View metadata, citation and similar papers at core.ac.uk brought to you by CORE provided by Elsevier - Publisher Connector JOURNAL OF ALGEBRA 148, 42-52 (1992) On the Lattice of S-Subnormal Subgroups* A. BALLESTER-BOLINCHES Departammt D’Algehm, Uniwrsitat de Val&cia, C/Doctor Moliner, 50, E-46100 Burjassot, Valencia, Spain K. DOERK Fachbereich Mnthematik, Johannes Gutenberg-linillersitiit, Postfach 3980, D-6500 Maim, Germmy AND M. D. PEREZ-RAMOS Departament D’AIgebra, Unioersitat de Val&ia, C/Doctor Moliner, 50, E-46100 Burjassot, Valencia, Spain Conmunicated by Gernot Stroth ReceivedMay 18, 1990 1. INTRODUCTION All groups considered are finite. It is well known that the set of all subnormal subgroups of a group G is a lattice. Now, assume that 5 is a subgroup-closed saturated formation containing the class of all nilpotent groups. It is known that the inter- section of two g-subnormal subgroups of a soluble group G is an S-subnormal subgroup of G (cf. [6, 5,2]). One might wonder if the set of all g-subnormal subgroups of a soluble group is a lattice. The answer is negative in general (see [2]), but there exist subgroup-closed saturated formations containing properly the class of all nilpotent groups for which the lattice property holds. In this paper, we obtain the exact description of the subgroup-closed saturated formations 5 of soluble groups such that the set of all g-subnormal subgroups is a lattice for every soluble group. 2. PRELIMINARIES In this section we collect some definitions and notations as well as some known results. * Researchpartially supported by DGICYT-proyecto No. PS87-005%CO2-02. 42 0021.8693192 $5.00 CopyrIght 0 1992 by Academic Press. Inc. Al, nghfs 01 reproductron ,n any form reserved S-SUBNORMALsuBGRouPs 43 First recall that if 5 is a saturated formation and G is a group, a maxi- mal subgroup A4 of G is said to be S-normal in G if the primitive group G/M, E 5, and B-abnormal otherwise (M, = n { Mg ( g E G} ). A subgroup H of a group G is called S-subnormal in G if either H = G or there exists a chain H=H,<H,-,< ... <H,=G such that Hi+, is a maximal g-normal subgroup of Hi, for every i = 0, .. n - 1. If M is a maximal subgroup of a group G such that M, = 1, we will say that M is a core-free maximal subgroup of G, and if X is a class of groups we denote char 3E= (pi P/C, E X}, where C, denotes the cyclic group of order p. Recall that the boundary b(X) of a class of groups X consists of all groups G satisfying G 6 3Zand G/N E X for all 1 # N u G. If rc is a set of prime numbers, let G and 6, denote the classesof soluble and soluble n-groups, respectively. ‘$2 denotes the class of all nilpotent groups. The following results will turn out to be crucial in the proof of our main result. (2.1) LEMMA [S, Theorem 1.3.111. Zf H is a subnormal subgroup of a finite group G, then Sot(G) normalizes H. (2.2) LEMMA[2]. Let G e !R& where 5 is a saturatedformation, and let E be an S-maximal subgroup of G satisfying G = EF(G). Then E is an B-normalizer of G. (2.3) LEMMA [3, 11. Let G be a group and let 3 be a saturatedforma- tion. If G5 is abelian, then G8 is complemented in G and any two com- plements in G of G5 are conjugate. The complements are the S-normalizers of G. (2.4) LEMMA[4, Hilfssatz 1.31. Let H be a group with a unique minimal normal subgroup M, where A4 is a q-group. If p is a prime distinct from q then H has a faithful irreducible representation over GF(p). (2.5) LEMMA [7, Lemma 1.11. Let 5 be a subgroup-closed saturated formation. Zf H is @subnormal in G and H < U < G, then H is S-subnormal in U. For details about formations the reader is referred to [4]. 