On half-factoriality of transfer Krull monoids

Abstract

Let H be a transfer Krull monoid over a subset G₀ of an abelian group G with finite exponent. Then every non-unit $a\in H$ can be written as a finite product of atoms, say $a=u_1·\dots ·u_k.$ The set $\mathsf{L}(a)$ of all possible factorization lengths k is called the set of lengths of a, and H is said to be half-factorial if $|\mathsf{L}(a)|=1$ for all $a\in H.$ We show that, if $a\in H$ is a non-unit and $|\mathsf{L}(a^{\lfloor (3\hspace{0.17em}\text{exp}\hspace{0.17em}(G)-3)/2\rfloor })|=1,$ then the smallest divisor-closed submonoid of H containing a is half-factorial. In addition, we prove that, if G₀ is finite and $|\mathsf{L}(\prod _{g\in G_0}{g}^{2\text{ord}(g)})|=1,$ then H is half-factorial.

Keywords:

• Half-factorial
• sets of lengths
• transfer Krull monoids

MATHEMATICS SUBJECT CLASSIFICATION:

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

1. Introduction

Let H be a monoid. If an element $a\in H$ has a factorization $a=u_1·\dots ·u_k,$ where $k\in \mathbb{N}$ and $u_1,\dots ,u_k$ are atoms of H, then k is called a factorization length of a, and the set $\mathsf{L}(a)$ of all possible k is referred to as the set of lengths of a. The monoid H is said to be half-factorial if $|\mathsf{L}(a)|=1$ for every $a\in H.$

The study of half-factoriality was pioneered by Leonard Carlitz in the setting of algebraic number theory: he proved in [4] that the ring of integers ${\mathcal{O}}_K$ of a number field K is half-factorial if and only if the cardinality of its class group is either 1 or 2. After this, the concept of half-factoriality seemed to remain dormant for more than a decade until papers by Abraham Zaks [32], Ladislav Skula [28], and Jan Śliwa [29] simultaneously appeared in 1976. In such papers, half-factoriality was studied in the context of Krull domains and c-monoids. Since then half-factoriality has been investigated in different classes of monoids (see [3, 6, 17, 18]) and integral domains (see [5, 7, 13, 19, 20, 23, 33]).

Given $a\in H,$ let $⟦a⟧=\{b\in H|b\mathrm{ }\text{divides}\mathrm{ }\text{some}\mathrm{ }\text{power}\mathrm{ }\text{of}\mathrm{ }a\}$ be the smallest divisor-closed submonoid of H containing a. Then $⟦a⟧$ is half-factorial if and only if $|\mathsf{L}(a^n)|=1$ for all $n\in \mathbb{N},$ and H is half-factorial if and only if $⟦c⟧$ is half-factorial for every $c\in H.$ It is thus natural to ask:

Does there exist an integer $N\in \mathbb{N}$ such that, if $a\in H$ and $|\mathsf{L}(a^N)|=1,$ then $⟦a⟧$ is half-factorial? (Note that, if $⟦a⟧$ is half-factorial for some $a\in H,$ then of course $|\mathsf{L}(a^k)|=1$ for every $k\in \mathbb{N}.$)

We answer this question affirmatively for transfer Krull monoids over finite abelian groups, and we study the smallest N having the above property (Theorems 1.1 and 1.2).

Transfer Krull monoids and transfer Krull domains are a recently introduced class of monoids and domains including, among others, all commutative Krull domains and wide classes of non-commutative Dedekind domains (see Section 2 and, for a survey, see [10]).

Let H be a transfer Krull monoid over a subset G₀ of an abelian group G. Then H is half-factorial if and only if the monoid $\mathcal{B}(G_0)$ of zero-sum sequences over G₀ is half-factorial (in this case, we also say that the set G₀ is half-factorial). It is a standing conjecture that every abelian group has a half-factorial generating set, which implies that every abelian group can be realized as the class group of a half-factorial Dedekind domain [11].

Suppose now that H is a commutative Krull monoid with class group G and that every class contains a prime divisor. It is a classic result that H is half-factorial if and only if $\left|G\right|\le 2,$ and it turns out that, also for $\left|G\right|\ge 3,$ half-factorial subsets (and minimal non-half-factorial subsets) of the class group G play a crucial role in a variety of arithmetical questions (see [12, Chapter 6.7], [15]). However, we are far away from a good understanding of half-factorial sets in finite abelian groups (see [25] for a survey, and [21, 22, 26]). To mention one open question, the maximal size of half-factorial subsets is unknown even for finite cyclic groups [22]. Our results open the door to a computational approach to the study of half-factorial sets.

More in detail, denote by $\mathsf{hf}(H)$ the infimum of all $N\in \mathbb{N}$ with the following property:

If $a\in H$ and $|\mathsf{L}(a^N)|=1,$ then $⟦a⟧$ is half-factorial.

(Here, as usual, we assume $\inf \varnothing =\infty .$) We call $\mathsf{hf}(H)$ the half-factoriality index of H. If H is not half-factorial, then $\mathsf{hf}(H)$ is the infimum of all $N\in \mathbb{N}$ with the property that $|\mathsf{L}(a^N)|\ge 2$ for every $a\in H$ such that $⟦a⟧$ is not half-factorial. In particular, if G is an abelian group with $\left|G\right|\ge 3,$ then $\mathsf{hf}(\mathcal{B}(G))$ is the infimum of all $N\in \mathbb{N}$ with the property that

For every sequence S over G, if $|\mathsf{L}(S^N)|=1,$ then $|\mathsf{L}(S^k)|=1$ for every $k\ge N.$

Theorem 1.1.

Let H be a transfer Krull monoid over a finite subset G₀ of an abelian group G with finite exponent. The following are equivalent.

1. H is half-factorial.

2. $\mathsf{hf}(H)=1.$

3. G₀ is half-factorial.

4. $|\mathsf{L}(\prod _{g\in G_0}{g}^{2\text{ord}(g)})|=1.$

We observe that in general if H is half-factorial, then $\mathsf{hf}(H)=1.$ But if H is a transfer Krull monoid over a subset of a torsion free group, then $\mathsf{hf}(H)=1$ does not imply that H is half-factorial (see Example 2.4.1). Furthermore, for every $n\in \mathbb{N},$ there exists a Krull monoid H with finite class group such that $\mathsf{hf}(H)=n$ (see Example 2.4.2).

Theorem 1.2.

Let H be a transfer Krull monoid over an abelian group G.

1. $\mathsf{hf}(H)<\infty$ if and only if $\text{exp}\hspace{0.17em}(G)<\infty .$

2. If $\text{exp}\hspace{0.17em}(G)<\infty$ and $\left|G\right|\ge 3$, then $\text{exp}\hspace{0.17em}(G)\le \mathsf{hf}(H)\le \frac{3}{2}(\hspace{0.17em}\text{exp}\hspace{0.17em}(G)-1).$

3. If G is cyclic or $\text{exp}\hspace{0.17em}(G)\le 6$, then $\mathsf{hf}(H)=\hspace{0.17em}\text{exp}\hspace{0.17em}(G).$

We postpone the proofs of Theorems 1.1 and 1.2 to Section 3.

2. Preliminaries

Our notation and terminology are consistent with [12]. Let $\mathbb{N}$ be the set of positive integers, let ${\mathbb{N}}_0=\mathbb{N}\cup \left\{0\right\},$ and let $\mathbb{Q}$ be the set of rational numbers. For integers $a,b\in \mathbb{Z},$ we denote by $[a,b]=\{x\in \mathbb{Z}|a\le x\le b\}$ the discrete, finite interval between a and b.

Atomic monoids. By a monoid, we mean a semigroup with identity, and if not stated otherwise we use multiplicative notation. Let H be a monoid with identity $1=1_H\in H.$ The set of invertible elements of H will be denoted by $H^{\times },$ and we say that H is reduced if $H^{\times }=\left\{1\right\}.$ The monoid H is said to be unit-cancellative if for any two elements $a,u\in H,$ each of the equations au = a or ua = a implies that $u\in H^{\times }.$ Clearly, every cancellative monoid is unit-cancellative.

