Cyclic-quotient characteristic implies upward-closed characteristic

This article gives the statement and possibly, proof, of an implication relation between two subgroup properties. That is, it states that every subgroup satisfying the first subgroup property (i.e., cyclic-quotient characteristic subgroup) must also satisfy the second subgroup property (i.e., upward-closed characteristic subgroup)
View all subgroup property implications | View all subgroup property non-implications
Get more facts about cyclic-quotient characteristic subgroup|Get more facts about upward-closed characteristic subgroup

Statement

Suppose $G$ is a group and $H$ is a cyclic-quotient characteristic subgroup of $G$. In other words, $H$ is a Characteristic subgroup (?) of $G$ that is also a Cyclic-quotient subgroup (?) of $G$, i.e., the quotient group $G/H$ is a cyclic group.

Then, $H$ is an upward-closed characteristic subgroup of $G$: every subgroup $K$ of $G$ containing $H$ is a characteristic subgroup of $G$.

Related facts

• Cyclic characteristic implies hereditarily characteristic
• Equivalence of definitions of upward-closed normal subgroup

Facts used

1. Cyclic implies every subgroup is characteristic
2. Characteristicity is quotient-transitive: If $A\le B\le C$ with $A$ characteristic in $C$ and $B/A$ characteristic in $C/A$, then $B$ is characteristic in $C$.

Proof

Given: A group $G$, a characteristic subgroup $H$ of $G$ such that $G/H$ is cyclic, a subgroup $K$ of $G$ containing $H$.

To prove: $K$ is characteristic in $G$.

Proof:

1. $K/H$ is characteristic in $G/H$: This follows from fact (1), and the given datum that $G/H$ is cyclic.
2. $K$ is characteristic in $G$: This follows from the previous step, the given datum that $H$ is characteristic in $G$, and fact (2).