ON; THE AUTOMDRPHISM GROUP OF AN QNEGRAL GROUP RING fiissertation for the Degree of Ph. D. MICHIGAN STATE UNIVERSETY GARY LEE PETERSON 1974 This is to certify that the thesis entitled On the Automorphism Group of an Integral Group Ring presented by Mr. Gary Lee Peterson has been accepted towards fulfillment of the requirements for Ph.D. Mathematics degree in Major professor 0-7 639 .. 1., _\ .auox'mNnéR . UBRARY l .l llgmsrssféguigéll , Hm” ll' .. ABSTRACT ON THE AUTOMORPHISM GROUP OF AN INTEGRAL GROUP Rmc By Gary Lee Peterson Let G be a finite group and let Z(G) denote the integral group ring of G. The primary purpose of this dissertation is to study A(G), the group of ring automorphisms of Z(G). If f 6 A(G), we say that f is a normalized automorphism if f(g) has augmentation one for all g 6 G. The set of normalized automorphisms of 2(6) form a subgroup of A(G) denoted by NA(G). Further, little generality is lost by studying NA(G) over A(G). In Chapter I, we deve10p some basic facts about NA(G). It is noted that the elements of NA(G) act as a permutation group on the class sums of G, the representations of G, and the characters of G. Next, some subgroups of NA(G) are introduced. First we let CP(G) denote the subgroup of elements of NA(G) which fix every class sum of G. It is well-known that the action of an element of CP(G) is equal to conjugation by a unit in the group ring of G over the rationals. We also let EA(G) denote the subgroup CP(G)Aut(G), Aut(G) the automorphism group of G, and let I(G) denote the group of inner automorphisms of 2K6). In addition, a normalized auto- morphism which lies in EA(G) is said to have an elementary repre- sentation and if EA(G) = NA(G) we say that G is an E.R. group. Gary Lee Peterson The questions of when NA(G), EA(G), CP(G) are each equal to Aut(G) are considered in Chapter I. It was previously known that NA(G) = Aut(G) is equivalent to (l) I(G) s Aut(G), (2) G is either abelian or a Hamiltonian 2-group, and (3) 2(6) has only trivial units of finite order. Here we extend the list of equi- valences to (4) G is the only group basis of Z(G), (5) [NA(G) : Aut(G)] is finite, and (6) I(G) is periodic. Necessary and sufficient conditions are determined for Aut(G) to equal EA(G), CP(G), and I(G). Another topic considered in Chapter I is the following. In his Ph.D. Thesis (Michigan State University, 1971), C.F. Brown showed that Aut(G) has a normal complement in NA(G) when G is metabelian. For N 4 G, let A(N) denote the kernel of the natural map from Z(G) to Z(G/N) and set W(G,N) = {f E NA(G)\f(g) '=‘ g mod A(N)A(G)} . Then W(G,N) is a subgroup of NA(G). Further, if G is metabelian, 'W(G,G') is exactly the complement Brown obtained. This led to the question of when is W(G,N) a complement for Aut(G) in NA(G)? It is shown that if one wants W(G,N) rlAut(G) = 1, one in general needs N abelian. Then, if N is abelian, we obtain W(G,N) is a complement for Aut(G) in NA(G) if and only if G/N is either abelian or a Hamiltonian Z-group. Finally in Chapter I, we consider the following. Let N <1G. If f(A(N)) = A(N) for all f in a subset S of NA(G), we say N is S-admissible. If N is NA(G)-admissible we say that N is NA-characteristic. The admissibility of some subgroups of C under Gary Lee Peterson various subgroups of NA(G) is studied. We also obtain that N <1G is NA-characteristic if and only if it is characteristic in G pro- vided that G is either an E.R. group or contains an abelian normal subgroup A such that W(G,A) is a complement for Aut(G) in NA(G). One of the major goals of this dissertation was to determine E.R. groups. That is, when is EA(G) = NA(G), or alternately, when can every element f of NA(G) be written in the form f(x)=u(o(x))u-1 for all x€Z(G) where aeAut(G) and u is a unit in the group ring of G over the rationals? In Chapter II, part of Chapter IV, and in Chapter V, we consider this problem. It was previously known that (1) class s 2 nilpotent groups, (2) groups containing a cyclic normal subgroup of index p, (3) groups G where \G" = 2 or 3, (4) groups with at most one non-linear character, and (5) Sn for n = l,...,10 are all E.R. groups. Some of the E.R. groups obtained in Chapter II and IV are (6) groups G of the form G = AB where A is a cyclic normal subgroup of G and B is an abelian subgroup of G, (7) groups G where G/Z(G) is metacyclic, Z(G) the center of G, (8) groups G where \G" = p, (9) p-groups containing a maximal abelian normal subgroup which is cyclic, and (10) p-groups G of the form G =.AB where A is an abelian normal subgroup of G with A 2 Z(G) and A/Z(G) elementary abelian of order p2 and B is an abelian subgroup of G. Most of these results are obtained by studying the action of W(G,G') on the irreducible characters of G. Included in this is a crucial lemma on the faithful irreducible characters of a metabelian group developed in Chapter II. Gary Lee Peterson Finally, in Chapter V, we show that Sn is an E.R. group for any positive integer n. One other problem considered in Chapter IV is when is W(G,G') S.CP(G) for a metabelian group G? This was known to be true in cases (1), (3), and (4) of the previous paragraph. In Chapter IV, we extend this to cases (6), (7), and (9) of the previous paragraph provided that G is a p-group and p > 2. Chapter III is concerned with normalized automorphisms of direct products. Suppose G = G1 X...x Gn' It is shown that NA(G1) x...x NA(Gn) has a normal complement in NA(G) which lies in CP(G) provided (\Gil,|Gj\) = 1 for i # j. A similar result is obtained for W(G,A) if W(G,A) is a complement for Aut(G) in NA(G) for an abelian normal subgroup A of G. In the above two cases we also obtain that G is an E.R. group if each Gi is an E.R. group. In the final chapter, Chapter VI, a technique is presented for extending the known groups for which the integral group ring problem holds. Using this technique we solve the group ring problem for 84. However, the technique used here involves lifting auto- morphisms. The remainder of the chapter involves a discussion of the problem of lifting automorphisms. w THE AUTOMORPHISM GROUP OF AN INTEGRAL GROUP RING By Gary Lee Peterson A DISSERTATION Submitted to Michigan State University in partial fulfillment of the requirements for the degree of DCI‘. TOR OF PHILOSOPHY Department of Mathematics 1974 ACKNOWLEDGEMENTS I would like to thank Professor Joseph E. Adney for his patience and guidance in the preparation of this dissertation. ii Chapter II III TABLE OF CONTENTS INDEX OF NOTATION INTRODUCTION PRELIMINARY RESULTS ABOUT AUTOMORPHISPS OF Z(G) Section Section Section Section Section Section U19 6. Facts about Group Bases of Z(G) Basic Facts about Automorphisms of Z(G) Some Subgroups of NA(G) and Conditions for Equality with Aut(G) Admissibility On a Complement for Aut(G) in NA(G) Some Further Results on Admissibility SOME METABELIAN E.R. GROUPS Section Section Section Section Section Introduction The Action of W(G,G') Sums and Characters A Lemma on Faithful Characters of Metabelian Groups Groups Containing a Cyclic Normal Subgroup with an Abelian Supplement Groups in which G/Z(G) is Metacyclic on Class NORMALIZED AUTOMORPHISMB OF DIRECT PRODUCTS Section Section Section Section Section Mb Introduction The Containment of NA(Gl) x...x NA(Gn) in NA(G) When the Summands have Relatively Prime Orders When W(G,A) The Converse is a Complement iii Page 10 15 18 24 27 27 27 30 30 34 4O 4O 40 42 45 48 Chapter IV VI ELEMENTARY REPRESENTATIONS IN p-GROUPS Section 1. Introduction Section 2. p-Groups which are E.R. Groups Section 3. When is W(G,G') SLCP(G)? ELEMENTARY REPRESENTATIONS IN Sn A.LOOK AT THE GROUP RING PROBLEM Section 1. Introduction Section 2. Lifting and a Generalization of Theorem 1.1.6 Section 3. Groups of Solvable Length Three Section 4. Lifting Units BIBLIOGRAPHY iv Page 50 SO SO 58 61 68 68 68 71 73 77 INDEX OF NOTATION I. Relations: 3 Is a subset of < Is a proper subset of d Is a normal subgroup of 9' Is isomorphic to h E g mod S Means h +’S = g +'S where h and g can either be elements or subsets of a ring and S is an ideal in the ring. II. Operations: S/T Quotient ring or group. X Direct product of groups. («9 )3 Direct sum of rings. < > The group generated by (a,b) Greatest common divisor of a and b. alb a divides b. ‘3‘ Number of elements of a set S. \81 ll [G:N] Index of N in G. EX.y] x 1y-1xy xu u-lxu Tu(x) The map x a uxu- . L0!) The augmentation map defined by L(x) = 2 a where x = 2 a g is an 366 5 566 3 element of the group ring R(G). III. Groups, Rings, and Modules: G C 8 E' a [M] N" (n) 1“,n (G) Z(G) zn(c> §(G) CG(N) R(G) A(S) R(e) A finite group. The conjugacy class of g 6 G. The class sum of g E G defined by C; = Z x. xECg <[n,m]\n e N, m e M) [N,N] [N'.N'] The nth term of the derived series of G starting with C(l) = G. th . The n term of the lower central series of G starting with P1(G) = G. The center of G. th . The n term of the upper central series of G starting with 21(6) = Z(G). The Frattini subgroup of G. The centralizer of a subset N of G. The symmetric group on n letters. The group ring of G over the ring R. The 2-sided ideal of R(G) generated by 8-1, s E S, for a subset S of R(G). The ring R with e adjoined. vi z..2.e.z n P ker Q ker x The integers, rationals, complexes, and Z/pnz reapectively. The induced C(G)~module where M is a CKN)-module for a Subgroup N of G. M viewed as a CKN)-module where N is a subgroup of G and M is a CKG)- module. The induced representation of G where F is a representation of a subgroup N of G. The kernel of a homomorphism (p. The kernel of a character x of G. vii INTRODUCTION The study of the automorphism group of an integral group ring Z(G) has previously received attention by Brown in [3], Hughes and Pearson in [7], Hughes and Wei in [8], and Sehgal in [14]. The primary purpose of this dissertation is to study the automorphism group of Z(G). Another problem in integral group rings that has received considerable attention is the group ring problem. In Chapter VI, we will see how knowledge of automorphisms of Z(G) may play a role in solving this problem. In Chapter I, we obtain some preliminary results about auto- morphisms of Z(G). In studying the automorphism group of Z(G), it suffices to study the group of normalized automorphisms of Z(G) denoted by NA(G). It was first noted by Sehgal in [14] that if f E NA(G) and g E G, then f(Cg) = 6S1 for some g1 E G. Hence NA(G) acts as a permutation group on the class sums of G. Further, £1658) = 5326's) for all g e c if and only if £1 = Tufz for some unit u in 2(G). In Section 2 of Chapter I, we note that NA(G) also acts as a permutation group on the representations