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Thread: 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 $\displaystyle S_n$ 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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    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 $\displaystyle \rho$ = (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 $\displaystyle \rho$ = (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 $\displaystyle \sigma = \left(\begin{array}{cccc}1 & 2 & 3 & 4 \\ 3 & 4 & 1 & 2\end{array}\right)$, which has order 2.

    $\displaystyle \sigma=(1\ 3)(2\ 4)$, 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 $\displaystyle \sigma$$\displaystyle \in$$\displaystyle S_n$ be such that |$\displaystyle \sigma$| = p where p is prime

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

    Now every permutation $\displaystyle \sigma$ 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 $\displaystyle \sigma$ 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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    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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    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

    |$\displaystyle \sigma$| = p $\displaystyle \Longrightarrow$ $\displaystyle \sigma$ 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 $\displaystyle \sigma$ $\displaystyle \in$$\displaystyle S_n$ be such that |$\displaystyle \sigma$| = p where p is prime

    Then we may that $\displaystyle \sigma$ 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 $\displaystyle \sigma=abc...$ where $\displaystyle a,\ b,\ c,\ ...$ are commutative p-cycles.

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

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

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

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

    then $\displaystyle |\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:


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

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

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

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

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

    Thus $\displaystyle k_1$ | p, $\displaystyle k_2$ | p, ... $\displaystyle k_r$ | p

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

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

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

    Is that OK?

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

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

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