# 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 $G$.

## 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 <mah>G[/itex], 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).