Abstract
We conjectured in Malle and Navarro (J Algebra 370:402–406, 2012) that a Sylow p-subgroup P of a finite group G is normal if and only if whenever p does not divide the multiplicity of \(\chi \in {{\text {Irr}}}(G)\) in the permutation character \((1_P)^G\), then p does not divide the degree \(\chi (1)\). In this note, we prove an analogue of this for p-Brauer characters.
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1 Introduction
In [3], we proved that if G is a finite group and \(P \in \mathrm{Syl}_p(G)\), then \(P\unlhd G\) if and only if p does not divide the degrees of the irreducible constituents of the permutation character \((1_P)^G\). Furthermore, we conjectured that this happens if and only if the irreducible constituents of \((1_P)^G\) with multiplicity not divisible by p had degree not divisible by p. This has turned out to be a difficult problem for symmetric groups (in the cases where \(p=2\) or \(p=3\)). In this short note, we aim for an analogous result on modular characters. By using the induction formula and [5, Theorem 2.13], notice that the permutation p-Brauer character \(((1_P)^G)^0\) decomposes as
where \(\Phi _\varphi \) is the projective indecomposable character associated with the irreducible Brauer character \(\varphi \).
We consider the following modular analogue of our conjecture in [3]:
Theorem A
Let p be a prime, let G be a finite group, and let \(P\in \mathrm{Syl}_p(G)\). Then \(P\unlhd G\) if and only if whenever \(\varphi \in {{\text {IBr}}}(G)\) is such that \(\Phi _\varphi (1)/|P|\) is not divisible by p, then \(\varphi (1)\) is not divisible by p.
This generalizes the well-known result of G. Michler [4, Theorem 5.5] that G has a normal Sylow p-subgroup if and only if all irreducible Brauer characters of G have degree prime to p.
Our proof (as in [4]) relies on the classification of finite simple groups for p odd only.
2 Proof of Theorem A
We use the notation in [5]. Let p be a fixed prime. We choose a maximal ideal of the ring of algebraic integers in \({{\mathbb {C}}}\) containing p, with respect to which we calculate the set \({{\text {IBr}}}(G)\) of irreducible (p-)Brauer characters for every finite group G. If \(N\unlhd G\) and \(\theta \in {{\text {IBr}}}(N)\), then \(\mathrm{IBr}(G|\theta )\) is the set of irreducible constituents of the induced character \(\theta ^G\), which, by [5, Corollary 8.7], is the set of irreducible Brauer characters \(\varphi \in {{\text {IBr}}}(G)\) such that \(\theta \) is a constituent of the restriction \(\varphi _N\). For \(\varphi \in {{\text {IBr}}}(G)\), we write \(c_\varphi =\Phi _\varphi (1)/|G|_p\). (This is an integer by Dickson’s theorem [5, Corollary 2.14].)
We start with the following observation:
Proposition 2.1
Let B be a p-block of a finite group G of non-maximal defect. Then B contains an irreducible Brauer character \(\theta \) of degree divisible by p but with \(c_\theta \) not divisible by p.
Proof
By a result of Brauer [5, Theorem 3.28], the p-part of the dimension of B is precisely \(p^{2a-d}\), where \(p^a\) is the order of a Sylow p-subgroup of G, and d is the defect of B. Now
and all \(\theta (1)\) are divisible by \(p^{a-d}\) as all \(\chi (1)\), for \(\chi \in {{\text {Irr}}}(B)\), are divisible by \(p^{a-d}\). It follows that there must be some \(\theta \in {{\text {IBr}}}(B)\) with \(\Phi _\theta (1)/p^a\) not divisible by p. \(\square \)
We shall need the first two parts of the following lemma. We keep part (c) for future use.
Lemma 2.2
Let \(N\unlhd G\), and let \(\theta \in {{\text {IBr}}}(N)\). Write
where \(a_\varphi \) are non-negative integers. Let T be the stabilizer of \(\theta \) in G, and let \(t=|G:T|\). Then for all \(\varphi \in \mathrm{IBr}(G|\theta )\), we have
-
(a)
$$\begin{aligned} a_\varphi tc_\theta =c_\varphi |G:N|_p \, ; \end{aligned}$$
-
(b)
if G/N is a p-group, then \(c_\varphi =c_\theta \); and
-
(c)
\(c_\theta \) divides \(c_\varphi \) and thus, \(|T:N|_p\) divides \(a_\varphi \).
Proof
By [5, Corollary 8.8], we have \(\Phi _\varphi (1)=a_\varphi t \Phi _\theta (1)\). Hence, (a) is clear.
To prove (b), notice that in this case, \(\mathrm{IBr}(G|\theta )=\{\varphi \}\) and \(\varphi (1)= t\theta (1)\) by [5, Theorem 8.11]. Now, \(\theta ^G=a_\varphi \varphi \) (using [5, Corollary 8.7]) and by part (a), we have \(c_\theta =c_\varphi \).
To prove (c), using the Clifford correspondence for Brauer characters [5, Theorem 8.9], we may assume that \(T=G\). We have \((\Phi _\varphi )_N=a_\varphi \Phi _\theta \) again by [5, Corollary 8.8]. Let \(H=PN\), where \(P\in \mathrm{Syl}_p(G)\). By [5, Problem 8.7], we have that \((\Phi _\varphi )_H\) is a positive sum of some \(\Phi _\tau \) for suitable \(\tau \in {{\text {IBr}}}(H)\). Using [5, Corollary 8.8] (and the fact that \(\{\Phi _\mu \mid \mu \in {{\text {IBr}}}(N)\}\) is linearly independent [5, Theorem 2.13]), we conclude that every such \(\tau \) necessarily lies over \(\theta \). However, there is a unique \({\hat{\theta }} \in {{\text {IBr}}}(H)\) over \(\theta \) (by [5, Theorem 8.11]). We conclude that
for some integer v. Then
using part (b). The last part now follows from (a). \(\square \)
We are now ready to prove our main result.
