"N“? “WNW" m GENERALIZED SYLOW TOWER GROUPS Thesis for the Degree of Ph. D. MICHIGAN STATE UNIVERSITY JAMES BERNARD DERR 1968 ,,¢. LIBRARY Tntwe This is to certify that the thesis entitled Generalized Sylow Tower Groups presented by James Bernard Derr has been accepted towards fulfillment of the requirements for PhoDo degree in Mgthfi‘matics Major professor Date September )+, 1968 0-169 y N BINDING BY 4...- "fi.¢~‘. ‘50"? .‘fi— V IIDAG & SONS’ ‘BIIUKBINDERY INC. LIBRARY em or: R5 1 menu. memes] III k.‘ - p .E§é§%€asififl I "I" m In“ V . r up P ..‘ u..- . 0 . r _ A ._ .. r C -., A V‘:" . Pva‘. *A...‘ ..‘ UV I ‘-\-. u ‘l‘ \ -\ ”‘9‘ v “-4 (I) ABSTRACT GENERALIZED SYLOW TOWER GROUPS By James B. Derr Both finite solvable groups and finite nilpotent groups can be characterized in terms of their Sylow structures. In this vein, P. Hall showed that a finite group is solvable if, and only if the group has a complete set of pairwise permutable Sylow subgroups. And it is well known that the Sylow subgroups of a finite nilpotent group centralize one another. The aim here is to study the structure of groups whose Sylow subgroups satisfy a normalizer condition (N). If S is a collection of subgroups of a finite group, we say S satisfies (N) if, for every pair of subgroups of relatively prime orders in S, at least one of the sub- groups normalizes the other. We first consider groups having a complete set of Sylow subgroups which satisfies (N). These groups are called generalized Sylow tower groups (GSTG). Sylow tower groups are GSTG's and GSTG's are solvable. All subgroups and factor groups of GSTG's are again GSTG's and the direct product of GSTG's having the same normalizing structure is a GSTG. The main result on GSTG's is the following. La - ., . at..- 'u-v-. I Q: r - Av u ”find: ‘Vio‘- " n {Yr-es r V-V‘» . 7-4" is q Draw“ Va. .1 “ . t“ ‘» H~U4L O r' 'YA James B. Derr ‘Ihegrem: If G is a GSTG, the nilpotent length of G is less than or equal to the number of distinct prime divisors of the order of G and, if equality holds, G is a STG. We next consider condition (N) for the set of all Sylow subgroups of a group. If the set of all Sylow sub- groups of a group satisfies (N), the group is called an N-group° An N-group is necessarily a GSTG and, if G is a GSTG whose order is divisible by at most 3 primes, then G is either an N-group or a STG. The inheritance prOper- ties of N—groups are identical with those of GSTG‘s and the classes of N-groups with similar normalizing structures are non-saturated formations. B. Huppert has studied groups with a permutability condition (V) on the set of all Sylow subgroups. A group satisfies (V) if any two Sylow subgroups of rela- tively prime orders permute as subgroups. We can then characterize N-groups as follows. .Ihggzem: Let G satisfy (V). Then G is an N- group if, and only if (1) G is a partially complemented extension of a nilpotent group H by a nilpotent group K, and (2) if p and q are distinct primes, then the Sylow p-subgroup of H normalizes the Sylow q-subgroup of K or the Sylow q-subgroup of H normalizes the Sylow . . . p 3 .5 u. L . . Y - r . _» “r o . . . .u. “A q . . ‘ C a: .w .3 M“ a: .. .u «v . . “Q I 5 . v . rt. "I“ o. . n . :u . ”L; n a as o 4 Ce K4 A: 3K. n e r... ;=\ ‘..¢ fi? In v.4 A- ._ V‘"r ,— ‘o. I James B. Derr p-subgroup of K. ‘We briefly consider condition (N) for Sylow systems of a group. A solvable group is a strongly Sylow towered group (SSTG) if some Sylow system satisfies (N). We give the following characterization of these groups. .Iheorem: G is a SSTG if, and only if G is an extension of a nilpotent group H by a nilpotent group K where H and K have relatively prime orders and either H or K is a p-group. The relative positions of various classes of solvable groups are shown by the following diagram. A line indi- cates that the lower class of groups lies in the higher class. solvrble GSTG .I. N-groups supersllvable SSTG nilpltent In order to give a decomposition of GSTG's in terms of N-groups, we examine the invariant series of a group Whose factor groups are N—groups. These series are called invariant N—series. James B. Derr A GSTG G has a unique descending invariant steries, whose length is denoted by m(G). m(G) is the minimal length of an invariant N-series of G and hence gives a measure of the deviation of the GSTG from an N-group. A GSTG G also has a unique ascending invariant N-series, whose length is denoted by e(G). We show by example that e(G) need not equal m(G). GENERALIZED SYLOW TOWER GROUPS By James Bernard Derr A THESIS Submitted to Michigan State University in partial fulfillment of the requirements for the degree of DOCTOR OF PHILOSOPHY Department of Mathematics 1968 prctl EBEEAQE I want to thank Professor W. E. Deskins for suggesting this problem and for his guidance throughout the preparation and writing of the dissertation. His patience and encourage- ment have served as a source of inapiration for the complen tion of this work. I should also like to thank Professors J. Adney and L. Sonneborn for stimulating discussions concerning the problem. ii .:;;Ji ‘7 Y‘vnr ‘A‘-Ol R" n Vo$n~ INDEI 4 BIBL. TABLE OF CONTENTS Page INTRODUCTION . . . . . . . . . . . . . 1 CHAPTER I. GENERALIZED SYLOW TOWER GROUPS . . . . 3 II. N-GROUPS . . . . . . . . . . . 28 III. STRONGLY SYLOW TOWERED GROUPS . . . . #3 IV. THE N-LENGTH 0F GENERALIZED SYLOW TOWER GRO UPS 0 o o o o o o o o o o 50 INDEX OF NOTATION o o e o o o o o o o o 70 BIBLIOGRAPIIY o o o o o o o o o o 9 o o 72 iii .l . .Q a -I .II. ,1. .5 .C . . h. vs. no a. “I. 3: o . v». A. "a. 4. 3. Ce ..q C. AV 9 .n e v f . . n . c . n...» urn A a L O G. n “ .nu "a be .. 4 "I. 5,. no .. .3 .. . t . n . a. «a a .3 . . Ce 15 Us. n5 rx. T; S . .. 3 9: to .4. ..i:>.. ,. #5.. ru. ’9 E 0‘; ‘ L;- au‘: CF \v L. n\v Q» +U INTRODUCTION Some well-known classes of finite groups can be described in terms of their Sylow structure. P. Hall [6] showed that finite solvable groups are characterized by the existence of a complete set of permutable Sylow subgroups. And finite nilpotent groups are known [13] to be the direct product of their Sylow subgroups. Recently Huppert [9,10] has characterized groups in which any two Sylow subgroups of different orders permute as groups. The aim here is to study groups which satisfy a normalizer condition (N) on a collection of subgroups. If S is a collection of subgroups of a group, we say S satisfies (N) if, for every pair of subgroups of rela- tively prime orders in S, at least one normalizes the other. In chapter I we take a complete set of Sylow sub- groups for our collection of subgroups. If some complete set of Sylow subgroups of a group satisfies (N), the group is called a generalized Sylow tower group (GSTG). The collection of subgroups considered in chapter II is the set of all Sylow subgroups of a group. A group G is called an N-group if the set of all Sylow subgroups of G satisfies (N). It is clear that every N-group is necessarily a GSTG. In chapter III we take a Sylow system of a group for our collection of subgroups. This choice is based on the work of P. Hall [6] on solvable groups. A group G is . io- n G C c n\ v A; - 2 called a strongly Sylow towered group (SSTG) if some Sylow system of G satisfies (N). In chapter IV we try to measure the deviation of an arbitrary GSTG from being an N-group. To accomplish this, we consider the normal series with N-factor groups of a GSTG. Among these series there is a unique descending invariant series, called the lower N-series of the group. This series is similar to the hypercommutator series of a solvable group. The length of the lower N-series of a GSTG gives the deviation of the group from being an N-group. There is also a unique ascending invariant series of a GSTG, called the upper N-series of G. An example shows that the length of the lower N-series may be shorter than the length of the upper N-series. v ~¢o. KII,‘ "haw“ (I) O 1101: Shem-.r Dr V~A~ CHAPTER I Solvable groups are characterized [6] by the existence of a complete set of permutable Sylow subgroups. We replace the permutability requirement by a normalizer condition (N) and examine groups with a complete set of Sylow sub- groups which satisfies (N). This class of groups includes the Sylow tower groups (STG) among others and so we call such groups generalized Sylow tower groups (GSTG). The