# Principal Blocks for different Primes, II

Gabriel Navarro  Departament de Matemàtiques, Universitat de València, 46100
Burjassot, València, Spain<EMAIL_ADDRESS>,  Noelia Rizo  Departamento de
Matemáticas, Universidad de Oviedo, 33007, Oviedo, Spain<EMAIL_ADDRESS>and  A. A. Schaeffer Fry  Department of Mathematics and Statistics,
Metropolitan State University of Denver, Denver, CO 80217, USA
<EMAIL_ADDRESS>
###### Abstract.

If $G$ is a finite group, we have proposed new conjectures on the interaction
between different primes and their corresponding Brauer principal blocks. In
this paper, we give strong support to the validity of these conjectures.

###### 2010 Mathematics Subject Classification:

20C20, 20C15

The first and second authors are partially supported by Grant
PID2019-103854GB-I00 funded by MCIN/AEI/10.13039/501100011033. The second
author also acknowledges support by Generalitat Valenciana AICO/2020/298 and
Grant PID2020-118193GA-I00 funded by MCIN/AEI/10.13039/501100011033. The third
author is partially supported by a grant from the National Science Foundation,
Award No. DMS-2100912.

To Pham Huu Tiep, on his 60th birthday

##  1\. Introduction

Meaningful interaction between the representation theory of finite groups from
the perspective of different primes is extremely rare. However, in [NRS21], we
proposed the following three plausible conjectures, which extended work of
several authors (see [LWXZ20], [NWi97], [BNOT07], [MN20]).

If $p$ is a prime and $G$ is a finite group, we denote by $B_{p}(G)$ the
principal $p$-block of $G$. The main subject of our work is the set ${\rm
Irr}_{p^{\prime}}(B_{p}(G))$ of the irreducible complex characters in the
principal $p$-block of $G$ whose degree is not divisible by $p$. This set
seems to possess remarkable properties.

######  Conjecture A.

Let $G$ be a finite group and let $p$ and $q$ be different primes. If

${\rm Irr}_{p^{\prime}}(B_{p}(G))\cap{\rm
Irr}_{q^{\prime}}(B_{q}(G))=\\{1_{G}\\},$

then there are a Sylow $p$-subgroup $P$ of $G$ and a Sylow $q$-subgroup $Q$ of
$G$ such that $xy=yx$ for all $x\in P$ and $y\in Q$.

######  Conjecture B.

Let $G$ be a finite group and let $p$ and $q$ be primes dividing the order of
$G$. If ${\rm Irr}_{p^{\prime}}(B_{p}(G))={\rm Irr}_{q^{\prime}}(B_{q}(G))$,
then $p=q$.

######  Conjecture C.

Let $G$ be a finite group, and let $p$ and $q$ be different primes. Then $q$
does not divide $\chi(1)$ for all $\chi\in{\rm Irr}_{p^{\prime}}(B_{p}(G))$
and $p$ does not divide $\chi(1)$ for all $\chi\in{\rm
Irr}_{q^{\prime}}(B_{q}(G))$ if and only if there are a Sylow $p$-subgroup $P$
of $G$ and a Sylow $q$-subgroup $Q$ of $G$ such that $xy=yx$ for all $x\in P$
and $y\in Q$.

The main result of [NRS21] was to reduce Conjecture A to a problem on almost
simple groups and to prove it in the case that one of the primes is $2$. To
prove Conjecture A for almost simple groups in the case where $p$ and $q$ are
both odd remains quite a challenge.

In the present paper, we focus on Conjectures B and C. Using the
Classification of Finite Simple Groups, in our first main theorem we prove the
following.

######  Theorem D.

Conjecture C implies Conjecture B.

After proving Theorem D, therefore, we concentrate our efforts in the
remainder of the paper towards Conjecture C.

######  Theorem E.

Conjecture C holds for finite simple groups.

Besides Conjecture C being true for simple groups, in Theorem 3.3 below, we
shall also prove that Conjecture C is true for $p$-solvable groups, assuming
the inductive Alperin–McKay condition. This gives strong support to the
validity of this conjecture.

Unfortunately, at the time of this writing, we still do not know how to reduce
Conjecture C to a question on almost simple groups.

##  2\. Theorem D

In this Section we prove that Conjecture C implies Conjecture B. This will
require the following result on simple groups.

######  Theorem 2.1.

Let $p,q$ be different primes and let $S$ be a non-abelian simple group with
$pq\mid|S|$. Assume that $[P,Q]=1$ for some Sylow $p$-subgroup $P$ of $S$ and
a Sylow $q$-subgroup $Q$ of $S$. Then one of the following holds:

  1. (a)

There exists $\alpha\in{\rm Irr}_{p^{\prime}}(B_{p}(S))-{\rm Irr}(B_{q}(S))$
which is ${\rm Aut}(S)_{p}$-invariant, where ${\rm Aut}(S)_{p}$ is some Sylow
$p$-subgroup of ${\rm Aut}(S)$.

  2. (b)

There exists $\alpha\in{\rm Irr}_{q^{\prime}}(B_{q}(S))-{\rm Irr}(B_{p}(S))$
which is ${\rm Aut}(S)_{q}$-invariant, where ${\rm Aut}(S)_{q}$ is some Sylow
$q$-subgroup of ${\rm Aut}(S)$.

###### Proof.

First, note that by [MN20, Lemma 3.1], the condition $[P,Q]=1$ implies that
$[\bar{P},\bar{Q}]=1$ for some Sylow $p$\- and $q$\- subgroups
$\bar{P},\bar{Q}$ of any covering group of $S$. If $S$ is one of the sporadic
groups $J_{1}$ or $J_{4}$, then we may use GAP and its Character Table Library
to see that the statement holds. Then [MN20, Propositions 3.2-3.4] further
imply that we may assume $S$ to be a simple group of Lie type that is not
isomorphic to a sporadic or alternating group and is defined in characteristic
$r_{0}\not\in\\{p,q\\}$.

So, let $S$ be of the form $S=G/{\bf Z}(G)$ for $G$ a group of Lie type of
simply connected type defined in characteristic $r_{0}\neq p,q$. In this case,
we may further assume that $p$ and $q$ are both odd and that the Sylow $p$\-
and $q$-subgroups of $G$ are abelian, using [MN20, Proposition 3.5].

Since $p$ and $q$ are odd, note that we may therefore assume without loss that
$p\geq 5$. Further, using [Ma14, Lemma 2.1 and Proposition 2.2], we have
$p\nmid|{\bf Z}(G)|$. Together, this implies that $p$ does not divide the
order of any diagonal or graph outer automorphism, so that
$\operatorname{Aut}(S)_{p}$ may be taken as a subgroup of $S\rtimes\langle
F\rangle$, where $F$ is a generating field automorphism.

Now, in [NRS21, Proposition 3.5], building off of [GSV19, Theorem 3.5], we
have shown that there is a character $\chi\in{\rm Irr}_{p^{\prime}}(B_{p}(S))$
such that $\chi$ is the deflation of a so-called semisimple character
$\chi_{t}$ of $G$ with $t\neq 1$ an element of the dual group $G^{\ast}$ of
order a power of $p$. In particular, $\chi_{t}$ lies in the Lusztig series
$\mathcal{E}(G,t)$. Since $B_{q}(G)$ contains only characters lying in Lusztig
series $\mathcal{E}(G,s)$ with $|s|$ a power of $q$ (see [CE04, Theorem
9.12]), we see that $\chi$ does not lie in ${\rm Irr}(B_{q}(S))$.

It now suffices to argue that $t$ can have been chosen so that $\chi_{t}$ is
invariant under $\operatorname{Aut}(S)_{p}$. Following the proof of [GSV19,
Theorem 3.5], we see that $t$ was chosen to lie in ${\bf Z}(P^{\ast})$, where
$P^{\ast}$ is a Sylow $p$-subgroup of the dual group $G^{\ast}$, and that we
may further choose $t$ to have order $p$ without losing the properties we have
already discussed.

Now, since $p\nmid|{\bf Z}(G)|$, we have ${\bf C}_{{\mathbf{G}}^{\ast}}(t)$ is
connected, where ${\mathbf{G}}^{\ast}$ is the ambient reductive group whose
fixed points under an appropriate Frobenius endomorphism yields $G^{\ast}$
(see [MT11, Exercise 20.16]). Then for $\varphi\in\langle F\rangle$, we have
$\chi_{t}^{\varphi}=\chi_{t^{\varphi^{\ast}}}$, where $\varphi^{\ast}$ is an
appropriate field automorphism of $G^{\ast}$ (see [NTT08, Corollary 2.5]).
Since $t^{\varphi^{\ast}}$ must be $G^{\ast}$-conjugate to another element of
order $p$ in ${\bf Z}(P^{\ast})$, we have $\chi_{t}^{\varphi}$ is also of the
form $\chi_{t^{\prime}}$ with $t^{\prime}\in{\bf Z}(P^{\ast})$ of order $p$,
is still trivial on ${\bf Z}(G)$, and still lies in $B_{p}(S)$ but not
$B_{q}(S)$ under deflation. Hence $\operatorname{Aut}(S)_{p}$ acts on such
characters and we let $\alpha$ be a fixed point under this action, since the
number of such elements $t^{\prime}$ is relatively prime to $p$. ∎

######  Lemma 2.2.

Suppose that $N$ is a minimal normal subgroup of $G$, which is a direct
product of the different $G$-conjugates of a non-abelian simple group $S$. Let
$P\in{\rm Syl}_{p}(G)$. Suppose that $\alpha\in{\rm Irr}(S)$ is ${\rm
Aut}(S)_{p}$-invariant, where ${\rm Aut}(S)_{p}$ is some Sylow $p$-subgroup of
${\rm Aut}(S)$. Then there are $g_{i}\in G,h_{i}\in P$ and $\sigma_{i}\in{\rm
Aut}(S^{g_{i}})$ such that
$(\alpha^{g_{1}})^{\sigma_{1}h_{1}}\times\cdots\times(\alpha^{g_{m}})^{\sigma_{m}h_{m}}\in{\rm
Irr}(N)$ is $P$-invariant.

