Brauer character degrees and Sylow normalizers

Abstract

If p and q are primes, and G is a p-solvable finite group, it is possible to detect that a q-Sylow normalizer is contained in a p-Sylow normalizer using the character table of G. This is characterized in terms of the degrees of p-Brauer characters. Some consequences, which include yet another generalization of the Itô–Michler theorem, are also obtained.

1 Introduction

If G is a finite group and $\pi$ is a set of primes, let $\mathrm{Irr}_{\pi }(G)$ be the subset of the irreducible complex characters $\chi$ of G such that all the primes dividing the degree $\chi (1)$ lie in $\pi$. It is fair to say that the interaction between $\mathrm{Irr}_{\pi }(G)$ and the structure of G is one of the recurrent problems in character theory. Recently, in [12], we have asked if it is possible to characterize group-theoretically when $\mathrm{Irr}_{\pi }(G)=\mathrm{Irr}_{\rho }(G)$ for sets of primes $\pi$ and $\rho$.

In this paper, we fix a prime p, and we turn our attention to p-Brauer characters, within the universe of finite p-solvable groups. (Degrees of modular representations in characteristic p outside p-solvable groups are usually deemed an intractable subject.) If G is a p-solvable finite group and $\mathrm{IBr}(G)$ is the set of the irreducible p-Brauer characters of G, we let $\mathrm{IBr}_{\pi }(G)$ be the subset of $\varphi \in \mathrm{IBr}(G)$ such that all the primes dividing $\varphi (1)$ lie in $\pi$. As usual, if r is a prime, $r'$ denotes the set of primes different from r.

The following is our first main result.

Theorem A

Let G be a finite p-solvable group, and let q be a prime. Then, $\mathrm{IBr}_{q'}(G) \subseteq \mathrm{IBr}_{p'}(G)$ if and only if there are $Q \in \mathrm{Syl}_q(G)$ and $P \in \mathrm{Syl}_p(G)$ such that $\mathbf{N}_{G}(Q) \subseteq \mathbf{N}_{G}(P)$.

Theorem A was the main result of [1], assuming that G is both p-solvable and q-solvable. To better understand our this new situation, we invite the reader to consider, for instance, $G=\mathrm{PSL}_2(3^5).5$, for $p=5$. If $q=2$, then $\mathbf{N}_{G}(P)=\mathbf{N}_{G}(Q)$ for some $p \in \mathrm{Syl}_p(G)$ and $Q\in \mathrm{Syl}_q(G)$. Of course, G is p-solvable, but not q-solvable. There are families of almost-simple groups like this, and we are able to deal with this new situation thanks to the main result of [6]. In particular, the proof of Theorem A depends on the Classification of Finite Simple Groups.

From Theorem A, and using McKay bijections for Brauer characters in p-solvable groups together with the recent proof of the divisibility of degrees between Glauberman correspondents by M. Geck in [2], we can prove our second main result.

Theorem B

Let G be a finite p-solvable group, and let q be a prime different from p. Then $\mathrm{IBr}_{p'}(G)=\mathrm{IBr}_{q'}(G)$ if and only if there are $P \in \mathrm{Syl}_p(G)$ and $Q \in \mathrm{Syl}_q(G)$ such that $\mathbf{N}_{G}(P)=P\mathbf{N}_{G}(Q)$ and Q is abelian.

As the reader can easily check, in the trivial case where p does not divide |G|, Theorem B is yet another restatement of the Itô–Michler theorem.

As happens with complex irreducible characters (see [12]), it does not seem easy to group-theoretically characterize when $\mathrm{IBr}_\pi (G)=\mathrm{IBr}_\rho (G)$ for arbitrary sets of primes $\pi$ and $\rho$, even if G is p-solvable. In the case, for instance, where $\pi =\mathbb P$ is the set of all primes and $\rho =q'$, where q is a prime, this constitutes Problem 3.2 of [9]. (This problem was studied long before in [8] and more recently in [5].) It has now been conjectured that $\mathrm{IBr}_{q'}(G)=\mathrm{IBr}_{\mathbb P}(G)$ if and only if the number of p-regular classes of G is the number of p-regular classes of $\mathbf{N}_{G}(Q)/Q'$, where $Q \in \mathrm{Syl}_q(G)$. This is a consequence of the Inductive McKay conjecture, and it seems difficult to obtain a direct proof. (See Conjecture D in [7].)

