WWNHINN)!ill!IHIMlHI||lilWIHWIHWIW SILTY LIIBRAR IE8 “"3“ MilWiHli\\|\\\\\\\\\W|\W|H1W11H“ 1 19430 This is to certify that the dissertation entitled SOME AMALGAMS IN CHARACTERISTIC 3 RELATED TO C01 presented by Panagiotis Papadopoulos has been accepted towards fulfillment of the requirements for Ph.D. degree“, Mathematics 44;,”‘M/ r Wprofe Date May 2, 1995 MS U is an Affirmative Action/Equal Opportunity Institution 0-12771 LIIRARY Mlchlgan State University PLACE II RETURN BOX to moon thIc chockout from your rocord. TO AVOID FINES rotum on or bdoro duo duo. DATE DUE DATE DUE DATE DUE MSU Io An Nfltmoflvo Mon/Emu Opportunlly Intuition W” 1 SOME AMALGAMS IN CHARACTERISTIC 3 RELATED TO Col By Panagiotis Papadopoulos A DISSERTATION Submitted to Michigan State University in partial fulfillment of the requirements for the degree of DOCTOR OF PHILOSOPHY Department of Mathematics 1995 ABSTRACT SOME AMALGAMS IN CHARACTERISTIC 3 RELATED TO 001 By Panagiotis Papadopoulos Col has parabolic subgroups of the shape 362M” and 31+4Sp4(3).2 Lyons’ simple group Ly has parabolics of the form 35(M11 x 2) and 32+4A5D8. We will characterize these parabolics and similarly the ones found in subgroups of C 01 using the amalgam method introduced by Goldschmidt. Dedicated to the memory of Amanda iii Acknowlegements I wish to express my sincere appreciation to my adviser, Professor Ulrich Meier- frankenfeld, for his invaluable guidance, advice and corrections at all stages of this project. I am very greatful to Professors R. Fintushel, J. Hall, R. Phillips and S. Schur for serving on my committee. I also wish to thank Dr. Sergey Shpectorov for his interest in my work and his valuable suggestions. A special word of thanks goes to my wife Maria, for her encouragement, under- standing and constant moral support. iv Contents 1 Introduction 2 Properties of (-9, \I! and their modules 3 Properties of the graph 1‘ 4 The case Z0: 5g Qa 5 The case b=2 6 The case Z0: 3 Q0, 7 The case b=1 and G E (2)M12 8 The case b=1 and (-3 E PSL2(9) or M11 11 19 31 35 41 46 1 Introduction Let G be a finite group, p a prime, S a Sylow p-subgroup of G and B = Na(S). A parabolic subgroup of G is a proper subgroup of G which contains a conjugate of B. Consider the set J of parabolic subgroups of G ordered by inclusion; then J becomes a partially ordered set called the parabolic geometry of G. In the case where G is a finite simple group of Lie type in characteristic p, J is the usual building given by Lie theory. The parabolic geometry may be viewed as a generalization of the concept of a building to an arbitrary group. In recent years the parabolic geometry (in particular for p=2) has been used to study, construct, characterize and prove uniqueness of many of the sporadic finite simple groups. The parabolic geometries (again for p=2) also play an important role in the ongoing revision of the classification of the finite simple groups, in particular in the so called quasi-thin and uniqueness cases. While parabolic subgroups have most intensively been studied for p=2, many interesting examples exist (besides the groups of Lie type) for arbitrary primes. In [RS], Roman and Stroth determined all the minimal parabolic geometries for all the 26 sporadic groups and all primes. One of the most interesting series of examples arises for the prime p=3 and G being the first Conway group 00;. C 01 has parabolic subgroups of the shape 362M” and 31+4 Sp4(3).2 (we will explain this notation later). Lyons’ simple group Ly has parabolics of the form 35(M11 x 2) and 32+4A5Dg. These parabolics have been used by M. Aschbacher and Y. Segev to prove the unique- ness of the Ly. It is the goal of this paper to characterize these parabolics and similarly the ones found in subgroups of C01. Let BG be the largest normal subgroup of G contained in B; then BC is contained in all the parabolic subgroups of G and thus acts trivially on the parabolic geometries and so the parabolic geometry carries out information only about G / 89. Also when B is contained in a unique maximal parabolic subgroup of G, the parabolic geometry becomes disconnected. So let us assume that BG = l and that P1 and P2 are parabolic subgroups of G containing B, with G = , for example two different maximal parabolic subgroups of G. Then we see that G, P1 and P2 fulfill the following statement: (A1) P1 and P2 are finite subgroups of G. (A2) G = . (A3) Let SESyIp(P1 (1 P2) and B = Np,np,(S); then B = Np,(S), i=1, 2. In particular SESylp(R), i=1, 2. (A4) No non-trivial normal subgroup of G is contained in B. If (G, P1, P2) fulfill (A1) - (A4), we say that G is an amalgamated product of P1 and P2. We remark that we allow G to be infinite in this definition in order to cover the case where G = P1 *3 P2, the free amalgamated product of P1 and P2 over B (see [S] for a precise definition). Notice that if (G, P1, P2) is an amalgamated product then also (P1 *3 P2, P1, P2) is an amalgamated product in our sense. To any amalgamated product (G, P1 , P2) we can associate a graph F whose vertices are the cosets of P1 and P2 in G and two cosets are adjacent if they are distinct and have non-empty intersection. We remark that if B = Ng(S) then the graph F can be embedded into the parabolic geometry of G. The amalgamated method introduced by Goldschmidt [G] and refined by Stell- macher [St], Delgado [DS] and Timmesfeld [T] uses I‘ as a tool to define important subgroups of B and as a book-keeping device to determine relations between the subgroups. This method has proven very succesful in determining the structure of P1 and P2 assuming the action of P1 and P; on their neighbours A(P1) and A(P2) respectively in the graph P is given. Let us assume for simplicity that P1 0 P2 = B (which will always be true for G = P1 *3 P2). Let Q.- = 0.00.), L.- = 0P’(P,-)= and P,“)/Q,- = CG,/Q,(L,-/Q,-). Then it is easy to see that 13,-“) is precisely the kernel of the action of P,- on A(P,-) and L,- acts transitively on A(P.-). Hence the group Lg/Q,‘ carries most of the information about the action of P,- on A(P,~) and we then refer to the pair (Ll/Q1, Lg/Qz) as the type of the amalgamated product (G, P1, P2). The main task of the amalgam method can now be described as determining (P1, P2) from the type (L1 /Q1, Lg/Qg). For example, the main part of Goldschmidt’s paper [G] determines the structure of (P1, P2) of type (Sym(3),Sym(3)) for p=2. For the remainder of this paper we will work under the following hypothesis: (P) (G, P1, P2) is an amalgamated product of type (9, \II) for p=3 so that: (Po) 9 E PSL2(9), M11, M12 or 2-M12, (P1) \II E PSL2(3), SL2(3), A5, 21%, 24/15, 2134/15, PSL2(9), SL2(9), 5194(3) or P5P4(3), (P2) Cp,(03(P,-)) _<_ 03(1),) for i=1, 2. Before we state the main theorem recall the following standard definitions: For a finite group X and a prime number p, 0,,(X) is the largest normal p-subgroup of X; 0”(X) is the smallest normal subgroup of X such that G / 0” (X ) is a p-group, or, equivalently, the group generated by all p’ elements; 0P'(X) is the smallest normal subgroup of X such that its index in X is not divisible by p, or, equivalently, the group generated by all Sylow p-subgroups of X. , Now introduce the following notation: G ~ 3d1+‘"+d"H means that there exists a normal series 1=H0_<.HIS"’SHnSGa so that for i=1, 2, - - - , n, H,- / H -_1 are elementary abelian minimal normal subgroups of G/H;-1 with |H;/H;_1| = 3d' and G/Hn E H. Also, by G ~ 2-H we mean that G/Z(G) g H, |Z(G)| = 2 and Z(H) _<_ H’. We are now able to state our main result. THEOREM P: Under hypothesis P the possible pairs (L1, L2) are as follows: (i) (34PSL2(9), 31+42-A5), (ii) (34PSL2(9),31+421_+4A5), (iii) (362-M12,31+1+1+2+2+13L2(3)), (iv) (362-M12,31+4Sp4(3)), (v) (35M11,31+1+2+2SL2(3)), (vi) (35M11,31+4SL2(9)), (vii) (34PSL2(9),31+2+2SL2(3)), (viii) (35M11,31+1+42-A5), (ix) (36PSL2(9), 31+1+4SL2(9)). Note that the examples for (i)-(ix) can be found in G ’5 M cL, C02, C01, Col, Suz, Cos, U4(3), Ly and PSp4(9) respectively. We will see later that all the cases occur when b=1 and [Zm Z01] = 1 where the notation in this remark will become apparent momentarily. 