# Isomorph-freeness is not transitive

This article gives the statement, and possibly proof, of a subgroup property (i.e., isomorph-free subgroup) not satisfying a subgroup metaproperty (i.e., transitive subgroup property).
View all subgroup metaproperty dissatisfactions | View all subgroup metaproperty satisfactions|Get help on looking up metaproperty (dis)satisfactions for subgroup properties
Get more facts about isomorph-free subgroup|Get more facts about transitive subgroup property|
This article gives the statement, and possibly proof, of a subgroup property (i.e., isomorph-containing subgroup) not satisfying a subgroup metaproperty (i.e., transitive subgroup property).
View all subgroup metaproperty dissatisfactions | View all subgroup metaproperty satisfactions|Get help on looking up metaproperty (dis)satisfactions for subgroup properties
Get more facts about isomorph-containing subgroup|Get more facts about transitive subgroup property|

## Statement

### Verbal statement

An isomorph-free subgroup of an isomorph-free subgroup need not be isomorph-free.

This also shows that an isomorph-containing subgroup of an isomorph-containing subgroup need not be isomorph-containing. Note that inside a finite group, the notions of isomorph-free and isomorph-containing are equivalent.

## Proof

### Example of the dihedral group

Consider the dihedral group:D8, the dihedral group acting on a set of size four, i.e., the dihedral group with eight elements, given explicitly by the presentation: $\langle a,x \mid a^4 = x^2 = e, xax^{-1} = a^{-1} \rangle$.

The cyclic subgroup generated by $a$ is a subgroup of order four, and since all the elements outside it have order two, it is an isomorph-free subgroup. Within this, the cyclic subgroup of order two generated by $a^2$ is an isomorph-free subgroup.

However, the cyclic subgroup of order two generated by $a^2$ is not isomorph-free in the whole group: it is isomorphic to the two-element subgroup generated by $x$.