This seems like a very fundamental thing, yet I simply cannot prove it. I have some ideas, but need help fleshing them out.
I want to prove:
Let $G$ be a finite abelian group, and let $G^\star$ be the set of all irreducible characters of $G$. Then, if $g\in G$ is not the identity, we have:
$$\sum\limits_{\chi\in G^\star}\chi(g)=0.$$
For what it is worth, everyone I've talked to has mentioned the row and column orthogonality relations for character tables, yet those are no real help here, as they require that I take inner products, and I have no way to manipulate this so that every character gets paired with itself. The only hint in the text is to multiply by $\chi'(g)$, where $\chi'\in G^\star$ and $\chi(g)\neq 1$.
This leads me to believe I must manipulate somehow to get this:
$$\chi'(g)\sum\limits_{\chi\in G^\star}\chi(g)\stackrel{?}=\sum\limits_{\chi\in G^\star}\chi(g),$$
which of course would give the result. But how do I do this? I know that since $G$ is abelian, all elements of $G^\star$ have degree $1$, so let's write:
$$\chi'(g)\sum\limits_{\chi\in G^\star}\chi(g)=\sum\limits_{\chi\in G^\star}\chi'(g)\chi(g),$$
which looks awfully like it will just amount to a reindexing of the sum to give the result I put with a question mark, but I have no idea how to justify that claim. That's what I need help with, I think. Or I need somebody to tell me I am barking up the wrong tree. Or something.
You're missing something very important: $G^{\ast}$, the set of group homorphisms $G\to\mathbb{C}^{\times}$, is a group under pointwise multiplication. That is, if $\chi$ and $\chi'$ are homomorphisms, then so too is their product $\chi\cdot\chi'$. It is easy to verify:
$$ \begin{array}{ll} (\chi\cdot\chi')(gh) & =\chi(gh)\chi'(gh) \\ & =\chi(g)\chi(h)\chi'(g)\chi'(h) \\ & =\chi(g)\chi'(g)\chi(h)\chi'(h) \\ & = (\chi\cdot\chi')(g)~(\chi\cdot\chi')(h). \end{array} $$
Therefore, multiplying $S=\sum_{\chi\in G^{\ast}}\chi(g)$ by $\chi'(g)$ doesn't actually change the value of $S$. But if we have $\chi'$ such that $\chi'(g)\ne1$, then how can multiplying $S$ by positive number besides $1$ not change its value? Only if $S=0$. (How you know there is a $\chi'\in G^{\ast}$ such that $\chi'(g)\ne1$ is another matter.)