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  pdftitle={Broue's abelian defect group conjecture for 2-blocks},
  pdfauthor={Qiyue Tang},
  pdfsubject={Blocks with abelian defect groups in characteristic two},
  pdfkeywords={Broue conjecture, abelian defect group, splendid Rickard equivalence, 2-blocks}}
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\title[Brou\'e's conjecture for 2-blocks]{Brou\'e's abelian defect group conjecture\protect\\ for $2$-blocks}
\author{Qiyue Tang}
\address{Zhili College, Tsinghua University, Beijing 100084, China}
\email{tangqy24@mails.tsinghua.edu.cn}
\date{November 10, 2026}
\subjclass[2020]{20C20, 18E30}
\keywords{Brou\'e's conjecture, abelian defect group, splendid Rickard equivalence, endopermutation source, inertial block}
\begin{document}
\begin{abstract}
We prove that every $2$-block of a finite group with an abelian defect
group is splendidly Rickard equivalent to its Brauer correspondent.
\end{abstract}
\maketitle

\section{Introduction}

Let $(K,\OO,k)$ be a splitting $2$-modular system: $\OO$ is a complete
discrete valuation ring of characteristic zero, $k$ is its algebraically
closed residue field of characteristic two, and $K$ is its field of
fractions, sufficiently large for the finite groups considered below.
A block of $\OO G$ is a primitive central idempotent $a$; its block
algebra is $\OO Ga$. If $P$ is a defect group of $a$, Brauer's first
main theorem associates to $a$ a block $a_P$ of $\OO N_G(P)$, called its
Brauer correspondent. Brou\'e's abelian defect group conjecture predicts
an equivalence
\[
 \Db(\OO Ga\text{-}\mathrm{mod})
 \simeq \Db(\OO N_G(P)a_P\text{-}\mathrm{mod})
\]
when $P$ is abelian. Here $\Db$ denotes the bounded derived category of
finitely generated modules.

The conjecture grew out of Brou\'e's study of perfect isometries and
their compatibility with local character theory. His 1990 paper
\cite{Broue1990} develops isotypies as compatible systems of perfect
isometries associated with Brauer pairs. Conjecture~6.1 concerns such a
local character correspondence; Question~6.2 asks for the corresponding
equivalence of derived categories. Section~3 relates derived
equivalences to perfect isometries, and the appendix treats principal
$2$-blocks of simple groups with abelian Sylow $2$-subgroups.
Thus the characteristic-two families already occur among the examples
that motivated the conjecture.

Rickard's theory expresses a derived equivalence by a complex of
bimodules with a two-sided tensor inverse. His splendid equivalences
\cite{RickardSplendid} impose a permutation condition on the terms.
This condition permits Brauer constructions at subgroups of the defect
group and connects the global equivalence with local block algebras.
The conjecture is known for cyclic defect groups, a case discussed
already in \cite[Section~6]{Broue1990}. Harris and Linckelmann
\cite{HL} proved the splendid form for blocks of finite $p$-solvable
groups with abelian defect groups. Source algebras and
endopermutation modules provide the connection between the Morita
equivalences arising in these reductions and the required complexes.

Two further developments are relevant to the present problem.
Chuang and Rouquier \cite{ChuangRouquier} constructed derived
equivalences by $\mathfrak{sl}_2$-categorification and proved
Brou\'e's conjecture for symmetric groups, with related results for
general linear groups. For groups of Lie type, Rickard's equivariant
cohomology construction \cite{Rickard} and Rouquier's \'etale chain
complexes \cite{RouquierEtale} associate permutation complexes to
Deligne--Lusztig varieties. Bonnaf\'e, Dat and Rouquier \cite{BDR}
established a splendid Jordan decomposition when the dual centralizer
is contained in the chosen Levi subgroup. Ruhstorfer \cite{Ruhstorfer}
developed equivariant and disconnected versions of this construction.
These results supply both the component equivalences and the
normalizer actions used below.

For $2$-blocks, Craven and Rouquier
\cite[Theorem~4.36]{CR} proved the principal-block case. The
classification of Klein four source algebras by Craven, Eaton, Kessar
and Linckelmann \cite{CEKL} identifies three source algebra types,
represented by $\OO V_4$, $\OO A_4$, and the principal block of
$\OO A_5$. The nonprincipal elementary abelian block of
$\mathrm{Co}_3$ was treated by Koshitani, M\"uller and Noeske
\cite{KMN}. Eaton, Kessar, K\"ulshammer and Sambale
\cite[Theorem~6.1]{EKKS} subsequently classified the $2$-blocks of
quasisimple groups with abelian defect groups up to Morita equivalence.
Their classification separates the elementary and Klein four cases,
certain rank-one Levi cases, and blocks covered by nilpotent blocks.
Its proof also identifies the groups and central quotients that occur
in the Levi cases.

This classification has led to results for arbitrary finite groups
with prescribed defect groups. Eaton and Livesey \cite{ELRank3},
together with Wu, Zhang and Zhou \cite{WZZ}, completed the
classification for abelian $2$-groups of rank at most three.
Ardito and Sambale \cite{AS32} proved Brou\'e's conjecture for
elementary abelian defect groups of order $32$. Eaton and Livesey
\cite{ELRank4} classified the Morita equivalence classes for abelian
defect groups of rank at most four and obtained the derived
equivalence with the Brauer correspondent. The rank here is the
minimal number of generators. Together these results cover all
abelian defect groups of order dividing $32$. The source and
permutation structure of the equivalence is an additional issue for
the splendid form.

There has also been recent progress without a bound on the rank.
Zhang and Zhou \cite{ZZInertial} proved that blocks with defect group
\[
 C_{2^{n_1}}\times\cdots\times C_{2^{n_t}},
 \qquad n_i\geq2,
\]
are inertial. Zhou and Zhang \cite{QZKZ} classified $2$-blocks with
abelian defect groups whose inertial quotient has prime order, and
deduced Brou\'e's conjecture for those blocks. For a maximal Brauer
pair $(P,e_P)$, the inertial quotient is
$N_G(P,e_P)/(PC_G(P))$, including its conjugation action on $P$.
Zhang and Zhou's extension theorem \cite{ZZ} treats extensions of
inertial blocks under a hypothesis on the block of $PH$.
Miller \cite{MillerEndosplit} characterized endosplit permutation
resolutions and studied their descent over coefficient fields.
Huang \cite{Huang} studies fusion-stable endosplit resolutions;
the abelian case gives the source resolution used in
Section~\ref{sec:sources}.

Passing from a quasisimple component to a block of an arbitrary
finite group requires compatibility with the automorphisms induced
by the ambient group. Zhou's equivariant reduction
\cite{ZhouEquivariant} formulates this requirement for principal
blocks. In the present argument the component complexes carry
actions of their marked stabilizers, and their restrictions to the
upper defect group come from specified upper complexes. This permits
tensoring the components over their shared centers and inducing to
the ambient group. The resulting theorem covers every abelian
$2$-group.

\Needspace{11\baselineskip}
\begin{theorem}\label{thm:main}
Let $G$ be a finite group and let $a$ be a block of $\OO G$ with abelian
defect group $P$. There is a bounded complex $X$ of
$(\OO Ga,\OO N_G(P)a_P)$-bimodules whose terms are $2$-permutation modules,
projective on both sides and relatively $\diag P$-projective, such that
\begin{align}
 X\otimes_{\OO N_G(P)a_P}X^\vee&\simeq\OO Ga,\label{eq:main-left}\\
 X^\vee\otimes_{\OO Ga}X&\simeq\OO N_G(P)a_P\label{eq:main-right}
\end{align}
in the respective homotopy categories of bimodule complexes. Here
$X^\vee=\Hom_{\OO}(X,\OO)$, with the dual differential. In particular,
these blocks are splendidly Rickard equivalent.
\end{theorem}

Reduction modulo the maximal ideal of $\OO$ gives the corresponding
statement over every algebraically closed field of characteristic two.
The theorem applies to arbitrary ranks and exponents of $P$.

We prove the theorem by induction on
\begin{equation}\label{eq:kappa}
 \kappa(G)=|G:O_{2'}(Z(G))|,
\end{equation}
where $O_{2'}(Z(G))$ is the odd-order subgroup of the center.
Section~\ref{sec:sources} resolves Morita bimodules with
endopermutation source and constructs their local return.
Section~\ref{sec:inertial} proves the central quotient and product
statements needed for the inertial case.
Sections~\ref{sec:components}--\ref{sec:geometry} construct the
component complexes, including their upper actions.
Section~\ref{sec:assembly} assembles one orbit of components using
transporters that preserve the common defect group.
Section~\ref{sec:global} completes the reduction to a proper normalizer.

At each reduction, let $R:A\longrightarrow C$ be the chosen
equivalence to a block of a smaller group, and let
$R_P:B_A\longrightarrow B_C$ be its maximal Brauer companion.
Induction gives $Z:C\longrightarrow B_C$. Its maximal companion
$Z_P$ is an autoequivalence of $B_C$, since the defect group is
normal in the group underlying $B_C$. The return is a reduced
integral representative of
\begin{equation}\label{eq:intro-return}
 R\otimes_C Z\otimes_{B_C}Z_P^\vee
       \otimes_{B_C}R_P^\vee.
\end{equation}
The Brauer-pair correspondence of Breland and Miller \cite{BM}
shows that $Z\otimes Z_P^\vee$ matches the chosen maximal roots,
and that the final complex has the fixed diagonal vertex bound.
In the inertial case, the component argument establishes inertia of
the block of $PF^*(G)$, where $F^*(G)$ is the generalized Fitting
subgroup, so the upper-block hypothesis of
\cite[Theorem~1.3]{ZZ} applies.

\section{Sources and local complexes}
\label{sec:sources}

\subsection{Conventions}
For a finite group $G$, let $O_2(G)$ and $O_{2'}(G)$ be its
largest normal $2$-subgroup and largest normal odd-order subgroup.
A quasisimple group is perfect with simple central quotient; a
component is a subnormal quasisimple subgroup. The layer $E(G)$
is the product of the components, the Fitting subgroup $F(G)$ is
the largest nilpotent normal subgroup, and $F^*(G)=F(G)E(G)$.
For a positive integer $n$, let $n_2$ be its largest power-of-two
divisor. For a finite abelian group $A$, write $A_2$ for its
unique Sylow $2$-subgroup.

We regard an $(\OO G,\OO L)$-bimodule as a module for $G\times L$ by
$(g,l)m=gml^{-1}$. For an isomorphism $\eta:P_C\to P$, put
\[
 \diag(P,\eta,P_C)=\{(\eta(u),u):u\in P_C\}.
\]
For the identity isomorphism we write $\diag P$. A marking
consists of the chosen maximal Brauer pairs and defect-group
identification. Conjugation transports these data simultaneously.

Fix a uniformizer $\pi$ of $\OO$. We use the same letter for a
block idempotent and its reduction modulo $\pi$, and interpret
its Brauer image after reduction. For an algebra $A$, write
$J(A)$ for its Jacobson radical and
$\Kb(A\text{-}\mathrm{mod})$ for the bounded homotopy category
of finitely generated $A$-modules, abbreviated to $\Kb(A)$ when
the module category is understood. We use analogous notation
for bimodules.

A lattice is a finitely generated free $\OO$-module. A $2$-permutation
lattice for a finite group is a direct summand of a permutation
lattice; the same definition over $k$ gives a $2$-permutation module. If $U$ is a
$kG$-module and $Q\leq G$ is a $2$-subgroup, its Brauer construction is
\[
 U(Q)=U^Q\Big/\sum_{R<Q}\Tr_R^Q(U^R),\qquad
 \Tr_R^Q(u)=\sum_{x\in Q/R}xu.
\]
For a lattice $M$, write $M(Q)=(k\otimes_{\OO}M)(Q)$; for a complex
apply this functor termwise. On a group algebra it gives the Brauer
homomorphism $\Br_Q:(kG)^Q\to kC_G(Q)$, which deletes the coefficients
of elements outside $C_G(Q)$. If $Q$ acts on a normal subgroup $H$,
the same formula defines the relative map
$\Br_Q:(kH)^Q\to kC_H(Q)$. Integral local complexes are lifted from
these constructions by Lemma~\ref{lem:integral-lift}.

An $a$-Brauer pair is a pair $(Q,e)$ with $e$ a block of $kC_G(Q)$ and
$e\Br_Q(a)\ne0$. We use the usual inclusion of Brauer pairs; beneath
a fixed maximal pair $(P,e_P)$ there is a unique pair $(Q,e_Q)$ for
each $Q\leq P$. Its block fusion system $\FF$ has the subgroups of
$P$ as objects and the conjugations in $G$ that preserve these pairs
as morphisms. A source idempotent is a primitive idempotent
$i\in(kGa)^P$ with $\Br_P(i)\ne0$; $i(kG)i$ is its source algebra.
The integral terminology is defined by reduction. A local point of
$Q$ on an algebra is a conjugacy class, under its $Q$-fixed units,
of primitive $Q$-fixed idempotents with nonzero Brauer image.

Write $A^e=A\otimes A^{\op}$, over the coefficient ring in use.
A complex is split if it is homotopy equivalent to its homology with
zero differential. An endosplit resolution of a module $V$ is a
bounded complex $S$ with $H^0(S)=V$, other homology zero, and
$S\otimes S^\vee$ split. An endopermutation $P$-module is a module
whose endomorphism algebra, with the conjugation action of $P$, is a
permutation module. It is capped if it has an indecomposable summand
with vertex $P$; the isomorphism class of such a summand is
uniquely determined and is called its cap. Here a vertex is a minimal
subgroup for relative projectivity, and a source is an indecomposable
module at that subgroup inducing the given summand. The Dade group
records capped endopermutation modules up to isomorphism of their
caps, with tensor product followed by taking the cap.

Write $D(P)$ for this group. For an injective homomorphism
$\varphi:Q\to P$, ${}_{\varphi}M$ denotes restriction along
$\varphi$, so that $q$ acts as $\varphi(q)$. The notation applies
termwise to complexes. A capped endopermutation module is
$\FF$-stable if restriction to $Q$ and restriction along
$\varphi$ have the same class in $D(Q)$ for every
$\varphi\in\Hom_{\FF}(Q,P)$. For an indecomposable capped
module this implies ${}_{\alpha}V\cong V$ for every
$\alpha\in\Aut_{\FF}(P)$.

For symmetric block algebras $A,B$, use the symmetrizing forms given
by the coefficient of the group identity. The dual and these forms
give the two adjunction evaluations
\begin{equation}\label{eq:evaluations}
 X\otimes_BX^\vee\longrightarrow A,
 \qquad X^\vee\otimes_AX\longrightarrow B.
\end{equation}
Whenever we call $X$ a Rickard complex, both maps are homotopy
equivalences in the indicated bimodule categories. For complexes
projective on both sides over symmetric algebras, $X^\vee$ is
the tensor functor's left and right adjoint. If it is a two-sided
inverse, the adjunction unit and counit are isomorphisms in the
homotopy categories. Thus the inverse identities also give these
canonical evaluations.
Relative to a fixed defect-group isomorphism $\eta:P_C\to P$,
a Rickard complex is splendid if its terms are $2$-permutation,
projective on both sides, and relatively
$\diag(P,\eta,P_C)$-projective. Tensor products use
the Koszul sign rule. A complex with an additional group action is always
a complex for that finite group, including its inner intersections
and multiplication relations. More explicitly, if finite extensions
$\widetilde G,\widetilde L$ of $G,L$ have a common quotient $\Gamma$,
the group of pairs is
\[
 \{(g,l)\in\widetilde G\times\widetilde L:
       gG=lL\text{ in }\Gamma\}.
\]
For a tensor through a middle extension $\widetilde H$, an outer
pair $(g,l)$ acts by
$(g,h)m\otimes(h,l)n$, where $h$ has their common image in
$\Gamma$. Changing $h$ by an element of $H$ leaves this tensor
unchanged by the balanced relation. This defines the tensor functors
in the operator categories used below.

\subsection{The source resolution}
We recall two source-theoretic inputs. If a Morita equivalence between
blocks $A=kGa$ and $B=kLc$ is induced by an indecomposable bimodule $M$
with endopermutation source and identified defect group $P$, the source
theorem supplies source idempotents $i,j$, a common fusion system $\FF$,
and a capped $\FF$-stable endopermutation $kP$-module $V$ such that $M$
is a summand of
\begin{equation}\label{eq:source-cover}
 U=Ai\otimes_{kP}T(V)\otimes_{kP}jB,
 \qquad T(V)=\Ind_{\diag P}^{P\times P}V.
\end{equation}
The graph of the given defect identification is understood. This is the
source theorem recalled in \cite[Section 5]{LinckelmannIntegral}; its
use in this form is also the starting point of \cite{Linckelmann}.
Moreover, every indecomposable summand of the source algebra $jBj$ as a
$kP$-$kP$-bimodule has the form
\begin{equation}\label{eq:fusion-graph}
 kP\otimes_{kQ}{}_{\varphi}kP,
 \qquad Q\leq P,\quad \varphi\in\Hom_{\FF}(Q,P).
\end{equation}
The integral version has the analogous permutation basis
\cite[Proposition 2.5]{Linckelmann}.

\begin{lemma}\label{lem:endosplit}
Suppose that $P$ is abelian and $V$ is the capped source in
\eqref{eq:source-cover}. There is a bounded endosplit
$2$-permutation resolution $S$ of $V$ such that ${}_{\alpha}S\cong S$
for every $\alpha\in E=\Aut_{\FF}(P)$.
\end{lemma}
\begin{proof}
The extension axiom for a saturated fusion system on an abelian group
extends every fusion morphism to $P$. The Sylow axiom gives
$|E|$ odd, since $\Aut_P(P)=1$.
The Dade group of an abelian $2$-group is generated by the inflated
relative syzygies associated with the quotients $P/Q$ for $Q<P$,
including $Q=1$. Its normal form has
integer exponents for noncyclic quotients and exponents modulo two for
cyclic quotients of order at least four. The quotient of order two
contributes no nontrivial generator. The exponents of the class of $V$
are constant on $E$-orbits.

Represent these factors by truncated permutation resolutions and use
their duals for negative exponents. Their tensor is endosplit and is
isomorphic to every $E$-transport. Its homology has $V$ as its cap.
For an endosplit resolution $S_0$, the canonical homology map gives
\[
 \End_{\Kb(kP)}(S_0)\cong\End_{kP}(H^0(S_0)).
\]
Choose a primitive projector with image $V$ on $H^0(S_0)$ and lift
it through this isomorphism. Splitting it gives an endosplit
resolution of $V$. For $\alpha\in E$, the transported projector,
viewed on the same $H^0(S_0)$ through the chosen transport of $S_0$,
also has image isomorphic to $V$. The two primitive projectors are
conjugate in $\End_{kP}(H^0(S_0))$. Their lifted images are therefore
homotopy equivalent. After removing contractible summands, a homotopy
equivalence of these minimal complexes is a chain isomorphism.
This proves the assertion and is the abelian construction of
\cite[Lemmas~4.1--4.2 and Proposition~4.3]{Huang}.
\end{proof}

\begin{proposition}\label{prop:source-upgrade}
Let $M$ be a Morita bimodule over $k$ as in \eqref{eq:source-cover}, with
$P$ abelian. There is a splendid Rickard complex $X$ over $k$ for the same
primitive blocks and defect identification, with $H^0(X)\cong M$ and
all other homology zero.
\end{proposition}
\begin{proof}
Use Lemma~\ref{lem:endosplit} and form
\begin{equation}\label{eq:resolved-cover}
 Y=Ai\otimes_{kP}T(S)\otimes_{kP}jB.
\end{equation}
Each term is a summand of an induced diagonal permutation module,
projective on both sides. The outside factors are projective over $kP$,
and $T(S)$ is free on either $P$-side. Thus $H^0(Y)=U$ and all other
homology vanishes. The bimodule $U$ is also projective on either side.

Decompose the middle factor $jBj$ in $Y\otimes_BY^\vee$ using
\eqref{eq:fusion-graph}. For its $(Q,\varphi)$-summand, the coefficient
complex on the diagonal $Q$ is
\begin{equation}\label{eq:twisted-square}
 \Res_Q^P S\otimes_k
 {}_{\varphi}\Res_{\varphi(Q)}^P S^\vee.
\end{equation}
Extend $\varphi$ to an element of $E$. The corresponding individual
isomorphism from Lemma~\ref{lem:endosplit} identifies
\eqref{eq:twisted-square} with a restriction of $S\otimes S^\vee$,
which is split. Inducing this splitting and tensoring with the outside
$Ai$ and $iA$ factors gives an $A$-$A$-bimodule splitting. The same
argument, with $iAi$ in the middle, gives a $B$-$B$-bimodule splitting.
Consequently
\begin{equation}\label{eq:split-squares}
 Y\otimes_BY^\vee\simeq (U\otimes_BU^\vee)[0],
 \qquad
 Y^\vee\otimes_AY\simeq (U^\vee\otimes_AU)[0].
\end{equation}
The splittings are in the full bimodule categories, since the outer
factors carry the given group actions throughout.

Symmetric adjunction identifies
$\Hom_B^\bullet(Y,Y)$ with $Y\otimes_BY^\vee$. The invariants
formula is
\begin{equation}\label{eq:correct-invariants}
 \End_{A-B}^\bullet(Y)=
 \Hom_{A^e}\bigl(A,\Hom_B^\bullet(Y,Y)\bigr).
\end{equation}
The argument of this additive functor is already split by bimodule
homotopies. Applying the functor preserves the given bimodule homotopies
and hence the splitting. The canonical homology map is an algebra isomorphism
\begin{equation}\label{eq:projector-lift}
 \End_{\Kb(A-B)}(Y)\xrightarrow{\sim}\End_{A-B}(U).
\end{equation}
Lift the idempotent whose image is the prescribed $M$ and split it in
the bounded homotopy category of $2$-permutation modules. This category
is idempotent complete. More concretely, remove contractible summands
from $Y$; null-homotopic endomorphisms then lie in the radical of its
finite chain-endomorphism algebra, so the idempotent has a strict lift.
Let $X$ be its termwise image.

