\documentclass[11pt,a4paper,reqno]{amsart} \usepackage{amsmath,amssymb,amsthm,mathtools} \usepackage[a4paper,margin=24mm]{geometry} \usepackage{lmodern} \usepackage{microtype} \usepackage{booktabs} \usepackage{tabularx} \usepackage{array} \usepackage{xcolor} \usepackage{hyperref} \definecolor{primary}{RGB}{20,60,120} \hypersetup{ colorlinks=true, linkcolor=primary, citecolor=primary, urlcolor=primary, pdftitle={Explicit Finite p-Group Counterexamples to the Amit-Ashurst Conjecture}, pdfauthor={Shihao Wang and Zhiwei Zheng} } \setlength{\parskip}{4pt plus 1pt minus 1pt} % Theorem environments \theoremstyle{plain} \newtheorem{theorem}{Theorem}[section] \newtheorem{proposition}[theorem]{Proposition} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{corollary}[theorem]{Corollary} \newtheorem*{amitconjecture}{Amit Conjecture} \newtheorem*{amitashurstconjecture}{Amit--Ashurst Conjecture} \theoremstyle{definition} \newtheorem{definition}[theorem]{Definition} \theoremstyle{remark} \newtheorem{remark}[theorem]{Remark} % Custom mathematical macros \newcommand{\F}{\mathbb{F}} \DeclareMathOperator{\cl}{cl} \DeclareMathOperator{\GL}{GL} \DeclareMathOperator{\Sp}{Sp} \DeclareMathOperator{\Mat}{Mat} \DeclareMathOperator{\Alt}{Alt} \DeclareMathOperator{\rank}{rank} \DeclareMathOperator{\Span}{span} \title[Counterexamples to the Amit--Ashurst Conjecture]{Explicit Finite $p$-Group Counterexamples to the Amit--Ashurst Conjecture} \author{Shihao Wang} \address{Shihao Wang, Qiuzhen College, Tsinghua University, Beijing, China} \email{wangshihao25@mails.tsinghua.edu.cn} \author{Zhiwei Zheng} \address{Zhiwei Zheng, Yau Mathematical Sciences Center, Tsinghua University, Beijing, China} \email{zhengzhiwei@mail.tsinghua.edu.cn} \date{} \subjclass[2020]{Primary 20F10; Secondary 20D15} \begin{document} \begin{abstract} For every prime $p$, we construct a finite $p$-group $G$ of order $p^{498}$ and nilpotency class two for which the word $w=\prod_{j=1}^{12}[x_{2j-1},x_{2j}]$ admits a value $g\ne1$ whose fibre has size strictly less than $|G|^{23}$. This gives counterexamples to the Amit--Ashurst conjecture for every prime. We determine the exact fibre size by reducing the word equation to matrix equations over $\F_p$ and counting their solutions. \end{abstract} \maketitle \section{Introduction} Let $F_r$ denote the free group of rank $r \ge 1$ on free generators $x_1, \dots, x_r$. An \emph{ordinary word} $w \in F_r$ induces on every group $G$ a word map $w \colon G^r \to G$ via substitution. When $G$ is finite, we denote the image of this map by $w(G) = \{w(g_1, \dots, g_r) : g_i \in G\}$, and for each $g \in G$ we write \[ N_{w,G}(g) = \bigl|\{(g_1, \dots, g_r) \in G^r : w(g_1, \dots, g_r) = g\}\bigr|, \qquad P_{w,G}(g) = \frac{N_{w,G}(g)}{|G|^r}, \] for the cardinality and induced probability measure of the fibre over $g$, respectively. Amit proposed the following conjecture; see \cite{Abert2006}. \begin{amitconjecture} For every finite nilpotent group $G$ and every word $w\in F_r$, \begin{equation}\label{eq:amit} N_{w,G}(1)\ge |G|^{r-1}. \end{equation} \end{amitconjecture} Ashurst \cite{Ashurst2012} proposed the following strengthening, called the Amit--Ashurst conjecture \cite{CaminaCockeThillaisundaram2023}. \begin{amitashurstconjecture} For every finite nilpotent group $G$ and every word $w\in F_r$, \begin{equation}\label{eq:amit-ashurst} N_{w,G}(g)\ge |G|^{r-1} \qquad\text{for every }g\in w(G). \end{equation} \end{amitashurstconjecture} Since every finite nilpotent group is the direct product of its Sylow $p$-subgroups \cite[Theorem~5.2.4]{Robinson1996} and word evaluation commutes with direct products, \[ N_{w, A \times B}\bigl((a,b)\bigr) = N_{w,A}(a) N_{w,B}(b), \] both