13.4. THE TREATMENT OF THE 5-DIMENSIONAL MODULE FOR A 6 899
We will now begin to produce subgroups Ho of G which are generated by
subgroups Hi and H2 in 1-l(T), such that (Ho, H 1 , H2) is a Goldschmidt triple in the
sense of Definition F.6.1. Proposition 13.4. 7.5 gives fairly strong information about
those pairs which also satisfy conditions (a)-( c) of the Proposition. In particular
subgroups satisfying (li) and (lii) of Proposition 13.4. 7 will eventually be identified
as the parabolics P2 and F3 of 8p5(2).
PROPOSITION 13.4.7. Assume Hi E 1-l(T), i = 1, 2, are distinct with Hi/0 2 (Hi)
~ 83. Let Ki:= 02 (Hi), Ho:= (H1, H 2 ), and Vo:= (zH^0 ). Assume:
(a) Either Ki or K2 has at least two noncentral 2-chief factors,
(b) IZ: Cz(Hi)I = 2, for i = 1 and 2, and
(c) If Ho E 1-l(T), then Ki :::l CH 0 (Cz(Ki)) Jori= 1and2, and
Kj = 0
31
(CH 0 (Cz(Kj)) for j := 1 or 2.
Then
(1) Ho E 1-l(T, M), Z = V1 = Cz(H1) x Cz(H2) with ICz(Hi)I = 2, and one
of the following holds:
(i) Ho= HiH2, [K1, K2] S 02(K1) n 02(K2), Ho/02(Ho) ~ 83 x 8s or
Z2/ E9, and Vo = V1 EB V2, where Vi := [Vo, Ki] = Cv 0 (K3-i) is of rank 2.
(ii) Ho= KoT where Ko E C(Ho) such that Ko E .Cj(G, T), Ko/02(Ko) ~
£ 3 (2), J(T) :::l H 0 , and Vo is either the sum of two nonisomorphic natural modules
for K 0 /02(K 0 ), or the core of the 7-dimensional permutation module.
(iii) 02(Ho) = CH 0 (Vo), Ho/02(Ho) ~ E4/31+^2 , m(Vo) = 6, and J(T) <I
Ho.
(iv) Ho = KoT, where Ko E C(Ho) with Ko/02(Ko) ~ A5, and Ji(T) <I
Ho for i = 0, 1.
(2) Assume conclusion (ii) holds, with Ki = [Ki, J1(T)] for some i, and X E
1-l(H 0 ). Then Vo= (Zx) and X = HoCx(Vo); so if Cx(Z) = T, then X =Ho.
(3) If K2 = £ 2 , then conclusion (iii) does not hold.
(4) Assume K 2 = £ 2. Then conclusion (iv) does not hold, and if conclusion
(ii) holds, then K2 = [K2, J1(T)].
PROOF. Let Qo := 02(Ho) and H 0 := Ho/CH 0 (Vo). Observe that the hy-
potheses say that (H 0 , H 1 , H 2 ) is a Goldschmidt triple in the sense of Definition
F.6.1, so
(H1/Qo, T/Qo, H2/Qo)
is a Goldschmidt amalgam by F.6.5.1, and hence is listed in F.6.5.2.
Assume that Q 0 = 1. By hypothesis (a), some Ki has at least two noncentral
2-chief factors, which eliminates cases (i)-(v) of F.6.5.2, and in case (vi) also elimi-
nates cases (1) and (2) of F.1.12. In cases (3), (8), (12), and (1~) of F.1.12, Z :::; Hi
for exactly one value of i, contrary to (b). Therefore Qo -f. 1, and hence Ho E 1-l(T).
Then Ho E ?-le by 1.1.4.6, so Vo E R2(Ho) by B.2.14, and hence 02(H 0 ) = 1.
By (b), [Z,Ki] -f. 1, so CK;(Vo) S 02(Ki) and Ki -f. 1 -f. K?,. Therefore
CT(Vo) S Qo by F.6.8, so as Vo E R2(Ho), Qo = CT(Vo). By (c),
CH 0 (Vo) S CH 0 (Z) S Na(K1) n Na(K2).
Also by (c), there exists an index j such that Kj = 031 (CH 0 (Cz(Kj)). Then
0
31
(CHo(Vo)):::; Kj, so in fact 0
31
(CHo(Vo)) = 1 since CKj(Vo):::; 02(Kj)• That
is, CHo (Vo) is a 3' -group.