Hadamard difference sets in groups with a cyclic odd quotient

Introduction

A Hadamard (or Menon) difference set is a $(v,k,\lambda)$ difference set with parameters

of order $n = k-\lambda= u^2$. Throughout, $u$ is an odd integer with $u > 1$. Then every group containing such a difference set has order $4u^2$, and its Sylow 2-subgroups have order 4; for abelian groups the Sylow 2-subgroup is $C_4$ or $C_2 \times C_2$. For background we refer to the surveys [1, 10] and to [4].

Lander’s conjecture [5] asserts that if an abelian group $G$ contains a $(v,k,\lambda)$ difference set and a prime $p$ divides $\gcd(v,n)$, then the Sylow $p$-subgroup of $G$ is not cyclic. For Hadamard parameters with $u$ odd, $\gcd(v,n) = u^2$. The conjecture is known when $n$ is a power of a prime $p > 3$ [6] and, for Hadamard difference sets, for all orders $n \le529$, i.e. $u \le23$, by Feng, Leung, Schmidt and Smith [4], who also obtain strong restrictions for $u = 2^a3^b$ (here $u$ is even) with a cyclic Sylow 3-subgroup. For individual values of $u$, classical criteria (the exponent bound under self-conjugacy and the field descent bounds of Leung and Schmidt [7]) exclude many abelian groups; see § 5.3. For groups of order $4p^{2f}$, $p \ge5$ prime, with a quotient $C_{p^f} \times C_2 \times C_2$, nonexistence was proved by AbuGhneim and Smith [2], extending results of Iiams and Wan; very recently Schmidt [9] showed that no group of order $4p^2$ with $p \equiv3 \pmod{4}$, $p \ge7$, contains a Hadamard difference set.

The cyclic case of Lander’s conjecture for Hadamard parameters is the circulant Hadamard conjecture of Ryser and Turyn [11]. A proof of it was announced in September 2026 [8]. The purpose of this note is to show that the method of [8] applies, with a simpler final step, to the other abelian 2-part $C_2 \times C_2$, and more generally to every group of order $4u^2$ admitting a suitable abelian quotient. We do not know of a previous result excluding $C_2 \times C_{2u^2}$ for all odd $u > 1$.

When $u/w = 1$ the self-conjugacy hypothesis is empty and $N = 1$. The case $G = C_4 \times C_{u^2}$ is the circulant Hadamard conjecture [8]. When $u/w > 1$ the group $N$, of order $(u/w)^2$, may be non-abelian, and $G$ need not be abelian.

Proof. Take $N = 1$ and $w = u$ in Theorem 2. □

Proof. Write $G = E \times P_T \times Q$, where $E$ is the Sylow 2-subgroup, $P_T$ is the product of the Sylow $p$-subgroups with $p \in T$ and $Q$ is the product of the other odd Sylow subgroups. Then

and the primes dividing $u/w$ are the primes $q \mid u$ with $q \notin T$. Theorem 2 applies with $N = Q$. □

Remark 5. The self-conjugacy condition of Corollary 4 concerns the primes outside $T$ (among them all primes with non-cyclic Sylow subgroups), and only modulo the radical $r_T$. It is of a different nature from the classical self-conjugacy hypotheses (Turyn’s exponent bound and its refinements [11, 1, 10]), which require a prime $p$ dividing $u$ to be self-conjugate modulo the $p$-free part of the exponent of $G$.

The proof follows [8] closely. Section 2 recalls the local comparison of two character values; Section 3 proves that alternating products of character values are roots of unity of odd order, under the self-conjugacy hypothesis at the primes of $u/w$; Section 4 contains the binary step for both possible 2-parts. Section 5 explains why the hypotheses cannot simply be dropped from the argument and describes the ancillary computations.

Group rings and local lemmas

For a finite group $H$ and a commutative ring $R \subset\mathbb{C}$ closed under complex conjugation, $R[H]$ carries the involution $x = \sum x_h h \mapsto x^* = \sum\overline{x_h}h^{-1}$. A subset $D$ of a group $G$ of order $4u^2$ which is a Hadamard difference set gives

(Conversely, such an $h$ with coefficients $\pm1$ comes from a Hadamard difference set or from the complement of one.) Indeed $D D^* = n + \lambda G$ with $n = u^2$, and $(G - 2D)(G - 2D)^* = (4u^2 - 4\kappa+ 4\lambda)G + 4n = 4u^2$. Every coefficient of $h$ is $\pm1$.

