Weights and Nilpotent Subgroups

TL;DR

This paper proves a pi-version of the Alperin Weight Conjecture for certain pi-separable groups, extending previous results and advancing understanding of weights in finite group theory.

Contribution

It introduces a pi-version of the Alperin Weight Conjecture applicable to pi-separable groups, generalizing earlier work by Isaacs and the first author.

Findings

Established a pi-version of the Alperin Weight Conjecture for specific groups

Generalized previous results by Isaacs and the first author

Enhanced understanding of nilpotent weights in finite groups

Abstract

In a finite group G, we consider nilpotent weights, and prove a pi-version of the Alperin Weight Conjecture for certain pi-separable groups. This widely generalizes an earlier result by I. M. Isaacs and the first author.

Equations

$$\Bigl{|}\bigcup_{Q\in{\mathcal{Q}}}{\bf I}_{\pi}(G|Q)\Bigr{|}=|{\bf I}_{\pi}({\bf N}_{G}(R)|R)|\,.$$

$$\phi (1{)}_{{\pi }^{\prime }}=|G:Q{|}_{{\pi }^{\prime }}.$$

$${\mathrm{Irr}}_S(G)\to \mathrm{Irr}(C),\chi \mapsto {\chi }^{\ast },$$

$${\mathbf{I}}_{\pi }(G|Q,\tau )={\mathbf{I}}_{\pi }(G|Q)\cap {\mathbf{I}}_{\pi }(G|\tau ).$$

$$|{\mathbf{I}}_{\pi }(G|Q,\tau )|=\sum _{U\in \mathcal{U}}|{\mathbf{I}}_{\pi }(G_{\tau }|U,\tau )|.$$

$$|{\mathbf{I}}_{\pi }(G|Q,\tau )|=|{\mathbf{I}}_{\pi }(G_{\tau }|Q,\tau )|.$$

$$\bigcup _{U\in \mathcal{U}}{\mathbf{I}}_{\pi }(G_{\tau }|U)$$

$${\mathbf{I}}_{\pi }(G|Q,\phi )=\bigcup _{\tau \in \mathcal{A}}{\mathbf{I}}_{\pi }(G|Q,\tau )$$

$${\mathbf{I}}_{\pi }({\mathbf{N}}_G(Q)|Q,\phi )=\bigcup _{\tau \in \mathcal{A}}{\mathbf{I}}_{\pi }({\mathbf{N}}_G(Q)|Q,{\tau }^{\ast })$$

$$|{\mathbf{I}}_{\pi }({\mathbf{N}}_G(Q)|Q,\phi )|=\sum _{\tau \in \mathcal{A}}|{\mathbf{I}}_{\pi }({\mathbf{N}}_{G_{\tau }}(Q)|Q,{\tau }^{\ast })|.$$

$$|{\mathbf{I}}_{\pi }(G_{\tau }|Q,\tau )|=|{\mathbf{I}}_{\pi }({\mathbf{N}}_{G_{\tau }}(Q)|Q,{\tau }^{\ast })|.$$

$${}^{\ast }:{\mathbf{I}}_{\pi }(G|\tau )\to {\mathbf{I}}_{\pi }({\mathbf{N}}_G(Q)|{\tau }^{\ast })$$

$$|{\mathbf{I}}_{\pi }(G|Q,\tau )|=|{\mathbf{I}}_{\pi }({\mathbf{N}}_G(Q)|Q,{\tau }^{\ast })|,$$

$$\Bigl{|}\bigcup_{Q\in{\mathcal{Q}}}{\bf I}_{\pi}(G|Q,\varphi)\Bigr{|}=|{\bf I}_{\pi}(M{\bf N}_{G}(R)|R,\varphi)|\,.$$

$$|{\mathbf{I}}_{\pi }(G|Q,\phi )|=|{\mathbf{I}}_{\pi }(G/K|Q/K,\hat{\phi }),$$

$$|{\mathbf{I}}_{\pi }({\mathbf{N}}_G(R)|R,\phi )|=|{\mathbf{I}}_{\pi }({\mathbf{N}}_G(R)K/K|RK/K,\hat{\phi })|=|{\mathbf{I}}_{\pi }({\mathbf{N}}_G(RK)/K|RK/K,\hat{\phi })|.$$

$$\Bigl{|}\bigcup_{Q\in{\mathcal{Q}}_{\tau}}{\bf I}_{\pi}(G_{\tau}|Q,\tau)\Bigr{|}=|{\bf I}_{\pi}(L{\bf N}_{G_{\tau}}(R)|R,\tau)|\,.$$

$$|{\mathbf{I}}_{\pi }(L{\mathbf{N}}_{G_{\tau }}(R)|R,\tau )|=|{\mathbf{I}}_{\pi }(L{\mathbf{N}}_G(R)|R,\tau )|.$$

$$|{\mathbf{I}}_{\pi }(L{\mathbf{N}}_G(R)|R,\phi )|=\sum _{\tau \in \mathcal{A}}|{\mathbf{I}}_{\pi }(L{\mathbf{N}}_G(R)|R,\tau )|,$$

$$|{\mathbf{I}}_{\pi }(L{\mathbf{N}}_G(R)|R,\phi )|=|{\mathbf{I}}_{\pi }({\mathbf{N}}_G(R)|R,\phi )|.$$

$$\sum_{\tau\in{{\mathcal{A}}}}\Bigl{|}\bigcup_{Q\in{\mathcal{Q}}_{\tau}}{\bf I}_{\pi}(G_{\tau}|Q,\tau)\Bigr{|}=|{\bf I}_{\pi}({\bf N}_{G}(R)|R,\varphi)|\,.$$

