Q:

Number of Orbits on ordered and unordered k-sets.

I do not understand several assertions in the proof below:
Let G be a permutation group on a set $\Omega$, and let $k$ be an integer s.t. $k<|\Omega|$. Then the following are equivalent:
(i) for all $\Delta$ with $|\Delta|=k$ the group $G_{(\Delta)}^{\Delta}=S_\Delta$, (where $G_{(\Delta)}^{\Delta}$ is the permutation group induced by the set-wise stabiliser of $\Delta$ on $\Delta$.)
(ii) the number of orbits of G as a permutation group  on the ordered subsets of k points of $\Omega$ is equal to the number of orbits of G as a permutation group on the unordered subsets of k points of $\Omega$.
Proof. The points of $\Delta$ form k! ordered sets of k points. Two of these ordered sets lie in the same orbit of G precisely when there exists an element of $G_{(\Delta)}^{\Delta}$mapping one onto the other. Consequently the number of different orbits of G in which the ordered sets of k points of $S_\Delta$ lie, is equal to the index of $G_{(\Delta)}^{\Delta}$ in $S_\Delta$. The result now follows.
The first two sentences are obvious, but why does the number of orbits equal the index in $S_\Delta$ ? Do the cosets also give orbits? And then how does the equivalence follow?
I have tried to use the orbit counting lemma, and simply to think about the structure, but something is not coming together.

A:

Fix $\Delta$ a set of $k$ points, and let $X$ be the set of $k!$ orderings of the elements of $\Delta$.  It is easy to check that the action of $G_{(\Delta)}^\Delta$ on $X$  is semiregular. (That is, all the point-stabilisers are trivial. Note that the points here are elements of $X$, not the original elements of $\Omega$ or $\Delta$.) It follows that all the orbits of $G_{(\Delta)}^\Delta$ have size $|G_{(\Delta)}^\Delta|$ and thus the number of orbits is $k!/|G_{(\Delta)}^\Delta|=|S_\Delta:G_{(\Delta)}^\Delta|$ .

