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

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.

1 Introduction

Let $m \ge 2$, 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

$$\begin{aligned} |F(y)|^2 = 2^n \end{aligned}$$

(1)

for all $y \in {\mathbb {Z}}_2^n$, where F(y) is defined as

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

(2)

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 [10], 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 et al. [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 bent functions, see Leung and Schmidt [6]. Schmidt [12] 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 [9, 11,12,13,14]). Very recently, Liu et al. [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 $m = 2m'$ with $m'$ odd.

The remainder of this paper is organized as follows. In Sect. 2, we introduce some basic tools and auxiliary results. In Sect. 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 Sect. 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. Suppose that R is a ring, and R[G] denotes 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 $r 1_G \in R [G]$. For the group ring element $D = \sum _{g \in G} d_g g \in R[G]$, its support is defined as

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

and we also define $|D| := \sum d_g$ and $|| D || := \sum |d_g|$ by convention when $R = {\mathbb {C}}$ or $R = {\mathbb {Z}}$ or R is the semiring ${\mathbb {N}}$ as we will use later. 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 an abelian group 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 $U^\perp := \{ \chi \in {\hat{G}}: \chi (g) = 1 \text { for all } g \in U \}$. If $\chi \in U^\perp$, we say that the character $\chi$ is trivial onU. 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

$$\begin{aligned} \sum _{\chi \in {\hat{G}}} \chi (g) = \left\{ \begin{array}{ll} 0 &{}\quad \text { if }g \ne 1_G,\\ v &{}\quad \text { if }g = 1_G, \end{array}\right. \end{aligned}$$

and

$$\begin{aligned} \sum _{g \in G} \chi (g) = \left\{ \begin{array}{ll} 0 &{}\quad \text { if } \chi \ne \chi _0, \\ v &{}\quad \text { if }\chi = \chi _0. \end{array}\right. \end{aligned}$$

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

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

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 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

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

Furthermore, we define an element $D_f$ in the group ring ${\mathbb {Z}}[C_m] [G]$ by

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

Remark 1

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

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

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

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

(3)

for all $y \in G$, where F(y) is defined in (2). This means that $\chi _y(B_f)$ is just the discrete Fourier transform of (m, n)-generalized bent functions. It then follows from (1) and (3) that f is an (m, n)-GBF if and only if

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

(4)

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

Proposition 1

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

$$\begin{aligned} B_f B_f^{(-1)} = 2^n. \end{aligned}$$

(5)

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

Proof

From (4) it follows that

$$\begin{aligned} |\chi (B_f)|^2 = \chi (B_f B_f^{(-1)}) = 2^n, \end{aligned}$$

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 $m'$-th root of unity. $\square$

Observe that we may write

$$\begin{aligned} D_f D_f^{(-1)} = \sum _{x \in G} \sum _{y \in G} g^{f(y + x)} g^{-f(y)} x = \sum _{x \in G} E_x x, \end{aligned}$$

(6)

where $E_x = \sum _{y \in G} g^{f(y + x)} g^{-f(y)} \in {\mathbb {Z}}[C_m]$. In fact, $E_x$ corresponds to the autocorrelation function of bent functions (for more details, see [1]).

Lemma 1

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 $x \ne 1_G$.

Proof

Note that $(D_f D_f^{(-1)})^{(-1)}=D_fD_f^{(-1)}$. Hence, we have $E_x=E_x^{(-1)}$ for all $x\in G$. Note that $E_{1_G}=2^n$. Thus, we may consider $x\ne 1_G$. Suppose that $x\ne 1_G$ and $(g_1x_1)(g_2x_2)^{-1}=g^{0}x$ for some $g_1,g_2\in C_m$ and $x_1,x_2\in G$. Note that $x_1\ne x_2$ and clearly, we have $(g_2x_2)(g_1x_1)^{-1}=g^{0}x$ 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

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

From (5) in Proposition 1, the conclusion follows. $\square$

The key in our study of (m, n)-GBFs is to investigate $E_x$. Lemma 1 (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 Sect. 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

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 3

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 $I \subsetneq S$, $\sum _{i \in I} a_i \mu _i \ne 0$.

