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\hypersetup{
 pdftitle={A proof of Brauer's k(B)-conjecture at the prime two},
 pdfauthor={Qiyue Tang},
 pdfsubject={The number of ordinary irreducible characters in a 2-block},
 pdfkeywords={Brauer conjecture, 2-block, defect group, ordinary character}}
\allowdisplaybreaks[2]
\setlength{\headheight}{14pt}
\title[Brauer's $k(B)$-conjecture at two]{A proof of Brauer's $k(B)$-conjecture\protect\\ at the prime two}
\author{Qiyue Tang}
\address{Zhili College, Tsinghua University, Beijing 100084, China}
\email{tangqy24@mails.tsinghua.edu.cn}
\date{November 13, 2026}
\subjclass[2020]{Primary 20C20; Secondary 20C15, 20C33}
\keywords{Brauer's $k(B)$-conjecture, $2$-block, defect group, character count}
\begin{document}
\begin{abstract}
We prove that every $2$-block of a finite group has at most as many
ordinary irreducible characters as the order of its defect groups.
\end{abstract}
\maketitle
\section{Introduction}\label{sec:introduction}

Let $p$ be a prime, let $B$ be a $p$-block of a finite group $G$,
let $D$ be a defect group of $B$, and write $k(B)$ for the number
of ordinary irreducible
characters belonging to $B$. Brauer's $k(B)$-conjecture asserts that
\[
                              k(B)\le |D|.
\]
The conjecture relates the number of characters in a block to the
order of its local defect group. In 1946 Brauer proved
$k(B)\le p^{d(d+1)/2}$, where $|D|=p^d$, and proposed the bound
$p^d$ in the paragraph following his Theorem~8 \cite{Brauer46}.
He also recorded the cases $d=0,1,2$. The question later appeared
as Problem~20 in his 1963 list \cite{Brauer63}. We prove the
conjecture at the prime two.

\begin{theorem}\label{thm:main}
Let $G$ be a finite group and let $B$ be a $2$-block of $G$ with
defect group $D$. Then $k(B)\le |D|$.
\end{theorem}

Brauer and Feit subsequently obtained the general estimate
$k(B)\le p^{2d}/4+1$ and the refinement
$k(B)\le p^{2d-2}$ for $d>2$
\cite[Theorems~1 and~1$^*$]{BrauerFeit59}.
Their Theorem~4 proves the conjectured bound when $D$ is cyclic.
For a nontrivial cyclic defect group, Dade's structure theorem
\cite[Theorem~1]{Dade66} gives the exact formula
\[
             k(B)=e+\frac{|D|-1}{e},\qquad l(B)=e\mid p-1,
\]
where $e$ is the inertial index and $l(B)$ is the number of
irreducible Brauer characters in $B$. Thus $e=1$ and $k(B)=|D|$
for cyclic defect at two. The study of noncyclic $2$-defect groups
includes the dihedral, semidihedral and quaternion cases and their
central extensions by cyclic $2$-groups.
Sambale's Cartan-matrix method proves the bound whenever the defect
group is a central extension of a metacyclic $2$-group by a cyclic
$2$-group \cite[Theorem~2]{SambaleCartan13}. This form is used in
the exceptional component calculations below.

The $p$-solvable case led to the coprime $k(GV)$ problem.
Nagao \cite{Nagao62} reduced the conjecture for $p$-solvable groups
to the assertion
\[
 k(V\rtimes H)\le |V|
\]
for a finite $p'$-group $H$ acting faithfully and irreducibly on
an elementary abelian $p$-group $V$. Here $k(X)$ denotes the
number of conjugacy classes of a finite group $X$.
Robinson and Thompson \cite{RT96} proved this assertion for
$p>5^{30}$. The remaining characteristics were resolved through
character-theoretic and orbit estimates; the final characteristic-five
case was settled by Gluck, Magaard, Riese and Schmid
\cite{GMRS04}. Their solution completed the $p$-solvable case for
every prime.
For arbitrary linear groups, the non-coprime problem was studied
by Guralnick--Tiep \cite{GT05} and Guralnick--Mar\'oti \cite{GM13}.
Their results on quasisimple and symplectic-type linear groups
provide representation-degree, support and class-number inputs
for our affine argument.

Recent work on the non-coprime problem also treats wreath products.
Hung, Mar\'oti and Mart\'inez Madrid \cite[Theorem~1.1]{HMM25}
prove $k(X\wr H)\le k(X)^n$ for a nontrivial finite group $X$
and a transitive semiprimitive group $H$ of degree $n$, whenever
$n$ or $k(X)$ is sufficiently large. Here semiprimitive means that
every normal subgroup is transitive or has trivial point stabilizers.
The theorem concerns the full wreath product. Our imprimitive
argument requires a bound for subgroups with relations between
their coordinate projections; these relations enter the weighted
permutation count in Section~\ref{sec:permutation}.

The classification of finite simple groups permits a separate
analysis of quasisimple blocks. Malle \cite{Malle18} studied whether
a block of a quasisimple group could be a minimal counterexample,
comparing with blocks of all finite groups having the same
defect-group isomorphism type and minimizing the index of the
center. His Theorems~1
and~2 exclude primes at least five and abelian defect groups.
His Theorem~3 confines a remaining quasisimple minimal
counterexample to an isolated block of a group of Lie type in
non-defining characteristic, at the prime two or three, with
nonabelian defect. At three the remaining block is nonunipotent.
Thus the exceptional isolated blocks at two are among the cases
requiring a full ordinary-character count. For classical groups,
the semisimple parametrization and generating functions give
such counts; Brenner's work \cite{Brenner20} supplies related
estimates for $2$-blocks of general linear and unitary groups.

Another line of progress concerns characters of height zero.
Write $k_0(B)$ for their number. Kessar and Malle proved that
abelian defect implies $k(B)=k_0(B)$ \cite{KM13}.
Ruhstorfer proved the Alperin--McKay conjecture for $2$-blocks
with almost abelian defect groups, in particular with abelian
defect groups \cite[Theorem~F]{Ruh22Sigma}.
These results already give Brauer's bound for every abelian-defect
$2$-block. Indeed, if $b$ is the Brauer correspondent of $B$ in
$N_G(D)$, they give
\[
                  k(B)=k_0(B)=k_0(b)=k(b)\le |D|.
\]
The last inequality follows from the normal-defect description
\cite{Kulshammer85} and the coprime $k(GV)$ theorem.
Ruhstorfer later proved the Alperin--McKay and height-zero
conjectures at two in full \cite{Ruh25}. For a block with
nonabelian defect, Brauer's bound also requires control of the
positive-height characters contributing to $k(B)$.

Jordan decomposition is central to the Lie-type analysis.
Bonnaf\'e--Rouquier \cite{BR03} and
Bonnaf\'e--Dat--Rouquier \cite{BDR17} construct equivalences
associated with semisimple centralizers and Levi subgroups.
Ruhstorfer's equivariant form \cite{RuhJordan22} provides
comparison maps compatible with diagonal automorphisms.
Kessar and Malle \cite{KM25} describe the block distribution
under Jordan correspondence, with the stated ambiguities for
exceptional twin blocks. Their result on Robinson's conjecture concerns individual
character defects. The block-distribution statements are the
input used here to count all characters in the exceptional
$2$-series.

The proof of Theorem~\ref{thm:main} combines an affine bound,
a fixed-point comparison for covering blocks, and a proper-Levi
descent. The affine bound is
\[
 k(V\rtimes H)\le |V|\,|H|_2
\]
for every faithful action of a finite group $H$ on a finite
$\mathbb F_2$-space $V$, where $|H|_2$ is the order of a Sylow
$2$-subgroup of $H$. Induction on the dimension handles arbitrary
modules. The primitive case uses the generalized Fitting subgroup,
support estimates and finite subgroup calculations. The imprimitive
case uses a support-weighted inequality for permutation groups,
with separate calculations over $\mathbb F_4$, $\mathbb F_8$ and
$\mathbb F_{32}$.

For a $G$-stable $2$-block $b$ of $X\lhd G$, let $M\le G$ consist
of the elements acting innerly on the integral block algebra of $b$.
When $M/X$ is solvable,
the character fibers in blocks of $M$ covering $b$ are bounded
by relative defect orders. Suppose there is an
$\mathbb F_2(G/M)$-module $W$ of cardinality equal to the defect
order of $b$, whose action has a $2$-group kernel and whose
fixed-point numbers dominate those of $\Irr(b)$ for every subgroup.
The affine inequality then bounds every covering block.
In a minimum counterexample, the component estimates supply
the required local modules. Their assembly proves the fixed-point
comparisons, solvability and kernel conditions for the ambient group.

For a component orbit satisfying the hypotheses on lifted ordinary
automorphisms in Theorem~\ref{thm:levi-descent}, with proper canonical
minimal enclosing Levi subgroups, we construct an integral Morita
equivalence with a block of an odd central extension of a proper
ambient subgroup.
The construction preserves the component-centralizer action, the
character count and the defect order, while decreasing the index
of the center. The remaining component families are counted
explicitly. In particular, the nonprincipal unipotent $E_8$
calculation includes every semisimple layer contributing to the
block.

Sections~\ref{sec:counting}--\ref{sec:inner} establish the
block-counting tools. The affine proof culminates in
Theorem~\ref{thm:affine}. Section~\ref{sec:geometry} proves the
proper-Levi descent. The classical counts and the component
analysis follow, and Section~\ref{sec:minimum} combines them by
minimizing, in order, the defect order, the index of the center
and the group order.
The finite calculations are specified by their complete input
ranges, algorithms and output tables.

\section{Character counts and fixed points}\label{sec:counting}

All groups in this paper are finite unless an algebraic group is explicitly
specified. We fix a splitting $2$-modular system $(K,\cO,k)$, where $\cO$ is
a complete discrete valuation ring, $K$ has characteristic zero, and $k$ is
algebraically closed of characteristic two. It may be enlarged when a finite
central extension is introduced. A block of $G$ means a primitive central
idempotent $b$ of $\cO G$, or the corresponding algebra $\cO Gb$. We write
$\Irr(b)$ and $\IBr(b)$ for its ordinary and irreducible Brauer characters,
$k(b)=|\Irr(b)|$, $l(b)=|\IBr(b)|$, and $\delta(b)=|D_b|$ for the
order of a defect group. For a group $H$, $k(H)=|\Irr(H)|$ is its
number of conjugacy classes, and $B_0(H)$ denotes its principal
$2$-block. For a nonzero integer $n$ and a prime $p$, let $n_p$
be the largest power of $p$ dividing $|n|$, put $n_{p'}=|n|/n_p$,
and write $v_p(n)=\log_p(n_p)$.
For $N\lhd G$ and $Y\subseteq\Irr(N)$, let $\Irr(G\mid Y)$ be the
set of ordinary irreducible characters of $G$ whose restrictions to
$N$ have a constituent in $Y$. For a block $c$ of $G$, put
$\Irr(c\mid Y)=\Irr(c)\cap\Irr(G\mid Y)$.
A block $b$ or a character $\theta$ in the right-hand argument means
$Y=\Irr(b)$ or $Y=\{\theta\}$, respectively.
We use the usual covering relation for blocks of normal subgroups.
The Cartan matrix $C_b$ is indexed by $\IBr(b)$; its largest
elementary divisor is $\delta(b)$.

We write $O_p(H)$ and $O_{p'}(H)$ for the largest normal
$p$-subgroup and normal $p'$-subgroup of $H$.
The Fitting subgroup $F(H)$ is its largest nilpotent normal subgroup.
A component is a subnormal quasisimple subgroup, where quasisimple
means perfect with nonabelian simple central quotient. The layer
$E(H)$ is the product of the components, and $F^*(H)=F(H)E(H)$
is the generalized Fitting subgroup. The solvable radical
$\operatorname{Rad}(H)$ is the largest solvable normal subgroup,
and $\Soc(H)$ is the product of the minimal normal subgroups.
We write $H^{\ab}=H/[H,H]$.

A normalized scalar factor set on $H$ is a map
$\alpha:H\times H\to K^\times$ satisfying
\[
 \alpha(h,j)\alpha(hj,r)=\alpha(j,r)\alpha(h,jr),\qquad
 \alpha(1,h)=\alpha(h,1)=1
 \quad(h,j,r\in H),
\]
with trivial $H$-action on the scalars.
For such a factor set, let $k_\alpha(H)$ be the number of irreducible
$\alpha$-projective characters. The center of the twisted group
algebra has a basis indexed by the $\alpha$-regular conjugacy classes,
where $x$ is $\alpha$-regular if $\alpha(x,y)=\alpha(y,x)$
for every $y\in C_H(x)$. Therefore
\begin{equation}\label{eq:projective-bound}
 k_\alpha(H)\le k(H).
\end{equation}
In Clifford theory the factor set records the obstruction to
extending the corresponding ordinary character.

For a finite $H$-set $Y$, write
$H_y=\{h\in H:hy=y\}$,
$Y^J=\{y\in Y:jy=y\text{ for all }j\in J\}$ for $J\le H$,
and $Y^h=Y^{\langle h\rangle}$ for $h\in H$.
Let $n(H,Y)=|Y/H|$ be the number of orbits.

\begin{lemma}\label{lem:class-counts}
If $H\le G$ and $N\lhd G$, then
\[
 k(H)\le |G:H|k(G),\qquad k(G)\le k(N)k(G/N).
\]
More generally, if a $G$-invariant block $b$ of $N$ is fixed, then
\begin{equation}\label{eq:cover-class-count}
 |\Irr(G\mid b)|\le k(b)k(G/N).
\end{equation}
\end{lemma}
\begin{proof}
The intersection of a $G$-class with $H$ is a union of at most $|G:H|$
$H$-classes: the classes are parametrized by suitable double cosets
of a centralizer and $H$. This proves the first inequality. For each
$G$-orbit representative $\theta$ in $\Irr(N)$, Clifford theory gives
$k_{\alpha_\theta}(G_\theta/N)$ irreducible characters above its orbit.
By \eqref{eq:projective-bound} and the first inequality this is at most
$|G:G_\theta|k(G/N)$. Summing the orbit sizes proves the second
inequality. Restricting the same sum to $\Irr(b)$ proves
\eqref{eq:cover-class-count}.
\end{proof}

For a finite $H$-set $Y$, put
\begin{equation}\label{eq:kappa-definition}
 \kappa_H(Y)=\sum_{y\in Y/H} k(H_y),
\end{equation}
where the sum is over orbit representatives. We call $|Y^J|$
the $J$-fixed-point number of $Y$.

\begin{lemma}\label{lem:kappa}
For every finite $H$-set $Y$,
\begin{equation}\label{eq:kappa-commuting}
 \kappa_H(Y)=\frac1{|H|}\sum_{\substack{x,z\in H\\xz=zx}}
 |Y^{\langle x,z\rangle}|.
\end{equation}
Consequently, for a finite $H$-set $Z$ and a real number $a\ge0$,
if $|Y^J|\le a|Z^J|$ for all two-generated abelian subgroups $J$ of $H$,
then $\kappa_H(Y)\le a\kappa_H(Z)$.
For a finite-dimensional $\F_2H$-module $V$, with dual module $V^*$,
\begin{equation}\label{eq:kappa-affine}
 \kappa_H(V)=k(V^*\sd H).
\end{equation}
\end{lemma}
\begin{proof}
For an orbit represented by $y$, the number of commuting ordered pairs
in $H_y$ is $|H_y|k(H_y)$. Count triples $(y,x,z)$ with $x,z\in H_y$
and $xz=zx$, first by orbits of $y$ and then by $(x,z)$, to obtain
\eqref{eq:kappa-commuting}. The comparison follows term by term.
The ordinary linear characters of the additive group $V^*$ form the
$H$-set $V$. Each such character extends to its inertia group in
$V^*\sd H$ by making it trivial on the complement. Clifford theory
therefore proves \eqref{eq:kappa-affine}.
\end{proof}

We shall prove the following affine inequality in
Section~\ref{sec:affine}:
\begin{equation}\label{eq:WA2}
 k(V\sd H)\le |V|\,|H|_2
 \quad\text{if $H$ acts faithfully on the $\F_2$-space $V$.}
\end{equation}
The following consequence permits a normal $2$-kernel.

\begin{lemma}\label{lem:affine-two-kernel}
Assume \eqref{eq:WA2}. If the kernel $R$ of the action of $H$ on $V$
is a $2$-group, then
\[
 \kappa_H(V)=k(V^*\sd H)\le |V|\,|H|_2.
\]
\end{lemma}
\begin{proof}
The subgroup $R$ is normal in $V^*\sd H$ and centralizes $V^*$.
Apply Lemma~\ref{lem:class-counts}, followed by \eqref{eq:WA2} for
$H/R$ on $V^*$, to obtain
\[
 k(V^*\sd H)\le k(R)k(V^*\sd(H/R))
 \le |R|\,|V|\,|H/R|_2=|V|\,|H|_2.
\]
\end{proof}

\begin{remark}
The dual in \eqref{eq:kappa-affine} is necessary. In characteristic
two the dimensions of common fixed spaces in $V$ and $V^*$ need not
agree for arbitrary subgroups. The fixed-point comparison uses the natural identification
$\Irr(V^*)\cong V$.
\end{remark}

\section{Linearization of normal \texorpdfstring{$2$}{2}-subgroups}\label{sec:normal-two}

This section retains individual blocks throughout the passage from a
normal $2$-subgroup to an elementary abelian group. Several of the
arguments work at an arbitrary prime $p$; in those statements the
coefficient system is understood to have residue characteristic $p$.

\begin{lemma}[Central character fibers]\label{lem:central-fiber}
Let $Z\le Z(X)$ be a $p$-subgroup, let $c$ be a block of $X$, and let
$\bar c$ be its dominated block in $X/Z$. For every $\lambda\in\Irr(Z)$,
\begin{equation}\label{eq:central-fiber}
 |\Irr(c\mid\lambda)|\le k(\bar c).
\end{equation}
If $D_\lambda$ consists of the decomposition rows belonging to
$\Irr(c\mid\lambda)$, with all columns of $\IBr(c)=\IBr(\bar c)$, then
\begin{equation}\label{eq:central-gram}
 D_\lambda^{\mathsf t}D_\lambda=C_{\bar c}.
\end{equation}
In particular its row-column incidence graph is connected and every
column occurs.
\end{lemma}
\begin{proof}
Multiplication by $Z$ permutes the conjugacy-class-sum basis of
$Z(\cO X)$. The direct summand $cZ(\cO X)$ is therefore an
$\cO Z$-permutation lattice.
Each transitive permutation lattice $\cO[Z/U]$ is indecomposable,
since its endomorphism ring is $\cO[Z/U]$, a local ring for a $p$-group
$Z/U$. Krull--Schmidt over the complete order $\cO Z$ implies that
every direct summand of a permutation lattice is again a sum of such
transitive lattices. Over $K$, the center of the block is the sum of
the idempotent lines indexed by $\Irr(c)$, and the $Z$-character on
the line of $\chi$ is the central character of $\chi$. Every
transitive permutation character of the abelian group $Z$ contains
the trivial character once and any other linear character at most
once. Thus the multiplicity of $\lambda$ is no larger than that of
$1_Z$, proving \eqref{eq:central-fiber}.

Column orthogonality proves \eqref{eq:central-gram} as follows;
compare \cite[Theorem~6.2]{Rizo18}.
Projection to $X/Z$ bijects the $p$-regular conjugacy classes. For a
$p$-regular $x$, one has
$C_X(x)/Z=C_{X/Z}(xZ)$: an element centralizing $xZ$ has a commutator
with $x$ in the central $p$-group $Z$, and the $p'$-order of $x$
forces that commutator to be trivial. Central-character column
orthogonality consequently gives the same column Gram matrix for
every $\lambda$. The ordinary character table restricted to this
central fiber is $D_\lambda$ times the common Brauer character table.
Cancel that invertible table and compare with the trivial fiber,
which is the character table of $X/Z$. This gives the Gram identity,
first for all blocks and then for the displayed block components.
The Cartan matrix of a block is indecomposable and has positive
diagonal entries. Its Gram realization by a nonnegative decomposition
matrix proves the last assertion.
\end{proof}

\begin{lemma}[Cartan matrices under linearization]\label{lem:cartan-linear}
Let $Q\lhd G$ be a $p$-group and choose a $G$-invariant central series
$1=Q_0<Q_1<\cdots<Q_t=Q$ with elementary abelian factors.
Let $V=\bigoplus_{i=1}^t Q_i/Q_{i-1}$ with its induced
$H=G/Q$-action, and put $L=V\sd H$.
Under the common inflation labeling by $\IBr(H)$, the full Cartan
matrices of $G$ and $L$ are equal. Their connected components give a
bijection of blocks $B\leftrightarrow B^{\lin}$ with
\[
 C_B=C_{B^{\lin}},\qquad \delta(B)=\delta(B^{\lin}).
\]
\end{lemma}
\begin{proof}
Every irreducible Brauer character is trivial on the normal
$p$-subgroup. The $p$-regular classes of $G$ and $H$ correspond
bijectively. If $h\in H$ is $p$-regular, choose a $p$-regular lift
$x\in G$. Coprime conjugacy and fixed-point lifting give
\[
 |C_G(x)|=|C_Q(x)|\,|C_H(h)|.
\]
Indeed, Schur--Zassenhaus identifies the $p'$-complements over
$\langle h\rangle$ and gives both uniqueness of the lifted class
and surjectivity of $C_G(x)\to C_H(h)$. Coprime fixed points are
exact on the chosen central series, so
\[
 |C_Q(x)|=\prod_i |(Q_i/Q_{i-1})^h|=|V^h|.
\]
The centralizer order at the corresponding $p$-regular element of
$V\sd H$ is therefore identical. The Gram matrix of the irreducible
Brauer characters for the scalar product on $p$-regular classes is
the inverse Cartan matrix. Its entries, expressed as sums with
denominators $|C_G(x)|$, are equal in the two groups. Thus their
Cartan matrices and their indecomposable components agree. Equality
of the largest elementary divisors gives equality of defect orders.
\end{proof}

\begin{lemma}[A block assignment from modular induction]\label{lem:induction-support}
Let $A\lhd G$ be an abelian $p$-subgroup, let
$\lambda\in\Irr(A)$, and put $T=G_\lambda$.
For a block $c$ of $T/A$, all characters in the $\lambda$-fiber of
its corresponding block of $T/\ker\lambda$, inflated to $T$,
induce to characters in one block of $G$.
That block is determined solely by the modular induction matrix
from $T/A$ to $G/A$ and by the block partition of $\IBr(G)$.
\end{lemma}
\begin{proof}
The group $A/\ker\lambda$ is a central cyclic $p$-subgroup of
$T/\ker\lambda$. Let $\hat c$ be the block corresponding to $c$.
Let $D_\lambda$ be its decomposition matrix in this central fiber,
and let $R$ be the nonnegative matrix of modular induction from
$T/A$ to $G/A$. Every row of $R$ is nonzero. Clifford theory says
that each character in $\Irr(\hat c\mid\lambda)$ induces
irreducibly to $G$, and its decomposition row is the corresponding
row of $D_\lambda R$. Such a row is supported on a single block.
By \eqref{eq:central-gram}, the columns of $D_\lambda$ are connected
by positive rows. Positivity now forces the rows of $R$ indexed by
$\IBr(c)$ all to be supported in the same block of $G$. This
description uses only $R$ and the indicated block partition.
\end{proof}

\begin{proposition}[Blockwise linearization]\label{prop:blockwise-linear}
In the notation of Lemma~\ref{lem:cartan-linear},
\begin{equation}\label{eq:blockwise-linear}
 k(B)\le k(B^{\lin})
\end{equation}
for every corresponding pair of blocks.
\end{proposition}
\begin{proof}
First let $A\lhd G$ be an abelian $p$-subgroup and compare $G$ with
$A\sd(G/A)$, writing $H=G/A$ for this comparison.
The Cartan block partitions agree by
Lemma~\ref{lem:cartan-linear}. Choose orbit representatives
$\lambda\in\Irr(A)/(G/A)$. The quotient inertia group is
$H_\lambda$ in both cases. For every block $c$ of $H_\lambda$,
Lemma~\ref{lem:induction-support} assigns the pair $(\lambda,c)$
to the same block in both groups. In $G$ its contribution is
$|\Irr(\hat c\mid\lambda)|\le k(c)$ by
Lemma~\ref{lem:central-fiber}. In the split group $\lambda$ has the
canonical extension, trivial on $H_\lambda$, and the contribution
is exactly $k(c)$. The extension has $p$-power order and is one on
$p$-regular elements, so its use changes neither the decomposition
rows nor the block assignment. Sum over precisely the pairs
assigned to $B$.

For general $Q$, put $A=Q_1\le Z(Q)$. The abelian case replaces $G$
by $A\sd(G/A)$, preserving the Cartan labels and increasing every
block count. In this split group the complement-side subgroup
$Q/A$ is normal, since it acts trivially on $A$. Apply induction to
$Q/A$. The quotient is $A\sd H$, and $A$ acts trivially on all
the remaining central factors of $Q/A$.
Let $V'$ be the direct sum of these remaining central factors.
The resulting group is $V'\sd(A\sd H)\cong(A\oplus V')\sd H$.
At every step the labels come from the same $\IBr(H)$, which proves
the assertion.
\end{proof}

\begin{proposition}[The kernel of an affine action]\label{prop:affine-block-kernel}
Let $L=V\sd H$, where $V$ is an elementary abelian $p$-group, and
put $K=C_H(V)$. For a block $B$ of $L$, after passing to the inertia
group of a covered block $b$ of $K$ by Fong--Reynolds, one has
\begin{align}
 \delta(B)&=|V|\delta(b)|H:K|_p,\label{eq:affine-defect}\\
 k(B)&\le k(b)k(V\sd(H/K)).\label{eq:affine-block-count}
\end{align}
The quotient $H/K$ in these formulas acts faithfully on $V$.
\end{proposition}
\begin{proof}
Fong--Reynolds preserves the count and the defect group, and the
inertia group is again split over $V$. Thus assume $b$ is
$H$-invariant. Set $N=V\times K=C_L(V)$ and let
$c=1_V\otimes b$ be its block. It has defect order $|V|\delta(b)$.
Write $e_B$ and $e_c$ for the corresponding block idempotents in the
residue-field group algebras. For a $p$-subgroup $P\le L$, the Brauer
map $\Br_P:(kL)^P\to kC_L(P)$ retains exactly the coefficients
supported on $C_L(P)$.
There is a unique block above $c$. In fact $V$, being a normal
$p$-subgroup, belongs to every block defect group of $L$. Hence
$\Br_V(e_B)\ne0$ for every covering block. The nonzero,
orthogonal idempotents $\Br_V(e_B)$ sum to $e_c$, since $e_c$ is
supported in $N=C_L(V)$. They are central in $kN$, and the
primitivity of $e_c$ forces uniqueness.

Let $P/N$ be a Sylow $p$-subgroup of $L/N$. The invariant block
$c$ has a unique covering block $d$ in $P$, whose defect order is
$\delta(c)|P:N|$. Its idempotent, viewed in $kL$, is $e_B=e_c$.
The Brauer map at a defect group of $d$ deletes the same basis
coefficients in $kP$ and in $kL$, so $\delta(B)\ge\delta(d)$.
The normal-block intersection theorem gives the reverse inequality.
This proves \eqref{eq:affine-defect}.

For $\lambda\in\Irr(V)/H$, ordinary Clifford theory identifies
the characters above its orbit and $b$ with the characters of
$H_\lambda$ above $b$. The canonical extension of $\lambda$ is
trivial on $K$. By \eqref{eq:cover-class-count},
\[
 k(B)=\sum_{\lambda\in\Irr(V)/H}|\Irr(H_\lambda\mid b)|
 \le k(b)\sum_{\lambda\in\Irr(V)/(H/K)}
 k((H/K)_\lambda).
\]
The last sum is $k(V\sd(H/K))$ by Clifford theory. All projective
factor sets in the first count have been bounded using
\eqref{eq:projective-bound}.
\end{proof}

\begin{corollary}\label{cor:normal-two-descent}
Assume \eqref{eq:WA2}.
If $B$ is a $2$-block of a group $G$ with $O_2(G)\ne1$ and
$k(B)>\delta(B)$, then a block of strictly smaller defect order
also violates Brauer's inequality. In particular a counterexample
of minimum defect order has $O_2(G)=1$.
\end{corollary}
\begin{proof}
Apply Proposition~\ref{prop:blockwise-linear} to $O_2(G)$, then
Proposition~\ref{prop:affine-block-kernel} to the resulting block.
Equation~\eqref{eq:WA2} gives
\[
 \frac{k(B)}{\delta(B)}\le
 \frac{k(B^{\lin})}{\delta(B^{\lin})}
 \le\frac{k(b)}{\delta(b)}.
\]
The exact defect formula gives
$\delta(b)=\delta(B)/(|V||H:K|_2)<\delta(B)$.
\end{proof}

\section{Integral inner action and relative character fibers}\label{sec:inner}

Let $X\lhd G$ and let $b$ be a $G$-invariant block of $X$. Define
\begin{equation}\label{eq:inner-subgroup}
 G[b]=\{g\in G:\text{conjugation by $g$ on $\cO Xb$ is inner}\}.
\end{equation}
This is a normal subgroup of $G$ containing $X$. It is Dade's
ramification subgroup in the integral formulation of
\cite[Lemma~2.1]{Murai13}. Here innerness is implemented by a unit of $\cO Xb$;
fixing ordinary characters concerns only the generic algebra.

\begin{lemma}[Central specialization]\label{lem:inner-specialization}
Suppose $G[b]=G$, and let $c$ be a block of $G$ covering $b$. For
every $\theta\in\Irr(b)$, the set $\Irr(c\mid\theta)$ is a single
$\theta$-block in the sense of \cite{Rizo18}, and is nonempty.
\end{lemma}
\begin{proof}
Put $A=\cO Xb$, $R=Z(A)$ and $H=G/X$. The commutative ring $R$
is finite over $\cO$, complete and local, with residue field $k$.
Choose representatives $g_h$ and units $u_h\in A^\times$
implementing their conjugation on $A$, with $g_1=1$ and $u_1=b$.
The homogeneous units
$v_h=u_h^{-1}g_hb$ centralize $A$, and
\[
 v_hv_j=\gamma(h,j)v_{hj},\qquad \gamma(h,j)\in R^\times.
\]
Associativity makes $\gamma$ a normalized $2$-cocycle, with trivial
$H$-action on $R$. Coefficientwise in the homogeneous decomposition,
\begin{equation}\label{eq:inner-centralizer}
 \cO Gb=\bigoplus_{h\in H}Av_h,\qquad
 C_{\cO Gb}(A)=R_\gamma H,\qquad
 Z(\cO Gb)=Z(R_\gamma H).
\end{equation}
The last two identities follow by comparing coefficients in the
homogeneous decomposition, since each $v_h$ centralizes $A$.

Let $\mathfrak m$ be the maximal ideal of $R$ and put
$C=R_\gamma H$. Central primitive idempotents of $C$ and of
$C/\mathfrak mC$ correspond. By completeness, lift an idempotent
modulo $\mathfrak mC$. If its image is central, then
\[
 eC(1-e)\subseteq\mathfrak mC,
 \qquad eC(1-e)=\mathfrak m\,eC(1-e).
\]
Nakayama's lemma gives $eC(1-e)=0$, and similarly $(1-e)Ce=0$.
Thus the lift is central. Central lifts are unique and preserve
primitivity.

For $\theta\in\Irr(b)$, its central character
$\omega_\theta:R\to\cO$ reduces to the residue map of $R$.
Specializing gives
\[
 C_\theta=\cO\otimes_{R,\omega_\theta}C
       =\cO_{\gamma_\theta}H,
 \qquad \gamma_\theta=\omega_\theta\gamma.
\]
The image $c_\theta$ of $c$ has the same nonzero central primitive
reduction for every $\theta$, and is consequently a primitive
idempotent of this twisted order. Over $K$, the $\theta$-part of
$A$ is a matrix algebra, so the $\theta$-part of $K Gc$ is
\[
 \operatorname{Mat}_{\theta(1)}(K)
 \otimes_K K_{\gamma_\theta}Hc_\theta.
\]
Thus $\Irr(c\mid\theta)$ is the irreducible-character set of one
twisted block.

We explain its agreement with the character-triple definition of a
$\theta$-block, including $2$-primary factor sets. Replace
$\gamma_\theta$ by a cohomologous cocycle with values in a finite
cyclic group of roots of unity. The cochain can be taken unit-valued:
the valuations of a cochain relating two unit-valued cocycles form
a homomorphism from the finite group $H$ to $\Q$, and hence vanish.
The resulting central representation group $\widehat H$ realizes
the twisted order as the central-character quotient of
$\cO\widehat H$. Its odd central part is selected by an integral
central idempotent. For its central $2$-part, reduction imposes
$z=1$ and the usual central-$2$ quotient preserves block idempotents.
The single twisted block above is therefore precisely the fixed
central-character fiber of one block of $\widehat H$.
This is the definition of a $\theta$-block; its independence of the
representation group is \cite[Theorem~4.3]{Rizo18}. This also proves
nonemptiness. The same compatibility is given by
\cite[Proposition~5.2]{MNS15} with the external coprime group trivial.
\end{proof}

\begin{lemma}[The relative fiber bound]\label{lem:relative-fiber}
Under the hypotheses of Lemma~\ref{lem:inner-specialization}, assume
that $G/X$ is $2$-solvable. Then, for every $\theta\in\Irr(b)$,
\begin{equation}\label{eq:relative-fiber}
 |\Irr(c\mid\theta)|\le\frac{\delta(c)}{\delta(b)}.
\end{equation}
\end{lemma}
\begin{proof}
We recall the precise defect input from relative block theory.
For a $\theta$-block contained in a block $c$ with defect group $D$,
its $\theta$-defect group $D_\theta/X$, defined by the central
representation group, is conjugate into $DX/X$
\cite[Theorem~5.1(a)]{Rizo18}. Since $D\cap X$ is a defect group
of $b$, this gives
\[
 |D_\theta/X|\le |DX/X|=\frac{\delta(c)}{\delta(b)}.
\]

The representation group of $G/X$ is $2$-solvable. Remove its
central $2$-subgroup. Lemma~\ref{lem:central-fiber} bounds the
selected central-character fiber by the entire corresponding
quotient block, and Brauer's inequality for $2$-solvable groups
applies to that block. This established case follows from the
coprime $k(GV)$ theorem \cite{GMRS04}.
Selecting the prescribed character of the remaining odd central
subgroup does not change the defect order. Equivalently, this
is the proof of \cite[Theorem~7.1]{Rizo18} restricted to
$2$-solvable representation groups. It bounds the size of this
single $\theta$-block by $|D_\theta/X|$. Now apply
Lemma~\ref{lem:inner-specialization}.
\end{proof}

\begin{lemma}[Full relative defect]\label{lem:full-relative-defect}
Let $M=G[b]$, and let $c$ be a $G$-invariant block of $M$ above $b$.
There is a unique block $B$ of $G$ covering $c$, and
\begin{equation}\label{eq:full-relative-defect}
 \delta(B)=\delta(c)|G:M|_2.
\end{equation}
\end{lemma}
\begin{proof}
The uniqueness is \cite[Theorem~3.5(i)]{Murai13}; its hypothesis is
the whole subgroup $G[b]$. Hence $e_B=e_c$. Choose a Sylow
$2$-preimage $P$ of $G/M$. The unique block of $P$ above $c$ has
defect order $\delta(c)|P:M|$. Its idempotent is again $e_c$.
The Brauer maps at its defect group delete the same coefficients
in $kP$ and in $kG$. Thus a defect group of $B$ has at least that
order. The normal-block intersection theorem bounds its order in
the other direction by $\delta(c)|G:M|_2$, proving equality.
\end{proof}

\begin{theorem}[A fixed-point criterion]\label{thm:inner-consumer}
Let $X\lhd G$, let $b$ be a $G$-invariant $2$-block of $X$, and
put $M=G[b]$. Suppose $M/X$ is $2$-solvable. Assume that
$Q=G/M$ has an $\F_2Q$-module $W$ satisfying
\begin{align}
 |W|&\le\delta(b),\label{eq:consumer-size}\\
 |\Irr(b)^J|&\le |W^J|
 \quad\text{for every two-generated abelian $J\le Q$},
 \label{eq:consumer-marks}\\
 \ker(Q\to\GL(W))&\text{ is a $2$-group}.
 \label{eq:consumer-kernel}
\end{align}
If \eqref{eq:WA2} holds, then every block $B$ of $G$ covering $b$
satisfies $k(B)\le\delta(B)$.
\end{theorem}
\begin{proof}
Choose a block $c$ of $M$ below $B$ and pass to its inertia group by
Fong--Reynolds. This preserves $k(B)$ and its defect order. It
also preserves the inner subgroup: if $I=G_c$, then
$I[b]=I\cap G[b]=M$, since $M\le I$. Restrict $W$ to $I/M$;
all three hypotheses remain valid. Relabel $I$ as $G$.

Now $c$ is $G$-invariant. Since $M$ acts innerly on $\cO Xb$,
each $\theta\in\Irr(b)$ is $M$-invariant. Restriction consequently
gives a $Q$-equivariant map
\[
 r:\Irr(c)\longrightarrow\Irr(b),
\]
sending a character to its unique irreducible constituent on $X$.
By Lemma~\ref{lem:relative-fiber} every fiber has size at most
$L=\delta(c)/\delta(b)$. Therefore
\[
 |\Irr(c)^J|\le L|\Irr(b)^J|\le L|W^J|
\]
for the subgroups $J$ occurring in \eqref{eq:kappa-commuting}.
Clifford theory, with \eqref{eq:projective-bound} for the
inertia factor sets, gives
\[
 k(B)\le\kappa_Q(\Irr(c))
       \le L\kappa_Q(W)
       \le L|W|\,|Q|_2.
\]
The last inequality is Lemma~\ref{lem:affine-two-kernel}.
Equations~\eqref{eq:consumer-size} and
\eqref{eq:full-relative-defect} finish the calculation:
\[
 k(B)\le\delta(c)|Q|_2=\delta(B).
\]
\end{proof}

\subsection{Tensor factors and component permutations}

We shall use the stronger form of \eqref{eq:consumer-marks} with all
subgroups $J$. The following elementary algebra fact ensures that
component modules descend through the group $G[b]$.

\begin{lemma}[Innerness on a tensor factor]\label{lem:tensor-inner}
Let $A$ and $B$ be finite free $\cO$-algebras with $Z(A)$ local.
If a unit of $A\otimes_\cO B$ induces an automorphism $\alpha$
on $A\otimes1$, then $\alpha$ is inner on $A$.
\end{lemma}
\begin{proof}
Write $u=\sum_i a_i\otimes b_i$ and
$u^{-1}=\sum_j c_j\otimes d_j$ using bases in the second tensor
factor. The identity $u(a\otimes1)=(\alpha(a)\otimes1)u$ gives
$a_i a=\alpha(a)a_i$, and the inverse identity gives
$c_j\alpha(a)=ac_j$. Hence $a_i c_j\in Z(A)$.
Choose an $\cO$-linear map $f:B\to\cO$ with $f(1)=1$; this is
possible because $1$ is primitive in the free lattice $B$. Apply
$1\otimes f$ to $uu^{-1}=1$ to obtain
\[
 \sum_{i,j}f(b_i d_j)a_i c_j=1.
\]
Since $Z(A)$ is local, some $a_i c_j$ is a unit. Thus $a_i$ has a
right inverse in $A$, and is a unit: this follows after scalar
extension to the finite-dimensional $K$-algebra, and its displayed
right inverse is integral. The equation for $a_i$ now shows
$\alpha(a)=a_i a a_i^{-1}$.
\end{proof}

\begin{proposition}[Assembly of component modules]\label{prop:component-assembly}
Suppose $O_2(G)=1$, $F(G)=Z(G)$, and
$C_G(E(G))=Z(G)$. Let $X=Z(G)E(G)$, and let $b$ be a
$G$-invariant block of $X$. Assume its component blocks $b_i$ on
the quasisimple components $L_i$ have positive defect and at least
two ordinary characters. Let $A_i$ be the image in $\Aut(L_i)$
of the stabilizer of $L_i$ in $G$, and let $T_i$ be the subgroup
acting innerly on $\cO L_i b_i$. Suppose an $\F_2(A_i/T_i)$-module
$W_i$ is given with
\begin{equation}\label{eq:component-local-marks}
 |W_i|=\delta(b_i),\quad
 |\Irr(b_i)^J|\le |W_i^J|\ (J\le A_i/T_i),\quad
 \ker(A_i/T_i\to\GL(W_i))\text{ a $2$-group}.
\end{equation}
Then $M=G[b]$ has solvable quotient $M/X$, and $Q=G/M$ has a
module $W$ satisfying all three conditions of
Theorem~\ref{thm:inner-consumer}, with equality in
\eqref{eq:consumer-size} and with all subgroup fixed-point numbers.
\end{proposition}
\begin{proof}
The intersections between the components and with $Z(G)$ have odd
order. In the central-character sector selected by $b$, the central
product block is canonically the tensor product of the component
block orders. Indeed the external tensor product descends through
the central kernel, which acts trivially because its prescribed
linear characters match. The additional central factor $Z(G)$ acts
by its chosen scalar character. Consequently
\begin{equation}\label{eq:component-products}
 \Irr(b)=\prod_i\Irr(b_i),\qquad
 \delta(b)=\prod_i\delta(b_i).
\end{equation}
These identifications are in the central-character sector selected
by $b$.

An element of $M$ fixes every character in \eqref{eq:component-products}.
It cannot permute component positions: each factor contains at least
two choices, so moving one position moves a suitable tuple.
Lemma~\ref{lem:tensor-inner} then shows that its automorphism on
each component block is inner. The image of the stabilizer $Q_i$
of a component in $Q$ therefore acts on $W_i$ through $A_i/T_i$.
For one representative of each component orbit define
\begin{equation}\label{eq:induced-component-module}
 W=\bigoplus_{i\in\{\text{component orbits}\}}
       \Ind_{Q_i}^{Q}W_i.
\end{equation}
Its cardinality is the product in \eqref{eq:component-products}.

Fix $J\le Q$. On each orbit of $J$ on the component positions,
a fixed character tuple is determined by one character fixed by
the corresponding position stabilizer, including all return
automorphisms. A fixed vector in \eqref{eq:induced-component-module}
is determined in exactly the same way by a vector in the relevant
$W_i$. Multiplying the local inequalities in
\eqref{eq:component-local-marks} over the position orbits proves
$|\Irr(b)^J|\le|W^J|$.

Let $R$ be the kernel on $W$. If $R$ had an element $q$ of odd
prime order $\ell$, choose an $\ell$-power-order lift $g\in G$
of $q$, using a Sylow subgroup of the preimage of $\langle q\rangle$.
The induced module has nonzero summands at all component positions,
so $q$ fixes those positions. On every component, the image of $g$
in $A_i/T_i$ has $\ell$-power order and lies in the local
$2$-kernel. It is trivial. Choose integral units implementing the
component actions; their tensor product implements the action on
$\cO Xb$, including its scalar central factor. Thus $g\in M$,
contrary to $q\ne1$. Cauchy's theorem proves that $R$ is a
$2$-group.

Finally, $M$ fixes the component positions and
$C_G(E(G))=Z(G)\le X$. The quotient $M/X$ therefore embeds in
a product of the component outer automorphism groups. These are
solvable by the Schreier theorem. This proves the remaining claim.
\end{proof}

\section{A weighted permutation inequality}\label{sec:permutation}

For a permutation $g$ of a finite set $\Omega$, write
$s_\Omega(g)=|\Omega|-|\Fix_\Omega(g)|$, and write $g=g_2g_o=g_og_2$
for its commuting $2$-part and odd part. Throughout this section
$t=2^{-1/2}$. If $f$ is a nonnegative class function on the odd-order
elements of a finite group $G$, set
\[
 \mathcal P(G;f)=\frac1{|G|}
   \sum_{\substack{a\in G\text{ odd}\\ b\in C_G(a)}}f(a)f(b_o),
 \qquad
 \mathcal P(G,\Omega)=\mathcal P(G;t^{s_\Omega}).
\]
Here and below ``odd'' includes the identity. The assertion needed for
the affine argument is the following.

\begin{theorem}\label{thm:permutation-weight}
For every faithful permutation action of a finite group $G$ on a finite
set $\Omega$,
\begin{equation}\label{eq:permutation-weight}
 \frac1{|G|}\sum_{\substack{a\in G\text{ odd}\\ b\in C_G(a)}}
  2^{-(s_\Omega(a)+s_\Omega(b_o))/2}\leq1.
\end{equation}
The same assertion holds with $2^{-1/2}$ replaced by any real number in
$(0,2^{-1/2}]$.
\end{theorem}

The almost simple input is stated in Proposition~\ref{prop:permutation-local}.

\subsection{Normal abelian subgroups}

\begin{lemma}\label{lem:permutation-two-kernel}
Let $L\unlhd G$ be a $2$-group. For an odd element $\bar a\in G/L$, let
$\bar f(\bar a)=f(a)$, where $a$ is any odd lift. This is well defined,
and
\begin{equation}\label{eq:permutation-two-kernel}
 \mathcal P(G;f)=\mathcal P(G/L;\bar f).
\end{equation}
\end{lemma}
\begin{proof}
Schur--Zassenhaus, applied to the inverse image of the cyclic group
$\langle\bar a\rangle$, shows that all odd lifts of $\bar a$ are
$L$-conjugate. Indeed their cyclic subgroups are complements, and a
complement contains exactly one element over $\bar a$. Thus odd
conjugacy classes correspond under the quotient map. Moreover
$C_G(a)\longrightarrow C_{G/L}(\bar a)$ is onto: if a lift $h$ centralizes
$\bar a$, conjugate $a^h$ back to $a$ by an element of $L$. Every fibre
has size $|C_L(a)|$. Quotients commute with taking odd parts. Consequently
\[
 \mathcal P(G;f)=
 \sum_{[a]\text{ odd}}\frac{f(a)}{|C_G(a)|}
                         \sum_{b\in C_G(a)}f(b_o)
 =\sum_{[\bar a]\text{ odd}}
   \frac{\bar f(\bar a)}{|C_{G/L}(\bar a)|}
                         \sum_{\bar b\in C_{G/L}(\bar a)}\bar f(\bar b_o).
\]
\end{proof}

\begin{lemma}\label{lem:permutation-abelian-descent}
If $G$ has a nontrivial abelian normal subgroup, a faithful action
$(G,\Omega)$ has a quotient $E$ with a faithful action on a strictly
smaller set $\Omega'$ such that
$\mathcal P(G,\Omega)\leq\mathcal P(E,\Omega')$.
\end{lemma}
\begin{proof}
It suffices to treat an abelian normal $r$-subgroup for a prime $r$.
First suppose $r=2$, and denote this subgroup by $K$. Its orbits inside
each $G$-orbit are blocks of a common size $b$. On such a block the
faithful image $K_0$ is regular abelian. The induced block stabilizer is
$K_0\sd A_0$, where $A_0$ is the stabilizer of an origin and acts
faithfully on $K_0$. Choose transports between the blocks; these embed
the action into the corresponding imprimitive wreath product.
For $b>1$, discard the translation base and replace each $K_0$-fibre by
$K_0\setminus\{1\}$, with its $A_0$-action. For $b=1$, retain the
original points and their action. The kernel $L$ of the resulting
homomorphism is a subgroup of the product of the translation
bases. Thus $L$ is abelian, normal and a $2$-group, and $K\leq L$.
The action of $G/L$ is faithful by construction and its degree decreases.

If an odd element moves a block, its old and new contributions to
support are $b$ and $b-1$. On a fixed block, its odd affine map is
translation-conjugate to its linear part, again by Schur--Zassenhaus.
Deleting the origin decreases the degree and the number of fixed
points by the same amount. Its support is unchanged. The result now
follows from Lemma~\ref{lem:permutation-two-kernel} and positivity.

Suppose next that $K$ has odd order. We supply the weighted estimate
required for this case. For every permutation subgroup
$D\le\operatorname{Sym}(\Omega)$ and every permutation
$h\in\operatorname{Sym}(\Omega)$, where $|\Omega|=n$,
\begin{equation}\label{eq:permutation-coset-maximum}
 \sum_{x\in Dh}t^{s(x)}\leq\sum_{x\in D}t^{s(x)}.
\end{equation}
To see this, expand
\[
 t^{s(x)}=\sum_{U\subseteq\Omega}
       t^{n-|U|}(1-t)^{|U|}\,1_{\{x\text{ fixes }U\text{ pointwise}\}}.
\]
The intersection of a coset with a pointwise stabilizer is empty or a
coset of the corresponding subgroup intersection. All coefficients are
nonnegative. If $D$ is odd abelian and faithful, then
\begin{equation}\label{eq:odd-abelian-support}
 \sum_{x\in D}t^{s(x)}\leq\sqrt{|D|}.
\end{equation}
For a transitive action its faithful image is regular of odd order $q$,
and $1+(q-1)t^q\leq\sqrt q$. At $q=3$, the square of the left side is
$3/2+\sqrt2<3$; for odd $q\geq5$, $(q-1)2^{-q/2}$ decreases in steps
of two and is at most $1/\sqrt2$. For several orbits, project to one
orbit and apply \eqref{eq:permutation-coset-maximum} to each coset of
the projection kernel on the remaining orbits. Induction gives
\eqref{eq:odd-abelian-support}, retaining all relations between orbits.

Write the $K$-orbits as $B_i$, identify $B_i$ with its regular odd
abelian image $K_i$, and put $q_i=|K_i|$. Let $L$ consist of the elements of $G$ which fix every fibre and act as translations on each.
It is normal abelian of odd order and contains $K$. The quotient $G/L$
acts by linear isomorphisms between the $K_i$. Replace a nontrivial
fibre by
\[
       Y_i=(K_i\setminus\{0\})/(v\sim-v),
       \qquad |Y_i|=(q_i-1)/2,
\]
and retain every singleton fibre. An automorphism of an odd abelian
group fixing all pairs $\{v,-v\}$ is either $1$ or $-1$: if it has
opposite signs on nonzero $v,w$, its value on $v+w$ cannot have either
allowed sign, since there is no $2$-torsion. Thus the kernel $R$ of
this action of $G/L$ is an elementary abelian $2$-group. Set
$E=(G/L)/R$. The new degree is strictly smaller, and its action is
faithful. For odd $\bar g\in G/L$ write
$w(\bar g)=t^{s_{\Omega'}(\bar g)}$.

We claim that for every subgroup $D\leq L$ and every odd $g\in G$,
\begin{equation}\label{eq:translation-coset-bound}
       \sum_{x\in Dg}t^{s_\Omega(x)}\leq\sqrt{|D|}\,w(gL).
\end{equation}
The subgroup $D$ need not be normalized by $g$. Let $F,M$ be the fixed
and moved fibres of $g$. On a fixed fibre let $u_i$ be the linear part,
$f_i=|C_{K_i}(u_i)|$, and $M_i=(1-u_i)K_i$, of order $q_i/f_i$.
The fixed-point equation for a translation times $g$ is solvable exactly
when the translation lies in a specified coset $c_i+M_i$; then it has
$f_i$ solutions. Its weight on this fibre is therefore
\[
 t^{q_i}+(t^{q_i-f_i}-t^{q_i})1_{c_i+M_i}.
\]
Expanding the product and using nonnegativity allows every $c_i$ to
be replaced by zero: each intersection with $D$ is empty
or a coset of a subgroup intersection. Let $D_0$ be the kernel of
$D\to\prod_{i\in F}K_i/M_i$. After division by
$t^{q_i-f_i}$, the fixed-fibre weight is $1$ at the zero coset
of $K_i/M_i$ and $t^{f_i}$ elsewhere. This is precisely the
support weight for its regular action of degree $f_i$.
Apply \eqref{eq:odd-abelian-support} to the image of $D$ on
the disjoint union of these quotient sets. Each image element
has $|D_0|$ preimages, giving
\[
 \sum_{x\in Dg}t^{s(x)}
 \leq\sqrt{|D|}\sqrt{|D_0|}
        t^{\sum_{i\in M}q_i+\sum_{i\in F}(q_i-f_i)},
 \qquad
 |D_0|\leq\prod_{i\in M}q_i\prod_{i\in F}(q_i/f_i).
\]
For a moved nontrivial fibre,
$\sqrt q\,t^q\leq t^{(q-1)/2}$, equivalently
$q\leq2^{(q+1)/2}$ for odd $q\geq3$. For a fixed fibre,
\[
       \sqrt{q/f}\,t^{q-f}\leq t^{(q-f)/2}
       \quad\Longleftrightarrow\quad
       q/f\leq2^{(q-f)/2}.
\]
Writing $q=fr$, this is immediate if $r=1$ or $f\geq3$. If $f=1$,
it holds for odd $q\geq7$; the apparent exceptions $q=3,5$ cannot
occur, since an odd automorphism of their abelian groups is trivial.
An odd automorphism fixes a pair $\{v,-v\}$ precisely when it fixes
$v$. Hence the new fixed-fibre support is $(q-f)/2$. A moved
singleton contributes one to each support. Multiplication proves
\eqref{eq:translation-coset-bound}.