Suppose that H is unit-cancellative. An element $u\in H$ is said to be irreducible (or an atom) if $u\notin H^{\times }$ and for any two elements $a,b\in H,$ u = ab implies that $a\in H^{\times }$ or $b\in H^{\times }.$ Let $\mathcal{A}(H)$ denote the set of atoms of H. 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 [9, Theorem 2.6]. If $a\in H\setminus H^{\times }$ and $a=u_1\dots u_k,$ where $k\in \mathbb{N}$ and $u_1,\dots ,u_k\in \mathcal{A}(H),$ then k is a factorization length of a, and $${\mathsf{L}}_H(a)=\mathsf{L}(a)=\{k\in \mathbb{N}|k\mathrm{ }\text{is}\text{ a }\text{factorization}\mathrm{ }\text{length}\mathrm{ }\text{of}\mathrm{ }a\}$$ denotes the set of lengths of a. It is convenient to set $\mathsf{L}(a)=\left\{0\right\}$ for all $a\in H^{\times }.$

Let H and B be atomic monoids. The homomorphism $\theta :H\to B$ is called a weak transfer homomorphism if it satisfies the following two properties.

(T1) $B=B^{\times }\theta (H)B^{\times }$ and ${\theta }^{-1}(B^{\times })=H^{\times }.$

(WT2) If $a\in H,\mathrm{ }n\in \mathbb{N},\mathrm{ }{v}_1,\dots ,v_n\in \mathcal{A}(B)$ and $\theta (a)=v_1·\dots ·v_n,$ then there exist $u_1,\dots ,u_n\in \mathcal{A}(H)$ and a permutation $\tau \in {\mathfrak{S}}_n$ such that $a=u_1·\dots ·u_n$ and $\theta (u_i)\in B^{\times }{v}_{\tau (i)}{B}^{\times }$ for each $i\in [1,n].$

A transfer Krull monoid is a monoid H having a weak transfer homomorphism $\theta :H\to \mathcal{B}(G_0),$ where $\mathcal{B}(G_0)$ is the monoid of zero-sum sequences over a subset G₀ of an abelian group G. If H is a commutative Krull monoid with class group G and $G_0\subset G$ is the set of classes containing prime divisors, then there is a weak transfer homomorphism $\theta :H\to \mathcal{B}(G_0).$ Beyond that, there are wide classes of non-commutative Dedekind domains having a weak transfer homomorphism to a monoid of zero-sum sequences ([31, Theorem 1.1], [30, Theorem 4.4]). We refer to [10, 16] for surveys on transfer Krull monoids. If $\theta :H\to \mathcal{B}(G_0)$ is a weak transfer homomorphism, then sets of lengths in H and in $\mathcal{B}(G_0)$ coincide [2, Lemma 2.7] and thus the statements of Theorems 1.1 and 1.2 can be proved in the setting of monoids of zero-sum sequences.

Monoids of zero-sum sequences. Let G be an abelian group and let $G_0\subset G$ be a non-empty subset. Then $〈G_0〉$ denotes the subgroup generated by G₀. In additive combinatorics, a sequence (over G₀) means a finite unordered sequence of terms from G₀ where repetition is allowed, and (as usual) we consider sequences as elements of the free abelian monoid with basis G₀. Let $$S=g_1·\dots ·g_{\ell }=\prod _{g\in G_0}{g}^{{\mathsf{v}}_g(S)}\in \mathcal{F}(G_0)$$ be a sequence over G₀. We call $$\begin{array}{cc}\text{supp}(S)\hfill & =\{g\in G|{\mathsf{v}}_g(S)>0\}\subset G\mathrm{ }\text{the}\mathrm{ }\mathit{support}\mathrm{ }\text{of}\mathrm{ }S,\hfill \\ \left|S\right|\hfill & =\ell =\sum _{g\in G}{\mathsf{v}}_g(S)\in {\mathbb{N}}_0\mathrm{ }\text{the}\mathrm{ }\mathit{length}\mathrm{ }\text{of}\mathrm{ }S,\hfill \\ \sigma (S)\hfill & =\sum _{i=1}^{\ell }{g}_i\mathrm{ }\text{the}\mathrm{ }\mathit{sum}\mathrm{ }\text{of}\mathrm{ }S,\hfill \\ \mathrm{ }\text{and}\mathrm{ }\Sigma (S)\hfill & =\left\{\sum _{i\in I}{g}_i\right|\varnothing \ne I\subset [1,\ell ]\}\mathrm{ }\text{the}\mathrm{ }\mathit{set}\mathrm{ }of\mathrm{ }\mathit{subsequence}\mathrm{ }\mathit{sums}\mathrm{ }\text{of}\mathrm{ }S.\hfill \end{array}$$

The sequence S is said to be

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

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

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

The set of zero-sum sequences $\mathcal{B}(G_0)=\{S\in \mathcal{F}(G_0)|\sigma (S)=0\}\subset \mathcal{F}(G_0)$ is a submonoid, and the set of minimal zero-sum sequences is the set of atoms of $\mathcal{B}(G_0).$ For any arithmetical invariant $*(H)$ defined for a monoid H, we write $*(G_0)$ instead of $*(\mathcal{B}(G_0)).$ In particular, $\mathcal{A}(G_0)=\mathcal{A}(\mathcal{B}(G_0))$ is the set of atoms of $\mathcal{B}(G_0)$ and $\mathsf{hf}(G_0)=\mathsf{hf}(\mathcal{B}(G_0)).$

Let G be an abelian group. We denote by $\text{exp}\hspace{0.17em}(G)$ the exponent of G which is the least common multiple of the orders of all elements of G. If there is no least common multiple, the exponent is taken to be infinity. Let $r\in \mathbb{N}$ 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 $e_i\ne 0$ for all $i\in [1,r]$ 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 $i\in [1,r].$ Suppose G is finite. The r-tuple $(e_1,\dots ,e_r)$ is said to be a basis of G if it is independent and $G=〈e_1〉\oplus \dots \oplus 〈e_r〉.$ For every $n\in \mathbb{N},$ we denote by C_n an additive cyclic group of order n. Since $G\cong C_{n_1}\oplus \dots \oplus C_{n_r},\mathrm{ }r=r(G)$ is the rank of G and $n_r=\hspace{0.17em}\text{exp}\hspace{0.17em}(G)$ is the exponent of G.

Let $G_0\subset G$ be a non-empty subset. For a sequence $S=g_1·\dots ·g_{\ell }\in \mathcal{F}(G_0),$ we call $$\begin{array}{c}\mathsf{k}(S)=\sum _{i=1}^l\frac{1}{\text{ord}(g_i)}\in {\mathbb{Q}}_{\ge 0}\hspace{1em}\text{the}\mathrm{ }\mathit{cross}\mathrm{ }\mathit{number}\mathrm{ }\text{of}\mathrm{ }S,\mathrm{ }\text{and}\\ \mathrm{ }\mathsf{K}(G_0)=\max \{\mathsf{k}(S)|S\in \mathcal{A}(G_0)\}\hspace{1em}\text{the}\mathrm{ }\mathit{cross}\mathrm{ }\mathit{number}\mathrm{ }\text{of}\mathrm{ }{G}_0.\end{array}$$ For the relevance of cross numbers in the theory of non-unique factorizations, see [22, 24, 27] and [12, Chapter 6].

The set G₀ is called

• half-factorial if the monoid $\mathcal{B}(G_0)$ is half-factorial;

• non-half-factorial if the monoid $\mathcal{B}(G_0)$ is not half-factorial;

• minimal non-half-factorial if G₀ is not half-factorial but all its proper subsets are;

• an LCN-set if $\mathsf{k}(A)\ge 1$ for all atoms $A\in \mathcal{A}(G_0).$

The following simple result [12, Proposition 6.7.3] will be used throughout the article without further mention.

Lemma 2.1.

Let G be a finite abelian group and $G_0\subset G$ a subset. Then the following statements are equivalent.

