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Calculating the order of an element in cycle notation

  1. Nov 22, 2014 #1
    1. The problem statement, all variables and given/known data
    Let ##g \in S_n## and suppose that we know the cycle notation for ##g##. How can one compute its order without repeatedly composing ##g## with itself?

    2. Relevant equations


    3. The attempt at a solution

    I took the group ##S_4## and tried formulating a conjecture. I composed several elements, such as (12), (13), (14), (1234), etc., and found that the order (the number of times one has to compose an element with itself to get the identity element) was equal to the length of the cycle minus one. I have been told that this is wrong, and have read elsewhere that the order is simply the length of the cycle.

    Why is that so?
     
    Last edited: Nov 22, 2014
  2. jcsd
  3. Nov 22, 2014 #2

    Dick

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    (12) has length 2. (12)(12)=(12)^2=identity. What's the order of (12)?
     
  4. Nov 22, 2014 #3
    I figured it would be one, because you only have to compose (12) with itself once.
     
  5. Nov 22, 2014 #4

    Dick

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    Composing a function with itself once is taking it to the second power.
     
  6. Nov 22, 2014 #5
    Oh, of course: the order is equal to the exponent.
     
  7. Nov 22, 2014 #6
    So, how would you go about proving something like this?
     
  8. Nov 22, 2014 #7

    Dick

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    Calculate. Suppose you have a k cycle ##\sigma=(n_1,n_2, ..., n_k )##. Then what is ##\sigma(n_1)##? What is ##\sigma^2(n_1)##? Etc.
     
    Last edited: Nov 22, 2014
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