# Principal blocks with six
ordinary irreducible characters

Nguyen N. Hung  Department of Mathematics, The University of Akron, Akron, OH
44325, USA<EMAIL_ADDRESS>,  A. A. Schaeffer Fry  Dept. Mathematics
and Statistics, MSU Denver, Denver, CO 80217, USA<EMAIL_ADDRESS>and
Carolina Vallejo  Dipartimento di Matematica e Informatica ‘Ulisse Dini’,
50134 Firenze (Italy<EMAIL_ADDRESS>
###### Abstract.

We classify Sylow $p$-subgroups of finite groups whose principal $p$-blocks
have precisely six ordinary irreducible characters.

###### Key words and phrases:

Principal block, defect group, irreducible character, Sylow subgroup

###### 2010 Mathematics Subject Classification:

Primary 20C20, 20C15, 20C33.

The first author is grateful for the support of an UA Faculty Research Grant.
The second-named author is grateful for the support of a grant from the
National Science Foundation, Award No. DMS-2100912. The third author, as part
of the GNSAGA, is grateful for the support of the Istituto Nazionale di Alta
Matematica (INDAM). The authors also thank Gunter Malle for comments on an
earlier draft.

In memory of Georgia Benkart, an inspiration to all mathematicians.

##  1\. Introduction

Classifying finite groups with a given (relatively small) number of conjugacy
classes is a classical and natural problem in finite group theory [VV85,
VS07]. Such classification not only provides a handy source for testing
several predictions/conjectures involving the class number, but also plays a
critical role in the proofs of theorems that require ad hoc arguments for
groups with a small number of conjugacy classes, see [HHKM11, Mar16] for
instance.

Given a positive integer $k$, it is expected that there are finitely many
isomorphism classes of groups that can occur as defect groups of blocks with
$k$ ordinary irreducible characters (this, often referred to as Brauer’s
Problem 21 [Bra63], has been shown by Külshammer and Robinson [KR96] to be a
consequence of the Alperin-McKay conjecture [Nav18, Conjecture 9.5]). The
problem of determining all possible structures of a defect group of a block
that has a given number of irreducible characters can be viewed as the modular
analogue of the above problem on conjugacy classes. As one should anticipate,
the modular problem is much harder and largely open – for instance, it is
still unknown whether a block has precisely $3$ ordinary characters if and
only if its defect group has order $3$ (see [KNST14, Section 4] and Table 1 at
the end of this paper).

Let $p$ be a prime. Recall that the principal $p$-block of a finite group $G$
is the one containing the principal character ${\bf 1}_{G}$ and that its
defect groups are the Sylow $p$-subgroups of $G$. Thanks to the recent work of
Koshitani-Sakurai [KS21] and of Rizo-Schaeffer Fry-Vallejo [RSV21], those
Sylow $p$-subgroups of finite groups with principal $p$-blocks having up to
five irreducible characters have been completely determined. This in turn has
contributed to the solution of the principal-block case [HS21] of Héthelyi-
Külshammer’s conjecture [HK00] and to obtaining a $p$-local lower bound for
the number of height-zero characters in principal blocks [HSV23].

The purpose of this paper is to advance on the determination of the structures
of defect groups of blocks with a given number of ordinary characters. Our
main result classifies the defect groups of principal blocks with six
characters.

######  Theorem A.

Let $G$ a finite group, $p$ a prime, and $P\in{\mathrm{Syl}}_{p}(G)$. Suppose
that the principal $p$-block of $G$ has precisely six irreducible ordinary
characters. Then $|P|=9$.

Let $B_{0}(G)$, or sometimes just $B_{0}$, denote the principal $p$-block of
$G$, and let $k(B_{0})$ denote the number of ordinary irreducible characters
of $B_{0}$. Our strategy for proving Theorem A, which is somewhat different
from the above-mentioned previous work on $k(B_{0})\in\\{4,5\\}$, is to
analyse the number $k_{0}(B_{0})$ of those characters in $B_{0}$ of _height
zero_. One of the main results of [HSV23] already classifies the Sylow
structure of finite groups with $k_{0}(B_{0})\leq 5$, and perhaps
surprisingly, none of these possibilities could occur when $k(B_{0})=6$.
Therefore, we are left to deal with $k(B_{0})=k_{0}(B_{0})=6$, in which case
the Sylow subgroup $P$ must be abelian, by the recent proof of Brauer’s height
zero conjecture for principal blocks by Malle and Navarro [MN21] (note that
the more general case of the conjecture has now also been proved [MNST22]).

Our proof makes use of the Classification of Finite Simple Groups. In
particular, we have to find lower bounds for the number of
$\operatorname{Aut}(S)$-orbits of irreducible characters in the principal
block of a non-abelian simple group $S$. This has been recently studied in
connection with other problems on character degrees and character bounds
[Mar21, HS21]. With this, we are able to restrict ourselves to studying the
$\operatorname{Aut}(S)$-orbits of characters in the principal $3$-blocks of
such an $S$ with an abelian Sylow $3$-subgroup, see Theorem 2.1.

There are two isomorphism classes of groups of order $9$, namely
${\sf{C}}_{9}$ and ${\sf C}_{3}\times{\sf C}_{3}$. Either of them can occur as
the defect group of a principal block with 6 ordinary characters, as shown in
semidirect products of ${\sf C}_{2}$ acting on ${\sf{C}}_{9}$ and ${\sf
C}_{3}\times{\sf C}_{3}$ by inversion. With that being said, the ‘full
inverse’ of Theorem A is not always true: if $|P|=9$ then $k(B_{0})$ could be
either $6$ or $9$ (see the proof of Theorem 3.9). The next result, which fully
characterizes finite groups with $k(B_{0})=6$ in terms of $p$-local structure,
offers a more complete version of Theorem A.

######  Theorem B.

Let $G$ a finite group, $p$ a prime, and $P\in{\mathrm{Syl}}_{p}(G)$. Let
$B_{0}$ denote the principal $p$-block of $G$. Then $k(B_{0})=6$ if and only
if precisely one of the following holds:

  * (i)

$P={\sf C}_{9}$ and $|{\bf N}_{G}(P):{\bf C}_{G}(P)|=2$.

  * (ii)

$P={\sf C}_{3}\times{\sf C}_{3}$ and either ${\bf N}_{G}(P)/{\bf
C}_{G}(P)\in\\{{\sf C}_{4},{\sf Q}_{8}\\}$ or ${\bf N}_{G}(P)/{\bf
C}_{G}(P)\cong{\sf C}_{2}$ acts fixed-point freely on $P$.

Proofs of the main theorems are contained in Section 3 and the necessary
results on finite simple groups are proved in Section 2. At the end of the
paper, we provide a summary of what is known about the structure of defect
groups of small blocks in Table 1.

##  2\. Principal blocks of simple groups

In this section, we prove the following statements on simple groups, which
will be needed for the proof of our main theorems.

######  Theorem 2.1.

Let $S$ be a non-abelian simple group. Let $p=3$ and let $B_{0}:=B_{0}(S)$ be
the principal 3-block of $S$. Assume that $Q\in{\rm Syl}_{p}(S)$ is abelian
and $|Q|\geq 9$. If $S\leq A\leq\operatorname{Aut}(S)$, then one of the
following holds:

  1. (a)

The action of $A$ defines at least $4$ orbits on ${\rm
Irr}(B_{0})\setminus\\{{\bf 1}_{S}\\}$.

  2. (b)

$S\in\\{{\rm PSL}_{2}(q),{\rm PSL}_{3}(q),{\rm PSU}_{3}(q)\\}$ with $(3,q)=1$,
$(|A:S|,3)=3$, and the action of $A$ induces $3$ orbits on ${\rm
Irr}(B_{0}(S))\setminus\\{{\bf 1}_{S}\\}$. In this case, $k(B_{0}(A))>6$.

  3. (c)

$S={\mathrm{Alt}}_{6}={\mathrm{PSL}}_{2}(3^{2})$, $A$ has a subgroup
isomorphic to ${\rm M}_{10}$, and the action of $A$ induces $3$ orbits on
${\rm Irr}(B_{0}(S))\setminus\\{{\bf 1}_{S}\\}$.

######  Theorem 2.2.

Let $S$ be a non-abelian simple group. Let $p$ be an odd prime and let
$B_{0}:=B_{0}(S)$ be the principal $p$-block of $S$. Assume that $P\in{\rm
Syl}_{p}(S)$ is abelian. If $k(B_{0})=6$, then $p=3$ and $|P|=9$.

We refer the reader to [Nav98, Chapter 9] for basics on the block theory
involving normal subgroups and quotient groups. Recall that if $N$ is a normal
subgroup of $G$ and $B$ and $b$ are blocks of $G$ and $N$ respectively, then
$B$ is said to _cover_ $b$ if there are $\chi\in{\rm Irr}(B)$ and
$\theta\in{\rm Irr}(b)$ such that $\theta$ is an irreducible constituent of
the restriction $\chi_{N}$. It is clear that $B_{0}(G)$ covers $B_{0}(N)$. For
$\theta\in{\mathrm{Irr}}(N)$, we write ${\mathrm{Irr}}(G|\theta)$,
respectively ${\mathrm{Irr}}(B|\theta)$, for the set of those characters of
$G$, respectively $B$, containing $\theta$ as a constituent when restricted to
$N$.

######  Lemma 2.3.

Let $G$ be a finite group, $N\trianglelefteq\,G$, and $p$ a prime. Let $B_{0}$
be the principal $p$-block of $G$.

  * (i)

${\rm Irr}(B_{0}(G/N))\subseteq{\rm Irr}(B_{0})$.

  * (ii)

If $N$ is a $p^{\prime}$-group, then ${\rm Irr}(B_{0}(G/N))={\rm Irr}(B_{0})$.

  * (iii)

For every $\theta\in{\rm Irr}(B_{0}(N))$, there exists $\chi\in{\rm
Irr}(B_{0}|\theta)$.

