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Need Help with Abstract Algebra Please

  1. Oct 15, 2004 #1
    My question is regarding abstract algebra.

    Suppost that B is a 10-cycle.
    For which integers i between 2 and 10 is
    B^(i) also a 10-cycle?

    I know that the answer is 3, 7, and 9 I just don't know
    how you arrive at these numbers. If someone could explain the
    process clearly to me I would greatly appreciate that.
    Thank you sooo much. :-)
  2. jcsd
  3. Oct 15, 2004 #2


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    Science Advisor
    Homework Helper

    The [tex]i[/tex] are values that are coprime (have no common factors other than 1) with 10.

    Here's a general proof:

    Let's say we have some element [tex]g[/tex] with order [tex]n[/tex]. Then [tex]g^i[/tex] has order [tex]\frac{n}{<n,i>}[/tex] where [tex]<n,i>[/tex] is the greatest common factor of [tex]n[/tex] and [tex]i[/tex].

    [tex]({g^{i}})^{\frac{n}{<n,i>}}=g^{i \times \frac{n}{<n,i>}}[/tex]
    [tex]i = k \times <n,i>[/tex]
    for some [tex]k[/tex] so
    [tex]g^{i \times \frac{n}{<n,i>}}=g^{k \times <n,i> \times \frac{n}{<n,i>}}=g^{k \times n}=g^{n\times k}=(g^{n})^{k}=e^k=e[/tex]
    so the order ok [tex]g{i}[/tex] is at most [tex]\frac{n}{<n,i>}[/tex]

    Now, let's say we have some [tex]j[/tex] so that
    then, since the order of [tex]g[/tex] is [tex]n[/tex]
    so [tex]n | ij[/tex] ([tex]n[/tex] divides [tex]ij[/tex])
    so [tex]\frac{n}{<n,j>} | j \Rightarrow j \geq \frac{n}{<n,i>}[/tex]

    So the order of [tex]g^i[/tex] is [tex]\frac{n}{<n,i>}[/tex].

    In this particular case, you have [tex]\frac{n}{<n,i>} = n[/tex] so [tex]<n,i>=1[/tex].
  4. Oct 15, 2004 #3
    Thank you so much I understand much better now. I am very grateful for all your help.
  5. Dec 5, 2009 #4
    I think that g has order 10 does not guarantee that g is a 10-cycle. If g is a product of two disjoint cycles of order 2 and 5, it can still have order 10. Is that right?
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