Powering-invariance is not quotient-transitive

This article gives the statement, and possibly proof, of a subgroup property (i.e., powering-invariant subgroup) not satisfying a subgroup metaproperty (i.e., quotient-transitive subgroup property).
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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 $n > 1$. 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, an element of $K$ whose image in $K/H$ generates the latter group does not have a $n^{th}$ root in $K$.

References

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