A CHARACTERIZATION OF CLASS GROUPS VIA SETS OF LENGTHS. 1. Introduction

Transcription

1 A CHARACTERIZATION OF CLASS GROUPS VIA SETS OF LENGTHS ALFRED GEROLDINGER AND WOLFGANG A. SCHMID Abstract. Let H be a Krull monoid with class group G such that every class contains a prime divisor. Then every nonunit a H can be written as a finite product of irreducible elements. If a = u 1... u k, with irreducibles u 1,... u k H, then k is called the length of the factorization and the set La) of all possible k is called the set of lengths of a. It is well-known that the system LH) = {La) a H} depends only on the class group G. In the present paper we study the inverse question asking whether or not the system LH) is characteristic for the class group. Consider a further Krull monoid H with class group G such that every class contains a prime divisor and suppose that LH) = LH ). We show that, if one of the groups G and G is finite and has rank at most two, then G and G are isomorphic apart from two well-known pairings). 1. Introduction Let H be a cancelative semigroup with unit element. If an element a H can be written as a product of k irreducible elements, say a = u 1... u k, then k is called the length of the factorization. The set La) of all possible factorization lengths is the set of lengths of a, and LH) = {La) a H} is called the system of sets of lengths of H. Clearly, if H is factorial, then La) = 1 for each a H. Suppose there is some a H with La) > 1, say k, l La) with k < l. Then, for every m N, we observe that La m ) {km + νl k) ν [0, m]} which shows that sets of lengths can become arbitrarily large. Under mild conditions on the ideal theory of H every nonunit of H has a factorization into irreducibles and all sets of lengths are finite. Sets of lengths together with parameters controlling their structure) are the most investigated invariants in factorization theory, in settings ranging from numerical monoids, noetherian domains, monoids of modules to maximal orders in central simple algebras. The focus of the present paper is on Krull monoids with finite class group such that every class contains a prime divisor. Rings of integers in algebraic number fields are such Krull monoids, and classical philosophy in algebraic number theory dating back to the 19th century) states that the class group determines the arithmetic. This idea has been formalized and justified. In the 1970s Narkiewicz posed the inverse question whether or not arithmetical phenomena in other words, phenomena describing the non-uniqueness of factorizations) characterize the class group [37, Problem 3; page 469]). Very quickly first affirmative answers were given by Halter-Koch, Kaczorowski, and Rush [33, 9, 40]). Indeed, it is not too difficult to show that the system of sets of factorizations determines the class group [18, Sections 7.1 and 7.]). All these answers are not really satisfactory because the given characterizations are based on rather abstract arithmetical properties which are designed to do the characterization and which play only a little role in other parts of factorization theory. Since on the other hand sets of lengths are of central interest in factorization theory it has been natural to ask whether their structure is rich enough to do characterizations. 010 Mathematics Subject Classification. 11B30, 11R7, 13A05, 13F05, 0M13. Key words and phrases. Krull monoids, maximal orders, seminormal orders; class groups, arithmetical characterizations, sets of lengths, zero-sum sequences, Davenport constant. This work was supported by the Austrian Science Fund FWF, Project Number P6036-N6, by the Austrian-French Amadée Program FR03/01, and by the ANR Project Caesar, Project Number ANR-1-BS

2 ALFRED GEROLDINGER AND WOLFGANG A. SCHMID Let H be a commutative Krull monoid with finite class group G and suppose that every class contains a prime divisor. It is classical that H is factorial if and only if G = 1, and by a result due to Carlitz in 1960 we know that all sets of lengths are singletons i.e., L = 1 for all L LH)) if and only if G. Let us suppose now that G 3. Then the monoid BG) of zero-sum sequences over G is again a Krull monoid with class group isomorphic to G, every class contains a prime divisors, and the systems of sets of lengths of H and that of BG) coincide. Thus LH) = L BG) ), and it is usual to set LG) := L BG) ). The Characterization Problem can be formulated as follows [18, Section 7.3], [, page 4], [44]). Given two finite abelian groups G and G such that LG) = LG ). Does it follow that G = G? The system of sets of lengths LG) for finite abelian groups is studied with methods from Additive Combinatorics it has been written down explicitly only for a handful small groups, see Proposition 4.). Zero-sum theoretical invariants, such as the Davenport constant, play a central role. Recall that, although the precise value of the Davenport constant is well-known for p-groups and for groups of rank at most two since the 1960s see Proposition.3), the precise value is unknown in general even for groups of the form G = C 3 n). Thus it is not surprising that all answers to the Characterization Problem so far have been restricted to very special groups including cyclic groups, elementary -groups, and groups of the form C n C n. Beyond the present paper there is recent progress towards groups of the form C r n when r is small with respect to n [4]). Apart from two well-known pairings see Proposition 4.) the answer is always positive. The goal of the present paper is to settle the Characterization Problem for groups of rank at most two. Here is our main result. Theorem 1.1. Let G be an abelian group such that LG) = LC n1 C n ) where n 1, n N with n 1 n and n 1 + n > 4. Then G = C n1 C n. Theorem 1.1 does not only apply to Krull monoids with class group G but also to certain maximal orders in central simple algebras and to certain seminormal orders in algebraic number fields. This will be outlined in Section see Proposition.1 and.). The proof of Theorem 1.1 is based substantially on Prior work on this problem as summarized in Propositions 6.1 and 6.), in particular on the recent paper [7]. The structure theorem for sets of lengths see Proposition 3.) and an associated inverse result [18, Proposition 9.4.9]; see the start of the proof of Proposition 6.5). The characterization of minimal zero-sum sequences of maximal length over groups of rank two Lemma 5.) which is crucial also for the above mentioned paper [7]. The difficulty of the Characterization Problem stems from the fact that most sets of lengths over any finite abelian group are arithmetical progressions with difference 1 see Proposition 3..4, or [18, Theorem ] for a density result of this flavor). Moreover, G and G are finite abelian groups with G G, then clearly LG) LG ). Thus in order to characterize a group G, we first have to find distinctive sets of lengths for G i.e., sets of lengths which do occur in LG), but in no other or only in a small number of further groups), and second we will have to show that certain sets are not sets of lengths in LG). These distinctive sets of lengths for rank two groups are identified in Proposition 6.5 which is the core of our whole approach, and Proposition 5.1 provides sets which do not occur as sets of lengths for rank two groups. In order to pull this through we proceed as follows. After gathering some background material in Section, we summarize key results on the structure of sets of lengths in Propositions , and 3.4. Furthermore, we provide some explicit constructions which will turn out to be crucial Propositions ). In Section 4 we characterize groups whose sets of lengths have a very special structure e.g., arithmetical progressions) which allows us to settle the Characterization Problem for small groups see Theorem 4.1). After that we are well-prepared for the main parts given in Sections 5 and 6.

