i4.3. FIRST STEPS; REDUCING (vG1) NONABELIAN TO EXTRASPECIAL 997By 14.3.18.4, A :'::) hQ, so that as L = 02 (! 2 ), K ::::; Na(A). Indeed from
14.3.18, Autr 2 (A) ~ 84, A= [A, L], and setting YA:= 02 (Ns(A)), AutRYA (A)~84 is of index at most 2 in the stabilizer in Auta(A) of Vi. We conclude from the
structure of L4(2) that AutK(A) is A 6 • Hence as Ca(A) is a 2-group, and as K =
02 (K) and K is Ca(A)-invariant by definition, it follows that K = 02 (KCa(A))
and K/02(K) ~ A5. Then as [R, Ca(A)] ::::; CR(A) = A, K = [K, R] centralizes
Ca(A)/A, so K is an A5-block. Next R = [R,L] and U = [U, YA], so RU::::; K.
Then as R/A ~ E4 is elementary abelian, K/A does not involve the double cover of
A5, so A= 02(K), and Na(K) ::::; Na(A), completing the proof of (4). As RU::::; K
and JRUJ = 2^7 = JKJ2, (8) is established. For u E U - Ran involution, u acts on
a complement B to Vin At, so B(u) is a complement to A in RU, and hence (9)holds using Gaschiitz's Theorem A.l.39. Let K 0 be a complement to A in K.
Let J := Aut(K) and A 0 := CJ(A). By (9), with 17.2and17.6 in [Asc86a], Ao
is elementary abelian with A 0 /A ~ Hi(K 0 , A). Hence Ao~ E 32 by I.1.6.1. Further
by 17.7 in [Asc86a] and a Frattini Argument, J = AoJo, where Jo := NJ(K 0 ), andof course Jo is the subgroup of Aut(K 0 ) stabilizing the representation of Ko on A,
so Jo ~ 86. Thus (10) holds.Recall NT(A) E 8yl2(Na(A)), so NT(A) E 8yl2(Na(K)). As L = 02 (P)
where P is the minimal parabolic of K with A = [A, P], CT(K) = CT(L) from
the structure of Aut(K) described in (10). Further CT(L) = 1, since Z ~ Z 2 by
14.2.2.6, while Z is not centralized by L. Thus CT(K) = 1, so (5) holds since
Ca(K) ::::; Cs(A) and we saw Cs(A) is a 2-group.
As Na(K) ::::; Na(A) with K transitive on A#, by a Frattini Argument, Na(K) =
KNs(A) = KQR. Thus Na(K) = KD.by (1) and (8). Now (6) follows from (10).
As NT(A) acts on K and is Sylow in Na(A), K ::::; Ki E C(Na(A)). Then
we conclude from the structure of GL 4 (2) and A.3.12 that either K = Ki or
AutK 1 (A) ~ A 7 , and the latter case is impossible as we saw Auts(A) is solvable.
Thus K :'::) Na(A) by 1.2.1.3, so (7) holds as we saw Na(K) ::::; Na(A). D
LEMMA 14.3.20. A= Ca(A).
PROOF. Let Ai := Ca(A) and suppose A < Ai. Let GA := Na(A). Then
GA ::::; Aut(K) by (5) and (7) of 14.3.19, so we. conclude from the structure of
Aut(K) described in (10) of 14.3.19 that E32 ~ Ai = 02(GA)· As the element
t defined in 14.3.18.4 acts on NT(A), Ai ::::; GA· As [A, At] =f l, [A, Ai] =f 1.
By B.3.2.4, K/0 2 (K) ~ A5 contains no FF*-offenders on Ai, so Ai = J(KAi).
Thus Ai f:. KAi, so JGA : KJ = 4, and hence GA = KAiAi ~ Aut(K) and
D = Cq(U) ~ Ds using 14.3.19.6.Next by 14.3.19.10, the action of GA/Ai on Ai is described in section B.3. In
the notation of that section, z = e5, 6 , so as UA/A = 02(CK(z)/A), D n Ai =
CA 1 (U) = (e5, e5). In particular d := e5 E Ai n D with Kd := CK(d) an A5-block,
and CaA(d) ~ S5/E 32. Further 02 (H) centralizes Das D ~ Ds. As [d,RQ] =Vi,
RQ = DCRq(d), so CRqo2(H) (d) = CRq(d)0^2 (H) ~ (83 x 83)/(Q~ x Z2). Let
Td := CT(d) and Sd := RQ n Td. As 02 (H) centralizes d, Td E 8yh(Cs(d)).
Further Z(Td) = Z(8d) = (z, d), and ITd : Sdl ::::; IT : RQJ = 2. As H is a 5'-