994 14. L 3 (2) IN THE FSU, AND L2(2) WHEN .Cr(G, T) IS EMPTY
U = VZu, contradicting U nonabelian. Thus L/0 2 (L) ~ L2(2)'. Here by 12.8.13.3,
m(Zu) = m(Zf;-):::; m 2 (if) = 1, so by Coprime Action, 02 (CH(U)) centralizes U,
and then by (2) and (4) of 12.8.4, CH(U) = CH(U) = Q. Then H* ~if~ 83, so
that IH : Tl = 3, and then 14.3.5 contradicts H i M. D
LEMMA 14.3.14. One of the following holds:
(1) m(WB) = d/2 - 1, so that the conclusions of 14.3.12 hold.
(2) d = 4 and m(W^9 ) 2 2. Further if contains A5 or 83 x 83.
(3) d = 6, if~ G 2 (2), and m(W^9 ) = 3.
PROOF. If m(WB) :::; d/2 - 1, then (1) holds by 14.3.12. Thus we may assume
m(WB) 2 d/2. But by 14.3.13, if and f) appear in one of the cases of G.11.2 otherthan (1), (4), (5), and (12). Thus as m 2 (if) 2 d/2, case (2), (6), or (7) of G.11.2
holds. Case (6) of G.11.2 gives conclusion (3), and case (2) gives conclusion (2)
as m 2 (if) 2 2 and if is a subgroup of 8p 4 (2) whose order is divisible by 10 or- Finally in case (7) of G.11.2, WB i E(if), so m(WB) :::; 3 < d/2, contrary to
assumption. D
LEMMA 14.3.15. Assume L/0 2 (L) ~ L2(2)'. Then U ~ Qg and H* ~ Ot(2)
with E totally singular.
PROOF. Suppose first that if is not solvable.· Then appealing to 14.3.13, and
inspecting the list of G.11.2, there exists a component K 1 of if isomorphic to
L2(4), A5, G2(2)', A1, L2(8), or M2 2. By 1.2.1.4 we may choose K E £(G, T)
with K/0 2 (K) quasisimple and K = K 1 , although K may not be in C(H); set
Ko := (KT). From G.11.2, either K = K 0 , or conclusion (7) of G.11.2 holds and
Ko/0 2 (Ko) 3:! nt(4). Further if K ~ A 6 , then from G.11.2, Tis trivial on the
Dynkin diagram of K/02(K). Finally if Ko~ nt(4), then KoT/02(KoT) is not
85 wr Z2 since Nsp(U) (Ko) is a proper subgroup of index 2 in 85 wr Z2. We
conclude using 14.3.6 that K =Ko 3:! L 2 (8), and KnM = T. However Out(L 2 (8))is of odd order, so NK(T) is a Borel subgroup of K. Then as Na(T) :::; M by
14.3.3.3, Kn M > T, contrary to the previous remark.
This contradiction shows that if is solvable. Thus in view of 14.3.13, if and
its action on U are described in conclusion (2) or (3) of G.11.2. Indeed if and U
are described in Theorem G.9.4 if His irreducible on U, and in G.10.5.2 if His not
irreducible on U.Suppose first that Vi = Zu. Then arguing as in the proof of ( 3) and ( 4) of
14.3.12, U is extraspecial with U = U, and if= H* preserves the quadratic form on
U in which Eis totally singular. In particular if d = 4 and if has order divisible by
9, then as 9 does not divide 104(2)1, U ~ Qg and so if= H* lies in ot(2); further
by 12.8.9.2, W^9 /Eis a natural L 2 (2)-module, so that WB = W*B has rank 2. So
since Hi M, 14.3.5 reduces cases (1)-(4) of G.9.4 and G.10.5.2 to H* ~ Ot(2),so that the lemma holds. Otherwise we have case (5) of G.9.4, with H* a subgroup
of 8D15/31+^2 acting irreducibly on 03 (H)/Z(0 3 (H)). Let X be the preimage in
Hof Z(03(H*)); again X:::; M by 14.3.5. This is impossible, since U = [U,X] in
G.9.4.5, so that X does not act on the subspace V of rank l.Thus Vi < Zu. Hence by 14.3.12.3, m(WB) 2 d/2, so case (2) of 14.3.14 holds
as if is solvable; that is, d = 4, m(WB) 2 2, and if contains 83 x 8 3. It follows
that if~ 83 x 8 3 or ot(2), and m(WB) = 2 = m 2 (0t(2)).