## Transitive group actions induce transitive actions on the orbits of the action of a subgroup

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 .

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## Comments

An easy way to show in 2: Since all orbits of on have the same cardinality, we see that (using the second isomorphism theorem). Also since is transitive, (using that subgroup index is multiplicative). Substituting this yields that .