# How to show that a surjective, non-injective homomorphism must have a finite codomain?

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 $G'$.

1. Show that $G'$ is also cyclic.
2. 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?

• Hint: If $G$ is finite then $G'$ is finite. There is only one infinite cyclic group up to isomorphism. If $f(a)=f(b)$ you can use what you know of its structure to conclude your proof. – Mark Bennet Feb 23 '15 at 23:13
• Okay - if $G$ is infinite, $G\simeq\mathbb Z$. However if $f(a) = f(b)$ (that is, there are two elements in $G$ that map to one element in $G'$), $G'$ cannot be isomorphic to $G$ and therefore must be finite. Correct or too simple? – bkaiser Feb 23 '15 at 23:25
• You have to use that $f$ is a homomorphism - I've put that bit in an answer. – Mark Bennet Feb 24 '15 at 7:46

Consider the kernel of $f$. If $f$ is not injective then this kernel is nonzero and $G/\ker f \simeq G'$. So show that $G/\ker f$ is finite when $\ker f \neq 1$.
To fill out the hint in comments, suppose $G=\mathbb Z$ and $f(a)=f(b)$ with $b\gt a$. Now call $c=b-a\neq 0$ and note that $f(c)=f(b-a)=f(b)-f(a)=0$ because $f$ is a homomorphism.
Now use the division algorithm to write an arbitrary number $n=cq+r$ with $0\leq r \lt c$, then show that $f(n)=f(r)$, and the options for $r$ are finite.
To complete the analysis of the image of $f$, rather than just the finiteness argument, work instead with the least integer $d$ for which $f(d)=0$ (it is not necessary to do this to prove finiteness, we just need some positive integer $c$ with $f(c)=0$).