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 $S_N$?


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...

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    $\begingroup$ From the insight of this answer and the other answer in the thread, it is seen that the cyclic group of order four is the smallest group which is not a normal subgroup of any $S_n$. $\endgroup$ – Jeppe Stig Nielsen Jun 28 '16 at 8:25
  • $\begingroup$ @JeppeStigNielsen Nice observation. Also it's interesting how $S_4$ is the only group "acting" differently in that regard. $\endgroup$ – Stefan4024 Jun 28 '16 at 8:52

In general (i.e. for $N \neq 4$), the only normal subgroups of $S_N$ are $S_N$ itself, $A_N$, and $1.$ Therefore no, because most $G$ will not map to one of these. ($S_4$ has an additional normal subgroup, the Klein $4$-group hiding in it.)


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