and characters of G. Finally, we use CP(G) to denote the subgroup {f e NA(G)\£((Tg) = 58 for all g e c}. In Section 3 of Chapter I, we introduce some subgroups of NA(G). First, the automorphism group of G, Aut(G), is naturally l embedded in NA(G) and hence is a subgroup of NA(G). Next, we say that f 6 NA(G) has an elementary representation if f = TuO for some unit u in 2(G) and some a 6 Aut(G). We use EA(G) to denote the set of all elements of NA(G) which have an elementary representation. Then EA(G) is a subgroup of NA(G) and in fact EA(G) = CP(G)Aut(G). Also, if EA(G) =NA(G), we say that G is an E.R. group. Finally, we let I(G) denote the group of inner auto- morphisms of Z(G). The question of when NA(G) equals Aut(G) was previously studied by Brown in [3] and Hughes and Wei in [8]. In Section 3 of Chapter I, this question is again studied and the previously known results are extended. We are also able to obtain necessary and sufficient conditions for Aut(G) to be equal to EA(G), CP(G) and I(G). Sections 4 and 6 of Chapter I focus on the question of when f(A(N)) = MN) for f E NA(G) and N <)G. This question received some attention by Brown in [3]. One of the major reasons for con- sidering this question here is that if f(ACN)) = A(N), then f in- duces a normalized automorphism on Z(G/N). For example, the above fact is useful in cases where one wishes to use induction on \G‘. In Section 5 of Chapter I, we consider generalizing another result of Brown's. Brown was able to show that Aut(G) has a normal complement in NA(G) when G is metabelian. Here we are able to generalize Brown's result by introducing some subgroups of NA(G), one of which is exactly Brown's complement when G is metabelian. Further, we are able to determine necessary and sufficient conditions for one of these subgroups to be a complement. Perhaps one of the outstanding questions in integral group rings is what groups are E.R. groups? This question was first studied by Sehgal in [14] and later by Brown in [3]. In Chapter II, part of Chapter IV, and Chapter V we study this question and determine several types of E.R. groups. For the most part, the previously known E.R. groups were obtained by studying the action of normalized automorphisms on class sums. In a similar manner, we use the action of NA(G) on class sums to show that Sn is an E.R. group in Chapter V. However, it is in general very hard to study the action of NA(G) on class sums. An alternate method is to study the action of NA(G) on the irreducible characters of G. This is the technique that we will use in Chapter II and part of Chapter IV in determining our list of E.R. groups. Included in this is a crucial lemma on the faithful irreducible characters of metabelian groups in Section 3 of Chapter II. It should be noted at this time that I know of no example of a group which fails to be an E.R. group. The other topic covered in Chapter IV consists of some cases in which the normal complement that we have for Aut(G) in NA(G) when G is metabelian lies in CP(G). Chapter III is devoted to studying normalized automorphisms in direct products. For the most part, the results of Chapter III were motivated by the following question: if G = G1 X...x Gn where each G1 is an E.R. group, then is G and E.R. group? Using the results of this chapter we can obtain a positive answer to the question when either (lGil,\Gj\) = l for i i j or when Aut(G) has a normal complement of the form described in Section 5 of Chapter I. We also show that the converse to this question always holds. Finally, we mention some open questions not considered in this dissertation: the relationships between the results concerning NA(G) and integral representations. For example, I(G) is always contained in CP(G), but there are examples where they are not equal. In [7], Hughes and Pearson obtain I(S3) = CP(S3) using integral representations. Perhaps integral representations may be used to study the question of how I(G) and CP(G) are related. Answers to this question may also have some connection with our remarks in Section 4 of Chapter VI towards solving the group ring problem. Con- versely, there may also be some applications of our results on NA(G) which would be useful in studying integral representations. CHAPTER I PRELIMINARY RESULTS ABOUT AUTOMORPHISMS OF Z(G) Section 1. Facts about Group Bases of Z(G). H is called a group basis of Z(G) if H is a group of units in Z(G) whose elements freely generate Z(G) and if ((h) = l for all h E H. In this section we record some results concerning group bases. The first result is due to Glauberman and its proof can be found in [12] or [16]. Theorem 1.1.1: Let H be a group basis of Z(G). Then for each h 6 H there exists a g E G such that 6% = C8 The next corollary follows easily from Theorem 1.1.1. Corollary 1.1.2: If H is a group basis of Z(G), then Z(H) = Z(G). Theorem 1.1.1, as shown in [12] and [16], yields a 1-1 correSpondence between the normal subgroups of G and those of a group basis H in the following manner. Let N14 G. Set K = U {h 6 H‘C£ = 6;}. Then it can be shown that K is a normal g€N subgroup of H. We state some facts concerning this correspondence from [16]. Theorem 1.1.3: Let H be a group basis of Z(G), N a normal sub- group of G, and let K be the correSponding normal subgroup of H. (i) Let n be the natural map from Z(G) to Z(G/N). Then K = {h E H\n(h) = l] and n(H) is a group basis of Z(G/N). (ii) A(N) ‘ A(K)- (iii) If M is another normal subgroup of G and if ‘L is the normal subgroup of H corresponding to M, then [L,K] correSponds to [M,N]. Another result concerning class Sums, whose proof can be found in [3] or [12], is the following: Theorem 1.1.4: Let H be a group basis of Z(G). Let h E H and let g E G such that 5% = Cg’ then an = an for every integer n. Finally, we state more results from [16]. Theorem 1.1.5: Let K be a normal subgroup of a group basis H of Z(G), then K/K' =- A(K)/A(K)A(H) under the mapping mk(k) = k - l + a(K)A(H) where k = kK'. Hence, if A is an abelian normal subgroup of G and if B is the correspond- ing normal subgroup of a group basis H of Z(G), then A.='B under the mapping qglqht Theorem 1.1.6: Let H be a group basis of Z(G), A an abelian normal subgroup of G, and let n denote the natural map from Z(G) to Z(G/A). If n(H) = n(G), then for each h E H there exists a unique gh E G such that h a gh mod A(A)A(G). Further, the mapping h a gh defines an isomorphism from H onto G. In the process of proving Theorem 1.1.6, Whitcomb actually showed the following result. Theorem 1.1.7: Let A be an abelian normal subgroup of G and let H be a group basis of Z(G). Suppose h ,h2 6 H such that l E d = , h1 h2 mo A(A)A(G), then h1 h2 Section 2. Basic Facts about Automorphisms of Z(G). We begin this section by reviewing some facts about automorphisms of Z(G). Let A(G) denote the group of automorphisms of Z(G). By NA(G) we will mean the subgroup of A(G) consisting of all f E A(G) such that L(f(g)) - l for all g E G. NA(G) is called the group of normalized automorphisms of Z(G). As remarked in [3], little generality is lost by studying NA(G) instead of A(G), since if f 6 A(G), the mapping g ~»L(f(g))f(g) for g E G extended linearly to Z(G) is in NA(G). Let f 6 NA(G). It then follows that f(G) is a group basis of Z(G) and that f(Cg) = 5' Thus, by Theorem 1.1.1, f(s)' f(Cg) = C; for some g1 6 G. Hence, we have that NA(G) acts as l a permutation group on the class sums of G. One of the basic facts about this permutation representation of NA(G) is the following result which is generalized from [14]. 1 g E G if and only if f1 = Tuf2 for some unit u in 2(G). Theorem 1.2.1: Let f ,f2 6 NA(G), then f1(5g) = f2(Cé) for all It should be noted that the unit u in Theorem 1.2.1 must necessarily normalize Z(G). That is, Z(G)u s Z(G). 'We also note that u need not be a unit in Z(G) as Whitcomb has given an example in [16] of an E E NA(G) such that f = Tu where u is a unit in 2(G) but u cannot be taken to be a unit in Z(G) when G is the dihedral group of order 8. Let CP(G) denote the kernel of this permutation representa- tion of NA(G) on the class sums of G, then by Theorem 1.2.1, CP(G) {f e NA(G)\f(C_g) = 58} {¢u\u is a unit in 216) normalizing Z(G)}- Also, note that \NA(G)/CP(G)\ is finite since NA(G)/CP(G) is isomorphic to a subgroup of SR where k is the number of conjugacy classes of G. We next state two lemmas on how NA(G) acts on class sums. Lemma 1.2.2: Suppose f e NA(G) and that f(Eg ) = 5g and -— ._ l 2 f(083) = Cg4 where g1,g2,g3, and g4 E G, then there exists x E G such that f(C ) = G x. 8182 8384 Proof: We have that f(C. CI ) = G- C- . g1 g2 g3 84 Now, the class suni f(Cg g ) will appear as a summand in l 2 f(C C ) when f(C- C. ) is written as a linear combination of g1 g2 g1 g2 class sums. Also, 6; C. is a linear combination of class sums of ._ 3 84 the form Cg3xlg4X2 where x1,x2 E G. Hence, there exist x1,x2 E G such that f(C ) = _ E '1 gls2 Cg3xlgax2 = g3s4x2x1 and so we have the result. The second lemma appears in [3] and follows directly from Theorem 1.1.4. ‘Lemma 1.2.3: Let f E NA(G) and suppose that f(Cg) ==Cg1 where 8,81 E C, then f(an) =:Cg n for every integer n. Further, l \gl = lgll' NA(G) can also be viewed as a permutation group on the representations or characters of G. For if F is a representation f of G, we can then define another representation F of G by setting Ff(g) = F(f(g)). (For the purposes of this dissertation we will assume that our representations are over the field of complex numbers.) Similarly, if X is a character, we can define another character Xf by setting X;(g) = X(f(g)). Further, it follows that Ff or Xf is irreducible if and only if F or X is irreducible. Lemma 1.2.4: If f(C?) ==Cé , then Xf(g) = X(g1) for any character 1 X of G. f l f ‘ l _ Proof: X (g) = -- "X (C ) = -——_'XKC ) = X(8 )- —— \c,\ g \c,\ s, 1 Lemma 1.2.5: If X is a faithful character of G, then Xf is also faithful. f f f ‘- - Proof: If g E ker X , then X (g) = X (1). Let f(Cg) = C8 , then 1 x(l). Hence g1 E ker X, so g1 = 1. Thus 3 = 1 f x(sl) = x (1) and we are done. Lemma 1.2.6: Let f ,f 6 NA(G), then 1 2 f f (i) f1(Cg) = f2(Cg) if and only if X 1(g) = X 2(g) for every irreducible character X and f f (ii) 13168) = £268) for all g e c if and only if x 1 = x 2 for every irreducible character X. g%oo_f: (of Let 51(ag)=e'g1 and £263) =c-gz, then X 1(g) = X 2(g) for every irreducible character X if and only if X(g1) = X(g2) for every irreducible character X. But X(g1) = X(g2) for every irreducible character X if and only if Cg1 ==ng and so we have (i). (ii) follows directly from (i). 