858 i2. LARGER GROUPS OVER F2 IN .Cj (G, T)LEMMA 12.9.3. Let 1::::; i < 5 when n = 5,, and i = 1 or 3 when n = 4. Then
Li ::S Ki E C(Na(Vi)) with Ki :::] Na(Vi), and one of the following holds:
(1) Li= Ki.
(2) i = 1, and Ki/02(Ki) ~ Ls(2), M24, or J4.
(3) i = 1, n = 4, and Kif0 2 (Ki) ~ L4(2), A1, A1, M23, HS, He, Ru, or
SL2(7)/E49.PROOF. The proof is of course similar to that of 12.5.3: First Li E £( G, T), so
the existence and normality of Ki follow from 1.2.4. If Ki > Li, the possibilities
for Ki/0 2 (Ki) are given by the sublist of A.3.12 for Lif 02(Li) ~ Lk(2) for asuitable choice of k := 3 or 4. When k = 4 we obtain the groups in conclusion (2).
When k = 3 we obtain the groups in conclusions (2) and (3), along with L2(49)and (S)L3(7)-but these last cases are out, since there T acts nontrivially on the
Dynkin diagram of Lif0 2 (Li), which is not the case by 12.8.3.1.
Thus when i = 1 the lemma is established, so we may assume i > 1 and
Li< Ki, and it remains to derive a contradiction. Set Ki_T* := KiT/02(KiT).Assume first that i = 3 or 4. Then Li/CL; (Vi) = GL(Vi), so Ki = LiCK; (Vi).
Hence Ki = Li if Ki/0 2 (Ki) is quasisimple, contrary to our assumption, so that
Ki/02(Ki) is not quasisimple. Then frQm the first paragraph, Ki/02(Ki) ~
SL2(7)/E49, Ki= XLi, where X := 'B1(Ki), and i = 3 since L4/02(L4) ~ L4(2)
is not involved in Ki. We argue much as in the proof of 12.5.3: Set Ki,3 :=
02 (CK 3 (Vi)). Then Ki,3T/02(Ki,3T) ~ SL2(3)/E49, since X ::S CK 3 (Vi) ::S
Ca(Vi). Further Ki,3 = YX where Y := 0
31(NL 1 nL 3 (Vi)) ::S K1, so that Ki,3 =
(Yx) ::::; Ki since Ki :::] Na(Vi). Now Ki_, 3 T* is a subgroup of Ki_T* contain-
ing T*. But from the structure of the overgroups of T* in the groups listed
in (2) and (3), no subgroup of these groups containing a Sylow 2-subgroup hasa GL2(3)/E49-section, except when Ki_ is also SL2(7)/E49. In this last case,
X = 071 (Ki,3) = 'B1(Ki) is normal in K 3 and Ki, so that L = (Li,L3)::::; Na(X).
Hence X ::::; Na(X) ::::; M = !M(LT), so that X = [X,£ 3 ] ::S L, contrary to
m1(L) = 1. This contradiction completes the proof that Ki= Li if i = 3 or 4.
Finally take i = 2. Thus n = 5 by our choice of i in the hypothesis, soL2/02(L2) ~ L3(2), and L 2 ::S Li with Li/02(Li) ~ L4(2). In particular m3(Li) =
2, and Li = 031 (Na(L1)) by A.3.18. We conclude G 2 1,. Na(Li), since G 2 contains
a subgroup X of order 3 faithful on 1;2, whereas if G 2 ::::; Na(L 1 ), then X ::::;
031 (Na(Li)) =Li ::S Gi. Similarly when L1 <Ki we conclude that G21,. Na(Ki).
We now claim
L2T < K2T < KiT and K2T =f. LiT.First as dim(V2) = 2, K 2 = Kf' ::::; Ca(V2) ::::; Ca(Vi). Then as L2 < Li ::::;
K1 :::] Na(V1), K2 = [K2, L2] ::S K1. As G2 does not act on Li or K1, K2 <Ki·
and K2 =f. Li. Finally by assumption, L 2 < K 2 , so the claim holds.
Now if Li = K1, then L2T is maximal in LiT = KiT, contrary to L2T <
K2T < K1T. Thus Li < Ki, so that Ki_ is in the list of (2). Observe in eachof those three groups that LiT* is determined (up to outer automorphism when
Ki is Ls(2)) as the unique overgroup of T* in Ki with LiT /02(LiT) ~ L4(2).
Suppose first that Ki_ ~ M24. Then from the list of overgroups of T*, L2 is normal
in each overgroup of L2T other than LiT, contradicting L 2 T < K 2 T < KiT
with L2 not normal in K2T =f. LiT. Therefore Ki_ ~ Ls(2) or J4, and a similar
argument shows that K 2 /02(K 2 ) is isomorphic to L 4 (2) in the former case, and