A characterization of Krull monoids for which sets of lengths are (almost) arithmetical progressions

TL;DR

This paper characterizes the class groups of Krull monoids where all sets of lengths form (almost) arithmetical progressions, using additive combinatorics to identify the structural conditions involved.

Contribution

It provides a new characterization of class groups ensuring sets of lengths are (almost) arithmetical progressions in Krull monoids.

Findings

Sets of lengths depend solely on the class group G.

Characterization of G where sets of lengths are (almost) arithmetical progressions.

Application of additive combinatorics methods.

Abstract

Let H be a Krull monoid with finite class group G and suppose that every class contains a prime divisor. Then sets of lengths in H have a well-defined structure which just depends on the class group G. With methods from additive combinatorics we establish a characterization of those class groups G guaranteeing that all sets of lengths are (almost) arithmetical progressions.

Equations

$$\rho(A)=\sup\Big{\{}\frac{m}{n}\mid m,n\in A\Big{\}}=\frac{\sup A}{\min A}\in\mathbb{Q}_{\geq 1}\cup\{\infty\}$$

$$\mathcal{L}(H)$$

$$\Delta (H)$$

$${\rho }_k(H)=\sup \{\sup L\mid L\in \mathcal{L}(H),k\in L\}\in \mathbb{N}\cup \{\infty \},\text{otherwise}.$$

$$\rho (H)=\sup \{\rho (L)\mid L\in \mathcal{L}(H)\}=\underset{k\to \infty }{\lim }\frac{{\rho }_k(H)}{k}\in {\mathbb{R}}_{\ge 1}\cup \{\infty \}$$

$$S=g_1\cdot \dots \cdot g_l=\prod _{g\in G_0}{g}^{{\mathsf{v}}_g(S)}\in \mathcal{F}(G_0),$$

$$\mathrm{supp}(S)$$

$$\sigma (S)$$

$$\mathcal{B}(G_0)=\{S\in \mathcal{F}(G_0)\mid \sigma (S)=0\}\subset \mathcal{F}(G_0)$$

$$\mathsf{D}(G_{0})=\sup\bigl{\{}|U|\,\bigm{|}\;U\in\mathcal{A}(G_{0})\bigr{\}}\in\mathbb{N}_{0}\cup\{\infty\}\,.$$

$${\Delta }^{\ast }(G)=\{\min \Delta (G_0)\mid G_0\subset G\text{with}\Delta (G_0)\ne \varnothing \}\subset \Delta (G).$$

$$\mathsf{L}(B)+\mathsf{L}(C)\subset \mathsf{L}(A).$$

$$ld\le \sum _{\nu =1}^l|V_{\nu }|=|A|=\sum _{\nu =1}^k|U_{\nu }|\le k\mathsf{D}(G_0)\text{whence}\frac{|A|}{\mathsf{D}(G_0)}\le \min \mathsf{L}(A)\le \max \mathsf{L}(A)\le \frac{|A|}{d}.$$

$$\Vert S{\Vert }_g=\frac{n_1+\dots +n_l}{n}.$$

$$\frac{\Vert A{\Vert }_g}{n-1}\le \min \mathsf{L}(A)\le \max \mathsf{L}(A)\le \Vert A{\Vert }_g.$$

$$L=y+(L^{\prime }\cup L^{\ast }\cup L^{\prime \prime })\subset y+\mathcal{D}+d\mathbb{Z}$$

$$y+L\in \mathcal{L}(G^{\ast })\text{ for all }y\ge y_L.$$

$$\bigl{[}1,\,\max\{\exp(G)-2,\,k-1\}\bigr{]}\subset\Delta(G)\subset\bigl{[}1,\mathsf{D}(G)-2\bigr{]}\ \text{ where }\ k=\sum_{i=1}^{\mathsf{r}(G)}\Bigl{\lfloor}\frac{n_{i}}{2}\Bigr{\rfloor}\,.$$

$$\mathsf{Z}(A)=\{U^{r+n\nu }{V}^{k-\nu }(-V{)}^{\ell -\nu }\mid \nu \in [0,\min \{k,\ell \}]\},$$

$$\mathsf{L}(A)$$

$$=r+k+\ell +(n-2)\cdot [0,\min \{k,\ell \}].$$

$$A_{k}=\begin{cases}g^{nk}(-g)^{nk}(2g)^{n}&\quad\text{if}\ n\ \text{is even},\\ g^{nk}(-g)^{nk}\big{(}(2g)^{(n-1)/2}g\big{)}^{2}&\quad\text{if}\ n\ \text{is odd}.\end{cases}$$

$$\Vert -W_0{\Vert }_g=n-1,\Vert U_2{\Vert }_g=\Vert V_1{\Vert }_g=2,\text{and}\Vert U_1{\Vert }_g=\Vert W_j{\Vert }_g=1\text{for all}j\in [0,n/2].$$

$$(W_0{U}_2{)}^{n-1}=(W_0{U}_2{)}^{n-1-\nu }(W_1{U}_1^2{)}^{\nu }\text{for all}\nu \in [0,n-1]$$

$$A_{k}=\begin{cases}W_{0}^{k}(-W_{0})^{k}W_{n/2}^{2}&\quad\ \text{if}\ n\ \text{is even},\\ W_{0}^{k}(-W_{0})^{k}\big{(}W_{(n-1)/2})^{2}&\quad\ \text{if}\ n\ \text{is odd},\end{cases}$$

$$W_0{W}_{n/2}=W_j{W}_{n/2-j}\text{for each}j\in [0,n/2],$$

$$W_0{W}_{(n-1)/2}=W_j{W}_{(n-1)/2-j}\text{for each}j\in [0,(n-1)/2].$$

$$\Vert A_k{\Vert }_g=k(n-1)+(k+2)=r(n-1)+2s+t,$$

$$|z^{\prime }|-|z|$$

$$=r+s+k(n-1)+(k+2)-r(n-1)-2s-(2k+2)$$

$$=(k-r)(n-2)-s.$$

$$(k-r)(n-2)-s\ge 2n-4-s\ge n-4>1.$$

$$(k-r)(n-2)-s\ge n-2-n/2=n/2-2>1.$$

Full text

This work was supported by the Austrian Science Fund FWF, Project Number P28864-N35.

1. Introduction

Let $H$ be a Krull monoid with class group $G$ and suppose that every class contains a prime divisor. Then every element has a factorization into irreducibles. If $a=u_{1}\cdot\ldots\cdot u_{k}$ with irreducibles $u_{1},\ldots,u_{k}\in H$, then $k$ is called the factorization length. The set $\mathsf{L}(a)\subset\mathbb{N}_{0}$ of all possible factorization lengths is finite and called the set of lengths of $a$. The system $\mathcal{L}(H)=\{\mathsf{L}(a)\mid a\in H\}$ is a well-studied means for describing the arithmetic of $H$. It is classic that $|L|=1$ for all $L\in\mathcal{L}(H)$ if and only if $|G|\leq 2$ and if $|G|\geq 3$, then there are arbitrarily large $L\in\mathcal{L}(H)$.

Sets of lengths in $H$ can be studied in the associated monoid of zero-sum sequences $\mathcal{B}(G)$. The latter is a Krull monoid again and it is well-known that \mathcal{L}(H)=\mathcal{L}\big{(}\mathcal{B}(G)\big{)} (as usual we write $\mathcal{L}(G)$ for \mathcal{L}\big{(}\mathcal{B}(G)\big{)}. Our results on $\mathcal{L}(G)$ also apply to the larger class of transfer Krull monoids over $G$. This will be outlined in Section 2.

If the group $G$ is infinite, then, by a Theorem of Kainrath, every finite set $L\subset\mathbb{N}_{\geq 2}$ lies in $\mathcal{L}(H)$ ([25], [16, Theorem 7.4.1]; for further rings and monoids with this property see [10] or [24, Corollary 4.7]). Now suppose that the group $G$ is finite. In this case sets of lengths have a well-defined structure. Indeed, by the Structure Theorem for Sets of Lengths (Proposition 2.2.1), the sets in $\mathcal{L}(G)$ are AAMPs (almost arithmetical multiprogressions) with difference in $\Delta^{*}(G)$ and some universal bound. By a realization result of the second author, this description is best possible (Proposition 2.2.2). By definition, the concept of an AAMP comprises APs (arithmetical progressions), AAPs (almost arithmetical progressions), and AMPs (arithmetical multiprogressions); definitions are gathered in Definition 2.1. The goal of this paper is to characterize those groups where all sets of lengths are not only AAMPs, but have one of these more special forms. We formulate the main result of this paper.

Theorem 1.1**.**

Let $G$ be a finite abelian group.

The following statements are equivalent :**

1. (a)

All sets of lengths in $\mathcal{L}(G)$ are APs with difference in $\Delta^{*}(G)$. 2. (b)

All sets of lengths in $\mathcal{L}(G)$ are APs. 3. (c)

The system of sets of lengths $\mathcal{L}(G)$ is additively closed, that is, $L_{1}+L_{2}\in\mathcal{L}(G)$ for all $L_{1},L_{2}\in\mathcal{L}(G)$. 4. (d)

$G$* is cyclic of order $|G|\leq 4$ or isomorphic to a subgroup of $C_{2}^{3}$ or isomorphic to a subgroup of $C_{3}^{2}$.* 2. 2.

The following statements are equivalent :**

1. (a)

There is a constant $M\in\mathbb{N}$ such that all sets of lengths in $\mathcal{L}(G)$ are AAPs with bound $M$. 2. (b)

$G$* is isomorphic to a subgroup of $C_{3}^{3}$ or isomorphic to a subgroup of $C_{4}^{3}$.* 3. 3.

For all finite sets $\Delta$ with $\Delta\supset\Delta^{*}(G)$ the following statements are equivalent :**

1. (a)

All sets of lengths in $\mathcal{L}(G)$ are AMPs with difference in $\Delta$. 2. (b)

$G$* is cyclic of order $|G|\leq 6$ or isomorphic to a subgroup of $C_{2}^{3}$ or isomorphic to a subgroup of $C_{3}^{2}$.*

A central topic in the study of sets of lengths is the Characterization Problem (for recent progress see [4, 17, 23, 31, 30]) which reads as follows:

1. Let $G$ be a finite abelian group with $\mathsf{D}(G)\geq 4$, and let $G^{\prime}$ be an abelian group with $\mathcal{L}(G)=\mathcal{L}(G^{\prime})$. Does it follow that $G\cong G^{\prime}$?

A finite abelian group $G$ has Davenport constant $\mathsf{D}(G)\leq 3$ if and only if either $|G|\leq 3$ or $G\cong C_{2}\oplus C_{2}$. Since $\mathcal{L}(C_{1})=\mathcal{L}(C_{2})$ and $\mathcal{L}(C_{3})=\mathcal{L}(C_{2}\oplus C_{2})$ (Proposition 3.1), small groups require special attention in the study of the Characterization Problem. As a consequence of Theorem 1.1 we obtain an affirmative answer to the Characterization Problem for all involved small groups.

Corollary 1.2**.**

Let $G$ be a finite abelian group with Davenport constant $\mathsf{D}(G)\geq 4$ and suppose that $\mathcal{L}(G)$ satisfies one of the properties characterized in Theorem 1.1. If $G^{\prime}$ is any abelian group such that $\mathcal{L}(G)=\mathcal{L}(G^{\prime})$, then $G\cong G^{\prime}$.

In Section 2 we gather the required tools for studying sets of lengths (Propositions 2.2, 2.3, and 2.4). The proof of Theorem 1.1 requires methods from additive combinatorics and is given in Section 3. Several properties occurring in Theorem 1.1 can be characterized by further arithmetical invariants. We briefly outline this in Remark 3.13 where we also discuss the property of being additively closed occurring in Theorem 1.1.1.(c).

