Difference Sets Disjoint from a Subgroup III: The Skew Relative Cases

Abstract

We study finite groups G having a subgroup H and $D \subset G {\setminus } H$ such that (i) the multiset $\{ xy^{-1}:x,y \in D\}$ has every element that is not in H occur the same number of times (such a D is called a relative difference set); (ii) $G=D\cup D^{(-1)} \cup H$; (iii) $D \cap D^{(-1)} =\emptyset$. We show that $|H|=2$, that H is central and that G is a group with a single involution. We also show that G cannot be abelian. We give infinitely many examples of such groups, including certain dicyclic groups, by using results of Schmidt and Ito.

1 Introduction

Here G will always be a finite group. We identify $X \subseteq G$ with the element $\sum _{x \in X} x \in {\mathbb {Q}} G$, and let $X^{(-1)}=\{x^{-1}:x \in X\}$. We write ${\mathcal {C}}_n$ for the cyclic group of order n. Let $H\le G$ and $h=|H|$. Then a $(v,k,\lambda )$-relative difference set (relative to H) is a subset $D \subset G{\setminus } H, |D|=k, v=|G|$, such that $DD^{(-1)}=\lambda (G-H)+k$, so that $g \in G{\setminus } H$ occurs $\lambda$ times in the multiset $\{xy^{-1}:x,y \in D\}$.

We now further assume

• (1) $D \cap D^{(-1)}=\emptyset$;

• (2) $G = D \cup D^{(-1)} \cup H$ (disjoint union).

A group having a difference set of the above type will be called a $(v,k,\lambda )$-skew relative Hadamard difference set group (with difference set D and subgroup H); or a $(v,k,\lambda )$-SRHDS group. Recall the following related concept: a group G is 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 [1,2,3,4,5,6,7,8].

In this paper we find infinitely many examples of such SRHDS groups. We also find groups that cannot be SRHDS groups, but which satisfy certain properties of a SRHDS group, as given in:

Theorem 1.1

For a $(v,k,\lambda )$ SRHDS group G with difference set D and subgroup H we have:

• (i) $|H|=2$;

• (ii) $H \triangleleft G$;

• (iii) G is a group having a single involution;

• (iv) $v \equiv 0 {\text {mod}}8$;

• (v) G is not abelian.

• (vi) A Sylow 2-subgroup is a generalized quaternion group.

For part (vi), suppose that G is a finite group with a unique involution. Then a Sylow 2-subgroup of G also has a unique involution. Now 2-groups with unique involution were determined by Burnside (see [9, Theorems 6.11, 6.12] and [10, 11]); they are cyclic or generalized quaternion groups. Corollary 4.5 shows they cannot be cyclic.

Groups with a single involution are studied in [12,13,14]. Dicyclic groups $\textrm{Dic}_{ {v}}$ are examples of such groups and we show that each $\textrm{Dic}_{8p}, 1\le p<9$ is an SRHDS group. However, we show that $\textrm{Dic}_{72}$ has no SRHDS (Proposition 8.1).

We now establish a connection between SRHDS groups and Hadamard groups. Recall that a Hadamard group is a group G containing $H\le Z(G)$ of order 2 such that there is an H-transversal $D, |D|=v/2,$ that is a relative difference set relative to H (so that $DD^{(-1)}=\lambda (G-H)+|D|$ and $HD=G$).

We show that if $D \subset G$ is a SRHDS, then G is also a Hadamard group (where $E=D+1$ is the relative difference set); see Proposition 2.5. Thus it is natural to try to obtain results for SRHDS groups that are similar to the results of Schmidt and Ito [15, 16] from the Hadamard group situation. For example Schmidt and Ito show that if $4p-1$ or $2p-1$ is a prime power, then the groups $\textrm{Dic}_{8p}$ or $\textrm{Dic}_{4p}$ (respectively) are Hadamard groups. For dicyclic SRHDS groups we show:

Theorem 1.2

If $p \in {\mathbb {N}}$ and $4p-1$ is a prime power, then $\textrm{Dic}_{8p}$ is a SRHDS group.

There is no analogous result when $2p-1$ is prime. Now Ito [16] determines a ‘doubling process’ that takes a Hadamard difference set for $\textrm{Dic}_{ {v}}$ and produces a Hadamard difference set for $\textrm{Dic}_{2 {v}}$. For us this doubling process gives:

Theorem 1.3

If $p \in {\mathbb {N}}$ and $4p-1$ is a prime power, then $\textrm{Dic}_{16p}$ is a SRHDS group.

We note that this doubling process does not work in general in the context of a SRHDS, however in our next paper we will show that it does work for a SRHDS under an additional hypothesis that we call doubly symmetric that is satisfied in the situation of Theorem 1.2, so that in this case we obtain an SRHDS in $\textrm{Dic}_{16p}$. This will allow us to prove, in the next paper, among other things:

Theorem 1.4

Let $G=\mathrm Dic_{8\cdot 2^u}$ be a generalized quaternion group for some $u\in {\mathbb {Z}}_{\ge 0}$. Then G contains a doubly symmetric SRHDS if and only if $2^{u+1}-1$ is either prime or 1.

Lastly, the following is a consequence of Proposition 8.2.

Theorem 1.5

Let $G = C_p \times \textrm{Dic}_{8n}$ with $p>2$ prime and n odd. Then G is not a SRHDS group.

2 $|H|=2$ and Normality of H

Recall that for $p \ge 2$ the dicyclic group of order 4p is

$$\begin{aligned} \textrm{Dic}_{4p}=\langle x,y|x^{2p}=y^2,y^4=1,x^y=x^{-1}\rangle . \end{aligned}$$

A generalized quaternion group, $Q_{2^a}$, is the dicyclic group $\textrm{Dic}_{2^a}$, $a\ge 3$.

Proposition 2.1

Let G be a SRHDS group with subgroup H. Then G has a single involution t, and $H=\langle t\rangle$. In particular $h=2, H\le Z(G)$ and $H \triangleleft G$.

Proof

Let $D\subset G$ be a SRHDS. Now D has no involutions since $D \cap D^{(-1)}=\emptyset$. Since $G-(D+D^{(-1)})=H$ all involutions are contained in H.

If $d_i \in D, h_i \in H,i=1,2$, with $h_1d_1=h_2d_2\in Hd_1 \cap Hd_2,$ then $h_2^{-1}h_1=d_2d_1^{-1} \in H$, so that $h_2^{-1}h_1=d_2d_1^{-1} =1$ (since $DD^{(-1)}=\lambda (G-H)+k$ implies that the only element of H of the form $d_2d_1^{-1}$ is 1). Thus $d_1=d_2$ and $h_1=h_2$.

Thus the cosets $Hd, d \in D,$ are disjoint and so $|\cup _{d \in D} Hd|=|H|\cdot |D|=hk$. Since $Hd \subset G - H$ for $d \in D$, we see that $hk=|\cup _{d \in D} Hd| \le |G{\setminus } H|= |D+D^{(-1)}|=2k$. Thus $h\le 2$ and so $h=2$ as $h>1$. The rest of the result follows. $\square$

This proves (i), (ii) and (iii) of Theorem 1.1. In what follows we will let $H=\langle t\rangle$, where $t \in Z(G)$ has order 2. Then:

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

(2.1)

These equations give: $v=2k+2, k^2=k+\lambda (v-2),$ and solving gives (i) of

Lemma 2.2

• (i) $v=2k+2, \quad \lambda =(k-1)/2=(v-4)/4$ and 4|v.

• (ii) $DH=HD=D^{(-1)}H=HD^{(-1)}=G-H.$

• (iii) $G,D,D^{(-1)},H$ all commute.

Proof

From $D \subset G-H$ we have $DH \cap H = \emptyset$, and $DH \subset G-H$; but $|G-H|=2k=|DH|$, so that

$$\begin{aligned} DH=HD=G-H=(G-H)^{(-1)}=D^{(-1)}H=HD^{(-1)}, \end{aligned}$$

giving (ii).

Since $D^{(-1)}=G-D-H$ and $H \le Z(G)$ it now follows that D and $D^{(-1)}$ commute. This shows that $G,D,D^{(-1)},H$ all commute. $\square$

Lemma 2.3

Let G be a SRHDS group with difference set D and subgroup $H=\langle t\rangle .$ Then $D^{(-1)}=tD$.

Proof

We have $D+Dt=(1+t)D=HD=G-H=D+D^{(-1)}$. $\square$

We now define Schur rings [17,18,19,20]. A subring ${\mathfrak {S}}$ of ${\mathbb {Z}} G$ is a Schur ring (or S-ring) if there is a partition ${\mathcal {K}}=\{C_i\}_{i=1}^r$ of G such that:

• 1. $\{1_G\}\in {\mathcal {K}}$;

• 2. for each $C\in {\mathcal {K}}$, $C^{(-1)}\in {\mathcal {K}}$;

• 3. ${C_i}\cdot {C_j}=\sum _k \lambda _{i,j,k} {C_k}$; for all $i,j \le r$.