2 Properties of 6), \II and their modules In this section we will list some of the properties of the groups G and ‘11 and their modules. A non-abelian p-group P (p a prime) is called extra-special if l‘1’(P)| = |Z(P)| =12- There are two extra-special groups of order 25; one, denoted by 23,“, contains an elementary abelian subgroup of order 8 and the other one, denoted by 21;”, does not. It is a well-known fact (see for example [Go; 5.5.2]) that 21.44 g Q8 * D83 where * denotes the central product. A Steiner system S(l, m, n) is a pair ((2,8), where Q is a set of size n, B is a set of subsets of size In called blocks and such that every subset of size 1 in Q lies in a unique member of B. By [W], there exists a unique, up to isomorphism, Steiner system of type S(5, 6, 12). Let 5=S(5, 6, 12). Define then the Mathieu group on 12 points to be the group M12 = Aut(8) = {7rESym(12)|B1r is a block for all blocks B}. Define M11 to be the stabilizer of a point in M12. Then M11 is 4-transitive on eleven points and its corresponding Steiner system is S(4, 5, 11). Lemma 2.1 (a) M12 is sharply 5-transitive on 12 points, i.e., M12 is 5-transitive on 12 points and the stabilizer of any five points in M12 is the identity group. (b) |M12| = 1211-1098 = 26-33-5-11. (c) The normalizer of a Sylow 3-subgroup of M12 has orbits of lengths 3 and .9 and therefore if an involution acts on these it has a fixed point. {(1) M12 has two classes of involutions, say D1 and 0;. Moreover :rEDl if and only ifzrfixes a point if and only ifa: fires four points if and only ifz belongs to a normalizer of a Sylow 3-subgroup of M12 if and only ifs: lifts to an involution in 2-M12. Proof: [A] and [Gr]. Notation 2.2 To avoid repetitions we will use the following notation throughout: X E (2)H means that either X E’ 2-H or X E H. Similarly, X g 2(1+)4A5 will denote a group X such that X/02(X) E A5, 02(X)/02(X)’ is the even permutation module on five letters for A5 and [02(X)’ [=1 or 2 respectively. Definition 2.3 Let X be a finite group. Slightly abusing the standard definition we will say that X is 3—stable provided that the following condition holds: If V is an irreducible GF{3)X-module and A _<_ X is such that [V,A, A] = 1 then [V, A] = 1. Lemma 2.4 Let Y be a finite group. Then: (a) The following statement is equivalent to Y being 3-stable: let V be any GF{3)- module and A S Y with [V,A,A] = 1. Then ACy(V)/Gy(V) S 03(Y/Cy(V)). (b) Y is 3-stable if and only if Y/03(Y) is 3-stable. (c) If every element of order 3 in Y lies in a perfect simple 3-stable subgroup of Y then Y is 3—stable. Proof: (a) Suppose first that Y is 3-stable and let V and A as in the statement. Let W be any compposition factor for Y and V. Then [W, A, A] = 1 and so by definition of 3-stable, [W, A] = 1. Hence ACy(V)/Cy(V) _<_ 03(Y/Cy(V)). Suppose next that the statement holds and let V be any irreducible GF(3)Y- module. Then 03(Y/Gy(V)) = 1 and so by the statement A S Cy(V) and Y is 3-stable. (b) It is clear, since 03(Y) acts trivially on every irreducible GF(3)Y—module. (c) Suppose that V is a GF(3)Y-module and aEV with [V,a,a] = 1. Then a3 = l and we may assume that [a] = 3. Then aEX S Y, where X is perfect and 3-stable. Then [W,a,a] = 1 for any composition factor W for X on V and so [W, a] = 1 and since X = , X is simple and [W,X]=1 we have [V,X,...,X| = 1 for some n (n-times) and since X is perfect, [V, X] = 1 and [V, a] = 1. Remark 2.5 It follows directly from [Go; p.111] that PSL2(3), A5 and PSL2(9) are all 3—stable. It is also easy to see that any element of order 3 in M11 or (2)M12 lies in a subgroup A5 of these groups and since A5 is 3-stable, 2.4 implies that so are M11 and (2)M12. Finally, 24A5 is 3-stable as it contains A5 which in turn it contains a Sylow 3-subgroup of 24A5. Definition 2.6 A GF{3)X-module V is called an FF-module for X ifo(V) = 1 and if there exists a non-identity 3—subgroup A ofX such that [VI/[CV(A)| S [A]. Lemma 2.7 IfX has an irreducible FF-module then X is not 3-stable. Proof: It follows from Thompson’s Replacement Theorem, see [Go; 8.2.4]. Lemma 2.8 Let X E 9, PSL2(3), (2)A5, 24/15, 21_'HA5 or PSp4(3). Then X does not have an FF-module. Proof: The proof for O, PSL2(3), A5 and 24A5 follows from 2.5 and 2.7 and the proof for PSp4(3) can be found in [M]. So we only worry about the cases 2-A5 and 2L+4A5, namely the cases where X / 02(X ) E A5. Let V be a faithful irreducible GF(3)X-module. Let A be a non-trivial 3—subgroup of X and suppose that IV/Gv(A)| S [A]. We want a contradiction. First, [A] = 3 since |X|3 = 3 where [X [3 denotes the 3-part of X. Second, since there exists an element of order 5 in X and since 5 does not divide We can choose d1,d2€L5 (where X E L5/Q5) of order 3 such that D := has a quotient A5. Since Cv(D) = Cv(d1) fl Cv(d2) has codimension less than or equal to two in V and since GL2(3) is solvable, D acts trivially on V, a contradiction since [V, d1] # 0. Hence X does not have an FF-module. Definition 2.9 (a) Let X E Sp4(3). A faithful GF(3)X-module W is called a natural Sp4(3)-module for X, if W carries the structure of a 4-dimensional symplectic space over GF(3) which is invariant under the action of X. (b) LetX E SL2(3’°) and Wafaithful GF(3)X-module. Then Wis called a natural SL2(3k)-module for X if W carries the structure of a 2—dimensional vector space over GF(3") invariant under the action of X. It is worth mentioning at this point that A, E PSL2(3) and 2A,, E 512(3), A5 g PSL2(5) and 2145 g 5112(5), A6 g PSL2(9) and 2146 21 514(9), and Remark 2.10 (i) SL2(3) has a unique faithful irreducible GF(3)-module; moreover, this module is an F F-module and its order is 32. (ii) PSL2(9) has four irreducible GF(3)-modules; their dimensions are: 1, 4, 6 and 9. (iii) Let X = SL2(3), SL2(9) or Sp4(3) and let V be an FF-module. Then v = [V, Z(X)] a CV(X) and [V,Z(X)] is a natural SL2(3), SL2(9) or Sp4(3)-module respectively. [M; p.469 and 470] (iv) M11 has two irreducible modules of dimension less than or equal to 8; more- over, both have dimension five and they are dual to each other. [J] (v) 2-M12 has a unique non-trivial irreducible GF(3)-module of dimension less than 10; moreover this module has dimension six and is faithful; in particular, M12 does not have any non-trivial module of dimension less than 10. [J] Lemma 2.11 Let G E (P)SL2(3), (P)SL2(9), M11 or (2)M12. Then G has no automorphism of order 2 centralizing a Sylow 3—subgroup. Proof: Well-known, see for example [A]. Lemma 2.12 Let X be any of our groups ('9 or \II, SIESy13(X) and B1 = NX(S,), Then B1 is irreducible on Z(Sl); in particular 81 is irreducible on 51 for X E (P)SL2(3), (2)A5, 24A5, 2(1+)4A5, (P)SL2(9) or M11. Proof: If |X|3 = 3 then [SI] 2 3 and the lemma holds trivially. If X E (2)M12 or (P)Sp4(3) then |Z(51)|= 3 and if X E PSL2(9)(E A6) then 51 = <(123),(456)> and Bl = Sl<(1425)(36)> and the lemma holds for PSL2(9) and so also for SL2(9). Since PSL2(9) S M11 and [Mlll3 = [PSL2(9)]3 it also holds for M11. Lemma 2.13 Let HEO, TESy13(H), tEAut(H) with |t| = 2, [NH(T),t] S T and T‘ = T. Then t is an inner automorphism. Proof: Suppose first that Z(H)=1. View H as a subgroup of Aut(H). Suppose H = PSL2(9). By 2.12, NH(T) is irreducible on H and so t either inverts T or centralizes T. Now the same is true for any involution z in N H(T). Hence by 2.11, both t and z invert T and so, again