142 960 __THS A STUDY OF THE AUTOMORPHISMS AND CHAINS OF A FENITE GROUP Thesis {or ”39 Degree 0{ Din D. MICHIGAN STATE UNIVERSITY Albert D. Polimem‘ 1965 «11111le mm m nmliiim { ' 3 129399 C" 1143.960 M; This is to certify that the thesis entitled "A Study of the Automorphisms and chains of a Finite group” presented by Albert D. Polimeni has been accepted towards fulfillment of the requirements for Ph.D. degree in Mathematics geéézév U Major pro ssor Date May 18. 1965 0-169 [M LIBRARY ichigan Stat University L13- 51.... e PLACE IN RETURN BOX to ram ave this checkout from your record. TO AVOID FINES return on or before date due. DATE DUE DATE DUE DATE DUE m _ ~;‘ .1/ 'd-QIIY h MSU Is An Aflirmativo ActiorVEquaI Opportunity Institution ammo ABSTRACT A STUDY OF THE AUTOMORPHISMS AND CHAINS OF A FINITE GROUP by Albert D Polimeni One of the main objects cf study in finite group thecr' is the grcup of automorphisms of a finite group. If one can say something about the automorphism group of a given finite group then conceivably he can say something about the group itself. he interest here is on how the automorphisms act 3 the chains of a group G, Given a chain s of subnormal subgroups of G, say 5: G = GO _ Gl _ . . , ; Gt—i 4 Gt = (l), we define recursively, the following: _ 9 _ 1 A _ 30(8) - +8- ~(GJI Gi - Gi for i - 0,1, ..t, tr, 8 (s) = {92 S {s} I 6; = 1 for l < k < t‘ k k—l‘ ‘ 'G. } — — K—l Gk t should be noted that Sfts) is the so—cailed stabiL;'y subgroup of s and has been shown by P. Hall to be nil— potent. Naturally one would like to know something abcu: the groups Skis) for a given subnormal chain s“ We see that 80(5) + 81(5) ; ... % St(s) and call this chain the stability series of s. In Chapters I and III we prove the ALBERT D. POLIMENI following theorems concerning the stability series: Sk(s)g So(s) for k= 0, l, ..., t. If 31 and 82 are subnormal chains of G such that $16 = 82 l for some as A(G), then 6 Sk(sl) e = Sk(s2). If s is a chief series of G, then 80(5) 1 if and only if lGI a 2. If s is a chief series of G, then St(s) 1 if and only if IGI = pl ... pt where pl, ..., pt are distinct primes. If s is a chief series of G then G is nilpotent if and only if IG :_St(s). If G is an A-group and s is a chief series of G, then 80(3) 8 St(s) if and only if G is a 2-group. If 30(8) is abelian for some normal series 5 of G, then G is nilpotent of class 1 or 2. The study of stability series becomes more significant under certain circumstances if we restrict ourselves to special classes of subnormal chains, for example chief series. If we let D denote the class of all chief series G of G and let Ak = QDG Sk(s), k = o, 1, ..., t , then we see that A 0 {ea A(G)| H9 = H for all H 3G} and :p u 1 {ea Aol dG/M =1 for all maximal normal subgroups M of G} ALBERT D. POLIMENI Similar statements can be made about the remaining Ak but we concentrate our attention on Just A0 and Al. To this end we prove the following theorems in Chapter III: If G is a p—group of class 2, g s G, and 6 8 A0, then a induces power automorphisms on Z1 and G/z and g6 = 1 gm zg for some postive integer m and some element 28 of 21. If G is nilpotent then Al ={95AOI dG = l}. W If G is nilpotent and all maximal subgroups of G are abelian then all elements of A1 are central and G is of class 1 or 2. 1'1 If [G] = p , n > 1, and |z = p, then all.elements of Al 1 l are central if and only if ¢(G) = 21 (i.e., G is an extra Special p—group). If {G} = pn, n > 1, IZlI = p, and all elements of A1 are central, then Al is an elementary abelian p—group. If IGI a pn, n > 1, Ile = p, all elements of A1 are central and Be 5 {x eGl x6 = x}, 62A tnen l’ A = I if and only if zl_< Z(Be) fer all 65 A l G 1' In the next situation if we let C denote the class G of all composition series of G and let B0 = é??? 80(5), 8 G then B0 = {esA(G)|He - H for all subnormal subgroups H of G}. A(G) ’ B Letting A (G) = we can;look at. EKG) as a permutation group on the class CG. We denote this group by ALBERT D. POLIMENI (A(G),C G). In Chapter II we prove the following theorems: If (A(G), CG) is transitive and 2IIGI, then: 1. The lower nilpotent series of G coincides with the then: derived series G 1 G' 1 .. :.G(r'l) ;.G(r) = (l),of G; Gun Each term of the derived series for G is com- plemented in G by the relative system normalizers of G(k-l) in G; There exist r subgroups H ..., Hr in G such that: 1’ a. Each H1 is either a cyclic pi-group or an elementary abelian pi-group for some prime p1; (k) _. 8 ' be G “ Hr Hr—l coo Hk+l for k 0,1, 000, r-1, 0. HiHJ = HJHi for all 1, J, and HlnnJ = (1) 11‘ 1+1; mm = (l); ¢(G) = (1) if r > 2; Gm 51111) is elementary abelian for k = 2, 3, ., r-l. If (K(G), CG) = (l), 2 I IGI and G is non-abelian, G is supersolvable; G is an A-group of lower nilpotent length 2; G'Z is the Fitting subgroup of G; a - G'K, G'nK = (1) and x =- NGCK) - com; G/G'Z is a generalized elementary abelian group; Either G has an abelian direct factor and this factor is a Hall subgroup of G, or Z 1 0(K). ALBERT D. POLIMENI It should be mentioned that in the process of proving the first of the above two theorems one is lead to consider groups all of whose subnormal subgroups are normal. Such groups have been studied by W. Gaschutz and are called tegroups. Some of the properties proved here have been proven by Professor Gaschutz but since the methods used are different we install them for the sake of completeness. A STUDY OF THE AUTOMORPHISMS AND CHAINS OF A FINITE GROUP By be Albert Di Polimeni A THESIS Submitted to Michigan State University in partial fulfillment of the requirements for the degree of DOCTOR OF PHILOSOPHY Department of Mathematics 1965 PREFACE The author is deeply indebted to his major professor, Dr. J. E. Adney for his patience and helpful guidance in the preparation of this thesis. The many discussions we had together during the research was of an immeasurable value to me. A special vote of thanks should go to Professors T. Yen and w. E. Deskins both of whom have con- tributed helpful suggestions. 11 TABLE OF CONTENTS Page INTRODUCTION . . . . . . . . . . . . . . 1 Chapter V I.. Basic Definitions and Properties . . . . . 3 II. The Group (A (G), CG ) . . . . . . . . . 11 III. Basic Properties of the Stability Series and the Groups A0 and A1 . . . . 13 APPENDIX . . . . . . . . . . . . . . . 5“ INDEX OF NOTATION. . . . . . . . . . . . . 60 BIBLIOGRAPHY . . . . '. . . . . . . . . . 62 iii INTRODUCTION In Chapter I the stability series of a chain 3 of subgroups of a finite group is defined. The motivation for this defintion comes from the stability subgroup of such a chain. This well known group has been studied by P. Hall [8] and shown to be nilpotent for an arbitrary chain of subgroups which terminates in the identity. The stability subgroup turns out to be the last term of the stability series of a given chain. Many natural questions arise concerning the stability series and we attempt to answer some of them here. However, since the ideas are very broad many very interesting questions must go unanswered. One of the main topics of investigation resulting from the definition of stability series is how the auto- morphisms of a finite group act on its chains of subgroups. In particular we investigate how the automorphisms of a finite solvable group G act on its composition series. The set B0 of all automorphisms of G which leave fixed every composition series of G formSa,norma1 subgroup of the auto- morphism group AKG) of G. We then see that the factor group EKG) = Aéglf- is a permutation group acting on the set of all composition series of G. In Chapter II we deter- mine what conditions are imposed on G if we assume first 1 that this permutation group is transitive and second that it is trivial. Similar discussions are made for the cases where G is abelian, nilpotent, and finally supersolvable. Upon completion of the above discussion we focus our attention on the chief series of a finite solvable group G and the following sets: 6 tx> ll {eeA(G)l s s for all chief series s of G}, :I> ll {eerl 1 for all maximal normal elG/M = subgroups M of G}. Both A0 and A are normal subgroups of A(G) and A0 consists l of all automorphisms of G which leave invariant every normal subgroup of G. In case G is nilpotent Al consists of all automorphisms of G which leave G/ elementwise fixed. ¢(G) In Chapter III we prove some theorems telling what the elements of A0 and A must be like for certain types of 1 groups. Finally, in Chapter III we prove some basic theorems relating the stability series of a given chain to the group itself. I The reader is asked to consult the index of notation for identification of symbolic notations of groups and relations. Further we have an appendix consisting of examples and theorems which are referred to throughout the thesis. CHAPTER I BASIC DEFINITIONS AND PROPERTIES In this chapter I would like to inroduce the main definitions and discuss some of the basic properties and problems associated with these definitions. Once and for all, the groups referred to are assumed finite. Definition 1.1: Let s ‘.G = G0: G1 ;_....: Gt-l 1 Gt-: (1) be a chain of subgroups for an arbitrary group G. We define so(s) = {BeA(G)|Gie = G1 for i = o, 1, ..., t}. ‘and'recursively Sk(s) {eeSk_l(s)| GIGk_l =‘1 } Gk for l i k i t. ‘It is easy to see that each Sk(s) is a subgroup of A(G) and that eeSk(s) if and only if 6280(3) and e induces the identity automorphism on the factor groups Gi/Gi+l, i = o, l, ...; k-l. From this it follows that if k i 2 , then 82(5) 1 Sk(s)‘and we get a descending chain of sub- groups of A(G): So(s) 1 81(3) 1 ... 