3
$\begingroup$

Suppose $G$ is a finite non-abelian p-group of nilpotent class $c$. Is there a subgroup $H$ of nilpotent class $c$ and size $p^{c+1}$?

If this is not true, is it possible to add some additional assumptions to satisfy the sentence?

$\endgroup$

1 Answer 1

5
$\begingroup$

The answer to the question is no for $p=2$. The $2$-groups of maximal class have been classified. There are only $3$ isomorphism types and they all have a normal cyclic subgroup of order $2^c$. But there are $2$-groups of exponent $4$ with arbitrary large class.

For a general method of constructing counterexamples, let $P$ be any $p$-group of maximal class $c >1$. Then $\Phi(P)=[P,P]$ and $|P/\Phi(P)|=p^2$.

Let $Q = C_{p^2}^2$, and let $G$ be a subdirect product of $P \times Q$ that contains $\Phi(P) \times \Phi(Q)$ such that $|G:(\Phi(P) \times \Phi(Q))|=p^2$. So $|G| = p^{c+1+2}$. Since $\Phi(P) = [G,G] \le \Phi(G)$ and $G^p\Phi(P) = \Phi(Q)$, we have $\Phi(G) = \Phi(P) \times \Phi(Q)$.

So any maximal subgroup of $G$ maps onto a proper subgroup of $G/(\Phi(P) \times \Phi(Q))$ and hence onto a proper subgroup of $P$, and hence has class less than $c$. So $G$ cannot contain a subgroup having maximal class.

$\endgroup$
2
  • $\begingroup$ Derek, thanks. Do you see any additional assumtions for subgroup $H$ such that $cl(G)=cl(H)$? $\endgroup$
    – maryam
    Jul 7, 2016 at 22:25
  • $\begingroup$ A finite nilpotent group is the direct product of its Sylow subgroups, so its nilpotence class is the maximum of their classes. In particular, your group not being a $p$-group suffices. $\endgroup$ Jul 8, 2016 at 6:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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