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Math Help - elements of symmetric groups and centralizers

  1. #1
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    elements of symmetric groups and centralizers

    Let P = (1, 2, 3, .... n) in S_n.
    a. Show there are (n-1)! distinct n-cycles in S_n.
    b. How many conjugates does P have in S_n?
    c. Let C(P)=N(P) be the centralizer of P in S_n. Using (b), find |C(P)|.
    d. Find the subgroup C(P).

    my thoughts:
    a -- this seems really obvious, but I never actually learned how to show it! I have no ideas.
    b. by part (a), wouldn't this be simply (n-1)!, since that's the property of symmetric groups?
    c & d, I really don't know what to do. Do I use the class equation somehow, or another theorem?

    Thanks so much for any help!
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  2. #2
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    Quote Originally Posted by kimberu View Post
    Let P = (1, 2, 3, .... n) in S_n.
    a. Show there are (n-1)! distinct n-cycles in S_n.
    b. How many conjugates does P have in S_n?
    c. Let C(P)=N(P) be the centralizer of P in S_n. Using (b), find |C(P)|.
    d. Find the subgroup C(P).

    my thoughts:
    a -- this seems really obvious, but I never actually learned how to show it! I have no ideas.
    b. by part (a), wouldn't this be simply (n-1)!, since that's the property of symmetric groups?
    c & d, I really don't know what to do. Do I use the class equation somehow, or another theorem?

    Thanks so much for any help!
    For (a), observe that (1 2 3 ... n )= (2 3 .. n 1) = (3, 4, .. n 1 2 ). To find the number of distinct n-cycles, you need to fix the first element in the n-cycle, let's say 1, and consider the permutation of the remaining elements. So there are (n-1)! distinct n-cycles in S_n.

    For (b), a conjugacy class in S_n has the same cycle type. So there are (n-1)! memebers in the conjugacy class of n-cycle in S_n.

    (c). The centralizer of (1 2 3 .. n) in S_n is simply a group generated by (1 2 3 .. n), which is <(1 2 3 .. n)>. If this is not immediate for you, cosider the permutation with two line notation.

    The members in the group <(1 2 3...n)> have a pattern in the two line notation. Try some members in the above group with two line notation and observe how they commute with (1 2 3 .. n).

    However, if you need to find the centralizer of (1 2 3 .. m) in S_n with m<n, the above needs a little bit of modification.

    For instance, the centralizer of (1 2 3 4 5 ) in S_8 is the group <(1 2 3 4 5), (6 8), (7 8)>.
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  3. #3
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    Quote Originally Posted by aliceinwonderland View Post

    (c). The centralizer of (1 2 3 .. n) in S_n is simply a group generated by (1 2 3 .. n), which is <(1 2 3 .. n)>. If this is not immediate for you, cosider the permutation with two line notation.

    The members in the group <(1 2 3...n)> have a pattern in the two line notation. Try some members in the above group with two line notation and observe how they commute with (1 2 3 .. n).
    Thanks a lot - I understand parts a&b, but I'm a little shaky on the centralizer. I don't think I know which unique pattern you mean in the two-line notation. (Also, is the order of <(1 2 3 .. n)> also (n-1)!, that is, is this group the same as the conjugacy class? I'm not sure how to find that either.)
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  4. #4
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    Quote Originally Posted by kimberu View Post
    Thanks a lot - I understand parts a&b, but I'm a little shaky on the centralizer. I don't think I know which unique pattern you mean in the two-line notation. (Also, is the order of <(1 2 3 .. n)> also (n-1)!, that is, is this group the same as the conjugacy class? I'm not sure how to find that either.)
    The order of <(1 2 3 ... n)> is n. It is the cyclic group generated by (1 2 3 .. n) in S_n. So it is an abelian group.

    For instance, if n=4 and let \tau = (1 2 3 4 ), then the cyclic group generated by \tau is \{1, \tau, \tau^2, \tau^3\}. I encourage you to verify this with the two line notation. All elements in this group commute with \tau, which implies \rho\tau\rho^{-1}=\tau where \rho \in \{1, \tau, \tau^2, \tau^3\}.
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