main result shows that the nilpotent length of a GSTG G is at most n(G), the number of distinct prime divisors of IGI. And if the nilpotent length of G equals v(G), then G is a STG. We conclude the section with a construction process which yields GSTG's of arbitrarily high nilpotent length. For the sake of completeness and easy reference, we include some definitions and basic theorems for solvable groups. W4 [1+] If G is a solvable group of order mn, where (m,n) = 1, then i.) G has at least one subgroup of order m ii.) any two subgroups of G of order m are conjugate iii.) any subgroup of G whose order divides m belongs to some subgroup of order m. 11; lov‘ .1 so I‘~ a R10 C\' i... r: nxw Q. n" ~y= TI. Cy a: L: D E' 1!. J 2 A subgroup H of G is called a Hall.suhgndup of G if (IH|,[G:H]) = 1. If r is a set of primes, then a Hall subgroup H of G is a Hall m;suhgrgnp of G if C(H) g n. For a set of primes a and a group G, G” will denote a Hall r-subgroup of G. A . . al 32 ar Let G be a finite group of order pl p2 °-°pr , where pl,p2,---,pr are distinct primes. If Si is a Sylow pi-subgroup of G (i = l,2,°-°,r), then a set J = {51’52’°'°’Sr} is called a mists set at 511m subgroups of G. A complete set of pairwise permutable Sylow subgroups of G is called a Sylow basis of G. The set of 2r Hall subgroups of G formed by taking all (group theoretic) products of members of a Sylow basis of G is called a Sylow system of G. .Definitidn_li& Let ‘29 and 7 be Sylow systems of a group G. We say a? and ‘7’ are coniugate if for some g E G, Tg E J for all T 6’5“. We denote this by 7g = A) . .Ihedrem_lI5 [5] A finite group G is solvable if, and only if G has a Sylow system. .Thenrem_lié [6] All Sylow systems of a solvable group G are conjugate. l‘]| . . . .- ! avuJ .W 2;" . . ..v . V L. 9‘ ‘ D‘ . . L x. . "A .w n . v a: — 5U S . . \ / "ct .- F.“ RD h. no ¢ ; V "TV do n "I.“ -~ I t. I. U! . Win» n» 031‘ S o I \ W [6] Let H be a subgroup of the solvable group G and let 7’ be a Sylow system of H. Then there is a Sylow system J of G such that ‘7 = {S (THIS 63/}. .Dfifiiniiifln_li5 If J is any Sylow system of G, the system mm- liner, NG( .3), associated with .3 is the set of elements in G which normalize every member of .3: NG(el) = {g 6 G ISg = S, for all S 6 3 }. One can easily show that the system normalizer NG(~l) is the intersection of the normalizers of the Sylow sub- groups which belong to J W [7] The index of a system normalizer in G is the number of distinct Sylow systems of G. W [7] A system normalizer NG(:J) is the direct product of . _ , _ lts Sylow subgroups Ni’ where Ni — Si n NG(Si ), Si — Sylow pi-subgroup of .9 and 81' = Sylow pi-complement of .4 . We are now ready to introduce the normalizer condition (N) which is used throughout this work. W Let I; be a collection of subgroups of a group G. 6 We say ._g£_satisfies__£fll if, for any two subgroups of relatively prime orders belonging to .3 , at least one of the subgroups normalizes the other. D E' 'I' J 12 If G has a complete set of Sylow subgroups which satisfies (N), G is called a_generalized_§ylgw_tgmer sump (GSTG). 2 'I' J 13 A GSTG G is necessarily solvable. moi: Let 2y = {81,82,---,Sr} be a complete set of Sylow subgroups of G which satisfies (N). Then 8.8. = 8.8. l J 3 l for 1,3 i.$ j,$ r and a? is a Sylow basis for G. By Theorem 1.5, G is solvable. C] W ' A finite group G is called a.fileW.LQE££.£ZQup (STG) if every nontrivial factor group of G has a nontrivial, normal Sylow subgroup. Equivalently a group G. is a Sylow tower group if G has an ascending series of normal subgroups S1 with S0 = l, Sk = G and Si/Si__l = the normal Sylow pi-subgroup of G/Si-l' This series is called a Sylow.tgwer of G. W If G is a STG, then G is a GSTG. I one o. a..' renn- any. ... 25‘ » u'd V " 4 ‘7 .LC . «A - - C:‘~'..- ~‘.i: K . ‘- - \ L: '1" 'Q‘Vutr- 7 .Ezggf: (Induction on W(G)). If r(G) = l, G is a p-group and the assertion is trivial. If W(G) = 2 and G is a STG, then G has a normal Sylow subgroup and G is clearly a GSTG. Now assume the theorem holds whenever r(G) S k and let G be a STG with w(G) = k + 1. Since G is a STG, G has a normal Sylow subgroup S. Let S‘ be any complement of S in G. S' is a STG and n(S') = k. By induction, 8' has a complete set of Sylow subgroups {T1,T2,°'°,Tk} which satisfies (N). Since S is a normal subgroup of G, {S,T1,T2,'°°,Tk} is a complete set of Sylow subgroups of G which satisfies (N) and so G is a GSTG. C] W If G is a GSTG, then every Sylow basis of G satisfies (N). EIOQ£= Since G is a GSTG, some complete set of Sylow sub- groups of G, .3 = {Sl,'°-,Sk}, satisfies (N). Then 23 is a Sylow basis of G which satisfies (N). If .73 is any other Sylow basis of G, then by theorem 1.6 II = 43g = {S g,"°,Skg} for some g 6 G. Since 81; NG(sj) implies sig 5 NG(Sjg), ’5’ = .83 satisifes (N). C] op:r 5....- 2. n3 wrhu ~\v «‘I‘ q 1H~ v «.1. ‘U 8 More generally, whenever a collection a: of subgroups of a group G satisfies (N), every conjugate 3g (g 6 G) of 93 satisfies (N). We now study the inheritence prOperties of the class of GSTG's. Lemma_l.._lZ If G is a GSTG and H is a Hall subgroup of G, then H is a GSTG. Emmi: Let as = {Sl,'°°,Sk} be a complete set of Sylow subgroups of G which satisfies (N). Let K = Si '°° Si be the product of those Sylow subgroups l m of 3 whose orders divide IHI. Then K is a Hall sub- group of G and IKI = |H|. By theorem l.l K is conju- gate to H and hence H = Kg for some g 6 G. Then {Sig,°° °°,Sig} is a complete set of Sylow subgroups of l m H which satisfies (N). U W If G is a GSTG and H is a subgroup of G, then H is a GSTG. .Ezggf: (Induction on IGI) Assume the theorem holds for all groups of order less than IGl. If w(H).§ n(G), then H lies in a prOper I‘Iall subgroup H of G. By the preceding lemma H is a GSTG. Induction then implies that H is a GSTG. I . . I . . AU ,1 a J r“ r: n n J 0: a J guy .r » \ I \. / OI ~I~ J n . a. . .3. "in VI. a. . a. . a. o u u g CI . .o; AU :e Il.\ I.I\ n A v do a O; . - «J. Fe . . he as n5. L win 9 Now let W(H) = V(G). If ()4 is any Sylow system of H, then, by theorem 1.7, there is a Sylow system :3 of G such that 74 = {H O S IS Gig }. Since W(H) = F(G), H n S is nontrivial for every 8 E El . In particular then, if {Sl,°° °°,Sk} is the set of all Sylow subgroups of .3 , {H n 31,” °°,H 0 SK} is a complete set of Sylow subgroups of H. By theorem 1.16 {Sl,°° °°,Sk} satisfies (N) and consequently {H n Sl,°° o»,H fl Sk} satisfies (N). Then H is a GSTG. ~ E] Lemma_l..13 [Um-131+] Let G be a finite group and let a be a homomorphism of G onto Ga. Then P is a Sylow p-subgroup of Go if, and only if P = G;’ for some Sylow p-subgroup Gp of G. W The homomorphic image of a GSTG G is again a GSTG. Email: Let a be a homomorphism of G onto GO and let .J = [Sl,°° °°,Sk} be a complete set of Sylow subgroups of G which satisfies (N). By lemma 1.19, .30 = {Sl°,°° ”’SkO} is a complete set of Sylow subgroups of GO. Since Sf, normalizes 85’ whenever Si normalizes s., .3“ satisfies (N) and GO is a GSTG. [:1 Suppose H is a normal subgroup of G and both H and Q/H are GSTG's. Can we conclude that G is a GSTG? . . . . 1. .2, h. wm A» to a: W. n. , E .r“ A» AC 2* wt. «T. I a he. “I. RD A‘U C a, 3 Q... .,. .pVU .. no and ..o . Aliv 10 The following examples show that we cannot answer "yes" even for simple cases. .Examnla_le AM is a GSTG, since the Sylow 2-subgroup of Ah is normal. And Sh/Au is a cyclic group of order 2. Yet SM is not a GSTG since W(Sh) = 2 and SM has no normal Sylow subgroup. Hence a GSTG extended by a cyclic group of prime order is not necessarily a GSTG. Example_2. S3 x A“ is not a GSTG since 1T(S3 x Ag) = 2 and S3 x Ah has no normal Sylow subgroup. Yet both C36 S3 and S3 x AH/C3 are GSTG's. This shows that an extension of a cyclic group of prime order by a GSTG is not necessar- ily a GSTG. Relative to this question however, we can show the following. W G is a GSTG if, and only if G/z(G) is a GSTG. £19m: Because of theorem 1.20, we need only show the necessity. Choose :8 = {G1,°° °°,Gk} to be a complete set of Sylow subgroups of G such that 3: {G12(G)/Z q and p';:> q. Then H N H K N K H K p.$ H( q), p'3 K( q) and consequently p x p normalizes Hq x Kq. Since p and q are arbitrary, 8! satisfies (N) and H x K is a GSTG. C] The previous result can be extended to central products. D E' 'I' J 28 A group G = H - K with H O K.g Z(G) and H.