On the Derived Length of Lie Solvable Group Algebras

3.2 THE BASIC GROUP IS NOT NILPOTENT 29
In the following we shall use freely these notations and facts and
the congruence
(x s − 1) ≡ s(x − 1) (mod ω 2 (F G)),
which is a simple consequence of (3.7) and is valid for all x ∈ G and
any integer s.
Lemma 3.2.5. Let G be a group with commutator subgroup G ′ =
〈x | xpn = 1〉 of odd order and let char(F ) = p. If G/C has order 2m with m > 1 then
for all l ≥ 0.
aω s(m)
l (F G ′ ) ⊕ a −1 ω s(m)
l (F G ′ ) ⊆ δ [l+1] (F G)
Proof. We shall prove this lemma by induction on l. Evidently,
δ [1] (F G) can be considered as a vector space over F , and according to
a result of R. Brauer [14] the element
αgg ∈ F G
g∈G
belongs to δ [1] (F G) if and only if
g∈Ch αg = 0 for all h ∈ G. The
binomial formula says that the support of ai (x − 1) j is a subset of the
coset aiG ′ and it is easy to check that aG ′ = Ca and a−1G ′ = Ca−1. Thus aω(F G ′ ), a−1ω(F G ′ ) ⊆ δ [1] (F G) and the statement follows for
l = 0.
The next step is to show the lemma for l = 1. Clearly, [a−1x, a] ∈
δ [1] (F G) and
[a −1 x, a] = x k − x = (x k − 1) − (x − 1) = (k − 1)(x − 1) + w
for some w ∈ ω 2 (F G ′ ). Keeping in mind that δ [1] (F G) is a vector space
over F and k − 1 is an invertible element of F , we have (x − 1) + w ′ ∈
δ [1] (F G) for some w ′ ∈ ω 2 (F G ′ ). Let zj = [(x−1)+w ′ , a(x−1) 1+j ] for
all j ≥ 0. Since the assertion is true for l = 0, zj ∈ δ [2] (F G). Evidently,
zj = a(k − 1)(x − 1) j+2 + awj+3 for some wj+3 ∈ ω j+3 (F G ′ ), so we can