1. G₀ is half-factorial.

2. $\mathsf{k}(U)=1$ for every $U\in \mathcal{A}(G_0).$

3. $\mathsf{L}(B)=\{\mathsf{k}(B)\}$ for every $B\in \mathcal{B}(G_0).$

Lemma 2.2.

Let G be a finite group, let $G_0\subset G$ be a subset, let S be a zero-sum sequence over G₀, and let A be a minimal zero-sum sequence over G₀.

1. If $\mathsf{k}(A)\ne 1$, then $|\mathsf{L}(A^{\hspace{0.17em}\text{exp}\hspace{0.17em}(G)})|\ge 2.$

2. If there exists a zero-sum subsequence T of S such that $|\mathsf{L}(T)|\ge 2$, then $|\mathsf{L}(S)|\ge 2.$

3. If $\mathsf{k}(A)<1$ and $\mathsf{k}(A)$ is minimal over all minimal zero-sum sequences over G₀, then $$\left|\mathsf{L}\left(A^{\lceil \frac{\text{ord}(g)}{{\mathsf{v}}_g(A)}\rceil }\right)\right|\ge 2,\hspace{1em}\mathit{\text{for}}\mathrm{ }\mathit{\text{all}}\mathrm{ }g\in \text{supp}(A).$$

Proof.

1. Suppose $\mathsf{k}(A)\ne 1$ and let $A=g_1·\dots ·g_{\ell },$ where $\ell \in \mathbb{N}$ and $g_1,\dots ,g_{\ell }\in G_0.$ Then $$A^{\hspace{0.17em}\text{exp}\hspace{0.17em}(G)}={\left(g_1^{\text{ord}\left(g_1\right)}\right)}^{\frac{\hspace{0.17em}\text{exp}\hspace{0.17em}\left(G\right)}{\text{ord}\left(g_1\right)}}·\dots ·{\left(g_{\ell }^{\text{ord}\left(g_{\ell }\right)}\right)}^{\frac{\hspace{0.17em}\text{exp}\hspace{0.17em}\left(G\right)}{\text{ord}\left(g_{\ell }\right)}},$$ which implies that $$\left\{\hspace{0.17em}\text{exp}\hspace{0.17em}\left(G\right),\sum _{i=1}^{\ell }\frac{\hspace{0.17em}\text{exp}\hspace{0.17em}\left(G\right)}{\text{ord}\left(g_i\right)}\right\}=\{\hspace{0.17em}\text{exp}\hspace{0.17em}(G),\hspace{0.17em}\text{exp}\hspace{0.17em}(G\left)\mathsf{k}\right(A\left)\right\}\subset \mathsf{L}\left(A^{\hspace{0.17em}\text{exp}\hspace{0.17em}\left(G\right)}\right).$$ It follows by $\mathsf{k}\left(A\right)\ne 1$ that $\left|\mathsf{L}\right(A^{\hspace{0.17em}\text{exp}\hspace{0.17em}\left(G\right)}\left)\right|\ge 2.$

2. Suppose T is a zero-sum subsequence of S with $\left|\mathsf{L}\right(T\left)\right|\ge 2.$ It follows by $\mathsf{L}\left(S\right)\supset \mathsf{L}\left(T\right)+\mathsf{L}\left(S{T}^{-1}\right)$ that $\left|\mathsf{L}\right(S\left)\right|\ge \left|\mathsf{L}\right(T\left)\right|\ge 2.$

3. Suppose $\mathsf{k}\left(A\right)<1$ and $\mathsf{k}\left(A\right)$ is minimal over all minimal zero-sum sequences over G₀. Let $g\in \text{supp}\left(A\right).$ Then there exist $s\in \mathbb{N}$ and minimal zero-sum sequences $W_1,\dots ,W_s$ such that $$A^{\lceil \frac{\text{ord}\left(g\right)}{{\mathsf{v}}_g\left(A\right)}\rceil }=g^{\text{ord}\left(g\right)}·W_1·\dots ·W_s.$$ Since $$k\left(A^{\lceil \frac{\text{ord}\left(g\right)}{{\mathsf{v}}_g\left(A\right)}\rceil }\right)=\lceil \frac{\text{ord}\left(g\right)}{{\mathsf{v}}_g\left(A\right)}\rceil \mathsf{k}\left(A\right)=1+\sum _{i=1}^s\mathsf{k}\left(W_i\right)>(1+s)\mathsf{k}\left(A\right),$$ we have $\lceil \frac{\text{ord}\left(g\right)}{{\mathsf{v}}_g\left(A\right)}\rceil \ne s+1$ and hence $\left|\mathsf{L}\left(A^{\lceil \frac{\text{ord}\left(g\right)}{{\mathsf{v}}_g\left(A\right)}\rceil }\right)\right|\ge 2.$ □

For commutative and finitely generated monoids, we have the following result.

Proposition 2.3.

Let H be a commutative unit-cancellative monoid. If $H_{\text{red}}$ is finitely generated, then $\mathsf{hf}\left(H\right)$ is finite.

Proof.

We may assume that H is reduced and not half-factorial. Suppose H is finitely generated and suppose $\mathcal{A}\left(H\right)=\{u_1,\dots ,u_n\},$ where $n\in \mathbb{N}.$ Set $A_0=\{\prod _{i\in I}{u}_i|\varnothing \ne I\subset [1,n]\}.$ Then A₀ is finite and hence there exists $M\in \mathbb{N}$ such that $\left|\mathsf{L}\right(a_0^M\left)\right|\ge 2$ for all $a_0\in A_0$ with $⟦a_0⟧$ not half-factorial. Let $a\in H\setminus H^{\times }$ such that $⟦a⟧$ is not half-factorial. It suffices to show that $\left|\mathsf{L}\right(a^M\left)\right|\ge 2.$ Suppose $a=u_1^{k_1}·\dots ·u_n^{k_n},$ where $k_1,\dots ,k_n\in {\mathbb{N}}_0.$ Set $I_0=\{i\in [1,n\left]\right|k_i\ge 1\}$ and $a_0=\prod _{i\in I}{u}_i.$ Then a₀ divides a and $⟦a_0⟧=⟦a⟧$ is not half-factorial, whence $\left|\mathsf{L}\right(a_0^M\left)\right|\ge 2$ and $\left|\mathsf{L}\right(a^M\left)\right|\ge 2.$ □

If G₀ is a finite subset of an abelian group, then $\mathcal{B}\left(G_0\right)$ is finitely generated [12, Theorem 3.4.2] and thus $\mathsf{hf}\left(G_0\right)<\infty .$ We refer to [8, Sections 3.2 and 3.3] and [14] for semigroups of ideals and semigroups of modules that are finitely generated unit-cancellative but not necessarily cancellative.

Example 2.4.

The following examples will help us to illustrate some important points.

1. Let (e₁, e₂) be a basis of ${\mathbb{Z}}^2$ and let $G_0=\{e_1,-e_1,e_2,-e_2,e_1+e_2,-e_1-e_2\}.$ Then $\mathcal{A}\left(G_0\right)=\left\{e_1\right(-e_1),e_2(-e_2),(e_1+e_2\left)\right(-e_1-e_2),e_1{e}_2(-e_1-e_2),(-e_1\left)\right(-e_2\left)\right(e_1+e_2\left)\right\}.$ Since $e_1(-e_1)·e_2(-e_2)·(e_1+e_2)(-e_1-e_2)=e_1{e}_2(-e_1-e_2)·(-e_1)(-e_2)(e_1+e_2),$ we obtain G₀ is not half-factorial. Furthermore, we have G₁ is half-factorial for every non-empty proper subset $G_1⊊G_0.$ Let $A\in \mathcal{B}\left(G_0\right).$ If $\text{supp}\left(A\right)=G_0,$ then $\left|\mathsf{L}\right(A\left)\right|\ge 2$ and $⟦A⟧=\mathcal{B}\left(G_0\right)$ is not half-factorial. If $\text{supp}\left(A\right)⊊G_0,$ then $⟦A⟧=\mathcal{B}\left(\text{supp}\right(A\left)\right)$ is half-factorial and $\left|\mathsf{L}\right(A\left)\right|=1.$ Therefore $\mathsf{hf}\left(G_0\right)=1.$

