Bilinear Forms on Finite Abelian Groups and Group-Invariant Butson Hadamard Matrices

TL;DR

This paper establishes conditions for the existence of group-invariant Butson Hadamard matrices using bilinear forms on finite abelian groups, providing both sufficient and necessary criteria in specific cases.

Contribution

It introduces new existence conditions for $BH(K,h)$ matrices based on bilinear forms and proves their necessity for cyclic groups of prime power order.

Findings

Existence of $BH(K,h)$ matrices under certain $p$-adic valuation conditions.

Necessary conditions for cyclic groups of prime power order.

Application of field descent method to establish necessity.

Abstract

Let $K$ be a finite abelian group and let $\exp(K)$ denote the least common multiple of the orders of the elements of $K$. A $BH(K,h)$ matrix is a $K$-invariant $|K|\times |K|$ matrix $H$ whose entries are complex $h$th roots of unity such that $HH^*=|K|I$, where $H^*$ denotes the complex conjugate transpose of $H$, and $I$ is the identity matrix of order $|K|$. Let $\nu_p(x)$ denote the $p$-adic valuation of the integer $x$. Using bilinear forms on $K$, we show that a $BH(K,h)$ exists whenever (i) $\nu_p(h) \geq \lceil \nu_p(\exp(K))/2 \rceil$ for every prime divisor $p$ of $|K|$ and (ii) $\nu_2(h) \ge 2$ if $\nu_2(|K|)$ is odd and $K$ has a direct factor $\mathbb{Z}_2$. Employing the field descent method, we prove that these conditions are necessary for the existence of a $BH(K,h)$ matrix in the case where $K$ is cyclic of prime power order.

Equations

$$\begin{array}{rl}f(g+h,k)& =f(g,k)+f(h,k)\text{ and }\\ f(g,h+k)& =f(g,h)+f(g,k)\end{array}$$

$$\begin{array}{rl}f(\alpha g,k)& =\alpha f(g,h)\text{ and }\\ f(g,\alpha h)& =\alpha f(g,h)\end{array}$$

$$s_{L}(g)=\left\{\begin{array}[]{ll}2G_{1}G_{2}^{T}&\text{ if }b=0,\\ 2G_{1}G_{2}^{T}+g_{c}&\text{ if }b=1.\end{array}\right.$$

$$\begin{array}{rl}\sum _{x\in L}{\zeta }_4^{s_L(x)-s_L(x\oplus g)}& =\sum _{x\in L}{\zeta }_4^{2X_1{X}_2^T+x_c-2(X_1{X}_2^T+X_1{G}_2^T+G_1{X}_2^T+G_1{G}_2^T)-(x\oplus g{)}_c}\\ & =(-1{)}^{G_1{G}_2^T}\sum _{x_c=0}^1{\zeta }_4^{x_c-(x_c\oplus g_c)}\sum _{X_1\in \{0,1{\}}^a}(-1{)}^{X_1{G}_2^T}\\ & \text{mbox}\sum _{X_2\in \{0,1{\}}^a}(-1{)}^{G_1{X}_2^T}.\end{array}$$

$$\sum _{x_c=0}^1{\zeta }_4^{x_c-(x_c\oplus g_c)}={\zeta }_4^{-1}+{\zeta }_4^1=0.$$

$$\sum _{x\in L}{\zeta }_4^{s_L(x)-s_L(x\oplus g)}=0$$

$$H_{y,x}={\zeta }_e^{f((x-y{)}_1/2,(x-y{)}_1)+\beta f((x-y{)}_1,(x-y{)}_2)}{\zeta }_4^{s_L(x_3\oplus y_3)}.$$

$$e_{1}=\left\{\begin{array}[]{ll}\exp(U)&\text{ if }L=\{0\},\\ {\rm lcm}(2,\exp(U))&\text{ if $L$ is of square order},\\ {\rm lcm}(4,\exp(U))&\text{ otherwise}.\end{array}\right.$$

$$(x-y{)}_2=x_2-y_2+u(x_2).$$

$$(x-y{)}_1=x_1-y_1-u(x_2),$$

$$\begin{array}{r}u(x_2)=0\text{ whenever }{y}_2=0.\end{array}$$

$$H_{y,x}={\zeta }_e^{f((x_1-y_1-u(x_2))/2,x_1-y_1-u(x_2))+\beta f(x_1-y_1-u(x_2),x_2-y_2+u(x_2))}{\zeta }_4^{s_L(x_3\oplus y_3)}.$$

$$\begin{array}{rl}A(y):& =\sum _{x\in K}{H}_{0,x}\overline{H_{y,x}}\\ & =\sum _{x_3\in L}\sum _{x_2\in R}\sum _{x_1\in U}{H}_{0,x_1+x_2+x_3}\overline{H_{y,x_1+x_2+x_3}}.\end{array}$$

$$\begin{array}{rl}A(y)& =\eta \sum _{x_3\in L}{\zeta }_4^{s_L(x_3)-s_L(x_3\oplus y_3)}\sum _{x_2\in R}{\zeta }_e^{T(x_2)}\sum _{x_1\in U}{\zeta }_e^{S(x_1,x_2)},\end{array}$$

$$\eta$$

$$T(x_2)$$

$$S(x_1,x_2)$$

$$A(y)=\eta |L|\sum _{x_2\in R}{\zeta }_e^{T(x_2)}\sum _{x_1\in U}{\zeta }_e^{S(x_1,x_2)}.$$

$${\zeta }_e^{S(z_1,x_2)}V(x_2)=\sum _{x_1\in U}{\zeta }_e^{S(x_1+z_1,x_2)}=\sum _{x_1\in U}{\zeta }_e^{S(x_1,x_2)}=V(x_2),$$

$$S(x_1,x_2)=f(x_1,y_1+\beta y_2+(1-\beta )u(x_2))=0$$

$$y_2=0.$$

$$u(x_2)=0$$

$$f(x_1,y_1)=0\text{ for all }{x}_1\in U$$

$$A(y)=\eta |L||U|\sum _{x_2\in R}{\zeta }_e^{T(x_2)}=\eta |L||U|\sum _{x_2\in R}{\zeta }_e^{\beta f(y_1,x_2)}.$$