2. Background on sets of lengths

Let $\mathbb{N}$ denote the set of positive integers, $\mathbb{P}\subset\mathbb{N}$ the set of prime numbers and put $\mathbb{N}_{0}=\mathbb{N}\cup\{0\}$. For real numbers $a,\,b\in\mathbb{R}$, we set $[a,b]=\{x\in\mathbb{Z}\mid a\leq x\leq b\}$. Let $A,B\subset\mathbb{Z}$ be subsets of the integers. We denote by $A+B=\{a+b\mid a\in A,b\in B\}$ their sumset, and by $\Delta(A)$ the set of $($successive$)$ distances of $A$ (that is, $d\in\Delta(A)$ if and only if $d=b-a$ with $a,b\in A$ distinct and $[a,b]\cap A=\{a,b\}$). For $k\in\mathbb{N}$, we denote by $k\cdot A=\{ka\mid a\in A\}$ the dilation of $A$ by $k$. If $A\subset\mathbb{N}$, then

[TABLE]

is the elasticity of $A$, and we set $\rho(\{0\})=1$.

Monoids. By a monoid, we mean a commutative cancellative semigroup with identity. Let $H$ be a monoid. Then $H^{\times}$ denotes the group of invertible elements of $H$, $\mathcal{A}(H)$ the set of atoms of $H$, and $H_{\rm red}=H/H^{\times}$ the associated reduced monoid. The factorization monoid $\mathsf{Z}(H)=\mathcal{F}(\mathcal{A}(H_{\rm red}))$ is the free abelian monoid with basis $\mathcal{A}(H_{\rm red})$ and $\pi\colon\mathsf{Z}(H)\to H_{\rm red}$ denotes the canonical epimorphism. For an element $a\in H$, $\mathsf{Z}(a)=\pi^{-1}(aH^{\times})$ is the set of factorizations of $a$ and $\mathsf{L}(a)=\{|z|=\sum_{u\in\mathcal{A}(H_{\rm red})}\mathsf{v}_{u}(z)\mid z=\prod_{u\in\mathcal{A}(H_{\rm red})}u^{\mathsf{v}_{u}(z)}\in\mathsf{Z}(a)\}\subset\mathbb{N}_{0}$ is the set of lengths of $a$ (note that $\mathsf{L}(a)=\{0\}$ if and only if $a\in H^{\times}$). We denote by

[TABLE]

For $k\in\mathbb{N}$, we set $\rho_{k}(H)=k$ if $H$ is a group, and

[TABLE]

Then

[TABLE]

is the elasticity of $H$.

Zero-Sum Theory. Let $G$ be an additive abelian group, $G_{0}\subset G$ a subset, and let $\mathcal{F}(G_{0})$ be the free abelian monoid with basis $G_{0}$. In Combinatorial Number Theory, the elements of $\mathcal{F}(G_{0})$ are called  sequences over  $G_{0}$. For a sequence

[TABLE]

we set $-S=(-g_{1})\cdot\ldots\cdot(-g_{l})$, and we call |S|=l=\sum_{g\in G}\mathsf{v}_{g}(S)\in\mathbb{N}_{0}\ \text{the \ {\it length} \ of \ S,}

[TABLE]

the set of subsequence sums  of  $S$. The sequence $S$ is said to be

• •

zero-sum free  if  $0\notin\Sigma(S)$,

• •

a zero-sum sequence  if  $\sigma(S)=0$,

• •

a minimal zero-sum sequence  if it is a nontrivial zero-sum sequence and every proper subsequence is zero-sum free.

The monoid

[TABLE]

is called the  monoid of zero-sum sequences  over  $G_{0}$. As usual we set, for all $k\in\mathbb{N}$, \rho_{k}(G_{0})=\rho_{k}\big{(}\mathcal{B}(G_{0})\big{)},\ \rho(G_{0})=\rho\big{(}\mathcal{B}(G_{0})\big{)},\ \mathcal{L}(G_{0})=\mathcal{L}(\mathcal{B}(G_{0})), \mathsf{Z}(G_{0})=\mathsf{Z}\big{(}\mathcal{B}(G_{0})\big{)}, and \Delta(G_{0})=\Delta\big{(}\mathcal{B}(G_{0})\big{)}. The atoms (irreducible elements) of the monoid $\mathcal{B}(G_{0})$ are precisely the minimal zero-sum sequences over $G_{0}$, and they will be denoted by $\mathcal{A}(G_{0})$. If $G_{0}$ is finite, then $\mathcal{A}(G_{0})$ is finite. The Davenport constant $\mathsf{D}(G_{0})$ of $G_{0}$ is the maximal length of an atom whence

[TABLE]

The set of minimal distances $\Delta^{*}(G)\subset\Delta(G)$ is defined as

[TABLE]

A tuple $(e_{i})_{i\in I}$ is called a basis of $G$ if all elements are nonzero and $G=\oplus_{i\in I}\langle e_{i}\rangle$. For $p\in\mathbb{P}$, let $\mathsf{r}_{p}(G)$ denote the $p$-rank of $G$, $\mathsf{r}(G)=\sup\{\mathsf{r}_{p}(G)\mid p\in\mathbb{P}\}$ denote the rank of $G$.

Transfer Krull monoids. A monoid is a Krull monoid if it is completely integrally closed and satisfies the ascending chain condition on divisorial ideals. An integral domain $R$ is a Krull domain if and only if its multiplicative monoid $R\setminus\{0\}$ is a Krull monoid whence every integrally closed noetherian domain is Krull. 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 ([16, Section 2.11]). Monoid domains and monoids of modules that are Krull are discussed in [5, 3, 1, 8]. Let $H$ be a Krull monoid with class group $G$ such that every class contains a prime divisor. Then there is a transfer homomorphism $\theta\colon H\to\mathcal{B}(G)$ which implies that $\mathcal{L}(H)=\mathcal{L}(G)$ ([16, Theorem 3.4.10]). A transfer Krull monoid $H$ over $G$ is a (not necessarily commutative) monoid allowing a weak transfer homomorphism to $\mathcal{B}(G)$ whence $\mathcal{L}(H)=\mathcal{L}(G)$. Thus Theorem 1.1 applies to transfer Krull monoids over finite abelian groups. Recent deep work, mainly by Baeth and Smertnig, revealed that wide classes of non-commutative Dedekind domains are transfer Krull ([29, 2, 28]). We refer to the survey [13] for a detailed discussion of these and further examples.

Sets of Lengths. Let $A\in\mathcal{B}(G_{0})$ and $d=\min\{|U|\mid U\in\mathcal{A}(G_{0})\}$. If $A=BC$ with $B,C\in\mathcal{B}(G_{0})$, then

[TABLE]

If $A=U_{1}\cdot\ldots\cdot U_{k}=V_{1}\cdot\ldots\cdot V_{l}$ with $U_{1},\ldots,U_{k},V_{1},\ldots,V_{l}\in\mathcal{A}(G_{0})$ and $k<l$, then

[TABLE]

For sequences over cyclic groups the $g$-norm plays a similar role as the length does for sequences over arbitrary groups. Let $g\in G$ with  $\operatorname{ord}(g)=n\geq 2$. For a sequence $S=(n_{1}g)\cdot\ldots\cdot(n_{l}g)\in\mathcal{F}(\langle g\rangle)$,  where $l\in\mathbb{N}_{0}$ and $n_{1},\ldots,n_{l}\in[1,n]$,  we define

[TABLE]

Note that $\sigma(S)=0$ implies that $n_{1}+\ldots+n_{l}\equiv 0\mod n$ whence $\|S\|_{g}\in\mathbb{N}_{0}$. Thus, $\|\cdot\|_{g}\colon\mathcal{B}(\langle g\rangle)\to\mathbb{N}_{0}$ is a homomorphism, and $\|S\|_{g}=0$ if and only if $S=1$. If $S\in\mathcal{A}(G_{0})$, then $\|S\|_{g}\in[1,n-1]$, and if $\|S\|_{g}=1$, then $S\in\mathcal{A}(G_{0})$. Arguing as above we obtain that

[TABLE]

Now we recall the concept of almost arithmetical multiprogressions (AAMPs) as given in [16, Chapter 4]. Then we gather results on sets of lengths and on invariants controlling their structure such as the set of distances and the elasticities (Propositions 2.2, 2.3, and 2.4). These results form the basis for the proof of Theorem 1.1 given in the next section.

Definition 2.1**.**

Let $d\in\mathbb{N}$, $l,\,M\in\mathbb{N}_{0}$, and $\{0,d\}\subset\mathcal{D}\subset[0,d]$. A subset $L\subset\mathbb{Z}$ is called an

• •

arithmetical multiprogression  (AMP  for short)  with  difference  $d$,  period  $\mathcal{D}$  and  length  $l$,  if $L$ is an interval of  $\min L+\mathcal{D}+d\mathbb{Z}$  (this means that $L$ is finite nonempty and $L=(\min L+\mathcal{D}+d\mathbb{Z})\cap[\min L,\max L]$),  and  $l$  is maximal such that  $\min L+ld\in L$.

• •

almost arithmetical multiprogression  (AAMP  for short)  with  difference  $d$,  period  $\mathcal{D}$,  length  $l$ and  bound  $M$,  if

[TABLE]

where  $L^{*}$  is an  AMP  with difference $d$ (whence $L^{*}\neq\emptyset$), period $\mathcal{D}$ and length $l$ such that  $\min L^{*}=0$,  $L^{\prime}\subset[-M,-1]$,  $L^{\prime\prime}\subset\max L^{*}+[1,M]$  and  $y\in\mathbb{Z}$.

• •

almost arithmetical progression  (AAP  for short)  with  difference $d$,  bound $M$  and  length  $l$,  if it is an  AAMP  with difference $d$, period  $\{0,d\}$, bound  $M$  and length  $l$.

Proposition 2.2** (Structural results on $\mathcal{L}(G)$).**

Let $G$ be a finite abelian group with $|G|\geq 3$.

There exists some $M\in\mathbb{N}_{0}$ such that every set of lengths $L\in\mathcal{L}(G)$ is an AAMP with some difference $d\in\Delta^{*}(G)$ and bound $M$. 2. 2.

For every $M\in\mathbb{N}_{0}$ and every finite nonempty set $\Delta^{*}\subset\mathbb{N}$, there is a finite abelian group $G^{*}$ such that the following holds :* for every AAMP $L$ with difference $d\in\Delta^{*}$ and bound $M$ there is some $y_{L}\in\mathbb{N}$ such that*

[TABLE] 3. 3.

Let $G_{0}\subset G$ be a subset. Then there exist a bound $M\in\mathbb{N}_{0}$ and some $A^{*}\in\mathcal{B}(G_{0})$ such that for all $A\in A^{*}\mathcal{B}(G_{0})$ the set of lengths $\mathsf{L}(A)$ is an AAP with difference $\min\Delta(G_{0})$ and bound $M$. 4. 4.

If $A\in\mathcal{B}(G)$ such that $\operatorname{supp}(A)\cup\{0\}$ is a subgroup of $G$, then $\mathsf{L}(A)$ is an AP with difference $1$.

Proof.

The first statement gives the Structure Theorem for Sets of Lengths ([16, Theorem 4.4.11]), which is sharp by the second statement proved in [27]. 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 ([16, Theorems 4.3.6 and 7.6.8]). ∎

Proposition 2.3** (Structural results on $\Delta(G)$ and on $\Delta^{*}(G)$).**

Let $G=C_{n_{1}}\oplus\ldots\oplus C_{n_{r}}$ where $r,n_{1},\ldots,n_{r}\in\mathbb{N}$ with $r=\mathsf{r}(G)$, $1<n_{1}\,|\,\ldots\,|\,n_{r}$, and $|G|\geq 3$.

