# 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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## Contents

## Statement

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

## 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

- Extending the action of quotient group on abelian normal subgroup to bigger abelian group gives rise to canonical bigger group
- Full invariance is transitive

## Proof

**Given**: A finite group , a homocyclic normal subgroup of .

**To prove**: There exists a finite group containing such that is a fully invariant subgroup of .

**Proof**: Let be the exponent of . The order of is thus for some ( is the number of copies of the cyclic group). Let be the order of . Let be the unique largest divisor of that is relatively prime to . Let be chosen such that divides .

Let be the homocyclic group of order , with exponent . In other words, is the direct product of copies of the cyclic group of order . sits inside as the set of elements whose order divides . The induced action by conjugation of on extends to an action of on . By fact (1), we get a group containing and with , , and the action of on equal to the chosen extension of the action on . Note that the order of is which has no prime factors other than those of .

Let be the subgroup of generated by all elements of the form . Then, is a verbal subgroup, hence a fully invariant subgroup. , because the order of divides which divides by assumption. Also, , because is the set . So, .

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