$$\begin{array}{rl}f(y_1,r+s)& =f(y_1,(r+s{)}_1+(r+s{)}_2)\\ & =f(y_1,(r+s{)}_2),\end{array}$$

$$\begin{array}{rl}{\zeta }^{\beta f(y_1,z_2)}W(y)& =\sum _{x_2\in R}{\zeta }_e^{\beta f(y_1,x_2+z_2)}\\ & =\sum _{x_2\in R}{\zeta }_e^{\beta f(y_1,(x_2+z_2{)}_2)}\\ & =\sum _{x_2\in R}{\zeta }_e^{\beta f(y_1,x_2)}=W(y).\end{array}$$

$$H_{y,x}=W_{y,x}{\zeta }_4^{s_L(x_3\oplus y_3)},$$

$$W_{y,x}^k={\zeta }_e^{f((x-y{)}_1/2,k(x-y{)}_1)+f(k(x-y{)}_1,(x-y{)}_2)}={\zeta }_e^0=1$$

$$\left|\sum _{x\in ({\mathbb{Z}}_2{)}^{2a}}(-1{)}^{F(x)+\alpha x^T}\right|=2^a$$

$$s_L(g_1,\dots ,g_c)=2F(g_1,\dots ,g_{2a})+\sum _{i=2a+1}^c{g}_i.$$

$${\nu }_p(h)\ge \lceil {\nu }_p(\exp (K))/2\rceil$$

Full text

• (i)

$\nu_{p}(h)\geq\lceil\nu_{p}(\exp(K))/2\rceil$ for every prime divisor $p$ of $|K|$ and

• (ii)

$\nu_{2}(h)\geq 2$ if $\nu_{2}(|K|)$ is odd and $K$ has a direct factor $\mathbb{Z}_{2}$.

Employing the field descent method, we prove that these conditions are necessary for the existence of a $\operatorname{{\rm BH}}(K,h)$ matrix in the case where $K$ is cyclic of prime power order.

1 Introduction

Let $U_{h}$ be the set of complex $h$th roots of unity. An $n\times n$-matrix $H$ with entries from $U_{h}$ is called a Butson Hadamard matrix if $HH^{*}=nI$, where $H^{*}$ is the complex conjugate transpose of $H$ and $I$ is the identity matrix of order $n$. We say that $H$ is a $\operatorname{{\rm BH}}(n,h)$ matrix. The Ph.D. thesis [15] of Szöllősi provides a good overview of most of the known results on Butson Hadamard matrices and [6, 4] contain more recent work and survey open problems in this area.

The focus of this paper is Butson Hadamard matrices invariant under abelian groups. Let $(G,+)$ be a finite abelian group of order $n$ with identity element [math]. An $n\times n$ matrix $A=(a_{g,k})_{g,k\in G}$ is $G$-invariant if $a_{g+l,k+l}=a_{g,k}$ for all $g,k,l\in G$. Such a matrix is sometimes also called group invariant or group developed. A $G$-invariant $\operatorname{{\rm BH}}(n,h)$ matrix is also called a $\operatorname{{\rm BH}}(G,h)$ matrix.

Remark 1.1**.**

For any multiple $h^{\prime}$ of $h$, every $\operatorname{{\rm BH}}(G,h)$ matrix is also a $\operatorname{{\rm BH}}(G,h^{\prime})$ matrix, as $U_{h}\subset U_{h^{\prime}}$.

By $\mathbb{Z}_{n}$ we denote the cyclic group of order $n$. Most existing work on group invariant Butson Hadamard matrices concerns circulant matrices, i.e., $\operatorname{{\rm BH}}(\mathbb{Z}_{n},h)$ matrices. Backelin [1] came up with the following result.

Result 1.2**.**

Let $p$ be a prime and let $n$ be a positive integer such that $n\equiv 0\ (\bmod\ p^{2})$ and $n\not\equiv 2\ (\bmod\ 4)$. Then there is a $\operatorname{{\rm BH}}(\mathbb{Z}_{n},n/p)$ matrix.

We remark that Backelin actually formulated his result in terms of so called cyclic $n$-roots; Result 1.2 translates his theorem into the language of group invariant Butson Hadamard matrices. We further remark that the condition $n\not\equiv 2\ (\bmod\ 4)$ is missing in the statement of the theorem in Backelin’s paper, but is necessary for his construction to work. In fact, for instance, it can be shown [9] that $\operatorname{{\rm BH}}(\mathbb{Z}_{2p^{2}},2p)$ matrices do not exist for any odd prime $p$. The special case $n=p^{2}$ of Result 1.2 was rediscovered in [7].

The main purpose of this paper is to provide a vast generalization and strengthening of Result 1.2. In fact, it turns out that any nondegenerate symmetric bilinear form on a finite abelian group can be used to construct group invariant Butson Hadamard matrices. Within our construction, given an abelian group $G$, there is ample freedom to choose “ingredients” (the bilinear form, a suitable subgroup of $G$, and a system of coset representatives).

There is a well developed theory of bilinear forms on finite abelian groups, see [10, 17, 18]. It is shown in these papers that, contrary to the case of bilinear forms over finite fields, in general there is quite a number of inequivalent nondegenerate symmetric bilinear forms on a finite abelian group. As any of these bilinear forms can be used in our construction, this theory turns out to be relevant for the existence of Butson Hadamard matrices and sheds light on the above mentioned flexibility of ingredients. We will not discuss the theory of bilinear forms on finite abelian groups in this paper though, and just focus on proving the correctness of our construction.

Finally, we will show that the conditions which are sufficient for our construction of $\operatorname{{\rm BH}}(G,h)$ matrices to work are also necessary for the existence of these matrices in the case where $G$ is cyclic of prime power order. The proof is an application of the field descent method developed in [13]. We remark that the field descent method relies on the fact that algebraic integers in a cyclotomic field $F$ whose squared modulus is an integer often are contained in proper subfields of $F$, which are, in fact, are also cyclotomic fields. This method was introduced in [13] and has, for instance, been used to obtain progress on the Circulant Hadamard Matrix Conjecture [13, 14] and Lander’s Conjecture [8].

