# A Characterization of class groups via sets of lengths II

Alfred Geroldinger and Qinghai Zhong

###### Abstract.

Let $H$ be a Krull monoid with finite class group $G$ and suppose that every
class contains a prime divisor. If an element $a\in H$ has a factorization
$a=u_{1}\cdot\ldots\cdot u_{k}$ into irreducible elements
$u_{1},\ldots,u_{k}\in H$, then $k$ is called the length of the factorization
and the set $\mathsf{L}(a)$ of all possible factorization lengths is the set
of lengths of $a$. It is classical that the system
$\mathcal{L}(H)=\\{\mathsf{L}(a)\mid a\in H\\}$ of all sets of lengths depends
only on the class group $G$, and a standing conjecture states that conversely
the system $\mathcal{L}(H)$ is characteristic for the class group. We verify
the conjecture if the class group is isomorphic to $C_{n}^{r}$ with $r,n\geq
2$ and $r\leq\max\\{2,(n+2)/6\\}$. Indeed, let $H^{\prime}$ be a further Krull
monoid with class group $G^{\prime}$ such that every class contains a prime
divisor and suppose that $\mathcal{L}(H)=\mathcal{L}(H^{\prime})$. We prove
that, if one of the groups $G$ and $G^{\prime}$ is isomorphic to $C_{n}^{r}$
with $r,n$ as above, then $G$ and $G^{\prime}$ are isomorphic (apart from two
well-known pairings).

###### Key words and phrases:

Krull monoids, maximal orders, seminormal orders; class groups, arithmetical
characterizations, sets of lengths, zero-sum sequences, Davenport constant

###### 2010 Mathematics Subject Classification:

11B30, 11R27, 13A05, 13F05, 20M13

This work was supported by the Austrian Science Fund FWF, Project Number
M1641-N26.

##  1\. Introduction and Main Result

Let $H$ be a Krull monoid with class group $G$ and suppose that every class
contains a prime divisor (holomorphy rings in global fields are such Krull
monoids and more examples will be given in Section 2). Then every nonunit
$a\in H$ can be written as a product of irreducible elements, say
$a=u_{1}\cdot\ldots\cdot u_{k}$, and the number of factors $k$ is called the
length of the factorization. The set $\mathsf{L}(a)$ of all possible
factorization lengths is the set of lengths of $a$, and
$\mathcal{L}(H)=\\{\mathsf{L}(a)\mid a\in H\\}$ is called the system of sets
of lengths of $H$ (for convenience we set $\mathsf{L}(a)=\\{0\\}$ if $a$ is an
invertible element of $H$). It is easy to check that all sets of lengths are
finite and, by definition of the class group, we observe that $H$ is factorial
if and only if $|G|=1$. By a result due to Carlitz in 1960, we know that $H$
is half-factorial (i.e., $|L|=1$ for all $L\in\mathcal{L}(H)$) if and only if
$|G|\leq 2$.

Suppose that $|G|\geq 3$. Then there is some $a\in H$ with
$|\mathsf{L}(a)|>1$. If $k,\ell\in\mathsf{L}(a)$ with $k<\ell$ and
$m\in\mathbb{N}$, then
$\mathsf{L}(a^{m})\supset\\{km+\nu(\ell-k)\mid\nu\in[0,m]\\}$ which shows that
sets of lengths can become arbitrarily large. The monoid $\mathcal{B}(G)$ 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 $\mathcal{B}(G)$ coincide. Thus
$\mathcal{L}(H)=\mathcal{L}\big{(}B(G)\big{)}$, and it is usual to set
$\mathcal{L}(G):=\mathcal{L}\big{(}\mathcal{B}(G)\big{)}$. In particular, the
system of sets of lengths of $H$ depends only on the class group $G$. The
associated inverse question asks whether or not sets of lengths are
characteristic for the class group. More precisely, the Characterization
Problem for class groups can be formulated as follows (see [8, Section 7.3],
[11, page 42], [22], and Proposition 2.1)).

  1. Given two finite abelian groups $G$ and $G^{\prime}$ such that $\mathcal{L}(G)=\mathcal{L}(G^{\prime})$. Does it follow that $G\cong G^{\prime}$?

The system of sets of lengths $\mathcal{L}(G)$ is studied with methods from
additive combinatorics. In particular, zero-sum theoretical invariants (such
as the Davenport constant or the cross number) and the associated inverse
problems play a crucial role. Most of these invariants are well-understood
only in a very limited number of cases (e.g., for groups of rank two, the
precise value of the Davenport constant $\mathsf{D}(G)$ is known and the
associated inverse problem is solved; however, if $n$ is not a prime power and
$r\geq 3$, then the value of the Davenport constant $\mathsf{D}(C_{n}^{r})$ is
unknown). Thus it is not surprising that affirmative answers to the
Characterization Problem so far have been restricted to those groups where we
have a good understanding of the Davenport constant. These groups include
elementary $2$-groups, cyclic groups, and groups of rank two (the latter were
recently handled in [13]; for a variety of partial results we refer to [20,
23, 21]).

The goal of the present note is to solve the Characterization Problem for
groups of the form $C_{n}^{r}$ if the exponent is large with respect to the
rank. Here is our main theorem.

######  Theorem 1.1.

Let $G$ be an abelian group such that $\mathcal{L}(G)=\mathcal{L}(C_{n}^{r})$
where $r,n\in\mathbb{N}$ with $n\geq 2$, $(n,r)\notin\\{(2,1),(2,2),(3,1)\\}$,
and $r\leq\max\\{2,(n+2)/6\\}$. Then $G\cong C_{n}^{r}$.

The groups $C_{n}^{r}$, where $r,n$ are as above, are the first groups at all
for which the Characterization Problem is solved whereas the Davenport
constant is unknown. This is made possible by a detailed study of the set of
minimal distances $\Delta^{*}(G)=\\{\min\Delta(G_{0})\mid G_{0}\subset G\
\text{is a non-half-factorial subset}\\}$ and the associated minimal non-half-
factorial subsets. Sets of minimal distances have been investigated by
Chapman, Grynkiewicz, Hamidoune, Plagne, Schmid, Smith, and others (see [8,
Section 6.8] for some basic information and [9, 19, 5, 21, 3, 15, 18] for
recent progress). In Section 2 we repeat some key facts on Krull monoids and
gather the required machinery, and in Section 3 we study structural properties
of (large) minimal non-half-factorial sets. The proof of Theorem 1.1 will be
provided in Section 4 where we also give a positive answer to the
Characterization Problem for all groups $G$ with Davenport constant
$\mathsf{D}(G)\in[4,11]$ (Proposition 4.1).

##  2\. Background on Krull monoids and their sets of minimal distances

Our notation and terminology are consistent with [12]. We denote by
$\mathbb{N}$ the set of positive integers, and for $a,b\in\mathbb{Q}$, we
denote by $[a,b]=\\{x\in\mathbb{Z}\mid a\leq x\leq b\\}$ the discrete, finite
interval between $a$ and $b$. If $A,B\subset\mathbb{Z}$ are subsets of the
integers, then $A+B=\\{a+b\mid a\in A,b\in B\\}$ denotes their sumset, and
$\Delta(A)$ 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 $[a,b]\cap
A=\\{a,b\\}$). Let $d,l\in\mathbb{N}$ and $M\in\mathbb{N}_{0}$. A subset
$L\subset\mathbb{Z}$ is called an almost arithmetical progression (AAP for
short) with difference $d$, length $l$, and bound $M$ if

$L=y+(L^{\prime}\cup L^{*}\cup L^{\prime\prime})\subset y+d\mathbb{Z},$

where $y\in\mathbb{Z}$, $L^{*}=\\{\nu d\mid\nu\in[0,l]\\}$ is an arithmetical
progression with difference $d$ and length $l$, $L^{\prime}\subset[-M,-1]$,
and $L^{\prime\prime}\subset\max L^{*}+[1,M]$.

By a monoid we mean a commutative semigroup with identity which satisfies the
cancellation laws. A monoid $F$ is called free abelian with basis $P\subset
F$, and we write $F=\mathcal{F}(P)$, if every $a\in F$ has a unique
representation of the form

$a=\prod_{p\in
P}p^{\mathsf{v}_{p}(a)}\quad\text{with}\quad\mathsf{v}_{p}(a)\in\mathbb{N}_{0}\
\text{ and }\ \mathsf{v}_{p}(a)=0\ \text{ for almost all }\ p\in P\,.$

A monoid $H$ is said to be a Krull monoid if it satisfies one of the following
two equivalent conditions (see [8, Theorem 2.4.8]).

  * (a)

$H$ is $v$-noetherian and completely integrally closed.

  * (b)

There exists a monoid homomorphism $\varphi\colon H\to F=\mathcal{F}(P)$ into
a free abelian monoid $F$ such that $a\,|\,b$ in $H$ if and only if
$\varphi(a)\,|\,\varphi(b)$ in $F$.

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 ([8, Section 2.11 and Examples
7.4.2]). Monoid domains and power series domains that are Krull are discussed
in [17, 4], and note that every class of a Krull monoid domain contains a
prime divisor. For monoids of modules that are Krull and their distribution of
prime divisors, we refer the reader to [6, 1].

Sets of lengths in Krull monoids can be studied in the monoid of zero-sum
sequences over its class group. Let $G$ be an additively written abelian group
and $G_{0}\subset G$ a subset. An element

$S=g_{1}\cdot\ldots\cdot g_{\ell}=\prod_{g\in
G_{0}}g^{\mathsf{v}_{g}(S)}\in\mathcal{F}(G_{0})$

is called a sequence over $G_{0}$, and we use all notations as in [16]. In
particular, $\sigma(S)=g_{1}+\ldots+g_{\ell}$ denotes the sum, $|S|=\ell$ the
length, $\mathsf{h}(S)=\max\\{\mathsf{v}_{g}(S)\mid g\in G_{0}\\}$ the maximal
multiplicity, $\operatorname{supp}(S)=\\{g_{1},\ldots,g_{\ell}\\}\subset
G_{0}$ the support, and
$\mathsf{k}(S)=\sum_{i=1}^{\ell}1/\operatorname{ord}(g_{i})$ the cross number
of $S$. The monoid

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

is the monoid of zero-sum sequences over $G_{0}$, and since the embedding
$\mathcal{B}(G_{0})\hookrightarrow\mathcal{F}(G_{0})$ satisfies Condition (b)
above, $\mathcal{B}(G_{0})$ is a Krull monoid. As usual, we write
$\mathcal{L}(G_{0})=\mathcal{L}\big{(}\mathcal{B}(G_{0})\big{)}$ for the
system of sets of lengths of $\mathcal{B}(G_{0})$ and
$\mathcal{A}(G_{0})=\mathcal{A}(\mathcal{B}(G_{0}))$ for the set of atoms (the
set of irreducible elements) of $\mathcal{B}(G_{0})$. Note that the atoms of
$\mathcal{B}(G_{0})$ are precisely the minimal zero-sum sequences over
$G_{0}$, and

$\mathsf{D}(G_{0})=\sup\\{|U|\mid
U\in\mathcal{A}(G_{0})\\}\in\mathbb{N}\cup\\{\infty\\}$

is the Davenport constant of $G_{0}$. The significance of the system of sets
of lengths $\mathcal{L}(G)$ (and hence of the Characterization Problem in the
formulation given in the Introduction) stems from its universal role which can
be seen from the following proposition.

