Primitive implies Fitting-free or elementary abelian Fitting subgroup

Statement

Suppose $G$ is a primitive group with a core-free maximal subgroup $M$. Then, there are two possibilities:

1. $G$ is a Fitting-free group (?): it has no nontrivial abelian normal subgroup
2. The Fitting subgroup of $G$ is an elementary Abelian normal subgroup of $G$, say $N$, and $N$ and $M$ are permutable complements. Further, this is the only nontrivial Abelian normal subgroup of $G$

Related facts

• Primitive implies innately transitive
• Abelian minimal normal subgroup and core-free maximal subgroup are permutable complements
• Abelian permutable complement to core-free subgroup is self-centralizing
• Plinth theorem

Facts used

1. Abelian minimal normal subgroup and core-free maximal subgroup are permutable complements
2. Abelian permutable complement to core-free subgroup is self-centralizing

Proof

'Given: A finite primitive group $G$ with core-free maximal subgroup $M$. $F(G)$ is the Fitting subgroup of $G$, and is nontrivial

To prove: $F(G)$ is elementary Abelian, and is the only nontrivial Abelian normal subgroup of $G$. If $N=F(G)$, then $NM=G$ and $N\cap M$ is trivial.

Proof: Since $F(G)$ is nontrivial, it is a nontrivial nilpotent characteristic subgroup of $G$. Consider the subgroup $Z(F(G))$, the center of $F(G)$, is thus a nontrivial Abelian characteristic subgroup. Hence, it is in particular an Abelian normal subgroup. Let $N$ be a minimal normal subgroup of $G$ contained in $Z(F(G))$.

Using fact (1), $NM=G$ and $N\cap M$ is trivial. Now using fact (2), we see that $C_G(N)=N$. But we know that since $N\le Z(F(G))$, So $N\le F(G)\le C_G(N)$. Thus, we're forced to conclude that $N=F(G)$.

Thus, the Fitting subgroup is itself a minimal normal subgroup. Since the Fitting subgroup, by definition, contains all Abelian normal subgroups, $N=F(G)$ is the unique Abelian normal subgroup.