I am trying to understand Conway's theorem on the number of orbits on the set of all ordered cycles in a $d$-valent graph. I quote it from Cycles in graphs and groups by Kantor.
Theorem $1$ (Conway) Let $G$ be a group of automorphisms of a $d$-valent graph $\Gamma (d \ge 2)$. Assume that $G$ acts transitively on the set of all ordered pairs of adjacent vertices. Let $X$ be the set of all (ordered) cycles in $\Gamma$ each of which is also a cycle occurring in some element of $G$. Then $G$ has exactly $d − 1$ orbits in its action on $X$.
I would like to write down my understanding up to the current level below.
I understand that a $d$-valent graph is a $d$-regular graph. $G$ acting transitively on the set of all ordered pairs of adjacent vertices means $G$ is acting on all the edges of $\Gamma$ in a way that there is only one orbit. This should be an obvious implication of the fact that $G = Aut (\Gamma)$. Now $X$ is constructed with all ordered cycles appearing in any all elements of $G$.
So, in this scenario $G$ acts on the following sets.
- Vertices of $\Gamma$
- Edges of $\Gamma$
- All cycles of $G$
This is where I am confused. If a permutation group acts on multiple sets, should not the cardinality of these sets be equal to each other? If the number of vertices in a $d$-regular (valent) graph $\Gamma$ is $n$, the number of its edges is $\frac{n d}{2}$. So, the number of elements in these three sets are not equal to each other.
So, how come $G$ is acting on all of them?