conjectures \eqref{eq:amit} and \eqref{eq:amit-ashurst} immediately reduce to the case of finite $p$-groups. Solomon's divisibility theorem \cite{Solomon1969} gives $N_{w,G}(1)\ge |G|$ for every finite group and every two-variable word. Ab\'ert \cite{Abert2006} and Nikolov--Segal \cite{NikolovSegal2007} showed that a finite group $G$ is solvable if and only if there exists a constant $\varepsilon_G>0$ such that $P_{w,G}(1)\ge\varepsilon_G$ for every $r\ge1$ and every word $w\in F_r$. For arbitrary numbers of variables, Levy \cite[Theorem~1.1]{Levy2011} and, independently, I\~niguez--Sangroniz \cite{IniguezSangroniz2017} proved the original Amit conjecture for finite groups of nilpotency class two. Levy's result also covers finite split metacyclic groups (semidirect products of two cyclic groups) \cite[Proposition~1.3]{Levy2011}. For nonidentity fibres, Ashurst \cite[Chapter~4]{Ashurst2012} established the Amit--Ashurst conjecture for dihedral $2$-groups and generalized quaternion groups. Camina--I\~niguez--Thillaisundaram \cite[Theorems~A and~C]{CaminaIniguezThillaisundaram2020} proved the Amit--Ashurst conjecture for two-variable words on finite $p$-groups of nilpotency class two and for products of disjoint commutators on finite $p$-groups with exactly two distinct irreducible complex character degrees. Camina--Cocke--Thillaisundaram \cite[Theorem~A]{CaminaCockeThillaisundaram2023} proved the full Amit--Ashurst conjecture for finite $p$-groups with a cyclic maximal subgroup. Kishnani--Kulshrestha \cite{KishnaniKulshrestha2024} proved that \[ N_{w,G}(g)\ge \frac{|G|^{r-1}}{|G'|} \qquad (g\in w(G)) \] for every finite group $G$ of nilpotency class two, where $G'=\langle[x,y]:x,y\in G\rangle$ and $[x,y]=x^{-1}y^{-1}xy$. For a group $G$, set \[ G_{(0)}=G,\qquad G_{(i)}=[G_{(i-1)},G] =\langle[x,y]:x\in G_{(i-1)},\ y\in G\rangle\qquad(i\ge1). \] A group $G$ has nilpotency class two, written $\cl(G)=2$, if \[ G_{(1)}\ne\langle1\rangle\qquad\text{and}\qquad G_{(2)}=\langle1\rangle. \] We obtain the following result. \begin{theorem}\label{thm:main} For every prime $p$, there exists an explicitly defined finite $p$-group $G$ of order $|G|=p^{498}$ and nilpotency class $\cl(G)=2$, such that for the product of twelve disjoint commutators \begin{equation}\label{eq:word-w} w(x_1, \dots, x_{24}) = \prod_{j=1}^{12} [x_{2j-1}, x_{2j}] \in F_{24}, \end{equation} there is an attained nonidentity value $g \in w(G) \setminus \{1\}$ whose fibre has exact cardinality \begin{align} N_{w,G}(g) &= p^{11298} \prod_{j=1}^{12} \bigl(p^{2j} - 1\bigr) \label{eq:exact-count} \\ &= |G|^{23} \prod_{j=1}^{12} \bigl(1 - p^{-2j}\bigr) < |G|^{23}. \label{eq:normalized-bound} \end{align} \end{theorem} Our construction uses the classical description of groups of nilpotency class two by bilinear maps \cite{Baer1938}, with quadratic maps describing central extensions of elementary abelian $2$-groups \cite[Section~2]{PakianathanYalcin2007}. We express commutators as alternating matrices and choose the relations using a nilpotent shift and an anisotropic binary quadratic form. The word equation then reduces to a symplectic matrix equation and a quadratic equation whose solutions are determined by the kernels of the shifts. Counting these solutions gives the fibre size in Theorem~\ref{thm:main}. \section{Ambient Linear Spaces and the Relation Subspace}\label{sec:linear} We first treat the case $p=2$. Unless otherwise stated, all vector spaces are over $\F_2$, and all vectors are column vectors. Let $H = \F_2^{11}$ be an $11$-dimensional vector space over $\F_2$ equipped with the ordered standard basis $e_0, e_1, \dots, e_{10}$. We define the nilpotent shift endomorphism $T \colon H \to H$ on