# Analytic Number Theory/Characters and Dirichlet characters

## Definitions, basic properties

Definition 4.1

Let $G$ be a finite group. A character of G is a function $f:G\to \mathbb {C}$ such that

1. $\forall \sigma ,\tau \in G:f(\sigma \tau )=f(\sigma )f(\tau )$ and
2. $\exists \rho \in G:f(\rho )\neq 0$ .

Lemma 4.2:

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

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

In particular, $\forall \sigma \in G:f(\sigma )\neq 0$ .

Proof:

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

$|f(\sigma )|=|f(\sigma ^{n+1})|=|f(\sigma )|^{n+1}\Rightarrow |f(\sigma )|=1$ .$\Box$ Lemma 4.3:

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

Proof:

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

since $\mathbb {C}$ is a field and thus free of zero divisors.$\Box$ Lemma 4.4:

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

Proof: Trivial, since $\forall \tau \in G:f(\tau )\neq 0$ as shown by the previous lemma.$\Box$ The previous three lemmas (or only the first, together with a few lemmas from elementary group theory) justify the following definition.

Definition 4.5

Let $G$ be a finite group. Then the group

$\{f:G\to \mathbb {C} |f{\text{ is a character }}\}=\mathrm {Hom} (G,\mathbb {C} ^{\times })$ is called the character group of $G$ .

## Required algebra

We need the following result from group theory:

Lemma 4.6

Let $G$ be a finite Abelian group, let $H\leq G$ be a subgroup of order $n\in \mathbb {N}$ , and let $\tau \in G\setminus H$ such that $k\in \mathbb {N}$ is the smallest number such that $\tau ^{k}\in H$ . Then the group

$N:=\{\tau ^{n}\sigma |\sigma \in H\}$ is a subgroup of $G$ containing $H$ of order $k\cdot n$ .

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 $\rho H=H\Leftrightarrow \rho \in H$ and $\tau ^{l}H=\tau ^{m}H\Leftrightarrow \tau ^{l-m}\in H\Leftrightarrow k|(l-m)$ . 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.$\Box$ 