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!
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.