# Characters and Sylow $3$-subgroup commutators

Eugenio Giannelli  Dipartimento di Matematica e Informatica ‘Ulisse Dini’,
50134 Firenze, Italy<EMAIL_ADDRESS>,  Noelia Rizo  Departament
de Matemàtiques, Universitat de València, 46100 Burjassot, València, Spain
<EMAIL_ADDRESS>,  A. A. Schaeffer Fry  Dept. Mathematics - University of
Denver, Denver, CO 80210, USA<EMAIL_ADDRESS>and  Carolina Vallejo
Dipartimento di Matematica e Informatica ‘Ulisse Dini’, 50134 Firenze, Italy
<EMAIL_ADDRESS>
###### Abstract.

We characterize when a finite group $G$ possesses a Sylow $3$-subgroup $P$
with abelianization of order $9$ in terms of the number of height zero
characters lying in the principal $3$-block of $G$, settling a conjecture put
forward by Navarro, Sambale, and Tiep in 2018. Along the way, we show that a
recent result by Laradji on the number of character of height zero in a block
that lie above a given character of some normal subgroup holds, without any
hypothesis on the group for blocks of maximal defect.

###### Key words and phrases:

Height zero character, principal block, Sylow subgroup commutator

The first and fourth-named authors are funded by the Italian national project
PRIN 2022- 2022PSTWLB - Group Theory and Applications and, as members of the
GNSAGA, both are grateful for the support of the Istituto Nazionale di Alta
Matematica. The second-named author is supported by Ministerio de Ciencia e
Innovación (Grant PID2019-103854GB-I00 and Grant PID2022-137612NBI00 funded by
MCIN/AEI/10.13039/501100011033 and “ERDF A way of making Europe”) and a
CDEIGENT grant CIDEIG/2022/29 funded by Generalitat Valenciana. The third-
named author gratefully acknowledges support from the National Science
Foundation, Award No. DMS-2100912, and her former institution, Metropolitan
State University of Denver, which holds the award and allows her to serve as
PI. The fourth-named author also acknowledges support from the Rita Levi-
Montalcini 2019 programme. This material is based upon work supported by the
National Science Foundation under Grant No. DMS-1928930, while the authors
were in residence at the Mathematical Sciences Research Institute in Berkeley,
California, during the summer of 2023.

##  1\. Introduction

A key role in the theory of finite groups is played by dihedral, semidihedral
and generalized quaternion $2$-groups. These can be characterized, for
example, as the non-abelian $2$-groups $P$ whose commutator subgroup
$P^{\prime}$ has index $4$ in $P$, by work of O. Taussky [Tau37]. Since the
beginning of the $20$th Century, the representation theory of finite groups
having a Sylow $2$-subgroup $P$ with $[P:P^{\prime}]=4$ (and more generally of
blocks having a defect group $P$ with $[P:P^{\prime}]=4$) has been largely
studied. Partly relying on this extensive study, G. Navarro, B. Sambale, and
P. H. Tiep showed in [NST18, Thm. A] that a finite group $G$ possesses a Sylow
$2$-subgroup $P$ whose commutator subgroup has index $4$ if, and only if, the
principal $2$-block of $G$ contains exactly $4$ irreducible characters of odd
degree.

Contrasting this situation, the representation theory of groups having a Sylow
$p$-subgroup $P$ with $[P:P^{\prime}]=p^{2}$ for an odd prime $p$ is far from
being understood. It is perhaps not so surprising, given the fact that the
structure of these $p$-groups for a general prime can be exceedingly
complicated. For instance, write $f_{p}(n)$ to denote the number of
isomorphism classes of groups $P$ of order $p^{n}$ with
$[P:P^{\prime}]=p^{2}$. For $p=2$, as mentioned above, we have that
$f_{2}(n)=3$ if $n\geq 3$, but already for $p=3$ it is not known whether the
function $f_{3}(n)$ is bounded or tends to infinity with $n$. In fact, proving
that a finite group has a Sylow $3$-subgroup $P$ with $[P:P^{\prime}]=9$ if,
and only if, its principal $3$-block contains exactly $6$ or $9$ irreducible
characters of degree coprime to $3$ has remained a challenge since this
statement was put forward in [NST18, Conj. B]. The main result of this article
is to prove the above mentioned conjecture.

######  Theorem A.

Let $G$ be a finite group and $P\in{\rm Syl}_{3}(G)$. Then $[P:P^{\prime}]=9$
if, and only if, there are exactly $6$ or $9$ irreducible characters of degree
coprime to $3$ lying in the principal $3$-block of $G$.

For a group $G$, we write $k_{0}(B_{0}(G))$ to denote the number of
irreducible characters of degree coprime to $3$ lying in the principal
$3$-block $B_{0}(G)$ of $G$ (as these are exactly those of $3$-height zero in
$B_{0}(G)$). Theorem A above, together with [NST18, Thm. A], imply that the
conditions $[P:P^{\prime}]\leq p^{2}$ and $k_{0}(B_{0}(G))\leq p^{2}$ are
equivalent for any finite group $G$ and $p\in\\{2,3\\}$. This statement does
not hold for arbitrary primes as the group $H=\verb|SmallGroup(1000,163)|$
satisfies $k(H)=k_{0}(B_{0}(H))=25$ with respect to the prime $p=5$ and its
Sylow $5$-subgroups are abelian of order $125$. We care to mention that, for a
general prime $p$, it has been proven very recently in [NS23] that the local
condition $[P:P^{\prime}]=p^{2}$ can be detected in the character table of the
group $G$. However, it is still not known whether there exists a simple
algorithm to do that for primes $p\geq 5$.

Our proof of Theorem A eventually reduces to showing that the statement of
Theorem A holds for almost simple groups $A$ with $[A:{\rm Soc}(A)]_{3}\leq
3$, where $n_{3}$ is the largest power of $3$ dividing an integer $n$ and
${\rm Soc}(A)$ is the socle of $A$. In order to perform such a reduction, we
have proven results of independent interest regarding the behavior of
characters in principal blocks with respect to normal subgroups and the
character theory of simple non-abelian groups with respect to the action of
group automorphisms on them. The former appear in Section 2 and the latter in
Section 3 of this work. Among the results briefly mentioned above, we would
like to highlight the following generalization of the main theorem of [Lar23].

######  Theorem B.

Let $p$ be a prime. Let $G$ be a finite group, $P\in{\rm Syl}_{p}(G)$ and $B$
a $p$-block of $G$ with defect group $P$. Let $N$ be a normal subgroup of $G$
such that $PN/N>1$. If $\theta\in{\rm Irr}(N)$, then the number
$k_{0}(B|\theta)$ of irreducible characters of degree coprime to $p$ in $B$
lying above $\theta$ is either 0 or $k_{0}(B|\theta)\geq 2$. If we further
assume that $p\leq 3$, then $k_{0}(B|\theta)$ is divisible by $p$.

The structure of the paper is as follows. In Section 2, we state some results
(both old and new) on principal blocks (and more generally, blocks of maximal
defect). In Section 3, we state our results on almost simple groups, which
will be proven in Sections 4 and 5. We address the alternating group in
Section 4 and the groups of Lie type in Section 5. Finally, we complete the
proof of Theorem A in Section 6.

##  2\. General statements on principal blocks

###  2.1. Known results on principal blocks

For a finite group $G$ and a prime $p$, we denote by $B_{0}(G)$ the principal
$p$-block of $G$. Recall that

${1}_{G}\in{\rm Irr}(B_{0}(G))=\\{\chi\in{\rm Irr}(G)\ |\ \sum_{{x\in
G}\atop{(p,o(x))=1}}\chi(x)\neq 0\\}\,.$

In particular, if $G=H_{1}\times\dots\times H_{t}$, then

$\operatorname{Irr}(B_{0}(G))=\\{\theta_{1}\times\dots\times\theta_{t}\mid\theta_{i}\in\operatorname{Irr}(B_{0}(H_{i}))\\}.$

Given a normal subgroup $N$ of $G$, we can view the irreducible characters of
$G/N$ as irreducible characters of $G$ containing $N$ in its kernel, and under
this identification

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

with equality in the case where the order of $N$ is coprime to $p$ by [Nav98,
Thm. 9.9(c)].

For any $\theta\in{\rm Irr}(B_{0}(N))$, there is some $\chi\in{\rm
Irr}(B_{0}(G))$ lying over $\theta$ (that is, with $[\chi,\theta^{G}]\neq 0$),
and for any $\chi\in{\rm Irr}(B_{0}(G))$, every constituent of $\chi_{N}$ lies
in $B_{0}(N)$ by [Nav98, Thm. 9.2, Cor. 9.3, and Thm. 9.4]. If moreover
$B_{0}(G)$ is the only block covering $B_{0}(N)$, then for any $\theta\in{\rm
Irr}(B_{0}(N))$ every constituent of $\theta^{G}$ lies in $B_{0}(G)$. This
happens, for instance, in the following case.

######  Lemma 2.1.

Let $G$ be a finite group and let $p$ be a prime. Suppose that $M$ is a normal
subgroup of $G$ containing ${\bf C}_{G}(Q)$, where $Q\in{\rm Syl}_{p}(M)$.
Then the principal $p$-block $B_{0}(G)$ of $G$ is the only block covering the
principal $p$-block $B_{0}(M)$ of $M$.

###### Proof.

Let $B$ be a $p$-block of $G$ covering $B_{0}(M)$. By [Nav98, Thm. 9.26], $Q$
is contained in some defect group $P$ of $B$. Since ${\bf C}_{G}(P)\leq{\bf
C}_{G}(Q)\leq M$, by [Nav98, Lem. 9.20] $B$ is regular with respect to $M$. By
[Nav98, Thm. 9.19], $B_{0}(M)^{G}$ is defined and hence
$B=B_{0}(M)^{G}=B_{0}(G)$ by Brauer’s Third Main Theorem [Nav98, Thm. 6.7]. ∎

Next is a well-known result of J. Alperin and E. C. Dade. We will use both the
notations ${\rm Irr}_{p^{\prime}}(B_{0}(G))$ and ${\rm Irr}_{0}(B_{0}(G))$ to
denote the set of irreducible characters of height zero ($p^{\prime}$-degree)
in the principal $p$-block of $G$, and $k_{0}(B_{0}(G))=|{\rm
Irr}_{0}(B_{0}(G))|$.

######  Theorem 2.2 (Alperin-Dade).

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 blocks of $G$ and $N$. In particular,
$k_{0}(B_{0}(G))=k_{0}(B_{0}(N))$.

###### Proof.

The case where $G/N$ is solvable was proved in [Alp76] and the general case in
[Dad77]. The last part of the statement follows from the first. ∎

The following is contained in [Mur94, Lem. 4.3].

######  Lemma 2.3 (Murai).

Let $p$ be a prime. Let $N$ be a normal subgroup of $G$ and let $\theta$ be a
character of height zero in a $p$-block $b$ of maximal defect of $N$. Let
$P\in{\rm Syl}_{p}(G)$. Suppose that $\theta$ extends to $PN$. If $B$ is a
$p$-block of $G$ of maximal defect covering $b$, then there exists
$\chi\in{\rm Irr}(B)$ of height zero lying over $\theta$.

For a normal subgroup $N$ of $G$ and given $\theta\in{\rm Irr}(B_{0}(N))$, we
will sometimes write ${\rm Irr}_{0}(B_{0}(G)|\theta)=\\{\chi\in{\rm
Irr}_{0}(B_{0}(G))\ |\ [\chi_{N},\theta]\neq 0\\}$ and
$k_{0}(B_{0}(G)|\theta)$ to denote the size of the above set. Lemma 2.3 above
guarantees that $k_{0}(B_{0}(G)|\theta)\neq 0$ if $\theta$ has height zero and
extends to $PN$, where $P\in{\rm Syl}_{p}(G)$.

###  2.2. New results on principal blocks

Here we collect some new properties of principal blocks that are key to our
reduction process and that we believe to be of independent interest.

If $\chi\in{\rm Irr}(B_{0}(G))$ and a normal subgroup $N$ is contained in the
kernel of $\chi$, then in general it is not true that $\chi$, seen as a
character of $G/N$, lies in $B_{0}(G/N)$. The following result was actually
proven within the proof of [NRSV21, Thm. 3.5]. We include it here for the
reader’s convenience. We recall that $\Phi(P)$ denotes the Frattini subgroup
of a $p$-group $P$.

######  Theorem 2.4.

Let $N$ be a normal subgroup of a finite group $G$. Suppose that
$N\subseteq\Phi(P)$, where $P\in{\rm Syl}_{p}(G)$. Write $B_{0}=B_{0}(G)$ to
denote the principal $p$-block of $G$. If $\chi$ is a height zero irreducible
character in $B_{0}$ and $N\subseteq{\rm Ker}(\chi)$, then $\chi$ seen as a
character of $G/N$ belongs to the principal $p$-block of $G/N$.

###### Proof.

Write $\bar{\chi}$ to denote the character $\chi$ seen as a character of
$G/N$. Since $\bar{\chi}$ has degree not divisible by $p$, it follows that
$\bar{\chi}$ lies in a block of $G/N$ of maximal defect $P/N$. By [Nav98,
Prob. 4.5], we only need to prove that if $Nx\in G/N$, with $x\in G$, is
$p$-regular with $P/N\subseteq{\bf C}_{G/N}(Nx)$, then

$\left({[G/N:{\bf
C}_{G/N}(Nx)]\bar{\chi}(Nx)\over\bar{\chi}(1)}\right)\equiv[G/N:{\bf
C}_{G/N}(Nx)]$

modulo any maximal ideal $\mathcal{P}$ of the ring $\bf R$ of algebraic
integers in $\mathbb{C}$ containing $p$. First notice that $Nx\in{\bf
N}_{G/N}(P/N)={\bf N}_{G}(P)/N$. Also, we may assume that $x$ is a $p$-regular
element, using that $Nx=(Nx)_{p^{\prime}}=Nx_{p^{\prime}}$. Notice that

${\bf C}_{G/N}(Nx)={\bf C}_{G}(x)N/N\,,$

using that $(|N|,o(x))=1$ (and [Isa08, Cor. 3.28]). We have shown that $x$ is
a $p$-regular element of ${\bf N}_{G}(P)$ centralizing $P/N$. Therefore $x$
centralizes $P/\Phi(P)$, and by [Isa08, Cor. 3.29] we have that $x$
centralizes $P$. In particular, $N\subseteq{\bf C}_{G}(x)$. Then $[G/N:{\bf
C}_{G/N}(Nx)]=[G:{\bf C}_{G}(x)]$. Since $\chi$ is in the principal block of
$G$, we have that

$\left({[G/N:{\bf
C}_{G/N}(Nx)]\bar{\chi}(Nx)\over\bar{\chi}(1)}\right)=\left({[G:{\bf
C}_{G}(x)]\chi(x)\over\chi(1)}\right)\equiv[G:{\bf C}_{G}(x)]\,,$

modulo any maximal ideal $\mathcal{P}$ of the ring of algebraic integers $\bf
R$ in $\mathbb{C}$ containing $p$, as desired. ∎

If $N\triangleleft G$ and $\theta$ is a character in the principal block of
$N$ that extends to $G$, in general it is not true that the principal block of
$G$ must contain some extension of $\theta$. The following guarantees the
existence of an extension in the principal block. The main idea of the proof
is taken from the proof of [Isa76, Thm. 6.26]. We denote by ${\rm Lin}(G)$ the
subset of ${\rm Irr}(G)$ of characters of $G$ of degree 1, often called linear
characters.

######  Lemma 2.5.

Let $N$ be a normal $p$-subgroup of $G$. Suppose that $\lambda\in{\rm Lin}(N)$
is $G$-invariant and extends to $P$, where $P\in{\rm Syl}_{p}(G)$. Then
$\lambda$ extends to some character in the principal block of $G$.

###### Proof.

Write $|G|=p^{a}m$ with $(m,p)=1$. Let $\delta\in{\rm Lin}(G)$ be an extension
of $\lambda$ (such an extension exists by [Isa76, Theorem 6.26]). Choose a
positive integer $b$ such that $m$ divides $b$ and $b\equiv
1\bmod\,o(\lambda)$. Notice then that $\delta^{b}\in{\rm Lin}(G)$. Write
$G^{0}=\\{g\in G\ |\ (o(g),p)=1\\}$ to denote the set of $p$-regular elements
of $G$. Since every $g\in G^{0}$ satisfies $o(g)$ divides $m$, we have that
$\delta(g)$ is an $m$th root of unity. In particular $\delta^{b}(g)=1$ for
every $g\in G^{0}$. Since then

$\displaystyle\sum_{x\in G^{0}}\delta^{b}(x)=|G^{0}|\neq 0\,,$

we have that $\delta^{b}$ lies in the principal $p$-block of $G$ by [Nav98,
Thm. 3.19]. ∎

We need to prove a relative version of classic results by Landrock [Lan81] and
Cliff–Plesken–Weiss [CPW87] (stated below for the reader’s convenience) for
principal blocks. In fact, we can do more than that. Below, $k_{0}(B)$ denotes
the number of height zero characters in a block $B$.

######  Theorem 2.6.

Let $B$ a $p$-block of nontrivial defect of a finite group $G$. Then

$k_{0}(B)\geq 2\,.$

If moreover $p\leq 3$, then $p$ divides $k_{0}(B)$.

###### Proof.

The first part appears in [CPW87] (see also [Nav98, Prob. 3.11]), and the
second is [Lan81, Cors. 1.3 and 1.6] (see also [NST18, Lem. 2.2] for a proof
for principal blocks). ∎

Below $k_{0}(B|\theta)=k_{0}(B)\cap{\rm Irr}(G|\theta)$ for a block $B$ and a
character $\theta$ of some normal subgroup of $G$. We next prove Theorem B
from the Introduction. We re-state it here in a way in which
$k_{0}(B|\theta)\neq 0$ (by using Lemma 2.3).

######  Theorem 2.7.

Let $G$ be a finite group, $p$ be a prime and $P\in{\rm Syl}_{p}(G)$. Let $N$
be a normal subgroup of $G$. Suppose that $PN/N>1$. Let $B$ be a $p$-block of
$G$ with defect group $P$ covering a $p$-block $b$ of $N$ with defect group
$N\cap P$. Let $\theta\in{\rm Irr}_{0}(b)$. Suppose that $\theta$ extends to
$PN$. Then $k_{0}(B|\theta)\geq 2$. If moreover $p\leq 3$, then $p$ divides
$k_{0}(B|\theta)$.

###### Proof.

Firs notice that, by induction on $[G:N]$, and using the Clifford
correspondence together with the Fong-Reynolds correspondence [Nav98, Thm.
9.14], we may assume that $\theta$ is $G$-invariant.

Define

$\psi=\sum_{\chi\in{\rm Irr}(G|\theta)}\chi(1)\chi\,.$

Notice that $\psi(x)=0$ if $x\notin N$ and $\psi(x)=|G/N|\theta(1)\theta(x)$
if $x\in N$ by a theorem of Knörr (see [Nav18, Thm. 5.21]). In particular,
$\psi^{*}\equiv 0$, where ${}^{*}\colon{\bf R}\to\mathbb{F}$ is the ring
homomorphism defined in [Nav98, Ch. 2] (recall that $\bf R$ is the ring of
algebraic integers in ${\mathbb{C}}$ and we choose a maximal ideal
$\mathcal{P}$ containing $p$ so that $\mathbb{F}$ is algebraically closed of
characteristic $p$ and ∗ is the natural epimorphism).

We now follow the proof of [Nav98, Thm. 5.10] to show that $(\psi_{C})^{*}$
vanishes on the set $G^{0}$ of $p$-regular elements of $G$ for every $p$-block
$C$ of $G$, where $\displaystyle\psi_{C}=\sum_{\chi\in{\rm
Irr}(C|\theta)}\chi(1)\chi$. Given $y\in G^{0}$ (so $x\in G$ such that
$(o(y),p)=1$), we have that

$\psi_{C}(y)=\sum_{\xi\in{\rm Irr}(C)}[\psi,\xi]\xi(y)=\sum_{\xi\in{\rm
Irr}(C)}[\psi,\xi]\left(\sum_{\varphi\in{\rm
IBr(G)}}d_{\xi,\varphi}\varphi(y)\right)\,.$

In particular,

$\psi_{C}(y)=\sum_{\varphi\in{\rm IBr}(G)}\left(\sum_{\xi\in{\rm
Irr}(C)}[\psi,\xi]d_{\xi,\varphi}\right)\varphi(y)\,.$

As $\psi=\sum_{C\in{\rm Bl}(G)}\psi_{C}$, we then have that

$0=\psi(y)^{*}=\sum_{\varphi\in{\rm IBr}(G)}\left(\sum_{\xi\in{\rm
Irr}(G)}[\psi,\xi]d_{\xi,\varphi}\right)^{*}\varphi(y)^{*}\,.$

As the trace functions $\\{\varphi^{*}\ |\ \varphi\in{\rm IBr}(G)\\}$ are
$\mathbb{F}$-linearly independent, by [Nav98, Thm. 1.19], we conclude that for
any $\varphi\in{\rm IBr}(G)$ we have that

$\left(\sum_{\xi\in{\rm Irr}(G)}[\psi,\xi]d_{\xi,\varphi}\right)^{*}=0\,.$

Now if $\varphi$ does not lie in $C$ then

$\sum_{\xi\in{\rm Irr}(C)}[\psi,\xi]d_{\xi,\varphi}=0\,.$

If $\varphi$ lies in $C$ then

$0=\left(\sum_{\xi\in{\rm
Irr}(G)}[\psi,\xi]d_{\xi,\varphi}\right)^{*}=\left(\sum_{\xi\in{\rm
Irr}(C)}[\psi,\xi]d_{\xi,\varphi}\right)^{*}\,.$

In particular, we have shown that $(\psi_{C}(1))^{*}=0$. In other words, the
positive integer

$\sum_{\chi\in{\rm Irr}(C|\theta)}\chi(1)^{2}$

is divisible by $p$, so is the positive integer

$\sum_{\chi\in{\rm Irr}_{0}(C|\theta)}\chi(1)^{2}\,,$

where we are using Lemma 2.3. This implies the first part of the statement. If
$p\leq 3$, then $m^{2}\equiv 1\bmod\,p$ for every integer $m$ coprime to $p$,
so the second part of the statement also follows. ∎

As mentioned in the Introduction, notice that above we have proven that, for
blocks of maximal defect, the main result of [Lar23] holds without the
$p$-solvability hypothesis on $G$.