3 A CHARACTERIZATION OF CLASS GROUPS VIA SETS OF LENGTHS 3. The arithmetic of Krull monoids: Background In this section we gather the required tools from the algebraic and arithmetic theory of Krull monoids. Our notation and terminology are consistent with the monographs [18,, 8]. Let N denote the set of positive integers, P N the set of prime numbers and put N 0 = N {0}. For real numbers a, b R, we set [a, b] = {x Z a x b}. Let A, B Z be subsets of the integers. We denote by A + B = {a + b a A, b B} their sumset, and by A) the set of successive) distances of A that is, d A) if and only if d = b a with a, b A distinct and [a, b] A = {a, b}). For k N, we denote by k A = {ka a A} the dilation of A by k. If A N, then { m } ρa) = sup n m, n A = sup A min A Q 1 { } is the elasticity of A, and we set ρ{0}) = 1. Monoids and Factorizations. By a monoid, we always mean a commutative semigroup with identity which satisfies the cancelation 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 H be a monoid. We denote by H the set of invertible elements of H, by AH) the set of atoms irreducible elements) of H, and by H red = H/H = {ah a H} the associated reduced monoid of H. A monoid F is free abelian, with basis P F, and we write F = FP ) if every a F has a unique representation of the form a = p P p vpa), where v p a) N 0 with v p a) = 0 for almost all p P, and we call a F = a = p P v p a) the length of a. The monoid ZH) = F AH red ) ) is called the factorization monoid of H, and the unique homomorphism π : ZH) H red satisfying πu) = u for each u AH red ) is the factorization homomorphism of H. For a H, Z H a) = Za) = π 1 ah ) ZH) is the set of factorizations of a, and L H a) = La) = { z z Za) } N0 is the set of lengths of a. Thus H is factorial if and only if H red is free abelian equivalently, Za) = 1 for all a H). The monoid H is called atomic if Za) for all a H equivalently, every nonunit can be written as a finite product of irreducible elements). From now on we suppose that H is atomic. Note that, La) = {0} if and only if a H, and La) = {1} if and only if a AH). We denote by LH) = {La) a H} the system of sets of lengths of H, and by H) = L) N the set of distances of H. L LH) For k N, we set ρ k H) = k if H = H, and ρ k H) = sup{sup L L LH), k L} N { }, if H H. Then ρ k H) ρh) = sup{ρl) L LH)} = lim R 1 { } k k is the elasticity of H. The monoid H is said to be half-factorial if H) = equivalently, ρh) = 1). If H is not half-factorial, then min H) = gcd H).

4 4 ALFRED GEROLDINGER AND WOLFGANG A. SCHMID decomposable if there exist submonoids H 1, H with H i H for i [1, ] such that H = H 1 H and H is called indecomposable else). For a free abelian monoid FP ), we introduce a distance function d: FP ) FP ) N 0, by setting { a b } da, b) = max, N 0 for a, b FP ), gcda, b) gcda, b) and we note that da, b) = 0 if and only if a = b. For a subset Ω FP ), we define the catenary degree cω) as the smallest N N 0 { } with the following property: for each a, b Ω, there are elements a 0,... a k Ω such that a = a 0, a k = b, and da i 1, a i ) N for all i [1, k]. Note that cω) = 0 if and only if Ω 1. For an element a H, we call c H a) = ca) := cz H a)) the catenary degree of a, and ch) = sup{ca) a H} N 0 { } is the catenary degree of H. The monoid H is factorial if and only if ch) = 0, and if H is not factorial, then + sup H) ch). Krull monoids and transfer homomorphisms. A monoid homomorphism ϕ: H F is said to be a divisor homomorphism if ϕa) ϕb) 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 following equivalent properties is satisfied [18, Theorem.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 = FP ) into a free abelian monoid such that for each p P there is a finite set E H with p = gcd ϕe) ). Suppose that H is a Krull monoid. Then a divisor theory ϕ: H F = FP ) is essentially unique. The class group CH) = qf )/q ϕh) ) depends only on H and it is isomorphic to the v-class group C v H). For a qf ), we denote by [a] = aq ϕh) ) CH) the class containing a. Thus every class g CH) is considered as a subset of qf ) and P g is the set of prime divisors lying in g. We use additive notation for the class group. An integral domain R is a Krull domain if and only if its multiplicative monoid R\{0} is a Krull monoid this generalizes to Marot rings: indeed, a Marot ring is a Krull ring if and only if the monoid of regular elements is a Krull monoid, [1]). 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 these domains are Krull monoids with finite class group such that every class contains a prime divisor [18, Section.11]). Monoid domains and power series domains that are Krull are discussed in [7, 35, 36]. For monoids of modules which are Krull we refer the reader to [6, 3, 11]. Much of the arithmetic of a Krull monoid can be studied in an associated monoid of zero-sum sequences. This is a Krull monoid again which can be studied with methods from Additive Combinatorics. To introduce the necessary concepts, let G be an additively written abelian group, G 0 G a subset, and let FG 0 ) be the free abelian monoid with basis G 0. In Combinatorial Number Theory, the elements of FG 0 ) are called sequences over G 0. If S = g 1... g l FG 0 ), where l N 0 and g 1,..., g l G 0, then σs) = g g l is called the sum of S, and the monoid BG 0 ) = {S FG 0 ) σs) = 0} FG 0 ) is called the monoid of zero-sum sequences over G 0. Since the embedding BG 0 ) FG 0 ) is a divisor homomorphism, Property b) shows that BG 0 ) is a Krull monoid. The monoid BG) is factorial if and only if G. If G =, then BG) is a Krull monoid with class group isomorphic to G and every class contains precisely one prime divisor. This is well-known and will also follow from a more general result given in Proposition.5. For every arithmetical invariant H) defined for a monoid H, it is usual to write G 0 ) instead of BG 0 )) although this is an abuse of language, there will be no danger of confusion). In particular, we set AG 0 ) = ABG 0 )) and LG 0 ) = LBG 0 )). Similarly, arithmetical properties of BG 0 ) are attributed to G 0. Thus, G 0 is said to be

5 A CHARACTERIZATION OF CLASS GROUPS VIA SETS OF LENGTHS 5 in)decomposable if BG 0 ) is in)decomposable, non-) half-factorial if BG 0 ) is non-)half-factorial. A monoid homomorphism θ : H B is called a transfer homomorphism if if it has the following properties: T 1) B = θh)b and θ 1 B ) = H. T ) If u H, b, c B and θu) = bc, then there exist v, w H such that u = vw, θv) b and θw) c. Transfer homomorphisms preserve sets of lengths and further arithmetical properties. We formulate the relevant results in the settings we need. Proposition.1. Let H be a Krull monoid with divisor theory ϕ: H FP ), class group G, and suppose that that each class contains a prime divisor. Let β : FP ) FG) be the homomorphism defined by βp) = [p] G for each p P. Then the homomorphism β = β ϕ: H BG) is a transfer homomorphism. In particular, we have 1. L H a) = L BG) βa) ) for each a H and LH) = LG).. If G 3, then ch) = c BG) ) i.e., the catenary degrees of H and of BG) coincide). Proof. See [18, Section 3.4]. There are recent deep results showing that there are non-krull monoids which allow transfer homomorphisms to monoids of zero-sum sequences. Proposition.. 1. Let O be a holomorphy ring in a global field K, A a central simple algebra over K, and H a classical maximal O-order of A such that every stably free left R-ideal is free. Then LH) = LG), 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 Ĥ such that the natural map XĤ) XH) is bijective and there is an isomorphism ϑ: C vh) C v Ĥ) between the v-class groups. Then LH) = LG), where G = C v H) is finite abelian. Proof. 1. See [50, Theorem 1.1], and [5, 48] for related results of this flavor. Note that H is a noncommutative Dedekind prime ring.. See [19, Theorem 5.8] for a more general result in the setting of weakly Krull monoids. Note, if H Ĥ, then H is not a Krull domain. Thus, beyond the Krull monoids occurring in Proposition.1, there are classes of objects H, where sets of lengths depend only on an abelian group G and where LH) = LG) holds. Hence all characterization results, such as Theorem 1.1, applies to them. To provide an example where the opposite phenomenon holds, consider the class of numerical monoids. There we can find infinitely many) non-isomorphic numerical monoids H and H with LH) = LH ) [1]). Zero-Sum Theory. Let G be an additive abelian group, G 0 G a subset, and G 0 = G 0 \ {0}. Then [G 0 ] G denotes the subsemigroup and G 0 G the subgroup generated by G 0. For a sequence S = g 1... g l = g G 0 g vgs) FG 0 ),

