14.7. FINISHING L 3 (2) WITH (VG1) ABELIAN 1053Let Y E Syh(K), set X := Y n Li, and let I consist of the ¥-invariant
subgroups I of Go with 3 E 7r(J). Then for IE I, there is a ¥-invariant Sylow 3-
subgroup Yr ofI, and Xr := 01(Z(YYr)nJ) = X since m3(Y) = 2 and m 3 (G 0 ):; 2.
Thus X :; Z(Yr) for each IE I.
Suppose next that 02(Go) < 02,3(Go). Then 02,3(Go) E I, so Xis in the
center of a Sylow 3-subgroup of 02,3(Go) by the previous paragraph. Then as
Li= X[02(Go),X], 02,F•(Go) S:: Na 0 (L1) = Hd using an earlier observation.
Hence as His a {2,3}-group by 14.7.15.3, 02,F•(Go) is a {2,3}-group. Then using
A.1.25.3, Go is a {2, 3}-group, so Go EI. Therefore Xis in the center of a Sylow3-group Yr of Go containing Y, so that Yr acts on X[X, 02(Go)] = L1. Then
Go= YrTd S:: Na 0 (L1) = Hd, contrary to Go E 9o. This contradiction shows that
02,3(Go) = 02(Go), so that 03(Go/02(Go)) = 1.
Now suppose J is a subnormal subgroup of Go contained in Hd. As Hd is a
{2, 3}-group, so is J, so as 02,3(J) :; 02,3(Go), J is a 2-group by the previous
paragraph. Hence 02(Go) is the largest subnormal subgroup of Go contained in
Hd.Suppose that Lo E C(Go) with 3 E 7r(Lo). Then Y = 02 (Y) acts on Lo by
1.2.1.3, so Lo EI, and hence Li= X[X, 02(Go)] :; Lo. Therefore Lo is the unique
member of C(Go) with 3 E 7r(Lo).
Suppose next that F*(Go) = 02(Go). Set J := 02,3'(Go). Then Td n J :;02 ,3'(Hd) = QH, so QH is Sylow in JQH. Therefore as Hypothesis C.2.3 holds in
G 0 , we conclude from C.2.5 that J:; Hd, so as J is normal in Go, J = 02(G 0 ) by
an earlier reduction. Thus 031(Go/02(Go)) = 1 = 03(Go/02(Go)), so 02,F• (Go)
is a product of 02 (G 0 ) with members of C(Go) whose order is divisible by 3. Then
we conclude from the previous paragraph that 02 (02,F• (Go)) =:Lo is the unique
member of C(Go) and L 1 :; L 0 • In particular Lo :::) G 0 , so that Lo is described
in C.2.7.3. As Li :; Lo and 03(Go/02(Go)) = 1, Y acts faithfully on Lo/02(Lo).
However no group K listed in C.2. 7.3 has a group of automorphisms A containing
Inn(K) and a subgroup HA of odd index in A with 02 (HA/0 2 (HA)) ~ 31 +2.
Therefore 02(Go) < F*(Go), so
Hd is maximal in {G+:; Gd: F(G+) = 02(G+)}. ()
Observe that by 1.1.6, Hypothesis 1.1.5 is satisfied with Gd, Td, H in the roles
of "H, S, M". However U = [U,L1] by 14.7.16.1, so U centralizes O(Gd) by A.l.26.
Then as z E U, O(Gd) = 1by1.1.5.2. Thus there is a component Ld of Gd, and
Ld i. H by 1.1.5.3..
Suppose first that Ld is a Suzuki group and set Lo := (L1jd). As Hd is a {2, 3}-
group, Hd n Lo = Td n Lo, so Hd acts on the Borel subgroup B := NLo (Td n Lo) of
L 0. Therefore as F(BHd) = 02(BHd), B :; Hd by (), impossible as Bis not a
{2, 3}-group.
Thus 3 E 1f(Ld), so by an earlier reduction, Ld is the unique component of Gd
and L 1 :; Ld. Similarly 03(Gd/02(Gd)) = 1 = 031(Gd/02(Gd)), so as before Y
acts faithfully on Ld. This time Ld is described in 1.1.5.3, and again no subgroup
A satisfying Inn(Ld) :; A :; Aut(Ld) contains a subgroup HA of odd index in A