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

[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

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

For convenience, we define the following notation.

Definition 4

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

$$\begin{aligned} \left\{ \sum _{g\in H} a_g g: a_g\in {\mathbb {Z}} \text{ and } a_g\ge 0 \right\} . \end{aligned}$$

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 5

[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 $\tau (\sum _{i = 0}^{m-1} b_i g^i)\ne 0$ whenever $0\le b_i \le a_i$ 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 5, then $\tau (D) = \sum _{i \in S} a_i \tau (g)^i$ is a minimal relation by Definition 3. We now define the reduced exponent of D as follows.

Definition 6

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 exponentk 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

If $D \in {\mathbb {N}}[C_m]$ is a minimal v-sum with reduced exponent k, then $D = D' h$ for some $D' \in {\mathbb {N}}[C_k]$ 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 6, 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. $\square$

In view of Proposition 2, we derive the following result.

Corollary 1

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

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

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 3

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 1. 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

$$\begin{aligned} D = \sum _{i=1}^9 h^i= & {} (1+h^5) + (1+h^5)h + (1+h^5)h^2 + (1+h^5) h^3 + (1+h^5) h^4 \quad \text { and } \\ D = \sum _{i=1}^9 h^i= & {} (1+h^2+h^4+h^6+h^8) + (1+h^2+h^4+h^6+h^8)h. \end{aligned}$$

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 7

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 4

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 $t \le s$ and $p_i$’s are distinct primes. Then we have the followings:

• (a) $||D|| \ge 2 + \sum _{i = 1}^t (p_i - 2)$;

• (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 3, we may assume that $D = \sum _{i=1}^t D_i$ such that each $D_i$ is a minimal v-sum. Hence, by Corollary 1, we have

$$\begin{aligned} ||D||&= \sum _{i=1}^t |D_i| \\&\quad \ge \sum _{i=1}^t \left[ 2+\sum _{q\in {{\mathcal {P}}}(k_i)} (q-2)\right] \\&\quad \ge 2+\sum _{q\in {{\mathcal {P}}}(k)} (q-2) \\&= 2+\sum _{i=1}^t (p_i-2) , \end{aligned}$$

because ${{\mathcal {P}}}(k)=\bigcup _{i=1}^t {{\mathcal {P}}}(k_i)$.

By Lemma 2, $D_i=E_i g_i$ 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 $E_i=\sum _{q\in {{\mathcal {P}}}(k_i)} Q_qF_q$, 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 $i=1,\ldots , t$. $\square$

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

Proposition 3

[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 of $C_m$, then $D=Ph$ for some $h\in C_m$.

• (b) If $k=\prod _{i=1}^t p_i$ with $t\ge 2$ and $p_1<p_2<\cdots <p_t$ are primes, then $t\ge 3$ and

  $$\begin{aligned} ||D|| \ge (p_1-1)(p_2-1)+(p_3-1). \end{aligned}$$

  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

It follows from Proposition 3 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 1 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 4

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 $1_G\ne x\in G$. Set

$$\begin{aligned} I=\{1 \le i \le s: p_i \not \mid k_x \ \forall x\in G\} \text{ and } {\overline{m}} = \prod _{i\notin I} p_i^{\alpha _i}. \end{aligned}$$

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

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

Recall that $E_x$ is a v-sum. By assumption $p_i$ does not divide $k_x$ for all $1_G \ne x \in G$. It follows from Lemma 4(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 4(a), $p_i$ does not divide $k_x$ for any $1_G\ne x\in G$.

$\square$

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

Lemma 5

Suppose that f is an (m, n)-GBF, and p, q are distinct primes that both divide m. Then there exist $y\ne 1_G$ 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. $\square$

Before we proceed, we need a technical result.

Lemma 6

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 $4\not \mid m$ and $\sum _{i=1}^t Q_ih_i=\sum _{i=1}^t Q_ih_i^{-1}$ for some $h_1,h_2,h_t\in C_m$ with $t \ge 2$.

