$p$-group and an normal subgroup, need help to start my solution

Question

Let $G$ be a $p$-group with |$G$|=$p^{n}$ with $p$ is prime and $n>1$. Let's say that $N$ is a strict normal subgroup of $G$.

1. Prove there is an $x\in Z(G/N)$ of order $p$, i.e. $x^{p}=e$ and $x \neq e$.
2. Prove that there exist a normal subgroup $M$ with $N\subset M\subset G$ and $[M:N]$=$p$.
3. Prove $G$ is soluble.

I wanted to solve the above but I'm really sorry i can't come further then definitions. I know it seems like there isn't a lot of effort in this post but I have really no clue how to start this. I looked here for some extra information but didn't find a post that could help me. I would appreciate it if you could help me out so i can understand similair questions or understand how stuff with a p-group works.

I'm really sorry i will edit the question with my solutions that i found with your help :)

1. I got this hint :Since N⊊G is a strict normal subgroup, G/N is a nontrivial p-group, i.e. |G/N|=$p^{k}$ for some k∈N. This implies Z(G/N)≠{e} ,because if $G/N$ is a p-group then |Z(G/N)|>1 ,and Z(G/N) is again a p-group. Now it should be easy to find an element x∈Z(G/N) of order p.

My solution: from the above we know that $Z(G/N)$ has a order $p^{k}$ for a $k\in \mathbb{N}$. Let's take a $x\in Z(G/N)$ with order $s\in \mathbb{Z}$. This means that $x^{s}=e$. Now it follows that $p^{k}|s$. It's trivial that if $s=p$, $p^{k}|p$ so there exist an element with order $p$.

Answer 1

I'll first post some hints here, but these are too long for a comment:

1. Since $N \subsetneq G$ is a strict normal subgroup, $G/N$ is a nontrivial $p$-group, i.e. $|G/N| = p^k$ for some $k \in \mathbb{N}$. This implies $Z(G/N) \neq \{e\}$ and $Z(G/N)$ is again a $p$-group. Now it should be easy to find an element $x \in Z(G/N)$ of order $p$.

2. Normal subgroups $M \subseteq G$ with $N \subseteq M$ and index $[M:N] = p$ correspond to normal subgroups $\tilde{M} \subseteq G/N$ with $|\tilde{M}| = p$.

3. Proceed by induction on $n$ and use the fact that for a normal divisor $M$ of $G$, $G$ is soluble if and only if both $M$ and $G/M$ are soluble.

EDIT: Some clarifications:

1. You should of course convince yourself why the given hints are true. The only nontrivial part of hint 1 is that $Z(G/N) \neq \{e\}$.

2. We have a bijection \begin{align*}\{M \subseteq G \text{ normal subgroup with } N \subseteq M\} &\to \{\tilde{M} \subseteq G/N \text{ normal subgroup}\} \\ M &\mapsto M/N. \end{align*}