A characterization of finite abelian groups via sets of lengths in transfer Krull monoids

ABSTRACT

Let H be a transfer Krull monoid over a finite abelian group G (for example, rings of integers, holomorphy rings in algebraic function fields, and regular congruence monoids in these domains). Then each nonunit a∈H can be written as a product of irreducible elements, say $a=u_1\dots u_k$, and the number of factors k is called the length of the factorization. The set L(a) of all possible factorization lengths is the set of lengths of a. It is classical that the system ℒ(H) = {L(a)∣a∈H} of all sets of lengths depends only on the group G, and a standing conjecture states that conversely the system ℒ(H) is characteristic for the group G. Let H^′ be a further transfer Krull monoid over a finite abelian group G^′ and suppose that ℒ(H) = ℒ(H^′). We prove that, if $G\cong C_n^r$ with r≤n−3 or (r≥n−1≥2 and n is a prime power), then G and G^′ are isomorphic.

KEYWORDS:

• Arithmetical characterizations
• class groups
• Davenport constant
• Krull monoids
• maximal orders
• seminormal orders
• sets of lengths
• zero-sum sequences

2010 MATHEMATICS SUBJECT CLASSIFICATION:

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

1. Introduction and main result

Let H be an atomic unit-cancellative monoid. Then each non-unit a∈H can be written as a product of atoms, and if $a=u_1\dots u_k$ with atoms $u_1,\dots ,u_k$ of H, then k is called the length of the factorization. The set L(a) of all possible factorization lengths is the set of lengths of a, and ℒ(H) = {L(a)∣a∈H} is called the system of sets of lengths of H (for convenience we set L(a) = {0} if a is an invertible element of H). Under a variety of noetherian conditions on H (e.g., H is the monoid of nonzero elements of a commutative noetherian domain) all sets of lengths are finite. Sets of lengths (together with invariants controlling their structure, such as elasticities and sets of distances) are a well-studied means for describing the arithmetic structure of monoids.

Let H be a transfer Krull monoid over a finite abelian group G. Then, by definition, there is a weak transfer homomorphism 𝜃:H→ℬ(G), where ℬ(G) denotes the monoid of zero-sum sequences over G, and hence ℒ(H) = ℒ(ℬ(G)). We use the abbreviation ℒ(G) = ℒ(ℬ(G)). By a result due to Carlitz in 1960, we know that H is half-factorial (i.e., |L| = 1 for all L∈ℒ(H)) if and only if |G|≤2. Suppose that |G|≥3. Then there is some a∈H with |L(a)|>1. If k,ℓ∈L(a) with k<ℓ and m∈ℕ, then L(a^m)⊃{km+ν(ℓ−k)∣ν∈[0,m]} which shows that sets of lengths can become arbitrarily large. Note that 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. In fact, we have the following conjecture (it was first stated in [6] and for a detailed description of the background of this problem, see [6], [7, Section 7.3], [8, page 42], and [24]).

Conjecture 1.1.

Let G be a finite abelian group with D(G)≥4. If G^′ is an abelian group with ℒ(G) = ℒ(G^′), then G and G^′ are isomorphic.

Note if D(G) = 3, then we have $\mathcal{ℒ}\left(C_3\right)=\mathcal{ℒ}(C_2\oplus C_2)$. The system of sets of lengths 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 D(G) is known and the associated inverse problem is solved; however, if n is not a prime power and r≥3, then the value of the Davenport constant $D\left(C_n^r\right)$ is unknown). Thus it is not surprising that most 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 two-groups, cyclic groups, and groups of rank 2 (for recent progress we refer to [1, 9]).

The first and so far only groups, for which the Characterization Problem was solved whereas the Davenport constant is unknown, are groups of the form $C_n^r$, where r,n∈ℕ and 2r<n−2 (this is done by Geroldinger and Zhong [12] and Zhong [27]), which use a deep characterization of the structure of Δ*(G). In this paper, we go on to study groups of the form $C_n^r$ and obtain the following theorem.

Theorem 1.2.

Let H be a transfer Krull monoid over a finite abelian group G with D(G)≥4. Suppose $G\cong C_n^r$ with r,n∈ℕ and ℒ(H) = ℒ(H^′), where H^′ is a further transfer Krull monoid over a finite abelian group G^′. Then

1. If r≤n−3, then G≅G^′.

2. If r≥n−1 and n is a prime power, then G≅G^′.

This is made possible by introducing new invariants ρ(G,d) and ρ*(G,d) which are only depending on ℒ(G) (see Definitions 3.1 and 3.3). In Section 2 we gather the required background both on transfer Krull monoids as well as on Additive Combinatorics. In Section 3, we provide a detailed study of ρ(G,d) and ρ*(G,d). The proof of Theorem 1.2 will be provided in Section 4. The final section is concluding remarks and conjectures.

Throughout the paper, let $G\cong C_{n_1}\oplus \dots \oplus C_{n_r}$ be a finite abelian group with D(G)≥4, where $r,n_1,\dots ,n_r\in \mathbb{N}$ and $1<n_1|\dots |n_r$.

2. Background on transfer Krull monoids and sets of lengths

Our notation and terminology are consistent with [7, 9]. For convenience, we set min∅ = 0. Let ℕ be the set of positive integers, let ℤ be the set of integers, let ℚ be the set of rational numbers, and let ℝ be the set of real numbers. For rational numbers a,b∈ℚ, we denote by [a,b] = {x∈ℤ∣a≤x≤b} the discrete, finite interval between a and b. If L⊂ℕ is a subset, then Δ(L) denotes the set of (successive) distances of L (that is, d∈Δ(L) if and only if d = b−a with a,b∈L distinct and [a,b]∩L = {a,b}) and ρ(L) = supL∕minL denotes its elasticity (for convenience, we set ρ({0}) = 1).

Let r∈ℕ and let $(e_1,\dots ,e_r)$ be an r-tuple of elements of G. Then $(e_1,\dots ,e_r)$ is said to be independent if e_i≠0 for all i∈[1,r] and if for all $(m_1,\dots ,m_r)\in {\mathbb{Z}}^r$ an equation $m_1{e}_1+\dots +m_r{e}_r=0$ implies that $m_i{e}_i=0$ for all i∈[1,r]. Furthermore, $(e_1,\dots ,e_r)$ is said to be a basis of G if it is independent and $G=⟨e_1⟩\oplus \dots \oplus ⟨e_r⟩$. For every n∈ℕ, we denote by C_n an additive cyclic group of order n. Since $G\cong C_{n_1}\oplus \dots \oplus C_{n_r}$, r = r(G) is the rank of G and n_r = exp(G) is the exponent of G.

2.1. Sets of lengths

By a monoid, we mean an associative semigroup with unit element, and if not stated otherwise we use multiplicative notation. Let H be a monoid with unit element 1 = 1_H∈H. An element a∈H is said to be invertible (or an unit) if there exists an element a^′∈H such that $a{a}^{\prime }=a^{\prime }a=1$. The set of invertible elements of H will be denoted by H^×, and we say that H is reduced if H^× = {1}. The monoid H is said to be unit-cancellative if for any two elements a,u∈H, each of the equations au = a or ua = a implies that u∈H^×. Clearly, every cancellative monoid is unit-cancellative.

Suppose that H is unit-cancellative. An element u∈H is said to be irreducible (or an atom) if u∉H^× and for any two elements a,b∈H, u = ab implies that a∈H^× or b∈H^×. Let 𝒜(H) denote the set of atoms, and we say that H is atomic if every non-unit is a finite product of atoms. If H satisfies the ascending chain condition on principal left ideals and on principal right ideals, then H is atomic [4, Theorem 2.6]. If a∈H∖H^× and $a=u_1\dots u_k$, where k∈ℕ and $u_1,\dots ,u_k\in \mathcal{𝒜}\left(H\right)$, then k is a factorization length of a, and

${\displaystyle \begin{array}{c}\multicolumn{1}{c}{L_H\left(a\right)=L\left(a\right)=\{k\mid k\text{is a factorization length of}a\}\subset \mathbb{N}}\end{array}}$

denotes the set of lengths of a. It is convenient to set L(a) = {0} for all a∈H^×. The family

${\displaystyle \begin{array}{c}\multicolumn{1}{c}{\mathcal{ℒ}\left(H\right)=\left\{L\right(a)\mid a\in H\}}\end{array}}$

is called the system of sets of lengths of H, and

${\displaystyle \begin{array}{c}\multicolumn{1}{c}{\rho \left(H\right)=sup\left\{\rho \right(L)\mid L\in \mathcal{ℒ}(H\left)\right\}\in {ℝ}_{\ge 1}\cup \left\{\infty \right\}}\end{array}}$

denotes the elasticity of H. We call

${\displaystyle \begin{array}{c}\multicolumn{1}{c}{\Delta \left(H\right)=\bigcup _{L\in \mathcal{ℒ}\left(H\right)}\Delta \left(L\right)\subset \mathbb{N}}\end{array}}$

the set of distances of H. Note that Δ(H) can be infinite, and by definition we have Δ(H) = ∅ if and only if ρ(H) = 1. If Δ(H)≠∅, then we have minΔ(H) = gcdΔ(H) [3, Proposition 2.9].

2.2. Monoids of zero-sum sequences

Let G₀⊂G be a non-empty subset. Then ⟨G₀⟩ denotes the subgroup generated by G₀. In Additive Combinatorics, a sequence (over G₀) means a finite sequence of terms from G₀ where repetition is allowed and the order of the elements is disregarded, and (as usual) we consider sequences as elements of the free abelian monoid with basis G₀. Let

${\displaystyle \begin{array}{c}\multicolumn{1}{c}{S=g_1\dots g_{\ell }=\prod _{g\in G_0}{g}^{v_g\left(S\right)}\in \mathcal{ℱ}\left(G_0\right)}\end{array}}$

be a sequence over G₀. We call

${\displaystyle \begin{array}{ccc}\multicolumn{1}{c}{supp\left(S\right)}& =\hfill & \{g\in G\mid v_g(S)>0\}\subset G\text{the}support\text{of }S,\hfill \\ \multicolumn{1}{c}{\left|S\right|}& =\hfill & \ell =\sum _{g\in G}{v}_g\left(S\right)\in {\mathbb{N}}_0\text{the}length\text{of }S,\hfill \\ \multicolumn{1}{c}{\sigma \left(S\right)}& =\hfill & \sum _{i=1}^{\ell }{g}_i\text{the}sum\text{of }S,\hfill \\ \multicolumn{1}{c}{\text{ and }\Sigma \left(S\right)}& =\hfill & \{\sum _{i\in I}{g}_i\mid \varnothing \ne I\subset [1,\ell ]\}\text{ the}setofsubsequencesums\text{of }S.\hfill \end{array}}$

The sequence S is said to be

• zero-sum free if 0∉Σ(S),

• a zero-sum sequence if σ(S) = 0,

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

The set of zero-sum sequences $\mathcal{ℬ}\left(G_0\right)=\{S\in \mathcal{ℱ}(G_0)\mid \sigma (S)=0\}\subset \mathcal{ℱ}\left(G_0\right)$ is a submonoid, and the set of minimal zero-sum sequences is the set of atoms of ℬ(G₀). For any arithmetical invariant ∗(H) defined for a monoid H, we write ∗(G₀) instead of ∗(ℬ(G₀)). In particular, $\mathcal{𝒜}\left(G_0\right)=\mathcal{𝒜}\left(\mathcal{ℬ}\right(G_0\left)\right)$ is the set of atoms of ℬ(G₀), $\mathcal{ℒ}\left(G_0\right)=\mathcal{ℒ}\left(B\right(G_0\left)\right)$ is the system of sets of lengths of ℬ(G₀), and so on. We denote by

${\displaystyle \begin{array}{c}\multicolumn{1}{c}{D\left(G_0\right)=max\left\{\right|S|\mid S\in \mathcal{𝒜}(G_0\left)\right\}\in \mathbb{N}}\end{array}}$

the Davenport constant of G₀.

2.3. Transfer Krull monoids

Let H be a atomic unit-cancellative monoid.

1. We say a monoid homomorphism 𝜃:H→B to an atomic unit-cancellative monoid B is a weak transfer homomorphism if it has the following two properties:

   • $B=B^{\times }𝜃\left(H\right)B^{\times }$ and ${𝜃}^{-1}\left(B^{\times }\right)=H^{\times }$.

   • If a∈H, n∈ℕ, $v_1,\dots ,v_n\in \mathcal{𝒜}\left(B\right)$ and $𝜃\left(a\right)=v_1\dots v_n$, then there exist $u_1,\dots ,u_n\in \mathcal{𝒜}\left(H\right)$ and a permutation τ∈𝔖_n such that $a=u_1\dots u_n$ and $𝜃\left(u_i\right)\in B^{\times }{v}_{\tau \left(i\right)}{B}^{\times }$ for each i∈[1,n].

