i3.9. CHAPTER APPENDIX: ELIMINATING THE Aw-CONFIGURATION 973Next by 13.9.8 and (*), Gz ::::; K and B n Q = (z, tG"). We conclude using
13.9.8.1 that the orbits of Gz on B# are E 0 := {z}, Ei := ri of length 6, and two
orbits Ei, i = 2, 3, on B - Q of length 4. Then since r 4 = B n .6.. 4 s;;; Q, appealing
to 13.9.4, we may choose notation so that E 2 = B n .6.. 3 and E 3 = B n .6.. 5.
By 13.9.2.3, K controls fusion in B, so it follows from (5) that the Gz-orbit E 3
is fused to z or t under K. In particular, Gz < K. Thus there are three possibilities
for zK: {z} U E3, {z} U E2, or {z} U E 2 U E3-of order 5, 5, or 9, respectively. Now
by 13.9.7, Cc(B) = CJ(T)(B) = B, so K/B::::; GL(B). As IGL4(2)J is not divisible
by 27, while 3 divides the order of CK(z) by 13.9.8, we conclude JzKI = 5 rather
than 9. Set K := K/B;·then CK(z) = G~ ~ 84 by 13.9.8. Further B = (Ei)
for i = 2, 3, so Bis the kernel of the action of CK(z) on E 2 and E 3. We conclude
K* acts faithfully as 85 on zK. As K has orbits of length 5 and 10 on E#, it
follows that Bis the natural module for K* ~ 0.4(2), with zK the singular points
of the orthogonal space B and tK the nonsingular points. This establishes (1) and
(2). Also if k EK - Gz then zzk E tK, while if zk E .6.. 5 , then zzk E .6.. 5. Thus
E 3 = B n .6.. 5 ~ zK, so zK = {z} u E 2 = {z} u (B n .6..3), and .6..5 n B = E 3 s;;; tK.
Now it follows using (*) that (3) and (4) hold. D
LEMMA 13.9.10. Let E :=Ti nA = 02 (Mi). Then E centralizes 02 (CK(t)) ~
A4·
PROOF. From the structure of K described in 13.9.9, and as J(T) E 8yl 2 (Gt)
by 13.9.2.6, CK(t) =Rix X, where t E Ri ~ Ds, and X ~ 84 with 02(X)# s;;; tK.
Let R2 := T n X. Then Ti x T2 = J(T) =Rix R2 with (t) = [Ri, Ri] =[Ti, Ti],
(t8) = [R 2 , R 2 ] = [T 2 , T 2 ], and z = tt^8 • Now by the Krull-Schmidt Theorem A.1.15,
Ti(z) = Ri(z) for each i = 1,2. Therefore E::::; Ti::::; Ri(z).
Suppose E t;_ R 1. Then there is e E E - (t) with ez E Ri. As En B = (t),
e rf B and hence ez rf B. Further ez E zM by 13.9.3.1, so ez = z9 E (Ri n zG) - B
for some g E G. Then as X centralizes z9, from the description of Gz in 13.9.8,
02 (X) = [0 2 (X), 02 (X)] ::::; Q9. So as 02 (X)# s;;; tK, it follows from (*) in the
proof of 13.9.9.1 that
U := (tG n Q^9 ) = (z^9 ) x 02(X).
By 13.9.8, (tGz n Q) = B n Q, while Cc(B n Q) ::::; Gz,t = J(T) by 13.9.7, so
we conclude U = B9 n Q^9 and Cc(U) ::::; CJ(T)9(U) = B^9. Now CR 1 o 2 (X)(U) =
CR 1 o 2 (x)(z^9 ) ~ Em ~ Cc(U), so B^9 = CR 1 o 2 (x)(U) ::::; K = Nc(B). Hence
Bg = B as B is weakly closed in T by 13.9.2.3, contradicting z9 tj. B.
ThiscontradictionshowsthatE::::; R 1. HenceEcentralizes0^2 (X) = 02 (CK(t))
~A4. D
We are now in a position to prove Theorem 13.9.1. The argument will be much
like that in the proof of 13.9.5.
Let E be as in 13.9.10, and recall GE = Cc(E). As J(T) E 8yb(Gt) by
13.9.2.6 and t EE :SI J(T), Ex T2 = CJ(T)(E) E 8yl2(GE)· Let H := 02 (GE),
H 1 := 02 (M 2 ), and H 2 := 02 (CK(t)). Thus Hi ~ A4 centralizes E, using 13.9.10
in the case of H2. Therefore Hi ::::; H. Furthermore 02(H1) = T2 n A, while
02 (H2) = H2 nB, with 02(Hi) n02(H2)·= (tz). Therefore as tz E 02(H2)# s;;; tG,
02 (H 2 )::::; T 2 , so as H 2 nB t;_ A, 02(H 2 ) normalizes but does not centralize, 02(Hi),
and then TH := 02 (Hi)02(H2) ~ Ds. As Hi ::::; H, TH ::::; H. As E ::::; Z(H), by
Thompson Transfer, EnH = 1, so that TH E 8yb(H). As all involutions in TH are