Comparable with all normal subgroups implies normal 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., Subgroup comparable with all normal subgroups (?)) must also satisfy the second subgroup property (i.e., Normal subgroup (?)). In other words, every subgroup comparable with all normal subgroups of nilpotent group is a normal subgroup of nilpotent group.
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Statement

If a subgroup of a nilpotent group is comparable with all normal subgroups, then it is a normal subgroup.

Related facts

• Subgroup comparable with all normal subgroups implies characteristic in finite nilpotent

Proof

Proof of normality

Given: A nilpotent group $G$, a subgroup $N$ of $G$ such that for every normal subgroup $H$ of $G$, either $N\leq H$ or $H\leq N$.

To prove: $N$ is a normal subgroup of $G$

Proof: Consider a central series of $G$:

$\{e\}=H_{0}\leq H_{1}\leq H_{2}\leq \dots \leq H_{n}=G$

Then, each $H_{i}$ is normal in $G$, so we can find some $i$ such that $H_{i}\leq N\leq H_{i+1}$. We have $H_{i+1}/H_{i}$ is central in $G/H_{i}$, so $N/H_{i}$ is central in $G/H_{i}$. In particular, $N/H_{i}$ is normal in $G/H_{i}$. Since $H_{i}$ is normal in $G$, and normality is quotient-transitive, we obtain that $N$ is normal in $G$.