   Let 𝜃:H→B be a weak transfer homomorphism between atomic unit-cancellative monoids. It follows that for all a∈H, we have $L_H\left(a\right)=L_B\left(𝜃\right(a\left)\right)$ and hence ℒ(H) = ℒ(B).

2. We say H is a transfer Krull monoid if one of the following equivalent conditions holds:

   • There exists a weak transfer homomorphism 𝜃:H→ℋ* for a commutative Krull monoid ℋ* (i.e., ℋ* is commutative, cancellative, completely integrally closed, and v-noetherian).

   • There exists a weak transfer homomorphism ${𝜃}^{\prime }:H\to \mathcal{ℬ}\left(G_0\right)$ for a subset G₀ of an abelian group.

   If the second condition holds, then we say H is a transfer Krull monoid over G₀. If G₀ is finite, then H is said to be a transfer Krull monoid of finite type.

3. We say a domain R is a transfer Krull domain (of finite type) if its monoid of cancelative elements R^∙ is a transfer Krull monoid (of finite type).

In particular, commutative Krull monoids are transfer Krull monoids. Rings of integers, holomorphy rings in algebraic function fields, and regular congruence monoids in these domains are commutative Krull monoids with finite class group such that every class contains a prime divisor [7, Section 2.11 and Examples 7.4.2]. Monoid domains and power series domains that are Krull are discussed in [17, 2], and note that every class of a Krull monoid domain contains a prime divisor. Thus all these commutative Krull monoids are transfer Krull monoids over a finite abelian group.

However, a transfer Krull monoid need neither be commutative nor v-noetherian nor completely integrally closed. To give a noncommutative example, let 𝒪 be a holomorphy ring in a global field K, A a central simple algebra over K, and H a classical maximal 𝒪-order of A such that every stably free left R-ideal is free. Then H is a transfer Krull monoid over a ray class group of 𝒪 [25, Theorem 1.1]. Let R be a bounded HNP (hereditary noetherian prime) ring. If every stably free left R-ideal is free, then its multiplicative monoid of cancelative elements is a transfer Krull monoid [26, Theorem 4.4]. A class of commutative weakly Krull domains which are transfer Krull but not Krull will be given in [10, Theorem 5.8]. Extended lists of commutative Krull monoids and of transfer Krull monoids, which are not commutative Krull, are given in [6].

Let G₀⊂G be a non-empty subset. For a sequence $S=g_1\dots g_{\ell }\in \mathcal{ℱ}\left(G_0\right)$, we call

${\displaystyle \begin{array}{ccc}\multicolumn{1}{c}{k\left(S\right)}& =\hfill & \sum _{i=1}^l\frac{1}{ord\left(g_i\right)}\in {\mathbb{Q}}_{\ge 0}\text{the}crossnumber\text{of}S\text{, and }\hfill \\ \multicolumn{1}{c}{K\left(G_0\right)}& =\hfill & max\left\{k\right(S)\mid S\in \mathcal{𝒜}(G_0\left)\right\}\text{the}crossnumber\text{of}{G}_0.\hfill \end{array}}$

They were introduced by U. Krause in 1984 (see [19]) and were studied under various aspects. For the relevance with the theory of non-unique factorizations, see [21, 20, 22, 23] and [7, Chapter 6].

Suppose $G\cong C_{q_1}\oplus \dots \oplus C_{q_{r^{\ast }}}$, where r* is the total rank of G and $q_1,\dots ,q_{r^{\ast }}$ are prime powers, and set

${\displaystyle \begin{array}{c}\multicolumn{1}{c}{K^{\ast }\left(G\right)=\frac{1}{exp\left(G\right)}+\sum _{i=1}^{r^{\ast }}\frac{q_i-1}{q_i}.}\end{array}}$

It is easy to see that K*(G)≤K(G) and there is known no group for which inequality holds. For further progress on K(G), we refer to [5, 15, 16, 18].

Lemma 2.1.

If G is a p-group, then K(G) = K*(G)<r(G).

Proof.

See [7, Theorem 5.5.9].

A subset G₀⊂G is called half-factorial if Δ(G₀) = ∅. Otherwise, G₀ is called non-half-factorial. Furthermore, the set G₀ is called

• minimal non-half-factorial if it is non-half-factorial and every proper subset $G_1⊊G_0$ is half-factorial.

• an LCN-set if k(A)≥1 for all A∈𝒜(G₀).

We collect some easy or well known results which will be used throughout the manuscript without further mention.

Lemma 2.2.

Let G₀⊂G be a non-empty subset. Then

1. G₀ is half-factorial if and only if k(A) = 1 for all A∈𝒜(G₀).

2. If G₀ is an LCN-set, then $min\Delta \left(G_0\right)\le \left|G_0\right|-2$.

3. ℬ(G₀) has accepted elasticity.

4. $\rho \left(G\right)=\frac{D\left(G\right)}{2}$.

5. Δ(G) is an interval with minΔ(G) = 1.

6. If B∈ℬ(G), then ρ(L(B^k))≥ρ(L(B)) for every k∈ℕ.

7. If A∈𝒜(G), then {exp(G),exp(G)k(A)}⊂L(A^exp(G)).

Proof.

1. follows from [7, Proposition 6.7.3] and 2. from [7, Lemma 6.8.6].

2. see [7, Proposition 6.7.3, Lemma 6.8.6].

3. follows from [7, Theorem 3.1.4] and 4. from [7, Section 6.3].

4. See [7, Theorem 3.1.4, Section 6.3].

5. See [11].

6. Let B∈ℬ(G) and k∈ℕ. Then maxL(B^k)≥kmaxL(B) and minL(B^k)≤kminL(B). It follows that $\rho \left(L\right(B^k\left)\right)=\frac{maxL\left(B^k\right)}{minL\left(B^k\right)}\ge \frac{kmaxL\left(B\right)}{kminL\left(B\right)}=\rho \left(L\right(B\left)\right)$.

7. Let A∈𝒜(G) and suppose $A=g_1\dots g_{\ell }$, where ℓ∈ℕ and $g_1,\dots ,g_{\ell }\in G$. Then

${\displaystyle \begin{array}{c}\multicolumn{1}{c}{A^{exp\left(G\right)}=\prod _{i=1}^{\ell }{\left(g_i^{ord\left(g_i\right)}\right)}^{\frac{exp\left(G\right)}{ord\left(g_i\right)}}.}\end{array}}$

Since A, $g_1^{ord\left(g_1\right)},\dots ,g_{\ell }^{ord\left(g_{\ell }\right)}$ are atoms, we obtain {exp(G),exp(G)k(A)}⊂L(A^exp(G)).

We need the following lemma.

Lemma 2.3.

Let $e_1,\dots ,e_r\in G$ be independent elements with the same order n, where r, n∈ℕ_≥2. Then

1. ${\displaystyle \begin{array}{c}\multicolumn{1}{c}{\Delta \left(\right\{e_1+\dots +e_r,e_1,\dots ,e_r\left\}\right)=\{r-1\}.}\end{array}}$

2. If n≠r+1, then

   ${\displaystyle \begin{array}{c}\multicolumn{1}{c}{\Delta \left(\right\{-(e_1+\dots +e_r),e_1,\dots ,e_r\left\}\right)=\left\{\right|n-r-1\left|\right\}.}\end{array}}$

3. If n = r+1, then

   ${\displaystyle \begin{array}{c}\multicolumn{1}{c}{min\Delta \left(\right\{-(e_1+\dots +e_r),e_1+\dots +e_r,e_1,\dots ,e_r\left\}\right)=r-1.}\end{array}}$

Proof.

1. See [7, Proposition 6.8.2].

2. See [7, Proposition 4.1.2.5].

3. Let $g=e_1+\dots +e_r$ and $G_0=\{g,-g,e_1,\dots ,e_r\}$. Let B∈ℬ(G₀) and assume

${\displaystyle \begin{array}{c}\multicolumn{1}{c}{B=U_1\dots U_k=V_1\dots V_{\ell },\text{ where }k,\ell \in N\text{ and }{U}_1,\dots ,U_k,V_1,\dots ,V_{\ell }\in \mathcal{𝒜}\left(G_0\right).}\end{array}}$

Let

${\displaystyle \begin{array}{ccc}\multicolumn{1}{c}{I_1}& =\hfill & \{i\in [1,k]\mid v_g(U_i)>0\text{ and }{v}_{-g}(U_i)=0\},I_2=\{i\in [1,k]\mid U_i=g(-g\left)\right\}\hfill \\ \multicolumn{1}{c}{J_1}& =\hfill & \{j\in [1,\ell ]\mid v_g(V_j)>0\text{ and }{v}_{-g}(V_j)=0\},J_2=\{j\in [1,\ell ]\mid V_j=g(-g\left)\right\}.\hfill \end{array}}$

Then for every i∈I₁, $k\left(U_i\right)=1+\frac{n-v_g\left(U_i\right)}{n}(n-2)$ and for every j∈J₁, $k\left(V_j\right)=1+\frac{n-v_g\left(V_j\right)}{n}(n-2)$.

Note that for every $i\in [1,k]\setminus (I_1\cup I_2)$ and every $j\in [1,\ell ]\setminus (J_1\cup J_2)$, $k\left(U_i\right)=k\left(V_j\right)=1$. Therefore

${\displaystyle \begin{array}{c}\multicolumn{1}{c}{k\left(B\right)=k-\left|I_1\right|-\left|I_2\right|+\sum _{i\in I_1}\left(1+\frac{(n-v_g(U_i\left)\right)(n-2)}{n}\right)+\sum _{i\in I_2}\frac{2}{n}=k+(n-2)\left|I_1\right|-\frac{n-2}{n}{v}_g\left(B\right)}\end{array}}$

and

${\displaystyle \begin{array}{c}\multicolumn{1}{c}{k\left(B\right)=\ell -\left|J_1\right|-\left|J_2\right|+\sum _{j\in J_1}\left(1+\frac{(n-v_g(V_j\left)\right)(n-2)}{n}\right)+\sum _{j\in J_2}\frac{2}{n}=\ell +(n-2)\left|J_1\right|-\frac{n-2}{n}{v}_g\left(B\right).}\end{array}}$

It follows that $k-\ell =(n-2)\left(\right|J_1|-|I_1\left|\right)$ and hence n−2 | minΔ(G₀). By 1., we obtain n−2 = r−1 = minΔ(G₀).

If d∈ℕ and ℓ,M∈ℕ₀, then a finite subset L⊂ℤ is called an almost arithmetical progression (AAP for short) with difference d, length ℓ, and bound M if

(1) ${\displaystyle \begin{array}{c}\multicolumn{1}{c}{L=y+(L^{\prime }\cup \{0,d,\dots ,\ell d\}\cup {L{}^{\prime }}^{\prime })\subset y+d\mathbb{Z}}\end{array}}$(1)

where y∈ℤ, L^′⊂[−M,−1], and ${L{}^{\prime }}^{\prime }\subset \ell d+[1,M]$.

Lemma 2.4.

There exist constants $M_1,M_2\in \mathbb{N}$ such that for every A∈ℬ(G) with Δ(supp(A))≠∅, the set $L\left(A^{M_1}\right)$ is an AAP with difference minΔ(supp(A)), length at least 1, and bound M₂.

Proof.

See [7, Theorem 4.3.6].

Next, we recall the definition of the invariants Δ*(G) and Δ₁(G) (see [7, Definition 4.3.12]) in the Characterization Problem.

Let

${\displaystyle \begin{array}{c}\multicolumn{1}{c}{{\Delta }^{\ast }\left(G\right)=\{min\Delta (G_0)\mid G_0\subset G\text{ is a subset with }\Delta (G_0)\ne \varnothing \}.}\end{array}}$

We define

${\displaystyle \begin{array}{ccc}\multicolumn{1}{c}{m\left(G\right)}& =\hfill & max\{min\Delta (G_0)\mid G_0\subset G\text{is an LCN-set with}\Delta (G_0)\ne \varnothing \},\hfill \end{array}}$

and we denote by Δ₁(G) the set of all d∈ℕ with the following property:

• For every k∈ℕ, there exists some L∈ℒ(G) which is an AAP with difference d and length ℓ≥k.

Lemma 2.5.

Let k∈ℕ be maximal such that G has a subgroup isomorphic to $C_{exp\left(G\right)}^k$. Then

1. ${\Delta }^{\ast }\left(G\right)\subset {\Delta }_1\left(G\right)\subset \{d\in \Delta (G)\mid \text{there exists }{d}^{\prime }\in {\Delta }^{\ast }(G\left)\text{ such that }d\right|d^{\prime }\}$.