2. Let G be a cyclic group with order n and let $g\in G$ with $\text{ord}\left(g\right)=n,$ where $n\in {\mathbb{N}}_{\ge 3}.$ Set $G_0=\{g,-g\}.$ Then G₀ is not half-factorial. Let $A_0=g(-g).$ Then $⟦A_0⟧$ is not half-factorial and $\left|\mathsf{L}\right(A_0^{n-1}\left)\right|=1,$ whence $\mathsf{hf}\left(G_0\right)\ge n.$ Let $A\in \mathcal{B}\left(G_0\right)$ such that $⟦A⟧$ is not half-factorial. Then $\text{supp}\left(A\right)=G_0$ and A₀ divides A, whence $\left|\mathsf{L}\right(A^n\left)\right|\ge 2.$ Therefore $\mathsf{hf}\left(G_0\right)=n.$ Let $G\cong C_2^2$ and let (e₁, e₂) be a basis of G. Set $G_1=\{e_1,e_2,e_1+e_2\}.$ Then G₁ is not half-factorial. Let $A_1=e_1{e}_2(e_1+e_2).$ Then $⟦A_1⟧$ is not half-factorial and $\left|\mathsf{L}\right(A_1\left)\right|=1,$ whence $\mathsf{hf}\left(G_1\right)\ge 2.$ Let $A\in \mathcal{B}\left(G_1\right)$ such that $⟦A⟧$ is not half-factorial. Then $\text{supp}\left(A\right)=G_1$ and A₁ divides A, whence $\left|\mathsf{L}\right(A^2\left)\right|\ge 2.$ Therefore $\mathsf{hf}\left(G_1\right)=2.$

3. Let H be a bifurcus monoid (i.e., $2\in \mathsf{L}\left(a\right)$ for all $a\in H\setminus (H^{\times }\cup \mathcal{A}(H\left)\right)$). For examples, see [1, Examples 2.1 and 2.2]. Since for every $a\in H\setminus H^{\times },$ we have $\{2,3\}\subset \mathsf{L}\left(a^3\right),$ it follows that $\mathsf{hf}\left(H\right)\le 3$ and $\mathsf{hf}\left(H\right)$ is the minimal integer $t\in \mathbb{N}$ such that $\left|\mathsf{L}\right(a^t\left)\right|\ge 2$ for all $a\in H\setminus H^{\times }.$ Therefore $\mathsf{hf}\left(H\right)=3$ if and only if there exists $a_0\in \mathcal{A}\left(H\right)$ such that $\mathsf{L}\left(a_0^2\right)=\left\{2\right\}.$

4. Let $H\subset F=F^{\times }\times [p_1,\dots ,p_s]$ be a non-half-factorial finitely primary monoid of rank s and exponent α (see [12, Definition 2.9.1]). For every $a=ϵp_1^{t_1}\dots p_s^{t_s}\in F,$ we define $\left|\mathrm{}\right|a\left|\mathrm{}\right|=t_1+\dots +t_s,$ where $t_1,\dots ,t_s\in {\mathbb{N}}_0$ and $ϵ\in F^{\times }.$ Let $a\in H\setminus H^{\times }.$ Since H is primary, we have $H=⟦a⟧$ is not half-factorial. Thus $\mathsf{hf}\left(H\right)$ is the minimal integer $t\in \mathbb{N}$ such that $\left|\mathsf{L}\right(a^t\left)\right|\ge 2$ for all $a\in H\setminus H^{\times }.$ Suppose $a_0\in H$ with $\left|\mathrm{}\right|a_0\left|\mathrm{}\right|=\min \left\{\right|\mathrm{}\left|a\right|\mathrm{}|:a\in H\setminus H^{\times }\}.$ Then $a_0\in \mathcal{A}\left(H\right)$ and $\mathsf{L}\left(a_0^2\right)=\left\{2\right\},$ whence $\mathsf{hf}\left(H\right)\ge 3.$

If $H\setminus H^{\times }={(p_1\dots p_s)}^{\alpha }F$ and $s\ge 2,$ then H is bifurcus and hence $\mathsf{hf}\left(H\right)=3.$ Suppose s = 1 and $H\setminus H^{\times }={\left(p_1\right)}^{\alpha }F.$ Let $b=ϵp^{\beta }\in H.$ Then $p^{3\alpha }$ divides b⁴. It follows by $p^{3\alpha }={\left(p^{\alpha }\right)}^3=p^{\alpha +1}{p}^{2\alpha -1}$ that $\left|\mathsf{L}\right(b^4\left)\right|\ge 2,$ whence $\mathsf{hf}\left(H\right)\le 4.$ If $3\beta \ge 4\alpha ,$ then $p^{3\alpha }$ divides b³ and hence $\left|\mathsf{L}\right(b^3\left)\right|\ge 2.$ If $\mathrm{(}\text{HTML}\mathrm{ }\text{translation}\mathrm{ }\text{failed}\mathrm{)},$ then b is an atom and $b^3={ϵ}^3{p}^{2\alpha -1}{p}^{3\beta -(2\alpha -1)},$ whence $\left|\mathsf{L}\right(b^3\left)\right|\ge 2.$ If $3\beta =4\alpha -1,$ then $\mathsf{L}\left(b^3\right)=\left\{3\right\}.$ Put all together, if $\alpha \equiv 1\mathrm{ }\mathrm{mod}\mathrm{ }3,$ then $\mathsf{hf}\left(H\right)=4.$ Otherwise $\mathsf{hf}\left(H\right)=3.$

3. Proof of main theorem

Proposition 3.1.

Let $G_0\subset G$ be a non-half-factorial subset and let S be a zero-sum sequence over G₀ with $\text{supp}\left(S\right)=G_0.$

1. If G₀ is an LCN-set, then $\left|\mathsf{L}\right(\prod _{g\in G_0}{g}^{\text{ord}\left(g\right)}\left)\right|\ge 2.$

2. If $\left|G_0\right|=2$, then $\left|\mathsf{L}\right(\prod _{g\in G_0}{g}^{\text{ord}\left(g\right)}\left)\right|\ge 2.$

3. If G₀ is a minimal non-half-factorial subset, then $\left|\mathsf{L}\right(S^{\hspace{0.17em}\text{exp}\hspace{0.17em}\left(G\right)}\left)\right|\ge 2.$

4. If $\left|\right\{g\in G_0\left|\text{ord}\right(g)/{\mathsf{v}}_g(S)=\hspace{0.17em}\text{exp}\hspace{0.17em}(G\left)\right\}|\le 1$, then $\left|\mathsf{L}\right(S^{\hspace{0.17em}\text{exp}\hspace{0.17em}\left(G\right)}\left)\right|\ge 2.$

Proof.

1. Suppose G₀ is an LCN-set. Since G₀ is not half-factorial, there exists a minimal zero-sum sequence T over G₀ such that $\mathsf{k}\left(T\right)>1.$ Note that T is a subsequence of $\prod _{g\in G_0}{g}^{\text{ord}\left(g\right)}.$ Then there exist $W_1,\dots ,W_l\in \mathcal{A}\left(G_0\right)$ such that $$\prod _{g\in G_0}{g}^{\text{ord}\left(g\right)}=T·W_1·\dots ·W_l.$$ Thus $\mathsf{k}\left(\prod _{g\in G_0}{g}^{\text{ord}\left(g\right)}\right)=\left|G_0\right|=\mathsf{k}\left(T\right)+{\sum }_{i=1}^l\mathsf{k}\left(W_i\right)>1+l.$ The assertion follows by $\left\{\right|G_0|,1+l\}\subset \mathsf{L}\left(\prod _{g\in G_0}{g}^{\text{ord}\left(g\right)}\right).$