$$\Bigl{|}\bigcup_{Q\in{\mathcal{Q}}}{\bf I}_{\pi}(G|Q,\varphi)\Bigr{|}=\sum_{\tau\in{{\mathcal{A}}}}\Bigl{|}\bigcup_{Q\in{\mathcal{Q}}_{\tau}}{\bf I}_{\pi}(G_{\tau}|Q,\tau)\Bigr{|}\,.$$

$$\bigcup _{Q\in \mathcal{Q}}{\mathbf{I}}_{\pi }(G|Q,\phi )=\bigcup _{Q\in \mathcal{R}}{\mathbf{I}}_{\pi }(G|Q,\phi )$$

$${ϵ}^{y^{-1}}=({\tau }^{gt}{)}^{y^{-1}}={\tau }^{x^{-1}xgt{y}^{-1}}={\tau }^{x^{-1}qv}={\tau }^{x^{-1}v}.$$

$$f:\bigcup _{Q\in \mathcal{R}}{\mathbf{I}}_{\pi }(G|Q,\phi )\to \bigcup _{\tau \in \mathcal{A}}\left(\left(\bigcup _{Q\in {\mathcal{Q}}_{\tau }}{\mathbf{I}}_{\pi }(G_{\tau }|Q,\tau )\right)\times \{\tau \}\right)$$

$$Q=Q(N\cap H)=QN\cap H◃N{\mathbf{N}}_G(Q)\cap H\le {\mathbf{N}}_H(Q).$$

$$\bigcup _{Q\in \mathcal{Q}}{\mathbf{I}}_{\pi }(G|Q)\to {\mathbf{I}}_{\pi }({\mathbf{N}}_G(R)|R).$$

Full text

The research of the first author is supported by MTM2016-76196-P and Prometeo/Generalitat Valenciana. The second author thanks the German Research Foundation (projects SA 2864/1-1 and SA 2864/3-1).

1. Introduction

Let $G$ be a finite group and let $p$ be a prime. The celebrated Alperin Weight Conjecture asserts that the number of conjugacy classes of $G$ consisting of elements of order not divisible by $p$ is exactly the number of $G$-conjugacy classes of $p$-weights. Recall that a $p$-weight is a pair $(Q,\gamma)$, where $Q$ is a $p$-subgroup of $G$ and $\gamma\in{\rm Irr}({\bf N}_{G}(Q)/Q)$ is an irreducible complex character with $p$-defect zero (that is, such that the $p$-part $\gamma(1)_{p}=|{\bf N}_{G}(Q)/Q|_{p}$).

In the main result of this paper, we replace $p$ by a set of primes $\pi$ as follows:

Theorem A**.**

Let $G$ be a $\pi$-separable group with a solvable Hall $\pi$-subgroup. Then the number of conjugacy classes of $\pi^{\prime}$-elements of $G$ is the number of $G$-conjugacy classes of pairs $(Q,\gamma)$, where $Q$ is a nilpotent $\pi$-subgroup of $G$ and $\gamma\in{\rm Irr}({\bf N}_{G}(Q)/Q)$ has $p$-defect zero for every $p\in\pi$.

Recall that a finite group is called $\pi$-separable if all its composition factors are $\pi$-groups or $\pi^{\prime}$-groups. Let us restate Theorem A in the (presumably trivial) case where $G$ itself is a (solvable) $\pi$-group. In this case, there is only one conjugacy class of $\pi^{\prime}$-elements of $G$. On the other hand, if $Q$ is a nilpotent subgroup of $G$, then $\gamma\in{\rm Irr}({\bf N}_{G}(Q)/Q)$ has $p$-defect zero for every $p\in\pi$ if and only if ${\bf N}_{G}(Q)=Q$. Amazingly enough, there is only one conjugacy class of self-normalizing nilpotent subgroups: the Carter subgroups of $G$ (see p. 281 in [R]).

Of course, if $\pi=\{p\}$, then Theorem A is the $p$-solvable case of the Alperin Weight Conjecture (AWC). As a matter of fact, AWC was proven for $\pi$-separable groups with a nilpotent Hall $\pi$-subgroup by Isaacs and the first author [IN]. Now we realize that the nilpotency hypothesis can be dropped if one counts nilpotent weights instead. The solvability hypothesis is still needed, as shown by $G={\sf A}_{5}$ and $\pi=\{2,3,5\}$.

There is a price to pay, however. The proof in [IN] relied on the so called Okuyama–Wajima argument, a definitely non-trivial but accessible tool on extensions of Glauberman correspondents. In order to prove Theorem A, however, we shall need to appeal to a deeper theorem of Dade and Puig (which uses Dade’s classification of the endo-permutation modules).

As it is often the case in a “solvable” framework, the equality of cardinalities in Theorem A has a hidden structure which we are going to explain now. For sake of convenience we interchange from now on the roles of $\pi$ and $\pi^{\prime}$ (of course, $\pi$-separable is equivalent to $\pi^{\prime}$-separable). Recall that in a $\pi$-separable group $G$, the set ${\bf I}_{\pi}(G)$ of irreducible $\pi$-partial characters of $G$ is the exact $\pi$-version (when $\pi$ is the complement of a prime $p$) of the irreducible Brauer characters ${\rm IBr}(G)$ of a $p$-solvable group (see next section for precise definitions). Each $\varphi\in{\bf I}_{\pi}(G)$ has canonically associated a $G$-conjugacy class of $\pi^{\prime}$-subgroups $Q$, which are called the vertices of $\varphi$. If ${\bf I}_{\pi}(G|Q)$ is the set of irreducible $\pi$-partial characters with vertex $Q$, unless $\pi=p^{\prime}$, it is not in general true that $|{\bf I}_{\pi}(G|Q)|=|{\bf I}_{\pi}({\bf N}_{G}(Q)|Q)|$. Instead we will prove the following theorem.