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

Lemma 2.4

$\{1\}, \{t\},D,D^{(-1)}$ are the principal sets of a commutative Schur ring.

Proof

Now $\{1\}, \{t\},D,D^{(-1)}$ partition G and $D^{(-1)}=tD,tD^{(-1)}=D,t^2=1,D^{(-1)}D=DD^{(-1)}=\lambda (G-H)+k= \lambda (D+D^{(-1)})+k, D^2=tDD^{(-1)}=t( \lambda (D+D^{(-1)})+k).$ This concludes the proof. $\square$

Proposition 2.5

If $D \subset G$ is a SRHDS, then G is a Hadamard group.

Proof

Now $DD^{(-1)}=\lambda (G-H)+k$. Let $E=D+1$, so that $EE^{(-1)}=DD^{(-1)}+D+D^{(-1)}+1= \lambda (G-H)+k+(G-H) =(\lambda +1)(G-H)+k+1,$ as required. $\square$

3 Intersection Numbers

Let $N\triangleleft G$ and let $g_1,g_2,\dots ,g_r$ be coset representatives for G/N. Then for each $1\le i \le r$ there is $1\le i'\le r$ such that $g_ig_{i'} \in N$ i.e. $Ng_i\cdot Ng_{i'}=N$ in G/N. If G is a SRHDS group with difference set D, then the numbers $n_i=|D \cap Ng_i|$ are called the intersection numbers. Standard techniques give (see Section 7.1 of [21]):

Lemma 3.1

Let $D \subset G$ be a SRHDS with subgroup $H=\langle t\rangle , t^2=1$. Let $N\triangleleft G$ have order s and index r in G. Let $g_1=1,g_2,\dots ,g_r$ be coset representatives for G/N and let $n_i=|D \cap Ng_i|, 1\le i\le r$. Then

$$\begin{aligned}{} & {} \sum _{i=1}^r n_i=k, \qquad \sum _{i=1}^r n_i^2=\lambda |N{\setminus } H|+k,\\{} & {} \sum _{i=1}^r n_in_{i'}=\lambda |N|+(\lambda +1)\cdot |H \cap N|-k. \end{aligned}$$

Lemma 3.2

Let $N \triangleleft G$ where $D \subset G$ is a SRHDS with subgroup H and $H \cap N=\{1\}$. Let $Ng_3,\cdots ,Ng_r$ be the cosets that don’t meet H, and let $n_i=|D \cap Ng_i|$. Suppose that we have distinct $i,i'>2$ where $g_ig_{i'} \in N$. Then $n_i+n_{i'}=|N|$.

Proof

We have $n_i=|D \cap Ng_i|=|D^{(-1)} \cap Ng_i^{-1}|=|D^{(-1)} \cap Ng_{i'}|.$ If $i\ge 3$, then $Ng_{i'} \subset G {\setminus } H=D+D^{(-1)}$, so that

$$\begin{aligned} |N|=|(D+D^{(-1)}) \cap Ng_{i'}|=|D \cap Ng_{i'}| + |D^{(-1)} \cap Ng_{i'}|=n_{i'}+n_i. \end{aligned}$$

$\square$

The next result concerns intersection numbers for subgroups that are not necessarily normal:

Proposition 3.3

Let G be a SRHDS group with difference set D and subgroup H. Let $K\le G$ be any subgroup where $t \in K$. Let $b=|G:K|$ and let $g_0=1,g_1,\ldots ,g_{b-1}$ be coset representatives for $K \le G$. Let $k_i=|D \cap Kg_i|, 0\le i<b$. Then $k_0=|K|/2-1 \text { and } k_i=|K|/2, \,\,\, 0<i<b.$

Let $D_i=D \cap Kg_i,i=0,\ldots ,b-1$. Then $\sum _{i=0}^{b-1}D_iD_i^{(-1)}=\lambda (K-H)+k.$

Proof

We have $D^{(-1)}=tD$. Let $D_i=D \cap Kg_i$; then $tD_i=t(D \cap Kg_i)=(tD) \cap tKg_i=D^{(-1)} \cap Kg_i,$ so that $D \cap tD=\emptyset$ and $i>0$ gives

$$\begin{aligned} D_i+tD_i&=(D \cap Kg_i)+(D^{(-1)} \cap Kg_i) =(D+D^{(-1)}) \cap Kg_i\\&=(G-H) \cap Kg_i =G \cap Kg_i=Kg_i. \end{aligned}$$

Taking cardinalities, again using $D \cap tD=\emptyset$, gives $2k_i=|K|,$ for $i>0$. Then $\sum _{i=0}^{b-1}k_i=k$ now gives

$$\begin{aligned} k_0+(b-1)|K|/2=k=v/2-1; \end{aligned}$$

but $v=b \cdot |K|,$ from which we obtain $k_0=|K|/2-1$.

Now from $DD^{(-1)}=\lambda (G-H)+k$ and $D=\sum _{i=0}^{b-1} D_ig_i$ we get $\sum _{i=0}^{b-1}D_iD_i^{(-1)}+\cdots =\lambda (G-H)+k$, so that $\sum _{i=0}^{b-1}D_iD_i^{(-1)}\subseteq \lambda (K-H)+k$. The last part will follow if we can show that both sides of this equation have the same size.

From $b=v/|K|$ and the first part, the size of the left hand side is

$$\begin{aligned} \sum _{i=0}^{b-1} |D_i|^2=\left( |K|/2-1\right) ^2+(b-1) |K|^2/4=2p|K|-|K|+1 \end{aligned}$$

and (since $H \subset K$) the number of elements of the right hand side is $\lambda (|K|-2)+k=2p|K|-|K|+1,$ and we are done. $\square$

4 Direct Products and G is not Abelian

Let $\zeta _n=\exp {2\pi i/n}, n \in {\mathbb {N}}.$ We first show

Theorem 4.1

Suppose that $N \unlhd G$, $G/N \cong {\mathcal {C}}_{2^a}, a\ge 2$, and $t \notin N$. Assume that $k=|G|/2-1$ is not a perfect square. Then G is not a SRHDS group.

Proof

Note that $a\ge 2$ means that k is odd. Now assume that G is a SRHDS group and that $G/N=\langle rN \rangle \cong {\mathcal {C}}_{2^a}, r \in G$. For $g \in G$ we have $g=r^ib, 0\le i<2^a, b \in N$. Then there is a linear character $\chi ':G/N\rightarrow {\mathbb {C}}^\times , \chi '(rN)=\zeta _{2^a}$ that induces $\chi :G\rightarrow {\mathbb {C}}^\times , \chi (r^ib)=\chi '(r^iN).$ Here $N =\ker \chi$. Then we can write

$$\begin{aligned} D=\sum _{j=0}^{2^a-1} r^jN_{j}, \text{ where } N_{j} \subseteq N. \end{aligned}$$

Since $t \notin N$ we have $\chi (t)=-1$ and so $\chi (H)=0.$ We certainly have $\chi (G)=0$. From $G=D+D^{(-1)}+H$ we get $\chi (D)+\chi (D^{(-1)})=0$, and from $DD^{(-1)}=\lambda (G-H)+k$ we get $\chi (D)\chi (D^{(-1)})=k.$ These give $\chi (D)^2=-k, \text{ and } \text{ so } \chi (D)=\pm \sqrt{-k}.$ But

$$\begin{aligned} \pm i\sqrt{k}&= \chi (D)=\chi \left( \sum _{j=0}^{2^a-1} r^jN_{j}\right) =\sum _{j=0}^{2^a-1} (\zeta _{2^a})^j |N_{j}|, \end{aligned}$$

(4.1)

which gives $\sqrt{k} \in {\mathbb {Q}}(i,\zeta _{2^a}) = \mathbb Q(\zeta _{2^a}),$ since $a\ge 2$. But the Galois group of $\mathbb Q(\zeta _{2^a})/{\mathbb {Q}}$ is ${\mathcal {C}}_2 \times \mathcal C_{2^{a-2}}$. These groups have at most three subgroups of index 2. The Galois correspondence tells us that $\mathbb Q(\zeta _{2^a})$ contains at most three quadratic extensions, the only possibilities being ${\mathbb {Q}}(i)/{\mathbb {Q}}, {\mathbb {Q}}(\sqrt{2})/{\mathbb {Q}}$ and ${\mathbb {Q}}(\sqrt{-2})/{\mathbb {Q}}$. But the hypothesis says that k is not a perfect integer square, so that $\sqrt{k} \notin {\mathbb {Z}}$. Now $k>1$ is also odd, and so $\sqrt{k} \notin {\mathbb {Q}}(i), {\mathbb {Q}}(\sqrt{2}), {\mathbb {Q}}(\sqrt{-2}).$ This contradiction gives Theorem 4.1. $\square$

Corollary 4.2

Suppose that $N \unlhd G$, $G/N \cong {\mathcal {C}}_{2^a}, a\ge 3$, and $t \notin N$. Then G is not a SRHDS group.

