592 3. DETERMINING THE CASES FOR LE Cj(G, T)
Definition C.1.31. As Y 2 /0 2 (Y 2 ) ~ £ 3 (2), case (5) of Theorem C.1.32 holds, so
that Y 2 T is described in C.1.34. Since T is Sylow in YzT, case (5) of C.1.34 does
not hold, so that one of cases (1)-(4) of C.1.34 holds.
Let Q := [0 2 (Y 2 T), Y 2 ] and U := Z(Q). By B.2.14, Z::::; S1 := S11(Z(02(Y2T))),
so [Y 2 , Z] ::::; Q n S1 = U and hence W 1 ::::; [Z, Yz] ::::; U and Y2 acts on UZ. Then
by 12.8 in [Asc86a], UZ = UZ 0 , where Zo := Cz(Y2), so Z = Zo(Z n U). On
the other hand CVi (Y 2 ) = 1 for each i, so Cv (Y2) = 1 and hence Zo = C z (L) by
3.3.7.4. Then as M = !M(LT), M = !M(Cc(zo)) for each zo E Z'/f,
Assume that c~se (4) of C.1.34 holds. Then U = UotJJU1, where Uo := Cu(Y2T)
is of rank 2 and U 1 is a natural module for Y2T/02(Y2T) ~ £3(2). Thus Un Z =
Uo ffi Z1, where Z1 := U1 n Z is of order 2, so as Z = Zo(U n Z), JZ : Zol = 2.
Further m(Z) ~ m(U n Z) = 3, so for each d E D, Zo n zg -:/:-1. Finally by an
earlier remark,
Md= !M(Cc(zd)) = M for some z E ztf with zd E Zo.
Thus D::::; Nc(M) =Mas ME M, contradicting 3.3.6.a. Hence case (4) of C.1.34
is eliminated.
Next Y 2 has m := 1 or 2 noncentral 2-chief factors in 02 (Y2), for n = 4 or
5, respectively, and Y 2 has r ~ 1 noncentral 2-chief factors in [V, L] ::::; 02(£).
Therefore Y 2 is not an £ 3 (2)-block, eliminating case (1) of C.1.34. Next the chief
factor(s) for Y 2 in 02 (Y 2 ) are isomorphic to W 1 ::::; U, so case (3) of C.1.34 is also
eliminated, since there the noncentral 2-chief factors of Y 2 other than U lie in Q /U
and are dual to U. Thus case (2) of C.1.34 holds, so Q = U = U 1 ffi U 2 is the
sum of two isomorphic natural modules Ui, and in particular Y 2 has exactly two
noncentral 2-chief factors. Thus m + r ::::; 2, so as m ~ 1 ::::; r, it follows that
m = r = 1, and therefore n = 4 and V = V 1 = [02(L), L]. We may choose notation
so that W1::::; U1. As V = [02(L),L], Lis an £4(2)-block, so P := 02 (Y1) ~ D~,
P/Z(P) = Pi/Z(P) ffi P 2 /Z(P) is the sum of two nonisomorphic natural modules
Pi/Z(P) for Yi/P, and we may choose notation so that Vi = P1. Thus as we
saw that D normalizes Y 1 , D normalizes 02 (Y 1 ) = P, and hence D = 02 (D)
also normalizes P1. Then as P 1 = Vi ::::) LT, D ::::; Nc(Pi) ::::; M = !M(LT),
contradicting 3.3.6.b. This completes the proof. D
LEMMA 3.3.19. L is not £5(2).
PROOF. Assume otherwise, and let Y := CL(Z)^00 • As V = (ZL), part (4) of
Theorem B.5.1 shows that V = [V, L] ffi Cz(L). Since 3.3.18 eliminates case (iv)
of B.5.1.1, either case (iii) of that result holds with [V, L] the sum of the natural
module and its dual, or case (i) there holds, with [V, L] irreducible. In the latter
case by Theorem B.4.2 and 3.3.18, [V, L] is a 10-dimensional irreducible.
Assume first that [V, L] is the sum of the.natural module and its dual. Then by
B.5.1.6, Y ~ £3(2)/2^1 +6, so YE .C(L,T). By 3.3.12.3, D acts on Y, and then also
on J(02(YT)). But again by B.5.1.6 (notice we can apply B.2.10 with 02 (YT) in
the role of "R"), we see that J(0 2 (YT)) ::::; CT(V) = 02 (LT), so J(0 2 (YT)) =
J(02(LT)) by B.2.3.3. Hence D::::; Nc(J(0 2 (LT)))::::; M = !M(LT), contradicting
3.3.6.b.
Therefore [V, L] is irreducible of dimension 10, and in particular is the exterior
square of a natural module. So this time (see e.g. K.3.2.3) Y is the parabolic
determined by the stabilizer of a 2-space in that natural module; again Y/0 2 (Y) ~
£3(2) so YE .C(L, T) and as before D normalizes Y by 3.3.12.3. Now 02 (Yf') does