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.

Related facts

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.