######  Proposition 2.1.

  1. 1.

If $H$ is a Krull monoid with class group $G$ such that each class contains a
prime divisor, then $\mathcal{L}(H)=\mathcal{L}(G)$.

  2. 2.

Let $\mathcal{O}$ be a holomorphy ring in a global field $K$, $A$ a central
simple algebra over $K$, and $H$ a classical maximal $\mathcal{O}$-order of
$A$ such that every stably free left $R$-ideal is free. Then
$\mathcal{L}(H)=\mathcal{L}(G)$, where $G$ is a ray class group of
$\mathcal{O}$ and hence finite abelian.

  3. 3.

Let $H$ be a seminormal order in a holomorphy ring of a global field with
principal order $\widehat{H}$ such that the natural map
$\mathfrak{X}(\widehat{H})\to\mathfrak{X}(H)$ is bijective and there is an
isomorphism
$\overline{\vartheta}\colon\mathcal{C}_{v}(H)\rightarrow\mathcal{C}_{v}(\widehat{H})$
between the $v$-class groups. Then $\mathcal{L}(H)=\mathcal{L}(G)$, where
$G=\mathcal{C}_{v}(H)$ is finite abelian.

###### Proof.

1\. See [8, Section 3.4].

2\. See [24, Theorem 1.1], and [2] for related results of this flavor.

3\. See [10, Theorem 5.8] for a more general result in the setting of weakly
Krull monoids. ∎

Next we discuss sets of distances and minimal sets of distances. Let

$\Delta(G_{0})=\bigcup_{L\in\mathcal{L}(G_{0})}\Delta(L)\ \subset\mathbb{N}$

denote the set of distances of $G_{0}$. Then $G_{0}$ is called half-factorial
if $\Delta(G_{0})=\emptyset$. Otherwise, $G_{0}$ is called non-half-factorial
and we have $\min\Delta(G_{0})=\gcd\Delta(G_{0})$. Note that $G_{0}$ is half-
factorial if and only if $\mathsf{k}(A)=1$ for all $A\in\mathcal{A}(G_{0})$.
Furthermore, the set $G_{0}$ is called

  * •

minimal non-half-factorial if it is half-factorial and every proper subset
$G_{1}\subsetneq G_{0}$ is half-factorial.

  * •

an LCN-set if $\mathsf{k}(A)\geq 1$ for all $A\in\mathcal{A}(G_{0})$.

The set $\Delta(G)$ is an interval and its maximum is studied in [7]. The
following two subsets of $\Delta(G)$, the set of minimal distances
$\Delta^{*}(G)$ and the set $\Delta_{1}(G)$, play a crucial role in the
present paper. We define

$\displaystyle\Delta^{*}(G)$ $\displaystyle=\\{\min\Delta(G_{0})\mid
G_{0}\subset G\ \text{with}\ \Delta(G_{0})\neq\emptyset\\}\subset\Delta(G)\,,$
$\displaystyle\mathsf{m}(G)$ $\displaystyle=\max\\{\min\Delta(G_{0})\mid
G_{0}\subset G\ \text{is an LCN-set with}\ \Delta(G_{0})\neq\emptyset\\}\,,$

and we denote by $\Delta_{1}(G)$ the set of all $d\in\mathbb{N}$ with the
following property:

  * For every $k\in\mathbb{N}$, there exists some $L\in\mathcal{L}(G)$ which is an AAP with difference $d$ and length $l\geq k$.

Thus, by definition, if $G^{\prime}$ is a further finite abelian group such
that $\mathcal{L}(G)=\mathcal{L}(G^{\prime})$, then
$\Delta_{1}(G)=\Delta_{1}(G^{\prime})$. The next proposition gathers the
properties of $\Delta^{*}(G)$ and of $\Delta_{1}(G)$ which are needed in the
sequel.

######  Proposition 2.2.

Let $G$ be a finite abelian group with $|G|\geq 3$ and $\exp(G)=n$.

  1. 1.

$\Delta^{*}(G)\subset\Delta_{1}(G)\subset\\{d_{1}\in\Delta(G)\mid d_{1}\
\text{divides some }\ d\in\Delta^{*}(G)\\}$. In particular,
$\max\Delta^{*}(G)=\max\Delta_{1}(G)$.

  2. 2.

$\max\Delta^{*}(G)=\max\\{\exp(G)-2,\mathsf{m}(G)\\}=\max\\{\exp(G)-2,\mathsf{r}(G)-1\\}$.
If $G$ is a $p$-group, then $\mathsf{m}(G)=\mathsf{r}(G)-1$.

  3. 3.

If $k\in\mathbb{N}$ is maximal such that $G$ has a subgroup isomorphic to
$C_{n}^{k}$, then

$\Delta_{1}(G)\subset[1,\max\\{\mathsf{m}(G),\lfloor\frac{n}{2}\rfloor-1\\}]\cup[\max\\{1,n-k-1\\},n-2]\,.$

and

$[1,\mathsf{r}(G)-1]\cup[\max\\{1,n-k-1\\},n-2]\subset\Delta_{1}(G)\,.$

###### Proof.

1\. follows from [8, Corollary 4.3.16] and 2. from [15, Theorem 1.1 and
Proposition 3.2].

3\. In [21, Theorem 3.2], it is proved that $\Delta^{*}(G)$ is contained in
the set given above. Since this set contains all its divisors, $\Delta_{1}(G)$
is contained in it by 1. The set
$[1,\mathsf{r}(G)-1]\cup[\max\\{1,n-k-1\\},n-2]$ is contained in
$\Delta_{1}(G)$ by [8, Propositions 4.1.2 and 6.8.2]. ∎

##  3\. Minimal non-half-factorial subsets

Throughout this section, let $G$ be an additive finite abelian group

with $|G|\geq 3$, $\exp(G)=n$, and $\mathsf{r}(G)=r$.

The following two technical lemmas will be used throughout the manuscript.

######  Lemma 3.1.

Let $G_{0}\subset G$ a subset.

  1. 1.

For each $g\in G_{0}$,

$\displaystyle\gcd\big{(}\\{\mathsf{v}_{g}(B)\mid
B\in\mathcal{B}(G_{0})\\}\big{)}=\gcd\big{(}\\{\mathsf{v}_{g}(A)\mid
A\in\mathcal{A}(G_{0})\\}\big{)}$ $\displaystyle=$
$\displaystyle\min\big{(}\\{\mathsf{v}_{g}(A)\mid\mathsf{v}_{g}(A)>0,A\in\mathcal{A}(G_{0})\\}\big{)}=\min\big{(}\\{\mathsf{v}_{g}(B)\mid\mathsf{v}_{g}(B)>0,B\in\mathcal{B}(G_{0})\\}\big{)}$
$\displaystyle=$ $\displaystyle\min\big{(}\\{k\in\mathbb{N}\mid kg\in\langle
G_{0}\setminus\\{g\\}\rangle\\}\big{)}=\gcd\big{(}\\{k\in\mathbb{N}\mid
kg\in\langle G_{0}\setminus\\{g\\}\rangle\\}\big{)}\,.$

In particular, $\min\big{(}\\{k\in\mathbb{N}\mid kg\in\langle
G_{0}\setminus\\{g\\}\rangle\\}\big{)}$ divides $\operatorname{ord}(g)$.

  2. 2.

Suppose that for any $h\in G_{0}$, we have that $h\not\in\langle
G_{0}\setminus\\{h,\ h^{\prime}\\}\rangle$ for any $h^{\prime}\in
G_{0}\setminus\\{h\\}$. Then for any atom $A$ with
$\operatorname{supp}(A)\subsetneq G_{0}$ and any $h\in\operatorname{supp}(A)$,
we have $\gcd(\mathsf{v}_{h}(A),\operatorname{ord}(h))>1$.

  3. 3.

If $G_{0}$ is minimal non-half-factorial, then there exists a minimal non-
half-factorial subset $G_{0}^{*}\subset G$ with $|G_{0}|=|G_{0}^{*}|$ and a
transfer homomorphism
$\theta\colon\mathcal{B}(G_{0})\to\mathcal{B}(G_{0}^{*})$ such that the
following properties are satisfied :

    1. (a)

For each $g\in G_{0}^{*}$, we have $g\in\langle
G_{0}^{*}\setminus\\{g\\}\rangle$.

    2. (b)

For each $B\in\mathcal{B}(G_{0})$, we have
$\mathsf{k}(B)=\mathsf{k}\big{(}\theta(B)\big{)}$.

    3. (c)

If $G_{0}^{*}$ has the property that for each $h\in G_{0}^{*}$,
$h\not\in\langle E\rangle$ for any $E\subsetneq G_{0}^{*}\setminus\\{h\\}$,
then $G_{0}$ also has the property.

###### Proof.

See [15, Lemma 3.4]. ∎

######  Lemma 3.2.

  1. 1.

If $g\in G$ with $\operatorname{ord}(g)\geq 3$, then
$\operatorname{ord}(g)-2\in\Delta^{*}(G)$. In particular,
$n-2\in\Delta^{*}(G)$.

  2. 2.

If $r\geq 2$, then $[1,\mathsf{r}-1]\subset\Delta^{*}(G)$.

  3. 3.

Let $G_{0}\subset G$ a subset.

