On the Derived Length of Lie Solvable Group Algebras

3.1 THE BASIC GROUP IS NILPOTENT 21
We shall show that
(3.3)
fk(l, m,s, t, z1, z2)
≡ (ms − lt)b l+s a m+t (x − 1) 2k+1 −1
(mod I(G ′ ) 2k+1
).
Lemma 2.2.1 ensures that the elements
l m 2
b a (x − 1) k
−1
, z2 , z1, z2 and
z1, b s a t (x − 1) 2k−1
belong to I(G ′ ) 2k+1,
thus
fk(l, m,s, t, z1, z2)
≡ b l a m (x − 1) 2k −1 , b s a t (x − 1) 2 k −1
(mod I(G ′ ) 2k+1
).
Furthermore, for p > 2 the inclusions
b l a m , (x − 1) 2 k −1 , (x − 1) 2k −1 , b s a t ∈ I(G ′ ) 2k +1
is guaranteed by Lemma 2.2.1 and it implies that
fk(l, m, s, t, z1, z2) ≡ [b l a m , b s a t ](x − 1) 2k+1 −2
(mod I(G ′ ) 2k+1
).
This congruence, together with (3.1), proves (3.3).
Define the following three series of elements of F G inductively by:
and, for k ≥ 0,
u0 = a, v0 = b, w0 = b −1 a −1 ,
uk+1 = [uk, vk], vk+1 = [uk, wk], wk+1 = [wk, vk].
Obviously, the k-th elements of these series belong to δ [k] (F G). By
induction on k we show for odd k that
(3.4)
uk ≡ ±ba(x − 1) 2k −1
vk ≡ ±b −1 (x − 1) 2k −1
wk ≡ ±a −1 (x − 1) 2k −1
(mod I(G ′ ) 2k
);
(mod I(G ′ ) 2k
);
(mod I(G ′ ) 2k
),