11.4. ELIMINATING THE REMAINING SHADOWS 777Suppose first that K/02(K) is not quasisimple. Then 000 (K) centralizes
R2(KT) by 3.2.14, and we conclude from A.4.11 that 02 ,F(K) centralizes VK and
hence also Vi. By 11.1.1, either K = 02,F(K)Li; or K/02,F(K) ~ SL 2 (p) for a
prime p with p = ±1 mod 5 and p = ±3 mod 8. However in the former case, K
centralizes Vi, contrary to an earlier remark, so the latter case holds.By 11.4.1.3, Ri S 000 (KRi). Thus Ri E Syl2(0 00 (KRi)), so by a Frattini
Argument, K = 02,F(K)KJ, where KJ := NK(Ri)^00 • Therefore· KJ/0 2 (KJ) ~
K/0 00 (K) ~ L2(p). As 02,F(K) centralizes Vi but K does not, KJ 1. Cc(Vi).
First XT S Nc(Ri) =: N. We claim that XT normalizes KJ. By 1.2.4, KJ S
Ko E C(N). The claim is immediate if KJ = K 0 , so we may assume that KJ < K 0 ,and hence the embedding is described in A.3.12. As Ri is Sylow in 000 (KRi)
and normal in N, Ko/02(Ko) is not SL2(p)/Ep2, so Ko/02(Ko) is quasisimple.
Hence asp 2: 11 since p = ±1 mod 5, we conclude that Ko/02(Ko) ~ L2(p^2 ).
But then T does not act on Ki, establishing the claim. Therefore as KJ S Cc(z)
while X S N N ( K J) by the claim, Vi = [z, X] S Cc ( K J), contrary to the previous
paragraph.
Therefore K/02(K) is quasisimple, so VK E R2(KT) and 02(KT) = Cr(VK)by 1.4.1.4. Set K* := K/CK(VK)· We saw 02(KT) S Ri. Thus if J(Ri) S
02(KT) then J(R1) = J(02(KT)). Hence KT S Nc(J(R1)), so Nc(J(R1)) S
Nc(K) by 1.2.7.3. Thus XS Nc(K), so as KS Cc(z) and Cv(L) = 1by11.4.1.1,
Vi= [z, X] S Cc(K), for our usual contradiction.
Therefore J(R 1 )* f=. 1, so we may apply B.2.10.2 with K, T, Ri in the roles of
"L, T, R" to conclude that VK is an FF-module for KT, with K = [K, J(R1)*] =
J(KT, VK ). Then Theorems B.5.1 and B.4.2 reduce the list in 11.l.1 to those
cases where either K is SL3(q), Sp4(q), or G2(q), or q = 4 and K is A1.
Assume K* is not A 7 , and let Y be a Hall 2'-subgroup of NK(R1). Then Y
centralizes z, and induces a group of order q - 1 on Z 1 := fh(Z(R1)) containing
Vi, so as Vi has order q, we conclude Vi < Zi. Hence Y S Nc(Ri) SM by 11.1.5.
Then by 11.0.4, Yi := 03 (Y)fh(0 3 (Y)) S L. As Y* hasp-rank 2 for primes p
dividing q - 1, so does Y. Therefore as V is the natural module for L/0 2 (L).
z E Cv 1 (Yi) = Cv(L), contrary to 11.4.l.l.
Thus K* is A1. As Li centralizes z E VK - CvK(K), [VK,K] is not a 4-
dimensional irreducible for K*. Therefore by B.4.2.5, [VK,K] is the natural 6-
dimensional module, J(Ri) is generated by a transposition, and q(KT*, VK) = 1,
so m 2 (T) = m 2 (Ri) by B.2.4.3. Thus conjugating in K, there is A E A(T) such
that A* induces a field automorphism on Li. However this is impossible since
J(T) s LCr(L) for the natural modules in 11.0.2.2 by (2)-(4) of B.4.2. This
contradiction finally shows that Ki = K, completing the proof. D
In the remainder of the section, fix z E Cv(T)#.
Set Gz := Cc(z), Mz := CM(z), and K := K(V, Vi). Then K = K(V, z) :::::! Gz
by 11.4.2. By 11.4.1.2, either K =Li, or K is described in case (1) or (2) of 11.1.2.
Furthermore if Gz SM, then as z E Vi, Cc(Vi) S Gz SM, so for HE H*(T, M),
[Z, H] ;:::: [z, H] f=. 1; thus Hypothesis 11.3.1 holds, so Cc(Vi) 1. M by 11.3.2:2, a
contradiction. Therefore
Gz 1. M.Define Gi := Nc(Vi) (as opposed to Cc(Vi) in earlier sections), and M1 .-
NM(Vi). Recall that XS G1.