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8 permutations that commute

  1. Dec 5, 2004 #1
    8 permutations that commute....

    Find 8 Permutations that commute with alpha=(1,2,4,5)

    I do not understand the concept of commuting with permutations.
  2. jcsd
  3. Dec 5, 2004 #2

    matt grime

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    Firstly in what permutation group?

    It simply means that the elements in the group commute.

    Eg (12) and (34) commute.

    In general transpositions commute iff they are disjoint (have no elements in common) or equal (are the same).

    Use that wisely...
  4. Dec 5, 2004 #3
    still not sure, is there any sites online that might explain this in detail?
  5. Dec 5, 2004 #4

    matt grime

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    do you know what it means fo elements of a group to commute?

    do you know how to compose cycles in S_n?

  6. Dec 5, 2004 #5

    Properties are said to commute under the following conditions:

    X*Y=Y*X; X+Y=Y+X
  7. Dec 5, 2004 #6
    alright cool then whatever gives me

    (1 2 3 4 5 6)
    (2 4 3 5 1 6)

    that's supposed to be two row form
  8. Dec 8, 2004 #7
    Could you PLEASE give sufficient detail of the problem for someone with zero knowledge and infinite intelligence to solve it?
    Reproduction of some more context and explanation of the notation would be helpful.
    So far I see whether head nor tail of your question, although I have a feeling that finding the answer may be much easier than understanding the question.
  9. Dec 8, 2004 #8

    matt grime

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    The onus should be on you if you want to learn it. All the notations used in this thread are absolutely standard and can be learned in any algebra textbook. But the answer is very easy, once you understand the definitions. But that's maths all over

    The OP still needs to answer 1 important question: in which permutation group is he/she working!

    Second, are you used to cycle notation? (1245) is the original cycle. Don't use the other notations since they won't help.

    If this is in S_7, then (67) commutes with it, for example. All powers of (1245) commute with (1245), the identity commutes with everything.
    Last edited: Dec 8, 2004
  10. Dec 14, 2004 #9
    Dr Think Deep is right
    I'm definitely sure that the problem is easier than it sounds,....that's my only problem

    look, there is only 8, i can't get a straight answer on how to do this.
    This is not HW or a test,...i just do not understand.

    so i will repeat the question as it is and if anybody can just spit out the 8 answers so i can figure out myself how to get there, and stop with all the concept garbage

    In S6 find "8" permutations that commute with alpha=(1,2,4,5)

    all i need is a list of answers
    if anybody is worried i'm cheating or sumthin stupid
    how bout doing it for (3,4,5,6) in S6 as well, i dont care, i just want to see the answer so i can figure out how to go from start to finish in my own way
    thanks in advance

    i know how to compute products of cycles and all that. i just do not under stand what this commuting stuff is, because it's not a+b=b+a here
    All i have found is that there is only 8 things and iota is one of them, a list of 1-7 for the rest, is all i ask,...do i multiply the original permutation, do i try to get the same thing,...i dont get it plain and simple.
  11. Dec 15, 2004 #10

    matt grime

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    For the second or third time two elements s and t in *ANY* group commute if


    where we are composing using the group operation. If you don't know the definitions you cannot do maths.

    That 's it, that's all.

    for instance (12) and (34) in S_n (n=>4) commute as (12)(34)=(34)(12)

    (12) and (23) do not commute as (12)(23) = (123) but (23)(12) = (132) =/=(123)

    (1245) commutes with all its powers, that is always true in any group:

    [tex] g^rg^s=g^{r+s}=g^sg^r[/tex]

    (1245) commutes with (36) notice that 3 and 6 are not in the cycle (1245).
  12. Dec 15, 2004 #11


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    Two operations are said to permute if it does not matter which order you do them in.

    For example, the identity permutation () commutes with all other permutatios. It commutes with yours since
    1 2 3 4 5 6-> 1 2 3 4 5 6-> 5 1 3 2 4 6
    1 2 3 4 5 6-> 5 1 3 2 4 6-> 5 1 3 2 4 6
    have the same result.

    Alternatively (36)
    1 2 3 4 5 6 -> 1 2 6 4 5 3 -> 5 1 6 2 4 3
    1 2 3 4 5 6 -> 5 1 3 2 4 6 -> 5 1 6 2 4 3

    Since there are only 6!=120 permuations in the group, you could grunt work your way through it, but there's a better way. The answer you were looking for is:
  13. Dec 17, 2004 #12
    SqrachMasda, you do know what S_6 is though, right? All you need to do is find other permutations (you can do this all in two-row form if you want, but remember it goes righ to left) in S_6 such that

    (alpha) (beta) = (beta) (alpha)

    where you defined alpha in your question. That is, do the operation on both sides and see if they're equal. There are tricks to find the answers quickly, but that's the gist of what you're trying to do. Also, when you write elements of S_n, do NOT put commas between the numbers. You wrote (1,2,4,5)...that's misleading as it looks like a 4-tuple of real numbers. Instead, write it with spaces like a row matrix (or juxtapose the numbers if dealing with single digits like how matt grime did it). That is, write (1 2 4 5). :smile: Have fun now.
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