Comparable with all normal subgroups implies normal in nilpotent group

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

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

Proof: Consider a central series of $$G$$:

$$\{ e \} = H_0 \le H_1 \le H_2 \le \dots \le H_n = G$$

Then, each $$H_i$$ is normal in $$G$$, so we can find some $$i$$ such that $$H_i \le N \le H_{i+1}$$. 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 $$G$$.