Cyclic characteristic implies hereditarily 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 characteristic subgroup) must also satisfy the second subgroup property (i.e., hereditarily characteristic subgroup)
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Statement

Suppose $G$ is a group and $K$ is a cyclic characteristic subgroup of $G$, i.e., $K$ is a cyclic group and is also a Characteristic subgroup (?) of $G$. Then, $K$ is also a hereditarily characteristic subgroup of $G$, i.e., every subgroup $H$ of $K$ is characteristic in $G$.

Related facts

• Cyclic normal implies hereditarily normal
• Hereditarily characteristic not implies cyclic in finite
• Cyclic-quotient characteristic implies upward-closed characteristic

Facts used

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

Proof

Given: A group $G$ with a cyclic characteristic subgroup $K$. A subgroup $H$ of $K$.

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

Proof:

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