# Characteristic not implies sub-isomorph-free in finite group

From Groupprops

(Redirected from Characteristic not implies sub-isomorph-free 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., characteristic subgroup) neednotsatisfy the second subgroup property (i.e., sub-isomorph-free subgroup)

View all subgroup property non-implications | View all subgroup property implications

## Contents

## Statement

We can have a finite group and a characteristic subgroup of such that is not a sub-isomorph-free subgroup of . In other words, there is no ascending chain of subgroups starting at and ending at with each member an isomorph-free subgroup of its successor.

## Related facts

- Characteristic not implies isomorph-free in finite
- Characteristic not implies sub-(isomorph-normal characteristic) in finite

## Proof

### The center of a non-abelian group of odd prime cube order

`Further information: Prime-cube order group:U3p, Prime-cube order group:p2byp`

Let be an odd prime. Let be the non-abelian group of order and exponent . Let be the center of . Then, we have:

- is characteristic in .
- is not sub-isomorph-free in : In fact, no proper subgroup of containing is isomorph-free. The subgroups of order are all elementary abelian, while itself is isomorphic to many other cyclic groups of prime order.

Another example is to take as the non-abelian group of order and exponent , and to be the center of . In this case:

- is characteristic in <mah>G</math>.
- is not sub-isomorph-free in : is not isomorph-free, because there are other cyclic groups of order . There is only one isomorph-free proper subgroup of properly containing , namely, an elementary abelian subgroup of order . But is not isomorph-free in this subgroup.