Minimal normal subgroup is contained in Centre of group for nilpotent groups

Question

This question is from Hungerford Algebra Chapter Structure of groups.

If $G$ is a finite nilpotent group, then every minimal normal subgroup of $G$ is contained in $C(G)$ and has prime order.

A minimal normal subgroup of a group $G$ is a nontrivial normal subgroup that contains no proper subgroup which is normal in $G$, where $C(G )$ is normal subgroup of $G$.

$G$ is nilpotent means there exists $n$ such that $C_n (G) =\langle e\rangle$, $C_n(G)$ is the inverse image of $C(G/ C_{n-1}(G))$.

Let $H$ be a minimal normal subgroup. I am at loss of ideas and not even able to start and would appreciate hints for it.

Also, I am unable to form some reasons on why minimal subgroup must have prime order?

Can you please help with that?

Any help would be much appreciated.

Answer 1

1. Any normal subgroup in the nilpotent group intersect the center non-trivially
2. the intersection of two subgroup is still a subgroup. In particular, the intersection of a subgroup with the center is still normal.