684 6. REDUCING L2(2n) TO n =^2 AND V ORTHOGONAL
Set LK := 02 (NK(RK)), so that LK/02(LK) ~ Z3. First suppose LK:::; M.
As K :S H = Ca(U), by 6.1.8 we obtain LK :S Kn M = NK(V). Then as
[LK, U] = 1 and U is of field type, [LK, V] = 1. But CuH(LK) = CuH(K), so
V:::; CuH(K):::; CuH(A), and then A:::; Na(V), contrary to paragraph one.
Therefore LK f:. M. By 6.2.4.2, Na(R) :::; M. If [Du, K*] # 1, then
R = 02(DuR) = 02(KR)(K n R) ~ LKR,
so LK :::; Na(R) :::; M, contradicting the reduction just obtained; hence [Du, K*] =
- Thus as A :::; K, [V K, Du] # 1 by our claim in paragraph two. Thus Du K*
acts on VK as GL 2 (4) with Du = Z(DuK*). As Na(R) :::; M but LK 1:. M,
R # RK_, so there is r E R inducing an involutory field automorphism on K.
This is impossible, as the field automorphism r* inverts the center Du of GL2(4),
whereas R ~ RDu. This contradiction completes the proof of 6.2. 7. D
For the remainder of the section, let z denote the generator of Z, set Gz :=
Ca(z), and Gz := Gz/Z. By 6.2.2.3, Gz 1:. M, so H1 # 0, where
H1 := {H:::; H(T) : H:::; Gz and H 1:. M}.
Consider any HE H 1 , and observe that Hypothesis F.7.6 is satisfied with LT, Hin
the roles of "G 1 , G 2 ". Form the coset graph r as in section F.7, and more generally
adopt the notational conventions of section F.7. By 6.2.3 and F.7.11.2, b := b(r, V)
is odd ..
LEMMA 6.2.8. V f:. 02(Gz)·PROOF. Choose H minimal in H1; then HE H*(T, M) n Gz. Thus n(H) = 1
by Hypothesis 6.1.1.2. We assume that V :::; 02 (Gz) and derive a contradiction.
Then V:::; 02 (H) so V:::; G~~) by F.7.7.2, and hence b > 1; thus b;::: 3 as we saw b
is odd. Let UH := (VH) :::; 02 (H). As b;::: 3, UH is abelian by F.7.11.4. As usual,
let 'Y E I' with d('Yo, 'Y) = b, and 'Yi at distance i from 'Yo on a fixed geodesic from
'Yo to 'Y· By F.7.11.6, [UH, u.,,] :::; UH nu.,,, where u.,, is the conjugate of UH defined
in section F.7.
As H :S Gz, H n M = NH(V) by 6.1.8. By 3.3.2.4, H n M is the unique
maximal subgroup of H containing T. Hence we may apply F.7.13 to UH in the
role of "A" to conclude there exists a E r('Y) such that m(UH/NuH(Va)) = 1.
As UH does not act on Vm there exists f3 E r('Y1) such that V,a does not
act on Va; we consider any f3 satisfying these two conditions. Notice that as
m(UH/NuH(Va)) = 1, also m(V,a/Nv/9(Va)) = 1. Let U,a := Cv/9(Va), so that
U,a :::; Nv/9(Va) < V,a. Then Va 1:. Na(V,a), since otherwise [Va, V,a] = 1 by 6.2.7,
contradicting U,a < V,a. As m(V,a/Nv/9(Va)) = 1, Ca(Nv/9(Va)) :::; Na(V,a) by6.1.10.1. So as Va centralizes U,a but does not normalize V,a, U,a < Nv/9 (Va); hence
Ua := [Nv/9 (Va), Va] is a noncyclic subgroup of Va. But Ua :S [UH, Va] :::; UH, so as
V,a :SUH which is abelian, Ua :S Cv"'(V,a). Now as V,a 1:. Na(Va), Ca(Cv"'(V,a)) 1:.Na(Va), so that m(Cv"'(V,a)) :::; 2 by 6.1.10.1; as Ua is noncyclic, we conclude
Ua = Cv"' (V,a) is a 4-group. Then Ua is of field type by 6.2.4.1. So as V,a centralizes
Ua, m(Nv/9 (Va)/U,a) = 1 by 6.2.4.2, with Nv/9 (Va) inducing a field automorphism
on Va· Then m(V,a/Nv/9(Va)) = 1 = m(Nv/9(Va)/U,a), so U,a is also a 4-group.Therefore as Va centralizes U,a but does not normalize V,a, Ca(U,a) 1:. Na(V,a), and
then U,a is also of field type by 6.2.4.1.