2 The Construction

For a finite abelian group $G$, we denote the least common multiple of the orders of the elements of $G$ by $\exp(G)$. For a positive integer $t$, write $\zeta_{t}=\exp(2\pi i/t)$. As before, we denote the cyclic group of order $t$ by $\mathbb{Z}_{t}$ and identify $\mathbb{Z}_{t}$ with $\{0,\ldots,t-1\}$, the group operation being addition modulo $t$.

Let $G$ be a finite abelian group and let $e$ be a positive integer. We say that a map $f:G\times G\to\mathbb{Z}_{e}$ is a bilinear form if

[TABLE]

for all $g,h,k\in G$. Note that (1) implies

[TABLE]

for all $g,h\in G$ and $\alpha\in\mathbb{Z}$. If $f(g,h)=f(h,g)$ for all $g,h\in G$, then $f$ is symmetric. If $f(g,h)=0$, then $g$ and $h$ are said to be orthogonal. We say that $f$ is nondegenerate if there is no $g\in G\setminus\{0\}$ such that $f(g,h)=0$ for all $h\in G$.

We will use the following conventions. For an abelian group $G$ and $g\in G$, we say that $h\in G$ is a square root of $g$ if $g=2h$ and we write $h=g/2$. Note that square roots are not unique in general; for our purposes, $g/2$ denotes any square root of $g$. In fact, the construction in Theorem 2.1 works no matter which square roots are chosen.

Let $L$ be an elementary abelian group of order $2^{2a+b}$ where $b\in\{0,1\}$ and write $c=2a+b$. We identify $L$ with $\{(g_{1},\ldots,g_{c}):g_{1},\ldots,g_{c}\in\{0,1\}\}$, the group operation being componentwise addition modulo $2$. For $g=(g_{1},\ldots,g_{c})\in L$, set $G_{1}=(g_{1},\ldots,g_{a})$ and $G_{2}=(g_{a+1},\ldots,g_{2a})$. Similarly, define $X_{1}$ and $X_{2}$ for $x=(x_{1},\ldots,x_{c})\in L$. For $x,y\in\{0,1\}^{a}$, write $xy^{T}=\sum_{i=1}^{a}x_{i}y_{i}$. Note that in this sum we use the addition of integers, not addition modulo $2$. For instance, if $x=y=(1,1)$, then $xy^{T}$ is the integer $2$. Define a function $s_{L}:L\to\mathbb{Z}$ by

[TABLE]

For $u,w\in\mathbb{Z}$, set $u\oplus w=0$ if $u+w$ is even and $u\oplus w=1$ if $u+w$ is odd. For $x,y\in L$, define $x\oplus y$ by $(x\oplus y)_{i}=x_{i}\oplus y_{i}$ for all $i$. First assume $b=1$. Let $g$ be any element of $L$. We have

[TABLE]

If $G_{1}\neq(0,\ldots,0)$, then the last sum in (2) vanishes and if $G_{2}\neq(0,\ldots,0)$, then the second last sum in (2) vanishes. If $g_{c}\neq 0$, then $g_{c}=1$ and

[TABLE]

In summary, we have

[TABLE]

whenever $g\neq 0$. If $b=0$, then (3) also holds and is proved in a similar way.

Theorem 2.1**.**

Let $K=G\times L$ be a finite abelian group, where either $L=\{0\}$ or $L$ is an elementary abelian $2$-group. Write $e=\exp(G)$. Let $U$ be a subgroup of $G$ such that every element of $U$ has a square root in $G$. Suppose that $f:G\times G\to\mathbb{Z}_{e}$ is bilinear, symmetric, and nondegenerate, and that no element of $G\setminus U$ is orthogonal to all elements of $U$. Let $R\subset G$ be a complete system of coset representatives of $U$ in $G$ with $0\in R$. For every $x\in K$, there are unique $x_{1}\in U$, $x_{2}\in R$, and $x_{3}\in L$ with $x=x_{1}+x_{2}+x_{3}$. Let $\beta$ be any integer coprime to $|G|$. Define a matrix $H=(H_{y,x})_{y,x\in K}$ by

[TABLE]

Then $H$ is a $\operatorname{{\rm BH}}(K,e_{1})$ matrix, where

[TABLE]

Proof.

We first make a remark on the assumption that every element of $U$ has a square root in $G$, as it will not be mentioned again in the proof. This assumption is necessary for the right hand side of (4) to be properly defined (note that $(x-y)_{1}/2$ must exist for all $x,y\in K$).

From the definition, it is clear that $H$ is $G$-invariant. Fix any $y\in K$. For every $x\in K$, there is a unique $u(x_{2})\in U$ with

[TABLE]

Note that $u(x_{2})$ depends on $y$, but we do not indicate this dependence, as we consider $y$ as fixed. We have

[TABLE]

by (5), as $x_{1}+x_{2}-y_{1}-y_{2}=(x-y)_{1}+(x-y)_{2}$. Note that $u(x_{2})$ only depends on $x_{2}$ (not on $x_{1}$), as $(x-y)_{2}=(x^{\prime}-y)_{2}$ whenever $x_{2}=x^{\prime}_{2}$. We claim that

[TABLE]

Indeed, if $y_{2}=0$, then $u(x_{2})=(x-y)_{2}-x_{2}=(x_{1}+x_{2}+x_{3}-y_{1}-y_{3})_{2}-x_{2}=x_{2}-x_{2}=0$.