2 Proof of theorem A

Our notation for Brauer characters follows [9]. The deepest part of the proof of Theorem A comes from the main result in [MN].

Theorem 2.1

Let G be a finite $\pi$-separable group. Let H be a Hall $\pi$-subgroup, let K be a $\pi$-complement of G, and let q be a prime. Then, every $\alpha \in \mathrm{Irr}_{q'}(H)$ extends to G if and only if there is $Q \in \mathrm{Syl}_q(H)$ such that $\mathbf{N}_{G}(Q) \subseteq \mathbf{N}_{G}(K)$.

Proof

This is Theorem A of [6]. $\square$

A very useful result to deal with the hypotheses in this paper appears in Suzuki’s book.

Theorem 2.2

Let G be a p-solvable group, and let q be a prime. Let $\pi =\{p,q\}$. Then, G has a unique conjugacy class of Hall $\pi$-subgroups, and every $\pi$-subgroup of G is contained in one of them.

Proof

This follows from 5.3.13 of [14]. $\square$

For the reader’s convenience, let us prove the following standard result.

Lemma 2.3

Let G be a finite group. Let Q be a Sylow q-subgroup of G and let N be a normal subgroup of G. If $\varphi \in \mathrm{IBr}_{q'}(G)$, then $\varphi _N$ has a Q-invariant irreducible constituent and any two of them are $\mathbf{N}_{G}(Q)$-conjugate.

Proof

Let $\varphi \in \mathrm{IBr}_{q'}(G)$ and let $\theta \in \mathrm{IBr}(N)$ be an irreducible constituent of $\varphi _N$. Let $G_\theta$ be the stabilizer of $\theta$ in G. By Clifford’s theorem (see Corollary 8.9 of [9], for instance), we have that $|G:G_\theta |$ divides $\varphi (1)$, and hence, it is a $q'$-number. It follows that there is an element $g\in G$ such that $Q^g$ is contained in $G_\theta$. Hence, $Q\subseteq G_\theta ^{g^{-1}}=G_{\theta ^{g^{-1}}}$ and $\theta ^{g^{-1}}$ is a Q-invariant irreducible constituent of $\varphi _N$.

Now, suppose that $\theta$ and $\mu$ are two Q-invariant irreducible constituents of $\varphi _N$. Again by Clifford’s theorem we have that there is an element $g\in G$ such that $\mu =\theta ^g$. It follows that Q and $Q^g$ are contained in $G_\mu$, and hence, there is an element $x\in G_\mu$ such that $Q=(Q^g)^x$. Therefore $gx\in \mathbf{N}_{G}(Q)$ and $\theta ^{gx}=(\theta ^g)^x=\mu ^x=\mu$, as wanted. $\square$

The following result implies Theorem A. In its proof, we shall use Fong characters of Brauer characters, and we refer the reader to Chapter 10 of [9] for their main properties. (The term Fong character was coined by I. M. Isaacs after some results of P. Fong.) We also use the fact that if G is p-solvable, H is a p-complement of G, and $\varphi \in \mathrm{IBr}(G)$ has degree not divisible by p, then $\varphi _H \in \mathrm{Irr}(H)$. (See Theorem 10.9 of [9].) A complication when dealing with Brauer characters is that we do not have any form of Frobenius reciprocity, even in favorable conditions. For instance, in the previous situation where H is a p-complement of a p-solvable group G, if $\alpha \in \mathrm{Irr}(H)$, and $\varphi$ is an irreducible constituent of $p'$-degree of the induced Brauer character $\alpha ^G$, then $\alpha$ needs not be the irreducible character $\varphi _H$; a fact that would simplify our proof below. (If $G=\mathsf{A}_4$, $p=2$, and $H=\mathsf{C}_3$, then the Brauer character $(1_H)^G=21_G + \lambda _1 + \lambda _2$, where $\lambda _i$ are distinct linear Brauer characters. Now take $\alpha =1_H$ and $\varphi =\lambda _i$.)

