Suppose acts transitively on and let be normal. Let be the distinct orbits of on .
- Prove that permutes the sets in the sense that for each and each there exists a such that . Prove that is transitive on . Deduce that all orbits of on have the same cardinality.
- Prove that if then and that . [Hint: Note that and use the Second Isomorphism Theorem.]
- Let and . Now is an arbitrary orbit of on . Moreover, for some . Then , since is normal. Thus permutes the orbits .
Now let be arbitrary; since the action of on is transitive, there exists such that . Then ; thus the action of on the orbits is transitive.
Let be arbitrary, and let such that . Because is normal, for every there exists a unique such that . We have a mapping given by . Because , this mapping is bijective, and . Because the action of is transitive on the orbits, all orbits have the same cardinality.
- Let . We can restrict the action of on the orbits of to ; with this action we have . Clearly ; thus . Considering now the (transitive) action of on the orbits of , we have .
We claim that . Proof of claim: Suppose . Now , so that in particular for some . Then , so that . Thus . Now suppose . Then for some and . Now , so that . Thus .