Cyclic groups and gcd

**Greg98** · March 12th 2011, 07:41 AM

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.

**tonio** · March 12th 2011, 08:32 AM

Originally Posted by Greg98

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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