Subgroups of $p$ -groups

It is a fact that a group of order $p^n$ has normal subgroups of order $p^k$ for any $k \le n.$ This can be proved by induction.

The base case $n=1$ is clear. Now suppose the result has been established for all group of order $<p^n,$ and let $G$ be a group of order $p^n.$ There are two key facts. The first is that the center $Z(G)$ is nontrivial (see the p-groups wiki for details). Then the quotient group $G/Z(G)$ is a smaller $p$ -group, so has normal subgroups of every possible size by the inductive hypothesis. The second key fact is that the normal subgroups of $G/Z(G)$ are in one-to-one correspondence with normal subgroups of $G$ containing $Z(G),$ by the third isomorphism theorem . So if $|Z(G)|=p^k,$ this shows that there are normal subgroups of order $p^k,p^{k+1},\ldots,p^n$ containing $Z(G).$

It remains to be shown that the center contains normal subgroups of every possible size. Note that every subgroup of the center is normal in $G,$ so it is only necessary to find subgroups of every possible size. And so the result follows from the structure theorem for abelian groups , and induction: any abelian group of order $p^n$ is isomorphic to a product of groups of the form ${\mathbb Z}_{p^i}$ , and a subgroup of order $p^{n-1}$ can be obtained simply by changing one of the ${\mathbb Z}_{p^i}$ to ${\mathbb Z}_{p^{i-1}}.$ That subgroup will have subgroups of every possible order by the inductive hypothesis.

(Note that the second paragraph of the proof is necessary: if $G$ is abelian, the center equals $G,$ so you can't use the original inductive hypothesis to get subgroups of it; you need to know the theorem is true for abelian groups.)

So the statements in the problem are all always true, and the answer is $\fbox{4}.$