# Morita equivalence classes of $2$-blocks with abelian defect groups of rank
$4$ 111This research was supported by the EPSRC grants no. EP/M015548/1 and
EP/T004606/1.

Charles W. Eaton222Department of Mathematics, University of Manchester,
Manchester M13 9PL. Email<EMAIL_ADDRESS>and Michael
Livesey333Department of Mathematics, University of Manchester, Manchester M13
9PL. Email<EMAIL_ADDRESS>
###### Abstract

We classify all $2$-blocks with abelian defect groups of rank $4$ up to Morita
equivalence. The classification holds for blocks over a suitable discrete
valuation ring as well as for those over an algebraically closed field. An
application is that Broué’s abelian defect group conjecture holds for all
blocks under consideration here.

Keywords: Morita equivalence; finite groups; block theory

##  1 Introduction

Let $p$ be a prime and $(K,\mathcal{O},k)$ be a $p$-modular system with $k$
algebraically closed, and let $P$ be a finite $p$-group. It is known by work
culminating in [19] that if $P$ is an abelian $2$-group, then there are only
finitely many Morita equivalence classes amongst all blocks of $\mathcal{O}G$
for all finite groups $G$ with defect groups isomorphic to $P$, i.e., that
Donovan’s conjecture holds for such $P$. This suggests the problem of
classifying the Morita equivalence classes of blocks that arise for given
abelian $2$-groups $P$. The Morita equivalence classes are already known
without the use of the classification when $P$ is abelian and $\mathop{\rm
Aut}\nolimits(P)$ is a $2$-group and when $P$ is a Klein four group. In the
former case every block must be nilpotent by [44] since $P$ controls fusion
for the block, and in the latter there are three Morita equivalence classes by
[36] (in fact these are the only source algebra equivalence classes by [13]).
In [20] the classification of finite simple groups was applied to describe the
$2$-blocks of quasisimple groups with abelian defect groups. This has been
used to classify the Morita equivalence classes of blocks with defect groups
isomorphic to $P$ in the following cases. When $P$ is elementary abelian of
order $16$ or $32$, there are $16$ or $34$ Morita equivalence classes
respectively by [18] and [3]. When $P$ is elementary abelian of order $64$,
there are $81$ Morita equivalence classes containing a principal block by [4].
Morita equivalence classes of blocks when $P$ is an abelian $2$-group of rank
at most three are determined in [22] and [51]. Morita equivalence classes have
been determined for a class of minimal nonabelian $2$-groups in [21]. Further
classifications are obtained when we place restrictions on the inertial
quotient of the blocks we consider, as in [42] and [5], where it is assumed
that the inertial quotient contains an element of maximal order, or is a
subgroup of a cyclic group of maximal order.

Here we determine the Morita equivalence classes of blocks when $p=2$ and $P$
is abelian of rank at most four. There are two main problems to overcome in
the reduction to quasisimple groups in order to apply [20]: the cases of a
normal subgroup of odd index and of a normal subgroup of index two. In the
former case we use Picard groups and crossed products to deduce possible
Morita equivalence classes for the block of the over group. In the latter we
use a combination of a method developed in [51] and one of our own.

Let $D\cong C_{2^{n_{1}}}\times C_{2^{n_{2}}}\times C_{2^{n_{3}}}\times
C_{2^{n_{4}}}$, where $n_{1},\ldots,n_{4}\geq 0$. The conjugacy classes of odd
order subgroups $E$ of $\mathop{\rm Aut}\nolimits(D)$ have representatives as
given in Table 1, and correspond to the possible inertial quotients for blocks
with defect group $D$ (where by inertial quotient, defined later, we consider
also the action on $D$). By [27, 5.2.3] we may write $D=[D,E]\times C_{D}(E)$.
To simplify notation we assume that labeling is chosen so that $[D,E]\cong
C_{2^{n_{1}}}\times\cdots\times C_{2^{n_{i}}}$ and $C_{D}(E)\cong
C_{2^{n_{i+1}}}\times\cdots\times C_{2^{n_{4}}}$ for some $i$. If $B$ is a
block with defect group $D$ and inertial quotient $E$, then we say that $B$ is
of type $E$, where we use the notation of Table 1 to distinguish between
isomorphic but non-conjugate subgroups. Note that the case $(C_{3})_{2}$
represents the simultaneous action of $C_{3}$ on $C_{2^{n_{1}}}\times
C_{2^{n_{2}}}$ and on $C_{2^{n_{3}}}\times C_{2^{n_{4}}}$, which for
$n_{1}=n_{2}=n_{3}=n_{4}$, also represents a subgroup of $C_{15}$.

Inertial quotient | Notes | Restrictions on $D$
---|---|---
$1$ |  | None
$(C_{3})_{1}$ |  $C_{D}(E)$ has rank $2$ | $n_{1}=n_{2}$
$(C_{3})_{2}$ | $C_{D}(E)=1$ |  $n_{1}=n_{2}$, $n_{3}=n_{4}$
$C_{3}\times C_{3}$ | $C_{D}(E)=1$ |  $n_{1}=n_{2}$, $n_{3}=n_{4}$
$C_{5}$ | $C_{D}(E)=1$ | $n_{1}=n_{2}=n_{3}=n_{4}$
$C_{15}$ | $C_{D}(E)=1$ | $n_{1}=n_{2}=n_{3}=n_{4}$
$C_{7}$ |  $C_{D}(E)$ cyclic | $n_{1}=n_{2}=n_{3}$
$C_{7}\rtimes C_{3}$ |  $C_{D}(E)$ cyclic | $n_{1}=n_{2}=n_{3}$
Table 1: Possible inertial quotients

We now give the classification. Throughout this paper, $G_{n}$ will denote the
nonabelian group $(C_{2^{n}}\times C_{2^{n}})\rtimes C_{3}$.

Following [46] a block is called _inertial_ if it is basic Morita equivalent
to a block with normal defect group.

######  Theorem 1.1.

main_theorem Let $G$ be a finite group and $B$ be a block of $\mathcal{O}G$
with defect group $D\cong C_{2^{n_{1}}}\times C_{2^{n_{2}}}\times
C_{2^{n_{3}}}\times C_{2^{n_{4}}}$ as above. Let $E$ be an inertial quotient
of $B$, with notation as in Table 1.

  1. (a)

singer If $n_{1}=n_{2}$, $n_{3}=n_{4}$ and $E$ acts as $1$, $(C_{3})_{2}$ or
$C_{5}$ then $B$ is inertial.

  2. (b)

    1. (i)

If $E$ acts as $(C_{3})_{1}$, then $B$ is Morita equivalent to one of
$\mathcal{O}(D\rtimes E)$ or $B_{0}(\mathcal{O}(A_{5}\times
C_{2^{n_{3}}}\times C_{2^{n_{4}}}))$.

    2. (ii)

If $E$ acts as $C_{7}$, then $B$ is Morita equivalent to $\mathcal{O}(D\rtimes
C_{7})$ or $B_{0}(\mathcal{O}(SL_{2}(8)\times C_{2^{n_{4}}}))$.

    3. (iii)

If $E$ acts as $C_{7}\rtimes C_{3}$, then $B$ is Morita equivalent to
$\mathcal{O}(D\rtimes(C_{7}\rtimes C_{3}))$, $B_{0}(\mathcal{O}(\mathop{\rm
Aut}\nolimits(SL_{2}(8))\times C_{2^{n_{4}}}))$ or
$B_{0}(\mathcal{O}(J_{1}\times C_{2^{n_{4}}}))$.

    4. (iv)

If $E$ acts as $C_{15}$, then either $B$ is inertial or $B$ is basic Morita
equivalent to $B_{0}(\mathcal{O}SL_{2}(16))$.

    5. (v)

Suppose $E$ acts as $C_{3}\times C_{3}$, with $n_{1}=n_{2}$ and $n_{3}=n_{4}$.
Then $B$ is Morita equivalent to one of $B_{0}(\mathcal{O}(G_{n_{1}}\times
G_{n_{3}}))$, $B_{0}(\mathcal{O}(A_{5}\times G_{n_{3}}))$,
$B_{0}(\mathcal{O}(A_{5}\times A_{5}))$ or a nonprincipal block of $D\rtimes
3_{+}^{1+2}$, where $Z(3_{+}^{1+2})$ acts trivially on $D$.

######  Remark 1.2.

(i) Nonprincipal blocks of $D\rtimes 3_{+}^{1+2}$ and $D\rtimes 3_{-}^{1+2}$
as above will be shown to be Morita equivalent in faithfulblocks.

(ii) Implicit in main_theorem is that for these defect groups Morita
equivalent blocks have the same fusion (i.e., the same inertial quotient).

(iii) Cases (a) and (b)(iv) of main_theorem have been proved in [42] and [5].

A consequence is the following Corollary that Broué’s abelian defect group
conjecture holds for the blocks under consideration:

######  Corollary 1.3.

broue Let $G$ be a finite group and $B$ be a $2$-block of $\mathcal{O}G$ with
abelian defect group $D$ of rank at most four. Let $b$ be the Brauer
correspondent of $B$ in $\mathcal{O}N_{G}(D)$. Then $B$ is derived equivalent
to $b$.

######  Remark 1.4.

Broué’s abelian defect group conjecture is often stated as requiring further a
splendid Rickard equivalence. Since main_theorem only provides a Morita
equivalence with no additional information on the bimodule affording the
equivalence, we make no claim to the existence of a splendid equivalence in
our result.

The structure of the paper is as follows. In Section 2 we briefly covered
necessary notation and general results. This includes: results on covering of
blocks of normal subgroups and related Morita equivalences; subpairs and
inertial quotients; a review of the different (stronger) versions of Morita
equivalence that we use, and their relationship with fusion; a summary of the
results of [20] that we use; and Picard groups. We prove a result on extending
Morita equivalences (in certain cases) from normal subgroups of index $2$ in
Section 4. This requires knowledge of some groups of perfect self-isometries,
which is the content of Section 3. In Section 5 we give results on extending
Morita equivalences from normal subgroups, which includes extensions using
crossed products where the index is odd, combined with techniques for index
$2$ from the previous section and the result developed in [51]. In Section 6
we present the proof of the main result. Finally in Section 7 we prove broue.

##  2 Notation and background

###  2.1 Morita equivalence

See [37, 2.8] for an introduction to Morita theory. It is not known whether or
not Morita equivalence of blocks of finite groups defined over $\mathcal{O}$
preserve the isomorphism type of the defect group (Morita equivalences defined
over $k$ are known not to necessarily preserve this by [26]). However, by [9]
Morita equivalence does preserve the isomorphism class of abelian defect
groups. It is also not known whether the inertial quotient (and its action)
are preserved. We will have to take some care in the proof of main_theorem to
check that this happens in our situation.

A Morita equivalence is _basic_ if it is induced by an endopermutation source
bimodule, and a block is _inertial_ if there is basic Morita equivalence with
the Brauer correspondent block of the normalizer of a defect group (see [45]
and [46]). A basic Morita equivalence preserves the isomorphism type of the
defect group and fusion.

We will also refer to source algebra equivalences (see [38, 6.4]). These occur
in the section on Picard groups. Also certain Morita equivalences that we use
are stated as source algebra equivalences (for example solvablequotient:lem),
but we do not use properties beyond that they are basic Morita equivalences.

Another point to mention here is the Bonnafé-Rouquier correspondence as in
[11]. Whilst correspondent blocks are only known to be Morita equivalences
(i.e, not basic), they are splendid Rickard equivalent, which implies they
have the same fusion.

###  2.2 Blocks and normal subgroups

We collect some background on blocks and normal subgroups that will be used
frequently and usually without further reference. A reference for this
material is [38, Section 6.8]

Let $G$ be a finite group and $B$ be a block of $\mathcal{O}G$ with defect
group $D$. We use ${\rm prj}(B)$ to denote the set of characters of projective
indecomposable $B$-modules. Let $N\lhd G$ and let $b$ be a block of
$\mathcal{O}N$ covered by $B$. We may choose a $G$-conjugate of $D$ such that
$D\cap N$ is a defect group for $b$. Write ${\rm Stab}_{G}(b)$ for the
stabiliser of $b$ under conjugation in $G$. There is a block of $\mathcal{O}G$
covering $b$ with a defect group $P$ such that $NP/N$ is a Sylow $p$-subgroup
of ${\rm Stab}_{G}(b)/N$.

If $[G:N]$ is a power of $p$, then $B$ is the unique block of $\mathcal{O}G$
covering $b$, and it follows that $b$ is $G$-stable if and only if $G=ND$. In
this case $B$ and $b$ share a block idempotent. We have the following crucial
fact in the case that $D$ is abelian:

######  Proposition 2.1.

Dabelian_inner:prop Let $G$ be a finite group and $B$ a block of
$\mathcal{O}G$ with abelian defect group $D$. Suppose that $N\lhd G$ with
$G=ND$ and that $b$ is a $G$-stable block of $\mathcal{O}N$ covered by $B$.
Then $D$ acts as inner automorphisms on $b$. Further,

(i) every irreducible character of $b$ is $G$-stable and extends to $p$
distinct irreducible characters of $B$,

(ii) induction gives a bijection between the projective indecomposable modules
for $b$ and those for $B$.

###### Proof.

The assertion that $D$ acts as inner automorphisms follows from [37, Corollary
6.16.3] or [24, Proposition 3.1]. The rest is [22, Proposition 2.6(i), (ii)].
∎

Note that the version of Proposition LABEL:Dabelian_inner:prop over $k$ is
also found within [30].

Recall that a block $B$ is _quasiprimitive_ if every block of every normal
subgroup covered by $B$ is $G$-stable. In particular $B$ covers a unique block
for each normal subgroup.

We will make frequent use of the following, especially in the case where the
quotient group is a subgroup of the outer automorphism group of a quasisimple
group.

######  Lemma 2.2 (Lemma 2.4 of [3]).

solvablequotient:lem Let $G$ be a finite group and let $N\lhd G$ with $G/N$
solvable. Let $b$ be a $G$-stable block of $\mathcal{O}N$ and let $B$ be a
quasiprimitive block of $\mathcal{O}G$ covering $b$ with defect group $D$.
Then $DN/N$ is a Sylow $p$-subgroup of $G/N$.

The normal subgroup $G[b]$ of $G$ is defined to be the group of elements of
$G$ acting as inner automorphisms on $b\otimes_{\mathcal{O}}k$ (see [33]).

######  Proposition 2.3.

innerautos Let $G$ be a finite group and $B$ a block of $\mathcal{O}G$ with
defect group $D$. Let $N\lhd G$ with $D\leq N$ and suppose that $B$ covers a
$G$-stable block $b$ of $\mathcal{O}N$. Let $B^{\prime}$ be a block of
$\mathcal{O}G[b]$ covered by $B$. Then

(i) $b$ is source algebra equivalent to $B^{\prime}$, and in particular has
isomorphic inertial quotient;

(ii) $B$ is the unique block of $\mathcal{O}G$ covering $B^{\prime}$.

