646 5. THE GENERIC CASE: L2(2") IN .Ct AND n(H) > i
Observe that Hypothesis D.3.1 is satisfied, with YT, Y*, Ur, W in the roles of
"M, M+, VM, V". So as ij < 2, we conclude from D.3.8 that Y* 1:-Sz(2k); hence
Y = K. By construction Ur plays the role of both "Vr" and "VM" in Hypothesis
D.3.2 and lemma D.3.4, so the hypotheses of D.3.10 are satisfied. Thus we conclude
from D.3.10 that W =Ur. Then B.4.2 and B.4.5 show that ij > 2, keeping in mind
that K* is L 3 (4) rather than SL 3 (4), and dim(W) -=/= 9 as T is nontrivial on the
Dynkin diagram of K*. This contradiction completes the proof of 5.1.23. D
We are now in a position to obtain a contradiction which will establish Theorem
5.1.14. We specialize to the case Hi = H. As L2 is transitive on vt and Zi is of
order 2, Gi is transitive on {Vi : Zi :::; Vi} by A.1.7.1. So by 5.1.23, [Vz, Vi] = 1
whenever Zi:::; Vi. Also CH(U) = 02 (H), since otherwise by Coprime Action, K
centralizes V2, contrary to 5.1.22.1 as Ki M. Further as D3 :::; L2, 02( (L2T, H)) =
1 by 5.1.20. Hence Hypothesis F.8.1 is satisfied with Zi , V2, L2 in the roles of
"Vi, V, L". As Zi is of order 2, Hypothesis F.9.8 is satisfied with Vz in the role of
"V+" by Remark F.9.9). Therefore by F.9.16.3 q(H, U) :::; 2. However we observe
that the argument at the end of the proof of 5.1.23, with H, U in the roles of "G!,
Ur", shows that q(H*, U) > 2.
The proof of Theorem 5.1.14 is complete.
5.2. Using weak EN-pairs and the Green Book
In this section, we continue to assume Hypothesis 5.1.8-in particular, n(H) >
1.
We work toward the goal of constructing a weak EN-pair of rank 2. This willbe accomplished by establishing Hypothesis F.1.1. In our construction, L plays the
role of "Li" in Hypothesis F.1.1, and we choose L2 to be a suitable subgroup of K.
To be precise, if Ki/02(Ki) is a Bender group in 5.1.10, we let L 2 :=Ki. Otherwise
K/0 2 (K) ~ (S)L 3 (2n) or Sp 4 (2n), in which case we let P+ be a maximal parabolic
of Kover TnK, and take L 2 E C(P+)· Notice in either case that TnL 2 E Syl 2 (L 2 ).
Further K = (L§/ and Hi M, so that L2 i M.
In any case, L2/02(L2) is a group of Lie type of Lie rank 1, and of course
L/02(L) ~ L2(2n) in this chapter. Next set S := 02 (MH) = 02 (BT). By 5.1.11,
SnK E Sylz(K), and SnL E Syl 2 (L). Then as SnK = TnK, SnL 2 E Syl 2 (L 2 )
by a remark in the previous paragraph. Further by 5.1.11.3:
LEMMA 5.2.1. If K/02(K) is not L3(4) then S acts on L 2.
Next the Cartan group B of K lies in M, and so normalizes L; therefore to
achieve condition (d) of F.1.1, we need to show that DL acts on L 2. To show DL
acts on L2, we first show that-modulo an exceptional case where we view L as
defined over lF2-DL acts on K. Then we deduce that DL acts on L 2. Eventually
it turns out that L2 = K.
LEMMA 5.2.2. Either
(1) DL :'.S Na(K), or
(2) K/02(K) ~ L/02(L) ~ L2(4), V is the sum of at most two copies of the
A5-module, and K:::; Kz := 031 (Ca(Z)), with Kz/02(Kz) ~ A1, J2, or Mz3.