i032 i4. L 3 (2) IN THE FSU, AND L 2 (2) WHEN .Cr(G, T) IS EMPTYvectors of UK. Then S := Cs(zY) for some Sylow 2-subgroup S of H containing
U-y. Now H:::; Gi with UH= (VH) :::; (V^01 ) = U; thus U-y = UiI:::; UY= (VY^0 i),
so since U is abelian by Hypothesis 14.5.1, [UY, Cs(zY)] :::; UY :::; Ca(U-y)· Now as
u; :::; S ·~ D 8 with U'fr,i = UH, 2 , [U-y, Cs(zY)] contains s withs = Z(S*). Then
s E [UY, Cs(zY)]:::; Ca(U-y), whereas s* does not centralize [UK, x]:::; U'Y"
Therefore U-y acts on UH,i· Suppose first that m(UK) = 4. Then by 14.6.12.2,
p = 3 and fj K is a 4-dimensional orthogonal space for H*. This time choose g
so that U 1 = UJ,, and choose notation so that [UH,i, UJ,] i= l; now UH :::; HY by
F.9.13.2, completing the verification that (1) holds.
Thus it remains to treat the case in 14.6.12 with m(UK) = 8. By the choice of
'Y:
As m(UK) = 8, if u; 1, H! for i = 1 or 2, then m(UK/CuK(u~)) = 4 for suitable
1 i= u~ Eu;; this is a contradiction as [DH, U-y] :::; Z-y by F.9.13.2, while m(Z-y) = 1
and m(UH/DH):::; 2 by(*). Therefore we may assume U-y:::; Hi, so that m(U;) = 1
from the structure of H in 14.6.12, and hence m(UH/DH) = 1 by(). Then since
[UH,i, U-y] has rank 2, 1 of= [DH rl UH,i, U-y], so that Z-y :::; UH,i by F.9.13.6. Also
m(U-y/D-y) = 1 = m(UH/DD), so our hypotheses are symmetric in 'Y and "fi, as
discussed in Remark F.9.17. Hence we may choose notation so that Z:::; UiI 11 so
that (2) holds, completing the proof of 14.6.13. ' D
Recall that Gi is a member of 7-i(T, M), so that the notational conventions of
section 14.5 apply also to Gi in the role of "H". Our convention in this subsection
is to define U := Ua 1 = (V^01 ), and set Qi := 02(Gi). Set Ci := Ci/Qi and
Kz := (KG1).
Now we further specify our choice of HE 7-i(T, M), so that the odd prime p E
7r(H) is maximal over odd primes such that Op(H 0 ) i= 1 for some Ho E 7-i(T, M);
that is, in view of 14.6.12.1, we choose H with p := 5 if 05 (H 0 ) i= 1 for some
Ho E 7-i(T, M), and otherwise p := 3.
LEMMA 14.6.14. One of the following holds:
(1) Kz = K, and if p = 3 then Gi is a {2, 3}-group.
(2) p = 3, Kz E C(Gi), and Ci~ Aut(Ln(2)) for n := 4 or 5.
(3) P = 3, Kz = KiKf for s E T - Nr(Ki) with Ki E C(Gi), and Ci !::::<
85 .wr Z2 or £3(2) wr Z2.
(4) p = 5, Kz = KiKf for s E T - Nr(Ki) with Ki E C(Gi), and G1 ~
Aut(L2(16)) wr Z2.
PROOF. First suppose H+ is a solvable overgroup of Hin Gi. If X :'.SJ H+ with
X/02(X) a q-group for some odd prime q, then XT E 7-i(T, M) by 14.6.1.4, and so
q:::; p:::; 5by14.6.12 and our maximal choice ofp. Thus setting H+ := H+/0 2 (H+),F*(H+) =II Oq(H+),
q5,_pwith mq(Oq(H+)) :::; 2 since H+ is an SQTK-group. Therefore using A.1.25 and
inspecting the order of GL 2 (q), we conclude H+ is a {2, 3}-group if p = 3, and a
{2, 3, 5}-group if p = 5.
We claim next that for J E C(Gi), J is not a Suzuki group: For if J ~ Sz(2m)
for some odd m > 1, then the T-invariant Borel subgroup B of J 0 := (Jr) has