A transitive permutation group acting on is called *doubly transitive* if for all , the subgroup is transitive on .

- Prove that is doubly transitive on for all .
- Prove that a doubly transitive group is primitive. Deduce that is not doubly transitive in its action on the four vertices of a square.

- We know that is transitive on . Now if and , we have a natural isomorphism ; this permutation group is also transitive in its action on . Thus is doubly transitive.
- Let act transitively on , and suppose further that the action is doubly transitive. Let be a proper block; then there exist elements and . By a previous exercise, we have . Thus if , we have . Suppose now that there exists an element with . Because is doubly transitive on , there exists an element such that . Thus , a contradiction. So no such element exists and we have . Now every block is trivial, thus the action of on is primitive.
We saw that the action of on the four vertices of a square is not primitive in the previous exercise. Thus this action is not doubly transitive.