There is a famous Theorem telling that:

For $n≥5$, $A_n$ is the only proper nontrivial normal subgroup of $S_n$.

For the proof, we firstly start with assuming a subgroup of $S_n$ which $1≠N⊲S_n$. We proceed until at the last part of proof's body, we assume $N∩A_n=\{1\}$. This assumption should be meet a contradiction with normality of $N$ in $S_n$. There; we get $N=\{1,\pi $} in which $\pi$ is an odd permutation of order $2$. Now for meeting desire inconsistency, I have two approaches:

(a) Since every normal subgroup, having two elements, lies in the center of $G$ so, our $N⊆ Z(S_n)=\{1\}$ for $n≥5$ and then $N=\{1\}$. enter image description here

(b) Clearly, $1≠N$ acting on set $\Omega=\{1,2,...,n\}$ is intransitive wherein $|\Omega|≥5$ and according to the following Proposition $S_n$ would be imprimitive. enter image description here

Proposition 7.1: If the transitive group $G$ contains an intransitive normal subgroup different from $1$, then $G$ is imprimitive (Finite Permutation Groups by H.Wielandt).

May I ask if the second approach is valid? I am fond of knowing new approach if exists. Thanks.

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    $\begingroup$ Yes it is valid, but by the time you know |N|=2, you know N is central. Perhaps use the primitivity result earlier to conclude that N is transitive, and so N does intersect An. $\endgroup$ – Jack Schmidt Jun 18 '12 at 18:31
  • $\begingroup$ @JackSchmidt: Thanks Jack. Honestly, I did the second one and wanted to give it to my Prof. :-) $\endgroup$ – mrs Jun 18 '12 at 18:39
  • $\begingroup$ Make sure you can easily prove the result in Wielandt (it is easy, orbits of a normal subgroup are blocks), and it sounds fine. (a) is even easier though :-) $\endgroup$ – Jack Schmidt Jun 18 '12 at 18:41

You are almost there. Try to prove that $Z(S_n)= 1$ for all $n \geq 3$. Then if $N$ is non-trivial and normal, you assume $N \cap A_n = 1$. This implies $N \subseteq Z(S_n)$. Why? Because in general, if $N \unlhd G$ and $N \cap [G,G] = 1$ then $N \subseteq Z(G)$.
We conclude that the normal subgroup $N \cap A_n \neq 1$. At this point I assume that you know that $A_n$ is a simple group for $n \geq 5$. Hence $N \cap A_n = A_n$, so $A_n \subseteq N \subseteq S_n$. Since $index[S_n:A_n] = 2$, it follows that $N=A_n$ or $N=S_n$.

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  • $\begingroup$ Thanks Nicky for the answer. $\endgroup$ – mrs Jun 18 '12 at 19:16
  • $\begingroup$ @Babak, you are welcome! Your question got me thinking about the general case: can one classify all finite groups $G$ possessing a single proper non-trivial normal subgroup $N$? Here $N$ must be characteristic and characteristically simple and hence a direct product of isomorphic simple groups. Can more be said here about the structure of $N$ and $G$? Have to think about this. $\endgroup$ – Nicky Hekster Jun 18 '12 at 21:16

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