Theorem B**.**

Suppose that $G$ is $\pi$-separable with a solvable Hall $\pi$-complement. Let $R$ be a nilpotent $\pi^{\prime}$-subgroup of $G$ and let $\mathcal{Q}$ be the set of $\pi^{\prime}$-subgroups $Q$ of $G$ such that $R$ is a Carter subgroup of $Q$. Then

[TABLE]

Since $|{\bf I}_{\pi}({\bf N}_{G}(R)|R)|$ is just the number of $\pi^{\prime}$-weights with first component $R$ (see Lemma 6.28 of [I2]), Theorem B implies Theorem A.

As happens in the classical case where $\pi=p^{\prime}$, and following the ideas of Dade, Knörr and Robinson, one can define chains of $\pi^{\prime}$-subgroups and relate them with $\pi$-defect of characters. This shall be explored elsewhere. Similarly, one can attach every weight to a $\pi^{\prime}$-block $B$ of $G$ by using Slattery’s theory [S]. In this setting we expect that the number of $\pi$-partial characters belonging to $B$ equals the number of nilpotent weights attached to $B$.

The groups described in Theorem A are sometimes called $\pi$-solvable. We did not find a counterexample in the wider class of so-called $\pi$-selected groups. Here, $\pi$-selected means that the order of every composition factor is divisible by at most one prime in $\pi$. P. Hall [H] has shown that these groups still have solvable Hall $\pi$-subgroups. Since every finite group is $p$-selected for every prime $p$, this version of the conjecture includes AWC in full generality.

Unfortunately, Theorem A does not hold for arbitrary groups even if they possess nilpotent Hall $\pi$-subgroups. It is not so easy to find a counterexample, though. The fourth Janko group $G=J_{4}$ has a cyclic Hall $\pi$-subgroup of order $35$ (that is, $\pi=\{5,7\}$). The normalizers of the non-trivial $\pi$-weights are contained in a maximal subgroup $M$ of type $2^{3+12}.(S_{5}\times L_{3}(2))$. However, $l(G)-k_{0}(G)=25\neq 30=l(M)-k_{0}(M)$, where $l(G)$ denotes the number of $\pi^{\prime}$-conjugacy classes and $k_{0}(G)$ is the number of $\pi$-defect zero characters of $G$. (The fact that $J_{4}$ was a counterexample for the $\pi$-version of the McKay conjecture for groups with a nilpotent Hall $\pi$-subgroup was noticed by Pham H. Tiep and the first author.)

We take the opportunity to thank the developers of [GAP]. Without their tremendous work the present paper probably would not exist.

The paper is organized as follows: In the next section we review $\pi$-partial characters which were introduced by Isaacs. In Section 3 we present two general lemmas on characters in $\pi$-separable groups. Afterwards we prove Theorem B. In the final section we construct a natural bijection explaining Theorem B in the presence of a normal Hall $\pi$-subgroup.

2. Review of $\pi$-theory

Isaacs’ $\pi$-theory is the $\pi$-version in $\pi$-separable groups of the $p$-modular representation theory for $p$-solvable groups. When $\pi=p^{\prime}$, the complement of a prime, then ${\bf I}_{\pi}(G)={\rm IBr}(G)$ and we recover most of the well-known classical results. In what follows $G$ is a finite $\pi$-separable group, where $\pi$ is a set of primes. All the references for $\pi$-theory can now be found together in Isaacs’ recent book [I2]. For the reader’s convenience, we review some of the main features. If $n$ is a natural number and $p$ is a prime, recall that $n_{p}$ is the largest power of $p$ dividing $n$. If $\pi$ is a set of primes, then $n_{\pi}=\prod_{p\in\pi}n_{p}$. The number $n$ is a $\pi$-number if $n=n_{\pi}$.

If $G$ is a $\pi$-separable group, then $G^{0}$ is the set of elements of $G$ whose order is a $\pi$-number. A $\pi$-partial character of $G$ is the restriction of a complex character of $G$ to $G^{0}$. A $\pi$-partial character is irreducible if it is not the sum of two $\pi$-partial characters. We write ${\bf I}_{\pi}(G)$ for the set of irreducible $\pi$-partial characters of $G$. Notice that if $\mu\in{\bf I}_{\pi}(G)$ by definition there exists $\chi\in{\rm Irr}(G)$ such that $\chi^{0}=\mu$, where $\chi^{0}$ denotes the restriction of $\chi$ to the $\pi$-elements of $G$. Also, it is clear by the definition, that every $\pi$-partial character is a sum of irreducible $\pi$-partial characters. Notice that if $G$ is a $\pi$-group, then ${\bf I}_{\pi}(G)={\rm Irr}(G)$.

Theorem 2.1** (Isaacs).**

Let $G$ be a finite $\pi$-separable group. Then ${\bf I}_{\pi}(G)$ is a basis of the space of class functions defined on $G^{0}$. In particular, $|{\bf I}_{\pi}(G)|$ is the number of conjugacy classes of $\pi$-elements of $G$.

Proof.

This is Theorem 3.3 of [I2]. ∎

We can induce and restrict $\pi$-partial characters in a natural way. If $H$ is a subgroup of $G$ and $\varphi\in{\bf I}_{\pi}(G)$, then $\varphi_{H}=\sum_{\mu\in{\bf I}_{\pi}(H)}a_{\mu}\mu$ for some uniquely defined nonnegative integers $a_{\mu}$. We write ${\bf I}_{\pi}(G|\mu)$ to denote the set of $\varphi\in{\bf I}_{\pi}(G)$ such that $a_{\mu}\neq 0$.

A non-trivial result is that Clifford’s theory holds for $\pi$-partial characters. If $N\triangleleft\,G$, it is then clear that $G$ naturally acts on ${\bf I}_{\pi}(N)$ by conjugation.