10 Section 3. figgg Subgroups of NA(G) and Conditions for Equality with Aut(G). Let Aut(G) denote the automorphism group of G, then Aut(G) is naturally embedded in NA(G) by extending every group automorphism linearly to Z(G). we will henceforth use Aut(G) to denote the image of Aut(G) in NA(G) under this embedding. Another type of normalized automorphism which will play an important role in what follows is contained in the following defini- tion first used in [3]. Definition: Let f 6 NA(G). We say that 'f has an elementary representation if f = Tug where o E Aut(G) and u is a unit in 2(6) normalizing Z(G). Let EA(G) denote the set of f 6 NA(G) whose elements have an elementary representation. Then EA(G) is a subgroup of NA(G) and EA(G) ==CP(G)Aut(G). If EA(G) = NA(G) we will say that G is an E.R. group. One basic result concerning elementary representations and group bases is contained in the following lemma. Lemma;l,3.l: Let H be a group basis of Z(G) such that H‘” G. Then the following are equivalent: (i) Every f 6 NA(G) such that f(G) = H is in EA(G). (ii) There exists f E EA(G) such that f(G) = H. (iii) There exists a unit u in 2(G) such that Gu = H- M: (i) =9 (ii) is clear. For (ii) a (iii), let f 6 EA(G) such that f(G) = H. Write f = Tug where u is a unit in .2(G) and a €.Aut(G), then -1 f(c) -= Tuo(G) = Tu(G) = c“ = a. 11 For (iii) =’(i), let f E NA(G) such that f(G) = H. Since G‘1 = H we have Tuf(G) = G. Hence, Tuf 6 Aut(G) and so f = Tu_1(Tuf) E EA(G). One final subgroup of NA(G) that we will introduce here is the group of inner automorphisms of Z(G) which we will denote by I(G). That is, I(G) = {Tulu is a unit in Z(G)}. Then I(G) SCP(G). But in general they are not equal, as we have seen in the remarks after Theorem 1.2.1. We now turn our attention to the question when are the sub- groups of NA(G) that we have so far defined equal to Aut(G). The first question that we consider is when does Aut(G) =‘NA(G)? This question was first partially solved in [3] and later completely solved in [8]. In fact, an even stronger statement can be made con- cerning this question than was made in [8]. In the next theorem, parts (1) - (4) appear in [8]. Theorem 1.3.2: The following are equivalent. (1) NA(G) =.Aut(G). (2) I(G) s Aut(G). (3) G is either abelian or a Hamiltonian 2-group. (4) jg,g 6 G, are the only finite units of Z(G). (5) G is the only group basis of Z(G). (6) [NA(G):Aut(G)] is finite. (7) I(G) is a periodic subgroup of NA(G). Since (1) - (4) are equivalent and since clearly (4) a (5) =1(6) = (7), it suffices to show (7) =1(3). To do this, we use basically the same proof as was used in [8] to show (2) a (3). 12 We begin with the following lemma. The technique used in this lemma first appeared in [6]. Lemma 1.3.3: Let g1,g2 6 G and n be a positive integer. Let 2 k- R = l +tg1 +g1 +3..+ g1 1 where k = lgll and let P = g2(l - g1). Then 1 - nRP is a unit in Z(G). Further, if T 82 for n 2 2, then g1 is a power of 31. l-nRP E Aut(G) Proof: Since PR = 0, it follows that l - nRP is a unit with in- verse 1 + nRP. Now, suppose 6 Aut(G) and n 2 2, then Tl-nRP (1 - nRP)g1(l +- nRP) 6 G. But since (1 - nRP)g1(1 +1 nRP) = 31 + n(RP - RPgl), it follows that RP - RPg1 = 0 since g1 + n(RP - RPgl) 6 G and since n 2 2. Thus k-l k-l (g2 + glgz +-. + g1 g2) - (3231 +...+ 81 3231) RP RPs 1 k-l 2 k-l 2 = ... - ...+ . (8281 + 313281 + + 31 szsl) (32g1 +- $1 szsl) It follows from the above equation that g2 = gigzg1 for some i. Thus gzglgg1 = g11 and we are done. we now prove Theorem 1.3.2 by showing (7) = (3). To do this, we assume G is not abelian and show that G is a Hamiltonian 2-group. We first show G is Hamiltonian. To do this it suffices 8 to show that g12 is a power of g1 for every g1,g2 6 G. Let P and R be as in Lemma 1.3.3. Since we are assuming I(G) is periodic, we have n 6 Aut(G) for some integer n. Further, we may (T1_RP) assume n 2 2. 13 Since PR = 0, it follows that (l - RP)n = 1 - .nRP. Hence 8 (T1 RP)n = Tl-nRP 6 Aut(G) and so g12 is a power of g1 by Lemma 1.3.3. Thus G is Hamiltonian. Now, suppose that G is Hamiltonian but is not a 2-group. Let a and b be generators of the quaternion group of order 8 where a4 = b4 = 1 and ab = a3. Let g E G of the form g = as where s is an element of G of odd prime power order p. Finally, for every positive integer d let ed denote a primitive dth root of unity. We have that 2() is isomorphic to 69 2 2(ed) under d‘4p the mapping 9 where e(g) = z e . Let R =<® 2 2(3 ). It d d d 4p d\4p follows that e(Z()) s R an we can find a positive integer m so that mR s e(Z()). By the Dirichlet Unit Theorem ([11], p. 128), we can find a unit v in Z(e4p) such that v1 is not in Z(eip) for any integer 1. Also, we can find an integer k so that l +-l +...+ 1'+ vk is a unit in 9(Z) since the ring R/mR is finite. Let u = e (l +...+ l + vk), then Tu 6 I(G). Since I(G) is periodic, we can find an integer n so that run is the identity. we will now show that this forces vkn to be in Z(gip), and hence obtain a contradiction which will prove the result. Since un 6 Z(), we may write un - a‘+ g5 where 0:9 6 Z(). But then b(a +'85) = (a'+ ge)b. It follows that (l - a2)3 = 0. Hence a = (l + a2)c where a 6 ZC<32>). Thus, un = a + a(1 + a2)os = f(gz) + (gp + g-p)h(g2) where f and h are polynomials over 2. 14 Thus, if ‘1 denotes the projection from 63 2. 2(e d‘4p d) onto Z(gap) , we have 2 - 2 Wu“) map) + (szp + 64:)8(34p) f(gip) vnk . Hence vnk 6 Z(eip) and we have our contradiction. We next treat the question of when does EA(G) = Aut(G)? Clearly, if NA(G) = Aut(G), then EA(G) = Aut(G). (Conversely, if EA(G) = Aut(G), then (2) of Theorem 1.3.2 holds and hence NA(G) = Aut(G). Thus we have Corollagy 1.3.4: EA(G) = Aut(G) if and only if NA(G) = Aut(G). Finally, we answer the question of when Aut(G) equals CP(G) or I(G). In fact, we can state the following. Corollary 1.3.5: The following are equivalent. (1) CP(G) =.Aut(G). (2) I(G) = Aut(G). (3) \G‘ = l or 2. 2522;; We first note that (l) = (3). For if CP(G) = Aut(G), then EA(G) = Aut(G). But then G is either abelian or a Hamiltonian 2-group by Corollary 1.3.4 and Theorem 1.3.2. But all abelian groups of order greater than 2 and all Hamiltonian 2-groups have non-class preserving automorphisms. Hence \G‘ = l or 2. Similarly (2) a (3), since if I(G) = Aut(G), (7) of Theorem 1.3.2 holds. Thus G is abelian or a Hamiltonian 2—group. But then again, \G\ = l or 2 since Aut(G) has non-class preserving automorphisms except when \G‘ = 1 or 2. 15 Conversely, if (3) holds, then both (1) and (2) hold since Aut(G) = 1, I(G) B l, and CP(G) = 1. Section 4. Admissibility. The purpose of this section is to define what we mean by a normal subgroup of a group being admissible under a set of normalized automorphisms. If S is a subset of Aut(G) and if N is a normal Subgroup of G admissible under 8, then every automorphism of S induces an automorphism on the quotient G/N. In defining admissibility for normal subgroups of G under a set of normalized automorphisms, we define it in a manner so that we induce normalized automorphisms on the quotient. Defigition: Let N + x) since f1 induces the identity on Z(G)/A(N)A(G). Hence, f-1f1f(g) = g + f") is isomorphic to (D Z .2(ed) where n - \gl under the mapping (9 d n defined by tp(g) 8 IL, ed. Similarly, 2() is isomorphic to d‘n (432 .2(sd) under the mapping 9 defined by 9(g) = 2 e . d d‘4p d‘4p Extend n toa mapping from 2(6) to 2(G/N). Then 1162()) = 2(<§>). Further, it follows that .2(ed) if d 5 4p 81129-1C2(€d)) = (*> 0 if d > 4p To see (it), first note that amp-1Q(ed)) is either some 2(ek) where k\4p or is 0 since it mist be a minimal ideal in 69(1):4 .2(ed) and since .‘2_( ed) , d‘4p, are the unique minimal ideals of'g) 8 2(ed). Further, if encp-1(.2(ed)) 9‘ 0, amp-'1 restricted to 2(2),?) is an isomorphism since .2(ed) is a minimal ideal. Thus, amp-laud» = 0 if d >4p. Also, amp.1 restricted to 69 2 2(3 ) d‘4p d is then an onto isomorphism. Then, since -1 - errep (2?. ed) = 911(8) = 9(8) = 8 ed . d‘n d\4p it must be the case that emp-1(.2(ed)) = .2(ed) when d s 4p. Now, let v be a unit in Z(eép) such that no power of v lies in Z(ez). Let u = 2 5 ,where 6 =1 if d$4p and 6d = v if d = 4p. Then, as in the proof of Theorem 1.3.2, we can find an integer 111 so that u“1 is a unit in tp(Z()). Then amp-I(um) will be an element of the form 1 +. ..+ l + v', where v' is a unit in Z(e4p) such that no power of v' lies in Z(‘:p)° Further, we have that l +...+ 1 + v' will lie in 23 8(Z()). Let w = 6-1(1 +3..+ 1 + v'). Then, as in the proof of Theorem 1.3.2, no power of Tw can be the identity or else some power of v' would lie in Z(eip)° Let w = m-1(um) and set H = Gw. Then n(H) = n(Gw) = 1'r(G)'3 # n(G), a contradiction. Hence, G must be a Hamiltonian 2-group. We now can state Corollary 1.5.5: If W(G,N)Aut(G) = NA(G) where N 4 G, then G/N is either abelian or a Hamiltonian Z-group. 2529;; Let H be a group basis of Z(G) such that H is isomorphic to G. Let f 6 NA(G) such that f(G) = H. Then, f = f o where 1 f1 6 W(G,N) and 0'6 Aut(G). Thus, f1(G) = H. Since f1(g) a g mod A(N)A(G), n(H) = n(G) where n is the natural map from Z(G) to Z(G/N). Thus, G/N is either abelian or a Hamiltonian 2-group. we can now state when W(G,A) is a complement for Aut(G) when A is abelian and normal. Corollary 1.5.6: Let A be an abelian normal subgroup of G. Then W(G,A) is a complement for Aut(G) in NA(G) if and only if CIA is either abelian or a Hamiltonian 2-group. 2322;; If W(G,A) is a complement for Aut(G), then G/A is either abelian or a Hamiltonian 2-group by the previous corollary. Conversely, if G/A is either abelian or a Hamiltonian 2- group, let f 6 NA(G). Set H = f(G). Then by Theorem 1.3.2, n(H) = n(G) where n is the natural map from Z(G) to Z(G/A). Hence by Theorem 1.1.6, there exists f1 6 NA(G) with f1(G) = H Such that f1(g) a g mod A(A)A(G). Then f1 6 W(G,A) and 24 filf 6 Aut(G). Hence f = f1(f;1f) 6 W(G,A)Aut(G). Thus W(G,A) is a complement by Corollary 1.5.3. Another question that arises here is when is W(G,N) a supplement for Aut(G) in NA(G)? That is, when is W(G,N)Aut(G) = NA(G)? Corollary 1.5.5 gives us necessary conditions on G/N for this to occur. However, I have been unable to determine necessary and sufficient conditions for W(G,N) to be a supplement for Aut(G), although we will see one sufficient condition in Section 3 of Chapter VI. Section 6. ngg Further Results on Admissibility. In Section 4 of this chapter we remarked that NA-characteristic implies characteristic and that the converse is true for E.R. groups. In this section we will see that the converse is also true if G contains an abelian normal subgroup A such that W(G,A) is a complement for Aut(G) in NA(G). Suppose G is a group containing an abelian normal subgroup A such that W(G,A) is a complement for Aut(G) in NA(G). Let f 6 NA(G) and write f = f o where o 6 Aut(G) and f1 6 W(G,A). 