A $p$-group of order $p^n$ has a normal subgroup of order $p^k$ for each $0\le k \le n$

This is problem 3 from Hungerford's section about the Sylow theorems.

I have already read hints saying to use induction and that $p$-groups always have non-trivial centres, but I'm still confused. This is what I have so far:

Suppose $|G| = p^n$. For $k = 0$, $\{e\}$ is a normal subgroup of order $p^0$. Suppose $N_1, ..., N_k$ are normal subgroups of $G$ with orders as described (ie. $|N_i| = p^i$) and $k < n$.

$G/N_k$ is a $p$-group, so it has a non-trivial centre $C(G/N_k)$. $\pi^{-1}(C(G/N_k))$ is a subgroup of $G$, where $\pi : G \to G/N_k$ is the quotient map. It is normal in $G$ because $\pi$ is a homomorphism and $C(G/N_k)$ is a normal subgroup of $G/N_k$.

I know it contains $N_k$ and some stuff not in $N_k$, so it has to have order at least $p^{k+1}$, but I do not know how to argue it must be equal, or indeed if it even is equal.

• Certainly you mean $k \leq log_p(|G|)$. – Yassine Guerboussa Nov 3 '13 at 8:41
• You are correct, I fixed it. – user104986 Nov 3 '13 at 17:38
• How are we assuming $N_1, \dots, N_k$ are normal subgroups of $G$ with orders $p^1, \dots, p^k$? This is part of what we need to prove. – user46372819 Jun 27 '19 at 19:03

Hint: Apply Cauchy's theorem to $C(G/N_k)$. This will give you a normal subgroup of $G$ of order $p^{k+1}$.