A GENERALIZED FRATTINI SUBGROUP OF A FINITE GROUP Thesis for the Degree of Ph. .D. MICHIGAN STATE UNIVERSITY HASSO CHELLARAM BHATIA 1970' TH?“ 0-169 Date This is to certify that the thesis entitled A GENERALIZED FRATTINI SUBGROUP OF A FINITE GROUP presented by Has so Che llaram Bhat ia has been accepted towards fulfillment of the requirements for Ph .1) 0 degree in M3 themat iCS Linux? Michigan State University August 7. l‘ g ‘ Bl NDlNG BY "DAG & SIJIIS' 500K BINDERY INC. LIBRARY BINDERS ABSTRACT A GENERALIZED FRATTINI SUBGROUP OF A FINITE GROUP BY Hasso Chellaram Bhatia The Frattini Subgroup and its influence on a group have been the objects of study by group theorists for many years. In recent years several mathematicians have inves- tigated various generalized Frattini subgroups. w. Gaschfitz and more recently J. Rose and H. Bechtellhave studied the family of non normal maximal subgroups of a finite group. w. Deskins has considered the family of maximal subgroups whose indices in the finite group are not divisible by a given preassigned prime. In the present investigation we consider another family of maximal subgroups of a finite group. Let 3(6) be the family of maximal subgroups of nonprime index of the group G. Let L(G) denote the intersection of the members of the family 3(6). In case 3(6) is empty, define L(G) = C. By a well-known theorem of B. Huppert, a finite group is supersolvable iff each of its maximal subgroups has prime index. It follows that L(G) = G iff G is supersolvable. Let {(6) be the family of all the non normal maximal subgroups of G and let A(G) denote the intersection of the Hasso Chellaram Bhatia members of 1K6). Since a group is nilpotent iff all its maximal subgroups are normal, it follows that A(G) = 6 iff 6 is nilpotent. The Subgroup A(G) and its generalizations have also been studied by, among others, J. Beidleman, T. Seo [Pacific Jour. of Math. 23 (1967)] and D. Dykes [Pacific Jour. of Math. 31 (1969)]. Our investigations show that the two families iKG) and 3(6) act in many respects as natural analogues of each other, the former with respect to the nilpotent structure and the latter with reSpect to the supersolvable structure of a group. We also show that L(G) is related to the hyperquasi- center [N. Mukherjee, Ph.D. Thesis, Michigan State University, 1968] in the same manner as A(G) is to the hypercenter. In Chapter I we also study the influence of 8(6) on the solv- ability of G improving some results of J. Rose. The main results of Chapter I are: (l) L(G) is supersolvable, (2) If G is solvable, L(G) n 6' is nilpotent, (3) The hyper- quasicenter of 6 is contained in 146), and (4) If every maximal subgroup of G of nonprime index is nilpotent, then G is solvable. In Chapter 11 some results of Beidleman, Seo and Dykes are generalized. It is shown that L(G) satisfies some Frattini-like properties described below. Definition (a). Let n be the set of primes dividing ‘6‘. Let n c n 1 such that for all elements p and q of n1 and n ~ n1 reSpectively, p > q. Then n1 is an upper set (UP-set) for Hasso Chellaram Bhatia 6. Definition (b). A proper normal subgroup is a special L-Subgroup of G if for every normal Subgroup N of 6 and A a Hall n-subgroup for n an arbitrary UP-set for N, G = HNG(A) implies 6 = NG(A). Definition (c). A proper normal subgroup H of G has property (9) in G if for every N A c with H _<. N, N/H n-closed for n a UP-set for N implies N is n-closed. Following R. Baer we also define the concept of a weakly hyperquasicentral subgroup. The main results of Chapter II are: (l) L(G) has property 0?) in G, (2) If H is a Special L-subgroup of G, then G/H has the Sylow tower prOperty if? G has that property, (3) For a nilpotent normal subgroup H, the definitions (b) and (c) are equivalent to each other and also to the concept of weakly hyper- quasicentral subgroups. So far we have not been able to confirm the necessity of the 'nilpotency' condition on the subgroup H above. We conjecture that this condition can be replaced by 'the Sylow tower property'. In Chapter III we conclude the present investigation by obtaining some conditions for the group G to be supersolvable; and for A(6) and L(G) to coincide. Following Bechtell [Pacific Jour. of Math. 14 (1964)], we define L-series and relative L-series (relative to the commutator subgroup) of the group G. The latter series turn out to be far more interesting for our consideration. Two of the results of this chapter are: (1) A(6) = L(G) if an L-series of 6 terminates in <1>. (2) 6 is supersolvable if and only if the upper relative L- series of G coincides with the descending central series of 6'. A GENERALIZED FRATTINI SUBGROUP OF A FINITE GROUP BY Hasso Chellaram Bhatia A THESIS Submitted to Michigan State University in partial fulfillment of the requirements for the degree of DOCTOR OF PHILOSOPHY Department of Mathematics 1970 TO MY FATHER ii ACKNOWLEDGEMENT I wish to express my gratitude to Professor W.E. Deskins for his guidance throughout the preparation and writing of the thesis. His patient understanding and constant interest were of singular importance in completion of this work. I also thank Professor J. Adney and Dr. I.N. Sinha for taking keen interest in my work and for many useful suggestions concerning the investigation. Special thanks are also due to my wife Champa whose patience and constant encouragement made this work possible and with whom I share this accomplishment. iii TABLE OF CONTENTS Page INDEX OF NOTATIONS ................................. v INTRODUCTION ....................................... 1 CHAPTER I ........... ...... . ...... ........... ...... ......... 8 1 1 Basic Concepts and Preliminary Results ........ 9 l 2 The Subgroup L(G) ............................ 12 1.3 L(G) Expressed as a Direct Product ........... 21 l 4 L(G): A Generalization of A(G) .............. 24 1 5 Relationship of L(G) with the Hyperquasicenter of G ........................ 29 1.6 A Condition for Solvability of a Group ........ 33 CI-{AHIER II O..COOOOOOOOOCCOOOCCOCOCCOOOO0.....00000000000000 41 1 L-SubgrOUPS Of a Gro‘lp G .‘C...’............ 41 2 Special L-8ubgroups of a Group G ............. 45 .3 Subgroups with Property (9) ................. 50 4 Weakly Hyperquasicentral Subgroups ............ 60 CHAPTER III O00......00000.00...00.......OOOOCOOOOOOOOOOOOOO 66 3.1 The L-series of the Group 6 .................. 67 3.2 Relative L-series of the Group G ............. 71 BIBLIOGRAPHY ............... . ....................... 78 iv INDEX OF NOTATIONS 1 Relations: c Is a subset of 3 Is a subgroup of S Is not a subgroup of < Is a proper subgroup of ($' for emphasis) A Is a normal subgroup of A Is not normal 6 Is an element of f 13 not an element of E Isomorphic to E Isomorphic to a subgroup of II Operations: ~ Set difference < > Subgroup generated by X Direct product of groups \GI The number of elements in G, the order of G \x‘ The order of the element x n Set of primes p(n) The prime divisors of the number n p(nl) < p(nz) The prime divisors of n1 are smaller than the prime divisors of n2 (m,n) = l m and n are two numbers relatively prime to each other Index of H in 6 Factor group X-lGx -l X 8X h'lx'lhk Subgroup generated by all [h,k]; h E H, k 6 K Set whose elements are Union Intersection For each The subgroup generated by nth powers of all the elements of 6. III Groups and Elements: n-element n'-e1ement n-group n'-group The identity of a group The identity of a factor group An element whose order is divisible by only the primes in the set n An element whose order is not diviSible by any prime in the set n A group whose order is divisible by only the primes in the set n A group whose order is not divisible by any prime in the set n Core K 2(6) Q(G) LP(G) Other: The largest normal subgroup of 6 con- tained in the subgroup K The center of the group G The quasicenter of the group G The hypercenter of the group G The hyperquasicenter of the group 6 Defined on page 24 Defined on page 12 The symmetric group of degree n The alternating group of degree n The Frattini subgroup of G The commutator subgroup of G The p-length of G x,y permute means , permute. vii INTRODUCTION In recent years a number of mathematicians have studied various generalized Frattini subgroups. W. Gaschfitz [1] and more recently J. Rose [16, 17] and H. Bechtell[7] have considered in detail the family of non-normal maximal subgroups and their intersection. W. Deskins [8] has con- sidered the family of maximal subgroups whose index in the group is not divisible by a given preassigned prime p, and studied the Structure of the subgroup ¢p which is the inter- section of the members of this family of maximal subgroups. More recently, J. Beidleman and T. Seo [5, 6] have generalized the work of H. Bechtell. In the present work we investigate another family of maximal subgroups of a finite group 6, namely that consisting of those maximal subgroups whose index in G is BEE a prime. Denote by L(G) the intersection of all the maximal subgroups of 6 whose index in G is not a prime. In case 6 has no maximal subgroup of non-prime index, then as is customary, we set L(G) = 6. It is well-known [13] that a finite group is supersolvable if and only if all its maximal subgroups have prime index. Thus L(G) = G if and only if G is super- solvable. In a natural way therefore, L(G) is a measure of how far 6 is from being supersolvable. Since for a finite group G a normal maximal subgroup has a prime index, the family of non-normal maximal subgroups considered by Gaschfitz contains all the maximal subgroups with non-prime index in G. If A(G) denotes the intersection of all the non-normal maximal subgroups, then A(G) is contained in L(G). Thus in part our work is a direct extension of the ideas of Gaschfitz. It is well-known (see for example Gaschfitz [11] and Deskins [8]) that A(G) is nilpotent. At the outset it is easy to see that L(G) in general need not be nilpotent. This is so because for any Supersolvable group G, L(G) and G coincide and there are finite Supersolvable groups which are not nilpotent as for example 83, the symmetric group on three symbols. It seems reasonable however, to conjecture that L(G) is Supersolvable. The verification of this con- jecture is one of the main results in Chapter I. It has been proved by, among others, Beidleman [6] that G is nilpotent if 6/A(G) is nilpotent. In Chapter I we have been able to Show that this result is still true when A(G) is replaced by L(G) and 'nilpotent' by 'supersolvable'. Incidentally our result also generalizes a well-known result, namely: 6 is Supersolvable iff G/§(6) is Supersolvable. Note that Q(G), the Frattini subgroup of a group G, is contained in L(G). The concept of the quasicenter as a generalization of the center was introduced by 0. Ore [15] in the following manner: An element x E 6 is quasicentral in G if for every element g G 6, the cyclic subgroups and permute. In other words = forms a subgroup. The subgroup generated by all the quasicentral elements of the group G is the quasicenter of G and is denoted by Q(6). Ore proved that the quasicenter of the group is a char- acteristic Subgroup. He studied the quasicenter as a quasi- normal subgroup for the symmetric and alternating groups. The structure of the quasicenter has been investigated in some detail by N. Mukherjee [14] who showed that the quasi- center of a group is nilpotent and that each of its Sylow sub- groups is generated by quasicentral elements of the group. Since every element in the center of a group commutes with every element of the group, it is in particular quasi- central in 6. Thus the center of a group is part of the quasi- center. This leads to the definition of the hyperquasicenter [14] which is a generalization of the concept of the hypercenter and is defined in an analogous manner. The hyperquasicenter, denoted by Q*(G), is a normal Supersolvable subgroup of C and contains every supersolvably embedded subgroup of G [14]. Also C is Supersolvable iff G = Q*(G). It is known [7, Theorem 2.2] that the hypercenter 2*(6) of a group is contained in A(6). In general A(G) does not contain the hyperquasicenter of 6. However the analogy between A(6) and L(G) suggests a relationship between the hyperquasi— center and L(G). We are able to establish this in Chapter I by proving that the hyperquasicenter of a group 6 is contained in L(G). J. Rose [16] proved that a group is solvable if all its non normal maximal subgroups are nilpotent. As a last result in Chapter I, we show that this theorem of Rose can be improved by replacing 'non normal maximal Subgroups' by only those 'maximal subgroups whose index in G is non—prime'. We have an example to show that our result is false if 'nilpotency' is replaced by 'supersolvability'. J. Rose extended his result by proving that a finite group is solvable if each of its non normal maximal subgroups is Supersolvable and has prime power index. Our example shows that this result of Rose is not true if 'non normal maximal subgroup' is replaced by 'maximal subgroup with non prime index'. This is not too surprising since maximal subgroups with non prime index may in general, form a very 'small' portion of all the non normal maximal subgroups of the group. However, we have so far not been able to obtain suitable con- ditions under which a finite group, with every maximal subgroup of non prime index Supersolvable, is solvable. In Chapter II we generalize the work of Beidleman and Seo [5, 6] and D. Dykes [10] and Show that the subgroup L(G) of a finite group 6 belongs to a class of normal subgroups called Special L-subgroups of 6. Following