On products of k atoms II
aa r X i v : . [ m a t h . N T ] M a r ON PRODUCTS OF k ATOMS II
ALFRED GEROLDINGER AND DAVID J. GRYNKIEWICZ AND PINGZHI YUAN
Abstract.
Let H be a Krull monoid with class group G such that every class contains a prime divisor(for example, rings of integers in algebraic number fields or holomorphy rings in algebraic functionfields). For k ∈ N , let U k ( H ) denote the set of all m ∈ N with the following property: There exist atoms u , . . . , u k , v , . . . , v m ∈ H such that u · . . . · u k = v · . . . · v m . Furthermore, let λ k ( H ) = min U k ( H ) and ρ k ( H ) = sup U k ( H ). The sets U k ( H ) ⊂ N are intervals which are finite if and only if G is finite. Theirminima λ k ( H ) can be expressed in terms of ρ k ( H ). The invariants ρ k ( H ) depend only on the classgroup G , and in the present paper they are studied with new methods from Additive Combinatorics. Introduction
Let H be a (commutative and cancelative) monoid. If an element a ∈ H has a factorization a = u · . . . · u k into atoms u , . . . , u k ∈ H , then k is called the length of the factorization, and the set L ( a ) ofall possible lengths is called the set of lengths of a . For k ∈ N , let U k ( H ) denote the set of all m ∈ N withthe following property: There exist atoms u , . . . , u k , v , . . . , v m ∈ H such that u · . . . · u k = v · . . . · v m .Thus U k ( H ) is the union of all sets of lengths containing k . Sets of lengths (and all invariants derivedfrom them, such as their unions) are the most investigated invariants in factorization theory. The sets U k ( H ) were introduced by S.T. Chapman and W.W. Smith in Dedekind domains ([14]) and since thenhave been studied in settings ranging from numerical monoids over Mori domains to monoids of modules([18, 10, 6, 24, 3]). Their suprema ρ k ( H ) = sup U k ( H ) and their minima λ k ( H ) = min U k ( H ) havereceived special attention. Indeed, the invariants ρ k ( H ) were first studied in the 1980s for rings ofintegers in algebraic number fields ([16, 36]). The supremum over all ρ k ( H ) /k is called the elasticity of H , whose investigation was a key topic in early factorization theory (see [1] for a survey, or to pick a fewfrom many, see [12, Problem 38] and [11, 2, 32, 13, 8]).In the present paper, we focus on Krull monoids having the property that every class in the classgroup contains a prime divisor. In Section 2 we present the necessary background and Proposition 2.4gathers the present state of the art. Among others, if H is such a Krull monoid with class group G and2 < | G | < ∞ , then U k ( H ) ⊂ N is a finite interval, hence U k ( H ) = [ λ k ( H ) , ρ k ( H )], and its minimum λ k ( H ) can be expressed in terms of ρ k ( H ). Moreover, ρ k ( H ) depends only on the class group G andhence it can be studied with methods from Additive Combinatorics. This is the starting point for thepresent paper. In Section 3 we discuss open problems, formulate two conjectures (Conjecture 3.3), andoutline the program of the paper. The main results are Theorems 4.1 and 5.1. The latter result is basedon the recent characterization of all minimal zero-sum sequences of maximal length in groups of rank two(see Main Proposition 5.4). Mathematics Subject Classification.
Key words and phrases. non-unique factorizations, sets of lengths, Krull monoids, zero-sum sequences.This work was supported by the Austrian Science Fund FWF, Project No. P26036-N26. Part of this manuscript waswritten while the first author visited South China Normal University in Guangzhou. He wishes to thank the School ofMathematics for their support and hospitality. Unions of sets of lengths in Krull monoids: Background
Let N denote the set of positive integers and set N = N ∪ { } . For real numbers a, b ∈ R , we denote by[ a, b ] = { x ∈ Z | a ≤ x ≤ b } the discrete interval. By a monoid, we mean a commutative semigroup withidentity which satisfies the cancellation law (that is, if a, b, c are elements of the monoid with ab = ac ,then b = c follows). The multiplicative semigroup of non-zero elements of an integral domain is a monoid.Let G be an abelian group, and let A, B ⊂ G be subsets. Then h A i ⊂ G is the subgroup generated by A , − A = {− a | a ∈ A } , and A + B = { a + b | a ∈ A, b ∈ B } is the sumset of A and B . Furthermore, A isa generating set of G if h A i = G , and A is a basis of G if all elements of A are nonzero and G = ⊕ a ∈ A h a i . Monoids and Sets of Lengths.
A monoid F is free abelian, with basis P ⊂ F and we write F = F ( P ),if every a ∈ F has a unique representation of the form a = Y p ∈ P p v p ( a ) with v p ( a ) ∈ N and v p ( a ) = 0 for almost all p ∈ P .
Let H be a monoid. We denote by H × the set of invertible elements of H and by q ( H ) a quotientgroup of H . For a subset H ⊂ H , we denote by [ H ] ⊂ H the submonoid generated by H . Let a, b ∈ H .We say that a divides b (and we write a | b ) if there is an element c ∈ H such that b = ac . We denote by A ( H ) the set of atoms (irreducible elements) of H . If a = u · . . . · u k , where k ∈ N and u , . . . , u k ∈ A ( H ),then k is called the length of the factorization and L ( a ) = { k ∈ N | a has a factorization of length k } ⊂ N is the set of lengths of a . For convenience, we set L ( a ) = { } if a ∈ H × . Furthermore, we denote by L ( H ) = { L ( a ) | a ∈ H } the system of sets of lengths of H .
Next we define the central concept of this paper. Let k ∈ N and suppose that H = H × . Then U k ( H ) = [ a ∈ H, k ∈ L ( a ) L ( a )is the union of all sets of lengths containing k . Thus, U k ( H ) is the set of all m ∈ N such that there areatoms u , . . . , u k , v , . . . , v m with u · . . . · u k = v · . . . · v m . Finally, we define ρ k ( H ) = sup U k ( H ) and λ k ( H ) = min U k ( H ) . Krull monoids.
A monoid homomorphism ϕ : H → F is said to be a divisor homomorphism if ϕ ( a ) | ϕ ( b )in F implies that a | b in H for all a, b ∈ H . A monoid H is said to be a Krull monoid if one of the followingequivalent properties is satisfied (see [23, Theorem 2.4.8] or [31]):(a) H is completely integrally closed and satisfies the ascending chain condition on divisorial ideals.(b) H has a divisor homomorphism into a free abelian monoid.(c) H has a divisor theory: this is a divisor homomorphism ϕ : H → F = F ( P ) into a free abelianmonoid such that for each p ∈ P there is a finite set E ⊂ H with p = gcd (cid:0) ϕ ( E ) (cid:1) .Let H be a Krull monoid. Then every non-unit has a factorization into atoms, and all sets of lengthsare finite. A divisor theory ϕ : H → F = F ( P ) is essentially unique, and the class group C ( H ) = q ( F ) / q ( ϕ ( H )) depends only on H . It will be written additively, and we say that every class contains aprime divisor if, for every g ∈ C ( H ), there is a p ∈ P with p ∈ g .An integral domain R is a Krull domain if and only if its multiplicative monoid R \ { } is a Krullmonoid, and Property (a) shows that a noetherian domain is Krull if and only if it is integrally closed.Rings of integers, holomorphy rings in algebraic function fields, and regular congruence monoids in thesedomains are Krull monoids with finite class group such that every class contains a prime divisor ([23,Section 2.11]). Monoid domains and power series domains that are Krull are discussed in [29, 33, 34].For monoids of modules which are Krull we refer the reader to [5, 3, 17]. N PRODUCTS OF k ATOMS II 3
Main portions of the arithmetic of a Krull monoid—in particular, all questions dealing with sets oflengths—can be studied in the monoid of zero-sum sequences over its class group. We provide the relevantconcepts and summarize the connection in the next subsection.
Transfer homomorphisms and Zero-sum sequences.
Let G be an additively written abelian group, G ⊂ G a subset, and let F ( G ) be the free abelian monoid with basis G . According to the tradition ofcombinatorial number theory, the elements of F ( G ) are called sequences over G . If S = g · . . . · g l ,where l ∈ N and g , . . . , g l ∈ G , then σ ( S ) = g + . . . + g l is called the sum of S , and the monoid B ( G ) = { S ∈ F ( G ) | σ ( S ) = 0 } ⊂ F ( G )is called the monoid of zero-sum sequences over G . Since the embedding B ( G ) ֒ → F ( G ) is a divisorhomomorphism, Property (b) shows that B ( G ) is a Krull monoid. The monoid B ( G ) is factorial if andonly if | G | ≤
2. If | G | ≥
3, then B ( G ) is a Krull monoid with class group isomorphic to G and every classcontains precisely one prime divisor.For every arithmetical invariant ∗ ( H ) defined for a monoid H , it is usual to write ∗ ( G ) instead of ∗ ( B ( G )) (although this is an abuse of language, but there will be no danger of confusion). In particular, weset A ( G ) = A ( B ( G )), L ( G ) = L ( B ( G )), U k ( G ) = U k ( B ( G )), ρ k ( G ) = ρ k ( B ( G )), and λ k ( G ) = λ k ( B ( G )).The next two propositions reveal the universal role of monoids of zero-sum sequences. Proposition 2.1.
Let H be a Krull monoid with class group G such that every class contains a primedivisor. Then there is a transfer homomorphism β : H → B ( G ) . In particular, for every k ∈ N , we have U k ( H ) = U k ( G ) , λ k ( H ) = λ k ( G ) , and ρ k ( H ) = ρ k ( G ) . Proof.
See [23, Theorem 3.4.10]. (cid:3)
Whereas the proof of the above result is quite straightforward, there are recent deep results showingthat there are non-Krull monoids (even non-commutative rings) which allow transfer homomorphisms tomonoids of zero-sum sequences.
Proposition 2.2. Let O be a holomorphy ring in a global field K , A a central simple algebra over K , and H a classicalmaximal O -order of A such that every stably free left R -ideal is free. Then U k ( H ) = U k ( G ) forevery k ∈ N , where G is a ray class group of O and hence finite abelian. Let H be a seminormal order in a holomorphy ring of a global field with principal order b H suchthat the natural map X ( b H ) → X ( H ) is bijective and there is an isomorphism ϑ : C v ( H ) → C v ( b H ) between the v -class groups. Then U k ( H ) = U k ( G ) for every k ∈ N , where G = C v ( H ) is finiteabelian.Proof.
1. See [41, Theorem 1.1], and [4] for related results of this flavor.2. See [24, Theorem 5.8] for a more general result in the setting of weakly Krull monoids. (cid:3)
We need some more notation for sequences over abelian groups (it is consistent with [23, 26, 30]). Asbefore, we fix an additive abelian group G and a subset G ⊂ G . Let S = g · . . . · g l = Y g ∈ G g v g ( S ) ∈ F ( G ) , be a sequence over G (whenever we write a sequence in this way, we tacitly assume that l ∈ N and g , . . . , g l ∈ G ). We set − S = ( − g ) · . . . · ( − g l ) and v G ( S ) = P g ∈ G v g ( S ) for a subset G ⊂ G . We ALFRED GEROLDINGER AND DAVID J. GRYNKIEWICZ AND PINGZHI YUAN call v g ( S ) the multiplicity of g in S , | S | = l = X g ∈ G v g ( S ) ∈ N the length of S , supp( S ) = { g ∈ G | v g ( S ) > } ⊂ G the support of S ,σ ( S ) = l X i =1 g i the sum of S , and Σ( S ) = n X i ∈ I g i | ∅ 6 = I ⊂ [1 , l ] o the set of subsums of S .
For a sequence T ∈ F ( G ), we write gcd( S, T ) ∈ F ( G ) for the maximal length subsequence dividing S and T . We write T | S to indicate that T is a subsequence of S , in which case ST − = T − S denotesthe subsequence obtained from S by removing the terms from T . The sequence S is said to be • zero-sum free if 0 / ∈ Σ( S ), • a zero-sum sequence if σ ( S ) = 0, • a minimal zero-sum sequence if it is a nontrivial zero-sum sequence and every proper subsequenceis zero-sum free.Clearly, the minimal zero-sum sequences are precisely the atoms of the monoid B ( G ), and they play acentral role in our investigations. Now suppose that G is finite. For n ∈ N , let C n denote a cyclic groupwith n elements. If | G | >
1, then we have G ∼ = C n ⊕ . . . ⊕ C n r , and we set d ∗ ( G ) = r X i =1 ( n i −
1) and D ∗ ( G ) = d ∗ ( G ) + 1 , where r = r ( G ) ∈ N is the rank of G , n , . . . , n r ∈ N are integers with 1 < n | . . . | n r and n r = exp( G )is the exponent of G . If | G | = 1, then r ( G ) = 0, exp( G ) = 1, and d ∗ ( G ) = 0. The Davenport constant D ( G ) of G is the maximal length of a minimal zero-sum sequence over G , thus D ( G ) = max (cid:8) | U | (cid:12)(cid:12) U ∈ A ( G ) (cid:9) ∈ N . (note that A ( G ) is finite). In other words, D ( G ) is the smallest integer ℓ such that every sequence S over G has a nontrivial zero-sum subsequence. We denote by d ( G ) the maximal length of a zero-sumfree sequence, and clearly we have 1 + d ( G ) = D ( G ). The next proposition gathers some facts on theDavenport constant which we will use without further mention. Proposition 2.3.
Let G be a finite abelian group. D ∗ ( G ) ≤ D ( G ) ≤ | G | . If G is a p -group or r ( G ) ≤ , then D ∗ ( G ) = D ( G ) . D ( G ) = 1 if and only if | G | = 1 , D ( G ) = 2 if and only if | G | = 2 , and D ( G ) = 3 if and only if G is cyclic of order | G | = 3 or isomorphic to C ⊕ C .Proof. See [23, Chapter 5]. Note that 1. is elementary and that 3. is a simple consequence of 1. and 2.There are more groups G with D ∗ ( G ) = D ( G ) (beyond the ones listed in 2.), but we do not have equalityin general ([25, 40]). (cid:3) The next proposition gathers the state of the art on unions of sets of lengths.
Proposition 2.4.
Let H be a Krull monoid with class group G such that every class contains a primedivisor. If | G | ≤ , then U k ( H ) = { k } for all k ∈ N . If < | G | < ∞ , then, for all k ∈ N , we have U k ( H ) = [ λ k ( G ) , ρ k ( G )] and λ k D ( G )+ j ( H ) = k for j = 02 k + 1 for j ∈ [1 , ρ k +1 ( G ) − k D ( G )]2 k + 2 for j ∈ [ ρ k +1 ( G ) − k D ( G ) + 1 , D ( G ) − , N PRODUCTS OF k ATOMS II 5 provided that k D ( G ) + j ≥ . If G is infinite, then U k ( H ) = N ≥ for all k ≥ .Proof.