By (4), (5), and (6), we have

[TABLE]

Note that the $(0,y)$ entry of $HH^{*}$ is

[TABLE]

Substituting (8) into the last expression and using the bilinearity of $f$, we get

[TABLE]

where

[TABLE]

From now on, we assume $A(y)\neq 0$. Our goal is to show that this implies $y=0$. First of all, note that $A(y)\neq 0$ implies $\sum_{x_{3}\in L}\zeta_{4}^{s_{L}(x_{3})-s_{L}(x_{3}\oplus y_{3})}\neq 0$ and thus $y_{3}=0$ by (3). Hence the first sum in the expression for $A(y)$ is equal to $|L|$ and

[TABLE]

Write $V(x_{2})=\sum_{x_{1}\in U}\zeta_{e}^{S(x_{1},x_{2})}$. Note that $S(x_{1}+z_{1},x_{2})=S(x_{1},x_{2})+S(z_{1},x_{2})$ for all $x_{1},z_{1}\in U$, since $f$ is bilinear. Suppose that there is $z_{1}\in U$ such that $S(z_{1},x_{2})\neq 0$ and thus $\zeta_{e}^{S(z_{1},x_{2})}\neq 1$. As

[TABLE]

we conclude $V(x_{2})=0$. So we see that $V(x_{2})$ is only nonzero if

[TABLE]

for all $x_{1}\in U$. As no element of $G\setminus U$ is orthogonal to all elements of $U$ by assumption, this implies $y_{1}+\beta y_{2}+(1-\beta)u(x_{2})\in U$ and thus $y_{2}\in U$, as $y_{1},u(x_{2})\in U$ and $\beta$ is coprime to $|G|$. This implies $y_{2}=0$, as $y_{2}\in R\cap U$ and $R\cap U=\{0\}$. Hence we have $y_{2}=0$ whenever $V(x_{2})\neq 0$. As $A(y)\neq 0$ by assumption, we have $V(x_{2})\neq 0$ for at least one $x_{2}$ and thus

[TABLE]

Hence

[TABLE]

for all $x_{2}\in R$ by (7). Combining (10), (11), and (12), we get

[TABLE]

and

[TABLE]

Hence $W(y):=\sum_{x_{2}\in R}\zeta_{e}^{\beta f(y_{1},x_{2})}\neq 0$.

Now suppose that there is $z_{2}\in R$ such that $f(y_{1},z_{2})\neq 0$. Note that the map $R\to R,\ x_{2}\mapsto(x_{2}+z_{2})_{2}$ is a bijection, as the elements of $R$ represent each coset of $U$ in $G$ exactly once. Moreover, for any $r,s\in R$, we have

[TABLE]

since $f(y_{1},(r+s)_{1})=0$ by (13). Thus

[TABLE]

As $\zeta_{e}^{\beta f(y_{1},z_{2})}\neq 1$, we conclude $W(y)=0$, a contradiction. Hence we have $f(y_{1},x_{2})=0$ for all $x_{2}\in R$. This, together with (13), implies $f(y_{1},x)=f(y_{1},x_{1}+x_{2})=f(y_{1},x_{1})+f(y_{1},x_{2})=0$ for all $x\in G$. Thus $y_{1}=0$, as $f$ is nondegenerate. So we have $y_{3}=0$, $y_{2}=0$, and $y_{1}=0$, that is, $y=0$, as desired.

In summary, we have shown that the $(0,y)$ entry of $HH^{*}$ is only nonzero if $y=0$. As $H$ is $K$-invariant, this shows that the $(x,y)$ entry of $HH^{*}$ is only nonzero if $x=y$. Hence $HH^{*}=|K|I$, where $I$ is the identity matrix of order $|K|$.

It remains to prove that the entries of $H$ are $e_{1}$th roots of unity, where $e_{1}$ is defined in the statement of the theorem. Recall that

[TABLE]

where $W_{y,x}=\zeta_{e}^{f((x-y)_{1}/2,(x-y)_{1})+f((x-y)_{1},(x-y)_{2})}$. Write $k=\exp(U)$. As $kx=0$ for all $x\in U$, we have $kf(x,y)=f(kx,y)=f(y,kx)=0$ for all $x\in U$ and $y\in G$ by the bilinearity of $f$. As $(x-y)_{1}\in U$ for all $x,y\in G$, this shows that

[TABLE]

for all $x,y\in K$. If $L=\{0\}$, then $e_{1}=k$ and $H_{y,x}=W_{y,x}$ and thus $H_{y,x}^{e_{1}}=1$ for all $x,y\in K$ by (16). If $L$ is of square order, then $e_{1}={\rm lcm}(2,k)$, $s_{L}(x_{3}\oplus y_{3})\equiv 0\ (\bmod\ 2)$ and thus $H_{y,x}^{e_{1}}=1$ for all $x,y\in K$ by (15) and (16). Note that, in any case, $H_{y,x}^{{\rm lcm}(4,k)}=1$ for all $x,y\in K$ by (15) and (16). Hence $H$ indeed is a $\operatorname{{\rm BH}}(G,e_{1})$ matrix. ∎

Remark 2.2**.**

In Theorem 2.1, there are alternative choices for the function $s_{L}$ for which the construction still works. Write $|L|=2^{2a+d}$ and $c=2a+d$, where $a$ and $d$ are any nonnegative integers. Recall that $F:(\mathbb{Z}_{2})^{2a}\to\mathbb{Z}_{2}$ is a bent function if

[TABLE]

for all nonzero $\alpha\in(\mathbb{Z}_{2})^{2a}$. Bent functions exist in abundance, see [11, 12], for instance. Let $F:(\mathbb{Z}_{2})^{2a}\to\mathbb{Z}_{2}$ be any bent function and set

[TABLE]

Then (4), with $s_{L}$ defined by (17), still is a $\operatorname{{\rm BH}}(K,e_{1})$ matrix. We omit the proof here, which is a straightforward extension of the proof of Theorem 2.1.

For a prime $p$ and an integer $t$, let $\nu_{p}(t)$ denote the $p$-adic valuation of $t$, that is, $p^{\nu_{p}(t)}$ is the largest power of $p$ dividing $t$. For groups $K$ and $W$, we say that $K$ has a direct factor $W$ if $K\cong W\times V$ for some group $V$.

Corollary 2.3**.**

Let $K$ be a finite abelian group and let $h$ be a positive integer such that

[TABLE]

Then there exists a $\operatorname{{\rm BH}}(K,h)$ matrix.