Beidleman and Seo, a prOper normal subgroup H of a finite group G is called a generalized Frattini subgroup of 6 if for every normal sub- group N of G and a Sylow subgroup P of N, G = HoNG(P) implies that G = NG(P). Generalized Frattini subgroups are normal, nilpotent subgroups of 6. Also A(C) is a gen- eralized Frattini Subgroup of 6. Furthermore, if H is a generalized Frattini subgroup of 6 such that G/H is nil- potent then 6 is nilpotent. Generalizing the above definition we call a normal subgroup H of 6 an L-subgroup of G if H satisfies the above definition only for the Sylow subgroups of N correSpond- ing to the largest prime divisor of the order of N. We show that an L-subgroup is p-closed, for p the largest prime divisor of its order. Moreover if p is the largest prime divisor of the order of G, then G/H p-closed implies 6 is p-closed. Also L(G) is an L-subgroup of 6. Some of the results of [5, 6] we have not been able to generalize. The main difficulties in this reSpect seem to be the following: (a) The concept of an L-subgroup seems to be too general to sufficiently restrict the behavior of Sylow subgroups for the various prime divisors. (b) It is known (see for instance Gaschfitz [11] and Deskins [8]) that A(G)/§(G) = Z(G/§(G)). An analogous relation connecting Q(C) and L(G) however, does not hold. A considerably restricted sub-class of the class of generalized Frattini subgroups, called Special generalized Frattini subgroups was considered by Dykes [10]. He proved that A(C) is a special generalized Frattini subgroup of the group 6. Moreover the concepts of Special generalized Frattini subgroups and the weakly hypercentral subgroups defined by R. Baer [4] are equivalent. Following these ideas we define special L-subgroups of the group 6; and prove that L(G) is one such Subgroup of G. In Chapter II we also define weakly hyperquasicentral subgroups and a certain property 6?). One of the main reSults we prove in this direction states that: For a normal nilpotent subgroup H of a group 6, the following are equivalent: (a) H is a special L-subgroup of G. (b) H is weakly hyperquasicentral in G. (c) H has property G?) in G. We also Show that if H is a Special L-Subgroup of G, then G has the Sylow tower property if G/H has that property. Our results on the above concepts are partial in the following two respects: (1) The equivalence of the definitions of Special L-subgroups and weakly hyperquasicentral Subgroups is shown only for nilpotent normal subgroups. However we believe this to be true for any normal subgroup satisfying the Sylow tower prOperty. ' (2) We have not been able to justify the term 'weakly hyperquasicentral' as we cannot yet prove that a hyperquasi- central Subgroup [14] is also weakly hyperquasicentral. How- ever we have shown this to be true for the nilpotent case. In Chapter III we conclude the present investigation by considering some conditions under which A(G) and L(G) coincide. FollowingBechtell[7] we define an L-series and the upper and lower L-series. It is shown that if the upper L-series of the group 6 terminates in <1>, then A(G) = L(G). We also define a relative L-series of 6 (relative to the commutator subgroup of 6). These series yield some inter- esting information about the group 6. Some of the results here are: (1) The upper relative L-Series coincides with the descending central series of 6' iff G is supersolvable. (2) The following are equivalent for a group G: (i) L*(G) = <1>, where L*(G) is the terminal member of the upper relative L-series. (ii) L(G) n c" = 2*(c') n c". (iii) If S is any subgroup of 6 generated by 3 elements one of which belongs to L(G), then L(G) n s' s 2*(3'). (3) If H is a solvable normal subgroup of G, then L*(G) = <1> implies 1*(H) = <1>. We have not been able to decide whether the solvability condition on H in (3) is necessary. CHAPTER I Let 8(6) be the family of those maximal subgroups of the finite group 6 which have nonprime index in 6. Let L(G) denote the Subgroup which is the intersection of all the members of 8(6). As is customary, define L(G) = G in case the set 8(6) is empty. It is obvious that L(G) is a characteristic subgroup of G. In this chapter we investigate the structure of the subgroup L(G) and study its influence on the structure of the group G. The main results proved here are the following: (1) The subgroup L(G) is supersolvable. (2) If G/L(G) is supersolvable, then 6 is super- solvable. (3) The hyperquasicenter of the group G is contained in L(G). (4) If every maximal subgroup of 6 of nonprime index is nilpotent, then G is solvable. For the sake of completeness and easy reference we include here some known results. Proofs or references are also given. Throughout, only finite groups will be considered. An index of notations and special symbols appears on a previous page C 1.1 Basic Concepts and Preliminary Results. We begin with the definition of the Frattini subgroup of a group and a lemma due to Frattini of which we shall make frequent use. Definition 1.1.1: Let Q(G) =11{MIM is a maximal sub- group of 6]. 6(6) is called the Frattini subgroup of 6. @(6) is a nilpotent characteristic Subgroup of 6. The argument used in the following lemma is due to Frattini. Lemma 1.1.2: Let H be a normal Subgroup of a group G. If P is a Sylow subgroup of H, then G = HNG(P). Proof: Since H is normal in G, for every g E 6, g-ng S H. Therefore P and g-1 Pg are both Sylow p-sub- groups of H, for the prime p. Consequently both P and g-ng are conjugate in H. Let x be an element of H such that x'lPx = g'1 -1) -1 Pg. Then (gx '1P(gx-1) = P, so that gx belongs to the normalizer of P in G. The element g there- fore belongs to the set NG(P)x of NG(P)H. This proves that 6 = HNG(P). We shall refer to the above result as 'Frattini's lemma'. Definition 1.1.3: A group 6 is p-closed if G has a normal Sylow p-subgroup. Definition 1.1.4 (Baer): If the group G is p-closed, and if G/P is abelian of exponent dividing p-l, where P is the Sylow p-subgroup of G, then G is strictly p-closed. It is easy to verify (see for example [1]) that 6 is Strictly p-closed if, and only if 6' and 6p.1 are p-subgroups 10 of 6. Definition 1.1.5: A group G has the Sylow tower pro- perty of supersolvableggroups, if for every homomorphic image H of G and p the maximal prime divisor of the order of H, H is p-closed. Note: We shall throughout refer to the above property as the Sylow tower prOperty. The above definition is easily seen to be equivalent to the following: Let p1 > p >...> pn be the natural ordering of prime 2 divisors of the order of 6, and let Pi be a Sylow pi-sub- group of G. Then for each k, P1P2'°°Pk is a normal sub- group of C. A group having the Sylow tower property is solvable. Subgroups and homomorphic images of 6 have the Sylow tower property if 6 has that property. Definition 1.1.6: A group 6 is supersolvable if every homomorphic image of G has a nontrivial normal Subgroup which is cyclic. It is well-known and easily verified that subgroups, homomorphic images and direct products of supersolvable groups are again supersolvable. An extension of a supersolvable group by a supersolvable group in general is not Supersolvable. How- ever, an extension of a cyclic group by a supersolvable group iS Supersolvable. By a well-known theorem of B. Huppert [13], we have, 11 Theorem 1.1.7: The following properties of a group G are equivalent: (1) G is supersolvable. (ii) 6/§(6) is supersolvable. (iii) Every maximal subgroup of G has index a prime. By a result of Baer [1, Theorem 2.1], the above conditions are also equivalent to, (iv) G has the Sylow tower prOperty; and NG(P)/CG(P) is, for every Sylow p-subgroup P of G, strictly p-closed. If 6 is supersolvable, then its commutator subgroup 6' is nilpotent; and 6 has the Sylow tower prOperty [13, p. 415]. Definition 1.1.8: A subgroup M of 6 is supersolvably eededded (SSE) in 6 if for each homomorphism of 6, the image of M contains a cyclic subgroup which is normal in the homo- morphic image of G. Remark 1. A supersolvably embedded subgroup is Super- solvable. A proof of this may be found in [14]. Remark 2. Every SSE subgroup M of 6 is normal in 6. This can be easily proved by induction on the order of Remark 3. If M is SSE in G and G/M is super- solvable, then G is supersolvable. This is true because there is a normal series of 6 up to M with cyclic factor groups. By the supersolvability of G/M, this chain can be extended to a normal series of G 12 with cyclic factor groups, which implies the supersolvability of 6. Next we state a result of Baer [1, Theorem 4.1] that we shall need. Theorem 1.1.9: The following properties of a normal subgroup K of a group 6 are equivalent: (i) K is SSE in 6. (ii) K/QCK) is sss in G/e(K). (iii) K has the Sylow tower prOperty; and if P is a Sylow p-subgroup of K, then NG(P)/CG(P) is strictly p-closed. 1.2 The Supgroupd L(G). Definition 1.2.1: Let 6 be a group. Define L(G) 0{M\M is a maximal Subgroup of 6 and [6:M] is not a prime} R(G) n[MIM is a maximal subgroup of 6 and [6:M] is a prime}. In case 6 has no maximal subgroup of nonprime index, we set L(G) = 6. Similarly R(6) = 6 if G has no maximal Subgroup of prime index. It is obvious that both L(G) and R(6) are characteristic subgroups of 6. Moreover R(6) n L(G) = 6(6), where @(G) is the Frattini subgroup of 6. By Theorem 1.1.7, a group is supersolvable if and only if each of its maximal subgroups has prime index. It follows therefore that G is supersolvable if and only if L(G) = 6. Proposition 1.2.2: If K is a normal subgroup of 6, then 13 (i) L(G)K/K s L(G/K). (ii) R(c)K/K s R(6/K). (iii) e(G)K/K s Q(G/K). In particular, if K A c and K s L(G), then L(G/K) = L(G)/K. Proof: (i) Let M; be a maximal Subgroup of G/K of nonprime index. There exists a maximal subgroup Mi of 6 containing K such that Mi = Mi/K° Also [G/K: Mi/K] = [6:Mi], so that Mi has nonprime index in 6. Then n(Mi/K) = L(G/K). Let J be the intersection of all the Mi's correSponding to M]; then J contains L(G) as well as K. Hence L(6)K/K s J/K. But now it is easy to Show that J/K S Q(Mi/K) = L(G/K). For if not, suppose there exists xK e J/K such that xK é n(Mi/K), where x E J. Then there exists some M; = Mi/K such that xK (Mi/K. This implies that X 6 Mi' But this is a contradic- tion since x 6 J 3 Mi. Thus L(G)K/K S J/K S L(G/K) and (i) is proved. (ii) This is proved in the same manner as (i). Let Mi be any maximal subgroup of G/K of prime index. There exists a maximal subgroup Mi of 6 containing K Such that Mi = Mi/K’ and [6/K: Mi/K] = [6:Mi]. Consequently R(6) 5 M1 and therefore R(6)K/K s T/K, where T is the intersection of all those maximal subgroups of G that correSpond to Mi's. Then as in (i) we can show that R(6)K/K s T/K sr1(Mi/K). (iii) Duplicating the argument of (i), let J be the intersection of all those maximal subgroups of 6 that contain K5 Then since 6(6) is contained in every maximal subgroup Of' G, Q(6)K/K S J/K. The maximal Subgroups of G/K are the 14 images under the homomorphism 9: G a G/K, of those maximal subgroups of 6 which contain K. Consequently, J/K s(.flMi)/K s n(Mi/K) = e(G/K). The proof of (i), (ii) and (iii) is thus complete. In the particular case, suppose K s L(G). By (i) L(G)/K s L(G/K). It only remains to show that L(G/K) s L(G)/K. For this assume that xK is an element of L(G/K) which does not belong to L(G)/K. Then x G L(G) and so there exists a maximal subgroup M of 6 of nonprime index such that x 4 M. Since by hypothesis K S L(G) < M, XK é M/K. But M/K is a maximal subgroup of G/K with LG/K: M/K] = [6:M] = nonprime. We thus arrive at a contradiction since xK is an element of L(G/K). Therefore L(G/K) = L(G)/K. Corollary 1.2.3: (a) L(G/e(c)) = L(G)/Q(6). (b) L(G/L(G)) = <1>. Remark: In general L(6)K/K < L(G/K). To confirm this consider 6 = A4. The 4-group V4 is a normal subgroup of 6. The maximal subgroups of 6 of nonprime index are cyclic of order 3. Since these maximal subgroups are non normal, their intersection is the identity subgroup. Thus L(G) = <1>. Hence L(CQVa/Va = <15. On the other hand since G/V4 E 63, we have L(G/Va) 9 C3. In general L(G) is not nilpotent. For example let 6 = 83. Since 8 is Supersolvable, L(G) = 8 As is well- 3 3' known S3 is not nilpotent. We also know that a supersolvable gzxmip has nilpotent commutator subgroup. Furthermore L(G) = 6 itif 6 is supersolvable. From these considerations it seems 15 natural to ask what the nature of the common Subgroup of L(G) and 6' would be. We answer this in the following result. Theorem 1.2.4: If G is solvable, then G' n L(G) is nilpotent. Proof: Let T be any maximal subgroup of G of prime index, say p. Then 6 has p cosets, say alT,azT,...,a T. Denote by S the set of these p cosets P of G. If g Q 6, then g(aiT) = gaiT E 3. Thus elements of 6 act as permutations of the set S. If aiT and ajT are any two arbitrary members of S, we can find an element g E G such that ai = gaj and hence aiT = ga T = g(ajT). Thus we J see that G acts as a transitive permutation group of degree p, on the set S. Also by hypothesis 6 is solvable. By [13, Satz 8], a solvable transitive permutation group of prime degree is metabelian. Let TG be the core of T in 6, which is the intersection of all the conjugates of T in G. If g 6 TG’ then g belongs to every c0njugate of T and con- sequently induces the identity permutation on the set S. Thus the group of permutations representing 6 is isomorphic with G/TG, which as remarked earlier is metabolism. This implies that 6" S TG S T. Since T was chosen arbitrarily, we see that G" S R(G). Now 6" S R(G) implies that G" n L(G) S R(G) n L(G) = Q(G). Also [6' n L(G), 6'] s [c',c'] = 6". Since G' n L(G) and 6' are both normal in 6, we have [6' n L(G), 6'] S L(G) fl 6". Thus, [0' n L(G), 6'] s L(G) n 6" 5 MG). 