I am attempting to solve the following problem:
Let $A$ be a finite abelian group, and let $φ:A\to \mathbb{C^\times}$ be a homomorphism that is not the trivial homomorphism. Prove that $\sum_{a\in A} φ(a)=0$.
I know by the structure theorem that $A$ is a direct product of cyclic groups. I have proven it for the special case that A is cyclic, but I need help to prove the general case.
Proof of special case:
Suppose $A$ is cyclic. Say $A=(\mathbb{Z}/n\mathbb{Z}, +)$. Since $φ$ is not trivial, $n\geq2$. Now, we have $φ(a+b)=φ(a)φ(b)$ and $φ(0)=1$. We also have $φ(a)=φ(1)^a$. It follows that $φ(1)\neq 1$, since otherwise $φ$ would be trivial. Now, $0=φ(1)^n-1=[φ(1)^0+φ(1)^1+\cdots+φ(1)^{n-1}][φ(1)-1]$, and therefore $\sum_{a\in A} φ(a)=φ(1)^0+φ(1)^1+\cdots+φ(1)^{n-1}=0,$ as needed. QED.
You can also take $b \in A$ such that $\phi (b) \neq 1$ Now $$\sum_{a\in A} \phi (a) = \sum_{a \in A} \phi (a\star b) = \phi (b) \sum_{a \in A} \phi (a)$$ So $$(1-\phi (b) ) \sum_{a\in A} \phi (a) = 0$$ You can conclude from that since $\phi (b) \neq 1$