=== Lemma === Let $H \le G$ be a subgroup. Let $\chi: H \to \C^\times$ be a character on $G$. Let $a \in G \mathrel \backslash H$ where $\backslash$ denotes divisibility. Then: :$\chi$ extends to $\left[{G : H}\right]$ distinct characters on $G$ where $\left[{G : H}\right]$ denotes the index of $H$ in $G$. === Proof === The proof proceeds by [[Second Principle of Mathematical Induction|stroing induction on $\left[{G : H}\right]$]]. If $\left[{G : H}\right] = 1$, then $H = G$ by Langrange's theorem. Hence (trivially) the result. Suppose that $\left[{G : H}\right] > 1$, and the result holds for all subgroups of smaller index in $G$. Let $a \notin H$ and let $n$ be the indicator of $a$ in $H$. Let $K = \left\langle{H, a}\right \rangle$ denote the subgroup generated by $H$ and $a$. From Subgroup Generated by Subgroup and Element, each element of $K$ has a unique representation in the form $x a^k$ with: :$x \in H$ and $0 \le k \le n-1$ and: :$\left|{K}\right| = n \left|{H}\right|$ Let $\tilde \chi$ be a character extending $\chi$. Then: :$(1): \quad \tilde \chi \left({x a^n}\right) = \tilde \chi \left({x}\right) \tilde \chi \left({a}\right)^n$ and: :$(2): \quad \tilde \chi \left({x a^n}\right) = \tilde \chi \left({x}\right) \tilde \chi \left({a^n}\right)$ Since $x, a^n \in H$, $\tilde \chi$ and $\chi$ agree on these values, so equating $(1)$ and $(2)$ we obtain: :$\tilde \chi \left({a}\right)^n = \tilde \chi \left({a^n}\right)$ That is: :$\tilde \chi \left({a}\right)$ is an $n$th root of $\tilde \chi \left({a^n}\right)$. So by nth roots of a complex number are distinct there are $n$ distinct possibilities. We choose one of these possibilities and define: : $\tilde \chi \left({x a^k}\right) = \chi \left({x}\right) \tilde \chi \left({a}\right)^k$ for $x a^k \in K$. We check that $\tilde \chi$ is multiplicative: For $x,y\in H$, {{begin-eqn}} {{eqn | l = \tilde \chi \left({x a^k y a^l}\right)       | r = \tilde \chi \left({x y a^{k + l} }\right) }} {{eqn | r = \chi \left({x y}\right) \tilde \chi \left({a}\right)^{k+l} }} {{eqn | r = \left[{\chi \left({x}\right) \tilde \chi \left({a}\right)^k}\right] \left[{\chi \left({y}\right) \tilde \chi \left({a}\right)^l}\right] }} {{eqn | r = \tilde \chi \left({x a^k}\right) \tilde \chi \left({y a^l}\right) }} {{end-eqn}} By Lagrange's theorem: : $n = \dfrac {\left|{K}\right|} {\left|{H}\right|} = \left[{K : H}\right]$ Also by Lagrange's Theorem: :$\left[{G : K}\right] < \left[{G : H}\right]$ since $H \subsetneq K$. So by the induction hypothesis, each $\tilde \chi$ extends to $\left[{G : K}\right]$ characters on $G$. Therefore by the Tower Law for Subgroups, $\chi$ extends to $\left[{K : H}\right] \left[{G : K}\right] = \left[{G : H}\right]$ characters on $G$. {{qed|lemma}} Since a character is a homomorphism, it must preserve the identity. Therefore there is only one character $\chi : \left\{{e}\right\} \to \C^\times : e \mapsto 1$ on the trivial subgroup. Moreover, by Langrange's theorem, the trivial subgroup has index $\left[{G : \left\{{e}\right\}}\right] = \left|{G}\right|$. So by the lemma there are $\left|{G}\right|$ distinct characters on $G$. {{qed}} Category:Analytic Number Theory bo9cl4unpag8bnu29nzn572d5idkni7