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Permutation Group

  1. Dec 4, 2009 #1
    1. The problem statement, all variables and given/known data

    Suppose G is a group generated
    by the two permutations (1 2 3 4 5) and
    (1 2)(3 4). Decide which group G is and
    prove your answer.


    2. Relevant equations



    3. The attempt at a solution
    So i crunched this out and I found

    identity
    15->2 cycles
    20->3 cycles
    and 24->5 cycles

    So i think this is the A5.

    Is there an easier way to determine this, other than crunching out the numbers.
     
  2. jcsd
  3. Dec 5, 2009 #2

    lanedance

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    how many elements do you think are in your group?
     
  4. Dec 5, 2009 #3
    I get 60 elements....took a while, I'm just assuming there is an easier way..
     
  5. Dec 5, 2009 #4

    Dick

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    Yes, there's somewhat easier ways. Both your generators are even permutations, so you know G will be a subgroup of A5. Now use Lagrange's theorem. Your first one has order 5, so you know 5 divides the order of G. The second has order 2. Now can you find subgroups of order 3 and 4? That would do it, right?
     
    Last edited: Dec 5, 2009
  6. Dec 5, 2009 #5
    wow. i've forgotten lots of group theory since last year. why is (12345) even? isn't (12345)=(12)(23)(34)(45)(51) i.e. the product of an odd no. of transpositions and thus the whole permutation is odd?

    even using lagrange's theorem,
    so if 5 divides 60 then there's a subgroup of size 12 and if 2 divides 60 then there's a subgroup of size 30, no?
     
  7. Dec 5, 2009 #6

    Dick

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    You've forgotten how to multiply permutations, that's for sure. (12345)=(15)(14)(13)(12). That's an even number of transpositions. Your product comes out to be (2345). And not necessarily for the second part. The order of a subgroup divides the order of the group. That's all.
     
  8. Dec 5, 2009 #7
    ok. checked my notes.
    a k-cycle is even if k is odd as it can be written as a product of k-1 transpositions
    (12345) is a 5-cycle and is therefore even as 5 is odd i.e. it can be written as (12)(23)(34)(45)
    the general case given in my notes was (x1 x2 x3 ... xk)=(x1 x2)(x2 x3) ... (xk-1 xk) so i just fitted that to this case.

    so the order of a subgroup divides the group. generators are always subgroups. so we know G is divisible by 5 and by 2. so #G is a multiple of 10. how can we deduce anything more?
     
  9. Dec 5, 2009 #8

    Dick

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    As I told carmine. Start taking products of the generators. See if you can find a subgroup of orders 3 and 4.
     
  10. Dec 5, 2009 #9
    i get (12)(12345)=(1543) which is of order 4
    and (12)(34)(12345)=(153) which is of order 3

    so that tells us #G is divisible by 2,3,4,5

    2x3x4x5=120
    but 60 is also divisible by 2,3,4,5
    nothing smaller than 60 is divisible by 2,3,4,5
    so G=A5
    yeah?
     
  11. Dec 5, 2009 #10

    Dick

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    The second one looks good. (1543) isn't in the group generated by the two given elements. It's an odd permutation. Notice, I didn't say 'find an element of order 4'. You aren't going to find one. I said 'find a subgroup of order 4'. It doesn't have to be cyclic.
     
  12. Dec 5, 2009 #11
    subgroup of order 4 is [itex]H=\{ (153),(145),(135),(235) \}[/itex]

    so now we can use lagrange to say 4 divides the size of the group.

    is the argument at the end of my last post ok for showing that G=A5?
     
  13. Dec 5, 2009 #12

    Dick

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    That's NOT a subgroup! It doesn't even have the identity in it. Think about it. A subgroup of order 4 is going to have to be generated by two elements of order 2. Right, the argument is fine. You just need to show the order of G is divisible by 4.
     
  14. Dec 5, 2009 #13
    depressing to forget identity

    so [itex]H=\{ e , (12) , (34) , (12)(34) \}[/itex] would do the trick.
     
  15. Dec 5, 2009 #14

    Dick

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    At least you've got a group there. But again (12) and (34) aren't in G. They are ODD. So it's not a SUBgroup.
     
  16. Dec 5, 2009 #15
    ok (12345)(12)(34)=(135)
    (34)(12345)(12)=(145)
    (12)(12345)(34)=(235)

    so H={(135),(145),(235)} looked ok until i tried combinations and found it wasn't closed under group action. any suggestions?
     
  17. Dec 5, 2009 #16

    Dick

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    Find the group generated by the two elements (12)(34) and (14)(32). What's it's order? You might ask how I knew (14)(32) was in G. Good question. I believe that carmine was correct in concluding the original two elements generate A5. So carmine knows how to check that (14)(32) is in G. I didn't bother to figure out how to generate it. Can you?
     
    Last edited: Dec 5, 2009
  18. Dec 5, 2009 #17
    ok well (12)(34)(14)(32)=(13)(24)

    so H={e,(12)(34),(13)(24),(14)(23)} which is my subgroup of order 4 and then i can use my argument to deduce G=A5, yes?

    so all that remains to be proven is that (14)(32) is in G

    i assume i just have to play about with combinations of the cycles
    (12)(34) and (12345) till it works?

    also, does lagrange's theorem work on elements of size 3 as well as subgroups of order 3? i didn't find a subgroup of size 3, all i found was an element of order 3.

    thanks for the revision tutorial in group theory. will be useful for my course in lie groups next semester hopefully!
     
    Last edited: Dec 5, 2009
  19. Dec 5, 2009 #18

    Dick

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    Yes, you just have to play around with it until you find one. If you didn't find a subgroup of order 3 you should probably find one. That should be easy. An element of order 3 will generate it. Good luck with Lie groups. That's a whole nother ball of wax. But if you didn't understand finite groups, Lie groups would be hopeless.
     
  20. Dec 6, 2009 #19
    rewinding about 7 months i was actually pretty competent at this stuff - clearly my christmas will need to involve some revision!

    so the subgroup of order 3 will be {e,(153),(135)}
     
  21. Dec 6, 2009 #20

    Dick

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    Yes, that's one of them.
     
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