720 8. ELIMINATING SHADOWS AND CHARACTERIZING THE J4 EXAMPLEPROOF. We begin with the proof of (1), although we will obtain (3) along the
way. Set H+ := H/0 2 (H) and Ho := (J, T). Then T:::;; Ho :::;; H but Ho i. M,
so Ho= H by minimality of HE H*(T, M). By 7.3.4 and Table 7.2.1, n(H):::;; 2.
Next H = (JT)T, so 02 (H) :::;; (JT) :::;; Ca(U) since I:::;; Ca(U) and U ::::! T by 8.2.5.Now I acts on Lu by 8.2.8.2, and hence so does H = (J, T); therefore m3(HLu) :'::'. 2
as HLu is an SQTK-group. We conclude from 8.2.8.2 and the description of Lu in
8.2.6 that 02 (H) centralizes Lu/0 2 (Lu), and m 3 (H):::;; 1.
Suppose that Vis the code module for L ~ M 2 4. Then Lu /02(Lu) ~ A5, sothe argument of the previous paragraph shows that H is a 3' -group. Therefore as
n(H) :::;; 2, and 5 E n(H) by 8.2.8.6, we conclude from E.2.2 and B.6.8.2 that H is
a {2, 5}-group. Then as m 5 (LuH) :'::'. 2, it follows that 02 (1) = 02 (H), whereas T
does not act on 02 (1) by 8.2.8.6. This establishes (3).
Suppose that L ~ M 24 , so that Vis the cocode module by the previous para-graph. Then n(H) = 2 = k = n(I) by 8.1.3 and 8.2.5, and I/02(J) ~ L2(4) by
8.2.8.3. As m 3 (H) :'::'. 1 by the first paragraph, inspecting the possibilities in E.2.2,
we conclude that 02 (H+) ~ L 2 (4) or U 3 (4). In the former case, H = IT and
02 (H) = 02 (1) so that (1) holds; so we may assume the latter. Then J+ ~ L 2 (4)
is generated by the centers of a pair of Sylow 2-groups of 02 (H+) and hence J+ is
centralized by a subgroup X of HnM of order 5. Recall HnM acts on V since Vis
a TI-set under M, so X acts on (V^02 (H)l) = (V^1 ) =I. Thus X acts on 02 (1) = P.
As m(U) = 3 by 8.2.5, GL(U) is a 5'-group, as is CGL(P)(AutLuI(F)) by 8.2.8.5.
Thus X centralizes P by Coprime Action, and then as m(V/V n P) = k = 2, Xcentralizes V. Then as I = 02 (J)C 1 (X), X centralizes (VCr(X)) = I. Therefore
I:::;; NH(X) :::;; H n M by Remark 4.4.2 and 4.4.3, impossible as we saw that Vis
normal in H n M but not in J.
Thus we may assume that L is L 3 (4) or M 22. Hence k = n(I) = 1 by 8.2.5,
and by 8.2.8, either
(i) L ~ L3(4), I/02(I) ~ D2m form:= 3 or 5, and B = [B, Lu], or
(ii) L ~ M22, I/02(1) ~ L2(2), and IE: [B,Lu]I = 2.Recall H acts on Lu and U, so that B :::;; 02 (Lu )U :::;; 02 (H) in case (i), and
similarly IB: B n 02 (H)I:::;; 2 in case (ii). As m(V/B) = 1 and Vi. 02 (H), either
(I) B = V n 02 (H), so that v+ = (v+) is of order 2, or(II) case (ii) holds and m(V+) = 2.
Suppose case (II) holds. As n(H) :'::'. 2, V :::;! HnM, and m 3 (H) = 1, we conclude
from E.2.2 that 02 (H+) ~ L 2 (4) or U 3 (4) and v+ is the center of 'f+ n 02 (H+).
This contradicts I E I(H, T, V) with n(I) = 1. The argument also shows that9.4. Eliminating n(H) =
Therefore case (I) holds and n(H) = 1. Thus for any g EH with 1 =f. lv+v+g\
an odd prime power, 11 := (V, V9) E I(H, T, V). Therefore by 8.2.8.4, \v+v+g\ En,where 7r := {3, 5} or {3}, in case (i) or (ii), respectively. Also we saw earlier that
m3(H) :'::'. 1, and as m5(LuH) :'::'. 2 while m 5 (Lu) = 1, m 5 (H) :'::'. 1. We conclude by
inspection of the list of possibilities for H with n(H) = 1 in B.6.8.2 and E.2.2 that
either H+ is L 2 (2) or Aut(L 3 (2)), or case (i) holds and 02 (H+) is Z 5 or L 2 (31).^2
(^2) In particular, we cannot have H+ ~ Us(2); thus in the first case we are eliminating the
shadow of U7(2), where N is not an SQTK-group-though the shadow of U6(2) still survives in
that first case.