###### Proof.

Suppose that $N$ is the direct product of $\Omega=\\{S^{g}|\,g\in G\\}$. Now,
write

$\Omega=\mathcal{O}(S_{1})\cup\ldots\cup\mathcal{O}(S_{t}),$

where $\mathcal{O}(S_{i})$ is the $P$-orbit of some $S_{i}=S^{x_{i}}$, for
some $x_{i}\in G$. Then $N=N_{1}\times\cdots\times N_{t}$, where $N_{i}$ is
the product of the elements in $\mathcal{O}(S_{i})$. Of course, $N_{i}$ is
$P$-invariant. Let $\alpha_{i}=\alpha^{x_{i}}\in{\rm Irr}(S_{i})$. Suppose
that $\\{S_{i}^{y_{1}},\ldots S_{i}^{y_{r}}\\}$ are the different
$P$-conjugates of $S_{i}$, where $y_{i}\in P$. Hence
$N_{i}=S_{i}^{y_{1}}\times\ldots\times S_{i}^{y_{r}}$, and

$P=\bigcup_{j=1}^{r}{\bf N}_{P}(S_{i})y_{j}$

is a disjoint union.

By hypothesis, $\alpha$ is invariant under $X\in{\rm Syl}_{p}({\rm Aut}(S))$.
Therefore $\alpha_{i}$ is invariant under $X_{i}=X^{x_{i}}\in{\rm
Syl}_{p}({\rm Aut}(S_{i}))$. Let $M_{i}={\bf N}_{G}(S_{i})$ and $C_{i}={\bf
C}_{G}(S_{i})$. We have that $M_{i}/C_{i}$ embeds into ${\rm Aut}(S_{i})$.
Then ${\bf N}_{P}(S_{i})C_{i}/C_{i}$ is a $p$-subgroup of ${\rm Aut}(S_{i})$,
and therefore, there is $\sigma_{i}\in{\rm Aut}(S_{i})$ such that $({\bf
N}_{P}(S_{i})C_{i}/C_{i})\subseteq X_{i}^{\sigma_{i}}$. Since $\alpha_{i}$ is
$X_{i}$-invariant, it follows that $\beta_{i}=(\alpha_{i})^{\sigma_{i}}\in{\rm
Irr}(S_{i})$ is $X_{i}^{\sigma_{i}}$-invariant, and therefore ${\bf
N}_{P}(S_{i})$-invariant. We claim that
$\gamma_{i}=\beta_{i}^{y_{1}}\times\cdots\times\beta_{i}^{y_{r}}\in{\rm
Irr}(N_{i})$ is $P$-invariant. Indeed, if $x\in P$, then
$y_{k}x=w_{k}y_{\sigma(k)}$ for some $w_{k}\in{\bf N}_{P}(S_{i})$, $1\leq
k\leq r$, and $\sigma$ a permutation of $\sf S_{r}$. Now, if $u\in S_{i}$,
then

$\gamma_{i}^{x^{-1}}(u^{y_{k}})=\gamma_{i}(u^{y_{k}x})=\alpha(1)^{r-1}\beta_{i}^{y_{\sigma(k)}}(u^{w_{k}y_{\sigma(k)}})=\alpha(1)^{r-1}\beta_{i}(u^{w_{k}})=\alpha(1)^{r-1}\beta_{i}(u)=\gamma_{i}(u^{y_{k}})\,.$

This proves that $\gamma_{i}$ is $P$-invariant. Hence
$\gamma=\gamma_{1}\times\cdots\times\gamma_{t}\in{\rm Irr}(N)$ is
$P$-invariant. ∎

We will need the following well-known result of J. Alperin and E. C. Dade.

######  Theorem 2.3.

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 principals blocks of $G$ and $N$.

###### Proof.

The case where $G/N$ is solvable was proved in [Al76] and the general case in
[Da77]. ∎

######  Theorem 2.4.

Assume that Conjecture C is true for all finite groups. Let $p$ and $q$ be
primes. Assume that $G$ is a finite group of order divisible by $p$ and $q$.
If ${\rm Irr}_{p^{\prime}}(B_{p}(G))={\rm Irr}_{q^{\prime}}(B_{q}(G))$, then
$p=q$.

###### Proof.

We argue by induction on $|G|$. Assume that $p\neq q$. By Conjecture C, we
know that $[P,Q]=1$ for some $P\in{\rm Syl}_{p}(G)$ and $Q\in{\rm
Syl}_{q}(G)$.

Step 0. If $1\neq N\lhd G$, then $p$ divides $|N|$ or $q$ divides $|N|$.

Otherwise by [Na98, Theorem 9.9(c)]

${\rm Irr}_{p^{\prime}}(B_{p}(G/N))={\rm Irr}_{p^{\prime}}(B_{p}(G))={\rm
Irr}_{q^{\prime}}(B_{q}(G))={\rm Irr}_{q^{\prime}}(B_{q}(G/N))$

and by induction we are done.

Step 1. Let $L$ be a proper normal subgroup of $G$. Then $G/L$ has order
divisible by $p$ or $q$.

Suppose that $G/L$ has $p^{\prime}$ and $q^{\prime}$-order. We claim that
${\rm Irr}_{p^{\prime}}(B_{p}(L))={\rm Irr}_{q^{\prime}}(B_{q}(L))$. Indeed,
let $\theta\in{\rm Irr}_{q^{\prime}}(B_{q}(L))$. Then there exists
$\chi\in{\rm Irr}(B_{q}(G))$ over $\theta$. Then $\chi$ has
$q^{\prime}$-degree by [I08, Theorem 11.29] and therefore $\chi\in{\rm
Irr}_{p^{\prime}}(B_{p}(G))$, by hypothesis. Therefore $\theta\in{\rm
Irr}_{p^{\prime}}(B_{p}(L))$. By symmetry, the claim is proved. Therefore, the
proof of Step 1 follows by using the inductive hypothesis.

Step 2. Let $N$ be a minimal normal subgroup of $G$ and suppose that $N$ is an
elementary abelian $p$-group. Then $G={\bf C}_{G}(N){\bf C}_{G}(Q)$.

Write $M={\bf C}_{G}(N)$ and $L=M{\bf C}_{G}(Q)$. We have that $Q\subseteq{\bf
C}_{G}(P)\subseteq{\bf C}_{G}(N)=M$ and $G/M$ is a $q^{\prime}$-group. Also by
the Frattini argument $G=M{\bf N}_{G}(Q)$ and $L\lhd G$. Notice that $G/L$ is
a $q^{\prime}$-group and a $p^{\prime}$-group (because $P\subseteq{\bf
C}_{G}(Q)\subseteq L$). Then we use Step 1.

Step 3. Let $N$ be a minimal normal subgroup of $G$ and suppose that $N$ is
abelian. Then $G={\bf C}_{G}(N)$.

By Step 0, we may assume that $N$ is a $p$-group or a $q$-group. By symmetry,
assume that $N$ is a $p$-group. Write $M={\bf C}_{G}(N)$. We prove first that
$B_{p}(G)$ is the only $p$-block of $G$ covering $B_{p}(M)$. Let $B$ be a
$p$-block of $G$ covering $B_{p}(M)$ and let $D$ be a defect group of $B$.
Then $N\subseteq D$ by [Na98, Theorem 4.8] and ${\bf C}_{G}(D)\subseteq M={\bf
C}_{G}(N)$. Then $B$ is regular with respect to $M$ ([Na98, Lemma 9.20]) and
hence by [Na98, Theorem 9.19] we have that $B=B_{p}(M)^{G}=B_{p}(G)$ by the
third main theorem (see [Na98, Theorem 6.7]). Hence $B_{p}(G)$ is the only
$p$-block of $G$ covering $B_{p}(M)$. In particular, we have that ${\rm
Irr}(G/M)\subseteq{\rm Irr}(B_{p}(G))$.

Next, we prove that $G=M$. Recall that $G/M$ is a $q^{\prime}$-group, because
$[Q,N]=1$. By Step 2, we have $G=M{\bf C}_{G}(Q)$. Hence by Theorem 2.3, we
have that the restriction map

${\rm res}:{\rm Irr}(B_{q}(G))\rightarrow{\rm Irr}(B_{q}(M))$

is a bijection. We claim that ${\rm Irr}_{p^{\prime}}(G/M)=\\{1_{G}\\}$.
Indeed, if $\chi\in{\rm Irr}_{p^{\prime}}(G/M)$, then $\chi\in{\rm
Irr}_{p^{\prime}}(B_{p}(G))={\rm Irr}_{q^{\prime}}(B_{q}(G))$. By Theorem 2.3,
$\chi_{M}$ is irreducible and since $\chi$ lies over $1_{M}$ we have
$\chi_{M}=1_{M}$. Hence $\chi=1_{G}$ by the uniqueness of the restriction map.
Thus ${\rm Irr}_{p^{\prime}}(G/M)=\\{1_{G/M}\\}$ and $G=M$ by [NRS21, Lemma
2.2].

Step 4. Let $N$ be a minimal normal subgroup of $G$, then $N$ is not abelian.

Suppose the contrary and assume without loss of generality that $N$ is an
elementary abelian $p$-group, so by Step 3 we have $G={\bf C}_{G}(N)$ and
$N\subseteq{\bf Z}(G)$. By [Na98, Theorem 9.10] we have that $B_{p}(G/N)$ is
the unique $p$-block of $G/N$ contained in $B$. We claim that ${\rm
Irr}_{p^{\prime}}(B_{p}(G/N))={\rm Irr}_{q^{\prime}}(B_{q}(G/N))$. Indeed, we
have that

${\rm Irr}_{p^{\prime}}(B_{p}(G/N))\subseteq{\rm
Irr}_{p^{\prime}}(B_{p}(G))={\rm Irr}_{q^{\prime}}(B_{q}(G))={\rm
Irr}_{q^{\prime}}(B_{q}(G/N))$

where we have used [Na98, Theorem 9.9] in the last equality. On the other
hand, let $\chi\in{\rm Irr}_{q^{\prime}}(B_{q}(G/N))$, so $\chi\in{\rm
Irr}_{q^{\prime}}(B_{q}(G))={\rm Irr}_{p^{\prime}}(B_{p}(G))$ and
$N\subseteq{\rm Ker}(\chi)$. Since $B_{p}(G/N)$ is the only $p$-block of $G/N$
contained in $B_{p}(G)$, we have that $\chi\in{\rm
Irr}_{p^{\prime}}(B_{p}(G/N))$ and the claim is proven. By using the inductive
hypothesis, we have that $p$ does not divide $|G/N|$. Therefore,
$\\{1_{G}\\}={\rm Irr}(B_{p}(G/N))={\rm Irr}_{p^{\prime}}(B_{p}(G/N))={\rm
Irr}_{q^{\prime}}(B_{q}(G/N))$ and $q\nmid|G/N|$ by [NRS21, Lemma 2.1]). Hence
$q\nmid|G|$ and this is a contradiction.

