Subgroup of abelian group not implies powering-invariant

Statement
It is possible to have an abelian group $$G$$ and a subgroup $$H$$ of $$G$$ that is not a powering-invariant subgroup of $$G$$: there exists a prime number $$p$$ such that $$G$$ is powered over $$p$$, but $$H$$ is not powered over $$p$$.

Related facts

 * Finite implies powering-invariant
 * Finite index implies powering-invariant

Thus, our example must be an infinite subgroup of infinite index.

Proof
The simplest example is as follows: $$G = \mathbb{Q}$$ is the additive group of rational numbers and $$H = \mathbb{Z}$$. In this case, $$G$$ is powered over all primes, but $$H$$ is not powered over any.