Difference sets disjoint from a subgroup

TL;DR

This paper investigates the structure of finite groups with a subgroup and a difference set disjoint from it, revealing normality conditions, size relations, and providing explicit examples and algebraic structures.

Contribution

It establishes that such groups must have a normal subgroup with order squared, characterizes the difference set distribution, and constructs a family of examples with specific algebraic properties.

Findings

H is normal in G

|G|=|H|^2

Existence of a 2-parameter family of examples

Abstract

We study finite groups $G$ having a subgroup $H$ and $D \subset G \setminus H$ such that the multiset $\{ xy^{-1}:x,y \in D\}$ has every non-identity element occur the same number of times (such a $D$ is called a {\it difference set}). We show that $H$ has to be normal, that $|G|=|H|^2$, and that $|D \cap Hg|=|H|/2$ for all $g \notin H$. We show that $H$ is contained in every normal subgroup of prime index, and other properties. We give a $2$-parameter family of examples of such groups. We show that such groups have Schur rings with four principal sets.

Equations

$$h=|H|,u=|G:H|.$$

$$\lambda =\frac{1}{4}h(h-2),k=\frac{1}{2}h(h-1).$$

$$\{1\},H\setminus \{1\},D,D^{-1},$$

$$\left(\begin{array}{cccc}1& h-1& \frac{1}{2}h(h-1)& \frac{1}{2}h(h-1)\\ 1& h-1& -\frac{1}{2}h& -\frac{1}{2}h\\ 1& -1& -\frac{1}{2}ih& \frac{1}{2}ih\\ 1& -1& \frac{1}{2}ih& -\frac{1}{2}ih\end{array}\right)$$

$$\mu (D)=(x-k)\left(x+\frac{h}{2}\right)\left(x^2+\frac{h^2}{4}\right).$$

$$1,h-1,h(h-1)/2,,h(h-1)/2$$

$${\mathfrak{G}}_{n,k}=⟨$$

$$a_2^{a_1}=a_2{b}_1,a_3^{a_1}=a_3{b}_2,\dots ,a_{k+1}^{a_1}=a_{k+1}{b}_k,$$

$$(a_1,a_{k+2})=(a_1,a_{k+3})=\cdots =(a_1,a_n)=1,$$

$$(a_i,a_j)=1,\text{ for }1<i,j\le n,$$

$$\text{ and }{b}_1,\dots ,b_n\text{ are central involutions}⟩.$$

$$\mathcal{H}(D)=2D-J_{h^2}$$

$$(x+h)(x^2+h^2).$$

$$\lambda =\frac{1}{4}h(h-2),k=\frac{1}{2}h(h-1).$$

$$A=H{g}_1\cup \cdots \cup H{g}_m,B=H{g}_1^{\prime }\cup \cdots \cup H{g}_m^{\prime },$$

$$|A|=|B|=hm,|D|=k=hm+|D_1|.$$

$$G=H+B+D_1+A+D_1^{-1}.$$

$$v=|G|=h+hm+|D_1|+hm+|D_1^{-1}|=h+2hm+2|D_1|=h+2k.$$

$$\lambda =\frac{1}{4}\frac{(hu-h)(hu-h-2)}{hu-1}.$$

$$a=\gcd (hu-h,hu-1),b=\gcd (hu-h-2,hu-1).$$

$$a=\gcd (h-1,u-1),b=\gcd (h+1,u+1).$$

$$ab\le \frac{1}{2}(h-1)(u+1)=\frac{1}{2}(hu-1+h-u)<\frac{1}{2}(hu-1).$$

$$ab\le \frac{1}{2}(u-1)(h+1)=\frac{1}{2}(hu-1+u-h)<\frac{1}{2}(hu-1).$$

$$\gcd \left(\frac{hu-h}{a},c\right)=\gcd \left(\frac{hu-h-2}{b},c\right)=1,$$

$$\lambda =\frac{1}{4}\frac{(hu-h)(hu-h-2)}{hu-1}=\frac{1}{4}\frac{(hu-h)}{a}\frac{(hu-h-2)}{b}\frac{1}{c},$$

$$\lambda =\frac{1}{4}\frac{(hu-h)(hu-h-2)}{hu-1}=\frac{1}{2}\frac{(hu-h)}{a}\frac{(hu-h-2)}{b}\frac{1}{c}.$$

$$\lambda =\frac{(h^2-h)(h^2-h-2)}{4(h^2-1)}=\frac{h(h-1)(h-2)(h+1)}{4(h-1)(h+1)}=\frac{h(h-2)}{4},$$

$$G=D+D^{-1}+H,D\cdot D^{-1}=\lambda G+(k-\lambda )\cdot 1.$$

$$(k-\lambda )(G-1)=D^2+DH.$$

$$(k-\lambda )(G-1)=(D^{-1}{)}^2+D^{-1}H.$$

$$(k-\lambda )(G-1)=(D^{-1}{)}^2+H{D}^{-1}.$$

$$DH=HD.$$

Full text

Keywords: Difference set, subgroup, skew Hadamard difference set, Schur ring.

1. Introduction

For a group $G$ we will identify a finite subset $X\subseteq G$ with the element $\sum_{x\in X}x\in\mathbb{Q}G$ of the group algebra. We also let $X^{-1}=\{x^{-1}:x\in X\}$. Also, write $\mathcal{C}_{n}$ for the cyclic group of order $n$.

A $(v,k,\lambda)$ difference set is a subset $D\subset G,|D|=k$, where $G$ is a finite group such that every element $1\neq g\in G$ occurs $\lambda$ times in the multiset $\{xy^{-1}:x,y\in D\}$. Further, $|G|=v$.

It is well-known that if $D\subset G$ is a difference set, then $gD=\{gd:d\in D\}$ and $\alpha(D)$ are also difference sets, for any $g\in G,\alpha\in\mathrm{Aut}(G)$. Thus in some sense, difference sets are spread out evenly over the group $G$. In this paper we seek to restrict the types of difference sets considered by imposing the following conditions:

We assume that $D\subset G$ is a $(v,k,\lambda)$ difference set where is a subgroup $1\neq H\leq G$ and $m\geq 0$ such that

(1) $D\cap D^{-1}=Hg_{1}\cup\dots\cup Hg_{m}$;

(2) $G\setminus(D\cup D^{-1})=H\cup Hg_{1}^{\prime}\cup\dots\cup Hg_{m}^{\prime}$.

Here $H,Hg_{1},\dots,Hg_{m},Hg_{1}^{\prime},\dots,Hg_{m}^{\prime}$ are distinct cosets of $H$. Let

[TABLE]

Then we have $h>1$. A group having a difference set of the above type will be called a $(v,k,\lambda)_{m}$ relative skew Hadamard difference set group (with difference set $D$ and subgroup $H$).

Recall that a group $G$ has a skew Hadamard difference set if it has a difference set $D$ where $G=D\cup D^{-1}\cup\{1\}$ and $D\cap D^{-1}=\emptyset$. Such groups have been studied in [2, 4, 6, 7, 8, 9, 11, 3].

Our first result is

Theorem 1.1**.**

Let $G$ be a $(v,k,\lambda)_{m}$ relative skew Hadamard difference set group with subgroup $H$ and difference set $D$. Then

(i) $h=u$ is even, $v=|G|=h^{2}$, and

[TABLE]

Now assume that $m=0$. Then we have:

(ii) $H\triangleleft G$;

(iii) each non-trivial coset $Hg\neq H$ meets $D$ in $h/2$ points;

(iv) $H$ contains the subgroup generated by all the involutions in $G$;

(v) the subgroup $H\leq G$ does not have a complement.