For a finite group $C$ and a class function $f$ on its odd elements, put
\[
 Q(C;f)=\sum_{\substack{a,u\in C\text{ odd}\\ {}[a,u]=1}}f(a)f(u).
\]
The unique decomposition $b=zu$, with $z$ a $2$-element, gives
\begin{equation}\label{eq:permutation-two-centralizers}
 \mathcal P(G;f)=\sum_{[z]\text{ a }2\text{-element}}
                       \frac{Q(C_G(z);f)}{|C_G(z)|}.
\end{equation}
Because $L$ is odd, Schur--Zassenhaus gives a bijection on $2$-element
classes under $G\to G/L$, and $C_G(z)\to C_{G/L}(zL)$ is onto, with
kernel $D=C_L(z)$. Every odd coset in this centralizer consists of
odd elements. For a commuting odd pair in the quotient, omit the
additional commutation condition on its lifts and apply
\eqref{eq:translation-coset-bound} twice. It follows that
\[
 Q(C_G(z);t^{s_\Omega})
       \leq |D|Q(C_{G/L}(zL);w).
\]
The factor $|D|$ cancels from the centralizer denominator in
\eqref{eq:permutation-two-centralizers}. Summing, and finally applying
Lemma~\ref{lem:permutation-two-kernel} to $R$, proves the claim.
\end{proof}

\subsection{Matrices for commuting pairs in prescribed cosets}

For $N\unlhd H$ and commuting $\alpha,\beta\in H/N$, with $\alpha$
odd, define
\begin{equation}\label{eq:two-coset-mass}
 U_N(\alpha,\beta;f)=\frac1{|N|}
   \sum_{\substack{aN=\alpha,\ a\text{ odd}\\ bN=\beta,\ [a,b]=1}}
                         f(a)f(b_o).
\end{equation}
Only $N$-conjugacy invariance of $f$ is required.

\begin{lemma}\label{lem:commuting-matrix}
If $\alpha=\beta_o=1$, let $I$ be the $\beta$-stable odd
$N$-classes, with identity class $0$. There is a symmetric nonnegative
matrix $K$, indexed by $I$, whose row sums are one and such that
\[
 U_N(1,\beta;f)=f^{\mathsf T}Kf,\qquad
 K_{0i}\geq |C_N(x_i)|^{-1}.
\]
Here $x_i$ represents the class $i$.
In general there is a rectangular nonnegative matrix on the odd
$N$-classes in $\alpha$ and $\beta_o$, with every row and column sum
in $\{0,1\}$, and with
$U_N(\alpha,\beta;f)=f_\alpha^{\mathsf T}Kf_{\beta_o}$,
where the two vectors restrict $f$ to the row and column class sets.
The number of nonzero rows equals the number of nonzero columns and
is at most $k(N)$.
\end{lemma}
\begin{proof}
Write $\beta=\beta_2\beta_o$. In both cases set
\[
 K_{ij}=\frac1{|N|}
  \sum_{\substack{a\in i,\ u\in j\\ {}[a,u]=1}}
  |\{z\in H:zN=\beta_2, z\text{ a }2\text{-element},
                              [z,a]=[z,u]=1\}|.
\]
The substitution $b=zu$ is the unique $2$/odd decomposition. For
fixed $a$, the relevant $b$ form either the empty set or a coset of
$C_N(a)$, proving that its row sum is zero or one. For columns use
$c=za$ instead. The image of $c$ is $\beta_2\alpha$; these commuting
factors have coprime orders, so the $2$-part and odd part of $c$ recover
$z$ and $a$ in their prescribed cosets. The same centralizer argument
proves the column assertion. Thus the two active counts agree.

For any $x$ normalizing $N$, the number of $N$-classes in $Nx$ equals
the number of $N$-classes in $N$ fixed by $x$. Indeed, Burnside's
formula for conjugation on $Nx$ sums $|C_{Nx}(n)|$ over $n\in N$;
an $x$-stable $N$-class contributes exactly $|N|$ to that sum and
an unstable class contributes zero. This proves the bound $k(N)$.

When both odd quotient labels are trivial, interchanging $a,u$
makes $K$ symmetric. The stable classes are exactly its nonzero
rows. Moreover
\[
 K_{0i}=
 \frac{|\{z\in C_H(x_i):zN=\beta, z\text{ a }2\text{-element}\}|}
      {|C_N(x_i)|}.
\]
The numerator is positive: take the $2$-part of any element of the
nonempty centralizer coset. This proves all the assertions.
\end{proof}

\begin{corollary}\label{cor:coset-quadratic}
Let $\beta\in H/N$ be a $2$-element. Suppose $f(1)=1$ and
$0\leq f\leq1$. If
\begin{equation}\label{eq:internal-quadratic}
       (\sqrt{|C_N(x)|}+1)f(x)\leq1
\end{equation}
for every nonidentity odd $x\in N$ whose $N$-class is $\beta$-stable,
then $U_N(1,\beta;f)\leq1$.
For general commuting labels $\alpha,\beta\in H/N$ with $\alpha$ odd,
if all nonidentity odd weights are at most $\epsilon\ge0$ and $R$ is
the active row count, then
\begin{equation}\label{eq:rectangular-bounds}
 U_N(\alpha,\beta;f)\leq
 \begin{cases}
 R\epsilon^2,&\alpha\ne1, \beta_o\ne1,\\
 \epsilon+(R-1)\epsilon^2,&\text{exactly one of }\alpha,\beta_o\ne1.
 \end{cases}
\end{equation}
In particular, for $q\geq k(N)$ and $d\ge0$, the conditions
\begin{equation}\label{eq:outer-quadratic}
 \epsilon\leq(\sqrt q+1)^{-1},\qquad
                   \sqrt q\,\epsilon\leq t^d
\end{equation}
make these two bounds at most $t^{2d}$ and $t^d$, respectively.
\end{corollary}
\begin{proof}
The symmetric stochastic matrix satisfies the exact energy identity
\[
 f^{\mathsf T}Kf=\sum_i f_i^2-
                       \frac12\sum_{i,j}K_{ij}(f_i-f_j)^2.
\]
Keeping the terms incident with the identity bounds this by
$1+\sum_{i\ne0}(f_i^2-K_{0i}(1-f_i)^2)$.
Each summand is nonpositive by \eqref{eq:internal-quadratic}.
For the rectangular assertion use Lemma~\ref{lem:commuting-matrix}.
If just one odd label is trivial, its identity row or column has
mass exactly one; all the remaining mass is at most $R-1$.
Finally
$1+(q-1)\epsilon\leq\sqrt q$ when
$\epsilon\leq(\sqrt q+1)^{-1}$, which gives the last assertion.
\end{proof}

These estimates persist with $t$ replaced by $t^c$, $c$ any positive
integer. In particular, if $\sqrt q\,\epsilon\leq t^d$, then
$\sqrt q\,\epsilon^c\leq(\sqrt q\,\epsilon)^c\leq t^{dc}$.
The weights are functions on odd classes, including when an even
return automorphism interchanges coordinate representations.

\subsection{Natural alternating groups}

\begin{lemma}\label{lem:natural-alternating-cosets}
Let $A=A_m$, $m\geq2$, in its natural action. For either sign
$\varepsilon\in\{0,1\}$ and $0<u\leq t$,
\begin{equation}\label{eq:alternating-sign-coset}
 \frac1{|A|}\sum_{\substack{x\in A\text{ odd},\ y\in S_m\\
                 \operatorname{sgn}(y)=(-1)^\varepsilon, [x,y]=1}}
                    u^{s(x)+s(y_o)}\leq1.
\end{equation}
For $m\geq5$, the analogous bound
$U_{A_m}(1,\beta;u^s)\leq1$ holds whenever $\beta\in\Aut(A_m)/A_m$
has $2$-power order, including the identity. It holds for each natural
representation, including either natural six-point representation of $A_6$.
\end{lemma}
\begin{proof}
We first take $u=t$. Put $P_m=\mathcal P(S_m,\{1,\ldots,m\})$
and $P_0=1$.
If the odd permutation $x$ has $r$ cycles of an odd length $a$, its
centralizer on these cycles is $C_a\wr S_r$. A $b$-cycle of the
permutation of these $a$-cycles has a total rotation $j\in\Z/a\Z$;
there are $a^{b-1}$ rotations with that total. On these $ab$ points
the second permutation has $\gcd(a,j)$ cycles of length
$ab/\gcd(a,j)$. Its odd part fixes all of them exactly when $b$ is a
power of two and $j=0$. Otherwise its odd part moves all $ab$ points.
The first permutation contributes weight $1$ if $a=1$, and $t^{ab}$
if $a>1$.

The cycle enumeration, including the averaging over rotations, therefore
gives the following identity of formal power series:
\begin{equation}\label{eq:symmetric-generating-function}
 F(z):=\sum_{m\geq0}P_mz^m
   =\exp\left(\sum_{n\geq1}\frac{h_nz^n}{n}\right),
 \quad
 h_n=\begin{cases}
 1,&n\text{ a power of }2,\\
 2t^n+(\sigma_1(n_{2'})-2)t^{2n},&\text{otherwise}.
 \end{cases}
\end{equation}
Here $\sigma_1(n)=\sum_{d\mid n}d$ is the divisor sum. To verify
the coefficient explicitly before collection, put $u_1=1$ and
$u_a=t^a$ for odd $a>1$. The logarithm is
\[
 \sum_{a\geq1\text{ odd}}\sum_{b\geq1}
 \frac{u_a^bz^{ab}}{ab}\sum_{j=0}^{a-1}v(a,b,j),
 \qquad
 v(a,b,j)=\begin{cases}1,&b\text{ a }2\text{-power}, j=0,\\
                                      t^{ab},&\text{otherwise}.
                       \end{cases}
\]
For a non-$2$-power $n$, the divisor $a=1$ contributes $t^n$,
the divisor $a=n_{2'}$ contributes $t^n+(n_{2'}-1)t^{2n}$, and each
other odd divisor $a>1$ contributes $at^{2n}$, proving
\eqref{eq:symmetric-generating-function}.

We have $0\leq h_n\leq1$. The small non-$2$-powers give
$h_3=1/\sqrt2+1/4<1$ and
$h_5=1/(2\sqrt2)+1/8<1$. For $n\geq6$, the bound
$\sigma_1(n_{2'})\leq n^2$ gives
$h_n\leq2^{1-n/2}+n^22^{-n}\leq13/16$; both terms decrease
thereafter. In particular $P_m\leq1$ coefficientwise by comparison
with $\exp(\sum z^n/n)=(1-z)^{-1}$.

To distinguish the two sign cosets, insert $\operatorname{sgn}(y)$
in the same enumeration. On the component above, this sign is
$(-1)^{ab-\gcd(a,j)}=(-1)^{b-1}=(-1)^{ab-1}$. The signed series is
therefore $R(z)=1/F(-z)$. Let $J=F+R$ and $K=F-R$.
Write $J(z)=\sum_{n\ge0}J_nz^n$ and $K(z)=\sum_{n\ge0}K_nz^n$.
Their coefficients are nonnegative, and, for $m\geq2$, are precisely
the two expressions in \eqref{eq:alternating-sign-coset}. Write
$H_{\rm odd}=\sum_{n\text{ odd}}h_nz^{n-1}$ and
$H_{\rm even}=\sum_{n\text{ even}}h_nz^{n-1}$. Then
\[
       J'=H_{\rm odd}J+H_{\rm even}K,
       \qquad K'=H_{\rm odd}K+H_{\rm even}J.
\]
The initial values are $J_0=J_1=2$, $K_0=K_1=0$, and
$J_2=K_2=1$. Inductively, in the coefficient recurrence for $nJ_n$
or $nK_n$, the terms with $1\leq j\leq n-2$ total at most $n-2$.
Of the last two terms exactly one has a nonzero initial coefficient,
equal to two. Thus $J_n,K_n\leq1$ for every $n\geq2$.
Nonnegativity proves the result for every $u\leq t$.

In applications a return automorphism, or holonomy, is the composite
of coordinate identifications around a closed coordinate path.
For $m\ne6$, $\Aut(A_m)=S_m$ for $m\geq5$, so this proves the
last assertion except at six. For $A_6$ the nonidentity odd classes
have the following data in either chosen natural action:
\[
\begin{array}{c|rrrr}
 \text{class}&3A&3B&5A&5B\\\hline
 \text{support}&3&6&5&5\\
 |C_{A_6}(x)|&9&9&5&5.
\end{array}
\]
The subgroup induced by $S_6$ is the stabilizer in $\Aut(A_6)$ of
the support-three class and has index two. Every automorphism outside
it interchanges $3A$ and $3B$. These finite facts can also be obtained
by the explicit permutation calculation given below. If the even
holonomy is induced by $S_6$, use the sign-coset result. Otherwise
neither order-three class is an index of the symmetric matrix in
Lemma~\ref{lem:commuting-matrix}. Every remaining nonidentity index
has support five and centralizer order five, and
\[
             (\sqrt5+1)2^{-5/2}<1.
\]
Corollary~\ref{cor:coset-quadratic} applies. This proof uses only the
stable classes and therefore also applies when the even holonomy
interchanges the two coordinate representation types.
\end{proof}

\subsection{The almost simple local estimates}

For each isomorphism type of nonabelian simple group $S$, fix a faithful
nonempty $\Out(S)$-set $\Psi_S$. We use its regular set, with one
point for the trivial outer group, except that for $S=\operatorname{PSL}_3(4)$
we use the faithful action of $\Out(S)\cong C_2\times S_3\cong D_{12}$,
the dihedral group of order twelve, on the six vertices of a hexagon.
Write $d_S=|\Psi_S|$. This choice is made once for each abstract $S$,
including occurrences in different
permutation degrees.

\begin{proposition}\label{prop:permutation-local}
Let $S\leq A\leq S_m$ be primitive almost simple, with
$A=N_{S_m}(S)$. Suppose the action is either natural alternating or
is outside the alternating subset actions, allowing abstract twists.
Identify $A$ with its conjugation image in $\Aut(S)$. The following
statements hold.
\begin{enumerate}
\item $d_S<m$, and the support of every odd element of $A$ is at
least the support of its image on $\Psi_S$.
\item $k(S)t^{3(m-d_S)}\leq1$.
\item Let $S\unlhd H\leq\Aut(S)$, and suppose every odd-order
element of $H$ belongs to $A$. For commuting $\alpha,\beta\in H/S$
with $\alpha$ odd, and every integer $c\geq1$, put
$f(x)=t^{cs(x)}$ on the odd elements of $H$, using their coordinate
action in $A$. Then
\[
 U_S(\alpha,\beta;f)
 \leq t^{c(s_{\Psi_S}(\alpha)+s_{\Psi_S}(\beta_o))}.
\]
Here $U_S$ is the prescribed-coset mass in \eqref{eq:two-coset-mass},
and $H/S$ acts through $\Out(S)$ on $\Psi_S$. The hypothesis permits
even return automorphisms that interchange coordinate representation types.
\end{enumerate}
\end{proposition}
\begin{proof}
For a natural $A_m$, the outer group is a $2$-group, so its odd
supports are zero, and (iii) is
Lemma~\ref{lem:natural-alternating-cosets}. Its outer degree is two,
except for $A_6$, where it is four. The general permutation class
bound of Garonzi--Mar\'oti \cite[Theorem~1.1]{GaM15} is
\begin{equation}\label{eq:permutation-class-bound}
       k(U)\leq5^{(n-1)/3}<2^{n-1}\qquad(U\leq S_n,\ n\geq4).
\end{equation}
Since $3(m-2)/2\geq m-1$ for $m\geq5$, this proves (ii) except
at six, where $k(A_6)=7\leq8=2^{3(6-4)/2}$.

Suppose $m\geq128$, and put $x=\log_2m$, $a=|\Out(S)|$ and
$q=k(S)$. The following four published bounds are used with the
indicated exceptions:
\begin{equation}\label{eq:large-as-inputs}
 \begin{aligned}
 s(u)&\geq m/2\qquad(1\ne u\in A\text{ odd}),\\
 |S|&<m^{1+\log_2m},\qquad
       a\leq3\log_2m,\qquad q\leq m^6.
 \end{aligned}
\end{equation}
The first follows from Burness--Guralnick
\cite[Corollary~2]{BG22}: outside the alternating subset/product
family, an element of odd prime order $r$ has fixed-point ratio at
most $1/\sqrt{r+1}\leq1/2$. Take a nonidentity prime-order power
of $u$. Mar\'oti's order theorem \cite[Theorem~1.1]{Mar02} gives
the second bound. Its alternating product alternative reduces here
to a subset action, and its Mathieu exceptions have degree at most
24. The bound on $a$ is Guralnick--Mar\'oti--Pyber
\cite[Lemma~7.7]{GMP17}, for any nontrivial simple-group permutation
representation. For the last bound, apply Garzoni--Gill
\cite[Theorem~1.2 and Table~1]{GG23} to the primitive ambient $A$. Outside the subset family, their projective
exception has $k(A)<100m$, and every entry of their finite table
satisfies this same bound; all other cases have $k(A)<m/2$.
An $A$-class in $S$ splits into at most $[A:S]$ $S$-classes. Hence
\[
       k(S)\leq[A:S]k(A)<100am\leq300m\log_2m\leq m^6.
\]

These bounds imply the three local inequalities
\begin{equation}\label{eq:large-as-thresholds}
 (\sqrt{|S|}+1)t^{m/2}\leq1,\qquad
 t^{m/2}\leq(\sqrt q+1)^{-1},\qquad
                 \sqrt q\,t^{m/2}\leq t^a.
\end{equation}
For completeness, at $x=7$ the difference
$2^x/4-(x^2+x)/2$ is four and its derivative
$(\log_e2)2^x/4-x-1/2$ is positive thereafter. Thus
$2^{m/4}\geq16m^{(1+\log_2m)/2}$, paying also the added one in the
first inequality. The difference $m/4-3\log_2m$ starts at eleven
and increases, proving the second. Finally
$m/2\geq9\log_2m\geq a+\log_2q$: the first comparison is
$64\geq63$ at 128 and its difference increases. The inequalities
remain true under $t\mapsto t^c$.
Corollary~\ref{cor:coset-quadratic} now proves (iii), and $d_S\leq a<m$
proves (i). For (ii), one may use
\eqref{eq:permutation-class-bound} and
$a\leq3\log_2m\leq(m+2)/3$, valid for $m\geq128$; this gives
$3(m-d_S)\geq2(m-1)$. The smaller outer action for
$\operatorname{PSL}_3(4)$ only improves these estimates.

It remains to give the complete finite range $5\leq m<128$.
Its classification input is the primitive group classification
\cite{RD05,CQRD11}, in the permutation-isomorphism models of
\texttt{PrimGrp} \cite{PrimGrp}. Select every primitive model in this
range of almost simple type; omit only the natural $A_m,S_m$ models
already proved. The selection has 292 models. The socle is computed
inside the primitive ambient, so the selection includes socles that
are imprimitive and the novelty maximal subgroups.

For every nonidentity odd socle class with centralizer order $C$
and support $s$, the required internal test is the exact integer test
\begin{equation}\label{eq:finite-internal-test}
       A_0=2^s-C-1\geq0,\qquad A_0^2\geq4C.
\end{equation}
This is equivalent to $(\sqrt C+1)2^{-s/2}\leq1$.
For 195 models, the test holds with $C=|S|$ and
$s=\lceil m/2\rceil$, using the first bound of
\eqref{eq:large-as-inputs}. All alternating labels are excluded from
this shortcut. The remaining 97 models give 887 odd-class
records and 267 distinct pairs $(C,s)$, all satisfying
\eqref{eq:finite-internal-test}. The $A_5\cong\operatorname{PSL}_2(5)$
label in degree six belongs to this exact branch. The exhaustive
algorithm, complete model incidence, and integer records are given
in the finite certificate accompanying this proof; the algorithm and
its data format are also displayed in
\S\ref{subsec:permutation-certificate}.

For odd outer labels the entire list in this range is
Table~\ref{tab:odd-outer-permutation}. Except in its first row, the
full primitive normalizer has minimum nonidentity odd-element support
at least the displayed $\mu=\lceil m/2\rceil$. The entries $Q$ are
upper bounds on $k(S)$;
the first four use $9,10,16,33$, respectively, and the other entries
are simply $|S|$. Each row other than the degree-nine exception
satisfies the integer tests
\begin{equation}\label{eq:finite-outer-tests}
 (2^\mu-Q-1)\geq0,\quad
 (2^\mu-Q-1)^2\geq4Q,\quad
 2^\mu\geq2^{d_S}Q,\quad \mu\geq d_S.
\end{equation}
They imply \eqref{eq:outer-quadratic} and (i). At higher degrees
of the same type the inequalities strengthen. If $\Out(S)$ is a
$2$-group there are no odd outer labels, and the internal test alone
suffices. Every selected model also satisfies
$3(m-d_S)\geq2(m-1)$, so \eqref{eq:permutation-class-bound} proves
(ii). The outer orders and four small class numbers can be recovered
by the explicit procedures in
\S\ref{subsec:permutation-certificate}.

\begin{table}[htbp]
\centering
\caption{The finite almost simple types with odd outer automorphisms.
The degree is the least selected degree.}
\label{tab:odd-outer-permutation}
\begin{tabular}{lrrrr}
\toprule
$S$ & $m$ & $d_S$ & $\mu$ & $Q\geq k(S)$\\
\midrule
$\operatorname{PSL}_2(8)$&9&3&6&9\\
$\operatorname{PSL}_3(4)$&21&6&11&10\\
$\operatorname{PSL}_2(27)$&28&6&14&16\\
$\operatorname{PSL}_2(32)$&33&5&17&33\\
$\operatorname{PSU}_3(5)$&50&6&25&126000\\
$\operatorname{PSL}_3(7)$&57&6&29&1876896\\
$\operatorname{Sz}(8)$&65&3&33&29120\\
$\operatorname{PSL}_2(64)$&65&6&33&262080\\
$\operatorname{PSL}_3(8)$&73&6&37&16482816\\
$\operatorname{P}\Omega_8^+(2)$&120&6&60&174182400\\
$\operatorname{PSL}_2(125)$&126&6&63&976500\\
\bottomrule
\end{tabular}
\end{table}

For $S=\operatorname{PSL}_2(8)$ in degree nine, the normalizer is
$S:3$, of order 1512. Each nontrivial outer coset has exactly two
odd $S$-classes, with $(\text{order},\text{support},\text{class size})$
equal to $(3,6,84)$ and $(9,9,168)$. Every nonidentity odd element
of the normalizer has support at least six. Thus the active row count
in \eqref{eq:rectangular-bounds} is at most two whenever a
nontrivial odd outer label is present. For the $c$th power weight
put $e=2^{-3c}\leq1/8$. The nonidentity weight in the regular
$C_3$ outer action is $\sqrt e$, and
\[
       e+e^2\leq\sqrt e,
       \qquad 2e^2\leq e.
\]
These prove (iii) for one or two nontrivial labels. The internal
test handles the remaining case. The class calculation is reproducible
directly from the fractional linear transformations of
$\F_8\cup\{\infty\}$ and the Frobenius map, as specified below.
This proves the proposition.
\end{proof}

\subsection{Smaller actions for the other primitive types}

If two faithful actions of the same group have
$s_{\rm new}(g)\leq s_{\rm old}(g)$ for all odd $g$, positivity gives
$\mathcal P(G,\mathrm{old})\leq\mathcal P(G,\mathrm{new})$.
We record replacements with this precise property.

\begin{lemma}\label{lem:whole-fibre-replacement}
Every primitive component of nonaffine, non-almost-simple type admits
a faithful action of strictly smaller degree with no greater support
on odd elements. The same is true of a non-natural alternating subset
action. These replacements can be made inside an arbitrary
block action, retaining all relations between its component kernels.
\end{lemma}
\begin{proof}
First, $\Aut(S)$ has a faithful primitive action of degree at most
$|S|/4$ for every nonabelian simple $S$. To construct one, let $P$
be a Sylow $2$-subgroup of $S$. The odd-order theorem \cite{FT63} implies that
$|S|$ is even. In fact $4\mid|S|$: otherwise the regular permutation
of an involution is odd, and its sign gives a nontrivial homomorphism
from the perfect group $S$ to $C_2$. The action of $\Aut(S)$ on
the Sylow $2$-subgroups is faithful. Its kernel intersects $S$
trivially and hence centralizes $S$, whereas $C_{\Aut(S)}(S)=1$.
Let $J=N_{\Aut(S)}(P)$ and take a maximal proper subgroup $M$
containing $J$. The Frattini argument gives $\Aut(S)=SJ$, so
$S\nleq M$. The core of $M$ again intersects $S$ trivially and is
trivial. The action on $\Aut(S)/M$ is therefore faithful primitive,
of degree
\begin{equation}\label{eq:aut-simple-small-action}
       d(S)\leq[S:N_S(P)]\leq |S|/|P|\leq|S|/4.
\end{equation}
Use one fixed such action for each abstract simple type.

Suppose a semisimple group $N=\prod_i S_i$ is regular and normal.
Choosing an origin puts the full acting group in
$\operatorname{Hol}(N)=N\sd\Aut(N)$.
Here $\operatorname{inn}_b(z)=bzb^{-1}$.
Since $N$ is centerless,
\[
 (b,\varphi)\longmapsto(\operatorname{inn}_b\varphi,\varphi)
        \quad\hbox{in }\Aut(N)\times\Aut(N)
\]
is injective. Each copy of $\Aut(N)$ acts faithfully on the disjoint
union of the sets in \eqref{eq:aut-simple-small-action}, including
its permutations of isomorphic factors. The two copies together have
degree
\[
       2\sum_i d(S_i)\leq\frac12\sum_i|S_i|
                         \leq\frac12\prod_i|S_i|=|N|/2.
\]
If an affine permutation $z\mapsto b\varphi(z)$ has a fixed point,
its fixed set is a translate of $C_N(\varphi)$. For a nonidentity
permutation this is a proper subgroup. A perfect group has no proper
subgroup of index at most four, since its nontrivial coset image would
be a perfect subgroup of the solvable group $S_4$. Thus a nonidentity
old permutation moves at least $4|N|/5$ points. The new action has
smaller degree and no greater support, even for elements of even order.

For a simple diagonal action put $N=S^r$, $r\geq2$,
$\Delta=\{(s,\ldots,s):s\in S\}$, and
$\Omega=\Delta\backslash N$. Its degree is $n=|S|^{r-1}$.
Centerlessness gives $N_N(\Delta)=\Delta$, so the elementary
centralizer formula for a transitive coset action gives
$C_{\operatorname{Sym}(\Omega)}(N)=1$. The full normalizer $W$
therefore embeds in $\Aut(S)\wr S_r$. Its stabilizer of $\Delta$
is $\Aut(S)\times S_r$: the coordinate automorphisms preserving
$\Delta$ must all agree. In particular this normalizer is primitive.
For a direct verification, a subgroup of $S^r$ containing $\Delta$
is a product of aligned full diagonal subgroups, and so corresponds
to a partition of the $r$ factors. The full $S_r$ preserves only the
two trivial partitions. This proves maximality of the point stabilizer.
The action is outside the alternating subset/product exceptions of
\cite[Corollary~2]{BG22}: its stabilizer intersects every individual
simple factor trivially, unlike a nontrivial subset stabilizer.
Thus every nonidentity odd element of $W$ moves at least $n/2$
points. On the other hand, the faithful imprimitive action induced
by \eqref{eq:aut-simple-small-action} has degree
\[
       rd(S)\leq r|S|/4\leq |S|^{r-1}/2=n/2.
\]
The last inequality follows from $|S|^{r-2}\geq r/2$.
Restriction to any subgroup of $W$ supplies the replacement.

For a product action $B\leq L\wr S_r$ with $r\ge2$, where $L$
acts faithfully on $m\ge2$ points, replace its $m^r$ points by the
$mr$ points of the imprimitive coordinate action. For an odd element, each top
cycle has odd length $a$. Its product-coordinate space has size
$m^a$ and at most $m$ fixed points. For $a=1$, its moved points
are exactly the local imprimitive support. For $a\geq3$,
$m^a-m\geq ma$, which pays all its imprimitive support. For several
cycles, the elementary inequality
\[
       \prod_i q_i-\prod_i f_i\geq\sum_i(q_i-f_i)
       \qquad(0\leq f_i\leq q_i, q_i\geq2)
\]
combines the estimates. For two factors it is equivalent to
$(q-1)(r-1)\geq(f-1)(g-1)$, including $f=0$ or $g=0$;
induction proves the displayed version. The degree decreases except
when $(m,r)=(2,2)$. In nonabelian primitive product types the base
degree is at least five. The compound diagonal types are product
actions with a diagonal base and are included here.

For the action of $S_m$ on $j$-subsets, $1\leq j\leq m-1$, send
the basis vector at a point to the sum of the basis vectors at the
$j$-subsets containing it. This incidence map from the natural real
permutation module has Gram matrix
\[
       \binom{m-2}{j-1}I+\binom{m-2}{j-2}J.
\]
Here $I,J$ are the $m\times m$ identity and all-ones matrices,
and binomial coefficients with a negative lower entry are zero.
The diagonal entries are $\binom{m-1}{j-1}$ and the off-diagonal
entries are $\binom{m-2}{j-2}$; Pascal's identity gives the formula.
It is positive definite, so the natural module embeds. In the
orthogonal complement the trace of a permutation is at most its
dimension; consequently subset support is at least natural support
for every element. This applies to all subgroups and through any
abstract identification of the component. A subset action inequivalent
to the natural action has $2\leq j\leq m-2$, so its degree decreases
under replacement; this excludes complementary singleton actions too.

The O'Nan--Scott alternatives are affine, almost simple, regular
nonabelian, simple diagonal, and product actions with almost simple
or diagonal base; equivalently the eight usual types
HA, AS, HS, HC, SD, CD, PA, TW, as in \cite{CQRD11}.
The constructions above cover every alternative claimed in the lemma.
Finally, embed the block action into its imprimitive component
wreath product and perform the replacement in that ambient action,
then restrict to the original group. A moved block contributes its
full old or new size; on a fixed block an odd element has an odd
local action. Thus support domination and faithfulness persist.
The weighted sum is then taken over the original subgroup.
\end{proof}

\subsection{Full diagonal subgroups with different coordinate actions}

A subdirect subgroup of a product of nonabelian simple groups is a
product of full diagonal subgroups on disjoint sets of isomorphic
factors. Here is the elementary induction behind this assertion.
Project away the last factor. The kernel in that simple factor is
either the whole factor, giving an independent factor, or trivial.
In the latter case Goursat's lemma identifies the last factor with a
simple quotient of the projected group. By induction that quotient
is one of its diagonal factors, to which the last coordinate is
attached by an isomorphism. We call the resulting full diagonal
factors \emph{strips}. Their coordinate representations need not
be equivalent.

\begin{lemma}\label{lem:heterogeneous-strip-transmission}
Let $G$ act faithfully on $\Omega$ and preserve a partition of $\Omega$.
On a nonempty invariant set
$I$ of blocks, suppose its primitive components have nonabelian simple
socles $S_i$ and satisfy Proposition~\ref{prop:permutation-local}.
Let $N\unlhd G$ be subdirect in $\prod_{i\in I}S_i$, acting trivially
on the other points. Replace each active block by $\Psi_{S_i}$ and
retain the inactive points. The quotient $E=G/N$ has a
faithful action on this new set $\Omega'$, and
\begin{equation}\label{eq:strip-transmission}
             \mathcal P(G,\Omega)\leq\mathcal P(E,\Omega').
\end{equation}
In particular $|\Omega'|<|\Omega|$.
\end{lemma}
\begin{proof}
Choose an abstract parametrization $\tau_i:S\to S_i$ on each
strip. An element normalizing $N$ gives coordinate permutations
and automorphisms between the parameters. Record their outer
classes and use the fixed representation of $\Out(S)$ on $\Psi_S$.
Changing a parameter conjugates the outer labels by an element of
$\Out(S)$, implemented by a permutation of $\Psi_S$. Hence these
labels define an action of $G/N$.

Each coordinate socle has trivial permutation centralizer. Indeed,
that centralizer is normal in the full primitive normalizer. If
nontrivial it is transitive, forcing the commuting transitive socle
to be regular, contrary to the almost simple hypothesis.

A subdirect product of centerless simple groups is self-normalizing
inside their direct product. On one strip an inner tuple normalizes
it precisely when all induced automorphisms of its parameter agree;
centerlessness then makes the tuple an element of that strip.
Apply this separately to every strip. Thus an element in the kernel
of the proposed action is in $N$: it fixes the active blocks, has
inner coordinate labels there, and is trivial on the inactive points.
The reverse inclusion is immediate. This proves faithfulness.

Fix a commuting pair $(\alpha,\beta)$ of $E$, with $\alpha$
odd. Decompose the strips into orbits of this pair. On one such orbit
there are $L$ strips of the same abstract type $S$. In either prescribed
lift coset, variation by the kernel varies the inner transports freely
through $S^L\cong\Inn(S)^L$. Regard the two permutations as the action of
$\Gamma=\Z^2$ on the $L$ strip positions. Choose a root and a rooted
spanning tree of its Schreier graph. Conjugate by the subgroup of $S^L$
whose root coordinate is one. Its action is free: the root value and
successive tree edges determine every other conjugating coordinate.
Each orbit therefore has $|S|^{L-1}$ elements and has a unique choice
whose tree transports are the prescribed representatives.
The transport of any remaining edge is then determined by the
holonomy of the closed path obtained by adjoining the tree paths.
The stabilizer $\Lambda\leq\Gamma$ of the root is a rank-two lattice.
Choose a basis of $\Lambda$. Its holonomies lie in two prescribed
$S$-cosets of $\Aut(S)$ and commute; these two values determine the
holonomies of every closed path. Conversely these assignments reconstruct
all edge transports satisfying the commuting-square relations.
Specified coordinate permutations or representation identifications may
impose further equations, which only reduce the possibilities.
For the full strip system, a commutator of prescribed lifts lies in
$N$. If its induced automorphism on every strip is trivial, it lies in
$N\cap C_G(N)=Z(N)=1$. Thus the commuting-square relations detect
commutation of the lifts.

For two prescribed automorphism cosets, the number of commuting pairs
divided by $|S|$ is at most $k(S)$. On any $S$-class in the first
coset, the allowed second elements are empty or a centralizer coset,
so that class contributes at most one. The number of classes in a
coset is at most $k(S)$ by the Burnside calculation in
Lemma~\ref{lem:commuting-matrix}. Therefore
\[
 \#\{\text{commuting lift pairs}\}
 \le |S|^{L-1}\#\{\text{commuting holonomy pairs}\}
 \le |S|^L k(S).
\]
After division by $|S|^L$, the
unweighted lift mass on the whole strip orbit is therefore at most
$k(S)$. This bound applies to lattice-basis holonomies of every order.

Suppose first that $\alpha$ or $\beta_o$ moves an original block
in this orbit. A moved block of degree $m_i$ gives excess
$m_i-d_{S_i}$ between old and new support. On a fixed block, the
old odd support dominates the new outer support by
Proposition~\ref{prop:permutation-local}(i). An odd moved cycle has
length at least three, and all its blocks have the same degree.
Consequently the old pair weight is at most
its quotient weight multiplied by $t^{3(m-d_S)}$, for one
such cycle. The factor $k(S)$ just obtained is paid by
Proposition~\ref{prop:permutation-local}(ii). Thus this entire strip
orbit has the required weighted bound, even if it also meets other
degrees and representation types.

Now suppose that both odd quotient elements fix every original block
on this orbit. Then $\alpha$ fixes every strip, and $\beta$ permutes
the $L$ strips in one cycle of $2$-power length. The preceding
transport calculation leaves an odd holonomy $x$ and a commuting
holonomy $y$ in two prescribed $S$-cosets. If a strip has $h$
original coordinates, let $s_j$ be odd support in its $j$th
representation. The pair weight on these $hL$ blocks is
\begin{equation}\label{eq:strip-weight-product}
          \prod_{j=1}^h t^{L(s_j(x)+s_j(y_o))}.
\end{equation}
These functions are defined even if the
even holonomy permutes the coordinate representation types.

Use the strip normalizer generated by the coordinate-fixing
subgroup and this cycle holonomy. Its permutation image on the
coordinates is a $2$-group. The conjugation map to $\Aut(S)$ has
a $2$-group kernel: a coordinate-fixing element centralizing $S$
on every coordinate is trivial by the coordinate-centralizer argument
above.

Every odd element of the automorphism image has an odd lift, obtained
by taking the odd part of a preimage. Such a lift fixes all coordinates.
It is unique, since two coordinate-fixing lifts with the same
automorphism differ by the coordinatewise trivial centralizer. This
defines all the support functions $s_j$, and they are $S$-class
functions. In each coordinate the odd automorphism images belong to
its coordinate normalizer, as required in
Proposition~\ref{prop:permutation-local}(iii).
The removed $2$-kernel meets $S$ trivially. Each prescribed
$S$-coset therefore maps bijectively to its image; a commutator of
two prescribed lifts lies in $S$ and vanishes exactly when its image
does. The first lift remains odd: its $2$-part would lie in both $S$
and the removed $2$-kernel. Thus the prescribed holonomy count is
preserved.

Also $(\beta^L)_o=\beta_o^L$. Since $L$ is a power of two, taking
this power does not change the support of the odd part, either in
an original coordinate or in its outer action. This proves
\eqref{eq:strip-weight-product} with these support functions.
Let $\mu$ be the common positive measure on the allowed holonomy
pairs, of mass $1/|S|$ on each pair. If additional relations remain,
enlarging to all commuting pairs in the prescribed cosets only
increases the sums below. Put
$F_j(x,y)=t^{L(s_j(x)+s_j(y_o))}$. H\"older's inequality gives
\begin{equation}\label{eq:strip-holder}
       \int\prod_{j=1}^h F_j\,d\mu
       \leq\prod_{j=1}^h\left(\int F_j^h\,d\mu\right)^{1/h}.
\end{equation}
Apply Proposition~\ref{prop:permutation-local}(iii) to each
coordinate with exponent $Lh$. Abstract changes of parameter
conjugate outer labels, so preserve their supports on the chosen
$\Psi_S$. Each factor on the right has the prescribed outer bound;
their product is exactly the required quotient weight on these
$hL$ blocks. For alternating types both odd outer labels are trivial,
and Lemma~\ref{lem:natural-alternating-cosets} includes every even
$A_6$ holonomy. It can therefore be combined in the same H\"older
inequality with non-natural representations of that simple group.

For the fixed quotient pair the equations on different strip orbits
involve different direct factors of $N$. Their bounds multiply.
On inactive points $N$ is trivial, and the support is already the
quotient support. Hence the normalized lift mass is at most
$t^{s_{\Omega'}(\alpha)+s_{\Omega'}(\beta_o)}$.
Averaging over the commuting quotient pairs proves
\eqref{eq:strip-transmission}. Every active degree decreases, proving
the final assertion.
\end{proof}

\subsection{The degree induction}

\begin{proof}[Proof of Theorem~\ref{thm:permutation-weight}]
Assume a counterexample exists and choose one of minimum degree.
Remove global fixed points, which changes neither faithfulness nor
the weighted sum. By Lemma~\ref{lem:permutation-abelian-descent}
its solvable radical is trivial: a nontrivial solvable radical contains
an elementary abelian minimal normal subgroup of $G$.

On each transitive orbit choose a smallest block with more than one
point and take all its translates; in a primitive orbit choose the
whole orbit. Each resulting block component is primitive, since a
smaller block in its stabilizer action would give a smaller original
block. Lemma~\ref{lem:whole-fibre-replacement} and minimality imply
that every component is affine or almost simple. They also remove
every larger alternating subset representation, including any
abstractly twisted instance.

Let $K$ be the kernel on this entire block set. If $K=1$, the block
action is faithful and has degree at most half the original degree.
Odd support can only decrease, contrary to minimality. Thus $K\ne1$.
Its solvable radical is characteristic in $K$ and normal in $G$,
so it is trivial. Consequently
$N=\Soc(K)$ is a nontrivial direct product of nonabelian simple
groups, characteristic in $K$ and normal in $G$.

Its image $N_i$ on any block is semisimple and normal in the
primitive component. In an affine component it must be trivial:
every nontrivial normal subgroup contains the unique elementary
abelian minimal normal subgroup, impossible in a semisimple group.
In an almost simple component with socle $S_i$ it is either trivial
or $S_i$. In fact a nontrivial $N_i$ contains $S_i$; this normal
simple subgroup is one direct factor of $N_i$, and all other
factors centralize it, whereas its component centralizer is trivial.

Let $I$ be the blocks on which $N_i=S_i$. The set $I$ is nonempty
and invariant. The faithful original action embeds $N$ subdirectly
in $\prod_{i\in I}S_i$, and $N$ is trivial on all inactive points.
The elementary subdirect description gives exactly the full strips
of Lemma~\ref{lem:heterogeneous-strip-transmission}, possibly across
different orbits, degrees and representations. Every active
component satisfies Proposition~\ref{prop:permutation-local}.
Lemma~\ref{lem:heterogeneous-strip-transmission} produces a strictly
smaller faithful quotient action with at least as large a weighted sum,
the final contradiction.

All terms are nonnegative and nondecreasing in $t$, so the assertion
for smaller parameters follows. For a $2$-group every odd part is
the identity and the weighted sum is one, showing sharpness of the
constant.
\end{proof}

\subsection{The finite permutation certificate}\label{subsec:permutation-certificate}

For reproducibility we specify the complete finite input to
Proposition~\ref{prop:permutation-local}. The primitive database used
here is GAP~4.14.0 with \texttt{PrimGrp}~3.4.4 and
\texttt{CtblLib} \cite{GAP,CTblLib,PrimGrp}.
Its completeness up to permutation isomorphism in
the present range is the classification of \cite{RD05}, incorporated
in the library described in \cite{PrimGrp}; \cite{CQRD11} extends
the range to 4095. We use only degrees 5 through 127. The official
GAP type string \texttt{"2"} denotes almost simple, and the socle
is obtained from the primitive model, including models with an
imprimitive socle action.

The following executable test gives every internal, outer and moving
inequality. For alternating groups, \texttt{OutOrder} uses
$|\Out(A_n)|=2$ for $n\ne6$ and $|\Out(A_6)|=4$.
The linear and unitary formulas are given in
\cite[\S7, proof of Lemma~7.7]{GMP17}; the symplectic and
orthogonal orders occur in \cite[proof of Theorem~2.9]{MNR21}.
For $M_{11},M_{12},M_{22},M_{23},M_{24},J_2,HS$ and
$\operatorname{Sz}(8)$, the respective orders are
$1,2,2,1,1,2,2,3$, as recorded in their entries in \cite{Atlas85}.
In the library's notation
\texttt{L} means $\operatorname{PSL}_n(q)$, \texttt{2A} with
parameter $[n,q]$ means $\operatorname{PSU}_{n+1}(q)$, and
\texttt{B} with parameter $[n,q]$ means
$\operatorname{PSp}_{2n}(q)$ for the types that occur here. The only
orthogonal labels present are $D_4(2)$ and ${}^2D_4(2)$.
The formulas below are used only for the types in
Table~\ref{tab:small-as-models}; unhandled series raise an error.

\begingroup\small
\begin{verbatim}
LoadPackage("primgrp");;
LoadPackage("ctbllib");;
ExactCriterion := function(c,s)
  local z;
  z := 2^s-c-1;
  return z >= 0 and z^2 >= 4*c;
end;;
OutOrder := function(st)
  local n,q,f,ser;
  ser := st.series;
  if ser="A" then
    if st.parameter=6 then return 4; else return 2; fi;
  elif ser="Spor" then
    if st.parameter in ["M(11)","M(23)","M(24)"] then
      return 1;
    else return 2; fi;
  elif ser="2B" then return LogInt(st.parameter,2);
  fi;
  n := st.parameter[1]; q := st.parameter[2];
  f := Length(FactorsInt(q));
  if ser="L" then
    if n=2 then return Gcd(n,q-1)*f;
    else return 2*Gcd(n,q-1)*f; fi;
  elif ser="2A" then return 2*Gcd(n+1,q+1)*f;
  elif ser="B" then
    if n=2 and q mod 2=0 then return 2*f; fi;
    return Gcd(2,q-1)*f;
  elif ser="D" and n=4 and q=2 then return 6;
  elif ser="2D" and n=4 and q=2 then return 2;
  fi;
  Error("Unrecognized simple type");
end;;
total := 0;; exact := 0;; triples := 0;; pairs := [];;
for m in [5..127] do
  for id in [1..NrPrimitiveGroups(m)] do
    g := PrimitiveGroup(m,id);
    if ONanScottType(g)="2" then
      st := SocleTypePrimitiveGroup(g);
      if not (st.series="A" and st.parameter=m) then
        total := total+1;
        s := Socle(g); sz := Size(s);
        key := [st.series,st.parameter];
        out := OutOrder(st); d := out;
        if key=["L",[3,4]] then d := 6; fi;
        mu := QuoInt(m+1,2);
        coarse := st.series<>"A" and ExactCriterion(sz,mu);
        if not coarse then
          exact := exact+1;
          for cl in ConjugacyClasses(s) do
            x := Representative(cl);
            if Order(x)>1 and IsOddInt(Order(x)) then
              c := sz/Size(cl); motion := NrMovedPoints(x);
              if not ExactCriterion(c,motion) then
                Error("Internal inequality failed");
              fi;
              triples := triples+1; AddSet(pairs,[c,motion]);
              Print("CLASS ",m," ",id," ",
                    [Order(x),motion,c],"\n");
            fi;
          od;
        fi;
        qbound := sz;
        if key=["L",[2,8]] then qbound := 9;
        elif key=["L",[2,27]] then qbound := 16;
        elif key=["L",[2,32]] then qbound := 33;
        elif key=["L",[3,4]] then qbound := 10; fi;
        if qbound<>sz and NrConjugacyClasses(s)<>qbound then
          Error("Small class number failed");
        fi;
        if out > 1 and ForAny(FactorsInt(out),r -> r<>2) then
          if not (key=["L",[2,8]] and m=9) then
            if not (ExactCriterion(qbound,mu) and
                    2^mu>=2^d*qbound and mu>=d) then
              Error("Outer inequality failed");
            fi;
          fi;
        fi;
        if not (d<m and 3*(m-d)>=2*(m-1)) then
          Error("Moving inequality failed");
        fi;
        Print("MODEL ",m," ",id," ",key," ",
              [sz,out,d,mu,qbound],"\n");
      fi;
    fi;
  od;
od;
Print([total,exact,triples,Length(pairs)],"\n");
# Output: [ 292, 97, 887, 267 ]
\end{verbatim}
\endgroup

The supplementary files
\path{verification/permutation/all_terminal_rows.json} and
\texttt{all\_socle\_labels.json} retain every $(m,\mathrm{id})$ and
its simple type, order, outer order, chosen outer degree and method.
The file \texttt{exact\_class\_certificate.json} contains, for each
of the 97 exact models, every triple
$[\operatorname{ord}(x),s(x),|C_S(x)|]$, with multiplicities.
The file \texttt{integer\_pairs.json} is its set of 267 distinct
$(|C_S(x)|,s(x))$ pairs. The full generating script and output are
included in the same directory. All comparisons use integers.

Table~\ref{tab:small-as-models} records the complete type and degree
incidence, with the number of models. The labels $L_2(5)$ and $A_5$
are kept separate to match the database; together they represent one
abstract simple type. Thus the 80 labels give 79 simple types.
Table~\ref{tab:small-as-integers} is a lossless reduction for checking
the internal inequalities: for each centralizer order $C$ occurring
among the 267 pairs it gives the smallest support $s_0$. The test
\eqref{eq:finite-internal-test} strengthens as $s$ increases, so these
109 boundary pairs suffice for its arithmetic. The full class records
provide the incidence with the selected models.

