Sylow Normalizers and Brauer Character Degrees

Journal of Algebra 229, 623–631 (2000)doi:10.1006/jabr.1999.8275, available online at http://www.idealibrary.com on

Sylow Normalizers and Brauer Character Degrees1

Antonio Beltran

Departament de Matematiques, Universitat Jaume I, Campus de Riu Sec, 12071Castello, Spain

E-mail: [email protected]

and

Gabriel Navarro

Departament d’Algebra, Facultat de Matematiques, Universitat de Valencia, 46100Burjassot, Valencia, Spain

E-mail: [email protected]

Communicated by George Glauberman

Received January 21, 1999

dedicated to professor martin isaacs on his 60th birthday

1. INTRODUCTION

Suppose that G is a finite group. In this note, we show that a localcondition about Sylow normalizers is equivalent to a global condition onthe degrees of certain irreducible Brauer characters of G.

Theorem A. Let G be a finite �p; q�-solvable group, and let Q ∈ Sylq�G�and P ∈ Sylp�G�. Then every irreducible p-Brauer character of G of q′-degree has p′-degree if and only if NG�Q� is contained in some G-conjugateof NG�P�.

Theorem A needs a solvability hypothesis. If p = 7, then the irre-ducible p-Brauer characters of the group G = PSL�2; 27� have degrees�1; 13; 26; 28�. If we set q = 2, then each q′-degree is also a p′-degree.

1This research was partially supported by DGICYT.

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624 beltran and navarro

However, if Q ∈ Sylq�G� and P ∈ Sylp�G�, we have that �NG�Q�� = 12 and�NG�P�� = 28. For the other direction, we let G =M11. If q = 2 and p = 3,there exist Q ∈ Syl2�G� and P ∈ Syl3�G� such that Q = NG�Q� ⊆ NG�P�.However, G has a p-Brauer irreducible character of degree 45 which is ofq′-degree and divisible by p. (For checking the Brauer character degrees,we have used [3].)

In [7], T. Wolf and the second author proved, with the same hypothesis,that every irreducible ordinary character of G of q′-degree has p′-degreeif and only if NG�Q� is contained in some NG�P� with CP ′ �Q� = 1. Notethat the hypothesis in our Theorem A here are weaker by the Fong–Swantheorem. Also, note that Theorem A above has content even when theprime q does not divide �G�.

One of the classical problems in character theory is determining whichproperties of a finite group can be read off from its character table.

Corollary B. Suppose G is a finite �p; q�-solvable group, and let Q ∈Sylq�G� and P ∈ Sylp�G�. Then the condition that NG�Q� is contained insome G-conjugate of NG�P� can be read off from the character table of G.

2. PRELIMINARIES

We start with some easy facts about Sylow normalizers.

(2.1) Lemma. Let G be a finite group, let P ∈ Sylp�G�, let Q ∈ Sylq�G�,and let N G G. If p and q do not divide �G x N�, then NG�Q� ⊆ NG�P� ifand only if NN�Q� ⊆ NN�P�.

Proof. We have that P;Q ⊆ N . Of course, if NG�Q� ⊆ NG�P�, thenNN�Q� ⊆ NN�P�. Assume now that NN�Q� ⊆ NN�P�. We prove thatNG�Q� ⊆ NG�P� by three applications of the Frattini argument. First,G = NNG�P� = NNG�Q�. Now, since Q ⊆ NN�P� G NG�P�, it follows thatNG�P� = NN�P��NG�P� ∩ NG�Q��. Therefore

G = N�NG�Q� ∩ NG�P��:

Hence, �G x NG�Q� ∩ NG�P�� = �N x N ∩ NG�Q� ∩ NG�P�� = �N x NN�Q��.Since �N x NN�Q�� = �G x NG�Q��, we deduce that �G x NG�Q� ∩ NG�P�� =�G x NG�Q��, and we are done.