by 2.11, tT=zT and tEH. If H 2 M11, Aut(H)=H and we are done. If H = M12, NH(T)/T E C2 x G2 and NAut(T)(T)/T E De (see A]); hence no element in NAu,(T)(T) \ Inn(H) centralizes NH(T)/T. If Z (H ) ¢ 1 then by the previous case, t induces an inner automorphism s‘ on H/Z(H). Pick sin H with sZ(H) = 3*. Then [s‘1t,H] S Z(H) so [s‘1t,H, H] = 1. Since H’ = H, the 3-subgroup lemma now implies [s’1t, H] = 1. Lemma 2.14 Let H = PSL2(9), M11 or (2)M12 and T S RESy13(H) with |T| = 9. Then: (a) H = for some gEH. (6) [fit = (2)M12 then H = = = <12, R’9> for some gEH. 10 Proof: Note first that (*) PSL2(9) E A6 = <(123), (125)(346)> and in particular (a) holds for H = PSL2(9). Clearly the statement for M12 implies the statement for 2-M12 and we may assume now that H = M11 or M12. Let (9,8) be a Steiner system of type S(4, 5, 11) and S(5, 6, 12) respectively with H = Aut(Q,B). Let D S T with ID] = 3 if H = M11 and D = R’ if H = M12. Then, in any case, D S T and D normalizes a block BEB. Hence NH(B) ESym(5) or Sym(6) respectively. In the M12 case, NH({B,Q \ B})(E Aut(A6)) interchanges the two conjugacy classes of elements of order 3 in N ”(B ) Hence, using A5 = <(123), (345)> and (*) respectively, NH(B)' = for some gEH. It is easy to see that N ”({B, Q \ B}) is the unique maximal subgroup of H con- taining NH(B)’ (see for example [A]). Since R S NH({B,Q \ B}) (by Lagrange’s Theorem), (a) is proved. Also, (b) holds unless T S NH(B). So assume H = M12 and T S NH(B). Then T has four orbits of length 3 on 9. Let X be a set of size two in It normalized by D. Then T S NH(X), NH(X) E Aut(As) and NH(X)’ = for some gENH(X) by (*). H = and the lemma is proved. Lemma 2.15 The normalizer of a Sylow 3-subgroup is maximal in SL2(3) and in SL2(9); for Sp4(3) the maximal overgroups of a normalizer of a Sylow 3-subgroup are N (E1) and N (E2) where Eg, i=1 or 2 is the i-dimensional singular subspace of W normalized by the Sylow 3-subgroups ( W any natural GF(3)-module for S p4(3) } Proof: [C; 8.3.2 and 11.3.2]. 11 3 Properties of the graph I‘ In this section we will define a graph 1‘ and we will list some of its properties. Definition 3.1 Let I‘ = {HzleG,i = 1,2}. From now on, small Greek letters will always denote elements of I‘. Make I‘ into a graph by defining a to be adjacent to ,3 if and only ifa 79 fl and a 0 fl 74 (0. Then G operates on I‘ by right multiplication. For aer let G. = StabG(6), G9” = largest normal subgroup of05 fixing all vertices of distance at most n from 6 and A(6) the set of all vertices adjacent to 6. Lemma 3.2 Let i=1, 2. Then: (a) Gm = Pf”, (b) The edge-stabilizers in G are conjugate to B, (c) Let 6,- = P,-. Then A(6,-) E P,/B as a G5,-set,' in particular, G5, is transitive on A(6,°), (d) Let (6,)\) be an edge; then G = , (e) G acts faithfully on P, { f ) P is connected. Proof: (a), (b) and (d) follow directly from the definitions; we will now prove the rest of the claims. (c) [DS; 2.1(c)]. (e) Let gEG be such that 79 = 7 for all 76F. Then Pig = P.- and therefore 963. Also, if heG then 79" = 7 for all '76P. Hence _<_ B and claim is proved by (A4)- (f) Let I}, be the connected component of F containing P1. Then also Pgero. By (a), S Ng(I‘o) and (A2) implies P = I‘o. Notation 3.3 Let d( , ) denote the usual distance on the graph P. For 661‘ and'i 2 1, 12 A“’(6) = {MP/(105A) S 2'}. Q6 = 03(05), 2. = <01Z(T)/T€Sy13(G,5)> , V; = , b. = mintendwa/zs z 09’}, b = min5:€p{b5:}, G5). = G5 0 GA and Q5; = Q5 0 Q; if 6€A(A). A pair of vertices (6,6’) such that Z5 S G9) and d(6, 6’) = b is called a critical pair. The bounding of the parameter b which we just introduced, will allow us to deduce a considerable amount of information about P1 and P2. Lemma 3.4 (a) G acts edge- but not vertex-transitively on F, (b) G5 is finite, (c) CGo(Q5) g Q5: (d) [fa is adjacent to B then Sy13(Ga (1 Ga) Q Syl3(Ga) fl Sy13(Gg). Proof: [DS; p.73]. Remark 3.5 Notice that as G acts edge-transitively, b = min{bo,,bg} for any pair of adjacent vertices 0, fl. Thus, we are allowed to choose a, ,8 such that be, = b S by and {GmGg} = {P1,P2}. In particular, Go, 0 G3 = B and SESyl3(Ga) fl Sy13(G5). Let 0/61" such that d(a,a’) = b and Z, S G9,). Let p be a path of length b from a to a’. We label the vertices of p by p=(a,a+1,...,a+b) = (a’-—b,---,a'—1,a’), i.e. a+i (respectively a’—i) is the unique vertex in p with d(a,a+i)=i (respectively d(a’ -i)=i). Furthermore, from 3.2 (c) we may assume that B=a+1ibeL 13 Note also that if Q5 = Q5 for some 6€A()\) then Q5 _<_] = G, a contradic~ tion. Hence Q6 5‘ QA WEAPO- Lemma 3.6 Let (6,/\) be an edge and N a subgroup of G55 such that NGH(N) acts transitively on A(p) for p6{6, A}. Then N=1. Proof: See [DS; (3.2)]. Lemma 3.7 For 661‘, (a) Q6 3 Gt”, (b) Z5 S Z(Q5) (1 V5; in particular, b 2 1, (c) Z5,, S Go, and [Z5,Zat] S Z5, 0 Z5,, (‘1) Z. # 912(T), T€5y13(Ge), (e) If SESy13(B) and (21(Z(S)) is centralized by a subgroup R of G5 which acts transitively on AM) then Z(La) = 1, (f) G§1’/Qa is a 5.5.0... Proof: (a) Let A€A(6) and TOESyl3(G5A). Then Q5 = 03(G5) S To S GA- Hence Q5eG. VAEA(6) and therefore Q5 3 G5,”. (b) Z5 S V5 is immediate from the definitions. Now show that Z5 S Z(Q5). Since C9,,(Q5) S Q5 it is enough to show Z5 S CG6(Q5). Let To be a Sylow 3-subgroup of G5 containing Q5. Then Q5 = Qg S T8 for every gEG5. Hence Q5 S T for every TESy13(G5). As T centralizes Z(T) so does Q5 for every TESyl3(G5). Thus, by the definition of Z5, [Z5,Q5] = 1 and Z5 S Cg,(Q5). In particular, by (a) Z, S G9) and soaaéa’andeI. (c) Minimality of b gives 25. g G331, g G... In particular, 25. normalizes Z5,. Hence [Z5, Z5,] S Z5,. Similarly, [Z5, Z5,] S Z5,. 14 ((1) Assume that Z, = 91Z(T) for some TESyl3(G5). Hence S = T9 for some gEGa and therefore 91(Z(5)) = 91(Z(Tg)) = (91(Z(T)))9 = Z0. where the last equality holds because 2,, g] G... Thus 2,, g Z5. Since 2,, 5 G53) we get that Z5 S GS.) and since d(a’,a) = b — 1, we get a contradiction to minimality of b. (e) Let R be a subgroup of G5 that acts transitively on AM) and such that 01(Z(S)) is centralized by R. Then 91(Z(L5)) S Ca.(Qe) S Q0: S 5. Hence Q1(Z(L5,)) S 01(Z(S)),and therefore 01(Z(La)) is also centralized by R. Thus, by 3'6, Q1(Z(Lcnr)):'1= Z(La) (f) Without loss of generality, 6€{oz, 5}. Since B = Na,(S) we get 03(B)ESy13(B); hence B/03(B) is a 3,-group. Let Q = 03(Ggl)). Then, as G551) S B, GP/Q is a 3,-group. Now Q is a normal characteristic subgroup of G5,” which is normal in G5 and so Q S G5. But Q is a 3-group. Thus Q S 03(G5) S G551) and we conclude Q = 03(G5). Thus Ggl)/03(G5) is a 3'-group. Remark 3.8 (i) By 3.7(b), Z, S Qal. (ii) Z5 3 G, V7€A(1)(6), B = Gag, Z5, 3 B, 25 g B. (iii) Also Sy13(B) Q Syl3(P1) fl Sy13(P2). (iv) Frattini argument gives L35 = L5 and for uEA(6), G5 = L5G5,,. Lemma 3.9 (1) [Z5, Z5., Z5,] = 1, (2) Vb S 06 V6EP) (3) Z, normalizes Vat. 15 PPOOf: (I) [Zen Zal, Z01] S [20 fl Zal, Zal] _<_ [201, Zat] = 1. (2) For €€A(6) and g€G5,Z§ = 2:9 S V5 as d(eg,6) = d(59,69) = d( G) On V II 5—: (3) It follows from V5,: S G5,: and Z, S Gav. Lemma 3.10 If b>2 then V5 is abelian. Proof: Let A,€€A(fl) U {B}; then d(/\,€) S 2 = 91Z(T) S 2),. Corollary 3.12 (a) 2,, S Z(Le), (b) If Z5: S Z(Lat) then a is not conjugate to a’. Proof: (a) follows from 3.11 as Z, S Z5. For (b) notice that if a were conjugate to 01’ then, since by (a) Z, S Z(L5), we get Z5, S Z(Lat) and we are done. Corollary 3.13 Z5 0 Q5. 