6 6 ALFRED GEROLDINGER AND WOLFGANG A. SCHMID we set ϕs) = ϕg 1 )... ϕg l ) for any homomorphism ϕ: G G, and in particular, we have S = g 1 )... g l ). We call supps) = {g G v g S) > 0} G the support of S, v g S) the multiplicity of g in S, S = l = v g S) N 0 the length of S, ks) = 1 Q the cross number of S, ordg) g G g G l { } σs) = g i the sum of S, and ΣS) = g i I [1, l] the set of subsequence sums of S. i=1 i I The sequence S is said to be zero-sum free if 0 / ΣS), a zero-sum sequence if σs) = 0, a minimal zero-sum sequence if it is a nontrivial zero-sum sequence and every proper subsequence is zero-sum free. Clearly, if S is zero-sum free, then σs))s is a minimal zero-sum sequence, and the minimal zero-sum sequences over G 0 are precisely the atoms of the monoid BG 0 ). Their study is basic for all arithmetical investigations of Krull monoids. The three invariants, the small) Davenport constant dg 0 ) = sup { S S FG 0 ) is zero-sum free } N 0 { }, the large) Davenport constant DG 0 ) = sup { U U AG 0 ) } N 0 { }, and the cross number KG 0 ) = sup { ku) U AG0 ) } Q { } are classical tools describing minimal zero-sum sequences all three of them are finite for finite subsets G 0 ). For n N, let C n denote a cyclic group with n elements. Suppose that G is finite. A tuple e i ) i I is called a basis of G if all elements are nonzero and G = i I e i. For p P, let r p G) denote the p-rank of G, rg) = max{r p G) p P} denote the rank of G, and let r G) = p P r pg) be the total rank of G. If G > 1, then G r = C n1... C nr, and we set d G) = n i 1), where r, n 1,..., n r N with 1 < n 1... n r, r = rg), and n r = expg) is the exponent of G. Similarly, we have G = C q1... C qs, and we set K 1 s G) = expg) + q i 1, qi where s = r G) and q 1,..., q s are prime powers. If G = 1, then rg) = r G) = 0, expg) = 1, and d G) = 0. We will use the following well-known results without further mention. Proposition.3. Let G be a finite abelian group d G) 1 + dg) = DG) G. If G is a p-group or rg), then dg) = d G).. K G) KG) 1 + log G, and the left inequality is an equality if G is a p-group or r G). Proof. See [18, Chapter 5]. There are more groups G with d G) = dg) but we do not have equality in general. On the other hand there is known no group G with K G) < KG). We refer to [8, 0, 49, 47, 10, 3, 34] for recent progress. We will make substantial use of the following result, which highlights the role of the Davenport constant for the arithmetic of Krull monoids. Proposition.4. Let H be a Krull monoid with finite class group G where G 3 and every class contains a prime divisor. Then ch) [3, DG)], and we have i=1 i=1

7 A CHARACTERIZATION OF CLASS GROUPS VIA SETS OF LENGTHS 7 1. ch) = DG) if and only if G is either cyclic or an elementary -group.. ch) = DG) 1 if and only if G is isomorphic either to C r 1 C 4 for some r or to C C n for some n. Proof. See [18, Theorem 6.4.7] and [5, Theorem 1.1]. We gather some simple facts on sets of lengths which will be used without further mention. A BG 0 ) and d = min{ U U AG 0 )}. If A = BC with B, C BG 0 ), then LB) + LC) LA). If A = U 1... U k = V 1... V l with U 1,..., U k, V 1,..., V l AG 0 ) and k < l, then ld l V ν = A = k U ν kdg 0 ), and hence A A min LA) max LA) DG 0 ) d. For sequences over cyclic groups the g-norm plays a similar role as the length does for sequences over arbitrary groups. Let g G with ordg) = n. For a sequence S = n 1 g)... n l g) F g ), where l N 0 and n 1,..., n l [1, n], we define S g = n n l n Note that σs) = 0 implies that n n l 0 mod n whence S g N 0. Thus, g : B g ) N 0 is a homomorphism, and S g = 0 if and only if S = 1. If S AG 0 ), then S g [1, n 1], and if S g = 1, then S AG 0 ). Arguing as above we obtain that A g n 1 min LA) max LA) A g. We will need the concept of relative block monoids as introduced by F. Halter-Koch in [30], and recently studied by N. Baeth et al. in [4]). Let G be an abelian group. For a subgroup K G let B K G) = {S FG) σs) K} FG), and let D K G) denote the smallest l N { } with the following property: Every sequence S FG) of length S l has a subsequence T with σt ) K. Clearly, B K G) FG) is a submonoid with BG) = B {0} G) B K G) B G G) = FG) and D {0} G) = DG). The following result is well-known [4, Theorem.]). Since there seems to be no proof in the literature and we make substantial use of it, we provide the simple arguments. Proposition.5. Let G be an abelian group and K G a subgroup. 1. B K G) is a Krull monoid. If G = and K = {0}, then B K G) = BG) is factorial. In all other cases the embedding B K G) FG) is a divisor theory with class group isomorphic to G/K and every class contains precisely K prime divisors.. The monoid homomorphism θ : B K G) BG/K), defined by θg 1... g l ) = g 1 +K)... g l +K) is a transfer homomorphism. If G/K 3, then c B K G) ) = c BG/K) ). 3. D K G) = sup{ U U is an atom of B K G)} = DG/K).. Let

