Eigenvalue Structure in Primitive Linear Groups Citation Huffman, William Cary (1974) Eigenvalue Structure in Primitive Linear Groups. Dissertation (Ph.D.), California Institute of Technology. doi:10.7907/1hnq-y922. https://resolver.caltech.edu/CaltechTHESIS:03092021-232801012 Abstract One approach to studying finite linear groups over the complex numbers is to classify those groups with an element possessing a certain eigenvalue structure. Let G be a finite group with a faithful, irreducible, primitive, unimodular complex representation X of degree n. Assume g ∈ G such that X(g) has eigenvalues ∈, ∈̅, 1, 1, . . ., 1 where ∈ is a primitive r th root of unity. H. F. Blichfeldt and J. H. Lindsey have classified G whenever r ⩾ 5. In this thesis r = 3 and 4 are handled. The main results are: Theorem 1: Let G be a finite group with a faithful, irreducible, primitive, unimodular complex representation X of degree n. Assume there is an element g ∈ G such that X(g) has eigenvalues i, -i, 1, 1, . . ., 1. Then n ⩽ 4 and G is a known group. Theorem 2: Let G be a finite group with a faithful, irreducible, primitive, unimodular complex representation X of degree n. Assume there is an element g ∈ G such that X(g) has eigenvalues ω, ω̅, 1, 1, . . ., 1 where ω = e 2πi/3 . Let N be the subgroup of G generated by all such elements. Then either 1. N ≅ A n+1 and G/Z(G) ≅ A n+1 or S n+1 · 2. n = 8, N = N', Z(N) has order 2, and N/Z(N) ≅ O 8 + (2); G/Z(G) is a subgroup of the automorphism group of O 8 + (2). 3. n ⩽ 7 and G is a known group. 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): Wales, David B. Thesis Committee: Unknown, Unknown Defense Date: 14 September 1973 Funders: Funding Agency Grant Number NSF UNSPECIFIED Ford Foundation UNSPECIFIED Record Number: CaltechTHESIS:03092021-232801012 Persistent URL: https://resolver.caltech.edu/CaltechTHESIS:03092021-232801012 DOI: 10.7907/1hnq-y922 Default Usage Policy: No commercial reproduction, distribution, display or performance rights in this work are provided. ID Code: 14100 Collection: CaltechTHESIS Deposited By: Benjamin Perez Deposited On: 11 Mar 2021 19:54 Last Modified: 29 Jul 2024 21:52 Thesis Files PDF - Final Version See Usage Policy. 38MB Repository Staff Only: item control page

EIGENVALUE STRUCTURE IN PRIMITIVE LINEAR GROUPS Thesis by William Cary Huffman In Partial Fulfillment of the Requirements for the Degree of Doctor of Philosophy California Institute of Technology Pasadena, California 1974 (Submitted September 14, 1973) ii to Mom, Dad, and Mike iii ACKNOWLEDGMENTS I would like to express my gratitude and thanks to my adviser, Dr. David B. Wales, not only for suggesting this thesis problem but also for his encouragement and patience. Dr. Wal es contributed many insights and ideas toward the solution; he also aided greatly in condensing and improving the proofs. I would also like to thank Dr. Michael Aschbacher, Dr. Marshall Hall, and Dr. Chris Landauer for their discussions with me. Dr. Robert Dilworth and Dr. Bill Iwan have been constant sources of encouragement throughout my graduate career. The faculty, staff, and students of the math department have contributed greatly in making Caltech a pleasant place to work. Also, I would like to thank Joyce Lundstedt for typing this thesis. I am grateful to the National Science Foundation and the Ford Foundation for providing me with graduate fellowships. I am also indebted to the California Institute of Technology for providing me with teaching assistantships. Words cannot express my thanks to my family for their devotion and love to me. It is to them that I dedicate this thesis. Any abilities in mathematics that I have are gifts from God. The certainty of the presence of Jesus Christ in my life has been my strength; He accomplishes everything for me. The constant care, encouragement, and prayers of other Christians have been invaluable. iv ABSTRACT One approach to studying finite linear groups over the complex numbers is to classify those groups with an element possessing a certain eigenvalue structure. Let G be a finite group with a faithful, irreducible, primitive, unimodular complex representation X of degree n. Assume g E G such that X(g) has eigenvalues E, E, 1, 1, ... , 1 where € is a primitive r th root of unity. H. F. Blichfeldt and J. H. Lindsey have classified G whenever r ? 5. In this thesis r = 3 and 4 are handled. The main results are: Theorem 1: Let G be a finite group with a faithful, irreducible, primitive, unimodular complex representation X of degree n. Assume there is an element g E G such that X(g) has eigenvalues i, -i, 1, 1, ... , 1. Then n ~ 4 and G is a known group. Theorem 2: Let G be a finite group with a faithful, irreducible, primitive, unimodular complex representation X of degree n. Assume there is an element g E. G such that X(g) has eigenvalues w, w, 1, 1, ... , 1 where w = e 21ri/3 . Let N be the subgroup of G generated by all such elements. Then either 1. N ~ An+i and G/Z(G) ~ An+i or Sn+i· 2. n = 8, N = N', Z(N) has order 2, and N/Z(N) ~ 0/(2); G/Z(G) is a subgroup of the automorphism group of o/(2). 3. n ~ 7 and G is a known group. V TABLE OF CONTENTS Page ACKNOWLEDGMENTS iii ABSTRACT . . iv INTRODUCTION 1 CHAPTER I The Constituents of X when Restricted to Subgroups . 6 II Building up of Certain Subgroups 19 III The Proof of Theorem 1 . . 35 IV A Special Case of Theorem 2 40 V The Proof of Theorem 2 . 57 RE FERENC ES . . . . . . . . . . . 86 1 INTRODUCTION Finite linear groups of degree n over the complex numbers can many times be classified according to the eigenvalue structure of an element in the group. As finite linear groups are subdirect products of irreducible linear groups, it is convenient to restrict the study to irreducible groups. For example, Mitchell [21] classified all irreducible linear groups containing an element with eigenvalues a, {3, J3, ... , {3 where a =1 {3. The next natural step is to consider the case where there exists an element of the group which has an eigenspace of dimension n - 2 corresponding to one of its eigenvalues. In this thesis, we will examine groups which contain an element with eigenvalues EC~, Ea, a, a, ... , a where E * 1. Let G be a finite group with a faithful, irreducible complex representation X of degree n over the vector space V. The representation X is said to be primitive if there does not exist a set of m ~ 2 proper, nontrivial subspaces Vi with V = V1 E9 V2 EB ... E9 Vm such that X(g) permutes {v.} for all g E G. An irreducible representation which is 1 not primitive is similar to one induced from a representation of some proper subgroup [ 5, Theorem 50. 2] . Thus it is not very restrictive to consider only primitive representations. The representation X is unimodular if X(g) has determinant 1 for all g E G. Any irreducible representation is projectively equivalent (i.e., as a collineation group) to a unimodular representation. Therefore by classifying unimodular, irreducible, primitive groups containing an element with eigenvalues 2 E, E, 1, 1, ... , 1, we are classifying, up to projective equivalence, those irreducible primitive groups containing an element with eigenvalues E:a, Ea, a, ... , a.. Some work has been done on this problem. If Xis irreducible and primitive, and if E is a primitive r th root of unity, the results are known for r ~ 5. A spec ial case of a theorem in Blichfeldt [2, p. 96] proves that if r ~ 6, then n ~2, r = 6, 8, or 10, and the groups are known. If r = 5, Lindsey [14, Lemma 2] proves n ~ 4 and the groups are known. In this thesis the cases r = 3 and 4 are handled. The results are: Theorem 1: Let G be a finite group with a faithful, irreducible, primitive, unimodular complex representation X of degree n. Assume there is an element g E G such that X(g) has eigenvalues i, -i, 1, 1, ... , 1. Then n ~ 4 and G is a known group. Theorem 2: Let G be a finite group with a faithful, irreducible, primitive, unimodular complex representation X of degree n. Assume there is an element g E G such that X(g) has eigenvalues w, w, 1, 1, ... , 1 2 ·/3 • Let N be the subgroup of G generated by all such where w = e 711 elements. 1. Then either N 9:: An + 1 and G/Z(G) 9:: An +1 or Sn +1 . 2. n = 8, N = N', Z(N) has order 2, and N/Z(N) 9:: ot(2); G/Z(G) is a subgroup of the automorphism group of 0:(2). 3. n ~ 7 and G is a known group. Note that all primitive linear groups of degree 7 or less are known (see Blichfeldt [2], Brauer [3], Lindsey [14, 16, 17], and Wales [ 27, 28]). 3 Some authors in recent years have become interested in classifying quasiprimitive groups rather than primitive ones. A representation X of G is quasiprimitive if for every normal subgroup N of G, X IN splits into isomorphic factors. By Clifford's theorem [ 5], a primitive repre- sentation is quasiprimitive. By examining the results of Chapter I, it is easy to see that a quasiprimitive irreducible representation X of G, where X(g) has eigenvalues i, -i, 1, 1, .. . , 1 or w, w, 1, 1, ... , 1 for some g E G, is indeed primitive. Therefore Theorems 1 and 2 are true if X is quasiprimitive rather than primitive. Chapters I and II give preliminary material used in the proofs of Theorems 1 and 2. The results help characterize the possible subgroup structure of G; in particular under certain conditions primitive subgroups of codimension 1 or 2 can be constructed. These results are useful for induction purposes to prove the main results. In Chapter III, Theorem 1 is proved. This proof is much easier than that of Theorem 2 because no primitive group of degree 5, 6, or 7 contains an element with eigenvalues i, -i, 1, ... , 1. In Chapter IV, the alternating groups of Theorem 2 are obtained by exhibiting appropriate generators and relations. The proof of Theorem 2 is completed in Chapter V. The powerful results of Aschbacher-Hall [1] and Stellmacher [26] on groups generated by elements of order 3 are used. The following corollary is also proved: Corollary: If G is a finite simple group containing An-i with a nontrivial representation of degree n ~ 10, then G ~ An-i ,An, or An+i· 4 The following notation is adopted. If H is a subgroup of a finite group K, NK(H) is the normalizer in K of H, CK(H) is the centralizer in K of H, Z(H) is the center of H, and H' is the derived group of H. If p is a prime, OP (H) denotes the largest normal p- subgroup of H and O 00 (H), the largest normal solvable subgroup of H. -1 If x, y E H, y xy is denoted by xY. Also m lH is the direct sum of m copies of the trivial representation of H. If Hand Lare subgroups of K, [H, L] = The symbol Zk denotes the cyclic group of order k. (h- 1 .f- 1 h.f lh E H, f EL). The order of H is denoted jH J; also H <l K means H is normal in K. Let H be a finite group with a faithful, irreducible, primitive complex representation Y of degree m. The term Blichf eldt refers to the result [2, p.96] that H\Z(H) does not contain an element h where Y(h) has an eigenvalue E such that all other eigenvalues are at most 60 ° away from E. The term Mitchell will refer to two results in [ 21] : If 2 h E H and Y(h) has eigenvalues a, f3, . .. , J3 such that h ri Z(H) but h4 E Z(H), then m ~ 2. If h EH and Y(h) has eigenvalues a, {3, •.. , {3 such that h rt. Z(H) but h 3 E Z(H), then m ~ 4. In the latter case if m = 3, H/Z(H) is a split extension of Z3 x Z3 by SL 2 (3) and if m = 4, H/Z(H) ~ Os{3). Which result the term Mitchell refers to will be clear from the context. If p is a prime, a p-element of H is an element in H of order a power of p. A special 4-element of H is an element h E H such that Y(h) has eigenvalues i, -i, 1, 1, ... , 1. A special 3-element of H is an element . h E H such that Y(h) has eigenvalues w, -w, 1, 1, ... , l where w = e 21ri/3 . The group G will be a finite group with a faithful, irreducible, primitive, unimodular complex representation X of degree n over the vector space ~- If v 1, ••• , vk E V, (vi, ... , vk) denotes the subspace of V generated by V 1, ••• ' Vk- 5 When working on this problem, it was often necessary to consult character tables of various . groups. Some of these tables are found in [7], [8], [13], [16], [17], [19], [20], [29], and [30]. General references for group theory and representation theory are [2], [ 5], (11], [ 19] , [ 23] , and [ 24] . 6 CHAPTER I THE CONSTITUENTS OF X WHEN RESTRICTED TO SUBGROUPS In this chapter, we give properties of the constituents of X when restricted to subgroups of G generated by special 3-elements or special 4-elements. In particular these constituents are shown to be either primitive or monomial. The possible monomial groups are investigated more closely, and conditions are given which guarantee the uniqueness up to scaling and ordering of the basis. These results are useful in constructing large subgroups in later chapters. Lemma 1. 1: Let N <J G and assume N contains a special 3element or special 4-element h. Then X IN is irreducible. Proof: By Clifford's theorem [ 5], X jN = X 1 EB ... EB~ where all the Xi's are equivalent irreducible representations of N. In particular the trace of ~ (h) is the trace of X1 (h). By the eigenvalue structure of X(h), t > 1 is impossible. Lemma 1. 2: Let H be a subgroup of G generated either by special 3-elements or special 4-elements. Assume X IH = X1 EB ~ where X1 is irreducible. Then either X1 is monomial or X1 is primitive. Proof: Let H = (hu ... , hr) where hi are special 3-elements or special 4-elements. Let X1 have degree m and act on the subspace V*. Assume the result is false. Then there exist l > 1 subspaces 7 V1 , ••• , V1 of V* all of dimension k > 1 with m = lk sue h that the X1 (hi) permute {Vu ... , v1}. We may assume m ~ 4. Assume X1 (hi) fixes exactly t subspaces, say Vu ... , Vt by renumbering if necessary. x /hi) be the trace of X1 Let Iv. (hi) for j = 1, ... , t. Then J m - 3 ~ !trace X1 (hi) t t j =1 j =1 I = I L x/hi) I ~ L Ix /hi) I ~ kt So kl- 3 ~ kt and hence t ~ £- 1 as k > 1. As X1 (hi) fixes i - 1 subspaces, it must fix all l subspaces; so X1 is reducible, a contradiction. Lemma 1.1 implies in particular that if V1 is a proper subspace of V, there exi sts a special 3-element or special 4-element g E G such that X(g) does not leave V1 invariant. It also implies that the subgroup generated by special 3-elements or special 4-elements could not be abelian. These facts along with Lemmas 1. 1 and 1. 2 are used without reference throughout this paper. We now want to look more closely at what happens when the hypothesis of Lemma 1. 2 holds. In particular we want to investigate H when X1 is monomial and ~ is of a special nature. Lemma 1. 3: Let H be a subgroup of G generated by special IH = X EB (n-r) lH where X is irreducible of degree r ~ 3. Suppose X acts on V and X IH is monomial in the basis 4-elements such that X 1 v 1 , ••• , vn of V . 1 1 1 Then 1. There exist special 4-elements hu ... , hr_ 1 EH such that when v1 , ••• , vn are properly scaled and ordered 8 Vi+1X(hi) = v.1 v .QX(hi) = v.Q for .Q El {i, i+l} Also (vi, ... , vr) = V 1 and (n-r)lH acts on (vr+u ... , vn). 2. XI (hu ... , hj) is irreducible on (vu ... , vj+ 1) for j ~ 2. 