Theorem 2.4

Let G be a p-solvable group, let $\mathrm{IBr}(G)$ be the set of irreducible Brauer characters of G, and let $q \ne p$ be a prime. Let $P \in \mathrm{Syl}_p(G)$ and $Q \in \mathrm{Syl}_q(G)$ such that $U=PQ$ is a subgroup of G. Suppose that H is a p-complement of G containing Q. Then, the following are equivalent.

• (a) $\mathrm{IBr}_{q'}(G) \subseteq \mathrm{IBr}_{p'}(G)$.

• (b) $\mathbf{N}_{G}(Q) \subseteq \mathbf{N}_{G}(P)$.

• (c) Every $\alpha \in \mathrm{Irr}_{q'}(H)$ extends to G.

Proof

Set $\pi =\{p,q\}$. By Theorem 2.2, notice that we can find $P \in \mathrm{Syl}_p(G)$ and $Q \in \mathrm{Syl}_q(G)$ such that $U=PQ$ is a Hall $\pi$-subgroup of G. (In fact, given a Hall $\pi$-subgroup U of G, then $U=PQ$ for every $P \in \mathrm{Syl}_p(U)$ and $Q\in \mathrm{Syl}_q(U)$.)

Assume (a). We prove (b) by induction on |G|. If K is a minimal normal subgroup of G, then we have that $\mathbf{N}_{G}(Q) \subseteq \mathbf{N}_{G}(P) K$, by using induction in G/K. Since G is p-solvable, then K is either a p-group or a $p'$-group. In the first case, $\mathbf{N}_{G}(P) K=\mathbf{N}_{G}(P)$ and we are done. So we assume that K is a $p'$-group.

Let V be any subgroup of G containing KQ and let $\delta \in \mathrm{IBr}_{q'}(V)$. By Lemma 2.3, there exists a Q-invariant irreducible constituent $\tau \in \mathrm{IBr}(K)$ of the restriction $\delta _K$ (using that $Q \in \mathrm{Syl}_q(V)$). We claim that $\tau$ is also P-invariant. By Corollary 8.7 of [9], we have that $\delta$ is an irreducible constituent of the induced Brauer character $\tau ^V$. Now consider the Brauer character $\delta ^G$, which has degree $|G:V|\delta (1)$, which is not divisible by q. Hence, there exists an irreducible constituent $\varphi \in \mathrm{IBr}(G)$ of $\delta ^G$ of degree not divisible by q. By hypothesis, we have that $\varphi$ has degree not divisible by p. Now, $\varphi$ is an irreducible constituent of $\tau ^G$, and therefore, $\tau$ is an irreducible constituent of the restriction $\varphi _K$, again by Corollary 8.7 of [9]. If $I=G_\tau$ is the stabilizer of $\tau$ in G, and $\mu \in \mathrm{IBr}(I)$ is the Clifford correspondent of $\varphi$ over $\tau$, we have that |G : I| is a $\pi '$-number. Therefore, using Theorem 2.2, we have that $U \subseteq I^g$ for some $g \in G$. Then, $\tau$ and $\tau ^g$ are Q-invariant constituents of $\varphi$, and thus, by Lemma 2.3, we have that $\tau ^g=\tau ^x$, for some $x \in \mathbf{N}_{G}(Q) \subseteq K\mathbf{N}_{G}(P)$. Hence, we may assume that $\tau ^g=\tau ^y$ for some $y \in \mathbf{N}_{G}(P)$. Then, $I^{gy^{-1}}=I$, $U^{y^{-1}} \subseteq I$, and we conclude that $P=P^{y^{-1}} \subseteq I$. In other words, $\tau$ is P-invariant, as claimed.