(2.2) Lemma. Suppose that P ∈ Sylp�G� and Q ∈ Sylq�G� are such thatPQ = QP . Let L G G and set H = PQL. Then NG�Q� ⊆ NG�P� if and onlyif NG�Q�L ⊆ NG�P�L and NH�Q� ⊆ NH�P�.

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Proof. One implication is clear. Assume now that NG�Q�L ⊆ NG�P�Land NH�Q� ⊆ NH�P� and we prove that NG�Q� ⊆ NG�P�. Let V = NG�H�.Since NG�Q� ⊆ NG�P�L = NG�PL� it follows that NG�Q� normalizes�PL�Q = H. Hence NG�Q� ⊆ V . Now, H G V , p and q do not divide�V x H�, and also we have that NH�Q� ⊆ NH�P�. By Lemma 2.1, we havethat NV �Q� ⊆ NV �P�. Since NV �Q� = NG�Q�, the proof of the lemma iscomplete.

(2.3) Lemma. Suppose that G is π-separable and let H be a Hall π-subgroup of G. If P is a Sylow p-subgroup of H, then NH�P� is a Hallπ-subgroup of NG�P�.

Proof. This follows from Proposition 3 of [7].

We are going to use several properties on characters and coprime ac-tion. If A acts on G, we denote by IrrA�G� the set of irreducible charactersof G which are fixed by A.

(2.4) Lemma. Suppose that A acts coprimely on G. Let B ⊆ A. ThenCG�A� = CG�B� if and only if IrrA�G� = IrrB�G�.

Proof. This is Lemma 2.2 of [5].

If G is a p-solvable group, we remind the reader that the set IBr�G� ofirreducible Brauer characters of G is canonically defined. Isaacs proved in[1] that there exists a canonical subset Bp′ �G� ⊆ Irr�G� such that the mapχ 7→ χ0 is a bijection Bp′ �G� → IBr�G� (where χ0 denotes the restrictionof χ to the p-regular elements of G). Since every χ ∈ Bp′ �G� is a canonicallifting of ϕ = χ0 ∈ IBr�G�, we have that χ and ϕ uniquely determine eachother. In particular, if A acts on G, then ϕ is A-invariant if and only if χis A-invariant.

(2.5) Lemma. Suppose that G is a p-solvable group. Assume that G =KH, L = K ∩H, where K;L G G and ��G x K�; �K x L�� = 1.

(a) If θ ∈ IBr�K� is H-invariant, then θL has an H-invariant irre-ducible constituent and all of them are C-conjugate, where C/L = CK/L�H�.

(b) If τ ∈ IBr�L� is H-invariant, then τK has some H-invariant irre-ducible constituent.

Proof. First we prove part (a). We have that H/L acts coprimely onK/L. Both of these groups act on the set � of irreducible constituents ofθL. By Clifford’s theorem for Brauer characters, K/L also acts transitivelyon �. Now, part (a) easily follows from Glauberman’s Lemma (Lemmas13.8 and 13.9 of [2]).

To prove part (b), we argue by induction on �K x L�. It is clear, usinginduction, that we may assume that K/L is a chief factor of G. Also, usinginduction and the Clifford correspondence for Brauer characters (Theorem


8.9 of [6]), we may assume that τ is G-invariant. If K/L is a p-group, thenIBr�K � τ� = �θ� by Theorem 8.11 of [6], and the lemma is certainly true inthis case. So we may assume that K/L is a p′-group. Now, let ψ ∈ Bp′ �L�be the Isaacs canonical lifting of τ. By uniqueness, we have that ψ is G-invariant. Now, by Theorem 13.31 of [2], there exists some H-invariantη ∈ Irr�K� lying over ψ. Since K/L is a p′-group, by Theorem 7.1 of [1]we have that η ∈ Bp′ �K�. Hence θ = η0 ∈ IBr�K� is H-invariant and liesover τ.

Next is a key fact in this paper.

(2.6) Theorem. Suppose that G is p-solvable. Let q 6= p be a prime.If q divides ϕ�1� for all nonlinear ϕ ∈ IBr�G�, then G has a normal q-complement.

