I have read somewhere (do know where) the following statement.
Alternating group $\operatorname{Alt}(n)$, $n>4$ has no subgroup of index less than $n$.
I want to prove it.
If there is a subgroup $H$ of $\operatorname{Alt}(n)$, $n>4$, of inex $d<n$ then the action of $\operatorname{Alt}(n)$ on the set of all left (or right) cosets of $H$ in $\operatorname{Alt}(n)$ gives rise to a nontrivial epimorphism from $\operatorname{Alt}(n)$ to $\operatorname{Alt}(d)$ contradicting the fact that $\operatorname{Alt}(n)$ is simple.
Is this proof correct?? If not, Please suggest a proof.