S56 iz. LARGER GROUPS OVER F2 IN Cf (G, T)12.8.10.6 that C z9 u (U) = Vf, so that Z[r ~ Zu /Vi ~ Zu, completing the proof of
(3)..
Suppose (4) fails; then Zu -:/= 1 but Z[r centralizes U. First n > 2 as there
Zlr ~ Zu by (3). Next by 12.8.10.6, Zlr = Czr)U) ::::; U; so by 12.8.10.5, Zu =
Vi x Z(lz) with [L, Z(Iz)] = 1. Then Nc(Z(lz)) ::::; M = !M(LT). Let J :=
(Lf) and suppose Li < J. Now Li ::::; CL(Vi) = CL(ViZ(Iz)) = CL(Zu), so
J::::; CH(Zu)::::; NH(Z(Jz))::::; Mi. If n > 3 then by 12.2.8, J::::; 0
31
(H n M) =Li.contrary to assumption. Hence n = 3, so Li/Oz(Li) ~ Z3. Then since M is an
SQTK-group and J = (Lf), J/Oz(J) ~ E 9 and J = Ldc, where Jc:= oz(C1(L).
As Li = [Li, T n L], Li and Jc are the only T-invariant subgroups Yi of J with
IYi : Oz(Yi)I = 3. Thus His not transitive on the four subgroups Y of J withJY: Oz(Y)J = 3, and we conclude IH: NH(Li)I = 3 and Jc :::! H. But Jc :::! LT,
so H::::; Nc(Jc) ::::; M = !M(LT), a contradiction. Therefore Li= J :::! H. Thus
by 12.8.5.1, Co(Li) = 1. However by hypothesis Zu -:/= 1, and we had seen that Licentralizes Zu. This contradiction establishes ( 4).
By 12.8.9.l, ]z :::) Gz, and a Sylow 3-subgroup of Iz is faithful on 1/2. Thus if(8) fails, then CH(V2) contains Y ~ Eg and m3(Gz) ~ m3(IzY) > 2, contradicting
Gz an SQTK-group. So (8) is established.
Define Hz := 031 (H n Gz) and observe that Hz= 0
31(CH(Vz)). We see that
if m 3 (C.H(Vz)) > 1, then by (8), 0
31(CH(Vz)) <Hz, so Hz does not centralize Zu
by Coprime Action. Hence Zu >Vi, so Z[r-:/= 1 by (4) under this assumption.
Asume the hypotheses of (5). Then by 1.2.1.1, there is K E C(H) with
k = F*(H), so by 1.2.1.4, K/Oz(K) ~As. Then K = 0
31(H) by A.3.18. Next
031 (C.H(Vz)) ~ E 9 /E 16 , so by the previous paragraph, [Zu,K]-:/= 1 and Zlr-:/= 1.
As Zlr-:/= 1, K = [K,Z[r]; so as [Zu,K]-:/= 1, Z[r i CH(Zu). Then 1-:/= [Zu,Zlr]::::;
Zu n Z[r = Z(Iz) by 12.8.10.2. Thus Zn := Zn Z(Iz) -:/= 1 by 12.8.10.3. There-fore n > 2 by (3), so Li -:/= 1 and Li ::::; 0
31(H) = K. Thus if [Zn, K] = 1, then
LT= (lz,LiT) centralizes Zn, so K::::; Cc(Zn)::::; M = !M(LT). This is impossi-
ble as Li :::) Mi, but Li is not normal in K as K/Oz(K) ~As. Thus [Zn, K]-:/= 1.
As Vn E Rz(KT) by B.2.13, KE £1(G, T), so (5) holds.
Assume that (7) fails; thus F* ( H) = k x kx for x E H -NH (K), U = [U, k] EB
[U, kx] with [U, k] the A5-module fork~ Lz(4), and X = (±)(X n kkx) ~Es.Then oz(Cj 1 (Vz)) is a Borel subgroup of E(H), and hence of 3-rank 2, so Z[r-:/= 1
by an earlier observation. On the other hand, in this case m(X) = 3 and m(U) = 8
so that m(E) ::::; 4 as E is totally isotropic by 12.8.11.2. Therefore by 12.8.11.5,
m(E) = 4 and m(W^9 / Z[r) = 3, contradicting Z[r -:/= 1. So (7) is established.
Finally assume that (6) fails; that is, F*(H) ~ A 7 and U the 6-dimensional
permutation module. Then U is described in section B.3, and we adopt the notationof that section. By 12.8.11.5: ·
m(E) = m(U) - m(X/Zlr) -1 ~ 5 - mz(H) ~ 2.
Thus E > Vz as m(Vz) = 1, so we conclude from 12.8.11.3 that 112 = [E, X] ::::;
[VzJ_, T] ::::; V;J_; it follows that a generator v for 112 is not of weight 2 or 6, so that v
is of weight 4. Hence, in the notation of section B.3, we may choose v = ei,z,3,4, soV;J_ = { e1: JJI and JJ n {1, 2, 3, 4}1 are even}.