Proof. This was proven in [4].

3. MAIN RESULTS

The strategy for proving our Theorem A was inspired by [7]. (The maindifference here is that we do not use the “McKay conjecture" type of argu-ment applied in [7] but our Theorem 3.3 below.) We fix a prime p, and forevery p-solvable group G we consider IBr�G� to be the set of irreducibleBrauer characters of G.

(3.1) Lemma. Suppose that G = PQ, where P ∈ Sylp�G� and Q ∈Sylq�G�. Then every ϕ ∈ IBr�G� of q′-degree has p′-degree if and only ifP G G.

Proof. Of course, we may assume that p 6= q. If P G G, then everyirreducible Brauer character of G is an ordinary character of the q-groupQ, and therefore it has degree not divisible by p. Now, if every ϕ ∈ IBr�G�of q′-degree has p′-degree, then note that every nonlinear ϕ ∈ IBr�G� hasdegree divisible by q (because G is a solvable group and ϕ�1� divides �G�).By Theorem 2.6, we conclude that P G G, as desired.

(3.2) Theorem. Suppose that G has a normal π-complement, where π =�p; q�. Let P ∈ Sylp�G� and Q ∈ Sylq�G� be such that PQ = QP . Then everyϕ ∈ IBr�G� of q′-degree has p′-degree if and only if NG�Q� ⊆ NG�P�.

Proof. Let K be the normal π-complement of G. Let H = PQ. Weclaim that, in either direction, we have that Q normalizes P . This is clearif we are assuming that NG�Q� ⊆ NG�P�. So let us assume the characterdegree condition on G. Since H is isomorphic to G/K, note that H alsosatisfies the same character degree condition. Hence, Q normalizes P byLemma 3.1, and this proves the claim.


Now, by Lemma 2.3, we have that NG�P� = NK�P�NH�P� =CK�P�NH�P� = CK�P�H and NG�Q� = CK�Q�NH�Q�. Hence, we havethat NG�Q� ⊆ NG�P� if and only if CK�Q� ⊆ CK�P�. By Lemma 2.4 ap-plied to the action of H and Q on K, we know that this happens if andonly if every θ ∈ IrrQ�K� is P-invariant.

We are reduced to proving that every ϕ ∈ IBr�G� of q′-degree has p′-degree if and only if every θ ∈ IrrQ�K� is P-invariant.

Assume that every θ ∈ IrrQ�K� is P-invariant. Let ϕ ∈ IBr�G� of q′-degree and let θ ∈ Irr�K� be under ϕ. (Recall that K is a p′-group, soIBr�K� = Irr�K�.) If η ∈ IBr�T � is the Clifford correspondent of ϕ overθ, since ϕ�1� is not divisible by q it follows that T contains some Sylowq-subgroup of G. Hence, replacing θ by some G-conjugate, we may assumethat θ is Q-invariant. Now, by hypothesis, we have that θ is P-invariant.Therefore, θ extends to G (by Theorem 8.13 of [6], for instance). Let ψ ∈IBr�G� be an extension of θ to G. Then ϕ = ψβ, where β ∈ IBr�G/K� byCorollary 8.20 of [6]. Now, β has p′-degree (because G/K has a normalSylow p-subgroup), ψ�1� = θ�1� is not divisible by p, and we conclude thatϕ�1� is not divisible by p.

Conversely, assume that every ϕ ∈ IBr�G� of q′-degree has p′-degree.Let θ ∈ IrrQ�K�. Let T be the inertia group and let γ ∈ Irr�T � be anextension of θ to T (by Theorem 8.13 of [6]). Now, γG ∈ IBr�G� has q′-degree (because Q ⊆ T and γ�1� is not divisible by q). By hypothesis, γG

has p′-degree and we conclude that G = T . Hence, θ is P-invariant, asdesired.

With the hypotheses and notation of Theorem 3.2, we have also shownthat NG�Q� ⊆ NG�P� if and only if CK�Q� ⊆ CK�P�.