\begingroup\small
\begin{longtable}{lrrrl}
\caption{Complete finite type and degree incidence. Here $L_n(q)=\operatorname{PSL}_n(q)$ and $U_n(q)=\operatorname{PSU}_n(q)$.}\label{tab:small-as-models}\\
\toprule
$S$ & $|\Out(S)|$ & $d_S$ & Models & Degrees\\
\midrule
\endfirsthead
\toprule
$S$ & $|\Out(S)|$ & $d_S$ & Models & Degrees\\
\midrule
\endhead
\bottomrule
\endfoot
$L_{2}(5)$ & 2 & 2 & 2 & 6\\
$L_{2}(7)$ & 2 & 2 & 5 & 7, 8, 21, 28\\
$L_{2}(8)$ & 3 & 3 & 6 & 9, 28, 36\\
$A_{5}$ & 2 & 2 & 2 & 10\\
$A_{6}$ & 4 & 4 & 13 & 10, 15, 36, 45\\
$L_{2}(11)$ & 2 & 2 & 7 & 11, 12, 55, 66\\
$M_{11}$ & 1 & 1 & 4 & 11, 12, 55, 66\\
$M_{12}$ & 2 & 2 & 2 & 12, 66\\
$L_{3}(3)$ & 2 & 2 & 3 & 13, 52, 117\\
$L_{2}(13)$ & 2 & 2 & 8 & 14, 78, 91\\
$A_{7}$ & 2 & 2 & 6 & 15, 21, 35, 120\\
$A_{8}$ & 2 & 2 & 9 & 15, 28, 35, 56, 105, 120\\
$L_{2}(16)$ & 4 & 4 & 9 & 17, 68, 120\\
$L_{2}(17)$ & 2 & 2 & 3 & 18, 102\\
$L_{2}(19)$ & 2 & 2 & 3 & 20, 57\\
$L_{3}(4)$ & 12 & 6 & 20 & 21, 56, 105, 120\\
$M_{22}$ & 2 & 2 & 4 & 22, 77\\
$M_{23}$ & 1 & 1 & 1 & 23\\
$M_{24}$ & 1 & 1 & 1 & 24\\
$L_{2}(23)$ & 2 & 2 & 2 & 24\\
$L_{2}(25)$ & 4 & 4 & 7 & 26, 65\\
$\operatorname{PSp}_{4}(3)$ & 2 & 2 & 10 & 27, 36, 40, 45\\
$U_{3}(3)$ & 2 & 2 & 8 & 28, 36, 63\\
$\operatorname{PSp}_{6}(2)$ & 1 & 1 & 4 & 28, 36, 63, 120\\
$L_{2}(27)$ & 6 & 6 & 4 & 28\\
$L_{2}(29)$ & 2 & 2 & 2 & 30\\
$L_{3}(5)$ & 2 & 2 & 1 & 31\\
$L_{5}(2)$ & 2 & 2 & 1 & 31\\
$L_{2}(31)$ & 2 & 2 & 2 & 32\\
$L_{2}(32)$ & 5 & 5 & 2 & 33\\
$A_{9}$ & 2 & 2 & 7 & 36, 84, 120, 126\\
$L_{2}(37)$ & 2 & 2 & 2 & 38\\
$L_{4}(3)$ & 4 & 4 & 4 & 40, 117\\
$L_{2}(41)$ & 2 & 2 & 2 & 42\\
$L_{2}(43)$ & 2 & 2 & 2 & 44\\
$A_{10}$ & 2 & 2 & 6 & 45, 120, 126\\
$L_{2}(47)$ & 2 & 2 & 2 & 48\\
$U_{3}(5)$ & 6 & 6 & 6 & 50, 126\\
$L_{2}(49)$ & 4 & 4 & 5 & 50\\
$L_{2}(53)$ & 2 & 2 & 2 & 54\\
$A_{11}$ & 2 & 2 & 2 & 55\\
$L_{3}(7)$ & 6 & 6 & 2 & 57\\
$L_{2}(59)$ & 2 & 2 & 2 & 60\\
$L_{2}(61)$ & 2 & 2 & 2 & 62\\
$L_{6}(2)$ & 2 & 2 & 1 & 63\\
$U_{3}(4)$ & 4 & 4 & 3 & 65\\
$\operatorname{Sz}(8)$ & 3 & 3 & 2 & 65\\
$L_{2}(64)$ & 6 & 6 & 4 & 65\\
$A_{12}$ & 2 & 2 & 2 & 66\\
$L_{2}(67)$ & 2 & 2 & 2 & 68\\
$L_{2}(71)$ & 2 & 2 & 2 & 72\\
$L_{3}(8)$ & 6 & 6 & 2 & 73\\
$L_{2}(73)$ & 2 & 2 & 2 & 74\\
$A_{13}$ & 2 & 2 & 2 & 78\\
$L_{2}(79)$ & 2 & 2 & 2 & 80\\
$L_{2}(81)$ & 8 & 8 & 8 & 82\\
$L_{2}(83)$ & 2 & 2 & 2 & 84\\
$\operatorname{PSp}_{4}(4)$ & 4 & 4 & 4 & 85, 120\\
$L_{4}(4)$ & 4 & 4 & 2 & 85\\
$L_{2}(89)$ & 2 & 2 & 2 & 90\\
$A_{14}$ & 2 & 2 & 2 & 91\\
$L_{3}(9)$ & 4 & 4 & 2 & 91\\
$L_{2}(97)$ & 2 & 2 & 2 & 98\\
$J_2$ & 2 & 2 & 2 & 100\\
$\operatorname{HS}$ & 2 & 2 & 2 & 100\\
$L_{2}(101)$ & 2 & 2 & 2 & 102\\
$L_{2}(103)$ & 2 & 2 & 2 & 104\\
$A_{15}$ & 2 & 2 & 2 & 105\\
$L_{2}(107)$ & 2 & 2 & 2 & 108\\
$L_{2}(109)$ & 2 & 2 & 2 & 110\\
$U_{4}(3)$ & 8 & 8 & 13 & 112, 126\\
$L_{2}(113)$ & 2 & 2 & 2 & 114\\
$\operatorname{P}\Omega_8^-(2)$ & 2 & 2 & 2 & 119\\
$\operatorname{PSp}_{8}(2)$ & 1 & 1 & 1 & 120\\
$\operatorname{P}\Omega_8^+(2)$ & 6 & 6 & 2 & 120\\
$A_{16}$ & 2 & 2 & 2 & 120\\
$L_{5}(3)$ & 2 & 2 & 1 & 121\\
$L_{2}(121)$ & 4 & 4 & 5 & 122\\
$L_{2}(125)$ & 6 & 6 & 4 & 126\\
$L_{7}(2)$ & 2 & 2 & 1 & 127\\
\end{longtable}
\endgroup

\begingroup\small
\begin{longtable}{rrrrrr}
\caption{The 109 exact internal boundary pairs.}\label{tab:small-as-integers}\\
\toprule
$C$&$s_0$&$C$&$s_0$&$C$&$s_0$\\
\midrule
\endfirsthead
\toprule
$C$&$s_0$&$C$&$s_0$&$C$&$s_0$\\
\midrule
\endhead
\bottomrule
\endfoot
3 & 6 & 90 & 60 & 3000 & 95\\
5 & 5 & 98 & 119 & 3240 & 90\\
6 & 9 & 105 & 102 & 3528 & 63\\
7 & 7 & 108 & 18 & 5400 & 76\\
9 & 9 & 126 & 104 & 5832 & 27\\
10 & 10 & 132 & 99 & 6480 & 51\\
11 & 11 & 150 & 75 & 7560 & 24\\
12 & 24 & 162 & 54 & 9720 & 81\\
13 & 13 & 180 & 15 & 12600 & 45\\
14 & 21 & 216 & 39 & 14580 & 105\\
15 & 15 & 252 & 85 & 17640 & 70\\
17 & 17 & 270 & 88 & 18000 & 105\\
18 & 9 & 300 & 35 & 22680 & 99\\
19 & 19 & 324 & 114 & 23328 & 114\\
20 & 40 & 375 & 105 & 37800 & 84\\
21 & 24 & 378 & 117 & 45360 & 57\\
23 & 23 & 405 & 91 & 58320 & 90\\
24 & 30 & 420 & 56 & 60480 & 27\\
25 & 30 & 450 & 117 & 100800 & 50\\
27 & 39 & 486 & 63 & 141120 & 77\\
30 & 25 & 500 & 25 & 181440 & 63\\
31 & 31 & 504 & 24 & 302400 & 92\\
33 & 88 & 540 & 81 & 362880 & 63\\
35 & 66 & 567 & 120 & 408240 & 99\\
36 & 12 & 600 & 85 & 544320 & 30\\
39 & 117 & 648 & 27 & 907200 & 55\\
42 & 21 & 660 & 110 & 1270080 & 84\\
45 & 52 & 810 & 119 & 3265920 & 69\\
49 & 63 & 900 & 68 & 5443200 & 33\\
50 & 65 & 972 & 66 & 9072000 & 60\\
54 & 9 & 1080 & 18 & 32659200 & 75\\
55 & 120 & 1176 & 56 & 59875200 & 36\\
60 & 20 & 1260 & 95 & 99792000 & 65\\
63 & 63 & 1800 & 40 & 718502400 & 39\\
75 & 78 & 1944 & 36 & 9340531200 & 42\\
81 & 36 & 2160 & 18 &  & \\
84 & 49 & 2520 & 63 &  & \\
\end{longtable}
\endgroup


There are two additional small automorphism computations. The data for
$A_6$ in Lemma~\ref{lem:natural-alternating-cosets} are obtained by
the following exact permutation procedure:
\begingroup\small
\begin{verbatim}
A := AlternatingGroup(6);;
cc := ConjugacyClasses(A);;
au := AutomorphismGroup(A);;
c3 := First(cc,c -> Order(Representative(c))=3 and
                        NrMovedPoints(Representative(c))=3);;
stab := Stabilizer(au,c3,function(c,a)
  return ConjugacyClass(A,Image(a,Representative(c)));
end);;
if Size(au)<>1440 or Size(stab)<>720 then
  Error("A6 automorphism assertion failed");
fi;
List(Filtered(cc,c -> IsOddInt(Order(Representative(c)))),
     c -> [Order(Representative(c)),
           NrMovedPoints(Representative(c)),Size(A)/Size(c)]);
# Up to order: [ [1,0,360], [3,3,9], [3,6,9], [5,5,5], [5,5,5] ]
\end{verbatim}
\endgroup
The permutation-induced subgroup has order 720 and fixes the
support-three class, so it is exactly this stabilizer. There are
precisely two order-three classes, proving the asserted interchange.
One may construct the exceptional automorphism independently through
the six one-factorizations of the complete graph on six vertices.

For $\operatorname{PSL}_2(8)$, take
$\F_8=\F_2[z]/(z^3+z+1)$ and the nine-point projective line.
The permutations
\[
       u\mapsto u+1,\qquad u\mapsto zu,
       \qquad u\mapsto u^{-1}
\]
(with $0,\infty$ interchanged by inversion) generate its 504-element
fractional linear group $S$. Adjoin $F:u\mapsto u^2$, fixing
$\infty$, to obtain the 1512-element normalizer. Close the three
generators under permutation multiplication; then on each of $SF$
and $SF^2$ partition the odd elements into orbits of conjugation by
$S$. The output in each coset is
\[
\begin{array}{c|rr}
 \operatorname{ord}(u)&s(u)&|u^S|\\\hline
 3&6&84\\9&9&168.
\end{array}
\]
This completely specified field and permutation construction is also
provided in \path{verification/permutation/verify_permutation.py},
with output in \texttt{permutation\_results.json}. It supplies the
two-class refinement; the matrix row and column identities themselves
were proved in Lemma~\ref{lem:commuting-matrix}.

Finally, the six-point outer action for $L_3(4)$ uses
$\Out(L_3(4))\cong D_{12}$; this standard automorphism structure
\cite{Atlas85} can be checked within the selected models by
\texttt{PrimitiveGroup(105,6)}. Its order is 241920, its socle has
order 20160, and the quotient has small-group identifier $[12,4]$,
the dihedral group of order twelve. Restrict this outer representation
to the outer subgroup of the degree-21 component.



\section{From permutation weights to affine groups}\label{sec:affine-transfer}

Write $\ell_2(H)=|\IBr(H)|$, equivalently the number of odd-order
conjugacy classes of $H$. For a finite $H$-set $Y$, define
\[
 \Lambda_2(H,Y)=\sum_{y\in Y/H}\ell_2(H_y).
\]
The analogue of \eqref{eq:kappa-commuting} is
\begin{equation}\label{eq:lambda-commuting}
 \Lambda_2(H,Y)=\frac1{|H|}
 \sum_{\substack{a\in H\text{ odd}\\z\in C_H(a)}}
 |Y^{\langle a,z\rangle}|.
\end{equation}
It follows by counting the same triples with the additional
condition that $a$ has odd order.

\begin{lemma}\label{lem:ordinary-spectrum}
Let $A\lhd G$, put $T=G/A$ and $Y=\Irr(A)$. Then
\begin{equation}\label{eq:ordinary-spectrum}
 k(G)\le |T|_2\Lambda_2(T,Y).
\end{equation}
Suppose $T$ acts faithfully on $\Omega$, and $s(a)$ is the number
of points moved by $a$. If there is a real number $M>0$ such that
\begin{equation}\label{eq:odd-character-decay}
 |Y^a|\le M2^{-s(a)}\qquad(a\in T\text{ odd}),
\end{equation}
then the permutation inequality of
Theorem~\ref{thm:permutation-weight} gives $k(G)\le M|T|_2$.
\end{lemma}
\begin{proof}
Grouping the conjugacy classes of a group $J$ by their odd parts
gives $k(J)\le |J|_2\ell_2(J)$. Over an odd class represented by
$a$, the remaining choices are $2$-element classes of $C_J(a)$;
each meets a fixed Sylow subgroup and hence there are at most
$|C_J(a)|_2$ of them. The number of irreducible characters above
a $T$-orbit in $Y$ is
$k_\alpha(T_y)\le k(T_y)\le|T|_2\ell_2(T_y)$.
Summing proves \eqref{eq:ordinary-spectrum}.

Write $z_o$ for the odd part of $z$. For every commuting pair
occurring in \eqref{eq:lambda-commuting},
\[
 |Y^{\langle a,z\rangle}|
 \le\sqrt{|Y^a|\,|Y^{z_o}|}
 \le M2^{-(s(a)+s(z_o))/2}.
\]
The normalized sum of these weights is at most one by
Theorem~\ref{thm:permutation-weight}. Now use
\eqref{eq:ordinary-spectrum}.
\end{proof}

\subsection{Ordinary character counts on coordinate kernels}

Throughout this subsection, $\pi_i$ denotes projection onto the
$i$th coordinate.
For an automorphism $y$ of $L$, let $f_L(y)=|\Irr(L)^y|$.
Equivalently this is the number of $L$-conjugacy classes in the
coset $Ly$, and twisted Burnside gives
\begin{equation}\label{eq:twisted-burnside}
 f_L(y)=\frac1{|L|}\sum_{x\in L}|C_L(xy)|.
\end{equation}
If $K\lhd L$ is $y$-invariant, projecting the coset classes to
$L/K$ gives
\begin{equation}\label{eq:twisted-filtration}
 f_L(y)\le f_{L/K}(y)\max_{x\in L}f_K(xy).
\end{equation}
Indeed, over one quotient coset class every lifted class meets
$Kxy$, whose $K$-classes give the asserted upper bound.

For a nonabelian simple group $S$, denote by $\widehat S$ its
universal central cover with the central $2$-part removed, and put
\begin{equation}\label{eq:odd-cover-capacity}
 \kappa^+(S)=\max\left\{1,\frac{k(\widehat S)}{|\widehat S|_2}\right\}.
\end{equation}
Every perfect central extension of $S$ with odd center is an
equivariant quotient of $\widehat S$. Automorphisms lift by the
universal property, and the central Sylow $2$-subgroup removed
from the universal cover is characteristic.

\begin{lemma}\label{lem:strip-capacity}
Let $M_i=L_iR_i$, where $L_i$ is a central product of quasisimple
groups with odd centers, $R_i=C_{M_i}(L_i)$ is solvable, and
$[L_i,R_i]=1$. Suppose $B$ is a subdirect subgroup of
$\prod_{i=1}^r M_i$ normalized by a coordinate automorphism $y$
that permutes the coordinates in one cycle.
Then, for every $i\in\{1,\ldots,r\}$,
\begin{equation}\label{eq:strip-capacity}
 f_B(y)\le |B|_2\,|R_i|_{2'}
 \prod_{L\text{ a component of }L_i}\kappa^+(L/Z(L)).
\end{equation}
The same bound holds when the coordinate images of $B$ are normal
subgroups of $M_i$.
\end{lemma}
\begin{proof}
Let $F=B^{(\infty)}$ be the last derived subgroup. Its projections
are $L_i$. Modulo the odd centers, it is a subdirect product of
nonabelian simple groups, and thus a product of full diagonal
simple subgroups on disjoint sets of factor positions. This
description follows directly by induction from Goursat's lemma:
adjoining a simple coordinate either adds an independent factor
or identifies it diagonally with one existing isomorphic factor.
The components of $F$ are therefore odd central quotients of the
corresponding $\widehat S$.

Put $C=C_B(F)$. An element of $B$ acts innerly on each coordinate
$L_i$ and preserves all factor positions. It consequently preserves
each diagonal component and acts innerly on its simple quotient.
An automorphism of a perfect central extension acting trivially
on its central quotient is trivial: its deviation from the
identity is a homomorphism to the center. Choose implementing
elements in the components of $F$. Their product shows that
\[
 B=FC,\qquad C\le\prod_i R_i.
\]
In particular $C$ is solvable and $y$-invariant. There is a
$y$-equivariant central cover
\[
 P=\left(\prod_j\widehat S_j\right)\times C\longrightarrow B
\]
with odd kernel. Its kernel has odd order because the component
cover kernels and $F\cap C=Z(F)$ have odd order. Inflation embeds
$\Irr(B)$ equivariantly in $\Irr(P)$ and $|P|_2=|B|_2$.

On an orbit of $m$ component factors under $y$, a fixed character
tuple is determined by one character, fixed by the composed return
automorphism. Its count is at most $k(\widehat S)$, whereas the
$2$-part of the order of all $m$ factors is $|\widehat S|_2^m$.
The normalized contribution is therefore at most $\kappa^+(S)$.
Each orbit of diagonal components meets a factor position in any
chosen coordinate; distinct orbits occupy disjoint positions.
Assigning one such position to each orbit proves that their
product is bounded by the product in \eqref{eq:strip-capacity}.
The maximum with one in \eqref{eq:odd-cover-capacity} permits
unused positions.

For $C$, each centralizer $C_C(cy)$ projects injectively into
one coordinate: commutation determines all other coordinates
successively around the cycle. Hence
\[
 f_C(y)\le |\pi_i(C)|\le |C|_2|R_i|_{2'}
\]
by \eqref{eq:twisted-burnside}. Multiplying proves the claimed
bound. If a coordinate image is only normal in $M_i$, it is the
product of those components that it contains and its intersection
with $R_i$. Indeed a noncentral normal image on a simple factor
contains the corresponding perfect component by taking
commutators. Apply the proved case to these images;
neither the odd order nor the product of the factors
$\kappa^+(S)\ge1$ increases.
\end{proof}

Suppose $O_2(E)=1$. Its Fitting subgroup $F(E)$ has odd order and
its components have odd centers. Define
\begin{equation}\label{eq:local-capacity-definition}
 c(E)=|E:F^*(E)|_{2'}\,|F(E)|
 \prod_{L\text{ a component of }E(E)}\kappa^+(L/Z(L)),
\end{equation}
with $c(1)=1$. The factors in $F^*(E)=F(E)E(E)$ commute, and
$C_{F^*(E)}(E(E))=F(E)$.

\begin{lemma}\label{lem:outer-cycle-capacity}
Let $K\le\prod_{i=1}^r E_i$ be subdirect, where $O_2(E_i)=1$,
and let $y$ cyclically transport the coordinates and normalize $K$.
Then $f_K(y)\le |K|_2c(E_i)$ for every $i\in\{1,\ldots,r\}$.
The same conclusion holds for a subgroup $K$ normal in a subdirect
group on the $E_i$, provided $y$ normalizes $K$.
\end{lemma}
\begin{proof}
Put $M_i=F^*(E_i)$ and $A=K\cap\prod_iM_i$. The images of
$A$ are normal in $M_i$, so Lemma~\ref{lem:strip-capacity}
applies to $A$ and every twisted automorphism $xy$ with $x\in K$.
The quotient $K/A$ embeds in $\prod_i E_i/M_i$. Again a
centralizer of a cyclic coordinate automorphism is determined by
one coordinate. Twisted Burnside gives
\[
 f_{K/A}(y)\le |\pi_i(K/A)|
 \le |K/A|_2|E_i:M_i|_{2'}.
\]
Combine the two bounds through \eqref{eq:twisted-filtration}; the
quotient $E_i/M_i$ enters only through its odd order part.
For the last assertion, normality in the subdirect overgroup
still makes every projection of $A$ normal in $E_i$; all the
same bounds therefore apply with the original $M_i$ and $c(E_i)$.
\end{proof}

\begin{proposition}\label{prop:affine-coordinate}
Let $H$ act faithfully on $V=\bigoplus_{i\in\Omega}W_i$ over
$\F_2$ and permute the summands. Let $N$ be the kernel
of that permutation action, and let $E_i$ be its image on $W_i$.
Suppose $O_2(E_i)=1$ and
\begin{equation}\label{eq:capacity-threshold}
 c(E_i)\le |W_i|^2/8.
\end{equation}
Assume also that $k(W_i\sd U)\le |W_i||U|_2$ for every subgroup
$U$ of the image in $\GL(W_i)$ of the stabilizer of $W_i$ in $H$.
Then
$k(V\sd H)\le|V||H|_2$.
\end{proposition}
\begin{proof}
Put $q_i=|W_i|$ and $T=H/N$. For $g\in H$, filter $V\sd N$
by the kernels of its projections onto successive unions of coordinate
cycles of $g$. Formula~\eqref{eq:twisted-filtration} applies at each
step. On a moved cycle $C$, choose $i\in C$ and put $q_C=q_i$ and
$E_C=E_i$, with $W_C=\bigoplus_{j\in C}W_j$.
Its successive affine image has the form $W_C\sd K_C$
and contains the full additive group $W_C$. The linear image $K_C$
is normal in the corresponding projection of $N$, which is subdirect
on the original $E_i$ in $C$. The last assertion of
Lemma~\ref{lem:outer-cycle-capacity} therefore uses the original
capacities. Any additive part of a coset twist is inner on $W_C\sd K_C$;
removing it preserves the character trace. The remaining twist $y$
normalizes $K_C$ and has the same coordinate cycle.
For $u\in W_C$ and $x\in K_C$, projection sends
$C_{W_C\sd K_C}(uxy)$ into $C_{K_C}(xy)$. Its kernel has at most
$q_C$ elements, since a fixed vector is determined by one
coordinate. Thus
\[
 |C_{W_C\sd K_C}(uxy)|\le q_C|C_{K_C}(xy)|.
\]
Summing over $u$ and $x$ and dividing by $|W_C||K_C|$ in
\eqref{eq:twisted-burnside} gives
\[
 f_{W_C\sd K_C}(y)\le q_C f_{K_C}(y)
 \le q_C|K_C|_2c(E_C).
\]
For a fixed coordinate use the assumed lower-dimensional affine
bound and $f_L(g)\le k(L)$. The orders of the successive
linear factors telescope to $|N|$, so their $2$-parts telescope
to $|N|_2$. We obtain
\begin{equation}\label{eq:affine-cycle-bound}
 |\Irr(V\sd N)^g|\le |N|_2
 \prod_{i\text{ fixed}}q_i
 \prod_{C\text{ moved cycle}}q_Cc(E_C).
\end{equation}
The parameters on a cycle agree because $g$ transports the coordinate
actions. The projection kernels preserve the subdirect relations and
the return automorphisms of the lift.

If $a\in T$ has odd order, all its moved cycles have odd length
$r\ge3$. By \eqref{eq:capacity-threshold},
\[
 \frac{q_Cc(E_C)}{q_C^r}
 \le\frac{q_C^{3-r}}8\le2^{-r}.
\]
At the identity \eqref{eq:affine-cycle-bound} also gives
$k(V\sd N)\le|V||N|_2$. Hence
\[
 |\Irr(V\sd N)^a|\le|V||N|_2\,2^{-s(a)}
\]
for every odd $a\in T$. Apply
Lemma~\ref{lem:ordinary-spectrum} to the normal extension
$V\sd N\lhd V\sd H$, with $M=|V||N|_2$.
\end{proof}

\section{Primitive linear groups and local capacities}\label{sec:primitive}

We prove two assertions needed in the affine reduction. Primitivity here
means the absence of a nontrivial direct-sum decomposition permuted by the
group. A module is quasiprimitive if its restriction to each normal
subgroup is homogeneous. An irreducible primitive module is quasiprimitive:
Clifford theory gives a permuted direct sum of isotypic components for each
normal subgroup, and primitivity makes that sum have one term.

\begin{proposition}\label{prop:primitive-affine}
If $E$ acts faithfully, irreducibly and primitively on a nonzero finite
$\F_2$-space $V$, then
\[
 k(V\sd E)\le |V|\,|E|_2.
\]
The same conclusion holds for a faithful irreducible quasiprimitive action.
\end{proposition}

\begin{proposition}\label{prop:local-capacity}
Let $B$ act faithfully, irreducibly and primitively on a finite
$\F_2$-space $W$, and let $E\lhd B$. Define $c(E)$ by
\eqref{eq:local-capacity-definition}. If $E(E)\ne1$ or $F(E)$ is
nonabelian, then
\[
 c(E)\le |W|^2/8.
\]
Here $O_2(E)=1$, and $W$ is homogeneous on every characteristic
subgroup of $E$, by faithfulness and primitivity.
\end{proposition}

The proofs retain the scalar field, component degrees, central
characters and homogeneous multiplicities. We first establish the
structural and counting estimates, then give the classification inputs
and the complete remaining numerical verification.

\subsection{The canonical field and the generalized Fitting subgroup}

\begin{lemma}\label{prim:field}
Let $E$ be a finite group acting faithfully on a nonzero
finite-dimensional $\F_2$-space $V$.
Assume that $V$ is homogeneous on every characteristic subgroup of $E$;
in applications $E\lhd B$ and $V$ is a faithful primitive irreducible
$B$-module. Put $R=F(E)$, $L=E(E)$, and $F^*=RL$. Then $O_2(E)=1$.
The center $K$ of $\End_{\F_2F^*}(V)$ is an $E$-normalized finite
field, say $K=\F_q$, $q=2^f$, and $V_{KF^*}$ is absolutely homogeneous.
Moreover
\begin{equation}\label{prim:fitting-form}
 R=ZR_1\cdots R_u,\qquad |R:Z|=n^2,\qquad
 n=\prod_{j=1}^u r_j^{a_j},
\end{equation}
where $Z=Z(R)$ is cyclic and acts by $K$-scalars, the $r_j$ are distinct
odd primes, and $R_j$ is a characteristic extraspecial subgroup of the
corresponding Sylow subgroup, of exponent $r_j$ and order
$r_j^{1+2a_j}$. The factors commute. If the components of $L$ are
$L_1,\ldots,L_t$, there are absolute component degrees $m_i\ge2$
and a positive integer $b$ such that
\begin{equation}\label{prim:tensor-degree}
 \dim_K V=bn\prod_{i=1}^t m_i.
\end{equation}
All the constituents have compatible central characters for the stated
central product. Finally, with $N=C_E(L)$, one has
\begin{equation}\label{prim:true-centralizer}
 C_N(R)=Z,\qquad
 H=E/N\le\Aut(S_1\times\cdots\times S_t),\qquad
 F^*(H)=S_1\times\cdots\times S_t,
\end{equation}
where $S_i=L_i/Z(L_i)$. If $R=Z$, then
$N\cap\GL_K(V)=Z$ and $N/Z$ is cyclic of order dividing $f$, acting
faithfully on $Z$.
\end{lemma}
\begin{proof}
Since $O_2(E)$ is characteristic in $E$, its restriction on $V$ is
homogeneous. Every irreducible module for a $2$-group in characteristic
two is trivial, so faithfulness gives $O_2(E)=1$. Thus $R$ has odd
order. The central $2$-parts of the components would generate a normal
$2$-subgroup of $E$, so their centers also have odd order.

Since $F^*$ is characteristic in $E$, write $V_{F^*}=U^{\oplus b}$
with $U$ irreducible. Then
$\End_{\F_2F^*}(V)=\operatorname{Mat}_b(K)$ by Schur's lemma and
Wedderburn's theorem. Its center is intrinsic and is normalized by $E$.
Centralizing this center leaves the same matrix algebra. Thus, when
$K$ is used as the scalar field, $U$ is absolutely irreducible and the
restriction remains homogeneous. The image of $KF^*$ on $U$ is
$\End_K(U)$. The algebras generated by the commuting factors $R,L_i$
are semisimple, by Clifford theory, and their centers commute with all
of $KF^*$. Their centers are therefore $K$. Each is a central simple
algebra over a finite field and hence a full matrix algebra. The usual
commuting-algebras decomposition gives a tensor product over this
\emph{same} $K$. Its factors are faithful on the corresponding
subgroups. This proves \eqref{prim:tensor-degree} and preserves the
central identifications.

Every characteristic abelian subgroup of a Sylow subgroup of $R$ is
normal in $E$. Its faithful homogeneous representation over $\F_2$
has field image, so the subgroup is cyclic. Hall's theorem in the
form \cite[Theorem~1.1]{Wolf82} says that an odd $r$-group with all
characteristic abelian subgroups cyclic is a central product of a
cyclic group and an extraspecial group of exponent $r$. The latter
is characteristic in each noncyclic Sylow subgroup $P_j$.
Indeed, write $P_j=Z(P_j)R_j$ with $R_j$ extraspecial of exponent
$r_j$. For $z\in Z(P_j)$ and $u\in R_j$ one has
$(zu)^{r_j}=z^{r_j}$. The unique subgroup of order $r_j$ in the
cyclic center lies in $R_j$, so the elements of order dividing
$r_j$ are exactly $R_j$. Thus
$R_j=\Omega_1(P_j):=\langle x\in P_j:x^{r_j}=1\rangle$
is characteristic. This proves \eqref{prim:fitting-form}. Its center
belongs to the center of the algebra generated by $F^*$, and so is
scalar over $K$. The faithful absolute degree of the extraspecial
product is $n$. Also $Z(L)\le R$, so $Z(RL)=Z(R)=Z$.

The generalized Fitting subgroup is self-centralizing. Therefore
$C_N(R)=C_E(RL)=Z$. Conjugation on $L/Z(L)$ has kernel $N$: an
automorphism of a perfect group acting trivially modulo its center
has central error a homomorphism to an abelian group and is the identity.
The image of $L$ is the full product of the $S_i$, with trivial
centralizer in $H$. A nilpotent normal subgroup or an additional
component of $H$ would centralize that product. This proves
\eqref{prim:true-centralizer}. If $R=Z$, its centralizer in $N$ is $Z$;
in particular every $K$-linear element of $N$ lies in $Z$. The remaining
quotient embeds in $\operatorname{Gal}(K/\F_2)$, and its action on $Z$
is faithful by the same centralizer equality.
\end{proof}

\subsection{Affine class counts, support and scalar centralizers}

For a linear group $X$ on a space $V$, put $v=|V|$. The split-extension
class formula and the ordinary little-groups formula are
\begin{equation}\label{prim:affine-formulas}
 k(V\sd X)=\sum_{x\in X/\text{conj}}n(C_X(x),V/(x-1)V)
          =\sum_{\lambda\in V^*/X}k(X_\lambda).
\end{equation}
The first formula uses a quotient space; it need not be equivariantly
identified with $C_V(x)$. Their cardinalities are equal.

\begin{lemma}\label{prim:support}
Let $r>0$, and suppose $|C_V(x)|\le v^{1-1/r}$ for every
$1\ne x\in X$. Then
\begin{equation}\label{prim:support-estimate}
 k(V\sd X)\le \frac v{|X|}
       +\left(k(X)-\frac1{|X|}\right)v^{1-1/r}.
\end{equation}
For $X\ne1$, writing $P=|X|_2$, a sufficient threshold for
$k(V\sd X)\le vP$ is
\begin{equation}\label{prim:threshold}
 v^{1/r}\ge\frac{|X|k(X)-1}{|X|P-1}.
\end{equation}
If $P\ge2$, the right side is at most $k(X)/P+1$.
For $X=1$ one has $k(V)=v$.
\end{lemma}
\begin{proof}
Burnside's lemma bounds the identity term of the first formula in
\eqref{prim:affine-formulas} by
$v/|X|+(1-1/|X|)v^{1-1/r}$. Each of the other $k(X)-1$ terms is at
most $v^{1-1/r}$. Rearrangement gives the assertions, including the
additive correction in \eqref{prim:threshold}.
\end{proof}

Two elementary auxiliary consequences will be useful. If a cyclic odd
central subgroup $Z$ of $X$ acts freely on $V\setminus\{0\}$, then
\begin{equation}\label{prim:free-center}
 k(V\sd X)\le\left(z+\frac{v-1}{z}\right)k(X/Z),
 \qquad z=|Z|.
\end{equation}
Indeed, $X_\lambda\cap Z=1$ for $\lambda\ne0$, and
$k(X_\lambda)\le[X/Z:X_\lambda]k(X/Z)$; summing these contributions
and using $k(X)\le z k(X/Z)$ proves the formula. If $v\ge(z+1)^2$,
its coefficient is at most $2v/(z+1)$. In particular it is at most $v$.

\begin{lemma}\label{lemma:soluble-affine}
If a solvable group $X$ acts faithfully on an $\F_2$-space $V$, then
$k(V\sd X)\le |V|\,|X|_2$, without a complete reducibility assumption.
\end{lemma}
\begin{proof}
A Hall $2'$-subgroup $J$ of a solvable $X$
meets every odd-order conjugacy class. The $X$-orbits of pairs
$(x,w)$ with $x$ odd and $xw=w$ are therefore images of the corresponding
$J$-orbits. Consequently
\[
 \Lambda_2(X,V)\le k(V\sd J)\le v.
\]
The last inequality is the coprime $k(GV)$ theorem \cite{GMRS04}.
Applying the same argument to $V^*$ and using
$k(X_\lambda)\le |X|_2\ell_2(X_\lambda)$ in every dual-vector
stabilizer gives
\[
 k(V\sd X)\le |X|_2\Lambda_2(X,V^*)\le v|X|_2.
\]
\end{proof}

\begin{lemma}\label{prim:scalar-reserve}
In Lemma~\ref{prim:field}, suppose $L\ne1$ and $R=Z\ne1$.
Write $z=|Z|$, $N/Z=C_e$, and $P=e_2$. Then
\[
 \frac{k(V\sd N)}{v|N|_2}\le\frac12,
 \qquad
 e>1\Longrightarrow\frac{k(V\sd N)}{v|N|_2}\le\frac5{16}<\frac13.
\]
Every element of $N\setminus Z$ fixes at most $v^{1/2}$ vectors.
\end{lemma}
\begin{proof}
Here $q\ge4$, $z\ge3$, and $v=q^m$ with $m\ge2$, since a nontrivial
perfect layer cannot act on a one-dimensional multiplicity space.
If $e=1$, direct orbit counting gives
\[
 \frac{k(VZ)}v=\frac1z+\frac{z^2-1}{zv}
 \le\frac2{z+1}\le\frac12.
\]
We give the calculation for the possibly nonsplit extension when $e>1$.
For $d\mid e$, let $z_d$ be the number of scalars fixed by the subgroup
of order $d$ in $N/Z$, and put
$J_2(d)=d^2\prod_{\ell\mid d,\ \ell\text{ prime}}(1-\ell^{-2})$.
Every character of $Z$ extends
to its inertia group, whose quotient over $Z$ is cyclic: the value on a
lift of a generator is obtained by taking a suitable complex root.
In ordinary Clifford theory a stabilizer of order $a$ in the cyclic
quotient contributes $a^2=\sum_{d\mid a}J_2(d)$. Averaging over
$\Irr(Z)$ gives
\begin{equation}\label{prim:scalar-k}
 k(N)=\frac1e\sum_{d\mid e}J_2(d)z_d
 \le\frac ze+\left(e-\frac1e\right)(\sqrt q-1).
\end{equation}
The action on $K$ has order $d$, so $z_d\le q^{1/d}-1$ for $d>1$.

Every nonzero dual-vector stabilizer in $N$ embeds in $C_e$ and is cyclic.
There are at most $z$ subgroups of order $d$ projecting isomorphically
onto the subgroup of order $d$ in $C_e$. If their generator $x$ fixes
a nonzero vector, then $x^d$ is a scalar fixing that vector, hence is one.
Galois descent gives $|C_V(x)|=v^{1/d}$. This descent is valid when
$d$ is even: if $b_i,c_i$ are trace-dual bases of $K/K^{\sigma}$, the
fixed vectors $\sum_{j=0}^{d-1}\sigma^j(c_i)x^j(w)$ span $V$ over
$K$, because $\sum_i b_i\sigma^j(c_i)$ is $1$ for $j=0$ and $0$
otherwise. Fixed vectors independent over $K^{\sigma}$ remain
independent over $K$, by applying $x$ to a minimal dependence.
There is no division by $d$.

The little-groups formula, with squared stabilizer orders retained, now
implies
\begin{equation}\label{prim:scalar-affine}
 k(VN)\le k(N)+\frac{v-1}{ze}
       +\frac1e\sum_{\substack{d\mid e\\d>1}}J_2(d)(v^{1/d}-1).
\end{equation}
Combining this with \eqref{prim:scalar-k} gives
\begin{equation}\label{prim:scalar-tail}
 \frac{k(VN)}{vP}\le\frac1{zeP}+\frac z{ePv}
               +e_{2'}(v^{-1/2}+\sqrt q/v).
\end{equation}
For $q\ge32$, use $eP\ge3$, $z\ge3$, $z<q$, $v\ge q^2$ and
$e_{2'}\le\log_2q$. The right side is at most
\[
 \frac19+\frac1{3q}+\frac{\log_2q}{q}
                   +\frac{\log_2q}{q^{3/2}}
 <\frac{89}{288}<\frac5{16}.
\]
Each nonconstant term decreases, and the displayed bound follows at
$q=32$ from $\sqrt{32}>5$. At $q=16$, $e=2$ or $4$, and
\eqref{prim:scalar-tail} is at most $17/96$. At $q=8$ the
parameters are $(z,e)=(7,3)$, and \eqref{prim:scalar-affine} gives
$k(VN)/v\le1/21+16/(7v)+(8/3)v^{-2/3}\le1/4$.
At $q=4$ they are $(z,e)=(3,2)$, and the bound is
$1/12+2/(3v)+3/(4\sqrt v)\le5/16$.
Finally a semilinear transformation with nontrivial field action has
fixed-space size at most $v^{1/2}$: commute it with a scalar of $K$
not fixed by that field action and intersect the two conjugate fixed
spaces. The scalar used for this argument need not belong to $N$.
\end{proof}

\subsection{An odd symplectic core}

Retain the notation of Lemma~\ref{prim:field}, and put $z=|Z|$.
Write $Q_0=\prod_j|\Sp_{2a_j}(r_j)|$ and $Q_o=(Q_0)_{2'}$.
The restriction to $R$ consists of $m$ copies of its faithful absolute
module of degree $n$, so $v=q^{nm}$. Let
$R\le X\le N_{\Gamma L_K(V)}(R)$ and assume
$C_{X\cap\GL_K(V)}(R)=Z$, where $\Gamma L_K(V)$ is the group of
invertible maps semilinear over $K$. Then
\begin{equation}\label{prim:core-orders}
 |R|=zn^2,\quad
 (X\cap\GL_K(V))/R\le\prod_j\Sp_{2a_j}(r_j),\quad
 |X/R|\le fQ_0,\quad |X/R|_{2'}\le f_{2'}Q_o.
\end{equation}
The kernel of the symplectic action consists of inner automorphisms of
$R$: a center-fixing central error is identified with a linear functional
on each symplectic quotient, and the commutator pairing is nondegenerate.
The scalar-centralizer hypothesis then gives the asserted kernel.

For every nonzero $\lambda\in V^*$,
\begin{equation}\label{prim:isotropic}
 |R_\lambda|\le n.
\end{equation}
The trace pairing identifies the additive dual with a $K$-dual. A
stabilizer injects into $R/Z$, because no nonidentity scalar fixes a
nonzero vector, and its image is isotropic for each extraspecial
commutator form. Its order is therefore at most $\prod r_j^{a_j}=n$.
This proof permits every homogeneous multiplicity.

Also
\begin{equation}\label{prim:core-support}
 |C_V(x)|\le v^{2/3}\qquad(1\ne x\in X).
\end{equation}
If a $K$-linear $x$ acts nontrivially on $R_j/Z(R_j)$, choose
$g\in R_j$ with $[x,g]$ noncentral. Such an extraspecial element has
all its $r_j$ eigenvalues with equal multiplicity. Working over an
algebraic closure $\overline K$ of $K$, let $U$ be a
$\lambda$-eigenspace of $x$ on $\overline K\otimes_K V$.
Its intersection with the corresponding eigenspace of $x^g$ lies in
the $1$-eigenspace of $[x,g]$. Hence
$2\dim_{\overline K}U-\dim_K V\le\dim_K V/r_j$, so every eigenspace
of $x$ has dimension at most $(r_j+1)\dim_K V/(2r_j)\le2\dim_K V/3$.
If all symplectic actions are trivial, remove an inner automorphism
to put $x$ in $R$;
a noncentral element has a nonidentity scalar commutator, and the same
argument gives the stronger bound $\dim_K V/2$. Nonidentity scalars
are fixed-point free. For a nontrivial field action commute with a
scalar of $K$; this also gives the bound $v^{1/2}$ over $\F_2$.

Burnside's lemma, \eqref{prim:isotropic}, and the little-groups formula
yield
\begin{equation}\label{prim:core-affine}
 k(VX)\le |X|+\frac n{|R|}v+n|X/R|v^{2/3}.
\end{equation}
For example, the number of nonzero $X$-orbits is at most
$v/|X|+v^{2/3}$, and each stabilizer has order at most $n|X/R|$.

\begin{lemma}\label{prim:nilpotent}
Under the explicit core hypotheses above, $k(VX)\le v|X|_2$.
This includes the scalar case $n=1$.
\end{lemma}
\begin{proof}
If $n=1$, $X$ is cyclic-by-cyclic and is solvable. Suppose $n>1$.
For $U=q^m$ put $T=fnQ_0/U^{n/3}$. Whenever
$q^{2nm/3-1}\ge n^2$, \eqref{prim:core-affine} gives
\begin{equation}\label{prim:core-unscaled}
 \frac{k(VX)}v\le\frac1n+(1+1/n)T.
\end{equation}
The absorption condition holds for $q\ge16,n\ge15,m\ge1$, and for
$q\ge4,n\ge5,m\ge2$: check the boundary and differentiate
$(2nm/3-1)\log q-2\log n$ in $n$.
The symplectic order formula gives
\begin{equation}\label{prim:symplectic-order}
 Q_0< n^{1+2\log_3n}.
\end{equation}
Indeed, put $x_j=a_j\log r_j$; use $\log r_j\ge\log3$ and
$\sum x_j^2\le(\sum x_j)^2$. For $U\ge16,n\ge35$,
\[
 T\le\frac{(\log_2U)n^{2+2\log_3n}}{U^{n/3}}
 \le\frac{4n^{2+2\log_3n}}{16^{n/3}}<\frac12.
\]
Both monotonicities follow by differentiation. In $n$, the logarithmic
derivative is $(2+4\log_3n)/n-(\log16)/3<0$ for $n\ge35$.
At $35$ the exact comparisons $35^4<3^{13}$ and $35^{51}<2^{262}$
prove the bound. Equation~\eqref{prim:core-unscaled} is then less than
one.

If there are at least two distinct core primes, $q\ge16$ and $n\ge15$.
Below $35$ the only values are $15,21,33$, with
$Q_0=2880,8064,31680$, respectively. Substitution in
\eqref{prim:core-unscaled}, using $T\le4nQ_0/16^{n/3}$, gives
bounds less than $243/1000,51/1000,31/1000$, respectively.
Thus this entire case satisfies $k(VX)<v$.

For one prime and $m\ge2$, $n=3$ gives a solvable group. For a prime
$n=r\ge11$, the condition $r\mid q-1$ gives
\[
 T<\frac{(\log_2(r+1))r^4}{(r+1)^{2r/3}}
 \le r^{5-2r/3}<\frac12.
\]
The remaining $n<35$ are covered by the following complete table.
The last column bounds $T$; larger $q$ or $m$ improve it.
\[
\begin{array}{c|r|r|c}
 n&q_{\min}&Q_0&T\text{ bound}\\\hline
 5&16&120&1/4\\
 7&8&336&441/1024\\
 9&16&51840&3645/32768\\
 25&16&9360000&1/2\\
 27&16&9170703360&1/2
\end{array}
\]
The nonintegral exponents are checked by cubing; for example the first
row is $9600^3<16^{10}$. This leaves $q=4,n=9$ or $27$.

For multiplicity one and one prime, apply the following precise part
of \cite[Theorem~1.1]{GM13}: if an odd group of symplectic type with
quotient order $r^{2a}$ has a faithful absolute module of degree $r^a$
over a field $K$ of characteristic two, every subgroup of its
$\F_2$-linear normalizer containing it satisfies $k(VX)\le v$, except
possibly $(r,a,|K|)=(3,2,4),(3,2,16),(3,3,4)$.
When $R$ has a larger cyclic scalar center, apply the theorem to its
characteristic extraspecial factor $R_0$ from
Lemma~\ref{prim:field}. The group $X$ normalizes $R_0$ and
contains it, so the theorem still applies to $X$ itself.
Thus only these three cases and their stated multiplicities remain.

Let $R_a$ be extraspecial of exponent three and order $3^{2a+1}$.
On $m$ copies of its degree $3^a$ module, put $U=q^m$ and
\[
 b_0(U)=(U+2)/3,\qquad
 b_a(U)=\frac{U^{3^a}+(3^{2a+1}-3)U^{3^{a-1}}+2}{3^{2a+1}}.
\]
The exact formula is
\begin{equation}\label{prim:extraspecial-exact}
 k(VR_a)=b_a(U)+2+(3^{2a}-1)b_{a-1}(U).
\end{equation}
Burnside proves the formula for $b_a$. There are two nonidentity
central classes, each contributing one. For a noncentral class,
$C_{R_a}(x)=\langle x\rangle\times R_{a-1}$, and its fixed eigenspace
is the faithful $R_{a-1}$-module; this gives the last term.
At $(q,a)=(4,2),(16,2),(4,3)$ the counts are
$1624$, $282813528$, and $8237037448664$.

The finite group data proved in Subsection~\ref{prim:finite} give,
for nonsolvable $J$,
\[
 \frac{k(J)}{|J|_2}\le
 \begin{cases}9/8&J\le\Sp_4(3),\\2187/4&J\le\Sp_6(3).
 \end{cases}
\]
The linear normalizer quotient has scalar kernel of order at most
$(q-1)/3$, and the field quotient in these three cases is a $2$-group.
If its symplectic image is solvable, so is $X$. Otherwise Nagao's
inequality and \eqref{prim:extraspecial-exact} give the normalized
bounds
\[
 \frac{1827}{262144},\qquad
 \frac{1590826095}{68719476736},\qquad
 \frac{2251800112528521}{9007199254740992},
\]
all less than one. The expression $k(VR_a)/U^{3^a}$ is a sum of
positive multiples of nonpositive powers of $U$, and hence decreases.
The same calculation therefore covers the multiplicities with
$q=4,n=9,27$ left above. This exhausts all cases.
\end{proof}

\begin{lemma}\label{prim:sixth}
In Lemma~\ref{prim:field}, if $L\ne1$ and $R$ is nonabelian, then
\begin{equation}\label{prim:sixth-bound}
 k(VN)<v|N|_2/6,\qquad N=C_E(L).
\end{equation}
\end{lemma}
\begin{proof}
The core multiplicity is at least two: $L$ centralizes $R$ and acts
nontrivially and perfectly on the multiplicity space. Thus
$m\ge2$, $q\ge4$, and $z\ge3$. Divide \eqref{prim:core-affine}
by the $v|N|_2$, and put
$T=f_{2'}nQ_o/q^{nm/3}$. Since
$q^{2nm/3-1}\ge n^2$, one obtains
\begin{equation}\label{prim:core-normalized}
 \frac{k(VN)}{v|N|_2}\le\frac1{3n}+(1+1/n)T.
\end{equation}
For $n\ge37$, \eqref{prim:symplectic-order} and $m\ge2$ give
\[
 T\le\frac{2n^{2+2\log_3n}}{16^{n/3}}<\frac18.
\]
The logarithmic derivative is negative on this range; at $37$ use
$37^3<3^{10}$ and $4096\cdot37^{26}<16^{37}$. Hence the right side
of \eqref{prim:core-normalized} is less than
$1/111+38/(8\cdot37)=61/444<1/6$.

If $n=r$ is prime, write $R=Z R_0$ with $|R_0|=r^3$ and put $U=q^m$.
Counting the scalar classes and the $r^2-1$ noncentral classes having
eigenvalue one gives
\begin{equation}\label{prim:prime-core-exact}
 k(VR)=\frac{z(r^2+r-1)}r+
 \frac{U^r+c_rU-c_r-1}{zr^2},\qquad
 c_r=r(r-1)(r+1)^2.
\end{equation}
Indeed the identity orbit count is
$[U^r+r(r^2-1)U+zr^2-1-r(r^2-1)]/(zr^2)$; each of those noncentral
classes contributes $(U+z-1)/z$. This proves the displayed formula.
For $N_1=C_N(Z)$, the exact kernel calculation gives
$N_1/R\le\SL_2(r)$. Every subgroup $J\le\SL_2(r)$ satisfies
$k(J)/|J|_2\le r$. Let $\overline J$ be its image in
$\operatorname{PSL}_2(r)$. If $-I\in J$, then
$k(J)\le2k(\overline J)$ and $|J|_2=2|\overline J|_2$.
If $-I\notin J$, projection is an isomorphism; moreover $J$ has
odd order, since $-I$ is the only involution of $\SL_2(r)$.
Dickson's classification \cite[Appendix~A]{GZ10} gives cyclic,
dihedral, Borel, $A_4,S_4,A_5$, or the full group. The Borel ratio is
bounded by $k(C_r:C_d)=d+(r-1)/d\le r$; all the other ratios are at
most $r$ directly. For $r=3$, intersect with the normal $Q_8$ in
$\SL_2(3)$ to obtain the sharper bound three.

The quotient $N/N_1$ acts faithfully on $Z$, so its odd order factor
is at most $\operatorname{ord}_z(2)_{2'}$. Here
$\operatorname{ord}_z(2)$ is the multiplicative order of two modulo
$z$, and $\phi$ is Euler's totient function. If $r=3$, this is at most
$\phi(z)/2\le z/3$, since $3\mid z$ forces the order of two modulo
$z$ to be even. Since $U\ge(z+1)^2$, Nagao and
\eqref{prim:prime-core-exact} give
\[
 \frac{k(VN)}{v|N|_2}
 \le\frac19+\frac{11z^2}{3(z+1)^6}
                  +\frac{32}{3(z+1)^4}
 \le\frac{5929}{36864}<\frac16.
\]
For $r\ge5$, use
$\operatorname{ord}_z(2)_{2'}\le\phi(z)\le z(r-1)/r$ instead. The
bound becomes
\[
 \frac{r-1}{r^2}+
 \frac{r(r-1)(r^2+r-1)}{(r+1)^{2r}}+
 \frac{(r-1)^2}{r(r+1)^{2r-4}}
 <\frac4{25}+\frac1{6^6}+\frac1{6^5}<\frac16.
\]
Here $z^2/(z+1)^{2r}$ decreases for $z\ge r$, and the two displayed
numerators are bounded by $(r+1)^4$ and $r(r+1)$, respectively.

The composite odd $n<37$ are exactly $9,15,21,25,27,33,35$.
Apart from $(n,q)=(9,4)$, the following table gives the endpoint
for the decreasing bound $T\le (\log_2q)nQ_o/q^{2n/3}$:
\[
\begin{array}{c|rrrrrrr}
 n&9&15&21&25&27&33&35\\\hline
 q_{\min}&16&16&16&16&4&16&16\\
 Q_o&405&45&63&73125&8955765&495&315
\end{array}
\]
At every entry $(100 f nQ_o)^3<q^{2n}$, so $T<1/100$ and
\eqref{prim:core-normalized} is less than $1/27+1/90<1/6$.
For $(n,q)=(9,4)$ the scalar order is three and the field quotient
is a $2$-group. The complete finite subgroup bound
$k(J)/|J|_2\le27$ for every $J\le\Sp_4(3)$ is proved in
Subsection~\ref{prim:finite}. Formula~\eqref{prim:extraspecial-exact}
at $U\ge16$ therefore gives
\[
 \frac{k(VN)}{v|N|_2}
 \le\frac{27\cdot282813528}{16^9}
 =\frac{954495657}{8589934592}<\frac16.
\]
All parameters have now been covered.
\end{proof}

\subsection{Component supports and the outer quotient}

For a nonabelian simple group $S$ write
\[
 \beta(S)=\max_{S\le A\le\Aut(S)}\frac{k(A)}{|A|_2}.
\]
If $F^*(H)=S_1\times\cdots\times S_t$ and $T$ is the permutation
image on these factors, then
\begin{equation}\label{prim:coordinate-budget}
 \frac{k(H)}{|H|_2}\le\frac{k(T)}{|T|_2}\prod_i\beta(S_i).
\end{equation}
Indeed, the kernel $J$ of that permutation action contains the full
socle and embeds in $\prod_i\Aut(S_i)$. Let $J_i$ be the subgroup of
$J$ acting trivially on the first $i$ factors. The quotient
$J_{i-1}/J_i$ is an almost simple group with socle $S_i$. Repeated
Nagao inequalities multiply their class numbers, while their two-parts
multiply \emph{exactly}. A final application to $J\lhd H$ proves the
formula, including coupled outer actions.

By \cite[Theorem~1.1]{GaM15}, a permutation group of degree $t\ge4$
has at most $5^{(t-1)/3}$ classes. Including the degrees $1,2,3$ gives
$k(T)\le2^{t-1}$. A sharper sufficient normalized bound for small
orbits is
\begin{equation}\label{prim:h}
 h(1)=h(2)=1,\quad h(3)=h(4)=3,\quad
 h(t)=2^{t-1}\ (t\ge5),\qquad \frac{k(T)}{|T|_2}\le h(t).
\end{equation}
For degree four, an intransitive group either has a three-point orbit
or is a $2$-group. A transitive group has two-part at least four,
so the class bound gives a ratio at most two. If components are
partitioned into $c$ nonempty unions of simple isomorphism types of
sizes $t_1,\ldots,t_c$, the same coordinate-kernel argument gives
$k(T)/|T|_2\le\prod_jh(t_j)\le2^{t-c}$.

\begin{lemma}\label{prim:component-support}
Let $L_i$ be a component, with absolute factor degree $m_i$
in \eqref{prim:tensor-degree}, and let $d=\dim_{\F_2}V$.
If $x$ normalizes $L_i$ but does not centralize it, then
\[
 \operatorname{codim}_{\F_2}C_V(x)\ge d/(2m_i).
\]
If $x$ moves a component, the lower bound is $d/4$.
Alternatively, suppose every prime-order automorphism of $S_i$ has at
most $r_i$ conjugates by $S_i$ generating a subgroup containing $S_i$.
An element with nontrivial induced action on $S_i$ has support at least
$d/r_i$.
\end{lemma}
\begin{proof}
For the first assertion choose $y=[x,a]\ne1$ in $L_i$. On every absolute
tensor constituent, $y-1$ has rank at least the total degree divided
by $m_i$, because the component representation is faithful. Frobenius
conjugate constituents have the same degree and kernel; summing over
them and over the homogeneous multiplicity gives
$\operatorname{rank}(y-1)\ge d/m_i$. On the other hand
$\operatorname{rank}(y-1)\le2\operatorname{rank}(x-1)$, since $y$
is a product of two conjugate matrices, one inverted. This argument
is over $\F_2$ and also applies to transformations semilinear over $K$.

For a moved component choose $a,b\in L_i$ mapping to generators of
$S_i$; they generate $L_i$, since it is perfect. With
$v(u)=u^{-1}u^x$ the group generated by $x,x^a,x^b,x^{ab}$ contains
$v(a),v(b),v(ab)$ and
\[
 v(a)v(b)v(ab)^{-1}=a^{-1}b^{-1}ab.
\]
The last element has nontrivial image in $S_i$. The intersection of
$\langle v(a),v(b),v(ab)\rangle$ with $L_i$ is normal under the
inner actions of $a,b$, and therefore maps onto $S_i$. Perfectness
puts $L_i$ itself in the subgroup. The four conjugate fixed spaces
have zero intersection, proving the asserted support.