2. $max{\Delta }^{\ast }\left(G\right)=max{\Delta }_1\left(G\right)=max\{exp(G)-2,r(G)-1\}$.

3. $m\left(G\right)\le max\left\{r\right(G)-1,\lfloor \frac{exp\left(G\right)}{2}\rfloor -1\}$.

4. $\{1,r(G)-1,exp(G)-2\}\subset {\Delta }^{\ast }\left(G\right)\subset {\Delta }_1\left(G\right)\subset [1,max\{r\left(G\right)-1,\lfloor \frac{exp\left(G\right)}{2}\rfloor -1\left\}\right]\cup [max\{1,exp\left(G\right)-k-1,exp\left(G\right)-2\left\}\right]$.

5. Δ₁(G) is an interval if and only if Δ*(G) is an interval if and only if r(G)+k≥exp(G)−1 or $G\cong C_{2r\left(G\right)+2}^{r\left(G\right)}$.

Proof.

1. Follows from [7, Corollary 4.3.16]

2. From [14, Theorem 1.1].

3. See [27, Proposition 3.7].

4. Follows from 1. and [27, Theorem 1.1].

5. If Δ₁(G) is an interval, then $exp\left(G\right)-k-2\le max\left\{r\right(G)-1,\lfloor \frac{exp\left(G\right)}{2}\rfloor -1\}$ by 4 which implies that Δ*(G) is an interval by Zhong [27, Theorem 1.1.2]. If Δ*(G) is an interval, then Δ₁(G) is an interval by 1.. It follows by Zhong [27, Theorem 1.1.2] that Δ*(G) is an interval if and only if r(G)+k≥exp(G)−1 or $G\cong C_{2r\left(G\right)+2}^{r\left(G\right)}$.

Lemma 2.6.

Let G₀⊂G with minΔ(G₀)≥⌊exp(G)∕2⌋. Then

1. Let A∈𝒜(G₀) with k(A)<1. Then $k\left(A\right)=\frac{exp\left(G\right)-min\Delta \left(G_0\right)}{exp\left(G\right)}$ and ord(h) = exp(G), v_h(A) = 1 for all h∈supp(A).

2. Let A∈𝒜(G₀) with k(A)≥1. Then supp(A) is an LCN-set and

   ${\displaystyle \begin{array}{c}\multicolumn{1}{c}{min\Delta \left(supp\right(A\left)\right)\le \left|supp\right(A\left)\right|-2.}\end{array}}$

3. If G₀ is a minimal non-half-factorial LCN-set with $min\Delta \left(G_0\right)=max{\Delta }^{\ast }\left(G\right)$, then |G₀| = r(G)+1 and for every h∈G₀, we have r(⟨G₀∖{h}⟩) = r(G).

4. Let B∈ℬ(G) with ρ(L(B)) = ρ(G). Suppose $B=U_1\dots U_k=V_1\dots V_{\ell }$, where k = minL(B), ℓ = maxL(B), and $U_1,\dots ,U_k,V_1,\dots V_{\ell }$ are atoms. Then k(U_i)≥1 for all i∈[1,k] and k(V_j)≤1 for all j∈[1,ℓ].

Proof.

1. Since {exp(G),exp(G)k(A)}⊂L(A^exp(G)), we have

${\displaystyle \begin{array}{c}\multicolumn{1}{c}{min\Delta \left(supp\right(A\left)\right)|exp(G)-exp(G\left)k\right(A)}\end{array}}$

and hence

${\displaystyle \begin{array}{c}\multicolumn{1}{c}{min\Delta \left(supp\right(A\left)\right)\le exp\left(G\right)-exp\left(G\right)k\left(A\right)\le exp\left(G\right)-2.}\end{array}}$

Since minΔ(G₀) | minΔ(supp(A)) and minΔ(G₀)≥⌊exp(G)∕2⌋, it follows that

${\displaystyle \begin{array}{ccc}\multicolumn{1}{c}{\frac{1}{2}(exp(G)-2)}& <\hfill & \lfloor exp\left(G\right)∕2\rfloor \le min\Delta \left(G_0\right)\le min\Delta \left(supp\right(A\left)\right)\hfill \\ \le \hfill & exp\left(G\right)-exp\left(G\right)k\left(A\right)\le exp\left(G\right)-2.\hfill \end{array}}$

Therefore minΔ(G₀) = minΔ(supp(A)) = exp(G)−exp(G)k(A) and hence $k\left(A\right)=\frac{exp\left(G\right)-min\Delta \left(G_0\right)}{exp\left(G\right)}$.

Let g∈supp(A) such that $\frac{ord\left(g\right)}{v_g\left(A\right)}=min\{\frac{ord\left(h\right)}{v_h\left(A\right)}\mid h\in supp(A\left)\right\}$. Then $A^{\lceil \frac{ord\left(g\right)}{v_g\left(A\right)}\rceil }=g^{ord\left(g\right)}\cdot B_1$, where B₁∈ℬ(supp(A)). If there exists an atom A₁ with k(A₁)<1 such that A₁ divides B₁, then $k\left(A_1\right)=\frac{exp\left(G\right)-min\Delta \left(G_0\right)}{exp\left(G\right)}=k\left(A\right)$. Therefore $1+maxL\left(B_1\right)<\lceil \frac{ord\left(g\right)}{v_g\left(A\right)}\rceil$ and hence $\lfloor exp\left(G\right)∕2\rfloor \le min\Delta \left(G_0\right)\le \lceil \frac{ord\left(g\right)}{v_g\left(A\right)}\rceil -2$. It follows by the minimality of $\frac{ord\left(g\right)}{v_g\left(A\right)}$ that ord(h) = exp(G) and v_h(A) = 1 for all h∈supp(A).

2. Assume to the contrary that supp(A) is not an LCN-set. Then there exists A₁∈𝒜(supp(A)) with k(A₁)<1. It follows by 1 that $v_h\left(A_1\right)=1$ for all h∈supp(A₁) which implies that A₁ | A, a contradiction. Thus supp(A) is an LCN-set and hence minΔ(supp(A))≤|supp(A)|−2.

3. By Geroldinger and Zhong [14, Lemma 4.2], we have |G₀| = r(G)+1 and r(⟨G₀⟩) = r(G). If h∈⟨G₀∖{h}⟩, then r(⟨G₀∖{h}⟩) = r(G). Otherwise let d = min{k∈ℕ∣kh∈⟨G₀∖h⟩} and hence (G₀∖{h})∪{dh} is also a minimal non-half-factorial LCN-set with $min\Delta \left(G_0\right)=max{\Delta }^{\ast }\left(G\right)$ by Geroldinger and Halter-Koch [7, Lemma 6.7.10]. It follows that r(⟨G₀∖{h}⟩) = r(G).

4. Assume to the contrary that there exists i∈[1,k], say i = 1, such that k(U₁)<1. Then exp(G)k(U₁)<exp(G). Since $exp\left(G\right)k\left(U_1\right)\in L\left(U_1^{exp\left(G\right)}\right)$, we infer

${\displaystyle \begin{array}{c}\multicolumn{1}{c}{minL\left(B^{exp\left(G\right)}\right)\le exp\left(G\right)k\left(U_1\right)+exp\left(G\right)(k-1)<exp\left(G\right)k.}\end{array}}$

It follows by maxL(B^exp(G))≥exp(G)ℓ that $\rho \left(G\right)=\rho \left(L\right(B\left)\right)\ge \rho \left(L\right(B^{exp\left(G\right)}\left)\right)>\frac{\ell }{k}=\rho \left(L\right(B\left)\right)$, a contradiction.

Assume to the contrary that there exists j∈[1,ℓ], say j = 1, such that k(V₁)>1. Then exp(G)k(U₁)>exp(G). Since $exp\left(G\right)k\left(U_1\right)\in L\left(U_1^{exp\left(G\right)}\right)$, we infer

${\displaystyle \begin{array}{c}\multicolumn{1}{c}{maxL\left(B^{exp\left(G\right)}\right)\ge exp\left(G\right)k\left(U_1\right)+exp\left(G\right)(\ell -1)>exp\left(G\right)\ell .}\end{array}}$

It follows by minL(B^exp(G))≤exp(G)k that $\rho \left(G\right)=\rho \left(L\right(B\left)\right)\ge \rho \left(L\right(B^{exp\left(G\right)}\left)\right)>\frac{\ell }{k}=\rho \left(L\right(B\left)\right)$, a contradiction.

3. The invariants ρ(G,d), ρ*(G,d), and K(G,d)

First, we introduce new invariants which play a crucial role in the proof of Theorem 1.2 (see Proposition 3.5).

Definition 3.1.

Let d∈Δ₁(G) and k∈ℕ. We define

${\displaystyle \begin{array}{c}\multicolumn{1}{c}{\rho (G,d,k)=sup\left\{\rho \right(L_k)\mid L_k\in \mathcal{ℒ}(G\left)\text{ is an AAP with difference }d\text{ and length at least}k\right\}\ge 1.}\end{array}}$

Then $(\rho (G,d,k){)}_{k=1}^{\infty }$ is a decreasing sequence of positive real numbers and hence converges. We denote by ρ(G,d) the limit of $(\rho (G,d,k){)}_{k=1}^{\infty }$.

It follows by Geroldinger and Zhong [13, Theorem 3.5] that ρ(G,1) = ρ(G) if and only if G is not a cyclic group of order 4,6 or 10.

Lemma 3.2.

Let G₀⊂G be a subset with Δ(G₀)≠∅. For every B∈ℬ(G₀) with minΔ(G₀)∈Δ(L(B)), we have ρ(G,minΔ(G₀))≥ρ(L(B)).

Proof.

Let B∈ℬ(G₀) with minΔ(G₀)∈Δ(L(B)). By definition, L(B) is an AAP with difference minΔ(G₀) and length at least 1. Therefore for every k∈ℕ, L(B^k) is an AAP with difference minΔ(G₀) and length at least k. Thus for every k∈ℕ, $\rho (G,min\Delta (G_0),k)\ge \rho \left(L\right(B^k\left)\right)\ge \rho \left(L\right(B\left)\right)$ by Lemma 2.2.6. Therefore ρ(G,minΔ(G₀))≥ρ(L(B)).

Definition 3.3.

Let d∈Δ₁(G). We define

${\displaystyle \begin{array}{ccc}\multicolumn{1}{c}{{\rho }^{\ast }(G,d)}& =\hfill & max\left\{\rho \right(G_0)\mid G_0\subset G\text{ is a non-half-factorial subset with }d\text{ divides }min\Delta (G_0\left)\right\},\text{ and}\hfill \\ \multicolumn{1}{c}{K(G,d)}& =\hfill & max\left\{K\right(G_0)\mid G_0\subset G\text{ is a non-half-factorial subset with }d\text{ divides }min\Delta (G_0\left)\right\}.\hfill \end{array}}$

Note that ρ*(G,1) = ρ(G) and K(G,1) = K(G). For every d∈Δ₁(G), there always exists G₀⊂G with G₀ non-half-factorial such that d | minΔ(G₀) by Lemma 2.5.1. Since ρ(G₀)>1 and K(G₀)≥1, we have ρ*(G,d)>1 and K(G,d)≥1. Furthermore, there exist $G_1,G_2\subset G$ with d | minΔ(G₁) and d | minΔ(G₂) such that ${\rho }^{\ast }(G,d)=\rho \left(G_1\right)$ and K(G,d) = K(G₂).

Lemma 3.4.

Let d∈Δ₁(G). Then

1. ρ*(G,d)≥ρ(G,d).

2. ${\rho }^{\ast }(G,d)\le \rho (G,k_{0}d)$ for some k₀∈ℕ with $k_{0}d\in {\Delta }_1\left(G\right)$.

3. ${\rho }^{\ast }(G,d)=max\left\{\rho \right(G,kd)\mid k\in \mathbb{N}\text{ and }kd\in {\Delta }_1(G\left)\right\}$.

Proof.

1. By the definition of Δ₁(G) and d∈Δ₁(G), for every k∈ℕ, we let B_k∈ℬ(G) be such that L(B_k) is an AAP with difference d and length at least k. Let

${\displaystyle \begin{array}{c}\multicolumn{1}{c}{{\ell }_k=min\{min{L}_k\mid L_k\in \mathcal{ℒ}(G\left)\text{ is an AAP with difference }d\text{ and length at least }k\right\}}\end{array}}$

and hence $\underset{k\to \infty }{lim}{\ell }_k=\infty$ by ρ(G) is finite.

