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How many elements of order 5 are in S7?

  1. Oct 5, 2004 #1
    Greetings, I just stumbled upon this site and thought someone might be able to help me. How many elements of order 5 are in S7? Are the elements permutations? I thought that S7 was a permutation of order 7. Also, are permutations cyclic groups? I know that permutations are a group and can be written as cycles ,but is this "cycle" the same as a "cyclic" group? Do these questions make any sense? :confused:
  2. jcsd
  3. Oct 6, 2004 #2

    matt grime

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    S_7 is the set of all permutations of 7 objects. I've never seen S7 written to be honest, they tend to use subscripts.

    'are permutations cyclic groups' is no, since one is an element of a group, and one is a group.

    A cyclic group is one generated by a single element.

    Given a cycle in S_n it generates a cyclic subgroup, which is a proper subgroup unless n=2.

    In usual notation, a cycle is something of the form (abc...d), and any permutation is writable as a product of disjoint cycles, that is cycles that have no common elements. eg (123)(45) is a product of disjoint cycles, but (12)(23) isn't. It is equal to (123).

    A cycle with n elements in it has order n.

    The product of two disjoint cycles of lengths p and q has order lcm(p,q)

    this is enough to find all the elements of S_7 that have order 5.
  4. Oct 6, 2004 #3
    Thank you, matt! your answer is really helpful for my present study! I hope people would feel the same..:niceteeth:
    Oh, sorry so unbearable that I have to :biggrin:

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