Class Two p Groups as Fixed Point Free Automorphism Groups Citation Berger, Thomas Robert (1967) Class Two p Groups as Fixed Point Free Automorphism Groups. Dissertation (Ph.D.), California Institute of Technology. doi:10.7907/NYAT-FG53. https://resolver.caltech.edu/CaltechTHESIS:11022015-081046019 Abstract Suppose that AG is a solvable group with normal subgroup G where (|A|, |G|) = 1. Assume that A is a class two odd p group all of whose irreducible representations are isomorphic to subgroups of extra special p groups. If p c ≠ r d + 1 for any c = 1, 2 and any prime r where r 2d+1 divides |G| and if C G (A) = 1 then the Fitting length of G is bounded by the power of p dividing |A|. The theorem is proved by applying a fixed point theorem to a reduction of the Fitting series of G. The fixed point theorem is proved by reducing a minimal counter example. IF R is an extra spec r subgroup of G fixed by A 1 , a subgroup of A, where A 1 centralizes D(R), then all irreducible characters of A 1 R which are nontrivial on Z(R) are computed. All nonlinear characters of a class two p group are computed. Item Type: Thesis (Dissertation (Ph.D.)) Subject Keywords: (Mathematics) Degree Grantor: California Institute of Technology Division: Physics, Mathematics and Astronomy Major Option: Mathematics Thesis Availability: Public (worldwide access) Research Advisor(s): Dade, Everett C. Thesis Committee: Unknown, Unknown Defense Date: 28 April 1967 Funders: Funding Agency Grant Number NSF UNSPECIFIED Woodrow Wilson Foundation UNSPECIFIED Caltech UNSPECIFIED Record Number: CaltechTHESIS:11022015-081046019 Persistent URL: https://resolver.caltech.edu/CaltechTHESIS:11022015-081046019 DOI: 10.7907/NYAT-FG53 Default Usage Policy: No commercial reproduction, distribution, display or performance rights in this work are provided. ID Code: 9262 Collection: CaltechTHESIS Deposited By: INVALID USER Deposited On: 02 Nov 2015 17:48 Last Modified: 15 Mar 2024 17:59 Thesis Files Preview PDF - Final Version See Usage Policy. 15MB Repository Staff Only: item control page

CLASS TWO p GROUPS AS FIXED POINT FREE AUTOMORPHISM GROUPS Thesis by Thomas Robert Berger In Partial Fulfillment of the Requirements For the Degree of Doctor of Philosophy California Institute of Technology Pasadena, California 1967 (Submitted April 28, 1967) ii ACKNOWLEDGMENTS This research was carriedout while on grants from the National Science Foundation, 1964 to 1967, and the Woodrow Wilson Foundation, 1963-64. The author was also assisted by the California Institute of Technology. Acknowledgment is due to Professor Marshall Hall, Jr. who, during the last year, advised the author on the many problems of just remaining a graduate student. Grateful thanks are due to Professor H. Nagao and ~ir. Edward Cline who read completely the initial drafts and offered many simplifying and helpful suggestions. An unwitting assistant has been E. Shult in his paper (8). The author is indebted to Professor E. being his adviser. c. Dade for It was his initial prompting that kept the author on this problem. He read through the many drafts of this thesis and provided major simplifications, especially in sections I and II. The method of approach in section III is largely due to his seminar notes on a theorem of Thompson. It was largely through his courses, seminars, articles, notes and discussions that the author learned about group theory. The author is 'also grateful to those who assisted him along the way, especially those who helped ease the last minute rush. iii Errors in this thesis are the sole responsibility of the author. Time limitations did not allow complete con- sideration of all suggestions. And unfortunately, as always, unfound misprints live eternally under the eyes of the author. iv ABSTRACT Suppose that AG is a solvable group with normal subgroup G where (!Al, !GI)= 1. Assume that A is a class two odd p group all of whose irreducible representations are isomorphic to subgroups of extra special p groups. If pc ~ rd + 1 for any c = 1,2 and any prime 2 r where r d+l divides !GI and if CG(A) = 1 the n the Fitting length of G is bounded by the power of p dividing !Al. The theorem is proved by applying a fixed point theorem to a reduction of the Fitting series of G. The fixed point theorem is proved by reducing a minimal counter example. If R is an extra special r subgroup of G fixed by A , a subgroup of A, where A centralizes 1 1 D(R), then all irreducible characters of A R which are 1 nontrivial on Z(R) are computed. All nonlinear characters of a class two p group are computed. v TABLE OF CONTENTS Acknowledgments ii Abstract iv Introduction 1 I Extensions of Group Characters 4 II Class Two p Groups 9 III Extensions of Extra Special Groups 15 IV Extra Special Extensions 32 v Reduction Lemmas 43 VI The Main Lemma 51 VII The Fitting Structure 67 VIII The Main Theorem 75 Bibliography 77 l INTRODUCTION Suppose that AG is a group with normal subgroup G where (!Al, !GI)= l. Assume that A is fixed point free on G, that is, CG(A) = l. Suppose that IAI is divisible by d primes, counting multiplicities. There is a conjecture that, not only is G solvable, but the Fitting length of G is bounded by d. The solvability half of the conjecture seems difficult and Thompson (10) made the first step by showing that for A of prime order G is nilpotent. Under · the assumption that G is solvable more progress has been made. Without the requirement that CG(A) = l Thompson has given a large upon the Fitting length of CG(A) (9). bound depending The cases where A is of order 4 have been handled (1, 4). A= s 3 has also been treated (7). The case where A large class of abelian groups for A have been handled by E. Shult in his thesis (8). The exceptions to his result relate to numbers like Fermat and Mersenne primes, but his exceptions add in certain composite numbers. These exceptions arise because of certain 11 bad 11 representations on extra special groups. As we allow A to become a nonabelian p group these same exceptions arise. However, in the cases treated here, except for one, these "bad" cases only arise from abelian representations of A. 2 (See (V. 8)). Thus it appears that, as far as exceptions to the conjecture go, if they exist at all, they probably exist for abelian groups. The main result of this thesis is contained in section VI. The proof proceeds by reducing a minimal counterexample. Several comments should be made here. The method of proof seems to be very general. Many of the reduction steps can be made without any hypothesis upon A. Others require the relative primeness and solvability of A. The real restrictions start at (VI. 10). Here a devious route is taken which depends upon I) a remarkable property of class two group characters (II. 2), and II) 7). an even more fortunate coincidence of inequalities (II. Section II then gets us past (VI. 10). All is fairly well ~£til the very end. In (VI. 15) we are forced to invoke the property (*). It is at this point that the exceptions enter. And it is (*) which requires the development of sections I, III, IV. section I are known. (The results of This is a simplification of the proof.) It is highly possible that the results of these sections hold for almost all choices of nonabelian A. It is this route which will be followed in further investigations. In fact, the case when A is an odd class two p group seems to be well within reach and requires only an 3 extension of the argument in section IV. This is now under investigation. The theorem proven here then is, Theorem: If A ~ ~ cla ss ~ 2 odd p gr oup all o f whose representations occur ~ subgroups of e x tra s ne c ial p groups ~ pc ~ rd + 1 f2£ c = 1, 2 a nd, !.££ prime r, . . d.ing r 2d+l d ivi !GI ~ the number S2i., p rimes d ivid i n g !Al bounds ~ Fitting length S2i., G provi d ed G ~ solvable. This includes all class two groups of exponent p (with exceptions on primes, of course). Originally, the result was attempted for A extra special. This choice was made first, because it seemed the next natural step above the abelian case and second, because of the important role played by these groups. Now, all special groups are included. As a sidelight, section VII contains a reduction of the Fitting series of a solvable group. A sequence of useful prime power factor groups is found on which the Fitting length depends. The complexity of the definition prompts the word edifice. But the situation is much less intricate than the words. This reduction in conjunction with the fixed point theorem of section VI is used to prove the theorem. 4 I. EXT ENS IONS OF GROUP CHARACT&-q_s Suppose that AG is a group with normal subgroup G and solvable subgroup A where (!Al, IGI) = 1. Let Q be the field of rational numbers and 8 a primitive IAG!th root of unity over Q. by 8. Let k = Q(8) be the field over Q generated In this section all characters will be k characters. For the remainder of this section we assume A is a fixed irreducible character of G which is stabilized by A, that for all x e G, y e A. is, We start by defining a function which maps characters Suppose that N is any group for which k into characters. • is a splitting field, and ~ any character of N. Then there is a representation by linear transformations A(x) over the field k such that tr A(x) = d(x) for all x e N where tr A(x) denotes the trace of A(x). The representaticn A by linear transformations is uniquely determined up to similarity by ~. In fact, if ~· 1 ~ is another character of N then an associated representation A' is not similar to A. Thus we may set ,i (Ol) ll'N = det A. This function is well defined since A is uniquely determined up to similarity. character on N. It is clear that r/>N (d.) is a linear Hence r/>N is a function mapping characters 5 of N onto linear characters of N. The object of this section is to determine, by use of ¢,all possible characters on AG which contain "A when restricted to G. {I. 1) There exists a uni que characte r i) e!G ="A ii) ~A{S!A) e o f AG satisfying: ~ = lA. We proceed by induction on !Al. Choose A0 A A of prime index p, and let ot be a faithful line ar character on AG/A G. 0 By induction there is a unique character e* of A G such that 0 i) e * IG = ~ ii) ~A {e*IA) = lA • 0 0 The charactere* is fixed by A. </> Ao {S*YI Ao ) = </; Ao <e*I Ao and )Y = 1 Ao 0 For let ye A. y = l Ao• Then Also, e*YIG = (e*la)Y = Ay =~so by the uniqueness of e* we find that e*Y = e*. Hence, e* IAG = [ e. is a sum of p irreducib le l. e. IA0 G = el,A0 G = S* for i = l, ••• , p. characters and l. Now ¢A(61 IA) is a linear character of A with A 0 kernel since ~A 0 <e 1 1A) =¢A {6*IA ) = lA • 0 0 0 0 ~A(6 1 I A) = ~j for some j. such that 0. f6 (l)+j l = lA. i n its And then There is a uni que chara cter We set 0 = Qfel. CJ...f Cl early then 6 ~A(e!A) = ~A((~felJIA) = ~te1<1>~ Ace 1 IA) = GIG= x, and o,.fSl(l)O(_j = lA. Since ei = cA.j(i)el for some j(i), e is unique. (I. 2) Theorem: Suppose ~ ~ any irreducible chara cter of AG ~ that pl G contains A.. Su ppose 8 ,i& ~ u n igue character of AG given i1! (1. l). Then there is~ unigue character ( .2f AG/G ~ that f3 = re Further, if ~ ~e is any irreducible character of AG/G then ,i& ~ unj_guely determined irreducible charac ter .2f AG. 