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I am self-studying Group Theory. I know that $A_{n}$ is a normal subgroup of $S_{n}$, but I've realized that I don't know another one, I mean, another subgroup of $S_{n}$ that is a normal subgroup of $S_{n}$. I would like to know if is known all the normal subgroups of $S_{n}$ and how many nonisomorphics of them there are in $S_{n}$, for all $n\geq 4$.

Thanks for your kindly help.

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8  
For $n \ge 5$ it's a classic result that $A_n$ is the only nontrivial normal subgroup of $S_n$. –  Qiaochu Yuan Feb 3 '12 at 22:54
1  
For $n=4$ there is also $\lbrace1,(12)(34),(13)(24),(14)(23)\rbrace$ –  Gerry Myerson Feb 3 '12 at 23:12

1 Answer 1

up vote 3 down vote accepted

Here are two hints:

  1. If $N\leq A_n$, then show that $N\unlhd A_n$.
  2. But $A_n$ is a simple group, for all $n\geq 5$. (This means that if $M\unlhd A_n$, then $M=1$ or $M=A_n$.)

If you're not there yet, here is one more hint:

  1. If $N\not\leq A_n$, then what is $A_nN$? Show that $A_n\cap N\unlhd A_n$.Use (2) again and determine $|N|$.
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The SE-sytem it a bit of a know-it-all. I want a "3" and I get a "1". –  Myself Feb 3 '12 at 23:01
    
My apologies for twice deleting and undeleting this post. –  Myself Feb 3 '12 at 23:18
    
This is an unfortunate gap in the Markdown syntax. The SE sites probably use some standard library for it. –  Dylan Moreland Feb 3 '12 at 23:22

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