Theorem 2.2** (Isaacs).**

Suppose that $G$ is $\pi$-separable and $N\triangleleft\,G$.

1. (a)

If $\varphi\in{\bf I}_{\pi}(G)$, then $\varphi_{N}=e(\theta_{1}+\cdots+\theta_{t})$, where $\theta_{1},\ldots,\theta_{t}$ are all the $G$-conjugates of some $\theta\in{\bf I}_{\pi}(N)$. 2. (b)

If $\theta\in{\bf I}_{\pi}(N)$ and $T=G_{\theta}$ is the stabilizer of $\theta$ in $G$, then induction defines a bijection ${\bf I}_{\pi}(T|\theta)\rightarrow{\bf I}_{\pi}(G|\theta)$.

Proof.

See Corollary 5.7 and Theorem 5.11 of [I2]. ∎

In part (b) of Theorem 2.2, if $\mu^{G}=\varphi$, where $\mu\in{\bf I}_{\pi}(T|\theta)$, then $\mu$ is called the Clifford correspondent of $\varphi$ over $\theta$, and sometimes it is written $\mu=\varphi_{\theta}$.

It is not a triviality to define vertices for $\pi$-partial characters (a concept that in classical modular representation theory has little to do with character theory). This was first accomplished in [IN] (generalizing a result of Huppert on Brauer characters of $p$-solvable groups).

Theorem 2.3**.**

Suppose that $G$ is $\pi$-separable, and let $\varphi\in{\bf I}_{\pi}(G)$. Then there exist a subgroup $U$ of $G$ and $\alpha\in{\bf I}_{\pi}(U)$ of $\pi$-degree such that $\alpha^{G}=\varphi$. Furthermore, if $Q$ is a Hall $\pi$-complement of $U$, then the $G$-conjugacy class of $Q$ is uniquely determined by $\varphi$.

Proof.

This is Theorem 5.17 of [I2]. ∎

The uniquely defined $G$-class of $\pi^{\prime}$-subgroups $Q$ associated to $\varphi$ by Theorem 2.3 is called the set of vertices of $\varphi$. If $Q$ is a $\pi^{\prime}$-subgroup of $G$, then we write ${\bf I}_{\pi}(G|Q)$ to denote the set of $\varphi\in{\bf I}_{\pi}(G)$ which have $Q$ as a vertex. By definition, notice in this case that

[TABLE]

Our last important ingredient is the Glauberman correspondence.

Theorem 2.4** (Glauberman).**

Let $S$ be a finite solvable group acting via automorphisms on a finite group $G$ such that $(|S|,|G|)=1$. Then there exists a canonical bijection, called the $S$-Glauberman correspondence,

[TABLE]

where $\operatorname{Irr}_{S}(G)$ is the set of $S$-invariant irreducible characters of $G$ and $C={\bf C}_{G}(S)$. Here, $\chi^{*}$ is a constituent of the restriction $\chi_{C}$. Also, if $T\triangleleft\,S$, then the $T$-Glauberman correspondence is an isomorphism of $S$-sets.

Proof.

See Theorem 13.1 of [I1]. ∎

3. Preliminaries

If $G$ is a finite group, $\pi$ is a set of primes, and $\chi\in{\rm Irr}(G)$, then we say that $\chi$ has $\pi$-defect zero if $\chi(1)_{\pi}=|G|_{\pi}$.

Lemma 3.1**.**

If $\chi\in{\rm Irr}(G)$ has $\pi$-defect zero, then ${\bf O}_{\pi}(G)=1$.

Proof.

Let $N\triangleleft\,G$, and let $\theta\in{\rm Irr}(N)$ be under $\chi$. Then we have that $\theta(1)$ divides $|N|$ and $\chi(1)/\theta(1)$ divides $|G:N|$ by Corollary 11.29 of [I1]. Thus $\chi(1)_{\pi}=|G|_{\pi}$ if and only if $\theta(1)_{\pi}=|N|_{\pi}$ and $(\chi(1)/\theta(1))_{\pi}=|G:N|_{\pi}$. The result is now clear applying this to $N={\bf O}_{\pi}(G)$. ∎

We shall use the following notation. Suppose $G$ is $\pi$-separable, $N\triangleleft\,G$, $\tau\in{\bf I}_{\pi}(N)$ and $Q$ is a $\pi^{\prime}$-subgroup of $G$. Then

[TABLE]

Lemma 3.2**.**

Suppose $G$ is $\pi$-separable and that $N\triangleleft\,G$. Let $Q$ be a $\pi^{\prime}$-subgroup of $G$.

1. (a)

Suppose that $\mu\in{\bf I}_{\pi}(G|Q)$. Then there is a unique ${\bf N}_{G}(Q)$-orbit of $\tau\in{\bf I}_{\pi}(N)$ such that $\mu_{\tau}\in{\bf I}_{\pi}(G_{\tau}|Q)$, where $\mu_{\tau}$ is the Clifford correspondent of $\mu$ over $\tau$. Every such $\tau$ is $Q$-invariant. 2. (b)

Suppose that $\tau\in{\bf I}_{\pi}(N)$ is $Q$-invariant. Let ${\mathcal{U}}$ be a complete set of representatives of the $G_{\tau}$-orbits on the set $\{Q^{g}\,|\,g\in G,Q^{g}\subseteq G_{\tau}\}$. Then

[TABLE]

Thus, if $G_{\tau}{\bf N}_{G}(Q)=G$, then

[TABLE]

Proof.

