Hereditarily characteristic not implies cyclic 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., hereditarily characteristic subgroup) need not satisfy the second subgroup property (i.e., cyclic characteristic subgroup)
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It is possible to have a finite group G, and a subgroup K of H such that:

  1. K is a 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


Further information: nontrivial semidirect product of Z4 and Z4, subgroup structure of nontrivial semidirect product of Z4 and Z4

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.