13.4. THE TREATMENT OF THE 5-DIMENSIONAL MODULE FOR A 6 909As L/02(L) ~ A6 by (3), R = Ri. Also either U E Irr +(K, R 2 (KT)), or
K* ~ L 3 (2) and U is a sum of two isomorphic natural modules. Suppose that the
latter case holds. Again R is the unipotent radical of the parabolic CK (Zn U)T!K
fixing a point in each summand of U, so we can finish much as in the proof of (1):
R contains no FF-offenders on U, so S ~ KTK by B.2.10.1. Then no nontrivial
characteristic subgroup of Sis normal in LT, so Lis an A 6 -block by 13.2.2.7, and
L 1 has exactly two noncentral 2-chief factors. This is a contradiction since L 1 :::; K
by (2), so that L 1 has a noncentral chief factor on each summand of U, plus one
more on 02(LJ:).
This contradiction shows that U E Irr +(K, R 2 (KT)). Thus from earlier re-
marks, VH = UCz(K) and U is the natural module for K* or a 5-dimensional mod-
ule for K* ~ A5. In particular, Z = (ZnU)Cz(K). Next CzL (K) = 1, as otherwise
K:::; Ca(CzL(K):::; M by 13.4.2.2. By 13.4.3.2, IZ: ZLI = 2, so ICz(K)I:::; 2, and
hence Cz(K) = (z). In particular if K* ~ A6 and m(U) = 5, then Cu(K) = (z),
establishing (4). Also IZ n U : Cznu(K)I = 2, so as Z = (Zn U)Cz(K) and
Cz(K) = (z), (5) and (6) hold.
Using (3) and 13.4.6.5, M = LT so Ca(Z) = LiT. Let WH := (ZH) and
UH := [WH,K]. By 13.4.5.4, 02,z(K) = CK(WK) = CK(UH) = CK(WH)· As
K = [K, J(T)], Theorems B.5.1 and B.5.6 say that either UH E Irr +(K, WH),
so that UH = U, or K* ~ L3(2) and UH is the sum of two isomorphic natural
modules for K* ~ L 3 (2). Assume the latter holds. Then as K is irreducible on
U and 02(H/CH(VH)) = 1 by B.2.14, AutH(UH) = L3(2) x L2(2) and UH is the
tensor product module. Then AutR(UH) contains no FF*-offenders, so as in earlier
arguments we obtain a nontrivial characteristic subgroup of R normal in KT and
M, a contradiction. Thus UH= U.
By A.1.41, CH(K/0 2 (K)):::; CH(U), so as Z = (ZnU)(z) and Out(K/0 2 (K))
is a 2-group, H = KTCH(K/02(K)) = KCH(U) = KCH(Z) = KL1T =KT since
L 1 :::; K by (2). Since Gz satisfies the hypotheses for H, we conclude Gz =KT= H.
Thus (7) holds since K f._ M.
Define H 2 and Ho as in (8), and let Hi := L2T· Observe that the hypotheses
of Proposition 13.4.7 are satisfied: For example (5) establishes part (b), with Zv =
Cz(L 2 ) and (z) = Cz(H2). Also if X E 'H(Ho), then Li f._ X: as otherwise
M = LT = (L 1 , L 2 T) :::; X whereas H2 :::; X but H 2 f._ M by (7). Therefore as
L 2 T is maximal in M = Ca(Zv) and H2 is maximal in H = Gz, we conclude
that L2T = Cx(Zv), so.L2 = 031 (CH 0 (Zv)); and H2 = Cx(z), so 02 (H2) =
031 ( C Ho ( z)). Thus part ( c) holds. Finally L2 has at least three noncentral 2-chief
factors, two on V and one on 02(L2), giving part (a). We conclude from 13.4.7.l
that Ho E 'H(T) and one of conclusions (i)-(iv) of that result holds. In applying
13.4.7, we interchange the roles of "H1" and "H2", so the hypothesis "K2 = L2" in
parts (3) and (4) of 13.4.7 also holds; hence conclusions (iii) and (iv) do not hold
here.
Suppose conclusion (ii) holds. Then J(T) ~ Ho. Further for any XE 'H(Ho),
Cx(Z) = Cx(Zv) n Cx(z) = L2T n H2 = T,
so we conclude from 13.4.7.2 that X = Ho. Thus Ho E M(T), and in particular
Ho= Na(J(T)). That is, conclusion (ii) of (8) holds.