16 Now consider the following two cases: Case (1): Suppose 6(6) = <1>, Then [6' O L(G), G'] = <1>, which implies that G' n L(G) commutes with every element of 6'. Since G' n L(G) s 6', we have 6' n L(G) s 2(6'). Hence 6' n L(G) is abelian and therefore nilpotent. Case (2): Suppose Q(G) # <1> and let Gl= G/®(G). Obviously s(c/e(c)) = <13. Then by Case (1), (E)' o L(E) is nilpotent. By Corollary 1.2.3, L(G) = L(G). Moreover by [12, p. 18], [c/e(G). G/¢(G)] = G :(:)G (E)' = (c/i(c))' = c'i(G)/e(c). Then we have that and so 6'(6)n L(G) = G'Mc) n L(G) e(c) Q(G) Q(G) GIQ(G) fl L(G) is a normal subgroup of 6 containing @(G) G'MG) n L(g) and (I) (C) H. Wielandt (see for instance [7, Corollary 2.3.23), we conclude is nilpotent. Now since is nilpotent, by a well-known theorem of that G'e(c) n L(G) and hence c' n L(G) is nilpotent. The proof is thus complete. Remark: As mentioned before, for any supersolvable group G the commutator subgroup G' is nilpotent. This re- sult can be easily deduced from the above theorem. For when 6 is supersolvable, it is solvable and L(G) = 6. Thus 6' n L(G) = 6', which is nilpotent by Theorem 1.2.4. One of our aims in this chapter is to prove that L(G) is supersolvable. Our next result is a step in this direction. It also shows that L(G) is solvable. Prqposition 1.2.5: In any group 6, L(G) has the Syrlow tower property. 17 Proof: We proceed by induction on the order of 6. Let p be the largest prime dividing the order of L(G) and P a Sylow p-subgroup of L = L(G). We claim that P is normal in 6. For this, suppose NG(P) # G and let S be a maximal subgroup of 6 containing NG(P). By Frattini's lemma, 6 = LNG(P) = LS. Since S is a proper subgroup of G, L S S. So S has prime index in 6, say q. Since q = [6:8] = [L:L{] S], we note that q divides [L(G)]. Moreover Since NL(P) S L n S, by the Sylow theorems and a result in Scott [18, 6.2.3] we have [LzL n S] = 1 + kp, where k is some integer. It follows that q = 1 + kp and hence p divides q-l. But this is impossible since both q,p divide [L(G)] and p is the largest such divisor. Hence we conclude that P A 6. Now by Proposition 1.2.2, L(G)/P = L(G/P) and by the induction hypothesis, L(G)/P has the Sylow tower property. It is now easy to see that L(G) itself has the Sylow tower preperty. Corollary 1.2.6: L(G) is solvable. This is true since the Sylow tower property implies solvability. Corollary 1.2.7: If 6/L(G) is solvable and (G/L(G))" is nilpotent, then 6" is nilpotent. Proof: (a) Obviously 6 is solvable. (b) By Theorem 1.2.4, 6" n L(G) S Q(6).. Then (6/L(6)" = E§§é%l'3 6"/6" n L(G) implies that 6" is nilpotent. Before proceeding to prove the supersolvability of L(G), we digress a bit to consider some generalizations and study the influence of L(G) on the group 6. 18 It is known that if 6/Q(G) has the Sylow tower pro- perty, then 6 has that property. In the following proposi- tion we improve this result. Proposition 1.2.8: If 6/L(6) has the Sylow tower prOperty, then G has the same property. Proof: We use induction on the order of 6. First let K be any normal subgroup of 6. We Show that G/K satisfies the hypothesis. By Proposition 1.2.2, L(6)K/K S L(6/K) and hence G/K/L(6/K) is isomorphic to a homomorphic image of G/K/L(G)K/K 5 G/L(G)K (by the isomorphism theorems). Since 6/L(6) and hence 6/L(6)K has the Sylow tower property, 6/K/L(6/K) has that property. Hence for any K A G, G/K satisfies the hypothesis. Now let p be the largest prime divisor of ‘6‘ and P be a Sylow p-Subgroup of 6. If P S L(G), then p is also the largest prime divisor of [L(G)] and P its Sylow p-sub- group. By PrOposition 1.2.5, P is normal in L(G) and hence normal in G. If P i L(G), then p/‘G/L(6)| and p is its largest prime divisor. By [18, 6.1.16], PL(6)/L(6) =IP is a Sylow p-Subgroup of 6/L(G). Since by hypothesis G/L(G) has the Sylow tower prOperty, P is normal in 6/L(6) and hence PL(6) A 6. Now P is also a Sylow p-subgroup of PL(6) and using Frattini's lemma, we have C = L(G)PNG(P) = L(6)NG(P). If P K 6, then 6 * NC(P). Let S be a maximal subgroup of 6 containing NG(P). So 6 = L(6)S and L(G) S S. Therefore S has prime index in 6, say q. Since NG(P) S S and P 19 is a Sylow subgroup of G, by the Sylow theorems as in Proposi— tion 1.2.5, q = [6:3] = 1 + kp, where k is some integer. This is impossible since p is the largest prime divisor of \6‘. Thus we conclude that P is normal in 6. By the first part of the proof, 6/P satisfies the hypothesis. This means that 6/P/L(6/P) has the Sylow tower property. By the induction hypothesis it follows that G/P has the same prOperty. Consequently G itself enjoys the Sylow tower property. This completes the proof. In the same vein we improve another well-known result. As remarked earlier (Theorem 1.1.7), a group 6 is Supersolv- able if and only if G/®(6) is supersolvable. We shall Show that this result still holds when Q(G) is replaced by L(G). The fact that L(G) is solvable (Corollary 1.2.6) will be used in the next theorem. Theorem 1.2.9: A group G is supersolvable if and only if G/L(6) is supersolvable. Proof: (a) Since every homomorphic image of a super- solvable group is supersolvable, clearly 6/L(6) is supersolv- able if 6 is supersolvable. (b) Conversely, assume that 6/L(C) is Supersolvable. Since L(G) and G/L(G) are solvable, 6 is solvable. Moreover, since G/L(6) and 6/6'§(6) are Supersolvable, G/L(G) x G/G'Q(G) is supersolvable. It follows that 6/L(6) n G'e(c) which is isomorphic to a subgroup of G/L(G) X G/G'§(G) is also supersolvable. Note that MG) s L(G) n G'MG). If Q(c) = L(G) n c'uc), then from the previous remark, 6/§(G) is supersolvable. This by Theorem 1.1.7 20 implies the supersolvability of 6. We may thus assume that Q(G) < L(G) n G'@(G). Next we Show that L(G) 2(2;¢§Gl is SSE in 6/§(G). Hereafter in the proof let 6 = 6(6) and L = L(G). As shown in the proof of Theorem 1.2.4 using the solvability of 6, LJ%-§:2 is nilpotent. Let P be the Sylow p-subgroup of 1.n G's/Q. Then P is normal in G/@. Let N/Q be a minimal normal subgroup of G/Q contained in P. Since Q(G/Q) = <1>, there exists a maximal subgroup S/Q of 6/6 such that N/é S 8/6. Also S is a maximal subgroup of 9, Then c/g = N/Q . 8/6 and since N/Q is solvable minimal normal in 6/6, we have N/é n S/Q = <15 [2, p. 118]. So 6 = NS, N n S = i and N i 3. Since N s L(G), it is contained in every maximal Subgroup of 6 of nonprime index. It follows that S has prime index in G and hence S/Q has prime index in G/i. Therefore IN/iI = [G/s: s/i] is a prime and hence N/Q is a subgroup of order p. Furthermore, since Q(G/Q) = <15 and P A G/T, it follows [11] that P is elementary abelian. From this it follows that P = C 1 are cyclic subgroups of order p and each 61 is normal in X C X... C , where the C,'s r l 2 6/Q(G). Thus there is a normal series of G/§(G) up to P with cyclic factor groups. By definition this implies that P is SSE in G/T. Hence every Sylow Subgroup of LJZEELS is SSE in G/Q. It can be easily shown that Product of two SSE sub- groups is again SSE. Since ILDEEIQ. is the product of its Sylow I Subgroups, we conclude that lLQE§_2 is SSE in G/T. 21 Now since G/Q/LIW G'Q/Q 5 G/1.n G'Q is supersolvable and 1.0 G'é/é is SSE in G/Q, it follows that G/é is super- solvable (Remark 3, Definition 1.1.8). Consequently G is supersolvable and the proof is complete. Corollary 1.2.10: If N A G and N s L(G), then G/N supersolvable implies that G is supersolvable. Proof: Since G/L(G) is isomorphic to a homomorphic image of G/N, it is Supersolvable. Hence by the theorem, G is supersolvable. Corollary 1.2.11: Let D(G) =n{N A G‘G/N is super- solvable}. Then either D(G) = <1> or D(G) £ L(G). Proof: Suppose D(G) # is part of L(G). By definition, G/D(G) is supersolvable and by Corollary 1.2.10, G is supersolvable. But this implies that D(G) = be- cause G/<1> is supersolvable. This contradiction proves the Corollary. 1.3 L(G) Expressed as a Direct Product. Next we study the relationship of L(G) with L(H) where H S G. Let H be a Subgroup of G. Gaschfitz [11] showed that if H §_G, then @(H) s Q(G). We see that no such relation holds in the case of L(G). For example, let G = A AS shown 4. earlier (Proposition 1.2.2), L(G) =<1>. But L(Va) = V4, where V 4 is the normal Sylow 2-subgroup of G. In the case of R(C) however, we do have such a re- lat: ion , 22 Proposition 1.3.1: If H A_G, then Q(H) S R(H) S R(G). Proof: If R(G) = G, we have nothing to prove. So let Esbe aumximal subgroup of G of prime index. If H s S, then R(H) s H s s. If H i s, then G = HS and so [CzS] = [HSzS] = [H:H n S]. Therefore H n S is a subgroup of H of prime index, hence it is also a maximal subgroup of H. Thus R(H) s H n S s S. We see that in any case R(H) S 8. Since 8 is arbitrary, R(H) s R(G). Since @(H) s R(H), the assertion is proved. The next two results concern L(G) when the group G is the direct product of two groups. If G = AB is an extension of a group A by a group B, it is not true in general that L(G) = L(A)-L(B). This is easily seen from the example of A However in case G is 4. a direct product we have, Proposition 1.3.2: If G = A x B and (|A|,\B\) = 1, then L(G) = L(A) X L(B). Proof: First we show that L(A) X L(B) s L(G). Suppose x E L(A) but x i L(G). Then there exists S, a maximal sub- group of nonprime index in G Such that x d S. Since G is the direct product of two groups of relatively prime orders, any subgroup of G has that property. Thus S = A1 X Bl’ where A1 S A, B1 S B. Since x E A and x Z S, A1 < A. Moreover if B1 < B, then S = A1 X B1 < A1 X B < C. This contradicts the maximality of 8. Hence B1 = B and S = A1 X B. Then S/B = A1 X B/B 5 A1 implies that A1 is a maximal sub- group of A of nonprime index. But x 6 A1 S S. This is a 23 contradiction. Therefore L(A) S L(G). Similarly L(B) S L(G). Next to Show that L(G) S L(A) X L(B), let x = ab 6 L(G), where a,b are unique elements of A,B reSpectively. We claim that a 6 L(A) and b E L(B). First Suppose a d L(A). Then there exists a maximal subgroup A1 of A of nonprime index such that a f A1. If S = A1 X B, then S is a maximal sub- group of G of nonprime index. For suppose S is not maximal in G and let T be a maximal Subgroup of G containing 3. Then T = A2 X B, which implies that A1 < A2 < A. This con- tradicts the maximality of A1. Therefore S is maximal and of nonprime index in G. Since x = ab 6 L(G), a 6 A1. There- fore a E L(A) and similarly b E l‘B). The result is thus proved completely. Remark: We have as yet, not been able to confirm that the condition (‘A\,‘B|) = 1 is necessary. The following is another useful condition for deter- mining L(G) when G is a direct product. Proposition 1.3.3: If G = A X B and A or B is supersolvable, then L(G) = L(A) X L(B). Proof: We proceed by induction on the order of G. Thus by the induction hypothesis if a group has order less than ‘6‘ and is the direct product of two groups one of them supersolvable, then the conclusion holds for that group. WIDG we may assume that A is supersolvable. Then there exists a minimal normal Subgroup M of A which is cyclic of prime order. Since A,B centralize each other, in particular M is normalized by B. Hence M is normal in 24 G and being of prime order it is a minimal normal subgroup of C. Now if S is a maximal Subgroup of G of nonprime index, then M S S. For otherwise M S S implies that G = MS and hence [G:S] = ‘M‘ is a prime. It follows there- fore that M S L(G). Now by Proposition 1.2.2, L(G/M) = L(G)/M and also L(A/M) = L(A)/M. Note that L(A) = A, since A is supersolvable. By the induction hypothesis, since G/M = A/M x B/<1>, we have L(G/M) = L(A/M) x L(B/<1>) and so L(G)/M = L(A)/M x L(B)/<1>. This implies that L(G) -- L(A) x L(B), 1.4 L(G): A generalization of A(Gl. In this section we show that our work is a generalization of some of the ideas of Gaschfitz. In [11] he considered the family 1K6) of all the non normal maximal Subgroups of a group G. Since in a nilpotent group all the maximal subgroups are normal, {(6) is empty for such a group. The family £(G) has also been studied from essentially two different standpoints by H. Bechte11[7] and J. Rose [16, 17]. Betchel has considered the basic Structure of A(C) and its relation to other subgroups of G; and Rose has studied the influence of the family iKG) on the group G, in particular on the solvability of G. We begin with a mention of some of the basic notions .and known results. Definition 1.4.1: For a group G define A(G) = n{M\M is a non normal maximal Subgroup of G}. IMefine A(G) = G is case G has no non normal maximal subgroup. 