Step 5. Let $N$ be a minimal normal subgroup of $G$, then $pq$ divides $|N|$.

Suppose that $N$ is a $p^{\prime}$-group. We claim first that $NQ$ does not
have a normal $q$-complement. Indeed, suppose the contrary and let $X\lhd NQ$
be a normal $q$-complement. Then $N\cap X$ is a normal $q$-complement of $N$
and by the minimality of $N$ we have that either $N\cap X=1$ or $N\cap X=N$.
If $N\cap X=N$, $N$ is $q^{\prime}$ and $p^{\prime}$, contradiction with Step
0. If $N\cap X=1$ then $N\cong XN/X$ is a $q$-group, which is a contradiction
with Step 4. Therefore $NQ$ does not have a normal $q$-complement. By [IS76,
Corollary 3] there is $\tau\in{\rm Irr}(B_{q}(QN))$ non-linear of
$q^{\prime}$-degree. Therefore $1\neq\tau_{N}\in{\rm
Irr}_{q^{\prime}}(B_{q}(N))$. By [Mu94, Lemma 4.3] we have that there is some
$\gamma\in{\rm Irr}_{q^{\prime}}(B_{q}(G))$ lying over $\tau_{N}$. By
hypothesis, we have that $\gamma$ is in the principal $p$-block of $G$, and
therefore $\tau_{N}$ is in the principal $p$-block of $N$, which is a
contradiction since $N$ is a $p^{\prime}$-group and $\tau_{N}\neq 1$.

Final Step. If $N$ is a minimal normal subgroup of $G$, then $N$ is semisimple
by Step 4. Suppose that $N$ is a direct product of all the different
$G$-conjugates of a simple group $S$ of order divisible by $pq$. Suppose that
(a) of Theorem 2.1 holds and let $\alpha$ be the character in ${\rm
Irr}_{p^{\prime}}(B_{p}(S))$ (not in ${\rm Irr}(B_{q}(S))$) which is ${\rm
Aut}(S)_{p}$-invariant, given by Theorem 2.1. Notice that any $G$-conjugate or
${\rm Aut}(G)$-conjugate of $\alpha$ is in the principal $p$-block and not in
the principal $q$-block os $S$. By Lemma 2.2, there exists $\tau\in{\rm
Irr}_{p^{\prime}}(B_{p}(N))$ which is $P$-invariant, and such that each of its
factors does not belong to the principal $q$-block. In particular, $\tau$ does
not belong to the principal $q$-block of $N$. Now $\tau$ extends to $PN$ by
[Is06, Corollary 8.16]. By [Mu94, Lemma 4.3] there is $\chi\in{\rm
Irr}_{p^{\prime}}(B_{p}(G))$ lying over $\tau$. Then $\chi\in{\rm
Irr}_{q^{\prime}}(B_{q}(G))$ and thus $\tau\in{\rm
Irr}_{q^{\prime}}(B_{q}(N))$, which is a contradiction. Assuming (b) in
Theorem 2.1 and reasoning analogously we get again a contradiction. ∎

##  3\. Conjecture C and $p$-solvable groups

As of the writing of this article, we can only prove Conjecture C for
$p$-solvable groups if the following (expected) consequence of the
Alperin–McKay conjecture is true.

######  Conjecture 3.1.

Suppose that $N$ is normal in $G$, and let $P\in{\rm Syl}_{p}(N)$. Then there
is a bijection

${}^{*}:{\rm Irr}_{p^{\prime}}(B_{p}(N))\rightarrow{\rm
Irr}_{p^{\prime}}(B_{p}({\bf N}_{N}(P)))$

such that for each $\theta\in{\rm Irr}_{p^{\prime}}(B_{p}(N))$, there is a
bijection $f_{\theta}:{\rm Irr}(B_{p}(G)|\theta)\rightarrow{\rm
Irr}(B_{p}({\bf N}_{G}(P))|\theta^{*})$ such that
$\chi(1)/\theta(1)=f_{\theta}(\chi)/\theta^{*}(1)$ for all $\chi\in{\rm
Irr}(G|\theta)$.

(In fact, it is expected that ∗ is ${\bf N}_{G}(P)$-equivariant, and that the
character triples $(G_{\theta},N,\theta)$ and $({\bf N}_{G}(P)_{\theta},{\bf
N}_{N}(P),\theta^{*})$ are isomorphic.)

In any case, we shall only need Conjecture 3.1 in the case where $G/N$ is a
$p^{\prime}$-group.

In our proof, we shall also need a McKay divisibility theorem, which was made
possible after M. Geck proved a remarkable conjecture on Glauberman
correspondents.

In the following, we follow the proof of [Ri19, Theorem A], and then use
Geck’s result.

######  Theorem 3.2.

Let $G$ be a $p$-solvable group and let $P\in{\rm Syl}_{p}(G)$. Then there is
a bijection

${}^{*}:{\rm Irr}_{p^{\prime}}(G)\rightarrow{\rm Irr}_{p^{\prime}}({\bf
N}_{G}(P))$

such that $\chi^{*}(1)$ divides $\chi(1)$ and $\chi(1)/\chi^{*}(1)$ divides
$|G:{\bf N}_{G}(P)|$.

###### Proof.

We argue by induction on $|G|$. As in the proof of [Ri19, Theorem A] we may
assume that ${\bf O}_{p}(G)=1$, and hence $K={\bf O}_{p^{\prime}}(G)>1$.

Let $S/K={\bf O}_{p}(G/K)$, and notice that $P_{0}=P\cap S$ is a Sylow
$p$-subgroup of $S$. By the Frattini argument we have that $G=K{\bf
N}_{G}(P_{0})$ and ${\bf N}_{G}(P_{0})<G$ since ${\bf O}_{p}(G)=1$. Let
$\theta_{1},\ldots,\theta_{s}$ be a complete set of representatives of the
orbits of the action of ${\bf N}_{G}(P)$ on the $P$-invariant irreducible
characters of $K$. By [Na18, Lemma 9.3], we have that

${\rm Irr}_{p^{\prime}}(G)={\rm
Irr}_{p^{\prime}}(G\,|\,\theta_{1})\cup\cdots\cup{\rm
Irr}_{p^{\prime}}(G\,|\,\theta_{s})$

is a disjoint union. Fix $\theta_{i}\in{\rm Irr}(K)$ and observe that
$\theta_{i}$ is also $P_{0}$-invariant. Let $\theta_{i}^{*}\in{\rm Irr}({\bf
C}_{K}(P_{0}))$ be the Glauberman correspondent of $\theta_{i}$ (see [Na18,
Theorem 2.9], for instance) and let $T_{i}$ be the stabilizer of $\theta_{i}$
in $G$. Since the Glauberman correspondence and the action of ${\bf
N}_{G}(P_{0})$ commute (see [Na18, Lemma 2.10]), it follows that ${\bf
N}_{T_{i}}(P_{0})=T_{i}\cap{\bf N}_{G}(P_{0})$ is the stabilizer of
$\theta_{i}^{*}$ in ${\bf N}_{G}(P_{0})$.

Again as in the proof of [Ri19, Theorem A] we obtain a bijection

${}^{*}:{\rm Irr}_{p^{\prime}}(T_{i}\,|\,\theta_{i})\rightarrow{\rm
Irr}_{p^{\prime}}({\bf N}_{T_{i}}(P_{0})\,|\,\theta_{i}^{*})$

satisfying $\psi(1)/\psi^{*}(1)=\theta_{i}(1)/\theta_{i}^{*}(1)$. Hence we
have that $\psi^{*}(1)$ divides $\psi(1)$ by the main result of [Ge20].
Moreover, since $G={\bf N}_{G}(P_{0})K$ we have that $|K:{\bf
C}_{K}(P_{0})|=|T_{i}:{\bf N}_{T_{i}}(P_{0})|$ and since
$\theta_{i}(1)/\theta_{i}^{*}(1)$ divides $|K:{\bf C}_{K}(P_{0})|$ we conclude
that $\psi(1)/\psi^{*}(1)$ divides $|T_{i}:{\bf N}_{T_{i}}(P_{0})|$.

The remaining part of the proof proceeds exactly as in the proof of [Ri19,
Theorem A]. ∎

######  Theorem 3.3.

Let $G$ be a finite $p$-solvable group. If Conjecture 3.1 is true, then
Conjecture C is true for $G$.

###### Proof.

By [MN20], we only need to prove that if all ${\rm
Irr}_{p^{\prime}}(B_{p}(G))$ have $q^{\prime}$-degree and all ${\rm
Irr}_{q^{\prime}}(B_{q}(G))$ have $p^{\prime}$-degree, then $[P,Q]=1$ for some
$P\in{\rm Syl}_{p}(G)$ and $Q\in{\rm Syl}_{q}(G)$. Let $N$ be a normal
subgroup of $G$. Since the hypothesis is satisfied by $G/N$, by induction, we
know that $[P,Q]\subseteq N$ for some $P\in{\rm Syl}_{p}(G)$ and $Q\in{\rm
Syl}_{q}(G)$.

Suppose that ${\bf O}_{p^{\prime}}(G)=1$. Then we know that ${\rm
Irr}(B_{p}(G))={\rm Irr}(G)$, by Theorem 10.20 of [Na98]. Hence, all the
irreducible characters in ${\rm Irr}_{p^{\prime}}(G)$ have
$q^{\prime}$-degree. Let $L={\bf O}_{p}(G)$ and let $P\in{\rm Syl}_{p}(G)$ and
$Q\in{\rm Syl}_{q}(G)$ such that $[P,Q]\subseteq L$. Therefore $Q$ normalizes
$P$ and $|G:{\bf N}_{G}(P)|$ is not divisible by $q$. By Theorem 3.2, we have
that all ${\rm Irr}({\bf N}_{G}(P)/P^{\prime})$ have $q^{\prime}$-degree. By
the Itô-Michler theorem we have that $QP^{\prime}$ is normal in ${\bf
N}_{G}(P)$, and hence $[QP^{\prime},P]\subseteq P^{\prime}$. Then $Q$ acts
trivially on $P/P^{\prime}$, and therefore $Q$ acts trivially on $P$ by
coprime action (see [I08, Corollary 3.29]). Thus $[Q,P]=1$ and we are done in
this case.

