1026 i4. L 3 (2) IN THE FSU, AND L 2 (2) WHEN .Cr(G, T) IS EMPTYRz = 02(Na 1 (Rz)) by A.1.6-that is Rz E B2(G1), so setting Qi := 02(Gi), we
conclude from C.2.1.2 that
(c) Qi S Rz.
But To= TuQi by 14.6.3.1, so as Rz S Tr, we conclude from (c) and 14.6.6.1 that
(d) Tr= Ta.
By (Ul), IT: Toi= 2, so T = Tr(r) by (d). Further as Gi E He, (c) says
(e) Zz := Di(Z(Rz)) S Di(Z(Qi)) =: Zi.
By 14.6.1.5 and (e):
(f) Y := 02 (0 2 ,F·(Gi)) centralizes Zi and Zz.
Next Kz =Xix X2 where Xi:= Kz n Ki, Ri := 02(Xi) ~ Q§, and IXi: Ril = 3.
Further as Tr is of index 2 in T, Hypothesis C.5.1 is satisfied with I, Tr, Tr, Tinthe roles of "H, TH, R, Mo". Similarly Hypothesis C.5.2 is satisfied using (b), as is
the hypothesis IT: Tri= 2 in C.5.6.7. So by C.5.6.7, D := CTr(K) S Z(Baum(Tr))
is elementary abelian, and 02 (I) = D0 2 (K). Hence setting Zo := Z(RiR2), we
·have
(g) 02(I) = D02(K) and Zz = DZo.
Observe since IT: Toi= 2 = IZI that z is diagonally embedded in Z(Ri) x Z(R2) =
Zo.We claim that D = 1. Suppose instead that D =/= 1. Then as D is normal in
KTr =I, (a) and (d) say that Iv:= Na(D) EI*, and Tr= To E Syl2(Iv).
Assume first that LT= LvTr02(LT), where Lv := 02 (L n Iv). As Tr= To
and
\02(L): 02(L) n To\ S \T: To\= 2,
Lv centralizes 02 (L)/(02(L) n To), and hence L = 02 (L) = Lv. Next K S
Ca(D) S Iv, and indeed Ki E £(Iv, To), so that Ki S Kt E C(Iv) with Kt de-
scribed in 1.2.8.2. Then using 1.2.2.a, LS 031 (Iv)= (KtT^0 ) S Ca(D). ThereforeCT(L) =/= 1, contrary to 14.2.2.6 as we mentioned at the start of the section.
This contradiction shows that LT > LvTr0 2 (LT). Next assume F*(Iv) =I=
02(Iv). Since O(Iv) = 1by14.6.7.3, Iv has a component Kv. By 14.6.7.1, Kvappears in the list of 1.1.5.3. As that list does not contain the possible proper
overgroups of KTr in 1.2.8.2, we conclude K centralizes Kv. But each component
in that list has order divisible by 3 or 5, so mp(KKv) > 2 for p = 3 or 5, contrary
to Iv an SQTK-group. Thus F*(Iv) = 02(Iv).
Since LvTr02(LT) < LT, 14.6.6.4 says that C(Iv, Tr) S Iv,z := Iv n Gi.Thus as F*(Iv) = 02(Iv), we may apply the local C(G,T)-Theorem C.1.29 to
conclude that Iv is the product of Iv,z with one or two xo-blocks. Since Iv
contains I = KTr, where K is the product of two A5-blocks not in Iv,z, and
no A5-block is contained in a larger xo-block, we conclude that the blocks in K
are the blocks in Iv, and K :::! Iv = Kiv,z· By (e) and (f), YQi centralizes
Zz, so YQi '.S Iv,z by (g). Then by A.4.4.1 applied with Gi, Iv, Iv,z, QiY in
the roles of "H, K, H n K, X", we conclude that Qi = 02(Iv,z)· Using A.1.6,
02(Iv) S 02(Iv,z) = Qi and 02(Iv) S 02(J). Further 02(I) = 02(K)D by (g),
and 02(K) S 02(Iv) as K :::! Iv, so we conclude that 02(Iv) = 02(1). Therefore02(I) = 02(Iv) :::; Qi S Rz by (c). As Ki is an A5-block, J(Rz) = J(02(I)),
so J(02(I)) = J(Qi) by B.2.3.3. Therefore IS Na(J(Qi)) = Gi as Gi EM by
14.6.1.1, contrary to IE I.