For any element x in X, let [x] denote the orbit of x under G. Since G acts transitively on X, it follows that [x] = X for any x in X. So the orbit equation implies:

|X| = |[x]| = |G| / |G_x| for any x in X (where G_x is the stabilizer at x).

Let [x]^H denote the orbit of x under H. Then the orbit equation implies:

|[x]^H| = |H| / |H_x| = |H| / |H int G_x| for any x in X (note that "int" denotes set-theoretic "intersection.")

Since H is normal in G, the set HG_x is a subgroup of G.

Let n = [G : HG_x]. Then:

n = |G| / |HG_x| = |G| |H int G_x| / (|H| |G_x|) = |X| / |[x]^H|.

So n |[x]^H| = |X|. Consequently, if |[x]^H| = |[y]^H| for every x and y in X, then the result follows.

Define an injection f: [x]^H --> [y]^H as follows:

f(hx) = g^-1hg.y, where g is chosen from G so that x = gy. Note that such a g exists since G acts transitively. Also note that the image of f lies in [y]^H since H is normal in G.

We claim that f is a well-defined injection:

Let h₁ and h₂ be elements of H. Then

h₁x = h₂x if and only if

h₁(gy) = h₂(gy) if and only if

g^-1h₁gy = g^-1h₂gy.

So f is a well-defined injection and hence |[x]^H| < |[y]^H|.

By interchanging the roles of x and y, it follows that |[y]^H| < |[x]^H|.

Thus, |[x]^H| = |[y]^H| and this completes the proof.