TITLE: Is there a neat way to see that $C_3$ is not a quotient of $S_n$?
QUESTION [5 upvotes]: Well, we know all normal subgroups of $S_n$, including the interesting case $n=4$. Using this one can conclude that the cyclic group of order 3 is not a quotient of $S_n$. Is there are more direct way to see this?
$C_3$ is simple, so if there were a surjective homomorphism from $S_n$ onto $C_3$, its kernel would be a maximal subgroup of index 3. The alternating group is also maximal as it has index 2. Can we combine these two fact together to get a contradiction?

REPLY [2 votes]: Suppose $N \unlhd S_n$, and $|S_n:N|=3$. Then $n \geq 3$ and $S_n/N$ is abelian. Hence, $A_n=[S_n,S_n] \subseteq N$, which is absurd because $|S_n:A_n|=2$.