i5.2. FINISHING THE REDUCTION TO Mr/CMf(V(Mf)) c:= 0!(2) 1111PROOF. Recall B = B(A) for some A E .6.'(V), and we may assume without
loss that A::::; Hi. Let X := (.6. n T n Hi) and Ji := (XH^1 ); then X ::::] Nr(Hi).
As A::::; Hi, Ki= [Ki, A]::::; Ii. Also Hi= QH(X,A), so Ii= (X,A).
We saw T = NH(V), so A does not normalize V, and hence [V, A] '=/:-1. But
B ::::; 02(H), so B does normalize V, and by 15.1.12.2, CuH(A) = CuH(B); so
[V, B] -=/:-1. Therefore (1) holds by 15.1.12.2, and then (3) follows from 15.2.10.
Recall A* is of order 2; thus 1 = m(A/B) = m(B/CB(UH)) by 15.1.12.2. 'Also
15.1.12.2 shows A is quadratic on UH, so Zs= [V,B] is centralized by A. Further
as .6. ~ A(H), X::::; J(T) ::::; Ca(V) by 15.1.9.1, so Zs is centralized by (X, A)= Ji.
Thus by (3), Zs ::::] (Ii, T) = H, so K = (Kf)::::; Ca(Zs), and hence (2) holds. D
5.
Recall from Notation 15.2.9 that H::::; Mc. Set Kc := (KMc).
LEMMA 15.2.12. Let BE I:'(V) and Zs:= [V, B]. Then
(1) KE S(G, T).(2) One of the following holds:
(i) Kc= K.
(ii) Kc E £*(G, T) = C(Mc) and KcT/02(KcT) ~ Aut(Ln(2)), n = 4 or
(iii) Kc = LLt with L E £*(G, T) = C(Mc) and L/02(L) ~ Lz(2n), n
even, or Lz(p) for some odd prime p.(3) Mc= !M(H) and Zs ::::] H, so Na(Zs)::::; Mc.
(4) Kc= 031 (Mc)::::; Ca(Z).
(5) Case (6) of 15.1.2 does not hold, so V = V(M).
PROOF. Part (1) is immediate from 15.2.8. By 1.3.4, either K =Kc, or Kc =
(L!') for some L E C(Mc) described in (1)-( 4) of 1.3.4. Suppose the latter holds.
By 14.1.6.2, LE £*(G, T). As Autr(K/02(K)) ~ Ds, we conclude from 1.3.4 that
either LT/0 2 (LT) ~ Aut(Ln(2)), n = 4 or 5, or L <Kc and L/02(L) ~ Lz(2n) or
L 2 (p). Therefore (2) is established.
By 15.2.5, Mc = !M(H), while by (2) and (3) of 15.2.11, H normalizes Zs,
so (3) holds. In case (i) of (2), as Autr(K/02(K)) ~ Ds and Aut(K/02(K)) =
GL 2 (3), it follows as TE Syl 2 (Na(K)) that Auta(K/02(K)) = Autr(K/02(K)),
so 031 (Mc)= K0
31
(CMc(K/02(K))). Therefore as K/02(K) ~ Eg and m3(Mc)::::;
2, we conclude Kc = K = 031 (Mc)· In cases (ii) and (iii) of (2), we obtain
Kc = 0
31
(Mc) using A.3.18 and 1.2.2.a. If K < Kc, then Kc = K'g° centralizes Z
by 14.1.6.3. If K =Kc, this follows from 15.1.9.3. This completes the proof of (4).
By ( 4), 031 (MnMc) ::::; CM(Z). However in case (6) of 15.1.2, IM : MnMcl = 3
and 02 (M) ~ E 9 with T = Cfil(Z n V). This contradiction establishes (5). D
LEMMA 15.2.13. (1) Either(i) case (1) or (4) of 15.2.1 holds, with M ~ 83, D10, or Sz(2), or
(ii) case (3) of 15.2.1 holds, and O(M) = [O(M), B] for each BE I:'(V).
(2) Let Bo be the unique subgroup of T of order 2 with O(M) = [O(M), Bo].
Then for each BE I:'(V), B =Bo, and Cv(B) = [V, B] = [V, Bo]= Cv(Bo).
PROOF. Assume conclusion (i) of (1) does not hold. Then either one of cases
(2) or (3) of 15.2.1 holds, or case (4) of 15.2.1 holds with F*(M) ~ Zi 5. Pick BE
I:'(V). By 15.2.11.1and15.2.10.1, Bis of order 2. Then either X := Co(M)(B) ~