1. (a)

Let $\nu\in{\bf I}_{\pi}(N)$ be under $\mu$, and let $\mu_{\nu}\in{\bf I}_{\pi}(G_{\nu}|\nu)$ be the Clifford correspondent of $\mu$ over $\nu$. If $R$ is a vertex of $\mu_{\nu}$, then $R$ is a vertex of $\mu$, by Theorem 2.3. Therefore $R=Q^{g}$ for some $g\in G$. If $\tau=\nu^{g^{-1}}$, then we have that $\mu_{\tau}$ has vertex $Q$. Suppose now that $\rho\in{\bf I}_{\pi}(N)$ is under $\mu$ such that $\mu_{\rho}$ has vertex $Q$. By Theorem 2.2(a), there exists $g\in G$ such that $\tau^{g}=\rho$. Thus $Q^{g}$ is a vertex of $\mu_{\rho}$. Then there is $x\in G_{\rho}$ such that $Q^{gx}=Q$. Since $\tau^{gx}=\rho$, the proof of part (a) is complete. 2. (b)

We have that induction defines a bijection ${\bf I}_{\pi}(G_{\tau}|\tau)\rightarrow{\bf I}_{\pi}(G|\tau)$. Notice that

[TABLE]

is a disjoint union. It suffices to observe, again, that if $\xi\in{\bf I}_{\pi}(G_{\tau}|\tau)$ has vertex $U$, then $\xi^{G}$ has vertex $U$.∎

4. Proofs

The deep part in our proofs comes from the following result.

Theorem 4.1**.**

Suppose that $L$ is a normal $\pi$-subgroup of $G$, $Q$ is a solvable $\pi^{\prime}$-subgroup of $G$ such that $LQ\triangleleft\,G$. Suppose that $M\subseteq{\bf Z}(G)$ is contained in $L$ and that $\varphi\in{\rm Irr}(M)$. Then $|{\bf I}_{\pi}(G|Q,\varphi)|=|{\bf I}_{\pi}({\bf N}_{G}(Q)|Q,\varphi)|$.

Proof.

Let ${\mathcal{A}}$ be a complete set of representatives of ${\bf N}_{G}(Q)$-orbits on ${\rm Irr}_{Q}(L|\varphi)$, the $Q$-invariant members of ${\rm Irr}(L|\varphi)$. Using Lemma 3.2, we have that

[TABLE]

is a disjoint union. Let ${\mathcal{A}}^{*}$ be the set of the $Q$-Glauberman correspondents of the elements of ${\mathcal{A}}$. Notice that ${\mathcal{A}}^{*}$ is a complete set of representatives of ${\bf N}_{G}(Q)$-orbits on ${\rm Irr}({\bf C}_{L}(Q)|\varphi)$. Moreover, ${\bf N}_{G_{\tau}}(Q)={\bf N}_{G_{\tau^{*}}}(Q)$. Then, as before,

[TABLE]

is a disjoint union. Thus

[TABLE]

Thus we need to prove that

[TABLE]

We may assume that $\tau$ is $G$-invariant.

Now, since $LQ\triangleleft\,G$ and $\tau$ is $G$-invariant, by Lemma 6.30 of [I2], we have that $Q$ is contained as a normal subgroup in some vertex of $\theta$, whenever $\theta\in{\bf I}_{\pi}(G)$ lies over $\tau$. Therefore $\theta\in{\bf I}_{\pi}(G|\tau)$ has vertex $Q$ if and only if $\theta(1)_{\pi^{\prime}}=|G:Q|_{\pi^{\prime}}$. Similarly, $\theta\in{\bf I}_{\pi}({\bf N}_{G}(Q)|\tau^{*})$ has vertex $Q$ if any only if $\theta(1)_{\pi^{\prime}}=|{\bf N}_{G}(Q):Q|_{\pi^{\prime}}=|G:Q|_{\pi^{\prime}}$ since $G=L{\bf N}_{G}(Q)$ by the Frattini argument and the Schur–Zassenhaus theorem.

Now we use the Dade–Puig theory on the character theory above Glauberman correspondents, which is thoroughly explained in [T]. By Theorem 6.5 of [T], in the language of Chapter 11 of [I1] (see Definition 11.23 of [I1]), we have that the character triples $(G,L,\tau)$ and $({\bf N}_{G}(Q),{\bf C}_{L}(Q),\tau^{*})$ are isomorphic. Write ${}^{*}:{\rm Irr}(G|\tau)\rightarrow{\rm Irr}({\bf N}_{G}(Q)|\tau^{*})$ for the associated bijection of characters. By Lemma 6.21 of [I2], there exists a unique bijection

[TABLE]

such that if $\chi^{0}=\phi\in{\bf I}_{\pi}(G|\tau)$ and $\chi\in{\rm Irr}(G)$ (which necessarily lies over $\tau$), then $(\chi^{*})^{0}=\phi^{*}$. Since $\chi(1)/\tau(1)=\chi^{*}(1)/\tau^{*}(1)$ (by Lemma 11.24 of [I1]), it follows that $\chi(1)_{\pi^{\prime}}=\chi^{*}(1)_{\pi^{\prime}}$. We deduce that

[TABLE]

as desired. ∎

In order to prove Theorem B, we argue by induction on the index of a normal $\pi$-subgroup $M$ of $G$. Theorem B follows from the special case $M=1$.

Theorem 4.2**.**

Suppose that $G$ is $\pi$-separable with a solvable Hall $\pi$-complement. Let $R$ be a nilpotent $\pi^{\prime}$-subgroup of $G$. Let $M\triangleleft\,G$ be a normal $\pi$-subgroup, and let $\varphi\in{\rm Irr}(M)$ be $G$-invariant. Let $\mathcal{Q}$ be the set of $\pi^{\prime}$-subgroups $Q$ of $G$ such that $R$ is a Carter subgroup of $Q$. Then

[TABLE]

Proof.

