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Induction on $n$:

If $n=1$, then $A_{n+2}=A_{3} \simeq \mathbb{Z}_3$ has subgroup $1 \simeq S_n = S_1$. Assume for some $n>1$, $S_n \simeq H$ for some $H \leq A_{n+2}$. I want to show that $S_{n+1} \simeq H'$ for some $H'\leq A_{n+3}$.

Note that $H \simeq S_n \leq S_{n+1}$, and by inductive hypothesis $H \leq A_{n+2} \leq A_{n+3}$. I guess if I can show that $S_{n+1} \simeq A_{n+2}$ then it will complete the proof, but I don't even know if that is a true statement.


marked as duplicate by user26857 abstract-algebra Jun 4 '17 at 20:15

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  • $\begingroup$ A subgroup of $A_{n+2}$ may contain odd permutations? $\endgroup$ – Sid Caroline Jun 2 '17 at 0:03
  • 1
    $\begingroup$ the "result" you are using is clearly false, just compare the orders of $A_{n+2}$ and $S_{n+1}$ $\endgroup$ – Jorge Fernández Hidalgo Jun 2 '17 at 0:07

Send $\sigma \mapsto \sigma$ if $\sigma$ is an even permutation. Let $\tau = (n+1 \, \, \, n+2)$ and send $\sigma \mapsto \sigma \tau $ for odd $\sigma$. This injects $S_n \to A_{n+2}$ and is a homorphism.


$A_{n+2}$ is the group of all even permutations of $\{1,2,\dots,n+2\}.$ Let $H$ be the subgroup of $A_{n+2}$ consisting of all permutations that leave the sets $\{1,2,\dots,n\}$ and $\{n+1,n+2\}$ fixed. Show that the obvious map $H\to S_n$ is an isomorphism.


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