Q: Prove that a subgroup of order $p^n$ has a normal subgroup of order $p^{k}$ for all $0\leq k \leq n$.
Attempt at a proof: We proceed by Induction. This is obviously true for $n=1, 2$. Suppose it is true for $m \leq n-1$. Now, take the group $G$ of order $p^{n}$. Since it is a $p$-group, it has a non-trivial center by the Conjugacy Class equation. This center then has a subgroup of $K$ order $p$ by Cauchy's theorem for example. Since this subgroup is contained in the center, it is normal in $G$.
Take the quotient $G/K$ which now has order $p^{n-1}$ hence by the induction hypothesis has subgroups $\bar{H_{k}}$ of order $p^{k}$ for $0\leq k \leq n-1$, which are normal in $G/K$. By the lattice isomorphism theorem, the lattice of $G$ has corresponding groups $H_{k}$ which are normal in $G$.
The index of each such subgroup is $|G:H_{k}|=|G/K: \bar{H_k}|=p^{n-1-k}$, hence $H_{k}$ has order $p^{k+1}$ in $G$ as desired.
Does this look okay?