On the Derived Length of Lie Solvable Group Algebras

40 CHAPTER 4
for suitable odd l. This implies that γm+1(G) ⊆ γm(G) 2 is a proper
subgroup of γm(G) while γm(G) = 1, which proves the assertion.
Assume in the sequel that n ≥ 3. It is well-known that the automorphism
group aut(G ′ ) of G ′ is a direct product of the cyclic group
〈α〉 of order 2 and the cyclic group 〈β〉 of order 2 n−2 where the action
of these automorphisms on G ′ is given by α(x) = x −1 , β(x) = x 5 . For
g ∈ G, let τg denote the restriction to G ′ of the inner automorphism
h ↦→ h g of G. The map G → aut(G), g ↦→ τg is a homomorphism
whose kernel coincides with the centralizer C = CG(G ′ ). Clearly, the
map ϕ : G/C → aut(G ′ ) given by ϕ(gC) = τg is a monomorphism.
The subset
Gβ = {g ∈ G | ϕ(gC) ∈ 〈β〉}
of G will play an important role in the sequel. It is easy to check that
Gβ is a subgroup of index not greater than two and g ∈ Gβ if and only
if xg = x5i for some i ∈ Z.
Lemma 4.1.1. Let G be a group with cyclic commutator subgroup of
order 2 n , where n ≥ 3 and let char(F ) = 2. Then
(i) (y, g) ∈ (G ′ ) 4 for all y ∈ G ′ and g ∈ Gβ;
(ii) ω m (F G ′ ), ω(F Gβ) ⊆ I(G ′ ) m+3 .
Proof. Let g ∈ Gβ and y ∈ G ′ .
(i) Clearly, (y, g) = y−1yg = y−1+5i for some i ≥ 0 and −1 + 5i ≡ 0
(mod 4). Therefore, (y, g) ∈ (G ′ ) 4 .
(ii) Using (i) we have that
[y − 1, g − 1] = [y, g] = gy (y, g) − 1 ∈ I(G ′ ) 4 ,
from which (ii) follows for m = 1. Now, assume that
ω m (F G ′ ), ω(F Gβ) ⊆ I(G ′ ) m+3
for some m ≥ 1. Then
ω m+1 (F G ′ ), ω(F Gβ)
⊆ ω m (F G ′ ) ω(F G ′ ), ω(F Gβ) + ω m (F G ′ ), ω(F Gβ) ω(F G ′ )
⊆ ω m (F G ′ )I(G ′ ) 4 + I(G ′ ) m+4 ω(F G ′ ) ⊆ I(G ′ ) m+4 ,