1. is classical, for 2. see [18, Theorem 4.1] and [22, Section 3.1], and 3. follows from [23, Theorem7.4.1]. (cid:3)
Let H be a Krull monoid with class group G such that every class contains a prime divisor, or any ofthe monoids in Proposition 2.2. Then Propositions 2.1, 2.2, and 2.4 show that, for a complete descriptionof the sets U k ( H ) of H , it remains to study the invariants ρ k ( G ) of an associated monoid of zero-sumsequences. This is the goal of the present paper.3. The extremal cases in the crucial inequality
We start with a simple and well-known lemma. For convenience, we provide its short proof.
Lemma 3.1.
Let G be a finite abelian group with | G | ≥ , and let k, l ∈ N . k + l ≤ ρ k ( G ) + ρ l ( G ) ≤ ρ k + l ( G ) . ρ k ( G ) = k D ( G ) and (1) 1 + k D ( G ) ≤ ρ k +1 ( G ) ≤ k D ( G ) + j D ( G )2 k . In particular, if D ( G ) = 3 , then ρ k +1 ( G ) = k D ( G ) + 1 . If ρ k +1 ( G ) ≥ m for some m ∈ N and l ≥ k , then ρ l +1 ( G ) ≥ m + ( l − k ) D ( G ) .Proof.
1. By definition, we have ρ m ( G ) ≥ m for each m ∈ N and hence k + l ≤ ρ k ( G ) + ρ l ( G ). Since U k ( G ) + U l ( G ) ⊂ U k + l ( G ), it follows that ρ k ( G ) + ρ l ( G ) = sup U k ( G ) + sup U l ( G ) ≤ sup U k + l ( G ) = ρ k + l ( G ) .
2. A simple counting argument shows that ρ k ( G ) ≤ k D ( G ) /
2; furthermore, if U = g · . . . · g D ( G ) ∈ A ( G ),then ( − U ) k U k = Q D ( G ) i =1 (cid:0) ( − g i ) g i (cid:1) k , whence k D ( G ) ≤ ρ k ( G ) and thus ρ k ( G ) = k D ( G ) (details can befound in [22, Theorem 2.3.1]). Using this and 1., we infer that1 + kD ( G ) = ρ ( G ) + ρ k ( G ) ≤ ρ k +1 ( G ) ≤ (2 k + 1) D ( G )2 . Clearly, D ( G ) = 3 implies that equality holds in both inequalities above.3. By 1. and 2., it follows that ρ l +1 ( G ) ≥ ρ k +1 ( G ) + ρ l − k ) ( G ) ≥ m + ( l − k ) D ( G ) . (cid:3) Our starting point is the crucial inequality (1). We conjecture that cyclic groups are the only groupswhere equality holds on the left hand side, whereas, for all noncyclic groups, there is a k ∗ ∈ N such thatequality holds on the right hand side for all k ≥ k ∗ . We are going to outline this in greater detail (seeConjecture 3.3 and Corollary 3.4). Proposition 3.2.
Let G be a finite abelian group with D ( G ) ≥ . If there exist U ∈ A ( G ) and S , S ∈ F ( G ) such that U = S S , | U | = D ( G ) and Σ( S ) ∪ Σ( − S ) = G \ { } , then ρ ( G ) > D ( G ) + 1 . If G is cyclic, then the property in 1. does not hold and ρ k +1 ( G ) = k D ( G ) + 1 for each k ∈ N . The following conditions are equivalent : ALFRED GEROLDINGER AND DAVID J. GRYNKIEWICZ AND PINGZHI YUAN (a)
There is a k ∗ ∈ N such that ρ k ∗ +1 ( G ) = k ∗ D ( G ) + j D ( G )2 k . (b) There is a k ∗ ∈ N such that ρ k +1 ( G ) = k D ( G ) + j D ( G )2 k for every k ≥ k ∗ . Proof.
1. Let S , S , and U have the above property. Then we choose an element g ∈ G \ (cid:0) Σ( S ) ∪ Σ( − S ) ∪ { } (cid:1) . Since g / ∈ Σ( S ), the sequence ( − S ) g is zero-sum free and U = ( − S ) g (cid:0) σ ( S ) − g (cid:1) ∈ A ( G ). Similarly,it follows that U = ( − S )( − g ) (cid:0) σ ( S ) + g (cid:1) ∈ A ( G ). Since − (cid:0) σ ( S ) − g (cid:1) = σ ( S ) + g in view of0 = σ ( U ) = σ ( S ) + σ ( S ), the product U U U has a factorization into | S | + | S | + 2 = D ( G ) + 2 atomsof length 2.2. Suppose that G is cyclic of order | G | = n . Then [20, Theorem 5.3] implies that ρ k +1 ( G ) = k D ( G )+1for each k ∈ N (see [22, Theorem 5.3.1] for a slightly modified proof). Clearly, every U ∈ A ( G ) of length | U | = | G | has the form U = g n for some g ∈ G with ord( g ) = n . Thus there are no S and S with thegiven properties.3. (a) ⇒ (b) If l ∈ N , then Lemma 3.1 implies that( k ∗ + l ) D ( G ) + j D ( G )2 k ≥ ρ k ∗ + l )+1 ( G ) ≥ ρ k ∗ +1 ( G ) + ρ l ( G )= (cid:18) k ∗ D ( G ) + j D ( G )2 k(cid:19) + l D ( G ) = ( k ∗ + l ) D ( G ) + j D ( G )2 k . (b) ⇒ (a) Obvious. (cid:3) Conjecture 3.3.
Let G be a noncyclic finite abelian group with D ( G ) ≥ . Then the following twoconditions hold : C1.
There exist U ∈ A ( G ) and S , S ∈ F ( G ) such that U = S S , | U | = D ( G ) , and Σ( S ) ∪ Σ( − S ) = G \ { } . C2.
There exists some k ∗ ∈ N such that ρ k +1 ( G ) = k D ( G ) + j D ( G )2 k for each k ≥ k ∗ . In Proposition 3.5, we show that Conjecture C1 holds for groups G with D ( G ) = D ∗ ( G ). All resultsof the present paper support Conjecture C2 . In particular, Theorem 4.1 provides groups satisfying C2 with k ∗ = 1, and Theorem 5.1 shows that C2 need not hold with k ∗ = 1.We start with some consequences of the above conjecture. The Characterization Problem is a centraltopic in factorization theory for Krull monoids (we refer to [23, Sections 7.1 - 7.3] for general information,and to [38, 37, 7, 27] for recent progress). The Characterization Problem studies the question whether ornot the system of sets of lengths of a Krull monoid, which has a prime divisor in every class, determinesthe class group. Thus, if G and G ′ are two finite abelian groups with D ( G ) ≥ L ( G ) = L ( G ′ ),does it follow that G and G ′ are isomorphic? The answer is affirmative (among others) for groups ofrank at most two, and there are no counter examples so far. Corollary 3.4 offers a simple proof in caseof cyclic groups which relies only on the ρ k ( · )-invariants. Corollary 3.4.
Suppose that Conjecture C1 holds. Let H be a Krull monoid with finite class group G such that every class contains a prime divisorand suppose that D ( G ) ≥ . Then the following statements are equivalent :(a) G is cyclic. N PRODUCTS OF k ATOMS II 7 (b) ρ k +1 ( H ) = k D ( G ) + 1 for every k ∈ N . (c) ρ ( H ) = D ( G ) + 1 . Let G be cyclic with D ( G ) ≥ . If G ′ is a finite abelian group with L ( G ) = L ( G ′ ) , then G ∼ = G ′ .Proof.
1. By Proposition 2.1, it suffices to consider ρ k ( G ) for all k ∈ N . The implication (a) ⇒ (b)follows from Proposition 3.2.2, and (b) ⇒ (c) is obvious.(c) ⇒ (a) If G would be noncyclic, then C1 and Proposition 3.2.1 would imply that ρ ( G ) > D ( G )+1.2. Suppose that L ( G ) = L ( G ′ ). Then D ( G ) = ρ ( G ) = ρ ( G ′ ) = D ( G ′ ) and D ( G ′ ) + 1 = D ( G ) + 1 = ρ ( G ) = ρ ( G ′ ) . Thus 1. implies that G ′ is cyclic, and since | G | = D ( G ) = D ( G ′ ) = | G ′ | , G and G ′ are isomorphic. (cid:3) For Conjecture C1 and for Corollary 3.4, the assumption D ( G ) ≥ C and C ⊕ C are the only groups (up to isomorphism) whose Davenport constant is equalto three. The group C ⊕ C does not satisfy C1 , ρ ( C ⊕ C ) = 4 (in contrast to Corollary 3.4.1), and L ( C ) = L ( C ⊕ C ) (see [23, Theorem 7.3.2]).The only groups G with D ( G ) > D ∗ ( G ), for which the precise value of D ( G ) is known, are groups ofthe form C ⊕ C n . We verify Conjecture C1 for them too. Proposition 3.5.
Let G be a noncyclic finite abelian group with D ( G ) ≥ . Let G = C n ⊕ . . . ⊕ C n r where < n | . . . | n r and suppose that there is some s ∈ [1 , r − suchthat n s < n s +1 = . . . = n r . Then ρ ( G ) ≥ D ∗ ( G ) + n s . If D ( G ) = D ∗ ( G ) , then Conjecture C1 holds. If G = C ⊕ C n with n ≥ , then Conjecture C1 holds.Proof. Let { e , . . . , e r } be a basis of G with ord( e i ) = n i for i ∈ [1 , r ] and n | . . . | n r . Set e = e + . . . + e r − .1. Let U = e n − · . . . · e n r − r ( e + e r ) ,U = ( − e ) n − · . . . · ( − e s − ) n s − − ( − e s + e r ) n s − ( − e s +1 ) n s +1 − · . . . · ( − e r ) n r − ( − e − n s e r ) ,U = ( − e s ) n s − ( e s − e r ) n s − ( − e − e r )( e + n s e r ) . Then the U i are each atoms, and clearly U U U is a product of12 | U U U | = 12 (2 D ∗ ( G ) + 2 n s ) = D ∗ ( G ) + n s atoms of length 2. The assertion follows.2. We consider the sequence U = e n − · . . . · e n r − r e , and distinguish two cases.First, suppose that n r >
2. We set S = e n r − r and S = S − U . Then − e − . . . − e r − + e r / ∈ Σ( S )and e + . . . + e r − − e r / ∈ Σ( S ) because − e r = e r .Second, suppose that n r = 2. Then G is an elementary 2-group and r ≥ d ( G ) ≥ S = e e and S = e · . . . · e r e . Then e + e / ∈ Σ( S ) ∪ Σ( − S ).3. Suppose that ord( e ) = . . . = ord( e ) = 2 and ord( e ) = 2 n with n ≥
70. If n is even, then D ( G ) = D ∗ ( G ) by [15, Theorem 5.8], and the assertion follows from 2. Suppose that n is odd. Then D ( G ) = D ∗ ( G ) + 1 by [15, Theorem 5.8]. By [28, Theorem 4], the sequence U = ( e + e )( e + e )( e + e )( e + e )( e − e )( e − e )( e − e )( e − e + e ) n − ( − e ) ALFRED GEROLDINGER AND DAVID J. GRYNKIEWICZ AND PINGZHI YUAN is a minimal zero-sum sequence of length | U | = D ( G ). We set S = ( e − e + e ) n − ( − e ) and S = S − U. Then the element e + e + e − e / ∈ Σ( S ), and we assert that its inverse—namely e + e + e + 2 e = e − e + e —does not lie in Σ( S ). If there would be a subsequence T of S with σ ( T ) = e + e + e + 2 e ,then we would have | T | = 2. But none of the subsequences of S of length two has sum e + e + e + 2 e ,a contradiction. (cid:3) Inductive Bounds
It is the aim of this section to prove the following result which confirms Conjecture C2 (with k ∗ = 1)for the groups G having the form below and satisfying D ( G ) = D ∗ ( G ). Theorem 4.1.
Let H be a Krull monoid with finite class group G such that every class contains a primedivisor. Suppose that G = C s n ⊕ . . . ⊕ C s r n r where < n | . . . | n r and s i ≥ for all i ∈ [1 , r ] . Then ρ k +1 ( H ) ≥ D ∗ ( G ) + j D ∗ ( G )2 k + ( k − D ( G ) for every k ≥ . In particular, if D ( G ) = D ∗ ( G ) , then ρ k +1 ( H ) = k D ( G ) + j D ( G )2 k for every k ≥ . Theorem 4.1 has the following straightforward consequences. Let n ≥
2. It is known that D ( C rn ) = D ∗ ( C rn ) for r ∈ [1 ,
2] and if D ( C rn ) = D ∗ ( C rn ) holds for some r ≥
3, then D ( C sn ) = D ∗ ( C sn ) for all s ∈ [1 , r ].The standing conjecture is that D ( C rn ) = D ∗ ( C rn ) for all r ∈ N . Corollary 4.2.
Let G = C rn , where n ≥ and r ≥ , and suppose that D ( G ) = D ∗ ( G ) . Then, for every k ≥ , we have ρ k +1 ( G ) = ( k D ( G ) + 1 r = 1 k D ( G ) + j D ( G )2 k r ≥ . Proof.
For r = 1, this follows from Proposition 3.2.2. For r ≥
2, it follows from Theorem 4.1. (cid:3)
Corollary 4.3.
Let G = C s q ⊕ . . . ⊕ C s r q r be a p -group where q , . . . , q r are powers of a fixed prime and s , . . . , s r ∈ N ≥ . Then ρ k +1 ( G ) = k D ( G ) + j D ( G )2 k for every k ≥ . Proof.
Since G is a p -group, we have D ( G ) = D ∗ ( G ) by Proposition 2.3, and hence the assertion followsfrom Theorem 4.1. (cid:3) We start with the preparations for the proof of Theorem 4.1. Let G be a finite abelian group. Theinequality ρ ( G ) ≥ ω means there are U , U , U , W , . . . , W ρ ∈ A ( G ) with(2) U U U = W · . . . · W ρ and ρ ≥ ω. For each W i , where i ∈ [1 , ρ ], we may write W i = T i, T i, T i, with the T i,j | U j subsequences such that Q ρi =1 T i,j = U j for each j ∈ [1 , | T i,j | ≤ i and j , then we say that the factorization (2) is weakly reduced . Let W ′ i be the sequenceobtained from W i = T i, T i, T i, by replacing each nonempty T i,j with the sum of its terms. Likewise, let U ′ j be the sequence obtained from U j = Q ρi =1 T i,j by replacing each nonempty T i,j with the sum of itsterms. The sequence X = Q ρi =1 | W ′ i | ∈ F ( Z ) is called a spread for the factorization (2), and it dependson the T i,j (so a given factorization (2) may have multiple spreads). It is readily seen that if U ∈ A ( G ) N PRODUCTS OF k ATOMS II 9 is an atom and T | U is nontrivial, then the sequence U T − σ ( T ), obtained by replacing the terms from T with their sum, is also an atom. Hence the W ′ i and U ′ i are atoms with U ′ U ′ U ′ = W ′ · . . . · W ′ ρ aweakly reduced factorization having | W ′ i | ≤ i . Thus, when ρ ( G ) ≥ ω , we may always assumeour factorization (2) is weakly reduced. We say that ρ ( G ) ≥ ω with spread X ∈ F ( { , , } ) if thereexists a factorization (2) having spread X , in which case, per the argument above, we may also assumethere is a weakly reduced factorization having spread X .If 1 ∈ supp( X ), then some W i , say W , has W = T ,j for some j , implying that the zero-sum sequence W = T ,j is a subsequence of U j . As U j is an atom, this is only possible if W = T ,j = U j , whichforces all other T i,j with i = 1 to be empty in view of Q ρi =1 T i,j = U j . In particular, | W ′ i | ≤ i when 1 ∈ supp( X ), meaning we cannot have both 1 , ∈ supp( X ). From this, we see that we have threemutually exclusive possibilities for a spread X :1 ∈ supp( X ) or supp( X ) = { } or 3 ∈ supp( X ) . Note there is a spread X with 1 ∈ supp( X ) precisely when W s = U t for some s ∈ [1 , ρ ] and t ∈ [1 , ∈ supp( U U U ), then 1 will be a term in any spread X . If supp( X ) = { } and (2) is weakly reduced, sothat | W i | = 2 for all i , then U must have a subsequence xy | U with − x ∈ supp( U ) and − y ∈ supp( U ),with similar statements holding for U and U . When 3 ∈ supp( X ), we refer to a W i with | W ′ i | = 3 as a traversal for the factorization (2). Lemma 4.4.