For the last assertion, first pass to a power whose induced automorphism
has prime order. Lift the conjugating elements to $L_i$. The resulting
generators lie in one coset modulo $L_i$, so their generated group has
cyclic quotient modulo its intersection with $L_i$. Its derived group
maps onto $S_i$; hence that intersection contains $L_i$. Their common
fixed space is zero. The intersection codimension inequality proves
the bound, and taking the power can only increase fixed vectors.
\end{proof}

\begin{lemma}\label{prim:profile-transfer}
Suppose $L\ne1$ in Lemma~\ref{prim:field}. Put $N=C_E(L)$ and
$H=E/N$, and let $B\ge k(H)/|H|_2$.
Let $0<M\le\prod_{i=1}^t m_i$, where the $m_i$ are the absolute
component degrees in \eqref{prim:tensor-degree}; thus $\dim_KV\ge nM$.
Suppose the support bounds just proved give
$|C_V(x)|\le v^{1-1/r}$ for $x\notin N$, with $r\ge3$.
Then the affine bound holds in either of the following situations:
\[
 B\le1,\qquad\text{or}\qquad M/r\ge\log_2(2B).
\]
For the second alternative with $B>1$, it is enough to use the
following overlapping ranges: $B\le6$ or $B\ge4$ when $R$ is
nonabelian; $B\le2$ or $B\ge2$ when $R\ne1$ is abelian; and the
ordinary support criterion when $N=1$.
\end{lemma}
\begin{proof}
Lemma~\ref{prim:nilpotent} applies to $N$: the core
centralizer is $Z$, and the homogeneous multiplicity is unrestricted.
Nagao gives $k(VE)\le k(VN)k(H)$, proving the first alternative.
Lemmas~\ref{prim:scalar-reserve} and \ref{prim:sixth} prove the stated
small-$B$ ranges.
All nonidentity elements of $N$ satisfy the needed support bound,
by \eqref{prim:core-support} or Lemma~\ref{prim:scalar-reserve}.

For an abelian nontrivial core, $z\le2^f-1$, $f\ge2$, and
\[
 \frac{k(E)}{|E|_2}\le z f_{2'}B\le(2^f-1)fB.
\]
If $B\ge2$, then $B^{f-1}\ge2^{f-1}\ge f$, and the profile gives
$v^{1/r}\ge(2B)^f\ge2^ffB>(2^f-1)fB+1$.
Use Lemma~\ref{prim:support}.

For a nonabelian core put $b=\log_2B\ge2$. Equations
\eqref{prim:core-orders} and \eqref{prim:symplectic-order} give
\[
 \frac{k(E)}{|E|_2}<2^f f n^2Q_oB,
 \qquad \log_2(n^2Q_o)<4n\quad(n\ge3).
\]
For the latter inequality, use
$\log n\le(2/e)\sqrt n<(3/4)\sqrt n$, $\log2>2/3$,
$\log3>1$, and $\sqrt n<(3/5)n$ for $n\ge3$; the resulting
coefficient is $297/80<4$. Therefore
\[
 f+\log_2(fn^2Q_o)+b<2f-1+4n+b.
\]
The profile gives $\log_2(v^{1/r})\ge fn(1+b)$, whose difference
from the last upper bound is at least
$2n(b-1)-b-3\ge1$. Thus $v^{1/r}>2\cdot2^ffn^2Q_oB$, which
exceeds $k(E)/|E|_2+1$. Finally, when $N=1$, the threshold follows
from $v^{1/r}\ge2B\ge B+1$. This proves the lemma.
\end{proof}

\subsection{Complete simple-group numerical inputs}

We now record the specific published representation and class estimates
used in the proof. They are stated for projective representations, so
odd central covers are included. Write $L_n(q)=\operatorname{PSL}_n(q)$,
$U_n(q)=\operatorname{PSU}_n(q)$, and use $\operatorname{PSp}$ and
$\operatorname{P\Omega}$ for the simple projective classical groups.
Isomorphic small groups are assigned to a single row.

\paragraph{The half-class families.}
Except for
\begin{equation}\label{prim:defining-exceptions}
 L_2(2^a),\quad L_3(2),\quad A_6,
\end{equation}
every simple group of Lie type in characteristic two has
$\beta(S)\le1/2$. In the generic ranges, the estimates used from
\cite[Propositions~2.5--2.8 and proof of Theorem~2.9]{MNR21} give
$k(A)\le k(U)$ for every almost simple $A$ and a unipotent subgroup
$U$ of the following order, strictly below the Sylow order:
\begin{center}\small
\begin{tabular}{lll}
\toprule Type & $|U|$ & $|S|_2$\\\midrule
$A_h$, $h\ge3$ & $q^{\lfloor(h+1)^2/4\rfloor}$ & $q^{h(h+1)/2}$\\
${}^2A_h$, $h\ge3$ & $q^{2h-1}$ & $q^{h(h+1)/2}$\\
$C_h$, $h\ge2$ & $q^{h(h+1)/2}$ & $q^{h^2}$\\
$D_h$, $h\ge4$ & $q^{h(h-1)/2}$ & $q^{h(h-1)}$\\
${}^2D_h$, $h\ge5$ & $q^{2+(h-1)(h-2)/2}$ & $q^{h(h-1)}$\\
${}^2D_4$ & $q^6$ & $q^{12}$\\
$G_2$, $q\ge4$ & $q^3$ & $q^6$\\
${}^3D_4$ & $q^5$ & $q^{12}$\\
$F_4$ & $q^9$ & $q^{24}$\\
${}^2F_4$, $q\ge8$ & $q^5$ & $q^{12}$\\
$E_6,{}^2E_6,E_7,E_8$ & elementary abelian & nonabelian Sylow\\
\bottomrule
\end{tabular}
\end{center}
In the last row the cited proof constructs an elementary abelian witness;
it is proper because adjacent simple root groups have a nontrivial
commutator. The small parameters in that comparison have the following
upper bounds for every $k(A)$:
\begin{center}\small
\begin{tabular}{lrr|lrr}
\toprule $S$ & bound & $|S|_2/2$ & $S$ & bound & $|S|_2/2$\\\midrule
$L_4(2)$&22&32&$L_4(4)$&168&2048\\
$L_5(2)$&54&512&$U_4(2)$&25&32\\
$U_4(4)$&256&2048&$U_4(8)$&3612&131072\\
$U_5(2)$&47&512&$\Sp_6(2)$&30&256\\
$\Sp_4(4)$&64&128&$O_8^+(2)$&68&2048\\
$O_8^-(2)$&60&2048&$O_{10}^-(2)$&230&524288\\
${}^3D_4(2)$&49&2048&&&\\\bottomrule
\end{tabular}
\end{center}
For $O_8^+(4)$ the bound is $<4^6<4^{12}/2$. For $\Sp_4(q)$,
$q\ge8$, the witness has order $q^3$. For $L_3(q),U_3(q)$,
$q\ge4$, the non-inner-diagonal bounds are $q^2$, apart from
$U_3(4).2$ with $20$ classes; the inner-diagonal formula is bounded
by $q^2+q+4\le q^3/2$. For $L_3(4)$ the all-extension bound is
$28<32$. For the Suzuki groups, $|U|=2q\le q^2/2$; the only extra
small field extension is $\operatorname{Sz}(8).3$, with $17<32$
classes. The Tits group has $(k,|S|_2)=(22,2048)$ and its extension
has $(29,4096)$. Under the convention $G_2(2)'=U_3(3)$, the two
class counts are $14,16$ and the two-parts $32,64$.
This proves the asserted half-class bound in every listed family.

For the three excluded types the exact sufficient bounds are
\begin{equation}\label{prim:residual-beta}
 \beta(L_2(2^a))\le11/8,\qquad
 \beta(L_3(2))=3/4,\qquad \beta(A_6)=7/8.
\end{equation}
Here $k(L_2(q))=q+1$; all proper extensions have class number at
most their two-part except $L_2(8).3$, whose ratio is $11/8$.
The additional exception $L_2(16).4$ has $17$ classes and two-part
$64$. For $A_6$, the $(k,2\text{-part})$ pairs are
$(7,8)$, $(11,16)$, $(11,16)$, $(8,16)$ and $(13,32)$.

Let $p(n)$ be the number of partitions of $n$ and $s_2(n)$ its
binary digit sum. For alternating groups $A_n$, $n\ge8$, the same
half-class bound follows from $k(A_n)<k(S_n)=p(n)$ and
\begin{equation}\label{prim:partition}
 p(n)\le2^{n-s_2(n)-2}=|A_n|_2/2.
\end{equation}
Indeed Euler's product gives
$p(n)<\exp(\pi\sqrt{2n/3})<2^{4\sqrt n}$. For $n\ge64$,
$n-s_2(n)-2\ge n-\log_2(n+1)-2\ge n/2\ge4\sqrt n$.
The complete range $8\le n\le63$ is covered by the integer
partition recurrence displayed in Subsection~\ref{prim:finite};
the minimum ratio of the right side of \eqref{prim:partition} to
$p(n)$ is $32/30$ at $n=9$.

For the sporadic groups the following are $k(S)/|S|_2$. Every nontrivial
outer extension, whose index is two, has no larger ratio.
\begin{center}\small
\begin{tabular}{lr|lr}
\toprule $S$&ratio&$S$&ratio\\\midrule
$M_{11}$&$5/8$&$M_{12}$&$15/64$\\
$M_{22}$&$3/32$&$M_{23}$&$17/128$\\
$M_{24}$&$13/512$&$J_1$&$15/8$\\
$J_2$&$21/128$&$J_3$&$21/128$\\
$J_4$&$31/1048576$&$Co_1$&$101/2097152$\\
$Co_2$&$15/65536$&$Co_3$&$21/512$\\
$Fi_{22}$&$65/131072$&$Fi_{23}$&$49/131072$\\
$Fi_{24}'$&$27/524288$&$HS$&$3/64$\\
$McL$&$3/16$&$He$&$33/1024$\\
$Ru$&$9/4096$&$Suz$&$43/8192$\\
$O'N$&$15/256$&$HN$&$27/8192$\\
$Ly$&$53/256$&$Th$&$3/2048$\\
$B$&$23/274877906944$&$M$&$97/35184372088832$\\\bottomrule
\end{tabular}
\end{center}
These are ordinary ATLAS character-table counts \cite{Atlas85,CTblLib}.
Thus all sporadic groups except $M_{11},J_1$ belong to the half-class
families.

\paragraph{Cross-characteristic degree and generation bounds.}
For odd $q$, the following table gives a lower bound $m$ for every
nontrivial absolute projective characteristic-two degree, a bound $r$
on the number of conjugates of a prime-order automorphism needed to
generate a subgroup containing the socle, and an upper bound $B$ on
$\beta(S)$. The degree and generation assertions are
\cite[Proposition~2.1, Table~2.1, Propositions~2.3--2.6]{Lee21},
which states the needed bounds including the small exceptions. The
class estimates are those of \cite[Propositions~2.5--2.8 and proof of
Theorem~2.9]{MNR21}: in particular $k(A)\le100q^h$ for untwisted
rank $h$, with the rank-one and rank-two refinements used below.
\begin{center}\small
\begin{tabular}{llll}
\toprule Socle and range & $m$ & $r$ & $B$\\\midrule
$L_3(q)$, $q\ge3$ & $q^2+q-1$&4&$q^3/16$\\
$U_3(q)$, $q\ge5$ & $q(q-1)$&3&$q^3/16$\\
$L_4(q),U_4(q)$, $q\ge5$&$q^2(q-1)$&6&$100q^3/8$\\
$L_n(q),U_n(q)$, $n\ge5$&$(q-1)q^{n-2}$&$n$&$100(q/2)^{n-1}$\\
$\operatorname{PSp}_4(q)$, $q\ge9$&$(q^2-1)/2$&5&$100q^2/64$\\
$\operatorname{PSp}_6(q)$, $q\ge5$&$(q^3-1)/2$&6&$100q^3/256$\\
$\operatorname{PSp}_{2h}(q)$, $h\ge4$&$(q^h-1)/2$&$2h$&$100q^h/2^{3h-1}$\\
$\operatorname{P\Omega}_{2h+1}(q)$, $h\ge3$&$q^{2h-3}$&$2h+1$&$100q^h/2^{3h}$\\
$\operatorname{P\Omega}^{\pm}_{2h}(q)$, $h\ge4$&$q^{2h-3}-2$&$2h$&$25q^h$\\
$G_2(q)$, $q\ge5$&$q^3-1$&5&$25q^2$\\
${}^3D_4(q)$, $q\ge3$&$(q^5-1)/2$&7&$25q^4$\\
$F_4(q)$, $q\ge3$&$q^5$&8&$25q^4$\\
$E_6(q),{}^2E_6(q)$, $q\ge3$&$q^5$&9&$25q^6$\\
$E_7(q)$, $q\ge3$&$q^5$&10&$25q^7$\\
$E_8(q)$, $q\ge3$&$q^5$&11&$25q^8$\\
${}^2G_2(q)$, $q\ge27$&$q(q-1)$&3&$25q^2$\\
$L_2(q)$, $q\ge37$&$(q-1)/2$&4&$(q+2)/4$\\\bottomrule
\end{tabular}
\end{center}
All unqualified odd fields in the table start at three. The bounds for
symplectic and orthogonal groups include the transvection and reflection
exceptions to generic generation bounds. In particular the $q=3$
odd-orthogonal degree is $27$ at rank three. For the even-orthogonal
minus type, the product bound in Lee's table implies the stated degree
because
\[
 (q^h+1)(q^{h-1}-q)-q^{2h-3}(q^2-1)
 =q^{h+1}(q^{h-4}-1)+q^{h-1}-q\ge0;
\]
the plus-type formulas give at least the same lower bound.

The two-parts used in this table also follow directly from the order
formulas. Use $\epsilon=1$ for $L_n(q)$ and $\epsilon=-1$ for
$U_n(q)$, and put
$a=v_2(q-\epsilon)$, $b=v_2(q+\epsilon)$, $j=\lfloor n/2\rfloor$.
Then
\[
 v_2(|S|)=(n-1)a+jb+v_2(j!)-\min(v_2(n),a)\ge n-1.
\]
At $n=3$ one has $|S|_2\ge16$. For symplectic and odd-orthogonal
rank $h$,
\[
 |S|_2=2^{h v_2(q^2-1)+v_2(h!)-1}
 \ge2^{3h+v_2(h!)-1}.
\]
For every other simple group $|S|_2\ge4$ suffices. The rank-two
class bound $q^3$ includes the inner-diagonal formula
$q^2+q+4\le q^3$ and the exception $k(L_3(3).2)=17<27$.
For $U_3(3)$ one instead uses its exact half-class bound.

The remaining odd rank-one profiles are
\begin{equation}\label{prim:rankone-profiles}
\begin{array}{c|rrrrrrr}
 q&17&19&23&25&27&29&31\\\hline
 m&8&9&11&12&13&14&15\\
 r&3&3&3&4&3&3&3\\
 B&19/16&3&25/8&27/8&29/4&31/4&33/32
\end{array}
\end{equation}
The refinement at $19$ uses $k(L_2(19))=12$, $|L_2(19)|_2=4$ and
$k(\operatorname{PGL}_2(19))=21$, with two-part eight. Odd field
exponent at $27$ permits $r=3$. The small fields $5,7,9$ are
$A_5,L_3(2),A_6$. For $q=11,13$ use the small table below.

Every row of the large tables satisfies
\begin{equation}\label{prim:large-property}
 m\ge8,\quad r\ge3,\quad m\ge2r,\quad
 m/r\ge\log_2(2B).
\end{equation}
Here is a full numerical verification of the unbounded ranges. For fixed
rank, write $F(q)=m(q)/r$ and $B=Cq^a$; then
$qF'(q)>3a/2$ at the first stated parameter and thereafter, proving
$F(q)-\log_2(2B)$ increases. The boundary comparisons are
$2^m\ge(2B)^r$ with the exact values in the tables. For the first
four rows the pairs $(m/r,2B)$ are
$(11/4,27/8)$, $(20/3,125/8)$, $(100/6,3125)$,
$(54/5,2025/2)$. The tight symplectic boundary is
$2^8>2025/8$ at $\operatorname{PSp}_4(9)$.
For variable rank it suffices to take $q=3$. The second difference of
$(q^j-c)/(aj+b)$, after multiplication by a positive denominator, has
numerator
\[
 q^j\{u^2(q-1)^2+au(q-1)(q-3)+2a^2\}-2ca^2,
 \qquad u=aj+b,
\]
which is positive in the ranges above. For orthogonal rows use the
geometric ratio $q^2$. Thus the first differences increase. At the
base rank the linear/unitary difference is $81/5>\log_2(3/2)$;
the symplectic and odd-orthogonal logarithmic slopes are negative;
the even-orthogonal first difference exceeds $2>\log_2 3$.
For rank one, $(q-1)/8-\log_2((q+2)/2)$ is positive at $37$ and
has positive derivative thereafter. The short exact-integer
verification in Subsection~\ref{prim:finite} includes every boundary
of these comparisons, including \eqref{prim:rankone-profiles}.

The small odd-field class data not assigned to these ranges are
\begin{center}\small
\begin{tabular}{lrr|rr}
\toprule $S$&$k(S)$&$|S|_2$&$k(S.2)$&$|S.2|_2$\\\midrule
$\operatorname{PSp}_4(3)$&20&64&25&128\\
$\operatorname{PSp}_4(5)$&34&64&47&128\\
$\operatorname{PSp}_4(7)$&52&256&77&512\\
$\operatorname{PSp}_6(3)$&74&512&76&1024\\
$G_2(3)$&23&64&28&128\\
$L_2(11)$&8&4&13&8\\
$L_2(13)$&9&4&15&8\\\bottomrule
\end{tabular}
\end{center}
The first, third, fourth and fifth rows have half-class bounds. The
second does not: its ratio is $17/32$, and its profile
$(m,r,B)=(12,5,17/32)$ satisfies \eqref{prim:large-property}.
The group $L_4(3)$ has the all-extension bound $56<128/2$.
For $U_4(3)$, $k(S)=20$, $|S|_2=128$ and $\Out(S)=D_8$;
Nagao's index factor cancels the two-part of every extension, so
$\beta(S)\le5/32$.

\paragraph{The remaining small profiles.}
After the half-class and large families, exactly the following seven
numerical classes remain:
\begin{equation}\label{prim:small-table}
\begin{array}{c|l|r|r|c}
\text{class}&\text{socle}&m&r&B\\\hline
 R&L_2(2^a)\ (a\ge2),\ L_3(2)&2&4&11/8\\
 A6&A_6&3&5&7/8\\
 A7&A_7&4&6&9/8\\
 J1&J_1&10&8&15/8\\
 M11&M_{11}&4&8&5/8\\
 L11&L_2(11)&5&3&2\\
 L13&L_2(13)&6&4&9/4
\end{array}
\end{equation}
For the $R$ and $A6$ generation bounds use
\cite[Lemma~3.1]{GS03}, equivalently \cite[Proposition~2.3]{Lee21};
the value five is only needed for the involutory field automorphism
at $q=9$. The projective degree for $L_3(2)$ and $A_6$, including
$3.A_6$, is at least three. For $A_7$ the characteristic-two degrees
are $1,4,4,6,14,20$; the additional faithful degrees of $3.A_7$ are
$6,6,15,15,24,24,24,24$. The central two-part acts trivially and
cannot supply another faithful case. The generation bound is six
by \cite[Lemma~6.1]{GS03}. Jansen's degree table \cite{Jansen05}
gives the stronger bounds $20,10$ for $J_1,M_{11}$; their multipliers
are trivial. Their prime-automorphism generation bounds are at most
four by \cite[Theorem~3]{RZ24}, so the displayed weaker eight is safe.
Dickson's classification implies that an absolute projective
characteristic-two representation of degree two has simple image
$L_2(2^a)$. Consequently every half-class component has degree
at least three, and Lemma~\ref{prim:component-support} gives it
support denominator $2m$ without any generation classification.

\subsection{One component, including the small almost simple quotients}

We first give the small-group argument for an almost simple central
quotient, over any field of definition of the module. For
$S\lhd H$ with $S$ simple put
\[
 \gamma_S(H)=\max_{J\le H,\ S\nleq J}\frac{|J|k(J)}{|H|}.
\]
If a complete list of overgroups $M$ covers the subgroups not containing
$S$, then
\begin{equation}\label{prim:capacity-overgroups}
 \gamma_S(H)\le\max_M\frac{|M|k(M)}{|H|},
\end{equation}
since $k(J)\le[M:J]k(M)$. Let $E/Z(E)=H$, put $z=|Z(E)|$, and
let $V$ be faithful irreducible. For a nonzero dual vector, its stabilizer intersects the
scalar center trivially and its image cannot contain $S$; otherwise
perfectness would put the lifted component in the stabilizer. Thus
\begin{equation}\label{prim:small-capacity}
 k(VE)\le z k(H)+\frac{v-1}{z}\gamma_S(H).
\end{equation}
Since $v\ge(z+1)^2$, the conditions
$\gamma_S(H)\le1$ and $k(H)\le2|H|_2+1$ imply $k(VE)\le v|E|_2$:
at the endpoint the difference is
$Pz^2-2z+P-2\ge2P-4\ge0$, with $P=|H|_2$.

The complete small capacities needed here are
\[
\begin{array}{l|rrrrrr}
 H&A_7&J_1&L_2(11)&\operatorname{PGL}_2(11)&L_2(13)&\operatorname{PGL}_2(13)\\\hline
 \gamma_S(H)&1&4/133&7/12&11/12&4/7&13/14
\end{array}
\]
For $J_1$ one may apply \eqref{prim:small-capacity} directly when
$z=1$: $15+(4/133)(v-1)\le8v$ for $v\ge2$. When $z>1$,
\eqref{prim:free-center} gives $30v/(z+1)\le15v/2<8v$.
The analogous argument for $A_7$ gives $v+8\le8v$ when $z=1$,
and $18v/(z+1)<8v$ otherwise. The overgroup lists establishing all
these capacities are given in Subsection~\ref{prim:finite}.
The conjugacy classes of $\mathfrak S_7$ are parametrized by the
partitions of seven, so $k(\mathfrak S_7)=15$ and
$|\mathfrak S_7|_2=16$. For $H=\mathfrak S_7$,
\eqref{prim:free-center} therefore gives
\[
 k(V\sd E)\le15\left(z+\frac{v-1}{z}\right)
             \le15v<16v=v|E|_2.
\]

For the rank-one defining-characteristic group $S=L_2(q)$, $q$ even,
its odd multiplier is trivial. A scalar central extension is therefore
$Z\times S$. Here a direct torus calculation supplies the affine
bound on every faithful module. If $T_-,T_+$ are the two torus types
of orders $q-1,q+1$, every nonidentity odd element of $S$ has torus
centralizer and is fused only with its inverse. Hence
\[
 \Lambda_2(S,V)=n(S,V)+\frac12\sum_{t\in T_-\setminus\{1\}}
 n(T_-,V^t)+\frac12\sum_{t\in T_+\setminus\{1\}}n(T_+,V^t)
 \le\frac{k(VT_-)+k(VT_+)}2\le v.
\]
For $Z\times S$, sum the same identity over $z\in Z$ and replace
$T_\pm$ by $Z\times T_\pm$. Both are faithful odd abelian acting
groups, so the coprime bound applies. This proves the weak affine
bound by applying $k(Y)\le|Y|_2\ell_2(Y)$ on the dual.
All proper almost simple extensions except $L_2(8).3$ have
$k(H)\le|H|_2$, as recorded above, and \eqref{prim:free-center}
applies. For $H=L_2(8).3$, its maximal subgroups have
$(|M|,k(M))=(504,9),(168,8),(54,10),(42,7)$; the first is $S$.
Since $[H:S]=3$, $S$ is the only proper overgroup containing the
socle. Its maximal subgroups have pairs $(56,8),(18,6),(14,5)$.
Thus a stabilizer avoiding $S$ has index at least nine and
$k(J)\le\sqrt{168\cdot8}<37$, hence $k(J)\le36$.
Each nonzero orbit has at least nine vectors, so the total
nonzero-orbit contribution is at most $36(v-1)/9=4(v-1)$.
It follows that
$k(VH)\le11+4(v-1)=4v+7\le8v$ for $v\ge2$.
For a nontrivial scalar center \eqref{prim:free-center} instead gives
$22v/(z+1)<8v$.

These arguments, the half-class bound and the large support table prove
the affine bound for every almost simple central quotient. For the
support-table assertion with a central scalar extension, write
$\End_E(V)=\F_{2^c}$.
The scalar center satisfies $z\le2^c-1$. The nontrivial component
has degree at least two over this field, so $v\ge2^{2c}\ge(z+1)^2$.
If $B\le1$, then \eqref{prim:free-center} gives
\[
 k(V\sd E)\le v\,k(H)\le v|H|_2=v|E|_2.
\]
For $B>1$, the inequality $m/r\ge\log_2(2B)$ gives
\[
 v^{1/r}\ge(2B)^c\ge2^cB\ge zB+1
               \ge k(E)/|E|_2+1.
\]
Thus Lemma~\ref{prim:support} applies over every finite field of
definition.

Now suppose the layer consists of one component with an arbitrary
Fitting subgroup. If the core is nonabelian, all non-large profiles
have $B\le9/4<6$, and all large profiles are covered by
Lemma~\ref{prim:profile-transfer}. If $R=Z\ne1$, every non-large
profile except $L_2(13)$ has $B\le2$. For $L_2(13)$ use $B=9/4$.
A nontrivial field quotient is paid by the factor-three reserve.
Otherwise $N=Z$, $m\ge6$, and
\[
 v^{1/4}\ge q^{3/2}\ge\frac94(q-1)+1\qquad(q\ge4).
\]
At four the values are $8$ and $31/4$, and the difference has
positive derivative. The large profiles again use
Lemma~\ref{prim:profile-transfer}. Finally $R=1$ implies $N=1$,
which is the already proved almost simple case. This completes every
one-component action.

\subsection{Tensor profiles and their five residual cases}

For $t\ge2$, let $m_i$ be the component degrees and
$r_i$ their support denominators. For a half-class component use
$r_i=2m_i$. For the other components use the published generation
bounds above. Put
\[
 M=\prod_i m_i,\quad R_0=\max(4,r_1,\ldots,r_t),\quad
 B=\frac{k(T)}{|T|_2}\prod_i\beta_i,
\]
where $\beta_i\ge1/2$ are the sufficient bounds just proved;
any smaller bound is replaced by $1/2$.
The factor four covers moved components. We will prove either $B\le1$
or $M/R_0\ge\log_2(2B)$, apart from explicitly resolved small
profiles. Lemma~\ref{prim:profile-transfer} will then apply.

Set $a_i=\log_2(2\beta_i)$. Partition simple types into the following
disjoint kinds: half-class factors $G$; large factors $P$ satisfying
\eqref{prim:large-property}; and the seven classes $F$ in
\eqref{prim:small-table}. The field ranges in a named simple type
are kept together when applying the permutation estimate. Splitting
into any of these disjoint unions of simple isomorphism types gives
\begin{equation}\label{prim:type-saving}
 \log_2(2B)\le\sum_i a_i-(c-1),
\end{equation}
where $c$ is the number of nonempty type classes used. A $G$ factor
has $a=0$, $m\ge3$, $r=2m$. A $P$ factor has $m\ge8$, $r\ge3$,
$m\ge2r$ and $a\le m/r$. Each $F$ factor has
\[
 m\ge2,\quad r\le2m,\quad a<9/4,\quad
 a\le m/2\ (m\ge3),\quad a<3/2\ (m=2).
\]
Every kind therefore satisfies $a\le(3/2)(m-1)$. Once a partial
product divided by the fixed $R_0$ is at least $3/2$, multiplying it
by a further degree $m$ increases it by at least the cost $a$.
Choose first a component with maximal $r_i$; all later attachments
leave the denominator $R_0$ unchanged.

If this component has kind $P$, put $u=m/R_0$. Then $u\ge2$ and
$a\le4u/3$; the slight weakening covers $r=3<R_0=4$. Pair it with
any other factor. If the other degree $m'\ge3$, the surplus is at least
\[
 u m'-4u/3-(3/2)(m'-1)\ge m'/2-7/6>0.
\]
If $m'=2$, the factors have different types, so the one-unit saving
in \eqref{prim:type-saving} gives surplus
$(2/3)u-1/2>0$. The pair has value at least four, and absorbs every
remaining factor. If a $P$ factor occurs but the maximizing component
is of kind $G$ or $F$, then $u\ge1/2$, its cost is less than $9/4$,
and pairing with the $P$ factor, of degree $m'\ge8$, gives surplus
\[
 m'/2-m'/3-9/4+1=m'/6-5/4\ge1/12>0.
\]
Again the pair has value at least four. Thus every layer containing a
large factor is covered.

If all factors have kind $G$, \eqref{prim:coordinate-budget} gives
$B\le1/2$. Otherwise, suppose a $G$ factor and an $F$ factor occur.
If a maximizing factor is $F$, pair it with a $G$ factor: the surplus
is greater than $3/2-9/4+1=1/4$, and the pair has value at least
$3/2$. If a maximizing factor is $G$, use
\begin{equation}\label{prim:small-product}
 \frac12\prod_{i\ne j}m_i\ge\sum_{i\ne j}a_i-1.
\end{equation}
For one factor this is immediate. For two degree-two factors it is
$2\ge3-1$; for degrees $2,m\ge3$ it is $m\ge m/2+1/2$;
for two degrees at least three it follows from
$m_1m_2\ge m_1+m_2-2$. Thereafter the value is at least two,
and the attachment inequality proves induction. Equation
\eqref{prim:type-saving} proves the required profile inequality.

Only the seven finite classes remain. For $t\ge7$, $R_0\le8$,
$M\ge2^t$, and $\beta_i\le9/4$, so
\[
 M/R_0\ge2^{t-3}\ge(9/4)t>t\log_2(9/2)\ge\log_2(2B).
\]
The middle inequality starts at $t=7$ and is preserved by doubling;
the strict logarithmic comparison is $(9/2)^4<2^9$.
For $2\le t\le6$, the complete rational enumeration of the multisets
in \eqref{prim:small-table}, using $h$ separately on their seven
multiplicities, gives
\begin{center}
\begin{tabular}{rrrrr}
\toprule $t$&total&$B\le1$&$2^M\ge(2B)^{R_0}$&residual\\\midrule
2&28&6&19&3\\3&84&13&70&1\\4&210&22&187&1\\
5&462&31&431&0\\6&924&33&891&0\\\midrule
Total&1708&105&1598&5\\\bottomrule
\end{tabular}
\end{center}
The complete algorithm, with integer rather than logarithmic
comparisons, is in Subsection~\ref{prim:finite}. Its five residuals are
\begin{equation}\label{prim:five-residuals}
 R^2,\quad R A6,\quad R A7,\quad R^3,\quad R^4.
\end{equation}
We resolve these uniformly in every field, multiplicity and core.

\paragraph{Residual layers of types $L_2(2^a),L_3(2),A_6$.}
It is convenient to prove the assertion for every number of such
components. For $t=2,3,4$ use
\[
 B_2=121/64<2,\quad B_3=3993/512<8,\quad
 B_4=43923/4096<11;
\]
for $t\ge5$ use $B_t=2^{t-1}(11/8)^t$. The support denominator is
at most four in the absence of $A_6$ and at most five otherwise.
Since an $A_6$ factor has degree at least three,
\begin{equation}\label{prim:residual-degree}
 M/r\ge2^{t-2}.
\end{equation}
For $t\ge5$, $2^{t-2}\ge t\log_2(11/4)=\log_2(2B_t)$;
the base is $2^{18}>11^5$, and doubling proves induction. Thus only
$t=2,3,4$ require more argument.

With a nonabelian core, $t=2$ is within the factor-six reserve. For
the other two values use the sharper elementary estimate
\begin{equation}\label{prim:three-n}
 (3+2\log_3n)\log_2n<3n\qquad(n\ge3).
\end{equation}
Here is a proof with explicit constants. Put $y=\log n\ge1$.
The inequalities $\log2>9/13$, $\log3>13/12$, obtained from the
positive series for $2\operatorname{atanh}(1/3)$ and
$2\operatorname{atanh}(1/2)$, bound the left side by
$(13/3)y+(8/3)y^2$. Taylor's series for $e^y$ bounds
$9e^y-13y-8y^2$ below by
$9-4y-(7/2)y^2+(3/2)y^3+(3/8)y^4$.
Writing $y=1+s$ gives
$27/8-5s+(13/4)s^2+3s^3+(3/8)s^4>0$, since the quadratic
has negative discriminant. This proves \eqref{prim:three-n}.
It gives
\[
 \log_2(k(E)/|E|_2)<2f-1+3n+\log_2B_t.
\]
For $t=4$, \eqref{prim:residual-degree} gives
$\log_2(v^{1/r})\ge4fn$; the sufficient comparison
$4fn\ge2f+3n+4$ holds already at $(f,n)=(2,3)$ and increases.
For $t=3,n\ge7$, use $2fn\ge2f+3n+3$, starting at $(2,7)$.
The remaining $n=3,5$ have $Q_o=3,15$; directly,
\[
 k(E)/|E|_2<216f(q-1),\quad v^{1/r}\ge q^6;
 \qquad
 k(E)/|E|_2<3000f(q-1),\quad v^{1/r}\ge q^{10}.
\]
The respective upper ratios
$216\log_2q/q^5+q^{-6}$ and $3000\log_2q/q^9+q^{-10}$ decrease
for $q\ge4$ and are less than one there. The additive unit in the
support criterion is thus retained in all cases.

If $N=1$, the dimension over $\F_2$ is at least $2^{t+1}$ for
$t\ge2$. To justify the field factor, a nontrivial characteristic-two
absolute module for $L_2(2^a)$ is, by Steinberg's tensor product
theorem, a tensor over a nonempty set of binary positions. If the
set has size $b$ and Frobenius period $e$, then $b\ge a/e$ and
$e2^b\ge e2^{a/e}\ge2a$. The Frobenius period of a tensor
constituent is a multiple of each labeled component period. Hence
any such component supplies dimension at least $a2^t\ge2^{t+1}$.
If none occurs, all degrees are at least three, and
$3^t\ge2^{t+1}$. The exact comparisons
\[
 2^{8/5}>B_2+1,\qquad2^{16/5}>B_3+1,\qquad
 2^{32/5}>B_4+1
\]
prove the support criterion. They follow by raising to the fifth power.

It remains to consider an abelian nontrivial core. For $t=2$ use
the factor-two reserve. For $t=4$,
$v^{1/r}\ge q^4>11(q-1)\log_2q+1$ for $q\ge4$;
dividing by $q^4$ proves monotonicity and reduces to four.
For $t=3$ one has $v^{1/r}\ge q^2$. If the field quotient
$e>1$, \eqref{prim:scalar-k} gives $k(N)/|N|_2\le3/2,5,6$ at
$q=4,8,16$, respectively; multiplying by $B_3<8$ and adding one
is still below $q^2$. For $q\ge32$ it gives at most
$q/3+f\sqrt q$, and
$8(q/3+f\sqrt q)+1<q^2$. After division by $q^2$ the terms decrease;
at $32$ use $\sqrt{32}>5$. If $e=1$ and $q\ge8$, simply use
$q^2>8(q-1)+1$. Only $q=4,z=3,e=1$ remains.

If one absolute component degree is at least three, then
$M/r\ge12/5$ and $v^{1/r}\ge4^{12/5}>25>3B_3+1$.
If all three degrees are two, absolute homogeneity over the canonical
$\F_4$ and Steinberg's field-period description force every socle to
be $A_5$. We now handle all the outer groups of this cube.

If $M\lhd J$ has $2$-group quotient $A$ and an automorphism $g$ of
order three preserves $M$, then
\begin{equation}\label{prim:fixed-label}
 |\Irr(J)^g|\le |A|\,|\Irr(M)^g|.
\end{equation}
Indeed each $g$-stable $A$-orbit in $\Irr(M)$ has size a power of
two. Its stabilizer in $A:C_3$ contains a Sylow-three subgroup;
Sylow conjugacy inside $A$ supplies a label fixed by the specified
$g$. Over that orbit there are at most $|A|$ projective irreducible
characters of the inertia quotient. This proves the formula with
the Clifford multiplier retained.

Now $A_5^3\le H\le S_5\wr S_3$. Let $U$ be the image in
$C_2^3:S_3$, $A=U\cap C_2^3$, and $J$ the preimage of $A$.
If the permutation image is a $2$-group, then
$k(H)/|H|_2\le(5/4)^3=125/64<2$. Otherwise first take its subgroup
above $C_3$ if needed. A Sylow-three subgroup of $U$ projects
isomorphically to the cycle group and is conjugate, by the base, to
the pure coordinate cycle. The standard section $C_2\le S_5$ lifts
this conjugation and this element. Since $H$ contains the full
$A_5^3$, it is the full preimage of $U$, so an order-three
pure cycle $g$ belongs to the resulting group. There are five
$g$-fixed labels on $\Irr(A_5^3)$. Thus
$k(J)\le125|A|$ and $|\Irr(J)^g|\le5|A|$.
Cyclic Clifford theory gives
\[
 k(J:C_3)=\frac{k(J)+8|\Irr(J)^g|}{3},\qquad
 \frac{k(H)}{|H|_2}\le\frac{125+40}{3\cdot64}=\frac{55}{64}.
\]
Passing to an index-two extension preserves this normalized upper
bound. The scalar reserve therefore closes every outer subgroup
of $A_5^3$. This completes all residual layers of the three named types.

\paragraph{The remaining $R A7$ pair.}
The types are distinct, so $T=1$ and $B\le99/64<2$. Both nontrivial
core cases are covered by their reserves. If $N=1$, the support
denominator is six. The $A_7$ degree is at least four. If the $R$
factor degree is at least three, $\dim_{\F_2}V\ge12$. Otherwise it
is a degree-two natural Frobenius twist for $L_2(2^a)$, $a\ge2$,
and its period forces $\dim_{\F_2}V\ge2\cdot2\cdot4=16$.
The group $L_3(2)$ has no degree-two projective constituent.
In either case $v^{1/6}\ge4>99/64+1$, proving the bound.

\begin{proof}[Proof of Proposition~\ref{prop:primitive-affine}]
Use Lemma~\ref{prim:field}; all its homogeneity hypotheses follow
from quasiprimitivity. If the layer is trivial,
Lemma~\ref{prim:nilpotent} applies with $X=E$. One component was
settled above. For at least two components the complete simple-group
classification assigns every factor to $G$, $P$, or one of the seven
classes in \eqref{prim:small-table}. The tensor argument handles every
profile containing $G$ or $P$, and every all-small profile of length
at least seven. The complete finite enumeration leaves exactly
\eqref{prim:five-residuals}, all of which have just been proved for
every core and field. Lemma~\ref{prim:profile-transfer} completes
the proof.
\end{proof}

\subsection{The capacity of one simple factor}\label{prim:one-factor}

Recall the universal odd central cover $\widehat S$ and
$\kappa^+(S)$ from \eqref{eq:odd-cover-capacity}. Put
\[
 \eta(S)=|\Out(S)|_{2'}\kappa^+(S).
\]
The maximum with one in this definition will be kept throughout.

\begin{lemma}\label{prim:one-factor-lemma}
If a nonabelian simple group $S$ has a nontrivial absolutely irreducible
projective representation of degree $a$ over $\F_Q$, where $Q=2^f$,
then
\begin{equation}\label{prim:one-factor-bound}
 \eta(S)\le Q^a/4.
\end{equation}
The representation may be a nonfaithful representation of $\widehat S$;
its projective image on $S$ is faithful.
\end{lemma}
\begin{proof}
For every such image $Q^a\ge8$. Indeed $a\ge2$, and the only smaller
possibility is $\operatorname{PGL}_2(2)$, a solvable group.
Let $\mu$ be the order of the odd part of the Schur multiplier and
$o=|\Out(S)|_{2'}$. For any upper bound $B\ge k(S)/|S|_2$, the
central-kernel inequality gives
\begin{equation}\label{prim:multiplier-bound}
 \eta(S)\le o\max(1,\mu B).
\end{equation}

\emph{Alternating and sporadic groups.}
The odd outer part is one for every alternating group. Its odd
multiplier is one except at $A_6,A_7$, where it is three. The class
estimates above therefore give
\[
 \eta(A_n)\le1\ (n\ge8),\qquad
 \eta(A_5)\le5/4,\quad \eta(A_6)\le21/8,\quad \eta(A_7)\le27/8.
\]
For $A_6,A_7$ one has $Q^a\ge16$, since
$|\operatorname{PGL}_3(2)|=168$ is divisible by neither $360$ nor
$2520$. This proves the required bounds. For all 26 sporadic groups
the odd multiplier is at most three and the odd outer part is one;
$M_{11},J_1$ have trivial multiplier. These exact metadata are the
ATLAS values \cite{Atlas85}. The complete class-ratio table above
then gives $\eta(S)\le15/8<8/4$.

\emph{Groups in defining characteristic two.}
Let $S$ be the simple quotient of a quasisimple simply connected
rational-point group $\widetilde S$ in characteristic two, with its
usual defining field of size $q_0=2^t$ and algebraic rank $h$.
The small simple derived groups are treated under their isomorphism
types below. For ordinary rational-point groups, the prime-to-two
part of the multiplier of $\widetilde S$ is trivial
\cite[Theorem, p.~235]{Prasad82}. For Suzuki and Ree groups in
characteristic two the same conclusion follows from the projective
lifting theorem \cite[\S12.4]{Steinberg63}: every class in
$H^2(\widetilde S,\overline{\F}_2^\times)$ has an irreducible
projective realization, so lifting all such representations makes
this group, which is dual to the odd multiplier, trivial.
If $U\to S$ is the universal central cover, the kernel of
$U\to\widetilde S$ is therefore a $2$-group. Since
$Z(\widetilde S)$ has odd order, that kernel is $O_2(Z(U))$.
Hence $\widetilde S\cong\widehat S$.
Let $D$ be the smallest
absolute degree of a nontrivial irreducible defining-characteristic
representation of $\widetilde S$.

The defining-characteristic restriction theorem
\cite[Theorems~7.4, 9.3 and~12.2]{Steinberg63}
realizes each irreducible module as a restriction from the simple
algebraic group. A nontrivial rational representation has finite
kernel and image of rank $h$, contained in $\SL_D$ because a connected
semisimple group has no nontrivial linear character. Thus $D\ge h+1$.
The precise field-sensitive input used here is the \emph{first}
assertion of \cite[Lemma~4.2]{GT05}: a nontrivial irreducible module
$U$ in defining characteristic for this finite group satisfies
$|U|\ge q_0^D$, over any finite field of definition. The later
$D\ge6$ and Suzuki--Ree restrictions in that lemma concern its later
alternatives, not this first inequality. Applied to the projective
module by pullback, it gives, with $n=fa$,
\begin{equation}\label{prim:field-degree}
 tD\le n,\qquad D\le a\le n.
\end{equation}
Write $z=|Z(\widetilde S)|$. The diagonal, field and graph formulas give
\begin{equation}\label{prim:outer-defining}
 z\le h+1\le D,
 \qquad |\Out(S)|_{2'}\le3zt.
\end{equation}
The factor three includes triality and the graph-field action in
${}^3D_4$.

For $S=L_2(q_0)$, $q_0\ge4$, one has
$\kappa^+(S)=1+1/q_0\le5/4$ and $o\le t$. Since $D=2$,
$n\ge2t\ge4$, so
\[
 \eta(S)\le5n/8\le2^n/4.
\]
The last inequality starts at $n=4$ and is preserved by induction.
For rank at least two, apart from $L_3(2)$, the half-class estimate
gives $\kappa^+(S)\le\max(1,z/2)\le D/2$. Consequently, if $n\ge9$,
\[
 \eta(S)\le(3/2)D^2t\le(3/2)n^2\le2^n/4.
\]
The last inequality is $6n^2\le2^n$, true at nine since
$486\le512$, and the ratio increases thereafter. For $L_3(2)$,
$\eta(S)=1$ directly.

It remains to check the entire boundary $n\le8$. Since $D\ge3$,
\eqref{prim:field-degree} forces $t\le2$, so the odd field factor is
one. The odd center has order at most three in this range. In linear
and unitary type it divides $q_0-1$ or $q_0+1$; a center of order
five at $q_0=4$ would force natural degree at least five, hence
$tD\ge10$. Other types have center at most three. Therefore
$\eta(S)\le3z\max(1,z/2)\le27/2\le2^n/4$ for $6\le n\le8$.
If $n\le5$, then $t=1$. A center of order three at $q_0=2$ could
only arise in unitary natural degree divisible by three or in an
$E_6$ type. The group $\operatorname{SU}_3(2)$ is not quasisimple;
unitary degree at least six forces $D\ge6$, and $E_6$ forces
$D\ge7$. Hence $z=1$. At $n=5$, $\eta(S)\le3\le8$; at $n=3,4$,
triality is impossible because $D_4$ has rank four and $D\ge5$, so
$\eta(S)=1\le2^n/4$. This completes all admissible defining-two types.
The Tits group is separate: its multiplier is trivial, its outer
group has order two, and $k(S)=22<2048=|S|_2$, so $\eta(S)=1$.

\emph{Odd-characteristic rank one.}
For $S=L_2(q)$, odd $q\ge11$, the multiplier has no odd part and
$\kappa^+(S)\le(q+5)/8$. The odd outer part is at most the defining
field exponent, hence at most $\log_3q\le q/3$. Every projective
characteristic-two degree is at least $(q-1)/2$. Thus
\[
 \eta(S)\le\frac{q(q+5)}{24}\le\frac{2^{(q-1)/2}}4\le\frac{Q^a}4.
\]
The middle inequality at eleven is $88/3\le32$; its logarithmic
right-to-left derivative is
$(\log2)/2-1/q-1/(q+5)>0$. The smaller simple cases are
$A_5,L_3(2),A_6$ and have already been treated.

\emph{The remaining odd-characteristic Lie groups.}
In this paragraph $q=r^t$ is the \emph{defining} odd field, distinct
from $Q$. The following exact sufficient multiplier and outer bounds
follow from \cite{Steinberg81,Atlas85} and the diagonal, graph and
field formulas:
\begin{center}\small
\begin{tabular}{llll}
\toprule Family & $\mu$ & $o$ & $\mu o$ bound\\\midrule
$L_n(q),U_n(q)$, except $U_4(3)$ & $\le n$ & $\le nt$ & $n^2t$\\
$\operatorname{PSp}_{2h}(q)$ & $1$ & $\le t$ & $t$\\
$\operatorname{P\Omega}_{2h+1}(q)$ & $\le3$ & $\le t$ & $3t$\\
$\operatorname{P\Omega}_{2h}^{\pm}(q)$ & $1$ & $\le3t$ & $3t$\\
Exceptional types & $\le3$ & $\le3t$ & $9t$\\\bottomrule
\end{tabular}
\end{center}
The odd factor three for $O_7(3)$ is necessary: its multiplier is six.
The exception $U_4(3)$ has odd multiplier nine and outer group $D_8$;
$G_2(3)$ has odd multiplier three and outer group of order two.
We retain these exceptions explicitly.

Use the projective degree bounds from the preceding table, with the
weaker $m=q(q-1)$ for both $L_3(q),U_3(q)$, including $U_3(3)$.
For $L_3(q)$ and $U_3(q)$, $B=q^3/16$ and $t\le q/3$ give
$4\eta(S)\le(3/4)q^4$. At $q=3$,
\[
 2^{q(q-1)}=64>243/4,
\]
and the logarithmic ratio has derivative
$(2q-1)\log2-4/q>0$. For $n=4,q\ge5$,
$4\eta(S)\le(800/3)q^4$; its base is $2^{100}>500000/3$ and the
derivative is $(3q^2-2q)\log2-4/q>0$.
At $L_4(3)$, $\eta(S)=1$ by $k(S)\le56$, $|S|_2=128$ and
$\mu=o=1$. At $U_4(3)$,
$\eta(S)\le\max(1,9\cdot20/128)=45/32<2\le Q^a/4$.
For $n\ge5$,
\begin{equation}\label{prim:capacity-linear}
 4\eta(S)\le \frac{400n^2q^n}{3\cdot2^{n-1}},\qquad
 m=(q-1)q^{n-2}.
\end{equation}
At $(n,q)=(5,3)$ this compares $2^{54}$ with $50625$.
The logarithm of the right-to-left ratio decreases on the full
rectangle: for the logarithm of $2^m$ divided by the right side,
the $q$-derivative multiplied by $q$ is
$\log2\{(n-2)m+q^{n-1}\}-n>0$, and the $n$-derivative is
$\log2\,m\log q-\log(q/2)-2/n>0$.

For the symplectic groups put
$m=(q^h-1)/2$ and $B=100q^h/2^{3h-1}$. Retaining the maximum
with one, the two required bounds are
\[
 4t\le4q/3\le2^m,\qquad
 4tB\le\frac{400q^{h+1}}{3\cdot2^{3h-1}}\le2^m.
\]
The group $\operatorname{PSp}_4(3)=U_4(2)$ is already covered.
All the rest lie in the rectangles $h\ge2,q\ge5$ or $h\ge3,q\ge3$.
At their bases the second right sides are $3125/6$ and $675/16$,
and the exponents are $12,13$. For the logarithmic ratio,
$q\partial_q=\log2\,h q^h/2-(h+1)>0$ and
$\partial_h=(\log2\,q^h/2-1)\log q+3\log2>0$.

For odd-orthogonal groups, $h\ge3,q\ge3$,
\[
 m=q^{2h-3},\qquad
 4\eta(S)\le\max\{4q,400q^{h+1}/2^{3h}\}.
\]
The first term is immediate from $m\ge q^3$. For the second the
base is $2^{27}>2025/32$; the two logarithmic derivatives are
$\log2(2h-3)m-(h+1)>0$ after multiplying the $q$-derivative by
$q$, and $(2\log2\,m-1)\log q+3\log2>0$ in $h$.
For even-orthogonal groups use $m=q^{2h-3}-2$ and
$4\eta(S)\le100q^{h+1}$. The base is $2^{241}>24300$ at $(4,3)$.
The derivatives are
$\log2(2h-3)q^{2h-3}-(h+1)>0$ and
$(2\log2\,q^{2h-3}-1)\log q>0$.

For exceptional groups, the class bound $B=25q^h$ and
$\mu o\le9t\le3q$ give $4\eta(S)\le300q^{h+1}$. The following
three rows exhaust the generic parameters:
\begin{center}\small
\begin{tabular}{lllll}
\toprule Family & range & $m$ & rank used & base right side\\\midrule
$G_2(q)$&$q\ge5$&$q^3-1$&2&$37500$\\
${}^3D_4,F_4,E_6,{}^2E_6,E_7,E_8$&$q\ge3$&$(q^5-1)/2$&8&$5904900$\\
${}^2G_2(q)$&$q\ge27$&$q(q-1)$&2&$5904900$\\\bottomrule
\end{tabular}
\end{center}
Their base exponents are $124,121,702$. The logarithmic derivative,
multiplied by $q$, is $\log2\,q m'(q)-(h+1)>0$ at the base and
increases thereafter. The common degree in the middle row is weaker
than each degree in Lee's table; for ${}^3D_4$ use
$q^5-q^3+q-1\ge(q^5-1)/2$. Finally at $G_2(3)$,
$\eta(S)\le69/64<2$, from $(\mu,o,k(S),|S|_2)=(3,1,23,64)$.
The small Ree derived group is $L_2(8)$, already covered. All
nonabelian finite simple groups have now been included.
\end{proof}

\subsection{Assembly of the local capacity}

\begin{lemma}\label{prim:fitting-capacity}
In Lemma~\ref{prim:field}, retain
$Q_o=\prod_j|\Sp_{2a_j}(r_j)|_{2'}$ and $n$ as the extraspecial
degree. If $n\ge3$, then
\begin{equation}\label{prim:fitting-eight}
 f(Q-1)n^2Q_o\le Q^{2n}/8,\qquad Q=2^f.
\end{equation}
For every $n\ge1$ the same assertion with denominator two holds,
where $Q_o=1$ if $n=1$.
\end{lemma}
\begin{proof}
For $n=3$, $Q_o=3$, and it is enough that
$8fn^2Q_o\le Q^{2n-1}$. At $f=2$ this is $432\le1024$;
the ratio $2^{5f}/f$ increases thereafter. For $n\ge5$,
\eqref{prim:symplectic-order} gives
$Q_o\le n^{2\log_3n+1}$, so it suffices to prove
\[
 8f n^{2\log_3n+3}\le2^{f(2n-1)}.
\]
At $(n,f)=(5,2)$, $\log_3 5<3/2$ gives left side
$<16\cdot5^6=250000<4^9=262144$. The logarithm of the
right-to-left ratio has derivatives
\[
 (2n-1)\log2-1/f>0,\qquad
 2f\log2-\frac{4\log_3n+3}{n}>0
\]
on $n\ge5,f\ge2$. The subtracted function in the second derivative
decreases; at five it is below $9/5$, whereas $4\log2>2$.
This proves the whole rectangle. If $n=1$, use $2f\le2^f=Q$
to obtain $f(Q-1)<Q^2/2$.
\end{proof}

\begin{proof}[Proof of Proposition~\ref{prop:local-capacity}]
Let $F^*(E)=RL$, with $R=F(E)$ and components $L_1,\ldots,L_t$.
Restrictions to characteristic subgroups of $E$ are homogeneous because
those subgroups are normal in $B$; also $O_2(E)=1$, as in
Lemma~\ref{prim:field}. That lemma supplies the normalized
field $K=\F_Q$, an extraspecial degree $n$, and absolute component
degrees $a_1,\ldots,a_t\ge2$. Put
$A=\prod_i a_i$ and $D=\sum_i a_i$. The module has
\begin{equation}\label{prim:local-size}
 |W|\ge Q^{nA},\qquad |R|\le(Q-1)n^2.
\end{equation}
Its homogeneous multiplicity can only increase the first lower bound.

