On the Derived Length of Lie Solvable Group Algebras

52 CHAPTER 5
Lemma 5.1.4. Let G be a group with commutator subgroup G ′ = 〈x |
x2n = 1〉, where n > 3, let char(F ) = 2 and assume that exp(G/C) ≤
2. Then dlL(F G) = 3 if and only if C is abelian and G/C = 〈aC〉,
where xa = x−1 .
Proof. Since exp(G/C) ≤ 2, only the following cases are possible:
Case 1: either G/C is trivial or G/C = 〈bC〉 where x b = x 2n−1 +1 .
Clearly, G has nilpotency class at most 3, therefore by Theorem 4.2.1
we have dlL(F G) = n + 1.
Case 2: G/C = 〈aC〉, where xa = x−1 . Then C ′ ⊆ G ′ ∩ ζ(G) =
〈x2n−1〉. If C ′ = 〈1〉 then C is an abelian subgroup of index two of G
and Theorem 5.1.2 implies that dlL(F G) = 3. Now, let C ′ = 〈x2n−1〉. Then we can choose b, c ∈ C such that
(c, a) = x, (c, b) = x 2n−1
, (a, b) ∈ 〈x 2 〉.
Indeed, let us consider the map ϕ : C → G ′ , where ϕ(c) = (c, a), which
is an epimorphism because G ′ = (a, C). Of course, H = ϕ−1 〈x2 〉 is
a proper subgroup of C. Let u ∈ C \ ζ(C) and c ∈ C \ H ∪ CC(u)
be such that (c, a) = x. Obviously, (c, u) = x2n−1. If (a, u) ∈ 〈x2 〉 then
set b = u, otherwise b = cu. It is easy to see that the elements b and
c satisfy the conditions stated. Then
[c, a],[c −1 a, c] , [c, a], [c −1 ba, c]
= [ac(x + 1), a(x −1 + 1)], [ac(x + 1), ba(x 2n−1 −1 + 1)]
= a 2 cx −1 (x + 1) 3 , ba 2 c (b, a)x −1 + 1 (x 2n−1 +1 + 1)(x + 1)
= a 4 bc 2 x −1 (b, a)x −1 + 1 (x 2n−1 +1 + 1)(x + 1) 2 n−1 +4
belongs to δ [3] (F G) and is not equal to zero, thus dlL(F G) > 3.
Case 3: G/C = 〈dC〉, where x d = x 2n−1 −1 . Since G ′ = (d, C),