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