Here is a problem I think I have a solution for, I am just not 100% sure on one of the detail
Let be a group that does not have a subgroup of index 2.
Show that every subgroup of index 3 is normal in
Let , first we will assume that is not trivial (if it is then it is normal).
By cayley's expanded theorem, there exists a homomorphism from to the permutation group on ( is the right coset of ). So this homomorphism is from to . now we will look at , according to cayley's expanded theorem is normal in , but since according to cayley's theorem every group is isomorphic to a subgroup of , then we can say that . Now since we assumed is not trivial and we get that . But the only normal subgroup of in this case is , then we get and thats a contradiction.
The only part I am not sure about is where I use and interchangeably, therfore saying that if then . Can I use this reasoning?
Appreciate your opinions.