On the Derived Length of Lie Solvable Group Algebras

5.1 PRELIMINARIES AND THE DESCRIPTION 51
Theorem 5.1.3. Let G be a group with cyclic commutator subgroup
of order 2 n and let char(F ) = 2. If G ′ β has order 2r , then
r + 1 ≤ dlL(F G) ≤ r + 3.
Proof. If G = Gβ then Lemma 4.1.2 and Theorem 4.2.1 say that
dlL(F G) = r+1. Otherwise, Gβ is of index two in G and we can apply
Theorem 5.1.2 to get dlL(F G) ≤ r + 3. Furthermore, Lemma 4.1.2
and Theorem 4.2.1 ensure that dlL(F Gβ) = r + 1. Since dlL(F Gβ) ≤
dlL(F G), the corollary is true.
Let char(F ) = 2 and H = 〈x | x2n = 1〉. We claim that if r > 0
and the kj’s are odd positive integers for 1 ≤ j ≤ r then the element
ϱ = (x k1 + 1)(x k2 + 1) · · · (x kr + 1) ∈ F H
is equal to zero if and only if r ≥ 2 n .
Indeed, ϱ ∈ ω r (F H) and if r ≥ 2 n then ϱ = 0, because t(H) = 2 n .
Assume now r < 2 n . Applying the identity
(x kj + 1) = (x kj−1 + 1)(x + 1) + (x (kj−1)/2 + 1) 2 + (x + 1)
for every 1 ≤ j ≤ r, we can write ϱ = (x + 1) r + ϱ1, where ϱ1 is the
sum of elements of weight greater than r. Clearly, (x+1) r ∈ ω r (F H)\
ω r+1 (F H) and ϱ1 ∈ ω r+1 (F H), hence ϱ ∈ ω r (F H) \ ω r+1 (F H) and
ϱ = 0.
In the sequel we shall use freely this fact.
In the proof of the next lemmas we will use that C ′ ⊆ G ′ ∩ ζ(G).
This inclusion is indeed valid, because for a, b, c ∈ G the Hall-Witt
identity states that
(a, b −1 , c) b (b, c −1 , a) c (c, a −1 , b) a = 1.
Evidently, if b, c ∈ C then this formula yields that (b, c, a) = 1, which
guarantees our statement.