Analytic Number Theory/Characters and Dirichlet characters

Definitions, basic properties

Lemma 4.2:

Let $G$ be a finite group and let $f:G\to ℂ$ be a character. Then

$\mathrm{\forall }\sigma \in G:|f(\sigma )|=1$.

In particular, $\mathrm{\forall }\sigma \in G:f(\sigma )\ne 0$.

Proof:

Since $G$ is finite, each $\sigma \in G$ has finite order $n:=\mathrm{o}\mathrm{r}\mathrm{d}(\sigma )$. Furthermore, let $\rho \in G$ such that $f(\rho )\ne 0$; then $f(\rho )=f(\sigma )f({\sigma }^{-1}\rho )$ and thus $f(\sigma )\ne 0$. Hence, we are allowed to cancel and

$|f(\sigma )|=|f({\sigma }^{n+1})|=|f(\sigma ){|}^{n+1}\Rightarrow |f(\sigma )|=1$.$◻$

Lemma 4.3:

Let $G$ be a finite group and let $f,g:G\to ℂ$ be characters. Then the function $h:G\to ℂ,h(\tau ):=f(\tau )\cdot g(\tau )$ is also a character.

Proof:

$h(\sigma \tau )=f(\sigma \tau )g(\sigma \tau )=f(\sigma )g(\sigma )f(\tau )g(\tau )=h(\sigma )h(\tau )\ne 0$,

since $ℂ$ is a field and thus free of zero divisors.$◻$

Lemma 4.4:

Let $G$ be a finite group and let $f:G\to ℂ$ be a character. Then the function $g:G\to ℂ,g(\tau ):=\frac{1}{f(\tau )}$ is also a character.

Proof: Trivial, since $\mathrm{\forall }\tau \in G:f(\tau )\ne 0$ as shown by the previous lemma.$◻$

The previous three lemmas (or only the first, together with a few lemmas from elementary group theory) justify the following definition.

Required algebra

We need the following result from group theory:

Proof:

Since $G$ is the disjoint union of the cosets of $H$, $N$ is the disjoint union $\bigcup _{j=0}^{n-1}{\tau }^{j}H$, as $\rho H=H\iff \rho \in H$ and ${\tau }^{l}H={\tau }^{m}H\iff {\tau }^{l-m}\in H\iff k|(l-m)$. Hence, the cardinality of $N$ equals $k\cdot n$.

Furthermore, if ${\tau }^l\sigma ,{\tau }^m\rho \in N$, then ${\tau }^l\sigma ({\tau }^m\rho {)}^{-1}={\tau }^{l-m}\sigma {\rho }^{-1}\in N$, and hence $N$ is a subgroup.$◻$