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, ...(adsbygoogle = window.adsbygoogle || []).push({});

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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# Homework Help: Abstract Algebra Problem (Group Isomorphisms)

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