By Lemma 2.4, there exists a constant M such that minΔ(supp(B))∈Δ(L(B^M)) for all B∈ℬ(G) with Δ(supp(B))≠∅. Since $\underset{k\to \infty }{lim}{\ell }_k=\infty$, we let k∈ℕ be large enough such that minL(B_k)≥M|𝒜(G)| and assume

${\displaystyle \begin{array}{c}\multicolumn{1}{c}{B_k=U_1^{t_1}\dots U_s^{t_s}=V_1\dots V_{maxL\left(B_k\right)},}\end{array}}$

where $s,t_1,\dots ,t_s\in \mathbb{N}$, $t_1+\dots +t_s=minL\left(B_k\right)$, $U_1,\dots ,U_s$ are pair-wise distinct atoms, and $V_1,\dots ,V_{maxL\left(B_k\right)}$ are atoms.

Since minL(B_k)≥M|𝒜(G)|, we infer there exists i∈[1,s], say i = 1, such that t₁≥M. Set I = {i∈[1,s]∣t_i≥M} and $G_0=supp\left({\prod }_{i\in I}{U}_i\right)$. It follows by the choice of M that

${\displaystyle \begin{array}{c}\multicolumn{1}{c}{min\Delta \left(G_0\right)\in \Delta \left(L\left(\prod _{i\in I}{U}_i^M\right)\right).}\end{array}}$

Since L(B_k) is an AAP with difference d and ${\prod }_{i\in I}{U}_i^M)$ is a subsequence of B_k, we obtain

${\displaystyle \begin{array}{c}\multicolumn{1}{c}{d|gcd\Delta (L\left(B_k\right)\left)\right|gcd\Delta \left(L\left(\prod _{i\in I}\underset{i}{\overset{M}{U}}\right)\right).}\end{array}}$

Therefore d | minΔ(G₀) and hence $\rho \left(G_0\right)\le {\rho }^{\ast }(G,d)$. Since $\left|{\prod }_{i\notin I}{U}_i^{t_i}\right|\le M\left|\mathcal{𝒜}\right(G\left)\right|D\left(G\right)$, there exists J⊂[1,maxL(B_k)] with |J|≥maxL(B_k)−M|𝒜(G)|D(G) such that ${\prod }_{j\in J}{V}_j$ divides ${\prod }_{i\in I}{U}_i^{t_i}$. It follows by $minL\left({\prod }_{i\in I}{U}_i^{t_i}\right)={\sum }_{i\in I}{t}_i$ that

${\displaystyle \begin{array}{ccc}\multicolumn{1}{c}{maxL\left(B_k\right)-M\left|\mathcal{𝒜}\right(G\left)\right|D\left(G\right)}& \le \hfill & \left|J\right|\le maxL\left(\prod _{i\in I}\underset{i}{\overset{t_i}{U}}\right)\hfill \\ \le \hfill & \left(\sum _{i\in I}{t}_i\right)\rho \left(G_0\right)\le minL\left(B_k\right){\rho }^{\ast }(G,d).\hfill \end{array}}$

Therefore

${\displaystyle \begin{array}{c}\multicolumn{1}{c}{\rho \left(L\right(B_k\left)\right)\le {\rho }^{\ast }(G,d)+\frac{M\left|\mathcal{𝒜}\right(G\left)\right|D\left(G\right)}{minL\left(B_k\right)}\le {\rho }^{\ast }(G,d)+\frac{M\left|\mathcal{𝒜}\right(G\left)\right|D\left(G\right)}{{\ell }_k}.}\end{array}}$

By definition, we infer $\rho (G,d,k)\le {\rho }^{\ast }(G,d)+\frac{M\left|\mathcal{𝒜}\right(G\left)\right|D\left(G\right)}{{\ell }_k}$ which implies that ρ(G,d)≤ρ*(G,d).

2. Let G₀⊂G with d | minΔ(G₀) and $\rho \left(G_0\right)={\rho }^{\ast }(G,d)$. Then there exists B∈ℬ(G₀) such that ρ(L(B)) = ρ*(G,d). Since supp(B)⊂G₀, we infer minΔ(G₀) divides minΔ(supp(B)) and hence d divides minΔ(supp(B)).

By Lemma 2.4, there exists a constant M such that minΔ(supp(B))∈Δ(L(B^M)). It follows by Lemma 3.2 and Lemma 2.2.6 that

${\displaystyle \begin{array}{c}\multicolumn{1}{c}{\rho (G,min\Delta (supp\left(B\right)\left)\right)\ge \rho \left(L\right(B^M\left)\right)\ge \rho \left(L\right(B\left)\right)={\rho }^{\ast }(G,d).}\end{array}}$

Thus the assertion follows by d divides minΔ(supp(B)).

3. For every k∈ℕ such that kd∈Δ₁(G), we have $\rho (G,kd)\le {\rho }^{\ast }(G,kd)\le {\rho }^{\ast }(G,d)$ by 1.. Therefore ${\rho }^{\ast }(G,d)\ge max\left\{\rho \right(G,kd)\mid k\in \mathbb{N}\text{ and }kd\in {\Delta }_1(G\left)\right\}$. It follows by 2. that ${\rho }^{\ast }(G,d)=max\left\{\rho \right(G,kd)\mid k\in \mathbb{N}\text{ and }kd\in {\Delta }_1(G\left)\right\}$.

Proposition 3.5.

Suppose ℒ(G) = ℒ(G^′) for some finite abelian group G^′ and let d∈Δ₁(G). Then $d\in {\Delta }_1\left(G^{\prime }\right)$, ρ(G,d) = ρ(G^′,d), and ${\rho }^{\ast }(G,d)={\rho }^{\ast }(G^{\prime },d)$.

Proof.

Since ℒ(G) = ℒ(G^′), it follows by definition that ${\Delta }_1\left(G\right)={\Delta }_1\left(G^{\prime }\right)$ and ρ(G,kd) = ρ(G^′,kd) for every k∈ℕ such that kd∈Δ₁(G). By Lemma 3.4.3, we obtain ${\rho }^{\ast }(G,d)={\rho }^{\ast }(G^{\prime },d)$.

Lemma 3.6.

Let d∈Δ₁(G). Then ρ*(G,d)≥K(G,d). In particular, if d∈[1,r−1], then ${\rho }^{\ast }(G,d)\ge K(G,d)\ge 1+\frac{(n_1-1)d}{n_1}\ge 1+\frac{d}{2}$.

Proof.

Suppose G₀⊂G with d dividing minΔ(G₀) and K(G₀) = K(G,d). Then there exists A∈𝒜(G₀) such that k(A) = K(G,d)≥1. Since

${\displaystyle \begin{array}{c}\multicolumn{1}{c}{\{exp(G),exp(G\left)k\right(A\left)\right\}\subset L\left(A^{exp\left(G\right)}\right),}\end{array}}$

we have

${\displaystyle \begin{array}{c}\multicolumn{1}{c}{{\rho }^{\ast }(G,d)\ge \rho \left(G_0\right)\ge \rho \left(L\right(A^{exp\left(G\right)}\left)\right)\ge \frac{exp\left(G\right)k\left(A\right)}{exp\left(G\right)}=k\left(A\right)=K(G,d).}\end{array}}$

In particular, if d∈[1,r−1], we let $e_1,\dots ,e_{d+1}$ be independent elements of order n₁. Set $e_0=e_1+\dots +e_{d+1}$ and $G_0=\{e_0,e_1,\dots ,e_{d+1}\}$. Then minΔ(G₀) = d by Lemma 2.3.1.

Since $A=e_0\cdot e_1^{n_1-1}\dots e_{d+1}^{n_1-1}$ is an atom with $k\left(A\right)=1+\frac{(n_1-1)d}{n_1}$, it follows that

${\displaystyle \begin{array}{c}\multicolumn{1}{c}{K(G,d)\ge K\left(G_0\right)\ge k\left(A\right)\ge 1+\frac{(n_1-1)d}{n_1}\ge \frac{1+d}{2}.}\end{array}}$

Lemma 3.7.

Suppose $r\ge \lfloor \frac{n_r}{2}\rfloor +1$. Let G₀⊂G be a subset with d | minΔ(G₀), where d∈Δ₁(G) satisfies $r-1\ge d\ge \lfloor \frac{n_r}{2}\rfloor$. If A∈𝒜(G₀) with k(A)>1, then $k\left(A\right)=1+\frac{sd}{n_{r-d}}$ for some s∈ℕ.

In particular,

1. $K(G,r-1)=1+\frac{s(r-1)}{n_1}$ for some $s\ge n_1(u-{\sum }_{i=1}^u\frac{1}{p_i^{k_i}})$, where $n_1=p_1^{k_1}\dots p_u^{k_u}$ with $u,k_1,\dots ,k_u\in \mathbb{N}$ and $p_1,\dots ,p_u$ are pairwise distinct primes.

2. if n_r is a prime power, then $K(G,r-1)=1+\frac{(n_1-1)(r-1)}{n_1}<r$.

3. if n₁ is not a prime power, then K(G,r−1)>r.

Proof.

Let t = r−d and we start with the following claim.

Claim A.

Let B∈ℬ(G₀) with $v_g\left(B\right)\equiv 0(mod{n}_t)$ for each g∈supp(B) and supp(B) is an LCN-set. Then there exists $B_0\in \mathcal{ℬ}\left(G_0\right)$ with B₀ is a product of atoms having cross number 1 and $v_g\left(B_0\right)\equiv 0(mod{n}_t)$ for each g∈supp(B₀) such that BB₀ is a product of atoms having cross number 1.

Proof of Claim A.

Assume to the contrary that there exists a B∈ℬ(G₀) with $v_g\left(B\right)\equiv 0(mod{n}_t)$ for each g∈supp(B) and supp(B) is an LCN-set, such that the assertion does not hold. Suppose |supp(B)| is minimal in all the counterexamples.

Set G₁ = supp(B). If for all g∈G₁, ord(g)∣v_g(B), then B is a product of atoms having cross number 1, a contradiction. Therefore there exits $g_0\in G_1$ such that $ord\left(g_0\right)\nmid v_{g_0}\left(B\right)$. Since $v_g\left(B\right)\equiv 0(mod{n}_t)$ for each g∈supp(B), we infer $v_{g_0}\left(B\right)g_0\in ⟨\{n_{t}g\mid g\in G_1\setminus \{g_0\left\}\right\}⟩$ and $n_t\ne n_r$.

Let $\frac{n_r}{n_t}=q_1^{s_1}\dots q_v^{s_v}$, where $v,s_1,\dots ,s_v\in \mathbb{N}$ and $q_1,\dots ,q_v$ are pairwise distinct primes. Let i∈[1,v] and $H_i=⟨\{\frac{n_r}{n_t{q}_i^{s_i}}{n}_{t}g\mid g\in G_1\setminus \{g_0\left\}\right\}⟩$. Then

${\displaystyle \begin{array}{c}\multicolumn{1}{c}{\frac{n_r}{n_t{q}_i^{s_i}}{v}_{g_0}\left(B\right)g_0\in H_i}\end{array}}$

and H_i is an q_i-group of rank r(H_i)≤r−t = d which implies that there exists $E\subset G_1\setminus \left\{g_0\right\}$ with |E|≤r(H_i)≤d such that

${\displaystyle \begin{array}{c}\multicolumn{1}{c}{\frac{n_r}{n_t{q}_i^{s_i}}{v}_{g_0}\left(B\right)g_0\in ⟨\left\{\frac{n_r}{n_t{q}_i^{s_i}}{n}_{t}g\mid g\in E\right\}⟩.}\end{array}}$

Therefore there exists $B_i\in \mathcal{ℬ}\left(G_1\right)$ such that

${\displaystyle \begin{array}{c}\multicolumn{1}{c}{\left|supp\right(B_i\left)\right|\le d+1,v_{g_0}\left(B_i\right)=\frac{n_r}{n_t{q}_i^{s_i}}{v}_{g_0}\left(B\right),\text{ and }{n}_t\left|v_g\right(B_i)\text{ for every }g\in G_1\setminus \{g_0\}.}\end{array}}$

Since $supp\left(B_i\right)\subset G_1$ is an LCN-set, we have

${\displaystyle \begin{array}{c}\multicolumn{1}{c}{min\Delta \left(supp\right(B_i\left)\right)\le \left|supp\right(B_i\left)\right|-2\le d-1.}\end{array}}$

Note that $d|min\Delta (G_0\left)\right|min\Delta \left(supp\right(B_i\left)\right)$. We infer minΔ(supp(B_i)) = 0 and hence supp(B_i) is half-factorial. Therefore B_i is a product of atoms having cross number 1.