Proof

Since $2^a\ge 8$ we see that $k=(|G|-2)/2$ satisfies $k \equiv 3 {\text {mod}}4$, and so the result follows from Theorem 4.1. $\square$

Corollary 4.3

If G is abelian with $|G|\equiv 0 {\text {mod}}8$, then G is not a SRHDS group.

Proof

Let G be an abelian SRHDS group, and write $G=A \times N$ where A is a Sylow 2-subgroup, and N is a subgroup of odd order. Since G has a single involution, we see that A is cyclic, say of order $2^a$. The results now follow from Corollary  4.2. $\square$

Corollary 4.4

If G is a SRHDS group, then $v=|G| \equiv 0 {\text {mod}}8.$

Proof

Assume that G is a SRHDS group with subgroup $H=\langle t\rangle$ and difference set D. Then we know that 4|v by Lemma 2.2, so suppose that $|G|=4n$ where n is odd. Then a Sylow 2-subgroup of G must be ${\mathcal {C}}_4=\langle r\rangle$ and $t=r^2$. Burnside’s theorem [9, Theorem 5.13] shows that $\langle r\rangle$ has a complement $N \triangleleft G, |N|=n, G=N \rtimes \langle r\rangle$. So we can write $D=D_0+D_1r+D_2r^2+D_3r^3, D_i \subset N$. Now $D+D^{(-1)}=G-H=N+Nr+Nr^2+Nr^3-H$ then gives

$$\begin{aligned} D_0+D_0^{(-1)}=N-1,\quad&D_1+\big (D_3^{(-1)}\big )^{r^3}=N,\quad D_2+\big (D_2^{(-1)}\big )^{r^2}=N-1\\&D_3+\big (D_1^{(-1)}\big )^r=N. \end{aligned}$$

Next, $D^{(-1)}=tD$ gives

$$\begin{aligned} D_0^{(-1)}=tD_0,\,\,\, \big (D_1^{(-1)}\big )^r=tD_3,\,\,\, \big (D_2^{(-1)}\big )^{r^2}=tD_2,\,\,\, \big (D_3^{(-1)}\big )^{r^3}=tD_1. \end{aligned}$$

Using $D_1+\big (D_3^{(-1)}\big )^{r^3}=N$ and $\big (D_3^{(-1)}\big )^{r^3}=tD_1$ we get $D_1(1+t)=N$. However $D_1(1+t)$ has an even number of elements (counting multiplicities), while |N| is odd. This contradiction gives the result. $\square$

Corollaries 4.3 and 4.4 now prove Theorem 1.1 (iv) and (v).

Corollary 4.5

If G is a SRHDS group, then a Sylow 2-subgroup of G is not cyclic.

Proof

Assume G is a SRHDS group with cyclic Sylow 2-subgroup $\langle r \rangle$. By Corollary 4.4, $|\langle r \rangle |\ge 8$. Again, Burnside’s theorem [9, Theorem 5.13] shows that $\langle r\rangle$ has a complement $N \triangleleft G, G=N \rtimes \langle r\rangle$. This now contradicts Corollary 4.2. $\square$

This concludes the proof of Theorem 1.1.

5 Construction of Some SRHDS Groups

We need the following set-up: For prime power $q=4p-1, p \in \mathbb N,$ we let ${\mathbb {F}}_{q^n}$ be the finite field of order $q^n$. Let $tr: {\mathbb {F}}_{q^2}\rightarrow {\mathbb {F}}_q, \beta \mapsto \beta ^q$ be the trace function. Let $\alpha \in {\mathbb {F}}_{q^2}$ satisfy $tr(\alpha )=0$. Let ${\mathbb {F}}_{q^2}^*=\langle z\rangle$. Let $Q=\{u^2: u \in {\mathbb {F}}_q, u\ne 0\}$. Then $-1\notin Q$ since $q\equiv 3 {\text {mod}}4$. Now choose $D \in {\mathbb {F}}_q {\setminus } (Q \cup \{0\})$. Then any $\beta \in {\mathbb {F}}_{q^2}$ has the form $\beta =a+b\sqrt{D},$ for some $c,d \in {\mathbb {F}}_q$ and $tr(c+ d\sqrt{D})=c-d\sqrt{D}.$ Write $\alpha =a+b\sqrt{D}$. Then $tr(\alpha )=0$ if and only if $a=0$, so we can choose $\alpha =\sqrt{D}.$

Let $U\le {\mathbb {F}}_{q^2}^*$ be the subgroup of order $(q-1)/2$, and let $\pi : {\mathbb {F}}_{q^2}^* \rightarrow W:={\mathbb {F}}_{q^2}^*/U$ be the natural map.

Theorem 5.1

Suppose that $4p-1$ is a prime power. Then $\textrm{Dic}_{8p}$ contains a SRHDS.

Proof

We follow [15, Theorem 3.3].

Let $q=4p-1$ and assume the above set-up. Let $g:=\pi (z)$ be a generator for W and note that $|W|=2(q+1)=8p$. Let $R=\{\pi (x): x \in {\mathbb {F}}_{q^2}^*, tr(\alpha x) \in Q \}.$ Then by [22, Thm 2.2.12], R is a relative $(q+1,2,q,(q-1)/2)$ difference set in W relative to the subgroup $H:=\langle g^{4p}\rangle$ of order 2.

Define $R_1,R_2 \subset W_2:=\langle g^2\rangle$ by $R=R_1+R_2g.$ Since R is a relative $(q+1,2,q,(q-1)/2)$ difference set, $RR^{(-1)}=\frac{q-1}{2} (W-H)+q$ from which we get

$$\begin{aligned} R_1R_1^{(-1)}+R_2R_2^{(-1)}=q+\frac{q-1}{2}\left( W_2-H\right) . \end{aligned}$$

If $d \in {\mathbb {F}}_{q^2}^*$ has order dividing $q+1$, then $d^q=d^{-1}$ and so

$$\begin{aligned} tr(\alpha d)=\alpha d+\alpha ^q d^q=\alpha d -\alpha d^{-1}=-tr(\alpha d^{-1}). \end{aligned}$$

Thus if tr$(\alpha d) \in Q$, then tr$(\alpha d^{-1}) \in -Q.$ But $q \equiv 3 {\text {mod}}4$ tells us that $g^{4p}=-1 \notin Q$, so that tr$(\alpha g^{4p} d^{-1})\in Q$. Thus $g^{4p}d^{-1} \in R_1$. Now the order of $g^{4p}d^{-1}$ is a divisor of $2(q+1)=|W|$. This gives a bijection, $Ud \leftrightarrow U g^{4p}d^{-1},$ between the elements of $R_1 \subset W_2$, which then gives $R_1^{(-1)}=g^{4p}R_1$.

Now let $G=\textrm{Dic}_{8p}$ $=\langle a,b|a^{2p}=b^2,b^4=1,a^b=a^{-1}\rangle$ and identify $\langle a\rangle$ with $W_2$, so that $a \leftrightarrow g^2$. From $R_1^{(-1)}=g^{4p}R_1$ we see that if $\gamma \in R_1 \cap R_1^{(-1)}$, then $g^{4p} \in R_1R_1^{(-1)}$, a contradiction to R being a relative difference set relative to H. It follows that $R_1 \cap R_1^{(-1)} =\emptyset .$ Now $1,-1=g^{4p} \notin R_1$ as tr$(\alpha 1)=0 \notin Q$, and so

$$\begin{aligned} R_1+R_1^{(-1)}=W_2-H. \end{aligned}$$

(5.1)

Then (5.1) and $R_1^{(-1)}=g^{4p}R_1$ gives

$$\begin{aligned} W_2-H=R_1(1+g^{4p})=R_1H, \end{aligned}$$

so that we have the first part of

Lemma 5.2

• (i) $R_1+1$ is a transversal for $W_2/H$.

• (ii) $R_2$ is a transversal for $W_2/H$.

Proof

(ii) We first show that R+1 is a transversal for W/H.

If $u \in W$, then tr$(\alpha u) \in Q$, and it follows that tr$(\alpha g^{4p} u) =- tr(\alpha u) \notin Q$. This sets up a bijection $u \leftrightarrow g^{4p} u$ of $W-H$ where the orbits of this bijection are the non-trivial H-cosets and a transversal corresponds to the elements of Q.

Since R+1 is a transversal for W/H and $R_1+1$ is a transversal for $W_2/H$ it follows that $R_2$ is a transversal for $W_2/H$. This concludes the proof. $\square$

Now if $\alpha =\sqrt{D}, \beta =a+b\sqrt{D}$, then tr$(\alpha \beta )=2bD \in Q$ if and only if $2b \in {\mathbb {F}}_q^* {\setminus } Q$.