(3.3) Theorem. Suppose that G = KH is p-solvable, where K G G, H ⊆G, K ∩H = L G G, and ��K x L�; �G x K�� = 1. Suppose that L ⊆ J ⊆ Hsatisfies CK/L�H� = CK/L�J�. Assume that �K/L� is not divisible by p. Letτ ∈ IBr�L� and θ ∈ IBr�K � τ� both be J-invariant. Then IH�θ� = IH�τ� andthere is a bijection∗x IBr�G � θ� → IBr�H � τ� such that

ϕ�1�θ�1� =

ϕ∗�1�τ�1�

for ϕ ∈ IBr�G � θ�.Proof. We argue by induction on �G x L�. Write C/L = CK/L�H� =

CK/L�J�.Suppose first that θ is H1-invariant, where J ⊆ H1 ⊆ H. Hence, C/L =

CK/L�H1�. By Lemma 2.5(a), we know that θL has an H1-invariant ir-reducible constituent τ1 and that all of them are C-conjugate. Also, byLemma 2.5(a), we have that the J-invariant irreducible constituents of θL


are C-conjugate. Since τ1 is J-invariant, we have that τc1 = τ for some c ∈ C.Hence, we see that τ is H1-invariant. This proves that IH�θ� ⊆ IH�τ�.

We let I = IG�τ�, so that J ⊆ IH�θ� ⊆ I ∩H, by the previous paragraph.Let M = K�I ∩ H�. Note that IG�θ� = KIH�θ� = IM�θ�. If M < G, byinduction we will have that IH�θ� = IH�τ� and that there is a bijection∗x IBr�M � θ� → IBr�I ∩H � τ� such that

ϕ�1�θ�1� =

ϕ∗�1�τ�1�

for ϕ ∈ IBr�M � θ�. Now, since M = IG�θ� because M ∩ H = I ∩ H =IH�θ�, by the Clifford correspondence for Brauer characters we will havethat induction of Brauer characters defines bijections IBr�M � θ� →IBr�G � θ� and IBr�I ∩H � τ� → IBr�H � τ�. Since �G x M� = �H x I ∩H�,the theorem easily follows in this case.

So we may assume that M = G or, equivalently, that H ⊆ I. Now,I ∩K = IK�τ� = U G I and KI = G. Also,

CU/L�H� = CK/L�H� ∩U/L = CK/L�J� ∩U/L = CU/L�J�;

so the hypotheses of the theorem are satisfied in I = UH with respectto L ⊆ J ⊆ H. Now, choose the unique ν ∈ IBr�I ∩ K � τ� with νK = θ.Since τ and θ are J-invariant, so is ν. Assume that I < G. By inductionwe have that ν (and hence θ) is H-invariant and that there is a bijection∗x IBr�I � ν� → IBr�H � τ� such that

ϕ�1�ν�1� =

ϕ∗�1�τ�1�

for ϕ ∈ IBr�I � ν�. Also, by the Clifford correspondence, we have thatcharacter induction is a bijection IBr�I � τ� → IBr�G � τ�. We claim thatinduction is a bijection IBr�I � ν� → IBr�G � θ�. If η ∈ IBr�I � ν�, then, byMackey, we have that �ηG�K = �ηU�K contains νK = θ. Since IBr�I � ν� ⊆IBr�I � τ� we have that induction defines an injection IBr�I � ν� → IBr�G �θ�. Now, let ψ ∈ IBr�G � θ�. Hence, ψ lies over τ and we let ζ ∈ IBr�I � τ�be such that ζG = ψ. To prove the claim, it suffices to show ζ ∈ IBr�I � ν�.Now, θ is a constituent of ψK = �ζG�K = �ζU�K and every constituent ofζU lies over τ. Now, θ is a constituent of some ρK for some ρ ∈ IBr�U � τ�which lies under ζ. But then ρK = θ and ρ = ν by the uniqueness inthe Clifford correspondence. So ζ ∈ IBr�I � ν�, and this proves the claim.Furthermore,