###### Proof.

Part (i) is [28, 2.2], noting that a source algebra equivalence over $k$
implies one over $\mathcal{O}$ by [44, 7.8]. Part (ii) follows from [15, 3.5].
∎

###  2.3 Subpairs, fusion and inertial quotients

Let $G$ be a finite group and $B$ be a block of $\mathcal{O}G$ with defect
group $D$. For convenience in stating definitions we assume that $D$ is
abelian. A $B$-subpair is a pair $(Q,B_{Q})$ where $Q$ is a $p$-subgroup of
$G$ and $B_{Q}$ is a block of $C_{G}(Q)$ with Brauer correspondent $B$. Such a
block $B_{Q}$ only exists when $Q$ is $G$-conjugate to a subgroup of $D$. For
a fixed $B$-subpair $(D,B_{D})$, for each $Q\leq D$ there is an unique
$B$-subpair $(Q,B_{Q})$ such that $B_{Q}$ is the Brauer correspondent of
$B_{D}$.

The $B$-subpairs define a fusion system
$\mathcal{F}=\mathcal{F}_{B}(D,B_{D})$, often called the _Frobenius category_
of the $B$ (see for example [7] or [37]). The _inertial quotient_ of $B$ is
$E=N_{G}(D,B_{D})/DC_{G}(D)$, together with the action of $E$ on $D$. Since
$D$ is abelian, $\mathcal{F}$ is determined by $E$. Consequently, every block
with abelian defect groups and trivial inertial quotient is nilpotent (we will
use this fact throughout this paper without further reference).

The following is well-known.

######  Proposition 2.4.

defect_group_factor:prop Let $B$ be a block of $\mathcal{O}G$ for a finite
group $G$ with abelian defect group $D$. Let $(D,B_{D})$ be a $B$-subpair.
Then $D=[D,N_{G}(D,B_{D})]\times C_{D}(N_{G}(D,B_{D}))$.

Suppose there is $N\lhd G$ such that $G=DN$, then $[D,N_{G}(D,B_{D})]\leq N$.

###### Proof.

The factorisation of $D$ follows from [27, Theorem 5.2.3]. For the second
part, note that $[D,N_{G}(D,B_{D})]=[D,DN_{N}(D,B_{D})]\leq N$. ∎

Most of the following is Lemma 5.3.5 of [41].

######  Proposition 2.5.

index_p_inertial:prop Let $B$ be a block of $\mathcal{O}G$ for a finite group
$G$ with abelian defect group $D$ and let $(D,B_{D})$ be a $B$-subpair. Let
$N\lhd G$ and suppose that $B$ covers a $G$-stable block $b$ of
$\mathcal{O}N$. Write $Q=N\cap D$, a defect group for $b$. Let $b_{Q}$ be a
block of $\mathcal{O}C_{N}(Q)$ covered by $B_{Q}$. Then $b_{Q}$ is
$C_{G}(Q)$-stable and $(Q,b_{Q})$ is a $b$-subpair. The blocks $B$ and $b$
have isomorphic inertial quotients and
$[N_{G}(D,B_{D}),D]=[N_{N}(Q,b_{Q}),Q]$.

###### Proof.

That $(Q,b_{Q})$ is a $b$-subpair forms part of the proof of the main theorem
of [30]. We note that due to the correspondence between $\mathcal{O}$-blocks
and $k$-blocks, the result may be proved for $k$-blocks, which is the setting
for [30]. Note also that this part of the proof of the main result of [30]
does not require that $G$ is a split extension of $N$.

We have $C_{G}(Q)=DC_{N}(Q)$ and $N_{G}(Q)=DN_{N}(Q)$. Hence
$N_{G}(Q,b_{Q})=DN_{N}(Q,b_{Q})$. Also, since $b_{Q}$ is $C_{G}(Q)$-stable, it
is the unique block of $C_{N}(Q)$ covered by $B_{Q}$ (and $B_{Q}$ is the
unique block of $C_{G}(Q)$ covering $b_{Q}$ since $C_{N}(Q)$ has index a power
of $p$ in $C_{G}(Q)$). Hence $N_{G}(Q,B_{Q})=N_{G}(Q,b_{Q})$. By definition we
have $N_{G}(D,B_{D})\leq N_{G}(Q,B_{Q})$.

Since $N_{G}(D,B_{D})$ controls fusion in $D$, by [1, Proposition 4.24] we
have $N_{G}(Q,B_{Q})=N_{G}(D,B_{D})C_{G}(Q)$. It follows that

$N_{N}(Q,b_{Q})/C_{N}(Q)\cong N_{G}(D,B_{D})/(C_{G}(Q)\cap N_{G}(D,B_{D})).$

Noting that $D=QC_{D}(N_{G}(D,B_{D}))$, we have
$C_{G}(D)=C_{N_{G}(D,B_{D})}(D)=C_{N_{G}(D,B_{D})}(Q)$, so $B$ and $b$ have
isomorphic inertial quotients.

Further

$[N_{G}(D,B_{D}),D]=[N_{G}(D,B_{D}),QC_{D}(N_{G}(D,B_{D}))]=[N_{G}(D,B_{D}),Q]$
$=[N_{G}(D,B_{D})C_{G}(Q),Q]=[N_{G}(Q,B_{Q}),Q]=[N_{G}(Q,b_{Q}),Q]=[N_{N}(Q,b_{Q}),Q]$

as required. ∎

The next result is essentially extracted from the proof of [51, Lemma 6.3],
and allows us to compare inertial quotients of blocks with those of blocks of
normal subgroups of $p^{\prime}$ index. We include the proof here for
convenience.

######  Lemma 2.6.

inertial_quotient_normal Let $B$ be a block of $\mathcal{O}G$ for a finite
group $G$, and let $b$ be a $G$-stable block of $\mathcal{O}N$ for $N\lhd G$
with $[G:N]=r$ a prime different to $p$. Let $E_{B}$, $E_{b}$ be the inertial
quotient of $B$, $b$ respectively. Let $D$ be an abelian defect group for $B$
(and for $b$). Then either $E_{b}$ is isomorphic to a subgroup of $E_{B}$ or
$E_{B}$ is isomorphic to a subgroup of $E_{b}$.

###### Proof.

We may take a $B$-subpair $(D,B_{D})$ and a $b$-subpair $(D,b_{D})$ with
$B_{D}$ covering $b_{D}$. If $C_{G}(D)\leq N$, then $B_{D}=b_{D}$ and
$N_{N}(D,b_{D})\leq N_{G}(D,B_{D})$, so the result is immediate in this case.
Hence suppose $C_{G}(D)$ is not contained in $N$, i.e,, $G=NC_{G}(D)$.

If $C_{G}(D)\leq N_{G}(D,b_{D})$, then $N_{G}(D,B_{D})\leq
N_{G}(D,b_{D})=N_{N}(D,b_{D})C_{G}(D)$. Hence $N_{G}(D,B_{D})/C_{G}(D)\leq
N_{G}(D,b_{D})/C_{G}(D)\cong N_{N}(D,b_{D})/C_{N}(D)$.

Suppose that $C_{G}(D)$ does not stabilize $b_{D}$. Since
$[C_{G}(D):C_{N}(D)]=r$ is a prime and $C_{N}(D)\leq C_{G}(D)\cap
N_{G}(D,b_{D})<C_{G}(D)$, we have $C_{N}(D)=C_{G}(D)\cap N_{G}(D,b_{D})$. Let
$T$ be a transversal of $C_{N}(D)$ in $C_{G}(D)$. Then $\\{b_{D}^{t}:t\in
T\\}$ is the set $r$ distinct conjugate blocks of $C_{N}(D)$ covered by
$B_{D}$ and $B_{D}$ is the unique block of $C_{G}(D)$ covering $b_{D}$. Hence
$N_{G}(D,B_{D})=N_{G}(D,b_{D})C_{G}(D)$. Then $N_{G}(D,B_{D})/C_{G}(D)\cong
N_{G}(D,b_{D})/C_{N}(D)$ and the result follows. ∎

###  2.4 Blocks of quasisimple groups

We extract the results of [20] necessary for this paper. Recall that a block
$B$ of a group $G$ is nilpotent covered if there is $H$ with $G\lhd H$ and a
nilpotent block of $H$ covering $B$. Properties of such blocks are considered
in [46].

######  Proposition 2.7 ([20]).

qsclassification Let $p=2$, and let $B$ be a block of $\mathcal{O}G$ for a
quasisimple group $G$ with abelian defect group $D$ of rank at most $4$. Then
one or more of the following occurs:

(i) $G\cong A_{5}$, $SL_{2}(8)$, $SL_{2}(16)$, $J_{1}$ or ${}^{2}G_{2}(q)$,
where $q=3^{2m+1}$ for some $m\in\mathbb{N}$, and $B$ is the principal block;

(ii) $G\cong Co_{3}$ and $B$ is the unique non-principal $2$-block of defect
$3$;

(iii) $B$ has inertial quotient of type $(C_{3})_{1}$ and is Morita equivalent
to a block $C$ of $\mathcal{O}L$ where $L=L_{0}\times L_{1}\leq G$ such that
$L_{0}$ is abelian and the block of $\mathcal{O}L_{1}$ covered by $C$ has
Klein four defect groups;

(iv) $B$ is nilpotent covered.

###### Proof.

This follows from Theorem 6.1 of [20] and its proof. The only point to address
is the inertial quotient in case (iii). Case (iii) arises in two ways. The
first is where $D\cong C_{2}\times C_{2}$, in which case either $B$ is
nilpotent (and so is covered by case (iv)), or has inertial quotient $C_{3}$.
The second way this arises is as in case (v) of [20, Proposition 5.3]. Here
the given Morita equivalence is given by the Bonnafé-Rouquier correspondence
as in [11]. Now the corresponding blocks in [11] are equivalent by a splendid
Rickard equivalence and so have the same fusion. Hence $B$ has the same
inertial quotient as $C$, which is of type $(C_{3})_{1}$. ∎

######  Corollary 2.8.

Gn_inertial Every $2$-block of a quasisimple group that is Morita equivalent
to $G_{n}$, for $n\in\mathbb{N}$, is inertial.

###### Proof.

For $n\geq 2$ we see from qsclassification that blocks of quasisimple groups
with defect groups $(C_{2^{n}})^{2}$ are nilpotent covered, and so inertial.
For $n=1$, by [13] every block that is Morita equivalent to $\mathcal{O}A_{4}$
is source algebra equivalent to $\mathcal{O}A_{4}$, and so inertial. ∎

###  2.5 Picard groups

Let $G$ be a finite group and $B$ be a block of $\mathcal{O}G$ with defect
group $D$. The Picard group $\mathop{\rm Pic}\nolimits(B)$ of $B$ has elements
the isomorphism classes of $B$-$B$-bimodules inducing $\mathcal{O}$-linear
Morita auto-equivalences of $B$. For $B$-$B$-bimodules $M$ and $N$, the group
multiplication is given by $M\otimes_{B}N$. For background and definitions we
follow [10].

We will use knowledge of $\mathop{\rm Pic}\nolimits(B)$ to refine Külshammer’s
analysis in [34] of the situation of a normal subgroup containing the defect
groups of a block. This involves the study of crossed products of a basic
algebra with a $p^{\prime}$-group, which in turn uses the outer automorphism
group of the basic algebra, a group that embeds into the Picard group. This
will be essential in reduction steps in our classification of Morita
equivalence classes of blocks.

Write $\mathcal{T}(B)$ for the subgroup of $\mathop{\rm Pic}\nolimits(B)$
consisting of bimodules with trivial source and $\mathcal{L}(B)$ for the
subgroup consisting of linear source modules.

$\mathcal{T}(B)$ and $\mathcal{L}(B)$ are described in [10], and we summarise
the relevant notation and results here. Let
$\mathcal{F}=\mathcal{F}_{B}(D,B_{D})$ be the fusion system for $B$ on $D$,
defined using a $B$-subpair $(D,B_{D})$, and let $E=N_{G}(D,B_{D})/DC_{G}(D)$
be the inertial quotient. Write $\mathop{\rm Aut}\nolimits(D,\mathcal{F})$ for
the subgroup of $\mathop{\rm Aut}\nolimits(D)$ of automorphisms stabilizing
$\mathcal{F}$ and $\mathop{\rm Out}\nolimits(D,\mathcal{F})=\mathop{\rm
Aut}\nolimits(D,\mathcal{F})/\mathop{\rm Aut}\nolimits_{\mathcal{F}}(D)$. By
[25], if $D$ is abelian, then $\mathop{\rm Out}\nolimits(D,\mathcal{F})\cong
N_{\mathop{\rm Aut}\nolimits(D)}(E)/E\cong\mathop{\rm Out}\nolimits(D\rtimes
E)$.

Let $A$ be a source algebra for $B$. Then $A$ is a $D$-algebra, so we have an
embedding of $D$ into $A$ and we may consider the fixed points $A^{D}$ under
the action of $D$. Write $\mathop{\rm Aut}\nolimits_{D}(A)$ for the group of
$\mathcal{O}$-algebra automorphisms of $A$ fixing each element of the image of
$D$ in $A$, and $\mathop{\rm Out}\nolimits_{D}(A)$ for the quotient of
$\mathop{\rm Aut}\nolimits_{D}(A)$ by the subgroup of automorphisms given by
conjugation by elements of $(A^{D})^{\times}$. By [44, 14.9] $\mathop{\rm
Out}\nolimits_{D}(A)$ is isomorphic to a subgroup of $\mathop{\rm
Hom}\nolimits(E,k^{\times})$.

By [10, Theorem 1.1] we have exact sequences

$\begin{array}[]{lllllll}1&\rightarrow&\mathop{\rm
Out}\nolimits_{D}(A)&\rightarrow&\mathcal{T}(B)&\rightarrow&\mathop{\rm
Out}\nolimits(D,\mathcal{F}),\\\ 1&\rightarrow&\mathop{\rm
Out}\nolimits_{D}(A)&\rightarrow&\mathcal{L}(B)&\rightarrow&\mathop{\rm
Hom}\nolimits(D/\mathfrak{foc}(D),\mathcal{O}^{\times})\rtimes\mathop{\rm
Out}\nolimits(D,\mathcal{F}),\\\ \end{array}$ (1)

where $\mathfrak{foc}(D)$ is the focal subgroup of $D$ with respect to
$\mathcal{F}$, generated by the elements $\varphi(x)x^{-1}$ for $x\in D$ and
$\varphi\in\mathop{\rm Hom}\nolimits_{\mathcal{F}}(\langle x\rangle,D)$. When
$D$ is abelian we have $\mathfrak{foc}(D)=[N_{G}(D,B_{D}),D]$, so that
$\mathop{\rm Hom}\nolimits(D/\mathfrak{foc}(D),\mathcal{O}^{\times})\cong
C_{D}(N_{G}(D,B_{D}))$ (see defect_group_factor:prop).

