On the Derived Length of Lie Solvable Group Algebras

16 CHAPTER 2
According to Lemma 2.2.1 the last three summands are in I(G ′ ) 2n.
Furthermore,
fi, gi
= x k
wi (1 − z2) · · · (1 − z2n−1), x k
vi (1 − z2n−1 +1) · · · (1 − z2n−1)
k
x wi, (1 − z2n−1 +1) · · · (1 − z2n−1) (1 − z2) · · · (1 − z2n−1) + x k vi
+ x k wi, x k
vi (1 − z2) · · · (1 − z2n−1) and the first two summands on the right-hand side belong to I(G ′ ) 2n
by Lemma 2.2.1. So,
fi + ϱ (i)
1 ,gi + ϱ (i)
2
≡ x k wi, x k
vi (1 − z2) · · · (1 − z2n−1) (mod I(G ′ ) 2n
),
for all 1 ≤ i ≤ s. Summing this for all possible i, we get
x j c(1 − z1)(1 − z2) · · · (1 − z2 n −1) + ϱ ∈ δ [n] (F G),
for some ϱ ∈ I(G ′ ) 2n,
as we claimed.
Let the abelian p-group G ′ be presented as G ′ = 〈c1〉 × · · · × 〈cr〉,
and assume that ci is of order pmi . The product
(1 − c1) i1 (1 − c2) i2 · · · (1 − cr) ir
with 0 ≤ ij ≤ p mj has s = i1 + i2 + · · · + ir factors and, according to
Jennings’ theory, it belongs to ω s (F G ′ ) \ ω s+1 (F G ′ ). It follows that if
2 n − 1 < tN(G ′ ) then the elements z1, z2, . . . , z2 n −1 of G ′ can be chosen
such that
x j c(1 − z1)(1 − z2) · · · (1 − z2 n −1) ∈ I(G ′ ) 2n −1 \ I(G ′ ) 2 n
.
We have that δ [n] (F G) has nonzero elements while 2 n − 1 < tN(G ′ ),
and therefore
dlL(F G) ≥ ⌈log 2 tN(G ′ ) + 1⌉.
The converse inequality follows immediately from the first part of the
theorem.