718 8. ELIMINATING SHADOWS AND CHARACTERIZING THE J4 EXAMPLE
subgroup I of H are solvable, so that 1 = n(I) = k by E.1.13; hence by 8.2.3, Vis
not the cocode module for M 2 4. Thus Lis L3(4) or M22·
Let Y E Sylp(X), so that also Y i. M. Then Y ~ Ep2 or p1+^2 by definition
of XE Bp(G, T). Suppose Y ~ p.1+^2. Then g>(Y) ::::; M by B.6.8.2, so asp> 3, Y
centralizes U from the action of AutM(U) on U in 8.2.6. Then asp> 3, [V, g>(Y)J =
1 by H.4.6.3 and H.16.3.4. Thus Y::::; Na(g>(Y))::::; M by 4.4.3 and Remark 4.4.2,
contradicting our observation that Yi. M. We conclude Y ~ EP2.
Let f:I := H/0 2 (H). As k = 1, H = 02 ,p, 2 (H) by B.6.8.2. Thus as we saw
02 (I) = P = [P, 02 (I)JU and H::::; N, P::::; 02 (H). Then as U =Zr ::::; Z(I) by
8.2.2.3, and I::::; 02 (H) = X, there is a chief factor W for Hon 02(X)U/U with
W = [W, YJ. As Vi. 02 (H) by 8.2.2.1, Vi. 02 (X); and V/B is of rank k = 1,
B = V n P = V n 02 (H) = V n 02 (X), so that Vis of rank 1. Therefore as
Tis irreducible on Y, V inverts Y, so m(W) = 2m([W, VJ). But [02(X)U, VJ ::::;
02 (X) n V = B, so [W, VJ::::; WE, where WE is the image of Bin W. Thus
m(W) = 2m([W, VJ)::::; 2m(WE)::::; 2m(B/U)::::; 10using 8.2.5. But this is impossible, as S L 2 ( q) / Ep2 for p > 3 has no faithful module
of dimension less than 52 - 1 = 24. D
PROPOSITION 8.2.8. (1) Lu =Ku <l N.
(2) [Lu, Ca(U)J::::; 02(Lu).
(3) .Either
(a) I/P ~ L2(2k), or
(b) Lis L 3 (4) or M 24 on the code module, and I/P ~ D10·(4) Lu acts on I and P with 02 (Lu I) = P02 (Lu) = CLu r(F).
(5) Let J E Irr +(I, F) and set F := F 2 in case (a) of (3), and F := F4 in case
(b). Then P, J, and B can be regarded as F-modules Fp, JF and Bp, for Lu I, I,
and Lu, respectively, and PF = J F l8l BF as an F Lu I -module.
(6) If Vis the code module for L ~ M 24 , then case (b) of (3) holds and T doesnot act on 02 (I).
PROOF. By 8.2.7, Ku/0 2 (Ku) is quasisimple, while Lu::::; Ku and CLu(U) =
02,z(Lu) by 8.2.6.3. Therefore CKu(U)::::; 02,z(Ku). But [Ku, Ca(U)J ::::; CKu(U),
so [Ku,Ca(U)J::::; 02(Ku). Hence (2) follows.
Choose has in 8.2.2.2. By 8.2.5 and (2), h EI i; Ca(U) ::::; Na(Lu0 2 (Ku )).
Therefore as Lu0 2 (Ku) acts on V, Lu0 2 (Ku) also acts on Vh, and hence on
(V, Vh) =I and on 02(I) = P.
Set Y :=I Lu and Y := Y/Cy(F). Since B is an Lu-submodule of rank m(A)
given in 8.2.5, in the various cases the Lu/0 2 (Lu)-module Bis identified as: the
natural module for L 2 ( 4) by H.4.6.2; the 5-dimensional indecomposable (with trivial
quotient) for L 2 (4) by H.16.3.3; a natural module for A 6 by H.16.1.3; the sum of
two isomorphic natural modules for L 3 (2) by H.15.6.3. Furthermore in each case
CLu (B) = 02 (Lu ). In particular, the number of Lu-constituents on Bis 1, 1, 1, 2,
and hence is equal to k by 8.2.5.
Now by E.2.10.2, P = BE9A is the sum of two I-conjugates of B, and P = Cr(F)
by E.2.14. Therefore as [Lu,IJ ::::; 02(Lu) = CLu(B) by (2), 02(Lu) = CLu(P)
and Lu ~ Lu/ 02 (Lu) is quasisimple and centralized by j ~ I/ P, so Y = j x Lu
and ( 4) holds.