New nonexistence results on $(m,n)$-generalized bent functions

TL;DR

This paper establishes new nonexistence results for $(m,n)$-generalized bent functions, especially for odd $n$ and specific $m$ congruences, using character sum techniques.

Contribution

It improves existing nonexistence results by deriving new conditions for the nonexistence of $(m,n)$-generalized bent functions, including explicit cases for $n=3$.

Findings

Nonexistence of $(m,n)$-generalized bent functions for odd $n$ and certain $m$

Explicit nonexistence of $(m,3)$-generalized bent functions for specific $m$

Use of character sums and group ring equations as main tools

Abstract

In this paper, we present some new nonexistence results on $(m,n)$-generalized bent functions, which improved recent results. More precisely, we derive new nonexistence results for general $n$ and $m$ odd or $m \equiv 2 \pmod{4}$, and further explicitly prove nonexistence of $(m,3)$-generalized bent functions for all integers $m$ odd or $m \equiv 2 \pmod{4}$. The main tools we utilized are certain exponents of minimal vanishing sums from applying characters to group ring equations that characterize $(m,n)$-generalized bent functions.

Equations

$$|F(y){|}^2=2^n$$

$$F(y):=\sum _{x\in {\mathbb{Z}}_2^n}{\zeta }_m^{f(x)}(-1{)}^{y\cdot x},$$

$$supp(D):=\{g\in G:d_g\ne 0\},$$

$$\sum_{\chi\in\hat{G}}\chi(g)=\left\{\begin{array}[]{cc}0&\textrm{ if $g\neq 1_{G}$},\\ v&\textrm{ if $g=1_{G}$},\end{array}\right.$$

$$\sum_{g\in G}\chi(g)=\left\{\begin{array}[]{cc}0&\textrm{ if $\chi\neq\chi_{0}$},\\ v&\textrm{ if $\chi=\chi_{0}$}.\end{array}\right.$$

$$d_g=\frac{1}{v}\sum _{\chi \in \hat{G}}\chi (D{g}^{-1}).$$

$$B_f:=\sum _{x\in G}{\zeta }_m^{f(x)}x.$$

$$D_f:=\sum _{x\in G}{g}^{f(x)}x.$$

$${\chi }_y(x)=(-1{)}^{y\cdot x},$$

$${\chi }_y(B_f)=\sum _{x\in G}{\zeta }_m^{f(x)}{\chi }_y(x)=\sum _{x\in G}{\zeta }_m^{f(x)}(-1{)}^{y\cdot x}=F(y),$$

$$|\chi (B_f){|}^2=2^n,$$

$$B_f{B}_f^{(-1)}=2^n.$$

$$|\chi (B_f){|}^2=\chi (B_f{B}_f^{(-1)})=2^n,$$

$$D_f{D}_f^{(-1)}=\sum _{x\in G}\sum _{y\in G}{g}^{f(y+x)}{g}^{-f(y)}x=\sum _{x\in G}{E}_{x}x,$$

$$\tau (D_f)\tau (D_f{)}^{(-1)}=B_f{B}_f^{(-1)}=2^n=\sum _{x\in G}\tau (E_x)x.$$

$$|S|\ge 2+\sum _{p\in \mathcal{P}(k)}(p-2).$$

$$\left\{\sum _{g\in H}{a}_{g}g:a_g\in \mathbb{Z}\text{mbox}and{a}_g\ge 0\right\}.$$

$$||D||\ge 2+\sum _{p\in \mathcal{P}(k)}(p-2).$$

$$D=\sum _{i=1}^9{h}^i$$

$$||D||=\sum _{i=1}^t|D_i|$$

$$||D||\ge (p_1-1)(p_2-1)+(p_3-1).$$

$$I=\{1\le i\le s:p_i\nmid k_x\forall x\in G\}\text{mbox}and\overline{m}=\prod _{i\notin I}{p}_i^{{\alpha }_i}.$$

$$\eta (D_f)\eta (D_f{)}^{(-1)}=2^n+\sum _{1_G\ne x\in G}\eta (E_x)x.$$

$$Q_1(h_1-h_1^{-1})=\sum _{i=2}^t{Q}_i(h_i^{-1}-h_i).$$

$$\sum _{i=2}^t{Q}_i(h_i^{-1}-h_i)=0.$$

$$p_1(p_2-1)+p_3-p_2\ge 2p_2+(p_3-2)\ge p_2+(5+7)\ge 3p_1+p_2$$

$$2^n=||E_x||\ge q_1+q_2+(t-2)p_1.$$

$$G_f:=\{x\in G:f(x)\text{ odd}\}.$$

$$\psi (D_f)\psi (D_f^{(-1)})$$

$$G_f^2=|G_f|+2\sum _{1_G\ne x\in G}{b}_{x}x.$$

Full text

Keywords: Exponent, generalized bent function, minimal relation, nonexistence, vanishing sum

1 Introduction

Let $m\geq 2$, $n$ be positive integers, and $\zeta_{m}=e^{\frac{2\pi\sqrt{-1}}{m}}$ be a primitive complex $m$-th root of unity. A function $f:\mathbb{Z}_{2}^{n}\rightarrow\mathbb{Z}_{m}$ is called an $(m,n)$-generalized bent function (GBF) if

[TABLE]

for all $y\in\mathbb{Z}_{2}^{n}$, where $F(y)$ is defined as

[TABLE]

and $y\cdot x$ denotes the usual inner product. In particular, when $m=2$, the generalized bent functions defined above are simply boolean bent functions introduced by Rothaus [9], whereas the function $F:\mathbb{Z}_{2}^{n}\rightarrow\mathbb{R}$ in fact becomes the Fourier transform of the boolean function $f$. In 1985, Kumar, Scholtz, and Welch [3] generalized the notion of boolean bent function by considering bent functions from $\mathbb{Z}_{m}^{n}$ to $\mathbb{Z}_{m}$. For recent nonexistence results on such generalized bend functions, see Leung and Schmidt [6]. Schmidt [11] investigated generalized bent functions from $\mathbb{Z}_{2}^{n}$ to $\mathbb{Z}_{m}$ for their applications in CDMA communications. For the boolean case, it is well known that bent function exists if and only if $n$ is even, and many constructions were reported (for a survey see [1]). In the literature, there exist constructions of generalized bent function from $\mathbb{Z}_{2}^{n}$ to $\mathbb{Z}_{m}$ for $m=4,8,2^{k}$ (for example, see [11, 10, 12, 13]). Very recently, Liu, Feng and Feng [7] presented several nonexistence results on generalized bent functions from $\mathbb{Z}_{2}^{n}$ to $\mathbb{Z}_{m}$. In this paper, we continue to investigate the nonexistence of such generalized bent functions, and present more new nonexistence results. If $m$ and $n$ are both even or $m$ is divisible by $4$, then there exists an $(m,n)$-generalized bent function [7]. Therefore, we restrict attention to the following two cases:

• (i)

$m$ is odd;

• (ii)

$n$ is odd and $m\equiv 2\pmod{4}$.

In the following, we always assume that $m$ is odd or $m=2m^{\prime}$ with $m^{\prime}$ odd.

The remainder of this paper is organized as follows. In Section 2, we introduce some basic tools and auxiliary results. In Section 3, we give several new nonexistence results of $(m,n)$-generalized bent functions, which improve the recent results in [7]. Furthermore, we show that no $(m,3)$-GBF exists for all $m$ odd or $m\equiv 2\pmod{4}$ in Section 4.

2 Basic tools and auxiliary results

In this section, we introduce some basic tools and auxiliary results, which will be used in later sections.

2.1 Group ring and character theory

It turns out that group ring and characters of abelian groups play an important role in the study of GBFs. Let $G$ be a finite group of order $v$, $R$ be a ring, and $R[G]$ denote the group ring of $G$ over $R$. For a subset $D$ of a group $G$, we may identify $D$ with the group ring element $\sum_{g\in G}d_{g}g\in R[G]$, also denoted by $D$ by abuse of notation, where $d_{g}\in R$ and these $d_{g}$’s are called coefficients of $D$. Let $1_{G}$ denote the identity element of $G$ and let $r$ be an element in $R$. For simplicity, we write $r$ for the group ring element $r1_{G}\in R[G]$. For the group ring element $D=\sum_{g\in G}d_{g}g\in R[G]$, its support is defined as

[TABLE]

and we also define $|D|:=\sum d_{g}$ and $||D||:=\sum|d_{g}|$. Let $t$ be an integer coprime to $m$. For $D=\sum_{g\in G}d_{g}g\in\mathbb{Z}[\zeta_{m}][G]$, we write $D^{(t)}=\sum d_{g}^{\sigma}g^{t}$, where $\sigma$ is the automorphism of $\mathbb{Q}[\zeta_{m}]$ determined by $\zeta_{m}^{\sigma}=\zeta_{m}^{t}$.

