880 1a. MID-SIZE GROUPS OVER FzTHEOREM 13.2.7. Assume n = 5 and
L+/0 2 (L+) ~ A5 for each L+ E .C1(G, T).Then 1-l*(T,M) ~ Ca(Z).
In the remainder of the section, we assume G is a counterexample to Theorem
13.2.7; thus there is HE 1-l(T, M) with Hi. Ca(Z). Hence:
Conclusion (b) of 13.2.4.2 holds.
In particular, Lf' ~ S 5 rather than A~. Let UH:= (zH), VH := [UH,H], LH :=
.0^2 (H), and H := H/CH(UH). As Hi. Ca(Z), by 5.1.7.2:
LH = [LH, J(T)] and L = [L, J(T)].
As LH = [LH, J(T)], we conclude from B.6.8.6d that [UH, J(T)] "I 1 Therefore
S := Baum(T) does not centralize UH, and UH is an FF-module for H*. Let
Q := 02(LT).
LEMMA 13.2.8. (1) H is solvable.
(2) UH= VH EB Cz(H).
(3) Either
(i) H* ~ Sa and m(VH) = 2, or
(ii) H = (Hi x H2)(t) ~Sa wr Z 2 and VH = U1 EB U2, where t* is an
involution with Hit= H2, Hi~ Sa, and U1 :=[UH, Hi]~ E4.
(4) S E Sylz(H) in {3i), and J(T) = S E Sylz(HiH2) in (3ii).
(5) SE Syl2(LHS).
(6) Let E := 01 (Z(J(T))); sets:= 1 and U1 := VH in case (i), and sets:= 2
in case (ii). Then E = CE(LH) EB E1 EB··· EB E 8 , where
Ei := (ei) = Cu; (S) ~ Z2.
PROOF. Assume H is not solvable. Then LH is the product of T-conjugates
of members of .C1(G, T), so by hypothesis (*), L'H ~ A 5 ; indeed it follows from(*)
that LH E .Cj(G, T). But then n(H) > 1, so that the hypothesis of Theorem 5.2.3 is
satisfied. Conclusions (2) and (3) of 5.2.3 are ruled out by Hypothesis 12.2.3, while
conclusion (1) of 5.2.3 does not hold as LH i. Ca(Z). This contradiction establishespart (1) of 13.2.8. Then as UH is an FF-module for H*, we conclude from Theorem
B.5.6 and B.2.14 that (2) holds, and from E.2.3.2 that (3)-(6) hold. D
We now adopt the notation of 13.2.8.6. Two cases appear in 13.2.8.3: s = 1
and s = 2. When s = 2, define Ht as in case (ii) of 13.2.8.3, and let Hi be the
preimage in H of Hi.LEMMA 13.2.9. 02(H) = CH(UH)· Thus LH/02(LH) ~ Eas.
PROOF. Set H := H/0 2 (H) and J := kerHnM(H). By 13.2.8 and B.6.8.2, LH
is a 3-group, j = Cf:i(LH), and Tis irreducible on LH/i. As Cf:i(L'H) = 1by13.2.8.3,
J = CH(UH)· Thus we may assume that X := 02 (J) f 1, and it remains to derive
a contradiction.
First X :::; M = Na ( L). If X centralizes L / 02 ( L), then L normalizes X =
02 (X02(L)), so H:::; Na(X):::; M = !M(LT), contrary to Hi. M. Therefore L =
[L,X]. Let R := 02 (XT). As XT =TX, X acts on TnL, so RE Syl 2 (LR). As
L = [L, X], R induces inner automorphisms on L/0 2 (L), and J(R) = J(02(LR)) '.::l