For an abelian group $P$ and a character $\chi$ of $P$ we write $\chi(x) = \sum_z x_z\chi(z)$; evaluation is a ring homomorphism and $\chi(x^*) = \overline{\chi(x)}$.

Proof. These are standard facts: $p$ is unramified in $\mathbb{Q}(\zeta_L)$ and totally ramified in $\mathbb{Q}(\zeta_L,\xi)/\mathbb{Q}(\zeta_L)$, with $\Phi_{p^k}(1+Y)$ Eisenstein over $A$; see e.g. [12 Ch. 2], or the self-contained proof in [8 Lemma 2.1]. The valuation statement holds because $p^\ell A[\xi] = (\xi-1)^{\ell D}A[\xi]$ is the set of elements of valuation at least $\ell$, and the last statement follows from $p^\ell A[\xi] = \bigoplus_{j}p^\ell A\,\xi^j$.

Proof. Here $F(1)$, $F(\rho)$ denote the images of $F$ under the ring homomorphisms determined by $X \mapsto1$ and $X \mapsto\rho$, $X$ the chosen generator. Since $F(1)K(1)=F(\rho)K(\rho)=p^e b \ne0$, all four values are nonzero, and the two ratios are inverse to each other; it suffices to prove the claim for one of them. We argue by induction on $e$.

If $e=1$, the valuations of $F(1)$ and $K(1)$ in $A$ sum to $1$, so one of them, say $F(1)$, is a unit. As $F(\rho)\equiv F(1)\pmod{\rho-1}$, the ratio $F(\rho)/F(1)$ is a unit congruent to $1$.

Let $e \ge2$ and let $\xi$ be a primitive $p^e$-th root of unity. In the discrete valuation ring $A[\xi]$, normalized by $v(p)=1$, we have $v(F(\xi))+v(K(\xi))=e \ge2$, so after exchanging $F$ and $K$ we may assume $v(F(\xi)) \ge1$, i.e. $F(\xi) \in pA[\xi]$. Write $F=Q\Phi_{p^e}+R$ in $A[X]$ with $\deg R<D$. Since $R(\xi)=F(\xi)$, Lemma 6 gives $R \in pA[X]$. Let $\pi:A[C_{p^e}] \to A[C_{p^{e-1}}]$ be the projection sending $X$ to a generator. Since $\pi(\Phi_{p^e})=\sum_{j<p}X^{jp^{e-1}} \mapsto p$, we get $\pi(F) \in pA[C_{p^{e-1}}]$. Put $F'=\pi(F)/p$ and $K'=\pi(K)$. Then $F'K'=p^{e-1}b$ (cancellation of $p$ is legitimate since $A[C_{p^{e-1}}]$ is a free $A$-module), and since evaluation at $1$ and at $\rho$ factors through $\pi$ (as $\rho^{p^{e-1}}=1$), the two ratios are unchanged. The induction hypothesis applies to $(F',K')$.

Remark 8. The hypothesis that the $p$-adic valuation of the scalar equals the exponent $e$ of the group order is essential: for an odd prime $p$, $J=\sum_{z\in C_p}z$ and $F=p-2J$ one has $FF^*=p^2$ in $\mathbb{Z}[C_p]$, but $F(\rho)/F(1)=-1$ [8 Remark 3.2].

Proof. (i) is classical. For (ii), if the order of $\zeta$ is $\ell^a m$ with $\ell\nmid m$, then $\eta=\zeta^{\ell^a}$ has order $m$ and $\eta\equiv1\pmod{\mathfrak{p}}$; if $m>1$, then $0=1+\eta+\cdots+\eta^{m-1}\equiv m\pmod{\mathfrak{p}}$, contradicting $\ell\nmid m$.

Alternating character products

Fix odd integers $u>1$ and $w>1$ with $w\mid u$ and $\gcd(w,u/w)=1$. Let $I$ be the set of primes dividing $w$, $r=\prod_{p\in I}p$, and let $P=C_{w^2}$, written as the product of its cyclic Sylow subgroups $P_p\cong C_{p^{e_p}}$ with $e_p=2v_p(w)=v_p(u^2)$. Fix $m_0\in\{1,4\}$ and put

For each $p \in I$ fix a generator $X_p$ of $P_p$ and a primitive $p$-th root of unity $\rho_p \in B$. For $S \subseteq I$ let $\chi_S$ be the character of $P$ with $\chi_S(X_p) = \rho_p$ for $p \in S$ and $\chi_S(X_p) = 1$ for $p \notin S$. For $x \in\mathbb{Z}[\zeta_{m_0}][P]$ put $x_S = \chi_S(x) \in B$ and $\varepsilon_S = (-1)^{|S|}$.