We close this part with an easy to prove yet useful observation.

######  Lemma 2.8.

Let $G$ be a finite group and $p$ be a prime. Let $N$ be a normal subgroup of
$G$. Let $\theta\in{\rm Irr}_{0}(B_{0}(N))$. Suppose that $p$ divides $[G:N]$
and $\theta$ extends to $G$. Then $k_{0}(B_{0}(G)|\theta)\geq k_{0}(\bar{B})$
for some block $\bar{B}$ of maximal defect of $G/N$. If $\theta$ extends to
some $\chi\in{\rm Irr}(B_{0}(G))$ then $k_{0}(B_{0}(G)|\theta)\geq
k_{0}(B_{0}(G/N))$.

###### Proof.

Write simply $B_{0}=B_{0}(G)$. Let $P/N\in{\rm Syl}_{p}(G_{\theta}/N)$. By
hypothesis, $P/N\in{\rm Syl}_{p}(G/N)$ and $\theta$ extends to $P$. By Theorem
2.7 $k_{0}(B_{0}|\theta)\geq 2$ and if $p\leq 3$ then $p$ divides
$k_{0}(B_{0}|\theta)$.

Let $\tilde{\theta}\in{\rm Irr}(G)$ be an extension of $\theta$ and let
$\chi\in{\rm Irr}_{p^{\prime}}(B_{0}|\theta)$. By Gallagher’s theory,
$\chi=\beta\tilde{\theta}$ for some $\beta\in{\rm Irr}_{p^{\prime}}(G/N)$. Now
the $p$-block $\bar{B}$ of $G/N$ containing $\beta$ is of maximal defect.
Using [Riz19, Lem. 2.4], for instance, we have that for every $\xi\in{\rm
Irr}_{p^{\prime}}(\bar{B})$ the character $\xi\tilde{\theta}\in{\rm
Irr}_{0}(B_{0}|\theta)$. Hence $k_{0}(B_{0}|\theta)\geq k_{0}(\bar{B})$. If we
can choose $\tilde{\theta}$ in $B_{0}$ then $\bar{B}=B_{0}(G/N)$ by [NV17,
Lem. 5.1], for instance. ∎

##  3\. Statements for almost simple groups

Our goal in the next several sections is to prove Theorems 3.1 and 3.3 stated
below. The latter is Theorem A from the Introduction for almost simple groups
$A$ with $[A:{\rm Soc}(A)]_{3}\leq 3$.

######  Theorem 3.1.

Let $S$ be a nonabelian simple group of order divisible by $3$. Then:

  1. (a)

Let $X\in{\rm Syl}_{3}({\rm Aut}(S))$. One of the following holds:

    1. (a1)

There exist $1_{S}\neq\theta_{1},\theta_{2}\in{\rm Irr}_{0}(B_{0}(S))$
nonconjugate in ${\rm Aut}(S)$ and $X$-invariant; or

    2. (a2)

There exist an $X$-invariant $1_{S}\neq\theta\in{\rm Irr}_{0}(B_{0}(S))$ such
that for every $S\leq T\leq{\rm Aut}(S)_{\theta}$, we have $\theta$ extends to
some character in $B_{0}(T)$.

  2. (b)

The set of degrees of $3^{\prime}$-irreducible characters in the principal
$3$-block of $S$ has size at least 3.

  3. (c)

If $S$ has order divisible by $9$, then there exist
$1_{S}\neq\theta_{1},\theta_{2},\theta_{3}\in{\rm Irr}_{0}(B_{0}(S))$ non-
conjugate in ${\rm Aut}(S)$.

######  Remark 3.2.

We will see that the only cases in which we rely on (a2) rather than (a1) are
when $S=\operatorname{PSL}_{2}(q)$ or when $3\nmid(q-\epsilon)$ and
$S=\operatorname{PSL}_{3}^{\epsilon}(q)$. In those (a2) cases,
${\mathrm{Out}}(S)$ has a normal cyclic Sylow $3$-subgroup. For $p\geq 5$, the
$p$-version of condition (b) above was shown to hold in [GRSS20]. Note that
condition (c) above holds if $S$ has abelian Sylow $3$-subgroups by [HSV24,
Thm. 2.1], so when proving (c) we may assume that $S$ has non-abelian Sylow
$3$-subgroups.

######  Theorem 3.3.

Let $A$ be an almost simple group with $[A:{\rm Soc}(A)]_{3}\leq 3$. Write
$k_{0}(B_{0})$ to denote the number of irreducible characters of degree
coprime to $3$ in the principal $3$-block of $A$ and $P\in{\rm Syl}_{3}(A)$.
Then $[P:P^{\prime}]=9$ if, and only if, $k_{0}(B_{0})\in\\{6,9\\}$.

######  Proposition 3.4.

Theorems 3.1 and 3.3 hold when $S$ is a simple sporadic group, a group of Lie
type with exceptional Schur multiplier, or one of:
$\operatorname{G}_{2}(2)^{\prime}$,
${}^{2}\\!\operatorname{G}_{2}(3)^{\prime}$,
${}^{2}\\!\operatorname{F}_{4}(2)^{\prime}$, $\operatorname{PSL}_{2}(27)$,
$\operatorname{PSL}_{2}(81)$, $\operatorname{PSL}_{3}^{\epsilon}(3)$,
$\operatorname{PSL}_{4}^{\epsilon}(3)$, or ${\mathrm{PSp}}_{4}(3)$.

###### Proof.

A list of exceptional Schur multipliers may be found, for example, in [MT11,
Table 24.3]. The result follows from computation in GAP, also taking into
account Propositions 5.17 and 5.18 below for some groups of Lie type with
exceptional Schur multipliers. ∎

The following appeared within the proof of [GRSS20, Thm. 2.4]. It shows how we
will use the statement of Theorem 3.1.(a2) in the context of minimal normal
non-abelian subgroups.

######  Theorem 3.5.

Suppose that $N\triangleleft\,G$ is a direct product
$N=S_{1}\times\cdots\times S_{t}$ of simple non-abelian groups of order
divisible by $p$ transitively permuted by $G$ by conjugation. Let
$\theta=\theta_{1}\times\cdots\times\theta_{t}\in{\rm Irr}(N)$ be
$G$-invariant. If $\theta_{1}$ extends to some irreducible character of degree
coprime to $p$ in the principal $p$-block $T$ for every $S_{1}\leq T\leq{\rm
Aut}(S_{1})_{\theta_{1}}$, then $\theta$ extends to some $\chi\in{\rm
Irr}_{p^{\prime}}(B_{0}(G))$.

###### Proof.

We may view $\theta_{1}$ as a character of $S_{1}{\bf
C}_{G}(S_{1})/S_{1}\triangleleft{\bf N}_{G}(S_{1})/{\bf C}_{G}(S_{1})$. By
hypothesis, $\theta_{1}$ extends to $\hat{\theta_{1}}\in{\rm
Irr_{p^{\prime}}}(B_{0}({\bf N}_{G}(S_{1})/{\bf C}_{G}(S_{1})))$. Let
$\displaystyle M=\bigcap_{i=1}^{t}{\bf N}_{G}(S_{i})\subseteq{\bf
N}_{G}(S_{1})$. We have that $\hat{\theta_{1}}|_{M}\in{\rm
Irr}_{p^{\prime}}(B_{0}(M))$. Consider $\chi=\hat{\theta_{1}}^{\otimes G}$ the
tensor induction of $\hat{\theta}_{1}$ as in [Nav18, Section 10.2]. By [Nav18,
Cor. 10.5], we have that $\chi_{N}=\theta$. Notice that $\chi_{M}$ is a
product of $G$-conjugates of $\hat{\theta_{1}}_{M}$ (see the proof of [GRSS20,
Thm. 2.4] for further details), and so $\chi_{M}\in{\rm
Irr}_{p^{\prime}}(B_{0}(M))$. Let $Q=P\cap M\in{\rm Syl}_{p}(M)$ and let
$Q_{i}=Q\cap S_{i}\in{\rm Syl}_{p}(S_{i})$. Since ${\bf C}_{G}(Q)\subseteq{\bf
C}_{G}(Q_{i})\subseteq{\bf N}_{G}(S_{i})$, we have that ${\bf
C}_{G}(Q)\subseteq M$ and then $B_{0}(G)$ is the only block covering
$B_{0}(M)$ by Lemma 2.1. Therefore, $\chi\in{\rm Irr}_{p^{\prime}}(B_{0}(G))$,
as wanted. ∎

The following result will come in handy when proving the case $[A:{\rm
Soc}(A)]_{3}=1$ of Theorem 3.3.

######  Proposition 3.6.

Let $p=3$, let $S$ be a nonabelian simple group of order divisible by $p$, and
let $S\leq A\leq{\rm Aut}(S)$ with $[A:S]_{3}=1$. Suppose that there exists a
nontrivial $\theta\in{\rm Irr}_{p^{\prime}}(B_{0}(S))$ $A$-invariant and that
there are at least 5 $A$-orbits of characters in ${\rm
Irr}_{p^{\prime}}(B_{0}(S))$. Suppose that $k_{0}(B_{0}(A))\in\\{6,9\\}$. Then
either $k_{0}(B_{0}(S))\leq 18$, or $[P:P^{\prime}]=9$, where $P\in{\rm
Syl}_{p}(A)$.

###### Proof.

In this case $P\in{\rm Syl}_{p}(S)$ and $A=S{\bf N}_{A}(P)$ by the Frattini
argument, so $M=S{\bf C}_{A}(P)$ is normal in $A$. By Lemma 2.1, $B_{0}(A)$ is
the only $p$-block of $A$ covering $B_{0}(M)$, so in particualr $k(A/M)\leq
k_{0}(B_{0}(A))\leq 9$, but since there are at least 5 orbits of characters in
${\rm Irr}_{p^{\prime}}(B_{0}(S))$ and $[A:M]=p^{\prime}$, then
$k(A/M)\in\\{1,2,4,5\\}$. If $A=M$, by Theorem 2.2 we have that
$k_{0}(B_{0}(S))=k_{0}(B_{0}(A))$ and we are done. So we may assume $A>M$. Let
$\tilde{\theta}\in{\rm Irr}_{p^{\prime}}(B_{0}(M))$ be the extension of
$\theta$ to $M$ given by Theorem 2.2, and notice that $\tilde{\theta}$ is
$A$-invariant because $\theta$ is $A$-invariant.

Suppose $k(A/M)=2$. In this case $A/M\cong C_{2}$. Since $1_{M}$ and
$\tilde{\theta}$ extend to $A$, we have at most 5 other $A$-orbits in ${\rm
Irr}_{p^{\prime}}(B_{0}(M))$ of size at most 2, so
$k_{0}(B_{0}(S))=k_{0}(B_{0}(M))\leq 12$.

Next, suppose that $k(A/M)=4$, then $A/M\in\\{C_{4},C_{2}\times
C_{2},D_{10}\\}$. Since $\tilde{\theta}$ is $A$-invariant, if $A/M=D_{10}$, or
$A=C_{4}$, we would have that $\tilde{\theta}$ extends to $A$, and we have too
many characters in ${\rm Irr}_{p^{\prime}}(B_{0}(A))$. So necessarily
$A/M=C_{2}\times C_{2}$ and ${\rm Irr}(A|\tilde{\theta})=\\{\chi\\}$, then
there are at most 4 other orbits of size at most 4, and then
$k_{0}(B_{0}(S))\leq 18$, as wanted.

Finally, suppose that $k(A/M)=5$, so
$A/M\in\\{C_{5},D_{8},Q_{8},D_{14},C_{5}\rtimes C_{4}\\}$. In this case, every
nontrivial $A$-orbit in ${\rm Irr}_{p^{\prime}}(B_{0}(M))$ has a unique
$\chi\in{\rm Irr}(A)$ over it, so in particular $\tilde{\theta}$ is fully
ramified in $A$. But then $[A:M]$ needs to be a square, a contradiction. ∎

##  4\. Alternating and Symmetric Groups

Here we first show that Theorem 3.1 holds for simple alternating groups. Then,
we show that Theorem 3.3 holds for symmetric and alternating groups. Given a
partition $\lambda$ of a natural number $n$, we let $\chi^{\lambda}$ be the
corresponding irreducible character of the symmetric group $S_{n}$. Moreover,
we let $C_{3}(\lambda)$ and $\lambda^{\prime}$ denote the $3$-core and the
conjugate partition of $\lambda$, respectively. We refer the reader to [Ols93]
for a complete account of the basic facts concerning the combinatorics of
partitions.

Let $n\geq 5$ be a natural number and let $A_{n}$ denote the alternating group
of rank $n$. We recall that $\mathrm{Aut}(A_{n})=S_{n}$ for all $n\neq 6$. On
the other hand, $\mathrm{Aut}(A_{6})/A_{6}\cong C_{2}\times C_{2}$. In
particular, if $X\in\mathrm{Syl}_{3}(\mathrm{Aut}(A_{n}))$ then $X\leq A_{n}$
and every irreducible character of $A_{n}$ is obviously $X$-invariant.

Let $a\in\\{0,1,2\\}$ be such that $n=3w+a$, for some $w\in\mathbb{N}$. Let
$\lambda_{1}$ and $\lambda_{2}$ be the partitions of $n$ defined as follows:

$\lambda_{1}=\begin{cases}(n-1,1)&\text{if }a=0,\\\ (n-2,2)&\text{if }a=1,\\\
(2,2,1^{n-4})&\text{if }a=2.\end{cases}\ \
\lambda_{2}=\begin{cases}(n-2,1^{2})&\text{if }a=0,\\\ (n-4,2,1^{2})&\text{if
}a=1,\\\ (2,1^{n-2})&\text{if }a=2.\end{cases}$

For any $i\in\\{1,2\\}$, it is easy to verify that $C_{3}(\lambda_{i})=(a)$
and that $\lambda_{i}\neq\lambda_{i}^{\prime}$. It follows that
$\chi^{\lambda_{i}}\in\mathrm{Irr}(B_{0}(S_{n}))$ and therefore that
$\theta_{i}:=(\chi^{\lambda_{i}})_{A_{n}}\in\mathrm{Irr}(B_{0}(A_{n}))$. An
easy application of the hook length formula [J78, 20.1] shows that:

$\theta_{1}(1)=\begin{cases}n-1&\text{if }a=0,\\\ n(n-3)/2&\text{if }a=1,\\\
n(n-3)/2&\text{if }a=2.\end{cases}\ \
\theta_{2}(1)=\begin{cases}(n-1)(n-2)/2&\text{if }a=0,\\\
(n^{2}-2n)(n-3)(n-5)/8&\text{if }a=1,\\\ n-1&\text{if }a=2.\end{cases}$

It follows that $\theta_{1}(1)$ and $\theta_{2}(1)$ are greater than $1$,
coprime to $3$, and distinct integers. This in turn implies that conditions
($a_{1}$) and (b) of Theorem 3.1 hold for simple alternating groups.

In order to prove that condition (c) holds as well, we can assume that $n\geq
6$ as $9$ should divide the order of $A_{n}$. We further assume that $n\geq
7$, as the case $n=6$ can be easily treated by hand (it is easy to see that
$\mathrm{Irr}_{0}(B_{0}(A_{6}))$ contains $3$ characters of degrees $5$, $8$
and $10$ respectively). We recall that Olsson’s theory of $3$-core towers
[Ols93, Cor. 11.8] allows us to give a precise formula for
$k_{0}(B_{0}(S_{n})))$. In particular, if $n=\sum_{k=0}^{t}a_{k}3^{k}$ is the
$3$-adic expansion of $n$ then we have that

$k_{0}(B_{0}(S_{n}))=\prod_{k=1}^{t}c(3^{k},a_{k}).$

Here $c(M,a)$ denotes the number of sequences of length $M$ of the form
$(\gamma_{1},\ldots,\gamma_{M})$ where $\gamma_{1},\ldots,\gamma_{M}$ are
$3$-core partitions whose sizes sum to $a$. In particular, we deduce that
$k_{0}(B_{0}(S_{n})))\geq c(3^{t},a_{t})\geq c(3^{t},1)=3^{t}$, this in turn
implies that $k_{0}(B_{0}(S_{n})))\geq 9$, whenever $n\geq 9$. Hence, for
$n\geq 9$ there are at least $5$ non $\mathrm{Aut}(A_{n})$-conjugate
irreducible characters of degree coprime to $3$ in the principal block of
$A_{n}$. For a sligthly different reason the same argument applies to the
cases $n\in\\{7,8\\}$. Here we have that $n=2\cdot 3+a$ for some
$a\in\\{1,2\\}$. It is easy to check that $c(3,2)=9$. Exactly as before we
deduce that there are at least $5$ non $\mathrm{Aut}(A_{n})$-conjugate
irreducible characters of degree coprime to $3$ in the principal block of
$A_{n}$. We conclude that condition (c) of Theorem 3.1 holds for $A_{n}$.

We end this section with the verification of Theorem 3.3 for symmetric and
alternating groups. For $k\in\mathbb{N}$ let $X_{k}$ denote a Sylow
$3$-subgroup of $S_{3^{k}}$. We know that $X_{k}\cong C_{3}\wr
C_{3}\wr\cdots\wr C_{3}$ the iterated wreath product of $k$ copies of the
cyclic group of order $3$. It is not difficult to show that
$[X_{k}:X_{k}^{\prime}]=3^{k}$ (it is very easy to show that $X_{k}$ has
exactly $3^{k}$ linear characters. A slightly more involved exercise is to
show that $X_{k}/X_{k}^{\prime}$ is isomorphic to the direct product of $k$
copies of $C_{3}$). Let $n=\sum_{k=0}^{t}a_{k}3^{k}$ be the $3$-adic expansion
of $n$ as above. Given $X\in\mathrm{Syl}_{3}(S_{n})=\mathrm{Syl}_{3}(A_{n})$,
we know from [JK81, 4.1.22] that

$X=\prod_{k=1}^{t}(X_{k})^{\times a_{k}}.$

Hence, we have that $[X:X^{\prime}]=\prod_{k=1}^{t}3^{k\cdot a_{k}}$. If we
now assume that $[X:X^{\prime}]=9$ we immediately deduce that $6\leq n\leq
11$. Using [Ols93, Cor. 11.8] and [Ols90, Prop. 4.4] we obtain that
$\\{k_{0}(B_{0}(S_{n}))\ |\ 6\leq n\leq 11\\}=\\{9\\}$ and that
$\\{k_{0}(B_{0}(A_{n}))\ |\ 6\leq n\leq 11\\}=\\{6,9\\}$. This proves one
implication of Theorem A. The converse is similar. The formulas provided in
[Ols93, Cor. 11.8] and [Ols90, Prop. 4.4] imply that if
$k_{0}(B_{0}(A_{n})),k_{0}(B_{0}(S_{n}))\in\\{6,9\\}$ then
$n\in\\{6,7,\ldots,11\\}$. As discussed above we know that in these cases we
have that $[X:X^{\prime}]=9$.

##  5\. Groups of Lie type

Let $p$ be a prime and let ${\mathbf{G}}$ be a connected reductive algebraic
group over $\bar{\mathbb{F}}_{p}$. Let
$F\colon{\mathbf{G}}\rightarrow{\mathbf{G}}$ be a Steinberg endomorphism. Then
the group of fixed points $G={\mathbf{G}}^{F}$ is a group of Lie type. In this
situation, we will write $G^{\ast}$ for the group $({\mathbf{G}}^{\ast})^{F}$
with $({\mathbf{G}}^{\ast},F)$ in duality with $({\mathbf{G}},F)$ as in [GM20,
Def. 1.5.17].

We summarize some relevant Deligne–Lusztig theory, for which we refer the
reader to, e.g. [GM20, Sec. 2.6]. The set ${\rm Irr}(G)$ is partitioned into
rational Lusztig series $\mathcal{E}(G,s)$ where $s$ ranges over the
semisimple elements of $G^{\ast}$, up to $G^{\ast}$-conjugacy. When $s=1$, the
characters in the set $\mathcal{E}(G,1)$ are known as unipotent characters.
The set $\mathcal{E}(G,s)$ is in bijection with the set of unipotent
characters of ${\bf C}_{G^{\ast}}(s)$, and we refer to those corresponding to
unipotent characters of ${\bf C}_{G^{\ast}}(s)$ lying above the trivial
character of $({\bf C}_{{\mathbf{G}}^{\ast}}(s)^{\circ})^{F}$ as semisimple
characters. In particular, we assume that a choice of this Jordan
decomposition has been made and will denote by $\chi_{s}\in\mathcal{E}(G,s)$
the semsisimple character corresponding to $1_{{\bf C}_{G^{\ast}}(s)}$.
(Although, when ${\bf C}_{{\mathbf{G}}^{\ast}}(s)$ is connected, no choice is
necessary.) The degree of $\chi\in\mathcal{E}(G,s)$ corresponding to
$\psi\in\mathcal{E}({\bf C}_{G^{\ast}}(s),1)$ is $[G^{\ast}:{\bf
C}_{G^{\ast}}(s)]_{p^{\prime}}\psi(1)$, so that $\chi_{s}(1)=[G^{\ast}:{\bf
C}_{G^{\ast}}(s)]_{p^{\prime}}$.

By [CE04, Prop. 8.26 and (8.19)], we have a bijection ${\bf
Z}(G^{\ast})\rightarrow{\rm Irr}(G/\textbf{O}^{p^{\prime}}(G))$, where if
$z\mapsto\hat{z}$, then $\mathcal{E}(G,s)\otimes\hat{z}=\mathcal{E}(G,sz)$.