The group ring notation is very useful when applying characters. A character $\chi$ of $G$ is a homomorphism $\chi:G\rightarrow\mathbb{C}^{*}$. The set of all such characters forms a group $\hat{G}$ which is isomorphic to $G$ itself, and the identity element of $\hat{G}$, denoted by $\chi_{0}$, which maps every element in $G$ to $1$ (i.e., $\chi_{0}(g)=1$ for all $g\in G$), is called the principal character of $G$. It is clear that the character group has the multiplication in $\hat{G}$ defined by $\chi\tau(g)=\chi(g)\tau(g)$ for $\chi,\tau\in\hat{G}$. For $D=\sum_{g\in G}d_{g}g\in\mathbb{C}[G]$ and $\chi\in\hat{G}$, we have $\chi(D)=\sum_{g\in G}d_{g}\chi(g)$. For a subgroup $U$ of the group $G$, we define a subgroup of $\hat{G}$ as U^{\perp}:=\{\chi\in\hat{G}:\chi(g)=1\textrm{ for all g\in U}\}. If $\chi\in U^{\perp}$, we say that the character $\chi$ is trivial on $U$. It is easy to see that $|U^{\perp}|=|G|/|U|$. The following two results are standard and well-known in character theory.

Fact 1** (Orthogonality relations).**

Let $G$ be a finite abelian group of order $v$ with identity $1_{G}$. Then

[TABLE]

and

[TABLE]

Fact 2** (Fourier inversion formula).**

Let $G$ be a finite abelian group of order $v$, let $D=\sum_{g\in G}d_{g}g\in\mathbb{C}[G]$ by abuse of notation and $\chi(D)=\sum_{g\in G}d_{g}\chi(g)$. Then the coefficients in $D$ are determined by

[TABLE]

2.2 Some auxiliary results

We now characterize $(m,n)$-generalized bent functions using the group ring equations. Instead of working with additive groups, we use multiplicative notation. We denote the cyclic group of order $m$ by $C_{m}$, and set $G=C_{2}^{n}$. Whenever $s|m$, we also denote the subgroup of order $s$ in $C_{m}$ by $C_{s}$.

Definition 2.1**.**

Let $f:G\rightarrow\mathbb{Z}_{m}$ be a function, and $g$ be a generator of $C_{m}$. We define an element $B_{f}$ in the group ring $\mathbb{Z}[\zeta_{m}][G]$ corresponding to $f$ by

[TABLE]

Furthermore, we define an element $D_{f}$ in the group ring $\mathbb{Z}[C_{m}][G]$ by

[TABLE]

Remark 2.2**.**

To study $(m,n)$-GBFs, we may assume that $C_{m}=\langle\{g^{f(x)}:x\in G\}\rangle$. By scaling if necessary, we may always assume $f(1_{G})=0$, i.e., $g^{f(1_{G})}=g^{0}$ is the identity element of $C_{m}$. From time to time, we may also interpret $\mathbb{Z}[C_{m}][G]$ as $\mathbb{Z}[C_{m}\cdot G]$, where $g^{0}$ and $1_{G}$ in $\mathbb{Z}[C_{m}\cdot G]$ both denote the identity element of $C_{m}\cdot G$.

Let $\tau$ be a character that maps $g$ to $\zeta_{m}$, then it is clear that $\tau(D_{f})=B_{f}$. Moreover, every element $y\in G$ determines a character $\chi_{y}$ of $G$ by

[TABLE]

for all $x\in G$. It is easily verified that every complex character of $G$ is equal to some $\chi_{y}$ with $y\in G$. Note that

[TABLE]

for all $y\in G$, where $F(y)$ is defined in (2). It then follows from (1) and (3) that $f$ is an $(m,n)$-GBF if and only if

[TABLE]

for all $\chi\in\hat{G}$. We now have the following characterization of $(m,n)$-GBFs.

Proposition 2.3**.**

Let $f$ be a function from $G$ to $\mathbb{Z}_{m}$. Then $f$ is an $(m,n)$-GBF if and only if

[TABLE]

Furthermore, if $f(G)=2\mathbb{Z}_{m}$, then $f$ can be regarded as an $(m^{\prime},n)$-GBF, where $m=2m^{\prime}$ with $m^{\prime}$ odd.

Proof.

From (4) it follows that

[TABLE]

for all characters $\chi$ of $G$. Using Facts 1 and 2, we are able to determine all the coefficients of $B_{f}B_{f}^{(-1)}$, i.e., (4) holds if and only if (5) is satisfied. The last statement follows from the fact that $\zeta_{m}^{f(x)}$ becomes an $m^{\prime}$-th root of unity. ∎

Observe that we may write

[TABLE]

where $E_{x}=\sum_{y\in G}g^{f(y+x)}g^{-f(y)}\in\mathbb{Z}[C_{m}]$.

Lemma 2.4**.**

Suppose that $f$ is a GBF from $G$ to $\mathbb{Z}_{m}$. Then

• (a)

$E_{x}=E_{x}^{(-1)}$* and the coefficient of $g^{0}$ in $E_{x}$ is even for all $x\in G$;*

• (b)

For each character $\tau$ of order $m$ on $C_{m}$, we have $\tau(E_{x})=0$ for all $x\neq 1_{G}$.

Proof.

Note that $(D_{f}D_{f}^{(-1)})^{(-1)}=D_{f}D_{f}^{(-1)}$. Hence, we have $E_{x}=E_{x}^{(-1)}$ for all $x\in G$. Note that $E_{1_{G}}=2^{n}$. Thus, we may consider $x\neq 1_{G}$. Suppose that $x\neq 1_{G}$ and $(g_{1}x_{1})(g_{2}x_{2})^{-1}=g^{0}x$ for some $g_{1},g_{2}\in C_{m}$ and $x_{1},x_{2}\in G$. Note that $x_{1}\neq x_{2}$ and clearly, we have $(g_{2}x_{2})(g_{1}x_{1})^{-1}=g^{0}x$ as well. This shows that the coefficient of $g^{0}$ in $E_{x}$ is even.

For any character $\tau$ of order $m$ on $C_{m}$, we obtain

[TABLE]

From (5) in Proposition 2.3, the conclusion follows. ∎

The key in our study of $(m,n)$-GBFs is to investigate $E_{x}$. Lemma 2.4 (b) allows us to define the notion of vanishing sum (v-sum), which was also studied in details in [4]. Another important notion to study v-sum is the idea of exponents and reduced exponents defined in [5]. In Section 3, we will use exponents to derive some new nonexistence results. To this end, we recall some notations defined in [5] and prove some preliminary lemmas.

Let $S$ be a finite index set, and we denote by $\mathcal{P}(k)$ the set of all prime factors of the integer $k$.

Definition 2.5**.**

Suppose that $X=\sum_{i\in S}a_{i}\mu_{i}$ where $\mu_{i}$’s are distinct roots of unity and all $a_{i}$’s are nonzero integers. We say that $u$ is the exponent of $X$ if $u$ is the smallest positive integer such that $\mu_{i}^{u}=1$ for all $i$. We say that $k$ is the reduced exponent of $X$ if $k$ is the smallest positive integer such that there exists $j$ with $(\mu_{i}\mu_{j}^{-1})^{k}=1$ for all $i$.

For example, the exponent of $\sum_{i=0}^{p-1}\zeta_{3}\zeta_{p}^{i}$ is $3p$, whereas the reduced exponent is $p$. To study vanishing sums, we consider those which are minimal.

Definition 2.6**.**

Suppose that $X=\sum_{i\in S}a_{i}\mu_{i}=0$ where $\mu_{i}$’s are distinct roots of unity and all $a_{i}$’s are nonzero integers. We say that the relation $X=0$ is minimal, if for any proper subset $I\subsetneq S$, $\sum_{i\in I}a_{i}\mu_{i}\neq 0$.

Based on the definition of minimal relation, we have the following restriction on the cardinality of the index set $S$, in terms of the reduced exponents of a minimal vanishing sum.

Proposition 2.7**.**

[2]** Suppose that $X=\sum_{i\in S}a_{i}\mu_{i}=0$ is a minimal relation with reduced exponent $k$ and all $a_{i}$’s are nonzero. Then $k$ is square free and

[TABLE]

For convenience, we define the following notation.

