# Upward-closed characteristic not implies cyclic-quotient in finite

This article gives the statement and possibly, proof, of a non-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., upward-closed characteristic subgroup) neednotsatisfy the second subgroup property (i.e., cyclic-quotient subgroup)

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## Contents

## Statement

It is possible to have a finite group and a subgroup of such that the following two conditions hold:

- is an upward-closed characteristic subgroup of : Every subgroup of containing is a Characteristic subgroup (?) of .
- is not a cyclic-quotient subgroup of , i.e., the quotient group is
*not*a cyclic group.

## Related facts

### Converse

## Facts used

## Proof

### Example of a group of order 16

`Further information: semidihedral group:SD16, subgroup structure of semidihedral group:SD16`

Consider the semidihedral group of order , given by the presentation:

.

The derived subgroup of euqals the Frattini subgroup of and it is the subgroup . The quotient by this subgroup is a Klein four-group, and is hence not cyclic. However, all subgroups of containing are characteristic -- in fact, all the subgroups of order eight containing are isomorph-free subgroups: the cyclic group of order eight , the dihedral group of order eight , and the quaternion group .

### Generalization

The Bryant-Kovacs theorem allows us to construct, for any prime number and any , a group such that (with the Frattini subgroup of ) is elementary abelian of order , and the image of the map is trivial. This in particular would imply that is upward-closed characteristic, but is not cyclic.