642 5. THE GENERIC CASE: L2(2n) IN Ct AND n(H) > i
This contradiction shows that D'Y < U'Y' Therefore there is /3 E r('Y) with
V,e i Qi, and d(/3,/'i) = b by minimality of b. Thus we have symmetry between
'Yo, 'Yi, 'Y and f3, 'Y, 'Yi; so reversing the roles of these triples if necessary, we may
assume that m(u:n = m(U'Y/D'Y)::::: m(U/DHJ· Thus if DH 1 :::; Cu-(U'Y), then U is
an FF-module for Hi, contrary to an earlier observation. Therefore [DH" U'Y] i=-l,
so there is g E G with Zf. = Z'Y (so that Vg:::; U'Y) and [DH" V^9 ] i=-1. By F.8.7.6,
[DH" U'Y] :::; Ai, so DH 1 acts on V9; then since Vis the natural module for Land
n is odd,
m(DH)CnH 1 (V^9 )):::; m2(1f') = n = m(V^9 /Cvg(DH,)).
Also V^9 n Qi:::; D'Y:::; Ca(DH,) by F.8.7.7. Thus
4 = m2(Hi)::::: m(V^9 +)::::: m(V^9 /Cvg(DHJ) = n,
son= 3 and
m(U /Cu-(V^9 )) :::; m(U/ DH,)+ m(DH 1 /CnH 1 (V^9 )) :::; m(U;;) + 3:::; 7,
contradicting m(Hi, U) ::::: 9. This contradiction completes the proof of 5.1.17. D
As n is even by 5.1.17, there is a unique subgroup D 3 of order 3 in DL.
LEMMA 5.1.18. If D 3 :::; M+, then D 3 = B, so that [Z, D3] = 1.
PROOF. Notice the final statement follows from the first, as B :::; H:::; Ca(Z)
by 5.1.14.2.
Assume D 3 :::; M+. It suffices to assume D 3 i=-Band establish a contradiction.
If D 3 induces inner automorphisms on K* then D 3 :::; K by 5.1.16.1. Then as BT
is the largest solvable subgroup of KT containing T, D 3 :::; BT and hence D 3 = B,
contrary to assumption. Therefore D 3 induces outer automorphisms on K*, and
K D3 ~ PGL 3 (4). Set D := DL n M+ and S := 02(DBT). Arguing as in 5.1.11,
TnL:::; Sand hence SnL E Syb(L); similarly SnK E Syb(K). From the structure
of Aut(L3(4)), CT·(BD3) = 1, so S = (TnK)Cs(K*) = (SnK)02(KS). Let P2
be a rank-1 parabolic of Kover SnK, and set K 2 := 02 (P 2 ). Then SD acts on K 2 ,
and as Ki M with T nontrivial on the Dynkin diagram of K/02(K), K 2 i M.
Thus 02(Go) = 1, where Go:= (LS,K2), since M = !M(L) by Theorem 4.3.2.
Suppose that DL :::; M+ = Na(K). Then DL = D acts on K 2 , so that
NL(S n L) = D(S n L) acts on K2. Now it is easy to verify the remainder of
Hypothesis F.1.1 with K 2 , Lin the roles of "Li, L 2 ": For example as 02 (M) :::;
S::::: 02(M+), LiSBD E 7-ie by 1.1.4.5. So by F.1.9, a:= (K2SD,BSD,BSL) is a
weak BN-pair of rank 2. Further by construction S :::;! SBD, so a is described in
F.1.12. Since K 2 /0 2 (K 2 ) ~ L2(4) and L/0 2 (L) ~ L 2 (2n) with n even, it follows
from F.1.12 that a is the amalgam of a (possibly twisted) group of Lie type over
F4. Then as K2 centralizes Z, a is the amalgam of G2(4) or U4(4). But now K 2
has only two noncentral chief factors, which is incompatible with the embedding of
K 2 in K with F*(K) = 02 (K).
Therefore DL i M+, so one of the last three cases of 5.1.15 must hold. However