Let G = G ⊕ G be a finite abelian group where G , G ⊂ G are nontrivial subgroups with D ( G ) ≤ D ( G ) . Then ρ ( G ) ≥ D ( G ) + D ( G ) − with spread X ∈ F ( { , } ) having v ( X ) = 1 .Proof. Let V = v · . . . · v d ( G ) ∈ A ( G ) and W = w · . . . · w d ( G ) ∈ A ( G )be atoms with maximal lengths | V | = d ( G ) + 1 = D ( G ) and | W | = d ( G ) + 1 = D ( G ). Since G and G are nontrivial, 0 / ∈ supp( V W ). Let V ′ = v · . . . · v d ( G ) and W ′ = w · . . . · w d ( G ) and define U = ( v + w ) V ′ W ′ d ( G ) Y i = d ( G )+1 w i = ( V v − )( W w − )( v + w ) ,U = ( − w )( − V ′ ) d ( G ) Y i =1 ( v i − w i ) d ( G ) Y i = d ( G )+1 ( − w i ) , and U = ( − v )( − W ′ ) d ( G ) Y i =1 ( w i − v i ) . It is easily seen that U , U , U ∈ A ( G ) are atoms with ( U ( v + w ) − )( U ( − w ) − )( U ( − v ) − ) havinga factorization into atoms of length 2, evidencing that ρ ( G ) ≥ d ( G )+2 d ( G )2 = 2 D ( G ) + D ( G ) − v + w )( − w )( − v ) the unique traversal. (cid:3) Lemma 4.5.
Let G = G ⊕ G be a finite abelian group where G , G ⊂ G are nontrivial subgroups suchthat ρ ( G ) ≥ ω and ρ ( G ) ≥ ω both hold with respective spreads X, Y ∈ F ( { , } ) . If v ( X ) + v ( Y ) ≥ , then ρ ( G ) ≥ ω + ω − holds with spread Z ∈ F ( { , } ) having v ( Z ) = v ( X ) + v ( Y ) − . If supp( X ) = supp( Y ) = { } , then ρ ( G ) ≥ ω + ω − holds with spread Z ∈ F ( { , } ) having v ( Z ) = 1 .Proof. For i ∈ { , } , let π i : G = G ⊕ G → G i denote the canonical projection.1. First suppose v ( X ) = r ≥ v ( Y ) = s ≥
1. Let V , V , V ∈ A ( G ) and W , W , W ∈ A ( G )be atoms having weakly reduced factorizations V V V = X · . . . · X ρ and W W W = Y · . . . · Y ρ with the X i ∈ A ( G ), the Y i ∈ A ( G ), ρ i ≥ ω i for i ∈ [1 , X i and Y j traversals in their respectivefactorizations for i ∈ [1 , r ] and j ∈ [1 , s ]. In particular, | X i | = | Y j | = 2 for i ≥ r + 1 and j ≥ s + 1, X = a a a with a i ∈ supp( V i ) for i ∈ [1 , Y = b b b with b i ∈ supp( W i ) for i ∈ [1 , V ′ i = V i a − i and W ′ i = W i b − i for i ∈ [1 , U = V ′ W ′ ( a + b ) = ( V a − )( W b − )( a + b ) ,U = V ′ W ′ ( a + b ) = ( V a − )( W b − )( a + b ) and U = V ′ W ′ ( a + b ) = ( V a − )( W b − )( a + b ) . It is easily observed that the U i are atoms. Moreover, V ′ V ′ V ′ = ( V V V ) X − = X · . . . · X ρ and W ′ W ′ W ′ = ( W W W ) Y − = Y · . . . · Y ρ . Thus U U U = X · . . . X ρ Y · . . . · Y ρ W with W =( a + b )( a + b )( a + b ) a traversal in view of a + a + a + b + b + b = σ ( X ) + σ ( Y ) = 0. Moreover, X , . . . , X s , Y , . . . , Y r also remain traversals in this factorization, while no X i nor Y j with i ≥ r + 1 or j ≥ s + 1 can be a traversal in view of | X i | = | Y j | = 2. Thus ρ ( G ) ≥ ρ + ρ − ≥ ω + ω − Z having v ( Z ) = v ( X ) + v ( Y ) − >
0, ensuring Z ∈ F ( { , } ) (noted before Lemma 4.4).Next suppose that either v ( X ) > v ( Y ) or v ( Y ) > v ( X ), say w.l.o.g. the former, so v ( X ) = r > Y ) = { } , the latter in view of Y ∈ F ( { , } ). Let V , V , V ∈ A ( G ) and W , W , W ∈ A ( G ) be atoms havingweakly reduced factorizations V V V = X · . . . · X ρ and W W W = Y · . . . · Y ρ with the X i ∈ A ( G ), the Y i ∈ A ( G ), ρ i ≥ ω i for i ∈ [1 , X i with i ∈ [1 , r ] traversals in theirfactorization, and | Y i | = 2 and | X j | = 2 for all i ∈ [1 , ρ ] and j ≥ r + 1. In particular, X = a a a with a i ∈ supp( V i ) for i ∈ [1 , / ∈ supp( XY ), we have 0 / ∈ supp( V V V W W W ) implying | V i | , | W i | ≥ i ∈ [1 , xy | W with − x | W and − y | W . Now we define U = ( V a − )( W x − y − )( x − a )( y + a + a ) ,U = ( V a − )( W ( − x ) − )( a − x ) and U = ( V a − )( W ( − y ) − )( a − y ) . Obviously, we have U , U ∈ A ( G ) and U ∈ B ( G ). Letting S = U ( y + a + a ) − and considering π ( S )and π ( S ) shows that S is zero-sum free, implying that U ∈ A ( G ). Since a + a + a = σ ( X ) = 0, wehave ( y + a + a ) + ( a − y ) = 0. Thus, since ( V a − )( V a − )( V a − ) = ( V V V ) X − = X · . . . · X ρ and since W W W has a factorization into ρ atoms of length 2, it is now clear that U U U has afactorization using ( ρ −
1) + ( ρ −
2) + 2 ≥ ω + ω − U U U = X · . . . · X ρ Z · . . . · Z ρ with | Z i | = 2 for all i . Hence ρ ( G ) ≥ ρ + ρ − ≥ ω + ω −
1. Moreover, each X i with i ∈ [2 , r ] remainsa traversal for (3), while this cannot be the case for X j with j ≥ r + 1 nor any Z i as | X j | = | Z i | = 2 for j ≥ s + 1 and all i . Thus (3) has a spread Z with v ( Z ) = v ( X ) − v ( X ) + v ( Y ) − . If v ( X ) ≥
2, this shows 3 ∈ supp( Z ), whence Z ∈ F ( { , } ) as discussed before Lemma 4.4. On theother hand, if v ( X ) = 1, then all atoms in the factorization (3) have length 2. Thus, since | U j | = | V j | + | W j | − ≥ j ∈ [1 , U j , meaning 1 / ∈ supp( Z ) for any spread Z for (3), also explained above Lemma 4.4. In this case,supp( Z ) = { } , completing the proof of Part 1. N PRODUCTS OF k ATOMS II 11
2. Suppose supp( X ) = supp( Y ) = { } . Let V , V , V ∈ A ( G ) be atoms such that V V V has aweakly reduced factorization into ρ ≥ ω atoms of length 2 and let W , W , W ∈ A ( G ) be atoms suchthat W W W has a weakly reduced factorization into ρ ≥ ω atoms of length 2. As explained beforeLemma 4.4, we may assume there is a length 2 subsequence xy | W with − x | W and − y | W and alength 2 subsequence ab | V with − a | V and − b | V . Now we define U = ( V ( − a ) − )( W x − y − )( x − a )( y ) ,U = ( V a − b − )( W ( − x ) − )( a − x )( b ) and U = ( V ( − b ) − )( W ( − y ) − )( − b − y ) . Obviously, we have U ∈ A ( G ) and U , U ∈ B ( G ). Letting S = U y − and considering π ( S ) and π ( S )shows that S is zero-sum free, implying that U ∈ A ( G ). Likewise, letting T = U b − and considering π ( T ) and π ( T ) shows that T is zero-sum free, implying that U ∈ A ( G ). Let c = y , c = b and c = − b − y . Since V V V and W W W both have factorizations into atoms of length 2, it is nowclear that ( U c − )( U c − )( U c − ) has a factorization into ( ρ −
2) + ( ρ −
2) + 1 = ρ + ρ − c c c gives a factorization of U U U into ρ + ρ − ρ ( G ) ≥ ρ + ρ − ≥ ω + ω − Z having v ( Z ) = 1, ensuring Z ∈ F ( { , } ) (noted before Lemma 4.4). (cid:3) Lemma 4.6.
Let G = C n with n ≥ . Then ρ ( G ) ≥ D ∗ ( G ) + ⌊ D ∗ ( G )2 ⌋ with spread X ∈ F ( { , } ) .Moreover, v ( X ) = 1 if D ∗ ( G ) is odd, and supp( X ) = { } if D ∗ ( G ) is even.Proof. Let { e , e , e } be a basis of G and for i ∈ [1 , π i : G = h e i ⊕ h e i ⊕ h e i → h e i i denote thecanonical projection. Note that D ∗ ( G ) = 3 n − ≡ n mod 2. We handle two cases.CASE 1: n is odd.Then D ∗ ( G ) = 3 n − ≥ U = e n − e n − ( e + e + e ) n − c with c = 2 e + 2 e + e ,U = ( − e ) n − e n − ( − e − e − e ) n − ( e − e ) n − c with c = − e − e + n + 12 e and U = ( − e ) n − ( − e ) n − ( − e − e − e ) n − ( − e + e ) n − c with c = − e − e + n − e . Clearly, U i ∈ B ( G ) for i ∈ [1 , π ( U c − ), π ( U c − ) and π ( U c − ), we infer that thesequences U i c − i are zero-sum free for every i ∈ [1 , U , U , U ∈ A ( G ), and it isnow easily seen that ( U c − )( U c − )( U c − ) has a factorization into atoms of length 2, which togetherwith the unique traversal c c c shows that ρ ( G ) ≥ n − = D ∗ ( G ) + ⌊ D ∗ ( G )2 ⌋ holds with spread X having v ( X ) = 1, so that X ∈ F ( { , } ) as noted before Lemma 4.4.CASE 2: n is even.Then D ∗ ( G ) = 3 n − ≥ U = e n − e n − ( e + e + e ) n − (2 e + e + e )( e + 2 e + e ) ,U = ( − e ) n − e n − ( − e − e − e ) n/ − ( e − e + e ) n/ − ( − e − e − e )( e − e + 2 e ) and U = ( − e ) n − ( − e ) n − ( − e − e − e ) n/ − ( − e + e − e ) n/ − ( − e − e − e )( − e + e − e ) . Clearly, U i ∈ B ( G ) for i ∈ [1 , π ( U ), π ( U ) and π ( U ), we infer that U , U , U ∈ A ( G ).By construction, U U U has a factorization into atoms of length 2, say U U U = Z · . . . · Z | U U U | ,implying that ρ ( G ) ≥ | U U U | = n − = D ∗ ( G )+ ⌊ D ∗ ( G )2 ⌋ . Moreover, since | U i | = 3 n − > | Z j | for all i and j , we see that 1 / ∈ supp( X ) in any spread X , whence supp( X ) = { } , completing the proof. (cid:3) Proof of Theorem 4.1.
By Proposition 2.1, we have ρ k ( H ) = ρ k ( G ) for all k ≥
1. By Lemma 3.1.1 andLemma 3.1.2, it suffices to prove the assertion for k = 1. By hypothesis, G can be written in the form G = C t m ⊕ . . . ⊕ C t α m α , where { m , . . . , m α } = { n , . . . , n r } with t i ∈ { , } . We proceed by induction on α to show that ρ ( G ) ≥ D ∗ ( G ) + ⌊ D ∗ ( G )2 ⌋ holds with spread X ∈ F ( { , } ) with v ( X ) = 1 when D ∗ ( G ) is odd and withsupp( X ) = { } when D ∗ ( G ) is even, which will complete the proof.Since n | . . . | n r with { m , . . . , m α } = { n , . . . , n r } , we have D ∗ ( G ) = d ∗ ( C t m ) + d ∗ ( C t m ⊕ . . . ⊕ C t α m α ) + 1 = D ∗ ( K ) + D ∗ ( L ) − , where K = C t m and L = C t m ⊕ . . . ⊕ C t α m α . If t = 2, then D ∗ ( K ) = D ∗ ( C m ) = 2 m − ρ ( K ) = ρ ( C m ) ≥ m − m −
1) + ⌊ m − ⌋ = D ∗ ( K ) + ⌊ D ∗ ( K )2 ⌋ with spread X having v ( X ) = 1, so that X ∈ F ( { , } ). If t = 3, then Lemma 4.6 implies that ρ ( K ) = ρ ( C m ) ≥ D ∗ ( K ) + ⌊ D ∗ ( K )2 ⌋ with spread X ∈ F ( { , } ). Moreover, v ( X ) = 1 if D ∗ ( K ) is odd,and supp( X ) = { } if D ∗ ( K ) is even. This completes the base case when α = 1. Thus we may assume α ≥
2, in which case the induction hypothesis ensures that ρ ( K ) ≥ D ∗ ( K ) + ⌊ D ∗ ( K )2 ⌋ and ρ ( L ) ≥ D ∗ ( L ) + ⌊ D ∗ ( L )2 ⌋ with respective spreads X, Y ∈ F ( { , } ).If D ∗ ( K ) and D ∗ ( L ) are both even, then D ∗ ( G ) = D ∗ ( K )+ D ∗ ( L ) − X ) = supp( Y ) = { } , whence Lemma 4.5.2 yields ρ ( G ) ≥ D ∗ ( K ) + D ∗ ( K )2 + D ∗ ( L ) + D ∗ ( L )2 − D ∗ ( G ) + D ∗ ( G ) − Z ∈ F ( { , } ) having v ( Z ) = 1, as desired. If D ∗ ( K ) and D ∗ ( L ) are both odd, then v ( X ) = v ( Y ) = 1 and D ∗ ( G ) = D ∗ ( K ) + D ∗ ( L ) − ρ ( G ) ≥ D ∗ ( K ) + D ∗ ( K ) −
12 + D ∗ ( L ) + D ∗ ( L ) − − D ∗ ( G ) + D ∗ ( G ) − Z ∈ F ( { , } ) having v ( Z ) = v ( X ) + v ( Y ) − D ∗ ( K ) and D ∗ ( L ) have different parities, then v ( X ) + v ( Y ) = 1, D ∗ ( G ) = D ∗ ( K ) + D ∗ ( L ) − ρ ( G ) ≥ D ∗ ( K ) + D ∗ ( K )2 + D ∗ ( L ) + D ∗ ( L )2 − − D ∗ ( G ) + D ∗ ( G )2with spread Z ∈ F ( { , } ) having v ( Z ) = v ( X ) + v ( Y ) − Z ) = { } . This completesthe induction. When D ∗ ( G ) = D ( G ), the needed upper bound comes from Lemma 3.1.2 (cid:3) Groups of rank two
The aim of this section is to prove the following characterization. It provides the first non-cyclic groups G at all for which ρ k +1 ( G ) is strictly smaller than the upper bound k D ( G ) + ⌊ D ( G ) / ⌋ for some k ∈ N . Theorem 5.1.
