I'm trying to prove that if $G$ is a simple group and $H$ is a proper subgroup of $G$ of index $n$ then $G$ is isomorphic to a subgroup of $S_n$.
To do this, I've been trying to find a homomorphism between $G$ and permutations on the set of right cosets of $H$, because this would give me a subgroup of $S_n$, but I've failed to find one with a trivial kernel(or at least one I can prove has a trivial kernel). Nor am I sure how to use the condition that $G$ is simple. I'm not even convinced this is the right approach, but I can't think of another way to get permutations on $n$ things.
Thanks!
[Edited the title]
Define $\phi_g(hH) = (gh)H$.
Show that this is independent of choice of h - if $h'\in hH$, then $(gh')H = (gh)H$.
Show that $\phi_{gg'}=\phi_{g}\circ\phi_{g'}$ and $\phi_e=id$, where $e\in G$ is the identity, and $id$ is the identity function on the set of left cosets of $H$.
Thus $g\rightarrow \phi_g$ is a homomorphism from $G$ to the set of permutations of the left cosets. There are $n$ left co-sets, to we have a homomorphism $G\rightarrow S_n$.
What can be said about the kernel of the homomorphism, given that $G$ is simple?