Abelian minimal normal subgroup and core-free maximal subgroup are permutable complements

Statement
Suppose $$G$$ is a primitive group with core-free maximal subgroup $$M$$. Then, if $$G$$ has an Abelian minimal normal subgroup $$N$$, the following are true:


 * 1) $$NM = G$$
 * 2) $$N$$ and $$M$$ intersect trivially
 * 3) $$M \cong G/N$$
 * 4) The induced action of $$N$$ on $$G/M$$ by left multiplication, is equivalent to the regular action of a group on itself.

Core-free subgroup
A subgroup of a group is termed core-free if its normal core in the whole group is trivial, or equivalently, if it does not contain any nontrivial normal subgroup of the whole group.

Maximal subgroup
A maximal subgroup of a group is a proper subgroup that is not contained in any bigger proper subgroup.

Primitive group
A primitive group is a group that possesses a core-free maximal subgroup.

Facts used

 * 1) Abelian normal subgroup and core-free subgroup generate whole group implies they intersect trivially

Plinth theorem
The plinth theorem is a similar result to the case of two distinct minimal nomral subgroups.

More results under similar hypotheses
Proceeding further along the lines of this proof, one can show that:


 * Abelian permutable complement to core-free subgroup is self-centralizing
 * Primitive implies Fitting-free or elementary Abelian Fitting subgroup: In other words, $$N$$ is maximal among Abelian normal subgroups, and also is the only nontrivial Abelian normal subgroup, and is the only nontrivial nilpotent normal subgroup.

Breakdown of similar results

 * Minimal normal subgroup and core-free maximal subgroup need not be permutable complements

Proof
Given: A primitive group $$G$$ with core-free maximal subgroup $$M$$, and an Abelian minimal normal subgroup $$N$$

To prove: $$N$$ and $$M$$ are permutable complements: $$NM = G$$ and $$N \cap M$$ is trivial

Proof: Since $$N$$ is normal, $$NM$$ is a subgroup of $$G$$ containing $$M$$. Since $$M$$ is maximal, either $$NM = M$$ or $$NM = G$$. If $$NM = M$$, then $$N \le M$$, but this cannot happen because $$M$$ does not contain any nontrivial normal subgroup of $$G$$. Hence $$NM = G$$.

Thus $$N$$ is an Abelian normal subgroup and $$M$$ is a core-free subgroup that together generate $$G$$. Applying the fact (1) stated above, $$N \cap M$$ is trivial, so $$N$$ and $$M$$ are permutable complements.