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

Statement

Suppose $G$ is a Primitive group (?), $M$ is a core-free maximal subgroup of $G$ (or, the stabilizer of a point for a faithful primitive group action of $G$), and $A$ is an abelian normal subgroup of $M$ that is a contranormal subgroup of $G$: $G$ is generated by the conjugates of $A$ in it. Then, the derived 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. Second isomorphism theorem
2. 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).

References

Textbook references

• Book:Cohn, Page 120, ^{More info}