Prove that: A group $G$ of order $p^n$ has normal subgroup of order $p^k$, for all $0\le k\le n$. Prove that: A group $G$ of order $p^n$ has normal subgroup of order $p^k$, for all $0\le k\le n$.
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 A: In fact we can show that $G$ has a normal series with quotients $\mathbb{Z}/p$. By induction on $n$. 
The step $n\to n+1$ is the only nontrivial one. Let $G$ be a group of order $p^{n+1}$.
It's enough to find a normal subgroup of $G$ of order $p$. 
It is a fact ( see the proof in the appendix below) that $Z(G)$ --the center of $G$, is nontrivial $|Z(G)| = p^k$ with $1\le k \le n+1$. Take any nontrivial  element $z$ of $Z(G)$, $\text{ord}z = p^l$. Then $z' \colon = z^{p^{l-1}}$ has order $p$. The subgroup $H$ generated by $z'$ has order $p$ and being inside $Z(G)$ is necessarily normal. Consider the group $G/H$ of order $p^n$. By induction, there exists an increasing  sequence $(\bar G_i)_{0\le i \le n}$ of normal subgroups of $G/H$  of orders $|\bar G_i| = p^i$. Their preimages $G_i$ in $G$ form an increasing sequence of normal subgroups of $G$ of orders $|G_i| = p^{i+1}$. Add to this $G_{-1} = (e)$ 
Proof that the center is nontrivial: Consider a finite $p$ group acting on a finite set $X$. Then cardinality of the set of fixed points of the action satisfies
$$|X_0| \equiv |X|\  ( \! \! \mod p\,)$$
This because the nontrivial orbits have cardinality $\equiv 0\  ( \! \! \mod p\,)$
Now consider the action of $G$ a finite $p$-group on $X=G$ by conjugation. The set of fixed points is $Z(G)$ and $|Z(G)|\equiv |G| (\mod p)$. But $|Z(G)| \ge 1$ since $e \in Z(G)$. 