Let H be a Krull monoid with finite class group G such that every class contains a primedivisor. Suppose that G = C m ⊕ C mn with n ≥ and m ≥ . Then ρ ( H ) = D ( G ) + j D ( G )2 k if and only if n = 1 or m = n = 2 . N PRODUCTS OF k ATOMS II 13
We start with two corollaries providing examples of groups G having rank two which show thatTheorem 5.1 is sharp in two aspects. Indeed, Corollary 5.2 shows that these groups G satisfy ρ ( G ) = D ( G ) + j D ( G )2 k − ρ k +1 ( G ) = k D ( G ) + j D ( G )2 k for all k ≥ . After that, we deal with groups of the form G = C ⊕ C n where n ≥
3. Since for cyclic groups G wehave ρ k +1 ( G ) = k D ( G ) + 1 for all k ≥
1, groups of the form C ⊕ C n are the canonical first choice fortesting Conjecture C2 . Indeed, we verify Conjecture C2 for them and show that there exists an integer k ∗ ∈ N (by Theorem 5.1 we must have k ∗ > n >
2) such that ρ k +1 ( G ) = k D ( G ) + j D ( G )2 k for all k ≥ k ∗ . Moreover, Corollary 5.3 provides the first example of a group where, for some odd k ∈ N , strict inequalitieshold in the crucial inequality (1). Corollary 5.2.
Let G = C m ⊕ C m with m ≥ . If m = 2 , then ρ k +1 ( G ) = k D ( G ) + ⌊ D ( G )2 ⌋ for every k ≥ . If m ≥ , then ρ ( G ) ≥ D ( G ) + ( m + 1) . If m ∈ { , } , then ρ ( G ) = D ( G ) + j D ( G )2 k − and ρ k +1 ( G ) = k D ( G ) + j D ( G )2 k for all k ≥ . Proof.
Let { e , e } be a basis of G with ord( e ) = m and ord( e ) = 2 m . Then D ( G ) = 3 m − U = e e ( e + e ) , U = e ( − e ) ( e − e ) and U = e ( e − e ) . Obviously, U U U may be written as a product of 7 which implies that ρ ( G ) = D ( G ) + ⌊ D ( G )2 ⌋ . Nowthe assertion follows from Lemma 3.1.3.2. We define U = e m − e m − ( e + e ) , U = ( − e ) m − e m − ( − e + e ) U = ( e + e ) m − ( − e ) m − ( e + me ) and U = ( − e − e ) m − ( − e + me ) ( e − e ) . Then U , U , U , U ∈ A ( G ) (note we need m ≥ U ∈ A ( G )), | U | = | U | = | U | = D ( G ) and | U | = 2 m + 2). By construction, U U U U has a factorization into atoms of length 2, which impliesthat ρ ( G ) ≥ | U U U U | D ( G ) + ( m + 1) .
3. Proposition 3.5.1 and Theorem 5.1 imply that D ( G ) + m ≤ ρ ( G ) ≤ D ( G ) + j D ( G )2 k − , which is an equality because m ∈ { , } . By Lemma 3.1.3, it suffices to show that ρ ( G ) ≥ D ( G ) + j D ( G )2 k , which follows from 2. above because m ∈ { , } ensures j D ( G )2 k = m + 1. (cid:3) Corollary 5.3.
Let G = C ⊕ C n with n ≥ . Then D ( G ) + 1 < ρ ( G ) < D ( G ) + j D ( G )2 k and ρ k +1 ( G ) = k D ( G ) + j D ( G )2 k for every k ≥ n − . Proof.
We have D ( G ) = D ∗ ( G ) = 2 n + 1. The left inequality follows from Proposition 3.5.2 and fromProposition 3.2.1, and the right inequality follows from Theorem 5.1.To prove the second statement, let { e , e } be a basis of G with ord( e ) = 2 and ord( e ) = 2 n . For i ∈ [1 , n ], we define U i = e n − (cid:0) e − ( i − e (cid:1)(cid:0) e + ie (cid:1) ∈ A ( G ) . Let V = ( e + e ) n − e e ∈ A ( G ) . Let W = e ( e + e )( e − e ). By construction, S = (cid:16) U ( − U ) (cid:17) · . . . · (cid:16) U n ( − U ) (cid:17)(cid:16) U ( − U ) V (cid:17) isa product of 4( n −
1) + 3 = 4 n − SW − has a factorization into atoms of length 2. Thisimplies that ρ n − ( G ) ≥ | S | −
32 = 1 + (4 n − n + 1) −
32 = (2 n − D ( G ) + j D ( G )2 k . The result now follows from Lemma 3.1.3. (cid:3)
The proof of Theorem 5.1 is based on the recent characterization of minimal zero-sum sequences ofmaximal length in groups of rank two, which will be formulated in Main Proposition 5.4. The proof ofthe characterization is obtained by combining the main results from [19], [21], [35], [39] with a few smallorder groups handled by direct computation [9]. The version below is derived from this original in a fewshort lines [7, Theorem 3.1] (apart from (e) and the fact that both parts of (d) hold when n = 2, whichwe will deduce from the rest of theorem in the explanations below). It eliminates some overlap betweentype I and II in the original statement. Main Proposition 5.4.
Let G = C m ⊕ C mn with n ≥ and m ≥ . A sequence S over G of length D ( G ) = m + mn − is a minimal zero-sum sequence if and only if it has one of the following two forms : • S = e ord( e ) −
11 ord( e ) Y i =1 ( x i e + e ) , where (a) { e , e } is a basis of G , (b) x , . . . , x ord( e ) ∈ [0 , ord( e ) − and x + . . . + x ord( e ) ≡ e ) .In this case, we say that S is of type I(a) or I(b) according to whether ord( e ) = m or ord( e ) = mn > m . • S = f sm − f ( n − s ) m + ǫ m − ǫ Y i =1 ( − x i f + f ) , where (a) { f , f } is a generating set for G with ord( f ) = mn and ord( f ) > m , (b) ǫ ∈ [1 , m − and s ∈ [1 , n − , (c) x , . . . , x m − ǫ ∈ [1 , m − with x + . . . + x m − ǫ = m − , (d) either s = 1 or mf = mf , with both holding when n = 2 , and (e) either ǫ ≥ or mf = mf .In this case, we say that S is of type II. We gather some simple consequences of the above characterization which will be used without furthermention. Let all notation be as in the Main Proposition 5.4.It is easy to see that | supp( S ) | ≥ N PRODUCTS OF k ATOMS II 15
When S has type II, it is always possible to find some f ′ ∈ G such that { f ′ , f } is a basis for G withord( f ′ ) = m and f = f ′ + αf for some α ∈ [1 , mn −
1] (see [7]). In particular, since mf = 0 (in viewof ord( f ) > m ), we have ord( f ) = tm for some t ≥ t | n . Moreover, it is now readily checkedthat, regardless of whether S has type I or II, every term of S must have its order being a multiple of m .When S has type II, it is clear that − x i f + f = − x i f ′ + (1 − αx i ) f = f in view of x i ∈ [1 , m − i ∈ [1 , m − ǫ ]. Likewise, a term − x i f + f = − x i f ′ + (1 − αx i ) f could only equal f = f ′ + αf if x i = m − − α ( m −
1) = 1 − αx i ≡ α mod mn , implying 1 ≡ αm mod mn , which is not possible.Consequently, we see that a term − x i f + f can never equal f or f . Likewise, since ord( f ) ≥ m , − x i f + f = − x j f + f is only possible if x i = x j ∈ [1 , m − S has type II, the condition x + . . . + x m − ǫ = m − x i ∈ [1 , m −
1] forces max x i ≤ ( m − − ( m − ǫ −
1) = ǫ . Thus we always have x i ≤ ǫ . In particular, if ǫ = 1, then x i = 1 for all i ∈ [1 , m − S has type II, then s ∈ [1 , n −
1] forces n ≥
2. Suppose n = 2. Then s = 1 and ord( f ) =ord( f ) = 2 m . Let f ′ ∈ G be such that { f ′ , f } is a basis for G with ord( f ′ ) = m . Let g = xf ′ + yf ∈ G with x, y ∈ Z . If y is odd, then mg = xmf ′ + ymf = ymf = 0, implying ord( g ) > m and thusord( g ) = 2 m . On the other hand, if ord( g ) = 2 m , then 0 = mg = ymf , implying y is odd. Consequently,the elements g ∈ G with ord( g ) = 2 m are precisely those g = xf ′ + yf with x, y ∈ Z and y odd, meaningany g ∈ G with ord( g ) = 2 m has mg = mf . In particular, mf = mf . This explains why bothconditions of (d) always hold when n = 2.If n >
1, then there are at most m − m in S . Indeed, if S has type I(a), then all termsof order m are contained in Q mi =1 ( x i e + e ). However, since m m P i =1 ( x i e + e ) = me = 0, they cannotall have order m , meaning there are at most m − S has type I(b), then it is clear thatall terms of the form x i e + e have order mn > m , leaving at most m − m , all equal to e .Finally, if S has type II, then we have ord( f ) ≥ m as remarked above. Thus only terms contained in Q m − ǫi =1 ( − x i f + f ) can have order m , meaning there are at most m − ǫ ≤ m − S has type II and contains precisely m − m , then we must have ǫ = 1with each term from Q m − i =1 ( − x i f + f ) having order m . However, since we have x i ∈ [1 , ǫ ] as remarkedabove, this is only possible if S = f sm − f ( n − s ) m +12 ( f − f ) m − with ord( f − f ) = m . In such case, { f , f − f } is also a generating set for G with ord( f ) = mn and ord( f − f ) = m , whichforces { f , f − f } to be a basis for G . Thus S has type I(b) (taking e = f − f and e = f ).In particular, if S had type II with mf = mf and ǫ = 1, then S would also have type I(b). Indeed ǫ = 1 forces x i = 1 for all i ∈ [1 , m −
1] in view of x i ∈ [1 , ǫ ], while each − x i f + f = − f + f has order m in view of mf = mf (and the fact that every term of S has its order being a multiple of m ). Thuswe would have m − m , so that the above argument shows that S has type I(b). Thisargument is what allows us to assume (e) in Main Proposition 5.4. In particular, if S has type II and n = 2, then ǫ ≥ m ≥ ǫ ∈ [2 , m − Lemma 5.5.
Let G = C m ⊕ C mn with n ≥ and m ≥ . Suppose S is a minimal zero-sum sequenceover G of length D ( G ) = m + mn − that is of type II, say S = f sm − f ( n − s ) m + ǫ m − ǫ Y i =1 ( − x i f + f ) with all notation as in Main Proposition 5.4. Suppose T | S is a subsequence with | T | ≥ m − . Then T contains a subsequence T | T with σ ( T ) = mf . Furthermore, if T has no proper subsequence with thisproperty, then T = f m − f ǫ Q m − ǫi =1 ( − x i f + f ) . Proof.
Since s ∈ [1 , n − n ≥
2. If s = 1, then v f ( T ) ≤ v f ( S ) = m −
1. On the otherhand, if s >
1, then mf = mf , in which case we must also have v f ( T ) ≤ m − f m | T will be aproper subsequence whose sum is mf = mf , as desired. Thus we may assume(4) v f ( T ) = m − − t for some t ∈ [0 , m − . Likewise, we must have v f ( T ) ≤ m − f m | T will be a proper subsequence whose sum is mf , asdesired. By re-indexing the − x i f + f appropriately, we may w.l.o.g. assume(5) ℓ Y i =1 ( − x i f + f ) = gcd m − ǫ Y i =1 ( − x i f + f ) , T ! , where ℓ ∈ [0 , m − ǫ ] . Hence, from the hypothesis | T | ≥ m −
1, we deduce that(6) v f ( T ) = | T | − v f ( T ) − ℓ ≥ m + t − ℓ. In particular, v f ( T ) ≤ m − ℓ ≥ t + 1 ≥ x + . . . + x m − ǫ = m − x i ∈ [1 , m −
1] for all i . Thus x + . . . + x ℓ = m − − x with x := m − ǫ X i = ℓ +1 x i ≥ m − ǫ − ℓ ≥ . Consequently, if t ≤ x , then the sequence f m − − t Q ℓi =1 ( − x i f + f ) contains at least ℓ disjoint subse-quences each having sum f and containing precisely one term of the form − x i f + f , while if t ≥ x ,then the sequence f m − − t Q ℓi =1 ( − x i f + f ) contains at least ℓ − (cid:16) ( m − − x ) − ( m − − t ) (cid:17) = ℓ − t + x disjoint subsequences each having sum f and containing precisely one term of the form − x i f + f . Ineither case, we have R · . . . · R w | f m − − t ℓ Y i =1 ( − x i f + f ) with σ ( R i ) = f for i ∈ [1 , w ] , where w = min { ℓ, ℓ − t + x } . Moreover, the subsequence R · . . . · R w of f m − − t Q ℓi =1 ( − x i f + f ) willbe proper unless m − − t = x + . . . + x ℓ = m − − x , i.e., unless t = x .Now, if t < x , then T = R · . . . · R ℓ f m − ℓ is a proper subsequence of T (in view of (4), (5), (6) and t = x ) with sum σ ( T ) = mf , as desired. On the other hand, if t ≥ x , then T = R · . . . · R ℓ − t f m − ℓ + t is a subsequence of T (in view of (4), (5) and (6)) with sum σ ( T ) = mf . Moreover, it will be a propersubsequence of T unless t = x = 0 and equality holds in (6). From x + . . . + x ℓ = m − − x = m − ℓ = m − ǫ in this case (recall that x + . . . + x m − ǫ = m − x i ∈ [1 , m −
1] for all i ),and now T = f m − − t f m + t − ℓ ℓ Y i =1 ( − x i f + f ) = f m − f ǫ m − ǫ Y i =1 ( − x i f + f ) , completing the proof. (cid:3) We are now ready to proceed with the proof of Theorem 5.1.
