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NUMBER of SYLOW SUBGROUPS in P-SOLVABLE GROUPS 1 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 SUBGROUPS IN p-SOLVABLE GROUPS GABRIEL NAVARRO (Communicated by Stephen D. Smith) Abstract. If G is a finite group and p is a prime number, let νp(G)bethe number of Sylow p-subgroups of G.IfH is a subgroup of a p-solvable group 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).ThenjC : C \ Hj divides jG : Hj. 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- 2 \ 2 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,jC \ P : C \ Q|≤jP : Qj, and therefore jC \ P : C \ Qj divides jP : Qj. We deduce that jCjq=jC \ Hjq divides jGjq=jHjq. Hence, jC : C \ Hj divides jG : Hj. 2 ⊆ ⊆ (2.2) Lemma. Suppose that G is p-solvable and let P Sylp(G).IfP M G, then jNG(P ):NM (P )j divides jG : Mj. Received by the editors April 25, 2002 and, in revised form, May 13, 2002. 2000 Mathematics 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 jGj. Suppose first that there is a proper normal subgroup N of G such that G=N is a p0-group. Let V = M \N. Now, by induction, we have that jNN (P ):NV (P )j divides jN : V j. Now, by the Frattini argument, we have that G = NNG(P ). Hence, G =(NM)NG(P ) and thus jG : NMj = jNG(P ):NNM(P )j. 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, jNG(P ):NM (P )j = jNG(P ):NNM(P )jjNNM(P ):NM (P )j = jG : NMjjNN (P ):NV (P )j divides jN : V jjG : NMj.Now,jN : V j = jNM : Mj, 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 jNN (P \N):NH (P \N)j divides jN : Hj = jG : Mj.Now,letG0 = NG(P \N). We have that NG(P ) ⊆ G0. \ j j Let M0 = G0 M. Suppose that G0 is proper in G.ThenNG0 (P ):NM0 (P ) = jNG(P ):NM (P )j divides jG0 : M0j.Now,sinceP ⊆ NM (P \ N)andPN = G, we have that jG0 : M0j = jNG(P \ N):NM (P \ N)j = jNN (P \ N):NH (P \ N)j divides jN : Hj = jG : Mj, 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 jNG(P ):NM (P )j = jC¯ : C¯ \ H¯ j : Also, jG : Mj = jN¯ : H¯ j : Now, the result follows by Lemma (2.1). Proof of Theorem A. We argue by induction on jG : Hj. Suppose that H<K<G. By induction, νp(H) divides νp(K), νp(K) divides νp(G), and we are done. So we may assume that H is a maximal subgroup of G.SinceG is p-solvable, we have that jG : Hj is a power of p or not divisible by p. j j 2 j Assume first that G : H is a power of p,andletP Sylp(G). Since ( G : j j j \ 2 H ; G : P ) = 1, we have that G = HP and Q = P H Sylp(H). Also, G = HNG(P )andjG : NG(P )j = jH : NH (P )j.Now,NH (P ) ⊆ NH (P \ H)and therefore νp(G)=jH : NH (P )j divides jH : NH (P \H)j = νp(H), and we are done in this case. j j 2 ⊆ Now, assume that G : H is not divisible by p and let P Sylp(H) Sylp(G). By Lemma (2.2), we have that jNG(P ):NH (P )j divides jG : Hj, and therefore jH : NH (P )j = νp(H) divides jG : NG(P )j = νp(G), as desired. Departament d'Algebra,` Facultat de Matematiques,` Universitat de Valencia,` 46100 Burjassot, Valencia,` Spain E-mail address: [email protected] License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use.
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