$\Delta(G)$* is an interval with*

[TABLE] 2. 2.

$1\in\Delta^{*}(G)\subset\Delta(G)$, $[1,\mathsf{r}(G)-1]\subset\Delta^{*}(G)$, and $\max\Delta^{*}(G)=\max\{\exp(G)-2,\mathsf{r}(G)-1\}$. 3. 3.

If $G$ is cyclic of order $|G|=n\geq 4$, then \max\big{(}\Delta^{*}(G)\setminus\{n-2\}\big{)}=\lfloor\frac{n}{2}\rfloor-1.

Proof.

The statement on $\max\Delta^{*}(G)$ follows from [22]. For all remaining statements see [16, Section 6.8]. A more detailed analysis of $\Delta^{*}(G)$ in case of cyclic groups can be found in [26]. ∎

Proposition 2.4** (Results on $\rho_{k}(G)$ and on $\rho(G)$).**

Let $G$ be a finite abelian group with $|G|\geq 3$, and let $k\in\mathbb{N}$.

$\rho(G)=\mathsf{D}(G)/2$* and $\rho_{2k}(G)=k\mathsf{D}(G)$.* 2. 2.

$1+k\mathsf{D}(G)\leq\rho_{2k+1}(G)\leq k\mathsf{D}(G)+\mathsf{D}(G)/2$. If $G$ is cyclic, then equality holds on the left side.

Proof.

See [16, Chapter 6.3], [12, Theorem 5.3.1], and [15]. ∎

We need the following technical lemma which follows from [9, Lemma 6.1].

Lemma 2.5**.**

Let $G$ be a finite abelian group, $A\in\mathcal{B}(G)$, $x=U_{1}\cdot\ldots\cdot U_{r}\in\mathsf{Z}(G)$, with $U_{1},\ldots,U_{r}\in\mathcal{A}(G)$ and $|U_{1}|=\ldots=|U_{r}|=2$, such that $U_{1}\cdot\ldots\cdot U_{r}\mid A$ in $\mathcal{B}(G)$. Then there exists a factorization $z\in\mathsf{Z}(A)$ with $|z|=\max\mathsf{L}(A)$ and $x\mid z$ in $\mathsf{Z}(G)$.

3. A characterization of extremal cases

The goal of this section is to prove Theorem 1.1. First we recall some cases where the systems of sets of lengths are completely determined.

Proposition 3.1**.**

\mathcal{L}(C_{1})=\mathcal{L}(C_{2})=\big{\{}\{m\}\mid m\in\mathbb{N}_{0}\big{\}}. 2. 2.

\mathcal{L}(C_{3})=\mathcal{L}(C_{2}\oplus C_{2})=\bigl{\{}y+2k+[0,k]\,\bigm{|}\,y,\,k\in\mathbb{N}_{0}\bigr{\}}. 3. 3.

\mathcal{L}(C_{4})=\bigl{\{}y+k+1+[0,k]\,\bigm{|}\,y,\,k\in\mathbb{N}_{0}\bigr{\}}\,\cup\,\bigl{\{}y+2k+2\cdot[0,k]\,\bigm{|}\,y,\,k\in\mathbb{N}_{0}\bigr{\}}. 4. 4.

\mathcal{L}(C_{2}^{3})=\bigl{\{}y+(k+1)+[0,k]\,\bigm{|}\,y\in\mathbb{N}_{0},\ k\in[0,2]\bigr{\}}* *

*   **  **        **  \cup\ \bigl{\{}y+k+[0,k]\,\bigm{|}\,y\in\mathbb{N}_{0},\ k\geq 3\bigr{\}}\cup\bigl{\{}y+2k+2\cdot[0,k]\,\bigm{|}\,y,\,k\in\mathbb{N}_{0}\bigr{\}}.* 5. 5.

$\mathcal{L}(C_{3}^{2})=\{[2k,l]\mid k\in\mathbb{N}_{0},l\in[2k,5k]\}$* *

*   **  **        **  $\cup\ \{[2k+1,l]\mid k\in\mathbb{N},l\in[2k+1,5k+2]\}\cup\{\{1\}\}$.*

Proof.

1. This is well-known. A proof of 2.,3., and 4. can be found in [16, Theorem 7.3.2]. For 5. we refer to [18, Proposition 3.12]. ∎

Let $G$ be a finite abelian group, $g\in G$ with $\operatorname{ord}(g)=n\geq 2$, $r\in[0,n-1]$, $k,\ell\in\mathbb{N}_{0}$, and $A=g^{kn+r}(-g)^{\ell n+r}$. We often use that

[TABLE]

where $U=g(-g)$, $V=g^{n}$, and that

[TABLE]

Proposition 3.2**.**

Let $G$ be a cyclic group of order $|G|=n\geq 7$, $g\in G$ with $\operatorname{ord}(g)=n$, $k\in\mathbb{N}$, and

[TABLE]

Then there is a bound $M\in\mathbb{N}$ such that, for all $k\geq n-1$, the sets $\mathsf{L}(A_{k})$ are AAPs with difference $1$ and bound $M$, but they are not APs with difference $1$.

Proof.

We set $G_{0}=\{g,-g,2g\}$, $U_{1}=(-g)g$, $U_{2}=(-g)^{2}(2g)$ and, if $n$ is odd, then $V_{1}=(2g)^{(n+1)/2}(-g)$. Furthermore, for $j\in[0,n/2]$, we define $W_{j}=(2g)^{j}g^{n-2j}$. Then, together with $-W_{0}=(-g)^{n}$, these are all minimal zero-sum sequences which divide $A_{k}$ for $k\in\mathbb{N}$. Note that

[TABLE]

It is sufficient to prove the following two assertions.

1. A1. 

There is an $M\in\mathbb{N}_{0}$ such that $\mathsf{L}(A_{k})$ is an AAP with difference $1$ and bound $M$ for all $k\geq n-1$. 2. A2. 

For each $k\in\mathbb{N}$, $\mathsf{L}(A_{k})$ is not an AP with difference $1$.

*Proof of  *A1.  By Proposition 2.2.1 there is a bound $M^{\prime}\in\mathbb{N}_{0}$ such that, for each $k\in\mathbb{N}$, $\mathsf{L}(A_{k})$ is an AAMP with difference $d_{k}\in\Delta^{*}(G)\subset[1,n-2]$ and bound $M^{\prime}$. Suppose that $k\geq n-1$. Then $(W_{0}U_{2})^{n-1}$ divides $A_{k}$. Since $W_{0}U_{2}=W_{1}U_{1}^{2}$, it follows that

[TABLE]

and hence \mathsf{L}\big{(}(W_{0}U_{2})^{n-1}\big{)}\supset[2n-2,3n-3]. Thus $\mathsf{L}(A_{k})$ contains an AP with difference $1$ and length $n-1$. Therefore there is a bound $M\in\mathbb{N}_{0}$ such that $\mathsf{L}(A_{k})$ is an AAP with difference $1$ and bound $M$ for all $k\geq n-1$.

*Proof of  *A2.  Let $k\in\mathbb{N}$. Observe that

[TABLE]

and it can be seen that $\min\mathsf{L}(A_{k})=2k+2$. We assert that $2k+3\notin\mathsf{L}(A_{k})$. If $n$ is even, then

[TABLE]

and similarly, for odd $n$ we have

[TABLE]

In both cases, all factorizations of $A_{k}$ of length $2k+2$ contain only atoms with $g$-norm $1$ and with $g$-norm $n-1$. Let $z^{\prime}$ be any factorization of $A_{k}$ containing only atoms with $g$-norm $1$ and with $g$-norm $n-1$. Then $|z^{\prime}|-|z|$ is a multiple of $n-2$ whence if $|z^{\prime}|>|z|$, then $|z^{\prime}|-|z|\geq n-2>1$.

Next we consider a factorization $z^{\prime}$ of $A_{k}$ containing at least one atom with $g$-norm $2$, say $z^{\prime}$ has $r$ atoms with $g$-norm $n-1$, $s\geq 1$ atoms with $g$-norm $2$, and $t$ atoms with $g$-norm $1$. Then $k>r$,

[TABLE]

and we study

[TABLE]

Note that $s\leq\mathsf{v}_{2g}(A_{k})\leq n$. Thus, if $k-r\geq 2$, then

[TABLE]

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 $2$ are equal to $U_{2}$. Since $\mathsf{v}_{-g}(U_{2})=2$, it follows that $s\leq\mathsf{v}_{-g}(-W_{0})/2=n/2$ whence

[TABLE]

Suppose that $V_{1}$ occurs among the $s$ atoms with $g$-norm $2$. Then $n$ is odd, $V_{1}$ occurs precisely once, and

[TABLE]

whence

[TABLE]

In order to handle the cyclic group with six elements (done in Proposition 3.8), we need five lemmas.

Lemma 3.3**.**

Let $G$ be a cyclic group of order $|G|=6$, $g\in G$ with $\operatorname{ord}(g)=6$, and $A\in\mathcal{B}(G)$ with $\operatorname{supp}(A)=\{g,2g,3g,-g\}$. Then $\mathsf{L}(A)$ is an AP with difference $1$.

Proof.

We set $G_{0}=\{g,2g,3g,-g\}$ and list all atoms of $\mathcal{B}(G_{0})$:

• $\|\cdot\|_{g}=1:$ $g^{6}$, $g^{4}(2g)$, $g^{3}(3g)$, $g^{2}(2g)^{2}$, $g(2g)(3g)$, $g(-g)$, $(2g)^{3}$, $(3g)^{2}$
• $\|\cdot\|_{g}=2:$ $U_{1}=(2g)^{2}(3g)(-g)$, $U_{2}=(2g)(-g)^{2}$,
• $\|\cdot\|_{g}=3:$ $U_{3}=(3g)(-g)^{3}$,
• $\|\cdot\|_{g}=5:$ $U_{4}=(-g)^{6}$.

We choose a factorization $z\in\mathsf{Z}(A)$ and we show that either $|z|=\max\mathsf{L}(A)$ or that $|z|+1\in\mathsf{L}(A)$. In each of the following cases it easily follows that $|z|+1\in\mathsf{L}(A)$.

• •

$U_{1}^{2}\mid z$ in $\mathsf{Z}(G)$, because $3\in\mathsf{L}(U_{1}^{2})$.

• •

$U_{1}U_{3}\mid z$ in $\mathsf{Z}(G)$, because $3\in\mathsf{L}(U_{1}U_{3})$.

• •

$U_{1}U_{4}\mid z$ in $\mathsf{Z}(G)$, because $3\in\mathsf{L}(U_{1}U_{4})$.

From now on we suppose that none of the above cases occurs. If $z$ is divisible only by atoms with $g$-norm one and by at most one atom having $g$-norm two, then we are done. Thus it is sufficient to handle the following four cases.

CASE 1: $U_{1}\mid z$.

If $z$ is divisible by an atom $W$ with $\mathsf{v}_{g}(W)\geq 1$ and $W\neq(-g)g$, then $3\in\mathsf{L}(U_{1}W)$ and we are done. Suppose this does not hold. Thus

[TABLE]

with $k_{2}\in\mathbb{N}$ and $k_{3},k_{4},k_{5}\in\mathbb{N}_{0}$. Then $|z|=1+k_{2}+\ldots+k_{5}$ and we assert that $|z|=\max\mathsf{L}(A)$. By Lemma 2.5 it is sufficient to show that

[TABLE]

Each atom $X\in\mathcal{B}(G_{0})$ with X\mid U_{1}U_{2}^{k_{2}}\Big{(}(2g)^{3}\Big{)}^{k_{3}} and $(2g)\mid X$ has length $|X|\geq 3$ whence the claim follows.

