Hereditarily characteristic not implies cyclic 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., hereditarily characteristic subgroup) need not satisfy the second subgroup property (i.e., cyclic characteristic subgroup)
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Statement

It is possible to have a finite group $G$, and a subgroup $K$ of $H$ such that:

1. $K$ is a Hereditarily characteristic subgroup (?) of $G$: For every subgroup $H$ of $K$, $H$ is a characteristic subgroup of $G$.
2. $K$ 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 $K=G$.
• 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:

$G:=⟨x,y\mid x^4=y^4=e,yx{y}^{-1}=x^3⟩$.

Let $K$ be the subgroup $⟨x^2,y^2⟩$. Then, $K$ is a Klein four-group but every subgroup of $K$ is characteristic in $G$. We explain here why each of the subgroups of order two is characteristic:

• The subgroup $⟨x^2⟩$ is the derived subgroup.
• The subgroup $⟨y^2⟩$ is the subgroup generated by the unique element that is a square but not a commutator.
• The subgroup $⟨x^2{y}^2⟩$ is the subgroup generated by the unique element that is a product of squares but not a square.

$K$ itself is ${\mho }^1(G)$ and is characteristic in $G$, and the trivial subgroup is characteristic in $G$. Thus, all subgroups of $K$ are characteristic in $G$.