Intersection of two isomorph-free subgroups need not be intermediately characteristic

This article gives the statement, and possibly proof, of a subgroup property (i.e., isomorph-free subgroup) not satisfying a subgroup metaproperty (i.e., intersection-closed 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 intersection-closed subgroup property|

This article gives the statement, and possibly proof, of a subgroup property (i.e., intermediately characteristic subgroup) not satisfying a subgroup metaproperty (i.e., intersection-closed 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 intermediately characteristic subgroup|Get more facts about intersection-closed subgroup property|

Statement

It is possible to have subgroups $H, K \leq G$ such that both $H$ and $K$ are isomorph-free (in other words, there is no other subgroup of $G$ isomorphic to $H$ and no other subgroup isomorphic to $K$ ) but $H \cap K$ is not isomorph-free, and in fact, not even an intermediately characteristic subgroup, i.e, there exists a subgroup $L$ of $G$ containing $H \cap K$ such that $H \cap K$ is not characteristic in $L$ .

In particular, this implies that an intersection of two isomorph-free subgroups need not be isomorph-free. Moreover, since isomorph-free subgroups are intermediately characteristic, it also implies that an intersection of intermediately characteristic subgroups need not be intermediately characteristic.

Proof

Example of a dihedral group of order 24

Let $G$ be the dihedral group of order $24$ : in other words, $G$ is the semidirect product of the cyclic group $C$ of order $12$ and the cyclic group of order two, acting via the inverse map. Let $H$ be the subgroup generated by multiples of $2$ in $C$ , and $K$ be the subgroup generated by multiples of $3$ .

Both $H$ and $K$ are isomorph-free subgroups of $G$ : $H$ is isomorphic to the cyclic group of order six and $K$ is isomorphic to the cyclic group of order four. To see that they are isomorph-free, observe that within the cyclic group of order twelve, there are no elements of order six (respectively four) other than those in $H$ (respectively $K$ ) while all elements outside $C$ have order two.

On the other hand, the intersection $H \cap K$ , which comprises the multiples of six, is a cyclic group of order two, and is isomorphic to the cyclic groups of order two generated by any element outside $C$ . Hence, $H \cap K$ is not isomorph-free.

In fact, if we take $L$ as the Klein four-group generated by $H \cap K$ and one of the elements outside $C$ , then $H \cap K$ is not a characteristic subgroup of $L$ .

Example of a semidihedral group of order 16

Further information: subgroup structure of semidihedral group:SD16

Let $G$ be semidihedral group:SD16, $H$ be the subgroup Z8 in SD16 and $K$ be the subgroup Q8 in SD16. Both $H$ and $K$ are characteristic maximal subgroups of $G$ , hence intermediately characteristic. The intersection $H \cap K$ is the derived subgroup of semidihedral group:SD16. This is not a characteristic subgroup inside $K$ (it is one of the three cyclic maximal subgroups of quaternion group).