On the Derived Length of Lie Solvable Group Algebras

14 CHAPTER 2
Theorem 2.2.2. Let G be a nilpotent group whose commutator subgroup
is a finite p-group, char(F ) = p and assume that γ3(G) ⊆ (G ′ ) p .
Then
dlL(F G) ≤ dl L (F G) = ⌈log 2(tN(G ′ ) + 1)⌉,
and if G is an abelian-by-cyclic p-group with p > 2 then
dlL(F G) = dl L (F G) = ⌈log 2 tN(G ′ ) + 1⌉.
Proof. Recall that δ (n) (F G) ⊆ (F G) (2n ) for all n ≥ 0. For n ≥ 1
Proposition 2.1.1 yields that
I(G ′ ) 2n −1 ⊆ δ (n) (F G)
and since γ3(G) ⊆ (G ′ ) p , by Proposition 2.1.4, we have
Combining this facts we get
(F G) (2n ) = I(G ′ ) 2 n −1 .
I(G ′ ) 2n −1 ⊆ δ (n) (F G) ⊆ (F G) (2 n ) = I(G ′ ) 2 n −1 .
Now it is easy to see that δ (n) (F G) = 0 if and only if 2n − 1 ≥ tN(G ′ ),
therefore n ≥ log2(tN(G ′ ) + 1), which implies the statement.
Let now G be an abelian-by-cyclic p-group with p > 2. Then
G = 〈A, x〉 for some abelian normal subgroup A of G and x ∈ G.
Clearly, G ′ = (A, x) is abelian and all z ∈ G ′ can be written in the
form z = (a1, x) · · · (as, x) for some a1, . . . , as ∈ A.
We are going to show that if c ∈ A, z1, z2, . . . , z2n−1 ∈ G ′ and j is
not divisible by p, then there exists ϱ ∈ I(G ′ ) 2n such that
x j c(1 − z1)(1 − z2) · · · (1 − z2 n −1) + ϱ ∈ δ [n] (F G).
We use induction on n. Let first n = 1 and let us choose an integer
k such that 2k ≡ j modulo the order of x. Then G ′ = (A, x k ) and
z1 = (a1, x k ) · · · (as, x k ) for some a1, . . . , as ∈ A. Applying the identity
1 − αβ = (1 − α) + (1 − β) − (1 − α)(1 − β)