16.4. INTERSECTIONS OF NG(L) WITH CONJUGATES OF CG(L) 1191

[To[ > 2, we showed z EL and L* is not L3(4) or M2~. So in any case z E Z(L);

it then follows by 16.1.2.2 that Z(L) = (z) is of order 2.

Suppose first that L ~ L3(4), and recall we are assuming that r does not

induce a graph-field automorphism on L. Assume that r induces a field auto-

morphism on L*, so that Lr~ L3(2) by 16.1.4.4. Let Lr,l be a maximal parabolic

of Lr; then there is an r-invariant maximal parabolic L 1 of L with Lr,l ~ L 1

and 02(Lr,1) ~ 02(L1). Then 02(L1) is transitive on r0 2 (Lr,1), contrary to (!).

Therefore by 16.1.4, r* induces a graph automorphism on L*, so Lr ~ L 2 (4)

by 16.1.4.6. Let U := T n Lr ~ E4 and V := EU; thus V ~ E 16 and as

Z(L) = (z), NrL (V)/V ~ E4 induces a group of F 2 -tranvections on V with axis

Vo:= (z,U) ~Es. Now R ~ T ~ Na(TL), so that [NrL(V),r] ==: W ~ U is a

hyperplane of Vo, and hence W = U. Then for 1 -/= v E W, rv E rL, contrary to

(!). This establishes the reduction to (ii) when L* ~ L 3 (4).

Next suppose that L* ~ G2(4),.M12, J2, or HS. Then [Z(L)[ = 2, so Z(L) =

(z); and from I.2.2.5b, r inverts y of order 4 in L with y^2 = z, so rY =.rz. Hence

the involutions in rZ(L) are fused under L; for example, from the description of

CL· (r) in 16.1.4 or 16.1.5, and observing that CL (r) ~ G 2 (2) when L ~ G 2 ( 4),

choose y E CL ( r) - L;, and observe y r E rL, so r inverts y as required. But

we showed earlier that rv = r^1 for some involution v EL; and l EL, and.now

from I.2.2.5a, we may choose a preimage v to be an involution. So as r^1 E rvZ(L)

and the involutions in rZ(L) are fused, rv E rL, again contrary to(!).

-This leaves the case L* ~ M22, where in view of 16.1.5 and 16.4.7, CL* (r*) =

L; ~ L 3 (2)/E 8. Choose an involutions* inducing an outer automorphism of the

type in 16.4.7, with r E s0 2 (Ls); then as we saw in the proof of 16.4.7, since

Z(L) = (z) is of order 2, the group 02(L~) ~ E15 splits over Z(L), so J(Cr(r)) is

the group V ofrank 6 in the proof of 16.4.7, now withs in the role of "r". Now the

argument in the final paragraph of that proof again supplies a contradiction. This

finally completes the proof of the reduction to (i) or (ii).

Thus to prove (2), it remains to eliminate the groups in parts (i) and (ii) of the
claim. Choose Tso that Nr(R) E Syb(Ns(R)). As in 16.4.2.4, choose g E G with
K^9 = K' and Nr(K') ~ T9...
S~ppose first that" (ii) hoids. Then Lr.~ Eg by 16.1.4.5, and NLE(E) = LrQ,
where Q/E ~ Qs and'(z,r) =.E:::,, CQ(Lr)·..
Assume that 02 (L) -/= ( Then since Tc= (z) is of order 2, (z) = Z(L). Also
CQ ( E) IE ~ Z4 is irreducible ori Lr' and Q induces a trarisvection on E with center
(z). By symmetry, Qg induces a transvection on E with center (r), so X :=;: (Q, Q^9 )
induces 83 on E. Since Nr(R) E Syb(Ns(R)), Cr(E) E Syl2(Ca(E)) by 16.4.2.6,
so by a Frattini Argument, Y := 02 (Na(E) n Na(Cr(E)}) is transitive on E#.
Hence as Lr ~ L',

Lr =0 "2(

n

yENa(E)

£Y) :::! Na(E),

•..

and then as Cr(E) is irreducible on Lr, either [Lri Y] = 1 or Y induces "SL 2 (3)

on Lr· In, the·former case as Cr(E) is irreducible on Lri there is an element

of order 3 in Y -:-Lr centralizing L,:, contradicting G quasithin. In the latter,

Nr(E)Y/E ~ GL 2 (3) has a unique Q 8 -subgroup, so that Q :::) Nr(E)Y, impossible

as Autq(E) is not normal in Autqy(E).