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Math Help - Cyclic groups and gcd

  1. #1
    Junior Member Greg98's Avatar
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    Cyclic groups and gcd

    Hello,
    the task is to deduce theorem (2) from theorem (1):

    (1) For cyclic group C_n=<c>: \ ord(c^m)=\frac{n}{gcd(n,m)} \ , \ (m \in \mathbb{Z}). Also <c^m>=\{g^m|g \in C_n\}.

    (2) \#\{g \in C_n|g^m=1\}=gcd(n,m)

    Group theory is fun, but in this case I can't get even started, though the puzzle seems trivial. I tried searching ProofWiki's group theory section for something useful, but I couldn't find anything. So, any help is welcome. Thank you.
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  2. #2
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    Quote Originally Posted by Greg98 View Post
    Hello,
    the task is to deduce theorem (2) from theorem (1):

    (1) For cyclic group C_n=<c>: \ ord(c^m)=\frac{n}{gcd(n,m)} \ , \ (m \in \mathbb{Z}). Also <c^m>=\{g^m|g \in C_n\}.

    (2) \#\{g \in C_n|g^m=1\}=gcd(n,m)

    Group theory is fun, but in this case I can't get even started, though the puzzle seems trivial. I tried searching ProofWiki's group theory section for something useful, but I couldn't find anything. So, any help is welcome. Thank you.


    Define f:C_n\rightarrow \langle c^m\rangle\,,\,\,f(g):=g^m ; since the group is abelian this is a homomorphism whose

    kernel is precisely K:=\{g \in C_n|g^m=1\} , so C_n/K\cong \langle c^m\rangle , according to (1).

    But \displaystyle{|\langle c^m\rangle |=\frac{n}{gcd\,(n,m)} , according to (1), again, so we get:

    \displaystyle{\left | C_n/K\right  |=\frac{n}{gcd\,(n,m)}} , and Lagrange's theorem now gives you the solution.

    Tonio
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