Define $S:=a^{2p} R_1+R_2b.$ First we show that $SS^{(-1)}=\lambda \left( G -H\right) +k$ where $k=(v-2)/2, \lambda =(k-1)/2$:

$$\begin{aligned} SS^{(-1)}&=(a^{2p}R_1+R_2b)\big (a^{2p}R_1^{(-1)}+b^{-1}R_2^{(-1)}\big ) \nonumber \\&=R_1R_1^{(-1)}+R_2R_2^{(-1)}+R_1R_2(1+a^{2p})b\nonumber \\&=R_1R_1^{(-1)}+R_2R_2^{(-1)}+R_1R_2Hb\nonumber \\&=R_1R_1^{(-1)}+R_2R_2^{(-1)}+R_1W_2b\nonumber \\&= q+\frac{q-1}{2}\left( W_2-H\right) + |R_1|W_2b\nonumber \\&= k+\lambda \left( W_2-H\right) + \lambda W_2 b\nonumber \\&= k+\lambda \left( W_2 +W_2b-H\right) = \lambda \left( G -H\right) +k, \end{aligned}$$

(5.2)

as desired. Next we need

Lemma 5.3

For S as above we have $S \cap S^{(-1)} =\emptyset .$

Proof

So assume that $r \in S \cap S^{(-1)}, S=a^{2p} R_1+R_2b$. Then there are two cases.

• (a) First assume that $r \in \langle a\rangle .$ Then there are $x^i, x^j \in R_1$ where $r=a^{2p}a^i=a^{2p}a^{-j}$ so we have $i=-j$. Since a corresponds to $g^2$ the elements $g^{2i},g^{-2j}$ satisfy tr$(\alpha g^{2i}),$ tr$(\alpha g^{-2j}) \in Q$. Let $g^{i}=c+b \sqrt{D}$. Then tr$(\alpha g^{2i}),$ tr$(\alpha g^{-2j}) \in Q$ (respectively) gives $4bcD \in Q,-\frac{4bcD}{(c^2-b^2D)^2} \in Q$ (respectively), which in turn gives $-1 \in Q$, a contradiction.

• (b) Next assume that $r \in \langle a\rangle b.$ Then there are i, j such that $r=a^ib=(a^jb)^{-1}=a^{j+2p}b,$ where $a^i, a^j \in R_2$. Thus $i=j+2p$. As in the first case this gives tr$(\alpha g^{2i+1}),$ tr$(\alpha g^{2j+1})=tr(\alpha g^{2i-4p+1})\in Q$. Since tr$(\alpha g^{2i-4p+1})=-$tr$(\alpha g^{2i+1})$, this gives $-1 \in Q$, a contradiction.

From $S \cap S^{(-1)} =\emptyset =S \cap H$ we get $G=S+S^{(-1)}+H$ and so Eq. (5.2) shows that S is a SRHDS, giving Theorem 5.1. $\square$

We next wish to show that we can double these examples (see Sect. 6 for the definition of this doubling process), and we will need the following symmetry results:

Symmetry proof for $R_1$ . Now $S=a^{2p}R_1+R_2b$ and if $a^i \in a^{2p}R_1$, then $i=2p+j$ where $tr(\alpha z^{2j}) \in Q$. We note that z, the generator of ${\mathbb {F}}_{q^2}^*$, has order $q^2-1$, and so $(z^{q})^q=z$, showing that the non-trivial Galois automorphism is given by $z \mapsto z^q$.

So from $tr(\alpha z^{2j}) \in Q$ we get $tr(\alpha ^q z^{2jq}) \in Q$. But $\alpha ^q=-\alpha =\alpha z^{(q^2-1)/2}$. Thus

$$\begin{aligned} tr(\alpha ^q z^{2jq}) = tr(\alpha z^{2jq+(q^2-1)/2})= tr(\alpha z^{2(jq+(q^2-1)/4)}) \in Q. \end{aligned}$$

This if $j'=(jq+(q^2-1)/4)$, then $a^{2p+j'} \in a^{2p}R_1$, and so $j \mapsto j'$ determines a function $R_1 \rightarrow R_1$ that one can show is an involution.

One can then check that $j=p+r$ is sent to $j'=p-r$ (recalling that j is defined mod 4p). This gives a ‘reflective’ symmetry for $R_1$.

Symmetry proof for $R_2$ . We now do a similar thing for $R_2$. So let $a^ib \in R_2b$, so that $tr(\alpha z^{2i+1}) \in Q$. Then acting by the Galois automorphism we get

$$\begin{aligned} tr(\alpha ^q z^{(2i+1)q})= tr(\alpha z^{(2i+1)q+(q^2-1)/2}) = tr(\alpha z^{ 2( iq+(q^2-1)/4+(2p-1) )+1 } ) \in Q. \end{aligned}$$

This similarly gives the involutive map

$$\begin{aligned} i \mapsto iq+(q^2-1)/4+(2p-1) \equiv -i-1 {\text {mod}}4p. \end{aligned}$$

(5.3)

$\square$

6 The Doubling Process

Lemma 6.1

Let $D\subset G=\textrm{Dic}_{ {v}}=\langle x,y\rangle , v=4n, k=2n-1, \lambda =n-1$. Let $K=\langle x\rangle$, $k_1=n-1,k_2=n$ and let $D=D_1+D_2y, D_i \subset K, k_i=|D_i|.$ Then the requirement that $D=D_1+D_2y$ is a SRHDS is equivalent to (a)–(d):

$$\begin{aligned}{} & {} (a) \, \, \, D_1H=K-H, \,\,\, (b) \, \, \, D_1^{(-1)}=tD_1, \,\,\, (c) \, \, \, D_2H=K, \,\,\, \\{} & {} (d) \, \, \,\lambda (K-H)+k=D_1D_1^{(-1)}+D_2D_2^{(-1)}. \end{aligned}$$

Proof

One checks that $D=D_1+D_2y$ is a SRHDS is equivalent to the conditions

• (i) $D_1 \cup \{1\}$ and $D_2$ are transversals for K/H (this comes from looking at $G-H=D+D^{(-1)}=D_1+D_2y+(D_1^{(-1)}+(D_2y)^{(-1)})$).

• (ii) $\lambda D_1H+k=D_1D_1^{(-1)}+D_2D_2^{(-1)}$;

• (iii) $\lambda Ky=D_2D_1y+D_1D_2y^{-1}$ (from $DD^{(-1)}=\lambda (G-H)+k$);

• (iv) $D_1^{(-1)}=tD_1$ and $D_i^y=D_i^{(-1)}.$

Now (iii) is equivalent to $D_1D_2(1+t)=\lambda K$ or $D_1 K=\lambda K$. But $D_1 K=\lambda K$ follows directly from $D_i \subset K,$ and $|D_1|=\lambda$. Thus (ii) and (iii) are equivalent to $\lambda D_1H+k=D_1D_1^{(-1)}+D_2D_2^{(-1)}.$ $\square$

Write $D = D_0 + D_1 y$. We construct the set $E\subseteq \textrm{Dic}_{16p}$ as

$$\begin{aligned} E:= E_0 + E_1 y \text { with } E_0:= D_0 + D_1 x \quad \text {and} \quad E_1:= D_1^{(-1)} x^{-1} t + D_0^{(-1)}+1. \end{aligned}$$

We show that if $D_1$ satisfies the symmetry: $x^{2i} \in D_1$ implies $x^{4p-2i-2} \in D_1$, then E is a $(v_2, k_2, \lambda _2)$-SRHDS with $v_2 = 16p$, $k_2 = 8p - 1$, and $\lambda _2 = 4p - 1$.

Theorem 6.2

Let $\textrm{Dic}_{16p}=\langle x,y$ | $x^{4p}=y^2,y^4=1, x^y=x^{-1}\rangle$, $t=y^2$. We let $\textrm{Dic}_{8p}=\langle x^2,y \rangle \le \textrm{Dic}_{16p}.$ Let D be a $(v_1, k_1, \lambda _1)$-SRHDS in $\textrm{Dic}_{8p}$, with $v_1 = 8p$, $k_1 = 4p - 1$, and $\lambda _1 = 2p - 1$. Then the unique involution t in $\textrm{Dic}_{16p}$ is the same as the unique involution in $\textrm{Dic}_{8p}$.

Write $D = D_0 + D_1 y, D_i \subset \langle x^2\rangle$, and let $E = E_0 + E_1 y \ \subseteq \textrm{Dic}_{16p}$ where:

$$\begin{aligned} E_0:= D_0 + D_1 x \quad \text {and} \quad E_1:= D_1^{(-1)} x^{-1} t + D_0^{(-1)}+1. \end{aligned}$$

Assume that $D_1$ satisfies the symmetry: $x^{2i} \in D_1$ implies $x^{4p-2i-2} \in D_1$. Then E is a $(v_2, k_2, \lambda _2)$-SRHDS with $v_2 = 16p$, $k_2 = 8p - 1$, and $\lambda _2 = 4p - 1$.