7Q Cza(Zat) if and only if Zat S Z(Lat). Proof: Assume first that ZaflQat 96 Cza(Zo,:). By 3.15(c), Z5,: S Q5, so [Z5, Zar] = 1 and 20,! S Z(Lal). Conversely, let Z5,. S Z(Lat). Then, since Z, S La, and since Z, S Q0", Za n Qa’ 7e CZa(Za’)° 16 Lemma 3.14 Let 6€{a,,8}. (a) Let A be a 3-subgroup of G5 with A S Q5. Then 03(L5) S and L5 = Q5. (b) Let N S G5 with 3 dividing [NQ5/Q5I; then NG55 = G5. Proof: Let E = L5/Q5. Since L5 = G or \II, L5/02(L5) is simple in all possible cases. Since A S Q5 and A is a 3-group, A S 02 (f5). We conclude that 02(L5) = .175. Thus L—5/ is a 3’-group. Since E is generated by 3-elements, L5 = and so Q5 2 L5. In particular, L5 / is a 3-group and (a) is proven. (b) By (a) applied to a Sylow 3-subgroup of N, L5 S N Q5, so G5 = L5Gag = (NQ5)Gafi 2 Name. Remark 3.15 (a) By 3.12(a), Z, S Z(L5) and so by 3.14(a) CGO(Z5,)/Q5 is a 3'- group. (b) If Z5,: S Q5, then [Z5, Z5,] = 1 and so Z, = CZO(ZO,I). Hence Z5, 0 Q5: # Z5, and Z5, (1 Q5: 5E Cza(Zat). (c) If Z5,: S Q5, then by (a) Cza(Za:) = Z5, 0 Q5: and since we have a complete symmetry between a and a’ in this case, we get that Cza,(Za) = Z5,, 0 Q5. Definition 3.16 (a) L5 = L5/03(L5). (b) Let K be a complement for S in B and K, = KnL, and K5 = ImLfi. 17 (c) Let 6€{a,fi}. Let t5 be an element of order 2 in K5 with t5Q5/Q5EZ(L5/Q5) if L5/Q5 is isomorphic to one of the groups SL2(3), 2'A5, 2L+4A5, SL2(9), 23112, 5114(3); otherwise let t5 = 1. Corollary 3.17 Let 66A().) and t5 at 1. If L,\/QA g (P)SL2(3), (P)SL2(9), M11 or (2)M12 then t5 does not centralize S/QA. Proof: Suppose t5 centralizes S/Ql. Then 2.11 implies that t5 centralizes LA/Qt. By Frattini argument L), = CL,(t5)Q,\. Similarly L5 = CL,(t5)Q5 and by 3.6 we get t5 = 1, a contradiction. Lemma 3.18 Q5 is not contained in Q), for any pair of adjacent vertices 6 and A. Proof: Without loss of generality assume {6,/\} = {0,fl}. Assume that Q5 is con- tained in Q,\. By 3.5 Q5 is properly contained in Q5. Hence Ql/Q5 is normal in B and as Q), 75 S we get that B is not irreducible on S/Q5. Hence by 2.12, L5/Q5 E PSL2(3), SL2(3), 24A5, 2L+4A5, A5, 2-A5, PSL2(9), 2PSL2(9) or M11 and therefore L5/Q5 E (P)Sp4(3) 0T (2)M12. For 76F let M(Q,) be the set of all maximal with respect to order abelian sub- groups of Q, and J(Q.,) = , namely . J(Q,) is the Thompson subgroup of Q.,. If J(Q,\) S Q5 then clearly M(Q5)=M(QA) and so l?é J(QA) = J(Q6) El , a contradiction to (A4). Hence, J(QA) S Q5 and there exists A6 M(Ql) with A S Q5. 18 Now notice that S G5 and therefore can not be normal in GA. Let Z; = <01Z(Q,\)G6>. Since Q5 is contained in Qy we get that. QIZ(Q,\) S C0,,(Q5) S Q5 and therefore Z; S Z(Q5). Hence ZgflAngflAflQ5. Let X = C0,,(Zg‘). Suppose CA(Zg) S Q5. Then 3.14 gives XG5,\ 2 G5. Since QlZ(Ql) is normalized by G ,\ and therefore by G5 A as well and since it is also normalized by X, we get that QIZ(Q),) is normalized by G5 and G,\, a contradiction. Hence A 0 Q5 = CA(Z6‘). Also, since zgnA=ZgnAnQ5 we now get 2; n A = z; n 05(25). Zg'(Afl Q5) is abelian. Hence |ZgCA(Zg)| S |A|. Then we have MI 2 IZECA(ZE)| = IZEIICA(Zi)l/|Zi fl CA(ZE)| = IZEIICA(Zi)l/|Zi 0 Al- Hence IZEI/IZE 0 Al 5 IAI/ICA(ZE)| and so IZE/Cz,-(A)l S IZi/Zi 0 Al S lAl/ICA(ZE)| = IAQol/lQol- Thus 2; is an FF-module for L5 /CL,(Zg') and 2.8 gives Lo/Qa E 5194(3)- Now t5 centralizes 5/ Q5 and as Q5, a contradiction to 3.6. 4 The case Za, S QC, In this section we work under the hypothesis Z5, S Q5. Notice that under this hypothesis, we have a complete symmetry between a and a’, so Zar S Z ( Lat). Lemma 4.1 (a) z, n Q... = 0242...); in particular b is even, (b) 2w D Qe = Czar(Za)r (C) 03'(CG.(Zo)) = Q01; ((0 03'(CG. -1 .(Zo_t)) = QM and (6) 03'(CG..(Z a— _1Z5, )) = Qo fl Qo_t. Proof: (a) follows from Z5,, S Z (L5,) and 3.13; (b) follows from (a) and symmetry between a and a’; (c) follows from 3.15; (d) is an immediate consequence of (c) and 3. 7(a); (c) and ((1) imply (e). Definition 4.2 5 =1 if Z5 # QIZ(S) and e = 2 if Zg = 5212(5). The main result in this section will be the following 20 Proposition 4.3 b = 2 and c = 2. Lemma 4.4 (a) Lat/Q0: ”2" Lat/Q0! E{SL2(3), 5122(9), Sp4(3)}. (b) Z5 is an FF—module for La/Qa, Z, = [Z5,La] EB 91Z(L5,) and [Z5,La] is the unique natural SL2(3), SL2(9) or Sp4(3) module for La/Qa respectively. Proof: (a) Since [Z5,Z5:,Z5:] = 1 and [Z5,Zal] yé 1 and as Z, S Q5: we get that Lat/Q0! cannot be 3-stable. Similarly L5/Qor is not 3-stable. Hence La/Qa E Loy/Q0! g SL2(3), 2°A5, 5142(9), 21+4A5 07‘ (P)Sp4(3). I want to exclude the possibility of 2-A5, 2L+4A5 and PSp4(3). Without loss of generality we may assume that lZaQa’/Qa’l S [ZO'Qa/Qal' Let V = Z5, and A = ZatQa/Qa. Then lV/Cv(A)| = IZo/Cz.(ZeI)| = IZo/Ze fl Qo'l = [ZaQa’/Qa’l S lZa’Qa/Qal : [A] Therefore Z5, is an FF-module for La/Qa. Since 2-A5, PSp4(3) and 2L+4A5 do not have an FF-module we conclude that La/Qa g La’/QO’E{SL2(3)> 5112(9), 5174(3)} (b) follows from 2.10. D By 4.4, L5, fixes some symplectic form on Z5, with 91Z(La) in its radical. In what follows “.1.” and “singular” is meant with respect to that form on Z, (or also on Z5,). Lemma 4.5 Let X S Gar. Then Cza,(X)L = [Zar,X] + QIZ(L5:). Proof: [As; 22.1]. 21 Definition 4.6 For La/Qo, 9: SL2(3) or SL2(9) let [\(afl') = A = A(01)\ {5} and for La/Qo, E 5194(3) let Egg be the 2-dimensional singular subspace of Z, nor- malized by S and define A(a,a’) = A by A = {a —1€A(a)/ZO_1 ,Z [ZmZm] and E040 (1 [Zn Z01] = 1} 4.7 A # 0. Proof: For La/Qa ’5 514(3) or SL2(9), this is clear. So suppose La/Qo, ’5 5194(3) and pick a singular 2-space in Z, whose intersection with [ZmZal] is 1 and is not perpendicular to [Za,Za:]. Call this space E and pick any l-space in E which is not perpendicular to [ZmZal], say W. Then NGO(W) fl NGO(E) is the normalizer of a Sylow 3-subgroup of Go, and so there exists a — 16A(a) with Z04 = W and E04“, 2: E. Then 01 ~16A. Lemma 4.8 If La/Qa ’.-_-‘_’ SL2(3) or 514(9) then Zn = 020(Za') = [20,, 2w] + Q12(Lc») = [ZaaQfil + 91Z(La) = Cz.(Qe) = 020(5)- Proof: By 4.10, [ZmLa] is 2-dimensional over GF(q), where q=3 or 9 respectively. Hence [ZmLa], Cza(Zo,.), [ZmZa’], [Zng] and C[ZmLa](Q,g) are all l—dimensional over GF(q). Moreover, [Z5, Q3] 2 1 = [23, 20;] and the lemma follows. Lemma 4.9 Leta — 16A. Then = Ga. Proof: If La/Qa ’3—1 SL2(3) or SL2(9), Lemma 4.8 implies [Za—la Za’] ié 1 and so Za: fl (30-14,. By 2.15, 00-13 is maximal in GO, and so = Ga. So suppose La/Qa E 3124(3) and ¢ G0,. By 2.15, normalizes 20-1 or E := Ea_1,a. 22 If Za: normalizes Z04 then [20-1, Z0:] 2 1 and 20-1 S CZa(Za—1) = [Zara Za'lla a contradiction. Thus Z0: does not normalize Z04. Hence Z0: normalizes E and [Za-nZa'l .<_ [1120'] S E r1[ZmZonl = 1 Therefore [Za_1, Zar] = 1, a contradiction since Z0: does not normalize Z04. Lemma 4.10 5 = 2. In particular Z0, is a natural SL2(3), SL2(9) or Sp4(3)-m0dule and Zg S Z0. Proof: Suppose 5 = 1. Let a — 16A. If 20-1 fl Qa:_1 then (a — 1, a’ — 1) has the same properties as (a, a'), which can’t happen as the vertices alternate in terms of 3-stability. Hence Z.-. _<_ 62.1.. s GEL. s a... and [20—13 Za’ 0 Q01» 20' n Q0] S [Go’s 201’, 201’] = 1 Now, 3-stability of 00-1 implies [20-1, 20,: 0 Q0] = 1 which gives CZ [(201) : 20’ 0 Q0: S CZOI(Za-l)- a Hence CZ :(Za-lli < CZar(Zcr)i and by 4.5, [20” Za—l] S [20” Zol- Hence Za_1Zo, is normalized by Za, and by G04 0 Ga we get by choice of a — 1 that Za_1Za 51 Ga and therefore CGa(Za—IZa) 3 Ga- 23 By 4.1(e) now 03'(CGO(ZO-IZO,)) = Q0 0 Q04 and so we conclude that Qa_1 fl QC, SJ Go, and Q5 0 Q0 S G... Let L = . As [Q.,Qa] g Q. n 62., 51 Ga, [L,Qa] g Q. D Q. g Q. Recall the definition of to, (see 3.16) now. Since Q3 S Qa, 3.14(a) implies that taEO3(La) S L. Hence [t., Q.] s Q. n Qe S (Q. fl Qe) :1 Q. and 02((Q. 0 Q.» s (Q. 0 Q.)- Thus [til S D Q. S 02(Q.) 0 Q. s ((Qa 0 Q3» 0 Q0 5. Q0: n Q10 5 Qfi' Hence to, centralizes S/Qg, a contradiction by 3.17. Thus 5 = 2. So Zg = 01Z(Lg) = 012(5) and by 3.7(e), 91Z(La) = 1. So the last statement of the lemma follows from 4.4(b). Notation 4.11 X, : 01Z(Qa), b=min{d(a,6)/Xa S Q5}. Lemma 4.12 b=b. Proof: Z, S Qa, and Z, S X0, give X0 S QC,“ Hence b S b. Suppose bQa and therefore [XmLa] S 20, = [tha]. Now: X0, = Cxa(ta) EB [tha]; but the first summand is normalized by LG and the second is [Xm L0] = Z0. Hence Cxa(ta) S CXa(La) S 912(La) =1 which implies Xa = Za. Remark 4.14 The following are equivalent: 0) 201—1 J: [ZaaZor’li (ii) Cz._.