25 It is obvious that A(G) is a characteristic subgroup of G and contains the Frattini subgroup Q(G). Also notice that A(G) is contained in L(G) since the maximal subgroups of G with nonprime index are non normal and hence are in the family £(G). The following theorem gives some of the results on A(G), due to Gaschiitz [ll], Deskins [8] and Bechtell[7]. Theorem 1.4.2: In a group G, (a) A(G) is nilpotent. (b) The hypercenter Z*(G) is contained in A(G). (c) A(G)/Q(G) = Z(G/Q(G)). To these we also add some results due to Beidleman and Seo [S] and Rose [16], (d) If G/A(G) is nilpotent, then G is nilpotent. (e) If all the non normal maximal subgroups of G are nilpotent, then G is solvable. (f) If all the non normal maximal subgroups of G are Supersolvable and have prime power indices in G, then G is solvable. From the definitions and some of the results we have obtained so far, it is apparent that there is a similarity in the nature and behaviour of the two families 8(6) and iKG); the former with respect to the supersolvable Structure of G and the latter with reSpect to the nilpotent structure of G. Thus L(G) acts in a natural way as a generalization of A(G). These considerations also motivate us to search for some suit- able form of 'generalized hypercenter' - one that is relevant to our family 8(6). We show that this role is effectively 26 played by the hyperquasicenter of the group G defined in the following pages. It is now natural to ask: Can all the results of Theorem 1.4.2 be generalized? We successfully answer this ques- tion before closing Chapter I. To begin in the chronological order then, we first prove the Supersolvability of L(G). For this we need the following lemma whose proof is analogous to that of Theorem 1.2.4, and uses the solvability of L(G). Lemma 1.4.3: In a group G, L'(G) is nilpotent. Proof: Since L'(G) S L(G), it is contained in every maximal subgroup of G of nonprime index. If L'(G) is also contained in every maximal Subgroup of G of prime index, then clearly L'(G) is contained in every maximal subgroup of G. This implies that L'(C) S Q(G), hence it is nilpotent. We may therefore assume that there exists a maximal subgroup S of G of prime index not containing L'(G). Observe that if S is normal in G, then G/S is a cyclic group and hence L'(G) s G' s s, which is a contradiction. Therefore 5 is not normal in C. Now since S is maximal in G and L'(G) S S, we have c = L'(G)S = L(G)S. Then, p = [G:S] = [L(G)S:S] = [L(G):L(G) n s], for p some prime. It follows that L(G) n S is a subgroup of L(G) of prime index and consequently it is maximal in L(G). Moreover by Corollary 1.2.6 L(G) is solvable. Now by the same argument as in Theorem 1.2.4, it follows that L(G) acting as a group of permutations on the cosets of L(G) n S 27 is represented by a metabelian group, and the kernel of the representation is the core of L(G) n S in L(G). Note: L(G) is not necessarily metabelian. The permutation representation of L(G) is metabelian. Hence L"(G) S Core (L(G) n S) S L(G) L(G) n S < 8. Thus we conclude that L"(G) is contained in every maximal subgroup of G. Note that L"(G) = implies that L'(C) is abelian and we are done. So L"(G) is non trivial normal subgroup of G contained in @(G). Also L'(G)/L"(G) is abelian. By a theorem of Wielandt (see Betchel [7]) we conclude that L'(G) is nilpotent. We are now ready to prove our main theorem. Theorem 1.4.4: L(G) is supersolvable, for any group Proof: (By induction on the order of G). Let p be the largest prime dividing ‘L(G)‘. Since L(G) has the Sylow tower prOperty (Proposition 1.2.5), the Sylow p-subgroup P of L(G) is characteristic in L(G) and hence normal in G. Since L(G)/P = L(G/P), by the induction hypothesis L(G)/P is supersolvable. Moreover, if G has a non trivial normal q-subgroup Q (q ¢ p) contained in L(G), then once again by the induction hypothesis L(G)/Q is supersolvable. This implies that L(G) E L(G)/P n Q is isomorphic to a subgroup of L(G)/P X L(G)flQ and hence L(G) is Supersolvable. Thus we can assume that G has no non trivial normal q-subgroup contained in L(G). Furthermore if Q(L(G)) f <1>, then by the induction hypothesis L(G)/@(L(G)) is supersolvable and hence by Huppert's theorem (Theorem 1.1.7) L(G) is Supersolvable. 28 Therefore we may also assume that Q(L(G)) = and this in turn implies that Q(P) = <1>, Hence P is elementary abelian. By a theorem of Gaschfitz [11, Satz 7] since P is an abelian normal subgroup of L(G) and Q(L(G))¢1 P =<1>, it follows that P is completely reducible in L(G). This means that every normal subgroup of L(G) in P has a complement in P which is normal in L(G). Since L'(G) is a normal nilpotent subgroup of G, by an earlier remark L'(G) S P. Hence L(G)/P is abelian. Also by the SchUr-Zassenhaus theorem [18, 9.3.6], L(G) = PD where D is a p-complement in L(G); and D is abelian. Now let Q be a Sylow q-subgroup of L(G), for q # p. WlDG we may take Q S D. Since G has no normal q-subgroup in L(G), NG(Q) # G. However since D is abelian, D S NG(Q). By Frattini's lemma, G = L(G)NG(Q). If s is a maximal Sub- group of G containing NG(Q) we have, C = L(G)S and L(G) S 3. Since D S S, P S S. Therefore G = PS. Also in G, S has a prime index which is p (observe that q/p-l). Let K = L(G) n s. Then p = [6:8] = [L(G):L(G) n s] = [L(G):K]. Since K has prime index in L(G), it is a maximal Subgroup of L(G) not containing P. Therefore L(G) = PK and [L(G):K] - [PzP n K] = p. Therefore P0 = Prfi K is normal in P and K and hence normal in L(G). Also the index of P0 in P is p. Since P is completely reducible in L(G), there exists P1 A L(G) such that P l is cyclic of order p. 29 * x x , , Now let P = H P = P P ...P . Each P 18 cyclic l 1 2 n l xEG * of order p and is contained in P. Therefore P is a normal subgroup of G contained in L(G). Moreover each Pi is normal in L(G), since P1 is normal in L(G) and L(G) A G. For let Pi = Pi, then for every g E L(G), xgx"1 6 L(G). So -1 X P1gx = P1 and hence Pig = Pi. Since P is elementary * . abelian, we have P = P1 X P2 X...X Pn. Consequently there * exists a normal series of L(G) up to P with cyclic factor * * groups. Also by the induction hypothesis L(G)/P = L(G/P ) is supersolvable. Hence L(G) is supersolvable and the theorem is proved. Corollary 1.4.5: If [L(G)] does not contain prime divisors p,q such that q/p-l, then L(G) is nilpotent. Proof: We use induction on ‘G‘ and then essentially the method of the theorem can be duplicated to Show that q/p-l for some p, whenever a Sylow q-subgroup of L(G) is not normal. Thus under the hypothesis, every Sylow subgroup of L(G) is normal and hence it is nilpotent. Remark: As mentioned before, in general L(G) is not nilpotent. For example, in S3 L(SB) = 83 is not nilpotent. However the hypotheses of the Corollary 1.4.5 are clearly not satisfied. 1.5 Relationship of L(G), with the Hyperquasicenter of G. Next we consider a generalization of the hypercenter and investigate its relationship with the subgroup L(G). 30 The concept of the quasicenter as a generalization of the center of a group was introduced by O. Ore [15] as follows: Definition 1.5.1: Let G be a group. An element x 6 G is called a 9pasicentral element (QC-element) in G if permutes with for every element y E G. The subgroup generated by all the quasicentral elements in G is the qgasicenter of G, denoted by Q(G). The quasicenter is a characteristic subgroup of G. Since every element in the center of a group permutes with every element in the group, it follows that the center is con- tained in the quasicenter of the grOup. Note that if x is a QC-element in G, then permutes with every subgroup of G. The structure of the quasicenter and its generalizations has been investigated in more detail by N. Mukherjee [14]. In the next few theorems we list some of the results obtained by him. On occasions we shall use some of these. Theorem 1.5.2: In a group G, (a) If x is a Qc-element of G, then Xr is a QC-element of G, for every integer r. (b) The quasicenter of G is nilpotent and each of its Sylow subgroups is generated by QC-elements of G. Proof: [14, Theorem.l.7, Lemma 1.9, Theorem 1.10]. Analogous to the ascending-central series we also have, Definition 1.5.3 [14]: For a group G let, * <1> S Q(G) "' Q1 S Q2 S...S Qn - Q (G) be a normal series of 6 Such that Q(G/Qi) =Qi+1/Qi. 31 This series is the ascendingequasicentral series of G. The terminal member Q*(G) of the series is the hyperquasicenter of G. Some properties of the hyperquasicenter are listed in the following: Theorem 1.5.4 [14]: For a group G, (a) The hyperquasicenter is Supersolvable. (b) If T is a normal subgroup of G contained in Q*(G). then Q*(G/T) =Q*(c)/T. (c) If G/Q*(G) is supersolvable, then G is super- solvable. Proof: [14, Theorems 2.5, 2.3] Remark: Mukherjee [14] has shown that the hyperquasi- center of a group G contains the largest SSE subgroup of G. However, during a communication he pointed out that the hyper- quasicenter is itself SSE in G and hence is the largest SSE subgroup of G. In the following we shall prove that the hyperquasi- center of G is contained in L(G) and thus slightly improve Theorem 1.5.4 in view of the fact that conclusions of that theorem have already been proved for L(G) in place of Q*(G). Also by the remark above, we observe that in general the hyper- quasicenter is not equal to L(G). For this we argue as follows. It can be shown that in general the Frattini subgroup Q(G) is not SSE in C. It is also easy to Show that a normal Subgroup of G contained in a SSE subgroup of G is SSE in G. Now * if L(G) =tQ (G), then L(G) is SSE in G and hence by the 32 previous comment ¢(G) is SSE in G. This, as we remarked * earlier is not in general true. Therefore L(G) #‘Q (G) in general. We now turn to the proof of our theorem. First we show the following: Proposition 1.5.5: The quasicenter of a group G is contained in L(G). Proof: Assume Q(G) is not contained in L(G). Since Q(G) is nilpotent (Theorem 1.5.2), there is a Sylow p-subgroup P of Q(G) which is not contained in L(G). Since P is generated by QC-elements of G, there is a QC-element x of G Such that x i L(G). Therefore there exists a maximal sub- group S of G of nonprime index such that x 4 S. Since x is a.QC-element, and S permute. Consequently G = S. Also [x] = pa, for some integer a. Let r T = S n and ‘TI = p . Then T is contained in the unique a-r-l a-r-l r+l subgroup of and ‘\ = p . Since a-r-l T is the intersection of S with , S S. More- a-r-l over by Theorem 1.5.2(a), x is again a QC-element of G. a-r-l Therefore G = 5 and the index of S in G is equal a-r a-r-l to the index of T = in which is p. But this is a contradiction since 3 was assumed to have nonprime index in G. Therefore we conclude that Q(G) S L(G). From the above result we now immediately deduce, Theorem 1.5.6: The hyperquasicenter of the group G is contained in L(G). 33 Proof: Lat <1> S Q(G) = Q1 S Q2 S...S Qn = Q*(G) be the ascending-quasicentral series of C. By Pr0position 1.5.5, Q(G) s L(G) and Q(G/Ql) s L(G)/Q1. By definition, QZ/Ql = Q(G/Ql) hence Q2 S L(G). Continuing in this manner we see that the terminal member of the series namely Q*(G), is contained in L(G). As mentioned earlier (Theorem 1.4.2) in the case of A(G), we have A(G)/Q(G) = Z(G/§(G)). But no Such relation holds between L(G) and the quasicenter of G. Example 1.5.7: Let G = 83. Then L(G) = S3 and Q(G) = <1>. But the quasicenter of S3 is the cyclic sub- group of order 3. The above example leads to the following observation: Proposition 1.5.8: If a group G is solvable and Q(G) = <1>, then L(G)flQ*(G) is abelian. Proof: As shown in Theorem 1.2.9 the solvability of G implies that L(G) n G' is SSE in G. Since Q*(G) the hyperquasicenter of G contains every SSE subgroup, L(G) n G' S Q*(G). Thus L(G)flQ*(G) is isomorphic to a homo- morphic image of L(G)/L(G) n G' g L£§l§:.3 G/G'. This implies GI * that L(G)/Q (o) is abelian. 1.6 A Condition for Solvabilipy of a Group. J. Rose [16] has proved that if every non normal maximal subgroup of a group is nilpotent, then the group is solvable. In this section we shall improve this result slightly by showing that 'every non normal maximal subgroup' can be replaced by 34 'every maximal subgroup of nonprime index'. We begin with some well-known definitions and preliminary results that we shall need. Definition 1.6.1: A group G is called pwsolvable, for a prime p if there exists a series <1> = G S G S... 6n = G such that 61 A 6 and each factor 0 1 Gi/Gi-l is either a p-group or a p'-group. The p-length LP(G) of the p-solvable group G is the least number of p-factors appearing in the series of the kind Specified above. Definition 1.6.2: A group G is p-nilpotent, for a prime p if G has a normal p-complement. Theorem 1.6.3 (Thompson [19]): Lat p be an odd prime and P a Sylow p-subgroup of a group G. If for every non trivial characteristic subgroup P of P, 6 induces in P 0 O a p-group of automorphisms, then G has a normal p-complement. Lemma 1.6.4 (Rose [16, Cor: Lemma 1]): A group G can have at most one conjugacy class of nilpotent non normal maximal subgroups. Lemma 1.6.5 (Baer [2, p. 181]): If S is a maximal subgroup of a group 6 whose Core SG = <1> and whose index [6:3] is a prime p, then q < p for every prime divisor q of ‘8‘. Definition 1.6.6: A subgroup N of G is Subnormal in G if there exists a series N =‘N0 A N1 A...A 6. We are now ready to prove our theorem. 