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Math Help - Commuting p-cycles in Sn

  1. #1
    Super Member Bernhard's Avatar
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    Commuting p-cycles in Sn

    Dummit and Foote Section 1.3 Symmetric Groups Exercise 15 is as follows:

    Let p be a prime. Show that an element has order p in S_n if and only if its cycle decomposition is a product of commuting p-cycles. Show by explicit example that this need not be the case if p is not prime.


    Would appreciate help with this problem

    Peter
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  2. #2
    MHF Contributor alexmahone's Avatar
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    Re: Commuting p-cycles in Sn

    Suggestion: Pick a small value of n and see if the statement is true.
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    Re: Commuting p-cycles in Sn

    sketch of proof:

    decompose σ in Sn into disjoint (non-trivial, that is, length > 1) cycles. now if |σ| = p, can any of these cycles have order < p or > p? why do disjoint cycles necesarily commute?

    for a counter-example, consider (1 2)(3 4 5) in S6.
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    Super Member Bernhard's Avatar
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    Re: Commuting p-cycles in Sn

    (Answer to AlexMahone)

    Good idea

    Take n = 3

    Consider the element \rho = (1 2 3) which is of order 3

    Then it should follow that it cycle decomposes in two commuting 3-cycles.

    But problems at this early stage - how to find them?

    I know (1 2 3) = (1 3) o (1 2) ... and these are not even commuting ... 3-cycles??

    How do I find the required 3-cycles in this case?

    Peter
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    MHF Contributor alexmahone's Avatar
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    Re: Commuting p-cycles in Sn

    Quote Originally Posted by Bernhard View Post
    (Answer to AlexMahone)

    Good idea

    Take n = 3

    Consider the element \rho = (1 2 3) which is of order 3

    Then it should follow that it cycle decomposes in two commuting 3-cycles.

    But problems at this early stage - how to find them?

    I know (1 2 3) = (1 3) o (1 2) ... and these are not even commuting ... 3-cycles??

    How do I find the required 3-cycles in this case?

    Peter
    In this case, (1 2 3) itself is the commutating 3-cycle.

    Take n = 4.

    Consider the element \sigma = \left(\begin{array}{cccc}1 & 2 & 3 & 4 \\ 3 & 4 & 1 & 2\end{array}\right), which has order 2.

    \sigma=(1\ 3)(2\ 4), which is the product of commutating 2-cycles.
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    Super Member Bernhard's Avatar
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    Re: Commuting p-cycles in Sn

    (Answer to Deveno)

    Let \sigma \in S_n be such that | \sigma| = p where p is prime

    Then | \sigma| = p \Longrightarrow | \sigma| is a p-cycle (is that correct?)

    Now every permutation \sigma of a finite set can be expressed as a product of disjoint cycles and, further, disjoint cycles are commutative - they do not have an effect outside their own elements.

    So \sigma is a product of commuting disjoint cycles BUT!!

    these cycles would be m-cycles where m is less than p (wouldn't they???)

    Surely if we express a cycle in terms of the product of a number of cycles the cycles in the product would be smaller - wouldn't they?

    Peter
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  7. #7
    Super Member Bernhard's Avatar
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    Re: Commuting p-cycles in Sn

    Thanks

    I assumed (wrongly) that the 'product' had to involve at least two cycles - hence was thrown!

    Peter
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  8. #8
    Super Member Bernhard's Avatar
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    Re: Commuting p-cycles in Sn

    I was quite wrong to say in the above post that

    | \sigma| = p \Longrightarrow \sigma is a p-cycle

    (I think I can make the claim that if a permutation can be expressed as a p-cycle then it has oder p)

    So removing the erroneous statement from the argument we now have the following:

    ===============================================

    Let \sigma \in S_n be such that | \sigma| = p where p is prime

    Then we may that \sigma can be expressed as a product of commuting cycles since

    (1) every permutation of a finite set can be expressed as a product of disjoint cycles
    (2) disjoint cycles are commutative

    But! Where to go from here??

    It must be the case (as Deveno implies in a post above) that these commuting cycles are p-cycles - but WHY?

    I have the intuition that each of the cycles must have order p - but ...

    Can anyone help me further with this proof

    (Thanks to Alexmahone and Deveno for help so far)

    Peter
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    MHF Contributor alexmahone's Avatar
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    Re: Commuting p-cycles in Sn

    Quote Originally Posted by Bernhard View Post
    (I think I can make the claim that if a permutation can be expressed as a p-cycle then it has oder p)
    Assume that \sigma=abc... where a,\ b,\ c,\ ... are commutative p-cycles.

    So, \sigma^p=(abc...)^p=a^pb^pc^p...=e

    Now try proving the converse.
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  10. #10
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    Re: Commuting p-cycles in Sn

    the "missing bit" is this lemma, which you need to prove:

    if we factor \sigma as a product of r distinct cycles of lengths k_1,k_2,\dots,k_r

    then |\sigma| = lcm(k_1,k_2,\dots,k_r) .
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  11. #11
    Super Member Bernhard's Avatar
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    Re: Commuting p-cycles in Sn

    OK, thanks for help.

    Will try to put together the proof that:


    \sigma has order p in S_n \Longrightarrow the cycle decomposition of \sigma is a product of commuting p-cycles
    ----------------------------------------------------------------------------------------------------

    So let \sigma \in S_n be such that | \sigma| = p where p is prime

    Then \sigma can be expressed as a product of disjoint, and hence commuting cycles

    Thus \sigma = a_1 a_2 ... a_r where the a_i are commuting cycles of lengths, say, k_1, k_2, ... k_r

    Then | \sigma| = p = lcm( k_1, k_2, ... k_r)

    Thus k_1 | p, k_2 | p, ... k_r | p

    Thus k_i = 1 or p with at least one k_1 = p since p prime

    [If p was not prime a number of the cycles could have lengths less than p]

    Thus cycle decomposition of \sigma is a product of commuting p-cycles

    Is that OK?

    If so I now have to show that if the cycle decomposition of \sigma is a product of commuting p-cycles then | \sigma| = p

    - but of course, reading through the posts again, Alexmahone has shown how to do this (thanks again!)

    Peter
    Last edited by Bernhard; October 16th 2011 at 02:50 AM.
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