Cyclic normal implies finite-pi-potentially verbal in finite

Statement
Suppose $$G$$ is a finite group and $$H$$ is a cyclic normal subgroup of $$G$$. Then, there exists a finite group $$K$$ containing $$G$$ such that every prime factor of the order of $$K$$ and also divides the order of $$G$$, and $$H$$ is a verbal subgroup of $$K$$.

Weaker facts

 * Abelian hereditarily normal implies finite-pi-potentially verbal in finite
 * Cyclic normal implies finite-pi-potentially characteristic in finite
 * Cyclic normal implies potentially verbal in finite

Other related facts

 * Homocyclic normal implies finite-pi-potentially fully invariant in finite
 * Homocyclic normal implies potentially fully invariant in finite
 * Central implies finite-pi-potentially verbal in finite

Facts used

 * 1) uses::Extending the action of quotient group on abelian normal subgroup to bigger abelian group gives rise to canonical bigger group
 * 2) uses::Verbality is transitive

Proof
Given: A finite group $$G$$, a cyclic normal subgroup $$H$$ of $$G$$.

To prove: There exists a finite group $$K$$ containing $$G$$ such that every prime factor of the order of $$K$$ divides the order of $$G$$, and $$H$$ is a verbal subgroup of $$K$$.

Proof: Let $$m$$ be the order of $$H$$ and $$n$$ be the order of $$G$$. let $$h$$ be the unique largest divisor of $$n$$ that is relatively prime to $$m$$. Further, let $$k$$ be a positive integer such that $$n/h$$ divides $$m^{k-1}$$.

Let $$M$$ be the cyclic group of order $$m^k$$, containing $$H$$ as the unique cyclic subgroup of order $$m$$. The action of the quotient group $$G/H$$ on $$H$$ extends to an action on $$M$$. Thus, by fact (1), we can construct a group $$K$$ containing $$G$$ and $$M$$, with $$G \cap M = H$$, $$GM = K$$, and the action of $$K/M \cong G/H$$ on $$M$$ the extension of the action we chose.

Now, consider the subgroup $$V$$ of $$K$$ generated by elements of the form $$x^{hm^{k-1}}$$. First, note that this subgroup is contained in $$M$$, because the order of $$G/H$$ divides the order of $$G$$ which in turn divides $$hm^{k-1}$$ by assumption. Thus, the verbal subgroup $$V$$ of $$K$$ is contained in $$M$$. Further, $$V$$ contains $$H$$ because the set of elements of the form $$x^{hm^{k-1}} : x \in M$$ itself equals $$H$$. Thus, $$H \le V \le M$$, with $$V$$ a verbal subgroup of $$K$$.

Since $$V$$ is a subgroup of $$M$$, it is cyclic. Every subgroup of $$V$$ is verbal in $$V$$, so $$H$$ is verbal in $$V$$. By fact (2), $$H$$ is verbal in $$K$$, completing the proof.