On the Derived Length of Lie Solvable Group Algebras

3.2 THE BASIC GROUP IS NOT NILPOTENT 27
and
vl+1
≡ t (l)
u t(l) w [ba(x−1 − 1) sl−2 sl−2+1 −1 sl−2 sl−2+1
(x − 1) , a(x − 1) (x − 1) ]
≡ t (l)
u t(l)
−1 2sl−2+1 2sl−2+1
w b(x − 1) (x − 1)
− aba(x −1 − 1) 2sl−2+1 2sl−2+1
(x − 1)
≡ t (l)
u t (l)
w b(x −1 − 1) sl−1+1 (x − 1) sl−1 (mod I(G ′ ) sl+2 ),
wl+1
≡ t (l)
w t(l) v [a(x−1 − 1) sl−2 sl−2+1 −1 −1 sl−2 sl−2+1
(x − 1) , b (x − 1) (x − 1) ]
≡ t (l)
w t(l)
−1 −1 2sl−2 2sl−2+2
v ab (x − 1) (x − 1)
− b −1 a(x −1 − 1) 2sl−2+1 2sl−2+1
(x − 1)
≡ t (l)
w t (l)
v (−2)b −1 a(x −1 − 1) sl−1+1 (x − 1) sl−1 (mod I(G ′ ) sl+2 ).
Supposing the truth of (3.10) for some even l, we apply Lemma 3.2.2
(with m = 1) and (3.8) to get the congruence
ul+1 ≡ t (l)
u t(l)
v [a(x−1 − 1) sl−2 (x − 1) sl−2 , b(x −1 − 1) sl−2+1 (x − 1) sl−2 ]
(mod I(G ′ ) sl+2 ).
Hence
ul+1 ≡ t (l)
u t(l)
−1 2sl−2+1 2sl−2
v ab(x − 1) (x − 1)
− ba(x −1 − 1) 2sl−2 2sl−2+1
(x − 1)
≡ t (l)
u t(l)
v (−2)ba(x−1 − 1) sl−1 (x − 1) sl−1+1
(mod I(G ′ ) sl+2 ).
Similar calculations give us the required congruences for vl+1 and wl+1.
So, (3.9) and (3.10) hold for all l ≥ 0.
Let d be the minimal integer such that sd ≥ p n . According to
the congruences (3.9) and (3.10), ud is a nonzero element of δ [d] (F G),
because sd−1 < p n . So dlL(F G) > d, from which the statement follows.
Let H = 〈h | hpn = 1〉 be a subgroup of a group G and char(F )=p.
Then ω(F H) is nilpotent and ωpn(F H) = 0. Evidently, for g ∈ G \ H