Subgroups of all orders dividing the group order not implies supersolvable

Statement with symbols
It is possible to have a finite group $$G$$ with subgroups of all orders dividing the order of $$G$$, with $$G$$ not a fact about::finite supersolvable group.

Example of the symmetric group of degree four
The symmetric group of degree four has subgroups of every order dividing its order. There are cyclic subgroups of order $$1,2,3,4$$; there is a dihedral Sylow subgroup of order $$8$$; there is a symmetric group on three elements of order $$6$$, and there is an alternating group of order $$12$$.

On the other hand, this group does not have any cyclic normal subgroups at all, hence it is not supersolvable.