1 St(S)° It should be noted that St(s) is the well-known stability subgroup of s and has been shown by P. Hall [8] to be nil- potent. Definition 1.2: The series of subgroups 80(3) 1 Sl(s) > ... 1 St(s) is called the stability series of s for G. In most discussions we will be considering various stability series for a given group G and shall refer to the stability series s for a given chain s, G being understood. Theorem 1.1: Let G be an arbitrary group and s : G = GO 1 G 1 G > > G 1 Gt = (l) *1 2-°°'— t-l be an arbitrary chain of subgroups of G. Then Sk(s).S‘SO(s) for k = l, 2, ... , t. roof: Let as 30(3) and 6e Sk(s), then for l i i i k and x GR 6 G i-l/Gz we have ‘ d-lea I afll 6a aul a (xGl) = (x G2) = \x G2) ==x G2 0-1 Gi-l —1 since x G1 2 —§—- and as Sk(s). Thus a Gas Sk(s) and 2 Sk(s) s So(s). In most instances we will take 5 to be a composition series or a chief series for G. It is to be noted that if s is a normal series for G then the group IG of inner auto- morphisms of G lies in 80(5). In particular, if G is a solvable group and s is a chief series for G, then IG 1 81(5). This follows from the fact that every maximal normal subgroup of a solvable group is of index a prime and IG 1 80(8). At this point I think it might be worthwhile to illustrate these definitions with an example. Example 1. G = > a3 = b2 = c2 = l, b_1 a b = a2, b_lcb = c, a_lca = c. 3 G _ SO Z(G) = , and IG - m implies IIGI- 6 AW) = <6, 0:, 8> 2 2 e: a >a a: a ———+ a. B: a ———+ a b -~+ab b "-—+ b b -———+ abc c ——+c c-~——+ c c ———+ c lol=3.-.o=l led-’2, a—I |B|=2 We consider all the compOSLtion series for G: G > (ac) > fic) > (1), 31 22 C} > (ac) :> > (l), 03 . G > (a) > (1), SA : G >>(a) e (1), then: 80(81) = 81(31) = <6,a,8> and 32(81) = 33(81) =(e), 80(52) = ul(82) = 52(s2) =\6,a,8> and 03(52) =<0), (e). 80(su) = 31(Su) = 82(84) = and S3(su) = (9), Further examples may be found in the Appendix and will be V II 82(83) =(e,s> and 83(83) referred to in what follows. ,) Definition 1.3: Let slzG = GO 1 G1 1--- “Gt“1 1 Gt = (l) and 32: G = H > H > ---v H 1 H = (1) be two chains of subgroups for G. We say that s1 and 52 are abstractly the same iff r = t and Hi & Gi for i = o, l,--o, t. We write s‘ l a s2 to denote this. If s1 and s2 are not abstractly the same then we write 51 i 52. . In the context of the above definition if 51 N 32 and there exists an automorphism e of G such that G19 = Hi for i = O, l,..., t, then we write 5: 52. We see that the relation % is an equivalence relation on the set of all chains of subgroups of G which terminate in (1). In particu— lar if we let CG’ NG’ and DG denote the classes of all com- position series, normal series, and chief series of G respectively then N is an equivalence relation on CG’ NG and DG' In Example 1 it is seen that 81 f 32 and 53 m s“. More— over we see that Sk(s3) = Sk(su) for k = O, l, 2, 3. Hence one is naturally led to ask about the stability series of a given group G for two chains 5 and s when s n 52. Example A l 2 l of the Appendix shows however that one cannot hOpe to prove that sl m 52 example also shows that it is possible for two chains not to implies Sk(sl) 5 Sk(s2) for all k. The sane be abstractly the same and yet they can have identical sta- bility series. However, we do get some results in this direction. Theorem 1.2: Let sl:G = GO 1 G1 1 1 Gt_l 1 Gt = (l) and s2zG = HO 1 H1 1 ... 1 Ht—l 1 Ht = (1) be two chains of subgroups for G such that $16 = s2 for some eeA(G). Then -1 _ - ... e Sk(sl)e - Sk(s2) for k — 0,1, , t. Proof: Let-a be any element of Sk(sl) and consider 9‘106. If xHg is any element of H III-‘1- '—l .1 9 a6 _ 6 -a6 _ 6 6 _ 6 (le) —.(x G£) — (x G2) - ng since x G2 5 -1 -l G£_l/G£. Hence a aeesk(s2) and we have thate Sk(sl) 6 i Sk(s2). In a similar manner one can show thate Sk(s2) £1l/Hl,‘l i 2: k, then , - —1 i Sk(sl), i.e, Sk(s2) < 6 Sk(sl)e' Thus it follows that — 6—1 -1 1 , e 'Sk(sl)e - Sk(s2). So one of the main problems is that of finding out when there exist automorphisms e for which 319 = 52 when slm 5 But this depends a great deal upon being able to con- 2. struct very special types of automorphisms of G and is essentially an extension problem. Since relatively little is known about such extensions and since this is not the main problem of this thesis I will not attempt to clarify this situation Completely. However in Chapter III I shall obtain some basic prOperties connecting stability series to the underlying group. I would like now to consider how the automorphisms of a given group G act on its chains which terminate in (1). Considering arbitrary chains may be difficult since the sub- groups in the chain need not be subnormal. In order to avoid this difficulty we further restrict ourselves to the (.0 class C, o? ull composition series of G. Let 31 5 CG’ say u = GO 1 G 1 ... 1 u _ 1 G = (l), H- H and let A(G) ){J t2SC(sl)(J ... Lj¢rSO(sl) be a ¢l o l coset decomposition of‘A(G) relative to 80(51) with ¢1 = l. q. ConSider the composition series sii‘ : n 1 G31 1 ... 1 i. s l i".1_'fl‘9 J“. _¢i Gt-l 1 Gt - II) for each 1, 1 i l i ,, and let si — s1 . -l rm ,e. ‘ ,5 ‘ S s, = ._ “ Then by Iiecrem 1.2 we hax that k( l) ¢i qk(bl) ¢i _. 1 _ ... Q o o ‘ o o ' ‘I and {ti -k(sl) till 1, ..., I} is the conJugate class 0: I! S gsj) in A(G) for k 0, J, ..., t. Moreover if we let 11 C, = is I i r l, ..., r} then the elements of C1 are all distin;t and are permuted transitively by the automorphisms o‘ C. he can repeat the above process until we have ex- J&U3t€d all the elements of CG and we will get a set of mutually disjoint subsets Cl, ..., an of CG such that C3 ‘3: : CG‘ 1‘1 J. r It should be noted that the subgroup (31 80(si) is a 0(3) U2 1 normal subgroup of A(&, so that the subgroup BO : f \ _ secp is also normal in A(G). Let A(G) = A(G)/BO. ’ In view of the above discussion it makes sense to consider the group A(G) as a permutation group acting on the elements of CG' If we denote this permutation group by (K(G), CG)’ then the subsets C CD will be the orbits l’ ... , o1'(ZKG), CG). In Chapter II we will investigate what happens to a solvable group G if we assume first that (K(G), is transitive and second that it is trivial. The reasons for requiring G to be solVable will become clear in Chapter II. Lgxt I would like to consider the subgroup Ar = f"; ecDg 80(3) of A(G). It is easy to see that Ad 3_ Ai?): in \r fact we have the following: Theorem 1.3: A0 = {eeA(G)| H8 = H for all H S G}. Proof; 1. Let Ger and let H 3 G. Since H :16 there exists a chief series 8 in which H appears as one of the term (11 Since a = s it fc110ws that H8 = H. 2. Suppose eeA(Gl and H6 = H for all H j P. Let s be any chief series for u. then since each term of s is‘a normal subgroup o: C we rust have that s = s. Thus FeAC. It is clear from the definition of A0 that IG 1 Ad. If we let A1 = 4:3 31(5) and assume G is solvable then one G " can show that Al 3 A(G) and IG i A]. should choose an arbitrary element eeA{G) and an arbitrary element aeAl and show that enlaeeAl. IG 1 A1 follows from the fact that every maximal normal subgroup of a solvable group To prove A, :3A(G) one is of index a prime. For if M is a maximal normal subgroup Of G, then G/M is abelian and hence centralized by every element of G. Using set inclusion arguments as in Theorem 1.5 one can show that A1 = {OeACI GIG/M =1, for all maximal normal subgroups M of G}. In Chapter III there will be some properties given roncerning this . 10 Definition 1.“: An automorphism eof G is called a -1X6 central automorphism iff x e Z(G) for all x e G. In examples A and B of the Appendix it is seen that all elements of A1 are central automorphisms and so in Chapter III we examine Al and determine conditions under which all of its elements are central automorphisms. CHAPTER II THE GROUP (A—(G), 0.) U As was mentioned earlier we wish to determine what restrictions are imposed on a group G if“we assume first that (EKG), CG) is transitive and second that it is trivial. Definition 2.1: Let G be an arbitrary group such that IGI = pl ... pr where pl, ..., pr are distinct primes. Let-< be an ordering of the set of all primes and assume pl.< p2‘< ... <§pr° Then the group G is called a §ylow tower group for the ordering -< iff G possesses a normal series G.= No.) Nl > N2 > ... > Nr-l > Nr = (1) such that O. 'Nk-l : Nkl = pk k for k = l, 2, ..., r. Lemma 2.1: Let G be an abelian group. Then (A(G),CG) is transitive iff G is an elementary abelian p—group or a cyclic p-group, for some prime p. 3319;: We prove the if part first. If G is a cyclic pwgroup then it is clear that (A(G), CG) is transitive since G has Just one composition series. Let G be an elementary abelian p—group, say G = ('x1, x2, ..., x5) , and let s1 and s, be two composition ser es for G. Any composition series C. for G can be obtained by selecting an apprOpriate basis for 11 12 G and then getting the terms of the series by deleting one basis element at a time: G = (xl,x2, ..., x5) > (x2, ..., x£> >...> (xr—l’ XE) > < xr> >° Thus, for the series 51 and 82 we can choose appropriate bases, say {Xl’ ..., Xr} and {y1, y2, ..., yr} for G such 31 and 32 are given as follows: 81: >' ...> > >(l): $2= 01,372. --.. yr.)> ..-> (yr—1J9) (yr) > x x be a direct decomposition of G in which the elements x1, x2, ..., xm are of maximal order in G. So m i r. We consider the two cases m < r and m = r separately. Case 1. l i m < r n Assume I x1 I = p for i = l, 2, ..., m-and Ixr I = pk. Thus 1 1 k < n. We can always get composition series for G 13 which begin as follows: G > (xl,x2, ..., x131, ..., Kr) > ...> (1) G»> ...> (1). Let G1 = I+II(Pn‘¢(9n))m-2J and the number of elements in G1 of order pn is given by ”2 = ¢(pn)[II (pn--¢(pn) )m'3]. Clearly n > n 1 so that Gl # H1. Thus this case cannot 2 arise. Case 2. m = r, n > 1, and r > 1. n fori = 1, 2, So G-=‘>> ...