14.3. FIRST STEPS; REDUCING (vG1) NONABELIAN TO EXTRASPECIAL 1003K1 is G2(2)' or A5, then Ki contains all elements of order 3 in H by A.3.18, so
L = (L1, L2) :::; (K1, h) :::; Ge, contrary to 14.3.25. On the other hand if iI ~ 85 ,
then iI contains no involution of type a 2 , contrary to 14.3.30.2.
Therefore case (i) of 14.3.23.2 holds. so iI ~ 83 x 83. Since a has type
a2, (J = [U,K], and since [U,a] = v, a centralizes v, so (a) = z&, k ~ 83 ,
iI = kLd', and K 1:. M by 14.3.23.2.
We saw earlier that Ke:= (KT,12T) :::; Ge; set Ue := (v'{e), Kt := Ke/CKJUe),
andKe := Kt/03'(Kt). Then0 2 (Kt) = 1 byB.2.14, soa := (I:J:T+,T+,K+T+)
is a Goldschmidt amalgam in the sense of Definition F.6.1. Observe that Vi =
(V/^2 ) :::; Ue, so U1 := (V2K) :::; Ue. Now K/02(K) ~ 83, U = [U, K], and
[Vi, U] = V1; so F*(K/CK(U1)) = 02(K/CK(U1)) and hence F*(K+) = 02 (K+).By 14.3.29.2, J(T) ~ LT. Hence J(T) :::; 02 (hT), and as K f:. M = !M(LT),
J(T) f:. 02(KT) in view ofB.2.3.3, so 02 (K) = [0^2 (K), J(T)]. Thus 02 (K+T+) =f.
02(I:}T+), and Ue is an FF-module for Kt. By F.6.11.1, 031(Kt) is of odd
order, so K+T+ ~ KT and rtT+ ~ 12 1', and hence F*(K) = 02 (K). Then
as 02(K+T+) =f. 02(ItT+), F.6.11.2 says Kt ~ Ke is described in Theorem
F.6.18. As F*(K) = 02(K), cases (1) and (2) of F.6.18 are ruled out. In the
remaining cases, Kt ~ Ke is not solvable, so Ko := K'i° E £1(G, T) by 1.2.10.
Then by 14.3.4.1, Ko/02(Ko) ~ L3(2) since A5 is not a composition factor of any
group in F.6.18. Then Ke appears in case (6) of F.6.18, so Ke = K 0 Y with Y
the preimage in Ke of 03 1(Kt). As 02 (K) = [0^2 (K),T n Ko] and 02 (K) is T-
invariant, 02 (K) :::; Ko. Similarly 02 (1 2 ) :::; Ko, so Ko = 02 (Ke) using F.6.6,
and hence Ke = KoT. Also Vi = (V/^2 ) and KT centralizes V 1 , so by H.5.5,
Ue = (V{e) is a 3-dimensional natural module for Kt~ L 3 (2). Thus Ue = (V2K).
We saw earlier that U = [U, K], K centralizes Zu, and CK(U) = 02 (K). Therefore
U = [U, K] EElZu. Now as iI = KL1T, U = (Vf^1 K), so Vi 1:. [K, U] and Ue = (Vl)
has rank greater than 3, contradicting m(Ue) = 3. D
u.
LEMMA 14.3.32. (1) U is the L2(4)-module for iI ~ 85.
(2) U ~ Q~ * Z4.
(3) Q = CH(U), so that iI ~ H*.
(4) H =KT with KE C(H), U = [02(K),K], and K acts indecomposably onPROOF. By 14.3.31, Z(h) = 1, so that by 12.8.10.6,
Czi (U) = Vf , so zg ~ Zu =f. I.
By 14.3.24, either case 14.3.14.3 holds with iI ~ G 2 (2), or iI is described in one
of cases (i)-(iii) of 14.3.23.2. Then d = 6 or 4, respectively. By 12.8.11.2, m(E) :::;
d/2. Then we can use 12.8.11.5 to show m(E) = d/2 and m(W9 I zg) = d/2-1: For
when d = 4, E = V ~ E4 by 14.3.23.1, and when d = 6, m(W9) = 3 by 14.3.14.3
and zg =f. 1 by (). This also shows m(Z&) = 1 when iI ~ G 2 (2). When d = 4,
m(W9/Z&) = 1by14.3.23.3, so as zg =f. l by(), either m 2 (iI) = 2, m(W9) = 2,
and m(Z&) = 1, or case (iii) of 14.3.23.2 holds with iI ~ 86 , m(W9) = 3, and
m(Z&) = 2.
Thus in view of (*), we have shown that either IZul = 2, or iI ~ 86 and
I Zu I = 4. In either case, H^00 centralizes Zu by Coprime Action, and in the former
H centralizes Zu. Thus as H = H^00 T in the latter case, (3) holds by 12.8.4.4.