Finite solvable not implies p-normal

Statement
It is possible to have a finite solvable group $$G$$ and a prime $$p$$ such that $$G$$ is not a p-normal group. In other words, there exists a $$p$$-Sylow subgroup $$P$$ of $$G$$ whose center $$Z(P)$$ is not weakly closed in it.

Example of the symmetric group of degree four
Let $$G$$ be the symmetric group on the set $$\{ 1,2,3,4 \}$$ and let $$p = 2$$. Then:


 * $$G$$ is solvable.
 * $$G$$ is not $$2$$-normal: The center of a $$2$$-Sylow subgroup is not weakly closed in it. For instance, consider the $$2$$-Sylow subgroup:

$$P = \{, (1,2,3,4), (1,3)(2,4), (1,4,3,2), (1,3), (2,4), (1,2)(3,4), (1,4)(2,3) \}$$.

The center of this is:

$$Z(P) = \{, (1,3)(2,4) \}$$.

This is not weakly closed in $$P$$. For instance, it is conjugate in $$G$$ to the subgroup generated by $$(1,2)(3,4)$$, which is another subgroup inside $$P$$.