1004 14. Ls(2) IN THE FSU, AND L 2 (2) WHEN £,f(G, T) IS EMPTY
Suppose next that case (i) of 14.3.23.2 does not hold; we will eliminate that
case at the end of the proof. Then there is K E C ( H) with K = F ( H*). As K
centralizes Zu and Tacts on Vz with [Vi, Q] = V1, CK(V2) = CK(Vz) by Coprime
Action. Then as H ~if by (3), CK(V2) acts on Z!f/ and W^9 * by 12.8.12.2. But
when H ~ G 2 (2), we saw Z{j has order 2, whereas CK(V2) is the stabilizer of a
4-subgroup of T, and in particular does not normalize Z(T) of order 2.
Thus d = 4, so V = E by 14.3.23.1. Further since fz :::] G2 by 12.8.9.1,
CK(V 2 ) = CK(Vz) normalizes E = Un U9. But in case (iii) of 14.3.23.2, the
maximal parabolic CK(V 2 )* does not normalize V.
Thus. we have reduced to case (ii) of 14.3.23.2, so that (1) holds, and also
[Zu[ = 4. If Zu ~ E 4 then iI preserves a quadratic form on U by 12.8.8.2, which
is not the case as here U is a natural £2(4)-module. Thus (2) holds.
Next as Q normalizes Vz with [Vi, U] = Vi, Q = UCq(V.f!). By 12.8.9.5,
W9nQ=E. Thus
[Q, W9] = [U, W9l[Cq(V.f!), Wg] ::; U(Wg n Q) = U.
Then as K = [K, W9], [Q, K] ::; U. If [U, K] < U, then [U, K] is extraspecial by
(2), impossible as fj is the £ 2 (4)-module fork. Thus U = [U, K] = [0 2 (K), K], so
K is indecomposable on U. By (1) and (3), H =KT, completing the proof of (4).
Finally we must eliminate case (i) of 14.3.23.2. Here L1 :::] if, so as iI ~ H*
by (3), £ 1 :::] H, and hence fj = [U, £ 1 ] by 12.8.5.1. This is a contradiction as we
saw H centralizes Zu. D
LEMMA 14.3.33. (1) P := 02(£) = (Zb) ~ Z~, with P/V isomorphic to V as
an £-module.(2) U = 02(K) and PUE Syl2(K).
(3) M =Land H =KT with U = 02 (H).
PROOF. By 14.3.32.2, Cu(V) = VZu, and Zu ~ Z4 is centralized by £ 1. By
14.3.23.1, E = V, so V::; Un Ug = E::; VZu and hence E = V(Zu n U9). By
(*) in the proof of 14.3.32 and symmetry, Zu n U9 =Vi, so E = V. By 12.8.8.4,
02(LU)/V is described in G.2.5; thus as E = V and m(W/V) = 1, we conclude
that 02(LU)/V is isomorphic to V as an £-module, and hence 02(LU) = (Zb)
and 02(LU) = [02(LU), L] = 02(£) = P. As Zu is a cyclic normal subgroup of
H = Ca(fh(Zu )), Zu is a TI-set in G. Further Zu::; CT(V), so [Zu, ztJ = 1 for
y EL by I.7.5, and hence (1) holds.
By (1),· V = f!i(02(L)) :::l M. By 14.3.32, H =KT, with KQ/Q ~ A 5 , so
H n M = L 1 T, and hence M =LT by 14.3.7.
From the structure of L, PU = 02(£ 1 ); so as £ 1 ::; 02 (H) = K, PU ::; K.
By 14.3.32.4, U = [0 2 (K), K], so if U < 02 (K), then K/U ~ 8£ 2 (5); but this is
impossible, as the central 2-chief factors of £ 1 are in Zu. by (1). Thus U = 02 (K).
Then [PU[ = [K[z, so (2) holds.
Now [K, CT(U)] ::; CK(U) = Zu with Zu centralized by K, so K = 02 (K)
centralizes CT(U) by Coprime Action. In particular CT(U) = CT(K) since U::; K.
Then by (2), CT(L) ::; CT(PU) ::; CT(U) = CT(K). But K i. LT= M, while
if CT(L) # 1, then Na(CT(L)) ::; M = !M(LT); so we conclude CT(L) = 1. By
(2), CT(K) centralizes PU; so as PU = 02(£1), from the structure of Aut(L),
CT(K)::; CT(PU)::; CT(L)Zu = Zu. Thus CT(U) = CT(K) = Zu.