Proof.

Write $K=G\times L$, where either $L=\{0\}$ or $L$ is an elementary abelian $2$-group, such that $G$ does not have a direct factor $\mathbb{Z}_{2}$. Without loss of generality, we can assume $G=\mathbb{Z}_{p_{1}^{a_{1}}}\times\cdots\times\mathbb{Z}_{p_{s}^{a_{s}}}$, where $p_{1},\ldots,p_{s}$ are (not necessarily distinct) primes and $a_{1},\ldots,a_{s}$ are positive integers. Note that $p_{i}^{a_{i}}\neq 2$ for all $i$, as $G$ has no direct factor $\mathbb{Z}_{2}$. Furthermore, $|L|>1$ if and only if $K$ has a direct factor $\mathbb{Z}_{2}$. We identify $G$ with

[TABLE]

where the group operation is componentwise addition, the $i$th component being taken modulo $p_{i}^{a_{i}}$. Let $U$ be the subgroup of $G$ given by

[TABLE]

Write $e=\exp(G)$ and define a map $f:G\times G\to\mathbb{Z}_{e}$ by

[TABLE]

It is straightforward to verify that $f$ is bilinear and symmetric. Let $g=(g_{1},\ldots,g_{s})$ be any nonzero element of $G$. Then $g_{i}\neq 0$ for some $i$. Let $t$ be the element of $G$ with $t_{i}=1$ and $t_{j}=0$ for all $j\neq i$. Then $f(g,t)=\frac{e}{p_{i}^{a_{i}}}g_{i}\neq 0$, since $g_{i}\not\equiv 0\ (\bmod\ p_{i}^{a_{i}})$. This shows that $f$ is nondegenerate.

Suppose that $g=(g_{1},\ldots,g_{s})$ an element which orthogonal to all elements of $U$. Let $i$ be arbitrary and let $s$ be the element of $U$ with $s_{i}=p_{i}^{\lfloor a_{i}/2\rfloor}$ and $s_{j}=0$ for all $j\neq i$. Then

[TABLE]

and thus $g_{i}\equiv 0\ (\bmod\ p_{i}^{\lceil a_{i}/2\rceil})$. This implies $g\in U$. Hence there is no element of $G\setminus U$ which is orthogonal to all elements of $U$.

Next, we show that every element $u=(u_{1},\ldots,u_{s})$ of $U$ has a square root in $G$. If $p_{i}=2$, then $a_{i}\geq 2$, as we are assuming $p_{i}^{a_{i}}\neq 2$ for all $i$. Thus $u_{i}\equiv 0\ (\bmod\ 2)$ by the definition of $U$, that is, $u_{i}=2y_{i}$ with $y_{i}\in\{0,\ldots,2^{a_{i}-1}-1\}$. On the other hand, if $p_{i}$ is odd, then the map $x\mapsto 2x\ (\bmod\ p_{i}^{a_{i}})$ is a bijection and thus there is $y_{i}\in\{0,\ldots,p^{a_{i}}-1\}$ with $2y_{i}\equiv u_{i}\ (\bmod\ p_{i}^{a_{i}})$. In summary, we have $u=2y$, i.e., $y$ is a square root of $u$.

We have shown that all assumptions of Theorem 2.1 are satisfied. Hence there exists a $\operatorname{{\rm BH}}(K,e_{1})$ matrix, where $e_{1}$ is defined in Theorem 2.1. In view of Remark 1.1, it suffices to show

[TABLE]

Note that

[TABLE]

and $h\equiv 0\ \left(\bmod\ p_{i}^{\lceil a_{i}/2\rceil}\right)$ by (18). Hence

[TABLE]

If $L=\{0\}$, then $e_{1}=\exp(U)$ and (20) follows from (22). Now assume $|L|>1$. If $\nu_{2}(\exp(K))\geq 3$, then $h\equiv 0\ (\bmod\ 4)$ by (18) and thus (20) holds by (22), as $e_{1}$ divides ${\rm lcm}(4,\exp(U))$. Hence we can assume $\nu_{2}(\exp(K))\leq 2$, i.e., $a_{i}=2$ whenever $p_{i}=2$. Note that this implies that $\nu_{2}(|G|)$ is even. Suppose that $\nu_{2}(|K|)$ is even. Then $\nu_{2}(|L|)$ is also even, that is, $L$ is of square order. Thus $e_{1}={\rm lcm}(2,\exp(U))$ and (20) follows from (18) and (22). Finally, suppose that $\nu_{2}(|K|)$ is odd. Note that $K$ has a direct factor $\mathbb{Z}_{2}$, as we are assuming $|L|>1$. Thus $h\equiv 0\ (\bmod\ 4)$ by (19), which implies (20). Thus (20) holds in every case and this completes the proof. ∎

Remark 2.4**.**

Corollary 2.3 only provides one possible choice of the bilinear form $f$ and the subgroup $U$. In general, there are numerous other choices, which produce group invariant Butson Hadamard matrices not equivalent to those constructed in Corollary 2.3.

Corollary 2.5**.**

If $v$ and $h$ are positive integers with

• $(i)$

$\nu_{p}(h)\geq\lceil\nu_{p}(v)/2\rceil$* for every prime divisor $p$ of $v$ and*

• $(ii)$

$\nu_{2}(h)\geq 2$* if $v\equiv 2\ (\bmod\ 4)$,*

then a (circulant) $\operatorname{{\rm BH}}(\mathbb{Z}_{v},h)$ matrix exists.