CASE 2: $U_{3}\mid z$ and $U_{1}\nmid z$.

If $z$ is divisible by an atom $W$ with $\mathsf{v}_{2g}(W)\geq 1$. If $W\in\{g^{4}(2g),g^{2}(2g)^{2},(2g)^{3},$ $g(2g)(3g)\}$, then $3\in\mathsf{L}(U_{3}W)$ and we are done. If this is not the case, then all atoms $W$ are equal to $U_{2}$. Thus $U_{2}U_{3}$ divides $z$. Thus $z$ is divisible by an atom $W$ with $\mathsf{v}_{g}(W)\geq 1$. If $W\in\{g^{6},g^{3}(3g)\}$, then $3\in\mathsf{L}(U_{2}W)$ and we are done. Thus we may assume that any atom $W$ with $\mathsf{v}_{g}(W)\geq 1$ is equal to $V=(-g)g$. This implies that $\mathsf{v}_{g}(A)=\mathsf{v}_{V}(z)$. We set

[TABLE]

and assert that $|z|=\max\mathsf{L}(A)$. By Lemma 2.5 it is sufficient to show that

[TABLE]

and this holds true.

CASE 3: $U_{2}\mid z$ and $U_{1}\nmid z$ and $U_{3}\nmid z$.

If $z$ is divisible by an atom $W$ with $\mathsf{v}_{2g}(W)\geq 2$, then $3\in\mathsf{L}(U_{2}W)$ and we are done. If $z$ is divisible by $W=g(2g)(3g)$, then U_{2}W=U_{1}\Big{(}g(-g)\Big{)}. Thus there is a factorization $z^{\prime}\in\mathsf{Z}(A)$ with $|z|=|z^{\prime}|$ and $U_{1}\mid z^{\prime}$ and the claim follows from CASE 1.

Suppose that this is not the case. Then the each atom dividing $z$ and containing $g$ is $V=(-g)g$ whence $\mathsf{v}_{g}(A)=\mathsf{v}_{V}(z)$. If $(2g)^{3}$ and $U_{4}$ divide $z$, then $|z|+1\in\mathsf{L}(A)$ because $3\in\mathsf{L}((3g)^{2}U_{4})$; otherwise $|z|=\max\mathsf{L}(A)$.

CASE 3: $U_{4}\mid z$ and $U_{1}\nmid z$ and $U_{2}\nmid z$ and $U_{3}\nmid z$.

If $W=(2g)^{3}$ divides $z$, then $3\in\mathsf{L}(U_{4}W)$ and we are done. Suppose this is not the case. If $W=g(2g)(3g)$ divides $z$, then $3\in\mathsf{L}(U_{4}W)$ and we are done. Suppose this is not the case. Then there are atoms $W_{1},W_{2}$ such that $2g\mid W_{1}$ and $3g\mid W_{2}$ such that $W_{1}W_{2}$ divides $z$. Then $W_{1}\in\{g^{4}(2g),g^{2}(2g)^{2}\}$ and $W_{2}\in\{(3g)^{2},g^{3}(3g)\}$. In each case we obtain that $W_{1}W_{2}=WW_{4}$ for some atom $W_{4}\in\mathcal{A}(G)$. Thus we obtain a factorization $z^{\prime}\in\mathsf{Z}(A)$ with $|z|=|z^{\prime}|$ with $WU_{4}\mid z^{\prime}$ and we are done. ∎

Lemma 3.4**.**

Let $G$ be a cyclic group of order $|G|=6$, $g\in G$ with $\operatorname{ord}(g)=6$, and $A\in\mathcal{B}(G)$ with $\operatorname{supp}(A)=\{g,3g,4g\}$. Then $\mathsf{L}(A)$ is an AP with difference $1$.

Proof.

We set $G_{0}=\{g,3g,4g\}$ and list all atoms of $\mathcal{B}(G_{0})$:

• $\|\cdot\|_{g}=1:$ $g^{6}$, $V_{3}=g^{3}(3g)$, $V_{5}=g^{2}(4g)$, $V_{4}=(3g)^{2}$
• $\|\cdot\|_{g}=2:$ $U_{1}=(4g)^{2}(3g)g$, $U_{2}=(4g)^{3}$,

We choose a factorization $z\in\mathsf{Z}(A)$ and we show that either $|z|=\max\mathsf{L}(A)$ or that $|z|+1\in\mathsf{L}(A)$. If $z$ is divisible by at most one atom having $g$-norm two, then the claim holds. Thus from now on we suppose that $z$ is divisible by at least two atoms having $g$-norm two. Since U_{1}^{2}=(3g)^{2}\Big{(}(4g)g^{2}\Big{)}\Big{(}(4g)^{3}\Big{)}, we may suppose that $U_{1}^{2}\nmid z$. Thus $U_{1}U_{2}$ divides $z$ or $U_{2}^{2}$ divides $z$.

Since V_{3}^{2}=(g^{6})\Big{(}(3g)^{2}\Big{)}, we may suppose that $\mathsf{v}_{V_{3}}(z)\leq 1$. If $z$ is divisible by $g^{6}$, then $3\in\mathsf{L}(U_{2}g^{6})$ shows that $|z|+1\in\mathsf{L}(A)$. Suppose that $g^{6}\nmid z$. It follows that

[TABLE]

where $\ell_{1}\leq 1$, $\ell_{3}\leq 1$. Since $U_{1}V_{3}=V_{4}V_{5}^{2}$ we may suppose that $\ell_{1}+\ell_{3}\leq 1$. We assert that $|z|=\max\mathsf{L}(A)$, and by Lemma 2.5 it suffices to show that

[TABLE]

Since $A^{\prime}$ is not divisible by an atom of length $2$, it follows that $\max\mathsf{L}(A^{\prime})\leq|A^{\prime}|/3$. Since $|U_{2}|=|V_{5}|=3$ and $\ell_{1}+\ell_{3}\leq 1$, we infer that $\ell_{1}+\ell_{2}+\ell_{3}+\ell_{5}=\lfloor|A^{\prime}|/3\rfloor$. ∎

Lemma 3.5**.**

Let $G$ be a cyclic group of order $|G|=6$, $g\in G$ with $\operatorname{ord}(g)=6$, and $A\in\mathcal{B}(G)\!$ with $\operatorname{supp}(A)=\!\{g,2g,3g,4g\}.\!$ Then $\!\mathsf{L}(A)\!$ is an AP with difference $1$.

Proof.

We set $G_{0}=\{g,2g,3g,4g\}$ and list all atoms of $\mathcal{B}(G_{0})$:

• $\|\cdot\|_{g}=1:$ $g^{6}$, $g^{4}(2g)$, $V_{3}=g^{3}(3g)$, $g^{2}(2g)^{2}$, $V_{5}=g^{2}(4g)$, $V_{1}=g(2g)(3g)$,

$\|\cdot\|_{g}=1:$ $V_{2}=(2g)(4g)$, $(2g)^{3}$, $V_{4}=(3g)^{2}$

• $\|\cdot\|_{g}=2:$ $U_{1}=(4g)^{2}(3g)g$, $U_{2}=(4g)^{3}$,

We choose a factorization $z\in\mathsf{Z}(A)$ and we show that either $|z|=\max\mathsf{L}(A)$ or that $|z|+1\in\mathsf{L}(A)$. If $z$ is divisible by at most one atom having $g$-norm two, then the claim holds. Thus from now on we suppose that $z$ is divisible by at least two atoms having $g$-norm two. Since U_{1}^{2}=(3g)^{2}\Big{(}(4g)g^{2}\Big{)}\Big{(}(4g)^{3}\Big{)}, we may suppose that $U_{1}^{2}\nmid z$. Thus $U_{1}U_{2}$ divides $z$ or $U_{2}^{2}$ divides $z$.

Since V_{3}^{2}=(g^{6})\Big{(}(3g)^{2}\Big{)}, we may suppose that $\mathsf{v}_{V_{3}}(z)\leq 1$. Since V_{1}^{2}=\Big{(}g^{2}(2g)^{2}\Big{)}(3g)^{2}, we may suppose that $\mathsf{v}_{V_{1}}(z)\leq 1$. Since V_{1}V_{3}=\Big{(}g^{4}(2g)\Big{)}(3g)^{2}, we may suppose that $\mathsf{v}_{V_{3}}(z)+\mathsf{v}_{V_{1}}(z)\leq 1$. Since $V_{1}V_{5}=V_{2}V_{3}$, we may suppose that $\mathsf{v}_{V_{5}}(z)+\mathsf{v}_{V_{1}}(z)\leq 1$. If $z$ is divisible by some $W\in\{g^{6},g^{4}(2g),g^{2}(2g)^{2},(2g)^{3}\}$, then $3\in\mathsf{L}(U_{2}W)$ shows that $|z|+1\in\mathsf{L}(A)$. Suppose that none of these four atoms divides $z$.

Since $\mathsf{v}_{2g}(A)\geq 1$, it follows that $V_{1}=g(2g)(3g)$ or $V_{2}=(2g)(4g)$ divides $z$. If $U_{1}V_{1}\mid z$, then $3\in\mathsf{L}(U_{1}V_{1})$ shows that $|z|+1\in\mathsf{L}(A)$. Since $U_{1}V_{2}=U_{2}V_{1}$, we may suppose that $U_{1}\nmid z$ whence $z$ has the form

[TABLE]

We assert that $|z|=\max\mathsf{L}(A)$. By Lemma 2.5 it is sufficient to show that

[TABLE]

If $k_{1}=1$, then $k_{3}=k_{5}=0$ and \mathsf{L}\Big{(}V_{1}^{1}U_{2}^{\ell}\Big{)}=\{\ell+1\}. Suppose that $k_{1}=0$, then $V_{3}^{k_{3}}V_{5}^{k_{5}}U^{\ell}$ is not divisible by an atom of length two whence (recall that $k_{3}\leq 1$)

[TABLE]

Lemma 3.6**.**

Let $G$ be a cyclic group of order $|G|=6$, $g\in G$ with $\operatorname{ord}(g)=6$, and $A\in\mathcal{B}(G)$ with $\operatorname{supp}(A)=\{g,2g,-g\}$. If $\mathsf{v}_{2g}(A)\geq 3$, then $\mathsf{L}(A)$ is an AP with difference $1$. If $\mathsf{v}_{2g}(A)=2$, then $\mathsf{L}(A)$ is an AMP with period $\{0,1,2,4\},\{0,1,3,4\}$, or $\{0,2,3,4\}$, and if $\mathsf{v}_{2g}(A)=1$, then it is an AMP with period $\{0,1,4\}$ or $\{0,3,4\}$.

Proof.

We set $G_{0}=\{g,2g,3g,4g\}$ and list all atoms of $\mathcal{B}(G_{0})$:

• $\|\cdot\|_{g}=1:$ $U_{1}=g^{6},U_{2}=g^{4}(2g),U_{3}=g^{2}(2g)^{2},U_{4}=(-g)g,U_{5}=(2g)^{3}$,
• $\|\cdot\|_{g}=2:$ $V_{1}=(2g)(-g)^{2}$,
• $\|\cdot\|_{g}=5:$ $V_{2}=(-g)^{6}$.

CASE 1: $\mathsf{v}_{2g}(A)=1$.

Then each factorization $z\in\mathsf{Z}(A)$ is divisible by $U_{2}$ or $V_{1}$. If $\mathsf{v}_{g}(A)<4$ or $\mathsf{v}_{-g}(A)<2$, it is easy to see that $\mathsf{L}(A)$ is a singleton. Thus, we assume that neither is the case and write $A=(2g)g^{v+4}(-g)^{w+2}$. Since $A$ is a zero-sum sequence, it follows that $v\equiv w+2\pmod{6}$.