###  5.1. Our Setting

We will work primarily in the following setting. Let $S$ be a simple group of
Lie type defined in characteristic $p$, so that there is a simple, simply
connected algebraic group ${\mathbf{G}}$ over $\bar{\mathbb{F}}_{p}$ and a
Steinberg endomorphism $F\colon{\mathbf{G}}\rightarrow{\mathbf{G}}$ such that
$S=G/{\bf Z}(G)$ with $G={\mathbf{G}}^{F}$. We can write
$\operatorname{Aut}(S)=\widetilde{S}\rtimes\mathcal{D}$, where $\widetilde{S}$
is the group of inner-diagonal automorphisms of $S$ and $\mathcal{D}$ is an
appropriate group generated by graph and field automorphisms. (See, e.g.
[MT11, Thm. 24.24].) In particular, we will write $F_{p}$ for a generating
field automorphism induced by the map $x\mapsto x^{p}$ on
$\bar{\mathbb{F}}_{p}$ and $F_{p^{i}}=F_{p}^{i}$ for a positive integer $i$.
Further, it will sometimes be useful to identify $\widetilde{S}$ with the
fixed points ${\mathbf{G}}_{\mathrm{ad}}^{F}$ of the adjoint group of the same
type as ${\mathbf{G}}$. (See [MT11, Prop. 24.21] and the subsequent
discussion.) If $\varphi\in\operatorname{Aut}(S)$, then a series
$\mathcal{E}(G,s)$ is mapped to a series $\mathcal{E}(G,s^{\varphi^{\ast}})$
for an appropriate dual automorphism $\varphi^{\ast}$ of $G^{\ast}$, by
[Tay18, Prop. 7.2].

We will also often use that, from [NT13, Lem. 4.4 and Rem. 4.6], we have the
characters in $\mathcal{E}(G,s)$ for $s\in[G^{\ast},G^{\ast}]$ are trivial on
${\bf Z}(G)$ as long as $S$ is not isomorphic to an alternating group or one
of the groups considered in Proposition 3.4.

For the moment, let $\ell$ be any prime and write $B_{0}(X)$ for the principal
$\ell$-block of a group $X$. Now, when $\ell\neq p$, that is $\ell$ is not the
defining characteristic of $G$, a unipotent block is one containing unipotent
characters, and the set of characters lying in unipotent blocks is exactly the
union of the series $\mathcal{E}(G,s)$ with $s\in G^{\ast}$ an $\ell$-element,
by [CE04, Thm. 9.12]. In particular, ${\rm Irr}(B_{0}(G))$ is contained in
this union. By [Hi90, Cor. 3.4] and [CE04, Thm. 21.13], the character
$\chi_{s}$ is contained in ${\rm Irr}(B_{0}(S))$ whenever $s$ is a semisimple
$\ell$-element and either ${\bf Z}({\mathbf{G}}^{\ast})$ is connected or
$\ell$ is good and ${\bf C}_{{\mathbf{G}}^{\ast}}(s)$ is connected.

On the other hand, when $\ell=p$ is the defining characteristic, we have ${\rm
Irr}(B_{0}(S))={\rm Irr}(S)\setminus\\{\mathrm{St}_{S}\\}$ by [CE04, Thm.
6.18].

The following observations regarding extensions will be useful.

######  Lemma 5.1.

Keep the setting above. Let $\ell\neq p$ be a non-defining prime dividing
$|S|$ but not dividing $|\widetilde{S}/S|$. Then restriction provides a
bijection between ${\rm Irr}(B_{0}(\widetilde{S}))$ and ${\rm Irr}(B_{0}(S))$.

###### Proof.

Let $\chi\in{\rm Irr}(B_{0}(S))$ and let $\widetilde{\chi}\in{\rm
Irr}(B_{0}(\widetilde{S})\mid\chi)$, which exists by [Nav98, Thm. 9.2]. Since
$\widetilde{S}/S$ is abelian and multiplicity-free (see e.g. [CE04, Thm.
15.11]), we have by Clifford theory that the number of constituents of
$\widetilde{\chi}|_{S}$ is the number of $\hat{z}\in{\rm
Irr}(\widetilde{S}/S)$ such that $\widetilde{\chi}\hat{z}=\widetilde{\chi}$
(see e.g. [RSV21, Lem. 1.4]). By [CE04, Thm. 9.12], the characters in
$B_{0}(\widetilde{S})$ lie in Lusztig series indexed by semisimple
$\ell$-classes of $({\mathbf{G}}_{\mathrm{ad}}^{\ast})^{F}$ and as discussed
above, the irreducible characters of
$\widetilde{S}/S={\mathbf{G}}_{\mathrm{ad}}^{F}/\textbf{O}^{p^{\prime}}({\mathbf{G}}_{\mathrm{ad}}^{F})$
are in bijection with elements $z$ of ${\bf
Z}({\mathbf{G}}_{\mathrm{ad}}^{\ast})^{F}$, and tensoring with $\hat{z}$ sends
the series $\mathcal{E}(\widetilde{S},s)$ to $\mathcal{E}(\widetilde{S},sz)$.
Since $\ell$ does not divide the order of any such $z\neq 1$, it follows that
$\widetilde{\chi}|_{S}=\chi$. Now, since $\widetilde{S}/S$ is abelian, every
character of $\widetilde{S}$ above $\chi$ is an extension, and hence of the
form $\widetilde{\chi}\hat{z}$ for some $z\in{\bf
Z}({\mathbf{G}}_{\mathrm{ad}}^{\ast})^{F}$. Again, since $\ell$ does not
divide the order of any such $z\neq 1$, we therefore see that
$B_{0}(\widetilde{S})$ contains only one extension of $\chi$. ∎

######  Lemma 5.2.

Keep the setting above and assume $S\neq\operatorname{D}_{4}(q)$ and let
$\ell$ be a prime dividing $|S|$. Suppose $\widetilde{\chi}\in{\rm
Irr}(B_{0}(\widetilde{S}))$ extends to
$\operatorname{Aut}(S)_{\widetilde{\chi}}$ and restricts irreducibly to
$\chi\in{\rm Irr}(S)$. Then for every every $S\leq T\leq{\rm
Aut}(S)_{\widetilde{\chi}}$, we have $\chi$ extends to some character in
$B_{0}(T)$.

###### Proof.

First, with our assumption on $S$, we have
$\operatorname{Aut}(S)/\widetilde{S}$ is abelian. Then every character in
$\operatorname{Irr}(\operatorname{Aut}(S)_{\widetilde{\chi}}\mid\widetilde{\chi})$
is an extension, and hence there is an extension $\hat{\chi}$ of
$\widetilde{\chi}$ in $B_{0}(\operatorname{Aut}(S)_{\widetilde{\chi}})$. Now,
let $S\leq T\leq\operatorname{Aut}(S)_{\widetilde{\chi}}$. Then
$\hat{\chi}|_{T\cap\widetilde{S}}=\widetilde{\chi}|_{T\cap\widetilde{S}}$ lies
in $B_{0}(T\cap\widetilde{S})$ since $B_{0}(\widetilde{S})$ must cover a
unique block of $T\cap\widetilde{S}$. By [Nav98, Thm. 9.4], $B_{0}(T)$
contains a character lying above $\hat{\chi}|_{T\cap\widetilde{S}}$. Since
$T/(T\cap\widetilde{S})$ is abelian, again this character is an extension,
completing the proof. ∎

For the remainder of the section, we will return to the case $\ell=3$, so that
$B_{0}(X)$ denotes the principal $3$-block of a group $X$.

###  5.2. Defining Characteristic

Here we deal with the case that $p=3$. That is, $S$ is defined in
characteristic $3$.

######  Proposition 5.3.

Let $q$ be a power of $3$ and let $S$ be one of the simple groups
$\operatorname{G}_{2}(q)$, $\operatorname{D}_{4}(q)$,
${}^{3}\\!\operatorname{D}_{4}(q)$, $\operatorname{F}_{4}(q)$,
$\operatorname{E}_{6}^{\pm}(q)$, $\operatorname{E}_{7}(q)$, or
$\operatorname{E}_{8}(q)$. Let
$X\in\operatorname{Syl}_{3}(\operatorname{Aut}(S))$. Then there exist at least
four $X$-invariant characters with distinct degrees in ${\rm
Irr}_{3^{\prime}}(B_{0}(S))$. Further, if
$S={}^{2}\\!\operatorname{G}_{2}(q^{2})$, with $q^{2}=3^{2n+1}>3$, then there
are six distinct character degrees in ${\rm Irr}_{3^{\prime}}(B_{0}(S))$ and
at least three not $\operatorname{Aut}(S)$-conjugate, nontrivial $X$-invariant
characters, two of which have distinct degrees. In particular, Theorem 3.1
(with (a1)), holds in these cases.

###### Proof.

Recalling that $B_{0}(S)=\operatorname{Irr}(S)\setminus\\{\mathrm{St}_{S}\\}$,
we have
$\operatorname{Irr}_{3^{\prime}}(B_{0}(S))=\operatorname{Irr}_{3^{\prime}}(S)$.
Keeping the notation of Sec. 5.1, we have $\widetilde{S}=S=G$ unless
$S=\operatorname{E}_{7}(q)$, in which case $|\widetilde{S}/S|=|{\bf Z}(G)|=2$
or $S=\operatorname{D}_{4}(q)$, in which case $|\widetilde{S}/S|=4$ and $|{\bf
Z}(G)|$ divides $4$. The character degrees of $\widetilde{S}$ and $G$ and
their multiplicities have been computed by Lübeck and are available at [Lub].
From this information, we see that in each case other than
${}^{2}\\!\operatorname{G}_{2}(q^{2})$, there are at least three nontrivial
$3^{\prime}$-character degrees of $\widetilde{S}$ whose multiplicities are not
divisible by $3$ and whose constituents on $S$ have distinct degrees of
multiplicity not divisible by $3$. When
$S={}^{2}\\!\operatorname{G}_{2}(q^{2})$ with $q^{2}>3$, we see five distinct
nontrivial character degrees, two of which have multiplicity prime to $3$. We
see directly from the character table in [Wa66] that there are at least three
nontrivial, non-$\operatorname{Aut}(S)$-conjugate, $X$-invariant characters in
this case. ∎

######  Lemma 5.4.

Let $S=\operatorname{PSL}_{n}^{\epsilon}(q)$, where $q$ is a power of $3$ and
$n\geq 2$. Then Theorem 3.1 holds for $S$. Further, (a1) holds for $n\geq 4$.

###### Proof.

Note that in this case, $XS/S$ must be generated by field automorphisms. As
before,
$\operatorname{Irr}_{3^{\prime}}(B_{0}(S))=\operatorname{Irr}_{3^{\prime}}(S)$
and we may assume by Proposition 3.4 that $S$ has a nonexceptional Schur
multiplier.

First suppose $S=\operatorname{PSL}_{2}(q)$ with $q\geq 9$. In this case, the
character tables for $S$ is well-known, even available in GAP4 [GAP]. The two
characters $\chi_{1},\chi_{2}$ of degree $\frac{q\pm 1}{2}$ must be invariant
under $X\in\operatorname{Syl}_{3}(\operatorname{Aut}(S))$ but fuse in
$\widetilde{S}$. Here $\mathcal{D}$ is cyclic, so
$\operatorname{Aut}(S)_{\chi_{1}}/S$ is cyclic. Then every character lying
above $\chi_{1}$ in $\operatorname{Aut}(S)_{\chi_{1}}$ is an extension, so
some extension $\hat{\chi}$ lies in $B_{0}(\operatorname{Aut}(S)_{\chi_{1}})$.
For $S\leq T\leq\operatorname{Aut}(S)_{\chi_{1}}$, we then have the
restriction of $\hat{\chi}$ to $T$ lies in $B_{0}(T)$ and extends $\chi_{1}$,
giving (a2). There are also characters of degree $q+1$ and $q-1$ in
$\operatorname{Irr}(S)$, giving (b) and (c).

We now assume $n\geq 3$. If $(n,q)\in\\{(3,3),(4,3)\\}$, we see the statement
using GAP, so we may further assume that $q\geq 9$ or $n\geq 5$. From [Mal07,
Thm. 6.8], we see that $\operatorname{Irr}_{3^{\prime}}(B_{0}(S))$ consists of
the deflations of the semisimple characters $\chi_{s}$ of
$G={\mathrm{SL}}_{n}^{\epsilon}(q)$ that are trivial on ${\bf Z}(G)$. Recall
that such a character has degree $[G^{\ast}:{\bf
C}_{G^{\ast}}(s)]_{3^{\prime}}$. Then for (b) and (c), it suffices by the
discussion at the beginning of the section to exhibit three nontrivial classes
of semisimple $s\in G^{\ast}$ satisfying $s\in[G^{\ast},G^{\ast}]$, and with
distinct $|{\bf C}_{G^{\ast}}(s)|_{3^{\prime}}$. We further write
$\widetilde{G}:={\mathrm{GL}}_{n}^{\epsilon}(q)$, so
$[\widetilde{G}^{\ast},\widetilde{G}^{\ast}]\cong G$. We will in fact exhibit
three nontrivial $\widetilde{s}\in[\widetilde{G}^{\ast},\widetilde{G}^{\ast}]$
such that $\chi_{\widetilde{s}}$ is irreducible on restriction to $G$ and with
distinct $|{\bf C}_{\widetilde{G}^{\ast}}(\widetilde{s})|_{3^{\prime}}$. Note
that the centralizer structures of semisimple elements are described in [FS82,
Prop. (1A)].

If $q\geq 9$, then there is some $\zeta\in C_{q-\epsilon}$ with $|\zeta|>2$.
Let $\widetilde{s}_{1}$ be the element
${\mathrm{diag}}(\zeta,\zeta^{-1},1,\cdots,1)$ in
$[\widetilde{G}^{\ast},\widetilde{G}^{\ast}]$. Then $|{\bf
C}_{\widetilde{G}^{\ast}}(\widetilde{s}_{1})|=(q-\epsilon)^{2}\times|{\mathrm{GL}}_{n-2}^{\epsilon}(q)|$
and the character $\chi_{\widetilde{s}_{1}}$ is irreducible on restriction to
$G$ since $\widetilde{s}_{1}$ is not $\widetilde{G}^{\ast}$-conjugate to any
$\widetilde{s}_{1}z$ for $z\in{\bf Z}(\widetilde{G}^{\ast})$. Similarly,
letting $\widetilde{s}_{2}:={\mathrm{diag}}(\zeta,\zeta,\zeta^{-2},1,\cdots
1)$, we have $\chi_{\widetilde{s}_{2}}$ is irreducible on restriction to $G$
and $|{\bf
C}_{\widetilde{G}^{\ast}}(\widetilde{s}_{2})|=(q-\epsilon)\times|{\mathrm{GL}}_{2}^{\epsilon}(q)|\times|{\mathrm{GL}}_{n-3}^{\epsilon}(q)|$.
Taking $\widetilde{s}_{3}$ to have eigenvalues $\\{\xi,\xi^{-1},1,\cdots,1\\}$
with $|\xi|>2$ and $|\xi|\mid(q+\epsilon)$, we again have
$\chi_{\widetilde{s}_{3}}$ is also irreducible on $G$ and now $|{\bf
C}_{\widetilde{G}^{\ast}}(\widetilde{s}_{3})|=(q^{2}-1)\times|{\mathrm{GL}}_{n-2}^{\epsilon}(q)|$.
Hence (b) and (c) hold.

Further, choosing $\zeta$ or $\xi$ to be a fourth root of unity $i$, depending
on whether $q\equiv\epsilon\pmod{4}$ or $q\equiv-\epsilon\pmod{4}$, the
corresponding character $\chi_{\widetilde{s}}=\chi_{\widetilde{s}_{1}}$, resp
$\chi_{\widetilde{s}_{3}}$, above is further invariant under field
automorphisms, as $F_{3}$ stabilizes the class of $\widetilde{s}$. Then this
character is $X$-invariant. Further, we see that this class is even
$\mathcal{D}$-invariant, so $\chi_{\widetilde{s}}$ extends to
$\operatorname{Aut}(S)_{\chi_{\widetilde{s}}}=\operatorname{Aut}(S)$ by
[Spa12, Prop. 3.4 and proof of Lem. 2.13], giving (a2) when combined with
Lemma 5.2. (Indeed, recall that ${\bf Z}(G)$ is a nonexceptional Schur
multiplier for $S$. Further, the factor set constructed in [Spa12, Lem. 2.13]
is trivial in this case since $\chi_{\widetilde{s}}$ is
$\mathcal{D}$-invariant.) Considering
$\widetilde{s}_{3}:={\mathrm{diag}}(i,i,-i,-i,1,...1)$ when $n\geq 4$, we
obtain a second $X$-invariant character not $\operatorname{Aut}(S)$-conjugate
to $\chi_{\widetilde{s}}$, yielding further (a1) when $n\geq 4$.

Then finally, we may assume $n\geq 5$ and $q=3$. Here (a1) is satisfied in
exactly the same way by the restriction of the same $\chi_{\widetilde{s}}$ and
$\chi_{\widetilde{s}_{3}}$. We obtain (b) and (c) in this case by using
$\widetilde{s}$, together with $\widetilde{s}_{3}$ and $\widetilde{s}_{4}$,
where $\widetilde{s}_{4}$ has nontrivial eigenvalues $\\{-1,-1\\}$. ∎

######  Lemma 5.5.

Let $q$ be a power of $3$ and let $S$ be a simple group
$S={\mathrm{PSp}}_{2n}(q)$ with $n\geq 2$, ${\mathrm{P\Omega}}_{2n+1}(q)$ with
$n\geq 3$, or ${\mathrm{P\Omega}}_{2n}^{\pm}(q)$ with $n\geq 4$. Let
$X\in\operatorname{Syl}_{3}(\operatorname{Aut}(S))$. Then there exist at least
four $X$-invariant characters with distinct degrees in ${\rm
Irr}_{3^{\prime}}(B_{0}(S))$. In particular, Theorem 3.1 holds for $S$ with
(a1).

###### Proof.

Again, we have
$\operatorname{Irr}_{3^{\prime}}(B_{0}(S))=\operatorname{Irr}_{3^{\prime}}(S)$.
The case $S=\operatorname{D}_{4}(q)={\mathrm{P\Omega}}_{8}^{+}(q)$ is covered
in Proposition 5.3, so we assume $S$ is not this group. Then note that $XS/S$
must be generated by field automorphisms. Let $H={\mathrm{Sp}}_{2n}(q)$,
${\mathrm{SO}}_{2n+1}(q)$, or ${\mathrm{SO}}_{2n}^{\pm}(q)$, respectively, and
write $\Omega:=\textbf{O}^{p^{\prime}}(H)$, so that $\Omega$ is perfect,
$S=\Omega/{\bf Z}(\Omega)$, and $H/\Omega$ is a $2$-group. If
$S={\mathrm{PSp}}_{4}(q)$, we can see the statement directly from the
character table constructed in [Sri68]. Hence, we may assume that $n\geq 3$.

Note that we can naturally embed ${\mathrm{GL}}_{n}(q)$ into
$H^{\ast}\cong{\mathrm{SO}}_{2n+1}(q)$, ${\mathrm{Sp}}_{2n}(q)$, or
${\mathrm{SO}}_{2n}^{+}(q)$, and can embed ${\mathrm{GL}}_{n-1}(q)$ into
${\mathrm{SO}}_{2n}^{-}(q)$. Then elements of ${\mathrm{SL}}_{n}(q)$ will
embed as elements of $[H^{\ast},H^{\ast}]$. In particular let
$s_{1},s_{2},s_{3}$ be elements of $H^{\ast}$ with nontrivial eigenvalues
$\\{-1,-1,-1,-1\\}$, $\\{i,-i,-i,i\\}$, and
$\\{\zeta,\zeta^{3},-1,\zeta^{-1},\zeta^{-3},-1\\}$, where
$\zeta\in\mathbb{F}_{q^{2}}^{\times}$ has order $8$. Then these lie in
$[H^{\ast},H^{\ast}]$, so that the corresponding semisimple characters of $H$
are trivial on ${\bf Z}(H)$. Further, the class of each $s_{i}$ is stable
under $F_{3}$, and hence $X$, so the $\chi_{s_{i}}$ are $X$-invarant. For each
$i$, let $\chi_{i}\in\operatorname{Irr}(\Omega\mid\chi_{s_{i}})$. Since
$|H/\Omega|\leq 2$, we see each $\chi_{i}$ is also $X$-invariant. Further, the
$|{\bf C}_{{H}^{\ast}}(\widetilde{s}_{i})|_{3^{\prime}}$ are distinct, using
the centralizer structures described in [FS89, (1.13) and discussion after].
Further, of these, only $s_{1}$ can potentially be conjugate to $s_{1}z$ for
$1\neq z\in{\bf Z}(H)$, so $\chi_{s_{1}}$ is the only of the three that can
possibly not restrict irreducibly. Since it also has the largest $|{\bf
C}_{{H}^{\ast}}(\widetilde{s}_{i})|_{3^{\prime}}$, we see that
$\chi_{1},\chi_{2},\chi_{3}$ also have distinct degrees. ∎

For a group $X$ with principal $3$-block $B_{0}$ and
$A\leq\operatorname{Aut}(X)$, we will write $k_{A,0}(B_{0})$ for the number of
distinct $A$-orbits in $\operatorname{Irr}_{3^{\prime}}(B_{0})$.

######  Proposition 5.6.

Let $S=G/{\bf Z}(G)$ be a simple group of Lie type such that
$G={\mathbf{G}}^{F}$ for ${\mathbf{G}}$ a simple simply connected algebraic
group and $F\colon{\mathbf{G}}\rightarrow{\mathbf{G}}$ a Frobenius
endomorphism defining $G$ over $\mathbb{F}_{q}$ with $q=3^{a}$. Assume that
$S$ is not isomorphic to an alternating group or one of the groups covered in
Proposition 3.4. Then $k_{\operatorname{Aut}(S),0}(B_{0}(S))\geq 10$.

###### Proof.

Write $k:=k_{\operatorname{Aut}(S),0}(B_{0}(S))$. Arguing as in [HS23, Sec.
(2D)], we have

$k\geq\frac{q^{r}}{|{\bf Z}(G)||{\mathrm{Out}}(S)|}$

where $r$ is the rank of ${\mathbf{G}}$. (Indeed, note that this gives a bound
for orbits of semisimple, hence $3^{\prime}$, characters, although this was
not needed in loc. cit.)