3 Necessary Conditions

From now on, we use the language of group ring equations to study group invariant Butson Hadamard matrices and write groups multiplicatively. Let $G$ be a finite abelian group, let $R$ be a ring and let $R[G]$ denote the group ring of $G$ over $R$. The elements of $R[G]$ have the form $X=\sum_{g\in G}a_{g}g$ with $a_{g}\in R$. The $a_{g}$’s are called the coefficients of $X$. Two elements $X=\sum_{g\in G}a_{g}g$ and $Y=\sum_{g\in G}b_{g}g$ in $R[G]$ are equal if and only if $a_{g}=b_{g}$ for all $g\in G$. A subset $S$ of $G$ is identified with the group ring element $\sum_{g\in S}g$. For the identity element $1_{G}$ of $G$ and $\lambda\in R$, we write $\lambda$ for the group ring element $\lambda 1_{G}$. For $R=\mathbb{Z}[\zeta_{h}]$ and $X=\sum_{g\in G}a_{g}g\in R[G]$, we write

[TABLE]

where $\overline{a_{g}}$ denotes the complex conjugate of $a_{g}$.

The group of complex characters of $G$ is denoted by $\hat{G}$. The trivial character $\chi_{0}$ is defined by $\chi_{0}(g)=1$ for all $g\in G$. For $D=\sum_{g\in G}a_{g}g\in R[G]$ and $\chi\in\hat{G}$, write $\chi(D)=\sum_{g\in G}a_{g}\chi(g)$. The following is a standard result and a proof can be found [2, Ch. VI, Lem. 3.5], for instance.

Result 3.1**.**

Let $G$ be a finite abelian group and $D=\sum_{g\in G}a_{g}g\in\mathbb{C}[G]$. Then

[TABLE]

for all $g\in G$. Consequently, if $D,E\in\mathbb{C}[G]$ and $\chi(D)=\chi(E)$ for all $\chi\in\hat{G}$, then $D=E$.

Lemma 3.2**.**

Let $p$ be a prime and let $m$ be a positive integer not divisible by $p$. Write $\zeta=\zeta_{p}$ if $p$ is odd and $\zeta=\zeta_{4}$ if $p=2$. If $X\in\mathbb{Z}[\zeta_{m}]$ satisfies $X\equiv 0\ (\bmod\ 1-\zeta)$, then $X\equiv 0\ (\bmod\ p)$.

Proof.

In this proof, we use basic algebraic number theory as covered in [5], for instance. Let $R=\mathbb{Z}[\zeta_{pm}]$ if $p$ is odd and $R=\mathbb{Z}[\zeta_{4m}]$ if $p=2$. The ideal $(1-\zeta)R$ of $R$ factorizes as

[TABLE]

where $k=\varphi(m)/{\rm ord}_{m}(p)$ and the $\mathfrak{p}_{i}$’s are distinct prime ideals. Furthermore, $p\mathbb{Z}[\zeta_{m}]=\prod_{i=1}^{k}\mathfrak{q}_{i}$, where $\mathfrak{q}_{i}=\mathfrak{p}_{i}^{p-1}$ and the $\mathfrak{q}_{i}$’s are prime ideals of $\mathbb{Z}[\zeta_{m}]$. Note that $X\equiv 0\ (\bmod\ \mathfrak{p}_{i})$ for all $i$, as $X\equiv 0\ (\bmod\ 1-\zeta)$ by assumption. Since $X\in\mathbb{Z}[\zeta_{m}]$, this implies $X\equiv 0\ (\bmod\ \mathfrak{q}_{i})$ for all $i$ and hence $X\equiv 0\ (\bmod\ p)$. ∎

Lemma 3.3**.**

Let $G$ be a finite abelian group, let $h$ be a positive integer, and let $a_{g}$, $g\in G$, be elements of $\{\zeta_{h}^{i}:i=0,\ldots,h-1\}$. Consider the element $D=\sum_{g\in G}a_{g}g$ of $\mathbb{Z}[\zeta_{h}][G]$ and the $G$-invariant matrix $H=(H_{g,k})$, $g,k\in G$ given by $H_{g,k}=a_{g-k}$. Then $H$ is a $\operatorname{{\rm BH}}(G,h)$ matrix if and only if

[TABLE]

Moreover, $(\ref{gr1})$ holds if and only if

[TABLE]

Proof.

Let $g\in G$ be arbitrary. The coefficient of $g$ in $DD^{(-1)}$ is

[TABLE]

On the other hand, the inner product of row $x+g$ and row $x$ of $H$ is

[TABLE]

Hence (23) holds if and only if any two distinct rows of $H$ have inner product [math], that is, if and only if $H$ is a $\operatorname{{\rm BH}}(G,h)$ matrix. Finally, the equivalence of (23) and (24) follows from Result 3.1. ∎

Result 3.4** ([14], Thm. 2.2.2).**

Let $p$ be a prime, let $a,b,h$ be positive integers with $(h,p)=1$, and write $v=p^{a}h$. Suppose that $X$ is an element of $\mathbb{Z}[\zeta_{v}]$ with $|X|^{2}=p^{b}$. Then there exist $Y\in\mathbb{Z}[\zeta_{h}]$, a root of unity $\eta\in\mathbb{Z}[\zeta_{h}]$, $\eta\neq 1$, of order dividing $p-1$, and an integer $j$ such that either

[TABLE]

where $\Theta=1-\zeta_{4}$ if $p=2$ and $\Theta=\sum_{i=0}^{p-2}\eta^{-i}\zeta_{p}^{t^{i}}$ if $p$ is odd, and $t$ is a primitive element modulo $p$.

Theorem 3.5**.**

Let $p$ be a prime and let $a,h$ be positive integers. If a $\operatorname{{\rm BH}}(\mathbb{Z}_{p^{a}},h)$ matrix exists, then $p$ divides $h$.

Proof.

