Powering-invariant and normal not implies quotient-powering-invariant

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 normal subgroup) need not satisfy the second subgroup property (i.e., quotient-powering-invariant subgroup)
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Statement

It is possible to have a group $G$ and a powering-invariant normal subgroup $H$ of $G$ that is not quotient-powering-invariant: in other words, $H$ is a normal subgroup of $G$ that is also a powering-invariant subgroup of $G$, but there exists a prime number $p$ such that $G$ is $p$-powered and $G/H$ is not $p$-powered.

We can in fact construct our example such that both $G$ and $H$ are rationally powered groups and $G/H$ contains a divisible group as subgroup where every element has infinitely many $n^{th}$ roots for all $n>1$.

References

• MathOverflow question: Normal subgroup that is invariant under powering such that the quotient group is not invariant