    1. (a)

If there exists an $U\in\mathcal{A}(G_{0})$ with $\mathsf{k}(U)<1$, then
$\min\Delta(G_{0})\leq\exp(G)-2$.

    2. (b)

If $G_{0}$ is an LCN-set, then $\min\Delta(G_{0})\leq|G_{0}|-2$.

###### Proof.

See [8, Proposition 6.8.2 and Lemmas 6.8.5 and 6.8.6]. ∎

######  Lemma 3.3.

Let $G_{0}\subset G$ be a subset, $g\in G_{0}$, and $s$ the smallest integer
such that $sg\in\langle G_{0}\setminus\\{g\\}\rangle$, and suppose that
$s<\operatorname{ord}(g)$. Then $\operatorname{ord}(sg)>1$ and for each prime
$p$ dividing $\operatorname{ord}(sg)$, there exists an atom
$A\in\mathcal{A}(G_{0})$ with
$2\leq|\operatorname{supp}(A)|\leq\mathsf{r}(G)+1$,
$s\leq\mathsf{v}_{g}(A)\leq\operatorname{ord}(g)/2$, and
$p\nmid\frac{\mathsf{v}_{g}(A)}{s}$. In particular,

  1. 1.

If $|G_{0}|\geq\mathsf{r}(G)+2$, then there exist
$s_{0}<\operatorname{ord}(g)$ and $E\subsetneq G_{0}\setminus\\{g\\}$ such
that $s_{0}g\in\langle E\rangle$.

  2. 2.

If $s=1$ and $\operatorname{ord}(g)$ is a prime power, then there exists a
subset $E\subset G_{0}\setminus\\{g\\}$ with $|E|\leq\mathsf{r}(G)$ such that
$g\in\langle E\rangle$.

###### Proof.

We set $\exp(G)=n=p_{1}^{k_{1}}\cdot\ldots\cdot p_{t}^{k_{t}}$, where
$t,k_{1},\ldots,k_{t}\in\mathbb{N}$ and $p_{1},\ldots,p_{t}$ are distinct
primes. Since $s<\operatorname{ord}(g)$, we have that
$\operatorname{ord}(sg)>1$. Let $\nu\in[1,t]$ with
$p_{\nu}\,|\,\operatorname{ord}(sg)$. Since $sg\in\langle
G_{0}\setminus\\{g\\}\rangle$, it follows that
$0\neq\frac{n}{p_{\nu}^{k_{\nu}}}sg\in
G_{\nu}=\langle\frac{n}{p_{\nu}^{k_{\nu}}}h\mid h\in
G_{0}\setminus\\{g\\}\rangle$. Obviously, $G_{\nu}$ is a $p_{\nu}$-group. Let
$E_{\nu}\subset G_{0}\setminus\\{g\\}$ be minimal such that
$\frac{n}{p_{\nu}^{k_{\nu}}}sg\in\langle\frac{n}{p_{\nu}^{k_{\nu}}}E_{\nu}\rangle$.
Since $\langle\frac{n}{p_{\nu}^{k_{\nu}}}E_{\nu}\rangle\subset G_{\nu}$ and
$G_{\nu}$ is a $p_{\nu}$-group, it follows that

$1\leq|E_{\nu}|=|\frac{n}{p_{\nu}^{k_{\nu}}}E_{\nu}|\leq\mathsf{r}(G_{\nu})\leq\mathsf{r}(G)\,.$

Let $d_{\nu}\in\mathbb{N}$ be minimal such that $d_{\nu}g\in\langle
E_{\nu}\rangle$. Since $0\neq\frac{n}{p_{\nu}^{k_{\nu}}}sg\in\langle
E_{\nu}\rangle$, it follows that $d_{\nu}<\operatorname{ord}(g)$. By Lemma
3.1.1, $d_{\nu}\,|\,\gcd(\frac{n}{p_{\nu}^{k_{\nu}}}s,\operatorname{ord}(g))$
and there exists an atom $U_{\nu}$ such that $\mathsf{v}_{g}(U_{\nu})=d_{\nu}$
and
$|\operatorname{supp}(U_{\nu})\setminus\\{g\\}|\leq|E_{\nu}|\leq\mathsf{r}(G)$.
Since $\mathsf{v}_{g}(U_{\nu})=d_{\nu}<\operatorname{ord}(g)$, it follows that
$|\operatorname{supp}(U_{\nu})|\geq 2$. By the minimality of $s$ and
$d_{\nu}\,|\,\frac{n}{p_{\nu}^{k_{\nu}}}s$, we have that $s\,|\,d_{\nu}$ and
$p_{\nu}\nmid\frac{d_{\nu}}{s}$.

If $|G_{0}|\geq\mathsf{r}(G)+2$, then
$|E_{\nu}|\leq\mathsf{r}(G)<|G_{0}\setminus\\{g\\}|$ implies that
$E_{\nu}\subsetneq G_{0}\setminus\\{g\\}$, and the assertion holds with
$E=E_{\nu}$ and $s_{0}=d_{\nu}$.

If $s=1$ and $\operatorname{ord}(g)$ is a prime power, then
$\operatorname{ord}(g)$ is a power of $p_{\nu}$ which implies that
$\gcd(\frac{n}{p_{\nu}^{k_{\nu}}}s,\operatorname{ord}(g))=1$ whence
$d_{\nu}=1$ and $g\in\langle E_{\nu}\rangle$. ∎

######  Lemma 3.4.

Let $G_{0}\subset G$ be a minimal non-half-factorial LCN-set with $|G_{0}|\geq
r+2$. Suppose that for any $h\in G_{0}$, $h\in\langle
G_{0}\setminus\\{h\\}\rangle$ but $h\not\in\langle G_{0}\setminus\\{h,\
h^{\prime}\\}\rangle$ for any $h^{\prime}\in G_{0}\setminus\\{h\\}$. Then
$|G_{0}|\leq r+\frac{n}{2}$. In particular, if each atom
$A\in\mathcal{A}(G_{0})$ with $\operatorname{supp}(A)=G_{0}$ has cross number
$\mathsf{k}(A)>1$, then $\min\Delta(G_{0})\leq\frac{5n}{6}-4$.

###### Proof.

We choose an element $g\in G_{0}$. If $\operatorname{ord}(g)$ is a prime
power, then there exists $E\subset G_{0}\setminus\\{g\\}$ such that
$g\in\langle E\rangle$ and $|E|\leq r<|G_{0}|-1$ by Lemma 3.3, a contradiction
to the assumption on $G_{0}$. Thus $\operatorname{ord}(g)$ is not a prime
power.

Let $s\in\mathbb{N}$ be minimal such that there exists a subset $E\subsetneq
G_{0}\setminus\\{g\\}$ with $sg\in\langle E\rangle$, and by Lemma 3.3.1, we
observe that $s<\operatorname{ord}(g)$. Let $E\subsetneq
G_{0}\setminus\\{g\\}$ be minimal such that $sg\in\langle E\rangle$. By Lemma
3.1.1, there is an atom $V$ with
$\mathsf{v}_{g}(V)=s\,|\,\operatorname{ord}(g)$ and
$\operatorname{supp}(V)=\\{g\\}\cup E\subsetneq G_{0}$. By Lemma 3.1.2, for
each $h\in\operatorname{supp}(V)$, $\mathsf{v}_{h}(V)\geq 2$. Since $G_{0}$ is
a minimal non-half-factorial LCN set, we obtain that

$1=\mathsf{k}(V)\geq\frac{2}{n}(|E|+1)\,,$

whence $|E|\leq\frac{n}{2}-1$.

Since $s\geq 2$, there is a prime $p\in\mathbb{N}$ dividing $s$ and hence
$p\,|\,s\,|\,\operatorname{ord}(g)$. By Lemma 3.3, there exists an atom
$U_{1}$ such that $|\operatorname{supp}(U_{1})|\leq r+1$ and
$p\nmid\mathsf{v}_{g}(U_{1})$, and therefore
$\operatorname{supp}(U_{1})\subsetneq G_{0}$.

Let $d=\gcd(s,\mathsf{v}_{g}(U_{1}))$ and
$E_{1}=\operatorname{supp}(U_{1})\setminus\\{g\\}$. Then $d<s$ and
$dg\in\langle sg,\mathsf{v}_{g}(U_{1})g\rangle\subset\langle E\cup
E_{1}\rangle\subset\langle G_{0}\setminus\\{g\\}\rangle$. The minimality of
$s$ implies that $E\cup E_{1}=G_{0}\setminus\\{g\\}$, and thus $|G_{0}|\leq
1+|E|+|E_{1}|\leq 1+r+\frac{n}{2}-1=r+\frac{n}{2}$.

Suppose that each atom $A\in\mathcal{A}(G_{0})$ with
$\operatorname{supp}(A)=G_{0}$ has cross number $\mathsf{k}(A)>1$. There exist
$x_{1}\in[1,\frac{\operatorname{ord}(g)}{s}-1]$ and
$x_{2}\in[1,\frac{\operatorname{ord}(g)}{\mathsf{v}_{g}(U_{1})}-1]$ such that
$dg=x_{1}sg+x_{2}\mathsf{v}_{g}(U_{1})g$. Thus
$d+y\operatorname{ord}(g)=x_{1}s+x_{2}\mathsf{v}_{g}(U_{1})$ with some
$y\in\mathbb{N}_{0}$. Let
$V^{x_{1}}U_{1}^{x_{2}}=(g^{\operatorname{ord}(g)})^{y}\cdot W$, where
$W\in\mathcal{B}(G)$ with $\mathsf{v}_{g}(W)=d$, and let $W_{1}$ be an atom
dividing $W$ with $\mathsf{v}_{g}(W_{1})>0$. Since $\mathsf{v}_{g}(W_{1})\leq
d<s$, the minimality of $s$ implies that $\operatorname{supp}(W_{1})=G_{0}$
and hence $\mathsf{k}(W_{1})>1$. Since $G_{0}$ is minimal non-half-factorial,
we have that $\mathsf{k}(V)=\mathsf{k}(U_{1})=1$. Therefore there exists
$l\in\mathbb{N}$ with $2\leq l<x_{1}+x_{2}$ such that
$\\{l,x_{1}+x_{2}\\}\subset\mathsf{L}(V^{x_{1}}U_{1}^{x_{2}})$. Then

$\min\Delta(G_{0})\leq
x_{1}+x_{2}-l\leq\frac{\operatorname{ord}(g)}{s}+\frac{\operatorname{ord}(g)}{\mathsf{v}_{g}(U_{1})}-4\leq\frac{5n}{6}-4\,.\qed$

For our next result we need the following technical lemma

######  Lemma 3.5.

Let $G_{0}\subset G$ be a non-half-factorial subset satisfying the following
two conditions:

  * (a)

There exists some $g\in G_{0}$ such that
$\Delta(G_{0}\setminus\\{g\\})=\emptyset$.

