Let $G$ be a group and $k$ be a field. A character of $G$ is a group homomorphism $G \to k^{\times}$.
Theorem (Artin). Distinct characters $\chi_1, \dots, \chi_n : G \to k^\times$ are linearly independent in $\operatorname{Map}(G,k)$.
(Special case in a different setting: Let $L/K$ be a finite separable field extension, $\tau : K \to \overline{L}$ an embedding, then there are exactly $n = [L:K]$ embeddings $\sigma_1, \dots, \sigma_n : L \to \overline{L}$ preserving $\tau$. These are linearly independent over $\Omega$.)
But why? I know the standard proof but I fail to really grasp why this statement should be true. The best I can come up with is the following: Take a linear combination $$ a_1 \chi_1 + \dots + a_n \chi_n = 0, $$ then by plugging in elements from $G$, we get a bunch of linear relations, e.g. by plugging in the identity element $e \in G$, we get $a_1 + \dots + a_n = 0$. Since those are a lot of relations, they should eventually give us $a_1 = \dots = a_n = 0$.
However, this is extremely vague. Does anyone have a better intuitive explanation of this result?
You meant $G$ is an arbitrary infinite group. Viewing the $\chi_j$ as characters on the abelian group $G/\cap_j \ker(\chi_j)$, they are assumed distinct thus they must disagree on some finitely generated subgroup $\langle s_1,\ldots,s_n\rangle$.
Assuming $a_1\ne 0$, take a polynomial $f(x)=\sum_{k=0}^K c_k x^k\in k[x]$ such that $f(\chi_1(s_1))=1$ and $f(\chi_j(s_1))=0$ for $\chi_j(s_1)\ne \chi_1(s_1)$. Then for all $g$, $$\sum_{l,\chi_l(s_1)=\chi_1(s_1)}a_l \chi_l(g)=\sum_{j=1}^n a_j f(\chi_j(s_1)) \chi_j(g)=\sum_{k=0}^K c_k \sum_{j=1}^n a_j \chi_j(s_1^k g)=0$$ Thus we get a linear dependence between the characters agreeing with $\chi_1$ at $s_1$, repeating with $s_2,s_3,\ldots$ we'll reach a contradiction, ie. that two characters are the same on $\langle s_1,\ldots,s_n\rangle$.