Now, let $V=\mathbf{N}_{G}(P) K$. We claim that $\mathrm{IBr}_{q'}(V) \subseteq \mathrm{IBr}_{p'}(V)$. Let $\delta \in \mathrm{IBr}_{q'}(V)$. By the claim in the previous paragraph, let $\tau \in \mathrm{IBr}(K)$ be PQ-invariant under $\delta$. By Theorem 8.11 of [9], there is a unique $\hat{\tau }\in \mathrm{IBr}(KP)$ over $\tau$, extending $\tau$, and we conclude that $\delta$ lies over $\hat{\tau }$. Now, $KP \triangleleft \,V$, V/KP is a $p'$-group, and we have that $\delta (1)/\hat{\tau }(1)$ divides |V : KP| by Theorem 8.30 of [9]. Since $\tau (1)$ is not divisible by p, we conclude that $\delta$ has $p'$-degree, as claimed. By induction, we may assume that $V=G$. That is, $KP \triangleleft \,G$. We want to use Theorem 2.1.

Next, we show that every $\alpha \in \mathrm{Irr}_{q'}(H)$ extends to G. By Lemma 2.3, let $\tau \in \mathrm{Irr}(K)$ be Q-invariant under $\alpha$. By the claim in the third paragraph, we have that $\tau$ is P-invariant too. By Corollary 6.28 of [3], we have that $\tau$ has a canonical extension $\gamma \in \mathrm{Irr}(KP)$. Using the uniqueness of $\gamma$, we easily check that $G_\gamma \cap H=H_\tau$. By Isaacs restriction theorem (Lemma 6.8(d) of [11]), the Clifford correspondence and Mackey’s theorem (Theorem 1.16 of [10]), we have that $\alpha$ extends to G. If $\pi _0=p'$ is the set of primes dividing |G| different from p, by Theorem 2.1 applied to $\pi _0$, we conclude that there is $Q_1 \in \mathrm{Syl}_q(H)$ such that $\mathbf{N}_{G}(Q_1) \subseteq \mathbf{N}_{G}(P)$. In particular, $P \triangleleft \,PQ_1$. By Theorem 2.2, $PQ_1$ and PQ are G-conjugate. Hence $P\triangleleft \,PQ$ and therefore $Q, Q_1 \in \mathrm{Syl}_q(\mathbf{N}_{G}(P))$. Hence, $Q_1^z=Q$ for some $z\in \mathbf{N}_{G}(P)$, and $\mathbf{N}_{G}(Q)=\mathbf{N}_{G}(Q_1)^z \subseteq \mathbf{N}_{G}(P)$.

We have that (b) implies (c), by Theorem 2.1 applied to $\pi _0=p'$.

Finally, we prove that (c) implies (a) by using Fong characters. Suppose now that every $\alpha \in \mathrm{Irr}_{q'}(H)$ extends to G. We show that $\mathrm{IBr}_{q'}(G) \subseteq \mathrm{IBr}_{p'}(G)$. Let $\varphi \in \mathrm{IBr}_{q'}(G)$, and let $\alpha \in \mathrm{Irr}(H)$ be an irreducible constituent of $\varphi _H$ such that $\alpha (1)=\varphi (1)_{p'}$ (using Theorem 10.18 of [9].) In other words, $\alpha$ is a Fong character for $\varphi$. Then, $\alpha (1)$ is not divisible by q, and by hypothesis, we have that $\alpha$ extends to some $\chi \in \mathrm{Irr}(G)$. Then, the Brauer character $\mu =\chi ^0 \in \mathrm{IBr}(G)$ extends $\alpha$. (Indeed, if $\chi ^0=\varphi _1+\varphi _2$ for Brauer characters $\varphi _i$ of G, then the irreducible character $\alpha$ would be written as $(\varphi _1)_H+(\varphi _2)_H$.) By Theorem 10.17 of [9], we have that $\varphi =\mu$, and $\varphi (1)=\mu (1)=\alpha (1)$ has degree not divisible by p. $\square$

