It is well known that finite $p$-groups have (normal) subgroups of all possible orders. Now, what can we say about subgroups containing a given non-normal subgroup? i.e.

Let $G$ be a group of order $p^n$ and let $H$ be a non-normal subgroup of $G$ of order $p^m$. Does there exist a (normal) subgroup of $G$ containing $H$ of order $p^i$, for $i=m,\ldots,n$? If not, can you show a counterexample?

Remark: I ask non-normality for $H$ because if it was normal, I could quotient out by it.

Thank you very much in advance!

  • 2
    $\begingroup$ I'm confused, for $i=m$ isn't the answer obviously no... $\endgroup$ Oct 19, 2012 at 15:21

1 Answer 1


You cannot require the containing subgroups to be normal ($i=m$ is an obvious problem, but $i<n$ can be a problem in general, for instance in the dihedral group).

However, you can certainly find subgroups above. This is more generally true because $p$-groups are supersolvable, and supersolvable groups have supersolvable subgroup lattices (so all subgroup maximal chains are the same length).

With $p$-groups you can prove this easily using an upper central series. If $H$ contains $Z(G)$, then mod out by $Z(G)$ and the problem has not really changed. Since $p$-groups are nilpotent and $H<G$, eventually we come to the case where $H$ does not contain $Z(G)$. Take $z \in \Omega(Z(G)) = \operatorname{Soc}(G)$ to be a central element of order $p$. Then $H_1 = \langle H, z_1 \rangle$ has the property that $[H_1:H]=p$. Now just repeat the process with $H_1$ instead. This gives a chain $H = H_0 < H_1 < \dots < H_n = G$ that stops when $H_n$ is no longer a proper subgroup, but at each step $[H_{i+1}:H_i]=p$.

  • $\begingroup$ Thank you very much Jack! A complete answer! $\endgroup$
    – azr
    Oct 19, 2012 at 23:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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