On the Derived Length of Lie Solvable Group Algebras

GROUP ALGEBRAS WITH ALMOST MAXIMAL... 63
Lemma 6.2.1. Let G be a nilpotent group with commutator subgroup
of order p n and let char(F ) = p. Then t L (F G) = p n − p + 2 if and
only if one of the following conditions holds:
(i) p = 2, n = 2 and d(2) = 2;
(ii) p = 2, n > 2, d (2 i +1) = d (2 n−1 ) = 1 and d(j) = 0,
where 0 ≤ i ≤ n − 2, j = 2 i + 1, j = 2 n−1 and j > 1;
(iii) p = 3, n = 2 and d(2) = 1 d(3) = 1;
(iv) p = 3, n ≥ 2, d (3 i +1) = d(3 n−1 ) = 1 and d(j) = 0,
where 0 ≤ i ≤ n − 2, j = 3 i + 1, j = 3 n−1 and j > 1.
Proof. If any one of the statements (i)-(iv) holds, then by (6.3), we
easily get t L (F G) = p n − p + 2.
Conversely, assume that t L (F G) = p n − p + 2. If n = 1 then
Proposition 6.1.1 states that t L (F G) = p n + 1, therefore n ≥ 2. For
n = 2 the equation (6.3) shows immediately that only (i) or (iii)
are possible. Now, we prove that p ≤ 3. Indeed, supposing that
p ≥ 5 and G ′ is not cyclic, we can apply Proposition 6.1.2 and (iii)
of Proposition 6.1.3 to get t L (F G) = tL(F G) ≤ p n−1 + 2p − 1. But
p n − p + 2 > p n−1 + 2p − 1, because (p n−1 − 3)(p − 1) > 0.
So, only the cases when n > 2 and either p = 3 or p = 2 remain.
First, we shall show that d (p i +1) > 0 for 0 ≤ i ≤ n − 2.
Suppose there exists 0 ≤ s ≤ n − 2 such that d(p s +1) = 0. From
(6.1) it follows at once that s = 0 and by (i) of Proposition 6.1.3
D(p s +1)(G) = 〈1〉, so d(r) = 0 for every r ≥ p s + 1. Moreover, if