3 Proof of theorem B

If N is a normal subgroup of G and $\theta \in \mathrm{IBr}(N)$, we write $\mathrm{IBr}(G\,|\,\theta )$ to denote the set of irreducible Brauer characters $\varphi$ of G such that $\theta$ is an irreducible constituent of $\varphi _N$. We write $\mathrm{IBr}_{p'}(G\,|\,\theta )=\mathrm{IBr}_{p'}(G)\cap \mathrm{IBr}(G\, |\, \theta )$.

Next is the version for Brauer characters and p-solvable groups of Theorem A of [13], which we shall need to prove Theorem B. It is worth mentioning that its proof uses the recent proof of the divisibility of the degrees of the Glauberman correspondence in [2].

Theorem 3.1

Let G be a p-solvable group, and $P\in \mathrm{Syl}_p(G)$, then there is a bijection $f: \mathrm{IBr}_{p'}(G)\rightarrow \mathrm{IBr}_{p'}(\mathbf{N}_{G}(P))$ such that $f(\varphi )(1)$ divides $\varphi (1)$ for all $\varphi \in \mathrm{IBr}_{p'}(G)$. Furthermore, $\varphi (1)/f(\varphi ) (1)$ divides $|G:\mathbf{N}_{G}(P)|$.

Proof

We argue by induction on |G|. Since $\mathbf{O}_{p}(G)$ is in the kernel of every $\varphi \in \mathrm{IBr}(G)$, by induction we may assume that $\mathbf{O}_{p}(G) =1$ and hence $K=\mathbf{O}_{p'}(G)>1$. Let $S/K=\mathbf{O}_{p}(G/K)$ and notice that $P_0=P\cap S$ is a Sylow p-subgroup of S. By the Frattini argument, we have that $G=K\mathbf{N}_{G}(P_0)$. Notice also that $\mathbf{N}_{G}( P_0)<G$ since $\mathbf{O}_{p}(G) =1$.

Let $\theta _1, \ldots , \theta _s$ be a complete set of representatives of the orbits of the action of $\mathbf{N}_{G}(P)$ on the P-invariant irreducible characters of K. By Lemma 2.3, we have that

$$\begin{aligned} \mathrm{IBr}_{p'}(G) = \ \mathrm{IBr}_{p'}( {G\,|\,\theta _1}) \cup \cdots \cup \mathrm{IBr}_{p'}( {G\,|\,\theta _s}) \end{aligned}$$

