Subgroups of all orders dividing the group order not implies supersolvable

This article gives the statement and possibly, proof, of a non-implication relation between two group properties. That is, it states that every group satisfying the first group property (i.e., group having subgroups of all orders dividing the group order) need not satisfy the second group property (i.e., supersolvable group)
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Statement

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 Finite supersolvable group (?).

Proof

Example of the symmetric group of degree four

Further information: symmetric group:S4, subgroup structure of symmetric group:S4

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.