1 If N is a characteristic subgroup of G, mm) = flo(A(N)) = f1(A(o(N))) = f1(ACN)) . <*) Hence, if we could show that every normal subgroup of G is W(G,A)- admissible, it would follow that every characteristic subgroup is NA-characteristic by equation (*), and so we show this. In fact, we will see that every solvable normal subgroup is W(G,A)-admissible regardless of whether or not W(G,A) is a complement for Aut(G). 25 We begin by showing that every solvable minimal normal sub- group is W(G,A)-admissible. Lemma 1.6.1: Let A be an abelian normal subgroup of G and let 'M be a solvable minimal normal subgroup of G, then M is W(G,A)- admissible. ‘ggggf: Since M is solvable, M is abelian. Further, M n.A = l or M $.A and hence MA is abelian. Now, let f 6 W(G,A), g 6 M, and let B be the abelian normal subgroup of G correSponding to f(M) in f(G). By Theorem 1.1.5, there is a b 6 B such that f(g) - 1 a b - 1 mod A(B)A(G) Also, B is a minimal normal subgroup of G and so BA is abelian. Thus, since b E f(8) mod A(BA)A(G) '=' 8 mod A(BA)A(G) , b = g by Theorem 1.1.7. Hence, M = B and so A(M) = ME) = A(f(M)) = f(A(M)) This completes the proof. we now can prove Lemma 1.6.2: Let A be an abelian normal subgroup of G and let N be a solvable normal subgroup of G. Then N is W(G,A)-admissible. 2592;; ‘Let f 6 W(G,A) and let M be a minimal normal subgroup of G contained in N. Set G = G/M, A = AM/M, and N = N/M. Then by the previous lemma, f induces an automorphism f of Z(G) with f 6 W(Glxb. Hence f(A(N)) = A(N) by induction on ‘6‘. Therefore, 26 MN) mod A(M) . A(N) + A(M) . f(MND or f(A(N)) IA Thus, f(A(N)) = A(N) since A(M) 5 MN). We now can prove Theorem 1.6.3: Suppose A is an abelian normal subgroup of G Such that W(G,A) is a complement for Aut(G) in NA(G) and let N <1G. Then N is NA-characteristic if and only if N is a characteristic subgroup of G. 2592;: By our previous remarks we only have to show that N is W(G,A)-admissible. Now, by Corollary 1.5.6, G is solvable so N is solvable. Hence, N is W(G,A)-admissible by Lemma 1.6.2. As an immediate corollary one should note Corollary 1.6.4: Let G be a metabelian group and let N 1, let N = ker X. Since N is W(G,G')- admissible, f induces an automorphism on Z(G) where G'= G/N which is in w(E,E‘). It then follows that Xf(g) = X(g) by in- duction on \G‘. If ker X = 1, let x 6 G such that gx # g. Then, X(g) = X(sx) = xx<[s.x1)/x(1> since [g,x] 6 Z(G). Hence X(g) = 0. Similarly, xf(g) = 0 since Xf is also faithful. Thus, Xf(g) = O = X(g) and we are done. It should be noted at this point that Lemma 2.2.1 immediately gives us a result of [16]; namely that W(G,G') s CP(G) when G has nilpotence class 5 2. Further, we then also have the result of [14] that class s 2 nilpotent groups are E.R. groups and so we state this as a corollary. Corollary 2.2.2: Let G be a class 5 2 nilpotent group. Then W(G,G') s.CP(G) and so G is an E.R. group. We also remark that if X is a linear character of G, then Xf = X for all f 6 W(G,G') since elements of W(G,G') in- duce the identity on Z(G/G') and since G' s ker x. From this observation, we get an easy proof of the following result of [3]. Theorem 2.2.3: If G has at most one non-linear irreducible char- acter, then G is an E.R. group. 29 nggf: By the results of [15], G is metabelian. Thus, if X is an irreducible character and if f 6 W(G,G'), Xf = X since all linear characters are fixed by f and since the non-linear character must be fixed if one exists. Thus, W(G,G') SLCP(G) by Lemma 1.2.6 and we are done. We conclude this section with two lemmas, the first of which also appears in [3]. Lemma 2.2.4: Let f e W(G,G'). If f(Cg) = Eaf then gl 6 gG'. nggf: Since f(C?) = C. a C; mod A(G')A(G), we have 81 C8 mod A(G'). But then, g1 E gx mod A(G') for some x 6 G 5' 81 and so g1 = gxy for some y 6 6'. Hence, g1 = g[g,x]y 6 gG'. The second lemma tells us that we can pick the g1 of Lemma 2.2.4 so that g1 6 gF3(G). Lemma 2.2.5: Let f 6 W(G,G'). Then there exists an x8 6 F3(G) such that f(C ) = C . 8 gxg Proof: Let G = G/F3(G). Since F3(G) is NA-characteristic, f induces an automorphism of Z(G) which will be in W(G;G‘). Let f(Cg) =:Cé . Since G- has nilpotence class s 2, we 1 have _ =_ 56 d . f(Cg) Cg1 g mo A(P3(G)) Therefore, g1 = gxy for some y 6 F3(G) and x 6 G. Then 6 =€x =6.*, 81 8 Y 8y x-l and so we have the result with xg = y 30 Section 3. A Lemma on Faithful Characters of Metabelian Groups. In this section we will obtain a useful lemma about normal subgroups from which a faithful character is induced in a metabelian group. Let A o G, let X be an irreducible character of G, and let M be an irreducible module affording X, As shown in §50 of is a homogeneous component of M and if 1 A * * A = {g 6 G‘ng = M1], then M1 is an irreducible CKA )-module [a], if M G If in addition, A is abelian and contains G' we can state * the following result about A when X is faithful. Lemma 2.3.1: Let X be a faithful irreducible character of G and let A be an abelian normal subgroup of G containing G'. Then A* st: C(A) ' * Proof: ‘Let F denote the irreducible representation of A afforded by M1. Since M1 is a direct sum of isomorphic CKA)-modules and since A is abelian, it follows that F(a) is a scalar matrix for * all a 6 A. Thus if g 6 A , F(ag) = P(ga) for any a 6.A. * , * , G G Since A 2 G , A <|G and it follows that F (ga) - F (ag) * for all g 6 A and a 6 A. But TC is a faithful representation * of G. Hence,A SCG(A). Section 4. Groups Containing a Cyclic Normal Subgroup with an Abelian Supplement. In this section we begin by showing that if G = BA where B is abelian and A is a cyclic normal subgroup of G, then G is an E.R. group. This result was originally motivated by trying to extend a result of [3], that any group with a cyclic normal aub- group of index p is an E.R. group, to metacyclic groups. 31 Theorem 2.4.1; Suppose G = BA where A is a cyclic normal sub- group of G and B is abelian. Then G is an E.R. group. Proof: Let A = u Note that G' s.A, and hence G is metabelian. Let f e W(G,G') , then f(C—a) = E S where as e aG' by a Lemma 2.2.4 and (s,\a‘) = l by Lemma 1.2.3. Let g 6 G and write g = ba , b 6 B. We define a mapping a of G by setting To see that o is well-defined, we first note that if ak 6 Z(G), then ak8 = ak. This follows since f(ak) = ak by Lemma 2.2.1 and so — — _ k _ k a a . i j Hence, if bla = bza where b1,b2 6 B, we have b1b21 = a"-1 6 B rlA s Z(G). Therefore, i _ is _ j-i is C(bla ) — bla - bZa a _ js-is is _ js _ j - bza a -— bza - g(b2a ) and so a is well-defined. o is also a homomorphism, since if b ,b2 6 B and if 1 ik+j b aulcms 1 2 C(blbza ) = b is s i bla bzaj = 0(b1a )o(b2aj) Finally, 0 is an automorphism of G, since if a(bai) = 1, then bais = 1. But then a18 6 B FlA 5,2(G), so a1 6 Z(G) since (s,\a\) = 1. Thus, a18 = a1 and hence bai = l. 32 If we can show that for any a defined in the above manner, X = X? for every irreducible character X ‘we will be done, so we show this. We also remark at this time that o is an automorphism of G which has been constructed to agree with f on the class sums of elements of A by Lemma 1.2.3. Suppose ker X > 1. Let M be a minimal normal subgroup of G contained in ker X. Set G.= G/M. It follows that f induces an automorphism on Z(G) by Lemma 1.6.2 which will be in W(G,Gw). We claim that 0' also induces an automorphism on G: To show this we show o(M) = M. If M slA, then clearly 0(M) = M. Suppose M t A, then M n A = 1. Further, [M,G] s M nA = 1 and so M s Z(G). Thus, if ba1 6 M where b 6 B, we have ai must commute with every element of B and hence a1 6 Z(G). Thus, a1 = a18 whence a(bai) = bais = bai. Hence, O'M = l and so 004) = M- Thus, f and a both induce automorphisms on Z(G). Further, if f and 5 denote these induced automorphisms on Z(G), we will C s where a have f(C?) aM and that 5 is defined in the same manner as a is on G. Hence, by induction on ‘G\, f(Cé) = 56g) for all g 6 G and so Xf = x0. Now, suppose ker X = 1. Let A? be as in the setting of Lemma 2.3.1. Then X is induced from an irreducible character of A* and A* sCG(A). Note that if g 6 c - A*, then x(g) = 0. Also, {(3) = o for g 6 G - A* for let f(Cg) = C; . Then 31 6 gG' and hence * l 31 6 A . Therefore, 33 f X(g) =x(31)= 0- * Similarly, one sees that X°(g) = O for g 6 G - A since o(g) 6 gA, * X0 on G - A . so 0(g) 6 A*. Hence, Xf Finally, suppose g 6 A*. If g 6 A, then X[(g) = Xo(g) since f and 0’ agree on class sums of elements of A. If g 6 A* - A, write g = bai where b 6 B. Since A* s.CG(A), we have [b,a] = l and hence b 6 Z(G). Thus, f(C 1L> = f((5 i)b) = (£6 1)» = (5 is» = o<€ i> ba a a a ba Therefore Xf(g) = X°(g), and the proof is complete. In the following corollary, we mention some groups which satisfy the hypothesis of Theorem 2.4.1. Corollary 2.4.2: The following are E.R. groups. (1) Any metacyclic group. (2) Any group which contains a cyclic Hall subgroup containing G'. (3) Any group which contains a cyclic Hall subgroup A such that CG(A) =A. (4) Any Frobenius group with cyclic Frobenius kernel. 3323;: (1) clearly satisfies the hypothesis of Theorem 2.4.1. For (2), let B be a complement of the cyclic Hall subgroup A 2 G'. Then, B' 5.3 FlG' = 1 and hence B is abelian. Thus, one can apply Theorem 2.4.1. In (3), let B be a complement for A. Then B is isomorphically contained in Aut(A) since CG(A) =.A. But Aut(A) is abelian since A is cyclic. Hence B is abelian, so 6' 51A. Now apply (2). 34 Finally in (4), one merely notes that if A is the Frobenius kernel, A = CG(A). Thus one can apply (3). In [3] Brown showed that if lG" = 2 or 3, then G is an E.R. group. This would lead one to wonder whether G is an E.R. group when ‘G'\ = p. By combining Corollary 2.2.2, Theorem 2.4.1, and the following lemma from [10], whose proof we include for the sake of completeness, we can obtain this result. Lemma 2.4.3: Suppose G' is a p-group. Let K be a p'—Hall sub- group of G. Then (i) K is abelian and (ii) there exists a subgroup Y of G such that G = KYG', Y is a p-group, and [K,Y] = l. nggf: (i) is easy since K' s K.F\G' = 1. (ii) Since K is a Hall subgroup of KG', = ' = ' = c NG(K)KG NG(K)G KP where P is the p-Sylow of G. Also, NG(K) = KCNG(K) nP). Let Y =NG(K) 0P. Then, G = KYG' and [K,Y] S K n? = 1. Corollary 2.4.4: Suppose \G" = p. Then G is an E.R. group. nggfg Let K. and Y be as in Lemma 2.4.3. If Y rlG' = 1, apply Theorem 2.4.1 with B = KY, A = c'. If Y (16' ii 1, then K s Z(G). It then follows that G is nilpotent of class s 2, so apply Corollary 2.2.2. Section 5. Groups in which G/Z(G) is Metacyclic. We have already seen that any metacyclic group is an E.R. group. In this section we will show that if G/Z(G) is metacyclic, then G is an E.R. group. 