(2.) = 1; Define now Y; and Yg by YE/ZB = and Y3 = CZ.(03(L3))- Note that [Yfifi Q3] S Z3. 4.15 If b>2 then Y5 S Z0. Proof: Let a — IEA. Since Y;_1 S V0-1 S 001-2 by minimality of b we have [Y;_1,Za,_2] S Za,_2. Now Z014 is centralized by Z0: since b>2 and therefore 25 [YJ—lazah-Z] S [Y;_1’Qa-ll S Za—l and [Y;_17Za’-2] S 020-1(Za’) : 1 BUt then Y;__1 S Qa,_2 S G014 and since b>2 implies that Va:_1 is abelian (see 3.10), we get [YOLD Val—1, VON-ll .<. [Gar-1, Val—1, VON—ll S (VaI—ll' = 1 and [16.1.2.0 Q... Z. 0 Q.] = 1- Look at now. For 6€A(a — 1) and gEVa:_1 S Ga we have d(69,a) = (1059,09) = d(6,a) S 2. Hence d(69,a — 1) S 3 and since b 2 4 we get that Z5. S Q0, (1 Q04 and V ;_ S Qa fl Qa—l- Therefore 1Y5-.. 1 3 [Y;..,Q.-1] s 2.-.. Hence [V.I-.,Y.:..] s [V.,-1,v.-1] s . Thus [Va’-13Y;_19Y;_1] S 20-1 m VON—1 S CZa_1(Za’) : 1 and 3-stability of La._1/Qa,_1 gives [Va;_1,Yo‘,"_1] = 1 whence Y;_1 S Qa’—1 S Go’- Now, if Z0, (1 Q0 S Qa:_1 then since Z0: 0 Q0, is quadratic on Yo:1 S Qa._2, we get [Y;_1,03(La_1)] = 1 by 3-stability. Hence CZa(Qa-l/Za—l) g 03(La-1)Ga,a-l = Ga-l SO Y.‘_1 = Cz.(Q.-1/Za-1) S Z. 26 and claim is proved. Hence, assume now that Za’ n Q0 S Qa’—l- Then [Y;_1,Za: 0 Q0] S Z0,_1 fl Vol-1 S Cza_,(Zaz) = 1 which by 4.5 implies [Y;_l, Zar] S Za. Hence, Y;_1Za _<_} = G... Therefore [Y;_1, Q0] S) GO, If [Y;_1,Qa] 7b 1 then C[Y;_,.Q.](Qa) 79 1 and since Z, = QIZ(QO) and Z0 is irreducible, Zor S [Y;_1,Qa]. So Y;_1 = Y;_1Za _<_] Go, and Y;_1 S1 , a contradiction. Hence [Y;_1, Q0] = 1 and Yof_1 S Z, by 4.13. Corollary 4.16 If b>2 then La/Qo, E Sp4(3). Proof: Suppose La/Qa g SL2(3) or SL2(9). Then, by 4.8, CZa/Z3(Qfi/ZB) = Za- Therefore, asmsa whence Z0 = Y; 3 a a contradiction. Remark 4.17 Suppose b>2. (1) Since Y; S Za, [Y", ] = 1 and so Y; S Y3 by 3.14. In particular, Y3 76 Z3 and since 003 normalizes Y3, E03 S Y3. Hence 1 Y3 S E05 S Y3: where Y; is the perp of Y3 in Za. So G3 = 03(L3)Ga3 normalizes Y; and Y; does only depend on B and not on 01. Moreover, Y; = Z3 if |Y3| = 33 and Y; = Y3 = E03 if |Y3| = 32. 27 Let a -— 16A. Note that if Yail = 0-1 then IC¢0_1(ZO,I)| = I and if Yoi1 = 3-1 then |C¢0_1(Za:)| S 3. (2) By 4.5, [Z,\,CQB(Y3)] S Y; VAEAW) and hence [V33 CQsO/fill S Yfi-L and [VfiaQfi 0 Q0] S Yfii' 4.18 Suppose b>2. Then Z3 = Cza(Q3). Proof: We have CZO(Q3) S Y; S Y3 so Cza(Q3) is centralized by 03(L3) and by Q3. By 3.14(a) and Q0, S Q3, L3 = Q3O3(L3)Qo, and hence L3 centralizes CZO(Q3). So Z3 S CZO(Q3) S QlZ(S) S Z3 and the claim is proved. 4.19 If b>2 then for all a — 16A, V0-1 S Qa,_2. Proof: Suppose we can pick a — IEA such that Va—l S Qa’—2- Then V0-1 S Qa,_2 S Ga,_1 and since b>2 we get that V0,-1 is abelian and therefore [Va-1, VON—19 VON-1] S (VO’-1), = 1 In particular, Va:_1 0 Q, S Q04 by 3-stability and 3.20. Hence by 4.17(2), [Va—la VON—1 (1 Q0] S Yell-1' Since b>2 and b is even, b 2 4. Let 6€A(a — 1) and gel/01-1 S Ga. Then d(69,a) = d(6,a) = 2 and so d(69,a — 1) S 3. Hence, by minimality of b, 269 S Qa-l fl an Thus S Q0, 0 Q04 and therefore we get using 4.17(2) that R == Iv.-. v.-. v.11 s v.-. n [, v.-.1 s Val—1 n Yer-L-l S CYOJ'_1(ZOI')' 28 Hence by 4.17(1) we get IRI =1 or 3. Suppose V0-1 S Qa,_1. By 4.9, 2 Ga and by (A4), V0-1 is not normal in Go. So Zai does not normalize V0-1 and since [Zm Zai][Va_1, Z0: 0 Q0] S V0-1 we get that [Va—19 Za’] z [Zara Za’][Va—ly Za’ 0 Q0]- Let W = [20, 20.][Va_1, 20,, n Q0]. Since W = [20,, Zar][Va_1, Z0! 0 Q0] S V0-1 n 20! S Cza,(Va-1) we get W S [Za’a Var—1] = CZa,(Va-l)-L S Wi- If |W| 2 32 then IWil S 32 and so [V0-1, Zar] S W, a contradiction. Hence ”Zora Za’][Va-la Za’ 0 Q0“ : 3 If [V0-1,Zai 0 Q0] 2 1 then by 4.5 [V0-1,Za:] S [ZmZa'] a contradiction and therefore [V0-1, Z0: (1 Q0] 75 1; since [Zm Zai] 75 1 we get [Zm Zai] = [V0-1, Za: 0 Q0]. But [V0-1,Za, 0 Q0] 3 Yail gives [20,204 g Yai, g 2;, a contradiction to the choice of a — 16A as [Zm Z01] S Z5,-4. Hence V0-1 S Qal_1. Suppose IRI = 1. Then 3-stability gives V0-1 S Qa:_1, a contradiction. Therefore |R| = 3 and so IYailI = 32 and |Ya_1| = 32. If Va:_1/Zar_1 has more than one non-central chief factor for L0,-1, say X2/X1 and X4/X3 with X1 and as 03(La) s by 3.14(a) we get that [20,,ng is centralized by 03(L3). Thus Z; = [Za,Q3] S Y3. But |Zfii| = 33 and, as seen above, |Y3| = 32 a contradiction. Therefore [Vat QB: 03(La)i 5E Za- So there is a noncentral chief factor in [V3, Q3]. Thus [V33 03(L5)i S [V37 Qfiizfi (otherwise we get another noncentral chief factor, but we should only have one). Hence Za[V3, Q3] S G3 and therefore V3 2 Za[V3, Q3] which implies Va/Za = [VatQalza/Za = [Va/2a,in- So we have a 3-group (Q3) acting on a 3-group (V3/Za) such that Va/Za = [Va/20,623]. Hence V3/Za = 1 and V3 = Z, S , a contradiction. Lemma 4.20 12:2. Proof: Assume that b>2. Suppose |Y5i : 33 01‘ IiZaaZoz’“ : 32; 30 then Z3 = Yfi‘L or Y3 2 E03 and [Zen Z03] = CZO(Z0,). Let a — 16A. Then, by the definition of A in both cases, Ya'L_l fl Cza(Zar) = 1. By 4.17(2), [VG-1: Za’—2] S Ya'L—l n CZa(ZOI') = 1? contradicting 4.19. So we can assume that |Y3| = 32 and |[ZO,,ZO,I]I = 3. Note that [Y3, Zai] = 1 and so Y3 S [Zm Za:]i. Look at Cza(Za:) \ Y3 = Z0 0 Q0: \ Y3 = [Zm Z0,]J' \ Y3. Pick l-spaces E17 E2, E3 S [Zara Za’]l so that they generate everything, (note that [Zm Z01]i is a 3-dimensional vector space) i.e. IE.) = 3, E.- S [Za,Zai]i, i=1, 2, 3 and E1'E2'E3 = [Za,Zai]i. Moreover, pick the above Ei’s so that E.- S Y3. Choose flteA in such a way that Z3,~ is perpendicular to E.- but not to [ZmZai]. Also, choose the 5:3 in such a way that Z3iE, = Y. (which implies that Cyfi.(Za:) = E,). Then, by 4.17(2) applied to 6.- in place of 5, [V3,,Zai_2] S Y; 2 Y3.. and since b>2, [V3,,Zo,i_2] S CY3'.(Za’) = E,. By 4.17(2) applied to a’ - 3 in place of 3, [V3, 201-2] S Yai_3. So, Z0, fl Q01 = _C_ YOU—3- But IZa fl Qai| = 33 and |Yai_3| = 32, a contradiction. Proof of the Proposition: It follows from 4.10 and 4.20. 31 5 The case b=2 In this section we assume that Z, S Q0. Recall from the previous section that La/Qa ’_—‘_’ 5112(3), 5142(9) 07' Sp4(3), Z0, is a natural SL2(3), SL2(9) or Sp4(3) — module, Lfi/Qfi g PSI/2(9), M11 07‘ (2)3412, €=b=2, (II—1:,B, Z3=QIZ(S') and a is conjugate to 01’. Proposition 5.1 The hypothesis in this section leads to a contradiction. Proof of the Proposition: Since [th] S Q0, 0 K = 1 we have [ta,1\'3] = 1 and the order of to, is 2. By 2.13, to induces an inner automorphism on L3/Q3. By 3.17 to, does not centralize L3/Q3. Also, as tor is an inner automorphism we can pick tEK3 which acts on the same way on L3/Q3 i.e. pick tEK3 so that .733 = tat and 2:3 centralizes L3 / Q3. I now claim that the order of t is 2 as well. By choice of t, ltl = tha/Qal and the image oft in L3/Qa g L3/Qa is to, which has order two. Hence the claim holds if t3 = 1 and so we are done for the cases PS'L2(9), M11 or M12. The only problem could appear in 2-M12 since when we lift M12 to 2-M12 the order of t could become 4. But this does not happen by 2.1(d). Moreover in any case 33 centralizes L3 / Q3 and the order of $3 is also one or two. Now to, acts non-trivially on Zo, which is irreducible for L0, so to, inverts Z0. KO, acts on Y3 faithfully and K3 centralizes Y3 so [K0, K3] = 1. We will distinguish two cases. 