We determine the set of lengths of $A$. Let $z$ be a factorization of $A$. If $z$ contains $U_{2}$, then $z$ is of the form $U_{2}z^{\prime}$ where $z^{\prime}$ is a factorization of $g^{v}(-g)^{w+2}$. If $z$ contains $V_{1}$, then $z$ is of the form $V_{1}z^{\prime\prime}$ where $z^{\prime\prime}$ is a factorization of $g^{v+4}(-g)^{w}$. Conversely, for each factorization $z^{\prime}$ of $g^{v}(-g)^{w+2}$ we have that $U_{2}z^{\prime}$ is a factorization of $A$, and for each factorization $z^{\prime\prime}$ of $g^{v}(-g)^{w+2}$ we have that $V_{1}z^{\prime\prime}$ is a factorization of $A$. Consequently $\mathsf{L}(A)=1+(\mathsf{L}(g^{v}(-g)^{w+2})\cup\mathsf{L}(g^{v+4}(-g)^{w}))$.

We determine $\mathsf{L}(g^{v}(-g)^{w+2})\cup\mathsf{L}(g^{v+4}(-g)^{w})$. If $v$ is odd, then each factorization contains of $g^{v}(-g)^{w+2}$ and $g^{v+4}(-g)^{w}$ contains $(-g)g$ and $\mathsf{L}(g^{v}(-g)^{w+2})\cup\mathsf{L}(g^{v+4}(-g)^{w})=1+(\mathsf{L}(g^{v-1}(-g)^{w-1+2})\cup\mathsf{L}(g^{v-1+4}(-g)^{w-1}))$. We thus focus on the case that $v$ is even.

If $v\equiv 0\pmod{6}$, then

[TABLE]

and

[TABLE]

Thus, the union is an AMP with difference $4$ and period $\{0,3,4\}$.

If $v\equiv 2\pmod{6}$, then

[TABLE]

and

[TABLE]

Thus, the union is an AMP with difference $4$ and period $\{0,1,4\}$.

If $v\equiv 4\pmod{6}$, then

[TABLE]

and

[TABLE]

Thus, the union is an AMP with difference $4$ and period $\{0,1,4\}$.

CASE 2: $\mathsf{v}_{2g}(A)=2$.

We set $A=(2g)^{2}g^{v+8}(-g)^{w+4}$ with $\mathsf{v}_{g}(A)=v+8>0$, $\mathsf{v}_{-g}(A)=w+4>0$, and $v,w\in\mathbb{Z}$. Then $v\equiv w+4\mod 6$. If $w$ is odd, then $\mathsf{Z}(A)=U_{4}\mathsf{Z}(A^{\prime})$ with $A^{\prime}=U_{4}^{-1}A$. Thus we may suppose that $w$ is even.

First suppose that $\mathsf{v}_{-g}(A)<4$. Then $\mathsf{v}_{-g}(A)=2$, $A=(-g)^{2}(2g)^{2}g^{6m+4}$ for some $m\in\mathbb{N}_{0}$, and $\mathsf{L}(A)=\{m+2,m+3\}$ is an AP with difference $1$. Now suppose that $\mathsf{v}_{-g}(A)\geq 4$. We discuss the cases where $\mathsf{v}_{g}(A)\in\{2,4,6\}$. If $\mathsf{v}_{g}(A)=2$, then $A=g^{2}(2g)^{2}(-g)^{6m}$ for some $m\in\mathbb{N}$ whence $\mathsf{L}(A)=\{m+1,m+2\}$. If $\mathsf{v}_{g}(A)=4$, then $A=g^{2}(2g)^{2}(-g)^{6m+2}$ for some $m\in\mathbb{N}$,

[TABLE]

whence $\mathsf{L}(A)=\{2+m,3+m,5+m\}$ is an AMP with period $\{0,1,3,4\}$. If $\mathsf{v}_{g}(A)=6$, then $A=g^{6}(2g)^{2}(-g)^{6m+4}$ for some $m\in\mathbb{N}_{0}$,

[TABLE]

where the last equations holds only in case $m\geq 1$. Thus $\mathsf{L}(A)=\{3,4,5\}$ or $\mathsf{L}(A)=\{3+m,4+m,5+m,7+m\}$ if $m\in\mathbb{N}$.

Now we suppose that $\mathsf{v}_{-g}(A)\geq 4$ and $\mathsf{v}_{g}(A)\geq 8$ whence $A=(2g)^{2}g^{v+8}(-g)^{w+4}$ with $v,w\in\mathbb{N}_{0}$ even. Then

[TABLE]

whence

[TABLE]

where the last equation holds because $1+\mathsf{L}(A_{1})\subset\mathsf{L}(g^{6}A_{1})$ and $g^{6}A_{1}=A_{2}$. The sets $\mathsf{L}(A_{2}),\mathsf{L}(A_{3})$ and $\mathsf{L}(A_{4})$ are APs with difference $4$. Thus in order to show that $\mathsf{L}(A)$ is an AMP with difference $4$, we study the minima and the lengths of $-1+\mathsf{L}(A_{2}),\mathsf{L}(A_{3})$ and $\mathsf{L}(A_{4})$. We distinguish three cases depending on the congruence class of $v$ modulo $6$ (recall that $v$ is even).

Suppose that $v\equiv 0\mod 6$. The set $-1+\mathsf{L}(A_{2})$ has minimum $a_{0}=(v+w+4)/6$ amd length $\ell_{0}=\min\{v+6,w+4\}/6$. The set $\mathsf{L}(A_{3})$ has minimum $a_{0}+2$ and length $\min\{v+6,w-2\}/6\in\{\ell_{0},\ell_{0}-1\}$. Finally $\mathsf{L}(A_{4})$ has minimum $a_{0}+3$ and length $\ell_{0}-1$. This implies that $\mathsf{L}(A)$ is an AMP with period $\{0,2,3,4\}$.

Suppose that $v\equiv 2\mod 6$. Then $\mathsf{L}(A_{4})$ has minimum $a_{2}=(v+w+6)/6$ and length $\ell_{2}=\min\{v+4,w+2\}/6$. The set $-1+\mathsf{L}(A_{2})$ has minimum $1+a_{2}$ and length $\ell_{2}$. The set $\mathsf{L}(A_{3})$ has minimum $a_{2}+3$ and length $\ell_{2}$ or $\ell_{2}-1$. This implies that $\mathsf{L}(A)$ is an AMP with period $\{0,1,3,4\}$.

Suppose that $v\equiv 4\mod 6$. The set $\mathsf{L}(A_{3})$ has minimum $a_{4}=(v+w+8)/6$ and length $\ell_{4}=\min\{v+8,w\}/6$. The set $\mathsf{L}(A_{4})$ has minimum $a_{4}+1$ and length $\ell_{4}$ or $\ell_{4}-1$. The set $-1+\mathsf{L}(A_{2})$ has minimum $a_{4}+2$ and the same length as $\mathsf{L}(A_{4})$. This implies that $\mathsf{L}(A)$ is an AMP with period $\{0,1,2,4\}$.

CASE 3: $\mathsf{v}_{2g}(A)\geq 3$.

We assert that $\mathsf{L}(A)$ is an AP with difference $1$. We choose a factorization $z\in\mathsf{Z}(A)$ and we show that either $|z|=\max\mathsf{L}(A)$ or that $|z|+1\in\mathsf{L}(A)$. Since $V_{2}U_{5}=V_{1}^{3}$, we may suppose that $\min\{\mathsf{v}_{V_{2}}(z),\mathsf{v}_{U_{5}}(z)\}=0$.

Suppose that $V_{1}\mid z$. If $U_{i}\mid z$ for some $i\in[1,3]$, then $|z|+1\in\mathsf{L}(A)$. If $\mathsf{v}_{U_{5}}(z)=0$, then $|z|=\max\mathsf{L}(A)$. Suppose that $\mathsf{v}_{V_{2}}(z)=0$. Then $z=U_{4}^{k_{4}}U_{5}^{k_{5}}V_{1}^{\ell_{1}}$ with $k_{4},k_{5},\ell_{1}\in\mathbb{N}_{0}$, and we claim that then $|z|=\max\mathsf{L}(A)$. By Lemma 2.5, it suffices to show that $k_{5}+\ell_{1}=\max\mathsf{L}(U_{5}^{k_{5}}V_{1}^{\ell_{1}})$. Since $U_{5}^{k_{5}}V_{1}^{\ell_{1}}$ is not divisible by an atom of length two, it follows that $\max\mathsf{L}(U_{5}^{k_{5}}V_{1}^{\ell_{1}})\leq|U_{5}^{k_{5}}V_{1}^{\ell_{1}}|/3=k_{5}+\ell_{1}$.

Suppose that $V_{1}\nmid z$. If $\mathsf{v}_{V_{2}}(z)=0$, then $|z|=\max\mathsf{L}(A)$ because all remaining atoms have $g$-norm one. From now on we suppose that $V_{2}\mid z$. This implies that $U_{5}\nmid z$. Since $\mathsf{v}_{2g}(A)\geq 3$, then $U_{5}\mid U_{2}^{\mathsf{v}_{U_{2}}(z)}U_{3}^{\mathsf{v}_{U_{3}}(z)}$ and we obtain a factorization $z^{\prime}\in\mathsf{Z}(A)$ with $|z^{\prime}|=|z|$ and with $U_{5}\mid z^{\prime}$ and we still have that $V_{2}\mid z^{\prime}$. Since $V_{2}U_{5}=V_{1}^{3}$, it follows that $|z^{\prime}|+1\in\mathsf{L}(A)$. ∎

Lemma 3.7**.**

Let $G$ be a cyclic group of order $|G|=6$, $g\in G$ with $\operatorname{ord}(g)=6$, and $A\in\mathcal{B}(G)$ with $\operatorname{supp}(A)=\{g,2g,4g,-g\}$. If $\mathsf{v}_{2g}(A)+\mathsf{v}_{4g}(A)\geq 3$, then $\mathsf{L}(A)$ is an AP with difference $1$ and otherwise it is an AMP with period $\{0,1,2,4\}$ or $\{0,1,3,4\}$ or $\{0,2,3,4\}$.

Proof.

We set $G_{0}=\{g,2g,4g,-g\}$ and list all atoms of $\mathcal{B}(G_{0})$:

• $\|\cdot\|_{g}=1:$ $U_{0}=g^{6}$, $U_{1}=g^{4}(2g)$, $U_{2}=g^{2}(2g)^{2}$, $U_{3}=(2g)^{3}$, $V_{2}=(2g)(4g)$,

$\|\cdot\|_{g}=1:$ $W=(4g)g^{2}$, $V_{1}=g(-g)$,

• $\|\cdot\|_{g}=2:$ $-U_{3}=(4g)^{3}$, $-W=(2g)(-g)^{2}$,
• $\|\cdot\|_{g}=3:$ $-U_{2}=(4g)^{2}(-g)^{2}$,
• $\|\cdot\|_{g}=4:$ $-U_{1}=(-g)^{4}(4g)$,
• $\|\cdot\|_{g}=5:$ $-U_{0}=(-g)^{6}$

First, suppose that $\mathsf{v}_{2g}(A)+\mathsf{v}_{4g}(A)\geq 3$. We choose a factorization $z\in\mathsf{Z}(A)$ and we show that either $|z|=\max\mathsf{L}(A)$ or that $|z|+1\in\mathsf{L}(A)$. We write $z$ in the form

[TABLE]

where $z^{+}$ is the product of all atoms from $\mathcal{B}(\{g,2g\})$ (these are $U_{0},U_{1},U_{2},U_{3}$), $z^{-}$ is the product of all atoms from $\mathcal{B}(\{-g,4g\})$, and $z^{0}$ is the product of all remaining atoms. Note that the sets in $\mathcal{L}(\{g,2g\})$ and $\mathcal{L}(\{-g,4g\})$ are singletons. If $U_{3}\mid z^{+}$ and $z^{-}\neq 1$, then $|z|+1\in\mathsf{L}(A)$. Similarly, if $-U_{3}\mid z^{-}$ and $z^{+}\neq 1$, then $|z|+1\in\mathsf{L}(A)$.