Proof. Applying $\chi_S$ to $xx^* = u^2$ gives $x_S\overline{x_S} = u^2$; in particular $x_S \ne0$ and $\Delta(x) \in K_0^\times$. Since $B$ is a Dedekind domain, it suffices to show $v_{\mathfrak p}(\Delta(x)) \ge0$ for every nonzero prime $\mathfrak p$ of $B$. Let $\ell$ be the rational prime below $\mathfrak p$.

Case $\ell\nmid u$. Then $u^2$ is a unit at $\mathfrak p$, hence so is every $x_S$, and $v_{\mathfrak p}(\Delta(x)) = 0$.

Case $\ell= q \mid u/w$. Then $q \nmid m_0r$, so $q$ is unramified in $K_0$, and complex conjugation fixes $\mathfrak p$ by the self-conjugacy hypothesis. Hence $v_{\mathfrak p}(\overline{x_S}) = v_{\mathfrak p}(x_S)$, and $x_S\overline{x_S} = u^2$ gives $v_{\mathfrak p}(x_S) = v_q(u)$ for every $S$. Since $\sum_{S \subseteq I}\varepsilon_S = (1-1)^{|I|} = 0$, we get $v_{\mathfrak p}(\Delta(x)) = 0$.

Case $\ell= p \in I$. Put $L = m_0r/p$, so $p \nmid L$ and $A_0 := \mathbb{Z}[\zeta_L] \subseteq B$ contains $\zeta_{m_0}$ and all $\rho_q$, $q \in I \setminus\{p\}$. Let $A$ be the localization of $A_0$ at $\mathfrak p \cap A_0$. Fix $S \subseteq I \setminus\{p\}$. Evaluating every component $P_q$, $q \ne p$, by $\chi_S$ and keeping $X_p$ as a variable defines a ring homomorphism $\mathbb{Z}[\zeta_{m_0}][P] \to A_0[P_p]$ compatible with the involutions; let $F$, $K$ be the images of $x$, $x^*$. Then

so $b \in A^\times$, and $P_p \cong C_{p^{e_p}}$ with $e_p \ge2$. By Lemma 7,

Since $A[\rho_p] \subseteq B_{\mathfrak p}$ and $\rho_p-1 \in\mathfrak p$, each such ratio is a unit of $B_{\mathfrak p}$ congruent to 1 modulo $\mathfrak p$. Pairing $S$ with $S \cup\{p\}$,

so $\Delta(x) \in B_{\mathfrak p}^{\times}$ and $\Delta(x) \equiv1 \pmod{\mathfrak p}$.

Thus $\Delta(x) \in B$. For every embedding $\sigma: K_0 \to\mathbb{C}$ we have $\sigma(\overline{y}) = \overline{\sigma(y)}$ ($K_0$ is abelian), hence $|\sigma(x_S)| = u$ for all $S$, and $|\sigma(\Delta(x))| = u^{\sum\varepsilon_S} = 1$. By Lemma 9(i), $\Delta(x)$ is a root of unity. For $p \in I$ choose a prime $\mathfrak p \mid p$; by the last case $\Delta(x) \equiv1 \pmod{\mathfrak p}$, so its order is a power of $p$ by Lemma 9(ii). $\square$

Remark 11. If $|I| \ge2$ the proposition gives $\Delta(x) = 1$. For $u = w = p^a$ the order can be $p$: $x = p^aX_p$ gives $\Delta(x) = \rho_p^{-1}$ [8 Section 3].

Proof of the main theorem

Let $D \subseteq G$ be a Hadamard difference set, $h = G - 2D$, and let $\bar{h} \in\mathbb{Z}[G/N]$ be the image of $h$. Then $\bar{h}\bar{h}^* = 4u^2$, and every coefficient of $\bar{h}$ is a sum of $|N| = (u/w)^2$ signs, hence an odd integer. Identify $G/N = E \times P$ with $P = C_{w^2}$, and let $I$, $r$, $\chi_S$, $\varepsilon_S$ be as in Section 3. Put

For $S \ne\varnothing$, $\chi_S$ is a nontrivial character, so $J_S=0$; and $J_{\varnothing}=w^2$.

We shall use the following elementary device [8 Lemma 4.1]. Let $O$ be a discrete valuation ring with maximal ideal $\mathfrak{n}$ and residue field $k=O/\mathfrak{n}$, and let $0 \ne t \in\mathfrak{n}$. Then $1+tO$ is a subgroup of $O^{\times}$ and

is a homomorphism, because $((1+tL)(1+tM)-1)/t=L+M+tLM$ and $((1+tL)^{-1}-1)/t=-L/(1+tL)$.

