15.4. COMPLETING THE PROOF OF THE MAIN THEOREM 1163F*(C1(z)) = 02(C1(z)) for each z E z#, and m(Z) = 2by15.4.14.2, we conclude
L ~ A 6 and L Z ~ 86. In particular Yi ::::; 031 (I) = L for i = 1, 2, so L = 02 (I)
using F.6.6. Now Tacts on Crr(Y) = Crr(L) x (ZnL), but Tacts on no nontrivial
subgroup of Crr (L) as Cz(X) n Cz(Y) = 1. Therefore as IT: T1I = 2, Crr(L) is of
order 2, and hence Crr(L) = Eo, so Gi ~ E4 x 84. Thus Y2 ~Ai, so X ~ A 4 x .Ai.
Then as Cz(X) =Ex is of order 2, m2(T1) 2: 5, contrary to Gi ~ E4 x 84.
Therefore F*(I) = 02 (I). Let Vr := (Z^1 ), so that Vr E R 2 (I) by B.2.14. But
Cr(Vr)::::; Cr(Z) =Tr using Theorem 15.4.8, so C1(Vr) = 02(I). Let i := I/0 2 (I).
Now Yi= [Yi, J(T)] as Yi E Y, and [Z, Yi] i-1by15.4.12.6, so Vr is an FF-module
for i. Then using Theorem B.5.6 to determin~ the FF-modules for the possible
groups in Theorem F.6.18, it follows as C1(Z) =Tr, that i ~ 83 x 83. But then Y 2
normalizes 02 (Y10 2 (I)) = Y 1 , so that Y2 ::::; Na(Y) = M 1 , contrary to an earlier
remark. D
We now define notation in force for the remainder of the subsection. By Hypoth-esis 15.4.1.2, we can pick distinct members Mi and M 2 of M(T), and by 15.4.12,
we can choose X E Y n Mi and Y E Y n M2. Thus Mi = Na(X) = !M(XT)
and M 2 = Na(Y) = !M(YT) by 15.4.12.2. Further s(X) = s(Y) = 2 by 15.4.18,
so that Y = Y1Y2 and X = X1X2 as in 15.4.12.6. Let To := Nr(Y1) n Nr(X1).
By 15.4.12.4, Sis Sylow in XS and YS, so as S ::::; T 0 by 15.4.12.6, To is Sylow
in XTo and YTo. Let Li := X1 or X2, and L2 := Y1 or Y2. Set Gi := LiTo,
and I := (G1, G2). Let 1li := [V(Mi), Li], so that 1li ~ E4 by 15.4.12.6. Observe
[T: Toi::::; 4 since IT: Nr(Yi)I = 2 =IT: Nr(Xi)[.
LEMMA 15.4.19. (1) 1 #-CE(I)::::; Z(I). In particular IE H.
(2) L3-i i Mi.
(3) Zn Z(I)Vi i-1.PROOF. If L2::::; M 1 then YT= (L2, T)::::; Mi, contrary to M2 = !M(YT) and
the choice of M 1 i-M 2. Thus (2) holds. Similarly Zn Z(I) = 1.
Let Er:= CE(T 0 ). Arguing as in the second paragraph of the proof of 15.4.18,
Er= Ei x Fi, where Ei := CE(Gi) and Fi:= Cv.(To) ~ Z2. Thus Eo := CE(I)
is of corank at most 2 in E1. As Cz(X) i-1by15.4.14, and CEn[z,xj(To) ~ E4
by 15.4.12.6, m(E1) 2: 3, so (1) holds. Further as Zn Z(I) = 1 and m(Z) = 2 by
15.4.14.2, Er = E 0 x Z, so 1 i-Zn EoF1 ::::; Z(I)Vi, and hence (3) holds. D
By 15.4.19, I E H, so that H(I) is nonempty.
LEMMA 15.4.20. To E Syb(Io) for each Io E H(I).
PROOF. Assume otherwise, and let To< Tr E Syl2(Io), and Ti:= TrnMinM2.
As S::::; To::::; Ti, S = Baum(Ti) by B.2.3.4, and hence Nrr(Ti)::::; Nrr(S)::::; T1 n
MinM 2 =Ti by 15.4.16. Thus T1 =Ti, so we may take Tr::::; T. Of course Tr< T,
as otherwise I contains XT and YT, contrary to !M(XT) =Mi i-M2 = !M(YT).Therefore IT: Tri = 2 since IT: Toi ::::; 4. Also Tr E Syl2(J) for any J E H(Io), and
in particular, To E Syl2(Nc(02(Io))).
As T1 > To, T1 does not normalize at least one of Li or L2, so we may assume
Tr does not normalize Li. Then X = (L'['r) ::::; I::::; Io and R := 02(XT1) ::::) XT,