8 8 ALFRED GEROLDINGER AND WOLFGANG A. SCHMID Proof. 1. If G = 1, then BG) = FG) is factorial, the class group is trivial, and there is precisely one prime divisor. If G = K =, then B K G) = FG) is factorial, the class group is trivial, and there are precisely two prime divisors. If G = and K = 1, then B K G) = BG) is factorial, and hence a Krull monoid with trivial class group. Suppose that G 3. Clearly, the embedding B K G) FG) is a divisor homomorphism. To verify that it is a divisor theory, let g G be given. If ordg) = n 3, then g = gcd g n, g g) ). If ordg) =, then there is an element h G \ {0, g} and g = gcd g, ghg h) ). If ordg) =, then g = gcd g)g, gg) 3g) ). The map ϕ: FG) G/K, defined by ϕs) = σs)+k for every S FG), is a monoid epimorphism. If S, S FG), then σs) + K = σs ) + K if and only if S [S] = Sq B K G) ). Thus ϕ induces a group isomorphism Φ: qfg))/qb K G)) G/K, defined by Φ[S]) = σs) + K, and we have [S] G = σs) + K. Thus the class [S] contains precisely K prime divisors.. If G, then θ is the identity map. If G 3, then this follows from 1. and from Proposition Since a sequence S FG) is an atom of B K G) if and only if S 1, σs) K and σt ) / K for all proper subsequences T of S, it follows that D K G) = sup{ U U is an atom of B K G)}. Since θ is a transfer homomorphism, we get θ AB K G)) ) = AG/K) and θ 1 AG/K) ) = AB K G)). Therefore U = θu) for all U B K G), and it follows that sup{ U U AB K G))} = sup{ V V AG/K)} = DG/K). 3. Structural results on LG) and first basic constructions Let G be an abelian group. Recall that all sets of lengths L LG) are finite indeed, max LA) A for all A BG)) and that G is half-factorial i.e., L = 1 for each L LG)) if and only if G. If G is infinite, then every finite subset L N is contained in LG) [18, Theorem 7.4.1]). If G is finite with G >, then sets of lengths have a well-studied structure which is the basis for all characterizations of class groups. First we repeat the results needed in the sequel and then we start with some basic constructions which will be used in all forthcoming sections. Definition 3.1. Let d N, l, M N 0 and {0, d} D [0, d]. A subset L Z is called an arithmetical multiprogression AMP for short) with difference d, period D and length l, if L is an interval of min L + D + dz this means that L is finite nonempty and L = min L + D + dz) [min L, max L]), and l is maximal such that min L + ld L. almost arithmetical multiprogression AAMP for short) with difference d, period D, length l and bound M, if L = y + L L L ) y + D + dz where L is an AMP with difference d whence L ), period D and length l such that min L = 0, L [ M, 1], L max L + [1, M] and y Z. almost arithmetical progression AAP for short) with difference d, bound M it is an AAMP with difference d, period {0, d}, bound M and length l. The subset G) of G), defined as plays a crucial role throughout this paper. G) = {min G 0 ) G 0 G with G 0 ) } G), Proposition 3. Structural results on LG)). Let G be a finite abelian group with G 3. and length l, if

9 A CHARACTERIZATION OF CLASS GROUPS VIA SETS OF LENGTHS 9 1. There exists some M N 0 such that every set of lengths L LG) is an AAMP with some difference d G) and bound M.. For every M N 0 and every finite nonempty set N, there is a finite abelian group G such that the following holds : for every AAMP L with difference d and bound M there is some y L N such that y + L LG ) for all y y L. 3. Let G 0 G be a subset. Then there exist a bound M N 0 and some A BG 0 ) such that for all A A BG 0 ) the set of lengths LA) is an AAP with difference min G 0 ) and bound M. 4. If A BG) such that suppa) {0} is a subgroup of G, then LA) is an arithmetical progression with difference 1. Proof. The first statement gives the Structure Theorem for Sets of Lengths [18, Theorem ]), which is sharp by the second statement proved in [45]. The third and the fourth statements show that sets of lengths are extremely smooth provided that the associated zero-sum sequence contains all elements of its support sufficiently often [18, Theorems and 7.6.8]). Proposition 3.3 Structural results on G) and on G)). Let G = C n1... C nr where r, n 1,..., n r N with r = rg), 1 < n 1... n r, and G G) is an interval with rg) [ ] [ ] 1, max{expg), k 1} G) 1, DG) where k =. 1 G) G), [1, rg) 1] G), and max G) = max{expg), rg) 1}. 3. If G is cyclic of order G = n 4, then max G) \ {n } ) = n 1. Proof. The statement on max G) follows from [6]. For all remaining statements see [18, Section 6.8]. A more detailed analysis of G) in case of cyclic groups can be found in [38]. i=1 ni. Proposition 3.4 Results on ρ k G) and on ρg)). Let G be a finite abelian group with G 3, and let k N. 1. ρ k G) = kdg) kdg) ρ k+1 G) kdg) + DG)/. If G is cyclic, then equality holds on the left side. 3. ρg) = DG)/. Proof. See [18, Chapter 6.3], [15, Theorem 5.3.1], and [17] for recent progress. In the next propositions we provide examples of sets of lengths over cyclic groups, over groups of rank two, and over groups of the form C r 1 C n with r, n N. All examples will have difference d = max G) and period D with {0, d} D [0, d] and D = 3, and we write them down in a form used in Definition 3.1 in order to highlight their periods. It will be crucial for our approach see Proposition 6.5) that the sets given in Proposition 3.5. do not occur over cyclic groups Proposition 3.6). It is well-known that sets of lengths over cyclic groups and over elementary -groups have many features in common, and this carries over to rank two groups and groups of the form C r 1 C n see Propositions 3.5., 3.7., and 6.5). Let G be an abelian group and L LG). Then there is a B BG) such that L = LB) and hence m + L = L0 m B) LG) for all m N 0. Therefore the interesting sets of lengths L LG) are those which do not stem from such a shift. These are those sets L LG) with m + L / LG) for every m N. Proposition 3.5. Let G = C n1 C n where n 1, n N with < n 1 n, and let d [3, n 1 ].

