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If we have $n\geq 5$ and let $N\lhd S_n$ with $N\neq 1$ and consider $N\cap A_n \lhd A_n$.

Now as $A_n$ is simple we have that $N\cap A_n=\{1\}$ or $N\cap A_n=A_n$

Now is $N\cap A_n=A_n$ then $N=A_n$ and we are done.

Else we have that $N\cap A_n=\{1\}$. Now if we have that $|N|\geq 3$ then we have a contradiction as the composition of two odd permutations is even so we assume that $|N|=2$

I am now unsure as how to proceed?

Nicky Hekster
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john smith
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2 Answers2

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In general, if $N \lhd G$ and $N \cap G'=\{1\}$, then $N \subseteq Z(G)$.

So if $N \cap A_n=\{1\}$, then $N \subseteq Z(S_n)=\{1\}$.

By the way, your step $N \cap A_n=A_n$, then $N=A_n$ is not correct. You can conclude $A_n \subseteq N$. Now use that index$[S_n:A_n]=2$ and conclude that either $A_n=N$ or $N=S_n$.

Nicky Hekster
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Hint: If $|N|=2$, then $N= \langle \tau \rangle$ for some $\tau \in S_n$ satisfying $\tau^2=\mathrm{Id}$. Now, $N$ is a normal subgroup of $S_n$ if and only if $\tau \in Z(S_n)$. What can you say about $Z(S_n)$?

Seirios
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