> (1.), 2 G > (xl,...,x§> >(xl,..é,xr_l,x: >> ...> (1). = p = Let H2 and G2 . By an argument similar to that in Case 1 we find that the number of elements in H of order pn exceeds 2 1“ the number of elements in G2of order pn. Hence it is impossible that H 5 G Hence we see that the only possi- 2 2' bilities are those stated in the conclusion of the lemma. Before proving the next lemma we need a definition. Definition 2.2: A group G is called a generalized n-l . , quaternion group iff G = (a,b> where a2 = bu 5'1, n 1 3, 2n-2 2 _ -1 a - b , and ab - b a . We denote G by Qn' Lemma 2.2: Let G be a nilpotent group. Then (A(G),CG) is transitive iff G is one of the following: 1. G is an elementary abelian p-group for some prime p; 2. G is a cyclic p—group for some prime p; 3. G is isomorphic to the quaternion group of order 8. 3392:: Assume G falls into one of the three categories listed. We already know by Lemma 2.1 that (A(G), CG) is transitive if G falls into category (1) or (2), so we need only consider the quaternion group Q . Q3 = (a,b> , a” = bu=1, a2 = b2, and ab = ba_l. 3 The automorphism group of Q is iso- 3 morphic to the symmetric group SA of degree A and has the following generators: a: a ——+ a3 B: a ——+ a Y: a ——+ ab e: a ——+ b b -——+ b b -—~+ a2b b -——+ a3 b -——+ a a = Ib, |a|=2 8=Ia, IS! = 2 |y| = 3 |¢| = 2 The composition series for Q are given by: 3 l:G>(a)>(a2>> (1), A2 : G > (b) > (a2) > (1), A3 : G > (ab)> (a2) > (1). S Moreover we have that <0: 8, “Y2¢> : <0: 89 87¢) : So(sl) a I 00‘82) 30(33) é:- _—:-— ._ , . AIQ3) - B 3 - ( K, ¢>and s: - s3, 3% = 32 Thus it o . follows that (5(Q3), CG ) is transitive. Conversely, suppose (A(G), CG) is transitive. Since G is nilpotent 'it fdllows from Theorem F of the Appendix that G is a~Sylow tower group for any ordering of the primes. Hence since (3(6), CG) is transitive it must follow that G is a p-group for some prime p. Since G is nilpotent any chief series for G must also be a composition series for G since all chief factors of G have rank 1. Hence if 31 is a chief series of G and s2 is a cOmposition series of G there exists an automorphism 9 of G such that slq 8 82. Thus we get that s is a chief series for G. So every compositiop 2 series for G is also a chief series for G and this implies that every subnormal subgroup of G is normal in G. By Theorem . C of the Appendix we get that every subgroup of G is sub- normal in G. Hence every subgroup of G is normal in G. Such groups are called Hamiltonian groups and have been completely characterized. In fact, according to Theorem I of the Appendix G = anA B where Qn is a generalized quaternion group, A is an abelian group of odd order and 16 B is an elementary abelian 2-group. Since G is a pvgroup it follows that either p is odd and G is abelian or p - 2 and G has form Qn x B. If p is odd then G being abelian implies that G is either cyclic or elementary abelian. If p = 2 then we claim that G i Q3. Given two subnormal sub— . groups H and K of G of the same order we know that H must appear in some composition series for G and likewise so must K. Moreover H and K must occupy the same relative position in the composition series in which they appear since IHI - IKI. Hence it follows that H O K since 2““1 _ u (1(0), CG) is transitive. Now Qn - (a, b) with a b - n-2 l, 32 . b2 and ab 8 ba-lo L813 B 8 (01,°°°,Or> , Ci . l for i - l, ---, r and consider Qn x B. We know that the n-l elements b, ab,"', a2 ‘1b are of order A, and B + (l) is of order A and is not cyclic. So 2n-2 implies (c1, a 2 n— (b) and (cl, a2) are two subnormal subgroups of Qn x B which are not isomorphic but are of the same order. Hence if G is of the form Qn x B then either B a (1) or Qn - (1). If Qn ' (l) and G s B then G falls into category (1) or category (2) of the conclusion. If B = (1) and G i Qn n— then suppose n>3. The element x =3r2 3 is of order A and so x” - bu - l, x2 2 and xb a bx-l. zn-u a b Thus (x,b) l Q3. n-A But a is of order 8 and hence (a ) and (x,b) are two subnormal subgroups of Qn of the same order which are not isomorphic. Hence it follows that the only possibility is n - 3. 17 Definition 2.3: A group G is called an A—group iff all of its Sylow subgroups are abelian. A—groups have been extensively studied by D. R. Taunt [11] and we will have occasion to utilize some of his results in what follows. Given two nilpotent normal subgroups of an arbitrary group G it is well known [10] that their product is a nil— potent normal subgroup of G. Hence G must possess a unique maximal nilpotent normal subgroup and this is called the Fitting subgroup of G. ’Theorem 2.1: If G is a supersolvable, non—nilpotent group and (A(G), CG) is transitive then G = Sp-Sq is a p—Sylow subgroup of G and S a p and q primes, where Sp q is a q-Sylow subgroup of G. Furthermore if q

’ permute the composition k series of gé?%¥l) transitively. Thus by Lemma 2.1 it follows that Ségél) is either a cyclic pk+l-group or an elementary abelian pk+l-group for some prime pk+l° Since G is a super- solvable group it follows from Theorem E of the Appendix that G is a Sylow tower group for the natural ordering of the primes. Hence G possesses a normal series G = H >H = (l) where IHi : H for i w 1 >"°> t-l t 1+1l = p1+1 - o, l, ..., t—l, and plH 0 that (A(G), CG) is transitive we see that IG and non-nilpotent, t>l. From Theorem D of the Appendix we have that G' is nilpotent. Moreover we know that the automorphisms of G, when restricted to G', permute the composition series of G' transitively. Hence, by Lemma 2.2 G' is either a cyclic p-group, an elementary abelian p—group, or iso— morphic to Q3, where p is a prime. Since pt IIG'I and pt>pl it follows that G' e Q Let IG'I = p*, so that G' 3. is either cyclic or elementary abelian and p is an odd prime. Also let IGzG'I = q*, then pq. If subgroup of G, Sq G' is cyclic and conditions l-6 of the conclusion of Theorem 2.1 are satisfied by G, then (3(G), CG) is transi- tive. 3333:: Since Sq and G' are abelian and Z(G) = 1, it follows that no element of Sq except 1 can centralize G'. Hence Sq is isomorphic to a subgroup of the automor— phism group of G'. Since G' is assumed cyclic and G' is a p—group A(G) must be cyclic. Hence Sq must be cyclic; since Sq is already elementary abelian it follows that it is cyclic of order q. Hence every composition series for G must have form G>G'>--->(l). Since G' is cyclic it 21 has just one composition series, hence G has Just one composi— tion series and so (K(G), CG) is transitive. Theorem 2.3. Let G = G'Sq with G' the p-Sylow suberUP of G, Sq a q—Sylow subgroup of G and G' elementary abelian. If Sq is cyclic and conditions l—6 of the conclusion of Theorem 2.1 hold, then (K(G), CG) is transitive. Proof: Since S is cyclic we must have IS q = q' ql As in the previous theorem every composition series for G must be of the form G > G'>. - ->(l). We know that (3(G'), CG.) is transitive by Lemma 2.1, thus if we can extend any automorphism of G' to an automorphism of G we will have shown that (A(G), CG) is transitive. As in the previous theorem we note that Sq is isomorphic to a subgroup of A(G'). Let yeS and consider the action of y on G' q under conjugation. Assume G' has rank r, say G' =‘(xi) x- - .x ( x5) . Then [xi] = p for i = l, 2, - - $, r. We know that every subgroup of G' is normal in G hence there exist integers :1, where l :igp. Let i>l. Now (xlxi)y = (xlxi)m for some integer m,ll m-il m—Ri xl ' = l = xi. , thus pim—ilio implies m = 21. Hence x. = xi 1 for i = l, 2, ---, r. Therefore the element y induces a power automorphism on G'. Now let a be any auto- morphism of G' and let g be any element of G. Since G = G'S7 22 a (I and Sq is cyclic, say Sq -(u) , g has form 3 8 x1 132 2... “r a — F “1 xr u . We define a mapping 9 on G by g '- (x1 ... 0:36 x u“ and claim that F is an automorphism of G. We r need only prove that 3 is a homomorphism of G since it is clearly one-to«one and onto. Let g1 and g2 be any two elements of G, say g1 - (x1 ---xr ) u“ and 52-’ 81 B r B F “1 (x1 o-oxr )u . _Then (3132) - [{xl ---xr B B 6 (x1 1 ...xr r) us] a We know that the element u‘ induces a power autbmorphism on us gr) u“ G‘ so there exists an integer 1 such that x - x‘1 for all Xe G'. Thus - a a B B 2 (8132)6 ' [(11 l°°'xr 1') x1 1 "‘X r) u 0+8] ll+8 1 a rr+8 z) ua+e a l_l+e 2 a +3 2 .1a3.[(x.oqxr 1°: .00 +8 _ [(x1°l)e a £8 .18 (xr P)O] [(x111)6 ...(xr r)0] “0+8 0 a 3 3 + - (x1 1 x, ">°- [ Kl> ~>Kt = (l) is a chain of normal subgroups of G then there exists a system normalizer NG of G and relative system normalizers NKl, NK2, - - -, NKt = G of K1, - ° ~, Kt respectively in G such that NG i NK 1 ° ° ° < NK '= G. l t Definition 2.6: Let G be a group. Let L1 be a sub— group of G which is minimal with respect to the prOperties Ll g.G and G/Ll is nilpotent and recursively, let Lk be a subgroup of Lkel which is minimal with respect to the Creperties Lk g.G and Lk/Lk—l is nilpotent. Then we call the chain G 1 L1 > - - -> Lk > - . - a lower nilpotent series of G. It is not hard to show that the groups L1, L2,: ° °, Lk’ ~ - . are unique with respect to the given prOperties so that one speaks about the lower nilpotent series of G. We are now in a position to state a theorem due to R. W. Carter [2] which will have a very significant bearing on Theorem 2.5. Theorem 2.