Upward-closed characteristic not implies cyclic-quotient 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., upward-closed characteristic subgroup) need not satisfy the second subgroup property (i.e., cyclic-quotient subgroup)
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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 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

• Hereditarily characteristic not implies cyclic in finite

Converse

• Cyclic-quotient characteristic implies upward-closed characteristic

Facts used

1. Bryant-Kovacs theorem

Proof

Example of a group of order 16

Further information: semidihedral group:SD16, subgroup structure of semidihedral group:SD16

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 $k\geq 2$, 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.