Abelian normal subgroup of core-free maximal subgroup is contranormal implies derived subgroup of whole group is monolith

Statement
Suppose $$G$$ is a fact about::primitive group, $$M$$ is a fact about::core-free maximal subgroup of $$G$$ (or, the stabilizer of a point for a faithful primitive group action of $$G$$), and $$A$$ is an fact about::abelian normal subgroup of $$M$$ that is a fact about::contranormal subgroup of $$G$$: $$G$$ is generated by the conjugates of $$A$$ in it. Then, the fact about::commutator subgroup $$[G,G]$$ is the monolith of $$G$$, i.e., every nontrivial normal subgroup of $$G$$ contains the commutator subgroup of $$G$$.

Facts used

 * 1) uses::Second isomorphism theorem
 * 2) uses::Abelianness is quotient-closed

Proof
Given: A group $$G$$, a core-free maximal subgroup $$M$$ of $$G$$. An abelian normal subgroup $$A$$ of $$M$$ that is contranormal in $$G$$. A nontrivial normal subgroup $$N$$ of $$G$$.

To prove: $$G/N$$ is abelian.

Proof:


 * 1) $$MN = G$$: Since $$M$$ is core-free and $$N$$ is nontrivial normal, $$M$$ does not contain $$N$$. Since $$M$$ is maximal, $$MN = G$$.
 * 2) $$AN$$ is normal in $$G$$: Clearly, $$N \le AN \le N_G(AN)$$. Also, $$A$$ is normal in $$M$$ and $$N$$ is normal in G, so $$M \le N_G(AN)$$. Thus, $$MN \le N_G(AN)$$, so $$N_G(AN) = G$$ by step (1).
 * 3) $$AN = G$$: By assumption, $$A$$ is contranormal. Thus, the only normal subgroup of $$G$$ containing $$A$$ is $$G$$. So, by step (2), $$AN$$ is normal in $$G$$.
 * 4) $$G/N \cong A/(A \cap N)$$ is abelian: This follows from facts (1) and (2).