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Math Help - Permutations and Conjugacy classes

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    Permutations and Conjugacy classes

    1) "Explain what is meant by \pi is a permutation of \bar{n} =(1,2...n). The permutation \pi is 123456 -> 635124, write in cycle notation. Find {\pi}^{-1}. How many permutations of {S}_{6 } are there in the conjugacy class of \pi"

    Ok for the first part I don't know how to word it. I know what a permutation does, but I don't know how to call it. \pi sends each element to another? I don't know how to define this. Cycle notation I make it (1 6 4)(2 3 5). Just need a confirmation on this one. Pi^-1, = 452631. The conjugacy classes part I have no idea. I think it means with the same cycle types? But I don't know how to do this?

    Also, how many permutations of 6 are there in total? Is it 6!?

    2) "The following are permutations of 9. a=(12)(345)(78), b=(1234)(6789), c=(197)(34)(58). Calculate bc. Calculate the conjugate c.b.c^-1.

    Does there exist a permutaiton oao^-1 = b, or oao^-1 = c."

    Now I remember a rule which says the conjugate sends the part of one to the other, the corresponding part. So a conjugate of something has to have the same cycle type? So for this there would be one for c but NOT a? And for oao^-1 = c, o must be... I forget how to do it?
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    Re: Permutations and Conjugacy classes

    Quote Originally Posted by TeaWithoutMussolini View Post
    1) "Explain what is meant by \pi is a permutation of \bar{n} =(1,2...n). The permutation \pi is 123456 -> 635124, write in cycle notation. Find {\pi}^{-1}. How many permutations of {S}_{6 } are there in the conjugacy class of \pi"

    Ok for the first part I don't know how to word it. I know what a permutation does, but I don't know how to call it. \pi sends each element to another? I don't know how to define this. Cycle notation I make it (1 6 4)(2 3 5). Just need a confirmation on this one. Pi^-1, = 452631. The conjugacy classes part I have no idea. I think it means with the same cycle types? But I don't know how to do this?

    Also, how many permutations of 6 are there in total? Is it 6!?

    2) "The following are permutations of 9. a=(12)(345)(78), b=(1234)(6789), c=(197)(34)(58). Calculate bc. Calculate the conjugate c.b.c^-1.

    Does there exist a permutaiton oao^-1 = b, or oao^-1 = c."

    Now I remember a rule which says the conjugate sends the part of one to the other, the corresponding part. So a conjugate of something has to have the same cycle type? So for this there would be one for c but NOT a? And for oao^-1 = c, o must be... I forget how to do it?

    1. If \pi(j)=k then  \pi^{-1}(k)=j

    For: "Also, how many permutations of 6 are there in total? Is it 6!?"
    The Answer is yes.


    I don't understand the second question.

    But if you need to find the composition of permutations try to put the permutations on matrices...
    Last edited by Also sprach Zarathustra; June 13th 2011 at 11:01 AM.
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    Re: Permutations and Conjugacy classes

    Quote Originally Posted by TeaWithoutMussolini View Post
    1) "Explain what is meant by \pi is a permutation of \bar{n} =(1,2...n).
    A permutation of a set A is a bijection from A to A.

    Cycle notation I make it (1 6 4)(2 3 5).
    Correct.

    Pi^-1, = 452631.
    Yes, if this is understood as the second row (under 123456) of a permutation, not as a cycle.

    How many permutations of {S}_{6 } are there in the conjugacy class of \pi
    Wikipedia says that conjugating preserves the cycle structure. So I think that the conjugates of (1 6 4)(2 3 5) are all possible products of two cycles of three elements.

    You may get a better response about conjugating in the Algebra forum.
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  4. #4
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    Re: Permutations and Conjugacy classes

    as emakarov points out, the term you are looking for is bijection, or as they are sometimes called "1-1 correspondences" (functions that are 1-1 AND onto). with a small finite set, it is sometimes easier to explicitly list the correspondence than to enumerate the "functional rule".

    another way to write the inverse \pi^{-1} is to "write the cycles in reverse order", so we would have (4 6 1)(5 3 2) = (1 4 6)(2 5 3).

    it is not that hard to prove that if (a b c)(d e f) is any pair of disjoint 3-cycles, (a b c)(d e f) is conjugate to (1 6 4)(2 3 5). let σ be the permutation:

    1-->a
    2-->d
    3-->e
    4-->c
    5-->f
    6-->b, then (a b c)(d e f) = σ(1 6 4)(2 3 5)σ^-1. on the other hand σ(1 6 4)(2 3 5)σ^-1 = (σ(1) σ(6) σ(4))(σ(2) σ(3) σ(5))

    for any permutation σ, so those are ALL the conjugates of (1 6 4)(2 3 5).

    so, how many 3-cycles are there in S6? we can pick a in 6 ways, and then b in 5 ways, and then c in 4 ways: 6(5)(4) = 120.

    but (a b c) = (b c a) = (c a b), so we must divide that total by 3, for 40 possible 3-cycles in S6.

    but given any 3-cycle (a b c), we can only make the remaining elements {d,e,f} into a 3-cycle 2 ways: (d e f) and (d f e).

    however, this does not give us 80 pairs of disjoint 3-cycles, because (d e f)(a b c) and (a b c)(d e f) are the same permutation.

    so we have 40 pairs of disjoint 3-cycles.

    for your second question, if you look closely i gave you a way to find a permutation σ such that σaσ^-1 = c, and since a and b have

    diferent cycle structure, there is no way they can be conjugate.
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