Hereditarily characteristic not implies cyclic in finite

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


 * 1) $$K$$ is a fact about::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 $$K = G$$.
 * SQ-dual::Upward-closed characteristic not implies cyclic-quotient in finite

Proof
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:

$$G := \langle x,y \mid x^4 = y^4 = e, yxy^{-1} = x^3 \rangle$$.

Let $$K$$ be the subgroup $$\langle x^2, y^2 \rangle$$. 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 $$\langle x^2 \rangle$$ is the derived subgroup.
 * The subgroup $$\langle y^2 \rangle$$ is the subgroup generated by the unique element that is a square but not a commutator.
 * The subgroup $$\langle x^2y^2 \rangle$$ 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 $$G$$.