# examples of primitive groups that are not doubly transitive

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

$\mathcal{D}_{2n}$ is clearly not doubly transitive for $n\geq 4$, since it preserves “adjacency” in the vertices. Thus, for example, clearly no element of $\mathcal{D}_{2n}$ can take $(1,2)$ to $(1,3)$. ($\mathcal{D}_{2\cdot 3}=\mathcal{D}_{6}$, the symmetry group of the triangle, is, however, doubly transitive).

We show that for $p$ prime, $\mathcal{D}_{2p}$ 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 $\{0,\ldots,p-1\}$, and let $r$ be the element of $\mathcal{D}_{2n}$ that takes each vertex into its successor $\pmod{p}$. Now, suppose a block contains two distinct elements $a,b$; assume wlog that $b\neq 0$. Iteratively apply $\displaystyle r^{b-a}$ to these elements to get

 $a$ $b$ $b$ $2b-a$ $2b-a$ $3b-a$ $\ldots$ $\ldots$

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

For nonprime $n$, $\mathcal{D}_{2n}$ 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.

Title examples of primitive groups that are not doubly transitive ExamplesOfPrimitiveGroupsThatAreNotDoublyTransitive 2013-03-22 17:22:37 2013-03-22 17:22:37 rm50 (10146) rm50 (10146) 7 rm50 (10146) Example msc 20B15