8.2. DETERMINING LOCAL SUBGROUPS, AND IDENTIFYING J 4 717are as described in the Table, then m(U) = m(Zr ), so Zr= U. Further the Table
says A :::! f', so U = Cv(A) :::! T. Hence it remains to verify the Table.
When Lis M24 on the cocode module, we verified the Table in 8.1.3. If LisL 3 (4), the Proposition follows from 8.2.4 modulo the following remark: As both
U = Cv(A) and A have rank 4, H.4.4.2 says that A= UL.
Thus we may assume L is M 22 or M 24 on the code module. By 8.2.3, k = w = l.
By 7.4.1 and the values in Table 7.2.1, r ;:::: 6.
Suppose L ~ M24 and r = 6. Then arguing as in the proof of (*) in the previous
lemma, Ca(Uo) i M for some subspace Uo of V of rank 5 which is the centralizer
of an element y of order 3 in Mv. By H.16.6, (y) = CMv (Uo) so that Uo Er. Then
by E.6.11.2, there is an element of order 63 centralizing U 0 , contradicting H.16.6.
Thus r ;:::: 7 when L ~ M 24 on the code module.
Now by E.3.28.3,
m(A) ;:::: r - w = r - 1,
so as r - 1 ;:::: 5, 6 = m2, we conclude m(A) = m2 = r - 1 = 5, 6. Then by 8.2.2.6,
m(A) = m(V/Zr) - k;:::: m(V/Cv(A)) -1, (*)
so m(V/Cv(A)) :::; m(A) + 1 = 6, 7. Since Vis not an FF-module, this inequality
is an equality, so the inequality in (*) is also an equality. Thus U = Zr is of rank
- Further it follows from H.16.5 that A= KQ, Ks, so the proof is complete. D
8.2.2. Constructing NG(U). We now use the results from the previous sub-
section to study the subgroup N := Na(U), where U is defined in 8.2.5. Let
N := N/U and Lu := NL(U)^00 • Recall from 8.2.5 that T :::; N, so N E He by
1.1.4.6.
As k :::; 2 by 8.2.5, 8.2.2.5 says I* ~ D 2 m or L2(4). Thus case (i) of E.2.14.2
holds, with P := 02 (J) = AB and A = Bh. By 8.2.5, U = Zr; it follows from
E.2.14 that P = [P, 02 (J)]U.
We first observe:LEMMA 8.2.6. (1) Lu E C(NM(U)).
(2) Lu acts naturally on U as A5, A5, A5, L3(2).
(3) Either 02 (Lu) = CLu(U), or Lis M24 on the code module, Lu/02(Lu) ~A5, and CLu(U) = 02,z(Lu).
PROOF. Part (1) follows from the definitions. Parts (2) and (3) follow from
H.4.6.2, H.16.3.2, H.16.1.2, and H.15.6.2. D
As T:::; NM(U), Tacts on Lu, so by 8.2.6.l and 1.2.4, Lu:::; Ku E C(N) with
T _::::; NN(Ku).
LEMMA 8.2.7. Ku/02(Ku) zs quasisimple.
PROOF. Assume not. Then by 1.2.1.4, Ku/02,F(Ku) ~ SL2(q) for some
odd prime q. Then as Lu :::; Ku, A.3.12 says that either Ku = Lu02,F(Ku)
or Lu/02(Lu) ~ L 2 (4) and q = ±1 mod 5. In any case (in the notation of chap-
ter 1) X := 'Bp(Ku) =!= 1 for some prime p > 3, and by 1.3.3, X E 'B(G,T).
By 1.2.l.4 either p = q and X = 02 (02,F(Ku)), or Ku = Lu02,F(Ku) and
Lu/0 2 (Lu) ~ L 2 (4). In particular Vis not the code module for L ~ M24, since
A 6 is not isomorphic to L 2 (p) for any odd prime p.
Now X = [X,Lu], so as Lu :::! NM(U), Xi M; hence XT E 1i*(T,M), so
replacing H by XT if necessary, we may take H = XT :::; N. Then Hand the