Both balanced squares of $X$ are retracts of the split squares in
\eqref{eq:split-squares}. Their degree-zero homology is respectively
$M\otimes_BM^\vee$ and $M^\vee\otimes_AM$. The evaluations
\eqref{eq:evaluations} induce the Morita evaluations, which are
isomorphisms onto $A$ and $B$. A map between split complexes that induces
an isomorphism on homology is a homotopy equivalence. This proves both
required evaluations on the same complex $X$.
\end{proof}

\begin{lemma}\label{lem:integral-lift}
Let $X$ be a splendid Rickard complex between blocks $A=kGa$ and
$B=kLc$, equipped with a compatible
action of a finite group of pairs. Suppose that every term is
$2$-permutation for this entire finite action group and that the two
evaluation cones are contractible in the corresponding full operator
categories. Then $X$ has an integral $2$-permutation lift with those
actions and evaluations. The same assertion holds in a specified
degree-preserving category.
\end{lemma}
\begin{proof}
Lift the terms to their integral $2$-permutation lattices for the whole
finite action group. Reduction on Hom spaces is surjective in that
group-module category: orbit sums prove it for permutation lattices,
and lifting summand projectors proves it for their summands. Without
extra operators, tensoring with the Rickard inverse gives
\begin{equation}\label{eq:homotopy-vanishing}
 \Hom_{\Kb(A-B)}(X,X[n])
 \cong\Hom_{\Kb(A-A)}(A,A[n])=0\qquad(n\ne0).
\end{equation}
With the prescribed operators, the tensor functors stay in the full
pair-group categories. In a presentation with a common operator quotient,
choose one middle-group lift of a quotient element and use it on both
tensor factors. Changing that lift is canceled by the balanced relation,
and multiplication retains the specified inner intersections. The stipulated
equivariant evaluations make these functors mutually inverse.
The inverse-tensor functor sends $X$ to the regular degree-zero bimodule with its
diagonal operator action, so the corresponding equivariant homotopy
Hom groups also vanish in nonzero degree. In the special case of an
odd operator quotient this follows equivalently by taking invariants,
which is exact. This homotopy vanishing supplies the differential lifts. The
whole-group permutation hypothesis supplies the termwise lifts
and surjectivity on Hom spaces.

Suppose that a differential has been lifted modulo $\pi^m$.
Its curvature modulo $\pi^{m+1}$ is a degree-two cocycle in the same
endomorphism complex. Equation~\eqref{eq:homotopy-vanishing} corrects it.
Completeness gives a differential on the fixed finite lattices. The
integral evaluation cones reduce to contractible complexes. Their terms
are again permutation summands for the full pair groups: before block
cuts, balanced permutation bases give balanced orbit sets, and tensor
products of the summand projectors descend to the corresponding
linearized orbit modules. The regular block terms are
invariant summands of the regular permutation bimodules. Thus the same
Hom-surjectivity applies to the cone terms. Lift a
reduced contraction $h$ and put $u=dh+hd$. Then $u$ is a chain
automorphism congruent to the identity modulo $\pi$, and $hu^{-1}$ is an
integral contraction.

In the graded applications, the grading is the group-coset grading:
its identity component is a module for the corresponding group of
pairs. Lift in that finite-group category and induce to recover the
graded terms. For a group grading, project a map on a homogeneous input of degree $g$
to its output of degree $g$. This projection is bimodule-linear: left
and right multiplication of degrees $h,k$ send $g$ to $hgk$ on both
sides. It commutes with the Hom differential and applies to the
homotopies as well. The same lifting argument therefore stays in the
degree-preserving category.
\end{proof}

\subsection{The local complex of an equivalence}
The anti-involution $*$ of a group algebra sends $g$ to $g^{-1}$.
For a complex $U$ of $2$-permutation bimodules between blocks
$A,B$ and an $A\otimes B^{\op}$-Brauer pair
$\lambda=(\diag(Q,\varphi,R),e\otimes f^*)$, put
\[
 U(\lambda)=eU(\diag(Q,\varphi,R))f.
\]
We say that $U$ supports $\lambda$ if this complex is
noncontractible, and write $\operatorname{BP}(U)$ for its set
of supported Brauer pairs. The centralizer block idempotents
$e,f$ are called the roots of the pair. The Lefschetz invariant
is the alternating sum $\sum_n(-1)^n[U^n]$ in the split
Grothendieck group of $2$-permutation modules.

\begin{proposition}\label{prop:companion}
Let $R:A\longrightarrow C$ be an integral splendid Rickard complex
between primitive blocks, with identified abelian defect groups.
Fix a supported maximal pair
\[
 \lambda_R=(\diag(P,\eta,P_C),e\otimes f^*).
\]
There is an integral maximal companion
$R_P:B_A\longrightarrow B_C$ between the prescribed Brauer
correspondents, supporting this same pair. If
$Z:C\longrightarrow B_C$ is splendid Rickard, it has a maximal
companion $Z_P:B_C\longrightarrow B_C$ such that a reduced integral
representative
\begin{equation}\label{eq:return}
 X\simeq R\otimes_CZ\otimes_{B_C}Z_P^\vee
                    \otimes_{B_C}R_P^\vee
\end{equation}
is splendid Rickard from $A$ to $B_A$. Its terms are relatively
$\diag P$-projective for the fixed diagonal, and
$(\diag P,e\otimes e^*)$ is a supported maximal pair.
\end{proposition}
\begin{proof}
Brauer constructions in this proof are over $k$; reduction of an
integral complex is understood. Write $A=\OO Ga$ and
$C=\OO G_Cc$, and put
\[
 I=N_G(P,e),\qquad I_C=N_{G_C}(P_C,f),\qquad
 W=N_{I\times I_C}(\diag(P,\eta,P_C)).
\]
With $\widehat e,\widehat f$ the block idempotents of the full
normalizers corresponding to the two maximal pairs, the modular
companion is
\begin{multline}\label{eq:companion-formula}
 \widehat e\,kN_G(P)e\otimes_{kI}
 \Ind_W^{I\times I_C}
       \bigl(eR(\diag(P,\eta,P_C))f\bigr)\\
 {}\otimes_{kI_C} f\,kN_{G_C}(P_C)\widehat f .
\end{multline}
Theorem~8.5 and Corollary~8.9 of \cite{BM} give its two Rickard
evaluation identities. Its endpoints are the prescribed Brauer correspondents.

To identify its maximal pair, apply
\cite[Corollary 7.11(b)]{BM} to the Lefschetz invariant of $R$
at the full defect groups. It says that the inertia-level complex
inside \eqref{eq:companion-formula} has $\lambda_R$ as a maximal
Brauer pair for its Lefschetz invariant. By
\cite[Corollary 8.8]{BM}, this is a Brauer pair of that complex.
The two Fong--Reynolds bimodules in
\eqref{eq:companion-formula} have, at their respective diagonal
maximal pairs, the regular centralizer blocks:
\[
 e\bigl(\widehat e\,kN_G(P)e\bigr)(\diag P)e
       \cong kC_G(P)e,
 \qquad
 f\bigl(f\,kN_{G_C}(P_C)\widehat f\bigr)(\diag P_C)f
       \cong kC_{G_C}(P_C)f.
\]
Indeed the fixed group elements in the regular permutation
bimodule are precisely the corresponding centralizer elements,
and the displayed block cuts retain the indicated root.
The blockwise Brauer--tensor formula
\cite[Theorem 6.7]{BM} therefore gives a noncontractible
root-cut Brauer complex at $\lambda_R$ for
\eqref{eq:companion-formula}.
Thus the full-normalizer companion supports the same
$\lambda_R$, including its two roots and its graph.
Remove its contractible summands in the ordinary bimodule category.
By \cite[Corollary 9.5]{BM}, its terms have vertices below the
graph in $\lambda_R$. Lift the resulting complex integrally by
Lemma~\ref{lem:integral-lift}. The lift has the same Brauer support.

Fix the left root $f$ for $Z$.
By \cite[Lemma 8.3 and Corollary 8.8(c)]{BM}, it extends to a
supported maximal pair
\[
 \lambda_Z=(\diag(P_C,\theta,P_C),f\otimes g^*)
\]
for some root $g$ of $B_C$ and some automorphism $\theta$ of
$P_C$. The right defect group is the chosen subgroup $P_C$:
it is normal in $N_{G_C}(P_C)$, so it is the unique defect
subgroup of $B_C$. Apply the preceding companion construction
to this pair. The left full-normalizer block is $B_C$.
The right full normalizer is again $N_{G_C}(P_C)$, and
$B_C$ is its own Brauer correspondent there. Hence the resulting
integral complex is an autoequivalence $Z_P$ of $B_C$
supporting $\lambda_Z$.

Let $Y$ denote the four-factor product on the right of
\eqref{eq:return}. The ordinary tensor-product assertion
\cite[Lemma 6.3]{BM} makes its terms permutation summands
with twisted diagonal vertices, meaning graphs of isomorphisms
between $2$-subgroups of the endpoints. The terms are projective
on both sides.
Associativity and the four given Rickard evaluations show that
the canonical evaluations of $Y$ are homotopy equivalences:
in the first square successively cancel
$R_P^\vee R_P$, $Z_P^\vee Z_P$, $ZZ^\vee$, and $RR^\vee$;
in the second square cancel
$R^\vee R$, $Z^\vee Z$, $Z_PZ_P^\vee$, and $R_PR_P^\vee$.

It remains to recover the fixed diagonal and root.
Put
\[
 E=kC_G(P)e,\qquad
 F=kC_{G_C}(P_C)f,\qquad
 F_g=kC_{G_C}(P_C)g .
\]
By \cite[Theorem 8.5]{BM}, the local complexes of $R,R_P$
at $\lambda_R$ are Rickard equivalences from $E$ to $F$,
and those of $Z,Z_P$ at $\lambda_Z$ are Rickard
equivalences from $F$ to $F_g$.
Consequently
\[
 R(\lambda_R)\otimes_F Z(\lambda_Z)
       \otimes_{F_g} Z_P(\lambda_Z)^\vee
       \otimes_F R_P(\lambda_R)^\vee
\]
is a Rickard autoequivalence of $E$, so it is
noncontractible. Iterating
\cite[Theorem 6.7 and Proposition 8.2]{BM} exhibits this
complex as a direct summand of the restriction of
$eY(\diag P)e$ to its centralizer bimodule category:
the chosen intermediate roots are $f,g,f$, and the graph
maps compose as
$\eta\theta\theta^{-1}\eta^{-1}=\mathrm{id}_P$.
Thus $(\diag P,e\otimes e^*)$ belongs to $\operatorname{BP}(Y)$.
It is maximal by \cite[Corollary 8.8(c)]{BM}.

Finally choose a reduced integral representative.
For permutation lattices and their summands, reduction on
group-module Hom spaces is surjective, and their finite
endomorphism algebras are complete. Therefore a summand
projector of a term after reduction lifts integrally. If the reduction
of the current complex has an invertible differential component
between two summands, lift their projectors. The corresponding
integral differential component is invertible by Nakayama's
lemma, and Gaussian elimination splits off an integral
contractible two-term complex. Repeating this finite process
produces $X\simeq Y$ such that $k\otimes_{\OO}X$ has no
contractible direct summands. Its evaluations and Brauer support
are unchanged. By \cite[Proposition 5.4]{BM}, its reduction is
indecomposable as a complex. Corollary~9.5 of \cite{BM},
applied to its supported maximal pair
$(\diag P,e\otimes e^*)$, now puts every term vertex below
a conjugate of the original $\diag P$. Reduction of
permutation lattices preserves vertices
\cite[Theorem 3.2(b)]{BM}, so the same bound holds
integrally. This proves the stated splendidness, with the
fixed marking and the prescribed target block.
\end{proof}

\section{Central quotients and inertial blocks}\label{sec:inertial}

A basic Morita equivalence is induced by an indecomposable
Morita bimodule whose vertex is the graph of the chosen
defect-group isomorphism and whose source is endopermutation.
For a maximal Brauer pair $(P,e_P)$, put $I=N_G(P,e_P)$.
The inertial block is the Fong--Reynolds correspondent in $I$
of the Brauer correspondent in $N_G(P)$.
A block is nilpotent if $N_G(Q,e)/C_G(Q)$ is a $2$-group for
each of its Brauer pairs $(Q,e)$, and is inertial if it is
basically Morita equivalent to its inertial block.

An interior $P$-algebra has a specified homomorphism
$P\to A^\times$. A primitive Dade $P$-algebra is the
endomorphism algebra of an indecomposable capped endopermutation
lattice, with this interior action and Brauer quotient $k$ at
$P$. It is $E$-stable if transport by every element of $E$
gives an isomorphic interior $P$-algebra. A determinant-one
interior structure means that every $p\in P$ has determinant
one on the corresponding lattice.
An inertial source algebra has the form
$S\otimes_{\OO}\OO_\alpha(P\rtimes E)$, where $S$ is a primitive
$E$-stable Dade $P$-algebra and $E$ is the odd inertial quotient.
Here $\alpha$ is a normalized $\OO^\times$-valued $2$-cocycle
on $E$, inflated to $P\rtimes E$.

We use two consequences of Puig's theory \cite{Puig}: a normal
subblock of an inertial block is inertial (Theorem~3.13), and
the rank criterion in Proposition~3.11. The latter says that an
$E$-stable Dade algebra $S$ gives inertia if the local corner
of $S^\op\otimes\OO G$ at the product of the Dade point and
the source point has rank $|P||E|$. Its proof shows that the
canonical interior embedding into that corner remains injective
after reduction; rank equality then makes it an isomorphism.

\subsection{Inertia and a central 2-quotient}
\begin{proposition}\label{prop:central-inertia}
Let $a$ be a block with abelian defect group $P$, and let
$Z\leq Z(G)\cap P$. The block $a$ is inertial if and only if the
dominated block of $G/Z$ is inertial.
\end{proposition}
\begin{proof}
Put $A=i\OO Gi$ for a source idempotent, and let
$E=\Aut_{\FF}(P)$. The group $E$ has odd order and fixes $Z$ pointwise.
The source-algebra permutation basis has stabilizers
$\diag_\varphi Q$, with $\varphi$ a fusion morphism. Since
$za=az$ for $z\in Z$, $\diag Z$ fixes every basis element. Hence
$Z\leq Q$ and $\varphi|_Z=1$ in every orbit. Left multiplication by
$Z$ is free on that basis.

Suppose that conjugation by $r\in P$ fixes the $Z$-orbit of a basis
element $\xi$. Then $r\xi r^{-1}=z\xi$ for some $z\in Z$.
The graph equation gives $r\in Q$ and $\varphi(r)=rz^{\pm1}$.
Extend $\varphi$ to $\alpha\in E$. On $\langle r,Z\rangle$, the
endomorphism $\delta=\alpha-1$ has image in $Z$, vanishes on $Z$, and
satisfies $\delta^2=0$. If $m$ is the odd order of this restriction,
$1=(1+\delta)^m=1+m\delta$; thus $\delta=0$ and $z=1$.
Passage to $Z$-orbits creates no new conjugation stabilizers.

Write $I_Z=(z-1:z\in Z)\OO Z$ and $\overline A=A/I_ZA$.
Orbit sums therefore lift with coefficient one, giving
\begin{equation}\label{eq:fixed-surjection}
 A^R\twoheadrightarrow\overline A^{RZ/Z}\qquad(R\leq P).
\end{equation}
The kernel is radical: $I_Z\subseteq J(\OO Z)$ and $A$ is finite
over the central local ring $\OO Z$, so $I_ZA\subseteq J(A)$.
For a fixed element, the inverse of a fixed unit is fixed; hence
$J(A)\cap A^R\subseteq J(A^R)$. In particular
$\overline A^{P/Z}$ is local.
Its identity has nonzero Brauer image: an expression by proper quotient
traces would lift, using \eqref{eq:fixed-surjection}, to proper traces
and radical elements in $A^P$. The proper-trace ideal is contained
in the proper Brauer kernel of the local algebra $A^P$, hence in
its radical, contradicting $1\notin J(A^P)$.
Thus $\overline A$ is the source algebra of the dominated block.

Quotienting each graph gives its quotient fusion graph. Every maximal
automorphism graph occurs in a source algebra, by
\cite[Proposition 2.5]{Linckelmann}. Since $Z\leq Q$, a quotient graph
can be maximal only when $Q=P$. The downstairs inertial quotient is
therefore the image of $E$. Its kernel is trivial: an automorphism
fixing $Z$ and $P/Z$ lies in a $2$-group automorphism kernel, whereas
$E$ is odd. Write this identification as $E\cong\overline E$.

Suppose that the dominated block is inertial. There is an
$\overline E$-stable primitive Dade algebra $\overline S$ with
\[
 \overline A\cong\overline S\otimes_{\OO}
 \OO_\beta((P/Z)\rtimes\overline E).
\]
Inflate $\overline S$ to $S$ and put $T=S^\op\otimes A$. A permutation
conjugation basis of $S^\op$, together with the preceding graph basis,
gives $T^P\twoheadrightarrow\overline T^{P/Z}$ with radical kernel.
Cancel the downstairs Dade factors and choose the resulting primitive
local idempotent $\overline j$, so that
\[
 \overline j\,\overline T\,\overline j
 \cong\OO_\beta((P/Z)\rtimes\overline E).
\]
Lift it to $j\in T^P$. Locality follows from the exact trace identity
\begin{equation}\label{eq:trace-central}
 \overline{\Tr_R^P(t)}=[RZ:R]
 \Tr_{RZ/Z}^{P/Z}(\overline t).
\end{equation}
For $R<P$, the right side is either a proper trace or is divisible by
two. Thus a nonlocal lift would have a nonlocal image.

The lifted idempotent belongs to the product local point. A primitive
Dade algebra satisfies $S(P)=k$, so
$T(P)\cong S^\op(P)\otimes A(P)\cong A(P)$. The last algebra is a
nonzero quotient of the local algebra $A^P$ and hence is local.
There is a unique local $P$-point of $T$, namely the product of its
Dade point and the chosen source point. The lifted $j$ belongs to it.
Now $T$ is free over the central local ring $\OO Z$, and $t\mapsto jtj$
is an $\OO Z$-linear projector. Thus $C=jTj$ is free over $\OO Z$ and
\[
 \rk_{\OO}C
 =|Z|\rk_{\OO}(C/I_ZC)
 =|Z|\,|P/Z|\,|\overline E|=|P|\,|E|.
\]
Puig's rank criterion proves ascent of inertia.

Conversely, write an inertial source as
$S\otimes\OO_\beta(P\rtimes E)$, using the determinant-one interior
structure on its Dade module. Centrality of $Z$ forces its action on
the Dade module to be scalar and forces $E$ to fix $Z$. A scalar of
$2$-power order with determinant one in odd dimension is one.
The source therefore descends to the indicated inertial source for
$P/Z$. This proves descent.
\end{proof}

\subsection{A supported nilpotent cover}

We specify the classification input used in this subsection.  In the
residual nonnilpotent Lie-type case, there is a simple simply connected
algebraic group $\mathbf X$ in odd characteristic, a Steinberg
endomorphism $F$, and a central subgroup $Z\leq X:=\mathbf X^F$ such
that $S=X/Z$.  The block of $X$ dominating the specified block of $S$
has a semisimple $2'$-label $s$ for which
$C_{\mathbf X^*}^{\circ}(s)$ is a torus.  This is the concrete case
in the proof of \cite[Theorem 6.1]{EKKS} to which
\cite[Lemma 4.2]{EKKS} is applied.  The other case needed here is that
the block of $S$ itself is nilpotent.  The toral input includes this
nonexceptional covering realization; the regular,
field, and graph automorphism construction below is made for this
realization.

We first record two block-theoretic facts in the forms that we use.
Suppose that $H=NP$, where $N\triangleleft H$ and $P$ is a defect
group of a block $B$ of $H$.  Then every block of $N$ covered by $B$
is $P$-stable.  Indeed, the covered blocks form one $H$-orbit, hence
one $P$-orbit.  If $b$ is one of them and its stabilizer $P_b$ is
proper in $P$, their sum is $\Tr_{P_b}^P(b)$ and has zero
$P$-Brauer image.  Multiplying this sum by $B$ contradicts
$\Br_P(B)\ne0$.  Thus the orbit has one member.  Since $H/N$ is a
$2$-group, $b$ has a unique covering block in $H$, and, as central
idempotents in $\OO H$, this block is $B=b$.  In addition,
$P\cap N$ is a defect group of $b$.  We use the usual
normal-subgroup theorem for defect groups and compatible Brauer
pairs, in the form recorded in \cite[Sections 3.2--3.4]{Puig}.

\begin{lemma}\label{lem:two-index-nilpotence}
Let $N\triangleleft H$, with $H/N$ a $2$-group, and let $B$ be a
block of $H$ covering a block $b$ of $N$.
\begin{enumerate}
\item If $B$ is nilpotent, then $b$ is nilpotent.
\item Suppose that $H=NP$ and that $P$ is an abelian defect group
of $B$.  If $b$ is nilpotent, then $B$ is nilpotent.
\end{enumerate}
\end{lemma}
\begin{proof}
For the first assertion, take a $b$-Brauer pair $(Q,d)$ and a
compatible $B$-Brauer pair $(Q,e)$, so that $e$ covers $d$.  Such
pairs exist by the normal-subgroup Brauer-pair theorem, after
conjugating a compatible choice of maximal pairs.  The quotient
$C_H(Q)/C_N(Q)$ is a $2$-group.  Consequently $e$ is the unique
block of $C_H(Q)$ covering $d$.  Every element of $N_N(Q,d)$
therefore stabilizes $e$, and conjugation gives an injection
\[
 N_N(Q,d)/C_N(Q)\ \longrightarrow\ N_H(Q,e)/C_H(Q).
\]
The latter group is a $2$-group by nilpotence of $B$.  This proves
the nilpotence condition for every $b$-Brauer pair.

For the second assertion, put $D=P\cap N$ and choose a maximal
$B$-Brauer pair $(P,e_P)$.  The compatible normal-subgroup pair
theorem supplies a $P$-stable maximal $b$-Brauer pair $(D,d)$ below
it.  Let $(D,e_D)$ be the unique $B$-subpair of $(P,e_P)$ with
first component $D$; it covers $d$.  Since $P$ is abelian and
$H=NP$, we have
\[
 C_H(D)=C_N(D)P.
\]
The $P$-stability of $d$ therefore makes it invariant under all of
$C_H(D)$.  Its unique block extension to $C_H(D)$ is $e_D$; in
particular, $e_D$ covers only $d$.  Now every
$y\in N_H(P,e_P)$ normalizes $D$ and stabilizes $e_D$, and hence
stabilizes $d$.