  * (b)

There exists some $U\in\mathcal{A}(G_{0})$ with $\mathsf{k}(U)=1$ and
$\gcd(\mathsf{v}_{g}(U),\operatorname{ord}(g))=1$.

Then $\mathsf{k}(\mathcal{A}(G_{0}))\subset\mathbb{N}$ and

$\min\Delta(G_{0})\mid\gcd\\{\mathsf{k}(A)-1\mid A\in\mathcal{A}(G_{0})\\}\,.$

Note that the conditions hold if $\Delta(G_{1})=\emptyset$ for each
$G_{1}\subsetneq G_{0}$ and there exists some $G_{2}$ such that $\langle
G_{2}\rangle=\langle G_{0}\rangle$ and $|G_{2}|\leq|G_{0}|-2$.

###### Proof.

The first statement follows from [8, Lemma 6.8.5]. If
$\Delta(G_{1})=\emptyset$ for all $G_{1}\subsetneq G_{0}$, then Condition (a)
holds. Let $G_{2}\subsetneq G_{1}\subsetneq G_{0}$ with $\langle
G_{2}\rangle=\langle G_{0}\rangle$. If $g\in G_{1}\setminus G_{2}$, then
$\langle G_{2}\rangle=\langle G_{0}\rangle$ implies that there is some
$U\in\mathcal{A}(G_{1})$ with $\mathsf{v}_{g}(U)=1$, and since
$G_{1}\subsetneq G_{0}$, it follows that $\mathsf{k}(U)=1$. ∎

######  Lemma 3.6.

Suppose that $\exp(G)=n$ is not a prime power. Let $G_{0}\subset G$ be a
minimal non-half-factorial LCN-set with $|G_{0}|\geq r+2$ such that
$h\in\langle G_{0}\setminus\\{h\\}\rangle$ for every $h\in G_{0}$. Suppose
that one of the following properties is satisfied :

  1. (a)

For any $h\in G_{0}$, $h\not\in\langle G_{0}\setminus\\{h,\
h^{\prime}\\}\rangle$ for any $h^{\prime}\in G_{0}\setminus\\{h\\}$ and there
exists an atom $A\in\mathcal{A}(G_{0})$ with $\mathsf{k}(A)=1$ and
$\operatorname{supp}(A)=G_{0}$.

  2. (b)

There is a subset $G_{2}\subset G_{0}$ such that $\langle G_{2}\rangle=\langle
G_{0}\rangle$ and $|G_{2}|\leq|G_{0}|-2$.

Then $\min\Delta(G_{0})\leq\frac{n+r-3}{2}$.

###### Proof.

Assume to the contrary that $\min\Delta(G_{0})\geq\frac{n+r}{2}-1$. Then Lemma
3.2.3.(b) implies that $|G_{0}|\geq\frac{n+r}{2}+1$. If Property (a) is
satisfied, then Lemma 3.1.2 implies that there exists some $g\in G_{0}$ such
that $\mathsf{v}_{g}(A)=1$. By Lemma 3.5, each of the two Properties (a) and
(b) implies that $\mathsf{k}(U)\in\mathbb{N}$ for each
$U\in\mathcal{A}(G_{0})$ and

$\min\Delta(G_{0})\,|\,\gcd\big{(}\\{\mathsf{k}(U)-1\mid
U\in\mathcal{A}(G_{0})\\}\big{)}\,.$

We set

$\Omega_{=1}=\\{A\in\mathcal{A}(G_{0})\mid\mathsf{k}(A)=1\\}\quad\text{and}\quad\Omega_{>1}=\\{A\in\mathcal{A}(G_{0})\mid\mathsf{k}(A)>1\\}\,.$

Thus for each $U_{1},U_{2}\in\Omega_{>1}$ we have

(3.1)
$\mathsf{k}(U_{1})\geq\frac{n+r}{2}\quad\text{and}\quad\big{(}\text{either}\
\mathsf{k}(U_{1})=\mathsf{k}(U_{2})\ \text{or}\
|\mathsf{k}(U_{1})-\mathsf{k}(U_{2})|\geq\frac{n+r}{2}-1\big{)}\,.$

Furthermore, for each $U\in\Omega_{=1}$ we have $\mathsf{h}(U)\geq 2$
(otherwise, $U$ would divide every atom $U_{1}\in\Omega_{>1}$). We claim that

  1. A1.

For each $U\in\Omega_{>1}$, there are $A_{1},\ldots,A_{m}\in\Omega_{=1}$,
where $m\leq\frac{n+1}{2}$, such that $UA_{1}\cdot\ldots\cdot A_{m}$ can be
factorized into a product of atoms from $\Omega_{=1}$.

Proof of A1. Suppose that Property (a) holds. As observed above there exists
some $g\in G_{0}$ such that $\mathsf{v}_{g}(A)=1$. Lemma 3.3 implies that
there is an atom $X$ such that
$2\leq|\operatorname{supp}(X)|\leq\mathsf{r}(G)+1$ and
$1\leq\mathsf{v}_{g}(X)\leq\operatorname{ord}(g)/2$. Since $g\not\in\langle
G_{0}\setminus\\{g,\ h\\}\rangle$ for any $h\in G_{0}\setminus\\{g\\}$, it
follows that $\mathsf{v}_{g}(X)\geq 2$, and $|G_{0}|\geq r+2$ implies
$\operatorname{supp}(X)\subsetneq G_{0}$.

Suppose that Property (b) is satisfied. We choose an element $g\in
G_{0}\setminus G_{2}$. Then $g\in\langle G_{2}\rangle$ and by Lemma 3.1.1,
there is an atom $A^{\prime}$ with $\mathsf{v}_{g}(A^{\prime})=1$ and
$\operatorname{supp}(A^{\prime})\subset G_{2}\cup\\{g\\}\subsetneq G_{0}$.
This implies that $A^{\prime}\in\Omega_{=1}$. Let $h\in G_{0}$ such that
$\mathsf{v}_{h}(A^{\prime})=\mathsf{h}(A^{\prime})$. Since
$\mathsf{h}(A^{\prime})\geq 2$, we obtain that
$A^{\prime\lceil\frac{\operatorname{ord}(h)}{\mathsf{h}(A^{\prime})}\rceil}=h^{\operatorname{ord}(h)}\cdot
W$ where $W$ is a product of
$\lceil\frac{\operatorname{ord}(h)}{\mathsf{h}(A^{\prime})}\rceil-1$ atoms and
$\mathsf{v}_{g}(W)=\lceil\frac{\operatorname{ord}(h)}{\mathsf{h}(A^{\prime})}\rceil$.
Thus there exists an atom $X^{\prime}$ with
$2\leq\mathsf{v}_{g}(X^{\prime})\leq\lceil\frac{\operatorname{ord}(h)}{\mathsf{h}(A^{\prime})}\rceil\leq\frac{n}{2}+1$.

Therefore both properties imply that there are $A,X\in\mathcal{A}(G_{0})$ and
$g\in G_{0}$ such that $\mathsf{k}(A)=\mathsf{k}(X)=1$, $\mathsf{v}_{g}(A)=1$,
and $2\leq\mathsf{v}_{g}(X)\leq\frac{n}{2}+1$. Let $U\in\Omega_{>1}$.

If
$\operatorname{ord}(g)-\mathsf{v}_{g}(U)<\mathsf{v}_{g}(X)\leq\frac{n}{2}+1$,
then

$UA^{\operatorname{ord}(g)-\mathsf{v}_{g}(U)}=g^{\operatorname{ord}(g)}S\,,$

where $S\in\mathcal{B}(G_{0})$ and
$\operatorname{ord}(g)-\mathsf{v}_{g}(U)\leq\frac{n}{2}$. Since
$\operatorname{supp}(S)\subsetneq G_{0}$, $S$ is a product of atoms from
$\Omega_{=1}$.

If $\operatorname{ord}(g)-\mathsf{v}_{g}(U)\geq\mathsf{v}_{g}(X)$, then

$UX^{\lfloor\frac{\operatorname{ord}(g)-\mathsf{v}_{g}(U)}{\mathsf{v}_{g}(X)}\rfloor}A^{\operatorname{ord}(g)-\mathsf{v}_{g}(U)-\mathsf{v}_{g}(X)\cdot\lfloor\frac{\operatorname{ord}(g)-\mathsf{v}_{g}(U)}{\mathsf{v}_{g}(X)}\rfloor}=g^{\operatorname{ord}(g)}S\,,$

where $S$ is a product of atoms from $\Omega_{=1}$ (because
$\operatorname{supp}(S)\subsetneq G_{0})$ and

$\displaystyle\lfloor$
$\displaystyle\frac{\operatorname{ord}(g)-\mathsf{v}_{g}(U)}{\mathsf{v}_{g}(X)}\rfloor+\operatorname{ord}(g)-\mathsf{v}_{g}(U)-\mathsf{v}_{g}(X)\cdot\lfloor\frac{\operatorname{ord}(g)-\mathsf{v}_{g}(U)}{\mathsf{v}_{g}(X)}\rfloor$
$\displaystyle\leq\frac{\big{(}\operatorname{ord}(g)-\mathsf{v}_{g}(U)\big{)}-\big{(}\mathsf{v}_{g}(X)-1\big{)}}{\mathsf{v}_{g}(X)}+\mathsf{v}_{g}(X)-1$
$\displaystyle\leq\frac{\operatorname{ord}(g)-\mathsf{v}_{g}(U)+1}{2}\leq\frac{n+1}{2}\,.\qquad\qquad\qquad\qed{\text{\rm(Proof
of {\bf A1})}}$

We set

$\Omega_{>1}^{\prime}=\\{A\in\mathcal{A}(G_{0})\mid\mathsf{k}(A)=\min\\{\mathsf{k}(B)\mid
B\in\Omega_{>1}\\}\\}\subset\Omega_{>1}\,,$

and we consider all tuples $(U,A_{1},\ldots,A_{m})$, where
$U\in\Omega_{>1}^{\prime}$ and $A_{1},\ldots,A_{m}\in\Omega_{=1}$, such that
$UA_{1}\cdot\ldots\cdot A_{m}$ can be factorized into a product of atoms from
$\Omega_{=1}$. We fix one such tuple $(U,A_{1},\ldots,A_{m})$ with the
property that $m$ is minimal possible. Let

(3.2) $UA_{1}\cdot\ldots\cdot A_{m}=V_{1}\cdot\ldots\cdot V_{t}\quad\text{
with}\quad t\in\mathbb{N}\quad\text{and}\quad
V_{1},\ldots,V_{t}\in\Omega_{=1}\,.$

We observe that $\mathsf{k}(U)=t-m$ and continue with the following assertion.