$ CG(K) is called the central_ngdnct of H and K. in W The central product of H and K is a GSTG if, and only if H and K are (N)-similar GSTG's. ,Erggf: Theorem 1.25 shows the sufficiency. To verify the necessity, let H and K be (N)-similar GSTG's. Since H.$ CG(K), M = H O K is normal in G = H ° K. Then H/M and K/M are (N)-similar GSTG's and hence 9/8 = H/M x 3/“ is a GSTG. Therefore .9 .11.. g M . z(g) /Z(G) is a GSTG and theorem 1.21 shows G is M a GSTG. ‘ C] The class of all (N)-similar GSTG's may be shown to be a formation in the sense of Gaschfitz [3]. Lamma_l_._3.0 If H and K are normal subgroups of G, then Q/HITK is isomorphic to a subgroup of 9/H x Q/K. Emmi: Define the mapping t : G'-> 9/H x Q/K by t(g) = gH x gK, for all g 6 G. t is a homomorphism into Q/H x G/k whose kernel is H n K. G E 'I' J 3] Let H and K be normal subgroups of G. If 9/H and G/k are (N)-simi1ar GSTGfs, then Q/HIlK is a GSTG. 15 limit: If Q/H and Q/K are (N)-similar GSTG's, then G/H x G/k is a GSTG. Since G/HITK is isomorphic to a subgroup of G/H x Q/k, Q/HITK is a GSTG. [3 D E' 'I' J 32 A formation F is a collection of finite solvable groups satisfying i) 1 e F ii) G e F implies every homomorphic image of G belongs to F iii) N l and N2 normal subgroups of G with 9/ 9/ . . G , 6 F lmplles / 6 F. The formation is saturated if 9/i(G) 6 F implies G E F. Let * be a relation on the set of all primes. A GSTG G is compatible_with__i if, for some complete set of Sylow subgroups 43 of G, p’tq implies the Sylow p-subgroup of :8 normalizes the Sylow q-subgroup of :8 . Then if H and K are GSTG's compatible with a relation *, H and K are (N)-similar. Consequently, theorems 1.26, 1.27 and 1.31 show that the set of all GSTG's compatible with a relation * is a formation. It is not known if the formation is saturated. The following definitions and theorems are needed for the main result of this section and for chapter IV. Ll 16 D E! 'I' J 3 Let G be a solvable group. i) The Eitting_suhgronp of G, denoted F(G), is the maximal normal nilpotent subgroup of G. It is well known [13, p.166] that F(G) is the product of all normal nilpotent subgroups of G. ii) The Eitting_series of G is defined inductively F _ i+l _ G _ iii) The length of the Fitting series of G, denoted z(G), is the nilpotent_length of G. iv) A normal series with nilpotent factors is called a.nilpntent_series- We combine several results of Spencer [12] in the next statement. .Ernnnsiiidn_li35 Let G be a solvable group. Then i) The length of any nilpotent series of G is greater than or equal to L(G). ii) For a normal subgroup K of G, £(Q/K) S L(G). iii) For H a subgroup of G, £(H) S £(G). iv) For normal subgroups H and K of G, £(Q/HI1K) .3 max [£(G/H), £(G/K)}. v) z(9/¢(G)) = £(G). 17 We are now able to state the main result of chapter I. W If G is a GSTG, then £(G) _<, 1r(G). Proof: (Induction on IGI). If F(G) g 2, then G has a normal Sylow subgroup and surely £(G) g F(G). Let G be a GSTG with F(G) = k.2 3 and take .J to be a complete set of Sylow subgroups which satisfies (N). Case 1. G has a normal Sylow subgroup, T (say). Since G is a GSTG, the factor group 'G = Q/T is a GSTG. W(G) = k-l and by the induction assumption k -1J Hence 'G has a nilpotent series 2(6) _<. m(E) ._ H H - l = T/T Q l/T 4 00.4 k l/T : G/T. Then 1‘4 T 4 H14 °°°.4 HK-l = G is a nilpotent series of G of length. k and theorem 1.35 (1) implies L(G) S R. Case 2. No Sylow subgroup of G is normal. Since G is a GSTG, G is solvable and has a minimal normal subgroup M, which has prime power order. Let IMI = pm and let P denote the maximal normal p-subgroup of G. If SD is the Sylow p-subgroup of G belonging to J , then Sp contains P. By assumption Sp,§ G. Then, for some prime q i p, the Sylow q-subgroup Sq of :3 fails to normalize Sp. 18 Since :8 satisfies (N), this means Sp normalizes Sq Then Sq,g CG(P) = C, because P normalizes Sq, P 4 G and (IPI, ISqI) 1. Notice that, since P 4 G, c = CG(P) .4. G. Now let P C O P denote the maximal normal 0 p-subgroup of C. Since P is a G p-group, Z(P) # 1 and 1:} Z(P)_§ PO. Since sC1 g c, C . . . _ /PO is a nonterlal C — CG(P) P solvable group and therefore S CITP = P0 has a minimal normal subgroup Y/P i'I. Suppose IY/P I is a power of the prime r. o 0 Now r # p, for r = p implies that V is a normal p-subgroup of C prOperly containing Po - contradicting the maximality of PO. Let W/P denote TG o the maximal normal r-subgroup of C/P . "CG(P) o W . . . . "W Then /P 18 characteristic in 0 hp 0 C/p‘a G/P and hence normal in Q/P . o o o By the isomorphism theorems, W 4 G. Since W centralizes P and W/P is an r-group, o W = W6 x PO where W0 is the Sylow r-subgroup of W. Since the normal Sylow r-subgroup WO of W is character- istic in W 4 G, W0 is normal in G. 19 . G G Now conSlder the factor groups /P and /w . By 0 the induction hypothesis, £(Q/p) S ”(Q/P) = k and £(9/w ).$ ”(Q/w ) = k. Since P n WO = l, PrOposition o o 1.3M iv) shows that £(G) = ‘(Q/prlw ).g k and the theorem 0 is proved. D Lemma_l.._12 Let G be a STG with £(G) = r(G). If s and T are Sylow subgroups of G having relatively prime orders, S does not centralize T. Ezggf: (by minimal counterexample) Assume that G is a STG of minimal order which satis- fies the hypothesis of the theorem but fails to satisfy the conclusion. Then £(G) = F(G) = n.2 3 and G has a Sylow tower l 4 S 4 S S ‘3 .. .. 4 S '°S S = G, l 2 l n 2 l where Si is a Sylow pi-subgroup of G (i = l,2,°°°,n). The factor group 9/8 is a STG with ”(9/8 ) = n-él. l I Since 9/8 is a STG and 2(6) 2 n, 1(9/8 ) = n-1 = r(9/S ). l l 1 Consequently, by the minimality of G, no Sylow subgroups of Q/ having relatively rime orders centralize one another. 31 p But, since we assumed G does not satisfy the conclusion of the theorem, some Sylow pj-subgroup T of G centra- lizes some Sylow pk-subgroup V of G. Then the Sylow TS pj-subgroup 1/S of 9/8 centralizes the Sylow l l 20 VS p -subgroup 1/ of 9/ . k S S l l . Isl ._ V31 ._ This is impossible unless /S = l or /S = 1. l 1 Consequently j = l or k = l and we may assume that the Sylow pk-subgroup V of G centralizes 81' Since S1 4 G and Sk = Vg for some g e G, Sk centralizes 81' Since 81‘4 G, CG(Sl) is a normal subgroup of G containing Sk' Let m be the smallest integer, 2.$ m.$ n, for which pm |cG(sl)| and put R = sIn -- 3251 n cG(sl). R is a normal subgroup of G and by the choice of m, p1 and pm are the only prime divisors of .|R|. Since R.S CG(Sl), any Sylow pm-subgroup Rm of R centralizes the normal Sylow pl-subgroup R1 = Sl n CG(Sl). Then R = R1 x Rm and Rm is a characteristic subgroup of the normal subgroup R of G and hence lea G. Suppose Rm is a Sylow subgroup of G. Then RmSl = Rm x S l is a normal nilpotent subgroup of G and 1 4 313m“ S251Rm“ " “Sm-1"S2Slfim‘ Sm+lSm--l' "(3251Rm""“G is a nilpotent series of G having length n-l. Then prOposition 1.35 1) implies 1(G).$ n - l, which is a contradiction. Therefore Rm is not a Sylow subgroup of S G and consequently we are assured that kRm/R is a m nontrivial Sylow pk-subgroup of Q/Rm. Since Sk.$ CG(Sl)’ S R S R k m/R centralizes l R/R . Then v(G/R ) = n and the m m m minimality of G imply 1(9/R ) < n. 'Since ‘(Q/S ) < n m l 21 and Rm 0 S1 = 1, prOposition 1.35 iv) implies that £(G) = £(Q/R OS ) < n - which is a contradiction. m l E] .Theorem_llld If G is a GSTG and £(G) = v(G), then G is a STG. Exggf: (Induction on IGI). If w(G).$ 2, G has a normal Sylow subgroup and clearly G is a STG. Let G be a GSTG with £(G) = F(G) = n.2 3. We distinguish two cases. Case 1. G has a normal Sylow subgroup K. Q/K is a GSTG and, by theorem 1.36, 2(G/k).s n(§/K) = n-l. Since £(G) = n, we must have 1(Q/K) = n-l. Then the induction assumption implies that G/K is a STG. Let G/K have Sylow tower S K S °°°S K —' K 1 -l l G l 2 /K4 /K4 oo 00 d n /K: /K, Where 81 is a Sylow pi-subgroup of G. Then 1 a K 4 le 4 ~ . . 4 sn_l---le = G is a Sylow tower of G and G is a STG. Case 2. No Sylow subgroup of G is normal. Then, as in the proof of theorem 1.36, there exist normal prime power order subgroups M1 and M2 of G with (|M1|,|M2|) = 1. Consider the factor groups Q/M l and G/M of G. If z(Q/M ) < n and L(9/M ) < n, 2 1 2 22 prOposition 1.35 iv) implies £(G) < n, a contradiction. Therefore we can assume £(Q/M ) = n. Since Ml cannot 1 be a normal Sylow subgroup of G, r(G/M ) = n = 2(9/M ) l l and the induction assumption implies G/M is a STG. , l M S M S °°°S M '— l l l n l l G M1 M1 M1 M1 be a Sylow tower of G/M , where Si is a Sylow pi-subgroup l of G (i = 1,2,--°,n). Furthermore, choose Sn to be the‘ Sylow pn-subgroup of a Sylow basis :3 of G which satis- fies (N) and contains Sl° Suppose M1.