Definition 2.8**.**

For any group $H$, by $\mathbb{N}[H]$ we denote

[TABLE]

Now we consider the corresponding notion of minimal relation in $\mathbb{N}[C_{m}]$. From now on, we assume that $g$ is a generator of $C_{m}$. We recall the notion of minimality defined in Section 4 of [4].

Definition 2.9**.**

[4]** Let $D=\sum_{i=0}^{m-1}a_{i}g^{i}\in\mathbb{N}[C_{m}]$. We say that $D$ is a v-sum if there exists a character $\tau$ of order $m$ such that $\tau(D)=\tau(\sum_{i=0}^{m-1}a_{i}g^{i})=0$. We say that $D$ is minimal if $\tau(\sum_{i=0}^{m-1}b_{i}g^{i})\neq 0$ whenever $0\leq b_{i}\leq a_{i}$ for all $i$ and $b_{j}<a_{j}$ for some $j$.

Suppose that $S\subseteq\{0,\ldots,m-1\}$ and $a_{i}>0$ for all $i\in S$. It is clear that if $D=\sum_{i\in S}a_{i}g^{i}$ is a minimal v-sum by Definition 2.9, then $\tau(D)=\sum_{i\in S}a_{i}\tau(g)^{i}$ is a minimal relation by Definition 2.6. We now define the reduced exponent of $D$ as follows.

Definition 2.10**.**

Suppose that $D=\sum_{i=0}^{m-1}d_{i}g^{i}\in\mathbb{N}[C_{m}]$ is a minimal v-sum. We define the reduced exponent $k$ of $D$ as the reduced exponent of the vanishing sum $\tau(D)=\sum_{i=0}^{m-1}d_{i}\tau(g)^{i}$.

Note that the reduced exponent defined above does not depend on the choice of the character $\tau$.

Lemma 2.11**.**

If $D\in\mathbb{N}[C_{m}]$ is a minimal v-sum with reduced exponent $k$, then $D=D^{\prime}h$ for some $D^{\prime}\in\mathbb{N}[C_{k}]$ and $h\in C_{m}$.

Proof.

Write $D=\sum_{i\in S}d_{i}g^{i}$ and $\tau(D)=\sum_{i\in S}d_{i}\tau(g^{i})$ where $S\subseteq\{0,\ldots,m-1\}$. Since $k$ is the reduced exponent of $D$, by Definition 2.10, the reduced exponent of $\tau(D)$ is also $k$. Thus, there exists a $j$ such that $(\tau(g^{i})\tau(g^{-j}))^{k}=1$ for all $i\in S$. It then follows that $Dg^{-j}\in\mathbb{N}[C_{k}]$. The proof is then completed. ∎

In view of Proposition 2.7, we derive the following result.

Corollary 2.12**.**

Suppose that $D=\sum_{i=0}^{m-1}a_{i}g^{i}\in\mathbb{N}[C_{m}]$ is a minimal v-sum with reduced exponent $k$. Then $k$ is square free and

[TABLE]

To deal with a v-sum $D\in\mathbb{N}[C_{m}]$ which is not minimal, we first decompose it into sum of minimal v-sums. It is straightforward to prove the following.

Lemma 2.13**.**

Let $D\in\mathbb{N}[C_{m}]$ be a v-sum. Then $D$ can be written as the form $D=\sum D_{i}$, where $D_{i}$’s are minimal v-sums in $\mathbb{N}[C_{m}]$.

We aim to find a lower bound of $||D||$ when $D$ is a v-sum. To do so, we need to extend the notion of reduced exponent and then apply Corollary 2.12. Suppose that $D=\sum_{i=1}^{t}D_{i}$ and $k_{i}$ is the reduced exponent of $D_{i}$ for each $i$. We may then define the exponent of $D$ to be $\mathrm{lcm}(k_{1},\ldots,k_{t})$. However, we note that such a decomposition is not necessarily unique. For example, if $m=10$ and $h$ is a generator of $C_{10}$, then we have

[TABLE]

Note that $(1+h^{5})h^{i}$ and $(1+h^{2}+h^{4}+h^{6}+h^{8})h^{j}$ are both minimal v-sums. If we use the notion of $\mathrm{lcm}$ of each decomposition, we will then get $2$ and $5$ as the reduced exponents, respectively. Thus, we need to modify the earlier definition of exponent as follows.

Definition 2.14**.**

Suppose that $D=\sum_{i=0}^{m-1}d_{i}g^{i}$ is a v-sum in $\mathbb{N}[C_{m}]$. We define the c-exponent of $D$ to be the smallest $k$ such that there exist $t$ minimal v-sums $D_{1},\ldots,D_{t}$ in $\mathbb{N}[C_{m}]$ with $D=\sum_{i=1}^{t}D_{i}$ and $k=\mathrm{lcm}(k_{1},\ldots,k_{t})$, where $k_{i}$ is the reduced exponent of $D_{i}$ for $i=1,\ldots,t$.

Note that in the example above, the c-exponent of $D$ is $2$.

Lemma 2.15**.**

Suppose that $D=\sum_{i=0}^{m-1}d_{i}g^{i}\in\mathbb{N}[C_{m}]$ is a v-sum with c-exponent $k$. Write $m=\prod_{i=1}^{s}p_{i}^{\alpha_{i}}$ and $k=\prod_{i=1}^{t}p_{i}$. Note that $t\leq s$ and $p_{i}$’s are distinct primes. Then we have the followings:

• (a)

$||D||\geq 2+\sum_{i=1}^{t}(p_{i}-2)$;

• (b)

$D=\sum_{i=1}^{t}P_{i}E_{i}$, where $P_{i}$ is the subgroup of order $p_{i}$ and $E_{i}\in\mathbb{Z}[C_{m}]$ for all $i$;

• (c)

Suppose that $\prod_{i=1}^{t}p_{i}^{\alpha_{i}}|d$ and $d|m$. If $\phi:\mathbb{Z}[C_{m}]\rightarrow\mathbb{Z}[C_{d}]$ is the natural projection, then $\chi(\phi(D))=0$ whenever $\mathrm{ord}(\chi)=d$.

Proof.

By Lemma 2.13, we may assume that $D=\sum_{i=1}^{t}D_{i}$ such that each $D_{i}$ is a minimal v-sum. Hence, by Corollary 2.12, we have

[TABLE]

because ${\mathcal{P}}(k)=\bigcup_{i=1}^{t}{\mathcal{P}}(k_{i})$.

By Lemma 2.11, $D_{i}=E_{i}g_{i}$ where $E_{i}\in\mathbb{N}[C_{k_{i}}]$ and $g_{i}\in C_{m}$. Clearly, $\tau(E_{i})=0$. Therefore, from [4, Theorem 2.2], it follows that $E_{i}=\sum_{q\in{\mathcal{P}}(k_{i})}Q_{q}F_{q}$, where $Q_{q}$ is the subgroup of order $q$ and $F_{q}\in\mathbb{Z}[C_{k_{i}}]$. Since $D=\sum D_{i}$, $D$ is of the desired form.

Finally, note that if $\phi$ and $\chi$ are defined as in (c), then $\chi(\phi(D))=0$ as $\chi(\phi(P_{i}))=\chi(P_{i})=0$ for $i=1,\ldots,t$. ∎

Next, we record a very useful result from [4, Theorem 4.8, Proposition 6.2].

Proposition 2.16**.**

[4]** Let $D\in\mathbb{N}[C_{m}]$ be a minimal v-sum with c-exponent $k$. Then we have the followings:

• (a)

If $k=p$ is prime and $P$ is the subgroup of order $p$, then $D=Ph$ for some $h\in C_{m}$.

• (b)

If $k=\prod_{i=1}^{t}p_{i}$ with $t\geq 2$ and $p_{1}<p_{2}<\cdots<p_{t}$ are primes, then $t\geq 3$ and

[TABLE]

Moreover, equality holds only if $D=(P_{1}^{*}P_{2}^{*}+P_{3}^{*})h$ for some $h\in C_{m}$. Here $P_{i}^{*}=P_{i}-\{e\}$, and $P_{i}$ is the subgroup of order $p_{i}$.

Remark 2.17**.**

It follows from Proposition 2.16 that either $k$ is a prime or $k$ has at least three prime factors.

3 New nonexistence results of $(m,n)$-GBFs

In this section, we derive some new necessary conditions on $(m,n)$-GBFs, and then give new nonexistence results accordingly. First we fix the following notation. As before, we assume that $g$ is the generator of $C_{m}$, and note that Remark 2.2 holds for any GBF $f$. To avoid confusion, we set $g^{0}$ as the identity element of $C_{m}$.