3. If XI H is monomial in a basis v 7, ... ,v~, by ordering and . v1* , . .. , vn* correctly, v1 = v1* , ... , vr = vr* and scalmg * * = (vr+u···,vn). <vr+u···,vn) Proof: As X1 is irreducible there is a special 4-element h1 E H such that X(h1) is not diagonal. So for some j * k, we have vjX(h 1) = -ovk vkX(h 1 ) v!X(h1 ) = ◊-1 v. J = vi for £ El {j, k} . By ordering v1, ... , vn correctly, we may assume j = 1, k = 2. Replacing v 2 by ov2 , we may assume o = 1. So X(h1) has the desired form. Note that if h is any special 4-element of H with j * k and vjX(h) = -Evk vkX(h) =E v .QX(h) = v1 for fJ/ {j, k} -1 v. J then the eigenspaces of X(h) corresponding to i and -i span (vj, vk) . Thus (v., vk) c V 1, the unique subspace of V on which X1 acts; in J - particular (vu v 2 ) i < r-1. ~ V 1. Assume we have constructed hu ... , hi where So (vu v 2 ) , (v2 , v 3) , ••• , (vi' Vi+ 1) ~ V1 and as X1 is irreducible 9 there is a special 4-element hi+i E H such that X(h1+i) does not leave (v 1 , ••• , vi+i) invariant. So there exist j, k with 1 ~ j ~ i+l and i+2 ~k ~ n such that * 1) vjX(hi+ = -ovk vkX(h1+J = 0-1 v. v _e_X(h1+1) = v J 1 for £ El {j, k} As (X(h1 ), ••• , X(l\)) acts transitively on (v1 ) , • • . , (vi+i) , we may choose -1 * h E (hi, ... , hi) such that vj X(h) = AVi+i · Letting hi+ 1 = h hi+ 1 h, Vi+iX(hi+i) = -1 -A ovk -1 vkX(hi+i) = AO v£ X(hi+i) = vl. vi+i for £ El {i+l, k} Replacing vk by ~ - ovk and rearranging vi+ 2 , ••• , vn, we may assume 1 1 A- 6 = 1 and k = i+2. We have inductively constructed hi, ... , hr_ 1 ; clearly (vi, ... , vr> = V 1 as (vi, vi+i) c V 1 • As (n-r)lH acts on a subspace of (vr+u ... , vn) and (Vr+u ... , vn) has dimension n - r, (n-r)lH acts on <vr+u ... , v n> , proving 1. The second assertion is proved by induction. Let j = 2. As h1 and~ do not commute, if XI (hi,~) is not irreducible on (vi, v2 , v3 ) , it must contain an eigenvector common to both X(h1 ) and X(h 2 ). This is not true because on (vu v 2 , v 3 ) , X(h1 ) has eigenspaces (v1 + iv 2 ) , (v1 - iv2 ) , (v 3 ) corresponding to i,-i, 1, respectively, while X(~) has eigenspaces (v2 + iv 3) , (v 2 - iv 3 ) , (v1 ) corresponding to i, -i, 1. Thus XI (h1 , !½) is irreducible on (v1 , v2 , v). Assume XI (h 1 , ••• , hj_ 1) is 10 irreducible on (vu ... , vj) for j-1 ~ 2. As XI (hu ... , hj) is invariant on (vu ... , vj+i> but not on (vj+i), it is irreducible on (vu ... , Vj+i), proving 2. To prove 3, we examine X(h1 ), • • • , X(hr_ 1 ). As a permutation on (vi), ... , (v~), each X(hi) acts trivially or as a transposition. As X I (h1 , h2 ) has an irreducible constituent of degree 3, the only possibility is that there exist i, j, k distinct with -ov.* v;x(h 1 ) = v;x(h1 ) = 0-1 v. v;x(h1 ) = v * for ] * and 1 1 vj*X(h2 ) = * vkX(h 2) = E -1 v. * l Et {i, j} V f X(l'½) -EVk* * J = v1* for i Et {j, k} . vj* by ovj* and vk* by EOVk, * we may assume o = E = 1. Replacmg By reordering Vi, ... , v~, we may assume i = 1, j = 2, k = 3. So assume under suitable ordering and scaling of v7, ... , v*n that we have (1) vi*X(hi) * = -Vi+l * vi+1X(hi) = v.* v* f X(hi) 1 = Vi* for l Et {i, i+l} * for 1 ~ i ~ j and some j ~ 2. Clearly (vu ... , vj+i) = (v1* , ••• , Vj+i), as XI (hi, ... , hj) is irreducible on (vu ... , Vj+i) and invariant on (vi', ... , Vj+i). As X(hj+J does not leave (vu ... , Vj+i) invariant, X(hj+i) must interchange (v;), (v~) for some 1 ::s; i ~ j+l and j+2 ~ k ~ n. By renumbering vj+ 2 , • • • , v~, we may assume k = j+2. As hj+i commutes with hu ... , hj-u the only possibility is that i = j+l. So 11 * -0Vj+2 -1 * * Vj+2X(hj+1) = 0 Vj+l v1*X(hj+i) = v; for l.. ti {j+ 1, j +2} Vj+i X(hj+i) = By replacing vj+ 2 by ovj+ 2 , which doesn't affect the form of X(h1 ), •• . . , X(hj), we may assume o = 1. By suitably ordering and scaling v:, ... , v~, X(~) has the form (1) for 1 ~ i ~ r-1. Clearly V1 = * ... , vn*> = (v r+i, ... , vn) . Let (v1*, ••• , v r*> = (vi, ... , vr ) an d (vr+P vis = vt; clearly S commutes with each hi. As X1 is irreducible, S must act as a scalar on V 1 • Hence by correctly scaling v~, ... ,v;, we have 3. Lemma 1. 4: Let H be a subgroup of G generated by special 3-elements such that X IH = X 1 EB~ EB (n-m-l)lH where X1 is irreducible of degree m ~ 5. Assume X1 is monomial on a subspace V1 of V, ~ is linear and acts on (v), and (n-m-l)lH acts on V2 • Let V1 have a basis vi, ... , vm such that X1 is monomial in that basis. Then there exist special 3-elements hi, ... , hm_ 2 E H such that by properly scaling and ordering v 1 , ••• , vm, we have viX(hi) = vi+ 1 Vi+iX(hi) = Vi+ 2 vi+2X(hi) = vi v1 X(hi) = v9- for f._ ti {i, i+l, i+2} Also ~(hi) = 1 for 1 ~ i ~ m-2. The elements h E H with (X1 EB ~)(h) diagonal in the basis v 1 , ••• , v m, v form a normal subgroup F of H with H = FA where A = (hi, ... , hm_ 2 ) . Also X1 IF splits into m distinct 12 linear representations all different from ~. Furthermore if v7, .. . , v~ is a basis for Vin which X IH is monomial and if Vi, ... ,v~ are properly scaled and ordered, v1 = v1* , ••• ,vm = vm. * If~= l H, (Vm+u * ... ,vn) * = (v) EB V2 • If ~ -J lH, by ordering and scaling v:X1+ 1 , ••• , v~ correctly, *> * 1 an d V2 = (Vm+ * 2 , ••• ,vn. v=vm+ Proof: As H is not abelian, there is a special 3-element h1 E H such that X1 (h1 ) is not diagonal. In particular, we must have i, j, k all distinct such that viX(h1 ) = av. J vjX(h1 ) = bvk abc = 1 vkX(h1 ) = cv.1 vf X(h1 ) = Vi for ! (1 {i, j, k} By reordering Vu ... , vn we may assume i = 1, j =2, k = 3. Replacing v2 by av2 and v 3 by abv 3 , we may assume a = b = c = 1 and we have h1 • Assume we have constructed hu ... , hi for some i ~ m-4. Then there is a special 3-element h E H which does not leave (vi, ... , vi+ 2 ) invariant, as X1 is irreducible. So in particular there exist r, s, t all distinct such that vrX(h) = av s V SX(h) = bvt abc = 1 vtX(h) = CVr v!X(h) = Vi for i (1 {r, s, t} where 1 ~ r ~ i+2 and at least one of s or t is greater than i+2. 13 Replacing h by h- if necessary, assume i + 3 ~ t ~ m. 1 Case a) i + 3 ~ s ~ m: By renumbering vi+ 3, ... , vm we may assume that s = i+3, t = i+4, which does not affect the form of X(h 1 ), ••• , X(hi). As XI (hi, ... , hi) is the alternating group on Vu ... , vi+ 2, choose g E (hi, ... , hi) with VrX(g) = Vi+2. Letting hi+2 = hg, Vi+2X(hi+2) = avi+3 Vi+3X(hi+2) = bviH Vi+4X(hi+2) = CVi+2 abc = 1 v1X(hi+2) = Vi.. for l (j_ {i+2, i+3, i+4} Replacing vi+ 3 by avi+ 3, vi+ 4 by abvi+4' which does not affect the form of X(h1 ), ••• , X(hi), we may assume a = b = c = 1. This is the desired hi+2hi . element hi+2 and hi 1s the element hi+i. Case b) 1 ~ s ~ i+2: By renumbering vi+ 3, ... ,vm, we may assume t = i+3. Then clearly XI (hi, ... , hi' h) acts as the alternating group on (v 1 ) , • • • , (vi+ 3). As X1 is irreducible and i+3 < m, there is a special 3-element g E H such that X1 (g) does not leave (vi, ... , vi+ 3 ) invariant. p, µ, v all distinct such that vpX(g) = Ci.V µ v~(g) = /3v V ct.{3y = 1 v~(g) = yvp v lX(g) = vi for l (j_ {p , µ , v} So there exist 14 where not all of p, µ, v are less than or equal to i+ 3, but at least one is. By replacing g by g- 1 if necessary, we may assume 1 ~ p ~ i+3 and i+4 ~ V ~ m. If i+4 ~ µ ~ m, then choose k E (h1 , ••• , h., h) such that 1 vpX(k) = XVi+2· Then 1 Vi+2X(k- gk) = X V µX(k- 1 V VX(k- 1 -1 gk) = {3v Ci.V µ 11 gk) = xyvi+2 v X(k- 1 gk) = v! for l. rJ {i+2, µ, 11} 1 We now have Case a, with k- gk in place of h. If 1 ~ µ ~ i+3, let 1 k E (hi, ... , hi' h) such that vPX(k) = xvi+ 2 and v µX(k) = YVi+ 3 • Therefore 1 -1 Vi+3X(k- gk) = y {3v V 1 Vl/X(k- gk) = xyvi+2 v X(k- 1 gk) = v for ! rJ {i+2, i+3, 11} 1 1 • 1 Again we have Case a, with k- gk in place of h. So by induction h1 , ••• , hm_ 2 are found. Clearly ~(hi) = 1 for 1 ~ i ~ m-2 as X1 (hi) is not diagonal in Vu ... , vm. Let F = {h E H I(X1 EB ~) (h) is diagonal in the basis v1 , . • • , vm, v} . 1 If g E H and f E F, (X1 EB ~)(g- fg) is still diagonal and so F <l H. Let g be a special 3-element in H. If (X1 EB ~)(g) is diagonal, then g E F. Assume (X1 EB E)(g) is not diagonal. vX(g) = v and Then for some i, j, k distinct, 15 viX(g) abc = 1 vjX(g) vkX(g) V = cv. 1 1 X(g) = v1 for I El {i, j, k} As X1 IA is the alternating group on Vi, ... ,vm, there is an hE A with viX(h) = vj vjX(h) = vk vkX(h) = vi v QX(h) = v! for f El {i, j, k} 2 Then gh E F and so g E FA. Hence H = FA and F O A = 1. Consider X1 IF = ~1 E9 ... EB ~m where ~i is linear on (vi) for 1 ~ i ~ m. Assume first that ~i = ~j for some i * j. Let f E F and by double transitivity, let gr E A such that viX(gr) = vi and v rX(gr) = vr 1 1 1 1 So ~/g; fgr)vj = vjX(g; fgr) = ~r(f)vj and ~/g; fgr)vi = viX(g; fgr) = 1 1 ~i (f)vi. This implies ~r(f) = ~/g; fgr) = ~i (g; fgr) = ~/f). Thus ~ 1 = ••• = ~m contradicting the irreducibility of X1 • So the ~i's are distinct. If ~ * lH, for some special 3-element g, ~(g) = w or w. So (X1 E9 ~)(g) is diagonal and ~i(g) * ~(g) for 1 ~ i ~ m. Thus ~ * ~i for 1 ~ i ~ m if ~ :1; lH. If ~ = lH, there is an f E F such that for some i, ~/f) * 1. 1 1 Let gr EA with viX(gr) = vr. Then ~r(g; fgr)vr = vrX(g; fgr) = 1 ~/f)vr. So ~r(g; fgr) * 1 and ~ * ~i for 1 ~ i ~ m. Let v~, ... , v~ be a basis of V in which XI H is monomial. We first consider X(h 1 ), ••• , X(hm_ 2 ) acting on this basis. As (hi, h 3 ) clearly the only possibility is that ~ A5 , 16 v*X(h1 ) r v;x(h 3 ) = avu* = av*s abc = 1 = bv; s v;x(h 1 ) = cv*r v1X(h1 ) = v 1* for l Et {r, s, t} v*X(h 1 ) and v:x(h 3 ) = /3v * Cl/3y w =1 * 3 ) = yvt* vwX(h v;x(h 3 ) = vl* for l Et{t,u,w} . vs* by av *s' vt* by abvt, * vu* by nabvu, * and vw* by a,Babvw' * we By replacmg may assume a = b = c = a = ,B = y = 1 . Also by renumbering we may assumer = 1, s = 2, t = 3, u = 4, w = 5. Note that~= h1 inductively assume we have reordered and rescaled h3hl • So Vi, ... ,v~ such that for some i, with 2 ~ i ~ m-4, v~X(h.) J J (1) * = Vj+i * vj+ 1 X(hj) = Vj+2 vj+2 X(hj) = v.* J v;x(hj) = v; for£ El {j,j+l,j+2} for 1 ~ j ~ i. As X(hi+2) does not commute with (X(h1 ), ••• , X(hi)) , 1 X(hi+ 2 ) must act trivially on at least n - (i + 4) of the vectors v +3 , ••• , y~. By reordering vi+ 3 , ••• , v~, assume X(hi+ 2 ) acts trivially on ViH, ... , v~. So XI (hi, ... , hi, hi+ 2 ) acts as a permutation gr oop on (v~) , ... , (vf+4 ) . As (hu ... , hi' hi+2> ~ AiH, which is simple, X l<hu ... , hi' hi+2) cannot act trivially on (vf+ 3 ) or (vf+4 ) • Thus there is a j with 1 ~ j ~ i+2 such that by ordering vi+ 3 , viH correctly vjX(hi+ 2 ) = avt+3 vi+3 X(hi+ 2 ) = bvf+4 abc = 1 vf+4 X(hi+ 2 ) = cvj v;x(hi+ 2 ) = v; for £ fl. {j, i+3, i+4} 17 · vi+ * 3 by avi+ * 3 an d viH * by abviH' * Replacmg we may assume a = b = c = 1. As hi+ 2 commutes with hu ... , hi-u j = i+2 is the only possibility. Noting h, h• that hi+i = hi 1+2 1 , we have by induction and properly scaling and ordering v:, ... , v* that X(h.) has the form (1) for 1 ~ j ~ m-2. n J Let D = {h E·H lx(h) is diagonal in Vi, ... , v~} <l H. If g is a special 3-element in H, by looking at (X1 EB ~)(g) in the basis Vu ... , v m , v, it is clear that g does not commute with A. Thus either g E D or for i, j, k distinct with i ~ m, av·* J v;x(g) = bvk* v~X(g) = abc = 1 vkX(g) = CV~1 = v; for l (l {i,j,k} In the latter case if j, k > m, by replacing vj by avj and v; by abv;, it is clear that (hu ... , hm- 2 , g) ~ Am+ 2 and XI (hi, ... , hm_ 2 , g) has an irreducible constituent of degree m+l, a contradiction. If only one of 1 j, k is greater than m, by replacing g by g- if necessary, we may assume j ~ m and k > m. Replacing v; by bv;, we may also assume c = a- 1 and b = 1. If a -:1= 1, xi (hu ... ,hm_ ,g) is irreducible on 2 * , a contradiction. So a = 1 and K = (h 11 ••• , hm_ 2 , g) (v1* , ••• , v * m, vk) ~ Am+i · As K n F is a nontrivial abelian normal subgroup of K, this is a contradiction. So i, j, k ~ m and as done previously, gh E D for some h EA. Therefore H = DA. As DF /F is a normal abelian sub- group of H/F ~Am, D c;_F. Hence D =Fas F n A= 1. Assume first that ~ = lH. As the only linear constituents of XI H are trivial, X In must be trivial on v~+u ... , v~; so ( v) EB V = 2 18 (vm+ * 1, • • • * ) =(vi, ... ,vm). Suppose~* lH. ,vn*> and (v1* , • • • ,vm As there is exactly one nontrivial linear constituent of X IH, X In must be trivial on n - m - 1 of the vectors v~+u ... , v~ while ~ acts on the remaining one. By reordering and rescaling, we may assume v = v~+ 1 ; so (vf, ... ,v~) =(vu ... ,vm) and V2 = (v~+ 2 , ••• ,v~). As X 1 IF = ~1 EB ... EB ~m = X1 In = µ EB ... EB µm where µi acts on (v;) 1 and each ~ i is unique, { (v;) I1 ~ i ~ m} = { (vj) jl ~ j ~ m} proving the lemma. If n ~ 5 and H is generated by special 3-elements such that X IH is irreducible, clearly Lemma 1. 4 holds when appropriately modified to avoid conclusions concerning ~ or (n-m-l)lH. So when necessary, Lemma 1. 4 will include this case. 19 CHAPTER II BUILDING UP OF CERTAIN SUBGROUPS In this chapter lemmas are developed which allow the building up of subgroups of G generated by special 3-elements and special 4elements. Also under certain conditions it is shown that there are subgroups of G which are primitive of smaller degree. These results are used for induction later. Lemma 2. 1: Let n ~ 4 and H be a subgroup of G generated by special 4-elements. Assume X !H = X1 EB (n-r)lH where X1 is irreducible such that 3 ~ r < n. h E G with XI (H, h) = ~ EB Then there exists a special 4-element (n-s)l (H, h) where X2 is irreducible of degree s = r + 1 or r + 2. Proof: Let X1 act on V 1. Choose a special 4-element h such that X(h) does not leave V 1 invariant. So XI (H, h) = ~ EB (n-r-2)1 (H, h) where either ~ is irreducible or X2 = X 3 EB i such that X:i is irreducible and t is linear. We are done unless X 2 = X 3 EB ~ where ~ =1= 1 (H, h) • Assuming this is the case, by Mitchell X3 is monomial. X !His monomial in some basis Vu ... , vn. In particular By Lemma 1. 3, there exist special 4-elements hu ... , hr_ 1 E H such that when Vu ... , vn are properly scaled and ordered vkX(hk) = -vk+1 Vk+1X(hk) = vk v _eX(hk) = vf_ for £ q {k, k+l} 20 for 1 ~k ~ r-1 and XI (hi, ... , hr_ 1 ) is irreducible on (vi, ... , vr). Thus V 1 = (vi, ... , vr). As ~(u) = 1 for u EH, ~(h) * 1. Since X(h) does not leave V 1 invariant, X(h) is not diagonal in Vu ... ,vn . So viX(h) = -ov. J -1 vjX(h) = o v.1 v1 X(h) = v 1 for /._ ti {i' j} where 1 ~ i ~ r and r+l ~ j ~ n. Thus XI (H, h) is irreducible on <vu ... , vr, vj) and trivial on <vr+i, ... , Vj-u vj+1' ... , vn) contradicting t(h) =1= 1. So the lemma is proved. Lemma 2. 