  1. A2.

For each $\nu\in[1,t]$, we have $V_{\nu}\nmid UA_{1}\cdot\ldots\cdot A_{m-1}$.

Proof of A2. Assume to the contrary that there is such a $\nu\in[1,t]$, say
$\nu=1$, with $V_{1}\,|\,UA_{1}\cdot\ldots\cdot A_{m-1}$. Then there are
$l\in\mathbb{N}$ and $T_{1},\ldots,T_{l}\in\mathcal{A}(G_{0})$ such that

$UA_{1}\cdot\ldots\cdot A_{m-1}=V_{1}T_{1}\cdot\ldots\cdot T_{\mathit{l}}\,.$

By the minimality of $m$, there exists some $\nu\in[1,l]$ such that
$T_{\nu}\in\Omega_{>1}$, say $\nu=1$. Since

$\sum_{\nu=2}^{l}\mathsf{k}(T_{\nu})=\mathsf{k}(U)+(m-1)-1-\mathsf{k}(T_{1})\leq
m-2\leq\frac{n-3}{2}\,,$

and $\mathsf{k}(T^{\prime})\geq\frac{r+n}{2}$ for all
$T^{\prime}\in\Omega_{>1}$, it follows that
$T_{2},\ldots,T_{l}\in\Omega_{=1}$, whence
$l=1+\sum_{\nu=2}^{l}\mathsf{k}(T_{\nu})\leq m-1$. We obtain that

$V_{1}T_{1}\cdot\ldots\cdot T_{\mathit{l}}A_{m}=UA_{1}\cdot\ldots\cdot
A_{m}=V_{1}\cdot\ldots\cdot V_{t}\,,$

and thus

$T_{1}\cdot\ldots\cdot T_{\mathit{l}}A_{m}=V_{2}\cdot\ldots\cdot V_{t}\,.$

The minimality of $m$ implies that $\mathsf{k}(T_{1})>\mathsf{k}(U)$. It
follows that

$\mathsf{k}(T_{1})-\mathsf{k}(U)=m-1-{\mathit{l}}\leq
m-2\leq\frac{n-3}{2}<\frac{r+n}{2}-1\leq\mathsf{k}(T_{1})-\mathsf{k}(U)\,,$

a contradiction. ∎(Proof of A2)

By Equation (3.2), there are $X_{1},Y_{1},\ldots,X_{t},Y_{t}\in\mathcal{F}(G)$
such that

$UA_{1}\cdot\ldots\cdot A_{m-1}=X_{1}\cdot\ldots\cdot X_{t},\
A_{m}=Y_{1}\cdot\ldots\cdot Y_{t},\ \text{and}\ V_{i}=X_{i}Y_{i}\ \text{ for
each}\ i\in[1,t]\,.$

Then A2 implies that $|Y_{i}|\geq 1$ for each $i\in[1,t]$, and we set
$\alpha=|\\{i\in[1,t]\mid|Y_{i}|=1\\}|$. If $\alpha\leq m+r$, then

$n\geq|A_{m}|=|Y_{1}|+\ldots+|Y_{t}|\geq\alpha+2(t-\alpha)=2t-\alpha\geq
2t-m-r\,,$

and hence $\min\Delta(G_{0})\leq t-1-m\leq\frac{r+n-3}{2}$, a contradiction.
Thus $\alpha\geq m+r+1$. After renumbering if necessary we assume that
$1=|Y_{1}|=\ldots=|Y_{\alpha}|<|Y_{\alpha+1}|\leq\ldots\leq|Y_{t}|$. Let
$Y_{i}=y_{i}$ for each $i\in[1,\alpha]$ and

(3.3) $S_{0}=\\{y_{1},y_{2},\ldots,y_{\alpha}\\}\,.$

For every $i\in[1,\alpha]$, $V_{i}\,|\,y_{i}UA_{1}\cdot\ldots\cdot A_{m-1}$
whence $\mathsf{v}_{y_{i}}(V_{i})\leq
1+\mathsf{v}_{y_{i}}(UA_{1}\cdot\ldots\cdot A_{m-1})$ and since $V_{i}\nmid
UA_{1}\cdot\ldots\cdot A_{m-1}$, it follows that

(3.4) $\mathsf{v}_{y_{i}}(V_{i})=\mathsf{v}_{y_{i}}(UA_{1}\cdot\ldots\cdot
A_{m-1})+1\,.$

Assume to the contrary that there are distinct $i,j\in[1,\alpha]$ such that
$y_{i}=y_{j}$. Then

$\mathsf{v}_{y_{i}}(UA_{1}\ \cdot\ldots\cdot
A_{m-1})+1=\mathsf{v}_{y_{i}}(V_{i})=\mathsf{v}_{y_{i}}(X_{i})+1=\mathsf{v}_{y_{i}}(V_{j})=\mathsf{v}_{y_{i}}(X_{j})+1\,.$

Since $X_{i}X_{j}\,|\,UA_{1}\ \cdot\ldots\cdot A_{m-1}$, we infer that

$\mathsf{v}_{y_{i}}(UA_{1}\ldots
A_{m-1})\geq\mathsf{v}_{y_{i}}(X_{i}X_{j})=\mathsf{v}_{y_{i}}(V_{i}V_{j})-2=2\mathsf{v}_{y_{i}}(UA_{1}\ldots
A_{m-1})\,,$

which implies that $\mathsf{v}_{y_{i}}(UA_{1}\ldots A_{m-1})=0$, a
contradiction to $\operatorname{supp}(U)=G_{0}$. Thus $|S_{0}|=\alpha$ and

(3.5) $|\operatorname{supp}(A_{m})|\geq|S_{0}|=\alpha\geq m+r+1\,.$

We proceed by the following two assertions.

  1. A3.

There exist $g^{\prime}\in G_{0}$ and $A^{\prime}\in\mathcal{A}(G_{0})$ with
$\mathsf{k}(A^{\prime})=1$ satisfying the following three conditions:

    1. (C1)

$\mathsf{v}_{g^{\prime}}(A^{\prime})<\operatorname{ord}(g^{\prime})$ is the
smallest positive integer $\gamma$ such that $\gamma
g^{\prime}\in\langle\operatorname{supp}(A^{\prime})\setminus\\{g^{\prime}\\}\rangle$;

    2. (C2)

$\mathsf{v}_{g^{\prime}}(A^{\prime})g^{\prime}\notin\langle E\rangle$ for any
$E\subsetneq\operatorname{supp}(A^{\prime})\setminus\\{g^{\prime}\\})$.

    3. (C3)

$UA_{1}\cdot\ldots\cdot A_{m-1}\cdot A^{\prime}$ can be factorized into a
product of atoms from $\Omega_{=1}$.

Proof of A3. Suppose that Property (a) is satisfied. As observed at the
beginning of the proof, there is a $g\in G_{0}$ such that
$\mathsf{v}_{g}(A)=1$. We choose $A^{\prime}=A$ and $g^{\prime}=g$, and we
need to prove that $UA_{1}\cdot\ldots\cdot A_{m-1}\cdot A$ can be factorized
into a product of atoms from $\Omega_{=1}$. Since
$S_{0}\subset\operatorname{supp}(A)=G_{0}$, then $V_{1}\cdot\ldots\cdot
V_{\alpha}\,|\,UA_{1}\cdot\ldots\cdot A_{m-1}\cdot A$ and hence
$\mathsf{k}(UA_{1}\cdot\ldots\cdot A_{m-1}\cdot A(V_{1}\cdot\ldots\cdot
V_{\alpha})^{-1})<\mathsf{k}(U)$. The minimality of $\mathsf{k}(U)$ implies
that $UA_{1}\cdot\ldots\cdot A_{m-1}\cdot A$ can be factorized into a product
of atoms from $\Omega_{=1}$.

Suppose that Property (b) satisfied. We choose $g^{\prime}=y_{1}$(see Equation
(3.3)) and distinguish two cases. First, suppose that there exists a subset
$E\subsetneq G_{0}\setminus\\{y_{1}\\}$ such that $y_{1}\in\langle E\rangle$.
Choose a minimal subset $E$ with this property. By Lemma 3.1.1, there exists
an atom $A^{\prime}$ satisfying the two conditions (C1) and (C2) with
$\mathsf{k}(A^{\prime})=1$ and $\mathsf{v}_{y_{1}}(A^{\prime})=1$. Since
$\mathsf{v}_{y_{1}}(V_{1})=\mathsf{v}_{y_{1}}(UA_{1}\cdot\ldots\cdot
A_{m-1})+1$ by Equation 3.4 and $V_{1}\,|\,UA_{1}\cdot\ldots\cdot A_{m-1}\cdot
y_{1}$, we obtain that $|\operatorname{supp}(UA_{1}\cdot\ldots\cdot
A_{m-1}\cdot A^{\prime}(V_{1})^{-1})|<|G_{0}|$ and hence
$UA_{1}\cdot\ldots\cdot A_{m-1}\cdot A^{\prime}$ can be factorized into a
product of atoms from $\Omega_{=1}$.

Now we suppose that $y_{1}\notin\langle E\rangle$ for any $E\subsetneq
G_{0}\setminus\\{y_{1}\\}$. Let $s_{0}\in\mathbb{N}$ be minimal such that
there exists a subset $E\subsetneq G_{0}\setminus\\{y_{1}\\}$ such that
$s_{0}y_{1}\in\langle E\rangle$, and by Lemma 3.3.1, we observe that
$s_{0}<\operatorname{ord}(g)$. Let $E$ be a minimal subset with this property.
Thus, by Lemma 3.1.1, there exists an atom $A^{\prime}$ with
$\mathsf{v}_{y_{1}}(A^{\prime})=s_{0}$ satisfying the two conditions (C1) and
(C2). Since $\operatorname{supp}(A^{\prime})\subsetneq G_{0}$, we have
$\mathsf{k}(A^{\prime})=1$. We distinguish two cases:

CASE 1: $|S_{0}\setminus\operatorname{supp}(A^{\prime})|\geq r+1$.

