$G$ is a $p$-group $\implies $ $G$ has a normal subgroup of $p^k \ \forall k$

Question

I am a beginner to $p$-groups, so please help me with the following qustion:

If $|G|=p^n$, where $p$ is prime, and $0\le k\le n$, then $G$ contains a normal subgroup of order $p^k$.

My work:

I have just been able to prove that

$G$ has a normal subgroup of order $p$.

Since $G$ is a $p$ group, its center $Z(G)\ne \{e\}$ $\therefore$ $Z(G)$ is $p$-subgroup of $G$ .By Cauchy's lemma, there is an element $a\in Z(G)$ such that $|a|=p$.Then $N=\langle a\rangle$ is a subgroup of order $p$.Moreover since every subgroup of $Z(G)$ is normal in $G$, $\therefore$ $N\triangleleft G$.

Now I dont know how to proceed. Maybe induction will help? Or is there some better way?

Answer 1

Hint If you know that $G$ has a normal subgroup $N$ of order $p^k$, then $G/N$ is a $p$ group, and hence has a normal subgroup of order $p$.

Deduce that $G$ has a normal subgroup of order $p^{k+1}$.

Answer 2

Recall the fourth isomorphism theorem: the subgroups of a quotient correspond to subgroups of the big group and moreover the image of a subgroup containing the kernel is normal in the quotient iff the subgroup itself was normal in the original group. Now if you have a normal subgroup of order $p^k$, $N$, consider $G/N$ having a normal subgroup of order $p$ (which you know exists by what you've already done). Then you may lift that subgroup to a normal subgroup of $G$ of order $p^{k+1}$ in $G$. This completes the induction.