“: If G is a solvable group and the lower nilpotent SGPiGS<1f G coincides with its derived series then each term G(k) of the derived series G = G(O) 1 G' °°' : G(r-l): G(P) = (l) is complemented in G by the relative system normalizers of G(k_l) in G. 25 We can now go on to consider the case that (K(G), CG) is transitive when G is solvable. It should be noted that the assumption of (K(G), CG) transitive is equivalent to the assumption that any two subnormal subgroups of G of the same order are isomorphic. In the proof of Theorem 2.1 it was shown that each factor G(K)/G(k+l) in the derived series for G was either a cyclic pk+l—group or an elementary abelian pk+l-group for some prime pk+1 and for k = 0,1,: ° ~,rel. The assumption of supersolvability was not used to proved this fact so that it still holds true if G is a solvable group. Moreover according to Theorem L of the Appendix it follows that no two consecutive factors of the series G‘: °'° i G(P) = (I) can be cyclic unless they collapse into one factor. Thus if all the derived factors are cyclic then the derived series for G must have form GiG': (1). In this case G is either cyclic or supersolvable. So we may as well assume that some derived factor is elementary abelian of rank greater than 1. G Lemma 2.3: If (I(G), CG) is transitive and 2 1 then any two consecutive derived factors must have relatively prime orders. Proof: Suppose |G(k): = p* and IG k+1z k+2| G = p* for some prime p. Then ET???) is a p-group and p is an odd prime. Moreover the automorphisms of G, when (k) restricced to ETEIE)’ permute the composition series of (k) (k) G -—-7 transitively so that by Lemma 2.2 :-; is abelian. Gang) 7G in) C) 26 (k+l) + G (k+2) then we have reached a contradic- Hence if G tion. So the lemma is proved. Lemma 2.4: Let H be a nilpotent normal subgroup of a group G and let G:G':° ' ':G(r'l) i G(r) = (I) be the derived series for G. If (K(G), CG) is transitive and 2*lGl, then H i G(r-l). E£22£= Since H‘g G and (I(G), CG) is transitive it follows that there exist atuomorphisms of G which permute the composition series of H transitively. Hence we have that (K(H), CH) is transitive. Therefore by Lemma 2.2 and the fact that 2X|H| we get that H is either a cyclic p—group or an elementary abelian p—group for some prime p. Hence from the transitivity of (K(G), CG) it follows that H i G(r-l). Corollary 2.1: Under the conditions of Lemma 2.4 (r-l) we have that G G (k+l) general Effil) is the Fitting subgroup of G/G . is the Fitting subgroup of G, and in Proof: Since every nilpotent normal subgroup of G lies in G(r-l) and G(r-l) is itself a p—group for some prime p it follows that G(P_l) G(k) G. The proof for _TKIl) is very much the same and will be G is the Fitting subgroup of omitted. Theorem 2.5: If (I(G), CG) is transitive and 2 1 [c] then the following are true: 27 l. The lower nilpotent series of G coincides with the derived series of G; 2. Each term G of the derived series for G is complemented in G by the relative system normalizers of G(k-l) in G; 3. There exist r subgroups H1, H2,- - -, Hr, r the length of the derived series for G, such that: a. Each H1 is either a cyclic pi—group or an elementary abelian pi-group for some prime pi; b. c = H H I: I'D-l. .Hk+l for k = O, l, . . ., P91; c. Hi HJ = HJ Hi for all 1,3 and HileJ = (1) if i + J; u. Z(G) = (l); 5- ¢(G) = (1) if P>2S Goo 6. ETEIT) is an elementary abelian group for k = 2, 3, °, r—l. Proof: Let L1 be the subgroup of G which is minimal With respect to the prOperties Ll'§_G and G/Ll is nilpotent. Clearly G' 3 L1. Moreover Ll is a characteristic subgroup Of G and so the automorphisms of G permute the composition series of G/Ll transitively. Thus by Lemma 2.2 G/Ll is an abelian p-group for some prime p. But this implies that L1 3 G', hence we get G' = Ll. We can continue this process until we finally get G(P-l) = L Hence the derived series r-l' of G coincides with the lower nilpotent series for G. a 28 By Theorem 2.“ it follows that each term G of the derived series for G is complemented by any relative system normalizer of G(k_l) in G. Let Kk be one such. In view of the discussion following Theorem 2.3 we can choose relative system normalizers K G(r«l) 1’ K23 ...’KI’-gl OfG: G'ans ..., reSpectively in G in such a way that K1G" >"' > G(P—l) > G(r) = (l) are cyclic and this is impossible. Thus G(r_l) is elementary abelian and ¢(G(r‘l)) = (1). Hence ¢(G) = (1). Finally let us consider G(k) G(k+l) where (r-l) > k >1. C(k) has a complement K in G, hence k 3 ' i ;o: .‘ . r , ism. . .‘.° .’ 3.5.5 sud-'1 ‘- -\ .uf'if. ' -,’) ‘-. 1,. .- ... a ' . . 1 (1‘1 ..." - 1?”: ”:32: \b‘gt', JO‘ 1 .i'.' ‘.'.I_‘.0i'#‘ . 1" Loy. '- 4 11- 39%.;1.‘ pjfipig c ., . . i-, '7‘ I . y'léa ’.."‘;I‘I. :. :"J' .e (DIHEIRZ... 1‘ng \I‘ 'l. ‘ ~ . ,1: ‘1’.” ..- L . 3: «L J ' t . .y _ g g , - ~.; 3*: 3.155,, . . . '- . WM, ‘1'“: "4...... ' . ‘ . . . 3O - \ G(k+l,K /G(k+l) k Using the same method of proof as in the preceding paragraph (k)/G(k+l) is a complement of Gkk)/G(k+l) in G/G(k+l). we can see that G must be elementary abelian. iG(r—l)l = k Corollary 2.2: If pr , then either \ '2 pr (prsl) OI" 2k-l lGll pr (pr-l) Proof: We know that G = G(r_l)Kr with G(r-l)r\K —l r—l (r-l) = (l) and we also know that G is the Fitting subgroup of G. Thus cG(c(r‘l)) = c(r‘l) and it follows that no element of Kr—l can centralize G; amua emfl . :11 Jada bé-':Q 13.110. an r.i..‘l:(£€! JI ” 3*; ”113 9'9; 3"} “5'3? r..'. (3‘? .xOY‘). .‘ . ”W I i" ‘ ' ;-; I . .'~‘ .' , '1 .I...1:IA’H9 3'. .LIL Ti" (3 '..|* I? ....._ a :1: '3 a !_ W“. .. 1:? - . . I. O 'J- — _ -. _ . “Er: C 96 n'rodss'! b=tf1ob 9V1“! . ' ‘ ‘ 1'3M'I‘113A01’A !- . r~19§dv¥%' ‘ ~ ‘ ‘ W? 'I!:- h'. “'3" ' .3:("'.. ‘3‘. L . l.‘ .1. ‘ - . .' ‘ J . ‘15-], "' - — o 9* It ,0- . . 9,. 3" .gQamA m1? ‘tc- '3 31:36“! 11.. 1‘31; a‘f‘u‘rgtl 3 .~ - . 1-1. 72w, " ‘E “3“)“. “(5.913133 {3 Was knit)“.x ~ of NP; ‘4'“. ~ “‘ - 45.; "'fi‘fi ma} ‘d SWJIG”; 51d? ‘:::-_5‘ ...- rwg: 9-: ;,p. "fix ‘ a 3"“ -3," H I? fifig 1‘ t ‘J-' Vii ' V. "-.l I -‘ H I 1.113“: ‘. 4" v‘ J“-"" -‘ ‘l 31.‘ . _-_. . 7‘ I ”I” at" '.~'.x-;;'U“,q‘t 33"...” ., _ — — 'qm4“w1w flgw~ ~ - ‘ I ‘ ' f. 11" 1. I 3 . 0 3 _ . ‘. . 3 _ ’3 .3 .LI __‘ 4 ‘ ’3', .9‘ l' '31. . 5'5 _ H" v .’ w ~ ..‘. ....“ {it ' ,, . _ . . ’ _ . a .3 3 . . ' ' I; . 3 _. . 11-3}.fll‘".' “ - . ,‘w . ‘ 3 . I. . 3 - V n .‘ l? I h ‘ if?“ - 5‘ r ‘ 3 .. 3 -- .rl 1:. u " . _ I. 3.3 . 33m 3 \- H 3‘ 3.3. “3.17, ; .. - - - ‘ 4.. _ - ‘ A . 7 Z A ~ . 1.0f; - ~ - _ - 7 --:§~ .. .; 3 31 consecutive derived factors are 2—groups then the conclusions of Theorem 2.5 are satisfied by G. In general however, we cannot say that the derived series for G coincides with the lower nilpotent series for G if 2 lGl. This completes the discussion for the case that (K(G), CG) is transitive. We next consider the case where (K(G), CG) is assumed trivial; ie, every automorphism of G fixes every composition series for G. It is to be noted that this is equivalent to assuming that every subnormal subgroup of G is characteristic. Groups G all of whose subnormal sub- groups are normal have been studied by W. Gaschutz [M] and are called t—groups. Many of the results proved in the theorems that follow are similar to those proved by Professor Gaschutz. However, we have that every subnormal subgroup of G is characteristic in G and the results in many cases are stronger. Since these results have been obtained independently using different techniques and since we are making stronger assumptions wehave-installed them for the sake of completeness. Lemma 2.5 If G is abelian then (X(G), CG) = (1) iff G is cyclic. grogf: If G is cyclic then for each divisor d of |G| there is exactly one subgroup of G of order d. Thus no two composition series of G are abstractly the same and so we must have (K(G), CG) = (l). I— - _I o O C O (i’fvy. a . ‘.] {‘3’3n ..i -. '4 A. o . I ' I 219.» Q . ‘9 3";r"l 4 I‘u "\ § ' 'a o -.. . 3 .4» I . _-,,,3 ‘3. I ,I:1;.,,3?‘,,.}3I 3 . _ . . ... I 1‘:- 1‘ II ‘Lk-’ t . -.-- . II‘ ‘I ..l. i; 1 d . ‘ A. -.. o ...- [.33 o". :‘gfa- .. I .n I 4. o< \— .n 7'... s .. _- iu - ‘ 07' n L :.- ff . \ . . VI ' t'-. ‘1. 0,1 l..~:” .3 , I 3 I I- ‘ “’3‘ 3'3m? with I ' I .Y .'|nl .‘J‘N' J, I - ‘3 . - L." :1, :33, Ha.;‘-.:WEW . *1 tn-gc. "' ’WILI‘. W";*‘Irj ‘ "0‘ I T. '5' '1’33”1"}..g'm9-r3f’c‘, { 1 . . . 3. ..J'r-A Hi“. '9; iff.— ‘ ' VII. '1‘ .. , )Ifr. I ."_II with I"; ...-fill; ' II. . 32 Assuming (HG), CG) = (l), suppose G = (x1) x (x232: ...x (x£)is a direct decomposition of G such that p is a prime divisor oflxll and |x2|. Let k = Ile/p and consider the mapping a: xl-——+ x1 xg, xi-——+ xi , i + l 3 e is an automorphism of G and moreover x1 x? ¢‘(x1) . Hence (xi) is not characteristic in this case. So if (3(G), CG) ' (1) then it must follow that G is cyclic. Lemma 2.6: If G is nilpotent then (K(G), CG) = (1) iff G is cyclic. 3333;: The if part of the lemma is clear in view of Lemma 2.5. For the only if part it suffices to show that G is abelian. Since (K(G), CG) = (l) we know that every subnormal subgroup of G is characteristic in G. Moreover by Theorem C of the Appendix we know that every subgroup of G is sub— normal in G. Hence every subgroup of G must be characteristic in G. So G is a Hamiltonian group, G = Qn x A x B where Q is a generalized quaternion group, A is an abelian group of odd order and B is an elementary abelian 2-group. All the automorphisms of G fix every composition series for Qn’ A, and B and moreover since A is a direct factor of G any automorphism of A can be extended to G. Hence it follows 3(38) .1 (I!) a 33' 2",: TI".‘.‘£"3 , -i -= 3'; ,M A) 3 ’ l :1 q Judi daua U lo,rciffetgm::eb Jos~£b 1 I3 ‘ 1151”” has q‘J'cx! .- 3"." . --. 2:... .r “10 so - _ j." - ' . ,3-1 I. - 3.x 6 ‘ 13.; fl ' . - ‘ é . a _ ' .— - “.1. 3f: . I I‘m ' -'l-'?- ..A' ; L * - 7‘ '. - v V‘- ‘t ‘3‘“. ' . .1 j '| “ ' 'v'.’ " . ". ' ‘ I? I. F, 3313‘- L . \ t u‘ .I «III. ?{I"-,._'3'.IL .J _ . _ N 3*; ‘. S3 IX '13V-L 3 A'Jl'f ;. H: L '.L - _nqnom. ..w .f-J ”_‘_ -.3 D Jan: wcliof 232m :1 mad: if? - (cfl'gA ~ II. <0 i 235%. “99‘ {(994) :1er 3..e 0 a: 9 ti :o.S '.