On the Derived Length of Lie Solvable Group Algebras

42 CHAPTER 4
Lemma 4.1.4. Let G be a group with commutator subgroup G ′ = 〈x |
x2n = 1〉, where n ≥ 3 and let char(F ) = 2. If G has nilpotency class
n + 1 then dlL(F G) ≤ n.
Proof. Clearly, the set of the Lie commutators [a, b] with a, b ∈ G spans
the F -space δ [1] (F G). Since [a, b] = g h +g with g = ba and h = b, while
of course g h + g = [a, b] with a = h −1 g and b = h whenever g, h ∈ G,
this spanning set for δ [1] (F G) can also be described as the set of the
elements g h + g with g, h ∈ G. It follows that the Lie commutators
[g1 h1 + g1, g2 h2 + g2], where g1, g2, h1, h2 ∈ G, span δ [2] (F G). We shall
compute these Lie commutators. It is easy to check that
(4.1)
[g1 h1 +g1, g2 h2 + g2] = g2g1
(g1, g2) + 1 (g2, h2) + 1 (g1, h1) + 1
+ (g2, h2) (g2, h2, g1) + 1 (g1, h1) + 1
+ (g1, g2)(g1, h1) (g1, h1, g2) + 1 (g2, h2) + 1
.
Firstly, if neither g1 nor g2 are in Gβ then
(4.2) [g h1
1 + g2, g2 h2 + g2] = bϱ3
for some b ∈ Gβ and ϱ3 ∈ ω 3 (F G ′ ). Indeed, it is clear from the
definition of Gβ that then g2g1 ∈ Gβ. Furthermore, the second factor
on the right-hand side of (4.1) always belongs to ω 3 (F G ′ ), because
γ3(G) ⊆ (G ′ ) 2 .
Secondly, if g1 or g2, say g1, belongs to Gβ, then we claim that
(4.3) [g h1
1 + g1, g2 h2 + g2] = gϱ4
for some ϱ4 ∈ ω 4 (F G ′ ) and g ∈ G.
For g1 ∈ Gβ, Lemma 4.1.1(i) asserts (g2, h2, g1) ∈ (G ′ ) 4 , therefore
(4.1) can be written as
(4.4)
[g1 h1 + g1,g2 h2 + g2] = g2g1
(g1, g2) + 1 (g1, h1) + 1
+ (g1, g2)(g1, h1) (g1, h1, g2) + 1 (g2, h2) + 1 + g2g1ϱ4