15.4. COMPLETING THE PROOF OF THE MAIN THEOREM 1159We come to the main result of this subsection:
THEOREM 15.4.8. Cc(Z) = T.The proof of Theorem 15.4.8 involves a short series of reductions. Until the
proof is complete, assume G is a counterexample. Let X consist of those X E
~(G, T) such that X ::::J Cc(Z). Recall by 15.4.5 that X/0 2 (X) ~ Z 3.
LEMMA 15.4.9. X -:/:-0.
PROOF. Set H := Ca(Z) and fl. := H/0 2 (H). By 15.4.5, H is a {2, 3}-group, so as G is a counterexample to Theorem 15.4.8, 03 (H) -:/:-1. Let P :=
D 1 (Z(0 3 (H))). If Pis of order 3, then X := 02 (P) EX, so we may assume that
IPI > 3. Then as m3(H) :S 2, Eg ~ P = D1(0s(H)), so that C:g(P) = Os(H)
by Coprime Action, and H/0 3 (H) is a subgroup of GL 2 (3). As H centralizes Z,
PT tj. 3( G, T) by 15.4.2.3, so Tis not irreducible on P. Therefore there is a normal
subgroup A of fl. of order 3, and so X := 02 (P 1 ) EX. D
LEMMA 15.4.10. For each XE X and each ME M(XT), X '.S CM(V(M)).
PROOF. Assume X, Mis a counterexample, and let V := V(M) and M :=
M/CM(V). In particular X =/= 1. If 02(X) = 1, then V = [V,X] E9 Cv(X) by
Coprime Action, and Zn [V, X] -:/:-1 as Xis T-invariant, contrary to X:::; Cc(Z).
Therefore 02(X) =/= l, so 02(X) i 02(M).
Let M := M/0 2 (M). Thus 1 -:/:- 02 (X) :S 02 (X). We claim next that
02 (X) centralizes 05 (M), so suppose not. Then by A.1.25, 02(X) acts non-
trivially on a supercritical subgroup P of Os(M), P* ~ Zs, E2s or 51 +^2 , and
Autx(P) is a subgroup of Aut(P)/0 5 (Aut(P)) ~ GL2(5). As 02(X) does
not centralize P and X = 02 (X), we conclude that P is not of order 5 and
Autx• (P) ~ Zsf Q 8. Thus 02(X) is irreducible on P /if!(P), and so the preim-
age P contains a member of B+(G,T), contrary to 15.4.2.3. This establishes the
claim.
As M is a solvable {2, 3, 5}-group by 15.4.4, F(M) = 03(M)Os(M), so
02 (X) is faithful on 03 (M) by the claim. Again by A.1.25, 02(X*) acts nontriv-
ially on a supercritical subgroup P of 03 (M), P ~Egor 31 +^2 , and 02(X) ~ Qs
is irreducible on P /if!(P). Let Y := 02 (P), so that YE 3(G, T) and Autx(P) ~
SL 2 (3). If P ~ 31+^2 , then as Autx·(P) ~ SL2(3), ms(XP) = 3, contrary to M
an SQTK-group; thus P ~ Eg.
Let H ;= YXT, W := (zH), and H+ := H/CH(W); then W = (ZH) E R2(H)
and W :::; 02 (M) by B.2.14. As [Z, Y] -:/:-1 by 15.4.2.3, and 02 (X*) is irreducible
on P*, Cy (W) = 02 (Y). Therefore H+ is the split extension of p+ ~ Eg by
either SL 2 (3) or GL 2 (3), so W conta.ins an 8-dimensional faithful irreducible H-
submodule I. Thus q(H+, W) ?:: q(H+, I) > 2.
By 15.4.2.3, Y E B*(G, T), so that N := Nc(Y) = !M(YT) by 1.3.7, and as
YT :SM, M = N. Of course YT :SH, so M = !M(H). Further 02(CH(W)) =
OT(W) since Cy(V) = 02 (Y). Thus we may apply 15.4.2.4 to conclude q(H+, W) :S
2, contrary to the previous paragraph. D
We are now ready to complete the proof of Theorem 15.4.8. By Hypothesis15.4.1.2, there exist distinct members M1 and M 2 of M(Cc(Z)). By 15.4.9, there
is X E X. Now X is not normal in both M 1 and M2, so we may assume X is
not normal in M 1. Let Y 1 := (XM^1 ), and set Mf := Mi/02(Mi) for i = 1, 2. By