Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
HINT: there are metabelian groups of arbitrarily large nilpotency class. – Arturo Magidin Sep 19 '11 at 3:29
FURTHER HINT: Restrict attention to p-groups. – user641 Sep 19 '11 at 3:31
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
@Steve D: the dihedral groups of order $2^n$ for any $n$! Thanks for the hint~ – ShinyaSakai Sep 19 '11 at 12:23
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 $$[x^{2^k},t] = x^{-2^k} x^{-2^k} = x^{-2^{k+1}} \qquad [N^{2^k},G]=N^{2^{k+1}}$$

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.