638 5. THE GENERIC CASE: L2(2n) IN .Ct AND n(H) > 1Suppose first that K is a block. Then so is k., and of the four subgroups of BD3
of order 3, B has three noncentral chief factors on 02(BD 3 T) and all others have
two such factors. Thus D 3 has at most three noncentral chief factors on 02(BD3T),
so Lis a £ 2 ( 4)-block. But then DL = D 3 has exactly three noncentral chief factors,
so D = B, contrary to D 1. B.
Consequently Lis a block. But if n = 2, then as Cz(L) = 1, Tis of order at
most 27 , so K is also a block, the case we just eliminated. Hence n > 2. Furtheras K is not a block, we saw that there is a C :::;! KT; then as C :::;! DLT,
KT= (KT, D 3 ):::; N 0 (C)-so that DL:::; Na(C):::; Na(K) = !M(KT) by 1.2.7.3,
since we chose k E C*(G, T). Now D 3 is inverted by t E T n k., sot induces a
nontrivial field automorphism on L/0 2 (L), and hence n is even. Then the subgroup
D of DL of order 2n/^2 + 1 satisfies D = [D, t] :::; DL n k as t E k.. As
K/0 2 (K) ~ A 7 , this forces D = D 3. But then n = 2, a case we eliminated at
the start of the paragraph. This contradiction shows that k:::; Ca(Z), establishing
( 4).
We have established (1), (4), and (5) and also showed k = 031 (X). As we
could take X = Ca(Z), it follows that (6) holds: for A7/ E 24 < M 23 is the only
proper inclusion in A.3.12 among the groups in (5).
As K :::; k :::; Ca(Z) by (4), as usual Cz(L) = 1 using 1.2.7.3. Hence 5.1.3
says either Vis the 04(2n/^2 )-module and indeed n/2 must be odd, or Vis the sum
of two 85 -modules. In the latter case, (2) and (3) hold. In the former case, the
subgroup D_ of DL of order 2n/^2 +1 centralizes Z. Now in each of the possibilities
for k in ( 5), D 3 is inverted by t E T n k. Then the final few sentences in the proofof (4) show that n = 2. This completes the proof of (2) and (3) and hence of the
lemma. D
5.1.3. More detailed analysis of the case K/0 2 (K) = L 3 (4). The re-
mainder of the section is devoted to an analysis of the subcase of 5.1.10.3 where
K/02(K) ~ £3(4). This case is the remaining major obstruction to applying the
Green Book [DGS85] and beginning the identification of G as a rank 2 group of
Lie type and characteristic 2 in Theorem 5.2.3 of the next section.
THEOREM 5.1.14. Let H* := Hj0 2 (H) and assume K* ~ £ 3 (4). Then
(1) KE C*(G, T), so Na(K) = !M(H) but K ¢'_ Cj(G, T).
(2) [Z, H] = 1 and Ca(z) :::; Na(K) for each z E z#.
(3) Cz(L) = 1.
(4) n = 2, V is the sum of one or two copies of the 85 -module for LT ~ 85,
and DL = B.
(5) Ca(K/0 2 (K)) is a solvable 31 -group.In the remainder of this section assume the hypotheses of Theorem 5.1.14, and
set H* := H/02(H). We will prove Theorem 5.1.14 by a series of reductions.
Note that B has order 3, since K ~ £ 3 (4), and B is a Cartan subgroup
of K. By 5.1.12, K E C(G, T). In particular Na(K) = !M(H) by 1.2.7.3.
On the other hand, K ¢'-C'j(G, T): For if KE Cj(G, T) then by 3.2.3, there exist
VK E R'2(KT) such that the pair K, V satisfies the FSU. By 5.1.10.3, Tis nontrivial
on the Dynkin diagram of K*, so case ( 4) of 3.2.9 in the FSU is excluded, while
£3(4) (as opposed to 8£3(4)) does not arise anywhere else in 3.2.8 or 3.2.9. This
contradiction establishes conclusion (1) of Theorem 5.1.14.