35 We begin with the following lemma in which parts (2) and (3) are generalized from [5] and part (4) appears in [5]. We will not need part (4) in this section, although it will be used in Chapter IV. Lemma 2.5.1: Let A be an abelian normal subgroup of G containing Z(G). (1) If G = AB where B is abelian, then CG(A) = A. (2) If CIA is cyclic, then \c'\ = \A/2(c)\. (3) If CIA and A/Z(G) are both cyclic, then G' = <[x,a]> where x generates CIA and a generates A/Z(G). (4) If, in addition to the hypothesis of (3), G is a p-group where p > 2, then \c'\ \A/Z(c)\ = \G/A‘ . Egggf: In (1), let g 6 CG(A). Write g = ba where b 6 B, a 6 A. Then b 6 CG(A), so b 6 Z(G). Hence b 6 A, so g 6 A. For (2), first note that if x generates G/A, the mapping g a [g,x] is a homomorphism of A onto G' ([9], Aufgabe 2, 8.259). Moreover, the kernel of this map is Z(G). Hence \A/Z(G)‘ = \G". In (3), one first notes that since a commutes with [x,a], [x,a]m = [x,am] for every integer m. Thus, \[x,a]\ = \A/Z(G)\ = ‘G'\ and we are done. Finally,for (4), let p“ = \c/A\ and am = lA/Z(G)\. Since G' is cyclic, G is a regular p-group ([9], Satz 10.2c.), S.322). m m Thus, [x,a]p = 1 = [xp ,a] ([9], Satz 10.6b.), 3.326) and hence 36 and hence pn 3 pm. Again by Satz lO.6b.), we have [xpn,a] = 1 = [x,a]pn. Thus pm s pn and we are done. Theorem 2.5.2: Suppose G/Z(G) is metacyclic. Then G is an E.R. group. 2529:; Let x, a 6 G such that x generates CIA and a generates A/Z(G) where A is a normal subgroup of G containing Z(G) with A/Z(G) and CIA cyclic. Also, suppose aX = arzx where 2x 6 Z(G). Then, [a 9x] = ar-lzx. Let f 6 W(G,G') and suppose f(C ) =»C' where a 6 G'. a aa1 1 Then, _ r-l k _ k(r-l) k a1 (a zx) a 2x for some integer k by Lemma 2.5.1(3). Also, note that if an 6 Z(G), n then 81 = 1 since f(Cn)=Cnn=f(a)=a a aa 1 Further, H CH f(Ci) =f((€i)z)=(511)z az a 8.31 aaz for any z 6 Z(G). Let g 6 G. Write g = xlajz where z 6 Z(G). We define a mapping a on G by setting c(g) = xiajaiz and claim that this is a well-defined automorphism of G. To see that o is well-defined, suppose xlajz1 = xLamz2 where e Z(G). Then xi" 6 A, so xi" 6 Z(G). It then zi’zz follows that am.J is also in Z(G). Thus, am-1 1 = l and so 37 o(xiajzl) = x a jaiz 21 = xiaja'i'z1 = xLamaTz2 = o(xLamzz) . It follows that o is a homomorphism, for let xiajz1 and xbamzz, 21 and 22 6 Z(G), be any elements of G. Then, i j L m i+L jr sz(rb 1+ ..+r+1) m o(x a zlx a 22) = o(x a 2x zla 22) = Xi+LaerHirLZi (r4‘1+...+r+1) jk a 2x z . L L L = ajk(r-l)r zikr = ajr ’ so 0 is indeed a homomorphism. Finally, a is an automorphism, for suppose q(x1an) = x iaja jz = 1. Then X1 6 A, so x1 6 Z(G). Thus, aja‘l' = (a1-'-k(r-1)z::)j 6 Z(G) , :i(""‘(r"))j 6 2(a). But (1 + k(r-l), \A/zmm = 1, for let f denote the automorphism f induces on Z(G/Z(G)). We have f(é'g) = 550+ k(r - 1)) 38 where a = aZ(G). Thus, (1 + k(r-1), \A/Z(G)\) = 1 by Lemma 1.2.3. Hence \A/Z(G)‘\j, so that aj 6 Z(G). Therefore, aj = l and hence 1 ii 13 o(x a z) = x a z = 1. Thus, a is 1-1 and we are done. Next, we claim that for any a defined in the above manner, f X = X0 for every irreducible character X, from which it follows that G is an E.R. group. If ker X > 1, let M be a minimal normal subgroup of G contained in ker X and let G'= G/M. Then f induces an auto- morphism on 2(5) by Lemma 1.6.2 which we will denote by E. It also follows that 0(M) = M. For if M,s G', then 'M is characteristic in G since 6' is a cyclic characteristic subgroup of G. If M i G, then M nG' = l and hence [M,G] SM (16' = 1. Thus, M s Z(G) so that O'M = 1. Hence a(M) = M. Therefore, 0' also induces an automorphism of G. which we will denote by 5. It will follow that Xf = X? by induction on ‘0‘ provided that 6 is defined in the same manner as a is on G. Let g denote gM for g 6 G and let A = <5,Z(G-)>. Since f(C-g) =C££ : _ 1 it suffices to show 6(xiajg) = £153; 3 where g 6 Z(G) in order 1. l to have 5 defined on G as o is on G. Write g = xLasz where z 6 Z(G). Then am 6 Z(G) so that (am)xM = armz:M = aWM . Thus, amM = akm(r-1)Z:WM = M . 1 Hence, 39 o(§ '5 g) = 5(i Wales; 5) -i-J-i-L-m - - i-j -j- xaaxaz=xaa 1 1g and we have that 5 is defined as 0- Thus we may assume ker X = 1. Since CG(A) =.A by Lemma 2.5.1(1), we have that A* =.A in Lemma 2.3.1. Hence, any faithful character is zero on G - A since the character is induced from A. Therefore, Xf = X? on G - A. Also, Xf = X? on A since f(C . ) = C . . = 0(6' ) and the proof is complete. 1 1 1 i a z a alz a 2 It is interesting to note at this point that Theorem 2.4.1 follows as a corollary to Theorem 2.5.2 when the cyclic normal sub- group A has odd prime power order by the next lemma. Lemma 2.5.3: Suppose that G = BA where B is abelian and A. is a cyclic normal subgroup of odd prime power order. Then G/Z(G) is metacyclic. 11199;: We have that Aut(A) is cyclic and that the mapping b —» Tb is a homomorphism from B into Aut(A). Further, the kernel of this map is B n Z(G). Hence, B Z(G)/Z(G) is cyclic and so G/Z(G) = (BZ(G)/Z(G))(AZ(G)/Z(G)) is metacyclic. It should be noted that the conclusion of Lemma 2.5.3 is not true if A does not have odd prime power order. For example, let A be a cyclic group of order 8 and let G be the holomorph of A. Then one can easily verify that G/Z(G) is not metacyclic. We will also find Lemma 2.5.3 useful in Chapter IV. CHAPTER III NORMALIZED AUTOMORPHISMS IN DIRECT PRODUCTS Section 1. Introduction. One question that naturally arises is the following: suppose G is a direct product of E.R. groups, then is C an E.R. group? Indeed, this would be a useful result to know if one wanted to determine whether nilpotent groups are E.R. groups, since one would then have to only consider p-groups. In this chapter, we will obtain a positive answer to this question when G contains an abelian normal subgroup A such that W(G,A) is a complement for Aut(G) in NA(G) or when the direct summands have pairwise relatively prime orders. Notice that the latter case will yield the sufficiency of studying only p-groups in the nilpotent case. We will also obtain the converse of this question (that is, if G is a direct product of groups and if G is an E.R. group, then each summand is an E.R. group) without any restriction on C. When we use the notation G] in this chapter, we will mean the subgroup G1 X...X Gi-l x G1+1 x...x Gn of G when Section 2. The Containment of NA(GI) x...x NA(Gn) in NA(G). Suppose G = C1 x...x Gn and let f1 6 NA(Gi). We then can extend fi to an element Fi of NA(G) by setting 40 41 Fi(3182"'gn) = gng°°°gi-lfi(gi)gi+l'°°gn where gj 6 Gj’ and then by extending Fi linearly to Z(G). Further, the mapping fi 4 F is an embedding of NA(Gi) into NA(G). i Also, if flf2°"fn’ where f1 6 NA(Gi), is an element of NA(Gl) x...x NA(Gn), it follows that the mapping m defined by ¢(f1f2 ... fn) = F1F2 ... Fn defines an embedding of NA(Gl) X...X.NA(Gn) into. NA(G). We summarize these remarks with the following theorem. Theorem 3.2.1: Suppose G = G1 x...x Gn’ then every f1 in NA(Gi) can be extended to an Fi in NA(G) such that F1 is the identity on Gi' Further, the mapping m given by tp(f1 ... fn) - F1 ... Fn defines an isomorphism of NA(Gl) x...x.NA(Gn) into NA(G). Because of the embedding, we will also let NA(GI) X...x NA(Gn) denote its image under m. We conclude this section with the following remark. Suppose G = G x...x Gn where (\G 1 "Gj') = 1 for i 6 j. ,l Then Aut(G) = Aut(Gl) x. . .x Aut (cm) This might lead one to conjecture that NA(G) = NA(GI) x...x NA(Gn) . However, this later equation is not true. For an example of this, let G = G x G2 where G is the l 1 quaterion group of order 8 and G2 is the cyclic group of order 3. 42 Then NA(G) # Aut(G) by Theorem 1.3.2. However, also by Theorem 1.3.2,NA(G1) =Aut(G and NA(GZ) =Aut(G2). Thus, 1) NA(Gl) x NA(GZ) = Aut(G) :4 NA(G) . Hence NA(GI) x...x NA(Gn) is always contained in NA(G), but they are not in general equal, even if the summands have relatively prime orders. Section 3. When the Summands have RelativelyiPrime Orders. In this section we will show that if G = C1 x...x Gn where (\Gi\,lGJ\) = l for i i j and if each G1 is an E.R. group, then G is an E.R. group. To accomplish this we first show the following: Theorem 3.3.1: Let G = C1 x...x Gn where (\G1\,\GJ‘) = l for i i j and let "1 denote the natural map from Z(G) to Z(Gi)° Then, for any f 6 NA(G), is in NA(Gi). Further, the f ”i 'Z(Gi) mapping W defined by W(f) = filf'Z(Gl)n2f'Z(G2) ... TTnle(Gn) gives rise to an exact sequence 0 a ker l ..NA(G) 1.NA(GI) x...x NA(Gn) a 0 and m is a splitting map for this sequence. In addition, ker ill 3 CP(G). Proof: In order to show nifl is in NA(Gi) we first note mi) that is clearly a homomorphism. Further, flif'Z(Gi) nif'Z(Gi) is 1-1, for suppose x 6 Z(Gi) such that nif(x) = 0, then 43 f(x) 6 A(G-i). But G1 is NA-characteristic by Lemma 1.4.4. Hence, -1 _ _ x 6 f (A(Gi)) = A(Gi) and so x = 0. Finally. we show nif'Z(Gi) is onto Z(Gi)' Let gi 6 Gi' Since Z(G)/A(GE) = Z(G)/A(f(A(G])) by Lemma 1.4.4, we can write 8. = z a f(g) + x 1 366 g i where ag 6 Z and x 6 A(Gi). Hence, n.f( Z: a g) =rr.(g. - x) =3 1 36G g 1 1 i and so "if‘ is onto. Thus we have established that Z(G i) fiif'Z(Gi) 6 NA(Gi) . Next, we show that the map W is a homomorphism. To do this it suffices to Show f - f' = ,ff' TTi 'Z(Gi) "i luci) "1‘ ”2(61) for any f,f' in NA(G). Let x 6 Z(Gi) and write f'(x) = x1 + x2 where x1 6 Z(Gi) and x2 6 A(Gi) . Then, I = = niff (X) 111150:1 + x2) Trif(x1) since f(xz) 6 A(Gi). Similarly, I .__ = Trifnif (x) nifni(x1 + x2) n1f(x1) , whence ‘l is a homomorphism. It now follows that (up = 1 from the definition of q; and q). Hence, it is onto and q) Splits the sequence. 44 Finally, we have to show f 6 CP(G) for all f 6 ker t. In order to do this it suffices to show f(Cg) = C' for any 8 ,- g 6 G.. For if g = glg2 ... gn where gi 6 G1, 1 CI = CD 6' ... C' and hence if the class sums of elements of each G1 are fixed, every class sum is fixed. Suppose f 6 ker u and g 6 G’. Let f(C ) = C; . Since 1 g 1 Ci is NA-characteristic, f(A(Gi)) = A(Gi) and hence f(Gi) is the normal subgroup of f(G) correSponding to Gi. Thus, since f f d _ = c , . (g) 6 (Ci) an since Cf(g) 81 we must have g1 6 Ci But then n,f C = C since f = 1 and n,f G = 11 C = C Thus C = C and we are done. 8 81 Since ker W is then a complement for NA(GI) we can state Corolla;yfi3.3.2: If G = G1 x...x Gn where (‘Gil,\Gjl) = l for 1 ¥ j, then NA(GI) x...x NA(Gn) has a normal complement in NA(G). Further, this normal complement can be taken to lie inside CP(G). As another corollary we get the result we seek concerning E.R. groups. Coroliagy 3.3.3: Let G = G1 x...x Gn where (\Gi"'Gj') = 1 for i i j and where each Gi is an E.R. group. Then G is an E.R. group. nggf: First note that if fi 6 NA(Gi) has an elementary representa- tion, then m(fi) also has an elementary representation. Thus, NA(Gl) X...X NA(Gn) S EA(G). Therefore, NA(G) = (NA(Gl) x...x NA(Gn))ker W S EA(G)CP(G) = EA(G) and hence NA(G) = EA(G). Thus G is an E.R. group. 