3
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ thank you for the reply. Im not familiar with the terminology, but what I think I am looking for is the following: the group is S2(d+1) and I am looking for an element of order 2 that interchanges at least d pair points - I dont care what happens to the others. For example when d=6 and I have a set of objects {e1,e2,e3,a1,a2,a3} that transform usnder some transitive subgroup of s6, I want there to always exist a group element that permutes (6=2(d+1)→d=2) at least 2 two pairs ei↔ai - it could permute all three pairs for instance. doesnt have to be order 2 as I dont care what happens to the rest $\endgroup$ – jdizzle Jun 20 '15 at 7:32
  • $\begingroup$ 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 $2$? $\endgroup$ – Derek Holt Jun 20 '15 at 8:27
  • $\begingroup$ Your right - thanks youve save me a lot of time! $\endgroup$ – jdizzle Jun 20 '15 at 9:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.