35 Theorem 1.6.7: Suppose every'uaximal subgroup of a group 6 of nonprime index is nilpotent. Then (i) G is solvable. (ii) If p is the largest prime divisor of ‘6‘, then either 6 is p-nilpotent or G has a normal p-sub- group P such that G/PO is p-nilpotent. More- 0 over, Lp(6) S 2. Proof: Let G be a minimal counter-example. Observe that if K A G, then G/K satisfies the hypothesis. (i) Let p be the largest prime divisor of ‘6‘ and P a Sylow p-subgroup of G. If P A G, then P and G/P are both solvable which implies that G is solvable. This con- tradicts,the assumption that G is a minimal counter-example. Therefore P A G. Let N(P) be the normalizer of P in G and let S be a maximal Subgroup of 6 containing N(P). The index of S in G is not a prime. For if q = [6:8] for some prime q, then by the Sylow theorems q = 1 + kp. This is impossible since q < p. Hence 8 has nonprime index in G and by hypothesis S is nilpotent. Since P is a Sylow Sub- group of S, N(P) = 8. Let PO be any non trivial char- acteristic Subgroup of P. By the same argument used for P, it follows that PO A 6. Consequently N(P) S N(PO) < 6. Since N(P) = S is maximal in 6, N(PO) = N(P). In a nil- potent group the elements of relatively prime order centralize each other. Hence N(Pb) induces only p-automorphisms in P0. Moreover, p is an odd prime. By Theorem 1J6.3, G has a normal p-complement D. Now 8 is a maximal subgroup of G 36 and is nilpotent of nonprime index. By Lemma 1.6.4, all the nilpotent non normal maximal subgroups of G are conjugate. Therefore all the maximal subgroups of G of nonprime index are conjugate to S. Since 6 is not simple, let M be a minimal normal subgroup of G. Since G/M is solvable, M cannot be solvable. Also M is the unique minimal normal subgroup of G. This is true because if N is another minimal normal subgroup of G distinct from M, then Mr] N = <1>» Since G/M and G/N are solvable, G/M n N‘E G/M X G/N is solvable. This contradicts the assumption that G is nonsolvable. Therefore M is unique. If M S S, then M is nilpotent. So M S S and hence G = MS. Let q be a prime divisor of [MzM n S] = [6:3] and Q a Sylow q-subgroup of M. Obviously Q < M since M is nonsolvable. Since M is minimal normal in 6, Q A G. Let T be a maximal subgroup of 6 containing NG(Q). Since D, the p-complement of G is normal in G, M is contained in D. This is so because of the uniqueness of M. Now by Frattini's lemma, (; = mam) = MT. So M at T and hence D at T. It follows that [6:T] = [DzD n T] is prime to p, since D is a p'-group. Suppose CoreGT # <1>. Since M S T, Mr] CoreGT < M. But this contradicts the minimality of M. Therefore CoreGT I“<1>. Again if [6:T] = r, a prime, then by Lemma 1.6.5, the prime divisors of ‘T‘ are smaller than r. But p/‘T‘ and it is the largest prime divisor of ‘6‘. This contradiction shows that T cannot have prime index. Consequently by an 37 earlier remark, T is conjugate to S. But this leads to a contradiction. To see this notice that [6:8] = [M:M n S] is a multiple of q and [6:T] = [M:M.n T] is prime to q since Q S M n T. But two conjugate Subgroups must have the same index in the group. Thus we arrive at a contradiction. Therefore 6 is solvable. (ii) If 6 does not possess a normal p-subgroup, then as in (i) by the application of Thompson's theorem, 6 has a normal p-complement and (ii) is proved. Assume therefore that G has a normal p-Subgroup and let P be the largest 0 such subgroup. Then consider G/PO. We note that G/P0 does not have a normal p-subgroup. For suppose there exists P, a non trivial normal p-subgroup of G/PO. Then there exists a p-subgroup P1 A G such that ‘P = Pl/PO. This contradicts the maximality of PO- Now since G/P0 satisfies the hypothesis and has no normal p-Subgroup, by the induction hypothesis G/P0 is p-nil- potent. Also if T/PO is the normal p-complement of G/PO’ then there is a normal series <1> S PO S T S 6. Hence the p ~1ength Lp(G) S 2. The theorem is thus proved. To continue further with the above theorem, let G’= G/P0 and let D. be the normal p-complement of Gl Let IE be a maximal subgroup of 6' containing D: Then 875' is maximal in 675.5 P. Since maximal subgroups of a nilpotent group are normal, S’A G. and [GT8] = p. 38 (a) Suppose now that every normal maximal subgroup of G of index p has nilpotent commutator subgroup. Consider P a Sylow p-subgroup of G. If ‘P‘ = p, then P is cyclic of order p. In the other case let P1 be any maximal sub- group of P. Since D A 6, 3,3' is a subgroup of 61 Moreover it is a maximal Subgroup in 6 containing D: By previous dis- cussion it follows that 1311? A E and is of index p. By hypothesis (a), ($15). is nilpotent. Now by [13, Satz 4] it follows that Pi A Fifi and hence is Subnormal in G. But it is easy to show (see for instance [16, Lemma 3]) that a group possessing a non trivial subnormal p-subgroup also possesses a non trivial normal p-subgroup. This is a contradiction since 6 has no non trivial normal p-subgroup.‘ Hence ‘Pi =and therefore every proper subgroup of P is abelian. Thus we see that under the hypothesis (a), P is either cyclic of order p or is minimal nonabelian. (b) Finally, suppose that every normal maximal subgroup of 6 of index p has the Sylow tower prOperty. Let P1 be a maximal Subgroup of P, where ‘P is a Sylow p-subgroup of G: Then as before, 5‘6' is a normal maximal Subgroup of G. of index p. Since p is the largest prime divisor of ‘PiD‘, by (b), Pi A Fifi. This implies that P1 is Subnormal p-sub- group of 6. As in (a) this is again a contradiction. There- fore we conclude that P' has no proper maximal subgroup and consequently it is cyclic of order p. The foregoing discussion can be summed up in the following proposition: 39 Proposition 1.6.8: Under the hypothesis of Theorem 1.6.7, if p is the largest prime divisor of ‘6‘ and PO the largest normal p-subgroup of G, then either PO is the Sylow p-subgroup of 6 or, (a) If every normal maximal subgroup of 6 of index p has nilpotent commutator subgroup, then a Sylow p-subgroup of G/PO is cyclic of order p or minimal nonabelian. (b) If every normal maximal subgroup of 6 of index p has the Sylow tower property, then a Sylow p-subgroup of G/PO is cyclic of order p. Rose [16, Theorem 4] extended his result by proving that if every non normal maximal subgroup of a group is supersolvable and has prime power index, then the group is solvable. By way of an example, we show that this result cannot be improved by replacing 'non normal maximal subgroups' by 'maximal subgroups with nonprime index'. Example 1.6.9: Let G B PSL (2,7), the simple group of order 168. It is well-known (see Scott [18, p. 336]) that the maximal subgroups of 6 have index 7 or 8. A maximal sub- group H of 6 of index 8 has order 21. Since H has two prime divisors, it is solvable and clearly it is supersolvable. H is not nilpotent since it can be shown [18, p. 336] that the normalizer of a Sylow 3-subgroup of C does not contain H. Thus the maximal subgroups of nonprime index are Super- solvable and have prime power index. To see how good a generalization of the family 1X6) is the family 3(6), we compare our results so far with 40 Theorem 1.4.2. We find that (a), (b) and (d) of that theorem have analogues for 3(6); (e) is slightly improved; while we have examples to Show that (c) has no counterpart and (f) can- not be improved in our context. This is the conclusion of Chapter I. CHAPTER II In this chapter we investigate some additional pro- perties of the subgroup L(G). It is Shown that L(G) belongs to a class of normal Subgroups we call the Special L-subgroups of the group 6. Following Beer [4] we define weakly hyperquasi- central subgroups of 6 and also a certain prOperty (9). Some of the work of Beidleman and Seo [5, 6] and Dykes [10] has also been generalized in this chapter. The main results here are: (l) L(G) is a Special L-Subgroup of the group 6. (2) If H A G is a Special L-subgroup of G and 6/H has the Sylow tower property, then 6 has the Sylow tower property. (3) The following statements are equivalent for a nilpotent normal subgroup H of the group 6: (a) H is a special L-subgroup of 6. (b) H has property (9) in 6. (c) H is weakly hyperquasicentral in G. 2.1 L-SUbgroups of the Group G. We begin with the following definition. Definition 2.1.1 [5]: Let H be a proper normal sub- group of a group 6. Then H is a generalized Frattini sub- gppgp of G if the following holds for every normal subgroup 41 42 N of 6: If P is any Sylow Subgroup of N, then G = HNG(P) implies 6 = NG(P). It is well-known that if G = Q(G)K for any K S G, then 6 = K. It follows therefore that the Frattini Subgroup of a group is a generalized Frattini subgroup. Some of the results on generalized Frattini subgroups are given in the next theorem. Theorem 2.1.2 [5, 6]: Let H be a generalized Frattini subgroup of 6. Then: (a) H is nilpotent. (b) If K is a normal subgroup of 6 containing H such that K/H is nilpotent, then K is nilpotent. (c) A(G) is a generalized Frattini subgroup of 6 if 6 is not nilpotent. As our investigations in Chapter I showed, the structure of L(G) and its related prOperties are very much influenced by the largest prime divisor of the order of the relevant sub- group. This leads naturally to the following generalization of Definition 2.1.1. Definition 2.1.3: Let H be a proper normal subgroup of a group G. Then H is an L-subgroup of G if the follow- ing holds for every normal subgroup N of 6: If p is the largest prime divisor of ‘N‘ and P is a Sylow p-subgroup of N, then G = HNG(P) implies G = NG(P) (i.e. P A 6). First we obtain some elementary properties of L-subgroups. 43 Proposition 2.1.4: Let G be a finite group. (i) If H is an L-subgroup of G, then (a) H is p-closed, where p is the largest prime divisor of ‘H‘, and S H, then H is also an L- (b) If H1 A G, H 1 l subgroup of 6. (ii) L(G) is an L-subgroup of 6. Proof: (i) (a) Let P be a Sylow p-subgroup of H for p the largest prime divisor of ‘H‘. Since H A G, by Frattini's lemma 6 = HNG(P). Since H is an L-subgroup of G, we have 6 = NG(P). Hence P A 6 and H is p-closed. (b) Let H1 S H and H1 A G. Let N be a normal subgroup of G and p the largest prime divisor of ‘N‘. Suppose G = HlNG(P)’ where P is a Sylow p-subgroup of N. Since H1 S H, 6 = HNG(P). Now since H is an L-subgroup of 6, 6 = NG(P). Hence H1 is an L-subgroup of 6. (ii) Let N A_6 and P a Sylow p-subgroup of N, where p is the largest prime divisor of ‘N‘. Suppose 6 = L(6)NG(P). Then we must Show that G = NG(P). Assume 6 # NG(P) and let M be a maximal subgroup of 6 containing NG(P). Then G = L(6)M and hence L(G) i M. It follows that M has prime index in 6. Let q = [6:M]. Now by Frattini's lemma since N A 6, 6 = NNG(P) and hence G = NM. This implies that q = [6:M] = [N: N n M]. Since NG(P) s M, by the Sylow theorems we have [6:M] = l + kp, where k is some integer. This means that p divides q - l, which is a contradiction since both p and q divide ‘N‘ and p is the largest such 44 divisor. Therefore 6 = NG(P) and L(G) satisfies the definition of an L-subgroup of 6. Corollapy 2.1.5: Q(G) and Q*(G) are L-Subgroups of 6. Remark: From the definition it is obvious that a generalized Frattini subgroup of a group is an L-subgroup. The converse of this however is not true, as the following example shows. Example 2.1.6: Let 6 = A4 X 83. By Proposition 1.3.3 since L(A4) = <1> and 83 is Supersolvable, L(G) = 83. By Theorem 2.1.2 however, a generalized Frattini Subgroup is nil- potent. Hence L(G) is not a generalized Frattini subgroup of 6 though it is an L-Subgroup of G by Proposition 2.1.4(ii). The next result shows the influence of L-subgroups on the group. Proposition 2.1.7: If H is an L-subgroup of 6 and p is the largest prime divisor of ‘6‘, then G/H is p-closed if and only if 6 is p-closed. Proof: If 6 is p-closed, then G/H being a homomorphic image of G is also p-closed. Conversely, assume G/H is p-closed. If p / ‘6/H‘, then a Sylow p-subgroup of G is also a Sylow p-subgroup of H. By 2.1.4, H is p-closed and since H A 6, 6 is p-closed. Suppose now that p/‘G/H‘ and let P1 be the normal Sylow p-subgroup of G/H. Then there exists P a Sylow p-subgroup of G such that P =PH/H. Since P AG/H, PHAG. More- 1 1 over, P is also a Sylow p-subgroup of PH. By Frattini's 45 lemma, 6 = PH.NG(P) = HNG(P). Now since H is an L-subgroup of G, we have 6 = NC(P). Thus 6 is p-closed. Unfortunately the concept of L-subgroup seems to be too general in that it does not sufficiently restrict the be- havior of the Sylow Subgroups for the various prime divisors. So in order to attain great control we Specialize the above concept to a certain degree. 2.2 Special L-subgroups of a Group, 6. We begin with a well-known definition. Definition 2.2.1: A subgroup H of a group G is called a Hall-subgroup if (‘H‘, [6:H]) = l. The Sylow subgroups of a group are well-known examples of Hall-subgroups. H is a Hall-subgroup of G and a n-subgroup for a set n of primes, then H is a Hall n-subgroup. Similarly a Hall subgroup which is also a n'-subgroup is a Hall n'-8ubgroup. Definition 2.2.2: Let n be the set of primes dividing the order of a group 6. Let n1 be a subset of n such that every element in the set n ~ n1 is smaller than every element in the set n1. Then we call the set a set of upperpprimes T‘1 for the group 6. As a notational convenience we denote by p(n) the set of prime divisors of the number n. Finally, we Shall call a set of upper primes for G a UP-set for 6. Now we are ready to make the following definition. 46 Definition 2.2.3: Let H be a prOper normal subgroup of a group 6. Then H is a Special L-subgroup of 6 (Sp-L- subgroup) if the following holds for every normal subgroup N of 6: If n is any UP-set for N and A any Hall n-subgroup of N, then G = H.NG(A) implies A A 6 (i.e., 6 = NG(A)). We note that every normal subgroup N of a group has at least one Hall n-Subgroup for n a UP-set; namely when n = {p}, where p is the largest prime divisor of ‘N‘. It is obvious that Q(G) is a sp-L-subgroup of 6. Moreover, every Sp-L-subgroup of G is also an L-subgroup of 6.- The converse of this, however, is not true as the follow- ing example shows: Example 2.2.4: Let 6 be the group with the follow- ing properties: (i) M = 22-3-7 (ii) 6 has a normal Sylow 7-subgroup P7. (iii) 6 has a normal Sylow 2-Subgroup V4, isomorphic to the 4-group. (iv) 6 has a Sylow 3-subgroup C which is self- 3 normalizing in 6. By Scott [18, 9.2.14] such a group 6 exists. Now 6 has a normal subgroup H = P7 X V4 of order 28. The only other normal subgroups K of G are: K = G,P7,V4. It is easy to verify that for each of these normal subgroups K, and P a Sylow p-subgroup of K correSponding to the largest prime divisor of ‘K‘, G = HNG(P) implies c = NG(P). In 47 other words H satisfies the definition of an L-subgroup of 6. However H is not a Sp-L-Subgroup of 6. To see this let N = G and A = P7°C3. Then A is a Hall {7,3}-subgroup of 6 and {7,3} is a UP-set for G. A is a.uaximal self- normalizing subgroup of 6 not containing H. Consequently, 6 = H-NG(A) = HA. But 6 # NG(A). Therefore H is an L-sub- group but not a sp-L-subgroup of 6. Remark: In case the group G has the Sylow tower pro- perty then every prOper normal subgroup of G is a Sp-L-Sub- group of 6. Next we have the following basic result on the structure of sp-L-subgroups. Proposition 2.2.5: Let H be a Sp-L-subgroup of a group 6. Then: (a) H has the Sylow tower property. (b) If H A G and H S H, then H l 1 group G. 1 is a sp-L-sub- Proof: (a) We prove by induction on n, where n is the number of distinct prime divisors of ‘H‘. > >00. .00 . Let p1 p2 pk > > pn be the natural ordering of the prime divisors of ‘H‘. Let Pi denote a Sylow pi-sub- group of H. Since H is a sp-L-subgroup, it is also an L- subgroup of 6. Therefore by Proposition 2.1.4, H is pl-Closed. 