\documentclass[11pt]{amsart} \usepackage[margin=1.1in]{geometry} \usepackage{amsmath,amssymb,amsthm,mathtools} \usepackage[hidelinks]{hyperref} \usepackage{enumitem} \newtheorem{theorem}{Theorem} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{lemma}[theorem]{Lemma} \theoremstyle{definition} \newtheorem{definition}[theorem]{Definition} \theoremstyle{remark} \newtheorem{remark}[theorem]{Remark} \newtheorem{example}[theorem]{Example} \newcommand{\Z}{\mathbb{Z}} \newcommand{\Q}{\mathbb{Q}} \newcommand{\C}{\mathbb{C}} \newcommand{\F}{\mathbb{F}} \newcommand{\m}{\mathfrak{m}} \newcommand{\p}{\mathfrak{p}} \newcommand{\n}{\mathfrak{n}} \newcommand{\eps}{\varepsilon} \DeclareMathOperator{\rad}{rad} \DeclareMathOperator{\Frac}{Frac} \title[Hadamard difference sets with a cyclic odd part]{Hadamard difference sets in groups\\ with a cyclic odd quotient} \author{David Niedbala Giraudin} \address{ORCID: 0009-0009-1526-1178} \date{November 2026} \begin{document} \begin{abstract} Let $u>1$ be odd. We prove that the abelian group $C_2\times C_{2u^2}$ contains no Hadamard difference set. Together with the recently announced proof of the circulant Hadamard conjecture, this shows that an abelian group of order $4u^2$ ($u>1$ odd) whose odd Sylow subgroups are all cyclic never contains a Hadamard difference set, so Lander's conjecture holds for Hadamard difference sets in these groups. More generally, let $G$ be a group of order $4u^2$ with a normal subgroup $N$ such that $G/N\cong E\times C_{w^2}$, where $|E|=4$, $w>1$, $w\mid u$ and $\gcd(w,u/w)=1$. If every prime divisor of $u/w$ is self-conjugate modulo $\rad(w)$ (modulo $4\rad(w)$ when $E$ is cyclic), then $G$ contains no Hadamard difference set. The proof adapts the alternating character products used for the cyclic case. \end{abstract} \maketitle \section{Introduction} A \emph{Hadamard} (or \emph{Menon}) difference set is a $(v,k,\lambda)$ difference set with parameters \[ (v,k,\lambda)=(4u^2,\;2u^2-u,\;u^2-u), \] 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 \cite{DavisJedwab,SchmidtUpdate} and to \cite{FLSS}. Lander's conjecture \cite{Lander} 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 $p>3$ \cite{LMS} and, for Hadamard difference sets, for all orders $n\le 529$, i.e.\ $u\le 23$, by Feng, Leung, Schmidt and Smith \cite{FLSS}, 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 \cite{LSfield}) exclude many abelian groups; see Section~\ref{rem:classical}. 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 \cite{AbuGhneimSmith}, extending results of Iiams and Wan; very recently Schmidt \cite{Schmidt4p2} 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 \cite{Turyn}. A proof of it was announced in September 2026 \cite{OAI}. The purpose of this note is to show that the method of \cite{OAI} 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 $u>1$. \begin{definition} Let $m\ge1$ and let $q$ be a prime with $q\nmid m$. We say that $q$ is \emph{self-conjugate modulo $m$} if $q^j\equiv-1\pmod m$ for some integer $j\ge0$. Equivalently, complex conjugation fixes every prime ideal above $q$ in $\Z[\zeta_m]$. \end{definition} \begin{theorem}\label{thm:main} Let $u>1$ be odd and let $G$ be a group of order $4u^2$. Suppose that $G$ has a normal subgroup $N$ with \[ G/N\;\cong\;E\times C_{w^2},\qquad |E|=4,\quad w>1,\quad w\mid u,\quad \gcd(w,u/w)=1, \] and put $r=\rad(w)$. Assume that every prime divisor of $u/w$ is self-conjugate modulo $r$ if $E\cong C_2\times C_2$, and modulo $4r$ if $E\cong C_4$. Then $G$ contains no Hadamard difference set. \end{theorem} 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 \cite{OAI}. When $u/w>1$ the group $N$, of order $(u/w)^2$, may be non-abelian, and $G$ need not be abelian. \begin{corollary}\label{cor:klein} For every odd $u>1$, the abelian group $C_2\times C_{2u^2}\cong C_2\times C_2\times C_{u^2}$ contains no Hadamard difference set. \end{corollary} \begin{proof} Take $N=1$ and $w=u$ in Theorem~\ref{thm:main}. \end{proof} \begin{corollary}\label{cor:lander} Let $G$ be an abelian group of order $4u^2$, $u>1$ odd, and let $I_c$ be the set of primes $p\mid u$ whose Sylow $p$-subgroup is cyclic. Suppose that there is a nonempty set $T\subseteq I_c$ such that every prime $q\mid u$ with $q\notin T$ is self-conjugate modulo $r_T=\prod_{p\in T}p$ (modulo $4r_T$ if the Sylow $2$-subgroup is cyclic). Then $G$ contains no Hadamard difference set; in particular, Lander's conjecture holds for Hadamard difference sets in $G$. This applies with $T=I_c$ whenever all odd Sylow subgroups of $G$ are cyclic, i.e.\ $G\cong C_{4u^2}$ or $G\cong C_2\times C_{2u^2}$. \end{corollary} \begin{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 $P_T\cong C_{w^2}$ with $w=\prod_{p\in T}p^{v_p(u)}>1$, $\gcd(w,u/w)=1$, $\rad(w)=r_T$, and the primes dividing $u/w$ are the primes $q\mid u$ with $q\notin T$. Theorem~\ref{thm:main} applies with $N=Q$. \end{proof} \begin{remark} The self-conjugacy condition of Corollary~\ref{cor:lander} 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 \cite{Turyn,DavisJedwab,SchmidtUpdate}), which require a prime $p$ dividing $u$ to be self-conjugate modulo the $p$-free part of the exponent of $G$. \end{remark} The proof follows \cite{OAI} closely. Section~\ref{sec:local} recalls the local comparison of two character values; Section~\ref{sec:alt} 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~\ref{sec:proof} contains the binary step for both possible $2$-parts. Section~\ref{sec:remarks} explains why the hypotheses cannot simply be dropped from the argument and describes the ancillary computations. \section{Group rings and local lemmas}\label{sec:local} For a finite group $H$ and a commutative ring $R\subseteq\C$ 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 \[ h:=G-2D\in\Z[G]\qquad\text{with}\qquad hh^*=4u^2 . \] (Conversely, such an $h$ with coefficients $\pm1$ comes from a Hadamard difference set or from the complement of one.) Indeed $DD^*=n+\lambda G$ with $n=u^2$, and $(G-2D)(G-2D)^*=(4u^2-4k+4\lambda)G+4n =4u^2$. 