The following result is very important, in the sense that it allows to eliminate all prime factors of $m$ greater than $2^{n}$ when deriving nonexistence results.

Proposition 3.1**.**

Suppose that $f$ is an $(m,n)$-GBF and $m=\prod_{i=1}^{s}p_{i}^{\alpha_{i}}$ where $p_{i}$’s are distinct primes. Let $k_{x}$ be the c-exponent of $E_{x}$ (as defined by (6)) for each $1_{G}\neq x\in G$. Set

[TABLE]

Then there exists an $(\overline{m},n)$-GBF. In particular, if $p_{i}|m$ and $p_{i}>2^{n}$, then there exists an $(m/p_{i},n)$-GBF.

Proof.

By induction, it suffices to show that if $p_{i}\in I$, then there exists an $(m/p_{i},n)$-GBF. Let $\eta:\mathbb{Z}[\langle g\rangle]\rightarrow\mathbb{Z}[\langle g^{p_{i}}\rangle]$ be the natural projection, it then follows that

[TABLE]

Recall that $E_{x}$ is a v-sum. By assumption $p_{i}$ does not divide $k_{x}$ for all $1_{G}\neq x\in G$. It follows from Lemma 2.15 (c) that $\tau(\eta(E_{x}))=0$ if $\tau$ is a character of order $m/p_{i}$. Therefore, $\tau(\eta(D_{f}))$ gives rise to an $(m/p_{i},n)$-GBF.

The last statement is now clear as if $p_{i}>2^{n}$, then by Lemma 2.15 (a), $p_{i}$ does not divide $k_{x}$ for any $1_{G}\neq x\in G$.

∎

We record the following result which will be used from time to time later.

Lemma 3.2**.**

Suppose that $f$ is an $(m,n)$-GBF, and $p,q$ are distinct primes that both divide $m$. Then there exist $y\neq 1_{G}$ and $h\in supp(E_{y})$ such that $pq|\circ(h)$.

Proof.

As $C_{m}=\langle\{g^{f(x)}:x\in G\}\rangle$, there exist $u,v\in G$ such that $p|\circ(g^{f(u)})$ and $q|\circ(g^{f(v)})$. Since $g^{f(1_{G})}=g^{0}\in C_{m}$, we know that $g^{f(u)}\in supp(E_{u})$ and $g^{f(v)}\in supp(E_{v})$. We are done if $q|\circ(g^{f(u)})$ or $p|\circ(g^{f(v)})$. Otherwise, $ug^{f(u)}(vg^{-f(v)})\in supp(E_{uv})$ and then clearly $pq|\circ(g^{f(u)-f(v)})$. The proof is completed. ∎

Before we proceed, we need a technical result.

Lemma 3.3**.**

Let $q_{1},q_{2},q_{3}$ be primes that divide $m$ and $Q_{1},Q_{2},Q_{3}$ be subgroups of order $q_{1},q_{2},q_{3}$, respectively. Suppose that $4\nmid m$ and $\sum_{i=1}^{t}Q_{i}h_{i}=\sum_{i=1}^{t}Q_{i}h_{i}^{-1}$ for some $h_{1},h_{2},h_{t}\in C_{m}$ with $t\geq 2$.

• (a)

If $q_{1}\neq q_{2}$ and $t=2$, then we may assume $h_{i}^{-1}=h_{i}$ for $i=1,2$.

• (b)

If $q_{1}\neq q_{2}=q_{3}$ and $t=3$, then we may assume $Q_{2}h_{2}+Q_{2}h_{3}=Q_{2}(h_{2}+h_{2}^{-1})$ and $h_{1}=h_{1}^{-1}$.

• (c)

If all $q_{i}$’s are distinct, then we may assume $h_{i}=h_{i}^{-1}$ for all $i$.

Proof.

By assumption, we have

[TABLE]

Suppose that $q_{1}^{\beta_{1}}||m$. Let $\phi:\mathbb{Z}[C_{m}]\rightarrow\mathbb{C}[C_{m}]$ be a ring homomorphism that fixes $g^{m/q_{1}^{\beta_{1}}}$ and sends $g^{q_{1}^{\beta_{1}}}$ to an $m/q_{1}^{\beta_{1}}$-primitive root of unity. Then, we have $\phi(Q_{i}(h_{i}-h_{i}^{-1}))=0$ for $i=2,\ldots,t$, which implies that $\phi(Q_{1}h_{1}-Q_{1}h_{1}^{-1})=0$. Write $h_{1}=g_{1}h^{\prime}$ with $g_{1}\in\langle g^{m/q_{1}^{\beta_{1}}}\rangle$ and $p_{1}\nmid\circ(h^{\prime})$. Then, we have $Q_{1}g_{1}\phi(h^{\prime})=Q_{1}g_{1}^{-1}\phi(h^{\prime-1})$. Hence $g_{1}^{2}\in Q_{1}$ and $\phi(h^{\prime})=\phi(h^{\prime-1})$. If $q_{1}$ is odd, then $g_{1}=g^{0}$. If $q_{1}=2$, then as $4\nmid m$, $g_{1}$ can be taken as $g^{0}$ as well. In both cases, we may assume $g_{1}=g^{0}$. It follows that $\phi(h^{\prime})^{2}=1$. As $\phi$ is of order $m/q_{1}^{\beta_{1}}$, $h^{\prime 2}=g^{0}$. Therefore, $g_{1}h^{\prime}=(g_{1}h^{\prime})^{-1}$. Furthermore, we have

[TABLE]

Now (a) follows easily by applying the same argument on $Q_{2}$.

If $t=3$ and $q_{2}=q_{3}$, we then obtain $Q_{2}(h_{2}+h_{3})=Q_{2}(h_{2}^{-1}+h_{3}^{-1})$. If $Q_{2}h_{2}=Q_{2}h_{2}^{-1}$, then we must have $Q_{2}h_{3}=Q_{2}h_{3}^{-1}$. Then, $h_{2}^{2}\in Q_{2}$ and $h_{3}^{2}\in Q_{2}$. Using a similar argument as before, we may assume that $h_{2}=h_{3}=g^{0}$. If $Q_{2}h_{2}=Q_{2}h_{3}^{-1}$, then clearly, we may take $h_{3}=h_{2}^{-1}$ and we are done.

To obtain (c), we set $t=3$. We then get our desired results by applying part (a) to Equation (7).

The proof is then completed. ∎

Now we are able to give the following necessary conditions on the existence of $(m,n)$ GBFs, where $m$ is odd.

Theorem 3.4**.**

Suppose that $m=\prod_{i=1}^{s}p_{i}^{\alpha_{i}}$, where $3\leq p_{1}<p_{2}<\cdots<p_{s}$ are odd primes and $\alpha_{i}$’s are all positive integers. If an $(m,n)$-GBF exists, then $s\geq 2$ and $3p_{1}+p_{2}\leq 2^{n}$.

Proof.

Recall that if $1_{G}\neq x\in G$ and $\chi$ is a character of order $m$, then $\chi(E_{x})=0$. If $s=1$, then by Lemma 2.15 (b), $E_{x}=P_{1}W$ where $P_{1}$ is a subgroup of order $p_{1}$ and $W\subseteq C_{m}$. In other words, $2^{n}=||E_{x}||=p_{1}||W||$. This is impossible as $p_{1}\neq 2$.

Next, we assume that $s\geq 2$. As $E_{x}\in\mathbb{N}[C_{m}]$, we may write $E_{x}=\sum D_{j}$ such that all $D_{j}$’s are minimal v-sums. Let $k_{j}$ be the reduced exponent of $D_{j}$. If $|{\mathcal{P}}(k_{j})|\geq 4$, then by Corollary 2.12, we have $||D_{j}||\geq 2+\sum_{i=1}^{4}(p_{i}-2)\geq 3p_{1}+p_{2}$. Thus, we may assume that $|{\mathcal{P}}(k_{j})|\leq 3$. But by Proposition 2.16, $|{\mathcal{P}}(k_{j})|=1$ or $3$. In case that $|{\mathcal{P}}(k_{j})|=3$, $||D_{j}||\geq q_{1}(q_{2}-1)+q_{3}-q_{2}\geq p_{1}(p_{2}-1)+p_{3}-p_{2}$. If $p_{1}\geq 5$, then clearly, $p_{1}(p_{2}-1)+p_{3}-p_{2}\geq 3p_{1}+p_{2}$. If $p_{1}=3$, it then follows that

[TABLE]

as $p_{2}\geq 5$ and $p_{3}\geq 7$.