+‘.‘. ' v , ' t 3:} I .C‘JEIO‘. .t . ’.. -" '. r --..-«r 63 asoitfiue It Jusq 11 {£40 and QU“ Inns ow (I) = ( J ,{c'fit 29618 A LI .0 n1 exactnesosusdo at D to w ‘ ‘- . f .Av" have 38:” worm sw xtbnoqd d it quoaadne t19ve eonefl .0 a! .” I .13 ’ r§r' “J'I' 11:5“ u r:*.——-———A 33 that (K(A), CA) = (1). So by Lemma 2.5 A must be cyclic. Likewise B must be cyclic, so |B| = 2. Hence AB is cyclic n—l of order 2 IA]. Consider on, we have on = Hl> --- > Hnel > Hn = (1) such -that IszHk+l| = p* for some prime pk+l and k = O,l,°°°, k+l n-l. Moreover ple yiyJ = x§9 y,yJ = x§l yiyJ = xiyixgy, = a? s3, Thus 5 is an automorphism of G and hence all automorphisms of G'Z can be extended to G. Since the automorphisms of G, when restricted to G'Z, fix all the composition series of G'Z it follows from Lemma 2.5 that G'Z is cyclic. We remark that since G'Z is cyclic and G'nz = (l), (IG'|,IZ|) = l. 'r ,. '— O- 21".“ I i" ' T I ' <2 - w x n b a." a; 3'5““. - ' .L '3! 3;? o 3! 113;.‘l96afi “.... . , - 10.39 331%». L .‘I I. y is :- x‘n 3 . . .. . ,. .. i ' ’ - r '. '- s‘-":5 9 ' . - 0‘. (1 .. 3“,? new .. - m. 1 ‘ ‘ c~ -.'._.-, I TIL. '3 “'3 ’3' .‘KICQIII :31 8, 33‘. has ' _ . . : 3:, WM q ‘ u‘~ [“5" in" dh“}1i\”.3’73::.:’_ II".Q 17.. , " ' 'T "‘4 -. - . .I‘k W’J‘ 7' 31"L r ' j'I, . . . '- - ' - - :1 ._ L. ”W 34‘. “:5” .- . .3. 4.3.3.1; I4II3I'Ii. .IIL$.‘I I o IIII ._ I ‘ ' I I I I .,~rt".ij..w\.~~ “35‘ M)“ '4-2 , _-' ’ ' * ‘ ' ' ”Writ! ‘Mc. my“ ..... is“: ‘4“ ' ' "" '1‘-IIv ‘IL’. I’I.4-..‘TM- 5'6“ LIM.‘ ,II‘Ig‘n‘ ~ I II'I,. 11".: Lin-2° 1.1 .I. _‘3 . - s H o .‘I; A ‘4’ o . L 3??! H . ' 7. 3! I.“ ”a 146 «3:31.31».me fius ms 3! '5 ~ #3,, .- ,. m-II. .— - ~-§-\,... 9 9; 5543379339 90 mo 8‘13 ' G‘I’i‘a 35.11! 3563 37 G' (l) LnK The Frattini subgroup ¢(G), of G is expressible in the form ¢(G) = ¢(G') ¢(K). For if M is any maximal subgroup of G then either G' i M or G'fl M is a maximal subgroup of G'. Moreover if MO is a maximal subgroup of G' then MOK is a maximal subgroup of G containing MO. Thus it follows that <1>(G') (\{G'F\M l M a maximal subgroup of G} = G'F\[(W{M I M a maximal subgroup of G}] = G'(\¢(G). Similarly we have ¢(K) = ¢(G)r\K. Since ¢(G) = (¢(G)f\G') f(4>(G)r'\K) the result follows. Moreover since ¢(G) is a nil- potent normal subgroup of G, ¢(G) v G'Z. Thus ¢(K)sZ and so K/Z 9 G/G'Z with G/G'Z a generalized elementary abelian group. G'L/ic' G' m L/xeL/K (1) Z Suppose that G has an abelian direct factor, say G = G B where G (\B = (1). Then B:K and K = NB with NfiB l l = (l) and N = GlfiK. Then G = (G'N)B; since 812 we have t ‘1: _o a v . _— . t 5'. ...~ it ....i d... ‘ ... . _M—-.r ..v . L10...— ......m. o.....} 3.1. T ._ “5:1. .....J. __ H... x .v . . ._ . . .. r .u. ...._ .0 Al. .L I . . . v y A . - .1 .L . r . A. . L.. I O . n l.'_. I . s ._ o O . .#p . . . .. , . a” .. I o I , O .l. . \ I.’ o. I‘”.. . . . _ IIk V A ‘ 4‘ '. J o. x . o . J ‘ I o r O w p t .... ..fl .., t c . ‘ _ .. . rl J I. .I’ f ‘5 .a . a. - v _ . .. A. . u '4 yr . l.‘ o . . ... .. . .4 . I . .. ..‘l a 1" '- . }‘.s0 . .0 . _ 3 _- . . o a .. 3... ....NH... . . uff.h. .- ... - I :11 .. .u.. . . 3 - en, L. . ._ .3 . 2%“ “hwy-(«n1 ....H...4 o ‘ .51.. O n t I on) t.‘ ...-II. .'.I.x | ..- . 'u. u _. o u: u‘ “0 .nH.qu l'J‘A-hrové -. ...." k ' c. - . . . .' ‘ . . . I .... ‘ . ; ._ I . .1" u .- . I .I u A - .. a. ..-..-23.-. .. - . .. - . - .- .... 0...... . .b. 38 (!B}, IG'§) = 1. Thus we need only show that (IBI, IN!) = l in order to see that B is a Hall subgroup of G. Since G'Z is cyclic so is B, say B = (bl) . Let {b2, b3,° ° °, bs—l} be a minimal generating system for N such that p leL where p is a supposed prime divisor of both [BI and INI. So {b2,' ' °, bsnl} is an irreducible generating system for K. Consider the following mapping: 6: bi-—-+ bi’ i + 2 t b2““* b1 b2 where lbltl = p. This is an automorphism of K which is not a power automorphism of K since if it were, then blt b2 = b2“ for some a and bit = bea’l. But BrXN = (1) so this is impossible and e is not a power automorphism of K. Now if we can show that every automorphism of G, when restricted to G/G', yields a power automorphism of G/G' and if we can extend a to an automorphism 3 of G then we will have a contradiction since ElG/G' cannot be a power automorphism of G/G'. Let ¢ be any automorphism of G and let H/G' be any subgroup of G/G'. Then H‘GG and so H is character— istic in G. Hence ¢ leaves H/G' invariant. So ¢ leaves every subgroup of the abelian group G/G' invariant. Hence ch/G' must be a power automorphism of G/G'. Let G' = (b3) , then G = (bl, b2,° ° ', bsel’ b8) and {bl,‘ ' ‘, bs} is an irreducible generating system for G. If g is C! any element of G then g is expressible in the form g = bl % a ' ' ' b 8. Define 3 as follows: S O I. . . . A _l . . a ... . . o I . .. . _ c 1:. . .v .0 “IL ... I _ fl 0. . I A . O a . F H . .. . . .. l ' ‘ . _ .. wa'..rtti syz.‘ .._ a ’9’ .v . “A. ._i . P7P: . : ... . o p . b. ....l .‘J . ‘ 3. O f; a”. J ly. . . .. - . r [Pi . . . h 5 .1 ._ Wu V7911; --...rv ...... dwwar. 7%..“ . I. N. . . . I . \ c 3.: . ... ._ I. ... . a e . . l :‘L . '1 - , J v to... a 4. u.“ ‘ wul v... - Viki ... o . . t . , —.rH ¢ .¢. ‘ I. .. 1 OT. . 4.4.. ... is». 38 ([8], IG'I) = 1. Thus we need only show that (IBI, IN!) = l in order to see that B is a Hall subgroup of G. Since G'Z is cyclic so is B, say B = (bl) . Let {b2, b3,. . ., bs—l} be a minimal generating system for N such that p |b2L where p is a supposed prime divisor of both [BI and |N|. So {b2,' ° °, bsel} is an irreducible generating system for K. Consider the following mapping: 9: bi—-+ bi’ i + 2 b2-—+ bi be where Ibltl = p. This is an automorphism of K which is not a power automorphism of K since if it were, then b1t b2 = “‘1. But BrNN = (1) so this b2“ for some a and blt = b2 is impossible and e is not a power automorphism of K. Now if we can show that every automorphism of G, when restricted to G/G', yields a power automorphism of G/G' and if we can extend e to an automorphism 3 of G then we will have a contradiction since alG/G' cannot be a power automorphism of G/G'. Let c be any automorphism of G and let H/G' be any subgroup of G/G'. Then H‘GG and so H is character- istic in G. Hence o leaves H/G' invariant. So ¢ leaves every subgroup of the abelian group G/G' invariant. Hence ch/G' must be a power automorphism of G/G'. Let G' = (b3) , then G a (bl, b2,' - -, bs_l, b8) and {bl,- - ~, b5} is an irreducible generating system for G. If g is any element of G then g is eXpressible in the form g a bl O. ’ ’ ' bS 5. Define e as follows: .r... _; ... . q . . QM It, I. .I W .. .v.»lfl1n....1 (“Mutinfi .xfiw. . . ‘ _ . «I ... . . . . . .1.....l|r. _”0 )« ....vho _vol‘.t_ I a D 0.. l o I a. r I I. I ... . 0r”. I‘ a. . .1. s l o ’ :1: . _ . .J m. -M’I. ... V. . u . - i u.» 1-1.»-.n M.- .. .wlrxnlfi xvi... . t‘l . ._ _ ... I: ‘ IN. (M ..n .r 0.. s. so ‘1‘... .... .0d..h.u’3.fl.~.ufl. 'w If... C. cau’. ..N - . . o s ... 0 {...}. I l . 0 0 Vt n g a d _ 6 a g . (bl 1- - «1331) b3 S. We need only show the homomorphism property for 8. Let gl (1 a and g2 be any two elements of G, say gl = bl 1 b2‘2 0 8 8 8 ' bs S and g2 = bl 1 b2.2 - - . bS 3. Then — a a a 8 8 8 9 l 2, . , s l. . . s (8182) = [Kbl b2 bS‘ )(b1 b8 3 a +8 a +8 a +8 - _ 1 1 2 2 8—1 s-l Bs "[b1 52. ° ' ° bsel b3 bs] a +8 a +8 0 +8 8 _ 1 1 t 2 2 5-1 8—1 * s ‘ b1 (b1 b2) ' ° ' bsel bs bs a a a 8 8 8 8 _ 1 t 2 s-1 8 a a a 8 B 8 t8 8 _ 1 t 2 3—1 1 2 . .. 8-1 a 2 s ‘ (b1 (b1 b2) ' ' ‘bs_1 )(b1 b2 .’ bS—l ) b8 b1 bs a a a a 8 8 8 t8 8 _ l t . 2, s—1 8 l~ 2 . s—1 2 s ' (b1 (b1 b2) bs_1 bs )(b1 b2‘ ' ‘bs_1 b1 bs ) a a a a 8 8 8 8 _ l t 2 3—1 s l t 2, , , 3-1 s - (b1 (b1 b2) '. . . bs_l bS )(b1 (b1 b2) bs—l bS ) .= 3 a g1 g2 Hence 3 is an automorphism of G and moreover 3 lG/G' is not a power automorphism of G/G' since we can write G/G' = (blG', b2G',- ° -, bS‘sl G'> . Thus we have a contradiction and so B is a Hall subgroup of G. Suppose now that G has no abelian direct factor. Then we wish to show Z i ¢(G). For this we need only show that ¢(K) = Z. Suppose Z i @(K), '99 n ' it . . a. ': ‘. H\ J "i ‘ i. . .' ' . C a. > o 0 ‘ f a . o - -. o O ’ q a ‘ . . r 1 ' .51. c d . I ' ‘ 'h- -. I. .. . r .' I. 0 r .' 1 ' - J. I . - -- . o . - ... , 2‘ .3 . v ‘ ‘—‘ c ' .0 J . . o. - .0 .. o a hi ‘ ‘ ' e .. , ‘ ' l‘ ‘7 - . - “ ‘ b 3 - a -Q'” - 't,‘ I" of s r’ :.i 'I “1 rdu‘“ . ' ' ‘ ' a ,_. '- ‘ o 'x‘“...‘~ ' ‘. ‘ ' .‘ I o --c 0"":‘ d '4?! " o. | I ‘0‘; .1'.6 ‘ . ..I‘ ‘ \.“.' MV‘ - I a—- " ‘ u. ' I ' r I ' ' ,‘ ‘i . v ; ”‘. ' n .4- 1 + ‘ r ' , on J Q! . , ‘ - In «JV ""1... 1' I ‘ ' o .‘ ‘ s d L', 1 , . , b I ‘ ,4 ’5. _ ,, ‘_‘ - - o A u v I' N r "- J L.- ' Q 1 . I . _ . . : § 4 .- ~. n ‘ - 'IWI: ‘,(: \| M1. “.5 b ' EM”4W" . ‘ ‘ l 1.3.?” ‘ ~-..; In ,v . . . o < ‘3 £48” . . Iiiefi 10 gnoxadss Lisfl a an 3967 some". .72 -' '17:", ..filnlq‘T- fjjilj.‘ . _ l ‘ a: h 'v‘ ' .. 'Pd". Huerfifl': "' -- ' 1" — " l‘ ‘ _ ‘ La Mil} L: ...-:2, C. '2'” . .1 ' " _-[;$5' “’1‘: 5 J ' _‘fi’ii‘f‘b, £0 —{1 £1: _ . - _ .iztj .rl‘ulo' . pgd-‘F -.‘ ..-- \o _‘..w10¢orow~, ,. _ -’ .1 . . '— . r - J r-lhr-JH‘: ‘fiLV".‘H_-" “ r‘ . '._' a , -_ -.-. 40 then there exists a maximal subgroup Ml of K such that Z i Ml' Thus there exists an element beZ such that b¢Ml. If IK: Mll = p then bpeMl and p lbl. If q + p and qllZl, then the q—Sylow subgroup Z, of Z lies in Ml' So q we can assume that (b) is the p—Sylow subgroup of Z since Z is cyclic. Let Kp be the p—Sylow subgroup of K, then we claim that Kp cannot be cyclic. For suppose Kp were cyclic, say Kp = (c) . Then cpeMl and ch1 since btml. But this implies that (b) = Kp and hence that G has an abelian direct factor. Since we are excluding this possibility it follows that Kp is not cyclic. We now consider two cases. G' mm ¢(G') (l) MK) Z Case a: Exponent (Ml)p > lbpl, where (Ml)p denotes the p—Sylow subgroup of Ml' . Let {b2,- ° °, bs—l} be an irreducible generating system for Ml where b2 is an element of maximal order in M1 such that Exp (Ml)p lbgl. So K = (b, b2,y- . o, bsvl). Moreover we can assume that the elements b, b2,- - -, bs-l form an irreducible generating system for K. Consider the mapping: r25!” 