Suppose that $z^{-}=1$. If $-W\nmid z$, then all atoms dividing $z$ have $g$-norm equal to $1$ whence $|z|=\max\mathsf{L}(A)$. Suppose that $-W\mid z$. If $(-W)Y\mid z$ with $Y\in\{U_{0},U_{1},U_{2},W\}$, then $|z|+1\in\mathsf{L}(A)$. Otherwise, $z$ is a product of the atoms $-W,U_{3},V_{2}$, and $V_{1}$ which implies that $|z|=\max\mathsf{L}(A)$. The case $z^{+}=1$ follows by symmetry.

Thus we may suppose that $z^{+}\neq 1$ and $z^{+}\neq 1$ and that $U_{3}\nmid z^{+}$ and that $(-U_{3})\nmid z^{-}$. Let $A^{+}$ resp. $A^{-}$ denote the zero-sum sequences corresponding to $z^{+}$ resp. $z^{-}$. If $\mathsf{v}_{2g}(A^{+})\geq 3$, then there is a factorization $z^{\prime}$ of $A^{+}$ with $|z^{\prime}|=|z^{+}|$ and with $U_{3}\mid z^{\prime}$ and we are back to an earlier case. Thus we may suppose that $\mathsf{v}_{2g}(A^{+})\leq 2$. By symmetry we may also suppose that $\mathsf{v}_{4g}(A^{-})\leq 2$.

Since $(-U_{1})^{2}=(-U_{0})(-U_{2})$, we may suppose that $\mathsf{v}_{-U_{1}}(z^{-})\leq 1$. By symmetry we infer that $\mathsf{v}_{U_{1}}(z^{+})\leq 1$. Now we distinguish two cases.

CASE 1:  $U_{2}\mid z^{+}$ or $(-U_{2})\mid z^{-}$.

By symmetry we may suppose that $U_{2}\mid z^{+}$. If $(-U_{2})\mid z^{-}$ or $(-U_{1})\mid z^{-}$, then $|z|+1\in\mathsf{L}(A)$. Otherwise $z^{-}$ is a product of $(-U_{0})$. Since $\mathsf{v}_{4g}(A)\geq 1$, it follows that $V_{2}\mid z$ or $W\mid z$. Since $(-U_{0})V_{2}=(-W)(-U_{1})$, we are back to a previous case. Since $(-U_{0})W=V_{1}^{2}(-U_{1})$, it follows that $|z|+1\in\mathsf{L}(A)$.

CASE 2:  $U_{2}\nmid z^{+}$ and $(-U_{2})\nmid z^{-}$.

If $(-W)\mid z_{0}$, then $|z|+1\in\mathsf{L}(A)$ because $z^{+}$ is a nonempty product of $U_{0}$ and $U_{1}$. If $W\mid z_{0}$, then $|z|+1\in\mathsf{L}(A)$ because $z^{-}$ is a nonempty product of $-U_{0}$ and $-U_{1}$. From now on we suppose that $W\nmid z_{0}$ and $-W\nmid z_{0}$. Since $\mathsf{v}_{2g}(A)+\mathsf{v}_{4g}(A)\geq 3$, it follows that $V_{2}\mid z_{0}$. If $U_{0}\mid z$, then $U_{0}V_{2}=WU_{1}$ leads back to the case just handled. If $U_{1}\mid z$, then $U_{1}V_{2}=U_{2}W$ and we are back to CASE 1.

Now suppose that $\mathsf{v}_{2g}(A)=\mathsf{v}_{4g}(A)=1$. Then $\mathsf{v}_{g}(A)\equiv\mathsf{v}_{-g}(A)\mod 6$, and we set $\mathsf{v}_{g}(A)=v=6m+r$ and $\mathsf{v}_{-g}(A)=w=6n+r$ with $v,w\in\mathbb{N},r\in[0,5]$, and $m,n\in\mathbb{N}_{0}$. Then

[TABLE]

where negative exponents are interpreted in the quotient group of $\mathcal{F}(G)$. Since each factorization of $A$ is divisible by exactly one of the following

[TABLE]

we infer that $\mathsf{L}(A)=L_{1}\cup L_{2}\cup L_{3}\cup L_{4}\cup L_{5}$, where

[TABLE]

Note, if exponents in the above equations are negative, then the associated sets $L_{i}$ are indeed empty. More precisely, if

• •

$v<6$, then $m=0$, $[0,\min\{m-1,n\}]=\emptyset$, and hence $L_{2}=\emptyset$;

• •

$w<6$, then $n=0$ and $L_{3}=\emptyset$;

• •

$r\in[0,1]$ and $\min\{m,n\}=0$, then $L_{4}=\emptyset$;

• •

$r\in[0,3]$ and $\min\{m,n\}=0$, then $L_{5}=\emptyset$.

We observe that $L_{2}\subset L_{1}$ and $L_{3}\subset L_{1}$.

If $r\in[4,5]$, then $\min L_{5}=r-2+m+n$, $\min L_{4}=2+\min L_{5}$, and $\min L_{1}=3+\min L_{5}$; since $L_{5}$, $L_{4}$, and $L_{1}$ are APs with difference $4$ and length $\min\{m,n\}$, $\mathsf{L}(A)$ is an AMP with period $\{0,2,3,4\}$.

If $r\in[2,3]$, then $\min L_{4}=r+m+n$, $\min L_{1}=1+\min L_{4}$, and $\min L_{5}=2+\min L_{4}$; since $L_{4}$, $L_{1}$, and $L_{5}$ are APs with difference $4$, $L_{4}$ and $L_{1}$ have length $\min\{m,n\}$, and $L_{5}$ has length $\min\{m,n\}-1$, $\mathsf{L}(A)$ is an AMP with period $\{0,1,2,4\}$.

If $r\in[0,1]$, then $\min L_{1}=1+r+m+n$, $\min L_{5}=1+\min L_{1}$, and $\min L_{4}=3+\min L_{1}$; since $L_{1}$, $L_{5}$, and $L_{4}$ are APs with difference $4$, $L_{4}$ and $L_{5}$ have length $\min\{m,n\}-1$, and $L_{1}$ has length $\min\{m,n\}$, $\mathsf{L}(A)$ is an AMP with period $\{0,1,3,4\}$. ∎

Proposition 3.8**.**

Let $G$ be a cyclic group of order $|G|=6$. Then every $L\in\mathcal{L}(G)$ has one of the following forms :**

• •

$L$* is an AP with difference in $\{1,2,4\}$.*

• •

$L$* is an AMP with one of the following periods: $\{0,1,4\}$, $\{0,3,4\}$, $\{0,1,2,4\}$, $\{0,1,3,4\}$ $\{0,2,3,4\}$.*

Each of the mentioned forms is actually attained by arbitrarily large sets of lengths.

Proof.

Let $A^{\prime}\in\mathcal{B}(G)$. If $A^{\prime}=0^{m}A$ with $m\in\mathbb{N}_{0}$ and $A\in\mathcal{B}(G\setminus\{0\})$, then $\mathsf{L}(A^{\prime})=m+\mathsf{L}(A)$. Thus it is sufficient to prove the assertion for $\mathsf{L}(A)$. Let $g\in G$ with $\operatorname{ord}(g)=6$.

If $\operatorname{supp}(A)=G\setminus\{0\}$, then $\mathsf{L}(A)$ is an AP with difference $1$ by Proposition 2.2.4. Suppose that $\operatorname{supp}(A)\subset G\setminus\{g,-g\}=\{2g,3g,4g\}=G_{0}$. Since $(3g)^{2}$ is a prime element in $\mathcal{B}(G_{0})$ and $\{0,2g,4g\}$ is a cyclic group of order three, it follows that $\mathsf{L}(A)$ is an AP with difference $1$ by Proposition 3.1. Thus from now on we may suppose that $g\in\operatorname{supp}(A)$. We distinguish two cases.

CASE 1:   $-g\notin\operatorname{supp}(A)$.

Then $\operatorname{supp}(A)\subset\{g,2g,3g,4g\}$ and we assert that $\mathsf{L}(A)$ is an AP with difference $1$. Since all atoms in $\mathcal{B}(\{g,2g,3g\}$ have $g$-norm one, the sets in $\mathcal{L}(\{g,2g,3g\})$ are singletons. Thus the claim holds if $\operatorname{supp}(A)\subset\{g,2g,3g\}$. If $\operatorname{supp}(A)=\{g,2g,3g,4g\}$, then the claim follows from Lemma 3.5. Thus it remains to consider the cases where $\operatorname{supp}(A)$ equals one of the following three sets:

[TABLE]

The multiplicity of $g$ in any zero-sum sequence $B\in\mathcal{B}(\{g,2g,4g\}$ is even, whence $\mathcal{L}(\{g,2g,4g\})=\mathcal{L}(\{2g,4g\})$. Since $\{0,2g,4g\}$ is a cyclic group of order three, all sets of lengths in $\mathcal{L}(\{0,2g,4g\}$ are APs with difference $1$ by Proposition 3.1. If $\operatorname{supp}(A)=\{g,3g,4g\}$, then the claim follows from Lemma 3.4.

CASE 2:   $-g\in\operatorname{supp}(A)$.

Then $\{g,-g\}\subset\operatorname{supp}(A)$ and we distinguish four cases.

CASE 2.1:  $\operatorname{supp}(A)\subset G_{0}=\{g,3g,-g\}$.

Since $\min\Delta(G_{0})=2$ by [16, Lemma 6.8.5] and $\max\Delta(G_{0})=4$, it follows that $\Delta(G_{0})=\{2,4\}$. If $4\in\Delta(\mathsf{L}(A))$, then $\mathsf{L}(A)$ is an AP with difference $4$ by [6, Theorem 3.2], and otherwise $\mathsf{L}(A)$ is an AP with difference $2$.

CASE 2.2:  $\operatorname{supp}(A)=\{g,2g,-g\}$ or $\operatorname{supp}(A)=\{g,4g,-g\}$.

By symmetry it suffices to consider the first case. Lemma 3.6 shows that $\mathsf{L}(A)$ has one of the given forms.

CASE 2.3:  $\operatorname{supp}(A)=\{g,2g,3g,-g\}$ or $\operatorname{supp}(A)=\{g,3g,4g,-g\}$.

By symmetry it suffices to consider the first case. Lemma 3.3 shows that $\mathsf{L}(A)$ is an AP with difference $1$.

CASE 2.4:  $\operatorname{supp}(A)=\{g,2g,4g,-g\}$.

Lemma 3.7 shows that $\mathsf{L}(A)$ has one of the given forms.

Finally, we note that the preceding lemmas imply that each of the mentioned forms is actually attained by arbitrarily large sets of lengths. ∎

The next proposition is used substantially in [19, Subsection 4.1], where the system $\mathcal{L}(C_{5})$ is written down explicitly.

Proposition 3.9**.**

Let $G$ be a cyclic group of order $|G|=5$. Then every $L\in\mathcal{L}(G)$ has one of the following forms :**

• •

$L$* is an AP with difference in $\{1,3\}$.*

• •

$L$* is an AMP with period $\{0,2,3\}$ or with period $\{0,1,3\}$.*

Each of the mentioned forms is actually attained by arbitrarily large sets of lengths.

Proof.

