The order of cyclic subgroups

1. Oct 21, 2008

SticksandStones

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. Oct 22, 2008

SticksandStones

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?

3. Oct 22, 2008

morphism

I really don't like the way the original proof is worded. What exactly is b?

4. Oct 22, 2008

SticksandStones

Some arbitrary integer as far as I can tell.

5. Oct 22, 2008

morphism

If that's the case, then a^kL = a^bn doesn't in general imply that kL = bn.

6. Oct 22, 2008

SticksandStones

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.