# Let $G$ be group with order $p^n$; does there then exist a sequence of normal subgroups?

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!

• Hint: $H\cap Z(G)$ is nontrivial, which gives you a subgroup of order $p$ of $H$ normal in $G$. Mod out by it. – Arturo Magidin Feb 4 at 17:17
• @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? – Tortuga Feb 4 at 17:25
• 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. – Arturo Magidin Feb 4 at 17:28
• Done it, thanks! – Tortuga Feb 4 at 17:42

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.
• 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 $p^i$..." – Arturo Magidin Feb 4 at 17:56
• Actually, if $K_i/N$ has order $p^i$, then $K_i$ has order $p^{i+1}$... – Arturo Magidin Feb 4 at 18:15