If $G={\mathrm{SL}}_{2}(q)$, this gives $k\geq\frac{3^{a}}{4a}$, which is
larger than 10 for $a\geq 5$. Note that our assumptions exclude the case
$a=1,2$ here. If $G={\mathrm{SL}}_{n}^{\epsilon}(q)$ with $n\geq 3$, we have
$|{\bf Z}(G)|=(n,q-\epsilon)$ and $|{\mathrm{Out}}(S)|=2a(n,q-\epsilon)$, so
we see from this bound that $k\geq 10$ unless possibly $q=3$ with $n=3,4$. If
$S=\operatorname{D}_{n}^{\pm}(q)$ with $n\geq 5$, we have $|{\bf Z}(G)|\leq 4$
and $|{\mathrm{Out}}(S)|\leq 8a$, so $k\geq 3^{an}/32a\geq 10$ for $n\geq 6$
or for $a\geq 2$ if $n=5$. For $S=\operatorname{D}_{4}^{\pm}(q)$, we instead
obtain $k\geq 3^{4a}/(96a)\geq 10$ for $a\geq 2$. If
$S=\operatorname{B}_{n}(q)$ or $\operatorname{C}_{n}(q)$ with $n\geq 2$, we
have $k\geq 3^{an}/(4a)\geq 10$ for $n\geq 4$ or for $a\geq 2$. In the cases
$S=\operatorname{D}_{5}^{\pm}(3),\operatorname{D}_{4}^{\pm}(3),$
$\operatorname{C}_{3}(3)$, or $\operatorname{B}_{3}(3)$, we see from GAP that
there are at least 10 distinct $3^{\prime}$-character degrees.

In the exceptional groups, the same bound for $k$ gives $k\geq 10$, except
possibly if $S=\operatorname{G}_{2}(3)$. In the latter case we can again see
the statement explicitly from the character table in GAP. ∎

######  Proposition 5.7.

Let $S$ be a simple group of Lie type defined in characteristic $3$. Then
Theorem 3.3 holds for $S$.

###### Proof.

We may assume that $S$ is not isomorphic to an alternating group or one of the
groups considered in Proposition 3.4. Again, recall that ${\rm
Irr}_{3^{\prime}}(S)={\rm Irr}_{3^{\prime}}(B_{0}(S))$.

So, we may assume that $S$ and $G$ are as in Proposition 5.6 or that
$S=G={}^{2}\\!\operatorname{G}_{2}(q^{2})$. Then $|Z(G)|$ is not divisible by
$3$ and the Sylow $3$-subgroup for $S$ is isomorphic to one of $G$, which is
of the form $U=\mathbf{U}^{F}$ with $\mathbf{U}$ the unipotent radical of an
$F$-stable Borel subgroup $\mathbf{B}=\mathbf{U}{\mathbf{T}}$ of
${\mathbf{G}}$. Here ${\mathbf{T}}$ is an $F$-stable maximal torus.

If $S={}^{2}\\!\operatorname{G}_{2}(q^{2})$ with $q^{2}=3^{2a+1}$ and $a\geq
1$, then a Sylow $3$-subgroup $U$ of $S$ satisfies $[U:U^{\prime}]=q^{2}\neq
9$ by [Wa66, Thm.(2)]. From the character table in [Wa66], we see that $|{\rm
Irr}_{3^{\prime}}(S)|>18$ and as discussed in Proposition 5.3, there are at
least six not $\operatorname{Aut}(S)$-conjugate characters in ${\rm
Irr}_{3^{\prime}}(B_{0}(S))$ and at least four of these are $X$-invariant,
where $X\in\mathrm{Syl}_{3}(A)$. Hence $k_{0}(B_{0}(A))>9$, using Theorem 2.7
when $|A/S|_{3}=3$, and Proposition 3.6 when $|A/S|_{3}=1$.

In the cases that $S$ is as in Proposition 5.6, since
$G\neq\operatorname{G}_{2}(3)$, we have by [DLM97, Lem. 2.2] that
$U/U^{\prime}$ is isomorphic to the direct product of the additive groups
$\mathbb{F}_{q^{|\omega|}}$, where $\omega$ runs over the orbits of the action
induced by $F$ on the fundamental roots for ${\mathbf{G}}$ with respect to
${\mathbf{T}}$. We see then that $U/U^{\prime}$ has size 9 exactly when
$(r,q)\in\\{(2,3),(1,9)\\}$ but $G\neq\operatorname{G}_{2}(3)$, where $r$ is
the rank of ${\mathbf{G}}$. These cases are covered in Proposition 3.4 or
Section 4. In the remaining cases, we therefore have $[U:U^{\prime}]\neq 9$,
and we see from Proposition 5.6 that
$k_{\operatorname{Aut}(S),0}(B_{0}(S))\geq 10$ (and hence $k_{0}(B_{0}(A))\geq
10$ if $|A/S|_{3}=1$). This completes the statement when $|A/S|_{3}=1$.

Now, assume $|A/S|_{3}=3$. A Sylow $3$-subgroup $P$ of $A$ can be identified
with $U\rtimes C$ where $C$ is a cyclic group of graph-field automorphisms of
size $3$ stabilizing $U$. From here, arguing exactly as in the last two
paragraphs of the proof of [NRSV21, Prop. 4.9] yields $P/P^{\prime}$ is not
$2$-generated, and hence $[P:P^{\prime}]\neq 9$, unless
$S=\operatorname{PSL}_{2}(27)$, which is again covered in Proposition 3.4.

From the discussion of ${}^{2}\\!\operatorname{G}_{2}(q^{2})$ above, note that
we may now assume that $S$ is as in Proposition 5.6. If $S$ is not
$\operatorname{PSL}_{n}^{\epsilon}(q)$, then Proposition 5.3 and Lemma 5.5
yield at least four $P$-invariant characters in ${\rm
Irr}_{3^{\prime}}(B_{0}(S))$ that are not $A$-conjugate. Then Theorem 2.7
yields at last 12 characters in ${\rm Irr}_{3^{\prime}}(B_{0}(A))$.

We are left to to argue that $k_{0}(B_{0}(A))>9$ in the case that
$S=\operatorname{PSL}_{n}^{\epsilon}(q)$ with $n\geq 2$ and $S$ is not covered
in Proposition 3.4. Let $G={\mathrm{SL}}_{n}^{\epsilon}(q)$,
$\widetilde{S}={\mathrm{PGL}}_{n}^{\epsilon}(q)$, and
$\widetilde{G}={\mathrm{GL}}_{n}^{\epsilon}(q)$. Since $A/S$ has order
divisible by $3$, it must be that $A$ contains a field automorphism $\varphi$
of order $3$, so $q=q_{0}^{3}$ for some $3$-power $q_{0}$. Note that by the
discussion at the beginning of the section, it suffices to show that there are
three nontrivial classes of semisimple elements $s\in G^{\ast}$ that are not
$\operatorname{Aut}(S)$-conjugate and such that $s\in[G^{\ast},G^{\ast}]$ and
$s$ is invariant under $\varphi^{\ast}$. This will yield four $P$-invariant,
not $A$-conjugate members of ${\rm Irr}_{3^{\prime}}(B_{0}(S))$, so that
$k_{0}({B_{0}(A)})>9$ by Theorem 2.7.

First, let $S=\operatorname{PSL}_{2}(q)$, so we assume $q_{0}\geq 9$. The
semisimple elements of $G^{\ast}$ whose preimages in $\widetilde{G}^{\ast}$
have eigenvalues $\\{i,-i\\}$, $\\{\zeta,\zeta^{-1}\\}$, and
$\\{\xi,\xi^{-1}\\}$ with $|\zeta|\mid(q_{0}-1)$ and $|\xi|\mid(q_{0}+1)$ but
$|\zeta|,|\xi|>4$ yield three nontrivial classes with the properties discussed
above. Next, consider $\operatorname{PSL}_{n}^{\epsilon}(q)$ with $n\geq 3$.
Let $\zeta\in\mathbb{F}_{q^{2}}^{\times}$ have order $8$. Then the semisimple
classes in $G^{\ast}$ whose preimages in $\widetilde{G}^{\ast}$ have
nontrivial eigenvalues $\\{-1,-1\\}$, $\\{i,-i\\}$, and
$\\{\zeta,\zeta^{3},-1\\}$ satisfy our properties, and the proof is complete.
∎

###  5.3. Non-Defining Characteristic

We now move to the case that $p\neq 3$. That is, $S$ is a simple group of Lie
type defined in characteristic distinct from $3$. We will keep the setting of
Section 5.1. By an abuse of terminology, we will say that $S$ is defined over
$\mathbb{F}_{q}$ if either $S={}^{2}\\!\operatorname{F}_{4}(q^{2})$ or if $F$
is a Frobenius map defining ${\mathbf{G}}$ over $\mathbb{F}_{q}$.

######  Proposition 5.8.

Let $S$ be a simple group of Lie type defined over $\mathbb{F}_{q}$ with
$3\nmid q$ and $3\mid|S|$. Then
$\operatorname{Irr}_{3^{\prime}}(B_{0}(\widetilde{S}))$ contains at least $4$
unipotent characters that extend to $\operatorname{Aut}(S)$ and restrict
irreducibly to $\operatorname{Irr}_{3^{\prime}}(B_{0}({S}))$, except

  * •

$S=\operatorname{PSL}_{4}^{\epsilon}(q)$ with $3\mid(q-\epsilon)$ or
${\mathrm{PSp}}_{4}(2^{a})$, in which case there are at least 3 such
characters;

  * •

$S=\operatorname{PSL}_{3}^{\epsilon}(q)$, in which case there are 3 such
characters if $3\mid(q-\epsilon)$ and $2$ if $3\mid(q+\epsilon)$; and

  * •

$S=\operatorname{PSL}_{2}(q)$, in which case there are $2$ such characters.

Further, these four, respectively 3, 3, 2, 2, characters can be taken with
distinct degrees. In particular, $B_{0}(\operatorname{Aut}(S))$ contains an
extension of each of these characters.

###### Proof.

The first part is proved in [RSV20, Propositions 3.9–3.11]. The Steinberg and
trivial characters are always two of these characters, and we refer the reader
to Tables 2-5 of loc. cit. for the additional explicitly chosen unipotent
characters. In fact, the nontrivial, non-Steinberg characters listed there
have distinct degree, using the formula in [GM20, Sec. 4.4]. The last
statement follows from the structure of $\operatorname{Aut}(S)/S$ and is
proven, for example, in [GMS22, Lemma 3.3]. ∎

######  Corollary 5.9.

Theorem 3.1 holds for $S$ a group of Lie type defined over $\mathbb{F}_{q}$
with $3\nmid q$. Further, (a1) holds unless $S=\operatorname{PSL}_{2}(q)$ or
$3\mid(q+\epsilon)$ and $S=\operatorname{PSL}_{3}^{\epsilon}(q)$.

###### Proof.

This follows from Proposition 5.8 if $S$ is not one of the excluded cases
there, and part (a) follows also in those cases. (Note that we achieve (a1)
except in the case $S=\operatorname{PSL}_{2}(q)$ and the case
$3\mid(q+\epsilon)$ with $S=\operatorname{PSL}_{3}^{\epsilon}(q)$, where we
obtain (a2) when combined with Lemma 5.2.)

Now let $S=\operatorname{PSL}_{2}(q)$ or let
$S=\operatorname{PSL}_{3}^{\epsilon}(q)={\mathrm{SL}}_{3}^{\epsilon}(q)$ with
$3\mid(q+\epsilon)$. In this case, Sylow $3$-subgroups of $S$ are abelian, and
(c) follows from [HSV23, Thm. 2.1]. For (b), it suffices to know that there is
at least one non-unipotent member in ${\rm Irr}_{3^{\prime}}(B_{0}(S))={\rm
Irr}(B_{0}(S))$. Let $G={\mathrm{SL}}_{2}(q)$, resp.
${\mathrm{SL}}_{3}^{\epsilon}(q)$. In this case, any semisimple $3$-element of
$G^{\ast}$ lies in $[G^{\ast},G^{\ast}]$ and has connected centralizer, so the
corresponding semisimple character of $G$ must be trivial on ${\bf Z}(G)$ and
lie in $B_{0}(G)$ by the discussion in Sec. 5.1.

In the remaining cases, it remains to show (c), so we need to exhibit one
additional character in ${\rm Irr}_{3^{\prime}}(B_{0}(S))$ not
$\operatorname{Aut}(S)$-conjugate to the three unipotent characters. Namely,
it suffices again to know there is a non-unipotent member of ${\rm
Irr}_{3^{\prime}}(B_{0}(S))$. If
$S={\mathrm{PSp}}_{4}(2^{a})={\mathrm{Sp}}_{4}(2^{a})=G$, then any $3$-element
$s$ of $G^{\ast}$ again yields a semisimple character in $B_{0}(S)$, for the
same reasoning. Choosing $s$ to be in ${\bf Z}(P^{\ast})$ with
$P^{\ast}\in\operatorname{Syl}_{3}(G^{\ast})$ further gives such a character
with $3^{\prime}$-degree.

Finally, assume $3\mid(q-\epsilon)$ and that
$S=\operatorname{PSL}_{4}^{\epsilon}(q)$ or
$S=\operatorname{PSL}_{3}^{\epsilon}(q)$. Let
$G={\mathrm{SL}}_{4}^{\epsilon}(q)$, respectively
${\mathrm{SL}}_{3}^{\epsilon}(q)$. Then $G$ has a unique unipotent $3$-block
using [FS82], and hence the semisimple character $\chi_{s}$ lies in ${\rm
Irr}(B_{0}(G))$ for any $3$-element $s\in G^{\ast}$. In the case
$G={\mathrm{SL}}_{4}^{\epsilon}(q)$, we have
$3\nmid[G^{\ast}:[G^{\ast},G^{\ast}]]$, so taking $s$ to be any element in the
center of a Sylow $3$-subgroup of $[G^{\ast},G^{\ast}]$, we again obtain
$\chi_{s}\in{\rm Irr}_{3^{\prime}}(B_{0}(G))$ is trivial on the center.
Finally, let $G={\mathrm{SL}}_{3}^{\epsilon}(q)$. The element
$(\omega,\omega^{-1},1)$ of ${\mathrm{SL}}_{3}^{\epsilon}(q)$, where
$|\omega|=3$, commutes with a Sylow $3$-subgroup of
${\mathrm{GL}}_{3}^{\epsilon}(q)$ up to multiplication by ${\bf
Z}({\mathrm{GL}}_{3}(q))$, and hence yields a $3$-element $s$ of
$[G^{\ast},G^{\ast}]$ centralizing a Sylow $3$-subgroup of $G^{\ast}$, again
yielding a semisimple character
$\chi_{s}\in\operatorname{Irr}_{3^{\prime}}(B_{0}(S))$. ∎

We remark that this completes the proof of Theorem 3.1.

###  5.4. Sylow Subgroups with $[P:P^{\prime}]=9$ in Non-Defining
Characteristic

Here we discuss when the condition $[P:P^{\prime}]=9$ holds for groups of Lie
type defined in characteristic $p\neq 3$. It is worth noting that when
$[P:P^{\prime}]=9$, we necessarily have $[P:\Phi(P)]\leq 9$, and $P$ is either
cyclic or $2$-generated. It will also be useful to note that when $G=H\wr K$
is a standard wreath product, we have $G/G^{\prime}\cong H/H^{\prime}\times
K/K^{\prime}$, for example from [Ro96, Prob. 1.6.20].

Throughout, for an integer $m$ with $3\nmid m$, we will write $d_{3}(m)$ for
the order of $m$ modulo $3$. If $3\mid m$, we write $m_{3}$ for the highest
power of $3$ dividing $m$.

####  5.4.1. $\operatorname{PSL}_{2}(q)$ and
$\operatorname{PSL}_{3}^{\epsilon}(q)$

We begin with the case of $\operatorname{PSL}_{2}(q)$ and
$\operatorname{PSL}_{3}^{\epsilon}(q)$.

######  Proposition 5.10.

Let $q$ be a power of a prime $p\neq 3$ and let $S=\operatorname{PSL}_{2}(q)$
or let $3\mid(q+\epsilon)$ and $\operatorname{PSL}^{\epsilon}_{3}(q)$. Let $A$
be an almost simple group $S\lhd A\leq\operatorname{Aut}(S)$ with
$|A/S|_{3}\leq 3$ and let $P\in\operatorname{Syl}_{3}(A)$. Then
$[P:P^{\prime}]=9$ if and only if $(q^{2}-1)_{3}=9$. Further, if
$3\nmid|A/S|$, then $A=S{\bf C}_{A}(P)$.

###### Proof.

Note that in this case $3\nmid|\widetilde{S}/S|$. First suppose that
$3\nmid|A/S|$. Let $q=p^{f}$ with $f=3^{a}\cdot m$ and $3\nmid m$. Then
$A\leq\widetilde{S}\rtimes\langle\tau,F^{\prime}\rangle$ with
$F^{\prime}=F_{p}^{3^{a}}$ and $\tau$ a graph automorphism in the case of
$\operatorname{PSL}_{3}(q)$ and trivial otherwise. Here $P$ is cyclic of order
$(q^{2}-1)_{3}$, and we can identify $P$ with a Sylow $3$-subgroup of a cyclic
Sylow $d$-torus of $G={\mathrm{SL}}_{2}(q)$ resp.
${\mathrm{SL}}_{3}^{\epsilon}(q)$, where $d=d_{3}(q)$, resp $d_{3}(\epsilon
q)=2$. The automorphism $F^{\prime}$ acts on this cyclic torus by $p^{d\cdot
3^{a}}$ powers, so we see $F^{\prime}$ acts trivially on $P$ or by inversion,
since we may write $(q^{2}-1)=\prod_{k\mid m}\Phi_{d}(p^{2\cdot 3^{a}})$ and
since $3\mid(p^{2\cdot 3^{a}}-1)$, we have $3\nmid\Phi_{k}(p^{2\cdot 3^{a}})$
for any $1\neq k\mid m$. In either case, we may find $g\in S$ such that
$F^{\prime}\in g{\bf C}_{A}(P)$. Similarly, $\tau$ acts by inversion on $P$
and $\widetilde{S}/S$ acts trivially on $P$, and we see $A=S{\bf C}_{A}(P)$ as
claimed. Further, in this case $P$ is abelian, so $[P:P^{\prime}]=9$ if and
only if $(q^{2}-1)_{3}=9$.

Now let $|A/S|_{3}=3$. Then $P\cong P_{1}\rtimes C_{3}$, where $C_{3}=\langle
F_{q_{0}}\rangle$ with $q=q_{0}^{3}$ and $P_{1}\in\operatorname{Syl}_{3}(S)$.
Since $P_{1}$ is abelian, note then that
$P/P^{\prime}=P_{1}/[P_{1},F_{q_{0}}]\times\langle F_{q_{0}}\rangle$. Let
$(q^{2}-1)_{3}=3^{b}$. Then $(q^{2}-1)=(q_{0}^{2}-1)(q_{0}^{4}+q_{0}^{2}+1)$
and the second factor is divisible by $3$ exactly once, so
$(q_{0}^{2}-1)_{3}=3^{b-1}$. Then we can see directly that
$P_{1}/[P_{1},F_{q_{0}}]$ has size $3^{b-1}$, so that $[P:P^{\prime}]=9$
exactly when $b=2$. ∎

We next complete the picture for $S=\operatorname{PSL}_{3}^{\epsilon}(q)$.

######  Proposition 5.11.

Let $q$ be a power of a prime and let $\epsilon\in\\{\pm 1\\}$ with
$3\mid(q-\epsilon)$. Let $G={\mathrm{SL}}_{3}^{\epsilon}(q)$,
$S=\operatorname{PSL}_{3}^{\epsilon}(q)$, and
$\widetilde{S}={\mathrm{PGL}}_{3}^{\epsilon}(q)$.

  1. (i)

Let $R\in\operatorname{Syl}_{3}(X)$, where
$X\in\\{S,\widetilde{S},G,G\langle\delta\rangle\\}$, where $\delta$ is a
diagonal automorphism of order $3$. Then $[R:R^{\prime}]=9$.

  2. (ii)

If $S\leq A\leq\operatorname{Aut}(S)$ with $|A/S|_{3}=3$, then
$R\in\operatorname{Syl}_{3}(A)$ satisfies $[R:R^{\prime}]\neq 9$ if and only
if the Sylow $3$-subgroup of $A/S$ induces a field automorphism on $S$.

  3. (iii)

If $S\leq A\leq\operatorname{Aut}(S)$ with $3\nmid|A/S|$ and
$R\in\operatorname{Syl}_{3}(A)$, then $[A:M]\leq 2$, where $M=S{\bf
C}_{A}(R)$.

###### Proof.

Throughout, we let $P$ denote a Sylow $3$-subgroup of $G$ and $\widetilde{P}$
a Sylow $3$-subgroup of $\widetilde{G}:={\mathrm{GL}}_{3}^{\epsilon}(q)$ with
$P=\widetilde{P}\cap G$. Let $Z={\bf Z}(G)$, so $|Z|=3$. Then
$Q:=P/Z=PZ/Z\in\operatorname{Syl}_{3}(S)$ and
$\widetilde{Q}:=\widetilde{P}{\bf Z}(\widetilde{G})/{\bf
Z}(\widetilde{G})\in\operatorname{Syl}_{3}(\widetilde{S})$.

(i) Let $(q-\epsilon)_{3}=3^{b}$. Then $\widetilde{P}\cong C_{3^{b}}\wr C_{3}$
and $\widetilde{P}^{\prime}=\\{(x,y,z)\in C_{3^{b}}^{3}\mid xyz=1\\}$. Then we
see $P=\widetilde{P}^{\prime}\rtimes C_{3}$ and
$P^{\prime}=[\widetilde{P}^{\prime},C_{3}]$. Since $P$ is not cyclic, we see
$[P:P^{\prime}]\neq 3$, and hence $\widetilde{P}^{\prime}\neq P^{\prime}$. We
claim that there is a subgroup $H\leq P^{\prime}$ with $|H|=3^{2b-1}$, which
will force $[P:P^{\prime}]=9$.

