9.2. REDUCING TO n = (^2 731)
PROOF. Recall that H centralizes Z. By 9.2.4.1, k := n(H) ;::: n - 1, so either
k > 1 or n = 2. Thus we may assume k > 1, and it remains to show that k = n.
As k > 1, K/02(K) is of Lie type over F 2 k by E.2.2.
If k f= 6, let p be a Zsigmondy prime divisor of 2k - 1; recall by Zsigmondy's
Theorem [Zsi92] that this means that a suitable element of order pin GLk(2) acts
irreducibly. If k = 6, let p = 3. Set BP:= Op(B). By Theorem 4.4.14, Bis faithful
on Lo, so as BT= TB, either some b E Bf induces an inner automorphism on L 0 ,
or IBPI divides n and Bp induces field automorphisms on L 0 • Assume the former.
If pis a Zsigmondy prime divisor of 2k - 1, then k divides n; while if k = 6, thenp = 3 so that n is even. Hence as k 2 n - 1, either n = k and the lemma holds, or
k = 6 and n = 2 or 4, impossible as then B 3 of order 9 is faithful on L 0. Therefore
we may assume that Bp induces field automorphisms on Land Lt, and IBPI divides
n. Then as k 2 n - 1, k f= 6. Thus pis a Zsigmondy prime divisor of 2k - 1, so
k divides p .,--l. Hence asp divides n and k ;::: n - 1, we conclude p = n = k + l.Then n is odd, and so Vi = Z ::::; Z(H) by 9.2.2.3, a contradiction as [V 1 , Bp] f= 1
since Bp induces field automorphisms on L 0. This establishes the lemma. D
LEMMA 9.2.6. If n(H) > 1, then B is contained in a Cartan subgroup D of Lo
acting on T n Lo.
PROOF. This is a consequence of 9.1.1 and 9.2.5. DLemma 9.2.6 has essentially eliminated the shadows of Aut(Lm(2n)) form:= 4
or 5, since in those groups B i. D: our argument above that B ::::; D assumes Gquasithin, whereas in those groups the parabolic M = Na(V) has 3-rank 3. So the
remainder of the proof (or more precisely, the reduction to n(H) = 1 in the next
section) can be viewed as showing that any embedding of B in D leads to a con-
tradiction. In the previous chapter 8, the road after eliminating the configurations
corresponding to shadows was typically fairly short; unfortunately in this case the
only route after that we know is fairly long and hard.
We can at least immediately eliminate all cases where n > 2:
PROPOSITION 9.2.7. (1) n = 2.(2) n(H) = 1 or 2.
(3) If n(H) = 2, then K/0 2 (K) ~ L 2 (4), B is cyclic of order 3, and B =
Cn(Vi) with B = [.D, a]..
PROOF. If n = 2 then (2) holds by 9.2.5, so it only remains to prove that (3)
holds; thus in this case we may assume n(H) = 2 = n. On the other hand if n > 2,
then n(H) = n by 9.2.5. So in any event we may assume that n(H) = n > l.By 9.2.6, B ::::; D, so as B ::::; K ::::; Ca(Z) and Cn(Z) = Cn(Vi) is cyclic of
order 2n -1, Bis cyclic of order at most 2n -1. Therefore as n(H) = n > 1, E.2.2
says K/02(K) ~ L2(2n) or Sz(2n) and IBI = 2n - 1, so B = Cn(Vi). By 9.1.2.2,
T contains f = a f o with fo a field automorphism of order a power of 2. Observe
a inverts B = C.oCVi) = [.D, a]. Pick a preimage t E NT(D). Then either t acts
nontrivially on B, or n = 2, fo f= 1, and t centralizes B. In the latter case the
lemma holds, so we may assume the former.
As B is not inverted by an inner automorphism of K/0 2 (K) in T, t induces
an outer automorphism on K/0 2 (K). Therefore n is even, and hence K/0 2 (K) ~
L 2 (2n) and t induces a field automorphism of some order 2i on K/0 2 (K). Therefore
n = 2im and ICB(t)I = 2m -1. If f =a, then t inverts B so m = 1 = i, and hence