32 Case 1: IZOI S |Y3|2. KO, acts on Y3 faithfully and K3 centralizes Y3 so [Km K3] = 1. Since KO, centralizes t and KO, centralizes to, we get that K0, centralizes 133. Thus [13, K0,] = 1. Now define Y3 = 020(02(L3)). Let A = 23 if La/Qa 2 514(3) or 512(9) and A = E03 if La/Qa ’5’ Sp4(3). Since t centralizes Y3 and to, inverts Y3, 2:3 inverts Y3 and so 933 inverts A. This means that if x3 is the image of .733 in Aut(A) then $3€Z(Aut(A)) and so [NCO (A), $5] centralizes A. Let L = NLa(A) and Q = CLa(A). Since Z0, is a natural SL2(3), SL2(9) or Sp4(3)- module, L/CL(A) 2’ GLF(A) where F = GF(3), GF(9) or GF(3) respectively and L acts irreducibly on A. Since A = Al, [ZmQ] S Al = A. Hence [Za,Q,Q] = 1 and Q is a 3-group. So Q = 03(L). Now [L,:z:3] S Q and so by Frattini argument L = CL(a:3)Q. Hence CL(:1:3) acts irreducibly on A and on Za/A (which is isomorphic to the dual of A). In particular 33 inverts or centralizes Za/A. Since C 1: Va = = we conclude that $3 inverts or centralizes V3 / A. Note that $3 inverts A so if x3 inverts V3/A, 2:3 inverts V3 and V3 is abelian, a contradiction to 1 S [Zm Z01] S V3. Ifz3 centralizes V3/A then V3 = Cvp(Z3)A = Cv,,(Z3)xA. Hence V3’ S (Cvfi(Z3))’ (as A s nae. 2: s 2%» and so V3, 0 Z3 S (Cvfi(Zg))' n A =1. Hence Cvé(S) = 1 and V3 = 1, again a contradiction. Case 22 IZQI>IY3]2. 33 As Z3 S Y3 and IZgl = |Zo,| for La/Qa E” SL2(3) or SL2(9), this implies that Lat/Q0: g 5114(3)- Then IZal = 34, Z, is a natural Sp4(3)-module for La/Qo, and [13] = 3. Thus CZG(O3(L3)) = Z3. Recall the definition of E03 in 4.6. Since |Y3| = 3 we have that [Eafia 03(Lfi)l # 1' Subcase 1: E03 S Z(V3). Choose a’EA(fl) such that [E03, Zar] 71E 1 (hence (a, 01’) is a critical pair). On the other hand we have [Zm Z0:, Z0:] = 1. Suppose |[Za, Zoi]| =2 3. Then [Zm Zai] 2 [E03, Z01] S E03 and so Cz.(Za') = [20,, Zeali 2 13$ = on, a contradiction to [E03203] : 1. Hence |[Za,Za,]| 7': 3. If IZGIQa/Qal = 3 then |Zai/Za: 0 Q0] = IZai/Cza,(Za)| S 3 and so [Zm Zai] = ICZO,(Za)i| = 3, a contradic- tion. Thus, lZa’Qa/Qa] 2 32° Since [Zm Zai, Zai] = 1 we have by the choice of a" that llZmZa’ll : 32 = CZa(Za’) ié E016- Now Q3 normalizes Z0: and hence it also normalizes Cza(Za:). Hence Q3Qo, normal- izes Cza(Za:). But the only S-invariant subgroup of order 32 in Z, is E03. Hence QaQ3 75 S which means (recall that from 3.19, QaQ3 aé 5' implies {La/Q0, L3/Q3} 2’ {(P)3P4(3),(2)M12)}) IQaQa/QaKlS/Qal = 33- Thus lQaQn/Qal S 32- Now look at Q0, 0 Q3; it centralizes [Zm Zai] = CZa,(Za) and the latter has order 32. Hence by 4.5 [62. 0 Q3, Zai] g [20,, 20m = [20,, 2...]. Let Q; = Qa/Za. Then |W, Zaiil = 1 and so lm/C-T,(Za')l S lQaQfi/Qfil S 32 S lZa’Qa/Qal 34 . So there exists a unique non-central composition factor and it is an FF-module isomorphic to Za (uniqueness of the FF-module). Now CE(ZGI) = m S 00,3 but on the other hand Cza(Zar) is not normal in 00,3. So we found one FF-module in which the centralizer of Z0: is normal in G03 and another F F-module in which the centralizer of Z0: was not normal in 003, a contradiction to the uniqueness of FF-modules. Subcase 2: E03 S Z(V3). Define W3 = S Z(V3). In particular, W3 S Q, and W3 is abelian. Also W3 has a non-central L3 chief factor since E03 is not centralized by 03(L3). Hence [W3, 03(L3)] # 1. Choose again a — 16A. Let’s also note that [WatQal = <[Ea3,Q3]G’3> S Za- If Wa_1 S Q3 then, as W3 S Q0, S 00-1, we get [W3, W04, W04] S [Wa_1, Wa_1] = 1 contradicting the 3-stability of ,6. Hence W04 S Q3 S Gal. So W04 normalizes Z0: 0 Q0 and therefore Z0. 0 Q0, is quadratic on Wed. Then 3-stability of La..1 gives Z0, 0 Q, S Q04. Since [Wa_1, QO,_1] S Z04 we now get [Wa_1, Z0: 0 Q0] S Z04. In particular, [Wa_1,Eai3] g 2..-, n 23 = 1. Hence by 4.5 [Za:,Wa_1] S E5153 2 E033. Thus [Wa_1, 2w n Qal S Za—l fl Ea’fl S Cza-i(Za') = 1 as E013 S Z(V3) and |Za_1| = 3. Hence [Wa_1, Z0: 0 Q0] = 1 which implies that [Za’aWa—I] S (20’ n Q01)-L : [ZaaZa’] .<_ Za- 35 This means that Wa-lZa 3 = Ga and therefore [War-la Q01] 3 Ga- But [W -1,Qa] 76 1 (since if WW1 centralizes Q0, Q0, S QO_1 and 3.14 imply [Wa_1, 03(La-1)] = 1, a contradiction). Now since |Z0_1| = 3 we get 20-1 S [Wa_1,Qo,]. On the other hand, [WG-17 Q01] 9. Ga and therefore 20 S W04. Hence Wa_1Za = WG—1 3 (Ga, Ga..1>, a contradiction. 6 The case Z0, S Q0, In this section we will deal with the case ZO,’ S Q0. We will show that b=1 and start the analysis of the case b=1. It follows from the hypothesis that there is no symmetry between a and a’ any more. Also [Zm 20:] S [20,, Q0] = 1 gives CZO(ZO,:) = Z0. Now notice that Z, 0 Q0: 7E 20, (otherwise get 20, S Q03, a contradiction). Hence, Cza(Zo,:) # Z0 0 Q0: and by 3.13, Za: S Z(Lat), a and a’ are not conjugate and b is odd. Therefore we have Z3 = QlZ(L3) and Z0. .—_ QIZ(LO,:). Lemma 6.1 [Q3, 20,, Zn] = 1. Proof: 3.7(b) gives Q3 S GS) S Ga. Hence, S = 20 which gives [QfiaZaa Za] S [Zm Z0] =1. 36 Lemma 6.2 Lori/Q0! '5 Lg/Qg '52" (P)Sp4(3), 5112(3), 5142(9), 21+4A5 07‘ 2A5. Proof: If b>1 then [Va:,Za,Za] S [Va:,V3,V3] S [V3,V3] = 1 by 3.10, so, since Z, S Q01, we conclude that L3 / Q3 is not 3-stable and the claim follows by 2.5. If b=1, 6.1 and Zor S Q3 again imply that L3/Q3 is not 3-stable and the claim follows by 2.5. Notation 6.3 For 76F let D, = CQ7(03(L7)). Lemma 6.4 Z(La) = Do, =1. Proof: Since 91(CQO(LO)) S Z3 S 91(CQO(S)) S QIZ(S) S Z(L3) we get that 91(CQO(LO)) is centralized by La and L3 and therefore CQO(LO) = 1. Hence CQO(O3(La)) = Do =1- Also, 912(La) S 00:1 and therefore Z(LO) = 1. Lemma 6.5 Q3 is not abelian. Proof: By 3.18, Q3 S Q0. So 3- stability of L0, gives [Za,Q3,Q3] # 1. Hence 1 75 [[ZaaQBlaQfil _<_ Q'g- Proposition 6.6 b=1. Proof of the Proposition: Assume that b>1. Since b is odd, I) Z 3. 6.6.1 [Vg fl Qa’, Val] 2' 1. Proof: Since b2 3, 3.10 implies [V3, V3] 2 1. Clearly V3 (1 Q0. centralizes 20.. Let 66A(a'). Since d(6,fl) S b we get Z; S G3 and since V3 S G3 we now have that 25 normalizes V3. Then [Z5, V3 0 Qa': Vfi n Qa’l S [Z59 VB, V3] S lVfiv Vfil = 1° But L5 is 3-stable as 6 is conjugate to a and Z5 S G5 and V3 {'1 Q0. S G; (since d(a’,6) = 1 so V3 0 Q0: S Q0, S G9,) S G5). Therefore, [25, V3 0 QOI] = 1. 37 6.6.2 L3/Q3 ’—‘_—’ 5124(3), SL2(3) or SL2(9) and V3 has a unique non-central L3- composition factor; moreover, this composition factor is the natural module for L3/Q3. Proof: 6.6.1 gives V3 (1 Q0: S Cvp(Va,) and by a similar argument we also have that Va, 0 Q3 S GVa,(V3). Without loss of generality, assume lVfiQa'/Qa’l S IVa'Qfi/Qal- Now let XzY/Z be a non-central chief factor in V3. As CY(Va')Z/Z S CY/Z(Voz’) we get that lX/CX(Va’)l = lY/Z/CY/Z(Va’)l S |Y/Z/CY(VO,.)Z/Z| = lY/CY(Va’)Zl S ly/CMVa'H = IY/Y fl CV3(Va’)l = lY'CV3(Va’)/CV3(V0’)I S lVfi/Cv3(Va')l S lVfiQa’/Qa’l S IVa'Qa/Qal so X is an FF-module; similarly, the direct sum of the L3 chief factors on V3 is still an FF-module for L3 / Q3 and lemma follows by 2.8. 6.6.3 [V3,Q3] S D3. Proof: Assume that [V3,Q3] S D3. Then by 6.6.2, Za[V3,Q3] is normalized by G0303(L3) = G3 and we get that Za[V3,Q3] = V3. Hence V3/Zo, = [V3/ZmQ3]. Since Q3 is a 3-group acting on the 3-group V3 / Z, in the above manner, we conclude that V3 /Zo, = 1. Therefore V3 = Z0, a contradiction. Hence [V3, Q3] S D3. Notation 6.6.4 Let Q}, = [Q3,03(L3)]. 6.6.5 Q}, g Q... 38 Proof: By 6.6.3, [V3,Qf3] S [V3,Q3] S D3. Note that Q}, S 03(L3) and therefore [V3,Q3,Qfal S [DanBl S [03,03(L3)l = 1- Hence [Zm Q5, Q3] = 1 and 3-stability of L0, gives that [Zm Qfil = 1 whence Q; S Q0. 