  * (iv)

Suppose that $B\in{\rm Bl}(G)$ is the only block covering $b\in{\rm Bl}(N)$.
Then for every $\theta\in{\rm Irr}(b)$, we have ${\rm
Irr}(G|\theta)\subseteq{\rm Irr}(B)$.

  * (v)

${\rm Irr}(B_{0})\cap{\rm Irr}(G/N)$ is a union of blocks of $G/N$.

###### Proof.

Part (i) and (v) follow as $B_{0}$ dominates $B_{0}(G/N)$, see the discussion
on [Nav98, pp. 198-199] on block dominance and inclusion. Part (ii) is [Nav98,
Theorem 9.9.(c)]. Part (iii) is [Nav98, Theorem 9.4]. Part (iv) can be found
in [RSV21, Lemma 1.2], for instance. ∎

We begin by proving Theorem 2.1 in various cases.

######  Lemma 2.4.

Theorem 2.1 holds if $S$ is a sporadic group, the Tits group
${}^{2}\operatorname{F}_{4}(2)^{\prime}$, a group of Lie type with exceptional
Schur multiplier, or an alternating group.

###### Proof.

First, if $S$ is a sporadic group, the Tits group
${}^{2}\operatorname{F}_{4}(2)^{\prime}$, a group of Lie type with exceptional
Schur multiplier (see [GLS98, Table 6.1.3] for a list), or
${\mathrm{Alt}}_{6}$, then the statement can be verified using GAP [GAP] and
the information available in its Character Table Library. In particular, we
see from this that case (c) occurs.

So, we now suppose that $S={\mathrm{Alt}}_{n}$ is an alternating group with
$n>6$. Then $A\in\\{{\mathrm{Alt}}_{n},{\mathrm{Sym}}_{n}\\}$, and it suffices
to show that ${\mathrm{Irr}}(B_{0}({\mathrm{Sym}}_{n}))$ contains at least 5
characters whose restrictions to ${\mathrm{Alt}}_{n}$ are distinct. Recall
that the characters of ${\mathrm{Sym}}_{n}$ are in bijection with partitions
of $n$, and non-conjugate partitions yield distinct restrictions to
${\mathrm{Alt}}_{n}$. The characters in $B_{0}({\mathrm{Sym}}_{n})$ are those
with $3$-core $(r)$, where $n=3m+r$, $0\leq r<3$ (see, e.g. [Ols93, Theorem
(11.1)]). Now, we see that there are at least 5 non-conjugate partitions with
$3$-core $(r)$ for $m\geq 2$, yielding the claim. ∎

We say that $\theta\in{\rm Irr}(N)$, where $N\trianglelefteq\,G$, _extends_ to
$G$ (or is _extendable_ to $G$) if there is some $\chi\in{\rm Irr}(G)$ such
that $\chi_{N}=\theta$. In that case,

${\rm Irr}(G|\theta)=\\{\beta\chi\ |\ \beta\in{\rm Irr}(G/N)\\}$

by a theorem of Gallagher [Isa76, Corollary 6.17].

The next observation will be useful in some of the remaining cases.

######  Lemma 2.5.

Let $X$, $Y$, $\widetilde{X}$, and $E$ be finite groups such that
$X\lhd\widetilde{X}\lhd\widetilde{X}E$; $X\lhd Y\leq\widetilde{X}E$; and $E$
is abelian. Suppose further that $p$ is a prime such that
$[\widetilde{X}:X]=p$ and $p\mid[Y:X]$. Let $\widetilde{\chi}\in{\rm
Irr}(B_{0}(\widetilde{X}))$ be a character in the principal $p$-block of
$\widetilde{X}$ that is extendable to $\widetilde{X}E$ and restricts to
$\chi\in{\rm Irr}(X)$. Then there exist at least two characters in ${\rm
Irr}(B_{0}(Y))$ lying above $\chi$.

###### Proof.

Note that our assumptions imply that $Y/(\widetilde{X}\cap Y)$ is abelian and
$\widetilde{X}\cap Y\in\\{\widetilde{X},X\\}$. Let $X\lhd Y_{p}\lhd Y$ such
that $Y_{p}=\widetilde{X}$ if $\widetilde{X}\cap Y=\widetilde{X}$ and such
that $|Y_{p}/X|=p$ if $\widetilde{X}\cap Y=X$. Then in either case, we have
$|Y_{p}/X|=p$ and $Y/Y_{p}$ is abelian.

Note that $\chi$ has $p$ distinct extensions $\chi_{1},\ldots,\chi_{p}$ to
$Y_{p}$, which all must lie in $B_{0}(Y_{p})$ by Lemma 2.3(iv) and the fact
that $[Y_{p}:X]=p$, so $B_{0}(Y_{p})$ is the unique $p$-block of $Y_{p}$ above
$B_{0}(X)$. Since $\chi$ extends to $\widetilde{X}E$, and hence $Y$, at least
one of these, say $\chi_{1}$, extends to $Y$. In the case that
$Y_{p}=\widetilde{X}$, we may even specify $\chi_{1}:=\widetilde{\chi}$. Since
$Y/Y_{p}$ is abelian, it follows that every character in ${\rm
Irr}(Y|\chi_{1})$ is an extension, by Gallagher’s theorem [Isa76, Corollary
6.17]. In particular, $\chi_{1}$ extends to some member of $B_{0}(Y)$ lying
above $\chi$, by Lemma 2.3(iii). But, note that by Lemma 2.3(iii), there is
also a member of ${\rm Irr}(B_{0}(Y)|\chi_{2})$, also lying above $\chi$, but
that this character cannot lie above $\chi_{1}$ since $\chi_{1}$ and
$\chi_{2}$ are not $Y$-conjugate. This yields at least two distinct members of
${\rm Irr}(B_{0}(Y)|\chi)$, as desired. ∎

In what follows, for $\epsilon\in\\{\pm 1\\}$, we will use
${\mathrm{PSL}}_{n}^{\epsilon}(q)$ to denote the group ${\mathrm{PSL}}_{n}(q)$
of type $\operatorname{A}_{n-1}$ if $\epsilon=1$ and the group
${\mathrm{PSU}}_{n}(q)$ of type ${}^{2}\operatorname{A}_{n-1}$ if
$\epsilon=-1$, and we will use analogous notation for the related groups
${\mathrm{SL}}_{n}^{\epsilon}(q)$, ${\mathrm{GL}}_{n}^{\epsilon}(q)$, and
${\mathrm{PGL}}_{n}^{\epsilon}(q)$.

######  Lemma 2.6.

Theorem 2.1 holds when $S$ is one of the groups ${\mathrm{PSL}}_{2}(q)$ or
${\mathrm{PSL}}_{3}^{\epsilon}(q)$ with $q=q_{0}^{f}$ a power of a prime
$q_{0}\neq 3$.

###### Proof.

Let $S={\mathrm{PSL}}_{n}^{\epsilon}(q)$ with $n\in\\{2,3\\}$ and
$q=q_{0}^{f}$ with $q_{0}\neq 3$ a prime. Write

$G:={\mathrm{SL}}_{n}^{\epsilon}(q),\widetilde{S}:={\mathrm{PGL}}_{n}^{\epsilon}(q),\text{
and }\widetilde{G}:={\mathrm{GL}}_{n}^{\epsilon}(q)$

for the appropriate choice of $n,\epsilon$. In this case, the dual group
$\widetilde{G}^{\ast}$ is isomorphic to $\widetilde{G}$, and we identify the
two groups. Note that $\operatorname{Aut}(S)=\widetilde{S}\rtimes D$, where
$D$ is an appropriate group generated by field and graph automorphisms. (See
e.g. [GLS98, Theorem 2.5.12].) In this case, $D$ is further abelian. Let
$e\in\\{1,2\\}$ be the order of $\epsilon q$ modulo $3$. The unipotent
characters of $\widetilde{G}$ (or $\widetilde{S}$, $S$) are in bijection with
partitions of $n$, and by [FS82], two such characters lie in the same block if
and only if they correspond to partitions with the same $e$-core. In the case
$e=1$ and $S={\mathrm{PSL}}_{3}^{\epsilon}(q)$, we see from this that there
are two nontrivial unipotent characters in $B_{0}(S)$, and these are
$\operatorname{Aut}(S)$-invariant (see e.g. [Mal08, Theorem 2.5]). In the
remaining cases, there is one nontrivial unipotent character in $B_{0}(S)$
(namely, the Steinberg character $\mathbf{St}_{S}$), which is again
$\operatorname{Aut}(S)$-invariant.

(I) Suppose first that $|Q|>9$.

(Ia) If $S={\mathrm{PSL}}_{3}^{\epsilon}(q)$ with $e=1$, then this means that
$9\mid(q-\epsilon)$ and all three unipotent characters lie in the principal
block. Let $a_{1},a_{2}\in C_{q-\epsilon}\leq{\mathbb{F}}_{q^{2}}^{\times}$
such that $|a_{1}|=3$ and $|a_{2}|=9$. Then for $i=1,2$, let

$s_{i}:={\mathrm{diag}}(a_{i},a_{i}^{-1},1)\in\widetilde{G}.$

Then each $s_{i}$ defines a semisimple character $\chi_{s_{i}}$ of
$\widetilde{G}$ that lies in $B_{0}(\widetilde{G})$ using [CE04, Theorem
9.12], is trivial on ${\bf Z}(\widetilde{G})$ since
$s_{i}\in[\widetilde{G},\widetilde{G}]=G$ (see e.g. [SFT23, Prop. 2.7]), and
such that $s_{1}^{\alpha}z$ is not $\widetilde{G}$-conjugate to $s_{2}$ for
any $z\in{\bf Z}(\widetilde{G})$ and $\alpha\in D$ (since semisimple classes
in $\widetilde{G}$ are determined by their eigenvalues and the eigenvalues of
$s_{1}^{\alpha}$ still have order 3). Further, there is an isomorphism
$z\mapsto\hat{z}$ between ${\bf Z}(\widetilde{G})$ and ${\rm
Irr}(\widetilde{G}/G)$, such that $\chi_{sz}=\chi_{s}\hat{z}$ in this
situation for $s\in\widetilde{G}$ semisimple. (See [CE04, (8.19) and
Proposition 8.26]). Then since $\chi_{s_{1}}^{\alpha}=\chi_{s_{1}^{\alpha}}$
by [NTT08, Corollary 2.5], we see $\chi_{s_{1}}^{\alpha}$ and $\chi_{s_{2}}$
must necessarily have distinct restrictions to $S=G/{\bf Z}(G)\cong G{\bf
Z}(\widetilde{G})/{\bf Z}(\widetilde{G})$, and we have obtained at least two
additional $\operatorname{Aut}(S)$-orbits in ${\mathrm{Irr}}(S)$ by
restriction.

