682 6. REDUCING L2(2n) TO n = 2 AND V ORTHOGONALU 0 is also of field type. Also (V; VB) S Ca(Uo), and by 6.2.4.4, VS 02(Ca(Uo)),
so (V, VB) is a 2-group and (1) holds.Thus we may assume that A is of order 4, so A 1:. L as we saw A tJ. Syl2(L).
From (1), our hypotheses are symmetric in V and VB, so also Autu(VB) is a 4-
group not contained in AutL9(VB). Let Q := UA and Q := Q/B. From the
action of A on V, B = Cv(A) is of order 2 and CA(V) is the centralizer in A of
each hyperplane of V. Also JVB : Bl = 8, so as JVBI = 16, it follows that B =
CA(V) = CA(U) = V n VB. Then we conclude Q ~ Q~ with B = Z(Q). Further
[[V,A],A] S Cv(A) =BS A, so [V,A] S Nv(A) S Nv(VB) = U by 6.2.5.6, andthus we conclude [V, A] = U as both groups have rank 3. Thus [V, A] S Q, so V
acts on Q, and then by symmetry, VB acts on Q. Hence Y := (V, VB) acts on Q.
Set Y := Y/Q, so that Y is dihedral, as V and VB are of order 2. We have
seen that [A, V] = U, so we conclude [Q, V] = U = CQ(V). Therefore V is
generated by an involution of type a2 in Out(Q) ~ Ot(2), Y* /Cy. (Q) ~ Ss with
Q :::;; 02 (Y), and the images of V and VB are conjugate in this quotient. Thus
U = CQ(V) is conjugate to A = CQ(VB) in Y, and hence U is conjugate to A
in Y. Therefore VB is conjugate to V in Y by 6.2.5.6. Thus V* is conjugate to
V*B in Y*, so that IY*I = 2 mod 4. Again by 6.2.5.6, Cy(Q) S Na(V), so as V*
inverts O(Y*), Cy(Q)* = Cy.(Q) = 1. Thus Cy(Q) = Z(Q) = B, so Y ~ S 3 /Qg,completing the proof of (2). D
LEMMA 6.2.7. Wo(T, V) S CT(V) = 02(LT), so that Na(W 0 (T, V)) SM.
PROOF. By E.3.34.2, it suffices the prove the first assertion. So assume by wayof contradiction that W 0 (T, V) 1:. CT(V). Then there is g E G such that V:::;; TB
but [V, VB] =/= 1. By 6.2.3, VB 1:. Na(V). Let U := Cv(VB). Then m(V/U) = 2
by 6.1.10.3, and as VB 1:. Na(V), Ca(U) 1:. Na(V), so the hypotheses of 6.2.4
are satisfied. Adopt the notation of that lemma (e.g., H = Ca(U), fI = H/U,UH = (VH), etc.) and let A := VB, B := Z~, and Du of order 3 in Lu. Then
V = [v; Du]. By 6.1.10.2 and E.3.10, VCa(A)/Ca(A) E A2(Na(A)/Ca(A), A), so
SB= VCsg(VB) and [A, VJ= B.We claim next that if K is a subgroup of CH·(D[;) with A S K*, then
[UH, K, D[;] =/= 1: For otherwise using the Three-Subgroup Lemma, A :::;; K* S
CH ([UH, D[;] S CH (V), contrary to the fact that A does not act on V.
Now ASH:= Ca(U) and VS UH, so B =[A, VJ SUH, which is abelian by
6.2.4.4. Thus UH S Ca(B) S Na(A) by 6.2.1, so we may take UH:::;; TB. Indeed
. as UH centralizes B, we have VS UH S CT9(Z~) =SB. Then UH= VCuH(A)
by the first paragraph of the proof, so [UH,A] = [V,A] = B, m(UH/CuH(A)) = 2,
and B = CA(Uo) for CuH(A) < Uo SUH.
We saw VS Ca(B), so BS NA(V). If B < NA(V), then as SB= VCsg(A),
B = [v; NA(V)] S V; but now Z~ = B s V =/= VB, contrary to 6.2.5.4. Hence
B = NA(V). We saw BS UH, so in particular B = CA(UH) as Ca(UH) S Na(V),
and hence A*~ E4.
Let B < Ai S A. Suppose that Ui := Cu-H(Ai) > CuH(A). We saw B =
CA(Uo) for CuH(A) < Uo S UH. Thus B = CA(Ui), so as [UH,A] = B, 1 =/=
[Ui, Ai] =: B1 S Un B. We will show that 1 =/= Un B leads to a contradiction.
For B is of rank 2, so m(B) S l. Then since [A, UH] = B, A* induces a 4-
group of transvections on UH with center B. Thus by G.3.1, there is K E C(H)