675Xt 9! 83, 85 or L3(2). In particular now 02,z(P) = 02(P) as the multiplier of
these groups is a 2-group. Thus P* = P/02(P).
Note that 02 (I) ::::; Nr(Xt) by G.6.4.3. Next if X;* is not 83, then DL
normalizes 02 (Xi) = Xf° by 1.2.1.3. On the other hand, if X;* 9! 83 , then
for d E DL, 02 (Xi)d ::::; 02 (I) ::::; Nr(Xt). Then recalling that m 3 (I) ::::; 2,
either 02 (Xi) = 0^2 (Xi)d, or else Xi0^2 (Xf)/0 2 (Xi0^2 (Xi)d) 9! 83 x Z 3 and
02 (Xi)0^2 (Xi)d = 03 ' (I) =: J. In the latter case, I/Or(J/02(J)) 9! 83 x 83
or 83 wr Z 2 , whose outer automorphism groups are 2-groups, so the former musthold. Thus in any case, DL and 02 (I) act on each Xi. So as m 3 (IDL) ::::; 2,
P =Xi and 02 (P) = 03 ' (I). If P* 9! 85 , then the T-invariant Borel subgroup of
Pis not contained in M-for otherwise, TP E H*(T, M) with n(PT) > 1, contrary
to 6.1.1.2. If P* is L 3 (2) then T induces inner automorphisms on P* by G.6.4.2a.
Thus in each case there exists a TDL-invariant parabolic subgroup Pi of P, with
Pi f:_ Mand TPi/02(P1) 9! 83. Then e := (LT, DLT, P1DLT) satisfies Hypothesis
F.1.1, and so by F.1.9 defines a weak EN-pair. Moreover the hypotheses of F.1.12
are satisfied by PiDLT, so that e is described in one of the cases of F.1.12.I. SinceL/02(L) 9! L2(4) and Pi/02(P1) 9! L2(2), the only possibility there is the U4(2)-
amalgam, which cannot occur here, since in that amalgam Vis the A5-module for
L/0 2 (L). This contradiction completes the proof of 6.1.19. D
Let U := (VI) and recall fI = H/Zs.LEMMA 6.1.20. (1) U::::; 02(I) and U::::; Z(02(f)).
( 2) U is nonabelian.
(3) For x EU - Z(U), [U, x] =Zs.
(4) U/Cu(V) s=! E 2 n. Further for g EI with [V, V9] #-1, U = VV9Cu(VV9),
and {V, V9} is the set of maximal elementary abelian subgroups of VV^9.
PROOF. Pick HE H*(T, M). If bis odd, then (2) holds by 6.1.19. On the otherhand, if bis even, then 1 #-[V, V'Y]::::; VnV'Y by F.7.11.2, so that V'Y::::; Na(V)::::; M,
and we may take V'Y ::::; T. Then by 6.1.11, Zs = [V, V'Y] and V'Y E vr. So (2) is
established in this case also.
Part (1) folows from 6.1.17.2 applied to IT in the role of "H". For x E U-Z(U),
x does not centralize all I-conjugates of V; so replacing x by a suitable I-conjugate,we may assume [x, V] #-1. Then' as x E 02(I) ::::; 8, [x, V] = Zs by 6.1.17.4, so (3)
holds. By (2) we may choose g E I with [V, V9] #-1; by (1), V^9 ::::; Ns(V). Then
by 6.1.10, m(V9 /Cvg (V)) = n = m(8/Cs(V)), so 8 = V^9 Cs(V), and hence also
U = V9Cu(V). Then we conclude that (4) holds from the symmetry between V
and V9. D
For the remainder of the section, we choose H := Nc(Zs); in contrast to our
earlier convention, this "H" is not in Hs. We also pick g E I with [V, V9] #-1;
such a g exists by 6.1.20.2. As NL(Zs) is irreducible on V/Zs, Hypothesis G.2.1
is satisfied with Zs in the role of "Vi". Recall from section G.2 that the condition
U nonabelian in 6.1.20.2 is equivalent to Cr #-1. Thus we have the hypotheses of
G.2.3, so we can appeal to that lemma.
LEMMA 6.1.21. Let l E L - H, and set L 1 .-(U, U^1 ), R := 02(L 1 ), and
E := UnU^1 • Then
(1) Li= (UMv) :":] Mv and Li= LU.
(2) R = Cu(V)Cui(V) and URE 8yl2(L1).