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 $\chi(x^*)=\overline{\chi(x)}$. \begin{lemma}\label{lem:local} Let $p$ be a prime, $L\ge1$ with $p\nmid L$, and let $A$ be the localization of $\Z[\zeta_L]$ at a prime ideal above $p$. Then $A$ is a discrete valuation ring with uniformizer $p$. For $k\ge1$ let $\xi$ be a primitive $p^k$-th root of unity and $D=(p-1)p^{k-1}$. Then $A[\xi]$ is a discrete valuation ring with uniformizer $\xi-1$, it is free over $A$ with basis $1,\xi,\dots,\xi^{D-1}$, and $p$ is a unit times $(\xi-1)^D$. Consequently, if $v$ denotes the valuation of $A[\xi]$ normalized by $v(p)=1$, then for $z\in A[\xi]$ and $\ell\ge0$ we have $v(z)\ge\ell\iff z\in p^\ell A[\xi]$, and for $R=\sum_{j1$, then $0=1+\eta+\dots+\eta^{m-1}\equiv m\pmod \p$, contradicting $\ell\nmid m$. \end{proof} \section{Alternating character products}\label{sec:alt} 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 \[ B=\Z[\zeta_{m_0r}],\qquad K_0=\Q(\zeta_{m_0r}). \] 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 $x\in\Z[\zeta_{m_0}][P]$ put $x_S=\chi_S(x)\in B$ and $\eps_S=(-1)^{|S|}$. \begin{proposition}[{cf.\ \cite[Proposition~3.4]{OAI}}]\label{prop:alt} Assume that every prime divisor of $u/w$ is self-conjugate modulo $m_0r$. Let $x\in\Z[\zeta_{m_0}][P]$ satisfy $xx^*=u^2$. Then \[ \Delta(x)=\prod_{S\subseteq I}x_S^{\eps_S} \] is a root of unity in $B$ whose order is a power of $p$ for every $p\in I$. In particular its order is odd. \end{proposition} \begin{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 $v_\p(\Delta(x))\ge0$ for every nonzero prime $\p$ of $B$. Let $\ell$ be the rational prime below $\p$. \smallskip\noindent\emph{Case $\ell\nmid u$.} Then $u^2$ is a unit at $\p$, hence so is every $x_S$, and $v_\p(\Delta(x))=0$. \smallskip\noindent\emph{Case $\ell=q\mid u/w$.} Then $q\nmid m_0r$, so $q$ is unramified in $K_0$, and complex conjugation fixes $\p$ by the self-conjugacy hypothesis. Hence $v_\p(\overline{x_S})=v_{\overline\p}(x_S)=v_\p(x_S)$, and $x_S\overline{x_S}=u^2$ gives $v_\p(x_S)=v_q(u)$ for every $S$. Since $\sum_{S\subseteq I}\eps_S=(1-1)^{|I|}=0$, we get $v_\p(\Delta(x))=0$. \smallskip\noindent\emph{Case $\ell=p\in I$.} Put $L=m_0r/p$, so $p\nmid L$ and $A_0:=\Z[\zeta_L]\subseteq B$ contains $\zeta_{m_0}$ and all $\rho_q$, $q\in I\setminus\{p\}$. Let $A$ be the localization of $A_0$ at $\p\cap A_0$. 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 $\Z[\zeta_{m_0}][P]\to A_0[P_p]$ compatible with the involutions; let $F,K$ be the images of $x,x^*$. Then \[ FK=u^2=p^{e_p}b,\qquad b=u^2/p^{e_p}\in\Z,\quad p\nmid b, \] so $b\in A^\times$, and $P_p\cong C_{p^{e_p}}$ with $e_p\ge 2$. By Lemma~\ref{lem:comparison}, \[ \frac{x_{S\cup\{p\}}}{x_S}=\frac{F(\rho_p)}{F(1)}\in A[\rho_p]^\times,\qquad \frac{x_{S\cup\{p\}}}{x_S}\equiv 1\pmod{\rho_p-1}. \] Since $A[\rho_p]\subseteq B_\p$ and $\rho_p-1\in\p$, each such ratio is a unit of $B_\p$ congruent to $1$ modulo $\p$. Pairing $S$ with $S\cup\{p\}$, \[ \Delta(x)=\prod_{S\subseteq I\setminus\{p\}} \Bigl(\frac{x_S}{x_{S\cup\{p\}}}\Bigr)^{\eps_S}, \] so $\Delta(x)\in B_\p^\times$ and $\Delta(x)\equiv1\pmod\p$. \smallskip Thus $\Delta(x)\in B$. For every embedding $\sigma:K_0\to\C$ we have $\sigma(\overline{y})=\overline{\sigma(y)}$ ($K_0$ is abelian), hence $|\sigma(x_S)|=u$ for all $S$, and $|\sigma(\Delta(x))|=u^{\sum\eps_S}=1$. By Lemma~\ref{lem:roots}(i), $\Delta(x)$ is a root of unity. For $p\in