PROCEEDINGS OF THE AMERICAN MATHEMATICAL SOCIETY Volume 131, Number 10, Pages 3019–3020 S 0002-9939(03)06884-9 Article electronically published on March 11, 2003

NUMBER OF SYLOW IN p-SOLVABLE GROUPS

GABRIEL NAVARRO

(Communicated by Stephen D. Smith)

Abstract. If G is a finite and p is a , let νp(G)bethe number of Sylow p-subgroups of G.IfH is a of a p- G,weprovethatνp(H) divides νp(G).

1. Introduction

Let G be a finite group, p aprimenumber,andwriteνp(G)forthenumber of Sylow p-subgroups of G.IfH is a subgroup of G, then it is well known that νp(H) ≤ νp(G). In this note, we prove the following elementary fact which perhaps has not been noticed up to now. Theorem A. Suppose that G is a finite p-solvable group. If H is a subgroup of G, then νp(H) divides νp(G). If G is not p-solvable, then Theorem A is simply not true. For instance, if G = A5, H = A4 and p =3,thenνp(H)=4andνp(G) = 10. 2. Proof We start with the following result. (2.1) Lemma. Suppose that A is a finite group acting coprimely on a finite group G,andletH be an A-invariant subgroup of G.LetC = CG(A).Then|C : C ∩ H| divides |G : H|. Proof. By standard arguments, recall that in any coprime action, if q is a prime, then every A-invariant q-subgroup of G is contained in an A-invariant Sylow q- subgroup of G.Furthermore,anytwoA-invariant Sylow q-subgroups are C-con- ∈ ∩ ∈ jugate and if P Sylq(G)isA-invariant, then P C Sylq(C). Now, let Q be an A-invariant Sylow q-subgroup of H.ThenQ∩C ∩H = Q∩C is a Sylow q-subgroup of C ∩ H.Now,Q is contained in an A-invariant Sylow q-subgroup P of G.Also, C ∩ P is a Sylow q-subgroup of C.Now,|C ∩ P : C ∩ Q|≤|P : Q|, and therefore |C ∩ P : C ∩ Q| divides |P : Q|. We deduce that |C|q/|C ∩ H|q divides |G|q/|H|q. Hence, |C : C ∩ H| divides |G : H|.  ∈ ⊆ ⊆ (2.2) Lemma. Suppose that G is p-solvable and let P Sylp(G).IfP M G, then |NG(P ):NM (P )| divides |G : M|.

Received by the editors April 25, 2002 and, in revised form, May 13, 2002. 2000 Subject Classification. Primary 20D20. The author’s research was partially supported by DGICYT.

c 2003 American Mathematical Society 3019

License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use 3020 GABRIEL NAVARRO

Proof. We argue by induction on |G|. Suppose first that there is a proper N of G such that G/N is a p0-group. Let V = M ∩N. Now, by induction, we have that |NN (P ):NV (P )| divides |N : V |. Now, by the Frattini argument, we have that G = NNG(P ). Hence, G =(NM)NG(P ) and thus |G : NM| = |NG(P ):NNM(P )|. Now, we claim that NNM(P )=NN (P )NM (P ). By the Frattini argument, we have that M =(M ∩ N)NM (P ). Thus, NM = NNM (P ). Now, NNM(P )=NNM(P ) ∩ NM (P )N = NM (P )NN (P ), as claimed. Now,

|NG(P ):NM (P )| = |NG(P ):NNM(P )||NNM(P ):NM (P )|

= |G : NM||NN (P ):NV (P )| divides |N : V ||G : NM|.Now,|N : V | = |NM : M|, and we are easily done. So we may assume that there is a proper normal subgroup N of G such that G/N is a p-group. Let H = N ∩M. By induction, we have that |NN (P ∩N):NH (P ∩N)| divides |N : H| = |G : M|.Now,letG0 = NG(P ∩N). We have that NG(P ) ⊆ G0. ∩ | | Let M0 = G0 M. Suppose that G0 is proper in G.ThenNG0 (P ):NM0 (P ) = |NG(P ):NM (P )| divides |G0 : M0|.Now,sinceP ⊆ NM (P ∩ N)andPN = G, we have that

|G0 : M0| = |NG(P ∩ N):NM (P ∩ N)| = |NN (P ∩ N):NH (P ∩ N)| divides |N : H| = |G : M|, and the result easily follows. Hence, we may assume that P ∩N/G.LetN¯ = N/P ∩N.ThenP acts coprimely on N¯.LetC = NG(P )∩N, ¯ ¯ and notice that C = CN¯ (P ). Then

|NG(P ):NM (P )| = |C¯ : C¯ ∩ H¯ | . Also, |G : M| = |N¯ : H¯ | . Now, the result follows by Lemma (2.1).  Proof of Theorem A. We argue by induction on |G : H|. Suppose that H

Departament d’Algebra,` Facultat de Matematiques,` Universitat de Valencia,` 46100 Burjassot, Valencia,` Spain E-mail address: [email protected]

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