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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!

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    $\begingroup$ I'm confused, for $i=m$ isn't the answer obviously no... $\endgroup$ – Noah Snyder Oct 19 '12 at 15:21
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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 $[H_{i+1}:H_i]=p$.

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  • $\begingroup$ Thank you very much Jack! A complete answer! $\endgroup$ – azr Oct 19 '12 at 23:09

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