3. THE LATTICEOF ~-SUBNORMALSUBGROUPS (3.1) LEMMA. Let G be a group and let H be an S-subnormal subgroup of G, where 5 is a subgroup-closed saturated formation. Then H5 is sub- normal in G. 44 BALLESTER-BOLINCHES, DOERK, AND PbREZ-RAMOS Proof: We argue by induction on the order of G. Let N be a minimal normal subgroup of G. Then HN/N is @subnormal in G, so that HsN is subnormal in G. If HN is a proper subgroup of G, then H n is subnormal in HN by Lemma 2.5. Therefore H” is subnormal in G and the lemma is true. So we can assume G = HN for each minimal normal subgroup N of G. This implies that H is contained in a core-free S-normal maximal subgroup of G. But then G is an S-group and H n = 1 is subnormal in G. (3.2) LEMMA. Let C%= {rc;: ie I} be a partition of z, a set of prime num- bers, and let 5 be the saturated formation of soluble groups locally defined by the formation function f given by f(p) = G=,, if p E 71, and ig 1, and f(q) = 0, if q $7t. Then G is an B-group if and only if G is a soluble n-group with a normal Hall ni-subgroup, for every i E 1. Proof Assume that GE 3. We see that G has a normal Hall rci-subgroup for each i E 1 by induction on [Gl. Let N be a minimal normal subgroup of G and let p E rc, the prime divisor of )Nj. If H is a Hall rci-subgroup of G, then Nd H and H/N is a Hall rci-subgroup of G/N. By induction we deduce that H is a normal subgroup of G. Now, let A be a Hall rcj-subgroup of G with j# i. Then A n N = 1 and since AN/N is a Hall nj-subgroup of G/N, we have that AN CI G. If AN < G, then A 4 AN and A a G. Therefore we can assume that G = AN and A is a maximal sub- group of G. Since GE 5, we have that G/A, E 6,, but then A = A, because AlA, E C,, that is, A is a normal subgroup of G. The converse is clear since the chief factors of a such group G are @central. (3.3) THEOREM. Let 3 be a subgroup-closed saturated formation of soluble groups containing ‘92, the class of all nilpotent groups, and let f be the full and integrated local formation function defining 5. Then 5 satisfies the following condition: (*) “If H, and H, are two @subnormal subgroups of GE G, then (H,, Hz) is an g-subnormal subgroup of G”, tf and only tf f can be described in the following way: “There exists a partition (zi}iSl of P, the set of all prime numbers, such that f (p) = 6,, for every prime number p E ni and for every i E I”. Proof Assume that the formation 5 = LF( f) as above satisfies (*). It is well known that f(p) is a subgroup-closed formation for every prime p (cf. [4, Hilfssatz 2.21). We split the first part of the proof into the following steps: (1) For each prime number p E P, every primitive group GE 5 n (b(f(p)) is cyclic. ~-SUBNORMAL sum3RouPs 45 It is clear that G has a unique minimal normal subgroup N, and evidently N must be a q-group, where p #q E P. Therefore there exists an irreducible and faithful G-module V, over GF(p). We claim that G has a unique maximal subgroup M such that MG = 1, which provides the result. Assume that M1 and M, are maximal subgroups of G, M, # M2 and (Mi)G = 1, i = 1, 2. Then Mi Ed. Consider now the semidirect product H = [V,,] G, with respect to the action of G on VP. Clearly H# 5, so H8 = V,, and G is not B-subnormal in H. But for i = 1,2, VPMi is g-normal maximal subgroup of H, and Mi is S-subnormal in V,M,, because 1/,M, E 6,f(p) = f(p) c 5, that is, Mi is G-subnormal in H. Since 8 satisfies (*), we have that G = (M,, M2) is $j-subnormal in H, which is a contradiction. (2) If p and q are prime numbers and q E char(f(p)), then P E char(f(q))- Assume that C,$f(q) and consider an irreducible and faithful C,-module VP over GF(p). Then the semidirect product [ V,]C,, with respect to the action of C, on VP, belongs to gnb(f(q)), which contradicts (1). (3) If p and q are prime numbers and p E char(f(q)), then char(.f(p)) = char(.f(q)). If r E P and r E char(f(q))\char(f(p)), then r # q and C, Ef(r), because of (2). Consider now an irreducible and faithful C,-module I/, over GF(r). Then [ V,] C, E i-j n b(f(p)), a contradiction with (1). (4) If p, qE P and p Echar(f(q)), then GPcf(q).