We record for later use that by [47], if $P$ is a $p$-group, then

${\mathop{\rm
Pic}\nolimits}(\mathcal{O}P)=\mathcal{L}(\mathcal{O}P)\cong{\mathop{\rm
Aut}\nolimits}(\mathcal{O}P)\cong{\mathop{\rm
Hom}\nolimits}(P,\mathcal{O}^{\times})\rtimes{\mathop{\rm Aut}\nolimits}(P).$

We gather together here the results regarding Picard groups that we will use
later on.

######  Proposition 2.9.

pic_list Let $P$ and $Q$ be abelian $2$-groups and
$n,n_{1},n_{2}\in\mathbb{N}$.

  1. (i)

$\mathop{\rm Pic}\nolimits(\mathcal{O}(G_{n}\times P))\cong
S_{3}\times(P\rtimes\mathop{\rm Aut}\nolimits(P))$. The subgroup of
$\mathop{\rm Pic}\nolimits(\mathcal{O}(G_{n}\times P))$ given by those self-
equivalences fixing the projective indecomposable modules is isomorphic to
$P\rtimes\mathop{\rm Aut}\nolimits(P)$.

  2. (ii)

$\mathop{\rm Pic}\nolimits(B_{0}(\mathcal{O}(A_{5}\times P)))\cong
C_{2}\times(P\rtimes\mathop{\rm Aut}\nolimits(P))$.

  3. (iii)

$\mathop{\rm Pic}\nolimits(B_{0}(\mathcal{O}(G_{n}\times A_{5})))\cong
S_{3}\times C_{2}$.

  4. (iv)

$\mathop{\rm Pic}\nolimits(\mathcal{O}(G_{n_{1}}\times
G_{n_{2}}))=\mathcal{T}(\mathcal{O}(G_{n_{1}}\times G_{n_{2}}))\cong$

    1. (a)

$S_{3}\wr S_{2}$ if $n_{1}=n_{2}$,

    2. (b)

$S_{3}\times S_{3}$ if $n_{1}\neq n_{2}$.

  5. (v)

If $Q\cong(C_{2^{n}})^{3}$, then $\mathop{\rm
Pic}\nolimits(\mathcal{O}((Q\rtimes C_{7})\times P))\cong(C_{7}\rtimes
C_{3})\times\mathop{\rm Aut}\nolimits(P)$.

###### Proof.

Throughout this proof, we denote by $D$ a defect group of the block in
question.

(i) The Picard group is described in [25, Theorem 1.1]. The $S_{3}$ factor
consists of (bimodules inducing) Morita equivalences corresponding to elements
of $C_{3}\rtimes\mathop{\rm Out}\nolimits(G_{n})$, where the $C_{3}$ is
generated by multiplying by a non-trivial linear character of $G_{n}$. The
$(P\rtimes\mathop{\rm Aut}\nolimits(P))$ factor is generated by equivalences
given by multiplication by a (linear) character and by automorphisms of $P$.
Since the projective indecomposable modules correspond in this case to the
irreducible characters with $D$ in their kernel, the remainder follows.

(ii)-(iv) are from [25, Theorem 1.1].

(v) Suppose $G=(Q\rtimes C_{7})\times P)$ where $Q\cong(C_{2^{n}})^{3}$. Then
$G=D\rtimes E$, where $E$ is the inertial quotient (note that $\mathcal{O}G$
has a unique block). By [39] $\mathop{\rm
Pic}\nolimits(\mathcal{O}G)=\mathcal{L}(\mathcal{O}G)$. By [25, Lemma 2.1]
$\mathop{\rm Out}\nolimits(D,\mathcal{F})\cong N_{\mathop{\rm
Aut}\nolimits(D)}(E)/E\cong\mathop{\rm Out}\nolimits(G)\cong
C_{3}\times(P\rtimes\mathop{\rm Aut}\nolimits(P))$. Also it follows from [25,
Lemma 2.2] that $\mathop{\rm Out}\nolimits_{D}(A)\cong E$, where $A$ is a
source algebra for the unique block $\mathcal{O}G$. We have $\mathop{\rm
Hom}\nolimits(D/\mathfrak{foc}(D),\mathcal{O}^{\times})\cong
C_{D}(N_{G}(D,B_{D}))=P$. The result follows from the description of
$\mathcal{L}(B)$ in (1) above, as it is clear that the elements of
$\mathop{\rm Out}\nolimits_{D}(A)$ cannot commute with the elements of
$\mathop{\rm Out}\nolimits(D,\mathcal{F})$ obtained as automorphisms of
$Q\rtimes C_{7}$. ∎

######  Remark 2.10.

In case (v), whilst a Sylow $3$-subgroup of the Picard group for the block in
question occurs as a conjugate of $\mathop{\rm Out}\nolimits(D,\mathcal{F})$,
this is not known to be the case for every block Morita equivalent to it. In
other words, it is theoretically possible for the Picard group of a Morita
equivalent block $C$ to have a subgroup $C_{7}\rtimes C_{3}$, but for
$\mathcal{T}(C)\not\cong C_{7}\rtimes C_{3}$. We will have to beware of this
inconvenience in Section 5.

##  3 Preliminaries on perfect isometries

We require a method for comparing the principal blocks of
$\mathcal{O}(A_{4}\times P)$ and $\mathcal{O}(A_{5}\times P)$ with those of
$\mathcal{O}(A_{4}\times Q)$ and $\mathcal{O}(A_{5}\times Q)$ respectively
when $Q$ is a subgroup of an abelian $2$-group $P$. We will do this in Section
4, but first require an analysis of their perfect self-isometries, which is
the content of this section. For a block $B$, write ${{\rm Perf}}(B)$ for the
group of perfect self-isometries of $B$, under composition. The results of
this section are an extension of those of part of Section 2 of [22].

Note that every perfect isometry $I$ between blocks $B_{1}$ and $B_{2}$ gives
rise to a bijection of character idempotents and so to a $K$-algebra
isomorphism between $Z(KB_{1})$ and $Z(KB_{2})$, and that by [12] this induces
an $\mathcal{O}$-algebra isomorphism $\phi_{I}:Z(B_{1})\rightarrow Z(B_{2})$.

By Dabelian_inner:prop, for a block with abelian defect groups, every
irreducible character in a block of a normal subgroup of index $p$ covered by
$B$ is $G$-stable and extends to $G$. The following Proposition tells us that
these extensions behave well with respect to perfect isometries.

######  Proposition 3.1.

prop:isomcent For $i=1,2$ let $G_{i}$ be a finite group and $N_{i}\lhd G_{i}$
with index $p$. Let $B_{i}$ be a block of $\mathcal{O}G_{i}$ with abelian
defect group $D$ and let $b_{i}$ be a $G_{i}$-stable block of
$\mathcal{O}N_{i}$ covered by $B_{i}$. For each $\chi\in\mathop{\rm
Irr}\nolimits(b_{1})$ write $\mathop{\rm
Irr}\nolimits(B_{1},\chi)=\\{\chi_{1},\ldots,\chi_{p}\\}$.

Suppose $I:\mathbb{Z}\mathop{\rm Irr}\nolimits(B_{1})\to\mathbb{Z}\mathop{\rm
Irr}\nolimits(B_{2})$ is a perfect isometry such that for each
$\chi\in\mathop{\rm Irr}\nolimits(b_{1})$ there is $\psi\in\mathop{\rm
Irr}\nolimits(b_{2})$ and $\epsilon_{\chi}\in\\{\pm 1\\}$ such that
$I(\chi_{i})=\epsilon_{\chi}\psi_{i}$ for $i=1,\ldots,p$ where $\mathop{\rm
Irr}\nolimits(B_{2},\psi)=\\{\psi_{1},\ldots,\psi_{p}\\}$. Then the isometry
$I_{N_{1},N_{2}}:\mathbb{Z}\mathop{\rm
Irr}\nolimits(b_{1})\to\mathbb{Z}\mathop{\rm Irr}\nolimits(b_{2})$ defined by
$I_{N_{1},N_{2}}(\chi):=\epsilon_{\chi}\psi$ is perfect and further
$\phi_{I_{N_{1},N_{2}}}=\phi_{I}|_{Z(b_{1})}$.

######  Remark 3.2.

We may restrict $\phi_{I}$ to $Z(b_{1})$ since, as by Dabelian_inner:prop
$B_{i}$ acts as inner automorphisms on $b_{i}$ and so $Z(b_{i})\subseteq
Z(B_{i})$.

###### Proof.

This is Proposition 2.6 and Lemma 2.7 of [22]. ∎

######  Lemma 3.3.

lem:perfP Let $P$ be an abelian finite $p$-group. Then ${{\rm
Perf}}(\mathcal{O}P)\cong{\mathop{\rm Aut}\nolimits}(\mathcal{O}P)\times
C_{2}$.

###### Proof.

Since there is only one indecomposable projective module for $\mathcal{O}P$,
every perfect self-isometry of $\mathcal{O}P$ must have all positive or all
negative signs. Now every perfect self-isometry induces a permutation of
$\mathop{\rm Irr}\nolimits(P)$, which induces an automorphism of $\mathop{\rm
Aut}\nolimits(Z(\mathcal{O}P))\cong\mathop{\rm Aut}\nolimits(\mathcal{O}P)$,
and the result follows. ∎

Now consider the character table of $A_{4}$. Let $\omega$ be a primitive $3$rd
root of unity. We set up the labelling of characters:

For the rest of the section we assume $p=2$.

######  Proposition 3.4.

prop:self_A4 The perfect self-isometries of $\mathcal{O}A_{4}$ are precisely
the isometries of the form:

$\displaystyle I_{\sigma,\epsilon}:\mathbb{Z}\mathop{\rm Irr}\nolimits(A_{4})$
$\displaystyle\to\mathbb{Z}\mathop{\rm Irr}\nolimits(A_{4})$
$\displaystyle\chi_{j}$
$\displaystyle\mapsto\epsilon\delta_{j}\delta_{\sigma(j)}\chi_{\sigma(j)}$

for $1\leq j\leq 4$, where $\sigma\in S_{4}$, $\epsilon\in\\{\pm 1\\}$ and
$\delta_{1}=\delta_{2}=\delta_{3}=-\delta_{4}=1$. Hence ${{\rm
Perf}}(B_{0}(\mathcal{O}A_{5})\cong{{\rm Perf}}(\mathcal{O}A_{4})\cong
C_{2}\times S_{4}$.

###### Proof.

This is [22, Proposition 2.8] together with the observation that by [12, A1.3]
$\mathcal{O}A_{4}$ and $B_{0}(\mathcal{O}A_{5})$ are perfectly isometric. ∎

######  Theorem 3.5.

thm:PA4 Let $P$ be a finite abelian $2$-group. Every perfect self-isometry of
$\mathcal{O}(P\times A_{4})$ is of the form $(J,I_{\sigma,\epsilon})$, where
$J$ is a perfect isometry of $\mathcal{O}P$ induced by an
$\mathcal{O}$-algebra automorphism, $\sigma\in S_{4}$ and $\epsilon\in\\{\pm
1\\}$.

###### Proof.

We proceed as in the proof of [22, Theorem 2.11]. The set of projective
indecomposable characters (characters of projective indecomposable modules) is

$\displaystyle{\rm prj}(\mathcal{O}(P\times
A_{4}))=\\{\chi_{P_{1}},\chi_{P_{2}},\chi_{P_{3}}\\},\text{ where
}\chi_{P_{j}}:=\left(\sum_{\theta\in\mathop{\rm
Irr}\nolimits(P)}\theta\right)\otimes\left(\chi_{j}+\chi_{4}\right).$

Let $I$ be a perfect self-isometry of $\mathcal{O}(P\times A_{4})$. Each
$I(\chi_{P_{i}})$ is an integer linear combination of projective
indecomposable characters. By counting constituents we see that

$\displaystyle
I(\chi_{P_{u}})=\pm\chi_{P_{1}},\pm\chi_{P_{2}},\pm\chi_{P_{3}},\pm(\chi_{P_{1}}-\chi_{P_{2}}),\pm(\chi_{P_{1}}-\chi_{P_{3}})\text{
or }\pm(\chi_{P_{2}}-\chi_{P_{3}}),$ (2)

for $1\leq u\leq 3$. Consider the set

$\displaystyle
X_{m}:=\left\\{j:\left\langle\zeta\otimes\chi_{j},I\left(\left(\sum_{\theta\in\mathop{\rm
Irr}\nolimits(P)}\theta\right)\otimes\chi_{m}\right)\right\rangle\neq 0,\text{
for some }\zeta\in\mathop{\rm Irr}\nolimits(P)\right\\},$

for $1\leq m\leq 4$. By (2) we have shown that $|X_{m}|=1$ or $2$ for every
$1\leq m\leq 4$. If $|X_{1}|=2$, then by considering (2) for $u=1$ we see that
$X_{4}=X_{1}$. Similarly by considering $I(\chi_{P_{2}})$, we get that
$X_{2}=X_{4}$. This is now a contradiction as then

$\displaystyle I\left(\left(\sum_{\theta\in\mathop{\rm
Irr}\nolimits(P)}\theta\right)\otimes\left(\chi_{1}+\chi_{2}+\chi_{4}\right)\right)$

has at most $2|P|$ constituents with non-zero multiplicity. Therefore
$|X_{1}|=1$ and so by considering $I(\chi_{P_{1}})$ we get that $|X_{4}|=1$
and then by considering $I(\chi_{P_{2}})$ and $I(\chi_{P_{3}})$ we get that
$|X_{2}|=|X_{3}|=1$. Moreover, $X_{1},X_{2},X_{3},X_{4}$ must all be disjoint.
By composing $I$ with the perfect isometry $({\rm Id}_{\mathbb{Z}\mathop{\rm
Irr}\nolimits(P)},I_{\sigma,1})$, for some appropriately chosen $\sigma\in
S_{4}$, we may assume $X_{m}=\\{m\\}$ for all $1\leq m\leq 4$. Therefore
$I(\chi_{P_{u}})=\pm\chi_{P_{u}}$ for $1\leq u\leq 3$ and by considering

$\displaystyle I\left(\left(\sum_{\theta\in\mathop{\rm
Irr}\nolimits(P)}\theta\right)\otimes\chi_{4}\right),$

we see that in fact all these signs are the same and we may assume, after
possibly composing $I$ with $({\rm Id}_{\mathbb{Z}\mathop{\rm
Irr}\nolimits(P)},I_{1,-1})$, that

$\displaystyle I\left(\left(\sum_{\theta\in\mathop{\rm
Irr}\nolimits(P)}\theta\right)\otimes\chi_{m}\right)=\left(\sum_{\theta\in\mathop{\rm
Irr}\nolimits(P)}\theta\right)\otimes\chi_{m},$

for $1\leq m\leq 4$. Next we note that

$\displaystyle\frac{1}{12}\theta\otimes(\chi_{1}+\chi_{2}+\chi_{3}+3\chi_{4})\in{\rm
CF}(P\times A_{4},\mathcal{O}(P\times A_{4}),\mathcal{O}),$

for each $\theta\in\mathop{\rm Irr}\nolimits(P)$. As $3$ is invertible in
$\mathcal{O}$, this implies

$\displaystyle\theta\otimes\left(\sum_{m=1}^{4}\delta_{m}\chi_{m}\right)\in
4{\rm CF}(P\times A_{4},\mathcal{O}(P\times A_{4}),\mathcal{O}),$

where $\delta_{m}$ is defined as in prop:self_A4, and so

$\displaystyle
I\left(\theta\otimes\left(\sum_{m=1}^{4}\delta_{m}\chi_{m}\right)\right)\in
4{\rm CF}(P\times A_{4},\mathcal{O}(P\times A_{4}),\mathcal{O}),$ (3)

for each $\theta\in\mathop{\rm Irr}\nolimits(P)$. Fixing for now
$\theta\in\mathop{\rm Irr}\nolimits(P)$, define
$\theta_{m}\otimes\chi_{m}:=I(\theta\otimes\chi_{m})$, for $1\leq m\leq 4$.
Let $x\in P$. Evaluating (3) at $(x,1)$, $(x,(123))$ and $(x,(132))$, gives

$\displaystyle\theta_{1}(x)+\theta_{2}(x)+\theta_{3}(x)+\theta_{4}(x)$
$\displaystyle\in 4\mathcal{O},$ (4)
$\displaystyle\theta_{1}(x)+\omega\theta_{2}(x)+\omega^{2}\theta_{3}(x)$
$\displaystyle\in 4\mathcal{O},$ (5)
$\displaystyle\theta_{1}(x)+\omega^{2}\theta_{2}(x)+\omega\theta_{3}(x)$
$\displaystyle\in 4\mathcal{O}.$ (6)

Proceeding as in the proof of [22, Theorem 2.11] we have
$\theta_{1}(x)=\theta_{2}(x)=\theta_{3}(x)=\theta_{4}(x)$ for all $x\in P$. In
other words $\theta_{1}=\theta_{2}=\theta_{3}=\theta_{4}$.