• (a) If $q_1\ne q_2$ and $t=2$, then we may assume $h_i^{-1}=h_i$ for $i=1,2$.

• (b) If $q_1\ne q_2=q_3$ and $t=3$, then we may assume $Q_2h_2+Q_2h_3=Q_2(h_2+h_2^{-1})$ 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

$$\begin{aligned} Q_1(h_1-h_1^{-1})=\sum _{i=2}^t Q_i(h_i^{-1}-h_i). \end{aligned}$$

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 $\phi (Q_1h_1-Q_1h_1^{-1})=0$. Write $h_1=g_1 h'$ with $g_1\in \langle g^{m/q_1^{\beta _1} }\rangle$ and $p_1\not \mid \circ (h')$. Then, we have $Q_1g_1\phi (h')=Q_1g_1^{-1}\phi (h'^{-1})$. Hence $g_1^2\in Q_1$ and $\phi (h')=\phi (h'^{-1})$. If $q_1$ is odd, then $g_1=g^0$. If $q_1=2$, then as $4\not \mid m$, $g_1$ can be taken as $g^0$ as well. In both cases, we may assume $g_1 = g^0$. It follows that $\phi (h')^2=1$. As $\phi$ is of order $m/q_1^{\beta _1}$, $h'^2=g^0$. Therefore, $g_1h'=(g_1h')^{-1}$. Furthermore, we have

$$\begin{aligned} \sum _{i=2}^t Q_i(h_i^{-1}-h_i)=0. \end{aligned}$$

(7)

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 $Q_2h_2=Q_2h_2^{-1}$, then we must have $Q_2h_3=Q_2h_3^{-1}$. 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 $Q_2h_2=Q_2h_3^{-1}$, 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 Eq. (7).

The proof is then completed. $\square$

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

Theorem 1

Suppose that $m = \prod _{i = 1}^s p_i^{\alpha _i}$, where $3 \le p_1< p_2< \cdots < p_s$ are odd primes and $\alpha _i$’s are all positive integers. If an (m, n)-GBF exists, then $s \ge 2$ and $3p_1 + p_2 \le 2^n$.

Proof

Recall that if $1_G\ne x \in G$ and $\chi$ is a character of order m, then $\chi (E_x) = 0$. If $s=1$, then by Lemma 4(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 $p_1\ne 2$.

Next, we assume that $s\ge 2$. 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 $|{{\mathcal {P}}}(k_j)|\ge 4$, then by Corollary 1, we have $||D_j||\ge 2+ \sum _{i=1}^4 (p_i-2) \ge 3p_1+p_2$. Thus, we may assume that $|{{\mathcal {P}}}(k_j)|\le 3$. But by Proposition 3, $|{{\mathcal {P}}}(k_j)|=1$ or 3. In case that $|{{\mathcal {P}}}(k_j)|=3$, $||D_j||\ge q_1(q_2-1)+q_3-q_2\ge p_1(p_2-1)+p_3-p_2$. If $p_1\ge 5$, then clearly, $p_1(p_2-1)+p_3-p_2\ge 3p_1+p_2$. If $p_1=3$, it then follows that

$$\begin{aligned} p_1(p_2-1)+p_3-p_2\ge 2p_2+(p_3-2)\ge p_2+(5+7)\ge 3p_1+p_2 \end{aligned}$$

as $p_2\ge 5$ and $p_3\ge 7$.

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 $E_x=Q_1Y$ for some $Y\in {\mathbb {Z}}[C_m]$. This is impossible as $q_1\not \mid 2^n$. In particular, it follows that $t\ge 2$ and we may assume $Q_1\ne Q_2$ without loss of generality. Recall that all $D_i\in {\mathbb {N}}[C_m]$. Therefore,

$$\begin{aligned} 2^n=||E_x|| \ge q_1+q_2+(t-2)p_1. \end{aligned}$$

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

We first study the case $t = 3$. As $q_1\ne q_2$, we may assume $q_1\ne q_3$ 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 $Q_2h_2+Q_2h_3=Q_2(h_2+h_2^{-1})$. Whereas if $Q_2\ne Q_3$, 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 1(a).

