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If a is even, prove a^(-1) is even

  1. Nov 12, 2012 #1

    Zondrina

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    1. The problem statement, all variables and given/known data

    If a is even, prove a-1 is even.

    2. Relevant equations

    We know that every permutation in [itex]S_n, n>1[/itex] can be written as a product of 2-cycles. Also note that the identity can be expressed as (12)(12) for this to be possible.



    3. The attempt at a solution

    Suppose a is a permutation made up of 2cycles, say [itex]a_1, ...,a_n[/itex].

    We know that :

    [itex]a^{-1} = (a_1, ...,a_n)^{-1} = a_{1}^{-1}, ..., a_{n}^{-1}[/itex]

    Now since we can write (ab) = (ba) for any two cycle, we know : [itex]a^{-1} = (a_1, ...,a_n)^{-1} = a_{1}^{-1}, ..., a_{n}^{-1} = a_1, ...,a_n = a[/itex]

    So if a is an even permutation, it means that |a| is even, say |a|=n. Then |a-1| is also even since |a| = |a-1| for 2cycles.

    Thus if a is even, then a-1 is also even.

    Is this correct?
     
  2. jcsd
  3. Nov 12, 2012 #2

    Dick

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    It's correct if you can get rid of all that unclearly defined symbolism and verbiage that's giving me a headache. What's the definition of 'even permutation' in simple english? Please don't use symbols!
     
  4. Nov 12, 2012 #3

    Zondrina

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    If a permutation 'a' can be expressed as a product of an even number of 2cycles, then every possible decomposition of a into a product of two cycles must have an even number of 2cycles.
     
  5. Nov 13, 2012 #4

    Dick

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    I'll just take the 'definition' part of that. a is a product, right? Write it as a product. So [itex]a=a_1 a_2 ... a_n[/itex] where the a's are tranpositions (2 cycles) and n is even. Now express [itex]a^{-1}[/itex] as a product of transpositions. Be careful about factor order.
     
    Last edited: Nov 13, 2012
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