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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.

Community
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Mikasa
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    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. – Jack Schmidt Jun 18 '12 at 18:31
  • @JackSchmidt: Thanks Jack. Honestly, I did the second one and wanted to give it to my Prof. :-) – Mikasa Jun 18 '12 at 18:39
  • 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 :-) – Jack Schmidt Jun 18 '12 at 18:41

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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 $N=S_n$.

Nicky Hekster
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  • Thanks Nicky for the answer. – Mikasa Jun 18 '12 at 19:16
  • @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. – Nicky Hekster Jun 18 '12 at 21:16
  • @NickyHekster Hi, Nicky, why if $[S_n:A_n]=2$, then $N=A_n or N=S_n$? – xyz Oct 02 '20 at 02:07
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    @xyz In general, if $p$ is a prime, $H \lt G$ with $|G:H|=p$, then for $H \subset K \subset G$, we have $H=K$ or $G=K$. This follows from the fact that indices are multiplicative: $|G:H|=|G:K| \cdot |K:H|$, and the factorization of a prime. – Nicky Hekster Oct 02 '20 at 09:41
  • @NickyHekster Thank you for your kindness. – xyz Oct 03 '20 at 14:26
  • @xyz No problem, you are welcome and ... never stop asking! – Nicky Hekster Oct 03 '20 at 17:12
  • @NickyHekster yes! Thank you! – xyz Oct 03 '20 at 23:53