'First, let ~o be any irreducible character of AG which is nontrivial on G. Then by the Clifford Theorems f30 IG = n(~ 1 + ••• +At) where~' ••• , At are all distinct nontrivial conjugate irreducible characters of G. So we have (lG, ~IG)G = o. Second, let ! be any irreducible character of AG/G. We want to compute (~,ofo0 )AG" thenD~0 = aa + ••• or (~fa0 ) IG and (lG, ~o!G)G>o. If it is greater than zero = o(l)~0 IG = aO(l)lG + ••• But, as above, allX.i :f lG so ( 0, 0 f3o) A G = 0 • Third, consider e. Now l = (e,e)AG = (lAG'ee)AG e§= lAG +A where (lAG' A)AG = Q. characters like ~o above. get S o ~ is a sum of Hence, (~, ~A)AG = o. l = (o,~)AG = (~,~(lAG +~))AG= (o,~Se) AG - And we so 7 (0 e, 0 S) AG" So 6 e is irreducible. Fourth, 69 is uniquely determined. For suppose o' is an irreducible character of AG/G such that ~·e = oe. If 0 :/: ~I then (3' e'0 •e ) AG (~,6 1 6)AG = O. This last is zero since, by the second part, to be nonzero, (lG, (o'6)jG)G>O; bu~ it isn't. Ther efore, 0 = DI a...'"'ld 6e is uniquely determine d. Finally, ~ (1) = (~(l)A, J...)G = ((oe) 'G' A )G = (oe, A. IAG)AG" Z:: 0 Further, = IAJA.(l) = AIAG(l). So 2 0(1)(oe)(l) = (60(1) ) "A(l) we get -x IAG = L: ~ a<i) ~ e where O ranges over irreducible characters of AG/G. Since (j3, A. !AG) AG = <13 IG' )... ) G ~ 0 we get the result that there is a unique t such that f3 = ~ e Suppose~ Ms Stab (Aut(AG),A). (I. 3) the stabilizer in Aut(AG) .Qi A., 2£ MS Stab (G(k/Q), A.), ~ stabilizer .2.f "AB! ~ Galois grou p .Q.f k/Q. .Ih.fil! e .Qf (I. l) is stabilized ,Ey M. Suppose x e: M. s6(6 J )x = l A follows, (1. 4) x A ~ lA Then exJG = /....x =A. = ~A(6jA). And ¢1A( exlA) = So by (1.1 ) t he result since Ax is conjugate to A in AG. Suppose ~ (3 is any irreducible charact:er 2£. AG ~ ~IG contains~ irreducible character~ 2f G. Suppose that A0 = Stab (A, n) =[x e: A l"t\x ="J is the 8 stabilizer £1 n in A. Then there is .fil! irreducible character ~ .Q!! AOG/G ~ ~ f1 = (~ 0) IAG where e is given in (I. 1). First we show that Stab (AG, Tr) = A0 G. stabilizes lT. Suppose x e AG stabilizes 1T. wh e r e y e A , z e G• stabilizes "Tr. = 'T"T'y • And lT = TIX = "ITzy " So y e A0 • 0 irreducible v Further = zy Hence y e A 0 ~on A0 G/G so that ~0 is Consider (39) IAG. This character is E 4 ~ ( l) 0 0) IAG = 1T IAG where ~ runs over the irreducible characters of A G/G. choice of~' Then x Therefore, z·y e A G. Next by (I. 2) we find irreducible on A G. Clearly A0 G (l0) IAG = p. 0 So for some 9 II. CLASS TWO p GROUPS In this section we compute the nonlinear irreducible characters of a class two p group. We then use this result to prove a fixed point theorem for a class two odd p group irreducible on a module over a prime Galois field. For the remainder of this section suppose that P is a class two p group, Q is the rational field, b is a primitive !Pith root of unity, and k = Q (o). Suppose that P has ~ faithful irreducible (11. 1) character{}. f3(x) = o · for ill x e P - Z(P). ~ Let x e P - Z(P). By the Clifford theorems f3 lz(P) = a multiple of a single linear faithful character of Z(P). Choose y so that {i,y] = x- 1 xY :/: 1. Then /3(x) = m~, ~(xY) = /3(x [x,y)) = /3(x) ol ( [x,y]) since [x, y] e Z(P). But ol is faithful on Z(P) so <X([x, y]) (II. 2) Theorem: character of P~ 1 1. Hence /3(x) =O. Suppose (3 is ~ faithful irreducible ~ [Pd~j~faithful linear 2!! Z(P) (3 and !Pl =Lo ; outside Z(P) = p 2dlz(P) I. Clearly ~IZ(P) = pd~ for some linear r:J... faithful on Z(P) and pd dividing IPI. Now 10 1 3 (/3' f3 )p = I~ I F.P /3 (x),B(x-1) = l~I P2d x~(P) ~(x)~(x-1) = l~I P2d lz(P) 1. This completes the proof. Suppose13 is~ faithful irreducible character of (II. 3) P. Assume that A is ~ subgroup ~ An Z(P) PIA= pmfA where pm= pd/IAI = 1. ~ ~PA i& the regular character 2f A. This is immediate from (II. 2). on 1 on AfF Suppose P ~ ~ faithful irreducible character of (II. 4) d Let s(P) be the number 2f subgroups A ,:s P of degree p • order p ~ ~ An Z(P) = l. ~ s(P) ,:S (~d-1) p p-1 Consider P/Z(P). p By (II. 2) this group has order 2d The largest number of subgroups of order p that P/Z(P) 2d can contain is p -1. This occurs precisely when P/Z(P) p-1 is of exponent p. Let B/Z(P) be a subgroup cyclic of order p. Then B is an abelian group of rank two or one. If it is of rank one then B ~ Z(P) and B is cyclic. If it is of rank two then B = AXZ(P) where A has order p. Further, B 2 contains p -l = p + 1 subgroups of order p. p-1 One of these 11 is <xlxeZ(P), xP = l>. Hence B contains at most p cyclic subgroups A of order p such that Ar\ Z(P) 2d 1 largest s(P) can be is (Pp-l ) p. i.d Remark: The only time s (P) = ( p - 1 p - 1 ) = 1. Hence the p occurs when P /Z(P) is of exponent p and B = AXZ(P), where A is of order p. This case is of special importance because P is the central product of cyclic group Z(P) and an extra special group of 2d+l order p • (II. 5) Suppose that p is ~ odd prime ~ r is a prime E£!. p. Suppose d ~ 2. ~ . p2d+l < rP d-1( p-1 ) unless p = 3, r = 2 and d = 2. In the latter ~ P2d+l < r2pd-l(p-l) 0 Clearly rP p 2d+l • d-1( p-l ) increases more rapidly in d than The extreme case of the inequality for d > 2 occurs at d = 3, r = 2 and p = 3. log 10 3 But here 7 < 7 x .48 = 3.36 ~ 5.4 = 18 x . 30 < loglO 23 x 2. Hence we may assumed= 2. Then we have the table : below. It i s clear that for fixed p extremal cases occur for small r. If the inequality holds for fixed p and r then it still holds if we enlarge p. table shows. Hence the inequality holds as the 12 d = 2 3 3 5 2.40 2.40 3.50 2 5 2 > 1.80 3.00 6.00 log r 2pd-l(p-l) > 3.60 p log p 2d+l < r log r (ll. 6) pd-l(p-l) Suppose p i!, .5!!! odd prime ~ r i!_ ~ prime !!2,! p. Assume d ~ 1. ~ 2. d -l)p p - 1 (p < rt\> d -1 t d-.l r P -1 ~ !11 t 2: 1, except ~ p = 3, r = 2 ~ d = 1,2 • . .!!! ~ latter ~ the inequality holds f2£ t ~ 2. d-1 and ..,..tp .. So ford~ 2 the result follows from (II. 5)G may assume d = 1. d <r-1) ....../ r :tpCl-1- 1 rtp -1 Hence, we That is, we want to prove rt?-1 ~4- l) · (p-1 p = p<p+l) < rt -1 The right hand side is obviously increasing in t. Again we can note that for fixed p the extremal cases occur for small r. Further, if the inequality holds for fixed p and r, it remains valid if we increase p. the table shows that (II. 6) holds if d = 1. Hence 13 p 3 3 5 p(p+l) 12 12 30 r 2 5 2 rP-1/r-1 2p 2 7 31 31 r (II. 7) r 21 -1/r -1 Theorem: 1 p, r ~ prime. Suppose ~ p !!. ~ ~ prime ~ Assume that V !,! ~ irreducible GF(r)[P] module faithful .2!! P. Then there exists ~vector v e v# which~ fixed .!2z !l2. element 91_ P#. We proceed by contradiction. Since r :;. p, ordinary character theory holds. we apply (II. 2) several times. Now Hence, IPI = p 2dlz(P)I so the Brauer character of V is a sum of t algebraic conjugates of the character of (II. 2). The number t = l if and only if V is absolutely irreducible. Hence dim v = tpd. d So there are rtp l vectors in v#. elementwise fixed point free on v. . Cp(v) :;. l then Cp(v) f\ Z(P) = 1. a cycl.ic subgroup of order p. We know that Z(P) is Hence, if v e v# and Further, Cp(v) contains So the l.argest number of vectors in v# which can be fixed by subgroups of order p will be s(P) times the maximum number of vectors in v# which can be fixed by a single subgroup of order p. Now by (II. 2) and (II. 3) we have dim Cy(A) = tpd-l where A is cyclic of order p and All Z(P) =le So the 14 latter number is rtp d-l - l. Hence, we must have d-1 d s(P) [rtp - tj~ rtp - 1. Using (II. 4) and (II. 6) we see we must have p = 3, r and d = 1,2 and t = 1. =2 But now vlz(P) is a multiple of a single linear Z(P) module since t = l implies V is absolutely irreducible. But this means 2 which is ridiculous for p = 3. Remarks: =l (mod IZ(P)j) Hence, (II. 7) holds. (II. 7) holds for p = 2 except possibly when P is the central product of d dihedral groups of order 8 1 2 3 3 3 3 • When d = l, p = 2, r = 3 and p is dihedral then this is a real exception to (II. 7). when d r However, all this requires proof. 15 III. EXTENSIONS OF EXTRA SPECIAL GROUPS In this section we investigate representations on symplectic spaces. We then combine this with results from section I to obtain all characters which are extensions over an extra special group. We assume that V is a nonsingular symplectic space over a field K = GF(r), r a prime, with pairing ( , ): VX.V ~> K+ • Suppose that A is a group represented upon V fixing the form ( , ). That is, (v , v )x = (xv , xv ) = (v ,v ) or 2 1 2 1 (xv , v 2 ) = (v , x· 1v 2 ) for all x E: G; v , v 2 e v. 1 1 1 1 2 Further, we assume that This done, we fix (!Al, r) = 1. ~: A ~> A as that unique antiautomorphism of A which sends x ---> x- 1 for all x E: A. Then Ol extends linearly to an antiautomorphism of K [A) • Assume that e E: K [A1 is an idempotent. . Then e"' is also an idempotent. Further, e is primitive if and only if e~ is primitive. And e is central if and only if e~ If e is central then eV is a left K [A1 module. is central. Let Ke = ker [ A - > Aut eV]. (III. 1) Suppose~ l = e 1 + ••• +et is~ decomoosition of l i!!!2, primitive central orthosonal idempotents 2£, K[A). ~,except possibly~ er= ej, ~ ~ 16 Choose any v , v 2 e V. 1 Then ei~ ej = O. So, (eiv1 , ejv 2 ) The symplectic space V is nonsingular. e l.. V ~ (O) then e.