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Here is my reasoning so far.

Let $H \leq A_{n+1}$, and suppose for a contradiction $H$ is isomorphic to $S_n$. If every permutation of $H$ fixes $n+1$, then $H \leq S_n$, but every permutation of $H$ is even, so $H \neq S_n$. Hence there must exist some permutation $h \in H$ that moves $n+1$. But then I'm kind of stuck. I tried to come up with an isomorphism from $H$ to $S_n$ and find a contradiction, but there is too little information available to gain any ground.

Sid Caroline
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  • The link has no solution – Asinomás Jun 02 '17 at 08:08
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    One elementary way to do this, at least for $ n \ge 5$ , is to note that the centralizer in $S_n$ of $(1,2)$ has order $2(n−2)!$, which is larger than the centralizer in $A_{n+1}$ of any element of order $2$ Of course it's easy for $n$ even, by Lagrange's Theorem. – Derek Holt Jun 02 '17 at 09:49
  • @JorgeFernándezHidalgo This is not true, see the "Proposition" in Arturo's answer. – Dietrich Burde Mar 31 '19 at 19:31
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    @DietrichBurde I don't detect any proof of the "proposition" in Arturo's answer. – YCor Mar 31 '19 at 20:35
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    @DerekHolt this is a nice approach (but the inequality requires some proof) – YCor Mar 31 '19 at 20:36

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