# Quotient groups of a finite symmetric group

What are the quotient groups of a finite symmetric group $S_n$? Can we classify them?

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See this (closed) MO post: mathoverflow.net/questions/91009/… –  Benjamin Dickman Nov 11 '12 at 22:08

For $n \neq 4$, the subgroup $A_n$ is simple. Thus its image in any quotient of $S_n$ is either trivial or all of $A_n$. In the former case the only quotient groups are the trivial group and $S_n/A_n \cong C_2$. In the latter case the only quotient group is $S_n$ (except for $n = 1, 2$ but nothing new happens here).
The only remaining case is $S_4$, and here one can just write down all normal subgroups by hand (e.g. by inspecting the character table, or just casework on cycle types). In this case the Klein four group $V_4$ is also a normal subgroup of $S_4$ so there is an exceptional quotient $S_4/V_4 \cong S_3$, but I think this is it. Algebraically this exceptional quotient is responsible for the existence of the resolvent cubic.
"In the latter case the only quotient group is $S_n$" Could you explain this? –  Makoto Kato Nov 11 '12 at 22:35
@Makoto: if $A_n$ injects into a quotient group of $S_n$, then the corresponding kernel has order at most $2$, and you can easily verify that $S_n$ has no normal subgroups of order at most $2$ except for very small $n$. –  Qiaochu Yuan Nov 12 '12 at 0:55
Thanks. Let $K$ be the kernel. By the assumption $K \cap A_n = 1$, Hence $|KA_n| = |K||A_n|$. Hence $|K| = 2$ if $K \neq 1$. As for the fact that there is no normal subgroups of order 2, I found the following proof. planetmath.org/NormalSubgroupsOfTheSymmetricGroups.html –  Makoto Kato Nov 12 '12 at 2:01