1148 15. THE CASE .Cf(G, T) =^02-chief factors, sos+ 2=r+1 by(*). Further m(U/X) = 1 and m(X/E) = r, so
using (2), ·m(E) = m(U) - (r + 1) = 2(s + 2) - (s + 2) = s + 2,
completing the proof of (3). Further F #-f!'t, so F n Ft:::;: P. Also [En P, Y 1 ] :::;:
[E, J] :::;: Vo:::;: V, so (En P)/V:::;: Cp;v(Y1) = P2/V by definition of P2, and then
En Et :::;: P2 n p~ = v.Now ( 4) holds, using an argument in the proof of 12.8.11.3; indeed the argument
is easier here since U is extraspecial. Next using 15.3.53.2, V n U = ZJ = Vo VJ,
and we saw earlier that [ Z J_, X] = Z, so X centralizes vnu. Therefore X acts on
Vo and Vt, so since E and Et are normal in Na(Vo) and Na(VJ), respectively, X
acts on E and Et. We saw earlier that En Et :::;: V, so
En Et :::;: u n UY n uyt n v = zj_ n zyj_ n zyu = z.Then as m(U) = 2m(E), U = EeoEt. Since the action of H* on U is self-dual, the
action of X on iJt is dual to its action on E, so (5) holds. By 15.3.53.2, V is of
order 2, so (4) and (5) imply (6) and (7). D
In the remainder of the proof of Theorem 15.3 .. 50, define V 0 , X, E, and y as in
lemma 15.3.56.
LEMMA 15.3.57. {1} His irreducible on U.
{2} If 1 #-K* = 02 (K*) '.:::l H* and the irreducibles of K* on U are of rank at
least 3, then [K*, V*] -:/:-1, and either
(a) K* is irreducible on U, or
{b) U = U1 ffi U2, where the Ui are irreducible K* -modules of rank s + 2,
and V* induces a transvection on each Ui.PROOF. Let Uo be a nonzero H-submodule of U. Then Cu 0 (T) -:/:-0, so by
- - - -H -
15.3.56.7, Zs:::;: Uo. Thus U =(Zs) :::;: Uo, so (1) holds.
Assume the hypothesis of (2). By (1) and Clifford's Theorem, U is a semisimple
K-module, and by hypothesis, each J E Irr +(K, U) is of rank at least 3. If
[K, V] = 1 then K acts on [U, V]; this is impossible. as [U, V] = vnu is
of rank 2 by 15.3.53.2, contradicting m(J) > 2 for J E Irr +(K, [U, V]). Thus
[K, V] -:/:-1.
Similarly if V does not normalize some J, then m([U, V]) 2: m(J) > 2 by
hypothesis, again contrary to 15.3.53.2. Thus we can write U = J 1 eo · · · eo Jk where
Ji E Irr +(K, U) and Ji is V-invariant. Again using 15.3.53.2,
k
2 = m([U, V]) = ~ m([Ji, V*]) 2: k,
i=l
so that (2) holds. D
The next lemma eliminates the shadow of Aut(L 4 (2)), and begins to zero in on
the shadows of Aut(L 5 (2)) and Aut(He).
LEMMA 15.3.58. {1} H* ~ Aut(L3(2)).( 2) s = 1 and U ~ D~.
{3) U = U1 ffi U{, fort E T - UCr(V), and some natural submodule U 1 for
02 (H*) ~ £ 3 (2), such that U{ is dual to U1.