Re: Orbitals of a transitive action

magidin@xxxxxxxxxxxxxxxxx (Arturo Magidin) writes:

In article <u4pu1zfhh.fsfhello@xxxxxxxxxxxxx>,
Rafael Villarroel <rvf0068@xxxxxxxxx> wrote:

It seems that a standard exercise is the following: If G acts
transitively on V, then G has a symmetric nondiagonal orbit on VxV if
and only if |G| is even (for example, Algebraic Graph Theory by Godsil
& Royle, p.31).

However, if I consider, say G=<(1,2,3)>x<(4,5)> acting on V={1,2,3},
(a transitive action where the elements of order 2 act trivially),
then the nondiagonal orbits on VxV are {(1,2),(2,3),(3,1)},
{(1,3),(2,1),(3,2)}, which are not symmetric, even though |G|=6.

What am I not seeing?

The action of G factors through through G/<(4,5)>, so you "really"
have only a group of order 3 acting on V. Are you sure the result does
not assume the action is faithful?

I think so, see for example Aschbacher's Finite Group Theory, Second
Edition, p. 59, 16.1(2). It is the first result that comes up when
searching ' aschbacher "finite group" "rank 3" ' in Google Books.

.

Relevant Pages

  • Re: Orbitals of a transitive action
    ... Rafael Villarroel wrote: ... It seems that a standard exercise is the following: If G acts ... I think so, see for example Aschbacher's Finite Group Theory, Second ... "In this section G is a transitive permutation group on a finite set ...
    (sci.math)
  • Re: Orbitals of a transitive action
    ... It seems that a standard exercise is the following: If G acts ... then G has a symmetric nondiagonal orbit on VxV if ... (a transitive action where the elements of order 2 act trivially), ... then the nondiagonal orbits on VxV are, ...
    (sci.math)