Ascendant not implies subnormal

Statement
An ascendant subgroup of a group need not be a subnormal subgroup.

Related facts

 * Stronger than::Normality is not transitive
 * Stronger than::There exist subgroups of arbitrarily large subnormal depth
 * Descendant not implies subnormal

Example of a generalized dihedral group
Let $$K$$ be the $$2$$-quasicyclic group. In other words, $$K$$ is the group of all $$(2^n)^{th}$$ roots of unity in $$\mathbb{C}$$ for all $$n$$, under multiplication. Consider $$G$$ the semidirect product of $$K$$ with a cyclic group $$H$$ of order two, where $$H$$ acts on $$K$$ by the inverse map. In other words, $$G$$ is the uses as intermediate construct::generalized dihedral group corresponding to the abelian group $$H$$. Then:


 * $$H$$ is an ascendant subgroup of $$G$$: Indeed, consider an ascending chain of subgroups whose $$n^{th}$$ member is the subgroup generated by $$x$$ and all the $$(2^n)^{th}$$ roots of unity. Each member of this ascending chain is normal in its successor, and the union of the ascending chain of subgroups is the whole group.
 * $$H$$ is not a subnormal subgroup of $$G$$: If $$H$$ were $$k$$-subnormal in $$G$$ for some $$k$$, $$H$$ would also be $$k$$-subnormal in every intermediate subgroup. However, the subnormal depth of $$H$$ in the semidirect product of the $$(2^n)^{th}$$ roots of unity with $$H$$ is $$n$$, and this grows arbitrarily large. Thus, $$H$$ cannot be $$k$$-subnormal for some finite $$k$$. (For more on why the subnormal depth grows arbitrarily large, refer the dihedral groups example in: there exist subgroups of arbitrarily large subnormal depth).