On the Derived Length of Lie Solvable Group Algebras

30 CHAPTER 3
apply Lemma 3.2.4 to conclude aω2 (F G ′ ) ∈ δ [2] (F G). Substituting
a−1 for a and thus k ′ for k, we can similarly get that a−1ω2 (F G ′ ) ∈
δ [2] (F G). Therefore the statement is indeed valid for l = 1.
Now, let l ≥ 2 and assume the truth of the assertion for all t < l.
Firstly, suppose that s (m)
l−1 is divisible by 2m . According to the
inductive hypothesis the element
belongs to δ [l] (F G). Clearly,
z = (x k′
z = [a(x − 1) s(m)
l−2, a −1 (x − 1) s(m)
l−2 +1 ]
− 1) s(m)
l−2(x − 1) s(m)
l−2 +1 − (x k − 1) s(m)
l−2 +1 (x − 1) s(m)
l−2
= (k ′ ) s(m)
l−2 − k s(m)
l−2 +1 (x − 1) s(m)
l−1 +1 + w
for some w ∈ ω s(m)
l−1 +2 (F G ′ ), and
(k ′ ) s(m)
l−2 − k s(m)
l−2 +1 = (k ′ ) s(m)
l−2(1 − k 2s(m)
l−2 +1 )
= (k ′ ) s(m)
l−2(1 − k s(m)
l−1 +1 ) = 0.
This implies that (x − 1) s(m)
l−1 +1 + w ′ ∈ δ [l] (F G) for suitable w ′ ∈
ω s(m)
l−1 +2 (F G ′ ). Let
zj = [(x − 1) s(m)
l−1 +1 + w ′ , a(x − 1) s(m)
l−1 +j ]
for all j ≥ 0. By the inductive hypothesis zj ∈ δ [l+1] (F G). At the
same time,
for some w s (m)
l
have that zj ∈ aωs(m) l
zj = a(k s(m)
l−1 +1 − 1)(x − 1) 2s(m)
= a(k s(m)
l−1 +1 − 1)(x − 1) s(m)
l
∈ ωs(m) l
+1+j
l−1 +1+j + aw s (m)
l
+j + aw s (m)
l
+1+j
+1+j
+1+j (F G ′ ), and since k s(m)
l−1 +1 − 1 = 0 we
+j (F G ′ )\aωs(m) l +1+j (F G ′ ) for all j ≥ 0. Applying
Lemma 3.2.4 we get aω s(m)
l (F G ′ ) ⊆ δ [l+1] (F G), and we can similarly