Let $\Phi(G)$ be the Frattini subgroup of $G$, the intersection of all the maximal subgroups of $G$. We also have the following result which concerns maximal subgroups of $G$:

Theorem 1.2**.**

Let $G$ be a group that is a $(v,k,\lambda)_{0}$ relative skew Hadamard difference set group with subgroup $H$ and difference set $D$. Then

(a) (i) every index $2$ subgroup of $G$ contains $H$ and $D$ meets each such subgroup in exactly $\lambda$ points.

(ii) if $N\triangleleft G$ has odd prime index $p$, then $H\leq N$. In this situation each non-trivial coset of $N$ meets $D$ in $\frac{1}{2p}h^{2}$ elements, while $|N\cap D|=\frac{1}{2p}h(h-p)$.

(b) Now assume that $G$ is also a $2$-group. Then $H\leq\Phi(G)$. Further, $D$ meets each maximal subgroup of $G$ in exactly $\lambda$ points.

Our original motivation for studying $(v,k,\lambda)_{0}$ relative skew Hadamard difference set groups was to produce examples of Schur rings with a small number of principal sets.

We now define Schur rings [14, 15, 16, 17].

A subring $\mathfrak{S}$ of the group algebra $\mathbb{C}G$ is called a Schur ring (or S-ring) if there is a partition $\mathcal{K}=\{C_{i}\}_{i=1}^{r}$ of $G$ such that the following hold:

$\{1_{G}\}\in\mathcal{K}$; 2. 2.

for each $C\in\mathcal{K}$, $C^{-1}\in\mathcal{K}$; 3. 3.

${C_{i}}\cdot{C_{j}}=\sum_{k}\lambda_{i,j,k}{C_{k}}$; for all $i,j\leq r$.

The $C_{i}$ are called the principal sets of $\mathfrak{S}$. Then we have:

Theorem 1.3**.**

Let $G$ be a $(v,k,\lambda)_{0}$ relative skew Hadamard difference set group with difference set $D$ and subgroup $H$. Then

[TABLE]

are the principal sets of a commutative Schur-ring over $G$.

Each Schur ring is also an association scheme over $G$. The $P$-matrix (where the columns correspond to the generators $1,H-1,D,D^{-1}$ and rows correspond to representations/eigenvalues of the Schur ring) of the association scheme corresponding to the above Schur ring is

[TABLE]

Here $i^{2}=-1$. Theorem 1.3 allows us to show

Theorem 1.4**.**

Let $G$ be a $(v,k,\lambda)_{0}$ relative skew Hadamard group with difference set $D$ and subgroup $H$. Then the minimal polynomial for $D$ is

[TABLE]

Further, the eigenvalues $k,-h/2,ih/2,-ih/2$ have multiplicities

[TABLE]

(respectively).

One can say something about the image of $D$ under an irreducible representation of $G$:

Theorem 1.5**.**

Let $G$ be a $(v,k,\lambda)_{0}$ difference set group with difference set $D$ and subgroup $H$. Let $\rho$ be a non-principal irreducible representation of $G$ of degree $d$. Then $\rho(G)=0I_{d},\rho(D^{-1})=\rho(D)^{*}$ and we have one of the following:

(i) $\rho(H)=0I_{d}$ and $\rho(D)=\mathrm{diag}\left(\varepsilon_{1}i\frac{h}{2},\varepsilon_{2}i\frac{h}{2},\dots,\varepsilon_{d}i\frac{h}{2}\right),$ for some $\varepsilon_{i}\in\{-1,1\}$;

(ii) $\rho(H)=hI_{d}$ and $\rho(D)=-\frac{h}{2}I_{d}$.

We next give examples of families of $(v,k,\lambda)_{0}$ relative skew Hadamard difference set groups. Let $n\geq 2,0\leq k<n-1$ and define the following bi-infinite family of groups:

[TABLE]

We will show:

Theorem 1.6**.**

For $n\geq 2,0\leq k<n-1,$ the group $\mathfrak{G}_{n,k}$ is a relative skew Hadamard difference set group.

Next, we note that, thinking of $D$ as a $h^{2}\times h^{2}$ matrix via the regular representation, the matrix

[TABLE]

is a matrix with entries $\pm 1$. In fact we have

Theorem 1.7**.**

The matrix $\mathcal{H}(D)$ is a Hadamard matrix and its minimal polynomial is

[TABLE]

We recall that an $n\times n$ matrix $M$ is a Hadamard matrix if its entries are $\pm 1$ and $MM^{T}=nI_{n}.$

We note that the difference sets that we study satisfy the parameter condition given by Kesava Menon in [13], and so (in this case, their complements) are examples of what are known as Menon difference sets. Thus the groups $\mathfrak{G}_{n,k}$ determine a $2$-parameter family of Menon difference sets.

Acknowledgements We are grateful to Pace Nielsen for useful conversations regarding this paper. All calculations made in the preparation of this paper were accomplished using Magma [1].

2. Results concerning the parameters

In this section we prove Theorem 1.1 (i):

Lemma 2.1**.**

Let $G$ be a $(v,k,\lambda)_{m}$ relative skew Hadamard group with difference set $D$ and subgroup $H$. Then $h=u$ is even. Further we have $|G|=h^{2}$ and

[TABLE]

Proof Let

[TABLE]

and $D=A+D_{1},D^{-1}=A+D_{1}^{-1},$ where $A\cap D_{1}=\emptyset$. Thus we have

[TABLE]

Then from (1) and (2) of $\S 1$ we obtain

[TABLE]

Thus we have

[TABLE]

Solving $v=hu,k(k-1)=\lambda(v-1),v=h+2k$ gives

[TABLE]

Let

[TABLE]

Then one can see that

[TABLE]

Thus $\gcd(a,b)|2$ since $a|(h-1),b|(h+1)$ and $h>1$.

We wish to show that $h=u$.

Now if we have $h<u$, then we cannot have $(u+1)|(h+1)$, so that we have $ab\leq(h-1)(u+1)/2$. This gives

[TABLE]

While if $h>u$, then we cannot have $(h+1)|(u+1)$, so that $ab\leq(u-1)(h+1)/2$, giving

[TABLE]

Thus in both cases we get $ab<\frac{1}{2}(hu-1).$ We show how this gives a contradiction. There are two cases:

Case 1: $\gcd(a,b)=1$. Then $a|(hu-1),b|(hu-1)$ and $\gcd(a,b)=1$ gives $ab|(hu-1)$. So let $hu-1=abc,c\in\mathbb{N}$. Then from (2.1) we have

[TABLE]

so that

[TABLE]

which implies that $c=1$. But then we have $ab=hu-1>\frac{hu-1}{2}$, a contradiction.

Case 2: $\gcd(a,b)=2$. Then $(ab/2)|(hu-1)$ and so that $hu-1=\frac{ab}{2}c,c\in\mathbb{N}$, where $\gcd\left(\frac{hu-h}{a},c\right)=\gcd\left(\frac{hu-h-2}{b},c\right)=1$. Then

[TABLE]

This again forces $c=1$, so that $\frac{ab}{2}=hu-1,$ which is a contradiction. Thus $h=u$ and $v=h^{2}$.

Now if $h=u$, then we have

[TABLE]

so that $h$ is even. ∎

3. $H$ is normal

Let $D$ denote the difference set where $G=D\cup D^{-1}\cup H,H\leq G,D\cap H=D\cap D^{-1}=\emptyset$.

Order the elements of $G$ according to the cosets $Hg_{1},Hg_{2},\dots,Hg_{h}$.

