How to construct a table of all the real-valued Dirichlet characters?

[Math100]

Homework Statement

Let $G$ be the group of reduced residue classes modulo $21$. Construct a table showing the values of all the real-valued, and one of the complex-valued, Dirichlet characters modulo $21$. (In your table, use $\omega$ for $e^{\pi i/3}$.)

Relevant Equations

Definition: Dirichlet characters. Let $G$ be the group of reduced residue classes modulo $k$. Corresponding to each character $f$ of $G$, we define an arithmetical function $\chi=\chi_{f}$ as follows:
$\chi(n)=f(\hat{n})$ if $(n, k)=1$,
$\chi(n)=0$ if $(n, k)>1$.
The function $\chi$ is called a Dirichlet character modulo $k$. The principal character $\chi_{1}$ has the properties $\chi_{1}(n)=1$ if $(n, k)=1$, and $\chi_{1}(n)=0$ if $(n, k)>1$.
Corresponding to each character $f_{i}$ of $G$, we define an arithmetical function $\chi_{i}$ as follows:
$\chi_{i}(n)=f_{i}(\hat{n})$ if $(n, k)=1$,
$\chi_{i}(n)=0$ if $(n, k)>1$.

\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|c|}
\hline n & 1 & 2 & 4 & 5 & 8 & 10 & 11 & 13 & 16 & 17 & 19 & 20 \\
\hline \chi_{1}(n) & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\
\hline \chi_{2}(n) & 1 & -1 & 1 & 1 & -1 & -1 & -1 & -1 & 1 & 1 & -1 & 1 \\
\hline \chi_{3}(n) & 1 & 1 & 1 & -1 & 1 & -1 & 1 & -1 & 1 & -1 & -1 & -1 \\
\hline \chi_{4}(n) & 1 & -1 & 1 & -1 & -1 & 1 & -1 & 1 & 1 & -1 & 1 & -1 \\
\hline \chi_{5}(n) & 1 & \omega & \omega^{2} & -\omega & -1 & -\omega^{2} & -\omega^{2} & -1 & -\omega & \omega^{2} & \omega & 1 \\
\hline \chi_{6}(n) & 1 & \omega^{2} & -\omega & \omega^{2} & 1 & -\omega & -\omega & 1 & \omega^{2} & -\omega & \omega^{2} & 1 \\
\hline \chi_{7}(n) & 1 & -\omega & \omega^{2} & -\omega & 1 & \omega^{2} & \omega^{2} & 1 & -\omega & \omega^{2} & -\omega & 1 \\
\hline \chi_{8}(n) & 1 & -\omega^{2} & -\omega & \omega^{2} & -1 & \omega & \omega & -1 & \omega^{2} & -\omega & -\omega^{2} & 1 \\
\hline \chi_{9}(n) & 1 & \omega & \omega^{2} & \omega & -1 & \omega^{2} & -\omega^{2} & 1 & -\omega & -\omega^{2} & -\omega & -1 \\
\hline \chi_{10}(n) & 1 & \omega^{2} & -\omega & -\omega^{2} & 1 & \omega & -\omega & -1 & \omega^{2} & \omega & -\omega^{2} & -1 \\
\hline \chi_{11}(n) & 1 & -\omega & \omega^{2} & \omega & 1 & -\omega^{2} & \omega^{2} & -1 & -\omega & -\omega^{2} & \omega & -1 \\
\hline \chi_{12}(n) & 1 & -\omega^{2} & -\omega & -\omega^{2} & -1 & -\omega & \omega & 1 & \omega^{2} & \omega & \omega^{2} & -1 \\
\hline
\end{array}

The table above is the answer/solution for this problem.
I know that $\varphi(21)=\varphi(3)\varphi(7)=2\cdot 6=12$, which means there are $12$ elements such that $G=\{1, 2, 4, 5, 8, 10, 11, 13, 16, 17, 19, 20\}$. But I do not know how to get $\chi_{2}(n)...$ except $\chi_{1}(n)$. May anyone please tell me how to get these values?

 

[andrewkirk]

Your notation has some problems that prevent us making sense of your question:

Corresponding to each character $f_{i}$ of $G$, ....

What do you mean by this? The elements of $G$ are not characters, they are effectively modulus classes, ie elements of $\mathbb Z_k$. It is the $\chi$ that are characters.

