Powering-invariant not implies 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., powering-invariant subgroup) need not satisfy the second subgroup property (i.e., divisibility-closed subgroup)
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Statement

It is possible to have a group $G$ and a subgroup $H$ such that:

1. $H$ is a powering-invariant subgroup of $G$: If $n$ is a natural number such that every element of $G$ has a unique $n^{th}$ root, then every element of $H$ has a unique $n^{th}$ root within $H$.
2. $H$ is not a divisibility-closed subgroup of $G$: There exists a natural number $n$ such that every element of $G$ has a $n^{th}$ root (not necessarily unique) but not every element of $H$ has a $n^{th}$ root within $H$.

Related facts

• Center not is divisibility-closed

Proof

Proof idea

The key fact is that any finite subgroup of a group must be powering-invariant, but it need not be divisibility-closed. We will construct an example where the subgroup is finite.

Proof details

For any prime number $p$:

• Let $G$ be the $p$-quasicyclic group.
• Let $H$ be the subgroup comprising the elements of order 1 or $p$.

Clearly:

• $H$, being finite, is powering-invariant (in fact, both $G$ and $H$ are powered over precisely the set of primes other than $p$).
• However, $H$ is not divisibility-closed: $G$ is $p$-divisible, but $H$ is not.