10 10 ALFRED GEROLDINGER AND WOLFGANG A. SCHMID 1. For each k N we have k + ) + {0, d, n } + {νn ) ν [0, k 1]} {kn + ) + d )} = k + ) + {0, d } + {νn ) ν [0, k]} LG).. For each k N we have ) k + 3) + {0, n 1, n } + {νn ) ν [0, k]} { } kn + 3) + n 1 ) + n ) LG). Proof. Let e 1, e ) be a basis of G with orde i ) = n i for i [1, ], and let k N. For i [1, ], we set U i = e ni i and V i = e i )e i. Then U i ) k Ui k = U i ) k ν U k ν i V νni i for all ν [0, k], and hence L U i ) k Ui k ) = k + {νni ) ν [0, k]}. 1. We set h = d 1)e 1, W 1 = e 1 ) d 1 h, and W = e n1 d 1) 1 h. Then ZU 1 W 1 ) = {U 1 W 1, V1 d 1 W } and LU 1 W 1 ) = {, d}. Therefore L U ) k U k ) U 1 W 1 = L U ) k U k ) ) + L U1 W 1 = {k + νn ) ν [0, k]} + {, d}. We define = k + ) + {νn ) ν [0, k]} + {0, d }. W 1 = e n1 1 1 e n 1 e 1 +e ), W = e 1 )e n 1 e 1 +e ), W 3 = e n1 1 1 e )e 1 +e ), W 4 = e 1 ) e )e 1 +e ), and B k = W 1 U 1 ) U )U k U ) k. Then any factorization of B k is divisible by precisely one of W 1,..., W 4, and we obtain that B k = W 1 U 1 ) U )U k U ) k = W V n1 1 1 U )U k U ) k = W 3 U 1 )V n 1 U k U ) k = W 4 V n1 1 1 V n 1 U k U ) k. Thus it follows that LB k ) = {3, n 1 + 1, n + 1, n 1 + n 1} + L U k U ) k) = k + 3) + {νn ) ν [0, k]} k + 3) + n 1 ) + {νn ) ν [0, k]} k + 3) + n ) + {νn ) ν [0, k]} k + 3) + n 1 ) + n ) + {νn ) ν [0, k]}. Thus max LB k ) = kn + 3) + n 1 ) + n ) and ) LB k ) = k + 3) + {0, n 1, n } + {νn ) ν [0, k]} {max LB k )}. Proposition 3.6. Let G be a cyclic group of order G = n 4, and let d [3, n 1]. 1. For each k N 0, we have k + ) + {0, d } + {νn ) ν [0, k]} LG).. For each k N 0, we set ) L k = k + 3) + {0, d, n } + {νn ) ν [0, k]} Then for each k N 0 and each m N 0, we have m + L k / LG). { } kn + 3) + d ) + n ).

11 A CHARACTERIZATION OF CLASS GROUPS VIA SETS OF LENGTHS 11 Proof. Let k N Let g G with ordg) = n, U = g n, V = g)g, W 1 = d 1)g ) g) d 1, W = d 1)g ) g n d 1), and B k = U)U ) k UW1. Then ZUW 1 ) = {UW 1, W V d 1 } and LUW 1 ) = {, d}. Since every factorization of B k is divisible either by W 1 or by W, it follows that LB k ) = L U) k U k) + L UW 1 ) = {k + νn ) ν [0, k]} + {, d} = k + ) + {νn ) ν [0, k]} + {0, d }.. Note that max L k = kn + 3) + d ) + n ) = k + 1)n + d 1). Assume to the contrary that there is a B k BG) such that LB k ) = L k. Then min LB k ) = k + 3 and, by Proposition 3.4, k + 1)n + d 1) = max LB k ) ρ k+3 G) = k + 1)n + 1, a contradiction. If m N 0 and B m,k BG) such that LB m,k ) = m+l k, then L0 m B m,k ) = L k LG). Thus m + L k / LG) for any m N 0. Proposition 3.7. Let G = C r 1 C n where r, n N and n is even. 1. For each k N 0 we have L k = k + ) + {0, n, n + r 3} + {νn ) ν [0, k]} LG). If r, then L k / LC n ).. For each k N 0, we have ) { } k + 3) + {0, r 1, n } + {νn ) ν [0, k]} kn + 3) + r 1) + n ) LG). Proof. Let e 1,..., e r 1, e r ) be a basis of G with orde 1 ) =... = orde r 1 ) = and orde r ) = n. We set e 0 = e e r 1, U i = e ordei) i for each i [1, r], U 0 = e 0 + e r )e 0 e r ), V r = e r )e r, Let k N 0. V = e 1... e r 1 e 0 + e r ) e r ), and W = e 1... e r 1 e 0 + e r )e n 1 r. 1. Obviously, L W )W ) = {, n, n + r 1} and L W )W U r ) k Ur k ) ) = L W )W + L Ur ) k Ur k ) = {, n, n + r 1} + {k + νn ) ν [0, k]} = k + ) + {0, n, n + r 3} + {νn ) ν [0, k]} Since min L k = k +, max L k = k + 1)n + r 1, and ρ k+ C n ) = k + 1)n by Proposition 3.4, r implies that L k / LC n ).. Let L k denote the set in the statement. We define B k = U 0 U 1... U r 1 U r ) k+1 U k+1 r and assert that LB k ) = L k. Let z be a factorization of B k. We distinguish two cases. CASE 1: U 1 z. ) k+1 νv νn Then U 0 U 1... U r 1 z which implies that z = U 0 U 1... U r 1 Ur )U r r for some ν [0, k+1] and hence z r + k + ) + {νn ) ν [0, k + 1]}. CASE : U 1 z.

12 1 ALFRED GEROLDINGER AND WOLFGANG A. SCHMID ) k νv Then either V z or W z. If V z, then z = V )V Vr n 1 νn Ur )U r r for some ν [0, k] and ) k νv νn hence z n + 1) + k + {νn ) ν [0, k]}. If W z, then z = W )W V r Ur )U r r for some ν [0, k] and hence z 3 + k + {νn ) ν [0, k]}. Putting all together the assertion follows. Proposition 3.8. Let G be a finite abelian group, g G with ordg) = n 5, and B BG) such that g)g ) n B. Suppose LB) is an AAMP with period {0, d, n } for some d [1, n 3] \ {n )/}. 1. If S A B g G) ) with S B, then σs) {0, g, g, d + 1)g, d + 1)g}.. If S 1, S A B g G) ) with S 1 S B, then σs i ) {0, g, g} for at least one i [1, ]. Proof. By definition, there is a y Z such that LB) y + {0, d, n } + n )Z. We set U = g n and V = g)g. 1. Let S A B g G) ) with S B and set σs) = kg with k [0, n 1]. If k {0, 1, n 1}, then we are done. Suppose that k [, n ]. Since S is an atom in B g G), it follows that W 1 = S g) k AG) and W 1 = Sg n k AG). We consider a factorization z ZB) with UW 1 z, say z = UW 1 y. Then z = W 1V k y is a factorization of B of length z = z + k 1. Since LB) is an AAMP with period {0, d, n } for some d [1, n 3] \ {n )/} it follows that k 1 {d, n d}.. Let S 1, S A B e G) ) with S 1 S B, and assume to the contrary σs i ) = k i e with k i [, n ] for each i [1, ]. As in 1. it follows that W 1 = S 1 g) k1, W 1 = S 1 g n k1, W = S g) k, and W = S g n k are in AG). We consider a factorization z ZB) with UW 1 UW z, say z = UW 1 UW y. Then z 1 = W 1V k1 1 UW y ZB) with z 1 = z + k 1 1 and hence k 1 1 {d, n d}. Similarly, z = UW 1 W V k 1 y ZB), hence k 1 {d, n d}, and furthermore it follows that k 1 = k. Now z 3 = W 1V k1 1 W V k 1 y ZB) is a factorization of length z 3 = z +k 1 +k. Thus, if k 1 1 = d, then d {n, n +d}, a contradiction, and if k 1 1 = n d, then n d) {n, n +n d)}, a contradiction. 4. Characterizations of extremal cases Let G be a finite abelian group. By the Structure Theorem for Sets of Lengths Proposition 3..1), all sets of lengths are AAMPs with difference in G) and some universal bound). By definition, the concept of an AAMP comprises arithmetical progressions, AAPs, and AMPs. The goal of this section is to characterize those groups where all sets of lengths are not only AAMPs, but have one of these more special forms. As a consequence we establish characterizations of all involved class groups. Since LC 1 ) = LC ) and LC 3 ) = LC C ) see Proposition 4. below), small groups require special attention in the study of the Characterization Problem. All results of this section are gathered in the following Theorem 4.1. Theorem 4.1. Let G be a finite abelian group. 1. The following statements are equivalent : a) All sets of lengths in LG) are arithmetical progressions with difference in G). b) All sets of lengths in LG) are arithmetical progressions. c) G is cyclic of order G 4 or isomorphic to a subgroup of C 3 or isomorphic to a subgroup of C 3.. The following statements are equivalent :