2: Let H be a subgroup of G generated by special 3elements such that XIH = X1 E9 ~ E9 (n-s)lH where X1 is a nonunimodular, irreducible, primitive representation of degree 4. Assume X2 is a representation of degree d with 1 ~ d ~ 4 ands = d+4. Suppose ~(H) = Hi. Then n = 8, X2 is irreducible and primitive of degree 4, and Hi ~ 6;(3) x Z3 where 6;(3) is the nonsplitting central extension of Z2 by 0 5 (3). If Li is the set of all elements of Hi which occur with component the identity of HJ. (j * i) in the subdirect product, then L. c 1 - 0 2 (Z(Hi)) for i = 1 and 2. Proof: By Mitchell, as X1 is not unimodular and X1 is primitive, H1 /Z(H 1 ) ~ 0 5 (3). ~ But Hf'Z(H 1 ) = H1 and so H{ = H{'. Thus H; / Z(H{) 0 5 (3). As 0 5 (3) does not have a nontrivial representation of degree 4, Z(H;) =1= 1. The Schur multiplier of 0 5 (3) is Z2 (see [ 6]); so Z(H:) ~ Z 2 • Because H1 has elements with determinant w and only elements with determinants 1, w, or w, 0 3 (Z(H 1 )) ~ Z 3 and 0 2 (Z(H 1)) £::: Z2 or Z4 • If 21 0 2 (Z(H 1 )) ~ Z4 , then K = tt; 0lZ(H 1 )) has index 2 in Hu contradicting the fact that H1 is generated by 3-elements. Thus 0 2 (Z(H 1 )) = Z(H 1') and __,,, H1 "' tt; x Z 3 where tt; ~ 0 5 (3), the nonsplitting central extension of Z2 by 0 5 (3). Let Li be as in the statement of the lemma. By Theorem 5. 5.1 of [11], L.1 4 H.1 and H1 /L 1 ~ H2 /L 2 • As L.1 consists of unimodular matrices, either L 1 = tt; or L 1 S 0 2 (Z(H 1 ) ) . In the first case L 1 contains an element with eigenvalues w, w, w, w and a central element with eigenvalues -1, -1, -1, -1. But then X( G) contains an element with eigenvalues -w, -w, -w, -w, 1, 1, ... , a contradiction to Blichfeldt. So ~ L 1 s_ 0 2 (Z(H 1 )) and H2 /L 2 ~ 0 5 (3) X Z 3 or 0 5 (3) X Z 3 • By examining the possible decompositions of ~ into its components and consulting Blichfeldt's list [2] carefully, it is easy to see that ~ is irreducible of degree 4 and is primitive. Carrying out the same analysis as on Hu we --..,/ see H2 ~ 0i3) x Z 3 and L 2 s_ 0 2 (Z(H 2 ) ) . Also in the isomorphism from H1 /L 1 onto H2 /L 2 , the central elements are mapped onto central elements. In particular if n ~ 9, X(G) contains an element with eigenvalues -w, -w, -w, -w, -w, -w, -w, -w, 1, ... , contradicting Blichfeldt. Son = 8. Lemma 2. 3: Let n ~ 5 and H a subgroup of G generated by special 3-elements. Assume X IH = X1 EB (n-r)lH where X1 is irreducible of degree r with 3 ~ r < n. Then there exists a subgroup K generated by spec ial 3-elements such that XI (H, K) = X 2 EB (n-s)l (H, K) where s = r+l or r+2 and ~ is irreducible of degree s. Proof: Let X1 act on the subspace V1 and let h be a special 3-element such that X(h) does not leave V1 invariant. So 22 XI (H, h) = ~ EB (n-r-2)1 (H, h) where either X2 is irreducible or X2 = X3 EB ~ such that X 3 is irreducible and ~ is linear. We are done unless X 2 = X 3 EB E and E=I= 1 (H, h). Assume this is the case. Suppose first that r = 3. X 3 is monomial. By Lemma 2. 2, X 3 is not primitive; so Let X 3 act on V3 :) V1 and let Vu v 2, v 3, v4 be a basis of V 3 such that X 3 is monomial in that basis. As X3 is irreducible, there exist special 3-elements hi, h 2 E (H, h) such that by scaling and ordering vi, ... , v4 correctly, v 1X(h1) = V2 v 1X (~) v 2 X(h1) = V3 v 2 X(h2 ) = av 3 and = V1 V3X(h1) = V1 v 3X(~) = V4 v 4X(h1) = V4 v 4X(~) = a V4 If a * 1, X 3 j (H, hu h 2) is irreducible and we are done. -1 The case a = 1 is handled in the same way as r ~ 4 is. Assume r ~ 4. ~(h) * 1. Let X 3 act on V 3. For g E H, ~(g) = 1; also By Mitchell, X 3 is monomial. Let Vu ... , vr+i be a basis of V 3 in which X 3 is monomial. By Lemma 1. 4, there exist hu ... , hr-i E (H, h) such that by scaling and ordering Vu ... , Vr+i properly, v.X(h.) 1 = Vi+i vi+1X(hi) = Vi+2 Vi+ 2 X(hi) = v.1 1 v 1 X(hi) = Vf for ltl{i,i+l,i+2} 1 Since ~(h) * 1, by replacing h by h- if necessary, v.X(h) = wv. for J J 23 some j and viX(h) = vi for i -:f. j. Let g E (hi, ... , hr_ 1 ) with vjX(g) = vk 1 1 for some k -:f. j. Leth* = g- h- gh. Then ~(h*) = 1 and Letting K = (hi, ... , hr-uh*), we are done since X3 jK is irreducible and~ jK = lK. Lemma 2. 4: Let H be a subgroup of G generated by special 4-elements (special 3-elements) such that X IH = X1 EB (n-r)lH where r ~ 4 (r ~ 5) and X1 is irreducible of degree r. Suppose X1 acts on V 1 and is monomial in a basis Vu ... , vr of V 1 • Let Hi = (h EH lviX(h) = vi and h is a special 4-element (special 3-element)) . Then X !Hi = x1 , i EB (n-r+l)lH, where X1 , i is irreducible and monomial on 1 (vi, ... ' vi-u Vi+u ... 'vr). Proof: Notice that multiplying vi by a does not change Hi. Let (n-r)lH act on V 2 • First take the case where His generated by special 4-elements. By Lemma 1. 3 after scaling and ordering Vu ... , vr cor- rectly, there exist special 4-elements hu ... , hr-i such that, 24 Hr = ( hgr lh E H.). Clearly H.1 has the desired properties if and only if 1 Hr does. But by Lemma 1. 3, X I (h1, ... , hr_ 2 ) is irreducible on (vi, ... , vr_ 1). As X !Hr clearly acts trivially on V2 EB (vr), we have X IHr = X , r EB (n-r+l)lH r where X , r is irreducible and monomial on 1 1 (vi, ... 'vr-1> . Now consider the case where H is generated by special 3-elements. By Lemma 1. 4, after scaling and ordering Vu ... , v r correctly, there exist special 3-elements hu ... , hr_ 2 such that viX(hi) = Vi+i Vi+iX(hi) = Vi+ 2 vi+2X(hi) = vi v f X(hi) = v f for £ E {i, i+l, i+2} Again as XI (h1 , ••• , hr_ 2 ) is transitive on Vu ... , vr' it suffices to show the results hold for Hr. Since X1 is irreducible, there exists h E H such that viX(h) = av. {viX(h) = J vjX(h) = bvk abc = 1 vkX(h) = cv.1 v fX(h) = vf for £ fl. {i, j, k} or WV1 vjX(h) = wv. J vfX(h) = vf for f q {i, j} In the first case not all of a, b, and c are 1. In either case, for some I.., vf X(h) = vf. By transitivity, there is a g E (hi, ... , hr) such that vf X(g) = vr. Replacing h by hg, we may assume i, j, k < r. In particular XI ( hu ... , hr_ 3 , h) is irreducible on (vi, ... , Vr_ 1) . Clearly X !Hr acts 25 trivially on V 2 EB (vr), and so X !Hr = X1 r EB (n-r+l)lH where X1 r is ' r ' irreducible and monomial on (vu ... , Vr_ 1 ) . We are now ready to prove that under certain circumstances there exist primitive subgroups of codimension 1 or 2. These are important for inductive purposes in the proofs of the main results. Lemma 2. 7 will also be useful in Chapter V to determine the possible subgroups generated by two special 3-elements. Lemma 2. 5: Let n ~ 6. Assume there exists a subgroup U of G gen~rated by special 4-elements such that X lu = X 1 EB rlu where r = 1 or 2 and X1 is irreducible. Then there exists a subgroup H of G generated by special 4-elements such that X IH = Y EB ylH where Y i.s irreducible and primitive, and y = 1 or 2. Proof: Assume the lemma is false. In particular X1 is monomial. By replacing U by another subgroup, we may assume r = 1 as follows. Let X1 act on V1 and let v 1 , ••• , Vn_ 2 be a basis of V1 such that X1 is monomial in that basis. Let 2 · lu act on V 2 • Choose a special 4- element h* such that X(h*) does not leave V 1 invariant. Then X(h*) does not leave both (vu ... , vn_) and (v2 , • • • , Vn_ 2 ) invariant. By reordering the vi's, we may assume X(h*) does not leave (vu ... , vn_) invariant. Let Un_ 2 = (h E U lvn_ 2 X(h) = Vn_ 2 and h is a special 4-element). By Lemma 2.4, xlun_ 2 = X1 ,n_ 2 EB 3 •lun_ 2 where Xun- 2 is irreducible on (vu ... , Vn_ 3 ) and monomial in the basis Vu ... , vn_ 3 • Also 3 · lu acts on V' = (vn_ 2 , V2 ) • So XI (Un_ 2 , h*) = ~ EB 1 (U n-2, n-2 h*) where ~ is irreducible or X2 = X 3 EB ~ with X3 irreducible of degree n - 2. If ~ is 26 irreducible, replace U by (Un_ 2 , h*). Suppose X2 = X3 EB ~. primitive, by Mitchell, ~ = 1 (U n-2, If X3 is h*) and the lemma is true, a con- tradiction. So X3 is monomial and there is a basis v7, ... , v~ of V on which XI (Un_ 2 , h*) is monomial. By Lemma 1. 3 applied to Un_ 2 , we * 3 may reorder and rescale v 1* , ••• , vn* such that v 1 = v *1 , ••• , vn_ 3 = Vn_ and V' = (v~_ 2 , v;_u v~). As (vi, ... , Vn_) is not left invariant by X(h*), in the basis v~, ... , v~, X(h*) is not diagonal. In particular ~ = 1 (U n-2, h*). So XI (Un_ 2 , h*) acts trivially on a subspace V3 of dimension 2 with V3 c V' . As V2 , V3 have dimension 2 and are in a subspace V' of dimension 3, V2 11 V3 =f:. {O }. Thus XI (U, h*) acts trivially on v2 n v3. Hence XI (U' h*) = X' EB 1 (U' h*) where X' is irreducible of degree n-1, as X(h*) does not leave V1 invariant. Replacing U by (U,h*), we may assume r = 1. So X lu = X1 EB lu where X1 is monomial and irreducible on V1 • Let V1 have a basis Vu ... , Vn-i in which X1 is monomial, and let lu act on (vn). First let g be any special 4-element such that X(g) leaves V1 invariant. Then XI (U, g) = X~ EB ~ where X~ is irreducible on V1 and ~ acts on (vn> . If ~ * 1 (U, g) , by Mitchell, X~ is monomial. If ~ = 1 (U, g) , as we are assuming this lemma is false, X~ is still monomial. If v7, ... , v~ is a basis of Vin which XI (U, g) is monomial, by Lemma 1. 3, after suitably scaling and ordering v1, ... , v~, we have v 1 = v:, ... , Vn_ 1 = v;_ 1 and (v~) = ( v n> . So X(g) is monomial in the basis vu ... , v n. Now let g be a special 4-element such that X(g) does not leave V1 invariant. Either X(g) does not leave (vi, ... , vn_ 3 ) or (v 3 , ••• , Vn_ 1 ) invariant. By reordering Vu ... , Vn-i if necessary, assume X(g) does 27 not leave (vu ... , vn_ 3 ) invariant. Let U = (h E U lh is a special 4- element and Vn_ 2 X(h) = Vn_ 2, Vn_ 1 X(h) = vn_ 1 ) • By Lemma 2. 4 applied twice, X lu = X EB 3 • lu where X is monomial and irreducible on <vu ... , vn_ 3 ) and has a monomial basis Vu ... , vn_ 3 on this subspace. Also 3 • lu acts on <vn-2, Vn-u vn). So XI (U' g) = ~ EB 1 (U' g) where either ~ is irreducible or X2 = X3 EB ~ such that X3 is irreducible and ~ is linear. By assumption that this result is false, ~ is monomial unless X2 = X3 EB ~ with ~ * 1 (U, g) . In that case, however, X3 is monomial by Mitchell. In any case XI (U, g) is monomial in some basis v7, ... , v~ of V. Applying Lemma 1. 3 to U, by ordering and scaling Vi, ... ,v~ cor- rectly, v 1* =Vu ... , vn_ * 3 = vn_ 3 , and <vn_ * 2, vn_u * vn> * = <vn-2' Vn-u vn). In particular vjX(g) = vj for some 1 ~ j ~ n-3. By reordering Vu ... , vn_ 3 , we may assume j = 1. Now let U1 = (h E U lv1 X(h) = v1 and h is a special 4-element). By Lemma 2. 4, X ju1 = Y1 EB 2 •lu where Y1 is irreducible and monomial 1 on (v2 , ••• , Vn_ 1 ) and Y1 has a monomial basis v2, ... ,vn-i. Also 2 · lu 1 acts on (vu v n>. As X(g) does not leave <Vu ... , Vn_ 1 ) invariant, X(g) does not leave (v2 , ••• , Vn_ 1 ) invariant. So XI (Uu g) = Y EB 1 (Uu g) - where Y acts irreducibly on V =:) (v2, ... , vn_ 1 ) and 1 (Ui, g) on (v 1 ) . By assumption that the result is false, Y is monomial with a basis v;, ... , v~. By Lemma 1. 3 applied to U1 , after reordering and rescaling v;, ... , v~, we may assume v1 = v2 , ••• , v~_ 1 = vn_ 1 and v~ E (vu vn). As X !u acts 1 trivially on (vuvn>, v~X(u) = v~ for u E U1 ; so as Y is irreducible, we have for some j with 2 ~ j ~ n-1, 28 By reordering v 2 , ••• , Vn- 1 , we may assume j = n-1. v .QX(g) = v f for 1 ~ f In particular ' ~ n-2. Let U2 = (h E U lv2 X(h) = v2 where h is a special 4-element). As in the analysis of (Uu g), there is v'n E (v2 , vn> such that = -EV' n = E -1 V. J v f for l Et {j , n} = The only possibility is that j = n-1 and (v~) = (vn). = (v~) :=. <vu vn> n (v2 , vn> So in fact X(g) is monomial in the basis Vu ... , Vn. Therefore if h is a special 4-element, X(h) is monomial in the basis v 1 , ••• , Vn· 4-elements. Let N be the subgroup of G generated by all special So N <l G. By Lemma 1. 3, the set { (vi) 11 ~ i ~ n} is the unique set of one dimensional subspaces of V permuted by X(N). 1 g E G, h E N, ghg- = h1 E N. For So ( (vi) X(g) )X(h) = ( (vi) X(h 1 ) )X(g) = (v.) X(g) for some j depending on i and h 1 • Thus X(h) for all h E N J permutes { (vi)X(g) jl ~ i ~ n}. So { (vi)X(g) 11 ~ i ~ n} = { (vi) 11 ~i ~n} and Xis monomial, a contradiction. Lemma 2. 6: Let n ?- 5. Let h 1 and h2 be noncommuting special 3-elements of G. Assume XI (hi, h2 ) is monomial in a basis Vu ... , vn such that XI (hu h2 ) leaves exactly n-5 of the subspaces (vi) fixed. Then 29 Proof: By ordering Vu ... , Vn correctly, we get · v 3X(h 2 ) = ClV4 v1 X(h1 ) = av2 v2 X(h 1 ) = bv 3 aj3y = 1 v4 X(h 2 ) = /3V5 abc = 1 and v 3 X(h 1 ) = CV1 v 5X(h 2 ) = yv3 v1 X(h 2 ) = vf for i q {3, 4, 5} v.e_X(h1 ) = v1 for fq{l,2,3} By replacing v2 by av2 , v 3 by abv 3, v4 by aabv4 , and v 5 by a{3abv 5 , we may assume a = b = c =a= J3 = y = 1. Clearly (hu ~) ~ A 5 , and X l<hu h 2 ) = X1 EB (n-4)1 (hu~> where X1 is irreducible. Lemma 2. 7: Let n ~ 8 and let K be a subgroup of G satisfying one of the following: 1. K = (g, h) where g, hare special 3-elements with X IK = X EB (n-r)lK such that Xis irreducible of degree r = 3 or 4. Assume for n = 8 and r = 4 that K ~ A 5 • 2. K = ( g, h, k) where g, h, k are special 3-elements with X jK = X EB (n-4)1K. Let X be monomial on a subspace V with a basis eu e 2 , e 3, e 4 such that in this basis 0 1 0 0 0 0 1 0 x(g) = 1 0 0 0 0 -1 0 X(h) = 1 0 0 0 0 0 1 0 , and X(k) = 1 0 0 0 0 0 0 1 0 0 0 1 0 0 0 1 0 -1 0 0 0 1 0 0 Then there exists a subgroup H of G generated by special 3-elements such that X IH = Y EB ylH where Y is irreducible and primitive with y = 1 30 or 2 and H contains a G-conjugate of K. Proof: First a few remarks are made about situation 2. The representation Xis irreducible and XI (g, h) = R1 EB R 2 where Ri are both irreducible with R1 acting on (e 1 + we 2 + we) EB (-e 2 + w e 3 + we 4 ) and R 2 on (e 1 + we 2 + we 3 ) EB (-e 2 + we 3 + we 4 ) . Also (g, h) (g, k) ~ A4 , and (h, k) ~ ~ SL 2 (3), A4 • Assume there is a basis u 1 , ••• , un on which XI (g, h, k) is monomial. By Lemma 2. 6, after ordering ui, ... , un correctly, we may assume X I (g, h) acts trivially on u 5 , • • • , un. As R 1 EB R2 acts on the same subspace as X does, (e 1 , e 2 , e 3 , e 4 ) = (ui, u 2 , u 3 , u 4 ) , and XI (g, h, k) acts trivially on u 5 , ••• , un. Let K satisfy either 1. or 2. By Lemma 2. 3 applied inductively, beginning with K, there exists a subgroup U generated by special 3elements with K ~ U such that X lu = X1 EB s • lu where s = 1 or 2 and X1 is irreducible. Assume the lemma is false. Then X1 is monomial. want to show that we may assume s = 1 . So assume s = 2. First we Let X1 act on V1 and let Vu •.• , vn_ 2 be a basis of V1 in which X1 is monomial. Let 2 • lu act on W. If K satisfies 2., we may assume X1 jK acts trivially on v 5 , • • • , vn_ 2 by ordering Vu •.• , vn_ 2 correctly. Suppose K satisfies 1. As g and h do not commute, we may assume by ordering correctly that X1 jK acts trivially on v 6 , ••• , vn_ 2 • By Lemma 2 .. 6, if n = 8 and r = 4, we may also assume that XI (g, h) acts trivially on v 5 • Thus in any case, we may assume X IK acts trivially on v O , ••• , v n- 2 where a = 6 if n > 8 and a = 5 if n = 8. Let h* be a special 3-element such that X(h*) does not leave V1 31 invariant. So X(h*) does not leave both (vi, ... , v o-u v o+u ... , vn_ 2) and (vi, ... , v O , v o+ 2, ... , vn_ 2 ) invariant. By renumbering v a, ... , vn_ 2, we may assume X(h*) does not leave (vi, ... , v o-v v o+v ... , vn_ 2 ) invariant. Let M = (u E U Iv 0 X(u) = v O where u is a special 3-element) . By Lemma 2. 4, X IM = X 1 ' 0 EB 3 · lM where X1 0 acts irreducibly on ' (vi,••·,vo-vVo+u···,vn- 2) and 3·1M on (v 0 ,W). Note K~M. Then XI (M, h*) = Y EB 1 (M, h*) where either Y is irreducible or Y = Y1 EB ~ with Y1 irreducible and ~ linear. If Y is irreducible, replace U by (M, h*). Assume then that Y = Y1 EB ~- If Y1 is primitive, ~ = 1 (M, h*) by Mitchell, contradicting the assumption that the result is false. So Y1 is monomial on the subspace V1 :J (vi, v 2, ... , v a-u v a+l , ••• , vn_ 2 ) . If Vu ... ,vn_ 2 is a basis of V1 in which Y1 is monomial, by Lemma 1. 