Theorem 2.3
Let p be a prime, let G be a finite group, and let \(P\in \mathrm{Syl}_p(G)\). Then \(P\unlhd G\) if and only if whenever \(\Phi _\varphi (1)/|P|\) is not divisible by p, then \(\varphi (1)\) is not divisible by p.
Proof
If \(P\unlhd G\), then p does not divide \(\varphi (1)\) for all \(\varphi \in {{\text {IBr}}}(G)\) since \(P \subseteq \ker \varphi \) ([5, Lemma 2.32]) and \(\mathrm{IBr}(G/P)=\mathrm{Irr}(G/P)\) (by [5, Theorem 2.12]).
To prove the converse in the case \(p=2\), we borrow an argument in [2]. First, recall that \(\varphi (1)\) is even if \(1 \ne \varphi \in {{\text {IBr}}}(G)\) is real-valued ([5, Theorem 2.30]) and \(c_1\) is odd ([5, Corollary 3.34]). By hypothesis, and using that \(c_\varphi =c_{{\bar{\varphi }}}\) since \((1_P)^G\) is real-valued, we can write
where \(\Delta \) is a Brauer character of G, \({\bar{\varphi }}\) is the complex conjugate of \(\varphi \), and \(\Lambda \) is a subset of non-trivial non-real-valued characters in \({{\text {IBr}}}(G)\). If P is not normal in G, then there exists a real element \(1\ne x\in G\) of odd order ([1, Proposition 6.4]). Then \(0=(1_P)^G(x)\), and we conclude that \(-c_1=2\alpha \) for some algebraic integer \(\alpha \). This is impossible since \(c_1\) is odd.
To prove the converse for p odd, we argue by induction on |G|. By Proposition 2.1, all blocks of G have maximal defect. Let N be a proper non-trivial normal subgroup of G.
First, we claim that p divides |G/N|. Assume the contrary. Let \(\theta \in {{\text {IBr}}}(N)\) be such that p does not divide \(c_\theta \). Let T be the stabilizer of \(\theta \) in G, and let \(t=|G:T|\). Let \(\varphi \in {{\text {IBr}}}(G)\) be over \(\theta \). By [5, Corollary 8.7], we have that \(\varphi (1)=et\theta (1)\), where e is the multiplicity of \(\theta \) in \(\varphi _N\). Also by [5, Corollary 8.7], this is the multiplicity of \(\varphi \) in \(\theta ^G\) (what we called \(a_\varphi \) in Lemma 2.2). By Lemma 2.2(a), we have \(c_\varphi ={\varphi (1) \over \theta (1)} c_\theta \). We conclude that \(c_\varphi \) is not divisible by p, (using Dade’s theorem [5, Theorem 8.30]). By hypothesis, we conclude that \(\varphi (1)\) is not divisible by p, and therefore neither is \(\theta (1)\). Hence N has a normal Sylow p-subgroup, and we are done in this case.
Suppose now that G/N is a p-group. Let \(\theta \in {{\text {IBr}}}(N)\) be such that p does not divide \(c_\theta \), and let \(\varphi \in \mathrm{IBr}(G|\theta )\). By Lemma 2.2(b), we have \(c_\theta =c_\varphi \). Using the hypothesis, we conclude that p does not divide \(\varphi (1)\), and therefore p does not divide \(\theta (1)\). Therefore, N has a normal Sylow p-subgroup, by induction.
Suppose now that \(c_{1_N}=1\). (By Fong’s dimensional formula [5, Corollary 10.14], this happens, for instance, if N is p-solvable.) By [6, Lemma 2.6], if \({\bar{\varphi }} \in \mathrm{IBr}(G/N)\) is the inflation of \(\varphi \in {{\text {IBr}}}(G)\), then \(c_\varphi =c_{{\bar{\varphi }}}\), and by induction, G/N has a normal Sylow p-subgroup. By the two previous paragraphs, we deduce that \(G=N\). In particular, G does not have proper non-trivial normal p-solvable subgroups.
Suppose now that N is a non-abelian minimal normal subgroup of G. Let \(S\unlhd N\) be simple, so that \(N=S^{x_1} \times \cdots \times S^{x_k}\) for some \(x_i \in G\). By [4, Theorem 5.3], there exists a block b of S that does not have maximal defect. Then the block \(e=b^{x_1} \times \cdots \times b^{x_k}\) does not have maximal defect. However, if B is any block of G covering e, then B has maximal defect, and therefore so does e by Knörr’s theorem [5, Theorem 9.26]. \(\square \)
A possible variation of Theorem A would be to consider \({{\text {Fix}}}_P(\varphi )\) instead of \(c_\varphi \), where \({{\text {Fix}}}_P(\varphi )\) is the dimension of the fixed points of P in any G-module affording the Brauer character \(\varphi \).
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Gunter Malle gratefully acknowledges financial support by SFB TRR 195. He thanks the Isaacs Newton Institute for Mathematical Sciences in Cambridge for support and hospitality during the programme “Groups, Representations and Applications: New Perspectives” when work on this paper was undertaken. This work was supported by: EPSRC grant number EP/R014604/1. The research of Gabriel Navarro is supported by Ministerio de Ciencia e Innovación PID2019-103854GB-I00 and FEDER funds.
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Malle, G., Navarro, G. On principal indecomposable degrees and Sylow subgroups. Arch. Math. 115, 489–493 (2020). https://doi.org/10.1007/s00013-020-01494-9
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DOI: https://doi.org/10.1007/s00013-020-01494-9