On the Derived Length of Lie Solvable Group Algebras

24 CHAPTER 3
since γ3(G) ⊆ (G ′ ) p , we have that k ≡ 1 (mod p). Therefore, aC has
p-power order and G/C is a p-group.
Since we have already closed the nilpotent case, by Lemma 3.2.1
we may assume that G/C has order 2 m p r with m > 0, r ≥ 0, or it is
divisible by some odd prime not equal to p. These two cases will be
discussed in the sequel.
Choose the element 〈aC〉 and the integer k such that G/C = 〈aC〉
and x k = x a . We claim that
(3.6) [(x − 1) 2l
, a] ≡ (k 2l
− 1)a(x − 1) 2l
for any l ≥ 0. Using the well-known equation
(3.7) (x s − 1) =
we have
as we claimed.
[(x − 1) 2l
, a] ≡ a
k
i=1
s
i=1
s
i
(x − 1) ,
i
(mod I(G ′ ) 2l +1 ),
2l k
i
(x − 1) −(x − 1) i
2l
≡ a k 2l
(x − 1) 2l
− (x − 1) 2l
≡ (k 2l
− 1)a(x − 1) 2l
(mod I(G ′ ) 2l +1 ),
Lemma 3.2.2. Let G be a group with cyclic commutator subgroup of
order p n , where p is an odd prime, and let char(F ) = p. If G/C has
order 2 m p r then
[I(G ′ ) 2m i , I(G ′ ) 2 m j ] ⊆ I(G ′ ) 2 m i+2 m j+1 .
Proof. With the notation introduced above, the assumption ensures
the congruence k2m ≡ 1 (mod p), thus we can apply (3.6) to conclude
the inclusion
[ω(F G ′ ) 2m
, F G] ⊆ I(G ′ ) 2m +1
.