I$ choose a prime $\p\mid p$; by the last case $\Delta(x)\equiv1\pmod\p$, so its order is a power of $p$ by Lemma~\ref{lem:roots}(ii). \end{proof} \begin{remark} If $|I|\ge2$ the proposition gives $\Delta(x)=1$. For $u=w=p^a$ the order can be $p$: $x=p^aX_p$ gives $\Delta(x)=\rho_p^{-1}$ \cite[Section~3]{OAI}. \end{remark} \section{\texorpdfstring{Proof of Theorem~\ref{thm:main}}{Proof of the main theorem}}\label{sec:proof} Let $D\subseteq G$ be a Hadamard difference set, $h=G-2D$, and let $\bar h\in\Z[G/N]$ 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 $I,r,\chi_S,\eps_S$ be as in Section~\ref{sec:alt}. Put \[ J=\sum_{z\in P}z\in\Z[P]. \] 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 \cite[Lemma~4.1]{OAI}. Let $O$ be a discrete valuation ring with maximal ideal $\n$ and residue field $k=O/\n$, and let $0\ne t\in\n$. Then $1+tO$ is a subgroup of $O^\times$ and \begin{equation}\label{eq:lambda} \lambda_t:1+tO\to(k,+),\qquad \lambda_t(1+tL)=L\bmod\n \end{equation} is a homomorphism, because $\bigl((1+tL)(1+tM)-1\bigr)/t=L+M+tLM$ and $\bigl((1+tL)^{-1}-1\bigr)/t=-L/(1+tL)$. \subsection{\texorpdfstring{The case $E\cong C_2\times C_2$}{The Klein case}} Write $E=\{1,y_1,y_2,y_1y_2\}$ and \[ \bar h=H_0+y_1H_1+y_2H_2+y_1y_2H_3,\qquad H_j\in\Z[P], \] where all coefficients of all $H_j$ are odd. For the four characters $\psi$ of $E$, put $x^{\psi}=\tfrac12\psi(\bar h)\in\Z[P]$: \[ c=\tfrac{H_0+H_1+H_2+H_3}{2},\quad d_1=\tfrac{H_0-H_1+H_2-H_3}{2},\quad d_2=\tfrac{H_0+H_1-H_2-H_3}{2},\quad d_3=\tfrac{H_0-H_1-H_2+H_3}{2}. \] 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 \begin{equation}\label{eq:klein-diff} d_i-c=-2b_i,\qquad b_1=\tfrac{H_1+H_3}{2},\ b_2=\tfrac{H_2+H_3}{2},\ b_3=\tfrac{H_1+H_2}{2}\in\Z[P], \end{equation} and \begin{equation}\label{eq:klein-sum} b_1+b_2+b_3=H_1+H_2+H_3=J+2V,\qquad V\in\Z[P], \end{equation} because each coefficient of $H_1+H_2+H_3$ is a sum of three odd integers. Apply Proposition~\ref{prop:alt} 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 $\m$ be a prime of $B=\Z[\zeta_r]$ above $2$, $O=B_\m$, $\n=\m O$, $k=O/\n$. As $r$ is odd, $2$ is unramified in $B$ and $\n=2O$. Each $c_S$ is a unit of $O$ since $c_S\overline{c_S}=u^2$ is odd. By \eqref{eq:klein-diff}, \[ r_{i,S}:=\frac{(d_i)_S}{c_S}=1+2L_{i,S},\qquad L_{i,S}=-\frac{(b_i)_S}{c_S}\in O, \] so $r_{i,S}\in1+2O$ and $R_i=\prod_S r_{i,S}^{\eps_S}\in 1+2O$. A root of unity of odd order congruent to $1$ modulo $\m$ equals $1$ by Lemma~\ref{lem:roots}(ii) (with $\ell=2$). Hence $R_i=1$, and applying $\lambda_2$, \[ \sum_{S\subseteq I}\eps_S\frac{(b_i)_S}{c_S}\equiv0\pmod\n\qquad(i=1,2,3). \] Summing over $i$ and using \eqref{eq:klein-sum}, \[ 0\equiv\sum_{S\subseteq I}\eps_S\frac{J_S+2V_S}{c_S} \equiv\sum_{S\subseteq I}\eps_S\frac{J_S}{c_S}=\frac{w^2}{c_\varnothing} \pmod\n. \] But $w^2/c_\varnothing$ is a unit of $O$. This contradiction proves the theorem when $E\cong C_2\times C_2$. \subsection{\texorpdfstring{The case $E\cong C_4$}{The cyclic case}} This is the argument of \cite[Section~4]{OAI}, 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 $H_j\in\Z[P]$ odd. Put \[ c=\tfrac{H_0+H_1+H_2+H_3}{2},\quad