On the Derived Length of Lie Solvable Group Algebras

3.3 SUMMARIZED RESULT 37
Indeed, according to the part (iii) of the theorem, if l = dl L (F G)
then s (1)
l−2 < pn . From (3.8) it follows that (2l −1)/3 < pn . Hence
l < log2(3pn /2 + 1/2) + 1, and therefore l ≤ ⌈log2(3pn /2 + 1/2)⌉.
Since ⌈log2(3pn /2+1/2)⌉ = ⌈log2(3pn /2)⌉, the proof is complete.
(ii) Since the order of G/C divides the order of U(Zpn), which is
equal to pn−1 (p−1), for primes p of the form 4k−1 the conditions
either (ii) of the theorem or (i) of this remark is satisfied, so we
can easily have the derived length and the strong Lie derived
length of F G.
(iii) Recall that p is called Fermat prime, if p has the form 22s + 1 for
some s ≥ 0. Let G be a non-nilpotent group with commutator
subgroup of order p > 3, where p is a Fermat prime, and let
char(F ) = p. Then
dlL(F G) = dl L
⌈log
(F G) =
2(2p)⌉ if G/C has order p − 1;
⌈log2(3p/2)⌉ otherwise.
Indeed, let us write p in the form 2r + 1 (r > 1). If G/C has
order p − 1 then the statement follows directly from Lemma
3.2.6. In the other case G/C has order 2m for some 0 < m < r.
Since ⌈log2(3p/2)⌉ = r + 1, by part (ii) of the theorem it is
enough to show that s (m)
r ≥ p. Indeed, from the definition it
follows that s (r−1)
r−1 = 2r−1 . Furthermore, for m = r − 1 we have
s (m)
r = s (r−1)
r = 2s (r−1)
r−1 + 1 = 2r + 1 = p, and if m < r − 1 then
s (m)
r−1 > s (r−1)
r−1 , which implies s (m)
r ≥ 2s (m)
r−1 > 2s (r−1)
r−1 = 2r = p − 1.
The proof is complete.
Finally, we note that some parts of the theorem can also be proved
using the Theorems A and B of [29]. Obviously, these theorems of
A. Shalev are not enough to determine the derived length under conditions
of our theorem. An example for such group algebra is the
following: let
G = 〈a, b, c | a 2 = b 9 = c 19 = 1, b a = b, c a = c 18 , c b = c 7 〉