Question is to :

Find all normal subgroups of $S_n$ for $n\geq 5$.

What I have done so far is :

We know that $A_n$ is one normal subgroup of $S_n$.

Suppose $H\neq (1)$ is another normal subgroup of $S_n$ then,

$H\cap A_n$ would also be a normal subgroup.

As $H\cap A_n \subseteq A_n$ we see that $H\cap A_n \unlhd A_n$.

But for $n\geq 5$, $A_n$ is a simple group so does not have proper normal subgroups.

Thus, $H\cap A_n =(1)$. So, we should have $A_n\leq H \leq S_n$ with $|A_n|=\frac{n!}{2}$ and $|S_n|=n!$

As there is no number in between $\frac{n!}{2}$ and $n!$ which is divisible by $\frac{n!}{2}$ and divides $n!$ we should end up with the case when $H=A_n$ or $H=S_n$.

I feel thankful and it would be appreciated if someone can proof read this solution

P.S : I have checked for this problem in this form and got two questions (one is tagged as duplicate) the original question do not have any proof for the Question "Find all normal subgroups of $S_n$ for $n\geq 5$". So, please do not tag it as duplicate as the question is not answered in any of the other OP.

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    $\begingroup$ Your proof seems fine up to the point where you conclude from $H\cap A_n=(1)$ that $A_n\leq H$: If $A_n\leq H$ then $H\cap A_n=A_n$, so this does not follow. $\endgroup$ – Servaes Sep 3 '13 at 10:15
  • $\begingroup$ I did not understand. What does not follow :O $\endgroup$ – user87543 Sep 3 '13 at 10:19
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    $\begingroup$ Despite what you say, this is an exact duplicate of math.stackexchange.com/questions/159994 - so I tagging it as duplicate! $\endgroup$ – Derek Holt Sep 3 '13 at 10:21
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    $\begingroup$ @DerekHolt : Another duplicate Question. :( I tried checking all questions but i am not sure how i missed it. I am brand new for this site (using from one month) so do not have much idea. anyways thank you for your reference :) $\endgroup$ – user87543 Sep 3 '13 at 10:24
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    $\begingroup$ Actually it isn't a duplicate. Even if the problem at hand is the same, this one clearly is a solution verfication while the other is not. They are different in nature and, for that reason, I believe they aren't duplicates. $\endgroup$ – Git Gud Sep 3 '13 at 10:48

Your step "$H \cap A_n = 1$ implies $A_n \leq H \leq S_n$" is too fast: you have to rule out the case $H \cap A_n = 1$. This can be done as follows: in general if $H \unlhd G$ with $H \cap G' = 1$, then $H \subset Z(G)$. Since $Z(S_n) = 1$ for $n \geq 3$, you can conclude $H=1$ in your case.
Further, in general if $K$ is a subgroup of $G$ with $index[G:K]=2$, then from $K \leq H\leq G$ it follows that $K=H$ or $H=G$. So the rest of your proof is correct!

  • $\begingroup$ I do not yet understand why $H \cap A_n = 1$ doesnot imply $A_n \leq H \leq S_n$ :( $\endgroup$ – user87543 Sep 3 '13 at 10:26
  • $\begingroup$ I understand that " in general if H⊴G with H∩G′=1, then H⊂Z(G)."In case of $G=S_n$ i understand that $Z(S_n)=1$ so, $H=(1)$ $\endgroup$ – user87543 Sep 3 '13 at 10:30
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    $\begingroup$ From the fact that $A_n$ is simple, you conclude $H \cap A_n=1$ or $A_n \subseteq H$. The first leads to $H=1$, which is of course also a (trivial) subgroup of $S_n$. $\endgroup$ – Nicky Hekster Sep 3 '13 at 11:07
  • $\begingroup$ please, what is $G'$? $\endgroup$ – Danilo Gregorin Oct 24 '17 at 18:27
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    $\begingroup$ It is the commutator subgroup of $G$: $\langle x^{-1}y^{-1}xy: x,y \in G \rangle$. This is the smallest normal subgroup $N$ of $G$, such that $G/N$ is abelian. It is also written as $[G,G]$,for $[x,y]:=x^{-1}y^{-1}xy$. It holds that $S_n'=A_n$. $\endgroup$ – Nicky Hekster Oct 24 '17 at 18:32

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