1178 16. QUASITHIN GROUPS OF EVEN TYPE BUT NOT EVEN CHARACTERISTICObserve also (cf. I.7.2.5):
REMARK 16.3.3. If Lis standard in G, then for each nontrivial 2-subgroup Xof Ca(L),
Na(X) :S Na(Ca(L)) = Na(L).
To show that L is terminal, we need to eliminate the proper inclusions of L in
K in parts (2) and (3) of 16.1.7. The first elimination makes use of an approach
suggested by Richard Lyons. Although the method could be applied without appeal
to Theorem 16.2.4, it goes more smoothly with such appeal. The method could also
be used to eliminate other proper containments in 16.1.7, but it is easier to use other
arguments like those in 16.3.9.LEMMA 16.3.4. Assume t is an involution in To, and L is a component of
CK(i) for some component K of Ca(t) and some involution i E Ca(L(t)) with
K = [K, i]. Then K is not U3(2n), M 12 , h, HS, or Ru.
PROOF. Assume K is one of the components we wish to eliminate. Inspectingthe list of possibilities for the pair L, K in 16.1. 7, we find that L is simple, so
TLTo =TL x To. By Theorem 16.2.4, Tacts on L, so IT: Toi:::; IAut(L)l2, while
by inspection of the pairs on our list, IKl2 > IAut(L)l2, so IKl2!Tol > ITI.When K ~ U 3 (2n) set V :=TL, and in the remaining cases choose V of order
2 in TL n Z(T). Thus in any case Ton V = 1 and T:::; Na(V), so as TE Syl2(G),
TE Syb(Na(V)). Thus for SE Syl 2 (Na(V)), S = T9 for some g E Na(V), and
setting SA:= Tl for A E {C, L}, So :::] S, IKl2IBol > ISi and Son V = 1.
Next we claim that there exists SK E Syb(K) such that V = Z(SK): This
is clear from the embedding of Lin K when K is U 3 (2n), while in the remainingcases we will show that 2-central involutions in L are 2-central in K, so the claim
holds there too. Choose SK so that Si:= CsK(i) E Syb(CK(i)). Then Si contains
an involution z in Z(SK), and by inspection of CK(i) for the pairs on our list,
this forces z E L: This is evident if Z(Si) :::; L, while in the remaining cases all
involutions in Csi (L) are not 2-central and all involutions in Z(Si) - L are fused
into CsJL) under Nx(Si)·
By the claim, Bx :::; Nx(V); let SK :::; S E Syl2(Na(V)). Then ISKllBol =
IKl2IBol > ISi, so SxnSo -=f. l. By the claim, Z(SK) = V, so as 1-=f. SxnSo ::::1 Bx,
V n (SK n 80 ) -=I-1, contrary to 80 n V = 1. D
LEMMA 16.3.5. Assume E is a subgroup of G of order 4, and K is a component
of Ca(e) for each e E E#. Let i be an involution in Ca(EK). Then one of the
following holds:(1) K is a component of Ca(i).
(2) K < I, where I is a component of Cc(i) such that E is faithful on I,
02(I)-=/:-1, E ~ CAut(I)(AutK(I)) ~ E4, and either
(a) K ~ A5 and I/02(I) ~ M12 or J2, or
(b) K ~ Sz(8) and I/02(I) ~Ru.PROOF. Assume that conclusion (1) fails, and set I := (KE(Oa(i)l). Then I
and the action of an involution t E E# on I are described in conclusion (2) or (3)
of 16.1.7, with K, I in the roles of "L, K ". Observe that CE(J) = 1 since K <I
and K is a component of Ca(e) for each e EE#, so Eis faithful on I.
Suppose the pair (t, I) satisfies case (2) of 16.1. 7. Then I= Iiii with Ii -=f. Ii,
for some component I1 of Ca(i) such that fi/Z(Ii) ~ K/Z(K). Thus NE(J 1 ) -=f. 1