Thus, we may assume $t = 2$ for all $x \ne 1_G$. 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 5.

The proof is then completed. $\square$

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 2

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 $s \ge 2$ 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 1. As for (c), it suffices to show that if $t\ge r+1$, then $p_t$ does not divide the c-exponent of $E_x$ for any $x\ne 1_G$. We follow the notation used in the proof of Theorem 1. 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 $E_x\ne D_1$ as otherwise $p_t|2^n$. Therefore, $||E_x||\ge ||D_1||+||D_2||\ge p_t+p_1>2^n$. On the other hand, if $k_1\ne p_t$, then as shown before, $k_1$ is a product of at least three primes. Hence, $||D_1||\ge p_t+p_1>2^n$, which is impossible. $\square$

Remark 3

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

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

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

Note that a $(2m', n)$ GBF is trivially an $(m',n)$ GBF if $G_f = \emptyset$ or G. Add f by m if necessary, we may always assume $|G_f|\le |G|/2$. 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

$$\begin{aligned} \psi (D_f)\psi (D_f^{(-1)})&= (G-2G_f)(G-2G_f^{(-1)}) \\&= (|G|-4|G_f|)G+4G_f^2 \\&= 2^n+\sum _{1_G \ne x\in G} \psi (E_x)x. \end{aligned}$$

Write

$$\begin{aligned} G_f^2 = |G_f|+ 2 \sum _{ 1_G \ne x\in G} b_x x. \end{aligned}$$

(8)

We denote $\psi (E_x)$ by $a_x$. It then follows that for $x \ne 1_G$,

$$\begin{aligned} a_x=|G|-4|G_f|+8b_x . \end{aligned}$$

(9)

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

Lemma 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 $(2m', n)$ GBFs.

Theorem 2

Let n be odd and $m = 2 p^\alpha$, 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 $n \le 3$, 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 $x \ne 1_G$, we conclude from Lemma 4(b) that $E_x = P_2 Y_x + P Z_x$ for some $Y_x, Z_x \in {\mathbb {N}}[C_m]$. Note that $\psi (E_x) \ne 0$ for some $x \ne 1_G$. Otherwise, the c-exponent of all $E_x$ is 2 and by Proposition 4, there exists a (2, 3)-GBF, which is impossible. Hence, $a_x = \psi (E_x) \ne 0$ for some $x \ne 1_G$. Therefore, we have $\psi (P) | \psi (E_x)$, i.e., $p | a_x$. Note that in view of Eq. (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 $x \ne 1_G$. 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 $G_f \ne \{1_G\}$. As $G_f$ is not a difference set, there exist two elements $x\ne 1_G$ and $x'\ne 1_G$ such that $b_x> b_{x'}\ge 0$. Since $p | a_x$ and $p | a_{x'}$, it follows that $p | (b_x-b_{x'})$ and $b_x - b_{x'} = tp$ for some positive integer t. To get our desired result, we need to find a bound on $b_x-b_{x'}$. Note that in view of Eq. (8), $b_x \le |G_f|/2 \le |G|/4$. Hence, we get our desired result if $t \ge 2$. Thus, we may assume that $t=1$, i.e., $b_x = p + b_{x'}$.

Suppose that $G = \langle x\rangle \cdot G'$, where $G'$ is a subgroup of order $2^{n-1}$ in G. As the coefficient of x in $G_f^2$ is $2b_x$, there are $2b_x = 2p + 2b_{x'}$ pairs (u, v) of elements in $G_f\times G_f$ such that $uv=x$. Therefore, there exists a set $Y \subseteq G'\cap G_f$ such that $Y\cup (Yx)\subseteq G_f$ with $|Y| = p + b_{x'}$. Write $G_f = (Y \cup Z_1) \cup (Yx \cup Z_2x)$ such that

$$\begin{aligned} Z_1\subseteq G', Z_2\subseteq G', Y\cap Z_1=\emptyset \text{ and } Y \cap Z_2=\emptyset . \end{aligned}$$

