# Thread: Cyclic groups and gcd

1. ## Cyclic groups and gcd

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

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

(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.

2. Originally Posted by Greg98
Hello,
the task is to deduce theorem (2) from theorem (1):

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

(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