I have the following exercise:
Problem. Group $G$ has order $p^n$, where $p$ is a prime number and $n$ is a positive integer. Prove that for any proper subgroup $H<G$ there is a subgroup $K\leq G$ such that $H\lhd K$ and $[K:H]=p$.
What I have done is consider center $Z(G)$ of $G$ and note that $Z(G)$ is nontrivial subgroup of $G$. For example, the statement is clear when $H\cap Z(G)=\{e\}$ or when $H\leq Z(G)$. However, it's not clear how to approach the problem in general case.
I think we should somehow induct on $n$ (maybe we can consider $G/Z(G)$), because the similar problem can be solved in this way:
Problem. Group $G$ has order $p^n$, where $p$ is a prime number and $n$ is a positive integer. Prove that for any positive integer $m\leq n$ there is a subgroup $H\leq G$ such that $|H|=p^m$.
So, does this approach lead to solution? Any advice would be appreciated.
Induct on $n$.
Take the quotient by ta subgroup of the center of size $p$ to obtain $G'$ and $H'$ and notice that there exists a subgroup $K'$ containing $H'$ of required index. And then reobtain $H$ by taking the union of the corresponding cosets. The subgroup is normal because of the lattice isomorphism theorem.
The base case when $n=1$ is clear.
In fact this is just an application of the lattic isomorphism theorem, you want to prove that for each dude there is a dude immediately above him, and this is clear because in the quotient lattice we have this.