Remove first the field action on $K$, then the component
permutation action. The remaining image acts on each extraspecial
symplectic quotient and each simple component. An element trivial on
all of these acts by inner automorphisms on $R$ and on the components.
Correct by those inner automorphisms; the remaining element centralizes
$F^*(E)$ and hence lies in it. This proves the kernel assertion in
\begin{equation}\label{prim:local-outer}
 |E:F^*(E)|_{2'}\le
 f_{2'}Q_o(t!)_{2'}\prod_{i=1}^t|\Out(S_i)|_{2'}.
\end{equation}
The component-permutation image is a subgroup of $\operatorname{Sym}(t)$,
so its order divides $t!$; the bound uses the successive kernels of
these actions.

If $t=0$ and $R$ is nonabelian, then $n\ge3$ and
\eqref{prim:local-outer}, \eqref{prim:local-size}, and
Lemma~\ref{prim:fitting-capacity} give $c(E)\le|W|^2/8$.
Suppose $t\ge1$. By Lemma~\ref{prim:one-factor-lemma},
\[
 c(E)\le f(Q-1)n^2Q_o\,t!\prod_i \eta(S_i)
 \le\frac{Q^{2n}}2\,t!\frac{Q^D}{4^t}.
\]
It remains to prove
\begin{equation}\label{prim:tensor-factorial}
 4t!\le4^t Q^{2n(A-1)-D}.
\end{equation}
The exponent is minimized by $n=1$ and every $a_i=2$:
increasing $a_i$ increases $2A-D$, and increasing $n$ increases it.
Its minimum is $2^{t+1}-2-2t\ge0$. Since $Q\ge2$, the desired
inequality follows from
$t!\le2^{2^{t+1}-4}$. Equality holds at $t=1$; the ratio on the
right from $t$ to $t+1$ is $2^{2^{t+1}}\ge t+1$, proving induction.
Thus $c(E)\le Q^{2nA}/8\le|W|^2/8$. All central characters and
multiplicities are those of the stated tensor factors.
\end{proof}

\subsection{Finite data and complete verification procedures}\label{prim:finite}

The finite checks concern the subgroup lists and bounded numerical
ranges specified below. The preceding monotonicity and tensor
inequalities cover the unbounded ranges. Character-table names in
the code are the identifiers of the GAP Character Table Library
\cite{CTblLib}; subgroup enumeration uses GAP \cite{GAP}. The group
orders and the numbers of subgroup classes are included as checks
on the groups being enumerated.

\paragraph{The symplectic finite inputs.}
The complete enumeration of subgroup conjugacy classes gives
\begingroup\small\setlength{\arraycolsep}{3pt}
\begin{equation}\label{prim:finite-sp4}
\begin{array}{l|r|r|c|c}
G&\text{subgroup classes}&\text{nonsolvable classes}&
 \max_{H\text{ nonsolvable}} k(H)/|H|_2&
 \max_{H\le G} k(H)/|H|_2\\\hline
\Sp_4(3)&162&8&9/8&27\\
\operatorname{PSp}_4(3)&116&8&5/4&\text{not needed}\\
\GL_3(3)&154&2&3/4&\text{not needed}
\end{array}
\end{equation}
\endgroup
The maxima in the nonsolvable column occur at groups isomorphic to
$\SL_2(5),A_5,L_3(3)$, respectively. The maximum 27 is attained by
an abelian subgroup of order 27. The following exact-arithmetic
algorithm enumerates every subgroup conjugacy class:
\begin{quote}\small
\begin{verbatim}
Part2 := n -> 2^Length(Filtered(FactorsInt(n),x->x=2));;
VecPerm := function(G,d)
  local vs,phi;
  vs:=Filtered(Elements(GF(3)^d),v->not IsZero(v));
  phi:=ActionHomomorphism(G,vs,OnRight);
  return Image(phi,G);
end;;
TestSubgroups := function(G,num,nonnum,bnd,allbnd)
  local cs,c,H,rat,best,allbest,non;
  cs:=ConjugacyClassesSubgroups(G);
  Assert(0,Length(cs)=num);
  best:=0; allbest:=0; non:=0;
  for c in cs do
    H:=Representative(c);
    rat:=NrConjugacyClasses(H)/Part2(Size(H));
    allbest:=Maximum(allbest,rat);
    if not IsSolvableGroup(H) then
      non:=non+1; best:=Maximum(best,rat);
    fi;
  od;
  Assert(0,non=nonnum and best=bnd);
  if allbnd<>fail then Assert(0,allbest=allbnd); fi;
end;;
G4:=VecPerm(Sp(4,3),4);;
Assert(0,Size(G4)=51840);;
TestSubgroups(G4,162,8,9/8,27);
P4:=FactorGroup(G4,Centre(G4));;
Assert(0,Size(P4)=25920);;
TestSubgroups(P4,116,8,5/4,fail);
G3:=VecPerm(GL(3,3),3);;
Assert(0,Size(G3)=11232);;
TestSubgroups(G3,154,2,3/4,fail);
\end{verbatim}
\end{quote}
The finite ambient group $\Sp_6(3)$ has eleven conjugacy classes of
maximal subgroups. The complete data needed are
\begin{center}\small
\begin{tabular}{rrrrl}
\toprule $|M|$&$k(M)$&$|\operatorname{Rad}(M)|$&
$|M/\operatorname{Rad}(M)|$&quotient or status\\\midrule
25194240&252&972&25920&$\operatorname{PSp}_4(3)$\\
2519424&264&2519424&1&solvable\\
8188128&124&1458&5616&$L_3(3)$\\
1244160&238&48&25920&$\operatorname{PSp}_4(3)$\\
82944&140&82944&1&solvable\\
22464&30&2&11232&\\
58968&29&2&29484&\\
48384&46&4&12096&\\
120&9&2&60&\\
2184&17&2&1092&\\
2184&17&2&1092&\\\bottomrule
\end{tabular}
\end{center}
For reproducibility, obtain the list by
\texttt{MaximalSubgroupClassReps(Sp(6,3))}; for each entry compute
\texttt{Size}, \texttt{NrConjugacyClasses}, and
\texttt{SolvableRadical}. Assert that the sorted triples are exactly
those in this table. For each of the three named nonsolvable radical
quotients, form \texttt{FactorGroup(M,SolvableRadical(M))} and test
\texttt{IsomorphismGroups} against \texttt{P4} above or
\texttt{VecPerm(SL(3,3),3)}, respectively. The latter group has
order $5616$, which is asserted before testing. The quotient
identifications are checked by explicit isomorphisms.

Here is why these finite data imply the all-subgroup bound needed in
Lemma~\ref{prim:nilpotent}:
\begin{equation}\label{prim:sp6-ratio}
 H\le\Sp_6(3),\ H\text{ nonsolvable}
 \quad\Longrightarrow\quad k(H)/|H|_2\le2187/4.
\end{equation}
For the full group use $k=141$ and two-part $1024$. Otherwise choose
a maximal overgroup $M$. In the first, third and fourth rows put
$K=H\cap\operatorname{Rad}(M)$. The quotient $H/K$ is nonsolvable. Since
$k(K)/|K|_2\le |\operatorname{Rad}(M)|_{2'}$, Nagao and
\eqref{prim:finite-sp4} give the respective bounds
$243(5/4)$, $729(3/4)$, and $3(5/4)$ for $k(H)/|H|_2$.
For the remaining nonsolvable rows,
\[
 k(H)\le\min\{|H|,[M:H]k(M)\}\le\sqrt{|M|k(M)}.
\]
A nonsolvable group has two-part at least four: groups of odd order
are solvable, and a group with Sylow two-subgroup of order two has
a normal odd complement. Each remaining product $|M|k(M)$ is less
than $2187^2$. Together these maximal-overgroup estimates prove
\eqref{prim:sp6-ratio} for every subgroup.

\paragraph{Maximal overgroups for the small capacities.}
The complete maximal-subgroup pairs $(|M|,k(M))$ for $A_7$ are
\[
 (72,9),\ (120,7),\ (360,7),\ (168,6),\ (168,6).
\]
For $J_1$ they are
\[
 (660,8),\ (168,8),\ (120,10),\ (114,9),\ (110,11),\
 (60,12),\ (42,7).
\]
Their respective isomorphism types are
$L_2(11)$, $2^3:7:3$, $2\times A_5$, $19:6$, $11:10$,
$D_6\times D_{10}$ and $7:6$. Formula
\eqref{prim:capacity-overgroups} gives $1$ and $4/133$.
For the two odd rank-one pairs the complete overgroup data are as
follows. Repeated conjugacy classes with the same pair are immaterial
to the maximum, but all subgroup types are covered.
\begin{center}\small
\begin{tabular}{lp{9.6cm}}
\toprule $H$&allowed maximal-overgroup pairs $(|M|,k(M))$\\\midrule
$L_2(11)$&$(60,5)$ twice, $(55,7)$, $(12,6)$\\
$\operatorname{PGL}_2(11)$&$(110,11)$, $(24,5)$, $(24,9)$,
$(20,8)$, and the preceding proper socle overgroups\\
$L_2(13)$&$(78,8)$, $(14,5)$, $(12,6)$, $(12,4)$\\
$\operatorname{PGL}_2(13)$&$(156,13)$, $(28,10)$, $(24,9)$,
$(24,5)$, and the preceding proper socle overgroups\\\bottomrule
\end{tabular}
\end{center}
Taking $|M|k(M)/|H|$ gives $7/12,11/12,4/7,13/14$ in order.
Here $[H:S]=2$ for each projective general linear group, so a
proper maximal overgroup containing $S$ is $S$ itself. Subgroups
inside it and avoiding $S$ are covered by the listed maximal
subgroups of $S$.

These lists can be reproduced using
\texttt{MaximalSubgroupClassReps(AlternatingGroup(7))} and the complete
\texttt{Maxes(CharacterTable("J1"))} attribute. The latter attribute
means the complete list of maximal-subgroup conjugacy classes in the
Character Table Library. Missing data must cause an error, not an
empty list. For $q=11,13$, construct the projective action on the
$q+1$ lines directly. For example:
\begin{quote}\small
\begin{verbatim}
ProjectivePair:=function(q)
  local F,pts,G,phi,S,H;
  F:=GF(q);
  pts:=Concatenation(List(Elements(F),x->[One(F),x]),
                    [[Zero(F),One(F)]]);
  G:=GL(2,q); phi:=ActionHomomorphism(G,pts,OnLines);
  H:=Image(phi,G); S:=Image(phi,SL(2,q));
  Assert(0,Size(H)=q*(q^2-1));
  Assert(0,Size(S)=q*(q^2-1)/2);
  return [H,S];
end;;
for q in [11,13] do
  pair:=ProjectivePair(q); H:=pair[1]; S:=pair[2];
  over:=Filtered(MaximalSubgroupClassReps(H),
                 M->not IsSubgroup(M,S));
  Append(over,MaximalSubgroupClassReps(S));
  cap:=Maximum(List(over,M->
       Size(M)*NrConjugacyClasses(M)/Size(H)));
  Assert(0,cap=([11/12,13/14][Position([11,13],q)]));
od;
\end{verbatim}
\end{quote}
For $L_2(8).3$, use the lines over $\F_8$, the projective image of
$\SL_2(8)$ of order $504$, and adjoin the permutation squaring all
line coordinates. Assert the resulting order $1512$ and the normality
of the socle. Its four maximal pairs and the three proper socle
maximal pairs are exactly those displayed in the preceding subsection.
This construction checks both the group and the required class-covering
list, including the subgroups lying inside the socle.

\paragraph{Ordinary tables and small projective degrees.}
All ordinary class counts in the sporadic and small Lie tables are
obtained by applying \texttt{NrConjugacyClasses} to
\texttt{CharacterTable(name)}, with order obtained by \texttt{Size}.
The sporadic names are
\begin{quote}\small
\begin{verbatim}
["M11","M12","M22","M23","M24","J1","J2","J3","J4",
 "Co1","Co2","Co3","Fi22","Fi23","Fi24'","HS","McL",
 "He","Ru","Suz","ON","HN","Ly","Th","B","M"]
\end{verbatim}
\end{quote}
The groups with outer order two are exactly
\[
 M_{12},\ M_{22},\ J_2,\ J_3,\ Fi_{22},\ Fi_{24}',\
 HS,\ McL,\ He,\ Suz,\ O'N,\ HN.
\]
Append \texttt{".2"} to each of these identifiers and require the
table to exist. For every one, its normalized ratio is at most that
of its simple socle. The small odd Lie table uses identifiers
\texttt{"U4(2)"}, \texttt{"S4(5)"}, \texttt{"S4(7)"},
\texttt{"S6(3)"}, \texttt{"G2(3)"}, \texttt{"L2(11)"}, and
\texttt{"L2(13)"}, together with their \texttt{".2"} extensions.
The characteristic-two $A_7$ degrees used above are reproduced by
the following commands, with \texttt{name} equal to \texttt{"A7"}
or \texttt{"3.A7"}:
\begin{quote}\small
\begin{verbatim}
ct := BrauerTable(CharacterTable(name),2);;
List(Irr(ct),x->x[1]);
\end{verbatim}
\end{quote}
These are complete Brauer
character tables, including the nontrivial odd central characters.

\paragraph{Exact profile and partition verification.}
The following Python program contains the entire finite seven-profile
calculation, the 24 support-table endpoints, and the partition
boundary. It uses only integers and rational numbers. Together with
the displayed monotonicity proofs, these assertions verify the
remaining finite inequalities in the tensor argument.
\begin{quote}\footnotesize
\begin{verbatim}
from fractions import Fraction as F
from itertools import combinations_with_replacement
from collections import Counter
from math import prod
P = [("R",2,4,F(11,8)), ("A6",3,5,F(7,8)),
     ("A7",4,6,F(9,8)), ("J1",10,8,F(15,8)),
     ("M11",4,8,F(5,8)), ("L11",5,3,F(2)),
     ("L13",6,4,F(9,4))]
def h(n):
    return 1 if n<=2 else 3 if n<=4 else 2**(n-1)
counts=Counter(); residual=[]; bylength={}
for t in range(2,7):
    bylength[t]=Counter()
    for ids in combinations_with_replacement(range(7),t):
        A=[P[i] for i in ids]
        B=prod(p[3] for p in A)*prod(h(v)
                                  for v in Counter(ids).values())
        M=prod(p[1] for p in A); R=max(4,*(p[2] for p in A))
        s="budget" if B<=1 else "support" if F(2**M)>=(2*B)**R \
          else "residual"
        counts[s]+=1; bylength[t][s]+=1
        if s=="residual": residual.append([p[0] for p in A])
assert counts=={"budget":105,"support":1598,"residual":5}
assert residual==[["R","R"],["R","A6"],["R","A7"],
                  ["R","R","R"],["R","R","R","R"]]
rows=[(18,4,F(39,4)),(8,3,F(19,16)),(9,3,F(3)),
 (11,3,F(25,8)),(12,4,F(27,8)),(13,3,F(29,4)),
 (14,3,F(31,4)),(15,3,F(33,32)),(11,4,F(27,16)),
 (20,3,F(125,16)),(100,6,F(3125,2)),(54,5,F(2025,4)),
 (40,5,F(2025,16)),(62,6,F(3125,64)),(40,8,F(2025,512)),
 (27,7,F(675,128)),(241,8,F(2025)),(124,5,F(625)),
 (121,7,F(2025)),(243,8,F(2025)),(243,9,F(18225)),
 (243,10,F(54675)),(243,11,F(164025)),(702,3,F(18225))]
for m,r,B in rows:
    assert F(2**m)>=(2*B)**r
assert F(2**12)>=(2*F(17,32))**5
parts=[0]*64; parts[0]=1
for j in range(1,64):
    for n in range(j,64): parts[n]+=parts[n-j]
for n in range(8,64):
    assert parts[n]<=2**(n-n.bit_count()-2)
assert min(F(2**(n-n.bit_count()-2),parts[n])
           for n in range(8,64))==F(32,30)
assert 37**3<3**10 and 4096*37**26<16**37
for n,q,Qo in [(9,16,405),(15,16,45),(21,16,63),
               (25,16,73125),(27,4,8955765),
               (33,16,495),(35,16,315)]:
    f=q.bit_length()-1
    assert (100*f*n*Qo)**3<q**(2*n)
for d,B in [(8,F(121,64)),(16,F(3993,512)),
            (32,F(43923,4096))]:
    assert F(2**d)>(B+1)**5
print(dict(counts),dict(bylength),residual)
\end{verbatim}
\end{quote}
The comparisons use exact integer and rational arithmetic. The support
tests retain their denominators and the additive unit in
\eqref{prim:threshold}.


\section{The small semilinear fields}\label{sec:small-semilinear}

The three scalar fields left by the capacity estimate are treated by
ordinary character counts. We prove a slightly more general statement
for odd prime field degree. The prime denoted by $r$ in this section
is a field-automorphism prime; the vector-space characteristic remains
two. We use
$\Gamma L_1(q)=\F_q^\times\rtimes\operatorname{Gal}(\F_q/\F_2)$
in its natural action on the additive group of $\F_q$.

\begin{proposition}\label{prop:small-semilinear}
Let $q=4$ or $q=2^r$ for an odd prime $r$, let $V=\F_q^n$, and let
$H\leq\Gamma L_1(q)\wr S_n$ act naturally on $V$. Then
\begin{equation}\label{eq:small-semilinear}
                    k(V\sd H)\leq q^n|H|_2.
\end{equation}
In particular this holds for $q=4,8,32$, for every scalar code,
field code and permutation image, including nonsplit lifts of the
permutation image.
\end{proposition}

\subsection{Scalar codes, norms, and shortening}

Put $m=q-1$ and identify the scalar torus with $(C_m)^n$.
For $A\leq(C_m)^n$, write $M_A=\F_q^n\sd A$. A coordinate
permutation $\pi$ normalizing $A$ has cycles of lengths
$\ell_1,\ldots,\ell_c$. A cycle is called good when
$\gcd(\ell_i,m)=1$, and bad otherwise.

\begin{lemma}\label{lem:scalar-code-trace}
Let $z\in(C_m)^n$ and let $z\pi$ normalize $A$. If $g,b$ are the
numbers of good and bad cycles of $\pi$, then
\begin{equation}\label{eq:scalar-code-trace}
 |\Irr(M_A)^{z\pi}|\leq q^g[2(q-1)]^b
                    \leq q^{n-s(\pi)/2}.
\end{equation}
The first inequality holds for every subgroup $A\leq(C_m)^n$.
For one cycle of length $\ell$ its bound is $q$ if
$\gcd(\ell,m)=1$, and $2m$ otherwise.
\end{lemma}
\begin{proof}
Since every scalar has odd order,
$V=C_V(s)\oplus[V,s]$ for $s\in A$. Conjugation by $V$ removes
the second summand. The ordinary classes of $M_A$ are therefore
the $A$-orbits of
\[
       Y_A=\{(v,s):s\in A, v\in V, sv=v\}.
\]
This parametrization is equivariant under every linear normalizer
of the displayed scalar group. The number of fixed ordinary
characters equals the number of fixed ordinary classes: the character
table is an invertible intertwining matrix for their two permutation
actions. We use this equality only for the trace of one automorphism.
The fixed-orbit form of Burnside's lemma consequently gives
\begin{equation}\label{eq:scalar-burnside}
       |\Irr(M_A)^{z\pi}|=
              \frac1{|A|}\sum_{a\in A}|Y_A^{az\pi}|.
\end{equation}

The fixed scalar labels form $S=C_A(\pi)$, identified with a
subgroup of $(C_m)^c$ by their constant values on cycles. Let
$R=N_\pi A\leq(C_m)^c$ be the image of the cycle-product map,
and let $u$ be the vector of cycle products of $z$. All norm fibres
have the same size. On cycle $i$ a simultaneous fixed vector has
$q$ possibilities if both its scalar label $s_i$ and cycle product
$w_i$ are one, and exactly one possibility otherwise. Thus
\begin{equation}\label{eq:scalar-norm-formula}
 |\Irr(M_A)^{z\pi}|=
   \frac1{|R|}\sum_{w\in uR}\sum_{s\in S}
                     \prod_{i=1}^c(1+m1_{\{w_i=s_i=1\}}).
\end{equation}
The fixed and norm codes satisfy the essential relation
\begin{equation}\label{eq:scalar-norm-relation}
                   (s_i^{\ell_i})_{i=1}^c\in R
                            \qquad(s\in S).
\end{equation}

For subgroups $P,Q\leq(C_m)^j$ and cosets $xP,yQ$, define
\[
 Z(xP,yQ)=\sum_{w\in xP,\ s\in yQ}
                         \prod_i(1+m1_{\{w_i=s_i=1\}}).
\]
Expanding over subsets $J$ of coordinates gives positive terms
$m^{|J|}|\{w\in xP:w_J=1\}|\,|\{s\in yQ:s_J=1\}|$.
Each cardinality is zero or the corresponding kernel size. Therefore
\begin{equation}\label{eq:scalar-coset-maximum}
                 Z(xP,yQ)\leq Z(P,Q).
\end{equation}
If $Q\leq P$, then
\begin{equation}\label{eq:scalar-nested-bound}
          Z(P,Q)\leq Z(P,P)=|P|k(\F_q^j\sd P)\leq|P|q^j.
\end{equation}
Here the last inequality is elementary. Project $P$ to one coordinate,
with cyclic image of order $d\mid m$, and use the actual kernel on
the other coordinates. The corresponding affine quotient has
$d+m/d\leq m+1=q$ classes. Induction and
Lemma~\ref{lem:class-counts} give $k(\F_q^j\sd P)\leq q^j$.
This proof also covers a trivial projection or a proper subdirect $P$.

By \eqref{eq:scalar-coset-maximum} the norm coset in
\eqref{eq:scalar-norm-formula} may first be replaced by $R$.
On the good coordinates apply the automorphism $s_i\mapsto
s_i^{\ell_i}$ to $S$, leaving the bad coordinates unchanged;
call the resulting code $S'$. The weights are unchanged. Let
$R_0,S'_0$ be their shortened codes on the good coordinates, that
is, kernels of projection onto the bad coordinates. Relation
\eqref{eq:scalar-norm-relation} gives $S'_0\leq R_0$.
Fix any projected bad-coordinate values. The remaining good-coordinate
fibres are cosets of $R_0,S'_0$, so
\eqref{eq:scalar-coset-maximum}--\eqref{eq:scalar-nested-bound}
bound their weighted sum by $|R_0|q^g$. Division by
$|R|=|R_0||\pi_{\rm bad}(R)|$ leaves
\[
 \frac{q^g}{|\pi_{\rm bad}(R)|}
 \sum_{w\in\pi_{\rm bad}(R),\ s\in\pi_{\rm bad}(S')}
                     \prod_{i\text{ bad}}(1+m1_{\{w_i=s_i=1\}}).
\]
For each $w$, enlarge the $s$-domain to the whole bad-coordinate
torus and replace $w$ by one. Each such coordinate costs
$q+(m-1)=2m$. This proves the first inequality of
\eqref{eq:scalar-code-trace}.

A bad cycle has length at least three because $m$ is odd, and
$2(q-1)\leq q^{3/2}$ for $q\geq4$. A nontrivial good cycle of
length $\ell$ costs $q\leq q^{\ell/2}$; a fixed coordinate costs
$q$. Multiplying gives the second bound. For $q=2$ the scalar group
is trivial and the same assertion follows directly.
\end{proof}

The coset in \eqref{eq:scalar-norm-formula} accounts for lifts with
nontrivial scalar parts. Bad cycles require the larger bound: at
$q=4$, the diagonal $C_3\leq C_3^3$ on a three-cycle has
$4+1+1=6$ fixed classes. The shortening argument includes both features.

\subsection{The field of four elements}

\begin{proof}[Proof of Proposition~\ref{prop:small-semilinear} for $q=4$]
Let $A=H\cap(C_3)^n$, $Q=H/A$, and $X=\Irr(V\sd A)$.
Put $\Omega=\{1,\ldots,n\}$ and recall $t=2^{-1/2}$.
The quotient $Q$ embeds in $C_2\wr S_n$. Let $T$ be its
image on $\Omega$ and $R$ its kernel; then $R$ is a normal $2$-group.
For every odd $a\in Q$ we claim
\begin{equation}\label{eq:q4-fixed-trace}
                         |X^a|\leq4^n2^{-s_\Omega(a)}.
\end{equation}
Choose a lift $g\in H$. The group $\langle A,g\rangle$ has
order $|A|\,|\langle a\rangle|$, which is odd, so $g$ is odd.
Every coordinate cycle of $g$ has odd length, and the product of
its field labels in $C_2$ is trivial, including on a fixed coordinate.
Successive diagonal field gauges around the cycle therefore conjugate
$g$ to a scalar times a pure permutation. The consistency condition
for the gauges is precisely that trivial product. This conjugation
transports $A$ to another scalar code. Applying
Lemma~\ref{lem:scalar-code-trace} to that code proves
\eqref{eq:q4-fixed-trace}; it also gives $|X|\leq4^n$ at the identity.

In the commuting formula \eqref{eq:lambda-commuting}, a common fixed
set is contained in $X^a$ and in $X^{b_o}$. Its size is at most their
geometric mean. Hence
\[
 \Lambda_2(Q,X)\leq4^n\,\mathcal P(Q;t^{s_\Omega}).
\]
The weight factors through $T=Q/R$, so
Lemma~\ref{lem:permutation-two-kernel} and
Theorem~\ref{thm:permutation-weight} give
$\mathcal P(Q;t^{s_\Omega})=\mathcal P(T,\Omega)\leq1$.
Lemma~\ref{lem:ordinary-spectrum}, applied to the extension
of $V\sd A$ by $Q$, yields
\[
       k(V\sd H)\leq|Q|_2\Lambda_2(Q,X)
                       \leq4^n|H|_2,
\]
since $A$ is odd. Thus the possible field $2$-kernel is retained
through the precise quotient identity.
\end{proof}

\subsection{Odd prime field degree and equivariant extensions}

Now let $q=2^r$ with $r$ an odd prime and $m=q-1$. Fermat's
theorem gives $m\equiv1\pmod r$, so $(r,2m)=1$.

\begin{lemma}\label{lem:semilinear-one-cycle}
Let $W=\F_q^\ell$, let $K\leq(C_m:C_r)^\ell$, and let $g$
be a semilinear monomial map normalizing $K$ whose coordinate
permutation is one $\ell$-cycle. Put
\[
 c_\ell=\begin{cases}q,&\gcd(\ell,m)=1,\\2m,&\gcd(\ell,m)>1.
          \end{cases}
\]
Then
\begin{equation}\label{eq:semilinear-one-cycle}
      |\Irr(W\sd K)^g|\leq c_\ell+2(r-1)\leq2(q+r-2).
\end{equation}
\end{lemma}
\begin{proof}
Put $A=K\cap(C_m)^\ell$. Its field image is a subgroup
$B\leq(C_r)^\ell$. Schur--Zassenhaus provides a complement to
$A$ in $K$. In the ambient inverse image $(C_m)^\ell:B$, all
complements are conjugate by the scalar torus. Thus, after a scalar
conjugation, we may write $K=A\sd B$ with $B$ a group
of pure field maps. Scalar conjugation does not alter $A$.
Since $g$ normalizes $K$, conjugacy of complements within $K$
allows multiplication of $g$ by an element of $A$ so that it
normalizes $B$. This inner change does not change its action on
$\Irr(W\sd K)$. It still has the same coordinate cycle.
Conjugation by $g$ on $B$ permutes its field exponents by this cycle.
Consequently
\begin{equation}\label{eq:field-fixed-code}
          D=C_B(g)\leq\text{the diagonal }C_r,
                          \qquad |D|\in\{1,r\}.
\end{equation}

Let $M=W\sd A$ and $X=\Irr(M)$. The little-group parametrization
over the abelian normal group $W$ shows that every $\theta\in X$
has degree $|A:A_\lambda|$ for some $\lambda\in\Irr(W)$: the
scalar stabilizer is abelian. Thus $r\nmid\theta(1)$. We now
justify the required extensions and their equivariance.

The obstruction to extending $\theta$ to $M\sd B_\theta$ is a
class $[\alpha]\in H^2(B_\theta,\C^*)$. It satisfies
$|B_\theta|[\alpha]=0$. If $T_b$ are
intertwining matrices, taking determinants in
$T_bT_c=\alpha(b,c)T_{bc}$ shows that the obstruction is killed
by $\theta(1)$, since
\[
 \alpha(b,c)^{\theta(1)}
       =\frac{\det T_b\det T_c}{\det T_{bc}}.
\]
These two coprime annihilators force it to vanish.
Thus $\theta$ extends. Its extensions differ by the linear characters
of the abelian $r$-group $B_\theta$. There is a unique extension
whose determinant restricted to this specified complement is trivial:
twisting changes that determinant by the $\theta(1)$st power, an
automorphism of its character group. This choice is preserved by
every automorphism normalizing $B$.

It follows that $\Irr(M\sd B)$ is parametrized equivariantly by
$B$-orbits of pairs $(\theta,\eta)$ with
$\theta\in X$, $\eta\in\Irr(B_\theta)$. A $g$-stable orbit of
$\theta$ has $|B:B_\theta|$ members, and for each member exactly
$|B_\theta|$ choices of $b\in B$ make $bg$ fix it. Since $B$ is
abelian, $bg$ and $g$ induce the same automorphism on $B_\theta$.
Its fixed linear-character count is $|C_{B_\theta}(g)|$, by
duality for finite abelian groups. Expanding this number over
$d\in D$ gives the exact identity
\begin{equation}\label{eq:field-code-joint-trace}
 |\Irr(M\sd B)^g|=
             \frac1{|B|}\sum_{b\in B}\sum_{d\in D}|X^{bg,d}|.
\end{equation}
The equivariant extension parametrization therefore gives the common
fixed-character count in \eqref{eq:field-code-joint-trace}.

We compute first the terms with $d\ne1$. Such a $d$ is the same
nontrivial Frobenius power on every coordinate. Its fixed field is
$\F_2$, so $C_A(d)=1$ and $d-1$ is an automorphism of $A$ and
of every $d$-stable subgroup and quotient of $A$.
Use the explicit little-group labels for $X$: they are the
$A$-orbits of pairs $(\lambda,\psi)$ with
$\lambda\in\Irr(W)$ and $\psi\in\Irr(A_\lambda)$. The extension
of $\lambda$ is the canonical character
$(v,a)\mapsto\lambda(v)$ on $W\sd A_\lambda$, so this
parametrization is equivariant under the displayed semilinear maps.
Each $d$-stable orbit on $\Irr(W)$ has a unique $d$-fixed member.
Indeed $A$ is abelian, so every point of the orbit has the same
stabilizer. Its $d$-stability therefore makes $A_\lambda$
$d$-stable. The fixed-point equation on the torsor $A/A_\lambda$
has a unique solution because
$d-1$ is invertible. For that member, the only $d$-fixed character
of $A_\lambda$ is the trivial one, since $C_{A_\lambda}(d)=1$.
We obtain a bijection
\begin{equation}\label{eq:field-fixed-bijection}
                         X^d\longleftrightarrow\Irr(W)^d.
\end{equation}
The uniqueness just proved makes this bijection equivariant for every
semilinear map normalizing $A$ and commuting with $d$, including
every $bg$ in \eqref{eq:field-code-joint-trace}.

The set $\Irr(W)^d$ is the product of $\ell$ binary fixed spaces,
one in each dual coordinate. Each has one nonzero vector. An
invertible coordinate map commuting with $d$ sends this vector to
the unique nonzero fixed vector of the next coordinate. Thus $bg$
acts on these binary letters as its coordinate $\ell$-cycle, and
\begin{equation}\label{eq:nonidentity-field-label}
                       |X^{bg,d}|=2\qquad(d\ne1).
\end{equation}

It remains to bound $|X^{bg}|$ for $d=1$. Put $y=bg$ and let
$\sigma$ be the total field map around its coordinate cycle.
If $\sigma\ne1$, a fixed scalar label is determined by an element
of $\F_q^*$ fixed by $\sigma$, so $C_A(y)=1$. Apply the ordinary
class parametrization $Y_A$ from the proof of
Lemma~\ref{lem:scalar-code-trace}. In the fixed-orbit average over
$ay$, the scalar label must be one. A scalar times a nontrivial
field map $u\mapsto u^{2^j}$ has exactly two fixed vectors, because
\[
             \gcd(2^j-1,2^r-1)=1\qquad(0<j<r).
\]
The cycle fixed vector is determined by one such vector, so every
summand in the average is two and $|X^y|=2$.
If $\sigma=1$, successive field gauges conjugate $y$ to a
scalar-permutation map and transport $A$ to another scalar
code. Lemma~\ref{lem:scalar-code-trace} gives $|X^y|\leq c_\ell$.
Combining these bounds with \eqref{eq:field-fixed-code},
\eqref{eq:field-code-joint-trace} and
\eqref{eq:nonidentity-field-label} proves
\eqref{eq:semilinear-one-cycle}.
\end{proof}

\subsection{Fixed coordinates and arbitrary permutation tops}

\begin{lemma}\label{lem:prime-semilinear-coordinate}
For every $E\leq\Gamma L_1(2^r)$, with $r$ an odd prime,
$k(\F_{2^r}\sd E)\leq2^r$.
\end{lemma}
\begin{proof}
Keep $q=2^r$ and $m=q-1$. If $E$ is cyclic scalar of order $d$,
then $k(\F_q\sd E)=d+m/d\leq q$. Otherwise its field image is
$C_r$, and a scalar conjugation puts $E=C_d:C_r$, with $d\mid m$.
Each nonidentity field element acts fixed-point-freely on
$C_d\setminus\{1\}$, so ordinary Clifford theory gives
$k(E)=r+(d-1)/r$. On the nonzero additive dual, every complement
$C_r$ has exactly one fixed vector, and distinct complements have
distinct fixed vectors because a nonidentity scalar fixes none.
The resulting $d$ vectors form one $E$-orbit with stabilizer $C_r$;
all other nonzero orbits have trivial stabilizer. This also holds
when $d=1$. The affine little-group formula therefore gives
\[
 \begin{split}
 k(\F_q\sd E)
   &=2r+\frac{d-2}{r}+\frac{m}{rd}\\
   &\leq2r+\frac{q-2}{r}\leq q.
 \end{split}
\]
The first comparison uses $d+m/d\leq m+1$; the last is equivalent
to $2(r+1)\leq2^r$, valid for $r\geq3$.
\end{proof}

\begin{proof}[Proof of Proposition~\ref{prop:small-semilinear} for odd prime field degree]
Let $\Omega=\{1,\ldots,n\}$, let $N$ be the kernel of the action of
$H$ on $\Omega$, and put $T=H/N$.
Since $\Gamma L_1(q)$ has odd order $mr$, $N$ is odd. Put
$Z=\Irr(V\sd N)$. Let $C_1,\ldots,C_s$ be the coordinate cycles
of a lift $g$ of an odd $a\in T$. Put $L_0=V\sd N$, and let
$L_j$ be the kernel of projection onto $C_1\cup\cdots\cup C_j$.
These kernels are $g$-invariant, $L_s=1$, and
\[
 L_{j-1}/L_j=W_j\sd K_j,\qquad
 W_j=\F_q^{C_j},\qquad
 K_j\le(C_m:C_r)^{|C_j|}.
\]
Every such image contains its full additive group $W_j$.
The twisted normal-subgroup inequality
\eqref{eq:twisted-filtration} takes the maximum over the prescribed
coset twists at each step. The additive part of a twist acts by inner
conjugation because the image contains the full $W_j$, so removing it
preserves the character trace. The remaining twist is semilinear with
the same coordinate cycle. Lemma~\ref{lem:semilinear-one-cycle}
therefore applies to every moved-cycle image.
For a fixed coordinate, its trace is at most its class number,
bounded by Lemma~\ref{lem:prime-semilinear-coordinate}. Consequently
\begin{equation}\label{eq:semilinear-cycle-filtration}
       |Z^a|\leq q^{\#\mathrm{fixed\ coordinates}}
                    \prod_{\text{moved cycles}}2(q+r-2).
\end{equation}
Every moved cycle has odd length $\ell\geq3$. Since $q\geq8$
and $r\leq q/2$,
\[
       2(q+r-2)\leq3q-4\leq q^3/8\leq(q/2)^\ell.
\]
For the middle inequality, $q^3-24q+32$ is positive at eight
and increasing thereafter. Thus
\begin{equation}\label{eq:prime-semilinear-fixed-trace}
                      |Z^a|\leq q^n2^{-s_\Omega(a)}.
\end{equation}
At the identity, the same coordinate filtration and
Lemma~\ref{lem:prime-semilinear-coordinate} give $|Z|\leq q^n$.
Now $T$ acts faithfully on the coordinate set.
Lemma~\ref{lem:ordinary-spectrum} and
Theorem~\ref{thm:permutation-weight} yield
\[
       k(V\sd H)\leq |T|_2\Lambda_2(T,Z)
                    \leq q^n|T|_2=q^n|H|_2,
\]
because $N$ is odd. This proves the proposition for all
subgroups stated, and in particular for the two remaining fields
of orders eight and thirty-two.
\end{proof}

\section{The affine inequality in characteristic two}\label{sec:affine}

\begin{theorem}\label{thm:affine}
Let $V$ be a finite-dimensional vector space over $\F_2$ and let
$H\le\GL(V)$. Then
\begin{equation}\label{eq:affine-theorem}
 k(V\sd H)\le |V|\,|H|_2.
\end{equation}
\end{theorem}
\begin{proof}
We induct on $\dim_{\F_2}V$. The zero-dimensional case has $H=1$.
Suppose first that $0<W<V$ is $H$-invariant, put $U=V/W$, and
let $C=C_H(U)$. The subgroup $W\sd C$ is normal in $V\sd H$
because $[C,V]\le W$, and its quotient is $U\sd(H/C)$ with
faithful action on $U$.

Let $J=C\cap C_H(W)$. Relative to $W$ and $V/W$, each element
of $J$ is the identity plus a map $U\to W$. Thus $J$ is a
$2$-group. It centralizes $W$, is normal in $W\sd C$, and
$(W\sd C)/J\cong W\sd(C/J)$ acts faithfully on $W$. Applying
Lemma~\ref{lem:class-counts} and the induction hypothesis gives
\[
 k(W\sd C)\le k(J)k(W\sd(C/J))
       \le |J|\,|W|\,|C/J|_2=|W|\,|C|_2.
\]
Apply the induction hypothesis also to $U\sd(H/C)$ and then
Lemma~\ref{lem:class-counts} to the normal subgroup $W\sd C$.
This proves \eqref{eq:affine-theorem} in the reducible case.

We may assume $V$ irreducible. The primitive case is
Proposition~\ref{prop:primitive-affine}. Otherwise choose an
$H$-invariant direct-sum decomposition
\[
 V=W_1\oplus\cdots\oplus W_t,\qquad t>1,
\]
with the smallest possible positive summand dimension. The action
on the summands is transitive. The full stabilizer image
$B_i\le\GL(W_i)$ is irreducible: an invariant proper subspace
would induce over a block transversal to an invariant proper
subspace of $V$. It is primitive, since any finer decomposition
would contradict the choice of the $W_i$.

Let $N$ be the kernel of the permutation action and $E_i$ its
image on $W_i$. Then $E_i\lhd B_i$. Every characteristic subgroup
of $E_i$ is normal in $B_i$. By Clifford theory its restriction
on $W_i$ is completely reducible, and its isotypic summands are
permuted by $B_i$. Primitivity forces a single isotypic summand;
the restriction is therefore homogeneous. Also $O_2(E_i)=1$:
this subgroup is normal in $B_i$, and a normal $2$-subgroup acts
trivially on an irreducible module in characteristic two.
Faithfulness of $B_i$ gives the assertion.

For every subgroup of $B_i$, the induction hypothesis applies on
$W_i$, whose dimension is smaller than that of $V$. Thus the
fixed-coordinate hypothesis of
Proposition~\ref{prop:affine-coordinate} holds without requiring
that a particular subgroup act irreducibly. By
Proposition~\ref{prop:local-capacity}, if $E_i$ has a nontrivial
layer or a nonabelian Fitting subgroup, then
$c(E_i)\le |W_i|^2/8$. The property is constant on the coordinate
orbit, so this case follows from
Proposition~\ref{prop:affine-coordinate}.

The remaining abelian Fitting subgroup is cyclic: its faithful
homogeneous representation has field image. Thus it remains to
treat $F^*(E_i)=F(E_i)$ cyclic. Write $E=E_i$ and
$W=W_i$. If $E\ne1$, put $R=F(E)$; the Fitting subgroup is
nontrivial because it is self-centralizing. The homogeneous
scalar algebra $Q=\F_2[R]$ is a finite field, say of order
$q=2^f$. The group $R$ embeds in $Q^\times$, and
$C_E(R)=R$. Consequently
\[
 E/R\hookrightarrow\operatorname{Gal}(Q/\F_2),\qquad
 c(E)=|E:R|_{2'}|R|\le f_{2'}(q-1).
\]
If $m=\dim_QW\ge2$, use $2f\le2^f=q$ to obtain
\[
 c(E)\le f(q-1)\le q^2/2\le q^4/8\le |W|^2/8.
\]
If $m=1$, the entire $B_i$ normalizes $R$ and its scalar
algebra $Q$, so $B_i\le\Gamma L_1(q)$. For $f=4$ the displayed
capacity is at most $15\le q^2/8$. For $f\ge6$, the elementary
inequality $f\le2^{f-3}$ gives the same conclusion. Since $R$
is nontrivial, $q\ne2$. Only $q=4,8,32$ remain. Transporting the
scalar fields along a block transversal embeds the original
$H$ in $\Gamma L_1(q)\wr\operatorname{Sym}(t)$ on its natural
direct sum. Proposition~\ref{prop:small-semilinear} is a theorem
for every subgroup of that wreath product and supplies
\eqref{eq:affine-theorem} for these three field sizes.

Finally, if $E=1$, then $c(E)=1$ and the capacity inequality
holds whenever $|W|\ge4$. The only remaining possibility is
$\dim_{\F_2}W=1$. Now $B_i\le\GL_1(2)=1$, so $H$ acts by
permutation matrices on these one-dimensional summands. It fixes
their nonzero all-ones vector. Since $t>1$, this contradicts
irreducibility. All cases have been covered, completing the
dimension induction.
\end{proof}

\section{Descent through a proper minimal Levi subgroup}
\label{sec:geometry}

Throughout this section the coefficient system is a sufficiently large
splitting $2$-modular system $(K,\cO,k)$, with $\cO$ complete and $k$
algebraically closed. Algebraic groups are defined over an algebraic closure
of a finite field of odd characteristic. For a finite group $T$, a block is
identified with its central idempotent in $\cO T$.
For a connected reductive group $\bH$ with Frobenius $F$, write
$H=\bH^F$ and let $H^*$ be its finite dual group. For an odd-order
semisimple element $s\in H^*$, put
\[
 \cE_2(H,s)=
 \bigcup_{\substack{t\in C_{H^*}(s)\\t\text{ a }2\text{-element}}}
 \cE(H,st),
\]
where $\cE(H,st)$ is the rational Lusztig series. The idempotent
$e_s^H$ is the sum of the block idempotents in this $2$-series.
For an algebraic group $\bH$, the superscript $\circ$ denotes its
identity component. For a torus $\bT$, write
$X^*(\bT)=\Hom(\bT,\mathbb G_m)$ and
$Y(\bT)=\Hom(\mathbb G_m,\bT)$ for its character and cocharacter
lattices.

An \emph{ordinary} finite group of Lie type in this section is a central
quotient of the fixed points of a simply connected algebraic group with
endomorphism $F=\gamma F_r^f$, where $\gamma$ is a pinned diagram
automorphism and $F_r$ is the standard field Frobenius. Its ordinary
automorphisms are the inner, diagonal, field and pinned diagram
automorphisms, with their relations on the finite group. This
terminology excludes the exceptional inseparable graph automorphism of
$G_2(3^f)$ and exceptional finite Schur covers. Those cases are treated by the component estimates.

\begin{theorem}[Proper minimal-Levi descent]\label{thm:levi-descent}
Let $D\lhd G$ be a central product of one $G$-orbit of ordinary quasisimple
finite groups of Lie type in odd characteristic, and let $b$ be a
$G$-stable block of $D$. Suppose the action induced by $G$ lifts to the
product of the ordinary simply connected covering groups, preserves its
original central kernel, and consists of ordinary automorphisms and
component permutations. For each component, let $\bS$ be its
simply connected algebraic covering group, let $\bS^*$ be the dual,
and let $s$ be the odd semisimple label of its covered block. Set
\begin{equation}\label{geo:minimal-levi}
 \bL^*=C_{\bS^*}\bigl(Z(C_{\bS^*}(s)^\circ)^\circ\bigr).
\end{equation}
Assume that each of these canonical minimal Levi subgroups is proper.

Then there exist a proper subgroup $H<G$ containing $C_G(D)$, a finite
central extension $\pi:\widehat H\longrightarrow H$ with odd kernel, and
an $\widehat H$-stable block $c$ of
$\widehat J=\pi^{-1}(D\cap H)$ such that
\begin{equation}\label{geo:ambient-equivalence}
                 \cO G b\ \simeq_{\mathrm{Mor}}\ \cO\widehat H c.
\end{equation}
The extension has a canonical copy of $C_G(D)$, and the equivalence
intertwines the action of every element of this copy with the action of
the same element of $C_G(D)$. Consequently, every block $B$ of $G$ covering
$b$ corresponds to a block $\widehat B$ of $\widehat H$ satisfying
\begin{equation}\label{geo:descent-invariants}
 k(\widehat B)=k(B),\qquad |D_{\widehat B}|=|D_B|,\qquad
 |\widehat H:Z(\widehat H)|<|G:Z(G)|.
\end{equation}
Any specified block of $C_G(D)$ covered by $B$ is retained.
\end{theorem}

The proof constructs the extension and its block after selecting a
primitive summand of a cohomology lattice. The geometric input is the
connected-centralizer Bonnaf\'e--Rouquier theorem and the full diagonal
comparison for different parabolics. The latter is the comparison of
\cite[Theorem~7.2]{BDR17}, in the diagonal form of
\cite[Lemma~3.1]{RuhJordan22}. Numbering for the latter reference is that
of arXiv:2010.04468v1. We give the remaining order and equivariance
arguments explicitly.

\subsection{The relative normalizer and the action}

We first work on an ordinary simply connected cover
$S=\bS^F$ and write $L=\bL^F$, where $\bL$ is dual to
\eqref{geo:minimal-levi}. Put
\[
              N=N_S(\bL,e_s^L),\qquad A=N/L.
\]
The normalizer here preserves both the algebraic Levi and its rational
series idempotent.

\begin{lemma}\label{geo:odd-component}
The group $C_{\bS^*}(s)/C_{\bS^*}(s)^\circ$ is abelian of odd order.
The group $A$ is abelian of odd order. A finite $2$-group of inner-diagonal
algebraic automorphisms which preserves $(\bL,e_s^L)$ acts trivially on
$Y(Z(\bL)^\circ)\otimes_{\Z}\mathbb R$.
\end{lemma}

\begin{proof}
Choose a finite central isogeny
$\nu:\bS^\sharp\longrightarrow\bS^*$ whose source has simply connected
derived subgroup. A point of an algebraic group over the algebraic
closure of a finite field has finite order. We may therefore lift $s$ to
an odd-order semisimple element $s^\sharp$: replace a finite-order lift
by its odd part. For $x\in C_{\bS^*}(s)$, choose a lift $x^\sharp$ and set
\[
 d(x)=x^\sharp s^\sharp(x^\sharp)^{-1}(s^\sharp)^{-1}.
\]
This lies in the central kernel of $\nu$, is independent of the lift,
and is multiplicative in $x$. Its order divides $|s^\sharp|$, since
$d(x)$ commutes with $s^\sharp$. Its kernel is the image of
$C_{\bS^\sharp}(s^\sharp)$, which is connected by the
connected-centralizer theorem for groups with simply connected derived
subgroup \cite[\S\S8--9]{Steinberg68}. This kernel is exactly
$C_{\bS^*}(s)^\circ$. The first assertion
follows.

An element of a relative normalizer preserving the $\bL^*$-class of $s$
has a representative in $C_{\bS^*}(s)$: multiply its dual representative
by a suitable element of $\bL^*$. Thus its relative Weyl image belongs to
\begin{equation}\label{geo:relative-image}
 \bigl(C_{\bS^*}(s)\cap N_{\bS^*}(\bL^*)\bigr)\bL^*/\bL^*.
\end{equation}
This is a quotient of a subgroup of the odd component group, because
$C_{\bS^*}(s)^\circ\leq\bL^*$. Rational Levi duality identifies $A$
with the corresponding rational series stabilizer in this relative
normalizer; see \cite[\S7.A]{BDR17}. Hence $A$ is abelian of odd order. In particular,
the relative image of a $2$-group of inner automorphisms is trivial.
Elements of $\bL$ act trivially on its center, giving the asserted
cocharacter action.

The inner-diagonal assertion can be checked on finite groups as well.
Take a compatible regular embedding
$\bS\hookrightarrow\widetilde{\bS}$ and put
$\widetilde{\bL}=\bL Z(\widetilde{\bS})$.
The quotient tori $\widetilde{\bS}/\bS$ and
$\widetilde{\bL}/\bL$ are naturally isomorphic. The Lang--Steinberg
theorem \cite[Theorem~10.1]{Steinberg68} for the connected kernel $\bL$ gives
\begin{equation}\label{geo:regular-factorization}
       \widetilde S=S\widetilde L,
       \qquad \widetilde S=\widetilde{\bS}^{F},\quad
       \widetilde L=\widetilde{\bL}^{F}.
\end{equation}
An inner-diagonal representative normalizing the pair can be written
$g\widetilde l$, with $g\in S$ and $\widetilde l\in\widetilde L$.
The diagonal Levi operator preserves $e_s^L$ and acts trivially on the
central cocharacter space. Thus $g\in N$, and the relative action is
that of its odd-order image in $N/L$. This proves the assertion.
\end{proof}

Automorphisms pass between central forms as follows. If
$T\twoheadrightarrow T/Z$ is a central quotient of a perfect group,
two lifts of the same automorphism differ by a homomorphism $T\to Z$.
Such a homomorphism is trivial. Consequently the existence of the
ordinary lifts implies their uniqueness and their compatibility with
multiplication. This observation applies to the whole product of
simply connected covering groups. In particular, an element of
$C_G(D)$ lifts to the identity automorphism of that product.

The field automorphisms admit the following algebraic realization.
Write $F=\gamma F_r^f$ and replace $\bS$ by $f$ copies. Set
\[
 \tau(x_0,\ldots,x_{f-1})
       =(\gamma(x_{f-1}),x_0,\ldots,x_{f-2}).
\]
Then
\begin{equation}\label{geo:product-frobenius}
 \tau^f=\gamma_{\mathrm{diag}},\qquad
 \overline F=\tau\,\operatorname{diag}(F_r).
\end{equation}
The inverse to projection onto the first coordinate is
\[
 \iota(x)=(x,F_r(x),\ldots,F_r^{f-1}(x))\qquad(x\in S).
\]
Use the same formula on the adjoint group. An inner-diagonal
automorphism represented by $x\in\bS_{\mathrm{ad}}^F$ is realized
by conjugation with $\iota(x)$. On the fixed-point group,
$F_r^f=\gamma^{-1}$. A compatible pinned diagram automorphism
$\delta$ satisfies $\delta\gamma=\gamma\delta$, whence
$\delta_{\mathrm{diag}}\tau=\tau\delta_{\mathrm{diag}}$ and
\[
 \tau^{-1}\iota(x)=\iota(F_r(x)),\qquad
 \delta_{\mathrm{diag}}\iota(x)=\iota(\delta(x)).
\]
Consequently these algebraic operators satisfy
\begin{align*}
 \tau^{-1}\operatorname{Ad}(\iota(x))\tau
   &=\operatorname{Ad}(\iota(F_r(x))),\\
 \delta_{\mathrm{diag}}\operatorname{Ad}(\iota(x))
       \delta_{\mathrm{diag}}^{-1}
   &=\operatorname{Ad}(\iota(\delta(x))),\qquad
 (\tau^{-1})^f=\gamma_{\mathrm{diag}}^{-1}.
\end{align*}
Together with the inner-diagonal and diagram multiplication laws,
these are the ordinary finite automorphism relations
\cite[\S2.1]{Harper24}. Thus the actual ordinary automorphism group
acts algebraically on the product, commuting with $\overline F$;
its field generator is $\tau^{-1}$.
For a component orbit, use these copies simultaneously and retain the
literal permutations of component factors.

