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If $G$ is a nilpotent group with positive class $c$, its derived length is at most $[\mathrm{log}_2c]+1$.

This statement can be proved by the inclusion of groups in the derived series and central series.

But I don't know how to prove

The class of a nilpotent group cannot be bounded by a function of the derived length.

I think I should find a sequence of nilpotent groups for which the derived lengths are equal, but the classes are not bounded. But I have no idea.

Thanks very much.

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    @ShinyaSakai: So, how about posting it with a proof as an answer? $T$hen you can accept it.2011-09-19

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(Community wiki summary of the answer to remove it from unanswered questions.)

The dihedral group $G$ of order $2^{n+1}$ has a cyclic subgroup $N$ of order $2^n$. Being index 2, this subgroup is normal, and the quotient, being order 2 is abelian. Hence the derived length of $G$ is 2; $G$ is metabelian.

The lower central series of $G$ is $G > N^2 > N^4 > \dots > N^{2^{n-1}} > N^{2^n} = 1$ so the nilpotency class of $G$ is $n$.

Explicitly: $G =\langle t, x : t^2 = x^{2^n} = 1, xt = tx^{-1} \rangle \qquad N =\langle x \rangle$ and $[x^{2^k},t] = x^{-2^k} x^{-2^k} = x^{-2^{k+1}} \qquad [N^{2^k},G]=N^{2^{k+1}}$

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    I am sorry, I should have done all this by myself. I didn't see Arturo Magidin's last comment and I thought it was a little bit strange to answer my own question. Thank you very much.2011-10-27