15.3. THE ELIMINATION OF Mr/CMr(V(Mr)) = S 3 wr Z 2 1149PROOF. Observe first that s > 0: For if s = 0, then by 15.3.54.4, Y ~ A 4 x A 4 ,
contrary to 15.3.9.
Next let Vu := V n U; as Vu = ZJ_ by 15.3.53.2, N 0 (Vu) ::; M by 15.3.46.2; soas Vu = [U, V] by 15.3.53.2, CH (V*) ::; MlI. Now using (4) and (5) of 15.3.49,
NH• (Vu), CH· (Zs), and CH* (V*) are 3'-groups.Let K be a minimal normal subgroup of H. As H* is faithful and irreduc.ible
on U using 15.3.57.1, U = [U, K]. If K is a 3-group, then as CH (V) is a
3'-group, V inverts K; so by 15.3.56.2 and 15.3.53.2,
2(s + 2) = m(U) = 2m([U, V*]) = 4,
contradicting s > 0.Therefore K is not a 3-group, so each irreducible for K on U has rank at least
- Thus by 15.3.57.2, [K, V] # 1, and K* satisfies one of the two conclusions of
15.3.57.2. Suppose K is solvable. Then K is a p-group for some prime p > 3.
As m([U, V]) = 2, it follows (cf. D.2.13.2) that [K, V*] ~ Z 5. However ass> 0,
there is a Ds-subgroup D of H with center V by 15.3.56.5. As V, and hence
also D*, is faithful on [K*, V*], this is a contradiction.Therefore K is not solvable, so as K is an SQTK-group, K is the direct
product of at most two isomorphic nonabelian simple groups.
Suppose first that conclusion (b) of 15.3.57.2 holds. Then U = U 1 EB U 2 is the
sum of two K*-irreducibles Ui ofrank s+2 with V* inducing a transvection on each
Ui· By G.6.4.4, K*V* ~ Ln(2), 3 ::; n::; 5, 85, or 87, and Ui is a natural module
for K*. Let (ui) = [Ui, V*]; then Vu= (u1,u2), so as NK·(Vu) is a 3'-group, we
conclude K* ~ L 3 (2), and sos= 1.Next as K* is irreducible on Ui, and Ui is not self-dual, Ui is totally singular,
and U2 is dual to U1. Thus Irr+(K,U) = {U1,U2} is permuted by H, and
as H is irreducible on U by 15.3.57.1, H is transitive on {U 1 , U 2 }. Further as
EndK (Ui) ~ F 2 , CH (K) = 1, so H ~ Aut(L3(2)). completing the proof of the
lemma in this case.Thus we may assume that K is irreducible on U. By 15.3.56.6, m 2 (H) ::'.'.: s+ 1,
· so using 15.3.56.2, m(U) = 2(s+2)::; 2(m 2 (H)+l). As [K, V*] # 1by15.3.57.2,
the hypotheses of Theorem G.9.3 are satisfied with K, U, X in the roles of "H, V,
A", so H* and its action on U are described in Theorem G.9.3. As m([U, V*]) = 2
with V* ::; Z(T*), we conclude: cases (0)-(2) and (15)-(17) do not hold (see e.g.chapter H of Volume I for the Mathieu groups); in cases (6)-(10), n ::; 2; and in
case (13), U is a natural module rather than a 10-dimensional module. As s > 0,
m(U) ::'.'.: 6; therefore case (3) does not hold, nor does (6) or (7) when n ::; 2,
completing the elimination of those cases. As Zs = Co(T) is of order 2, and
CH• (Zs) is a 3' -group, the remaining cases are eliminated. DLet K := 02 (H) and TK := T n K, so that K* ~ L 3 (2) by 15.3.58.1.
LEMMA 15.3.59. (1) U = QH.
(2) TK E Syb(YU).
(3)[T:TK[=2.
(4) M= YT.PROOF. Let Qc := CT(U). As QH = CT(U) by 15.3.48, and U is extraspecial,
QH = UQc. Now [Qc, V] ::; Cv(U) = Z, so [Qc, V*] = 1. Then as K = [K, VJ