We have shown that we may assume $I$ is of the form

$\displaystyle I(\theta\otimes\chi_{m})\mapsto J(\theta)\otimes\chi_{m}$

for all $\theta\in\mathop{\rm Irr}\nolimits(P)$, where $J$ is a permutation of
$\mathop{\rm Irr}\nolimits(P)$. In particular the $\mathcal{O}$-algebra
automorphism of $Z(\mathcal{O}(P\times A_{4})$ induced by $I$ leaves
$\mathcal{O}P$ invariant. Therefore the permutation $J$ of $\mathop{\rm
Irr}\nolimits(P)$ must induce an automorphism of $\mathcal{O}P$ and the
theorem is proved. ∎

We need two technical lemmas before we continue. Set

$\displaystyle A$ $\displaystyle=k[X,Y,Z]/(X^{2},Y^{2},Z^{2},XY,XZ,YZ)\cong
Z(kA_{4}),$ $\displaystyle A_{(m_{1},\dots,m_{s})}$
$\displaystyle=k[X_{1},\dots,X_{s}]/(X_{1}^{2^{m_{1}}},\dots,X_{s}^{2^{m_{s}}})\cong
k(C_{2^{m_{1}}}\times\cdots\times C_{2^{m_{s}}}),$

for $s\in\mathbb{N}$, $(m_{1},\dots,m_{s})\in\mathbb{N}^{s}$.

######  Lemma 3.6.

Let $s,t\in\mathbb{N}$, $(m_{1},\dots,m_{s})\in\mathbb{N}^{s}$, with
$m_{1}\geq\dots\geq m_{s}$ and $(n_{1},\dots,n_{t})\in\mathbb{N}^{t}$, with
$n_{1}\geq\dots\geq n_{t}$. If $A_{(m_{1},\dots,m_{s})}\otimes_{k}A\cong
A_{(n_{1},\dots,n_{t})}\otimes_{k}A$, then $s=t$ and
$m_{1}=n_{1},\dots,m_{s}=n_{s}$.

###### Proof.

Throughout this proof we use $X_{1},\dots,X_{s},X,Y,Z$ to denote the images of
the elements of the same name in $A_{(m_{1},\dots,m_{s})}\otimes_{k}A$.

We first note that

$\displaystyle\mathcal{B}:=\Bigg{\\{}\Bigg{(}\prod_{i=1}^{s}X_{i}^{u_{i}}\Bigg{)}X^{\epsilon_{X}}Y^{\epsilon_{Y}}Z^{\epsilon_{Z}}|0\leq
u_{i}$ $\displaystyle<2^{m_{i}},\text{ for all }1\leq i\leq s,$
$\displaystyle\epsilon_{X},\epsilon_{Y},\epsilon_{Z}\in\\{0,1\\},\text{ with
}\epsilon_{X}+\epsilon_{Y}+\epsilon_{Z}\leq 1\Bigg{\\}}$

forms a basis for $A_{(n_{1},\dots,n_{t})}\otimes_{k}A$ and, setting
$J:=J(A_{(m_{1},\dots,m_{s})}\otimes_{k}A)$ (the Jacobson radical of
$A_{(m_{1},\dots,m_{s})}\otimes_{k}A$), we have that

$\\{X_{1}+J^{2},\dots,X_{s}+J^{2},X+J^{2},Y+J^{2},Z+J^{2}\\}$

forms a basis of $J/J^{2}$.

Suppose some $W\in J\setminus J^{2}$ has image

$\left(\sum_{i=1}^{s}\lambda_{i}X_{i}\right)+(\lambda_{X}X+\lambda_{Y}Y+\lambda_{Z}Z)+J^{2}$

in $J/J^{2}$, for some $\lambda_{i},\lambda_{X},\lambda_{Y},\lambda_{Z}\in k$.
Then $W$ has order of nilpotency at least $2$. If $\lambda_{i}\neq 0$ for some
$1\leq i\leq s$, then, by looking at the coefficients of powers of $X_{i}$
with respect to the basis $\mathcal{B}$, we can see that $W$ has order of
nilpotency at least $2^{m_{i}}$.

Now let $W_{1},\dots W_{s+3}\in J$ be such that
$\\{W_{1}+J,\dots,W_{s+3}+J\\}$ forms a basis of $J/J^{2}$. We set $o_{i}$ to
be the order of nilpotency of $W_{i}$, for $1\leq i\leq s+3$. By reordering,
we may assume that $o_{1}\geq\dots\geq o_{s+3}$. As a consequence of the
previous paragraph $o_{i}\geq 2^{m_{i}}$, for each $1\leq i\leq s$ and
$o_{s+1},o_{s+2},o_{s+3}\geq 2$. Moreover, setting $W_{i}=X_{i}$, for each
$1\leq i\leq s$ and $W_{s+1}=X,W_{s+2}=Y,W_{s+3}=Z$, these lower bounds on the
$o_{i}$’s can all be achieved. We have now shown that the tuple
$(m_{1},\dots,m_{s})$ can be retrieved from the isomorphism type of
$A_{(m_{1},\dots,m_{s})}\otimes_{k}A$ and the result is proved. ∎

######  Corollary 3.7.

lem:notPI Let $P$ and $Q$ be finite abelian $2$-groups. If
$\mathcal{O}(P\times A_{4})$ is perfectly isometric to $\mathcal{O}(Q\times
A_{4})$, then $P\cong Q$.

###### Proof.

Suppose $\mathcal{O}(P\times A_{4})$ is perfectly isometric to
$\mathcal{O}(Q\times A_{4})$. Then certainly $Z(k(P\times A_{4}))\cong
Z(k(Q\times A_{4}))$. We may assume that $P,Q\neq 1$. Note that
$Z(kA_{4})\cong A$, $kP\cong A_{(m_{1},\dots,m_{s})}$ and $kQ\cong
A_{(n_{1},\dots,n_{t})}$, where $P\cong C_{2^{m_{1}}}\times\dots\times
C_{2^{m_{s}}}$ and $Q\cong C_{2^{n_{1}}}\times\dots\times C_{2^{n_{t}}}$. By
Lemma 3.6, we have $\\{m_{1},\dots,m_{s}\\}=\\{n_{1},\dots,n_{t}\\}$ and so
$P\cong Q$. ∎

##  4 Normal subgroups of index 2

######  Proposition 4.1 (Theorem 3.15 of [23]).

Let $G$ be a finite group and $N$ a normal subgroup of $G$ of index $p$. Now
let $B$ be a block of $\mathcal{O}G$ with abelian defect group $D$ such that
$G=ND$. Then there exists a block $b$ of $\mathcal{O}N$ with the same block
idempotent as $B$ and defect group $D\cap N$. Moreover there exists a
$G/N$-graded unit $a\in Z(B)$, in particular $B=\bigoplus_{j=0}^{p-1}a^{j}b$.

######  Theorem 4.2.

index2theorem Let $G$, $N$, $B$, $b$ and $D$ be as in Proposition 4.1. Suppose
further that $D\cong P\times(C_{2})^{2}$, for some finite abelian $2$-group
$P$, $D\cap N\cong Q\times(C_{2})^{2}$, for some subgroup $Q\leq P$ of index
$2$, and that $b$ has inertial quotient $C_{3}$ and is Morita equivalent to
the principal block of $\mathcal{O}(Q\times A_{4})$ (respectively
$\mathcal{O}(Q\times A_{5})$). Then $B$ is Morita equivalent to the principal
block of $\mathcal{O}(P\times A_{4})$ (respectively $\mathcal{O}(P\times
A_{5})$).

###### Proof.

We follow the proof of [22, Theorem 2.15].

Suppose that $b$ is Morita equivalent to the principal block of
$\mathcal{O}(Q\times A_{4})$ or $\mathcal{O}(Q\times A_{5})$ and has inertial
quotient $C_{3}$. By Proposition LABEL:Dabelian_inner:prop $D$ acts as inner
automorphisms on $b$, so every irreducible Brauer character of $b$ is fixed
under conjugation in $G$. Since $G/N$ is a cyclic $2$-group, each irreducible
Brauer character extends uniquely to $G$ and lies in $B$ (the unique block of
$G$ covering $b$), so $l(B)=l(b)=3$. By index_p_inertial:prop $B$ and $b$ have
isomorphic inertial quotients $(C_{3})_{1}$. Since $l(B)$ is equal to the
order of the inertial quotient, by the main result of [50] and its proof we
have a perfect isometry

$\displaystyle\mathbb{Z}\mathop{\rm Irr}\nolimits(B)\to\mathbb{Z}\mathop{\rm
Irr}\nolimits(\mathcal{O}(P\times A_{4})).$

Similarly there is a perfect isometry $\mathbb{Z}\mathop{\rm
Irr}\nolimits(B_{0}(\mathcal{O}(P\times A_{5})))\to\mathbb{Z}\mathop{\rm
Irr}\nolimits(\mathcal{O}(P\times A_{4}))$.

Write $\mathcal{A}$ for $A_{4}$ or $A_{5}$. From the above we have a perfect
isometry

$\displaystyle I:\mathbb{Z}\mathop{\rm
Irr}\nolimits(B)\to\mathbb{Z}\mathop{\rm
Irr}\nolimits(B_{0}(\mathcal{O}(P\times\mathcal{A}))),$

for $\mathcal{A}$ as according to the Morita equivalence class of $b$.

Now $I$ induces an isomorphism of groups ${\rm Perf}(B)\cong{\rm
Perf}(B_{0}(\mathcal{O}(P\times\mathcal{A})))$ via $\beta\mapsto
I\circ\beta\circ I^{-1}$ for $\beta$ any perfect self-isometry of $B$, and we
denote this isomorphism by $I_{\operatorname{PI}}$. Consider the perfect self-
isometry

$\displaystyle L:\mathbb{Z}\mathop{\rm Irr}\nolimits(B)$
$\displaystyle\to\mathbb{Z}\mathop{\rm Irr}\nolimits(B)$ $\displaystyle\chi$
$\displaystyle\mapsto\operatorname{sgn}_{N}^{G}.\chi,$

where $\operatorname{sgn}_{N}^{G}$ is the linear character of $G$ with kernel
$N$. So for each $\theta\in\mathop{\rm Irr}\nolimits(b)$, $L$ swaps the two
extensions of $\theta$ to $G$. We know that $L$ is indeed a perfect isometry
as it is induced by the $\mathcal{O}$-algebra automorphism of $\mathcal{O}G$
given by $g\mapsto\operatorname{sgn}_{N}^{G}(g)g$ for all $g\in G$.

Note that $L$ is a perfect self-isometry of order $2$ and that it induces the
trivial $k$-algebra automorphism on $Z(kB)$. Furthermore, since by Proposition
LABEL:index_p_inertial:prop induction gives a bijection between ${\rm prj}(b)$
and ${\rm prj}(B)$, each element of ${\rm prj}(B)$ is fixed under
multiplication by $\operatorname{sgn}_{N}^{G}$ and so $L$ is the identity on
$\mathbb{Z}{\rm prj}(B)$. Therefore $I_{\operatorname{PI}}(L)$ must be of
order $2$, induce the identity $k$-algebra automorphism on
$Z(B_{0}(k(P\times\mathcal{A})))$ and be the identity on $\mathbb{Z}{\rm
prj}(B_{0}(\mathcal{O}(P\times\mathcal{A})))$. We claim that
$I_{\operatorname{PI}}(L)$ is induced by multiplication by the sign character
of $G^{\prime}:=P\times\mathcal{A}$ with respect to the subgroup
$N^{\prime}:=R\times\mathcal{A}\leq G^{\prime}$, for some index $2$ subgroup
$R\leq P$, with $R\cong Q$.

We first deal with the $\mathcal{A}=A_{4}$ case. Adopting the notation of
thm:PA4, set $I_{\operatorname{PI}}(L)=(J,I_{\sigma,\epsilon})$, where $J$ is
a perfect self-isometry of $\mathcal{O}P$ induced by an $\mathcal{O}$-algebra
automorphism, $\sigma\in S_{4}$ and $\epsilon\in\\{\pm 1\\}$. Then the fact
that $I_{\operatorname{PI}}(L)$ is the identity on $\mathbb{Z}{\rm
prj}(\mathcal{O}(P\times A_{4}))$ forces $\sigma$ to be the identity
permutation and $\epsilon=1$. So $J$ is induced by an element of
$\alpha\in\mathop{\rm Aut}\nolimits(\mathcal{O}P)$ that has order $2$ and
induces the identity on $kP$. Recall that $\mathop{\rm
Aut}\nolimits(\mathcal{O}P)\cong\mathop{\rm
Hom}\nolimits(P,\mathcal{O}^{\times})\rtimes\mathop{\rm Aut}\nolimits(P)$. The
fact that $\alpha$ induces the identity on $kP$ forces $\alpha$ to be given by
multiplication by $\lambda_{\alpha}\in\mathop{\rm
Hom}\nolimits(P,\mathcal{O}^{\times})$ of order $2$. In other words $\alpha$
is induced by multiplication by the sign character of $P$ with respect to some
normal subgroup $R$ of index $2$. Hence $I_{\operatorname{PI}}(L)$ is induced
by multiplication by the sign character of $P\times A_{4}$ with respect to the
subgroup $N^{\prime}:=R\times A_{4}\leq G^{\prime}$. (Note that we do not know
yet that $R\cong Q$.)

