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I am asked to show there exists a subgroup of $A_n$ that is isomorphic to $S_{n-2}$ for $n \ge 3$. Here are some of my thoughts:

Certainly there exists a homomorphism $f : A_n \rightarrow S_n$---namely, take $f$ to be the inclusion map, $f(\sigma) = \sigma$. Because of this, $f^{-1}(S_n)$ will be a subgroup of $A_n$, and I suspect I can extend $f$ in someway to an isomormphism by defining $\varphi : S_{n-2} \rightarrow f^{-1}(S_n)$

I am not really sure how to move forward, but does this seem to be moving in the right direction? It seems that I want to take all even permutation $S_{n-2}$ and map them to themselves, and then take every odd permutation and "correct" it to become an even permutation, but I am not sure why this is right--or if it is even remotely correct. Can $\varphi$ map an odd permutation to an even permutation and still be an isomorphism? It doesn't seem that it could.

user193319
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  • The setwise stabilizer in $A_n$ of two points is isomorphic to $S_{n-2}$. – Derek Holt Oct 02 '16 at 13:04
  • @DerekHolt I really haven't yet dealt with stabilizers of groups, especially of the alternating group, so I would prefer to deal with the problem without reference to that. – user193319 Oct 02 '16 at 13:06
  • I don't think there is any other way of doing it, because the only subgroups of $A_n$ that are isomorphic to $S_{n-2}$ are the stabilizers of ses of two points. To be expicit, assuming that $A_n$ is permuting the set ${1,2,\ldots,n}$, the subgroup ${ g \in A_n : g(1),g(2) \in {1,2}}$ is isomorphic to $S_{n-2}$. – Derek Holt Oct 02 '16 at 15:56

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If $\sigma\in S_{n-2}$ is even, map it to itself. Otherwise, map it to $\sigma\circ(n-1,n)$.

ajotatxe
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  • And $\sigma \circ (n-1,n)$ is an even permutation, right? So was my suspicion right, that $\varphi$ couldn't be an isomorphism if it mapped odd permutations to even permutations? – user193319 Oct 02 '16 at 13:09