It remains to consider the case $|{\mathcal{P}}(k_{j})|=1$, i.e., $D_{j}=Q_{j}h_{i}$ where $h_{i}\in C_{m}$ and $Q_{j}$ is a subgroup of order $q_{j}$. Note that $q_{j}$’s need not be distinct. Therefore, $E_{x}=\sum_{j=1}^{t}Q_{j}h_{j}$. If all $Q_{j}$’s are the same, then $E_{x}=Q_{1}Y$ for some $Y\in\mathbb{Z}[C_{m}]$. This is impossible as $q_{1}\nmid 2^{n}$. In particular, it follows that $t\geq 2$ and we may assume $Q_{1}\neq Q_{2}$ without loss of generality. Recall that all $D_{i}\in\mathbb{N}[C_{m}]$. Therefore,

[TABLE]

Hence, we are done if $t\geq 4$.

We first study the case $t=3$. As $q_{1}\neq q_{2}$, we may assume $q_{1}\neq q_{3}$ as well. Since $E_{x}^{(-1)}=E_{x}$ and $m$ is odd, we may then assume $h_{1}=g^{0}$. Moreover, if $Q_{2}=Q_{3}$, then $Q_{2}h_{2}+Q_{2}h_{3}=Q_{2}(h_{2}+h_{2}^{-1})$. Whereas if $Q_{2}\neq Q_{3}$, then $h_{2}=h_{3}=1_{G}$ as $m$ is odd. Therefore, the coefficient of $g^{0}$ is either $1$ or $3$ in both cases. This contradicts Lemma 2.4 (a).

Thus, we may assume $t=2$ for all $x\neq 1_{G}$. Moreover, as $m$ is odd, $E_{x}$ is of the form $Q_{1}+Q_{2}$. In particular, each non-identity element in $supp(E_{x})$ is of prime order. This contradicts Lemma 3.2.

The proof is then completed. ∎

The theorem above provides an alternative proof of [7, Corollary 2], from which we can have an improved result on the case $s=2$.

Corollary 3.5**.**

Suppose that $m=\prod_{i=1}^{s}p_{i}^{\alpha_{i}}$, where $p_{1}<p_{2}<\cdots<p_{s}$ are odd primes and $\alpha_{i}$’s are all positive integers.

• (a)

There is no $(m,n)$-GBF when $s=1$.

• (b)

There is no $(m,n)$-GBF if $s\geq 2$ and $3p_{1}+p_{2}>2^{n}$.

• (c)

There is no $(m,n)$-GBF if there is no $(\prod_{i=1}^{r}p_{i}^{\alpha_{i}},n)$-GBF where $p_{r+1}$ is the smallest prime such that $p_{1}+p_{r+1}>2^{n}$.

Proof.

(a) and (b) follow directly from Theorem 3.4. As for (c), it suffices to show that if $t\geq r+1$, then $p_{t}$ does not divide the c-exponent of $E_{x}$ for any $x\neq 1_{G}$. We follow the notation used in the proof of Theorem 3.4. We write $E_{x}=\sum D_{j}$ such that all $D_{j}$’s are minimal v-sums. Again, we denote by $k_{j}$ the reduced exponent of $D_{j}$. Suppose that $p_{t}|k_{1}$. If $k_{1}=p_{t}$, then $E_{x}\neq D_{1}$ as otherwise $p_{t}|2^{n}$. Therefore, $||E_{x}||\geq||D_{1}||+||D_{2}||\geq p_{t}+p_{1}>2^{n}$. On the other hand, if $k_{1}\neq p_{t}$, then as shown before, $k_{1}$ is a product of at least three primes. Hence, $||D_{1}||\geq p_{t}+p_{1}>2^{n}$, which is impossible. ∎

Remark 3.6**.**

For $s=2$, our result is stronger than  [7, Corollary 2].

Now we consider the case when $m=2m^{\prime}$ with $m^{\prime}$ odd. If $f$ is a $(2m^{\prime},n)$ GBF, then we define

[TABLE]

Note that a $(2m^{\prime},n)$ GBF is trivially an $(m^{\prime},n)$ GBF if $G_{f}=\emptyset$ or $G$. Multiply $f$ by $-1$ if necessary, we may always assume $|G_{f}|\leq|G|/2$. Note that $G_{f}^{(-1)}=G_{f}$ as $G$ is $2$-elementary. Apply a homomorphism $\psi:\mathbb{Z}[G\cdot C_{m}]$ such that $\psi$ fixes every element in $G$ and maps the generator $g$ of $C_{m}$ to $-1$, then we have

[TABLE]

Write

[TABLE]

We denote $\psi(E_{x})$ by $a_{x}$. It then follows that for $x\neq 1_{G}$,

[TABLE]

The following is a consequence of [8, Theorem 1].

Lemma 3.7**.**

If $n$ is odd, then $G_{f}$ is a difference set in $G$ if and only if $G_{f}=\{1_{G}\}$.

We now give the following nonexistence results on $(2m^{\prime},n)$ GBFs, which are weaker than those in Theorem 3.4.

Theorem 3.8**.**

Let $n$ be odd and $m=2p^{\alpha}$, where $\alpha$ is a positive integer. Suppose that an $(m,n)$-GBF exists. Then $p<2^{n-3}$ unless $p=2^{n-2}-1$ is a Mersenne prime. In particular, if $n\leq 3$, there is no $(m,n)$-GBF if $m=p^{\alpha}$ or $m=2p^{\alpha}$.

Proof.

Let $P_{2}$ be the subgroup of order $2$ and $P$ be a subgroup of order $p$. For any $x\neq 1_{G}$, we conclude from Lemma 2.15 (b) that $E_{x}=P_{2}Y_{x}+PZ_{x}$ for some $Y_{x},Z_{x}\in\mathbb{N}[C_{m}]$. Note that $\psi(E_{x})\neq 0$ for some $x\neq 1_{G}$. Otherwise, the c-exponent of all $E_{x}$ is $2$ and by Proposition 3.1, there exists a $(2,3)$-GBF, which is impossible. Hence, $a_{x}=\psi(E_{x})\neq 0$ for some $x\neq 1_{G}$. Therefore, we have $\psi(P)|\psi(E_{x})$, i.e., $p|a_{x}$. Note that in view of Equation (9), $4p|a_{x}$ if $|G_{f}|$ is odd and $8p|a_{x}$ if $|G_{f}|$ is even. We are done if $8p|a_{x}$ as $|a_{x}|<2^{n}$. We may therefore assume that $|G_{f}|$ is odd.

Suppose that $G_{f}=\{1_{G}\}$. Then, $a_{x}=2^{n}-4$ if $x\neq 1_{G}$. Hence, $4p|a_{x}$. It follows that $p_{1}<2^{n-3}$ unless $4p=2^{n}-4$ which implies that $p=2^{n-2}-1$ is a Mersenne prime.

Suppose that $G_{f}\neq\{1_{G}\}$. As $G_{f}$ is not a difference set, there exist two elements $x\neq 1_{G}$ and $x^{\prime}\neq 1_{G}$ such that $b_{x}>b_{x^{\prime}}\geq 0$. Since $p|a_{x}$ and $p|a_{x^{\prime}}$, it follows that $p|(b_{x}-b_{x^{\prime}})$ and $b_{x}-b_{x^{\prime}}=tp$ for some positive integer $t$. To get our desired result, we need to find a bound on $b_{x}-b_{x^{\prime}}$. Note that in view of Equation (8), $b_{x}\leq|G_{f}|/2\leq|G|/4$. Hence, we get our desired result if $t\geq 2$. Thus, we may assume that $t=1$, i.e., $b_{x}=p+b_{x^{\prime}}$.