13 A CHARACTERIZATION OF CLASS GROUPS VIA SETS OF LENGTHS 13 a) There is a constant M N such that all sets of lengths in LG) are AAPs with bound M. b) G is isomorphic to a subgroup of C 3 3 or isomorphic to a subgroup of C The following statements are equivalent : a) All sets of lengths in LG) are AMPs with difference in G). b) G is cyclic with G 5 or isomorphic to a subgroup of C 3 or isomorphic to a subgroup of C Suppose that DG) 4 and that LG) satisfies the property in 1.,., or 3. If G is a finite abelian group such that LG) = LG ), then G = G. We proceed in a series of lemmas. The proof of Theorem 4.1 will be given at the end of this section. Proposition LC 1 ) = LC ) = { {m} m N 0 }.. LC 3 ) = LC C ) = { y + k + [0, k] y, k N0 }. 3. LC 4 ) = { y + k [0, k] y, k N 0 } { y + k + [0, k] y, k N 0 }. 4. LC 3 ) = { y + k + 1) + [0, k] y N0, k [0, ] } { y + k + [0, k] y N 0, k 3 } { y + k + [0, k] y, k N 0 }. 5. LC 3) = {[k, l] k N 0, l [k, 5k]} {[k + 1, l] k N, l [k + 1, 5k + ]} {{1}}. Proof. 1. This is straightforward and well-known. A proof of.,3., and 4. can be found in [18, Theorem 7.3.]. For 5. we refer to [3, Proposition 3.1]. Lemma 4.3. Let G be a cyclic group of order G = n 7, g G with ordg) = n, k N, and { g nk g) nk g) n if n is even, A k = g nk g) nk g) n 1)/ g ) if n is odd. Then there is a bound M N such that, for all k n 1, the sets LA k ) are AAPs with difference 1 and bound M, but they are not arithmetical progressions with difference 1. Proof. We set G 0 = {g, g, g}, U 1 = g)g, U = g) g) and, if n is odd, then V 1 = g) n+1)/ g). Furthermore, for j [0, n/], we define W j = g) j g n j. Then, together with W 0 = g) n, these are all minimal zero-sum sequences which divide A k for k N. Note that W 0 g = n 1, U g = V 1 g =, and U 1 g = W j g = 1 for all j [0, n/]. It is sufficient to prove the following two assertions. A1. There is a bound M N 0 such that LA k ) is an AAP with difference 1 and bound M for all k n 1. A. For each k N, LA k ) is not an arithmetical progression with difference 1. Proof of A1. By Proposition 3..1 there is a bound M N 0 such that, for each k N, LA k ) is an AAMP with difference d k G) [1, n ] and bound M. Suppose that k n 1. Then W 0 U ) n 1 divides A k. Since W 0 U = W 1 U 1, it follows that W 0 U ) n 1 = W 0 U ) n 1 ν W 1 U 1 ) ν for all ν [0, n 1] and hence L W 0 U ) n 1) [n, 3n 3]. Thus LA k ) contains an arithmetical progression of difference 1 and length n 1. Therefore there is a bound M N 0 such that LA k ) is an AAP with difference 1 and bound M for all k n 1.

14 14 ALFRED GEROLDINGER AND WOLFGANG A. SCHMID Proof of A. Let k N. Observe that { W k A k = 0 W 0 ) k Wn/ if n is even, W0 k W 0 ) k W n 1)/ ) if n is odd, and it can be seen that min LA k ) = k +. We assert that k + 3 / LA k ). If n is even, then and similarly, for odd n we have W 0 W n/ = W j W n/ j for each j [0, n/], W 0 W n 1)/ = W j W n 1)/ j for each j [0, n 1)/]. In both cases, all factorizations of A k of length k + contain only atoms with g-norm 1 and with g-norm n 1. Let z be any factorization of A k containing only atoms with g-norm 1 and with g-norm n 1. Then z z is a multiple of n whence if z > z, then z z n > 1. Next we consider a factorization z of A k containing at least one atom with g-norm, say z has r atoms with g-norm n 1, s 1 atoms with g-norm, and t atoms with g-norm 1. Then k > r, and we study A k g = kn 1) + k + ) = rn 1) + s + t, z z = r + s + t k + ) = r + s + kn 1) + k + ) rn 1) s k + ) = k r)n ) s. Note that s v g A k ) n. Thus, if k r, then k r)n ) s n 4 s n 4 > 1. Suppose that k r = 1. Then we cancel W 0 ) k 1, and consider a relation where W 0 occurs precisely once. Suppose that all s atoms of g-norm are equal to U. Since v g U ) =, it follows that s v g W 0 )/ = n/ whence k r)n ) s n n/ = n/ > 1. Suppose that V 1 occurs among the s atoms with g-norm. Then n is odd, V 1 occurs precisely once, and whence s 1 v g A k ) n + 1 = n 1) n + 1 k r)n ) s n ) n 1 = n + 1 = n 3, > 1. Lemma 4.4. Let G be a cyclic group of order G = 6, g G with ordg) = 6 and, for each k N, A k = g 6k g) 6k 4g) g) 4 3g)g 3. Then there is a bound M N such that, for all k N, the sets LA k ) are AAPs with difference 1 and bound M, but they are not arithmetical progressions with difference 1. Proof. We set U = g 6, W 1 = 4g) g) 4, and W = 3g)g 3. Then, for each k N, we have A k = U k U) k W 1 W. By Proposition 3.3, we obtain that G) = {1,, 4}. By Proposition 3..1, there is a bound M N such that, for every k N, LA k ) is an AAMP with difference d k G) and bound M. We show that k + 4, k + 5, k + 6, k + 7 LA k ) which implies that there is a bound M N such that, for every k N, LA k ) is an AAP with difference 1 and bound M. Let k N. We set V = g)g, W 3 = 4g)3g) g), W 4 = 4g)g, and obtain that A k = U k U) k W 1 W = U k U) k W 3 V 3 = U k 1 U) k W 4 W V 4 = U k 1 U) k 1 W 1 W V 6 = U k 1 U) k 1 W 4 W )V 7,

