If $\phi$ is a homomorphism between cyclic groups $G$ and $F$, why must $|\phi(G)|$ divide $|G|$?

Question

I am reading a solution to another problem that counts the homomorphisms. I don't get why $|G|/|\phi(G)|$, I get that $|F|/|\phi(G)|$ by Lagrange's theorem.

I guess it has to do with if $|G| = m$, then $\phi(g^m) = \phi(g)^m$, so there are $m$ copies which divide $|G|$?

Answer 1

Think about the first theorem of homorphism. It tells us that $G/ker(\phi)\cong Im(\phi)$ so if the group $G$ is finite so you' ll get the answer.

Answer 2

If $\phi :G\to F$ then $G/\operatorname{ker \phi}\cong \operatorname{Im\phi}=\phi(G)\implies o(G/\operatorname{ker \phi})=o(\operatorname{Im \phi})\implies \dfrac{o(G)}{o(\operatorname {Im \phi)}}=o(\operatorname{ker \phi})$

and $\operatorname{ker \phi}$ is a subgroup of $G$