Finite solvable not implies supersolvable

Statement
A finite solvable group need not be a supersolvable group. In particular, a solvable group need not be supersolvable.

Note that for a finite group, being solvable is equivalent to being a fact about::polycyclic group, so this also gives an example of a fact about::polycyclic group that is not supersolvable.

Proof
Consider the alternating group of degree four on the set $$\{1,2,3,4 \}$$. This is a group of order 12:


 * The group is solvable and polycyclic: The derived subgroup, V4 in A4, is isomorphic to Klein four-group, which is a finite abelian group, and hence the group is solvable. We can also construct a polycyclic series as follows: $$\{ \} \le  \{, (1,2)(3,4) \} \le \{ , (1,2)(3,4), (1,3)(2,4), (1,4)(2,3) \} \le G$$. This is a subnormal series where all the successive quotients are cyclic: cyclic group:Z2, cyclic group:Z2, and cyclic group:Z3.
 * This group is not supersolvable: The group has no cyclic normal subgroup, so there is no way it can have a normal series where all successive quotients are cyclic.