(Ib) Now let $S={\mathrm{PSL}}_{3}^{\epsilon}(q)$ with $e=2$ or
$S={\mathrm{PSL}}_{2}(q)$, so that $Q$ is cyclic and the condition $|Q|>9$
means $27\mid(q^{2}-1)$. Let
$a_{1},a_{2},a_{3}\in{\mathbb{F}}_{q^{2}}^{\times}$ with orders
$|a_{i}|=3^{i}$ for $i=1,2,3$. Then considering $s_{i}\in\widetilde{G}$ whose
nontrivial eigenvalues are $\\{a_{i},a_{i}^{-1}\\}$, we again obtain
$\chi_{s_{i}}\in{\rm Irr}(B_{0}(\widetilde{G}))$, trivial on ${\bf
Z}(\widetilde{G})$, and hence $\chi_{s_{i}}$ may be identified with a
character in ${\rm Irr}(B_{0}(\widetilde{S}))$. Suppose now for a
contradiction that $\chi_{s_{i}}^{\alpha}\hat{z}$ restricts to the same
character of $S$ as $\chi_{s_{j}}$ for $i\neq j\in\\{1,2,3\\}$, some
$\alpha\in D$, and $z\in{\bf Z}(\widetilde{G})$. This means that
$s_{i}^{\alpha}z$ is $\widetilde{G}$-conjugate to $s_{j}$ and that the
character $\hat{z}$ must be trivial on ${\bf Z}(\widetilde{G})$. Then we see
that the corresponding $z$ (and hence $\hat{z}$) must have order a nontrivial
power of $3$, contradicting that $3$ does not divide $|\widetilde{S}/S|$ in
the cases being considered. This yields our additional 3
$\operatorname{Aut}(S)$-orbits in this case.

(II) Finally, assume that we are in the last situation:
$S={\mathrm{PSL}}_{2}(q)$ or ${\mathrm{PSL}}_{3}^{\epsilon}(q)$ and $|Q|=9$,
and further assume that $A$ defines fewer than 4 orbits on ${\rm
Irr}(B_{0}(S))\setminus{\bf 1}_{S}$.

(IIa) First, if $e=1$ and $S={\mathrm{PSL}}_{3}^{\epsilon}(q)$, this means
that $3\mid\mid(q-\epsilon)$. In this case, we still have two nontrivial,
$\operatorname{Aut}(S)$-invariant, unipotent characters in ${\rm Irr}(B_{0})$.
Taking $s_{1}=\mathrm{diag}(a_{1},a_{1}^{-1},1)$ as before with $|a_{1}|=3$,
the corresponding semisimple character $\chi_{s_{1}}$ of $\widetilde{G}$ is
trivial on ${\bf Z}(\widetilde{G})$, lies in $B_{0}(\widetilde{S})$, and
restricts to the sum of three characters in $B_{0}(S)$. So, if $S\lhd
A\leq\operatorname{Aut}(S)$ with $3\nmid[A:S]$, these characters must also be
invariant under $A$, giving more than 4 $A$-orbits on
${\mathrm{Irr}}(B_{0}(S))$. So, assume that $3\mid[A:S]$. By [Mal08, Theorems
2.4 and 2.5], the unipotent characters extend to $\operatorname{Aut}(S)$. Now,
by applying Lemma 2.5 with $(X,\widetilde{X},Y,E)=(S,\widetilde{S},A,D)$ to
each unipotent character, we obtain at least 6 characters just from those
above the three unipotent characters, and hence more than 6 in total. Hence we
are in the situation of (b).

(IIb) We are left with the case that $9\mid\mid(q^{2}-1)$, so that
$9\mid\mid(q-\eta)$ for some $\eta\in\\{\pm 1\\}$ and $Q$ is cyclic of size
$9$. (In particular, we have $e=2$ and $\eta=-\epsilon$ in case
$S={\mathrm{PSL}}_{3}^{\epsilon}(q)$.) Here the only nontrivial unipotent
character in $B_{0}({S})$ is the Steinberg character $\mathbf{St}$, and there
is a unique unipotent block of $\widetilde{G}$ with positive defect. Let
$a_{1},a_{2}\in{\mathbb{F}}_{q^{2}}^{\times}$ and
$s_{1},s_{2}\in\widetilde{G}$ be defined exactly as in the case (Ib) above.
Then $\chi_{s_{i}}\in{\mathrm{Irr}}(B_{0}(\widetilde{S}))$ and $\chi_{s_{1}}$
and $\chi_{s_{2}}$ lie in distinct $\operatorname{Aut}(S)$-orbits, as before.
Note that each $\chi_{s_{i}}$ is irreducible on $S$ using the same arguments
as before, and that $\chi_{s_{1}}$ is further $D$-invariant, since any element
of $D$ either inverts or stabilizes the eigenvalues of order $3$. From the
restrictions of $\chi_{s_{1}}$ and $\chi_{s_{2}}$ to $S$, in addition to
$\mathbf{St}$, this yields three $\operatorname{Aut}(S)$-orbits on ${\rm
Irr}(B_{0}(S))\setminus\\{{\bf 1}_{S}\\}$. Then since we have assumed we do
not have four $A$-orbits on this set, the remaining characters in ${\rm
Irr}(B_{0}(S))$ must be $A$-conjugate to the restriction of $\chi_{s_{2}}$ to
$S$. But note this means the three choices of pairs $\\{a_{2},a_{2}^{-1}\\}$
with $|a_{2}|=9$ must yield $A$-conjugate characters, say
$\chi_{s_{2}},\chi_{s_{2}^{\prime}},\chi_{s_{2}^{\prime\prime}}$, and hence
$3$ divides $|A/S|$.

Hence, we see that if $A$ is an almost simple group with socle $S$ permitting
only 3 orbits on ${\rm Irr}(B_{0})\setminus\\{{\bf 1}_{S}\\}$, then
$3\mid|A/S|$ and $A/(\widetilde{S}\cap A)$ is abelian. Another application of
Lemma 2.5 applied to $\mathbf{1}_{S}$, $\mathbf{St}$, and $\chi_{s_{1}}$ now
forces at least 6 characters in ${\rm Irr}(B_{0}(A))$, along with at least one
more above $\chi_{s_{2}}$. ∎

######  Corollary 2.7.

Theorem 2.1 holds when $S$ is one of the groups ${\mathrm{PSL}}_{2}(q)$ or
${\mathrm{PSL}}_{3}^{\epsilon}(q)$ with $q=q_{0}^{f}$ a power of a prime
$q_{0}$.

###### Proof.

From Lemma 2.6, we may assume that $q_{0}=3$. Then the condition that $Q$ is
abelian and $|Q|\geq 9$ means that $S={\mathrm{PSL}}_{2}(3^{f})$ with $f\geq
2$. (See, e.g. [SZ16, Theorem].) Since the case $f=2$ is covered by Lemma 2.4,
we assume that $f>2$.

Here $B_{0}(S)$ contains all irreducible characters, aside from the Steinberg
character (see [Cab18, Theorems 1.18 and 3.3]). Then we see from the well-
known character table for $S$ that there are three distinct character degrees
in ${\rm Irr}(B_{0}(S))\setminus\\{{\bf 1}_{S}\\}$, and it suffices to show
that there are two semisimple characters of the same degree that are not
$\operatorname{Aut}(S)$-conjugate. We will employ similar strategies to the
second paragraph of the proof of Lemma 2.6.

Note that since $q=3^{f}\geq 27$, at least one of $q-\eta$ for $\eta\in\\{\pm
1\\}$ is a composite number of the form $4m$ with $m\geq 7$. Then
$C_{q-\eta}\leq{\mathbb{F}}_{q^{2}}^{\times}$ contains two elements
$\zeta_{1},\zeta_{2}$ of order larger than 4 and satisfying
$|\zeta_{1}|\not\in\\{|\zeta_{2}|,2|\zeta_{2}|\\}$. For $i=1,2$, let $s_{i}$
be a semisimple element of ${\mathrm{GL}}_{2}(q)$ with eigenvalues
$\\{\zeta_{i},\zeta_{i}^{-1}\\}$. Then the semisimple characters
$\chi_{s_{i}}$ of ${\mathrm{GL}}_{2}(q)$ corresponding to the $s_{i}$ will
restrict irreducibly to ${\mathrm{SL}}_{2}(q)$ (since $s_{i}z$ cannot be
conjugate to $s_{i}$ for any $1\neq z\in{\bf Z}({\mathrm{GL}}_{2}(q))$) and be
trivial on the center (since
$s_{i}\in{\mathrm{SL}}_{2}(q)=[{\mathrm{GL}}_{2}(q),{\mathrm{GL}}_{2}(q)]$).
Further, the restrictions of $\chi_{s_{1}}$ and $\chi_{s_{2}}$ to
${\mathrm{PSL}}_{2}(q)$ cannot be conjugate under field automorphisms since
$\chi_{s_{1}}^{\alpha}=\chi_{s_{1}^{\alpha}}$ for $\alpha\in D$, where we
write $\operatorname{Aut}(S)=\widetilde{S}\rtimes D$ as in the proof of Lemma
2.6, using [NTT08, Corollary 2.5], and ${s_{1}}^{\alpha}$ cannot be conjugate
to $s_{2}z$ for $z\in{\bf Z}({\mathrm{GL}}_{2}(q))$ of order dividing $2$.
(Recall that the $z\in{\bf Z}({\mathrm{GL}}_{2}(q))$ are in bijection with
$\hat{z}\in{\mathrm{Irr}}({\mathrm{GL}}_{2}(q)/{\mathrm{SL}}_{2}(q))$, and
that $\chi_{s_{2}z}=\chi_{s_{2}}\hat{z}$. Further, if $|z|>2$, then $\hat{z}$
is not trivial on ${\bf Z}({\mathrm{GL}}_{2}(q))$.) This shows that the
restrictions of $\chi_{s_{1}}$ and $\chi_{s_{2}}$ to $S$ are not
$\operatorname{Aut}(S)$-conjugate, as desired. ∎

We now complete the proof of Theorem 2.1, which essentially follows from the
observations in [RSV21, Section 3].