ξG�1�θ�1� =

ξ�1�ν�1�


for ξ ∈ IBr�I � ν�. Hence, we easily deduce in this case that there exists abijection ∗x IBr�G � θ� → IBr�H � τ� such that

ϕ�1�θ�1� =

ϕ∗�1�τ�1�

for ϕ ∈ IBr�G � θ�. So we now assume that τ is G-invariant.By use of a modular character triple isomorphism (Theorem 8.28 of

[6]) we may assume that L is a central p′-subgroup of G. Hence, we havethat K is a normal p′-subgroup of G. Let Z be the Hall π-subgroup of Lwhere π is the set of prime divisors of �H x L�. Then K = K0 × Z whereK0 is the normal Hall π ′-subgroup of G. We let L0 = L ∩K0, so that L =L0 ×Z. Now H = L0 ×A where A is the Hall π-subgroup of H and we setB = J ∩A. Set λ = τZ and write θ = θ0 × λ and τ = τ0 × λ for uniquelydefined θ0 and τ0 irreducible characters of K0 and L0 (respectively). Now,we have that CK/L�A� = CK/L�H� = CK/L�J� = CK/L�B�. Also, since K/Land K0/L0 are A-isomorphic, we see that CK0/L0

�A� = CK0/L0�B�. Since

��A�; �K0�� = 1 and L0 ⊆ Z�G�, by coprime action we deduce that

CK0�A� = CK0

�B�:Now, τ0 is A-invariant (because L0 ⊆ Z�G�) and θ0 is B-invariant (since θis). Since CK0

�A� = CK0�B�, it follows by Lemma 2.4 that θ0 is A-invariant.

Thus θ = θ0 × λ is invariant in LA = H. This proves the first part of thetheorem. Now, since K/Z and L/Z are π ′-groups and G/K and H/L areπ-groups, there exist extensions θ ∈ IBr�G� and τ ∈ IBr�H� of θ0 × 1Zand τ0 × 1Z (respectively) by Theorem 8.13 of [6]. Now, by Corollary 8.20of [6], we have that β 7→ βθ is a bijection IBr�G/K0� → IBr�G � θ0� andβ 7→ βHτ is a bijection IBr�G/K0� → IBr�H � τ0�. (Of course, we are usingthat restriction defines a bijection IBr�G/K0� → IBr�H/L0�.) Thus thereis a bijection

∗x IBr�G � θ0� → IBr�H � τ0�such that

ζ�1�θ�1� =

ζ∗�1�τ�1�

for all ζ ∈ IBr�G � θ0�. Now, the theorem follows because βθ lies over θ ifand only if βA lies over λ if and only if βHτ lies over τ.

Theorem A of the Introduction easily follows from the next result.

(3.4) Theorem. Suppose that G is p-solvable and q-solvable. Let P ∈Sylp�G� and Q ∈ Sylq�G� be such that PQ = QP . Then NG�Q� ⊆ NG�P�if and only if every irreducible p-Brauer character of G of q′-degree has p′-degree.


Proof. We argue by induction on �G�. Let π = �p; q� so that G is π-separable. Let M = Oπ ′ �G�. Let ϕ ∈ IBr�G� and let θ ∈ IBr�M� be underϕ. By Theorem 8.30 of [6], we have that ϕ�1�

θ�1� divides �G xM�, and we deducethat ϕ has p′-degree (or q′-degree) if and only if θ has p′-degree (or q′-degree). Since NM�Q� ⊆ NM�P� if and only if NG�Q� ⊆ NG�P�, by Lemma2.1, arguing by induction, we may assume that M = G.

