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I would like to show the following statement:

Let $p$ be a prime. Let $G$ be group with order $p^n$. Let $H$ be a normal in $G$ with order $p^k$. Then prove $H$ has subgroups $K$ such that $K$ has order $1,p,p^2,\ldots,p^k$ and $K$ is normal in $G.$

I was trying to prove this by induction on $k$. When $k=0$ or $1$, this is clear, since $H=\{e\}$ or $H\le Z(G)$. Suppose this is true for $k-1$. Let $H$ be a normal subgroup of $G$ with order $p^k$, then by Sylow theorem, $H$ has (in fact normal) subgroups of order $1,p,\ldots,p^{k-1}$. However how can I show they are normal in $G$?

Thanks for any hints or helps!

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    $\begingroup$ Hint: $H\cap Z(G)$ is nontrivial, which gives you a subgroup of order $p$ of $H$ normal in $G$. Mod out by it. $\endgroup$ – Arturo Magidin Feb 4 at 17:17
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    $\begingroup$ @ArturoMagidinThanks for your hint! Let $N$ be the intersection you mentioned. If we mod it out from $G$, then $H/N$ is normal in $G/N$ by correspondence theorem. Then $H/N$ is a group with order smaller than $p^k$, thus, by induction it has subgroups $K/N$ normal in $G/N$. Therefore by correspondence theorem again, $K$ normal in $G$, and the order of $K$ is $1,\ldots,p^{k-1}$ automatically. Does this sound correct? $\endgroup$ – Tortuga Feb 4 at 17:25
  • $\begingroup$ Except that you don't want to mod out by $H\cap Z(G)$, you want to mod out by a subgroup of $H\cap Z(G)$ of order $p$. Now, write it out in detail an as "answer" to your own question. That way, it can be upvoted. $\endgroup$ – Arturo Magidin Feb 4 at 17:28
  • $\begingroup$ Done it, thanks! $\endgroup$ – Tortuga Feb 4 at 17:42
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I would thank @ArturoMagidin for the hints

Consider $H\cap Z(G)$. This is nontrivial, since $H$ is normal in a $p$-group $G$. Also since $H\cap Z(G)$ is normal subgroup in $G$, it has order $p^t$ for some $t=1,\ldots,k-1$. Thus by Cauchy's lemma (or Sylow theorem with $m=1$), it has a subgroup $N$ of order $p$, and $N$ is necessarily normal in $G$.

Mod it out from $G$, then $H/N$ is normal in $G/N$ by correspondence theorem.

Therefore $H/N$ is a group with order $p^{k-1}$. Thus, by induction it has subgroups $\{K_1/N,\ldots,K_{k-1}/N\}$ which are normal in $G/N$, where $K_i/N$ has order $p^{i}$ for each $i=1,\ldots,k-1$. Therefore by correspondence theorem again, $K_i$ normal in $G$, and the order of $K_i$ is $p^{i+1}$, which completes the proof by adding the trivial group to the set.

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    $\begingroup$ It would be better if you did not use the same letter to represent all the subgroups of $G/N$. So, "it has subgroups $K_0/N$, $K_1/N,\ldots,K_{k-1}/N$, where $K_i/N$ has order $p^i$..." $\endgroup$ – Arturo Magidin Feb 4 at 17:56
  • $\begingroup$ already edited :) $\endgroup$ – Tortuga Feb 4 at 18:10
  • $\begingroup$ Actually, if $K_i/N$ has order $p^i$, then $K_i$ has order $p^{i+1}$... $\endgroup$ – Arturo Magidin Feb 4 at 18:15
  • $\begingroup$ my bad, thanks:) $\endgroup$ – Tortuga Feb 4 at 18:39

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