Each minimal normal subgroup of is contained in the center Let $G$ be finite nilpotent. I want to show that each minimal normal subgroup of $G$ is contained in $Z(G)$ and has order $p$. 
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We have that $Z(G)=\{g\in G\mid ga=ag , \ \forall a\in G\}$. 
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$G$ 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? 
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Could you give me some hints how we could show that each minimal normal subgroup of $G$ is contained in $Z(G)$ ? 

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