Powering-invariance is not finite-join-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., finite-join-closed group property).
View all subgroup metaproperty dissatisfactions | View all subgroup metaproperty satisfactions|Get help on looking up metaproperty (dis)satisfactions for subgroup properties
Get more facts about powering-invariant subgroup|Get more facts about finite-join-closed group property|

Statement

It is possible to have a group $G$ and subgroups $H$ and $K$ of $G$ such that both $H$ and $K$ are both powering-invariant subgroups of $G$ but the join of subgroups $\langle H,K\rangle$ is not a powering-invariant subgroup of $G$.

Related facts

• Powering-invariance is strongly intersection-closed
• Divisibility-closedness is not finite-join-closed
• Divisibility-closedness is not finite-intersection-closed

Nilpotent case

• Powering-invariance is strongly join-closed in nilpotent group
• Divisibility-closedness 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).
• $\langle H,K\rangle$ is isomorphic to the infinite dihedral group. It is not powered over any primes, and in particular it is not powering-invariant in $G$.