7-” 04m 1"” .l "H“: ..-,,-__ "...".o‘a: h, -:"'§‘*fi*'-FL 1.1 e: b. ————> bi, i # 2 --> bb2 k-l b -———> bp +1 in case b2 is the only element of p-power order in the set k {b2, °--, bs-l} and b5 = bp. Otherwise we can assume that n n l)p = x and bp = b8 2 b§ 3. In this case we define a mapping a as follows: (M a: b. —-—-> bi’ i # 2 or 3 -—-—> bb2 t -———> b b3 1 + - b -——-> bpn2 tn3 +1 where Ibtl = |b3 . Both a and e are automorphisms of K which can be extended to G and which do not induce power auto- morphisms on G/G’. So this case cannot arise. Case b: Exp (Ml)p = Ibpl In this case we have that bp is an element of (Ml)p of maximal order and hence is a direct factor of (Ml)p. Thus it follows that is a direct factor of Ml' So we can write M1 = x <02) x- - - x ° Consider relative to any . We have n = (1) since n = (1). Hence is a direct factor of K since K = x x- . . 2 x ‘ Then must be an abelian direct factor of G since bez, and we have reached a contradiction. Hence we have 2 §(K) and it follows that G’Z is the intersection of all the maximal normal subgroups of G. D , rr 3.. II I - .)‘ - I 9 ‘r_" . s . . “in ‘ Q "-"n ‘ - 1 ‘ . 'I r " ‘ u '2 r _ !~ Ir fi.’ V .. ‘9’” ’_ o O a". .' c’ . Of 1 “cu. " ‘ ‘ . q ' “ - -.. “r' a. ..., I - ’ f{{;:1;‘)\t'-':'.H't\o _.|:Qd:fil#_‘xl‘£ 0 ..-'¢.. "3". - ,' LL,'.'(‘_...'3:- ‘ I‘.‘ " ‘ ‘ ' _ y, OJ.' 4mg :33t» 'H‘ with O a ':':;'. ;,b;f;.. - i t v, “2 Corollary 2.3: If Z = ¢(K) then G/G'Z is cyclic iff no two maximal normal subgroups of G are of the same prime index. 3399:: We know that G/G'Z is a generalized elementary abelian group so the corollary follows from the fact that G‘Z is the intersection of all the maximal normal subgroups of G. Corollary 2.4: G can be expressed in the form G = M'U where U<:G, M is abelian, (IUI, IMI) l, G':U and 23M. Proof: Since G is solvable it can be expressed as a product of permutable Sylow subgroups, say G = Sp ° Sp 1 2, 0k an .Spn. Assume IG'I = pk ° ° ° pn and let U = Spk. . - Spn. Since the Sylow subgroups spk, . . .’ Spn are permutable U is a group. Moreover since G'<.G, and G'0. Let x be an element of Gr such that xe+ x, then -1 since (xG )6: xG x-lxee G Assume x6 = x where r r’ r' - gr P gre Gr‘ Then it follows that x = x6 = x grp. Thus since x6 + x,gr is of order p. So p |G|. 43 .42...- ' 1“”; _ fl: ' ...- .1 .. ,' * ' , 5“; 1K! ..9 u ... . V H . . “ Alum ’ ’9 .‘ "I L :I . ‘ v “'1 5. 13 1| 1 15w ,3 ‘. Mr W vy‘ 3‘ l3 0". u‘“ 3? . “N"? *‘2' fjm ‘1 ' 1773a“. Ht. "-‘w 1..| ‘I: A -;‘.e :1 ’ n 1 . 1 ."I u. \ 0 2...: ' v9"! -, n 1. I \ ~o .. l. a 1,. )n "i . 1 "r h‘ 1.3! A A w- afié; .. A \t: 1 1 -. J t ' 1. ”H" i. \O‘. o ' 3 v I“ 1.,“ w i 1 iff: CHAPTER III Basic Properties of the Stability Series and the Groups A0 and Al. Recall that for a given subnormal series s:G = GO 1 G1 :4 - oi-Gt-l: Gt = (l) of G we defined 80(3) = {eeA(G)|Gi = GE for i = O,l,° ° °, t} and Sk(s) = {eesk¢l(s) el Gkvl/Gk = l} for likit. Theorem 3.1: Let G be a solvable group and s:G = GoiGli' ' ° :Gt_l:Gt= (l) be‘a chief series for G. If p is a prime divisor of lst(s)1 then p lGl- Proof: Let 6 be any element of St(5) of order p, then 9 fixes Gt—l elementwise. Let Gr be the largest term of s which is elementwise fixed by 9; since 6 + l r>O. Let x be an element of Gr-l such that xe+ x, then x-lxee Gr' Assume x6 = xgr where P gre Gr‘ Then it follows that x = x6 = x grp. Thus since 9- since (xGr) - XGr’ x6 + x,gr is of order p. So p |G|. 43 .l’.1 . ‘Et'f d . 1"; . 3 an: 5118.55: e' ' . .. on 1' ..’. I. I" - -. -.. - -h i 1. f3}. ' oél'i’vc .: - , -2 - " II“* :v- . 1‘1 . :1 a-.nm - ° ’. f I « ,.. " 1 4:. b- O 'm‘ 7“} a 1... o! 1 .3 H " H D n v a 0" 3 , .. -r. l: d O '0 53 WW 9.1158110: .9 1 - _._v 1 3 ' .1r -0' o ' . --_"ia.£::9 a so .- a . . . -< . l5.” " a u {3'33‘ 17 “I” . f' ‘~3‘.;°:’!C1‘ 8 .t {If ‘ t 10 mangle 1.15 1 a 9. --‘x 303333 .' .i‘ .39; “ ... 2 K'.“r* '. . . $54.1“ .. ..éiflft‘ 1...: 1.31 5:311 '3' 9mm scluhmua 21 tom": 5'10?" 3 . .. . .0 30 Juan-2-- .‘ is: I 38:1 Me . 33:11:5-n-zf- '- 3 ‘ “z" ’ . 1:51:31; $1“: 1 1" ' '3‘ ‘ e r w 1 1”.”4'1’,’ '. .. I. - _.7 . I . O . I’M” ' '1! :‘LJ ‘3 . _t‘ ' . . h 3 ‘ s ' {110*} :3" :‘- -. ' ' ”I '2 ‘2‘" :1) “with “"21?" . ‘ 35.1?inth anciio? :1 M16? -q ‘zshflc 10. -a.‘ 133 ...\ _ ..‘. .1 ‘ . '. ; .. . _ g _ ‘ I I ‘ 1 J I , ‘ v " . .‘ I ... _I _ _ _— . +.. 11.3"“;‘11‘nwt : ~ 1‘ 1 ,'.. I .r .. 1' . _ .' ~ T h _ "’1‘”! ' “.13.: 27- ;:.,w , - - , TTN" 3! ' '11.-” :1... . r1 » . Maw-ii . l ‘1': ' ' ' TIL“ , 4 “2111111111”. -1 . -- 1‘13Igilafi! .l ‘I , .. ' .. ... 1 :1. .' '1'. ‘ I _ ~ 7 "1}.11. .mp1 _ flew..- .. 7 — ° {hf-1-13.33 ‘ ’ - _1 #5 i... 44 Corollarl 3.1: If G is a p-group, p a prime, then St(s) is a p—group. Theorem 3.2: Let G be a pegroup, p a prime, and let s:G = GoiGli' ° ~> Gt_l_>_Gt = (1) be a chief series for G. — Then St(s) is a p-Sylow subgroup of St—l(s) and 6e St-1(3) implies [6| ISt(s)l(p-l). Proof: Let ¢ be any element of S (s) of p—power tml order. Since G is a p-group and s is a chief series of G, [G Hence by Theorem A of the Appendix A(G ) is . tnl | = p- t 1 ¢p—l Cyclic of order p—l and x = x for all xe Gtel° But (|¢I, p—l) 5 1, thus x¢ = x for all x e Gt—l and ¢e St(s). So St(s) must be the p—Sylow subgroup of St_1(s)' Next let as S Since I AG _ t_l:Gt = (1). Then 30(5) = 1 iff [cl = 2. 3399:: If IGI 2 then A(G) = (1) so that 30(8) = (1). Assume 80(8) = 1. Since 8 is a chief series for G we know that IG 1 80(3), so that G must be abelian. Suppose G is not an elementary abelian 2~group, then the mapping 6: x-———->x'1 for all Xe G, is an automorphism of G. Moreover ’%3 ~ ,. ;1: . -- rsfia ,3.‘:1 ‘ ' - ... T l’- 1 - ~*h— - 2 391 fins-,smtfiq n 1 ,.3 1 .1... - i ‘. H !.'_ _' ’ c '1 ' ' '-'- . '. , 'vb ~' ‘ - a . .L,N1fi . ' .9,1U$ 561%5; 18i3 * .x - 4 ~1 ‘ - '- ‘ . -' . 1. ‘7'. :. _‘M, U -." ‘ - ‘3-1m..£'1k‘..' ' J . '1' I I‘ .g‘. ‘ t ‘ '>.T~. “3.17,. ($5.1...3 fie URL, . c. o - . - ‘3' J . ‘. ‘ . .. ‘v. “'5- " ‘-' 3 . 1" 1.57.31. .‘. | :1 (I. .I ‘— "n -’.--‘l "A“! ."l', l -‘ IJ” -..:H‘de ' ' . 1.41-".3. ' emu-1.11.11.- ,, . .3 y 1 1 .: . 1 ' . ,1‘ 11.1511 .' 1.- .___‘ f 313- ‘. , 1 1 ."5‘5‘ .mw— 1:- '- .3 -- —- - 3 t .I'._ 1 ‘1 :1 ‘d 1-‘11 L .'- L - ‘t-vx‘t 5 :21. .W‘ i q . b _ 1: -1... ‘9; ‘ 3....’311‘?“‘" “V -— «1" lb" . ,3 1 11‘115151'31 M. x “2:42.: _ . v 1. _ ,- _ . . : .: ".0”. ...“. 1 ‘ 41 ' .. ' ’- _ .t 1'?" . f‘ —. ' ’ Lf _. . 1‘ 3 r 1 L | l 1. . I . - - 3L: ”J 'F D I ..- . ~33 , v. ‘ ' 71:11"! 113.: L‘ ~1' 1 -, 11“} '4»... 3.13139 '4 f.» W111 11:11, ,1 1 1. :1 3 . [_WV: ‘:jv 1 ‘2‘ “:11 11.1 115.11%” "‘1’ 1,1: 1 ' ‘. . "'31 I}: .. ’khl - _ A?“ “a (1' ‘7 1 ,Q J I J 3 cl L asnla . Kv‘.“ 4‘— —L,3 1...“; ' v I 1~é13:1{.1:l:_‘ k?» f :4 It’»'136:“' . :9-7 [_"f' ,‘I .’ :uE '” ”.3 ‘ ."" -: b - If ' V...’. . ' A- -?v F‘_ o c , . _'.1 1 ' . 3142. ; 3“ 38.3113 - ~GI’N 0:: "_h§-al .fl )2 116 ’ M‘ Q I ‘1 11 ‘- v ,4‘ _. é.“ abfi‘w'afi‘ m". a.” ' :‘7-‘J- . _' 337‘." J 1 1 .. - L , , ...-W - “5 e + l and as So(s), so we have reached a contradiction. Hence G is an elementary abelian 2—group, say G = l, and as 80(5). Hence we must have that t = l and [G] = 2. Theorem 3.4: Let G be a nilpotent group and let s:G GOiGli° ° 'iGt-liet = (1) be a chief series for G. Then St(s) = (1) iff IGI pl' ' 'pt where p1,° ° °, pt are distinct primes Proof: Assume St(s) = (1). Since G is a nilpotent group evenychief factorcfi‘G is centralized by every element of G. This follows from Theorem B of the Appendix, Hence IG 1 St(s). But since St(s) = (l), IG = (1) and so G is abelian, say G = S - 3 3 is the ‘ —S low sub- pl x szpx x Spn where Spi‘ pi y group of G. If some Spi has an element x of order p1k where k>l, then the mapping . _____, p+l o 9. y y 3 y E: Qpi y_"'y 9 y 5 Sp 3 J 1: i 3 J is an automorphism of G which is not trivial and lies in St(s). But this is impossible since St(s) = (1), so each Sp must be elementary abelian. Suppose Sp were of rank 1 l ;.'_..-l ' ‘N? u . , w ' 1 - . - .9— 2 J .1734: -o A :c 1‘. ' I 2%: «fire-my 1 "1", &.5“‘ .' 31 5‘! ‘ o ‘ ‘k‘; 4 .151 cC-‘I‘rv "1 " .“ {3011? 1111?“. ..1' .3 1.1; 3:3 ' 1 1- 31 ‘3 l. - . -121 1 '.;-:3-33,,,3.333; 3 - ' - . 3 :3. -- 11:11:13.. .3, 1.11:3, 13:14 111133”11{*131'.'7"3' 1 , ‘ . 153111-3411?” - 3371111161111; 341151, 1113;111‘1111110311111!‘ ‘1 1 1‘ - . ' 1111. , ‘J '- ’ 11"11‘115' "‘ ..| ‘1 . ‘,.I‘ f . 1' ”MI ’3: ‘ V _ .133 SAL. [fr-1"“: ' '9’1 11 ‘ {$‘1h? 11mg. I" 111 1‘1111I"1‘l‘|; I 1 .. _qN 1 .- I W'- b. 1 I u ‘ 'Joflau 16331.1; "J :z: myf‘l’lyl‘é "‘1 | 1 1 $411; ' x'dgrthvll“‘ 1.111; '1'W1“ '. ‘ '1‘ $11.1£31 133“ 11113 211:?- ‘ “11\:i”" ”1m"! '"‘ yr“? ”11'"; 3113:. II ‘I 1‘1““ 1431.“ 11m 4 - . 2:: _?:-'fi . 31-13913 1 cut" 109‘ 46 greater than 1, say S = (yl) x- ° °x (ym) and m>l. We p1 can assume without loss of generality that there is a positive integer k such that Gk :_ but y2 i Gk+l' So Gk/Gk+1 = ° The mapping 0“ y2_“*y1y2, yi—eyi . 