668 6. REDUCING L2(2n) TO n = 2 AND V ORTHOGONAL
each Li. Choose numbering so that L 0 := Li · · · L; is the product of those factors
Li upon which some X-conjugate of K projects nontrivially; in particular K =
[K, J(T)] :::; L 0 , 1 :::; r :::; 2, and by construction L 0 :::! X. Thus X =
(K, D£T) = L 0 D£T and DL acts on each Li,.
Now for 1 :::; i :::; r, [U, Li] is an FF-module for Li, and we claim Li is on
the following list: Lk (2), k = 2, 3, 4, 5; Bk, k = 5, 6, 7, 8; Ak, k = 6, 7, 8; A5, or
G2(2). For no Li can be isomorphic to L2(2m), BL3(2m), Bp4(2m), or G2(2m)
with m > 1, acting on the natural module, since in those cases J(T)* induces inner
automorphisms on Li, whereas Tacts on the solvable group Kand K = [K, J(T)].
Thus the claim follows from B.5.6 and B.4.2. Furthermore Li is not isomorphic to
L 2 (2) for all i::::; r, since DL does not normalize K* by a previous reduction.
As DLT = TDL and the groups Li do not appear in A.3.15, we conclude
03 (D£) centralizes Li,. So as D£ does not normalize K :::; L 0 , 03 (D£) < D£. As
L/0 2 (L) ~ L 2 (2n), it follows that 3 divides 2n - 1, so that n is even. As Out(Li)
is a 2-group for each Li, DL induces inner automorphisms on L 0. Then as DL is
cyclic and Li has no element of order 9, DL/CDt (Li··· L;) is of order 3.
Set Do := 02 (DLT) and let Ai be the projection of D 0 on Li. By the previous
paragraph, 1 f=. Ai for some i, and Ai = 02 (Ai)B for B of order 3. As Do is
invariant under the Sylow group T, we conclude by inspection of the possibilities
for Li listed above that Ai = 02 (P), where P* is either a rank one parabolic
over T n Li, or a subgroup isomorphic to B3 or B4 containing T n Li in case
02 (Li) ~ A1. Let Li denote the preimage of Li,. In each case Ai = [T n Li, Ai],
so 0
31
(Do) = [0
31
(Do), T n Li] :::; Li. It follows as DL is cyclic that Ai f=. 1 for a
unique i, and T n Li centralizes a subgroup of index 3 in D 0 /0 2 (Do). We concludefrom the structure of Aut(L/02(L)) that n = 2; hence DL = 03(DL) :S Li and
DoT/02(DoT) ~ B3· We may choose notation so that i = 1.
As T acts on D 0 , T acts on L 1 , so as 02 (X) normalizes each Li, Li :::! X.
Recall by definition that the projection A of K on Li is nontrivial. As A* is
T-invariant with A /02(A) ~ Z3 or Eg, arguing as in the previous paragraph, we
conclude that A = [A, T n L 1 ]. Then as T acts on K, A n K f=. 1, so as Tis
irreducible on K/0 2 (K), K =A :::; Li. Now as X acts on L 1 , and.DL and K
are contained in Li, X = (DL, KT) = LiT.
Assume Li is L 2 ( 2) or B 5. Then there is a unique T* -invariant nontrivial
solvable subgroup Y = 02 (Y) of Li. Hence K = Y = D 0 , impossible as D'L
does not act on K*. Therefore Li is Lk(2), 3:::; k :S 5, Bk or Ak, 6:::; k:::; 8, A5, or
G2(2).
Suppose that H/02(H) ~ B 3 wr Z 2. Then as K* :::; Li and X = LiT,
X* ~ Aut(Lk(2)), k = 4 or 5, and K* a rank-2 parabolic determined by a pair of
non-adjacent nodes. As T normalizes D 0 , with D 0 /0 2 (D 0 ) of order 3, k = 4. Thenas [Kj, Z] f=. 1 for j = 1 and 2, Theorems B.5.1 and B.4.2 show that [U, L 1 ] is the
sum of the natural module and its dual. But then J(T)* = 02 (K*), contrary toK = [K, J(T)].
This contradiction shows that H/0 2 (H) ~ B 3. Recall also D 0 T/0 2 (DoT) ~
B 3. Now X = (H,D 0 T), so that 02 (X*) is generated by K* and D 0. We conclude
02 (Li) is L 3 (2), U 3 (3), A 6 , A 7 , or A 6 • Further neither Do nor K centralizes Z, so
we conclude X* ~ B 7 and [U, Li] is the natural module for X*. From the description
of offenders in B.3.2.4, J(T)* is generated by the three transpositions in T*, so as
J(T) :::! D 0 T, it follows that D 0 permutes these transpositions transitively, and