Isomorph-freeness is not transitive

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.

Example of the dihedral group
Consider the particular example::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 $$x$$.