Each minimal normal subgroup of is contained in the center

Question

Let $G$ be finite nilpotent. I want to show that each minimal normal subgroup of $G$ is contained in $Z(G)$ and has order $p$.

$$$$

We have that $Z(G)=\{g\in G\mid ga=ag , \ \forall a\in G\}$.

$$$$

$G$ is nilpotent iff $\exists$ a series of normal subgrousp $N_i$: $$1\leq N_1\leq N_2 \leq \dots \leq N_k=G$$ such that $N_{i+1}/N_i\subseteq Z(G/N_i)$, right?

$$$$

Could you give me some hints how we could show that each minimal normal subgroup of $G$ is contained in $Z(G)$ ?

Answer 1

Since $\;G\;$ is nilpotent, any non-trivial normal subgroup intersects non-trivially the center, and then for minimal non-trivial normal $\;N\lhd G\;$ we get

$$1<|N\cap Z(G)|\stackrel{\text{minim.}}\implies N\cap Z(G)=N\iff N\le Z(G)$$

Answer 2

Suppose that $N$ is a minimal normal subgroup of $G$.

1.) Since $N$ is nontrivial and normal, $N ∩ Z(G)$ is nontrivial by nilpotency of $G$. Since $N ∩ Z(G)$ is a normal subgroup of $G$, by minimality of $N$ we have $N = N ∩ Z(G)$, so $N\subseteq Z(G)$.

2.) Since $N$ is in the center of $G$, every subgroup of $N$ is a normal subgroup of $G$, so $N$ contains no proper nontrivial subgroups. It follows that $N$ is abelian and simple, so it has prime order $p$.