# Hereditarily characteristic not implies cyclic in finite

From Groupprops

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., hereditarily characteristic subgroup) neednotsatisfy the second subgroup property (i.e., cyclic characteristic subgroup)

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## Statement

It is possible to have a finite group , and a subgroup of such that:

- is a Hereditarily characteristic subgroup (?) of : For every subgroup of , is a characteristic subgroup of .
- is
*not*a cyclic group.

## Related facts

- Finite group implies cyclic iff every subgroup is characteristic: This just states that we cannot get a counterexample when .
- Upward-closed characteristic not implies cyclic-quotient in finite

## Proof

`Further information: nontrivial semidirect product of Z4 and Z4, subgroup structure of nontrivial semidirect product of Z4 and Z4`

Consider the semidirect product of cyclic group:Z4 and cyclic group:Z4 where the generator of the acting group acts via the inverse map on the other group. In other words, it has the presentation:

.

Let be the subgroup . Then, is a Klein four-group but every subgroup of is characteristic in . We explain here why each of the subgroups of order two is characteristic:

- The subgroup is the derived subgroup.
- The subgroup is the subgroup generated by the unique element that is a square but not a commutator.
- The subgroup is the subgroup generated by the unique element that is a product of squares but not a square.

itself is and is characteristic in , and the trivial subgroup is characteristic in . Thus, all subgroups of are characteristic in .