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Either all the permutations in H are even or

  1. Nov 29, 2007 #1
    1. The problem statement, all variables and given/known data
    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.


    2. Relevant equations



    3. 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.
     
  2. jcsd
  3. Nov 29, 2007 #2

    morphism

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    Consider H[itex]\cap[/itex]A(n) and HA(n). Use the fact that A(n) is an index 2 normal subgroup of S(n).
     
  4. Nov 29, 2007 #3

    Dick

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    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.
     
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