Either all the permutations in H are even or

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Homework Statement


Show that for every subgroup H of S(n) [the symmetric group on n letters] for n>=2 either all the permutations in H are even or exactly half of them are even.


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The Attempt at a Solution



I didn't really know how to do this but i thought maybe since H is closed it has something to do with the fact that an odd permutation times an odd permutation produces an even permutation and an odd permutation times an even permutation produces an odd permutation.
 
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Consider H\capA(n) and HA(n). Use the fact that A(n) is an index 2 normal subgroup of S(n).
 
Even simpler, if H contains an odd permutation 'o', then H*o=H. But multiplying by o turns even permutations into odd, and vice versa.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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