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Q: Prove that a subgroup of order $p^n$ has a normal subgroup of order $p^{k}$ for all $0\leq k \leq n$.

Attempt at a proof: We proceed by Induction. This is obviously true for $n=1, 2$. Suppose it is true for $m \leq n-1$. Now, take the group $G$ of order $p^{n}$. Since it is a $p$-group, it has a non-trivial center by the Conjugacy Class equation. This center then has a subgroup of $K$ order $p$ by Cauchy's theorem for example. Since this subgroup is contained in the center, it is normal in $G$.

Take the quotient $G/K$ which now has order $p^{n-1}$ hence by the induction hypothesis has subgroups $\bar{H_{k}}$ of order $p^{k}$ for $0\leq k \leq n-1$, which are normal in $G/K$. By the lattice isomorphism theorem, the lattice of $G$ has corresponding groups $H_{k}$ which are normal in $G$.

The index of each such subgroup is $|G:H_{k}|=|G/K: \bar{H_k}|=p^{n-1-k}$, hence $H_{k}$ has order $p^{k+1}$ in $G$ as desired.

Does this look okay?

TheManWhoNeverSleeps
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    Do it by induction. First consider the abelian case. Then consider the non abelian case, for this one after the base case, split by the center(I mean the quotient group $G/Z(G)$) and apply correspondence theorem. – Diego Robayo Dec 11 '14 at 06:17
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    Is there an error in what I wrote up there? The way I set it up it doesn't seem I need to split it into an abelian and non abelian case, but I might be wrong. – TheManWhoNeverSleeps Dec 11 '14 at 06:21
  • I don't think so. It seems quite well. – Diego Robayo Dec 11 '14 at 06:24
  • @DiegoRobayo Okay, I see what you mean though. If I take the center to be say order $p^{k}$ in there I have all the subgroups of order less $p^{m}, m\leq k$ since for Abelian groups it's easy to construct the subgroups, and then I still need to apply the correspondence theorem to $G/Z$ in the same way. Thanks :) – TheManWhoNeverSleeps Dec 11 '14 at 06:31
  • @TheManWhoNeverSleeps It looks just fine to me. Consider writing an answer to your own question so that it will not be contained in the "Unanswered Questions" section – Timbuc Dec 13 '14 at 11:31

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It seems everyone agrees that this is fine, so I'm posting this in order for the question to be removed from the unanswered section.

TheManWhoNeverSleeps
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  • You could make this post CW. – Pedro Dec 13 '14 at 18:35
  • Well, by induction each $\bar{H_{i}}$ is normal in $G/K$. The 4th isomorphism theorem says that the lattices for $G$ above $K$ and that of $G/K$ are the same, and $\bar{H_{i}}$ is normal if and only if $H_{i}$ is normal. For example pg 99 of http://zeth.ciencias.uchile.cl/~cortiz/Apuntes/ebooks/Abstract%20algebra%20-%20Dummit%20and%20Foote.pdf And sorry, what is CW? – TheManWhoNeverSleeps Dec 13 '14 at 18:43
  • Right. You can do this more generally for a group of any finite order and look at its Sylow subgroups. Then you can claim that the subgroups are normal in the Sylow, but not in all of $G$. My bad. – Pedro Dec 13 '14 at 18:49
  • No worries, I'm glad you asked. :) – TheManWhoNeverSleeps Dec 13 '14 at 18:52