Then thinking of $D$, $H$ and $G$ as matrices via the regular representation (relative to the above order of $G$) we have

[TABLE]

Note that the fact that $D^{-1}$ is also a difference set [10, p. 57], together with the last equation of (3.1), gives $DD^{-1}=D^{-1}D$.

For $m\in\mathbb{N}$ let $J_{m}$ be the all $1$ matrix of size $m\times m$. Then we have ordered the elements of $G$ so that $H=\mathrm{diag}(J_{h},J_{h},\dots,J_{h})$. So solving for $D^{-1}$ from the first equation of (3.1), and using $DG=kG$, the second equation gives

[TABLE]

However (since $D^{-1}$ is also a difference set) we can interchange $D$ and $D^{-1}$ so as to obtain

[TABLE]

Now taking the inverse of equation (3.2) we have

[TABLE]

Thus from equations (3.3) and (3.4) we must have $D^{-1}H=HD^{-1}$; taking inverses gives

[TABLE]

Thus $D$ commutes with $G,H,D^{-1}$. Now multiplying $(k-\lambda)(G-1)=D^{2}+HD$ by $H$ we obtain

[TABLE]

Multiplying by $H$ again we have

[TABLE]

We now find the minimal polynomial for $DH$, by first finding the minimal polynomial for $hG-H$. A calculation shows that

[TABLE]

Thus $(hG-H),(hG-H)^{2},(hG-H)^{3}$ are in the span of $H,G$ and so are linearly dependent. Define

[TABLE]

Then one finds that

[TABLE]

Now let $\Delta=DH$. Then from (3.5) we have

[TABLE]

It follows from (3.6), (3.7) that $\Delta$ satisfies the polynomial

[TABLE]

For $n\in\mathbb{N}$ we let $1_{n}=(1,1,\dots,1),0_{n}=(0,0,\dots,0)\in\mathbb{R}^{n}$. Now from equation (3.5) and the definition of the function $F$ we see that:

(A) the matrix $hG-H$ has eigenvalue $\mu=h^{3}-h$ with an eigen space containing $1_{h^{2}}$.

(B) the matrix $hG-H$ has eigenvalue $\mu=-h$ with corresponding eigenspace containing the span of

[TABLE]

so that this eigenspace has dimension at least $h-1$.

(C) Lastly, $hG-H$ has eigenvalue $\mu=0$ with corresponding eigenspace containing the span of all vectors of the form $(v_{1},v_{2},\dots,v_{h})$, where $v_{i}\in\mathbb{R}^{h}$ satisfies $J_{h}v_{i}=0.$ Thus this eigenspace has dimension at least $h^{2}-h$.

Since $1+(h-1)+(h^{2}-h)=h^{2}$ we see that (A), (B), (C) describe all the eigenspaces, and we conclude that $hG-H$ is diagonalizable.

The eigenvalues for $(k-\lambda)h(hG-H)$ are thus

[TABLE]

with corresponding eigenspaces $E_{\mu^{\prime}}$, as given in (A), (B), (C).

Let $g_{1}=1,g_{2},\dots,g_{h}$ be coset representatives for $G/H$. Let $d_{i}=|D\cap Hg_{i}|$, so that $d_{1}=0$. Let $D=(D_{ij})$, where the blocks are of size $h\times h$ and are $\{0,1\}$ matrices.

Now from $DH=HD$ we see that $J_{h}D_{ij}=D_{ij}J_{h}$ for all $1\leq i,j\leq h$.

Lemma 3.1**.**

Let $A$ be an $h\times h$ matrix whose entries are $0,1$, and such that $J_{h}A=AJ_{h}$. Then every row and column of $A$ has the same number of $1$s in it.

Proof Note that the $k$th column of $J_{h}A$ is $u(1,1,\dots,1)^{T}$, where $u$ is the number of $1$s in the $k$th column of $A$. Similarly, the $k$th row of $AJ_{h}$ is $u(1,1,\dots,1)$, where $u$ is the number of $1$s in the $k$th row of $A$.

Let $a_{i},1\leq i\leq h,$ be the number of $1$s in the $i$th row of $A$. Then the $i$th row of $AJ_{h}$ is $(a_{i},a_{i},\dots,a_{i})$. Thus

[TABLE]

Let $b_{i},1\leq i\leq h$ be the number of $1$s in the $i$th column of $A$. Then the $i$th column of $AJ_{h}$ is $(b_{i},b_{i},\dots,b_{i})^{T}$, so that we have

[TABLE]

Since $AJ_{h}=J_{h}A$ we see from the first columns of these matrices that

[TABLE]

and from the first rows that

[TABLE]

But this shows that $a_{i}=a_{j}=b_{r}=b_{s}$ for all $1\leq i,j,r,s\leq h$, and the result follows. ∎

Thus from $HD_{ij}=D_{ij}H$ we see that each row and column of $D_{ij}$ has the same number of $1$s in it. Let this number be $d_{ij}$, so that $d_{ii}=0$. Thus $DH=HD=(d_{ij}J_{h})$.

Now $DD^{-1}=D^{-1}D$ with $D^{-1}=D^{T}$ shows that $D$ is a normal matrix. Clearly $H$ is a normal matrix. Thus we have

Lemma 3.2**.**

The matrices $D,H,G$ are commuting normal matrices and are simultaneously diagonalizable. ∎

In particular $DH$ is diagonalizable. Next: if $\alpha$ is an eigenvalue for $\Delta=DH$ with eigenvector $v$, then

[TABLE]

But $\Delta^{2}+h^{2}\Delta=(k-\lambda)h(hG-H)$, which shows that $v$ is also an eigenvector for $(k-\lambda)h(hG-H)$ with eigenvalue $\alpha^{2}+h^{2}\alpha$. However we know the eigenvalues and eigenvectors for $(k-\lambda)h(hG-H)$. Thus there are three cases:

(A) Here $\alpha^{2}+h^{2}\alpha=(k-\lambda)h^{2}(h^{2}-1)$, in which case we solve for $\alpha$:

[TABLE]

Here the eigenvector is $1_{h^{2}}$. Since $\Delta^{2}+h^{2}\Delta$ is a matrix with non-negative entries it follows that $\frac{1}{2}\left(-h^{3}-h^{2}\right)$ is not possible with this eigenvector. Thus we only have $\frac{1}{2}\left(h^{3}-h^{2}\right)$ as an eigenvalue in this case.

(B) Here $\alpha^{2}+h^{2}\alpha=-(k-\lambda)h^{2},$ so that $\alpha=-h^{2}/2$.

(C) Here $\alpha^{2}+h^{2}\alpha=0$, so that $\alpha=0,-h^{2}$.

Since $DH$ is diagonalizable the dimensions of the eigenspaces in cases (A), (B), (C) must be $1,h-1,h^{2}-h$ (respectively). In particular, each eigenvector for $hG-H$ as in (B) is also an eigenvector for $DH$. Thus we have

[TABLE]

which, since $J_{h}1_{h}=h1_{h}$, gives

[TABLE]

Similarly, using

[TABLE]

we obtain

[TABLE]

Continuing we see that

[TABLE]

This shows that $|D\cap gH|=\frac{h}{2}$ for all $g\notin H$, and so also gives

[TABLE]

Proposition 3.3**.**

Let $H\leq G,|H|=h,|G|=n,$ and order the elements of $G$ according to the cosets of $H$ as in the above. Represent elements of $G$ using the regular representation relative to this ordering. Then for $g\in G$ we write $g=(g_{ij})$, where each $g_{ij}$ is a $0,1$ matrix of size $h\times h$. Then $H\triangleleft G$ if and only if for all $g\in G$ and all $1\leq i,j\leq n/h$ each $g_{ij}$ is either the zero matrix or is a permutation matrix.