By Proposition 2.3 we obtain that $\Delta^{*}(G)=\{1,3\}$. Let $A^{\prime}\in\mathcal{B}(G)$. If $A^{\prime}=0^{m}A$ with $m\in\mathbb{N}_{0}$ and $A\in\mathcal{B}(G\setminus\{0\})$, then $\mathsf{L}(A^{\prime})=m+\mathsf{L}(A)$. Thus it is sufficient to prove the assertion for $\mathsf{L}(A)$. If $|\operatorname{supp}(A)|=1$, then $|\mathsf{L}(A)|=1$. If $|\operatorname{supp}(A)|=4$, then $\mathsf{L}(A)$ is an AP with difference $1$ by Proposition 2.2.4. Suppose that $|\operatorname{supp}(A)|=2$. Then there is a nonzero $g\in G$ such that $\operatorname{supp}(A)=\{g,2g\}$ or $\operatorname{supp}(A)=\{g,4g\}$. If $\operatorname{supp}(A)=\{g,2g\}$, then $\mathsf{L}(A)$ is an AP with difference $1$ (this can be checked directly by arguing with the $g$-norm). If $\operatorname{supp}(A)=\{g,4g\}$, then $\mathsf{L}(A)$ is an AP with difference $3$.

Thus it remains to consider the case $|\operatorname{supp}(A)|=3$. We set $G_{0}=\operatorname{supp}(A)$. Then there is an element $g\in G_{0}$ such that $-g\in G_{0}$. Thus either $G_{0}=\{g,2g,-g\}$ or $G_{0}=\{g,3g,-g\}$. Since $\{g,3g,-g\}=\{-g,2(-g),-(-g)\}$, we may suppose without restriction that $G_{0}=\{g,2g,-g\}$. We list all atoms of $\mathcal{B}(G_{0})$:

• $\|\cdot\|_{g}=1:$ $g^{5}$, $g^{3}(2g)$, $g(2g)^{2}$, $g(-g)$,
• $\|\cdot\|_{g}=2:$ $(2g)^{5}$, $(2g)^{3}(-g)$, $(2g)(-g)^{2}$,
• $\|\cdot\|_{g}=4:$ $(-g)^{5}$.

If $\Delta(\mathsf{L}(A))\subset\{1\}$, then $\mathsf{L}(A)$ is an AP with difference $1$. If $3\in\Delta(\mathsf{L}(A))$, then $\Delta(\mathsf{L}(A))=\{3\}$ by [6, Theorem 3.2], which means that $\mathsf{L}(A)$ is an AP with difference $3$. Thus it remains to consider the case where $2\in\Delta(\mathsf{L}(A))\subset[1,2]$. We show that $\mathsf{L}(A)$ is an AMP with period $\{0,2,3\}$ or with period $\{0,1,3\}$. Since $2\in\Delta(\mathsf{L}(A))$, there exist $k\in\mathbb{N}$, $A_{1},\ldots,A_{k},B_{1},\ldots,B_{k+2}\in\mathcal{A}(G_{0})$ such that

[TABLE]

We distinguish two cases.

CASE 1:  $(-g)^{5}\notin\{A_{1},\ldots,A_{k}\}$.

Then $\{A_{1},\ldots,A_{k}\}$ must contain atoms with $g$-norm $2$. These are the atoms $(2g)^{5},(2g)(-g)^{2},(2g)^{3}(-g)$. If $g^{5}$ or $g^{3}(2g)$ occurs in $\{A_{1},\ldots,A_{k}\}$, then $k+1\in\mathsf{L}(A)$, a contradiction. Thus none of the elements $(-g)^{5},g^{5}$, and $g^{3}(2g)$ lies in $\{A_{1},\ldots,A_{k}\}$, and hence

[TABLE]

Now we set $h=2g$ and obtain that

[TABLE]

Since the $h$-norm of all these elements equals $1$, it follows that $\max\mathsf{L}(A)=k$, a contradiction.

CASE 2:  $(-g)^{5}\in\{A_{1},\ldots,A_{k}\}$.

If $(2g)^{5}$, or $g(2g)^{2}$, or $(2g)^{3}(-g)$ occurs in $\{A_{1},\ldots,A_{k}\}$, then $k+1\in\mathsf{L}(A)$, a contradiction. Since $\Delta(\{-g,g\})=\{3\}$, it follows that

[TABLE]

Since \Big{(}g^{3}(2g)\Big{)}\Big{(}(2g)(-g)^{2}\Big{)}=\Big{(}(-g)g\Big{)}^{2}\Big{(}g(2g)^{2}\Big{)} and $k+1\notin\mathsf{L}(A)$, it follows that $|\Omega|=1$. We distinguish two cases.

CASE 2.1:  $\{A_{1},\ldots,A_{k}\}\subset\{g^{5},(-g)^{5},g(-g),(2g)(-g)^{2}\}$.

We set $h=-g$, and observe that

[TABLE]

Since $(-h)^{5}$ is the only element with $h$-norm greater than $1$, it follows that $(-h)^{5}\in\{A_{1},\ldots,A_{k}\}$. Since $\Delta(\{h,-h\})=\{3\}$, it follows that $h^{2}(3h)\in\{A_{1},\ldots,A_{k}\}$. Since \Big{(}(-h)^{5}\Big{)}\Big{(}h^{2}(3h)\Big{)}=\Big{(}h(-h)\Big{)}^{2}\Big{(}(3h)(-h)^{3}\Big{)}, we obtain that $k+1\in\mathsf{L}(A)$, a contradiction.

CASE 2.2:  $\{A_{1},\ldots,A_{k}\}\subset\{g^{5},(-g)^{5},g(-g),g^{3}(2g)\}$.

Since \Big{(}g^{3}(2g)\Big{)}^{2}\Big{(}(-g)^{5}\Big{)}=\Big{(}g^{5}\Big{)}\Big{(}g(-g)\Big{)}\Big{(}(2g)(-g)^{2}\Big{)}^{2} and $k+1\notin\mathsf{L}(A)$, it follows that

[TABLE]

and hence $\mathsf{v}_{2g}(A)=1$. Thus every factorization $z$ of $A$ has the form

[TABLE]

where $z_{1},z_{2}$ are factorizations of elements $B_{1},B_{2}\in\mathcal{B}(\{-g,g\})$. Since $\mathsf{L}(B_{1})$ and $\mathsf{L}(B_{2})$ are APs of difference $3$, $\mathsf{L}(A)$ is a union of two shifted APs of difference $3$. We set

[TABLE]

where $m_{1}\in\mathbb{N}_{0}$, $m_{2}\in\mathbb{N}$, and $m_{3}\in[0,4]$. Suppose that $m_{1}\geq 1$. Note that

[TABLE]

and hence $\mathsf{L}(A^{\prime})=\{3,5,6\}$. We set $A=A^{\prime}A^{\prime\prime}$ with $A^{\prime\prime}\in\mathcal{B}(\{g,-g\})$. The above argument on the structure of the factorizations of $A$ implies that $\mathsf{L}(A)$ is the sumset of $\mathsf{L}(A^{\prime})$ and $\mathsf{L}(A^{\prime\prime})$ whence

[TABLE]

Since $\mathsf{L}(A^{\prime\prime})$ is an AP with difference $3$, $\mathsf{L}(A)$ is an AMP with period $\{0,2,3\}$. Suppose that $m_{1}=0$. If $m_{3}\in[2,4]$, then $\mathsf{L}(A)=\{m_{2}+m_{3},m_{2}+m_{3}+1,m_{2}+m_{3}+3\}$ is an AMP with period $\{0,1,3\}$. If $m_{3}=1$, then $\mathsf{L}(A)=\{m_{2}+2,m_{2}+4\}$. If $m_{3}=0$, then $\mathsf{L}(A)=\{m_{2}+1,m_{2}+3\}$. ∎

Proposition 3.10**.**

Let $G$ be a finite abelian group, $e_{1},e_{2}\in G\setminus\{0\}$ with $\langle e_{1}\rangle\cap\langle e_{2}\rangle=\{0\}$, $\operatorname{ord}(e_{2})=n\geq 4$, and $U=(e_{1}+e_{2})(-e_{1})e_{2}^{n-1}$. Then, for every $k\in\mathbb{N}$, we have

[TABLE]

Proof.

Let $k\in\mathbb{N}$ and A_{k}=U(-U)\big{(}e_{2}(-e_{2})\big{)}^{kn}. If $X\in\mathcal{A}(G)$ with $X\mid A_{k}$, then

[TABLE]

Thus

[TABLE]

and we obtain the claim by using (3.1). ∎

Proposition 3.11**.**

Let $G=C_{2}^{4}$, $(e_{1},e_{2},e_{3},e_{4})$ be a basis of $G$, $e_{0}=e_{1}+\ldots+e_{4}$, $U_{4}=e_{0}\cdot\ldots\cdot e_{4}$, $U_{3}=e_{1}e_{2}e_{3}(e_{1}+e_{2}+e_{3})$, and $U_{2}=e_{1}e_{2}(e_{1}+e_{2})$.

For each $k\in\mathbb{N}$, \mathsf{L}\big{(}(U_{3}U_{4})^{2k}\big{)}=\{4k\}\cup[4k+2,9k]. 2. 2.

For each $k\in\mathbb{N}$, $\mathsf{L}(U_{4}^{2k}U_{2})=(2k+1)+\{0,1,3\}+3\cdot[0,k-1]$.

Proof.

1. Let $k\in\mathbb{N}$ and $A_{k}=(U_{3}U_{4})^{2k}$. Then

[TABLE]

Since $\max\mathsf{L}(A_{k})\leq|A_{k}|/2=9k$, it follows that $\max\mathsf{L}(A_{k})=9k$. We set $z=U_{3}^{2k}U_{4}^{2k}\in\mathsf{Z}(A_{k})$. Since $U_{4}$ is the only atom of length $\mathsf{D}(G)=5$ dividing $A_{k}$ and since $2k=\mathsf{v}_{e_{0}}(A_{k})=\mathsf{v}_{U_{4}}(z)=\max\{\mathsf{v}_{U_{4}}(z^{\prime})\mid z^{\prime}\in\mathsf{Z}(A_{k})\}$, it follows that $\min\mathsf{L}(A_{k})=|z|=4k$.

Next we assert that $4k+1\notin\mathsf{L}(A_{k})$. For $\nu\in[0,4]$, we set $V_{\nu}=e_{\nu}^{2}$ and $V_{5}=(e_{1}+e_{2}+e_{3})^{2}$. Since $\mathsf{Z}(U_{3}^{2})=\{U_{3}^{2},V_{1}V_{2}V_{3}V_{5}\}$, $\mathsf{Z}(U_{4}^{2})=\{U_{4}^{2},V_{1}V_{2}V_{3}V_{4}V_{0}\}$, and $\mathsf{Z}(U_{3}U_{4})=\{U_{3}U_{4},V_{1}V_{2}V_{3}W\}$ where $W=(e_{1}+e_{2}+e_{3})e_{0}e_{4}$, it follows that $\min(\mathsf{L}(A_{k})\setminus\{4k\})=4k+2$.

Since $A_{k}$ has precisely one factorization $z$ of length $|z|=\max\mathsf{L}(A_{k})=9k$, and $z$ is a product of atoms of length two with $V_{0}V_{4}V_{5}\mid z$. Since $V_{0}V_{4}V_{5}=W^{2}$, it follows that $A_{k}$ has a factorization of length $9k-1$.

