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 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

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:

$M N = G$ : Since $M$ is core-free and $N$ is nontrivial normal, $M$ does not contain $N$ . Since $M$ is maximal, $M N = G$ .
$A N$ is normal in $G$ : Clearly, $N \leq A N \leq N_{G} (A N)$ . Also, $A$ is normal in $M$ and $N$ is normal in <mah>G</math>, so $M \leq N_{G} (A N)$ . Thus, $M N \leq N_{G} (A N)$ , so $N_{G} (A N) = G$ by step (1).
$A N = G$ : By assumption, $A$ is contranormal. Thus, the only normal subgroup of $G$ containing $A$ is $G$ . So, by step (2), $A N$ is normal in $G$ .
$G / N ≅ A / (A \cap N)$ is abelian: This follows from facts (1) and (2).

References

Textbook references

Book:Cohn, Page 120, ^{More info}