2. Suppose $\left|G_0\right|=2$ and let $G_0=\{g_1,g_2\}.$ If G₀ is an LCN-set, the assertion follows by 1. Suppose there exists a minimal zero-sum sequence T over G₀ with $\mathsf{k}\left(T\right)<1.$ Let $T_0=g_1^{l_1}·g_2^{l_2}$ be the minimal zero-sum sequence over G₀ such that $\mathsf{k}\left(T_0\right)$ is minimal. If $\min \{\frac{\text{ord}\left(g_1\right)}{l_1},\frac{\text{ord}\left(g_2\right)}{l_2}\}\le 2,$ say $\frac{\text{ord}\left(g_1\right)}{l_1}\le 2$ then $$T_0^2=g_1^{\text{ord}\left(g_1\right)}·W, \text{where }W \text{is an on}‐\text{empty zero}‐\text{sum sequence}.$$ Thus $\mathsf{k}\left(W\right)=2\mathsf{k}\left(T_0\right)-1<\mathsf{k}\left(T_0\right),$ a contradiction to the minimality of $\mathsf{k}\left(T_0\right).$ Therefore $\min \{\frac{\text{ord}\left(g_1\right)}{l_1},\frac{\text{ord}\left(g_2\right)}{l_2}\}>2$ and hence $$g_1^{\text{ord}\left(g_1\right)}·g_2^{\text{ord}\left(g_2\right)}=T_0^2·V, \text{where}\mathrm{ }V\mathrm{ }\text{is}\mathrm{ }\text{non}‐\text{empty}\mathrm{ }\text{zero}‐\text{sum}\mathrm{ }\text{sequence}.$$ It follows that $\left|\mathsf{L}\right(g_1^{\text{ord}\left(g_1\right)}·g_2^{\text{ord}\left(g_2\right)}\left)\right|\ge 2.$

3. Suppose that G₀ is a minimal non-half-factorial set. If S has a minimal zero-sum subsequence A with $\mathsf{k}\left(A\right)\ne 1,$ then the assertion follows by Lemma 2.2. If G₀ is an LCN-set, then the assertion follows from 1 and Lemma 2.2.2. Therefore we can suppose $\mathsf{L}\left(S\right)=\left\{\mathsf{k}\right(S\left)\right\}$ and suppose there exists a minimal zero-sum sequence T over G₀ with $\mathsf{k}\left(T\right)<1.$

Let $T_0={\prod }_{i=1}^{\left|G_0\right|}{g}_i^{l_i}$ be the minimal zero-sum sequence over G₀ such that $\mathsf{k}\left(T_0\right)$ is minimal. The minimality of G₀ implies that $l_i\ge 1$ for all $i\in [1,|G_0\left|\right].$ After renumbering if necessary, we let $$\frac{\text{ord}\left(g_1\right)}{l_1}=\min \left\{\frac{\text{ord}\left(g_i\right)}{l_i}\right|i\in [1,|G_0\left|\right]\}.$$ By Lemma 2.2.3, $\left|\mathsf{L}\left(T_0^{\lceil \frac{\text{ord}\left(g_1\right)}{l_1}\rceil }\right)\right|\ge 2.$ If $T_0^{\lceil \frac{\text{ord}\left(g_1\right)}{l_1}\rceil }$ divides $S^{\hspace{0.17em}\text{exp}\hspace{0.17em}\left(G\right)},$ then the assertion follows by Lemma 2.2.2. Suppose $T_0^{\lceil \frac{\text{ord}\left(g_1\right)}{l_1}\rceil }\nmid S^{\hspace{0.17em}\text{exp}\hspace{0.17em}\left(G\right)}.$ Let $$I=\{i\in [1,\left|G_0\right|]|\lceil \frac{\text{ord}\left(g_1\right)}{l_1}\rceil l_i>\hspace{0.17em}\text{exp}\hspace{0.17em}\left(G\right){\mathsf{v}}_{g_i}\left(S\right)\}.$$ Thus for each $i\in I,$ we have $$2\text{ord}\left(g_i\right)>l_i\lceil \frac{\text{ord}\left(g_i\right)}{l_i}\rceil \ge l_i\lceil \frac{\text{ord}\left(g_1\right)}{l_1}\rceil >\hspace{0.17em}\text{exp}\hspace{0.17em}\left(G\right){\mathsf{v}}_{g_i}\left(S\right)\ge \hspace{0.17em}\text{exp}\hspace{0.17em}\left(G\right),$$ which implies that $\text{ord}\left(g_i\right)=\hspace{0.17em}\text{exp}\hspace{0.17em}\left(G\right),\mathrm{ }{\mathsf{v}}_{g_i}\left(S\right)=1,$ and $\lceil \frac{\text{ord}\left(g_1\right)}{l_1}\rceil >\frac{\text{ord}\left(g_i\right)}{l_i}=\frac{\hspace{0.17em}\text{exp}\hspace{0.17em}\left(G\right)}{l_i}.$

Let $i_0\in I$ such that $l_{i_0}=\max \left\{l_i\right|i\in I\}.$ Therefore for every $j\in [1,|G_0\left|\right]\setminus I,$ we have $$l_j\le \frac{\hspace{0.17em}\text{exp}\hspace{0.17em}\left(G\right){\mathsf{v}}_{g_j}\left(S\right)}{\lceil \frac{\text{ord}\left(g_1\right)}{l_1}\rceil }\le \frac{\hspace{0.17em}\text{exp}\hspace{0.17em}\left(G\right){\mathsf{v}}_{g_j}\left(S\right)}{\frac{\hspace{0.17em}\text{exp}\hspace{0.17em}\left(G\right)}{l_{i_0}}}=l_{i_0}{\mathsf{v}}_{g_j}\left(S\right).$$ Note that for every $i\in I,$ we have $l_i\le l_{i_0}=l_{i_0}{\mathsf{v}}_{g_i}\left(S\right).$ It follows by ${\mathsf{v}}_{g_{i_0}}\left(T_0\right)=l_{i_0}=l_{i_0}{\mathsf{v}}_{g_{i_0}}\left(S\right)={\mathsf{v}}_{g_{i_0}}\left(S^{l_{i_0}}\right)$ that $$S^{l_{i_0}}=T_0·W,\mathrm{ }\text{where}\mathrm{ }W\mathrm{ }\text{is}\text{ a }\text{zero}-\text{sum}\mathrm{ }\text{sequence}\mathrm{ }\text{over}\mathrm{ }{G}_0\setminus \left\{g_{i_0}\right\}.$$ By the minimality of G₀, we have $G_0\setminus \left\{g_{i_0}\right\}$ is half-factorial which implies that $\mathsf{k}\left(W\right)\in \mathbb{N}.$ Therefore $\mathsf{k}\left(T_0\right)=l_{i_0}\mathsf{k}\left(S\right)-\mathsf{k}\left(W\right)$ is an integer, a contradiction to $\mathsf{k}\left(T_0\right)<1.$

4. Let $G_1=\{g\in G_0|\text{ord}\left(g\right)=\hspace{0.17em}\text{exp}\hspace{0.17em}\left(G\right){\mathsf{v}}_g\left(S\right)\}.$ Suppose $G_0\setminus G_1$ is not half-factorial. If $G_0\setminus G_1$ is an LCN-set, then the assertion follows by Proposition 3.1.1 and Lemma 2.2.2. Otherwise there exists a minimal zero-sum sequence A over $G_0\setminus G_1$ such that $\mathsf{k}\left(A\right)<1.$ We may assume that $\mathsf{k}\left(A\right)$ is minimal over all minimal zero-sum sequences over $G_0\setminus G_1$ and that $\min \left\{\frac{\text{ord}\left(g\right)}{{\mathsf{v}}_g\left(A\right)}\right|g\in \text{supp}\left(A\right)\}=\frac{\text{ord}\left(g_0\right)}{{\mathsf{v}}_{g_0}\left(A\right)}$ for some $g_0\in \text{supp}\left(A\right)\subset G_0\setminus G_1.$ Thus by Lemma 2.2.3, we have $|\mathsf{L}\left(A^{\lceil \frac{\text{ord}\left(g_0\right)}{{\mathsf{v}}_{g_0}\left(A\right)}\rceil }\right)|\ge 2.$ The definition of G₁ implies that $$A^{\lceil \frac{\text{ord}\left(g_0\right)}{{\mathsf{v}}_{g_0}\left(A\right)}\rceil }\hspace{1em}\mathrm{ }\text{divides}\mathrm{ }\hspace{1em}{S}^{\hspace{0.17em}\text{exp}\hspace{0.17em}\left(G\right)}$$ and hence the assertion follows.

