Proof of Artin's Theorem (linearly independent functions)

Solution 1:

Suppose there are nontrivial linear relations between the maps $f_1,\dots,f_n$ seen as elements of the vector space $K^G$; among them choose one with the minimum number of nonzero coefficients. Upon a reordering, we can assume it is $$ \alpha_1f_1+\dots+\alpha_kf_k=0 $$ with all $\alpha_i\ne0$. This means that, for every $x\in G$, $$ \alpha_1f_1(x)+\dots+\alpha_kf_k(x)=0 $$ Note that $k>1$ or we have a contradiction.

Fix $y\in G$; then also $$ \alpha_1f(yx)+\dots+\alpha_kf_k(yx)=0 $$ and, since the maps are homomorphisms, $$ \alpha_1f_1(y)f_1(x)+\dots+\alpha_kf_k(y)f_k(x)=0\tag{1} $$ for every $x\in G$ and $$ \alpha_1f_1(y)f_1(x)+\dots+\alpha_kf_1(y)f_k(x)=0\tag{2} $$ By subtracting $(2)$ from $(1)$ we get $$ \alpha_2(f_2(y)-f_1(y))f_2(x)+\dots+\alpha_k(f_k(y)-f_1(y))f_k(x)=0 $$ for all $x$, hence $$ \alpha_2(f_2(y)-f_1(y))f_2+\dots+\alpha_k(f_k(y)-f_1(y))f_k=0 $$ which would be a shorter linear relation, so we conclude that $$ f_2(y)=f_1(y),\quad \dots,\quad f_k(y)=f_1(y) $$ Now, choose $y$ such that $f_1(y)\ne f_2(y)$ and you have your contradiction.