basis vectors by \begin{equation}\label{eq:shift-T} T e_i = e_{i+1} \quad (0 \le i < 10), \qquad T e_{10} = 0. \end{equation} For each integer $1 \le s \le 6$, the $s$-fold iterate $T^s$ satisfies \begin{equation}\label{eq:ker-Ts} \ker T^s = \langle e_{11-s}, \dots, e_{10} \rangle, \qquad \dim_{\F_2} \ker T^s = s. \end{equation} Next, let $P = \F_2^{24}$ have ordered standard basis $f_1, \dots, f_{24}$, and form the $35$-dimensional direct sum \[ D = P \oplus H \cong \F_2^{35}, \] where we order the basis of $D$ by placing all $24$ basis vectors of $P$ before the $11$ basis vectors of $H$. In characteristic two, we denote by $\Alt_n = \Alt_n(\F_2)$ the $\F_2$-vector space of symmetric $n \times n$ matrices with zero diagonal: \[ \Alt_n = \bigl\{ M \in \Mat_{n \times n}(\F_2) : M^T = M \text{ and } M_{ii} = 0 \text{ for all } 1 \le i \le n \bigr\}. \] Here $\Mat_{a\times b}(\F_2)$ denotes the space of $a\times b$ matrices. Clearly $\dim_{\F_2} \Alt_n = \binom{n}{2}$. The canonical bilinear assignment \begin{equation}\label{eq:wedge-ident} v \wedge u \longmapsto v u^T + u v^T \qquad (v, u \in \F_2^n) \end{equation} identifies the exterior square $\bigwedge^2 \F_2^n$ isomorphically with $\Alt_n$, sending each elementary wedge $e_i \wedge e_j$ ($i < j$) to the elementary alternating matrix having $1$ at entries $(i,j)$ and $(j,i)$ and $0$ elsewhere. For each \[ z=(z_{1,1},z_{1,2},\ldots,z_{6,1},z_{6,2})\in H^{12}, \] define $C(z)\in\Mat_{24\times11}(\F_2)$ by \begin{align} C(z)_{2s-1,*}&=z_{s,1}^{T}, & C(z)_{2s,*}&=z_{s,2}^{T},\label{eq:C-top}\\ C(z)_{12+2s-1,*}&=\bigl(T^s(z_{s,1}+z_{s,2})\bigr)^T, & C(z)_{12+2s,*}&=(T^s z_{s,2})^T,\label{eq:C-bot} \end{align} for $1\le s\le6$, where $C(z)_{i,*}$ denotes the $i$-th row of $C(z)$. Define \begin{equation}\label{eq:K-W-def} K=\left\{\begin{pmatrix}0_{24\times24}&C(z)\\C(z)^T&0_{11\times11}\end{pmatrix}:z\in H^{12}\right\} \subseteq\Alt_{35}. \end{equation} Since each $T^s$ is linear, the map \[ H^{12}\longrightarrow\Alt_{35},\qquad z\longmapsto\begin{pmatrix}0&C(z)\\C(z)^T&0\end{pmatrix} \] is linear. By the construction of $C(z)$, this map is injective. Hence $K$ is a linear subspace and \[ \dim_{\F_2}K=\dim_{\F_2}H^{12}=12\cdot11=132. \] Set $W=\Alt_{35}/K$, and write $\pi:\Alt_{35}\to W$, $A\mapsto A+K$, for the quotient map. Then \begin{equation}\label{eq:dim-W} \dim_{\F_2}W=\binom{35}{2}-\dim_{\F_2}K=595-132=463. \end{equation} \begin{remark} \label{rem:coords-W} One can realize $W$ explicitly as a direct sum of matrix spaces. For $R\in\Mat_{24\times11}(\F_2)$, let $R_+$ consist of its first twelve rows and $R_-$ of its last twelve rows. Thus \[ R=\begin{pmatrix}R_+\\R_-\end{pmatrix}, \qquad R_+,R_-\in\Mat_{12\times11}(\F_2). \] Define the linear operator $L \colon \Mat_{12 \times 11}(\F_2) \to \Mat_{12 \times 11}(\F_2)$ row-wise for $1 \le s \le 6$ by \[ L(R_+)_{2s-1, *} = \bigl(R_{+, 2s-1, *} + R_{+, 2s, *}\bigr)(T^s)^T, \qquad L(R_+)_{2s, *} = R_{+, 2s, *}(T^s)^T, \] where $R_{+, i, *}$ denotes row $i$ of $R_+$. Then $\binom{R_+}{R_-}=C(z)$ for some $z\in H^{12}$ if and only if $R_- = L(R_+)$. Consequently, the linear map \[ \Alt_{35} \longrightarrow \Alt_{24} \oplus \Mat_{12 \times 11}(\F_2) \oplus \Alt_{11}, \qquad \begin{pmatrix} A & R \\ R^T & B \end{pmatrix} \longmapsto \bigl(A,\, R_- + L(R_+),\, B\bigr), \] is surjective (as seen by restricting to $R_+ = 0$) and has kernel precisely equal to $K$, inducing an explicit coordinate isomorphism $W \cong \Alt_{24} \oplus \Mat_{12 \times 11}(\F_2) \oplus \Alt_{11}$. \end{remark} \section{Construction of the \texorpdfstring{$2$}{2}-Group \texorpdfstring{$G$}{G}}\label{sec:group} To construct a central extension of $D = P \oplus H \cong \F_2^{35}$ by $W$, we