$ Sk’ for some k < n° Then H = Sk"SlMl = sk-osl since £(G) = n, L(H) = H(H) = k, By the induction assump- is a normal Hall subgroup of G and, tion H is then a STG and consequently has a normal Sylow subgroup W. Since H is a normal Hall subgroup of G, ‘W is then a normal Sylow subgroup of G - a contradiction. Consequently, since a normal p-subgroup of a group lies in every Sylow p-subgroup of the group, we must have Ml.$ Sn' Since S1 and Sn were chosen to belong to a Sylow 'basis 3 cd‘ G- which satisfies (N), either Sl normalizes Sn or Sn normalizes 81' Suppose first that Sn.$ NG(Sl). Then M normalizes S1 and SlMl 2 S1 x Ml’ since 1 [Sl’Ml]'S S1 0 M1 = 1. Consequently S1 is a characteristic . subgroup of SlMl a G- and hence S1 is a normal Sylow pl—subgroup of G - a contradiction. Now suppose 23 81.3 NG(Sn). Since SlMl'd G, we then have that the Sylow S pl-subgroup 1 1/M of 9/M centralizes the Sylow l l S M S _ n l _ n G pn subgroup /M1 . /Ml Of /Ml. Consequently, since G /M1 is a STG, z(9/M ).$ n-l - which is a contradiction. 1 Therefore this case cannot occur and the theorem is proved. C] The remainder of the chapter is devoted to showing that GSTG's of arbitrarily high nilpotent length do exist. The construction process given utilizes wreath products, which we now describe. Let A and B be finite groups and let AIBl denote the direct product of IBI COpies of A. Suffix the IBI IBI by the elements of B and construct direct factors of A a group W = A wr B, which is the extension of AlBl by B. To complete the definition of W, we Specify the automorphisms induced on AIBI by the elements b e B. If a 6 Ab , i -161b as the element corresponding to a in -1 define b IBI. Ab b“ This determines b rlb for any element n 6 A i The group W = A wr B is called the.mneaih.pzndnnt.nf .A .hy .B and its order is IAIIBI ° lBI. W If A and B are GSTG's of relatively prime orders, A wr B is a GSTG. 2% Email: Let _O-= {A1,°°,Ak} and Q> = {Bl’°°’BL} be Sylow bases of, A and B (reSpectively) which satisfy (N). Let AilB' denote the subgroup of AIBI which is the direct product of IBI c0pies of A1 (i = l,°°,k). Then A1IBI is a Sylow subgroup of AIBI and every element of B normalizes AiIBI (i = l,°°,k). Furthermore, A1 3 NA(A.) implies AiIBI normalizes Aj IBI. Therefore WV'= {A1|B|,°-,Ak|BI,Bl,°°,B£} is a Sylow basis of W = A wr B which satisfies (N). D LemmaJIItO Let A and B be finite groups and let K;g A. If w = A wr B, then W/KIBI '5' A/K wr B. Proof: KIBI is normal in w = AIBI . B, since KIBI <_i AIBI and by the definition of multiplication in W, B normali- zes KIBI. We will show that the mapping o : boal-°°alBl'—> b°alK°°a|B|K for b E B, a1 6 Abi is a homomorphism from W to 4/K wr B with kernel KlBl and image A/K wr B. Clearly the mapping is well-defined and has image A IBI ,[K wr B. Let c,d 6 B and ni,gi 6 Abi S A for i = 1,°°,|B|. (l) m is a homomorphism. 25 r 00 o .0 —— w L(C g1 g|B|) (d nl n|B|)] - m (c m [cd (gl°°g|B|)d(nl°°nlB|)] = d m [Cd (hlnl°°h|B|n|Bl)]: ht : gs 6 Ab 3 S: t : 1929°°lBl Cd (hlanoohlBlnlBlK) = cd (th°°h|B|K)(an°°nlBlK) u .. d .. — cd (glK gIBIK) (an nlBlK) : [C(glK°°g|B|K)][d(an°°n|BIK)] o (cgl°°g|B|) ° w (d nl°°n|B|). (2) Ker co=K|B|. oo __- lBl oo — lBl g1 ng|)-l—K ecglK nglK—K e c = 1 and g. 6 K i = l °° IBI l bi 3 9 B G C gl°°ng| E Kl ‘. U W [13. p.167] If G is a finite group, then Fit(G) = e in | K(p), P G where K(p) is the intersection of all Sylow p-subgroups of G and Fit(G) denotes the Fitting subgroup of G. As a corollary of theorem 1.Hl, we have the following. Lemma_li&2 If G = G x Gk is the direct product of finite l Xoo STOUps, then Fit(G) = Fit(Gl) X°°X Fit(Gk). .Ezanasiiion_lla3 If G is a finite group of nilpotent length k and 26 C = is a cyclic group of order 2, (IG|,£) = 1, then G wr C has nilpotent length k + l. .Eroof= (1) Fit (G wr c) = (Fit(G))lCI. By lemma l.h2 Fit(G|C|) = (Fit(G))|C| and consequently (Fit(G))|C| is a normal nilpotent subgroup of G wr C. Hence (Fit(G))ICI S Fit(G wr c). SUppose (Fit(G))ICI < Fit (G wr C) and let d°n e Fit(G wr c)\\(Fit(G))|C|, where d e c and n e GICI. Then d = c1 for 1.3 i < A. dt’1 d . n - o n 11 for any integer t, Since (dn)t = dt d-n has order j-z' where jllGIlCl and l # L"£. Then (dn)j has order 1' and (dn)j 6 Fit(G wr C). Since G wr C is solvable, there exists x 6 G wr C such that (dn)j 6 0x and therefore (dn)j = (cw)x for some w, l.$ w.; 2. Then cw e Fit(G wr C) and Fit(G wr C) has a nontrivial Hall subgroup H whose order divides ICI. Then, H.g C and hence H.$ Z(G wr C) - since C is cyclic and (IH|,|G|) = 1. This is impossible and there- fore we must have (Fit(G))|C| = Fit(G wr C). Now use induction on £(G) to complete the argument. If £(G) = 1, then surely L(G wr C).$ 2. If £(G wr C) = 1, then G wr C = G x C, which is nonsense. Therefore £(G wr C) = 2 in this case. Let £(G) = m and put F = Fit(G). Then 2(9/F) = m-l 27 and by induction 1(9/F wr C) = m. Applying lemmas 1.HO G gGWI’C :GwrC and 1.H2 we have /F wr C — /F|Cl — /Fit(Glm7C)° G wr C _ Therefore L( /Fit(G wr C)) — m and consequently L(G wr C) = m + l - and the proof is completed. I] 1’ 2,°°,Cn denote cyclic groups whose orders are relatively prime in pairs. Then, by the previous results, Let C C we see that the repeated wreath product W = ((°(C1 wr 02)wr°°)wr Cn is a GSTG of nilpotent length n. Hence if k and n are any positive integers and k,2 n, we can construct a GSTG G with £(G) = n and 1T(G) = k. Chapter II In this chapter we consider condition (N) for the collection of all Sylow subgroups of a group. If the set of all Sylow subgroups of G satisfies (N), we call G an N-group. Clearly every N-group is a GSTG. The inheri- . tance prOperties of N-groups are identical with those of [ GSTG's and certain subclasses of N-groups are (non—satura‘ ted) formations. r Using some results of Huppert [10], it is shown that the nilpotent length of an N-group is at most 2. A characterization of N-groups as a Special type of product of nilpotent groups is then obtained. The remaining re- sults describe the Sylow structure of N-groups or the structure of groups which just fail being N-groups. We begin by formally defining an N-group. D E’ 'Ii 2 1 Let Syl(G) denote the set of all Sylow subgroups of G. A finite group G is an N;grgnp if Syl(G) satis- fies (N). Several observations follow immediately. If G is an N-group, then every complete set of Sylow subgroups of G satisfies (N). In particular, every N-group is a GSTG and consequently is solvable. And since a finite nilpotent group is the direct product of its Sylow subgroups, a finite nilpotent group is clearly an N-group. Furthermore, every group which is a nilpotent group extended by a p-group 29 is an N-group. S3 is an example of such a group. The next two examples further describe the position of N-groups re- ..- -. -~ lative to other classes of solvable groups. We. The following is an N-group which is not a STG. Let s 2 3 3 H==®1r x®1r x®1r i=1 1 i=1 k=l and k = g Aut(H), where x_, _ x__ _ ai ai+l’ bjx- bj’ ck - ck (a6 — a1) bjy“ bj+l’ aiy= ai, cky= Ck (b3 = b1) z_ z_ z_ _ ck — Ck+l’ ai — ai, bj — bj (CH - cl). Define G to be the Split extension of H by K - = [H]°K. Then G is an N-group whose Sylow structure is described by ,(’5*\ . Since G has no normal -——> Sylow subgroup, G is not a STG. .Example_21 A supersolvable group which is not an N-group. Let G be the Split extension of C7 = < x Ix7 = l > by .Aut(C7) = < y Ixy = x3 > 5 C6’ and put G2 = , G3 = . G is supersolvable since 1 4 C7 4 07G3 4 G is an invariant series of G with cyclic factors. And G is not an N-group because 3.3 NG (G2 x) and G: .j N G).(G3 E 'I' 2 2 If G is a GSTG and v(G).S 3, then G is either a 41ll.rn‘ll..ombl.< . t -4 3O STG or an N-group. Proof: If F(G).S 2, then G has a normal Sylow subgroup and G is a STG and an N-group. Suppose H(G) = 3 and let {Gp,G Gr} be a Sylow basis for G which satisfies (N). q, If G has a normal Sylow subgroup, then G is a STG. Otherwise we may assume Gp,$ NG(Gq), Gq.