Homocyclic normal implies potentially fully invariant in finite

This article gives the statement and possibly, proof, of an implication relation between two subgroup properties, when the big group is a finite group. That is, it states that in a Finite group (?), every subgroup satisfying the first subgroup property (i.e., Homocyclic normal subgroup (?)) must also satisfy the second subgroup property (i.e., Finite-potentially fully invariant subgroup (?)). In other words, every homocyclic normal subgroup of finite group is a finite-potentially fully invariant subgroup of finite group.
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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 finite group. That is, it states that in a Finite group (?), every subgroup satisfying the first subgroup property (i.e., Homocyclic normal subgroup (?)) must also satisfy the second subgroup property (i.e., Potentially fully invariant subgroup (?)). In other words, every homocyclic normal subgroup of finite group is a potentially fully invariant subgroup of finite group.
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View all subgroup property non-implications in finite groups

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Statement

Suppose $G$ is a finite group and $H$ is a homocyclic normal subgroup of $G$: in other words, $H$ is a normal subgroup of $G$ as well as a homocyclic group: it is a direct product of isomorphic cyclic groups. Then, there exists a finite group $K$ containing $G$ such that $H$ is a Fully invariant subgroup (?) of $K$.

Facts used

1. Extending the action of quotient group on abelian normal subgroup to bigger abelian group gives rise to canonical bigger group
2. 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 $K</mathis<math>m^{dk}(n/m)$ which has no prime factors other than those of $G$.

Let $V$ be the subgroup of $K$ generated by all elements of the form $x^{h{m}^{k-1}}$. 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 $h{m}^{k-1}$ by assumption. Also, $H\le V$, because $H$ is the set $\{x^{h{m}^{k-1}}\mid x\in K\}$. 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 $K$.