5.2. USING WEAK EN-PAIRS AND THE GREEN BOOK 65i(b) a is an extension of the U4(q) amalgam of degree 2 and 02 (KS) is the
extension of 02(K) by an involution t such that CK(t) ~ P^00 for P a maximal
parabolic of Sp4(q).
PROOF. We have already verified Hypothesis F.1.1, so by F.1.9, a is a weak BN-pair of rank 2. By 5.2.4.2, B2 ::::; Na(S), so that we may apply F.l.12 to determine
a. As Li and L 2 are not solvable, cases (8)-(13) of F.l.12.I are ruled out. Together
with F.l.12.II, this shows that S::::; Li and hence also 02 (Li) = 02 (Gi) for i = 1
and 2, unless possibly a is the extension of the U4(q) amalgam of degree 2. In the
latter case by F.4.29.5, (II.i) fails only weakly, in the sense that 02 (L) = 02 (LS)
and IS : s n LI = IS : s n L21 = I02(L2S) : 02(L2)I = 2. Further by F.4.29.4,
02(Gi) = 02(Li)· Now the remaining amalgams in cases (1)-(7) of F.l.12.I arethose given in 5.2.6; notice that the numbering convention for Li and L 2 in F.l.12
differs in some cases from that used here in 5.2.6. We are using the facts that
L/02(L) ~ L2(2n) and 1 -I [Z, L].
We next show that L2 = K; that is, we eliminate cases (1) and (3) of 5.1.10.
First suppose L2 =Ki < K. Then fort ET - NT(Ki), 02 (L 2 S) contains Sn L~
with 02(L2) n L~ ::::; 02(L~) and IS n L~ : 02(L~)I > 2; therefore I02(L2S) :02(L2)I > 2, contrary to an earlier observation. So we may suppose instead that
K/02(K) is (S)L 3 (2k) or Sp4(2k). We recall in this case that L 2 = Pf for a
maximal parabolic P+ of K. Thus L 2 /0 2 (L 2 ) ~ L 2 (2k), 02 (L 2 )0 2 (K)/0 2 (K)has a natural chief factor, and there is at least one more noncentral 2-chief factor
for L 2 in 02 (K). Thus L 2 has at least two noncentral 2-chief factors, so that a
is not the L3(q) or Sp4(q)-amalgam. As L2/02(L2) ~ L2(2k), rather than Sz(q)
or SU 3 (q), a is not the^2 F4(q) or U 5 (4)-amalgam. If a is the amalgam for G2(q)
or^3 D 4 ( q), then L 2 D has just one noncentral 2-chief factor, and that factor is not
natural. This leaves the U 4 (q)-amalgam, where L 2 has two natural 2-chief factorson the Frattini quotient of 02 (L 2 ), but L 2 D is irreducible on the Frattini quotient.
However D acts on 02(K) and hence on the 2-chief factor for 02 (L 2 ) in 02(K).
This contradiction shows that L 2 = K, completing the proof of the claim.
Recall S = 02 (MH ), so that 02 (KT) ::::; S by A.1.6, and hence 02 (KT) =
02 (KS). By F.l.12.II, 02(L) = 02(LS), and either 02(K) = 02(KS) or a is
the extension of the U 4 (q)-amalgam of degree 2. In the latter case by F.4.29.5,
02 (KS) = 02 (K)(t), where t induces a graph-field automorphism on U4(q), and
hence Cu 4 (q)(t) ~ Sp4(q), so that CK(t) is as claimed. This completes the proof of
5.2.6. D
In the remainder of this section, if a is the extension of degree 2 of the U4(q)-
amalgam, we replace a by its subamalgam of index 2. Thus in effect, we are
replacing S = 02 (BT) by Sn L of index 2 in S. Subject to this convention:
LEMMA 5.2.7. (1) a:= (Gi, Gi,2, G2) is the amalgam of L3(q), Sp4(q), G2(q),(^3) D4(q), (^2) F4(q), U4(q), with q > 2, or U5(4).
(2) Gi = LiBD and Gi,2 = SBD, where Li= L, L2 = K; and S = T n Li=
T n L2 = 02(Gi,2).
(3) 02(Li) = 02(GiT).PROOF. Parts (1) and (2) are immediate from 5.2.6 and the convention for
U4(q). Let So := 02(BT). By 5.2.6, 02(GiSo) = 02(Li) for i = 1, 2. Further as
B::::; Gi, 02(GiT) ::::; 02(BT) =So using A.1.6, so 02(GiT) ::::; 02(GiSo) = 02(Li)
and hence 02 (GiT) = 02(Li), establishing (3). D