6.6.6 The hypothesis that b>1 gives a contradiction. Proof: By 6.6.5, Q2, centralizes Z0, and so it centralizes = V3 as well. Since [t3,Q3] S Q5, t3 is the unique involution in t3Q3/Qg and so t3Q5EZ(L3/Q3). In particular, L3 normalizes [V3,t3]. By 6.6.2, [V3,t3] 75 1 and so C[vfi,tfi](S) 7t 1. Hence Z3 fl[V3,t3]7£ 1. On the other hand, since by 3.10 V3 is abelian, Va = CvfiUa) X [Vain] and [Z3,t3] S [Z3, L3] = 1, a contradiction. Notation 6.7 For 76F let F, be a normal 3-subgroup of L, minimal with respect to the property F, S D,. Remark 6.8 As F, is a 3-group we get F, S Q, and F4 # F,. Also, the definition implies F, S 1. Since Q, is a 3-group acting on the 3-group F,, F, ¢ [F,,Q,] and by minimality of F,, [F,,Q,] S D,. Also it is clear from the definitions that F3 = [F3,03(L3)] S 03(L3) and therefore [D3,F3] S [D3,03(L3)] = 1. Lemma 6.9 F3 S Q0 and D3 S Qa. Proof: If F3 S Qa, [F3, Za] = 1 and by 3.14 [F3, 03(L3)] = 1, a contradiction. By 6.8 we have [D3, F3] = 1 and since [Zm D3] S D3, [Zm D3, F3] =1 39 and [20,193, ] = 1. Suppose now that D3 S Q0. By 2.12, B is irreducible on Z(S/Qa) and so Z(S/Qa) S DfiQa/Qa S LaQa/Qa- Similarly, Z(S/Qa) S Qm Hence [Za,Z(S/Qa),Z(S/Qa)] =1, a contradiction to the 3-stability of La. Thus D3 S Qa. Lemma 6.10 Q, is elementary abelian, [Qa,03(LO,)] is an irreducible La-module and F0, = Z, = [Qm 03(La)]. Proof: Step 1: F,, (1 F3 S D3. Proof of step 1: g g g F,,. Similarly, 5 F3. Hence [Fm F3] S F0 (1 F3. Assume now that F0, 0 F3 S D3. Then, since Fri = [Fna03(La)l S 03(Lfi) we get [Fa,F3,F3] S [D3,F3] S [D3,O3(L3)] = 1. But F3 S Q3 S Go, and by 6.9 F3 S Qa. Hence [Fa,F3,F3] = 1 and 3-stability of L0, gives [Fa,03(La)l = 1, a contradiction to the definition of F0. Step 2: [F,, n F3173] S La and F3 _<_ F,,. Proof of step 2: F,,, n F3 S D3 implies <(Fo, fl F3)Lfi> S D3. And since <(Fa n F3)Lfi> 51 L3 minimality of F3 gives F3 S <(FaflF3)Lfi>. Clearly the other inclusion is also true so F3 = <(Fo, fiF3)Lfl>. Now F5 S L3 and F5 ¢ F3 so minimality of F3 gives Ff, S D3. 40 This now means that F,, = [F3,F3] = <[FanF3, F3]’+B> g D,. = CQ,(03(L,3)). Hence [For n F3,F3] is centralized by 03(L3). Since [F0 0 F3, Fol _<_] S and 03(L3)S 2 L3 we get [Fa n F3, F3] 51 L3. Thus F,, = [F,, n F3, F3] 3 Fa. Step 3: Q, is elementary abelian and [Qo, 03(Lall = F,,. Proof of step 3: Since S = F3 we get [Qananol S F3 S Fo- Then 3-stability of a gives [Qa/Fa,03(La)] = 1 i.e. 03(La) centralizes Qa/Fa. So, Q, has a unique non-central chief factor. By the properties of the Frattini group (for example see [Go; p.173]) we get that @(Qa) Q DcJr = 1. Hence Q, is elementary abelian. Step 4: [Qm 03(La)] is irreducible La-module and F0 = Z0. Proof of step 4: Since D0, = 1, Z, is the unique non-central chief factor for L, on Q0; moreover, by Gaschiitz’ Theorem, Z, = Z3Fo, = QIZ(LO)FO, 2 Fa. Corollary 6.11 Note that from 6.9, D3 S Q0, and so (D3) = 1. Corollary 6.12 CGO(Q0) = Q0. In particular, ifX S G, then X D QC, = CX(QO). Proof: By (P2), CGO(QO,) S Q0. But as Q0, is abelian we get Qo S CG.(Qo) and therefore the claim follows. Lemma 6.13 If L3/Q3 2’ (P)Sp4(3) then La/Qa ’5 (2)M12. Proof: Q, is abelian implies QOQ3 / Q3 is abelian. If L3/Q3 ’5’ (P)Sp4(3) then the group S/Q3 is not abelian and we conclude S S QaQ3. But then 3.19 gives La/Qa E’ (2)M12. 41 7 The case b=1 and 92(2)M12 Proposition 7.1 US 2 QaQ3 then (L0,,L3) ~ (362-M12,31+1+1+2+2+ISL2(3)). Proof: Suppose that QaQ3 = S. Then by 6.10 S' / Q3 is abelian and therefore La/Qa $73 (P)5P4(3)- Hence by 6.2 L3/Q3 ’5 SL2(3),2'A5,21_+4A5, SL2(9). Also from anfi = S we get [FoQo/Qe. 5] = [FoQo/Qo. Qo] and as [F3, Q3] 5 D3 5 Q. (see 6.8 and 6.9) we conclude that [FoQo/QmQfil = 1- Hence F3Qa/Qa S Z(S/Qa). Since IS/Qal = 33 and S/Qo, is not abelian we get that |Z(5/Qo)| = 3 and therefore F3Qa/Qo, = Z(S/Qa). But F3 S Go, and therefore 6.12 gives lFfi/CF3(Qo)l = 3- In particular F3/D3 is an FF-module for L3/Q3 so 2.8 implies that L3/Q3 $3 2oA5,21_+4A5. If L3/Q3 E’ SL2(9), then by 2.10, F3/D3 is a natural SL2(9)-module, a contradiction to IFB/CF3(Qa)l : 3° SO, Lfi/Qfi '5 514(3)- 42 Now [Q3, Q0, Q0] 2 1 and PSL2(3) is 3-stable imply that t3 acts non-trivially on each non-central chief factor of L3 in Q3 and therefore it inverts every non-central chief factor of L3 in Q3. Also t3 inverts F3Qa/Qo, = Z(S/Qa); in particular t3 acts non-trivially on S = S/Qa. Then by [Go; p.173], t3 has to act non-trivially on S/¢(S). If t3 completely inverts S/(S) then, since t3 also inverts (NS) = Z(S), it completely inverts S. Since a fixed point free automorphism of order 2 of a group implies that the group is abelian we get that S is abelian, a contradiction. Therefore |[S,t3]Qo,/Qa| = 32. Recall that by 6.6.4, Q; = [Q3, 03(L3)]. Then, as t3 inverts each of the non-central chief factors we get that lQEQa/Qal = 32 and so |Q5/GQ73(QO)| = 32. Hence L3 has exactly two non-central chief factors in Q3. Moreover, @(Qg) S Q0, and so [@(Qg),Za] = 1 and (Qg) S D3 by 3.14(a) applied to . Put 627, = (23/03. Since Q2} acts trivially on a; and (see proof of 6.6.6) t3Q3/Q36Z(L3/Qg) we have Q}: = C5303) x [Q, t3] and L3 normalizes 05;;(t3). Now since t3 inverts all the non-central chief factors in Q3, 052%) S C5;(03(L6))=1- Thus Q} = [Q3, t3] has order 34. Let E = 04,463). 7.1.1 CE(t3) S D3. Proof: First notice that C'Qp(t3) normalizes CE(t3). Now if the claim is not true, pick F S CE(t3) with [F,Cqfl(t3)] S D3 and F S D3. Since by Frattini argument L3 = Q3CLfi(t3), a composition series for L3 in Q3 is also a composition series for CLfl(t3) in Q3 and we conclude that [CQfi(tfi),03(CLB(tfi)] = 1. Hence FD3/D3 is centralized by C'Qfl(t3) (by choice of F), Q}: (by choice of E) and 03(CLfi(t3)). 43 As Q3 2 QBCQfi(t3) and L3 = Q3CL6(t3), we conclude 03(La) S QECQe(t6)03(CLe(t6))- Hence [Fa03(LB)l S- DH and [Fa03(L6)a03(Lo)l S [DB,O3(LB)1= 1- Now we have a group generated by 3’ elements (03(L3)) acting quadratically on a 3-group (F); thus [Fa03(Lfi)l = 1 which implies F S D3, a contradiction. Hence CE(t3) S D3. 7.1.2 Q0, (1 Q3 S E. Proof: Recall that by 6.10 QC, is abelian and therefore We 0 QoaQo n QEl =1 which gives [(QO, (1 Q3)Qg, Q33] = 1 (where Q33 = Q0 (1 Q5). But (Qo n QolQZ} S Go603(Lfi) = Go; thus [(62, n Q3)Q;,, = 1. Since 22?, = [ageing] = 4523,6246» and [Q3,Qol S Q33 we get = Q73 and therefore we get [(QO, (1 (wk-3,55 : 1. Then [Q0 0 Q3,Qvf3] = 1 and the claim follows. D Now E = CQfi(Qg) = CE(t3)Qg = D3Qg by 7.1.1 and so by 7.1.2, Q0 0 Q6 S DfiQE 44 Since [Q3, Q0, Q0] = 1, [Q3, Q0] S C5;(QO). As Q} has two non-central chief factors. IQE/C;,(Q.)I 2 32 and “63.6.1 2 32. From ngl = 34 and [QéQa] S Q30, S C5:§(Qa) we conclude [Q3,Q0]D3 = Q0, 0 Q3 and IQa fl Q3/D3I = 32. Since IQa/Qa fl Q3| = 3 we finally get that IQa/D3l = 33. Hence IQa/CQO(Q3)| S 33. Now since by 2.14 L0, = Qa for some gEGa, we get IQa/CQO(LO)| S 36. By 2.10, only 2-M12 has an irreducible module of dimension less than or equal to six. Moreover, this module is unique and its dimension is actually six. Hence, La/Qo, E 2-M12 and [Q0] = 36 and therefore we also get that |D3] = 33, ISI 2 39 and |Q3| = 38. It is clear now that since Q0, is an irreducible elementary abelian normal subgroup of L, of order 36, Lo, ~ 362M”. Reviewing, |D3I = 33 and D3 is central for 03(L3). Also IQEI = 34 and Q}, has two composition factors each of dimension 2. Finally, [Q3] = 38 and so |Q3/Qf3l = 3. Thus, L3 ~ 31+1+1+2+2+15L2(3). Proposition 7.2 US S QOQ3 then (L0,,L3) ~ (362-M12,31+4Sp4(3)). Proof: Suppose that S S QaQ3. Then by 3.19 and 6.13, La/Qo, 9.“ (2)M12 and L3/Q3 E“ (P)S'p4(3). Therefore IS/Qal = 33 and [S/Q3I = 34. Hence, lQol/lQal = 3- Then lQfi/CQ6(Qa)l = lQa/Qo n Qal = IQoQo/Qol S lQaQfl/Qfil' Hence all composition factors for L3 in Q3 are FF -modules for L3 / Q3. Thus by 2.8, L3 / Q3 ~ 5134(3) and L3 has a unique non-central composition factor in Q3; moreover, 45 this composition factor is a natural module. In particular, (Q3) S D3 and so by 2.10 [Q3/D3, t3] is a natural Sp4(3)-module for L3/Q3 and CQfi/Dfl(t3) = CQfi/DB(L3). Hence [CQeUBLLmLol = 1, 005056) = Do and [QB/Dfil = 34- Thus, as t3€03(L3), Q6 = [Qm 03(LollDe and since L3 = 03(L3)Q3 and [Q3,03(L3)] S 03(L3) we get that L6 = 03(LellQ6, 03(LollDa = 03(LfilDB- By 6.11, D3 is elementary abelian and so [DaaLel = [D6s03(Lfi)Dfil = 1- Hence D3 = QIZ(L3) = Z3. NOW lQa/Zfil = 3'lQfi/Zfil = 3'lQa/Dnl = 3'34 = 35- Pick Z3 for some gEG and so lQa/CQo(Lo)l S 35-34 = 39. Since CQa(La) = 1, IQal S 39. By 6.10, Z0, is the unique non-central chief factor for L0, in Q0. From 2.10 now we get that La/Qo, ’5 2-M12 and [Z0] 2 36. Furthermore, Q0, 2 CQa(ta) x [tha]. But CQa(to,) S D0, = 1. Since Z0, is the unique non-central chief factor for LC, in Q0, [Qo/Zo,03(Le)l = 1 46 and IS [tha] S 20,. Irreducibility of Z, yields now that [Q09 to] S Za- As Q0, is an irreducible elementary abelian normal subgroup of LC, of order 36 we now get that L0, ~ 362-M12. Also, |Q3| = IQOI/3 = 35 and as [Q3/D3l = 3“, |D3| = 3 and so we get L3 ~ 31+4Sp4(3). 