Descendant 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., descendant subgroup) need not satisfy the second subgroup property (i.e., subnormal subgroup)
View a complete list of subgroup property non-implications | View a complete list of subgroup property implications
Get more facts about descendant subgroup|Get more facts about subnormal subgroup

EXPLORE EXAMPLES YOURSELF: View examples of subgroups satisfying property descendant subgroup but not subnormal subgroup|View examples of subgroups satisfying property descendant subgroup and subnormal subgroup

Statement

A descendant subgroup of a group need not be subnormal.

Related facts

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

Definitions used

Descendant subgroup

Further information: Descendant subgroup

Subnormal subgroup

Further information: Subnormal subgroup

Proof

Example of the dihedral group corresponding to a 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. Then:

• $H$ is a descendant subgroup of $G$: Indeed, consider a descending 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 descending chain is normal in its predecessor, and the intersection of all these members is precisely $H$.
• $H$ is not a subnormal subgroup of $G$: In fact, the descending chain constructed above is precisely the one obtained where each member is the normal closure of $H$ in its predecessor. Since this chain has infinite length, $H$ is not subnormal.