Suppose that $G=\langle x\rangle\cdot G^{\prime}$, where $G^{\prime}$ is a subgroup of order $2^{n-1}$ in $G$. As the coefficient of $x$ in $G_{f}^{2}$ is $2b_{x}$, there are $2b_{x}=2p+2b_{x^{\prime}}$ pairs $(u,v)$ of elements in $G_{f}\times G_{f}$ such that $uv=x$. Therefore, there exists a set $Y\subseteq G^{\prime}\cap G_{f}$ such that $Y\cup(Yx)\subseteq G_{f}$ with $|Y|=p+b_{x^{\prime}}$. Write $G_{f}=(Y\cup Z_{1})\cup(Yx\cup Z_{2}x)$ such that

[TABLE]

Since $b_{x}=|Y|$, it follows that $Z_{1}\cap Z_{2}=\emptyset$. Moreover, we have

[TABLE]

Note that the support of $[2Y^{2}+2Y(Z_{1}+Z_{2})+Z_{1}^{2}+Z_{2}^{2}]$ is in $G^{\prime}$ and the support of $[2Y^{2}+2Y(Z_{1}+Z_{2})+2Z_{1}Z_{2}]x$ is in $G^{\prime}x$. We now consider the coefficients of the following group elements

[TABLE]

For any $1_{G}\neq v\in G^{\prime}$, the coefficient of $v$ in $Z$ is equal to $2(b_{v}-b_{vx})$. Clearly, the absolute value of the coefficient of $v$ in $Z$ is less than $|Z_{1}|+|Z_{2}|$ as $Z_{1}$ and $Z_{2}$ are disjoint. Thus, if there exists $v\neq 1_{G}$ in $G^{\prime}$ such that $b_{v}-b_{vx}$ is nonzero, then $p|(b_{v}-b_{vx})$ and we obtain

[TABLE]

Hence, we get $4p\leq|G_{f}|\leq|G|/2$ and $p\leq 2^{n-3}$. Thus, it remains to deal with the case $(Z_{1}-Z_{2})^{2}=|Z_{1}|+|Z_{2}|$.

If both $Z_{1}=Z_{2}=\emptyset$, then $2b_{x}=|G_{f}|$. Hence, $|G_{f}|$ is even and as remarked earlier, we are done in this case. Note that as $G=C_{2}^{n}$, all character values of $Z_{1}-Z_{2}$ are integers. Thus, $|Z_{1}|+|Z_{2}|$ is a square. Since $Z_{1}\cap Z_{2}=\emptyset$, all nonzero coefficients of $Z_{1}-Z_{2}$ is $\pm 1$. On the other hand, if $q$ is an odd prime divisor or $|Z_{1}|+|Z_{2}|$, then $q$ divides the all nonzero coefficients of $Z_{1}-Z_{2}$ by applying Fourier inversion formula. This is impossible. It follows that $|Z_{1}|+|Z_{2}|=2^{t}$. Again, we are done if $t\geq 1$ as then $|G_{f}|=2b_{x}+|Z_{1}|+|Z_{2}|$ is even. Hence, we may assume that $t=0$, i.e., $|Z_{1}|+|Z_{2}|=1$. Note that the coefficient of $1_{G}$ in $[2Y^{2}+2Y(Z_{1}+Z_{2})+Z_{1}^{2}+Z_{2}^{2}]$ is $|G_{f}|$ and the coefficient of $1_{G}$ in $[2Y^{2}+2Y(Z_{1}+Z_{2})+2Z_{1}Z_{2}]$ is the same as the coefficient of $x$ in $G_{f}^{2}$. As $Z=1$, it follows that $2b_{x}=|G_{f}|-1$. Hence, $a_{x}=|G|-4|G_{f}|+4(|G_{f}|-1)=2^{n}-4$. Recall that $4p|a_{x}$. Hence either $p=2^{n-2}-1$ or $p<2^{n-3}$.

The proof is then completed. ∎

Corollary 3.9**.**

Let $n$ be odd and $m=2\prod_{i=1}^{s}p_{i}^{\alpha_{i}}$, where $p_{1}<p_{2}<\cdots<p_{s}$ are odd primes and $\alpha_{i}$’s are all positive integers.

• (a)

If $s=1$, then there is no $(m,n)$-GBF if one of the following conditions is satisfied:

• (i)

$p_{1}>2^{n-2}$;

• (ii)

$p_{1}$* is not a Mersenne prime and $p_{1}>2^{n-3}$;*

• (iii)

$p_{1}\equiv 3,\ 5\pmod{8}$.

• (b)

If $s\geq 2$, and $r$ is the least integer such that $p_{r+1}+p_{1}>2^{n}+2$, then there is no $(m,n)$-GBF if there is no $(2\prod_{i=1}^{r}p_{i}^{\alpha_{i}},n)$-GBF. In particular, there is no $(m,n)$-GBF if $p_{1}>2^{n-2}$ and $p_{1}+p_{2}>2^{n}+2$.

Proof.

It is easily seen that (i) and (ii) of (a) directly follow from Theorem 3.8. If (iii) holds, it is known that no $(2p_{1}^{\alpha_{1}},n)$-GBF exists.

To prove (b), it is sufficient to show that for $i\geq r+1$, $p_{i}$ does not divide the c-exponent of any $E_{x}$ for $x\neq 1_{G}$. As before, we wirte $E_{x}=\sum D_{j}$ and $k_{j}$ the reduced exponent of $D_{j}$. We may assume that $p_{i}$ divides $k_{1}$. If $k_{1}$ consists of at least three prime factors, then $||D_{i}||\geq 2+(p_{1}-2)+(p_{i}-2)$. Thus, $2^{n}\geq p_{1}+p_{i}-2\geq p_{1}+p_{r+1}-2>2^{n}$. This is impossible. Therefore, we have $k_{1}=p_{i}$.

Otherwise, we assume that $p_{i}$ divides the reduced exponent $k_{x}$ of $\tau(E_{x})$. If $k_{x}=p_{i}$, it follows from the argument in (a) that $4p_{i}\leq 2^{n}$. This is impossible as $2^{n}<p_{1}+p_{i}<4p_{i}$. Therefore, $p_{j}|k_{x}$ for some $j\neq i$. But then by Proposition 2.16, $2^{n}\geq p_{j}+p_{i}-2>p_{r+1}+p_{1}-2$. This is impossible. ∎

Remark 3.10**.**

When compared with [7, Theorem 2], our result in Corollary 3.9 is stronger in all cases quoted in Table 2 [7] therein except for the case that $p=191$.

4 Nonexistence results for $n=3$

In this section, we show that there in no $(m,3)$-GBF for all $m$ odd or $m\equiv 2\pmod{4}$. By Proposition 3.1, we may assume that all prime factors of $m$ are less than or equal to $7$. According to Corollary 3.5, we conclude that there is no $(m,3)$-GBF if $m$ is odd. Therefore, we may write $m=2\cdot 3^{a}5^{b}7^{c}$. For convenience, we fix the following notation. Let $g_{2},g_{3},g_{5},g_{7}$ be elements of order $2,3,5,7$, respectively. Let $P_{2},P_{3},P_{5},P_{7}$ be subgroups of order $2,3,5$ and $7$, respectively.

We assume that $f$ is an $(m,3)$-GBF. We first determine what $E_{x}$ is if $x\neq 1_{G}$. As seen before, $\tau(E_{x})=0$ for any character of order $m$. Recall that $\mathcal{P}(k)$ denotes the set of all prime factors of the integer $k$.

Lemma 4.1**.**

For any $x\neq 1_{G}$, write $E_{x}=\sum D_{i}$ where each $D_{i}$ is a minimal v-sum with reduced exponent $k_{i}$. Then ${\mathcal{P}}(k_{i})=\{2\},\{3\},\{5\},\{7\}$ or $\{2,3,5\}$ or $\{2,3,7\}$. Moreover,

• (a)

If ${\mathcal{P}}(k_{i})=\{j\}$ for some $j\in\{2,3,5,7\}$, then $D_{i}=P_{j}h_{j}$ for some $h_{j}\in C_{30}$.

• (b)

If ${\mathcal{P}}(k_{i})=\{2,3,7\}$, then $E_{x}=g_{2}^{\alpha}(P_{7}^{*}+g_{2}P_{3}^{*})$ for some integer $\alpha$.

Proof.

Let $k_{x}$ be the reduced exponent of $E_{x}$. Note that $k_{x}\neq 2\cdot 3\cdot 5\cdot 7$, $3\cdot 5\cdot 7$, or $2\cdot 5\cdot 7$ as $||E_{x}||>(7-2)+(5-2)+2>8$. Therefore, either $|{\mathcal{P}}(k_{i})|=1$, ${\mathcal{P}}(k_{i})=\{2,3,7\}$ or $\{2,3,5\}$. (a) then follows from Lemma 2.11.

For (b), note that $||D_{i}||\leq 8$. Hence, by Proposition 2.16 (b), $D_{i}=h(P_{7}^{*}+g_{2}P_{3}^{*})$ for some element $h\in C_{m}$. As $||E_{x}||=8$, $E_{x}=D_{i}$. As $E_{x}=E_{x}^{(-1)}$, we have $h=g_{2}^{\alpha}$ for some integer $\alpha$. ∎

Corollary 4.2**.**

If $7|k_{x}$, then $E_{x}=g_{2}^{\alpha}(P_{7}^{*}+g_{2}P_{3}^{*})$.