4, applied to M, when Vi, ... , "n- 2 are ordered and scaled correctly, -V1 =Vu ... ,vo-v = Va-uVa+1 = Va+1' ... ,vn-2 = vn-2 an d,.,,, va E ( va, W) . As X(h*) does not leave (v1, ... , v o-u v a+i' ... , vn_ 2 ) invariant, Y1(h*) is not diagonal. In particular ~(h*) = 1 and so XI (M, h*) = Y1 EB 2 · 1 (M, h*) · Let 2 · 1 (M, h*) act on W*. Then W* c (v 0 , W). As Wand W* have dimension 2, W n W*=!: {O }, and so XI (U, h*) acts trivially on W n W*. Hence XI (U, h*) = X EB 1 (U, h*) where Xis irreducible of degree n - 1, because X(h*) does not leave V 1 invariant. So without loss of generality, we may assume s = 1 and X1 acts on V1 • Let Vu ... , Vn-i be a basis of V1 in which X1 is monomial. Suppose lu acts on (vn). Assume first that h is a special 3-element such that X(h) leaves V1 invariant. Then XI (U, h) = X EB ~ where X acts on V1 and~ on (vn). If Xis primitive, by Mitchell, ~(h) = 1, and we have a 32 contradiction to the assumption that the result is false. - So X is monomial and by Lemma 1. 4, X(h) is monomial in Vi, ... , vn. Now let h be a special 3-element such that X(h) does not leave V 1 invariant. As earlier, we may order Vi, ... , vn_ 1 correctly so that X IK acts trivially on Va, ... , vn-i where a = 5 if n = 8 and a = 6 if n > 8. Also X(h) does not leave all three of (vu ... , v a-u v a+ 2 , ••• , vn_ 1 ) , (vi, ... , Va-u Va+u ... , vn_ 2 ) , and (vu ... , v a, Va+ 3 , • • • , Vn_ 1 ) invariant. By numbering v a, ... , Vn-i correctly, we may assume X(h) does not leave (vu ... , v a-u v a+z, ... , vn_ 1 ) invariant. Let M = (u E U lvaX(u) = v a , v a+ 1 X(u) = v a+i where u is a special 3-element). By Lemma 2. 4 applied twice, X IM = X EB 3 · lM where Xis irreducible and monomial on (vi, ... , v a-v Va+z' ... , vn_ 1 ) and 3 · lM acts on (v a, v a+u vn). But K ~ M and XI (M, h) = Y EB 1 (M, h) where either Y is irreducible or Y = Y1 EB ~ with Y1 being irreducible and ~ being linear. By assumption that the result is false, if Y is irreducible, Y is monomial. Also if Y is reducible, Y1 is monomial by Mitchell and the assumption that the result is false. In either case there is an i < n with viX(h) = vi by Lemma 1. 4. If 1 ~ i ~ a-1, by Lemma 1.4, choose u EU such that vaX(u) = vi. If i ~ a, let u = 1. Let Ni = (g E U lviX(g) Then X !Ku acts trivially on vi. =viand g is a special 3-element). So Ku~ Ni and by Lemma 2. 4, X !Ni = Yi EB 2 · lN, where Yi is irreducible and acts with monomial basis l Vu ... ,vi-vVi+u ... ,vn_ 1 • Also 2 ·lN. acts on (vi,vn). As X(h) leaves l vi invariant but not Vu X(h) does not leave (vu ... , vi-u vi+i, ... , vn_ 1) invariant. Thus XI (N., h) = R. EB 1 (N h) where R. acts irreducibly on l V 2 and 1 ( N., h) acts on (vi). l i, l As the result is assumed false, Ri is l monomial in some basis Vu ... ,vn-1 of V 2 • By Lemma 1. 4 applied to Ni, 33 we can renumber and rescale ,7i, ... ,Vn-i so that v 1 =Vu ... , Vi-i = ="i-1' Vi+1 = vi+1' • • •, Vn-1 = "n-1 and "i E (vi, vn). As X(h) does not leave (Vi, ... , vi-1' Vi+u ... , Vn-1> invariant, viRi(h) (/_ (\\). So for some j, k < n with i, j, k all distinct, vjX(h) = avk vkX(h) = b\ abc = 1 ,\X(h) = cvj v.e_ X( h) = vf for 1 ~ £ ~ n-1, f * j , k . Choosing £ with 1 ~ .e. ~ n-1 but distinct from i, j, k, we may apply the same argument to get v X(h) p vqX(h) = avq v;x(h) = yvp = {3v; a.f3y = 1 vtX(h) = Vt for 1 ~ t ~ n-1, t -:/= p, q where p, q, .Q < n are distinct and v; E (vf, vn). By comparing the two expressions for X(h), the only possibility is j = p and k = q. So (t\) = (v!) ~ (vi' vn) n (v.e_, vn) = (vn). In particular, X(h) is monomial in the basis Vu •.. , vn. The ref ore if N is the normal subgroup of G generated by the special 3-elements of G, X IN is monomial. As in the concluding para- graph of the proof of Lemma 2. 5, this is a final contradiction. Corollary 2 .1: Let K be as in Lemma 2. 7 and n ~ 8. Then there exists a subgroup U of G generated by special 3-elements such that 34 X lu = R EB rlu where R is primitive and irreducible with r = 1 or 2, and K cU. Proof: Let H be as in the conclusion of Lemma 2. 7. Then -1 Kg~ H for some g E G. Letting U = Hg , the desired result follows. Corollary 2. 2: Let H be a subgroup of G generated by special 3-elements such that X IH = X EB (n-r)lH where Xis irreducible of degree r and 3 ~ r < n. Then there exists a subgroup U of G such that XI U = Y EB ylu where Y is irreducible and primitive with y = 1 or 2. Proof: By Lemma 2. 3 and induction, we may assumer = n-2 or n-1. If X is primitive, we are done. So assume X is monomial. By Lemma 1. 4, there exist special 3-elements hu h2 such that XI {h h 1, 2) = S EB (n-3)1 {h h ) where S is irreducible of degree 3. We 1' are done by Corollary 2 .1. 2 35 CHAPTER III THE PROOF OF THEOREM 1 Theorem 1 will now be proved. or less are known. The primitive groups of degree 7 By examining Brauer [3], Lindsey [14, 16, 17] , and Wales [27, 28] very carefully, we see there are no primitive, irreducible, unimodular linear groups of degree 5, 6, or 7 containing special 4elements. We may assume n ~ 8. Let g and h be special 4-elements which do not commute. Then XI (g,h) = Y EB (n-4)1 (g,h) where Y has an irreducible constituent of degree at least 2. Case A: Y = Y1 EB Y2 where Y1 is irreducible of degree 2. Let Hi = Y/ (g, h)). Then Y( (g, h)) =His a subdirect product of H1 and H2 • By examining Blichfeldt [2], we check the various possibilities for H1 • Subcase 1: H1 /Z(H 1 ) ~ A5 • (This proof is as in Lemma 2 of [ 14]). Then H' is a subdirect product of H~ and H~, which are unimodular groups of 2 x 2 matrices. Let Mi be the set of all elements in Hi which occur with component the identity of Hj (j =1= i) in the subdirect product. By Theorem 5. 5.1 of Hall [11], Mi <l Hi and H~/M 1 ~ H~/M 2 • As H~/Z(H~) ~ A 5 , either M1 = H~ = H1 or M1 ::: Z(H~). As A 5 has no IZ(H~) I = 2. If M = H~, there is an element of order 3 in Mu which must have eigenvalues w, w. Multiplying by the representation of degree 2, 1 36 nontrivial element in Z(H~) gives an element in X(G) with eigenvalues -w,-w,1,1,1, ... , contradicting Blichfeldt. If M1 ~Z(H~), the only possibility is that H~/Z(H~) ~ A 5 • In any case there are elements h.l E H~l with eigenvalues -w, -w which are paired in H. So X(G) contains an element with eigenvalues -w, -w, -w, -w, 1, 1, ... , contradicting Blichf eldt. Subcase 1. is therefore impossible. Then H' is isomorphic to a subdirect product of H~ and H~, which again are uni modular. Let Mi be as in subcase 1. Then H~ /Z (H~) ~ A4 • The only element of order 2 in a 2 dimensional unimodular group is (-J _~). So IZ(H~) I 2 and the Sylow 2-subgroup of H~, which has order = 8, is either cyclic or quaternion. As A4 has no elements of order 4, the Sylow 2- subgroup of H~ must be quaternion. If M1 contains the Sylow 2-subgroup of H~, there exist special 4-elements gi, g 2 such that (gi, g 2 ) is the quaternion group of order 8 and XI (g g = x* EB (n-2)1 ( ) g1, g2 where X*( g 1 ) = (-i0 0) i and X*( g 2 ) = (01 -10 ) in some basis. If M1 does not contain the Sylow 2-subgroup of H~, as HUZ(H~) ~ A4 and M1 <l H~, 1, 2) M1 ::: Z(H~). As H~/M 2 ~ H~/Mu the only possibility, by looking at Blichfeldt's list of 2 dimensional groups is H~/Z(H;) ~ A4 • As in sub- -w, which are paired in H. So X( G) contains an element with eigenvalues - w, - w, -w, -w, 1, 1, ... , contradicting Blichfeldt. 1 case 1, there are elements hi E H with eigenvalues -w, This subcase is impossible as 2-elements do not generate A4 • 37 Subcase 4: Y1 is monomial and unimodular. As Y1 is irreducible and unimodular, Y1 (g) and Y1 (h) both have eigenvalues i and -i. So Y2 = 2 · 1 (g, h). Subcase 5: Y1 is monomial and not unimodular. Let Yi act on Vi and let Vu v 2 be a basis of V1 in which Y1 is monomial. As both Y1 (g), Y1 (h) could not be diagonal, only one of Y1 (g), Y1 (h) could have eigenvalue structure 1, i or 1, -i. As Y1 is not unimodular, 1 we may assume by replacing g by g- if necessary that Y1 (g) has eigenvalues 1, i. So Y2(h) is the identity and hence Y2 = ~1 EB ~2 • Let ~1 act on v 3 , ~2 on v 4 • By ordering Vu v 2 and v 3 , v 4 correctly we get in the basis Vu . . . , v 4 that (1 0 0 0 0 i 0 0 and (Y 1 EB y 2)(g) = 0 0 -i 0 0 0 0 1 (Y1 EB Y2)(h) = -0 0 0 0 0 0 0 1 0 0 0 1 But 1 (Y l EB Y2)(h- ghg- 1 ) = -1 and letting g 1 = h, g 2 = h ghg XI (gi, g 2) = Y EB (n-2)1 (gi, g 3 -1 ) 2 0 0 0 -i 0 0 0 1 0 0 0 1 , which a.re special 4-elements, where Y3 is monomial and irreducible. 38 Therefore if case A holds, we may choose special 4-elements gu g 2 such that XI ( gu g 2 ) and monomial. = X 1 EB (n-2) 1 (gu g ) where X1 is irreducible 2 Let X1 act on V 1 and choose a special 4-element g 3 such that X(g 3 ) does not leave V 1 invariant. So X I ( gu g 2 , g 3 ) = X 2 EB (n-4)1 (gi, g , g ) where ~ is irreducible or ~ = X3 EB ~ such that 2 3 X 3 is irreducible of degree 3. Assume the latter is the case with ~ :t 1( g1' g2, g ) . Then by Mitchell, X 3 is monomial on some basis 3 Vu v 2 , v 3 • Consider X/gi) in this basis; X 3 (gi) must act trivially on at least one vr Assume first that X j(g1 , g 2 ) does not leave any (vi) invariant. Then by ordering Vu v2 , v 3 correctly, we get 0 -0 0 0 0 0 1 0 and X3 (g 2 ) = 1 0 0 0 0 -E 0 E- But then X 3 j (gu g 2 ) is irreducible, a contradiction. leaves some (vi) invariant. 1 0 So XI (gu g 2 ) But as ~(g 1 ) = ~(g 2 ) = 1, ~(g 3 ) :t 1, X3 (g 3) is diagonal and so also leaves (vi) invariant, a contradiction. any case xi (gug ,g 2 3) = Thus in X EB (n-s)l (g 1 ,g 2 ,g 3 ) wheres= 3 or 4 andX is irreducible when case A holds. Case B: Y = Y1 EB ~ where Y1 is irreducible of degree 3. If Y1 is primitive, ~ = 1 (g, h) by Mitchell. Assume Y1 is monomial on V 1 ; let Vu v 2 , v 3 be a basis of V 1 in which Y1 is monomial. If Y1 (g) or Y1 (h) have eigenvalues 1, 1, -i or 1, 1, i, they are diagonal in this basis. In any case Y1 (g) and Y1 (h) each fix one of the subspaces <v/, 39 j depending on g and h. So neither Y1 (g) nor Y1 (h) is diagonal and hence ~ = 1 (g, h). Therefore in any case G contains a subgroup H generated by special 4-elements such that X jH = X 1 EB (n-r)lH where X1 is irreducible and r = 3 or 4. Assume Theorem 1 is false. Let n ~ 8 be minimal such that there is a counterexample. Thus, recalling the remark at the beginning of this section, there does not exist irreducible, primitive unimodular groups of degree n-1 or n-2 which contain special 4-elements. Applying Lemma 2.1 inductively, starting with H, there exists a subgroup U generated by special 4-elements such that X /u = Y EB y • lu where Y is irreducible and y = 1 or 2. By Lemma 2. 5, we may assume Y is primitive, contradicting the minimality of n. So Theorem 1 holds. 40 CHAPTER IV A SPECIAL CASE OF THEOREM 2 We return to the proof of Theorem 2. In this chapter we classify G when n ~ 8 and a certain hypothesis holds, which will be useful for induction purposes. Define hypothesis (A) as fallows: (A) If U is any subgroup of G generated by special 3-elements such that X ju = S EB s • lu where S is irreducible and primitive of degree n - s withs = 1 or 2, then U ~ An-s+i · We first prove some results on the irreducible characters of the alternating and symmetric groups. By the work of Frobenius [ 9] , the characters of Sn are all related to the partitions of n into integers. Let (A) ={Au ... , Ak} be a partition of the integer n into nonnegative integers where A1 ~ A2 ~ . • • ~ Ak > 0 . Let f i = \ + k - i. So n ~ .f. 1 > £ 2 > . . . > f k > 0. Then the degree d(A) of the irreducible character corresponding to (A) is All irreducible characters come from such partitions. Lemma 4 .1: The group Sn for n ~ 9 has two irreducible characters of degree 1 due to the partitions (A) = {n} and (A) = {1, 1, ... , 1}, two irreducible characters of degree n - 1 due to the partitions 41 (A) = {n-1, 1} and (A) = {2, 1, 1, ... , 1}. All other irreducible characters have degree greater than 2(n+l). Proof: By checking the chara cter tables, we see that the r esult holds for n = 9, ... , 14 (see [13, 19, 29, 30]). So assume n ~ 15 and proceed by induction. We split this into several cases. The case k = 1: The only partition is (A) = {n} . So d(A) = n ! The case k = 2: Then (A) = {Au AJ and A1 ~ ~- Jr = 1 . If A1 = n - 1, thenf.. 1 =n, f.. 2 =1, andd(A) =n-1. IfA 1 =n-2, thenf.. 1 =n-1, f.. 2 =2, and d(A) = n(n2- 3) > 2 (n+l ) for n ~ 15. So assume n ~ A. ~ n - 3. Let A = m. 2 l. 2 = n - m. 1 (m+l+m-n) So d(A) = n I m ! (n-m) I ' m'!(n-m')I n! m,m ' >=; .-- n 2'"w1· th m > m, and n ;;,. 15, m !(~-Im)! The case k = n: 1 Then ! 1 = m + 1 and > m ! (n-m) n! ! . Notice that if > m!(n-m)! n! Al so f or m =n- 3 > 2(n+l). So d(A) > 2(n+l). The only partition is (A) = {1, 1, ... , 1}. So £. = 1 + n - i and l = 1 The case (A) = {n-k+l, 1, 1, ... , 1} for 3 ~ k ~ n-1: and f i = k - (i-1) for i ~ 2 . So Then f.. 1 = n 42 = {(n-(k-1) ) ... (n-1 )} {(k-2) I (k-3) ! ... 1 !} (k-1) l(k-2) ! ... 1 ! If k =n-1, (A) ={2, 1, 1, ... , 1} and d(A) =n-1. = (n-1\ k - 1) Because(~= i) =( ~ = fl , fl > 2(n+l) for n ~ 15 and 3 ,;; k,;; n-2, it suffices to prove it for 3 ~ k ~ ~- But if 3 ~ k ~ ~' for n ;?: 15, 2(n+l) < (n-l~(n- 2) = to prove (~ = (n ;1} ,; (~ = fl The remaining cases: In the remaining cases we have k ;?: 3 and we do not have the partition (A) = {n-k+l, 1, ... , 1} . Let >tk = r. As k ;?: 3, ~ ;?: r. Consider the partition (A') = {Au ... , Ak_J of n - r. 1'. =A . + (k-1) - i =/.. . - 1. 1 1 1 Let Then ep -!'q =l p -i q and But is the degree of a character of S n-r corresponding to the partition (A'). 43 By induction, noting that we have enough initial cases (as r ~ i ), d(A') > 2(n-r+l) because (A') is not of the form {1, ... , 1}, {n-r}, {n-r-1,1}, or{2,1, ... ,1}. Assume first that k - 1 > r. As n ~ f 1 > 1 2 > ... > ik > 0, Qp-r ~ Qp+r for p ~ k- 1 - r. So .fp-r /.e.p+r ~ 1 for p ~ k - 1 - r. As ik = r, .fk-r-r > ... > .fk_ 1 -r > 0 and so k-1 TT p=k-r (t -r) p r! Thus k-1 (~\ n (i p -r) p=k-r p+r} As n ~ £ 1 > £ 2 . > • • • > £r, n+l-i 1i ~ 1 for 2 ~ 1 ~ r. r! ( n So d(A) > 2 n-r+l) r; . If r = 1, A1 ~ n - (k-1) with equality holding only if (A) = {n-k+l, 1, 1, ... , 1} which we are excluding. So £1 = A1 + k - 1 ~ n - 1 and d(A) > n~l 2n > 2(n+l). If r > 1, A1 ~ n - r(k-1) and £ 1 = A1 + k- 1 ~ n - (r-l)(k-1) ~ n - 2(r-l) ask~ 3. So d(A) > n- 2(r-l) 2(n-r+l). But n- 2(r-l) 2(n-r+l) ~ 2(n+l) is equivalent to O ~ n(r-2) + 2(r-1), which is true for r ~ 2. Assume that k - 1 ~ r. Then 44 The middle product is vacuous if k-1 = r. As l 1 > f 2 > ... > .e.k-i > lk = r, l... - r ~ k - i for i ~ k - 1 . So 1 k-1 (l...-rj 11' _1___ i=l ~ 1 k-1 As~~ r ~ k-1, n-(k-1) ~ rand n-(k-1)-i ~ r-i for O ~ i ~ r - k. Thus { (n-(k-lwn-k) ... (n-r+l)~ " l r(r- ... (r-(r-k)) ~ j As n ~ f 1 > f 2 > ... > l..k_u ~~i 1+1 ~ 1. Therefore d(A) > 2(n-r+l) -J!: 1 But A1 ~ n-r(k-1) and l.. 1 = A1 + k- 1 ~ n - (r-l)(k-1) ~ n - 2(r-1). So d(A) > n- 2(r-l) 2(n-r+l). Proceeding as above, d(A) > 2(n+l), and the lemma is proved. Lemma 4. 