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Math Help - Prove that Disjoint cycles commute

  1. #1
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    Prove that Disjoint cycles commute

    Question: if f and g in Sn are disjoint cycles fg = gf

    Answer:
    f = (a1 a2 a3... an)
    g = (b1 b2 b3 ... bm)
    where ai != bj for any i or j

    gf(ai) where i = 1 to n-1
    gf(ai) = g(ai+1) = ai+1
    fg(ai) = f(ai) = ai+1
    thus f and g commute for ai where i = 1 to n-1 this same method can be used to show that f and g commute for an and for bi.

    Is there a better way to prove this or to think about it?
    Thanks!
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  2. #2
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    Quote Originally Posted by temp31415 View Post
    Question: if f and g in Sn are disjoint cycles fg = gf
    It is trivial. Say f=(a1,a2,a3) and g=(b1,b2) for simplicity sake. For fg = gf we require that they are the same function. We want to show fg (x) = gf (x). Now if x ! = a_1,a_2,a_3,b_1,b_3 then it says fixed. If x=a1 then fg(a1)=a_2 and gf(a_1)=a_2 because they are disjoint. And the same idea in general.
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  3. #3
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    This not intended to answer the original question. It is an interesting observation I think. Several years ago I saw this question posed a different way. Prove that two permutations commute iff their active sets are disjoint.
    Example
    f = \left( {\begin{array}{*{20}c}   1 & 2 & 3 & 4 & 5  \\<br />
   1 & 4 & 3 & 5 & 2  \\\end{array}} \right)\quad \& \quad g = \left( {\begin{array}{*{20}c}<br />
   1 & 2 & 3 & 4 & 5  \\   3 & 2 & 1 & 4 & 5  \\ \end{array}} \right).

    The idea being that if an element is active in f it is not in g, and visa versa.
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  4. #4
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    Exclamation

    Quote Originally Posted by Plato View Post
    This not intended to answer the original question. It is an interesting observation I think. Several years ago I saw this question posed a different way. Prove that two permutations commute iff their active sets are disjoint.
    Example
    f = \left( {\begin{array}{*{20}c}   1 & 2 & 3 & 4 & 5  \\<br />
   1 & 4 & 3 & 5 & 2  \\\end{array}} \right)\quad \& \quad g = \left( {\begin{array}{*{20}c}<br />
   1 & 2 & 3 & 4 & 5  \\   3 & 2 & 1 & 4 & 5  \\ \end{array}} \right).

    The idea being that if an element is active in f it is not in g, and visa versa.
    Sorry I don't think it's true. Just take permutations f,g such that f=g.
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