Let $a$ be a generator for $C_{3^{b}}$ and $c$ a generator for $C_{3}$, chosen
such that $[x,c]=(x_{1}x_{2}^{-1},x_{2}x_{3}^{-1},x_{3}x_{1}^{-1})$ for
$x=(x_{1},x_{2},x_{3})$ in the base group $C_{3^{b}}^{3}$. The elements
$g_{1}:=(a,a^{-1},1)$ and $g_{2}:=(a,a,a^{-2})$ lie in
$\widetilde{P}^{\prime}$, and $P^{\prime}$ contains
$h_{1}:=[g_{1},c]=(a^{2},a^{-1},a^{-1})$ and
$h_{2}:=[g_{2},c]=(1,a^{3},a^{-3})$. But note that $|h_{1}|=3^{b}$,
$|h_{2}|=3^{b-1}$, and $\langle h_{1}\rangle\cap\langle h_{2}\rangle=1$, so
$H=\langle h_{1},h_{2}\rangle$ is the desired subgroup. Hence we have shown
that $[P:P^{\prime}]=9$.

Now, we have $[Q:Q^{\prime}]=[P:P^{\prime}Z]$. But note that $Z=\\{(x,x,x)\in
C_{3^{b}}^{3}\mid x^{3}=1\\}={\bf Z}(P)$. Then $Z\leq P^{\prime}$ since $Z\cap
P^{\prime}\neq 1$, and we therefore have $[Q:Q^{\prime}]=[P:P^{\prime}]=9$.

Now write $\widetilde{Z}_{3}:={\bf Z}(\widetilde{G})\cap\widetilde{P}={\bf
Z}(\widetilde{P})$ and note
$\widetilde{Q}\cong\widetilde{P}/\widetilde{Z}_{3}$. We have
$[\widetilde{Q}:\widetilde{Q}^{\prime}]=[\widetilde{P}:\widetilde{P}^{\prime}\widetilde{Z}_{3}]=[\widetilde{P}:\widetilde{P}^{\prime}]/[\widetilde{Z}_{3}:\widetilde{Z}_{3}\cap\widetilde{P}^{\prime}]=[\widetilde{P}:\widetilde{P}^{\prime}]/[\widetilde{Z}_{3}:Z]=[\widetilde{P}:\widetilde{P}^{\prime}]/3^{b-1}=3^{b+1}/3^{b-1}=9$.
Now note that
$P\langle\delta\rangle\in\operatorname{Syl}_{3}(G\langle\delta\rangle)$ and
$\widetilde{Q}\cong P\langle\delta\rangle/Z$. Recalling that $Z\leq
P^{\prime}\leq(P\langle\delta\rangle)^{\prime}$, we have from before,
$9=[\widetilde{Q}:\widetilde{Q}^{\prime}]=[P\langle\delta\rangle:(P\langle\delta\rangle)^{\prime}Z]=[P\langle\delta\rangle:(P\langle\delta\rangle)^{\prime}]$
as claimed.

(ii) Now suppose that $|A/S|_{3}=3$. If $A$ contains $\widetilde{S}$, then we
are done by (i). So, we may assume $3\nmid[A\cap\widetilde{S}:S]$. Here, the
statement can be seen using [NS23, Lem. 5.1]. Alternatively, we may argue as
follows. A Sylow $3$-subgroup of $A$ is of the form $\hat{Q}=Q\langle
F_{q_{0}}\rangle=P\langle F_{q_{0}}\rangle/Z$ or $\hat{Q}=Q\langle\delta
F_{q_{0}}\rangle=P\langle\delta F_{q_{0}}\rangle/Z$, where $P$, $Z$, $Q$ and
$\delta$ are as before and $q=q_{0}^{3}$. In the case that
$\hat{Q}=Q\langle\delta F_{q_{0}}\rangle$, we can compute that
$(P\langle\delta F_{q_{0}}\rangle)^{\prime}=\widetilde{P}^{\prime}$, giving
$[\hat{Q}:\hat{Q}^{\prime}]=9$. If instead $\hat{Q}=Q\rtimes\langle
F_{q_{0}}\rangle$, we can compute that $(P\langle
F_{q_{0}}\rangle)^{\prime}=P^{\prime}$, yielding that
$[\hat{Q}:\hat{Q}^{\prime}]=27$.

(iii) Let $q=p^{f}$ with $f={3^{a}\cdot m}$ and $(m,3)=1$ and let
$F_{0}:=F_{p^{3^{a}}}$. In this case, $|\widetilde{S}/S|=3$, so since
$3\nmid|A/S|$, we have $A\leq S\rtimes\langle\tau,F_{0}\rangle$, where $\tau$
denotes a graph automorphism if $\epsilon=1$ and $A\leq S\rtimes\langle
F_{0}\rangle$ if $\epsilon=-1$. First suppose $\epsilon=1$. We choose $\tau$
to be induced by the transpose-inverse automorphism on $G$. Note that
$\langle\tau,F_{0}\rangle=\langle\tau,\tau F_{0}\rangle\cong C_{2}\times
C_{m}$. Let $M=S{\bf C}_{A}(Q)$. We compute that $F_{0}$ centralizes $P$ if
$p\equiv 1\pmod{3}$ and $\tau F_{0}$ centralizes $P$ if $p\equiv 2\pmod{3}$,
yielding the claim in this case. Now, if $\epsilon=-1$, then $\langle
F_{0}\rangle\cong C_{2m}$. Here we calculate that $F_{0}$ centralizes $P$ if
$p\equiv 2\pmod{3}$ but acts by inversion on $\widetilde{P}^{\prime}$ if
$p\equiv 1\pmod{3}$. But in any case, $F_{0}^{2}$ centralizes $P$, completing
the proof. ∎

####  5.4.2. Other Linear and Unitary Groups

Throughout this subsection, we let $G={\mathrm{SL}}_{n}^{\epsilon}(q)$,
$\widetilde{G}:={\mathrm{GL}}_{n}^{\epsilon}(q)$,
$S=\operatorname{PSL}_{n}^{\epsilon}(q)$, and
$\widetilde{S}:={\mathrm{PGL}}_{n}^{\epsilon}(q)$. We will write
$\widetilde{P}\in\operatorname{Syl}_{3}(\widetilde{G})$,
$P\in\operatorname{Syl}_{3}(G)$ with $P=\widetilde{P}\cap G$,
$\widetilde{Q}\in\operatorname{Syl}_{3}(\widetilde{S})$ with
$\widetilde{Q}=\widetilde{P}{\bf Z}(\widetilde{G})/{\bf Z}(\widetilde{G})$,
and $Q\in\operatorname{Syl}_{3}(S)$ with $Q=\widetilde{Q}\cap S\cong P{\bf
Z}(G)/{\bf Z}(G)$.

The situation for ${\mathrm{GL}}_{n}^{\epsilon}(q)$ will be useful for the
remaining classical groups, so we next discuss this case.

######  Proposition 5.12.

Let $n\geq 1$ and let $\widetilde{P}$ be a Sylow $3$-subgroup of
${\mathrm{GL}}_{n}^{\epsilon}(q)$.

If $3\mid(q-\epsilon)$, then $[\widetilde{P}:\widetilde{P}^{\prime}]=9$ if and
only if one of the following holds:

  * •

$n=1$ and $(q-\epsilon)_{3}=9$; or

  * •

$n\in\\{2,3\\}$ and $(q-\epsilon)_{3}=3$.

If $3\mid(q+\epsilon)$, then $[\widetilde{P}:\widetilde{P}^{\prime}]=9$ if and
only if one of the following holds:

  * •

$n\in\\{2,3\\}$ and $(q+\epsilon)_{3}=9$; or

  * •

$4\leq n\leq 7$ and $(q+\epsilon)_{3}=3$.

###### Proof.

First, suppose $3\mid(q-\epsilon)$. Let
$n=a_{0}+3a_{1}+3^{2}a_{2}+\cdots+3^{t}a_{t}$ be the $3$-adic decomposition of
$n$. Then by [We55], we have
$\widetilde{P}\cong\prod_{i=0}^{t}\widetilde{P}_{i}^{a_{i}}$, where
$\widetilde{P}_{i}\in\operatorname{Syl}_{3}({\mathrm{GL}}_{3^{i}}^{\epsilon}(q))$.
From this we already see that $P$ is at least $3$-generated if
$\sum_{i=0}^{t}a_{i}\geq 3$, so we assume $\sum_{i=0}^{t}a_{i}\leq 2$.

Now, we have $\widetilde{P}_{0}\in\operatorname{Syl}_{3}(C_{q-\epsilon})$ is
isomorphic to $C_{3^{b}}$, where $3^{b}=(q-\epsilon)_{3}$. So, we see if
$n=1$, we have $[\widetilde{P}:\widetilde{P}^{\prime}]=|\widetilde{P}|=9$ if
and only if $(q-\epsilon)_{3}=9$, giving the first item.

Again by [We55], for $i\geq 1$, we have
$\widetilde{P}_{i}=\widetilde{P}_{i-1}\wr C_{3}=C_{3^{b}}\wr C_{3}\wr
C_{3}\wr\cdots\wr C_{3}$, with $i$ copies of $C_{3}$ in this iterated wreath
product. If $\sum_{i=0}^{t}a_{i}=2$, then the condition that $\widetilde{P}$
is generated by at most two elements forces that each $\widetilde{P}_{i}$ with
$a_{i}\neq 0$ is cyclic. This only happens for $i=0$, and hence this is the
case $n=2$. Since the Sylow $3$-subgroup of ${\mathrm{GL}}_{2}^{\epsilon}(q)$
is of the form $C_{3^{b}}\times C_{3^{b}}$, we have
$[\widetilde{P}:\widetilde{P}^{\prime}]=9$ if and only if $b=1$ in this
situation.

Hence we may assume $n=3^{i}$ for some $i\geq 1$, and
$\widetilde{P}=\widetilde{P}_{i}$. Then by the comment about wreath products
at the beginning of the section, we have
$\widetilde{P}_{i}/\widetilde{P}_{i}^{\prime}\cong\widetilde{P}_{i-1}/\widetilde{P}_{i-1}^{\prime}\times
C_{3}\cong\widetilde{P}_{0}/\widetilde{P}_{0}^{\prime}\times C_{3}^{i}\cong
C_{3^{b}}\times C_{3}^{i}$, and the statement follows.

Now suppose $3\mid(q+\epsilon)$. Then letting $w=\lfloor\frac{n}{2}\rfloor$,
we can instead write $w=a_{0}+3a_{1}+\cdots+3^{t}a_{t}$. The work of [We55]
again shows $P$ is isomorphic to $\prod_{i=0}^{t}\widetilde{P}_{i}^{a_{i}}$,
where now $\widetilde{P}_{i}\in\operatorname{Syl}_{3}({\mathrm{GL}}_{2\cdot
3^{i}}^{\epsilon}(q))$. In this case, each $\widetilde{P}_{i}$ is of the same
form as before, where now $3^{b}=(q+\epsilon)_{3}$, and the result follows. ∎

######  Proposition 5.13.

Let $n\geq 3$ and let $P$ be a Sylow $3$-subgroup of
$G={\mathrm{SL}}_{n}^{\epsilon}(q)$ with $3\mid(q-\epsilon)$. Then
$[P:P^{\prime}]=9$ if and only if $n=3$ or if $n=4$ and $(q-\epsilon)_{3}=3$.

###### Proof.

From the discussion in the proof of Proposition 5.12, we have
$\widetilde{P}/\widetilde{P}^{\prime}\cong\prod_{i=0}^{t}(C_{3^{b}}\times
C_{3}^{i})^{a_{i}}$. So $[\widetilde{P}:\widetilde{P}^{\prime}]=3^{b\sum
a_{i}+\sum ia_{i}}$.

Further, we have $[\widetilde{P}:P]=3^{b}$,
$P^{\prime}\leq\widetilde{P}^{\prime}\leq P\leq\widetilde{P}$, and
$[P:P^{\prime}]=3^{-b}[\widetilde{P}:\widetilde{P}^{\prime}][\widetilde{P}^{\prime}:P^{\prime}]$.
Then $[P:P^{\prime}]$ is divisible by $3^{m}$ with $m={b(\sum a_{i}-1)+\sum
ia_{i}}$. Note that $m>2$ unless we are in one of the following situations:
$n=2$; $n=3$; $n=4$ and $b=1$; or $n=9$.

In the latter case, we can see explicitly that $P=P_{T}\rtimes P_{W}$ with
$P_{T}\cong C_{3^{b}}^{8}$ and $P_{W}\cong C_{3}\wr C_{3}$ has a minimal
generating set of three elements, and hence $[P:P^{\prime}]\neq 9$. The cases
that $n\in\\{2,3\\}$ are covered in Propositions 5.10 and 5.11. Finally, if
$n=4$ with $b=1$, then $P$ is isomorphic to the Sylow $3$-subgroup for
${\mathrm{SL}}_{4}(4)$, and we see explicitly that $[P:P^{\prime}]=9$ in this
case, for example in GAP. ∎

######  Lemma 5.14.

Let $n\geq 3$ and let $P$ be a Sylow $3$-subgroup of
$G={\mathrm{SL}}_{n}^{\epsilon}(q)$ with $3\mid(q-\epsilon)$. Let
$Z\in\operatorname{Syl}_{3}({\bf Z}(G))$. Then $Z\leq\widetilde{P}^{\prime}$.

###### Proof.

First, assume $n=3^{i}\geq 3$. For $i=1$, we see explicitly that $Z={\bf
Z}(P)$, so that $Z\cap P^{\prime}\neq 1$, and hence $Z\leq
P^{\prime}\leq\widetilde{P}^{\prime}$ since it has size $3$. We now induct on
$i$. Write $P_{i}=\widetilde{P}_{i}\cap G$, where
$\widetilde{P}_{i}\in\operatorname{Syl}_{3}({\mathrm{GL}}_{3^{i}}^{\epsilon}(q))$.
We have
$\widetilde{P}_{i}^{\prime}=\\{(x_{1},x_{2},x_{3})\in\widetilde{P}_{i-1}^{3}\mid
x_{1}x_{2}x_{3}\in\widetilde{P}_{i-1}^{\prime}\\}$ by [Ols76, Lem. (1.4)].
Further, we can write $Z=\langle z\rangle$ where $z=\zeta\cdot
I_{3^{i}}=(x,x,x)$ with $x=\zeta\cdot I_{3^{i-1}}$ and $\zeta\in
C_{q-\epsilon}$ such that $(\zeta^{3})^{3^{i-1}}=\zeta^{3^{i}}=1$. Note that
we have $x\in\widetilde{P}_{i-1}$ with $x^{3}\in{\bf
Z}({\mathrm{SL}}_{3^{i-1}}^{\epsilon}(q))$. So
$x^{3}\in\widetilde{P}_{i-1}^{\prime}$ by induction and we conclude
$z\in\widetilde{P}_{i}^{\prime}$.

Now, suppose that $n=a_{0}+a_{1}3+a_{2}3^{2}+\cdots+a_{t}3^{t}=\sum
a_{i}3^{i}$ is the $3$-adic expansion of $n$, and let $j$ be the smallest such
that $a_{j}\neq 0$. Then $Z$ is generated by some $\zeta I_{n}$ with
$|\zeta|=\gcd(3^{j},3^{b})=\gcd(n,q-\epsilon)$, and we see $\zeta^{3^{i}}=1$
for each $i$ such that $a_{i}\neq 0$. That is, $\zeta
I_{n}\leq\prod_{i=0}^{t}Z_{3^{i}}^{a_{i}}$, where $Z_{3^{i}}$ is a Sylow
$3$-subgroup of ${\bf Z}({\mathrm{SL}}_{3^{i}}^{\epsilon}(q))$. By above, we
have $Z_{3^{i}}\leq\widetilde{P}_{i}^{\prime}$, so we have
$Z\leq\prod(\widetilde{P}_{i}^{\prime})^{a_{i}}=\widetilde{P}^{\prime}$ as
desired. ∎

######  Corollary 5.15.

Let $n\geq 3$ and let $Q$ be a Sylow $3$-subgroup of
$S=\operatorname{PSL}_{n}^{\epsilon}(q)$ with $3\mid(q-\epsilon)$. Then
$[Q:Q^{\prime}]=9$ if and only if $n=3$ or if $n=4$ and $(q-\epsilon)_{3}=3$.

###### Proof.

Let $G={\mathrm{SL}}_{n}^{\epsilon}(q)$ and write $Z:={\bf Z}(G)$. Let
$Z_{3}\in\operatorname{Syl}_{3}(Z)$. Then $Q=PZ/Z\cong P/Z_{3}$ with
$P\in\operatorname{Syl}_{3}(G)$. We have
$[Q:Q^{\prime}]=[P/Z_{3}:P^{\prime}Z_{3}/Z_{3}]=[P:P^{\prime}Z_{3}]$. If
$Z_{3}\leq P^{\prime}$, we are done by Proposition 5.13. Write $n=\sum
a_{i}3^{i}$ and $(q-\epsilon)_{3}=3^{b}$ as before. If $n$ is not a power of
$3$, then $\sum a_{i}\geq 2$ and we have $[Q:Q^{\prime}]\geq 3^{m-b}\geq
3^{\sum ia_{i}}$, where $m$ is as in the proof of Proposition 5.13. For $n>9$,
we have $\sum ia_{i}\geq 3$. Hence we are left to consider the cases $n=6$ and
$n=3^{i}$. If $n=3$, we saw in the proof of Lemma 5.14 that $Z_{3}\leq
P^{\prime}$. If $n=6$, then we again have $Z_{3}\leq P^{\prime}$, since
$P\cong P_{3}\times P_{3}$, and we have $Z_{3}\leq P_{3}^{\prime}\times
P_{3}^{\prime}\leq P^{\prime}$.

So let $n=3^{i}\geq 9$. By Lemma 5.14 we have
$P^{\prime}Z_{3}\leq\widetilde{P}^{\prime}$. Then exactly as in the proof of
Proposition 5.13, we see $[P:P^{\prime}Z_{3}]$ is divisible by $3^{m}$, and we
reduce to the same cases: $n=3$; $n=4$ and $b=1$; or $n=9$. Since $Z_{3}\leq
P^{\prime}$ in the first two cases, these follow from Proposition 5.13. Then
we are left with the case $n=9$, in which case we see again that $P/Z_{3}$ has
a minimal generating set of size three, so $[Q:Q^{\prime}]\neq 9$. ∎

######  Corollary 5.16.

Let $n\geq 3$ and let $S=\operatorname{PSL}_{n}^{\epsilon}(q)$ with
$3\mid(n,q-\epsilon)$. Let $S\leq A\leq\operatorname{Aut}(S)$ such that
$|A/S|_{3}=|(A\cap\widetilde{S})/S|_{3}=3$. Let
$R\in\operatorname{Syl}_{3}(A)$ with $[R:R^{\prime}]=9$. Then $n=3$.

###### Proof.

We may write $R=(P\langle\delta\rangle)/Z\leq\widetilde{P}/Z$, where $P$ and
$\widetilde{P}$ are Sylow $3$-subgroups of $G$ and $\widetilde{G}$ as before,
$\delta$ is a diagonal automorphism of order $3$, and
$Z\in\operatorname{Syl}_{3}({\bf Z}(G))$. Recall from Lemma 5.14 that
$Z\leq\widetilde{P}^{\prime}$. Further, notice that
$(P\langle\delta\rangle)^{\prime}\leq(P\langle\delta\rangle)^{\prime}Z\leq\widetilde{P}^{\prime}\leq
P\leq P\langle\delta\rangle\leq\widetilde{P}$, so that
$[R:R^{\prime}]=[P\langle\delta\rangle:(P\langle\delta\rangle)^{\prime}Z]\geq[P\langle\delta\rangle:\widetilde{P}^{\prime}]=3^{1-b}[\widetilde{P}:\widetilde{P}^{\prime}]\geq
3\prod_{i=0}^{t}3^{ia_{i}}$, where the last equality is from the calculations
in the proof of Proposition 5.12. Then if $[R:R^{\prime}]=9$, we must have
$\sum_{i}ia_{i}=1$, and hence $n\in\\{3,4,5\\}$. Then since $3\mid n$, we must
have $n=3$ as claimed. ∎

####  5.4.3. The Remaining Cases

We now summarize (when combined with Propositions 5.10 and 5.11) when
$[P:P^{\prime}]=9$ holds for simple groups of Lie type in non-defining
characterstic.

######  Proposition 5.17.

Let $S$ be a simple group of Lie type defined in characteristic $p\neq 3$ and
let $P\in\operatorname{Syl}_{3}(S)$. Assume that
$S\not\in\\{\operatorname{PSL}_{2}(q),\operatorname{PSL}_{3}^{\epsilon}(q)\\}$.
Then $[P:P^{\prime}]=9$ if and only if $S$ is one of:

  1. (i)

$\operatorname{PSL}_{4}^{\epsilon}(q)$ when $3\mid\mid(q-\epsilon)$;

  2. (ii)

$\operatorname{PSL}_{n}^{\epsilon}(q)$ with $4\leq n\leq 7$ and
$3\mid\mid(q+\epsilon)$;

  3. (iii)

${\mathrm{PSp}}_{2n}(q)$ or ${\mathrm{P\Omega}}_{2n+1}(q)$ with
$n\in\\{2,3\\}$ and $3\mid\mid(q^{2}-1)$;

  4. (iv)

${\mathrm{P\Omega}}_{8}^{-}(q)$ when $3\mid\mid(q^{2}-1)$;

  5. (v)

$\operatorname{G}_{2}(q)$, ${}^{3}\\!\operatorname{D}_{4}(q)$, or
${}^{2}\\!\operatorname{F}_{4}(q^{2})$.

In (ii)-(v), we further have $\widetilde{S}A=\widetilde{S}{\bf C}_{A}(P)$ if
$S\leq A\leq\operatorname{Aut}(S)$ with $3\nmid|A/S|$.