The product Levi is
\[
 \bL\times F_r(\bL)\times\cdots\times F_r^{f-1}(\bL).
\]
It is $\overline F$-stable because $F(\bL)=\bL$, and the first-coordinate
projection identifies its fixed points with $L$. Its native dual label
represents the original rational series. This compatibility follows from the fixed-point
projection and series comparison of \cite[\S\S8.2--8.8]{Taylor19}; the
same comparison applies to the Levi itself. We apply cohomology directly
to this product datum. Component permutations act on this product variety. Under
K\"unneth they are the graded tensor permutations, which satisfy
the symmetric-group relations.

\begin{lemma}\label{geo:fixed-parabolic}
In the product realization \eqref{geo:product-frobenius}, an actual
Sylow $2$-subgroup of the stabilizer of the Levi, its rational series,
and the selected block fixes a parabolic whose Levi is the selected
product Levi.
\end{lemma}

\begin{proof}
Let $P$ be that Sylow subgroup and let $P_j$ be the stabilizer of an
algebraic factor. A field power which returns to that factor is already
a diagram holonomy by $\tau^f=\gamma_{\mathrm{diag}}$. Thus $P_j$ acts
on the factor by inner-diagonal and pinned diagram automorphisms;
there is no remaining inseparable field map in this factor action.
It preserves the geometric class of the label inside the selected
dual Levi, since it preserves the product Levi series.

Write $Y_j=Y(Z(\bL_j)^\circ)\otimes\mathbb R$. The kernel of the diagram
projection of $P_j$ acts trivially on $Y_j$ by
Lemma~\ref{geo:odd-component}. If the diagram group has odd order, the
whole $P_j$ acts trivially on $Y_j$, and an integral point outside the
finitely many relative-root hyperplanes gives the desired parabolic
on this factor.

The ordinary diagram groups with even order occur in types $A$, $D$
and $E_6$ (including the full diagram group $\mathfrak S_3$ of $D_4$).
In these cases properness and minimality imply
\begin{equation}\label{geo:central-minimal}
                    C_{\bS_j^*}(s_j)^\circ=\bL_j^*.
\end{equation}
Indeed, in type $A$ every semisimple connected centralizer is a Levi.
For an odd-order semisimple element in an orthogonal or symplectic
realization, there is no $-1$-eigenspace: paired nontrivial eigenvalues
give general linear factors, and the $1$-eigenspace gives the one
residual classical factor. This is the standard Levi form. The
assertion passes through central isogenies by taking an odd-order
lift. Every proper Levi in $E_6$ has only $A$- and $D$-type derived
components. Therefore
$C_{\bL_j^*}(s_j)^\circ=C_{\bS_j^*}(s_j)^\circ$ is a Levi of $\bL_j^*$,
hence of $\bS_j^*$; the canonical minimal choice forces
\eqref{geo:central-minimal}.

Work at this point with the simply connected primal group and its
adjoint dual. The center of a dual Levi is connected: its character
lattice is the quotient of the adjoint root lattice by the sublattice
generated by a subset of the simple roots, and this quotient is
torsion-free. Equation~\eqref{geo:central-minimal} places $s_j$ in this
connected center. Its Levi-conjugacy class is a singleton, so $P_j$
fixes it as a point.

Choose a Weyl-and-diagram-invariant positive definite form
$(-,-)_j$ on the real cocharacter space of a maximal torus
$\bT_j\le\bL_j$. Under $X^*(\bT_j^*)=Y(\bT_j)$, the real
character space of $Z(\bL_j^*)$ is the quotient by the span of
the Levi coroots. The space $Y_j$ is the orthogonal complement
of that span. Thus
\[
 \langle y,\overline u\rangle_j=(y,u)_j
 \quad\bigl(y\in Y_j,\ 
       \overline u\in X^*(Z(\bL_j^*))\otimes\mathbb R\bigr)
\]
is a nondegenerate invariant pairing. Suppose an outside root
$\alpha$ vanished on $Y_j^{P_j}$. Averaging its restriction over
$P_j$ gives zero on $Y_j$. The root-coroot proportionality and
this pairing then give
\begin{equation}\label{geo:root-norm}
       \sum_{g\in P_j}{}^g\!\alpha^\vee
             \bigm|_{Z(\bL_j^*)}=0
       \quad\text{in }X^*(Z(\bL_j^*))\otimes\mathbb R.
\end{equation}
The proportionality factor between a root and its coroot is constant
on its orbit; in the diagram-primary cases here the root system is
simply laced. Different torus-normalizer representatives differ by
Levi Weyl elements, which act trivially on the central quotient and
on $Y_j$. Thus these are genuine actions on the two central spaces.
The sum in \eqref{geo:root-norm} is integral, and the character lattice
of the connected center is torsion-free, so it is the zero integral
character. Evaluating at $P_j$-fixed $s_j$ yields
\[
                         \alpha^\vee(s_j)^{|P_j|}=1.
\]
The left-hand base has odd order and $|P_j|$ is a power of two. Hence
$\alpha^\vee(s_j)=1$, which would put the outside dual root in
$C_{\bS_j^*}(s_j)^\circ=\bL_j^*$. This contradiction proves that no
outside-root hyperplane contains $Y_j^{P_j}$.

The fixed space is rational. Choose an integral fixed cocharacter
$y_j$ outside all these hyperplanes. More explicitly, for a basis
$v_1,\ldots,v_t$ of its fixed lattice, each outside-root value on
$v_1+u v_2+\cdots+u^{t-1}v_t$ is a nonzero polynomial. Some integer
$u$ therefore avoids their finite sets of zeros. For each $P$-orbit
of algebraic factors choose such a representative and put
$y_{gj}=g y_j$. The stabilizer property makes this well defined.
The product cocharacter $y$ is $P$-fixed and regular relative to the
selected product Levi. Its parabolic $\mathbf P(y)$ has precisely
that Levi and is $P$-stable. Frobenius stability of this parabolic is
not required.
\end{proof}

\subsection{Cohomology and the full prescribed diagonal subgroup}

Suppress the product notation and write $\bS,F,\bL,S,L$ for the data
just obtained. For a parabolic $\mathbf P=\bL\bU$, put
\[
 Y_{\bU}=\{g\bU\in\bS/\bU:
                   g^{-1}F(g)\in\bU F(\bU)\},\qquad
 d=\dim Y_{\bU},\qquad
 M=H_c^d(Y_{\bU},\cO)e_s^L.
\]
Here $H_c^d$ denotes compactly supported \'etale cohomology in the
middle degree. It has commuting left $S$
and right $L$ actions and the prescribed full diagonal Levi action.
For a regular embedding as in \eqref{geo:regular-factorization}, write
\begin{equation}\label{geo:prescribed-X}
       X=(S\times L^{\op})\Delta\widetilde L.
\end{equation}
Here $\Delta\widetilde L=\{(t,t^{-1}):t\in\widetilde L\}$
inside $\widetilde S\times\widetilde L^{\op}$.
The regular-embedding center acts trivially diagonally. After selecting
a left block $b_S$, let $X_{b_S}$ denote its diagonal stabilizer.

The comparison theorem \cite[Theorem~7.2]{BDR17}, together with
\cite[Lemma~3.1 and its proof]{RuhJordan22}, supplies an isomorphism
of the full prescribed diagonal modules for any two parabolics with
this Levi. Its hypothesis is
$C_{\bS^*}(s)^\circ\leq\bL^*$. Applying a pair-preserving automorphism
first sends $Y_{\bU}$ to $Y_{a(\bU)}$, after which this comparison gives
an intertwiner for $M$. We establish the multiplication law for these intertwiners below.

For the parabolic of Lemma~\ref{geo:fixed-parabolic}, $P$ and the
prescribed diagonal Levi operators act on the same variety.
Let $H_0\le G$ be the stabilizer of the product datum
$(\bL,e_s^L,b_S)$, allowing its component permutations, and let
$\rho_0:H_0\to\Aut(S)$ be its uniquely lifted action on the
covering product. In this automorphism group put
\[
 \mathcal A=\rho_0(H_0),\qquad
 D_0=\{\operatorname{Ad}(t):t\in\widetilde L,
                                  {}^tb_S=b_S\},\qquad
 \Gamma=D_0\mathcal A.
\]
Intrinsically, $D_0$ consists of the rational adjoint-Levi
automorphisms preserving $b_S$. Every element of $\mathcal A$
preserves the Levi and the block, so it normalizes $D_0$.
Thus $D_0$ is normal in $\Gamma$. To record their
intersection relations, form the finite groups
\begin{equation}\label{geo:labels}
 \mathcal U=(S\times L^{\op})\rtimes\Gamma,
 \quad
 K_0=\{(t^{-1},t,\operatorname{Ad}(t)):t\in L\},
 \quad
 \mathcal Z=\mathcal U/K_0.
\end{equation}
The group $K_0$ is normal: it centralizes $S\times L^{\op}$, and an
actor sends its $t$-element to its $a(t)$-element. Its geometric
operator is the identity. Let $\overline X_{b_S}$ be the subgroup
of $\mathcal Z$ above $D_0$. The homomorphism
\[
 X_{b_S}\longrightarrow\overline X_{b_S},\qquad
 (s,\ell)\Delta t\longmapsto[(s,\ell,\operatorname{Ad}(t))]
\]
is onto and has kernel $\Delta Z(\widetilde S)$, the trivially acting
diagonal regular-embedding center. The relations in $K_0$ make this
map independent of the chosen expression $(s,\ell)\Delta t$.
The action on $M_b=b_SM$ factors through $\overline X_{b_S}$,
and the two groups give the same endomorphism order. Thus
\begin{equation}\label{geo:R-quotient}
                  R=\mathcal Z/\overline X_{b_S}=\Gamma/D_0,
\end{equation}
and the actor maps onto $R$. The internal subgroup $A=N/L$
is a normal subgroup of $R$.

Let $\overline P$ be the image in $R$ of a Sylow $2$-subgroup
$P$ of $\mathcal A$. It is a Sylow $2$-subgroup of $R$. The groups
$S\times L^{\op}$, $D_0$ and $P$ act on the same variety, and $K_0$
acts trivially. Hence the prescribed $\overline X_{b_S}$-module extends to
\emph{the whole inverse image} of $\overline P$ in $\mathcal Z$.
Indeed, the formula
\[
                  \rho(xp)=\rho_X(x)\rho_P(p)
\]
is well defined because the two actions coincide on their intersection
and satisfy the geometric conjugation relation. The extension obstruction therefore restricts to zero on the
Sylow subgroup $\overline P$ of $R$.

\subsection{Endomorphisms before a normalizer extension}

\begin{proposition}\label{geo:endomorphism-order}
The lattice $M$ is projective on both the $S$ and $L$ sides and is a
left $\cO S e_s^S$-progenerator. For each $a\in A$, choose
$n_a\in N$ representing $a$ and a full-$X$-compatible diagonal
intertwiner $U_a$ for $\operatorname{Ad}(n_a)$. Write
$L_x(m)=xm$ and $R_y(m)=my$ for the left and right operators,
and put $T_a=L_{n_a^{-1}}U_a$. Then $T_a$ is $S$-linear,
$T_aR_\ell=R_{{}^{n_a}\ell}T_a$ for $\ell\in L$, and
\begin{equation}\label{geo:generic-span}
       \End_{KS}(KM)
           =\bigoplus_{a\in A}(KL e_s^L)T_a.
\end{equation}
The integral order $E=\End_{\cO X}(M)$ is commutative. For a selected
left block $b_S$, put
$M_b=b_SM$ and $E_b=\End_{\cO X_{b_S}}(M_b)$. This is commutative,
and $E_b^A$ is local. Consequently $A$ acts transitively on the
primitive idempotents of $E_b$, whose number is odd.
\end{proposition}

\begin{proof}
We recall the regular-embedding calculation, including the parts
needed over $\cO$. Put $\widetilde N=N\widetilde L$ and
$\widetilde X=\widetilde S\times\widetilde L^{\op}$. The series
idempotents above $e_s^L$ have free $A$-orbits
\cite[Lemma~7.4]{BDR17}. Let $e_0$ be the sum of one representative
from each such orbit. The lifted dual centralizers are connected,
so the connected-centralizer Bonnaf\'e--Rouquier theorem makes
$\widetilde M e_0$ an integral Morita lattice
\cite[Theorem~B$'$]{BR03}. Here
\[
       \widetilde M=\Ind_X^{\widetilde X}M
                   =H_c^d(Y_{\bU}^{\widetilde{\bS}},\cO)e_s^L.
\]
Induction along the free orbit extends this upstairs module to
$\widetilde N$ and gives an upstairs Morita equivalence with
$\cO\widetilde N e_s^L$. The idempotents are
orthogonal elements of $\cO\widetilde L$; the connected-centralizer
Morita theorem holds over $\cO$; and the induction along the orbit is
an identity of integral modules. In particular the calculation in
\cite[\S7.C]{BDR17} gives
\begin{equation}\label{geo:upstairs-commutative}
 \End_{\cO\widetilde X}(\widetilde M)
       \cong Z(\cO\widetilde L e_0)^{[N:L]}.
\end{equation}
It is obtained by separating the free right idempotents and using
the Morita equivalence on each representative summand.

Write $d_{\mathrm{reg}}=[\widetilde S:S]=[\widetilde L:L]$. On the left,
\begin{equation}\label{geo:induction-restriction}
 \widetilde M\cong\Ind_S^{\widetilde S}M,
 \qquad \Res_S^{\widetilde S}\widetilde M\cong M^{\oplus d_{\mathrm{reg}}}.
\end{equation}
The second identity uses the prescribed diagonal $\widetilde L$
action and \eqref{geo:regular-factorization}. Restriction of the
upstairs projective module makes $M^{\oplus d_{\mathrm{reg}}}$ projective, hence
so is $M$. The upstairs regular module
$\cO\widetilde S e_s^S$ is a direct summand of finitely many copies
of $\widetilde M$. Its restriction is
$(\cO S e_s^S)^{\oplus d_{\mathrm{reg}}}$, so $\cO S e_s^S$ is a direct summand
of finitely many copies of $M$. This proves the left generator
property. On the right, restriction
of the upstairs projective module to $L$, followed by the analogous
direct-sum decomposition, proves right $L$-projectivity. These
statements also imply that $M$ is an $\cO$-lattice.

Frobenius reciprocity and the upstairs Morita equivalence give the
following dimension calculation:
\begin{align}\label{geo:dimension}
 d_{\mathrm{reg}}\dim_K\End_{KS}(KM)
 &=\dim_K\End_{K\widetilde S}(K\widetilde M)\notag\\
 &=\dim_K(K\widetilde N e_s^L)
   =d_{\mathrm{reg}}\dim_K(KN e_s^L).
\end{align}
Use the common left-induced model
\[
 K\widetilde M=K\widetilde S\otimes_{KS}KM,\qquad
 \widetilde T_a(g\otimes m)=g\otimes T_am.
\]
Each operator is well defined because $T_a$ is $S$-linear.
Let $\delta(t)$ be the prescribed diagonal operator for
$t\in\widetilde L$. In this model the right action is
\[
                (g\otimes m)t=gt\otimes\delta(t)^{-1}m.
\]
For $n=n_a$, put $u=ntn^{-1}$ and $h=t^{-1}u\in L$.
Full-$X$ compatibility gives
$U_a\delta(t)^{-1}=\delta(u)^{-1}U_a$. Since
$h u^{-1}n^{-1}u=n^{-1}$, it follows that
\begin{align*}
 (g\otimes T_am)u
 &=gt\otimes L_h\delta(u)^{-1}L_{n^{-1}}U_am\\
 &=gt\otimes L_{n^{-1}}\delta(u)^{-1}U_am
   =\widetilde T_a\bigl((g\otimes m)t\bigr).
\end{align*}
Thus $\widetilde T_a$ permutes the free right lifted idempotents
by $a$. If $\sum_a R_{u_a}T_a=0$ with $u_a\in KL e_s^L$,
induction of these $S$-linear maps gives
$\sum_a\widetilde R_{u_a}\widetilde T_a=0$; here
$\widetilde R_{u_a}$ is the upstairs right action of $u_a$.
For each lifted source idempotent separately,
the terms have mutually orthogonal target idempotents. Projection
to those targets shows that every induced $u_aT_a$ vanishes on
that source summand. Doing this for all source summands makes
each induced $u_aT_a$ zero. Faithfulness of the upstairs right
Levi action, and invertibility of $T_a$, give $u_a=0$.
The resulting independent summands have total dimension
$|A|\dim_K(KL e_s^L)=\dim_K(KN e_s^L)$; \eqref{geo:dimension}
proves \eqref{geo:generic-span}.

Induction of endomorphisms is faithful, so it embeds $E$ into
\eqref{geo:upstairs-commutative}. Hence $E$ is commutative.
If $b_S$ is not fixed by the whole diagonal group, decompose $M$
by the orbit of $b_S$ under that group. Induction from its diagonal
stabilizer identifies the orbit endomorphism order with
$\End_{\cO X_{b_S}}(b_SM)$; different left blocks are orthogonal.
Thus $E_b$ is the corresponding direct factor and is commutative.
Two full-$X_{b_S}$ intertwiners for the same automorphism differ
by a unit of $E_b$. Their conjugation actions on $E_b$ therefore
agree. The prescribed subgroup acts trivially on its endomorphism
order, so these actions define an action of $R=\Gamma/D_0$ on
$E_b$.
The projection of \eqref{geo:generic-span} onto $b_S$ still spans
the full generic left endomorphism algebra.

An $A$-fixed element of $E_b$ commutes with right $L$ and with all
internal intertwiners. It therefore centralizes the full algebra
in \eqref{geo:generic-span}, after projection onto $b_S$.
Intersecting integrally, and using that $M_b$ is a progenerator,
gives
\begin{equation}\label{geo:invariant-center}
 E_b^A\cong
 Z(\cO S b_S)^{\text{diagonal stabilizer}}.
\end{equation}
Conversely every fixed central element has the stated commutation
properties, proving equality. The center of a block order over
$\cO$ is local; an invariant unit has invariant inverse, so the
invariant subring in \eqref{geo:invariant-center} is local as well.
The sum of an $A$-orbit of primitive idempotents is an idempotent
of this local ring. There can therefore be only one such orbit.
Its size divides the odd order $|A|$.
\end{proof}

\subsection{One scalar extension of the full stabilizer}

We isolate the obstruction argument in a form which keeps the
prescribed subgroup action.

\begin{lemma}\label{geo:scalar-cover}
Let $X_0\lhd Z_0$ be finite groups, let $R_0=Z_0/X_0$, and let
$U_0$ be a $Z_0$-stable $\cO X_0$-lattice. Suppose
$E_0=\End_{\cO X_0}(U_0)$ is a commutative complete local order.
If the prescribed $X_0$-action extends to the full preimage of a
Sylow $2$-subgroup of $R_0$, then $U_0$ extends to a finite central
odd cover of $Z_0$ pulled back from $R_0$. Its central kernel acts
by a single scalar character.
\end{lemma}

\begin{proof}
Choose intertwiners for the elements of $R_0$. Because $E_0$ is
commutative, their conjugation action on $E_0$ is independent of
the choices and is a genuine group action. Their multiplication
obstruction is
\[
                       \omega\in H^2(R_0,E_0^\times).
\]
The residue field of $E_0$ is $k$, and $R_0$ acts trivially on it
because all operators are $\cO$-linear. Put
$V_0=1+\mathfrak m_0$, where $\mathfrak m_0$ is the maximal ideal
of $E_0$, and put $m=|R_0|_{2'}$. The image of $\omega$ in
$H^2(R_0,k^\times)$ is killed by $m$: positive-degree cohomology
is killed by $|R_0|$, while squaring is an automorphism of
$k^\times$ and hence of its cohomology. Let $\mu_m$ be the group
of $m$th roots of unity in $k^\times$. The exact sequence
\[
 1\longrightarrow\mu_m\longrightarrow k^\times
       \xrightarrow{\,x\mapsto x^m\,}k^\times\longrightarrow1
\]
therefore supplies a $\mu_m$-valued cocycle representing that image.
Lift the roots to $\cO^\times$ and choose this one normalized
scalar cocycle on the whole $R_0$. Its pullback defines the required
finite central odd cover.

Subtract its class from $\omega$. The remainder comes from a
class $v\in H^2(R_0,V_0)$. Let $P_0\in\Syl_2(R_0)$. The original
obstruction restricts to zero on $P_0$ by the stipulated extension
on its whole preimage. The scalar cocycle also has zero restriction,
since its coefficient group has odd order. Moreover
$H^1(P_0,k^\times)=0$, so the exact coefficient sequence makes
\[
 H^2(P_0,V_0)\longrightarrow H^2(P_0,E_0^\times)
\]
injective. Thus $v$ restricts to zero in $H^2(P_0,V_0)$.
Restriction followed by corestriction multiplies $v$ by
$[R_0:P_0]$. Hensel's lemma makes the power map for this odd
integer an automorphism of $V_0$, and hence of its cohomology.
It follows that $v=0$. An integral cochain adjustment now gives
a coherent action on the cover, extending the originally
prescribed $X_0$-action.
\end{proof}

Apply Proposition~\ref{geo:endomorphism-order} to the whole component
product, with its block and diagonal stabilizer. The strict
Sylow action constructed above permutes the odd number of primitive
idempotents of $E_b$. Hence some primitive idempotent $e$ is fixed.
Put
\begin{equation}\label{geo:primitive-corner}
 M_e=eM_b,\qquad E_e=eE_be,
 \qquad R_e=\operatorname{Stab}_R(e),\qquad A_e=A\cap R_e.
\end{equation}
The ring $E_e$ is commutative and local. The chosen Sylow image
$\overline P$ lies in $R_e$ and is still Sylow there. The module
$M_e$ retains the strict action on its whole preimage. Thus
Lemma~\ref{geo:scalar-cover} constructs one scalar odd cover of
its full stabilizer, including all component permutations.

All primitive summands of $M_b$ are isomorphic as left block
modules, since the internal intertwiners giving $A$-transitivity
commute with the left group. Therefore $M_e$ is still a left
block progenerator. Cornering the generic span gives
\begin{equation}\label{geo:corner-span}
 \End_{KSb_S}(KM_e)
       =\sum_{a\in A_e}(KL e_s^L)eT_a.
\end{equation}
Indeed, $eT_ae=0$ if $a(e)\ne e$, and $e$ commutes with right $L$.
This remains true after the cochain adjustments. A unit of $eE_be$
extends by the identity to $E_b$ and then over its diagonal block
orbit to a unit of the original full-$X$ endomorphism order.
Consequently the adjusted operators are restrictions of a new
arbitrary full-$X$ intertwiner family, to which
\eqref{geo:generic-span} again applies.

\subsection{An integral support criterion}

The following symmetric-order argument constructs the target block.
Its conclusion includes the integrality of the generic support.
The proof uses the adjunction argument of
\cite[Theorems~1.1 and~2.4]{Broue90}.

\begin{lemma}\label{geo:partial-morita}
Let $A_0$ and $B_0$ be symmetric $\cO$-orders with split semisimple
generic fibers. Let ${}_{A_0}M_0{}_{B_0}$ be projective of finite
type on both sides. Suppose $KM_0$ is faithful as a $KA_0$-module
and the right-action homomorphism
\[
          (KB_0)^{\op}\longrightarrow\End_{KA_0}(KM_0)
\]
is surjective. There is a central idempotent $c\in B_0$ with
$M_0=M_0c$, and $M_0$ induces a Morita equivalence between $A_0$
and $B_0c$. If $A_0$ is indecomposable, $c$ is primitive.
\end{lemma}

\begin{proof}
Put $N_0=\Hom_{\cO}(M_0,\cO)$, $P_0=M_0\otimes_{B_0}N_0$ and
$Q_0=N_0\otimes_{A_0}M_0$. Symmetry identifies $N_0$ both with
$\Hom_{A_0}(M_0,A_0)$ and with $\Hom_{B_0}(M_0,B_0)$.
The functors $\mathcal F=M_0\otimes_{B_0}-$ and
$\mathcal G=N_0\otimes_{A_0}-$ are therefore adjoint in both orders.
Denote the unit and counit for $\mathcal F\dashv\mathcal G$ by
\[
 \eta_B:B_0\longrightarrow Q_0,\qquad
 \epsilon_A:P_0\longrightarrow A_0,
\]
and those for $\mathcal G\dashv\mathcal F$ by
\[
 \eta_A:A_0\longrightarrow P_0,\qquad
 \epsilon_B:Q_0\longrightarrow B_0.
\]
All four are bimodule maps and satisfy their triangle identities.
The bimodules $P_0,Q_0$ are $\cO$-free: projectivity expresses each
as a direct summand of finitely many copies of $M_0$ or $N_0$.

The generic hypotheses give a semisimple Morita equivalence between
$KA_0$ and a central summand $KB_0c_K$. Thus $K\eta_A$ and
$K\epsilon_A$ are isomorphisms. Tensor $\eta_A$ on the left by
$N_0$ and on the right by $M_0$. The resulting map
\[
 N_0\eta_A M_0:Q_0\longrightarrow Q_0\otimes_{B_0}Q_0
\]
has left inverse $\epsilon_B\otimes1_{Q_0}$ by the triangle
identity for $\mathcal G\dashv\mathcal F$. Its generic fiber is
an isomorphism, so its complementary integral summand is free of
rank zero and vanishes. It is therefore an integral isomorphism.
Similarly
\[
 N_0\epsilon_A M_0:Q_0\otimes_{B_0}Q_0\longrightarrow Q_0
\]
has right inverse $\eta_B\otimes1_{Q_0}$ by the triangle identity
for $\mathcal F\dashv\mathcal G$, and is an integral isomorphism.

The composite $\epsilon_A\eta_A$ is multiplication by an element
$z\in Z(A_0)$. Its action on $Q_0$ is the composite of the preceding
two isomorphisms. Consider
\[
 Z(A_0)\longrightarrow
 T_0=\End_{B_0\text{-}B_0}(Q_0),
 \qquad z\longmapsto[n\otimes m\mapsto nz\otimes m].
\]
This map becomes an isomorphism over $K$, by the generic Morita
equivalence, and is injective. The order $T_0$ is commutative since
its generic fiber is commutative; it is finite over $\cO$, hence
integral over $Z(A_0)$. The image of $z$ is a unit, so $z$ is a
unit already in $Z(A_0)$. Indeed, a monic relation
\[
 z^{-r}+a_{r-1}z^{-(r-1)}+\cdots+a_0=0,
 \qquad a_i\in Z(A_0),
\]
gives $z^{-1}=-(a_{r-1}+a_{r-2}z+\cdots+a_0z^{r-1})\in Z(A_0)$.
It follows that $z^{-1}\epsilon_A$ is a left inverse of $\eta_A$.
The same free-rank-zero argument proves that $\eta_A$, and then
$\epsilon_A$, is an integral isomorphism.

Set $T=\eta_B\epsilon_B\in\End_{B_0\text{-}B_0}(Q_0)$.
The triangle identities and the isomorphisms just proved show
that $M_0\eta_B$ and $M_0\epsilon_B$ are isomorphisms.
Furthermore $Q_0=\mathcal G(M_0)$ belongs to the essential image
of the fully faithful functor $\mathcal G$. On this image
$\mathcal F$ detects isomorphisms. Hence $T$ is an automorphism;
its inverse is also a bimodule map. The maps
\[
 p=\eta_B:B_0\longrightarrow Q_0,
 \qquad j=\epsilon_BT^{-1}:Q_0\longrightarrow B_0
\]
satisfy $pj=1_{Q_0}$. The projector $jp$ on the regular
$B_0$-bimodule is multiplication by an integral central idempotent
$c$. Thus $Q_0\cong B_0c$.

Under the canonical identification
$Q_0\cong\End_{A_0}(M_0)^{\op}$, $p$ is the right-action map;
hence $B_0c\cong\End_{A_0}(M_0)^{\op}$ as algebras.
The isomorphism $M_0\eta_B$ also gives $M_0c=M_0$.
Finally $P_0\cong A_0$ and left $B_0$-projectivity of $N_0$
show that $A_0$ is a direct summand of finitely many copies of
$M_0$. Thus $M_0$ is an $A_0$-progenerator, proving Morita
equivalence. The induced center isomorphism makes $c$ primitive
when $A_0$ is indecomposable.
\end{proof}

Let $J$ be the inverse image of $A_e$ in the internal normalizer
$N$, and restrict the same cover to $\widehat J$. Write
$\pi:\widehat J\to J$ for the projection and $\rho$ for its
coherent action on $M_e$. For $\widehat n\in\widehat J$, put
$n=\pi(\widehat n)$ and $T_{\widehat n}=L_{n^{-1}}\rho(\widehat n)$.
Since $\rho(\widehat n)$ commutes with $L_n$, the right operator is
\[
 R_{\widehat n}=L_n\rho(\widehat n)^{-1}
              =T_{\widehat n}^{-1}.
\]
For $\widehat u\in\widehat J$, put $u=\pi(\widehat u)$.
Semilinearity gives
\[
 T_{\widehat n}T_{\widehat u}
 =L_{n^{-1}}L_{nu^{-1}n^{-1}}\rho(\widehat n\widehat u)
 =T_{\widehat n\widehat u}.
\]
Hence $R_{\widehat n}R_{\widehat u}=R_{\widehat u\widehat n}$,
as required for a right action.
The right-action image contains these operators and their inverses;
thus \eqref{geo:corner-span} gives its generic surjectivity.
The cover is canonically trivial on $L$, and $[\widehat J:L]$ is odd. Since
$M_e$ is right $L$-projective, averaging over this odd index shows
that it is right $\widehat J$-projective. It is left block
projective as a summand of $M_b$. Let $\lambda$ be the scalar
character of the covering kernel on the right. In the notation
of Lemma~\ref{geo:partial-morita}, take
\[
             A_0=\cO S b_S,
       \qquad B_0=\cO\widehat J e_\lambda e_s^L.
\]
These are symmetric orders. Generic left faithfulness follows
from the progenerator property. The lemma therefore supplies one
block $c$ of $\widehat J$ and an equivalence
\begin{equation}\label{geo:local-morita}
                 \cO S b_S\ \simeq_{\mathrm{Mor}}\
                 \cO\widehat J c.
\end{equation}
The block $c$ is stable under the whole covering actor. Indeed,
a semilinear operator sends $M_e=M_ec$ to
$M_e=M_e{}^{\widehat h}c$. If the two primitive central
idempotents differed, their orthogonality would force $M_e=0$.
Thus the target block in \eqref{geo:local-morita} is the same
block throughout the ambient construction.

\subsection{The subgroup, the central quotient, and ambient transport}

The construction on the simply connected component product carries
the unique lifts of the $G$-action. The subgroup $H_0\le G$
defined above stabilizes the tuple of Levi and Levi-series data.
Then
\begin{equation}\label{geo:first-frattini}
                              G=DH_0.
\end{equation}
Here is the rational conjugacy justification. An ambient automorphism
preserves the global series, so its dual sends $s$ to a rationally
conjugate element and sends the canonical minimal Levi to a
rationally conjugate Levi. Rational Levi duality allows an inner
adjustment by an element of the original finite group which
normalizes the primal Levi. After this adjustment, a rational
conjugator of the two labels normalizes the dual Levi, because
both canonical minimal Levis are now the same. The relative
normalizer correspondence supplies the matching primal normalizer
element. A second inner adjustment therefore preserves the
Levi-series idempotent itself. This is the rational conjugacy
argument of \cite[Lemma~2.10, Corollary~2.13, and the initial
argument of Proposition~4.8]{RuhJordan22}. Apply it to every
component, with the same component permutation, to obtain
\eqref{geo:first-frattini}.

Let $N$ now denote the image in $D$ of the product normalizer;
then $H_0\cap D=N$. The homomorphism
\[
 \eta:H_0\xrightarrow{\rho_0}\mathcal A
       \lhook\joinrel\longrightarrow\Gamma
       \longrightarrow R=\Gamma/D_0
\]
satisfies $\eta(N)=A$ and $C_G(D)\le\ker\eta$.
Put $H=\eta^{-1}(R_e)$ and $J=D\cap H$. Thus
$\eta(J)=A_e$, and $H$ is the stabilizer of the chosen primitive
idempotent $e$. The conjugation action on $E_b$ of an internal
right normalizer intertwiner and of the corresponding actual
diagonal normalizer element is the same: their ratio is a
prescribed left group operator, which commutes with $E_b$.
Transitivity of $A$ on the primitive idempotents therefore gives
\begin{equation}\label{geo:second-frattini}
                H_0=NH,\qquad G=DH,\qquad J=D\cap H.
\end{equation}
The local groups used above are their full inverse images in the
simply connected product. The identity $G=DH$ retains every
ambient coset. The centralizer $C=C_G(D)$ acts trivially on the
data and lies in $H$.

The group $H$ is proper. For one factor, $N/L$ is abelian. If
$N=S$, perfectness would give $S=[S,S]\leq L$. But if $r$ is
the defining characteristic and $q=r^f$, let $\Phi^+$ be a
positive root system of $\bS$ and $\Phi_L^+$ its Levi subset.
The group-order formulas give
\[
        |L|_r=q^{|\Phi_L^+|}<q^{|\Phi^+|}=|S|_r
\]
for a proper Levi, a contradiction. Hence $N<S$; the same argument
on a proper factor applies to the product. Thus $J<D$, and
\eqref{geo:second-frattini} implies $H<G$.

We next impose the original central kernel $Z_0$ of the simply
connected product mapping onto $D$. Write $(Z_0)_2$ and
$(Z_0)_{2'}$ for its Sylow $2$-subgroup and odd Hall subgroup.
If $\zeta_i$ is the prescribed central character on the $i$th
component, select their tensor-product sector with
\[
       \prod_i\zeta_i(z_i)=1
       \qquad\bigl((z_i)_i\in(Z_0)_{2'}\bigr).
\]
The geometric lattice has identical left and
right action of each $z\in Z_0$:
\begin{equation}\label{geo:same-center}
                              zm=mz.
\end{equation}
The primitive cut and the scalar extension preserve this identity.
The cocycle is trivial on the prescribed Levi and hence on $Z_0$.
The local Morita equivalence consequently matches the two central
ideals generated by $z-1$, $z\in Z_0$. Quotienting by those
corresponding ideals yields a Morita equivalence of the actual
quotient group-block orders. To see this directly, restrict the
Morita functors to the full subcategories annihilated by these
ideals. They are exactly the module categories of the quotient
orders. The quotient equivalence makes the quotient bimodule projective
and a generator on both sides; it is therefore $\cO$-free.

For the odd part of $Z_0$, the selected trivial sector is already
the quotient group algebra. For its central $2$-part, the usual
central-$2$ block bijection gives one quotient block on each side.
Their defect orders both lose the same factor $|(Z_0)_2|$.
An actor need only normalize $Z_0$: it sends $(z-1)m$ to
$({}^{h}z-1)\rho(h)m$, and so preserves the quotient module.
The covers descend as well, with odd scalar kernel. Properness
is retained because $Z_0\leq L\leq J$. From this point on, $M$ denotes this quotient Morita lattice,
$J=D\cap H$, and $\widehat J$ is the corresponding covered subgroup
of the group. The lattice is a Morita bimodule for
$\cO D b$ and one block $\cO\widehat Jc$.


With $n_{\mathrm{odd}}=|R_e|_{2'}$, let
$\omega:R_e\times R_e\to\mu_{n_{\mathrm{odd}}}$ be the normalized scalar
cocycle defining the cover above. Its pullback to $H$ is
\[
                (h,h')\longmapsto\omega(\eta(h),\eta(h')).
\]
Since $\eta(C)=1$, this cocycle is $1$ on every pair with one
entry in $C$. Consequently the resulting central odd cover
$\widehat H\to H$ has a canonical embedded copy of $C$, and its
coherent operator satisfies $\rho(c_0)=1$ for every $c_0\in C$.
Its restriction to $\widehat J$ is exactly the descended cover used
in the local Morita equivalence.

The resulting local Morita lattice $M$ carries semilinear operators.
For $\widehat h\in\widehat H$, $a\in\cO Db$, $m\in M$ and
$j\in\widehat J$, they satisfy
\begin{align}\label{geo:literal-relations}
 \rho(\widehat h)(a m j)
   &={}^{\pi(\widehat h)}a\,\rho(\widehat h)m\,
                         {}^{\widehat h}j,\notag\\
 \rho(\widehat j)m
   &=\pi(\widehat j)m\widehat j^{-1}
                 \qquad(\widehat j\in\widehat J).
\end{align}
The second identity is the original intersection relation in
\eqref{geo:labels}. Equivalently, the right operator is
$R_{\widehat j}=L_{\pi(\widehat j)}\rho(\widehat j)^{-1}$.
Thus the scalar kernel acts by $\lambda$ on the right and by
$\lambda^{-1}$ under $\rho$.

For a left $\cO G b$-module $V$, define
\begin{equation}\label{geo:hom-functor}
 \Phi(V)=\Hom_{\cO D b}(M,V),\qquad
 (\widehat h f)(m)=\pi(\widehat h)
                      f\bigl(\rho(\widehat h)^{-1}m\bigr).
\end{equation}
For $\widehat j\in\widehat J$, relation
\eqref{geo:literal-relations} reduces this to $f(m\widehat j)$,
the ordinary local Morita action. Thus $\Phi(V)$ is an actual
$\cO\widehat Hc$-module in the prescribed scalar sector.
Conversely, for such a module $W$, put
\begin{equation}\label{geo:tensor-functor}
 \Psi(W)=M\otimes_{\cO\widehat J c}W,
 \qquad
 h(m\otimes w)=\rho(\widehat h)m\otimes\widehat h w,
                   \quad\pi(\widehat h)=h.
\end{equation}
The two scalar kernel actions cancel, so the formula is independent
of the lift. On $J=D\cap H$ it agrees with the left $D$-action by
\eqref{geo:literal-relations}. It also respects conjugation by $H$.
The resulting representation of $D\rtimes H$ kills the exact kernel
$\{(j^{-1},j):j\in J\}$ of multiplication onto $DH$. It is
therefore a representation of the group $G=DH$.

Local Morita evaluation is equivariant. Explicitly,
$m\otimes f\mapsto f(m)$ sends
$\rho(\widehat h)m\otimes\widehat h f$ to
$\pi(\widehat h)f(m)$. The inverse coevaluation is equivariant
by the same formulas. Thus \eqref{geo:hom-functor} and
\eqref{geo:tensor-functor} are inverse integral equivalences.
For $c_0\in C$, the operator $\rho(c_0)$ is $1$; both functors
therefore retain its original operator on the module factor.
In particular, every $C$-block idempotent retains its action.


Finally, integral Morita equivalence of the sums in
\eqref{geo:ambient-equivalence} bijects their primitive central
idempotents. It sends the specified $B$ to one $\widehat B$.
Extension of scalars preserves the number of generic simple
modules, giving $k(\widehat B)=k(B)$. The Cartan elementary divisors
are preserved, and their largest member is the defect order of a
group block; hence $|D_{\widehat B}|=|D_B|$.
Since $Z(G)\leq C$ and the cocycle has value $1$ on every mixed
pair involving $C$, the full preimage of $Z(G)$ is central in
$\widehat H$. Therefore
\[
        |\widehat H:Z(\widehat H)|
          \leq |H:Z(G)|<|G:Z(G)|.
\]
This proves Theorem~\ref{thm:levi-descent}.

\begin{remark}\label{geo:minimality-essential}
The canonical minimal Levi in \eqref{geo:minimal-levi} makes the
odd label central in types $A$, $D$ and $E_6$ in
Lemma~\ref{geo:fixed-parabolic}. For types with odd-order diagram
group, the proof uses the trivial relative action of the factor
stabilizer.
\end{remark}

\section{Principal blocks in type \texorpdfstring{$A$}{A}}\label{sec:type-a-counts}

We count all constituents on restriction through the determinant
quotient. A character with scalar stabilizer of order $s$
contributes $s^2$ to the orbit sum.
Let $q$ be an odd prime power, let $\epsilon\in\{1,-1\}$, and set
\[
 T_n=\begin{cases}\GL_n(q)&\epsilon=1,\\
                       \operatorname{GU}_n(q)&\epsilon=-1,\end{cases}
 \qquad
 N_n=\begin{cases}\SL_n(q)&\epsilon=1,\\
                       \operatorname{SU}_n(q)&\epsilon=-1.\end{cases}
\]
We also write $\GL_n(-q)=\operatorname{GU}_n(q)$
and $\SL_n(-q)=\operatorname{SU}_n(q)$.
Write $a=v_2(q-\epsilon)$, $b=v_2(q+\epsilon)$, and $A=2^a$.
Exactly one of $a,b$ is one and the other is at least two.
Let $K(n)=k(B_0(T_n))$, with $K(0)=1$, and let
$P(t)=\prod_{i\ge1}(1-t^i)^{-1}$.

\begin{lemma}\label{lem:type-a-series}
One has
\begin{equation}\label{eq:type-a-series}
 F_{a,b}(t):=\sum_{n\ge0}K(n)t^n
 =P(t)^A P(t^2)^{2^{a-1}(2^b-1)}
   \prod_{r\ge2}P(t^{2^r})^{2^{a+b-2}}.
\end{equation}
For $n\ge1$, put $m=\min\{a,v_2(n)\}$. Then
\begin{equation}\label{eq:type-a-restriction}
 k(B_0(N_n))=\frac{S(n)}A,
 \qquad
 S(n)=K(n)+\sum_{j=1}^{m}3\cdot4^{j-1}K(n/2^j).
\end{equation}
\end{lemma}
\begin{proof}
Ordinary Jordan labels in $B_0(T_n)$ are partitions attached to
the orbits of $\epsilon q$ on the roots of unity of $2$-power
order. A root orbit of length $d$ carries one partition and gives
the factor $P(t^d)$. The lifting-the-exponent formula gives $A$
orbits of length one, $2^{a-1}(2^b-1)$ of length two, and
$2^{a+b-2}$ of each length $2^r$ for $r\ge2$. This proves
\eqref{eq:type-a-series}; see also \cite[Lemma~1.2]{Brenner20}.

Here are the block facts needed to pass to $N_n$. For an odd
semisimple label $s$, the set $\cE_2(T_n,s)$ is a single block.
The block with $s=1$ covers just the principal block of $N_n$,
and the other blocks covering that block are its odd scalar
twists. These are the ordinary covering statements in
\cite[Sections~4 and~6, Theorem~6.8 and Remark~6.9]{FLZ23}.
Consequently, if $X=\Irr(B_0(T_n))$ and $C$ is the cyclic group
of determinant characters of $2$-power order, then $|C|=A$,
each full determinant-twist orbit meeting $X$ meets it in one
$C$-orbit, and every stabilizer is contained in $C$.

Restriction through a cyclic quotient is multiplicity free.
An orbit with stabilizer order $s$ has $A/s$ members and supplies
$s$ distinct constituents on $N_n$. Hence
\[
 k(B_0(N_n))=\frac1A\sum_{\chi\in X}|C_\chi|^2.
\]
For the subgroup $C_j\le C$ of order $2^j$, write $X_j=X^{C_j}$.
Counting ordered pairs in a cyclic stabilizer by the subgroup
they generate gives
\begin{equation}\label{eq:type-a-square}
 \sum_{\chi\in X}|C_\chi|^2
 =|X|+\sum_{j=1}^a3\cdot4^{j-1}|X_j|.
\end{equation}
We prove that the generating function for $|X_j|$ is
$F_{a,b}(t^{2^j})$.

Suppose first $a\ge2$, so $b=1$. Scalar multiplication of order
$2^j$ has orbits of length $2^j$ on the $A$ root labels of
degree one. For each $r\ge1$, the $A/2$ root orbits of degree
$2^r$ arise from primitive roots of order $2^{a+r}$. The
subgroup generated by $\epsilon q$ on these roots is the group
of units congruent to one modulo $2^a$. Thus scalar multiplication
has orbit length $2^{\max\{j-r,0\}}$ on their Frobenius orbits.
Equal partitions are required along a scalar orbit. The resulting
exponent of $P(t^{2^j})$ is
\[
 \frac A{2^j}+\sum_{r=1}^{j-1}2^{a+r-j-1}+\frac A2=A,
\]
and that at every subsequent weight $2^r$, $r>j$, is $A/2$.
This is exactly $F_{a,1}(t^{2^j})$.

If $a=1$, only $j=1$ occurs. Put $B=2^b$. The labels $1,-1$
are interchanged, giving $P(t^2)$. Among the $B-1$ degree-two
root orbits, precisely the orbit of primitive fourth roots is
fixed by multiplication by $-1$; the others occur in pairs.
Their contribution is $P(t^2)P(t^4)^{B/2-1}$.
Every root orbit of degree $2^r$, $r\ge2$, is fixed by $-1$:
on roots of order $2^{b+r}$ the even powers of $\epsilon q$
generate the units congruent to one modulo $2^{b+1}$, including
the exponent change for multiplication by $-1$. They contribute
$\prod_{r\ge2}P(t^{2^r})^{B/2}$. Combining the factors again
gives $F_{1,b}(t^2)$. Thus
\[
 |X_j|=\begin{cases}K(n/2^j)&2^j\mid n,\\0&2^j\nmid n.
                 \end{cases}
\]
Substitution in \eqref{eq:type-a-square} proves the lemma.
\end{proof}

\subsection{A uniform coefficient estimate}

We give the elementary partition estimates used below. If $p_2(n)$
counts partitions into powers of two, then
\begin{equation}\label{eq:type-a-binary}
 p_2(n)\le2^{\lfloor n/2\rfloor}.
\end{equation}
Indeed $p_2(2u+1)=p_2(2u)$ and
$p_2(2u)=p_2(2u-2)+p_2(u)\le2p_2(2u-2)$ for $u\ge2$.
The assertion follows from $p_2(0)=p_2(1)=1$, $p_2(2)=2$.
If $k(c,n)=[t^n]P(t)^c$, then
\begin{equation}\label{eq:type-a-multipartition}
 k(c,n)\le c^n\quad(c\ge3),\qquad k(2,n)\le3^n.
\end{equation}
The quantity $k(c,n)$ is a polynomial in $c$ with nonnegative
coefficients and degree at most $n$. Therefore $k(c,n)/c^n$
is nonincreasing for positive $c$, and it suffices to take $c=3$.
With $\sigma_1(j)=\sum_{d\mid j}d$, logarithmic differentiation gives
\[
 n k(3,n)=3\sum_{j=1}^n\sigma_1(j)k(3,n-j).
\]
The cases $n=1,2$ are $3,9$. For the induction at $n\ge3$, use
\[
 \sum_{j\ge1}\frac{\sigma_1(j)}{3^j}
 =\sum_{d\ge1}\frac{d3^{-d}}{1-3^{-d}}
 \le\frac12+\frac98\left(\frac34-\frac13\right)
 =\frac{31}{32}<1.
\]
This proves \eqref{eq:type-a-multipartition}.

The order formula and the lifting-the-exponent formula give
\begin{equation}\label{eq:type-a-sylow}
 \Delta_n:=|T_n|_2
 =2^{an+(b-1)\lfloor n/2\rfloor+n-s_2(n)},
\end{equation}
where $s_2(n)$ is the binary digit sum. We claim
\begin{equation}\label{eq:type-a-budget}
 2^m S(n)<\Delta_n\qquad(n\ge3).
\end{equation}

If $a\ge2$, expand \eqref{eq:type-a-series} by binary
decompositions of $n$ and use \eqref{eq:type-a-multipartition}.
Every term is at most $A^n$, so $K(n)\le p_2(n)A^n$.
For $n\ge4$ this yields
\[
 S(n)\le p_2(n)A^n\bigl(1+(n^2-1)A^{-n/2}\bigr)
 \le2p_2(n)A^n,
\]
because $n^2\le2^n$. Since $2^m\le n$ and
$2^{s_2(n)}\le n+1$,
\begin{equation}\label{eq:type-a-large-plus}
 \frac{2^mS(n)}{\Delta_n}\le2n(n+1)2^{-n/2}<1
 \quad(n\ge20).
\end{equation}
The last expression is $840/1024$ at $20$ and decreases thereafter.
For $3\le n\le19$, enlarge $m$ to $v_2(n)$. Every coefficient
$K(w)$ is a polynomial in $A$ with nonnegative coefficients and
degree at most $w$. Thus the resulting ratio is largest at $A=4$.

If $a=1$, put $B=2^b\ge4$, $M=\lfloor n/2\rfloor$, and
enlarge the exponent $B-1$ of $P(t^2)$ to $B$. Denote the enlarged
coefficients by $\widetilde K(n)$. Their dependence on $B$ is
polynomial with nonnegative coefficients and degree at most $M$.
For a binary decomposition with $w_0$ parts of size one, the
other multiplicities $w_r$ satisfy
$\sum_{r\ge1}w_r\le M-\lfloor w_0/2\rfloor$. Equations
\eqref{eq:type-a-binary}--\eqref{eq:type-a-multipartition} give
\[
 \frac{K(n)}{\Delta_n}
 \le\frac{p_2(n)3^n}{2^{2n+M-s_2(n)}}
 \le(n+1)(3/4)^n.
\]
Here we used
$3^{w_0}/4^{\lfloor w_0/2\rfloor}\le3^n/4^M$,
since $w_0$ and $n$ have the same parity. The sequence $K(n)$
is nondecreasing, so $S(n)\le4K(n)$ and $2^m\le2$. In particular
\begin{equation}\label{eq:type-a-large-minus}
 \frac{2^mS(n)}{\Delta_n}
 \le8(n+1)(3/4)^n<12(n+1)(3/4)^n<1
 \quad(n\ge20).
\end{equation}
The last inequality at $20$ is $252\cdot3^{20}<4^{20}$,
and the expression decreases thereafter. For $3\le n\le19$,
replace $K$ by $\widetilde K$ and set $B=4$: after division by
$B^M$ each polynomial term is nonincreasing in $B$.

The remaining finite calculation is displayed
in Table~\ref{tab:type-a-boundary}. The plus columns mean
$K_+(n)=[t^n]F_{2,1}(t)$ and
$S_+(n)=K_+(n)+\sum_{j=1}^{v_2(n)}3\cdot4^{j-1}K_+(n/2^j)$.
The minus columns use $\widetilde K_-(n)$ at $a=1,B=4$ and
$\widetilde S_-(n)=\widetilde K_-(n)+3\widetilde K_-(n/2)$
for even $n$, with no correction for odd $n$. The inequalities
to check in each row are precisely
\[
 \frac{2^{v_2(n)}S_+(n)}{2^{3n-s_2(n)}}<1,
 \qquad
 \frac{2^{\min\{1,v_2(n)\}}\widetilde S_-(n)}
 {2^{2n+\lfloor n/2\rfloor-s_2(n)}}<1.
\]
Their largest value is $9/16$. The entries can be reproduced
using integer arithmetic alone: if
$F(t)=\prod_{d\ge1}(1-t^d)^{-e_d}=\sum K(n)t^n$, put
$h_j=\sum_{d\mid j}de_d$ and use
$nK(n)=\sum_{j=1}^nh_jK(n-j)$, starting from $K(0)=1$.

\begin{table}[htbp]
\centering\small
\caption{The complete finite boundary for \eqref{eq:type-a-budget}.}
\label{tab:type-a-boundary}
\begin{tabular}{rrrrr}\toprule
$n$&$K_+(n)$&$S_+(n)$&$\widetilde K_-(n)$&$\widetilde S_-(n)$\\\midrule
3 & 48 & 48 & 18 & 18 \\
4 & 140 & 236 & 56 & 83 \\
5 & 360 & 360 & 108 & 108 \\
6 & 896 & 1040 & 273 & 327 \\
7 & 2080 & 2080 & 510 & 510 \\
8 & 4696 & 5500 & 1145 & 1313 \\
9 & 10132 & 10132 & 2076 & 2076 \\
10 & 21344 & 22424 & 4281 & 4605 \\
11 & 43536 & 43536 & 7560 & 7560 \\
12 & 87024 & 90288 & 14665 & 15484 \\
13 & 169736 & 169736 & 25290 & 25290 \\
14 & 325376 & 331616 & 46761 & 48291 \\
15 & 611680 & 611680 & 78948 & 78948 \\
16 & 1132852 & 1150156 & 140526 & 143961 \\
17 & 2064664 & 2064664 & 232758 & 232758 \\
18 & 3714192 & 3744588 & 401470 & 407698 \\
19 & 6591664 & 6591664 & 653570 & 653570 \\
\bottomrule\end{tabular}
\end{table}

\begin{proposition}\label{prop:type-a-principal}
Every principal $2$-block of $N_n/Z$, where $Z\le Z(N_n)$,
satisfies $k(B_0(N_n/Z))\le|N_n/Z|_2$.
\end{proposition}
\begin{proof}
An odd central quotient retains exactly the principal central
sector and does not change the defect order. We may therefore
first consider a central $2$-subgroup $Z$. For $n\ge3$ its order
is at most $2^m$, while $|N_n|_2=\Delta_n/A$.
Inflation and \eqref{eq:type-a-budget} give
\[
 k(B_0(N_n/Z))\le k(B_0(N_n))
 =\frac{S(n)}A<\frac{\Delta_n}{A2^m}\le|N_n/Z|_2.
\]
The case $n=1$ is trivial. We include $n=2$ explicitly, since
discarding the central-character condition is too costly there.