$ NG(Gr) and Gr,$ NG(Gp). We will Show that every conjugate of Gp in G normalizes Gq. Consider Ggg, for any g E G. Since G = GpGqu, g = xyz for some x E Gp, y E Gr’ 2 E Gq. 8 = XYZ 2 YZ = Z Then Gp Gp Gp Gp , since y E Gr,$ NG(Gp). Since Gp normalizes Gq and z 6 Gq, Gpg = sz norma- lizes Gq(= qu). Therefore every Sylow p-subgroup of G normalizes every Sylow q-subgroup of G. The argument may be repeated to Show qu.$ NG(Gr) and Grg g NG(Gp), for all g E G. D W Let G be an N-group and let some Sylow p-subgroup of G normalize some Sylow q-subgroup of G, where p # q. Then every Sylow p-subgroup of G normalizes every Sylow q-subgroup of G. Proof: Let P E Sy1;)(G), Q 6 Syl<1(G) and suppose P normalizes Q. Since all Sylow subgroups of G of the Same order are conjugate in G, it is sufficient to show that Fx normalizes Qy, for all x,y E G. 31 Since every complete set of Sylow subgroups of G forms a Sylow basis, PX and Qy belong to some basis 7’ of G. Likewise P and Q belong to some Sylow basis .J of G. By theorem 1.6 a? and 7' are conjugate in G and so there exists g E G such that PX 2' Pg, Qy = Qg. Pl Since P normalizes Q, Pg normalizes Qg and we are 7 done. [I] We now examine the inheritance prOperties of N-groups. Lemmaiiit t Let H be a subgroup of G. If P is a Sylow p- subgroup of H, then P = H O'P for some Sylow p-subgroup P of G. Proof: By theorem 1.1 P lies in some Sylow p-subgroup P of G. Then P g H n ”p. Since H n p is a p-subgroup of H which contains a Sylow p-Subgroup P of H, we must have P = H n'E. W If G is an N-group and H.g G, then H is an N-group. Proof: Let P e Syl p(H), Q 6 Syl q(H) where p +q. By lemma 2.# there exist subgroups P E Sy113(G) and GGSqu(G) such that P=HOP and Q=Hn§. Since G is an N-group we may assume IQ normalizes P. Then Q = H O Q normaliZeS P = H HIP. g 32 E 'I' 2 5 The homomorphic image of an N-group G is again an N-group. Ihxmdh Let a be a homomorphism of G onto Go. Let P E Syl})(G°) and Q 6 Syl<1(G°) where p # q. By lemma 1.19 there exist Sylow p- and q-subgroups Gp and G of G such that P = Gp“ and Q = sq“. Since G q is an N-group we may assume Gq normalizes Gp. Then qu normalizes GpO and the assertion follows. C] The next definition is crucial for the study of direct products of N-groups. D E' 'l' 2 2 The N-groups H and K are said to be similar if for any pair of distinct primes p, q 6 c(H) n c(K), either i) p > q and p > q Syl(H) Syl(K) or ii) q > p and q > p . Syl(H) Syl(K) The proofs of prOpositions 2.5 and 2.6 have established the next results. W If H and K are any subgroups of an N-group, then H and K are similar N-groups. E 'I' 2 9 If H and K are any normal subgroups of an N-group G, then 9/H and 9/K are Similar N-groups. 33 The next theorem is the analogue of theorem 1.27 for N-groups. B 'I' 2 10 H x K is an N-group if, and only if H and K are Similar N-groups. Proof: The sufficiency follows from PrOposition 2.8. To Show the necessity, let H and K be Similar N-groups. Let p and q be distinct primes and suppose p >q and p >q. If P€Sylp(HxK) and Syl(H) Syl(K) Q E Sy1<1(H x K), then P = Hp x Kp and Q = Hq x Kq for some HI) 6 Sy1p(H), HQ 6 Squ(H), Kp 6 Sy1p(K) and Kq 6 Syl<1(K). Consequently P = Hp x Kp normalizes Q = Hq x Kq and the assertion follows. C] The groups SH and 83 x AH’ which were discussed in the first chapter, Show that an N-group extended by a cyclic group need not be an N-group and that a cyclic group extended by an N-group need not be an N-group. Using the same argument as for GSTG's [theorem 1.21], we can prove that a group G is an N—group if, and only if 9/Z(G) is an N-group. Consequently prOposition 2.10 can be extended to central products using the argument for theorem 1.29. Similarly the proof of prOposition 1.31 carries over to the present case and establishes the next result. 3% E 'I' 2 J] Let H and K be normal subgroups of G. If 9/H and 9/K are similar N-groups, then Q/HITK is an N-group. We are now in a position to see what collections of N-groups are formations. D E' 'I’ 2 12 Let * be a relation on the set of all primes. An N-group G is ' * if p * q implies that every Sylow p-subgroup of G normalizes every Sylow q- Subgroup of G. Since any two N-groups compatible with a relation * are Similar, it follows from prOpositionS 2.6 and 2.11 that the set of all N-groups compatible with a given relation * is a formation. The next example shows these formations are not necessarily saturated. Example: Let S be the nonabelian group of order 73 and eXponent 7 defined by S = . S is generated by y and z and S' = . Let D be the group of automorphisms of S defined by D = <. Take G to be the Split extension of S by D. Then Q(G) = <:x> = S' and 2 3 G _ (Z d ,x> . G /i(G) ‘ ’ / x ’ / ' Since /i(G) is the direct product of similar N-groups, Q/Q(G) is an 35 N-group. Yet G is not an N-group because x.4 NG() and i NG(x). Huppert [10] has studied groups which satisfy (V): every pair of Sylow subgroups of different orders permute as subgroups. He proves the following result. .Theorem_2113 If G satisfies (v), then 9/§(G) is isomorphic to a subgroup of a direct product of groups of orders a b pi i qi 1, where pi and q1 are primes (not necessarily distinct). As a consequence of this we have the following. luxuxundfirulJLGUT If G is an N-group, then L(G).S 2. Taxman Since an N-group clearly satisfies (V), theorem 2.13 shows /§(G) — H.§ ® izl Ti’ where ITiI — pi qi . For each integer k, 1.3 k.$ n, define the projection wk of H into Tk by wk (t1'°tk--tn) = tk, where ti ETHQ” Since H is isomorphic to a factor group of G, H is an N-group. Consequently the homomorphic image of H under ”k’ denoted Im(rk), is an N-group for k = l,---,n. Since Im(wk) S Tk’ at most two primes divide IIm(nk)I and so £(Im(nk)).$ 2. Since n H.$ e 'r Im(wk) proposition 1.35 implies k=1 3.- . u! __. 36 n £(H).S £( ® v Im (Wk)) 3 max {Z(Im(vk)},$ 2. Conse- k=l l$k$n quently £(G).$ 2. C] We now examine N-groups having nilpotent length exactly 2 more closely. . .. : I Let G be a group with normal subgroup H. We say t Kisapamialrdmflememw if G=HoK and L K is a prOper subgroup of G. The subgroup H O K is E normal in K, but is not necessarily the identity. W [8] If Q/H is any nilpotent factor of G, then some system normalizer NG(-?) of G is a partial complement Of H in G. Let the hypercommutatgr_gf__fi_ be the intersection of all normal subgroups K of G such that g/K is nilpotent. Q/fi (G) is nilpotent and Hm(G) is the minimal (normal) subgroup with this prOperty. If £(G) = 2 and G is an A-group (i.e. all Sylow subgroups of G are abelian), Taunt [1%] shows that any system normalizer of G is a complement of Hm(G). We can characterize N-groups as a Special product of nilpotent groups. .Iheglem_2LlZ Let G satisfy (v). Then G is an N-group if, and 37 only if (1) G is a partially complemented extension of a nil- potent group H by a nilpotent group K and (2) if p and q are distinct primes, then Hp norma- lizes Kq or Hq normalizes Kp. .EIQQ£= Let G be an N-group of nilpotent length 2. By theorem 2.16, G = Hm(G) ~ NG(-!) for some system normalizer NG(.J) of G. Since Z(G) = 2, Hm(G) is nilpotent and, by theorem 1.10, NC( J) is always nilpotent. Consequently (1) holds. To verify (2), let p and q be any distinct prime divisors of (IHm(G)I,IG/fim(G)|). Put H = Hm(G), K = NG( 3 ) and consider the subgroup L1 = Hp(Kqu) of G. L1 is an N-group of order paqb and consequently Kq normalizes Hpr or H Kp normalizes Kq. In the latter P case Hp normalizes Kq. In the first case, we may assume Kq.