###### Proof of Theorem 2.1.

From Lemma 2.4 and Corollary 2.7, we may assume that $S$ is a simple group of
Lie type defined over ${\mathbb{F}}_{q}$ with nonexceptional Schur multiplier,
where $q$ is a power $q=q_{0}^{f}$ of a prime $q_{0}$ and that
$S\not\in\\{{\mathrm{PSL}}_{2}(q),{\mathrm{PSL}}_{3}^{\epsilon}(q)\\}$.
Further, $q_{0}\neq 3$, as otherwise $Q$ is not abelian by [SZ16, Theorem],
since $S\neq{\mathrm{PSL}}_{2}(q)$.

Assume first that $S$ is of exceptional type, including
${}^{2}\operatorname{F}_{4}(q)$ and ${}^{3}\operatorname{D}_{4}(q)$. Then the
proof of [RSV21, Lemma 3.7] yields more than 4 orbits under
$\operatorname{Aut}(S)$ in ${\rm Irr}(B_{0}(S))$, and we are done in this
case. (Note that we do not need to consider the cases
${}^{2}\operatorname{G}_{2}(q)$ and ${}^{2}\operatorname{B}_{2}(q)$, since we
would have $q_{0}=3$ in the first case and $3$ does not divide $|S|$ in the
second.)

We are left with the case that $S$ is of classical type
$\operatorname{A}_{n-1}$ or ${}^{2}\operatorname{A}_{n-1}$ with $n\geq 4$,
$\operatorname{B}_{n}$ with $n\geq 3$, $\operatorname{C}_{n}$ with $n\geq 2$,
or $\operatorname{D}_{n}$ or ${}^{2}\operatorname{D}_{n}$ with $n\geq 4$.
Here, the proof of [RSV21, Lemma 3.8] shows that we can assume $me\leq 4$ when
$S={\mathrm{PSL}}_{n}^{\epsilon}(q)$, where $e\in\\{1,2\\}$ is the order of
$\epsilon q$ modulo 3 and $n=me+r$ with $0\leq r<e$. Similarly, the proof of
[RSV21, Lemma 3.10] shows that there are at least 5
$\operatorname{Aut}(S)$-orbits in $B_{0}(S)$ in the remaining classical cases,
except possibly if $S=\operatorname{C}_{2}(q)$. (Note that since $p=3$, we
have $e=1$ and $n=m$ for these remaining classical cases in the notation of
loc. cit.) If $S=\operatorname{C}_{2}(q)$, then there is at worst one pair of
unipotent characters in $B_{0}(S)$ that are interchanged by elements of
$\operatorname{Aut}(S)$ (see [Mal08, Theorem 2.5]). Then the same arguments as
in [RSV21, Lemma 3.10] again finish this case, since $k(2,2)=5$ (in the
notation of loc. cit).

Finally, we assume $S={\mathrm{PSL}}_{4}^{\epsilon}(q)$ or
${\mathrm{PSL}}_{5}^{\epsilon}(q)$ with $em=4$. Recall that by [FS82],
unipotent characters lie in the same block if and only if they correspond to
partitions with the same $e$-core. We see that there are 5 partitions of $4$
with trivial $e$-core and similarly 5 partitions of $5$ with $e$-core (1), and
hence there exist 5 unipotent characters in $B_{0}(S)$, which are
$\operatorname{Aut}(S)$-invariant by [Mal08, Theorem 2.5]. ∎

We are now ready to prove Theorem 2.2.

###### Proof of Theorem 2.2.

First, assume $P$ is cyclic. Then by Dade’s cyclic-defect theory [Dad66,
Theorem 1], we have $k(B_{0})=f+\frac{|P|-1}{f}$, where $f:=[{\bf
N}_{S}(P):{\bf C}_{S}(P)]$. This forces $|P|=9$ when $k(B_{0})=6$.

Hence we assume that $P$ is non-cyclic, and for contradiction, we assume that
$k(B_{0})=6$ and $|P|\neq 9$. By [HS21, Theorem 1.1], we know that
$k(B_{0})\geq 2\sqrt{p-1}$, and hence we may further assume that
$p\in\\{3,5,7\\}$, so that $|P|\geq 25$.

As before, the cases of sporadic groups, the Tits group,
${}^{2}\operatorname{G}_{2}(3)^{\prime}$, and groups of Lie type with
exceptional Schur multiplier can be seen using GAP.

Next suppose $S={\mathrm{Alt}}_{n}$ is a simple alternating group and let
$n=pm+r$ with $m,r$ integers satisfying $m\geq 1$ and $0\leq r<p$. Then

$k(B_{0})\geq\frac{1}{2}k(B_{0}({\mathrm{Sym}}_{n}))=\frac{1}{2}k(B_{0}({\mathrm{Sym}}_{pm}))$

where the equality comes from [MO83, Theorem (1.10)]. Note that our assumption
$|P|\geq 25$ forces $m\geq 2$ if $p\geq 5$ and $m\geq 3$ if $p=3$. In these
cases, we can explicitly find at least 13 partitions of $pm$ with trivial
$p$-core, so that $k(B_{0}({\mathrm{Sym}}_{pm}))\geq 13$ and $k(B_{0})>6$.

Finally, let $S$ be a simple group of Lie type defined over $\mathbb{F}_{q}$
with $q$ a power of some prime. If $p\mid q$, then the condition $P$ is
abelian again leaves only $S={\mathrm{PSL}}_{2}(q)$, and the condition
$|P|\geq 25$ forces $q\geq 25$. Now, by [Cab18, Theorem 3.3], we have
$k(B_{0})=k(S)-1>6$, using the well-known character table for $S$.

Hence we assume $S$ is a simple group of Lie type defined over
$\mathbb{F}_{q}$ with $p\nmid q$. We remark that our proof in this remaining
case follows the work in [RSV21] closely, although we need to exhibit one more
character here than we needed there.

Assume first that $S$ is of exceptional type. Note that
${}^{2}\operatorname{B}_{2}(q)$ and ${}^{2}\operatorname{G}_{2}(q)$ have
cyclic Sylow $p$-subgroups, $B_{0}({}^{2}\operatorname{F}_{4}(q))$ has more
than 6 characters, using [Mal90] and $B_{0}({}^{3}\operatorname{D}_{4}(q))$
has more than 6 characters using [DM87], so we further assume that $S$ is not
of Suzuki, Ree, or triality type. If the order $d_{p}(q)$ of $q$ modulo $p$ is
not a so-called regular number, then (with our previous assumptions) we see
from [BMM93, Table 2] that $B_{0}(S)$ contains more than 6 characters. So,
assume that $d_{p}(q)$ is regular, so that every $p^{\prime}$-degree unipotent
character lies in $B_{0}(S)$ (see e.g. [RSV21, Lemma 3.6]). For
$S=\operatorname{G}_{2}(q)$, we see using [Sha07] that $B_{0}(S)$ contains
more than 6 characters. In the remaining exceptional groups
$\operatorname{F}_{4}(q)$, $\operatorname{E}_{6}^{\pm}(q)$,
$\operatorname{E}_{7}(q)$, and $\operatorname{E}_{8}(q)$, we see using [Car85,
Section 13.9] that there are more than 6 $p^{\prime}$-degree unipotent
characters when $P$ is noncyclic and abelian and $d_{p}(q)$ is regular, so we
are done with the exceptional groups.

We are left with the case that $S$ is a finite classical group. If $S$ is
${\mathrm{PSL}}_{n}^{\epsilon}(q)$ with $\epsilon\in\\{\pm\\}$ and $n\geq 2$,
let $e$ be $d_{p}(\epsilon q)$. If $S$ is
${\mathrm{P\Omega}}_{2n}^{\epsilon}(q)$ with $n\geq 4$,
${\mathrm{P\Omega}}_{2n+1}(q)$ with $n\geq 3$, or ${\mathrm{PSp}}_{2n}(q)$
with $n\geq 2$, let $e$ be $d_{p}(q^{2})$. Arguing as in [HS21, Sections 6 and
7], we have the number of unipotent characters in $B_{0}(S)$ is at least
$k(e,m)$, $k(2e,m)/2$, $k(2e,m)$, and $k(2e,m)$, in the cases if
$S={\mathrm{PSL}}_{n}^{\epsilon}(q)$, ${\mathrm{P\Omega}}_{2n}^{\epsilon}(q)$,
${\mathrm{P\Omega}}_{2n+1}(q)$, or ${\mathrm{PSp}}_{2n}(q)$, respectively.
Here $m$ is such that $n=em+r$ with $0\leq r<e$ and $k(s,t)$ can be computed
as in [Ols84, Lemma 1]. Further, our assumption that $P$ is abelian but not
cyclic forces $p\geq m>1$. With this and our assumption that $P$ is not
cyclic, we have the number of unipotent characters in $B_{0}(S)$ is at least 7
unless possibly if $(e,m)\in\\{(1,3),(1,4),(2,2)\\}$ if
$S={\mathrm{PSL}}_{n}^{\epsilon}(q)$ or $(e,m)\in\\{(1,2),(1,3)\\}$ otherwise.
In the latter cases, since $e=1$, we have $n=m$, so that
$S={\mathrm{PSp}}_{2n}(q)$ or ${\mathrm{P\Omega}}_{2n+1}(q)$, and the case
$(e,m)=(1,3)$ gives 10 unipotent characters in $B_{0}(S)$. Then we may assume
$S={\mathrm{PSL}}_{n}^{\epsilon}(q)$ with $(e,m)\in\\{(1,3),(1,4),(2,2)\\}$ or
$S={\mathrm{PSp}}_{4}(q)$ with $(e,m)=(1,2)$. Let
$G={\mathrm{SL}}_{n}^{\epsilon}(q)$ and
$\widetilde{G}={\mathrm{GL}}_{n}^{\epsilon}(q)$ in the first cases and
$G=\widetilde{G}={\mathrm{Sp}}_{4}(q)$ in the latter.

