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.
The key argument should be G/K is isomorphic to a subgroup of . Thus, |G/K|=[G:K] must divide |S_3|=6 by Lagrange's theorem. The possible choice of |G/K| is 1,2,3, and 6. Since G has no subgroup of index 2 by hypothesis and (verify this), the only possible choice of |G/K| is 3 (verify this). Since [G:K]=[G:H][H:K], we conclude that H=K. Thus, H is normal in G.