Proof of Theorem 5.1.
By Proposition 2.1, we have ρ ( H ) = ρ ( G ). We study ρ ( G ) and recall that D ( G ) = D ∗ ( G ) = m + mn −
1. If n = 1, then G = C m ⊕ C m , and the theorem follows from Corollary 4.2.If m = n = 2, then G = C ⊕ C , and the theorem follows from Corollary 5.2.1. We now assume n ≥ N PRODUCTS OF k ATOMS II 17 m ≥ n = 2. In particular, D ( G ) ≥
7. It remains to show ρ ( G ) < ρ := ⌊ D ( G ) / ⌋ = ⌊ m +3 mn − ⌋ in this case. Assume by contradiction that there are U , U , U , V , . . . , V ρ ∈ A ( G ) such that U U U = V · . . . · V ρ . Without loss of generality, we may assume | U | ≥ | U | ≥ | U | and | V | ≥ . . . ≥ | V ρ | . We continue byshowing we can assume the following assertion holds true. Note that | U | = D ( G ) − D ( G ) is odd and | V | = 2. Assertion A. | U | = | U | = D ( G ) and D ( G ) − ≤ | U | ≤ D ( G ) with the U i satisfying either U = AB, − U = AC, U = ( − B ) C, | A | = (cid:24) D ( G )2 (cid:25) and | V | = 2 or U = ABw , − U = ACw , U = ( − B ) C ( w − w ) , | A | = D ( G ) −
12 and | V | = 3 , where A = gcd( U , − U ) , B = gcd( U , − U ) , C = gcd( − U , U ) , | B | = | C | = (cid:22) D ( G )2 (cid:23) and w , w ∈ G. Proof of Assertion A.
We trivially have | U U U | = | U | + | U | + | U | ≤ D ( G ). Also, | V i | ≥ i (as 0 cannot divide any U i , else ρ ≤ D ( G ) + 1), implying3 D ( G ) ≥ | U U U | = | V · . . . · V ρ | = ρ X i =1 | V i | ≥ ρ = 2 ⌊ D ( G ) / ⌋ ≥ D ( G ) − , with equality in the latter estimate only possible when D ( G ) is odd. It follows that, if D ( G ) is even, then | U | = | U | = | U | = D ( G ) with | V i | = 2 for all i , while if D ( G ) is odd, then either | U | = | U | = | U | = D ( G ) with | V | = 3 and | V i | = 2 for all i ≥ | U | = | U | = D ( G ) and | U | = D ( G ) − | V i | = 2for all i .When | V i | = 2 for all i , then S = U U U has a factorization into length 2 atoms. Thus U = AB , − U = AC and U = ( − B ) C for some A, B, C ∈ F ( G ). Since | A | + | B | = | U | = D ( G ) = | U | = | A | + | C | ,it follows that | B | = | C | . But now 2 | B | = | B | + | C | = | U | ∈ { D ( G ) , D ( G ) − } , implying | B | = | C | = j D ( G )2 k and | A | = | U | − | B | = D ( G ) − j D ( G )2 k = l D ( G )2 m . If there is some g ∈ supp( B ) ∩ supp( C ),then U will contain both g and − g . However, since U is an atom, this is only possible if | U | = 2,contradicting that | U | ≥ D ( G ) − ≥
6. Therefore we instead conclude that supp( B ) ∩ supp( C ) = ∅ ,implying gcd( U , − U ) = A . Similar arguments show that B = gcd( U , − U ) and C = gcd( − U , U ),completing the proof of Assertion A in this case. It remains to consider the case when | V | = 3 with | V i | = 2 for i ≥
2, which is only possible when | U | = | U | = | U | = D ( G ) is odd.If some U i , say w.l.o.g. U , contains two terms from V , say g g | gcd( V , U ), then replacing U by U ′ = U ( g g ) − ( g + g ) and replacing V by V ′ = V ( g g ) − ( g + g ) yields atoms U , U , U ′ ∈ A ( G )having a factorization U U U ′ = V ′ V . . . V ρ with | U | = | U | = D ( G ), | U ′ | = D ( G ) − | V ′ | = | V | = . . . = | V ρ | = 2. These atoms also provide a counter-example to the theorem and satisfy the previouslyhandled case of Assertion A. Thus we may assume (for the purpose of proving the theorem) that thisdoes not occur: no length two subsequence of V divides any U i . In consequence, precisely one of each ofthe three terms of V occurs in each U i while ( U U U ) V − has a factorization into length 2 atoms (inview of | V i | = 2 for i ≥ U = ABw , − U = ACw and U = ( − B ) C ( w − w ) forsome A, B, C ∈ F ( G ), where V = w ( − w )( w − w ).Since | A | + | B | + 1 = | U | = D ( G ) = | U | = | A | + | C | + 1, it follows that | B | = | C | . But now2 | B | + 1 = | B | + | C | + 1 = | U | = D ( G ) follows, implying | B | = | C | = D ( G ) − = j D ( G )2 k and | A | = | U | − | B | − D ( G ) − D ( G ) − − D ( G ) − .Suppose there were some g ∈ supp( Bw ) ∩ supp( Cw ). Note w = w , else V would contain a length 2zero-sum subsequence, contradicting that V is an atom. Consequently, if g = w , then w = g ∈ supp( C ), in which case U contains the two term subsequence w ( w − w ) of V , contrary to assumption. Likewise,if g = w , then w ∈ supp( B ), in which case U contains the two term subsequence ( − w )( w − w ) of V ,once more contrary to assumption. On the other hand, if g ∈ supp( B ) ∩ supp( C ), then U will containboth g and − g , yielding the contradiction 2 = | U | ≥ D ( G ) − | V i | = 2 for all i . Sowe instead conclude that supp( Bw ) ∩ supp( Cw ) = ∅ , implying gcd( U , − U ) = A . Similar argumentsshow that B = gcd( U , − U ) and C = gcd( − U , U ), completing the proof of Assertion A. (cid:3) In view of Assertion A, we see that we can apply Main Proposition 5.4 to U and − U to characterizethe possible structures for U and − U . Since the roles of U and U are symmetric, this gives us sixcases.CASE 1: U and − U are both of type I(b), say U = e m − mn Y i =1 ( x i e + e ) and − U = f m − mn Y i =1 ( y i f + f ) , where { e , e } and { f , f } are bases for G with ord( e ) = ord( f ) = m and ord( e ) = ord( f ) = mn > n .Let H = h e , f i . Since ord( e ) = ord( f ) = m , we conclude that H is isomorphic to a subgroupof C m . In particular, D ( H ) ≤ D ( C m ) = 2 m −
1. Since m, n ≥ n ≥ m = 2, we have | B | = | C | ≥ D ( G ) − = mn + m − > m . Likewise | A | ≥ D ( G ) − > m . Any element of the form xe + e or yf + f , where x, y ∈ Z , has order mn > m = ord( e ) = ord( f ) and thus cannot be equal to e nor f . Since | A | ≥ m + 1, we conclude that A must contain a term from U of the form xe + e , whichmust, by the previously mentioned order restriction, be equal to a term from − U of the form yf + f .Hence f − e ∈ H . But now it is clear that difference between any two terms of the form x ′ e + e and y ′ f + f , where x ′ , y ′ ∈ Z , must also be an element from H .If e = f , then H ∼ = C m and D ( H ) = D ( C m ) = m . In this case, B = b · . . . · b ℓ consists entirelyof terms of the form xe + e while C = c · . . . · c ℓ consists entirely of terms of the form yf + f ,where ℓ = | B | = | C | ≥ m + 1. Consequently, ( − b + c ) · . . . · ( − b m + c m ) ∈ F ( H ) is a sequence of m = D ( H ) terms from H , meaning ( − B ) C contains a nontrivial zero-sum subsequence of length at most2 m < ℓ = | B | + | C | ≤ | U | . But this contradicts that U is an atom with ( − B ) C | U . Therefore wemay now assume e = f .In view of e = f and the previously mentioned order restriction, neither e nor f can be a termfrom A . Thus every term equal to e in U must be contained in B except possibly one such term equalto w . Likewise, every term equal to f in − U must be contained in C except possibly one such termequal to w . It follows that m − ≤ v e ( B ) ≤ m − m − ≤ v f ( C ) ≤ m −
1. Consequently, in viewof | B | = | C | ≥ m + 1, there must be subsequences b · b | B and c · c | C with each term b i of the form b i = x ′ i e + e and each term c i of the form c i = y ′ i f + f . Moreover, if v e ( B ) = v f ( C ) = m −
2, thenthere exists a third term b from B also of the form b = x ′ e + e and a third term c from C also ofthe form c = y ′ f + f so that b · b · b | B and c · c · c | C . Observe that v f ( C ) < m −
1, as well as v e ( B ) < m −
1, is only possible if U = ( − B ) C ( w − w ).If v e ( B ) = v f ( C ) = m −
1, then ( − e ) m − f m − ( − b + c ) ∈ F ( H ) is a sequence of terms from H oflength 2 m − ≥ D ( H ), meaning ( − B ) C contains a nontrivial zero-sum subsequence of length at most2 m < ℓ = | B | + | C | ≤ | U | . But this contradicts that U is an atom with ( − B ) C | U .If v e ( B ) = v f ( C ) = m −
2, then ( − e ) m − f m − ( − b + c )( − b + c )( − b + c ) ∈ F ( H ) is a sequenceof length 2 m − ≥ D ( H ), meaning ( − B ) C contains a nontrivial zero-sum subsequence, contradictingthat U is an atom since U = ( − B ) C ( w − w ).If v e ( B ) = m − v f ( C ) = m −
2, then ( − e ) m − f m − ( − b + c )( − b + c ) ∈ F ( H ) is a sequenceof length 2 m − ≥ D ( H ), meaning ( − B ) C contains a nontrivial zero-sum subsequence, contradictingthat U is an atom since U = ( − B ) C ( w − w ). N PRODUCTS OF k ATOMS II 19 If v e ( B ) = m − v f ( C ) = m −
1, then ( − e ) m − f m − ( − b + c )( − b + c ) ∈ F ( H ) is a sequenceof length 2 m − ≥ D ( H ), meaning ( − B ) C contains a nontrivial zero-sum subsequence, contradictingthat U is an atom since U = ( − B ) C ( w − w ), which completes CASE 1.CASE 2: U and − U are both of type I(a), say U = e mn − m Y i =1 ( x i e + e ) and − U = f mn − m Y i =1 ( y i f + f ) . where { e , e } and { f , f } are bases for G with ord( e ) = ord( f ) = mn > m and ord( e ) = ord( f ) = m .Since m, n ≥ n ≥ m = 2, we have mn − > mn + m = D ( G )+12 ≥ | A | . If e = f , thengcd( U , − U ) = A implies | A | ≥ v e ( U ) = mn −
1, contrary to what we just noted. Therefore e = f .On the other hand, since v e ( U ) = v f ( − U ) = mn − > D ( G )+12 ≥ D ( G ) − | A | = | U | − | A | = | U | − | A | ,we must have e , f ∈ supp( A ). It follows that e = yf + f and f = xe + e for some x, y ∈ Z . Since U contains at most m terms not equal to e , we deduce that v e ( A ) ≥ | A | − m ≥ D ( G ) − − m = mn − m −
1. However, since e = f with the highest multiplicity of a term in − U other than f being m −
1, we have v e ( A ) ≤ v yf + f ( − U ) ≤ m − . Hence mn − m − ≤ v e ( A ) ≤ m −
1, implying n ≤ n = 3. Then D ( G ) = 4 m − n ≤ | A | = D ( G ) − , forcing the case corresponding to | V | = 3 in Assertion A, and all m terms of U not equal to e must be contained in A . Arguing as in the previous paragraph, we must alsohave 12 mn − m − ≤ | A | − m ≤ v f ( A ) ≤ v xe + e ( U ) ≤ m − , implying n ≤
3. Once more, equality must hold in all these estimates, meaning all m terms of − U notequal to f must be contained in A . Consequently, U = ( − B ) C ( w − w ) = ( − e ) m − f m − ( f − e ) = ( − e ) m − ( xe + e ) m − (( x − e + e ) . Since σ ( U ) = 0, we see that x ≡ − e ) m ( xe + e ) m is a properzero-sum subsequence of U , contradicting that U is an atom. So we may instead assume n = 2.Since n = 2, it follows that D ( G ) = 3 m − m ≥
3. Recall that e = yf + f and f = xe + e .Thus, since ord( e ) = ord( f ) = 2 m , we conclude that x and y are both odd, whence(7) me = myf = mf = mxe with ord( me ) = ord( mf ) = 2 . If v − e ( − B ) ≥ m and v f ( C ) ≥ m , then ( − e ) m f m is a zero-sum subsequence of U (in view of (7)) oflength 2 m < m − D ( G ) − ≤ | U | , contradicting that U is an atom. Therefore we may assumeeither v − e ( − B ) < m or v f ( C ) < m , say w.l.o.g. v − e ( − B ) < m (the role of e in U is identical to thatof f in − U ).As noted earlier, v e ( A ) ≤ m −
1. Consequently, if | V | = 2, then v − e ( − B ) = v e ( U ) − v e ( A ) ≥ m − − ( m −