$ NG(Hpr) and Hpr.i NG(kq) - for otherwise we are l is Similar to G, H K must norma- done. Then, Since L q q l' H K . lze p p Now consider the subgroup L2 = Hq(Kqu) of G. L2 is an N-group of order paqb and so Kp normalizes H K H K ' K . q q or q q normalizes p In the latter case, Hq.S NG(Kp) and we are done. Otherwise we may assume Kp normalizes HqKq and HqKq i NG(Kp). Since L2 is 38 similar to G, we must have Hpr normalizes HqKq. Therefore H K and H K normalize one another and, P P q q since they have relatively prime orders, H K H K H K n H K = 1. Th' b ' h th I p p’ q q].S p p q q is esta lis es e sufficiency. We now show the necessity. Suppose G satisfies (V) and conditions (1) and (2) hold. We proceed by induction on w(G). If G is a p-group, the theorem is trivial. Suppose G=ab d G=H°K=HH KK.B 2 I I p q an let ( p q)( p q) y ( ) we may assume Hp normalizes K . Since K is nilpotent, q Kp normalizes Kq. And Since H is a nilpotent normal subgroup of G, Hq 4 G. Then Hpr normalizes HqKq and G is an N-group. Suppose F(G) = k.2 3 and let p and q be any distinct prime divisors of IGI. Let Gp and Gq be any Sylow p- and q-subgroups of G (reSpectively). By Sylow arguments we then have G = (H K )x = H K x for some x E G P P P P P G: Ky=HKy G. q (Hq q) q q for some y 61 Since G satisfies (V), L = (Hpr)(HqKqyx ) is a sub- group of G, for all x,y E G. L is a prOper subgroup of G which satisfies the induction hypothesis and so L is an N-group. Hence Hpr normalizes Hqqux or HqK yx K. = KX HKy G normalizes Hp p Then Gp Hp p normalizes q q q 39 or Gq = Hqqu normalizes HprX = Gp. Since Gp and Gq were chosen to be arbitrary Sylow subgroups of G, we have shown that G is an N-group. D Owing to Taunt's result on A-groups, we see that con- dition (l) of theorem 2.17 can be improved to complements (in place of partial complements) for N-groups with abelian Sylow subgroups. It is not known if this is generally true for N-groups. The following examples Show the conditions of theorem 2.17 cannot be relaxed. Example—l. Let G be the Split extension of C7 by Aut (C7) 3 C6' We already know G is not an N-group. G satisfies (1) and (2), but fails to satisfy (v). Let W = e _ X_ X_ V__ and let H — < x,v|ai - ai+l, b. - b., b. — b a1v = ai, an = a1, b3 = bl >.g Aut (w). Put G = [W]-H, the Split extension of W by H. G satisfies (V) and condition (1) but is not an N-group - since F(G) = 2 and G has no normal Sylow subgroup. In view of theorems 2.16 and 2.17, we try to describe the system normalizers of an N-group. E 'I' 2 13 Let G be an N-group. If .3 = {Sl,°°,Sk} isa Sylow basis for G, then the system normalizer associated ’M‘w‘w uni-fig #0 n withw3 isN(28)=®1rN,whereN=C G k=1 k for T1’ °° O’TL those members of .3 \.[Sk} which norma- lize Sk' Proof: n By theorem 1.10, NG(a8) = e 'H Li’ where i=1 L. = S. n N (S.') S.‘ = W S.. G ’ . . l l 1 1 «Ph- J _ I 000 Let x 6 Lk — Sk n NG(Sk ) and let y 6 T1 T1' l— — 00 Then [y,x] E Sk n Sk — 1 and so x 6 NK — CSk(Tl T£)' Conversely, let w 6 CSK(Tl°-T£) = Nk' Then w E Sk and by our choice of T1’ ."T£’ w e NG(Sk') - since Sk normalizes all members of :3‘\ {T1,---,TL}, D The order of a system normalizer can also be computed using theorem 1.9 : [G : NG(.3)] = the number of distinct Sylow bases of G. Since the number of Sylow bases for an N-group G is the number n of complete sets of Sylow F(G) subgroups of G, we see n = w [G :NG(Sk)] and k=1 ING(v8)I = |G|/n. Although we gave an example to show that G need not be an N-group whenever Q/§(G) is an N-group, we can make some progress in this direction. .Erflnnsition_2119 _ a b G . If IGI — p q and /i(G) is an N-group, then G is an N-group. #1 3225212: ‘We may assume that 9/§(G) has normal Sylow p-subgroup G §(G) _ 3 p /,(G). Let W—NG(Gp). If w G, then Gpa G and we are done. Otherwise W lies in some maximal sub- Gpi(G) G PI group S of G. Let x 6 G\(S. Since /§(G)‘¢ ’/§(G)’ ,11 G XMG) G 6(G) G MG) S we have p /i(G) =< /t(G)>x z p /i(G)$ /i(G)° Hence pr.3 pr§(G).3 S and there exists 5 6 S with I x s . xs”1 Gp = Gp . Since Gp = Gp, xs 6 W = NG(Gp) and (xs-l)s = x 6 S - a contradiction. G .Eropnsition_2120 If 9/§(G) is an N-group and all maximal (prOper) subgroups of G are N-groups, then G is an N-group. .Eroof: Let Gp 6 Sy113(G) and Gq 6 Syl<1(G) for p t q. G unG) Since ‘/§(G) is an N-group, we may assume /§(G) G MG) ' p : : normalizes /i(G)° Let W NG(Gp). If W G, then G 4 G and we are done. Otherwise W lies in some maximal P G §(G) subgroup S of G. Since q /§(G) normalizes G 6(G) G XMG) G MG) p /6(G)’Wesee p /t.3 S, either Gp q normalizes Gq or Gq normalizes Gp. D we mention a result of Rose [11] for classes of groups closed under homomorphic images (i.e. Q-closed). a W Let (3 be any Q-closed class ofgroups. Suppose G is a finite solvable but non-nilpotent group in which every abnormal maximal subgroup is a e -group. Then G has a normal subgroup W of prime power order such that 9/W is a B -group. Since N-groups form a Q-closed class, the theorem applies if we let G = class of all N-groups. Chapter III The idea of a Sylow system is basic to P. Hall's in- vestigation of solvable groups. In this section we consider condition (N) for Sylow systems of a group. If some Sylow system of G satisfies (N), we call G a strongly Sylow towered group (SSTG). The inheritance prOperties of SSTG'S are derived and we give a characterization of these groups. 'We begin with a formal definition. .Definition_3il Let G be a solvable group. If some Sylow system .8 of G satisfies (N), G is calledastronalLsxlow W (SSTG). It follows immediately that a SSTG is necessarily a GSTG. The next examples show that SSTG'S are unrelated to N-groups. Examples Let G = [C7]°C6, the Split extension of C7 by its automorphism group. Write Aut(C7) = C6 as the product of its Sylow subgroups, C2 x C3. G is a SSTG since 3 = {l,C7,C3,C2,C702,C7C3,0203,G} is a Sylow system of G which satisfies (N). By a previous remark, G is not an N-group. Example—2~ We will Show [PrOposition 3.H] that a SSTG is necessarily w. a STG. Therefore the example of an N-group which is not a STG, given in chapter II, is an N-group which is not a SSTG. E 'I' 3 2 If G is a SSTG, then every Sylow system of G satis- fies (N). Ifixxfl? This follows at once from the fact that all Sylow sys- tems of a group are conjugate [Theorem 1.6]. C] If G is a SSTG, then every subgroup H of G is a SSTG. 229.5212: Since G is solvable, the subgroup H is solvable. Let TI be any Sylow system of H. By theorem 1.7 there is a Sylow system J of G such that 'H = {H nsls 6 '8 }. Since :3 satisfies (N), ‘H satisfies (N). D E 'I' 3 2 If G is a SSTG, then G is a STG. ,Ezggf: (Induction on F(G)) If n(G).3 2, then G has a normal subgroup and clearly G is a STG. Suppose Tr(G) = n2 3 and let :3 be a Sylow system of G. Let T and' W be complementary (prOper) subgroups of G belonging to :8 - i.e. G=T~W and Tin W = 1. Since :3 satisfies (N), we may assume T normalizes W. Then W 4 G = T -W. #5 By induction, both T and W are STG's. Let ldTldw °°4Tk°'Tl=T mm Il4Wi4°- ” 4W -°W =w be Sylow towers of T and W. Then 14W< H 4W~W==W4TW4-° ~41?--TW=G 1 L l l k l is a Sylow tower of G and G is a STG. I] W Let G be a finite solvable group and o a homomor- phism of G onto G°. Then L is a Hall w-subgroup of GO if, and only if L = HO, for some Hall v-subgroup H of G. Email: Let V be any Hall n-subgroup of G. Then Va is a Hall w-subgroup of Go, since [G9 :Vo] [G :V] and |V°| ||V|. Since G is solvable, GO is solvable and theorem 1.1 implies Va and L are conjugate in GO. Therefore L = (V0)Xo = (Vx)°, some x 6 G. [3 .Bzapnsitinn_3lfi If G is a SSTG and o is a homomorphism of G onto G“, then G“ is a SSTG. EIQQ£= Let 23 be a Sylow system of G. Then 28° = {SOIIS 6.3} is, by the lemma, a Sylow system of G°. And So normali- zes To whenever S normalizes T (S,T e :8). Therefore 3 0 satisfies (N) and G0 is a SSTG. [3 46 In order to study direct products of SSTG's, we need to define similarity for SSTG's. .Definiiicn_3lz Let H and K be SSTG's with Sylow systems 34 = {Hnlw.sic(H)} and 7< = {KWIW.§10(K)]. Let Z and A be arbitrary disjoint sets of primes. If either (1) Hz normalizes HA and K2 normalizes KA or (ii) HA normalizes H2 and KA normalizes K2, we say H and K are,similaz_SSIGLs, It is easy to check [PrOpositions 3.3 and 3.6] that all subgroups and factor groups of a SSTG are similar SSTG's. And by adapting the argument used for the direct product of GSTG's [theorem 1.27] to the present situation, we have the following. E 'l' 3 8 HI x K is a SSTG if, and only if H and K are similar SSTG's. SSTG's can be characterized as follows. W G is a SSTG if, and only if G is a Split extension of a nilpotent group A by a nilpotent group B with (IA|,|B|) = 1 and either A or B a p—group. .EIQQ£= Let :3 be any Sylow system of G. Let S and T be any distinct non-normal Sylow subgroups of a? and let S' and T' be (reSpectively) the complements of S and T ‘43-; -. L+7 in .3 . Since .3 satisfies (N) and SéG, TflG, we must have S'.s G and T';s G. Furthermore, we may assume S normalizes T. Now 8.3 T'14 G implies [S,T] S T' n T = l and hence, any distinct non-normal Sylow subgroups of a? centralize one another. If we let A be the product of all the normal Sylow subgroups of G and B be the product of the non-normal Sylow subgroups of G in A , then G: [AJ-B and (IA|,|BI) = 1. We now show