5.1. PRELIMINARY ANALYSIS OF THE L 2 (2n) CASE 643Recall Sn LE Syl2(L), so SDs ::::1 TDs, and hence SoD 5 is a subgroup of G
for each subgroup So of T containing S. As B acts on D 5 and S, it acts on SD 5.
As Sn K E Syb(K) and S ::::1 T, 1 i- 02 ( (Na(S), H)) by Theorem 3.1.1. Then
Na(S) ~ M+ by 5.1.14.1, and hence Ds i. Na(S).Let X := (SBDs, K2)· Suppose first that 02(X) = l. We just saw D 5 does
not act on S, so SDs/02(SD 5 ) ~ D10 or Sz(2). Therefore Hypothesis F.1.1 is
satisfied with K2, SDs in the roles of "£1, £ 2 ". Thus f3 := (K 2 S, BS, BSD 5 ) is a
weak BN-pair of rank 2 by F.1.9, and as S is self-normalizing in SD 5 , f3 is on the
list ofF.l.12. But D10 or Sz(2) occur as factors of Li/0 2 (Li) only in the amalgams
of^2 F4(2)' and^2 F4(2), where the rank-1 parabolic over S other than K 2 in those
amalgams is solvable, a contradiction as K 2 is not solvable.This contradiction shows that 02 (X) -=f-1. Set T 0 := NT(K 2 ). We saw earlier
that T acts on SDs and similarly T acts on SB. Thus T acts on SBD 5 , so To
acts on X, and hence on 02(X). Embed To ~ Ti E Syl2(XT0); as IT : Toi = 2,
IT1 : Toi ~ 2. As 02(KT0) ~ To and K ¢ £-J(G, T) by 5.1.14.1, [Z(To), K] = lusing B.2.14; hence Na(To) ~ M+ by 5.1.14.l. Also by 4.3.17, Na(To) ~ M, so
Ti ~ Mn M+. Thus if To < Ti we may take Ti = T. However if Ti = T, then
KT = (K2, T) ~ XTo E H, so that D5 ~ X ~ M+ using 5.1.14.1, contrary to an
earlier reduction. Hence To E Syb(X).We claim Ds acts on K 2 ; assume otherwise. As K 2 E C(X, To) and To E
Syb(X), K 2 < Kx E C(X) by 1.2.4, with the embedding described in A.3.14.
Let Y := KxToDs and Y* := Y/Cy(Kx/02(Kx)). Arguing as in the beginning
of the proof of 5.1.13, CD 5 (Kx/0 2 (Kx)) normalizes K 2 ; so as we are assuming
Ds i. Na(K2), Df, i- 1.. As S ~ To, D5 permutes with To and so DsTo is a
subgroup of G by an earlier remark; therefore Kx appears on the list of A.3.15.Comparing that list to the list of A.3.14, we conclude that case (3) of A.3.14 holds
with Kx ~ L 2 (p), p^2 = 1 mod 5 and p = ±3 mod 8. But B acts on DL, so that
[B,D 5 ] = 1. Then Df, permutes with the subgroup (Ton Kx)*B* ~ A 4 of KX:,
which is not the case in Aut(L2(P)).This contradiction establishes our claim that Ds acts on K 2 • By symmetry,
D 5 also acts on K 3 := 02 (P 3 ), where P 3 is the second rank one parabolic of K
over T n K. Therefore D 5 acts on K = (K2, K3), a contradiction as we· showed
D 5 i. M+. This completes the proof of 5.1.18. D
From this point on, we assume H is a counterexample to Theorem 5.1.14.
Under this assumption we show:LEMMA 5.1.19. One of the following holds:
(1) D3 i. M+, and either{i) n = 2, and V is the direct sum of two natural modules for L, or
{ii) n = 2 or 4, and [V, L] is a natural module for L.
{2) n = 4 or 8, V is the f2"4(2nf^2 )-module for L, and D 3 i. M+.
{3) n = 2 mod 4, n > 2, 3 does not divide n, D 3 = B ~ M+, and V is therr4(2nf^2 )-module for L.
PROOF. First suppose D 3 ~ M+. Then by 5.1.18, [Z,D3] = 1 and D3 = B.
This forces one of cases (2), (4), or (5) of 5.1.3 to hold, with n = 2 mod 4 in (4).
Assume first that n = 2, so that DL = D 3 = B. As Cz(L) = 1 by part (3) of
Theorem 5.1.14, Vis the sum of at most two copies of the Ss-module, so part (4)