Since $b_x = |Y|$, it follows that $Z_1\cap Z_2=\emptyset$. Moreover, we have

$$\begin{aligned} G_f^2 = [2Y^2+2Y(Z_1+Z_2)+Z_1^2+Z_2^2]+[2Y^2+2Y(Z_1+Z_2)+2Z_1Z_2]x. \end{aligned}$$

Note that the support of $[2Y^2+2Y(Z_1+Z_2)+Z_1^2+Z_2^2]$ is in $G'$ and the support of $[2Y^2+2Y(Z_1+Z_2)+2Z_1Z_2]x$ is in $G'x$. We now consider the coefficients of the following group elements

$$\begin{aligned} Z=[2Y^2+2Y(Z_1+Z_2)+Z_1^2+Z_2^2]-[2Y^2+2Y(Z_1+Z_2)+2Z_1Z_2]=(Z_1-Z_2)^2. \end{aligned}$$

For any $1_{G}\ne v\in G'$, 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 $v \ne 1_G$ in $G'$ such that $b_v-b_{vx}$ is nonzero, then $p | (b_v - b_{vx})$ and we obtain

$$\begin{aligned} 2p \le 2|b_v-b_{vx}| \le |Z_1|+|Z_2| \le (|G_f|-2b_x) \le |G_f| - 2p. \end{aligned}$$

Hence, we get $4p \le |G_f| \le |G|/2$ and $p \le 2^{n-3}$. 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 $t \ge 1$ 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 $[2Y^2+2Y(Z_1+Z_2)+2Z_1Z_2]$ 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. $\square$

Corollary 3

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 $s \ge 2$, 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 2. If (iii) holds, it is known that no $(2 p_1^{\alpha _1}, n)$-GBF exists.

To prove (b), it is sufficient to show that for $i\ge r + 1$, $p_i$ does not divide the c-exponent of any $E_x$ for $x \ne 1_G$. 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 $|| D_i || \ge 2 + (p_1 -2) + (p_i - 2)$. Thus, $2^n \ge p_1 + p_i - 2 \ge p_1 + p_{r+1} - 2 > 2^n$. 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 $4p_i \le 2^n$. This is impossible as $2^n< p_1 + p_i < 4p_i$. Therefore, $p_j|k_x$ for some $j\ne i$. But then by Proposition 3, $2^n \ge p_j+p_i-2 > p_{r+1}+p_1-2$. This is impossible. $\square$

Remark 4

When compared with [7, Theorem 2], our result in Corollary 3 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 4, we may assume that all prime factors of m are less than or equal to 7. According to Corollary 2, 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 $x\ne 1_G$. 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 8

For any $x\ne 1_G$, 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 $k_x\ne 2\cdot 3 \cdot 5 \cdot 7$, $3 \cdot 5 \cdot 7$, 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.

For (b), note that $||D_i|| \le 8$. Hence, by Proposition 3(b), $D_i = h (P_7^* + g_2P_3^*)$ 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 $\alpha$. $\square$

Corollary 4

If $7|k_x$, then $E_x = g_2^{\alpha }(P_7^* + g_2P_3^*)$.

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 $j\ne i$. This is impossible as then $\tau (D_j)\ne 0$. Hence, $k_i$ is not a prime and therefore, $k_i=2 \cdot 5 \cdot 7$. By Lemma 8 (b), our desired result follows. $\square$

Let $\psi$ be as defined in Sect. 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 9

If $7\not \mid k_x$, then $E_x$ is in one of the forms below:

• (a) $E_x=P_2W$ 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 8. If all $k_i$’s are prime, then in view of Lemma 8,

$$\begin{aligned} E_x=P_2X+P_3Y+P_5Z, \end{aligned}$$

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

$$\begin{aligned} (||X||,||Y||,||Z||)=(4,0,0), \ (1,2,0), \ \text{ or } (0,1,1). \end{aligned}$$