~ V ~ (O). l. e . ~ V = (O). So if Further, e.V = (O) implies l. By choosing complementary bases we see that l. din\z eiv = di.n\z e .<>- V. Further, e.V + e l..ol.. V is a nonsingular i x- 1 e K Since x e K we also e.l. implies ei l. ' subspace of v. have Ke. =Ke.«· l. In ~ notation of (III. 1) ~ have, for all i, (Ill. 2) a) K e. = K e.cJ. l. b) c) So, (III. 1) has the following corollary: l. l. ~ dill\z eiv = d~ e.l. V, and °' e.V + e. V j& ~ nonsingular subspace of v l. l. if e.V ~ (O). - l. For each primitive central idempotent e, eK[A1 is a left K[A] module. Let SA be the collection of all subgroup5 Hof A such that: There exists a primitive central idempotent e e K(A] with eV ~ (O) and H = ker lA ~> Aut e K(A)). For H e SA let EH be the set of all primitive central idempotents e e K[A] such that i) H = ker [A~> Aut e K[A11 and ii) eV ~ (O). 17 So, SA is the collection of kernels of irreducible K(A1 submodules of V. And EH r ¢ is the collection of idempotents associated with irreducible K[Alsubmodules of V having k~rnel H. We then set eH = Le e: Err e. From (111. 2) c) we conclude that VH subspace of V and EH o. = eHV is a nonsingulai:- = EH. And so we have V = L H e: SA + VH. Next we define some numbers. (III. 3) 2m(x) = diflKCv(x), x e: A n(x) = the nu~ber of nontrivial irreducible K [<x>] modules in a complete decomposition of V. Note that if B ~ A is a subgroup of A then all the preceding discussion holds for B also. Further, the numbers in (III. 3) may also be defined for spaces other than v. In particular, we may consider nonsingular subspaces of V. messy. In these situations notation gets a bit Distinguishing notation is added in these cases. We have set up the apparatus needed from symplectic spaces. We proceed now to develop apparatus for extra special groups expressed as central products. Suppose AR is a group with normal extra special r 2sn+l subgroup R of order r • Assume that A centralizes D(R). 18 Consider the K vector space R/D(R) R/D(R) choose x e v = V. If v 1 , v 2 e V = 1 = xD(R) and ye v 2 = yD(R). Then set (v , v ) = [x,y] e D(R) = GF(r)+ = K+. Using the 2 1 identification of D(R) with GF(r)+ = K+, ( , ) becomes a nonsingular symplectic pairing on V = RID(R) into K+. For x e R, y e A we set y(xD(R)) = (yxy- 1 )D(R) = xY -1 D(R). With this conjugation as action V becomes a left K[A] module. Further, A centralizes D(R) so A fixes the pairing ( , ). We now may apply all the notation and machinery developed in the first part of this section. We define RH for H e SA to be the inverse image in R of VH. CR(A) That is, RH is "the part of R" with kernel H. = D(R) then CR(H) ~ RH precisely. If Because VH is nonsingular, RH is an extra special group. (III. 4) R is ~ central product of ~ R8 Since each~~ D(R), ff H e SA. ' ~H = M ~ D(R). HeSA Further, M/D(R) = L + VH = V = R/D(R). Hence M = R. HeSA Next, if H, R* e SA then (~, RH*) = 1. This is immediate since (Ill. 1) applies to show that VH is orthogonal and disjoint to VH*· or equivalently [RH, RH*1 That is, (VH, VH*) = 1. Therefore, R is the central product of the R-u h =0 19 We now reintroduce the field of section I. that Q is the rational field and S is a primitive We let k = Q(S). root of unity over Q. distinct from K = GF(r). Suppose th IARl This field is ln what follows we will be discussing k characters. For the following lemma, the construction of the Let R0 = TT XRH. Also, set HeSA M equal to the subgroup of all 1txy.H e R such tha t the 0 central product is important. product in R TryH = l. is in TI X D(RH). HeSA Since V = L This subgroup is normal in R 0 Further, R ~ R0 /M in a natural way. + VH for y e R, yD(R) = '[::; vH uniquely. zH e vH so that the product in R TI z.H = y. </>(y) = TC and Choose Then setting xy gives the desired isomo-cphi sm. In fact, this is an A isomorphism as is easily verified. (III. 5) Suppose ~ 0H ~ .5!,!! irreducible c haracter o f ~ which i.2, nontrivial fill D(R) = D(RH). Suppose ~ for every H e s A' eH ID(R ) con t ains ~ fixed linear c haracter H X ~'f ~ D(R) = D(R__). _!i Ass thL t x__ • . bl e _ _ m... m... e _""H 1:.2. ~ i• r re d u ci. c haracter .2.{ ~ ~ Xa:IRxi = eH. /3 = I.hfill ~ d irec t product TI --H x.._ HeS A .a& i rreduci ble fill AR ~ A~ R0 /M where A~ g ~ diagon a l subgroup of lTx A. HeS A 20 It is sufficient to note that ~IR irreducible character of R0 = TT eH is an He:SA with M in its kernel. o Hence, p, considered as a character on AR, is irreducible. (III. 6) = H. Suopose ~ A0 = CA(R). ~ ~ ~ CA(Rs) Further, assume ~ (3 is ~ chara cter of AR constructed ~ !!1 (III. 5). Suppose ~ (~IA, 0) > 0 £.2!: every irreducible character ~ 21 A/H. ~ <p IA' (J)A > 0 !2£. every irreducible character a -of A/ A0 • First let us prove the following statement: n isomorphic to a subgroup of B = We know that A0 = .( \ H. map of A into B. X A/H. He:SA So consider the following For y e A let ¢(y) = llx(yH) e B. Clearly ¢ is a homomorphism of A. = l. A/A0 is Assume then that ¢(y) Since y e H for every H e SA, y e A0 = (\ H. Therefore, ¢(A) ~ A/A0 • Conversely, x e ker ¢ if x e A0 • Second, we prove that if YH is the character of A which is the sum of everr.J character of A/H, and TIYH is considered as a character of AA then fjyH contains every character of A/A0 • Now YH is a character of A/H. 11 subgl:-oup 11 of Tix A/H. character of B = Suppose nx A/H. Then <5 Further, A/A0 is a is any irreducible 21 0 n ~H = where OH is an irreducible character of A/H. But YH = OH+ OH'· Hence, 11YH = l1<oH + O'H) = <TloH) + 01 = ~ + o'. Therefore, IT YH contains every char.s:.ct er of B. Finally, assume that character of A/A0 • ~is an arbitrary irreducible ~1B = ~l + ••• + bt where Oi is Then irreducible. · Also, ~l IA/A = <S + • • • • 0 in But lf 1 appears n YH on B so a appears in TI YH restricted to A/Ao. The result is immediate since YH is contained in XrIIA by hypothesis. Character Values From (111. 5) and (Ill. 6) it is evident that, in order to compute the character values on AR, we need only consider the spaces VH. In other words, we need only consider submodules of V which are faithful on A/H. The next few le:mmas are technical in nature and are used to compute actual character values. Look again at A represented on the nonsingular symplectic space V fixing the form ( , ). irreducible K[A] submodule of V. Let W be an Let W0 be the sum of all irreducible K[A] submodules of V which are isomorphic to W. ·.J' , That is, for some HeSA and some e e ~' WO = eV. By (Ill. 2) we know that wo°' = e"v is complementary to W and 0 W0 + W~ is a nonsingular symplectic space. 0 22 First consider the case that e~ ~ e. = o. • Since rad W in W + W 0 d.. Then (W0 ,W0 ) is K(AJ invariant, we may, 0 by choosing appropriate complementary bases, split W0 (W as + W*) + W' a W 0 + W0 + W~ where OI. w + W* and w• are nonsingular symplectic spaces and also K(A1 modules. In particular, if W0 sum of t copies of W then w: = Wl + • • • + wt is a may be = W1* .f. • • • + W* t Wi + Wi* written as the sum of t copies of W* satisfying: is a nonsingular symplectic space. Second, we consider the case that ed = e. is a bit more complex. Here W ~ = W0 , so W0 is a 0 nonsingular symplectic space. It may be nonsingular. This case Two cases can arise for W. In which case W = W + W' where W, 0 W' are nonsingular and K[A] invariant. In the other case W is isotropic since W(\ rad W ~ (O) is a K(A) submodule of W. As above then we may choose K [A] invariant submodules so that W = (W i- W*) .f. W1 where W .f. W* and W1 0 are nonsingular. In any case . where all Wf, Wi* are irreducible K[A] modules isomorphic to W, the Wi have symplectic complement Wi -,'", and the Wj are nonsingulal:; Notice then what the situation is. If W is an irreducible K[A] module of dimension g then V contains an irreducible K[A] module isomorphic to W which is nonsingular, or WO + w0cA. is a sum of complementary paired 23 K[A] modules isomorphic to W or w«. If V contains nonsingular W of dimension g then we say situation (i) arises. If W0 is a sum of paired modules as described we say situation (ii) arises. (Ill. 7) Suppose A~ cyclic. consider VH. Assume that H e SA~ Suppose that W is ,!!! irreducible K[A] .!B situation (mod [A:H]) where g 12, ~' submodule £1.. VH of dimension g. (i) rg/Z,:: -1 (ii) r 8 ,:: 1 (mod [A:H]) Further, every irreducible K(A) submodule .Qi VH ~ dimension g. §.2 g dim VH = hg ~ (iii) rhg/Z,:: (-l)h (mod (A:H] ). If [A:H] = l the result is trivial. If [A:H] = 2 then ((A:H], r) = l by hypothesis so r is odd and rj,:: l ,::~lf (mod 2) for all i,j. So we may assume (A:H] > 2. Let t be the smallest positive integer such that rt.:: 1 (mod [A:H] ). Consider the collection EH of all primitive idempotents of K[A] such that eVH 1: (O). If we take e e EH then eK[A] is an extension field of K = GF(r) since A is cyclic. As a left K[A) module, eK[A) is faithful on A/H, so eK[A)::::;: GF(rt). diIIJ< eK[A]= din\.c. GF(rt) = t. In particular, This holds for every e e EH. Therefore, t = g is the dimension of every irreducible K[A] submodule of VH• 24 ln situation (i), there is an irreducible submodule W of VH which is nonsingular and hence of even dimension g. =l Now rg (mod [A:H] ) and [A:H] > 2 so rg/ 2 =-l (mod [A:H] ). In situation (ii) we obviously have rg = l (mod CA:H]). For (iii) we consider situation (i) first. Here we just raise the congruence of (i) to the h power. we consider situation (ii). modules so h is even. =1 rhg/ 2 Second Here VH is a sum of pairs of That is, (-l)h = 1 or = (-l)h (mod [A:H] ). This completes the proof of (111. 