2. Setting $W=(e_{1}+e_{2})e_{3}e_{4}e_{0}$ we infer that $U_{4}^{2}U_{2}=U_{4}(e_{1}^{2})(e_{2}^{2})W=U_{2}(e_{0}^{2})\cdot\ldots\cdot(e_{4}^{2})$ and hence $\mathsf{L}(U_{4}^{2}U_{2})=\{3,4,6\}$. Thus for each $k\in\mathbb{N}$ we obtain that

[TABLE]

Proposition 3.12**.**

Let $G=C_{3}^{r}$ with $r\in[3,4]$, $(e_{1},\ldots,e_{r})$ a basis of $G$, $e_{0}=e_{1}+\ldots+e_{r}$, and $U=(e_{1}\cdot\ldots\cdot e_{r})^{2}e_{0}$.

If $r=3$, then for each $k\in\mathbb{N}$ we have

[TABLE] 2. 2.

If $r=4$ and $V_{1}=e_{1}^{2}e_{2}^{2}(e_{1}+e_{2})$, then for each $k\in\mathbb{N}$ we have

[TABLE]

Proof.

1. Let $r=3$ and $k\in\mathbb{N}$. We set $A_{k}=U^{6k+1}(-U)$ and $L_{k}=\mathsf{L}(A_{k})$. For $\nu\in[0,3]$, we set $U_{\nu}=e_{\nu}^{3}$, $V_{\nu}=(-e_{\nu})e_{\nu}$, and we define $X=e_{0}^{2}e_{1}e_{2}e_{3}$.

First, consider $\mathsf{L}(U^{6k})$. We observe that $\mathsf{Z}(U^{2})=\{U^{2},U_{1}U_{2}U_{3}X\}$ and $\mathsf{Z}(U^{3})=\{U^{3},UU_{1}U_{2}U_{3}X,$ $U_{0}U_{1}^{2}U_{2}^{2}U_{3}^{2}\}$. Furthermore, $\min\mathsf{L}(U^{6k})=6k$, $\max\mathsf{L}(U^{6k})=14k$, $\Delta(\{e_{0},\ldots,e_{3}\})=\{2\}$, and hence

[TABLE]

Next, consider \mathsf{L}\big{(}(-U)U\big{)}. For subsets $I,J\subset[1,3]$ with $[1,3]=I\uplus J$, we set

[TABLE]

Since

[TABLE]

it follows that

[TABLE]

which implies that

[TABLE]

It remains to show that $14k+6\notin L_{k}$. For this, we consider the unique factorization $z\in\mathsf{Z}(A_{k})$ of length $|z|=14k+7$ which has the form

[TABLE]

Assume to the contrary that there is a factorization $z^{\prime}\in\mathsf{Z}(A_{k})$ of length $|z^{\prime}|=14k+6$. If $V_{0}\,|\,z^{\prime}$, then $V_{0}V_{1}^{2}V_{2}^{2}V_{3}^{2}\,|\,z^{\prime}$ and $z^{\prime}=V_{0}V_{1}^{2}V_{2}^{2}V_{3}^{2}x$ with $x\in\mathsf{Z}(U^{6k})$. Whence $|x|\in\mathsf{L}(U^{6k})$ and $|z^{\prime}|\in 7+\mathsf{L}(U^{6k})$, a contradiction. Suppose that $V_{0}\nmid z^{\prime}$. Then there are $I,J\subset[1,3]\ \text{ with}\ [1,3]=I\uplus J$ such that $W_{I}(-W_{I})\prod_{j\in J}V_{j}\,|\,z^{\prime}$ and hence z^{\prime}=W_{I}(-W_{I})\big{(}\prod_{j\in J}V_{j}\big{)}x with $x\in\mathsf{Z}(U^{6k})$. Thus $|z^{\prime}|\in[2,5]+\mathsf{L}(U^{6k})$, a contradiction.

2. Let $r=4$ and $k\in\mathbb{N}$. We have $\mathsf{L}(U^{2})=\{2,5\}$ and $\mathsf{L}(U^{3k})=3k+3\cdot[0,2k]$. We define

[TABLE]

and observe that

[TABLE]

whence $\mathsf{L}(U^{3}V_{1})=\{4,5,7,8\}$. Clearly, each factorization of $U^{3k}V_{1}$ contains exactly one of the atoms $V_{1},V_{2},V_{3}$, and it contains it exactly once. Therefore we obtain that

[TABLE]

Proof of Theorem 1.1.

1. (a)  $\Rightarrow$  (b)  This implication is obvious.

(b)  $\Rightarrow$  (d)  Suppose that $\exp(G)=n$, and that $G$ is not isomorphic to any of the groups listed in (d). We have to show that there is an $L\in\mathcal{L}(G)$ which is not an AP. We distinguish four cases.

(d)  $\Rightarrow$  (a)   Proposition 3.1 shows that, for all groups mentioned, all sets of lengths are APs. Proposition 2.3 shows that all differences lie in $\Delta^{*}(G)$.

(c)  $\Leftrightarrow$  (d)  This is the special case for finite groups of [18, Theorem 1.1] (see Remark 3.13.1).

CASE 1:  $n\geq 5$.

Then [17, Proposition 3.6.1] provides examples of sets of lengths which are not APs.

CASE 2:  $n=4$.

Since $G$ is not cyclic, it has a subgroup isomorphic to $C_{2}\oplus C_{4}$. Then [16, Theorem 6.6.5] shows that $\{2,4,5\}\in\mathcal{L}(C_{2}\oplus C_{4})\subset\mathcal{L}(G)$.

CASE 3:  $n=3$.

Then $G$ is isomorphic to $C_{3}^{r}$ with $r\geq 3$, and Proposition 3.12.1 provides examples of sets of lengths which are not APs.

CASE 4:  $n=2$.

Then $G$ is isomorphic to $C_{2}^{r}$ with $r\geq 4$, and Proposition 3.11.1 provides examples of sets of lengths which are not APs.

2. (b)  $\Rightarrow$  (a)  Suppose that $G$ is a subgroup of $C_{4}^{3}$ or a subgroup of $C_{3}^{3}$. Then Proposition 2.3.2 implies that $\Delta^{*}(G)\subset\{1,2\}$, and hence Proposition 2.2.1 implies the assertion.

(a)  $\Rightarrow$  (b)  Suppose that (b) does not hold. Then $G$ has a subgroup isomorphic to a cyclic group of order $n\geq 5$, or isomorphic to $C_{2}^{4}$, or isomorphic to $C_{3}^{4}$. We show that in none of these cases (a) holds.

If $G$ has a subgroup isomorphic to $C_{n}$ for some $n\geq 5$, then [17, Proposition 3.6.1] shows that (a) does not hold. If $G$ has a subgroup isomorphic to $C_{2}^{4}$, then Proposition 3.11.2 shows that (a) does not hold. If $G$ has a subgroup isomorphic to $C_{3}^{4}$, then Proposition 3.12.2 shows that (a) does not hold.

3. Let $\Delta$ be finite with $\Delta^{*}(G)\subset\Delta\subset\mathbb{N}$. If $G$ is cyclic with $|G|\leq 6$, then all sets of lengths are AMPs with difference in $\Delta^{*}(G)$ by Propositions 3.1, 3.8, and 3.9. If $\exp(G)\geq 7$, then Proposition 3.2 shows that there are arbitrarily large sets of lengths that are not AMPs with difference $d\in\Delta$.

Suppose that $G$ has rank $r\geq 2$ and $\exp(G)=n\in[2,6]$. If $n\geq 4$, then Proposition 3.10 shows that there are arbitrarily large sets of lengths that are not AMPs with difference $d\in\Delta$. Thus it suffices to consider elementary $2$-groups and elementary $3$-groups.

Suppose that $G=C_{2}^{r}$. If $r\leq 3$, then the assertion follows from 1. If $r\geq 4$, then the assertion follows from Proposition 3.11.1.

Suppose that $G=C_{3}^{r}$. If $r\leq 2$, then the assertion follows from 1. If $r\geq 3$, then the assertion follows from Proposition 3.12.1. ∎

Proof of Corollary 1.2.

Let $G^{\prime}$ be an abelian group such that $\mathcal{L}(G)=\mathcal{L}(G^{\prime})$. Then $G^{\prime}$ is finite by Proposition 2.2 and by [16, Theorem 7.4.1]. By Proposition 2.4, we have $\mathsf{D}(G)=\rho_{2}(G)=\rho_{2}(G^{\prime})=\mathsf{D}(G^{\prime})$, and $\mathcal{L}(G)$ satisfies one of the properties given in Theorem 1.1 if and only if the same is true for $\mathcal{L}(G^{\prime})$. We distinguish three cases.

CASE 1:  $\mathcal{L}(G)$ satisfies the property in Theorem 1.1.1.

Then $G$ is cyclic of order $|G|\leq 4$ or isomorphic to a subgroup of $C_{2}^{3}$ or isomorphic to a subgroup of $C_{3}^{2}$. Moreover, the same is true for $G^{\prime}$. Since $\mathsf{D}(G)\geq 4$, the assertion follows from Proposition 3.1.

CASE 2:  $\mathcal{L}(G)$ satisfies the property in Theorem 1.1.2.

By CASE 1, we may suppose that $\mathcal{L}(G)$ and $\mathcal{L}(G^{\prime})$ do not satisfy the property in Theorem 1.1.1. Thus $G$ and $G^{\prime}$, are isomorphic to one of the following groups: $C_{3}^{3}$, $C_{2}\oplus C_{4}$, $C_{2}^{2}\oplus C_{4}$, $C_{2}\oplus C_{4}^{2}$, $C_{4}^{2}$, or $C_{4}^{3}$. Since $C_{3}^{3}$ and $C_{4}^{2}$ are the only non-isomorphic groups having the same Davenport constant, it remains to show that $\mathcal{L}(C_{3}^{3})\neq\mathcal{L}(C_{4}^{2})$. Since $\max\Delta(C_{4}^{2})=3$ (by [21, Lemma 3.3]) and $\max\Delta(C_{3}^{3})=2$ (by [14, Proposition 5.5]), the assertion follows.

CASE 3:  $\mathcal{L}(G)$ satisfies the property in Theorem 1.1.3.

By CASE 1, we may suppose that $G$ and $G^{\prime}$ do not satisfy the property in Theorem 1.1.1. Thus $G$ and $G^{\prime}$ are cyclic of order five or six. Since $\mathsf{D}(C_{5})=5$ and $\mathsf{D}(C_{6})=6$, the claim follows. ∎

Remark 3.13**.**

1. Let $H$ be an atomic monoid. The system of sets of lengths $\mathcal{L}(H)$ is said to be additively closed if the sumset $L_{1}+L_{2}\in\mathcal{L}(H)$ for all $L_{1},L_{2}\in\mathcal{L}(H)$. Thus $\mathcal{L}(H)$ is additively closed if and only if $(\mathcal{L}(H),+)$ is a commutative reduced semigroup with respect to set addition. If this holds, then $\mathcal{L}(H)$ is an acyclic semigroup in the sense of Cilleruelo, Hamidoune, and Serra ([7]). $\mathcal{L}(H)$ is additively closed in certain Krull monoids stemming from module theory ([1, Section 6.C]). Examples in a non-cancellative setting can be found in [20, Theorem 4.5] and a more detailed discussion of the property of being additively closed is given in [18].

2. Several properties occurring in Theorem 1.1 can be characterized by further arithmetical invariants such as the catenary degree $\mathsf{c}(G)$ and the tame degree $\mathsf{t}(G)$ (for background see [16, Sections 6.4 and 6.5]). For example, the properties (a) - (d) given in Theorem 1.1.1. are equivalent to each of the following properties (e) and (f):

• (e)

$\mathsf{c}(G)\leq 3\quad\text{or}\quad\mathsf{c}(G)=4\ \text{and}\ \{2,4\}\in\mathcal{L}(G)$.

• (f)

$\mathsf{c}(G)\leq 3\quad\text{or}\quad\mathsf{t}(G)=4$.

(use [21, Theorem A], [11, Theorem 4.12], and [16, Theorem 6.6.3]).