45 Section 4. When W(G,A), is a Complement. In the previous section we were successful in showing that a direct product of E.R. groups of relatively prime order is an E.R. group since we could construct the map 1' If G contains an abelian normal subgroup of G such that W(G,A) is a complement for Aut(G) in NA(G), we can restrict ourselves to W(G,A) when studying whether G is an E.R. group. In this case we will show that a direct product of E.R. groups is an E.R. group by carrying out the same procedure as in the last section on W(G,A). Here our success will hinge upon the fact that the summands are W(G,A)-admissible when W(G,A) is a complement for Aut(G). We first state two lemmas about the map m. Lemma 3.4.1: Suppose G = G x...x Gn and let N = N x...x N l l n where Ni <1 G then i, tp(W(G1,N1) x...x W(Gn,Nn)) s W(G,N) Proof: Let fi 6 W(G1,Ni). IL suffices to show m(fi) 6 W(G,N). If g 6 61’ then fi(s) =‘-— 3 mod A(N1)A(Gi) in Z(Gi)' Hence by the definition of W: o<£1> a g mod some) . If a e 51, then being) SO ll 00 coins) -=- a mod ss 46 The result now follows since it is true on a set of generators for G. Lemma 3.4.2: Let G = G X...X Gn’ n denote the natural map from l i G to G1, and let A be an abelian normal subgroup of G such that W(G,A) is a complement for Aut(G) in NA(G). Then W(G,Al x...x An) = W(G,A) and m(w(c1,A1) x...x W(Gn,An)) s W(G,A) where A. = fi.(A). l 1 Proof: Since W(G,A) is a complement for Aut(G), G/A is either abelian or a Hamiltonian 2-group. If CIA is abelian, certainly G/A1 X...X An is abelian. If G/A is a Hamiltonian 2-group, then G/A1 x...x An is either an abelian 2—group or a Hamiltonian 2-group since every subgroup of G/A1 x...x AD is normal. Thus, W(G,A1 x...x A“) is a complement for Aut(G) and so W(G,Al x...x An) = W(G,A). Also, ¢(W(GI,A1) x...x W(Gn,An)) s W(G,A) now follows from the previous lemma. We next construct a map n similar to 1- Theorem 3.4.3: Let G = G1 x...x Gn’ "i denote the natural map from Z(G) to Z(Gi)’ A be an abelian normal subgroup of G, and A1 = ni(A). Further suppose W(G,A) is a complement for Aut(G) in NA(G). Then for every f 6 W(G,A), nifl 6 W(G,Ai). Moreover, the Z(Gi) mapping n defined by . nnf‘ “m I "1'E ‘ z(c:1)"2f 'Z(G2) Z(Gn) 47 for f 6 W(G,A) gives rise to an exact sequence .... —O I] X... w ,A "'9 O ker n W(G,A) W(Gl’Al) X (Gn n) 0 such that m restricted to W(Gl,A1) x...x W(Gn’An) is a Splitting map. Also ker Tl s CP(G). nggf: First note that G is solvable by Corollary 1.5.6. Hence every normal Subgroup of G is W(G,A)-admissible by Lemma 1.6.2. Let f e W(G,A). Then tune-1)) = A(Ei). Thus, going through the same Steps as in the proof of Theorem 3.3.1, nifl 6 NA(Gi). Further, if g 6 Gi’ f(g) = g + x where z X 6 A(A)A(G) and so nif(g) = g + ni(X) E 8 “Mi A(A1)A(Gi) in Z(Gi)' Thus 6 W(Gi’Ai)' Also, one can again mimic "if \zmi) the proof of Theorem 3.3.1 to obtain that n is a homomorphism. It then follows that nm = 1 on W(G1,A1) x...x W(Gn’An) and hence n is onto and m is the splitting map. Finally, it follows that ker n S.CP(G). For if f 6 ker n, one can again go through the same Steps as in Showing ker t SLCP(G) in Theorem 3.3.1 to conclude f 6 CP(G) since f(A(Gi)) = A(Gi)° CorreSponding to Corollary 3.3.2 we can State Corollary 3.4.4: Let G = C1 x...x Gn and suppose G contains an abelian normal Subgroup A Such that W(G,A) is a complement for Aut(G) in NA(G). Then, if n denotes the natural map from Z(G) i to Z(Gi) and if A1 = ni(A), W(G1,A1) x...x W(Gn,An) has a normal complement in W(G,A). Further, this normal complement can be taken to lie inside CP(G). 48 Finally corresponding to Corollary 3.3.3, we have Corollary 3.3.5: Let G = G1 x...x Gu and suppose that G contains an abelian normal subgroup A such that W(G,A) is a complement for Aut(G) in NA(G). Then, if each Gi is an E.R. group, G is an E.R. group. Proof: It suffices to Show W(G, A) s.EA(G). Since each Gi is i="i(A) and "i is the natural map from Z(G) to Z(Gi)' Hence in NA(G), W(G,,Ai) S EA(G). an E.R. group, W(G,,Ai) s EA(Gi) where A Thus, W(G,A) = (W(G1,A1) x...x W(Gn,An))ker n IA EA(G)CP(G) EA(G) and we are done. In concluding this section, one should also note that Theorem 3.3.3 and Corollary 3.3.4 will hold under the assumptions that G is solvable and A is equal to A1 x...x An instead of assuming W(G,A) is a complement. Section 5. The Converse. we conclude this chapter by showing that if G = G1 x...x Gn and if G is an E.R. group then each Gi is an E.R. group. Theorem 3.5.1: Let G 8 G1 x...x Gu and suppose that G is an E.R. group, then each G1 is an E.R. group. Proof: Let f 6 NA(Gi) and set H = f(Gi)' By Lemma 1.3.1, it i suffices to Show the existence of an f' in NA(Gi) such that f'(Gi) = H I- =— 1 and f (cg) C8 in Z(Gi) for all g 6 oi. 49 Set H = HiGi’ Then H is a group basis of Z(G). Further, H is isomorphic to G. Hence by Lemma 1.3.1, there exists f' 6 CP(G) such that f'(G) = H since G is an E.R. group. Then, Since H1 is the normal subgroup of H corresponding to G1 and since f'(A(Gi)) = A(Gi) by Theorem 1.4.3, f'(Gi) = H by Lemma 1.4.1. 1 Hence f' restricted to Z(Gi) lies in NA(Gi), f'(Gi) = H , and i f'(C ) = C. for all g 6 G since the class sum of an element of 8 g i G1 in Z(Gi) is also its class sum in Z(G). This completes the proof. CHAPTER IV ELEMENTARY REPRESENTATIONS IN p-GROUPS Section 1. Introduction. We consider two problems in this chapter. In Section 2 we will see some p-groups which are E.R. groups, al- though all the E.R. groups obtained here will also be metabelian. We also emphasize again that knowledge of p-groups which are E.R. groups yields knowledge about nilpotent groups which are E.R. groups by Corollary 3.3.3. In Section 3 we will consider the question of when W(G,G') s.CP(G) for p-groups. Section 2. peGroups which are E.R. Groups. One useful fact about p-groups is the following. Let G be a metabelian p-group and let A be a maximal abelian normal subgroup of G containing G'. Since A is a maximal abelian normal subgroup of G, CG(A) = A. Hence, if X is a faithful character of G, X is induced from a linear character on A by Lemma 2.3.1. Also, X is then zero on G - A. One immediate result we can obtain from the above observation on faithful characters is Theorem 4.2.1: Suppose G is a p-group such that Z(G) is cyclic, \P3(G)[ = p, and G contains a maximal abelian normal subgroup such that G' S.A s 22(G). Then W(G,G') stCP(G) and hence G is an E.R. group. 50 51 (2522;: Let X be an irreducible character of G and let f 6 W(G,G'). It suffices to Show Xf = X. If ker X > 1, let N = ker X. Then N 2 F3(G) and N is W(G,G')-admissible by Lemma 1.6.2. Hence, f induces an automorphism on Z(G) which will be in W(G,G') where G=G/N. Further, X; = X since 5- has class 5 2. If ker X = 1, then X? = X since both X; and X. are zero on G - A and since f fixes the class sums of elements of A. This completes the proof. In attempting to prove that a metabelian p-group is an E.R. group, one might try to construct an automorphism of G that agrees with a given f 6 W(G,G') on class sums of a maximal abelian normal subgroup .A containing G', as this would be similar to the technique used in proving Theorems 2.4.1 and 2.5.2. One case in which one might first try the above technique is when A is cyclic. However, by the next lemma we will see that we have already treated this case in Chapter II. Lemma 4.2.2: Let G be a p-group such that G contains a maximal abelian normal subgroup A which is cyclic. Then G BA where B is abelian. m: Let g denote gA for g6G. Since CG(A) =A, it follows that G/A is isomorphically contained in Aut(A) under the mapping 8 r 7g. We may assume G/A is not cyclic, or else we are done. Hence, we have that p = 2 and that \A‘ 2 23 since Aut(A) is cyclic when p is odd or \A‘ S 22. 52 Let .A = and Suppose \A\ = 2m, m 2 3. Then 5 - Aut(A) = X ’ where a(a) = a and 5(a) = a 1. Thus since G/A is not cyclic, we can choose d and c in G such that d and c generate G/A and so that T is some power of a and d Claim: There exists b 6 G such that b and c generate G/A and [b,c] = 1. Note that once we have established the claim, we are done with B I . Proof of the claim: Let [d,c] = ar. We first Show that 2 g r. Case 1: If \dA‘ = 2. 2 i Let d = a Then Z‘j, or else : would be an abelian normal subgroup of G contradicting the maximality of A. We also have (d2>° - (aj>° - a‘3 2 2 (I) = (dc) = (da‘> . . m-Z Further, Slnce \dA\ = 2 and since ‘0‘ = 2 , -3 2m m-l ad = a5 = al+2 . (2) Thus from (1) and (2), m-l m-l -° 2 a j = dar dar = dZar(2+2 ) = aj+r( +2 ) Therefore, r(2 + 2m-1) E -2j mod 2m. Since 4\-2j and since 4 I 2 + 2m~l, we have er. 53 Case 2: If \dA\ > 2. Suppose 2 Y r in this case. Then a 6 G' and hence 6(G) = . Also, Z(§(G)) SCG(A) = A so that Z(§(G)) is cyclic. But then 6(G) is cyclic by Satz 7.8 c.), 8.306 of [9], a contradiction. Hence 2\r. To complete the proof of the claim, let b = da . Then b and c generate GAA and b and c commute since bc = (dar/2)c = dara-r/2 = dar/2 = b. Thus we get as a corollary to Theorem 2.4.1 Corollary 4.2.3: Let G be a p-group containing a maximal abelian normal subgroup which is cyclic. Then G is an E.R. group. Another type of p-group which is an E.R. group is contained in the following result. Theorem 4.2.4: Let G be a p-group and suppose that G = BA where B is abelian and A is an abelian normal Subgroup of G containing Z(G). Further, suppose A/Z(G) is elementary abelian of order p2. Then G is an E.R. group. nggfig Let x and y be elements of A which generate A/Z(G). We may further assume that y is chosen so that y 6 22(G). Since x lies in the center of G/, G/ is abelian and hence G has nilpotence class s 3. Further, if the class of G is 3 then x 6 22(G). Let b 6 B. Since G/ is abelian, be(G) = xij(G) for some 0 s j < p. Thus, if the class of G is 3, we can find bx 6 B such that b . x XZ(G) = nyZ(G) 54 where O < j < p. In the remainder of the proof we will assume j = l, for if not replace y by yj. Let f 6 W(G,G'). Since y 6 ZZ(G), we have f(C;) = Cy by Lemma 2.2.1. We also know that r(E’x) = c"xz for some 2x 6 F3(G) s Z(G) by Lemma 2.2.5. In addition? ‘zx\ s p Since xp 6 Z(G) and so f(Cxp) = f(xp) = xp = cxpzp . X J Let g 