1016 14. L 3 (2) IN THE FSU, AND L2(2) WHEN .Cf(G, T) IS EMPTYor p1+^2 by A.1.25. As m 2 (Aut(R)) ::::; 2 and X = [X,J1(T)] by 14.5.10, the
hypothesis of D.2.17 is satisfied for each indecomposable pair in a decomposition
of (R X J 1 (T), VH)· So asp> 3 and R* is not cyclic, we conclude from D.2.17
that p = 5, and that there are two indecomposable components: that is, R* =
Ri x R2 with Ri, ~ Z5, [VH,R] = VH,1 EB VH,2, and V;, := [VH,Ri] is of rank 4.
But by definition of the decomposition, X* acts on each component, contradicting
[R*,X*] # 1.
Therefore [03'(H*),X*] = 1, so (1) follows from F.6.9. Of course (1) implies
(2). DLEMMA 14.5.12. (1) H+ is described in Theorem F.6./8.
(2) 02 (XT) # 02 (YT); in particular, case (2) of F.6.18 holds.PROOF. By 14.5.11, H is a quotient of H+, and by 14.5.7.1, X # 1 # Y*.
Thus if H+ ~ 83 , then H ~ 83, so that Y = X*, contrary to 14.5.6. Thus (1)
follows from 14.5.7.5. As H+ is solvable by 14.5.8, case (1.) or (2) of F.6.18 holds.
As 03'(H+) = 1 by definition, H+ is a {2,3}-group by F.6.9.
Assume (2) fails; then 02(XT) = 02(YT) = QH, so x+T+ ~ Y+T+ ~ 83.
As T+ is of order 2, case (1) of F.6.18 holds, and we may apply Cyclic Sylow2-Subgroups A.1.38 and F.6.6 to conclude that
(x+, y+) = 02 (H+) = O(H+).
Then as H+ is a {2, 3}-group, 02 (H+) =: p+ is a 3-group. Furthermore p+ is
noncyclic in case (1) of F.6.18, so that m3(P+) = 2 as His an SQTK-group.
We claim p+ ~ 31 +2; the proof will require several paragraphs. By 14.5.6,
p+ is nonabelian with x+ and y+ of order 3, so fh(P+) is nonabelian. Thus as
we saw m 3 (P+) = 2, if p+ is of symplectic type (cf. p. 109 in [Asc86a]), then
fh(P+) ~ 31+^2 and the claim holds.
So assume p+ is not of symplectic type. Then p+ has a characteristic subgroup
E+ ~ E 9. If x+ or y+ is contained in E+, say x+, then p+ = (X+, y+) =
E+y+ ~ 31+^2 , and again the claim holds, so we may assume neither x+ nor y+is contained in E+. Now p+ := Cp+(E+) is of index 3 in p+, and E+ = fh(F+).
Let T+ = (t+). Then t+ inverts x+, so as x+E+ ~ 3i+^2 , B+ := CE+(t+) ~
Z3, and hence NE(T)+ # 1. But Na(T) ::::; M by 14.3.3.3, ·so either E =
(NE(T)x) ::::; M, or B+ = fh(Z(P+)). The former case is impossible, as X s:J
H n M, whereas E+ does not normalize x+. Thus the latter case holds, and
we let Bo E 8yl 3 (B n M), where B is the preimage of B+, and set BM :=
02 (BoQH)· Observe that 02(BM) # 1 since HE He. By a Frattini Argument,
H = 031(H)NH(Bo), so H+ = NH(BM)+. As X 1. BM, with X/0 2 (X) inverted
in T n Land TBM =EMT, we conclude BM ::::; CM(L/0 2 (L)), so L normalizes02 (BM02(L)) =BM. Hence Na(BM)::::; M = !.Ait(LT). As H+ = NH(BM)+ and
X s:J H n M but E+ 1. NH+(X+), this is a contradiction.
This establishes the claim that p+ ~ 31+^2. Thus t+ inverts p+ /Z(P+) as t+
inverts x+ and y+. Hence t+ centralizes Z(P+). Then we obtain a contradiction
as in the previous paragraph. DLEMMA 14.5.13. (1) (X, Y) = P = 03 (H) ~ 31+^2 and H = PT, where
PE 8yiJ(H).