Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Even Permutations

  1. Apr 5, 2009 #1
    Changed to a different question:
    Can anyone provide a proof for the order of alternating groups |An| = ½(n!)?
     
    Last edited: Apr 5, 2009
  2. jcsd
  3. Apr 6, 2009 #2

    matt grime

    User Avatar
    Science Advisor
    Homework Helper

    It is almost immediate from the definition that it is a group of index 2 inside S_n.
     
  4. Apr 6, 2009 #3
    The way i like to show this goes something along these lines:

    THe whole idea is to show that there are as many odd permutations as there are even permutation in Sn. So, let


    [tex]S_n= ( \alpha_i,\beta_j)[/tex] which represents the set of even and odd permutations.

    where [tex]\alpha_i, i=1,2,3,...,r[/tex] and [tex]\beta_j,j=1,2,3,...,k[/tex] are even and odd permutations respectively.

    Now, let's consider the following:


    [tex]\beta_1\beta_j, j=1,2,3,....,k[/tex] (there is some extra work here to show that all these elements are indeed unique, but it is not difficult to establish it. a proof by contradiction would work)

    So, we know that the multiplication of odd permutations is an even permutation, so we know that in our set Sn, we have the following relation:

    [tex]|k|\leq|r|------(1)[/tex]

    Now, consider the following:


    [tex]\beta_1\alpha_i,i=1,2,...,r[/tex]

    so all these permutations now are odd. From this we get the following relation:

    [tex]|r|\leq |k|-----(2)[/tex]


    From (1) &(2) we get the following:


    [tex]|k|\leq |r| \leq |k|=>|k|=|r|[/tex]

    Which means that the number of even and odd permutations in Sn is equal.

    Now, since |Sn|=n! => |An|=n!/2
     
  5. Apr 6, 2009 #4

    matt grime

    User Avatar
    Science Advisor
    Homework Helper

    There is no work in showing this: that multiplication by a group element is a bijection is almost the definition of a group.

    Apart from that, your proof is 'the correct one': there exists an injection from the set of odd elements to even elements, and vice versa, hence they have the same cardinality (this has no dependence on their being a finite number of them, which is always nice).
     
  6. Apr 6, 2009 #5
    I see. The proof was written in good clarity, so I'm thinking I got a good hold from it. Thanks.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Even Permutations
  1. Permutation problem (Replies: 10)

  2. Permutation Matrices (Replies: 4)

Loading...