is a disjoint union. Fix $\theta _i \in \mathrm{Irr}(K)$, P-invariant, and observe that $\theta _i$ is also $P_0$-invariant. Let $\theta _i^*\in \mathrm{Irr}(\mathbf{C}_{K}(P_0))= \mathrm{IBr}(\mathbf{C}_{K}(P_0))$ be the Glauberman correspondent of $\theta _i$ and let $T_i=G_{\theta _i}$ be the stabilizer of $\theta _i$ in G. Since the Glauberman correspondence and the action of $\mathbf{N}_{G}(P_0)$ commute (see Lemma 2.10 of [10]), it follows that $\mathbf{N}_{T_i}(P_0) =T_i\cap \mathbf{N}_{G}(P_0)$ is the stabilizer of $\theta _i^*$ in $\mathbf{N}_{G}(P_0)$. By Dade’s theorem (see Theorem (6.5) of [15]), we have that $(T_i, K, \theta _i)$ and $(\mathbf{N}_{T_i}(P_0), \mathbf{C}_{K}(P_0), \theta _i^*)$ are isomorphic character triples. In particular, there is a bijection $\Delta :\mathrm{Irr}(T_i|\theta _i)\rightarrow \mathrm{Irr}(\mathbf{N}_{T_i}(P_0)|\theta _i^*)$ such that $\chi (1)/\theta _i(1)=\Delta (\chi )(1)/\theta _i^*(1)$ for all $\chi \in \mathrm{Irr}(T_i|\theta _i)$. Now using $\Delta$ and Lemma 3.12 of [4] we can construct a bijection $^*: \mathrm{IBr}(T_i\, |\, \theta _i) \rightarrow \mathrm{IBr}( \mathbf{N}_{T_i}(P_0) \, |\, \theta _i^*)$ such that $\psi (1)/\theta _i(1)= \psi ^*(1)/\theta _i^*(1)$ for all $\psi \in \mathrm{IBr}(T_i\, |\, \theta _i)$, and since $\theta _i(1)$ and $\theta _i^*(1)$ are $p'$-numbers we have that $^*: \mathrm{IBr}_{p'}({T_i\, |\, \theta _i}) \rightarrow \mathrm{IBr}_{p'}( { \mathbf{N}_{T_i}(P_0) \, |\, \theta _i^*})$ is a bijection.

Let $\chi \in \mathrm{IBr}_{p'}(G)$ and let $\theta _i\in \mathrm{IBr}(K) =\mathrm{Irr}(K)$ be such that $\chi \in \mathrm{IBr}_{p'}( {G\,|\,\theta _i})$. By the Clifford correspondence (Theorem 8.9 of [9]), we have that there is $\psi \in \mathrm{IBr}_{p'}(T_i\, |\, \theta _i)$ such that $\psi ^G=\chi$. Since $\mathbf{N}_{T_i}(P_0)$ is the stabilizer of $\theta _i^*$ in $\mathbf{N}_{G}(P_0)$, again by the Clifford correspondence we have that $(\psi ^*)^{\mathbf{N}_{G}(P_0)}\in \mathrm{Irr}(\mathbf{N}_{G}(P_0))$. Furthermore, since $P\subseteq \mathbf{N}_{T_i}(P_0)$, we have that

$$\begin{aligned} (\psi ^*)^{\mathbf{N}_{G}(P_0)}(1)=|\mathbf{N}_{G}(P_0) : \mathbf{N}_{T_i}(P_0)| \psi ^*(1) \end{aligned}$$

is a $p'$-number. Hence, we define

$$\begin{aligned} g: \mathrm{IBr}_{p'}(G) \rightarrow \mathrm{IBr}_{p'}({\mathbf{N}_{G}(P_0)}) \end{aligned}$$

by $g(\chi )=(\psi ^*)^{\mathbf{N}_{G}(P_0)}$.

Since $\theta _1, \ldots , \theta _s$ is a complete set of representatives of the action of $\mathbf{N}_{G}(P)$ on the P-invariant characters of K and the Glauberman correspondence commutes with the action of $\mathbf{N}_{G}(P)\subseteq \mathbf{N}_{G}(P_0)$, we have that $\theta _1^*, \ldots , \theta _s^*$ is a complete set of representatives of the action of $\mathbf{N}_{G}(P)$ on the P-invariant irreducible characters of $\mathrm{Irr}(\mathbf{C}_{K}(P_0))=\mathrm{IBr}(\mathbf{C}_{K}(P_0))$. By Lemma 2.3, we have that

$$\begin{aligned} \mathrm{IBr}_{p'}({\mathbf{N}_{G}(P_0)})= \mathrm{IBr}_{p'}( {\mathbf{N}_{G}(P_0)\, |\,\theta ^*_1})\cup \cdots \cup \mathrm{IBr}_{p'}( {\mathbf{N}_{G}(P_0)\, |\,\theta ^*_s}) \end{aligned}$$

is a disjoint union and it follows that g is a bijection.