First let $(e,m)\neq(1,3)$, so that there are at least 5 unipotent characters
in $B_{0}(S)$. We may consider two semisimple characters $s_{1},s_{2}$ of
$\widetilde{G}$ corresponding to semisimple elements of
$\widetilde{G}^{\ast}\in\\{{\mathrm{GL}}_{n}^{\epsilon}(q),{\mathrm{SO}}_{5}(q)\\}$
with nontrivial eigenvalues $\\{a,a^{-1},b,b^{-1}\\}$, where $a,b\in
C_{(q^{2}-1)_{p}}\leq{\mathbb{F}}_{q^{2}}^{\times}$ satisfy $a\neq b$ in the
case of $s_{1}$ and $a=b$ in the case of $s_{2}$. These necessarily give
distinct characters on restriction to $G$ since $s_{1}z$ is not conjugate to
$s_{2}$ for any $1\neq z\in{\bf Z}(\widetilde{G}^{\ast})$ and are trivial on
${\bf Z}(G)$, as they lie in $[\widetilde{G}^{\ast},\widetilde{G}^{\ast}]$,
which yields two more members of ${\rm Irr}(B_{0}(S))$.

Finally, let $(e,m)=(1,3)$, so we have $e=1$, $p>3$, and
$S={\mathrm{PSL}}_{3}^{\epsilon}(q)$. Here $B_{0}(S)$ contains all 3 unipotent
characters and therefore $B_{0}(G)$ contains all characters lying in Lusztig
series indexed by semisimple elements $s\in G^{\ast}$ with $|s|$ a power of
$p$ (see [CE04, Theorem 9.12]). We may obtain three semisimple elements
$s_{1},s_{2},s_{3}$ with eigenvalues $\\{a,a,a^{-2}\\}$,
$\\{a,b,(ab)^{-1}\\}$, and $\\{a,a^{-1},1\\}$ with $a,b\in C_{q-\epsilon}$;
$|a|=p=|b|$; and $a\neq b$, whose Lusztig series give distinct restrictions to
$G$ and are trivial on ${\bf Z}(G)$ arguing as before. Further, the series
$\mathcal{E}(\widetilde{G},s_{1})$ contains two characters, corresponding to
the two unipotent characters of ${\bf
C}_{\widetilde{G}}(s_{1})\cong{\mathrm{GL}}_{2}^{\epsilon}(q)\times{\mathrm{GL}}_{1}^{\epsilon}(q)$.
This yields at least 4 additional characters in $B_{0}(S)$, and we are done. ∎

##  3\. Principal blocks with six irreducible characters

The purpose of this section is to prove Theorems A and B.

###  3.1. Preliminaries

When $p$ is a prime, we write $k_{p}$ for the $p$-part of an integer $k$.
Also, $l(B)$ denotes the number of irreducible Brauer characters in a block
$B$.

######  Lemma 3.1.

Let $G$ be a finite group and $p$ a prime. Let $B_{0}$ denote the principal
$p$-block of $G$.

  * (i)

$k(B_{0})\geq|{\bf Z}(G)|_{p}l(B_{0})$.

  * (ii)

$l(B_{0})=1$ if, and only if, $G$ has a normal $p$-complement.

###### Proof.

Part (i) is a direct consequence of [Nav98, Theorem 5.12]. Part (ii) is
[Nav98, Corollary 6.13]. ∎

######  Lemma 3.2.

Let $M\trianglelefteq\,G$ and $P\in{\rm Syl}_{p}(G)$. If $P{\bf
C}_{G}(P)\subseteq M$, then $B_{0}(G)$ is the only block of $G$ covering
$B_{0}(M)$. In particular, $k(G/M)<k_{0}(B_{0}(G))$ as long as $P>1$.

###### Proof.

See [HSV23, Lemma 2.3]. ∎

We will also make use of Alperin-Dade’s theory of isomorphic principal blocks.

######  Theorem 3.3 (Alperin-Dade correspondence).

Suppose that $N$ is a normal subgroup of $G$, with $G/N$ a $p^{\prime}$-group.
Let $P\in{\rm Syl}_{p}(G)$ and assume that $G=N{\bf C}_{G}(P)$. Then
restriction of characters defines a natural bijection between the irreducible
characters of the principal $p$-blocks of $G$ and $N$. In particular,
$k(B_{0}(G))=k(B_{0}(N))$.

###### Proof.

The case where $G/N$ is solvable was proved in [Alp76] and the general case in
[Dad77]. ∎

Recall that by Gallagher’s theorem, if $\theta\in{\rm Irr}(N)$, where
$N\trianglelefteq\,G$, extends to $G$, then $|{\rm Irr}(G|\theta)|=k(G/N)$.
The following is well-known and guarantees the existence of an extension.

######  Lemma 3.4.

Let $N\trianglelefteq\,G$ and $\theta\in{\rm Irr}(N)$ be $G$-invariant. If the
Sylow subgroups of $G/N$ are cyclic (for all primes dividing $|G/N|$), then
$\theta$ extends to $G$.

###### Proof.

The statement follows by [Isa76, Theorem 11.7 and Corollary 11.21]. ∎

We also need a version of [HSV23, Lemma 2.2]. We write ${\rm
Irr}_{p^{\prime}}(G)$ to denote the subset of ${\rm Irr}(G)$ consisting of
those characters with degree coprime to $p$.

######  Lemma 3.5.

Let $N\trianglelefteq\,G$ and let $B_{0}=B_{0}(G)$ be the principal $p$-block
of $G$. If $\theta\in{\rm Irr}(B_{0}(N))$ extends to $\chi\in{\rm Irr}(B_{0})$
with $(\chi(1),p)=1$, then

$|{\rm Irr}_{p^{\prime}}(B_{0}|\theta)|\geq k_{0}(B_{0}(G/N)),$

where ${\rm Irr}_{p^{\prime}}(B_{0}|\theta)={\rm Irr}_{p^{\prime}}(G)\cap{\rm
Irr}(B_{0}|\theta)$.

###### Proof.

By Gallagher’s theorem, ${\rm
Irr}_{p^{\prime}}(B_{0}|\theta)=\\{\beta\chi\in{\rm Irr}(B_{0})\ |\
\beta\in{\rm Irr}_{p^{\prime}}(G/N)\\}$. By [GRSS20, Lemma 2.3] we have
$\beta\chi\in{\rm Irr}(B_{0})$ whenever $\beta\in{\rm Irr}(B_{0}(G/N))$, and
the lemma follows. ∎

We note that the example of $G={\mathrm{Sym}}_{3}$ and
$N={\mathrm{Alt}}_{3}=\langle(1\,2\,3)\rangle$ for $p=3$ shows that the
inequality in Lemma 3.5 is not always an equality.

###  3.2. Proof of Theorem A

As we will see later, known results on height zero characters in principal
blocks essentially reduce Theorem A to the following.

######  Theorem 3.6.

Let $G$ a finite group of order divisible by $p\in\\{3,5,7\\}$. Suppose that
$P\in{\mathrm{Syl}}_{p}(G)$ is abelian and the principal $p$-block $B_{0}$ of
$G$ has precisely $6$ irreducible characters. Then $p=3$ and $|P|=9$.

###### Proof.

Assume that the statement is false and let $G$ be a counterexample of minimal
order. In particular, $k(B_{0})=6$ and $|P|\neq 9$. If $|P|<9$ then $P$ would
be ${\sf C}_{3}$, ${\sf C}_{5}$, or ${\sf C}_{7}$, and Dade’s cyclic-defect
theory [Dad66, Theorem 1] quickly shows that $k(B_{0})\neq 6$. We therefore
indeed have

$|P|>9.$

Furthermore, by Lemma 2.3(ii), we have ${\bf O}_{p^{\prime}}(G)=1$.

We now show that $G$ is not $p$-solvable. Assume that $G$ is $p$-solvable.
Then, by Fong’s theorem (see [Nav98, Theorem 10.20]), $G$ has a unique
$p$-block - the principal one, and so $k(G)=k(B_{0})=6$. An inspection of the
list of finite groups with 6 conjugacy classes ([VV85, Table 1]) reveals no
counterexamples. Therefore $G$ is in fact not $p$-solvable.

Let $N\trianglelefteq\,G$ be a minimal normal subgroup of $G$. As ${\bf
O}_{p^{\prime}}(G)=1$, we have that $p$ divides the order of $N$, which
therefore is either an elementary abelian $p$-group or a semisimple group.

(A) We claim that $p$ does not divide $|G:N|$.

Assume otherwise. Let $\bar{B}_{0}:=B_{0}(G/N)$ be the principal block of
$G/N$. On one hand we have $k(\bar{B}_{0})\leq 5$ by Lemma 2.3 (i) and (iii);
on the other hand, $k(\bar{B}_{0})\geq\lceil 2\sqrt{p-1}\rceil\geq 3$ by
[HS21, Theorem 1.1]. In summary,

$k(\bar{B}_{0})\in\\{3,4,5\\}.$

Also notice that, by Lemma 2.3, we have that ${\rm Irr}(B_{0})\cap{\rm
Irr}(G/N)$ is a union of blocks of $G/N$ including $\bar{B}_{0}$. Hence, we
can write

${\rm Irr}(B_{0})=({\rm Irr}(B_{0})\cap{\rm
Irr}(G/N))\cup\\{\chi_{1},\ldots,\chi_{l}\\}\,,$

with $l\in\\{1,2,3\\}$, where the union is disjoint.

