652 5. THE GENERIC CASE: L2(2n) IN .C.1 AND n(H) > i
Recall the notion of a completion of an amalgam from Definition F.1.6. Let
G(a) denote the simple group of Lie type for which there is a completion~: a->
G(a); that is, a is an amalgam of type G(o:). To establish conclusion (3) of Theorem
5.2.3, we must show that G ~ G(a). Let 2m(o:) be the order of the Weyl group of
G(a).
LEMMA 5.2.8. Either
(1) KE .C*(G,T), or
( 2) a is the L 3 ( 4 )-amalgam, and K < k E C* ( G, T) with k an exceptional
A1-block.
PROOF. Assume K < k E C*(G,T) and let Q := 02 (K) andK* := K/02(K).Then K/Q is not SU 3 (4), since in that event K E .C*(G, T) by 1.2.8.4. Thus we
may assume a is not of type U5(4).
Recall k E 1-ie by 1.1.3.1, so 1 # [0 2 (K), K] :::; Kn 02 (K). If K/Q ~ Sz(q),
then a appears in case (5) of 5.2.6, and K ~^2 F 4 (q) by 1.2.4 and A.3.12. But then
K is isomorphic to its image K in K*, so Kn 02 (K) = 1, contrary to our earlier
observation. Thus we have eliminated the case where a is the^2 F4(q)-amalgam.
If a is the L 3 (q) or Sp 4 (q) amalgam, then K is an L 2 (q)-block, so it has a
unique noncentral 2-chief factor, and hence the same holds for k, with Q :::; 02 ( K).
By 5.2.6, Q = 02 (KT), so Q = 02 (K). Therefore K* ~ L 2 (q) is a T-invariant
quasisimple subgroup of K*, so by A.3.12, q = 4; and then by A.3.14, K* is A1,
A 7 , Ji, L 2 (25), or L 2 (p), p = ±3 mod 8 and p^2 = 1 mod 5. As a is of typeL 3 (4) or Sp 4 (4), Q is an extension of a natural module for K/Q ~ L2(4) and
m(Q) = 4 or 6. Ask E 1-{e and K is quasisimple, K is faithful on Q, so that
K :::; GL(Q). Comparing the possiblities for K listed above to those in G.7.3,
we conclude from G.7.3 that K* ~ A1, and then as m(Q) = 4 or 6, k is an A1-
block or an exceptional A 7 -block. In the former case, the noncentral chief factor
for Kon Q is not the L2(4)-module, so the latter case holds, forcing a to be the
L 3 (4)-amalgam. Thus (2) holds in this case.
Suppose a is the U 4 (q)-amalgam. From 5.2.6, K/Q ~ L 2 (q) for q = 2n/^2 > 2
and Q is special of order qi+4 with K trivial on Z(Q). Further by 5.2.6, either
Q = 02(KT), or 02 (KT) = Q(t) where tis an involution with CQ(t) ~ Eqs.
We claim Q :::] k, so assume otherwise. Suppose first that Q:::; R := 02 (K).Then as R :::; 02 (KT), and Q < R by assumption, R = 02 (KT) = Q(t). But
now Z(Q) = Z(R) :::] k, and Q/Z(Q) = J(R/Z(Q)), so Q :::] k, contrary to
assumption. Thus Q 1:. R, so as K has two natural chief factors on Q/Z(Q) and
[R,K] # 1, we conclude (Q n R)Z(Q)/Z(Q) is one of these chief factors. Thus
Z(Q) = [QnR, Q]:::; Rand QnR ~ Eqs. Again as R:::; 02 (KT), either R = QnR
or IR: Q n RI= 2. In the latter case Q n R = [Q, t] = CQ(t), so R = (Q n R)(t).
In any case K is an L 2 (q)-block with I0 2 (K)I = q^2. The only possibilities
for such an embedding in A.3.12 are that K* ~ (S)L 3 (q), or q = 4 and K* ~ M 22 ,
M22, or M2 3. The last three cases are impossible, as those groups are of order
divisible by 11, a prime not dividing the order of GL 7 (2). Thus K* ~ SL 3 (q) and
[R, K] is the natural module for K*, so [R, K] = [R, K] = Q n K. However as a isthe U4(q)-amalgam, J(T) = 02 (L) is normal in LT, so N 0 (J(T)):::; M = !M(LT).
From the action of k on R, Ki:= Nf<(J(T)) is the scond maximal parabolic of k
over knT. Thus as TnL = TnK by 5.2.7.2, K'{' = [K'{',KnT]:::; L, and then