2: If n ~ 7, An has only one irreducible character of degree 1, which is the trivial character, and only one irreducible character of degree n-1, which is the nontrivial constituent of the permutation character. All other irreducible characters have degree greater than n+l. Proof: By Frobenius [ 10] , all irreducible characters of S0 remain irreducible or split into two conjugate irreducible characters, 45 obviously of equal degree, when restricted to An. All characters of A are obtained in this way. n The result is true for n = 7, 8 by checking the character tables. Assume n ~ 9. Let p 1 be the permutation char- acter of Sn on n points and let µ be t he nontrivial linear character of S . Two irreducible characters of Sn of degree n-1 are p 1 -ls and n n µ(p 1 -ls ) . They are equal and irreducible when restricted to An. As n all other irreducible characters of Sn have degree greater than 2(n+l) by Lemma 4.1, all other irreducible characters of An have degree greater than n+ 1 . Notice that if H ~ Am for m ~ 7 is a subgroup of G such that X jH = X 1 EB (n-m+l)lH, X1 is irreducible and the 3-cycles of H correspond precisely to the special 3-elements of H. Also X 1 is primitive by Lemma 1. 4 as Am is simple. These facts will be used without reference in the rest of the paper. We now give some results on generators and relations of An. Lemma 4. 3: Let Uk = {fu ... , fk_ 2 ) for k ~ 5. Suppose the following relations hold: 3 (1) f 1 2 = 1, (f afd-i ... f 1 ) = 1 for d = 2, ... , k-2 (2) f d+ 1 = 1 for d = 1 , . . . , k- 3 ( 3) ( (f d . . . f 1 )(f e . . . f 1 ) ) 2 = 1 for d = 1, . . . , k- 4 and e = d+ 2 , . . .. , k-2. Then either Uk = 1 or Uk ~ Ak. Also in Ak if we let fd = (d, d+l, d+2), then f u ... , fk_ 2 satisfy (1), (2), and (3). 46 Proof: Let hd = fafd-i ... f 1. Then (1) is equivalent to ( 1 ') h~ = 1, h~ = 1 for d = 2 , . . . , k- 2 Also (2) is equivalent to (2') (hdhd+1) 3 = 1 3 ( h- 1 ) 3 ( ( ( -1 _1 3 _1 3 because ( hdhd+i ) = hd d+i = f d ... f 1 ) f 1 ••• f d+i) ) = (f d+i) . In addition, (3) is equivalent to 2 (3') (hdhe) = 1 for d=l, ... , k-4 and e=d+2, ... , k-2. By Moore [ 22] , Uk = 1 or Uk ~ Ak as Uk = (hu ... , hk_ 2 ) . Now let fd = (d, d+l, d+2). d ~ 2. Then fd ... f 1 = (1, 2)(d+l, d+2) for 2 Thus (1) and (2) hold. Also (f 1fe ... f 1 ) = ( (1, 2, 3)(1, 2)(e+l, e+2)) = 1 for e ?: 3 and for d > 1, e ?: d+2, (fd ... f 1f e ... f 1 ) 2 = 2 ( (1, 2)(d+l, d+2)(1, 2)(e+l, e+2) ) = 1. So ( 3) holds. Lemma 4. 4: Let U be a group with a subgroup H ~ Ak for k ?: 5. Let H = (fu ... , fk_ 2 ) where fd corresponds to (d, d+l, d+2) on {1, ... , k}. Assume fk-i E U such that fk-i = 1, (fk_ 2 fk_ 1)2 = 1, fk-i commutes with fu ... ,fk_ 4 , and fk_ 1 fk_ 2 fk-ik-i = fk_ 2 fk_ 3 • Then (H,fk_ 1 ) ~ Ak+i· Proof: It suffices to show (1), (2), and (3) hold in Lemma 4. 3. By Lemma 4. 3, these relations hold for the f i's with i ~ k - 2. (1) we only need to consider d = k - 1. So for But Thus (1) holds. Clearly (2) holds. We only need to consider e = k-1 in (3). Assume first that d ~ k-4. Then 2 47 Now assume d = k-3. (fk-ik-1) 2 First as fk_ 1fk_ 2fk-ik_ 1 = fk_ 2fk_ 3 and 2 = 1, fk-ik-1 = fk_1fk_2fk_3• So Lemma 4. 5: Let Ube a group containing a subgroup H = (h 1 , •• . . , hk_2) ~ Ak for some k ~ 6. Assume H acts on {1, ... , k} with hi = (i, i+l, i+2) for 1 ~ i ~ k-2. Let g E U such that (H, g) ~ Ak+s where s = 1 or 2. Assume (H, g) acts on {b 1 , ••• , bk+J and that hu ... , hk_ 2, g are 3-cycles in (H, g). Then by numbering bu ... , bk+s correctly, hi = (bi, bi+i' bi+ 2). Also there is an h E H such that gh or 1 (g- )h is (bk+s- 2 , bk+s-u bk+s); if g commutes with hi, we can choose h E (h4 , • • • , hk- 2 ) • Proof: Because (hu h 3 ) ~ A 5 and hu h 3 are 3-cycles in (H, g), by correct ordering of bu ... , bk+s' h1 = (bu b 2, b 3 ) and h 3 = (b 3 , b 4 , b 5 ). h3h1 As h 2 = h1 , ~ = (b 2 , b 3, b 4 ). Assume we have numbered bi, ... , bk+s correctly so that hj = (bj, bj+i' bj+ 2 ) for 1 ~ j ~ m where m ~ 2. If m = k-3, omit hm in the latter classification and so assume m ~ k-4. 48 As (hi, ... , hm, hm+ 2) ~ Am+ 4 and each hi is a 3-cycle, by numbering bm+ 3 , ••• , bk+s correctly, hm+ 2 = (bk, bm+ 3 ' bm+ 4 ) for some k ~ m+2. Since hm+ 2 commutes with hu ... , hm-u the only possibility is k = m+2. hm+2hm Also ~+ 1 = hm = (bm+i, bm+ 2, bm+ 3 ) • Therefore by induction hi, ... , hk_ 2 have the desired form. If s = 1, g = (bi, bj, bk+ 1 ) where i, j ~ k, and if s = 2, g = (bj, bk+i, bk+ 2) where j ~ k upon ordering bk+i, bk+ 2 correctly. By double transitivity, choose h E H with bih = bk_ 1 and bjh = bk. Then g h = (bk+s- 2, bk+s-u bk+s). Suppose g com- mutes with h 1 • Then i,j ~ 4. So we could have chosen h E (h4 , ••• ,hk_ 2) unless (h4 , • • • , hk_ 2) is not doubly transitive, which occurs only if k = 6. If k = 6, however, g h or (g -l)h where h E (h4 ) is of the desired form. Lemma 4. 6: Let U be a group with a subgroup H = (h 1 , ••• , hk_ 2) ~ Ak where k ~ 7. Assume H acts on {1, ... , k} and that hi = (i, i+l, i+2) for 1 ~ i ~ k-2. Let g E U such that g commutes with h1 and h2 • Let H1 = (h2 , ••• , hk_ 2, g) ~ Ak and assume H1 acts on {b 1 , ••• , bk} . Furthermore assume h2, ... , hk_ 2, g are 3-cycles in H1 . Proof: Then (H, g) r-J Ak+i. By Lemma 4. 5, we may assume hi = (bi-v bi' bi+ 1 ) for 2 ~ i ~ k-2 and for some h E (h 5 , ••• ,hk_ 2), gh or (g- )h is 1 g 1 = (bk-1' bk, bk+ 1 ). Also g1 commutes with h 1 as g, h do. Thus g: = 1, 2 (hk_ 2 g 1 ) = 1, g 1 commutes with hu ... , hk_ 4 , and g 1 hk_ 2 hk_ 3 g 1 = hk_ 2 hk_ 3 • By Lemma 4. 4, ( H, g) = (H, g 1 ) ~ Ak+i. Lemma 4. 7: Assume hypothesis (A) holds and n ~ 8. Let M be a subgroup of G generated by special 3-elements such that X /M = X 1 EB (n-m)lM where X1 is irreducible of degree m ~ 5 (including the 49 possibility that m = n) or X IM = X 1 EB ~ EB (n-m-l)lM where X1 is irreducible of degree m ~ 5 and ~ is linear. Then X1 is primitive and if X IM is of the latter form, ~ = lM. Proof: Let K be as in the hypothesis of Lemma 2. 7. By Corollary 2. 1 and hypothesis (A), there is a subgroup U generated by special 3-elements such that K 5= U ~ An-s+i where U and s are as in hypothesis (A). As two special 3-elements of U, which must be 3-cycles of U, either commute, generate A4 , or generate A 5 , K can only satisfy 1. of Lemma 2. 7 and K ~ A4 or A 5 • Let X1 act on the subspace V 1 and assume X1 is monomial. Vu ... , vm be a basis of V1 in which X1 is monomial. Let By Lemma 1. 4, after rescaling and reordering Vu ... , v m, there exist special 3-elements hu ... 'hm-2 EM such that v.X(h.) 1 1 = Vi+l vi+iX(hi) = vi+2 vi+2X(hi) = v.1 v f X(hi) = vi for itl_{i,i+l,i+2} for 1 ~ i ~ m-2. Assume first that X IM = X 1 EB ~ EB (n-m-l)lM where ~ * lM. Therefore there is a special 3-element g E M with ~(g) = w. So X1 (g) must be diagonal in the basis Vu ... , vm and for some i, viX(g) = wvi. Let gv g 2 E (hi, ... , hm_ 2 ) such that viX(g 1 ) = v1 and viX (g 2 ) = v 2 • Then 1 1 1 h = g~ gg 1 g; g- g 2 is a special 3-element and 50 (1) { V 1 X(h) = WV1 V2X(h) = WV2 vfX(h) = vl for f > 2 But then XI (huh) = Y EB (n-3)1 (huh) where Y is irreducible and (h1 , h) i A4 or A 5 , a contradiction. Therefore ~ = lM and we may assume X IM = X1 EB (n-m)lM. As X1 is irreducible, there is a special 3-element g EM with X(g) not leaving (v1 + ... +v m> invariant. basis vi, ... , vm. First assume X1 (g) is diagonal in the Then viX(g) = wvi and vjX(g) = wvj for some i and j. Let g 1 E (hu ... , hm_ 2 ) such that viX(g 1 ) = v1 and vjX(g 1 ) = v2 , by double transitivity. 1 Letting h = g~ ggu we get X(h) as in (1), which is again a contradiction. Therefore X1 (g) is not diagonal. Hence there exist distinct i, j, and k such that viX(g) = av. J vjX(g) = bvk abc = 1 vkX(g) = cv.1 V f X(g) = Vf for ftl_{i,j,k} where not all of a, b, c are 1. Let g 1 E (hu ... , hm_ 2 ) with viX(g 1 ) = v 2 , 1 vjX(gi) = v 3 , and vkX(g 1 ) = v4 • Replacing g by g~ gg 1 , we may assume i =2, j =3, andk=4. If g commuteswithh2 , then a =b =cE {w,w}; however XI ( hi, g) = Y EB (n-4)1 (hi, g) where Y is irreducible but (hi, g) -;/. A4 or A 5 , a contradiction. So g does not commute with h 2 • Hence XI (I½, g) = Y EB (n-3)1 (h , g) where Y is irreducible. The only 2 possibility is that (~, g) ~ A4 and {a, b, c} = {-1, -1, 1} . By conjugating 51 -1 g by ~ or h 2 if necessary, we may assume a = -1, b = 1, and c = -1. But then situation 2. of Lemma 2. 7 occurs, a contradiction. Therefore we can only conclude X1 is primitive. If X IM = X 1 EB ~ EB (n-m-l)lM, by Mitchell ~ = lM. Lemma 4. 8: Assume hypothesis (A) holds and n ~ 8. Let H be a subgroup of G generated by special 3-elements. Assume X IH = X 1 EB (n-m) lH where X1 is irreducible of degree m ~ 5. special 3-element which does not commute with H. Let g be a Then X I (H, g) = Y EB (n- s) 1 (H, g) where s = m, m + 1, or m + 2 and Y i s irreducible. Proof: We must have XI (H, g) = R EB (n-m-2)1 (H, g) where either 1 . R is irreducible. 2. R = R1 EB ~ where R 1 is irreducible and ~ is linear. 3. R = R 1 EB ~1 EB ~2 where R 1 is irreducible and ~1 , ~ 2 are both linear. We are done if 1. holds. If 2. holds, ~ = 1 (H,g) by Lemma 4. 7. If 3. holds, since g does not commute with H, one ~i' say ~2 , is 1 ( H, g). By Lemma 4. 7, t 1 is also 1 (H, g) and the lemma is proved. Lemma 4. 9: Assume hypothesis (A) holds and n ~ 8. Let H be a subgroup of G generated by special 3-elements such that X !H = X1 EB 2 · lH where X1 is irreducible. such that Let g E G be a special 3-element XI (H, g) is irreducible. Then (H, g) ~ An+ Proof: H ~ An_ 1 . 1• By Lemma 4. 7, X 1 is primitive; so by hypothesis (A), Let H act on the set {av ... , au-J. The special 3-elements 52 in Hare 3-cycles and so define hi= (ai,ai+vai+ 2 ) for 1 ~ i ~ n-3. H = (hi, ... , hn_ 3 ) . As Then XI (H, g) is irreducible, g does not commute with H. Thus g does not commute with both (hi, ... , hn_ 4 ) and -1 1 ~ i ~ n-1. . Let gi = hn-i- 2 for 1 ~ 1 ~ n-3 and bi = an-i for (h2 , • • • , ~ - 3 ) . Then gi = (bi, bi+v bi+ 2 ) and (gv ... , gn_ 4 ) = (hn_ 3 , ••• , ~ ) . If necessary, replacing hi by gi and an-i by bi, we may assume notation is chosen so that g does not commute with (hi, ... , hn_ 4 ) . First consider XI (hi, ... , hn_ XI (hi, ... , hn_ g) . By Lemma 4. 2, 4, = X 1 I (hi, ... , hn_ 4 ) EB 2 · 1 (h h ) = 1, ... , n-4 X 2 EB 3 · 1 (h h ) where ~ is irreducible. By Lemma 4. 8, 1, ••• , n-4 xl(h1 , ••• ,hn_ 4 ,g) =Y EB s·l(h h g) wheres= 1, 2, or 3 and v ... ' n-4' Y is irreducible. If s = 3, clearly (hi, ... , ~- 4 , g, ~- 3 ) is not 4) XI irreducible, a contradiction. So s = 1 or 2. By Lemma 4. 7, Y is primitive and hence by hypothesis (A), (hi, ... , ~- 4 , g) ~ An-s+ 1 and hu ... , hn_ 4 , g represent 3-cycles. By Lemma 4. 5 there is an element g 1 = g h for some h E (hi, ... , hn_ 4 ) such that g 1 commutes with hv .. . . , hn_ 6 • As (H, g) = (H, g 1 ) , by replacing g by gv we may assume g commutes with hu ... , hn_ 6 • XI (H, g) is irreducible, g does not commute with ( h . . , hn-). By Lemma 4. 2, XI (h hn_ = X I (h hn_ EB As 2 , •• 2 , ••• , 2 · 1 (h h 2, ... , n-3 ) = X3 EB 3 · 1 (h Therefore by Lemma 4. 8, XI (h h 2, ... , n-3 2 , ••• , 3) 1 hn_ 3 , g) = R EB r · 1 (h XI h g) 2,· • ., n-3' (h 2 , ••• , hn_ 3 , g, h 1 ) By Lemma 4. 7, R is primitive. If r = 2, by hypothesis (A), (~, ... , hn_ 3 , g) An; however 3) ) where X3 is irreducible. where r = 1, 2, or 3 and R is irreducible. If r = 3, is reducible, a contradiction. 2 , ••• , ~ An_ 1 • By Lemma 4. 6 (H, g) XI (H, g) could not be irreducible by Lemma 4. 2. ~ 53 Therefore r = 1 and by hypothesis (A), (h2 , ••• , ¾- 3 ' g) (h2 , • • • , hn_ 3 , g) act on {bu ... , bJ. ~ An. Let By Lemma 4. 5, we may assume hi = (bi-u bi, bi+ 1 ) for 2 ~ i ~ n-3 and g = (bi, bn-u bn). Also there is . g = an element h E ( h 5 , ••• , hn_ 3 ) such that g h or (g - l)h 1s 1 (bn_ 2 , bn-u bn). As g and h commute with hi, so does g 1 • Let g1hn-3 hn_ 2 = hn_ 3 • Then by Lemma 4~ 6, (H, hn_ 2 ) ~ An. But hu ... , hn_ 2 represent 3-cycles in (H, ~- 2 > and if (H, hn_ 2 ) acts on {cu ... , en}, by Lemma 4. 5, we may assume hi = (ci, ci+u Ci+ 2 ) for 1 ~ i ~ n-3 and hn_ 2 = (cj, ck, en). Looking in (~, ... , ~- 2 >, the only possibility i s hn_2 = (cn_ 2 ,Cn_ucn). Thus by Lemma 4.6, (H,hn_ 2 ,g 1 ) = (H,g) ~ An+i· Lemma 4.10: Assume hypothesis (A) holds and n ?: 8. Let H be a subgroup of G generated by special 3-elements such that X IH = X1 EB lH where X1 is irreducible. Let g E G be a special 3-element such that XI (H, g) is irreducible. Then (H, g) ~ An+i · Proof: By Lemma 4. 7, X1 is primitive. Therefore by hypothesis (A), H ~ An. Assume H acts on {au ... , an} . The special 3-elements in H are precisely the 3-cycles and so let hi = (ai, ai+i, ai+ 2 ). Hence H = (h11 ••• , hn_ 2 ) and as XI (H, g) is irreducible, g does not commute with H. So g could not commute with both (hu ... , hn_) and (h2 , • • • , hn_ 2 ) • As in the proof of Lemma 4. 9, we may assume g does not commute with (hu ... , hn_) . By Lemma 4. 2, XI (hu ... , hn_ 3 ) = X 1 I (hu ... , hn_ 3 ) EB l(h1, ... , hn-3 ) =Xz EB 2 ·l(h1, ... , hn-3 ) . By Lemma 4.8, (hu ... , hn_ 3 , g) = Y EB s · 1 (h h g) where Y is irreducible and 1, ••• , n-3, s = 0, 1, or 2. By Lemma 4.7, Y is primitive. Ifs= 2, by hypothesis XI 54 (A), (hu ... , hn_ 3 , g) irreducibility of ~ An_ 1 ~ (hu ... , hn_) and g E H, contradicting the XI (H, g). Thus s = 0 or 1 and by Lemma 4. 9 or hypothesis (A), respectively, (hi, ... , hn_ 3 , g) ~ An-s+ 1 • The elements hu ... , ~- 3 ,g must all represent 3-cycles and by Lemma 4. 5, there is a conjugate g 1 of g by an element in (hu ... , ¾-) such that g 1 commutes with hu ... , hn_ 5 • Without loss of generality, we may replace g by g 1 and hence assume g commutes with hu ... , hn_ 5 • As • • ,~_ 2 ). XI (H, g) is irreducible, g does not commute with (h But XI (h ,hn_ = X I (h ,hn_ EB 1 (h h ) 2, · · · , n-2 2 , •• 2 , ••• = X 3 EB 2 · 1 (h h 2,·· ·, n-2 by Lemma 4. 8, XI (h 2) 1 2 , ••• 2) ) where X3 is irreducible by Lemma 4. 2. 2 , ••• , hn_ 2 ,g) = R EB r -1 (h is irreducible and r = 0, 1, or 2. n-2, g ) where R By Lemma 4. 7, R is primitive. If r = 2, by hypothesis (A), ( h2 , ••• , ~ - 2 , g) ~ A0 _ 1 ~ so g E H, contradicting the irreducibility of Lemma 4. 9, (h2 , ••• , ~ - 2 , g) h 2, • • · , So ~ An+i · (h 2 , ••• , ~ - 2 > and XI (H, g). If r = 0, by As h 2 , ••• , hn_ 2 , g are special 3-elements, they represent 3-cycles. Assume (h 2 , • • • , ¾-z, g) acts on {bu ... , bn+J. By Lemma 4. 5, we may assume hi = (bi-u bi, bi+ 1 ) and g = (bj, bn, bn+ 1 ) for some j. By Lemma 4. 5, there is an . g = (bn_u bn, bn+ ) • As g h E (h5 , ••• , ~ - 2 ) such that g h or (g -1 )h 1s 1 1 . g1hn-2 and h commute with hu so does g 1 • Let ~- 1 = hn_ 2 = (bn_ 2 , bn-u bn). This commutes with h 1 and~; so by Lemma 4. 6, (H, hn_ 1 ) ~ An+i· Again hu ... , hn_ 1 represent 3-cycles in (H, hn_ 1 ) and if (H, hn_ 1 ) acts on {cu ... , cn+J, by Lemma 4. 5, we may assume hi= (ci,ci+uci+ 2 ) for 1 ::$ i ::$ n-2 and hn-i =(ci,cj,cn+ 1 ). Examining (h 2 , ••• , hn_ 1 ) , the only possibility is ~- 1 = (cn-u en, cn+J. 4. 