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 substracting 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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