648 5. THE GENERIC CASE: L2(2n) IN .CJ AND n(H) >^1again K = £ 2. Then B s K = £ 2 , so as S = 02 (BT), a is not the Aut(J2)-
amalgam. Now either a is of type^3 D 4 (2) and 02 ' (Aut(K)) = Inn(K), or a is of
type Jz and 02 ' (Aut(K)) = Aut(K) ~ S 5 /E 16 • So either T = S s K; or a is
the J 2 -amalgam, IT: SI = 2, and (D3TB, TB, TK) is a weak EN-pair extendinga, and hence is the Aut(Jz)-amalgam. Therefore if a is of type J2, then T is a
Sylow 2-subgroup of either J 2 or Aut(Jz), so m 2 (T) = 4 and T has no normal E15-
subgroup. This is impossible as m(V) ::'.'.'. 4 from 5.1.6. Thus a is of type^3 D4(2)
with S = T, so K/0 2 (K) ~ £ 2 (8), B is of order 7, and Tis a Sylow 2-subgroup
of^3 D 4 (2). We are free to choose V to be (zL); thus Zs V, so V2 := (ZD^3 ) s V.
From the structure of a, V 2 s CT(B) ~ D 8 • As B acts on Land Z, B acts on
· (ZL) = V. Therefore Vz = Cv(B) and in particular [B, V] =fa l, so Bis faithful on
L/0 2 (£). This is impossible as n = 2, 4, or 8 and B acts on Sn L = T n L with
IBI =7.
This contradiction shows that 02(G 0 ) =fa 1. Let To := NT(L2). As TB acts
on DPS and S, and T 0 B acts on L2, T 0 B acts on Go; hence 02(GoToB) =I 1.
Thus as 02(Xvp) = 1 and Dp s Go, H 1:. GoTo; hence To < T and £2 < K.
Therefore either case (1) of 5.1.10 holds with £ 2 =Ki < K, or case (3) holds with
£ 2 < Ki = K. In either case £2 < K, T 0 < T, and K = (£2, £~) for t E T - To.
Furthermore as Tacts on DpTo, (L~)DpTo = (L~pTo)t, so as Dp 1:. Na(K) it follows
that Dp 1:. Na(L2).
Embed To in Ti E Sylz(GoToB). As IT: Toi = 2, ITi : Toi S 2. As SS To,
Na(To) s M by 4.3.17; hence Ti acts on T 0 nL, and then as Dp(TonL) ::;i·NM(Ton
£),Ti acts on DpTo.
By Theorem 3.1.1, applied with T 0 , Na(T 0 ), Hin the roles of "R, M 0 , H",
we conclude 02 (X) =fa l, where X := (Na(To), H). Now Ki E .C(X, T) and TE
Syl 2 (X), so by 1.2.4, Ki s Kx E C(X), and we set K+ := (K'i:). Recallingthat £ 2 < K, we conclude from A.3.12 and 1.2.8 that either K = K+ :SI X,
or Ki/02(Ki) ~ L2(4) and Ki < Kx, with Kx =I K1: for t E T - To and
Kx/02(Kx) ~Ji or L 2 (p). In any case, T n K+ =Sn K+ =Tin K+.
Suppose that T 0 < Ti. Set Ho := (£ 2 ,Ti) and Ko := (Lf^1 ). As T,Ti E
Sylz(X), and T n K+ =Ti n K+, K = (Ln = (Lf^1 ) from the structure of K+T.
Thus KE .C(Ho,Ti), K = (Lf^1 ) s GoToB, and applying 5.1.13 with G 0 ToB, Ti,Ho in the roles of "X, T, H", we conclude that either K/02(K) ~ £2(4), or Dp acts
on K. The first case is impossible as we saw £ 2 < K, and the second is impossible
as we chose p so that Dp does not act on K.
This contradiction shows that Ti = T 0 E Sylz(GoTo). Now we can repeat
parts of the proof of 5.1.13 with G 0 ToB, L 2 T 0 , Dp in the roles of "X, H, D" to
obtain a contradiction: We know GoToB E 1i(L 2 T 0 ) and Dp 1:. Na(L 2 ) from earlier
reductions. Then £2 < L2 E C(GoTo) using 1.2.4, and arguing as in 5.1.12 with £ 2
in the role of "Ki", one of conclusions (2)-(5) of that result must hold. Indeed as
L2 is normalized by the Sylow group To, conclusion (2) of that result cannot arise.
Then the argument in the second paragraph of the proof of 5.1.13 shows L 2 /0 2 (L)
is not of Lie type in characteristic 2 of Lie rank 2, so that conclusions (3) and ( 4) of
5.1.12 are ruled out. Hence we are reduced to case (5) of 5.1.12, and in particular,L2/02(L2) ~ £ 2 (4), with the embedding £ 2 < L 2 described in A.3.14. We saw
K/02(K) '¥-£ 3 (4), so by 5.1.10, K/02(K) is Sp 4 (4) or £ 2 (4) x £ 2 (4), and in either
case B ~ Eg. Next proceeding as in the proof of 5.1.13 with Dp in the role of "D",