###### Proof.

First let $S=\operatorname{PSL}_{n}^{\epsilon}(q)$. If $3\mid(q-\epsilon)$, we
are done by Proposition 5.15. So assume $3\mid(q+\epsilon)$. Then $P$ can be
identified with a Sylow $3$-subgroup of
$\widetilde{G}={\mathrm{GL}}_{n}^{\epsilon}(q)$ and the statement about
$[P:P^{\prime}]$ is Proposition 5.12.

Now suppose that $S$ is of one of the remaining classical types, so that
$S={\mathrm{PSp}}_{2n}(q)$ with $n\geq 2$, ${\mathrm{P\Omega}}_{2n+1}(q)$ with
$n\geq 3$, or ${\mathrm{P\Omega}}_{2n}^{\pm}(q)$ with $n\geq 4$. In these
cases, $P$ can be identified with a Sylow $3$-subgroup of the corresponding
symplectic or orthogonal group ${\mathrm{Sp}}_{2n}(q)$,
${\mathrm{SO}}_{2n+1}(q)$, or ${\mathrm{SO}}_{2n}^{\pm}(q)$. Weir [We55] has
again described the Sylow subgroups in these cases. In the first two cases,
$P$ can be identified with a Sylow $3$-subgroup of ${\mathrm{GL}}_{n}(q)$ if
$3\mid(q-1)$ and of ${\mathrm{GL}}_{2n}(q)$ if $3\mid(q+1)$. The Sylow
$3$-subgroups of ${\mathrm{SO}}_{2n}^{\pm}(q)$ can be identified with those of
either ${\mathrm{SO}}_{2n+1}(q)$ or ${\mathrm{SO}}_{2n-1}(q)$. Note that in
the latter situation when $n=4$, the order polynomials show that $P$ is in
${\mathrm{SO}}_{7}(q)$ in the case $-$ and ${\mathrm{SO}}_{9}(q)$ in the case
$+$. Together with Proposition 5.12, this now gives the first statement for
the classical groups. Arguments just like in Proposition 5.10 and using the
explicit structure of $P$ yield that $\widetilde{S}A=\widetilde{S}{\bf
C}_{A}(P)$ in cases (ii)-(iv).

Now let $S$ be an exceptional group of Lie type, including
${}^{3}\\!\operatorname{D}_{4}(q)$ and the Ree groups
${}^{2}\\!\operatorname{F}_{4}(q^{2})$. (Recall that the sizes of the Suzuki
groups ${}^{2}\\!\operatorname{B}_{2}(q^{2})$ are not divisible by $3$ and
that the Ree groups ${}^{2}\\!\operatorname{G}_{2}(q^{2})$ are defined in
characteristic $3$.) If $S=\operatorname{G}_{2}(q)$ and $\epsilon\in\\{\pm
1\\}$ such that $3\mid(q-\epsilon)$, then we have
$P\in\operatorname{Syl}_{3}(S)$ is isomorphic to a Sylow $3$-subgroup of
${\mathrm{SL}}_{3}^{\epsilon}(q).2$, where the extension is by the graph
automorphism $\tau$ if $\epsilon=1$ and the graph-field $F_{0}^{m}$ if
$\epsilon=-1$ (in the notation of the proof of Proposition 5.11). Similarly,
when $S={}^{3}\\!\operatorname{D}_{4}(q)$, we see from [DZ87, Prop. 2.2] that
$P$ is an extension of a Sylow $3$-subgroup of
${\mathrm{SL}}_{3}^{\epsilon}(q).2$ by a diagonal automorphism of order three,
where $3\mid(q-\epsilon)$. Next, let $S={}^{2}\\!\operatorname{F}_{4}(q^{2})$,
with $q^{2}=2^{2a+1}$. Note that $3\mid(q^{2}+1)$. Here by [Mal91, Main Thm.],
$S$ has a maximal subgroup ${\mathrm{SU}}_{3}(q^{2}):2$ and $P$ is isomorphic
to a Sylow $3$-subgroup of ${\mathrm{SU}}_{3}(q^{2})$. Then in all of these
cases, $[P:P^{\prime}]=9$ by Proposition 5.11. Now, note that in these cases
$A/S$ is generated by field automorphisms. Further, recalling from the proof
of Proposition 5.11 that either $F_{0}$ or $\tau F_{0}$ centralizes $P$, we
obtain that $A=S{\bf C}_{A}(P)$ in these cases.

Now consider the remaining cases $S=\operatorname{F}_{4}(q)$,
$\operatorname{E}_{6}^{\epsilon}(q)$, $\operatorname{E}_{7}(q)$, and
$\operatorname{E}_{8}(q)$ and let $G$ be as in our setting 5.1. If
$S\neq\operatorname{E}^{\epsilon}_{6}(q)$ with $3\mid(q-\epsilon)$ we may
identify $P$ with a Sylow $3$-subgroup $\hat{P}$ of $G$. If
$S=\operatorname{E}^{\epsilon}_{6}(q)$ with $3\mid(q-\epsilon)$, we have
$P=\hat{P}/Z$, where $\hat{P}\in\operatorname{Syl}_{3}(G)$ and $Z={\bf Z}(G)$
has size 3. By [GLS98, Thm. 4.10.2], we may write $\hat{P}=P_{T}\rtimes
P_{W}$, where $P_{T}$ is homocyclic of exponent $3^{b}$ with
$3^{b}=(q^{2}-1)_{3}$ and $P_{W}\in\operatorname{Syl}_{3}(W_{d})$, where
$W_{d}={\bf N}_{G}(T)/T$ is the relative Weyl group for a
Sylow-$d_{3}(q)$-torus $T$ of $({\mathbf{G}},F)$. Further, note that $Z\leq
P_{T}$ in the case of $\operatorname{E}_{6}^{\epsilon}(q)$. In these cases,
the relative Weyl groups are listed in [GM20, Table 3.2]. In each case, we see
that $[P_{W}:P_{W}^{\prime}]\geq 9$ and $P_{W}$ is generated by at least two
elements. From this structure, we see $P$ cannot be $2$-generated and
$[P:P^{\prime}]>9$. ∎

Finally, we complete the picture when $|A/S|_{3}=3$.

######  Proposition 5.18.

Let $S$ be a simple group of Lie type defined in characteristic distinct from
$3$ such that $S$ is not $\operatorname{PSL}_{3}^{\epsilon}(q)$ nor
$\operatorname{PSL}_{2}(q)$. Let $A$ be an almost simple group $S\leq
A\leq\operatorname{Aut}(S)$ such that $|A/S|_{3}=3$. Let
$P\in\operatorname{Syl}_{3}(A)$. Then $[P:P^{\prime}]>9$.

###### Proof.

Let $S=G/{\bf Z}(G)$ where $G={\mathbf{G}}^{F}$ with ${\mathbf{G}}$ simple of
simply connected type, and let $d:=d_{3}(q)\in\\{1,2\\}$ if $F$ is a Frobenius
map. If $S={}^{2}\\!\operatorname{F}_{4}(q^{2})$, we make the slight
alteration on $d$ according to [GM20, 3.5.3]. Let
$P_{1}\in\operatorname{Syl}_{3}(S)$ and
$\hat{P}_{1}\in\operatorname{Syl}_{3}(G)$ such that $P_{1}=\hat{P}_{1}/Z$ with
$Z\in\operatorname{Syl}_{3}({\bf Z}(G))$. Then as before, $\hat{P}_{1}$ is of
the form $\hat{P}_{1}=P_{T}\rtimes P_{W}$ by [GLS98, Thm. 4.10.2], where
$P_{T}\in\operatorname{Syl}_{3}(T)$ for a Sylow $d$-torus $T$ and
$P_{W}\in\operatorname{Syl}_{3}(N_{G}(T)/T)$. Then $P_{1}=(P_{T}/Z)\rtimes
P_{W}$.

First, assume that $P=P_{1}\rtimes\langle\varphi\rangle$, where
$\varphi=F_{q_{0}}$ is a field automorphism of order $3$ (in this case
$q=q_{0}^{3}$ for some prime power $q_{0}$) or $S=\operatorname{D}_{4}(q)$ and
$\varphi\in\\{F_{q_{0}},\tau,\tau F_{q_{0}}\\}$ where $\tau$ is a triality
graph automorphism. Write $\hat{P}=\hat{P}_{1}\rtimes\langle\varphi\rangle$.
We can choose the decomposition so that $\varphi$ centralizes $P_{W}$ and
normalizes $P_{T}$. Then we calculate that $\hat{P}^{\prime}\leq
P_{T}P_{W}^{\prime}$. In particular, we have $[P:P^{\prime}]\geq
3[P_{W}:P_{W}^{\prime}]$, and we may assume that $P_{W}$ is cyclic of size
$3$. However, since $P_{1}/Z$ is not cyclic, we still see from the structure
above that $P$ cannot be $2$-generated, completing the proof in this case.

Then we are left with the case that $S=\operatorname{PSL}_{n}^{\epsilon}(q)$
with $3\mid(n,q-\epsilon)$ or $\operatorname{E}_{6}^{\epsilon}(q)$ with
$3\mid(q-\epsilon)$. First, suppose $S=\operatorname{PSL}_{n}^{\epsilon}(q)$.
If $3\mid|\widetilde{S}/S|$ and $q=q_{0}^{3}$ for some prime power $q_{0}$,
then [NS23, Thm. 5.1] yields that if $[P:P^{\prime}]=9$, then $n=3$. If
$3\mid|A\cap\widetilde{S}/S|$, then Corollary 5.16 also tells us that $n=3$.
Hence in this case, we may assume that $P=P_{1}\rtimes\langle
F_{q_{0}}\rangle$, and we are done by the first part of the proof.

Finally, suppose that $S=\operatorname{E}_{6}^{\epsilon}(q)$ with
$3\mid(q-\epsilon)$. If $P$ is not of the form $P=P_{1}\rtimes\langle
F_{q_{0}}\rangle$ as above, then $P$ is either a Sylow $3$-subgroup
$\widetilde{P}$ of the adjoint version
$\widetilde{S}=\operatorname{E}_{6}^{\epsilon}(q)_{\mathrm{ad}}$, or
$P=P_{1}\langle\delta F_{q_{0}}\rangle$, where $\delta$ is a diagonal
automorphism of order $3$ induced by $\widetilde{S}/S$. In the latter case, we
have $P$ is an index-3 subgroup of $\widetilde{P}\rtimes\langle
F_{q_{0}}\rangle$. Now, by [GLS98, Thm. 4.10.2], we also have
$\widetilde{P}=\widetilde{P}_{T}\rtimes P_{W}$, and we note that by the same
reasoning as Proposition 5.17, $|\widetilde{P}/\widetilde{P}^{\prime}|>9$.
(See also the proof of [NS23, 5.2].) Now let $P=P_{1}\langle\delta
F_{q_{0}}\rangle\leq\widetilde{P}\langle F_{q_{0}}\rangle$. Note that $P_{1}$
has index 9 in $\widetilde{P}\langle F_{q_{0}}\rangle$, and therefore $P_{1}$
contains $(\widetilde{P}\langle
F_{q_{0}}\rangle)^{\prime}=\widetilde{P}^{\prime}[\widetilde{P},\langle
F_{q_{0}}\rangle]$, which contains $P^{\prime}$. But here, we see similar to
before that $P_{1}/(\widetilde{P}\langle F_{q_{0}}\rangle)^{\prime}$ has a
quotient $P_{W}/P_{W}^{\prime}$, which has size 9. But then
$[P:P^{\prime}]=3[P_{1}:P^{\prime}]\geq 27[(\widetilde{P}\langle
F_{q_{0}}\rangle)^{\prime}:P^{\prime}]$, completing the proof. ∎

###  5.5. Theorem 3.3 in Non-Defining Characteristic

In this section, we let $p\neq 3$ and let $q$ be a power of $p$. With the
classification of when $[P:P^{\prime}]=9$ now complete, here we turn our
attention to the “character side” of proving Theorem 3.3. We begin by
completing the cases of $\operatorname{PSL}_{2}(q)$ and
$\operatorname{PSL}_{3}^{\epsilon}(q)$.

######  Proposition 5.19.

Let $S=\operatorname{PSL}_{2}(q)$ or $\operatorname{PSL}_{3}^{\epsilon}(q)$
with $3\nmid q$. Then Theorem 3.3 holds for $S$.

###### Proof.

Let $S\leq A\leq\operatorname{Aut}(S)$ such that $|A/S|_{3}\leq 3$. Let
$G:={\mathrm{SL}}_{2}(q)$, resp. ${\mathrm{SL}}_{3}^{\epsilon}(q)$.

First, suppose that $S=\operatorname{PSL}_{2}(q)$ or that $3\mid(q+\epsilon)$
and $S=\operatorname{PSL}^{\epsilon}_{3}(q)$. Recall that the “Sylow side” has
been discussed in Proposition 5.10, and we keep the notation from that proof.
Since $\operatorname{Aut}(S)=\widetilde{S}\rtimes\langle\tau,F_{p}\rangle$ and
the only outer automorphisms of order $3$ are field automorphisms, we can
write $A$ as $M\rtimes\langle F_{q_{0}}^{i}\rangle$ where $q=q_{0}^{3}$,
$i\in\\{0,1\\}$, and $3\nmid|M/S|$. Let $P=P_{1}\rtimes\langle
F_{q_{0}}^{i}\rangle$ with $P_{1}\in\operatorname{Syl}_{3}(S)$. By Proposition
5.10, we have $M=S{\bf C}_{M}(P_{1})$ and $\widetilde{S}=S{\bf
C}_{\widetilde{S}}(P_{1})$ since $3\nmid|\widetilde{S}/S|$, so that
$k_{0}(B_{0}(M))=k_{0}(B_{0}(S))=k_{0}(B_{0}(G))$ by the Alperin–Dade
correspondence Theorem 2.2 and the fact that $3\nmid|{\bf Z}(G)|$. We see
using [Mal20, Thm. 5.12] and [MO83, Prop. 2.13] that $k_{0}(B_{0}(S))\leq 9$
exactly when $|P_{1}|\leq 9$ in this case.

Then suppose that $|A/S|_{3}=3$. The set ${\rm Irr}_{3^{\prime}}(B_{0}(S))$
consists of the unipotent characters $1_{S}$ and $\mathrm{St}_{S}$, together
with the deflation of the semisimple characters $\chi_{s}$ where $s\in
G^{\ast}={\mathrm{PGL}}_{n}^{\epsilon}(q)$ (for the appropriate $n,\epsilon$)
lifts to an element of $\widetilde{G}^{\ast}={\mathrm{GL}}_{n}^{\epsilon}(q)$
with nontrivial eigenvalues $\\{\zeta,\zeta^{-1}\\}$ with
$|\zeta|\mid(q^{2}-1)_{3}:=3^{b}$. Here we see the characters
$1_{S},\mathrm{St}_{S}$, and the deflation of $\chi_{s}$ with $|\zeta|$
dividing $(q_{0}^{2}-1)_{3}=3^{b-1}$ are invariant under $F_{q_{0}}$ and are
further the only characters in ${\rm Irr}_{3^{\prime}}(B_{0}(S))$ invariant
under $F_{q_{0}}$. If $b=2$, this yields only $1_{S},\mathrm{St}_{S}$, and
$\chi_{s}$ with $|\zeta|=3$, so there are exactly 9 characters in ${\rm
Irr}_{3^{\prime}}(B_{0}(A))$, lying above the three corresponding characters
in ${\rm Irr}_{3^{\prime}}(B_{0}(M))$. On the other hand, if $b\geq 3$, we
obtain at least a fourth not $\operatorname{Aut}(S)$-conjugate character in
$B_{0}(S)$ using $|\zeta|=9$, giving more than $9$ characters in ${\rm
Irr}_{3^{\prime}}(B_{0}(A))$ using Theorem 2.7. When combined with Proposition
5.10, this completes the proof in this case.

Now let $S=\operatorname{PSL}_{3}^{\epsilon}(q)$ with $3\mid(q-\epsilon)$. We
keep the notation and setting of the proof of Proposition 5.11. The set ${\rm
Irr}_{3^{\prime}}(S)$ and action of $\operatorname{Aut}(S)$ on ${\rm
Irr}_{3^{\prime}}(S)$ is described in [Mal08, Secs. 3.1, 3.2].

We have $|{\rm Irr}_{3^{\prime}}(S)|=6$, with three of these characters coming
from the deflation of restrictions of unipotent characters of $\widetilde{G}$,
and the remaining three coming from the deflation of the three constituents of
${\rm Irr}(G\mid\widetilde{\chi}_{s})$, where $\widetilde{\chi}_{s}$ is a
semisimple character of $\widetilde{G}$ corresponding to a semisimple element
$s$ in $\widetilde{G}^{\ast}$ of order $3$. From [FS82], we see that
$\widetilde{\chi}_{s}$ and the unipotent characters all lie in
$B_{0}(\widetilde{G})$, so these six characters all lie in $B_{0}(S)$ and
$k(B_{0}(S))=6$.

Let $P\in\operatorname{Syl}_{3}(S)$ and suppose we have $3\nmid|A/S|$. Write
$M:=S{\bf C}_{A}(P)$ as before. From the considerations in [Mal08, Secs. 3.1,
3.2] and discussion in the proof of Proposition 5.11, we see that in the cases
that $[A:M]=2$, the generator of $A/M$ interchanges two of the characters in
${\rm Irr}(G\mid\widetilde{\chi}_{s})$ and fixes the remaining four characters
in ${\rm Irr}_{3^{\prime}}(S)$. Using Theorem 2.2, this yields that
$k_{0}(B_{0}(A))=6$ if $A=M$ and $k_{0}(B_{0}(A))=9$ if $[A:M]=2$.

Now, let $|A/S|_{3}=3$. Let $Y$ be a Sylow $3$-subgroup of $A/S$. If $Y$
induces a field automorphism of order $3$ on $S$, then again, we have the
three unipotent characters are $Y$-invariant. We further see by adapting the
descriptions of the characters in [Mal08, Sec. 3.1] that $F_{q_{0}}$ will
stabilize the characters in ${\rm Irr}(G\mid\widetilde{\chi}_{s})$, so they
are $Y$-invariant. Since these four characters are not
$\operatorname{Aut}(S)$-conjugate, we have $k_{0}(B_{0}(A))\geq 12$ by Theorem
2.7.

Hence, we may assume that $Y$ induces $\delta$ or $\delta F_{q_{0}}$ on $S$.
Then the three characters in ${\rm Irr}(G\mid\widetilde{\chi}_{s})$ fuse in
$A$, and the only characters in ${\rm Irr}_{3^{\prime}}(B_{0}(A))$ are those
lying above the three unipotent characters, which extend to $A$ by [Mal08,
Thms. 2.4, 2.5]. Let $\bar{S}:=SY$. We have ${\rm
Irr}_{3^{\prime}}(B_{0}(\bar{S}))$ contains the 9 extensions of the three
unipotent characters of $S$. Let $\beta\in{\rm Irr}(\bar{S}/S)$ be of order
$3$. Note that the graph automorphism $\tau$ commutes with $F_{q_{0}}$ and
inverts $\delta$, and therefore inverts $\beta$. Now, let $M:=\bar{S}{\bf
C}_{A}(P)$ with $P\in\operatorname{Syl}_{3}(A)$, so $k_{0}(B_{0}(M))=9$ by
Theorem 2.2. We have $[A:M]\leq 2$ by the same proof as in Proposition
5.11(iii). Then when $|A/M|=2$, we have $A$ fuses two of the characters of $M$
above each unipotent character, corresponding to the fusion of $\beta$ with
$\beta^{-1}$. This yields that $k_{0}(B_{0}(A))=9$, as desired. ∎

The following follows from Proposition 5.8 above.

######  Corollary 5.20.

Let $S$ be a simple group of Lie type defined in characteristic distinct from
$3$ such that $S$ is not $\operatorname{PSL}_{3}^{\epsilon}(q)$ nor
$\operatorname{PSL}_{2}(q)$. Let $A$ be an almost simple group $S\leq
A\leq\operatorname{Aut}(S)$ such that $|A/S|$ is divisible by 3. Then
$k_{0}(B_{0}(A))>9$.

###### Proof.

If $S$ is not in the list of exceptions in Proposition 5.8, then we have at
least four unipotent characters in $\operatorname{Irr}_{3^{\prime}}(B_{0}(S))$
that extend to $\operatorname{Aut}(S)$. Then by Theorem 2.7, we have at least
12 characters in $\operatorname{Irr}_{3^{\prime}}(B_{0}(A))$ lying above these
four characters of $S$.

Next, suppose that $S=\operatorname{PSL}_{4}^{\epsilon}(q)$ with
$3\mid(q-\epsilon)$ or ${\mathrm{PSp}}_{4}(2^{a})$. Then Proposition 5.8 and
Theorem 2.7 yield at least 9 characters in
$\operatorname{Irr}_{3^{\prime}}(B_{0}(A))$ lying above unipotent characters
in $S$. It suffices to show that there is at least one additional
$3^{\prime}$-degree character. Here the class of $s$ discussed in the proof of
Corollary 5.9 can be chosen to be invariant under field automorphisms of
$3$-power order. Using Lemma 2.3, this completes the proof since the only
outer automorphisms of $3$-power order in this case are field automorphisms. ∎

######  Proposition 5.21.

Let $q$ be a power of a prime distinct from $3$ and let $S$ be a simple group
of Lie type $\operatorname{F}_{4}(q)$, $\operatorname{E}_{6}^{\epsilon}(q)$,
$\operatorname{E}_{7}(q)$, or $\operatorname{E}_{8}(q)$. Then
$k_{\operatorname{Aut}(S),0}(B_{0}(S))\geq 10$.