Suppose $G_0\setminus G_1$ is half-factorial. Then G₁ is non-empty and hence $G_1=\left\{g_0\right\}$ for some $g_0\in G_0.$ If G₀ is an LCN-set, then the assertion follows by Proposition 3.1.1 and Lemma 2.2.2. Otherwise there exists a minimal zero-sum sequence A over G₀ such that $\mathsf{k}\left(A\right)<1.$ We may assume that $\mathsf{k}\left(A\right)$ is minimal over all minimal zero-sum sequences over G₀ and that $\min \left\{\frac{\text{ord}\left(g\right)}{{\mathsf{v}}_g\left(A\right)}\right|g\in \text{supp}\left(A\right)\}=\frac{\text{ord}\left(g_1\right)}{{\mathsf{v}}_{g_1}\left(A\right)}$ for some $g_1\in \text{supp}\left(A\right)\subset G_0.$ Thus by Lemma 2.2.3, we have $|\mathsf{L}\left(A^{\lceil \frac{\text{ord}\left(g_1\right)}{{\mathsf{v}}_{g_1}\left(A\right)}\rceil }\right)|\ge 2.$ For every $g\in G_0\setminus G_1,$ we obtain $${\mathsf{v}}_g\left(A\right)\lceil \frac{\text{ord}\left(g_1\right)}{{\mathsf{v}}_{g_1}\left(A\right)}\rceil \le {\mathsf{v}}_g\left(A\right)\lceil \frac{\text{ord}\left(g\right)}{{\mathsf{v}}_g\left(A\right)}\rceil <2\text{ord}\left(g\right)\le \hspace{0.17em}\text{exp}\hspace{0.17em}\left(G\right){\mathsf{v}}_g\left(S\right).$$ If ${\mathsf{v}}_{g_0}\left(A\right)\lceil \frac{\text{ord}\left(g_1\right)}{{\mathsf{v}}_{g_1}\left(A\right)}\rceil \le \text{ord}\left(g_0\right)=\hspace{0.17em}\text{exp}\hspace{0.17em}\left(G\right),$ then $$A^{\lceil \frac{\text{ord}\left(g_1\right)}{{\mathsf{v}}_{g_1}\left(A\right)}\rceil }\hspace{1em}\mathrm{ }\text{divides}\mathrm{ }\hspace{1em}{S}^{\hspace{0.17em}\text{exp}\hspace{0.17em}\left(G\right)}$$ and hence $\left|\mathsf{L}\right(S^{\hspace{0.17em}\text{exp}\hspace{0.17em}\left(G\right)}\left)\right|\ge 2.$

Otherwise for every $g\in G\setminus G_1,$ we have $$\frac{\hspace{0.17em}\text{exp}\hspace{0.17em}\left(G\right)}{{\mathsf{v}}_{g_0}\left(A\right)}<\lceil \frac{\text{ord}\left(g_1\right)}{{\mathsf{v}}_{g_1}\left(A\right)}\rceil \le \lceil \frac{\text{ord}\left(g\right)}{{\mathsf{v}}_g\left(A\right)}\rceil \le \lceil \frac{\hspace{0.17em}\text{exp}\hspace{0.17em}\left(G\right){\mathsf{v}}_g\left(S\right)}{2{\mathsf{v}}_g\left(A\right)}\rceil \le \frac{\hspace{0.17em}\text{exp}\hspace{0.17em}\left(G\right){\mathsf{v}}_g\left(S\right)}{{\mathsf{v}}_g\left(A\right)}.$$ Therefore ${\mathsf{v}}_g\left(A\right)<{\mathsf{v}}_{g_0}\left(A\right){\mathsf{v}}_g\left(S\right)$ for all $g\in G_0\setminus G_1$ which implies that A divides $S^{{\mathsf{v}}_{g_0}\left(A\right)}.$ Thus there exists a zero-sum sequence W over $G_0\setminus G_1$ such that $S^{{\mathsf{v}}_{g_0}(A)}=A·W.$ Since $G_0\setminus G_1$ is half-factorial, we obtain $k(A)={\mathsf{v}}_{g_0}(A)k(S)-k(W)$ is an integer, a contradiction to $k(A)<1.$ □

Proof of Theorem 1.1.

By the definition of transfer Krull monoid, it suffices to prove the assertions for $H=\mathcal{B}(G_0)$ and hence H is half-factorial if and only if G₀ is half-factorial. If G₀ is half-factorial, it is easy to see that $\mathsf{hf}\left(G_0\right)=1$ and $\left|\mathsf{L}\right(\prod _{g\in G_0}{g}^{2\text{ord}\left(g\right)}\left)\right|=1.$ Therefore we only need to show that (b) implies (c) and that (d) implies (c).

(b) $\Rightarrow$ (c) Suppose $\mathsf{hf}\left(G_0\right)=1$ and assume to the contrary that G₀ is not half-factorial. Then there exists $A\in \mathcal{A}\left(G_0\right)$ such that $\mathsf{k}\left(A\right)\ne 1,$ whence $\text{supp}\left(A\right)$ is not half-factorial. Therefore $\mathsf{hf}\left(\text{supp}\right(A\left)\right)\ge 2,$ a contradiction.

(d) $\Rightarrow$ (c) Suppose $\left|\mathsf{L}\right(\prod _{g\in G_0}{g}^{2\text{ord}\left(g\right)}\left)\right|=1$ and assume to the contrary that G₀ is not half-factorial. If G₀ is an LCN set, then Proposition 3.1.1 implies that $\left|\mathsf{L}\right(\prod _{g\in G_0}{g}^{\text{ord}\left(g\right)}\left)\right|\ge 2,$ a contradiction. Thus there exists an atom $A\in \mathcal{A}\left(G_0\right)$ with $\mathsf{k}\left(A\right)<1$ and we may assume that $\mathsf{k}\left(A\right)$ is minimal over all atoms of $\mathcal{B}\left(G_0\right).$ Let $g_0\in \text{supp}\left(A\right).$ Then by Lemma 2.2.3, we have $\left|\mathsf{L}\left(A^{\lceil \frac{\text{ord}\left(g_0\right)}{{\mathsf{v}}_{g_0}\left(A\right)}\rceil }\right)\right|\ge 2,$ a contradiction to $A^{\lceil \frac{\text{ord}\left(g_0\right)}{{\mathsf{v}}_{g_0}\left(A\right)}\rceil } | \prod _{g\in G_0}{g}^{2\text{ord}\left(g\right)}.$ □

Proof of Theorem 1.2.

By the definition of transfer Krull monoid, it suffices to prove all assertions for $H=\mathcal{B}\left(G\right).$

1. Suppose $\text{exp}\hspace{0.17em}\left(G\right)<\infty .$ If $\left|G\right|\ge 3,$ then 2 implies that $\mathsf{hf}\left(G\right)<\infty .$ If $\left|G\right|\le 2,$ then $\mathcal{B}\left(G\right)$ is half-factorial and hence $\mathsf{hf}\left(G\right)=1.$

Suppose $\text{exp}\hspace{0.17em}\left(G\right)=\infty .$ If there exists an element $g\in G$ with $\text{ord}\left(g\right)=\infty ,$ then $A_n=\left(\right(n+1\left)g\right)(-ng)(-g)$ is an atom for every $n\in \mathbb{N}.$ Since $\left\{\right(n+1)g,-ng,-g\}$ is not half-factorial and $\left|\mathsf{L}\right(A_n^n\left)\right|=1$ for every $n\ge 2,$ we obtain that $\mathsf{hf}\left(G\right)\ge n$ for every $n\ge 2,$ that is, $\mathsf{hf}\left(G\right)=\infty .$ Otherwise G is torsion. Then there exists a sequence ${\left(g_i\right)}_{i=1}^{\infty }$ with $g_i\in G$ and $\underset{i\to \infty }{\lim }\text{ord}\left(g_i\right)=\infty .$ It follows by 1 that $\mathsf{hf}\left(G\right)\ge \mathsf{hf}(〈g_i〉)\ge \text{ord}\left(g_i\right)$ for all $i\in \mathbb{N},$ that is, $\mathsf{hf}\left(G\right)=\infty .$