The Klein case

Write $E=\{1,y_1,y_2,y_1y_2\}$ and

where all coefficients of all $H_j$ are odd. For the four characters $\psi$ of $E$, put $x^\psi=\frac{1}{2}\psi(\bar{h})\in\mathbb{Z}[P]$:

Each coefficient is half a sum of four odd integers, hence an integer. Since $\psi$ is real-valued, evaluation at $\psi$ commutes with ${}^{*}$, so $cc^{*}=d_id_i^{*}=u^2$. We have the exact identities

and

because each coefficient of $H_1+H_2+H_3$ is a sum of three odd integers.

Apply Proposition 10 with $m_0=1$ (the hypothesis is that every prime divisor of $u/w$ is self-conjugate modulo $r$): $\Delta(c)$ and $\Delta(d_i)$ are roots of unity of odd order, hence so is $R_i=\Delta(d_i)/\Delta(c)$.

Let $\mathfrak{m}$ be a prime of $B=\mathbb{Z}[\zeta_r]$ above $2$, $O=B_{\mathfrak{m}}$, $\mathfrak{n}=\mathfrak{m}O$, $k=O/\mathfrak{n}$. As $r$ is odd, $2$ is unramified in $B$ and $\mathfrak{n}=2O$. Each $c_S$ is a unit of $O$ since $c_Sc_S^{*}=u^2$ is odd. By (2),

so $r_{i,S}\in1+2O$ and $R_i=\prod_{S\subseteq I}r_{i,S}^{\varepsilon_S}\in1+2O$. A root of unity of odd order congruent to $1$ modulo $\mathfrak{m}$ equals $1$ by Lemma 9(ii) (with $\ell=2$). Hence $R_i=1$, and applying $\lambda_2$,

Summing over $i$ and using (3),

But $w^2/c_{\varnothing}$ is a unit of $O$. This contradiction proves the theorem when $E\cong C_2\times C_2$.

The cyclic case

This is the argument of [8 Section 4], which we include for completeness, now with $P=C_{w^2}$ and the scalar $u^2$. Let $Z$ generate $E$ and write $\bar{h}=H_0+ZH_1+Z^2H_2+Z^3H_3$ with all coefficients of the $H_j\in\mathbb{Z}[P]$ odd. Put

the halves of the evaluations of $\bar{h}$ at $Z=1,-1,i$. They lie in $\mathbb{Z}[P]$, $\mathbb{Z}[P]$, $\mathbb{Z}[i][P]$ and satisfy $cc^{*}=dd^{*}=gg^{*}=u^2$. With $b=(H_1+H_3)/2$ one has

with $U \in\mathbb{Z}[i][P]$ (each coefficient $\alpha+i\beta$ of $H_2+iH_3$ has $\alpha,\beta$ odd). Apply Proposition 10 with $m_0=4$ (the hypothesis is that every prime divisor of $u/w$ is self-conjugate modulo $4r$; note $\mathbb{Z}[P]\subseteq\mathbb{Z}[i][P]$). Then $R=\Delta(d)/\Delta(c)$ and $W=\Delta(g)/\Delta(c)$ are roots of unity of odd order in $B=\mathbb{Z}[\zeta_{4r}]$.

Let $\mathfrak{m}$ be a prime of $B$ above 2, $O=B_{\mathfrak{m}}$, $\mathfrak{n}=\mathfrak{m}O$. Here $1+i$ is a uniformizer of $O$, the image of $i$ in $k$ is 1, and every $c_S$ is a unit. Using $i-1=i(1+i)$ and $2=(1+i)(1-i)$,

Hence $R,W\equiv1\pmod{\mathfrak{m}}$, so $R=W=1$ by Lemma 9(ii). Applying $\lambda_2$ and $\lambda_{1+i}$,

Since $L_{r,S}+L_{w,S}=(i-1)\frac{b_S}{c_S}-\frac{J_S}{c_S}-(1-i)\frac{U_S}{c_S}\equiv-\frac{J_S}{c_S}\equiv\frac{J_S}{c_S}\pmod{\mathfrak{n}}$ (the residue field has characteristic 2), adding gives $w^2/c_{\varnothing}\equiv0\pmod{\mathfrak{n}}$, a contradiction. This completes the proof of Theorem 2. □