Write $y=xp$ with $x\in N$ and $p\in P$.  Both $y$ and $p$
stabilize $(D,d)$, so $x\in N_N(D,d)$.  Nilpotence of $b$ and
abelianness of its defect group $D$ give
$N_N(D,d)=C_N(D)$: its action on $D$ is both a $2$-group, by
nilpotence, and a $2'$-group, by the maximal-pair inertial-quotient
theorem.  Thus $y$ acts trivially on $D$.  It also acts trivially
on $P/D$, since this is the image of $P$ in the abelian group
$H/N=P/D$.

For an abelian $2$-group $P$, the automorphisms that fix $D$ and
$P/D$ pointwise form a $2$-group.  Indeed, writing the group law
additively, every such automorphism is $1+\delta$, where
$\delta\in\operatorname{Hom}(P/D,D)$; these maps compose by
addition because the product of any two such endomorphisms is
zero.  The inertial quotient of $B$ is a $2'$-subgroup of
$\Aut(P)$, so it is trivial.  Fusion in a block with abelian
defect group is controlled by its maximal Brauer pair.  It follows
that $B$ is nilpotent.
\end{proof}

\begin{lemma}\label{lem:upper-inertia}
Let $S$ be a quasisimple component, let $b$ be its specified block,
and suppose that $b$ is in the toral case described above or is
nilpotent.  Let $P$ be an abelian $2$-subgroup normalizing $S$, and
let $c$ be a block of $G_0=SP$ covering $b$ whose defect group is
the subgroup $P$.  Then $c$ is inertial.
\end{lemma}
\begin{proof}
Put $D=P\cap S$.  The preceding observation shows that $b$ is
$P$-stable, that $D$ is a defect group of $b$, and that $c=b$ as
idempotents in $\OO G_0$.  In particular,
$\Br_P(b)=\Br_P(c)\ne0$, with the first expression interpreted
as the relative Brauer map on $(\OO S)^P$.  If $b$ is nilpotent,
Lemma~\ref{lem:two-index-nilpotence}(2) proves that $c$ is
nilpotent.  We henceforth treat the toral case.

First we justify the central reductions.  Let $R=O_2(G_0)$.
Since $R$ is contained in every defect group of $c$, we have
$R\leq P$.  Also
$[R,S]\leq R\cap S\leq O_2(S)\leq Z(S)$.  For a fixed element
of $R$, the resulting central commutator map on the perfect group
$S$ is trivial.  Thus $R$ centralizes $S$, and it centralizes $P$
because $P$ is abelian.  Conversely, $C_P(S)$ centralizes both
$P$ and $S$ and is a normal $2$-subgroup of $G_0$.  Hence
\[
 R=C_P(S)\leq Z(G_0),\qquad R\cap S=O_2(Z(S)).
\]
Apply Proposition~\ref{prop:central-inertia} to $c$ and pass to
$G_0/R$.  We retain the notation $G_0,S,P,b,c$.  Now
$O_2(G_0)=1$ and $Z(S)$ has odd order.  The group
$C_{G_0}(S)/Z(S)$ is a $2$-group and $Z(S)$ is central in
$C_{G_0}(S)$.  Schur--Zassenhaus gives a unique Sylow
$2$-subgroup of this centralizer.  It is normal in $G_0$ and is
therefore trivial.  Consequently
\[
 C_{G_0}(S)=Z(S),\qquad C_P(S)=1.
\]

Let $\theta$ be the odd central character of $b$ and put
$K_0=\ker(\theta)\leq Z(S)$.  The $P$-stability of $b$ makes
$K_0$ normal in $G_0$.  The averaging idempotent of $K_0$ acts as
the identity on $c=b$, so quotienting by $K_0$ gives an
isomorphism of block algebras, with the subgroup $P$
identified with its image.  This preserves the source algebra
and inertia.  The centralizer is preserved: if an element
centralizes $S/K_0$, its commutators with $S$ lie in $K_0$, and
perfection of $S$ again makes that central commutator map trivial.
Thus
\[
 C_{G_0/K_0}(S/K_0)=Z(S)/K_0.
\]
In particular $O_2(G_0/K_0)=1$.  We may therefore retain the
notation and assume that $\theta$ is faithful.  Its invariance
implies that $P$ centralizes $Z(S)$.  The same block of the
simply connected group $X$ dominates the successive central
quotients of $b$, so the toral classification input is retained.

Choose a regular embedding $\mathbf X\hookrightarrow
\widetilde{\mathbf X}$ with connected centre, compatible with
$F$, and put $\widetilde X=\widetilde{\mathbf X}^{F}$.
Let $\Gamma_Z$ denote the suitable field and graph automorphisms
stabilizing $Z$, so that $\widetilde X\rtimes\Gamma_Z$ induces
the automorphisms of $S=X/Z$.  This is the regular automorphism
realization used in the proof of \cite[Proposition 4.2 and
Lemma 4.5]{JZZ}.  Set
\[
 J_0=Z\,O_2(Z(\widetilde X)),\qquad
 \check G=\widetilde X/J_0,\qquad
 \check A=(\widetilde X\rtimes\Gamma_Z)/J_0.
\]
Since $Z(S)$ has odd order,
$O_2(Z(X))\leq Z$, and hence $J_0\cap X=Z$.
The original $S$ therefore embeds in $\check G$.  Conjugation
gives a homomorphism
$\rho:\check A\longrightarrow\Aut(S)$ whose kernel is the
odd-order group $Z(\widetilde X)/J_0$.

Let $\tau:G_0\longrightarrow\Aut(S)$ be the original action.
The restriction $\tau|_P$ is injective and
$\tau(P)\cap\Inn(S)=\tau(D)$.  Choose a Sylow $2$-subgroup
$\check P$ of $\rho^{-1}(\tau(P))$ containing the already
embedded $D$.  Then $\rho|_{\check P}$ is an isomorphism onto
$\tau(P)$.  Thus
\[
 \lambda=(\rho|_{\check P})^{-1}\tau|_P:P\longrightarrow\check P
\]
is an isomorphism, equals the identity on $D$, and satisfies
$\check P\cap S=D$.  Since $\lambda(x)$ induces the same
automorphism of $S$ as $x$, the map
\begin{equation}\label{eq:actual-envelope}
 G_0\longrightarrow\check A,\qquad sx\longmapsto s\lambda(x)
 \quad(s\in S,\ x\in P)
\end{equation}
is a well-defined homomorphism.  If its value is $1$, then
$\lambda(x)\in S$, hence $x\in D$, and $s\lambda(x)=sx$.
It is therefore injective and is the identity on $S$.  We now
identify $G_0$ and $P$ with these embedded subgroups.

Every block of the regular subgroup $\check G$ covering $b$
is nilpotent, as follows.  Let $b_X$ be the block of $X$
dominating $b$.  The proof of \cite[Lemma 4.2]{EKKS} starts with
an arbitrary block $\widetilde B$ of $\widetilde X$ covering
$b_X$.  Its semisimple label has toral centralizer in the dual
regular group, and the Bonnaf\'e--Rouquier equivalence in that
proof makes $\widetilde B$ nilpotent.  Thus every such $\widetilde B$ is nilpotent.
The odd part of $Z$ is in its kernel; quotienting that part and
the central $2$-group $O_2(Z(\widetilde X))$ preserves
nilpotence.  Conversely, every block of $\check G$ covering
$b$ lifts to one of these blocks of $\widetilde X$.
Consequently every block of $\check G$ covering $b$ is
nilpotent.

The quotient $\check G/S$ is abelian.  Let $\check K$ be the
inverse image of its odd-order subgroup.  Then
$\check K/S$ is odd and abelian, $\check G/\check K$ is a
$2$-group, and $P$ normalizes $\check K$.  Every block of
$\check K$ covering $b$ is covered by a block of $\check G$
covering $b$, and is nilpotent by
Lemma~\ref{lem:two-index-nilpotence}(1).

Let $e$ be the sum of all blocks of $\check K$ covering $b$.
Then $e$ is $P$-stable and $eb=b$, so
\begin{equation}\label{eq:supported-cover}
 \Br_P(e)\Br_P(b)=\Br_P(b)\ne0.
\end{equation}
The sum of a nonfixed $P$-orbit of covering blocks is a relative
trace from its proper stabilizer, and its $P$-Brauer image is
zero.  Therefore there is a $P$-fixed nilpotent block $f$ of
$\check K$ such that
$\Br_P(f)\Br_P(b)\ne0$.

Put $U=P\check K$.  The invariant block $f$ is the unique block
$F$ of $U$ covering it, with $F=f$ as an idempotent.  Its
$P$-Brauer image is nonzero, so some defect group of $F$
contains $P$.  Every defect group of $f$ has order $|D|$:
$\check K/S$ is odd, and the intersection theorem for defects
applied to $S\triangleleft\check K$ gives this assertion.
Also $P\cap\check K=D$.  The normal-extension defect bound,
for a defect group $Q$ of $F$, therefore gives
\[
 |Q|
 \leq |D|\,|U:\check K|
 =|D|\,|P:D|=|P|.
\]
It follows that the subgroup $P$ is a defect group of
$F$.  Lemma~\ref{lem:two-index-nilpotence}(2), with
$N=\check K$, now proves that $F$ is nilpotent.

Choose a block $e_P$ of $kC_U(P)$ satisfying
\[
 e_P\Br_P(F)=e_P,\qquad e_P\Br_P(b)\ne0.
\]
This is possible by the chosen nonzero product of Brauer images,
and $(P,e_P)$ is a maximal $F$-Brauer pair.  Put
$N=N_U(P)$ and $L=SN$.  Brauer's first main theorem first
gives the correspondent $F_N$ of $F$ in $N$, and then gives
the unique block $E_L$ of $L$ with defect $P$ whose correspondent
in $N$ is $F_N$.  Indeed,
\[
 N_L(P)=N_U(P)=N,\qquad C_L(P)=C_U(P).
\]
All three blocks use the same maximal pair $(P,e_P)$.
Their inertial quotients are therefore equal.  The inertial
quotient of $F$ is trivial; hence so is that of $E_L$, and the
abelianness of $P$ makes $E_L$ nilpotent.

The same maximal pair also retains the original covering
relation.  Namely
\[
 e_P\Br_P(E_Lc)
 =e_P\Br_P(c)
 =e_P\Br_P(b)\ne0,
\]
so $E_Lc\ne0$.  Finally, $G_0=SP$ is normal in $L$: the
subgroup $S$ is normal and every element of $N$ normalizes
both $S$ and $P$.  Thus $c$ is a normal subblock of the
nilpotent block $E_L$.  Puig's theorem
\cite[Theorem 3.13]{Puig}, including the stabilizer reduction
in its Section 3.2, proves that $c$ is inertial.  The normal
odd-kernel block isomorphism and
Proposition~\ref{prop:central-inertia} restore the original
group and original block.
\end{proof}

\subsection{Products of the component extensions}

\begin{lemma}\label{lem:product-inertia}
Let $G=PE$, where $P$ is abelian and
$E=S_1\cdots S_m$ is a central product of normal quasisimple
components of $G$.  Let $a$ be a block of $G$ with defect group
$P$, and let $b_i$ be the component blocks covered by $a$.
Suppose that, for each $i$, the block $c_i$ of $PS_i$ covering
$b_i$ is inertial.  Then $a$ is inertial.
\end{lemma}
\begin{proof}
We first establish stability and the defects of the blocks
in this statement.  Let $b$ be a block of $E$ covered by $a$.
The support observation above, applied to $G=EP$, shows that
$b$ is $P$-stable, is the unique block of $E$ covered by $a$,
and satisfies $a=b$ as an idempotent.  The block $b$ covers a
unique block $b_i$ of each component $S_i$, and
\[
 b=b_1\cdots b_m
\]
inside $\OO E$.  This is the block correspondence obtained by
inflating to $\prod_iS_i$ and then taking the central character
sector of the central product.  Since each $S_i$ is normal in
$G$, the $P$-stability of $b$ makes each $b_i$ $P$-stable.
The normal-subgroup defect theorem gives
$D_i:=P\cap S_i$ as a defect group of $b_i$.

Brauer multiplicativity gives
\[
 0\ne\Br_P(a)=\Br_P(b)
       =\prod_i\Br_P(b_i).
\]
Thus every $\Br_P(b_i)$ is nonzero.  The unique block $c_i$ of
$G_i:=PS_i$ covering $b_i$ is the idempotent $b_i$ in
$\OO G_i$.  A defect group of $c_i$ contains $P$, while its
order is at most
$|D_i|\,|G_i:S_i|=|D_i|\,|P:D_i|=|P|$.
Consequently $P$ is a defect group of every $c_i$.

Let $R=O_2(G)$.  We have $R\leq P$.  For every $i$,
$[R,S_i]\leq O_2(S_i)\leq Z(S_i)$, so perfection of $S_i$
makes $R$ centralize $S_i$.  It also centralizes $P$.
Hence
\[
 R=C_P(E)\leq Z(G).
\]
Apply Proposition~\ref{prop:central-inertia} simultaneously
to $a$ and to all $c_i$ and quotient by $R$.  The component
images remain quasisimple, $O_2(G)=1$, and $Z(E)$ has odd
order.  The last assertion follows since
$O_2(Z(E))\leq R$ before quotienting and a central quotient
of a perfect group does not enlarge its centre.

Let $\theta$ be the character of $Z(E)$ belonging to $b$ and
let $K=\ker(\theta)$.  Stability of $b$ makes $K$ normal in
$G$, and the averaging idempotent of $K$ acts as the identity
on $a=b$.  Thus quotienting by $K$ gives the corresponding
block-algebra isomorphism, identifying $P$ with its image.
For each $i$, the kernel of the map $G_i\to G/K$ is
$K\cap S_i$: it is odd and $G_i/S_i$ is a $2$-group.  It is
contained in the kernel of $c_i$, so the same quotient
preserves the source algebra and inertia of every $c_i$.

The quotient also has trivial $2$-core.  Before taking it,
$C_G(E)/Z(E)$ is a $2$-group with central odd kernel $Z(E)$.
Its unique Sylow $2$-subgroup would be normal in $G$, so
$C_G(E)=Z(E)$.  Since $E$ is perfect and $K\leq Z(E)$, an
element centralizing $E/K$ already centralizes $E$; hence
\[
 C_{G/K}(E/K)=Z(E)/K.
\]
A normal $2$-subgroup of $G/K$ centralizes $E/K$, since
$O_2(E/K)=1$.  The displayed centralizer has odd order, so
that normal $2$-subgroup is trivial.  We retain
the notation and assume from now on that
\[
 O_2(G)=1,\qquad C_G(E)=Z(E),\qquad
 \theta\text{ is faithful}.
\]
The $P$-invariance of the faithful character $\theta$ forces
$P$ to centralize $Z(E)$, and in particular every $Z(S_i)$.

For $G_i=PS_i$ put $R_i=O_2(G_i)$.  The same commutator
argument gives $R_i=C_P(S_i)\leq Z(G_i)$.  Since $Z(S_i)$
has odd order, $R_i\cap S_i=1$.  Moreover,
$C_{G_i}(S_i)/Z(S_i)$ is a $2$-group with central odd kernel.
Its unique Sylow $2$-subgroup is normal in $G_i$, and is
exactly $R_i$.  Thus
\begin{equation}\label{eq:component-centralizer}
 C_{G_i}(S_i)=Z(S_i)\times R_i.
\end{equation}
Put $\overline G_i=G_i/R_i$.  The original $S_i$ embeds
elementwise in $\overline G_i$, its centralizer is $Z(S_i)$,
and the block $\overline c_i$ dominated by $c_i$ is inertial
by Proposition~\ref{prop:central-inertia}.  Its defect
group is $P_i=P/R_i$, with $P_i\cap S_i=D_i$.

Let
\[
 A=\ker\bigl(\prod_iS_i\longrightarrow E\bigr).
\]
This is an odd subgroup of $\prod_iZ(S_i)$.  Because $P$
centralizes each component centre, $A$ is central in
$\prod_i\overline G_i$.  Form
\[
 \widehat G=(\prod_i\overline G_i)/A.
\]
It contains the original $E$, and
\[
 \widehat G/E\cong\prod_i(\overline G_i/S_i)
\]
is an abelian $2$-group.

We now construct an embedding of the original $G$ which is
the identity on $E$.  Let
$\tau:G\to\prod_i\Aut(S_i)$ be its componentwise action,
and let $\rho:\widehat G\to\prod_i\Aut(S_i)$ be the
componentwise action of the constructed product.  The image
of $\rho$ contains $\tau(G)$, its restriction to $E$ is
$\tau|_E$, and
\[
 \ker(\tau)=Z(E),\qquad \ker(\rho)=Z(E).
\]
The latter kernel is central and has odd order.  The actions
are well defined on the central product because every factor
fixes the shared centres pointwise.

Choose a Sylow $2$-subgroup $T$ of $G$ containing
$P$.  The group $\rho^{-1}(\tau(T))$ is a central extension
of a $2$-group by the odd group $Z(E)$ and therefore has a
unique Sylow $2$-subgroup $\widehat T$.  It contains the
already embedded $T\cap E$.  The map
\[
 \lambda_T=(\rho|_{\widehat T})^{-1}\tau|_T
       :T\longrightarrow\widehat T
\]
is an isomorphism and is the identity on $T\cap E$.
Also $G=ET$, and $\lambda_T(t)$ induces the original action
of $t$ on $E$.  Hence
\[
 \Phi:G\longrightarrow\widehat G,\qquad
 et\longmapsto e\lambda_T(t)
\]
is a well-defined homomorphism.  If $\lambda_T(t)\in E$,
then $\tau(t)\in\tau(E)$, so
$t\in E\ker(\tau)=E$ and $\lambda_T(t)=t$.
This proves that $\Phi$ is injective.  In particular it
is the identity on $E$.  For $p\in P$,
its image is the class of the tuple $(pR_i)_i$: that tuple
is a $2$-element with action $\tau(p)$, and belongs to the
unique Sylow $2$-subgroup of
$\rho^{-1}(\tau(\langle p\rangle))$.

It remains to keep the specified block and central character.
On $Z(S_i)$ the block $\overline c_i$ has character
$\theta_i=\theta|_{Z(S_i)}$.  Thus the tensor block
$\widetilde a=\bigotimes_i\overline c_i$ of
$\prod_i\overline G_i$ has character
\[
 (z_i)_i\longmapsto\prod_i\theta_i(z_i)
          =\theta\bigl(\prod_i z_i\bigr),
\]
which is trivial on $A$.  It therefore descends to a block
$\widehat a$ of $\widehat G$ via the central odd-kernel
block-algebra isomorphism.

The tensor block is inertial.  Explicitly, if the source of
$\overline c_i$ is
$\mathcal D_i\otimes_{\OO}\OO_{\alpha_i}(P_i\rtimes I_i)$,
then its product source is
\[
 \Bigl(\bigotimes_i\mathcal D_i\Bigr)
 \otimes_{\OO}
 \OO_{\prod_i\alpha_i}
 \Bigl((\prod_iP_i)\rtimes(\prod_iI_i)\Bigr).
\]
The product Dade algebra is primitive and stable under the
product inertial quotient.  The odd central quotient by $A$
preserves this source with its defect-group identification,
so $\widehat a$ is inertial.  In fact $\overline c_i=b_i$ as
idempotents in $\OO\overline G_i$, because $c_i=b_i$ and the
quotient by $R_i$ embeds $S_i$ identically.  The product and
the compatible central character therefore give
$\widehat a=b$ as idempotents in $\OO\widehat G$.

Finally, $G$ is normal in $\widehat G$ because it contains
$E$ and $\widehat G/E$ is abelian.  The block of $G$ covering
$b$ is uniquely $a=b$.  Hence $\widehat a$ covers the block $a$.  Puig's normal-subblock theorem
\cite[Theorem 3.13]{Puig} now proves that $a$ is inertial.
The preceding odd-kernel block isomorphism and
Proposition~\ref{prop:central-inertia} restore all the
given groups and blocks.
\end{proof}

\subsection{The strong inertial extension theorem}
The following extension theorem applies when the indicated block
of $PH$ is inertial.

\begin{theorem}[Zhang--Zhou, strong form]\label{thm:strong-extension}
Let $H\triangleleft G$, let $b$ be a $G$-stable block of $\OO H$, and
let $P$ be a maximal $2$-subgroup with nonzero relative Brauer support
for $b$. Put $Q=P\cap H$. Suppose that $P$ is abelian and that the
unique block of $PH$ covering $b$ is inertial. Put $J=N_G(Q)$,
$H_0=N_H(Q)$, and let $b_0$ be the Brauer correspondent of $b$ in
$H_0$. There is a module for $(H\times H_0)\diag J$
whose lower restriction is a basic Morita bimodule and whose induction
to $G\times J$ gives a Morita equivalence between
$\OO Gb$ and $\OO Jb_0$. Every corresponding primitive summand is a
basic Morita equivalence with the chosen defect identification.
\end{theorem}

This is \cite[Theorem 1.3]{ZZ}. We describe the graded source
comparison and its consequence for the diagonal action. Put
\[
 A=\OO Gb,\quad B=\OO Hb,\quad
 A'=\OO Jb_0,\quad B'=\OO H_0b_0,
 \qquad \Gamma=G/H\cong J/H_0.
\]
The last identification is the pointed Frattini equality. The algebras
$A,A'$ have their group-coset $\Gamma$-gradings, with identity
components $B,B'$.

