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