# Group $G\hat{}$ of characters of finite abelian group $G$

I am reading the book "Finite Fields" by Lidl, Niederreiter and I ran into the moment which is not so clear to me.

Suppose that $$G$$ is a finite multiplicative abelian group and let $$U=\{z\in \mathbb{C}: |z|=1\}$$. Homomorphism $$\chi:G\to U$$ is called character of group $$G$$. Then we see that $$\chi(g_1g_2)=\chi(g_1)\chi(g_2)$$ for all $$g_1,g_2\in G$$ and it follows that $$\chi(1_G)=1$$.

By Lagrange's theorem we see that $$(\chi(g))^{|G|}=\chi(g^{|G|})=\chi(1_G)=1.$$

Character $$\chi_0:G\to U$$ such that $$\chi_o(g)=1$$ for all $$g\in G$$ is called trivial character of $$G$$.

If we have finite number of characters $$\chi_1,\dots, \chi_n$$ we can define their product and it is also a character which is defined by formula: $$(\chi_1\cdot\dots\cdot \chi_n)(g)=\chi_1(g)\cdot \dots \cdot \chi_n(g)$$.

Denote by $$G\hat{}$$ the set of all characters $$\chi$$ of $$G$$. Then it is easy to see that $$G\hat{}$$ is multiplicative group. Since $$G$$ is finite then $$G\hat{}$$ is also finite.

Question: It is not so clear to me why the group $$G\hat{}$$ is finite. Can anyone give more details, please?

Let $$n = \lvert G \rvert$$ be the order of $$G$$. Let $$T = \{z \in \mathbb{C} : z^n = 1\}$$ be the set of $$n$$th roots of unity. Since $$\mathbb{C}$$ is a field, $$T$$ is finite -- indeed you might know that $$T = \{e^{2\pi i k/n} : k \in \{0, \dots, n-1\}\}$$.
Now for any character $$\chi$$ and any $$g \in G$$, we have $$\chi(g)^n = \chi(g^n) = \chi(1_G) = 1$$ by Lagrange's Theorem. Thus, $$\chi(g) \in T$$. This means that every character of $$G$$ is actually a function $$G \to T$$! Since $$G$$ and $$T$$ are finite, there are only finitely many such functions.
• Thanks for your reply. Let me ask you question: Let denote by $F(G;T)$ the set of all functions from $G$ to $T$. So you basically showed that $G\hat{}\subset F(G;T)$, right?
• Correct! And since $F(G;T)$ is finite, so is $G\hat{}$. Commented Apr 1, 2021 at 17:09