Since $s_{0}\geq 2$, there is a prime $p$ dividing $s_{0}$. Since by
assumption, $y_{1}\in\langle G_{0}\setminus\\{y_{1}\\}\rangle$, Lemma 3.3
implies that for each prime $p$ dividing $\operatorname{ord}(y_{1})$, there
exists an atom $A_{p}^{\prime}$ such that
$|\operatorname{supp}(A_{p}^{\prime})|\leq r+1<|G_{0}|$,
$1\leq\mathsf{v}_{y_{1}}(A_{p}^{\prime})\leq\operatorname{ord}(y_{1})/2$, and
$p\nmid\mathsf{v}_{y_{1}}(A_{p}^{\prime})$.

Let $d=\gcd(s_{0},\mathsf{v}_{y_{1}}(A_{p}^{\prime}))$. Then $d<s_{0}$ and
$dy_{1}\in\langle
s_{0}y_{1},\mathsf{v}_{y_{1}}(A_{p}^{\prime})y_{1}\rangle\subset\langle(\operatorname{supp}(A^{\prime})\cup\operatorname{supp}(A_{p}^{\prime}))\setminus\\{y_{1}\\}\rangle$.
By the minimality of $s_{0}$, we have
$G_{0}\setminus\\{y_{1}\\}=(\operatorname{supp}(A^{\prime})\cup\operatorname{supp}(A_{p}^{\prime}))\setminus\\{y_{1}\\}$.
It follows that

$\displaystyle|\operatorname{supp}(A^{\prime})|+r$
$\displaystyle\geq|\operatorname{supp}(A^{\prime})|+|\operatorname{supp}(A_{p}^{\prime})|-1\geq|G_{0}|$
$\displaystyle\geq|\operatorname{supp}(A^{\prime})|+|S_{0}\setminus\operatorname{supp}(A^{\prime})|\geq|\operatorname{supp}(A^{\prime})|+r+1\,,$

a contradiction.

CASE 2: $|S_{0}\setminus\operatorname{supp}(A^{\prime})|\leq r$.

Therefore $|\operatorname{supp}(A^{\prime})\cap S_{0}|\geq m+1$ by Equation
(3.5), and we may suppose that
$\\{y_{1},\ldots,y_{m+1}\\}\subset\operatorname{supp}(A^{\prime})\cap S_{0}$.
Then $V_{1}\cdot\ldots\cdot V_{m+1}\,|\,UA_{1}\cdot\ldots\cdot
A_{m-1}A^{\prime}$ and $\mathsf{k}(UA_{1}\cdot\ldots\cdot
A_{m-1}A^{\prime}(V_{1}\cdot\ldots\cdot V_{m+1})^{-1})<\mathsf{k}(U)$. By the
minimality of $\mathsf{k}(U)$, we have that $UA_{1}\cdot\ldots\cdot
A_{m-1}A^{\prime}$ can be factorized into a product of atoms from
$\Omega_{=1}$. ∎(Proof of A3)

  1. A4.

Let $g^{\prime}\in G_{0}$ and $A^{\prime}\in\mathcal{A}(G_{0})$ with
$\mathsf{k}(A^{\prime})=1$ satisfying the following three conditions:

    1. (C1)

$\mathsf{v}_{g^{\prime}}(A^{\prime})<\operatorname{ord}(g^{\prime})$ is the
smallest positive integer $\gamma$ such that $\gamma
g^{\prime}\in\langle\operatorname{supp}(A^{\prime})\setminus\\{g^{\prime}\\}\rangle$;

    2. (C2)

$\mathsf{v}_{g^{\prime}}(A^{\prime})g^{\prime}\notin\langle E\rangle$ for any
$E\subsetneq\operatorname{supp}(A^{\prime})\setminus\\{g^{\prime}\\})$.

    3. (C3)

$UA_{1}\cdot\ldots\cdot A_{m-1}\cdot A^{\prime}$ can be factorized into a
product of atoms from $\Omega_{=1}$.

If $|\operatorname{supp}(A^{\prime})|\geq m+r+1$, then there exists an atom
$A^{\prime\prime}\in\mathcal{A}(G_{0})$ with $\mathsf{k}(A^{\prime\prime})=1$
and
$|\operatorname{supp}(A^{\prime\prime})|<|\operatorname{supp}(A^{\prime})|$
such that (C1), (C2), and (C3) hold.

Suppose that A4 holds. Iterating A4 we find an atom $A^{*}$ with
$|\operatorname{supp}(A^{*})|\leq m+r$ such that $UA_{1}\cdot\ldots\cdot
A_{m-1}\cdot A^{*}$ can be factorized into a product of atoms from
$\Omega_{=1}$, a contradiction to (3.5).

Proof of A4. For simplicity of notation, we suppose that $A^{\prime}=A_{m}$.

Let $s_{0}\in\mathbb{N}$ be minimal such that there exists a subset
$E\subsetneq\operatorname{supp}(A_{m})\setminus\\{g^{\prime}\\}$ such that
$s_{0}g^{\prime}\in\langle E\rangle$. By (C1) and
$|\operatorname{supp}(A^{\prime})|\geq m+r+1\geq r+2$, Lemma 3.3 implies that
$s_{0}<\operatorname{ord}(g^{\prime})$. Let $E$ be a minimal subset with this
property. Thus, by Lemma 3.1.1, there exists an atom $A^{\prime\prime}$ with
$\mathsf{v}_{g^{\prime}}(A^{\prime\prime})=s_{0}$ satisfying the two
conditions (C1) and (C2). Since
$\operatorname{supp}(A^{\prime\prime})\subsetneq G_{0}$, we have
$\mathsf{k}(A^{\prime\prime})=1$. We distinguish two cases:

CASE 1: $|S_{0}\setminus\operatorname{supp}(A^{\prime\prime})|\geq r+1$.

We set
$s^{\prime}=\mathsf{v}_{g^{\prime}}(A_{m})<\operatorname{ord}(g^{\prime})$.
Since $A_{m}$ satisfies condition (C1), Lemma 3.1.1 implies that
$s^{\prime}\,|\,s_{0}$ and $\frac{s_{0}}{s^{\prime}}>1$. Let $p$ be a prime
dividing $\frac{s_{0}}{s^{\prime}}$. Since
$s^{\prime}\,|\,\operatorname{ord}(g^{\prime})$ and
$s_{0}\,|\,\operatorname{ord}(g^{\prime})$, it follows that
$p\,|\,\frac{s_{0}}{s^{\prime}}\,|\,\frac{\operatorname{ord}(g^{\prime})}{s^{\prime}}=\operatorname{ord}(s^{\prime}g^{\prime})$.
Lemma 3.3 (applied to the subset $\operatorname{supp}(A_{m})\subset G$)
implies that there exists an atom
$A_{p}^{\prime}\in\mathcal{A}(\operatorname{supp}(A_{m}))$ such that
$|\operatorname{supp}(A_{p}^{\prime})|\leq r+1<|\operatorname{supp}(A_{m})|$,
$s^{\prime}\leq\mathsf{v}_{g^{\prime}}(A_{p}^{\prime})\leq\operatorname{ord}(g^{\prime})/2$,
and $p\nmid\frac{\mathsf{v}_{g^{\prime}}(A_{p}^{\prime})}{s^{\prime}}$.

Let
$d=\gcd(\frac{s_{0}}{s^{\prime}},\frac{\mathsf{v}_{g^{\prime}}(A_{p}^{\prime})}{s^{\prime}})$.
Then $d<\frac{s_{0}}{s^{\prime}}$ and

$ds^{\prime}g^{\prime}\in\langle
s_{0}g^{\prime},\mathsf{v}_{g^{\prime}}(A_{p}^{\prime})g^{\prime}\rangle\subset\langle(\operatorname{supp}(A^{\prime\prime})\cup\operatorname{supp}(A_{p}^{\prime}))\setminus\\{g^{\prime}\\}\rangle\,.$

Thus by minimality of $s_{0}$, we have
$\operatorname{supp}(A_{m})\setminus\\{g^{\prime}\\}=(\operatorname{supp}(A^{\prime\prime})\cup\operatorname{supp}(A_{p}^{\prime}))\setminus\\{g^{\prime}\\}$.
It follows that

$\displaystyle|\operatorname{supp}(A^{\prime\prime})|+r$
$\displaystyle\geq|\operatorname{supp}(A^{\prime\prime})|+|\operatorname{supp}(A_{p}^{\prime})|-1\geq|\operatorname{supp}(A_{m})|$
$\displaystyle\geq|\operatorname{supp}(A^{\prime\prime})|+|S_{0}\setminus\operatorname{supp}(A^{\prime\prime})|\geq|\operatorname{supp}(A^{\prime\prime})|+r+1\,,$

a contradiction.

CASE 2: $|S_{0}\setminus\operatorname{supp}(A^{\prime\prime})|\leq r$.

Therefore $|\operatorname{supp}(A^{\prime\prime})\cap S_{0}|\geq m+1$ by
Equation (3.5), and we may suppose that
$\\{y_{1},\ldots,y_{m+1}\\}\subset\operatorname{supp}(A^{\prime\prime})\cap
S_{0}$. Then $V_{1}\cdot\ldots\cdot V_{m+1}\,|\,UA_{1}\cdot\ldots\cdot
A_{m-1}A^{\prime\prime}$ and $\mathsf{k}(UA_{1}\cdot\ldots\cdot
A_{m-1}A^{\prime\prime}(V_{1}\cdot\ldots\cdot V_{m+1})^{-1})<\mathsf{k}(U)$.
By the minimality of $\mathsf{k}(U)$, we have that $UA_{1}\cdot\ldots\cdot
A_{m-1}A^{\prime\prime}$ can be factorized into a product of atoms from
$\Omega_{=1}$. This completes the proof of (A4) and thus Lemma 3.6 is proved.
∎

######  Proposition 3.7.

We have
$\mathsf{m}(G)\leq\min\\{\frac{n}{2}+r-2,\max\\{r-1,\frac{5n}{6}-4,\frac{n+r-3}{2}\\}\\}$.