Suppose that $L={\bf O}_{p^{\prime}}(G)>1$ and let $P\in{\rm Syl}_{p}(G)$ and
$Q\in{\rm Syl}_{q}(G)$ with $[Q,P]\subseteq L$. By Hall–Higman 1.2.3 Lemma
(see [I08, Theorem 3.21]) we have that ${\bf C}_{G/L}({\bf
O}_{p}(G/L))\subseteq{\bf O}_{p}(G/L)$. Since $QL/L\subseteq{\bf
C}_{G/L}(PL/L)$ we conclude that $QL/L\subseteq{\bf O}_{p}(G/L)$ and hence
$Q\subseteq L$. Thus $G/L$ is $q^{\prime}$ and $G=L{\bf N}_{G}(Q)$, in
particular $|G:{\bf N}_{G}(Q)|$ is not divisible by $p$. We claim that all the
irreducible characters in ${\rm Irr}_{q^{\prime}}(B_{q}({\bf N}_{G}(Q)))$ have
$p^{\prime}$-degree. Indeed, let $\chi^{*}\in{\rm Irr}_{q^{\prime}}(B_{q}({\bf
N}_{G}(Q)))$ and let $\theta^{*}\in{\rm Irr}_{q^{\prime}}(B_{q}({\bf
N}_{L}(Q)))$ under $\chi^{*}$. Let $\theta\in{\rm Irr}_{q^{\prime}}(B_{q}(L))$
be the pre-image of $\theta^{*}$ given by the bijection in Conjecture 3.1.
Again using Conjecture 3.1, let $\chi\in{\rm Irr}(B_{q}(G)|\theta)$ be such
that $f_{\theta}(\chi)=\chi^{*}$, so we know that

$\frac{\chi(1)}{\theta(1)}=\frac{\chi^{*}(1)}{\theta^{*}(1)}.$

Then $\chi(1)/\theta(1)$ is not divisible by $q$ and thus $\chi\in{\rm
Irr}_{q^{\prime}}(B_{q}(G))$. By hypothesis, $\chi(1)$ is not divisible by
$p$. Hence $\chi^{*}(1)/\theta^{*}(1)$ is not divisible by $p$, and therefore,
since $\theta^{*}$ is of $p^{\prime}$-degree we have that $\chi^{*}(1)$ is not
divisible by $p$, and the claim follows.

Let $X={\bf O}_{q^{\prime}}({\bf N}_{G}(Q))$. Then all the elements in ${\rm
Irr}({\bf N}_{G}(Q)/Q^{\prime}X)={\rm Irr}_{q^{\prime}}(B_{q}({\bf
N}_{G}(Q)))$ have degree not divisible by $p$. By the Itô-Michler theorem, we
have that this group has a normal Sylow $p$-subgroup (which is a Sylow
$p$-subgroup of $G$, since $|G:{\bf N}_{G}(Q)|$ is $p^{\prime}$), and
therefore $P$ centralizes $Q/Q^{\prime}$. By coprime action, $[P,Q]=1$. ∎

##  4\. Conjecture C and Simple Groups

Note that the “if” direction for simple groups follows from the work of
Malle–Navarro in [MN20] and their proof of Brauer’s height-zero conjecture for
principal blocks [MN21]:

######  Proposition 4.1.

Let $S$ be a simple group. If there exists $P\in{\operatorname{Syl}}_{p}(S)$
and $Q\in{\operatorname{Syl}}_{q}(S)$ such that $[P,Q]=1$, then $q$ does not
divide $\chi(1)$ for all $\chi\in{\rm Irr}_{p^{\prime}}(B_{p}(S))$ and $p$
does not divide $\chi(1)$ for all $\chi\in{\rm Irr}_{q^{\prime}}(B_{q}(S))$.

###### Proof.

In [MN20, Section 3], the condition that $[P,Q]=1$ for some
$P\in{\operatorname{Syl}}_{p}(S)$ and $Q\in{\operatorname{Syl}}_{q}(S)$ is
studied. In particular, we see from the results there that this condition
implies that $P$ and $Q$ are abelian, and hence ${\rm
Irr}_{p^{\prime}}(B_{p}(S))={\rm Irr}(B_{p}(S))$ and ${\rm
Irr}_{q^{\prime}}(B_{q}(S))={\rm Irr}(B_{q}(S))$ by the principal block case
of Brauer’s height-zero conjecture [MN21]. Hence the statement follows from
[MN20, Theorem 3.6]. ∎

Next we consider the cases easily dealt with in GAP:

######  Proposition 4.2.

Conjecture C holds for sporadic simple groups, alternating and symmetric
groups $\mathfrak{A}_{n}$ and $\mathfrak{S}_{n}$ with $n\leq 8$, the Tits
group ${}^{2}\\!\operatorname{F}_{4}(2)^{\prime}$, and groups of Lie type with
exceptional Schur multipliers.

###### Proof.

This can be seen using GAP and its Character Table Library. ∎

###  4.1. Conjecture C for Alternating and Symmetric Groups

Here we prove Conjecture C in the case of alternating groups
$\mathfrak{A}_{n}$ and symmetric groups $\mathfrak{S}_{n}$. Note that it
follows from [MN20, Proposition 3.3] that $[P,Q]\neq 1$ for every Sylow
$p$-subgroup $P$ and Sylow $q$-Subgroup $Q$ of $\mathfrak{S}_{n}$ or
$\mathfrak{A}_{n}$.

######  Proposition 4.3.

Let $G$ be an alternating or symmetric group $\mathfrak{A}_{n}$ or
$\mathfrak{S}_{n}$ with $n\geq 9$ and let $p,q$ be primes dividing $|G|$. Then
either there exists $\chi\in{\rm Irr}_{p^{\prime}}(B_{p}(S))$ with degree
divisible by $q$ or there exists $\chi\in{\rm Irr}_{q^{\prime}}(B_{q}(S))$
with degree divisible by $p$.

The strategy: We first recall some facts and give the basic idea of the proof.
The set ${\rm Irr}(\mathfrak{S}_{n})$ is indexed by partitions of $n$, and two
characters $\chi_{\lambda},\chi_{\mu}$ corresponding to partitions
$\lambda,\mu$ lie in the same $p$-block if $\lambda,\mu$ have the same
$p$-core (and similar for $q$). In particular, writing $n=pm+b$ with $0\leq
b<p$, the set ${\rm Irr}(B_{p}(\mathfrak{S}_{n}))$ consists of the characters
$\chi_{\lambda}$ such that $\lambda$ has $p$-core $(b)$. Further, recall that
the degree of the character $\chi_{\lambda}$ is given by the hooklength
formula $\chi_{\lambda}(1)=\frac{n!}{\prod h_{\lambda}}$, where the
denominator is the product of all hooklengths in the tableaux corresponding to
the partition $\lambda$. Further, if $\lambda$ is not self-conjugate, then the
corresponding character restricts irreducibly to $\mathfrak{A}_{n}$.

So, our strategy will be (up to switching $p$ and $q$) to illustrate a non-
self-conjugate partition $\lambda$ of $n$ with $p$-core $(b)$ such that the
numerator in the hooklength formula has the same $p$-part as the denominator
and larger $q$-part than the denominator. Our proof will require several
technical cases and analysis of the degrees given by the hooklength formula.

Setting notation: Throughout our proof, we will assume without loss that
$q<p$. We will write $mp=wq+r$ with $0\leq r<q$, and let $n=mp+b$ with $0\leq
b<p$. Note that we may assume that $m>1$, since otherwise a Sylow $p$-subgroup
is abelian, and the result follows from [GMV19, Theorem 3.5], together with
the principal block version of Brauer’s height zero conjecture [MN21]. In
studying the degrees of the characters that we construct, several expressions
will appear repeatedly. Hence we define once and for all:

$Y:=\frac{(mp+b)\cdots(mp+1)}{b!},\quad\quad
Y^{\prime}:=\frac{(mp+b-1)\cdots(mp+1)}{(b-1)!},$
$Z:=\frac{(mp+r)\cdots(mp+1)}{r!},\quad\hbox{ and }\quad
Z^{\prime}:=\frac{(mp-1)\cdots(mp-r)}{r!}$

which we see are each relatively prime to $p$.

It will also be useful to set the $p$-adic and $q$-adic expansions of $mp$:
Let

$mp=a_{1}q^{t_{1}}+\cdots a_{k}q^{t_{k}}=b_{1}p^{s_{1}}+\cdots
b_{k^{\prime}}p^{s_{k^{\prime}}}$

with $t_{i}<t_{i+1}$, $s_{j}<s_{j+1}$, $0<a_{i}<q$, and $0<b_{j}<p$ for each
appropriate value of $i,j$. With these established, we further define

$X:=\frac{\prod_{i=1}^{a_{1}q^{t_{1}}}(mp-i)}{\prod_{i=1}^{a_{1}q^{t_{1}}}i}\quad\hbox{
and }\quad
X^{\prime}:=\frac{\prod_{i=1}^{b_{1}p^{s_{1}}}(mp-i)}{\prod_{i=1}^{b_{1}p^{s_{1}}}i}.$

Note that $mp\neq a_{1}q^{t_{1}}$, since $p\nmid a_{1}$. In the situation that
the expression $X^{\prime}$ becomes relevant, we will see that also $mp\neq
b_{1}p^{s_{1}}$. Finally, for an integer $x$, we will write $(x)_{p}$ (or just
$x_{p}$ if it is clear) for the $p$-part of $x$.

###### Proof of Proposition 4.3.

Keep the notation above.

(I) First, suppose that $r>0$. If $b=0$, so that $n=mp$, then consider
$\lambda=(1^{mp-r-1},1+r)$. Then $\chi_{\lambda}\in{\rm
Irr}(B_{p}(\mathfrak{S}_{n}))$ and $\chi_{\lambda}(1)=Z^{\prime}$, which is
$p^{\prime}$ but divisible by $q$ since $mp-r=wq$ and $r<q$. If $b\neq 0$ and
$0<r<b$, consider $\lambda=(1^{mp-r-1},1+r,b)$. If $0<b<r$, consider
$\lambda=(1^{mp-r-1},1+b,r)$. In these cases, $\chi_{\lambda}\in{\rm
Irr}(B_{p}(\mathfrak{S}_{n}))$ and $\chi_{\lambda}(1)=\displaystyle{Y\cdot
Z^{\prime}\cdot\frac{|b-r|}{mp-r+b}}$. Note that the $q$-part of the numerator
of $Y$ must be at least as large as the $q$-part of the denominator. (Each
remainder modulo $b$ appears once as a factor in the numerator.) Hence, this
character still has degree that is $p^{\prime}$ and divisible by $q$, with the
possible exception of if $q\mid b$ and $b!(wq+b)$ has larger $q$-part than
$(mp+1)\cdots(mp+b)$. In the latter case, $(1^{mp},b)$, giving degree
$Y^{\prime}$, works instead.