Inertiality of the $PH$-block supplies a determinant-one Dade
$P$-algebra. The extension
construction of \cite{PuigZhou}, followed by the cancellation and
comparison in \cite[Sections 3--5]{ZZ}, gives idempotents
$i\in B^P$, $i'\in(B')^P$ and compatible graded interior-$P$
isomorphisms, with $V,W$ the $\OO P$-lattices of the two
Dade factors,
\begin{align}
 iAi&\cong\End_{\OO}(V)\otimes_{\OO}R,&
 iBi&\cong\End_{\OO}(V)\otimes_{\OO}R_1,\label{eq:strong-corners}\\
 i'A'i'&\cong\End_{\OO}(W)\otimes_{\OO}R,&
 i'B'i'&\cong\End_{\OO}(W)\otimes_{\OO}R_1.\notag
\end{align}
The idempotents contain source idempotents for the lower blocks and
are full in $B,B'$. The matrix factors have degree $1$.
The algebra $R$ is the twisted group algebra of the local extension
group, graded by the original quotient $\Gamma$; it contains a unit
in every homogeneous component. Both occurrences use the same
central extension and the same embedding of $P$. In
\cite[Section 3.9 and Lemma 5.15]{ZZ}, the central extensions are
identified over the identity of the local extension group. Their
splittings over the $2$-subgroup agree by uniqueness: the central
kernel $k^\times$ has no $2$-torsion, and powering by a power of
two is an automorphism of $k^\times$. Equivalently, choose a finite
odd central realization of the cocycle and its unique $2$-subgroup
splitting. This supplies the stated interior identification.

The fixed-algebra condition used in this comparison follows from the
upper inertia hypothesis. If $E$ is its odd inertial quotient, then
$E$ acts trivially on $P/Q$. Choose $n$ with
$n|E|\equiv1\pmod{\exp(P)}$. For $p\in P$, the element
\[
 c=\left(\prod_{\alpha\in E}\alpha(p)\right)^n
\]
belongs to $C_P(E)$ and satisfies $cQ=pQ$. Thus $P=QC_P(E)$.
After the Dade cancellation, $R_1=\OO_\beta(Q\rtimes E)$.
The chosen $2$-subgroup splitting shows that $C_P(E)$ commutes with
$Q$ and with the lifted elements of $E$. Conjugation by $p=qc$ on
$R_1$ is consequently conjugation by $q$, and is trivial on
$(R_1)^Q$. This is the fixed-algebra condition in
\cite[Sections 5.9--5.15]{ZZ}, under the hypothesis of its
Theorem~1.3.

\begin{lemma}\label{lem:graded-extension}
The common graded source corners \eqref{eq:strong-corners} give a
Morita bimodule $D$ for $A,A'$ whose identity component is a Morita
bimodule for $B,B'$. This component is a module for
$(H\times H_0)\diag J$, and its induction to $G\times J$ is $D$.
\end{lemma}
\begin{proof}
Put $S=\End_{\OO}(V)$ and $S'=\End_{\OO}(W)$ and form
\[
 F=Ai\otimes_{S\otimes R}(V\otimes R),\qquad
 F'=(R\otimes W^\vee)\otimes_{S'\otimes R}i'A',
 \qquad D=F\otimes_R F'.
\]
Fullness of $i,i'$ gives the full-idempotent Morita equivalences;
the natural column and row modules give the two matrix-algebra
Morita equivalences. Their composition is $D$.
All tensor products have the displayed $\Gamma$-gradings.

Multiplication gives graded isomorphisms
\[
 F\cong F_1\otimes_{R_1}R,
 \qquad F'\cong R\otimes_{R_1}F'_1.
\]
Indeed, choose a unit $r_t\in R_t$. Its inverse image $u_t\in iA_ti$, under the corner
isomorphism applied to $1\otimes r_t$, is a unit of $iAi$. For $a\in A_ti$,
$a u_t^{-1}\in Bi$, which proves the first formula in degree $t$;
the second follows by right multiplication at the other endpoint.
The inverses use $r_t^{-1}$, so the formulas are independent of the
chosen homogeneous units. Hence
\[
 D\cong F_1\otimes_{R_1}R\otimes_{R_1}F'_1,
 \qquad M:=D_1=F_1\otimes_{R_1}F'_1.
\]
The two identity-component corner isomorphisms in
\eqref{eq:strong-corners} show that $F_1,F'_1$ are Morita
bimodules. Thus $M$ gives the lower Morita equivalence.

The existing $G\times J$ action on $D$ is
$(g,j)d=gdj^{-1}$. It sends degree $t$ to
$(gH)t(jH)^{-1}$. Its subgroup preserving degree $1$ is
\[
 \{(g,j):gH=jH\}=(H\times H_0)\diag J.
\]
Restricting this action to $M$ gives all multiplication and
intersection relations. For each $t\in\Gamma$, choose
$g_t\in G$ above $t$. Then $D_t=g_tM$, since $g_t$ is invertible
on $D$. The map
\[
 \Ind_{(H\times H_0)\diag J}^{G\times J}M
 \longrightarrow D,\qquad \sigma\otimes m\longmapsto\sigma m,
\]
is an isomorphism: its coset summands map bijectively to the
homogeneous components $D_t$.
\end{proof}

For a primitive correspondence with defect subgroup $P_a\leq P$,
the interior-$P$ comparison gives the capped summand of
$\Res_{P_a}^{P}(V\otimes W^\vee)$ as its endopermutation source.
In the application below, the selected primitive block has
$P_a=P$.
Reducing the primitive bimodule modulo the maximal
ideal, applying Proposition~\ref{prop:source-upgrade}, and then
Lemma~\ref{lem:integral-lift} gives the integral splendid equivalence
used below.

\section{Quasisimple blocks and the upper defect action}
\label{sec:components}

Let $S$ be a quasisimple component of the ambient finite group
$G$, let $P$ be the abelian defect group of the specified block
of $SP$, and let $b$ be the block of $S$ that it covers.
Its defect group $D=P\cap S$ is assumed noncentral.
Automorphisms are induced by the ambient stabilizers.

\subsection{The classification in the form used here}
The classification \cite[Theorem~6.1 and Propositions~3.4,
5.3 and~5.4]{EKKS}, together with its proof, gives the following
cases. We first assign all nilpotent blocks, and all Lie-type
cases in which $C_{\mathbf X^*}^{\circ}(s)$ is a torus, to (iv).
Among the remaining cases, the type-$A$ blocks in (ii) are
precisely those of \cite[Proposition~5.3(iii)]{EKKS}, in both
linear and unitary forms. The cases of
\cite[Proposition~5.3(v)]{EKKS} are assigned to (iii), including
their Klein-defect overlaps.

\begin{enumerate}[label=(\roman*),leftmargin=*]
\item The principal blocks of $\SL_2(2^a)$, with $a\geq2$, the Ree groups
${}^2G_2(3^{2a+1})$, with $a\geq1$, and $J_1$, and the nonprincipal block of
$\mathrm{Co}_3$ with defect $C_2^3$.
\item The nonnilpotent Klein four blocks in the specified nontoral
type-$A$ cases, in alternating groups, and in the sporadic and
central-cover cases described below. Their possibilities are
retained from \cite[Section 8]{CEKL} and the proof in \cite{EKKS}; the
classified $D_{2t}/E_7$ Levi cases are assigned to (iii).
\item The rank-one Levi cases in split type $D_{2t}$, with $t\geq3$
odd, and in type $E_7$, with $t=3$. Here $q$ is an odd prime
power with $q\equiv\pm3\pmod8$.
The full centralizer of the semisimple odd-order label is the
classification Levi; its derived root system is a single Frobenius
orbit of $t$ factors of type $A_1$. After the prescribed central
quotient, its finite group has the form
$\PSL_2(q^t)\times A_0$ with $A_0$ abelian.
\item The nilpotent blocks and the cases in which the connected
component $C_{\mathbf X^*}^{\circ}(s)$ of the dual centralizer
is a torus. The latter have nilpotent regular-embedding covers
by \cite[Lemma~4.2]{EKKS}.
\end{enumerate}

The central-defect components will be removed before this list is
applied. The classification branches enter the list as follows.
In defining characteristic two, Proposition~8.7 and Lemma~8.8
of \cite{CEKL}, together with the proof of
\cite[Theorem~6.1]{EKKS}, leave the $\SL_2$ family and the
listed small exceptions after the central-defect cases are removed. For alternating and sporadic
groups, \cite[Section 8]{CEKL} gives (i) and (ii).
In odd characteristic, the toral case of \cite[Lemma 4.2]{EKKS}
belongs to (iv). In \cite[Proposition 5.3]{EKKS}, cases (i) and
(iv) give nilpotent blocks, as observed in the proof of
\cite[Theorem~6.1]{EKKS}, and thus belong to (iv) here; case (iii)
gives the type $A$ quaternion preimage treated in (ii); case (v)
gives precisely the $D_{2t}$ and $E_7$ families in (iii).
The remaining quasi-isolated
$E_7$ case with dihedral defect $D_8$ before a central involution
quotient belongs to (iv). An element normalizing the quotient
defect lifts to an element normalizing its full preimage $D_8$.
Its action on $V_4$ therefore lies in the image of $\Aut(D_8)$,
a $2$-group. The odd inertial quotient is trivial, and abelian
defect then makes the quotient block nilpotent. These are all the alternatives in the
proof of \cite[Theorem 6.1]{EKKS}.

The central-cover cases are as follows. In odd characteristic,
the proof of \cite[Theorem~6.1]{EKKS} excludes faithful blocks
with noncentral abelian defect on exceptional covers.
A nonnilpotent block with toral connected dual centralizer can
have disconnected full centralizer only in type $A$ or $E_6$,
by \cite[Proposition~5.4]{EKKS}. In odd characteristic the
exceptional multipliers in type $A$ occur for $\PSL_2(9)$ and
$\operatorname{PSU}_4(3)$ \cite[Sections~9--10]{CEKL}.
Their standard simply connected centers are $2$-groups.
An odd-order semisimple label therefore has connected
centralizer: lift it with odd order through the central
$2$-isogeny; a conjugate differing by a central $2$-element
must be equal, and the centralizer upstairs is connected.
A toral case for these groups is consequently nilpotent by
\cite[Lemma~4.2]{EKKS}. The odd-characteristic $E_6$ and
${}^2E_6$ cases have their standard centers of orders
$\gcd(3,q-1)$ and $\gcd(3,q+1)$, respectively
\cite[Lemmas~12.5 and~12.7]{CEKL}.
A nonfaithful odd central cover has the same block algebra
after quotienting its block kernel. A central $2$-quotient
preserves the inertial quotient for abelian defect, as proved
in Proposition~\ref{prop:central-inertia}; it therefore
preserves nilpotence in this situation. Thus each nonnilpotent
case in (iv) has the nonexceptional simply connected realization
used in Lemma~\ref{lem:upper-inertia}.

In defining characteristic two, Proposition~8.7 of \cite{CEKL}
gives full defect after the central reductions. The abelian
Sylow case is the $\SL_2$ family. Its exceptional double
cover at $\SL_2(4)$ has quaternion Sylow preimage and gives
no further abelian-defect case. The Tits group has defect-zero
blocks and a principal block of nonabelian defect; the remaining
small groups are $\operatorname{PSp}_4(2)'\cong A_6$ and
$G_2(2)'\cong\operatorname{PSU}_3(3)$, covered by the
alternating and odd-characteristic unitary cases
\cite[Lemma~8.8]{CEKL}. The exceptional odd covers in the
alternating and sporadic Klein cases are treated below.

Cases (i)--(iii) provide the component complexes used in the
assembly. In case (iv), Lemma~\ref{lem:upper-inertia} proves
that the block of $SP$ is inertial.

\Needspace{5\baselineskip}
\subsection{An inner-block test for upper 2-elements}
For an algebra $A$, $\Pic(A)$ is the group of isomorphism
classes of invertible $A$-$A$-bimodules under tensor product.
For an interior $D$-algebra $A$, $\Out_D(A)$ is the quotient
of the algebra automorphisms fixing the specified image of $D$
pointwise by conjugation with units in $(A^D)^\times$.

\begin{lemma}\label{lem:block-inner}
Let a finite $2$-group $T$ act on $L$ by automorphisms. Suppose
that it stabilizes a block of $L$,
centralizes its defect group $D$, and preserves a supported maximal
Brauer pair. Then its image in the Picard group of that block is
trivial. In particular it fixes every ordinary irreducible character
of the block.
\end{lemma}
\begin{proof}
The twist of the regular block bimodule by an element of $T$ is an
invertible trivial-source bimodule with diagonal vertex $\diag D$.
Its fusion automorphism is the identity. By
\cite[Theorem 1.1(ii) and Remark 1.2]{BKL}, its Picard class belongs
to $\Out_D$ of a source algebra, which embeds in
$\Hom(E,k^\times)$ for the inertial quotient $E$. This is an odd
group. The homomorphism from $T$ is therefore trivial. An inner
algebra automorphism fixes all ordinary characters.
\end{proof}

The stable root required here is available from the given upper
defect pair. Indeed $\Br_T(b)\ne0$. In $\Br_D(b)$ a nontrivial
$T$-orbit of primitive root idempotents is a proper relative trace
and has zero $T$-Brauer image. Choose a fixed summand supported by
the original pair. This gives the required supported maximal pair.

\subsection{Klein four blocks and upper 2-actions}
\begin{proposition}\label{prop:klein-inner}
In case (ii), the original abelian upper defect group $P$ acts
on $S$ by inner automorphisms. Moreover $O_2(Z(S))=1$.
\end{proposition}
\begin{proof}
The inertial quotient of a nonnilpotent Klein block is $C_3$, acting
transitively on $D\setminus\{1\}$. A central involution would be fixed
by this action, so $O_2(Z(S))=1$.

First consider the type-$A$ cases of
\cite[Proposition~5.3(iii)]{EKKS}. Here $q$ is an odd prime
power, $\varepsilon\in\{1,-1\}$, and
$\GL_n^+(q)=\GL_n(q)$, $\GL_n^-(q)=\operatorname{GU}_n(q)$,
$\SL_n^+(q)=\SL_n(q)$, $\SL_n^-(q)=\operatorname{SU}_n(q)$.
The natural dimension is $2t$, with $t$
odd, and $q\equiv\pm3\pmod8$. An odd-order lift of the semisimple
label in $\GL_{2t}^{\varepsilon}(q)$ has $t$ distinct eigenvalues,
each of multiplicity two, forming one orbit of
$f:\lambda\mapsto\lambda^{\varepsilon q}$.
If an inner-diagonal graph automorphism stabilizes the label, then
\[
 j:\lambda\longmapsto c\lambda^{-1}
\]
permutes these eigenvalues for an $f$-fixed scalar $c$. It is an
involution commuting with the $t$-cycle $f$. The centralizer of that
cycle is cyclic of odd order, so $j$ is the identity. Every eigenvalue
then has square $c$. Squaring is injective on their odd-order group,
so $t=1$. For $t>1$ the graph component is excluded. For $t=1$ the
group is rank one and has the same conclusion below. This also
treats the graph component in the unitary outer group. Since
$q\equiv3$ or $5\pmod8$, the exponent of its defining prime is odd;
there is no remaining field $2$-part.

It remains to exclude the diagonal $2$-part. The inverse image of
$D$ in $\SL_{2t}^{\varepsilon}(q)$ has Sylow $2$-subgroup $Q_8$.
Its natural module is the sum of $t$ copies of the faithful
two-dimensional quaternion module. If a general linear or unitary
element induces an automorphism centralizing $D$, it normalizes this
$Q_8$: the central $2$-part consists of $\{\pm1\}$, and an odd scalar
cannot occur in the conjugate of an element of order at most four.
The kernel of $\Aut(Q_8)\to\Aut(Q_8/Z(Q_8))$ is inner. After
multiplication by an element of $Q_8$, our element centralizes
$Q_8$. Its centralizer on the natural module is $\GL_t(q)$ or
$\GU_t(q)$ on the multiplicity space, and
\begin{equation}\label{eq:quaternion-determinant}
 \det_{2t}(u)=\det_t(u)^2.
\end{equation}
The diagonal outer quotient has $2$-part of order two. A square
determinant has trivial image there. Hence the upper $2$-action is
inner. The rank-one cases are included in this calculation.

For alternating groups, \cite[Lemma 8.5]{CEKL} makes the Klein block
$S_n$-stable. Its unique cover has defect of order eight and weight
two, hence has five ordinary characters. If the four characters of
the lower block comprise $a$ fixed characters and $b_0$ exchanged
pairs, Clifford theory gives
\[
 a+2b_0=4,\qquad 2a+b_0=5.
\]
Thus $b_0=1$: the outer involution moves a character and is excluded
by Lemma~\ref{lem:block-inner}. The exceptional odd covers give no
additional faithful Klein case: $A_6$ and the faithful blocks of
$3.A_6$ have no Klein defect, and the faithful blocks of $3.A_7$
have defect groups of order eight or one. These are the cover cases in the
classification and its modular tables \cite{CEKL,MOC}.

The sporadic Klein four cases in (ii) are $M_{12},J_2,HS,\mathrm{Fi}_{24}'$
\cite[Section 8]{CEKL}. In the first three, the outer involution
exchanges, respectively, the following two character rows in the
Klein block:
\[
\begin{array}{c|c|c}
 S&\text{rows}&\text{common degree}\\ \hline
 M_{12}&\chi_4,\chi_5&16\\
 J_2&\chi_{16},\chi_{17}&224\\
 HS&\chi_{14},\chi_{15}&896.
\end{array}
\]
The upper character tables identify their induced characters of
twice those degrees using the conjugate-pair notation in
\cite{MOC}. Apply
Lemma~\ref{lem:block-inner}.

For $S=\mathrm{Fi}_{24}'$, choose $u\in D\setminus\{1\}$. The
classification gives
$C=C_S(u)\cong2.\mathrm{Fi}_{22}.2$ with a unique block $c$ of
defect $D$ \cite[Lemma 8.3]{CEKL}. Put
$H_0\cong2.\mathrm{Fi}_{22}\triangleleft C$ and $z=u$.
The block $c$ contains two extensions each of the characters
$\chi_{56}$ and $\chi_{104}$ of $H_0$, all of degree $1441792$;
the latter pair is faithful on $z$ \cite{MOC}.
The full automorphism centralizer is
\[
 C_{\Aut(S)}(u)\cong(2\times2.\mathrm{Fi}_{22}):2,
\]
with a normal $V_4=\langle a,z\rangle$, where $a,az$ are outside
$S$ \cite[Section 2.4-10]{CTbl}. The perfect subgroup $H_0$
centralizes $V_4$, whereas $v\in C\setminus H_0$ satisfies
$a^{-1}va=zv$. Conjugation by $a$ therefore exchanges the two
extensions of $\chi_{104}$: on the $v$-coset it changes their
representations by the sign of the faithful central involution.
Every element of the outer coset has this same character action.
The original $P$ centralizes $u$ and stabilizes the supported block
$c$ with defect $D$, so Lemma~\ref{lem:block-inner}, now applied to
$C$, excludes that outer coset.

Faithful $3.\mathrm{Fi}_{24}'$ blocks contribute no Klein case
\cite[Lemma 8.3]{CEKL}. Nonfaithful blocks of odd central covers
are handled in their chosen character sector. Finally an
automorphism of a perfect central cover inducing an inner
automorphism of its simple quotient is itself inner: after
correcting by the inner automorphism, the central correction is a
homomorphism from a perfect group to its center and vanishes.
\end{proof}

\begin{lemma}\label{lem:actual-direct-product}
Suppose that $O_2(Z(S))=1$ and that the original abelian $P$ acts on $S$
by inner automorphisms. Then, inside the given group,
\begin{equation}\label{eq:actual-product}
 SP=S\times R,\qquad P=D\times R,
 \qquad R=P\cap C_G(S).
\end{equation}
\end{lemma}
\begin{proof}
The central odd extension of the abelian $2$-image in $\Inn(S)$ has
a unique Sylow $2$-subgroup. Hence the inner action has a unique
homomorphic lift $\delta:P\to S$ with $2$-group image and
$\delta|_D=1_D$. Set $r(p)=\delta(p)^{-1}p$. This is a homomorphism
into $C_{SP}(S)$, with kernel $D$. Its image $R_0$ meets $S$
trivially, and $SP=S\times R_0$.

The specified upper
block is the tensor of the lower block with $\OO R_0$, so its defect
groups have form $D'\times R_0$, where $D'$ is an $S$-conjugate of
$D$. Apply this to $P$ and use $P\cap S=D$.
Then $P=D\times R_0$, giving $R_0=R$. Uniqueness of $\delta$
makes this construction natural under further marked operators.
\end{proof}

\section{Klein four and elementary abelian component complexes}
\label{sec:elementary}

\subsection{The relative operator group}
Let $N=N_S(D)$, let $c$ be the original Brauer correspondent of $b$,
and let $\mathcal A\leq\Aut(S)_b$ contain $\Inn(S)$.
Suppose that $\mathcal A/\Inn(S)$ is odd and $O_2(Z(S))=1$.
Write $\mathcal A_D=N_{\mathcal A}(D)$ and $\delta(n)=(n,n)$,
and let $c_n(x)=nxn^{-1}$ denote conjugation by $n$.
The finite relative group is
\begin{equation}\label{eq:relative-operator}
 \mathcal T=\bigl((S\times N)\rtimes\mathcal A_D\bigr)
 \big/\{(\delta(n)^{-1},c_n):n\in N\}.
\end{equation}
The denominator centralizes the base and is normal. Its intersection
with $S\times N$ is $\diag Z(S)$; the resulting base is
$(S\times N)/\diag Z(S)$ and its quotient is
$\mathcal A/\Inn(S)$, by the marked Frattini equality.
The corner $b\OO Sc$ is a module for \eqref{eq:relative-operator},
since $c_n$ acts there by the inner diagonal action. The original
central character is the same at both ends. An ambient stabilizer
acts by pullback; its centralizer kernel acts as the identity on
this factor, while remaining in the ambient group.

\begin{proposition}\label{prop:klein-complex}
For a Klein four defect block under the preceding relative-operator
hypotheses, the block and its Brauer
correspondent have an integral splendid Rickard complex carrying
the action \eqref{eq:relative-operator}. Its inner operators are the
literal diagonal operators. In the situation of
Lemma~\ref{lem:actual-direct-product}, it has an upper complex
\begin{equation}\label{eq:klein-upper}
 C^+=C\otimes_{\OO}\OO R
\end{equation}
for the given groups $SP$ and $NP$, with the same lower identity
component and the same $P$-action.
\end{proposition}
\begin{proof}
Let $M$ be the unique marked full-$\diag D$-vertex summand of
$b\OO Sc$. Green correspondence gives this summand with
multiplicity one; all complementary summands have proper vertices.
Any competing vertex of full
order in the restriction of the regular block has right projection
contained in the normal defect group $D$ of the right block. That
projection has order $|D|$, so it equals $D$; its right transporter
already belongs to $N$. Thus this vertex is conjugate to the marked
$\diag D$ inside $S\times N$, and Green uniqueness excludes a
second such summand.
It is invariant under the relative group. In the
endomorphism algebra of the corner, its isotype projector modulo
the radical is invariant. Odd averaging lifts the projector to an
invariant element, and iteration of $t\mapsto3t^2-2t^3$ gives an
invariant idempotent. Thus the action on $M$ is extracted from the
original corner.

