Supersolvable implies every nontrivial normal subgroup contains a cyclic normal subgroup

This article gives the statement and possibly, proof, of an implication relation between two group properties. That is, it states that every group satisfying the first group property must also satisfy the second group property
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This article gives the statement and possibly, proof, of an implication relation between two subgroup properties, when the big group is a [[{{{group property}}}]]. That is, it states that in a "{{{group property}}}" is not a number., every subgroup satisfying the first subgroup property must also satisfy the second subgroup property . In other words, every is a .
[[:Category:Subgroup property implications in {{{group property}}}s|View all subgroup property implications in {{{group property}}}s]] $|$ [[:Category:Subgroup property non-implications in {{{group property}}}s|View all subgroup property non-implications in {{{group property}}}s]] $|$ View all subgroup property implications $|$ View all subgroup property non-implications

[[Category:Subgroup property implications in {{{group property}}}s]]

Statement

Property-theoretic statement

The group property of being a supersolvable group is stronger than, or implies the property of being a group in which every nontrivial normal subgroup contains a cyclic normal subgroup.

Statement with symbols

Let $G$ be a supersolvable group. Then, if $N$ is a nontrivial normal subgroup of $G$, there exists a (nontrivial) cyclic normal subgroup of $G$ contained in $N$.

Proof

Given: A supersolvable group $G$, with a nontrivial normal subgroup $N$

To prove: There exists a nontrivial cyclic normal subgroup of $G$ contained in $N$

Proof: By assumption, there exists a normal series for $G$:

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

where each $K_{i}/K_{i-1}$ is a cyclic group and each $K_{i}$ is normal in $G$.

Now consider the series:

$\{e\}=K_{0}\cap H\leq K_{1}\cap H\leq \dots \leq K_{n}\cap H=H$

It's clear that each of the $K_{i}\cap H$ is normal in $G$ (since normality is intersection-closed), and moreover, each $(K_{i}\cap H)/(K_{i-1}\cap H)$ is cyclic (since it can be identified with a subgroup of the cyclic group $K_{i}/K_{i-1}$). Pick the first $i$ such that $K_{i}\cap H$ is nontrivial. Such an $i$ exists, because $K_{n}\cap H=H$ is nontrivial. Then, this is a cyclic normal subgroup of $G$ contained in $H$.