1547845830-Classification_of_Quasithin_Groups_-_Volume_II__Aschbacher_

13.4. THE TREATMENT OF THE 5-DIMENSIONAL MODULE FOR A 6 905

This establishes the claim that [VH, J(R 1 )] i= 1. By the first paragraph of the

proof, GR 1 (VH) = 02(H), so we may apply B.2.10.1 to conclude that

PRi,H = {A*h i= 1: A E A(R1), h EH}

is a nonempty stable subset of P(H, VH ). Hence by B.1.8.5, I = (J(R 1 )*H) =

Ii x · · · x I; with It ~ 83, and [VH, I] is the direct sum of the subgroups Ui :=

[VH, Ii] ~ E4. Further s::::; 2 by Theorem B.5.6.

Recall that L = B(M), Li= 031 (GL(z)), and H::::; Gz. Thus L 1 = B(GM(z)) =

B(H n M). Similarly if L/02(L) ~ A 6 , then L 1 = 031 (H n M) using 13.4.2.5.

Next by B.1.8.5, J(R1) E 8yl2(I); thus J(R1) is self-normalizing in I. We

claim that 02 (I*)nLi = 1: If L/0 2 (L) ~ A5, then Li normalizes R 1 , so this follows

from the previous observation. So suppose L/02(L) ~ A 6 • Then L 1 ,+ normalizes

Ri, so 02 (I*) n Li is trivial or L 0. Assume the latter case holds. Then as Lo is

T-invariant, L 0 = 02 (It) for some i, and then L 0 :::;] H* since s ::::; 2. In case (i)

of 13.4.4.2, GH(VH) = 02 (H) acts on Lo, so Lo= 02 (LoGH(VH)) :::;] H. In case

(ii) of 13.4.4.2, Li,+= 02 (GH(VH)), so Li= Li,+Lo = 02 (LoGH(VH)) ::9 H. In

either case H::::; M by 13.2.2.9, contrary to Ht_ M. This contradiction completes

the proof of the claim that 02 (I*) n Li= 1.

_Since L 1 = B(H n M), it follows from the claim that I t. M. Furthermore

02 (I*) = 0

31

(NaL([VH,I])(I*)), so the claim says I*Lj_ =I* x Li. Thus when

Li i= 1, it follows from A.1.31.1 applied in the quotient I* Li/02(Li) that s = 1.

We first treat case (i) of 13.4.4.2, where GH(VH) = 02(H). Then m3(L 1 ) =

m3(Li) = 1, sos= 1 by the previous paragraph and L/0 2 (L) ~ A 6. Thus (1) and

(2) hold. By 13.4.3.2, IZ: ZLI = 2, so as z E Z(H) does not lie in U1,

1 i= ZL n (z)(Z n U1) =: Z1

and H = IGH(U1) = IGH(Z1) = I(H n M), where the final equality holds as
Ga(Z1) ::::; M = !M(LT). As GH(VH) ::::; M, IH : H n Ml = II : 02(I)I = 3,

so 0{^2 •^3 }(H) ::::; GM(z). Then applying 13.3.9 to 0{^2 •^3 }(H) in the role of "Y", we

conclude that His a {2, 3}-group. So as L 1 = 0
31
(HnM), H = I(HnM) = IL 1 T,
with H/0 2 (H) ~ 83 x 83, since Ri ::::; GH(L1/02(L1)). Thus (3) holds.

We must treat case (ii) of 13.4.4.2, where 02 (H) < GH(VH) with 02 (GH(VH))

= L+ = L 1 or Li,+, when L/02(L) ~ A 6 or A5, respectively, and R 1 is Sylow in

the normal subgroup H 1 := GH(L+/0 2 (L+)) of H. Thus I= (J(R1)H)::::; H1, and

hence Ri E 8yl2(IR1).

Assume that L/0 2 (L) ~ A5. As Li,+ = 02 (GH(VH)), Li_ = L 0 is of order
3, and hence s = 1 and Li/0 2 (L 1 ) ~ E 9 ~ 02 (!*) x Li by an earlier remark.
Therefore as m 3 (H) ::::; 2, 02 (I)Lif0 2 (0^2 (I)L 1 ) ~ 31+^2. Then 02 (I) normalizes
02 (L 102 (0^2 (I)L 1 )) = L 1 , so that I ::::; Na(L1) ::::; M by 13.2.2.9, contrary to
It. M.
Therefore L/0 2 (L) ~ A 6 , so (1) holds. If s = 1, then (2) holds, and an
argument above shows that (3) holds. Thus we may assume that s = 2. Then
a8 Li = L+ = 02 (GH(VH)) and m3(H) ::::; 2, I/02(I) ~ E4/31+^2 with Li =
02 (0 2 ,(I)). This is impossible, since R 1 E 8ylz(IR 1 ), and J(R 1 ) centralizes
Lif0 2 (L1). This completes the proof of 13.4.8. D