Proof We note that $H\triangleleft G$ if and only if for all $g\in G$ we have $Hg=gH$, where $H=\mathrm{diag}(J_{h},J_{h},\dots,J_{h})$.

Assume that $H\triangleleft G,g\in G,g=(g_{ij})_{1\leq i,j\leq h}$, where each $g_{ij}$ is an $h\times h$ matrix. Then $gH=Hg$ implies that $g_{ij}J_{h}=J_{h}g_{ij}$ for all $1\leq i,j\leq n/h$. By Lemma 3.1 this is true if and only if all the rows and columns of $g_{ij}$ have the same number of $1$s in them. Since each row and column of $g$ has exactly one $1$ in it (the rest of the entries being [math]) we see that if $g_{ij}\neq 0$, then each row and column of $g_{ij}$ has exactly one $1$ in it. Thus, for fixed $i,j$, no other $g_{ik},k\neq j,$ or $g_{kj},k\neq i,$ can be non-zero. In particular, each $g_{ij}$ is a permutation matrix.

Now assume that for all $g\in G$ and all $1\leq i,j\leq n/h$ each $g_{ij}$ is either the zero matrix or is a permutation matrix. We wish to show that $H\triangleleft G$ i.e. that $g_{ij}J_{h}=J_{h}g_{ij}$ for all $1\leq i,j\leq n/h.$ This is certainly true if $g_{ij}=0$, while if $g_{ij}$ is a permutation matrix, then $g_{ij}J_{h}=J_{h}=J_{h}g_{ij}$, and so we are done.∎

Let $D$ denote a difference set where $G=D\cup D^{-1}\cup H,H\leq G,D\cap H=D\cap D^{-1}=\emptyset$. We now wish to show that $H\triangleleft G$.

From the above we know that $|G|=h^{2},h=|H|$, where $h$ is even and $D=(d_{ij})$, where either $D_{ij}=0$ or $D_{ij}$ is a $0,1$ matrix that has $h/2$ $1$s in each row and column. We wish to show that $gH=Hg$ for all $g\in G$. This is certainly true if $g\in H$, so assume that $g\notin H$. Write $g=(g_{ij})$ as in the above. Since $g\notin H$ we either have $g\in D$ or $g\in D^{-1}$. Without loss of generality we can assume that $g\in D$. Now either $D_{ij}=0$ or $D_{ij}$ is a $0,1$ matrix that has $h/2$ $1$s in each row and column, so either $g_{ij}=0$ or $g_{ij}$ is a $0,1$ matrix that has one $1$ in each row and column. It follows that $g_{ij}J_{h}=J_{h}g_{ij}$, and so $H\triangleleft G$.∎

We have thus proved the following, which includes proofs of Theorem 1.1 (ii), (iii):

Theorem 3.4**.**

Let $D$ denote a difference set where $G=D\cup D^{-1}\cup H,H\leq G,D\cap H=D\cap D^{-1}=\emptyset$. Then

(i) $|G|=h^{2}$ where $h=|H|$ and $h$ is even;

(ii) $H\triangleleft G$;

(iii) each coset $Hg\neq H$ meets $D$ in $h/2$ points;

(iv) if we write $D=(D_{ij})$, with $h\times h$ blocks $D_{ij}$, then each $D_{ij}$ is either a zero matrix or is a $0,1$ matrix with $h/2$ $1$’s in each row and column. ∎

For Theorem 1.1 (iv) we note that if $g\in G$ is an involution that is not in $H$, then we must have $g\in D$ or $g\in D^{-1}$. In either case we have $g\in D\cap D^{-1}$, a contradiction.

For Theorem 1.1 (v) we show that $H\triangleleft G$ does not have a complement. So suppose that $L\leq G$ is a complement to $H$. Now since $L$ is a complement to $H$ we have $|L|=|G|/|H|=h$, which is even. Thus $L$ contains an involution that is not in $H$, a contradiction.

This now concludes the proof of Theorem 1.1. ∎

4. $H$ and subgroups of index $2$

We prove the following result, which gives Theorem 1.2 (i):

Theorem 4.1**.**

Let $G$ be a $(v,k,\lambda)_{0}$ relative skew Hadamard difference set group with difference set $D$ and subgroup $H$.

Then the subgroup $H$ is contained in every index $2$ subgroup of $G$. Further, $D$ meets each such subgroup in exactly $\lambda$ points.

If $G$ is a $2$-group, then $H$ is a subgroup of the Frattini subgroup of $G$.

Proof From Lemma 2.1 we know that $|G|=h^{2},k=\frac{h(h-1)}{2},\lambda=\frac{h(h-2)}{4}$. Let $M\leq G$ be a subgroup of index $2$ and let $\pi:G\to G/M=\langle t:t^{2}=1\rangle$ be the quotient map. Let

[TABLE]

so that

[TABLE]

Let $d_{2}=k-d_{1},h_{2}=h-h_{1}.$ Then we have the equations

[TABLE]

Now from equations (3.2) and (3.8) we deduce that

[TABLE]

Taking the image of this under $\pi$, and using the fact that $\pi(D)=d_{1}1+d_{2}t$, we obtain two equations (by looking at the coefficients of $1$ and $t$):

[TABLE]

Now $D+D^{-1}=G-H$ gives (by acting by $\pi$)

[TABLE]

which gives

[TABLE]

Solving equations (4.1), (4.2), (4.3) we find that

[TABLE]

Thus $D$ meets $M$ in $\lambda$ points, and all of $H$ is in $M$. Since this is true for any maximal subgroup $M$ we see that $H$ is contained in the Frattini subgroup if $G$ is a $2$-group, since every maximal subgroup of such a group has index $2$. ∎

5. Proof of the $p$ odd case

Let $N\triangleleft G$ be of index $p$, a prime. Let

[TABLE]

be the quotient map. Then we have

[TABLE]

We let

[TABLE]

where $x_{i},y_{i}\geq 0$ are integers, and

[TABLE]

We may represent elements of $Q$ as matrices where the generator $t$ corresponds to $P=\begin{pmatrix}0&1&0&0&\dots\\ 0&0&1&0&\dots\\ \vdots&\vdots&\vdots&\ddots&\vdots\\ 0&0&0&\dots&1\\ 1&0&0&\dots&0\end{pmatrix}$. Now we can simultaneously diagonalize $\pi(D),\pi(D^{-1}),$ $\pi(G),$ $\pi(H)$ using the matrix $R=(\zeta^{-(i-1)(j-1)})$, where $\zeta=\exp{2\pi i/p}$. We first note from [5, §3.2] that

[TABLE]

From this it follows that

[TABLE]

We also have

[TABLE]

From $H^{2}=hH$ we see that the minimal polynomial of $H$ is $x(x-h)$, and so the minimal polynomial of $\tilde{H}$ is a divisor of $x(x-h)$. In particular, the eigenvalues of $\tilde{H}$ are either [math] or $h$. Now we know that $\sum_{i=0}^{p-1}y_{i}=h$, and for $1\leq j\leq p-1$ we must also have

[TABLE]

We rewrite equations (5.2) in matrix form as

[TABLE]

Let $T$ denote the matrix in (5.3). Then $T^{-1}=\frac{1}{p}T^{*}$. Thus

[TABLE]

Now, since $p$ is a prime, the minimal polynomial for $\zeta$ is $\sum_{i=0}^{p-1}x^{i}$. We also have $y_{0},\dots,y_{p-1}\in\mathbb{Z}$. Thus each choice of [math] or $h$ in the above equation must be the same. Call this choice $\eta$. Thus we have two possibilities:

(a) $\eta=h$ so that $y_{0}=h,y_{1}=y_{2}=\dots=y_{p-1}=0$, or

(b) $\eta=0$, so that $y_{0}=y_{1}=y_{2}=\dots=y_{p-1}=\frac{h}{p}$.

