On the Derived Length of Lie Solvable Group Algebras

3.2 THE BASIC GROUP IS NOT NILPOTENT 23
Assume that k < ⌈log 2(p n +1)⌉. Then 2 k −1 < p n and the elements
uk, vk, wk are nonzero in δ [k] (F G), thus dlL(F G) ≥ ⌈log 2(p n + 1)⌉.
At the same time, Theorem 2.2.2 says that
The proof is complete.
dl L (F G) ≤ ⌈log 2(p n + 1)⌉.
3.2 The basic group is not nilpotent
Let G be a group with commutator subgroup G ′ = 〈x | xpn = 1〉 of
odd order, and let C denote its centralizer in G. As it is well-known,
the automorphism group of G ′ is isomorphic to the unit group U(Zpn) of Zpn. Since p is odd, U(Zpn) is cyclic, so the factor group G/C, which
is isomorphic to a subgroup of it, is cyclic, too. In our observation the
order of the group G/C will play an important role.
It is easy to check that
Sylp U(Zpn) = k ∈ Zpn | k ≡ 1 (mod p) ,
from which it follows for a ∈ G that the automorphism x ↦→ xa of G ′
has p-power order if and only if xa = xk
, where k ∈ Sylp U(Zpn) . In
other words, the element aC has p-power order if and only if xa = xk for some k ≡ 1 (mod p).
Lemma 3.2.1. Let G be a group with cyclic commutator subgroup of
order p n , where p is an odd prime. Then G is nilpotent if and only if
G/C is a p-group.
Proof. Assume first that G/C is a p-group. The normal subgroup
G ′ can be written as union of different conjugacy classes of G, moreover,
the assumption ensures that every class has p-power order. Since
G ′ has order pn , there are at least p classes which contain only one
element, that is, G ′ contains nontrivial central elements of G. Consequently,
G is nilpotent.
Conversely, assume that G is nilpotent. Set G ′ = 〈x | xpn = 1〉,
G/C = 〈aC〉 and xk = xa . Then (x, a) = x−1+k belongs to γ3(G) and