1058 14. L 3 (2) IN THE FSU, AND L 2 (2) WHEN .Cr(G, T) IS EMPTY
Yo:= (QH,02(Ga)) induces GL(Vy) with kernel 02(Yo) = CqH(Vy)Co 2 (Ga)(Vy),
and
Na(Vy) ::::; Na(Yo).
Set Ty:= UaQHCTnKoQH(U~) and Y :=(Ty, 02(Ga)). Then Ty centralizes
U~ and preserves the F 4 -structure on U, so Ty centralizes the F4-point [U, U~]
containing A}, and hence acts on Vy. Then by(**), Yo :::] Y = YoTy, and Y acts
on Vy.
As Ua ::; R 1 , from the structure of H*:(t*L~T*) = (t*T*) = (t*)Co2(Li)(t*) =TY.;that is, Ty :::l LiT.
Recall that 02 (Yo) = Cy 0 (Vy), so 02(Yo) ::; 02(CH(Vy)) by (**), while
02 (CH (Vy)) = U~ from the action of H on the orthogonal module U, so 02 (Yo) ::::;
QHUa::; Ty. Thus Ty E Syb(Y), and as Yo induces GL(Vy) on Vy, Cy(Vy) =
02 (Y). Further Vy ::; U, so Vy ::::; UY for each y E Y. This completes the verifi~cation of the hypotheses for part (2) of 14.7.31, so we conclude from 14.7.31.2 that
Wo(Ty, V) :::l Y.
Set I:= (L 1 T, Y). We saw earlier that LiT acts on Ty, so I acts on Wo(Ty, V).Set Vr := (Vl) and J+ := I/Cr(Vr); as usual Vr E R2(J) by B.2.14. Also Vy::::; Vr
as Y::::; I. We claim that Lt is not subnormal in J+: For otherwise 02(L1)+ = 1,
so that 02 (Li) centralizes Vy. This is impossible, as 02(Li) does not act on Vy
since 02 (Li) E Syb(K[)) and A.} is nonsingular. This completes the proof of the
claim. By the claim, Lt-=/= 1 and also Yo i Na(Li).Now Lo := CL 1 (Vy) ::::; Na(Yo) by(**), so [Yo, Lo] ::::; Cy 0 (Vy) = 02(Yo) ::::;
Ty ::; Na(L 0 ). Thus Yo acts on 02 (L 0 ) =: Ly. Also Li = L 0 02(Li) from the
action of Hon U, so Li = Ly02(Li).Suppose next that Yo ::; M. Then as Yo normalizes Ly, we conclude from the
structure of Aut(L 3 (2)) that 02 (Y 0 ), and hence also 02 (Y 0 )Ty = Y, acts on Li,
whereas we saw that Yo i Na(Li).
Therefore Yo i M. We claim next that [Vr, J(T)] -=/= 1. For otherwise J(T) ::::;CT(Vr)::; CT(Vy)::::; Ri from the action of H* on U. Then J(T) = J(02(Y)) by
B.2.3.3, so that Yo ::::; Na(J(T)) ::::; M = !M(LT) using 14.3.9.2, a contradiction
establishing the claim.
By the claim, J(J)+ -=/= 1. If J(J)+ is solvable, then by Solvable ThompsonFactorization B.2.16, J(J)+ has a direct factor Kt~ 83 , and there are at most two
such factors by Theorem B.5.6, so that Kt is normalized by 02 (J+) and Lt. If
J(I)+ is nonsolvable, then there is Kr E C(J(J)) with Kt-=/= 1, so Kr E LJ(G,T)
by 1.2.10-and then by parts (1) and (2) of 14.3.4, Kt is A 5 or L 3 (2), and Kr :::] J.
We saw that Ly= 02 (L 0 ) contains a Sylow 3-subgroup PL of Li, and that Lo
acts on Yo. Since Vy= [Vy, Yo], PL::::; PE Syls(YoLy) with P ~ Eg. As Lt-=/= 1,
Pt-=/= 1, and then as Pt= Cp+(Vy), p+ ~ E 9 • From the previous paragraph, p+
normalizes Kt and Out( Kt) is a 2-group, so P =PK x Pc, where PK := P n Kr
and Pc:= Cp(Kt). Now Px has order at most 3 by the structure of Kt, and Pchas order at most 3 by A.1.31.1, so we conclude both PK and Pc are of order 3.
As Y = (P n Y 0 )Ty, I = (L 1 T, Y) = (L 1 T, P), so as Lt is not normal in J+, p+
does not act on Lt. Finally one of the following holds:
(a) Lt ::::; Kf.