# transitive subgroups of the symmetric group on $2(d+1)$ elements: can I always do at least $d$ permutations?

I have a set of $2(d+1)$ elements which are labelled as pairs $\{e_i, a_i\}_{i=1}^{d+1}$, transforming under some transitive subgroup of the symmetric group. This can be thought of as a regular $2(d+1)$ sided polygon and the $i^\text{th}$ opposite vertexes as the pairs $e_i$ and $a_i$, but the group of transforamtions can be more general than the dihedral group. For any transitive subgroup, can I always achieve a transformation permuting at least $d$ pairs of elements (I dont care what happens to the other elements)?

For example, if I have $d=2$ the elements are $\{ e_1, e_2, e_3, e_4, e_5, e_6 \}$. I would be looking for a transitive group that has an element that permutes 2 or 3 of these pairs. I dont care what happens to the other (in the case of permuting 2 pairs), do for general $d$ the transformation doesnt have to be order 2.

The reason for asking is the following: I have a state space comprising of $2d+2$ "pure state" vectors, which can be divided into $d+1$ pairs of antipodal vectors. I have a group of transformations that maps the state space to its self. I want to show that for any transitive group I can realise a transformation that permutes $d$ pairs of antipodal vectors. I.e. if my state space is $\{e_1,e_2,e_3,a_1,a_2,a_3\}$, where $e_i$ and $a_i$ are the $i^{th}$ pair of mutually antipodal vectors, I want to show that any transitive group can reach something like $\{a_1,a_2,e_3,e_1,e_2,a_3\}$, i.e. permuting at least $d$ (2) antipodal elements.

I am not sure if I am understanding your question correctly. I think it is the following. Let $G$ be a transitive subgroup of $S_n$ with $n$ even. Does there necessarily exist an element of order $2$ in $G$ that fixes at most $2$ points?
I am afraid that the answer to that is no. In the terminology of the GAP and Magma computer algebra systems, the group $\mathtt{TransitiveGroup}(12,31)$ is a transitive subgroup of $S_{12}$ of order $48$. The only elements of order $2$ in this group interchange $4$ pairs of points and fix $4$ points. There are several other transitive groups of order $12$ that are counterexamples to your question.
• I am finding this hard to follow, because first you said $d=6$, which would mean there were $14$ points being permuted, but then you said $d=2$. But if there are $2(d+1)$ points altogether and $d$ pairs are interchanged, then there are only two remaining points anyway, and those two would have to be either interchanged or fixed. So surely the permutation must have order $2$? – Derek Holt Jun 20 '15 at 8:27