I am studying this proof in my algebra notes, and I would like some help regarding the requirements of the proof.
The statement of the proof is:
For each finite abelian group $G$ and each $h \in G$ with $h \neq 1$, we have $\chi(h) \neq 1$ for some $\chi$ in character group of $G$.
The proof is:
Let $H = \langle h\rangle$, the cyclic subgroup generated by $h$. Since $H \neq 1$, we have character group of $H \neq 1$ so there is a ${\phi}$ in the character group of $H$ with ${\phi}(h) \neq 1$. Let $\chi$ extend ${\phi}$. Then $\chi(h) \neq 1$.
Why is it necessary for $G$ to be finite abelian?
Because if the group is not abelian, it does not necessarily have any non-trivial characters at all. For instance, if the group is simple, i.e. it does not have any normal subgroups (for instance, the alternating group $A_5$), then it only has the trivial character, and then of course you cannot have $\chi_0(x)\neq 1$.
A non-abelian simple group $G$ only has the trivial character because if it had a non-trivial character $\chi\neq \chi_0$, its kernel would not be the whole group, so it would be a proper normal subgroup; that subgroup cannot be trivial, either, because then $G$ would be abelian. Therefore, no nontrivial character exists.
Edit: This does not mean that characters are useless for non-abelian groups; on the contrary, it is extremely useful in many other classes of groups, plus it plays a big role in the decomposition of some groups with respect to maximal abelian subgroups inside them. But it is abelian groups that are most closely linked to their character groups (and give the simplest correspondence possible), and this is why you study their characters first.
Edit2: I am assuming the OP is talking about group characters, i.e. homomorphisms to $\mathbb{C}^*$; if not please clarify.