# 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.
View all subgroup property implications in nilpotent groups $|$ View all subgroup property non-implications in nilpotent groups $|$ View all subgroup property implications $|$ View all subgroup property non-implications

## Statement

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

## Proof

### Proof of normality

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