7). We now build a character. Assume that H e SA and consider RH, the inverse image in the extra special group R of VH. Suppose that dim VH = hg where g is the dimension of an irreducible K LA] submodule and A is cyclic. (111. 8) Suppose ~ A ~ cyclic ~A is ~ nontrivial linear character~ D(RH). ~ rhg/ 2A(z); x = yz, ye H, z e D(~) X)l.(x) = (-l)h).(z); X"'YZ, ye A-H, z e D(RH) elsewhere 0 l l !!.!1 irreducible character .2.f ARii· By (II. 2) rohg/2). (x) (3)..(x) = { is an irreducible character of x e D(RH) elsewhere Ra· The character (3).. 25 extends to HXRH so that ~Ae is trivial on H. Set [A:H]rhg/Z/..(z); x = yz, yeH, zeD(Ra) 0 elsewhere The character A extends to a linear character ~ of AXD(~) which is trivial on A. .Set rhgA(z); x=yz, yeH, zeD(RH) AR 1'1)..(x) = )..e I A(z); X"" yz, yeA-H, zeD(RH) H(x) = 0 By (III. 7) 1-(-l)hrhg/ 2 elsewhere is an integer. Hence, [A:it) xA = is a generalized character of ARH. X A (1) = rhg/ 2 > 0. X ) = rhg [ ). ARH But ~ (A) = D(RH) H )..e(y) A6(y-l) + L )..e(z) i\.e(z-1) yeHXD(RH) y rv ze [A-H)XD(~) = rhg+l IHI + r IA - HI <IRHl/l~(A)I). so = rhg+l IHI + rhg+l IA - HI = IARHI. Therefore (III. 8) holds. 26 Assume~ condition s ££_ (III. 8). (III. 9) 1£. (A:H] character 2f A/H. x e H X").I A (x) = x e A-H [ rhg~~~~-l)h = where Jp A/H + (-l)hlA f A/H is the regular character of A/H. We still consider A to be cyclic, but: now we want to find a character on all of R rather than just RH. (III. 10) Assume~ A~ cycl ic. nontri vial linear character of D(R). Supoose ~ {\~ ~ ~ x e A~ consider m{x) ~ n(x) ~defined ,1!! (III. 3). ~ rm(x)(-l)n(x)X{z); yrvx.z, x e A, z e D{R) 0 ~ ~ elsewhere irreducible character of AR. We apply {III. 5) and (III. 8). From (Ill. 8) for each H e SA we get hH and gH dependent upon H. (III. 5) we form the product character. difficult to see that TI rm(x) = xeHeSA And in the same fashion From It is not 27 n(x) :: L hH (mod 2) • x ~HeSA So that (Ill. 5) yields, using the character of (Ill. 8), t:he values given for Y-;.... · (III. 11) Assume the conditions !2.f (III. 10). 1.f. A0 = ker [A---> Aut R/D(R~ ~ Y~IA contains every character ££. A/A0 provided that (-l) ~ = 1 for every H e SA. Here we apply (III. 9) and (III. 6) in much the same manner as in (III. 10) above. The inequality hypothesis of t his lemma may be improved under certain restricted hypotheses. (III. 12) Assume that A is cyclic ~ A* ~ ~ s ubgroun . Suppose that f A is ~ regular character of A a."'ld = PA - 1 A ~ fA/A-1/f = PA/A·k - lA . l l f3 ~ any linear character 2i, A and 0 = (>A#{i/A*# then (A:A~'') - l ~ := lA [A:A*] -2 f3 :f. 1A, ~I A* = pA# (A:A*] -l ~IA* =/: 1 A*• We treat the three cases separately. 28 (IA! -1) ([A:A-i•) - 1) a (x) = 1 - (A:A')'~ 1 We treat = o = 0 - lA. 0 elsewhere. Then IA I ( i 0 , lA) A [ (IA I - l) ( [A:A')"]-1)-1 )+ (-[A:A*]) (IA>'• I - l) = lAI ( [A:A*] - 2). Suppose f3 ':f: lA, p 'A* = lA*. IA I ( 0° ' /3 )A = [(I A I -1) ( [A: A*] (-[A:A-J~) ·!· (- - l) -1] + (IA*I - l) = !Al ([A:A*] - 2). {31 A* :f lA-;\-• Then !Al (o 0 Then Finally take ,(3) A= [< !Al - 1) ( [A:A"i•] - l) - ~ (I ker ~IA* I - l) + I ker (3 IA7, I [A:A*] [A:A*]) = IA I ( [A:A')"] If in each case we note the value of 1). 10 (lA, ~)A then the proof is complete. (Ill. 13) Suppose that A 1& ~ cycli c p group for .QS1£. p. Assu.me the hypotheses of (Ill. 10). Y"'IA contains every characte r = .J[R:~(A)] £i A/A0 Let A - 0 = CA(R). ~ e x cept when [A:A01 + l .s!!£ R/~(A) 1& ~ irreducible K[A)module. Note that A 0 e SA since A is a cyclic p group. subgroups of A form a chain. So suppose there is an A"i• e SA such that A0 < A* < A. The character Yr.. is constructed from (III. 6) and (Ill. 9). The But since both A*, A0 e SA from (Ill. 12) we get Y~IA containing every character of A/A0 • Since R/CR(A) Therefore, only A and A0 can be in SA. ~A VA the result follows directly from 0 (III. 11). For if A 0 =A the result is trivial. If 29 A0 :f A then an exception can only occur if [A :A0l where [R:CR(A)] = r 2 f = rf + l and R/CR(A) is an irreducible A module. We are now in a position to prove the main theorem of this section. The previous results apply for cyclic A. We remove this restriction now. We apply t:he previous results to cyclic subgroups of A. Before the main theorem we prove a simple lemma concerning the . Galois group of k/Q. ( : 11. 14) Assume that a,b ~ positive integer s ~ (a, b) = 1. Suppose that ~ Sg Q, ~rational field. ~ nrimitive ab £22! !21. unity Then there is x e G*, ~ Galois group .2f Q(S)/Q which fixes Sb ~ t akes &a into 0-a • The automorphisms of k/ Q are given by where (n,ab) = l. n ; 1 (mod a) and n S :__> gn So we need onl y find n so that = -1 (mod b). But a;b = 1 (mod a) = l (mod b) are solvable f or a• and b 8 so n = a 1 b - b 1 a works since (n,ab) = l. and b 1 a Assume ~ AR 1§. .§; solvable group 2m+l with normal extra special subgroup R S2.f o=de r r ~ (Ill. 15) (!Al,r) =l. ~ Theorem: Suppose~ A cen tral izes D(R) . Assume A.ll !! nontrivial linear chara cter 2!1 D(R). ~ 30 rm(x) (-l)n(x) ~ (x)).(z); y""' xz, xeA, zeD(R) elsewhere, 0 where ~ : A - > {l,-l) .!! ~ class function, is .!!!!. irreducible character 2f AR. Further icx) = l whenever l<x>I !!. ~LetA. 0 be the irreducible character of R lying over ~. Then ~o is fixed by A. By (l.l), (l. 3) we may choose an extension of -X 0 on AR, 0 , such that i) e IR = Ao ii) G0 *, the subgroup of G* fixing all rth roots and of unity of the field k = Q(6) where S is a primitive IARI root of unity over Q, fixes e. e of (l.l) is a good choice. This choice of 9 is unique. Further, if A* ~ A is a subgroup then 9IA*R is the corresponding unique character of (1. l) on A*R also satisfying i) and ii). Let x e A. By (III. 10) and (I. 2) 9 l<x>R = Y}. ~ for some linear character f3 of <x>R/R. character. But f3 is an r• By (III. 14) G0 * contains an element taking · But both e and Y~ are fixed by G0 * • So p is a character of <x>R/<x2>R. That is, pmaps x into fl,-1}. ra - > r(-\-1. Therefore~9 = *>.. has the values of (III. 15). 31 Remark: ~(y) = l. If x e A and x 2 = y and [<x>:<y>) = 2 then This follows by looking at *~l<x>R• 32 IV. EXTRA SPECIAL EXTENSIONS Following (III. 10) we proved (III. 11) which concerns itself with which characters appear in Y~IA. Ideally, we would like an analogous result to follow (111.15). For purposes of the main theorem it would be sufficient to have such a result for A a class two odd p group. For such a case, the computations here are incomplete. So we settle for the case where A is the central product of a cyclic group of order p or p 2 and an extra special group. We carry the argument as far as possible for the class two group. Assume that p, r are distinct primes and p is odd. Suppose that Pis a class two p group of order p 2 dlZ(P)I where IZ(P)I =pa. Assume that PR is a group with normal extra special r subgroup R of order r2m+l. Suppose that every irreducible P submodule of R/D(R) = V is faithful, and P centralizes D(R). Let K = GF(r) and k = Q(8) where Q is the rational field and Sis a primitive IPRlth root . of unity. All characters are k characters untess otherwise specified. Recall that V is a symplectic space. The Brauer character of P on v (pjE r) is a sum of t characters as in (II. 2). Let~ Hence, dilIJz V = tpd. We must find out what t is. be the smallest positive integer such that 33 Then for b = l, (IV. ~ r = ml = l (mod p). r 1 (mod pb). SUJ2J20Se c !.§. the largest positive integer ~ l) ml r~ = l - c · (mod p ). Ill!'m mb = ml l l b :s c~ mb = ml Pb-c l l b > c. If b :S c the result is obvious. Suppose b~ c, and n = ml p b-c ' and rn =. 1 (mod pb) but rn i l (mod p b+l) • Clearly mb+l = np. The question then is, what about rnp = 2 1 (mod pb+ )7 That is, what about rnp - 1 2 - - - ;: 1 (mod p )7 rn - l Now (rnp - l) = (rn - l)(rn(p-l) + rn(p- 2 ) + • • • and rn(p-j) - 1 (mod pb) exactly • . Therefore, (rn(p-l) + ••• + l);: p (mod pb). If b > l then we are done. b = c = le for j = O, l, ••• , p-1. Hence, n = ~· 2 jfp + f .p. J So we may assume that Now rnj =l (mod p) exactly Sorn= l + fp and rnj = l + Therefore ~ ~:5 rnj .=. p + fp [: ~:5 j = p(l + fp(2-l)) 2 p (mod p ) since p is odd. (IV. 2) m8 GF(r ) a IZ(P) I = p • Hence the result. 1! ~ splitting field 12£ P ,2E V where 34 The character of an absolutely irreducible P module over an extension of GF(r) is given by (II. 2) and takes ma = p 2d then ) exactly. lf m an irreducible GF(r a)[P] module has dimension pd over IPI its values in GF(r lz(P)I some finite division algebra by the Wedderburn Structure Theorems. So by the Wedderburn theorem on finite division ma algebras, GF(r ) is the splitting field for P . (IV. 3) 1" jz(P)I =pa~ t =man where n ~ ~ number 2£. irreducible GF(r)[P] modules in~ decomposition 21 v. The dimension over GF(r) of V is tpd. By (IV. 1) and (IV. 2) every irreducible GF(r)[P] submodule must have dimension mapd. There are n of them so tpd = manpd. Hence the result. Next we compute information concerning (IV. 5) a) n(l) .= 0 m(x) and n(x). (mod 2) m(l) = m b) ll x e P ~ n(x) .=n m(x) c) <x> (\ Z(P) ~ 1 (mod 2) =0 l l x e P, <x> (\ Z(P) = 1, ~ ~ ~ n{x) m(x) =0 {mod 2) = m/pf. l<x>I = Pf 35 The K dimension of V is 2m. m{l) = m. immediately that n(l) Hence {lII.3) shows Further, n(l) = 2m so =0 {mod 2). Next, Z(P) is fixed point free elementwise on v. <x> So if x GP <x>n Z(P) · ~ l then and is fixed point free elementwise on v. m(x) = O. l<x>I = pf then an irreducible K(<x>] If submodule is faithful of dimension mf. n(x) Hence = 2 m/mf =t/mf = m1pa-cn/m1pf-c : n pis odd. Therefore, Finally, for x s P, <x>n Z(P) {mod 2) since = l, and l<x>I = pf we find from (11. 