On the Derived Length of Lie Solvable Group Algebras

3.2 THE BASIC GROUP IS NOT NILPOTENT 31
verify the inclusion a−1ωs(m) l (F G ′ ) ⊆ δ [l+1] (F G) too. So, the statement
is proved when s (m)
l−1 is divisible by 2m .
Now, suppose that s (m)
l−1 is equal to 2rq, where r < m and q is an
odd integer. We claim that
(3.11) (x − 1) s(m)
l−1 + w ∈ δ [l] (F G) for some w ∈ ω s(m)
l−1 +1 (F G ′ ).
First assume that r = 0. By the inductive hypothesis
furthermore
[a(x − 1) s(m)
l−2, a −1 (x − 1) s(m)
l−2] ∈ δ [l] (F G),
[a(x − 1) s(m)
l−2,a −1 (x − 1) s(m)
l−2]
= (x k′
= (k ′ ) s(m)
for some w ′ ∈ ω s(m)
l−1 +1 (F G ′ ). Since
− 1) s(m)
l−2 (x − 1) s(m)
l−2 − (x k − 1) s(m)
l−2(x − 1) s(m)
l−2
l−2 − k s(m)
l−2
s
(x − 1) (m)
l−1 + w ′
(k ′ ) s(m)
l−2 − k s(m)
l−2 = (k ′ ) s(m)
l−2(1 − k 2s(m)
l−2 ) = (k ′ ) s(m)
l−2(1 − k s(m)
l−1) = 0,
(3.11) is true. Now suppose r = 0. Clearly,
[a(x − 1) s(m)
l−2, a −1 (x − 1) s(m)
l−2 +1 ]
= (x k′
− 1) s(m)
l−2(x − 1) s(m)
l−2 +1 − (x k − 1) s(m)
l−2 +1 (x − 1) s(m)
l−2
= (k ′ ) s(m)
l−2 − k s(m)
l−2 +1 (x − 1) s(m)
l−1 + w ′
for some w ′ ∈ ω s(m)
l−1 +1 (F G ′ ) and
(k ′ ) s(m)
l−2 − k s(m)
l−2 +1 = (k ′ ) s(m)
l−2(1 − k 2s(m)
l−2 +1 ) = (k ′ ) s(m)
l−2 (1 − k s(m)
l−1 ) = 0,
so (3.11) follows again from the inductive hypothesis.
For j ≥ 0 let zj = [(x − 1) s(m)
l−1 + w, a(x − 1) s(m)
l−1 +j ]. By the inductive