Results are out! Find what you need...fast. Get quick advice or join the chat
Hey there Sign in to join this conversationNew here? Join for free

Product of Disjoint Cycles

Announcements Posted on
    • Thread Starter
    • 0 followers
    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?
    • 4 followers
    Offline

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

    ReputationRep:
    The first one is correct. What is your method?
    • Thread Starter
    • 0 followers
    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!
    • Thread Starter
    • 0 followers
    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!
    • 0 followers
    Offline

    ReputationRep:
    Why are you using such a complicated method?!
    • 10 followers
    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?
    • 4 followers
    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.
    • Thread Starter
    • 0 followers
    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 followers
    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 .
    • 4 followers
    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.)

Reply

Submit reply

Register

Thanks for posting! You just need to create an account in order to submit the post
  1. this can't be left blank
    that username has been taken, please choose another Forgotten your password?
  2. this can't be left blank
    this email is already registered. Forgotten your password?
  3. this can't be left blank

    6 characters or longer with both numbers and letters is safer

  4. this can't be left empty
    your full birthday is required
  1. By joining you agree to our Ts and Cs, privacy policy and site rules

  2. Slide to join now Processing…

Updated: April 15, 2012
New on TSR

Find out what year 11 is like

Going into year 11? Students who did it last year share what to expect.

Article updates
Reputation gems:
You get these gems as you gain rep from other members for making good contributions and giving helpful advice.