Every Abelian characteristic subgroup is central implies every characteristic maximal subgroup is centralizer-dense

Statement
Suppose $$G$$ is a group with the property that every Abelian characteristic subgroup of $$G$$ is central. Then, if $$K \le G$$ is a maximal subgroup that is also a characteristic subgroup (in other words, $$K$$ is a characteristic subgroup of prime index), then $$C_G(K) = Z(G)$$, i.e., $$K$$ is a centralizer-dense subgroup of $$G$$.

Facts used

 * 1) Maximal implies cocentral or self-centralizing
 * 2) Cocentral implies centralizer-dense
 * 3) Characteristicity is transitive

Proof
Given: A group $$G$$ in which every Abelian characteristic subgroup is central. A characteristic subgroup $$K \le G$$ that is also maximal in $$G$$.

To prove: $$C_G(K) = Z(G)$$.

Proof: By fact (1), either $$K$$ is cocentral in $$G$$ or $$K$$ is self-centralizing in $$G$$.

If $$K$$ is cocentral in $$G$$, then fact (2) completes the proof.

If $$K$$ is self-centralizing in $$G$$, then $$C_G(K) = Z(K)$$ is Abelian, and is characteristic in $$K$$. By fact (3), $$Z(K)$$ is an Abelian characteristic subgroup of $$G$$, hence $$C_G(K) = Z(K) \le Z(G)$$ by the assumption that every Abelian characteristic subgroup is central. On the other hand $$Z(G) \le C_G(K)$$ is clear, so $$Z(G) = C_G(K)$$.