On the Derived Length of Lie Solvable Group Algebras

64 CHAPTER 6
d(q+1) = 0, then q ≤ p s . According to (6.3), it follows that
t L s−1
(F G) = 2 + (p − 1) p i s−1
+ p i (d (pi +1) − 1) +
i=0
s−1
< 2 + (p − 1) p i +
i=0
i=0
i=0
s−1
q=p i
(d (pi +1) − 1) +
i=0
s−1
= 2 + (p − 1) p i + (n − s) · p s
< 1 + p n−2 + (p − 1)(n − (n − 2)) · p n−2 .
q=p i
qd(q+1)
s
d(q+1) p
Hence, for p = 2 we have t L (F G) < 1 + 3 · 2 n−2 < p n , and if p = 3
then t L (F G) < 1 + 5 · 3 n−2 < p n − 1, which contradicts the assumption
t L (F G) = p n − p + 2.
Therefore d (p i +1) > 0 for 0 ≤ i ≤ n − 2 and by (6.4) there exists
α ≥ 2 such that d(α) = 1. Then by (6.3),
t L n−2
(F G) = 2 + (p − 1) p i + (p − 1)(α − 1)d(α)
i=0
= 1 + p n−1 + (p − 1)(α − 1).
Since t L (F G) = p n − p + 2, we must have α = p n−1 and the proof is
done.
Lemma 6.2.2. Let G be a nilpotent group with commutator subgroup
of order 2 n and let char(F ) = 2. If t L (F G) = 2 n then one of the
following conditions holds:
(i) G is of class 2 and G ′ is noncyclic of order 4;
(ii) G is of class 4 with one of the following properties:
(a) G ′ ∼ = C4 × C2, γ3(G) ∼ = C2 × C2;
(b) G ′ ∼ = C2 × C2 × C2.