Let K = Oπ�G� and let L = Oπ ′ �K�. Let H = PQL. Note that wemay assume that H < G, since otherwise G is a π-group and the theoremis true in this case by Lemma 3.1. Now, by Theorem 3.2, we have thatNG�Q�L ⊆ NG�P�L if and only if every irreducible p-Brauer character ofG/L of q′-degree has p′-degree. Hence, in either direction, we have thatNG�Q�L ⊆ NG�P�L. By the remark after the proof of Theorem 3.2, wealso have that CK/L�QL� = CK/L�H�. Assuming this, we prove that everyirreducible p-Brauer character of G of q′-degree has p′-degree if and onlyif every irreducible p-Brauer character of H of q′-degree has p′-degree.

Assume that every irreducible p-Brauer character of G of q′-degree hasp′-degree. Let η ∈ IBr�H� be of q′-degree. Let τ ∈ IBr�L� under η. If Iis the inertia group of τ in H, by the Clifford correspondence (and usingthat η�1� is not divisible by q) we have that I contains a Sylow q-subgroupof H. Hence, by replacing τ by some H-conjugate, we may assume that τis Q-invariant. By Lemma 2.5(a), let θ ∈ IBr�K� be a Q-invariant characterover τ. By Theorem 3.3 (with J = QL), there exists ϕ ∈ IBr�G � θ� suchthat

ϕ�1�θ�1� =

η�1�τ�1� :

Now, θ�1�τ�1� divides �K x L� (by Theorem 8.30 of [6]), and hence p and q do

not divide θ�1�τ�1� . Now,

ϕ�1� = η�1�θ�1�τ�1�

is not divisible by q. Therefore, by hypothesis, ϕ�1� is not divisible by p,and we conclude that η�1� is not divisible by p. The other direction followsfrom a similar argument.

Assume now that NG�Q� ⊆ NG�P�. Then NH�Q� ⊆ NH�P� and, arguingby induction, we have that every irreducible Brauer character of H of q′-degree has p′-degree. By the previous paragraph, the same happens withG. Conversely, suppose that every irreducible Brauer character of G of q′-degree has p′-degree. Then the same happens with H and, by induction,we have that NH�Q� ⊆ NH�P�. Since NG�Q�L ⊆ NG�P�L, we obtain thatNG�Q� ⊆ NG�P� by Lemma 2.2. This proves the theorem.

Next is Corollary B from the Introduction.


(3.5) Corollary. Suppose that G is a finite �p; q�-solvable group, andlet Q ∈ Sylq�G� and P ∈ Sylp�G�. Then the condition that NG�Q� is con-tained in some G-conjugate of NG�P� can be read from the character tableof G.

Proof. By Higman’s Theorem 8.21 of [2], we know in the character ta-ble which are the conjugacy classes of p-regular elements. Since G is p-solvable, by the Fong–Swan Theorem ((10.1 of [6]) we have that IBr�G� =�χ0 � χ ∈ Irr�G� and χ0 is not of the form α0 + β0 for characters α;βof G� is determined by the character table of G. Now, by Theorem 3.4,we only have to check whether or not every ϕ ∈ IBr�G� of q′-degree hasp′-degree.

REFERENCES

1. M. Isaacs, Character of π-separable groups, J. Algebra 86 (1984), 98–128.2. M. Isaacs, “Character Theory of Finite Groups,” Dover, New York, 1994.3. C. Jansen, K. Lux, R. Parker, and R. Wilson, “An Atlas of Brauer Characters,” London

Mathematical Society Monographs, New Series, Vol. 12 11, Clarendon, Oxford, 1995.4. G. Navarro, Two groups with isomorphic group algebras, Arch. Math. 55 (1990), 35–37.5. G. Navarro, Actions and invariant character degrees, J. Algebra 160 (1993), 172–178.6. G. Navarro, “Characters and Blocks of Finite Groups,” London Math Society Lecture

Note Series, Vol. 10 250, Cambridge Univ. Press, Cambridge, UK, 1998.7. G. Navarro and T. Wolf, Character degrees and local subgroups of π-separable groups,

Proc. Amer. Math. Soc. 126 (1998), 2599–2605.

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Sylow Normalizers and Brauer Character Degrees