We argue by induction on $|G:M|$.

By Lemma 3.11 of [I2], let $(G^{*},M^{*},\varphi^{*})$ be a character triple isomorphic to $(G,M,\varphi)$, where $M^{*}$ is a central $\pi$-subgroup of $G^{*}$. If $Q$ is a $\pi^{\prime}$-subgroup of $G$, notice that we can write $(QM)^{*}=M^{*}\times Q^{*}$, for a unique $\pi^{\prime}$-subgroup $Q^{*}$ of $G^{*}$. If $R$ is contained in a $\pi^{\prime}$-subgroup $Q$, then $R$ is a Carter subgroup of $Q$ if and only if $RM/M$ is a Carter subgroup of $QM/M$, using that $Q$ is naturally isomorphic to $QM/M$. This happens if and only if $(RM/M)^{*}$ is a Carter subgroup of $(QM/M)^{*}$, which again happens if and only if $R^{*}$ is a Carter subgroup of $Q^{*}$. Notice further that if $R$ is a Carter subgroup of $Q$, then $R$ is a Carter subgroup of every Hall $\pi$-complement $Q_{1}$ of $QM$ that happens to contain $R$ (again using the isomorphism between $QM/M$ and $Q$). We easily check now that the set of $\pi^{\prime}$-subgroups of $G^{*}$ that contain $R^{*}$ as a Carter subgroup is exactly $\mathcal{Q^{*}}=\{Q^{*}|Q\in\mathcal{Q}\}$.

By the Frattini argument and the Schur–Zassenhaus theorem, notice that ${\bf N}_{G}(MR)=M{\bf N}_{G}(R)$. By Lemma 6.21 and the proof of Lemma 6.32 of [I2], there is a bijection ${}^{*}:{\bf I}_{\pi}(G|\varphi)\rightarrow{\bf I}_{\pi}(G^{*}|\varphi^{*})$ such that $\eta$ has vertex $Q$ if and only if $\eta^{*}$ has vertex $Q^{*}$. From all these arguments, it easily follows that we may assume that $M$ is central. In particular, $M\leq{\bf N}_{G}(R)$.

Let $K={\bf O}_{\pi^{\prime}}(G)$. Suppose that there exists some $\mu\in{\bf I}_{\pi}(G|Q,\varphi)$ for some $Q\in\mathcal{Q}$. By Lemma 6.30 of [I2] (in the notation of that lemma, $K$ is $1$ and $Q$ is $K$), we have that $K$ is contained in $Q$. Hence, it is no loss if we only consider $Q\in\mathcal{Q}$ such that $K\subseteq Q$.

Suppose that ${\bf N}_{K}(R)$ is not contained in $R$. Then there cannot be weights $(R,\gamma)$, where $\gamma\in{\rm Irr}({\bf N}_{G}(R)/R)$ has $\pi$-defect zero by Lemma 3.1. So the right hand side is zero. Suppose that there exists some $\mu\in{\bf I}_{\pi}(G|Q,\varphi)$ for some $Q\in\mathcal{Q}$ (with $K\subseteq Q$). Since $R$ is a Carter subgroup of $Q$, then $R$ is a Carter subgroup of $KR$, and therefore ${\bf N}_{K}(R)$ is contained in $R$. Therefore may assume that ${\bf N}_{K}(R)$ is contained in $R$. We claim that $R$ is a Carter subgroup of $Q$ if and only if $RK/K$ is a Carter subgroup of $Q/K$. One implication is known (see 9.5.3 in [R]). Suppose that $RK/K$ is a Carter subgroup of $Q/K$. Since ${\bf N}_{Q}(R)$ normalizes $RK$, it is contained in $RK$. Hence ${\bf N}_{Q}(R)={\bf N}_{KR}(R)=R$, and $R$ is a Carter subgroup of $Q$. In this situation the Frattini argument yields ${\bf N}_{G}(R)K={\bf N}_{G}(RK)$.

Next, we will replace $G$ by $G/K$. By Lemma 6.31 of [I2] (the roles of $K$ and $M$ are interchanged in that lemma),

[TABLE]

where $\hat{\varphi}\in{\rm Irr}(MK/K)$ corresponds to $\varphi$ via the natural isomorphism. Similarly,

[TABLE]

Hence, for the remainder of the proof we may assume that ${\bf O}_{\pi^{\prime}}(G)=K=1$.

Suppose now that $L={\bf O}_{\pi}(G)$. Let ${\mathcal{A}}$ be a complete set of ${\bf N}_{G}(R)$-representatives of the $R$-invariant characters in ${\rm Irr}(L|\varphi)$. If $L=M$, then $L=G$ by the Hall-Higman Lemma 1.2.3, and $G$ is a $\pi$-group. In this case, $R=1=Q$, and there is nothing to prove. Thus, we may assume that $|G:L|<|G:M|$.

For each $\tau\in{\mathcal{A}}$, let ${{\mathcal{Q}}}_{\tau}$ be the set of $\pi^{\prime}$-subgroups $Q$ of $G_{\tau}$ such that $R$ is a Carter subgroup of $Q$. By induction,

[TABLE]

Since $L$ is a $\pi$-group and $R$ is a $\pi^{\prime}$-subgroup, we have that $L{\bf N}_{G}(R)={\bf N}_{G}(LR)$. Also, ${\bf N}_{G}(LR)_{\tau}=L{\bf N}_{G}(R)_{\tau}$. By Lemma 3.2, we have that

[TABLE]

Also,

[TABLE]

by using the first paragraph of the proof of Theorem 4.1. By Theorem 4.1,

[TABLE]

Therefore,

[TABLE]

We are left to show that

[TABLE]

Let ${\mathcal{R}}$ be a complete set of representatives of ${\bf N}_{G}(R)$-orbits in ${\mathcal{Q}}$, and notice that

[TABLE]

is a disjoint union. Indeed, if $\mu\in{\bf I}_{\pi}(G|Q_{1},\varphi)\cap{\bf I}_{\pi}(G|Q_{2},\varphi)$ for $Q_{i}\in{\mathcal{Q}}$, then we have that $Q_{1}=Q_{2}^{g}$ for some $g\in G$ by the uniqueness of vertices. Hence $R^{g}$ and $R$ are Carter subgroups of $Q_{1}$, and therefore $R^{gx}=R$ for some $x\in Q_{1}$. It follows that $Q_{1}=Q_{2}^{gx}$ are ${\bf N}_{G}(R)$-conjugate.

