Divisibility-closed not implies local divisibility-closed

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

EXPLORE EXAMPLES YOURSELF: View examples of subgroups satisfying property divisibility-closed subgroup but not local divisibility-closed subgroup|View examples of subgroups satisfying property divisibility-closed subgroup and local divisibility-closed subgroup

Statement

It is possible to have a group $G$ and a subgroup $H$ satisfying the following:

1. $H$ is a divisibility-closed subgroup of $G$: If $n$ is a natural number such that every element of $G$ has a $n^{th}$ root in $G$, then every element of $H$ has a $n^{th}$ root in $H$.
2. $H$ is not a local divisibility-closed subgroup of $G$: There exists a natural number $n$ and an element $g\in H$ such that $x^{n}=g$ has solutions for $x\in G$ but no solution for $x\in H$.

Related facts

• Derived subgroup not is local divisibility-closed in nilpotent group whereas derived subgroup is divisibility-closed in nilpotent group

Proof

We can take any example of a subgroup of finite group where the subgroup is not local divisibility-closed. Some simple examples are below:

• Z2 in Z4: Let $G$ be cyclic group:Z4 and $H$ be Z2 in Z4, the unique cyclic subgroup of order two. The non-identity element of $H$ has square roots in $G$ but not in $H$, so $H$ is not local divisibility-closed. However, being a subgroup of a finite group, it is divisibility-closed.
• Center of dihedral group:D8
• Center of quaternion group