If $b=r>0$, note that $r+1<wq$, as otherwise we would have $r=q-1$ and
$w=1=m$, contradicting our assumption that $m>1$. Then let
$\lambda=(1^{r},r+1,wq-1)$, so that $\chi_{\lambda}(1)=Z\cdot
Z^{\prime}\cdot({wq-r-1})/({2r+1})$, which is $p^{\prime}$ and divisible by
$q$ unless $2r+1=p$ and $p\mid(m-1)$ or if $2r+1=q$ and $q\nmid w$. In the
latter cases, the partition $(1^{wq-2},1+r,1+r)$ works, unless we were in the
case $2r+1=p$ with $p\mid(m-1)$ and $r+1=q$ with $q\nmid w$. In this case, if
$q\neq 2$ (and hence $r\neq 1$), take $\lambda=(1^{mp},r)$, which lies in
$B_{p}(\mathfrak{S}_{n})$ and has degree
$Y^{\prime}=(mp+1)\cdots(mp+r-1)/(q-2)!$. This is relatively prime to $p$, and
is divisible by $q$ since there must be a number between $mp$ and
$mp+q-1=mp+r$ divisible by $q$, but neither $mp=(w+1)q-1$ nor $n=mp+r=wq+2q-2$
can be divisible by $q$. If $q=2$, we have $r=1,q=2,p=3$, and the character
corresponding to $(1^{n-2},2)$ lies in $B_{2}(\mathfrak{S}_{n})$ and has
degree $n-1=3m=2w+1$, which is odd and divisible by $3$.

From now on, we may therefore assume that $r=0$, so that $mp=wq$.

(II) First, assume that $a_{1}q^{t_{1}}<b_{1}p^{s_{1}}$. If $b=0$, so
$n=mp=wq$, consider $\lambda=(1^{mp-a_{1}q^{t_{1}}-1},1+a_{1}q^{t_{1}})$. Then
the corresponding degree is $X$, which we see is equal to

$X=\frac{\prod_{i=0}^{a_{1}q^{t_{1}}-1}(i+a_{2}q^{t_{2}}+\cdots
a_{k}q^{t_{k}})}{\prod_{i=1}^{a_{1}q^{t_{1}}}i}=\frac{\prod_{i=1}^{a_{1}q^{t_{1}}}(b_{1}p^{s_{1}}-i+b_{2}p^{s_{2}}+\cdots
b_{k^{\prime}}p^{s_{k^{\prime}}})}{\prod_{i=1}^{a_{1}q^{t_{1}}}i}.$

Note that the $q$-part of this is $q^{t_{2}}/q^{t_{1}}$, which is divisible by
$q$. Further, the $p$-part is $1$, since $a_{1}q^{t_{1}}<b_{1}p^{s_{1}}$
implies that the $p$-part of
$-i+b_{1}p^{s_{1}}+\cdots+b_{k^{\prime}}p^{s_{k^{\prime}}}$ is the same as
that of $i$ for $1\leq i\leq a_{1}q^{t_{1}}$.

Now suppose that $b>0$, so $n=mp+b=wq+b$. If $a_{1}q^{t_{1}}\neq b$, consider
either $(1^{mp-a_{1}q^{t_{1}}-1},1+a_{1}q^{t_{1}},b)$ or $(1^{mp-
a_{1}q^{t_{1}}-1},1+b,a_{1}q^{t_{1}})$, depending on whether $b$ is larger or
smaller than $a_{1}q^{t_{1}}$. Then the corresponding character lies in
$B_{p}(\mathfrak{S}_{n})$ and has degree $X\cdot
Y\cdot\displaystyle{\frac{|b-a_{1}q^{t_{1}}|}{n-a_{1}q^{t_{1}}}}$. Note that
since $a_{1}q^{t_{1}}<b_{1}p^{s_{1}}$, the $p$-part of
$|b-a_{1}q^{t_{1}}|_{p}\leq p^{s_{1}}$, and from this we see
$|b-a_{1}q^{t_{1}}|$ and $n-a_{1}q^{t_{1}}=mp+b-a_{1}q^{t_{1}}$ have the same
$p$-part. Hence this characters is a member of ${\rm
Irr}_{p^{\prime}}(B_{p}(\mathfrak{S}_{n}))$. Further, it is still divisible by
$q$, except possibly if $|b-a_{1}q^{t_{1}}|_{q}<(n-a_{1}q^{t_{1}})_{q}$. This
can only happen if $(b)_{q}\geq q^{t_{2}}$. In the latter case, consider again
the partition $(1^{mp},b)$. The degree is $Y^{\prime}$ and hence we have
removed $\frac{mp+b}{b}$ from the expression $Y$, which is $p^{\prime}$ and,
from before, has $q$-part of the numerator at least as large as that of the
denominator. Since in our situation
$(mp+b)_{q}=q^{t_{1}}<q^{t_{2}}\leq(b)_{q}$, this degree $Y^{\prime}$ is also
divisible by $q$.

Now assume $b=a_{1}q^{t_{1}}$. If $b+1\neq p$ or $p\nmid(m-1)$, we take
$\lambda=(1^{mp-b-2},b+1,b+1)$, with
$\displaystyle{\chi_{\lambda}(1)=Y\cdot\frac{(mp-2)\cdots(mp-b-1)}{(b+1)!}}.$
Then $\chi_{\lambda}\in{\rm Irr}(B_{p}(\mathfrak{S}_{n}))$ and
$p\nmid\chi_{\lambda}(1)$ due to the assumption $b+1\neq p$ or $m-1$ is not
divisible by $p$. Further, $\chi_{\lambda}(1)$ is divisible by $q$ since the
$q$-part of the numerators of each of the two fractions is at least that of
the denominators, as before, and in this case, the factor $(mp-b)$ is divisble
by $q^{t_{2}}$, but no factor in the denominator is. Now, if
$p=b+1=1+a_{1}q^{t_{1}}$ and $p\mid(m-1)$, this forces also
$b_{1}p^{s_{1}}=p$, as $mp-p=\sum{b_{i}p^{s_{i}}}-p$ must be divisible by
$p^{2}$. Here consider the partition $(1^{mp-p},b+p)$, which gives a character
with degree ${(mp+b-1)\cdots(mp-p+1)}/{(p+b-1)!}$ in
$B_{p}(\mathfrak{S}_{n})$. Note that the only factor in the numerator
divisible by $p$ is $mp$, which is divisible by $p$ eactly once. Then since
the denominator is divisible by $p$, we see this character lies in ${\rm
Irr}_{p^{\prime}}(B_{p}(\mathfrak{S}_{n}))$. Further, since
$mp-p+1=\sum_{i\geq 2}a_{i}q^{t_{i}}$ in this case, we see
$(mp-p+1+j)_{q}\geq(j)_{q}$ for $1\leq j\leq p+b-2$, and that
$\displaystyle{\frac{mp-p+1}{p+b-1}=\frac{\sum_{i\geq
2}a_{i}q^{t_{i}}}{2a_{1}q^{t_{1}}}}$ is divisible by $q$ except possibly if
$q=2$ and $t_{2}=t_{1}+1$. In the latter case, the character corresponding to
$(1^{mp-1},b+1)$, which has degree
$\displaystyle{\frac{mp\cdot\prod_{i=1}^{2^{t_{1}}-1}(mp+i)}{2^{t_{1}}!}}$
lies in ${\rm Irr}_{2^{\prime}}(B_{2}(\mathfrak{S}_{n}))$ and has degree
divisible by $p$.

(III) Finally, suppose that $a_{1}q^{t_{1}}>b_{1}p^{s_{1}}$. Note here that
$mp\neq b_{1}p^{s_{1}}$, as $mp\geq a_{1}q^{t_{1}}$. If $b=0$, then reversing
the roles of $p$ and $q$ in the corresponding case in (II) above yields a
character in $B_{q}(\mathfrak{S}_{n})$ with degree $X^{\prime}$, which is
relatively prime to $q$ but divisible by $p$.

Hence we assume $b>0$, so $n=mp+b=wq+b$. Note here that
$b=w^{\prime}q+b^{\prime}$ for some integers $w^{\prime},b^{\prime}$ with
$0\leq b^{\prime}<q$, and $B_{q}(\mathfrak{S}_{n})$ consists of those
characters whose corresponding partitions have $q$-core $(b^{\prime})$.

Now, the partition $(1^{mp-b_{1}p^{s_{1}}-1},b+1,b_{1}p^{s_{1}})$ gives a
character in $B_{q}(\mathfrak{S}_{n})$ with degree $\displaystyle{Y\cdot
X^{\prime}\cdot\frac{b_{1}p^{s_{1}}-b}{n-b_{1}p^{s_{1}}}}$. Note that the
third factor is not divisible by $p$ nor $q$, since $p\nmid b$ and
$b_{1}p^{s_{1}}-b<a_{1}q^{t_{1}}$ so $b_{1}p^{s_{1}}-b$ and
$n-b_{1}p^{s_{1}}=mp-(b_{1}p^{s_{1}}-b)$ have the same $q$-part. Further,
$p\nmid Y$, and also $q\nmid Y$ as long as $b<(q-a_{1})q^{t_{1}}$. So if
$b<(q-a_{1})q^{t_{1}}$, this character lies in ${\rm
Irr}_{q^{\prime}}(B_{q}(\mathfrak{S}_{n}))$ with degree divisible by $p$.

So, we now assume that $q\mid Y$, so $b\geq(q-a_{1})q^{t_{1}}$. Then the
partition $(b+1,mp-1)$ corresponds to a character in $B_{p}(\mathfrak{S}_{n})$
with degree $\displaystyle{Y\cdot\frac{mp-(b+1)}{b+1}}$, which is divisible by
$q$ and is relatively prime to $p$ if $b+1\neq p$ or $p\nmid(m-1)$. Hence we
may now assume that further $p=b+1$ and $p\mid(m-1)$. Then setting
$\lambda=(1^{mp},b)$ yields $\chi_{\lambda}\in{\rm
Irr}(B_{p}(\mathfrak{S}_{n}))$ and $\chi_{\lambda}(1)=Y^{\prime}$, which is
prime to $p$. Since $q\mid Y$, we have $q\mid Y^{\prime}$ unless $q\mid b$ and
$(mp+b)_{q}>(b)_{q}$, which forces $b=(q-a_{1})q^{t_{1}}$.

