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Product of Disjoint Cycles

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  1. Offline

    ReputationRep:
    Ive found many a theory of how to find the product of disjoint cycles, and when I've applied each one I've got many a different answer!

    The cycles are (15)(1243)(12)

    I've got the answers of

    (1435)(2)
    (15)(243)
    (15423)

    Are any of these actually correct?
  2. Offline

    The first one, (1435)(2), is correct. What method are you using to put it in disjoint cycle form?
  3. Offline

    ReputationRep:
    The first one is correct. What is your method?
  4. Offline

    ReputationRep:
    I put them all into 2 line forms,

    (12345). (12345). (12345)
    (52341). (24135). (21345)

    Then starting from the right, I put the middle perm, underneath the right one with the rows matching ie

    (12345)
    (21345)

    (21345)
    (42135)

    Cancelled out the two middle rows to get the
    (12345)
    (42135)

    Then put the left perm underneath

    (12345)
    (42135)

    (42135)
    (42531)

    Canceled the middle rows again

    (12345)
    (42531)

    Then turned this into the one line cycle.

    I'm sure there must be an easier way, but I understand this!!

    Thank you!
  5. Offline

    ReputationRep:
    I put them all into 2 line forms,

    (12345). (12345). (12345)
    (52341). (24135). (21345)

    Then starting from the right, I put the middle perm, underneath the right one with the rows matching ie

    (12345)
    (21345)

    (21345)
    (42135)

    Cancelled out the two middle rows to get the
    (12345)
    (42135)

    Then put the left perm underneath

    (12345)
    (42135)

    (42135)
    (42531)

    Canceled the middle rows again

    (12345)
    (42531)

    Then turned this into the one line cycle.

    I'm sure there must be an easier way, but I understand this!!

    Thank you!
  6. Offline

    ReputationRep:
    Why are you using such a complicated method?!
  7. Offline

    ReputationRep:
    (Original post by Aslsh)
    I put them all into 2 line forms,

    (12345). (12345). (12345)
    (52341). (24135). (21345)

    Then starting from the right, I put the middle perm, underneath the right one with the rows matching ie

    (12345)
    (21345)

    (21345)
    (42135)

    Cancelled out the two middle rows to get the
    (12345)
    (42135)

    Then put the left perm underneath

    (12345)
    (42135)

    (42135)
    (42531)

    Canceled the middle rows again

    (12345)
    (42531)

    Then turned this into the one line cycle.

    I'm sure there must be an easier way, but I understand this!!

    Thank you!
    How did you get your other two answers?
  8. Offline

    A much easier method is to "chase the numbers". That is, find where 1 goes to, then find where that goes to, and so on until you get back to 1. Then choose the smallest number not yet accounted for, and do the same.

    For instance, say you have (123)(145)(25)(34).

    What happens to 1? Well working from right to left we see that
    1 \xrightarrow{(34)} 1 \xrightarrow{(25)} 1 \xrightarrow{(145)} 4 \xrightarrow{(123)} 4
    so 1 goes to 4. What happens to 4? Similarly we get
    4 \to 3 \to 3 \to 3 \to 1
    so 4 goes back to 1 and this completes a cycle.

    We haven't yet accounted for 2. We find 2 \to 2 \to 5 \to 1 \to 2, so 2 is fixed.

    Next up is 3. We find 3 \to 4 \to 4 \to 5 \to 5, so 3 goes to 5. And then 5 \to 5 \to 2 \to 2 \to 3, so 5 goes back to 3 and we get another cycle.

    So (123)(145)(25)(34) = (14)(2)(35), or we could just write it as (14)(35).

    You don't even need to write anything down to this steps, you can do the bit with all the arrows in your head and just fill in the disjoint cycles as you go along.
  9. Offline

    ReputationRep:
    That I understand! Thank you!

    My middle answer I got by doing the same method I used above but going left to right, and then third answer I got, by the method of 'chasing number' which appears I did wrong!
    But how it's been shown above I get, and using that on my question I got the same answer as my more complicated way.

    One question for u though, as it's asked for the product does it matter that I haven't got one cycle with all 5 elements in? And that I get the 2 on its own?
  10. Offline

    ReputationRep:
    (Original post by nuodai)
    A much easier method is to "chase the numbers". That is, find where 1 goes to, then find where that goes to, and so on until you get back to 1. Then choose the smallest number not yet accounted for, and do the same.

    For instance, say you have (123)(145)(25)(34).

    What happens to 1? Well working from right to left we see that
    1 \xrightarrow{(34)} 1 \xrightarrow{(25)} 1 \xrightarrow{(145)} 4 \xrightarrow{(123)} 4
    so 1 goes to 4. What happens to 4? Similarly we get
    4 \to 3 \to 3 \to 3 \to 1
    so 4 goes back to 1 and this completes a cycle.

    We haven't yet accounted for 2. We find 2 \to 2 \to 5 \to 1 \to 2, so 2 is fixed.

    Next up is 3. We find 3 \to 4 \to 4 \to 5 \to 5, so 3 goes to 5. And then 5 \to 5 \to 2 \to 2 \to 3, so 5 goes back to 3 and we get another cycle.

    So (123)(145)(25)(34) = (14)(2)(35), or we could just write it as (14)(35).

    You don't even need to write anything down to this steps, you can do the bit with all the arrows in your head and just fill in the disjoint cycles as you go along.
    I was waiting for someone else to post this so I didn't have to .
  11. Offline

    (Original post by Aslsh)
    One question for u though, as it's asked for the product does it matter that I haven't got one cycle with all 5 elements in? And that I get the 2 on its own?
    It asks for a product of disjoint cycles. They could be 1-cycles, 2-cycles, 3-cycles, or whatever. The fact that you don't have a 5-cycle quite simply means that it's not a 5-cycle; for instance (123)(45) can't be written as a 5-cycle, and nor can (123) or (123456). (This is to do with conjugacy classes in the symmetric group, see here.) Getting the 2 on its own isn't a problem, but it's usual to omit 1-cycles from the notation since they have no effect. (It's not wrong to include them though.)

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Updated: April 15, 2012
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