13.2. SOME PRELIMINARY RESULTS ON A 5 AND A 6 883Next z = ele2 = el,2e3,4 generates Z, and as K 2 = ((1, 2), (1, 5)), (zK^2 ) =: F ~
Es, with each coset f E2 of E2 = (e2) = (e3,4) in F distinct from E 2 containinga K2-conjugate ZJ of z. Therefore Co( Go) (J) = Co( Go) (zt) = 1 using the previ-
ous paragraph. Thus no hyperplane of F centralizes an element of O(G 0 ), so by
Generation by Centralizers of Hyperplanes A.1.17, 0 (Go) = 1.Now el E Z(To), so el centralizes 02(G 0 ), but e 1 fj. 02 (G 0 ) by assumption.
Hence as 0 (Go) = 1, Lo = [Lo, el] for some component Lo of G 0 • Thus Lo = E (Go)
is described in one of cases (3)-(13) of Theorem F.6.18. As [Ki, e 1 ] # 1 for i = 1, 2,
and el E Z(Qi), Ki does not centralize Z(Qi)· Therefore G 0 must satisfy conclusion(6) or (8) of F.6.18. But then Ki ~ A4, contrary to 13.2.11.6.
This contradiction finally completes the proof of 13.2.12. DBy 13.2.12, Ho E H(H). Let U := (zHo), so that (ZH) = UH :S U, and let
Ho:= Ho/CHo(U).
LEMMA 13.2.13. 02 (Ho/031(Ho)) is not a 3-group.PROOF. Assume that 02 (H 0 /0 3 1(H 0 )) is a 3-group. Then
02(LH) :S 031 ,3(Ho) nT :S CT(Li/02(L1)) = 02(L1T) = Rl,
so Rl E Sylz(R1LH)· By 13.2.4.1, B := Baum(R1) = Baum(Q). Thus as LH i M,
J(R1) is not normal in RlLH, so as [Z,LH] # 1, BE Sylz(BLH) by E.2.3.2. Thus
Q E Sylz(QLH ), so by Theorem 3.1.1 applied to LT, Qin the roles of "Mo", "R",
02( (LT, H)) # 1, and hence we obtain our usual contradiction to Hi M. DLEMMA 13.2.14. s = 1.PROOF. Assume that s = 2. By 13.2.10, LH E Sj(G, T), so LH :::'.! Ho by
1.3.5. Therefore Ho = LHL1T = L 1 H. Recall we are in case (b) of 13.2.4.2, so
that [Z,L 1 ] = 1, and hence U = (zHo) = (zL^1 H) = (ZH) = UH. By 13.2.8.6,
UH = U1 E9 U 2 E9 Cz(H). Now L 1 = 02 (L1) fixes the two subgroups 02 (Hi) with
image of index 3 in LH/02(LH) such that CuH(LH) < CuH(0^2 (Hi)) <UH. Hence
Ll acts on U1, U2, and Cz(H). Therefore as [Z, Ll] = 1 and Hi induces GL(Ui)
on Ui, we conclude [UH, Ll] = 1. Therefore [L1, LH] :S CLH(UH) = 02(LH), so as
Ho= LlH, Ho/031(Ho) is a 3-group, contrary to 13.2.13. D
We are now ready to complete the proof of Theorem 13.2.7.
Ass = 1 by 13:2.14, H/02(H) ~ 83 by 13.2.9. Hence (H 0 ,L 1 T,H) is a
Goldschmidt triple. As 02 (H 0 ) # 1 by 13.2.12, Ho is an SQTK-group. Let Ho :=
H 0 /03'(Ho) and a := (L1T, T, H). By 13.2.13 and F.6.11.2, a is a Goldschmidt
amalgam; hence as Ho is an SQTK-group, Ho is described in Theorem F.6.18.
Let Lo := H 0. By 13.2.13, neither conclusion (1) nor (2) of F.6.18 holds, so
Lo is quasisimple and described in one of cases (3)-(13) of F.6.18. By F.6.11.1,
031(Ho) is 2-closed, so Lo E C(Ho) by A.3.3. Thus Lo E .C(G, T); so if [Z, Lo]# 1,then Lo/02(Lo) ~ As by hypothesis (*) of Theorem 13.2.7. As Lo is not As in
any of the conclusions of F.6.18, we conclude [Z, Lo] = 1. Thus LH i L 0 , so case
(3) of F.6.18 holds; that is, 02 (H 0 ) = iJ x L 0 , where Lo ~ L 2 (q), q = 11 or 13
mod 24, and iJ ~ Z3. Let D be a Sylow 3-subgroup of the preimage of iJ which
permutes with T. Then D does not centralize Z as 02 (H) = LH. does not. Further
Ll :S Ca2(.H)(Z) =Lo, so LlT ~ D2~ and Li/02(L1) is inverted in CT(D). Thus