###### Proof.

First, we remark that all unipotent characters of $3^{\prime}$-degree lie in
$B_{0}(S)$, as in [RSV21, Lem. 3.6]. Then from the character degrees of
unipotent characters available in [Car85, Section 13], together with the
knowledge of the action of $\operatorname{Aut}(S)$ on unipotent characters
(see [Mal08, Thm. 2.5]), we see there are at least 10 unipotent characters of
$3^{\prime}$-degree of $S$ in different $\operatorname{Aut}(S)$-orbits in
these cases, except if $S=\operatorname{E}_{6}^{\epsilon}(q)$ with
$3\mid(q-\epsilon)$. In the latter case, there are 9 such unipotent
characters, and the proof of [NST18, Prop. 4.7] additionally yields at least
one non-unipotent character in $\operatorname{Irr}_{3^{\prime}}(B_{0}(S))$,
which cannot be $\operatorname{Aut}(S)$-conjugate to a unipotent character. ∎

######  Proposition 5.22.

Let $q$ be a power of $p\neq 3$ and let $S$ be one of the simple groups
$\operatorname{PSL}_{n}^{\epsilon}(q)$ with $n\geq 2$;
${\mathrm{PSp}}_{2n}(q)$ with $n\geq 2$; ${\mathrm{P\Omega}}_{2n+1}(q)$ with
$n\geq 3$; or ${\mathrm{P\Omega}}_{2n}^{\pm}(q)$ with $n\geq 4$. Let
$B_{0}(S)$ be the principal $3$-block of $S$. Then there are at least $9$
unipotent characters in ${\rm Irr}_{3^{\prime}}(B_{0}(S))$ in distinct
$\operatorname{Aut}(S)$-orbits, unless:

  * •

$S=\operatorname{PSL}_{n}^{\epsilon}(q)$ with $3\mid(q-\epsilon)$ and $n\leq
5$;

  * •

$S=\operatorname{PSL}_{n}^{\epsilon}(q)$ with $3\mid(q+\epsilon)$ and $n\leq
7$;

  * •

$S={\mathrm{PSp}}_{4}(q)$, ${\mathrm{PSp}}_{6}(q)$, or
${\mathrm{P\Omega}}_{7}(q)$; or

  * •

$S={\mathrm{P\Omega}}_{8}^{\pm}(q)$.

###### Proof.

Let $H={\mathrm{GL}}_{n}^{\epsilon}(q)$, ${\mathrm{Sp}}_{2n}(q)$,
${\mathrm{SO}}_{2n+1}(q)$, or ${\mathrm{SO}}^{\pm}_{2n}(q)$, respectively, and
let $d=d_{3}(q)$. In the case of $\operatorname{PSL}_{n}^{\epsilon}(q)$,
further define $e:=d_{3}(\epsilon q)$. In the other cases, let
$e:=d_{3}(q^{2})$, so that $e=1$. Write $n=ew+r$ with $0\leq r<e$. Note that
since $e\in\\{1,2\\}$ and $e=1$ in the cases of symplectic and orthogonal
groups, the case $w=1$ only happens for $\operatorname{PSL}_{2}(q)$ or
$\operatorname{PSL}_{3}^{\epsilon}(q)$ when $e=2$.

Using the results of [BMM93, Mal07] and arguing as in part (IV) of the proof
of [HSV23, Prop. 3.2], we see the number of unipotent characters in ${\rm
Irr}_{3^{\prime}}(B_{0}(S))$ and of ${\rm Irr}_{3^{\prime}}(B_{0}(H))$ is at
least $|{\rm Irr}_{3^{\prime}}(W_{d})|$, where $W_{d}$ is the relative Weyl
group of a Sylow $d$-torus of $H$. Moreover, from [BMM93, Sec. 3A], we have
$W_{d}=C_{e}\wr S_{w}$ if $S=\operatorname{PSL}_{n}^{\epsilon}(q)$, $C_{2}\wr
S_{w}$ if $S={\mathrm{PSp}}_{2n}(q)$ or ${\mathrm{P\Omega}}_{2n+1}(q)$, and is
either $C_{2}\wr S_{w}$ or is the index-two group $G(2,2,n)$ in the case
$S={\mathrm{P\Omega}}_{2n}^{\pm}(q)$.

From this, we see that the number of unipotent characters in $B_{0}(S)$ of
height zero is at least $9$ unless we are in one of the stated exceptions.
Further, by [Mal08, Thm. 2.5], these are $\operatorname{Aut}(S)$-invariant
unless $S=\operatorname{D}_{n}(q)$ or ${\mathrm{Sp}}_{4}(2^{a})$, in which
case the graph automorphisms permute certain unipotent characters. Except for
$\operatorname{D}_{4}(q)$, the graph automorphisms have order 2. If
$S=\operatorname{D}_{n}(q)$, then the second-to-last paragraph of the proof of
[HSV23, Prop. 3.2] yields that there are at least $2|{\rm
Irr}_{3^{\prime}}(S_{n})|$ unipotent characters in ${\rm
Irr}_{3^{\prime}}(B_{0}(S))$, which is therefore at least 18 for $n\geq 6$.
Then we see we still have at least 9 $\operatorname{Aut}(S)$-orbits of
unipotent characters, except possibly if $S=\operatorname{D}_{5}(q)$. However,
in this case we see explicitly that the Weyl group $G(2,2,5)$ has 15
characters of $3^{\prime}$-degree. Hence there are 15 unipotent characters in
${\rm Irr}_{3^{\prime}}(B_{0}(S))$. Since at least two of these (namely,
$1_{S}$ and $\mathrm{St}_{S}$) are $\operatorname{Aut}(S)$-invariant, this
yields again at least 9 $\operatorname{Aut}(S)$-orbits of unipotent characters
in this case. ∎

Finally, we complete the proof of Theorem 3.3.

######  Theorem 5.23.

Theorem 3.3 holds whenever $S$ is a simple group of Lie type.

###### Proof.

By Propositions 5.7 and 5.19, we may assume $S$ is defined in characteristic
$p\neq 3$ and that $S$ is not one of $\operatorname{PSL}_{2}(q)$ or
$\operatorname{PSL}_{3}^{\epsilon}(q)$. If $|A/S|_{3}=3$, then the result now
follows from Proposition 5.18 and Corollary 5.20. Hence, we may further assume
that $S\leq A\leq\operatorname{Aut}(S)$ with $3\nmid|A/S|$.

First, suppose $S$ is of exceptional type. Since $3$ does not divide the order
of ${}^{2}\\!\operatorname{B}_{2}(q^{2})$, note that $S$ is not a Suzuki
group. If $S$ is one of ${}^{2}\\!\operatorname{F}_{4}(q^{2})$,
$\operatorname{G}_{2}(q)$, ${}^{3}\\!\operatorname{D}_{4}(q)$, we have
$k_{0}(B_{0}(S))\leq 9$ using the results of [Mal90], [HiSh90], and [DZ87],
respectively. Since $A=S{\bf C}_{A}(P)$ by Proposition 5.17, we see
$k_{0}(B_{0}(A))\leq 9$ in these cases. In the remaining exceptional cases, we
have $k_{0}(B_{0}(A))\geq 10$ by Proposition 5.21, and hence we are done when
combined with Proposition 5.17.

Now let $S$ be of classical type such that $S$ is not one of the listed
exceptions in Proposition 5.22. Then there are at least 9 orbits under $A$ in
${\rm Irr}_{3^{\prime}}(B_{0}(S))$ from unipotent characters. It suffices in
these cases to show there is at least one non-unipotent member of ${\rm
Irr}_{3^{\prime}}(B_{0}(S))$, as such a character cannot be
$\operatorname{Aut}(S)$-conjugate to a unipotent character. Here [GSV19, Lem.
3.3 and Thm. 3.5] and their proofs yield a character $\chi\in{\rm
Irr}_{3^{\prime}}(S)$ that is the deflation of a semisimple character
$\chi_{s}\in{\rm Irr}(G)$, for $s$ a semisimple $3$-element of $G^{\ast}$.
Then by [CE04, Thm. 21.13], we have $\chi_{s}\in{\rm
Irr}_{3^{\prime}}(B_{0}(G))$, so $\chi\in{\rm Irr}_{3^{\prime}}(B_{0}(S))$.

We now deal with the various exceptions pointed out in Proposition 5.22. The
distribution of the unipotent characters of the relevant groups into blocks
can be determined by their combinatorial data by [FS82, FS89]. We note that
the centralizer structures of semisimple elements in relevant classical groups
are also given in loc. cit. Throughout, we will let
$\omega\in\mathbb{F}_{q^{2}}^{\times}$ denote an element of order $3$. We will
also let $F_{0}$ denote the Frobenius map $F_{p}^{3^{a}}$, where $q=p^{f}$
with $f=3^{a}m$ and $3\nmid m$.

Suppose $S=\operatorname{PSL}_{n}^{\epsilon}(q)$ for some $n\geq 4$ in the
list. We will write $G={\mathrm{SL}}_{n}^{\epsilon}(q)$,
$\widetilde{S}={\mathrm{PGL}}_{n}^{\epsilon}(q)$, and
$\widetilde{G}={\mathrm{GL}}_{n}^{\epsilon}(q)$ as before. In each of the
relevant cases, we have $3\nmid|{\bf Z}(G)|$, so that
$\operatorname{Irr}_{3^{\prime}}(B_{0}(S))=\operatorname{Irr}_{3^{\prime}}(B_{0}(G))$.
In this case, we further recall that by [Mal08, Thm. 2.4, 2.5], every
unipotent character extends to $\operatorname{Aut}(S)$. Further, in this case
$A\leq\widetilde{S}\rtimes\langle\tau,F_{0}\rangle$, where $\tau$ denotes a
graph automorphism. (When $\epsilon=-1$, $\tau$ may be identified with the
element $F_{0}^{m}$ of $\langle F_{0}\rangle$ of order $2$.) Let
$F_{0}^{\prime}$ denote $F_{0}$ if $p\equiv\epsilon\pmod{3}$; $\tau F_{0}$ if
$p\equiv-1\pmod{3}$ and $\epsilon=1$, and $F_{0}^{2}$ if $p\equiv 1\pmod{3}$
and $\epsilon=-1$. Then $F_{0}^{\prime}$ has index two in the abelian group
$\langle\tau,F_{0}\rangle$, and we write $\tau^{\prime}$ for a representative
of the nontrivial coset.

First, let $S=\operatorname{PSL}_{4}^{\epsilon}(q)$ with $3\mid(q-\epsilon)$.
Here we see that $B_{0}(S)$ contains all unipotent characters, and hence
$B_{0}(S)$ is the unique unipotent block by [CE04, Thm. 9.12]. Hence ${\rm
Irr}_{3^{\prime}}(B_{0}(S))$ contains three unipotent characters. We have
$\widetilde{S}/S$ and ${\bf Z}(G)$ are cyclic groups of order prime to $3$, so
${\rm Irr}(B_{0}(S))={\rm Irr}(B_{0}(G))$ and Lemma 5.1 applies to any
$\widetilde{S}$-invariant character in ${\rm Irr}(B_{0}(S))$. Let
$s^{\prime}\in\widetilde{G}^{\ast}$ be the element $(\omega,\omega,\omega,1)$
and let $s$ be its image in $G^{\ast}$ under the natural projection map. Then
the class of $s$ and of $s^{-1}$ yield Lusztig series of $G$ that are stable
under $\widetilde{S}$ and $F_{0}^{\prime}$ and are interchanged by
$\tau^{\prime}$. Further, since ${\bf
C}_{\widetilde{G}^{\ast}}(s^{\prime})={\bf
C}_{\widetilde{G}^{\ast}}((s^{\prime})^{-1})\cong{\mathrm{GL}}_{3}^{\epsilon}(q)\times
C_{q-\epsilon}$ and the three unipotent characters of
${\mathrm{GL}}_{3}^{\epsilon}(q)$ have (distinct) degrees prime to $3$, we see
the six characters in these series all lie in ${\rm
Irr}_{3^{\prime}}(B_{0}(G))={\rm Irr}_{3^{\prime}}(B_{0}(S))$. Now, write
$M:=\widetilde{S}{\bf C}_{A}(P)$. Then exactly as in Proposition 5.11(iii), we
have $[\widetilde{S}A:M]\leq 2$, with the nontrivial coset (when it exists)
acting on $M$ via $\tau^{\prime}$. If $(q-\epsilon)_{3}=3$, then the classes
of $s$ and $s^{-1}$ are the only classes of $3$-elements in $G^{\ast}$ that
yield Lusztig series with $3^{\prime}$-degree characters. Hence in this case,
we have $k_{0}(B_{0}(A))=9$ using Lemma 5.1, Theorem 2.2, and the above
analysis about $\tau^{\prime}$ on these characters. Now suppose that
$(q-\epsilon)_{3}=3^{b}\geq 9$. In this case, we obtain three additional
characters in ${\rm Irr}_{3^{\prime}}(B_{0}(G))$, lying in the Lusztig series
corresponding to the image in $G^{\ast}$ of the class representative
$(\zeta,\zeta,\zeta,\zeta^{-3})\in\widetilde{G}^{\ast}$, where
$|\zeta|=3^{b}$. Since these characters cannot be
$\operatorname{Aut}(S)$-conjugate to any of the other 9 characters discussed,
we see $k_{0}(B_{0}(A))>9$ in this case.

Now, in the remaining cases, we could make similar explicit arguments.
Instead, we will use [Mal20, Thm. 5.12] and [MO83, Prop. 2.13] to obtain
$k_{0}(B_{0}(S))$, and apply Proposition 3.6. In each case, we have at least
five unipotent characters in $\operatorname{Irr}_{3^{\prime}}(B_{0}(S))$,
yielding 5 characters that extend to $A$. Here [MO83, Prop. 2.13] gives us
$k_{0}(B_{0}(\widetilde{G}))$. Then by [Mal20, Thm. 5.12] we have
$k_{0}(B_{0}(G))=k_{0}(B_{0}(\widetilde{G}))/3^{b}$ if $3\mid(q-\epsilon)$. If
$3\mid(q+\epsilon)$, we instead see that
$k_{0}(B_{0}(S))=k_{0}(B_{0}(\widetilde{S}))=k_{0}(B_{0}(\widetilde{G}))$
since $3\nmid|{\bf Z}(\widetilde{G})|$ and using Lemma 5.1, since the
characters in series indexed by $3$-elements of $\widetilde{G}^{\ast}$ must
restrict irreducibly to $G$. From this (and using Proposition 5.17 and Theorem
2.2), we obtain that $k_{0}(B_{0}(S))=k_{0}(B_{0}(A))\leq 9$ whenever
$[P:P^{\prime}]=9$, and further that $k_{0}(B_{0}(S))>18$ whenever
$[P:P^{\prime}]>9$, unless $S=\operatorname{PSL}_{6}^{\epsilon}(q)$ or
$\operatorname{PSL}_{7}^{\epsilon}(q)$ with $(q+\epsilon)_{3}=9$. Then we are
done by Proposition 3.6 unless we are in the latter case.

So now suppose $S=\operatorname{PSL}_{6}^{\epsilon}(q)$ or
$\operatorname{PSL}_{7}^{\epsilon}(q)$ with $(q+\epsilon)_{3}=9$. We see that
$B_{0}(G)$ is the unique unipotent block of maximal defect and contains six
unipotent characters of $3^{\prime}$-degree. Recall that $P$ can be identified
with a Sylow $3$-subgroup of ${\mathrm{GL}}_{6}^{\epsilon}(q)$ in this case,
as can a Sylow $3$-subgroup of $G^{\ast}$. Let $s$ be a semisimple element of
$G^{\ast}$ corresponding to an element whose pullback to
$\widetilde{G}^{\ast}$ has nontrivial eigenvalues
$\\{\omega,\omega,\omega,\omega^{-1},\omega^{-1},\omega^{-1}\\}$. Then $s$
centralizes a Sylow $3$-subgroup. Since ${\bf
C}_{\widetilde{G}^{\ast}}(s)\cong{\mathrm{GL}}_{3}(q^{2})\times V$ with $V$ a
torus, this series contains three characters of distinct $3^{\prime}$-degrees.
Further, the class of $s$ is invariant under $\operatorname{Aut}(S)$, so these
characters are $\operatorname{Aut}(S)$-invariant. We then have additional
characters in ${\rm Irr}_{3^{\prime}}(B_{0}(G))$ in a series corresponding to
eigenvalues $\\{\zeta I_{3},\zeta^{-1}I_{3}\\}$ with $|\zeta|=9$, yielding
characters not $\operatorname{Aut}(S)$-conjugate to the 9 discussed above, and
hence $k_{0}(B_{0}(A))>9$.

Together with Proposition 5.17, we see the proof is complete when $S$ is one
of the groups $\operatorname{PSL}_{n}^{\epsilon}(q)$.

Now let $S={\mathrm{PSp}}_{4}(q)$ and $G={\mathrm{Sp}}_{4}(q)$. In this case
there is again a unique unipotent block of maximal defect, and ${\rm
Irr}_{3^{\prime}}(B_{0}(S))={\rm Irr}_{3^{\prime}}(B_{0}(G))$ contains 5
unipotent characters, which lie in distinct $\operatorname{Aut}(S)$-orbits
unless $q=2^{a}$, in which case they lie in 4 distinct orbits. Since a non-
unipotent character cannot be $\operatorname{Aut}(S)$-conjugate to a unipotent
character, we will have at least 5 $A$-orbits once we know that
$k_{0}(B_{0}(S))>4$. Here arguing like above, but now using [Mal20, Thm. 5.17]
to obtain $k_{0}(B_{0}(G))$, yields $k_{0}(B_{0}(A))\leq 9$ if
$[P:P]^{\prime}=9$ and $k_{0}(B_{0}(A))>18$ if $[P:P^{\prime}]>9$, yielding
the statement using Proposition 3.6.

Next, consider the case $S={\mathrm{PSp}}_{6}(q)$. Again we have ${\rm
Irr}_{3}(B_{0}(S))={\rm Irr}_{3}(B_{0}(G))$ where $G={\mathrm{Sp}}_{6}(q)$.
This case is very similar to that of $\operatorname{PSL}_{6}^{\epsilon}(q)$
with $3\mid(q+\epsilon)$ above, again using [Mal20, Thm. 5.17] to obtain
$k_{0}(B_{0}(S))$. We again obtain $k_{0}(B_{0}(A))\leq 9$ if
$(q^{2}-1)_{3}=3$, $k_{0}(B_{0}(S))=18$ if $(q^{2}-1)_{3}=9$, and
$k_{0}(B_{0}(S))>18$ if $(q^{2}-1)_{3}>9$. If $(q^{2}-1)_{3}=9$, we may argue
similar to the case $\operatorname{PSL}_{6}^{\epsilon}(q)$ with
$3\mid(q+\epsilon)$, now using semisimple elements in
$G^{\ast}={\mathrm{SO}}_{7}(q)$ with eigenvalues
$\\{\omega,\omega,\omega,\omega^{-1},\omega^{-1},\omega^{-1},1\\}$ and
$\\{\zeta,\zeta,\zeta,\zeta^{-1},\zeta^{-1},\zeta^{-1},1\\}$, whose series
yield three characters each in ${\rm Irr}_{3^{\prime}}(B_{0}(S))$, now
yielding six $\operatorname{Aut}(S)$-orbits in addition to the six from
unipotent characters. Hence again $k_{0}(B_{0}(A))>9$.

Now let $S={\mathrm{P\Omega}}_{7}(q)$. Here note that an order-2 diagonal
automorphism is induced by the action of $H:={\mathrm{SO}}_{7}(q)$. Arguing
exactly as in the case of ${\mathrm{Sp}}_{6}(q)$ above but with corresponding
semisimple elements of $H^{\ast}={\mathrm{Sp}}_{6}(q)$, we obtain that there
are more than 9 $A$-orbits in ${\rm Irr}_{3^{\prime}}(B_{0}(H))$ when
$(q^{2}-1)_{3}\geq 9$, and we further can argue similarly to above to see
$k_{0}(B_{0}(A))<10$ when $3\mid\mid(q^{2}-1)$. Their construction further
shows that these 10 characters restrict to distinct $A$-orbits as well in
${\rm Irr}_{3^{\prime}}(B_{0}(S))$, so that $k_{0}(B_{0}(A))>9$ when
$(q^{2}-1)_{3}\geq 9$.

Finally, consider the cases $S={\mathrm{P\Omega}}_{8}^{+}(q)$ or
$S={\mathrm{P\Omega}}_{8}^{-}(q)$ and let $H$ be the corresponding
${\mathrm{SO}}_{8}^{\pm}(q)$ and $\Omega=[H,H]$ so that $S=\Omega/{\bf
Z}(\Omega)$. In these cases, there are six unipotent characters in ${\rm
Irr}_{3^{\prime}}(B_{0}(S))$, and again note that ${\rm
Irr}_{3^{\prime}}(B_{0}(\Omega))={\rm Irr}_{3^{\prime}}(B_{0}(S))$ since
$3\nmid|{\bf Z}(\Omega)|$. The unipotent characters moved by the graph
automorphisms (see [Mal08, Thm. 2.5]) have degree divisible by $3$, so the six
unipotent characters in ${\rm Irr}_{3^{\prime}}(B_{0}(S))$ are $A$-invariant.

If $H={\mathrm{SO}}_{8}^{+}(q)$, we may consider the classes in $H^{\ast}$ of
the semisimple elements $s_{1}$ with eigenvalues
$\\{\omega,\omega^{-1},I_{6}\\}$ and $s_{2}$ with eigenvalues $\\{\omega\cdot
I_{3},\omega^{-1}\cdot I_{3},1,1\\}$. These each give series of at least three
characters with $3^{\prime}$-degree. (Note that ${\bf C}_{H^{\ast}}(s_{i})$ is
of the form $H^{\ast}\cap({\mathrm{GL}}_{n}^{\epsilon}(q)\times
O_{m}^{\epsilon}(q))$ where $m$ is the multiplicity of the eigenvalue $1$ and
$n$ is the multiplicity of $\omega$, since $n$ is odd.) These two classes
cannot be $\operatorname{Aut}(S)$-conjugate and the characters restrict to
not-$\operatorname{Aut}(S)$-conjugate characters in ${\rm
Irr}_{3^{\prime}}(B_{0}(\Omega))={\rm Irr}_{3^{\prime}}(B_{0}(S))$ as well
since $s_{1}$ cannot be $\operatorname{Aut}(S)$-conjugate to $s_{2}z$ for any
nontrivial element $z$ in the $2$-group ${\bf Z}(H^{\ast})$. This gives at
least 12 $\operatorname{Aut}(S)$-orbits in ${\rm Irr}_{3^{\prime}}(B_{0}(S))$,
so $k_{0}(B_{0}(A))>9$.

