Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 4.1 Exercise 4.1.8
A transitive permutation group $G \leq S_A$ acting on $A$ is called doubly transitive if for all $a \in A$, the subgroup $\mathsf{stab}(a)$ is transitive on $A \setminus \{a\}$.
(1) Prove that $S_n$ is doubly transitive on $\{1,2,\ldots,n\}$ for all $n \geq 2$.
(2) Prove that a doubly transitive group is primitive. Deduce that $D_8$ is not doubly transitive in its action on the four vertices of a square.
Solution:
(1) We know that $S_n$ is transitive on $A = \{1,2,\ldots,n\}$. Now if $n \geq 2$ and $k \in A$, we have a natural isomorphism $\mathsf{stab}(k) \cong S_{A \setminus \{k\}}$; this permutation group is also transitive in its action on $A \setminus \{k\}$. Thus $S_n$ is doubly transitive.
(2) Let $G \leq S_A$ act transitively on $A$, and suppose further that the action is doubly transitive. Let $B \subseteq A$ be a proper block; then there exist elements $b \in B$ and $a \in A \setminus B$. By a previous exercise, we have $\mathsf{stab}(b) \leq \mathsf{stab}(B)$. Thus if $\sigma \in \mathsf{stab}(b)$, we have $\sigma[B] = B$. Suppose now that there exists an element $c \in B$ with $c \neq b$. Because $G$ is doubly transitive on $A$, there exists an element $\tau \in \mathsf{stab}(b)$ such that $\tau(c) = a$. Thus $\tau[B] \neq B$, a contradiction. So no such element $c$ exists and we have $B = \{b\}$. Now every block is trivial, thus the action of $G$ on $A$ is primitive.
We saw that the action of $D_8$ on the four vertices of a square is not primitive in the previous exercise. Thus this action is not doubly transitive.
