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 fact about::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) uses::Abelian minimal normal subgroup and core-free maximal subgroup are permutable complements
 * 2) uses::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.