Upward-closed characteristic not implies cyclic-quotient in finite

Statement
It is possible to have a finite group $$G$$ and a subgroup $$H$$ of $$G$$ such that the following two conditions hold:


 * 1) $$H$$ is an upward-closed characteristic subgroup of $$G$$: Every subgroup $$K$$ of $$G$$ containing $$H$$ is a fact about::characteristic subgroup of $$G$$.
 * 2) $$H$$ is not a cyclic-quotient subgroup of $$G$$, i.e., the quotient group $$G/H$$ is not a cyclic group.

Related facts

 * SQ-dual::Hereditarily characteristic not implies cyclic in finite

Converse

 * Cyclic-quotient characteristic implies upward-closed characteristic

Facts used

 * 1) uses::Bryant-Kovacs theorem

Example of a group of order 16
Consider the semidihedral group of order $$16$$, given by the presentation:

$$G := \langle a,x \mid a^8 = x^2 = e, xax^{-1} = a^3 \rangle$$.

The derived subgroup $$H$$ of $$G$$ euqals the Frattini subgroup of $$G$$ and it is the subgroup $$\langle a^2 \rangle$$. The quotient by this subgroup is a Klein four-group, and is hence not cyclic. However, all subgroups of $$G$$ containing $$H$$ are characteristic -- in fact, all the subgroups of order eight containing $$H$$ are isomorph-free subgroups: the cyclic group of order eight $$\langle a \rangle$$, the dihedral group of order eight $$\langle a^2, x \rangle$$, and the quaternion group $$\langle a^2, ax \rangle$$.

Generalization
The Bryant-Kovacs theorem allows us to construct, for any prime number $$p$$ and any $$k \ge 2$$, a group $$G$$ such that $$G/\Phi(G)$$ (with $$\Phi(G)$$ the Frattini subgroup of $$G$$) is elementary abelian of order $$p^k$$, and the image of the map $$\operatorname{Aut}(G) \to \operatorname{Aut}(G/\Phi(G))$$ is trivial. This in particular would imply that $$\Phi(G)$$ is upward-closed characteristic, but $$G/\Phi(G)$$ is not cyclic.