Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    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?
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Need Help with Abstract Algebra Please
  1. Help abstract algebra (Replies: 2)

  2. Abstract Algebra (Replies: 6)

  3. Abstract Algebra (Replies: 0)

  4. Abstract Algebra (Replies: 9)