2. If G is an elementary 2-group and e₁, e₂ are two independent elements, then $\{e_1,e_2,e_1+e_2\}$ is not a half-factorial set and $\left|\mathsf{L}\right(e_1{e}_2(e_1+e_2)\left)\right|=1$ which implies that $\mathsf{hf}\left(G\right)\ge 2=\hspace{0.17em}\text{exp}\hspace{0.17em}\left(G\right).$ Otherwise there exists an element $g\in G$ with $\text{ord}\left(g\right)=\hspace{0.17em}\text{exp}\hspace{0.17em}\left(G\right)\ge 3.$ Since $\{g,-g\}$ is not half-factorial and $\left|\mathsf{L}\right(g^{\text{ord}\left(g\right)-1}{(-g)}^{\text{ord}\left(g\right)-1}\left)\right|=1,$ we obtain $\mathsf{hf}\left(G\right)\ge \text{ord}\left(g\right)=\text{exp}\hspace{0.17em}\left(G\right).$

Let S be a zero-sum sequence over G such that $\text{supp}\left(S\right)$ is not half-factorial. In order to prove $\mathsf{hf}\left(G\right)\le \lfloor \frac{3\hspace{0.17em}\text{exp}\hspace{0.17em}\left(G\right)-3}{2}\rfloor ,$ we show that $$\left|\mathsf{L}\right(S^{\lfloor \frac{3\hspace{0.17em}\text{exp}\hspace{0.17em}\left(G\right)-3}{2}\rfloor }\left)\right|\ge 2.$$ Set $G_0=\text{supp}\left(S\right).$ If G₀ is an LCN-set, the assertion follows by Proposition 3.1.1. Suppose there exists an atom $A\in \mathcal{A}\left(G_0\right)$ with $\mathsf{k}\left(A\right)<1.$ Let $A_0\in \mathcal{A}\left(\text{supp}\right(S\left)\right)$ be such that $\mathsf{k}\left(A_0\right)$ is minimal over all minimal zero-sum sequences over G₀ and set $A_0=g_1^{l_1}·\dots ·g_y^{l_y},$ where $y,l_1,\dots l_y\in \mathbb{N}$ and $g_1,\dots ,g_y\in \text{supp}\left(S\right)$ are pairwise distinct elements. If there exists $j\in [1,y]$ such that $2l_j\ge \text{ord}\left(g_j\right),$ then $g_j^{\text{ord}\left(g_i\right)}$ divides $A_0^2$ and hence $A_0^2=g_j^{\text{ord}\left(g_j\right)}·W$ for some non-empty sequence $W\in \mathcal{B}\left(\text{supp}\right(S\left)\right).$ Thus $\mathsf{k}\left(W\right)=2\mathsf{k}\left(A_0\right)-1<\mathsf{k}\left(A_0\right),$ a contradiction to the minimality of $\mathsf{k}\left(A_0\right).$ Therefore $$2l_i\le \text{ord}\left(g_i\right)-1\mathrm{ }\text{for}\mathrm{ }\text{all}\mathrm{ }i\in [1,y].$$ After renumbering if necessary, we assume $\frac{\text{ord}\left(g_1\right)}{l_1}=\min \left\{\frac{\text{ord}\left(g_i\right)}{l_i}\right|i\in [1,y]\}.$ Then $$l_i\lceil \frac{\text{ord}\left(g_1\right)}{l_1}\rceil \le l_i\lceil \frac{\text{ord}\left(g_i\right)}{l_i}\rceil \le l_i\frac{\text{ord}\left(g_i\right)+l_i-1}{l_i}\le \text{ord}\left(g_i\right)+\frac{\text{ord}\left(g_i\right)-1}{2}-1\le \frac{3\hspace{0.17em}\text{exp}\hspace{0.17em}\left(G\right)-3}{2},$$ which implies that $A_0^{\lceil \frac{\text{ord}\left(g_1\right)}{l_1}\rceil }$ divides $S^{\lfloor \frac{3\hspace{0.17em}\text{exp}\hspace{0.17em}\left(G\right)-3}{2}\rfloor }.$ The assertion follows by Lemma 2.2.3.

3(a). Suppose that G is cyclic and that $g\in G$ with $\text{ord}\left(g\right)=\left|G\right|\ge 3.$ We will show that $\mathsf{hf}\left(G\right)=\hspace{0.17em}\text{exp}\hspace{0.17em}\left(G\right).$

Let S be a zero-sum sequence over G such that $\text{supp}\left(G\right)$ is not half-factorial. It suffices to show that $\left|\mathsf{L}\right(S^{\hspace{0.17em}\text{exp}\hspace{0.17em}\left(G\right)}\left)\right|\ge 2.$ If $\left|\right\{g\in \text{supp}\left(S\right)\left|\text{ord}\right(g)=|G\left|{\mathsf{v}}_g\right(S\left)\right\}|\le 1,$ then the assertion follows from Proposition 3.1.4. Suppose $\left|\right\{g\in \text{supp}\left(S\right)\left|\text{ord}\right(g)=|G\left|{\mathsf{v}}_g\right(S\left)\right\}|\ge 2.$ Then there exist distinct $g_1,g_2\in \text{supp}\left(S\right)$ such that $\text{ord}\left(g_1\right)=\text{ord}\left(g_2\right)=\left|G\right|.$ We may assume that g₁ = kg₂ for some $k\in {\mathbb{N}}_{\ge 2}$ with $\gcd (k,|G\left|\right)=1.$ It follows by $\mathsf{k}(g_1^{\left|G\right|-k}·g_2)<1$ that $G_0=\{g_1,g_2\}=\{g_1,k{g}_1\}$ is not half-factorial. By Proposition 3.1.2, we obtain that $\left|\mathsf{L}\right(S^{\hspace{0.17em}\text{exp}\hspace{0.17em}\left(G\right)}\left)\right|\ge 2.$

3(b). Suppose G is a finite abelian group with $\text{exp}\hspace{0.17em}\left(G\right)\le 6.$ We need to prove that $\mathsf{hf}\left(G\right)=\hspace{0.17em}\text{exp}\hspace{0.17em}\left(G\right).$ Let S be a zero-sum sequence over G such that $\text{supp}\left(G\right)$ is not half-factorial. It suffices to show that $\left|\mathsf{L}\right(S^{\hspace{0.17em}\text{exp}\hspace{0.17em}\left(G\right)}\left)\right|\ge 2.$

If $\text{supp}\left(S\right)$ is an LCN-set, the assertion follows by Proposition 3.1.1. Thus there is a minimal zero-sum sequence W over $\text{supp}\left(S\right)$ such that $$\mathsf{k}\left(W\right)<1.$$ By Proposition 3.1.2 and Lemma 2.2.2, we have $$\left|\text{supp}\right(W\left)\right|\ge 3.$$ Suppose $W | S.$ Since $\mathsf{k}\left(W\right)<1,$ it follows by Lemma 2.2 that $\left|\mathsf{L}\right(W^{\hspace{0.17em}\text{exp}\hspace{0.17em}\left(G\right)}\left)\right|\ge 2$ and hence $\left|\mathsf{L}\right(S^{\hspace{0.17em}\text{exp}\hspace{0.17em}\left(G\right)}\left)\right|\ge 2.$ Therefore we may assume that $W\nmid S,$ whence $\left|W\right|\ge \left|\text{supp}\right(W\left)\right|+1\ge 4.$ It follows that $6\ge \hspace{0.17em}\text{exp}\hspace{0.17em}\left(G\right)\ge \frac{\left|W\right|}{\mathsf{k}\left(W\right)}>\left|W\right|\ge 4.$

We distinguish two cases according to $\text{exp}\hspace{0.17em}\left(G\right)\in \{5,6\}.$

