678 6. REDUCING L2(2°) TO n =^2 AND V ORTHOGONALbe in the list of F.l.12. As n = 3 and k = 1, the only possibility is the^3 D4(2)
amalgam of F.l.12.I.4. However, in that case Z is central in the parabolic Li with
Li/0 2 (Li) ~ L2(8), contradicting V the natural module for L/02(L) ~ Lz(S).
This contradiction shows that B i M. In particular K* is not^3 D4(2), soKi E .C(G,T). Next as R = [R,DL], and Out(K) is 2-nilpotent for each K,
R induces inner automorphisms on K, so that R ::::; 02 (B R) := C. Then
RU ::::; So := Sn C E Sylz(C), and as Ki/02(K1) is quasisimple, So = 02(C).
However No(So)::::; M by 6.1.23, contradicting Bi M.This contradiction shows K* is not of Lie type and characteristic 2. Thus by
our earlier discussion, either n = 2 and K* ~ L 2 (p) for p = ±3 mod 8 or Ji, or
n = 3 and K ~Ji. In each case as R = [R,DLl, R::::; 02(NK(Sk)R) := C*;
then the argument of the previous paragraph shows N K* (Bk) ::::; Mk.
Suppose K ~Ji. Then NK·(Sk) ~ Frob2i/Es is maximal in K, so Mk=
NK*(Sk)· Now DLTL :'.SJ MK, so we conclude DL is of order 7 rather than 3, and
DL :S [DL,MKl :SK :S Co(Zs)-impossible, as [Zs,DLl =Zs.
Therefore K* ~ Lz(p) with p = ±3 mod 8 and n = 2. As K* is not L 2 (4) byan earlier reduction, p 2 11. Therefore (1) and (2) are established.
As n = 2, DL is of order 3, so as m3(DLI) ::::; 2, m 3 (J) = 1 and hence
K == 0
31(I). As DL is not inverted in DLS and DL is faithful on K*, S in-
duces inner automorphisms on K*. As K = 031 (I), if Ko E C(J) with Ko =f=. K,
then Ko/02(Ko) ~ Sz(2k). As DL = 02 (DL), DL acts on each member of KJ by
1.2.1.3, and hence so does R* = [R*,DLl· The case [R*,K 0 l =f.1 was eliminated in
our earlier treatment of groups of Lie type in characteristic 2. Therefore R* cen-
tralizes K 0 I, so R* centralizes CF*(I*)(K*) in view of 6.1.24.3. Recall S* inducesinner automorphisms on K, so as 02 (1) = 1 by 6.1.24.4, we conclude R ::::; K.
Thus R::::; SK:, so as R = [R,DLl, we conclude R =SK:. In particular R* is of
order 4, so by 6.1.24.2, s = 1 and hence (3) holds. DLEMMA 6.1.26. If there exists e E E - V, then:
(1) R is transitive on eV.
(2) IE : VI ::::; 4.PROOF. Set Lo := (V9, V9^1 ), where l E Lis as in 6.1.21. Then ifg = tJ by
6.1.20.4, and so V^91 = tP. Therefore by 6.1.21.1, L = L 1 = L 0 and L ::::; L 1 =LoR. By 6.1.20.4, m(U/Cu(V^9 ))) = 2, so m(E/CE(V9)) ::::; 2 = m(Zs). Then as
Cz 8 (V^91 ) = 1 = Czi s (V^9 ) and L acts on Eby 6.1.21.3,
E = ZsCE(V^91 ) = Z1CE(V^9 ),so that E = VCE(Lo).
If E = V then the lemma is trivial, so assume e EE - V. As E = VCE(L 0 )
there is f E eV n CE(Lo). If [R, fl = 1, then f is centralized by R and Lo, so
L ::::; LoR ::::; Co(!), a contradiction as CT(L) = 1 by 6.1.6.l. This contradiction
shows [R, f] i= l. But by 6.1.21.3, [R, fl ::::; V, so as Lo is irreducible on V,
[R, fl = V. Therefore (1) holds, and we may take e = f E CE(Lo) =: F. Now
V^9 E::::; Cu(F) and V^9 = tJ, so IU: Cu(F)I::::; IU: V^9 EI::::; IU n R: El. But n = 2
by 6.1.25.1, and R/ E is the natural module for Li/ R by 6.1.25.3, so we conclude
IU: Cu(F)I ::::; 4. We saw R does not centralize f, so as Lo centralizes F and acts
irreducibly on R/ E, [Un R, Fl i= l. Thus there is u E (Un R) - Cu(F), and for
each such u, [F,ul::::; Zs by 6.1.17.2. Then IF/Cp(u)I::::; IZsl = 4 by Exercise 4.2.2