Upward-closed characteristic not implies cyclic-quotient in finite

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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

Converse

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.