i3.2. SOME PRELIMINARY RESULTS ON A 5 AND A 6 88iLT by 13.2.4.1, so Na(J(R)):::; M. To complete the proof we show H acts on J(R),
contrary to H i M.Assume H does not act on J(R). Then 02 (H) < R, so by E.2.1, LH ~ Z 3 , E 9 ,
of 31+^2 , and as X f= 1, the last case must hold. Then byt 13.2.8.3, H* ~ 83 wr Z 2 ,
and as R = Cy.(X), R ~ Z 4. But then from the action of H* on UH, J(R) =J ( 02 ( H)), contrary to assumption. D
LEMMA 13.2.10. Ifs= 2, then LH E 3j(G, T).
PROOF. Assume s = 2. Then by 13.2.8.3 and 13.2.9, T is irreducible on
LH/02(LH) ~ Eg, so that LH E 3(G, T). As [Z,LH] f= 1, LH E 2t(G,T). Furtherif LH.::; Lo for some L 0 := (L~) with L+ E C(G,T), then L+ E Ct(G,T). There-
fore by hypothesis (*), L+/02(L+) ~ A 5 and L+ E Cj(G, T), so Lo = L+ since
conclusion (3) of Theorem 12.2.2 holds by Hypothesis 12.2.3; but this contradicts
m 3 ( L H) = 2. Thus no such Lo exists, so by definition L H E 3j ( G, T). D
LEMMA 13.2.11. Assume Z(H) = 1. Then
{1} Cr(L) = Cr(LH) = CE(L) = CE(LH) = 1.
{2} J(T) = S = ((1, 2), (3, 4)) ~ E4.
{3} s = 2, E = (ei,2, e3,4) = (ei, e2) ~ E4, and Z = (eie2) is of order 2, and
{interchanging Hi and H2 if necessary) we may take ei = ei,2 and e 2 = e3,4·
(4) To:= Nr(Hi) = Nr(H2) = Cr(ei) = Cr(e2) = Q8 = 02(H)8.
{5} L is not an A 5 -block.{6} 02 (H 2 ) is not an A 3 -block.
PROOF. As Tacts on CE(LH) and Z(H) = 1, CE(LH) = 1 = Cr(LH), and
hence E ~ E2• by 13.2.8.
Next as we saw that L = [L, J(T)], (2) follows from 13.2.4.2. Thus V n E
contains (ei, 2 , e3,4) ~ E4, so as E ~ E2•, we concludes= 2 and E = V n E.::; V.
As Z :::; E :::; V, Z = Cv(T) has order 2 and is generated by z := ei,2e3,4. As
Z.::; V, Cr(L) = 1, completing the proof of (1). Further E = (ei,2, e3,4) = EiE2.
Then Z = (z) = (eie 2 ), so (3) holds. Most of the equalities in (4) are clear; observe
To= 02(H)8 by 13.2.8.4, and To= Q8 by (2) and (3).
If Lis an A 5 -block, then by C.1.13.c, Q = 02(LT) = V x Cr(L). Thus Q = V
by (1). Now f' 0 = ((1, 2), (3, 4)) by (2) and (4), so as Q = V, To~ D8 x D8. Thus
LHTo ~ 84 x 84, so 02 (H2) =: K2 is an A3-block.
Therefore if (5) fails, then so does (6); so to prove both parts of the lemma, wemay assume that K 2 is an A 3 -block. Thus Ki = K~ is also an A3-block; and again
by C.1.13.c, Ki~ A4 and 02(H) = Cr(LH) x VH. Thus 02(H) = VH by (1), so
H~S4 wr Z2..
By 13.2.10, LH E 3*(G, T), so Mi:= Na(LH) = !M(H) by 1.3.7. As 02(H) =
VH = 02(LH), 02(H) = 02(M1) using A.1.6. Then as F*(Mi) = 02(Mi),
CM 1 (VH) = VH so that Mi/VH :S GL(VH)· Then as H/VH ~ Ot(2) is a maximal
subgroup of L 4 (2) with Sylow group T/VH ~ Ds, we conclude that Mi= H. But
now Theorem 13.9.1 contradicts the simplicity of G. D
Set Ho := (H, Li).
LEMMA 13.2.12. 02(Ho) f= 1.
PROOF. Assume that 02 (H 0 ) = 1. Since Li = 02 (NL(T n L)), we conclude