introduce the bilinear map $F \colon D \times D \to \Alt_{35}$ by \begin{equation}\label{eq:F-def} F(v,u)_{ij} = \begin{cases} v_i u_j, & \text{if } 1 \le i < j \le 35, \\ v_j u_i, & \text{if } 1 \le j < i \le 35, \\ 0, & \text{if } i = j. \end{cases} \end{equation} It is immediate from \eqref{eq:F-def} that $F(v,u)$ is a symmetric matrix with zero diagonal and $F$ is bilinear in $(v,u)$. Moreover, $F$ satisfies the fundamental identity \begin{equation}\label{eq:F-sym} F(v,u) + F(u,v) = v u^T + u v^T \in \Alt_{35} \qquad (v, u \in D). \end{equation} Composing $F$ with the quotient map $\pi \colon \Alt_{35} \to W$, we set \[ f(v,u) = \pi\bigl(F(v,u)\bigr) \in W \] and endow the Cartesian product $G = D \times W$ with the binary operation \begin{equation}\label{eq:group-law} (v,a)(u,b) = \bigl(v + u,\, a + b + f(v,u)\bigr) \qquad (v,u \in D,\; a,b \in W). \end{equation} Write $Z(G)=\{z\in G:zx=xz\text{ for every }x\in G\}$ for the center of $G$. The exponent $\exp(G)$ of a finite group $G$ is the least positive integer $n$ such that $x^n=1$ for every $x\in G$. \begin{proposition}\label{prop:group-props} The multiplication rule \eqref{eq:group-law} makes $G$ into a finite $2$-group with the following properties: \begin{enumerate} \item\label{item:group-invariants} $|G|=2^{498}$, $\cl(G)=2$ and $\exp(G)=4$. \item\label{item:commutator} For all $(v,a),(u,b)\in G$, \begin{equation}\label{eq:commutator} [(v,a),(u,b)] =\bigl(0,\pi(vu^T+uv^T)\bigr). \end{equation} \item\label{item:subgroups} $G'=Z(G)=\{0\}\times W$. \end{enumerate} \end{proposition} \begin{proof} We first verify the group law, and then prove (\ref{item:commutator}), (\ref{item:subgroups}) and (\ref{item:group-invariants}). Since $f\colon D\times D\to W$ is bilinear, we have for all $v,u,t\in D$ the $2$-cocycle identity \[ f(v,u)+f(v+u,t) =f(v,u)+f(v,t)+f(u,t) =f(u,t)+f(v,u+t), \] which guarantees that the multiplication \eqref{eq:group-law} is associative. Because $f(v,0)=f(0,v)=0$, the element $1_G=(0,0)$ is the two-sided identity, and every element of $\{0\}\times W$ satisfies \[ (v,a)(0,b)=(v,a+b)=(0,b)(v,a), \] so $\{0\}\times W\le Z(G)$. Moreover, since $2v=0$ in $D$ and $2a=0$ in $W$, we have \[ (v,a)\bigl(v,a+f(v,v)\bigr) =\bigl(2v,2a+2f(v,v)\bigr) =(0,0) =\bigl(v,a+f(v,v)\bigr)(v,a), \] establishing the inverse formula \[ (v,a)^{-1}=\bigl(v,a+f(v,v)\bigr). \] By \eqref{eq:dim-W}, \[ |G|=|D|\cdot|W| =2^{35}\cdot2^{463} =2^{498}, \] so $G$ is a finite $2$-group. \emph{Proof of (\ref{item:commutator}).} For $x=(v,a)$ and $y=(u,b)$ in $G$, applying \eqref{eq:group-law} and the inverse formula above gives \[ \begin{aligned} x^{-1}y^{-1}x &=\bigl(v,a+f(v,v)\bigr) \bigl(u,b+f(u,u)\bigr)(v,a)\\ &=\bigl(u,b+f(u,u)+f(v,u)+f(u,v)\bigr), \end{aligned} \] and multiplying on the right by $y=(u,b)$ yields, via \eqref{eq:F-sym}, \[ [x,y] =\bigl(0,f(v,u)+f(u,v)\bigr) =\bigl(0,\pi(vu^T+uv^T)\bigr), \] proving \eqref{eq:commutator}. \emph{Proof of (\ref{item:subgroups}).} Since the elementary wedges $vu^T+uv^T$ span $\Alt_{35}$ as $v,u$ range over $D$ and $\pi\colon\Alt_{35}\to W$ is surjective, we obtain \[ G' =\{0\}\times \Span_{\F_2}\bigl\{\pi(vu^T+uv^T):v,u\in D\bigr\} =\{0\}\times W. \] To determine the center, write $v=p+h$ with $p\in P$ and $h\in H$. If $(v,a)\in Z(G)$, then $vu^T+uv^T\in K$ for all $u\in D$. Since every element of $K$ has zero diagonal blocks, choosing $u\in P$ and $u\in H$ gives, respectively, \[ pu^T+up^T=0\quad(u\in P), \qquad hu^T+uh^T=0\quad(u\in H). \] Thus $p=h=0$, and $Z(G)=\{0\}\times W$. \emph{Proof of (\ref{item:group-invariants}).