We have now shown that

$\displaystyle
I(\operatorname{sgn}_{N}^{G}.\chi)=\operatorname{sgn}_{N^{\prime}}^{G^{\prime}}.I(\chi),$

for all $\chi\in\mathop{\rm Irr}\nolimits(B)$. By Proposition
LABEL:prop:isomcent, $b$ is then perfectly isometric to
$\mathcal{O}N^{\prime}$. However, $b$ is Morita equivalent to
$B_{0}(\mathcal{O}(Q\times A_{4}))$ and so lem:notPI implies that $R\cong Q$
as desired.

For the $\mathcal{A}=A_{5}$ case we fix a perfect isometry
$I_{A}:\mathbb{Z}\mathop{\rm
Irr}\nolimits(B_{0}(\mathcal{O}A_{5}))\to\mathbb{Z}\mathop{\rm
Irr}\nolimits(\mathcal{O}A_{4})$. As above, we can then show that

$({\rm Id}_{\mathbb{Z}\mathop{\rm Irr}\nolimits(P)},I_{A})\circ
I_{\operatorname{PI}}(L)\circ({\rm Id}_{\mathbb{Z}\mathop{\rm
Irr}\nolimits(P)},I_{A})^{-1}:\mathbb{Z}\mathop{\rm
Irr}\nolimits(\mathcal{O}(P\times A_{4}))\to\mathbb{Z}\mathop{\rm
Irr}\nolimits(\mathcal{O}(P\times A_{4}))$

is induced by multiplication by the sign character of an appropriate subgroup.
That $I_{\operatorname{PI}}(L)$ is of the desired now follows immediately.

Composing the perfect isometry induced by the Morita equivalence between $b$
and $B_{0}(\mathcal{O}(Q\times\mathcal{A}))$ with that given by the
isomorphism between $Q\times\mathcal{A}$ and $N^{\prime}$, we obtain a perfect
isometry $I_{\operatorname{Mor}}:\mathbb{Z}\mathop{\rm
Irr}\nolimits(b)\rightarrow\mathbb{Z}\mathop{\rm
Irr}\nolimits(B_{0}(\mathcal{O}N^{\prime}))$.

Denote by $I_{N,N^{\prime}}$ the perfect isometry $\mathbb{Z}\mathop{\rm
Irr}\nolimits(b)\rightarrow\mathbb{Z}\mathop{\rm
Irr}\nolimits(B_{0}(\mathcal{O}N^{\prime}))$ induced by $I$ as in Proposition
LABEL:prop:isomcent.

Write $I_{N,N^{\prime}}\circ
I_{\operatorname{Mor}}^{-1}=(J^{\prime},I_{\tau,\delta})$ in the notation of
thm:PA4 applied to $B_{0}(\mathcal{O}(R\times\mathcal{A}))$, where
$J^{\prime}$ is a perfect self-isometry of $\mathcal{O}R$ induced by an
$\mathcal{O}$-algebra automorphism $\alpha^{\prime}$, $\tau\in S_{4}$ and
$\delta\in\\{\pm 1\\}$. By post-composing $I$ with the perfect self-isometry
$({\rm Id}_{\mathbb{Z}\mathop{\rm Irr}\nolimits(P)},I_{\tau,\delta})^{-1}$ of
$B_{0}(\mathcal{O}G^{\prime})$ and post-composing the Morita equivalence
$b\sim_{\operatorname{Mor}}B_{0}(\mathcal{O}N^{\prime})$ with that induced by
$\alpha^{\prime}\otimes{\rm Id}_{B_{0}(\mathcal{O}A)}$, we may assume that
$I_{N,N^{\prime}}=I_{\operatorname{Mor}}$.

Let $\phi_{I}:Z(B)\to Z(B_{0}(\mathcal{O}G^{\prime}))$ be the isomorphism of
centres induced by $I$ as in Lemma LABEL:prop:isomcent and let $M$ be the
$B_{0}(\mathcal{O}N^{\prime})$-$b$-bimodule inducing the above Morita
equivalence $b\sim_{\operatorname{Mor}}B_{0}(\mathcal{O}N^{\prime})$. Since
$I_{N,N^{\prime}}=I_{\operatorname{Mor}}$, by Proposition LABEL:prop:isomcent
we have that $\phi_{I}|_{Z(b)}=\phi_{I_{N,N^{\prime}}}:Z(b)\to
Z(B_{0}(\mathcal{O}N^{\prime}))$ is the isomorphism of centres induced by the
Morita equivalence. In other words,

$\displaystyle\phi_{I}(\mathsf{b})m=m\mathsf{b},\text{ for all }\mathsf{b}\in
b,m\in M.$ (7)

Let $a\in B$ be a graded unit as described in Proposition 4.1 and set
$a^{\prime}:=\phi_{I}(a)$. Since $\phi_{I}$ respects the $G/N$ and
$G^{\prime}/N^{\prime}$-gradings, $a^{\prime}$ is also a graded unit. We now
give $M$ the structure of a module for

$\displaystyle(B_{0}(\mathcal{O}N^{\prime})\otimes_{\mathcal{O}}b^{\operatorname{op}})\oplus(a^{\prime-1}B_{0}(\mathcal{O}N^{\prime})\otimes_{\mathcal{O}}(ab)^{\operatorname{op}})$

by defining $a^{\prime-1}.m.a=m$, for all $m\in M$, where (7) ensures that
this does indeed define a module. Now by [40, Theorem 3.4] we have proved that
$B$ is Morita equivalent to $B_{0}(\mathcal{O}(P\times\mathcal{A}))$.

∎

##  5 Extensions of blocks

In this section we give the possible Morita equivalence classes of blocks
covering a block of a normal subgroup in some relevant Morita equivalence
classes.

Let $G$ be a finite group and $N\lhd G$. Let $b$ be a $G$-stable block of
$\mathcal{O}N$ covered by a block $B$ of $G$ with abelian defect group $D$.
Then $b$ has defect group $Q=D\cap N$. Let $(D,B_{D})$ be a maximal
$B$-subpair.

We first extract and summarize two results of [51] and [52].

The following is a weaker version of Theorem 5.10 of [51] that is sufficient
for our purposes. We use it in some cases where there is a normal subgroup of
index $p$.

######  Theorem 5.1 ([51]).

WZZextendp With the notation above, let $E=N_{G}(D,B_{D})/C_{G}(D)$ suppose
that $E$ is cyclic and acts freely on $[N_{G}(D,B_{D}),D]\setminus\\{1\\}$
(that is, all orbits have length $|E|$). Suppose that $b$ is inertial, i.e.,
there is a basic Morita equivalence with $\mathcal{O}(Q\rtimes E)$. Then $B$
is Morita equivalent to $\mathcal{O}(D\rtimes E)$.

The following is the main result of [52], and is particularly relevant to the
case that $b$ is a nilpotent covered block.

######  Theorem 5.2 ([52]).

inertialp’extend Suppose that $N$ has $p^{\prime}$-index. If $b$ is inertial,
then $B$ is inertial.

We now apply Külshammer’s analysis in [34] of the situation of a normal
subgroup containing the defect groups of a block, which involves the study of
crossed products of a basic algebra with an $p^{\prime}$-group. In the general
setting he finds finiteness results for the possible crossed products, but in
our situation we are able to precisely describe the possibilities.

Background on crossed products may be found in [34], but we summarize what we
need here. Let $X$ be a finite group and $R$ an $\mathcal{O}$-algebra. A
crossed product of $R$ with $X$ is an $X$-graded algebra $\Lambda$ with
identity component $\Lambda_{1}=R$ such that each graded component
$\Lambda_{x}$, where $x\in X$, contains a unit $u_{x}$. Given a choice of unit
$u_{x}$ for each $x$, we have maps $\alpha:X\rightarrow\mathop{\rm
Aut}\nolimits(R)$ given by conjugation by $u_{x}$ and $\mu:X\times
X\rightarrow U(R)$ given by
$\alpha_{x}\circ\alpha_{y}=\iota_{\mu(x,y)}\circ\alpha_{xy}$, where $U(R)$ is
the group of units of $R$ and $\iota_{\mu(x,y)}$ is conjugation by $\mu(x,y)$.
The pair $(\alpha,\mu)$ is called a parameter set of $X$ in $R$. In [34] an
isomorphism of crossed products respecting the grading is called a weak
equivalence. By the discussion following Proposition 2 of [34] weak
isomorphism classes of crossed products of $R$ with $X$ are in bijection with
pairs consisting of an $\mathop{\rm Out}\nolimits(R)$-conjugacy class of
homomorphisms $X\rightarrow\mathop{\rm Out}\nolimits(R)$ for which the induced
element in $H^{3}(X,U(Z(R)))$ vanishes, and an element of $H^{2}(X,U(Z(R)))$.

We adapt Proposition in Section 3 of [34], and include a proof for
completeness (as given in [18]). Note that $\alpha:X\rightarrow\mathop{\rm
Aut}\nolimits(R)$ restricts to a map $X\rightarrow\mathop{\rm
Aut}\nolimits(Z(R))$. Hence we also have homomorphisms
$X\rightarrow\mathop{\rm Aut}\nolimits(Z(R)/J(Z(R)))$ and
$X\rightarrow\mathop{\rm Aut}\nolimits(U(Z(R)/J(Z(R))))$.

######  Lemma 5.3.

vanishing_cohomology Suppose that $X=\langle x\rangle$ is a cyclic
$p^{\prime}$-group. Then $H^{i}(X,U(Z(R)))=0$ for each $i\geq 1$.

###### Proof.

Let $i\in\mathbb{N}$. Following [34], $U(Z(R))\cong
U(Z(R)/J(Z(R)))\times(1+J(Z(R))$ and $H^{i}(X,U(Z(R)))\cong
H^{i}(X,U(Z(R)/J(Z(R))))\times H^{i}(X,1+J(Z(R)))$. We have
$H^{i}(X,1+J(Z(R)))=0$ since $X$ is a $p^{\prime}$-group. Now $Z(R)/J(Z(R))$
is a commutative semisimple $k$-algebra, which we denote by $A$, and note as
above we have a homomorphism $X\rightarrow\mathop{\rm Aut}\nolimits(A)$. Write
$A=A_{1}\times\cdots\times A_{r}$, where each $A_{j}$ is a product of simple
algebras constituting an $X$-orbit. We have $H^{i}(X,U(A))\cong
H^{i}(X,U(A_{1}))\times\cdots\times H^{i}(X,U(A_{r}))$. Now each
$H^{i}(X,U(A_{j}))$ vanishes, for as a $kX$-module $A_{j}$ is induced from the
trivial module of $kY$ for some $Y\leq X$, and so by Shapiro’s Lemma
$H^{i}(X,U(A_{j}))\cong H^{i}(Y,k^{\times})$ (see [8, 2.8.4]), which vanishes
since $X$ is cyclic. Hence $H^{i}(X,U(Z(R)))=0$ for each $i$. ∎

We apply the above with $X=G/N=\langle x\rangle$, where $G/N$ is a
$p^{\prime}$-group. Let $f$ be an idempotent of $b$ such that $fbf$ is a basic
algebra for $b$. By [3, Lemma 4.2] $fBf$ is a crossed product of $fbf$ with
$X$ and $fBf$ is Morita equivalent to $B$. Hence we may take $R=fbf$ in the
above. By vanishing_cohomology weak isomorphism classes of crossed products of
$fbf$ with $X$ are in bijection with $\mathop{\rm
Out}\nolimits(fbf)$-conjugacy classes of homomorphisms
$X\rightarrow\mathop{\rm Out}\nolimits(fbf)$. Note however, that such crossed
products may be isomorphic as algebras but not weakly isomorphic as crossed
products. Indeed, given $\alpha:X\rightarrow\mathop{\rm Out}\nolimits(fbf)$,
the same algebra gives rise to parameter sets associated to
$\alpha\circ\varphi$ for each $\varphi\in\mathop{\rm Aut}\nolimits(X)$.

Now $\mathop{\rm Out}\nolimits(fbf)$ embeds in $\mathop{\rm
Pic}\nolimits(fbf)\cong\mathop{\rm Pic}\nolimits(b)$ and since $fbf$ is a
basic algebra $\mathop{\rm Out}\nolimits(fbf)\cong\mathop{\rm
Pic}\nolimits(fbf)$, so we may apply the descriptions of Picard groups in
pic_list. The strategy will be to limit the number of possible Morita
equivalence classes for $B$ (given $b$), and to identify examples where all
such Morita equivalence classes are realised.

A special case that will arise frequently, and that demonstrates the
phenomenon of crossed products isomorphic as algebras but not weakly
isomorphic as crossed products is as follows:

######  Lemma 5.4.

Picard_argument_special_case With the notation above, suppose $G/N$ has prime
order $r$ different to $p$ and that $\mathop{\rm Out}\nolimits(fbf)$ has
cyclic Sylow $r$-subgroups of order $r$. Then there are precisely two
possibilities for Morita equivalence class of $B$, one of which is that $B$ is
source algebra equivalent to $b$.

###### Proof.

The trivial homomorphism $G/N\rightarrow\mathop{\rm Out}\nolimits(fbf)$
corresponds to the case that $B$ is source algebra equivalent to $b$ by
innerautos, since $G=G[b]$. Consider nontrivial
$\alpha:G/N\rightarrow\mathop{\rm Out}\nolimits(fbf)$, and consider a block
$B$ such that $fBf$ is a crossed product of $fbf$ with $G/N$ corresponding to
$\alpha$. Then for each $\varphi\in\mathop{\rm Aut}\nolimits(G/N)$ we have
that $\alpha\circ\varphi$ also realises $fBf$ as a crossed product. This
accounts for all possible homomorphisms $\alpha:G/N\rightarrow\mathop{\rm
Out}\nolimits(fbf)$. ∎

######  Example 5.5.

Puig_example As observed in [3], there are examples of nilpotent blocks
covering non-nilpotent blocks constructed in [46, Remark 4.4] that will be
useful in the arguments that follow. Let $P$ be an abelian $2$-group on which
$C_{r}$ acts regularly for a prime $r$. Define $N=(P\rtimes C_{r})\times
C_{r}$ with Sylow $r$-subgroup $H$. Note that $N$ has $r$ $2$-blocks,
corresponding to $\mathop{\rm Irr}\nolimits(Z(N))$. There is a group $T\cong
C_{r}$ acting on $N$ fixing the elements of $Z(N)$ and $H/Z(N)$. Define
$G=N\rtimes T$. Then each nontrivial element of $\mathop{\rm
Irr}\nolimits(Z(N))$ is covered by a nilpotent block of $G$ (with defect group
$P$). These cover block of $N$ Morita equivalent to $\mathcal{O}(P\rtimes
C_{r})$. We in particular require the cases $N=G_{n}=(C_{2^{n}})^{2}\rtimes
C_{3}$ and $N=(C_{2^{n}})^{3}\rtimes C_{7}$.

