2.5. ELIMINATING THE SHADOWS WITH rg EMPTY 555or by B.2.7 and B.4.2.1, AutA(V) is Sylow in AutALi (V). In the former case
V :::; A since A E A(S), so as V is self-centralizing in Aut(K), A = VOA(K),
where OA(K) :::; Ra. In the latter case A induces an elementary abelian group of
inner automorphisms on K not centralizing V, and hence A centralizes vs, so by
symmetry between V and vs, A= vsoA(K). Thus the claim holds if R:::; Ns(L1),
so we may assume there is a E A - Ns(L1). Then m2(GK/Z(K)(a)) = n, so
m(OA(K)) 2 m(A) - (n + 1). Hence as A E A(S), and n > 1 by hypothesis, weconclude that
m(A) 2 m(VOA(K)) 2 2n + m(OA(K)) 2 m(A) + n - l > m(A),
since we are assuming that n > 1. This contradiction completes the proof of the
claim.Next suppose that <'P(Rc) = 1. Set Q := 02(L1Sc) :::;:: VSa. By the claim,
QR := Q n R = V(Sc n R) = VRa. Then QRSo = Q and Ns(QR) = Ns(Q).
Since A(SK) = {V, vs}, and we are assuming that Re is elementary abelian, QR=
VRc E A(S), and A(S) ={QR, Q:R} is of order 2. Hence IS: Ns(QR)I = 2, and forTq := NT(S)nNT(QR), NT(S) = Tq(s). Also ITql 2 ISi, since S < NT(S) because
S < T. As RSa = SKSc :::; LiSc normalizes 02(L1Sc) = Q, RSc :::; Ns(Q) =
Ns(QR)· Thus we have shown that IS : Ns(QR)I = 2 and both J(S) = Rand
Sa= 02(H) lie in Ns(QR)· Also Os(Ns(QR)) :::; Os(QR) :::; Ns(QR)· Thereforeapplying 2.3.9.8 to Ns(QR) in the role of "R", we conclude that Ns(QR) E (3. So as
Ns(QR) :::; Tq, Tq E f3 by 2.3.2.1. We saw earlier that Ns(QR) = Ns(Q). Further
Ns(Q) = LiNs(Q), so Q = 02(Ns(Q)) and NH(Q) E He. Also Ns(Q) fo M
since Ko nM is a Borel subgroup of Ko. Therefore (Ns(Q), Ns(Q)) E U(Na(Q)),
and henceNa(Q) Er. Then by 2.3.8.2, Q = 02(Ns(Q)) E Si,(G), so Na(Q) E He.
Since we saw above that Tq E /3, (Tq,Na(Q)) E U(No(Q)). However ITql 2 ISi 2
IU1I for each U1 EU by 2.3.6. Hence by the maximality of IUI and/or ISi in the
definitions of HE r* or r* in Notation 2.3.5, Na(Q) Ero, and therefore Na(Q) E
r 0 , contrary to our assumption in this section that r 0 = 0. This contradiction
shows that <'P (Re) # 1.
By 2.5.5.1, Ren So = 1, while Re :::; RX = R; so as R = SKRc, Re isisomorphic to a subgroup of SK/ Z ( K). Indeed we further claim that the members of
A(S) are of the formAaxAK with Ax E A(Rx) for eachX E {O, K}: If Z(K) = 1,
then R =Rex SK, so the second claim is clear in this case. Otherwise K/0 2 (K) ~
L 3 (4), and as <P(Z(K)) = 1, from the structure of the covering group Kin I.2.2.3b,each elementary subgroup of SK/Z(K) lifts to an elementary subgroup of SK,
completing the proof of the second claim.Hence as <P(Ro) =f:. 1 and Re is isomorphic to a subgroup of SK/Z(K), which
has exactly two maximal elementary subgroups V/Z(K) and vs /Z(K), we conclude
that A(Sc) = {A1,A 2 }, where Ai and A2 are the two maximal elementary abelian
subgroups of Re.Now suppose that [V, vx] =f:. 1. Then as V:::; A E A(S), m(Vx /Ovx(V)) = n =
m(R/OR(V)). Similarly m(V/Ov(Vx)) = m(R/OR(Vx)), so R = VVxOR(VVx)
with <'P(OR(vvx)):::; Re, and as V and vx are normal in R, [V, vx] = V n vx =Ovx(V) = vx n Z(R). By symmetry, <'P(OR(vvx)) :::; Re, so <P(OR(vvx)) = 1
by 2.5.5.1. Further for v E V - Z(R), m([v, R]) = n and [v, R] n Re = 1; so for
u E vs - Z(R), m([u, R]) = n and [u, R] n Ra = 1. Now for w E vx - Z(R),