} The order of $G$ was computed above. By (\ref{item:subgroups}), $G'=\{0\}\times W\subseteq Z(G)$. Since $\dim_{\F_2}W=463>0$, we have $G'\ne\{1_G\}$, and hence $\cl(G)=2$. For any $(v,a)\in G$, squaring via \eqref{eq:group-law} gives $(v,a)^2=(0,f(v,v))\in\{0\}\times W$, and squaring again gives $(v,a)^4=(0,2f(v,v))=(0,0)=1_G$. To see that $\exp(G)=4$, take $v=f_1+f_2\in P$, where $f_i$ is the standard basis vector; then $F(v,v)$ has $(1,2)$ entry $v_1v_2=1$ in the upper-left block, so $F(v,v)\notin K$. Hence $(v,0)^2=(0,\pi F(v,v))\ne1_G$, confirming $\exp(G)=4$. \end{proof} \section{The Attained Target Value and the Anisotropic Shift Lemma}\label{sec:target} On $P$, we fix the standard $24 \times 24$ nondegenerate alternating block matrix \[ J = \begin{pmatrix} 0 & I_{12} \\ I_{12} & 0 \end{pmatrix} \in \GL_{24}(\F_2), \] where $I_n$ is the $n\times n$ identity matrix and $\GL_n(\F_2)$ is the group of invertible $n\times n$ matrices. The matrix $J$ satisfies $J^T = J$ and $J^2 = I_{24}$, so that $J^{-1} = J$. We define the target matrix $A_0 \in \Alt_{35}$ and the corresponding central group element $g \in G' = \{0\} \times W$ by \begin{equation}\label{eq:A0-g-def} A_0 = \begin{pmatrix} J & 0 \\ 0 & 0 \end{pmatrix} \in \Alt_{35}, \qquad g = (0, \pi A_0) \in G. \end{equation} Because the upper-left $24 \times 24$ block of $A_0$ is $J \ne 0$, whereas every matrix in $K$ has zero upper-left $24 \times 24$ block by Remark~\ref{rem:coords-W}, we have $A_0 \notin K$ and hence $g \ne 1_G$. To verify that $g$ is attained by the word $w = \prod_{j=1}^{12} [x_{2j-1}, x_{2j}] \in F_{24}$, evaluate $w$ on the specific $24$-tuple of basis lifts \begin{equation}\label{eq:witness} x_{2j-1} = (f_j, 0), \qquad x_{2j} = (f_{12+j}, 0) \qquad (1 \le j \le 12). \end{equation} By the commutator formula \eqref{eq:commutator} and the centrality of $G'$, we have \[ w(x_1, \dots, x_{24}) = \left(0,\, \pi\!\left(\sum_{j=1}^{12} \bigl(f_j f_{12+j}^T + f_{12+j} f_j^T\bigr)\right)\right) = (0, \pi A_0) = g. \] Thus $g \in w(G)$, confirming that the fibre $w^{-1}(g)$ is nonempty. Before parameterizing the full fibre $w^{-1}(g)$, we study the equation $C(z)^T J C(z) = 0$ for $z\in H^{12}$, with $C(z)$ defined in \eqref{eq:C-top}--\eqref{eq:C-bot}. Under the identification \eqref{eq:wedge-ident} on $H = \F_2^{11}$, expanding the block product $C(z)^T J C(z) \in \Alt_{11}$ across the rows \eqref{eq:C-top}--\eqref{eq:C-bot} yields \begin{align} C(z)^T J C(z) &= \sum_{s=1}^6 \Bigl( z_{s,1} \wedge T^s(z_{s,1} + z_{s,2}) + z_{s,2} \wedge T^s z_{s,2} \Bigr) \nonumber \\ &= \sum_{s=1}^6 Q_s(z_{s,1}, z_{s,2}), \label{eq:CTJC-sum} \end{align} where for each $1 \le s \le 6$ and $a, b \in H$, the quadratic map $Q_s \colon H \times H \to \Alt_{11}$ is defined by \begin{equation}\label{eq:Qs-def} Q_s(a,b) = a \wedge T^s a + a \wedge T^s b + b \wedge T^s b. \end{equation} Set $q(a,b)=a^2+ab+b^2$ for $a,b\in\F_2$. It is direct to show that $q(a,b)=0$ if and only if $(a,b)=(0,0)$. The following lemma is the heart of the characteristic-two counting argument: it exploits a degree grading on $\bigwedge^2 H$ together with the anisotropy of $q(c_1, c_2) = c_1^2 + c_1 c_2 + c_2^2$ over $\F_2$ to show that $C(z)^T J C(z)$ vanishes if and only if every parameter vector $z_{s,1}, z_{s,2}$ lies in the kernel of $T^s$. \begin{lemma}\label{lem:CTJC} For every $z\in H^{12}$, the matrix $C(z)$ defined by \eqref{eq:C-top}--\eqref{eq:C-bot} satisfies \begin{equation}\label{eq:CTJC-equiv} C(z)^T J C(z) = 0 \quad \Longleftrightarrow \quad z_{s,1}, z_{s,2} \in \ker T^s \quad \text{for every } s = 1, \dots, 6. \end{equation} Consequently, the solution set in $H^{12}$ has exact cardinality \begin{equation}\label{eq:CTJC-count} \bigl|\{z \in H^{12} : C(z)^T J C(z) = 0\}\bigr| = 2^{42}. \end{equation} \end{lemma} \begin{proof} Expand any vector $z \in H$ in