6, (H, hn-u g 1 ) = (H, g) ~ By Lemma An+ 2 , contradicting Lemma 4. 2. So r = 1 55 and by hypothesis (A), (~, ... , ~ _2 , g) ~ An. By Lemma 4. 6, (H, g) ~ An+i · We are now ready to classify G when n ~ 8 and hypothesis (A) holds. Lemma 4.11: Assume hypothesis (A) holds and n ~ 8. Assume there is a subgroup H of G generated by special 3-elements such that X IH = Y EB (n-s)lH where Y is irreducible of degree s with 3 ~ s < n. Then G contains a normal subgroup N generated by special 3-elements such that N ~An+i and G/Z(G) ~ An+i or Sn+i· Proof: or n-1. By Lemma 2. 3 and induction, we may assume s = n-2 Let Y act on the subspace V1 of dimension s. Let g be a special 3-element such that X(g) does not leave V1 invariant. If s = n-1 XI (H, g) is irreducible and by Lemma 4.10, (H, g) ~ An+i · Ifs= n-2 andXl(H,g) is irreducible, byLemma4.9, (H,g) ~An+i· Ifs=n-2 and XI (H, g) = X 1 EB ~ where X1 is irreducible of degree n-1, by Lemma 4. 7, ~ = 1 (H, g). In this case, let X1 act on V2 and let g 1 be a special 3-element such that X(g 1 ) does not leave V2 invariant. So XI (H,g,g1 ) is irreducible and by Lemma 4.10, (H, g, g1 ) ~ An+i. Hence in any case, G contains a subgroup N generated by special 3-elements such that X IN is irreducible and N ~ An+i · Let N act on {a1 , ••• , ~+J and choose special 3-elements hu ... , hn-i such that hi = (ai, ai+u ai+ 2 ). not in N. Let h be a special 3-element By Lemma 4. 2, X I (h 1 , ••• , ~ - 2 ) = X1 EB 1 (h , ••• , ¾- ) where 1 2 X1 is irreducible. Let X1 act on the subspace W. If X(h) leaves W 56 invariant, by Lemma 4. 7, where X2 is primitive. XI (hi, ... , ~- ,h) = ~ EB 1 (h u · · · , hn-2, h) 2 But by hypothesis (A), h E ( hu ... , l\i_ 2 ) ~ An, a contradiction. So X I (h 1 , ••• , ~ - 2 , h) is irreducible and by Lemma 4.10, (hi, ... , hn_ 2 , h) ~ An+i· By Lemma 4. 5, there is a g 1 = hg for some g E (hi, ... , hn_ 2 ) such that g 1 commutes with hv ... , hn_ 5 • Also g 1 is not in N. As above XI (h 2 , ••• , Lemma 4.10, (h2 , ••• , ~-u g 1 ) hn-u g 1 ) is irreducible and so by ~ An+i and h2 , ••• , hn-v g 1 are 3-cycles. By Lemma 4. 6, ( hu ... , hn-v g 1 ) ~ An+2 which contradicts Lemma 4. 2. So h does not exist and N is the subgroup of G generated by all special 3-elements of G. Therefore N 4 G and as X IN is irreducible, CG(N) = Z(G). Thus G /Z( G) is a subgroup of the automorphism group of An+i. G/Z( G) ~ An+i or Sn+i. (See [24]). So 57 CHAPTER V THE PROOF OF THEOREM 2 In this chapter we complete the proof of Theorem 2. We first begin by considering what groups could be generated by two special 3- elements. Lemma 5 .1: Let h 1 and h 2 be special 3-elements such that h 1 and h2 do not commute. 1. Let n ~ 6. Then one of the following holds: XI (hi,~) = Y EB Y EB (n-4)1 (h h) where Yu Y are 1 2 irreducible of degree 2. 1' 2 2 Let Yi ( (h1 , h2 )) = Hi and let Mi be the set of all elements in Hi which occur with component the identity of H/j * i) in the subdirect product. Then Hi ~ SL 2 (3) for i = 1 and 2, IZ(Hi) I = 2, and either M.1 = Z(H.) for i = 1 and 2, or M1. = 1 for i = 1 and 2. Also l Yi (hj) are not unimodular for i = 1, 2 and j = 1, 2. 2. XI (hi, h 2) = Y1 EB ~ EB (n-4) 1 (h h ) where Y1 is irreducible 1' 2 of degree 3. 3. X I (h1 , h2 ) = Y EB (n-4) 1 (h , ~ ) where Y is irreducible of 1 degree 4. Proof: As h1 and h2 do not commute, XI (hi, h 2) = Y EB (n-4) 1 (h where Y has a constituent of degree at least 2. So we have three possibilities: a. Y = Y1 EB Y2 where Y1 is irreducible of degree 2. b. Y = Y1 ED ~ where Y1 is irreducible of degree 3. c. Y is irreducible. h) 1' 2 58 As b. and c. give 2. and 3., respectively, we only need to show that a. gives 1. So assume YI (hu h 2 ) 2. Let Yi ( (h1, h 2 ) ) = Hi. = Y1 EB Y 2 where Y1 is irreducible of degree So Y( (h1, ~)) is a subdir ect product of H1 and H2 • If Y1 is monomial in some basis, then Y1(h1) and Y1(h2 ) would both be diagonal in that basis as they have odd order. This contradicts the irreducibility of Y1 • So Y1 is primitive. We now examine the possibilities for Y1 by examining Blichfeldt's list [2]. This case is impossible as in case A of Chapter 3. This case is also impossible as S4 is not generated by its 3e lements. Assume first that Y2 is reducible. abelian and so Y( (h1, h 2 ) )' = So Y2 = ~1 EB ~2 • Thus H2 is H: EB 1 EB 1 and H:/(Z(H1) n H;) is elementary abelian of order 4. As Hi' consi(s:~ of ~ x) 2 unimodular matrices, the only element of order 2 in H{ is an element x of order 4. _1 . In particular H{ contains O As x is unimodular, x has eigenvalues i and -i. So G contains a special 4-element, a contradiction by Theorem 1. So Y2 is irreducible also. As previously, Y2 is primitive; also H2 /Z(H 2 ) ~ A 5 or S4 are eliminated as in cases A and B. So 59 H2 /Z(H 2 ) ~ A4 • Note that if any one of Yi(hj) is the identity, then Y1 or Y2 is reducible, a contradiction. So each Y/hj) is not unimodular . Let Mi be as defined in the statement of the lemma. By Theorem 5 .5 .1 of [11], Mi <3 Hi and H1 /M 1 ~ H2 /M 2 • Let Ci = Z(Hi). Assume first that for some i, MiC/Ci contains the Sylow 2-subgroup of Hi/Ci. Then Mi contains a unimodular subgroup of order 4. As earlier Mi could only have one involution, the central involution, and so has an element with eigenvalues i and -i. Thus G contains a special 4-element, a contradiction to Theorem 1. So for i = 1 and 2, M.1c- C.. As elements of H1 1 and H2 have only determinant 1, w, or w, Ci could(~rly bt)l, Z 2 , Z 3 , or Z 6 • As H.' contains an involution, which must be Q l -1 , Z 2 c C. - l for i = 1 and 2. Assume Ci = Z 6 for some i, say i = 1. As M1 contains unimodular matrices, M1 = Z 2 or M1 = 1. In either case, as H1 /M 1 ~ H2 /M 2 , we would have C2 = Z 6 and M2 ~ M1 • The isomorphism between H1 /M 1 and H2 /M 2 maps the centers onto one another; so X(G) contains an element with eigenvalues -w, -w, -w, -w, 1, 1, ... , a contradiction to Blichfeldt. Soc. ~ Z 2 and H.' => C. for i = 1 and 2. l l - l By Schur [25], the only nonsplitting central extension of Z 2 by A4 is SL 2 (3). So H1 ~ H2 ~ SL2 (3) and either Mi = Z(Hi) for i = 1 and 2 or M.l = 1 for i = 1 and 2. Lemma 5. 2: The only nonabelian linear unimodular group P of degree 3 and order 27 has exponent 3 and is given by P = (g, h) where and 0 w 0 in some basis. 60 Proof: The representation is monomial ( 52 .1 of [ 5]). Let vu v2 , v 3 be a basis for the space on which the representation is defined. Assume P has an element t of order 9. Assume first that t is not diagonal. By ordering Vu v2 , v 3 correctly, 0 a t = 0 0 ( C 0) b O 0 But t has determinant 1. So abc = 1 implying t has order 3, a con1 4 9 3 tradiction. So t is diagonal. Also P = (s, t ls- ts = t , t = 1, s = 1). (See Chapter 4 of [11] . ) As s could not be diagonal, by scaling and ordering Vu v 2 , v 3 correctly, s = G: D and t = ( a O 0 b 0 0 0) 0 C where abc = 1 . As t 3 E Z ( P), a 3 = b ~ = c 3 E { w, -w} . Let a = E where E is a primitive ninth root of unity. If E 3 = w either b = c = EW or {b, c} = {E, Ew}. If E3 = W either b = C = EW or {b, c} = {E, Ew}. -1 s But (~ ts=~ and none of the combinations of a, b, So P has exponent 3. scaling Vu v2 , v 3 correctly, and c work. Let s be nondiagonal. By ordering and 61 Let t be an element of order 3 not commuting with s. If t is diagonal, 1 by replacing t by C if necessary and permuting Vi, v2 , v 3 cyclically, we may assume If t is not diagonal, by replacing t by t -1 if necessary where abc = 1. But 2 As s and s 2 t do not commute and s t has order 3, {a, b, c} = {1, w, w} , and the result holds. Lemma 5. 3: Let h 1 and h2 be special 3-elements. Assume XI (hu h 2) = Y1 EB ~ EB (n-4)1 (hi,~) where Y1 is irreducible of degree 3 and ~ * 1 (h h). 1' Then there exist special 3-elements h( and l½' 2 contained in (hi, I½) such that X I (h:, h;} = Y EB (n-3) 1 (h:, h{) where Y 62 is irreducible and (h{, h;) is the nonabelian group of order 27 and exponent 3. Proof: Assume first that Y1 is monomial. Let Y1 act on V1 and ~ on (v4 ) . Let Vu v 2 , v 3 be a basis of V1 in which Y1 is monomial; since Y1 (h 1 ) and Y1 (h2 ) cannot both be diagonal, we may assume Y1 (h 1 ) 1 is not diagonal. By replacing h2 by h; if necessary and scaling and ordering v 1 , v 2 , v 3 correctly, 0 0 1 0 (Y1 EB ~) (h1 ) = -1 -1 Let h* = h1 h 2 h1 h2 • 1 0 0 0 0 1 0 0 and (Yl EB ~)(h2) = 1 0 0 0 0 1 0 0 w 0 0 0 0 0 0 Then 0 0 0 w 0 0 0 w 0 0 0 1 and h: = hv h; = h* gives the desired group by Lemma 5. 2. Assume now that Y1 is primitive. Let Y1 ( (hi,~)) = H1 and ~( (h1 , ~ ) ) = H2 • Then H = (Y EB ~)( (h1 , ~ ) ) is a subdirect product of H1 and H2 • By Mitchell, H1 /Z(H 1 ) is an extension of Z 3 x Z 3 by SL 2 (3). Now IH; /(Z(H n H;) I = 72. If S is the Sylow 3- subgroup of H:, S ., tt; and so S char tt; <3 H. Thus H' is a subdirect product of Hf and H; = 1; also 1) S <J Hand by Clifford's theorem (see [ 5)), as S <t Z(H 1 ), Sis nonabelian. As Z (H 1 ) n tt; consists of nonsingular uni modular scalar matrices, jz(H 1 ) n H; I ~· 3. So as Sis nonabelian, Z(H 1 ) n H( has 63 order 3 and S has or de r 27. By Lem ma 5. 2, S is of exponent 3 and the result follows . The next t hree lemmas deal with special 3- e lements whic h satisfy conclusion 1. of Le m ma 5.1. The possibility of conclusion 1. occurring is one reason why Theor e m 2 is mor e difficult than Theorem 1 . Induction techniques do not work quite as ea s ily in t his situation as one would hope. Lemma 5. 4: Let h 1 a nd ~ be s pecia l 3- e lements such that they satisfy 1. of Lemma 5.1. Let n ~ 6 and let Yi act on the subspace Vi . Let h 3 be a special 3-element such that X(h 3 ) does not leave V1 EB V2 invariant. 1. Then one of the fallowing holds: XI <hu h h 2, 3) = U EB (n-s) l (hi, h , h~) where U is irreducible 2 of degree s withs = 5 or 6. 2. There are special 3-elements hf, h; E (hi,~' h:) such that XI (h{, 11;) = U EB (n-3) 1 (h, h ') where U is irreducible of 1 , 2 degree 3 and (h;, hf) is t he nonabelian group of order 27 and exponent 3. 3. XI (hi, h2' h 3 ) = U1 EB U2 EB (n-6) 1 (h h h ) where both Ui are 1' 2, 3 irreducibl e and primitive of degree 3. Also U/hj) are nonunimodular for i = 1, 2 and j = 1, 2, 3. Let Gi = U1. ( (h1 , h2 , h 3 )) and N. be the set of all elements in G. which 1 1 occur with component the identity of Gj (i =t- j) in the subdirect product. by SL 2 (3). Then G.1 /Z (G.) is an extension of Z 3 x Z 3 l If N. ct Z(G.) for some i, 2. also holds. 1- 1 64 Proof: We have XI (hu h2 , h 3 ) = U EB (n-6)1 (h h h). As 1' 2, 3 X(h 3 ) does not leave V1 EB V2 invariant, X(h 3 ) does not leave both V 1 and V2 invariant. Hence U has an irreducible constituent of degree at least 3. So we have four possibilities for U: A. U is irreducible. B. U = U1 EB ~ where U1 is irreducible of degree 5. C. U = U1 EB U2 where U1 is irreducible of degree 4. D. U = U1 EB U2 where U1 is irreducible of degree 3. First, case A gives 1. Assume case B holds; then 1. holds if we show ~ = 1 (h h h ) . 1' ~ = 1 (h 1' 2, 3 By Mitchell, if U1 is primitive, h h). Assume U1 acts on a subspace Wand is monom ial in 2, ~ a bas is vu ... , v :',. As U1 I (h u h2 ) = Y1 EB Y2 EB 1 (h h ) , if ~ * 1 (h h h ) , 1' 1' 2 2, 3 U1 (h 3 ) must be diagonal in the basis Vu ... , v 5 • Hence as U1 is irreducible, U1 I (h 1 , h 2 ) does not leave any (vi) invariant. (hu ~) ~ By Lemma 2. 6, A 5 , a contradiction. So ~ = 1 (h h h ) and 1. holds. 1' 2, 3 Assume case C holds. Let Ui act on Wi. If V1 EB V2 .:: Wu V1 EB V2 = W1 and X(h 3 ) leaves V1 EB V2 invariant, a contradiction. Hence as Vu V2 are unique subspaces, we may assume V1 c W1 and V2 = W2 • So U1 (h 1 ) and U1 (h2 ) have eigenvalues 1, 1, 1, w or 1, 1, 1, w. If U1 is monomial in some basis, U1 (h 1 ) and U1 (~) would have to be diagonal , contradicting U1 I (h1 , h 2 ) = Y1 EB 2 · 1 (h h ) . Therefore U1 is primitive. 1' But this is impossible by Lemma 2. 2. Finally assume case D holds. Ui act on Wi. 2 So case C does not occur. Let Gi = Ui ( (hu h 2 , h:)) and let By ordering Y1 and Y2 correctly, we may assume Vi ~ Wi. Therefore U2 has a constituent of degree at least 2, and U/h J and Ui (h2 ) are nonunimodular for i = 1 and 2. We have two possibilities: 65 (i) U2 = U3 EB ~ where U 3 is irreducible of degree 2 and ~ -:f:. 1 (hu h2' h3) . (ii) U2 is irreducible or U2 = U 3 EB 1 (h h h) where U3 is 1' 2, 3 irreducible of degree 2. If (i) holds, then ~(h 1 ) = ~(~) = 1 and ~(h 3) -:t 1. As U1 is irreducible, h 3 does not commute with both h1 and h 2 • Without loss of generality assume h 3 does not commute with h 1 • As U1 (h 3) is not trivial, U2(hJ is trivial. I So U1 (hi, h 3) = R EB 1 (h h ) where R is irreducible of degree 2 and 1' 3 XI (hi, h~) = R EB ~ EB t EB (n~4)1 (h h ) . 1 2 1' 3 But by Lemma 5. 1, this is impossible. So (ii) holds. If either Ui is monomial, in some basis U/h 1 ) and U/h2) are both diagonal as they have eigenvalues 1, 1, w or 1, 1, w. contradicts the irreducibility of Yi. So U1 is primitive; if U2 is irreducible, it is primitive and if U2 = U 3 EB 1 (h itive. This h 1 , .142 , h ) , U3 is prim3 Let Gi = U/ (hi,~, h 3) ) . So U( (h1 , l½, h 3 )) is a subdirect product of G1 and G2. By Mitchell, G1 /Z(G1 ) is Z 3 X Z 3 extended by SL 2 (3). Let Ni be the set of all elements in Gi which occur with the identity of Gj (j =f:: i) in the subdirect product. By Theorem 5. 5.1 of [11], Ni <1 Gi and G1 /N1 ~ G2 /N 2. All nontrivial normal subgroups of G1 /Z( G1 ) contain S, a normal subgroup of order 9. Let S be the inverse image of Sin G1 • Assume first that N1 ~ Z(G1 ). Then Sc N1 = N1 Z(G1 )/Z(G1 ). Also S .n N1 <J G1 and Is I = 9 · !Z(G1 ) 1- As N1 contains only unimodular Is n N I = 9 or 27. By Clifford's theorem [5] as G is primitive and Sn Nl <t_ Z(Gl), Is n Nl I= 27. So by Lemma 5.2, we matrices, have 2. 1 1 Now assume N1 c Z( G1 ). As G1 /N1 ~ G2 /N 2 , by looking at Blichfeldt's list of primitive groups of degree 2 and 3, U2 must be 66 irreducible. As it is primitive G2 /Z(G2 ) ~ G1 /Z(G1 ). If N2 ~ Z(G2 ), we have 2. as above. We must have U/h 3 ) nonunimodular as both U1 and U2 are irreducible, and so 3. holds. Lemma 5. 5: One of the following occurs for n ~ 6. 1. There exists a subgroup H of G generated by special 3elements such that X IH = Y EB (n-r)lH where Y is irreducible of degree r for some r with 3 ~ r ~ 6. 2. Let h 1 and h 2 be any two special 3-elements. Then either a) h1 and h2 commute b) (hu h2 ) ~ SL/3) and XI ( hu h2 ) satisfies 1. of Lemma 5 .1. For any special 3-element h 3 satisfying the hypothesis of Lemma 5. 4, conclusion 3. holds in Lemma 5. 4 but 2. doesn't. Proof: By Lemma 1. 1, not all special 3-elements commute. So let h1 and h2 be special 3-elements which don't commute. If conclusion 3. of Lemma 5.1 holds, we have 1. If conclusion 2. of Lemma 5.1 holds, we have 1. ·by Lemma 5. 3. So assume conclusion 1. holds of Lemma 5.1. Let Yi act on Vi. There is a special 3-element h 3 such that X(h 3 ) does not leave V1 EB V2 invariant. Hence, if for this h 3 , conclusion 1. or 2. of Lemma 5. 