Suppose that a $\operatorname{{\rm BH}}(\mathbb{Z}_{p^{a}},h)$ exists and that $p$ does not divide $h$. Write $G=\mathbb{Z}_{p^{a}}$ and let $g$ be generator of $G$. By Lemma 3.3, there is $D\in\mathbb{Z}[\zeta_{h}][G]$, $D=\sum_{i=0}^{p^{a}-1}\zeta_{h}^{a_{i}}g^{i}$, with

[TABLE]

for all $\chi\in\hat{G}$. Let $\chi$ be the character of $G$ with $\chi(g)=\zeta_{p^{a}}$. By Result 3.4, we have $\chi(D)=\zeta_{p^{a}h}^{k}Y$ or $\chi(D)=\zeta_{p^{a}h}^{k}\Theta Y$ for some integer $k$, where $Y\in\mathbb{Z}[\zeta_{h}]$ and $\Theta$ is defined in Result 3.4. Replacing $D$ by $\zeta_{h}^{c}g^{d}D$ with suitable integers $c,d$, if necessary, we can assume $k=0$. Hence

[TABLE]

We first assume $p=2$. In this case, $h$ is odd and we have $\chi(D)\in\mathbb{Z}[\zeta_{4h}]$ by (26). If $a=1$, it is easy to show that $h$ is divisible by $4$. Hence we can assume $a\geq 2$. Write $D=\sum_{i=0}^{2^{a-2}-1}g^{i}\sum_{j=0}^{3}\zeta_{h}^{a_{i,j}}g^{2^{a-2}j}$ with $a_{i,j}\in\mathbb{Z}$. We have

[TABLE]

As $\chi(D)\in\mathbb{Z}[\zeta_{4h}]$ and $\{1,\zeta_{2^{a}},\ldots,\zeta_{2^{a}}^{2^{a-2}-1}\}$ is linearly independent over $\mathbb{Q}(\zeta_{4h})$, we conclude

[TABLE]

where $A=\zeta_{h}^{a_{0,0}}-\zeta_{h}^{a_{0,2}}$ and $B=\zeta_{h}^{a_{0,1}}-\zeta_{h}^{a_{0,3}}$. We conclude

[TABLE]

As $\{1,\zeta_{4}\}$ is linearly independent over $\mathbb{Q}(\zeta_{h})$, this implies $-A\bar{B}+B\bar{A}=0$ and $A\bar{A}+B\bar{B}=2^{a}$. As $h$ is odd, we have $|A|,|B|<2$ and thus $A\bar{A}+B\bar{B}<8$. Hence $a=2$ and $A\bar{A}+B\bar{B}=|\chi(D)|^{2}=4$. A quick computation shows that this implies $\eta+\bar{\eta}+\gamma+\bar{\gamma}=0$, where $\eta=\zeta_{h}^{a_{0,0}-a_{0,2}}$ and $\gamma=\zeta_{h}^{a_{0,1}-a_{0,3}}$. Hence ${\rm Re}(\eta)=-{\rm Re}(\gamma)$, where ${\rm Re}(z)$ denotes the real part of $z\in\mathbb{C}$. Write $\eta=\zeta_{h}^{c}$ and $\gamma=\zeta_{h}^{d}$ with $c,d\in\mathbb{Z}$. We have ${\rm Re}(\eta)=\cos(2\pi ic/h)$ and ${\rm Re}(\gamma)=\cos(2\pi id/h)$ and thus ${\rm Re}(\eta)=-{\rm Re}(\gamma)$ implies $2\pi d/h=\pm 2\pi c/h+k\pi$ where $k$ is an odd integer. This is only possible if $h$ is even. The proof is complete for $p=2$.

Now assume that $p$ is odd. We rewrite $D$ in the form

[TABLE]

where the $a_{i,j}$’s are integers with $0\leq a_{i,j}\leq p-1$. Note that

[TABLE]

As $\{1,\zeta_{p^{a}},\ldots,\zeta_{p^{a}}^{p^{a-1}-1}\}$ is linearly independent over $\mathbb{Q}(\zeta_{ph})$ and $\chi(D)\in\mathbb{Z}[\zeta_{ph}]$ by (26), we conclude that $\sum_{j=0}^{p-1}\zeta_{h}^{a_{i,j}}\zeta_{p}^{j}=0$ for all $i>0$ and thus

[TABLE]

where $b_{j}=a_{0,j}$. This implies $\left|\chi(D)\right|<p$ and thus $a=1$ by (25). In particular, $g$ has order $p$ and we have

[TABLE]

According to (26), it suffices the consider the following two cases.

Case 1 $\chi(D)=\Theta Y$. Recall that $\Theta=\sum_{i=0}^{p-2}\eta^{-i}\zeta_{p}^{t^{i}}$ and note that

[TABLE]

since $\eta\neq 1$ and the order of $\eta$ divides $p-1$. We conclude

[TABLE]

By Lemma 3.2, this implies

[TABLE]

However, for the trivial character $\chi_{0}$ of $G$, we have $\chi_{0}(D)=\sum_{j=0}^{p-1}\zeta_{h}^{b_{j}}$ by (29) and $|\chi_{0}(D)|^{2}=p$ by (25). But (30) implies $|\chi_{0}(D)|^{2}\equiv 0\ (\bmod\ p^{2})$, a contradiction.

Case 2 $\chi(D)\in\mathbb{Z}[\zeta_{h}]$. By (28), we have $\chi(D)=\sum_{j=0}^{p-2}(\zeta_{h}^{b_{j}}-\zeta_{h}^{b_{p-1}})\zeta_{p}^{j}$ and since $\chi(D)\in\mathbb{Z}[\zeta_{h}]$ and $\{1,\zeta_{p},\ldots,\zeta_{p}^{p-2}\}$ is linearly independent over $\mathbb{Q}(\zeta_{h})$, we conclude $b_{1}=\cdots=b_{p-1}$ and $\chi(D)=\zeta_{h}^{b_{0}}-\zeta_{h}^{b_{p-1}}$. As $|\chi(D)|^{2}=p$ and $p$ is odd, this implies $p=3$. Replacing $D$ by $\zeta_{h}^{-b_{0}}D$, if necessary, we can assume $\chi(D)=1-\zeta_{h}^{u}$ for some integer $u$. Note that $\zeta_{h}^{u}\neq 1$. We have $3=|\chi(D)|^{2}=2-\zeta_{h}^{u}-\zeta_{h}^{-u}$ and thus $\zeta_{h}^{u}+\zeta_{h}^{2u}+1=0$, which implies $\zeta_{h}^{3u}=1$. As $\zeta_{h}^{u}\neq 1$, we conclude $h\equiv 0\ (\bmod\ 3)$, contradicting the assumption that $p$ does not divide $h$. ∎

Theorem 3.6**.**

Let $v$ be a power of a prime $p$ and let $h$ be a positive integer. A (circulant) $\operatorname{{\rm BH}}(\mathbb{Z}_{v},h)$ matrix exists if and only if

[TABLE]

Proof.

