# Plinth theorem

From Groupprops

## Statement

Suppose is a finite group with a faithful group action on a set . Suppose and are two distinct minimal normal subgroups of that act transitively on under the induced action. Then, the following are true:

- and .
- is isomorphic to .
- There exists an element such that is involutive, normalizes , and conjugates to .
- The action of on is a primitive group action.

## Related facts

- Abelian minimal normal subgroup and core-free maximal subgroup are permutable complements
- Abelian permutable complement to core-free subgroup is self-centralizing
- Primitive implies Fitting-free or elementary Abelian Fitting subgroup

## Proof

**Given**: A finite group with a faithful group action on a set . and are two distinct minimal normal subgroups of that act transitively on under the induced action.

**To prove**:

- and .
- is isomorphic to .
- There exists an element such that is involutive, normalizes , and conjugates to .
- The action of on is a primitive group action.

**Proof**:

- and intersect trivially: Since an intersection of normal subgroups is normal (Fact (1)), and and are
*distinct*minimal normal subgroups, so their intersection must be trivial. - Every element of commutes with every element of : Since and are normal subgroups, the commutator is contained in the intersection . Thus, is trivial, and thus, every element of commutes with every element of .
- The induced action of on is regular: Suppose there exists a non-identity element and such that . Now, for every , we have , so , so fixes . Since is transitive on , we obtain that acts as the identity on , a contradiction to the assumption that the action was faithful. Thus, the action of on is regular.
- The induced action of on is regular: This follows from the same reasoning as the previous step, interchanging the roles of and .
- The centralizer of in is isomorphic to : Let be any chosen element. This can be seen in two ways:
- Suppose commutes with the action of . Then, we must have for all . Thus, is completely determined by the value of . Further, if and , then . Thus, there is a bijection between the centralizer of in and the elements of , that reverses the order of multiplication. By fact (2) (every group is isomorphic to its opposite group), we obtain that the centralizer of in is isomorphic to .
- A shorter way is to observe that the action of on is equivalent to the left-regular group action on itself. Now, we use the fact that the only permutations that commute with all left multiplications are the right multiplications, which form the right-regular group action. Thus, the centralizer of in is an isomorphic group acting via right multiplication.

- is a subgroup of the centralizer of in , hence isomorphic to a subgroup of : This combines step (2) with the previous step.
- is a subgroup of the centralizer of in , hence isomorphic to a subgroup of : The same argument as the previous step, reversing the roles of and .
- and are isomorphic, and equal each other's centralizers in : This follows from the previous two steps, and the fact that they are both finite groups. Note that this settles parts (1) and (2) of what we need to prove.
- We now construct the involution in that conjugates to : Note that we can think of as a set identified with both and , where acts by left multiplication and acts via right multiplication by the
*inverse element*. Consider now a bijection from to itself that sends each element of to its*inverse*(when identified with either or ). This map conjugates left multiplication by an element of with right multiplication by the inverse of its corresponding element in . This settles part (3) of what we need to prove. - The action is primitive: We prove this by showing that if the action is not primitive, we can construct strictly smaller nontrivial normal subgroups of contained in .