On the Derived Length of Lie Solvable Group Algebras

54 CHAPTER 5
Case 2: G/C = 〈dC〉, where x d = x −5 . Then G ′ = (d, C) and, as
before, we can choose c ∈ C such that (c, d) = x and
[c, d], [d −1 c, d] , [c, d], [c, dc]
dc(x
2 −5
= + 1), c(x + 1) , dc(x + 1), dc (x + 1)
= dc 2 (x −4 + 1)(x + 1), d 2 c 3 (x −5 + 1)(x + 1) 2
= b 3 c 5 x 6 (x −5 + 1)(x 9 + 1)(x + 1) 9
belongs to δ [3] (F G) and is not zero.
Case 3: G/C = 〈aC, bC〉, where x a = x −1 and x b = x 5 . Then by
similar arguments as in the last case of the previous lemma we can
restrict ourselves to the case when (a, b) = x. Then
[a, b], [b −1 a, b] , [a, b], [b, ab]
which was to be proved.
ba(x 2 −5
= + 1), a(x + 1) , ba(x + 1), ab (x + 1)
= ba 2 (x 10 + 1)(x + 1), b 3 a 2 (x 10 + 1)(x 7 + 1)
= b 4 a 4 x 3 (x 5 + 1) 4 (x + 1) 6 = 0,
Theorem 5.1.6. Let G be a group with cyclic commutator subgroup
of order p n and let F be field of characteristic p. Then dlL(F G) = 3
if and only if one of the following conditions holds:
(i) p = 7, n = 1 and G is nilpotent;
(ii) p = 5, n = 1 and either x g = x −1 for all x ∈ G ′ and g ∈ CG(G ′ )
or G is nilpotent;
(iii) p = 3, n = 1 and G is not nilpotent;
(iv) p = 2 and one of the following conditions is satisfied: