Powering-invariance is not commutator-closed

Statement
It is possible to have a group $$G$$ and subgroups $$H,K$$ of $$G$$ such that both $$H$$ and $$K$$ are powering-invariant subgroups of $$G$$ but the commutator $$[H,K]$$ is not powering-invariant.

Related facts

 * Powering-invariance is commutator-closed in nilpotent group
 * Powering-invariance is not finite-join-closed
 * Powering-invariance is strongly join-closed in nilpotent group

Proof
Suppose $$G$$ is the generalized dihedral group corresponding to the additive group of rational numbers. Let $$H$$ and $$K$$ both be subgroups of order two generated by different reflections. Then, the following are true:


 * $$G$$ is powered over all primes other than 2.
 * $$H$$ and $$K$$ are both powering-invariant subgroups on account of being finite subgroups (see finite implies powering-invariant).
 * $$[H,K]$$ is an infinite cyclic group, isomorphic to the group of integers. It is not powered over any prime, hence is not powering-invariant in $$G$$.