6 G. Write g = bxly z where b 6 B and z 6 Z(G). . . i i We define a mapping a by setting a(g) = bx yjzxz and claim that o is a well-defined automorphism of G. To see that o is well-defined, suppose blxiyjz1 = bzxkybz2 where b1,b2 e B, 21,22 6 Z(G). Then, blA = bZA and so blbgl e Z(G). Thus, xi-kyj-L 6 Z(G) and hence p‘i-k. Therefore zi-k = 1. Hence o(b1xiyjzl) = blxiyj :21 = blxiyjz:z1 = bzxkyLz:z2 = 0(b2xkyLzz) , so a is well-defined. o is a homomorphism, for if blxiyjz1 and bzxkyLz2 are b i 2 i any elements of G, we have (x yj) = x ymz for some integer m and some 2 6 Z(G). Then 55 j k L i+k th zlbzx y 22) 5(b1b2x y 21222) i C(blx y _ i+k i+k,m+L — blbzx zx y zlzzz ij i 1:11: (blx y lexflbzx y 2x22) ll ' k o(blxlyjzl)o(bzx yLzz) and g is a homomorphism. Finally, a is an automorphism. For if i j i i j _ - 0(bx y z) = bx zxy z-lq thenb 6 A and hence b 6 Z(G). Thus, xiyj 6 Z(G) so that x1 6 Z(G). Then, 2: = 1 and so bxiyjz = 1. At this time, we remark that A is a maximal abelian normal subgroup by Lemma 2.5.1(1) and that A 2 G'. Further, if G has class 3, f and 0 agree on class sums of elements of A. For 1et g 6 A. Write g xiyjz where z 6 Z(G), O s i < p, and O s j < p. If i = 0, then g 6 22(G) so that f(c‘>=5 =o<5) 8 8 8 If j = 0, then f(Cg)=f(Ci)=Cii =°(Ci) x z x zxZ x 2 Finally, if both i i O and j # 0, let k be an integer such that ki E j mod p. Then bk x i ik i j ) Z(G) = X Y Z(G) = X y Z(G) bk (xi Hence xlyj = (x1) X21 for some 21 6 Z(G). Hence, f(Cij) f(Ci )=Cii x y z x z z x z z z 1 x 1 =6 =6 i j i °( i j ) ° x y zxz x y z 56 We claim that Xf = X0 for every irreducible character X and for any a defined as above, from which it will follow that G is an E.R. group. Suppose our claim is false. Let G be a minimal counter- example and let X be an irreducible character such that Xf i x?. We first Show that X is not faithful, for Suppose it is. If G has class 3 2, then X, Xf, and X0 are all zero on G - Z(G) since they are all faithful. From this observation it follows that Xf = X0 since f and 0 agree on Z(G). If G has class 3, then the faithful characters X, Xf, and X0 are all induced from A. But then Xf = X0 Since X5 and X? are both zero on G - A and since f and 0 agree on class Sums of A. Hence, X cannot be faithful. Thus we have ker X n Z(G) >.1. Let ‘N = ker X n Z(G). Then f and 5 are both the identity on N. Therefore, f and o induce automorphisms on Z(G) where G'= G/N. Let f and 5 de- note the induced automorphisms, g denote gN for g 6 G, X: the character X induces on G; B'= BN/N, and let A'= <§,y, Z(G)>. Also, note that f 6 W(G,G‘). Since X; # X9, Xf # Xv. we will obtain a contradiction by Showing X? = XO- Suppose i 6 Z(G). Since not) “52;. IC’ ’ x Ex = l and so 5 = 1. Also, G. has class 2 since G7Z(G) would have to be abelian. Thus, 57 If y 6 Z(G), then again G. would have class 5 2. Thus, since f(G)?) =C;£ =C‘ , X there exists a 61 6 B such that x = xfx. We also must have f(Cé) = C; for g 6 G. since G has class 5 2. Hence, f(C__i_j_) E;_i_j = (C i j ) bx y z bx y , -f so again X = X Finally, suppose neither x nor y lies in Z(G). Since f(C-) = C;— , if we can Show x xzx --i_j- __i-j-_i o(bx y g) = bx y szx (*) where g 6 Z(G), we will again have 'Xf =';U since G is a minimal counter-example. To Show (*), it suffices to Show that 5(g) = g. Write - - -k-L- g = blx y z where b1 6 B, z 6 Z(G), O s k < p, and O s L < p. Then, and so 61 6 Z(G). But then ikyf 6 Z(G), so k = L = 0. Hence, 5(g) = g and so we have (*). This completes the proof. 58 Theorem 4.2.4 was originally motivated by the following corollary. Corollary 4.2.5: Let G be a p-group where p > 3. Suppose that every normal abelian subgroup of G is generated by at most two elements. Then G is an E.R. group. 2392;: By Satz 12.4, 8.343 of [9], G is one of the following types of groups: (1) G is metacyclic. (2) G . n n-l x S X «1),:z‘xp "' Yp zp = [y,2] 1: Y = yz P 92 = 2y) (3) G where s = 1 or is a quadratic non-residue mod p. We already know that (l) is an E.R. group. (2) is easily seen to have class 2 and hence is also an E.R. group. In (3), let B = and A = . Also in (3), Z(G) = and A is an abelian normal subgroup. Thus one can apply Theorem 4.2.4. Finally we mention one more result. Corollary 4.2.6: Suppose G contains an abelian normal subgroup A such that G/A is cyclic, Z(G) s.A, and \A/Z(G)‘ 5 p2. Then G is an E.R. group. 2329;; Note that A/Z(G) is either cyclic or elementary abelian of order p2. If A/Z(G) is cyclic, apply Theorem 2.5.2. If A/Z(G) is elementary abelian of order p2, apply Theorem 4.2.4. Section 3. When is W(G,G') s.CP(G)? An interesting question re- lated to whether a metabelian group is an E.R. group is when is W(G,G') sCP(G)? Certainly, if W(G,G') sCP(G) where G is metabelian, then G is an E.R. group. In addition, one can obtain 59 the following generalization of Lemma 2.2.5 by replacing F3(G) by N in the proof of Lemma 2.2.5 when G is metabelian. Lgmma 4.3.1: Let N be a normal subgroup of the metabelian group G. Further, suppose that W(G,G') sCP(G) where G = G/N. Then for each f 6 W(G,G') and for each g 6 G, there exists an xg in N such that f(Eg) = ch . Thus we see that knowledge of when W(G,G') $.CP(G) will also increase our knowledge on how W(G,G') acts on class sums. Up to this point, we have seen that W(G,G') s CP(G) when- ever G haS nilpotence class 5 2, G has at most one non-linear char- acter, and when G satisfies the hypothesis of Theorem 4.2.1. In this section we will obtain some metabelian p-groups where this is true. Before obtaining some p-groups where W(G,G') s.CP(G), one should note that we do have the following result on direct products as a corollary to Theorem 3.4.3. Corollary 4.3.2: Let G = G1 x...x Gn be a metabelian group. Then, if W(Gi’Gi) SCP(G1) for each i, W(G,G') 5CP(G). Proof: If W(Gi’ci) SCP(Gi), then W(Gi’ci') SCP(G). Hence, W(G,G') =(W(cl,ci) x...x W(Gn,Gt"))ker n sCP(G) . Then in particular, studying the question of when W(G,G') s CP(G) for metabelian nilpotent groups reduces to studying this question for p-groups. The major case in which we can get p-groups for which W(G,G') $.CP(G) is the following. . slilli 60 Theorem 4.3.4: Let G be a p-group where p > 2 such that G/Z(G) is metacyclic. Then W(G,G') s CP(G). Proof: Let x, a 6 G such that A = : is a normal subgroup of G and such that x generates G/A. By Lemma 2.5.1, we know that \G/Aj = \G". Hence, ‘Ca' = \G/CG(a)\ = ‘G/Aj = \G", so that C = aG'. a In the proof of Theorem 2.5.2, for an f 6 W(G,G'), we took an a1 6 G' such that f(Cg) = Eaa , constructed an automorphism J 11 1 o by setting C(Xia z) = x a a z where z 6 Z(G), and showed that J l f(C?) = 5(Cé) for all g 6 G. In this case we can actually take a1 = 1 Since C8 = aG'. Then a - l and we are done. Theorem 4.3.4 gives us some additional cases of when W(G,G') s CP(G) which are contained in the following corollary. Corollary 4.3.5: Let G be a p-group, p > 2. Suppose either (1) G = AB where A is a cyclic normal subgroup of G and B is abelian or (2) G contains a maximal abelian normal subgroup A where A is cyclic. Then W(G,G') some). LEQQL; (1) follows since Lemma 2.5.3 implies G/Z(G) is metacyclic. (2) follows since, as we noted in the proof of Lemma 4.2.2, G is metacyclic. 1‘ \I'I. 1'1 .1 it. \I CHAPTER V ELEMENTARY REPRESENTATIONS IN Sn In [3], Brown showed that Sn’ n = 1,...,10, are E.R. groups. In this chapter, we will Show that Sn is an E.R. group for any positive integer n. We begin by recording two lemmas about Sn’ the first of which is Exercise 11.4.11 of [13]. The second lemma is a well- known result about the order of conjugacy classes of Sn and can be found, for instance, in [2]. Lemma 5.1.1: Let n > 2, n i 6, then x 6 Sn is a 2-cycle if and only if \x‘ = 2 and max‘xxy\ = 3, where y 6 Sn. Lemma 5.1.2: Let g 6 Sn and suppose that g is the product of disjoint a1 l-cycles, a2 2-cycles,..., an n-cycles. Then n! CY 0' 0' 1 2 \c,\ = In showing that Sn is an E.R. group, we will show that NA(Sn) = CP(Sn) for n > 2, n # 6. (Note that this would have to. be the case if SD is to be an E.R. group since Aut(Sn) = Inn(Sn) for n f 6.) We first Show that every normalized automorphism of Z(Sn) fixes the class sums of elements of order 2 for n > 2, n H 6. Lemma 5.1.3: Let f be a normalized automorphism of Z(Sn) and suppose that n > 2, n 6 6. Let g 6 Sn be a product of disjoint tranSpositions. Then f(Cg) = Cg. 61 A lit.‘ 62 Proof: Let t denote the number of tranSpositions appearing in g. we may assume that g has the form g = (12)(34)...(2t-1,2t). Let f(G) = C- . To Show 6 = G we proceed by induction on t. g 81 81 g If t = 1, let y 6 S . Then by Lemma 1.2.2, we can find cu m x 6 S such that f(C. ) = n x y 88 8181 Lemma 5.1.1. Next, let x 6 Sn Such that \ggx\ = 3. Again by . Thus, ‘glgi‘ = ‘ggx‘ S 3 by Lemma 1.2.2, there exists y 6 Sn such that f(C- x) = C. . gs glgi Hence 181811 = \ggxl = 3. Therefore, max‘glgil = 3, so g1 is a tranSposition by Lemma 5.1.1 and we are done when t = 1. In the general case we have that f(C(12)(34)...(2t-3.2t-2)) ‘ C(12)<34>-~(2t-3:2t‘2> and f(C' Hence, (2t-l,2t)) = C(2t-l,2t)° f(Cg) = C(12)(34)...(2t-3,2t-2)(2t-1,2t)x for some x 6 Sn by Lemma 1.2.2. If (2t-l,2t)x is disjoint from (12)...(2t-3,2t-2) we are done. Suppose this is not the case. If (2t-l,2t)x has one letter in common with (12)...(2t-3,2t-2), it follows that (12)...(2t-3,2t-2)(2t-l,2t)x is a product of disjoint cycles which are tranSpositions and a 3-cycle. But then 3|lgll which is impossible. Next, Suppose (2t-l,2t)x has two letters in common with one tranSposition of (12)...(2t-3,2t-2). It follows that g1 is a product of disjoint transpositions and has one less trans- position than g. But then f fixes C. So that f(Cg) # C; 8l l The final possibility would be for (2t-1,2t)x to have one letter 63 in common with two different transpositions of (12)...(2t-3,2t-2). But then (12)...(2t-3,2t-2)(2t-1,2t)x is a product of disjoint cycles which are tranSpositions and a 4-cycle. Therefore “1131‘ which is impossible. Hence, (2t-l,2t)x is disjoint from (12)...(2t-3,2t-2) so that C? = 6E1. Theorem 5.1.4: Sn is an E.R. group for every positive integer n. EEQQEF By the results of [3] we may assume n > 2, n i 6. Let f e NA(G) and let N = {g e sn\£(6é) a 6;). We will show that N = ¢, which proves the theorem. Suppose N # g. We pick a "minimal element" g of N, which satisfies the following properties in the order that they are listed: (1) Suppose that g has its largest cycle of smallest length among the elements of N. Let h denote the length of its largest cycle. (2) Suppose that g has the fewest number of cycles of length h among the elements of N satisfying (1). (3) Suppose that g has the fewest number of cycles of length greater than or equal to two among the elements of N satisfying (2). Note that we have h 2 3 by Lemma 5.1.3. Write g = 3192 ... Br where the Bi are disjoint cycles and 2 3 'Bi' s \6 Also, assume that er = (l,2,...,h). Let r+1“ I ._.. - s ' Br (l,2,...,h 1). Then g 51 ... Br_18r(1,h). Also, rd? .) = 5' a1"'Br-19r Bl°°°Br-la' r by the minimality of g and f(C = C(1,h) (1 .h)’ 64 Hence by Lemma 1.2.2, _. =6 ' f(Cg) erueflerum)?