Abstract Algebra Problem (Group Isomorphisms)

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Hi. My latest question concerns the following. I must prove that the alternating group A_n contains a subgroup that is isomorphic to the symmetric group S_{n-2} for n = 3, 4, ...

So far, here's what I have (not much). The cases for n = 3 and n = 4 are elementary, since the group lattices are very easy to visualize. But, once n = 5, this becomes quickly unwieldy ( A_5 has 60 elements.)

I also know, by Lagrange, that this subgroup must divide the order of A_n. Since the order of S_{n-2} is (n-2)! and the order of A_n is \frac{n!}{2} I took the ratio of these and got \frac{n(n-1)}{2} which is always a natural number for n = 2, 3, ... But that just shows that the isomorphism is always possible in terms of the orders of the groups involved. It doesn't guarantee that such a subgroup exists.

The other thing I know is that A_n is a non-abelian simple group for n > 4. This means that it contains no proper normal subgroups. What this could possibly do for me, I'm not sure. But is seems to eliminate the isomorphism theorems from my arsenal, since they require subgroups to be normal.

Any suggestions for an approach here? Thanks.
 
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How would you change the elements of Sn-2 into elements of Sn such that they're all even? Answer the question in two parts: a) how would you "change" an even element of Sn-2 into an even element of Sn, and b) how would you change an odd element of Sn-2 into an even element of Sn? Start with a reasonable guess, and try to prove that this procedure for changing elements of one group into elements of another is an injective homomorphism. If it doesn't work out, tweak your guess a little until you find one.
 
Take a set of n elements. Sn-2 can act on n-2 of these as you'd expect. For the other two, if you can think of something to do with them that ensures that for each element in Sn-2, the total associated permutation of all n elements is even, then you'll be able to associate each element of Sn-2 with an element of An. Can you do this in such a way that the result is an injective homomorphism, and so an isomorphism between Sn-2 and its image in An?
 
Thanks to both of you, I think I've got it.

I send the even elements to themselves and I multiply the odd elements by the transposition made up of the elements n and n-1. This guarantees that each permutation in the image is even. I checked and confirmed that this is an injective homomorphism.

Thanks!
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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