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.
View all subgroup property implications in finite groups $|$ View all subgroup property non-implications in finite groups $|$ View all subgroup property implications $|$ View all subgroup property non-implications

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

Proof

The result follows directly from fact (1), which is a stronger version.