For $a\ge2$, $K(2)=A(A+4)/2$ and
$S(2)=A(A+10)/2\le2A^2=\Delta_2$.
For $a=1$, $K(2)=B+4$ and $S(2)=B+10\le4B=\Delta_2$.
This proves the unquotiented assertion. The unique nontrivial
central $2$-subgroup has order two. Let $x_0$ be the number of principal
$T_2$ labels trivial on this central involution and let $x_1$
be the number among them fixed by the scalar involution. The same
restriction calculation gives $(x_0+3x_1)/A$ characters on the
central quotient. The ordinary torus pairing identifies the
central character of a Jordan label with its determinant label;
this follows already from the common central action on the two
sides of Deligne--Lusztig induction. Thus the following count is
on the required central sector.

If $a\ge2$, the degree-one labels are the $A$ roots with
exponents modulo $A$. The unordered pairs of distinct exponents
with even sum number $A(A-2)/4$; the repeated labels each carry
two partitions, adding $2A$. Degree-two root orbits have odd
determinant charge and do not occur. Hence
\[
 x_0=A(A+6)/4,\qquad x_1=A/2.
\]
The second equality counts pairs $\{\alpha,-\alpha\}$; their
determinant charge is even because $4\mid A$. Therefore
$2(x_0+3x_1)=A(A+12)/2\le2A^2$.
If $a=1$, the two repeated degree-one labels contribute four
characters, and the allowed degree-two root orbits contribute
$B/2-1$. Exactly the primitive-fourth-root orbit is scalar fixed.
Thus
\[
 x_0=B/2+3,\qquad x_1=1,\qquad
 2(x_0+3x_1)=B+12\le4B.
\]
Division by $A$ proves the bound against the quotient defect
$\Delta_2/(2A)$ in both cases. Finally take any odd central
quotient, which cannot increase the count or change its $2$-part.
\end{proof}

\begin{lemma}\label{lemma:type-a-small-ranks}
For either sign $\epsilon$ and every odd $q$,
\[
 \frac{k(B_0(N_3))}{|N_3|_2}\le\frac12,
 \qquad
 \frac{k(B_0(N_6))}{|N_6|_2}\le\frac1{16}.
\]
The same estimates hold after an odd central quotient.
\end{lemma}
\begin{proof}
For $n=3$, the quotient is $K(3)/\Delta_3$. If $a\ge2$, then
$K(3)=A(A^2+12A+8)/6$ and $\Delta_3=2A^3$; the ratio is
largest at $A=4$ and is at most $1/2$.
If $a=1$, the two quantities are $2B+8$ and $8B$, again giving
at most $1/2$ for $B\ge4$.

For $n=6$ the numerator is $K(6)+3K(3)$, including the scalar
stabilizer term. For $a\ge2$, polynomial positivity shows that
the ratio is largest at $A=4$. There $K(6)=896$, $K(3)=48$,
and $\Delta_6=16\cdot4^6$, giving $65/4096<1/16$.
For $a=1$, use the enlarged series above. Its corresponding
coefficients at $B=4$ are $273$ and $18$, and
$\Delta_6=128\cdot4^3$. The ratio is at most
$327/8192<1/16$. An odd central quotient only removes characters.
\end{proof}

\section{Principal blocks of classical groups in odd characteristic}
\label{sec:classical-counts}

We prove the principal-block bounds for classical components,
including the stronger estimate for $P\Omega_8^+(q)$ needed for
triality.

\begin{proposition}\label{prop:classical-principal}
The principal $2$-blocks of the simple symplectic and orthogonal
groups in odd characteristic satisfy $k(b)\le\delta(b)$.
For $S=P\Omega_8^+(q)$, $q$ an odd prime power, the stronger estimate
\begin{equation}\label{eq:D4-margin}
 k(B_0(S))\le\frac{149}{1024}|S|_2<\frac14|S|_2
\end{equation}
holds. The same Brauer inequality holds for every quasisimple
central quotient of the corresponding simply connected finite groups.
\end{proposition}

\subsection{The series to be counted}

Put $A=\max\{(q-1)_2,(q+1)_2\}\ge4$, and define the following
formal power series:
\begin{align}
 P(t)&=\prod_{m\ge1}(1-t^m)^{-1},&
 H_A(t)&=P(t)^{A/2-1}\prod_{r\ge1}P(t^{2^r})^{A/4},
 \label{eq:classical-H}\\
 \Psi(t)&=\sum_{j\ge0}t^{j(j+1)},&
 \Theta(t)&=\sum_{j\in\Z}t^{j^2},\nonumber\\
 B(t)&=P(t)^2\Psi(t),& O(t)&=P(t)^2\Theta(t),\nonumber\\
 F(t)&=P(t^2).&&\nonumber
\end{align}
The factor $H_A$ records all primary factors other than the
eigenvalues $1$ and $-1$ in a semisimple $2$-element. A reciprocal
pair of polynomials gives a general linear centralizer factor; a
self-reciprocal polynomial gives a unitary factor. In both cases
the unipotent characters are indexed by partitions of the
multiplicity. There are $A/2-1$ inverse--Frobenius orbits of rank
weight one, and $A/4$ of rank weight $2^r$ for each $r\ge1$.
Indeed, if $A=2^c$, the subgroup generated by $q$ and $-1$ on
the primitive $2^j$-roots, $j>c$, has order $2^{j-c+1}$.
This proves \eqref{eq:classical-H}. At weight one the factor is
linear for $q\equiv1\pmod4$ and unitary for $q\equiv-1\pmod4$.
The latter changes the orthogonal sign by the parity of the
multiplicity. Thus the signed primary series is $H_A(t)$ or
$H_A(-t)$, respectively.

The standard symbol parametrization of unipotent characters gives
$B(t)$ for either symplectic or odd special orthogonal factors:
a symbol of odd defect $2j+1$ has a bipartition after a staircase
of rank $j(j+1)$. Ordered even-defect symbols similarly give
$O(t)$ for the sum of the two full orthogonal types. These are
the parametrizations in \cite[Section~13.8]{Carter85} and
\cite[Sections~4--6]{Vinroot18}.
Writing $O_+$ and $O_-$ for the two types, Jacobi's product identity
gives
\begin{equation}\label{eq:orthogonal-signs}
 O_++O_-=O,
 \qquad O_+-O_-=P(t)^2\sum_{j\in\Z}(-1)^jt^{j^2}=F.
\end{equation}
The determinant-twist-fixed full orthogonal characters arise from
the pairs of degenerate special orthogonal symbols. Their series
is $F$, entirely of plus type. At rank zero its formal value is one.

For two full orthogonal factors with unipotent counts $x_1,x_2$
and determinant-twist-fixed counts $f_1,f_2$, restriction to the
determinant-one subgroup gives
\begin{equation}\label{eq:disconnected-count}
 \frac{x_1x_2+3f_1f_2}{2}.
\end{equation}
In fact a nonfixed pair of characters has one irreducible
restriction, whereas a determinant-fixed character splits into
two. When one rank is zero this is the familiar special orthogonal
formula $(x+3f)/2$. When both ranks vanish the formal value two
counts the two special orthogonal semisimple classes into which
the full orthogonal class splits. Thus the same expression counts
the entire Jordan series, including that boundary case. By
\eqref{eq:orthogonal-signs}, its sum over the two total signs is
$(O^2+3F^2)/2$, and its signed difference is $2F^2$.

Jordan decomposition, including the disconnected centralizers in
\eqref{eq:disconnected-count}, now gives the following exact
counts of the full $2$-series at the identity:
\begin{align}
 K_C(n,A)&=|\cE_2(\Sp_{2n}(q),1)|=[t^n]BOH_A,\nonumber\\
 K_B(n,A)&=|\cE_2(\mathrm{SO}_{2n+1}(q),1)|=[t^n]B^2H_A.
 \label{eq:BC-series}
\end{align}
For the even special orthogonal group of sign $\eps$, call the
sign good when $q^n\equiv\eps\pmod4$ and bad otherwise. For
$n\ge1$ the two series counts are
\begin{align}
 K_g(n,A)&=[t^n]\frac{H_A}{4}(O^2+7F^2),\nonumber\\
 K_b(n,A)&=[t^n]\frac{H_A}{4}(O^2-F^2).
 \label{eq:D-series}
\end{align}
For $q\equiv-1\pmod4$, the sign at degree $n$ is multiplied by
$(-1)^n$, because $F^2$ contains only even powers. This gives the
stated convention for the good sign. The constant terms are
formal boundary terms; the formulas are used for $n\ge1$.

\subsection{Uniform coefficient bounds}

Let $s_2(n)$ be the number of nonzero binary digits of $n$, and
$n_2=2^{v_2(n)}$. The Sylow order for the symplectic and odd
special orthogonal groups is
\begin{equation}\label{eq:classical-delta}
 \Delta_n=\prod_{i=1}^n(q^{2i}-1)_2
         =\frac{(4A)^n}{2^{s_2(n)}}.
\end{equation}
We prove
\begin{align}
 2K_C(n,A)&\le\Delta_n,&4K_B(n,A)&\le\Delta_n
 &&(n\ge2),\label{eq:BC-budgets}\\
 16K_g(n,A)&\le\Delta_n,&4A n_2K_b(n,A)&\le\Delta_n
 &&(n\ge4).\label{eq:D-budgets}
\end{align}

For $B$ and $C$, the coefficients in \eqref{eq:BC-series} are
polynomials in $A$ with nonnegative coefficients and degree at
most $n$. To see the positivity, combine the four factors $P$
from $B,O$ with $P^{A/2-1}$, obtaining $P^{A/2+3}$, and use the
nonnegative logarithmic coefficients of the other Euler factors.
Therefore $K_C(n,A)/A^n$ and $K_B(n,A)/A^n$ decrease with $A$.

For $D$, replace $H_A$ coefficientwise by
\[
 \widehat H_A=P H_A
 =P^{A/2}\prod_{r\ge1}P(t^{2^r})^{A/4},
\]
and denote the resulting upper counts by $\widehat K_g$ and
$\widehat K_b$. They also have nonnegative polynomial coefficients
in $A$. Moreover $(O^2-F^2)/4=O_+O_-$ has zero constant term,
so the degrees of these two polynomials are at most $n$ and
$n-1$, respectively. Hence both
$\widehat K_g(n,A)/A^n$ and
$A\widehat K_b(n,A)/A^n$ decrease with $A$.
It suffices to establish all four inequalities at $A=4$.

Set $t=1/8$ and $J(t)=\prod_{r\ge1}P(t^{2^r})$.
The elementary bounds
\[
 \log P(x)\le\frac{x}{(1-x)^2},\qquad
 P(1/8)<\frac65,\qquad J(1/8)<\frac{48}{47}
\]
follow by expanding the logarithm. For the last bound, use
$\log J(t)\le t^2/(1-t^2)^3<1/48$ and
$e^x\le(1-x)^{-1}$. Also
\[
 \Psi(1/8)\le\frac{64}{63},\qquad
 \Theta(1/8)\le\frac97,\qquad F(1/8)<\frac65.
\]
At $A=4$ the four series in question are
\[
 P^5\Psi\Theta J,\quad P^5\Psi^2J,\quad
 \frac J4(P^6\Theta^2+7P^2F^2),\quad
 \frac J4(P^6\Theta^2-P^2F^2).
\]
The displayed rational bounds show that all four have value less
than eight at $t=1/8$. Their coefficients are nonnegative, so
each degree-$n$ coefficient is less than $8\cdot8^n$. Since
$2^{s_2(n)}\le n+1$ and $n_2\le n$, every normalized
left side of \eqref{eq:BC-budgets}--\eqref{eq:D-budgets} is at most
\[
 \frac{128n(n+1)}{2^n}.
\]
At $n=15$ this is $15/16$, and the ratio of consecutive values
is $(n+2)/(2n)<1$. This proves the inequalities for $n\ge15$.

The remaining coefficients are obtained by multiplying the
displayed Euler products modulo $t^{15}$:
\begin{center}
\begin{tabular}{r r r r r}
\toprule
$n$&$K_C(n,4)$&$K_B(n,4)$&$\widehat K_g(n,4)$&$\widehat K_b(n,4)$\\
\midrule
2&32&23&28&12\\
3&119&80&88&56\\
4&386&256&298&208\\
5&1134&731&866&694\\
6&3089&1968&2492&2092\\
7&7918&4963&6632&5896\\
8&19302&11985&17168&15632\\
9&45098&27701&42274&39542\\
10&101587&61927&101144&95864\\
11&221642&134048&233464&224344\\
12&470124&282575&525162&508480\\
13&972372&580875&1148968&1120888\\
14&1966088&1168480&2458196&2408996\\
\bottomrule
\end{tabular}
\end{center}
For example, the coefficients of an Euler product
$U(t)=\prod_{m\ge1}(1-t^m)^{-a_m}$ may be generated from
$u_0=1$ by
\[
 n u_n=\sum_{j=1}^n\left(\sum_{d\mid j}d a_d\right)u_{n-j}.
\]
Together with the sparse series $\Psi,\Theta$, this recurrence
independently specifies every entry of the table. Substitution in
\eqref{eq:classical-delta} proves the remaining 48 inequalities
(the $D$ columns are only used for $n\ge4$). The largest ratios
are $1/2$, $23/32$, $149/1024$ and $13/32$, respectively.

\begin{proof}[Proof of Proposition~\ref{prop:classical-principal}]
The principal block is contained in the counted $2$-series.
For $P\Sp_{2n}(q)$ the Sylow order is $\Delta_n/2$, and
quotient characters form a subset of the symplectic characters.
For $\Omega_{2n+1}(q)$ it is again $\Delta_n/2$. The index-two
inclusion $\Omega\lhd\mathrm{SO}$ gives a unique block of
$\mathrm{SO}$ covering $B_0(\Omega)$; it contains the trivial
character, so it is $B_0(\mathrm{SO})$. Every character of
$B_0(\Omega)$ occurs in a restriction from this block, and each
such restriction has at most two distinct constituents. Thus
restriction costs at most a factor two, giving
\eqref{eq:BC-budgets}.

For even orthogonal groups,
\[
 |P\Omega_{2n}^{\eps}(q)|_2=
 \begin{cases}
 \Delta_n/8,&q^n\equiv\eps\pmod4,\\
 \Delta_n/(2A n_2),&q^n\equiv-\eps\pmod4.
 \end{cases}
\]
The same unique-covering-block argument makes restriction to
$\Omega$ cost at most two, and the subsequent
central quotient only removes characters. Thus
\eqref{eq:D-budgets} proves the assertion. For $n=4$, plus type
is good and the same comparison, using the table and the
monotonicity in $A$, gives the ratio $149/1024$ in
\eqref{eq:D4-margin}. Lower ranks are covered by the standard
type-$A$ isomorphisms. Finally, a central $2$-extension multiplies
the defect order by its kernel size, and
Lemma~\ref{lem:central-fiber}, summed over its central characters,
increases the block count by at most that same factor. Odd central
quotients in the principal sector preserve the count. This proves
the assertion for these central quotients.
\end{proof}

The principal block is the only unipotent $2$-block for a classical
derived component in odd characteristic; the precise statement
used here is recorded in \cite[Section~5]{FLLMZ19} from the
Cabanes--Enguehard classification. Noncentral odd semisimple
labels are handled by the proper-Levi descent, not by the
principal-series estimate alone.

\section{Local modules and integral ramification}
\label{sec:component-modules}

Throughout this section $L$ is quasisimple with center of odd order,
$b$ is a $2$-block of $L$, and $P$ is a nonabelian defect group of $b$.
An algebraic central cover is a central quotient of the fixed points
of a simply connected algebraic group; for Suzuki and Ree groups
we use the defining Steinberg endomorphism. Exceptional odd covers
are the additional odd parts of the finite Schur cover.
Write $\delta=|P|=2^d$ and
$\Aut(L,b)=\{a\in\Aut(L):{}^ab=b\}$. If
\[
 \Inn(L)\le A\le\Aut(L,b),\qquad
 T=\{a\in A:a\text{ is inner on }\cO Lb\},\qquad R=A/T,
\]
then $R$ acts on $\Irr(b)$. Innerness is implemented by a unit of
the integral block algebra $\cO Lb$.

We shall construct $\F _2R$-modules $W$ with
\begin{equation}\label{eq:component-marks}
 |W|=\delta,\qquad |\Irr(b)^J|\le |W^J|\quad(J\le R),\qquad
 \ker(R\longrightarrow\GL(W))\text{ a }2\text{-group}.
\end{equation}
The use of every subgroup in this condition includes subgroups of
even order and mixed field--graph subgroups. By Schreier's theorem,
$R$ is solvable, since $T$ contains $\Inn(L)$.

\begin{lemma}[The intrinsic ramification bound]\label{comp:murai-index}
With the notation above, let $E_b$ be the inertial quotient of a
maximal $b$-Brauer pair. Then
\begin{equation}\label{eq:intrinsic-index}
 |R|_{2'}\le
 \frac{|\Aut(P)|_{2'}}{|E_b|}\,|E_b^{\ab}|
 \le |\Aut(P)|_{2'}.
\end{equation}
\end{lemma}

\begin{proof}
We give the argument for an invariant block $b$ of an arbitrary
normal subgroup $N\lhd H$. Fix a maximal pair $(P,e)$ and put
\[
 J=N_H(P,e),\quad J_0=N_N(P,e),\quad C=C_H(P)_e,
 \quad C_0=C_N(P),\quad H_1=NC.
\]
Conjugacy of maximal pairs gives $H=NJ$. If $I,I_0$ denote the
images of $J,J_0$ in $\Aut(P)$, respectively, then
\[
 H/H_1\cong I/I_0,\qquad |I_0|_{2'}=|E_b|,
 \qquad E_b=J_0/(PC_0).
\]
In particular,
\[
 |H:H_1|_{2'}\le |\Aut(P)|_{2'}/|E_b|.
\]
The canonical character of $e$ is the inflation of the unique
character of the defect-zero block of $C_0/Z(P)$ dominated by $e$.
It determines $e$, so an element centralizing $P$ stabilizes this
character exactly when it stabilizes $e$. Thus $C$ retains the
canonical-character stabilizer required in Murai's pairing.
Murai's pairing theorem, in the form of
\cite[Lemma~3.12 and Theorem~3.13]{Murai13}, gives a homomorphism
\[
 C\longrightarrow\Hom(E_b,K^\times)
\]
with kernel $C_!$ such that $C_0\le C_!$ and
\[
 H[b]=NC_!\le H_1.
\]
Here $H[b]$ is the integral ramification group. Since
$N\cap C=C_0\le C_!$, we have
$H_1/H[b]\cong C/C_!$, the image of this homomorphism. Its
order is odd and at most $|E_b^{\ab}|$. Multiplication of the two
indices proves the first inequality; the second follows from
$|E_b^{\ab}|\le |E_b|$.

Apply this argument to the semidirect product
$H=L\sd A$ and $N=L$. An element $(x,a)$ acts on $\cO Lb$ as
the composition of conjugation by $x$ and $a$. Consequently
$H[b]=L\sd T$ and $H/H[b]\cong A/T$. This proves the assertion
for every central character of $b$, as well as for centreless $L$.
\end{proof}

\begin{lemma}[Two elementary module constructions]\label{comp:small-odd}
The following conditions each imply \eqref{eq:component-marks}.
\begin{enumerate}
\item $R$ is a $2$-group and $k(b)\le\delta$.
\item $|R|_{2'}\le3$ and $k(b)\le\delta/4$.
\end{enumerate}
\end{lemma}

\begin{proof}
In the first case take the trivial module $\F _2^d$. For the second
case put $U=R/O_2(R)$. If $U\ne1$, solubility implies
$F(U)\ne1$. Since $O_2(U)=1$ and $|U|_{2'}\le3$, its Fitting
subgroup is $C_3$. The self-centralizing property of the Fitting
subgroup yields $U/C_3\le\Aut(C_3)=C_2$. Hence
\[
 U\in\{1,C_3,S_3\}.
\]
Use the natural two-dimensional module $E=\F _2^2$ for
$S_3=\GL_2(2)$, restricted to $C_3$ when appropriate, and inflate
it to $R$. If $U\ne1$, take $W=E\oplus\F _2^{d-2}$.
Its kernel is $O_2(R)$, and every fixed space has order at least
$2^{d-2}=\delta/4$. This dominates $|\Irr(b)^J|\le k(b)$.
For $U=1$ use the first construction.
\end{proof}

\begin{lemma}[Defect groups with small odd automorphism part]
\label{comp:defect-aut}
The following assertions hold.
\begin{enumerate}
\item If $Z(P)$ is cyclic and $P/Z(P)$ is nonabelian dihedral,
then $\Aut(P)$ is a $2$-group. In particular this applies to
$C_{2^a}\wr C_2$ and $\mathrm{SD}_{2^{a+2}}$, $a\ge2$.
\item If $P$ is a product of $r\le3$ wreathed or semidihedral
groups of these orders, then $|\Aut(P)|_{2'}\le(r!)_{2'}$.
\item Suppose $P$ contains a normal subgroup
$D\cong C_{2^a}\times C_{2^a}$, $a\ge2$, and
$P/D\cong V_4=C_2\times C_2$
acts faithfully on $D$ by the matrices $-I$ and
$\left(\begin{smallmatrix}0&1\\1&0\end{smallmatrix}\right)$.
Then $\Aut(P)$ is a $2$-group.
\end{enumerate}
\end{lemma}

\begin{proof}
For (i), the map
\[
 \Aut(P)\longrightarrow\Aut(Z(P))\times\Aut(P/Z(P))
\]
has $2$-group image. Indeed, an automorphism of the dihedral
group of order $2^n$, $n\ge3$, maps its characteristic rotation
generator to an odd power and a reflection to a rotation times
a reflection. Its automorphism group has order $2^{2n-3}$.
An element $\alpha$ of the displayed kernel has the form
$\alpha(x)=x f_\alpha(xZ(P))$, where
$f_\alpha\in\Hom(P/Z(P),Z(P))$. This embeds the kernel in a
$2$-group. For the wreathed group
$\langle x,y,t\mid x^{2^a}=y^{2^a}=t^2=1,[x,y]=1,x^t=y\rangle$,
the center is $\langle xy\rangle$ and the quotient is dihedral
of order $2^{a+1}$. The semidihedral presentation
\[
 \langle x,t\mid x^{2^{a+1}}=t^2=1,
             x^t=x^{-1+2^a}\rangle
\]
has center $\langle x^{2^a}\rangle$ and the same dihedral
quotient.

Write $\Phi(P)$ for the Frattini subgroup, the intersection of the
maximal subgroups of $P$. For (ii), each factor is directly indecomposable because its
center is cyclic, and its center lies in its Frattini subgroup.
For a wreathed factor this follows from
$xy\in\langle x^2,y^2,xy^{-1}\rangle=\Phi(P)$; the
semidihedral assertion follows immediately from its presentation.
The central-automorphism form of the Remak theorem
\cite[Theorem~12.1 and the proof of Theorem~12.8]{MarinMichel10}
says that automorphisms of the product are generated by central
automorphisms, factor automorphisms and permutations of isomorphic
direct factors. Central automorphisms form a normal subgroup and
act trivially on $P/\Phi(P)$, so they belong to the $2$-group kernel
in Burnside's basis theorem. Modulo this normal $2$-subgroup, the
automorphism group is an image of the product of the factor
automorphism groups extended by permutations of isomorphic factors.
The factor automorphism groups are $2$-groups by (i). Thus only
factor permutations contribute to the odd part, giving the claimed bound.

For (iii), let $B$ be an abelian subgroup not contained in $D$
and write $V=BD/D\ne1$. Then
\[
 B\cap D\le C_D(V),\qquad |B|=|B\cap D|\,|V|.
\]
For $V=\langle-I\rangle$ this gives $|B|\le8$; for either
reflection subgroup it gives $|B|\le2^{a+1}$; for $V=V_4$ it
again gives $|B|\le8$. Each bound is strictly smaller than
$|D|=2^{2a}$. Thus $D$ is the unique abelian subgroup of largest
order and is characteristic. The restriction image in
$\GL_2(\Z/2^a\Z)$ normalizes the given $V_4$. Modulo $2$ it
normalizes a nontrivial transvection subgroup of $\GL_2(2)$,
whose normalizer has order $2$. The congruence kernel is a
$2$-group, so the restriction image is a $2$-group. An
automorphism fixing $D$ pointwise acts trivially on $P/D$,
because its action on $D$ is faithful. Its errors on a fixed
transversal form a $1$-cocycle in $Z^1(V_4,D)$. The restriction
kernel therefore embeds in a $2$-group. This proves (iii).
\end{proof}

We shall also use the following published defect-group statements
with their exact scope. A block with defect a central extension
of a metacyclic $2$-group by a cyclic $2$-group satisfies
$k(b)\le|P|$ \cite[Theorem~2]{SambaleCartan13}. A metacyclic
$2$-group has $2$-group automorphism group unless it is homocyclic
abelian or $Q_8$ \cite[Lemma~1]{SambaleMeta12}. Thus dihedral,
wreathed and semidihedral defects in the preceding lemma give
the first construction of Lemma~\ref{comp:small-odd} for every
block stabilizer.

\section{Components in defining characteristic two}
\label{sec:defining-two}

Let $S$ be a simple group of Lie type in characteristic $2$, written
over $\F_q$ with $q=2^f$, and let $L$ be a perfect algebraic central
cover with odd center. The three simple derived groups arising
from the nonsimple fixed-point groups $\Sp_4(2)$, $G_2(2)$ and
$ {}^2F_4(2)$ are treated separately in the last table below;
until that point they are excluded. The defining-characteristic block theorem
of Humphreys, recalled in \cite[Proposition~8.1]{MalleNavarro11},
has the following precise form. The Steinberg character gives a
defect-zero block; every other block has full Sylow defect, and
there is one such block for each character of $Z(L)$. For $L=S$
this says
\begin{equation}\label{eq:defining-principal-characters}
 \Irr(B_0(S))=\Irr(S)\setminus\{\mathrm{St}\}.
\end{equation}
We first check the integral quotient on which our modules must act.

\begin{lemma}\label{comp:defining-inner}
For every positive-defect block $b$ of $L$ under the preceding
hypotheses, the subgroup of $\Aut(L,b)$ acting
innerly on $\cO Lb$ is $\Inn(L)$.
\end{lemma}

\begin{proof}
If $L=S$ is simple, an integral-inner automorphism fixes every
character of $b$. The Steinberg character is fixed by every
automorphism, so \eqref{eq:defining-principal-characters} shows
that it fixes all ordinary irreducible characters. The character
table then shows that it fixes every conjugacy class. The theorem
that every class-preserving automorphism of a finite simple group
is inner \cite[Theorem~C]{FeitSeitz89} proves the assertion.

A nontrivial algebraic odd center occurs only in types
$A_r,{}^2A_r,E_6,{}^2E_6$. The following argument applies to every
central-character sector.
Apply Murai's theorem to $L\sd A$, where $A=\Aut(L,b)$. Since
a defect group $P$ is Sylow, that theorem gives
\[
 (L\sd A)[b]\le L C_{L\sd A}(P).
\]
Thus every element of $T/\Inn(L)$ has a representative which
centralizes $P$ pointwise. We prove that such a representative
is inner.

Work first with a simply connected algebraic group $\bS$ and
Frobenius $F$, with a pinned Borel subgroup $\mathbf B=\bT\bU$.
Let $\bU_2$ be the
connected subgroup generated by positive root groups of height
at least two. The commutator relations imply
$[\bU,\bU]\le\bU_2$, and
\[
 \bU/\bU_2=\prod_{\alpha\text{ simple}}\bU_\alpha.
\]
Lang's theorem for the connected group $\bU_2$ gives a
surjection $\bU^F\to(\bU/\bU_2)^F$. We use this height quotient.

An automorphism normalizing $\bU^F$ has the standard pinned
form $\operatorname{Int}(ut)\sigma$, with $u\in\bU^F$,
$t$ in the rational adjoint torus, and $\sigma$ an ordinary
field--diagram operator. This follows from the standard generators
\cite[\S2.1]{Harper24} and
$N_{\bS_{\mathrm{ad}}^F}(\bU^F)=\mathbf B_{\mathrm{ad}}^F$.
In the split case the simple-root
coordinates of the height quotient are copies of $\F_q$.
Pointwise triviality first forces the diagram permutation to
be trivial and then gives
\[
 a_\alpha x^{2^e}=x\qquad(x\in\F_q)
\]
in each coordinate. Taking $x=1$ yields $a_\alpha=1$; the
remaining equality yields $e\equiv0\pmod f$. The simple roots
form a basis of the character lattice of the adjoint torus, so
$t=1$. The factor $u$ is inner.

In a twisted group write $F=\gamma F_2^f$. On the finite group
the basic field map $v=F_2$ satisfies
\[
 v^f=\gamma^{-1},\qquad v^{2f}=1.
\]
The height quotient contains a coordinate $\F_{q^2}$ from a
pair of diagram-interchanged roots. Surjectivity from Lang's
theorem includes the adjacent-root pair in type $A_2$. Trivial
action on this coordinate gives
$a x^{2^e}=x$ for all $x\in\F_{q^2}$, hence $a=1$ and
$e\equiv0\pmod{2f}$. Triviality on all the remaining
coordinates again forces the adjoint torus factor to be one.

Finally, let $\pi:\widetilde L\to L$ be the simply connected
presentation, with odd central kernel, and let $a\in\Aut(L)$.
The pinned inner-diagonal, field and diagram generators lift
its induced automorphism on $S$ to an automorphism
$\widetilde a$ of $\widetilde L$. The homomorphisms
$\pi\widetilde a$ and $a\pi$ agree modulo $Z(L)$, so their
pointwise quotient is a homomorphism from the perfect group
$\widetilde L$ to $Z(L)$. It is trivial. Hence
$\pi\widetilde a=a\pi$, which proves both kernel preservation
and existence of the lift; the same argument proves uniqueness.
If $a$ centralizes $P$ pointwise and $\widetilde P$ lies above
$P$, then
$\pi^{-1}(P)=\widetilde P\times\ker\pi$. Its unique Sylow
$2$-subgroup $\widetilde P$ is $\widetilde a$-stable, and
$\widetilde P\to P$ is an isomorphism. Thus $\widetilde a$
centralizes $\widetilde P$ pointwise. The preceding root-coordinate
argument proves the assertion for every algebraic odd central quotient.
\end{proof}

For a block in the central character $\zeta$, the complex
$\zeta$-sector of $L$ is a twisted group algebra of $S=L/Z(L)$.
Only cocycle-regular conjugacy classes contribute to its center.
Consequently
\begin{equation}\label{eq:defining-sector-bound}
 k(b)\le k_\alpha(S)\le k(S).
\end{equation}
The bound is independent of $|Z(L)|$. We use the simple-group
class estimate
\begin{equation}\label{eq:FG-bound}
 k(S)\le\frac{76}{5}q^r,
\end{equation}
where $r$ is the absolute Lie rank
\cite[Corollary~1.2(1)]{FulmanGuralnick12}. The small ranks
where this estimate is too coarse will be calculated separately.

\subsection{Outer modules}

Write $f=2^a m$, with $m$ odd. For odd $h$, put
\[
 U_h=\{(x_0,\ldots,x_{h-1})\in\F_2^h:
                         \textstyle\sum_i x_i=0\}.
\]
Cyclic rotation gives a faithful $C_h$-module of dimension
$h-1$: the permutation module is the direct sum of $U_h$ and
its trivial line, so an element trivial on $U_h$ is trivial
on the whole faithful permutation module. We allow $U_1=0$.
Let $E$ be the natural module of $S_3=\GL_2(2)$, with generators
\[
 c=\begin{pmatrix}0&1\\1&1\end{pmatrix},\qquad
 t=\begin{pmatrix}0&1\\1&0\end{pmatrix},\qquad c^t=c^{-1}.
\]

Here are two modules that retain diagonal automorphisms. For
an odd divisor $d$ of $q-1$, the split group
$C_d\sd(C_f\times C_2)$ has a faithful representation on
$\F_q\oplus\F_q$ given by
\begin{equation}\label{eq:split-outer-module}
 z(u,v)=(\zeta u,\zeta^{-1}v),\quad
 \phi(u,v)=(u^2,v^2),\quad \tau(u,v)=(v,u),
\end{equation}
where $\zeta$ has order $d$. The field map raises $z$ to its
square and $\tau$ inverts it. A transformation acting trivially
cannot interchange the two summands; evaluation at $1$ and
then at all elements of $\F_q$ forces both its scalar and
field parts to be trivial. For $d\mid q+1$ the twisted group
$C_d\sd C_{2f}$ acts faithfully on $\F_{q^2}$ by
\begin{equation}\label{eq:twisted-outer-module}
 z(u)=\zeta u,\qquad \phi(u)=u^2.
\end{equation}
The relation $\phi^f z\phi^{-f}=z^{-1}$ is the actual
field--graph relation. These representations both have
$\F_2$-dimension $2f$.

If an active module has dimension $e\le D$ and kernel a
$2$-group, adjoining $D-e$ trivial dimensions gives a module
$W$ of order $2^D$ for which
\begin{equation}\label{eq:fixed-floor}
 |W^J|\ge2^{D-e}\qquad\text{for every subgroup }J.
\end{equation}
By Lemma~\ref{comp:defining-inner}, these are modules on the
integral quotient; restricting from the full outer
group to any block stabilizer preserves the kernel condition.
The standard outer-group descriptions used below are realized
by pinned diagonal, field and diagram operators
\cite[Section~2.1 and Lemma~2.1]{Harper24}.

\subsection{Ranks one, two and three}

For $S=\mathrm{PSL}_2(q)$, $q\ge4$, take $W=\F_q$ with
its Frobenius action of $C_f$. For $J\le C_f$, write
$\F_{q_0}=\F_q^J$. Every nonzero trace labels one conjugacy
class in $\SL_2(q)$; trace zero gives the identity and the
single nonidentity unipotent class. Thus there are exactly
$q_0+1$ classes fixed by a generator of $J$. The permutation
representations on conjugacy classes and on irreducible
characters have equal traces, by the invertibility of the
character table. Since $J$ is cyclic, exactly $q_0+1$
irreducible characters are $J$-fixed. Removing the Steinberg
character gives
\[
 |\Irr(B_0(S))^J|=q_0=|W^J|.
\]
The action on $W$ is faithful, and $|W|=|S|_2=q$.

For rank two of type $A$, write
$d=(3,q-\eps)$, with $\eps=1$ in the linear case and
$\eps=-1$ in the unitary case. The exact class numbers are
\begin{equation}\label{eq:A2-classcounts}
\begin{array}{c|cc}
 &d=1&d=3\\ \hline
 \mathrm{PSL}_3(q)&q^2+q&(q^2+q+10)/3\\
 \mathrm{PSU}_3(q)&q^2+q+2&(q^2+q+12)/3.
\end{array}
\end{equation}
These formulas are given in \cite[Proposition~2.6]{MNR21} away
from its excluded linear cases $q=2,4$. In those cases the
ordinary character tables give $k(\mathrm{PSL}_3(2))=6$ and
$k(\mathrm{PSL}_3(4))=10$ \cite{CTblLib}, agreeing with
\eqref{eq:A2-classcounts}.
They also follow from the degree-three Macdonald formula:
the general linear and unitary coefficients are respectively
$(q-1)(q^2+q)$ and $(q+1)(q^2+q+2)$, and the two divisor
terms for $3$ add $2+8=10$ before division by the central
factor. In particular the unitary numerator is $q^2+q+12$.

If $d=1$, project the field part to $C_m$ and use $U_m$,
ignoring the graph involution. The kernel is a $2$-group,
whether the outer group is $C_f\times C_2$ or
$C_{2f}$. Here $D=3f$, and
\[
 2^{D-(m-1)}\ge2q^2\ge k(S).
\]
For the linear $d=3$ case, $f$ is even. Map the diagonal
generator to $c$, and both the field generator and graph
involution to $t$. Also map the field generator to a generator
of $C_m$. The resulting module $E\oplus U_m$ has dimension
$m+1$. Its kernel is generated by $\phi^m\tau$ and has
order $2^a$. Since $m\le f-1$,
\[
 2^{3f-m-1}\ge q^2\ge(q^2+q+10)/3\qquad(q\ge4).
\]
For the unitary $d=3$ case, $f$ is odd. The outer group is
$C_3\sd C_{2f}$ with the field generator inverting $C_3$.
Send this generator to $t$ and to a generator of $C_f$, and the
diagonal generator to $c$. On $E\oplus U_f$ this action is
faithful. Since $q=2$ is nonsimple, $q\ge8$, and
\[
 2^{3f-(f+1)}=q^2/2\ge(q^2+q+12)/3,
\]
because $q^2-2q-24\ge0$ for $q\ge8$.
Equation~\eqref{eq:defining-sector-bound} shows that these
same estimates include every odd central sector of the
algebraic central covers.

For $A_3$ and ${}^2A_3$ the diagonal center has no odd part
when $q$ is even. Use $U_m$ and pad to $D=6f$. For $q\ge4$
its fixed-space lower bound is at least $2q^5$, exceeding
$(76/5)q^3$. At $q=2$ the exact class numbers are $14$ and
$20$, respectively, both less than $2^6$. More generally the
class formulas here are $q^3+q^2+q$ and
$q^3+q^2+3q+2$; only their values at two are needed.

\subsection{The remaining families}

For $A_r,{}^2A_r$ with $r\ge4$, use
\eqref{eq:split-outer-module} or
\eqref{eq:twisted-outer-module}, with the diagonal
order $d=(r+1,q-\eps)$, and pad to
$D=fN$, $N=r(r+1)/2$. Since $N-r-2\ge4$,
\[
 2^{D-2f}=q^{N-2}\ge16q^r>(76/5)q^r.
\]
The same two constructions work for $E_6,{}^2E_6$, where
$r=6$, $N=36$ and $d=(3,q-\eps)$; the lower bound is
$q^{34}$.

In characteristic two, $B_r(q)\cong C_r(q)$. In the following
table the outer group has a cyclic odd
quotient $C_m$ and a $2$-group kernel. Use $U_m$ and pad to
$fN$ dimensions.
\begin{center}
\begin{tabular}{l c c c}
\toprule
Type&Range&$N$&Field--diagram group\\
\midrule
$C_r$&$r\ge3$&$r^2$&$C_f$\\
$D_r$&$r\ge5$&$r(r-1)$&$C_f\times C_2$\\
${}^2D_r$&$r\ge4$&$r(r-1)$&$C_{2f}$\\
$G_2$&$q\ge4$&$6$&$C_f$\\
$F_4$&&$24$&$C_{2f}$\\
$E_7$&&$63$&$C_f$\\
$E_8$&&$120$&$C_f$\\
\bottomrule
\end{tabular}
\end{center}
For every row, $N-r-1\ge3$. Its fixed-space floor therefore
satisfies
\[
 2^{fN-m+1}\ge2q^{N-1}\ge16q^r>(76/5)q^r.
\]
The cyclic $C_{2f}$ entry records the exceptional graph--field
relation in $F_4$.

For $C_2$ with $q\ge4$, the outer group is likewise the actual
cyclic group $C_{2f}$. For $q\ge8$ the construction with $U_m$
has floor $2q^3\ge(76/5)q^2$. At $q=4$, $m=1$, the outer
group is a $2$-group and the trivial module of order $q^4=256$
suffices, since $(76/5)q^2<256$.

For $D_4$, the outer group in even characteristic is
$S_3\times C_f$. Use $E\oplus U_m$, of dimension $m+1$,
and pad to $12f$. Its kernel is the $2$-part of the field
group, and its fixed-space floor is at least
$q^{11}/2>(76/5)q^4$. For ${}^3D_4$, the outer group
is cyclic of order $3f$. Put $h=(3f)_{2'}$ and use $U_h$.
Here $h\le3f$, so the floor in $12f$ dimensions is at least
$2q^9>(76/5)q^4$. This keeps the relation between triality
and field automorphisms.

For $S={}^2B_2(q)$, $q=2^f\ge8$ with $f$ odd, take $U_f$
and pad to $2f$ dimensions. The fixed-space floor is $2q$,
whereas $k(B_0(S))=q+2\le2q$. For
$S={}^2F_4(q)$, $q=2^f\ge8$ with $f$ odd, use $U_f$
and pad to $12f$ dimensions. The full class formulas give
\[
 k(B_0(S))\le q^2+4q+16\le2q^{11}.
\]
These class counts are included in the small-rank exceptional
class data used in \cite[Section~3]{Malle18} and
\cite{FulmanGuralnick12}. The remaining small derived groups
have the following principal-block counts:
\begin{center}
\begin{tabular}{l c c c}
\toprule
$S$&$|S|_2$&$k(B_0(S))$&$\Out(S)$\\
\midrule
$\Sp_4(2)'\cong A_6$&$8$&$5$&$V_4$\\
$G_2(2)'\cong\mathrm{PSU}_3(3)$&$32$&$12$&$C_2$\\
${}^2F_4(2)'$&$2048$&$20$&$C_2$\\
\bottomrule
\end{tabular}
\end{center}
In each group the principal block is the only positive-defect
$2$-block; the other two blocks have defect zero. The complete
lists of defect exponents and ordinary character counts are
\[
 (3,0,0),\ (5,1,1);\qquad
 (5,0,0),\ (12,1,1);\qquad
 (11,0,0),\ (20,1,1),
\]
respectively. They are obtained from the ordinary character tables
\cite{CTblLib} by the following commands; the last output identifies
the principal block by its trivial character.
\begingroup\small
\begin{verbatim}
LoadPackage("ctbllib");;
for nam in ["A6", "U3(3)", "2F4(2)'"] do
  t := CharacterTable(nam);;
  r := PrimeBlocks(t,2);;
  counts := List([1..Length(r.defect)],
                 i -> Number(r.block,j -> j=i));;
  principal := r.block[Position(Irr(t),TrivialCharacter(t))];;
  Print(nam," ",r.defect," ",counts," ",principal,"\n");
od;
\end{verbatim}
\endgroup
Thus every positive-defect block has full defect and satisfies
$k(b)\le\delta(b)$. Since each outer group is a $2$-group, the
trivial modules of dimensions $3,5,11$ satisfy
\eqref{eq:component-marks}.
Equations~\eqref{eq:defining-sector-bound} and
\eqref{eq:fixed-floor}, together with all the preceding
cases, prove the following statement.

\begin{proposition}\label{comp:defining-marks}
Every positive-defect block of a quasisimple algebraic central
cover in characteristic two with odd center satisfies
\eqref{eq:component-marks} for its integral quotient.
The assertion includes every nontrivial central character
sector and every block stabilizer. It also holds for the three
small simple derived groups in the preceding table.
\end{proposition}

\section{Principal blocks in odd characteristic}
\label{sec:odd-principal-marks}

We first establish integral innerness for the odd diagonal and
field automorphisms of a principal block. Let $N\lhd H$, assume
$H/N$ has odd order, and let $P\in\Syl_2(N)$. If
\begin{equation}\label{eq:principal-centralizer-extension}
 H=N C_H(P),
\end{equation}
the Alperin--Dade theorem gives a restriction bijection
\[
 \Irr(B_0(H))\longrightarrow\Irr(B_0(N))
\]
\cite{Alperin76,Dade77}. The implication (iii)$\Rightarrow$(i)
of \cite[Theorem~4.1]{Murai13} then gives
\begin{equation}\label{eq:principal-integral-inner}
 H=H[B_0(N)].
\end{equation}
We use the integral conclusion \eqref{eq:principal-integral-inner}.

Here is the diagonal verification of
\eqref{eq:principal-centralizer-extension}. Suppose $H=NT$,
where $T$ is abelian, $P$ normalizes $T$ and $H/N$ has odd
order. The odd Hall subgroup $T_{2'}$ maps onto $H/N$. Its quotient
action by $P$ is trivial, because $P\le N$. Coprime averaging
on the abelian group $T_{2'}$, or equivalently
$H^1(P,T_{2'}\cap N)=1$, shows that
\[
 C_{T_{2'}}(P)\longrightarrow H/N
\]
is surjective. For a finite reductive group, a rational maximal
torus in the adjoint group maps onto the diagonal quotient:
this follows from the central-isogeny cohomology sequence and
Lang's theorem for its covering torus. Choose a Sylow torus
normalized by $P$. The preceding argument supplies
\eqref{eq:principal-centralizer-extension} for the odd
diagonal extension.

For an odd field extension $q=q_0^h$, the identities
\begin{equation}\label{eq:odd-field-valuations}
 (q_0^{hi}-1)_2=(q_0^i-1)_2,\qquad
 (q_0^{hi}+1)_2=(q_0^i+1)_2\qquad(h\text{ odd})
\end{equation}
show that the corresponding same-type smaller-field group
contains a full Sylow $2$-subgroup. The central divisors in
the order formula have unchanged $2$-parts as well. Thus an
odd field subgroup centralizes such a $P$. For an order-two
twist, the same argument uses $F=F_0^h$, with the twist
unchanged because $h$ is odd. The torus used for the diagonal
step can be chosen over this smaller field. Consequently the
whole odd diagonal--field subgroup satisfies
\eqref{eq:principal-centralizer-extension}, including when
it is nonabelian.

The cyclic field--triality group of ${}^3D_4(q)$ requires a
separate order check. Write $q=r^f$, $f=2^a u$ with $u$ odd.
Its full odd outer subgroup has order $3u$ and a fixed subgroup
containing $G_2(r^{2^a})$. Since
\[
 |{}^3D_4(q)|_2=(q^2-1)_2^2
              =|G_2(r^{2^a})|_2,
\]
it also centralizes a full Sylow subgroup. Here the full
field--triality group is cyclic of order $3f$. For small Ree groups the analogous fixed
subgroup ${}^2G_2(3)$ has the full Sylow order $8$; all their
defects are abelian in any case.

For $G_2(3^f)$ the exceptional graph--field generator $\tau$
satisfies $\tau^2=F_3$, and the full outer group is $C_{2f}$.
If $f=2^a u$ with $u$ odd, its odd subgroup is generated by
$\tau^{2^{a+1}}=F_3^{2^a}$. It is therefore an ordinary odd
field subgroup, already contained in the integral-inner kernel.

For every ordinary simple group in odd characteristic other
than split $D_4$, the outer group has a normal odd
diagonal--field subgroup with $2$-group quotient. The preceding
argument puts that subgroup in the integral-inner kernel.
The principal block bounds of
Propositions~\ref{prop:type-a-principal} and
\ref{prop:classical-principal}, and the exceptional principal
block theorem \cite[Section~6, introduction and Proposition~6.1]{Malle18}, therefore give
\eqref{eq:component-marks} by the trivial-module construction.
The exceptional theorem cited here is specifically the
principal-block theorem; the nonprincipal $E_8$ calculation
will be given separately below.

\subsection{The split triality module}

Let $S=P\Omega_8^+(q)$ with $q$ odd. The diagonal--graph part
of its outer group is $V_4\sd S_3\cong S_4$, and the field
part commutes with it. Let $H\le S_3$ be the graph image of
the block stabilizer modulo inner automorphisms.
Odd field automorphisms are integral inner by the preceding
argument. All unipotent characters belong to the principal
$2$-block; diagonal and field automorphisms fix them individually,
whereas triality has two orbits of length three
\cite[p.~1138, before Theorem~5.4]{HungSchaefferFry23}.

If $H$ has no element of order three, then $A/T$ is a
$2$-group. Otherwise $H=C_3$ or $S_3$. The graph image of
$T$ is normal in $H$ and contains no element of order three,
because such an element would move a unipotent character.
Neither $C_3$ nor $S_3$ has a nontrivial normal subgroup of
$2$-power order, so this image is trivial. The natural
$E=\F_2^2$ representation of $H\le\GL_2(2)$ therefore
descends to $R=A/T$. Its kernel is a $2$-group: the remaining
diagonal group is $V_4$ and the remaining field group has
$2$-power order. Pad $E$ to dimension $d=\log_2|S|_2$.
The fixed-space floor is $\delta/4$, while
\eqref{eq:D4-margin} gives
\[
 k(B_0(S))\le(149/1024)\delta<\delta/4.
\]
This proves every subgroup inequality. For an algebraic central
cover with odd center, the principal sector is canonically the
principal block algebra of the simple quotient; this
identification is equivariant for the stabilizer and
preserves the defect order.

\begin{proposition}\label{comp:principal-marks}
Every principal $2$-block of a quasisimple algebraic central cover
of a simple group of Lie type in odd characteristic, with odd
center and nonabelian defect,
satisfies \eqref{eq:component-marks}.
\end{proposition}

\section{Isolated odd labels}
\label{sec:isolated-marks}

Let $\bG$ be a simple simply connected algebraic group
with Frobenius $F$ in odd characteristic, let
$(\bG^*,F^*)$ be dual to $(\bG,F)$, and let
$s\in(\bG^*)^{F^*}$ be semisimple of odd order.
The element $s$ is isolated if
$C_{\bG^*}(s)^\circ$ is not contained in any proper
Levi subgroup of $\bG^*$.
Put $A(s)=C_{\bG^*}(s)/C_{\bG^*}(s)^\circ$.
The isolated-element classification
\cite[Proposition~4.9 and Tables~II--III]{Bonnafe05}, together
with the $2$-block tables in \cite{KM13} and their correction,
gives the following noncentral odd isolated types:
\begin{equation}\label{eq:isolated-types}
\begin{array}{c|l}
 G_2&A_2\\
 F_4&A_2A_2\\
 E_6,{}^2E_6&A_2^3\quad(A(s)\cong C_3)\\
 E_7&A_5A_2\\
 E_8&A_8, A_4A_4, E_6A_2.
\end{array}
\end{equation}
The rational forms, including factor permutations, are retained.
The $E_6$ rational three-cycle entries consist of three blocks,
as recorded in the correction
\cite[Supplement, Section~2.1]{KM13supp}, also recorded in
\cite[Remark~6.8]{KM25}. In the other rows the centralizer is
connected. For $E_7$ this follows also because its component
group has exponent dividing both the odd label order and the
$2$-power algebraic fundamental group. For $G_2,F_4,E_8$ the
dual group is simply connected. The odd quasi-isolated $D_4$
centralizer in $E_6$ is not isolated and belongs to the
proper-Levi alternative.

\subsection{A bound for the full two-series of type \texorpdfstring{$E_6$}{E6}}

Let $(\bH,F)$ and $(\bH^*,F^*)$ be dual connected reductive
groups over $\F_q$, where $q$ is odd, and put
$H=\bH^F$ and $H^*=(\bH^*)^{F^*}$. Assume that the
centralizer in $\bH^*$ of every semisimple $2$-element is
connected. Let $n_2(H^*)$ be the number of semisimple
$2$-classes in $H^*$. Choose an $F^*$-stable maximal torus
$\bT^*$ of $\bH^*$, let
$W=N_{\bH^*}(\bT^*)/\bT^*$, and write $F^*=q\varphi$
on its character lattice, with $\varphi$ of finite order.
Then
\begin{equation}\label{eq:torus-mass}
 n_2(H^*)=\frac1{|W|}
      \sum_{w\in W}|\det(q\varphi w-I)|_2.
\end{equation}
Indeed, geometric semisimple $2$-classes are $W$-orbits
on the $2$-power torsion of $\bT^*$. An $F^*$-stable
orbit contributes exactly $|W|$ pairs $(w,x)$ with
$wF^*(x)=x$, and a nonstable orbit contributes none.
The fixed sets are finite; Smith normal form gives their
orders as the determinant $2$-parts in
\eqref{eq:torus-mass}. Connectedness of the centralizers
and Lang's theorem give one rational class in each
stable geometric class. Suppose that
$C_{\bH^*}(t)^{F^*}$ has at most $u$ unipotent characters
for every semisimple $2$-element $t\in H^*$.
Jordan decomposition then gives
\begin{equation}\label{eq:torus-jordan}
 |\cE_2(H,1)|\le u\,n_2(H^*)
 \le u\max_{w\in W}|\det(q\varphi w-I)|_2.
\end{equation}

For type $E_6$ one can take $u=30$. Here are details of the
finite root-system calculation. Label the affine diagram by
the edges
\[
 (0,2),(2,4),(1,3),(3,4),(4,5),(5,6),
\]
with highest-root marks
$(1,1,2,2,3,2,1)$ at nodes $0,1,\ldots,6$.
The alcove description of a semisimple centralizer
\cite[Proposition~3.13(a)]{Bonnafe05} says that its roots are
generated by the zero-coordinate nodes. For a $2$-element,
the sole omitted node cannot be node $4$, since its mark is
three. Thus one examines the $126$ subsets other than the
full affine diagram and the subset obtained by omitting
only node $4$. Deleting vertices in the displayed tree gives
the following maxima:
\begin{center}
\begin{tabular}{c c l}
\toprule
Semisimple rank&Maximum unipotent count&Types attaining it\\
\midrule
$0$&$1$&$1$\\
$1$&$2$&$A_1$\\
$2$&$4$&$A_1^2$\\
$3$&$8$&$A_1^3$\\
$4$&$16$&$A_1^4$\\
$5$&$20$&$D_5, A_3A_1^2$\\
$6$&$30$&$E_6$\\
\bottomrule
\end{tabular}
\end{center}
Let $p(n)$ be the number of partitions of $n$.
To reproduce this table, multiply the factor counts
$u(A_i)=p(i+1)$, $u(D_4)=14$, $u(D_5)=20$ and
$u(E_6)=30$ for each connected component of an induced
subgraph. The only allowed rank-six types are $E_6$ and
$A_5A_1$, the latter having $11\cdot2=22$ characters.
At rank five $D_4A_1$ does not occur in this tree; the
other possible products are bounded by $20$. All lower
ranks are bounded by $2^r$. These statements can also be
checked by choosing the omitted vertices successively
along each of the three arms of the displayed tree.
The values are the usual partition and symbol counts;
in particular $D_4$ triality has eight, the nonsplit $D_i$
counts do not exceed the split counts, and both $E_6$ and
${}^2E_6$ have thirty unipotent characters
\cite{Carter85,DudasMalle20}. Frobenius permutation of equal
factors replaces a product by one factor count and cannot
increase it. This proves $u=30$ for every rational form.