If $(||X||,||Y||,||Z||)=(1,2,0)$, then $E_x=P_2(h_1+h_2)+P_3h_3$. 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 $E_x=P_3h_1+P_5h_2$. By Lemma 6(a), $h_i=g_2^{\alpha _i}$. Note that $\psi (E_x)=\pm 2\ne \pm 4$ if $\alpha _1\ne \alpha _2 \bmod 2$. 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 8, $k_1=2\cdot 3 \cdot 5$. But then by Proposition 3(b), $||D_1||\ge 6$. If $E_x\ne D_1$, then $||D_2||\le 2$. Hence $D_2=P_2 h'$ for some $h'\in C_m$ 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,

$$\begin{aligned} 8=\sum _{i=0}^4 ||A_i||. \end{aligned}$$

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 $k_1 \ne 30$. This is impossible. Hence, $||A_i||\ge 1$ for each i.

Claim$||A_j|| \le 3$ for all $j = 0, \ldots , 4$.

Otherwise, we assume that $||A_{\ell }||\ge 3$ for some $\ell$. It then follows that $||A_j||\le 2$ if $j\ne \ell$. Since $E_x = E_x^{(-1)}$, we have $\ell =0$. On the other hand, if $||A_j|| = 2$ for some j, then again $||A_t||\ne 2$ whenever $t \ne j$. 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 $\tau (A_0 + g_2h) = 0$. Note that $||A_0 + g_2 h|| = 5$. Since $\tau (A_0 + g_2h) = 0$, we may apply a similar argument as in Lemma 8 to conclude that $A_0 + g_2h = P_2 h_1 + P_3 h_2$ for some $h_1, h_2\in C_6$. Therefore, $A_0 = P_2h_1 + h_3 + h_4$ or $A_0 = h_3 + P_3h_2$ 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 $||A_j|| \le 2$ 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 $i=0, 1, 4$, $||A_i + g_2h||=3$ and $\tau (A_i + g_2h)=0$. Therefore, $A_i+g_2h = P_3g_2h$ as $g_2h$ is in the support of all $A_i + g_2h$. In other words, $A_i = P_3^*(g_2h)$ for $i=0, 1, 4$. It is now clear that $E_x$ is of desired form. This shows that (c) holds. $\square$

Theorem 3

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 $x \ne 1_G$. By Lemma 9, we see that $E_x = h^{\alpha }(P_7^* + hP_3^*)$ and $\psi (E_x)=\pm 4$. It follows from Eq. (8) that $a_v=\pm 4$ for any $v\ne 1_G$. Therefore, $E_v$ is of the form in Corollary 4 or Lemma 9(c). That means there is no element in $supp(E_v)$ of order a multiple of 21 for any v. This contradicts Lemma 5. Thus, we may assume that 7 does not divide the c-exponent of $E_x$ for all $x\in G$. By Proposition 4, it remains to show that $(2\cdot 3^a \cdot 5^b, 3)$-GBF does not exist.

In view of Lemma 9, $E_x\in {\mathbb {N}}[C_{30}]$ for all $x\ne 1_G$. It follows that $supp(D_f)\subset G\cdot C_{30} h'$ for some $h'\in C_m$. After multiplying $D_f$ with $h'^{-1}$, we may assume $D_f\in {\mathbb {N}}[G\cdot C_{30}]$. Recall that we may assume that $1 \le |G_f| \le 4$. 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 0. It follows that $a_v = 8$ and $a_x = 0$ if $x \ne 1_G, v$. By Lemma 9, 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 $x\ne 1_G, v$.

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 $E_x = P_2W_x$ for some $W_x\in {\mathbb {N}}[C_{30}]$. Thus, we get

$$\begin{aligned} \eta (D_f)\eta (D_f)^{(-1)} = 11+P_5. \end{aligned}$$

Write $\eta (D_f)=\sum a_ig_5^i$ 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 $|a_i| \ge 2$ 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 0. 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 $1 \ne \beta \ne \gamma \ne 1$. 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 $|G_f| \ne 2$.

Case (2)$|G_f| = 4$. We may assume that $G_f=\{1_G, v_1, v_2, v_3\}$.

