910 lS. MID-SIZE GROUPS OVER F2
normalize each other. Thus L 2 centralizes U 2 := (Z{f^2 ), and K2 centralizes U1 :=
(zL^2 ). But as conclusion (i) of 13.4.7.1 holds, Cv 0 (L2) =: U~ ~ E4, so that U2 =
u~ ~ E4; in particular, Zv :::; U2 :::; VK. Similarly U1 ~ E4, so it follows that z is
of weight 4 rather than 2 in V. Similarly Zv is of weight 4 in VK when VK is the
5-dimensional module for K* ~ A 6 • Thus conclusion (i) of (8) holds, and the proof
of 13.4.10 is complete. D
LEMMA 13.4.11. (1) L/a2(L) ~ A5.
(2) M= LT.
{3) 'Hz= {Gz}.
PROOF. Part (1) follows from 13.4.10.3, and (2) and (3) follow from 13.4.10.7.
DLEMMA 13.4.12. (1) Z = V 1 has rank 2, and there exists a unique z E z# such
that Ca(z) i. M.
(2) There is a unique member H2 of 'H* (T, M) contained in G z.
PROOF. By 13.4.10.5, Z = V 1 is of rank 2. Recall z E vt with Gz i. M, and
z has weight 2 or 4 in V while a generator of Zv is of weight 6. Let Zk denote the
element of Vi of weight k and choose m with Gzrn i. M. In this subsection Gzrn is
not solvable, so by parts (1) and (7) of 13.4.10, Gzrn = KzrnT for Kzrn E C(Gzrn),
and there is a unique Hm,2 E 'H*(T,M) contained in Gzrn· Set Km,2 := a^2 (Hm,2).
As Hm, 2 is the unique member of 'H* (T, M) contained in Gzrn, (2) holds. More-
over by 13.4.10.5, Cz(Hm,2) = (zm)· As z2 # Z4, H2,2 # H4,2·
It remains to prove the final statement in (1), so we assume that Gz= i. M for
both m = 2 and 4. Set Hm,o := (L2T, Hm,2). Then by 13.4.10.8, Hm,o E 'H(T),
and Hm,o satisfies conclusion (i) or (ii) of both 13.4.7.1 and 13.4.10.8. As z2 has
weight 2 in V, H 2 , 0 satisfies conclusion (ii) rather than (i) of 13.4.10.8, and hence
H2,o = Na(J(T)) E M(T).
Suppose that H4,0 also satisfies conclusion (ii) of both results. Then by 13.4.10.8,
H2,0 = Na(J(T)) = H4,0 and Hm,o contains a unique member Hm,2 of 'H*(T, M).
Therefore H2,2 = H4,2, contrary to an earlier observation. Hence H4,0 satisfies
conclusion (i) of both results.
Next let Ho := (H2,2, H4,2). We check that the hypotheses of Proposition
13.4.7 are satisfied: We already observed that m(Z) = 2 and (zk) = Cz(Hk, 2 ),
establishing (b). We saw that Hk,2 does not centralize z 6 -k, so Ho i. Gzk and
hence as' (CHo(zk))T < GZk" Now GZk = KzkT for k = 2, 4, and in each case
Kk,2T is a maximal subgroup of Gzk' so we conclude Kk,2 =as' (CH 0 (zk)), giving
( c). Finally by 13.4.10.4, Kzk has at least two noncentral 2-chief factors, one in
Vazk and one in Kzk/CKzk (Vazk), giving (a).
So we may apply 13.4.7. Assume first that Ho satisfies one of conclusions (ii)-
(iv) of 13.4.7.1. Then Ho :::; Na(J(T)) = H2,0· Recall however that Ho is generated
by distinct members Hk, 2 of 'H*(T, M), whereas H 2 , 2 is the unique member of
'H*(T, M) contained in H2,0·
Therefore Ho satisfies conclusion (i) of 13.4.7.1. Thus K 2 , 2 :::; Na(K 4 , 2 ) and
hence H2,0 = (K2,2, L2T) :::; Na(K4,2), since H4,o also satisfies conclusion (i) of