Non-central subgroup which is maximal among abelian normal subgroups is self-centralizing?

Question

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 $Z(G)\neq A$.)

Answer 1

Here is a counterexample of minimal order. Consider the binary octahedral group $2\mathcal{O}$ and let $\langle x \rangle$ be the cyclic subgroup of order $4$ which contains $Z(2\mathcal{O})$ but is not contained in the derived subgroup $2\mathcal{O}^\prime$. (So, $x^2$ generates the center.)

There is a $\sigma\in\operatorname{Aut}\left(2\mathcal{O}\right)$ of order $2$ defined by $\sigma:x\mapsto x^{-1}$. We construct a semidirect product based on that automorphism: let $\langle z \rangle = C_2$ and consider $G=2\mathcal{O}\rtimes \langle z \rangle$ formed by $z\mapsto \sigma$.

We observe that $z$ fixes $x^2$, so this group the same center as $2\mathcal{O}$. Therefore, $A=\langle x^2,z \rangle=Z(2\mathcal{O})\times \langle z \rangle$ is maximal among abelian normal subgroups. However, its centralizer has order $48$ (in fact it is isomorphic to $\operatorname{SL}_2(\mathbb{F}_3)\times \langle z \rangle$). So, this is a counterexample.

To see that this counterexample has minimal order, notice that the binary octahedral group is the smallest counterexample where $A$ is central.