On the Derived Length of Lie Solvable Group Algebras

3.2 THE BASIC GROUP IS NOT NILPOTENT 35
Proof. The upper bound ⌈log 2(2p n )⌉ on dl L (F G) is a consequence of
Proposition 2.1.1.
The statement (i) follows directly from Lemma 3.2.6. In order to
prove the inequality
⌈log 2(3p n /2)⌉ ≤ dlL(F G) ≤ dl L (F G) ≤ ⌈log 2(2p n )⌉,
it remains to show that if G/C has order 2 m p r , then ⌈log 2(3p n /2)⌉ ≤
dlL(F G). Since G is not nilpotent, Lemma 3.2.1 ensures m > 0, so
there is an element bC ∈ G/C of order 2. Let H = 〈x, b〉. Clearly,
b 2 ∈ ζ(H) and x b = x −1 , therefore the factor group H = H/ζ(H) is
isomorphic to the dihedral group of order 2p n , so by Lemma 3.2.3, if
d is the minimal integer such that s (1)
d ≥ pn then we have
d + 1 ≤ dlL(F H) ≤ dlL(F H) ≤ dlL(F G).
At the same time, by (3.8) we have (2 d+2 − 1)/3 ≥ s (1)
d ≥ pn , whence
d + 1 ≥ ⌈log 2(3p n /2 + 1/2)⌉ follows. Since ⌈log 2(3p n /2 + 1/2)⌉ =
⌈log 2(3p n /2)⌉, the the required inequality is guaranteed.
(ii) Let d be the integer such that s (m)
d
≥ pn , but s (m)
d−1 < pn . To
prove that d + 1 is an upper bound on dl L (F G) it is sufficient to show
that
δ (l+1) (F G) ⊆ I(G ′ ) s(m)
l for all l ≥ 0.
This is clear for l = 0, and assuming that δ (l) (F G) ⊆ I(G ′ ) s(m)
l−1, by
Lemma 3.2.2 we obtain
δ (l+1) (F G) = [δ (l) (F G), δ (l) (F G)]F G
⊆ [I(G ′ ) s(m)
l−1 , I(G ′ ) s(m)
l−1]F G ⊆ I(G ′ ) s(m)
l .
Therefore, dl L (F G) ≤ d + 1. Let us choose an element aC of order
2 m in G/C, set x k = x a and consider the group H = 〈x, a〉. Since
(x, a) = x −1+k ∈ H ′ and k ≡ 1 (mod p), we have that H ′ has order
p n . Moreover, H/CH(H ′ ) has also order 2 m . If m = 1 then, as
we have already seen before, H = H/ζ(H) is isomorphic to the dihedral
group of order 2p n , so the statement is a consequence of Lemma