Proof

We note that $D^{(-1)}=tD$ implies that $E^{(-1)}=tE$. We also observe that the map $x^{2i}\rightarrow x^{4p-2i-2}$ is an involution. Using Lemma 6.1, to show E is a SRHDS it suffices to show that E satisfies

$$\begin{aligned}&(1) \,\,\,E \cup E^{(-1)} = \textrm{Dic}_{16p} - \langle t \rangle ;\\&(2)\,\,\,E \cap E^{(-1)} = \emptyset ;\\&(3)\,\,\,E_0 E_0^{(-1)} + E_1 E_1^{(-1)} = \lambda _2 (\langle x \rangle - \langle t \rangle ) + k_2. \end{aligned}$$

This is sufficient because conditions (1) and (2) along with $E^{(-1)}=tE$ imply conditions (a) and (c) of Lemma 6.1. First we note that E does not contain t or the identity, as this would imply that $D_0$ contains these. We now show (2), which will imply (1). We split condition (2) into cases by considering the intersection of E with each coset of $\langle x^2 \rangle$, all of which cosets are their own inverses. There are four such cosets: $\langle x^2 \rangle , \langle x^2 \rangle x,\langle x^2 \rangle y,$ and $\langle x^2 \rangle xy$.

$\langle x^2\rangle :$ For $E \cap \langle x^2 \rangle =D_0$, we know that $x^{2i} \in D_0$ implies $x^{-2i} \not \in D_0$ since $D_0\cap D_0^{(-1)}=\emptyset$.

$\langle x^2\rangle x:$ We have $E \cap \langle x^2 \rangle x = D_1 x$. We show $D_1 x\cap (D_1x)^{(-1)}=\emptyset$.

$$\begin{aligned} x^{2i+1} \in D_1 x&\iff x^{2i} \in D_1 \iff x^{4p - 2i - 2} \in D_1 \nonumber \\&\iff x^{4p - 2i - 2}y \in D_1y \iff tx^{4p-2i-2}y \not \in D_1y \nonumber \\&\iff x^{-2i-2}\not \in D_1 \iff x^{-2i-1}\not \in D_1x. \end{aligned}$$

(6.1)

Here we used the symmetry and the fact that $(D_1y)\cap (D_1y)^{(-1)}=\emptyset$ where $(D_1y)^{(-1)}=tD_1y$.

$\langle x^2\rangle y:$ Here we have $E \cap \langle x^2 \rangle y = D_0^{(-1)} y + y$. First we check that $D_0^{(-1)} y$ doesn’t contain any of its inverses:

$$\begin{aligned} x^{-2i} y \in D_0^{(-1)} y&\iff (x^{-2i} y)^{-1}=t x^{-2i} y \not \in D_0^{(-1)} y. \end{aligned}$$

We also check the additional y doesn’t have an inverse in $D_0^{(-1)} y$:

$$\begin{aligned} t \not \in D_0^{(-1)}&\iff y^{-1} =t y \not \in D_0^{(-1)} y. \end{aligned}$$

$\langle x^2\rangle xy:$ Here we have $E \cap \langle x^2 \rangle x y = D_1^{(-1)} x^{-1} t y$, and

$$\begin{aligned} x^{-2i-1} t y \in D_1^{(-1)} x^{-1} t y&\iff x^{2i} \in D_1 \iff t x^{2i} \not \in D_1 \\ \iff tx^{-2i} \not \in D_1^{(-1)}&\iff x^{-2i-1} y = tx^{-2i}x^{-1}ty \not \in D_1^{(-1)} x^{-1} t y. \end{aligned}$$

Thus $E \cap E^{(-1)} = \emptyset$. This concludes (2) and implies (1), since both E and $E^{(-1)}$ don’t intersect $\langle t \rangle$ and $|E|=k_2=8p-1.$

Now we prove (3): we have

$$\begin{aligned} E_0 E_0^{(-1)} + E_1 E_1^{(-1)}&= \left( D_0 + D_1x \right) \left( D_0^{(-1)} + D_1^{(-1)} x^{-1} \right) \nonumber \\&\quad + \left( D_1^{(-1)} x^{-1} t + D_0^{(-1)}+1 \right) \left( D_1 x t + D_0 + 1 \right) \nonumber \\&= 2 D_0 D_0^{(-1)} + 2 D_1 D_1^{(-1)} \nonumber \\&\quad + (1 + t) D_0 D_1^{(-1)} x^{-1} + (1 + t) D_1 D_0^{(-1)} x \nonumber \\&\quad + D_1 x t + D_0 + D_1^{(-1)} x^{-1} t + D_0^{(-1)} + 1. \end{aligned}$$

(6.2)

For E to be a SRHDS we need (6.2) to be equal to $\lambda _2(\langle x \rangle - \langle t \rangle )+k_2$. Looking at just the even powers of x, we need

$$\begin{aligned} 2 D_0 D_0^{(-1)} + 2 D_1 D_1^{(-1)} + D_0 + D_0^{(-1)} + 1 \end{aligned}$$

to be equal to $\lambda _2(\langle x^2 \rangle - \langle t \rangle ) + k_2$. We note that $D_0 + D_0^{(-1)} = \langle x^2 \rangle - \langle t \rangle$, and $D_0 D_0^{(-1)} + D_1 D_1^{(-1)}=\lambda _1(\langle x^2 \rangle -\langle t \rangle )+k_1$ since D is a SRHDS for $\langle x^2,y \rangle$. Since $\frac{k_2 - 1}{2} = \lambda _2$, we have

$$\begin{aligned}&2(D_0 D_0^{(-1)} + D_1 D_1^{(-1)}) + (D_0 + D_0^{(-1)}) + 1\\&\quad =2(\lambda _1(\langle x^2 \rangle -\langle t \rangle )+k_1)+(\langle x^2 \rangle - \langle t \rangle )+1\\&\quad =(2\lambda _1+1)(\langle x^2 \rangle -\langle t \rangle )+(2k_1+1) =\lambda _2(\langle x^2 \rangle - \langle t \rangle ) + k_2, \end{aligned}$$

as desired. We now look at the odd powers of x in (6.2), which must equal $\lambda _2 \langle x^2 \rangle x$. We see that

$$\begin{aligned}&(1 + t) D_0 D_1^{(-1)} x^{-1} + (1 + t) D_1 D_0^{(-1)} x + D_1 x t + D_1^{(-1)} x^{-1} t\nonumber \\&\quad = (1 + t) \left( D_0 + 1 \right) D_1^{(-1)} x^{-1} + (1 + t) \left( D_0 + 1 \right) ^{(-1)} D_1 x \nonumber \\&\qquad - \left( D_1 x\right) ^{(-1)} + D_1 x. \end{aligned}$$

(6.3)

Looking at the first two terms of (6.3), $D_0 + 1$ is a transversal of $\langle t \rangle$ in $\langle x^2 \rangle$, so $(1 + t)\left( D_0 + 1 \right) = \langle x^2 \rangle$ and $(1 + t)\left( D_0 + 1 \right) ^{(-1)} = \langle x^2 \rangle$. So we can reduce (6.3) to

$$\begin{aligned} \langle x^2 \rangle D_1^{(-1)} x^{-1} + \langle x^2 \rangle D_1 x - \left( D_1 x\right) ^{(-1)} + D_1 x. \end{aligned}$$

To evaluate the last two terms of (6.3), we note that (6.1) gives us: if $x^{2i} \in D_1$, then $x^{-2i-2} \not \in D_1$. Thus $D_1$ and $\left( D_1 x^2 \right) ^{(-1)}$ are disjoint, so their sum is $\langle x^2 \rangle$ since $|D_1|=4p$. Thus $\left( D_1 x\right) ^{(-1)} + D_1 x = \left( \left( D_1 x^{-2}\right) ^{(-1)} + D_1\right) x = \langle x^2 \rangle x.$ So the sum of the odd powered terms is

$$\begin{aligned}&\langle x^2 \rangle \left( D_1 \right) ^{(-1)} x^{-1} + \langle x^2 \rangle D_1 x - \langle x^2 \rangle x =D_1^{(-1)}\langle x^2 \rangle x^{-1} +(D_1-1)\langle x^2 \rangle x\\&\quad =|D_1|\langle x^2 \rangle x +(|D_1|-1)\langle x^2 \rangle x =\lambda _2 \langle x^2 \rangle x \end{aligned}$$

as desired. Therefore we have shown (3), and E is a SRHDS. $\square$

Corollary 6.3

The set $E=E_0+E_1y$ as defined above is an SRHDS in $\textrm{Dic}_{16p}$ if $D = D_0 + D_1 y$ is an SRHDS in $\textrm{Dic}_{8p}$ and $x^{2i} \in D_1$ implies $x^{- 2i - 2} \in D_1$.

