Analytic Number Theory/Characters and Dirichlet characters

Definitions, basic properties
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
 * $$|f(\sigma)| = |f(\sigma^{n+1})| = |f(\sigma)|^{n+1} \Rightarrow |f(\sigma)| = 1$$.

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.

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.

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 $$\rho H = H \Leftrightarrow \rho \in H$$ 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.