14.6. ELIMINATING L 2 (2) WHEN (vG1) IS ABELIAN 1039(4) Ca(u) EI, so that I* is nonempty.
PROOF. Set To := CT(u). To prove (1) we must verify (Ul), (U2), and (U3).
By construction To is NT(UK,1) or NT(K), and so is of index 2 in T. Then as
To E Syb(CH(u)), (Ul) holds. By 14.6.21, NH=To (To)= T 0 , so as G 1 = H = H^00 T
by 14.6.23.1, Na 1 (To) = T, establishing (U3). As u E UK - Z, [K, u] =f. 1, and for
t ET-To, ut lies in either UK, 2 or UK•, so 1 =f. [K,uut]. By 14.6.1.5, K centralizes
D1(Z(02(G1))), so as F*(G1) = QH, neither u nor uut centralizes QH, establishing
(U2). This completes the proof of (1).In view of 14.6.22.1 and 14.6.23, we conclude from Theorem B.5.6 that for any
KE C(H),
U is not an FF-module for H* , and UK is not an FF-module for AutH(UK)·
(a)In particular, no member of H* induces a transvection on U, so by 14.5.18.1,
D'Y < Uy. Therefore by 14.5.18.4, we can choose 'Y as in 14.5.18.4; in particular
u; E Q(H*' U), and from that choice and (a):(b)In view of (b), [D, u;J =f. 1 by (a), so z'Y = [D, u;J by F.9.13.6. Then z'Y:::; [U, H^00 ]
by 14.6.23.3. Set g := gb, so that "(1g = 'Y and Z'Y := Z^9 plays the role of "A1" of
section F.9.Now we begin the proof of (3), which will be lengthier. Thus we assume that
K < H^00 and derive a contradiction. Observe that case (2) or (3) of 14.6.21 holds,
so that C(H) = {K, K+} with K/02(K) ~ K+/02(K+) ~ L3(2). By 14.6.22.1 and
14.6.23.3, U = f~h EB U+ EB Cu-(H), where U+ := [K+, U]. By 14.6.22.1, UK and U+
have rank 6.
Assumethatsomea* Eu; doesnotnormalizeK*. ThenCH•(a*) ~ Z2XL3(2),
andm(U/Co(U'Y))?: m(U/Cu-(a)) = m(UK) = 6 = 2m2(CH·(a*)),
with {a) the kernel of the action of CH·(a) on Cu-(a) of corank 6 in U. Thus
m(U;):::; m 2 (CH·(a)) = 3, while U /Cu-(U'Y) is of rank 6 if u; ={a) is of rank 1,
and rank greater than 6 if m(U'Y) = 2 or 3, contrary to u; E Q(H, U).
Thus u'Y normalizes Kand K+, and hence also UK and U+· So as u; is faithful
on F(H) = K K.f. we may choose notation so that K = [K, u;J.
We claim that z'Y :::; UK or U+. Suppose otherwise. Then as [D, U] :::; z'Y by
F.9.13.6, u'Y centralizes jj n UK and jj n U+. Then by (a),
m(UK/(UK n D))?: m(UK/Cu-K(U'Y))?: m2(Autu'Y(K*)) + 1. (c)
Set u+ := U/(UK + Cu-(H)). As u; is faithful on K* K.f. and normalizes both
factors,
(d)so using (b)-(d):
m := m(Autu'Y(K.f.))?: m(U;)-m(Autu'Y(K*))?: m(U/D)-( m(UK/(UKnD))-1)
?: m(u+ jD+) + 1?: m(u+ /Cu+(U'Y))-m + 1, (e)