######  Remark 5.6.

conj_perm In the crossed product construction, we are considering
homomorphisms $G/N\rightarrow\mathop{\rm Pic}\nolimits(b)$ induced by
congugation by elements of $G$. Conjugation on $\mathcal{O}N$ affords a
permutation module, and so conjugation on an invariant summand (in our case
$b$) affords a trivial source module. It follows that actually we have
$G/N\rightarrow\mathcal{T}(b)$. In general this is a less useful observation
than we might hope, as $\mathcal{T}(b)$ is not (known to be) preserved under
Morita equivalence. However, it does allow us to keep some control over
inertial quotients as we move from $b$ to $B$. We will use this observation in
the following proofs.

We now apply the above to every extension of a Morita equivalence from a
normal subgroup of odd prime index that we will need.

######  Proposition 5.7.

Picard_application Let $G$ be a finite group and $N\lhd G$ with $[G:N]=r$,
where $r$ is an odd prime. Let $B$ be a $2$-block of $\mathcal{O}G$ covering a
$G$-stable block of $b$ of $\mathcal{O}N$.

  1. (i)

Suppose that $b$ is Morita equivalent to $\mathcal{O}(((C_{2^{n}})^{3}\rtimes
C_{7})\times P)$ where $n\in\mathbb{N}$ and $P$ is a cyclic $2$-group. Suppose
$r=3$. Then $B$ is either source algebra equivalent to $b$ or Morita
equivalent to $\mathcal{O}(((C_{2^{n}})^{3}\rtimes(C_{7}\rtimes C_{3}))\times
P)$. Further, if $b$ is known to have inertial quotient $C_{7}$, then in the
latter case $B$ has inertial quotient $C_{7}\rtimes C_{3}$. Suppose $r=7$.
Then $B$ is either source algebra Morita equivalent to $b$ or is nilpotent. If
$r$ is an odd prime other than $3$ and $7$, then $B$ is source algebra
equivalent to $b$.

  2. (ii)

Suppose that $b$ is Morita equivalent to $\mathcal{O}(G_{m}\times P)$, where
$m\in\mathbb{N}$ and $P$ is an abelian $2$-group with rank at most $2$. Let
$D$ be the defect group of $b$. If $r\neq 3$, then $B$ is source algebra
equivalent to $b$. If $r=3$ and $P\cong(C_{2^{n}})$ for some $n\in\mathbb{N}$,
then $B$ is either nilpotent, source algebra equivalent to $b$, or Morita
equivalent to $\mathcal{O}(G_{m}\times G_{n})$ or a nonprincipal block of
$\mathcal{O}(D\rtimes 3_{+}^{1+2})$. Further, if $b$ is known to have inertial
quotient $(C_{3})_{1}$, then in the latter two cases the inertial quotient of
$B$ is $C_{3}\times C_{3}$. If $r=3$ and $P$ is cyclic or a product of two
cyclic groups of different orders, then $B$ is either nilpotent or source
algebra equivalent to $b$.

  3. (iii)

Suppose that $b$ is Morita equivalent to $B_{0}(\mathcal{O}(A_{5}\times P))$
where $P$ is an abelian $2$-group with rank at most $2$. If $P$ is cyclic or a
product of two cyclic groups of different orders, then $B$ is source algebra
equivalent to $b$. If $P\cong(C_{2^{n}})$ for some $n\in\mathbb{N}$, then $B$
is either source algebra equivalent to $b$ or $B_{0}(\mathcal{O}(A_{5}\times
G_{n}))$, with the latter case only occurring when $r=3$.

  4. (iv)

Suppose that $b$ is Morita equivalent to $B_{0}(\mathcal{O}(A_{5}\times
G_{n}))$. Then $B$ is Morita equivalent to
$B_{0}(\mathcal{O}(A_{5}\times(C_{2^{n}})^{2}))$ or source algebra equivalent
to $b$, with the former case only occurring when $r=3$.

###### Proof.

(i) By pic_list $\mathop{\rm Pic}\nolimits(b)\cong(C_{7}\rtimes
C_{3})\times(C_{2^{n}}\rtimes\mathop{\rm Aut}\nolimits(C_{2^{n}})$. By
Picard_argument_special_case, for both $r=3$ and $r=7$ there are two
possibilities for the Morita equivalence class of $B$, and one is that $B$ is
source algebra equivalent to $b$. For $r=7$ the second is that $B$ is
nilpotent as realised in Puig_example. For $r=3$, the second case is realised
by the group given in the statement. For all other odd primes, since they do
not divide the order of the Picard group, $B$ is source algebra equivalent to
$b$. It remains to prove the statement regarding inertial quotients. Suppose
that $b$ is known to have inertial quotient $C_{7}$ and that $r=3$. It follows
from inertial_quotient_normal that $B$ has inertial quotient $C_{7}$ or
$C_{7}\rtimes C_{3}$. By conj_perm we observe that we may consider elements of
$\mathcal{T}(b)$. By the description of $\mathcal{T}(b)$ in Section 2.5 we
know that $\mathcal{T}(b)$ maps to a subgroup of $\mathop{\rm
Out}\nolimits(D,\mathcal{F})\cong C_{3}$. If this subgroup is trivial, then we
may not construct any nontrivial homomorphism $G/N\rightarrow\mathcal{T}(b)$
and we are done. Hence suppose the subgroup has order three. Then the action
of $G$ induces an element of order three in $\mathop{\rm
Aut}\nolimits(D,\mathcal{F})$, so that $B$ must have inertial quotient
$C_{7}\rtimes C_{3}$.

(ii) By pic_list $\mathop{\rm Pic}\nolimits(b)\cong
S_{3}\times(P\rtimes\mathop{\rm Aut}\nolimits(P))$. If $P$ is cyclic or a
product of two cyclic groups of different orders, then $\mathop{\rm
Aut}\nolimits(P)$ is a $2$-group, and so by Picard_argument_special_case $B$
is either Morita equivalent to $b$ or is nilpotent, with this case realised as
in Puig_example. Suppose that $P\cong(C_{2^{n}})$ for some $n\in\mathbb{N}$.
Then $\mathop{\rm Pic}\nolimits(b)\cong S_{3}\times(P\rtimes S_{3})$. There
are four conjugacy classes of homomorphisms $G/N\rightarrow\mathop{\rm
Pic}\nolimits(b)$, and so at most four possibilities for the Morita
equivalence class of $B$. These are account for by the four cases listed in
the statement, with the cases that $B$ is Morita equivalent to $b$ or
$\mathcal{O}(G_{m}\times G_{n})$ requiring no further explanation. The case
that $B$ is nilpotent is again realised as in Puig_example. Note that
$D\rtimes 3_{+}^{1+2}$ has a normal subgroup $M$ of index $3$ isomorphic to
$N\times C_{3}$. Now $B$ covers a nonprincipal block of $\mathcal{O}M$, which
is Morita equivalent to $\mathcal{O}N$. Hence the final case is realised. It
remains to prove the statement regarding inertial quotients. Suppose that $b$
is known to have inertial quotient $(C_{3})_{1}$. It follows from
inertial_quotient_normal that $B$ has inertial quotient $(C_{3})_{1}$ or
$C_{3}\times C_{3}$. as above, by conj_perm we observe that we may consider
elements of $\mathcal{T}(b)$. By the description of $\mathcal{T}(b)$ in
Section 2.5 we know that $\mathcal{T}(b)$ maps to a subgroup of $\mathop{\rm
Out}\nolimits(D,\mathcal{F})\cong C_{3}$. If this subgroup is trivial, then we
may not construct any nontrivial homomorphism $G/N\rightarrow\mathcal{T}(b)$
and we are done. Hence suppose the subgroup has order three. Then the action
of $G$ induces an element of order three in $\mathop{\rm
Aut}\nolimits(D,\mathcal{F})$, so that $B$ must have inertial quotient
$C_{3}\rtimes C_{3}$.

(iii) By pic_list $\mathop{\rm Pic}\nolimits(b)\cong
C_{2}\times(P\rtimes\mathop{\rm Aut}\nolimits(P))$. If $P$ is cyclic or a
product of two cyclic groups of different orders, then $\mathop{\rm
Pic}\nolimits(b)$ is a $2$-group, and so $B$ is Morita equivalent to $b$.
Suppose that $P\cong(C_{2^{n}})$ for some $n\in\mathbb{N}$. Then $\mathop{\rm
Pic}\nolimits(b)\cong C_{2}\times(P\rtimes S_{3})$. Hence by
Picard_argument_special_case the result follows.

(iv) By pic_list $\mathop{\rm Pic}\nolimits(b)\cong C_{2}\times S_{3}$. By
Picard_argument_special_case there are two possibilities for the Morita
equivalence class of $B$, one of which is the class containing $b$. The case
$B_{0}(\mathcal{O}(A_{5}\times(C_{2^{n}})^{2}))$ is realised by taking a
product of $A_{5}$ with a group as in Puig_example.

∎

######  Remark 5.8.

since (a) source algebra equivalence preserves the inertial quotient, and (b)
for abelian defect groups a block is nilpotent if and only if the inertial
quotient is trivial, we have shown that if the inertial quotient of $b$ is
isomorphic to that of the given Morita equivalence class representative, then
$B$ also has inertial quotient isomorphic to that of the given Morita
equivalence class representative.

Putting the reduction techniques of the section together, we have the
following that is in a form directly applicable to the proof of main_theorem.
Note that we do not need to consider all forms for $b$ here: only those that
will arise later.

######  Proposition 5.9.

covers_inertial Let $G$ be a finite group and $N\lhd G$ with $G/N$ solvable.
Let $B$ be a quasiprimitive $2$-block of $\mathcal{O}G$ with abelian defect
group $D$ of rank at most $4$. Suppose that $B$ covers a block $b$ of
$\mathcal{O}N$.

  1. (i)

If $b$ is inertial with inertial quotient $C_{7}$, then $B$ is Morita
equivalent to $\mathcal{O}D$, $\mathcal{O}(D\rtimes C_{7})$,
$\mathcal{O}(D\rtimes(C_{7}\rtimes C_{3}))$ or $\mathcal{O}(G_{n}\times P)$
for some $n$ and some abelian $2$-group $P$, and $B$ has inertial quotient
$1$, $C_{7}$, $(C_{7}\rtimes C_{3}$ or $(C_{3})_{1}$ respectively.

  2. (ii)

If $b$ is inertial with inertial quotient $C_{3}\times C_{3}$, then $B$ is
inertial with inertial quotient $1$, $(C_{3})_{1}$ or $C_{3}\times C_{3}$.

  3. (iii)

If $b$ is inertial with inertial quotient $(C_{3})_{2}$, $C_{5}$ or $C_{15}$,
then $B$ is inertial with inertial quotient $1$, $(C_{3})_{2}$, $C_{5}$ or
$C_{15}$.

  4. (iv)

If $b$ is Morita equivalent to a block with normal defect group and has
inertial quotient $(C_{3})_{1}$, then $B$ is Morita equivalent to one of:

    1. (a)

$\mathcal{O}D$;

    2. (b)

$\mathcal{O}(G_{n}\times P)$, where $D\cong(C_{2^{n}})^{2}\times P$, and $B$
has inertial quotient $(C_{3})_{1}$;

    3. (c)

$\mathcal{O}(G_{n_{1}}\times G_{n_{2}})$, where
$D\cong(C_{2^{n_{1}}})^{2}\times(C_{2^{n_{2}}})^{2}$, and $B$ has inertial
quotient $C_{3}\times C_{3}$;

    4. (d)

a non-principal block of $\mathcal{O}(D\rtimes 3_{+}^{1+2})$ and $B$ has
inertial quotient $C_{3}\times C_{3}$;

  5. (v)

If $b$ is Morita equivalent to $B_{0}(\mathcal{O}(A_{5}\times Q))$ for some
$Q\leq D$ and has inertial quotient of type $(C_{3})_{1}$, then $B$ is Morita
equivalent to one of:

    1. (a)

$B_{0}(\mathcal{O}(A_{5}\times P))$, where $D\cong(C_{2})^{2}\times P$, and
$B$ has inertial quotient $(C_{3})_{1}$;

    2. (b)

$B_{0}(\mathcal{O}(A_{5}\times G_{n}))$, where
$D\cong(C_{2})^{2}\times(C_{2^{n}})^{2}$, and $B$ has inertial quotient
$C_{3}\times C_{3}$.

  6. (vi)

If $b$ is Morita equivalent to $B_{0}(\mathcal{O}(A_{5}\times G_{n}))$ for
some $n\geq 1$, then $B$ is Morita equivalent to $b$ or
$B_{0}(\mathcal{O}(A_{5}\times(C_{2^{n}})^{2}))$, where
$D\cong(C_{2})^{2}\times(C_{2^{n}})^{2}$, and $B$ has inertial quotient
$C_{3}\times C_{3}$ or $(C_{3})_{1}$ respectively.

###### Proof.

Since $b$ is $G$-stable and $G/N$ is solvable, it follows from Lemma
LABEL:solvablequotient:lem that $DN/N$ is an abelian Sylow $2$-subgroup of
$G/N$, which must then have $2$-length at most one. By Proposition
LABEL:Dabelian_inner:prop $D\leq G[b]$, which must also have $2$-length at
most one. Hence there are normal subgroups $N_{i}$ of $G$ such that
$N=N_{0}\lhd N_{1}\lhd N_{2}\lhd N_{3}=G[b]$ with $N_{1}/N_{0}$, $N_{3}/N_{2}$
of odd order and $N_{2}/N_{1}$ a $2$-group, so in particular $N_{2}=N_{1}D$.
Write $b_{i}$ for the unique block of $\mathcal{O}N_{i}$ covered by $B$. Let
$(D,(b_{2})_{D})$ be a $b_{2}$-subpair. By defect_group_factor:prop
$D\cong[D,N_{N_{2}}(D,(b_{2})_{D})]\times C_{D}(N_{N_{2}}(D,(b_{2})_{D}))$
with $[D,N_{N_{2}}(D,(b_{2})_{D})]\leq N_{1}$ (and so
$[D,N_{N_{2}}(D,(b_{2})_{D})]\leq N$).

By innerautos $b$ is source algebra equivalent to $b_{1}$, with the same
inertial quotient as $b$. If furthermore $b$ is inertial, then $b_{1}$ is also
inertial.

