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

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 $$k . Induction should start by assuming for every $$p$$-group of order $$p^k$$ where $$0\le k , 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 $$0 \le i \le r$$.

• 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.) Commented Jun 27, 2019 at 18:43
• @the_fox That's exactly what I wrote. Commented Jun 27, 2019 at 18:44
• Do you want me to comment on the other "solutions" or how to give a proof? Commented Jun 27, 2019 at 18:47
• Actually, you haven't asked anything. Commented Jun 27, 2019 at 18:49
• @the_fox I'm interested in knowing how to go about solving this. Commented Jun 27, 2019 at 18:50

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 $$|\{1\}|=p^0, |H_0|=p^1, |H_1|=p^2, \dots, |H_{r-1}|=p^r.$$

• You don't really need the third iso. theorem. Immediately, $|H_i|=p^i|N|$. Commented Jun 27, 2019 at 19:40
• Ahh, you're right. Edited. Thanks. Commented Jun 27, 2019 at 19:44
• Good! Now you don't need to look elsewhere :) Commented Jun 27, 2019 at 19:54