534 2. CLASSIFYING THE GROUPS WITH IM(T)I = 1If Z(L) = 1 then z = zx by (2); hence (zx)L = zL gives the partition of the
natural module U by its I-dimensional F q-subspaces, so (3) holds in this case. If
Z(L) =f. 1 then Z(R) = zzx and (Z(R)/z)L is the corresponding partition of U/Z,
so again (3) holds as Lis indecomposable on U. D
LEMMA 2.4.17. {1) Either R = CT(Z); or case (1) of 2.4.14 holds, so that
Z(L) = 1 and CT(Z) = R(t) for some involution tin T-S with Z = Z(R) = CR(t).
{2) If Z(L) = 1 then T/CT(Z) is cyclic.PROOF. By 2.4.12.1, Cs(Z) = R, since the field automorphisms in F do not
centralize Z. Assume CT(Z) > R. Then CT(Z) = R(u) for some u E CT(Z) - S;
so u^2 E R and hence 2.4.14 completes the proof of (1).
Assume Z(L) = 1, so that Z = Z(R). By 2.4.8.6, AutB(Z) is cyclic and
regular on z#, so AutaL(z)(AutB(Z)) is the multiplicative group of Fq extended
by Aut(F q)· Since AutB(Z) is normal in AutBT(Z) by 2.4.8.5, we conclude AutT(Z)
is cyclic, so that (2) holds. D
LEMMA 2.4.18. (1) Z is a TI-subgroup of G.
(2) If Z(L) = 1 then Na(Z) = M.
{3) if Z(L) =f. 1 then Na(Z) = GQ.
PROOF. If Z(L) =f. 1 then (1) and (3) hold by 2.4.11.3. Thus we may assume
Z(L) = 1, so Z = Z(R) from Notation 2.4.15. Set P := 02 (M). As T normalizes R
by 2.4.8.1, there is an involution z in ZnZ(T). As F*(M) = 02 (M), z E CM(P) =
Z(P). Then as D :::; M and D is irreducible on Z, Z:::; Z(P). It suffices to show
that Z :::l M: For then M = Na(Z) since ME M, so that (2) holds. Further as
M = !M(T), Ca(z):::; M, and hence as as Dis transitive on z#, Z is a TI-set inG by I.6.1.1, so that (1) also holds.
Thus it remains to show that Z :::l M. If R :::; P, then as R = J(T), also
R = J(P) by B.2.3, so that Z = Z(J(P)) :::l M. Thus we assume that R 1: P.
Now for x E T - S, R = uux by 2.4.8.3, so U 1: P. Then as Z :::; P and D
is irreducible on U/Z, Z =Un P, and then also Z = ux n P. So since U and
ux are the maximal elementary subgroups of R by 2.4.8.7, Z = fh(R n P). We
now assume Z is not normal in M, and it remains to derive a contradiction. We
saw Z :::; Z(P), so that Z < Zp := D 1 (-Z(P)) and hence there is an involution
t E Zp - Z. As Z = D 1 (R n P), t tj. R, so as t centralizes Z, the second case of
2.4.17.1 holds. Therefore Z = CR(t) and tis described in case (1) of 2.4.14. But
[R, t] :::; [R, Zp] :::; Rn Zp :::; CR(t) = Z, impossible as [R, t] > Z in case (1) of
2.4.14.. DLEMMA 2.4.19. R is the weak closure of Z in T.
PROOF. By 2.4.16.3, Q = U = ((zx)L), and R = QQx by 2.4.8.3. Hence R is
contained in the weak closure of Z in T. Thus we may assume that there is g E G
with zg :::; T but zg 1: R, and it remains to derive a contradiction. By 2.4.16.1,
[ZJ = 2n > 2 = [T : Sf ~ [T : NT(Z)J, so that Nzg(Z) =f. 1. Then as Z is a
TI-subgroup of G by 2.4.18.1, and (Z, zg) is a 2-group, zg :::; CT(Z) by I.6.2.1. As
zg f: R, there is an involution t E zg - S with CT(Z) = R(t) by 2.4.17.1. Then t
satisfies conclusion (1) of 2.4.14 with CR(t) = Z. Hence as [Vf > 2 = JCT(Z) : RJ,
Rn zg =f. 1. But Rn zg :::; CR(t) = Z, so as Z is a TI-subgroup of G, zg = Z:::; R,