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I'm trying to prove that $A_n$ is the only subgroup of index $2$ in $S_n$ by a contradiction argument, which have led me to the equality

$$A_nK=S_n,\tag{$*$}$$

where $K$ is a subgroup of index $2$ in $S_n$, distinct from $ A_n$. Why is $(*)$ a contradiction (if it actually is)?

Shaun
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    We need more details about your proof. It's possible for a finite group to have distinct subgroups of index $2$, so there must be something about your particular construction that leads to the contradiction. – Robert Shore Jun 15 '22 at 21:44
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    If we are allowed to use the fact that $A_n$ is simple for $n \ge 5$ then we know that $A_n$ has no subgroup of index $2$, so this cannot happen, since this would imply $|A_n:A_n \cap K|=2$. – Derek Holt Jun 15 '22 at 21:50

1 Answers1

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This reasoning is possible.

Since every cycle of odd length lies in $K$ (why?) and $(123)(234)=(12)(34)$ it follows that $A_n\leq K$.

kabenyuk
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    Yes that's better than my solution in the comments, because there is no need to assume the simplicity of $A_n$. – Derek Holt Jun 16 '22 at 07:52