Let $\chi \in \mathrm{IBr}_{p'}(G|\theta _i)$ and let $\psi \in \mathrm{IBr}_{p'}(T_i|\theta _i)$ with $\psi ^G=\chi$. Then $\psi (1) = \psi ^*(1)\theta _i(1)/\theta ^*_i(1)$ and, since $\theta ^*_i(1)$ divides $\theta _i(1)$ (by the recent main theorem of [2]), we have that $\psi ^*(1)$ divides $\psi (1)$. Since $G=\mathbf{N}_{G}(P_0) K$ we have that $g(\chi )(1)= |\mathbf{N}_{G}(P_0) :\mathbf{N}_{T_i}(P_0) |\psi ^*(1)$ divides $|G:T_i|\psi (1)=\chi (1)$.

Finally, since $\mathbf{N}_{G}(P_0)<G$, we apply induction to obtain a bijection

$$\begin{aligned} h:\mathrm{IBr}_{p'}(\mathbf{N}_{G}(P_0))\rightarrow \mathrm{IBr}_{p'}(\mathbf{N}_{G}(P)) \end{aligned}$$

such that $h(\psi )(1)$ divides $\psi (1)$ and $\chi (1)/h(\chi )(1)$ divides $|\mathbf{N}_{G}(P_0): \mathbf{N}_{G}(P)|$ for all $\psi \in \mathrm{IBr}_{p'}(\mathbf{N}_{G}(P_0))$. Let $f=gh=h\circ g$. Clearly, f is a bijection and $f(\chi )(1)$ divides $\chi (1)$ for all $\chi \in \mathrm{IBr}_{p'}(G)$. Now, since $g(\chi )(1)/h(g(\chi ))(1)$ divides $|\mathbf{N}_{G}(P_0): \mathbf{N}_{G}(P)|$, we have that $\chi (1)/f(\chi )(1)$ divides $|\mathbf{N}_{G}(P_0): \mathbf{N}_{G}(P)|\psi (1)/\psi ^*(1)=|\mathbf{N}_{G}(P_0): \mathbf{N}_{G}(P)|\theta _i(1)/\theta _i^*(1)$. By Problem 13.2 of [3], $\theta _i(1)/\theta _i^*(1)$ divides $|K:\mathbf{C}_{K}(P_0)|=|G:\mathbf{N}_{G}(P_0)|$. Hence, $\chi (1)/f(\chi )(1)$ divides $|G:\mathbf{N}_{G}(P)|$. $\square$

The following is Theorem B.

Theorem 3.2

Suppose that G is a p-solvable finite group and let q be a prime different from p. Then

$$\begin{aligned} \mathrm{IBr}_{p'}(G)=\mathrm{IBr}_{q'}(G) \, \end{aligned}$$

if and only if there is a Sylow p-subgroup P of G and a Sylow q-subgroup Q of G, such that $\mathbf{N}_{G}(P) =P\mathbf{N}_{G}(Q)$ and Q is abelian.

Proof

Suppose that $\mathrm{IBr}_{q'}(G)=\mathrm{IBr}_{p'}(G)$. By Theorem 2.4, we have that there is a Sylow p-subgroup of G and a Sylow q-subgroup of G such that $\mathbf{N}_{G}(Q)\subseteq \mathbf{N}_{G}(P)$. We claim that

$$\begin{aligned} \mathrm{IBr}_{p'}(\mathbf{N}_{G}(P))=\mathrm{IBr}_{q'}(\mathbf{N}_{G}(P)) \, . \end{aligned}$$

First, we notice that Theorem 2.4 applied to $\mathbf{N}_{G}(P)$ shows $\mathrm{IBr}_{q'}(\mathbf{N}_{G}(P) )\subseteq \mathrm{IBr}_{p'}(\mathbf{N}_{G}(P))$. Let $\mu \in \mathrm{IBr}_{p'}(\mathbf{N}_{G}(P))$. By Theorem 3.1, there is $\varphi \in \mathrm{IBr}_{p'}(G)$ such that $\mu (1)$ divides $\varphi (1)$ which is a $q'$-number. Hence, we conclude $\mu \in \mathrm{IBr}_{q'}(\mathbf{N}_{G}(P))$. Then $\mathrm{IBr}_{q'}(\mathbf{N}_{G}(P) )= \mathrm{IBr}_{p'}(\mathbf{N}_{G}(P))$, as claimed.