Subcase (a)$v_3 = v_1v_2$ and $G_f$ is a subgroup of order 4. Hence, $G_f^2 = 4G_f$ and $(G-2G_f)^2 = 8G_f - 8G_fv$ for some nonzero $v\in G$. Therefore, $a_x = \pm 8$ for all $x\in G$. In view of Lemma 9, $E_x = g_2^{\alpha } (P_3+P_5)$ for all nonzero $x\in G$. By Lemma 5, this is impossible as there is no element in $supp(E_v)$ which is divisible by 15 for any v.

Subcase (b)$v_3 \ne v_1 v_2$. Let $H = \{1_G, v_1, v_2, v_1v_2\}$ be the subgroup of order 4. Then $G_f^2 = 2G + 2 - 2v_1v_2v_3$. For convenience, we write $v=v_1v_2v_3$. Thus, $a_v=-8$ and $a_x=0$ if $x\ne 1_G$ or v. As $v\notin H$, there exists a ring homomorphism $\eta '$ that maps $H\cdot P_3$ to identity, and $\eta '(g_2)=\eta '(v) = -1$. Then as before $\eta '(E_x)=0$ if $a_x=0$. Hence, we obtain

$$\begin{aligned} \eta '(D_f)\eta '(D_f)^{(-1)} = 8+(-1)(-3-P_5) = 11 + P_5. \end{aligned}$$

Write $\eta '(D_f) = \sum a_ig_5^i$. Observe that $\sum a_i=\eta '(G-2G_f)=-4$. 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 $x \ne 1_G$ in G. Therefore, by Lemma 9(c), for any $E_x$ with $1_G\ne x\in G$,

$$\begin{aligned} E_x= & {} g_2^{\alpha }[(g^0+g_5+g_5^4)(g_3+g_3^2)+g_2(g_5^2+g_5^3)] \text{ or } \\&g_2^{\alpha }[(g^0+g_5^2+g_5^3)(g_3+g_3^2)+g_2(g_5+g_5^4)]. \end{aligned}$$

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 $||A_{ij}||\ge 2$, 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 $A_{ij}A_{ij}^{(-1)} = 2 + x_1x_2 h_1 h_2 + \cdots$. Hence, $supp(E_{x_1})\cap \{g^0, g_2\}\ne \emptyset$. This contradicts Lemma 9(c). Thus, $|A_{ij}|\le 1$ and $||B_i||\le 3$. Note that

$$\begin{aligned} ||Z_0|| = 8 + \sum _{x \ne 1_G} ||W_{x0}|| = 8 + 2\times 7 = 22 = \sum _{i=0}^4 ||B_i||^2. \end{aligned}$$

Observe that not all $||B_i||\le 2$. 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

$$\begin{aligned} B_i = \sum _{t=0}^2 u_t g_2^{\alpha _i}g_3^t = u_0g_2{\alpha _0} \left( 1+ u_0u_1g_2^{\alpha _1-\alpha _0}g_2g_3+ u_0u_1g_2^{\alpha _2-\alpha _0}g_3^2\right) . \end{aligned}$$

Let $\phi$ be a character on $G\cdot C_{30}$ such that $\phi (u_0u_1)=(-1)^{\alpha _1-\alpha _0}$ and $\phi (u_0u_2)=(-1)^{\alpha _2-\alpha _0}$. Note that such a $\phi$ exists as $u_0u_1\ne u_0u_2$. 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

$$\begin{aligned} |\phi (B_j)|^2+|\phi (B_k)|^2+ \phi (B_j)\overline{\phi (B_k)}\zeta _5^{j-k}+ \phi (B_k)\overline{\phi (B_j)}\zeta _5^{k-j}=8. \end{aligned}$$

This is impossible unless $\phi (B_j)=0$ or $\phi (B_k)=0$. But then $||B_k||=2$ and $||A_{kj}||\le 1$ imply that $\phi (B_k)\ne 0$. Thus $\phi (B_j)=0$ and $|\phi (B_k)|^2=8$. This is impossible as $||B_k||=2$. This finish showing that $|G_f|\ne 1$ or 3.

The proof is then completed. $\square$