2.4. THE CASE WHERE r~ IS NONEMPTY 545Set Gz := Cc(Z) and Gz := Gz/Z. Then Pi'c = J(S) is x-invariant, so
PTc :::;! (x,KS) =: Mi. Observe T ~Mi ~ Gz and TcZ = Z(PTc) :::;! Mi.
By 2.4.27.3, Tc n T 0 = 1, so as x normalizes Z(PTc) = ZTc, and Tc is ofindex 2 in TcZ, ITcl ~ 2 with [x, Tc] = Z in case of equality. As IZTcl ~ 4
and Mi ~ Ncz(ZTc), 02 (Mi) centralizes ZTc. As S centralizes Tc, H =LS
. centralizes Tc.
Next PTc ~ Q~ or Q~ x Z2, so AutAut(PTc)(PTc/ZTc) ~ Ot(2). Let Mt:=
Mi/CM 1 (PTc/ZTc). Then Mt ~ Ot(2) and K+R+ ~ Ss x Z2 with u+ =
02(K+ R+). As ux 1. 02(KR) and x E Mi, U 1. 02(Mi); then as Mt ~ Ot(2),Mt~ Ot(2). In particular, Mi is irreducible on PTc/ZTc.
Suppose first that Tc =/= 1. As H = LS ~ C := Cc(Tc), we conclude from
2.3.7.2 that CE I'o and SE Sylz(C). By 1.2.4, L ~Le E C(C), and the embedding
of Lin Le is described in A.3.14. From the previous paragraph, 02 (Mi) ~ C but
U 1. 02(Mi), so U 1. 02(C). Hence as Lis irreducible on U, 02 (0) =Tc~ Z(C).
Therefore as m2(Lc) 2::: m2(L) > 1, Le is quasisimple by 1.2.1.5, and so Le is a
component of C. But the list of A.3.14 contains no embedding Le > L with Lan
A5-block.Therefore Tc = 1, so Q = Tc x U = U, and hence U = 02(Nc(U)) =
F*(Nc(U)) by 2.4.3.1. ·rn particular, Cc(U) = U, so Cc(L) = Cu(L) = 1. By
2.4.6, L :::;! Nc(U), so as LS= Aut(L), H =LS= Nc(U).
As T ~ Mi ~ Gz and M = !M(T), Mi ~ Gz ~ M. As G is of even type,M E He, so Z ~ CM(02(M)) ~ 02(M) =: PM. Also Q~ ~ P = CT(F), so as
Mt ~ Ot(2), P = 02(Mi). Therefore as T ~ Mi ~ M, PM ~ P by A.1.6.
Then as Mi is irreducible on P/Z, PM is either P or Z. As M E He, the latter
is impossible, so P = PM. Then as Z = Z(P), M ~ Gz, so that M = Gz as
ME M. Since Mi/P ~ Ot(2) ~ Out(P), M =Mi= Gz = Cc(z). In particular,
M is solvable.Let u E Z(R) - Z. As U is the S 5 -module for H/U, we can adopt the notation
of section B.3 to describe U, and choose u = ei,2· Then z = ei,2es,4 = UU^8 for a
suitable s E S-R. Set Gu:= Cc(u), Hu:= CH(u), etc. Then as H/U ~ S5, Hu~
Ds x S4, so R = Su is of index 2 in S. Further Cu ( 02 (Hu)) # = { u, ei,s,4,5, ez,s,4,5}with ei,s,4,5 and e2,S,4,5 in zL. As (u, z) = Z(R) :::;! T and U^8 = uz, there is
x E Tu - S, and Tu = (x)R is of index 2 in T with Tu = Mu- As Tu :::;! T,
Nc(Tu) ~ M = !M(T). Then as Tu = Mu, Tu E Sylz(Gu), and in particular
u 1. zG. Also Hu 1. M with ITul = ISi = ITl/2, so by 2.3.7.1, Gu Er*.
Suppose first that F*(Gu) = 02(Gu), so that Gu E r5. Then we may apply the
results of this section to Gu in the role of "H". By 2.3.8.4, Gu= MuKi ···Kt is a
product of blocks Ki, where Ki is an A5-block or As-block, since 2.4.7 eliminated
the case where some Ki is an L2(2n)-block. Indeed as we saw Mu= Tu is a 2-group,
and Ki n M is a Borel subgroup of Ki in 2.3.8.4, each Ki is in fact an As-block.
Then as Tu is of order 27 and 2-rank 4 with 1 =/= u E Z(Gu), t = 1. But nowKi = 02 (Gu) ~ A4 contains 02 (Hu) ~ A4, so that 02 (Gu) = 02 (Hu)· As Su
is of index 2 in Tu, Hu is of index 2 in Gu, and hence is normal in Gu. Then as
U ~ 02 (Hu) and x E Tu, ux ~ 02(Hu), which is not the case.
Thus F*(Gu) =/= 02(Gu)· As 02 (Hu) ~ A4, 02(0^2 (Hu)) centralizes O(Gu)by A.1.26, so z E (u)02(0^2 (Hu)) ~ Ccu (O(Gu)); hence O(Gu) = 1, as z inverts
O(Gu) by 2.3.9.5. Therefore Gu has a component K, which must appear in 2.3.9.7.
We further restrict the list of 2.3.9.7 using the facts that Mu= Ccu (z) is a 2-group