644 5. THE GENERIC CASE: L2(2n) IN .CJ AND n(H) >^1of Theorem 5.1.14 holds. Hence by our remark after 5.1.16, Theorem 5.1.14 holds,contrary to our assumption that H is a counterexample to that Theorem.
Son> 2, and then case (4) of 5.1.3 holds, with n = 2 mod 4. Thus DL:::; M+
by 5.1.15. Further 3 does not divide n, or otherwise DL contains a cyclic subgroupof order 9, which must be faithful on K* as D 3 = B is faithful. However this is
impossible as Aut(K) has no cyclic subgroup of order 9 permuting with T. So
conclusion (3) holds when D3 :::; M+.
Therefore we may assume D 3 1:. M+. Then one of the last three cases of 5.1.15
must hold. Cases (2) and (3) give conclusion (1), and case ( 4) gives conclusion(2). D
LEMMA 5.1.20. D3 1:. M+, so 02( (H, D3)) = 1.
PROOF. If D 3 1:. M+, then 02 ( (H, D 3 )) = 1 by 5.1.13.1. Thus it suffices to
assume D 3 :::; M+, and derive a contradiction. As D 3 :::; M+, case (3) of 5.1.19
holds; thus n = 2 mod 4, n > 2, and 3 does not divide n, so n 2:: 10. Set
S := (T n L)02(LT); then S E Syl 2 (LS). Also S = 02(D3T), so as D3 :::; M+,
SE Syl2(KS).
Next as case (3) of 5.1.19 holds, J(T) :::] LT by 5.1.2, so J(T) :::; 02(LT) :::; S
and hence J(T) = J(S) by B.2.3.3. As K 1:. M = !M(LT), J(S) is not normal in
KS. By B.5.1 and B.4.2, KS has no FF-modules, so as m 2 (KS) = 4, E.5.4
says E := D 1 (Z(J4(S))) :::! KS. Therefore as K 1:. Mand M = !M(L), J4(S) 1:.
02(LS) = Cs(V). By E.5.5, there is A E A^2 (S) with m(V/Cv(A)) - m(A) :::; 4.
But by construction S:::; L, so by H.1.1.3 applied with n/2 in the role of "n",
n/2:::; m(V/Cv(A)) - m(A):::; 4.
Thus n:::; 8, whereas we saw earlier that n 2:: 10. This contradiction completes the
~~ D
By 5.1.20, D 3 1:. M+. So by 5.1.19, case (1) or (2) of 5.1.19 holds. In particular,
n = 2, 4, or 8. However by 5.1.14.1 we may apply Theorem 3.3.1 to K, to conclude
Na(T) :::; M+; hence D 3 1:. Na(T). Therefore LT~ Aut(L 2 (2n)).
By B.2.14, Vz := (ZL) E R 2 (LT), so we can apply the results of this section
to Vz in the role of "V". In particular as Lf' = Aut(L), from the structure of
the modules in case (1) or (2) of 5.1.19, either Z is of order 2, in which case we
set Z1 := Z; or Vz is the sum of two natural modules for L ~ L 2 (4), where we
take Z1 := Zn V1 for some Vi E Irr +(L, Vz). Thus in any case Z1 is of order
2, and Vi := (Zf^3 ) ~ E4. Set G1 := Ca(Z1), G2 := Na(Vz), and consider any
H -
H1 with H :::; H1 :::; G1. Set U := (V 2 1 ), Qi := 02(H1), G1 := Gi/Z1, and
L2 := (D'[) = D3[02(D3T), D3]. Observe Hypothesis G.2.1 is satisfied with L 2 ,
Vi, Z1, H1 in the roles of "L, V, V 1 , H", so by G.2.2 we have:
LEMMA 5.1.21. fj:::; Z(Q 1 ) and <l?(U) :::; Z1.
LEMMA 5.1.22. (1) Ca(V2) = CT(V2)B:::; M.
(2) n = 2 or 4, and [V, L] is the sum of at most two natural modules for L.
(3) [Vi, 02(K)] = Z1 and D302(Ca(Vz)) :'::! G2.
PROOF. Notice (1) implies (2), since if case (2) of 5.1.19 holds, then 1 =l-
CDL(V2) is a 3'-group.
If K normalizes Vi, then by 5.1.14.1, D 3 :::; G 2 :::; M+, contradicting 5.1.20.