###### Proof.

Let $G_{0}\subset G$ be a non-half-factorial LCN set. We have to prove that

$\min\Delta(G_{0})\leq\min\\{\frac{n}{2}+r-2,\max\\{r-1,\frac{5n}{6}-4,\frac{n+r-3}{2}\\}\\}\,.$

If $G_{1}\subset G_{0}$ is non-half-factorial, then
$\min\Delta(G_{0})=\gcd\Delta(G_{0})\,|\,\gcd\Delta(G_{1})=\min\Delta(G_{1})$.
Thus we may suppose that $G_{0}$ is minimal non-half-factorial. By Lemma
3.1.3.(a), we may suppose that $g\in\langle G_{0}\setminus\\{g\\}\rangle$ for
all $g\in G_{0}$.

If $n$ is a prime power, then $\mathsf{m}(G)=r-1$ by Proposition 2.2, and the
assertion follows. Suppose that $n$ is not a prime power. If $|G_{0}|\leq
r+1$, then $\min\Delta(G_{0})\leq|G_{0}|-2\leq r-1$ by Lemma 3.2.3. Thus we
may suppose that $|G_{0}|\geq r+2$ and we distinguish two cases.

CASE 1: There exists a subset $G_{2}\subset G_{0}$ such that $\langle
G_{2}\rangle=\langle G_{0}\rangle$ and $|G_{2}|\leq|G_{0}|-2$.

Then Lemma 3.6 implies that $\min\Delta(G_{0})\leq\frac{n+r-3}{2}$.

CASE 2: Every subset $G_{1}\subset G_{0}$ with $|G_{1}|=|G_{0}|-1$ is a
minimal generating set of $\langle G_{0}\rangle$.

Then for each $h\in G_{0}$, $G_{0}\setminus\\{h\\}$ is half-factorial and
$h\notin\langle G_{0}\setminus\\{h,h^{\prime}\\}\rangle\ \text{for any }\
h^{\prime}\in G_{0}\setminus\\{h\\}$. Thus Lemma 3.4 and Lemma 3.6 imply that
$\min\Delta(G_{0})\leq\max\\{\frac{5n}{6}-4,\frac{n+r-3}{2}\\}$. By Lemma 3.4,
we obtain that $|G_{0}|\leq r+\frac{n}{2}$. Therefore Lemma 3.2.3 implies that
$\min\Delta(G_{0})\leq\min\\{r+\frac{n}{2}-2,\max\\{\frac{5n}{6}-4,\frac{n+r-3}{2}\\}\\}$.
∎

######  Proposition 3.8.

Let $G^{\prime}$ be a finite abelian group with
$\mathcal{L}(G)=\mathcal{L}(G^{\prime})$. If $r\in[2,(n-2)/4]$, then
$n=\exp(G^{\prime})>\mathsf{r}(G^{\prime})+1$.

###### Proof.

Let $k\in\mathbb{N}$ be maximal such that $G$ has a subgroup isomorphic to
$C_{n}^{k}$. Then $k\leq r\leq\frac{n-2}{4}\leq\frac{n-3}{2}$. By Proposition
3.7, we obtain that

$\mathsf{m}(G)\leq\frac{n}{2}+r-2\leq n-2\leq n-k-3\,$

and hence

$\max\\{\mathsf{m}(G),\lfloor\frac{n}{2}\rfloor-1\\}\leq n-k-3\,.$

By Proposition 2.2.3, we have

$\Delta_{1}(G)\subset[1,\max\\{\mathsf{m}(G),\lfloor\frac{n}{2}\rfloor-1\\}]\cup[n-k-1,n-2]\,,$

and thus $n-k-2\not\in\Delta_{1}(G)$. Thus
$n-k-2\not\in\Delta_{1}(G^{\prime})$ and Proposition 2.2 implies that

$n-2=\max\\{\mathsf{m}(G),n-2\\}=\max\Delta_{1}(G)=\max\Delta_{1}(G^{\prime})=\max\\{\mathsf{r}(G^{\prime})-1,\exp(G^{\prime})-2\\}\,.$

If $n-2=\mathsf{r}(G^{\prime})-1$, then $\Delta_{1}(G^{\prime})=[1,n-2]$ by
Lemma 3.2.2, a contradiction to $n-k-2\not\in\Delta_{1}(G^{\prime})$.
Therefore it follows that $n=\exp(G^{\prime})>\mathsf{r}(G^{\prime})+1$. ∎

##  4\. Proof of the Main Result and groups with small Davenport constant

###### Proof of Theorem 1.1.

Let $G$ be an abelian group such that $\mathcal{L}(G)=\mathcal{L}(C_{n}^{r})$
where $r,n\in\mathbb{N}$ with $n\geq 2$, $(n,r)\notin\\{(2,1),(2,2),(3,1)\\}$,
and $r\leq\max\\{2,(n+2)/6\\}$.

First we note that $G$ has to be finite and that
$\mathsf{D}(C_{n}^{r})=\mathsf{D}(G)$ and
$\Delta_{1}(C_{n}^{r})=\Delta_{1}(G)$ (see [8, Proposition 7.3.1 and Theorem
7.4.1]). If $r=1$, then the assertion follows from [8, Theorem 7.3.3]. If
$r=2$, then the assertion follows from [21], and hence we may suppose that
$r\in[3,(n+2)/6]$.

Let $k\in\mathbb{N}$ be maximal such that $G$ has a subgroup isomorphic to
$C_{n}^{k}$. If $k\geq r$, then
$\mathsf{D}(C_{n}^{r})=\mathsf{D}(G)\geq\mathsf{D}(C_{n}^{k})$ implies that
$k=r$ and that $G\cong C_{n}^{r}$. Suppose that $k<r$. By Proposition 3.8, we
obtain that $n=\exp(G)>\mathsf{r}(G)+1$. By Proposition 2.2.3 (applied to
$C_{n}^{r}$) we infer that
$[n-r-1,n-2]\subset\Delta_{1}(C_{n}^{r})=\Delta_{1}(G)$. By Proposition 2.2.3
(applied to $G$), we obtain that

$[1,\mathsf{r}(G)-1]\cup[n-r-1,n-2]\subset\Delta_{1}(G)\subset\big{[}1,\max\big{\\{}\mathsf{m}(G),\lfloor\frac{n}{2}\rfloor-1\big{\\}}\big{]}\cup[n-k-1,n-2]\,,$

which implies that $\mathsf{m}(G)\geq n-r-1$. By Proposition 3.7, we have that

$n-r-1\leq\mathsf{m}(G)\leq\max\Big{\\{}\mathsf{r}(G)-1,\frac{5n}{6}-4,\frac{n+\mathsf{r}(G)-3}{2}\Big{\\}}\,.$

If $n-r-1\leq\frac{5n}{6}-4$, then $r\geq\frac{n}{6}+3$, a contradiction. Thus
$n-r-1\leq\max\big{\\{}\mathsf{r}(G)-1,\frac{n+\mathsf{r}(G)-3}{2}\big{\\}}$
which implies that $\mathsf{r}(G)\geq n-2r+1$. Therefore

(4.1)
$[1,n-2r]\subset[1,\mathsf{r}(G)-1]\subset\Delta_{1}(G)=\Delta_{1}(C_{n}^{r})\,.$

Since $r\leq\frac{n+2}{6}$, we have that
$n-2r-1\geq\frac{n}{2}+r-2\geq\lfloor\frac{n}{2}\rfloor-1$. By Proposition
3.7, we obtain that $\mathsf{m}(C_{n}^{r})\leq\frac{n}{2}+r-2\leq n-2r-1$.
Therefore

$\max\\{\mathsf{m}(C_{n}^{r}),\lfloor\frac{n}{2}\rfloor-1\\}<n-2r<n-r-1\,.$

By Proposition 2.2.3, $n-2r\notin\Delta_{1}(C_{n}^{r})$, a contradiction to
Equation (4.1). ∎

Our proof of Theorem 1.1, to characterize the groups $C_{n}^{r}$ with $r,n$ as
above, uses only the Davenport constant and the set of minimal distances.
Clearly, there are non-isomorphic groups $G$ and $G^{\prime}$ with
$\mathsf{D}(G)=\mathsf{D}(G^{\prime})$,
$\Delta^{*}(G)=\Delta^{*}(G^{\prime})$, and
$\Delta_{1}(G)=\Delta_{1}(G^{\prime})$. We meet this phenomenon in Proposition
4.1. Indeed, since $\mathcal{L}(C_{1})=\mathcal{L}(C_{2})$ and
$\mathcal{L}(C_{3})=\mathcal{L}(C_{2}\oplus C_{2})$ ([8, Theorem 7.3.2]),
small groups definitely deserve a special attention when studying the
Characterization Problem. Clearly, the groups $C_{1},C_{2},C_{3}$, and
$C_{2}\oplus C_{2}$ are precisely the groups $G$ with $\mathsf{D}(G)\leq 3$.
In our final result we show that for all groups $G$ with
$\mathsf{D}(G)\in[4,11]$ the answer to the Characterization Problem is
positive.

Suppose that $G\cong C_{n_{1}}\oplus\ldots\oplus C_{n_{r}}$ where
$r\in\mathbb{N}_{0}$ and $1<n_{1}\,|\,\ldots\,|\,n_{r}$ and set
$\mathsf{D}^{*}(G)=1+\sum_{i=1}^{r}(n_{i}-1)$. Then
$\mathsf{D}^{*}(G)\leq\mathsf{D}(G)$. If $\mathsf{r}(G)=r\leq 2$ or if $G$ is
a $p$-group, then equality holds.

######  Proposition 4.1.

Let $G$ be a finite abelian group with $\mathsf{D}(G)\in[4,11]$. If
$G^{\prime}$ is a finite abelian group with
$\mathcal{L}(G)=\mathcal{L}(G^{\prime})$, then $G\cong G^{\prime}$.

###### Proof.

Suppose that $G^{\prime}$ is a finite abelian group with
$\mathcal{L}(G)=\mathcal{L}(G^{\prime})$. Then
$\mathsf{D}(G)=\mathsf{D}(G^{\prime})$. If $\mathsf{D}(G)\in[4,10]$, then the
assertion follows from [21, Theorem 6.2].

