14.4. FINISHING THE TREATMENT OF (VG1) NONABELIAN 1009axis W, or [W, DJ = Vf = V2 with m([U, d]) S 2 for each d ED. This contradicts
14.4.6, so D centralizes U, and hence [D, W] s V 1 n V:f = 1. Recall that W9* is
elementary abelian of rank d/2 - 1 > 1 by 14.4.2, and this forces K = [K, W9*]
in each of cases (3)-(6) of 14.4.2. Thus by symmetry K9 = [K9, W] s Ca(D). But
K^9 i Mand M = 1M(LT9), so as T9 acts on CD(L), it follows that CD(L) = 1.
Next we saw D centralizes U so that [D, U] S V 1 S V, and hence by symmetry,
[D, ux] S V{ S V for each x EL. Thus LS (Ux: x EL)=: I centralizes DV/V.Further by 12.8.8.4, I is described by G.2.5, so S := U9WCui (V) is Sylow in LS, for
l E L-L2T. As W centralizes D, so does Cui (V) by symmetry, so that S centralizes
D; then we conclude from Gaschiitz's Theorem A.1.39 that DV = V x CD(L) with
CD(L) a complement to V[ in D. Then as CD(L) = 1, D = Vf. Then Q9 = U9,
so Q = U, completing the proof of the lemma. DWe now define certain {2, 3}-subgroups X of H, which are analogous to £ 1 : for
example, 14.4.8 will show that (X, £ 2 ) =: Lx satisfies the hypotheses of L. Then14.4.13 will show that (LT,Lx) ~ £ 4 (2), leading to our final contradiction.
So let X consist of the set of T-invariant subgroups X = 0^2 (X) of H such that
IX: 02(X)I = 3. Let Y consist of those XE X such that Vx := [Vi,X] is of rank
3 and contained in E, and set Lx := (£2, X).LEMMA 14.4.8. (1) L1 E Y, with VL 1 =[Vi, L1] = V and LL 1 = L.
(2) If XE Y thenLx E Cj(G,T), Lx/02(Lx) ~ £3(2), LxT inducesGL(Vx)
on Vx with kernel 02(LxT), and I2 and XT are the maximal parabolics of LxTover T.
PROOF. By construction, L 1 E X with V = [Vi, L1], and V s E by 12.8.13.l.
Thus (1) holds.
Assume XE Y. Then Vis Vx SES Un U9, so U and U9 act on Vx, and
hence also I2 = (U, U^9 ) acts on Vx. Then Auth(Vx) is the maximal subgroup of
GL(Vx) stabilizing the hyperplane V 2 of Vx, and X does not act on that hyperplane
as Vx = [Vx,X], so Lx/CLx(Vx) = GL(Vx). Thus there is L+ E C(Lx) with
L+CLx(Vx) = Lx, so L+ E Ct(G, T). Then by 14.3.4.1, L+ E .lj(G, T) and
L+/02(L+) ~ £3(2). The projection P of L2 on L+ satisfies P = [P, T n L+J, so
as Tacts on L2, L2 = [£2,TnL+] SL+. Similarly XS L+, so Lx = L+, and (2)
holds. D
The shadow of the Harada-Norton group F 5 is eliminated in the proof of the
next lemma. We obtain a contradiction in the 2-local which would correspond to
the local subgroup rr5(2)/E 2 a in F5.
LEMMA 14.4.9. Case (3) of 14.4.2 does not hold.
PROOF. Assume case (3) of 14.4.2 holds. Then we can view U as a 4-dimensional
orthogonal space over F 4 preserved by K. In particular V 2 ~ Co(W9) lies in some
totally singular F4-point U2 of U. Further X = {X 1 , X2}, where a subgroup of or-
der 3 in each x; is diagonally embedded in K*, and we may choose notation so
that [X2, U2] = 1 and U2 = [X1, U2]. Thus L1 = X1 and V = U2 by 14.4.8.1.
Therefore the subspace f7 -1^2 orthogonal to f7 in the Frorthogonal space U is
the same as the subspace V-1^4 =: W1 orthogonal to f7 in F4-orthogonal space u.
Choose k E K so that f/k i W 1. As W 1 is an Frhyperplane of W and L1 is
transitive on fl#, we can choose k so that sk E W (recall s is the generator of V[).