No nontrivial group homomorphism $ \mathbb{Z}/n\mathbb{Z} \rightarrow\mathbb{Z} $

Question

I need to show that there is no nontrivial group homomorphism between $$ \mathbb{Z}/n\mathbb{Z} \rightarrow\mathbb{Z} $$ with $n \in \mathbb{N}, n\geq1$.

Now I know that $\mathbb{Z}/n\mathbb{Z} = 0+n\mathbb{Z},1+n\mathbb{Z},...,(n-1)+n\mathbb{Z}$ and that this is a group with "+", the same goes for $(\mathbb{Z},+)$. But I have no idea how to prove that there is no other group homomorphism. if $f$ is a homomorphism, then $f(0+n\mathbb{Z})=0$ since $0$ and $0+n\mathbb{Z}$ are the neutral elements in both groups. But where do I go from here? It is clear that $$ f: \mathbb{Z}/n\mathbb{Z} \rightarrow\mathbb{Z}, f(a+n\mathbb{Z})=a $$ is no group homomorphism because $$ f(0+n\mathbb{Z})=f((1+n\mathbb{Z})+(n-1)+n\mathbb{Z})= 1+n-1=n\neq0 $$ so this is wrong. For the same reason it also cannot be $f(a+n\mathbb{Z})=x$ for $x\neq 0$. But from here on I find no more argumentation to show the rest and would appreciate your help very much.

EDIT// with the help from below: Assume $f(1+n\mathbb{Z})=a\neq0.$ $$ 0=f(0+n\mathbb{Z})=f(n+n\mathbb{Z})=f((1+...+1)+n\mathbb{Z})=f((1+n\mathbb{Z})+...+(1+n\mathbb{Z}))= f(1+n\mathbb{Z})+...+f(1+n\mathbb{Z})=n\cdot a\neq0 $$ so $f(1+n\mathbb{Z})=0$ but then because all $b\in \mathbb{Z}_+, b\neq 0$ can written as a sum of 1 we get the same result as above, which is zero.

Answer 1

Hint: Homomorphisms have the property that $h(a)$ has order dividing the order of $a$ for each $a$, if the order of $a$ is finite.

Answer 2

More generally, if $G$ is a finite group, then there is no nontrivial group homomorphism $\phi: G \to \mathbb Z$.

Indeed, the image of $\phi$ is a finite subgroup of $\mathbb Z$ and so must be the trivial subgroup.