I found a way to generalize Herstein's problem.
Note in this proof we do not really need that fact the group is abelian. We just need the fact that all sub-group products are subgroups. As a result, let me give this group a name in honor of Grothendieck.
Defintion: A "Grothendieck group" is a group such that if are subgroups then is a subgroup.
Note, Grothendieck groups do not need to be abelian. For instance, Dedekind groups are Grothendieck groups because all its subgroups are normal. So the quaternion group which is Dedekind but not abelian is Grothendieck.
Here is the generalization:
"If is a finite Grothendieck group and has subgroups of orders then it will have a subgroup of order ".
Not only does it generalize Herstein's problem it generalizes the problem of the converse of the theorem of Lagrange. By extending the converse to a larger class or groups.