3) that <X> acts as tpd-f regular representations on v. Therefore, m(x) = tpd-f /'2 = m/pf. Now (v,<x>] has dimension 2m - (2m/pf) In other words, if f is the regular = {2m/pf) (pf -l). representation of <x> then <x> is represented upon [v,<x>) as 2m/pf times p- l. We notice that for an appropriate extension field f contains pg - pg-l absolutely irreducible representations with kernel precisely Hence [v,<x>] contains {2m/pfm ){pg-pg-l) of order pf-s. g irreducible submodules with kernel precisely of order pf-g. from But l to n(x) f - is the sum of these numbers as l. Fix g. Then g<f .S d .S a l<x> I .S Pd and f [<X>,P] I = pa. g runs since Therefore, (2m/pfmg)pg-l = {ma/m )npd+g-f-l isan integer; so n{x) is even since p -1 8 is even. This completes (IV. 5). (IV. 6) a) r[m{p-l)] /p =l (mod pd+a-l) 36 b) f2E s > 0 = sp2d+2a-l > 0 rm _ (-l)n _ p(rm/p _ (-l)n] unless = n = l, d = a ~ m = p = 3, r = t = 2. To do this we require (Ill. 7). representation of Z(P) on v. faithful Z(P) module over We examine the Since an irreducible K always has dimension ma and since vlz(P) is a S\41.~ of such modules, vlz(P) must contain tpd/ma = npd irreducible Z(P) modules. case p is odd. a) In our If n is even then follows by a simple computation. If n is odd we are in situation of (Ill. 7) and rma/ 2 = -l {mod pa). i) We raise this expression to the npd-l{p-1) power to get a) since [m(p-1)]/p = (ma/2) (npd-l(p-l)). For b) we rewrite rm - (-l)n - p(rm/p - (-l)nJ = rm/p(r[m{p-l)]/p_p) + (p-1)(-l)n. q this becomes Using a), for some rm/p(l + qpd+a-l_p) + (p-1)(-l)n. We assume this number is less than or equal to zero. rm/p(l + qpd+a-l - p) ~ (p-1)(-l)n+l. Hence, But the left hand side is positive so n + l is even. Further, the left hand side is greater than p-l = 1 and d + a - l = 1. This forces d = a = l. unless q And [m{p-l)]/p become~ So rt(p-1)/2 = l + p. So r = 2. Now t = m n 1 ml r = l + fp for some f. But m1 ~ t(p-1)/2 so and m1 = m1 n{p-l)/2 = t(p-1)/2. Therefore, n t(p-1)/2. and f = 1 = 1 and 37 p = 3. From here we get d = a = n = l, r = m1 = m1 n = t = 2, and p = tpd/2 = m = 3. We argue on congruences for the rest. As above. when m n/2 is even, r a (-l)n {mod pa). And when n is odd = n m n/2 r a ={-l)n {mod pa). In particular. rm/ P = Therefore, rm/p = (-l)n + fpd+a-l. (-l) n (mod pd+a-l). J l:'m = [(-lf + fpd+a-1 p = (-l)n + fpd+a + L: J.~ 2 (~) (fpd+a-l)j (-l)n(p-j)• J And finally rm - (-l)n - P [rm/p - {-l)n] = [: j~ (~)(fpd+a-l)j (-l)n(p-j) = 0 2 (mod p2d+2a-l). From this, and the above argument, b) follows. We now put strong hypotheses upon P. We assume that P is the central product of an extra special group of order pZd+l and a cyclic group of. order pa where (IV. 7) a = 1,2. ~ lz(P)j = p ~ P conta i n s a) p 2 d+l -1 elements 2! o r d e r p .Q.£ p 2d - l e leme nts 2! o r d er p ~ p 2d(p- l) b) 2 e lements .Q.! order p • If !z(P)I = p 2 .!hfil! P c ontains pZd+l - l elements o f order p ~ p 2d+l(p-l) e lemen ts 2! order p 2 • Here P' is of order p so x ---> xP i s a homomorphism of P whose kernel contains all elements o f orde r p plus l , the i dentity. In case a) P is of exponent p and in t he remai ning cases P is of exponent p 2 • Using t h e order o f . 38 P equal to p 2dlz(P)I, the result follows easily. P is ~ central product 2£ Z(P) ~ Assume~ (IV. 8) ~ extra special p group. ~ .QB PR has ~ property ~ (*,._lp,1-1.)p > 0 !!?.£ ill characte r *.,._of (Ill. 15) irreducible .QB P except~ IPI = 27 ..fil1£ IRI = 2 M- 7 ~ R/D(R) ~~faithful irreducible K[P]module. We compute the inner product directly. First suppose that ~Ip• Then ~IZ(P) :f lp•• for a faithful linear character M 0 of Z(P). _rmd a" p+pl-J// (-l)n(x).M.o(x -1). By (IV. 5) xeZ(P) n(x) =n {mod 2) = pd~o IPl(*~lp,J.L) b) so = rmpd - (-l)npd 1 0 Therefore Second, suppose that ~Ip• = lp•• IP I (*;a.I p,M.) = rm + L ff (-l)n(x)rm(x)~(x- 1 ). xeP For our group P, for every x e ?. xP e Z(P) So for every element x of order p 2 , <X>C\ Z(P) :f 1 m(x) = 0 and n(x) = = n order p, <x> (\ Z(P) = 1 x A) t Z(P) then n(x) 0 (mod 2). unless For the elements of x e Z(P). If xP = 1 and = m/p. Here a = l. (mod 2) and m(x) Suppose that P is of exponent p. i) and so Assume that J..\ = lp• = ~ _ (-l)n + p(-l)n + rm/p (p2d+l -p) > o. 39 order p 2d In this case ker µ.. is of p elements not in contains p 2d :f lp• Assume that )J. ii) . So Ptff'\ ker - JJ.. Z(P) and p - 1 elements i n Z(P)ii. There are then p 2d+l - p 2 d elements on which µ. is nontrivial. Thus = rm _ (-l)n + p(-l)n + prm/p(p2d-l _ l) - prm/p(p = rm 2d - p p - 2d-l l ) (-l)n - p ( rm/p - (-l)n] except when r > 0 = 2, m = p = 3, and n = 1. Suppose P is of exponent p 2 • Here a = l, 2. 2 contains p 2d+a - p 2d+a-l elements of order p and B) Also, P p 2d+a-l - l elements of order p. Assume that i) = lp• M = rm _ (-l)n + p(-l)n ~ rm/p(p2d+a-l _ p) + (-l)n(p2d+a _ p2d+a-l) = rm (-l)n - p[rm/p - (-l)n] + P2d+a-l(rm/p _ (-l)n) + (-l)np2d+a = sp2d+2a-l + p2d+a-l(rm/p _ (-l)n) + (-l)np2d+a > 0 since 2a - l ~ a. :f lp• ii) Suppose thatµ. a) Assume ker}J. has exponent p. The n (kerA )fl is precisely the set of all elements of order p. = rm _ (-l)n + p(-l)n + rm/ p( p2d+a-l _ p) (-l)n(p 2d+a - P p - 1 2d+a-l ) 40 (-l)n - p [rm/p - (-l)n] + p2d+a-l [rm/p _ (-l)n] > o. Assume that ker u contains an element of order Then it contains pZdTa-Z(p-1) elements of order p 2 b) p2• 2 2 and p d+a- - p noncentral elements of order p. Outside of ker)..l there are (p 2d+a-l - p 2d+a- 2 )(p-l) elements of order p 2 and p 2d+a-Z(p-l) of order p. = rm _ (-l)n + p(-l)n _ rm/pp2d+a-2 _ (-l)n(p2d+a-l _ P2d+a-2) + r m/p( p 2d+a-2 - p ). + (-l)n(p2d+a-~ _ p2d+a-2) = rm - (-l)n - p[rm/p - (-l)n] > 0 except for the noted cases. It is always true that P - ker .u. contains an element 2 of order p2 • Suppose x e ker J..J.. and l<x>I = p • Suppose y e p - keru and yP = 1. Then (P) (xy)P = xPyP [ x, y] 2 = xP :j: l. But )..l(xy) = .U(y) :j: l. Hence xy e P .:. kerJ.l • So we have treated all possible cases and (IV. 9) Sunpose ~ P .!.§..~extra special 29,S p group. Assume ~ ·that P 1 ~ P, ~ P 0 A P 1 , ~ P /P 0 ~ 1 a faithful irreducible ·k[P /P 1 module. Then P /P is 1 o1 0 2 cycli c of exponent p .2£ p , ~ P /P0 12, ~ central 1 41 S2.£. exponent or p x e p If then so the exponent is r. is nonabelian. > D(f). 2 p • Xp€D(P) = Z(?) p 2 or p • p = Pl/Po• Set Then D(E) = D(P)P /P 0 But Z(P) is cyclic. since f./Z(f.) must lzCP)I = p; and 0 • Suppose Hence, Z(P) The result follows easily be of exponent and be a nonsingular p symplectic space. (IV. 10) odd Suppose extra special c = 1,2. Suppose iQ!: any ~ rdlrm '*)..' for every character~.Q! a11 J.t I i 21 character~ P PR P P0 = Cp(R). Suppose~ Ass ume~ ~ IRI = r2m+l where P p' .1.l ) p > 0 ~ P/P 0 ~ ~ ll Ip (*)I.Ip,~ )p = 0 ~ i, 0 il i2.£ *" ~ ~ given,!!! (III. 15). As in section III, let subgroups of i§. ~ group ~ n orma l PR R (r ~ p). r subgroup centralizes D(R). pc ~rd + l is ~subgroup 21 ~extra spe c ial Assume ~ group. p P Sp be the collection of all which appear as kernels of irreducible K[P] submodules of V = R/D(R). For He Sp l a t EH be the collection of all primitive central idempotents of K[P] where eV I (O) and ker [P - > Aut eK[P]= H. Let VH be the corresponding subspace of V and RH the inverse 42 image in R of VH. Note that (P/H)RH is a group just as was considered in this section. That is, VH is a sum of By (IV. 9) the faithful irreducible K[P/H} modules . character *A of (III. 15) on PRH is either that considered in (lV. 8) or that in (111. 13). Applying these two results with (III. 6) gives the conclusion. Assume~ (IV. 11) odd p group. extra special P ,;!;§. ~ subgroup 2f ~extra spe cial Assume ~ PR ,;!;§. ~ group with norma l r s ubgroup R (r ~ p) £i_ order r 2m-:·1 • Suppose Cp(R) = 1. Ass'Uffie that Assume !hfil:. pc ':f rd + 1 £.2.E. P centralizes D(R). c = l,2 ~ d :;: m. Su ppose X ,;!;§.~irreducible character .Q.f PR nontrivial .Q!1 D(R). ~ For ~ irreducible on P, (o ~, lp) P > 0. By (I. 2) and (III. 15) *~· Hence the result. X = ~ */\· But by (IV. 10) a~ is in 43 V. REDUCTION LEMMAS Suppose that AG is a solvable group with normal (!Al, IGI) = 1. subgroup G where qmq 0 where (q, q 0 = 1, ) Suppose that m ~ 0, and q is a prime. Suppose that Q = GF(q) or the rational field primitive q 0 !GI = ~ndS is a IAI, (q, !Al) = l, root of unity and k = Q(S). Suppose that V is a k [AG] module. (V. l) Suonose ~ J Suppose A' .S A* .S A ~ Al .S A. i& .sm irreducible k CA.1) L = ker [A I= C . JI 1 mo dule . Suooose ~ - > Aut J) 2: Al(\ A*. ~ ~ A(A*). ker (A-> Au1; I] = LA*. JIA1 A')" (A*) = J o' has kernel LA*. First suppose C A A* Jl = (A*, JI 1 ]. Then JI A1 A* Set: = J 0 ~ J 1 as a k[A 1A*) Let J' be an irreducible component of J • 1 Then [A*, J'] = J'. Hence, [A*, J' IA]= J' IA. So module. must be contained wholly in = -rrf 1 A*"' 'lt!liJ0 ILA*" is both a trivial LA* J 0 IA. But Now A*, LA* A A and A* module. J 0 I IAILA* so 1t l'&J 0 ILA* Hence, J 0 jA = I. So we may assume that A A* = A and prove the lemma 1 in that case. 44 A A* ~LA* A ---------- l l~L~ ~A* ~ ~ A*('\ A 1 ~ I ILA* = J IL ILA* since -LA* J A /\LA* 1 A1LA* = A1 A* and L ~ A1(\ LA*~ L(A1 f'\A*) trivial on JIL so JI = L. But Lis A A..,': 1 ILA* '::::LA* (dim J) l L ILA* where lL is the trivial L module of d imension 1. Nex t , LA* dim Ho141<(LA~(lLA*, lLI ) =dim Ho111<[L)(lLA*IL, lL) = 1. So dim C A A*(LA*) = dim J. JI i contained in I. Clearly C A A*(LA*) is JI i But . LA* =dun Ho~[L](lA*I IL, lL) = 1. A J I l A* ILA-I: IA* And, in addition, ~Ai• (dim J) 1L ILA*I Ai( • Therefore dim l A A*(LA*). Hence C A A-lc(LA*) JI 1 JI i LA* is in the kernel of I. dim J = dim C Suppose B >LA*. = l. = So Then since A1 B = A1A*, [ A1B: B] = [A1 : A1 (\ B] and A1 (\ B 2: L we must have A f\ B > L. 1 Now 45 is a sum of faithful irreducible submodules. So the kernel of l is LA*. (V. 2) Suppose A' ~A*~ A~ A1 ~A. k(A1 G] module~ V .-v:AG ulAG. i) Suppose U !.! ~ ~ Cy(A*) = (0) if ~ only i4, SJ(Al (\A*) = (0) • .ll Cv(A*) :/- {0) ~ ii) CACv{A*) = A*CA Cu(Al()A*). l We know that vlA* ~* ulAGIA* IVA* But then Cul A G f\A* 1 (A*). IA* dim Ho~[A (\ A*] {lA* IA "A*' U IA "A*)• l 1 l Hence Cu(A1 f'\A*) = (O) if and only if Cv(A*) = (O). Remark: With A* = A this says, c0 {A1 ) = (O) if and only if 46 Cy(A) = (O). To get ii) we apply i) and (V. l). Suppose ~Hi§.