Since $gcd\left(\right\{v_{g_0}\left(B_i\right)\mid i\in [1,v]\left\}\right)=v_{g_0}\left(B\right)$, there exist $x_1,\dots ,x_v\in \mathbb{N}$ such that

${\displaystyle \begin{array}{c}\multicolumn{1}{c}{v_{g_0}(B{B}_1^{x_1}\dots B_v^{x_v})\equiv 0(modord(g_0\left)\right).}\end{array}}$

Therefore $B{B}_1^{x_1}\dots B_v^{x_v}=g_0^{yord\left(g_0\right)}C$ for some y∈ℕ and $C\in \mathcal{ℬ}(G_1\setminus \{g_0\left\}\right)$. Note $n_t\left|v_g\right(B_i)$ for every i∈[1,v] and every $g\in G_1\setminus \left\{g_0\right\}$. Thus $n_t\left|v_g\right(C)$ for each g∈supp(C). Since |supp(C)|<|supp(B)|, it follows by the minimality of |supp(B)| that there exists $C_0\in \mathcal{ℬ}\left(supp\right(G_0\left)\right)$ satisfying C₀ is a product of atoms having cross number 1 and $n_t\left|v_g\right(C_0)$ for each g∈supp(C₀), such that CC₀ is a product of atoms having cross number 1. Let $B_0=C_0{B}_1^{x_1}\dots B_{t_1}^{x_{t_1}}$. Then BB₀ is a product of atoms having cross number 1, a contradiction to our assumption.

Let A∈𝒜(G₀) be with k(A)>1. Then Lemma 2.6.2 implies that supp(A) is an LCN-set. Set $B=A^{n_t}$ and hence Claim A implies that there exist atoms $W_1,\dots ,W_{\ell }\in \mathcal{𝒜}\left(G_0\right)$ having cross number 1, where ℓ∈ℕ₀, such that $A^{n_t}{W}_1\dots W_{\ell }$ is a product of atoms having cross number 1. Therefore $\left\{n_{t}k\right(A)+\ell ,n_t+\ell \}\subset L(A^{n_t}{W}_1\dots W_{\ell })$. Since d | minΔ(G₀), we infer $d\left|\right(n_{t}k\left(A\right)-n_t)$. It follows that $k\left(A\right)=1+\frac{sd}{n_t}$ for some s∈ℕ.

Now we begin to prove the “in-particular’’ parts.

1. For every j∈[1,u] and every m∈ℤ, we denote by ∥m∥_j the least positive residue of m modulo $p_j^{k_j}$, that is, $\parallel m{\parallel }_j\in [1,p_j^{k_j}]$ and $\parallel m{\parallel }_j\equiv m(mod{p}_j^{k_j})$.

By definition of K(G,r−1), we have $K(G,r-1)=1+\frac{sd}{n_1}$ for some s∈ℕ. Let H be a subgroup of G with

${\displaystyle \begin{array}{c}\multicolumn{1}{c}{H\cong C_{n_1}^r\cong C_{p_1^{k_1}}^{r-1}\oplus \dots \oplus C_{p_u^{k_u}}^{r-1}\oplus C_{n_1}}\end{array}}$

and let $(e_{1,1},\dots ,e_{1,r-1},\dots ,e_{u,1},\dots ,e_{u,r-1},e)$ be a basis of H with ord(e) = n₁ and $ord\left(e_{j,i}\right)=p_j^{k_j}$ for all i∈[1,r−1], j∈[1,u]. Set $e_0=e+{\sum }_{j=1}^u{\sum }_{i=1}^{r-1}{e}_{j,i}$ and $G_2=\{e_{1,1},\dots ,e_{1,r-1},\dots ,e_{u,1},\dots ,e_{u,r-1},e,e_0\}$. For every W∈𝒜(G₂), we have

${\displaystyle \begin{array}{c}\multicolumn{1}{c}{W=e_0^{v_{e_0}\left(W\right)}{e}^{n_1-v_{e_0}\left(W\right)}\prod _{i\in [1,r-1],j\in [1,u]}{e}_{j,i}^{p_j^{k_j}-\parallel v_{e_0}\left(W\right){\parallel }_j}}\end{array}}$

and

${\displaystyle \begin{array}{c}\multicolumn{1}{c}{k\left(W\right)=1+\sum _{j=1}^u\frac{p_j^{k_j}-\parallel v_{e_0}\left(W\right){\parallel }_j}{p_j^{k_j}}(r-1).}\end{array}}$

We suppose that $U_1\dots U_{{\ell }_1}=V_1\dots V_{{\ell }_2}$ with ${\ell }_2-{\ell }_1=min\Delta \left(G_2\right)$, where ${\ell }_1,{\ell }_2\in \mathbb{N}$ and $U_1,\dots ,U_{{\ell }_1}$, $V_1,\dots ,V_{{\ell }_2}\in \mathcal{𝒜}\left(G_2\right)$. Then

${\displaystyle \begin{array}{c}\multicolumn{1}{c}{k(U_1\dots U_{{\ell }_1})={\ell }_1+\sum _{x=1}^{{\ell }_1}\sum _{j=1}^u\frac{p_j^{k_j}-\parallel v_{e_0}\left(U_x\right){\parallel }_j}{p_j^{k_j}}(r-1)={\ell }_2+\sum _{y=1}^{{\ell }_2}\sum _{j=1}^u\frac{p_j^{k_j}-\parallel v_{e_0}\left(V_y\right){\parallel }_j}{p_j^{k_j}}(r-1).}\end{array}}$

Since $v_{e_0}(U_1\dots U_{{\ell }_1})=v_{e_0}(V_1\dots V_{{\ell }_2})$, we infer

${\displaystyle \begin{array}{c}\multicolumn{1}{c}{\sum _{x=1}^{{\ell }_1}\parallel v_{e_0}\left(U_x\right){\parallel }_j\equiv \sum _{y=1}^{{\ell }_2}\parallel v_{e_0}\left(V_y\right){\parallel }_j(mod{p}_j^{k_j})\text{ for each }j\in [1,u].}\end{array}}$

Therefore

${\displaystyle \begin{array}{cc}\multicolumn{1}{c}{}& \sum _{x=1}^{{\ell }_1}\frac{p_j^{k_j}-\parallel v_{e_0}\left(U_x\right){\parallel }_j}{p_j^{k_j}}-\sum _{y=1}^{{\ell }_2}\frac{p_j^{k_j}-\parallel v_{e_0}\left(V_y\right){\parallel }_j}{p_j^{k_j}}\hfill \\ ={\ell }_1-{\ell }_2-\frac{\sum _{x=1}^{{\ell }_1}\parallel v_{e_0}\left(U_x\right){\parallel }_j-\sum _{y=1}^{{\ell }_2}\parallel v_{e_0}\left(V_y\right){\parallel }_j}{p_j^{k_j}}\in \mathbb{Z}\hfill \end{array}}$

for each j∈[1,u], whence

${\displaystyle \begin{array}{c}\multicolumn{1}{c}{\sum _{x=1}^{{\ell }_1}\sum _{j=1}^u\frac{p_j^{k_j}-\parallel v_{e_0}\left(U_x\right){\parallel }_j}{p_j^{k_j}}-\sum _{y=1}^{{\ell }_2}\sum _{j=1}^u\frac{p_j^{k_j}-\parallel v_{e_0}\left(V_y\right){\parallel }_j}{p_j^{k_j}}\in \mathbb{Z}.}\end{array}}$

Since $min\Delta \left(G_2\right)={\ell }_2-{\ell }_1$, we infer r−1 | minΔ(G₂) which implies that K(G,r−1)≥K(G₂). Let $W_1\in \mathcal{𝒜}\left(G_2\right)$ be the atom with $v_{e_0}\left(W_1\right)=1$. Then $k\left(W_1\right)=1+(u-{\sum }_{i=1}^u\frac{1}{p_i^{k_i}})(r-1)$.

Since $K(G,r-1)=1+\frac{s(r-1)}{n_1}\ge k\left(W_1\right)=1+(u-{\sum }_{i=1}^t\frac{1}{p_i^{k_i}})(r-1)$ for some s∈ℕ, it follows that $s\ge n_1(u-{\sum }_{i=1}^u\frac{1}{p_i^{k_i}})$.

2. If n_r is a prime power, then K(G)<r by Lemma 2.1. It follows by 1. that $r>K(G,r-1)=1+\frac{s(r-1)}{n_1}\ge 1+\frac{(n_1-1)(r-1)}{n_1}$ for some s∈ℕ. Therefore $K(G,r-1)=1+\frac{(n_1-1)(r-1)}{n_1}$.

3. If n₁ is not a prime power, then u≥2 and $u-{\sum }_{i=1}^u\frac{1}{p_i^{k_i}}>u-\frac{u}{2}\ge 1$. It follows by 1. that K(G,r−1)>1+r−1 = r. □

Proposition 3.8.

Let s∈ℕ be maximal such that G has a subgroup isomorphic to $C_{n_r}^s$. Suppose d∈Δ₁(G) with $d\ge max\{\lfloor \frac{n_r}{2}\rfloor ,\lfloor \frac{r+1}{2}\rfloor \}$.

1. If d≥r, then ${\rho }^{\ast }(G,d)=\frac{n_r}{n_r-d}$.

2. If d≥n_r−1, then ρ*(G,d) = K(G,d).

3. Suppose r = n_r−1≥3. If $n_1=n_r=p^k$ for some prime p and k∈ℕ, then $K(G,r-1)<{\rho }^{\ast }(G,r-1)=1+\frac{(r+1)(r-1)}{r+2}<r$. Otherwise, K(G,r−1) = ρ*(G,r−1).

Proof.

1. Since $d\ge max\{\lfloor \frac{n_r}{2}\rfloor ,\lfloor \frac{r+1}{2}\rfloor \}$ and d≥r, it follows by Lemma 2.5(items 3 and 4) that d>m(G) and $d\in [n_r-s-1,n_r-2]$.

Let $e_1,\dots ,e_{n_r-d-1}$ be independent elements with order n_r and let $e_0=-e_1-\dots -e_{n_r-d-1}$. Then $min\Delta \left(\right\{e_0,e_1,\dots ,e_{n_r-d-1}\left\}\right)=|n_r-(n_r-d-1)-1|=d$ by Lemma 2.3.2. Since $A=e_0{e}_1\dots e_{n_r-d-1}$ is an atom, we infer $L\left(A^{n_r}\right)=\{n_r-d,n_r\}$. Therefore

${\displaystyle \begin{array}{c}\multicolumn{1}{c}{{\rho }^{\ast }(G,d)\ge \rho \left(\right\{e_0,e_1,\dots ,e_{n_r-d-1}\left\}\right)\ge \rho \left(L\right(A^{n_r}\left)\right)=\frac{n_r}{n_r-d}.}\end{array}}$

Let G₀⊂G with d | minΔ(G₀) be such that $\rho \left(G_0\right)={\rho }^{\ast }(G,d)$. Then there exists B∈ℬ(G₀) with ρ(L(B)) = ρ*(G,d). Set

${\displaystyle \begin{array}{c}\multicolumn{1}{c}{B=U_1\dots U_k=V_1\dots V_{\ell },}\end{array}}$

where k = minL(B), ℓ = maxL(B), and $U_1,\dots ,U_k,V_1,\dots ,V_{\ell }\in \mathcal{𝒜}\left(G_0\right)$ are atoms.

If there exists i∈[1,k] such that k(U_i)>1, then Lemma 2.6.2 implies that supp(U_i) is an LCN-set and hence minΔ(supp(U_i))≤m(G). Since

${\displaystyle \begin{array}{c}\multicolumn{1}{c}{d|min\Delta (G_0\left)\right|min\Delta \left(supp\right(U_i\left)\right),}\end{array}}$

we get a contradiction to d>m(G). Therefore k(U_i)≤1 for all i∈[1,k].

If there exists i∈[1,ℓ] such that k(V_i)<1, then Lemma 2.6.1 implies that $k\left(V_i\right)=\frac{n_r-d}{n_r}$. Therefore $k\left(V_i\right)\ge \frac{n_r-d}{n_r}$ for all i∈[1,ℓ]. It follows that

${\displaystyle \begin{array}{c}\multicolumn{1}{c}{k\ge \sum _{i=1}^{k}k\left(U_i\right)=k\left(B\right)=\sum _{i=1}^{\ell }k\left(V_i\right)\ge \ell \frac{n_r-d}{n_r}.}\end{array}}$

Then ${\rho }^{\ast }(G,d)=\rho \left(L\right(B\left)\right)=\frac{\ell }{k}\le \frac{n_r}{n_r-d}$ and hence ${\rho }^{\ast }(G,d)=\frac{n_r}{n_r-d}$.

