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The title is the question. Is $A_n$ characteristic in $S_n$?

If $\phi \in \operatorname{Aut}(S_n)$, Then $[S_n : \phi(A_n)]$ (The index of $\phi(A_n)$) is 2. Maybe the only subgroup of $S_n$ of index 2 is $A_n$?

Thanks in advance.

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    Homework question?2012-04-18
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    $A_n$ is commutator of $S_n$.2012-04-18
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    In fact, since $A_n=[S_n,S_n]$, $A_n$ is *fully invariant*: if $f\colon S_n\to S_n$ is any endomorphism, then $f(A_n)\subseteq A_n$.2012-04-18

2 Answers 2