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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?

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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.

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    $\begingroup$ 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? $\endgroup$
    – RFZ
    Commented Apr 1, 2021 at 17:07
  • $\begingroup$ Correct! And since $F(G;T)$ is finite, so is $G\hat{}$. $\endgroup$ Commented Apr 1, 2021 at 17:09
  • $\begingroup$ Nice proof! Thanks a lot for your help :) I'll accept your answer as the best one. $\endgroup$
    – RFZ
    Commented Apr 1, 2021 at 17:10

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