4 hold, we have 1. So assume conclusion 3. holds but 2. doesn't. Then N. c Z(G.). As elements of U1 have determinant 1, w, 1 - 1 w, IZ(Gi) I 19. Also I(huh ,h I = IG I· IN I = 216 · IZ(Gi) I IN I= 2 3 3a for some a ~ 1. So in particular 48 ~ I (hi, h h I and hence or 2 3 3) 1 2 2 3 • • by cone lusion 1. of Lemma 5. 1, (h1 , h2 ) 2, ~ SL 2 ( 3). 3) 67 Lemma 5. 6: Assume n ~ 8 and that conclusion 2. of Lemma 5. 5 holds. Let hu h 2 , h 3 be special 3-elements of G that satisfy conclusion 3. of Lemma 5. 4 where Ui acts on Wi. Let h 4 be a special 3-element of G such that X(h4 ) does not leave W1 EB W2 invariant. Then one of the following holds: 1. XI (hi,~' h h 3, 4) = X EB (n-s)l (h1, h-2 , h1, h) where Xis 4 irreducible and primitive of degree s = 7 or 8. 2. n = 8 and XI (hi, h2 , h 3 , h4 ) = R 1 EB R 2 where R 1 and R 2 are irreducible and primitive of degree 4. If Fi = R/ (hi, h2 , h 3 , h4 ) ), ___,. ,,,--..-, then Fi ~ 0 5 (3) x Z 3 where 0~(3) is the nonsplitting central extension of Z 2 by 0 5 (3). If Li is the set of all elements in Fi which occur with component the identity of F j (i * j) in the subdirect product, then Li := O2 (Z(F i)) for i = 1 and 2. Proof: As X(h4 ) does not leave both W1 and W2 invariant, we have the following possibilities: (i) XI ( hv h2 , h 3 , h4 ) = X EB (n-8)1 (h1' h2, h 3, h4 ) where Xis irreducible. (ii) XI (hi, h2 , h 3 , h4 ) = X EB Y EB (n-8)1 (h h h h) where X 2, 1' 3' 4 is irreducible of degree r with 4 ~ r ~ 7 and Y has degree 8-r. First assume either (i) holds or (ii) holds with r = 7. - act on V. - Let X - Then by Lemma 4, Suppose that ,.., X is monomial on V. 1. there exist special 3-elements generating A~, a contradiction to the assumption that 2. holds in Lemma 5. 5. So X is primitive and by Mitchell, if (ii) holds with r = 7, Y = 1 ( h h h h ) . So in these 1' cases, 1. holds. 2, 3' 4 68 We now assume (ii) holds for 4 ~ r ~ 6. If r = 6, X must be acting on W1 EB W2 , a contradiction that X(h4 ) does not leave W1 EB W2 invariant. If r = 5, the only possibility by correctly ordering U1 and U2 is that X I ( hu ~, h3 ) = U1 EB 2 · 1 (h h h ) and Y I (hi, ~, h) = U2 • 1' 2, 3 As X(hi) for 1 ~ i ~ 3 have eigenvalues 1, 1, 1, 1, w or 1, 1, 1, 1, Mitchell, Xis not primitive; so Xis monomial. w, by But then in some basis of the space on which X acts, X(hi) are all diagonal for 1 ~ i ~ 3, contradicting the fact that U1 is irreducible. So r = 4. Let X = R 1 and Y = R 2 • By correctly ordering U1 and U2 , the only possibility is I Ri (hu h2 , h 3 ) = Ui EB 1 (h h h ) . If either Ri is monomial, in some 1' 2' 3 basis Ri (h 1 ), Ri (h 2 ), and Ri (h 1 ) are all diagonal because they have eigenvalues 1, 1, 1, w or 1, 1, 1, w. of Ui. This contradicts the irreducibility So R 1 is primitive and the result follows by Lemma 2. 2. The next lemma eliminates a possibility which occurs in [15]. Using the powerful results of [ 1] , Lemma 5. 8 shows that condition 1. of Lemma 5. 5 holds. This allows us to construct primitive subgroups of codimension 1 or 2. Lemma 5. 7: If X1 is an irreducible representation of a group H containing a special 3-element and X1 has degree 8, then X1 is not the tensor product of two representations of smaller degree. Proof: Assume X1 is the tensor product of Y1 and Y2 each of degree less than 8. degree 2. Then we may assume Y1 has degree 4 and Y2 has There exist elements Yu y 2 such that Y1 (y 1 ) 0 Y2 (y 2 ) has eigenvalues w,w,1,1,1,1,1,1. Let Y1 (y 1 ) have eigenvalues n;_, ~' a:P a 4 and Y2 (y 2 ) have eigenvalues /3 11 {3 2 • So Y1 (y 1 )@ Y2 (y2 ) has 69 eigenvalues l\/3j for 1 ~ i ~ 4 and 1 ~ j ~ 2. We may assume a 1 {3 1 = w . So /3 1 * /3 2 and ~ * a 2 • Thus { n;_/3 2 , a 2 {3J = {w, 1}. So {a 3p2 , a 4 {3J = {1, 1} and hence a 3 = a 4 • So ~/3 1 = OJ. 4 /3 1 = 1 and hence p1 = /3 2 , a contradiction. Lemma 5.8: Let n ;?:- 8. Then 1. of Lemma 5.5 holds. Proof: Assume 1. of Lemma 5. 5 fails. By Lemma 1. 1 not all special 3-elements commute. Choose special 3-elements h 1 and h 2 which do not commute . So (hi,~) ~ SL 2 (3) and by replacing h2 by Ii;" 1 if necessary, we may assume h1 and ~ are conjugate in (hi, h2 ) . Also XI (h h 1, 2) satisfies 1. of Lemma 5. 1. There is a special 3- element h 3 satisfying the hypothesis to Lemma 5. 4. So (hi, h 3 ) = SL 2 (3) 1 for either i = 1 or i = 2 and in particular, replacing h 3 by h; if necessary, we may assume h 3 is conjugate to h1 in (h 1 , h 2 , h:) . Also 3. holds in Lemma 5. 4 but 2. doesn't. We may choose h 4 as in the hypothesis of Lemma 5. 6, and as above we may assume h4 is conjugate to h1 in (hi, h2 , h3 , h4 ) . Assume 2. holds in Lemma 5. 6. Then n = 8 and by Lemma 1. 1 there is a special 3-element h5 such that XI (hu h h h hi) is irreducible. By Lemma 1. 4 and our hypothesis, 2, 3, 4, XI (hi, ... , hs) is primitive. As before we may assume h 5 is conjugate to h 1 in ( hv ... , h!j). Therefore in any case we may assume that there is a subgroup Hof G generated by an H-conjugate class n of special 3-elements such that X jH = X 1 EB (n-r)lH where r = 7 or 8 and X1 is irreducible and primitive. Any two elements in n either commute or generate SL2 (3). By Aschbacher-Hall [1], H/O 00 (H) ~ SPk:(3), Uk(3), or PGUk(2). By examination of Wales [27, 28], we see that r = 7 is impossible . So 70 assume r = 8. First consider the case when O 00(H) > Z (H). By Lemma 5. 7 and inspection of the theorem and proof in Lindsey [15], there is a 2-group Q 4 H with X1 IQ irreducible. Consider the group T = (hu h2 ) Q which has order 2a3. Let XI (hu h2 ) = Y1 EB Y2 EB (n-4)1 (h h) and let Y. act on v.. Assume there is a conu 2 l l jugate h E T of h1 by an element in T sue h t hat h ti. (h1, h2 ) • Then as 3 2 JIT I, by Lemma 5. 5, X(h) must leave V1 EB V invariant. If h 2 2 commutes with ( hu h2 ) , 3 j IT I, a contradiction. Thus X J ( hu h 2 , h) = Y EB ~ EB (n-5) 1 (h h h) where Y acts on V1 EB V2 and ~ is linear. 1' for one of h 1 or h2 , But 2, say hu x!(hi,h) =U 1 EB U2 EB (n-4)1 (huh) where Uu U2 are irreducible of degree 2, by Lemma 5.1. So ~ = 1 (hu ~' h) and by assumption that 1. of Lemma 5. 5 fails, X(h) must leave both V1 and V2 invariant. Hence xi (huh2 ,h) = R 1 EB R 2 EB (n-4)1 (h h h) where 1' 2, R. acts on ~. In a manner analogous to case C of Lemma 5. 1, 1 l I ( h1, ~, h) I = 24 or 48. But (h1, h2 ) has four Sylow 3- subgroups and as I (hu h2 , h) I = 24 or 48, so does (hu h2 , h) , contradicting h (/_ (hu ~) . Thus all conjugates of h 1 by elements in T lie in (h1, ~) . In particular Q normalizes (h1, h2 ) • For either i = 1 or i = 2, (h., h 3 ) l paragraph, Q normalizes (hi' h 3 ) . ~ SL 2(3). As in the preceding So Q normalizes H1 = (hu h 2 , h). By examining the groups listed in section 3 of Aschbacher-Hall [1], IH 1 I = 648 and 0 2 (H 1) = 1. As Q and H1 normalize each other, [ Q 'H1] ~ Q n H1 <J QHl. As 02(H1) = 1, Q n Hl = 1 and hence Q and Hl centralize each other. This is clearly impossible as X1 /Q is irreducible. Thus O 00(H) = Z(H). By 2D of Brauer [ 3] , we may assume the highest prime dividing IHI is 7. Then H/Z(H) ~ Ofi(3), U3 (3), or U4 (3). 71 I (hu ~' h~) I = 648, a contradiction. Let K ~ 0 (3) or U (3) and let 5 L be a group with Z(L) 4 a 2-group and L/Z(L) ~ K. An element of order 5 in K is self-centralizing, and so CL((7T 5 )) = (1T 5 ) x Z(L) where 1T 5 is a 5-element of L. All 5-elements of Kare conjugate and so all 2 5-elements of Lare conjugate. As 5 f IL I, by Brauer [4, I, Theorem 10], every 5-block of defect 1 has characters of degree z = ± 1 (mod 5). So L does not have an irreducible character of degree 8, and hence H does not exist. This is a final contradiction and so 1. of Lemma 5. 5 holds. The next lemma allows us to use the results of Stellmacher [26] when n = 8. Lemma 5.10 shows that when hypothesis (A) of Chapter IV fails in the case n = 8, we get a large primitive subgroup of degree 7. The final two lemmas complete the proof of Theorem 2. Lemma 5.9: Let n = 8. Let h1 and h 2 be special 3-elements. Then either h 1 and h2 commute or (hu I½> is isomorphic to A4 , SL2 (3), Proof: Assume there is a subgroup H of G generated by special 3-elements such that X IH = X 1 EB (8-s)lH where X1 is irreducible and primitive of degree s = 6 or 7. a subgroup exists. By Lemma 5. 8 and Corollary 2. 2, such By examination of the primitive groups in Lindsey [14, 16, 17] and Wales [27, 28] and by applying Blichfeldt, we get that if s = 6, H ~ A7 or 0 5 (3) and if s = 7, H ~ A8 or PSp 6 (2). (If s = 6, the group H with H/Z(H) ~ U4 (3) has a special 3-element (see I 16]); but IZ(H) I = 6 in this case, which gives a contradiction by Blichfeldt.) In 72 particular if p j jG I where p is a prime we must have p ~ 7 (see Brauer [ 3] ) . Also in these cases X1 has a rational character. Let h1 and h2 be noncommuting special 3-elements. We study XI (hi, h given in Lemma 5. 1. Consider the case XI (hi, h = Y EB ~ EB 4 · 1 ¢1u ~) where Y is the possibilities for 2) 2) 1 1 irreducible of degree 3. If ~ * 1 (hi,~), by Lemma 5. 3, there exist special 3-elements h{, h; such that XI (h{, h;) = Y EB 5 · 1 <h{, h;) where ( h{, ~) is the nonabelian group of order 27 and exponent 3. Then by Corollary 2 .1, there is a subgroup H of G generated by special 3elements such that (h{, h;) ~ H and X jH = X1 EB (8-s)lH where X1 is irreducible and primitive of degree s = 6 or 7. By the first paragraph of this proof, X1 has a rational character, contradicting the existence of an element z E Z( (h{, h;)) such that X(z) has eigenvalues w,w,w,1,1,1,1,1. So~ =l(h h)' 1' ThusbyCorollary2.1, there is 2 a subgroup H of G generated by special 3-elements such that (hi, h2 ) :::. H and X jH = X1 EB (8-s)lH where X1 is irreducible and primitive of degree s = 6 or 7. As above Z( (hi, h2 )) = 1. If Y1 is primitive, the only possibility is (h 1 , h2 ) ~ (hi, h 2 ) in X1 jH. ~ SL 2 (7) or A 5 by looking at Blichfeldt's list. If SL/7), the eigenvalue structure of a 7-element is incorrect So if Y1 is primitive, (hu h 2 ) ~ Af',. Now assume Y1 acts on V 1 and is monomial in some basis Vu v 2 , Vu v 2 , v 3 correctly and numbering hu h2 correctly, we may assume 73 As (hi, h 2 ) cannot be the nonabelian group of order 27 and exponent 3, 1 as above, by replacing h2 by h; if necessary, we may assume If H is A 7 or A 8 , the special 3-elements correspond to 3- cycles and so two of them commute, generate A4 , or generate A 5 • H ~ 0 5 ( 3) or PSp 6 (2). So we may assume 2 But X1 (h 1 ~ ) has eigenvalues a, b, c, 1, 1, . . . . By examining X1 IH, the only possibilities are {a, b, c} = { 1, -1, -1} or {1, w, w}. The first case gives (hu h2 ) ~ A4 and the second case gives (hi, h2 ) the nonabelian group of order 27 and exponent 3, a contradiction. So if case 2. of Lemma 5 .1 holds, ( hu hJ ~ A4 or A 5 • Now assume X I (h 1 , h2 ) = Y EB 4 · 1 ( h , ~ ) where Y is irreducible 1 of degree 4. Let Y act on V 1 • Assume fir st that Y is monomial on V 1 • Let Vu v 2 , v 3 , v 4 be a basis of V 1 in which Y is monomial. By ordering 1 and scaling Vi, v 2 , v 3 , v4 correctly and replacing h 2 by h; if necessary, we may assume ·o 1 1 0 0 0 1 0 0 0 0 0 Y(h1 ) = where a * 1. 0 0 0 1 and Y(~) = 1 0 0 0 0 0 0 a -1 0 a 0 0 0 1 0 0 By Corollary 2. 1, there is a subgroup H of G generated by special 3-elements such that (bu h 2 ) ~Hand X IH = X 1 EB (8-s)lH where X1 is irreducible and primitive of degree s = 6 or 7. first paragraph H ~ A 7 , A 8 , 0 5 (3), or PSp 6 (2). By the If H ~ A 7 or A 8 , h1 and h2 represent 3-cycles and generate A4 or A':, contradicting Y being 74 irreducible and monomial. So H ~ 0 5 (3) or PSp 6 (2). 2 Now X1 ( (h1 h2 ) ) 1 has eigenvalues a, a- , a, a - 1 , 1, 1, . . . . By looking in the character table of 0 5 (3) or PSp 6 (2), the possibilities for a are -1, w, w, i, or -i. If a = -1, Y has an invariant subspace (v 1 + wv2 + wv 3 ) EB (-v2 + wv 3 + -wv4 ) , a contradiction. Let g = h 1 ( h 1 ~ ) 2 h 1-1( h2-1 h 1-1 ) 2 and 2 2 h = h 1 gh; 1 g- 1 • Then X1 (h) has eigenvalues a 4, a- , a - , 1, 1, 1, ... which is impossible if a = w or -w. has eigenvalues a -1 ,a - 1 3 Let h = h 1-1( h2-1 h 1-1 ) 2( h2 h1 h 2 ) 2 • ,a ,a -1 Then X1 ( h) , 1, 1, ... which is impossible if a = ± i. Thus Y must be primitive. There is a special 3-element h 3 such that X(h 3 ) does not leave V 1 invariant. So XI (hu h 2 , h 3 ) = R EB 2 · 1 (h h h ) where either R is ir1' 2' 3 reducible or R = R 1 EB ~ where R 1 is irreducible of degree 5. Assume first that R acts on V 2 and is monomial. Then there is a basis Vu ... , v 6 of V 2 in which R is monomial. As Y is not monomial, R I (hu h2 ) can fix at most one (vi) trivially. by Lemma 2. 6 is that (h1 , h2 ) ~ But the only possibility A 5 • So we may assume that if R is irreducible it is primitive, and if R = R 1 EB ~' R 1 is primitive and ~ = 1 (h h h) by Mitchell. 1' 2, G is abelian. By Lindsey [18], the Sylow 5-subgroup of 3 Using this fact and the list of primitive groups of degree 5 in Brauer [ 3] , if R is reducible (h1 , ~ , h 3 ) ~ A 6 or 0 5 ( 3) . Now if (hu h 2 , h 3 ) ~ 0 5 (3), R 1 doesn't have a rational character. Let R 1 act on V 3 and let h 4 be a special 3-element such that X(h4 ) does not leave V 3 invariant. So XI (hu h h h 2, 3, 1) = S EB 1 (h 1, h h h ) where either S 2, 3, 4 is irreducible or S = S1 EB ~ with S1 irreducible of degree 6. By Lemma 1. 4, as 0 5 (3) is simple and not contained in A7 , S cannot be monomial. So S is primitive if it is irreducible, and if S = S1 EB ~, · ~ = 1 (h h h h ) 1' 29 3' 4 75 and S1 is primitive. By the first paragraph, S has a rational character, a contradiction. So if R is reducible, ( h1 , h2 , h 3 ) possibility is that (hu h2 ) ~ ~ A 6 and the only A 5 • So we assume R is irreducible. the first paragraph, (h1 , ~ ' h) ~ A7 By or 0 5 (3). Noting that in either group a 5-element is self-centralizing, by examining Blichf eldt' s list [2] of primitive groups of degree 4, and by applying Blichfeldt, the only possibility is that (hu ~) ~ (hu h2 ) ~ A 5 • So if 3. holds of Lemma 5 .1, A5 • Now assume 1. holds in Lemma 5 .1 . In the notation of conclusion 1. of Lemma 5. 1, it suffices to prove M.1 = 1 for i = 1 and 2. Let h 3 be a special 3-element such that X(h 3 ) does not leave V1 EB V 2 invariant by Lemma 1. 1. Then the possibilities for X I ( h1 , ~ , h:1) are listed in Lemma 5. 4. Assume Mi * 1 for both i = 1 and i = 2, and hence I(h I = 48. Assume fir st that X I(h h h) = U EB 1 , ~) 1, 2, (8-s)l (h h h) where U is irreducible of degree s with s = 5 or 6. 1' 2, 3 Suppose U is primitive. By what was done in the previous paragraph, if s = 5, ( hu ~' h 3 ) ~ A 6 and if s = 6, (hu h2 , h 3 ) ~ A 7 or 0 5 (3). But as M. =1= 1 for i = 1 and i = 2, there is an element g E (h1 , h 2 ) with 1 Y1 (g) = Y 1 (h1 ), Y2 (g) = '( - 1 0) Y2 (h1 ). Then U(g) does not have a 0 -1 rational trace, a contradiction as U must have a rational character. So U is monomial. Thus in some basis vu ... , vs of the subspace on which U acts, by ordering and scaling v 1 , ••• , vs correctly and replacing -1 h 2 by ~ if necessary, we may assume 76 V 1 X(h1 ) = V 2 v2 X(~) = av 3 V 2 X(h 1 ) = V 3 V 3 X(~) = V4 and 1 V 3 X(h1 ) = V1 v 4 X(~) = a- v 2 V £X(h1) = v.f for l > 3 v.fX(~) = v.f for £ i {2, 3, 4} As U I (hu h 2 ) = Y1 EB Y2 EB (s-4)1 (h h ) and as U( (h1h 2 ) ) = 2 1' 1 2 1 So a = ± 1. If diag{a, a- , a, a- , 1, ... } , the only possibility is a = a - i . a= 1, (h 1,h2 ) ~ A4 , ~ a contradiction. If a= -1, (h1,h2 ) SL 2 (3), a contradiction. Thus if Mi * 1 for i = 1 and i = 2, 1. of Lemma 5. 