~ group~ normal subgroup N {V. 3) of index n. U l l ~ H module Suppose~ .Qi. characteristic~~ prime !2 n. !§. completely reducible. Let 1 = "Kl' • • • , ~ ™ ~ f i e ld K As s ume tha t UjN U is completely reducible. lfn be coset representatives of Then 1\.l. n j -- 1\" (i, j)nij where nij e N. Let J be an irreducible submodule of u. Then UjN = Nin G. . JIN + I where is an I N module. UjN is compl,etely reducible. Let uniquely where y e J and z e I. We may find 1 since w e u. Let </> : Then w = y + z w - > y be the usual projective K[N]homomorphism of U onto J . Set cf = n-l Ci ~i;, ni-1 • Note that n-l exists in K. This is clearly a K(H1 homomorphism of U into U since </> is a K [N] homomorphism. For 1t. -lz. 1 l. J. w e U we have uniquely ~i-lw = ~i~lyi' + where 11"'J.. ·ly.l. ' e J and 1T.J. -lz.l. 1 e I. an H module so y.J. 1 e J. ~ (w) = n -1 L i J. L . ir. "' 1\ . -1y. - n -l L . -rr "i ~i -1Y· , = n-1 L J..y. , "" J • ''ii" 1t.J. -1w -- n -1 .J.. ( l. J. I l. l. _ l. l. In other words, if w e J then ~(w) = w. Hence, q is idempotent. l. Hence, "7T" ,I. + .,.,. "· -1 z. 1 ) Now J is l. So the kernel of ~ is a K[H)module ~ 47 complementary to J in u. Therefore, U is completely reducible. (V. 4) Suppose~ V l l ~ completely reducible k(AG] module. Assume M ~ G .!E,. normal in AG. Suppose A1 ~ A. ~ vlAM il completely reducible. 1 By Clifford's Theorems VIM is completely reducible. Hence (V. 3) applies to VIA M• l (V. 5) Suppose .th§!.! H .!.§.~group~ normal subgroup N .2.f index n. Assume ~ U l l ~ completely reducible N module ~ ~ field K .2.f characteristic 0 .2!: prime !£ n. ~ ulHIN i§. completely reducible .§..2 ulH 12, completely reducible. n 1 = 1, ••• , Let N in H. Then ujHjN ~n be coset representatives of = 1'"1~u ~ it2~u e ••• Q itnW. Suppose U = U1 + ••• +Us where the Ui are irreducible N modules. 1T ~ = Then Tii~Uj i ""·l.i&Us where the are irreducible N modules since N A G. ulHIN is completely reducible. So So by (V. 3) ulH is completely reducible. (V. 6) Suppose V i l !!!,l irreducible K [AG] module .§E£ VIA Gil !!Q.! homogeneous :£.2.E. A0 G A AG. Assume~ All 0 nilpotent. ~ index n ~ ~ there ,i& ~ subgroup A0 ~ A* A A of prime 48 ... .... un We know that A G is normal in AG since A is nilpotent. 0 So by Clifford's Theorems VIA G is completely reducible. 0 So VIA G = vl + ... +Ve where the vi are homogeneous 0 Let A = Stab (A,V ). Since A0 G A AG, 1 1 A G = Stab(AG,V ). So v is an irreducible A1 G module 1 1 1 and v (A G)IAG e:!AGV. But A is nilpotent so there ·is 1 1 Al S A* A A maximal of prime index n so that VIA*G components. . ... + Un where the = ul + A ';ltG with ul (V. !::t A*G and so UllAG ~AG V. Suppose~ YAX S G ~A invaria nt subgroups 7) in AG. vl (Al G) I ui are irreducible AitG modules 1f A fixes~ coset xY . for x e X ~A fixes ~element xy e xY. Further CX/Y(A) = Cx(A)Y/Y. Suppose that A fixes the coset xY. Let AY act upon xY by ay: h e xY - > ayha-l where a e A and y e Y. Since A fixes xY and Y A X we see that AY fixes xY permuting the elements. Further, Y acts transitively as the regular representation of Y. The subgroup of AY fixing h e xY is then of order (AY: Y] = IAI. Therefore, it is a Hall !Al subgroup of AY and is conjugate in AY to A. Therefore A fixes an element of xY. obvious. The rest is 49 (V. 8) Suppose M .::S G ~ normal iE AG. ~ ~ may Let A0 = A f\ (Af-'l)1t. 1 = Af\ (AM)lt. Now n-M e CG/M(A0 ) . choose n• e ~M so that n' e CG(A0 ). CA(n') 9) n- e G. choose n' e: 1t M ~ ~ CA(7t') = A(\ (AM)l\" (V. Assume So we may Then = A() (AM)n' =A(\ (AM)n = A0 • Suppose p I I G J. Then A fixes P, ~ p Sylow subgroup .2£. G. Let N = NAG(P Choose P 0 a p Sylow subgroup of G. Then NG = AG. Suppose x e AG. 0 !NI. words, N contains a Hall IA! subgroup, A0 • 0 0 • So Hence AG .::S NG. Next (IGl,!A!) = l so IAI divides z e G so that A z =A. In other There then is By putting P = P 0 z we get A~ NAG(P). (V. 10) l.f H .::S CG(A) fill9. N = NG(H) ~ N = CN(A)CN(H). We apply the Three Subgroup Lemma here which says that for subgroups H,J,L of a group [H,J,L] = l and [J,L,H] = l implies [L,H,J] = l. Clearly [A,H) = l and ·. ). Now P 0 x is a p. Sylow subgroup of G so there is y e: G with P xy = P xy e N or x e Ny-l .::S NG. 0 [H,N) ~ H. Therefore 50 (A,H,N) S CG(H). = (H,N,A) = l. And so (N,A,H) = 1 So by (V. 7) N = CN(A)CG(H). CG(H) = CN(H). or ~~,A) But obviously 51 VI. THE MAIN LEl1t-'i.A In this section we prove the major res ult of this thesis. It is the lemma which makes everything work. The familiar technique of reducing a minimal counter example is used. The pattern was set in E. Shult's work (8). The object is to reduce the minimal counterexample so that the character lemmas of sections Ill and IV may be applied. The result is carried out for a general class two odd p group. Only section IV prevents us from getting the strong result. In the beginning reduction steps are fairly complete. As arguments are repeated often they become shorter with use. Suppose that A is a group and CIA!, r) = l for a primer. Assume that pis a prime and (!Al, p) = l, p1r. Suppose that AR is a group with extra special normal r subgroup R. We assume that A is irreducible on R/D(R) and trivial on D(R) with A0 = CA(R). Suppose that Q is GF(p) or the rational field and Sis a primitive !AR.Ith root of unity and k = Q(S). Assume that V is an irreducible k[A..~) module which is nontrivial on D(R) and Cy(A0 ) suppose that for any n r 2c+l I IRI. (*) = v. We also I exp A, n ~ re + 1 for any c if A1 is called a (*) group if VIA contai ns the 52 trivial A module for any R, V, and A~ A satisfying the 1 above conditions. By (III. (IV. 11) abelian groups are (*) groups. By 11) an odd p group P is a (*) group if all of its irreducible representations are cyclic or the central product of an extra special p group with a cycl i c group of 2 exponent p or p • That is, all irreducible r epresentations are subgroups of extra special groups. and factor groups of Note that subgroups (*) groups are (*) groups. In particular, all class two groups of exponent p are (*) groups. We do not know if all those of exponent p 2 are (*) groups since there is a class two group P of o rder p 6 of exponent p 2 where P/Z(P) is of rank two and exponent p 2 and P has a faithful irreducible character. not covered in section IV. The lemma is stated on the next page. This group is 53 (VI. 1) Suppose ~ A ~ ~ p group -2.f class Theorem: _s 2 for £££ p. Suppose A l2, ~ (*) group. ~!!solvable group with normal subgroup Suppose~ = 1. ~ (q,q 0 ) = l. As su."!le ~ AG Gwhere (!Al, !GI) !GI = qmq0 f2E. ~prime q f p (m ~ 0) Assume k = Q(S) where Q = GF(q) .2E the rational field~ S 1.2. !! primitive 1Alq0 ~ of unitv . Sunoose V l l ~ k[AG] module faithful 2£ G. Assu.me . ~ i) V ~ ~ ~ of eauiv alent irreduc i ble k[AG] rrodules ii) !! exp A = pa ~ pb ~ re+ l f£E. l ,S b < a ~any primer~~ r 2 c+ll IGI. ~ 1) Cv(A) 1- (0) 2E 2) Cv(A') = (O) 2!: 3) Cv(A') 1- (0) implies there l l cycl ic D < A~ a) CV(A'D) = (0) - ~ b) CG(A'D) > CG(A'). The proof begins on the next page. 54 We assume that (VI. 1) is false and choose a counterexample (A, G, V) minimizing IAI + !GI +dim V. So we have the following: (VI. l') Cv(A) = (0) and 2') Cv(A') :f: (0) and 3 I) for any cyclic D ,!S A 2) a•) Cv(A 1 D) =f: (O) b') CG(A 1 D) 'f:. CG(A 1 ) . or V !§.. ~ irreducible l<[AG] module. Here V = v 1 + •.. i- Vt. is a sum of equivalent irreducible k[AG] modules. Hence, (A,G,V 1 ) is a counterexample if and only if (A,G,V) is also. So t = l. (VI. 3) VIG is a multiple of a single irre ducible A G il\.b-- - -- - 0 module i..2.£ every A0 l::i. A. 1!! p articular , VIG i& homogenev~. Suppose not. 6) there is A0 _!S A1 l::i. A of pri~ By (V. index p so that ••• where the Ui are irreducible A1 G modules and V NAG u 1 1AG. Let Gi = ker[G - > Aut Ui), Qi= G/Gi. Clearly (A1 ,Q1 ,u1 ) satisfies the hypotheses of (VI. l). Hence, (VI. Now vlA l) 1) holds in this case by induction. ~ u1 1AGIA ~A u1 1A IA. l So by (V. 2) Cu (Al) = (O) if and only if . Cy(A) = l (O)~ 55 2) Also by (V. 2) we have, since A .2: A' 1 <o) :f cu ( A1 f\ A• ) = c l u1 (A• ) ;S Cu (Al'). l Hence we find 3) D ;:: A1 cyclic so that there is = (O) CU (A1 1 D) a") l and CG (A1 1 D) > CG (Alt)• -1 -1 Using the fact that A ~A' .2: A 1 , from 1 1 b") au) we get CU (A'D) < cu (A1 • D) = ( o). al) 1 1 - And CG (D) .2: CG (A1 1 D) > CG (Al') > C,.., (A') 1..:1 -l -l of Al in A. . -l -l l = -nl, . .. . , 1\p as cos et representatives We may arrange the j{'. Further, x e: Al acts upon ui as x y -1 -~:l 1\'i l. G 1 it7" l. so that u.l. = it"iul • = "Ti. -lxl\". acts upon l. l. 1t·1 acts upon Q with the isomorphism 1 of Q onto Q1 .. Since A' ;:: A and A' A A, and upon G. as x ~> 1 1 A'D A A we get ai) CU. (A 1 D) = (0) and l. bi) so -1 CG (A'D) > CG (A'). bl) Choose -1 1 1 CG. ( A D) > CG. (A ) . -i -i So finally a) CV(A 1 D) = (O) b) CG(A'D) > CG(A') and Therefore, VIA G is homogeneous . 0 by (V. 7). 56 (VI. 4) ~every A0 <A~~ Cy(A0 ) Suppose A 0 A 0 ~ = (O). <A and Cv(A0 ) :f (O). Hence we may choose 1 A A and A1 < A since A is nilpotent, and Cv(A1 ) A = (O). Clearly A1' ~A'. So Cv(Ao') ~ Cv(A 1 ) ~ (0). So by (VI. 3), VIA G is homogeneous. Hence, using inGuction, 1 we may apply (VI. 1) to (A , G, V). From the foregoing, 1 it is clear that we have 3) a I) Cv(A1 1 D) = (O) b') CG(A1 1 D) > CG(A1') and for a cyclic D ~ A • So 1 a) Cv(A'D) ~ cv<A •n)::: (O) 1 CG(D) > CG(A1 1 D) ~ CG(A1 1 ) ~ CG(A 1 ) b) B.l'ld or CG(A'D) ~ CG(A'). Hence the conclusion. (VI. 5) A!