that either A or B is a p-group. Assume this is not the case and suppose w(A) = k.2 2, w(B) = £.2 2. Let Nl,'°',Nk denote the normal Sylow subgroups of G and let S °°°,S£ denote the non-normal l, Sylow subgroups of G belonging to :8 . Without loss of generality we may assume that Nl does not normalize 81' If some normal Sylow subgroup Ni (i,2 2) fails to norm- alize some Sj (j.2 2), consider the subgroups N183 and Nisl' Now lej and Nisl belong to 93 and (Ilej|,|NiSl|) = 1. Hence lej normalizes NiS1 or NiS1 normalizes lej' Then [Nl’SIJ'S N1 n N131 = l or [Ni’Sj]-S Ni n lej = l - a contradiction. Therefore we see Ni normalizes Sj whenever i,j 2 2. Consequently Nl must not normalize Sj for j = 2,°°°,£. In particu- lar, Nl does not normalize S2. Now suppose Ni normalizes S1 whenever i.2 2. Then Ni normalizes every non-normal Sylow subgroup Sj 1+8 k L and H = V Ni ' w S. is a nilpotent subgroup of G. i=2 j=l Therefore G = [N1] . H - which is a contradiction. Hence, we may assume that N2 fails to normalize 81. Consider the subgroups lel and N282 of G. lel and N282 belong to .3 and have relatively prime orders. I‘_‘ f ‘ Therefore lel normalizes N282 or N282 normalizes lel. Then [N152] 5 N1 n N282 = or if . [NZ’SlJ'S N2 0 N1 1 = l - which is a contradiction. This J establishes the sufficiency. Let G be the split extension of a p-group A by a nilpotent group B where (p,|BI) = 1. Let B -°-,Bn 1: denote the Sylow subgroups of B. Then the Sylow system of G generated by the Sylow basis {A,Bl,---,Bn} clearly satisfies (N) and G is a SSTG. Now let G be the Split extension of a nilpotent group A by a p-group B with (|A|,p) = 1. Let {Al,"',Ak} be the complete set of Sylow subgroups of A. Then the Sylow system of G generated by the Sylow basis {Al,'--,Ak,B} of G clearly satisfies (N) and G is a SSTG. D We now give an example of a supersolvable group which is not a SSTG. Let G be the Split extension of C3 x Cll by its automorphism group C2 x C10° Then 1 4034 C3 x C11 4 C2 (C3 x Cu) 4 G is a cyclic invariant series of G, and G is supersolvable. By the preceeding N9 theorem, G is not a SSTG. The group G = C3 wr 02 is an example of a SSTG which is not supersolvable. The following diagram Shows the relative positions of various classes of solvable groups. A line indicates the class at the lower end lies in the class at the upper end. All containments are prOper. solvable GJFG STG N-groups supersolvable SSTG nilpotent Chapter IV If G is a solvable group, the Fitting length of G is a measure of the nilpotency of G. We ask if there is a measure of the deviation of a GSTG from an N-group. This leads to an examination of invariant Series whose factor groups are N-groups. Such series will be called invariant N:Ssrieso A GSTG G has a unique descending invariant N-series, called the lower N-series for G. The length of the lower N-series, denoted by m(G), is the minimal length of an invariant N-series of G. A GSTG G also has a unique ascending invariant N-series, called the upper N-series for G. The length of this ascending series is denoted by e(G). The nilpotent length of G is greater than or equal to e(G). An invariant N-series l 4 L14 . - . 4 Lk = G with Li+1 G /L a maximal normal N-subgroup of /L i i (0.3 i.$ k-l) is called a.maxima1_inyaniant_n;sezies of G. If k is the length of a maximal invariant N-series of G, then m(G).: k.$ e(G). We show by example that k can be less than e(G), but it is not known if k must equal m(G). 'We begin with the following observation. Ifimmm_&il Let H and K be (N)-similar GSTG's. If H and K 51 are N-groups, they are Similar N-groups. £2991: Let p and q be distinct primes. Since H and K are (N)-similar GSTG's, we may assume that some Sylow p-subgroup of H normalizes some Sylow q-subgroup of H and, some Sylow p-subgroup of K normalizes some Sylow q-subgroup of K. If H and K are N-groups, pr0position 2.3 then implies that every Sylow p-Subgroup of H normalizes every Sylow q-subgroup of H and, every Sylow p-subgroup of K normalizes every Sylow q-subgroup of K. [3 Our study of ascending invariant N-series of a GSTG is motivated by Baer's work [1] on supersolubly immersed subgroups. D E' 'I’ 2 2 A subgroup H of a group G is N-embedded in G if for every pair of distinct primes (p,q), either Gq g NG(Hp) or Hp g NG(Gq) for all Hp e Sylp(H), Gq € Syl<1(G). A normal p-Subgroup of a group is an example of an N-embedded subgroup. An N—embedded subgroup need not be normal however. For example, a Sylow 2-subgroup of S3 is an N-embedded subgroup which is not a normal subgroup. E 'I' 3 3 If H is N-embedded in G, then H is an N-group. 1" .. :_ 52 moi: Let HD and Hq be arbitrary Sylow subgroups of H corresponding to distinct primes p and q. Let G be q a Sylow q-subgroup of G containing Hq. Then Gq.S NG(Hp) or Hp.$ NG(Gq). Consequently we see Hq normalizes H or H normalizes G n H = H . [3 p P q q W t If H and K are N-embedded in G, then H n K is N- embedded in G. £10212: Let p and q be distinct primes and consider (H n K)p 6 Sy113(H n K) and Gq 6 Syl<1(G). By lemma 2.M (H n K)p = Hp n Kp for some Hp 6 Sy11)(H), Kp e Syl p (K). Now if HI) 5 Nqu) or Kp s Nqu), then (H n K) = H n K normalizes G . Otherwise G P P P q q normalizes both Hp and Kp. Then Gq normalizes H n K = H n K . P P ( )P C] WSW If K is N-embedded in G and H is a Hall sub- group of K, then H is N-embedded in G. Email: This follows at once Since Sylow subgroups of H are also Sylow subgroups of K. 53 An arbitrary subgroup of an N-embedded subgroup is not necessarily an N-embedded subgroup. For example, let G be the Split extension of A x B by C, where A = < aIa3 = 1 >, B = < bIb3 = l> and C=§Aut(AxB). Then AxB is _ N-embedded in G but neither A nor B is N-embedded in he G. W If H is N-embedded in G and K3 H, Kg G, then EL . K is N-embedded in G. hoof: Let p and q be distinct primes and Kp E Sy1}3(K), G E Sy1<1(G). Since K.$ H, Kp = K n HD for some q H S H.I H NG K=Kn pe ylp() 1‘ pg G( q), then p Hp normalizes Gq. Otherwise Gq $,ING(Hp) and, since K S.G, Gq normalizes K n Hp = Kp. D W If H is an N-embedded (normal) subgroup of G and o is a homomorphism of G onto G°, then H0 is an N-embedded (normal) subgroup of Go. 2292:: Let p i q and let P e Sylp(H"), Q e Squ(G°). By [13, p.13h], P’= (Hp)° and Q = (Gq)° for some Hp E Sylp(H), Gq G, either Hpg Nqu) or Gq_<. NG(Hp). Then P e Syl q(G). Since H is N-embedded in 51+ normalizes Q or Q normalizes P. C] In particular, every conjugate of an N-embedded sub- group is an N-embedded subgroup. All the previous results hold for arbitrary finite groups. From this point on, we restrict our attention to GSTG' s . “1 E 'I' 3 8 P Let G be a GSTG with N-embedded subgroup H. Then every Sylow p-subgroup of H normalizes every Sylow E‘s q-subgroup of G or every Sylow q-subgroup of G norma- lizes every Sylow p-subgroup of H, for p and q dis- tinct primes. .EIQQfl3 Let p ‘Jt q and consider Hp E Syl p(H), G E Syl q(G). q Since H is N-embedded in G, Hp qu is a subgroup of G, for all x 6 G. Since G is a GSTG, all the subgroups Hqux are (N)-similar. Therefore, either qu normalizes Hp or Hp normalizes qu, for all x E G. D W Let H be an N-embedded (normal) subgroup of G- and let K be an N-embedded normal subgroup of G. Then the product H - K is an N-embedded (normal) subgroup of G. moi: * Let p # q and consider (H-K)p e Sy113(HK) and Gq E Syl<1(G). 55 We first Show (H°K)p = HpXKpX for some Hp 6 Sy113(H), Kp E Syl p(K) and x E H-K. Let P be any Sylow p-subgroup of H. Then P.$ G for some Gp 6 Sle)(G) and P P = H n Gp. Since K e_G, K n Gp = Kp is a Sylow p-sub- group of K and Kp 9 GP. Then PoKp is a p-subgroup of H°K and IP-Kp| = |P| IKPI, the order of a Sylow subgroup D1 ___,. , a |P n K | P of HoK. Therefore (H-K)p = (P°Kp)x = PXKpX for some x e H°K. g n Since G is a GSTG, all subgroups of G are (N)- similar GSTG's and hence p‘:?> q or q':r> p, where .3 is a Sylow basis of any subgroup of G. Suppose first that p':?