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Mathematics 18 Online
OpenStudy (anonymous):

Prove that if you compose two even permutations, you get an even permutation.

OpenStudy (watchmath):

even+even=even

OpenStudy (anonymous):

permutation formula nPr = n!/(n-r)! where n is total object taken, and r= objects taken at a time proof: nPr =n!/(n-r)! 2P2= 2!/(2-2)! = 2!/0! = 2 where 2 is even

OpenStudy (anonymous):

therefore, if you compose two even permutation you'll always get an even answer

OpenStudy (anonymous):

try other given like 4P2 and you will also get an even answer.

OpenStudy (watchmath):

I think what he meant by even permutation are the elements in permutation group that can be written as a product of even cycles.

OpenStudy (anonymous):

Sorry back, ye what watchmath said, logically it makes sense since you need a even number of transpositions to compose a even permutation, but i wasn't sure of the generic proof.

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