Proof

This follows by applying the automorphism $\varphi (x)=x, \varphi (y)=x^{2p}y$ to $\textrm{Dic}_{16p}$ in the preceding theorem. We have that D is a SRHDS for $\textrm{Dic}_{8p}$ if and only if $\varphi (D)$ is, and similarly E is a SRHDS for $\textrm{Dic}_{16p}$ if and only if $\varphi (E)$ is. The condition $x^{2i} \in \varphi (D_1)$ implies $x^{- 2i - 2} \in \varphi (D_1)$ is equivalent to the condition $x^{2i} \in D_1$ implies $x^{4p-2i-2} \in D_1$. $\square$

Many other equivalent symmetries can be obtained by using a different automorphism that fixes $\langle x \rangle$. The one we have used is that obtained at the end of Theorem 5.1. In the SRHDS $S=a^{2p}R_1+R_2b$ of $\textrm{Dic}_{8p}$ from Theorem 5.1, we showed that $a^i\in R_2$ implies $a^{-i-1}\in R_2.$ See (5.3). As a subgroup of $\textrm{Dic}_{16p}$, this is the necessary symmetry condition for Corollary 6.3 to apply. Thus $\textrm{Dic}_{16p}$ is a SRHDS group when $4p-1$ is a prime power. This proves Theorem 1.3. $\square$

7 D and Cosets of $Q_8$

Let G be a SRHDS group with subgroup H and difference set D. Suppose that $Q\le G$ has even order and that $g_0=1,\ldots ,g_{p-1}$ is a transversal for $Q \le G$. Then we can write

$$\begin{aligned} D=F_0g_0+F_1g_1+\cdots +F_{p-1}g_{p-1}, \quad F_i \subset Q. \end{aligned}$$

(7.1)

Lemma 7.1

Let $Q\le G$ be as above. For all subsets $F \subseteq Q$ of size greater than |Q|/2, the multiplicity of t in $FF^{(-1)}$ is greater than zero. $\square$

Proof

Now $t \in Q$, so $H \le Q$ and if $|F| >|Q|/2$, then some coset of $H \le Q$ meets F in two elements and so $t \in FF^{(-1)}$. $\square$

Now $DD^{(-1)}=\lambda (G-H)+k$ and a part of the left hand side is $\sum _{i=0}^{p-1} F_iF_i^{(-1)}$. Thus $|F_i|\le |Q|/2$ when D is written as in Eq. (7.1).

Now let $f_i=|F_i|,0\le i<p-1,$ so that

$$\begin{aligned} \sum _{i=0}^{p-1} f_i=|D|=k=\frac{(|G|-2)}{2}=\frac{(|Q|p-2)}{2}=\frac{|Q|}{2}p-1. \end{aligned}$$

Since $f_i \le |Q|/2$ we must have $f_i=|Q|/2$ for all $0\le i\le p-1$ except one. To see that $f_0=|Q|/{2}-1$ we just note that $Q-H$ has $|Q|-2$ elements that come in inverse pairs. Thus $f_0=|Q|/{2}-1$.

Next note that $DD^{(-1)}=\lambda (G-H)+k$ and $F_iF_i^{(-1)} \subseteq Q$. We want to show

$$\begin{aligned} \sum _{i=0}^{p-1} F_iF_i^{(-1)}=\lambda (Q-H)+k. \end{aligned}$$

(7.2)

Now, $v=8p, k=\frac{|Q|}{2}p-1,\lambda =\frac{|Q|}{4}p-1$ and so $\lambda (Q-H)+k$ has $(\frac{|Q|}{4}p-1)(|Q|-2)+(\frac{|Q|}{2}p-1)=\frac{|Q|^2}{4}p-|Q|+1$ elements, while $\sum _{i=0}^{p-1} F_iF_i^{(-1)}$ has $\left( \frac{|Q|}{2}-1\right) ^2+(p-1)\left( \frac{|Q|}{2}\right) ^2=\frac{|Q|^2}{4}p-|Q|+1$ elements, so we must have Eq. (7.2).

For $Q=Q_8,$ considering those $F_i$ of size $|Q|/2=4$ a Magma [12] calculation gives the following result by finding all those subsets $F \subset Q_8$ such that $FF^{(-1)}$ does not contain t:

Lemma 7.2

Suppose that $Q=Q_8\le G.$ Then each $F_i$ of size 4 is one of the following 16 sets:

$$\begin{aligned}&\{1,x ,y,x y\};\quad \{1,x ,y,x^{3}y\};\quad \{x ,x^{2},x^{2}y,x^{3}y\};\quad \{1,x ,x^{2}y,x^{3}y\};\\&\{1,x^{3},x^{2}y,x^{3}y\};\quad \{1,x^{3},y,x y\};\quad \{x ,x^{2},y,x^{3}y\};\quad \{x^{2},x^{3},y,x^{3}y\};\\&\{x ,x^{2},x y,x^{2}y\};\quad \{x^{2},x^{3},x y,x^{2}y\};\quad \{x^{2},x^{3},y,xy\};\,\,\,\, \{1,x ,x y,x^{2}y\};\\&\{x ,x^{2},y,x^{2}y\};\quad \{x^{2},x^{3},x^{2}y,x^{3}y\};\,\,\, \{1,x,x y,x^{2}y\};\,\,\, \{1,x^{3},y,x^{3}y\}.\square \end{aligned}$$

Each of these is a relative difference set for $Q_8$. Thus each $F_i, i>0,$ is a relative difference set for $Q_8$. It follows then from Eq. (7.2) that $F_0$ is a SRHDS for $Q_8$. Thus $F_0$ is determined by

Lemma 7.3

The following sets are equal:

• (i) The set of all SRHDS for $Q_8=\langle i,j,k \rangle$.

• (ii) The set of all conjugate (by elements of $Q_8$)-translates (by elements of H) of $\{i,j,k\}$.

• (iii) The set of all $\{a,b,c\} \subset Q_8 {\setminus } H$ where $|\{a,b,c\}|=3$ and $t \notin \{uv^{-1}:u,v \in \{a,b,c\}\}.$ $\square$

Call this common set ${\mathcal {S}}$ and note that $|{\mathcal {S}}|=8$.

Now any $F_0$ must satisfy (iii), so $F_0 \in {\mathcal {S}}$. Further, we can choose $F_0$ to be any element of ${\mathcal {S}}$ by applying the operations in (ii) to D, which still result in a SRHDS.

Assume that $G=\textrm{Dic}_{8p}$ so that a transversal of $Q_8\le G$ is $1,x,\ldots , x^{p-1}$. Now we can write $D=F_0+F_1x+F_2x^2+\cdots +F_{p-1}x^{p-1}$ where $F_i \subset Q_8$ and $F_0\in {\mathcal {S}}$.

Here each $F_i, i>0,$ is one of the 16 subsets of $Q_8$ in Lemma 7.2 and $F_i=(1+x^p)(a+by)=a+by+x^pa+x^pby, \text{ where } a,b \in \langle x^p\rangle .$

Now $D^{(-1)}t=D$ and so if $F_ix^i \subset D$, then $t(F_ix^i)^{(-1)}=tx^{-i}F_i^{(-1)} \subset D$. Here $F_i^{(-1)}=a^{-1}+bty+x^{-p}a^{-1}+x^pbty,$ and so

$$\begin{aligned} t(F_ix^i)^{(-1)}&=tx^{-1}F_i^{(-1)}=tx^{-i}( a^{-1}+bty+x^{-p}a^{-1}+x^pbty )\\&=ta^{-1}x^{-i}+tx^{-p}a^{-1}x^{-i} + byx^{i} + x^pbyx^{i}. \end{aligned}$$

Thus $F_i$ and $t(F_ix^i)^{(-1)}$ have $byx^{i} + x^pbyx^{i}$ in common and so

$$\begin{aligned} F_ix^i\cup t(F_ix^i)^{(-1)}=ax^i+byx^i+x^pax^i+x^pbyx^i + ta^{-1}x^{-i}+tx^{-p}a^{-1}x^{-i}. \end{aligned}$$

We denote this by $J_i(a,b)$, so that D is a union of $D_0$ and some of the $J_i(a,b)$.

Now $J_i(a,b)$ has four elements in $Q_8x^i$ and has two elements in $Q_8x^{-i}$. Since we know that each non-trivial coset of $Q_8$ has to contain four elements of D we know that D has to contain some $J_{-i}(c,d)$ so that

$$\begin{aligned} (a+x^pa)x^i+(a^{-1}+x^{-p}a^{-1})tx^{-i} = (c+x^pc)x^{-i} + (b^{-1}+x^{-p}b^{-1})tx^i. \end{aligned}$$

This is true if and only if we have $a+x^pa=b^{-1}t+x^{-p}b^{-1}t$ and $(a^{-1}+x^{-p}a^{-1})t=b+x^pb$. However these equations are equivalent and we note that for any choice of $a \in \langle x^p\rangle$ there is a $b \in \langle x^p\rangle$ that solves the first equation.

Thus we now obtain eight element sets by taking the union of these two $J's$. We denote these by $L_i(a,b,c)$:

$$\begin{aligned}&(a+x^pa)x^i + (a^{-1}+x^{-p}a^{-1})tx^{-i} +(by+x^pby)x^i + (cy+x^pcy)x^{-i}\\&\quad =(1+x^p)(a+by)x^i+(1+x^p)(x^pa^{-1}+cy)x^{-i}. \end{aligned}$$

We note that $L_i(a,b,c)=L_j(a',b',c')$ if and only if $i=j, a=a', b=b', c=c'.$ For $1\le i\le p-1$ let $\mathcal L_i=\{L_i(a,b,c):a,b,c \in \langle x^p\rangle \}$. Then $|\mathcal L_i|=64$.

