12.7. THE TREATMENT OF A 6 ON A 6-DIMENSIONAL MODULE S45
Let a+ := M 24. Arguing as in the proof of Theorem 12.7.7, Mis determined
up to isomorphism, so as a+ satifies the hypotheses of this Theorem, there is an
isomorphism cp : Af+ ---+ M. As Qt = J(T+ ), cp(Qt) = Qz. Now either U is the
socle of Kon Qz, or Qz is the sum of U and its dual as a K-module. As in the final
three paragraphs of the proof of 12.7.7, any LtT+-submodule of Qt isomorphic
to fl+ splits over fl+, so applying cp the same holds in K, and hence again U is
a semisimple K-module. Thus AutGL(Qz)(K*) ~ Aut(L3(2)), so as K ::::l Gz and
T* ~ Ds, K* = Autc.(Qz). Therefore K = Gz. As r+ splits over Qt, applyingcp, T splits over Qz; so K splits over Qz, and hence Gz = K is determined up to
isomorphism. We have seen that z is not weakly closed in Qz with respect to G, so
that we may apply Theorem 44.4 in [Asc94]. This time as Gt::::; M by 12.7.8, we
conclude that G ~ M24. D
By 12.7.16, to complete the proof of Theorem 12.7.14, we may assume that
U f=. R 1 , and it remains to derive a contradiction. In particular U is not abelian by
12.7.15.3, and <J?(U) = (z) by 12.7.15.1. Let U1 := Z(L1T). By 12.7.15.2, Mv ~ S5,
so 02(L1T) = U1 x R1 ~Es, and either Cr= U1, or Cr= 02(L1T) contains 01.
In any case Es ~ [V, U 1 ] ::::; Un V, and Li is irreducible on V/V1[V, U1] ~ E4, so
V n U = V1 [V, U1], and hence:
LEMMA 12.7.17. V* ~ E4.
LEMMA 12.7.18. U = 02(L1T) ~Es.
PROOF. If not, by the discussion before 12.7.17, U = 01 is of order 2. ThenV induces a group of transvections on U -with axis U ---------n Q, so using the dual of
G.3.1 as in the proof of 12.7.16, Li ::::; K ~ Ln(2) with n = 3, 4, or 5. This time
since we are arguing in the dual of U, [U, K] is the natural module for K. Then
U = [U, K*] EB Cu(K*) as K* is genrated by m([U, K*]) transvections. Next as
[V, U 1 ] is of rank 3 and contains z,[U, V] = [U 1 , V]/(z) ~ U1, V,L1/(z) = [U, V,Li]
is of rank 2. Thus in its action on the natural module [U, K], LiT is the rank
one parabolic stabilizing the line [U, V] and centralizing [U, K]/[U, V*]. In par-
ticular LiT fixes no point in the natural module, so Vi ::::; Cu(LiT) = Cu(K*),
contradicting U = (V{). D
We may represent LT on S1 := {1, ... , 6} so that P2T is the global stabilize
of {1,2}. Then by 12.7.18, U = ((1,2),(3,4),(5,6)). Pick g EL with (jg=
((1, 6), (2, 3), (4, 5)). Then
L = (0, Og) = (0, x)
for each 1 f=. x E (jg which is not a transposition. Let I := (U, Ug). Arguing as in
the the proof of G.2.3, L::::; I and
[02(I), J] =: p = (P n U)(P n Ug)
with [I, Un Ug]::::; V, and setting I/(U n Ug)V =: J+,
p+ = (Un P)+ EB (Ug n P)+
with Gp+ (U) = (Un P)+. Indeed if 1 f=. x E (jg such that xis not a transposition,
then I= (U,x), so