On the Derived Length of Lie Solvable Group Algebras

3.2 THE BASIC GROUP IS NOT NILPOTENT 25
Hence, for i = j = 1 we have
[I(G ′ ) 2m
, I(G ′ ) 2m
] = [ω 2m
(F G ′ )F G, ω 2m
(F G ′ )F G]
⊆ ω 2m+1
(F G ′ )[F G, F G] + ω 2m
(F G ′ )[ω 2m
(F G ′ ), F G]F G
⊆ I(G ′ ) 2m+1 +1 ,
and by induction on i we obtain
[I(G ′ ) 2m i , I(G ′ ) 2 m
] = I(G ′ ) 2m
[I(G ′ ) 2m (i−1) , I(G ′ ) 2 m
]
+ [I(G ′ ) 2m
, I(G ′ ) 2m
]I(G ′ ) 2m (i−1)
⊆ I(G ′ ) 2m i+2 m +1 .
Now we can finish the proof by an easy induction on j. Indeed,
[I(G ′ ) 2m i , I(G ′ ) 2 m j ] = I(G ′ ) 2 m
[I(G ′ ) 2m i , I(G ′ ) 2 m (j−1) ]
+ [I(G ′ ) 2m i , I(G ′ ) 2 m
]I(G ′ ) 2m (j−1)
⊆ I(G ′ ) 2m i+2 m j+1 .
Definition. For m ≥ 0 let us define the series (s (m)
l
⎧
⎪⎨ 1 if l = 0;
s (m)
l
= 2s
⎪⎩
(m)
l−1
2s (m)
l−1
+ 1 if s(m)
l−1 is divisible by 2m ;
otherwise.
) as follows:
It is easy to check that for m = 1 this series can also be given in
the form
(3.8) s (1)
l =
(2l+2 − 1)/3 if l is even;
(2l+2 − 2)/3 if l is odd.
Lemma 3.2.3. Let G = D2pn = 〈a, b | a2 = bpn = 1, ba = ab−1 〉
be the dihedral group of order 2pn , where p is an odd prime, and let
char(F ) = p. If d is the minimal integer such that s (1)
d ≥ pn , then
dlL(F G) ≥ d + 1.