‘ x . for some x 6 Sn' Let g1 = 51...Br_15£(l,h) . We wlll now Show that Eg = cg, from which it follows that N = s and so we will 1 be done. . X. I Case 1. Suppose (1,h) ls dLSJOlnt from 81"'Br-lar° . x . . . u — If (1,h) ls dlSjoint from 31...Br_13r, then Cg1 is fixed by f Since it has fewer cycles of length h with its largest cycle of length 5 h. But then f(C?) # C' , so this case cannot gl OCCUI‘ 0 Case 2: If (1,h)X has one letter in common with Bl"°Br-laf° Suppose that (1,h)x has one letter in common with I 31...B ler and that (n1,n2,...,nt) is the cycle of 61"°Br-16f where the common letter occurs. Then (1,h)x = (nj,a) where a does not appear in 61°'°Br-lef' Also, note that (n1,...,nt)(nj,a) = (n1,...,nj_1,a,nj,...,nt) . If t < h-l, then g1 would have fewer cycles of length h with its largest cycle having length s h. Thus C8 is fixed ._ ._ l by f, so f(Cg) f Cg . Hence, we must have t 2 h-l. 1 If t > h-l, then t = h. But then g1 has one more cycle of length h-l than g does. Thus, g: has one more cycle of length h-l than gh does since (h,h-l) = 1. But E'h is fixed by f 8 Since all cycles in gh have length less than h. Thus, 65 f(C h) “Ch‘ch g s 81 which is impossible. Thus we must have t = h-l. But then g1 has the same cycle structure as g does. Hence f(Cé) = C. = 6E1. Case 3: If (1,h)X has two letters in common with Bl'°'Br-lef° First, Suppose that (1,h)x has two letters in common with one cycle of. 81"'Br-lar' Let (n1,n2,...,nt) denote this cycle and Suppose (1,h)x = (nj,ns) where j < s S t. Then (n1,...,nt)(nj,ns) = (n1,...,n, ,n ,n ...,n ). j-l s s+l’°°°’nt)(nj’nj+l’ 8-1 But then g1 has fewer cycles of length h with its largest cycle having length S h. Thus, f fixes 6g and so f(Cg) 6 G 1 1 Therefore, (1,h)x must have its letters in common with two cycles of 61"°Br-1Bf' Let (n1,n2,...,nt) and (ni,né,...,n;) denote these cycles where t S s. Then (1,h)x has the form I (ni,nj) and I I I = I I ... I, I,..., (n1,...,nt)(n1,...,ns)(ni,nj) (n1,...,ni_1,nj,nj+1, ,n8 n1 nj-1,ni,...,nt) If s + t S h, then g1 has its largest cycle of length S‘h, has at most as many cycles of length h as g, and has fewer cycles of length 2 2 than g. Thus f fixes C so that f(C ) i C. . g1 g 81 Hence, we must have S + t >:h. If neither 3 nor t is h-l, then g1 has one more cycle h of length h-l than g. Therefore, g1 has at least one more cycle of length h-l than g does and so Eh :4 Eh' But f(C- h) = (T 81 8 8 8 h 66 since gh has its largest cycle of length < h. Thus, f(C h) I C h I C h ’ s a s1 a contradiction. Hence either 3 or t must be h-l. Suppose that g has a l-cycles, a2 2-cycles,...,ah h-cycles. 1 Then, t \c,\ = “,2 ,h all (12! 2 ah! h If s = h-l and t < s, then C = n! /( ! 12m2 l(t-1)atI1( -1)!ta't-1 !(t+ at“ \ 81' 0 0’1 a2 oooat-1 at at+1 ... ah-l O‘h'1 ! - - I 01h_1 (h 1) (oh 1) h > Thus since \C \ = \C \, we have 81 8 S + t = attahh . But th > s-+ t, a contradiction. If t = h-l and s = h-l, then 0’2 °‘h-1'1 "’h'1 \cg1\ — ni/(oltoziz °'°’S + t, so we have a contradiction. Finally, if t = h-l and s = h, then _ “2 O‘h-l "‘h‘2 \cgll — nl/(alla212 ...ah_1!(h-l) (ah-2)!h (s +-t)) 67 Again, since \Cgll = \Cg" we have 2 S + t = (ah " 1)ahh 2 But h > S + t, so we again have a contradiction and case 3 cannot OC cur . Thus we have now shown f(G) = Cg, so N = 95 . 8 CHAPTER VI A LOOK AT THE GROUP RING PROBLEM Section 1. Introduction. Let R be.a ring and let G and H be two groups. A question which has received considerable attention is when does R(G) ='R(H) imply G==‘H? This is known as the group ring problem or isomorphism problem. In the particular case when R = Z, the best general result known at this time is that Z(G) =- Z(H) implies G 3‘ H when G is a metabelian group. This result was shown by Whitcomb in [16] by using Theorem 1.1.6. It Should be noted at this time that in [10], Jackson claims to have obtained a positive answer to the group ring problem when G is either metabelian or nilpotent. However, the nilpotent result depends on Lemma 4 of [10], which is false, and I have been unable to follow Jackson's proof in the metabelian case. In this chapter, we will give some procedures which may prove fruit- ful in extending Whitcomb's result. Finally, we remark, as noted in [16], that in order to prove the group ring problem when R = Z, it suffices to show that every group basis of Z(G) is isomorphic to G. Section 2. Lifting and g Generalization of Theoremfil.1.6. We begin this section with the following definition. Jackson uses Lemma 4 in his proof of the metabelian result, but the use of it can be omitted. 68 69 Definition: Let N 4G and let f 6 NA(G/N). We say that f can be lifted if there exists an f 6 NA(G) which induces f on ZKG/N) (that is, n(f(x)) = f(n(x)) for all x 6 Z(G) where n is the natural map from Z(G) to Z(G/N)). We will also say that f is a lift of f. Note that in the above definition it is implicit that f(A(N)) = MN) if f is to be a lift of If. Next, we state a generalization of Theorem 1.1.6. Theorem 6.2,L5 Let A be an abelian normal subgroup of G, let n denote the natural map from Z(G) to Z(GIA), and let H be a group basis of Z(G). Suppose that there exists an f 6 NA(G/A) such that f(n(H)) = n(G) which possesses a lift f. Then for each h 6 H, there is a unique gh 6 G such that f(h) a gh mod A(A)A(G) Further, the mapping h a gh defines an isomorphism of H onto G. Proof: Let h 6 H, then there exists g 6 G such that f(h) a g mod A(A). Thus, f(h) = g +: 2 (a-l)t(a) where t(a) 6 Z(G). Then, computing a6A as Whitcomb did in [16], f(h) = g +' EA(a-l)t(a) E gAaL(t(a))g mod A(A)A(G). a a n aL(t(a)) 86A The mapping h a gh will be a homomorphism since f is a Letting gh = g, we have gh is unique by Theorem 1.1.7. homomorphism and since A(A)A(G) is an ideal. Further, h a gh is an isomorphism. For if ghl = ghz, then f(hl) a f(hz) mod A(A)A(G) so that f(hl) = f(hz) by Theorem 1.1.7. But then h1 = h2' 70 Theorem 6.2.1 can be used to solve the group ring problem for S To see this, first note that S4 = S3V4 where 40 v4 = <(l,2)(3,4),(1,3)(2,4)>. Thus, 2(54/v4)=~ 2(33). Now, in [7] it is shown that NA(S3) = 1(33). Hence if f 6 NA(S4/V4), we can write f = T; where 6 is a unit in Z(S4/V4). Since Z(S3) S 2(84), we can then find a unit u in 2(33) such that n(u) = d where n is the natural map from 2(84) to Z(S4/V4). Therefore, Tu will be a lift of Ta. Thus every element of NA(Sa/V can be lifted. 4) Finally, let H be a group basis of 2(84)' Then, n(H)== “(34) since S is metabelian. Hence, since V is abelian 3 and Since every element of NA(S4/V 4 4) can be lifted, we can apply Theorem 6.2.1 to conclude H=~ 54. More generally, one might attempt to use Theorem 6.2.1 in the following manner to solve the group ring problem for solvable groups. Let G be a solvable group, let A i 1 be an abelian normal subgroup of G, and let n be the natural map from ZKG) to Z(G/A). Suppose H is a group basis of Z(G). Then by induction on \G\, there exists f 6 NA(G/A) such that f(n(H)) = n(G). Then, if f can be lifted, we can apply Theorem 6.2.1 to conclude H‘='G. Unfortunately, the success of the above process depends on f having a lift, and in general it is not true that every normalized automorphism of Z(G/A) can be lifted from Z(G). For an example of this, let G be the dihedral group of order 8 with generators a and b where a4 = b2 = l and ab = a3. Let 2 - A = ' and let f denote the normalized automorphism of Z(G/A) defined by f(5) = 6 and f(h) = a where a = aA and B = bA. 71 Then f cannot be lifted, for suppose that f did have a lift f. Then, f(a) a b mod A(A), so that f(a) = b +(a2 - 1)t(a) where t(a) 6 2(6). But then b +-(a2 - 1)t(a) E a2L(t(a))b mod A(A)A(G) f(a) Hence f(az) 1 mod A(A)A(G), so that f(az) = l by Theorem 1.1.7. But \f(a)‘ = 4, a contradiction. Thus f does not have a lift. However, it Should be noted that we do not have to be able to lift every f such that f(n(H)) = n(G) to prove the group ring problem for solvable grOUpS. Rather, it would suffice to have the existence of one f mapping n(H) to n(G) which has a lift. It is also interesting to note, that although not every normalized automorphism of Z(G/N) can be lifted when N 4 G, they can be lifted from .2(G). By this we mean the following. Let f 6 NA(G/N). Extend f to .2(G/N) and let n be the natural map from {A(G) to .2(G/N). Then, since 2(G) a-'.2,(G/N) 63 ker n, it follows that any automorphism of the form f(Q f', where f' is an automorphism of ker n, will induce f on .2(G/N). Let us at this time introduce some notation. For a group G and a normal subgroup N of G, let L(G/N) denote the normalized automorphisms of Z(G/N) which can be lifted. Section 3. Groups of Solvable Length Three. Let G be a group such that G" is abelian and set G'= G/G". Let H. be a group basis of Z(G). Then there exists f 6 W(G;GH) such that f(H) = G. by Theorem 1.1.6. Hence, if W(G;GH) S'L(G/G"), we can apply Theorem 6.2.1 to obtain that the group ring problem holds for G. 72 Whether every element of W(G,G‘) can be lifted is to my knowledge unknown. However, we can state a necessary condition for this to occur. Theorem 6.3.1: Let G be a group such that G" is abelian and such that W(G;G‘) S L(G/G") where G'= G/G". Then W(G,G')Aut(G) = NA(G). nggj; Let f 6 NA(G) and set H = f(G). Let f1 6 W(G;GH) Such that f1(n(H)) = n(G) where n is the natural map from Z(G) to z(c/c") and let f be a lift of f1. Note that f1 6 W(G,G'), l for let g 6 G. Then n(f1(g)) a n(g) mod A(GH)A(G). Hence f1(g) 6 g'+ A(G')A(G) + A(G"). But applying Theorem 1.1.5 with K = G', we see A(G") S A(G')A(G). Thus, f1(g) E g mod A(G')A(G) or f1 6 W(G,G'). Also, note that f induces a normalized auto- morphism f on Z(G) since G' is NA-characteristic. Further, nf1f(G) = n(G). ‘Now, we have f1f(g) a g1 mod A(G") for some g1 6 G, or f1f(g) = g1 +' z (a-1)t(a) where t(a) 6 Z(G). Again computing as a6A Whitcomb did in [16], L(t(a))g f1f(g) = a, + z (a-1)t(a> n a mod A(G")A(G) 86A a6G" 1 Thus for each g 6 G, we can find a gf f in G Such that 1 f f(g) = g mod A(G")A(G). Further, g is unique by Theorem 1 flf flf 1.1.7. Let a be the mapping 5(g) = gf f' Since gf f. is unique, 1 1 flf is an isomorphism, and since A(G")A(G) is an ideal, it follows that o 6 Aut(G). Also, o"f,f a 3 mod ss