If we have (a), then $\tilde{H}=hI_{p}$, so that the equation $\tilde{D}\tilde{H}=\frac{h}{2}(\tilde{G}-\tilde{H})$ gives

[TABLE]

Using the expression for $\tilde{D}$ given in equation (5.1) we see that the above is expressible as a matrix equation:

[TABLE]

As before, if we denote the matrix in (5.4) by $U$, then $U^{-1}=\frac{1}{p}U^{*}$, and we obtain

[TABLE]

This concludes consideration of (a).

We now show that (b) is not possible. So assume (b). Then as noted we get $y_{0}=y_{1}=y_{2}=\dots=y_{p-1}=h/p.$ Thus $H=\frac{h}{p}J_{p}$, so that $\tilde{H}=\begin{pmatrix}h&0&0&\dots&0\\ 0&0&0&\dots&0\\ \vdots&\vdots&\vdots&\dots&\vdots\\ 0&0&0&\dots&0\end{pmatrix}.$

Now from $\tilde{D}\tilde{D}^{-1}=\lambda\tilde{G}+(k-\lambda)$ we get

[TABLE]

Now the $x_{i}\in\mathbb{Z}$ and so from the (2,2) entry of (5.5) we see that (taking the rational part) we have

[TABLE]

Using $\sum_{i=0}^{p-1}x_{i}=k$, the fact that $x_{i}\in\mathbb{Z},$ and (5.6) gives

[TABLE]

so that $\lambda\leq 0,$ a contradiction. This proves Theorem 1.2 (a) (i) and (ii). ∎

Since we only have (a) we see that $|N\cap D|=\frac{1}{2p}h(h-p)$, so that in the case $p=2$ we have $|N\cap D|=\frac{1}{4}h(h-4)=\lambda,$ which gives Theorem 1.2 (b).∎

6. The Schur ring and minimal polynomials

We have $(G-H)^{-1}=G-H,(H-1)^{-1}=H-1,(D^{-1})^{-1}=D$, and so we just need to show that $D,D^{-1},H-1,1$ commute and span the ring that they generate. We have already seen in Lemma 3.2 that they commute.

We have $H\cdot G=hG,D\cdot G=kG=D^{-1}\cdot G.$

Using equations (3.2) and (3.8) we get

[TABLE]

We collect together the rest of the products that we need:

[TABLE]

Since $k=h(h-1)/2,\lambda=h(h-2)/4,k-\lambda=h^{2}/4\in\mathbb{Z},$ one can check that all the coefficients in the above sums are non-negative integers. This proves that $D,D^{-1},H-1,1$ commute and span the ring that they generate. Theorem 1.3 follows. ∎

One can show that

[TABLE]

For a matrix or an element $M$ of an algebra we let $\mu(M)$ denote the minimal polynomial of $M$.

To help us find $\mu(D)$ we have the equations

[TABLE]

Thinking of $D,D^{-1},G,H$ as variables in a polynomial ring over $\mathbb{Q}(h)$ we can consider the ideal generated by the above elements. This has dimension zero, and calculating a Gröbner basis for this ideal and doing an elimination, one finds that $D$ satisfies the polynomial $(x-k)\left(x+\frac{h}{2}\right)\left(x^{2}+\frac{h^{2}}{4}\right)$. Thus $\mu(D)$ divides this polynomial.

We note that $\frac{1}{k}D$ is a stochastic matrix, and since $D^{2}=(k-\lambda)(G-1)-\frac{h}{2}(G-H)$ it follows that

Lemma 6.1**.**

The matrix $\frac{1}{k}D$ is an irreducible doubly stochastic matrix.∎

Further, we know that $\mu(D)$ factors as a product of distinct linear factors $(x-\kappa)$, where $\kappa$ is an eigenvalue (since $D$ is diagonalizable by Lemma 3.2).

Next we note that $k$ is an eigenvalue of $D$, since each row sum and column sum of $D$ is $k$. Next we show that $-h/2$ is an eigenvalue of $D$: for $g\notin H$ we have $H-Hg\neq 0$ and

[TABLE]

Thus $-\frac{h}{2}$ is an eigenvalue for $D$.

Since $D$ is a matrix with real entries it follows that the eigenspaces for eigenvalues $\pm ih/2$ have the same dimension, and that either $\mu(D)=(x-k)(x+h/2)$ or $\mu(D)=(x-k)(x+h/2)(x^{2}+\frac{h^{2}}{4}).$ If $\mu(D)=(x-k)(x+h/2)$, then, since $D$ is diagonalizable, Lemma 6.1 and the Perron Frobenius theorem show that $D$ has eigenvalue $k$ with multiplicity one, and $-h/2$ with multiplicity $h^{2}-1$. Now, since $D\cap H=\emptyset,$ we see that $D$ has trace zero. Thus we must have

[TABLE]

but the lefthand side of this expression is $-h^{2}(h-1)$, which gives a contradiction. Thus $\mu(D)=(x-k)(x+h/2)(x^{2}+\frac{h^{2}}{4}).$ In fact it easily follows from Trace$(D)=0$ that the eigenvalue $-h/2$ has multiplicity $h-1$. ∎

This gives a proof of Theorem 1.4.∎

Using the above ideal it is easy to show that $G\setminus D$ is another difference set, where

[TABLE]

Further, we have

[TABLE]

as required for a part of Theorem 1.7. As we have seen $D,G$ are simultaneously diagonalizable, and $2D-G$ is similar to

[TABLE]

and it follows that $\mu(2D-G)=(x+h)(x^{2}+h^{2})$. This gives Theorem 1.7.∎

7. Examples of relative skew Hadamard difference set groups

The groups $\mathfrak{G}_{n,k}$ have been defined in the introduction. We now show that they are relative skew Hadamard difference set groups. We let

[TABLE]

Then a transversal for $H$ in $G$ is the set of products $a_{i_{1}}a_{i_{2}}\cdots a_{i_{u}}$, where these are indexed by the sequences $i_{1}<i_{2}<\cdots<i_{u}$ of $1,2,\dots,n$, or in other words, indexed by the subsets $X=\{i_{1},i_{1},\dots,i_{u}\}$ of $\{1,2,\dots,n\}$. We let $a_{X}=a_{i_{1}}a_{i_{2}}\cdots a_{i_{u}}$ denote the corresponding element of $G$. Here $a_{\emptyset}=1$. We may also employ a similar notation for the elements $b_{X}=b_{i_{1}}b_{i_{2}}\cdots b_{i_{u}}$.

We note that for any $g\in G$ we have $g^{2}\in H$. We are interested in the hypothesis

(H1): there is a set of distinct maximal subgroups $M_{1},\dots,M_{2^{n}-1}$ of $H$, and an ordering $S_{1},\dots,S_{2^{n}-1}$ of the non-empty subsets of $\{1,\dots,n\}$ so that $a_{S_{i}}^{2}\notin M_{i}$.

Proposition 7.1**.**

The groups $\mathfrak{G}_{n,k}$ satisfy (H1).

Proof We first show that the squares of the coset representatives $a_{S},S\subseteq\{1,2,\dots,n\}$, are distinct.

We note that the subgroup $J=\langle a_{2},a_{3},\dots,a_{n}\rangle$ is isomorphic to $\mathcal{C}_{4}^{n-1}$. We also have $J\triangleleft\mathfrak{G}_{n,k}$, so that $\mathfrak{G}_{n,k}=J\rtimes\langle a_{1}\rangle=J\rtimes\mathcal{C}_{4}$.