So we are reduced to the case $q\nmid Y^{\prime}$, $b=(q-a_{1})q^{t_{1}}=p-1$,
and $p\mid(m-1)$. Then the partition $(1^{mp-1},1+b)$ gives a character in
$B_{q}(\mathfrak{S}_{n})$ with degree $Y^{\prime}\cdot{mp}/{b}$, which is
divisible by $p$ but is not divisible by $q$ since the $q$-part of both $b$
and $mp$ are $q^{t_{1}}$.

Finally, note that in all cases, the $\lambda$ described is not self-
conjugate, and hence the characters restrict irreducibly to
$\mathfrak{A}_{n}$, completing the proof. ∎

###  4.2. Conjecture C for Simple Groups of Lie Type

Let $r_{0}$ be a prime and $r:=r_{0}^{a}$ be some power of $r_{0}$. For $p$
another prime, we denote by $d_{p}(r)$ the order of $r$ modulo $p$,
respectively modulo $4$, if $p$ is odd, respectively $p=2$. Here we will prove
the remaining direction of Conjecture C for simple groups of Lie type $S$
defined over ${\mathbb{F}}_{r}$.

Several cases here also follow quickly from [MN20]:

######  Proposition 4.4.

Let $S$ be a simple group of Lie type defined over ${\mathbb{F}}_{r}$, and let
$p,q$ be primes such that $[P,Q]\neq 1$ for every Sylow $p$-subgroup $P$ and
Sylow $q$-subgroup $Q$ of $S$ but that at least one of the following
conditions holds:

  1. (1)

$r_{0}\in\\{p,q\\}$;

  2. (2)

a Sylow $p$-subgroup or a Sylow $q$-subgroup of $S$ is abelian; or

  3. (3)

$d_{p}(r)\neq d_{q}(r)$;

Then either there exists $\chi\in{\rm Irr}_{p^{\prime}}(B_{p}(S))$ with degree
divisible by $q$ or there exists $\chi\in{\rm Irr}_{q^{\prime}}(B_{q}(S))$
with degree divisible by $p$.

###### Proof.

First suppose that $r_{0}\in\\{p,q\\}$ and without loss, say $r_{0}=p$. Then
${\rm Irr}(B_{p}(S))={\rm Irr}(S)\setminus\\{\mathrm{St}_{S}\\}$. Let $G$ be a
quasisimple group of Lie type of simply connected type such that $G/{\bf
Z}(G)=S$. As in [MN20, Theorem 5.1], we let $\chi\in{\rm Irr}(S)$ be of
positive $q$-height, so that it must lie in $B_{q}(S)$. But then it must be
that $\chi$ is the deflation of some $\bar{\chi}\in{\rm Irr}(G)$ in a Lusztig
series $\mathcal{E}(G,s)$, where $s$ is a semisimple element of the dual group
$G^{\ast}$ that does not centralize a Sylow $q$-subgroup of $G^{\ast}$ and
which lies in $O^{p^{\prime}}(G^{\ast})$. Considering a semisimple character
$\chi_{s}$ in this series instead, we see that $\chi_{s}(1)=[G^{\ast}:{\bf
C}_{G^{\ast}}(s)]_{p^{\prime}}$, so $\chi_{s}$ still has degree divisible by
$q$ but lies in ${\rm Irr}_{p^{\prime}}(B_{p}(S))$ as a character of $S$.

Next, suppose that $r_{0}\not\in\\{p,q\\}$ and that a Sylow $p$-subgroup of
$S$ is abelian. Then ${\rm Irr}_{p^{\prime}}(B_{p}(S))={\rm Irr}(B_{p}(S))$ by
Brauer’s height zero conjecture for principal blocks [MN21], and hence the
proof of [MN20, Theorem 5.1] yields the desired character.

Finally, assume that $r_{0}\not\in\\{p,q\\}$, that no Sylow $p$\- or $q$\-
subgroup of $S$ is abelian, and that $d_{p}(r)\neq d_{q}(r)$. Note that the
assumption $[P,Q]\neq 1$ for every choice of Sylow $p$\- and Sylow
$q$-subgroups of $S$ (and hence analogously for $G$) implies that also
$[P^{\ast},Q^{\ast}]\neq 1$ for any Sylow $p$\- and $q$\- subgroups of the
dual group $G^{\ast}$. (Again, this is pointed out already in [MN20, Theorem
5.1].) Now, let $d_{p}(r)\leq d_{q}(r)$ and suppose that there exists $1\neq
s\in{\bf Z}(Q^{\ast})$ for some
$Q^{\ast}\in{\operatorname{Syl}}_{q}(G^{\ast})$ such that $s$ centralizes a
Sylow $p$-subgroup of $G^{\ast}$. Then the same argument as in the first two
paragraphs of [MN20, Proposition 3.5], but now applied to $G^{\ast}$, yields
that $d_{p}(r)=d_{q}(r)$, a contradiction. Hence, given $1\neq s\in{\bf
Z}(Q^{\ast})$, we have ${\bf C}_{G^{\ast}}(s)$ does not contain a Sylow
$p$-subgroup $P^{\ast}$ of $G^{\ast}$. Then the corresponding semisimple
character $\chi_{s}$ of $G$ has degree divisible by $p$ but lies in ${\rm
Irr}_{q^{\prime}}(B_{q}(G))$. Further, arguing as in [GSV19, Theorem 3.5]
shows that such $s$ can be chosen so that $\chi_{s}$ is trivial on ${\bf
Z}(G)$, completing the proof. ∎

Our task is now to prove Conjecture C in the case that neither $p$ nor $q$ is
the defining characteristic, $d_{p}(r)=d_{q}(r)$, and no Sylow $p$\- or
$q$-subgroup of $S$ is abelian. We begin with the case of exceptional groups
of Lie type, by which we mean the groups $S=\operatorname{G}_{2}(r)$,
$\operatorname{F}_{4}(r)$, $\operatorname{E}_{6}^{\epsilon}(r)$,
$\operatorname{E}_{7}(r)$, $\operatorname{E}_{8}(r)$,
${}^{3}\\!\operatorname{D}_{4}(r)$, ${}^{2}\\!\operatorname{G}_{2}(r)$,
${}^{2}\\!\operatorname{F}_{4}(r)$, and ${}^{2}\\!\operatorname{B}_{2}(r)$.

######  Proposition 4.5.

Let $S$ be an exceptional simple group of Lie type defined over
${\mathbb{F}}_{r}$, and let $p,q$ be primes such that $[P,Q]\neq 1$ for every
Sylow $p$-subgroup $P$ and Sylow $q$-subgroup $Q$ of $S$. Then either there
exists $\chi\in{\rm Irr}_{p^{\prime}}(B_{p}(S))$ with degree divisible by $q$
or there exists $\chi\in{\rm Irr}_{q^{\prime}}(B_{q}(S))$ with degree
divisible by $p$.

###### Proof.

By Proposition 4.4, we may assume that $r$ is not a power of $p$ nor $q$, that
no Sylow $p$\- or $q$\- subgroup of $S$ is abelian, and that
$d_{p}(r)=d_{q}(r)$. Let $d:=d_{p}(r)=d_{q}(r)$. With these constraints, we
see that $S$ is not of Suzuki or Ree type, that $p$ and $q$ are at most 7, and
that $d$ is a regular number. Hence we see that the principal blocks
$B_{p}(S)$ and $B_{q}(S)$ are the the unique blocks of $S$ containing
$p^{\prime}$-, respectively, $q^{\prime}$-degree unipotent characters (see,
e.g., [RSV21, Lemma 3.6]). Under these conditions, we see by observing the
explicit list of unipotent character degrees in [Car85, Section 13.9] that
there exists a unipotent character $\chi$ satisfying either $\chi\in{\rm
Irr}_{p^{\prime}}(B_{p}(S))$ and $q\mid\chi(1)$ or $\chi\in{\rm
Irr}_{q^{\prime}}(B_{q}(S))$ and $p\mid\chi(1)$. ∎

We next consider the case of linear and unitary groups.

######  Proposition 4.6.

Let $S={\operatorname{PSL}}_{n}^{\epsilon}(r)$ with $n\geq 2$, and let $p\neq
q$ be primes such that $[P,Q]\neq 1$ for every Sylow $p$-subgroup $P$ and
Sylow $q$-subgroup $Q$ of $S$. Then either there exists $\chi\in{\rm
Irr}_{p^{\prime}}(B_{p}(S))$ with degree divisible by $q$ or there exists
$\chi\in{\rm Irr}_{q^{\prime}}(B_{q}(S))$ with degree divisible by $p$.

###### Proof.

By Proposition 4.2, we may assume $S$ is not isomorphic to a sporadic or
alternating group and that $S$ has a nonexceptional Schur multiplier. Further,
by Proposition 4.4, we may assume that $r$ is not a power of $p$ nor $q$, that
no Sylow $p$\- or $q$\- subgroup of $S$ is abelian, and that
$d_{p}(r)=d=d_{q}(r)$.

Write $G:={\operatorname{SL}}_{n}^{\epsilon}(r)$,
$\widetilde{G}:=\operatorname{GL}_{n}^{\epsilon}(r)$, and $e:=d_{p}(\epsilon
r)=d_{q}(\epsilon r)$. Note that $\widetilde{G}^{\ast}\cong\widetilde{G}$, so
we will make this identification. Let
$\widetilde{P}\in{\operatorname{Syl}}_{p}(\widetilde{G})$ and
$\widetilde{Q}\in{\operatorname{Syl}}_{q}(\widetilde{G})$. It suffices to show
that (up to switching $p$ and $q$) there is some $s\in{\bf
Z}(\widetilde{P})\cap G$ such that ${\bf C}_{\widetilde{G}}(s)$ is not
divisible by $|\widetilde{Q}|$ and $sz$ is not $\widetilde{G}$-conjugate to
$s$ for any $1\neq z\in{\bf Z}(\widetilde{G})$. (Indeed, that $s\in
G={\operatorname{O}}^{r_{0}^{\prime}}(\widetilde{G})$ implies that the
corresponding semisimple character $\chi_{s}$ is trivial on ${\bf
Z}(\widetilde{G})$ and that $sz$ is not conjugate to $s$ for nontrivial
$z\in{\bf Z}(\widetilde{G})$ implies that $\chi_{s}$ is irreducible on
restriction to $G$ using e.g. [SFT22, Proposition 2.7] and [RSV21, Lemma 1.4];
the remaining conditions imply $\chi_{s}\in{\rm Irr}(B_{p}(\widetilde{G}))$
using [Hi90, Corollary 3.4] and $\chi_{s}(1)$ is $p^{\prime}$ but divisible by
$q$ since $\chi_{s}(1)=[\widetilde{G}:{\bf
C}_{\widetilde{G}}(s)]_{r_{0}^{\prime}}$.)

