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Given some integer $n$, is there an easy way to construct a finite solvable group of derived length $n$?

It would seem that given a solvable group of length $n-1$, one should be able to form the semidirect product with a suitably chosen abelian subgroup of its automorphism group. But I don't see an easy way to ensure that the derived length actually increases this way.

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It can be done with the wreath product...have you studied this beast? – DonAntonio Jan 30 '13 at 19:55
Yes, I have seen those. – Tobias Kildetoft Jan 30 '13 at 20:14
Well, then try to read here: – DonAntonio Jan 30 '13 at 20:17
Thank you, that is a very nice way to do it. – Tobias Kildetoft Jan 30 '13 at 20:30
@SteveD In the paper linked, the wreath product is taken with an arbitrary cyclic group. It is also interesting to note that this actually shows that the $p$-Sylow subgroup of $S_{p^n}$ has derived length $n$. – Tobias Kildetoft Jan 31 '13 at 13:36
up vote 3 down vote accepted

A handy example is provided by the group of $n \times n$ upper triangular unipotent matrices. If $n =2^{t-1}+1$, then its derived length is $t$.

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Can you provide a proof or a link to a proof of this result, please? – DonAntonio Jan 30 '13 at 20:17
I can see that the derived length is at most $t$. Is there an easy argument for why it is exactly $t$? – Tobias Kildetoft Jan 30 '13 at 20:17
@DonAntonio see the reference on top of page 4 of… – Andreas Caranti Jan 30 '13 at 20:37
Thanks a lot, @AndreasCaranti . Huppert's first book in finite groups is imo one of the most important books that still waits to be translated into english (mainly to avoid those annoying germanic letters...yucks!), but it certainly has contributed a lot to enhance the skills in german language of many algebraists I know. +1 – DonAntonio Jan 30 '13 at 20:42
@DonAntonio, I completely agree, my copy is seriously worn out, but still precious. +1 – Andreas Caranti Jan 30 '13 at 20:44

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