14.6. ELIMINATING L 2 (2) WHEN (VG1) IS ABELIAN 1029In the remaining conclusions of F.6.18, there is K E C(I) with K ::::] I, and
either I= KT1, or case (3) of F.6.18 holds with KT1 of index 3 in I. Since 03 ,(I)
is 2-closed by F.6.11.1, K/0 2 (K) is quasisimple by 1.2.1.4. Next by (1), C(I, R 1 ) ::::;
M1 := InM. Further L1::::] M1 and Ri = T1n02(L1T1) E Syl2(CMr(Lr/02(L1))),
so Ri E B2(M1) by C.1.2.4; then as N1(R1) :S M1, Ri E B2(I). Now Hypothesis
C.2.3 is satisfied with I, Ri, M1 in the roles of "H, R, MH". As K appears in
F.6.18, K/02(K) is not L2(2n), so that K is not a xo-block. Now as K is Tr
invariant and K/02(K) is quasisimple, we may apply C.2.7 to conclude that K is
described in C.2.7.3. Comparing the lists of C.2.7.3 and F.6.18, we conclude that
02(I) = 031(1), I* = I/02(I) ~ La(2), A5, A1, 85, 81, or G2(2), and except
possibly in the first case, K is a block. In particular case (3) of F.6.18 is now
ruled out, so I= KT1. Then again using F.6.6, K = 02 (!) = (K 1 , K 2 ), where
Ki := 02 (Ii)· Thus L1 = K1 ::::; K, so
To prove (5), we must show that Vo := (V^12 ) is of rank 3, so we assume
m(Vo) =/= 3, and it remains to derive a contradiction.
Suppose first that K* ~ L 3 (2). Then case (g) of C.2. 7.3 occurs, so we may apply
C.1.34 to conclude that Wis either a natural module, the sum of two isomorphic
natural modules, or a 4-dimensional indecomposable module with a 1-dimensional
submodule. As V = [V, L 1 ] is a Ti-invariant projective line in W, it follows that
m(W) =/= 4, and that (VK) is an irreducible K-submodule of W of rank 3, so
Vo= (V^12 ) = (VK) is of rank 3, contrary to assumption. Therefore K is a block.
Suppose first that K is an A5-block. Then since K = (K1, K2), K1 = L1 1:.
X := 02 (0 2 ,z(K)), and of course Xis normalized by K 1 =Ji. Thus X::::; CM(V)
by (3), impossible as Cw(X) = 1 in an A5-block.
Next V = [V, L1] is a Ti-invariant line and I2 stabilizes the point Z on that
line. In particular if K is a G 2 (2)-block then V is a doubly singular line in the
language of [Asc87], and so Vo is of rank 3, contrary to assumption. Similarly
when m(W) = 4 and K* ~ A 6 or A 7 , we compute that Zand Vo have ranks 1 and
3, respectively-again contrary to assumption.
Thus K is an An-block for n := 6 or 7, I* ~ An or Sn, with m(W) =. 5
when n = 6, and we can represent I on n := {1, ... , n} ·as in section B.3, so
that W is the core of the permutation module on 0. Further M1 is the stabilizer
in I of the T 1 -invariant line V. So when n = 6, Mj = Ii is the stabilizer of
the partition A := {{1, 2}, {3, 4}, {5, 6}}, V = (e1,2,a,4, ei,2,5,6), z = ei,2,a,4, and
Vo = {eJ : [Jn {1,2,3,4}[ = 0 mod 2}, while I2 =I; is the stabilizer of the
partition { {1, 2, 3, 4}, {5, 6} }. Next assume for the moment that n = 7. Then Ii
and I2 are (in some order) the stabilizers of the partitions A' :=AU {7} and e :=
{ {1, 2}, {3, 4}, {5, 6, 7}}. However if Ii is the stabilizer of e then V = (e5,5, e5,7) and
z = e 5 , 6 , impossible as I 2 centralizes z but the stabilizer of A' does not. Thus Mj =
Ii is the stabilizer of A', I2 is the stabilizer of e, and as before V = (e1,2,a,4, ei,2,5,6)
and z = e 1 , 2 , 3 , 4 , while now Vo = (V, e5,6, e5,7)· Observe in this case that I2 is a
proper subgroup of the stabilizer I; of the partition { {1, 2, 3, 4}, {5, 6, 7} }.
Suppose first that T1 = To. We saw earlier that [LT : L1T1[ = [T : T1[, so
as IT: Toi = 2, L1 = 02 (L1T1) = 02 (LT) = L. Further Cr(L) = 1 by 14.2.2.6.