Orbits of Transitive Group Actions

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Homework Statement


Suppose G is a finite group, which acts transitively on a set X and let H be a normal subgroup of G. Show that the size of the orbits of the action of H on X are of the same size.


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The Attempt at a Solution


I haven't been able to get very far with this. The only fact I've gotten is that |H(x)| divides |G|, but I am not sure that is even relevant in this case.
 
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First think about this example.

Let G=group of all rotations in the plane about the origin.

Let H=some subgroup, e.g. H=\{ R_0, R_{120}, R_{240} \}.

Let X=the unit circle with center at origin.

Let x=x_0= some fixed point p_0 on the unit circle, e.g. p_0=(1,0).

Let y= some other point on the unit circle.

Draw a picture of Hx.

Now, what if you want to find Hy, but for some crazy reason you don't know how to find it directly. You only know how to apply the rotations of H to the point x, not any other point. Geometrically, describe how to find Hy. Use the word "rotate."
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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