Proof.

We will follow the notation used above. By assumption, $7|k_{i}$ for some $i$. If $k_{i}=7$, then $D_{i}=P_{7}h_{i}$. Since $||E_{x}||=8$, it follows that $||D_{j}||=1$ if $j\neq i$. This is impossible as then $\tau(D_{j})\neq 0$. Hence, $k_{i}$ is not a prime and therefore, $k_{i}=2\cdot 5\cdot 7$. By Lemma 4.1 (b), our desired result follows. ∎

Let $\psi$ be as defined in Section 3. As we have seen before, $a_{x}=\psi(E_{x})\equiv 0\bmod 4$. With the condition $E_{x}=E_{x}^{(-1)}$, this allows us to narrow down the possibilities of $E_{x}$ when $7$ does not divide the c-exponent of $E_{x}$.

Lemma 4.3**.**

If $7\nmid k_{x}$, then $E_{x}$ is in one of the forms below:

• (a)

$E_{x}=P_{2}W$* and $a_{x}=0$.*

• (b)

$E_{x}=(P_{3}+P_{5})g_{2}^{\alpha}$* and $a_{x}=\pm 8$.*

• (c)

$E_{x}=g_{2}^{\alpha}[g_{2}(g^{0}+g_{5}+g_{5}^{4})(g_{3}+g_{3}^{2})+(g_{5}^{2}+g_{5}^{3})]$* or $g_{2}^{\alpha}[g_{2}(g^{0}+g_{5}^{2}+g_{5}^{3})(g_{3}+g_{3}^{2})+(g_{5}+g_{5}^{4})]$ and $a_{x}=\pm 4$. In particular, $supp(E_{x})\cap P_{2}=\emptyset$. [Recall that $g^{0}$ is the identity of $C_{m}$. ]*

Proof.

We continue with the notation used in Lemma 4.1. If all $k_{i}$’s are prime, then in view of Lemma 4.1,

[TABLE]

where $X,Y,Z\in\mathbb{N}[C_{m}]$. As $||E_{x}||=8$ and $8=2||X||+3||Y||+5||Z||$. It is clear that

[TABLE]

If $(||X||,||Y||,||Z||)=(1,2,0)$, then $E_{x}=P_{2}(h_{1}+h_{2})+P_{3}h_{3}$. In this case, $\psi(E_{x})=\pm 3$. This is impossible. Next, if $(||X||,||Y||,||Z||)=(4,0,0)$, then (a) holds. If $(||X||,||Y||,||Z||)=(0,1,1)$, then $E_{x}=P_{3}h_{1}+P_{5}h_{2}$. By Lemma 3.3 (a), $h_{i}=g_{2}^{\alpha_{i}}$. Note that $\psi(E_{x})=\pm 2\neq\pm 4$ if $\alpha_{1}\neq\alpha_{2}\bmod 2$. Since $a_{x}\equiv 0\bmod{4}$, (b) holds.

As $k_{i}$’s are not all prime, we may assume that $k_{1}$ is not a prime. Then by Lemma 4.1, $k_{1}=2\cdot 3\cdot 5$. But then by Proposition 2.16 (b), $||D_{1}||\geq 6$. If $E_{x}\neq D_{1}$, then $||D_{2}||\leq 2$. Hence $D_{2}=P_{2}h^{\prime}$ for some $h^{\prime}\in C_{m}$ and $||D_{1}||=6$. Thus, $D_{1}=(P_{2}^{*}P_{3}^{*}+P_{5}^{*})h$ for some $h\in C_{30}$. Since $||E_{x}||=8$, $E_{x}=D_{1}+D_{2}$. But $\psi(E_{x})=\psi(D_{1}+D_{2})=\pm 2$. This is impossible as $4|a_{x}$. Hence, $E_{x}$ is a minimal v-sum and $E_{x}=D_{1}=Dh$ for some $D\in\mathbb{N}[C_{30}]$. As $E_{x}=E_{x}^{(-1)}$, we have $h\in C_{30}$. So, $E_{x}\in\mathbb{N}[C_{30}]$. We may write $E_{x}=\sum_{i=0}^{4}A_{i}g_{5}^{i}$, where $A_{i}\in\mathbb{N}[C_{6}]$. Clearly,

[TABLE]

Let $\tau$ be a character of order $30$. If $A_{i}=0$ for some $i$, then $\tau(A_{j})=0$ for all $j$ as $\tau(E_{x})=0$. Then, $E_{x}$ is not a minimal v-sum unless $E_{x}=A_{j}$ for some $j$. So, $k_{1}|6$ and $k_{1}\neq 30$. This is impossible. Hence, $||A_{i}||\geq 1$ for each $i$.

Claim. $||A_{j}||\leq 3$ for all $j=0,\ldots,4$.

Otherwise, we assume that $||A_{\ell}||\geq 3$ for some $\ell$. It then follows that $||A_{j}||\leq 2$ if $j\neq\ell$. Since $E_{x}=E_{x}^{(-1)}$, we have $\ell=0$. On the other hand, if $||A_{j}||=2$ for some $j$, then again $||A_{t}||\neq 2$ whenever $t\neq j$. Using the condition $E_{x}=E_{x}^{(-1)}$ again, we have $j=0$. This is impossible. Hence, all other $||A_{j}||=1$. Thus we conclude, $||A_{0}||=4$ and $||A_{i}||=1$ if $i=1,2,3,4$. Write $A_{1}=h$, where $h\in C_{6}$. As $\tau(A_{0})=\tau(h)$, we have $\tau(A_{0}+g_{2}h)=0$. Note that $||A_{0}+g_{2}h||=5$. Since $\tau(A_{0}+g_{2}h)=0$, we may apply a similar argument as in Lemma 4.1 to conclude that $A_{0}+g_{2}h=P_{2}h_{1}+P_{3}h_{2}$ for some $h_{1},h_{2}\in C_{6}$. Therefore, $A_{0}=P_{2}h_{1}+h_{3}+h_{4}$ or $A_{0}=h_{3}+P_{3}h_{2}$ for some $h_{3},h_{4}\in C_{6}$. In either case, it contradicts the assumption that $E_{x}$ is a minimal v-sum.

Hence, we conclude that $||A_{j}||\leq 2$ for all $j$. Using the assumption that $E_{x}=E_{x}^{(-1)}$ again, we then obtain two possible cases.

(i) $||A_{0}||=||A_{1}||=||A_{4}||=2$ and $||A_{2}||=||A_{3}||=1$ or

(ii) $||A_{0}||=||A_{2}||=||A_{3}||=2$ and $||A_{1}||=||A_{4}||=1$.

It remains to show that $E_{x}$ is of the desired form when (i) holds. We may assume that $A_{i}=h_{i}$ for some $h_{i}\in C_{6}$ for $i=2,3$. Since $\tau(E_{x})=0$ for any character $\tau$ of order $30$, we set $h_{2}=h_{3}=h$. As $E_{x}=E_{x}^{(-1)}$, we see that $h=g_{2}^{\alpha}$.

Note that for $i=0,1,4$, $||A_{i}+g_{2}h||=3$ and $\tau(A_{i}+g_{2}h)=0$. Therefore, $A_{i}+g_{2}h=P_{3}g_{2}h$ as $g_{2}h$ is in the support of all $A_{i}+g_{2}h$. In other words, $A_{i}=P_{3}^{*}(g_{2}h)$ for $i=0,1,4$. It is now clear that $E_{x}$ is of desired form. This shows that (c) holds. ∎

Theorem 4.4**.**

There is no $(m,3)$-GBF for any integer $m$ odd or $m\equiv 2\pmod{4}$.

Proof.

Recall that by earlier discussion of this section, we may assume that $m=2\cdot 3^{a}\cdot 5^{b}\cdot 7^{c}$. We first remove the case $7|m$.

We may assume that $7$ divides the c-exponent of $E_{x}$ for some $x\neq 1_{G}$. By Lemma 4.3, we see that $E_{x}=h^{\alpha}(P_{7}^{*}+hP_{3}^{*})$ and $\psi(E_{x})=\pm 4$. It follows from Equation (8) that $a_{v}=\pm 4$ for any $v\neq 1_{G}$. Therefore, $E_{v}$ is of the form in Corollary 4.2 or Lemma 4.3 (c). That means there is no element in $supp(E_{v})$ of order a multiple of $21$ for any $v$. This contradicts Lemma 3.2. Thus, we may assume that $7$ does not divide the c-exponent of $E_{x}$ for all $x\in G$. By Proposition 3.1, it remains to show that $(2\cdot 3^{a}\cdot 5^{b},3)$-GBF does not exist.

