13.4. THE TREATMENT OF THE 5-DIMENSIONAL MODULE FOR A 6 911that H2,o/02(H2,o) ~ L3(2), and hence 02,3(H 2 ,o) = 02 (H 2 , 0 ) is a 2-group. This
final contradiction completes the proof of 13.4.12. DBy 13.4.12, there is a unique z E Vt = z# with Gz i M. For the remainder
of the section, set H := Gz and VH := [(ZH), 02 (H)]. By 13.4.10.7 there is a
unique H2 E 1-l*(T, M) contained in H. Let K 2 := 02 (H 2 ), Ho:= (L 2 T, H 2 ), and
Vo:= (zHo).
LEMMA 13.4.13. (1) H =KT with KE .Cj(G,T), K/0 2 (K) ~ £ 3 (2) or A 6 ,
and VH is the natural module for K/02(K) or a 5-dimensional module for A 6 •
(2) (ZH) = VH(z), so VH E 'R.2(H).PROOF. By 13.4.10, (1) holds and Z = Zv(z) where (z) = Cz(K). So by
B.2.14, (zH) = [Z, K]Cz(K) = VH(z) and VH E R.2(H). D
LEMMA 13.4.14. Ho satisfies conclusion (i) or (ii) of 13.4. 7.1, ·and:
( 1) If Vo is semisimple then z is of weight 4 in V and Zv :::; [ Z, K 2] :::; V .ff. If
further VH is the 5-dimensional module for A5, then Zv is of weight 4 in VH.
(2) If Ho/02(Ho) ~ L3(2) and Vo is the core of the permutation module, then
either:or else(i) Zv i Soc(Vo), z is of weight 4 in V, and either
(a) Zv i VH, or
(b) V H is a 5-dimensional module for A6 and Zv is of weight 2 in V H;(ii) Zv :::; Soc(Vo), z is of weight 2 in V, Zv :::; VH; and if VH is a
5-dimensional module for A5 then Zv is of weight 4 in VH.PROOF. The initial statement follows from 13.4.10.8. In the remainder of the
proof, we extend arguments used in the last few lines of the proof of that result:
First z is of weight 4 in V iff z E [Z, L2]. Further Zv :::; [Z, K2] iff Zv :::; VH with
Zv of weight 4 in VH when VH is of dimension 5. Thus the subcase of conclusion
(1) where Ho is solvable can be treated exactly like the subcase in the earlier proof
corresponding to 13.4.10.Si.
So assume Ho/02(Ho) ~ L3(2). Then Zv = Cv 0 (L2T) and (z) = Cv 0 (K2T).In case (1), where Vo is semisimple, Cv 0 (L2T) :::; [Z, K2] and Cv 0 (K2T) :::; [Z, L2],
completing the proof of (1) in view of the equivalences in the previous paragraph.
It remains to prove (2), so we assume Vo is the core of the permutation module.
Suppose first that Zv i Soc(Vo). Then L2 centralizes the generator for Zv, which
lies in V 0 -Soc(Vo). Thus we may apply section H.5 with L2, K2 in the roles of "Lp,
Lz": Then (z) = Cv 0 (K2T) :::; Soc(Vo) by H.5.2.5 and H.5.3.3, and z E [Z, L 2 ] by
H.5.4.2, so that z is of weight 4 in V. Further Zv = Cv 0 (L2T) i [Z, K2] by H.5.4.l.
Therefore if Zv :::; VH, then z E Z:::; Zv[Z, K 2 ] :::; VH, so VH is a 5-dimensional
module for A 6 , and hence Zv is of weight 2 in VH by our earlier equivalences. Thus
either (a) or (b) of (2i) holds in this case.
On the other hand if Zv :::; Soc(Vo), then the roles of £ 2 and K2 are reversed
in the application of section H.5. Thus Zv = Cv 0 (L 2 T) :::; [Z, K 2 ] and z ~ [Z, £ 2 ],
so that (2ii) holds. D
Recall since L/0 2 (£) ~ A5 by 13.4.11 that R 2 = 02 (L 2 T).
LEMMA 13.4.15. Assume for some g E G that Vcf :::; R2 and V:::; Rg. Then
1 = [V,Vcf].