15.1. INITIAL REDUCTIONS WHEN L'.r(G, T) IS EMPTY 10972-group, <P(S) induces inner automorphisms on L. Then as S '.::J T by 15.1.18.3,
1 i-Zn <P(S) induces inner automorphisms, completing the proof of the claim.
Notice the claim eliminates case (d) of 1.1.5.3 where L ~ A 7 , as there each
z E z# induces outer automorphisms on L. Thus L ~ L 3 (2) or A 6 •
Next Y := 02 (Ca(Z)) ::::; Mz by 15.1.18.1, and so Y acts on L by 1.2.1.3.
However as Lis L3(2) or A5, CA.ut(L)(Z(SL)) is a 2-group, so as ZL induces Z(SL)
on L, 02 (CAut(L)(Z))::::; 02 (CAut(L)(Z(SL))) = 1, and hence Y::::; Ca 1 (L). Then asMc= !M(Ca(Z)), L::::; Na(Y)::::; Mc, contrary to our choice of L. This completes
the treatment of cases (a)-(d) of 1.1.5.3.
Next suppose case (f) of 1.1.5.3 holds. Then L/0 2 (L) is sporadic, Z induces in-
ner automorphisms on L, and Z(SL0 2 (L)/0 2 (L)) is of order 2. Thus by paragraph
one, Z induces Z(SL) on Land CL(Z(SL)) ::::; ML. Indeed if L/Z(L) is not M22,
M23, or M24, then CL(Z(SL)) is a maximal subgroup, so that CL(Z(SL)) =ML.
Hence in these cases either conclusion (d) of (3) holds, or L ~ J 4 , a case we post-pone temporarily. Next assume L/Z(L) ~ M22, M23, or M24· To complete our
treatment of case (f) in these cases, we assume that CL(Z(SL)) <ML, and derive
a contradiction. Here since F*(ML) = 02 (ML) 'from paragraph one, the subgroup
structure of L determines ML uniquely as a block of type A5, exceptional A 7 , or
L4(2), respectively. Therefore [ML, Z] f- 1 as CL(Z(SL)) = CL(ZL) = CL(Z).
Then ML::::; M:;o, so 1.2.1.1 says M:;o contains a member of .C1(G,T), contrary to
Hypothesis 14.1.5.1.To complete our treatment of case (f), we may assume L ~ J 4. Then there is
KE £(G1,S) with K::::; L, F*(K) = 02(K), and K ~ M24/E 2 1i. Now K::::; ML
by (2). But then L = (K, CL(Z(SL))) ::::; ML, contrary to our choice of Li. Mz.Finally suppose case (e) of 1.1.5.3 holds. We have already treated the cases
where L ~ L 2 (4) ~ L2(5) and L ~ L3(2) ~ L2(7). Thus Lis either L3(3), or L2(P)
for p > 7 a Fermat or Mersenne prime. By 15.1.19.7, X := 02 (CM(Z 1 )) acts on
L, and Vi acts nontrivially on L. Thus X is nontrivial on L since V2 = [V2, X]
by construction of Vi in 15.1.16. This is impossible since SL acts on X, whereas
neither L 3 (3) nor L 2 (p) has a subgroup of odd index in which an element of odd
order acts nontrivially on a normal elementary abelian 2-subgroup. (Cf. Dickson'sTheorem A.1.3 in the case of L 2 (p)). D
In the remainder of this section, we will eliminate cases from 15.1.20 until we
have reduced to case (2c), at which point we will derive our final contradiction.
We begin by eliminating cases (2a) and (3ab) of 15.1.20:LEMMA 15.1.21. Assume L E C(G 1 ). Assume further that either F*(L)
02 (L) with L an L 2 (2n)-block or an A 5 -block, or L is a component of G1 with
L/0 2 (L) a Bender group, Ls(2n) or Sp4(2n). Then L::::; Mz.
PROOF. Set SL := SnL, ML := LnMz, and assume that Li. Mz. By 15.1.20,
ML is a either the Borel subgroup BL of Lover SL, or a maximal parabolic of L.
Set Lo := (L^8 ); then Mz n Lo is either the Borel subgroup B := (Bf) of Lo over
Sn L 0 , or a maximal parabolic of Lo. So in any case, B::::; Mz, and S normalizes
B.
When L is a block, L = [L, J(S)] by 15.1.20.2, so the action of J(S) on L is
described in Theorem B.4.2. When L is a component, n > 1 by 15.1.20.3. Then