How to show that the orbits of the action of Gs on S \ {s} have lengths that are equal in pairs. Question:Let G be a group of odd order acting transitively on a set S. Fix s ∈ S. Show that the orbits of the action of Gs on S \ {s} have lengths that are equal in pairs. 
My idea: set a point a$\in$S \ {s}, then the order of the stabilizer of a in Gs equals to another one b$\in$S \ {s}'s order of its stabilizer. But how to use the odd  order?
 A: For every orbit of $G_s$ on $S \setminus \{ s \}$, there is a so-called paired orbit, which has the same length.
It's easiest to understand this by considering the orbits of $G$ in the induced action of $G$ on the set $S^2 = S \times S$. There is a bijection between these orbits and the orbits of $G_s$ on $S$, where an orbit $O$ of $G$ on $S^2$ corresponds to the orbit $\{ t \in S : (s,t) \in O \}$ of $G_s$ on $S$.
The paired orbit of the orbit of $(a,b) \in S^2$ is defined to be the orbit of $(b,a)$, which has the same size. Then the paired orbit of an orbit of $G_s$ on $S$ is the orbit corresponding to the paired orbit of the corresponding orbit of $G$ on $S^2$. (Sorry that sounds complicated, but it's not really!)
In general, the paired orbit might be the same orbit (in which case it is called self-paired) and the orbit of $(a,b) \in S^2$ is self-paired if and only if it contains $(b,a)$, which is the case if and only if there is an element of $G$ interchanging $a$ and $b$, which is not possible when $a \ne b$ and $|G|$ is odd.
A: Let $G$ act on a set $S$.  The orbits of the induced action of $G$ on $S \times S$ are called the orbitals of $G$.  Fix $s \in S$. It can be shown that if the action of $G$ on $S$ is transitive, then there is a 1-1 correspondence between the orbits of the action of $G_s$ on $S$ and the orbitals of $G$.  The orbital $\Delta:=(s,t)^G$ corresponds to the $G_s$-orbit $\{t: (s,t) \in \Delta \}$.  One way to visualize this is using the tabular representation of relations on $S$: each orbital corresponds to a union of permutation matrices, and the columns corresponding to the nonzero coordinates in row $s$ of an orbital form the corresponding $G_s$-orbit.  
An orbital $(s,t)^G$ is said to be self-paired if it equals its transpose $(t,s)^G$.  If this is the case, then some element of $G$ interchanges $s$ and $t$. This element contains the 2-cycle permutation $(s,t)$ and hence has even order.  Thus $G$ also has even order.  Conversely, if $G$ has even order, then by Cauchy's theorem it has an element of order 2, which contains a 2-cycle $(x,y)$. Thus, there exists an orbital $(x,y)^G$ that equals its transpose.  We have shown that $G$ has even order if and only if some orbital is self-paired.  Take the contrapositive, and we get that $G$ has odd order if and only if every orbital is not equal to its transpose.  In particular, for each orbital $\Delta$ of $G$, there is another orbital $\Delta^*$ which is disjoint from $\Delta$ but which has the same order as $\Delta$.  Due to the 1-1 correspondence, these two orbitals give rise to two distinct $G_s$-orbits of the same order.  For example, if $\Delta$ is a union of $k$ permutation matrices, then so is $\Delta^*$, and the two corresponding $G_s$-orbits have $k$ elements each.
