278, Chapman & Hall/CRC, 2006.[7] A. Geroldinger,
Additive group theory and non-unique factorizations , to ap-pear.[8] 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. 215-222.[9] A. Pixton,
Sequences with small subsum sets , J. Number Theory (2009),806-817. 1010] F. Sun,
On subsequence sums of a zero-sum free sequence , the ElectronicJournal of Combinatorics (2007), ♯ R52.[11] P.Z. Yuan,
On the index of minimal zero-sum sequences over finite cyclicgroups , J. Combin. Theory Ser. A (2007), 1545-1551.[12] P.Z. Yuan,