This exercise is from Lang's Algebra.
Let $n\geq 3$, and let $H$ be a subgroup of the alternating group $A_n$. Suppose that $H$ has index $n$ in $A_n$. Show that the action of $A_n$ on $A_n/H$ by translation induces an isomorphism between $A_n$ and the alternating group $Alt(A_n/H)$ on $A_n/H$.
The action by left-translation gives a group homomorphism $\phi$ from $A_n$ to $Sym(A_n/H)$. If we can prove that every member of $Alt(A_n/H)$ is in the image of $\phi$, then we immediately obtain that $\phi$ is an isomorphism: if the image of $\phi$ contains $Alt(A_n/H)$ then it is exactly $Alt(A_n/H)$ by cardinality and then $\phi$ is injective, being a surjective map between two sets of the same cardinality. I don't manage to prove this fact, though. I have tried considering the cycle structure of an arbitrary element $\sigma\in Alt(A_n/H)$ but couldn't find an element $a\in A_n$ such that $\sigma$ is left-translation by $a$.
When $n\geq 3$ and $n\neq 4$, $A_n$ is simple. The kernel of $\phi$ is included in $H<A_n$, so that by simplicity $\ker(\phi)=\{e\}$. If $g\in im(\phi)$, then $g^2\in im(\phi)\cap Alt(A_n/H)$ since the sign of a square is $1$. It follows that $|im(\phi)\cap Alt(A_n/H)|\geq \frac1{2}|im(\phi)|$. Rearranging and using the fact that $|A_n|=|Alt(A_n/H)|$, we obtain $[Alt(A_n/H) : im(\phi)\cap Alt(A_n/H)]\leq 2$. The index cannot be two, for a subgroup of index $2$ is normal and $Alt(A_n/H)$ is simple. Thus the index is one and $im(\phi)=Alt(A_n/H)$.
It remains to treat the case $n=4$. In this case $H$ is a cyclic subgroup of order $3$ of $A_4$, and the order of $\ker(\phi)$ is $1$ or $3$. $A_4$ does not have a normal subgroup of order $3$ (there are $4$ Sylow $3$-subgroups and all of them are conjugated). Thus $\phi$ is injective. Finally $|Alt(A_4/H)\cap im(\phi)|\geq 6$ but $A_4$ does not have a subgroup of order $6$ (this subgroup would be abelian, thus having a normal subgroup of order $3$ but none exists in $A_4$), so that $im(\phi)$ is again $Alt(A_4/H)$.