We also need a bound on the multiplicity of $-1$ in the
reflection representation of $W(E_6)$. It is at most four.
For if that multiplicity were at least five, orthogonality
would make the remaining subspace, of dimension at most one,
have eigenvalue $1$ or $-1$. The element would be an involution.
Carter's involution lemma \cite[Lemmas~2 and~5]{Carter72}
expresses it as the product of as many mutually orthogonal
root reflections as the dimension of its $-1$-space. The
roots perpendicular to any root of $E_6$ form $A_5$:
for the highest root this is seen by deleting its adjacent
node $2$ in the displayed affine diagram, and all roots are
Weyl conjugate. In the realization
$A_5=\{e_i-e_j:1\le i\ne j\le6\}$, orthogonal roots
have disjoint pairs of indices, so there are at most three.
Thus $E_6$ has at most four mutually orthogonal roots, a
contradiction.

Put $Q=\eps q$, where $q$ is an odd prime power and
$\eps=1$ in the split case and $\eps=-1$ in the
twisted case. Write $E_6(Q)_{\mathrm{sc}}$ for
$E_6(q)_{\mathrm{sc}}$ or ${}^2E_6(q)_{\mathrm{sc}}$,
respectively, and let
$A=\max\{(Q-1)_2,(Q+1)_2\}\ge4$. The diagram action in
the twisted case is $-w_0$, where $w_0$ is the longest Weyl-group
element. Let $\Phi_j$ denote the $j$th cyclotomic polynomial.
Replacing $q$ by $Q$ gives
the same torus determinants. For odd $Q$, only
$\Phi_1(Q),\Phi_2(Q)$ and $\Phi_{2^j}(Q)$ contribute to
the $2$-part; the latter have $2$-part two for $j\ge2$.
If $Q\equiv1\pmod4$, every torus has $2$-part at most
$A^6$. If $Q\equiv-1\pmod4$, the multiplicity bound just
proved gives at most $A^4\cdot2^2=4A^4$. The group orders are
\begin{equation}\label{eq:E6-orderparts}
 |E_6(Q)_{\mathrm{sc}}|_2=
 \begin{cases}
 128 A^6,&Q\equiv1\pmod4,\\
 512 A^4,&Q\equiv-1\pmod4.
 \end{cases}
\end{equation}
For example these follow at once from degrees $2,5,6,8,9,12$
in the order formula and the lifting-the-exponent identity.
The central divisors have odd order. Equations
\eqref{eq:torus-jordan} and \eqref{eq:E6-orderparts} prove
\begin{lemma}\label{comp:E6-series}
Let $\bH$ be a connected semisimple algebraic group of
type $E_6$, and let $F$ define a split or twisted
$\F_q$-structure, where $q$ is an odd prime power.
Then
\[
 |\cE_2(\bH^F,1)|\le\frac{15}{64}|\bH^F|_2.
\]
\end{lemma}
The central isogenies between these forms have odd degree.
Their dual semisimple
$2$-centralizers are connected, since a component group would
have exponent dividing both a power of two and the odd
fundamental group. The rational $2$-classes and the
unipotent-character counts are therefore unchanged.

\subsection{The three rational configurations in type \texorpdfstring{$E_6$}{E6}}

Let $G=E_6(\eps q)_{\mathrm{sc}}$, $q$ odd, and let $b$
have a noncentral isolated odd label. Its connected dual
centralizer has geometric type $A_2^3$ and component group
$C_3$. The rational action on the three factors is the
identity, a three-cycle, or a transposition. Each rational
$A_2$ factor has a wreathed or semidihedral Sylow subgroup
of order at least sixteen.

We spell out how restriction through a regular embedding is
counted. Choose a one-dimensional regular embedding
$G\le X$ with connected center. The group $X/G$ is cyclic
of order $q-\eps$; the map $Z(X)\to X/G$ has odd kernel
and odd cokernel, each of order dividing three. Let
$Z_2=O_2(Z(X))$ and let $Y$ be the preimage of the odd Hall
subgroup of $X/G$. Then
\begin{equation}\label{eq:E6-regular-split}
 X=Z_2\times Y,\quad |Y:G|\text{ odd},\quad
 |Y:Z(Y)G|\le3.
\end{equation}
Indeed $Z_2$ maps isomorphically onto the Sylow $2$-subgroup
of $X/G$. If $\widetilde b$ is the connected-center block
over $b$, then
\[
 \widetilde b=\cO Z_2\otimes c,\qquad
 k(\widetilde b)=|Z_2|k(c),\qquad
 \delta(\widetilde b)=|Z_2|\delta(b).
\]
Connected-center Jordan decomposition identifies its full
ordinary count with the principal $2$-series count of the
connected centralizer. Multiplication of its central torus
and its simply connected $A_2$ factors has odd degree, so
the same connected-centralizer argument as above identifies
the $2$-labels and their counts. Removing $Z_2$ leaves
precisely the product of the rational $A_2$ principal counts.

By Lemma~\ref{lemma:type-a-small-ranks}, each rational
factor has ratio at most $1/2$. For a factor over
$\F_{q_0}$, put $Q=q_0$ in the linear case and
$Q=-q_0$ in the unitary case. Writing
$a=v_2(Q-1)$ and $b_1=v_2(Q+1)$, its general
linear or unitary count is
\[
 K_3(Q)=
 \begin{cases}
 A(A^2+12A+8)/6,&a\ge2, A=2^a,\\
 2B+8,&a=1, B=2^{b_1},
 \end{cases}
\]
and its Sylow orders are $2A^3$ and $8B$, respectively.
Since no nontrivial scalar $2$-element fixes a degree-three
label, the special linear ratio is the same. The respective
ratios are at most $3/8$ and $1/4+1/B\le1/2$.

In the three-cycle case, the corrected block table gives
three $G$-blocks, permuted transitively by diagonal
automorphisms. Fix one, and fix the central character
$\mu$ of $c$ on $Z(Y)$. A character of $b$ extends uniquely
to $Z(Y)G$ in the sector $\mu$; its inertia group in $Y$
is exactly $Z(Y)G$. Induction gives a bijection with
$\Irr(c)$. Hence $k(b)/\delta(b)\le1/2$. In the
transposition case the diagonal quotient in
\eqref{eq:E6-regular-split} is trivial; the same
central-product argument gives $k(b)=k(c)$ and ratio
at most $1/4$.

In the separate-factor case there is one invariant block
and $Y/Z(Y)G=C_3$. Write
\[
 K=k(B_0(\SL_3(\eps q))),\qquad
 D=|\SL_3(\eps q)|_2.
\]
After fixing $\mu$ and removing $Z_2$ we have
$k(c)=K^3$ and $\delta(b)=D^3$. The quotient linear
characters of $Y/Z(Y)G$ fix $c$: under central duality
their order-three labels preserve the lifted odd class
and cyclically permute the three $A_2$ factors. The
cube of a rational conjugating element is in the
connected centralizer, so the returning action on a
factor's ordinary Jordan labels is inner. More explicitly,
identify the three label sets with a set $\Omega$ of size $K$.
The action has the form
\[
 (x_1,x_2,x_3)\longmapsto
       (f_3(x_3),f_1(x_1),f_2(x_2)),\qquad f_3f_2f_1=1.
\]
Its fixed set is
$\{(x,f_1(x),f_2f_1(x)):x\in\Omega\}$.
The one-dimensional central torus is fixed: the induced
automorphism has order dividing three, whereas the automorphism
group of a one-dimensional torus has order two. Thus there are
exactly $K$ fixed characters of $c$. Before removing $Z_2$ the
number is $|Z_2|K$, which explains why that removal was
necessary.

If a block $c$ in a cyclic extension
$Y/N=C_3$ is fixed by quotient linear characters, it is
the unique block over its invariant block $b$ of $N$.
Indeed the sum of the idempotents in any linear-twist
orbit belongs to $\cO N$; primitivity of $b$ makes that
orbit the whole covering set. If $a$ is the number of
invariant characters of $b$ and $t$ the number of its
orbits of length three, cyclic Clifford theory gives
\[
 k(c)=3a+t,\qquad k(b)=a+3t.
\]
The $t$ induced characters are exactly those fixed by
a nontrivial quotient linear character. Here $t=K$, so
\begin{equation}\label{eq:E6-separate-count}
 k(b)=\frac{K^3+8K}{3},\qquad
 \frac{k(b)}{D^3}\le\frac1{24}+\frac4{3D^2}
                 \le\frac3{64},\qquad D\ge16.
\end{equation}
The count uses the central-twist and conjugation compatibility
of connected-center Jordan decomposition
\cite[Theorem~1.1(3),(7) and Theorem~1.2]{Ruh22pre}.

The defects in the three configurations are products
of three, one and two wreathed or semidihedral factors,
respectively. Lemmas~\ref{comp:defect-aut} and
\ref{comp:murai-index} show that their integral quotients
have odd parts at most $3,1,1$. In the first case,
\eqref{eq:E6-separate-count} gives $k(b)<\delta/4$, so
Lemma~\ref{comp:small-odd}(ii) applies; in the other cases
use its trivial-module construction. Every odd central quotient
of an algebraic central cover is included: the corresponding central
sector has unchanged defect and its ordinary characters
are those of the covering block with the quotient kernel
in their kernel.

\subsection{The connected isolated cases}

Consider the connected centralizers in
\eqref{eq:isolated-types}, excluding the $E_6$ row. For
$E_8:E_6A_2$, this subsection concerns the maximal-defect
semisimple block. Its other blocks will be treated below.
The block distribution and defect statements from
\cite{KM13,KM25} have the following consequences. The
$G_2:A_2$, $F_4:A_2^2$, $E_7:A_5A_2$,
$E_8:A_8$ and $E_8:A_4^2$ series each consist of one
block; the maximal $E_8:E_6A_2$ block has full Sylow
defect in its associated connected centralizer. Ordinary
Jordan decomposition counts the indicated full series,
so its count also bounds any selected block in it.

For factors of type $A_2,A_4,A_8,E_6$, the fundamental
groups have odd order. Thus all semisimple $2$-centralizers
are connected, and odd central isogenies identify their
rational $2$-classes and unipotent-character counts.
This is the odd-isogeny case of
\cite[Proposition~4.6]{Malle18}; it also follows directly
by lifting the unique $2$-part through the odd kernel
and applying Lang's theorem to the connected centralizer.
It follows that the relevant series-to-defect ratios
are products of the type-$A$ principal ratios and, when
needed, Lemma~\ref{comp:E6-series}. Factor permutations
are treated as single rational factors over an extension
field. Thus
\begin{equation}\label{eq:connected-isolated-counts}
\begin{array}{c|c}
 \text{type}&k(b)/\delta(b)\text{ at most}\\ \hline
 G_2:A_2&1/2\\
 F_4:A_2^2&1/2\\
 E_8:A_8&1\\
 E_8:A_4^2&1\\
 E_8:E_6A_2\text{ (maximal)}&15/128.
\end{array}
\end{equation}
For $F_4$ the two factors give $1/4$, which implies the displayed
$1/2$ bound. The last $E_8$ row uses the complete $E_6$ two-series, so it
bounds every allocation among cuspidal or twin blocks.

For $E_7:A_5A_2$ the even center of $A_5$ must be
retained. Work first with the simply connected group
$G=E_7(q)_{\mathrm{sc}}$, and let $\bC$ be the
connected reductive group \emph{dual to}
$C_{\bG^*}(s)$ in the adjoint dual group. Put $C=\bC^F$.
Jordan decomposition on $G$ and on $\bC$ gives
$k(b)=k(B_0(C))$ and $\delta(b)=|C|_2$; this is
\cite[Lemma~4.4]{Malle18}, which allows disconnected
centralizers of subsequent $2$-elements. The character
lattice of $\bC$ is the $E_7$ weight lattice. Write $Q(\Phi)$
and $P(\Phi)$ for the root and weight lattices of a root system
$\Phi$, and $C_\Phi$ for its Cartan matrix. With the common
root-length normalization,
\[
 [Q(E_7):Q(A_5A_2)]^2
   =\frac{\det C_{A_5}\det C_{A_2}}{\det C_{E_7}}
   =\frac{6\cdot3}{2}=9.
\]
Consequently
\[
 [P(E_7):Q(A_5A_2)]=2\cdot3=6,\qquad
 [P(A_5A_2):P(E_7)]=\frac{18}{6}=3.
\]
Hence the simply connected covering of
$\bC$ has kernel of order three, and its rational
points are $\SL_6(\eps q)\times\SL_3(\eps q)$.
Its image $H$ has index at most three in $C$.
Both this kernel and index are odd. This identifies
the group being counted; the adjoint-dual centralizer
itself can have an additional even covering kernel.
The principal block of $H$ has the product character
count. For a cyclic quotient of order at most three,
Clifford theory gives
\[
 k(B_0(C))\le3\,k(B_0(H)),\qquad |C|_2=|H|_2.
\]
Lemma~\ref{lemma:type-a-small-ranks} gives
\[
 \frac{k(B_0(\SL_6(\eps q)))}{|\SL_6(\eps q)|_2}
 \le\frac1{16},\qquad
 \frac{k(B_0(\SL_3(\eps q)))}{|\SL_3(\eps q)|_2}
 \le\frac12.
\]
The first estimate includes the scalar-stabilizer correction in
\eqref{eq:type-a-restriction}. The Jordan count therefore gives
ratio at most $3/32$ for the simply connected $E_7$ block. Its
quotient by the central involution has
no more characters and half the defect order, so
\begin{equation}\label{eq:E7-isolated-budget}
 k(b)/\delta(b)\le3/16
\end{equation}
for the ordinary simple group relevant to odd-centred
components.

\begin{lemma}[Integral innerness for connected isolated labels]
\label{comp:isolated-inner}
Let $L$ be an algebraic central cover with odd center
in one of the connected rows of \eqref{eq:isolated-types},
and let $b$ be a block of maximal defect in that series.
For every $\Inn(L)\le A\le\Aut(L,b)$, put
$T=\{a\in A:a\text{ acts innerly on }\cO Lb\}$.
Then $A/T$ is a $2$-group.
\end{lemma}

\begin{proof}
Work first with $G=\bG^F$, where $\bG$ is simply
connected, and let $s\in(\bG^*)^{F^*}$ be the odd
label of the covering block. As in
\cite[Section~2.2]{Ruh22pre}, choose a compatible identification
of $\bG_{\mathrm{ad}}$ with $\bG^*$, let
$\pi:\bG\to\bG_{\mathrm{ad}}$ be the adjoint
quotient, and put
\[
 G(s)=\bigl(\pi^{-1}(C_{\bG^*}(s))\bigr)^F.
\]
The local structure theorem
\cite[Proposition~6.1]{Ruh22pre} supplies a maximal
pair $(P,e)$ for which $P\in\Syl_2(G(s))$ and
\begin{equation}\label{eq:isolated-normalizer}
 N_G(P,e)=N_{G(s)}(P).
\end{equation}
The construction uses the odd linear torus character
$\widehat s$ and the uniquely corresponding block
at $C_G(P)$. In particular it is compatible with
automorphisms fixing that torus and $s$. The two
exceptions to the cited proposition, the rational
$A_2(\eps q^3).3$ centralizer and the triality case,
are absent from the present list.

The simply connected rational factors of $G(s)$ have
type $A$ of rank at least two, or type $E_6$. For
these factors
\cite[Corollary~2.4 and Remark~2.2]{BroughRuhstorfer22}
gives
\[
 N(P)=P C(P),
\]
and the property descends through a central quotient.
Any further rational diagonal index is odd. A maximal
torus normalized by $P$ maps onto it, and the averaging
argument preceding
\eqref{eq:principal-centralizer-extension} shows that
its $P$-fixed part still maps onto that quotient.
To see the normalizer equality explicitly, let $H_0$ be the
image of the product of the simply connected rational factors
and put $H=G(s)$. The averaging argument gives
$H=H_0C_H(P)$, whence
\[
 N_H(P)=N_{H_0}(P)C_H(P)
       =PC_{H_0}(P)C_H(P)=PC_H(P).
\]
Combining this with
\eqref{eq:isolated-normalizer} gives
\begin{equation}\label{eq:isolated-Eb}
 E_b=1.
\end{equation}

Let an odd field subgroup of order $h$ stabilize the
block. Write $F=F_0^h$ and $F^*=(F_0^*)^h$ for
the corresponding primal and dual Frobenius maps.
We may assume that $s$ is $F_0^*$-fixed. Indeed,
an $F_0^*$-stable geometric semisimple class has an
$F_0^*$-fixed representative by Lang's theorem in
the connected ambient dual group. Since
$C_{\bG^*}(s)$ is connected, this representative
is conjugate to the original one by an element
of $(\bG^*)^{F^*}$. In dual
adjoint type $E_7$ its rational conjugator can be
chosen in the image of the simply connected group:
an $F$-stable maximal torus of $C(s)$ maps onto the
diagonal quotient, by the torus cohomology sequence;
multiply the conjugator by an appropriate element
of this centralizing torus. Thus the change is
implemented by an inner automorphism on the original
ordinary presentation.

All diagram twists and factor permutations in these
centralizers have $2$-power order; none has triality
or an odd factor cycle. Their central covering
divisors are odd. Equation~\eqref{eq:odd-field-valuations}
therefore shows that $C_{G(s)}(F_0)$ and $G(s)$ have
equal $2$-parts. Choose $P$ in the smaller-field
Sylow torus normalizer. It is a full Sylow subgroup
of the larger group and is centralized pointwise
by $F_0$.

We next determine the second component of the Brauer pair. In the torus construction of
\eqref{eq:isolated-normalizer}, the torus is defined
over the smaller field and its $2$-torsion is
unchanged by the odd extension. Its canonical odd
character $\widehat s$ is $F_0$-fixed because $s$ is.
The uniquely corresponding block $e$ of $C_G(P)$
is therefore $F_0$-fixed. For the field
extension $H$ this proves
\[
 H=G C_H(P)_e.
\]
By \eqref{eq:isolated-Eb}, Murai's pairing is trivial,
so $H=H[b]$ integrally. The odd part of the ordinary
outer group in each of these types is precisely its
normal odd field subgroup; the remaining quotient
has $2$-power order. Hence $A/T$ is a $2$-group.

For the $E_7$ simple quotient the conclusion descends
through the central involution. A unit implementing
an automorphism of the covering integral block maps
to a unit implementing it in the dominated quotient
block. Thus the argument gives integral innerness
on the quotient, in addition to the count in
\eqref{eq:E7-isolated-budget}.
\end{proof}

Equations~\eqref{eq:connected-isolated-counts} and
\eqref{eq:E7-isolated-budget}, followed by
Lemma~\ref{comp:isolated-inner}, give trivial modules
for all these maximal blocks.

\subsection{Nonmaximal blocks in the isolated \texorpdfstring{$E_8$}{E8} series}

The remaining nonabelian-defect rows in the isolated
$E_6A_2$ series are rows $4$ and $6$ of the
$2$-block portion of \cite[Table~5]{KM13}, with their
Ennola forms. Row $4$ has a wreathed defect group.
Lemma~\ref{comp:defect-aut}(i), the intrinsic bound
and the cited central-extension theorem of Sambale
give its trivial module.

Row $6$ has a metacyclic defect group
\cite[Proposition~7.5 and Table~11]{KM25}. We check
that it is neither abelian nor $Q_8$, the two
exceptions to the automorphism lemma. Write $\bL$ for its
cuspidal Levi in the ambient algebraic group $\bG$ of type $E_8$.
Its derived subgroup has type $E_7$, its center is connected
of dimension one, and its relative Weyl group has type $A_1$.
Connectedness of its center follows because
the seven Levi simple roots span a primitive
summand of the $E_8$ character lattice. Its
intersection with the simply connected derived
$E_7$ is that group's central subgroup of order
two. For $q\equiv\eps\pmod4$, put
$A=(q-\eps)_2\ge4$. Let $\lambda$ be the cuspidal character in
this row, and put
$W_{\bG^F}(\bL,\lambda)=N_{\bG^F}(\bL,\lambda)/\bL^F$.
The defect formula \cite[Theorem~1.2(b) and the proof of
Proposition~6.4, case~6]{KM13} gives
\begin{equation}\label{eq:E8-metacyclic-defect}
 |P|=|Z(\bL)^F|_2
 \left|\bL^F/(O_2(Z(\bL)^F)[\bL,\bL]^F)\right|_2
 |W_{\bG^F}(\bL,\lambda)|_2.
\end{equation}
The first factor is $A$. Lang's theorem for the
derived subgroup makes the map to the rational
one-dimensional quotient torus surjective. The
image of $Z(\bL)^F$ has index two there, as its
kernel is the central involution. Thus the middle
factor is two, in both the split and nonsplit
rank-one torus cases. The last factor is two,
so
\[
 |P|=4A\ge16.
\]
The nonabelianness assertion in
\cite[Theorem~1.2(c)]{KM13} applies since the
relative Weyl group has even order. Hence $P$
is neither homocyclic abelian nor $Q_8$.
By \cite[Lemma~1]{SambaleMeta12}, $\Aut(P)$
is a $2$-group. The metacyclic block bound and
Lemma~\ref{comp:murai-index} once again give the
trivial module for the quotient, including
every field stabilizer.

\section{The remaining component blocks}
\label{sec:remaining-components}

\subsection{Nonprincipal unipotent blocks}

The complete list of nonprincipal unipotent $2$-blocks of
exceptional groups with nonabelian defect is
\cite[Section~7, immediately before Table~10]{KM25}.
Only $E_7$ and $E_8$ occur, for both congruences modulo four.
The $E_7$ defects are nonabelian dihedral. Their automorphism
groups are $2$-groups by \cite[Lemma~1]{SambaleMeta12}.
The intrinsic bound and the dihedral block theorem therefore
give the required trivial module.
The $E_8$ case requires a full ordinary-character estimate,
which we now prove.

\begin{proposition}\label{comp:E8-unipotent}
Let $b$ be a nonprincipal unipotent $2$-block of $E_8(q)$
with nonabelian defect, where $q$ is an odd prime power. Put
$A=\max\{(q-1)_2,(q+1)_2\}\ge4$. Then
\begin{equation}\label{eq:E8-full-count}
 k(b)\le\frac23 A^2+8A+\frac{10}{3}<4A^2=\delta(b).
\end{equation}
Its integral quotient is a $2$-group, so the
trivial module of order $\delta(b)$ satisfies
\eqref{eq:component-marks}.
\end{proposition}

\begin{proof}
Let $\bG$ be the algebraic group of type $E_8$ with
Frobenius $F$ defining $E_8(q)$, and identify it with its dual.
The quasi-central cuspidal source has derived type
$E_6$ (the twisted rational form for $q\equiv-1\pmod4$),
a connected central torus of rank two, and one of the
two cuspidal characters labelled $\theta,\theta^2$, where
$\theta$ is a primitive cube root of unity.
Its geometric relative Weyl group, and its rational
relative Weyl group, is $S_3\times C_2$, acting faithfully
on that central torus. The defect order is $4A^2$.
These assertions and the following single-source
unipotent counts are the local data of
\cite[Lemma~7.1(ii), the proof of Lemma~7.2, and the first
paragraph of the proof of Lemma~7.4]{Ruh22pre}.
For a semisimple $2$-element $t\in\bG^F$, the contributions are
\begin{center}
\begin{tabular}{l c}
\toprule
Rational derived type of $C_{\bG}(t)$&Contribution from one source\\
\midrule
$E_6$ or ${}^2E_6$&$1$\\
$E_6A_1$ or ${}^2E_6A_1$&$2$\\
$E_7$&$2$\\
$E_7A_1$&$4$\\
$E_8$ ($t=1$)&$6$\\
\bottomrule
\end{tabular}
\end{center}
The $E_7A_1$ contribution is four per source. The cited
calculation counts all ordinary characters in these layers before
restricting to height zero. Moreover,
\cite[Section~4.2 and Table~1]{KM25} retains a twin-block
ambiguity when the order of $q$ modulo four is two.
We therefore allow both sources for every nonidentity
$t$: their union contributes at most eight. At $t=1$
the unipotent distribution gives six in the selected
individual block.

We next justify the source assertion needed to count
the possible $t$. If
$\chi\in\Irr(b)\cap\cE(\bG^F,t)$, let $c_t$ be
the unipotent block containing its Jordan correspondent
in $C_{\bG}(t)^F$. The block/source comparison
\cite[Theorem~1, Lemma~3.4, Definition~3.5,
Propositions~3.6 and~4.1]{KM25} gives a quasi-central
cuspidal pair $(\bL_t,\lambda_t)$ for $c_t$ whose
Levi is also an ambient Levi of $\bG$. Uniqueness of
the quasi-central cuspidal source of a unipotent block,
as used in the proof of Proposition~4.1(b) of that
paper, identifies its derived source with that of
$b$ or its indicated twin. Definition~3.5 and
Proposition~3.6(a) preserve this derived subgroup
up to rational conjugacy; the exception in
Proposition~3.6(b) is only at the block prime three.
Both present twins have geometric derived source
$E_6$. Thus $\bL_t$ is an $F$-stable ambient
$E_6$ Levi. Since it is a Levi of $C_{\bG}(t)$,
it contains the center of that connected group.
In particular $t\in\bL_t$, and because $t$
centralizes it,
\begin{equation}\label{eq:E8-source-central}
 t\in Z(\bL_t).
\end{equation}
This membership uses the Levi property, in addition
to the fact that $\bL_t\le C_{\bG}(t)$.

Fix a split $E_6$ Levi $\bL$ and put $\bS=Z(\bL)$.
Every ambient Levi with derived type $E_6$ is geometrically
conjugate to $\bL$: the $E_8$ Dynkin diagram has a unique
subdiagram of type $E_6$.
The six Levi simple roots form a primitive summand
of the $E_8$ character lattice, whose simple roots
are a basis. Thus $\bS$ is a connected torus of
rank two. For a torus $\bT$, write $\bT[2^\infty]$
for its subgroup of elements of $2$-power order.
Consider geometric conjugacy classes of
marked pairs
\[
 (t,\bL'),\qquad
 t\in Z(\bL')[2^\infty],\quad
 \bL'\text{ conjugate to }\bL.
\]
After conjugating $\bL'$ to $\bL$, the residual
equivalence relation is the action of
\[
 W=N_{\bG}(\bL)/\bL=S_3\times C_2
\]
on $\bS[2^\infty]$. This action is faithful,
because $C_{\bG}(\bS)=\bL$. It commutes with
Frobenius. By \eqref{eq:E8-source-central}, each
rational semisimple class contributing to $b$
is the image of an $F$-stable geometric marked-pair
class. All semisimple centralizers in simply
connected $E_8$ are connected; consequently Lang's
theorem gives exactly one rational class in each
$F$-stable geometric semisimple class. Hence the
number $n_b$ of contributing rational $2$-classes
is at most the number $N$ of $F$-stable $W$-orbits
on $\bS[2^\infty]$. Thus $N$ counts stable geometric orbits
with all rational torus twists.

The twisted Burnside argument used in
\eqref{eq:torus-mass} gives
\[
 N=\frac1{12}\sum_{w\in W}|\det(qw-I)|_2.
\]
It remains valid on the infinite torsion set,
since every fixed-point set is finite and each
stable orbit contributes twelve pairs. The
faithful rank-two representation of $W$ is the
dihedral reflection representation of order
twelve. Let $\omega$ have order three. The spectra and multiplicities are
\[
\begin{array}{c|c|c}
 &\text{number}&\text{eigenvalues}\\ \hline
 1&1&(1,1)\\
 -I&1&(-1,-1)\\
 \text{reflections}&6&(1,-1)\\
 \text{order-three rotations}&2&(\omega,\omega^{-1})\\
 \text{order-six rotations}&2&(-\omega,-\omega^{-1}).
\end{array}
\]
Their determinant
$2$-parts are $A^2,4,2A,1,1$, respectively,
with the first two possibly interchanged.
Replacing $q$ by $-q$ for the twisted source
does not change the sum, because $-I\in W$.
Thus
\begin{equation}\label{eq:E8-relative-torus-mass}
 N=\frac{A^2+4+12A+4}{12}
   =\frac{A^2}{12}+A+\frac23.
\end{equation}
There is one marked orbit above $t=1$. The
single-block identity contribution and the
two-source bound at nonidentity classes give
\[
 k(b)\le6+8(n_b-1)\le6+8(N-1)
       =\frac23A^2+8A+\frac{10}{3}.
\]
Writing $A=4+x$, $x\ge0$, the difference from
$4A^2$ is
\[
 18+\frac{56}{3}x+\frac{10}{3}x^2>0.
\]
This proves the count for every valuation, and
includes positive-height characters.

Return to the $F$-stable cuspidal source Levi $\bL_b$ of
$b$. It has rational type $E_6\Phi_1^2$ when
$q\equiv1\pmod4$ and ${}^2E_6\Phi_2^2$ when
$q\equiv-1\pmod4$. Put
$D=O_2(Z(\bL_b)^F)\cong C_A^2$.
Choose a defect group $P$ of $b$ containing $D$.
The proof of \cite[Lemma~7.2]{Ruh22pre} gives
$D\lhd P$, $P/D\cong V_4$, and generators of $P/D$
acting on $D$ by inversion and interchange of its two
cyclic factors. The odd central intersection with the
derived $E_6$ subgroup does not change this $2$-torsion.
Lemma~\ref{comp:defect-aut}(iii) applies without
assuming that the extension splits. Thus
$\Aut(P)$, and by Lemma~\ref{comp:murai-index}
also $R$, are $2$-groups. The trivial module
completes the proof.
\end{proof}

\subsection{The exceptional graph in \texorpdfstring{$G_2$}{G2}}

We prove \eqref{eq:component-marks} directly for every
nonprincipal $2$-block of $G_2(q)$ with nonabelian defect,
where $q$ is an odd prime power. This includes the
exceptional graph action when $q$ is a power of three.

The block list in \cite[Appendix~B.3.1]{Schulte15}
is
\[
 B_0;\quad B_3\ (3\nmid q);\quad
 B_{1a},B_{1b},B_{2a},B_{2b};\quad
 BX_1,BX_2,BX_a,BX_b;\quad\text{defect-zero blocks}.
\]
Section~12.2 of that source expressly includes
$q=3$. The $B_1,B_2$ blocks have two Brauer
characters, $B_3$ has three, and the $BX$ blocks
have one. The $BX_a,BX_b$ defects are cyclic.
Write $q\equiv\eps\pmod4$ and
$(q-\eps)_2=2^a$, $a\ge2$. The defect
groups in \cite[Remark~12.22]{Schulte15} are
\begin{equation}\label{eq:G2-small-defects}
 P\cong
 \begin{cases}
 C_{2^a}\wr C_2,&B_{1a},B_{1b},\ \eps=1,
                    \text{ or }B_{2a},B_{2b},\ \eps=-1,\\
 \mathrm{SD}_{2^{a+2}},&\text{the other congruence combinations}.
 \end{cases}
\end{equation}
We show that $BX_1$ and $BX_2$ also have abelian
defect. First observe the following rank-one fact.
Let $\bM$ be connected reductive with connected center and
derived type $A_1$, and let $F$ define an $\F_q$-structure
with $q$ odd. Then
\begin{equation}\label{eq:rank-one-brauer}
 l(B_0(\bM^F))\ge2.
\end{equation}
Indeed Lang's theorem for the center gives a
surjection onto $\mathrm{PGL}_2(q)$. Inflate
its permutation action on $\mathbb P^1(q)$.
The ordinary augmentation character is the
irreducible Steinberg character of degree $q$.
Modulo two its augmentation lattice contains
the constant vector, since $q+1$ is even, so
its reduction has the trivial composition
factor and belongs to the principal block.
It also has a nontrivial composition factor:
an odd-order nonidentity root element fixes
exactly one point of $\mathbb P^1(q)$ and has
Steinberg value zero. If all composition
factors were trivial, the Brauer value there
would be $q$. This proves
\eqref{eq:rank-one-brauer}.

By the unipotent list stated above, a
nonprincipal unipotent block of $G_2(q)$ has
abelian defect. For a non-unipotent block
let $s\ne1$ be its odd dual label. The dual
centralizer is connected. Its only noncentral
isolated possibility for an odd label is
$A_2$, with an order-three label; the
$A_1A_1$ centralizer requires an even label.
The isolated series is a single block
\cite[Lemma~6.13]{KM13}, and it is exactly
$B_3$ in the above notation. One can check
this identification from the character
degrees: if $q\equiv\eta\pmod3$ with $\eta\in\{1,-1\}$, the Jordan
semisimple character has degree
\[
 (q+\eta)\Phi_6(\eta q),
\]
which for $\eta=-1$ is $(q-1)\Phi_3(q)$.
This is the degree of the character denoted $X_{32}$ in
\cite[Table~B.2]{Schulte15}; Appendix~B.3.1
places it in $B_3$. For $q\ge5$ this degree
is unique in that table. In fact every
nontrivial unipotent degree is at least
$q(q-1)^2\Phi_6(q)/6$, which is larger
than $(q+1)\Phi_3(q)$ because
\[
 \Phi_6(q)>\tfrac23\Phi_3(q),\qquad
 q(q-1)^2>9(q+1).
\]
The other non-unipotent families, apart from
the two strict multiples $X_{31},X_{33}$,
have degree at least $\Phi_3(q)\Phi_6(q)$,
also larger. In characteristic three
there is no such semisimple order-three
label, so $B_3$ is absent.

The remaining odd centralizers are proper
Levis of semisimple rank one or zero. A
rank-one Levi in $G_2$, and its dual, have
connected center: the Cartan determinant
is one, every root and coroot is
Weyl-conjugate to a simple one, and hence
is primitive in its lattice. Ordinary
Bonnaf\'e--Rouquier equivalence sends the
block to a block of that dual Levi. Its
central odd label makes it a linear twist
of a unipotent block; the principal
$2$-block is the only unipotent block for
a classical derived component, as recalled
in \cite[the proof of Lemma~2.6]{KM25}.
Morita equivalence and linear twisting
preserve $l(b)$, so
\eqref{eq:rank-one-brauer} excludes this
case for $BX_1,BX_2$. In rank zero the
correspondent is a torus block. The
abelian-defect preservation theorem
\cite[Theorem~1.3]{KM13} for this
correspondence gives abelian defect for the original block.
Thus $BX_1$ and $BX_2$ have abelian defect.

It follows that every nonprincipal
nonabelian-defect block is either the
isolated $B_3$ case already treated or
one of \eqref{eq:G2-small-defects}.
For the latter, the intrinsic bound and
Lemma~\ref{comp:defect-aut}(i) make
$R$ a $2$-group, and the wreathed or
semidihedral block bound gives its
trivial module. This argument includes
the full exceptional graph--field
stabilizer of $G_2(3^f)$.

\subsection{Nonprincipal triality blocks}

For $S={}^3D_4(q)$ with $q$ odd, put
$(q^2-1)_2=2^{a+1}$, $a\ge2$. Every nonabelian defect group
is isomorphic to one of the following:
\begin{equation}\label{eq:triality-defects}
 S_2,\qquad C_{2^a}\wr C_2,\qquad
 \mathrm{SD}_{2^{a+2}},\qquad S_2\in\Syl_2(S),
\end{equation}
by \cite[p.~274]{An95}. The same source
uses $C_S(S_2)=Z(S_2)$ on p.~287. This
centralizer is a $2$-group and has just
one block. Thus every block with defect
$S_2$ has the same maximal Brauer pair
as the principal block and is principal.
For a nonprincipal block only the last
two groups in \eqref{eq:triality-defects}
remain. Their automorphism groups are
$2$-groups by Lemma~\ref{comp:defect-aut}(i),
and the central-extension block theorem
gives $k(b)\le|P|$. Lemmas
\ref{comp:murai-index} and
\ref{comp:small-odd} provide the trivial
module for every stabilizer. The construction applies to the
full stabilizer with its cyclic field--triality action.

\subsection{Alternating, sporadic and exceptional covering groups}

Let $S$ be nonabelian simple and let $L$ be quasisimple with
$L/Z(L)\cong S$. The map $\Aut(L)\to\Aut(S)$ is injective.
An automorphism trivial on $S$ has a
central error homomorphism from $L$
to its center and is therefore the
identity. An automorphism inducing an
inner automorphism of $S$ is inner
on $L$ after lifting the conjugating
element. Hence $\Out(L)$ embeds in
$\Out(S)$. If this latter group is a
$2$-group, every quotient $A/T$ is a
$2$-group.

The scalar block inequalities for
alternating groups, sporadic covers,
the Tits group and exceptional covers
are \cite[Proposition~2.3 and
Theorem~2.4]{Malle18}. The outer groups of alternating and
sporadic simple groups are $2$-groups.
After removing exceptional multiplier
$2$-parts, the exceptional odd-cover list is
\begin{center}
\begin{tabular}{l c c}
\toprule
Simple quotient&Exceptional odd multiplier&Outer group\\
\midrule
$A_6$&$C_3$&$V_4$\\
$A_7$&$C_3$&$C_2$\\
$\mathrm{PSU}_4(3)$&$C_3\times C_3$&$D_8$\\
$\Omega_7(3)$&$C_3$&$C_2$\\
$G_2(3)$&$C_3$&$C_2$\\
\bottomrule
\end{tabular}
\end{center}
The multiplier list is
\cite[Introduction]{SuleimanWilson99};
the outer orders are the corresponding
entries of \cite{Atlas85}. Every
quotient of the displayed multiplier
is allowed, including both triple
covers of $\mathrm{PSU}_4(3)$.
The exceptional multiplier parts of
$\mathrm{PSL}_3(4)$,
$\mathrm{PSU}_6(2)$ and ${}^2E_6(2)$
are $2$-primary. Their remaining odd central covers arise
from the simply connected algebraic presentations and were included
in Proposition~\ref{comp:defining-marks}.
In every displayed exceptional case
the scalar bound and the outer
$2$-group give the trivial module.

\subsection{The local alternative}

\begin{theorem}[Component alternative]\label{thm:component-marks}
Let $L$ be a quasisimple group with odd center and let
$b$ be a $2$-block of $L$ with nonabelian defect. Let
$\Inn(L)\le A\le\Aut(L,b)$, and put
\[
 T=\{a\in A:a\text{ acts innerly on }\cO Lb\},
 \qquad R=A/T.
\]
At least one of the following holds.
\begin{enumerate}
\item There is an ordinary presentation
$L=\bS^F/Z_0$ in odd characteristic, with $\bS$
simply connected and $Z_0\le Z(\bS^F)$, such that
the $A$-action lifts to $\bS^F$ by ordinary
automorphisms preserving $Z_0$. Let $b_S$ be the
block of $\bS^F$ dominating $b$, let $\bS^*$ be
dual to $\bS$, and let $s$ be the odd semisimple
label of $b_S$. The canonical minimal Levi
\[
 \bM^*=C_{\bS^*}
       \bigl(Z(C_{\bS^*}(s)^\circ)^\circ\bigr)
\]
is proper.
\item The group $R$ has an $\F_2R$-module $W$
satisfying \eqref{eq:component-marks}.
\end{enumerate}
\end{theorem}

\begin{proof}
The preceding propositions cover
algebraic central covers in defining
characteristic two, their principal blocks in odd characteristic,
all noncentral isolated odd labels,
nonprincipal unipotent blocks with
nonabelian defect, all $G_2$ and triality endpoints, and
all alternating, sporadic and
exceptional odd covers.
The small Ree groups have elementary
abelian Sylow $2$-subgroups of order
eight and have no nonabelian-defect
case. The nonsimple small
presentations are treated through
their simple derived groups, already
listed above.

For completeness we check the
remaining ordinary-action boundary.
Every block of a finite group of
Lie type in nondefining
characteristic belongs to an odd
semisimple $2$-series. At a central
label the resulting unipotent
cases are precisely those already
listed, allowing their central
linear twists in a regular
presentation. At a noncentral
non-isolated label the canonical
minimal Levi in the statement is
proper by definition. In classical
types the connected odd centralizer
is a proper Levi. In $E_6$ its
possible component group of order
three is retained in the actual
normalizer of that Levi. The
classification in
\cite{Bonnafe05,KM13,KM25} gives
exactly \eqref{eq:isolated-types}
when the canonical Levi is the
whole group. Thus these alternatives
exhaust all remaining labels.

The standard automorphism theorem
realizes the remaining finite
automorphisms by adjoint
conjugations, field maps and pinned
diagram maps
\cite[Section~2.1]{Harper24}.
Adjoint conjugation acts canonically
on the simply connected cover,
independently of a central choice
of lift. Field and pinned diagram
maps act on its Chevalley generators
with their relations. The
original quotient kernel is
preserved: in type $A$ and $E_6$
it is a subgroup of a cyclic
center; for the odd-centred
orthogonal and $E_7$ forms the
entire $2$-primary center has
already been removed. In twisted
$E_6$, if $F=\gamma F_r^f$ and
$u=F_r$ on fixed points, retain
$u^f=\gamma^{-1}$ and $u^{2f}=1$.
Perfectness makes the lift unique, so the ambient relations
are preserved. On a component orbit
fixed identifications give the
literal permutation action, and
the original central gluing is
preserved by the actual
conjugation action.

The exceptional inseparable graph
families are $B_2$ and $F_4$ in
characteristic two and $G_2$ in
characteristic three
\cite[Lemma~2.1]{Harper24}. The
first two have direct defining-two
modules; the last has the direct
$G_2$ endpoint proved above.
The triality nonprincipal cases
also have their direct endpoint.
Thus none of these operators is
inserted into an ordinary Levi
descent. For every other remaining
proper Levi, the hypotheses of
Theorem~\ref{thm:levi-descent} are
exactly the properties just
verified. This proves the first
alternative, and completes the
exhaustion.
\end{proof}



\section{The minimum counterexample}\label{sec:minimum}

We now combine the affine inequality, the component estimates and the
block descent. The order of the minimization is relevant:
the normal-$2$ argument decreases the defect order even when an
intermediate Morita equivalence changes the size of the ambient group.

\begin{lemma}[Reduction of normal subblocks]\label{lem:reduced-minimum}
Assume \eqref{eq:WA2}, and choose a counterexample $(G,B)$ for which
\begin{equation}\label{eq:minimum-triple}
 \bigl(\delta(B),\ |G:Z(G)|,\ |G|\bigr)
\end{equation}
is lexicographically minimal. Then $B$ is quasiprimitive, meaning
that every block of a normal subgroup covered by $B$ is $G$-invariant.
Moreover
\begin{equation}\label{eq:reduced-group}
 O_2(G)=1,\qquad O_{2'}(G)=F(G)=Z(G),\qquad
 C_G(E(G))=Z(G).
\end{equation}
The center is cyclic of odd order. If $B$ covers a nilpotent block
of $N\lhd G$, then $N\le Z(G)$.
\end{lemma}
\begin{proof}
Corollary~\ref{cor:normal-two-descent} gives $O_2(G)=1$.
If a covered normal block has proper inertia group $I$, the
Fong--Reynolds correspondent has the same character count and defect
group. Since $Z(G)\le I$,
\[
 |I:Z(I)|\le |I:Z(G)|<|G:Z(G)|,
\]
contrary to minimality. Thus $B$ is quasiprimitive.

We spell out the use of the nilpotent-block reduction
\cite{KP90}. Suppose $B$ covers a nilpotent block of $N\lhd G$.
The K\"ulshammer--Puig construction, applied also to the central
factor, gives a Morita equivalent block $\widehat B$ with the same
defect group in a finite group $\widehat L$. There is a central odd
subgroup $\widehat Z$ and a normal subgroup
$\widehat R\cong D_B\cap Z(G)N$ such that
\begin{equation}\label{eq:KP-quotient}
 \widehat L/(\widehat Z\widehat R)\cong G/Z(G)N.
\end{equation}
This is the usual reduced-pair construction; its precise form for
arbitrary defect groups is also recorded in
\cite[Proposition~6.1 and its proof]{AnEaton25}.
If $\widehat R\ne1$, apply
Corollary~\ref{cor:normal-two-descent} to $\widehat B$. It produces
a counterexample of smaller defect order, which is impossible.
Thus $\widehat R=1$. If $N\not\le Z(G)$,
\eqref{eq:KP-quotient} gives
\[
 |\widehat L:Z(\widehat L)|\le |G:Z(G)N|<|G:Z(G)|,
\]
again contradicting \eqref{eq:minimum-triple}. This proves the
assertion about nilpotent covered blocks.

Every block of the odd normal subgroup $O_{2'}(G)$ has defect zero
and is nilpotent. It follows that $O_{2'}(G)\le Z(G)$. Conversely
$Z(G)$ is odd because $O_2(G)=1$, so equality holds. The nilpotent
Fitting subgroup is therefore $F(G)=Z(G)$. The generalized Fitting
subgroup is $F^*(G)=Z(G)E(G)$ and is self-centralizing. Moreover
$Z(E(G))$ is an abelian normal subgroup, so
$Z(E(G))\le F(G)=Z(G)$. Hence
\[
 C_G(E(G))\le Z(G)E(G)\cap C_G(E(G))
             =Z(G)Z(E(G))=Z(G),
\]
and equality follows.

The block selects one linear character of the odd center. If its
kernel $Z_0$ is nontrivial, quotienting by $Z_0$ preserves the
block's character count and defect group, does not increase the
second entry of \eqref{eq:minimum-triple}, and decreases the third.
Its central character is thus faithful, so $Z(G)$ is cyclic.
\end{proof}

\begin{lemma}[Abelian normal subblocks]\label{lem:abelian-normal}
In the minimum counterexample of Lemma~\ref{lem:reduced-minimum},
a covered block of a noncentral normal subgroup cannot have
abelian defect.
\end{lemma}
\begin{proof}
The defect-zero case is covered by the last assertion of
Lemma~\ref{lem:reduced-minimum}. Let $b$ be a covered block of
$N\lhd G$ with nontrivial abelian defect group $P$. Write
$\Irr_0(b)=\{\chi\in\Irr(b):\chi(1)_2=|N|_2/\delta(b)\}$
for the height-zero characters. Brauer's height-zero theorem
\cite{KM13} gives $\Irr(b)=\Irr_0(b)$.
The inductive Alperin--McKay condition at the prime two holds for
all nonabelian simple groups \cite{Ruh25}. We may therefore apply
the normal-subgroup theorem \cite[Theorem~B]{NS14}.

Let $b^*$ be the Brauer correspondent of $b$ in $N_N(P)$.
It has the same abelian defect group, so
$\Irr(b^*)=\Irr_0(b^*)$. Conjugacy of the defect groups of
the $G$-invariant block $b$ gives $G=N N_G(P)$.
The normal-subgroup theorem gives the equivariant bijection on the lower
height-zero characters and, for each corresponding lower character,
a bijection of the \emph{full} ordinary upper block fibers.
The identity $G=N N_G(P)$ identifies the lower orbit sets.
Let $\Theta$ be a set of $G$-orbit representatives in $\Irr(b)$,
and let $\theta^*$ correspond to $\theta\in\Theta$. Since all
lower characters have height zero, the full upper fibers give
\[
 \Irr(B)=\coprod_{\theta\in\Theta}\Irr(B\mid\theta)
 \ \longleftrightarrow\
 \coprod_{\theta\in\Theta}\Irr(B^*\mid\theta^*)=\Irr(B^*),
\]
where $B^*$ is the corresponding block of $N_G(P)$.
The two upper blocks have a common defect group:
this is the Harris--Kn\"orr correspondence used in
\cite[Corollary~3.10]{NS14}. Hence
\[
 k(B^*)=k(B),\qquad \delta(B^*)=\delta(B).
\]
But $P\lhd N_G(P)$ is a nontrivial $2$-subgroup.
Corollary~\ref{cor:normal-two-descent} again produces a
counterexample of smaller defect order, a contradiction.
\end{proof}

\begin{proof}[Proof of Theorem~\ref{thm:main}]
The affine inequality \eqref{eq:WA2} has been proved in
Theorem~\ref{thm:affine}. Suppose the conclusion is false and
choose $(G,B)$ as in \eqref{eq:minimum-triple}.
Lemma~\ref{lem:reduced-minimum} applies. Put
$X=Z(G)E(G)$, and let $b$ be the unique block of $X$ covered by
$B$. Write $L_1,\ldots,L_t$ for the components of $G$, and $b_i$
for the corresponding component blocks. There is at least one
component: otherwise the self-centralizing property of $F^*(G)$
would give $G=Z(G)$, whose $2$-blocks have defect zero.

For any $G$-orbit of components, their central product is a normal
subgroup. If a component block in that orbit had defect zero,
the product block would be nilpotent, contrary to
Lemma~\ref{lem:reduced-minimum}. If its defect were nontrivial
abelian, the product defect would be abelian as well, contrary to
Lemma~\ref{lem:abelian-normal}. Thus every $b_i$ has nonabelian
defect. All central overlaps have odd order.

Since $b$ is $G$-stable, $N_G(L_i)$ stabilizes $b_i$.
For each $i$, let $A_i$ be its image in $\Aut(L_i,b_i)$, and apply
Theorem~\ref{thm:component-marks} to $(L_i,b_i,A_i)$.
If its first alternative holds for one component, it holds
throughout its $G$-orbit. Fix a representative $L_0$ of this orbit
and its presentation $\pi_0:S_0\twoheadrightarrow L_0$ from that
alternative. For each component $L_i$ in the orbit choose $g_i\in G$
with ${}^{g_i}L_0=L_i$, take a copy $S_i$ of $S_0$, and set
$\pi_i=c_{g_i}\pi_0$, where $c_g$ denotes conjugation by $g$.
If ${}^{g}L_i=L_j$, define the map $S_i\to S_j$ by the unique
ordinary lift of the automorphism of $L_0$ induced by
$g_j^{-1}gg_i\in N_G(L_0)$. Two lifts differ by a homomorphism
from the perfect group $S_0$ to $\ker\pi_0$, so the lift is unique.
For ${}^{h}L_j=L_k$, the identity
\[
 (g_k^{-1}hg_j)(g_j^{-1}gg_i)=g_k^{-1}hgg_i
\]
therefore makes these maps a $G$-action on $\prod_iS_i$.
By the defining relation for a lift, the product map onto the
original central product is $G$-equivariant, so its central kernel
is $G$-stable. The lifted action satisfies the hypotheses of
Theorem~\ref{thm:levi-descent}. Applied to the central
product of that orbit, the theorem supplies a block
$\widehat B$ of a finite central odd cover $\widehat H$
with
\[
 k(\widehat B)=k(B),\qquad
 \delta(\widehat B)=\delta(B),\qquad
 |\widehat H:Z(\widehat H)|<|G:Z(G)|.
\]
This contradicts the second entry of
\eqref{eq:minimum-triple}. Thus every component satisfies
the second alternative for its induced group $A_i$.
Let $T_i$ be the subgroup of $A_i$ acting innerly on
$\cO L_i b_i$, and choose the resulting
$\F_2(A_i/T_i)$-module $W_i$. These modules satisfy
\eqref{eq:component-local-marks}.
Each $b_i$ has at least two ordinary characters.
Indeed, Brauer's subsection formula expresses $k(b_i)$
as the sum of $l(c)$ over conjugacy classes of $b_i$-subsections
$(u,c)$, where $c$ is a block of $C_{L_i}(u)$ belonging to $b_i$.
The identity subsection and a subsection at a nonidentity element
of the center of a defect group each contribute at least one.
Proposition~\ref{prop:component-assembly}
constructs a module $W$ for $Q=G/G[b]$ with
\[
 |W|=\delta(b),\qquad
 |\Irr(b)^J|\le|W^J|\quad(J\le Q),\qquad
 \ker(Q\to\GL(W))\text{ a $2$-group},
\]
and shows that $G[b]/X$ is solvable.
Theorem~\ref{thm:inner-consumer} therefore gives
$k(B)\le\delta(B)$, the desired contradiction.
\end{proof}

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