2. Let G₀⊂G be such that d | minΔ(G₀) and $\rho \left(G_0\right)={\rho }^{\ast }(G,d)$. If there exists an atom A∈𝒜(G₀) such that k(A)<1, then Lemma 2.6.1 implies that $k\left(A\right)=\frac{n_r-d}{n_r}\le \frac{1}{n_r}$, a contradiction to |A|≥2. Thus G₀ is an LCN-set. Let B∈ℬ(G₀) such that $\rho \left(L\right(B\left)\right)=\rho \left(G_0\right)={\rho }^{\ast }(G,d)$. Then

${\displaystyle \begin{array}{c}\multicolumn{1}{c}{minL\left(B\right)K(G,d)\ge k\left(B\right)\ge maxL\left(B\right)}\end{array}}$

which implies that ρ*(G,d) = ρ(L(B))≤K(G,d).

Let G₀⊂G be such that d | minΔ(G₀) and K(G₀) = K(G,d). Then there exists an atom A∈𝒜(G₀) such that k(A) = K(G,d)≥1. Since $\{n_r,n_{r}k(A\left)\right\}\subset L\left(A^{n_r}\right)$, we infer

${\displaystyle \begin{array}{c}\multicolumn{1}{c}{{\rho }^{\ast }(G,d)\ge \rho \left(G_0\right)\ge \rho \left(L\right(A^{n_r}\left)\right)\ge k\left(A\right)=K(G,d)}\end{array}}$

and hence ρ*(G,d) = K(G,d).

3. Let r = n_r−1≥3 and we proceed to prove the following claim.

Claim B.

Suppose ρ*(G,r−1)>K(G,r−1) and let G₀⊂G be such that (r−1) | minΔ(G₀) and ρ(G₀)>K(G,r−1).

1. There exists g∈G₀ with ord(g) = n_r such that −g∈G₀.

2. Let $G_2\subset G_0\setminus \{g,-g\}$ with |G₂| = r. If there exists a∈[1,n_r−1] such that ag∈⟨G₂⟩, then $g={\sum }_{h\in G_2}h$, $G\cong C_{n_r}^r$, and G₂ is a basis of G.

3. $G\cong C_{n_r}^r$, $G_0=\{e_1,\dots ,e_r,g,-g\}$, where $g=e_1+\dots +e_r$ and $(e_1,\dots ,e_r)$ is a basis of G, and $\rho \left(G_0\right)=1+\frac{n_r(r-1)}{n_r+1}$. In particular, ${\rho }^{\ast }(G,r-1)=1+\frac{n_r(r-1)}{n_r+1}$.

Proof of Claim B.

By Lemma 2.5.2, we infer that $min\Delta \left(G_0\right)=r-1=n_r-2=max{\Delta }^{\ast }\left(G\right)$.

a. If G₀ is an LCN-set, then for every B∈ℬ(G₀), we have

${\displaystyle \begin{array}{c}\multicolumn{1}{c}{minL\left(B\right)K(G,r-1)\ge k\left(B\right)\ge maxL\left(B\right)}\end{array}}$

which implies that ρ(L(B))≤K(G,r−1). Therefore ρ(G₀)≤K(G,r−1), a contradiction. Thus there exists A∈𝒜(G₀) such that k(A)<1. Lemma 2.6.1 implies that A = g(−g) for some g∈G₀ with ord(g) = n_r. Hence {g,−g}⊂G₀.

b. Let E⊂G₂ be minimal such that there exists a∈[1,n_r−1] such that ag∈⟨E⟩ and let $d_g\in [1,n_r-1]$ be minimal such that d_gg∈⟨E⟩. Then there exists an atom V∈𝒜(E∪{g}) with $v_g\left(A\right)=d_g$ and |supp(V)|≤r+1. Let $V=g^{d_g}T$, where T∈ℱ(E). Then

${\displaystyle \begin{array}{c}\multicolumn{1}{c}{V{(-g)}^{n_r}={\left(g\right(-g\left)\right)}^{d_g}\left({(-g)}^{n_r-d_g}T\right),\text{ where }L\left({(-g)}^{n_r-d_g}T\right)\subset [1,n_r-d_g].}\end{array}}$

Note that for each $\ell \in L\left({(-g)}^{n_r-d_g}T\right)$, we have r−1 | d_g+ℓ−2. Therefore $L\left({(-g)}^{n_r-d_g}T\right)=\{n_r-d_g\}$ or (d_g = 1 and $L\left({(-g)}^{n_r-1}T\right)=\left\{1\right\}$). We distinguish two cases.

Suppose $L\left({(-g)}^{n_r-d_g}T\right)=\{n_r-d_g\}$. Let $T=T_1\dots T_{n_r-d_g}$ such that (−g)T_i, $i\in [1,n_r-d_g]$, are atoms, where $T_1,\dots ,T_{n_r-d_g}\in \mathcal{ℱ}\left(E\right)$. Thus −g∈⟨E⟩ which implies that d_g = 1 by the minimality of d_g. The minimality of E implies that supp(T_i) = E for each i∈[1,n_r−1]. Then for every h∈E, $v_h\left(V\right)\ge n_r-1$. Therefore ord(h) = n_r and $v_h\left(V\right)=n_r-1$. It follows that

${\displaystyle \begin{array}{c}\multicolumn{1}{c}{T_1=\dots =T_{n_r-1}=\prod _{h\in E}h.}\end{array}}$

If k((−g)T₁)<1, then Lemma 2.6.1 implies that T₁ = g, a contradiction. Therefore

${\displaystyle \begin{array}{c}\multicolumn{1}{c}{k\left(\right(-g\left)T_1\right)=\frac{1}{n_r}+\sum _{h\in E}\frac{1}{ord\left(h\right)}=\frac{1+\left|E\right|}{n_r}\ge 1.}\end{array}}$

Since |E|≤|G₂|≤r and r = n_r−1, we have |E| = |G₂| = r. Let $G_2=\{e_1\dots ,e_r\}$. Then $V=g{e}_1^{n_r-1}\dots e_r^{n_r-1}$ implies that $(e_1,\dots ,e_r)$ is a basis of G, $G\cong C_{n_r}^r$, and $g=e_1+\dots +e_r$.

Suppose d_g = 1 and $L\left({(-g)}^{n_r-1}T\right)=\left\{1\right\}$. Note that V = gT and hence |T|≥2. We infer $k\left({(-g)}^{n_r-1}T\right)>1$. It follows by Lemma 2.6.2 that {−g}∪E is an LCN-set and minΔ({−g}∪E)≤|E|−1. Since (r−1) | minΔ({−g}∪E) and |E|≤|G₂| = r, we have

${\displaystyle \begin{array}{c}\multicolumn{1}{c}{\left|E\right|=\left|G_2\right|=r\text{ and }min\Delta \left(\right\{-g\}\cup G_2)=r-1.}\end{array}}$

Let $E_1\subset \{-g\}\cup G_2$ be a minimal non-half-factorial LCN-set. Then

${\displaystyle \begin{array}{c}\multicolumn{1}{c}{min\Delta \left(E_1\right)=min\Delta \left(\right\{-g\}\cup G_2)=r-1=max{\Delta }^{\ast }\left(G\right).}\end{array}}$

It follows by Geroldinger and Zhong [14, Lemma 4.2] that |E₁| = r+1 and hence $E_1=\{-g\}\cup G_2$. Therefore

${\displaystyle \begin{array}{c}\multicolumn{1}{c}{\{-g\}\cup G_2\text{ is a minimal non-half-factorial LCN-set.}}\end{array}}$

Let $G_2=\{e_1,\dots ,e_r\}$ and assume

${\displaystyle \begin{array}{c}\multicolumn{1}{c}{V=g{e}_1^{k_1}\dots e_r^{k_r},V_1={(-g)}^{n_r-1}{e}_1^{k_1}\dots e_r^{k_r},}\end{array}}$

where $k_i\in [1,ord(e_i)-1]$ for each i∈[1,r].

If k(V)>1, then Lemma 2.6.2 and [14, Lemma 4.5] imply

${\displaystyle \begin{array}{c}\multicolumn{1}{c}{\left\{g\right\}\cup G_2\text{ is a minimal non-half-factorial LCN-set with }min\Delta \left(\right\{g\}\cup G_2)=max{\Delta }^{\ast }\left(G\right)=r-1.}\end{array}}$

By the minimality of E = G₂, we have for every m∈[1,n_r−1] and every h∈G₂, mg∉⟨G₂∖{h}⟩. Note that n_r≥4. Let $I=\{i\in [1,r]\mid k_i\ge \frac{ord\left(e_i\right)}{2}\}$. Then [14, Lemma 4.4.1] implies that

${\displaystyle \begin{array}{c}\multicolumn{1}{c}{W_1=g^2\prod _{i\in I}{e}_i^{2k_i-ord\left(e_i\right)}\prod _{i\in [1,r]\setminus I}{e}_i^{2k_i}}\end{array}}$

and

${\displaystyle \begin{array}{c}\multicolumn{1}{c}{W_2={(-g)}^{n_r-2}\prod _{i\in I}{e}_i^{2k_i-ord\left(e_i\right)}\prod _{i\in [1,r]\setminus I}{e}_i^{2k_i}}\end{array}}$

are both atoms. Since $V^2={\prod }_{i\in I}{e}_i^{ord\left(e_i\right)}{W}_1$ and $V_1^2={(-g)}^{n_r}{\prod }_{i\in I}{e}_i^{ord\left(e_i\right)}{W}_2$, we infer r−1 | |I|−1 and r−1 | |I|. Hence r = 2, a contradiction to r≥3.

Thus k(V) = 1 which implies that $k_1=\dots =k_r=1$ and $ord\left(e_i\right)=n_r$ for each i∈[1,r]. It follows by k(V₁)>1 and [14, Lemma 4.3.2] that $(-g)e_1^{n_r-1}\dots e_r^{n_r-1}$ is also an atom. Therefore $e_1,\dots ,e_r$ are independent and $g=e_1+\dots +e_r$. Since r(G) = r and exp(G) = n_r, we infer $(e_1,\dots ,e_r)$ is a basis of G and $G\cong C_{n_r}^r$.

c. If for all W∈𝒜(G₀), k(W)≤1, then for every B∈ℬ(G₀), we have

${\displaystyle \begin{array}{c}\multicolumn{1}{c}{minL\left(B\right)\ge k\left(B\right)\ge maxL\left(B\right)\frac{2}{n_r}}\end{array}}$

which implies that $\rho \left(G_0\right)\le \frac{n_r}{2}$. It follows by Lemma 3.6 that $K(G,r-1)\ge 1+\frac{r-1}{2}=\frac{n_r}{2}$, a contradiction to ρ(G₀)>K(G,r−1).

Let W∈𝒜(G₀) with k(W)>1. Then supp(W) is an LCN-set with minΔ(supp(W)) = r−1 by Lemma 2.6.2. Let G₁⊂supp(W) be a minimal non-half-factorial subset. Then minΔ(G₁) = r−1 and hence |G₁| = r+1 by Geroldinger and Zhong [14, Lemma 4.2.1]. Since {g,−g}⊄supp(W), we choose h∈G₁ such that {g,−g}∩(G₁∖{h}) = ∅. Lemma 2.6.3 implies r(⟨G₁∖{h}⟩) = r. Thus there exists a∈[1,n_r−1] such that ag∈⟨G₁∖{h}⟩. Since |G₁∖{h}| = r, it follows by a and b that $G\cong C_{n_r}^r$ and there exists a basis $(e_1,\dots ,e_r)$ of G such that $g=e_1+\dots +e_r$ and $\{e_1,\dots ,e_r,g,-g\}\subset G_0$.

Assume to the contrary that there exists $h_0\in G_0\setminus \{e_1,\dots ,e_r,g,-g\}$. After renumbering if necessary, we may assume that $h_0=k_1{e}_1+\dots +k_t{e}_t$, where t∈[1,r], $k_i\in [1,n_r-1]$ for each i∈[1,t] and $k_1=min\{k_1,\dots k_t\}$. Thus $k_{1}g\in ⟨\{h_0,e_2,\dots ,e_r\}⟩$. It follows by b that $g=h_0+e_2+\dots +e_r$ and hence $h_0=e_1$, a contradiction. Therefore $G_0=\{e_1,\dots ,e_r,g,-g\}$.

We only need to prove $\rho \left(G_0\right)=1+\frac{n_r(r-1)}{n_r+1}$ which immediately implies that ${\rho }^{\ast }(G,r-1)=1+\frac{n_r(r-1)}{n_r+1}$.