The integral source algebra classification \cite{CEKL} identifies
the source algebra of the left block with $\OO D$, $\OO A_4$, or the principal
$\OO A_5$ block. These are interior-$D$ identifications. The
normal-defect right block has source $\OO D$ or $\OO A_4$:
normal $D$ acts trivially on every simple right module, excluding
the nontrivial two-dimensional simple sources in the $A_5$ case.
The induced Puig tensor functors preserve projective covers,
indecomposability, permutation summands, and the marked full
vertex. For the last assertion, a proper vertex after transport
would remain proper after transport back, contradicting the
original full vertex.

To identify the underlying bimodule $M$, use the source models.
For the $A_5$--$A_4$ and $A_4$--$A_4$ models,
\[
 N_{A_5\times A_4}(\diag D)/\diag D\cong A_4,
 \qquad N_{A_4\times A_4}(\diag D)/\diag D\cong A_4.
\]
The Brauer image of the standard Green module is
$kD\cong\Ind_{C_3}^{A_4}k$, the indecomposable projective cover of
the trivial character. Its three inertial-character twists exhaust
the projective indecomposables of $kA_4$. The permutation-module
Brauer correspondence therefore shows that the transported $M$
is exactly one of these three twists. In the $D$--$D$ case there
is one possibility. Consequently $M$ is already a Morita
equivalence in the $D$ and $A_4$ cases; in the $A_5$ case it is
the standard Green stable equivalence with its specified twist.

In that last case $\Pic(\OO A_5b_0)\cong C_2$, where $b_0$
is the principal block idempotent of $\OO A_5$
\cite[Theorem~1.5]{BKL}, so the odd operator quotient fixes the
three left simple classes. The Green equivalence intrinsically
identifies the distinguished right simple whose image is simple.
It is fixed, and an odd group cannot exchange the other two.
In the $A_5$--$A_4$ model, label the two nontrivial left simples
$U_1,U_2$ over $k$. Each
$(k\otimes_{\OO}M)^\vee\otimes_{kA_5b_0}U_i$ is an
indecomposable module of dimension two for the right block;
let $V_i$ be its unique simple submodule. The standard correction of
\cite[Sections~5.1 and~6.3.5]{Rouquier} uses the two
projective bimodules
\[
 P(U_1)\otimes_{\OO}P(V_1)^\vee,
 \qquad P(U_2)\otimes_{\OO}P(V_2)^\vee,
\]
where $P(U)$ denotes the projective $\OO$-lattice lifting the
projective cover of the simple $k$-module $U$. The retained inertial
twist transports these pairings together. Thus the two required
isotypes of the projective cover of $M$ are fixed by the odd
operator quotient.

Take a projective cover $q:P_M\twoheadrightarrow M$ in the actual
relative group. Its restriction is the lower projective cover. Here the normal
base is $K_0=(S\times N)/\diag Z(S)$ and $\Lambda=\mathcal T$.
For a normal subgroup $K_0\triangleleft\Lambda$ of odd
index,
\begin{equation}\label{eq:radical-odd}
 J(\OO\Lambda)=J(\OO K_0)\OO\Lambda:
\end{equation}
the right side is radical modulo the uniformizer, and its quotient
is a crossed product of a semisimple algebra by a group of
invertible order, hence semisimple by averaging splittings.
The two selected isotypes give an invariant idempotent modulo the
radical of $\End_{\OO K_0}(P_M)$. Lift an invariant element $e_0$
by odd averaging, and iterate
\[
 e_{n+1}=3e_n^2-2e_n^3.
\]
The error satisfies
\[
 e_{n+1}^2-e_{n+1}
 =-(e_n^2-e_n)^2(3-2e_n)(1+2e_n).
\]
Completeness gives an invariant idempotent $e$. The required
complex is
\begin{equation}\label{eq:klein-two-term}
 C=[eP_M\xrightarrow{q|_{eP_M}}M],
 \qquad \deg(eP_M)=-1,\quad\deg M=0.
\end{equation}
Under the Puig recognition this is the standard correction with
the retained inertial twist. Projective-cover uniqueness and
conjugacy of idempotent lifts preserve its underlying Rickard
property. Its terms are permutation summands for the actual
relative group: the correction term is projective there, and the
Green term is a summand of the permutation corner $b\OO Sc$.
Alternatively, the odd base index places every $2$-subgroup in the
base, where the restriction is $2$-permutation. The evaluations are equivariant, and their lower
contractions become equivariant after averaging over the odd
quotient. This proves both evaluations for
\eqref{eq:klein-two-term}.

Finally \eqref{eq:actual-product} makes
\eqref{eq:klein-upper} an upper splendid complex. Its
identity component is $C$, with the fixed diagonal $P$-action.
The regular factor retains the whole centralizer kernel and its
group multiplication.
\end{proof}

\begin{lemma}\label{lem:green-companion}
The full maximal companion of the complex constructed in
Proposition~\ref{prop:klein-complex} is the regular right block
in degree zero, with its operator action.
\end{lemma}
\begin{proof}
If $D\triangleleft N$ and $c$ is a central block idempotent of
$kN$, then $\Br_D(c)=c$. Indeed the quotient map to $k(N/D)$
kills each nontrivial $D$-orbit sum. Thus $c$ and $\Br_D(c)$
have the same image modulo the nilpotent ideal $J(kD)kN$;
both are central idempotents and are equal. The maximal roots
of $\Br_D(b)$ form the same full $N$-orbit as those of $c$.
Consequently
\[
 \Br_D(b)=c,\qquad
 M(\diag D)=kC_S(D)c.
\]
The proper-vertex complementary terms and the projective correction
vanish at $\diag D$. Moreover
\[
 N_{S\times N}(\diag D)=(C_S(D)\times C_S(D))\diag N.
\]
The full companion is identified with $kNc$ by the multiplication
map $[(x,y),m]\mapsto xmy^{-1}$. It is a bijection on the
original coset decomposition and commutes with all original
operators. Uniqueness of the integral lift of a permutation summand
in the full finite action group gives $\OO Nc$ with this action.
\end{proof}

\subsection{The other elementary cases}
\begin{proposition}\label{prop:elementary-complex}
For a relative operator group with odd image modulo
$\Inn(S)$, case (i) has the same component data as
Proposition~\ref{prop:klein-complex}: a full relative-operator
splendid complex and an upper complex on $P$ with
the same lower identity component.
\end{proposition}
\begin{proof}
For $S=\SL_2(2^a)$, the construction in
\cite[Sections 4.2.6 and 6.3]{CR} is in the full relative
automorphism module category. We restrict it to the stipulated odd
outer image. In the global application, oddness of the entire
relevant image follows from the full relative-defect equality proved
in Section~\ref{sec:global}.
To see the exclusion of an upper outer $2$-action, write the
Sylow unipotents as $u(x)$, $x\in\mathbb F_{2^a}$. Let $F$ square matrix entries. An element $gF^r$ normalizing
this Sylow subgroup has $g$ in its Borel normalizer; write
$t,t^{-1}$ for its diagonal entries. If it centralizes every
$u(x)$, its action on parameters is
$x\mapsto t^2x^{2^r}$. Evaluating at $1$ gives $t=1$, and
evaluating at all $x$ gives $r=0$ modulo $a$. The remaining
representative is inner. Lemma~\ref{lem:actual-direct-product}
supplies the upper regular factor. Restriction to an odd outer
quotient preserves the permutation property and allows averaging
of both contractions. Apply Lemma~\ref{lem:integral-lift} if the
initial representative is modular.

For Ree groups, the lower integral splendid kernel is the one
recorded in \cite[Theorem 1.6]{KMN} and
\cite[Section 6.2]{CR}. The Picard computation
\cite[Proposition~5.4]{EL} gives $\Pic(\OO Sb)\cong C_3$;
its nonidentity elements cycle three unit rows of the decomposition
matrix, including the ordinary trivial character. Every actual
group automorphism fixes that character, so its Picard class is
trivial. The outer group is odd. The following construction gives
its action on the chosen kernel.

Work first over $k$. A block-inner automorphism gives an isomorphism
between the regular block bimodule and its twist. Their full vertices
are the diagonal of $D$ and the graph of that automorphism on $D$.
Conjugacy of these vertices shows that the latter action belongs to
the normalizer fusion. Put $\Omega=\mathcal A/\Inn(S)$ and
choose representatives $a_g\in\mathcal A_D$, $g\in\Omega$.
Correct them by inner normalizer elements so that they fix $D$
pointwise. Since $S$ is centerless, there are unique
$n(g,h)\in C_S(D)$ satisfying
\[
 a_ga_h=c_{n(g,h)}a_{gh}.
\]
Choose units $u_g\in(kSb)^D$ implementing $a_g$. Then
\[
 u_gu_h=\alpha(g,h)n(g,h)u_{gh},\qquad
 \alpha(g,h)\in Z(kSb)^\times.
\]
The center is a finite-dimensional local commutative $k$-algebra,
and
\[
 Z(kSb)^\times=k^\times\times(1+J(Z(kSb))).
\]
If $J(Z(kSb))^m=0$ and $2^r\geq m$, then
$(1+x)^{2^r}=1$ for every $x$ in this radical. The principal-unit
group therefore has bounded $2$-power exponent. The carry relation
for the representatives, and the fact that $u_g$ implements their
action, give the cocycle identity for $\alpha$. Inner automorphisms
fix the center. Multiplication by the odd order of the outer group
is invertible on its principal-unit subgroup, so its positive-degree
cohomology vanishes. Rescale the $u_g$ by central principal units
to obtain $\alpha(g,h)\in k^\times$.
Set $v_g=\Br_D(u_g)$ in the right block. These are units,
because $\Br_D(b)=c$ and the Brauer images of the inverses are
their inverses. They implement the automorphism on the full right
normalizer block: for $n\in N_S(D)$,
the element $a_g(n)n^{-1}$ centralizes $D$, and applying the Brauer
map to
$u_g({}^{n}u_g^{-1})=a_g(n)n^{-1}b$ gives the corresponding
identity for $v_g$. They therefore satisfy the same product equation
with the same scalar $\alpha(g,h)$. Hence
\begin{equation}\label{eq:same-scalar}
 \tau_g(x)=u_gxv_g^{-1},\qquad
 \tau_g\tau_h=\delta_{n(g,h)}\tau_{gh}.
\end{equation}
Here $\delta_n(x)=nxn^{-1}$ is the diagonal bimodule action.
On the dual use $y\mapsto v_gyu_g^{-1}$. The middle factors
cancel in both balanced products. Every bimodule differential,
evaluation, and contraction commutes with this action.
It therefore gives the finite relative action with its inner
intersections. Every $2$-subgroup of the relative group is conjugate
into the base, since the quotient is odd. Restriction to the base
is $2$-permutation, so the terms are $2$-permutation for the whole
relative group. The two contractions are equivariant by the
displayed unit action. Lift the resulting complex integrally using
Lemma~\ref{lem:integral-lift}.

For $J_1$, use the explicit splendid construction in
\cite[Section 6.2.3]{CR}. For the nonprincipal $C_2^3$-block of
$\mathrm{Co}_3$, \cite[Theorems 1.3, 1.5--1.6 and Section 7]{KMN}
uses the original marked Green module for
$(\SL_2(8)\rtimes C_3)\times S_3$. Its Puig equivalence,
defect-zero factor and Brauer-normalizer identification are integral
and retain the original defect group. Both sporadic groups have
trivial outer automorphism group. Their relative action is just
the prescribed inner action, pulled back with the ambient
centralizer acting trivially.

The Ree groups and these sporadic groups are centerless, and the
Ree outer group is odd. Thus the given upper $P$ has inner
image in every case. Lemma~\ref{lem:actual-direct-product} gives
the upper complex by the same regular tensor factor.
Proposition~\ref{prop:companion} supplies its local return.
\end{proof}

\section{The rank-one Levi components}\label{sec:geometry}

We treat case (iii). We construct a complex for $S$ and $N_S(D)$
with the marked component stabilizer action, as the identity
component of a splendid complex for $SP$ and $N_S(D)P$.
The operator groups are the marked stabilizers in the reduced
ambient situation of Section~\ref{sec:global}. In particular, the
image of $P$ is Sylow at the $2$-part of their quotient
by the lower inner operators. This condition will be used for the
rank-one factor, whose outer operator image must be odd.
Let $\mathcal A$ be the automorphism image on $S$ of the group
generated by $S$ and the ambient marked component stabilizer,
and put $\mathcal A_D=N_{\mathcal A}(D)$.

\subsection{The full-centralizer Levi and field operators}
Let $\mathbf G$ be the simply connected algebraic group giving the
component before its central quotient, with Frobenius $F$.
Write $S_{\mathrm{sc}}=\mathbf G^F$ and
$S=S_{\mathrm{sc}}/Z_0$, where $Z_0$ is central; $Z_0=1$
is allowed. A regular embedding is an inclusion
$\mathbf G\leq\widetilde{\mathbf G}$ with connected center
and $\widetilde{\mathbf G}=\mathbf GZ(\widetilde{\mathbf G})$.
The proof of \cite[Proposition 5.3(v)]{EKKS} gives, in the present
$D_{2t}$ and $E_7$ branches,
\begin{equation}\label{eq:full-centralizer}
 C_{\mathbf G^*}(s)=C_{\mathbf G^*}^{\circ}(s)=\mathbf L^*
\end{equation}
for the chosen odd-order semisimple label $s$ and the
classification Levi. 

Here is also a direct reason for connectedness in this setting.
The simply connected covering of the dual group has a central
$2$-kernel. An odd-order $s$ has an odd-order lift $\widehat s$.
If $\widehat g\widehat s\widehat g^{-1}=z\widehat s$ for a central
$2$-element $z$, taking the odd order of $\widehat s$ gives $z=1$.
The connected centralizer upstairs therefore maps onto the full
centralizer downstairs.

For type $D_{2t}$, the two possible $A_1^t$ root systems can be
written as paired difference roots, or as the same pairs with one
pair replaced by a sum root. The parity of the number of sum pairs
is invariant under $W(D_{2t})$, whose signed permutations change
an even number of coordinate signs. The nontrivial graph
automorphism exchanges these two classes. An automorphism
stabilizing the block stabilizes its rational-series
idempotent and hence the full-centralizer Levi class. Thus its
graph component is trivial. There is no triality case, since
$t\geq3$ is odd. Type $E_7$ has no nontrivial diagram
automorphism. Also $q\equiv\pm3\pmod8$ forces its field exponent
to be odd. The given upper $2$-group consequently acts by
inner-diagonal automorphisms.

We choose the Levi compatibly with the field part. Let $F_0$ be
the pure field root representing the relevant cyclic field
subgroup, so $F=F_0^r$ with $r$ odd. Its geometric semisimple
class is stable. Lang's theorem gives an $F_0$-fixed representative of this
class. Connectedness of its centralizer implies that the
geometric class contains a single rational $F$-class, so this
representative is $(\mathbf G^*)^F$-conjugate to $s$.
Transport the label, the dual Levi, and all block and defect
markings by that rational conjugation, retaining the notation
$s,\mathbf L^*$. Choose the dual primal Levi $\mathbf L$ and put
$L_{\mathrm{sc}}=\mathbf L^F$, $L=L_{\mathrm{sc}}/Z_0$.


Choose a parabolic radical $\mathbf U$ with Levi $\mathbf L$,
and put
\begin{equation}\label{eq:restriction-scalars}
 \mathbb G=\mathbf G^r,\quad
 \mathbb L=\mathbf L^r,\quad
 \mathbb U=\mathbf U^r,\quad
 \mathbb F=\tau(F_0,\ldots,F_0),
\end{equation}
where $\tau$ is cyclic permutation. First-coordinate projection
identifies $\mathbb G^{\mathbb F}$ with $\mathbf G^F$; its inverse
is $\iota(g)=(g,F_0^{r-1}g,\ldots,F_0g)$. The finite-order
operator $\tau$ is $F_0^{-1}$ on these finite points and preserves
$\mathbb U$ literally. The dual tuple uses the reverse cycle.
The rational-series comparison in \cite[Section 5]{Ruhstorfer}
retains the chosen label. The Deligne--Lusztig variety is
\[
 Y_{\mathbb U}^{\mathbb G}
 =\{g\mathbb U\in\mathbb G/\mathbb U:
       g^{-1}\mathbb F(g)\in\mathbb U\mathbb F(\mathbb U)\}.
\]
Its left and right actions are multiplication by
$\mathbb G^{\mathbb F}$ and $\mathbb L^{\mathbb F}$.
The same definition is used for the disconnected upper group
\cite[Section~2.E]{BDR}.

We use the point-stabilizer form of Rickard's cochain construction.
Let $\Lambda$ be the whole finite group acting on the chosen variety:
it includes the left and right finite groups and all the stipulated
regular-embedding and finite field operators. If a primitive block
cut is not $\Lambda$-invariant, first restrict to its inertia
group; full block orbits are recovered by finite induction.
For a $\Lambda$-variety $Z$ put
\[
 \mathscr P_\Lambda(Z)
 =\operatorname{add}\{\OO[\Lambda/\Lambda_z]:z\in Z(\overline{\mathbb F}_q)\}.
\]
Here $\operatorname{add}$ denotes finite direct sums and direct
summands in the category of $\OO\Lambda$-lattices. We write
$\Omega_c(Z,\Lambda,\OO)$ for the resulting compact-support
cochain object in that category; the group parameter is omitted
when its action has been fixed. There are finitely many subgroup
types in the displayed expression.
Choose the bounded cochain complex in
$K^b(\mathscr P_\Lambda(Z))$ supplied by
\cite[Theorems 3.2 and 3.5]{Rickard}; the integral point-stabilizer
form is also explicit in \cite[Section 2.5.1]{RouquierEtale}.
Apply the prescribed block projectors to this complex.
This specifies the terms, differential, and finite operator
action at the initial choice.

The integral lower equivalence is supplied by
\cite[Theorem 7.7]{BDR}, equivalently
\cite[Theorem 2.28]{Ruhstorfer} in the full-centralizer case.
Our specified complex represents the same geometric object in the
homotopy category of the base double group. Thus its canonical
evaluations are homotopy equivalences there as well, by naturality
of duality and evaluation under a homotopy equivalence.
Denote the chosen lower block complex by $C$. For a central
quotient, this choice is made on the quotient variety below.

Two consequences will be used. For a subgroup $B\leq\Lambda$,
Mackey restriction of $\OO[\Lambda/\Lambda_z]$ is a sum of
$\OO[B/B_{\lambda z}]$. Thus restriction to a smaller action group
still has a point-stabilizer term bound. The same statement
holds for pullback through a homomorphism of finite action groups.
Also, every operator acting pointwise as the identity on $Z$
acts as the identity on every term of this representative:
it belongs to every point stabilizer.
In particular, the diagonal central kernels and the
inner-overlap relations hold on the chosen terms.

\subsection{Finite induction of the chosen cochain model}
\begin{lemma}\label{lem:finite-induction}
Let $T\leq\Lambda$ be finite groups, and let $Y$ be a
quasi-projective variety with $T$-action. With coefficients of
characteristic different from the defining characteristic,
\begin{equation}\label{eq:finite-induction}
 \Omega_c(\Lambda\times_TY,\Lambda,\OO)
 \simeq\Ind_T^\Lambda\Omega_c(Y,T,\OO)
\end{equation}
in the bounded $2$-permutation homotopy category. The comparison
is natural for additional finite operators preserving the data.
\end{lemma}
\begin{proof}
The induced variety is a finite disjoint union of open-and-closed
copies of $Y$. Godement stalk products, and subsequently
compact-support sections, decompose into the corresponding finite
sums. The action of $\Lambda$ permutes them with the original
coset carries. This gives induction at the cochain-construction
level. The bounded permutation replacement is homotopy equivalent
to this object \cite[Section 2.5.1]{RouquierEtale}.

Equivalently, put $B=Y/T=(\Lambda\times_TY)/\Lambda$.
The finite quotient pushforwards satisfy the constructible-sheaf
identity
\[
 (\pi_{\Lambda\times_TY})_*\OO
 \cong\OO\Lambda\otimes_{\OO T}(\pi_Y)_*\OO.
\]
On each geometric stalk this is the permutation-module identity
for the induced fiber, compatible with restriction maps.
Rickard's additive-functor compatibility
\cite[Lemma 2.8 and Theorem 3.5]{Rickard} gives
\eqref{eq:finite-induction}, including the integral limit.
In particular, restrict a representative chosen first for the
whole lower finite action, and then induce that representative.
It realizes the upper object in the homotopy category.
\end{proof}

\subsection{The disconnected upper group}
First suppose that the component is not centrally quotiented.
Let $D=S\cap P$. Use the algebraic inner action of $P$ in
the restriction-of-scalars group and form
\begin{equation}\label{eq:disconnected-group}
 \widehat{\mathbb G}
 = (\mathbb G\rtimes P)
   /\{(\iota(d)^{-1},d):d\in D\}.
\end{equation}
The graph subgroup centralizes $\mathbb G$ and is normalized by
$P$, so it is normal. The Frobenius acts as $\mathbb F$ on
$\mathbb G$ and trivially on the finite factor. If a class of
$(g,p)$ is fixed, then
\[
 (\mathbb F(g),p)=(g,p)(\iota(d)^{-1},d)
\]
for some $d\in D$. Its $P$ coordinate forces $d=1$.
Thus the finite fixed-point group is exactly
$(S\rtimes P)/\diag(D^{-1},D)\cong SP$, with the original
multiplication and powers. 

