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Permutations form subgroup.

  1. Apr 29, 2008 #1
    1. The problem statement, all variables and given/known data
    Consider the group D4 (rigid motions of a square) as a subgroup of S4 by using
    permutations of vertices. Identify all the even permutations and show that they form a subgroup of D4.

    3. The attempt at a solution
    I think I have the permutations of correct. They are: (1,2,3,4), (1,3)(2,4), (1,4,3,2), (1), (2,4), (1,3), (1,2)(3,4), (1,4)(2,3). If this is correct, then the only one that is not even is (1).

    Can someone check my work thus far? I know how to go about proving it is a subgroup assuming the rest is correct.
     
  2. jcsd
  3. Apr 29, 2008 #2

    Dick

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    Your list of group elements looks correct. The conclusion that (1) is not even is dead wrong. Better reread the definition of an 'even' permutation. There are four even permutations in there and four odd ones.
     
  4. Apr 29, 2008 #3
    The four even then would be (13)(24), (12)(34), (14)(23), and (1), correct?

    Is 1 even because it is 0 transpositions?
     
  5. Apr 29, 2008 #4

    Dick

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    Yes.
     
  6. Apr 29, 2008 #5

    Dick

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    Yes.
     
  7. Apr 29, 2008 #6
    I'm having some problems completing the proof that the set of even elements forms a group under D4. I do know that if G is any group of permutations then the set of all even permutations G form a subgroup of G, but I'm not sure how to prove that. Does that seem the like the easiest way to go about it?
     
  8. Apr 29, 2008 #7

    Dick

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    If a*b is the product of the two permutations a and b, what can you say about whether a*b is even or odd in terms of the even or oddness of a and b? Then to show the evens are a subgroup, show it's closed, has inverses, has an identity etc.
     
  9. Apr 29, 2008 #8
    OK, that makes sense. I've completed this problem. Thanks!
     
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