Minimal normal implies central in nilpotent group

This article gives the statement and possibly, proof, of an implication relation between two subgroup properties, when the big group is a nilpotent group. That is, it states that in a Nilpotent group (?), every subgroup satisfying the first subgroup property (i.e., Minimal normal subgroup (?)) must also satisfy the second subgroup property (i.e., Central subgroup (?)). In other words, every minimal normal subgroup of nilpotent group is a central subgroup of nilpotent group.
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Statement

In a Nilpotent group (?), any Minimal normal subgroup (?) is contained in the center (i.e., is a central subgroup). In fact, a minimal normal subgroup must be a cyclic group of prime order contained in the center.

Facts used

1. Nilpotent implies center is normality-large: In a nilpotent group, any nontrivial normal subgroup intersects the center nontrivially.

Proof

Given: A nilpotent group $G$, a minimal normal subgroup $N$. Let $Z(G)$ denote the center of $G$.

To prove: $N$ is contained in $Z(G)$. Further, $N$ must be cyclic of prime order.

Proof: By fact (1) stated above, we see that since $N$ is nontrivial, so is $N \cap Z(G)$. Also, $N \cap Z(G)$ is normal (being the intersection of two normal subgroups). Since $N$ is a minimal normal subgroup, this forces $N = N \cap Z(G)$, so $N \le Z(G)$.

Since any subgroup of the center is normal, any minimal normal subgroup contained in the center, must literally be a minimal subgroup. The only possibility for a minimal subgroup is a cyclic group of prime order, so $N$ is cyclic of prime order.