# Is there a simple construction of a finite solvable group with a given derived length?

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: hindawi.com/journals/ijmms/2011/245324 –  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

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 $t$.
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