1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Unique Elements of Relatively Prime Order

  1. Oct 12, 2008 #1
    The problem statement, all variables and given/known data
    Let G be a group and let g be an element G such that |g| = mn, where gcd(m, n) = 1. Show that there are unique elements a, b in G such that g = ab = ba and |a| = m and |b| = n.

    The attempt at a solution
    Since gcd(m, n) = 1, there are integers s and t such that ms + nt = 1. Hence g = gms+nt = gmsgnt. Let a = gnt and b = gms. Then g = ab = ba. We also have that |a| ≤ m and |b| ≤ n.

    I'm having trouble showing that |a| = m and |b| = n. Also, the fact that s and t are not unique proves that a and b are not unique right? This is as my thought process has taken me. Any tips?
     
  2. jcsd
  3. Oct 12, 2008 #2

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Hi e(ho0n3! :smile:

    s and t are unique (to prove it, assume they're not).

    And |a| must be a factor of mn, so … ? :smile:
     
  4. Oct 12, 2008 #3
    I don't think so: Take m = 7 and n = 11. Then gcd(7, 11) = 1 and 7(-3) + 11(2) = 7(8) + 11(-5) = 1.

    Why must it be a factor of mn?
     
  5. Oct 12, 2008 #4

    morphism

    User Avatar
    Science Advisor
    Homework Helper

    Notice that |g^m| = |g|/gcd(|g|, m) = n. So |g^(ms)|=<fill blank>. Similarly, |g^(nt)|=<fill blank>.

    So now you've found the appropriate a and b. To show uniqueness, suppose that g=a'b'=b'a' and |a'|=m and |b'|=n, then show that a'=a and b'=b. The fact that |a'|=|a|=m and |b'|=|b|=n is important here.
     
  6. Oct 12, 2008 #5
    We have that |b| = mn / gcd(mn, ms) = n / gcd(n, s) and |a| = mn / gcd(mn, nt) = m / gcd(m, t). Are you suggesting that gcd(n, s) = gcd(m, t) = 1 because ms + nt = 1?

    Concerning uniqueness, I've managed to get bm = (b')m from (ab)m = (a'b')m and similarly an =(a')n. How do I continue.
     
  7. Oct 12, 2008 #6
    If gcd(n,s)>1, what could you say about ms + nt?

    Use your existing information.
     
  8. Oct 12, 2008 #7
    I could say that gcd(n,s) divides ms + nt since it divides ms and nt, but this means it would divide 1, which is impossible. I get it know.

    Somehow I need to get rid of the n in an = (a')n. I'm thinking of the following: Consider the cyclic group <a>. Since <a> has order m and n is relatively prime to m, we must have <an> = <a>. We also get that <(a')n> = <a'>, hence <a> = <a'> and a = a'?
     
  9. Oct 12, 2008 #8

    morphism

    User Avatar
    Science Advisor
    Homework Helper

    You already used the trick required here: b = bnt bms.
     
  10. Oct 12, 2008 #9
    Ah, right! That hadn't occured to me. Thanks a lot.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Unique Elements of Relatively Prime Order
  1. Order of element (Replies: 5)

Loading...