Let $$f: G \to C^{×}$$ is homomorphism, whereas $C^{×}$ is the multiplicative group of non-zero complex numbers, |G|=n and G is abelian.
Prove that \begin{align} \sum_{g \in G}f(g)&=0,\quad \text{or}\\[1em] \sum_{g \in G}f(g)&=n \end{align}
I know how to break abelian group into modulo $n$, but after that I could not imagine a homomorphism.
Hint: Either all elements are sent to $1$ by $f$, or not. If not all elements are sent to $1$, pick an $h\in G$ such that $f(h)\neq 1$. Then compare the two sums $$ \sum_{g\in G}f(g)\\ \sum_{g\in G}f(gh) $$
The last two lines of the solution: $$ \sum_{g\in G}f(g) = \sum_{g\in G}f(gh)\\ = \sum_{g\in G}f(g)f(h) = f(h)\left(\sum_{g\in G}f(g)\right) $$ Look at the first and last term in this chain of equalities, use that $f(h) \neq 1$, and conclude something about $\sum_{g\in G}f(g)$.