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#1 2009-11-10 02:04:59
- Pythagorean12
- Guest
Groups
.
I cant really get anywhere with this. Could someone give me a hint please?
Thanks
#2 2009-11-10 07:58:27
- JaneFairfax
- Member
- Registered: 2007-02-23
- Posts: 6,868
Re: Groups
Suppose there are odd permutations, .
Consider the list
. Each is an even permutation, being a product of two odd permutations. They are also distinct, for by the cancellation law we have . Hence there are at least even permutations.Conversely, if
is an even permutation, then is an odd permutation and so is one of the permutations . Hence is one of the permutations .From this, we conclude that there are exactly
even permutations, i.e. there are as many odd permutations as even ones.Last edited by JaneFairfax (2009-11-10 07:59:03)
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#3 2009-11-10 13:41:33
- Ricky
- Moderator
- Registered: 2005-12-04
- Posts: 3,791
Re: Groups
An alternate proof is to let g be an odd permutation, and note that {g, A_n} generates S_n. Now we have by the diamond (second) isomorphism theorem:
The left side is 2, giving the right side is 2, and you're done.
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
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