1+ 2 1+1, X———-—+X , X e S pi, is an automorphism of G which is non—trivial and which leaves each Gi/G1+ elementwise fixed. Thus aeSt(s) 1 and we have reached a contradiction. Hence 13p.1 = pi l for i = l,° ' ', n and n = t. Conversely if G is abelian of order pl- - -pt then no non—trivial automorphism of S can leave every Gk/Gk+1 pi elementwise fixed, 1 i i t and l i k i t. Thus | A St(s) = (1). Theorem 3.5: Let s:G = G01G11° - -: Gtith = (1) be a chief series for G. Then G is nilpotent iff IG 1 St(s). Proof: The proof is a direct application of Theorem B of the Appendix. If G is nilpotent then each Gk/Gk+1 is centralized by every element of G and so it must follow that IG : St(s). On the other hand if IG 1 St(s) then each chief factor of G in s is centralized by every element of G. Hence it follows that G is nilpotent. 4‘64 5 NJ. . ‘. ' . n ~-.‘ ' . . '1...” . . u . " o. 4 ‘ ‘.' 7 p . . .P . ‘ Y” ‘ o' '43 a" ”:fi I I :‘I‘l _'.."“r '.‘ r". l a. 3 {junvz'uy'pn‘ . " . T” ..I ' "I ”..., :. \ "" u v ‘V. .. M103 I‘ ‘o,. . ,. '~.- °; --...-l; g Q - .1 3 ’ .. ‘ _ .- I'm ‘ '.~ . g , T." - . H" . J— ‘ V I . _ | ' effing” ‘, “F . u _ j _‘ "'-_ . ~:;J ,' 5.58;. :m - y :‘-‘.’=:' -. - w I, 5' '-' ”Hi-7143f”! ,ji . ‘ . ‘ a . . P"«w.‘$""".+.' . ‘ 3 . L; 1 ‘ ‘ "' ' (:3 - .- ‘ I ' an “-1 - U '8 a“, 'H ~' 3.1 I ,- ....- uh! “7 Corollary 3.1: If So(s) = St(s), then G is nilpotent. Proof: This follows from the fact that IG 1 30(5). Theorem 3.6: Let G be an A-group and s:G = GoiGl: ‘:_Gt—1:Gt = (1) be a chief series for G. Then So(s)’ = St(s) iff IGI = 2*. Proof: If IGI= 2* then it is clear that any auto- morphism of G which leaves each Gk invariant, likit, must fix each Gk/Gk+1 elementwise since Gk/Gk+1| = 2. Thus So(s) = St(s). , Conversely if 80(3) = St(s) then by Corollary 3.1 G is nilpotent. Hence since G is already an A—group it must be abelian, say G = Sp1 x SPZ-X. ' ' x Spn. If n>l, then some pi is odd, say pl, and the mapping 9: x————+x_l for all x e G, is an automorphism of G which leaves s invariant and is non-trivial. So Be 80(5) but at St(s). Contradiction, thus |G|= 2*. Corollary 3.2: If G is arbitrary, s a chief series for G and So(s) = St(s), then G can contain no abelian Sylow subgroup of odd order. Proof: Since G is nilpotent G = SpoQ where Sp is any Sylow subgroup of G and Q is a complement of Sp in G. In fact Q g G. If Sp were abelian of odd order then the mapping 9: x————+x_l , x e S y-—-—+y , y s Q “53953 5;L\ “gi'1519'1.3?.1'5 ”112:1. 5 t “555131 . 32"331153541'5 V 1 :1. ‘1 55 ”I“ 1 .. .5 1'31\11111"11 11 ‘15 3155 11111513.? '115 '”. ' , M .‘1111'1 p131 H A‘55 , 1119'.th5 5511' 1 5 t "'13 1:135”; 5. gm; 3W131:.3133‘:3:“‘;:1333n-3 . 5.15515 “5.55555" '31:“ M11 7.5‘1l1513'11‘ ’1 11“11'1‘1111 (H. 111' :11. 111-1 51. NL’ 1".‘1'1'1Mw‘111 _ . 3(5- . 5 :15: 5 . ‘ 15-3" 11“.: 5M 3| 5- . 3 ‘ 3 3156'. 53"1“|I.: 33‘ 5.53.35 ...; 555 51539 5 5551' "1': 11173111315911?" 531' " [dJ‘l'P ' 5{.51 -1 ‘ 15.1”:5155537. ' - 5 .55... 5555:3151 5'51 .55 ’3'”qu 3 . 3 '5'; P T141“ INT-o W” 73" gm“ “5,5: {511.15. ‘34:: ,- ’8 ' . 1 L35 .. . '5. ’ 3" .5555 55 «E 5.45;. 55“” '5'555135r133; . I * 4 . F333 . 3.333» 133335451 35:1: 5:353:48 ‘ fimf.’ “W3 ”5”“ 1315;115:3111“ ,~'~:-:.1'. 5.5.. 5,55 544355" ”“55.“ 5; “13...: " 'b"_ ' 13";95.‘ Fifi . .5535551553 "1.,- ' ‘535 .5: W74“ - 5 55. 55-. . 55 «5.555 55-— ‘ 15- .3. j. '11.:‘1111t’1'n ":3. {“l‘ij 3B3 [3.1-.3 -.--fL-I 333-! 3.33333 3 3 33:3.43_ 3 , 3‘ 333-— 15 11.131: .u3 5 ‘ '1'"! 1‘ "33:5"1 f3. :3% ‘ . 1")" 14 w ?t:'"’ 3‘ '31:. 5535!$1 111 "5113 ' 11.. 1 ' ."31 , '3 an». o 55 5 :1 ._~.. {3_ y -r- :(P': I J“ _ W11“ 0 1“.1:f3“3*1 5" 3 .54 5|. , ..I 3., >1|5 .h I 5". 33 “8 is an automorphism of G such that 6e 80(5) but a; St(s). Hence we have a contradiction and so the theorem follows. Theorem 3.7: If 80(5) is abelian for some normal series s of a group G, then G is nilpotent of class 1 or 2 . I 3333:: Since 3 is a normal series for G, 1G1 80(3). Thus IG is abelian. Since IG 5 G/Z(G) it follows that G/Z(G) is abelian and hence that G is nilpotent of class 1 or 2. We now investigatescmm properties of the subgroups A0 and Al which were defined in Chapter I. Particular emphasis is placed on when the elements of A1 are central automorphisms. In what follows let Zl denote the center of G and in general ZR = Z(G/Zk-l)° Theorem 3.8: Let G be a p—group of class 2, g be any element of G and a any element of A0. Then 89 a ngE; for some positive integer m and some 2821, and 6induces power automorphisms on 21 and G/Zl. £3993: Since 6 e Ao,eleaves every normal subgroup of G invariant. Hence 6 leaves every subgroup of Zl invariant, therefore by a previous argument 6 must induce a power automorphism on 21, say 29 = Zn for all z: Zl and some positive integer n. Likewise 9 must induce a power morphism on G/Zl’ say (g Zl)e = ng for all g Z a G/Zl l l and some positive integer m. Let g be any element of G, then (3 Z1)e = gm Z implies that g6 = gng for some z 521. l g .1 - ‘ J _ ‘ . ‘ .: c ' 1 I '0 I 1 ‘1 u . I 0 .k“ I: . - . In??? 1' ' .1'" N . '11 '0. "0;" 1 , 1 1 1 ‘11. ... .t r 1",} I'a' . I .. r.. --‘ ' ‘ 11 .3 ." 11:31 I k "A" ‘1. II. _11 '. m: 1.4:? ‘ ” ' 1 . ; 4 . ‘5 ' . 1 ,1 - “ . .1 ' .‘ '~ t >1 , .1 'L .3 1 6 '- - VI H I I I“’" '¢| 1 ‘ A r‘ ' ;t-:“a‘ -‘ 0 O. R‘- , Um .d‘41'v‘l 1‘" H. . l' u" '4- ) l V J , J' ' . H’. 1"‘.‘ ', . 1) 1H l'll 'Vl “ . 4r}. " ' ' 1 I "' I ' L ' rv‘ v' 1 1 I . ‘ "It.};Y:3:'1::h .1,IA.1‘, 1.1;. ‘. LI'. . . " ‘ . I ‘ 1k 1 .1; :4. " I '| d L . . 1"r19.u : .11’:: . - ' .- . 1,“* \.I11‘iti‘o I\-.’+ " .7‘1 , " .1; I? ,. 1 ‘ .."1 I ‘11.: .l .1. , .1 I. - 1 l . a-‘h-‘k”: d "It. . I 1. , E1" .6 .r Ifijg1.':fj"lr ' 41‘": "'0‘. ‘ ‘ I V.7101 V " . I U"? .11 111.11 "f“; ‘-. .v.’ 1- ”1‘1 10-1» I ll ‘ y , 1 M : ' II I 1 "1" '1‘ '1'“an '1I.1 1:1 ' .. ~1lr 1. Thin. 1 . .111“ -~ ';1.'- . , '11. .2"; I an ”“1 g ‘- ’1' If: ' '4'. 1 1 7:); ‘ ' .1. . ' no ‘ 1 'I-A'Z' I "’1‘." I 1: 7 ‘ ‘.-*.-V" I! . ; I ‘.' «I‘l‘l ‘. Id. ‘ ,bII.‘ - ' ‘ 1:91“ 1-‘~‘ . C .. . _ I'm. ..9 , bu; g! 31.1. . ' IN”, ' '11.. 'f 1 . '. n—Ilv - ”.1.” 1 ' k I .‘ _ 2‘“ ... . f .‘ _ 1“ 1. . 1 ‘ 1‘.‘ I‘1-‘1t'tI 31,, , "" » ~,‘ "" 'A~ .' ' ‘. " 1"“ 3'; i a". t: ’91. 1- 1 1 w' :21 .7». .2 :‘Iggmjlww .1 Ufim : W Ei'fiflwsiiir : 1 1 ~ .111 ""“. ~- 1 -..J -..":é~ -, I ’0' ::¥:1%109'1ll‘3*!': ' w m :é’L'.” “‘1," ””0” ’. I ~ , 1 , _ ~ I . .. U.,...II1. . .1 - '1I‘IWfllfi 13pm ;-11"-".E‘- ‘fq f: fa“wl.'"i:- I 1 I11" . 41.11 1.12% 1 .“ £11.. [11' ”"‘”':-:-. 'fihmvflt ' 41311.1" ’"' n" . ‘41:."4115'1: "' . - :L. ~ . 1"“! ‘Lu'M‘. 1 ~-.1 . _fl ' w. -1 - A 111 4' (“I“). .1") gnu-‘19 1. . 7. ‘- ° '°'" III" .3)" a.,“r,b. OQ-“W’mfi an 11 “13‘ ‘-“’-' -' ' r: .. WUfim1- #11 ~.. llé IN ' u *1. .- ., L. E I: _ ‘N 11.311 ‘1'. .K '13“: 1’ ‘ "F913. - ' ""111 wins, 1211;»: ...;L;. 3:1“ ‘35 1" “f. ‘ zlgrl‘1'l 11.1 1f. t«?-1 L! {9' '1 f ... Io“. ' I ”if ‘ .‘V".".1-‘: ‘_. -ut.. .. ( II -1 '- a} 11H ‘ :1‘ - E'1v'1'" 1‘1" ;. 1.- . '.'.;" a: 4.11»! -‘ I‘é. ; ‘ +9.1 v'O-" «1 J. | "1.1 - | .9 . “1H 1' 1'. ..I hi .. 1v —o‘ 111.5 H ..“1 1 I. 1 PH, "1-‘.'&‘I"’f“ .. _ 1'1“ ’ 1 ‘1 ‘ Q “9 We recall that an inner automorphism Ia of G is a central automorphism of G if and only if a e Z2. and the number of inner automorphisms of G which are central is given by |Z2:Zl| . Thus every inner automorphism of G is central if and only if Z2 = G; ie, G' s Z1. Lemma 3.1: If G is nilpotent, then A1 = {68 Aol 6|G/¢(G) = 1} Proof: as AI if and onlyii‘elG/M = l for every 1 maximal subgroup M of G since G is nilpotent. But this is true if and only if x‘1 x6e M for all maximal subgroups M of G. Equivalently x‘l xee ¢CG)cn"6lG/¢(G) = 1. Thus the lemma is proved. Lemma 3.2: If G is a solvable group and all elements of Al_are central, then G is nilpotent of class 1 or 2. Proof: Let [G,Al] a (xgl Xel x e G, e e Ai> . Since as A1 is central x‘l x9e 21 for all x e G. Hence EG, Al] : 21' Moreover since G is solvable IGffll and thus G':_[G,A1] i 21' So G/Zl is abelian and G is nilpotent of class 1 or 2. Theorem 3.9: Let G be a nilpotent group all of whose maximal subgroups are abelian. Then all elements of A1 are central and G is of class 1 or 2. 3399:: Let ng denote the intersection of all the maximal subgroups of G; we claim that 21 = qng. Clearly Zl = Q B if G is abelian. Assume G is non-abelian, then it is clear that Z1 1 B for any maximal subgroup B of G '4 Nut” ' WT“: .¢:I;IL. I5“. 1",: .Hlxc':‘ —0, v-‘- , 1‘ H f'hiy“ V' ' .LL" I4LT?I1-r'"' .' . .Z't’ ,4“. .HJ".b UH, J; ...JL ‘5 \ I!” .'..A.|. :1". «4‘ L'” 3:» " 1's, 50 since B is abelian. So Z1 :_/;\B. Let x be any element of I§\B and let g be any element of G, then there exists a maximal subgroup BO containing g. Since x 5 BO and B0 is abelian we have that xg = gx. Thus x a Z1 and 21 = 493. But <1>(G)= @B = Z1 and so by Lemma 3.1 x"1 Xee Zl for all 6 5 Al and for all x a G. Hence all elements of A1 are central and by Lemma 2.2 G is of class 1 or 2. n Lemma 3.3: Let IGI = p , n>l, [21] p and let aeG, ll aéZl. If all elements of Al are central then Ca = CG(a) is a maximal subgroup of G. Brogfi: Since G is nilpotent IG 1 Al and so every inner automorphism of G is central. Thus G' i 21' Let G = x1 Ca LJ x2‘Ca\J° ° ° ler Ca be a coset decomposition of G relative to Ca vith x1 : 1. Consider the elements 1, x51 Xea, ' ' ', Xr‘l Xras these . _ . -l a _ -l a ) -l are all distinct. For if Xi' Xi — xj xj , then (Xi Xj 2a -1 _ -l . _ . - xi Xj and so Xi xj 6 Ca‘ Thus xiCa — ij and Xi = xj, so all of the above elements are distinct. Moreover since G' 5 z1 and |Z1| = p it follows that r = p and G = CatJ x2-CaLJ' ° .kagp-l Ca for some x2 5 G. Thus Ca is a maximal subgroup of G. Theorem 3.10: Let c = p“, n > 1 and Izll = p. Then all elements of A1 are central iff ¢(G) = Z1 (ie, G is an extra-special p—grwup). - U I. '}.1 .I‘d .“.‘I.' - or “I I , “It!" ' 1. ' ' 71‘?" a. .. x ”J. . n ' . _~ I 0 ‘1'" 1"! ‘ - 3‘; t‘ '\ l f V . ' - 31:." , ". 5-; :5 . .‘o‘nlt'l 1,“; 1': ‘ ‘4 '. ‘ .‘J-' "' ‘l " "Ir . :. . '3: 30;.“ “-147.- .' ‘ ' ‘j' . JHC '3 . " I .- 'v~ “{er ’ .4 M . m , ‘ .: v z . ' 'g’ -_ 1:; «3.: .‘ :t‘a , . . .‘ ‘. '0 I‘ ‘ ' o . I- ' ' i ' I.._ WW I ... fin:- 'fii’fi _. 