If $S\subseteq\{1,2,\dots,n\}$ and $m\in\mathbb{Z}$ we let $S+m$ be the set $\{u+m:u\in S\}$, where we take numbers mod $n$ so that $S+m\subseteq\{1,2,\dots,n\}$.

Now for a coset representative $a_{S},S=\{i_{1},i_{2},\dots,i_{u}\}\subseteq\{2,\dots,n\},$ we have $a_{S}\in J$ and so from the relations in $\mathfrak{G}_{n,k}$ we have

[TABLE]

We note that in this situation, since $1\notin S$, we have $1+k\notin S+k$.

Now for a coset representative $a_{S}$ that is not in $J$ we can write $S=\{1,i_{1},i_{2},\dots,i_{u}\}$, where $a_{S\setminus\{1\}}\in J$. So if we let $K=S\setminus\{1\}$, then $a_{S}=a_{1}a_{K}$.

Now write $K=K_{1}\cup K_{2}$, where the elements $a_{m},m\in K_{2},$ commute with $a_{1}$, and those $a_{m},m\in K_{1},$ do not. Note that

[TABLE]

Then from the relations in $\mathfrak{G}_{n,k}$ we have:

[TABLE]

Thus we have

[TABLE]

We next show that $b_{1+k}$ has non-zero exponent in (7.1). But from the above we know that $K_{1}\subseteq\{2,3,\dots,k+1\},$ so that $1+k\notin K_{1}-1.$ If $1+k\in K_{i}+k,i=1,2,$ then $1\in K_{i},$ a contradiction. This shows that $b_{1+k}$ has non-zero exponent in (7.1).

Note that in the above we have also shown (i) of

Lemma 7.2**.**

With the above definitions we have:

(i) the element $b_{1+k}$ occurs with non-zero coefficient in $a_{S}^{2}$ if and only if $1\in S$.

(ii) The squares of the coset representatives $a_{S},S\subseteq\{1,2,\dots,n\}$, where $1\in S,$ are distinct.

Proof (ii) We need to show that the map $S\mapsto b_{K_{1}-1}b_{S+k}$ is injective.

We represent $S$ as a (column) vector $v_{S}\in V=\mathbb{F}_{2}^{n}$, where the $i$th coordinate of $v_{S}$ is $1$ if and only if $i\in S$. Then the action on $V$ of replacing $S$ by $S+1$ is determined by the $n\times n$ permutation matrix

[TABLE]

Thus for any $i\in\mathbb{Z}$ we have

[TABLE]

Let $0_{k,m}$ denote the $k\times m$ zero matrix, and let $0_{k}=0_{k,k}$. If $k\leq 0$ or $m\leq 0$, then $0_{k,m}$ will denote the empty matrix. Then, the map $S\mapsto K_{1}$, is determined by the $n\times n$ matrix

[TABLE]

so that

[TABLE]

Thus the map $S\mapsto b_{K_{1}-1}b_{S+k}$ is represented by the matrix $P^{-1}A+P^{k},$ and we will be done if we can show that $P^{-1}A+P^{k}$ is a non-singular matrix in GL$(2,\mathbb{F}_{2})$. But this is the same as showing that $A+P^{k+1}$ is non-singular, where

[TABLE]

We note that since $k<n-1$ the second matrix is not a diagonal matrix, and that the submatrix $I_{k}$ in the second matrix of equation (7.2) occurs to the right of the diagonal. (This shows that $A+P^{k+1}$ is singular when $k=n-1$.) Thus the $I_{k}$ in the second matrix of equation (7.2) can be used to column-reduce the $I_{k}$ in the first matrix to zero. This shows that $A+P^{k+1}$ column-reduces to $P^{k+1}$, which is non-singular, and we are done.∎

Let $V^{\times}=\mathbb{F}_{2}^{n}\setminus\{0\}$. Then non-empty subsets of $S$ correspond bijectively to elements of $V^{\times}$, as explained above. Further, maximal subgroups of $H$ correspond to subspaces of $V$ of dimension $n-1$, which, in turn, are determined by elements of $V^{\times}$: a vector $v\in V^{\times}$ determines the subspace $M_{v}=\{u\in V|u\cdot v=0\}$, where $\cdot$ is the usual dot-product on $V$ taking values in $\mathbb{F}_{2}$. Since $V$ is a vector space over $\mathbb{F}_{2}$ the correspondence $v\leftrightarrow M_{v}$ is bijective. Further, given a maximal subgroup (or subspace) $M$ we let $v_{M}$ denote the corresponding vector.

Thus the correspondence of subsets with maximal subgroups that we require is $S\leftrightarrow M_{S}$ where $v_{S}\leftrightarrow v_{M_{S}}$, with $v_{S}\notin M_{S}$ i.e. $v_{S}\cdot v_{M_{S}}=1$. But this correspondence determines a function

[TABLE]

Conversely, such a function determines the correspondence that we want. We now show how to construct such a function:

Lemma 7.3**.**

For all $n\in\mathbb{N},V=\mathbb{F}_{2}^{n},$ there is a function $\mu:V^{\times}\to V^{\times}$ such that $u\cdot{\mu(u)}=1$ for all $u\in V^{\times}.$

Proof We will show that there is a function $\mu$ that is an involution i.e. where we have $\mu(\mu(v))=v$ for all $v\in V^{\times}$. For $0\leq k\leq n$ we let

[TABLE]

where there are $k$ $1$s (so for $k=0$ we have the zero vector of $V$).

Write $v\in V^{\times}$ as $v=(v_{1},v_{2},\dots,v_{n}),v_{i}\in\mathbb{F}_{2}$. If $1\leq k\leq n$ where $v_{k}=1$ and $v_{m}=0$ for $k+1\leq m\leq n$, then we let

[TABLE]

This satisfies $\mu(v)\cdot v=1$, as required. Further, since the same $k$ works for $\mu(v)$, we have

[TABLE]

This defines a function $\mu$ that is an involution.∎

Lemma 7.3 determines the pairing for hypothesis (H1) for the groups $\mathfrak{G}_{n,k}$, and concludes the proof of Proposition 7.1.∎

We will next show

Proposition 7.4**.**

The groups $\mathfrak{G}_{n,k}$ are relative skew Hadamard difference set groups.

Proof We first note that since $b_{1},\dots,b_{n}$ are central idempotents, all the maximal subgroups of $H$ are normal subgroups of $G$.

As usual, subsets $S$ of $G$ will correspond to elements $\sum_{s\in S}s$, of the group algebra. We define $D$ as follows:

[TABLE]

Let $a_{i}=a_{S_{i}}$. We first show that $(a_{i}M_{i})^{-1}=a_{i}(H-M_{i}).$ But this is true if and only if $a_{i}^{-1}M_{i}=a_{i}(H-M_{i})$ if and only if $M_{i}=a_{i}^{2}(H-M_{i})$ if and only if $M_{i}=H-a_{i}^{2}M_{i}$. But this latter equation is true since $a_{i}^{2}\in H$ and $a_{i}^{2}\notin M_{i}$.

Thus we have:

[TABLE]

Let $1\leq i\neq j\leq 2^{n}-1$; then, since $M_{i},M_{j}$ are distinct maximal subgroups of $H\cong\mathcal{C}_{2}^{n}$, we have

[TABLE]

so that for $1\leq i\neq j\leq 2^{n}-1$ we have

[TABLE]

We use this to obtain:

[TABLE]

Since $|\mathfrak{G}_{n,k}|=2^{2n},h=|H|=2^{n}$ we have

[TABLE]

Returning to equation (7.3), in particular looking at the first sum of equation (7.3), we see that every non-trivial coset of $H$ occurs $2^{n}-2$ times in equation (7.3). Thus from equation (7.3) we see that the coefficient in $DD^{-1}$ of each element of that coset is

[TABLE]

as we desire.

