640 5. THE GENERIC CASE: L2(2n) IN .Cf AND n(H) > i
composition factor. By F.1.12.II, T:::; K+. But then T:::; K+ S CK, contradicting
TnK-f:.CK. D
Notice now that to complete the proof of Theorem 5.1.14, it suffices to prove
part (4) of 5.1.14: Namely we have already established the first three parts of
Theorem 5.1.14. Further if part (4) holds then DL = B S M+, which by 5.1.16
forces CK to be a solvable 3'-group, establishing part (5) of 5.1.14.
LEMMA 5.1.17. n is even.PROOF. Assume n is odd, so in fact n ?: 3 as n > 1. Let F := F2n· Then
T induces inner automorphisms on L, so T :::; L. By 5.1.15, DL :::; M+; then
as DL is a {2, 3}'-group acting on T and K/0 2 (K) ~ L 3 (4), we conclude that
DL s CK:= Ca(K/02(K)).
We now specialize our choice of V to be the module "V" in the Fundamental
Setup (3.2.1) for L, as we may by 3.2.3. As L/0 2 (L) ~ L 2 (2n), case (1) or (2)
of Theorem 3.2.5 holds, so Lis irreducible on V/Cv(L) and V is a TI-set under
M. Since n is odd, V/Cv(L) is the the natural module for L by 5.1.3; then as
Cz(L) = 1 by 5.1.14.3, Vis a natural module. Let Zi :=Zn V. Notice as T:::; L,
Zi is the I-dimensional F-subspace of V stabilized by T. In particular Zi is a
TI-set under NM(V), so as Vis a TI-set under M, Zi is a TI-set under M.
Observe also that Lis not a block: For if it were, then as Cz(L) = 1, CT(DL) =
1, contradicting DL :::; CK. Also CK is a solvable 3'-group by 5.1.16, since we saw
DL SM+.
Let S := Baum(T), and recall from 5.1.7.1 that Na(S) :::; M.
We claim Zi is a TI-set in G. For let Z 0 := (Z°K); then Zo E 1?'2(CKT)
by B.2.14. As CK is a solvable 3'-group, by Solvable Thompson Factorization
B.2.16, [Z 0 , J(T)] = 1, so that S = Baum(CT(Z 0 )) using B.2.3. Now by a Frattini
Argument, CK = CcK(Zo)NcK(S). Then as Zi :::; Zo while NcK(S) :::; Mand
Zi is a TI-set under M, Zi is a TI-set under CK. Now n # 6 since n is odd, so
by Zsigmondy's Theorem [Zsi92], there is a Zsigmondy prime divisor p of 2n - 1,
namely such that a suitable element of order p is irreducible on Zi. Let P E
Sylp(CK)· As DL S CK= CcK(Zo)NcK(S) with NcK(S) SM, we may choose P
so that P = Cp(Zo)(P n M) and PL := P n DL E Sylp(DL)· By the choice of p,
PnM =PL x CPnM(Zi), so P = PLCp(Zi), and Pis irreducible on Zi. Therefore
Zi is a TI-set under NM+(P). Further by a Frattini Argument, M+ = CKNM+(P),
so as Zi is a TI-set under CK, Zi is a TI-set under M+. Finally by 5.1.14.2,
Ca (z) :::; M+ for each z E Zf, so as D L :::; M+ is transitive on Zf, Zi is a TI-set
under G by I.6.1.1, and hence the claim holds.
Let Gi := Na(Zi) and Gi := Gi/Zi. Recall by 5.1.14.2 that H:::; Ca(Zi), so
Gi s M+ by 5.1.14.1.
Consider any Hi with HDL S Hi :::; Gi, and set Qi := 02(Hi) and U :=
(VH^1 ). Observe that Hypothesis G.2.1 is satisfied with Zi and Hi in the roles of
"Vi" and "H". Therefore U:::; Z(Qi) and q?(U):::; Zi by G.2.2.
Suppose by way of contradiction that q?(U) # 1. Then U = (VH^1 ) is not
elementary abelian, so U i. CT(V). Thus fJ # 1, and hence the hypotheses of G.2.3
are satisfied. Therefore fJ E Syl 2 (L) by G.2.3.1. Set I := (UL) and W := 02 (!).
By G.2.3.4, there exists an I-series
1 = Wo s Wi s W2 s W3 = W,