By Corollary 2.5, condition (31) is sufficient for the existence of a $\operatorname{{\rm BH}}(\mathbb{Z}_{v},h)$ matrix. It remains to prove the necessity of (31). Assume that a $\operatorname{{\rm BH}}(\mathbb{Z}_{v},h)$ exists and write $v=p^{a}$, where $a$ is a positive integer. Write $G=\mathbb{Z}_{p^{a}}$ and let $g$ be generator of $G$. By Lemma 3.3, there is $D\in\mathbb{Z}[\zeta_{h}][G]$ with $D=\sum_{i=0}^{p^{a}-1}\zeta_{h}^{a_{i}}g^{i}$ and $DD^{(-1)}=v$.

We first show that we can assume $a\geq 5$ if $p=2$. Suppose $p=2$. Then $h$ is even by Theorem 3.5. If $a=1$, then $2=DD^{(-1)}=2+(\zeta_{h}^{a_{0}-a_{1}}+\zeta_{h}^{a_{1}-a_{0}})g$ and hence $\zeta_{h}^{a_{0}-a_{1}}+\zeta_{h}^{a_{1}-a_{0}}=0$. This implies $\zeta_{h}^{2a_{0}-2a_{1}}=-1$ and thus $h\equiv 0\ (\bmod\ 4)$. This shows that (31) is necessary in the case $a=1$. As $h$ is even, condition (31) is also necessary in the case $a=2$. Now suppose $a\in\{3,4\}$. To prove the necessity of (31), it suffices to show that $\nu_{2}(h)=1$ is impossible. Thus assume $\nu_{2}(h)=1$. Similar to the case $p=2$ in the proof of Theorem 3.5, we see that we can assume $\chi(D)\in\mathbb{Z}[\zeta_{2h}]$ where $\chi$ is the character of $G$ with $\chi(g)=\zeta_{2^{a}}$ (note that we have $\chi(D)\in\mathbb{Z}[\zeta_{2h}]$ and not only $\chi(D)\in\mathbb{Z}[\zeta_{4h}]$, as $h$ is even now). Furthermore, as in the proof of Theorem 3.5, we have $D=\sum_{j=0}^{3}\zeta_{h}^{a_{0,j}}g^{j}$, $A\bar{A}+B\bar{B}=2^{a}$ where $A=\zeta_{h}^{a_{0,0}}-\zeta_{h}^{a_{0,2}}$ and $B=\zeta_{h}^{a_{0,1}}-\zeta_{h}^{a_{0,3}}$. This implies $a=3$ and $|A|=|B|=2$. This is only possible if $\zeta_{h}^{a_{0,2}}=-\zeta_{h}^{a_{0,0}}$ and $\zeta_{h}^{a_{0,1}}=-\zeta_{h}^{a_{0,3}}$. But then $\chi_{0}(D)=\sum_{j=0}^{3}\zeta_{h}^{a_{0,j}}=0$ for the trivial character $\chi_{0}$ of $G$, a contradiction. In summary, we have shown that (31) is necessary if $p=2$ and $a\leq 4$.

By what we have shown, we can assume $a\geq 5$ if $p=2$. Suppose that condition (31) is not satisfied. Set $b=\nu_{p}(h)$ if $p$ is odd and $b=\max\{2,\nu_{2}(h)\}$ if $p=2$. By Theorem 3.5, we have $b\geq 1$. Moreover, $\nu_{p}(h)<a/2$, as we assume that (31) does not hold. Note that, for $p=2$, this implies $b<a/2$, as $a\geq 5$ and thus $a/2>2$ is this case. In summary, we have

[TABLE]

Let $\chi$ be the character of $G$ with $\chi(g)=\zeta_{p^{a}}$. Note that $|\chi(D)|^{2}=p^{a}$ by Lemma 3.3. Write $h=p^{b}h^{\prime}$ where $(h^{\prime},p)=1$. By Result 3.4, there is an integer $s$ such $\chi(D)\zeta_{p^{a}}^{s}\in\mathbb{Z}[\zeta_{ph^{\prime}}]$ if $p$ is odd and $\chi(D)\zeta_{p^{a}}^{s}\in\mathbb{Z}[\zeta_{4h^{\prime}}]$ if $p=2$. In view of (32), this implies $\chi(D)\zeta_{p^{a}}^{s}\in\mathbb{Z}[\zeta_{h}]$. Replacing $D$ by $Dg^{-s}$, if necessary, we can assume

[TABLE]

Write $h=p^{b}m$, where $(p,m)=1$. Every integer in $\{0,\ldots,p^{a}-1\}$ has a unique representation as $i+p^{a-b}j$ with $0\leq i\leq p^{a-b}-1$ and $0\leq j\leq p^{b}-1$. Hence we can write

[TABLE]

with $a_{i,j},b_{i,j}\in\mathbb{Z}$ and we have

[TABLE]

In view of (33), and since $\{1,\zeta_{p^{a}},\ldots,\zeta_{p^{a}}^{p^{a-b}-1}\}$ is linearly independent over $\mathbb{Q}(\zeta_{p^{b}m})$, we have $\sum_{j=0}^{p^{b}-1}\zeta_{p^{b}}^{a_{i,j}+j}\zeta_{m}^{b_{i,j}}=0$ for all $i>0$. We conclude $\chi(D)=\sum_{j=0}^{p^{b}-1}\zeta_{p^{b}}^{a_{0,j}+j}\zeta_{m}^{b_{0,j}}$ and hence $|\chi(D)|^{2}\leq p^{2b}$. Thus $|\chi(D)|^{2}<p^{a}$ by (32), contradicting $|\chi(D)|^{2}=p^{a}$. ∎

Acknowledgement We thank the anonymous referees for their careful reading of the manuscript and their suggestions that helped to improve the readability of the paper.