Primitive implies innately transitive

This article gives the statement and possibly, proof, of an implication relation between two group properties. That is, it states that every group satisfying the first group property (i.e., primitive group) must also satisfy the second group property (i.e., innately transitive group)
View all group property implications | View all group property non-implications
Get more facts about primitive group|Get more facts about innately transitive group

Statement

In the language of group actions

Suppose $G$ is a finite group with a primitive group action with a faithful primitive group action $\alpha$ on a set $S$. Then, there exists a minimal normal subgroup $N$ of $G$ such that the restriction of $\alpha$ to $N$ is still transitive to $S$.

In fact, the above statement holds for any minimal normal subgroup (and in fact, any nontrivial normal subgroup) of $G$. (This is the stronger fact that any primitive group is quasiprimitive).

In the language of group theory

Suppose $M$ is a core-free maximal subgroup of a group $G$. Then, $G$ has a minimal normal subgroup $N$ such that $NM=G$.

Related facts

• Primitive implies quasiprimitive, Quasiprimitive 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
• Primitive implies Fitting-free or elementary abelian Fitting subgroup
• Primitive solvable group acts on a set iff the set has prime power size

Proof

In the language of group theory

Since the action of $G$ is primitive, the stabilizer of any point is a core-free maximal subgroup. Call this $M$. In particular, $M$ does not contain any nontrivial normal subgroup. Hence, its product with any nontrivial normal subgroup $N$ is strictly bigger than $M$, so $MN=G$.