1052 i4. L 3 (2) IN THE FSU, AND L2(2) WHEN .Cf(G, T) IS EMPTYc_ is the preimage in p_ of Cp_ (Li). As case (ii) of 14.7.20.2 holds, P_/C_ is
a 6-dimensional faithful irreducible module for H*. This time we work modulo Vrather than modulo R, so we let Ro denote the preimage in R of Soc(R/V). Since
we are in case (2) of 14.7.23, R 0 /V is isomorphic to V as an L-module, so that
R 0 /V = [R 0 /V, Li]. Since R is generated by the L-conjugates of E, E f; Ro, so
from the analysis of case (2) in the proof of 14.7.23, the noncentral Li-chief factor
in R 0 /V is (Ro nP)j(Ro n C). Thus Ro= PoV, where Po:= [PnRo, Li] SP.
This time we set Co:= Pon C_, so that CoV/V = CR 0 ;v(Li) is of rank 1. Now
V s C 0 , but as Li is fixed-point-free on U, Co j; U and hence Co =/= l. As K
is trivial on C, K acts on CoU, and hence again K acts on [CoU,P] = [Co,P].
Now [Co, P] S Vas Co V/V is T-invariant of rank 1, so as K is irreducible on U,
we conclude P centralizes Co. Let Do denote the preimage in CoU of C 00 -v(Li);
just as at the end of the proof of 14.7.24.2, Do s Z(K), so as Co =I= l, Vi < Do,
completing the proof. D
As H =KT, by 14.7.25 there is a subgroup D of order 4 in Z(K) containing
Vi and normal in H.LEMMA 14.7.26. D is a Tl-subgroup of G.
PROOF. If D is cyclic, then Vi = fh(D), so as D :SJ H = Gi, the lemma
holds. Thus we may assume D ~ E4. If D s Z(T) then D s Z(H), so asH E M by 14.7.21.1, the lemma follows from I.6.1.2. Therefore we may assume
that [D, T] =Vi.
Let d ED - Vi, and set Gd:= Ca(d), and Hd := H n Gd. Then Td := T n Gd
is Sylow in Hd and of index 2 in T, so that Hd := KTd is of index 2 in KT= H;
hence Hd :SJ H, and so Hd E He by 1.1.3.l. Then Z(Td) S Z(0 2 (Hd)) =: Zd. As
P = 02(K) and K :SJ Hd, P s 02(Hd), so [Zd, K] s Zd n P S Z(P) S Z(K) by
14.7.21.2. Therefore K centralizes Zd by Coprime Action, and so Z(Td) :SJ KT= H.
Thus Na(Td) S Na(Z(Td)) =Has HEM, so that Td E Syh(Gd)· In particulard tf. za, so that H controls fusion in D. So appealing to I.6.1.1, it suffices to show
that Gd SH. Thus we assume Gd j;_ H, and it remains to derive a contradiction.
As Gd j;_ H,
go:= {Gos Gd: Hd <Go}
is nonempty. The bulk of the proof consists of an analysis of g 0 •
Let Go Ego; then Go E 1-l(Td) as d E 02(Go). As Li :SJ HEM, H = Na(Li),
so Hd = Nad(Li) and in particular Li is not normal in G 0.
Suppose first that Td is irreducible on K/Li. Then Hd E B(Go,Td), so the
conclusions of 1.3.2 hold with Td in the role of "T", and we may apply the proof
of 1.3.4 to Go in the role of "H" (as that argument uses only 1.3.2 and the fact
that Go is an SQTK-group, and does not actually require T to be Sylow in G 0 )
to conclude since K/02(K) is not elementary abelian that K :SJ G 0 , and hence
Li = 0^2 (02,z(K)) :SJ Go, contrary to the previous paragraph.
Thus Td is reducible on K/Li, so as Ri is irreducible on K/Li by 14.7.15.2,
Ri f;_ TJ, and hence Ri f;_ TdQH. So TdQH < T, and then as IT : Tdl = 2,
QH S Td. Therefore QH = 02(Hd)· Further H = Na(QH) as H E M, sothat Hd = Na 0 (QH), and hence C(Go,QH) = Hd. Therefore Hypothesis C.2.3 is
satisfied with Go, Hd, QH in the roles of "H, MH, R".