On the Derived Length of Lie Solvable Group Algebras

2.2 THE GENERALIZED RESULT 15
a sufficient number times, we have
(2.1) x j c(1 − z1) ≡
s
x j c 1 − (ai, x k )
i=1
(mod I(G ′ ) 2 ).
Since p is an odd prime, we can choose the elements ui, vi such that
u 2 i
= ca−1
i
and v 2 i = cai. Then ui, vi ∈ A, (uivi) 2 = c 2 and (u −1
i vi) 2 =
a 2 i which implies uivi = c and u −1
i vi = ai. Setting wi = x k ui(x k ) −1 we
have
[x k wi, x k vi] = x j (w xk
i vi − wiv xk
i )
= x j w xk
i vi
−1
1 − (wi vi, x k ) = x j c 1 − (ai, x k ) ,
because (w −1
i vi, x k ) = (u −1
i vi, x k ) = (ai, x k ). Now by (2.1) it follows
that
(2.2) x j c(1 − z1) ≡
s
[x k wi, x k vi] (mod I(G ′ ) 2 ),
i=1
which proves our statement for n = 1.
Now, assume that j, c, z1, z2, . . . , z2 n −1 have already been given, and
let 2k ≡ j modulo the order of x. We can apply the method above to
find elements wi, vi ∈ A such that the congruence (2.2) holds. Set
and
fi = x k wi(1 − z2) · · · (1 − z2 n−1)
gi = x k vi(1 − z2 n−1 +1) · · · (1 − z2 n −1).
for 1 ≤ i ≤ s. By the induction hypothesis there exist ϱ (i)
1 , ϱ (i)
2 ∈
I(G ′ ) 2n−1 such that fi + ϱ (i)
1 , gi + ϱ (i)
2 ∈ δ[n−1] (F G). Evidently,
fi + ϱ (i)
1 , gi + ϱ (i)
2
= [fi, gi] + [fi, ϱ (i)
2 ]
+ [ϱ (i)
1 , gi] + [ϱ (i)
1 , ϱ (i)
2 ] ∈ δ [n] (F G).