Suppose that $\mathsf{D}(G)=\mathsf{D}(G^{\prime})=11$. If $\mathsf{r}(G)\leq
2$ or $\mathsf{r}(G^{\prime})\leq 2$, then the assertion follows from [12,
Theorem 1.1]. If $G$ or $G^{\prime}$ is an elementary $2$-group, then the
assertion follows from [8, Theorem 7.3.3].

Thus we suppose that $\mathsf{r}(G)\geq 3$, $\mathsf{r}(G^{\prime})\geq 3$,
$\exp(G)\in[3,8]$, and $\exp(G^{\prime})\in[3,8]$. If $G\cong C_{4}^{3}$ or
$G^{\prime}\cong C_{4}^{3}$, then the assertion follows from [12, Theorem
4.1]. Thus we may suppose that all this is not the case. Since there is no
finite abelian group $H$ with $\mathsf{D}(H)=11$ and $\exp(G)\in\\{5,7\\}$, it
remains to consider the following groups:

$C_{2}^{4}\oplus C_{4}^{2},\ C_{2}^{7}\oplus C_{4},\ \ C_{2}^{r}\oplus C_{6},\
C_{2}^{3}\oplus C_{8},\ C_{3}^{5}\quad\text{for some}\ r\in\mathbb{N}\,.$

By [14, Corollary 2], $\mathsf{D}(C_{2}^{r}\oplus
C_{6})=\mathsf{D}^{*}(G)=r+6$ if and only if $r\leq 3$. Thus
$\mathsf{D}(C_{2}^{4}\oplus C_{6})\geq 11$, and this is the only group for
which $\mathsf{D}(C_{2}^{r}\oplus C_{6})=11$ is possible. Thus we have to
consider

$G_{1}=C_{2}^{4}\oplus C_{4}^{2},\ G_{2}=C_{2}^{7}\oplus C_{4},\ \
G_{4}=C_{2}^{4}\oplus C_{6},\ G_{5}=C_{2}^{3}\oplus C_{8},\ \text{and}\
G_{6}=C_{3}^{5}\,.$

Since $\max\Delta^{*}(G_{1})=5$, $\max\Delta^{*}(G_{2})=7$,
$\max\Delta^{*}(G_{4})=4$, $\max\Delta^{*}(G_{5})=6$, and
$\max\Delta^{*}(G_{6})=4$, it remains to show that
$\mathcal{L}(C_{2}^{4}\oplus C_{6})\neq\mathcal{L}(C_{3}^{5})$. Note that
Proposition 2.2 implies that $\Delta^{*}(C_{2}^{4}\oplus
C_{6})=[1,4]=\Delta^{*}(C_{3}^{5})$. By [8, Theorem 6.6.2], it follows that
$\\{2,8\\}\in\mathcal{L}(C_{2}^{4}\oplus C_{6})$, and we assert that
$\\{2,8\\}\notin\mathcal{L}(C_{3}^{5})$.

Assume to the contrary that $\\{2,8\\}\in\mathcal{L}(C_{3}^{5})$. Then there
exists $U,V\in\mathcal{A}(C_{3}^{5})$ such that $\mathsf{L}(UV)=\\{2,8\\}$. We
choose the pair $(U,V)$ such that $|U|$ is maximal and observe that
$11\geq|U|\geq|V|\geq 8$. There exists an element $g\in G$ such that $g\,|\,U$
and $-g\,|\,V$. Then $\mathsf{v}_{g}(U)\leq 2$ and $\mathsf{v}_{-g}(V)\leq 2$.
If $\mathsf{v}_{g}(U)=\mathsf{v}_{-g}(V)=2$, then $gV(-g)^{-2}$, $(-g)Ug^{-2}$
and $g(-g)$ are atoms and hence $3\in\mathsf{L}(UV)$, a contradiction.
Therefore $\mathsf{v}_{g}(U)+\mathsf{v}_{-g}(V)\in[2,3]$ and we set

(4.2)
$\\{g\in\operatorname{supp}(U)\mid\mathsf{v}_{g}(U)+\mathsf{v}_{-g}(V)=3\\}=\\{g_{1},\ldots,g_{\ell}\\}\quad\text{
where}\quad\ell\in\mathbb{N}_{0}\,.$

We continue with the following assertion.

  1. A1.

For each $i\in[1,\ell]$ we have $\mathsf{v}_{g_{i}}(U)=2$.

Proof of A1. Assume to the contrary that there is an $i\in[1,\ell]$ with
$\mathsf{v}_{g_{i}}(U)=1$. Then $g_{i}V((-g_{i})^{2})^{-1}$ is an atom and
$(-g_{i})^{2}Ug_{i}^{-1}$ is an atom or a product of two atoms. Since
$3\notin\mathsf{L}(UV)$, we obtain that $(-g_{i})^{2}Ug_{i}^{-1}$ is an atom
but $|(-g_{i})^{2}Ug_{i}^{-1}|>|U|$, a contradiction to our choice of $|U|$.
∎(Proof of A1)

Now we set

$U^{\prime}=(-g_{1})\cdot\ldots\cdot(-g_{\ell})U(g_{1}^{2}\cdot\ldots\cdot
g_{\ell}^{2})^{-1}\quad\text{and}\quad V^{\prime}=g_{1}^{2}\cdot\ldots\cdot
g_{\ell}^{2}V((-g_{1})\cdot\ldots\cdot(-g_{\ell}))^{-1}\,.$

Using the above argument repeatedly we infer that $U^{\prime}$ and
$V^{\prime}$ are atoms. Clearly, we have
$\mathsf{L}(U^{\prime}V^{\prime})=\mathsf{L}(UV)=\\{2,8\\}$ whence
$|V|+\ell=|V^{\prime}|\leq|U|$ and thus $\ell\leq 3$. We consider a
factorization

$UV=W_{1}\cdot\ldots\cdot W_{8}\,,$

where $W_{1},\ldots,W_{8}\in\mathcal{A}(C_{3}^{5})$ such that
$|\\{i\in[1,8]\mid|W_{i}|=2\\}|$ is maximal under all factorization of $UV$ of
length $8$. We set $U=U_{1}\cdot\ldots\cdot U_{8}$, $V=V_{1}\cdot\ldots\cdot
V_{8}$ such that $W_{i}=U_{i}V_{i}$ for each $i\in[1,8]$, and we define
$W=\sigma(U_{1})\cdot\ldots\cdot\sigma(U_{8})$. We continue with a second
assertion.

  1. A2.

There exist disjoint non-empty subsets $I,J,K\subset[1,8]$ such that

$I\cup J\cup K=[1,8]\quad\text{ and}\quad\sigma\big{(}\prod_{i\in
I}U_{i}\big{)}=\sigma\big{(}\prod_{j\in
J}U_{j}\big{)}=\sigma\big{(}\prod_{k\in K}U_{k}\big{)}\,.$

Proof of A2. First we suppose that $\mathsf{h}(W)\geq 2$, say
$\sigma(U_{1})=\sigma(U_{2})$. Then $I=\\{1\\}$, $J=\\{2\\}$, and $K=[3,8]$
have the required properties. Now suppose that $\mathsf{h}(W)=1$. Since the
tuple $(\sigma(U_{1}),\ldots,\sigma(U_{7}))$ is not independent and the
sequence $\sigma(U_{1})\cdot\ldots\cdot\sigma(U_{7})$ is zero-sum free, there
exist disjoint non-empty subset $I,J\subset[1,8]$, such that $\sum_{i\in
I}\sigma(U_{i})=\sum_{j\in J}\sigma(U_{j})$. Therefore, $I$, $J$, and
$K=[1,8]\setminus(I\cup J)$ have the required properties. ∎(Proof of A2)

We define

$X_{1}=\prod_{i\in I}U_{i}\,,X_{2}=\prod_{j\in J}U_{j}\,,\text{ and
}X_{3}=\prod_{k\in K}U_{k}\,,$ $Y_{1}=\prod_{i\in I}V_{i}\,,Y_{2}=\prod_{j\in
J}V_{j}\,,\text{ and }Y_{3}=\prod_{k\in K}V_{k}\,.$

By construction, we have $X_{1}Y_{1}=\prod_{i\in I}W_{i}$,
$X_{2}Y_{2}=\prod_{j\in J}W_{j}$, $X_{3}Y_{3}=\prod_{k\in K}W_{k}$,
$\sigma(X_{1})=\sigma(X_{2})=\sigma(X_{3})$,
$\sigma(Y_{1})=\sigma(Y_{2})=\sigma(Y_{3})$, and hence
$X_{i}Y_{j}\in\mathcal{B}(G)$ for all $i,j\in[1,3]$.

We choose a factorization of $X_{1}Y_{2}$, a factorization of $X_{2}Y_{3}$,
and a factorization of $X_{3}Y_{1}$, and their product gives rise to a
factorization of $UV$, say $UV=W_{1}^{\prime}\cdot\ldots\cdot W_{8}^{\prime}$,
where all the $W_{i}^{\prime}$ are atoms, and we denote by $t_{1}$ the number
of $W_{i}^{\prime}$ having length two. Similarly, we choose a factorization of
$X_{1}Y_{3}$, a factorization of $X_{2}Y_{1}$, and a factorization of
$X_{3}Y_{2}$, obtain a factorization of $UV$, and we denote by $t_{2}$ the
number of atoms of length $2$ in this factorization. If $g\in G$ and
$i,j\in[1,3]$ are distinct with $g(-g)\,|\,X_{i}Y_{j}$ and $g\,|\,X_{i}$, then
the choice of the factorization $UV=W_{1}\cdot\ldots\cdot W_{8}$ implies that
$g(-g)\,|\,X_{i}Y_{i}$ or $g(-g)\,|\,X_{j}Y_{j}$ whence
$\mathsf{v}_{g}(U)+\mathsf{v}_{-g}(V)\geq 3$. Therefore Equation (4.2) implies
that $g\in\\{g_{1},\ldots,g_{\ell}\\}$ whence $t_{1}+t_{2}\leq\ell\leq 3$, and
we may suppose that $t_{1}\leq 1$. Therefore we infer that

$2+3\times 7\leq\sum_{i=1}^{8}|W_{i}^{\prime}|=|UV|\leq
2\mathsf{D}(C_{3}^{5})=22\,,$

a contradiction. ∎

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