80 P1 A H and hence P1 A 6. Suppose now that A = P P ...P 1 2 k is a normal Hall subgroup of H. Consider H/A A G/A. Let ml k+l be a Sylow Pk+1-8ubgroup of H/A. Then there exists _ - =P AA. Pk+1 a Sylow pk+1 subgroup of H such that Pk+1 k+1 / 48 Now by Frattini's lemma, G/A = H/A-NG/A(Pk+IA/A). We show that G = HNG(Pk+1A). For this let x be an arbitrary element of G. Then xA (hA)(yA), where h E H and y 6 G such that yA normalizes Pk+1A/A. So (hy)-1x E A. If (hy)-1x = t E A, -1 _ -l -1 then x - hyt. Also (yt) (Pk+1A)(yt) — t (y Pk+lAy)t. Now yA normalizes +1A/A hence y normalizes P A. Since Pk k+l t E A, it is clear that x E HNG(Pk+1A). Thus 6 = HNG(Pk+1A). Note that nk+1 = [p1,p2,...,pk+1] 18 a UP-set for H and A ‘ - , ' ' - - - Pk+1 is a Hall nk+1 subgroup of H Since H is a Sp L Sub group of 6 it follows that A A 6. Thus H has normal Pk+l Hall n-Subgroup for every UP-set n. Hence H has the Sylow tower prOperty. (b) Let N A_6, n a UP-set for N and A a Hall n-subgroup of N. Suppose 6 = HING(A)' Since H1 S H, 6 = HNG(A). Since H is a sp-L-subgroup of 6, 6 = NG(A). Thus H1 is a Sp-L-subgroup of 6. Corollary 2.2.6: If H is a Sp-L-subgroup of 6, then H¢(6) and HZ(G) are also Sp-Ldsubgroups of 6. Proof: Since G = Q(G)K implies G = K for any K S 6, it is clear that Q(G)H is a sp-L-subgroup when H is a sp-L-subgroup of 6. Next, let N be a normal subgroup of G, n a UP-set for N and A a Hall n-subgroup of N. Suppose G = HZ(G).NG(A). Since 2(6) commutes with every element of 6, it normalizes A. Hence 6 = HNG(A) and since H is a sp-L-Subgroup, G = NG(A). Remark: In general a Sp-L-subgroup need not be super- solvable. This is so because there are groups with the Sylow 49 tower preperty which are not supersolvable (see for example [18, 9.2.13]). Let H be Such a group and let 6 = H X K, where K is any group having the Sylow tower prOperty. Then 6 has the Sylow tower property. By an earlier remark, every prOper normal subgroup of G is a sp-L-subgroup of 6. Con- sequently H is a Sp-L-Subgroup of 6 but H is not super- solvable. The next result gives a condition for a group 6 to have the Sylow tower prOperty. Proposition 2.2.7: Let H be a Sp-L-subgroup of G. Then G has the Sylow tower property if and only if G/H has the Sylow tower property. Proof: If 6 has the Sylow tower property, then G/H, being a homomorphic image of 6, has that property. Conversely, assume that G/H has the Sylow tower pro- perty. We shall use induction on ‘6‘. Let p be the largest prime divisor of ‘6‘ and P be a Sylow p-subgroup of G. If P S H, then H is p-closed since H has the Sylow tower prOperty (Proposition 2.2.5). Hence P A 6. Suppose P S H. Then p/‘G/H‘ and by hypothesis G/H is p-closed. By Proposition 2.1.7 since H is an L-Subgroup, 6 is p-closed. Thus in any case P A 6. Now we Show that HP/P is a Sp-L-subgroup of G/P. Suppose that N/P A_G/P and A/P is a Hall n-subgroup of N/P, where n is an arbitrary UP-set for N/P. Let n1 3 n U [p]. Then A/P is also a Hall nl-subgroup of N/P. HP . Q = __. Moreover n1 18 a UP-set for N. Suppose P P NG/P(A/P). 50 Then as shown in Proposition 2.2.5, 6 = HP.NG(A) = H'NG(A)'. Now [NzA] = [N/PzA/P] is a ni-number (because p does not divide ‘N/P‘). Also A is a nl-number. Therefore A is a Hall nl-subgroup of N and n is a UP-set for N. Since 1 H is a Sp-L-subgroup of 6, it follows that 6 = NG(A). 80 A A G and hence A/P A G/P. Thus we have shown that HP/P is a Sp-L-subgroup of G/P. Furthermore, G/P/HP/P 5 G/HP has the Sylow tower property. By the induction hypothesis we conclude that G/P has the Sylow tower property. It is now clear that 6 itself has that property. The above result does not hold if instead of being a sp-L-subgroup, H is merely an L-subgroup of 6. To see this consider the following: Example 2.2.8: Let 6 be the group defined in Example 2.2.4. Then H = P X V is an L-subgroup of 6 but not a 7 4 sp-L-subgroup. However, G/H E C has the Sylow tower property. 3 But 6 does not have that property since we have seen that P7-C3 is not normal in 6. Our objective in the following pages is to Show that L(G) is a sp-L-subgroup of G and also to attempt to char- acterize the sp-L—subgroups of the group. In order to achieve our purpose we shall proceed by first considering some defini- tions and intermediate results which are also of some independent interest. 2.3 Subgroups with Property 0?). Definition 2.3.1: Let n be a set of primes. A group 6 is n-closed if G has normal Hall n-subgroup. 51 If the primes in the set u do not divide ‘6‘, then 6 is trivially n-closed. It is well-known and easily verified that every Subgroup and homomorphic image of a n-closed group is again n-closed. A n-group is obviously n-closed. If H is a normal n-subgroup of 6 such that G/H is n-closed, then G is n-closed. If however, H and G/H are both n-closed then G is not necessarily n-closed. Even in the particular case when n is a UP-set for G and H is an L-subgroup of 6 Such that both H and G/H are n-closed, 6 is not always n-closed. An example to illustrate this is the following: Example 2.3.2: Let G be the group defined in Example 2.2.4. Let H = P7 X V4 A G, which is an L-Subgroup of 6. Hence V4 is also an L-Subgroup of 6. Let n = [7,3], which is a UP-set for 6. Now G/VA is a n-group, hence n-closed. Also V4 is trivially n-closed (note that V is not a n-group). 4 But 6 is not n-closed as we have shown before. The above example leads us to make the following definition. Definition 2.3.3: Let H be a proper normal subgroup of a group G. Then H has pr0per§y_ (9), in 6 if the follow- ing holds for every normal subgroup K of 6 containing H: If n is a UP-set for K, then K/H n-closed implies K is n-closed. The next two results establish the relationship between Sp-L-subgroups of G and the Subgroups with property 0?) in G. 52 Proposition 2.3.4: If H A G has property (9) in G, then H is a sp-L-subgroup of 6. Proof: Let N be a normal subgroup of 6, n be an arbitrary UP-set for N and A a Hall n-subgroup of N. Suppose G = HNG(A). We must Show that G = NG(A). First observe that HA A 6. Next, HA/H 5 A/A.n H is a n-group since A is a n-group. Also [N:A] is a n'-number. Now let HI = n U {all primes dividing ‘HA‘ larger than n}. Then in particular, HA/H is a nl-group and hence it is nl-closed. Since n is a UP-set for HA and H has the l prOperty G?) in 6, HA is n -closed. Hence HA.n N is l nl-closed. But n is a UP-set for N and the primes in n1 ~ n are larger than the primes in n. This means that HA.n N is n-closed. Since A is a Hall n-Subgroup of N and hence a Hall n-subgroup of HA.n N, it follows that A A HA.n N. So A is normal in 6 as we wanted to Show. The following is a partial converse of the above. Proposition 2.3.5: If H is a nilpotent Sp-L-sub- group of G, then H has the property (9) in 6. Proof: Let K be a normal Subgroup of 6 containing H and n be an arbitrary UP-set for K. Suppose that K/H is n-closed and let L/H be the Hall n-subgroup of K/H. Then L/HAG/H and LAG. If H is arr-subgroup, then L is a n-subgroup. Also [K:L] = [K/HzL/H] is a n'-number. Thus L is a normal Hall n-subgroup of K and this implies the n-closure of K. 53 Assume therefore that H is not a n-subgroup. By hypothesis H is nilpotent and hence it has normal Hall n'-Subgroup A. Consider [L:A] = [L:H][H:A]. Since [L:H] and [HzA] are both n-numbers, [L:A] is a n—number. There- fore L has normal Hall n'-subgroup A. By the Schfir- Zassenhaus theorem [18, 9.3.6], L has a n'-complement 6. Moreover since A is solvable, any two such complements are conjugate in L. Now since L A G, for any x E G if x 6 NG(C)’ then X-ICX S L. This means that all conjugates of C in G are also conjugates in L. Therefore 6 = ING(C) AC.NG(C) = ANG(C). Since A A G and A S H, A is also a sp-L-subgroup of 6. Moreover, C is a Hall n-subgroup of L and n is a UP-set for L A 6. So 6 = NG(C) i.e., c A c. Now [K:C] = [K:L][L:C] = [K/H:L/H][L:C] = (n'-number)(n'-number). Therefore C is normal Hall n-subgroup of K. Thus K is n-closed as we were required to Show. Rpmggk: We are as yet unable to decide whether the condition that H be nilpotent in the above proposition is necessary. We know however, that 'nilpotent Sp-L-subgroup' cannot be replaced by 'nilpotent L-subgroup'. Once again Example 2.3.2 confirms this. The next two results will be useful in the sequel. Proposition 2.3.6: Let K be a normal subgroup of 6 having prOperty 09) in G and H be a normal subgroup of 6 contained in K. Then, 54 (i) H has property 6?) in 6. (ii) K/H has property (9) in G/H. Proof: (i) Let N be a normal subgroup of 6 con- taining H. Let n be an arbitrary UP-set for N and suppose N/H is n-closed. Since H S N n K, N/N n K is isomorphic to a homomorphic image of N/H and hence it is n-closed. Con- sequently NK/K E N/N n K. is n-closed. Let n* = n U {811 primes dividing ‘K‘ larger than n}. Since n is a UP-set for N, there are no primes dividing ‘N‘ and larger than the primes in the set w. Hence n*-closure of NK/K is essentially n-closure of NK/K. Now K has property (9) in G, n* is a UP-Set for NK A G and NK/K is n*-closed. It follows that NK is n*-closed. In particular N is n*-closed, which in effect means that N is n-closed. Hence H has the property G?) in 6. (ii) Let L/H be a normal subgroup of G/H such that K/H S L/H. Suppose L/H/K/H is n-closed, for n any UP-set for L/H. ‘We must show that L/H is “-closed. Notice that L/H/K/H i=- L/K implies that L/K is n-closed. Moreover the primes dividing ‘L/H‘ also divide ‘L‘. As before, let n1 = n U {all the primes dividing ‘L‘ and larger than n}. Clearly n1 is a UP-set for L. Also, L/K being isomorphic to a homomorphic image of L/H implies that the primes dividing ‘L/K‘ are not larger than the primes dividing ‘L/H‘ hence no larger than the elements in n. This means that L/K n-closed implies it is n -closed. Now by hypothesis, K has property 1 G?) in 6. Hence L is nl-closed. Therefore L/H is 55 nl-closed, which means it is n-closed. Thus we Show that K/H has the property (9) in G/H. The following is a converse of the above. Proposition 2.3.7: If H has property (9), in 6 and K/H has property (9) in G/H, then K has prOperty Q?) in 6. Proof: Let L be a normal subgroup of 6 containing K. Suppose L/K is n-closed, n any UP-Set for L. Then L/K 5 L/H/K/H is also n-closed and n is a UP-set for L/H. By hypothesis, it follows that L/H is n-closed. Again H too has the property 6?) in G and therefore L is n-closed. This proves that K has the property 69) in 6. We now show that the hyperquasicenter of a group 6 has the property (9) in G and hence by Proposition 2.3.4 is a Sp-L-subgroup of 6. This will eventually enable us to prove that L(G) is a sp-L-subgroup of G - our main objective in this section. We begin with the following: Theorem 2.3.8: The quasicenter of a group G has pro- perty (9) in 6. Proof: Let N be a normal subgroup 6 containing Q = Q(6). Let n be an arbitrary UP-set for N and suppose N/Q is n-closed. let AflQ be the normal Hall n-subgroup of NAQ. We must Show that N has normal Hall n-subgroup. Note that A A G. In case Q is a n-Subgroup then A is a n-Sub- group. Moreover, [NzA] - [NflQ:AflQ] is a n'-number. Thus A is normal Hall n-subgroup and we are done. Assume therefore that Q is not a n-subgroup. Since the quasicenter of a group 56 is nilpotent (Theorem 1.5.2), Q has normal Hall n'-Subgroup Q1. Consider [AJQI] = [AJQ][Q:Q1] = (n-number)(n-number). Thus O1 is a normal Hall n'-subgroup of A. By the Schfir- Zassenhaus theorem, A has a n'-complement B. So A ==Q1B. Since Q1 is solvable and A A 6, all the conjugates of B in 6 are actually conjugates by elements of A. This implies that 6 = ANG(B) ==Q1NG(B). Note that Q1? is a n'-subgroup and B is a n-subgroup. Both Q1, B lie in N and hence the prime divisors of ‘Q1‘ are smaller than the prime divisors of ‘B‘. In the following discussion we show that Q1 S NG(B). Suppose Q1 S NG(B). Since Q is nilpotent and all its Sylow subgroups are generated by QC-elements of 6 (Theorem 1.5.2), we can choose an element x E 6 satisfying the following: (a) x is a p-element for some p dividing ‘6‘. (b) x is a.QC-element of 6. (c) x i NGCB), and (d) x 6 Q1 and therefore p is smaller than the elements in n. Let y be an arbitrary element in B. Then y is a n-element. Since x is a quasicentral element of 6, it permutes with y. Hence T = = is a subgroup. By Scott [18, 13.3.1], T is supersolvable. Also is a Sylow p-subgroup of T and is contained in Q(T), the quasicenter of T. Therefore i is a Sylow subgroup of Q(T). Consequently by Theorem 1.5.2, A T. On the other hand is a Hall n-subgroup Of T and is in fact a p-complement of T. Since p is the 57 smallest prime divisor of ‘T‘, T is penilpotent. Thus we conclude that A T. Consequently we see that x,y centralize each other. Since y was an arbitrary element of B, it follows in particular that x E