BSMSMSTMSPHD
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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.
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.