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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]

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    And personally, I suggest you pick the problem from the first sentence, since $A_n \lhd S_n$.2011-05-30

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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?