d=\tfrac{H_0-H_1+H_2-H_3}{2},\quad g=\tfrac{H_0-H_2}{2}+i\,\tfrac{H_1-H_3}{2}, \] the halves of the evaluations of $\bar h$ at $Z=1,-1,i$. They lie in $\Z[P]$, $\Z[P]$, $\Z[i][P]$ and satisfy $cc^*=dd^*=gg^*=u^2$. With $b=(H_1+H_3)/2$ one has \[ d-c=-2b,\qquad g-c=(i-1)b-(H_2+iH_3),\qquad H_2+iH_3=(1+i)J+2U, \] with $U\in\Z[i][P]$ (each coefficient $\alpha+i\beta$ of $H_2+iH_3$ has $\alpha,\beta$ odd). Apply Proposition~\ref{prop:alt} with $m_0=4$ (the hypothesis is that every prime divisor of $u/w$ is self-conjugate modulo $4r$; note $\Z[P]\subseteq\Z[i][P]$). Then $R=\Delta(d)/\Delta(c)$ and $W=\Delta(g)/\Delta(c)$ are roots of unity of odd order in $B=\Z[\zeta_{4r}]$. Let $\m$ be a prime of $B$ above $2$, $O=B_\m$, $\n=\m O$. 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 $2=(1+i)(1-i)$, \[ \frac{d_S}{c_S}=1+2L_{r,S},\quad L_{r,S}=-\frac{b_S}{c_S};\qquad \frac{g_S}{c_S}=1+(1+i)L_{w,S},\quad L_{w,S}=i\frac{b_S}{c_S}-\frac{J_S}{c_S}-(1-i)\frac{U_S}{c_S}. \] Hence $R,W\equiv1\pmod\m$, so $R=W=1$ by Lemma~\ref{lem:roots}(ii). Applying $\lambda_2$ and $\lambda_{1+i}$, \[ \sum_S\eps_S L_{r,S}\equiv0,\qquad \sum_S\eps_S L_{w,S}\equiv0\pmod\n . \] Since $L_{r,S}+L_{w,S}=(i-1)\frac{b_S}{c_S}-\frac{J_S}{c_S}-(1-i)\frac{U_S}{c_S} \equiv-\frac{J_S}{c_S}=\frac{J_S}{c_S}\pmod\n$ (the residue field has characteristic $2$), adding gives $w^2/c_\varnothing\equiv0\pmod\n$, a contradiction. This completes the proof of Theorem~\ref{thm:main}. \qed \section{Remarks on the hypotheses and computations}\label{sec:remarks} \subsection{The exponent condition} The argument needs the $p$-part of the cyclic quotient to have order exactly $p^{v_p(u^2)}$ (Remark~\ref{rem:exact}). 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 $R_1=-1\neq1$: some alternating products have even order, and the argument correctly breaks down. \subsection{The self-conjugacy condition} At a prime $\p$ 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~\ref{prop:alt}, where the scalar is $u^2$ and $P=C_{w^2}$) illustrates this local phenomenon: in $\Z[i][C_3]$ 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 $\Z[i]$; the quotient $x(1)/x(\rho)$ is not integral. Here $5$ is not self-conjugate modulo $12$ (it is modulo $3$). \subsection{Comparison with classical criteria}\label{rem:classical} For an individual $u$, many of the groups covered by Corollary~\ref{cor:lander} 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. \begin{enumerate}[label=\textup{(\roman*)}] \item The exponent bound under self-conjugacy (Turyn \cite{Turyn}; Arasu, Davis and Jedwab \cite{ADJ}; see \cite[Thm.~2.5]{SchmidtUpdate}): 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. \item The field descent bound of Leung and Schmidt \cite[Thm.~4.3]{LSfield} 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 $F=\gcd\bigl(u^2,F(2u^2,u^2)\bigr)$. \end{enumerate} 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 $u\le 2\cdot10^5$ 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 $u\varphi(u)\le F(u^2,u)$ of \cite[Cor.~4.5]{LSfield}; the same three values survive, in agreement with \cite[Cor.~4.5]{LSfield}, which excludes circulant Hadamard matrices of order $4