1076 14. Ls(2) IN THE FSU, AND L 2 (2) WHEN Cf(G, T) IS EMPTY
14.7.70.3, VU 0 =Un Ua = [U, Ua]V. Thus Ua centralizes [U, Q], so R:::; Z(S) by
14.7.13.3. Also <I>(S) = [S, SJ= R by 14.7.13.4. In particular U i Ras S =(UL),
so as L 1 is irreducible on U/UoV, UoV = RnU. Therefore as R:::; Z(S):::; GH(U),
we conclude from 14.7.71.1 that [R,L1]:::; RnU = UoV, so [R,L1] = [UoV,L1] = V
. in view of 14.7.65. Thus [R,L]:::; V, so R = UoV. Further UR/R = [UR/R,L1] ~
E4, so by H.6.5:
(*) S/R is one of: the Steinb~rg module, the dual of V, the core (denoted Gore)
of the permutation module for LT on LT/ L2T, or the sum of the Steinberg module
with either the dual of V or Gore.
Suppose first that U 0 = Vi, so that R = V. Then by 14.7.71.2, Li has three
noncentral chief factors on S /V, so that S / R = S /V must be the Steinberg module,
since by 14. 7.22.2, the Steinberg module is the only module listed in (*) with
this property. It follows that V = Z(S), and then the rest of the lemma holds:
For example, S = [Q,L] by 14.7.71.3, and then as QH n 02(L1) - Q contains an
involution H-conjugate to an involution in Um the double cover of L 3 (2) is not
involved in L/S, so that S = [0 2 (L),L] = 02 (L).
Thus we assume that V1 < U 0 , and it remains to derive a contradiction. By
14.7.70.2, m(Uo) = 2, so as R = U 0 V, m(R/V) = 1. Therefore as Lis irreducible on
V, either R:::; Z(Q) or [R, Q] = V, and the latter is impossible as IT: GT(Uo)I :::; 2.
Thus R:::; Z(Q) and m(R/V) = 1, but IT: GT(Uo)I :::; 2 so R is not the extension
in B.4.8.3; thus R = V 61 GR(L) where GR(L) = Rn Z(L) is of rank 1. Hence
GR(L)Vi = GR(T) = GR(L1), so as Uo:::; GR(L1), there exists u E Gu 0 (LT) - V1.
But now by 14.7.68.2, L:::; Ga(u):::; H, contrary to Hi M = !M(LT). D
Recall M1 = H n M = LiT = NH(V).
LEMMA 14.7.73. (1) S02(K) = 02(L1) E Sy[z(K).
(2) ITnL: TnKI = 2.
(3) Letk E K-M1. ThenK = (S,Sk), 02 (K) = (Sn0 2 (K))(Skn0 2 (K)) is of
order 211 , SnSk = Go 2 (K)(U), and 02 (K)/U is the 6-dimensional indecomposable
for K/02(K) with Go 2 (K)/u(K) = (Sn Sk)/U ~ E 4 and 02 (K)/(S n Sk) the
L2 ( 4 )-module.
PROOF. By 14.7.72.2 and H.6.3.5, S/V = [S/V,L 1 ]. Then as V = [V,L1],
S = [S,L1]:::; 02(L1):::; K. We saw in() in the proofof 14.7.71that0 2 (Li):::; S,
so we conclude that 02 (Li) = S E Sy[z(K).
We can now argue much as in the proof of G.2.3: Let k E K -M 1 and set Ko :=
(S,Sk). Now K* = K 0 , so K:::; KoQH; therefore as QH:::; Na(S), 3K = 3Ko, so
K:::; (SK^0 ) =Ko. Then as S:::; K ::::1 H, K =Ko. Let P :=(Sn QH)(Sk n QH)·
Then [P, SJ :::; S n QH :::; P and similarly [P, Bk] :::; P, so P ::::J K; then as
PS/P ~ S/S n P ~ S* E Syl2(K*), P = 02(K).
Next U:::; Sn Bk, and [S, SJ= <I>(S) = V:::; U by 14.7.72.4, so (Sn Sk)/U:::;
Z(K/U). Further setting p+ := P/S n Sk,
p+ =(Sn P)+ 61 (Bk n P)+.
For each s ES - P, [P+, s]:::::; (Sn P)+:::; Gp+(s) again since [S, S] = V:::; Sn Sk
using 14. 7. 72.4. So by G.1.5.3 and Theorem G.1.3, p+ is the sum of natural modules
for K/02(K). Hence as U:::; SnSk, we conclude from 14.7.71.1 that p+ is a natural
module and Sn Sk = Gp(U). Therefore P/(S n P) = [P/(S n P),L 1 ]. Thus as
S:::; 02(L1) :::; SP, P = [P, Li](S n P) :::; 02(L1) and S02(K) =SP= 02(L1) E
