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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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3  
HINT: there are metabelian groups of arbitrarily large nilpotency class. –  Arturo Magidin Sep 19 '11 at 3:29
3  
FURTHER HINT: Restrict attention to p-groups. –  user641 Sep 19 '11 at 3:31
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FURTHER FURTHER HINT: ... and restrict further to $p=2$. –  Arturo Magidin Sep 19 '11 at 3:54
    
@Arturo Magidin: the dihedral groups of order $2^n$ for any $n$! Thanks for the hints~ –  ShinyaSakai Sep 19 '11 at 12:23
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@Steve D: the dihedral groups of order $2^n$ for any $n$! Thanks for the hint~ –  ShinyaSakai Sep 19 '11 at 12:23

1 Answer 1

up vote 2 down vote accepted

(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. –  ShinyaSakai Oct 27 '11 at 13:14

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