Originally Posted by

**skyking** Here is a problem I think I have a solution for, I am just not 100% sure on one of the detail

Let $\displaystyle G$ be a group that does not have a subgroup of index 2.

Show that every subgroup of index 3 is normal in $\displaystyle G$

My proof

Let $\displaystyle H\leq G, [G:H]=3$, first we will assume that $\displaystyle H$ is not trivial (if it is then it is normal).

By cayley's expanded theorem, there exists a homomorphism from $\displaystyle G$ to the permutation group on $\displaystyle S$ ($\displaystyle S$ is the right coset of $\displaystyle H$). So this homomorphism $\displaystyle h$ is from $\displaystyle G$ to $\displaystyle S_{3}$. now we will look at $\displaystyle K=kerh$, according to cayley's expanded theorem $\displaystyle K$ is normal in $\displaystyle G$, but since according to cayley's theorem every group is isomorphic to a subgroup of $\displaystyle S_{n}$, then we can say that $\displaystyle K\lhd S_{n}$. Now since we assumed $\displaystyle H$ is not trivial and $\displaystyle [G:H]=3$ we get that $\displaystyle n>5$. But the only normal subgroup of $\displaystyle S_{n}$ in this case is $\displaystyle A_{n}$, then we get $\displaystyle [G:K]=2$ and thats a contradiction.

The only part I am not sure about is where I use $\displaystyle S_{n}$ and $\displaystyle G$ interchangeably, therfore saying that if $\displaystyle K\lhd G$ then $\displaystyle K=A_{n}$. Can I use this reasoning?

Appreciate your opinions.

SK