Therefore, we may assume, arguing by induction that $P\triangleleft \,G$. Then, we have that $\mathrm{IBr}(G)=\mathrm{IBr}(G/P)=\mathrm{Irr}(G/P)$ and

$$\begin{aligned} \mathrm{Irr}_{q'}(G/P)=\mathrm{IBr}_{q'}(G)=\mathrm{IBr}_{p'}(G)=\mathrm{Irr}_{p'}(G/P)=\mathrm{Irr}(G/P). \end{aligned}$$

By the Itô-Michler theorem, we know that G/P has a normal and abelian Sylow q-subgroup PQ/Q. Hence, Q is abelian and $PQ\triangleleft \,G$. Then $G=P\mathbf{N}_{G}(Q)$ by the Frattini’s argument.

Conversely, suppose that there is $P\in \mathrm{Syl}_p(G)$ and $Q\in \mathrm{Syl}_q(G)$ such that $\mathbf{N}_{G}(P) =P\mathbf{N}_{G}(Q)$ and Q is abelian. Notice that since $\mathbf{N}_{G}(Q) \subseteq \mathbf{N}_{G}(P)$ by Theorem 2.4 we have that $\mathrm{IBr}_{q'}(G)\subseteq \mathrm{IBr}_{p'}(G)$. We only need to prove the reverse containment. If $\mathbf{N}_{G}(P)<G$, arguing by induction, we have that $\mathrm{IBr}_{q'}(\mathbf{N}_{G}(P) )= \mathrm{IBr}_{p'}(\mathbf{N}_{G}(P))$. Let $\varphi \in \mathrm{IBr}_{p'}(G)$ and let $f:\mathrm{IBr}_{p'}(G)\rightarrow \mathrm{IBr}_{p'}(\mathbf{N}_{G}(P))$ be the bijection given in Theorem 3.1. Hence $\varphi (1)/ f(\varphi )(1)$ divides $|G:\mathbf{N}_{G}(P)|$ which is not divisible by q. Since $\mathrm{IBr}_{q'}(\mathbf{N}_{G}(P) )= \mathrm{IBr}_{p'}(\mathbf{N}_{G}(P))$, we have that $f(\varphi )(1)$ is not divisible by q and thus $\varphi \in \mathrm{IBr}_{q'}(G)$.

Hence, we may assume that P is a normal subgroup of G and then $\mathrm{IBr}(G)=\mathrm{Irr}(G/P)$. Also $G=\mathbf{N}_{G}(Q) P$ and hence PQ is normal in G. Then, PQ/P is an abelian normal Sylow q-subgroup of G/P. By Itô’s theorem, we have that $\mathrm{Irr}(G/P)=\mathrm{Irr}_{q'}(G/P)$. It follows that

$$\begin{aligned} \mathrm{IBr}_{q'}(G)=\mathrm{Irr}_{q'}(G/P)=\mathrm{Irr}(G/P)=\mathrm{IBr}_{p'}(G), \end{aligned}$$

and we are done. $\square$

To prove the assertion in the Abstract, it is enough to notice that in p-solvable groups, the ordinary character table of G uniquely determines the Brauer characters of G, by using the Fong–Swan theorem. (See Theorem 10.1 and Corollary 10.4 of [9].)

Finally, using Isaacs $\pi$-characters, the Glauberman–Isaacs correspondence, and some ad hoc arguments, it is possible to replace in Theorems A and B of this paper $p'$ by $\pi$, p-solvable groups by $\pi$-separable groups, and Sylow p-subgroups by Hall $\pi$-complements.