556 2. CLASSIFYING THE GROUPS WITH IM(T)I = 1
is of rank n, and hence [R, w] = [V, u] and [R, w] n Re = [V, u] n Re = 1. Thus
[Re, w] = 1, so [Re, vx] = 1, and hence 'P(Re) S 'P(GR(vvx)) = 1, contrary to an
earlier reduction. This contradiction shows that vx S GR(V) = VRc, and hence
vx S V Ai for i = 1 or 2 using the second claim.
Next suppose that x normalizes Ns(V). Set I := 01(Z(T))vvx. Then
I :'::l N 8 (V) = Ns(Vr using our assumption. Further as J(S) = R = SKRc,
01 (Z(T)) S VRe. Therefore as vx S VRe, IS VRc with [VRc,L1] =VS I,
and hence
I :'::l L1Ns(V).
Also arguing as above using 2.3.9.8, Ns(V) E fl. As 01 (Z(T)) SI, IE 85(G) by
1.1.4.3. Hence as No(I) contains Li 1. M, (Ns(V), No(I)) E U(Na(I)) and thus
Na(I) Ere. However S1 := (Ns(V), x) s NT(I) with IS1I = ISi, so again by 2.3.6,
IS 11 ~ IU1I for each U1 EU. Hence again from the maximality of IUI and/or ISi in
the definitions of HE I' or r in Notation 2.3.5, Na(I) E I'o. Then Na(I) E r5,
contrary to our assumption in this section that r5 = 0.
Therefore x does not normalize Ns(V). Set W := Ns(V) nNs(vr and Tw :=
S(x). As IS: Ns(V)I = 2 and Ns(V) =I-Ns(Vx), S/W ~ E4, Tw/W ~ Ds, and
we can choose x withs:= x^2 ES - Ns(V). Thus (Vx, vx-
1) = (V, vs)x. Hence
-1
setting D := [V, V^8 ], DX= [vx, vx ]. We showed [V, vx] = 1, and by symmetry
between x and x-^1 , vx-
1
also centralizes V, so (Vx, vx-
1
) centralizes V. Thus
conjugating bys,
(Vx' vx-l) s Gs(VV^8 ) = RcD.
Therefore nx S 'P(RcD) S Re. Also Dx :'::l S, so as K is not A5 since n > 1,
K :'::l Na(Dx) by 2.5.6.
Let p be a prime divisor of 2n - 1, and for J s G, let e(J) := QP' (J) if p > 3,
and e(J) := (j E J : IJI = 3) if p = 3. By A.3.18, either K = e(Na(Dx)), or
p = 3 and e(Na(Dx)) / 03 , (e( Na(Dx) )) ~ PGL 3 (2n). Thus, except possibly
in the exceptional case, as x^2 E Ns(D) and nx S Re, we have e(NK(D)) S
KX s Ga(D), impossible as [D, e(NK(D))] =I- 1. Thus K/Z(K) ~ L3(2n); 3
is the only prime divisor of 2n - 1, so that n = 2; and K/Z(K) ~ £ 3 (4) and
a subgroup X of order 3 in NK(D) induces outer automorphisms on Kx. Now
X s Y E Syl3 ( Na(D) n N 0 (Dx) n No(R) ) with Y = X(Y n Kx) ~ E 9 •
By a Frattini Argument, we may assume x acts on Y. Now Re = GR(X) from
the structure of K, so as Re n RC: = 1 by 2.5.5.1, R 0 = [R 0 , X] s K. Now Y
normalizes R and Kx, so Y normalizes R 0 ; then as R 0 is not elementary abelian,
R 0 = SK. This is impossible, as xx centralizes R 0 , but is faithful on SK. This
contradiction completes the proof of 2.5. 7. D
As a consequence of 2.5.7, the groups remaining in cases (2)-( 4) of 2.5.2 have
the following common features:
LEMMA 2.5.8. (1) Out(K) is a 2-group.
(2) K is simple so KoSe =Ko x Sc.
(3) Either SK is dihedral of order at least 8 or SK semidihedral of order 16.
(4) Z(S) = (Z(S) n SKo) x (Z(S) n Sc) and Z(S) n SKo = (zK) is of order 2,
where ZK is the projection on Ko of z.
(5) For each 4-subgroup F of K, NK(F) ~ S4; and furthermore if F S SK,