8 Groups that are not SRHDS Groups

Proposition 8.1

The dicyclic group $\textrm{Dic}_{72}$ is not a SRHDS group.

Proof

Suppose it is and that D is the SRHDS. Let $G=\textrm{Dic}_{72}$ = $\langle x,y|x^{36}=1,y^2=x^{18}, x^y=x^{-1}\rangle .$ Then by the above section there are $D_i \in {\mathcal {L}}_i, 1\le i\le 4,$ such that $D=D_0+\sum _{i=1}^4 D_i$. There are $64=|L_i|$ choices for each $D_i, 1\le i\le 4$. Using the standard irreducible representation $\rho : \textrm{Dic}_{72}\rightarrow \textrm{GL}(2,{\mathbb {C}})$ given by $\rho (x)=\begin{bmatrix} \zeta _{36}&{}0\\ 0&{}\zeta _{36}^{-1}\end{bmatrix}, \rho (y)=\begin{bmatrix} 0&{}-1\\ 1&{}0\end{bmatrix},\zeta _{36}=e^{2\pi i/36},$ we have $\rho (G)=\rho (H)=0$. From $D+D^{(-1)}=G-H$ we then have $\rho (D)+\rho (D^{(-1)})=0$. By $DD^{(-1)}=\lambda (G-H)+k$ we have $\rho (D)\rho (D^{(-1)})=kI_2=35I_2$. Therefore, $35I_2=\rho (D)\rho (D^{(-1)})= -\rho (D)^2$. A Magma calculation determines that of the $64^4$ possibilites for D, only 648 have $\rho (D)^2=-35I_2$. Another Magma [23] calculation verifies that none of these 648 give a SRHDS, completing the proof. $\square$

Proposition 8.2

Let G be a group where $Q_8 \le G$. Suppose that there is an epimorphism $\pi :G\rightarrow {\mathcal {C}}_p \times Q_8$ for p prime where $\pi (Q_8)=\{1\} \times Q_8$ and $|\ker \pi |$ is odd. Then G is not a SRHDS group.

Proof

So suppose that G is a SRHDS group with difference set D and subgroup $H=\langle t\rangle$. Let $Q_8=\langle x,y|x^4,x^2=y^2,x^y=x^{-1}\rangle \le G$, so that $t=x^2, \pi (x)=x, \pi (y)=y$. First note that p must be odd since G has a unique involution. Let $N=\ker \pi$. Put ${\mathcal {C}}_p =\langle \pi ( r) \rangle , r \in G,$ so that we can write

$$\begin{aligned} D=\sum _{i=0}^{p-1} \sum _{j=0}^3 r^i x^j D_{0,i,j} + \sum _{i=0}^{p-1} \sum _{j=0}^3 r^i x^jyD_{1,i,j}, \quad D_{ k,i,j}\subset N. \end{aligned}$$

We note that $|D_{i,j,k}|\le |N|$.

Let $p_2=(p-1)/2.$ We can also write $D =\sum _{i=0}^{p-1} r^iD_i, D_i \subset \langle x,y,N\rangle$ so that

$$\begin{aligned} D_i= \sum _{j=0}^3 x^j D_{0,i,j} +\sum _{j=0}^3 x^jy D_{1,i,j} \end{aligned}$$

From $D^{(-1)}=tD$ we get $D_i^{(-1)}r^{-i}=tr^{p-i}D_{p-i}, 0\le i<p,$ so that $D_{p-i} = tr^{-p}(D_{i}^{(-1)})^{r^{-i}}.$ Thus $D=D_0+\sum _{i=1}^{p_2} r^iD_i + r^{-i} t(D_{i}^{(-1)})^{r^{-i}}.$

Now let $\rho :Q_8 \rightarrow \textrm{GL}(2,{\mathbb {Q}}( {i})), \textit{i}$ = $\sqrt{-1},$ be an irreducible faithful unitary representation of $Q_8$ where $\rho (x)=\begin{bmatrix} i&{}0\\ 0&{}-i\end{bmatrix}, \rho (y)=\begin{bmatrix} 0&{}-1\\ 1&{}0\end{bmatrix}.$ Then the $\mathbb Q$-span of the image of $\rho$ has basis

$$\begin{aligned} B_1=I_2,\,\,\, B_2=\rho (x) = \begin{bmatrix} i&{}0\\ 0&{}-i\end{bmatrix},\,\,\, B_3=\rho (y) = \begin{bmatrix} 0&{}-1\\ 1&{}0\end{bmatrix},\,\,\, B_4=\rho (xy)= \begin{bmatrix} 0&-i\i&0\end{bmatrix} \end{aligned}$$

since $\rho (x^2)=-B_1$. We note from Lemma 7.3 that we may assume $D_0=\{x,y,xy\},$ so $\rho (D_0)=\begin{bmatrix} i&{}-i-1\\ 1-i&{}-i\end{bmatrix}=B_2+B_3+B_4$.

Let $\omega =\exp {2 \pi i/p}$. Then $\pi ,$ $\rho$ and $r \mapsto \omega I_2$ determine an irreducible unitary representation of G that we also call $\rho$. Then $\rho (r^iD_i)=\omega ^i \sum _{j=1}^4 a_{ij}B_j$, where $a_{ij} \in {\mathbb {Z}}$, so that

$$\begin{aligned} \rho \big ( r^{-i} t\big (D_{i}^{(-1)}\big )^{r^{-i}} \big )= -\omega ^{-i} \rho \big ( D_{i}^{(-1)}\big )^{r^{-i}} \big ) = -\omega ^{-i} \rho \big ( D_{i}^{(-1)} \big ) = -\omega ^{-i} \sum _{j=1}^4 a_{ij}B_j^*. \end{aligned}$$

Here $B_1^*=B_1, B_2^*=-B_2, B_3^*=-B_3, B_4^*= -B_4$.

This gives

$$\begin{aligned} \rho (D)&=\begin{bmatrix} i&{}-i-1\\ 1-i&{}-i\end{bmatrix} + \sum _{i=1}^{p_2} \rho \big (D_ir^i + r^{-i} t\big (D_{i}^{(-1)}\big )^{r^{-i}} \big )\nonumber \\&=\begin{bmatrix} i&{}-i-1\\ 1-i&{}-i\end{bmatrix} + \sum _{i=1}^{p_2} \sum _{j=1}^4\big (a_{ij}B_j\omega ^i - a_{ij} B_j^*\omega ^{-i}\big ). \end{aligned}$$

(8.1)

We can write this as

$$\begin{aligned} \rho (D)=\begin{bmatrix} i&{}-i-1\\ 1-i&{}-i\end{bmatrix} + \sum _{u=1}^4 a_u B_u, \text{ where } a_u \in {\mathbb {Z}}[\omega ]. \end{aligned}$$

(8.2)

From $DD^{(-1)}=\lambda (G-H)+k$ and $D^{(-1)}=tD$ we get $D^2=\lambda (G-H)+kt$. Now if $\rho (D)^2=(e_{ij})$, then from $(e_{ij})=\rho (D^2)=\rho (\lambda (G-H)+tk)=-kI_2$ and Eq. (8.2) we get

$$\begin{aligned}&0=e_{11}-e_{22}=4ia_1(1+a_2),\,\,\, 0=e_{12}=2a_1(i+1+a_3+ia_4),\\&0=e_{21}=2a_1(-1+i-a_3+ia_4). \end{aligned}$$

Solving, we must have either

$$\begin{aligned} (i) \,\,\, a_1=0; \text{ or } (ii) \,\,\, a_2=-1,\,\, a_3=-1,\,\,\, a_4=-1. \end{aligned}$$

Now we find $a_1,\cdots ,a_4$ in terms of the $a_{ij}$. From (8.1) and (8.2) we have

$$\begin{aligned} \sum _{u=1}^4 a_uB_u&=\sum _{i=1}^{p_2} \sum _{j=1}^4 a_{ij}B_j\omega ^i - a_{ij}B_j^*\omega ^{-i}\\&=\sum _{i=1}^{p_2} a_{i1}B_1\omega ^i-a_{i1}B_1\omega ^{-i} + a_{i2}B_2\omega ^i+a_{i2}B_2\omega ^{-i}\\ {}&\qquad + a_{i3}B_3\omega ^i+a_{i3}B_3\omega ^{-i}+ a_{i4}B_4\omega ^i+a_{i4}B_4\omega ^{-i}. \end{aligned}$$

From this we get

$$\begin{aligned}&a_1=\sum _{i=1}^{p_2} a_{i1}(\omega ^i-\omega ^{-i});\quad a_2=\sum _{i=1}^{p_2} a_{i2}(\omega ^i+\omega ^{-i});\\&a_3=\sum _{i=1}^{p_2} a_{i3}(\omega ^i+\omega ^{-i});\quad a_4=\sum _{i=1}^{p_2} a_{i4}(\omega ^i+\omega ^{-i}). \end{aligned}$$

Now if we have (i) $a_1=0$, then $p>2$ is a prime means that the $\omega ^i-\omega ^{-i},i=1,2,\cdots ,p_2$ are linearly independent over ${\mathbb {Q}}$, so that we must than have $a_{i1}=0$ for all i.