By index_p_inertial:prop $b_{2}$ has inertial quotient isomorphic to that of
$b$, with the same action (this last statement uses the fact that in this case
there is a unique action given the isomorphism type of the inertial quotient
and the order of commutator part $[D,N_{N_{2}}(D,(b_{2})_{D})]$ of $D$).

(i) Suppose that $b$, and so $b_{2}$, has inertial quotient $C_{7}$ and that
$b$ is inertial. By WZZextendp $b_{2}$ is Morita equivalent (but not
necessarily basic Morita equivalent) to $\mathcal{O}(D\rtimes C_{7})$. Note
that $C_{D}(N_{N_{2}}(D,(b_{2})_{D}))$ is cyclic. By pic_list $\mathop{\rm
Pic}\nolimits(b_{2})\cong(C_{7}\rtimes
C_{3})\times(C_{D}(N_{N_{2}}(D,(b_{2})_{D}))\rtimes\mathop{\rm
Aut}\nolimits(C_{D}(N_{N_{2}}(D,(b_{2})_{D})))$. Now consider $G[b_{2}]$.
Again applying innerautos, $b_{2}$ is source algebra equivalent to the unique
block $c$ of $G[b_{2}]$ covered by $B$. Since $G/G[b_{2}]$ is an odd order
subgroup of $\mathop{\rm Pic}\nolimits(b_{2})$, it is isomorphic to a subgroup
of $C_{7}\rtimes C_{3}$. Suppose that $G/G[b_{2}]\cong C_{7}$. By
Picard_application either $B$ is nilpotent or it is Morita equivalent to $c$
with inertial quotient $C_{7}$. If $G/G[b_{2}]\cong C_{3}$, then again by
Picard_application $B$ is Morita equivalent to $c$ or
$\mathcal{O}(D\rtimes(C_{7}\rtimes C_{3}))$, with the appropriate inertial
quotient. Suppose $G/G[b_{2}]\cong C_{7}\rtimes C_{3}$. Write $G_{1}$ for the
preimage of $O_{7}(G/G[b_{2}])$ in $G$ and $B_{1}$ for the unique block of
$G_{1}$ covered by $B$. Then by Picard_application $B_{1}$ is either nilpotent
or Morita equivalent to $c$. Applying Picard_application again, in the first
case $B$ is either nilpotent or Morita equivalent to a block with normal
defect group and inertial quotient $(C_{3})_{1}$. In the second, $B$ is either
Morita equivalent to $c$ or to $\mathcal{O}(D\rtimes(C_{7}\rtimes C_{3}))$. In
each of these cases by Picard_application the inertial quotient is as stated.

(ii) and (iii) Suppose that $b$, and so $b_{2}$, has inertial quotient
$(C_{3})_{2}$, $C_{5}$, $C_{15}$ or $C_{3}\times C_{3}$, and that $b$ is
inertial. We have $C_{D}(N_{N_{2}}(D,(b_{2})_{D}))=1$, so
$D=[D,N_{N_{2}}(D,(b_{2})_{D})]\leq N_{1}$, i.e., $N_{1}=N_{2}$ and $G/N$ has
odd order. The result then follows from inertialp’extend.

(iv) Suppose that $b$, and so $b_{1}$, $b_{2}$, has inertial quotient
$(C_{3})_{1}$. Hence $b_{1}$ is Morita equivalent to $\mathcal{O}(G_{n}\times
Q)$ for some $n$ and $Q\leq D$. Recalling that
$[D,N_{N_{2}}(D,(b_{2})_{D})]\leq N_{1}$, by index2theorem $b_{2}$ is Morita
equivalent to $\mathcal{O}(G_{n}\times Q)$, where $D\cong(C_{2^{n}})^{2}\times
P$. Applying innerautos $b_{2}$ is source algebra equivalent to the unique
block $c$ of $G[b_{2}]$ covered by $B$. By pic_list $\mathop{\rm
Pic}\nolimits(b_{2})\cong S_{3}\times(P\rtimes\mathop{\rm Aut}\nolimits(P))$.
Suppose that $P$ is not homocyclic. Since $G/G[b_{2}]$ is an odd order
subgroup of $\mathop{\rm Pic}\nolimits(b_{2})$ and $|\mathop{\rm
Pic}\nolimits(b_{2})|_{2^{\prime}}=3$, it is isomorphic to a subgroup of
$C_{3}$. By Picard_application $B$ is either Morita equivalent to $c$ or is
nilpotent. Suppose that $P$ is homocyclic. Then $\mathop{\rm
Pic}\nolimits(b_{2})\cong S_{3}\times(P\rtimes S_{3})$ and $G/G[b_{2}]$ is
isomorphic to a subgroup of $C_{3}\times C_{3}$. Let $H$ be the kernel of the
action of $G$ on the irreducible Brauer characters of $c$ and let $B_{H}$ be
the unique block of $\mathcal{O}H$ covered by $B$. By pic_list $[G:H]$ divides
$3$. Suppose that $[G:G[b_{2}]]=3$. By Picard_application the possible Morita
equivalence classes for $B$ are represented by the four blocks listed. Suppose
that $[G:G[b_{2}]]=9$. If $G=G[c]$, then by innerautos $B$ is Morita
equivalent to $c$ and we are done. If $[G:G[c]]=3$, then $B_{1}$ is Morita
equivalent to $c$, and again there are four possibilities for the Morita
equivalence class of $B$ and these are as listed. Hence suppose that
$G[c]=G[b_{2}]$, so that $G[b_{2}]<H<G$. The subgroup of $Pic(c)$ of self-
equivalences that preserve all irreducible Brauer characters is isomorphic to
$S_{3}$. Hence there is one possibility for the Morita equivalence class for
$B_{H}$ (as we are excluding the case that $H$ acts as inner automorphisms),
that $B_{1}$ is Morita equivalent to $\mathcal{O}(G_{n_{1}}\times G_{n_{2}})$.
One may check as in (i) that the inertial quotients are as stated.

(v) Suppose that $b$, and so $b_{1}$, $b_{2}$, is Morita equivalent to
$B_{0}(\mathcal{O}(A_{5}\times Q))$ for some $Q\leq D$. By index2theorem
$b_{2}$ is Morita equivalent to $B_{0}(\mathcal{O}(A_{5}\times P))$, where
$D\cong(C_{2})^{2}\times P$. By innerautos $b_{2}$ is source algebra
equivalent to the unique block of $G[b_{2}]$ covered by $B$. By pic_list
$\mathop{\rm Pic}\nolimits(b_{2})\cong C_{2}\times(P\rtimes\mathop{\rm
Aut}\nolimits(P))$. Since $G/G[b_{2}]$ is an odd order subgroup of
$\mathop{\rm Pic}\nolimits(b_{2})$ and $P$ has rank at most $2$, it is
isomorphic to a subgroup of $C_{3}$. If $P$ is not homocyclic, then
$G=G[b_{2}]$ and we are done. Suppose $P$ is homocyclic, i.e.,
$P\cong(C_{2^{n}})^{2}$ for some $n$. We may suppose $[G:G[b_{2}]]=3$. The
result then follows by Picard_application.

(vi) Suppose that $b$ is Morita equivalent to $B_{0}(\mathcal{O}(A_{5}\times
G_{n}))$ for some $n\geq 1$. As in (ii) $G/N$ has odd order, so $N_{1}=N_{2}$
and $b$ is source algebra equivalent to $b_{3}$. By pic_list $\mathop{\rm
Pic}\nolimits(b)\cong C_{2}\times S_{3}$. Since $G/G[b]$ is an odd order
subgroup of $\mathop{\rm Pic}\nolimits(b)$, it is isomorphic to a subgroup of
$C_{3}$. If $G[b]=G$, then $b$ is source algebra equivalent to $B$ and we are
done. If $[G:G[b]]=3$, then we are done by Picard_application.

∎

######  Remark 5.10.

We have not treated the case that $b$ is Morita equivalent to
$B_{0}(\mathcal{O}(A_{5}\times A_{5})$ here since we do not at present know
the Picard group of this block. This case will be treated in the main part of
the reduction, where we will make use of additional hypotheses.

Finally, the methods above may also be used to prove the following.

######  Proposition 5.11.

faithfulblocks Let $D\cong(C_{2^{n_{1}}})^{2}\times(C_{2^{n_{2}}})^{2}$ for
some $n_{1},n_{2}\in\mathbb{N}$ and consider $G=D\rtimes 3_{+}^{1+2}$, where
the centre of $3_{+}^{1+2}$ acts trivially. The $2$-blocks of $\mathcal{O}G$
correspond to the simple modules of $Z(3_{+}^{1+2})$, and the two non-
principal blocks are Morita equivalent. Further, these blocks are Morita
equivalent to the two non-principal blocks of $\mathcal{O}(D\rtimes
3_{-}^{1+2})$.

###### Proof.

Let $B$ be any faithful $2$-block of $G=D\rtimes 3_{+}^{1+2}$ or $D\rtimes
3_{-}^{1+2}$. Then $l(B)=1$. Take a maximal subgroup $N$ of $G$ and a block
$b$ of $N$ covered by $B$. Then $N\cong(D\rtimes C_{3})\times C_{3}$ or
$D\rtimes C_{9}$ and without loss of generality $b$ is Morita equivalent to
$\mathcal{O}(G_{n_{1}}\times(C_{2^{n_{2}}})^{2})$ for some $n_{1},n_{2}$. By
Picard_application(ii) there is only one possibility for the Morita
equivalence class of $B$ under the restriction that there is just one simple
module. ∎

##  6 Proof of the main theorem

We first recall the case where the defect group is normal.

######  Lemma 6.1.

normaldefect Let $B$ be a block of $\mathcal{O}G$ for a finite group $G$ with
abelian normal defect group $D$ with rank at most $4$ and inertial quotient
$E$. Then $B$ is source algebra equivalent to a block of $D\rtimes\hat{E}$,
where $E$ is the inertial quotient of $B$ and $\hat{E}/Z(\hat{E})=E$. If
$E\cong C_{3}\times C_{3}$, then $\hat{E}\cong E$ or $3^{1+2}$ with
$Z(3^{1+2})$ acting trivially on $D$. Otherwise $\hat{E}=E$.

###### Proof.

See [38, Theorem 6.14.1]. In all cases except $E\cong C_{3}\times C_{3}$, all
Sylow subgroups of $E$ are cyclic, so the Schur multiplier is trivial. For
$C_{3}\times C_{3}$ the Schur multiplier is $3$ and the result follows. ∎

We now state a result used in previous reductions for results concerning
Morita equivalence classes of blocks, that encapsulates the use of Fong-
Reynolds reductions and [35]. It appears in [2] in full, but was extracted
from the first part of the proof of [19, Proposition 4.3].

######  Lemma 6.2.

reduced_block Let $G$ be a finite group and $B$ a block of $\mathcal{O}G$ with
defect group $D$. Then there is a finite group $B$ and a block $C$ of
$\mathcal{O}H$ such that $B$ is basic Morita equivalent to $C$, a defect group
$D_{H}$ of $C$ is isomorphic to $D$ such that:

  1. (R1)

$C$ is quasiprimitive, that is, if $N\lhd H$, then $C$ covers a unique block
of $\mathcal{O}N$;

  2. (R2)

If $N\lhd H$ and $C$ covers a nilpotent block of $\mathcal{O}N$, then $N\leq
O_{p}(H)Z(H)$ with $O_{p^{\prime}}(N)\leq[H,H]$ cyclic. In particular
$O_{p^{\prime}}(G)\leq Z(G)$;

  3. (R3)

$[H:O_{p^{\prime}}(Z(H))]\leq[G:O_{p^{\prime}}(Z(G))]$.

Note that $B$ and $C$ have the same Frobenius categories.

###### Proof.

This is [2, Proposition 6.1], in which Fong-Reynolds and Külshamme-Puig
reductions are applied repeatedly. It is noted in the proof of [2, Proposition
6.1] that application of these reductions reduces $[G:O_{p^{\prime}}(Z(G))]$,
hence (R3).

∎

We call the pair $(H,C)$, where $C$ is a block of $\mathcal{O}H$, _reduced_ if
it satisfies conditions (R1) and (R2) reduced_block. If the group is clear,
then we just say $C$ is reduced.

Before proceeding we recall the definition and some properties of the
generalized Fitting subgroup $F^{*}(G)$ of a finite group $G$. Details may be
found in [6]. A _component_ of $G$ is a subnormal quasisimple subgroup of $G$.
The components of $G$ commute, and we define the _layer_ $E(G)$ of $G$ to be
the normal subgroup of $G$ generated by the components. It is a central
product of the components. The _Fitting subgroup_ $F(G)$ is the largest
nilpotent normal subgroup of $G$, and this is the direct product of $O_{r}(G)$
for all primes $r$ dividing $|G|$. The _generalized Fitting subgroup_
$F^{*}(G)$ is $E(G)F(G)$. A crucial property of $F^{*}(G)$ is that
$C_{G}(F^{*}(G))\leq F^{*}(G)$, so in particular $G/F^{*}(G)$ may be viewed as
a subgroup of $\mathop{\rm Out}\nolimits(F^{*}(G))$.

Proof of main_theorem Let $B$ be a $2$-block of $\mathcal{O}G$ for a finite
group $G$ with defect group $D$ of rank $4$ with $[G:O_{2^{\prime}}(Z(G))]$
minimised such that $B$ is not Morita equivalent to any of the blocks listed
in the statement of the theorem, or that the inertial quotient of $B$ is not
as stated. In the remainder of the proof, the reader may check that the
inertial quotients are respected at each step. By reduced_block we may assume
that $(G,B)$ is reduced.

If $D\lhd G$ and $B$ has inertial quotient $E_{B}$, then by the main result of
[32] $B$ is source algebra equivalent to either: (i) $D\rtimes E_{B}$ when
$E_{B}$ has cyclic Sylow $r$-subgroups for every $r$, that is, for every
inertial quotient other than $C_{3}\times C_{3}$; a block of $D\rtimes
3_{+}^{1+2}$ or $D\rtimes 3_{-}^{1+2}$ when $E_{B}\cong C_{3}\times C_{3}$,
where $Z(3^{1+2})$ acts trivially on $D$. By faithfulblocks the non-principal
blocks of $D\rtimes 3_{+}^{1+2}$ and $D\rtimes 3_{-}^{1+2}$ form a single
Morita equivalence class. Hence $B$ is Morita equivalent to a block listed in
the statement of the theorem, a contradiction. Hence $D$ is not normal in $G$.

Let $b^{*}$ be the unique block of $\mathcal{O}F^{*}(G)$ covered by $B$.

Write $L_{1},\ldots,L_{t}$ for the components of $G$, so $E(G)=L_{1}\cdots
L_{t}\lhd G$. Note that $G$ permutes the $L_{i}$. There must be at least one
component, since otherwise $b^{*}$ is nilpotent and so
$F^{*}(G)=Z(G)O_{2}(G)$. But $O_{2}(G)\leq D$ and $D$ is abelian, so we would
have $D\leq C_{G}(F^{*}(G))\leq F^{*}(G)=Z(G)O_{2}(G)$, so that $D\lhd G$, a
contradiction.

