Lemma 9.13.1. Let L be a field. Let G be a monoid, for example a group. Let \chi _1, \ldots , \chi _ n : G \to L be pairwise distinct homomorphisms of monoids where L is regarded as a monoid by multiplication. Then \chi _1, \ldots , \chi _ n are L-linearly independent: if \lambda _1, \ldots , \lambda _ n \in L not all zero, then \sum \lambda _ i\chi _ i(g) \not= 0 for some g \in G.
Proof. If n = 1 this is true because \chi _1(e) = 1 if e \in G is the neutral (identity) element. We prove the result by induction for n > 1. Suppose that \lambda _1, \ldots , \lambda _ n \in L not all zero. If \lambda _ i = 0 for some, then we win by induction on n. Since we want to show that \sum \lambda _ i\chi _ i(g) \not= 0 for some g \in G we may after dividing by -\lambda _ n assume that \lambda _ n = -1. Then the only way we get in trouble is if
\chi _ n(g) = \sum \nolimits _{i = 1, \ldots , n - 1} \lambda _ i\chi _ i(g)
for all g \in G. Fix h \in G. Then we would also get
\begin{align*} \chi _ n(h)\chi _ n(g) & = \chi _ n(hg) \\ & = \sum \nolimits _{i = 1, \ldots , n - 1} \lambda _ i\chi _ i(hg) \\ & = \sum \nolimits _{i = 1, \ldots , n - 1} \lambda _ i\chi _ i(h) \chi _ i(g) \end{align*}
Multiplying the previous relation by \chi _ n(h) and subtracting we obtain
0 = \sum \nolimits _{i = 1, \ldots , n - 1} \lambda _ i (\chi _ n(h) - \chi _ i(h)) \chi _ i(g)
for all g \in G. Since \lambda _ i \not= 0 we conclude that \chi _ n(h) = \chi _ i(h) for all i by induction. The choice of h above was arbitrary, so we conclude that \chi _ i = \chi _ n for i \leq n - 1 which contradicts the assumption that our characters \chi _ i are pairwise distinct. \square
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