8 The case 13:1 and GE’PSL2(9) or M11 In this section, 6') ’_—‘_’ PSL2(9), M11 and b=1. Notice that by 6.13 ‘11 S (P)S'p4(3) and therefore by 6.2, \II ’5 SL2(3), 2-A5, 21_+4A5, SL2(9). Recall also from 3.19 that S' = QOQ3. Moreover [Zm Z03] = 1. Remark 8.1 Since a Sylow 3—subgroup of G) is elementary abelian we have ¢(Q6) S Qa- Similarly (Qo,) S Q3. Lemma 8.2 IfN S S,N S B,6E{a,fl} then N S Q5 or NQ5 = 5'. Proof: It follows from irreducibility of B on 5/Q5 (see 2.12). Corollary 8.3 S = ZQQ3. Proof: It is an immediate consequence of 8.2. 47 Lemma 8.4 Let Xg = n6EA(fi) Q6. Then: (a) Q3/X3 is an irreducible G3-module, (b) lQa/DaJal = Qfi/DB and CQa/D3(tfi) = 1, (C) 00.903) S De and (d) X3 = 03. Proof: Let X3QeaQ6l S A. Let Q3 = Q3/X3. Then Q3 is abelian. Now Q; = €73,303) x [Q3,t3] and both parts are normalized by L3. If 05:3(t3) S 1, we may assume A = CQB(t3)X3 (since then A S Q3, A S G3 and as Ca‘;(t3) S 1 we also have X3 S A). Hence 4 = Cam) and get [[Q3,t3],t3] S [[L3,Q3],t3] S [A,t3,t3] = 1. Hence (element of order 2 acting on a 3-group) [Q3,t3] = 1 a contradiction to [Q3,QmQa] = 1 and the 3- stability of L3/. Therefore C673(t3) = 1 and Q3 = [Q3, t3] = [@,L3]. Thus Q3 S [L3, Q3] S A which implies A = Q3 and Q3/X3 is an irreducible G3-module. ‘ Now by 6.9, D3 S Q0, and as D3 S G3 we get D3 S X3. But [XfiaZal S [Q07 Zol = 1 and Z, S Q3 give X3 S D3. Hence X3 2 D3. 48 Lemma 8.5 There is gEG3 such that t3EQ3. Proof: If ‘1! E SL2(3) or 2-A5 it is clear since in these cases L3 = Q3 for some gEG3 and t3EL3 by definition. Since inside 3112(9) we can generate a 2-A5 this case is also clear. The case 2L+4A5 is left. Let a, b be two elements in L3 / Q3 of order three and H = be such that 21+4H = 21+‘A5. The possibilities for H then are 21+4A5, 2-A5 or A5. In the first two cases t3Q3EH and we are done and the last case can not happen as A5 is 3-stable and Z0, acts quadratically on Q3. Notation 8.6 Q: = Q,/D,. Lemma 8.7 IQ—3l = 3“. Proof: By 8.5, pick g 6G3 such that t3EQ3. Since lQe/CQ,3(Za)l = lQaQa/Qal = lS/Qal = 32, we get lax/0530a» s 34. By 8.4(b), Ca(t3) = 1 and therefore IQEI S 34. Suppose lQ—3I<34. Since 5 does not divide IGL3(3)| we conclude that Lfi/Qfi '5 SL2(3)- From 8.4(a) and 2.10, |Q—3| = 32 and so |Q3/[Q3, QO]D3| S 3. Since [Q3, Q0]D3 3 Q0, lQfiQa/Qal S 3 and S 7é QaQB since IS/Qal = 32, a contradiction. Hence Iml = 3“. Lemma 8-8 llzaffill = lQaal = lQa fl Zal = 9- 49 Proof: If |[ZO,Q_3]| = 3, then, with same argument as before, we get IGEI = HQ—stsll s 32, a contradiction. Hence 9 s HZmQ—ell s |an Zal 3 I323}! s 9 and lemma is proved. Lemma 8.9 D3 = Z3. Proof: First, show 03 g 23. Let L = . Then by 3.14(a), 03(Ls) _<_ L and L3 = LQ3. Since Q3 is irreducible for G3 we get [Q3, L] = 1 or Q3. If [Q3, L] = 1 then [Q3,L] 3 D3 so [Q3,03(L3)] = 1, a contradiction. Therefore [Q—3,L] = :27, which gives [Q3, L]D3 = Q3. Also, as L S G3, we have Q3 S NGfi(L). Hence [Q3,L] Q L, Q3 S D3L and L3 = LD3. But from 6.11 now, [D3,D3] S (D3) = 1. As D3 S Q0, [L,D3] = 1 so 03 and L both centralize D3 But then, we also get [D3, L3] = [03, L03] = 1. Thus D3 S Z(L5) _<_ Z3. Therefore D3 S Z3. Since Z3 2 QIZ(L3) S CQ5(03(L3) 2 D3 the lemma follows. Lemma 8.10 Q0, 0 Q3 = Z0, 0 Q3. Proof: It is enough to show that Q0, 0 Q3 S Z, 0 Q3. Let xEQa 0 Q3. Then xD3EQa fl Q3/D3 = Q; = Z0, 0 Q3 = Z0, 0 Q3/D3. Therefore, xD3 = yD3, where yEZo, 0 Q3. Then a: = yd, dED3. 3.11 gives Z3 S Za. By 8.9, D3 = Z3 S Z0. Therefore :cEZa and hence xGZo, fl Q3. Corollary 8.11 Q0, = Z0. 50 Proof: Since Q0, C S = ZaQ3 we get Q0, 9; ZaQ3 fl Q0 2 Za(Qa 0 Q3) and hence Q0 = Za(Qa n QB) : Za- Lemma 8.12 (1) Q: = Z. is irreducible, (2) ma 9—1 PSL2(9) and 11: g 51.2(3), 2-A5 or 21_+4A5 then |Za| = 34, |Z3| = 3 and lQal = 35, (3)1f9 g PSL2(9) and \II ’—_‘—’ SL2(9) then IZal = 36, |Z3| = 32 and |Q3| = 36, (4) If(-) 2:“ M11 and \II E“ SL2(3), 2-A5 or 2L+4A5 then IZOI = 35, IQ3I = 36 and lZBl = 32, (5)1f0 2 M11 and \II 9:” SL2(9) then then IZal = 35, |Q3| = 35 and |Z3| = 3. Proof: 3.11 and 8.9 give D3 = Z3 S 20,. Hence lZa/Zfi: = IZaQa/Qallza n Qfi/Za n Dfil = lZaQfi/Qfillza fl QB/Zfil- Recall now 8.8 to get IZO, fl Q3/Z3I = 32 and hence lZa/Zfil = 32lZaQfi/Qfil = 32|5/Qa|o Since S/Q3ESyl3(\II) we get that IS/Qal = 3 if ‘1’ S 5L2(9) and |S/Q3l = 32 if ‘1! E” SL2(9). Hence if ‘1! S SL2(9) then IZa/Z3I = 33 and if \11 E“ SL2(9) then IZa/Z3I = 34; in particular, IZa/Z3l S 34. Since by 2.14 we can generate LG by two Sylow 3-subgroups we get IZOI S 38. By 6.10, Za is irreducible as La-module. Case 9 ’5’ PSL2(9): Then by 2.10 IZGI = 34 or 36. Moreover, if |Za| = 34 then |Z3| = 3 and therefore IZa/Z3I = 33 and \II S SL2(9) and if |Za/Z3I = 36 then 51 |Z3| = 32 and therefore |Za/Z3I = 3“ and \II S SL2(9). Case 6 3 M11: 2.10 gives that IZOI = 35 and |Z3| = 3 or 32. If |Z3| = 3 then IZa/Z3I = 3“ and ‘II ’5 SL2(9) and if |Z3| = 32 then |Za/Z3I = 33 and ‘II S SL2(9). Corollary 3.13 (1) ma z PSL2(9) and 111 g SL2(3) then (L0,, L3) ~ (34PSL2(9),31+2+2SL2(3)). (2) Us a PSL2(9) and 111 g 2-A5 then (La, L3) ~ (34PSL2(9),31+42-A5). (3) Us 2 PSL2(9) and \p 9: 254/15 then (Lan) ~ (3‘PSL2(9),31+421_+4A5). (4) Us g 1951,2(9) and 1p 2 SL2(9) then (LmLfi) ~ (36PSL2(9),31+1+4SL2(9)). (5) Us 2 M11 and q: a SL2(3) then (LmLfi) ~ (35M11,31+1+2+2SL2(3)). {6) rm 2 M11 and 11: g 2-A5 then (L0,, L3) ~ (35M11,3‘+1+42-A5). (7) ma ;—v. M11 and 111 2 514(9) then (L0,, L3) ~ (35M11,31+4PSL2(9)). Proof: By 8.12, Q0, = Z0, is an irreducible elementary abelian normal subgroup of L0,. Moreover, lQal = 341'}r o g PSL2(9) and 11: a: SL2(9), IQQI = 36 if e g PSL2(9) and q: 2 514(9) and lQol = 35 2if 6 ’5 M11- Thus, the structure of L0, is as given in the corollary. Notice now that in all the above cases, D3 is central as by 8.9 we have D3 = Z3. Moreover in cases (1), (2), (3) and (7), |D3| = 3 and for the rest of the cases we have IDal = 32- Finally, in all the cases, |Q3/Z3I = 3“ and hence Q3/Z3 is an irreducible L3- module whenever ‘1! S SL2(3) which proves (2), (3), (4), (6) and (7). By 8.4 though, t3 inverts Q3 / Z3. Since by 2.10 SL2(3) has a unique faithful irreducible GF(3)-module which is of order 32, (1) and (5) follow. 52 Lemma 8.14 The case (L0,, L3) ~ (35M11,21_+4A5) is impossible. Proof: Let L3 = L3/Q3. Since 02(L—3)/<5> is the even permutation module, S centralizes a group D of order 8 in 02(L—3). It is easy to see that D S D8. Let D' be the inverse image of D in L3. Then [D',.5'] S Q3 S S and so D S N045) = B. Now recall the definition of K from 3.16 and let D = K D D‘ and pick tED \ with |t| = 2. Since t3 inverts Q0, 0 Q3/Z3 and = [t,D], t neither centralizes nor inverts Q0, 0 Q3/Z3. Since IQa fl Q3/Z3l = 32, HQ, 0 Q3/Z3,t]| = 3. Now 12m] 5 12M] = 1 and [est] 3 [omBinlsz D] s QeflQa and we get llQeJJI = 3. Similarly, llthfill = ”Q: n Qfi/Zfiatfill = 32- Since M11 has no outer automorphism and only one class of involutions, there exists gELa so that [t9t3, L0,] S Q0. Since Q, is an irreducible La-module, t9t3 centralizes or inverts Qa. 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