Remarks on the hypotheses and computations

The exponent condition

The argument needs the $p$-part of the cyclic quotient to have order exactly $p^{v_p(u^2)}$ (Remark 8). This is not a technicality: Hadamard difference sets exist in $C_2\times C_2\times C_3\times C_3$ and in $C_4\times C_3\times C_3$ (explicit $(36,15,6)$ difference sets are verified in the ancillary script). Projecting the first one onto $C_2\times C_2\times C_3$, whose 3-part has order $3<9=u^2$, one finds $\Delta(c)=-1$, $\Delta(d_1)=1$, $\Delta(d_2)=\Delta(d_3)=-1$, so $R_1=-1\ne1$: some alternating products have even order, and the argument correctly breaks down.

The self-conjugacy condition

At a prime $\mathfrak{n}$ above $q\mid u/w$ the comparison lemma is not available, and without self-conjugacy the values $x_S$ may lie above different primes over $q$. The following example (which is not itself in the setting of Proposition 10, where the scalar is $u^2$ and $P=C_{w^2}$) illustrates this local phenomenon: in $\mathbb{Z}[i][C_3]$ the element $x=-i+(1+i)X+(1+i)X^2$ satisfies $xx^*=5$, while $x(1)=2+i$ and $x(\rho)=-1-2i$ generate the two distinct primes above 5 in $\mathbb{Z}[i]$; the quotient $x(1)/x(\rho)$ is not integral. Here 5 is not self-conjugate modulo 12 (it is modulo 3).

Comparison with classical criteria

For an individual $u$, many of the groups covered by Corollary 4 were already excluded by classical results. We mention two criteria for $G=C_2\times C_{2u^2}$, $u$ odd. When $u$ is a prime power the first one alone excludes $G$, since the $p$-free part of $\exp G$ is 2 and every odd prime is self-conjugate modulo 2; we therefore assume that $u$ is not a prime power.

1. The exponent bound under self-conjugacy ([11]; [3]; see [10 Thm. 2.5]): if a prime $p\mid u$ with $p^a\Vert u$ is self-conjugate modulo the $p$-free part $2(u/p^a)^2$ of $\exp G$, the cyclic Sylow $p$-subgroup, of exponent $p^{2a}>p^a$, is impossible.

2. The field descent bound of Leung and Schmidt [7 Thm. 4.3] with $A=C_{u^2}$ and $H=C_2\times C_2$ (so that $\gcd(|A|,|H|)=\gcd(n,|H|)=1$): $n=u^2\le|H|F^2/(4\varphi(F))=F^2/\varphi(F)$, where $F=\gcd(u^2,F(2u^2,u^2))$.

By our computation (ancillary script, Part F), these two criteria exclude $C_2\times C_{2u^2}$ for every odd $u<11715$ that is not a prime power, and among $u\le2\cdot10^5$ exactly $u\in\{11715,16401,82005\}$ survive. For the cyclic group $C_{4u^2}$ we combined the exponent bound (with $p$-free part $4(u/p^a)^2$) with the bound $\psi(u)\le F(u^2,u)$ of [7 Cor. 4.5]; the same three values survive, in agreement with [7 Cor. 4.5], which excludes circulant Hadamard matrices of order $4<v<548\,964\,900=$ 4 · $11715^{2}$. We have not applied other known criteria, so the novelty of Corollary 3 lies in its uniformity in $u$ rather than in any particular small case.

Ancillary computations

The script anc/verify_hds.py (exact integer arithmetic, except for a floating-point pre-filter whose survivors are all checked exactly; about one minute in total; it requires numpy)

• checks the identities (2), (3) and those of the cyclic case on random sign arrays;

• enumerates all $x \in\mathbb{Z}[C_{9}]$ and all $x \in\mathbb{Z}[i][C_{9}]$ with $xx^{*}=9$ (there are 108 and 432 of them) and verifies Proposition 10 for each of them: $\Delta(x)$ is integral, is a root of unity of order 1 or 3, and is congruent to 1 above 3;

• verifies the two $(36,15,6)$ difference sets and the computations of § 5;

• counts the abelian groups of order $4u^{2}$, $u$ odd, $u \le105$, covered by Corollary 4: of 424 groups, 204 have no cyclic odd Sylow subgroup, 104 have all odd Sylow subgroups cyclic, 53 others are covered through the self-conjugacy hypothesis (for some choice of $T$), and 63 satisfy the hypothesis of Lander’s conjecture but are not covered;

• carries out the comparison of § 5.3.

References

ORCID: 0009-0009-1526-1178