Alternating Groups: Even Permutations in Sn for n > 2

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Alternating groups encompass all even permutations in Sn for n > 2, but for n = 2, the only transposition permutation is (w x), which is classified as an odd permutation. Consequently, the order of the alternating group A_2 is calculated as 2!/2, resulting in a trivial group that contains only the identity element. This means that A_2 does not include any permutations of the form (wx), as they are odd. The discussion highlights the unique nature of A_2 and its distinction from larger alternating groups. Ultimately, A_2 consists solely of the identity element.
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Alternating groups apply to all even permutations in Sn for n > 2. Since n = 2 is inclusive, what got me wondering is that for such a case there are only 2 elements in S (say w and x); wouldn't that mean that the only transposition permutation would be (w x), which is an odd permutation?
 
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If you go by the fact that the order of any alternating group is n!/2 then you would have that the order of A_2 is 2!/2=1 and therefore it's just the trivial group consisting of the identity element. Anything in the form of (wx) would be an odd permutation and therefore not in A_2
 
jeffreydk said:
If you go by the fact that the order of any alternating group is n!/2 then you would have that the order of A_2 is 2!/2=1 and therefore it's just the trivial group consisting of the identity element. Anything in the form of (wx) would be an odd permutation and therefore not in A_2

I see...so it would simply imply the identity element...Thanks
 
No problem
 
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