Let $\mathbf Z_1=C_{\widehat{\mathbb G}}(\mathbb G)$.
Algebraic innerness gives
$\widehat{\mathbb G}=\mathbb G\mathbf Z_1$. Put $\widehat{\mathbb L}=\mathbb L\mathbf Z_1$ and extend the
parabolic by $\mathbf Z_1$. Define the upper finite Levi by
$L^+=\widehat{\mathbb L}^{\mathbb F}$. Lang's theorem in the
connected Levi implies that $L^+$ meets every component, so
$SP=S L^+$. Each component preserves the chosen rational series.
For example, if an algebraically inner representative $g$ commutes
with $F$, then $g^{-1}F(g)$ is central. In a rational torus solve
$a^{-1}F(a)=g^{-1}F(g)$; the rational element $ga^{-1}$ transports
the same torus-character pair as $g$. The same argument applies
inside the Levi. This gives the series-inertia equality required
in \cite[Corollary 3.8]{Ruhstorfer}.

The full-centralizer hypothesis gives almost-superregularity.
Propositions 3.6--3.7 and Corollary 3.8 of \cite{Ruhstorfer}
therefore give the upper splendid complex and its full
bimodule-homotopy endomorphism splitting. Those statements allow
disconnected groups with even component group. The upper
splendid correspondence puts a conjugate of the marked $P$
inside $L^+$. Since $SP=S L^+$, a conjugation by an element of
$S$ suffices. Conjugate the Levi, radical, field operators and
all markings together. We now have $L^+=LP$ and the original
$P$ preserves the same lower variety.

\subsection{The central quotient and its coarse identity component}
For a component $S=S_{\mathrm{sc}}/Z_0$, let
$\alpha:SP\to\Aut(S)$ be conjugation and let
$\rho:\widetilde{\mathbf G}^{F}\to\Aut(S)$ be the
regular-embedding action. Form the finite fiber product
\[
 \{(u,g)\in SP\times\widetilde{\mathbf G}^{F}:
                      \alpha(u)=\rho(g)\}.
\]
The kernel is central:
commutators of an element inducing the identity on $S$ with
$S_{\mathrm{sc}}$ lie in $Z_0$ and define a homomorphism from a
perfect group to a central group, so vanish. Choose a Sylow $2$-subgroup $T_0$ of the inverse image
of $P$ in this fiber product, and let
\[
 E_0=S_{\mathrm{sc}}T_0\twoheadrightarrow SP
\]
be the resulting finite extension, with central kernel
$A$ and $A\cap S_{\mathrm{sc}}=Z_0$. The construction
\eqref{eq:disconnected-group} applies to this inner-diagonal
extension as well. Denote its finite upper Levi by $L_0^+$.
The central kernel satisfies $A\leq L_0^+$; put
$L^+=L_0^+/A$. The same correspondence and simultaneous
conjugation place the marked $P$ in $L^+$.

Let $Y_0=Y_{\mathbb U}^{\mathbb G}$ be the variety for the
connected lower group with the chosen radical, and let $Y_0^+$
be the corresponding variety for the upper group. Before quotienting, coarsen the component grading to
\begin{equation}\label{eq:coarse-grading}
 E_0/(S_{\mathrm{sc}}A)\cong SP/S.
\end{equation}
Its identity piece is $AY_0$, and the quotient map gives
\begin{equation}\label{eq:coarse-quotient}
 (AY_0)/A\cong Y_0/Z_0=:\overline Y_0.
\end{equation}
An $A$-orbit meeting $Y_0$ meets it in exactly one $Z_0$-orbit.
The finite quotient universal property gives the inverse as a
map of varieties, which proves the variety isomorphism.
Two lifts of the same original $p\in P$ differ by $A$, whose
diagonal conjugation on the variety is pointwise trivial.
Thus \eqref{eq:coarse-quotient} is equivariant for the original
$P$, with all its power relations.

After placing $P$ in the upper Levi, the stabilizer of the
identity component consists of pairs with equal quotient degrees.
Multiplying by one common original $P$-lift turns them into
lower pairs. Consequently, with $K=SP$,
\begin{equation}\label{eq:upper-variety}
 \overline Y_0^+
 \cong(K\times L^+)\times_{(S\times L)\diag P}
                \overline Y_0.
\end{equation}
The map is $[(g,l),[y]]\mapsto[gyl^{-1}]$; the same degree
calculation proves its well-definedness, injectivity and
surjectivity. The original stable primitive block idempotents
lie in identity degree: a stable block in a normal
$2$-extension has a unique covering block, represented by that
same idempotent. This applies at both endpoints and preserves
the chosen cuts in \eqref{eq:upper-variety}.

Choose the lower point-stabilizer cochain representative directly on
$\overline Y_0$ with its whole field and regular-embedding
action; these operators preserve the central quotient. The action of $(S\times L)\diag P$ on the lower cochains is
obtained from its base action and the conjugation action of $P$;
their intersection acts diagonally. Restriction to this action
is denoted by $\Res C$. Lemma~\ref{lem:finite-induction}
gives the representative
\begin{equation}\label{eq:same-identity}
 C^+=\Ind_{(S\times L)\diag P}^{K\times L^+}\Res C,
 \qquad (C^+)_1=C.
\end{equation}
This equality is literal for the displayed induced model.

Put $T=(S\times L)\diag P$ and $\Gamma=K\times L^+$.
After restricting each of the specified point-stabilizer summands
and then inducing, its permutation overmodule is
$\OO[\Gamma/T_z]$. This is the permutation module of a point
stabilizer on $\Gamma\times_T\overline Y_0$. Thus the chosen
induced complex has the point-stabilizer term bound for the
upper variety.

We verify that the two indicated $P$-actions are equal.
A nontrivial field power cannot be inner-diagonal on the
finite central quotient. If it were, its implementing
regular-embedding element, preserving the opposite standard
Borels, would be toral. Since a field map fixes every
$x_\alpha(1)$, all root multipliers would be one, contradicting
its nontrivial action on some $x_\alpha(t)$. A central quotient
does not identify distinct root-group elements. Thus the kernel
of the regular/field finite operator group over its automorphism
image has no field part. The remaining kernel consists of central regular-embedding
elements and acts pointwise trivially on the variety. The two
lifts of an original $P$-operator in
\eqref{eq:same-identity} consequently induce the same cochain
action. The kernel of $P\to\Aut(S)$ acts through its homogeneous
units in the upper grading.

\begin{lemma}\label{lem:geometric-deck-stabilizers}
Let $K_0$ contain $L_0$, let $A\leq Z(K_0)\cap L_0$, and let $Z$
be a $(K_0,L_0)$-variety on which the left and right actions of $A$
agree and the resulting deck action of $A$ is free.
Put $K=K_0/A$, $L=L_0/A$, and $\overline Z=Z/A$.
Suppose every $2$-subgroup of a point stabilizer on $Z$ is
$(K_0\times1)$-conjugate to a subgroup of $\diag L_0$.
Then every $2$-subgroup of a point stabilizer on $\overline Z$
is $(K\times1)$-conjugate to a subgroup of $\diag L$.
If $P\in\Syl_2(L)$, it is $(K\times L)$-conjugate to a subgroup
of the same literal $\diag P$.
\end{lemma}
\begin{proof}
Let $T\leq(K\times L)_{\bar z}$ be a $2$-subgroup and choose
$z\in Z$ above $\bar z$. For the inverse image $\widetilde T$
in $K_0\times L_0$ there is an exact sequence
\[
 1\longrightarrow A\times A
 \longrightarrow\widetilde T\longrightarrow T\longrightarrow1.
\]
Choose a Sylow $2$-subgroup $Q$ of $\widetilde T$.
Since $A\times A$ is central, $Q$ contains $A_2\times A_2$
and maps onto $T$.

For each $u\in Q$, freeness supplies a unique $\chi(u)\in A$
with $uz=\chi(u)z$. Centrality of the deck action implies
$\chi(uv)=\chi(u)\chi(v)$, so $\chi:Q\to A_2$ is a homomorphism.
With the bimodule convention $(a,b)z=azb^{-1}$, it satisfies
\[
 \chi(a,b)=ab^{-1}\qquad(a,b\in A_2).
\]
Every $u$ can therefore be multiplied by
$(\chi(u)^{-1},1)$, without changing its image in $T$, to obtain
an element fixing $z$. Hence $Q_z=\ker\chi$ is a $2$-subgroup
of the upstairs point stabilizer and
\[
 1\longrightarrow\diag A_2
 \longrightarrow Q_z\longrightarrow T\longrightarrow1
\]
is exact. The stipulated upstairs conjugacy applies to $Q_z$.
Its image gives the first assertion. Finally conjugate the
right projection inside $L$ into $P$ and use the same element
diagonally in $K\times L$.
\end{proof}

In the present application $Z=Y_0^+$ and
$(K_0,L_0)=(E_0,L_0^+)$. The deck kernel lies in the central
part used to extend the Levi, so $A\leq L_0^+$.
It acts freely: an element of this central deck group fixing
$y\mathbb U$ would lie in the unipotent radical, whose
intersection with the central deck group is trivial.
The upstairs point-stabilizer assertion is
\cite[Lemma 3.7]{BDR}. That lemma is stated in the section
allowing disconnected reductive groups, and its conjugator
belongs to the finite left group. For the extended parabolic
and Levi used here their simultaneous normalizer is the
extended Levi, so its diagonal is the stated $\diag L_0^+$.
The lemma therefore applies to the central quotient,
including its additional fixed-point components.

Combining Lemma~\ref{lem:geometric-deck-stabilizers} with the
initial point-stabilizer choice proves the following precise
term assertion once $P\in\Syl_2(L^+)$ has been established:
\begin{equation}\label{eq:actual-C-diagonal-bound}
 (C^+)^i\in
 \operatorname{add}
 \{\OO[(K\times L^+)/\diag R]:R\leq P\}
 \quad\hbox{for every }i.
\end{equation}
Indeed, for a point stabilizer $B$ and $R\in\Syl_2(B)$,
$\OO[\Gamma/B]$ is a direct summand of $\OO[\Gamma/R]$:
the coset projection and transfer have composite $[B:R]$,
a unit of $\OO$. The preceding lemma conjugates $R$ into
$\diag P$. Block cuts preserve this term assertion.
The conjugations establish relative projectivity upon restriction
to the base double group and preserve the chosen operator action.

\begin{lemma}\label{lem:central-evaluations}
The central quotient above preserves the two upper evaluations,
their homotopies, and the relative dual on the specified
geometric complexes.
\end{lemma}
\begin{proof}
The central deck action is free. Use Rickard's point-stabilizer
representative: its terms are direct summands of permutation
modules whose restriction to the deck group is free. In particular
the terms, including their block cuts, are finite projective over
$\OO A$. On each free orbit the norm identifies the basis of
coinvariants with the orbit-sum basis of invariants with
coefficient one. Thus coinvariants and invariants are isomorphic
on these free terms and on their direct summands, without dividing
by $|A|$. The quotient theorem
\cite[Theorem 4.1]{Rickard}, or
\cite[Theorem 2.28]{RouquierEtale}, identifies these invariants
with quotient cochains in the homotopy category, retaining the
remaining finite operators.

Central balance is termwise: the left and right central
operators agree. Tensoring the quotient complexes over the
quotient middle algebra is therefore the quotient of the
original balanced tensor. For a finite projective $\OO A$-module
$V$, the coefficient-of-identity Frobenius trace gives
\begin{equation}\label{eq:frobenius-dual}
 \Hom_{\OO A}(V,\OO A)\cong\Hom_{\OO}(V,\OO).
\end{equation}
The inverse sends $f\in\Hom_{\OO}(V,\OO)$ to
\[
 v\longmapsto\sum_{a\in A}f(a^{-1}v)a.
\]
To check the evaluations, put $R=\OO A$ and let $B_0$ be an
upstairs middle block algebra. Its relative symmetrizing trace is
\[
 t^R_{B_0}\Bigl(\sum_g b_g g\Bigr)=\sum_{a\in A}b_a a,
\]
restricted to the block. Composing with the coefficient of $1$
in $R$ gives the ordinary symmetrizing trace; composing with
augmentation $R\to\OO$ gives the symmetrizing trace of the
quotient block. Since $V$ is finite projective over $R$,
\[
 \OO\otimes_R\Hom_R(V,R)
 \cong\Hom_{\OO}(\OO\otimes_RV,\OO).
\]
The relative tensor-dual adjunction and these trace identities
identify the quotient of each evaluation with the downstairs
evaluation. Their coefficients are unchanged. Homotopies
descend under the additive quotient functor. The upper
contractions supplied by
\cite[Propositions~3.6--3.7]{Ruhstorfer} therefore descend to
the specified quotient complexes. When $A$ is a $2$-group,
the finite projective $R$-terms are free because $R$ is local.
\end{proof}

\subsection{Rank one and the upper direct product}
The lower quotient Levi is $L=S_0\times A_0$ with
$S_0=\PSL_2(q^t)$, $A_0$ central, and a principal Klein block
on $S_0$. Here $t\geq3$, also in the $E_7$ case where $t=3$;
thus $q^t\geq27$ and $S_0$ is perfect and centerless.
Let $V\cong V_4$ be its marked defect group. The projective
centralizer of $V$ in the algebraic $\PGL_2$ is $V$: diagonalize
one involution and impose commutation with a second swapping
involution; precisely four projective matrices remain.

The original abelian $P$ centralizes
$D=V\times(A_0)_2$ and acts algebraically inner on the Levi.
It therefore acts on $S_0$ by conjugation with a unique
$v(p)\in V$. Uniqueness makes $v:P\to V$ a homomorphism,
identical on $V$. Put $P_0=\ker v$. Then
$P=V\times P_0$. Algebraic innerness fixes the connected
central torus, so $A_0$ centralizes $P$. Hence, inside the
given upper Levi,
\begin{equation}\label{eq:rank-one-upper}
 L^+=S_0\times A',\qquad A'=A_0P_0,
 \qquad P=V\times A'_2.
\end{equation}
The intersection $A'\cap S_0$ is trivial by centerlessness.
The factors are
the derived subgroup and center and are preserved by the
relevant operators.

\begin{lemma}\label{lem:rank-one-transporter}
In the decomposition
\[
 L^+=S_0\times A',\qquad
 P=V\times A'_2,\qquad D=V\times(A_0)_2,
\]
where $V\in\Syl_2(S_0)$ and $A'$ is central abelian, put
\[
 H^+=N_{L^+}(D)=N_{S_0}(V)\times A'.
\]
Then
\[
 H^+=N_{L^+}(P),\qquad
 \{l\in L^+:l^{-1}Dl\leq P\}=H^+.
\]
Moreover $H^+$ controls the fusion of all subgroups of the
abelian Sylow subgroup $P$ in $L^+$.
\end{lemma}
\begin{proof}
Write $l=(s,a)$. The condition $l^{-1}Dl\leq P$ says
$s^{-1}Vs\leq V$, hence $s^{-1}Vs=V$.
All central factors are fixed under conjugation.
This proves the two normalizer equalities and the transporter
assertion.

For the fusion assertion, let $Q\leq P$ and
$lQl^{-1}\leq P$. Both $P$ and $lPl^{-1}$ centralize
$lQl^{-1}$, because $P$ is abelian.
They are Sylow $2$-subgroups of $C_{L^+}(lQl^{-1})$.
Choose $c$ in that centralizer with
$c(lPl^{-1})c^{-1}=P$. Then $h=cl\in H^+$ and
$hqh^{-1}=lql^{-1}$ for every $q\in Q$.
\end{proof}

Here $q^t\equiv\pm3\pmod8$ gives $|S_0|_2=4$, so the
marked $V$ is a Sylow subgroup and the displayed $P$ is
indeed a Sylow $2$-subgroup of $L^+$.

The Klein four complex of Proposition~\ref{prop:klein-complex}
on $S_0$ has its odd outer action and the literal inner
restriction $v(p)$. In the global application the relevant
outer image is odd: the given upper $P$ represents its
Sylow $2$-image and has just been shown inner. Tensor that
complex with the original regular $\OO A'$ block in its
original odd central character sector. This gives $Y^+$,
whose identity part is the corresponding complex $Y$ on $L$.
All central powers in $A'$ are retained.

\subsection{The companion after quotienting and the full overlap}
Write
\[
 N=N_S(D),\quad H=N_L(D),\quad
 N^+=N_K(D)=NP,\quad H^+=N_{L^+}(D)=HP.
\]
Choose a maximal supported pair
$(\diag P,e_P^+\otimes(f_P^+)^*)$ of $k\otimes C^+$, and let $(D,e_D^+)$ and $(D,f_D^+)$ be its two
subpairs at $D$. The block $e_D^+$ of
$C_K(D)=C_S(D)P$ covers a single $P$-orbit of blocks of
$C_S(D)$ and is their orbit sum, by the normal $2$-extension
block theorem and Fong--Reynolds correspondence. Since
\[
 \Br_P(e_D^+)e_P^+=e_P^+\ne0,
\]
this orbit has size one: a nontrivial orbit sum is a proper
relative trace from a subgroup of $P$ and has zero Brauer image
at $P$. Write its unique member as $e$; thus $e_D^+=e$.
The same argument gives $f_D^+=f$ on the right. In particular
$e,f$ are $P$-stable lower roots and are also the upper roots.
We use the same symbols for their integral lifts.

These lower roots are supported by $k\otimes C$.
Indeed, downward closure of the supported-pair set
\cite[Corollary~8.8]{BM} gives the supported upper pair at
$\diag D$. Its cut is a Rickard equivalence of nonzero
centralizer blocks by \cite[Theorem~8.5]{BM}, and hence has
nonzero ordinary homology. In the induced model
\eqref{eq:same-identity}, its homogeneous pieces are the
$p$-translates of $eC(\diag D)f$, for $pD\in P/D$:
$p$ centralizes $D$ and fixes $e,f$. Nonzero homology of
the upper cut therefore implies nonzero homology of this
identity piece. The lower pair is supported and maximal,
since $D$ is a defect group at both lower endpoints.

Let $\widetilde E$ be the finite group of regular-embedding
Levi and field operators preserving the block, the chosen
geometric data, and $D$, whose automorphism image lies in
$\mathcal A_D$. Denote this image by $E$. We keep its finite action on the cochains.
Set
\[
 \begin{aligned}
 I&=N_N(e),& T&=N_H(f),&
 J_0&=N_{I\times T}(\diag D),\\
 I^+&=IP,&T^+&=TP,&
 J_{e,f}^+&=N_{I^+\times T^+}(\diag D),\\
 W_0&=N\times H,&W^+&=N^+\times H^+,&
 J&=N_{K\times L^+}(\diag D).
 \end{aligned}
\]
Since $P$ centralizes $D$ and fixes $e,f$, direct multiplication
in the given groups gives
\begin{equation}\label{eq:companion-inertia-groups}
 J_{e,f}^+=J_0(P\times P),\qquad
 (J_{e,f}^+)_{\mathrm{degree}\,1}=J_0\diag P.
\end{equation}

Put $\mathcal J=N_{S\times L}(\diag D)$ and choose
\[
 \Omega_D=\Omega_c(\overline Y_0^{\diag D},\OO)
\]
with the point-stabilizer construction for the whole finite
$\mathcal J\rtimes\widetilde E$ action. All supported root
pairs at $\diag D$ form one $\mathcal J$-orbit, by maximal-pair
conjugacy in \cite[Corollary~8.8]{BM}. This joint orbit is
$\widetilde E$-stable. Consequently
\[
 \pi_D=\sum_{(e',f')\in\mathcal J(e,f)}e'\otimes(f')^*,
 \qquad V_D=e\Omega_D f
\]
give an invariant complex $\pi_D\Omega_D$ and an isomorphism
\begin{equation}\label{eq:joint-root-induction}
 \Ind_{J_0}^{\mathcal J}V_D\ \xrightarrow{\ \sim\ }
 \pi_D\Omega_D,\qquad [j,v]\longmapsto jv.
\end{equation}
Distinct cosets give orthogonal root summands; this proves
injectivity and surjectivity term by term and commutes with
the differential. The selected $V_D$ carries the action of
$J_0\diag P$, because $P$ fixes the roots.

Define, at this initial choice,
\begin{equation}\label{eq:fixed-model-upper-induction}
 V_D^+=\Ind_{J_0\diag P}^{J_{e,f}^+}V_D .
\end{equation}
This is the fixed-point cochain object for
$(\overline Y_0^+)^{\diag D}$ with the specified cuts.
Indeed, the cosets of the component grading have
representatives $(p,1)$ with $p\in P$, and these representatives
centralize $\diag D$. Thus taking fixed points in the induced
variety gives the same finite induction.
This also follows modulo the maximal ideal from
\cite[Theorem 4.2]{Rickard}, or
\cite[Theorem 2.29]{RouquierEtale}, including the normalizer action.

Let
\[
 \widehat e=\Tr_I^N(e),\qquad \widehat f=\Tr_T^H(f).
\]
These are the blocks on the full lower normalizers.
The same sums, in degree one, are the upper block idempotents:
coset representatives for $N/I$ and $H/T$ also represent
$NP/IP$ and $HP/TP$.
The complete companions are the following complexes:
\begin{align}
 U_D={}&
 \widehat e\,\OO N e\otimes_{\OO I}
 \Ind_{J_0}^{I\times T}(\Res_{J_0}V_D)
 \otimes_{\OO T}f\,\OO H\widehat f,
 \label{eq:full-lower-companion}\\
 U_D^+={}&
 \widehat e\,\OO N^+ e\otimes_{\OO I^+}
 \Ind_{J_{e,f}^+}^{I^+\times T^+}(V_D^+)
 \otimes_{\OO T^+}f\,\OO H^+\widehat f .
 \label{eq:full-upper-companion}
\end{align}
These are precisely the two Fong--Reynolds factors and the
graph-normalizer induction of
\cite[Theorem 8.5 and Corollary 8.9]{BM}.
Their subpair form applies also when $D<P$.
The lower formula also equals
\[
 \widehat e\,\Ind_{\mathcal J}^{N\times H}
                (\pi_D\Omega_D)\,\widehat f
\]
by \eqref{eq:joint-root-induction} and induction transitivity.
This expression supplies its $\widetilde E$-action; the selected
root formula supplies its compatible $P$-action.