the standard basis as $z = \sum_{i=0}^{10} z_i e_i$, and endow the basis elements $e_i \wedge e_j$ ($0 \le i < j \le 10$) of $\bigwedge^2 H \cong \Alt_{11}$ with the additive degree \[ \deg(e_i \wedge e_j) = i + j. \] For a given parameter tuple $z = (z_{s,1}, z_{s,2})_{s=1}^6 \in H^{12}$, consider the active index set \[ S = \bigl\{ s \in \{1, \dots, 6\} : (T^s z_{s,1}, T^s z_{s,2}) \ne (0,0) \bigr\}. \] If $S = \varnothing$, then $T^s z_{s,1} = T^s z_{s,2} = 0$ for every $1 \le s \le 6$, so each summand $Q_s(z_{s,1}, z_{s,2})$ in \eqref{eq:CTJC-sum} vanishes identically and $C(z)^T J C(z) = 0$. Suppose conversely that $S \ne \varnothing$. For each $s \in S$, define the leading nonvanishing coordinate index \[ a_s = \min\bigl\{ i \in \{0, \dots, 10\} : \bigl((z_{s,1})_i, (z_{s,2})_i\bigr) \ne (0,0) \bigr\}, \] and set $c_1 = (z_{s,1})_{a_s} \in \F_2$ and $c_2 = (z_{s,2})_{a_s} \in \F_2$. Therefore, $(c_1,c_2)\ne (0,0)$ in $\F_{2}^{2}$. Since $(T^s z_{s,1}, T^s z_{s,2}) \ne (0,0)$, at least one of $z_{s,1}, z_{s,2}$ has a nonzero coordinate at an index $\le 10 - s$, which implies that $a_s + s \le 10$. Expanding $Q_s(z_{s,1}, z_{s,2})$ via \eqref{eq:Qs-def} and \eqref{eq:shift-T}, every nonzero basis term in $Q_s(z_{s,1}, z_{s,2})$ is of the form $e_i \wedge e_{j+s}$ with $i, j \ge a_s$. Its degree therefore satisfies \[ \deg(e_i \wedge e_{j+s}) = i + j + s \ge 2a_s + s, \] with equality $i + j + s = 2a_s + s$ holding if and only if $i = j = a_s$. At this unique minimal-degree basis element $e_{a_s} \wedge e_{a_s + s}$, the coefficient in $Q_s(z_{s,1}, z_{s,2})$ is precisely \[ c_1^2 + c_1 c_2 + c_2^2 = q(c_1, c_2) = 1 \in \F_2, \] because $(c_1, c_2) \in \F_2^2 \setminus \{(0,0)\}$. Thus $Q_s(z_{s,1}, z_{s,2})$ has homogeneous degree-$(2a_s + s)$ component equal to $e_{a_s} \wedge e_{a_s + s}$, and all of its other nonzero terms have strictly larger degree. Now let $\delta = \min_{s \in S}(2a_s + s)$ be the minimum leading degree among all active indices $s \in S$. Extracting the homogeneous degree-$\delta$ component of the sum $\sum_{s=1}^6 Q_s(z_{s,1}, z_{s,2})$ in \eqref{eq:CTJC-sum} gives \[ \Biggl(\sum_{s=1}^6 Q_s(z_{s,1}, z_{s,2})\Biggr)_{\!\delta} = \sum_{\substack{s \in S \\ 2a_s + s = \delta}} e_{a_s} \wedge e_{a_s + s}. \] Crucially, the basis wedges $e_{a_s} \wedge e_{a_s + s}$ appearing in this sum are pairwise distinct. We conclude that $\sum_{s=1}^6 Q_s(z_{s,1}, z_{s,2}) \ne 0$ whenever $S \ne \varnothing$. This proves \eqref{eq:CTJC-equiv}. Finally, since $\dim_{\F_2} \ker T^s = s$ for $1 \le s \le 6$ by \eqref{eq:ker-Ts}, the number of tuples $z\in H^{12}$ satisfying $C(z)^T J C(z) = 0$ is \[ \prod_{s=1}^6 \bigl|\ker T^s\bigr|^2 = \prod_{s=1}^6 2^{2s} = 2^{2(1+2+3+4+5+6)} = 2^{42}, \] which establishes \eqref{eq:CTJC-count}. \end{proof} \section{Bijective Parameterization and Exact Fibre Enumeration}\label{sec:fibre} We now determine the exact cardinality of the fibre $w^{-1}(g)$ in $G^{24}$. Let $x_i=(v_i,a_i)\in G$, with $v_i\in D$ and $a_i\in W$, for $1\le i\le24$, and arrange their $D$-coordinates $v_i \in \F_2^{35}$ as the columns of a $35 \times 24$ matrix $X$ with the odd-indexed columns first, followed by the even-indexed columns: \[ X = ( v_1 \ v_3 \ \cdots \ v_{23} \ \ v_2 \ v_4 \ \cdots \ v_{24} ) = \begin{pmatrix} U \\ Y \end{pmatrix} \in \Mat_{35 \times 24}(\F_2), \] where $U \in \Mat_{24 \times 24}(\F_2)$ records the $P$-coordinates and $Y \in \Mat_{11 \times 24}(\F_2)$ records the $H$-coordinates. By \eqref{eq:commutator} and $(0,a)(0,b)=(0,a+b)$, we have\begin{align*} w(x_1,\ldots,x_{24}) &=\prod_{j=1}^{12}[x_{2j-1},x_{2j}]=\prod_{j=1}^{12} \left(0,\pi\left( v_{2j-1}v_{2j}^{T}+v_{2j}v_{2j-1}^{T} \right)\right)\\ &=\left(0,\pi\left( \sum_{j=1}^{12} \left(v_{2j-1}v_{2j}^{T}+v_{2j}v_{2j-1}^{T}\right) \right)\right). \end{align*} Since \[XJX^T=\sum_{j=1}^{12} \left(v_{2j}v_{2j-1}^{T}+v_{2j-1}v_{2j}^{T}\right),\] we obtain \begin{equation}\label{eq:w-XJX} w(x_1,\ldots,x_{24})=\left(0,\pi(XJX^T)\right). \end{equation} Notice that the $W$-coordinates $a_1, \dots, a_{24} \in W$ do not appear in \eqref{eq:w-XJX}, so they may be chosen arbitrarily in $W^{24}$. Let $\Sp_{2m}(2)$ denote the symplectic group of $2m \times 2m$ matrices over $\F_2$: \[ \Sp_{2m}(2) = \bigl\{ U \in \GL_{2m}(\F_2) : U J_{2m} U^T = J_{2m} \bigr\}, \qquad J_{2m}=\begin{pmatrix}0&I_m\\I_m&0\end{pmatrix}. \] \begin{proposition}\label{prop:bijection} The set of matrices $X = \binom{U}{Y} \in \Mat_{35 \times 24}(\F_2)$ satisfying $w(x_1, \dots, x_{24}) = g$ in \eqref{eq:A0-g-def} is in explicit bijection with the Cartesian product \[ \Sp_{24}(2) \times \bigl\{ z \in H^{12} : C(z)^T J C(z) = 0 \bigr\} \] via the map \begin{equation}\label{eq:bijection-maps} (U, z) \longmapsto \begin{pmatrix} U \\ C(z)^T U^{-T} J \end{pmatrix}. \end{equation} The inverse sends $\binom{U}{Y}$ to $(U,z)$, where \[ z_{s,1}=\bigl((UJY^T)_{2s-1,*}\bigr)^T,\qquad z_{s,2}=\bigl((UJY^T)_{2s,*}\bigr)^T\qquad(1\le s\le6). \] \end{proposition} \begin{proof} By the definitions of $K$ in \eqref{eq:K-W-def}, $g = (0, \pi A_0)$ in \eqref{eq:A0-g-def}, and $w(x_1, \dots, x_{24})$ in \eqref{eq:w-XJX}, a matrix $X = \binom{U}{Y}$ satisfies $\pi(X J X^T) = \pi(A_0)$ if and only if $X J X^T - A_0 \in K$. Expanding the block product \[ X J X^T - A_0 = \begin{pmatrix} U J U^T - J & U J Y^T \\ Y J U^T & Y J Y^T \end{pmatrix} \] and comparing with the block structure of $K$ in \eqref{eq:K-W-def}, we see that $X J X^T - A_0 \in K$ is equivalent to the following three matrix equations \begin{equation}\label{eq:coupled-system} U J U^T = J, \qquad U J Y^T = C(z)\quad(z\in H^{12}), \qquad Y J Y^T = 0. \end{equation} The first equation $U J U^T = J$ implies $U \in \Sp_{24}(2) \le \GL_{24}(\F_2)$. For each fixed $U \in \Sp_{24}(2)$ and each $z\in H^{12}$, the second equation $U J Y^T = C(z)$ has the unique solution \begin{equation}\label{eq:YT-sol} Y^T = J^{-1} U^{-1} C(z) = J U^{-1} C(z) \quad \Longleftrightarrow \quad Y = C(z)^T U^{-T} J, \end{equation} since $J^{-1} = J = J^T$. Substituting \eqref{eq:YT-sol} into the third equation of \eqref{eq:coupled-system} then gives \begin{align*} Y J Y^T &= \bigl(C(z)^T U^{-T} J\bigr) J \bigl(J U^{-1} C(z)\bigr)\\ &= C(z)^T \bigl(U^{-T} J U^{-1}\bigr) C(z) = C(z)^T J C(z), \end{align*} where we used $J^3 = J$. Thus $Y J Y^T = 0$ holds if and only if $C(z)^T J C(z) = 0$. Thus the map \eqref{eq:bijection-maps} is a bijection, with inverse given by the formulas above. \end{proof} For completeness, we record the classical order formula for $\Sp_{2m}(2)$. \begin{lemma} \label{lem:sp-order} For every integer $m\ge1$, the symplectic group $\Sp_{2m}(2)$ has order \[ |\Sp_{2m}(2)|=2^{m^2}\prod_{j=1}^{m}(2^{2j}-1). \] \end{lemma} \begin{proof} See \cite[p.~70]{Taylor1992}. \end{proof} We can now complete the proof of \eqref{eq:exact-count} and \eqref{eq:normalized-bound} in Theorem~\ref{thm:main} for $p=2$. \begin{proof}[Proof of Theorem~\ref{thm:main} for $p=2$] A tuple $(x_1, \dots, x_{24}) \in G^{24}$ with $x_i = (v_i, a_i) \in D \times W$ satisfies $w(x_1, \dots, x_{24}) = g$ if and only if its matrix of $D$-coordinates $X = \binom{U}{Y}$ satisfies $\pi(X J X^T) = \pi(A_0)$, while the twenty-four coordinates $(a_1, \dots, a_{24}) \in W^{24}$ are completely unrestricted. Combining Proposition~\ref{prop:bijection}, Lemma~\ref{lem:CTJC}, and Lemma~\ref{lem:sp-order} (for $m = 12$) with $|W| = 2^{463}$ from \eqref{eq:dim-W}, we