This is a homework question for an abstract algebra class. As such, I don't expect a complete answer but a hint in the right direction would be great. The question is as follows:
Suppose that $f: (G, ∗) \rightarrow (G', ∗')$ is a surjective homomorphism from a cyclic group $G$ to a group $G'$.
- Show that $G'$ is also cyclic.
- Show that if $f$ is not injective, then $G'$ must be finite.
The proof for part 1 seems quite simple: because $f$ is surjective, $G' = f(G)$. Since $G$ is cyclic, $G = \langle x \rangle$ so $G' = f(G) = f(\langle x \rangle) = \langle f(x) \rangle$. Since $G$ is generated by $\langle f(x) \rangle$, it is cyclic.
I am unclear on how to show the link between $f$ not being one-to-one and the size of $G'$. I have considered dividing it into two cases: $G$ is either finite or infinite. If $G$ is finite, $G'$ must be as well. For the other case, I tried to assume by way of contradiction that $f$ is a non-injective, surjective homomorphism and $G'$ is infinite, but was unable to produce the contradiction.
Can someone provide an intuition about how to tackle this problem?