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Let $p$ be a prime number, and let $G$ be a $p$-group: $|G| = p^r$. Prove that $G$ contains a normal subgroup of order $p^k$ for every nonnegative $k \le r$.

The answers here and here use induction but they assume $G$, where $|G|=p^r$, has normal subgroups of order $p^k$ for $k <r$. Induction should start by assuming for every $p$-group of order $p^k$ where $0\le k <r$, there exists normal subgroups of order $p^i$ where $0 \le i \le k$.

We have to show there exists normal subgroups of order $p^i$ where $0 \le i \le r$.

user5826
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  • Figure out your own solution then. My suggestion: use induction and the fact that $p$-groups have non-trivial centres. (The induction would be on $r$, so for $s<r$ the inductive hypothesis will give you that a $p$-group of order $p^s$ has normal subgroups of each possible order.) – the_fox Jun 27 '19 at 18:43
  • @the_fox That's exactly what I wrote. – user5826 Jun 27 '19 at 18:44
  • Do you want me to comment on the other "solutions" or how to give a proof? – the_fox Jun 27 '19 at 18:47
  • Actually, you haven't asked anything. – the_fox Jun 27 '19 at 18:49
  • @the_fox I'm interested in knowing how to go about solving this. – user5826 Jun 27 '19 at 18:50
  • Let me give a concrete example (it's not a proof). Say you have $G$ with $|G|=p^5$ and that the claim holds for $p$-groups of order $p^4$. Take $N$ to be a subgroup of the centre of $G$ of order $p$. Look at $G/N$. It has normal subgroups of order $1, p, p^2, p^3, p^4$. – the_fox Jun 27 '19 at 18:53
  • If the above makes sense to you, you can easily turn it into a proper proof. If there's a sticky point somewhere, let me know. – the_fox Jun 27 '19 at 18:57
  • @the_fox I think I got it. Would you mind if I post the solution? – user5826 Jun 27 '19 at 19:00
  • Not at all. Go ahead. – the_fox Jun 27 '19 at 19:02

1 Answers1

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If $|G|=p^0=1$, then vacuous. If $|G|=p$, then $\{1\}$ and $G$ are normal subgroups of order $p^0$ and $p^1$.

Suppose the statement is true for $p$-groups of order $p^k$ where $k < r$. Let $|G|=p^r$.

Since $G$ is a $p$-group, it has a non-trivial center, $Z(G)$. So, $Z(G)$ is a $p$-group. Since the center is abelian, then, by Cauchy's theorem, there exists an element of order $p$ and thus a subgroup of order $p$, say $N$. Since $N \subset Z(G)$, then $N$ is normal in $G$.

Consider $G/N$. Then $|G/N|=p^{r-1}$. By the induction hypothesis, there exists normal subgroups of order $p^i$ for $0\le i \le r-1$. By the correspondence theorem, these normal subgroups have the form $H_0/N, H_1/N, \dots, H_{r-1}/N$, where $H_i$ is a normal subgroup of $G$ containing $N$, and where $|H_i/N|=p^i$.

So, $|H_i|/|N|=p^i$ and thus $|H_i|=p^i|N|=p^ip=p^{i+1}$. So, there exists a normal subgroup of order $$|\{1\}|=p^0, |H_0|=p^1, |H_1|=p^2, \dots, |H_{r-1}|=p^r.$$

user5826
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  • You don't really need the third iso. theorem. Immediately, $|H_i|=p^i|N|$. – the_fox Jun 27 '19 at 19:40
  • Ahh, you're right. Edited. Thanks. – user5826 Jun 27 '19 at 19:44
  • Good! Now you don't need to look elsewhere :) – the_fox Jun 27 '19 at 19:54
  • Why do you mention the fact that the center is abelian in the third paragraph? Cauchy's theorem applies regardless whether the group is abelian or not? – Jonathan Huang Oct 20 '24 at 03:46
  • @JonathanHuang You're right. But I guess user5826 wanted to use a weaker version for the argument. In fact we do not need Cauchy's theorem for this argument. We could have noted that $Z(G)$ is normal in $G$ and that $G/Z(G)$ is a $p$-group with $|G/Z(G)|<|G|$. So we could have let $N=Z(G)$ instead, while keeping in mind that any subgroup normal in $Z(G)$ is a normal $p$-subgroup of $G$. – 19021605 May 26 '25 at 08:43