15 A CHARACTERIZATION OF CLASS GROUPS VIA SETS OF LENGTHS 15 and hence {k +, k + 4, k + 5, k + 6, k + 7} LA k ). Furthermore, min LA k ) = k +, and z = U k U) k W 1 W is the only factorization of A k of length k +. From this we see that there is no factorization of length k + 3, and hence LA k ) is not an arithmetical progression with difference 1. Lemma 4.5. Let G be a cyclic group of order G = 5. Then every L LG) has one of the following forms : L is an arithmetical progression with difference 1. L is an arithmetical progression with difference 3. L is an AMP with period {0,, 3} or with period {0, 1, 3}. Proof. By Proposition 3.3 we obtain that G) = {1, 3}. Let A BG). If A = 0 m A with m N 0 and A BG ), then LA ) = m + LA). Thus it is sufficient to prove the assertion for LA). If suppa) = 1, then LA) = 1. If suppa) = 4, then LA) is an arithmetical progression with difference 1 by Proposition Suppose that suppa) =. Then there is a g G such that suppa) = {g, g} or suppa) = {g, 4g}. If suppa) = {g, g}, then LA) is an arithmetical progression with difference 1 this can be checked directly by arguing with the g-norm). If suppa) = {g, 4g}, then LA) is an arithmetical progression with difference 3. Thus it remains to consider the case suppa) = 3. We set G 0 = suppa). Then there is an element g G 0 such that g G 0. Thus either G 0 = {g, g, g} or G 0 = {g, 3g, g}. Since {g, 3g, g} = { g, g), g)}, we may suppose without restriction that G 0 = {g, g, g}. If LA)) {1}, then LA) is an arithmetical progression with difference 1. If 3 LA)), then LA)) = {3} by [9, Theorem 3.], which means that LA) is an arithmetical progression with difference 3. Thus it remains to consider the case where LA)) [1, ]. We show that LA) is an AMP with period {0,, 3} or with period {0, 1, 3}. Since LA)), there exist k N, A 1,..., A k, B 1,..., B k+ AG 0 ) such that A = A 1... A k = B 1... B k+, and k + 1 / LA). For convenience we list the elements of AG 0 ), and we order them by their lengths: g 5, g) 5, g) 5, g 3 g), g) 3 g), gg), g) g), g g). Clearly, { S g S AG 0 )} = {1,, 4}, and g) 5 is the only atom having g-norm 4. We distinguish two cases. CASE 1: g) 5 / {A 1,..., A k }. Then {A 1,..., A k } must contain atoms with g-norm. These are the atoms g) 5, g) g), g) 3 g). If g 5 or g 3 g) occurs in {A 1,..., A k }, then k + 1 LA), a contradiction. Thus none of the elements g) 5, g 5, and g 3 g) lies in {A 1,..., A k }, and hence Now we set h = g and obtain that {A 1,..., A k } {g) 5, g) 3 g), gg), g) g), g g)}. {A 1,..., A k } {g) 5, g) 3 g), gg), g) g), g g)} = {h 5, h 3 h), h 3h), hh), h)3h)}. Since the h-norm of all these elements equals 1, it follows that max LA) = k, a contradiction. CASE : g) 5 {A 1,..., A k }. If g) 5, or gg), or g) 3 g) occurs in {A 1,..., A k }, then k + 1 LA), a contradiction. Since { g, g}) = {3}, it follows that Ω = {A 1,..., A k } {g 3 g), g) g) }.

16 16 ALFRED GEROLDINGER AND WOLFGANG A. SCHMID ) Since g 3 g) )g) g) = two cases. CASE.1: {A 1,..., A k } {g 5, g) 5, g g), g) g) }. We set h = g, and observe that ) ) g)g gg) and k+1 / LA), it follows that Ω = 1. We distinguish {A 1,..., A k } {g 5, g) 5, g g), g) g) } = {h 5, h) 5, h h), h 3h)}. Since h) 5 is the only element with h-norm greater than 1, it follows that h) 5 {A 1,..., A k }. Since ) ) ) ) {h, h}) = {3}, it follows that h 3h) {A 1,..., A k }. Since h) 5 h 3h) = h h) 3h) h) 3, we obtain that k + 1 LA), a contradiction. CASE.: {A 1,..., A k } {g 5, g) 5, g g), g 3 g)}. ) ) ) ) ) Since g 3 g) g) 5 = g 5 g g) g) g) and k + 1 / LA), it follows that {i [1, k] A i = g 3 g)} = 1, and hence v g A) = 1. Thus every factorization z of A has the form z = g)g 3) z 1 or z = g) g) ) z, where z 1, z are factorizations of elements B 1, B B{ g, g}). Since LB 1 ) and LB ) are arithmetical progressions of difference 3, LA) is a union of two shifted arithmetical progression of difference 3. We set A = g 5) m 1 g) 5 ) g)g ) m 3 g)g 3 ), where m 1 N 0, m N, and m 3 [0, 4]. Suppose that m 1 1. Note that A = g 5) g) 5) g)g 3) = g)g ) 3 g) g) ) g 5) = g)g ) 5 g)g 3 ), and hence LA ) = {3, 5, 6}. We set A = A A with A B{g, g}). The above argument on the structure of the factorizations of A implies that LA) is the sumset of LA ) and LA ) whence LA) = LA ) + LA ) = 3 + {0,, 3} + LA ). Since LA ) is an arithmetical progression with difference 3, LA) is an AMP with period {0,, 3}. Suppose that m 1 = 0. If m 3 [, 4], then LA) = {m + m 3, m + m 3 + 1, m + m 3 + 3} is an AMP with period {0, 1, 3}. If m 3 = 1, then LA) = {m +, m + 4}. If m 3 = 0, then LA) = {m + 1, m + 3}. Lemma 4.6. Let G = C n1 C n where n 1, n N with 4 n 1 n, e 1, e ) be a basis of G with orde i ) = n i for i [1, ], and set W = e n1 1 1 e n 1 e 1 + e ). Then there is a bound M N such that, for all sufficiently large k, the sets L W k W ) k) are AAPs with difference 1 and bound M, but they are not arithmetical progressions with difference 1. Proof. We set e 0 = e 1 + e, G 0 = {e ν, e ν ν [0, ]}, U ν = e ordeν) ν and V ν = e ν )e ν for ν [0, ]. For k N we set A k = W k W ) k and L k = LA k ). Since gcd G 0 ) gcd{n 1, n, W = n 1 + n 3}) = 1, it follows that min G 0 ) = 1. Thus, by Proposition 3..3, there are M, k 0 N such that for all k k 0, the set L k is an AAP with difference 1 and bound M. Let k N. We assert that 1 + min L k / L k. This implies that L k is not an arithmetical progression with difference 1. Since W = W = DG), it follows that min L k = k, and clearly W k W ) k is the only factorization of A k having length k. If S = e 1 )e n 1 e 1 + e ), then W W ) = S S)V n1 1, k + n 1 L k, and this is the second shortest factorization length of A k. Lemma 4.7. Let G = C 4, e 1, e, e 3, e 4 ) be a basis of G, e 0 = e e 4, U 4 = e 0... e 4, U 3 = e 1 e e 3 e 1 + e + e 3 ), and U = e 1 e e 1 + e ).

