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The order of cyclic subgroups

  1. Oct 21, 2008 #1
    Today we learned about subgroups of cyclic groups G = <a>. During the discussion we reached this point:

    |<a^k>| = minimum L, L > 0, such that a^(kL) = 1.
    |G| = n.
    Then a^kL = a^bn, thus kL = bn, and thus L = n/gcd(k, n).

    However, I don't understand the bolded. My number theory is terrible, and I don't really see where the gcd(k, n) comes from.

    I understand that if gcd(k, n) = 1 that <a^k> = <a>, but the connection to L = n/gcd(k,n) just isn't apparent to me. Can someone shine some light on this for me?

    Thanks, I appreciate it!
  2. jcsd
  3. Oct 22, 2008 #2
    Bleh, I think I figured it out. kL is a multiple of n and L is the lowest integer that makes kL a multiple of N. If gcd(k,n)=1 then L is just n, otherwise it has to be a multiple of n. If gcd(k,n)=d then k=sd and L=n/gcd(k,n) so kL=sn and s=b.

    Is this correct?
  4. Oct 22, 2008 #3


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    I really don't like the way the original proof is worded. What exactly is b?
  5. Oct 22, 2008 #4
    Some arbitrary integer as far as I can tell.
  6. Oct 22, 2008 #5


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    If that's the case, then a^kL = a^bn doesn't in general imply that kL = bn.
  7. Oct 22, 2008 #6
    Ok then I'm really confused. a^kL=a^bn = 1, how can I then generalize this to find L?

    EDIT: Normally I'd just ask the professor, but it was a "substitute professor" if you will and I honestly don't know who the man is/was.
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