In view of Lemma 4.3, $E_{x}\in\mathbb{N}[C_{30}]$ for all $x\neq 1_{G}$. It follows that $supp(D_{f})\subset G\cdot C_{30}h^{\prime}$ for some $h^{\prime}\in C_{m}$. After multiplying $D_{f}$ with $h^{\prime-1}$, we may assume $D_{f}\in\mathbb{N}[G\cdot C_{30}]$. Recall that we may assume that $1\leq|G_{f}|\leq 4$. We may assume that $1_{G}\in G_{f}$ instead of $1_{G}\in G\setminus G_{f}$. We now discuss by cases.

Case (1) $|G_{f}|=2$.

As $1_{G}\in G_{f}$, we write $G_{f}=\{1_{G},v\}$. Note that $a_{x}=8$ or [math]. It follows that $a_{v}=8$ and $a_{x}=0$ if $x\neq 1_{G},v$. By Lemma 4.3, we have $E_{v}=P_{5}+P_{3}$ and $E_{x}=P_{2}W_{x}$ for some $W_{x}\in\mathbb{Z}[C_{30}]$ if $x\neq 1_{G},v$.

Let $\eta:\mathbb{Z}[G\cdot C_{30}]\rightarrow\mathbb{Z}[G\cdot C_{5}]$ be a ring homomorphism such that $\eta(g_{2})=-1$ and $\eta(g_{3})=1$ and $\eta(g_{5})=g_{5}$; and $\eta(x)=1$ for all $x\in G$. Note that $\eta(E_{x})=0$ if $a_{x}=0$ as $E_{x}=P_{2}W_{x}$ for some $W_{x}\in\mathbb{N}[C_{30}]$. Thus, we get

[TABLE]

Write $\eta(D_{f})=\sum a_{i}g_{5}^{i}$ where $a_{i}\in\mathbb{Z}$. Observe that if we further map $g_{5}$ to $1$, then the resulting map is just $\psi$. Thus, we have $\sum a_{i}=\psi(D_{f})$. Then as $|G_{f}|=2$, $\psi(D_{f})=\sum a_{i}=8-2\cdot 2=4$. By considering the coefficient of identity of $11+P_{5}$, we get $\sum a_{i}^{2}=12$. Thus $|a_{i}|\geq 2$ for some $i$. If the maximum value of $|a_{i}|$ is $2$, then there must be two more $a_{j}$’s with $|a_{j}|=2$ and the rest is [math]. That is impossible as then $2$ divides $\eta(D_{f})$ but $2$ does not divide $11+P_{5}$ in $\mathbb{Z}[P_{5}]$.

Hence, the maximum value of $|a_{i}|$ is $3$. Then there are exactly three $a_{j}$’s with $|a_{j}|=1$. Since $\sum a_{i}=4$, exactly one $a_{i}$ is $-1$. So we may assume that $\eta(D_{f})=3+g_{5}+g_{5}^{\beta}-g_{5}^{\gamma}$ with $1\neq\beta\neq\gamma\neq 1$. Clearly, we may assume either $\beta=4$ or $\gamma=4$.

If $\beta=4$, then we may take $\gamma=2$ or $3$. Then, the coefficient of $g_{5}^{\gamma}$ is $-2$, which is impossible. If $\gamma=4$, then $\beta=2$ or $3$. In that case, the coefficient of $g_{5}^{\beta}$ is $-2$, which is also impossible. Therefore, we have $|G_{f}|\neq 2$.

Case (2) $|G_{f}|=4$. We may assume that $G_{f}=\{1_{G},v_{1},v_{2},v_{3}\}$.

Subcase (a) $v_{3}=v_{1}v_{2}$ and $G_{f}$ is a subgroup of order $4$. Hence, $G_{f}^{2}=4G_{f}$ and $(G-2G_{f})^{2}=8G_{f}-8G_{f}v$ for some nonzero $v\in G$. Therefore, $a_{x}=\pm 8$ for all $x\in G$. In view of Lemma 4.3, $E_{x}=g_{2}^{\alpha}(P_{3}+P_{5})$ for all nonzero $x\in G$. By Lemma 3.2, this is impossible as there is no element in $supp(E_{v})$ which is divisible by $15$ for any $v$.

Subcase (b) $v_{3}\neq v_{1}v_{2}$. Let $H=\{1_{G},v_{1},v_{2},v_{1}v_{2}\}$ be the subgroup of order $4$. Then $G_{f}^{2}=2G+2-2v_{1}v_{2}v_{3}$. For convenience, we write $v=v_{1}v_{2}v_{3}$. Thus, $a_{v}=-8$ and $a_{x}=0$ if $x\neq 1_{G}$ or $v$. As $v\notin H$, there exists a ring homomorphism $\eta^{\prime}$ that maps $H\cdot P_{3}$ to identity, and $\eta^{\prime}(g_{2})=\eta^{\prime}(v)=-1$. Then as before $\eta^{\prime}(E_{x})=0$ if $a_{x}=0$. Hence, we obtain

[TABLE]

Write $\eta^{\prime}(D_{f})=\sum a_{i}g_{5}^{i}$. Observe that $\sum a_{i}=\eta^{\prime}(G-2G_{f})=-4$. As shown above, there is no solution in $\mathbb{Z}[P_{5}]$.

Case (3) $|G_{f}|=1$ or $3$. Ten $a_{x}=\pm 4$ for all $x\neq 1_{G}$ in $G$. Therefore, by Lemma 4.3 (c), for any $E_{x}$ with $1_{G}\neq x\in G$,

[TABLE]

Observe that if we write $E_{x}=\sum_{i=0}^{4}W_{xi}g_{5}^{i}$, then $||W_{x0}||=2$ and $P_{2}\cap supp(E_{x})=\emptyset$.

Write $D_{f}=\sum_{i=0}^{4}B_{i}g_{5}^{i}$ where $B_{i}\in\mathbb{Z}[G\cdot C_{6}]$ and $D_{f}D_{f}^{(-1)}=\sum_{i=0}^{4}Z_{i}g_{5}^{i}$ with $Z_{i}\in\mathbb{N}[G\cdot C_{6}]$. For each $i$, $B_{i}=A_{i0}+A_{i1}g_{3}+A_{i3}g_{3}^{2}$ where $A_{ij}\in\mathbb{N}[G\cdot P_{2}]$. If $||A_{ij}||\geq 2$, i.e., $A_{ij}=x_{1}h_{1}+x_{2}h_{2}+\cdots$, where $x_{1},x_{2}\in G$ and $h_{1},h_{2}\in P_{2}$, then $A_{ij}A_{ij}^{(-1)}=2+x_{1}x_{2}h_{1}h_{2}+\cdots$. Hence, $supp(E_{x_{1}})\cap\{g^{0},g_{2}\}\neq\emptyset$. This contradicts Lemma 4.3 (c). Thus, $|A_{ij}|\leq 1$ and $||B_{i}||\leq 3$. Note that

[TABLE]

Observe that not all $||B_{i}||\leq 2$. Using the equation above, we may assume that $||B_{i}||=||B_{j}||=3$ and $||B_{k}||=2$ for some distinct $i,j,k$. Then we have

[TABLE]

Let $\phi$ be a character on $G\cdot C_{30}$ such that $\phi(u_{0}u_{1})=(-1)^{\alpha_{1}-\alpha_{0}}$ and $\phi(u_{0}u_{2})=(-1)^{\alpha_{2}-\alpha_{0}}$. Note that such a $\phi$ exists as $u_{0}u_{1}\neq u_{0}u_{2}$. Then, it is clear that $\phi(B_{i})=0$. Thus, $|\phi(D_{f})|^{2}=|\phi(B_{j})\zeta_{5}^{j}+\phi(B_{k})\zeta_{5}^{k}|^{2}=8$. In other words, we have

[TABLE]

This is impossible unless $\phi(B_{j})=0$ or $\phi(B_{k})=0$. But then $||B_{k}||=2$ and $||A_{kj}||\leq 1$ imply that $\phi(B_{k})\neq 0$. Thus $\phi(B_{j})=0$ and $|\phi(B_{k})|^{2}=8$. This is impossible as $||B_{k}||=2$. This finish showing that $|G_{f}|\neq 1$ or $3$.

The proof is then completed. ∎