On the Derived Length of Lie Solvable Group Algebras

44 CHAPTER 4
Similarly, we can obtain that
Hence
y k−5l
+ 1 = (y k−5l −2 + 1)(y 2 + 1) + (y k−5 l −2 + 1) + (y 2 + 1)
≡ y 2 + 1 (mod ω 3 (F G ′ )).
ϑ = y k−5l
+ 1 + y(y k−1 + 1) ≡ 2(y 2 + 1) ≡ 0 (mod ω 3 (F G ′ )),
which completes the checking of (4.3).
Let S be the additive subgroup generated by all elements of the
form gϱ4 and bϱ3, where g ∈ G, b ∈ Gβ and ϱ3 ∈ ω 3 (F G ′ ), ϱ4 ∈
ω 4 (F G ′ ). We claim that [S, S] ⊆ I(G ′ ) 8 . Indeed, the additive subgroup
[S, S] can be spanned by some Lie commutators of the forms
[gϱ4, hϱ3] and [b1ϱ3, b2η3] with g ∈ G, b1, b2 ∈ Gβ, ϱ3, η3 ∈ ω 3 (F G ′ ),
ϱ4 ∈ ω 4 (F G ′ ). Furthermore, by Lemma 2.2.1,
[gϱ4, hϱ3] = g[ϱ4, hϱ3] + [g, hϱ3]ϱ4
= g[ϱ4, h + 1]ϱ3 + hg (g, h) + 1 ϱ3ϱ4
+ h[g + 1, ϱ3]ϱ4 ∈ I(G ′ ) 8 ,
and by Lemma 4.1.1(ii) and Lemma 4.1.3,
[b1ϱ3, b2η3] =b1[ϱ3, b2η3] + [b1, b2η3]ϱ3
=b1[ϱ3, b2 + 1]η3 + b2[b1 + 1, η3]ϱ3 + b1b2 (b1, b2) + 1 η3ϱ3
also belongs to I(G ′ ) 8 . Therefore, [S, S] ⊆ I(G ′ ) 8 .
From (4.2) and (4.3) we get δ [2] (F G) ⊆ S, so we have
δ [3] (F G) = δ [2] (F G), δ [2] (F G) ⊆ [S, S] ⊆ I(G ′ ) 8 .
Now, we use induction on k to show that
(4.5) δ [k] (F G) ⊆ I(G ′ ) 2k
for all k ≥ 3.
Indeed, assuming the validity of (5) for some k ≥ 3 we have
δ [k+1] (F G) = δ [k] (F G), δ [k] (F G) ⊆ I(G ′ ) 2k
, I(G ′ ) 2k ′ 2
⊆ I(G ) k+1
and this proves the truth of (4.5) for every k ≥ 3.
Keeping in mind that G ′ has order 2 n , (4.5) implies that δ [n] (F G) =
0. Hence dlL(F G) ≤ n and the proof is complete.