540 2. CLASSIFYING THE GROUPS WITH IM(T)I = 1
We can now argue much as in the proof of 2.4.8.5, but using M 0 , Lo in the roles
of "T, L", to show that B := DDx is abelian of odd order, omitting details except
to point out where the argument differs slightly: Notice this time that D normalizes
Q and the unique member Qx of A(S) with R = QQx. Further Dx = 02 (Dx), so
Dx normalizes each of the two conjugates L 1 and L2 of Lin LoS.
Now Gq is an SQTK-group, so mp(Gq) :::; 2 for pa prime divisor of JDJ; then
as mp(D) = 2 it follows that B = D = Dx.
Next x E T - S acts on B and hence on CR(B). As B = D and L is an
L 2 (2n)-block, CR(B) = CR(L 0 ), and as Q is abelian, CR(Lo) :::; Z(R) so that
CR(B) :::'.) LoNs(B) = LoS = H. Hence 1 = CR(B) = CR(D) by 2.4.4. Then
Cu 1 (L) = 1 and Q = CUo = Uo :::; Lo, so that Gq = Na(Lo), just as in the proof of
2.4.11.1. Also as RQ is Sylow in LoQ, R = QQx is Sylow in Lo. Then Lo= L1 x L2
so R = R 1 x R2, where Ri :=Rn Li E Syl2(Li) is of order q^3 , and D = D1 x D2,
where Di := D n Li, and D 1 and D 2 are the subgroups of D maximal subject to
CR(Di) =/= 1. Therefore as Ns(D) interchanges D 1 and D 2 , we may choose x in
Mo - S so that x normalizes D 1 and D 2 , and hence x acts on CR(D 3 -i) = Ri·
As L2 = [L2, Qx], x ~ Na(L2). Thus L2 < K := (L2, L~) :::; Ca(R1D1), and
S1 := (x)Ns(R1) = NM 0 (R1) normalizes K. Observe that JS: Ns(R1)I = 2 with
R = J(S) :::; Ns(R1), Q = 02(H) :::; Ns(R1), and HE r5. Thus Ns(Ri) E (3 by
2.3.8.5b. Then as L2Ns(R1) E He and L2 i. M, from the definitions in Notation
2.3.4 and Notation 2.3.5, (Ns(R1),L2Ns(R1)) E U(KS1), so that KSi Er.
We claim next that R = J(M 0 ): For suppose A E A(Mo) with Ai. R. By 2.4.3,
S = Nr(Q), so as R = J(S), there is an involution a EA-S; hence Qa = Qx,
since Mo = S(x) = S(a) and S acts on Q. If R't = R2 then CR(a) ~ R1 is of
rank 2n, while if R't = R 1 , then as Qa = Qx, fh(CRJa)) :::; Z(Ri), and so again
m2(CR(a)) :::; 2n. Now S/R is contained in the wreath product of a cyclic group
of field automorphisms of L 2 (2n) by Z 2 , so that m 2 (S / R) :::; 2; hence
4n:::; m(A):::; m(Mo/S) +m(S/R) +m(AnR):::; 1+2+m(CR(a)):::;3+2n < 4n
since n 2: 2. This contradiction establishes the claim that R = J(M 0 ).
Next from the proof of C.5.6.7, JA(R)J = 4, and Mo - Nr(R) induces a 4-
subgroup on A(R) generated by a pair of commuting transpositions. Thus either
Mo= Nr(R) and QNT(R) = {Q, Qx} is of order 2, or Mo< Nr(R) with QNT(R) =
A(R) and Nr(R) inducing Ds on A(R).
Assume that the latter case holds. Now D acts on each member of A(R),
so for each y E Nr(R), D :::; G~ = Na(L'b), and by 1.2.2, D :::; L'b. It follows
that Nr(R) normalizes the intersection RD of the groups Lo and L5; hence RD :::'.)
Nr(R)D, so Nr(R) = R(Nr(R)nNr(D)) by a Frattini Argument. Then arguing as
above, Nr(R) permutes the subgroups Di maximal subject to CR(Di) =I= 1, and so
permutes their fixed spaces {R1, R2}· Therefore Ns(R 1 ) is of index 2 in a subgroup
S2 :::; Nr(R) such that S2 acts on R 1 and U 2. We have seen that N 8 (R 1 ) E (3, so
S2 E (3 by 2.3.2.1. Next R1U2 = QQ^8 for s E S2 - S with A(R1U2) = {Q, Q^8 }, soN := Na(R1U2) = (Na(Q) n Na(Q^8 ))S2. By 2.4.3.1, Q E Si_(G), so by 1.1.4.1,
N E He. Then as L 2 :::; N with L 2 i. M, (S 2 , N) E U(N), so N E r. But
JS2I = ISi, so by 2.4.3.2, NE r5 and S2 E Syh(N). Now H1 := (S2, L2) :::; N and
as S2 E Syh(N) and NE He, Hi E He by Ll.4.4. Thus (S2,H1) E U(H 1 ), so