On the Derived Length of Lie Solvable Group Algebras

28 CHAPTER 3
and s > 0 the set gω s (F H) = {gy | y ∈ ω s (F H)} is a vector space
over F , and by [16], the set {g(h − 1) s+j | j ≥ 0} forms an F -basis of
gω s (F H).
Lemma 3.2.4. Let H = 〈h | hpn = 1〉 be a subgroup of a group G
and char(F ) = p. Then for all g ∈ G \ H the vector space gω s (F H)
can be spanned by the elements zi ∈ gωs+i (F H) \ gωs+i+1 (F H), where
0 ≤ i ≤ pn − s − 1.
Proof. Denote by V the vector space spanned by the zi’s. It is enough
to prove that g(h − 1) s+j ∈ V for all j ≥ 0. This is clear when
j ≥ p n − s. Let t be an integer such that 0 ≤ t < p n − s and assume
that g(h−1) s+j ∈ V for all j > t. Since zt ∈ gω s+t (F H)\gω s+t+1 (F H),
it can be written as
where α (zt)
r
∈ F and α (zt)
t
zt =
r≥t
α (zt)
r g(h − 1) s+r ,
= 0. Hence
α (zt)
t g(h − 1) s+t = zt −
r≥t+1
α (zt)
r g(h − 1)s+r .
According to the inductive hypothesis, the right-hand side of this
equality belongs to V , and since α (zt)
t = 0, we have g(h − 1) s+t ∈ V .
The proof is complete.
As before, let G be a group with commutator subgroup G ′ =
〈x | xpn = 1〉, where p is an odd prime, but now assume that G/C has
order 2m , with m > 1. Since G/C is cyclic, we can choose an element
a ∈ G such that 〈aC〉 = G/C. Clearly, 〈a−1C〉 = 〈aC〉 = G/C.
Setting xk = xa and xk′ = xa−1, it follows that k ≡ 1 (mod p),
k ′ ≡ 1 (mod p), furthermore k2m ≡ (k ′ ) 2m ≡ 1 (mod pn ) and kk ′ ≡ 1
(mod p). Now, we show that ki ≡ 1 (mod p) if and only if 2m divides
i. Indeed, if ki ≡ 1 (mod p) then ki
∈ Sylp U(Zpn) and (k) ipr = 1
holds in U(Zpn) for some r. Since k is of order 2m we have that 2m divides ipr and therefore 2m divides i. The converse statement is trivial.