Every doubly transitive group action is primitive

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.

  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.

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