Now fix $Q\in{\mathcal{R}}$. For each $\mu\in{\bf I}_{\pi}(G|Q,\varphi)$, we claim that there is a unique $\tau\in{\mathcal{A}}$ such that $\mu_{\tau}\in{\bf I}_{\pi}(G_{\tau}|Q^{x},\tau)$, for some $x\in{\bf N}_{G}(R)$. We know that there is $\nu\in{\rm Irr}(L|\varphi)$ such that $\mu_{\nu}\in{\bf I}_{\pi}(G_{\nu}|Q,\nu)$ by Lemma 3.2(a). Now, $\nu^{x}=\tau$ for some $x\in{\bf N}_{G}(R)$ and $\tau\in{\mathcal{A}}$, and it follows that $\mu_{\tau}\in{\bf I}_{\pi}(G_{\tau}|Q^{x},\tau)$. Suppose that $\mu_{\epsilon}\in{\bf I}_{\pi}(G_{\epsilon}|Q^{y},\epsilon)$, for some $y\in{\bf N}_{G}(R)$ and $\epsilon\in{\mathcal{A}}$. Now, $\epsilon=\tau^{g}$ for some $g\in G$, by Clifford’s theorem. Thus $Q^{xgt}=Q^{y}$ for some $t\in G_{\epsilon}$, by the uniqueness of vertices. Thus $xgty^{-1}\in{\bf N}_{G}(Q)$. Since $R$ is a Carter subgroup of $Q$, by the Frattini argument we have that $xgty^{-1}=qv$, where $q\in Q$ and $v\in G$ normalizes $Q$ and $R$. Since $Q^{x}$ fixes $\tau$, then $Q$ fixes $\tau^{x^{-1}}$. Now

[TABLE]

So $\epsilon$ and $\tau$ are ${\bf N}_{G}(R)$-conjugate, and thus they are equal.

Now we define a map

[TABLE]

given by $f(\mu)=(\mu_{\tau},\tau)$, where $\tau\in{\mathcal{A}}$ is the unique element in ${\mathcal{A}}$ such that $\mu_{\tau}\in{\bf I}_{\pi}(G_{\tau}|Q^{x},\tau)$, for some $x\in{\bf N}_{G}(R)$. Since $\mu_{\tau}^{G}=\mu$, we have that $f$ is injective. If we have that $\gamma\in\bigcup_{Q\in{\mathcal{Q}}_{\tau}}{\bf I}_{\pi}(G_{\tau}|Q,\tau)$ then $\gamma^{G}\in\bigcup_{Q\in{\mathcal{Q}}}{\bf I}_{\pi}(G|Q,\varphi)$, so $f$ is surjective. ∎

Some of the difficulties in Theorem 4.2 are caused by the fact that Clifford correspondence does not necessarily respect vertices, even in quite restricted situations. Suppose that $N$ is a normal $p^{\prime}$-subgroup of $G$, $\tau\in{\rm Irr}(N)$, $Q$ is a $p$-subgroup of $G$ and $\tau$ is $Q$-invariant. Then it is not necessarily true that induction defines a bijection ${\rm IBr}(G_{\tau}|Q,\tau)\rightarrow{\rm IBr}(G|Q,\tau)$. For instance, take $p=2$ and $G={\tt SmallGroup}(216,158)$. This group has a unique normal subgroup $N$ of order 3. The Fitting subgroup $F$ of $G$ is $F=N\times M$, where $M$ is a normal subgroup of type $C_{3}\times C_{3}$, and $G/F=D_{8}$. Let $1\neq\tau\in{\rm Irr}(N)$. Then $G_{\tau}\triangleleft\,G$ has index 2, and $G_{\tau}/N=S_{3}\times S_{3}$. Now $\tau$ has a unique extension $\hat{\tau}\in{\rm IBr}(G_{\tau})$. The group $G_{\tau}$ has three conjugacy classes of subgroups $Q$ of order 2. Take $Q_{1}$ that corresponds to $C_{2}\times 1$ and $Q_{2}$ that corresponds to $1\times C_{2}$. Then $Q_{1}$ and $Q_{2}$ are not $G_{\tau}$-conjugate but $G$-conjugate. So $|{\rm IBr}(G_{\tau}|Q_{1},\tau)|=1$ and $|{\rm IBr}(G|Q_{1},\tau)|=2$.

5. A canonical bijection

If $G$ has a normal Hall $\pi$-subgroup, then we have a canonical bijection in Theorem 4.2. This seems worth to be explored.

Lemma 5.1**.**

Suppose that $G=NH$ where $N$ is a normal $\pi$-subgroup and $H$ is a $\pi^{\prime}$-subgroup. Then ${\bf N}_{G}(Q)={\bf C}_{N}(Q){\bf N}_{H}(Q)$ for every $Q\leq H$.

Proof.