> q. Since H is N-embedded in G, every conjugate Hx of H is N-embedded in G. Then H quv is a subgroup of G and consequently pr normalizes Gq (x 6 G). Likewise prGq is a subgroup of G and pr normalizes G (x 6 G). Then q prpr = (HK)p normalizes Gq' . x x — . G Now suppose q 18> p Slnce Hp Gq and Kp q are subgroups of G, Gq normalizes both pr and pr Cx e G). Then Gq normalizes prpr = (HK)p and the theorem is proved. U .Gcrollanx_&ilQ The product of all normal N-embedded subgroups of G is a.characteristic N-embedded subgroup of G. We denote 56 this maximal normal N-embedded subgroup of G by E(G). QQIQllaI¥_&ill The characteristic subgroup E(G) of G is the inter- section of all maximal N-embedded subgroups of G. 2mm: Let {MX}XEA be the set of maximal N-embedded sub- Imus ‘1 groups of G. By theorem h.h the intersection 0 M1 REA is an N-embedded subgroup of G. And, by theorem h.7, I t p j. n MX is a characteristic subgroup of G. Consequently *‘h ASA n MX.S E(G). leA By the preceding theorem we know Mx-E(G) is an N-embedded subgroup of G, for each A GA. Since MA is a maximal N-embedded subgroup, E(G) S Ml for each X EA and E(G)_<. n Mk. [:1 le/\ Using the maximal normal N-embedded subgroup E(G) of G we can define a unique ascending invariant N-series of the GSTG G. Before proceeding to this, we examine the subgroup E(G) in more detail. 2 'I' 2 12 If G is a GSTG, then Fit(G) s E(G). Emflnfit The Sylow subgroups of Fit(G) are normal p-subgroups of G. Since a normal p-subgroup is N-embedded,the asser- tion is proved. C] 57 Since S3 is an N-group which is not nilpotent, Fit(S3).i S3 = E(S3). Therefore we cannot hOpe for equality in prOposition 4.12. TL3:T h 13 Let K be a normal N-subgroup of G. If H is a E‘ normal N-embedded subgroup of G, then the product H °K 1‘1 is an N-group. .EIQQf= Let p # q and consider (HK)p E Sylp(HK) and (HK)q E Syl<1(HK). Let (HK)p.$ Gp and (HK)q.S Gq where Gp 6 Sle)(G), G E Syl<1(G). Since H and K q are normal in G, we have the following relations: H n G = = H p Hp e Syl}3(H), H n Gq Hq e Syl<1( ), K n G = K S K K n G = K s l K d p p E yl.p( ), q q E y <1( ) an HK = HK n G = H K HK = HK G = H K . ()p p pp’()q nc1 qq By lemma h.l we may assume that p':;> q for a Sylow basis .8 of any N-subgroup of G. Since H is N- embedded in G, Hqu is an N-subgroup of G and hence Hp 5 NG(Gq). Since HKfl G, this implies that Hp norma- 1izes HK n Gq = (HK)q. Hqu is also an N-subgroup of G and so Gp.$ HG(Hq). Therefore Kp = K n Gp normalizes Hq. Since K is an N-subgroup of G, Kp also normalizes ' K = . Kq and hence Kp normalizes the product Hq q (HK)q Therefore both Hp and Kp normalize (HK)q and the assertion follows. E 58 The following example Shows we cannot drOp the normality of the subgroup K in theorem H.13. G = [C7]°C6 is the product of a normal N-embedded subgroup C7 and a (non- normal) cyclic subgroup C6, but G is not an N-group. W A maximal normal N-subgroup of G contains every normal N-embedded subgroup of G. In particular, the subgroup E(G) lies in every maximal normal N-subgroup of G. The next example shows that E(G) is not necessarily the intersection of all maximal normal N-subgroups of G. Let S3 = 2, 05 ‘normalizes M2. Consequently td 6 M2 and ttd = (ba,bba,b,l,l) e M ‘1 is a 3-element and ttd a _ 2. But bb — a is therefore not a 2-element - 60 which is nonsense. E 'I' 3 15 If H and K are (N)-similar GSTG's, then E(H x K) = E(H) x E(K). Roof: Since E(H) and E(K) are normal N-embedded subgroups h of H x K, E(H) x E(K) S E(H x K). “1 Let E1 = {h e H|hk e E(H x K) for some k E K} and g _ E2 = [k E thk. E E(H x K) for some h E H}. By the :3 isomorphism theorems El 2 El/Ean g Elg/K = E(Hx K)°K/K. :m Since E(HXK)‘K/K is a normal N-embedded subgroup of H)(K/K, E1 is a normal N-embedded subgroup of H and so El,$ E(H). Similarly E2 3 E(K). Therefore, E(HxK) s Elx E2 g .S E(H) x E(K). U Baer [1] has shown that a normal supersolvably immersed subgroup K of a group G satisfies the following prOperties: (1) If K g S _<. G and S/K is supersolvable, then s is supersolvable. (2) The elements of G induce a supersolvable group of automorphisms on K. The corresponding prOpertieS do not hold for normal N-embedded subgroups of a group. NZ (1) Let -G = [C7]°C6. E(G) = C7 and 9/0 C6 but 7 G .is not an N-group. (2) Let G = C5 wr 8% and let G; be the Sylow 5-subgroup 61 of G. Since SM is not an N—group, the automorphism group induced on G; by the elements of G is not an N-group. Consequently we expect the ascending invariant N-series of a GSTG which arises in connection with the subgroup E(G) to have some Shortcomings. D E' 'l' 3 15 If G is a GSTG, the.nPPfiZ_N:§&£ififi_Q£__fi is the invariant series of G defined inductively by E0 = l, E 1+1/H. = E(Q/E.)° The length of this series, denoted l l by e(G), is the number of distinct nontrivial terms in the series. Since the Fitting subgroup of G lies in E(G), the next result is clear. 2 'I' 2 12 If G is a GSTG, then e(G) S Z(G), the nilpotent length of G. ‘We now compare the upper N-series of a GSTG G with the maximal invariant N-series of G. W Let t be the length of a maximal invariant N-series of G. Then e(G).2 t. 62 M: Let 1 = Moro MlM11> .. >M£=l be the lower N- Series for G. Then, by PrOposition h.23, M M K M K G _ O b l D ooob‘ __K /K— /K— /K— -— /K-/K is the lower N-series of Q/K. C] 67 The next result shows that the lower N-series of G is an invariant N-series of G having minimal length. .Theorem_&l2fi Let G be a GSTG. Then any invariant N-series of G has at least m(G) distinct nontrivial terms. £19912: Let G=LO|> le -~>Lk=l be any invariant N-series of G and let G = MO > Mlb -- > MI. = 1 be the lower N-series of G. Since 9/L is an N-group, Ml.$ Ll' Next, 1 M1 ~ M1L2 Ll /fi 0L = /L .S /L , which is an N-group. Hence 1 2 2 2 M1 /MlflL2 is an N-group and M2.g Ml n L2.$ L2. Proceedlng M. ~ L. inductively, 1/M 0L c 1/L implies that 1 1+1 1+1 Mi+l'$ Mi n Li+1'S Li+l' It follows that k 2.1 = m(G). C] The preceding argument has established the following relation between the terms of the lower N-series and the terms of the upper N-series. CQrQJJary 3.29 Let l=EO<3 E14 °° 4En=G be the upper N- series of G and let G=MO> M19 .. >M£=l be 68 the lower N-series of G. Then Lk.$ En-k’ 0,3 k.$ L. We have Shown that the length t of any maximal invariant N-series of a GSTG G satisfies the relation m(G)_$ t_$ e(G). An example was given to Show that t need not be e(G). However, it is not known if t must equal m(G). There is a refinement theorem for N-series which should be mentioned, although we have not been able to use it. W Let 1=HO n 1,.. The number of elements in G Set of prime divisors of |G| The number of prime divisors of |G| The Fitting (nilpotent) length of G Defined on page 61 . r "V"! U.”- III. ,ch32: Z(G) Z.(G) HmCG) F(G) 71 Defined on page 66 Index of H in G Factor group X.1 G x -l x a x x.1 y-1 x y Subgroup generated by all [h,k]; h e H, k E K Center of G Hypercenter of G Hypercommutator of G Fitting subgroup of G (also denoted Fit(G)) Frattini subgroup of G Derived group of G Defined on page 56 Defined on page 6% Automorphism group of G Symmetric group of degree n Alternating group of degree n Set of all Sylow subgroups of G Set of all Sylow p-subgroups of G. Split extension of H by K Wreath product of H by K 72. BIBLIOGRAPHY l. R. Baer, Supersoluble immersion, Canadian Journal of Mathematics, 11(1959), pp. 353—369. 2. ,_______, Sylowturmgrup en II, Mathematische Zeitschrift, 92(1966), pp. 25 -268. 3. w. Gaschfitz, Zur theorie der endlichen aurlosbaren Gruppen, Mathematische Zeitschrift, 80(1963), pp. 300-305. #. P. Hall, A note on soluble groups, Journal of the London Mathematical Society, 3(1928), PP. 98-105. 5- __+_____, A characteristic pr0perty of soluble groups, Journal of the London Mathematical Society, 12(1937). pp. 198-200. 6. ._______, On the Sylow systems of a soluble group, Proceedings of the London Mathematical Society, h3(l937), pp» 316-323. 7. _______, On the system normalizers of a soluble group, Proceedings of the London Mathematical Society, Lt3(1937), pp. 507-528. 8. ._______, The construction of soluble groups, Journal fur die reine und angewandte Mathematik, 182 (19h0), pp. 206-21%. 9. B. Huppert, Zur Sylowstruktur auflSsbarer Gruppen, Arkiv der Mathematik, 12(1961), pp. 161-169. 10. , Zur Sylowstruktur auflgsbarer Gruppen II, Arkiv der Mathematik, 15(196H), pp. 251-257. 11. J. 8. Rose, Finite groups with prescribed Sylow tower subgroups, Proceedings of the London Mathematical Society, 16(1966), pp. 577—589. 12. A. E. Spencer, Maximal chains in solvable groups, Ph.D. thesis (1967). _ 13. ‘W. R. Scott, Group theory, (Englewood Cliffs, New Jersey: Prentice Hall, 19 H). 1%. D. R. Taunt, On A-groups, Proceedings of the Cambridge PhilOSOphical Society, h5(l9h9), pp. 2h-h2. WIN 3071 0564 .1111 llHlIHlm