(A1) Assume that $l=3$. Then ${\rm Irr}(B_{0})\cap{\rm Irr}(G/N)={\rm
Irr}(\bar{B}_{0})$ and $k(\bar{B}_{0})=3$. It follows from [KS21, Theorem 3.1]
that

$p=3\text{ and }|PN:N|=3.$

Moreover, the action of $G$ on ${\rm Irr}(B_{0}(N))\setminus\\{{\bf 1}_{N}\\}$
defines at most 3 orbits.

Suppose first that $N$ is elementary abelian. Then

$P{\bf C}_{G}(P)={\bf C}_{G}(P)\subseteq{\bf C}_{G}(N)=:M\trianglelefteq\,G.$

By Lemma 3.2, $B_{0}$ is the only block of $G$ covering $B_{0}(M)$ and $1\leq
k(G/M)\leq 3$, as ${\rm Irr}(G/M)\subseteq{\rm Irr}(B_{0})\cap{\rm
Irr}(G/N)={\rm Irr}(\bar{B}_{0})$. Since $G/M$ is a $3^{\prime}$-group, we
necessarily have $|G/M|\leq 2$. If $G=M$, then $N\subseteq{\bf Z}(G)$ and by
Lemma 3.1, $6=k(B_{0})\geq|N|l(B_{0})$, implying that $|N|$ must be 3, so that
$|P|=3|N|=9$, a contradiction. We therefore must have $|G/M|=2$. As the action
of $G/M$ on the nontrivial elements of $N$ defines at most 3 orbits, we have
$|N|\leq 7$, and thus $|N|=3$, which implies that $|P|=9$, a contradiction
again.

Now suppose that $N\cong S^{t}$ for some non-abelian simple group $S$ of order
divisible by $p=3$ and $t\in{\mathbb{Z}}^{+}$. By [Mar21, Theorem 2.2],
$\operatorname{Aut}(S)$ produces at least 2 orbits on ${\rm
Irr}(B_{0}(S))\setminus\\{{\bf 1}_{S}\\}$. Since the action of $G$ on ${\rm
Irr}(B_{0}(N))\setminus\\{{\bf 1}_{N}\\}$ defines at most 3 orbits, we deduce
that $t=1$ and $N=S$ is simple (otherwise, assuming that $\alpha,\beta\in{\rm
Irr}(B_{0}(S))\setminus\\{{\bf 1}_{S}\\}$ lie in different
$\operatorname{Aut}(S)$-orbits, then $\alpha^{t}$, $\alpha\times{\bf
1}_{S}^{t-1}$, $\beta^{t}$ and $\beta\times{\bf 1}_{S}^{t-1}$ would lie in
different $G$-orbits). We then have $|P:(P\cap S)|=|PS:S|=3$. The fact that
$G$ is a counterexample then implies that $Q:=P\cap S\in{\rm Syl}_{3}(S)$ has
order at least 9.

We now use Theorem 2.1, with the almost simple group $G/{\mathbf{C}}_{G}(S)$
in place of $A$, to arrive at one of the following cases.

  1. (a)

The action of $G$ on ${\rm Irr}(B_{0}(S))\setminus\\{{\bf 1}_{S}\\}$ defines
at least 4 orbits, a contradiction.

  2. (b)

$S\in\\{{\rm PSL}_{2}(q),{\rm PSL}_{3}(q),{\rm PSU}_{3}(q)\\}$ with $(3,q)=1$,
and $3$ divides $|G:S{\bf C}_{G}(S)|$. Since $|G:S|_{3}=3$, ${\bf C}_{G}(S)$
has order not divisible by 3. Then ${\bf C}_{G}(S)=1$, because ${\bf
O}_{3^{\prime}}(G)=1$ and $G=A$ is almost simple. Theorem 2.1(b) yields
$k(B_{0})>6$, a contradiction.

  3. (c)

$S={\mathrm{Alt}}_{6}$ and $A:=G/{\bf C}_{G}(S)$ has a subgroup isomorphic to
${\rm M}_{10}$. It follows that $A$ is either $M_{10}$ or the full
automorphism group $\operatorname{Aut}(S)$. In fact, as $k(B_{0}(A))\leq
k(B_{0})=6$ and $k(B_{0}(\operatorname{Aut}(S)))>6$, we have $A\cong{\rm
M}_{10}$, in which case $k(B_{0}(A))=k(B_{0})=6$. We conclude that 3 does not
divide the order of ${\bf C}_{G}(S)$ (otherwise $B_{0}({\bf C}_{G}(S))$
contains a nontrivial character, say $\theta$, and by Lemma 2.3(iii),
$B_{0}(G)$ would have a character lying above $\theta$, implying that
$k(B_{0}(A))<k(B_{0})$). In particular, $|G|_{3}=|{\mathrm{Alt}}_{6}|_{3}=9$
and $G$ is not a counterexample.

(A2) Assume that $l=2$. If ${\rm Irr}(\bar{B}_{0})\subsetneq{\rm
Irr}(B_{0})\cap{\rm Irr}(G/N)$ then ${\rm Irr}(B_{0})$ would contain a block
of $p$-defect zero of $G/N$, but that is impossible as every character in
${\rm Irr}(B_{0})$ has degree coprime to $p$ by [KM13]. Thus ${\rm
Irr}(\bar{B}_{0})={\rm Irr}(B_{0})\cap{\rm Irr}(G/N)$. In particular
$k(\bar{B}_{0})=4$ and, by [KS21, Theorem 1.1],

$p=5\text{ and }|PN:N|=5.$

In this case the action of $G$ on ${\rm Irr}(B_{0}(N))\setminus\\{{\bf
1}_{N}\\}$ defines at most 2 orbits.

Suppose that $N$ is elementary abelian. Then, as above, $P\subseteq{\bf
C}_{G}(N):=M\trianglelefteq\,G$. By Lemma 3.2, $B_{0}$ is the only block
covering $B_{0}(M)$. Consequently, Lemma 2.3(iv) implies that ${\rm
Irr}(G/M)\subseteq{\rm Irr}(\bar{B}_{0})$ and so $1\leq k(G/M)\leq 4$.
Moreover, using Lemma 2.3 (iii) and (iv), considering a non-principal
character $\theta\in{\rm Irr}(B_{0}(M/N))\subseteq{\rm Irr}(B_{0}(M))$, we in
fact have that $k(G/M)<4$. If $G=M$ then $N\subseteq{\bf Z}(G)$ and, by Lemma
3.1(i), $6=k(B_{0})\geq 5l(B_{0})$. It follows that $l(B_{0})=1$ and, by Lemma
3.1(ii), $G$ would be $p$-solvable, a contradiction. Hence the the action of
the nontrivial group $G/M$ on $N$ is faithful and defines at most 2 orbits on
the nontrivial elements of $N$. The only possibility is that $|N|=5$ and
$|G/M|=2$. Since ${\rm Irr}(B_{0}(M))$ consists of the irreducible
constituents of $\psi_{M}$ for $\psi\in{\rm Irr}(B_{0})$, the fact $|G/M|=2$
forces $k(B_{0}(M))\in\\{3,6,9\\}$. As $p=5$, the main result of [HS21]
implies that $k(B_{0}(M))\geq 2\sqrt{p-1}=4$. Furthermore, $k(B_{0}(M))$
cannot be 6 neither by the minimality of $G$. So $k(B_{0}(M))=9$. Lemma 3.1
then implies that $l(B_{0}(M))=1$ and $M$ is $p$-solvable. Thus $G$ is
$p$-solvable as well, a contradiction.

Suppose that $N\cong S^{t}$ for some non-abelian simple group $S$ of order
divisible by $p=5$ and $t\in{\mathbb{Z}}^{+}$. If $t>1$, then, as before,
[Mar21, Theorem 2.2] shows that $G$ defines more than 2 orbits when acting on
${\rm Irr}(B_{0}(N))\setminus\\{{\bf 1}_{N}\\}$, violating the fact that
$l=2$. Hence $N=S\trianglelefteq\,G$.

Assume first that $S\neq P\Omega_{8}^{+}(q)$. By [GRSS20, Proposition 2.1(i)],
there are some ${\bf 1}_{S}\neq\theta\in{\rm Irr}(B_{0}(S))$ that extends to
$\hat{\theta}\in{\rm Irr}(B_{0}(G))$ and ${\bf 1}_{S}\neq\varphi\in{\rm
Irr}(B_{0}(S))$ with $\varphi(1)\nmid\theta(1)$. Furthermore, both $\theta$
and $\varphi$ have degree coprime to $p$. Recall that
$k(B_{0}(G/S))=k(B_{0}(G/N))\geq 3$. We therefore can write

${\rm Irr}(B_{0})=\\{{\bf 1}_{G},\alpha,\beta,\gamma,\chi_{1},\chi_{2}\\}$

so that $\alpha,\beta,\gamma$ contain $S$ in their kernels,
$\chi_{1}=\hat{\theta}$, and $\chi_{2}$ lies over $\varphi$. In particular,
$\chi_{1}$ is the only character in ${\rm Irr}(B_{0})$ that is above $\theta$.
This together with the facts that $p\nmid\theta(1)$ and $p\mid|G:S|$ would
contradict Lemma 3.5.

We are left with the case $S={\rm P}\Omega_{8}^{+}(q)$. Then by [GRSS20,
Proposition 2.1.(ii)], the set ${\rm Irr}(B_{0}(S))\setminus\\{{\bf 1}_{S}\\}$
contains two $\operatorname{Aut}(S)$-invariant members. Since $k(B_{0}(S))\geq
2\sqrt{p-1}=4$ by [HS21, Theorem 1.2.(i)], it follows that the action of $G$
on ${\rm Irr}(B_{0}(S))\setminus\\{{\bf 1}_{S}\\}$ defines at least 3
different orbits, which is again a contradiction.