17 A CHARACTERIZATION OF CLASS GROUPS VIA SETS OF LENGTHS There is a bound M N such that, for all sufficiently large k, the sets L U 3 U 4 ) k) are AAPs with difference 1 and bound M, but they are not arithmetical progressions with difference 1.. For each k N, we have LU k 4 U ) = k + 1) + {0, 1, 3} + 3 [0, k 1]. Proof. 1. We set G 0 = suppu 3 U 4 ), A k = U3 k U4 k and L k = LA k ) for each k N. Since gcd G 0 ) gcd{ U 3 =, U 4 = 3}, it follows that min G 0 ) = 1. Thus, by Proposition 3..3, there are M, k 0 N such that for all k k 0, the set L k is an AAP with difference 1 and bound M. Let k N. Then min L k = 4k, and we assert that 1 + 4k / L k. For ν [0, 4], we set V ν = e ν and V 5 = e 1 + e + e 3 ). Since ZU3 ) = {U3, V 1 V V 3 V 5 }, ZU4 ) = {U4, V 1 V V 3 V 4 V 0 }, and ZU 3 U 4 ) = {U 3 U 4, V 1 V V 3 W } where W = e 1 + e + e 3 )e 0 e 4, it follows that minl k \ {4k}) = 4k +.. Setting W = e 1 + e )e 3 e 4 e 0 we infer that U4 U = U 4 e 1)e )W = U e 0)... e 4) and hence LU4 U ) = {3, 4, 6}. Thus for each k N we obtain that LU4 k U ) = {1} + LU4 k ) ) LU4 k ) + LU4 U ) ) = k [0, k] ) k + 3 [0, k 1] + {3, 4, 6} ) = k [0, k] ) k [0, k 1] ) k [0, k 1] ) = k + 1) + {0, 1, 3} + 3 [0, k 1]. Lemma 4.8. Let G = C r 3 with r [3, 4], e 1,..., e r ) a basis of G, e 0 = e e r, and U = e 1... e r ) e If r = 3, then there is a bound M N such that, for all k N, the sets L U 6k+1 U) ) are AAPs with difference 1 and bound M, but they are not arithmetical progressions with difference 1.. If r = 4 and V 1 = e 1e e 1 + e ), then for each k N we have LU 3k V 1 ) = 3k + 1) + {0, 1, 3} + 3 [0, k 1]. Proof. 1. Let r = 3 and k N. We set A k = U 6k+1 U) and L k = LA k ). For ν [0, 3], we set U ν = e 3 ν, V ν = e ν )e ν, and we define X = e 0e 1 e e 3. First, consider LU 6k ). We observe that ZU ) = {U, U 1 U U 3 X} and ZU 3 ) = {U 3, UU 1 U U 3 X, U 0 U 1 U U 3 }. Furthermore, min LU 6k ) = 6k, max LU 6k ) = 14k, {e 0,..., e 3 } = {}, and hence LU 6k ) = 6k + [0, 4k]. Next, consider L U)U ). For subsets I, J [1, 3] with [1, 3] = I J, we set W I = e 0 e j ). Since Z U U) ) = { } V 0 V1 V V3 e i i I j J { W I W I ) } V j I, J [1, 3] with [1, 3] = I J, j J it follows that L U)U ) = { } 7 { } + J I, J [1, 3] with [1, 3] = I J = {, 3, 4, 5, 7}. This implies that [6k +, 14k + 5] {14k + 7} = L U)U ) + LU 6k ) LA k ) [6k +, 14k + 7], and we claim that [6k +, 14k + 5] {14k + 7} = LA k ). Then the assertion of the lemma follows. To prove this, we consider the unique factorization z ZA k ) of length z = 14k + 7 which has the form z = U 0 U1 U U3 ) k V 0 V1 V V3 ).

18 18 ALFRED GEROLDINGER AND WOLFGANG A. SCHMID Assume to the contrary that there is a factorization z ZA k ) of length z = 14k + 6. If V 0 z, then V 0 V1 V V3 z and z = V 0 V1 V V3 x with x ZU 6k ), whence x LU 6k ) and z 7 + LU 6k ), a contradiction. Suppose that V 0 z. Then there are I, J [1, 3] with [1, 3] = I J such that W I W I ) j J V j z and hence z = W I W I ) j J V j) x with x ZU 6k ). Thus z [, 5] + LU 6k ), a contradiction.. Let r = 4 and k N. We have LU ) = {, 5} and LU 3k ) = 3k + 3 [0, k]. We define and observe that V = e 1 + e )e 1 e e 3e 4e 0, V 3 = e 1 + e )e 3 e 4 e 0, and W = e 1... e 4 e 0, U 3 V 1 = U V e 3 1)e 3 ) = UV 3 e 3 1) e 3 ) e 3 3)e 3 4) whence LU 3 V 1 ) = {4, 5, 7, 8}. Clearly, each factorization of U 3k V 1 contains exactly one of the atoms V 1, V, V 3, and it contains it exactly once. Therefore we obtain that LU 3k V 1 ) = {1} + LU 3k ) ) LU 3 V 1 ) + LU 3k 3 ) ) = 3k + 1) + 3 [0, k] ) {4, 5, 7, 8} + 3k 3) + 3 [0, k ] ) = 3k + 1) + 3 [0, k] ) 3k + 1) + {0, 1, 3, 4} + 3 [0, k ] ) = 3k + 1) + {0, 1, 3} + 3 [0, k 1]. Proof of Theorem c) a) Proposition 4. shows that, for all groups mentioned, all sets of lengths are arithmetical progressions. Proposition 3.3 shows that all differences lie in G). a) b) Obvious. b) c) Suppose that expg) = n, and that G is not isomorphic to any of the groups listed in c). We have to show that there is an L LG) which is not an arithmetical progression. We distinguish four cases. CASE 1: n 5. Then Proposition provides examples of sets of lengths which are not arithmetical progressions. CASE : n = 4. Since G is not cyclic, it has a subgroup isomorphic to C C 4. Then [18, Theorem 6.6.5] shows that {, 4, 5} LC C 4 } LG). CASE 3: n = 3. Then G is isomorphic to C r 3 with r 3, and Lemma provides examples of sets of lengths which are not arithmetical progressions. CASE 4: n =. Then G is isomorphic to C r with r 4, and Lemma provides examples of sets of lengths which are not arithmetical progressions.. b) a) Suppose that G is a subgroup of C 3 4 or a subgroup of C 3 3. Then Proposition 3.3. implies that G) {1, }, and hence Proposition 3..1 implies the assertion. a) b) Suppose that b) does not hold. Then G has a subgroup isomorphic to a cyclic group of order n 5, or isomorphic to C 4, or isomorphic to C 4 3. We show that in none of these cases a) holds. If G has a subgroup isomorphic to C n for some n 5, then Proposition shows that a) does not hold. If G has a subgroup isomorphic to C 4, then Lemma 4.7. shows that a) does not hold. If G has a subgroup isomorphic to C 4 3, then Lemma 4.8. shows that a) does not hold. 3. Suppose that G is cyclic. If G 4, then all sets of lengths are arithmetical progressions with difference in G) by 1. and hence they are AMPs with difference in G). If G 5, then the assertion follows from the Lemmas 4.3, 4.4, and 4.5.