i5.2. FINISHING THE REDUCTION TO Mr/CMf(V(Mf)) ~ Oj:"(2) 1107Suppose Li is not quasisimple. Then by 1.2.1.4, fA := F(Li) = F*(Li) is a 3'-
group, so the preimage Fi of fA lies in Mi by C.2.6.2. Then Y:::; Fi: for otherwise
[Fi, Y] :::; Fin Y :::; Fin 02(Y) :::; 02(Fi), and then Li = [Li, Y] centralizes Fi,
contradicting F*(L1) = F1. Therefore fr·::; 05 (L 1 ), so L 1 ~ 8L 2 (5)/E 25 by1.2.1.4.
In particular 8 is irreducible on Fi, impossible as 8 acts on Y and Y <Fi.Therefore Li is quasisimple, so as Li :::;I G 0 , Li is described in C.2.7.3. Further
Y is an S-invariant subgroup of Li with iY: 02 (Y)I = 5, so we may apply A.3.15;
comparing the list of A.3.15 with the list of C.2.7.3, we conclude Li/0 2 (Li) ~
L2(2n) or 8L3(2n) with n = 0 mod 4. This is impossible, as Ki E C(L 1 8, 8) with
Ki/02(K 1 ) ~ A 5. This contradiction completes the proof of 15.2.4. DWe are now able to obtain the analogue of 14.2.2.5:LEMMA 15.2.5. M(T) = {M, Mc} ..
PROOF. We assume M 1 E M(T) - {M, Mc}, and derive a contradiction. Set
H :=Mn Mi and Zv :=Zn V. As Mc= !M(Ca(Z)), CM 1 (V(M1)) ~ Ca(Z) :::;Mc ;:::: Na(Zv ). By 15.1.9.2, Mi ::, M, so that Mi = HCM 1 (V(M1)), and hence
as CM 1 (V(M1)) :::; Mc but Mi i. Mc, also Hi. Mc. Thus Hi. NM(Zv), so
as CM(V) :::; Mc by 15.1.5.2, it follows that fI i. Mc, so that f' < fl. On theother hand if 02 (M 0 ) :::; HNM(Zv), then M = !M(H) by 15.2.2, contrary to
H:::; M 1 =/= M. Thus 02 (M 0 ) i. HCM(V), so that fI < M.
Now if either case (1) or (2) of 15.2.1 holds, then IM: NM(Zv )I= 3 is prime,
so as Hi. NM(Zv ), M = NM(Zv )H, which is contrary to the previous paragraph.
Similarly in case (4) of 15.2.1, as 02 (M 0 ) i. fI and M > fI > T, F*(M) has order
15, and fI = 03 (M)f'. But M =Mn Mc0^2 (M 0 ) by 15.2.3.4, so fI = 03 (M)f' =
M n Mc, contrary to the previous paragraph.
Thus case (3) of 15.2.l holds, so as M > fI > T, fI ~ Z 2 x 83. Set M* :=M/02(M) and R := 02(H).
Suppose for the moment that V < V(M). Then from Notation 15.1.4, case (6)
of 15.1.2 holds. Let MJ deno.te the preimage in M of J(AutM(V(M)), V(M)), and
VJ:= Cv(M)(MJ); by 15.1.4, V = [V(M), MJ], and by 15.1.2.6, V(M) = V x VJ
with VJ =/= 1, CM(V)CM(VJ )T = Mc, and IM : M n Mel = 3 is prime. So as
H i. Mc, M = H(M n Mc) = HCM(V)CM(VJ) in this case. We now drop the
assumption that V < V(M).
Suppose that R = 1. Observe that hypothesis (a) of 14.1.17 is satisfied with
V(M), 02 (M) in the roles of "V, Y 0 " so as R = 1, we conclude V < V(M)
from 14.1.17.1, and we adopt the notation of the previous paragraph. As R =
1, R:::; CM(V), so [CM(VJ),R]:::; CM(V(M)). Also R = 02(RCM(V(M))) by14.1.17.5 applied to V(M) in the role of "V", so CM(VJ) :::; N M(R). From the
previous paragraph, M = HCM(V)CM(VJ), so M = CM(V)NM(R), and henceM = !M(NM(R)) by 15.2.2. Therefore C(G,R) :::; M, so C(M1,R) = H, and
hence M 1 = !M(H) by 14.1.16, contrary to H :::; M =/=Mi.
Therefore R =/= 1. Since R:::; 02(fI) with fI ~ 83 x Z2, R = 02(fI) is of order
- Let Yo denote the preimage in M of O(H); then fI = YoT. As 02 (M) is abelian
and T :::; H, Yo of order 3 is normal in M.
Set Ri := 02(M1 n Mc), Vi := V(M1), M1 := Mi/CM 1 (Vi), and M{ :=
Mi/02(M1). Recall M1 = fl and CM 1 (Vi) :::; Mc, so Mi = (M1 n Mc)H, and
2 A