Abelian characteristic is not join-closed

Statement
It is possible to have a group $$G$$ with Abelian characteristic subgroups (i.e., subgroups that are both Abelian and characteristic) $$H$$ and $$K$$ such that the join $$\langle H, K \rangle$$ is not an Abelian characteristic subgroup.

Since the join is always a characteristic subgroup, the particular thing that can fail is that the join need not be Abelian.

Related facts

 * Characteristicity is strongly join-closed
 * Abelian normal is not join-closed
 * Cyclic normal is not join-closed