Ordinary induction in the two factors and transitivity identify
\eqref{eq:full-upper-companion} with
\begin{equation}\label{eq:FR-as-single-induction}
 U_D^+\cong
 \widehat e\,
 \Ind_{J_{e,f}^+}^{W^+}(V_D^+)
 \,\widehat f
 =
 \widehat e\,
 \Ind_{J_0\diag P}^{W^+}(V_D)
 \,\widehat f .
\end{equation}
For example, the adjacent $e$ and $f$ can be removed from the
outer tensor factors because they act as the identity on $V_D^+$;
the remaining outer factors are the ordinary induction
bimodules. This proves the formula as a complex, with its
given differential and coset carries.

Let $Q=P/D$. The original quotient maps identify
$K/S$ and $L^+/L$ with $Q$; the degree of a pair is the left
quotient degree times the inverse right quotient degree.
The subgroup of $W^+$ preserving degree one is
$W_0\diag P$. Restricting the degree-one part of
\eqref{eq:FR-as-single-induction} gives
\[
 \widehat e\,
 \Ind_{J_0\diag P}^{W_0\diag P}V_D
 \,\widehat f
 \cong
 \widehat e\,
 \Ind_{J_0}^{W_0}\Res_{J_0}V_D
 \,\widehat f
 = U_D.
\]
The middle isomorphism sends $[w\delta(p),v]$ to
$[w,\delta(p)v]$; its ambiguity is exactly
$\diag(P\cap S)=\diag D\leq J_0$.
Using these coset normal forms in the definitions gives the identity
of complexes with their specified $P$-actions
\begin{equation}\label{eq:companion-identity}
 (U_D^+)_1=U_D.
\end{equation}
All block idempotents used here have degree one.

\begin{lemma}\label{lem:full-companion-diagonal-bound}
Every term of the specified integral complex $U_D^+$ belongs to
\[
 \operatorname{add}
 \{\OO[(N^+\times H^+)/\diag R]:R\leq P\}.
\]
\end{lemma}
\begin{proof}
First note the useful finite-set calculation
\begin{equation}\label{eq:fixed-diagonal-cosets}
 ((K\times L^+)/\diag R)^{\diag D}
 =
 \begin{cases}
 J/\diag R,&D\leq R,\\
 \varnothing,&D\nleq R,
 \end{cases}
 \qquad R\leq P.
\end{equation}
If $(g,l)\diag R$ is fixed, then
\[
 (g,l)^{-1}\diag D(g,l)\leq\diag R,
 \qquad l^{-1}Dl\leq P.
\]
Lemma~\ref{lem:rank-one-transporter} gives $l\in H^+$ and
$l^{-1}Dl=D$. The same displayed inclusion then gives
$g\in N^+$, $D\leq R$, and
$g^{-1}dg=l^{-1}dl$ for every $d\in D$.
Thus $(g,l)\in J$. Conversely $P$ is abelian, so
$\diag R\leq J$, and every coset of $J/\diag R$ is fixed
when $D\leq R$. This proves \eqref{eq:fixed-diagonal-cosets}.
It explains directly why full normalizer induction has right
endpoint $H^+$ and preserves the literal diagonal in the
modular Brauer model.

For the preselected integral fixed-point model a separate
point-stabilizer argument gives the same term assertion.
Let $z\in(\overline Y_0^+)^{\diag D}$ and let $R$ be a Sylow
$2$-subgroup of its stabilizer in $J_{e,f}^+$.
The normal subgroup $\diag D$ acts trivially on this variety,
so $\diag D\leq R$.
The preceding global point-stabilizer bound supplies
$x=(g,l)\in K\times L^+$ with
$x^{-1}Rx\leq\diag P$.
Applying the transporter argument just given to
$x^{-1}\diag D x\leq\diag P$ shows that $x\in J\leq W^+$.
Hence $R$ is already $W^+$-conjugate to a subgroup of
$\diag P$.

Restrict the lower point-stabilizer model to $J_0\diag P$
and then induce as in \eqref{eq:fixed-model-upper-induction}.
A permutation overmodule $\OO[(J_0\diag P)/B_z]$ becomes
$\OO[J_{e,f}^+/B_z]$, the module of the corresponding
upper fixed-variety point stabilizer. Thus $V_D^+$ has terms
in the additive category of these permutation modules.
Passing to a Sylow $2$-subgroup is a split operation on their
permutation overmodules, as above.
Formula~\eqref{eq:FR-as-single-induction} then proves the bound
after both Fong--Reynolds factors and full normalizer induction.
Primitive root cuts and full orbit induction preserve the term bound.
\end{proof}

The reductions of these particular fixed-point complexes are
the Brauer cochain objects, so the modular local evaluations
come from \cite[Theorem 8.5 and Corollary 8.9]{BM}.
For these integral models lift the
contractions of their evaluation cones. The Hom-surjectivity for integral
permutation summands and the correction $h\mapsto
h(dh+hd)^{-1}$ in Lemma~\ref{lem:integral-lift} apply.
The grading projector preserves the bimodule equations, so
these contractions can be taken degree-preserving.

For a variety $\mathcal Z$ with free central $2$-deck group
$A$, let $R$ be the full inverse image of a quotient subgroup.
Its quotient fixed-point variety has the decomposition
\[
 (\mathcal Z/A)^{\diag(R/A)}
 =\coprod_{a\in\Hom(R/A,A)}
 \mathcal Z^{\{(r,a(rA)r):r\in R\}}/A,
\]
with empty components omitted. Freeness supplies a unique deck
change for each fixed orbit; centrality makes this change a
homomorphism. The displayed fixed loci are disjoint and closed,
and there are finitely many, so they are open and closed in
their union. The maps $r\mapsto a(rA)r$ are automorphisms:
they fix $A$, induce the identity on $R/A$, and have inverses
$r\mapsto a(rA)^{-1}r$. Each component contributes to the
quotient Brauer construction.

Define
\begin{equation}\label{eq:geometric-sandwich}
 X_S=C\otimes Y\otimes U_D^\vee,
 \qquad
 X_S^+=C^+\otimes Y^+\otimes(U_D^+)^\vee,
\end{equation}
with the forced intermediate block algebras. The endpoints
are the block of $S$ and its block in $N_S(D)$,
and their given upper extensions.

\begin{lemma}\label{lem:geometric-sandwich-diagonal-bound}
For the specified models, every term of
\[
 C^+\otimes_{\OO L^+}Y^+
       \otimes_{\OO H^+}(U_D^+)^\vee
\]
is $2$-permutation and relatively $\diag P$-projective as
an $(\OO K,\OO N^+)$-bimodule.
The assertion includes the indicated block cuts.
\end{lemma}
\begin{proof}
The terms of $C^+$ satisfy
\eqref{eq:actual-C-diagonal-bound}. The Green term in the
rank-one complex has vertex $\diag V$, and its correction
terms are projective. Tensoring with the regular central
$\OO A'$ block therefore puts each term of $Y^+$ in
\[
 \operatorname{add}
 \{\OO[(L^+\times H^+)/\diag T]:T\leq P\}.
\]
Lemma~\ref{lem:full-companion-diagonal-bound} gives the same
bound for $(U_D^+)^\vee$, with its factors reversed.

It suffices to tensor the displayed permutation overmodules:
split inclusions and retractions remain split after balanced
tensor product. For the first tensor, the double cosets
$R\backslash L^+/T$ give stabilizers
\[
 \{(a,l^{-1}al):a\in R\cap lTl^{-1}\},\qquad l\in L^+ .
\]
Both projections lie in $P$.
By the fusion assertion of
Lemma~\ref{lem:rank-one-transporter} there is $h\in H^+$
with $h^{-1}ah=l^{-1}al$ on this intersection.
Conjugation by $(1,h)$ puts the displayed graph in the
literal $\diag P$. Thus the first tensor has the required
$(K,H^+)$ term bound.

For the second tensor the double-coset representative belongs
to $H^+$, which is a subgroup of the right endpoint
$N^+$. Its graph stabilizer is
\[
 \{(a,h^{-1}ah):a\in R'\cap hT'h^{-1}\},
 \qquad h\in H^+,\quad R',T'\leq P .
\]
Now $(1,h)\in K\times N^+$ itself absorbs the graph.
Its resulting diagonal subgroup is contained in $P$.

Linearization of a balanced product of finite bisets is a
permutation module. Block cuts and the given summand
projectors give its direct summands.
Tensoring over a block algebra is the same balanced tensor
as over its group algebra on these block-supported modules.
This proves the assertion for every term of the total
complex. The conjugations witness relative projectivity and preserve
the chosen complex and its operator action.
\end{proof}

The same proof applies to the lower complex, with
$(K,L^+,P,N^+,H^+)$ replaced by $(S,L,D,N,H)$.
Here $D$ is Sylow in $L=S_0\times A_0$ and
$N_L(D)$ controls its fusion by the same centralizer argument.


 For the graded tensor identity, let $B=\bigoplus_{g\in Q}B_g$
be an intermediate upper block algebra, with $Q=P/D$,
and choose a homogeneous unit $u_g\in B_g$.
For the adjacent graded bimodule complexes
$M=\bigoplus M_g$ and $N=\bigoplus N_g$, multiplication by
these units gives
\begin{equation}\label{eq:homogeneous-tensor}
 (M\otimes_B N)_1\ \xrightarrow{\ \sim\ }
 M_1\otimes_{B_1}N_1,\qquad
 x_g\otimes y_{g^{-1}}\longmapsto
 x_gu_g^{-1}\otimes u_gy_{g^{-1}}.
\end{equation}
Its inverse is induced by the inclusions of $M_1,N_1$.
For $b_h\in B_h$, the relation
$u_gb_hu_{gh}^{-1}\in B_1$ verifies balancing and independence
of the chosen units. These units are group elements multiplied
by the relevant block idempotent, so their inverses retain the
group multiplication relations.
Equations~\eqref{eq:same-identity} and
\eqref{eq:companion-identity} give
$(X_S^+)_1=X_S$, with the same operators and both
evaluations. Degree-preserving projection of the upper
homotopies is bimodule-linear even for a nonabelian grading;
it retains the specified contractions.

Write $c_n$ for conjugation by $n$ on $S$, and, for
$U\leq S$, put $c(U)=\{c_u:u\in U\}\leq\Aut(S)$.
The kernel calculation following \eqref{eq:same-identity}
shows that the finite operator presentation $\widetilde E$
has a pointwise-trivial kernel over $E$ consisting of central
regular-embedding elements.
The point-stabilizer cochains, their fixed-point models and
inductions kill this kernel termwise. Thus the action factors
to $\rho:E\to\Aut_{\OO}(X_S)$.

With $N=N_S(D)$ and $H=N_L(D)$ the precise overlap is
\begin{equation}\label{eq:full-geometric-overlap}
 \mathcal A_D=c(N)E,\qquad
 E\cap c(N)=c(H).
\end{equation}
For the intersection, an inner representative in $S$
preserving the chosen Levi and its block preserves the unique
rational-series idempotent containing that block.
Its dual relative normalizer takes $s$ to an $\mathbf L^*$-conjugate.
Correcting by that conjugator puts it in
$C_{\mathbf G^*}(s)=\mathbf L^*$, so its relative Weyl element
is trivial and its representative lies in $L$.
This is the normalizer calculation
\cite[Section~7.A, equation~(7.1)]{BDR}, applied with
$C_{\mathbf G^*}(s)=\mathbf L^*$.
The same statement descends through the central quotient,
whose kernel acts trivially. Imposing the marking $D$ gives
$H$. Conversely every $h\in H$ supplies this inner operator.

For the product equality, choose a regular embedding
$\widetilde{\mathbf G}=\mathbf GZ(\widetilde{\mathbf G})$
and put
$\widetilde{\mathbf L}=\mathbf LZ(\widetilde{\mathbf G})$.
Then $\widetilde{\mathbf L}/\mathbf L\cong
\widetilde{\mathbf G}/\mathbf G$. Lang's theorem for the
connected groups $\mathbf L$ and $\mathbf G$ gives
\[
 \widetilde{\mathbf L}^{F}/\mathbf L^{F}
 \cong\widetilde{\mathbf G}^{F}/\mathbf G^{F},
 \qquad
 \widetilde{\mathbf G}^{F}
   =\mathbf G^{F}\widetilde{\mathbf L}^{F}.
\]
Together with the finite field operator $\tau$, these regular
Levi elements represent every relevant inner-diagonal-field
outer class and preserve $\mathbb U$ in
\eqref{eq:restriction-scalars}. The same assertions hold after
the simultaneous conjugation used to place $P$ in the Levi.
The equivariance and the identification of the block
labels are given by \cite[Corollary~4.2, Proposition~4.9,
Corollary~5.5 and Lemma~5.6]{Ruhstorfer}; comparison of block
bijections uses \cite[Theorem~7.2]{BDR}.
A representative in the same outer class as a block stabilizer
fixes that block, since inner automorphisms fix blocks. Its
geometric action then fixes the corresponding Levi block.
Conjugating by an element of $L$ returns its defect group to
$D$. The resulting representative belongs to $E$, and differs
from the original marked automorphism by $c_n$ with $n\in N$.
This proves \eqref{eq:full-geometric-overlap}.

On $c(H)$ the constructed action is the literal
diagonal action: the geometric action of $h$ is conjugation
on the varieties, the rank-one action has this property by
construction, and the induction and Fong--Reynolds formulas
preserve it. Also $\delta_z=1$ for every $z\in Z(S)$,
because the original complexes are strictly balanced over
that center. Therefore
\begin{equation}\label{eq:normalizer-overlap}
 \rho_{\mathrm{full}}(c_ne)=\delta_n\rho(e),
 \qquad n\in N,\ e\in E,
\end{equation}
is well-defined even when $S$ has a nontrivial center.
Indeed, two factorizations differ by $c_h$ with $h\in H$;
the corresponding representatives in $N$ differ by $hz$
with $z\in Z(S)$, and $\delta_z=1$.
It is multiplicative because
\[
 \rho(e)\delta_n\rho(e)^{-1}=\delta_{e(n)}.
\]

Equivalently $X_S$ is a module complex for the finite
relative group
\[
 \mathcal T_S=
 \bigl((S\times N)\rtimes\mathcal A_D\bigr)
 \Big/
 \{(\delta(n)^{-1},c_n):n\in N\}.
\]
The displayed normal subgroup centralizes the base, and its
intersection with $S\times N$ is exactly $\diag Z(S)$.
Thus the base of $\mathcal T_S$ is
$(S\times N)/\diag Z(S)$.
More precisely, the pre-existing Levi-operator group
\[
 \mathcal T_E=
 ((S\times N)\rtimes E)
 /\{(\delta(h)^{-1},c_h):h\in H\}
\]
maps isomorphically to $\mathcal T_S$.
Surjectivity follows from $\mathcal A_D=c(N)E$, since an
inner operator is identified with its base diagonal operator.
For the kernel, if an $E$-operator represents $c_n$, then
$c_n\in E\cap c(N)=c(H)$; write $n=hz$ with
$h\in H$ and $z\in Z(S)\leq H$. Its relation is already in
the displayed denominator of $\mathcal T_E$.
The isomorphism of relative groups transfers the chosen
permutation terms and the action on their differentials.
Pull back this action through the original ambient
automorphism map. Its kernel $C_{\mathrm{ambient}}(S)$
acts as the identity on this lower component.
On $P$, the action agrees with the geometric
action used in the identity-component inductions, because
the two regular/field lifts differ only by the pointwise
central kernel already identified.
This assertion concerns the diagonal operator on the lower
complex; the original kernel elements of $P$ remain as
homogeneous units with their powers in the upper
groups and upper complexes.

\begin{proposition}\label{prop:geometric-component}
For case (iii) with the ambient marked stabilizers
specified above, there is an integral splendid component
complex $X_S$ with its genuine full marked stabilizer
action and literal inner restriction. It is the
identity component of the upper splendid complex
$X_S^+$ for $SP$ and $N_S(D)P$. Both evaluations admit
homogeneous upper contractions, and the complexes are
strictly balanced over their shared centers.
\end{proposition}

\section{Assembly of component complexes}\label{sec:assembly}

Let $H=S_1\cdots S_mC\triangleleft G$ be a central product,
let $b=b_1\cdots b_m b_C$ be a $G$-stable block of $\OO H$,
and let $a$ be a block of $\OO G$ covering $b$, with abelian
defect group $P$. Assume
\begin{equation}\label{eq:full-relative-assumption}
 D=P\cap H,\qquad |P:D|=|G:H|_2.
\end{equation}
The factors $S_i$ constitute one chosen $G$-orbit of components
in cases (i)--(iii); the remaining factors form $C$.
Assume that $P$ fixes each $S_i$ and that $N_G(P)$ is transitive on
this component orbit. Put $D_i=P\cap S_i$ and
\begin{equation}\label{eq:partial-normalizer}
 Q=P\cap(S_1\cdots S_m),\quad J=N_G(Q),\quad H'=H\cap J.
\end{equation}
The factor blocks and defect intersections give
\[
 H'=N_{S_1}(D_1)\cdots N_{S_m}(D_m)C.
\]
Let $c_i$ be the Brauer correspondent of $b_i$ in $N_{S_i}(D_i)$,
and write $b'=c_1\cdots c_m b_C$ for the resulting block of $H'$.
The marked Frattini argument on the normal component product
gives $G=HJ$. Hence
$\Gamma=G/H\cong J/H'$, $P\leq J$, and $P\cap H'=D$.

For every chosen factor use the data established in
Sections~\ref{sec:elementary} and \ref{sec:geometry}:
an integral splendid complex $X_i$ with the marked stabilizer
action and diagonal inner action; an upper complex $X_i^+$
for $S_iP$ and $N_{S_i}(D_i)P$ whose identity component is
$X_i$ with that $P$-action; degree-preserving evaluations and
contractions; and balance over the shared centers in the chosen
central-character summands. On $C$ use the identity complex
$\OO Cb_C$. Choose the data on one representative $S_1$.
For each $i$, choose $g_i\in N_G(P)$ with ${}^{g_i}S_1=S_i$, taking
$g_1=1$, and transport its blocks, Brauer pairs, lower and upper
complexes, evaluations, and contractions by this same $g_i$.
Since the component product is normal, $N_G(P)\leq J$.
Thus every transported upper complex still uses the specified group
$P$.

\begin{proposition}\label{prop:assembly}
These data give a splendid Rickard complex between the
block-sum algebras on $G$ and $J$. A primitive cut
at the block gives a primitive target block
with the same marked defect group $P$.
\end{proposition}
\begin{proof}
First construct the full action. With the transporters $g_i$ fixed
above, for $g\in J$ the operator on the $i$th factor is
the given operator of
\[
 h(g,i)=g_{g(i)}^{-1}gg_i.
\]
The equality
\begin{equation}\label{eq:carry}
 h(g_2g_1,i)=h(g_2,g_1(i))h(g_1,i)
\end{equation}
proves the group law. For $p\in P$, the equality $p(i)=i$ gives
$h(p,i)=g_i^{-1}pg_i\in P$. Hence the action on each transported
lower complex is the identity-component action of its specified
upper complex. Tensor these maps with the Koszul
permutation signs and with the original conjugation action
on the regular $C$ factor. Inner elements of $H'$ act by
the literal diagonal bimodule action. Balance over the
shared centers and retain the chosen odd
character sector. This gives a lower complex $X$
for
\begin{equation}\label{eq:pair-groups}
 T=(H\times H')\diag J,
 \qquad T_P=(H\times H')\diag P.
\end{equation}

We prove that its restriction to $T_P$ is a permutation
summand with the required vertex bound. Temporarily keep
$P$ as a separate acting group:
\[
 \widetilde T_P=
 \left(\prod_i(S_i\times N_{S_i}(D_i))
              \times(C\times C)\right)\rtimes P.
\]
For the $i$th factor the inverse image of the original
$\diag P$ is
\begin{equation}\label{eq:inverse-diagonal}
 J_i=\diag D_i\times P.
\end{equation}
Indeed $(sp,tp)$ is diagonal on $P$ precisely when
$s=t\in D_i$. The product is direct here because $P$ is
abelian. Restrict the upper complex $X_i^+$ to its
identity stabilizer and take its identity summand.
Mackey intersections remain below the diagonal bound:
an upper pair is a base pair times $(p,q)\in P\times P$,
and the latter centralizes $\diag P$.
Thus the pulled-back $X_i$ is $2$-permutation and
relatively $J_i$-projective.

The corresponding witness for the remainder is the
regular block of $CP$. Its original $P$-Brauer support
is nonzero, and the normal-extension defect bound is
$|P\cap C|\,|P:P\cap C|=|P|$; hence its defect is
$P$. It has identity component
$\OO Cb_C$ and relative inverse-image bound
$\diag(P\cap C)\times P$.

Permutation summands relatively projective for $J_i$
are summands of sums of permutation modules with
stabilizers $Q_i\leq J_i$. The stabilizers in their
external tensor are intersections of conjugates of
the inverse-image $Q_i$. Remove the $P$-coordinate
from each conjugator, since $P$ normalizes $J_i$.
The remaining conjugators belong to independent base
factors and combine into one base conjugation.
Every stabilizer is consequently conjugate into
\begin{equation}\label{eq:common-bound}
 \left(\prod_i\diag D_i\times\diag(P\cap C)\right)
 \times P.
\end{equation}

Now impose the central-product relations.
Strict central balance makes the left relation and
the corresponding right relation act identically.
For a quotient kernel $K$, the coinvariants of
$\OO[\widetilde T_P/Q_0]$ are the permutation module
on the orbit set, with stabilizer $Q_0K/K$.
Summand projectors descend. Next impose the original
graph identifying $d\in D$ with its inner diagonal
action. This graph already acts trivially, by the
literal inner restriction and the identity action
of off-factor centralizers. Under these two
quotients the image of \eqref{eq:common-bound}
is first $\diag D\cdot P$ and then exactly the
original $\diag P$. This proves the whole
$T_P$ permutation and relative-vertex assertions.
One-sided projectivity is preserved by the same
balanced tensor: tensoring summands of free factor
modules gives summands of free modules for the
central-product algebra.

