5.2. USING WEAK BN-PAIRS AND THE GREEN BOOK 647PROOF. Assume that neither (1) nor (2) holds. In particular DL i B as (1)
fails. For D :::; DL let Xn := (D, H). Let 1J consist of those D :::; DL such
that 02(Xn) = l. If D E 1J then D i Na(K) as 02(K) -=/= 1 and K ::::) H. If
02(Xn) -=/= 1, then by 5.1.13, either D:::; N 0 (K) or the various conclusions of 5.1.13
hold, and the latter contradicts our assumption that (2) fails. Thus 02 (X.b) -=/= 1
iff D :::; Na(K) iff D r:J_ 1J. Finally if~ is a collection of subgroups generating DL,
then as (1) fails, D i Na(K) for some D E ~' so that ~ n 1J-=/= 0.
In particular DL E 7J. We conclude from 5.1.6 that one of the four cases of
5.1.6 holds. Now in the first three cases of 5.1.6, n = 2 or 4. If case (4) holds, then
we may take~ to consist of D and D+. However 1-=/= Z:::; 02 (Xn) in that case
by 5.1.6, so that D+ E 7J. Therefore n = 4 or 8 by 5.1.6.4.
So in any case, we haven= 2, 4, or 8. Next let Dp denote the subgroup of DL
of order p. When n = 2, D3 = DL so D3 E 7J. When n = 4, DL = (D 3 , D5), soDp E 1J for p = 3 or 5. Finally when n = 8, D+ = (D 3 , D 5 ), and we saw D_ acts
on K, so again Dp E 1J for p = 3 or 5. Thus in each case, Dp E 1J for p = 3 or 5;
choose p with this property during the remainder of the proof.
As DL i B, K/02(K) is not L3(4) by part (4) of Theorem 5.1.14. Hence S
acts on L2 by 5.2.1, and, as we observed at the beginning of this section, Sn KE
Syb(K) and Sn L E Syl2(L). Recall B normalizes 02 (BT) = S and L 2. SetGo:= (Dp, L2S).
We first suppose that 02(Go) = l. This gives part (e) of Hypothesis F.1.1,
with DPS and L2 in the roles of "L1" and "L 2 ". Part (f) follows from 1.1.4.5, as
Mand Hare in He and S contains 02(H) and 02 (M). To check part (c), we only
need to prove that Sis not normal in DpS, since then DpS/02(DpS) ~ L2(2), D10,
or Sz(2). But if S ::::! SDp, then as S ::::! T and Sn KE Syl 2 (K), Theorem 3.1.1
says 1-=/= 02( (DpT, H)) :::; 02(H) :::; 02(BT) = S:::; Go using A.1.6, contrary to our
assumption that 02(Go) = 1. The remaining parts of Hypothesis F.1.1 are easily
verified.
Now by F.1.9, a := (DpSB2, SB2, SL2) is a weak BN-pair of rank 2, where
B 2 := B n L2. Indeed since Sis self-normalizing in SDp, a is described in F.1.12.
As we saw in 5.1.18, when p = 5 the amalgams in F.1.12 have solvable parabolics,
and so are ruled out as L2 is not solvable. Sop= 3 and D3S/02(D3S) ~ L2(2);
then as L 2 is not solvable, we conclude that a is of type J 2 , Aut(J 2 ),^3 D 4 (2), or
U 4 (2). In each case, Z(S) is of order 2, and is centralized by one of the parabolics
in the amalgam.
Suppose first that a has type U4(2). Then D3S is the solvable parabolic cen-
tralizing Z(S), with [02(SD3),D3] ~ Q~, and L2 is an A5-block with 02 (L 2 ) =
F*(L2S). Thus 02(L2) is the unique 2-chief factor for L2S, so K = L2. Also
Cs(L2) = 1, so Z(H) = l. As Z(H) = 1, from the discussion above we are
in case (3) of 5.1.6, so that [V, L] is the A 5 -module for L/0 2 (L); in particular
n = 2 and DL = D3 i B. As [D3, 02(D3S)] ~ Q~, L also is an As-block. But
then as D3 < D3B and S = 02(BT), 1 -=/= CBn 3 (L) :::; O(LTB), contradicting
F*(LTB) = 02 (LTB)).
Thus we may suppose a is of type J2, Aut(J2), or^3 D4(2). In each case L 2 S is
the parabolic centralizing Z(S), so as Z = 01 (Z(T)):::; Z(S) and Z(S) is of order 2,
we conclude Z(S) = Z centralizes (L 2 , T) = H. Again in each case Q := 02 (L 2 S)
is extraspecial and L 2 is irreducible on Q/Z; so as HE He, Q = 02 (H) using A.1.6.