Now, if $2\not\in\\{p,q\\}$, we see using the results of Weir [We55] that
$\widetilde{P}$ and $\widetilde{Q}$ are naturally isomorphic to the
corresponding Sylow subgroups of $\operatorname{GL}_{we}^{\epsilon}(r)$,
embedded naturally into
$\operatorname{GL}_{we}^{\epsilon}(r)\times\operatorname{GL}_{b}^{\epsilon}(r)\leq\operatorname{GL}_{n}^{\epsilon}(r)$
where $n=we+b$ with $0\leq b<e$. The results of Carter–Fong [CF64] yield the
same when $p$ or $q$ is $2$, except that if $(e,b)=(2,1)$, then
$|\operatorname{GL}_{b}^{\epsilon}(r)|$ is divisible by $2$ exactly once, and
the Sylow $2$-subgroup of $\operatorname{GL}_{n}(r)$ in this case is that of
$\operatorname{GL}_{we}^{\epsilon}(r)\times\operatorname{GL}_{b}(r)$.

Let
$w=a_{1}p^{t_{1}}+a_{2}p^{t_{2}}+\cdots+a_{k}p^{t_{k}}=b_{1}q^{m_{1}}+\cdots+b_{k^{\prime}}q^{m_{k^{\prime}}}$
be the $p$-adic and $q$-adic expansions of $w$, with $t_{1}\leq\cdots\leq
t_{k}$; $m_{1}\leq\cdots\leq m_{k^{\prime}}$; $1\leq a_{i}<p$ for each $1\leq
i\leq k$; and $1\leq b_{j}<q$ for each $1\leq j\leq k^{\prime}$. By [We55,
CF64], we have $\widetilde{P}\cong P_{1}^{a_{1}}\times\cdots\times
P_{k}^{a_{k}}\times X$, where $P_{i}$ is a Sylow $p$-subgroup of
$\operatorname{GL}_{ep^{t_{i}}}^{\epsilon}(r)$ (which, if $p$ is odd, is a
Sylow $p$-subgroup of $\operatorname{GL}^{\epsilon}_{p^{t_{i}}}(r^{e})$
embedded naturally) for each $1\leq i\leq k$ and
$X\in{\operatorname{Syl}}_{p}(\operatorname{GL}_{b}^{\epsilon}(r))$ is
isomorphic to $C_{2}$ if $(p,e,b)=(2,2,1)$ and is trivial otherwise. Here we
view $\prod\operatorname{GL}_{ep^{t_{i}}}^{\epsilon}(r)$ as the natural
diagonally-embedded subgroup. Similarly, $\widetilde{Q}\cong
Q_{1}^{b_{1}}\times\cdots\times Q_{k^{\prime}}^{b_{k}}\times Y$ where
$Q_{j}\in{\operatorname{Syl}}_{q}(\operatorname{GL}_{eq^{m_{j}}}^{\epsilon}(r))$
for $1\leq j\leq k^{\prime}$ and
$Y\in{\operatorname{Syl}}_{q}(\operatorname{GL}_{b}^{\epsilon}(r))$.

Without loss, assume $a_{1}p^{t_{1}}<b_{1}q^{m_{1}}$. (Note that we cannot
have $a_{1}p^{t_{1}}=b_{1}q^{m_{1}}$, as this would contradict that either
$a_{1}<p<q$ or $b_{1}<q<p$.) Let $x\in{\bf Z}(P_{1})$ have no eigenvalues
equal to $1$ (indeed, this can be done by taking $x$ as an element of a Sylow
$p$-subgroup of ${\bf Z}(\operatorname{GL}_{p^{t_{1}}}^{\epsilon}(r^{e}))$
embedded naturally into $\operatorname{GL}_{ep^{t_{1}}}^{\epsilon}(r)$) and
consider the element $s=\mathrm{diag}(x,\ldots,x,I_{n-ea_{1}p^{t_{1}}})\in{\bf
Z}(\widetilde{P})$, with $a_{1}$ copies of $x$. In fact, taking $x$ (and hence
its eigenvalues) to have order $p$, we obtain $\det(x)=1$ and hence $s\in G$.
Here ${\bf C}_{\widetilde{G}}(s)={\bf
C}_{\operatorname{GL}_{a_{1}ep^{t_{1}}}^{\epsilon}(r)}(\mathrm{diag}(x,\ldots,x))\times\operatorname{GL}_{n-ea_{1}p^{t_{1}}}^{\epsilon}(r)$.
Now, by considering the structure, and hence size, of $\widetilde{Q}$ (namely,
each $Q_{j}$ is a wreath product $Q^{\prime}\wr C_{q}$ where $Q^{\prime}$ is a
Sylow $q$-subgroup of $\operatorname{GL}_{eq^{m_{j}-1}}^{\epsilon}(r)$), we
see
$\operatorname{GL}_{a_{1}ep^{t_{1}}}^{\epsilon}(r)\times\operatorname{GL}_{n-ea_{1}p^{t_{1}}}^{\epsilon}(r)$
is not divisible by $|\widetilde{Q}|$. Further, by considering the block
sizes, we see that $sz$ and $z$ cannot have the same eigenvalues (and hence
they cannot be $\widetilde{G}$-conjugate) for any nontrivial scalar matrix
$z\in{\bf Z}(\operatorname{GL}_{n}^{\epsilon}(r))$. This completes the proof.
∎

We next consider the remaining classical types, for which the proof is very
similar to the linear and unitary case.

######  Proposition 4.7.

Let $S=\operatorname{PSp}_{2n}(r)$ with $n\geq 2$,
${\operatorname{P\Omega}}_{2n+1}(r)$ with $n\geq 3$, or
${\operatorname{P\Omega}}_{2n}^{\epsilon}(r)$ with $n\geq 4$. Let $p\neq q$ be
primes such that $[P,Q]\neq 1$ for every Sylow $p$-subgroup $P$ and Sylow
$q$-Subgroup $Q$ of $S$. Then either there exists $\chi\in{\rm
Irr}_{p^{\prime}}(B_{p}(S))$ with degree divisible by $q$ or there exists
$\chi\in{\rm Irr}_{q^{\prime}}(B_{q}(S))$ with degree divisible by $p$.

###### Proof.

As before, we may assume $S$ is not isomorphic to a sporadic or alternating
group, that $S$ has a nonexceptional Schur multiplier, $r$ is not a power of
$p$ nor $q$, no Sylow $p$\- or $q$\- subgroup of $S$ is abelian, and that
$d_{p}(r)=d=d_{q}(r)$.

First we set some notation. We define $H_{n}:=\operatorname{Sp}_{2n}(r)$,
$\operatorname{SO}_{2n+1}(r)$, and $\operatorname{SO}_{2n}^{\epsilon}(r)$ in
the cases $S=\operatorname{PSp}_{2n}(r)$,
${\operatorname{P\Omega}}_{2n+1}(r)$, and
${\operatorname{P\Omega}}_{2n}^{\epsilon}(r)$, respectively. Let $H:=H_{n}$
and let $\Omega:={\operatorname{O}}^{r_{0}^{\prime}}(H)$ so that $\Omega$ is
perfect and $S=\Omega/{\bf Z}(\Omega)$. Note that the dual groups are
$H_{n}^{\ast}=\operatorname{SO}_{2n+1}(r)$, $\operatorname{Sp}_{2n}(r)$, and
$\operatorname{SO}_{2n}^{\epsilon}(r)$, respectively, and we will write
$H^{\ast}:=H_{n}^{\ast}$. Note that ${\bf Z}(\Omega)\leq{\bf Z}(H)$ and that
$H/\Omega$ and ${\bf Z}(H)$ are $2$-groups. Let $\widetilde{P}$ and
$\widetilde{Q}$ be Sylow $p$\- and $q$-subgroups of $H^{\ast}$.

In this situation, it suffices to show that (up to switching $p$ and $q$)
there is some $s\in{\bf Z}(\widetilde{P})$ such that ${\bf
C}_{\widetilde{G}}(s)$ is not divisible by $|\widetilde{Q}|$, using similar
reasoning to the above case. Indeed, if $p$ is odd, then $\widetilde{P}$ may
be considered as a Sylow $p$-subgroup of
${\operatorname{O}}^{r_{0}^{\prime}}(H^{\ast})$ and $sz$ cannot be
$H^{\ast}$-conjugate to $s$ for any $1\neq z\in{\bf Z}(H^{\ast})$ since ${\bf
Z}(H^{\ast})$ is a 2-group, so a corresponding semisimple character $\chi_{s}$
of $H$ is trivial on ${\bf Z}(H)$ and restricts irreducibly to $\Omega$. If
instead $p=2$, then such a character $\chi_{s}$ would have odd degree, and
therefore restrict irreducibly to $\Omega$ since $H/\Omega$ is a $2$-group.
Then since $\Omega$ is perfect, ${\bf Z}(\Omega)$ is a $2$-group, and
$\chi_{s}$ has odd-degree, this forces $\chi_{s}$ to be trivial on ${\bf
Z}(\Omega)$. Further, as before, $\chi_{s}\in{\rm Irr}(B_{p}(H))$ in either
case.

