# Cyclic normal implies finite-pi-potentially verbal 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., Cyclic normal subgroup (?)) must also satisfy the second subgroup property (i.e., Finite-pi-potentially verbal subgroup (?)). In other words, every cyclic normal subgroup of finite group is a finite-pi-potentially verbal subgroup of finite group.

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## Statement

Suppose is a finite group and is a cyclic normal subgroup of . Then, there exists a finite group containing such that every prime factor of the order of and also divides the order of , and is a verbal subgroup of .

## Related facts

### Weaker facts

- Abelian hereditarily normal implies finite-pi-potentially verbal in finite
- Cyclic normal implies finite-pi-potentially characteristic in finite
- Cyclic normal implies potentially verbal in finite

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

## Facts used

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

## Proof

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

**To prove**: There exists a finite group containing such that every prime factor of the order of divides the order of , and is a verbal subgroup of .

**Proof**: Let be the order of and be the order of . let be the unique largest divisor of that is relatively prime to . Further, let be a positive integer such that divides .

Let be the cyclic group of order , containing as the unique cyclic subgroup of order . The action of the quotient group on extends to an action on . Thus, by fact (1), we can construct a group containing and , with , , and the action of on the extension of the action we chose.

Now, consider the subgroup of generated by elements of the form . First, note that this subgroup is contained in , because the order of divides the order of which in turn divides by assumption. Thus, the verbal subgroup of is contained in . Further, contains because the set of elements of the form itself equals . Thus, , with a verbal subgroup of .

Since is a subgroup of , it is cyclic. Every subgroup of is verbal in , so is verbal in . By fact (2), is verbal in , completing the proof.