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Find a permutation of order 15 in S8.

  1. Sep 5, 2009 #1
    Hello, can anyone tell me how to find order 15 in S8.

    I only know. Permutation (abc)(defgh) have order 15.

    Next, I would think about 8*7*6*5*4*3*2*1 = 13440

    Number of permutations, for order 15 in s8. would be

    8*7*6*5*4*3*2*1 / 3*5 = 896.

    There are 896 permutation of order 15 in s8.

    But how to find just one permutation in s8?

  2. jcsd
  3. Sep 5, 2009 #2
    Well, you already presented one. Do you mean, you want to find more? That's easy.

    if ab is order n > 1 with a and b commuting, then

    [tex](ab)^{n} = a^{n}b^{n} = e[/tex]

    so n is divisible by the order of a and the order of b. combined with the example that you have, you should be able to get the idea. It would be a good exercise to prove that if a is order m and b is order n, then ab has order [m, n] which is just the least common multiple of m and n. Assume that a and b commute for that last sentence.

    What condition can guarantee that two permutations commute?
    Last edited: Sep 5, 2009
  4. Sep 6, 2009 #3
    We say two permutations f and g commute if fg = gf.

    I only know. Permutation (abc)(defgh) have order 15. (123)(45678)

    have order 15. I don't know the permutation (123)(45678) in s8 or not.

    how to verify it?
    (123)(456) is a permutation has order 3 in s6?

    Thanks for your help again :biggrin:
  5. Sep 6, 2009 #4

    Same as A group G said to be commutative, or ablian if operation *, in addition to be above four axioms, satisfies.

    Commutativity, For all x,y belongs to G, we have x*y = y*x


    Can we use this definition to satisfy S8 is also ablian?
  6. Sep 6, 2009 #5
    The http://mathworld.wolfram.com/SymmetricGroup.html" [Broken] says:
    "The symmetric group S_n of degree n is the group of all permutations on n symbols."
    So you have to show that (123)(45678) is a permutation of 8 symbols.
    Recall that a permutation is a bijective function from a set X to X.

    The order tells you how many times you have to apply the permutation to get the "identity" element. For example applying (456) three times yields the identity element. Calculate (456)(456)(456) and you should get (4)(5)(6).
    Last edited by a moderator: May 4, 2017
  7. Sep 6, 2009 #6

    That's the definition of commutativity. I asked for a condition on two permutations that would guarantee that they satisfy that definition. They are two distinct concepts.

    Do (1 2 3) and (4 5 6) commute? What about (1 2 3) and (4 2 5)? Why do the two that commute do so? What is the obstruction to commutativity in the two that do not?

    What do you think? Is it abelian? Remember that two elements of a non-abelian group can commute with each other. For an easy example, see the permutations earlier in this post.

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