Ascendant not implies subnormal

This article gives the statement and possibly, proof, of a non-implication relation between two subgroup properties. That is, it states that every subgroup satisfying the first subgroup property (i.e., ascendant subgroup) need not satisfy the second subgroup property (i.e., subnormal subgroup)
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Statement

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

Related facts

• Normality is not transitive
• There exist subgroups of arbitrarily large subnormal depth
• Descendant not implies subnormal

Definitions used

Ascendant subgroup

Further information: Ascendant subgroup

Subnormal subgroup

Further information: Subnormal subgroup

Proof

Example of a generalized dihedral group

Further information: generalized dihedral group of 2-quasicyclic 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 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).