1) = m , contrary to our assumption above. Thus we must have | V | = 3, which is onlypossible (in view of Assertion A) if | U | = D ( G ) = 3 m − | m and m ≥ | V | = 3, we again obtain the contradiction v − e ( U ) ≥ m unless v e ( A ) = m − w = e . It follows that there are at most | A | − v e ( A ) = m terms of A not equal to e . Hence, since f = e , we conclude that v f ( A ) ≤ m , implying(8) v f ( C ) ≥ m − − m − m − , with equality only possible if w = f and w − w = f − e = ( x − e + e . Since v e ( A ) = m − w = e , we have − B = ( − e ) m − m/ Y i =1 ( − x i e − e ) , where we have appropriately re-indexed the terms x i e + e in U so that the first m terms correspondto those from B . Thus U = ( − e ) m − m/ Y i =1 ( − x i e − e ) f m − g g = ( − e ) m − m/ Y i =1 ( − x i e − e ) ( xe + e ) m − g g with w.l.o.g. g ∈ { f , y f + f } (by re-indexing the y i f + f appropriately) and g = w − w = w − e .If g = f , let g = g = f = xe + e . If g = f , the equality must hold in (8). In this case, let g = g = f − e = ( x − e + e . Regardless, we see that g = g j = ze + e for some z ∈ { x, x − } and j ∈ [1 , − e ) m − ( − x e − e )( ze + e ) , which would contradict that U is an atom, we must have x / ∈ { z, z − , . . . , z − ( m − } modulo 2 m . Onthe other hand, in view of (7), we have σ (( xe + e ) m ) = me , so that to avoid a zero-sum subsequenceof ( − e ) m − ( xe + e ) m ( − x e − e )( ze + e ) , which would contradict that U is an atom in view of m − ≥ m , we must have x / ∈ { m + z, m + z − , . . . , m + z − ( m − } modulo 2 m . However, this leaves no possibilities left for the value of x modulo2 m , which is a contradiction that concludes CASE 2.CASE 3: Either U is of type I(b) and − U is of type I(a) or else U is of type I(a) and − U is of typeI(b), say w.l.o.g. the former with U = e m − mn Y i =1 ( x i e + e ) and − U = f mn − m Y i =1 ( y i f + f ) , where { e , e } and { f , f } are bases of G with ord( e ) = ord( f ) = m and ord( e ) = ord( f ) = mn > m .Since m, n ≥ n ≥ m = 2, we have v f ( − U ) = mn − > D ( G )+12 ≥ D ( G ) −| A | = | U |−| A | ,implying f ∈ supp( A ). Consequently, since f cannot equal e due to ord( f ) = mn > m = ord( e ), itfollows that f = xe + e for some x ∈ Z . Let(9) y = v e ( B ) ∈ [0 , m − . Then v e ( A ) = m − − y − ǫ , where ǫ = 1 if | V | = 3 and w = e , and ǫ = 0 otherwise. Since f = e , itfollows that v f ( A ) ≤ | A | − v e ( A ) = | A | − m + 1 + y + ǫ , implying(10) v f ( C ) ≥ mn − − δ − | A | + m − − y − ǫ ≥ mn + 12 m − − y, where δ = 1 if | V | = 3 and w = f , and δ = 0 otherwise. Moreover, the estimate on the far right of(10) improves by 1 unless w = e and w = f , in which case w − w = ( x − e + e is a term of U . As a result, we see that U ( − B ) − contains at least mn + m − − y terms from e + h e i , say c · . . . · c s | U ( − B ) − with(11) s ≥ mn + 12 m − − y and c i ∈ e + h e i for all i. N PRODUCTS OF k ATOMS II 21
On the other hand, per definition of y , we see that − B contains | B | − y ≥ mn + m − − y terms from − e + h e i , say b · . . . · b t | − B with(12) t ≥ mn + 12 m − − y and b i ∈ − e + h e i for all i. Now e ∈ h e i and b i + c i ∈ h e i for all i ∈ [1 , min { s, t } ], while D ( h e i ) = D ( C m ) = m . Moreover,( mn + m − − y ) + y > m − m, n ≥ n ≥ m = 2. Consequently, weconclude from (9), (11) and (12) that U contains a nontrivial zero-sum subsequence of length at most2 ⌈ mn + m − − y ⌉ + y ≤ mn + m − < D ( G ) − ≤ | U | , contradicting that U is an atom.CASE 4: U and − U are both of type II, say U = f s m − f ( n − s ) m + ǫ m − ǫ Y i =1 ( − y i f + f ) and − U = g s m − g ( n − s ) m + ǫ m − ǫ Y i =1 ( − z i g + g ) , where { f , f } and { g , g } are generating sets for G such that ord( f ) = ord( g ) = mn > m andord( f ) , ord( g ) ≥ m , where s , s ∈ [1 , n − ǫ , ǫ ∈ [1 , m −
1] and y i , z i ∈ [1 , m −
1] for all i , andwhere y + . . . + y m − ǫ = z + . . . + z m − ǫ = m −
1. Moreover, either s = 1 or mf = mf and either s = 1 or mg = mg .Per the remarks after Main Proposition 5.4, let { f ′ , f } and { g ′ , g } be bases for G with ord( f ′ ) =ord( g ′ ) = m such that f = f ′ + αf and g = g ′ + βg for some α, β ∈ Z . We distinguish two subcases.CASE 4.1: n ≥ n ≥
3, we have | A | ≥ | B | = | C | ≥ mn + m − ≥ m −
1. Thus v { f , f } ( A ) ≥ | A | − ( m − ǫ ) ≥| A | − m + 1 ≥ m > m − ǫ , implying(13) { f , f } ∩ { g , g } 6 = ∅ . Also, applying Lemma 5.5 to B | U and C | − U , we conclude that there exist subsequences T | B and T | C with σ ( T ) = mf , σ ( T ) = mg and | T | , | T | ≤ m − mf = mg , so that σ ( T ) = σ ( T ). Then ( − T ) T is a zero-sum subsequence of ( − B ) C | U ,which contradicts that U is an atom unless ( − T ) T = ( − B ) C = U with | B | = | C | = | T | = | T | =2 m −
1, implying n = 3. However, in view of the equality conditions in Lemma 5.5, this is only possibleif U = ( − B ) C = ( − f ) m − ( − f ) ǫ m − ǫ Y i =1 ( y i f − f ) · g m − g ǫ m − ǫ Y i =1 ( − z i g + g ) . In particular, the terms − f , − f , g and g all occur in U in view of m ≥ ǫ , ǫ ≥
1. But then(13) ensures that U contains a zero-sum subsequence of length 2, contradicting that U is an atom with | U | = 4 m − >
2. So we instead conclude that(14) mf = mg . If s > s >
1, then mf = mf and mg = mg , which combined with (13) yields mf = mf = mg = mg , contrary to (14). Therefore we may w.l.o.g. assume s = 1 and v g ( − U ) = m − . Since | A | ≥ m − > v g ( − U ) + m − ǫ , we conclude that g ∈ supp( A ). Observe that g = f , for g = f would contradict (14). In consequence, we find that g = f or g = − yf + f for some y ∈ [1 , m − . This gives two further subcases.CASE 4.1.1: g = − yf + f for some y ∈ [1 , m − g = f as already remarked. Also, g = − yf + f = f as remarked after Main Proposition 5.4.Thus (13) ensures that we must have g = f or g = f . If g = f , then f can have multiplicity at most v g ( − U ) = m − A , meaning f must also becontained in A in view of | A | ≥ m −
1. By an analogous argument, if g = f , then f must be containedin A . In other words, in both cases, we have f , f ∈ supp( A ) . Suppose that g = f . Then, as f ∈ supp( A ) but f = f = g and f = g , it follows that f = − zg + g for some z ∈ [1 , m − f = − zg + g = − zf + g = − zf − yf + f , implying that ( z + y ) f = 0 with z + y ∈ [2 , m − f ) ≥ m , this is not possible.So we instead conclude that g = f . Now f ∈ supp( A ) but g = f = f and g = − yf + f = f as remarked after Main Proposition 5.4.In consequence, f = − zg + g for some z ∈ [1 , m − f ′ + αf = f = − zg + g = − zf + g = − zf − yf + f = − yf ′ + (1 − z − αy ) f , which, in view of y ∈ [1 , m − y = m − α ≡ − z − αy ≡ − z − α ( m −
1) mod mn.
The above congruence implies that z ≡ − αm mod mn , which, in view of z ∈ [1 , m − z = 1 and αm ≡ mn . Thus mf = m ( f ′ + αf ) = mf ′ + αmf = 0, contradicting thatord( f ) ≥ m for type II.CASE 4.1.2: g = f .If s >
1, then mg = mf = mf , contrary to (14). Therefore s = 1 and v f ( U ) = m − . Since | A | ≥ m − > v f ( − U ) + m − ǫ , we conclude that f ∈ supp( A ). As already remarked, we have f = g . Consequently, f = g or f = − zg + g for some z ∈ [1 , m − . Observe, however, that the roles of U and − U are now symmetric (we have the same information about − U that we did about U before CASE 4.1.1). Thus, if f = − zg + g for some z ∈ [1 , m − U and − U , we fall under the hypotheses of CASE 4.1.1, and the proof is completeby those prior arguments. So, combined with the subcase hypothesis, we may instead assume(15) f = g and f = g . Now v f ( A ) ≤ m − v f ( A ) = v g ( A ) ≤ m − s = s = 1. Consequently, since | A | ≥ m −
1, we conclude from (15) that − yf + f = − zg + g = − zf + f for some y, z ∈ [1 , m − y ) f − (1 + z ) f = (1 + y ) f ′ + ( α (1 + y ) − − z ) f , which, in view of y ∈ [1 , m − y = m − ≡ α (1 + y ) − − z ≡ αm − − z mod mn. N PRODUCTS OF k ATOMS II 23
The above congruence implies that z ≡ αm − mn , which, in view of z ∈ [1 , m − z = m − αm ≡ m mod mn . Thus mg = mf = m ( f ′ + αf ) = mf , contradicting (14) andcompleting CASE 4.1.CASE 4.2: n = 2.Since s , s ∈ [1 , n −
1] = [1 , s = s = 1. We also have m ≥ ǫ , ǫ ≥ n = 2 (the latter per (d) and (e) in Main Proposition 5.4). Now U = f m − f m + ǫ m − ǫ Y i =1 ( − y i f + f ) and − U = g m − g m + ǫ m − ǫ Y i =1 ( − z i g + g )with ord( f ) = ord( f ) = ord( g ) = ord( g ) = 2 m and (as remarked after Main Proposition 5.4)(16) mf = mf = mg = mg . Observe that 32 m − ≤ | A | ≤ m and 32 m − ≤ | B | = | C | ≤ m − . If neither f nor g is a term from A , then ( − f ) m g m will be a subsequence of U which is zero-sum (inview of (16)) and has length 2 m < m − ≤ | U | , contradicting that U is an atom. Therefore(17) f ∈ supp( A ) or g ∈ supp( A ) . We handle several subcases.CASE 4.2.1: f = g .We may w.l.o.g. assume ǫ ≤ ǫ . Then v f ( A ) = m + ǫ and v f ( B ) = 0. As remarked afterMain Proposition 5.4, we have y i ≤ ǫ and z j ≤ ǫ for all i and j . Also, since there are precisely2 m > m − − ( m − ≥ D ( G ) − | B | terms of U of the form − xf + f with x ∈ [0 , m − v f ( B ) = 0, it follows that yf − f ∈ supp( − B ) for some y ∈ [1 , ǫ ] ⊆ [1 , m − . Now, since v f ( B ) = 0, we have v f ( B ) ≥ | B | − ( m − ǫ ) ≥ m − ǫ ≥ ǫ . Thus ( − f ) y ( yf − f ) is asubsequence of − B . If f = g ∈ supp( C ), then ( − f ) y ( yf − f ) f would be a zero-sum subsequence of U of length y + 2 ≤ m + 1 < m − ≤ | U | , contradicting that U is an atom. Therefore we may insteadassume v g ( C ) = 0. But now, repeating the prior arguments for − U instead of U , we find that − zg + g = − zg + f ∈ supp( C ) for some z ∈ [1 , ǫ ] ⊆ [1 , m − v g ( C ) ≥ | C | − ( m − ǫ ) ≥ m − ǫ ≥ ǫ . Thus g z ( − zg + f )( − f ) y ( yf − f ) is a zero-sumsubsequence of U of length z + y + 2 ≤ ǫ + ǫ + 2 ≤ m < m − ≤ | U | (in view of m ≥ U is an atom and completing the subcase.CASE 4.2.2: f = g or g = f .By symmetry, we may w.l.o.g. assume f = g . Then v f ( A ) = v g ( A ) = m −
1, meaning v g ( C ) = v f ( C ) = 0 and ǫ + 1 ≥ v − f ( − B ) ≥ ǫ . Suppose g = f . Then v f ( A ) = v g ( A ) = m −
1, yielding v g ( C ) = v f ( C ) ≥ ǫ and m ≥ | A | ≥ v f ( A ) + v f ( A ) = 2 m −
2, which is only possible if 3 ≤ m ≤ | A | = 2 m − | V | = 2. In thiscase, ( − f ) ǫ f ǫ m − ǫ Y i =1 ( y i f − f ) m − ǫ Y i =1 ( − z i f + f ) | U . Thus, if ǫ = ǫ = m −
1, then Q m − ǫ i =1 ( y i f − f ) Q m − ǫ i =1 ( − z i f + f ) is a proper subsequence of U withsum ( m − f − f − ( m − f + f = mf − mf = 0 (in view of (16)), contradicting that U is an atom.Therefore we may w.l.o.g. assume that ǫ ≤ m −
2. Hence, since ǫ i ∈ [2 , m −
1] for n = 2, it follows that m = 4 with 2 ≤ ǫ ≤ m −
2, so that ǫ = 2. Consequently, since y + y = m − y i ∈ [1 , ǫ ] = [1 , y = 1 and y = 2. Likewise, if ǫ = 2, then w.l.o.g. z = 1 and z = 2, while if ǫ = 3 = m −
1, then z = m − − f + f )( f − f )(2 f − f ) is a properzero-sum subsequence of U (in view of (16) and m = 4), while in the latter case, ( − f + f )(2 f − f ) f is a proper zero-sum subsequence of U (again, in view of (16) and m = 4), both contradicting that U is an atom. So we instead conclude that g = f . Suppose next that f , g ∈ supp( A ). In view of g = f and g = f , this is only possible if f = − zg + g = − zf + g and g = − yf + f for some y ∈ [1 , m −
1] and z ∈ [1 , m − . Thus f ′ + αf = f = − zf + g = − zf − yf + f = − yf ′ + (1 − z − αy ) f . However, since y ∈ [1 , m − y = m − − z − αy = 1 − z − α ( m − ≡ α mod 2 m . Hence z ≡ − αm mod 2 m , which, in view of z ∈ [1 , m − z = 1 with αm ≡ m , implying mf = mf ′ + αmf = 0. Since this contradicts that ord( f ) = 2 m > m , wemay now assume f / ∈ supp( A ) or g / ∈ supp( A ) . Suppose f / ∈ supp( A ). Then A | f m − Q m − ǫ i =1 ( − y i f + f ). If | supp( A ) | ≥
3, then, since g = f , wemust have − yf + f = − zg + g and − y ′ f + f = − z ′ g + g for some distinct y, y ′ ∈ [1 , m −
1] anddistinct z, z ′ ∈ [0 , m − y − y ′ ) f ′ + α ( y − y ′ ) f = ( y − y ′ ) f = ( z − z ′ ) g = ( z − z ′ ) f . Thus y ≡ y ′ mod m , which, in view of y, y ′ ∈ [1 , m − y = y ′ , contrary to assumption. Thereforewe must instead have | supp( A ) | = 2 . Let − yf + f be the element of supp( A ) \ { f } , where y ∈ [1 , m − − yf + f has multiplicityat least v − yf + f ( A ) = | A | − v f ( A ) = | A | − ( m − ≥ m . Consequently, m y ≤ v − yf + f ( A ) y ≤ y + . . . + y m − ǫ = m −
1, which together with y ∈ [1 , m −
1] ensures that y = 1, so that − f + f ∈ supp( A ).Since − f + f ∈ supp( A ) with g = f , it follows that − f + f = − zg + g = − zf + g ∈ supp( A ) for some z ∈ [0 , ǫ ] . If z = 0, then mg = − mf + mf = 0 (in view of (16)), contradicting that ord( g ) = 2 m . Therefore wemust have z ∈ [1 , m − − zg + g = − f + f has multiplicity at least v − zg + g ( A ) = v − f + f ( A ) = v − yf + f ( A ) ≥ m . Consequently, m z ≤ v − zg + g ( A ) z ≤ z + . . . + z m − ǫ = m −
1, which together with z ∈ [1 , m −
