1028 14. L 3 (2) IN THE FSU, AND L 2 (2) WHEN .Cr(G, T) IS EMPTYSet I*:= I/0 2 (I). Then U]r :::1 TJ, while m(UH/CuH(QH)) 2 4 by hypothesis.
Therefore by 14.6.8, m 2 (I*) 2 m(UlI) 2 4, so from the previous paragraph, Ko >
Kand K/0 2 (K) ~ L 2 (p) for p;::::: 7 a Fermat or Mersenne prime, with Aut1(K*) ~PGL 2 (7) if p = 7. But in these groups TJ has no normal elementary abelian
subgroup of rank at least 4. This contradiction completes the proof of 14.6.9. D
This subsection culminates in the technical lemma 14.6.10. In each of the
subsequent two subsections, the final contradiction will be to part (5) of 14.6.10.
LEMMA 14.6.10. Assume the hypotheses of 14.6.9 and let L1 := 02 (L n I).
Assume that I= (Ii,1 2 ), where Ii:= L1T1 and T1::; I2::; H with I2/02(lz) ~ 83.
Set Ri := 02(Ii)· Then
(1) C(G, Ri) ::; M.
(2) Ri -j. R2.
(3) If P is an Ii-invariant subgroup of I, then either L1 ::; P or P:::; CM(V).
(4) F*(I) = 02(I).
(5) IfT1 =Tu, assume further that I::; Ca(u). Then m((V^12 )) = 3.
PROOF. As the hypotheses of 14.6.9 hold, by that result LT= L1T102(LT) =
Ii02(LT). In particular L1 f:. G1, li/02(li) ~ 83, LJ/02(L1) ~ Z3, and Ri =
02 (LT)nT 1. By 14.6.6.1, T1 < T, so T1 < NT(T1)::; NT(R1) since Ri = 02(LT)n
T1. Then as IE I* and NLT(R1) contains Ii f:. G1, we conclude from 14.6.4 that
M = !M(NLT(R 1 )), so (1) holds. Since If:. M but Ji :::; M, I2 f:. M, so (1) implies
(2).
Assume P is a counterexample to (3). If P ::; G 1 , t!J.en as P is Ii-invariant,
P centralizes (zli) = V, so that P::; Ca(V) = CM(V), contrary to the choice of
Pas a counterexample; thus Pf:. G 1. Set M+ := M/0 2 (M). By 14.2.2.4, M+ =
L+Rt x CM(V)+, where Re:= 02(MnG1). As L1 -j."1 while L = [L,Co 2 (a,)(u)]
by 14.6.3.3, L +Rt = T;l", where Ie := Ii n LRe· As we are assuming that P isI 1 -invariant vyith L 1 f:. P, P n LJ ::; 02 (L1), so as 02 (L n P) ::; 02 (L n I) = L1,
02 (L n P) = 1. If P ::; M then [P, Ie] ::; P n LRe ::; 02(L n P)Re = Re, so
p+::; CM+(0^2 (It))::; CM(V)+, again contrary to the choice of P since 02 (M)::;
CM(V). Therefore Pis contained in neither M nor G 1 , and as PT1 ::; I, PT1 E
H(Tu)· Hence PT1 EI* by 14.6.4. Thus we may apply 14.6.9 to PT1 in the roleof "I", to conclude that 02 (L n P) -::f. 1, contrary to an earlier observation. So (3)
is established.Set I* := I/031(I). Observe as T1 E Syl 2 (I) by 14.6.6.2, that (I,I1,I2) is
a Goldschmidt triple in the sense of Definition F.6.1. In view of (2), case (i) of
F.6.11.2 holds, so I* is a Goldschmidt amalgam, and hence as I is an SQTK-group,
I* is described in Theorem F.6.18.
To prove ( 4), we assume F* (I) -::f. 02 (I), and derive a contradiction. By hypoth-esis m(UH/CuH(QH)) 2 4, so since 02(I) E Syl2(031(I)) by F.6.11.1, m(UlI) 2 4
by 14.6.8. Now the only case of Theorem F.6.18 in which m 2 (I*) 2 4 is case (13),
where I*~ Aut(M12). Thus II;: TJI = 3 = II2: TJI, so I2 =I;. Thus as Iz::; H,
U]r :::1 I; with m(UlI) ;::::: 4, whereas in Aut(M 12 ) (as we saw during the proof of
14.5.5), I; has no such normal subgroup. This contradiction establishes ( 4).
Assume the hypotheses of (5). By (2), conclusion (1) of Theorem F.6.18 does
not hold. If either case of conclusion (2) of F.6.18 holds, then there is a normal
subgroup P of I with I= PI 1 and P n L = 02 (L). But then by (3), P::; CM(V),
so I= IiP::; M, contrary to IE I.