The second sum of equation (7.3) gives the contributions to the trivial $H$-coset. We rewrite it as

[TABLE]

But we are assuming that $a_{S_{i}}^{2}\notin M_{i}$, so we must have $H-a_{S_{i}}^{2}M_{i}=M_{i}.$ Thus equation (7.4) is

[TABLE]

Now since the $M_{i}$ are distinct maximal subgroups, and there are $2^{n}-1$ of them, we see that every maximal subgroup of $H\cong\mathcal{C}_{2}^{n}$ is in the list $M_{1},\dots,M_{2^{n}-1}$, and so one has

[TABLE]

Thus if $h^{\prime}\in H,h^{\prime}\neq 1,$ then the coefficient of $h^{\prime}$ in equation equation (7.5) is

[TABLE]

as required.

The coefficient of $1$ in $D\cdot D^{-1}$ is then

[TABLE]

which is equal to $k$, as required. Thus we have $D\cdot D^{-1}=\lambda(G-1)+k\cdot 1.$ ∎

8. Another criterion for the existence of a relative skew Hadamard difference set group

Theorem 8.1**.**

Given $H\lhd G$, $H\cong\mathcal{C}_{2}^{h},[G:H]=h,$ let $t_{1}=1,t_{2},\dots,t_{h}$ be a transversal for $H$ in $G$. Then for each $1\leq i\leq h$ there is a $j=j(i)$ such that $t_{i}^{-1}\in Ht_{j}$. Suppose there exist distinct maximal subgroups $H_{2},\dots,H_{h}$ of $H$ such that

(1) $(H_{j})^{t_{i}^{-1}}=H_{i},$ and

(2) $t_{i}t_{j}\in H_{i}$,

for all $2\leq i\leq h$ where $j=j(i)$. Then $D=\sum_{i=2}^{h}H_{i}t_{i}$ is a difference set such that $D=D^{-1}$.

Proof.

We note that $H_{2},\dots,H_{h}$ are all the maximal subgroups of $H$. First, to show that $D=D^{-1}$, we show that the identity occurs $k$ times in the product $D^{2}$:

[TABLE]

Fix $2\leq i\leq h$ and let $j=j(i)$. If $\ell\neq j$, then $Ht_{i}t_{\ell}\neq H$ so that $1\notin H_{i}t_{i}H_{\ell}t_{\ell}$. If $\ell=j$, then using (1) and (2) we have

[TABLE]

There are $h-1$ values of $i$, so the identity occurs $\frac{h}{2}(h-1)=k$ times in $D^{2}$. Thus, $D=D^{-1}$.

Now we will show that $D\cdot D^{-1}=k+\lambda(G-1)$. We have just shown that the identity occurs $k$ times in $D\cdot D^{-1}=D^{2}$. Consider the product

[TABLE]

When $t_{i}t_{j}\in H$, we have, as above,

[TABLE]

Each element of $H$ that is not the identity occurs in $\frac{h-2}{2}$ of the $h-1$ maximal subgroups $H_{i},2\leq i\leq h,$ of $H\cong\mathcal{C}_{2}^{n}$. Thus, the coefficient, in $D^{2}$, of each element of $H$ that isn’t the identity is $\frac{h}{2}(\frac{h-2}{2})=\lambda$.

Now, when $t_{i}t_{\ell}\notin H$, we have,

[TABLE]

We observe that conjugation by $t_{i}^{-1}$ permutes $H_{2},H_{3},\dots,H_{h}$. Thus, since $t_{i}H_{j}t_{i}^{-1}=H_{i}$, and $j(i)\neq\ell$, we have that $t_{i}H_{\ell}t_{i}^{-1}\neq H_{i}$, so

[TABLE]

For $g\notin H$, each non-trivial coset $Hg$ occurs $h-2$ times in the form $Ht_{i}t_{\ell}$. So $g$ will occur $\frac{h}{4}(h-2)=\lambda$ times in $D^{2}$. Thus

[TABLE]

which shows that $D$ is a difference set. ∎

Remark A computer calculation [1] shows that, of the $267$ groups of order $64$, at least $50$ of them satisfy the hypotheses of Theorem 8.1. These include groups with nilpotency classes $1,2,3,4$. There are also four groups of order $16$ that satisfy the hypotheses of Theorem 8.1.

Proposition 8.2**.**

The group $G=\mathcal{C}_{4}^{n}$ satisfies the hypotheses of Theorem 8.1.

Proof Let $a_{1},a_{2},\dots,a_{n}\in G$ be the generators of order $4$, and let $b_{i}=a_{i}^{2},i=1,\dots,n$. Let $H=\langle b_{1},b_{2},\dots,b_{n}\rangle$. Let $t_{0}=1,t_{1},t_{2},\dots,t_{2^{n}-1}$ be a transversal for $H$. We clearly have hypothesis (1) of Theorem 8.1, since $G$ is abelian and $j(i)=i$. Thus we must order the maximal subgroups $M_{1},M_{2},\dots,M_{2^{n}-1}$ of $H$ so that we have (2): $t_{i}^{2}\in M_{i}$ for $1\leq i\leq 2^{n}-1$. Note that $t_{1}^{2},t_{2}^{2},\dots,t_{2^{n}-1}^{2}$ are all the non-trivial elements of $H$.

Now we think of $H$ as the additive group $V=\mathbb{F}_{2}^{n}$, so that maximal subgroups correspond to maximal subspaces of $V$. Further, each such maximal subspace $M$ is determined by a vector that we denote $v_{M}$, so that we have $M=\{v\in V:v\cdot v_{M}=0\}$. Here $\cdot$ denotes the standard dot product on $V$, taking values in $\mathbb{F}_{2}$.

Each element $t_{i}^{2}$ is then a vector that we denote by $w_{i}$, and we need to find a correspondence $w_{i}\leftrightarrow v_{M_{i}}$ where $w_{i}\in M_{i}$, i.e. where $v_{i}\cdot v_{M_{i}}=0$ This is thus equivalent to finding a bijection $\kappa:V^{\times}\to V^{\times}$ such that $v\cdot\kappa(v)=0$ for all $v\in V,v\neq 0$. We show how to do this. In fact we will define $\kappa$ to be an involution.

Let $(1)\in V$ denote the all $1$ vector.

Assume first that $n$ is even, and let

[TABLE]

Since $n$ is even we have $(1)\cdot(1)=0$, while if $v\neq(1)$, then $v\cdot\kappa(v)=0$. This does the $n$ even case.

Now assume that $n$ is odd. Let

[TABLE]

Then for $v\in V\setminus S$ we let $\kappa(v)=(1)-v,$ noting that $(1)-v\in V\setminus S$. We define $\kappa$ on the elements of $S$ as follows:

[TABLE]

This defines $\kappa$ (as an involution) and one checks that, since $n$ is odd, this gives $v\cdot\kappa(v)=0$ for a $v\neq 0$.∎

9. Normal subgroups

Suppose that $G$ is a relative skew Hadamard difference set with difference set $D$ and subgroup $H,h=|H|,h^{2}=|G|$.

Let $N\leq G,n=|N|,$ be normal, with $\pi:G\to K=G/N$ the quotient map. Let $g_{0}=1,g_{1},\dots,g_{u-1}$ be coset representatives for $K=G/N,u=|K|$. Then $\pi(G)=\frac{h^{2}}{u}K$.