$\chi_{i}(n)=f_{i}(\hat{n})$ if $(n, k)=1$,

What do you mean by this? You have not defined $\hat n$, and $f$ is not a function, as you have presented it, but a modulus number.

 

[martinbn]

It depends on what you already know about characters/representations of finite groups (in your case abelian). One thing you can do is find normal subgroups (in the abelian case that is all subgroups). Say $H$ is a normal subgroup of $G$, then consider the group $G/H$ it is smaller and it is easier to find characters of it. If $f$ is a character of $G/H$, when you compose it with the projection $G\rightarrow G/H$ you get a character of $G$, which is trivial on $H$.

For example look at the group generated by $5$, then $H=\langle 5\rangle = \{1,4,5,16,17,20\}$. The the factor $G/H$ has two elements. It has two characters. One is the trivial (all values are 1). This will pull back to $\chi_1$. The other is the character that gives value -1 to the non-identity element of the group. When you pull back that one you get $\chi_2$.

Playing around like that you can get many characters, may be all. There are also the orthogonality relations the characters satisfy. So, if you already know some you can use easily these relations to find the others.

 

[Math100]

Your notation has some problems that prevent us making sense of your question:

What do you mean by this? The elements of $G$ are not characters, they are effectively modulus classes, ie elements of $\mathbb Z_k$. It is the $\chi$ that are characters.

What do you mean by this? You have not defined $\hat n$, and $f$ is not a function, as you have presented it, but a modulus number.

This is my question, too. I do not know what they mean, I just posted them under the relevant equation(s) just because my book has these definitions. Since these definitions are preventing people to make sense of my question, then please ignore them. How should I find those values then, starting from $\chi_{2}(n)...$?

 

[martinbn]

Your notation has some problems that prevent us making sense of your question:

What do you mean by this? The elements of $G$ are not characters, they are effectively modulus classes, ie elements of $\mathbb Z_k$. It is the $\chi$ that are characters.

What do you mean by this? You have not defined $\hat n$, and $f$ is not a function, as you have presented it, but a modulus number.

The notations are fine. It says that $f_i$ is a character of $G$. For some reason you think it is a group element, but it doesnt say that.

 

[martinbn]

How should I find those values then, starting from $\chi_{2}(n)...$?

I told you! Did you not read my post?

 

[Math100]

I told you! Did you not read my post?

I already did. But I still don't understand. How should I find normal subgroups?

 

[martinbn]

My be it would help if you told us what you already know. For example do you know what the structure of finite abelian groups is? Can you figure out how the group of your example decomposes as a product of cyclic groups? Do you know what the characters of cyclic groups are? For instance your group $G$ is isomorphic to $C_6\times C_2$, two cyclic groups of order 6 and 2. Does this help?

 

[Math100]

My be it would help if you told us what you already know. For example do you know what the structure of finite abelian groups is? Can you figure out how the group of your example decomposes as a product of cyclic groups? Do you know what the characters of cyclic groups are? For instance your group $G$ is isomorphic to $C_6\times C_2$, two cyclic groups of order 6 and 2. Does this help?

No. I do not know the structure of finite abelian groups. I do not know the characters of cyclic groups are. You said that my group of $G$ is isomorphic to $C_{6}\times C_{2}$, which are the two cyclic groups of order $6$ and $2$. But how did you get these?

 

[martinbn]

No. I do not know the structure of finite abelian groups. I do not know the characters of cyclic groups are. You said that my group of $G$ is isomorphic to $C_{6}\times C_{2}$, which are the two cyclic groups of order $6$ and $2$. But how did you get these?

It would be easier if you told us what you do know. Why are looking at this question?

 

[Math100]

It would be easier if you told us what you do know. Why are looking at this question?

Since $\varphi(21)=\varphi(3)\varphi(7)=2\cdot 6=12$, there are $12$ elements such that $G=\{1, 2, 4, 5, 8, 10, 11, 13, 16, 17, 19, 20\}$. So $G$ can be generated by order $2$ or $6$. And we have $\chi(k)^{2}=\chi(k)\cdot \chi(k)=\chi(k\cdot k)=\chi(k^{2})=\chi(1)=1$ and $\chi(k)^{6}=\chi(k^{6})=\chi(1)=1$. But how should I find out which elements are of order $2$ and which ones are of order $6$ from group $G$?