1] ensures that z = 1. Thus − f + f = − zg + g = − zf + g = − f + g . Hence g = − f + 2 f and supp( A ) = { f , − f + f } = { g , − g + g } .Since f / ∈ supp( A ), we have(18) v − f ( − B ) ≥ v f ( U ) − m − , with equality only possible if | V | = 3 with w = f . Since f = g with v f ( A ) = v g ( A ) = m −
1, wehave v − f ( − B ) ≥ v f ( U ) − − v f ( A ) = ǫ ≥ ǫ ≥ n = 2). Since g = − f + 2 f / ∈ {− f + f , f = g } = supp( A ), we have v − f +2 f ( C ) = v g ( C ) ≥ v g ( − U ) − m + ǫ − ≥ m + 1 . Since − yf + f = − f + f ∈ supp( A ), we know y k = 1 for some k ∈ [1 , m − ǫ ]. If y i = 1 for all i ∈ [1 , m − ǫ ], then y + . . . + y m − ǫ = m − ǫ = 1, contradicting that ǫ ≥ n = 2. Therefore N PRODUCTS OF k ATOMS II 25 we may instead assume there is some y j ≥ j ∈ [1 , m − ǫ ]. Then, since y + . . . + y m − ǫ = m − y k = y = 1, we conclude that 2 ≤ y j ≤ m −
2, implying m ≥ . Since supp( A ) = { f , − f + f } , we must either have y j f − f ∈ supp( − B ) or w = − y j f + f . In theformer case, ( y j f − f )( − f + 2 f )( − f ) y j − ( − f )is a zero-sum subsequence of U of length y j + 2 ≤ m < m − ≤ | U | , contradicting that U is an atom.In the latter case, | V | = 3 and we have strict inequality in (18), in which case( − f ) m − ( − f ) ( − f + 2 f ) m +1 is a zero-sum subsequence of U having length 2 m + 2 < m − | U | (in view of m ≥ U is an atom. So we may now assume f ∈ supp( A ) and g / ∈ supp( A ) . Since g / ∈ supp( A ), we have A | g m − Q m − ǫ i =1 ( − z i g + g ) = f m − Q m − ǫ i =1 ( − z i f + g ). Thus, since f ∈ supp( A ), we have f = − xg + g = − xf + g for some x ∈ [1 , m − . If supp( A ) = { f , f } , then − yf + f = − zg + g for some y, z ∈ [1 , m − − yf ′ + (1 − αy ) f = − yf + f = − zg + g = − zf + ( xf + f ) = f ′ + ( x − z + α ) f . Thus, since y ∈ [1 , m − y = m − x − z ≡ − αm mod 2 m . Since x − z ∈ [ − ( m − , m −
2] (in view of x, y ∈ [1 , m − x − z = 1 with αm ≡ m . Butthis means mf = mf ′ + αmf = 0, contradicting that ord( f ) = 2 m . Therefore, we instead concludethat supp( A ) = { f , f } , whence v − xg + g ( A ) = v f ( A ) = | A | − v f ( A ) = | A | − m + 1 ≥ m . Thus f = − xg + g = − g + g in view of m x ≤ v − xg + g ( A ) x ≤ z + . . . + z m − ǫ = m − x ∈ [1 , m − mf = − mg + mg = 0 (in view of (16)), contradicting that ord( f ) = 2 m , which completes CASE4.2.2.CASE 4.2.3: f = g In this case, v f ( A ) = v g ( A ) = m − v f ( B ) = v g ( C ) = 0. In view of (17), we may w.l.o.g.assume f ∈ supp( A ). We have f = f = g while we can assume f = g else CASE 4.2.1 completesthe proof. Therefore(19) f = − xg + g = − xf + g for some x ∈ [1 , m − . Likewise, if g ∈ supp( A ), then g = − yf + f for some y ∈ [1 , m − f = − xf + g = − xf − yf + f , in which case ( x + y ) f = 0 with x + y ∈ [2 , m − f ) = 2 m . Therefore weconclude that g / ∈ supp( A ). As a result, all elements in supp( A ) \ { g } have the form − z i g + g = − z i f + g with z i ∈ [1 , m − − zg + g ∈ supp( A ) \ { g } be arbitrary. Let us show that z ≥ x . If z = x , this is trivial, sosuppose z = x . Then − yf + f = − zg + g = − zf + g for some y ∈ [1 , m − − yf + f = − zf + g = − zf + ( xf + f ) , yielding ( x − z + y ) f = 0. Consequently, since ord( f ) = 2 m with x − z + y ∈ [ − ( m − , m − −
1] =[ − m + 3 , m − z = x + y ≥ x + 1, as claimed. All terms of A not equal to g = f have the form − z i g + g . There are at least | A | − v g ( A ) = | A |− m +1 ≥ m such terms all with z i ≥ x as shown above. Consequently, m x ≤ z + . . . + z m − ǫ = m − x = 1. Hence f = − xf + g = − f + g , which yields mf = − mf + mg = 0 (in view of (16)), contradicting that ord( f ) = 2 m and completingthe subcase.CASE 4.2.4: { f , f } ∩ { g , g } = ∅ .Let a i = v − if + f ( A ) and b i = v − ig + g ( A ) for i ∈ [1 , m − . Let c = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) gcd m − ǫ Y i =1 ( − z i g + g ) , m − ǫ Y i =1 ( − y i f + f ) !(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) . Thus c counts the number of terms of A simultaneously equal to some − y i f + f as well as some − z j g + g . In view of the hypothesis { f , f } ∩ { g , g } = ∅ , we see that every term of A is either equalto some − y i f + f or to some − z j g + g . As a result, the inclusion-exclusion principle gives(20) m − X i =1 a i + m − X i =1 b i − c = | A | ≥ m − . Note − f + f and − g + g have order m (in view of (16)), meaning {− f + f , − g + g }∩{ f , f , g , g } = ∅ . Consequently, if − f + f occurs in A , then it must be equal to some − zg + g with z ∈ [1 , m − c ≥ a . Likewise, if − g + g occurs in A , then it must be equal to some − yf + f , so that c ≥ b . Averaging these estimates, we obtain c ≥ a + b . Applying this estimate in (20) along with the pigeon-hole principle, we conclude that either12 a + m − X i =2 a i ≥ m −
12 or 12 b + m − X i =2 b i ≥ m − , and we w.l.o.g. assume the former:(21) 12 a + m − X i =2 a i ≥ m − . By definition of the a i , we have(22) a + m − X i =2 a i ≤ a + 2 a + 3 a + . . . + ( m − a ≤ y + . . . + y m − ǫ = m − . Combining (21) and (22) yields 32 m − ≤ a m − X i =2 a i ! ≤ m − , which is a contradiction, concluding CASE 4.If | U | = D ( G ), then it possible to also apply Main Proposition 5.4 to U and (by symmetry) re-indexthe U i with i ∈ [1 ,
3] in any fashion. Consequently, if one of U , U or U has the same type from amongI(a), I(b) and II, then we may w.l.o.g. re-index the U i so that U and − U have the same type and applyCASE 1, 2 or 4 to yield the desired conclusion (note U i and − U i have the same type). On the otherhand, if U , U and U have distinct types I(a), I(b) and II, then we my re-index the U i so that U has N PRODUCTS OF k ATOMS II 27 type I(b) and − U has type I(a), in which case CASE 3 completes the proof. In summary, the proof isnow complete when | U | = D ( G ), so we instead assume | U | = D ( G ) − . By Assertion A, this is only possible if | V | = 2 , | A | = D ( G ) + 12 and | B | = | C | = D ( G ) −
12 with D ( G ) = mn + m − , which we now assume for the final two cases of the proof, where by symmetry we now assume − U hastype II.CASE 5: U is of type I(b) and − U is of type II, say U = e m − mn Y i =1 ( x i e + e ) and − U = f sm − f ( n − s ) m + ǫ m − ǫ Y i =1 ( − y i f + f ) , where { e , e } is a basis for G with ord( e ) = mn > m and ord( e ) = m , where { f , f } is a generatingset for G with ord( f ) = mn and ord( f ) ≥ m , and where y + . . . + y m − ǫ = m − y i ∈ [1 , m − ǫ ∈ [1 , m −
1] and s ∈ [1 , n − s = 1 or mf = mf , with both holding when n = 2.Since | A | ≥ m > m −
1, we must have f ν ∈ supp( A ) for some ν ∈ [1 , f ν ) ≥ m > m =ord( e ), we cannot have f ν = e . Thus f ν ∈ h e i + e . It is easily noted that any g ∈ h e i + e hasord( g ) = ord( e ) = mn . Moreover, U will also have type I(b) using the basis { e , g } replacing each x i with x i − α , where g = αe + e . Consequently, since f ν ∈ h e i + e for some ν ∈ [1 , f = e or f = e . CASE 5.1: n ≥ mf = me . If s >
1, then this follows from Main Proposition 5.4 and (23). If s = 1, then | A | = mn + m ≥ m > m − n ≥ f ∈ supp( A ). Hence, by the argument above CASE 5.1, we may w.l.o.g.assume f = e , implying mf = me in this case as well. Thus (24) is established.Let H = h me i . Then G/H ∼ = C m . Since n ≥
3, we have | C | = | B | = mn + m − ≥ m − D ( G/H ).Let B ′ | B be a subsequence with | B ′ | = 2 m − B ′ = e t · b · . . . · b m − − t , where v e ( B ′ ) = t ∈ [0 , m − b i = x i e + e for i ∈ [1 , m − − t ]. Since 2 m − − t ≤ m − t ≤ m −
1, it is readily seen that the only way B ′ can contain a nontrivial subsequence T | B ′ with σ ( T ) ∈ H = h me i is if T contains precisely m terms from b · . . . · b m − − t , in which case σ ( T ) = me .Consequently, since | B ′ | = 2 m − D ( G/H ), we conclude that there exits a subsequence T | B ′ with σ ( T ) = me = mf and | T | ≤ m + t ≤ m − . Moreover, T will be a proper subsequence of B unless n = 3 (so that 2 m − | B | = | B ′ | = | T | ) and(w.l.o.g. re-indexing the x i e + e ) B = e m − m Y i =1 ( x i e + e ) with m X i =1 x i ≡ m. Since | C | ≥ m −
1, Lemma 5.5 ensures that there is a subsequence R | C with σ ( R ) = mf and | R | ≤ m −
1. Moreover, R will be a proper subsequence unless n = 3 and C = f m − f ǫ m − ǫ Y i =1 ( − y i f + f ) . Now ( − T ) R is a nontrivial zero-sum subsequence of ( − B ) C = U . Since U is an atom, this is onlypossible if T = B and R = C . Thus n = 3 and U = ( − e ) m − m Y i =1 ( − x i e − e ) f m − f ǫ m − ǫ Y i =1 ( − y i f + f ) , where m P i =1 x i ≡ m and m − ǫ P i =1 y i = m −
1. Since v f ( − U ) = ( n − s ) m + ǫ ≥ m + ǫ > ǫ , we concludethat f ∈ supp( A ), whence (as argued before CASE 5.1) we may w.l.o.g. assume e = f . As a result,we see that ( − x e − e )( − y f + e ) f y ( − e ) z , where z ∈ [0 , m −
1] is the integer such that z + x ≡ m , will be a zero-sum subsequence of U of length 2 + y + z ≤ m < m − | U | , contradictingthat U is an atom.CASE 5.2: n = 2.Similar to CASE 4.2, we now have ord( e ) = ord( f ) = ord( f ) = 2 m , s = 1, ǫ ≥ m ≥ D ( G ) = 3 m − mf = mf = me , with U = e m −
11 2 m Y i =1 ( x i e + e ) and − U = f m − f m + ǫ m − ǫ Y i =1 ( − y i f + f ) . We handle several subcases.CASE 5.2.1: e = − f + f .Let t be the number of terms from C of the form − yf + f with y ∈ [1 , m − e = − f + f ,we see that v e ( A ) ≤ m − ǫ − t , so that(26) v − e ( − B ) ≥ ǫ − t. By (23), we have f = e or f = e . In either case, the hypothesis e = − f + f ensures that f , f ∈ h e i + e . Thus there are | C | − t = m − − t terms of C from h e i + e , say c · . . . · c ℓ | C with c i ∈ h e i + e and ℓ = m − − t , and there are (by (26)) ℓ := | B | − v − e ( − B ) ≤ m − − ( ǫ − t ) = 32 m − ǫ − t ≤ m − − t terms of − B from h e i − e , say b · . . . · b ℓ | − B with b i ∈ h e i − e and ℓ < ℓ . Consequently,( − e ) v − e ( − B ) ( b + c ) · . . . · ( b ℓ + c ℓ ) ∈ F ( h e i ) is a sequence of terms from h e i ∼ = C m of length | B | = m − ≥ m = D ( h e i ). Thus the proper (in view of ℓ > ℓ ) subsequence ( − e ) v − e ( − B ) b · . . . · b ℓ · c · . . . · c ℓ of ( − B ) C = U contains a nontrivial zero-sum subsequence, contradicting that U is an atom.CASE 5.2.2: f ∈ supp( A ).In this case, we may assume f = e per the argument before CASE 5.1.First suppose that e / ∈ supp( A ). Then v − e ( − B ) = m −
1. Since | B | = m − > m −
1, theremust be some − xe − e ∈ supp( − B ). If v e ( C ) = v f ( C ) >
0, then ( − xe − e ) e ( − e ) z , where z ∈ [0 , m −
1] is the integer with z + x ≡ m , will be a zero-sum subsequence of U of length z + 2 ≤ m + 1 < m − | U | , contradicting that U is an atom. Therefore we instead assume v f ( C ) = 0. Thus m − ≥ v f ( C ) ≥ | C | − ( m − ǫ ) = m − ǫ ≥ m , implying ǫ ≤ m , and thereare at least | C | − m + 1 ≥ m > C of the form − y i f + e with y i ∈ [1 , ǫ ] ⊆ [1 , m ]. Let − yf + e ∈ supp( C ) with y ∈ [1 , m ] be one such term. Then f y ( − yf + e )( − xe − e )( − e ) z , where z ∈ [0 , m −
1] is the integer such that x + z ≡ U of length N PRODUCTS OF k ATOMS II 29 y + z + 2 ≤ m + 1 < m − | U | , contradicting that U is an atom. So we instead conclude that e ∈ supp( A ). As a result, since ord( e ) = m < m = ord( f ) = ord( f ) = ord( e ), we must have(27) e = − yf + f = − yf + e for some y ∈ [1 , ǫ ] . Furthermore, since me = 0, we conclude from ord( e ) = 2 m that y is odd, and in view of CASE 5.2.1,we can assume y ≥ f ∈ supp( A ). Then f = xe + e for some x ∈ Z . Combining this with (27) yields − e + e = yf = xye + ye , which implies y ≡ m . Hence, since y ∈ [1 , m − y = 1, which is contrary to our above assumption. So we may instead assume f / ∈ supp( A ), implying v f ( C ) = m − . Each term of A equal to e = − yf + f = − yf + e is also equal to some − y i f + f . Thus3 v e ( A ) ≤ v e ( A ) y ≤ y + . . . + y m − ǫ = m −
1, implying v e ( A ) ≤ m − and v − e ( − B ) ≥ m − ≥ m . Since v f ( C ) = m −
