Powering-invariance is not quotient-transitive

Statement
It is possible to have groups $$H \le K \le G$$ such that $$H$$ is a powering-invariant normal subgroup of $$G$$ and $$K/H$$ is a powering-invariant subgroup of the quotient group $$G/H$$, but $$K$$ is not powering-invariant in $$G$$.

Related facts

 * Powering-invariant over quotient-powering-invariant implies powering-invariant
 * Quotient-powering-invariance is quotient-transitive

Proof
The proof idea is follows: use the construction in the reference for $$H$$. Now take $$K$$ as a subgroup containing $$H$$ such that $$K/H$$ is a finite cyclic group of order greater than 1. Now:


 * $$H$$ is powering-invariant in $$G$$ by construction, since both $$G$$ and $$H$$ are rationally powered groups.
 * $$K/H$$ is powering-invariant in $$G/H$$ since $$G/H$$ is not powered over any prime.
 * $$K$$ is not powering-invariant in $$G$$: For instance, any element in the coset of $$H$$ in $$K$$ other than $$H$$ cannot have a square root in $$K$$, even though $$G$$ is 2-powered.