Case 1. $\text{exp}\hspace{0.17em}\left(G\right)=5.$

Then, $G\cong C_5^r$ and for all $W\in \mathcal{A}\left(\text{supp}\right(S\left)\right)$ with $\mathsf{k}\left(W\right)<1,$ we have that $$W\nmid S,\mathrm{ }\left|\text{supp}\right(W\left)\right|=3,\mathrm{ }\mathrm{ }\text{and}\mathrm{ }\mathrm{ }\left|W\right|=4.$$

Let W₀ be an atom over $\text{supp}\left(S\right)$ with $\mathsf{k}\left(W_0\right)<1.$ Then $$W_0=g_1^2{g}_2{g}_3\hspace{1em}\text{and}\hspace{1em}S=T{g}_1{g}_2{g}_3,$$ where $g_1,g_2,g_3\in \text{supp}\left(S\right)$ are pairwise distinct and $T\in \mathcal{F}\left(\text{supp}\right(S)\setminus \{g_1\left\}\right)$ with $\sigma \left(T\right)=g_1.$

We may assume T is zero-sum free. Otherwise $T=T_{0}T\prime ,$ where T₀ is a zero-sum sequence and $T\prime$ is zero-sum free. We can replace S by $T\prime g_1{g}_2{g}_3,$ since $\left|\mathsf{L}\right({(T\prime g_1{g}_2{g}_3)}^5\left)\right|\ge 2$ implies that $\left|\mathsf{L}\right(S^5\left)\right|\ge 2.$ Therefore S is a product of at most three atoms and every term of S has order 5.

Assume to the contrary that $\left|\mathsf{L}\right(S^5\left)\right|=1,$ that is, $\mathsf{L}\left(S^5\right)=\left\{\right|T|+3\}.$ Since $g_1^5{g}_2^5{g}_3^5=W_0^2\left(g_1{g}_2^3{g}_3^3\right)$ is a zero-sum subsequence of S⁵, we obtain that $g_1{g}_2^3{g}_3^3$ is an atom. Note that $${\left(g_1^2{g}_2{g}_3\right)}^{2}S=\left(g_1^{4}T\right)\left(g_1{g}_2^3{g}_3^3\right)\mathrm{ }\text{is}\text{ a }\text{zero}-\text{sum}\mathrm{ }\text{subsequence}\mathrm{ }\text{of}\mathrm{ }{S}^5.$$

Suppose S is an atom. Then $\mathsf{L}\left(g_1^{4}T\right)=\left\{2\right\}$ and hence $\left|g_1^{4}T\right|\ge 2\times 4=8,$ that is, $\left|T\right|\ge 4.$ It follows by $\left\{5\right\}=\mathsf{L}\left(S^5\right)=\left\{\right|T|+3\}$ that $\left|T\right|=2,$ a contradiction.

Suppose S is a product of two atoms. Then $\mathsf{L}\left(g_1^{4}T\right)=\left\{3\right\}$ and hence $\left|g_1^{4}T\right|\ge 3\times 4=12,$ that is, $\left|T\right|\ge 8.$ It follows by $\left\{10\right\}=\mathsf{L}\left(S^5\right)=\left\{\right|T|+3\}$ that $\left|T\right|=7,$ a contradiction.

Suppose S is a product of three atoms. Then $\mathsf{L}\left(g_1^{4}T\right)=\left\{4\right\}$ which implies that $T=T_1{T}_2{T}_3{T}_4$ such that $g_1{T}_i$ is zero-sum for all $i\in [1,4].$ Since $g_1{T}_i | S,$ we obtain $\mathsf{k}\left(g_1{T}_i\right)\ge 1$ and hence $\left|g_1{T}_i\right|\ge 5.$ Therefore $\left|g_1^{4}T\right|\ge 4\times 5=20,$ that is, $\left|T\right|\ge 16.$ It follows by $\left\{15\right\}=\mathsf{L}\left(S^5\right)=\left\{\right|T|+3\}$ that $\left|T\right|=12,$ a contradiction.

Case 2. $\text{exp}\hspace{0.17em}\left(G\right)=6.$

Let W be an atom over $\text{supp}\left(S\right)$ with $\mathsf{k}\left(W\right)<1.$ If $\left|W\right|=4,$ then $\left|\text{supp}\right(W\left)\right|=3$ and hence W must be of the form $$W=g_1^2{g}_2{g}_3$$ where $g_1,g_2,g_3\in \text{supp}\left(S\right)$ are pairwise distinct. Since $\left(g_1^6\right)\left(g_2^6\right)\left(g_3^6\right)=W^3\left(g_2^3{g}_3^3\right),$ we obtain that $\left|\mathsf{L}\right(g_1^6{g}_2^6{g}_3^6\left)\right|\ge 2.$ It follows from the fact that $g_1^6{g}_2^6{g}_3^6$ divides $S^{\hspace{0.17em}\text{exp}\hspace{0.17em}\left(G\right)}$ that $\left|\mathsf{L}\right(S^{\hspace{0.17em}\text{exp}\hspace{0.17em}\left(G\right)}\left)\right|\ge 2.$

If $\left|W\right|=5,$ then $\left|\text{supp}\right(W\left)\right|=3$ or 4 and hence W can be written in one of the following ways.

1. $W=g_1^3{g}_2{g}_3,$ where $g_1,g_2,g_3\in \text{supp}\left(S\right)$ are pairwise distinct.

2. $W=g_1^2{g}_2^2{g}_3,$ where $g_1,g_2,g_3\in \text{supp}\left(S\right)$ are pairwise distinct.

3. $W=g_1^2{g}_2{g}_3{g}_4,$ where $g_1,g_2,g_3,g_4\in \text{supp}\left(S\right)$ are pairwise distinct.

Suppose (i) holds. Then $0=2\sigma \left(W\right)=6g_1+2g_2+2g_3=2g_2+2g_3.$ Since $\left(g_1^6\right)\left(g_2^6\right)\left(g_3^6\right)=W^2{\left(g_2^2{g}_3^2\right)}^2,$ we obtain that $\left|\mathsf{L}\right(g_1^6{g}_2^6{g}_3^6\left)\right|\ge 2.$ It follows from the fact that $g_1^6{g}_2^6{g}_3^6$ divides $S^{\hspace{0.17em}\text{exp}\hspace{0.17em}\left(G\right)}$ that $\left|\mathsf{L}\right(S^{\hspace{0.17em}\text{exp}\hspace{0.17em}\left(G\right)}\left)\right|\ge 2.$

Suppose (ii) holds. Then $0=3\sigma \left(W\right)=6g_1+6g_2+3g_3=3g_3.$ Thus $\text{ord}\left(g_3\right)=2$ and hence $\mathsf{k}\left(W\right)\ge 1/2+4/6>1,$ a contradiction.

Suppose (iii) holds. Then $0=3\sigma \left(W\right)=6g_1+3g_3+3g_3+3g_4=3g_2+3g_3+3g_4.$ Therefore $W_0=g_2^3{g}_3^3{g}_4^3$ is zero-sum. If $W_0=g_1^6{g}_2^6{g}_3^6{g}_4^6{\left(W\right)}^{-3}$ is not a minimal zero-sum sequence, then $\left|\mathsf{L}\right(g_1^6{g}_2^6{g}_3^6{g}_4^6\left)\right|\ge 2$ and hence $\left|\mathsf{L}\right(S^{\hspace{0.17em}\text{exp}\hspace{0.17em}\left(G\right)}\left)\right|\ge 2.$ If W₀ is minimal zero-sum, then $g_1^6{g}_2^6{g}_3^6{g}_4^6=\left(g_1^6\right)W_0^2$ implies that $\left|\mathsf{L}\right(g_1^6{g}_2^6{g}_3^6{g}_4^6\left)\right|\ge 2$ and hence $\left|\mathsf{L}\right(S^{\hspace{0.17em}\text{exp}\hspace{0.17em}\left(G\right)}\left)\right|\ge 2.$ □

Acknowledgments

We thank the referee for very careful reading and for providing valuable suggestions.

Additional information

Funding

This work has been supported in part by the National Science Found of China with grant no. 11671218 and by the Austrian Science Fund with grant no. P33499.