Finally, let $H={\mathrm{SO}}_{8}^{-}(q)$, and note that ${\bf
C}_{H^{\ast}}(s_{i})$, with $s_{i}$ as above, is of the form
$H^{\ast}\cap({\mathrm{GL}}_{n}^{\epsilon}(q)\times O_{m}^{-\epsilon}(q))$,
where $m$ is multiplicity of the eigenvalue $1$ and $n$ is the multiplicity of
$\omega$, since $n$ is odd. In this case, we see $s_{1}$ above yields
characters of degree divisible by $3$, but $s_{2}$ still yields three
not-$\operatorname{Aut}(S)$-conjugate characters in ${\rm
Irr}_{3^{\prime}}(B_{0}(H))$, and hence their restrictions yield three
not-$\operatorname{Aut}(S)$-conjugate characters in ${\rm
Irr}_{3^{\prime}}(B_{0}(S))$ as before. If $(q^{2}-1)_{3}\geq 9$, we may in
addition consider the class corresponding to $\\{\zeta\cdot
I_{3},\zeta^{-1}\cdot I_{3},1,1\\}$ with $|\zeta|=(q^{2}-1)_{3}$ to obtain
three additional $\operatorname{Aut}(S)$-orbits of characters in ${\rm
Irr}_{3^{\prime}}(B_{0}(S))$. Then it suffices to show that
$k_{0}(B_{0}(A))\leq 9$ if $(q^{2}-1)_{3}=3$. Recall that the Sylow
$3$-subgroup of $S$ or of $H\cong H^{\ast}$ can be identified with one of
${\mathrm{SO}}_{7}(q)$, and hence of ${\mathrm{GL}}_{3}(q)$ or
${\mathrm{GL}}_{6}(q)$. From here, the proof is almost exactly as in the case
of $\operatorname{PSL}_{6}^{\epsilon}(q)$ with $(q+\epsilon)_{3}=3$ above. ∎

##  6\. Proof of Theorem A

We divide the proof in three parts. We first analyse the character theory of
groups whose Sylow 3-subgroup have abelianization of size 9, relying mostly on
[NS23].

######  Theorem 6.1.

Let $G$ be a finite group and $P\in{\rm Syl}_{3}(G)$. Write $B_{0}=B_{0}(G)$
to denote the principal $3$-block of $G$. If $[P:P^{\prime}]=9$, then
$k_{0}(B_{0})\in\\{6,9\\}$.

###### Proof.

We will sometimes write $p=3$. If $P$ is abelian then $P\cong
C_{9},C_{3}\times C_{3}$. In either case $k_{0}(B_{0})=k(B_{0})\in\\{6,9\\}$
by work of Dade and of Sambale (see [Sam14, Thm. 8.6 and Cor. 8.11], for
instance). We may thus assume that $P$ is not abelian. In particular
$P^{\prime}=\Phi(P)$.

Arguing by induction on $|G|$ we may assume that ${\bf O}_{3^{\prime}}(G)=1$
by [Nav98, Thm. 9.9.(c)]. We can also assume that ${\bf N}_{G}(P)<G$, as
otherwise the group $G$ would be $p$-solvable and by Fong’s Theorem ([Nav98,
Thm. 10.20]) $k_{0}(B_{0})=k_{0}(B_{0}(G/P^{\prime}))=k(G/P^{\prime})$. Note
that the group $G/P^{\prime}$ has abelian Sylow $3$-subgroup.

We want to show that also ${\bf O}_{3}(G)$ may be assumed to be trivial.
Otherwise, let $N$ be a minimal normal subgroup of $G$ contained in ${\bf
O}_{3}(G)$. Then $N<P$. Notice that
$(P/N)^{\prime}=P^{\prime}N/N=\Phi(P)N/N=\Phi(P/N)$. As $[P/N:P^{\prime}N/N]$
divides $[P:P^{\prime}]=9$ and $P/N>1$ we conclude that either $P^{\prime}\leq
N$ and $|P/N|=3$ or $N\leq P^{\prime}=\Phi(P)$.

In the case where $|P/N|=3$ then $N={\bf O}_{3}(G)$ and $P\cap P^{g}=N$
whenever $g\in G\setminus{\bf N}_{G}(P)$. By An-Eaton (see [AE05, Cor. 1.6] or
[Sam14, Thm. 7.17]) $k_{0}(B_{0}({\bf N}_{G}(P)))=k_{0}(B_{0})$ and we are
done by induction. In the case where $N\leq P^{\prime}$, by induction
$k_{0}(B_{0}(G/N))\in\\{6,9\\}$. We claim that
$k_{0}(B_{0}(G/N))=k_{0}(B_{0})$. Let $\chi\in{\rm Irr}_{0}(B_{0})$. Then
$\chi_{P}$ contains some linear character $\lambda$ of $P$. Since
$\lambda_{N}=1_{N}$, we conclude that $\chi$ lies over $1_{N}$ and
$N\subseteq{\rm ker}(\chi)$. By Theorem 2.4, $\chi\in{\rm
Irr}_{0}(B_{0}(G/N))$.

We have shown that if $N$ is a minimal normal subgroup of $G$, then $N$ is
semisimple of order divisible by $p$. If $G$ is almost simple, write $S={\rm
Soc}(G)$. Recall that $P^{\prime}=\Phi(P)$. By [NRSV21, Lem. 2.2] we have that
either $[G:S]_{3}=1$ or $[G:S]_{3}=3$ and we are done by Theorem 3.3. So we
may assume that $G$ is not almost simple. In this case, [NS23, Thm. 3.2 and
3.3] yields $k_{0}(B_{0})\leq 9$. As we know that $3$ divides $k_{0}(B_{0})$
(by [NST18, Lem. 2.2], for instance) and $k_{0}(B_{0})>3$ by [NST18, Thm. C],
we are done in this case. ∎

We continue by analyzing the structure of $P/P^{\prime}$ for groups $G$
satisfying $k_{0}(B_{0}(G))=6$ and $P\in{\rm Syl}_{3}(G)$.

######  Theorem 6.2.

Let $G$ be a finite group with $k_{0}(B_{0}(G))=6$, where $B_{0}$ is the
principal $3$-block. Then $[P:P^{\prime}]=9$ where $P\in{\rm Syl}_{3}(G)$.

###### Proof.

Write $p=3$ and $B_{0}=B_{0}(G)$. We proceed by induction on $|G|$.

Working by induction, we may assume that ${\rm\textbf{O}}_{p^{\prime}}(G)=1$
by [Nav98, Thm. 9.9.(c)]. We may also assume that $P$ is not normal in $G$.
Indeed, if $P\triangleleft G$, then $G$ is $p$-solvable and since
${\rm\textbf{O}}_{p^{\prime}}(G)=1$ we have that ${\rm
Irr}_{p^{\prime}}(B_{0}(G))={\rm Irr}_{p^{\prime}}(G)={\rm
Irr}(G/P^{\prime})$. Analysing the structure of the finite groups with an
abelian normal Sylow 3-subgroup and 6 conjugacy classes, we would arrive to
$[P:P^{\prime}]=9$.

Let $N$ be a minimal normal subgroup of $G$. Then $p$ divides the order of
$N$.

(1) Suppose first that $N$ is an elementary abelian 3-group. Notice that
$P/N>1$ as ${\bf N}_{G}(P)<G$, so $3\mid k_{0}(B_{0}(G/N))\leq 6$ by Theorem
2.6. Write $\bar{B}_{0}=B_{0}(G/N)$. The only options are then
$k_{0}(\bar{B}_{0})\in\\{3,6\\}$.

(1.a) Suppose that $k_{0}(\bar{B}_{0})=3$, then $|P/N|=3$ by [NST18, Thm. C].
Notice that $N={\bf O}_{p}(G)$ and that if $g\in G\setminus{\bf N}_{G}(P)$
then $P\cap P^{g}=N$. By [AE05, Cor. 1.5] (see also [Sam14, Thm. 7.17]), the
block $B_{0}$ satisfies the Alperin-McKay conjecture. In particular,
$k_{0}(B_{0}({\bf N}_{G}(P)))=6$ and we are done by induction (as $P$ is not
normal in $G$).

(1.b) Suppose that $k_{0}(\bar{B}_{0})=6$. Then ${\rm
Irr}_{p^{\prime}}(B_{0}(G))={\rm Irr}_{p^{\prime}}(B_{0}(G/N))$, so every
height zero character in $B_{0}(G)$ lies over $1_{N}$. By induction
$[P/N:(P/N)^{\prime}]=9$. Notice that $(P/N)^{\prime}=P^{\prime}N/N$. We want
to show that actually $N\subseteq P^{\prime}$ and hence $[P:P^{\prime}]=9$.
Given any linear character $\lambda\in{\rm Irr}(P)$, we have that
$\tau=\lambda_{N}\in{\rm Irr}(N)$. By Lemma 2.3 there is some $\chi\in{\rm
Irr}(B_{0})$ of height zero lying over $\tau$. Then $\tau=1_{N}$. Therefore
$N$ is contained in the kernel of every linear character of $P$, giving us

$N\subseteq\bigcap_{\lambda\in{\rm Lin}(P)}{\rm ker}(\lambda)=P^{\prime},$

as wanted.

(2) Suppose now that $N$ is semisimple, and write $N=S_{1}\times
S_{2}\times\cdots\times S_{t}$, where $S_{i}\cong S$ nonabelian simple of
order divisible by 3. Since $p=3$, we have that
$k_{0}(B_{0}(G/N))\in\\{1,3,6\\}$ again by Theorem 2.6.

(2.a) Suppose first that $k_{0}(B_{0}(G/N))=1$. Then 3 does not divide $[G:N]$
and $P\leq N$. Also, since ${\rm\textbf{O}}_{p^{\prime}}(G)=1$, we have that
${\bf C}_{G}(N)=1$. By the Frattini argument, we have that $G=N{\bf
N}_{G}(P)$, so $M=N{\bf C}_{G}(P)$ is normal in $G$. By Lemma 2.1 we have that
$B_{0}(G)$ is the only block covering $B_{0}(M)$. In particular, since $3$
does not divide $[G:M]$, we have that ${\rm Irr}(G/M)\subseteq{\rm
Irr}_{0}(B_{0}(G))$. More generally ${\rm Irr}(G|\theta)\subseteq{\rm
Irr}_{0}(B_{0})$ for every $\theta$ of height zero in the principal block of
$M$. If $t=1$ then $G$ is almost simple and we are done by Theorem 3.3. Hence
we may assume that $t\geq 2$. By Theorem 3.1(b), the set of degrees of ${\rm
Irr}_{p^{\prime}}(B_{0}(N))$, has size at least 6. Notice that for every
$\psi\in{\rm Irr}_{p^{\prime}}(B_{0}(M))$, there is $\chi\in{\rm
Irr}_{p^{\prime}}(B_{0}(G))$ lying over it and hence the set of character
degrees of characters in ${\rm
Irr}_{p^{\prime}}(B_{0}(M))\setminus\\{1_{M}\\}$ has size at most $6-k(G/M)$.
By Theorem 2.2 so has the set of character degrees of characters in ${\rm
Irr}_{p^{\prime}}(B_{0}(N))\setminus\\{1_{N}\\}$. These facts force $M=G$.
Then again by Theorem 2.2 we have that
$6=k_{0}(B_{0}(N))=k_{0}(B_{0}(S))^{t}$, absurd as $t\geq 2$.

(2.b) Suppose now that $k_{0}(B_{0}(G/N))=3$. Then $|PN/N|=3$. We first claim
that ${\bf C}_{G}(N)=1$. Indeed, consider $k_{0}(B_{0}(G/{\bf C}_{G}(N)))$. As
$[G:{\bf C}_{G}(N)]$ is divisible by 3, we have that $k_{0}(B_{0}(G/{\bf
C}_{G}(N)))\in\\{3,6\\}$ by Theorem 2.6. If $k_{0}(B_{0}(G/{\bf C}_{G}(N)))=3$
then $[G:{\bf C}_{G}(N)]_{3}=3$ by [NST18, Thm. C] and hence $|N|_{3}=3$,
which implies $|G|_{3}=9$, and the result follows. Finally, suppose that
$k_{0}(B_{0}(G/{\bf C}_{G}(N)))=6$, so all the characters in $\chi\in{\rm
Irr}_{0}(B_{0}(G))$ lie over $1_{{\bf C}_{G}(N)}$. Since $[G:N]_{3}=3$ and 3
divides $|{\bf C}_{G}(N)|$, we necessarily have that 3 does not divide
$[G:N{\bf C}_{G}(N)]$. Now take $\varphi\in{\rm Irr}_{0}(B_{0}({\bf
C}_{G}(N)))$ nontrivial and consider $1_{N}\times\varphi\in{\rm
Irr}_{0}(B_{0}(N\times{\bf C}_{G}(N)))$, then there is $\chi\in{\rm
Irr}_{0}(B_{0}(G))$ lying over $\varphi$, a contradiction.

We now claim that $P$ (and hence of $PN/N$) acts transitively on
$\\{S_{1},\ldots,S_{t}\\}$. In particular, $t\in\\{1,3\\}$ and by Theorem 3.3,
that implies $t=3$. Suppose that the action of $P$ on
$\\{S_{1},\ldots,S_{t}\\}$ is not transitive and write $N=R\times T$, where
$R$ is the product of one $P$-orbit. Since $p$ divides $k_{0}(B_{0}(R))$, then
necessarily $p$ divides the size of the set ${\rm
Irr}_{p^{\prime},P}(B_{0}(R))$ of $P$-invariant characters of height zero in
${\rm Irr}(B_{0}(R))$. As $1_{R}\in{\rm Irr}_{p^{\prime},P}(B_{0}(R))$, there
exists $1_{R}\neq\theta\in{\rm Irr}_{p^{\prime}}(B_{0}(R))$ $P$-invariant.
Analogously, let $1_{T}\neq\xi\in{\rm Irr}_{p^{\prime}}(B_{0}(T))$ be
$P$-invariant. We have that $1_{R}\times\xi$ and $\theta\times\xi$ are
$P$-invariant, so they extend to $PN$ using [Isa76, Cor. 8.16]. By Theorem
2.7, there are at least 3 characters in ${\rm Irr}(B_{0}(G))$ of height zero
above them. As $1_{R}\times\xi$ and $\theta\times\xi$ are not $G$-conjugate,
this implies that $k_{0}(B_{0}(G))\geq 9$, a contradiction.

Therefore $N=S_{1}\times S_{2}\times S_{3}$, $|PN/N|=3$ and $P$ acts
transitively on $\\{S_{1},S_{2},S_{3}\\}$. We claim that from every character
$\varphi$ in ${\rm Irr}_{p^{\prime}}(B_{0}(S))$ we can construct a
$P$-invariant character $\theta$ in ${\rm Irr}_{p^{\prime}}(B_{0}(N))$ with
$\theta(1)=\varphi(1)^{3}$. Indeed, if $xN$ generates $PN/N$, then $xN$ seen
as an element of ${\rm Out}(N)$ acts on characters of $N$ like
$(\alpha_{1},\alpha_{2},\alpha_{3})\sigma$ with $\alpha_{i}\in{\rm Out}(S)$
and $\sigma\in{S}_{3}$ of order $3$, satisfying
$\alpha_{1}\alpha_{\sigma(1)}\alpha_{\sigma(2)}=1$. The character
$\theta=\varphi\times\varphi^{\alpha_{1}}\times\varphi^{\alpha_{1}\alpha_{\sigma(1)}}$
is $P$-invariant whenever $\varphi$ is a character of $S$, and hence it does
extend to $PN$ using [Isa76, Cor. 8.16]. Now, by Theorem 2.7, there are at
least $3$ characters in ${\rm Irr}_{p^{\prime}}(B_{0}(G)|\theta)$. With this
method, and using Theorem 3.1(b), we can construct at least $2$ non-trivial
$P$-invariant characters $\theta\in{\rm Irr}_{p^{\prime}}(B_{0}(N))$ with
different degrees, but again that is not possible.

(2.c) Finally suppose that $k_{0}(B_{0}(G/N))=6$, thus every irreducible
character in ${\rm Irr}_{0}(B_{0}(G))$ lies over $1_{N}$. Since 3 divides
$k_{0}(B_{0}(N))$, we have that there are at least 3 irreducible characters in
${\rm Irr}_{0}(B_{0}(N))$ fixed by $P$. Let $\varphi$ one of them, so
$\varphi$ extends to $PN$, using [Isa76, Cor. 8.16]. Then by Lemma 2.3 there
exists $\chi\in{\rm Irr}_{0}(B_{0}(G))$ lying over $\varphi$, a contradiction.
∎

We now finish the proof of Theorem A.

######  Theorem 6.3.

Let $G$ be a finite group, $p=3$ and $P\in{\rm Syl}_{p}(G)$. Suppose that
$k_{0}(B_{0}(G))=9$, then $[P:P^{\prime}]=9$.

###### Proof.

We proceed by induction on $|G|$. Working as in the proof of Theorem 6.2, we
may assume that ${\rm\textbf{O}}_{p^{\prime}}(G)=1$ and ${\bf N}_{G}(P)<G$.
Write $B_{0}=B_{0}(G)$.

(1) We next want to prove that we may also assume that
${\rm\textbf{O}}_{p}(G)=1$. Otherwise, let $N$ be a minimal normal subgroup of
$G$ contained in ${\rm\textbf{O}}_{p}(G)=1$. Notice that $p$ divides $|G/N|$
as otherwise $N=P\triangleleft G$. By Theorem 2.6, $3$ divides
$k_{0}(\bar{B_{0}})=k_{0}(B_{0}(G/N))\leq k_{0}(B_{0})=9$. The only
possibilities are then $k_{0}(\bar{B_{0}})\in\\{3,6,9\\}$.

(1.a) Suppose that $k_{0}(\bar{B_{0}})=9$. In this case, ${\rm
Irr}_{0}(B_{0}(G))={\rm Irr}_{0}(B_{0}(G/N))$, so every character of height
zero in $B_{0}(G)$ lies over $1_{N}$. Moreover, by induction,
$[P:P^{\prime}N]=9$. We claim that $N\subseteq P^{\prime}$. Let
$\lambda\in{\rm Lin}(P)$. Then $\theta=\lambda_{N}\in{\rm Irr}(N)$ extends to
$\lambda$. By Lemma 2.3 we have that there exists $\chi\in{\rm
Irr}_{0}(B_{0}(G))$ over $\theta$. Hence $\theta=1_{N}$. Then

$N\subseteq\bigcap_{\lambda\in{\rm Lin}(P)}{\rm ker}(\lambda)=P^{\prime},$

as wanted.

(1.b) Suppose that $k_{0}(\bar{B_{0}})=6$. In this case, by Theorem 6.2,
$[P:P^{\prime}N]=9$. We work to prove that $N\subseteq P^{\prime}$. Let
$\lambda\in{\rm Lin}(P)$. If $\theta=\lambda_{N}\neq 1_{N}$, then by Lemma 2.3
some $\chi\in{\rm Irr}_{p^{\prime}}(B_{0})$ lies over $\theta$. By Theorem
2.7, we have that $3$ divides $k_{0}(B_{0}|\theta)$. In fact,
$k_{0}(B_{0}|\theta)=3$ as $k_{0}(B_{0})=9$ and $k_{0}(\bar{B}_{0})=6$.
Moreover $k_{0}(B_{0}|\theta)=k_{0}(B_{0}(G_{\theta})|\theta)$, by the
Clifford correspondence and the Third Main Theorem ([Nav98, Thm. 6.7]). By
Lemma 2.5, we have that $\theta$ extends to some $\psi\in{\rm
Irr}_{0}(B_{0}(G_{\theta}))$. Then, the equality
$k_{0}(B_{0}(G_{\theta})|\theta)=k_{0}(B_{0}(G_{\theta}/N))$ follows from
Gallagher’s Theorem and [NV17, Lem. 5.1]. But $k_{0}(B_{0}(G_{\theta}/N))>3$
by [NST18, Thm. C] because $[P/N:(P/N)^{\prime}]=9$, a contradiction. Hence
$\theta=1_{N}$, and we conclude $N\subseteq P^{\prime}$, as in (1.a).

(1.c) Suppose that $k_{0}(\bar{B_{0}})=3$. Then $|P/N|=3$. Recall that $P$ is
not normal in $G$, so $N={\rm\textbf{O}}_{p}(G)$. Let $g\in G\setminus{\bf
N}_{G}(P)$. Then $P\cap P^{g}=N$. Again, by Corollary 1.5 of [AE05], the
Alperin–McKay conjecture holds for $B_{0}(G)$. Hence $k_{0}(B_{0}({\bf
N}_{G}(P))=9$. Since ${\bf N}_{G}(P)<G$, we are done by induction.

(2) Now let $N$ be a minimal normal subgroup of $G$. By (1), $N$ is a
semisimple group of order divisible by $p$. Then $N=S_{1}\times\cdots\times
S_{t}$, where $S_{i}\cong S$ are transitively permuted by the action of $G$,
and $S$ is a simple non-abelian group of order divisible by $p$. We now
analyse the different possibilities for $k_{0}(B_{0}(G/N))\in\\{1,3,6,9\\}$.

(2.a) $k_{0}(B_{0}(G/N))=1$. In this case $(|G/N|,p)=1$ and $P\subseteq N$.
Since ${\rm\textbf{O}}_{p^{\prime}}(G)=1$, we have that ${\bf C}_{G}(N)=1$, so
we may assume that $t>1$ by Theorem 3.3. Consider $M=N{\bf C}_{G}(P)$. Then
$M$ is normal in $G$, by the Frattini argument, and $B_{0}$ is the only block
covering $B_{0}(M)$ by Lemma 2.1. In particular $k(G/M)\leq 8$. Notice that by
Theorem 3.1 (b) and using that $t\geq 2$ we obtain that there are at least 6
different character degrees of irreducible characters in ${\rm
Irr}_{p^{\prime}}(B_{0}(N))$. Using Theorem 2.2, there are at least 6