# 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.

## 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 $k \ge 2$, 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.