Maximal among abelian normal not implies self-centralizing in solvable

Statement
It can happen that $$G$$ is a solvable group, and $$H$$ is maximal among Abelian normal subgroups of $$G$$, but $$H$$ is not a self-centralizing subgroup of $$G$$. In fact, it can even happen that the center of $$G$$ is maximal among Abelian normal subgroups of $$G$$.

This is significant because if we replace solvability by somewhat stronger conditions, like being nilpotent or supersolvable, then any subgroup maximal among Abelian normal subgroups is self-centralizing.

Proof
The simplest example is the group $$GL(2,3)$$. This is a group of order 48, where the center, a subgroup of order 2, is the only nontrivial normal Abelian subgroup. Clearly, it is not self-centralizing because its centralizer is the whole group.

The reason why the proof that works for supersolvable groups fails in this setting is because the quotient by the center, which is isomorphic to the symmetric group on four elements, fails the property that every nontrivial normal subgroup contains a cyclic normal subgroup. (In fact, there is no nontrivial cyclic normal subgroup).