A simple group such that $[G:H]=n$ can be embedded into $A_n$

Question

Let $G$ be a finite simple group and $H$ be a proper subgroup of $G$ such that $|G:H|=n$. Then, how do I prove that $G$ can be embeded into $A_n$?

I can prove that $G$ can be embedded into $S_n$ by using the group action by left multiplication, but don't know how to prove it for $A_n$. Thank you in advance.

Answer 1

Wlog, assume that $|G|\geq 3$.

Let $f:G\rightarrow S_n$ be a permutation representation induced by $G$ acting on the left cosets of $H$. Since $G$ is simple and its kernel is contained in $H$ and $H$ is proper, its kernel must be trivial. Hence $f$ is injective.

Now note that $A_n$ is normal in $S_n$, hence $f^{-1}(A_n)$ is normal in $G$. Suppose $f^{-1}(A_n)=1$. Then, $f(G)\cap A_n$ is trivial, but this is impossible since then $|f(G)A_n|\geq 3|A_n|>|S_n|$. Since $G$ is simple, it must be the case $f^{-1}(A_n)=G$.

Answer 2

If the image is not in $A_n$, then the image must have a non-trivial normal subgroup, so apply the correspondence thm, we get a normal subgroup in $G$, which is a contradiction.