First note that

[TABLE]

Let $xh\in{\bf N}_{G}(Q)$ where $x\in N$ and $h\in H$. Then $h=x^{-1}(xh)\in N{\bf N}_{G}(Q)\cap H\leq{\bf N}_{H}(Q)$. This shows ${\bf N}_{G}(Q)={\bf N}_{N}(Q){\bf N}_{H}(Q)={\bf C}_{N}(Q){\bf N}_{H}(Q)$. ∎

Lemma 5.2**.**

Suppose that $G=NH$ where $N$ is a normal $\pi$-subgroup and $H$ is a solvable $\pi^{\prime}$-subgroup. Let $R\leq H$, and let $\tau\in{\rm Irr}({\bf C}_{N}(R))$ be such that ${\bf N}_{G}(R)_{\tau}={\bf C}_{N}(R)\times R$. Let $\gamma\in{\rm Irr}_{R}(N)$ be the Glauberman correspondent of $\tau$. Then $R={\bf N}_{H_{\gamma}}(R)$.

Proof.

Suppose that $R<S\leq H_{\gamma}$, where $R\triangleleft\,S$. Then $S$ acts on the $R$-Glauberman correspondence. Since $S$ fixes $\gamma$, therefore it fixes $\gamma^{*}=\tau$. But this gives the contradiction $S\subseteq{\bf N}_{G}(R)_{\tau}={\bf C}_{N}(R)\times R$. ∎

Theorem 5.3**.**

Suppose that $G=NH$ where $N$ is a normal $\pi$-subgroup and $H$ is a solvable $\pi^{\prime}$-subgroup. Let $R$ be a nilpotent subgroup of $H$. Let $\mathcal{Q}$ be the set of subgroups $Q\subseteq H$ such that $R$ is a Carter subgroup of $Q$. Then there is a natural bijection

[TABLE]

Proof.

Let $Q\in\mathcal{Q}$. By the Frattini argument, notice that ${\bf N}_{G}(Q)=Q({\bf N}_{G}(Q)\cap{\bf N}_{G}(R))$, and that $Q\cap({\bf N}_{G}(Q)\cap{\bf N}_{G}(R))=R$.

Let $\phi\in{\bf I}_{\pi}(G|Q)$. By Lemma 3.2, there exists a $Q$-invariant $\theta\in{\rm Irr}(N)$ under $\phi$. Then $T=G_{\theta}=QN$ using Corollary 8.16 in [I1] for instance. If $\theta_{1}$ is another such choice, then $\theta_{1}=\theta^{g}$ for some $g\in{\bf N}_{G}(Q)$. Thus, we may assume that $g\in{\bf N}_{G}(Q)\cap{\bf N}_{G}(R)$. Let $\theta^{*}\in{\bf C}_{N}(R)$ be the $R$-Glauberman correspondent of $\theta$. Now, by Lemma 5.1 applied in $T$, we have that ${\bf N}_{T}(R)={\bf C}_{N}(R){\bf N}_{Q}(R)={\bf C}_{N}(R)\times R$. We claim that ${\bf N}_{T}(R)$ is the stabilizer of $\theta^{*}$ in ${\bf N}_{G}(R)$. If $x\in{\bf N}_{G}(R)$ fixes $\theta^{*}$, then $x$ fixes $\theta$, and thus $x\in{\bf N}_{T}(R)$, as claimed. Now $\phi^{*}:=(\theta^{*}\times 1_{R})^{{\bf N}_{G}(R)}$ is irreducible, and belongs to ${\bf I}_{\pi}({\bf N}_{G}(R)|R)$. Since $\theta_{1}$ is ${\bf N}_{G}(Q)\cap{\bf N}_{G}(R)$-conjugate to $\theta$, $\phi^{*}$ is independent of the choice of $\theta$.

Suppose that $\phi^{*}=\mu^{*}$, where $\phi\in{\bf I}_{\pi}(G|Q_{1})$ and $\mu\in{\bf I}_{\pi}(G|Q_{2})$, where $R$ is a Carter subgroup of $Q_{i}$ and $Q_{i}\subseteq H$. Suppose that we picked $\theta$ for $\phi$ and $\epsilon$ for $\mu$, so that $\phi^{*}=(\theta^{*}\times 1_{R})^{{\bf N}_{G}(R)}$ and $\mu^{*}=(\epsilon^{*}\times 1_{R})^{{\bf N}_{G}(R)}$. Then ${\bf N}_{G}(R)={\bf C}_{N}(R){\bf N}_{H}(R)$, and $\theta^{*}$ and $\epsilon^{*}$ are ${\bf N}_{H}(R)$-conjugate, say $(\theta^{*})^{x}=\epsilon^{*}$. Then $\theta^{x}=\epsilon$. By replacing $(Q_{1},\theta)$ by $(Q_{1}^{x},\theta^{x})$, we may assume that $\theta=\epsilon$. But then $Q_{1}=H_{\theta}=H_{\epsilon}=Q_{2}$. Since $\pi$-partial character are determined on the $\pi$-elements, we must have $\phi=\mu$ now.

Suppose conversely that $\tau\in{\bf I}_{\pi}({\bf N}_{G}(R)|R)$. Then $\tau$ is induced from ${\bf C}_{N}(R)\times R$. Let $\mu\in{\rm Irr}({\bf C}_{N}(R))$ such that $\mu\times 1_{R}$ induces $\tau$. Then the stabilizer of $\mu$ in ${\bf N}_{G}(R)$ is ${\bf C}_{N}(R)\times R$. If $\rho\in{\rm Irr}_{R}(N)$ is the $R$-Glauberman correspondent of $\mu$, then by Lemma 5.2 we know that $R$ is a Carter subgroup of $Q=H_{\rho}$, where $Q$ is the stabilizer in $H$ of $\rho$. Thus with the notation of the first part of the proof we obtain $\tau=\mu^{*}$ where $\mu$ is induced from $G_{\rho}=QN$. ∎