examples of primitive groups that are not doubly transitive

The group ${𝒟}_{2n},n\ge 3$, the dihedral group of order $2n$, is the symmetry group of the regular $n$-gon. (Note that we use the more common notation ${𝒟}_{2n}$ for this group rather than ${𝒟}_n$).

${𝒟}_{2n}$ is clearly not doubly transitive for $n\ge 4$, since it preserves “adjacency” in the vertices. Thus, for example, clearly no element of ${𝒟}_{2n}$ can take $(1,2)$ to $(1,3)$. (${𝒟}_{2\cdot 3}={𝒟}_6$, the symmetry group of the triangle, is, however, doubly transitive).

We show that for $p$ prime, ${𝒟}_{2p}$ is primitive. To prove this, we need only verify that any block containing two distinct elements is the entire set of vertices. Number the vertices consecutively $\{0,\mathrm{\dots },p-1\}$, and let $r$ be the element of ${𝒟}_{2n}$ that takes each vertex into its successor $(modp)$. Now, suppose a block contains two distinct elements $a,b$; assume wlog that $b\ne 0$. Iteratively apply $r^{b-a}$ to these elements to get

$a$	$b$
$b$	$2b-a$
$2b-a$	$3b-a$
$\mathrm{\dots }$	$\mathrm{\dots }$

Since blocks are either equal or disjoint, we see that the block in question contains $a,b$, and $nb-a$ for each $n$. But $a\ne b$, so $nb-a$ runs through all residues (http://planetmath.org/ResidueSystems) $(modp)$ and thus the block contains each vertex. Thus $D_{2p}$ is primitive.

For nonprime $n$, ${𝒟}_{2n}$ is not primitive. In this case, if $d$ is a divisor of $n$, then the set of vertices that are multiples of $d$ form a block.