 

[andrewkirk]

Since $\varphi(21)=\varphi(3)\varphi(7)=2\cdot 6=12$, there are $12$ elements such that $G=\{1, 2, 4, 5, 8, 10, 11, 13, 16, 17, 19, 20\}$. So $G$ can be generated by order $2$ or $6$. And we have $\chi(k)^{2}=\chi(k)\cdot \chi(k)=\chi(k\cdot k)=\chi(k^{2})=\chi(1)=1$ and $\chi(k)^{6}=\chi(k^{6})=\chi(1)=1$. But how should I find out which elements are of order $2$ and which ones are of order $6$ from group $G$?

Any group has only one element of order one, which is its identity. It's easy to see which element of G that is.
Only one element of $C_6\times C_2$ can have order two: $(1_6,(-1)_2)$ where $1_6$ is the identity of $C_6$ and $(-1)_2$ is the sole non-identity element of $C_2$. So you only need to find one element out of the eleven non-identity elements of G, that has order two. The other ten elements all have order six.

Which of the elements of G gives 1 modulo 21 when you square it?

 

[Math100]

Any group has only one element of order one, which is its identity. It's easy to see which element of G that is.
Only one element of $C_6\times C_2$ can have order two: $(1_6,(-1)_2)$ where $1_6$ is the identity of $C_6$ and $(-1)_2$ is the sole non-identity element of $C_2$. So you only need to find one element out of the eleven non-identity elements of G, that has order two. The other ten elements all have order six.

Which of the elements of G gives 1 modulo 21 when you square it?

$1, 8, 13, 20$
$1^{2}\equiv 1\pmod {21}, 8^{2}\equiv 1\pmod {21}, 13^{2}\equiv 1\pmod {21}, 20^{2}\equiv 1\pmod {21}$

 

[andrewkirk]

$1, 8, 13, 20$
$1^{2}\equiv 1\pmod {21}, 8^{2}\equiv 1\pmod {21}, 13^{2}\equiv 1\pmod {21}, 20^{2}\equiv 1\pmod {21}$

Correct.
BTW, that identifies an error in my post. I should have said that three elements in $C_6\times C_2$ have order two:$$(1_6,g_2),\ \ (b, 1_2),\ \ (b, g_2)$$ where $g_2$ is the sole non-identity element of $C_2$. and $b$ is the sole non-identity element of $C_6$ whose square is $1_6$. You can map those three order-two elements of $C_6\times C_2$ to 8, 13 and 20 in G in any way you like.
My apologies.

 

[Math100]

Correct.
BTW, that identifies an error in my post. I should have said that three elements in $C_6\times C_2$ have order two:$$(1_6,g_2),\ \ (b, 1_2),\ \ (b, g_2)$$ where $g_2$ is the sole non-identity element of $C_2$. and $b$ is the sole non-identity element of $C_6$ whose square is $1_6$. You can map those three order-two elements of $C_6\times C_2$ to 8, 13 and 20 in G in any way you like.
My apologies.

So now we have that $\chi(k)^{2}=\chi(k)\cdot \chi(k)=\chi(k\cdot k)=\chi(k^{2})=\chi(1)=1$ for all $k\{8, 13, 20\}$ and $\chi(k)^{6}=\chi(k^{6})=\chi(1)=1$ for all $k\{2, 4, 5, 10, 11, 16, 17, 19\}$. And this implies that $\chi(n)=8, 13, 20$ can either be $-1, 1$. But how should I find out the order of $-1, 1$ for each $\chi_{1}(n),...\chi_{12}(n)$ in $n=8, 13, 20$? For example, why does $\chi_{1}(8)=1, \chi_{2}(8)=-1, \chi_{3}(8)=1, \chi_{4}(8)=-1, \chi_{5}(8)=-1, \chi_{6}(8)=1, \chi_{7}(8)=1, \chi_{8}(8)=-1, \chi_{9}(8)=-1, \chi_{10}(8)=1, \chi_{11}(8)=1, \chi_{12}(8)=-1$? And how should I find other values in $\chi(2), \chi(4), \chi(5), \chi(10), \chi(11), \chi(16), \chi(17), \chi(19)$? The question asks to use $\omega$ for $e^{\pi i/3}$ but how should I do this?