§. faithful .Q.!! V. Suppose not. = ker [A ~> Aut V] • Let A 0 Since G is faithful and V is an irreducible AG module we must have [A0 , G] = l. Hence (VI. Let ~ = A/A0 and t:.1 CV(~) 1) applies to (A/A0 , G, V). = A A /A 1 0 0 • Then A1 =A 1 so clearly = Cv(A) = (O) and CV(~•) = Cv(A') ~ (0). so there is Q ~ ~ cyclic with - a") cv<t:.' Q) = (O) b") CG(t:,' Q) > CG(a•) • and Let D < A be cyclic such that DA /A 0 0 = Q. Then since 57 A'Ao/Ao = /!', a) And Cv(A 1 D) = (O). CG(D) ~ CG(A'D) ~ CG(A') so b) CG(A 1 D) ~ CG(A 1 ) . Therefore A is faithful on v. Choose M < G as a maximal AG invariant subgroup of G. The group G/M is an irreducible A module, where the action, for x e A and ~M e G/M, is -l x(nM) = nx M = (xlrx- 1 )M. From each A orbit on G/M choose a representative tt1M. that ll'lM, ••• So , 7tmM form a complete set of A orbit representatives. 8) we may choose iri, i = l, "Jr. -l = 'tt -l = A()A l. Af\(AM) i - Ai. By By (V. ... , choosing A conjugates of 1\1 = l, ••• , -rrm we get a complete set of coset representatives of M in G; rr = l, 1 • • • ' ltm' • • • , l\"':l AA (AM) J = Aj' j = l, ••• , e. Further, A permutes the l \ j if we specify for x s A that, x(~jM) = lfj(x)M. Now v IG is homogeneous. Therefore, v IM= vi+ ••• + vf with homogeneous components Vi. Further, G is transitive on the Vi's and M fixes each one. That is, f divides IG/MI. 58 (VI. 6) .ll f 1 1 !llim f = e = IG/MI ~~Vi may ~ numbered~ that A fixes V1 , itiVl = Vi, ~A permutes !h2 Vi exactly .!!. ll. permutes ~ n1 • Consider the permutation representation f> of AG on Now M is in the kernel of f>. the Vi's • Further G f'\ ker r/, is a proper AG invariant subgroup of G containing M, so it must be M. Since G/M is abelian, Gf\ker f> is the And now f = e = IG/MI. subgroup of G fixing every v 1 • But ~ is a transitive representation of A(G/M) given on the cosets of some subgroup B of order IA(G/M)j/e = IA!. So Band A are Hall IAI subgroups of A(G/M). they are conjugate in A(G/M). In other words, the representation is given on the cosets of A. fixes, say, v • Hence Therefore A Setting Vi= n 1 v 1 , for x e A we get 1 -1 xvi= x(~ivl) = (xlri.x-l)vl =~ix vl = ~i(x)vl = vi(x)• So this step is complete. l! f ~ l !llim !2£. ~ (A,AM) coset representa- (VI. 7) tives we have 1'ri = l, ••• , 1(m -vJA ~ V "VA Li~lti> vllA IA, i ~AG Vl(AM)jAG. Since AM stabilizes vl and lstab(AG,Vl)I = IAGl/e = IAMI we have AM= Stab(AG,V1 ). Now M ~AG so V ~AG 59 Vl(AM)jAG. By the Mackey Decomposition we get vlA ::YA Vl(AM) IAGIA Cl.A -1 since (AM)...,.i n A Remark: ci:l$ 1\iVll 1t'-l IA 1 . (AM) f\ A = CA(1\'i) = Ai. If v11A. contains the trivial Aj module then J contains the trivial A module by (V. 2). So (0) implies that Cv (A.) = (0) for each j = l, • • • l , m. (Hence also for j = 1, ••• , e.) ~ = ker[A ---> Aut G/M] • Let (VI. J 8) l l v 1 1Ar1 ~ !l£.! contain ~ trivial Ar1 submodule ~ f = 1. (i.e. VIM~ homogeneous.) Suppose v 1 1~ does not contain the trivial~ submodule. Now ~ ~ AG since [~,G] ,::: M and ~ ~ A. So v 1 1Ax1'1 is a homogeneous A.M1- module by (VI. 3) and (VI. l) applies to (A.i~,M/M ,v ) 1 M 1 1 where = kar[M - > Aut v 1) . by induct ion. By assumption Cy (~) = (O). l Next ~ ,::: Aj for every j. j. So ~· ,::: Aj ()A' for every If Cv (~') = (O) then Cy (Ajf\A 1 ) = (0) for every j. 1 1 60 I Hence by (V. 2) Cv l IA (A') = (O) for every j. Hence Aj So we must have Cv (~ 1 ) ~ (O). l This means that when we apply (VI. 1) to 1 1 (~,M/M ,V ) we have 3) a") Cv (~ 1 0) = (O) and l b") CM/M (~'O) ~ CM/M (~'). 1 1 Set Mi = ker[M - > Aut Vi), !:li = M/Mi . ~'D ~~so ~'D Now is centralized by every 1'1 • Hence conjugation of ~'D by ni1 fixes '\r'D elementwise. Therefore, So by (V. 8) That is, And Again, since every 1ri. centralizes ~· Cv(~'D) = (O). That is, a) Hence f = 1. Cv(A 1 0) ~ CV(~ 1 0) = (O). 61 (VI. 9) If f .Y. A/P;v1 1&, abelian ~ f = l. ~ l then A is cyclic and irreducible on G/M. ~i ~ n = l then the orbit [n~ I x e 1 A/~. A} If is faithful on Hence it is the regular representation of A/~. That is, Ai = ~· By the remark preceding (VI. 8) and (VI. 8) we are done. (VI. 10) 1f A/~ = A ~ nonabelian ~ f Now G/M is a.n r group for a prime r. = l. But ~ is a class two p group which is faithful and irreducible on the GF(r) module G/M. So we apply (II. 7) to get a 7\~M which .L In other words, C~(n ) = 1 1 is fixed by no element of A#· or CA(TTi) =Ai= ~· So again by (VI. 8) and the remark preceding it we get f = l. Under the hypothesis of (VI. l) this means VIM is homogeneous and f = l. Now G/M is an r section. So by (V. 9) we may choose an r Sylow subgroup R0 of G fixed by A. Next let R be chosen in R0 minimal such that i) ii) R is A invariant, and RM = G. We eventually show that R is extra special. Next consider VIAM = v1 homogeneous components. + ... +Vt where the Vi are Since VIM is homogeneous, each 62 Vi is faithful and is a multiple of a single fixed irreducible M module. v 1 Since Cv(A') ~ (O) we may choose so that Cy (A 1 ) ~ (0). Clearly Cy(A) = (0) implies l Cy.(A) = (O), i = l, ••• , t. So we apply (Vl. l) to l. (A, M, v ) and obtain D $ A cyclic so that 1 a") Cy (A 1 D) = (O) and l b") (VI • 11) ~(A'D) ,2:: CM(A'). l l A 12, abelian ~ CA(M) = A* rf l. In this case, A1 = l so ~(A 1 D) = CM(D) ,2:: CM(A') = M. Hence D $ CA(M). But Cv (A'D) l 1 ~ D $A*. (VI. 12) = Cv (D) = (O) so l CA(M) =A*~ 1. We may assume that A is nonabelian. Let U be a homogeneous component of v 1 1A'DM• Since VIM is homogeneous, U is faithful on M. Now (A'D)' = 1 since A is class two, A' $ Z(A), and D is cyclic. Since Cy (A 1 D) = (0), Cu(A'D) = (O). = u. Further, cu[<A'D) •] applying (VI. l) to (A'D, M, U) we get 3) and D l So in 1 $ A'D cyclic so that Also since a*) we have D1 ~ l. c0 ((A 1 D) 1 D1] = Cu(D 1 ) = (0), Hence n1 $ CA(M) = A*. 63 (VI. 13) A*() .°'M A = 1 -and CG(A·lc) = M. Suppose A* n '\I = A0 :f l. Now A0 A A so we may take A1 = Z(A)nA0 ~ 1 since A is nilpotent. We know that A* centralizes Mand~ centralizes G/M. Hence by (V. 7) A 1 centralizes A and G. So A1 s Z(AG). But Vis irreduci a l e so A1 is cyclic and acts as scalar multiplication on V by (VI. 5). Hence CV(A1 ) = (O). By (VI. 4) Al= A. But then A is cyclic and a) Cv(A 1 A) b) CG(A) = Cv(A) = (O) and = G ~ CG(A 1 ) = G. Hence A*r\ ~ = 1. But then A*~/~ A A/~1 so A*Ay/~n Z(A/A~) :/: 1 and CG/M(Aic) = l. Hence CG(A*) = M. (VI. 14) ~Em choose R ~~RS CG(M), R is e x tra special, and RA AG. Now G = NG(M). Further, D(R) SM, D(R) S CG(AG). But M = CG(A*) so by (V. 10) G = CG(M)CG(A*) = CG(M)M. Now CG(M) is A invariant so R may be chosen so that RS CG(M). Let R1 Ri.~ = G. = Z(R). Now R1 S CG(M) so R1 ~ Z(G), since Further VIG is homogeneous and faithful so R1 is cyclic and acts as scalar multiplication on v. In particular, because AG is faithful, R s Z(AG). 1 Mand R1 S CG(AG). So R < 1 By the minimal choice of R we must have M\\R = D(R) 64 as the unique maximal A invariant normal subgroup of R. Let R0 be any characteristic abelian subgroup of R. R/D(R) ~A G/M so if Ro <R then Ro ~ D(R). D(R) and so Z(R) = Rl < R. Ro ~ M. 0 ~ But CR (A*) Hence Ro ~ D(R). We already know that R Now ~ But then < CG(M) ('\ M = Z(M) and VIM - Z(R) = R is homogeneous. So R0 and R0 is cyclic. So R 1 is the central product of a cyclic r group and an extra special r group. By the minimality of R, this means R is extra special. Finally, R ~ CG(M) normalizes itself and is normalized by A. (VI. 15) Hence R b AG. VIR ~homogeneous; CV(~) ~ (O). Here VIG is homogeneous. So, since RA G, v!R is completely reducible and the homogeneous components are permuted transitively by M since MR = G. But M centralizes R so vjR is homogeneous. G/M Suppose next that cv<Ar1 ) ~ (O). Now~ centralizes A A. "'A R/D(R), so it centralizes R. Further, ~ iL Hence CV(~) is a k[AR] submodule of v. be an irreducible k[AR] submodule. ~ Let v 0 -< Cv(AM) l Since Z(R) = D(R) Z(AG) it acts as scalar multiplication (nontrivially) on V hence also on V • 0 as A/An· Further, on V , A is represented Now ~< A since Cv(A) k((A/~)PJ irreducible module. 0 = (O). Therefore V0 is a Also A/~ is faithful and 65 irreducible on R/D(R). 2 Now !RI = r c+l I !GI. Further, by hypothesis, pb I re + 1 for any e ~ c and any pb ~ exp A. Hence we may apply the fact that A is a ( * ) group to find that (O) ~ Cv (A)~ Cv(A). But Cy(A) = (O). 0 Hence Cv(i\i) = (O). (VI. 16) (VI. 1) holds. By (VI. 12) A*~ 1. Hence Ai1 <A. And by (VI. 13) A*(\Ar~ = l. So by (VI. 4) CV(~) ~ (O). contradicts (VI. 15). This Therefore (VI. 1) holds. We now curtail the hypotheses on k. WI. 17) Corollary: any subfield 2£. Q(S). 1£ (VI. 1) ~ ma y a s s ume t hat k l l 1£l particular, ~ may~ k = GF(q). Suppose U is a homogeneous K[AQ)module satis f ying all of (VI. 1) except that KS Q(S) is a subf ield of Q(S). Let K(S) = k = Q(S). Let g = ~(U. of Q. Then k is a finite extension of K. Let V be any irreducible l<[AG] submodule Then V is a K[AG] module isomorphic to m copies of U for some integer dividing the degree of the extension [k: r<]. We apply the theorem to (A,G,V). v ~K[AG] ,u e ...,.---e--u1 • m Suppose 66 It is clear that Cv(L) !:!K(AG] lCU(L) 0 ••• e Cu(L~ v m for any L ~ A. Also G is faithful on V since it is on U. The two isomorphisms give (VI. 17). (VI. 18) conclusion 2) arises. 2) Sunpose ~ !,n (VI. 17) Corollary: ~ i.§, CV(A') = (O). t: D ~A' ~ CV(D) = (0) im£ CG(D) = G. ~ there i l 1 a) b) A1 G = V 1 :r ••• +Vt where the Vi are (in the case of (VI. l)) homogeneous components. Let Now VI G.i = ker[G - > Aut V..), • G. -i = G/Gi. • Then we apply (VI. l) to (A•,g1 ,v1 ). Since A11 = 1, and CV (A') = (O) we get by (VI. 1) a cyclic D ~A' so that 1 a') CV (D) = (O) and 1 b') CG (D) = !h· -1 a) CV(D) = (0) b) CG(D) = G. Now D ~A' < Z(A). Remark: So and Again it is no trouble to extend this by the - argument of (VI. 17) to the field K < k • . 67 VII. THE FITTING STRUCTURE Suppose AG is a group with normal solvable subgroup G, and (IAl,lal) = 1. i Consider an AG invariant series = s 0 < s 1 < ••• < st .:s G satisfying, for j = l, 2, ••• , t, b) c) d) s ,., = l 1 s 1 /s 1 * is a nontrivial s(l) group s 1 > S 1 o -> S 1* , S 1 o is AG invariant and ,,rrinue -~ maximal containing s 1* 2) ° is an irreducible AG module, e) ~l = s 1 /s 1 a) s 2 > s1 s 2* = ker [s 2 - > Aut ~ 1] s 2 /s 2* is a nontrivial s(2) group s 2 > s 2° ~ s 2*, s 2° is unique maximal AG invariant containing s 2* liz = s 2 /s 2° is a irreducible AG module, b) c) d) e) • j) a) s.J > s.J- 1 b) Sj* = ker [SJ. - > Aut -j-1 S ) c) S./S .