Homocyclic normal implies finite-pi-potentially fully invariant in finite

Statement
Let $$G$$ be a finite group and $$H$$ be a homocyclic normal subgroup of $$G$$. In other words, $$H$$ is a normal subgroup of $$G$$ and is also a homocyclic group: it is a direct product of pairwise isomorphic cyclic groups. Then, there exists a finite group $$K$$ containing $$G$$ such that all the prime factors of the order of $$K$$ also divide the order of $$G$$, and $$H$$ is a fully invariant subgroup of $$K$$.

Related facts

 * Cyclic normal implies finite-pi-potentially verbal in finite, cyclic normal implies potentially verbal in finite
 * Central implies finite-pi-potentially verbal in finite, central implies potentially verbal in finite
 * Normal not implies finite-pi-potentially characteristic

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::Full invariance is transitive

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

To prove: There exists a finite group $$K$$ containing $$G$$ such that $$H$$ is a fully invariant subgroup of $$K$$.

Proof: Let $$m$$ be the exponent of $$H$$. The order of $$H$$ is thus $$m^d$$ for some $$d$$ ($$d$$ is the number of copies of the cyclic group). Let $$n$$ be the order of $$G$$. Let $$h$$ be the unique largest divisor of $$n$$ that is relatively prime to $$m$$. Let $$k$$ be chosen such that $$n/h$$ divides $$m^{k-1}$$.

Let $$M$$ be the homocyclic group of order $$m^{dk}$$, with exponent $$m^k$$. In other words, $$M$$ is the direct product of $$d$$ copies of the cyclic group of order $$m^k$$. $$H$$ sits inside $$M$$ as the set of elements whose order divides $$m$$. The induced action by conjugation of $$G/H$$ on $$H$$ extends to an action of $$G/H$$ on $$M$$. By fact (1), we get a group $$K$$ containing $$G$$ and $$M$$ with $$G \cap M = H$$, $$GM = K$$, and the action of $$K/M \cong G/H$$ on $$M$$ equal to the chosen extension of the action on $$H$$. Note that the order of $$K$$ is $$m^{dk}(n/m)$$ which has no prime factors other than those of $$G$$.

Let $$V$$ be the subgroup of $$K$$ generated by all elements of the form $$x^{hm^{k-1}}$$. Then, $$V$$ is a verbal subgroup, hence a fully invariant subgroup. $$V \le M$$, because the order of $$K/M \cong G/H$$ divides $$n$$ which divides $$hm^{k-1}$$ by assumption. Also, $$H \le V$$, because $$H$$ is the set $$\{ x^{hm^{k-1}} \mid x \in K \}$$. So, $$H \le V \le M$$.

Further, $$H$$ is a fully invariant subgroup of $$V$$, since it is the set of elements whose order divides $$m$$. Thus, by fact (2), $$H$$ is fully invariant in $$K$$.