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

My proof

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.

SK