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Parity of a Permutation

  1. Sep 11, 2014 #1
    I'm asked to show that a permutation is even if and only if the number of cycles of even length is even. (And also the odd case)

    I'm having trouble getting started on this proof because the only definitions of parity of a permutation I can find are essentially this theorem. And obviously I can't use this theorem to prove this theorem.. (If only). So what is the most basic, abstract definition of parity of a permutation that I might use, for a permutation of a set of size, n, that is even.

    And as a side note, I've not gotten my text book yet, I've been mooching off my class mates because, well, books are expensive and I'm broke atm. So I'm sure a relevant definition for parity of a permutation is in there but I don't have access to it right now.

    Edit: We've been using what we called double row notation for permutations I.e.

    (1 2 3 4 5)
    (2 4 3 5 1)

    And this is the question. Statement up until "Show that.."

    Let ##{\sigma}{\epsilon}S_n## and suppose that ##{\sigma}## can be written as a product of disjoint cycles. Show that..
    Last edited: Sep 11, 2014
  2. jcsd
  3. May 9, 2015 #2
    You just have to remember that the "signature map" is a group morphism from the permutation group to the multiplicative group {-1,1}.
    If you have an odd number of odd cycles, your permutation will be odd.
    If you have an even number of odd cycles, your permutation will be even.
  4. May 11, 2015 #3
    So the permutation is even if and only if there are an even number of odd cycles.

    You have to show that a cycle is odd if and only if it has even length.

    - Assume you have an odd cycle ##c## with length ##l(c)##.
    You get ## -1 = \epsilon(c) = \epsilon(c^{l(c)+1}) = \epsilon(c)^{l(c)+1} = (-1)^{l(c)+1} ##. You get that the length of ##c## must be even.

    - Cycles of even length are odd :
    a- 2 - cycles are transpositions, a transposition is odd.
    b- Assume 2k - cycles are odd. Let ##c = (x_1,...., x_{2(k+1)}) ## be a 2(k+1) - cycle. Then, unless I'm mistaking, ##c = \tau_{x_{2k+1},x_{2k+2}}\circ \tau_{x_{2k},x_{2k+1}} \circ (x_1,...., x_{2k})##. By induction hypothesis, ##c## is odd. ​
  5. May 11, 2015 #4
    Thanks for responding, though I asked this toward the first semester of Abstract Algebra, I just took the final for the second semester of abstract algebra. I had this figured out at this point. But hopefully someone else finds this useful.
  6. May 11, 2015 #5
    Yes, you were on the unanswered thread list. I don't know why old unanswered threads are not deleted.

    Also I think I've made a mistake at the end of my last reply,
    I'd say ## c = \tau_{x_1,x_{2k+2}}\circ \tau_{x_1,x_{2k+1}} \circ (x_1,...., x_{2k})##

    Looks better ?
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