By Cayley's theorem, we know that for any finite group $G$, there exists $N \in \mathbb{N}$ such that $G$ is isomorphic to a subgroup of $S_N$, the symmetric group on $N$ letters. Can we prove that for every finite group $G$ there is some symmetric group $S_N$ such that $G$ is isomorphic to a $normal$ subgroup of $S_N$?
[Math] Is every finite group a normal subgroup of a symmetric group
group-theory
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Best Answer
HINT: Try to prove that for $n \ge 5$, $A_n$, the alternating group of $n$ elements is the only proper and nontrivial normal subgroup of $S_n$.
UPDATE: This has to do something with the fact that $A_n$ is simple for $n \ge 5$. After proving this and checking the cases $n \le 4$ we can conclude that a normal subgroup of symmetric group has order $1, 4, \frac{n!}{2}$ or $n!$. Hence...