1040 14. L 3 (2) IN THE FSU, AND L 2 (2) WHEN .Cr(G, T) IS EMPTY
where the last inequality follows from the fact that Uy induces a group of transvec-
tions on v+ with center zt.
By (e),
2m ~ m(u+ /Cu+(U'Y)) + 1. (!)
In particular m > 0, so that u; is nontrivial on U+, and hence also on Kf.. There-
fore m(U+ /Cu+(U'Y)) > 1 by (a), so that (f) now gives m > 1. Hence m = 2 as
m 2 (AutH(Kf.)) = 2. As m = 2, we conclude from (e) and the structure of U+ that
u; induces inner automorphisms on K.'f-, and
(g)Therefore ( f) is an equality, and hence all inequalities in ( c )-( f) are equalities. Then
(d) becomes:
m(U;) = m(Autu-r(K*)) + m(Autu-r(Kf-)). (h)As the inequalities in ( c) are equalities,
fJ n UK= CuK(U'Y) is of codimension m2(Autu-r(K*)) + 1 in UK, (i)
and since m = 2 and the inequalities in (e) are equalities, we obtain m(u+ / D+) =
m(u+ /Cu+(U'Y)) - 2. Thus by (g):
v+ is a hyperplane of u+. (j)
We had chosen notation so that K = [K, u;J, but we also saw after (f) that
K.'f- = [K.'f-, u;J. Thus we have symmetry between Kand K+, so we conclude
m(Autu-r(K)) = 2 and u; induces inner automorphisms on K. Then by (h),
u; =A xA+, where A and A+ are 4-subgroups of K* and Kf., respectively. Since
U'Y induces a group of transvections on D+ with center zt, and we are assuming
that z'Y is not contained in fjK or U+, it follows from (j) that z'Y is generated by
zB = z1z2, where 1-:/=-z1 E tJK,l and 1-:/=-z 2 E tJK+,l· Therefore from the structure
of UK in 14.6.22, CH(zB) = Ca(ZZ'Y) has a Sylow 3-subgroup P isomorphic to
Eg. However by 14.5.21.2 and 14.5.15.1, QH and Q51 induce transvections on ZZ'Ywith centers Z and Z'Y, respectively, so that m 3 (Ca(V)) :::; 1 by A.1.14.4. This
contradiction finally completes the proof of the claim.
By the claim we may choose notation so that z'Y :::; fj K' and hence z'Y = ZB
centralizes K+. Set HB := HB /Q51; then K+ :::; Ca(ZB) = HB = Na(U'Y) since
H = G1 E M by 14.6.1.1. Therefore u; centralizes K.'f-, and hence m(U;) :::;
m2(CH*(K.'j..)) = 2. Also HB = KBK+TB by 14.6.23.1, so either K+ is KB or k+,
or else K+ is a full diagonal subgroup of KB x k+. Suppose this last case holds.
Then as K+ also acts on U, fj = (w) for w an involution interchanging KB and
K+, and hence Uk:w = u+. Then as [D'Y, U] :::; Z by F.9.13.6 and Cu-r(w) is of
codimension 6 in U'Y, m(U;) = m(U'Y/D'Y) ~ m(UK) - 1 = 5 > 2, contradicting
m(U;) :::; 2. Thus K+ = J where J :=KB or K+. Hence K+ :::; JQ51, so that
K+ = K'f:::; (JQ5iI)oo = J.
Suppose K+ = J. Then by 14.6.22.3, zB E [UB, J] :=:;; 02 (J). Then UK =
(zBNa(K)) :=:;; 02 (J) :=:;; Q51, so by 14.5.15.1, [UK, U'Y] :=:;; (zB), contrary to (a). Hence
K+ < J, so in particular JK+I < JJJ, and hence J = KB. Further K and K+
have different orders and so are normal in H, so that case (3) of 14.6.21 holds.
As K+ < J = KB with Jj0 2 (J) ~ L 3 (2) ~ K/0 2 (K) by 14.6.21.3, K+ has a