1192 16. QUASITHIN GROUPS OF EVEN TYPE BUT NOT EVEN CHARACTERISTICTherefore 02 (L) = 1, so L ~ L 3 (4). Then Q centralizes E, and there is a
complement P to E in Q. Let fl' := H'/K'. Each Q 8 -subgro\lp of C.H,(z) is
contained in L'(z), so that P::::; T9 n K'L'E::::; L'E. Hence 1 -:f. (P)::::; L'. This
contradicts(+) as (P) = (v) where vis an involution in CL(r) with rv E rL.
Thus case (i) holds, so L ~ A6 or A 8. Since R = (r), Tc= (z), and Nr(K') ::::;
T9, r = z9. Further r induces an outer automorphism on L, and RL is not PGL2(9)
by 16.4.6, so we conclude that zG nLT ~LE and LE= LR x (z) ~Sn x Z 2 , with
n := 6 or 8. Represent LE on n := {1, ... , n} with kernel Tc= (z). Observe that
either LT= LE and T = TLE, or H ~ Aut(A6) and ILT : LEI = IT : TLEI = 2.
By (), zG n L(z) = {z}, so since zG n LT ~ LE, we conclude zG n LT ~
{ z} Ur LU rzL. If n = 6, we may choose notation so that r induces a transposition
on n, and if n = 8, r is either a transposition, or of type 23 , 12. In any case,
setting m := n/2 + 1, there is a subgroup Ao of TLR generated by a set of m - 1
commuting transpositions, and by the m - 1 conjugates of r under L in Ao. Thus
there is E 2 ,,, ~A::::; LE with a:= {z} U (rL n A) of order m.
Let a, b, c be a triple from a. Then c := z"' for suitable x E G; set X := L"'(z"').
By the previous paragraph,
zGnX={z"'}. (**).
Then a, b E L'1' A-X, so as zGnL"'T"' ~ L"' E"' and l(LE)"': XI= 2, ab EX. Henceby (**), neither ab nor abz"' is in zG. Thus no product of two or three members of
a is in zG.
Now assume n = 8, let a= {a 1 , ... ,a5}, and take a 5 = z"'. By the previousparagraph, aia2 and a3a4 are in X, so aia2a3a4 E X. Hence by (**), neither
aia2a3a4 nor aia2a3a4z"' = aia2a3a4a5 is in zG. Thus no product of four or five
members of a is in zG, so zG n A = a. But each involution in T is fused into A
under L, so we conclude that zG n (rz)L = 0, and hence z ¢. 02 (G) by Thompson
Transfer,^1 contrary to the simplicity of G.
Therefore n = 6. Set a = { ai, ... , a 4 } and zG n T =: (3. First assume H* ~
Aut(A6)· Then LT= LE= LA and T = TLR x (z). If t := a 1 a 2 a 3 a 4 ¢, za, thenthe argument of the previous paragraph supplies a contradiction, so t E zG. Hence
f3 = a U { t} is of order 5. But A(T) is of order 2, while the member A of A(T) is
normal in T, so A is weakly closed in T with respect to G. Hence by Burnside's
Fusion Lemma A.1.35, Na(A) is transitive on (3. Then as Naz (A)/A induces 83 on
(3, Auta(A) is a subgroup of 85 of order 30, a contradiction.
Therefore H* ~ Aut(A6).·Thus A(T) = {A,A^8 } for s E T-ATL, so Ca(z) is
transitive on A(Ca(z)); hence by A.1.7.1, Na(A) is transitive on (3. In particular
as INr(A) : Al = 2, l/31 -:f. 4, so t E f3 and l/31 = 5, for the same contradiction as at
the end of the previous paragraph.
This finally completes the proof of 16.4.9. DIn view of 16.4.9:
In the remainder of this chapter, we choose K' E ~ 0. Thus n1(R) ::::; KL,
where RE Syb(NK1(K)).
LEMMA 16.4.10. RE Syb(K').
PROOF. This is more or less the argument on page 101 of [Asc76]: By 16.4.2.1,
R ~ Nr 0 (K') = Cr 0 (R) =: S. Assume R ¢. Syl2(K'); then also S <Tc, so in(^1) Here we are in particular eliminating the shadow of 810.