$p$-groups have normal subgroups of each order [duplicate]

Suppose $|G|=p^n$. Then $G$ has a normal subgroup of order $p^m$ for every $0\le m\le n$.

By induction. It is clearly true for $n=0$. Now suppose $k<n$ and $H_i$ is a normal subgroup of $G$ of order $p^i$ for $i=0,...,k$.

I read in another thread that if we consider $Z(G/H_k)\neq 1$ and the Cauchy's theorem, this will provide us a group of order $p^{k+1}$. I don't see why yet.

By that theorem there exists $gH_k\in Z(G/H_k)$ with order $p$. How can we construct a group of order $p^{k+1}$ from here?

Thank you.

Lemma 1: If $G$ is a group, every subgroup of $Z(G)$ is normal in $G$.

Proof: If $H\subset Z(G)$ is a subgroup of the center, then $gHg^{-1}=H$ for every $g\in G$ since every $g$ commutes with every element of $H$.

Lemma 2: If $N$ is a normal subgroup of $G$ and $H\subset G/N$ is a normal subgroup of $G/N$, then the preimage in $G$ of $H$ is a normal subgroup of $G$.

Proof: this is just the correspondence theorem.

Okay, now to your situation. You have a subgroup $H_k$ in $G$ of order $p^k$. You have found yourself an element $gH_k\in G/H_k$ of order $p$ that is contained in the center of $G/H_k$. This means that the order-$p$ subgroup it generates in $G/H_k$ is contained in the center of $G/H_k$. By the first lemma, this subgroup is normal in $G/H_k$. By the second lemma, its preimage in $G$ is normal in $G$. And its order is $p$ times the order of $H_k$, which is $p^{k+1}$.