NG(B). But this is a contradiction since x was chosen such that x é NG(B). Therefore we conclude that Q1 S NG(B). This implies that B A 6. Further, [NzB] = [NzA][A:B] = (n'-number)(n'-number). So B is a normal Hall n-Subgroup of N and hence N is n-closed. We have thus Shown that Q(G) has the property 6?). Corollary 2.3.9: The hyperquasicenter Q*(G) has pro- perty 09) in 6. Proof: Let <1> S Q(G) = Q1 S Q2 S...S Qr = Q*(G) be the ascending-quasicentral series of 6. By definition QZ/Ql = Q(G/Ql). Since Q1 has property (0) in G and 02/01 =‘Q(6/Q1) has property 69) in G/lQ1 by Proposition 2.3.7, Q2 has property 09) in 6. By repeating this argu- ment we eventually see that Q*(G), the terminal member of the above series has property (9) in 6. This proves the corollary. A hyperquasicentral subgroup of a group is defined by Mukherjee [14] as follows: Definition 2.3.10: A normal subgroup H of 6 is hyperqoasicentral in 6 if for every M A G and M_$_H, H/M n Q(G/M) f <1>. It is shown in [14, Theorem 2.18] that every hyperquasi- central subgroup of 6 is contained in the hyperquasicenter 58 Q*(G). This leads to the following rather obvious corollary. Corollary 2.3.11: A hyperquasicentral subgroup of a group G has the property (9) in 6. Moreover every SSE subgroup of G has property 6?) in G. We are now in a position to derive the main result of this section. Theorem 2.3.12: In a group G, L(G) has the property 6?). Proof: Case (1). Suppose L'(G) ¥ <1>, i.e. L(G) is not abelian. We show that L'(G/s(c)) is SSE in c/s(c). Notice that if G is solvable, then as in Theorem 1.2.9 L(Gis(c)) o (G/o(c))' is SSE in c/s(c). This implies that L'(6/§(6)) is SSE in 6/§(6). In the general case, we know L'(G) is nilpotent. Suppose Q(G) = Q = <1>. Then @(L'(6)) S 6(6) = <1>. Let P be the Sylow p-subgroup of L'(G), where p is an arbitrary prime divisor of ‘L'(G)‘. Then P A 6 and P is elementary abelian. Let N S P be a minimal normal subgroup of G. Since Q(G) = <1>, there exists a maximal Subgroup M of 6 not containing N. Since N S P is solvable, c = MN and ‘N‘ = [6:M]. But N s P s L'(G) s L(G). Therefore M must have a prime index in 6. Consequently ‘N‘ = p. Since P is abelian normal in G and Q(G) =<1>, by [11, Satz 7] P is completely reducible in G. This means that P = C1 X C2 x... Cr, where the Ci's are minimal normal subgroups of 6 of order p. Therefore 6 induces in P a Strictly p-closed group of automorphisms. By Theorem 1.1.9 we conclude that L'(G) is SSE in G. 59 If 6(6) ¢.<1>, then Q(G/Q) = <1} and from above, L'(C/Q) is SSE in G/Q. By Corollary 2.3.11 every SSE Sub- L'(G) gs) L' (clap) = Q has Q(G) preperty (9) in 6/§(6). Since Q(G) has property 0?) group has property 6?). Hence in G, by Proposition 2.3.7 L'(6)@(G) has property (9) in 6. Now L'(G) A G and L(6/L'(6)) = L(G)/L'(G). By the induction hypothesis L(G)/L'(G) has property (9) in G/L'(6). Application of Proposition 2.3.7 now Shows that L(G) has the property (9) in 6. Case (2). Suppose L'(G) = <1>, i.e. L(G) is abelian. As in Case (1) it is not difficult to Show that L(G)/Q(G) is SSE in 6/Q(6) and hence has the property G?) in 6/¢(G). This by Proposition 2.3.7 implies that L(G) has property 09) in 6. The theorem is now completely proved. Corollary 2.3.13: L(G) is a sp-L-subgroup of 6. In an earlier remark we mentioned that if a group 6 has the Sylow tower property, then each of its proper normal subgroups is a Sp-L-subgroup of G. The following result con- siders the case when 6 does not have the Sylow tower pro- perty but every proper subgroup of 6 has that property. It is seen that in such a group G the sp-L-subgroups are severely restricted. Proposition 2.3.14: If a group 6 does not have the Sylow tower property but each of its proper subgroups has that pr0perty, then Q(G) is the largest sp-L-subgroup of 6. Moreover Q(G) = L(G). 60 Proof: Let K be a maximal sp-L-subgroup of 6 (i.e., K is not properly contained in any sp-L-Subgroup of 6). By Corollary 2.2.6, K§(G) is a sp-L-subgroup of 6. From the maximality of K, we have Q(G) S K. Suppose on the other hand that K S Q(6). Then there exists M, a maximal subgroup of 6 such that K S M. Hence 6 = KM. By hypothesis M has the Sylow tower property. Hence G/K E MIMI] K has the Sylow tower property. Since K is a sp-L-subgroup of 6, by Proposition 2.2.7 6 has the Sylow tower property. This contradicts the hypothesis. So 6(6) = K. Finally Since L(G) is a sp-L-sub- group of G, we have Q(G) = L(G). We continue our investigation of Special L-subgroups and give one more characterization of nilpotent sp-L-subgroups. 2.4 Weakly Hyperqpasicentral Subgpoups. In [4] Beer introduced the concept of a weakly hyper- central subgroup as follows: Definition 2.4.1: Let H be auproper normal subgroup of a group G. Then H is weakly hypercentral in G if the follow- ing holds for every normal Subgroup K of 6 containing H: For every pair of elements x,y belonging to H,K reSpectively if (‘x‘,‘y‘) = l and (‘x‘,[K:H]) = 1, then x and y commutes Since a group is nilpotent if and only if elements of relatively prime orders permute, it follows by setting K = H above that a weakly hypercentral subgroup is nilpotent. 61 Two characterizations of weakly hypercentral Subgroups based on the concepts of generalized Frattini subgroups are given by D. Dykes [10]. We shall begin by first stating these. Definition 2.4.2 [10]; ,AprOper normal subgroup H of a group 6 is a Special generalized Frattini subgroup of 6 provided that for every N A'G and A any Hall subgroup of N, 6 = HNG(A) implies G = NG(A). It is obvious that a Sp-generalized Frattini Subgroup is already a generalized Frattini subgroup and hence is nil- potent. Moreover a Sp-generalized Frattini Subgroup is also a sp-L-Subgroup of the group. The converse Of this however, is not necessarily true. This can be easily seen from the fact that Sp-generalized Frattini Subgroups are nilpotent; while sp-L-Subgroups in general are not. For instance L(G) is a Sp-L-subgroup of the group G which in general is not' nilpotent (see Example 2.1.6). Definition 2.4.3 [10]: Let H be a proper normal sub- group of a group 6. Then H satisfies property (NU) in 6 if the following holds for every normal Subgroup K of 6 containing H: If n is any set of primes, then K/H n-closed implies K is n-closed. The following theorem shows the equivalence of the con- cepts defined above. Theorem 2.4.4 [10]: The following statements for a normal subgroup H of the group 6 are equivalent: (i) H is a Special generalized Frattini subgroup of G. 62 (ii) H satisfies the property (Np) in 6, for any set of primes n. (iii) H is a weakly hypercentral Subgroup of 6. We shall attempt to generalize the above theorem in the present section. For this we consider the following gen- eralization of the Definition 2.4.1: Definition 2.4.5: A proper normal Subgroup H of a group G is weakly hyperquasicentral (WHQC) in 6 if the follow- ing holds for every normal subgroup K of 6 containing H: Forevery pair of elements x,y belonging to ‘H and K reSpectively if (‘x‘,‘y‘) = 1. and p(‘x‘) < p([KzH]), then and . permute. It is obvious from the definition that every weakly hypercentral subgroup of 6 is already weakly hyperquasicentral in 6. Next, we list some of the elementary properties of weakly hyperquasicentral subgroups. The proofs of these are rather straightforward and therefore will be omitted. Proposition 2.4.6: Let H be a.WHQC subgroup of 6. If H1 A 6 and H S H, then H 1 1 is WHQC 1n 6 and h/H1 is WHQC in 6/H1- Proposition 2.4.7: Let K be a WHQC subgroup of G. If H A G and p(‘K‘) < p(‘H‘), then KH/H is WHQC in G/H. It seems unlikely that in general the product of two WHQC subgroups of G is again WHQC in 6. So let T* be the intersection of all the maximal.WHQC subgroups of G. Then * T satisfies the following property. 63 Proposition 2.4.8: The following are equivalent for a normal Subgroup N of a group G: (a) N ST*. (b) If M is WHQC in c, then MN is WHQC in 6. We now turn to our main objective in this section and establish the relationship of WHQC Subgroups with Sp-L-Subgroups and the subgroups satisfying property 6?). Proposition 2.4.9: Let H be a normal subgroup of G such that, (i) H has the Sylow tower property, and (ii) H is WHQC in 6. Then H is a Sp-L-subgroup of 6. Proof: Let N be a normal Subgroup of 6 and n a UP-set for N. Let A be any Hall n-subgroup of G such that 6 = HNG(A). Then we must Show that A A 6. It is obvious that HA A 6. Let n1 = n U [all prime divisors of ‘H‘ larger than primes in n}. Then n1 is a UP-set in HA. Since H has the Sylow tower property, it has normal Hall nl-subgroup H1. Moreover by the Schfir-Zassenhaus theorem, H has a n -complement 1 H2 and H = Hl'HZ. Now let x E H2 and y E A S HA A 6. Then (‘x‘,‘y‘) = 1. Also p(‘x‘) E "1 and p([HAzH] E n g ml. Since n1 is a UP-set for HA, it follows that p(‘x‘) < p([HA:H]). Since x E H, y E HA and H is WHQC in 6, we conclude that x,y permute. Thus =. = T. Since T is a product of two cyclic subgroups, it is Super- solvable. Hence T has the Sylow tower property. Since 64 is a Hall nl-subgroup of T and n1 is a UP-set for T, A T. Furthermore if y1 E S HA, then y1 is a n1- element. So , permute. Thus x permutes with every 1 element of T. This implies that x belongs to Q(T), the quasi- center of T. Since Q(T) is normal nilpotent and is a Sylow subgroup of T, it follows that A T. Hence x and y centralize each other. Since y is an arbitrary element of A, we see that x E NG(A). 80 H S NG(A). Then 2 G(A) = H1N6(A)' Therefore HlA A 6. Also H1 is a nl-group and HlA/H1 is a nl-group. Hence HlA is G = HNG(A) = HIHZN nl-closed. In particular HlA n N is fll-closed. But n is a UP-set for N and hence for HlA.n N. This means that A is the characteristic Hall n-subgroup of H A n N. 80 A A G, l as we were required to Show. Proposition 2.4.10: If H is a nilpotent sp-L-subgroup of 6, then H is WHQC in 6. Proof: Since H is a nilpotent Sp-L-subgroup of 6, by PrOposition 2.3.5 H has the property 0?) in G. Let K be a normal subgroup of 6 containing H and let n be the set of primes dividing [K:H]. Then K/H is a n-group. Let HI = n U {all the primes dividing ‘K‘ larger than the smallest prime in n}. It is evident that n1 is a UP-set for K. By the prOperty 0?) it follows that K is nl-closed. Let A be the normal Hall nl-subgroup of K. Since H is by hypothesis nilpotent, H = H 1 X H2, where H1, H2 are n and l ni Hall subgroups of H. Since A contains every n1 hence n-element of K, we have K = AH = A.(H1 X H2) = A X H2. Now 65 let x e H, y e x such that (‘x‘,‘y‘) = 1 and P(‘X‘) < P(1K=H])- Then p(‘x‘) < n C n1 . So x is a ni-element of H, hence belongs to H Also y = ah = has 2. where h E H2, a E A. Notice that (‘h‘,‘x‘) = 1. For if some prime q divides (‘h‘,‘x‘), then since ‘ha‘ = ‘al-‘h‘: it follows that q divides ‘ha‘ = ‘y‘. But this is a con- tradiction since (‘x‘,‘y‘) = 1. Thus (‘h‘,‘x‘) = 1. Now since h,x are elements of H2 which is nilpotent, they commute. Thus xy = xah = xha = (hx)a = ahx = yx. So x,y commute. Therefore H is WHQC in G. We derive from above one of our main results of this chapter. Theorem 2.4.11: The following statements are equi- valent for a nilpotent normal subgroup H of the group G: (i) H is an sp-L-Subgroup of 6. (ii) H has property (9) in 6. (iii) H is weakly hyperquasicentral in 6. Moreover, every nilpotent hyperquasicentral Subgroup of G is weakly hyperquasicentral in G. CHAPTER III In this chapter we investigate the conditions under which the subgroups L(G) and A(G) coincide. Following Bechtell[7] we define an L-series of the group G and also the upper and the lower L-series of G. We also define the relative L-series (relative to the commutator subgroup of G) and study some properties of this series and also its relation to the group 6. Some of the results obtained here are the follow- ing: Let G be a group. (1) If L*(6), the terminal member of the upper L-series, is the identity subgroup, then L(G) = A(G). (2) The upper relative L-series of G coincides with the lower central series of 6' if and only if 6 is super- solvable. (3) The following are equivalent: (i) 1*(6) = <1>, where 1*(6) is the terminal member of the upper-relative L-series of 6 . (ii) L(G) n 6" = 2*(0') n 6". (iii) If S is any subgroup of 6 generated by 3 elements one of which belongs to L(G), then L(G) n S' S 2*(3')- (4) If H A G and H is solvable, then 1*(6) = <1> implies 15(H) = <1>. 