'd' *ko‘ ‘Tmriyt‘vhf. l...“ ,J‘ . 5" - 09.,- . , . .l v. .4 ‘L ' . ' 'L . ’1‘]wa ,, ' “Hf. "U'if’flf ‘h. .7 Grip?“ L -. ....«n, I. " ‘ ‘3! . 5 I ‘. - _ 0' 2 : o .9: r1,‘ -._—."' v. . .- cw. v .4 'v, "Y?uo:vdl 'H'I‘I r ”:5“ 41,}; 51 Proof: Assume that all elements of A1 are central automorphisms. We know that Zl = égb CG(a) and also that CG(a) is a maximal subgroup of G if a e Z1. Hence Z1 is an intersection of maximal subgroups and as such contains ¢(G). Since G is nilpotent G' i ¢(G) and since lzll = p we must have that G' = @(G) = Z1. Assuming ¢(G) = Z1 we see that x“1 xee ¢(G) = 21 for all xe G and all 6 6 Al. Hence 9 is a central automorphism of G and so all elements of A1 are central. I'l Theorem 3.11: Let [G] = p , n > 1, z 1' = p and let B6 = {xeGlxe = x} where Ge A(G). If all elements of A1 are central then A1 = IG iff Zl < Z (Be) for all as Al. Proof: Assume A = I . For each a.eG B = .__—_- l G Ia CG(a) = C If a i Z1, then by Lemma 3.3 C8 is a maximal a' subgroup of G. Hence since a e 2 (Ca) and a ¢ Zl it follows that Zl < Z(Ca). Conversely suppose Zl < Z (Be) for all e 8 Al' Using the same sort of argument as was used in Lemma 3.3 one can Show that Be is a maximal subgroup of G for each Ge Al' Hence G = Betlx B6 L1x2 BeLJ°°° uxp_lBe for some x e G. By hypothesis there exists an element yeZ(Be) such that y¢Zl. Hence Zl =(x-l y_lxy)= {x_1 y-2 x y2 2: = O,l,-:', p-l}. Since x-l xee Zl we have that x_1 x9 = x-ly-£ x y2 for some t. We can assume without loss of generality that 2 = 1. Thus we get xe =|y—l x y. Now let g be any element of G, say g = xm bl for some m and some .II ; . 1 9‘ 4‘“ I d I. 0 ‘. 3W :"c ' 1 .49: [3' L :1; fl1'l3¢.~,l' 'L’ . 1555-55“ fmh ‘C'JILLZI: ‘, 3 "I3" ' f | ‘ ' III 3 i, I I.. ‘ ‘ 1333,13.“ “1.3" 5v” I 1 '_ 5‘5555“.I.“:' .0 “I "' A. ‘ yg‘t_“ II II‘ 555.1. 9 3’ ‘ ' P N. “f" I ' .f ‘ ’ ' "‘ ~“"‘{ .Iutllf .h f“ , o: ‘ I. . 5 v 5 3 V 5 5: .55 ...-I199” In. M“) $4.35,... 5.33.5»: ... I]. l...‘ 9:?" «L :1.. 3‘; l, I 3&3? Wm" “'33.‘ . 1:13" ' “f. fl‘éi‘éfi "“fi ..‘atfv‘ifi‘ . . I :33. 'MW 52 -l xm -1 b1 5 89' Now g6 = (xmbl)e = (xm)e bl = y 3:" bi y xmbl y = y-;gy since y e 2 (Be). Thus as IG and so Al 3 IG' But since G is nilpotent IG 1 Al, hence A1 = IG. .One can say a little more about the groups 80’ as Al, in case G is an arbitrary finite group and e is a central automorphism. Theorem 3.12: Let G be an arbitrary group and let as A . If all elements of A1 are central then Be<<}. Proof: Let x a B6 and g e G. Then since x6 = x 6-9 6 we have that (g x 3'1)6 =.ge x g = $9 x 3' Since 6 is a central automorphism g'lges Zl so (g-l.ge)-l X (gulge) = x. Hence 8 x 8'1 = s9 x 8'9 = (a x s" and g x g-1 5 Be. So BeId G. l)9 Theorem 3.13: Let |G| = pn and |[G, Aljl = p. Then 0(G) i 21' £3393: If G' =(l)then the theorem is obvious. So assume G' +(1L Then IG 1 A1, so that G' i [G,Al]. But IEG,A1]| a p, thus G' = [G,Al]. Now since IG 1 Al and since 21 = 5:5 CG(a) it follows that 8A1Be 5.21- Let 0 be an arbitrary element of A.l and consider Be. Let G= = x1 B6L1x2 BBL) ° ° 'LJxr Be , x1 = 1, be a coset decomposi- tion of G relative to Be and consider the elements 1, x2fl x3 , - - -, XI,-l xre. These are all distinct and are con- tained in G'. Hence r = p and [Bel = pn-l, so ¢(G): Be for all as A1. Thus @(G) i 21' I o .‘ . I" - ‘ ..- O 'J ‘- a» - o ‘ '0 “-3 ‘{- O c ’0 ° 0 v. , 11 l - k . 6 ‘4 ;:r, , . ‘ L I: \_. 7' - '. Q l.‘ - - ‘ .4. o' _ 7“ } :9 1 r. 11 ‘Mo. A lil‘ c , ‘ ~ 0 -' ...1- ' | “ .I’. .‘I.:;. c _. - J . ‘ ‘ "'r»: .. ' _ .> . -. _._ 1 >’~_ _ — t, “ A- ,r- . | -z{—' -wr,; .. ‘4 v-Ij ..1 ‘h \é -1;:{..‘ - . . . . . . k . " ‘ t” . . ’ 1*. . ‘ ----‘l!"“. '1‘ .T‘ _ Lr’ 't?‘ ‘l '9,” *1 »,r ' r u .""b .“ . -.. . 1.. s ' 14% - ° ““1 1'“ ."v I ...; .IVI I 5 i2; nit-3w; - .. . 7 _ ‘.‘ . .;- ".ul” t ‘0 o. - ‘u ‘ . .iL‘ ..—- _ _ . ... .0 _ V .-,v‘.‘" ’.{.";.‘L.:' ”I." 35‘!" “‘1‘ _ . ~ ,_ _.' ' ” ‘ " " - " ‘ . ""‘if‘kr -.' . v 1. ,. .‘. . «W‘fltn A "" ‘--’i «'3 53 n Theorem 3.14: If IGI = p , n > 1, |zl| = p and all elements of A1 are central then Al is an elementary abelian p—group. Proof: Let a 5 Al and G = Ba!) x Bel) ' - - U xp-l Be , then x6 = xb for some b s 21 since 0 is a central 2 x bl for some Q and some automorphism. If g e G then g bl a B6 and 6 9. g = (xgbl)e = . . 9. 9. (x2)6 bl xfib bl = (xlbl) b = g b 6p 1p Thus it follows that g = g b = g and la] = p. So every element of Al is of order p. Let a be any other element of Al, say ga g H with b e 21' Then gae = (85 )6 = 895 = g bib = g 6 b“ = gab2 = (g bk)“ 6 Hence a6 = 6a and Al is elementary abelian. . _‘I!_» I ; . A. ”I . JIM" "P~O‘M,U.w‘ ’.'--'_'-;d'..'fi&§:’3-1 * -‘ waft-4 1...... .l a»... {~71 .43 .1. a. a I .... Wain 1L.» {fol-41!.1’ lit... ‘bH ._ if.‘ ‘1‘, L fi. ‘ .0 1'... .0- 3 - APPENDIX I. Examples Example A: G = , a27 = b3 = l , b-l ab = alo. Consider the chains 51: G > (a3, b) > (a3) >(a9) > (1) and 52: G > (a3, b) > (a3 b)> (a9) > (1). We have that A(G) = <¢, a, B,K>, where: ¢: a——+au a: a——+ab B: a——+a Y: a—-—>a26 b———+b b-—~+b b——>a9b b——*b From this we get that SO (51) = A(G) and 80(52) = (¢, a, 8) while sl m 52. We note that G also has the following chain: 8 G > (a3, b) > (a9,b) > (a9) > (1), 3: and 80(33) = A(G), Sk(s3) = (¢, a, 8) for k = l, 2, 3. More- over Sk(sl) = (¢, a, B) for k = l, 2, 3 , so Sk(sl) = Sk(83) for k = O, l, 2, 3. Finally it should be noted that all elements of A are 1 central automorphisms. l u Example B: G = (a, b) , a9 = b3 = l, b— ab = a The composition series of G are: G > (ab) > (a3) > (1), S 1: s2: G > (a) > (a3) > (1), 33: G > (a3) > (1), 54 55 G > > > <1). S“! SS: G > > (1). 86: G > (a3, b) > (a6b) > (1), 87: G > (a3, b) > (a3b> > (1). We have that A(G) = (Ia, lb, 0, ¢), where: Ia: a——+a Ib: a——+au a: a-+a5b ¢: a-—+ab b——+a6b b+—+b 'b——+b b——+b It follows that AC =‘(IB). Since Z(G) = (a3) and a"1 b-la follow that all elements of A are central since A i A b = a , 3 lb is a central automorphism. Thus it must 1 l o' 4 Example C: G = (a, b, x) , an = b = 3 = l, a2 = b (a, b) 2 III x ba_l, x_lax= ab, )5le = a3. ab We note that Q (x) + G and (a, b)<1 G. The composition series ‘0 3 for G are: 812 S22 S 3: G > (a, b) > (a) > (a2) > (1), G > (a, b) > (b) > (a2) > (1), G > (a, h) > (ab) > (a2) > (1), By using the inner automorphism of G induced by the element x we see that (K (G), CG) is transitive. Moreover the ’3 derived series is given by G, G' = (a, b) , G" = S X°°°x S > S x'°°x S > °'° > S x S > S > (1) is a normal series for G. Hence it follows that G is a Sylow tower group for the ordering < of the primes. Assuming G to be a Sylow tower group for any ordering of the primes immediately results in all Sylow subgroups of G being normal in G. Thus G is nilpotent. Theorem GElO]: If A is a subgroup of G such that A 3 G and (IAI, IG:AI) = 1, then A has a complement in G and all such complements are conjugate. Theorem H[ll]: If.G is an A—group then G'f\ Z(G) = (l). Theorem IE5]: If G is a Hamiltonian group then. G = Qn X A X B where Qr1 is a generalized quaternion group, A is an abelian group of odd order and B is an elementary abelian 2-group. “LgrO‘ #1.“. a). I .- . . " ’;}!ll'f ".1 . -l. I _ "t‘ ‘1“.' Ht: 'A“ 4 J. ‘ 1.10;"; a? * " I... 3.3- :1“! JP». ("‘3 l' 1“ figflfi‘l 'T;':i: 58 Theorem KLll]: If G is an A-group of derived length 2 then the Fitting subgroup of G is G'»Z(G). Theorem LES]: Given a group G no two consecutive factors of the series G' i G" : --- : G(r—l) i G = (1) can be cyclic unless they collapse into one factor. Remark A: Let G be an abelian p—group, p a prime, and let_ G = (x1) x "' x (xm) x °'° xm [ll

> + + n n _ ll (p -¢(p ))IT1 2] a where r is assumed greater than 1. Proof: Let y 5 (x2, -°' , xi) and let k be an integer such that l i k < pn and (k, pn) = 1. There are ¢ 0| ~‘ L. ' . 1" ._ ‘ 4 .- l‘( o I“ . b . .‘ O 'A— .' u!» 0‘" ‘J -' ' o :I v - .~ 1 ' V fq.i'M,. fiQ§$§1 avzsuosanoo ~2 . . e“:r " - A. . a ' \— . I. , I . Ir ‘ . ' . , .. . "' ,J .‘ . . . . 1 J , . -, .. .. . .. b J I ( I t v , ' " I. J l‘ ' v I Q, .i" ‘l‘ ‘ ' 8 - .' "ca -.‘i a ‘ -. - F . I" . ‘ . ‘ i 'J ~| . l'x- ‘ . k 6 .- ‘ . ' r, . ‘ ‘ . I. - I . , ' ‘ l..',,'f.1*""~ .1 I‘LL. I 1' ' '- ,Il (VII, - ' . ‘L . J" ' . " ‘ I... p" 71“ ,..' niv' ‘ " . .. - u '- * ' 1.— ‘ ‘. '— !:' 1' ' I i. . ' at“, II ' I I '1“: ' ‘ '5 ‘ ’—L'. ’9. '- ~~ an 10"?! f. gr - -3 r. ‘v- ' ' .‘- ' . - ' '3 ‘ b w” . ‘ ~vl- . I ._J 0‘: I 1 : ' ' ~ ’ I . . I ‘ 0' o | -1 ‘. yo; {.01" . Il Q ‘ . ‘ , — o . 0 .- _"r-\I|1 U,.. ‘ , 11!)“ . ' 1‘. ' . a .. 3“.“ a 98 at . My: “Mum" \ r I P "I g 3 . m (43 r . ' ' \ ' x) - If. ‘33»... :mo‘; . c- 4 > . ‘ ' _'. . "‘..1‘* ' '. . n . _ . (.34. .i..;bi ’ at; .: :nr- '1 1‘4 .. l 3953' . x . I!!! 9'19“? a; .1 1st. :.;-:-.‘ I A. 3:“) L16 3“. Vex? I-‘I";‘ ‘C ’1‘]? If! JLE-{;?:C¥":.“ 3““{3 “IIILI .J. v. ,3» C " I. 4' ...»I‘. " ' H. 0. : I ‘ ‘ A ”ml '-. . , - .O' I . r .'-K * ... ‘ : - Lira ...' .. Lu; ‘, .~ 'L-l‘. “Hid "' A r9“! ‘1‘ JV‘ 4”: "3 ,_v‘ . ,~ I . . . . ‘1 1 i . " ' _. F- * I." 'i‘.‘ 'A ‘ T‘" ' ' .. . . '.: “I it ‘ ”3% 3 gt‘ j I*{: Qfi‘: _ "In ‘ l' '1 - " ’_'\ . ' ' L " : _ 7-- WW”? _0 ' u ”i. "Hf-Q " , " _ :6.” 5"l"|1'. :’ .Ilo: " f In “I ‘- I 'l :‘ . ANTHIJH- , a I'M E IV cl}; “ II . . . . ' .. . _. ... . ,. u ,1 .' ' I ‘; .r: ".‘l O 0- VI : O . " :7 v .. "~"“‘ ' 4.. _ ‘ .‘2 a". ., I . .‘i- ,1. A 3.3 : a A 3'15: ..a ‘ 59 and consider the element x: y', there are ¢(pn)°|(x3, °°°, xi)l such elements and they are all of maximal order in H. Clearly they are not of the form xi y. Also, if h is any integer such that l < h < pn and (h, pn) + 1, then x? x; is of maximal order in H and there are (pn-¢(pn))¢(pn) such elements in H. More- k t I over these are not of the form xl y or x2 y Hence we see that there are exactly ¢(pn)-l