Minimal normal implies central in nilpotent group

Statement
In a fact about::nilpotent group, any fact about::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.

Other facts in nilpotent groups

 * Minimal normal implies contained in Omega-1 of center for nilpotent p-group, socle equals Omega-1 of center in nilpotent p-group
 * Minimal characteristic implies central in nilpotent
 * Minimal characteristic implies contained in Omega-1 of center for nilpotent p-group
 * Formula for number of minimal normal subgroups of group of prime power order
 * Congruence condition relating number of normal subgroups containing minimal normal subgroups and number of normal subgroups in the whole group

Facts in other kinds of groups

 * Normal of order equal to least prime divisor of group order implies central
 * Minimal normal implies characteristically simple
 * Minimal normal implies elementary Abelian in finite solvable

Generalizations

 * Normal of prime power order implies contained in upper central series member corresponding to prime-base logarithm of order in nilpotent

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.