Write $b_{E}$ for the unique block of $E(G)$ covered by $B$ and $b_{i}$ for
the unique block of $L_{i}$ covered by $b_{E}$. We claim that no $b_{i}$ can
be nilpotent (in our minimal counterexample).

Let $Z=O_{2}(Z(E(G)))$. Write $\bar{b}_{E}$ for the unique block of
$\overline{E(G)}:=E(G)/Z$ corresponding to $b_{E}$ and $\bar{b}_{i}$ for the
unique block of $\bar{L}_{i}$ corresponding to $b_{i}$. Writing
$M:=\overline{L}_{1}\times\cdots\times\overline{L}_{t}$, where
$\overline{L}_{i}:=L_{i}Z/Z$, there is a $2^{\prime}$-group $W\leq Z(M)$ and a
block $b_{M}$ of $M$ with $W$ in its kernel such that $\overline{E(G)}=M/W$
and $b_{M}$ is isomorphic to $\bar{b}_{E}$. Then $D\cap E(G)$ is a defect
group for $b_{E}$, $(D\cap E(G))/Z$ is a defect group for $\bar{b}_{E}$ and
$b_{M}$ has defect groups isomorphic to $(D\cap E(G))/Z$. Then $\bar{b}_{i}$
has defect group $D_{i}=((D\cap E(G))Z)\cap\bar{L}_{i}$. We have that
$b_{M}=\bar{b}_{1}\otimes\cdots\otimes\bar{b}_{t}$ and $b_{M}$ has defect
group $D_{1}\times\cdots\times D_{t}$. Suppose $b_{j}$ is nilpotent for some
$j$. Let $J\subseteq\\{1,\ldots,t\\}$ correspond to the orbit of $L_{j}$ under
the permutation action of $G$ on the components. Define $L_{J}\lhd G$ to be
the product of the $L_{i}$ for $i\in J$, and write $b_{J}$ for the unique
block of $L_{J}$ covered by $b_{E}$. For $i\in J$, since $\beta_{j}$ is
nilpotent, so is $b_{i}$, hence so is $\bar{b}_{i}$. Hence the unique block
$\bar{b}_{J}$ of $L_{J}/Z$ corresponding to $b_{J}$ is also nilpotent (to see
this, observe that as above $L_{J}/Z$ is the quotient of
$\mathop{\mathchoice{\hbox{\sf\Large\lower
1.20007pt\hbox{X}}}{\hbox{\sf\large\lower
1.20007pt\hbox{X}}}{\hbox{\sf\normalsize\lower
1.20007pt\hbox{X}}}{\hbox{\sf\tiny\lower 1.20007pt\hbox{X}}}}_{i\in
J}\bar{L}_{i}$ by a central $2^{\prime}$-group and that products of nilpotent
blocks are nilpotent). It follows that $b_{J}$ is nilpotent by [49] (where the
result is stated over $k$, but follows over $\mathcal{O}$ immediately), a
contradiction. Hence no $b_{i}$ is nilpotent. In particular, it follows that
no $D_{i}$ can have rank one, and so $t\leq 2$.

Now it follows from Schreier’s conjecture that $G/E(G)$ is solvable. By Lemma
LABEL:solvablequotient:lem $DE(G)/E(G)$ is a Sylow $2$-subgroup of $G/E(G)$.
Since $DE(G)/E(G)$ is abelian, it follows that $G/E(G)$ has $2$-length at most
one, so there are normal subgroups $N_{i}$ of $G$ such that $E(G)=N_{0}\lhd
N_{1}\lhd N_{2}\lhd N_{3}=G[b_{E}]$ with $N_{1}/N_{0}$, $N_{3}/N_{2}$ of odd
order and $N_{2}/N_{1}$ a $2$-group. Further $D\leq N_{2}$ by Proposition
LABEL:Dabelian_inner:prop. Write $b_{N_{i}}$ for the unique block of
$\mathcal{O}N_{i}$ covered by $B$.

Suppose that $t=2$. Write $\bar{D}_{i}:=D_{i}/O_{2}(Z(L_{i}))$, which must
have rank $2$ for each $i$. Note that $O_{2}(G)\leq Z(E(G))$, otherwise a
quotient of $D$ would have rank greater than $4$. Further
$\bar{D}_{i}\cong(C_{2^{n_{i}}})^{2}$ for some $n_{i}$, otherwise
$\bar{b}_{i}$, and so $b_{i}$, would be nilpotent. Now
$D_{i}\cong(C_{2^{n_{i}}})^{2}\times O_{2}(Z(L_{i}))$, again since otherwise
$b_{i}$ would be nilpotent. It follows that $O_{2}(Z(L_{i}))=1$ for each $i$,
hence $O_{2}(G)=1$. Further note that each $L_{i}$ is normal in $G$, since
$N_{G}(L_{i})$ has index at most $2$ and $G=DN_{G}(L_{i})$. If $G\neq
N_{G}(L_{i})$, then some element of $D$ must permute $L_{1}$ and $L_{2}$,
contradicting our hypothesis that $D$ is abelian. It follows from [20, Theorem
1.1] and [36] that $b_{i}$ is Morita equivalent to $\mathcal{O}G_{n}$ for some
$n\geq 1$ or to $B_{0}(\mathcal{O}A_{5})$. Further, $b_{E}$ is isomorphic (in
fact basic Morita equivalent) to a block $b_{1}\otimes b_{2}$ of $L_{1}\times
L_{2}$.

Suppose $b_{i}$ is Morita equivalent to $B_{0}(\mathcal{O}A_{5})$ for each
$i$. Then by [10, Theorem 1.5] $\mathop{\rm Pic}\nolimits(b_{i})\cong C_{2}$,
and so $[G:G[b_{i}]]$ divides $2$. But then $G=DG[b_{i}]=G[b_{i}]$. It follows
that $G=G[b_{E}]$. Since the inertial quotient of $b_{E}$ is $C_{3}\times
C_{3}$, it follows that $D=[D,N_{N_{2}}(D,(b_{N_{2}})_{D})]$, which is
contained in $N_{1}$ by Proposition LABEL:defect_group_factor:prop. We have
shown that $[G:E(G)]$ is odd, hence by innerautos $B$ is Morita equivalent to
$b_{E}$, and so to $B_{0}(\mathcal{O}(A_{5}\times A_{5}))$, a contradiction.

Suppose that $b_{1}$ is Morita equivalent to $B_{0}(\mathcal{O}A_{5})$ and
$b_{2}$ to $\mathcal{O}G_{2^{n}}$ for some $n$. Then by covers_inertial(vi)
$B$ is Morita equivalent to a block in our list, a contradiction.

Suppose that $b_{1}$ and $b_{2}$ are Morita equivalent to
$\mathcal{O}G_{2^{n_{1}}}$ and $\mathcal{O}G_{2^{n_{2}}}$ for some
$n_{1},n_{2}\in\mathbb{N}$. Then by Gn_inertial $b_{1}$ and $b_{2}$ are both
inertial. Hence $b_{E}$ is inertial, and so by covers_inertial $B$ is
inertial, a contradiction.

We have ruled out $t=2$, so now suppose that $t=1$, so that $E(G)=L_{1}$ is
quasisimple. We refer to qsclassification for the possibilities for $L_{1}$
and $b_{1}$. For ease of notation, write $N=E(G)$ and $b=b_{E}$. Note that
$F^{*}(G)\cong O_{2}(G)O_{2^{\prime}}(G)N$. Here we have already established
that $O_{2^{\prime}}(G)\leq Z(G)$, however it is not immediately the case that
$O_{2^{\prime}}(G)\leq N$, since in principal $G/O_{2^{\prime}}(G)$ may have
Schur multiplier larger than that of $N/Z(N)$. Since $F^{*}(G)$ is self-
centralizing, $G/F^{*}(G)$ is isomorphic to a subgroup of $\mathop{\rm
Out}\nolimits(F^{*}(G))$.

We first consider the cases $N/Z(N)\cong SL_{2}(8)$, $SL_{2}(16)$, $J_{1}$,
$Co_{3}$ and ${}^{2}G_{2}(3^{2m+1})$, where $m\in\mathbb{N}$. In each case
$D\cap N$ has rank $3$ or $4$, so $F^{*}(G)\cong NC_{2^{n}}$ for some $n$.
Also $N/Z(N)$ has cyclic odd order outer automorphism group and trivial Schur
multiplier. It follows from [16, Lemma 3.4] that $\mathop{\rm
Aut}\nolimits(N/Z(N)$ also has trivial Schur multiplier in each of these
cases. Hence $G\cong C^{2^{n}}\times H$ for some $H$ with $N\leq
H\leq\mathop{\rm Aut}\nolimits(N)$.

Suppose that $N\cong SL_{2}(8)$ and $b$ is the principal block, which has
elementary abelian defect group of order $8$ and inertial quotient $C_{7}$.
Note that $\mathop{\rm Out}\nolimits(SL_{2}(8))\cong C_{3}$. Then by the above
$G\cong H\times C_{2^{n}}$, where $SL_{2}(8)\leq H\leq\mathop{\rm
Aut}\nolimits(SL_{2}(8))$. Hence $B$ is Morita equivalent to the principal
block of $SL_{2}(8)\times C_{2^{n}}$ or $\mathop{\rm
Aut}\nolimits(SL_{2}(8))\times C_{2^{n}}$.

Suppose that $N\cong SL_{2}(16)$ and $b$ is the principal block, which has
elementary abelian defect group of order $8$ and inertial quotient $C_{15}$.
Note that $\mathop{\rm Out}\nolimits(SL_{2}(16))=1$. Hence $G=N$ and we are
done in this case.

Suppose that $N\cong J_{1}$ and $b$ is the principal block, which has
elementary abelian defect group of order $8$ and inertial quotient
$C_{7}\rtimes C_{3}$. Note that $\mathop{\rm Out}\nolimits(J_{1})=1$, so
$G\cong J_{1}\times C_{2^{n}}$ and we are done in this case.

Suppose that $N\cong\ {}^{2}G_{2}(3^{2m+1})$ for some $m\in\mathbb{N}$ and $b$
is the principal block, which has elementary abelian defect group of order $8$
and inertial quotient $C_{7}\rtimes C_{3}$. Then $G\cong H\times C_{2^{n}}$,
where ${}^{2}G_{2}(3^{2m+1})\leq H\leq\mathop{\rm
Aut}\nolimits({}^{2}G_{2}(3^{2m+1})$. It follows from [17, Proposition 3.1]
that $B$ is Morita equivalent to $b$, and these blocks have the same inertial
quotient. In turn $b$ is Morita equivalent to $B_{0}(\mathcal{O}(\mathop{\rm
Aut}\nolimits(SL_{2}(8))\times C_{2^{n}}))$ by [43, Example 3.3].

Suppose that $N\cong Co_{3}$ and $b$ is the non-principal block with
elementary abelian defect group of order $8$. Note that $\mathop{\rm
Out}\nolimits(Co_{3})=1$, so $G\cong Co_{3}\times C_{2^{n}}$. By [31] $b$ is
Morita equivalent to $B_{0}(\mathcal{O}\mathop{\rm Aut}\nolimits(SL_{2}(8)))$,
so $B$ is Morita equivalent to $B_{0}(\mathcal{O}(\mathop{\rm
Aut}\nolimits(SL_{2}(8))\times C_{2^{n}})$.

We now move on to case (iii) of qsclassification. Then $b$ is Morita
equivalent to the principal block of $A_{5}\times P$ or $A_{4}\times P$ for
some abelian $2$-group $P$ of rank at most $2$. Since $G/N$ is solvable,
covers_inertial(iv,v) apply, and $B$ is Morita equivalent to a block on our
list, a contradiction.

It remains to consider the case that $b$ is nilpotent covered. By [46, 4.3]
$b$ is inertial. Since $F^{*}(G)=NZ(G)O_{2}(G)$ and $O_{2}(G)$ has rank at
most $2$, it follows that $\mathop{\rm Out}\nolimits(O_{2}(G))$ and
$\mathop{\rm Out}\nolimits(N)$, and so $\mathop{\rm Out}\nolimits(F^{*}(G))$
are solvable. It follows from covers_inertial that $B$ is Morita equivalent to
a block in our list, a contradiction. We have covered each possibility for $b$
as given in qsclassification, so we have established the Morita equivalences
in all cases.

Finally we observe that parts (a) and (b)(iv) are proved in [41] and [5].
$\Box$

##  7 Derived equivalences and Broué’s abelian defect group conjecture

In this section we prove broue.

We first recall Külshammer-Puig classes of blocks. Let $D$ be a defect group
for a block $B$ with Frobenius category $\mathcal{F}$. Following the
presentation in [38, Section 8.14], a Külshammer-Puig class is an element of
$H^{2}(\mathop{\rm Aut}\nolimits_{\mathcal{F}}(Q))\cong H^{2}(E,k^{\times})$.
As we have used in normaldefect, by [38, Theorem 6.14.1] the Morita
equivalence class of a block with normal defect group is determined by the
inertial quotient and the Külshammer-Puig class. If $E$ has cyclic Sylow
subgroups for all primes, then $H^{2}(E,k^{\times})$ is trivial, so in this
case the Morita equivalence class is determined just by $E$. This is the case
for all blocks considered in this paper except for those with inertial
quotient $C_{3}\times C_{3}$. Hence, since inertial quotients are preserved by
the Morita equivalences in main_theorem, in order to prove broue for inertial
quotients other than $C_{3}\times C_{3}$ it suffices to show that in each of
(b)(i)-(iv) all blocks are derived equivalent when the defect groups are
isomorphic (the result is trivial for blocks in main_theorem(a)). This follows
from [14, Theorem 4.36] since for the groups in each of (b)(i)-(iv), the
principal block of the normalizer of a Sylow $2$-subgroup $P$ is Morita
equivalent to $P\rtimes E$, where $E$ is the inertial quotient. This proves
broue when $E\not\cong C_{3}\times C_{3}$.

Now suppose that $B$ has inertial quotient $E\cong C_{3}\times C_{3}$. We must
distinguish whether the Brauer correspondent $b$ of $B$ in $N_{G}(D)$ is
Morita equivalent to $\mathcal{O}(D\rtimes E)$ or to $\mathcal{O}(D\rtimes
3_{+}^{1+2})$. A nonprincipal block of $D\rtimes 3_{+}^{1+2}$ has just one
simple module whilst all other blocks in main_theorem(b)(v) have nine simple
modules. However by [48] and [29, Proposition 5.5] $l(B)=l(b)$, so we may
distinguish the Morita equivalence class of $b$ by $l(B)$. By [14, Theorem
4.36] the principal blocks occurring in (b)(v) are derived equivalent, so
broue follows.

Acknowledgment

We thank Benjamin Sambale for a useful observation.

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