Let

[TABLE]

so that we can write

[TABLE]

It follows that

[TABLE]

Then from $D\cdot D^{-1}=\lambda G+(k-\lambda)1$ we get

[TABLE]

We also have

[TABLE]

From $DH=\frac{h}{2}(G-H)$ we obtain

[TABLE]

Lastly, from $G=D+D^{-1}+H$ we get

[TABLE]

Each of equations (9.1)-(9.4) determines $u=\frac{h^{2}}{n}$ elements of $R=\mathbb{Q}[x_{1},\dots,x_{u},$ $y_{1},\dots,y_{u}]$, one for each element of $K$.

Let $\mathcal{I}_{K}\subseteq R$ denote the ideal generated by the elements in equations (9.1)-(9.4).

Example 9.1: $K=\mathbb{C}_{2}^{2}$. So suppose that $K=\mathcal{C}_{2}^{2}$. Then finding a Gröbner basis for $\mathcal{I}_{K}$ gives

[TABLE]

Since $y_{0}=h$ this shows that $H\leq N$.

A slightly more involved example is given by

Example 9.2: $K=\mathcal{C}_{4}$. So suppose that $K=\mathcal{C}_{4}=\langle t\rangle,$ where the ordering of the elements is $g_{1}=1,g_{2}=t,g_{3}=t^{2},g_{4}=t^{3}$. Then finding a Gröbner basis for $\mathcal{I}_{K}$ gives:

[TABLE]

This ideal has dimension zero. The fifth element of this basis factors as

[TABLE]

and the eighth as $y_{2}\left(y_{2}-\frac{1}{2}h\right)$. By an elimination process one finds that the variable $x_{3}$ satisfies

[TABLE]

Thus one of the factors in each of these three products must be zero. This gives $12$ cases to check, and a quick analysis shows that we have one of

[TABLE]

Example 9.3: $K=\mathcal{C}_{6}$: For $G=\mathcal{C}_{6}=\langle t\rangle$ a similar argument shows that we have

[TABLE]

Since $y_{0}=h$ we have $H\leq N$.

Example 9.4: $K=\mathcal{C}_{9}=\langle t\rangle$: One further idea occurs in this case: Calculating $\mathcal{I}_{K}$ and doing an elimination shows that the variable $y_{1}$ (corresponding to the element $t$) satisfies

[TABLE]

Here $\zeta=\zeta_{36}$ is a primitive $36$th root of unity. Thus we see that $y_{1}$ has to be a root of one of these linear factors. However we know that $y_{1}\in\mathbb{Z}$, so that the only possibilities are $y_{1}=0,y_{1}=\frac{h}{9}$, thus greatly restricting the number of cases that we have to consider. One can apply this idea to many of the variables $x_{i},y_{j}$ so as to reduce the number of cases that one has to consider. In this way we conclude that $y_{0}=h$.

One can use the above techniques to show:

Proposition 9.1**.**

If $N\triangleleft G$, where $G/N$ is one of

[TABLE]

then $H\leq N$. ∎

10. Representations

Suppose that $G$ is a relative skew Hadamard difference set with difference set $D$ and subgroup $H,h=|H|$. We recall that $D,D^{-1},G,H$ satisfy the equations

[TABLE]

Let $\rho$ be a non-principal irreducible representation of $G$ with irreducible character $\chi$ and $d=\chi(1)$. We assume that $\rho$ is unitary. Since $\chi$ is not principal we see from orthogonality of the character table that

[TABLE]

Since $\rho$ is unitary we see that $\rho(D^{-1})=D^{*}$. Now we know from Lemma 3.2 that $D,D^{-1},G,H$ pairwise commute, and so $\{\rho(D),\rho(D^{-1}),\rho(H),\rho(G)\}$ is a set of commuting normal matrices. Thus they are simultaneously diagonalizable, and we may assume that in fact they are diagonal matrices.

Since $H\triangleleft G$ and $\rho$ is irreducible it follows that $\rho(H)$ is a scalar matrix, which we write as

[TABLE]

Since $H^{2}=hH$ we have $\rho(H)^{2}=h\rho(H)$, which shows that either $\rho(H)=0$ or $\rho(H)=h$; i.e. $h_{0}\in\{0,h\}$. From (10.1) and (10.2) we see that

[TABLE]

CASE 1: If $h_{0}=0$, then these give (where $i^{2}=-1$)

[TABLE]

Here $\varepsilon_{i}\in\{-1,1\}$. In this case $\mu(\rho(D))$ divides $x^{2}+\frac{h^{2}}{4}.$

CASE 2: If $h_{0}=h$, then (10.3) gives

[TABLE]

But then we have $\rho(D^{-1})=D^{*}=D$. In this case $\mu(\rho(D))=x+\frac{h}{2}.$ This gives Theorem 1.5.∎

11. Some tests for relative skew Hadamard difference set groups

We will say that a group $K$ is $H$-swallowing, if whenever we have a relative skew Hadamard difference set group $G$ with subgroup $H$ and there is a normal subgroup $N$ of $G$ with $G/N\cong K$, then $H\leq N$.

For a group $G$ let $\mathcal{N}(G)$ denote the intersection of all normal subgroups of $G$ that have prime index or whose quotients are $H$-swallowing. Thus if $G$ is a $p$-group, then $\mathcal{N}(G)$ is just the Frattini subgroup of $G$. Then by Theorem 1.2 we see that $H\leq\mathcal{N}(G)$. Thus we have our first test:

(T1) $|\mathcal{N}(G)|\equiv 0\operatorname{mod}h$.

Other tests that we can use are (see Theorem 1.1):

(T2) $H$ contains the subgroup generated by all the involutions.

(T3) $G$ contains a normal subgroup of order $h$ and index $h$.

(T4) $H$ does not have a complement in $G$.

Using tests (T1), (T2), (T3), (T4) enables us to eliminate various groups from being relative skew Hadamard difference sets groups:

The case $h=6$ Of the $14$ groups of order $36$, only the group $G_{36,1}$ passes the above tests. One can show by a short computer calculation that $G_{36,1}$ is not a relative skew Hadamard difference set group. Here

[TABLE]

This gives:

Theorem 11.1**.**

For $h=6$ there are no relative skew Hadamard difference set groups. ∎

12. Can we have $m>0$?

Let $G$ be a relative skew Hadamard difference set group with subgroup $H$ and difference set $D$. Assume, as in $\S 1$, that (1) $D\cap D^{-1}=Hg_{1}\cup\dots\cup Hg_{m}$; (2) $G\setminus(D\cup D^{-1})=H\cup Hg_{1}^{\prime}\cup\dots\cup Hg_{m}^{\prime}$.

We note that $0\leq m\leq(h-1)/2,$ but we can do better:

Lemma 12.1**.**

$m\leq\frac{h-1}{4}.$**

Proof Let $\pi:G\to Q=G/H$ be the quotient map. Then $|Q|=h,\pi(G)=hQ,\pi(H)=h$. Let $Q=\{q_{0},q_{1},\dots,q_{h-1}\}$ and put

[TABLE]

Now we know that $\sum_{i=0}^{h-1}x_{i}=k$ and by Theorem 7.1 of [10] we have:

[TABLE]

But $D$ contains $m$ cosets of $H$ and so $\sum_{i=0}^{h-1}x_{i}^{2}\geq mh^{2}.$ This gives the result.∎

Corollary 12.2**.**

For $h=4$ we have $m=0$ and for $h=6,8$ we can only have $m=0,1$.∎

13. Open questions

1. Are there relative skew Hadamard difference set groups that are not $2$-groups.

2. Are there relative skew Hadamard difference set groups with $m>0$?