(A3) Assume that $l=1$. If ${\rm Irr}(\bar{B}_{0})\subsetneq{\rm
Irr}(B_{0})\cap{\rm Irr}(G/N)$ then arguing as in the second sentence of case
(A2) we see that ${\rm Irr}(B_{0})$ contains a block $\bar{B}_{1}$ of $G/N$
with $k(\bar{B}_{1})=2$. But $k(\bar{B}_{1})=2$ would force $p=2$ by [Bra82,
Theorem A], contradicting our hypothesis on $p$. Hence $k(\bar{B}_{0})=5$.
Using the main result of [RSV21], we then have

$p\in\\{5,7\\}\text{ and }|PN:N|=p.$

In this case all the characters in ${\rm Irr}(B_{0}(N))\setminus\\{{\bf
1}_{N}\\}$ must be $G$-conjugate (and lie under $\chi_{1}$). In particular,
the set of character degrees of ${\rm Irr}(B_{0}(N))$ has size 2, and it
follows from [Mar21, Theorem A] that $N$ is $p$-solvable. We therefore
conclude that $N$ is an elementary abelian $p$-group. Also, $N\subseteq P$ and
$|P/N|=p$.

As before, $B_{0}$ is the only block covering $B_{0}(M)$, where $M:={\bf
C}_{G}(N)\trianglelefteq\,G$. Moreover, $k(G/M)<5$. If $G=M$, then
$N\subseteq{\bf Z}(G)$ and Lemma 3.1(i) implies that
$6=k(B_{0})\geq|N|l(B_{0})$, forcing $l(B_{0})=1$ as $p\geq 5$, and so $G$
would be $p$-solvable by Lemma 3.1(ii), a contradiction. Therefore the
$p^{\prime}$-group $G/M\in\\{{\sf C}_{2},{\sf C}_{3},{\mathrm{Sym}}_{3},{\sf
C}_{4},{\sf C}_{2}\times{\sf C}_{2},{\sf D}_{10},{\mathrm{Alt}}_{4}\\}$ acts
faithfully on $N$ and transitively on $N\setminus\\{1\\}$. (See [VV85] for the
list of finite groups of relatively small class number.) There are only two
possible scenarios: $|G/M|=4$ and $|N|=5$ or $G/M\cong{{\mathrm{Sym}}}_{3}$
and $|N|=7$. The latter case in fact cannot happen as $\operatorname{Aut}({\sf
C}_{7})\cong{\sf C}_{6}$. In the former case, since ${\rm
Irr}(G/M)\subseteq{\rm Irr}(\bar{B}_{0})$ and $k(\bar{B}_{0})=5$, the
principal block $\bar{B}_{0}$ of $G/N$ would have only 2 different character
degrees. It again follows from the main result of [Mar21] that $G/N$, and
therefore $G$, would be $p$-solvable, a contradiction.

(B) We have shown that $p$ does not divide $|G:N|$. Therefore we have $P\leq
N$. In particular, as $G$ is not $p$-solvable, $N$ is isomorphic to a direct
product $S^{t}$ of $t$ copies of a non-abelian simple group $S$ of order
divisible by $p$.

Let $M:=N{\bf C}_{G}(P)$. Then $M\trianglelefteq\,G$ by the Frattini argument,
and

$k(B_{0}(M))=k(B_{0}(S))^{t}$

by Theorem 3.3. Moreover, $G/M$ transitively permutes the simple factors of
$N$. By Lemma 3.2, $B_{0}$ is the only block of $G$ covering $B_{0}(M)$. In
particular, ${\rm Irr}(G|\phi)\subseteq{\rm Irr}(B_{0})$ for every
$\phi\in{\rm Irr}(B_{0}(M))$. Moreover, as $k(B_{0})=6$, we have $1\leq
k(G/M)<6$.

If $G=M$ then $t=1$ and $k(B_{0}(S))=k(B_{0}(G))=6$, and it follows from
Theorem 2.2 that $|P|=9$, contradicting the choice of $G$ as a counterexample.
If $k(G/M)=5$ then all members of ${\rm Irr}(B_{0}(M))\setminus\\{{\bf
1}_{M}\\}$ lie under the same character in ${\rm Irr}(B_{0})$. In particular,
[Mar21, Theorem A] implies that $M$ is $p$-solvable, and so would be $G$,
which is not the case. We have shown that

$k(G/M)\in\\{2,3,4\\}.$

Recall that $|P|>9$. Therefore, by previous results on possible Sylow
structure of finite groups with up to $5$ characters in the principal block
(see Theorems 4.1–4.5 of [HSV23]), we have $k(B_{0}(M))\geq 6$. Note that
$P\in{\mathrm{Syl}}_{p}(M)$, and so the minimality of $G$ as a counterexample
further implies that

$k(B_{0}(M))\geq 7.$

Let us first handle the special case $(S,p)=({\rm PSL}_{2}(3^{m}),3)$ for some
$m\geq 3$. If $S={\rm PSL}_{2}(27)$ then $\operatorname{Aut}(S)$ has $5$
orbits on ${\rm Irr}(B_{0}(S))\backslash\\{\textbf{1}_{S}\\}$ ([Atl85]), so
$k(B_{0})\geq 5+k(G/M)\geq 7$ and we would be done. For $S={\rm PSL}_{2}(q)$
with $q\geq 81$ we have $k(B_{0}(S))=(q+3)/2\geq 42$ (the ordinary characters
of $B_{0}(S)$ are ${\mathrm{Irr}}(S)\backslash\\{{\mathsf{St}}_{S}\\}$ and
$k({\rm PSL}_{2}(q))=(q+5)/2$ for odd $q$). We therefore have at least $41$
nontrivial members in ${\mathrm{Irr}}(B_{0}(M))$. With $k(G/M)\leq 4$, we have
$|G/M|\leq 12$. So $G$ produces at least $\lceil 41/12\rceil=4$ orbits on
${\mathrm{Irr}}(B_{0}(M))\backslash\\{\textbf{1}_{M}\\}$. As $k(B_{0})=6$, it
follows that $k(G/M)\leq 2$, which would imply that $G$ now defines more than
$41/2$ orbits on ${\mathrm{Irr}}(B_{0}(M))\backslash\\{\textbf{1}_{M}\\}$, a
contradiction.

We may assume from now on that $(S,p)\neq({\rm PSL}_{2}(3^{m}),3)$ for every
$m\geq 3$. Then ${\rm Irr}(B_{0}(S))\setminus\\{{\bf 1}_{S}\\}$ contains an
$\operatorname{Aut}(S)$-invariant member $\phi$ by [GRSS20, Proposition 2.1]
for $p\geq 5$ and by [Mar21, Proposition 3.7] for $p=3$ and $S\neq{\rm
PSL}_{2}(3^{n})$ for $n\geq 2$. (Note that the irreducible character of degree
10 of $S={\rm Alt}_{6}\cong{\rm PSL}_{2}(9)$ is
$\operatorname{Aut}(S)$-invariant and lies in the principal 3-block.) In
particular, by Theorem 3.3, there exists some

${\bf 1}_{M}\neq\psi\in{\rm Irr}(B_{0}(M))$

that is $G$-invariant. More precisely, under the Alperin-Dade correspondence,
$\psi$ corresponds to $\phi^{t}$ where $\phi\in{\rm
Irr}(B_{0}(S))\setminus\\{{\bf 1}_{S}\\}$ is
$\operatorname{Aut}(S)$-invariant.

(B1) Assume that $k:=k(G/M)\in\\{2,3\\}$. Then all the Sylow subgroups of
$G/M$ are cyclic and $\psi$ extends to $G$ by Lemma 3.4. Therefore
${\mathrm{Irr}}(B_{0})$ has $2k$ members lying over the characters
$\textbf{1}_{M}$ and $\psi$ in ${\mathrm{Irr}}(B_{0}(M))$. As $k(B_{0}(M))\geq
7$, the remaining at least $5$ characters in ${\mathrm{Irr}}(B_{0}(M))$
produce at least $\lceil 5/k\rceil$ additional irreducible characters in
$B_{0}$. We arrive at

$k(B_{0})\geq 2k+\lceil 5/k\rceil,$

which is greater than 6 when $k\in\\{2,3\\}$, a contradiction.

(B2) Finally we consider $k(G/M)=4$. Now ${\mathrm{Irr}}(B_{0})$ contains four
members lying above $\textbf{1}_{M}$ and at least one member above $\psi$.
Furthermore, the set
${\mathrm{Irr}}(B_{0}(M))\backslash\\{\textbf{1}_{M},\psi\\}$, which again has
cardinality at least 5, would cover at least two $G$-orbits. As a result, we
have

$k(B_{0})\geq 4+1+2=7,$

and this contradiction completes the proof. ∎

The following result on $2$-blocks with metacyclic defect groups will be
useful. The weaker case of maximal-class defect groups, which is what we
really need, is due to Brauer [Bra72] and Olsson [Ols75].

######  Lemma 3.7.

Let $B$ be a $2$-block of a finite group with metacyclic defect group $D$ of
order at least $8$. Then either $k(B)\geq 7$ or one of the following holds:

  1. (i)

$B$ is nilpotent and $k(B)=k(D)$.

  2. (ii)

$D={\sf D}_{8}$ and $k(B)=5$.

###### Proof.

This follows from [Sam14, Theorem 8.1]. ∎

######  Theorem 3.8.

Let $G$ a finite group, $p$ a prime, and $P\in{\mathrm{Syl}}_{p}(G)$. Suppose
that the principal $p$-block of $G$ has precisely six irreducible characters.
Then $p=3$ and $|P|=9$.

