# 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.

## Definitions used

### Core-free subgroup

Further information: 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

Further information: Maximal subgroup

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

### Primitive group

Further information: 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

## Related facts

### Plinth theorem

Further information: 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:

## 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.