14.6. ELIMINATING L2(2) WHEN (vG1) rs ABELIAN 1041least 3. However K+ = Cc((z,zg})^00 is invariant under SE Syb(Cc((z,zg}), so
Sn J ¢. Syl2(J) and hence z does not centralize J, and IQ~S : SI 2". 8, so thatICH(zg)l2 = ISi:::; ITl/8.
Suppose first that m(U/D) = m(u;). Then, as in Remark F.9.17, we have
symmetry of hypotheses between 'Yl and')', so there exists a unique J 1 E C(Hg) such
that J1 = CJ 1 (z)02(J1). Thus as z centralizes K+ :::; J, J = J 1 , whereas we saw
[J,z] -=f-1. Therefore m(U/D) < m(U;), and we saw earlier that m(U;):::; 2, so we
conclude from (b) that m(U;) = 2 and m(U/ D) = 1. Thus m(UK,i/(DnUK,i)) :::; 1,
and as u; is of rank 2, u; does not centralize a hyperplane of both U K,l and itsdual UK,2· Therefore as [D, U1'] :::; Z1' by F.9.13.6, we may take zg E UK,l· But
then as K ::::! H, ICH(zg)l2 = ITl/4, contrary to the previous paragraph. This
contradiction finally completes the proof of (3).
Suppose next that (2) fails. Then K/02(K) ~ A 6 by 14.6.21.1, and m(UK,i) =4 or 5 by 14.6.22.1. Then for i an involution in H, m([UK,l, i*]) 2". 1, and in case
of equality, i* induces a transposition on K*. But also by 14.6.22.1, U :K, 2 = Uk, 1 for
s E Nr(K) nontrivial on the Dynkin diagram of K*, so if i* acts as a transvection on
UK,1, then m([UK, 2 ,i*]) = 2. We conclude m(U;) > 1, since u; E Q(H*, U). Next
as u; is quadratic on UK,1, from the action of NH(UK,1) on UK,1, u; is contained
in a 2-subgroup of H* generated by transpositions. Then again as U K, 2 = UK, 1 and
u; is quadratic on u K,2' u; is a 4-group F* generated by a transposition and the
product of three commuting transpositions. Then m(U/Cu(U;)) = 4 = 2m(U;).This contradicts 14.5.18.2, as F* does not induce a group of transvections on any
subspace of UK of codimension 2. This establishes (2).By (2) and (3), H^00 = K with H* ~ Aut(L 3 (2)), so by 14.6.22.1, UK =
U K,l E9 U K, 2 with U K,l natural and U K, 2 its dual. Thus H has three orbits on
UK - Z: ut 1uut2 =UH plus two diagonal classes, one of which is 2-central in
' '
fI. Denote this latter 2-central class by C. Recall that u E U(H) by (1), so that
u (j!:_ zG by 14.6.3.4. As CK(u) is not a 2-group, CK(u) -j;_ M since H n M = T by
14.6.23.2. Thus to prove (4), we must also show that Ca(u) -j;_ G1 = H.
Now u; is of rank 1 or 2 as m 2 (H*) = 2. Suppose first that m(U;) = 1.
Then [UK,U1'] = (u 1 ,u 2 ) with ui E UK,i and u 1 u2 EC. Conjugating in H, wemay take u = u 1. Recall Z1' :::; [U, U1'] :::; (u 1 , u2, z}, with Ui ¢. zG as u r/:-zG,
so that z1' is generated by U1U2 or U1U2Z, and hence c ~ zG. Since m(u;) = 1,
m(U / D) = 1 by (b ), and so our hypotheses are symmetric between 'Y and 'Yl· Thus
[U,U1'] = (ui,u~,zg}, with u~ E Uk,i and uiu~ E Cg. As C ~ zG, Cg~ z^0 , so
u fj. Cg, and hence u E Uf<: 1 or Uf<: 2. So since H is transitive on U# 1 U U# 2 , we
may take g E Cc(u). Thu~ as Z /v, Cc(u) -j;_ H. Hence Cc(u) EI, and ~o (4)
holds.So suppose instead that m(U;) = 2. Then [UK, U1'] = (U1, fi2) where U1 is a
hyperplane of UK,l and u2 E UK, 2 , with Ufu2 ~ C. If m(U/D) = 2, we again
have symmetry between 'Y and 'Yl, so the argument of the previous paragraph
establishes (4) in this case also. Thus by (b) we may take m(U/D) = 1. As u; is
of rank 2, u; does not centralize a hyperplane of both UK,1 and its dual UK,2, so
z1' = [U1', D n UK,i] :::; UK,i for i = 1 or 2, contrary tour/:-zG. This contradiction
completes the proof of (4), and of 14.6.24. D