Observe from previous definitions that $a_{i1}=|D_{0,i,0}|-|D_{0,i,2}|.$ From $D^{(-1)}=tD$ and $D \cup D^{(-1)}=G-\langle t \rangle$ we have $|D_{0,i,0}|+|D_{0,i,2}|=|N|$. So $|D_{0,i,0}|=|D_{0,i,2}|=|N|/2$. Thus |N| is even, which contradicts our assumption on $\ker \pi$.

So now assume (ii), so that

$$\begin{aligned} \rho (D)&=\begin{bmatrix} i&{}-i-1\\ 1-i&{}-i\end{bmatrix} + \sum _{i=1}^4 a_iB_i\\ {}&= \begin{bmatrix} i&{}-i-1\\ 1-1&{}-i\end{bmatrix}+a_1 B_1 -B_2-B_3-B_4=a_1I_2. \end{aligned}$$

But $-\rho (D^2)=\rho (DD^{(-1)})=kI_2$ then gives $a_1^2=-k.$ Here $a_1 \in {\mathbb {Q}}[\omega ]$. Recall that $\omega =e^\frac{2\pi i}{p}$, so the Galois group of $[{\mathbb {Q}}(\omega ):{\mathbb {Q}}]$ is cyclic of even order $p-1$. By the Galois correspondence, $\mathbb Q(\omega )$ has a unique quadratic subfield. In particular, we can verify that the subfield is exactly ${\mathbb {Q}}(\sqrt{p})$ if $p\equiv 1 \pmod {4}$, and ${\mathbb {Q}}(\sqrt{-p})$ if $p\equiv 3 \pmod {4}$. This follows from the Gauss sum:

$$\begin{aligned} \left( \sum _{n=0}^{p-1}\left( \frac{n}{p}\right) \omega ^n\right) ^2 = (-1)^{\frac{p-1}{2}}p \end{aligned}$$

Note that $k \equiv 3 \pmod {4}$ so k is not an integer square. Therefore $a_1^2=-k$ implies $k=px^2$ for some $x \in {\mathbb {Z}}$. However, $k=4p|N|-1$ so we have a contradiction, as k must be congruent to both 0 and $-1 \pmod {p}$. $\square$

9 Groups of Order Less Than or Equal to 72

Here are the non-dicyclic groups (using magma notation) of order at most 72 that meet the following requirements: (i) they are not abelian; (ii) their Sylow 2-subgroups are generalized quaternion groups; (iii) they have a single involution.

$$\begin{aligned}&G_{24, 3},\quad G_{24, 11},\quad G_{40, 11},\quad G_{48, 18},\quad G_{48, 27},\quad G_{48, 28},\quad G_{72, 3}, \\ {}&G_{72, 11},\quad G_{72, 24},\quad G_{72, 25},\quad G_{72, 26},\quad G_{72, 31},\quad G_{72, 38} \end{aligned}$$

We note that all of the dicyclic groups of order less than 72 and divisible by 8 are SRHDS groups by Theorems 1.2 and 1.3, while $\textrm{Dic}_{72}$ is not by Proposition 8.1.

We will determine whether the remaining groups have a SRHDS. If they have a SRHDS then we give a SRHDS explicitly. If not, then we give a proof that the group is not a SRHDS group.

In the cases of $G_{72, 3},\,\,\, G_{72, 11}, \,\,\,G_{72, 24},\,\,\, G_{72,25},$ and $G_{72,31},$ we use the following process to show they are not SRHDS groups: Given one of the four groups G, we take a right transversal $g_0=1, \dots , g_8$ for $Q_8\le G$. Assuming there is an SRHDS D, we write D as in (7.1). We can assume $F_0=\{x,y,xy\}$ by Lemma 7.3. By Lemma 7.2, there are 16 possibilities for each $F_i$, and a Magma [23] calculation verifies that none of these combinations give a SRHDS.

• (1) $G_{24,3}=\textrm{SL}(2,3)=\langle a,b,c,d| a^3=1, b^2 = d, c^2 = d, d^2, b^a = c, c^a = b c, c^b = c d\rangle .$ Here $D=\{a^2 c d, a b c d, a c d, c d, a^2 b d, a^2 d, a^2 b c, a, b c, a b, b \}.$

• (2) $G_{24,11}={\mathcal {C}}_3 \times Q_8$. This is not a SRHDS group by Proposition  8.2.

• (3) $G_{40,11}={\mathcal {C}}_5 \times Q_8$. This is not a SRHDS group by Proposition  8.2.

• (4) $G_{48,18}={\mathcal {C}}_3 \rtimes \textrm{Dic}_{16}$ $= \langle a,b,c,d,e|d^2=e^3=1, a^2=b^2=c^2=d, b^a=bc, c^a=c^b=cd, d^a=d^b=d^c=d, e^a=e^2, e^b=e^c=e^d=e\rangle$ and let D be

  $$\begin{aligned}&\{ a d e^2, d e^2, a e, e, a b c e^2, a b c, b c e^2, a b d e^2, b d e^2, b c e, a c d, a c d e^2, a b d,\\ {}&c d e^2, c d, a c d e, c d e, b d e, b c d, a, a b c d e, b, a b e\}. \end{aligned}$$

• (5) $G_{48,27}={\mathcal {C}}_3 \times \textrm{Dic}_{16}$. We show $G_{48,27}$ is not a SRHDS group. Let ${\mathcal {C}}_3=\langle r \rangle$. Then $D=D_0+D_1r+D_2r^2, D_i \subset \textrm{Dic}_{16}$. Now $D^{(-1)}=tD$ gives $D_0^{(-1)}=tD_0$ and $D_2=tD_1^{(-1)}$. Also Lemma 3.1 shows that the sizes of $D_0,D_1,D_2$ are 7, 8, 8 (in some order). By replacing D by $r^iD$ if necessary we may assume that $|D_0|=7$ and that $D_0+1,D_1,D_2$ are transversals for G/H. Using $D_0^{(-1)}=tD_0$ one sees that there are 64 possible $D_0$s and 256 possible $D_1$s. Further, $D_2$ is determined by $D_2=tD_1^{(-1)}$. There are thus $64 \cdot 256$ possibilities for D and one checks that none of these give a SRHDS.

• (6) Let $G_{48,28} = \langle a,b,c,d,e|b^3=e^2=1, a^2=c^2=d^2=e, b^a=b^2, c^a=d, c^b=de, d^a=c, d^b=cd, d^c=de, e^a=e^b=e^c=e^d=e\rangle$. Here one D is

  $$\begin{aligned}&\{ a b^2 d e, a b^2 c d e, b^2 c d e, c e, a b c, b^2 c, b c, d, a d e, a b^2 c e, a c, a b^2, a c d, c d, \\ {}&b^2 d, b^2 e, a b d e, b d e, b c d, a, a b, a b c d e, b\}. \end{aligned}$$

• (7) $G_{72,3} =Q_8 \rtimes {\mathcal {C}}_9=\langle i,j,b|i^4=j^4=b^9=1, i^j=i^{-1}, i^2=j^2, i^b=j, j^b=ij\rangle$. The Magma search described at the beginning of this section shows this is not an SRHDS group.

• (8) $G_{72,11} ={\mathcal {C}}_9 \times Q_8$. The Magma search described at the beginning of this section shows this is not an SRHDS group.

• (9) $G_{72,24} ={\mathcal {C}}_3^2 \rtimes Q_8=\langle a,b,i,j|a^3=b^3=i^4=j^4=1, ab=ba, i^j=i^{-1}, i^2=j^2, a^i=a, b^i=b^2,a^j=a^2, b^j=b\rangle$. The Magma search described at the beginning of this section shows this is not an SRHDS group.

• (10) $G_{72,25} ={\mathcal {C}}_3 \times SL(2,3)$. The Magma search described at the beginning of this section shows this is not an SRHDS group.

• (11) $G_{72,26} ={\mathcal {C}}_3 \times Dic_{24}$. This is not an SRHDS group by Proposition 8.2.

• (12) $G_{72,31} ={\mathcal {C}}_3^2 \rtimes Q_8=\langle a,b,i,j|a^3=b^3=i^4=j^4=1, ab=ba, i^j=i^{-1}, i^2=j^2, a^i=a^2, b^i=b^2,a^j=a, b^j=b\rangle$. The Magma search described at the beginning of this section shows this is not an SRHDS group.

• (13) $G_{72,38} ={\mathcal {C}}_3^2 \times Q_8$. This is not an SRHDS group by Proposition 8.2.