Powering-invariance is not commutator-closed

This article gives the statement, and possibly proof, of a subgroup property (i.e., powering-invariant subgroup) not satisfying a subgroup metaproperty (i.e., commutator-closed subgroup property).
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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 $G$.