Divisibility-closed implies powering-invariant

This article gives the statement and possibly, proof, of an implication relation between two subgroup properties. That is, it states that every subgroup satisfying the first subgroup property (i.e., divisibility-closed subgroup) must also satisfy the second subgroup property (i.e., powering-invariant subgroup)
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Statement

Suppose $G$ is a subgroup and $H$ is a divisibility-closed subgroup of $G$, i.e., if $n$ is a natural number such that every element of $G$ has an $n^{th}$ root in $G$, then every element of $H$ has a $n^{th}$ root in $H$. Then, $H$ is also a powering-invariant subgroup of $G$, i.e., if $p$ is a prime number such that every element of $G$ has a unique $p^{th}$ root, then every element of $H$ has a unique $p^{th}$ root in $H$.

Related facts

Converse

• Powering-invariant not implies divisibility-closed

Proof

This statement is pretty direct. Note that existence follows from being divisibility-closed, and uniqueness in the whole group implies uniqueness in the subgroup as well.