Assume first that $p$ and $q$ are odd. In these cases, the work of Weir [We55]
again describes the structure of $\widetilde{P}$ and $\widetilde{Q}$, building
off of the case of linear groups. If $H^{\ast}=\operatorname{SO}_{2n+1}(r)$ or
$\operatorname{Sp}_{2n}(r)$, we have Sylow $p$\- and $q$-subgroups are already
Sylow subgroups of $\operatorname{GL}_{2n+1}(r)$ (and hence of
$\operatorname{GL}_{2n}(r)$) when $d$ is even, and are Sylow subgroups of the
naturally-embedded $\operatorname{GL}_{n}(r)$ if $d$ is odd. For these cases,
let $e:=d_{p}(r^{2})=d_{q}(r^{2})$, write $n=ew+b$ with $0\leq b<e$, and let
$w=a_{1}p^{t_{1}}+a_{2}p^{t_{2}}+\cdots+a_{k}p^{t_{k}}=b_{1}q^{m_{1}}+\cdots+b_{k^{\prime}}q^{m_{k^{\prime}}}$
be the $p$-adic and $q$-adic expansions of $w$ as before. Again without loss,
we assume $a_{1}p^{t_{1}}<b_{1}q^{m_{1}}$. In particular, $\widetilde{P}$ and
$\widetilde{Q}$ are again isomorphic to Sylow subgroups of $H_{ew}^{\ast}$ and
of the form $\widetilde{P}\cong P_{1}^{a_{1}}\times\cdots\times
P_{k}^{a_{k}}$, where each $P_{i}$ is a Sylow $p$-subgroup of
$\operatorname{GL}_{dp^{t_{i}}}(r)$ and can be identified with a Sylow
$p$-subgroup of $H_{ep^{t_{i}}}^{\ast}$, and similar for $\widetilde{Q}$. As
before, let $x\in{\bf Z}(P_{1})$ with no eigenvalues equal to 1 and let
$s=(x,\cdots x,1,\ldots 1)\in{\bf Z}(\widetilde{P})$ with $a_{1}$ copies of
$x$. Then we can see from the centralizer structure of semisimple elements
that ${\bf C}_{H^{\ast}}(s)\cong{\bf
C}_{H^{\ast}_{a_{1}ep^{t_{1}}}}((x,\ldots,x))\times
H^{\ast}_{n-ea_{1}p^{t_{1}}}$. Since the Sylow $q$-subgroups of
$H_{a_{1}ep^{t_{1}}}^{\ast}$ and $H_{n-ea_{1}p^{t_{1}}}^{\ast}$ can be
identified with Sylow subgroups of linear groups in an analogous way as for
$H^{\ast}$, depending on whether $d$ was even or odd, we have
$|\widetilde{Q}|\nmid|{\bf C}_{H^{\ast}}(s)|$ for the same reason as the case
of linear groups above.

If $H^{\ast}=\operatorname{SO}_{2n}^{\epsilon}(r)$, then we have embeddings
$\operatorname{SO}_{2n-1}(r)\leq H^{\ast}\leq\operatorname{SO}_{2n+1}(r)$, and
$\widetilde{P}$ and $\widetilde{Q}$ are both Sylow subgroups of either
$\operatorname{SO}_{2n-1}(r)$ or $\operatorname{SO}_{2n+1}(r)$. In this case,
letting $m\in\\{n,n-1\\}$ so that $\widetilde{P},\widetilde{Q}$ are Sylow
subgroups of $\operatorname{SO}_{2m+1}(r)$ and now writing $m=ew+b$ with $w$
written with $p$\- and $q$-adic expansions as before, $\widetilde{P}$ can
again be written $\widetilde{P}\cong P_{1}^{a_{1}}\times\cdots\times
P_{k}^{a_{k}}$ with each $P_{i}$ a Sylow subgroup of
$\operatorname{GL}_{dp^{t_{i}}}(r)$, which in this case can also be identified
with a Sylow $p$-subgroup of either $\operatorname{SO}_{2ep^{t_{i}}}^{+}(r)$
or $\operatorname{SO}_{2ep^{t_{i}}}^{-}(r)$. From here, arguing similar to
before, we obtain an element $s\in{\bf Z}(\widetilde{P})$ with ${\bf
C}_{H^{\ast}}(s)$ isomorphic to a subgroup of ${\bf
C}_{\operatorname{GO}_{2a_{1}ep^{t_{1}}}^{\pm}(r)}(\mathrm{diag}(x,\ldots,x))\times\operatorname{GO}^{\pm}_{2(n-ea_{1}p^{t_{1}})}(r)$.
Since $q$ is odd, we again see in the same way as above that $|\widetilde{Q}|$
does not divide $|{\bf C}_{H^{\ast}}(s)|$.

We are finally left with the case that $2\in\\{p,q\\}$. Let $\hat{H}^{\ast}$
denote the group $\operatorname{GO}_{2n+1}(r),\operatorname{Sp}_{2n}(r)$, or
$\operatorname{GO}_{2n}^{\epsilon}(r)$ respectively, so that
$[\hat{H}^{\ast}:H^{\ast}]$ divides $2$. Note that if $p=2$ and
$H^{\ast}\neq\hat{H}^{\ast}$, then $\widetilde{P}$ is index-2 in a Sylow
$2$-subgroup $\hat{P}$ of $\hat{H}^{\ast}$, which are again described by
Carter–Fong [CF64]. Here in the case of $\operatorname{GO}_{2n+1}(r)$ or
$\operatorname{Sp}_{2n}(r)$, writing $n=2^{t_{1}}+\cdots+2^{t_{k}}$ for the
$2$-adic expansion with $t_{1}\leq\ldots\leq t_{k}$, we have $\hat{P}\cong
P_{1}\times\cdots\times P_{k}$, where $P_{i}$ is a Sylow $2$-subgroup of
$\operatorname{GO}_{2\cdot 2^{t_{i}}+1}(r)$, respectively,
$\operatorname{Sp}_{2\cdot 2^{t_{i}}}(r)$. In the case
$\hat{H}^{\ast}=\operatorname{GO}_{2n}^{\epsilon}(r)$, we have $\hat{P}$ is
either a Sylow $2$-subgroup of $\operatorname{GO}_{2n+1}(r)$, embedded as
before, or of the form $P_{0}\times C_{2}\times C_{2}$, where $P_{0}$ is a
Sylow $2$-subgroup of $\operatorname{GO}_{2n-1}(r)$. From here, we may argue
analogously to before, keeping in mind that when $p=2$, choosing $x\in{\bf
Z}(P_{1})$ to have $2^{t_{1}+1}$ eigenvalues $-1$ yields an element of
determinant 1, and hence an element of $\widetilde{P}=\hat{P}\cap H^{\ast}.$ ∎

Conjecture C for simple groups (and Theorem E) now follows from Propositions
4.1-4.7.

## References

  * [Al76] J. L. Alperin,  Isomorphic blocks.  J. Algebra 43 (1976), 694–698.
  * [BNOT07] C. Bessenrodt, G. Navarro, J. B. Olsson, Pham H. Tiep, On the Navarro-Willems conjecture for blocks of finite groups. _J. Pure Appl. Algebra 208_ (2007), 481–484.
  * [CE04] M. Cabanes and M. Enguehard, _Representation theory of finite reductive groups_ , New Mathematical Monographs 1, Cambridge University Press, Cambridge, 2004.
  * [Car85] R. W. Carter, _Finite groups of Lie type. Conjugacy classes and complex characters_ , Wiley and Sons, New York et al, 1985, 544 pp.
  * [CF64] R. Carter and P. Fong, The Sylow $2$-subgroups of the finite classical groups. _J. Algebra_ 1 (1964), 139–151.
  * [Da77] E. C. Dade,  Remarks on isomorphic blocks.  J. Algebra 45 (1977), 254–258.
  * [GAP] The GAP group, ‘GAP \- groups, algorithms, and programming’, Version 4.10.0, 2018, http://www.gap-system.org.
  * [Ge20] M. Geck, Green Functions and Glauberman Degree-divisibility. Annals of Mathematics 192, no. 1 (2020), 229–49.
  * [GMV19] E. Giannelli, G. Malle, and C. Vallejo, Even degree characters in principal blocks. _J. Pure Appl. Algebra_ 223 (2019), 900–907.
  * [GSV19] E. Giannelli, A. A. Schaeffer Fry, and C. Vallejo, Characters of $\pi^{\prime}$-degree, _Proc. Amer. Math. Soc._ 147:11 (2019), 4697–4712.
  * [Hi90] G. Hiss, Regular and semisimple blocks of finite reductive groups, _J. London Math. Soc._ 41 (1990), 63–68.
  * [IS76] I. M. Isaacs and S. D. Smith, ‘A note on groups of $p$-length 1’, J. Algebra, 38 (1976), 531–535.
  * [Is06] I. M. Isaacs, Character theory of finite groups, AMS Chelsea Publishing, Providence, RI, 2006.
  * [I08] I. M. Isaacs, ‘Finite Group Theory’, Grad. Stud. Math., vol.92, American Mathematical Society, Providence, 2008.
  * [LWXZ20] Y. Liu, W. Willems, H. Xiong, J. Zhang, Trivial intersection of blocks and nilpotent subgroups, _J. Algebra 559_ (2020) 510–528.
  * [Ma14] G. Malle, On the inductive Alperin-McKay and Alperin weight conjecture for groups with abelian Sylow subgroups, _J. Algebra_ 397 (2014), 190–208.
  * [MN20] G. Malle, G. Navarro, Brauer’s height zero conjecture for two primes, _Math. Z. 295_ (2020), no. 3-4, 1723–1732.
  * [MN21] G. Malle, G. Navarro, Brauer’s height zero conjecture for principal blocks, _Journal für die reine und angewandte Mathematik (Crelles Journal)_ 778 (2021), 119–125.
  * [MT11] G. Malle, D. Testerman, _Linear Algebraic Groups and Finite Groups of Lie Type_ , Cambridge Studies in Advanced Mathematics 133, Cambridge University Press, 2011.
  * [Mu94] M. Murai, Block induction, normal subgroups and characters of height zero, _Osaka J. Math_ 31, (1994) 9–25.
  * [Na98] G. Navarro, _Characters and Blocks of Finite Groups_. LMS Lecture Note Series 250, Cambridge University Press, 1998.
  * [Na03] G. Navarro, Number of Sylow subgroups in p-solvable groups. _Proc. Am. Math. Soc._ 131(10), (2003) 3019–3020.
  * [Na18] G. Navarro, Character Theory and the McKay Conjecture, Cambridge University Press, 2018.
  * [NRS21] G. Navarro, N. Rizo, A. A. Schaeffer Fry, Principal blocks for different primes I, submitted.
  * [NTT08] G. Navarro, P. H. Tiep, and A. Turull, Brauer characters with cyclotomic field of values, _J. Pure App. Algebra_ 212 (2008), 628–635.
  * [NWi97] G. Navarro, W. Willems When is a $p$-block a $q$-block? _Proc. Amer. Math. Soc. 125_ (1997), no. 6, 1589–1591.
  * [Ri19] N. Rizo, Divisibility of degrees in McKay correspondences. Arch. Math.  112 (2019), 5–11.
  * [RSV21] N. Rizo, A. A. Schaeffer Fry, and C. Vallejo, Principal blocks with 5 irreducible characters. _J. Algebra_ 585 (2021), 316–337.
  * [SFT22] A. A. Schaeffer Fry and J. Taylor, Galois automorphisms and classical groups. _Transformation Groups_ , To appear (2022).
  * [We55] A. J. Weir, Sylow $p$-subgroups of the classical groups over finite fields with characteristic prime to $p$. _Proc. Amer. Math. Soc._ 6:4 (1955), 529–533.