1, we find that there are precisely | C | − m + 1 = m terms of C either equal to e = f or − y i f + e for some i . Hence, since y + . . . + y m − ǫ = m − v f ( C ) with the y i ∈ [1 , m − T · . . . · T m/ | C with each T i ∈ F ( G ) a subsequence having σ ( T i ) = e . There are at least | B | − m + 1 = m terms of − B of the form − xe − e , say b · . . . · b m/ | − B with b i ∈ h e i− e for all i . Now ( σ ( T )+ b ) · . . . · ( σ ( T m/ )+ b m/ )( − e ) m/ ∈ F ( h e i ) is a subsequence ofterms from h e i of length m = D ( h e i ). Consequently, the subsequence T · . . . · T m/ · b · . . . · b m/ · ( − e ) m/ of ( − B ) C = U contains a nontrivial zero-sum subsequence of length at most | T | + . . . + | T m/ | + m ≤| C | + m = m − < m − | U | , contradicting that U is an atom and completing CASE 5.2.1.CASE 5.2.3: f / ∈ supp( A ).Since f / ∈ supp( A ), all terms of A not equal to f are equal to some − y i f + f , and there are at least | A | − m + 1 = m + 1 such terms of A . If y i ≥ m + 2 = ( m + 1)2 ≤ y + . . . + y m − ǫ = m −
1. Thus − f + f ∈ supp( A ). If − f + f = xe + e forsome x ∈ Z , then (25) implies 0 = − mf + mf = xme + me = me , contradicting that ord( e ) = 2 m .Therefore we instead conclude that − f + f = e , so that CASE 5.2.1 completes the proof of CASE 5.CASE 6: U is of type I(a) and − U is of type II, say U = e mn − m Y i =1 ( x i e + e ) and − U = f sm − f ( n − s ) m + ǫ m − ǫ Y i =1 ( − y i f + f ) , where { e , e } is a basis for G with ord( e ) = m and ord( e ) = mn > m , where { f , f } is a generatingset for G with ord( f ) = mn and ord( f ) ≥ m , and where y + . . . + y m − ǫ = m − y i ∈ [1 , m − ǫ ∈ [1 , m −
1] and s ∈ [1 , n − s = 1 or mf = mf , with both holding when n = 2.CASE 6.1: n ≥ | A | = mn + m > m , we must have e ∈ supp( A ). If e = − y i f + f for some y i ∈ [1 , m − | A | ≤ v e ( A ) + m ≤ ( m − ǫ ) + m ≤ m − < mn + m = | A | (in view of n ≥ e = f or e = f . Suppose s = 1. If e = f , then mn + m = | A | ≥ v e ( A ) = ( n − m + ǫ ≥ mn − m + 1, contradictingthat n ≥
3. If e = f , then | A | ≤ v e ( A ) + m ≤ v f ( − U ) + m = 2 m − < mn + m = | A | , again in viewof n ≥
3, which is a contradiction. So (in view of (28)) we may instead assume s >
1, whence(29) mf = mf = me , where the first equality follows from Main Proposition 5.4 and the second from (28). The argument is now similar to CASE 5.1. Let H = h me i . Then G/H ∼ = C m . Since n ≥
3, wehave | C | = | B | = mn + m − ≥ m − D ( G/H ). Let B ′ | B be a subsequence with | B ′ | = 2 m −
1. If v e ( B ′ ) ≥ m , then B ′ will contain a subsequence T = e m with σ ( T ) = me . If v e ( B ′ ) < m , then this isonly possible if B ′ = e m − m Y i =1 ( x i e + e ) with m X i =1 x i ≡ mn, in which case it is easily seen that T = B ′ is a subsequence of B ′ with σ ( T ) = me . Since | C | ≥ m − R | C with σ ( R ) = mf and | R | ≤ m −
1. Moreover, R will be a proper subsequence unless n = 3 and C = f m − f ǫ m − ǫ Y i =1 ( − y i f + f ) . Now ( − T ) R is a nontrivial zero-sum subsequence of ( − B ) C = U (in view of (29)). Since U is an atom,this is only possible if T = B ′ = B and R = C . Thus n = 3 and U = ( − e ) m − m Y i =1 ( − x i e − e ) f m − f ǫ m − ǫ Y i =1 ( − y i f + f ) . As a result, since v f ( − U ) = ( n − s ) m + ǫ ≥ m + ǫ > m > ǫ , we conclude that f ∈ supp( A ). Moreover, v f ( A ) ≥ m + ǫ − v f ( C ) = m . Hence, as the only term in U with multiplicity at least m is e (recall | supp( U ) | ≥ e = f , in which case( − e )( f ) = ( − e )( e ) is a proper zero-sum subsequence of U , contradicting that U is an atom.CASE 6.2: n = 2.Similar to CASE 5.2, we now have ord( e ) = ord( f ) = ord( f ) = 2 m , s = 1, ǫ ≥ m ≥ D ( G ) = 3 m − mf = mf = me , with U = e m − m Y i =1 ( x i e + e ) and − U = f m − f m + ǫ m − ǫ Y i =1 ( − y i f + f ) . Since | A | = m > m , we must have e ∈ supp( A ). We have three possibilities for e .Suppose e = f . Then v e ( A ) = v f ( A ) = m −
1, implying v − e ( − B ) = m and v f ( C ) = 0. Let T = c · . . . · c m | C be any length m subsequence of C . As v f ( C ) = 0, each c i = − z i f + f = − z i e + f for some z i ∈ [0 , m −
1] with0 ≤ z := z + . . . + z m ≤ y + . . . + y m − ǫ = m − . Then σ ( T ) = − zf + mf = ( m − z ) e (in view of (30) and e = f ), in which case ( − e ) m − z T is azero-sum subsequence of ( − B ) C = U of length m − z + | T | ≤ m < m − | U | , contradicting that U is an atom.Suppose e = f . Then v e ( A ) = v f ( A ) = m + ǫ ≤ | A | = m , implying ǫ ≤ m , v − e ( − B ) = m − − ǫ ≥ m − > v f ( C ) = 0. Since v f ( A ) = m + ǫ , it follows that there are at most | A | − v f ( C ) = m − ǫ terms of A of the form − y i f + f = − y i f + e , meaning there are at least m − ǫ − ( m − ǫ ) = m terms of C ofthis form, say b · . . . · b ℓ | C with w.l.o.g. b i = − y i f + f = − y i f + e for i ∈ [1 , ℓ ] and ℓ ≥ m . If b i ≥ N PRODUCTS OF k ATOMS II 31 for all i ∈ [1 , ℓ ], then we obtain the contradiction m ≤ ℓ ≤ b + . . . + b ℓ ≤ y + . . . + y m − ǫ = m − y i = 1 for some i ∈ [1 , ℓ ], meaning − f + e ∈ supp( C ) . Since v f ( A ) = m + ǫ , there are also at most | A | − m − ǫ = m − ǫ terms of A equal to f , whence v f ( C ) ≥ m − − ( m − ǫ ) = m − ǫ >
0. Hence f ( − e )( − f + e ) is a zero-sum subsequence of( − B ) C = U of length 3 < m − | U | , contradicting that U is an atom.It remains to consider the case when e = − yf + f for some y ∈ [1 , m − e ) = 2 m , we must have y even, whence y ≥
2. Thus 2 v e ( A ) ≤ y + . . . + y m − ǫ = m − v e ( A ) ≤ m − . But now m = | A | ≤ v e ( A ) + m ≤ m − + m , which is a proof concludingcontradiction. (cid:3) References [1] D.F. Anderson,
Elasticity of factorizations in integral domains : a survey , Factorization in Integral Domains, Lect.Notes Pure Appl. Math., vol. 189, Marcel Dekker, 1997, pp. 1 – 29.[2] D.F. Anderson, P.-J. Cahen, S.T. Chapman, and W.W. Smith, Some factorizations properties of the ring of integer-valued polynomials , Zero-Dimensional Commutative Rings, Lect. Notes Pure Appl. Math., vol. 171, Marcel Dekker,1995, pp. 95 – 113.[3] N.R. Baeth and A. Geroldinger,
Monoids of modules and arithmetic of direct-sum decompositions , Pacific J. Math. (2014), 257 – 319.[4] N.R. Baeth and D. Smertnig,
Factorization theory from commutative to noncommutative settings , J. Algebra, toappear.[5] N.R. Baeth and R. Wiegand,
Factorization theory and decomposition of modules , Am. Math. Mon. (2013), 3 – 34.[6] P. Baginski, S.T. Chapman, N. Hine, and J. Paixao,
On the asymptotic behavior of unions of sets of lengths in atomicmonoids , Involve, a journal of mathematics (2008), 101 – 110.[7] P. Baginski, A. Geroldinger, D.J. Grynkiewicz, and A. Philipp, Products of two atoms in Krull monoids and arith-metical characterizations of class groups , Eur. J. Comb. (2013), 1244 – 1268.[8] M. Banister, J. Chaika, S.T. Chapman, and W. Meyerson, A theorem on accepted elasticity in certain local arithmeticalcongruence monoids , Abh. Math. Semin. Univ. Hamb. (2009), 79 – 86.[9] G. Bhowmik, I. Halupczok, and J.-C. Schlage-Puchta, The structure of maximal zero-sum free sequences , Acta Arith. (2010), 21 – 50.[10] V. Blanco, P. A. Garc´ıa-S´anchez, and A. Geroldinger,
Semigroup-theoretical characterizations of arithmetical invariantswith applications to numerical monoids and Krull monoids , Illinois J. Math. (2011), 1385 – 1414.[11] P.-J. Cahen and J.-L. Chabert, Elasticity for integer-valued polynomials , J. Pure Appl. Algebra (1995), 303 – 311.[12] S.T. Chapman and S. Glaz,
One hundred problems in commutative ring theory , Non-Noetherian Commutative RingTheory, Mathematics and Its Applications, vol. 520, Kluwer Academic Publishers, 2000, pp. 459 – 476.[13] S.T. Chapman and B. McClain,
Irreducible polynomials and full elasticity in rings of integer-valued polynomials , J.Algebra (2005), 595 – 610.[14] S.T. Chapman and W.W. Smith,
Factorization in Dedekind domains with finite class group , Isr. J. Math. (1990),65 – 95.[15] F. Chen and S. Savchev, Long minimal zero-sum sequences in the groups C r − ⊕ C k , Integers (2014), Paper A23.[16] A. Czogala, Arithmetic characterization of algebraic number fields with small class number , Math. Z. (1981), 247– 253.[17] A. Facchini,
Krull monoids and their application in module theory , Algebras, Rings and their Representations (A. Fac-chini, K. Fuller, C. M. Ringel, and C. Santa-Clara, eds.), World Scientific, 2006, pp. 53 – 71.[18] M. Freeze and A. Geroldinger,
Unions of sets of lengths , Funct. Approximatio, Comment. Math. (2008), 149 – 162.[19] W. Gao and A. Geroldinger, On zero-sum sequences in Z /n Z ⊕ Z /n Z , Integers (2003), Paper A08, 45p.[20] , On products of k atoms , Monatsh. Math. (2009), 141 – 157.[21] W. Gao, A. Geroldinger, and D.J. Grynkiewicz, Inverse zero-sum problems III , Acta Arith. (2010), 103 – 152.[22] A. Geroldinger,
Additive group theory and non-unique factorizations , Combinatorial Number Theory and AdditiveGroup Theory (A. Geroldinger and I. Ruzsa, eds.), Advanced Courses in Mathematics CRM Barcelona, Birkh¨auser,2009, pp. 1 – 86.[23] A. Geroldinger and F. Halter-Koch,
Non-Unique Factorizations. Algebraic, Combinatorial and Analytic Theory , Pureand Applied Mathematics, vol. 278, Chapman & Hall/CRC, 2006.[24] A. Geroldinger, F. Kainrath, and A. Reinhart,
Arithmetic of seminormal weakly Krull monoids and domains , J.Algebra, to appear. [25] A. Geroldinger, M. Liebmann, and A. Philipp,
On the Davenport constant and on the structure of extremal sequences ,Period. Math. Hung. (2012), 213 – 225.[26] A. Geroldinger and I. Ruzsa, Combinatorial Number Theory and Additive Group Theory , Advanced Courses in Math-ematics - CRM Barcelona, Birkh¨auser, 2009.[27] A. Geroldinger and Wolfgang A. Schmid,
A characterization of class groups via sets of lengths , submitted.[28] A. Geroldinger and R. Schneider,
On Davenport’s constant , J. Comb. Theory, Ser. A (1992), 147 – 152.[29] R. Gilmer, Commutative Semigroup Rings , The University of Chicago Press, 1984.[30] D.J. Grynkiewicz,
Structural Additive Theory , Developments in Mathematics, Springer, 2013.[31] F. Halter-Koch,
Ideal Systems. An Introduction to Multiplicative Ideal Theory , Marcel Dekker, 1998.[32] F. Kainrath,
Elasticity of finitely generated domains , Houston J. Math. (2005), 43 – 64.[33] H. Kim, The distribution of prime divisors in Krull monoid domains , J. Pure Appl. Algebra (2001), 203 – 210.[34] H. Kim and Y. S. Park,
Krull domains of generalized power series , J. Algebra (2001), 292 – 301.[35] C. Reiher,
A proof of the theorem according to which every prime number possesses property B , PhD Thesis, Rostock,2010 (2010).[36] L. Salce and P. Zanardo,
Arithmetical characterization of rings of algebraic integers with cyclic ideal class group , Boll.Unione. Mat. Ital., VI. Ser., D, Algebra Geom. (1982), 117 – 122.[37] W.A. Schmid, Arithmetical characterization of class groups of the form Z /n Z ⊕ Z /n Z via the system of sets of lengths ,Abh. Math. Semin. Univ. Hamb. (2009), 25 – 35.[38] , Characterization of class groups of Krull monoids via their systems of sets of lengths : a status report , NumberTheory and Applications : Proceedings of the International Conferences on Number Theory and Cryptography (S.D.Adhikari and B. Ramakrishnan, eds.), Hindustan Book Agency, 2009, pp. 189 – 212.[39] , Inverse zero-sum problems II , Acta Arith. (2010), 333 – 343.[40] D. Smertnig,
On the Davenport constant and group algebras , Colloq. Math. (2010), 179 – 193.[41] ,
Sets of lengths in maximal orders in central simple algebras , J. Algebra (2013), 1 – 43.
Institute for Mathematics and Scientific Computing, University of Graz, NAWI Graz, Heinrichstraße 36,8010 Graz, Austria
E-mail address : [email protected] Department of Mathematical Sciences, University of Memphis, Memphis, TN 38152, USA
E-mail address : [email protected] School of Mathematics, South China Normal University, Guangzhou 510631, P.R. China
E-mail address ::