4 does not hold. hold. By what we have already done, 2. of Lemma 5. 4 could not So assume 3. of Lemma 5. 4 holds. Ni£ Z(Gi) for i = 1 and 2 and In the notation of that lemma, I (hi,~' h3 ) I = 2 3 • 3a for some integer a. But then I (hi,~) I =t- 48, a contradiction. So in any case, we must conclude Mi = 1 for both i = 1 and i = 2, and hence (hi, h2 ) ~ SL 2 (3). Lemma 5.10: Let n = 8. Assume the special 3-elements of G do not generate A 9 • Then there is a subgroup H of G generated by special 3-elements such that X jH = R EB lH where R is irreducible and primitive of degree 7 with H ~ PSp 6 ~). Proof: By Lemma 5. 8 and Lemma 4.11, hypothesis (A) of Chapter 4 does not hold. Thus there is a subgroup K of G generated by special 3-elements such that X IK = X1 EB (8- s) lK where X1 is irreducible and primitive of degree s = 6 or 7 and K ~ As+i. By the first para- graph of the proof of Lemma 5. 9, if s = 6, K ~ 0 5 (3) and if s = 7, K ~ PSpi2). If s = 7, we are done and so assume s = 6. There is a subgroup K1 c K with K1 ~ A 6 • This subgroup is the derived group of the stabilizer of a point in the permutation representation of 0 5 (3) on 36 letters [12] . So X IK1 = X1 IK 1 EB 2 · lK = ~ EB 1 3 · lK 1 77 where X2 is irreducible and primitive, by examining the character table of A 6 • Also K1 is generated by special 3-elements. Let Xi act on Vi for So V2 cV 1 • Let 2 ·lK act on v; and 3 ·lK act on v;. Hence i = 1 and 2. 1 v; c v;. Assume the lemma is false. Let g be a special 3-element such that X(g) does not leave V1 invariant. mute with K. In particular g does not com- Let Ki, ... , Kk be all the K-conjugates of K1 • As K is simple, K = <Ku ... , Kk). Thus g does not commute with some Ki and by renumbering, we may assume g does not commute with K1 • Assume first that X(g) leaves V2 invariant. So XI (Ki, g) = X3 EB ~1 EB ~2 EB 1 (Ki, g) where X3 acts on V2 • As g does not commute with Ku ~1 (g) * 1, >. ~2 (g) * 1 is impossible. ~1 (g) = 1, X(g) leaves V1 invariant, a contradiction. So ~1 (g) * 1. So one~-, say for i = 2, is 1 (K 1 1'g this is a contradiction by Mitchell as X 3 jK1 = X2 is primitive. If But So X(g) does not leave V2 invariant. We have three possibilities for XI (Ku g): A. XI <Ku g) = R EB 1 (Ki, g) where R is irreducible and primitive. B. XI <Ku g) = R EB 1 (Ku g) where R is irreducible and monomial. C. XI (Ki, g) = R EB ~ EB l(Ki, g) where R is irreducible of degree 6. We show that in any case there is a special 3-element h such that XI (K h) = Y EB 2 · 1 (K h) where X(h) does not leave V invariant. 1, , 1 1 Assume case A holds. As we are assuming the lemma is false, (Ki, g) ~ A 8 • As all special 3-elements of (Ki, g) represent 3-cycles, by Lemma 4. 5, K1 is the stabilizer of two points. Choose a special 3-element h such that <Kuh) is the stabilizer of one point, and hence so that <Kuh) ~ A7 • So XI (Ki, h) = Y EB 2 · 1 (Ki, h) where Y is irreducible. If X(h) leaves V1 invariant, then Y acts on V1 and 78 XI (K, h) = S EB 2 • l (K, h) where S is primitive on V1. So (K, h) ~ 0 5 (3) or A 7 • But (K, h) contains subgroups isomorphic to both 0 5 (3) and A 7 , which is impossible. Assume case B holds. So X(h) does not leave V1 invariant. Let R act on Wand 1 ( Ku g) on (v 8 ) . Let Vi, ... , v 7 be a basis of Won which R is monomial. Choose special 3-elements gu g 2, g 3 , g 4 in K 1 with gi corresponding to the 3-cycle (i, i+l, i+2). By ordering and scaling v 1, ... , v 7 correctly we may assume vkX(gk) = Vk+1 vk+1X(gk) = vk+2 and viX(g) = av. J vjX(g) = V7 -1 V7X(g) =a v. vk+2X(gk) = vk l v f_ X(gk) = vf_ for I._ Et {k, k+l, k+2} V X(g) = Vf for £ El{i,j,7} 1 In the expression for X(g), a =1= 1 and i =1= j with i, j < 7. by an element of Ki, we may assume i = 5 and j = 6. By conjugating -1 Leth = gg 3 gg 3 g -1 . Then v4 X(h) v 5 X(h) v6 X(h) Now XI <Kuh) = Y EB 2 · 1 (Ki, h), where Y is irreducible on (vu ... , v 6 ) . By examining (g 4 , h), by Lemma 5. 9, a = -1 is the only possibility. Assume X(h) leaves V1 invariant. Then V1 = (vi, ... , v 6 ) and XI (K, h) = S EB 2 • 1 (K, h). So as S jK = Xu h EK and (K, h) ~ 0 (3). 5 Let f = gfh. Then viS(f) = -vi for i = 4 and 6 and vjS(f) = vj for j = 1, 2, 3, and 5. Conjugating f by elements of K1, we get elements 79 hv I½, h 3 with V 2 i_ 1 X(h 1.) = -V2 i-v V 2 iX(h.) = -V 2 i and v.X(h.) = v. for 1 J 1 J j * 2i-1, 2i. But then S(h1 l½h 3 ) is a nontrivial scalar matrix, con- tradicting Z(0~(3)) = 1. Thus X(h) does not leave V1 invariant. Assume case C holds. Suppose ~ =t- 1 (K , g) ; so ~(g) 1 * 1. But there is some special 3-element g 1 E K 1 with g 1 not commuting with g. The only possibility is that XI (gu g) = Y1 EB ~1 EB ~2 EB 4 · 1 ( g1' g Y1 is irreducible of degree 2 or XI (gv g) = Y2 EB /J. EB 4 · 1 ( g1' g ) where ) where Y2 is irreducible of degree 3 and µ{g) =t- 1. By Lemma 5 .1, the first case is impossible. possible. By Lemmas 5. 3 and 5. 9, the second case is im- Hence ~ = 1 (Ku g) and letting g = h, we have the desired result. So there is a special 3-element h such that XI (Ki, h) = Y EB 2 -1 <Kuh) where X(h) does not leave V1 invariant. Thus either XI (K, h) is irreducible or XI (K, h) = R EB ~ where R is irreducible of degree 7. Let Y act on W1 and 2 · 1 (K , h) on W; . Note that W{ c V;. 1 As V{ and W{ are subspaces of dimension 2 contained in the subspace V; of dimension 3, V; n W{ has dimension at least 1 and XI (K, h) acts trivially on V{ n W{. So V{ n w; has dimension 1 and XI (K, h) = R EB~ where ~ = 1 (K, h). As 0 5 (3) is simple and not contained in A 7 , R is primitive by Lemma 1. 4. As 0 5 (3) is not contained in A 8 , the only possibility is that (K, h) ~ PSp 6 (2). This contradicts the assumption that the lemma is false and so the result follows. Lemma 5 .11: Let n = 8 and let N be the subgroup of G generated by all special 3-elements. 1. Then one of the following holds: N ~ A 9 and G/Z(G) ~ A 9 or S 9 • 80 N/Z(N) ~ 0:(2) where Z(N) has order 2 and N = N'. 2. Also G/Z( G) is a subgroup of the automorphism group of 0 8+ (2). Proof: If N ~ A 9 , then X jN is irreducible and so CG(N) = Z(G). As N 4 G, G/Z(G) = G/CG(N) is contained in the automorphism group of A9 • Thus G/Z(G) ~ A 9 or S 9 and 1. holds. Assume N is not isomorphic to A 9 • Then by Lemma 5. 10, there is a subgroup H generated by special 3-elements such that H ~ PSp 6 (2) and X !H = R EB lH where R is irreducible and primitive of degree 7. All special 3-elements of H are conjugate in H and the special 3- elements generate H. Also by Ase hbacher- Hall [ 1] and Lem ma 5. 9, there are two special 3-elements in H which generate A 5 • Let R act on V1 and let g be a special 3-element in G such that X(g) does not leave V 1 invariant. So g does not commute with H; hence g does not commute with some special 3-element h 1 E H. h 1 or h1- 1 in (g, h 1 ) • By Lemma 5. 9, g is conjugate to Thus g is conjugate to the special 3-elements of H by an element in (H, g). Let h be any special 3-element of G. Then h I does not commute with <H, g) as X (H, g) is irreducible and so h does not commute with some special 3-element h 2 E (H, g). By Lemma 5. 9, h is conjugate to ~ or h; in (g, h 2 ) and we can cone lude that all special 1 3-elements of Gare conjugate in the group N that they generate. By Lemma 5. 9 and Stellmacher [26], N/O 00 (N) is isomorphic to PSp2 k(2) fork ~ 3, O2±k (2) for k ~ 3, Ak fork~ 5, HJ, G2 (4), Sz, or C 0 1 • As H is simple, H n O 00 (N) = 1 and N /O 00 (N) contains a subgroup isomorphic to PSpi2). If p is a prime and p I IG I, then p ~ 7 ( see [ 3]). Therefore using these two facts, we can conclude that N/O 00 (N) ~ PSpi2) or o: (2). 81 Con sider first the case that 0 00 (N) > Z (N). By Lemma 5. 7 and inspection of the theorem and proof in Lindsey [ 15] , there is a 2-group Q ~ 0 00 (N) such that Q 4 N, X /Q is irreducible, and N /Q is a subgroup of PSpi2). From the fact that N/0 00 (N) ~ PSp 6 (2) or 0:(2), we could only have 0 00(N) = Q and N /0 00 (N) ~ PSp 6 (2) . Therefore as 0 00 (N) n H = 1, N = HQ. By Frame [7], H has a subgroup H1 with H1 ~ 0 5 (3). As X jH has a rational character, the only possibility is that X jH 1 = R 1 EB 2 · lH where R 1 is irreducible and primitive of degree 6. H1 is generated by special 3-elements. As Q 4 N, consider the group N1 = H1 Q. jN1 I = IH 1 I· 2a for some integer a. Let R 1 act on V2 . Then Let h 1 be any special 3-element of H1 and h any conjugate of h1 by an element of N1 • Assume X(h) does not leave V2 invariant. Then XI <Huh) is irreducible or X I ( H11 h) = S EB ~ where S is irreducible of degree 7. As 7 ~ jN1 I, the latter is impossible. So X I (H 1 , h) is irreducible; as 0 5 ( 3) is simple and not a subgroup of A8 , by Lemma 1. 4, X I (H 1 , h) must be primitive. But this contradicts Lemma 5.10. So X(h) leaves V2 invariant. Hence XI (H 11 h) = S EB ~1 EB ~2 where S is irreducible of degree 6. As S jH 1 is primitive, so is S. If ~1 (h) =1- 1 and ~2 (h) -t- 1, h commutes with H1 and h El H1 • But then the Sylow 3-subgroup of N1 is larger than the Sylow 3-subgroup of H11 a contradiction. So at least one ~i is 1 (Huh) and by Mitchell, both must be. By the first paragraph of the proof of Lemma 5. 9, the only possibility is h E H1 • Therefore H1 <l N1 and so [H 1 , Q] ~ H1 n Q = 1. So H1 centralizes Q, a contradiction as X jQ is irreducible. Thus 0 00 (N) = Z(N). Hence if N/0 00 (N) ~ PSp 6 (2), N = HZ(N), a contradiction as X jN is irreducible. So N/Z(N) ~ 0 8+(2). By 1 82 examining Frame [8], it is easy to see that 0:(2) has no irreducible representation of degree 8. So Z(N) is cyclic of order 2, 4, or 8. Also N'Z(N) = N as N/Z(N) is simple. Hence N/N' is a 2-group. As N is generated by elements of order 3, N = N'. The Schur multiplier of 0 8+(2) by Steinberg (see [6]) is Z 2 x Z2 • As Z(N) is cyclic, we must have ]Z(N) I = 2. Frame [ 8] exhibits a group, the Weyl group of E 8 , whose derived subgroup has the properties of N described here. As N <1 G and X /N is irreducible, CG(N) = Z(G). So G/Z(G) is a subgroup of the automorphism group of N. But the automorphism group of N is clearly isomorphic to a subgroup of the automorphism group of o/(2). Lemma 5. 12: Let n ?:. 9 and let N be the subgroup of G generated by all special 3-elements. Then N ~ An+i and G/Z( G) ~ An+ 1 or Sn+ 1 • Proof: We first consider the case n = 9. Assume hypothesis (A) of Chapter 4 does net hold. Then there is a subgroup K of G generated by special 3-elements such that X /K = X1 EB (9-s) lK where X1 is primitive and irreducible of degree s = 7 or 8 and K ';f. As+i. If s = 8, by Lemma 5.11, K/Z(K) ~ 0 8+(2) and Z(K) * 1. But then X(G) contains an element with eight eigenvalues equal to -1 and the other one equal to 1. By Mitchell [21], G/Z(G) ~ S10 , clearly a contradiction. So s = 7 and by examining Wales [ 27, 28] K ~ PSp 6 (2). Let X1 act on V1 and 2 · lK act on V(. By Frame [7], there is a subgroup K1 of K with K1 ~ Os{3) and X jK1 = ~ EB 3 · lK where ~ is irreducible and primitive of degree 6. 1 Also K1 is generated by its special 3-elements. Let g be a special 3-element in G such that X(g) does not leave V1 invariant. In particular 83 g does not commute with K. As in the proof of Lemma 5.10, by replacing K1 by an appropriate K-conjugate, we may assume g does not commute with K1 • Let X2 act on V2 and 3 · lK on V;. As in the 1 proof of Lemma 5.10, X(g) does not leave V2 invariant. Therefore XI ( Ku g) = S EB 1 <Kv g) where S is irreducible or S = Y EB ~ such that Y is irreducible of degree 7. Assume first that Sis irreducible. By Lemma 1. 4, as 0 5 (3) is simple and not a subgroup of A 8 , S is primitive. But by Lemma 5. 11, (Ki, g) /Z( (Ki, g)) ~ o: (2) is the only possibility. This is a contradiction, as in the first paragraph. So S = Y EB ~. By Lemma 1. 4, Y must be primitive. Hence XI (K 1 , g) = Y EB 2 · 1 (K ) i,g by Mitchell. Arguing as in Lemma 5.10, XI (K, g) = R EB 1 (K, g) where R is irreducible. By Lemma 1. 4, since PSp 6 (2) is simple and is not contained in A8 , R is primitive. By Lemma 5.11, (K, g) / Z( (K, g)) ~ 0 8+(2) is the only possibility, a contradiction as in the first paragraph. So hypothesis (A) holds and by Lemmas 5. 8 and 4.11, the result holds for n = 9. We now consider n = 10. Again assume hypothesis (A) fails. Then there is a subgroup K of G generated by special 3-elements such that X IK = X1 EB (10-s)lK where X1 is irreducible and primitive of degree s = 8 or 9 with K ~ As+i· s = 8. From the case n = 9, we must have By Lemma 5.11, K/Z(K) ~ 0 8+(2). By Lemma 5.10, K contains a subgroup K1 generated by special 3-elements such that X jK1 = ~ EB 3· lK where K1 ~ PSpi2) and X2 is primitive and irreducible. Let X1 act onV 1 and2·1KactonV{. Let~actonV 2 and3·1K onv;. 1 Let g be a special 3-element such that X(g) does not leave V1 invariant. As in the proof of Lemma 5.10, X(g) does not leave V2 ] 84 invariant. Therefore X I (K1 , g) = S EB 1 (Ku g) where S is irreducible or S = Y EB ~ such that Y is irreducible and ~ is linear. By Lemma 1. 4, as PSp 6 (2) is simple and not a subgroup of A 9 , if Sis irreducible, it is primitive, and if S is reducible, Y is primitive. From the result for n = 9, S cannot be irreducible as PSpi2) is not a subgroup of A 10 • So XI <Ku g) = Y EB 2 · 1 <Ki, g) by Mitchell. Arguing as in Lemma 5.10, XI (K, g) = R EB 1 (K, g) where R is irreducible. But again by Lemma 1. 4, R is primitive and so (K, g) ~ A10 • But K 1 ~ PSpi2), a contradiction. So hypothesis (A) holds. Therefore by Lemmas 5. 8 and 4. 11, the result holds for n = 10. Assume the result fails for some n ? 11 and let n be minimal so that a counterexample exists. By Lemmas 5. 8 and 4.11, hypothesis (A) fails and so there exists a subgroup H of G generated by special 3elements such that X IH = X1 EB (n- s) lH where X1 is irreducible and primitive of degree s = n-1 or n-2 but H ~ As+i. As n ? 11, this contradicts the minimality of n and the lemma holds. Combining Lemmas 5.11 and 5.12, the proof of Theorem 2 is now complete. The corollary is an easy application of Theorem 2. Let G be simple and X a nontrivial complex representation of G of degree n ? 10. Let H be a subgroup of G isomorphic to An-i · By Lemma 4. 2, X IH = X1 EB 2 · lH where X1 is irreducible and primitive of degree n-2. In particular G contains a special 3-element, and as G is simple, it is generated by its special 3-elements. Also X = Y1 EB Y2 where Y1 is irreducible of degree n-2, n-1, or n. If Y1 has degree n-2, it is primitive as X 1 is. So special 3-elements g E G '\H have the property that 85 either Y1 (g) or Y2 (g) is trivial; as G is simple, Y2 (g) must be trivial and hence Y2 = 2 · lG. By Theorem 2, G ~ An-i· n -1, Y2 = lG as G is simple. Theorem 2, G ~ An. If Y1 has degree By Lemma 1. 4, Y1 is primitive and by If X is irreducible, by Lemma 1. 3 it is primi- tive and by Theorem 2, G ~ ¾+i· The corollary is proved. 86 REFERENCES [1] Aschbacher, M. and M. 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