Abstract Algebra Problem (Group Isomorphisms)

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Homework Help Overview

The discussion revolves around proving that the alternating group A_n contains a subgroup isomorphic to the symmetric group S_{n-2} for n = 3, 4, and beyond. Participants explore the properties of these groups and the implications of their orders.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to understand the relationship between the orders of A_n and S_{n-2}, noting that while the order condition is satisfied, it does not guarantee the existence of the subgroup. Participants question how to map elements of S_{n-2} to A_n while ensuring the resulting permutations are even.

Discussion Status

Participants are actively engaging with the problem, with some suggesting methods to construct an injective homomorphism from S_{n-2} to A_n. The original poster expresses progress in their understanding and has formulated a potential solution, although the discussion remains open for further exploration.

Contextual Notes

There is an emphasis on the challenge posed by the non-abelian nature of A_n for n > 4, which complicates the use of certain group theory theorems. The discussion also reflects on the need for careful consideration of even and odd permutations in the context of subgroup isomorphisms.

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Hi. My latest question concerns the following. I must prove that the alternating group [tex]A_n[/tex] contains a subgroup that is isomorphic to the symmetric group [tex]S_{n-2}[/tex] 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 ( [tex]A_5[/tex] has 60 elements.)

I also know, by Lagrange, that this subgroup must divide the order of [tex]A_n[/tex]. Since the order of [tex]S_{n-2}[/tex] is [tex](n-2)![/tex] and the order of [tex]A_n[/tex] is [tex]\frac{n!}{2}[/tex] I took the ratio of these and got [tex]\frac{n(n-1)}{2}[/tex] 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 [tex]A_n[/tex] 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!
 

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