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I want show that

If $ n \geq 3$ , A subgroup $H$ of $S_n$ having index 2 is $A_n$.

Say $H$ is such a subgroup. Since $A_n$ is a normal subgroup of $S_n$,$H$ normalizes $A_n$ .

By 2nd isomorphism theorem , $HA_n$ is a group such that $$ HA_n / A_n \mbox{ is isomorphic to } H/ H \cap A_n$$

What shall i do?

Seongqjini
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1 Answers1

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For $n>4$: a subgroup of index $2$ is normal. Show any (non-trivial) normal subgroup $H$ (non-trivially) intersects $A_n$. Then $H \cap A_n$ is normal in $A_n$ and by simplicity of $A_n$ the claim follows.

For $n=4$ you can look at the subgroups directly.

GiantTortoise1729
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