We next retain both evaluations. The homogeneous
identity comparison \eqref{eq:homogeneous-tensor}
gives $P$-linear contractions for the chosen
lower factor evaluations. Tensor them and the
regular remainder identity, and descend through
the central relations. The duality here is the
relative Frobenius dual. For a common central
subgroup $Z$, put $R=\OO Z$. Finite projective
$R$-modules satisfy
\begin{align*}
 \Hom_R(U,R)&\cong\Hom_{\OO}(U,\OO),\\
 \Hom_R(U\otimes_RV,R)
 &\cong\Hom_R(U,R)\otimes_R\Hom_R(V,R).
\end{align*}
The first map composes with the coefficient-of-identity trace
$R\to\OO$. Its inverse sends $f\in\Hom_{\OO}(U,\OO)$ to
\[
 u\longmapsto\sum_{z\in Z}f(z^{-1}u)z.
\]
The second is the ordinary dual-tensor isomorphism for finite
projective modules over the commutative ring $R$. These maps are
natural under automorphisms of $Z$.
Thus for $A=A_1\otimes_R A_2$, $B=B_1\otimes_R B_2$
and $X=X_1\otimes_R X_2$,
\begin{equation}\label{eq:balanced-evaluation}
 X\otimes_BX^\vee
 \cong(X_1\otimes_{B_1}X_1^\vee)
          \otimes_R(X_2\otimes_{B_2}X_2^\vee),
\end{equation}
and likewise on the other side. The resulting
evaluations are the quotient-algebra
evaluations. On a coset basis the displayed inverse has coefficient
one at the identity, including when $Z$ has even order.

By \eqref{eq:full-relative-assumption},
\begin{equation}\label{eq:odd-index}
 [T:T_P]=[\Gamma:PH/H]
\end{equation}
is odd. Restriction and induction across this
index show that the already constructed $T$-module
terms are permutation summands with vertices
below $\diag P$. For either evaluation, let $\mathcal U$ be its operator
group, $(H\times H)\diag J$ or $(H'\times H')\diag J$,
and let $\mathcal U_P$ be the subgroup obtained by replacing
$J$ with $P$. If $h_P$ is its supplied
$\mathcal U_P$-linear contraction, put
\begin{equation}\label{eq:relative-trace}
 h_{\mathcal U}=\frac{1}{[\mathcal U:\mathcal U_P]}
       \sum_{u\in\mathcal R}u h_Pu^{-1},
\end{equation}
where $\mathcal R$ represents the left cosets of
$\mathcal U_P$ in $\mathcal U$. This index is odd by
\eqref{eq:full-relative-assumption}. Changing a representative
by a right element of $\mathcal U_P$ leaves its summand
unchanged. Left multiplication permutes the cosets with these
relations, proving equivariance. Linearity of $dh+hd=1$
proves that $h_{\mathcal U}$ is a contraction.

Write $\tau_j$ for the diagonal action of $j\in J$ on $X$ and
$\tau_j^\vee(f)=f\circ\tau_{j^{-1}}$ for its action on the dual.
Finally induce $X$ from $T$ to $G\times J$.
Equivalently use $\OO Gb\otimes_{\OO Hb}X$,
with right action
\begin{equation}\label{eq:induced-right}
 (a\otimes x)j=aj\otimes\tau_{j^{-1}}(x).
\end{equation}
The inner intersection relation makes this well-defined. Put
$A=\OO Hb$, $A^+=\OO Gb$, $B=\OO H'b'$, and
$B^+=\OO Jb'$, where $b'$ is the retained block of $H'$.
Write $Y=A^+\otimes_A X$. Its dual is
$X^\vee\otimes_A A^+$, with
\[
 j(f\otimes a)=\tau_j^\vee(f)\otimes ja
 \qquad(j\in J).
\]
Choose representatives $j\in J$ for $G/H$. Every homogeneous
element of $A^+$ has the form $ja_0$ with $a_0\in A$. The two
evaluation identifications are
\begin{align*}
 Y\otimes_{B^+}Y^\vee
 &\longrightarrow A^+\otimes_A(X\otimes_BX^\vee),\\
 (a\otimes x)\otimes(f\otimes ja_0)
 &\longmapsto aj\otimes
   \bigl(\tau_{j^{-1}}x\otimes\tau_{j^{-1}}^\vee f\bigr)a_0,
\end{align*}
and
\begin{align*}
 X^\vee\otimes_A A^+\otimes_A X
 &\longrightarrow B^+\otimes_B(X^\vee\otimes_A X),\\
 f\otimes ja_0\otimes x
 &\longmapsto j\otimes
   \bigl(\tau_{j^{-1}}^\vee f\otimes a_0x\bigr).
\end{align*}
On the first target put $Z=X\otimes_BX^\vee$ and
$\tau_j^Z=\tau_j\otimes\tau_j^\vee$. Its upper right action is
\[
 (a\otimes z)j=aj\otimes\tau_{j^{-1}}^Z(z).
\]
The second target has the analogous action, with
$Z=X^\vee\otimes_A X$ and
$\tau_j^Z=\tau_j^\vee\otimes\tau_j$.
Their inverses send $a\otimes(x\otimes f)$ to
$(a\otimes x)\otimes(f\otimes1)$ and
$j\otimes(f\otimes x)$ to $\tau_j^\vee f\otimes j\otimes x$,
respectively. The relation
$\tau_j(hxh')=jhj^{-1}\tau_j(x)jh'j^{-1}$ verifies balancing;
changing a coset representative uses this same relation. The maps
intertwine both endpoint actions and carry the upper evaluations
to the induced lower evaluations. The contractions from
\eqref{eq:relative-trace} therefore give both upper contractions.

The primitive-block correspondence
\cite[Theorem~8.4]{BM} gives a unique primitive idempotent
$c$ of $B^+$ such that $aYc$ is noncontractible. For $c'\ne c$,
the complex $aYc'$ is contractible, and the two evaluations
restrict to the regular blocks at $a,c$.
Apply \cite[Lemma~8.3]{BM} to the chosen left maximal pair
$(P,e_P)$. It gives a supported graph pair for $aYc$;
\cite[Corollary~8.8(c)]{BM} makes its right pair maximal,
of order $|P|$. Every term already has vertices below
conjugates of $\diag P$. A supported graph of this full order
is therefore conjugate to $\diag P$ in $G\times J$.
Its right projection is a $J$-conjugate of $P$, so conjugating
the right pair gives the chosen defect subgroup $P$ without
changing $c$. Since $N_G(P)\leq J$, a further right normalizer
conjugation also matches the graph with the chosen left
marking. Thus this primitive cut has the required defect
group and diagonal bound.
\end{proof}

\section{Reduction and proof of the theorem}\label{sec:global}

\subsection{Strict reductions}
We use induction on the positive integer
$\kappa(G)$ in \eqref{eq:kappa}. Every
equivalence used in a reduction is an actual
splendid complex with its specified defect
graph, so Proposition~\ref{prop:companion}
returns the inductive result to the original
Brauer correspondent.

If a block $d$ of a normal subgroup is covered
and its inertia group $I=I_G(d)$ is proper,
Fong--Reynolds gives a source-algebra
equivalence to the corresponding
primitive block of $I$. Transport the
maximal pair simultaneously so its defect
is the selected $P\leq I$. Put
$Z_0=O_{2'}(Z(G))$. Since $Z_0\leq I\cap Z(I)$,
\begin{equation}\label{eq:fr-descent}
 \kappa(I)\leq|I:Z_0|<\kappa(G).
\end{equation}
Thus we may assume quasiprimitivity: every
covered block of every normal subgroup is
$G$-stable.

\begin{lemma}\label{lem:kp-descent}
Suppose that a covered block of
$N\triangleleft G$ is nilpotent and
$N\nsubseteq Z(G)O_2(G)$. There is an actual
splendid equivalence to a primitive block
of a group $G_1$ with $\kappa(G_1)<\kappa(G)$.
\end{lemma}
\begin{proof}
Put $M_0=Z(G)N$. Its covered block is
nilpotent and stable. Its defect is
$Q=P\cap M_0$. The nilpotent-extension
theorem \cite[Theorem 3.5 and Corollary
3.15]{PuigZhou}, with the source-corner
correction in its Section 3.16, gives a
model group $L$, a finite central odd cover
$\widetilde L\to L$, and basic
Morita correspondences on primitive
summands, with
\begin{equation}\label{eq:kp-order}
 |L|=|G:M_0|\,|Q|.
\end{equation}
Choose a maximal relative $2$-subgroup $R\geq P$ for
the covered $M_0$-block. Normal defect intersection gives
$R\cap M_0=Q$, since $P\cap M_0$ is already a defect group of that
block. The construction has an exact sequence
\[
 1\longrightarrow Q\longrightarrow L
   \longrightarrow G/M_0\longrightarrow1.
\]
It embeds $R$ with its original intersection $Q$ and identifies the
defect of the selected primitive model block with the given subgroup
$P\leq R$. The source resolution is applied to this primitive
correspondence and hence to its abelian defect group $P$.

Here is the source-corner passage in this correspondence.
In \cite[Section~3.1]{PuigZhou} the chosen source idempotent
$j$ of the lower block satisfies $j\leq i$. Since $j$ is full
in that lower block algebra, $i$ is a full idempotent in the
upper block-sum algebra. Corollary~3.15 identifies its corner,
as an $R$-interior graded algebra, with a matrix factor given
by a Dade $R$-algebra tensored with the local-extension algebra.
The full-idempotent equivalence followed by matrix cancellation
is therefore a Morita equivalence of the block-sum algebras.
The induced center isomorphism selects the corresponding
primitive block. Under the compatible pointed-subgroup
embeddings, its source at the defect $P\leq R$ is the
capped summand of the restriction of the Dade $R$-source.
This is an endopermutation $P$-source with the given defect
identification, so the primitive correspondence is basic.
The central $k^\times$-extension has a finite odd central
realization in its prescribed character sector, giving the
finite group $\widetilde L$ used above.

Apply Proposition~\ref{prop:source-upgrade}
to this primitive Morita bimodule,
whose capped endopermutation source and
fusion stability are supplied by the
source theorem. Lift integrally by
Lemma~\ref{lem:integral-lift}. The central
odd cover gives
\begin{equation}\label{eq:kp-kappa}
 \kappa(\widetilde L)
 \leq|L|=|G:M_0|\,|Q|\leq|G:Z_0|.
\end{equation}
Equality in the last inequality would
imply $M_0=Z_0Q$. This is an internal
direct product; $Q$ is its unique Sylow
$2$-subgroup and is characteristic in
$M_0$, hence $Q\leq O_2(G)$.
It would follow that
$N\leq Z(G)O_2(G)$, contrary to the
hypothesis on $N$. The inequality is therefore
strict. Take $G_1=\widetilde L$ with
the selected primitive block and its
chosen central-character summand.
\end{proof}

After these reductions, for every $N\triangleleft G$ we have
\begin{equation}\label{eq:weak-reduced}
 \begin{gathered}
 \text{if a block of $\OO N$ covered by $a$ is nilpotent, then}\\
 N\leq Z(G)O_2(G).
 \end{gathered}
\end{equation}

\subsection{The generalized Fitting subgroup and relative defect}
\begin{lemma}\label{lem:reduced-layer}
Let $a$ be quasiprimitive and satisfy
\eqref{eq:weak-reduced}. Put $H=F^*(G)$.
Every component has noncentral defect,
$P$ fixes each component, and, for
$D=P\cap H$,
\begin{equation}\label{eq:full-relative}
 |P:D|=|G:H|_2.
\end{equation}
In particular, $P$ is maximal among
$2$-subgroups with nonzero relative
Brauer support for the covered $H$-block.
\end{lemma}
\begin{proof}
Every block of $O_{2'}(G)$ has defect zero
and is nilpotent, so
\eqref{eq:weak-reduced} puts
$O_{2'}(G)$ in $Z(G)$. The normal
$2$-subgroup lies in abelian $P$.
Thus $H$ is the central product of its
components and its original abelian
factor $O_2(G)Z_0$.

We justify the component intersections
before using them. Apply the normal
defect-intersection theorem first to
$E(G)\triangleleft G$, then to
$S_i\triangleleft E(G)$. It gives
that $P\cap E(G)$ is a defect group
of the covered layer block and
$D_i=P\cap S_i$ is a defect group
of its component block.
For a central product these factor
defects give its defect by the
group multiplication map, with
the shared centers retained.

If a whole orbit of components had
central-defect blocks, its covered
central-product block would be
nilpotent. Its normal perfect group
would, by \eqref{eq:weak-reduced},
lie in the abelian group
$Z(G)O_2(G)$, a contradiction.
Thus every component defect is
noncentral. An element of $P$
centralizes each $D_i$. If it
moved $S_i$ to a different
component, $D_i$ would lie in
their central intersection,
again a contradiction. Hence
$P$ fixes every component.

We also verify the upper-defect hypothesis of the
component constructions. Write $b_E$ for the covered
$E(G)$-block and $b_i$ for its retained component block.
Since $P$ fixes $S_i$ and stabilizes $b_E$, uniqueness
of the factor blocks makes $b_i$ $P$-stable. In the
group algebra, $ab_E=a$ and $b_Eb_i=b_E$,
so $ab_i=a$. Applying the Brauer homomorphism at $P$
gives $\Br_P(a)\Br_P(b_i)=\Br_P(a)\ne0$.
The unique covering block of the normal $2$-extension
$PS_i/S_i$ is represented by $b_i$ and has nonzero
$P$-support. Its defect order is at most
$|D_i|\,|P:D_i|=|P|$ by the normal-extension bound.
It therefore has $P$ as a defect group,
as required in all four component cases.

Let $K_0$ be the kernel of the
component permutation action,
intersected with $C_G(O_2(G))$.
Then
\[
 P\leq K_0,\qquad C_G(P)\leq K_0,
 \qquad H\leq K_0.
\]
The middle inclusion uses the same
noncentral-defect argument and
$O_2(G)\leq P$. The covered $K_0$-block $d$ has defect $P$.
Since $ad=a$ and $C_{K_0}(P)=C_G(P)$, the original root satisfies
\[
 e_P\Br_P(d)=e_P\Br_P(a)\Br_P(d)=e_P.
\]
It is therefore the same maximal root for $d$. Quasiprimitivity
and conjugacy of maximal pairs
therefore give
\begin{equation}\label{eq:marked-frattini}
 G=K_0N_G(P,e_P).
\end{equation}
The quotient $G/K_0$ is an image
of the odd inertial quotient
$N_G(P,e_P)/(PC_G(P))$, hence
is odd. The componentwise outer-automorphism map has kernel $H$
on $K_0$. Indeed an element in its kernel can be multiplied by
elements of the components so as to centralize every component.
It already centralizes $O_2(G)$ and $Z_0$, and hence centralizes $H$.
The generalized-Fitting centralizer theorem gives $C_G(H)\leq H$.
Conversely every element of $H$ induces inner automorphisms on the
components. Thus $K_0/H$ embeds in the product of their outer
automorphism groups and is solvable by the Schreier theorem.
Consequently $G/H$ is
$2$-solvable.

Pull back a $2$/$2'$ series of normal subgroups to
$H=H_0\triangleleft H_1\triangleleft\cdots\triangleleft H_t=G$
with every $H_j\triangleleft G$, and retain the covered blocks.
They are stable under the next group,
by quasiprimitivity, and have defect groups $P_j=P\cap H_j$.
If $H_j/H_{j-1}$ is a $2$-group, the normal-extension defect theorem
gives $|P_j:P_{j-1}|=|H_j:H_{j-1}|$. If it has odd order, the
defect groups have equal order and $P_j=P_{j-1}$.
Multiplication gives \eqref{eq:full-relative}.
If $R\geq P$ has nonzero
relative support, then
$|R\cap H|\leq|D|$ and
$|R:R\cap H|\leq|G:H|_2$.
The equality for $P$ forces
$R=P$.
\end{proof}

The lemma also gives the odd outer
image used in the component constructions.
For the partial normalizer $J$ in
\eqref{eq:partial-normalizer}, the
stabilizer $J_i$ of a component
contains $P$ and $H'$. The image
$PH'/H'$ is a Sylow $2$-subgroup
of $J/H'$, hence also of
$J_i/H'$. Its image in the
outer automorphism group is
trivial for the elementary
and direct Klein cases,
by Propositions~\ref{prop:klein-inner}
and \ref{prop:elementary-complex}.
Therefore the entire relevant
outer image is odd.
For the rank-one factor inside the geometric Levi,
let $E$ be its marked Levi/block stabilizer after the
simultaneous conjugation, so that $P\leq E\leq J_i$.
The projection $E\to J_i/H'$ has kernel $E\cap H'$.
By the full-centralizer inner-overlap calculation,
this kernel acts on $S_0$ through $N_L(D)$, with
the other component factors acting trivially; its
action is therefore inner. The image of $P$ is
a Sylow $2$-subgroup of $E/(E\cap H')$, because
it is already Sylow in $J_i/H'$. Its outer
action on $S_0$ is trivial by
\eqref{eq:rank-one-upper}. Consequently the
entire relevant outer image of $E$ on $S_0$
is odd.

\subsection{Completion of the induction}
\begin{proof}[Proof of Theorem~\ref{thm:main}]
Induct on $\kappa(G)$. A proper
Fong--Reynolds reduction lowers
it by \eqref{eq:fr-descent};
Lemma~\ref{lem:kp-descent} lowers it whenever a covered
nilpotent normal block satisfies that lemma's hypothesis.
For either reduction, use
induction on the actual
primitive target and return
by Proposition~\ref{prop:companion}.
We may therefore assume the
hypotheses of
Lemma~\ref{lem:reduced-layer}.

If there are no components,
$H=O_2(G)Z_0$ is centralized
by $P$. The generalized-Fitting
centralizer theorem gives
$P\leq C_G(H)\leq H$,
so $P=O_2(G)$ is normal.
Its normalizer is $G$ and
the identity bimodule is
the required splendid
equivalence.

Suppose there are components
and at least one active
case (i), (ii), or (iii)
from Section~\ref{sec:components}.
Take its complete original
$G$-orbit and set $Q,J,H'$
as in
\eqref{eq:partial-normalizer}.
Equation~\eqref{eq:marked-frattini} shows that $N_G(P,e_P)$ is
transitive on the chosen component orbit, because $K_0$ fixes each
component. Its elements belong to $J$ and normalize the same $P$.
Choose the transporters and all transported component data as in
Section~\ref{sec:assembly}. The component propositions provide the full
action, common-$P$ identity
component, central balance,
and upper evaluations required
by Proposition~\ref{prop:assembly}.
The complementary layer
keeps its identity
bimodule. Hence there is a
splendid Rickard complex
\[
 R:\OO Ga\longrightarrow\OO Jc
\]
to a primitive block with
the original defect $P$.
The subgroup $J$ is proper:
if $Q$ were normal in $G$,
its intersection with each
component of the orbit would
be an abelian normal subgroup
of that quasisimple group,
hence central, contradicting
Lemma~\ref{lem:reduced-layer}.
Also $Z_0\leq Z(J)$, so
\[
 \kappa(J)\leq|J:Z_0|<\kappa(G).
\]
Because the component product
is normal in $G$, every
element of $N_G(P)$ normalizes
its intersection $Q$ with
$P$. Thus $N_G(P)\leq J$.
Apply induction
to the selected block of
$J$ and return using the
maximal companion of this
same $R$.

Finally suppose that all
components use case (iv).
Lemma~\ref{lem:upper-inertia}
makes every original
$PS_i$-block inertial.
Write $b_E$ for the covered block of $E(G)$. The equality
$ab_E=a$ gives $\Br_P(b_E)\ne0$. Its defect is $P\cap E(G)$,
and the unique block of $PE(G)$ covering $b_E$ is represented
by that idempotent. Its defect contains $P$ and has order at most
$|P\cap E(G)|\,|P:P\cap E(G)|=|P|$, so its defect is $P$.
Lemma~\ref{lem:product-inertia} therefore makes this
$PE(G)$-block inertial. The remaining $O_2(G)$ is already in $P$. To include
$Z_0$, put $K=PE(G)$, $I=K\cap Z_0$, let $d$ be its covered block,
and let $\lambda$ be the central character of the block
on $Z_0$. Put
\[
 e_\lambda=|Z_0|^{-1}\sum_{z\in Z_0}\lambda(z^{-1})z.
\]
The matching character on $I$ gives the interior-$P$
algebra isomorphism
\[
 \OO Kd\otimes_{\OO I}\OO Z_0e_\lambda
 \xrightarrow{\sim}\OO Kd,\qquad
 x\otimes ze_\lambda\longmapsto\lambda(z)x.
\]
Consequently
the block of the original
$PH$, $H=F^*(G)$, is inertial.
Lemma~\ref{lem:reduced-layer}
gives the relative maximality
of this same $P$. The strong
hypotheses of
Theorem~\ref{thm:strong-extension}
are now satisfied.

Take its primitive
basic Morita bimodule from
$\OO Ga$ to the corresponding
block of $J=N_G(D)$, with
$D=P\cap H$.
Proposition~\ref{prop:source-upgrade}
and Lemma~\ref{lem:integral-lift}
give a splendid Rickard complex $R$
with the original defect
graph. There is a component,
and its defect is noncentral;
therefore $D$ is not normal
in $G$, so $J<G$.
Again $Z_0\leq Z(J)$,
$N_G(P)\leq J$, and
$\kappa(J)<\kappa(G)$.
Apply induction and
Proposition~\ref{prop:companion}.

Every branch either ends
with the identity complex
or strictly lowers the
same positive integer
$\kappa$. Formula~\eqref{eq:return}
and its two evaluation
identities
complete every return.
All complexes remain
bounded, integral,
$2$-permutation and
projective on both sides,
with vertices below the
fixed defect diagonal.
This proves
\eqref{eq:main-left} and
\eqref{eq:main-right}.
\end{proof}

\begin{corollary}
Let $k$ be an algebraically closed
field of characteristic two.
Every block $kGa$ with abelian
defect group $P$ has a splendid
Rickard equivalence with its
Brauer correspondent in
$kN_G(P)$. In particular,
Brou\'e's abelian defect
group conjecture holds at
the prime two.
\end{corollary}
\begin{proof}
Choose a sufficiently large splitting
$2$-modular system with residue field
$k$, and lift the block idempotent
to $\OO G$. Reduce the complex and
both integral contractions of
Theorem~\ref{thm:main} modulo the
maximal ideal of $\OO$.
Reduction preserves the two identities,
the contractions, and the permutation
summands.
\end{proof}

\begingroup
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