1188 16. QUASITHIN GROUPS OF EVEN TYPE BUT NOT EVEN CHARACTERISTIC03 ,(Lr) = 1, Lr is faithful on L' and Lr ::l 02 (Cb,(z)). We conclude from
16.1.4 first that z induces an outer automorphism on L', and second as 03 ' (H') is
PGL3(2n) or L~'^0 (2n) that either Lr :SL', or 02 (H') ~ PGL3(4) and z induces a
graph-field automorphism on L' with 02 (Cu(z)) ~ Eg ~Lr. In the former case
conclusion (1) holds, so we may assume the latter. Then ICL·(r*) : CL(r)*I :SIZ(L)I =: m, and by 16.1.2.2, m :S 4. On the other hand, CL (r) contains a
Q 8 -subgroup faithful on L;, so if m < 4, then r centralizes x EL with x^2 inverting
Lr; but then x^2 E L' by 16.4.2.5, so that Lr = [Lr, x^2 ] :S L', and conclusion
(1) holds again. Finally if m = 4, then r centralizes Z(L) by I.2.2.3b, so that
Z(L) :S Ca(r) :SH' using 16.4.2.5. Then as m2(CAut(L'J(Lr)) = 1, Z(L) nK' =/.1,
contradicting K tightly embedded in G.
Next we treat the subcase where r induces an inner automorphism on L. Recall
here that ICL(r)/02(CL(r))I = (2n - E)/3. Assume that n > 3. Then there are
prime divisors p > 3 of ICL(r)I, and mp(L) > 1 for each such p; hence L =OP' (H)by 16.4.5, so that OP' (Lr) :SOP' (H') = L'. Thus 03 (Lr) :SL', so that conclusion
(2) holds. Finally suppose that n = 3, so that L ~ U3(8), ILr : TLI = 3, and
02 (Lr) centralizes Z(TL)· Let x be of order 3 in.Lr; we may assume (1) fails, sox ~ L'. By 16.4.5, 031 (H') ~ PGU 3 (8) or U3(8), and CK'(r) is a 3'-group, so as
x ~ L', x ~ L'. As 1 =f. z E CH,(x), CH,(x) is of even order, and xis of order 3; so
from the structure of Aut(U 3 (8)), and as 031 (H') ~ PGU 3 (8) or U3(8), x EL', a
contradiction.This completes the treatment of the case where 031 (H') is described in case (2)
of 16.4.5, so we may assume that A(Lr) :S L9. In particular if Lr = A(Lr), then
conclusion (1) holds, so we may assume that A(Lr) <Lr; thus Lr =f. 1.
Suppose first that L ~ L 2 (q), q odd. Then q is a Fermat or Mersenne prime or 9
by (E2), and r does not induce an outer automorphism in PGL 2 (q) on L by 16.4.6.
If q = 9 and r induces an outer automorphism in 86 , then Lr ~ A4, contrary to
our assumption that A(Lr) < Lr. Thus we conclude from 16.1.4 that r induces an
inner automorphism of L. But then CL(r) is a 2-group, so Lr= 1, again contrary
to assumption.
Next suppose L* = X(2n) is of Lie type. We can assume by the previous
paragraph that Lis not L2(4) ~ L 2 (5), L 3 (2) ~ L 2 (7), or Sp4(2)' ~ L2(9). Hence
as A(Lr) <Lr, we conclude from 16.1.4 and 16.1.2.1 that L ~ L 2 (2n) for n > 2
even, and r induces a field automorphism on L. For each involution i E CT 0 (r),Lr ~ L2(2n/^2 ) is a component of CH' ( (i, r)) using 16.4.2.5. Therefore either Lr :SL'
so that conclusion (1) holds, or Lr::::; K', and we may assume the latter. Then using
16.4.2.5,
L = (CL(j) : j is an involution in Lr) :SH',
contrary to 16.4.2.2.
It remains only to consider the cases in (E2) where L ~ Ls(3) or L* is sporadic.
Inspecting centralizers of involutory automorphisms of L using 16.1.5, we conclude
that Lr = A(Lr), except in the situation which we already eliminated in 16.4.7.
This completes the proof of 16.4.8. DWe now focus on those members of ~ with the property that involutions of R
induce inner automorphisms on L. Set~o = ~o(K) := {K' E ~: fh(R) :S KL for RE Syb(NK'(K))}.