16.3. SHOWING L IS STANDARD IN G 1181by I.2.2.7b, contrary to(!!). Thus K ~ G 2 (4). By I.2.2.5a, 2-central involutions of
K lift to involutions of K; so since the uni potent radical of the stabilizer in K of
a point in the natural representation contains a product of two long roots groups
with elements permuted transtivitely by a subgroup L 2 (4) of a Levi complement,we conclude that m2(K) 2". 5. Therefore Z(K) = (t) by (!!).
Next from I.2.2.5b, short-root involutions in K lift to elements of order 4 in
K squaring to t. We may choose such a u of order 4 to normalize L. But now
as Tis Sylow in Na(L), if necessary replacing T by a Sylow 2-subgroup of Na(L)
containing (u, z), we may assume that u lies in T and so normalizes Tc. But by
16.3.8, Tc is dihedral or semidihedral of order at least 8 and Cr 0 (t) = (z, t), while
as v? = t, t centralizes the characteristic subgroup of Tc isomorphic to Z 4. D
LEMMA 16.3.10. Neither t nor tz is in zG.
PROOF. Suppose z9 = t. Then as L9 and K are distinct components of Gtdescribed in 16.3.9, m3(KL9) > 2, contrary to G quasithin.
Therefore t ¢:. zG. But by 16.3.8, tz E tTo, so also tz ¢:. zG. D
LEMMA 16.3.11. (1) (t,z) E Syb(Ca,(L)).
(2) (t) E Syl2(Ca,(K)).
(3) K ~As, L ~ A 6 , and z induces a transposition on K.
PROOF. By 16.3.8, (t,z) =:EE Syb(Ca,(L(z)), and by 16.3.10, tz tj:. z^0 , so z
is weakly closed in E with respect to Gt. Hence (1) holds, and of course (1) implies(2) since z does not centralize K.
Assume (3) fails. Then K appears in one of cases (2)-(4) of 16.3.9. Thus
1 -=f. 02 (K), so by (2), 02 (K) = (t), and if z induces an inner automorphism on K,
then z EK. Let K := K/(t).
Suppose z induces an inner automorphism on K. Then z EK by the previous
paragraph, so as z centralizes L, we conclude from 16.1.5 that z is a non-2-central
involution of R. Then from I.2.2.5b, the lift in K of z is of order 4, a contradiction.Thus z induces an outer automorphism on K. Again using 16.1.5, z centralizes
a non-2-central involution u in R. Thus a preimage u of u in K is of order 4, and
u acts on L = 02 (C.K(z)), sou acts on L. Now the argument in the last paragraph
of the proof of 16.3.9 supplies a contradiction. DLet Tt := Cr(t); as T E Syb(Gz) and L ::::! Gz, we may choose t so that
Tt E Syh(Ca,(z)). Let Tt:::; PE Syb(Gt)·
As K(z) ~ 88 by 16.3.11.3, we can represent K(z) as the symmetric group on
n := {l, ... , 8} with z := (1, 2). Then there is an involution u E CK(z) acting as
(1, 2)(3, 4) on n and inducing a transposition on L ~ A5. Let y denote a generatorfor the characteristic cyclic subgroup Y of index 2 in Tc provided by 16.3.8. Choose
w E {u, tu} with ICy(w)I maximal.
LEMMA 16.3.12. J(T) =Rx TL x (w), where either
(1) w centralizes Tc, R :=Tc if Tc is dihedral, and R is the dihedral subgroup
of Tc of index 2 if Tc is semidihedral; or ·
(2) IYI > 4, yw = yz, and R = (y^2 , t)".
PROOF. First (u, t) acts on Y with yt = y-^1 or y-^1 z for Tc dihedral or semidi-