If $G$ is a supersolvabe group, and $A$ is a maximal among abelian normal subgroups of $G$, then the centralizer of $A$ in $G$ is $A$ itself (see link).
My question is about the importance of the hypothesis that "$G$ is supersolvable".
Question: Let $G$ be any group, $Z(G)$ be its center, and $A$ be a maximal among abelian normal subgroups of $G$, such that $Z(G)\neq A$ (i.e. $Z(G)<A$). Prove or disprove: the centralizer of $A$ in $G$ is $A$ itself?
(The examples I found were direct product of an abelian group with a non-abelian simple group; but here we can not have a maximal abelian normal subgroup $A$ such that $Z(G)\neq A$.)