Doubly transitive groups (D&F 4.1.8)

A transitive permutation group \(G\) on a set \(A\) is doubly transitive if for all \(a \in A\) the subgroup \(G_a\) is transitive on \(A-\{a\}\).

(a) Prove that \(S_n\) is doubly transitive on \(A = \{1,2,\cdots,n\}\) for all \(n \ge 2\).

Let \(G = S_n\). Clearly \(G\) is transitive on \(A\). Furthermore, for any \(i \in A\), \(G_i \cong S_{n-1}\) is also transitive on \(A-\{i\}\).

(b) Prove that a doubly transitive group is primitive. Deduce that \(D_8\) is not doubly transitive in its action on the 4 vertices of a square.

Let \(a \in A\), let \(B\) be any proper subset of \(A\) containing \(a\), and let \(x \in A-B\), \(y \in B-\{a\}\). By double-transitivity, there is some \(\sigma \in G_a\) such that \(\sigma(x) = y\). Since both \(a = \sigma^{-1}(a)\) and \(x\) are in \(\sigma^{-1}(B)\) but \(x \not\in B\) while \(a \in B\), it follows that \(\sigma^{-1}(B)\) is neither equal nor disjoint from \(B\). So \(B\) cannot be a block. Since every block must contain some \(a \in A\), we can conclude that the only blocks are the singletons and \(A\).