10i4 i4. L 3 (2) IN THE FSU, AND L2(2) WHEN .Cf(G, T) IS EMPTY
Next we saw V < (V^01 ) = U :::;l YT, so m 2 (U) ~ 4 since m(V) = 3 by
14.5.4.2. As the 2-rank of G2(2)', G2(2), and Mi2 is at most 3, it follows that a is the
Aut(M12)-amalgam and m(U) = 4. Thus A:= Aut(Mi2) is a faithful completion of
a, so identifying YT with its image under this completion, we may assume YT :::; A.
As m2(Mi2) = 3, there is an involution u EU - Mi2· Thus CA(u) ~ Z2/(E4 x J)
where J := Ca(u)^00 ~ A 5. Therefore U = J(CT(u)) = 02 (CA(u)) x (UnJ). Then
from the structure of A, NA(U) = T(NA(U) n CA(u)), and Vi= Z = Cu(T):::; J.
Thus IYTI = 2^7 ·3 = INA(U)I, so NA(U) =YT as U :::;l YT. This is a contradiction
as YT centralizes Vi but Z(NJ(U)) = 1. D
By 14.5.5 and 1.1.4.6, HE H(T) ~He.
LEMMA 14.5.6. Y does not act on X.
PROOF. Assume otherwise. Then ViXY = vr X =Vix* as y :::; Gi. There-
fore Y acts on (V{) =Vi, and hence [Y, Vi] = 1 by Coprime Action. Thus Y is
not a counterexample to conclusion (1) or (3) of 14.5.3, so Y must be a counterex-
ample to conclusion (2). Therefore Y Li = Li Y, and hence. Y acts on (V 2 L^1 ) = V,
contradicting 14.5.4.1. D
Set H+ := H/0 3 1(H).
LEMMA 14.5.7. (1) Cx(VH) '.S 02(X) and Cy(VH) '.S 02(Y).
(2) QH = CT(VH)·
(3) VH :S Z(QH) and 02(H*) = l.
(4) QH E Syb(031(H)), so 031(H) is 2-closed and in particular solvable.
(5) Either(i) H+ is described in Theorem F. 6.18, or
(ii) 02(XT) = 02(YT) = QH, and H+ ~ 83.
PROOF. We saw H E He, so as Vi :::; Z, part (3) follows from B.2.14. Next
Cx(VH) :::; 02(X) as X 1:. Gi. Thus if Y:::; CH(VH), then Y =-1, so Y acts on
X*, contrary to 14.5.6. Hence (1) holds. By (3), QH:::; CT(VH), while by (1), we
may apply F.6.8 to CH(VH) in the role of "X" to conclude that CT(VH) :::; QH,
so (2) holds. Similarly F.6.11.1 implies (4), and F.6.11.2 implies (5) as His an
SQTK-group. D
LEMMA 14.5.8. H is solvable.PROOF. Assume His nonsolvable. Then by 1.2.1.1 there is KE C(H), and by
14.5.7.2, CH(VH) is 2-closed and hence solvable, so K* =f. 1. Then KE £ 1 (G, T) by
1.2.10, so by 14.3.4.1, K/02(K) ~ A5 or L3(2). Now 02(K) = 03 1(K) = CK(VH),
so K+ ~ K/02(K) ~ K*. By 14.5.7.5, H+ is described in F.6.18, so we conclude
that case (6) of F.6.18 holds, with H+ = K+ ~ L 3 (2). Hence K = 031 (H) =
(X, Y). Then K = 02 (H) by F.6.6.3, so that H = KT. Now as [Vi, Y] = 1
and \Vix) = Vi ~ E4, VH is the natural module for K* by H.5.5. In particular
Vi= \Vix):::; VH and VH = (Vl).
By 14.3.4.2, K E Cj(G, T), so by our discussion after Hypothesis 14.3.1, part
(1) of that Hypothesis holds with Kin the role of "L". Thus by Theorem 14.4.14,
either (Vfj^1 ) is abelian, or G ~ G 2 (3) or HS. However in the latter two cases, L