15.3. THE ELIMINATION OF Mf/CMr(V(Mr)) = Ss wr Z2 1141fused in G. Recall weak closure parameters r(G, V) and w(G, V) from Definitions
E.3.3 and E.3.23.LEMMA 15.3.46. (1) M controls G-fusion of involutions in V.
(2) For g E G - M, V n VB is totally singular. In particular if 1 < U < V with
Na(U) i M, then U is totally singular:
(3) r(G, V) > 1.
(4) W 0 (T, V) centralizes V, so that w(G, V) > 0, v^0 n M ~ Ca(V), andNa(Wo(T, V)) 5:. M.
(5) Na(Zs) 5:. M 2: Ca(v) for v EV nonsingular.
PROOF. Let (v) =Zs n V 1 ; as we just observed, v and z are representatives for
the orbits of Mon V#. Now SE Syh(CM(v)), and Ca(v) 5:. M by 15.3.45.3, so v
is not 2-central in G, and hence is not fused to the 2-central involution z. Thus (1)
holds. As Ca(v) 5:. M = Na(V), (1) and A.1.7.2 say that Vis the unique member
of v^0 containing v, so (2) holds. As no hyperplane of V is totally singular, (2)
implies (3). Similarly Zs is not totally singular, so (5) holds.
It remains to prove (4), so suppose g E G - Mand A:= VB 5:. T. We must
show [V, A] = 1, so assume otherwise.
Assume first that V 5:. Na(A). Then we have symmetry between V and A,
1 #- [V, A] 5:. V n A, and [V, A] is totally singular by (2). As [V, A] is totally
singular, m(A) = 1 and A= (a) with v 1 a =Vi. But as m(A) = 1, V centralizes the
hyperplane CA (V) of A, so that V induces a group of transvections on A, contrary
to Vt = V2 and symmetry.
Therefore V i Na(A). In the notation of Definition F.4.41, by (3), U :=
r 1 ,A(V) 5:. Nv(A), so U < V. Hence m(A) > 1 so m(A) = m2(M) = 2, and so A is
one of the-two 4-subgroups off'. As V = r 1 ,s(V), A is the 4-subgroup distinct from
S, so U = z1-and Cu(A) = Z. Let B := CA(V); then m(B) = 2 = m(Autu(A)).
As V i Na(A), Ca(B) i Na(A), so B is totally singular in A by (2). This is
impossible, as U centralizes B and m(Autu(A)) = 2, whereas the centralizer of a
totally singular line is of 2-rank 1.
Therefore W 0 (T, V) centralizes V, and hence w(G, V) > 0 and v^0 n Na(V) ~
Ca(V). Then by a Frattini Argument, M = CM(V)NM(Wo(T, V)), and it follows
that N 0 (W 0 (T, V)) 5:. M by 15.3.2.4. D
LEMMA 15.3.47. If x E Ca(Zs), then either [V,x] = 1 or v^0 n Na([V,x]) ~
Ca(V).
PROOF. Assume [V, x] #-1. As x E Ca(Zs), x EM by 15.3.46.5. Therefore as
[V, x] #- 1 and x centralizes Zs, [V, x] is not totally singular, so Na ( [V, x]) 5:. M by
15.3.46.2. But then V^0 n Na([V, x]) ~ Ca(V) by part (4) of 15.3.46. D
Observe that Mc E 'H(T, M), and in particular 'H(T, M) is nonempty.
In the remainder of the section, H denotes a member of 'H(T, M).
LetMH := MnH, VH := (VH), UH= (Zff), QH := 02(H), andH* := H/QH.
By 15.3.2.3 and 15.3.4, H 5:. Mc= Ca(Z), so we can form fI := H/Z.
LEMMA 15.3.48. If case (2) of 15.3. 7 holds, assume that Cy(V) 5:. H. Then
(1) Hypothesis F.9.1 is satisfied with Y, Zs, Z in the roles of "L, V+, Vi".
(2) Zs 5:. Z(T), UH 5:. n1(Z(QH)), and <P(UH) 5:. z.
{3) QH = CH(UH).