On the Derived Length of Lie Solvable Group Algebras

32 CHAPTER 3
hypothesis and (3.11), zj ∈ δ [l+1] (F G) and
for some w s (m)
zj ∈ aωs(m) l
l
zj = a(k s(m)
l−1 − 1)(x − 1) 2s(m)
= a(k s(m)
l−1 − 1)(x − 1) s(m)
l
∈ ωs(m) l
+1+j
l−1 +j + aw s (m)
l
+j + aw s (m)
l
+1+j
+1+j
+1+j (F G ′ ). Since k s(m)
l−1 = 1, we have that
+j (F G ′ ) \ aωs(m) l +j+1 (F G ′ ) for all j ≥ 0. According to
Lemma 3.2.4, aωs(m) l (F G ′ ) ⊆ δ [l+1] (F G) and we can similarly verify
that a−1ωs(m) l (F G ′ ) ⊆ δ [l+1] (F G). The proof is complete.
Lemma 3.2.6. Let G be a group with cyclic commutator subgroup
of order p n and let char(F ) = p. Assume that one of the following
conditions holds:
(i) the order of G/C is divisible by an odd prime q = p;
(ii) p can be written in the form 2 r + 1 for some r > 1, n = 1 and
G/C has order 2 r .
Then dlL(F G) ≥ ⌈log 2(2p n )⌉.
Proof. Let G ′ = 〈x | xpn = 1〉. Assume first that condition (i) is
satisfied. Let us choose an element bC ∈ G/C of order q and set
xk = xb . Evidently, k2m ≡ 1 (mod p) for all m. Under condition (ii)
let bC be of order 2r and let again xk = xb . Then k2m for all 0 ≤ m ≤ r − 1.
≡ 1 (mod p)
Set H = 〈b, C〉. In both cases xk−1 = (x, b) ∈ H ′ is of order pn ,
so H ′ too has order pn . Since H ′ = (b, C) and the map c ↦→ (b, c)
is an epimorphism of C onto H ′ , we can choose c from C such that
(b, c) = x. Define the following three series in F G:
and, for l > 0,
u0 = b, v0 = c, w0 = c −1 b −1 ,
ul+1 = [ul, vl], vl+1 = [ul, wl], wl+1 = [wl, vl].