Since ${(g{e}_1^{n_r-1}\dots e_r^{n_r-1})}^{n_r}{(-g)}^{n_r}={\left(g\right(-g\left)\right)}^{n_r}{\left(e_1^{n_r}\right)}^{n_r-1}\dots {\left(e_r^{n_r}\right)}^{n_r-1}$, we obtain

${\displaystyle \begin{array}{c}\multicolumn{1}{c}{\rho \left(G_0\right)\ge \frac{n_r+r(n_r-1)}{n_r+1}=1+\frac{n_r(r-1)}{n_r+1}.}\end{array}}$

Let B∈ℬ(G₀) such that ρ(L(B)) = ρ(G₀) and assume that

${\displaystyle \begin{array}{c}\multicolumn{1}{c}{B=U_1\dots U_k=V_1\dots V_{\ell },\text{ and }\frac{\ell }{k}=\rho \left(G_0\right),}\end{array}}$

where k,ℓ∈ℕ and $U_1,\dots ,U_k,V_1,\dots ,V_{\ell }$ are atoms.

Note that $\mathcal{𝒜}\left(G_0\right)=\mathcal{𝒜}(G_0\setminus \{-g\left\}\right)\cup \left\{g\right(-g),(-g)e_1\dots e_r,{(-g)}^{n_r}\}$. By Lemma 2.6.4, we have k(U_i)≥1 for all i∈[1,k] and k(V_j)≤1 for all j∈[1,ℓ]. Since ${\left(\right(-g)e_1\dots e_r)}^{n_r}={(-g)}^{n_r}{\prod }_{i\in [1,r]}{e}_i^{n_r}$ and $\rho \left(B^{n_r}\right)=\rho \left(G_0\right)$, substituting B by $B^{n_r}$, if necessary, we can assume that $U_i\ne (-g)e_1\dots e_r$, $V_j\ne (-g)e_1\dots e_r$ for each i∈[1,k] and each j∈[1,ℓ].

Since there must exist j₀∈[1,ℓ] such that $V_{j_0}=g(-g)$, there must exists i₀∈[1,k] such that $U_{i_0}={(-g)}^{n_r}$ which implies that $V_j\ne {(-g)}^{n_r}$ for all j∈[1,ℓ]. If there exists i₁∈[1,k] such that $U_{i_1}=g^{n_r}$, then $maxL\left(B\right)=n_r+maxL\left(B{\left(U_{i_0}{U}_{i_1}\right)}^{-1}\right)$ which implies that

${\displaystyle \begin{array}{c}\multicolumn{1}{c}{\rho \left(L\right(B\left)\right)=\frac{n_r+maxL\left(B{\left(U_{i_0}{U}_{i_1}\right)}^{-1}\right)}{2+minL\left(B{\left(U_{i_0}{U}_{i_1}\right)}^{-1}\right)}=\rho \left(G\right).}\end{array}}$

Therefore

${\displaystyle \begin{array}{c}\multicolumn{1}{c}{\rho \left(L\right(B\left)\right)=\frac{n_r}{2}=\rho \left(L\right(B{\left(U_{i_0}{U}_{i_1}\right)}^{-1}\left)\right).}\end{array}}$

It follows by Lemma 3.6 that $K(G,r-1)\ge 1+\frac{r-1}{2}=\frac{n_r}{2}$, a contradiction to ρ(G₀)>K(G,r−1). If there exists j₁∈[1,ℓ] such that $V_{j_1}=g^{n_r}$, then $maxL\left(B{(-g)}^{n_r}\right)=n_r+\ell -1$ which implies that

${\displaystyle \begin{array}{c}\multicolumn{1}{c}{\rho \left(L\right(B{(-g)}^{n_r}\left)\right)\ge \frac{n_r+\ell -1}{k}>\frac{\ell }{k}=\rho \left(G_0\right),}\end{array}}$

a contradiction.

To sum up, we obtain

${\displaystyle \begin{array}{c}\multicolumn{1}{c}{\{U_i\mid i\in [1,k\left]\right\}\subset \left(A\right(G_0\setminus \{-g\})\setminus \{g^{n_r}\left\}\right)\cup \left\{{(-g)}^{n_r}\right\}}\end{array}}$

and

${\displaystyle \begin{array}{c}\multicolumn{1}{c}{\{V_j\mid j\in [1,\ell \left]\right\}\subset \mathcal{𝒜}(G_0\setminus \{g,-g\left\}\right)\cup \left\{\right(-g\left)g\right\}.}\end{array}}$

Let $I_1=\{i\in [1,k]\mid U_i={(-g)}^{n_r}\}$, $I_2=\{i\in [1,k]\mid v_g(U_i)\ge 1\}$, and J = {j∈[1,ℓ]∣V_j = g(−g)}. Then

${\displaystyle \begin{array}{c}\multicolumn{1}{c}{n_r\left|I_1\right|=\left|J\right|=v_{-g}\left(B\right)=v_g\left(B\right)=\sum _{i\in I_2}{v}_g\left(U_i\right)\ge \left|I_2\right|}\end{array}}$

and

${\displaystyle \begin{array}{ccc}\multicolumn{1}{c}{\frac{2}{n_r}\left|J\right|+\ell -\left|J\right|=k\left(B\right)}& =\hfill & \sum _{i\in I_2}k\left(U_i\right)+k-\left|I_2\right|\hfill \\ =\hfill & \sum _{i\in I_2}(1+\frac{n_r-v_g\left(U_i\right)}{n_r}(n_r-2\left)\right)+k-\left|I_2\right|\hfill \\ \\ \\ =\hfill & k+(n_r-2)\left|I_2\right|-\frac{n_r-2}{n_r}\sum _{i\in I_2}{v}_g\left(U_i\right)\hfill \\ =\hfill & k+(n_r-2)\left|I_2\right|-\frac{\left|J\right|(n_r-2)}{n_r}.\hfill \end{array}}$

Therefore $k\ge \left|I_1\right|+\left|I_2\right|\ge \frac{n_r+1}{n_r}\left|I_2\right|$ and $\frac{\ell }{k}=1+\frac{\left|I_2\right|}{k}(n_r-2)\le 1+\frac{n_r(n_r-2)}{n_r+1}$. It follows that $\rho \left(G_0\right)=1+\frac{n_r(r-1)}{n_r+1}$.

We distinguish two cases to finish the proof.

Suppose that $G\cong C_{p^k}^{p^k-1}$ for some prime p and k∈ℕ with p^k≥4. Then $K(G,p^k-2)=1+\frac{(p^k-1)(p^k-2)}{p^k}$ by Lemma 3.7.2. Let $(e_1,\dots ,e_{p^k-1})$ be a basis of G and $G_0=\{e_1,\dots ,e_{p^k-1},g,-g\}$, where $g={\sum }_{i\in [1,p^k-1]}{e}_i$. By Lemma 2.3.3, we have $p^k-2=min\Delta \left(G_0\right)$. Since

${\displaystyle \begin{array}{c}\multicolumn{1}{c}{{(g{e}_1^{p^k-1}\dots e_r^{p^k-1})}^{p^k}{(-g)}^{p^k}={\left(g\right(-g\left)\right)}^{p^k}{\left(e_1^{p^k}\right)}^{p^k-1}\dots {\left(e_r^{p^k}\right)}^{p^k-1},}\end{array}}$

we obtain

${\displaystyle \begin{array}{c}\multicolumn{1}{c}{{\rho }^{\ast }(G,p^k-2)\ge \rho \left(G_0\right)\ge \frac{p^k+(p^k-1)r}{p^k+1}=1+\frac{p^k(p^k-2)}{p^k+1}>K(G,p^k-2).}\end{array}}$

It follows by Claim B that

${\displaystyle \begin{array}{c}\multicolumn{1}{c}{K(G,p^k-2)<{\rho }^{\ast }(G,p^k-2)=1+\frac{p^k(p^k-2)}{p^k+1}<p^k-1.}\end{array}}$

Suppose that $G\ncong C_{p^k}^{p^k-1}$ for any prime p and any k∈ℕ. Assume to the contrary that K(G,r−1)<ρ*(G,r−1). Then Claim B implies that $G\cong C_{n_r}^r$ and ρ(G,r−1)<r. Since n_r is not prime power, it follows by Lemma 3.7.3 that K(G,r−1)>r, a contradiction. □

4. Proof of main theorems

From now on, we assume G^′ is a further finite abelian group with $G^{\prime }\cong C_{n_1^{\prime }}\oplus \dots \oplus C_{n_r^{\prime }}\prime$, where $r^{\prime },n_1^{\prime },\dots ,n{r}^{\prime }\prime \in \mathbb{N}$ and $1<n_1^{\prime }|\dots |n_r^{\prime }\prime$. For convenience, we collect some necessary results which will be used all through the following two sections without further mention.

Lemma 4.1.

Suppose ℒ(G) = ℒ(G^′) and d∈Δ₁(G). Then

1. If G is isomorphic to a subgroup of G^′, then G≅G^′.

2. $max\left\{r\right(G)-1,exp(G)-2\}=max\left\{r\right(G^{\prime })-1,exp(G^{\prime })-2\}$.

3. $d\in {\Delta }_1\left(G^{\prime }\right)$ and ${\rho }^{\ast }(G,d)={\rho }^{\ast }(G^{\prime },d)$.

Proof.

1. By Geroldinger and Halter-Koch [7, Proposition 7.3.1.3], we have D(G) = D(G^′). It follows from [7, Proposition 5.1.3.2 and 5.1.11.1] that G≅G^′.

2. follows from [7, Corollary 4.3.16] and [14, Theorem 1.1.3].

3. See Proposition 3.5.

Theorem 4.2.

Suppose ℒ(G) = ℒ(G^′). Then

1. If r≥n_r−1 and n₁≠2, then $r\left(G\right)=r\left(G^{\prime }\right)\ge exp\left(G^{\prime }\right)-1$.

2. If r≥n_r−1, n₁≠2, and n_r is a prime power , then r(G) = r(G^′) and $n_1=n_1^{\prime }$.

3. If r≤n_r−3, then exp(G) = exp(G^′) and $r\left(G^{\prime }\right)\le n_r-3$.

4. If $\lfloor \frac{n_r}{2}\rfloor +1\le r\le n_r-3$, then exp(G) = exp(G^′) and r(G) = r(G^′).

5. If r≤n_r−3 and Δ*(G) is an interval, then exp(G) = exp(G^′) and r(G) = r(G^′).

Proof.

1. Assume to the contrary that r(G^′)≠r(G) = r. Since $r-1=max\left\{r\right(G)-1,exp(G)-2\}=max\left\{r\right(G^{\prime })-1,exp(G^{\prime })-2\}$, it follows that $r\left(G^{\prime }\right)-1<exp\left(G^{\prime }\right)-2=r-1$. Let d = r−1. Then $d\in {\Delta }_1\left(G\right)={\Delta }_1\left(G^{\prime }\right)$. Since ${\rho }^{\ast }(G,d)\ge 1+\frac{(n_1-1)d}{n_1}$ by Lemma 3.6 and ${\rho }^{\ast }(G^{\prime },d)=\frac{exp\left(G^{\prime }\right)}{2}$ by Proposition 3.8.1, it follows that ${\rho }^{\ast }(G^{\prime },d)=\frac{exp\left(G^{\prime }\right)}{2}={\rho }^{\ast }(G,d)\ge 1+\frac{(n_1-1)(exp(G^{\prime })-2)}{n_1}$, a contradiction to n₁≠2. Thus $r-1=r\left(G^{\prime }\right)-1=max\left\{r\right(G^{\prime })-1,exp(G^{\prime })-2\}$ and hence $r\left(G^{\prime }\right)\ge exp\left(G^{\prime }\right)-1$.

2. Note that r≥n_r−1≥2. If r = 2, then $G\cong C_3\oplus C_3$. Therefore D(G) = D(G^′) = 5 and $max\{3-2,2-1\}=max\{exp(G^{\prime })-2,r(G^{\prime })-1\}$. It follows that G≅G^′. Thus we can assume r≥3.

By 1., we have $r\left(G\right)=r\left(G^{\prime }\right)\ge exp\left(G^{\prime }\right)-1$. Since n_r is a prime power and n_r≥3, it follows by Lemma 3.7.2 that $K(G,r-1)=1+\frac{n_r(r-1)}{n_r}<r$. Then r≥3 and Propositions 3.8.2 and 3.8.2 imply that ρ*(G,r−1)<r and hence ${\rho }^{\ast }(G^{\prime },r-1)={\rho }^{\ast }(G,r-1)<r$. Therefore

${\displaystyle \begin{array}{c}\multicolumn{1}{c}{1+\frac{(n_1^{\prime }-1)(r-1)}{n_1^{\prime }}\le K(G^{\prime },r-1)\le {\rho }^{\ast }(G^{\prime },r-1)<r}\end{array}}$

by Lemma 3.6. It follows by Lemma 3.7.1 that $K(G^{\prime },r-1)=1+\frac{(n_1^{\prime }-1)(r-1)}{n_1^{\prime }}$.