*is a nontrivial s(j) group d) S. > s . J J 0 J J > S .*, s. 0 is unique maximal AG - J J invariant containing S.* J 68 S. = S./s. 0 is an irreducible AG module, e) -J J J = 1, 2, ••• , tare primes. and s(i), i Such a series is called a t- e d i fice in G. (VII. 1) Suppose AG is a group with normal Hypo~hesis. solvable subgroup G and (IAI, !GI)= 1. E = {si, Si*' si , ~i = 1, •.• , t} 0 G. I i Suppos e that is at-edifice for Suppose 1 = Fo < F 1 < ••• < Fn = G is the Fitting series of G. 2) (VII. Assume (VII. 1). Suppose~ l ;::: Go .:£ Gl .:£ • • • .:£ Gr = G 1§.. ~normal series 2:f G and Gi+l/Gi' i = O, ••• , r-1 i s nilpotent. Ihfil! We proceed by induction on j. For j = 0 the res u l t is Suppose F. > G. • Then G . / (F . ~ [\ G . ) is a J J- J. J J- 1 J- 1 homomorphic image of G./G. since F. nG. > G . , and J J- 1 J- 1 J J- 1 h e nce is nilpotent. But G./(F. 1 1\G.) ~ G.F . /F. b J JJ J J- 1 J- 1 G/F. • Therefore trivial. J- 1 G.F. 1 /F. l < F./F. l J JJ- J JGj _< G.F. l < F . • J J- - J or (VII. ~ 3) Suppose H ,:£ G ~ Fitting length m. Fitting length n. ~ m ,:£ n. Sup pose G 69 Suppose l = F0 s F 1 S ... S F n = G is the Fitting series of G. We apply (VII. 2) to the normal series of H given by 1 = FoAH = Go .:s Fl(\ H = Gl .:s (VII. 4) Suppose (VII. 1). ker [ G - > Aut s i] , i = l, 2, Then F.i . < K.1. = - - ... , t. Hen c e t: .:S n. We proceed by L~duction on i. Since s 1 Suppose p ••• .:s Fnf\ H = Gn == H. First consider i = 1 . A G and is an s(l) group, we know that s 1 .:s F 1 • = s(l). Now Op 1 (F ) centralizes s 1 and h e nce ~ • 1 1 But then OP(F )K /K 1 is a normal p subgroup o f G/K • On 1 ~l' 1 1 G/K 1 is completely reducible since G A AG a nd therefore Op(F1 ) S K1 • < K. • J- 1 - J- 1 Suppose that F. - So F 1 = Op 1 (F 1 ) X Op(F1 ) < K~.... • We notice that F .K . l/K . 1 ,..., F . I ( F . " K . l) and F . n K . 1 > F . l so J J- JJ J JJ J- JF .K. 1 /K. 1 is nilpotent and normal in AG/K . 1 • JJ JJF .K. 1 /K. l < F(AG/K. 1 ). J JJ- J- So Further S.K. l/K. l~ s./(S . l\K. 1) and S.(\K . 1 =SJ·*· J JJJ J JJ JAnd so S.K. 1 /K. 1 is normal and nilpotent i n AG/K. 1 • J JJJTheref ore S .K. /K. l < F(AG/K. ). J J- 1 J- J- 1 Finally AG is irreducible on ~j' a section of S .K . 1 /K. 1 , J JJhence, as in j = 1, F.K. /K. centralizes S.. And -J .) .J- 1 J- 1 therefore, FJ. < F.K . l < K .• J .)- J (VII. P. 5) Suppose ~ HP is ~ group ~ norma l p s ubg;yuu --- - - Assume P has a.'1 HP s e ction X/Y on 'C·J hich HP i s - 70 irreducible im£ H ~ H 12, non trivial . Then P con tains ~ 1 . . b . HP HP invariant su ~roup p 0 wh ere p 0 > p 1 ~ D(D ... 0 )·, ...D 1 ~ i nvaria..'t'lt ~ ~HP X./Y i£ ~ ir:.:-e dnc iblc HP module , i) P /P ii) P /D(P ) is a n i n dGcomp os able HP module , iii) pl .1.§. ~ uniaue maximal HP inv ari ant subgr oup .2.f 0 1 0 0 -- Po, ~ P0 !.§.minimal satisfying i), ii), and iii). Let J be the class o~ all subgroups 1) P 2 is HP invariant satisfying 2) P 2 contains an HP invariant subgroup P ·:c 2 * ~HP such that P 2/P 2 Clearly X e J so J f ~- Since P 0 e J there is P So P 0 P2 ~ P X/Y. We choose P0 e J of minimal order. 1 ~ D(P ) 0 so that P /P 0 1 ~HP X/Y. satisfies i). is a ma ximal HP Since X/Y is HP irreducibl e, P invariant subgroup of P 0 • 1 Suppose P 1 maximal HP invariant subgroup of P • 0 * ~ D(P 0 ) is &is o a Then as an HP module where now pl~'c/(Pl I\ Pl*) So P1 * e J. ~HP · plpl~\'/Pl =- Po/Pl ~HP This contradicts the minimali~y of P X/Y. 0 e J. Hence iii) holds. Suppose that as an HP module P /D(P 0 0 ) = P 0 */D(P0 ) ~- P 0 "/D(P0 ) is decomposable. 71 Choose P 1* maximal HP invariant in P 0 invariant in P0 11 • * and P1 11 maximal HP Then P1*P0 11 and P1 11 P0 * are dist:inc\: maximal HP invariant subgroups of P 0 contradicting iii). . Hence ii) holds. Clearly P 0 (VII. 6) is minimal satisfying i), ii), and iii). Suppose that AG is ~ group ~ normal solvable subgroup G ~ (jA!,IG!) = 1. Fitting subgroup of G. Suppose that Fis~ Suppose ~ S ~ G ~ S/F ~ ~ normal p subgroup !2.:f AG/F. Then there is ~prime r :f p .2!19. !l section g! Or(F) .Q!! which AG i2, irreducible ~ S i s nontrivial. Consider an r Sylow subgroup R of F. Let P be a. p Sylow subgroup of S. for each such r. Suppose r p. Suppose that [R, P) = 1 Then S = P X OP 1 (F) A G so S ~ F. means there is some r for which [R, Pis nontrivial on R/D(R). ~ J3 :f:. 1. This In particular, Now let D(R)/D(R) = Ro/D(R) < Rl/D(R) < ••• < Re/D(R) = R/D(R) be an AG composition series of R/D(R). irreducible AG module. some i, say i = j. Then Ri+l/R1 is an Since p :f r, P is nontrivial for Then X = Rj+l' Y = Rj is the desired section. (Vll. 7) Assume (VII. l). ~ I< = ker [G - > Aut §.i} , Q = G/K, Ti= si+lK/K, Ti*= si+l*K/K, Tio= si+l°K/K, .imS! 72 , t-1. T. = Ti/Ti 0 ~ i = l, 2, ••• -i ~= ~ T. I i = l, • • • , t-1) * , Ti ' -J. 0 {Ti, Ti edifice 12£ Q. We just verify the definition. Then since Kr\ s course, TO = l. Further S2* ,:SK so T 1* = l. We do just 1). Of 2 = s 2* < Sz, Tl > To• Now T1 /T 1 * S2/S2* OlAG proving the rest. (VII. Suppose AG !.§. ~ group ~ norma l solvable 8) subgroup G ~ (IAI, ~ length sll_ G !.[ n. Suppose~ !h§:. Fit ting jGj) = l. G~ E n edifice. Proof is by induction on n. For n = l we take a minimal AG invariant subgroup of G for s • 1 So suppose n > 1. induction. Hence G/F satisfies (VII. So we get an n-1 edifice for G/F. ~ = {Ti/F, Ti*/F, Ti /F, Ii 0 We apply (VII. with Ti 8) by I i = 1, ••• , n-1). 6) to obtain s 1 • 5) and (VII. Then = S i-1' Ti * = S.l.-1*, T.l.0 = S i-1 0 we get the desired n edifice. (VII. 9) Set K. = ker l,G - > Aut s.j i Assume (VII. l). .fil1£ -Gi. = G/Ki. - ~~all l. j > i, Kj > K. and S. is - l. - -J AG isomorphic !2 ~ section of G.• - -i We first prove that Kj ~ Ki for j > i. by induction on j - We proceed i and save the nf i r st case" of 73 - = j i So assume the result holds for all 1 for last. - numbers less than j > f > i. j By induction Hence we need J - - i = l. for all f ~ i, we may assume si 0 0 =1 f ~ i. But in this case = l, j = 2, simplifying the notation. We want to prove by considering G/Si i Then choose i > l. K. > Kf -> K.l. • only prove the result for j Since Kf ~ si - i and j x ' and Kf/si is nontrivial on §. 2 • S2/ S2* = S2/K1() S2. prime to s(2), x 0 , x e K1 has order prime to s (2) Suppose that K2 ~Kl. and 0 Then x Therefore, since is nontrivial on is nontrivial on x is of order s 1 = ~1 . This contradiction forces K K2 /K 2 to be a normal s(2) subgroup 1 of AG/K 2 which is faithful and irreducible on the GF(s(2)) Therefore, K1 ~ K K2 = K2 completing the proof module ~ 2 • of: 1 K.J -> K.l. for all j > i. .*. The rest is easy since S. (\K . 1 = S J JJ (VII. 10) Assume (VII. 1). and Qi = G/Ki. i) Suppose I<..1. = ker L' rG - > Jut -J:: SJ Assume ~A i2_ ~ class Ei2 p grouo. D ~A' CG.(D) = -G. i. -1. 12!: ill ii) j > i. 1f D~ A, and CG. (A'D) ~ CG:. (A') -i. j . ~ for every -i > i ~~Cs . <A') = (O) ~ ~ (D,Cs _(A')] -J Consider i) first. -J By (VII. 9), ~j for j > i = (O). 74 is a section of -Gi.• . Therefore, D centralizes this section. Next consider ii). The condition that CG (A 1 D) ~ -i CG. (A') -i. says that D centralizes every section of G.i which admits A'D and is centralized by A'. But c 5 .<A•) -J is just such a section, by (VII. 9) so (p,c5 . (A')] -J = (O). 75 VIII. THE MAIN THEOREM We are now in a position to prove the final theorem of the thesis. (VIII. l) Theorem: 2:& ~ odd p Eroun ~ assume that A .£f class ~ 2. Further, A ~ be a in section VI. Assume ~ AG is solvable with normal subgroup r 2 c+l divides Suppose Suppose~ expA =pa G where (!Al, !GI)= 1. £m2. for every prime r ~ (*) group ~ defined every integer c ~ ~ !GI, ~every l ~ b ~ a~~ pb ='f re·:· 1. IAI = pd .fil!£ G ~ Fitting length n. Ass~L~e ~ A 1,a fixed point ~ £!! G (J?.y, ~ ~ ~ only !.h.ru:. CG(A) = 1). ~ d ,2:: n. = [si, s. 0 , Si*, s.l. I i = 1, ' .. , n} be an n edifice of G. Let K. = ker [G - > Aut s.1 and -i. G. We apply (VI. 17), (VI. 18), and (VII. 10) - i = G/Ki. Let E ]. ]. to obtain descending chains of subgroups in A. Set Ano = ker [A' - > Aut ~]. If CS (A 1 ) = ( 0) set A:0,'r = A • -n If Cs (A') t= (0) -n set An* = ker [ A - > Aut c8 (A')]. . -n Continuing inductively we set A . 0 J If c5 .(A') = (0) set A.*= A~+i· -J J J = ker [A' ->Aut%1 If c5 .(A') ~ (O) -J 76 set AJ.* = ker [Aj;r-> Aut Cs. (A') -J Now either Aj+lo > Aj 0 J. Aj+l* > Aj*• or Suppose not. and A .. - .,,_ = AJ.*. Then by (VII. 10) we get Aj~lo = A~ J JVJ.. we must investigate the representation of AQj Suppose first that Cs .<A') ~ (O). So ~j. on Then by (VI. 17) -J applied to (A, G . , S.) there is a D _<A so that -J -J If for some CS . (A'D) = (0) but cG_(A 1 D) ~ cG.(A 1 ) . -J -J -J c5 . (A• ) ~· (0) then we may choose . -1 so that c5 .(A 1 ) ~ (O). Fix this i. i > j , -i definition. Now D s Aj+l* =Ai* > j i but D ~ Aj*• minimal Therefore, Aj+l.," > Aj*• So we may assume that c .(A') = (0) for all 5 -i i > j. But then A *=A*= A_> D and again A.* j+l n J < Aj+l • * Hence we assume that c5 .(A') = (O). Now by (VI. 18) -J applied to (A, G., S.) we get l < D S A' with -J -J c5 .<o) = (O) and CG (D) = G .• In particular, Di AJ. 0 -J -j J but D < Aj+l 0 so Aj+l 0 0 > Aj • Therefore we get a chain (Ai or A.J.* < Ai+l*'• 0 , Ai*) where It is easy to see that the length of this chain is bounded by d where The length is obviously n. d ,::: n. Therefore p d • 77 BIBLIOGRAPHY 1) Bauman, s., 11 The Klein Group As a Fixed Point Free Automorphism Group," Phd. thesis, University of Illinois, 1962. 2) Curtis, Charles W., and Irving Reiner, Re presentation Theory of Finite Grouus .§ill_q Associative Algebras, New York, c. 1962. 3) Dade, E. c., Group Theor;z Seminmz 196L;. - 1965, California Institute of Technology, Pasadena, 1966. 4) Gorenstein, D., and I. N. 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