ok, we have a map from from the orbit of a (that is all possible images g.a) to the set of left cosets of .

that is, given b in the set A, such that b is in the equivalence class of a, then b~a, so b = g.a (g permutes A in such a way as to take a to b).

so we send b (=g.a) to the coset of that contains g, which is .

now, given an arbitrary coset of , say , we know that x.a is some element of the orbit of a,

and our mapping takes x.a to , so x.a is a pre-image of (pretty much because that's the way wedefinedthe orbit of a).

now, suppose two elements of [a], let's call them b and c, get mapped to the same coset. we can write b = g.a, and c = h.a, so we have:

. this means that .

but what is ? it's the stabilizer of a; that is, the set of g in G such that g.a = a. so, .

therefore:

, that is, b = c.

perhaps this will be more clear if we see it "in action" (excuse the terrible pun). let's have A = {1,2,3,4}, and let G be the group D4 (where we are considering A as the set of vertices (corners) of the square).

now, just from looking at the rotations alone, we can see that the orbit of 1, is all of A. let's write the elements of D4 in permutation form:

{e, (1 2 3 4), (1 3)(2 4), (1 4 3 2), (1 2)(3 4), (1 4)(2 3), (1 3), (2 4)}.

what is Stab(1)?

well clearly e stabilizes (or "fixes") 1, and the only other element to do so is (2 4).

so Stab(1) = {e, (2 4)}. the left cosets are as follows:

Stab(1) = {e, (2 4)}

(1 2 3 4)Stab(1) = {(1 2 3 4), (1 2)(3 4)}

(1 3)(2 4)Stab(1) = {(1 3)(2 4), (1 3)}

(1 4 3 2)Stab(1) = {(1 4 3 2), (1 4)(2 3)}, and our bijection is:

1<--->Stab(1)

2<--->(1 2 3 4)Stab(1) (every element of (1 2 3 4)Stab(1) takes 1-->2)

3<--->(1 3)(2 4)Stab(1) (every element of (1 3)(2 4)Stab(1) takes 1-->3)

4<--->(1 4 3 2)Stab(1) (every element of (1 4 3 2)Stab(1) takes 1-->4).

and indeed we see that |[1]| = |A| = [D4:Stab(1)] = 4.

you can look at it this way: if g,h take a to different elements of A, then give rise to different cosets of .