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Question is to prove that :

$A_n$ contains an isomorphic copy of $S_{n-2}$

I have no idea from where to start.

I do not even believe this can be true (expect for cardinality conditions).

But as this is an exercise.it has to be true.

please provide some hints/suggestions to attack this problem.

Thank you :)

1 Answers1

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Hint: Suppose $\;S_{n-2}\;$ acts on $\;\{1,2,...,n-2\}\;$ and $\;A_n\;$ on $\;\{1,2,..,n\}\;$ , then check the map

$$S_{n-2}\to A_n\;,\;\;\sigma\mapsto \begin{cases}\sigma\cdot(n-1\;\;n)&,\;\;if\;\;\sigma\notin A_{n-2}\\\sigma&,\;\;if\;\;\sigma\in A_{n-2}\end{cases}$$

with $\,\sigma\in S_{n-2}\;$ any permutation and $\,(n-1\;\;n)\;$ is the trasnposition.

DonAntonio
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  • I see this map is an injection. So, we have (assuming it is a homomorphism) $S_{n-2}\cong H\leq A_n$... Am i correct?? –  Sep 07 '13 at 11:19
  • Yes, indeed...and it is a homomorphism. – DonAntonio Sep 07 '13 at 11:22
  • yes, yes, I have now checked it, it is a homomorphism... I wonder "all of a sudden " how can i think of such a strange map between $S_{n-2}$ and $A_n$... could you please specify if there is something more to think of such map.. or it is just like that... –  Sep 07 '13 at 11:25
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    I think there's all there is to it: we need a difference of two between $,n-2,,,n;$ in order to be able to stick that permutation there to make the map work (i.e., to "turn" odd permutations in $;S_{n-2};$ into even ones in $;A_n;$). That's all. – DonAntonio Sep 07 '13 at 11:33
  • ok,ok. I understood :) –  Sep 07 '13 at 11:35
  • So, all we are trying to do is Convert (give a map) an "odd permutation" in $S_{n-2}$ to "even permutation" in $A_n$ so that every element in $S_{n-2}$ has image in $A_n$ and make sure it is injective :) –  Sep 07 '13 at 11:41