If K(G,r−1) = ρ*(G,r−1) and $K(G^{\prime },r-1)={\rho }^{\ast }(G^{\prime },r-1)$, then $1+\frac{(n_r-1)(r-1)}{n_r}=1+\frac{(n_1^{\prime }-1)(r-1)}{n_1^{\prime }}$ which infers $n_1=n_r=n_1^{\prime }$.

If K(G,r−1)<ρ*(G,r−1) and $K(G^{\prime },r-1)={\rho }^{\ast }(G^{\prime },r-1)$, then $G\cong C_{n_r}^r$, r = n_r−1, and ${\rho }^{\ast }(G,r-1)=1+\frac{\left(n_r\right)(r-1)}{n_r+1}$ by Proposition 3.8.3. Therefore $1+\frac{\left(n_r\right)(r-1)}{n_r+1}=1+\frac{(n_1^{\prime }-1)(r-1)}{n_1^{\prime }}$ which infers $r+2=n_r+1=n_1^{\prime }\le exp\left(G^{\prime }\right)$, a contradiction.

If K(G,r−1) = ρ*(G,r−1) and $K(G^{\prime },r-1)<{\rho }^{\ast }(G^{\prime },r-1)$, then $G^{\prime }\cong C_{exp\left(G^{\prime }\right)}^r$, r = exp(G^′)−1, and ${\rho }^{\ast }(G^{\prime },r-1)=1+\frac{exp\left(G^{\prime }\right)(r-1)}{exp\left(G^{\prime }\right)+1}$ by Proposition 3.8.3. Therefore $1+\frac{(n_1-1)(r-1)}{n_1}=1+\frac{(exp(G^{\prime }\left)\right)(r-1)}{exp\left(G^{\prime }\right)+1}$ which infers $n_r\ge n_1=exp\left(G^{\prime }\right)+1=r+2$, a contradiction.

If K(G,r−1)<ρ*(G,r−1) and $K(G^{\prime },r-1)<{\rho }^{\ast }(G^{\prime },r-1)$, then $G\cong C_{r+1}^r\cong G^{\prime }$ by Proposition 3.8.3.

3. Note that n_r≥4. Assume to the contrary that $r\left(G^{\prime }\right)\ge n_r-2$. Let $d=n_r-3\in [1,r(G^{\prime })-1]$. Therefore $d\in {\Delta }_1\left(G^{\prime }\right)={\Delta }_1\left(G\right)$ and Lemma 3.6 implies that ${\rho }^{\ast }(G^{\prime },d)\ge 1+\frac{d}{2}$. Since d≥max{r,⌊n_r∕2⌋}, Proposition 3.8.1 implies that ${\rho }^{\ast }(G,d)=n_r∕3<1+\frac{d}{2}$, a contradiction to ${\rho }^{\ast }(G,d)={\rho }^{\ast }(G^{\prime },d)$.

Thus $r\left(G^{\prime }\right)\le n_r-3$. Since $n_r-2=max\left\{r\right(G)-1,exp(G)-2\}=max\left\{r\right(G^{\prime })-1,exp(G^{\prime })-2\}$, it follows that $n_r-2=exp\left(G^{\prime }\right)-2$.

4. By 3, exp(G) = exp(G^′) and $r\left(G^{\prime }\right)\le exp\left(G^{\prime }\right)-3$. Assume to the contrary that r(G)≠r(G^′).

Suppose that r(G)>r(G^′). Choose d = r−1. Therefore $d\in {\Delta }_1\left(G\right)={\Delta }_1\left(G^{\prime }\right)$ and Lemma 3.6 implies that $\rho (G,d)\ge 1+\frac{r-1}{2}$. Since $d\ge max\left\{r\right(G^{\prime }),\lfloor exp(G^{\prime })∕2\rfloor \}$, Proposition 3.8.1 implies that $\rho (G^{\prime },d)=\frac{exp\left(G^{\prime }\right)}{exp\left(G^{\prime }\right)-d}\le \frac{n_r}{4}<1+\frac{r-1}{2}$, a contradiction to ${\rho }^{\ast }(G,d)={\rho }^{\ast }(G^{\prime },d)$.

Suppose that r(G)<r(G^′). Choose d = r∈[1,r(G^′)−1]. Then $d\in {\Delta }_1\left(G^{\prime }\right)={\Delta }_1\left(G\right)$ and Lemma 3.6 implies that ${\rho }^{\ast }(G^{\prime },d)\ge 1+\frac{r}{2}$. Since d≥max{r,⌊n_r∕2⌋}, Proposition 3.8.1 implies that ${\rho }^{\ast }(G,d)=\frac{n_r}{n_r-r}=1+\frac{r}{n_r-r}<1+\frac{r}{2}$, a contradiction to ${\rho }^{\ast }(G,d)={\rho }^{\ast }(G^{\prime },d)$.

5. By 3, we have $r\left(G^{\prime }\right)+3\le exp\left(G^{\prime }\right)=n_r$ and by 4, we can assume that $r\le \lfloor \frac{n_r}{2}\rfloor$ and $r\left(G^{\prime }\right)\le \lfloor \frac{n_r}{2}\rfloor$. Since Δ*(G) is an interval, we obtain that ${\Delta }^{\ast }\left(G^{\prime }\right)={\Delta }_1\left(G^{\prime }\right)={\Delta }_1\left(G\right)={\Delta }^{\ast }\left(G\right)$ is an interval by Lemma 2.5.5. Let k be maximal such that there exists a subgroup H of G with $H\cong C_{n_r}^k$ and let k^′ be maximal such that there exists a subgroup H^′ of G^′ with $H^{\prime }\cong C_{n_r}^{k^{\prime }}$. Then 2r≥r+k≥n_r−2 and $2r\left(G^{\prime }\right)\ge r\left(G^{\prime }\right)+k^{\prime }\ge n_r-2$ by Lemma 2.5.5.

Assume to the contrary that r≠r(G^′) and by symmetry, we can assume r<r(G^′). Thus $\frac{n_r}{2}-1\le r<r\left(G^{\prime }\right)\le \lfloor \frac{n_r}{2}\rfloor$ which implies that n_r is even, n_r = 2r+2, and r(G^′) = r+1. Since Δ*(G) and ${\Delta }^{\ast }\left(G^{\prime }\right)$ are intervals, it follows by Lemma 2.5.5 that $G\cong C_{n_r}^r$ and k^′≥r. Then G is a subgroup of G^′ which implies that G≅G^′ by Lemma 4.1.1, a contradiction.

Proof of Theorem 1.2.

By definition of transfer Krull monoids, it follows that $\mathcal{ℒ}\left(G\right)=\mathcal{ℒ}\left(H\right)=\mathcal{ℒ}\left(H^{\prime }\right)=\mathcal{ℒ}\left(G^{\prime }\right)$.

1. Let r≤n−3. If Δ*(G) is not an interval, then [27, Theorems 1.1 and 1.2] implies that G≅G^′.

Suppose Δ*(G) is an interval. Then Theorem 4.2.5 implies that n = exp(G^′) and r = r(G^′). Therefore G^′ is isomorphic to a subgroup of G which implies that G≅G^′ by Lemma 4.1.1.

2. If n = 2, then G is an elementary two-group and the assertion follows by Geroldinger and Halter-Koch [7, Theorem 7.3.3]. We assume n≥3 is a prime power. Then Theorem 4.2.2 implies that r = r(G^′) and $n=n_1^{\prime }$. Therefore G is isomorphic to a subgroup of G^′ which implies that G≅G^′ by Lemma 4.1.1.

5. Concluding remarks and conjectures

Throughout this section, let $G\cong C_n^r$ with D(G)≥4 and $n=p_1^{k_1}\dots p_u^{k_u}$, where $u,k_1,\dots ,k_u\in \mathbb{N}$ and $p_1,\dots ,p_u$ are pair-wise distinct primes.

Conjecture 5.1.

Suppose r≥n−1. Then the following hold:

1. $K(G,r-1)=1+\frac{s(r-1)}{n}$ for some s∈ℕ with gcd(s,n) = 1.

2. $K(G,r-1)=1+\left({\sum }_{i=1}^u\frac{p_i^{k_i}-1}{p_i^{k_i}}\right)(r-1)$.

It is easy to see that C2 implies C1. By Lemma 3.7.1, we have $K(G,r-1)=1+\frac{s(r-1)}{n}$ for some s∈ℕ and if n is a prime power, then C2 holds.

Proposition 5.2.

Let G^′ be a finite abelian group with ℒ(G^′) = ℒ(G). If r≥n−1 and C1 holds, then G≅G^′.

Proof.

If n is a prime power, then the assertion follows by Theorem 1.2.

Suppose n is not a prime power. Then r≥n−1≥5. Let ℒ(G) = ℒ(G^′) and let $G^{\prime }\cong C_{n_1^{\prime }}\oplus \dots \oplus C_{n_r^{\prime }}\prime$ with $r^{\prime },n_1^{\prime },\dots ,n_r^{\prime }\prime \in \mathbb{N}$ and $1<n_1^{\prime }|\dots |n_r^{\prime }\prime$. Then Theorem 4.2.1 implies that $r=r^{\prime }\ge n_r^{\prime }\prime -1$. By Lemma 3.7.1 and Proposition 3.8(items 2. and 3.), we have ${\rho }^{\ast }(G,r-1)=K(G,r-1)=1+\frac{s(r-1)}{n}$ for some s∈ℕ. If ${\rho }^{\ast }(G^{\prime },r-1)=K(G^{\prime },r-1)$, then Lemma 3.7.1 implies that ${\rho }^{\ast }(G^{\prime },r-1)=1+\frac{s^{\prime }(r-1)}{n_1^{\prime }}$ for some s^′∈ℕ. Thus ${\rho }^{\ast }(G,r-1)={\rho }^{\ast }(G^{\prime },r-1)$ implies that $\frac{s}{n}=\frac{s^{\prime }}{n_1^{\prime }}$ and hence $n|n_1^{\prime }$ by gcd(s,n) = 1. Therefore G is isomorphic to a subgroup of G^′ which implies that G≅G^′ by Lemma 4.1.1.

Suppose ${\rho }^{\ast }(G^{\prime },r-1)\ne K(G^{\prime },r-1)$. Then Lemma 3.6 implies that ${\rho }^{\ast }(G^{\prime },r-1)>K(G^{\prime },r-1)$ and hence Proposition 3.8(items 2. and 3.) implies that $n_r^{\prime }=r+1$ and ${\rho }^{\ast }(G,r-1)=1+\frac{(r+1)(r-1)}{r+2}$. Thus ${\rho }^{\ast }(G,r-1)={\rho }^{\ast }(G^{\prime },r-1)$ implies that $\frac{s}{n}=\frac{r+1}{r+2}$. Since gcd(s,n) = 1, we have n = r+2, a contradiction.

Recall that $K^{\ast }\left(G^{\prime }\right)\le K\left(G^{\prime }\right)$ for all finite abelian group G^′ and there is known no group G^′ with $K^{\ast }\left(G^{\prime }\right)<K\left(G^{\prime }\right)$.

Proposition 5.3.

Let G^′ be a finite abelian group with ℒ(G^′) = ℒ(G). If r≥max{(u−1)n+1,n}≥3 and K(G) = K*(G), then G≅G^′.

Proof.

If u = 1, then the assertion follows by Theorem 1.2.2. Suppose u≥2. Note $K\left(G\right)=K^{\ast }\left(G\right)=\frac{1}{n}+r\left({\sum }_{i=1}^u\frac{p_i^{k_i}-1}{p_i^{k_i}}\right)=1+\frac{s}{n}(r-1)+{\sum }_{i=1}^u\frac{p_i^{k_i}-1}{p_i^{k_i}}-\frac{n-1}{n}$, where $s=n\left({\sum }_{i=1}^u\frac{p_i^{k_i}-1}{p_i^{k_i}}\right)$. Since r≥(ω(n)−1)n+1, we have ${\sum }_{i=1}^u\frac{p_i^{k_i}-1}{p_i^{k_i}}-\frac{n-1}{n}<u-1\le \frac{r-1}{n}$. It follows by Lemma 3.7.1 that $1+\frac{s}{n}(r-1)\le K(G,r-1)=1+\frac{s^{\prime }}{n}(r-1)\le 1+\frac{s+1}{n}(r-1)$, where s^′∈ℕ. Therefore $1+\frac{s}{n}(r-1)=K(G,r-1)$. Since $s=n\left({\sum }_{i=1}^u\frac{p_i^{k_i}-1}{p_i^{k_i}}\right)$, we have gcd(s,n) = 1. The assertion follows by Proposition 5.2.