904 13. MID-SIZE GROUPS OVER Fzof Baum(R 1 ) is normal in Y, and hence not normal in LT as Yi. M = !M(LT).
Therefore L is an A 6 -block by 13.2.2.7, and in particular L 1 has two noncentralchief factors. This is impossible, as Li has two noncentral chief factors on Vy and
one on 02 (Li). So the proof of (3), and hence of Proposition 13.4.7, is finally
complete. D13.4.2. The case Gz solvable, leading to Sp 6 (2). Recall the definitions
of z and Hz given before 13.4.3, and recall that Gz := Ca(z). In the next lemma,
we begin to identify Gz and a suitable 2-local Ho with the parabolics P2 and F3 of
G = 8p6(2).
LEMMA 13.4.8. Assume HE Hz is solvable, choose VH E R2(H) with Zv S
VH, and let I:= (J(R1)H). Then(1) L/02(L) ~ A6.
(2) I/02(I) ~ 83 with m([VH,I]) = 2.
(3) H = L 1 IT and H/0 2 (H) ~ 83 x 83. In particular IT is the unique memberofH*(T,M) in H.
PROOF. Let H := H/CH(VH)· As usual, 02(H) = 1 by B.2.14. As Zv S
VH, we may apply 13.4.4.2, to conclude that
CR1 (VH) = 02(H).
For in case (i), CH(VH) = 02(H) S Ri; and in case (ii), Ri E 8yh(CH(L+/02(L+)),
where L+ = 02 (CH(VH)).We claim that [VH, J(R1)] =I-1, so we assume that [VH, J(R1)] = 1 and derive
a contradiction. ThenB := Baum(R1) =Baum( CR~ (VH)) = Baum(02(H)),
by B.2.3.5 and the previous paragraph. Hence as Ht;. M = !M(LT), no nontrivial
characteristic subgroup of B is normal in LT, so by 13.2.2.7, Lis an A6-block. In
particular, L/02(L) ~ A6 rather than .A.6.
Calculating in the core V of the permutation module:
Vi= [V,L1] = [V,TnL1] = [V,02(L1)] = {eJ: J s;;; {1,2,3,4} and IJI is even},
and [Vi, 02(L1)] = (e1,2,3,4). Further if M ~ 86, then also Zv =(en) S [V, Ri].
By 13.2.2.6, V s J(R1), so by (*), V S 02(H) S NH(V) and V centralizes
· VH. Hence U := (VH) s 02(H) and [V, Vh] S V n Vh for each h E H. If
US Cr(V) =: Q, then L normalizes U because [Q, L] = V since Lis a block. But
then HS Na(U) SM= !M(LT), contrary to Hi. M. So we conclude instead
that [V, Vh] =f. 1 for some h E H.
Suppose that Li ::::J H. Then as V3 = [Vi, Li], V3h = [V3\ Li] S 02(L1). Thus
either V 3 h = [Co 2 (L 1 )(V), Li]= Vi, or V:f' n Q =Vi, 02(L1)::::; V 3 hCo 2 (L 1 )(V) and
fTh = R 1 i. L, since Vi/V 1 is the unique minimal Li-invariant subgroup of V/Vi.Assume the former case holds. Then Vh centralizes Vi, so fTh = ((5, 6)) and hence
[V, Vh] = (e 5 , 6 ). But then Zv S V 3 (e 5 , 6 ) S V n Vh, contrary to 13.4.2.3. In the
latter case, Zv S [V, Ri] = [V, Vh], for the same contradiction.
This contradiction shows that L 1 is not normal in H. Hence [VH, Li] =I- 1 by
13.4.4.2. We saw VHS Q, so 1 =f. [VH,L1] S [Q,L] = V, and hence [VH,L1] =
[V, Li] = Vs. By C.1.13.d, 02(L) S Cr(Q) S CH(VH), so Li is a quotient of
L 1 ~ A4. Then by A.1.26, 02(Li) centralizes F*(H*) of odd order, so 02(Li) = 1,
and hence 02(L1) S CH(VH) S CH(V3), whereas we saw [Vi, 02(L1)] =I-1.