###### Proof.

Recall that $B_{0}$ denotes the principal $p$-block of $G$. First, as
$k(B_{0})=6>1$, we know that $G$ has order divisible by $p$. As before, let
$k_{0}(B_{0})$ denote the number of height zero characters in $B_{0}$.
Obviously $k_{0}(B_{0})\leq 6$ and moreover $k_{0}(B_{0})\geq 2$ by [Nav98,
Problem 3.11]. By [HSV23, Theorems 4.2 and 4.3] we further have
$k_{0}(B_{0})\geq 4$. The case $k_{0}(B_{0})=5$ cannot happen since otherwise,
by [HSV23, Theorem 1.2(C)], $P$ is cyclic, and thus $k(B_{0})=k_{0}(B_{0})=5$
by the main result of [KM13], which is a contradiction. We now have

$k_{0}(B_{0})\in\\{4,6\\}.$

Suppose that $k_{0}(B_{0})=4$. Then it follows from [HSV23, Theorem 1.2(B)]
that $p=2$ and $P$ has maximal class (so $P$ is either dihedral, semidihedral,
or generalized quaternion). Since $k_{0}(B_{0})<k(B_{0})$, $P$ is nonabelian
by again [KM13]. In particular, $P$ is a metacyclic group of order at least
$8$. Using Lemma 3.7, we deduce that $B_{0}$ is nilpotent and
$6=k(B_{0})=k(P)$. However, according to [VV85, Table 1], there is no
$p$-group with precisely six conjugacy classes.

We are left with the case $k_{0}(B_{0})=k(B_{0})=6$, and so $P$ is abelian by
[MN21]. Using [Lan81, Corollary 1.3.(i)], we deduce that $p$ must be odd. On
the other hand, the main result of [HS21] implies that $p\leq
k(B_{0})^{2}/4+1=10$, and we conclude that $p\in\\{3,5,7\\}$. The result now
follows from Theorem 3.6. ∎

###  3.3. Proof of Theorem B

######  Theorem 3.9.

Let $G$ a finite group, $p$ a prime, and $P\in{\mathrm{Syl}}_{p}(G)$. Let
$B_{0}$ denote the principal $p$-block of $G$. Then $k(B_{0})=6$ if and only
if precisely one of the following happens:

  * (i)

$P={\sf C}_{9}$ and $|{\bf N}_{G}(P):{\bf C}_{G}(P)|=2$.

  * (ii)

$P={\sf C}_{3}\times{\sf C}_{3}$ and either ${\bf N}_{G}(P)/{\bf
C}_{G}(P)\in\\{{\sf C}_{4},{\sf Q}_{8}\\}$ or ${\bf N}_{G}(P)/{\bf
C}_{G}(P)\cong{\sf C}_{2}$ acts fixed-point freely on $P$.

###### Proof.

In either implication of the statement, by Theorem A, we have $p=3$ and
$|P|=9$. In particular, $P$ is abelian and $k_{0}(B_{0})=k(B_{0})$ by [KM13].

If $P$ is cyclic then ${\bf N}_{G}(P)/{\bf C}_{G}(P)$ is a
$3^{\prime}$-subgroup of $\operatorname{Aut}(P)\cong{\sf C}_{6}$. By Dade’s
cyclic-defect theory [Dad66, Theorem 1], we have $k(B_{0})=8/f+f$, where
$f:=|{\bf N}_{G}(P)/{\bf C}_{G}(P)|$. The only possibilities are that
$k(B_{0})=6$ with $f=2$ or $k(B_{0})=9$ with $f=1$.

If $P$ is elementary abelian, then Broué’s abelian defect conjecture holds for
$B_{0}$ by [KK02]. In particular, we have
$k_{0}(B_{0})=k_{0}(B_{0}({\mathbf{N}}_{G}(P)))$. Since every irreducible
character of ${\mathbf{N}}_{G}(P)$ has $p^{\prime}$-degree,

$k_{0}(B_{0}({\mathbf{N}}_{G}(P)))=k(B_{0}({\mathbf{N}}_{G}(P)))=k({\mathbf{N}}_{G}(P)/{\mathbf{O}}_{p^{\prime}}({\mathbf{N}}_{G}(P))),$

where the second equality follows from Lemma 2.3(ii). In summary,

$k(B_{0})=k({\mathbf{N}}_{G}(P)/{\mathbf{O}}_{p^{\prime}}({\mathbf{N}}_{G}(P))).$

Note that, as $P$ is abelian, we have
${\mathbf{C}}_{G}(P)\cong{\mathbf{O}}_{p^{\prime}}({\mathbf{N}}_{G}(P))\times
P$, so that
${\mathbf{N}}_{G}(P)/{\mathbf{O}}_{p^{\prime}}({\mathbf{N}}_{G}(P))$ is a
semidirect product of the $3^{\prime}$-group

$A:={\mathbf{N}}_{G}(P)/{\mathbf{C}}_{G}(P)$

acting faithfully (and coprimely) on $P$. Since
${\mathrm{Aut}}(P)\cong{\mathrm{GL}}_{2}(3)$, there are eight possibilities
for the $3^{\prime}$-group $A\leq{\mathrm{GL}}_{2}(3)$:

$1,{\sf C}_{2},{\sf C}_{4},{\sf C}_{2}\times{\sf C}_{2},{\sf C}_{8},{\sf
D}_{8},{\sf Q}_{8},\text{ and }{\sf SD}_{16}.$

(Here ${\sf SD}_{16}$ is the semi-dihedral group of order $16$.) While ${\sf
C}_{2}$ can either invert every element of $P$ or invert just elements in one
factor ${\sf C}_{3}$ and fix elements of the other, each of the other
possibilities can only act on $P$ in a unique way. Using [GAP], we easily
check that the corresponding semidirect products have 6 or 9 conjugacy
classes. Indeed, a straightforward inspection of [VV85, Table 1] reveals that
$k(PA)=6$ if and only if $A\in\\{{\sf C}_{4},{\sf Q}_{8}\\}$ or $A={\sf
C}_{2}$ acts on $P$ by inversion. This completes the proof. ∎

We conclude the paper with a summary of what is now known on the problem of
classifying defect groups of blocks with a “relatively small” number of
irreducible ordinary characters. In Table 1, as usual, we have used $B$ for an
arbitrary block and $B_{0}$ for a principal one. Also, $k(B)$ and $l(B)$
denote the numbers of irreducible ordinary and Brauer characters,
respectively, in $B$. A defect group of $B$ is denoted by $D$ and, in the case
of principal blocks, by $P$. Moreover, a root of $B$ is denoted by $b$ – a
block of $D{\mathbf{C}}_{G}(D)$ with defect group $D$ that induces $B$, see
[Nav98, Theorem 9.7 and p. 198]. The second column presents all the possible
defect-group structures (for a defect group $D$) of a block with corresponding
given values of $k(B)$ and $l(B)$ in the first column. The third column
presents the “if and only if” conditions for those values to be achieved, with
a note that a missing item indicates no conditions being needed; as usual
$T(b)$ denotes the stabilizer of the block $b$ as in [Nav98, p. 193].

Table 1. Defect groups of small blocks | $k(B),l(B)$
---
| Defect group
---
| Local condition
---
Reference
| $k(B)=1$
---
| $1$
---
| well-known
| $k(B)=2$
---
| ${\sf C}_{2}$
---
| [Bra82]
| $k(B_{0})=3$
---
| ${\sf C}_{3}$
---
| [Bel90, KS21]
|  $k(B)=3$,
---
$l(B)=1$
| ${\sf C}_{3}$
---
| [Kul84]
|  $k(B)=3$,
---
$D\trianglelefteq\,G$
| ${\sf C}_{3}$
---
| [KNST14]
| $k(B_{0})=4$
---
| ${\sf C}_{2}\times{\sf C}_{2}$
---
${\sf C}_{4}$
${\sf C}_{5}$
| $|{\mathbf{N}}_{G}(P)/{\mathbf{C}}_{G}(P)|=2$
---
[KS21]
|  $k(B)=4$,
---
$l(B)=1$
| ${\sf C}_{2}\times{\sf C}_{2}$
---
${\sf C}_{4}$
| [Kul84]
|  $k(B)=4$,
---
$D\trianglelefteq\,G$
| ${\sf C}_{2}\times{\sf C}_{2}$
---
${\sf C}_{4}$
${\sf C}_{5}$
| $|T(b)/{\mathbf{C}}_{G}(D)|=2$
---
[MRS22]
| $k(B_{0})=5$
---
| ${\sf C}_{5}$
---
${\sf C}_{7}$
${\sf D}_{8}$
${\sf Q}_{8}$
| $|{\mathbf{N}}_{G}(P)/{\mathbf{C}}_{G}(P)|\in\\{1,4\\}$
---
$|{\mathbf{N}}_{G}(P)/{\mathbf{C}}_{G}(P)|\in\\{2,3\\}$
$|{\mathbf{N}}_{G}(P)/P{\mathbf{C}}_{G}(P)|=1$
[RSV21]
|  $k(B)=5$,
---
$l(B)=1$
| ${\sf C}_{5}$
---
${\sf D}_{8}$
${\sf Q}_{8}$
|  $B$ is nilpotent.
---
$B$ is nilpotent.
[CK92]
| $k(B_{0})=6$
---
| ${\sf C}_{9}$
---
${\sf C}_{3}\times{\sf C}_{3}$
| $|{\mathbf{N}}_{G}(P)/{\mathbf{C}}_{G}(P)|=2$
---
${\mathbf{N}}_{G}(P)/{\mathbf{C}}_{G}(P)\in\\{{\sf C}_{4},{\sf Q}_{8}\\}$ or
${\mathbf{N}}_{G}(P)/{\mathbf{C}}_{G}(P)={\sf C}_{2}$
acts on $P$ by inversion.
Theorem B
|  $k(B_{0})=7$,
---
$l(B_{0})=6$
| ${\sf C}_{7}$
---
| $|{\mathbf{N}}_{G}(P)/{\mathbf{C}}_{G}(P)|=6$
---
[HST21]

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