66 67 3.1 The L-series of the Grogp 6. Definition 3.1.1: For a group G, (i) an L-Series of G is a series L(G) = c0 2 c1 2..., where Ci/Ci+1 s 2(G/ci+ 1> or [01.0] s c1+1 (ii) the upper L-series (UL-series) is the series L(G) = LO 2 L1 2 L2 2..., where L1+1 = [L1,6] (iii) the lower L-Series is defined as the ascending central series of G, i.e. =z SZ 0 s2 S...,where z, /zi=2(c/zi). l 2 1+1 A Let L (6) denote the terminal member of the UL-series. The terminal member of the lower L-series is the hypercenter of * G, Z (G). Some elementary properties of the above series are the following: PrOposition 3.1.2: For a group 6: (i) An L-series is a normal series of 6. (ii) The upper and lower L-series are characteristic. (iii) If G is solvable, L.'s are all nilpotent, for l i 2 1. (iv) There is no normal subgroup H of 6 contained * * in L (6) such that L (G)/H s Z(6/H). * 7': (v) If 9 is a homomorphism of 6, then L (6)9 S L (69). Proof: (i) and (ii) are obvious by definition. (iii). If 6 is solvable, by 1.2.4 L(G) n 6' is nil— potent. Now L1 = [L(6),6] s L(G) n 6', since L(G) A G. So 68 L is nilpotent. Thus L1 is nilpotent for every i 2 l. 1 (iv). Suppose there exists H A G and H S L*(6) such that L*(6)/HSZ(6/H). Then [L*(6),6] SH. But L*((;) is the terminal member of the UL-series hence [L*(G),G] = L*(c). Thus L*(c) s H and this implies that * H = L (6). (v). Let 9 be a homomorphism of 6. By PrOposition 1.2.3, L(G)9 S L(Ge). Moreover, [L(6),6]e = [L(6)9,Ge]. From this it follows that L19 S [L(Ge),69] = L1(Ge). By the same argument it is easy to see that Li(66) S Li(66)r V i. As with many series, it is the terminal member of the series which yields some information about the group. We shall say that the group 6 possesses an L-series if the terminal member of the L-series is the identity subgroup. For such a group we have the following: Proposition 3.1.3: In a group possessing an L-series, i.e. L(G) = C 2 C 2 C 2... C 0 1 2 k i = O,l,...,k, and Ck-j S Zj’ j = O,l,...,k-l. = <1> we have Li S 61 for Remark: The proof of the above is analogous to the proof of similar results for the ascending and descending central series of the group (see for example Scott [18, 6.4.1]). However for completeness we outline the proof here. Proof: (i) Obviously L S C = L(G). By the induc- 0 0 tion hypotheSLS we can assume that Li S Ci. Since [61.6] 5 Ci+1’ we have [L1,G] S Ci+1° Thus L1+1 = [L1,G] S Ci+1° 80 Li S Ci’ V i. 69 (ii) To Show 6 S Z Show that when j = 0, k'j J Ck = <1> S ZO = <1>. Now assume that for j = r, the assertion holds. Then C S Z . Let T = G/Z ; then T is a homo- k-r r r morphic image of G/Ck r’ and the kernel Of this homOmorphism S Z and in particular Z . Ck-r-i r r+1 Ck-(r+1) S Zr+1 Corollary 3.1.4: If there exist integers j and k * such that Lj S Zk, then L S 2 (G). 0 Proof: By exactly the same method as in the Proposi- tion 3.1.3 we see that Lj S Zk implies that Lj-l S zk+l° Now if j < k, then by repeating the above process, we get * * L S 2 S Z (6). If j > k, then L, S 2 (6) and repeat- 0 k+l j-k * ing the process we have LO S 2 (6). Corollary 3.1.5: In a group G possessing an L-series, * L(G) = Z (6). Proof: By 3.1.3, Li S Ci; and in particular * * C S Z = Z (G). Then L S C S Z = Z (G). On the other 0 k 0 0 k * * hand we know 2 (G) S A(G) S L(G). Thus Z (G) = L(G). Proposition 3.1.6: In a group possessing an L-series, A(G) = L(G). Proof: First we Show that the existence of L-series implies the existence of A-series (defined by replacing L(G) by A(G)). Also if Ar s Lr’ then Ar.” = [Ar,6] a [Lr,6] = Lr+1. Since the terminal member of L-series is the identity Subgroup * of 6, A (6) = 1. Now by the same argument as used for L(G), 70 it is easily seen that A*(6) = <1> implies A(G) = 2*(6). Therefore when 6 possesses the L-series, it also possesses the A—series and A(G) = L(G). Also note that since 2*(c) s 0*(6) s L(G), it follows that Q*(G) = L(G). The above result is false if 6 does not possess an L-Series. This can be easily confirmed from the following example. Example 3.1.7: Let G be the group of order 84 des- cribed in Example 2.2.4. Then 6 has, (i) 28 Sylow 3-Subgroups (ii) 1 Sylow 7-subgroup P7 (iii) 1 Sylow 2-subgroup, V4. Obviously G is solvable. G has maximal subgroups of order 21 and hence index 4. Every such maximal Subgroup has a normal Sylow 7-subgroup which must be P Hence P7 S L(G). 7. Since L(G) is normal in 6 and P7-C3 is not normal in 6, it follows that L(G) = P . It is not difficult to verify 7 that 6(6) = A(G) = <1>. But 6 does not possess an L-series. To see this first notice that if L1 = [L(G),6] = <1>, then L(G) S 2(6) = <1>, which is a contradiction. So L1 # <1>, hence L1 S L(G) implies L1 = P7 = 1&6). Thus we see that an L-series of 6 does not terminate in <1>. The example also shows that A*(G) = <1> does not guarantee that Lf(G) =<1>. Proposition 3.1.8: If every proper subgroup of a group G has the Sylow tower property (but not the group 6 itself), then G has an L-series. 71 Proof: By Proposition 2.3.14, Q(G) = L(G). Since Ma) s 13(6) S Ma). A(G) -= L(G), By J. Rose [17, p. 588], under the above hypothesis 6 is an SRI-group. An SRI-group has the form 6 = R3, P A G and Q is a cyclic Sylow q-subgroup of 6. Moreover 663) S 2(6), so it is normal in 6. By [7, Theorem 4.1], A(G) = 2*(6) and so ¢(6) S Z*(6). Since Q(G) = L(G) and 2*(6) = A(G) S L(G), it follows that 2*(6) = A(G) = L(G). By Proposition 3.1.6, 6 possesses an L-series. The converse of above is not true. This is easily seen by considering any simple group G. 6 cannot satisfy the hypothesis of Proposition 3.1.8, otherwise 6 is solvable. Moreover for a simple group, L(G) = 2*(6) = <1> and an L-series exists trivially. More interesting however, from our viewpoint is the concept of relative L-series of 6 defined in the following. 3.2 Relative L-series of the Group G. We have noticed in Chapter I that the structure of the group 6 depends in large measure on the relation between 6' and L(G). For example we know that 6 is supersolvable if and only if 6' S L(G). Also when G is solvable 6' n L(G) is nilpotent. This leads to the study of an L-series of 6 relative to the commutator Subgroup of 6. Definition 3.2.1: For a group 6 define B0 = L(G) n 6' 2 B1 2 B2 2..., such that Bi/Bi+1 s Z(6'/Bi+1) i = 0,1,... . This is a relative L-series (RL-series) of G. 72 The upper relative L-series (URL-series) of 6 is the series 0 The lower relative L-Series (IRIrseries) of 6 is the ascending '=A It A =A ' L(G) n G L 2 L1 2..., where L1+1 [L1,G ]. central series of 6', i.e. 1>= sz S...SZ S... wh = ' . < 20 1 r , ere zi/zi_1 Z(6 /zi_ ) The terminal members of URL-series and URL-series are denoted A* * respectively, by L (G) and Z (6'). The following is easily verified. Proposition 3.2.2: In a group 6, (i) RL-series are normal series of 6'; (ii) the URL-series and LRL-series are characteristic series of 6'; (iii) the URL-series coincides with the upper central series of 6' if and only if 6 is supersolvable. (iv) If 6 is solvable, then Lt S Q(G), for every r 2 1. Moreover Lf(6) S Q(6). (v) For any homomorphism e of 6, 1%(6)9 S L*(69). Proof: (i) and (ii) are obvious from the definitions of these series. (iii) If 6 is supersolvable, then L(G) = G and L(G) n 6' = 6' is nilpotent. Hence in this case URL-series defines precisely the upper central series for 6' which terminates in <1>, Since 6' is nilpotent. Conversely, suppose URL-series coincides with the upper central series of 6'. This implies that L(G) n G' = 6' and so 6' S L(G). By Theorem 1.2.5 it follows that 6 is Super- solvable. 73 (iv) L1 = [L(G) n c',c'] S L(G) n [c',c']. Since G is solvable, 6" is contained in every maximal subgroup of prime index (Proposition 1.2.4). Then L(G) n 6" S 6(6) and con- sequently L S Q(6). Since Lt S L V'r 2 1, we have Lr S Q(G), l l .* , , , V r 2 1. Now to Show that L (C) is 1n Q(G) first notice A* A 5* A that if L (6) # L0, then L (c) S L A 1 S Q(6). In case L (G) = LO’ we have LO = L(G) n c' = L, = [L(G) n c',c'] s L(G) n c" s s(c). Thus in any case L*(G) is contained in Q(6). (v) Let 9 be any homomorphism of 6, then (L(G) n 6')e S L(G)e n 6'9. Since L(6)e S L(ce) and G'e = (ce)', we have Lo(c)e S io(ce). Similarly, Ll(G)e = [ib,c'je = [LO(G)S,G'9] s [10(69),6'e] = 11(69). This argument leads to (v). We shall say that a group possesses a relative L-series if the series terminates in the identity subgroup. For such a group we have the following. Proposition 3.2.3: In a group G possessing a RL-series, ' ' = ... = a 1,9, L(g) n 6 BO 2 B1 2 2 Bk <1>' we have (a) Li S Bi’ i = 0,1,...,k and (b) B zj, j = O,l,...,k-l. k-J 5 Proof: The method of proof is essentially a duplication of the correSponding results for the L-series. Therefore we only outline the proof here. (a) Notice that In S B0 = L(G) n 6'. Suppose now that L, S B, for a fixed i < R. Then by definition B,/B. l l 1 1+ S Z(G'/Bi+ ). l l . . . I “ = “ I This implies that [B1,6 ] S Bi+1° Hence Li+l [L1,6 ] S [B.,G'] S B. So we conclude that L S B. for i = O,l,...,k. l 1 1 +1' 1 (b) Now to show that Bk-j 5 zj’ observe that when j = 0, Bk = <1> = 20. Again assume that Bk-i S Zi for a flxed 74 integer i. Let T = 6'/Zi be the homomorphic image of ' o 6 /BK-i With kernel zi/Bk-i° By definition, B /B S Z(G'/BK_i). Hence under K-i-l K-i the homomorphism the image of BK-i-llBK-i must lie in the center of T. But this image is BK-izi/Zi which lies in Z(T) = 2(c'/zi) = zi+1/zi. From this it follows that z. 1 This proves (b). BK-(i+1) S +1“ As a consequence of the above results we have the following. Proposition 3.2.4: In a group G, the following are equivalent: (a) 1*(c) = <1>. (b) 10(6) S 2*(0'). (c) L(G) n c" = z* n 6". Proof: (a) ='(b). Let Lt = LT(G) = Z0 = <1>. By definition and the hypothesis, ir = [Lr_1,6'] = <1>. So 1. If Z1 = <1>, then 2*(6') = <1> and hence r-l’ir-2"'°’10 are all equal to <1> and (b) is proved. 80 A ' = Lr_1 SZ(G) Z ti) assume ir-l S Z1 # <1> and let T = G'IZl. T is a homomorphic . g " . A image of c /Lr_1 Wlth kernel 21/Lr_1. Also Lr_1 = [Lr_2,6'], so Lr_2/Lr_1 S Z(6'/Lr_1) [12, p. 18]. Consequently, the image of ir-Zlir-l lies in the center of A ' = A T. Thus Lr_zzl/z1 S 2(c /zl) 22/21 and hence Lr_2 S 22. a * By repeating this process we eventually have LO S Zt S Z (6'), if r < t (t is the length of LRL-series). On the other a * hand if Lr-t S Z (6'), then by repeating the above process A * we reach L0 S Z (6'). 75 . * (b) a (c). Suppose L0 S Z (6'). By [7, Theorem 2.2], 2*(6') n 6" S Q(G'). Therefore 2*(6') n 6” S Q(G') S 6(6) 0 6" S L(G) n 6". Also by hypothesis LO = L(G) n 6' S 2*(6') and so L(G) n G" S 2*(6') n 6". Thus, L(G) n G" = 2*(6') n G" and this is (c). Finally assume (c). Then since [L(G) o 6',6'] S L(G) n G", Ll s 1(a) n c" = 2*(6') n c". Thus L S Z*(G') = Zt’ where t is the length of LRL-series. 1 a * O ' = ' _1 < zt, we have L1 zt_1/zt_1 S 2 (c )/zt_1 2(c /zt_1). This implies that [Llozt_1,c'] S zt_1. Hence Since Zt .* This argument leads to L (G) S 2 = . A = A ' L2 [L1,G ] S Z 0 t-l' Thus (a), (b) and (c) are equivalent. Remark: In the above proposition 10(6) may actually be smaller than 2*(6'). For example consider 6 = A4. As we know already, L(G) e <1> and so 10(6) = L(G) n 6' = <1>. However 6' 5 V4, the 4-group and 2*(6') 3 V4. Notice that "3(a) = <1>. Corollary 3.2.5: The conditions in Proposition 3.2.4 are also equivalent to the following: If S is any Subgroup of 6 generated by 3 elements one of which belongs to L(G), then L(G) n 8' S Z*(8'), where 2*(8') is the hypercenter of 8'. Proof: This can be immediately deduced from a result of R. Baer [3, p. 177] which states that any normal subgroup N of 6 satisfies the hypothesis of the corollary (in place of L(G)) if and only if G' n N S 2*(6'). Remark: The above may have some interesting consequences, and we hope to return to its investigation on a later occasion. 76 a* * a Corollary 3.2.6: If L (G) = <1>, then Z (6')/LO = U 2(0 /LO). * Proof: By Proposition 3.2.4, L(G) n 6" = Z (6') n G". * * A Then [2 (c'),c'] S 2 (6') n c" = L(G) n c" S L(G) n 6' = LO. . A * * A ' A Moreover Since Lo S Z 63') we have, Z (6')/LO S 2(6 /Lo). A * * 0n the other hand since LO S Z (6'), G'/Z (6') is a homo- A * A morphic image of 6'/L0 with kernel 2 (6')/L0. Under this homomorphism e, Z(G'/LD) is mapped into the center of ‘k * — A c‘lz (6'). But Z(G/Z (6')) = <1>. Therefore Z(6-'/LO) is in the kernel of the homomorphism 6. Consequently, I " * I " " * I " Z(6 /L0) S 2 (6 )/LO. Thus we conclude that Z(6'/L0) = z (6 )/L0. We close the chapter and the present investigation with the following. Proposition 3.2.7: If H is a solvable normal sub- a* ..-k group of 6 and L (G) = <1>, then L (H) B<1>. Proof: If H is Supersolvable, then by Pr0position 3.2.2 the URL-series of H coincides with the descending 5* central series of H' and hence L (H) "<1>w We may assume therefore, that H and hence 6 is not supersolvable. Since .* A * L (6) = <1>, by Proposition 3.2.4, L0(G) = L(G) n 6' S Z (6'). Also since H A 0, MN) S Q(G). Then H' 0 6(6) s c' n L(G) S * Z (6'). Suppose H' n Q(H) = <1>. Since H is solvable, by .* Proposition 3.2.2 (iv), L (H) S H' n §(H) -.<1>s Therefore assume that H' n Q(H) # <1>, in particular Q(H) # <1>. It is known that the hypercenter of a group is the intersection of the normalizers of all the Sylow Subgroups of the group. In the following discussion we shall make use of this definition. 77 If P1 is the Sylow p-subgroup of H' n Q(H), then * P1 A 6'. Furthermore from above P S H' n Q(H) S Z (6'). 1 Therefore P1 S NG,(Q), where Q is any Sylow q-subgroup of 6'. Now if p i q, then Pf} = P1 X Q and hence every p- element of H' n Q(H) centralizes every p'-element of 6'. In particular, every p-element of H' n Q(H) centralizes every p'-element of H'. Let ‘P be a Sylow p-Subgroup of H' and P a Sylow p-subgroup of 6' containing P. Since H' A 6', P=PnH'. If xEP1,then Px=(PnH')x=Pan'= P 0 H' = P' since x E NG.(P). This shows that x normalizes every Sylow p-subgroup of H'. Hence x E 2*(H') and so P1 S 2*(H'). Since H' n Q(H) is nilpotent, we conclude that H' n @(H) S Z*(H'). Since H is solvable, L(H) n H" S Q(H). Thus L(H) n H" S Q(H) n H' S 2*(H'). Also by [7, Theorem 2.2], 2*(H') n H" S Q(H') S §(H) n H". 80 we conclude that 2*(H') n H” = L(H) n H". By Proposition 3.2.4 it follows that .* L (H) = <1>. Remark: We are not able to decide whether the solv- ability condition in the above prOposition is necessary. BIBLIOGRAHiY 10. 11. 12. l3. l4. BIBLIOGRAPHY R. Baer; Supersolvable immersion, Canadian Journal of Mathematics, 11(1959), pp. 353-369. ________j Classes of finite groups and their prOperties, Illinois Journal of Mathematics, Vol. 1 (1957), pp. 115-187. ; The hypercenter of functorily defined subgroups, Illinois Journal of Mathematics, 8(1964), pp. 177-230. ; Nilpotent characteristic subgroups of finite groups, American Journal of Mathematics, 75(1953), pp. 633-664. J. Beidleman and T. 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