# Homocyclic normal implies finite-pi-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-pi-potentially fully invariant subgroup (?)). In other words, every homocyclic normal subgroup of finite group is a finite-pi-potentially fully invariant subgroup of finite group.
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## 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 $K$.

## 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 $K$.