obtain \begin{align*} N_{w,G}(g) &= |W|^{24} \cdot \bigl|\{z \in H^{12} : C(z)^T J C(z) = 0\}\bigr| \cdot \bigl|\Sp_{24}(2)\bigr|\\ &= 2^{24 \cdot 463} \cdot 2^{42} \cdot 2^{144} \prod_{j=1}^{12} \bigl(2^{2j} - 1\bigr). \end{align*} Summing the powers of $2$ in the prefactor gives \[ 24 \cdot 463 + 42 + 144 = 11112 + 186 = 11298, \] which proves \eqref{eq:exact-count}. Since $|G|^{23} = (2^{498})^{23} = 2^{11454}$ and $\sum_{j=1}^{12} 2j = 156$, we have \[ N_{w,G}(g) = 2^{11298 + 156} \prod_{j=1}^{12} \bigl(1 - 2^{-2j}\bigr) = |G|^{23} \prod_{j=1}^{12} \bigl(1 - 2^{-2j}\bigr). \] Because $0 < 1 - 2^{-2j} < 1$ for every $1 \le j \le 12$, the product is strictly less than $1$, proving $N_{w,G}(g) < |G|^{23}$ in \eqref{eq:normalized-bound}. Together with the explicit witness \eqref{eq:witness}, this establishes that $G$ and $w$ form a counterexample to the Amit--Ashurst conjecture \eqref{eq:amit-ashurst}. \end{proof} The proof for odd primes is similar to the case $p=2$, with slight changes to the construction. We briefly give the details. \begin{proof}[Proof of Theorem~\ref{thm:main} for odd $p$] Let $p$ be an odd prime and choose a nonsquare $d\in\F_p^\times$. Use $H=\F_p^{11}$, $P=\F_p^{24}$, $D=P\oplus H$, and the same shift $T$ as in \eqref{eq:shift-T}. In this proof, write \[ \Alt_n=\{A\in\Mat_{n\times n}(\F_p):A^T=-A,\ A_{ii}=0\}, \qquad J=\begin{pmatrix}0&I_{12}\\-I_{12}&0\end{pmatrix}, \] and identify $u\wedge v$ with $uv^T-vu^T$. For $z\in H^{12}$, define $C(z)$ by \[ \begin{aligned} C(z)_{2s-1,*}&=z_{s,1}^T,& C(z)_{2s,*}&=z_{s,2}^T,\\ C(z)_{12+2s-1,*}&=(T^sz_{s,1})^T,& C(z)_{12+2s,*}&=-d(T^sz_{s,2})^T \end{aligned} \qquad(1\le s\le6). \] Set \[ K=\left\{\begin{pmatrix}0&C(z)\\-C(z)^T&0\end{pmatrix}:z\in H^{12}\right\}, \qquad W=\Alt_{35}/K, \] and let $\pi:\Alt_{35}\to W$ be the quotient map. As before, $\dim K=132$ and $\dim W=463$. Define $G_p=D\times W$ by \[ (v,a)(u,b)=\left(v+u,a+b+\tfrac12\pi(vu^T-uv^T)\right), \] where $1/2$ is the inverse of $2$ in $\F_p$. The bilinearity of $(v,u)\mapsto\pi(vu^T-uv^T)$ implies that the multiplication is associative, and \[ (v,a)^{-1}=(-v,-a),\qquad (v,a)^k=(kv,ka),\qquad [(v,a),(u,b)]=(0,\pi(vu^T-uv^T)). \] Thus $|G_p|=p^{498}$, $G_p'=\{0\}\times W\le Z(G_p)$, $\cl(G_p)=2$, and $\exp(G_p)=p$. Expanding $C(z)^TJC(z)$ gives \[ C(z)^TJC(z)=\sum_{s=1}^6 \left(z_{s,1}\wedge T^sz_{s,1}-d z_{s,2}\wedge T^sz_{s,2}\right). \] In the proof of Lemma~\ref{lem:CTJC}, the lowest-degree coefficient is now $c_1^2-dc_2^2$, which is nonzero whenever $(c_1,c_2)\ne(0,0)$. The corresponding basis wedges remain pairwise distinct. Hence \[ C(z)^TJC(z)=0 \iff z_{s,1},z_{s,2}\in\ker T^s\ (1\le s\le6), \qquad \bigl|\{z:C(z)^TJC(z)=0\}\bigr|=p^{42}. \] Put $A_0=\begin{pmatrix}J&0\\0&0\end{pmatrix}$ and $g_p=(0,\pi A_0)$. The tuple \eqref{eq:witness}, interpreted over $\F_p$, shows that $g_p\in w(G_p)\setminus\{1\}$. Define $\Sp_{24}(p)=\{U\in\GL_{24}(\F_p):UJU^T=J\}$. Since $J^T=J^{-1}=-J$ and $UJU^T=J$, the block equations give \[ UJY^T=C(z)\iff Y=C(z)^TU^{-T}J, \qquad YJY^T=C(z)^TJC(z). \] Proposition~\ref{prop:bijection} therefore applies with $\Sp_{24}(p)$ in place of $\Sp_{24}(2)$, with the same inverse formulas. Using $|\Sp_{24}(p)|=p^{144}\prod_{j=1}^{12}(p^{2j}-1)$ \cite[p.~70]{Taylor1992}, we obtain \begin{align*} N_{w,G_p}(g_p) &=p^{24\cdot463+42+144}\prod_{j=1}^{12}(p^{2j}-1)\\ &=|G_p|^{23}\prod_{j=1}^{12}(1-p^{-2j})<|G_p|^{23}.\qedhere \end{align*} \end{proof} \begin{remark}\label{rem:dimension-choice} The dimensions are chosen by balancing the powers of $p$ in the normalized fibre size. More generally, take $k\ge1$ pairs of parameter vectors, $\dim P=4k$, and $\dim H=n\ge k$, with the shift $Te_i=e_{i+1}$ for $0\le i