Fourier transform of the Klein group

Question

Obviously we have the elements of $G = \lbrace e, a, b, c\rbrace\;$ where $\;c=ab$

Define:

$$\chi : G \rightarrow \mathbb{Z}$$ How would I construct the transform for each of the elements of $G$?

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What I mean is, for example take another group $G$ as $G = \mathbb{Z_5} = \lbrace0,1,2,3,4\rbrace$

As defined above, let $\chi : G \rightarrow \mathbb{Z} $

Define: $$\chi(0)=1$$ $$\chi(1)=a$$ $$\chi(2)=a^2$$ $$\chi(3)=a^3$$ $$\chi(4)=a^4$$

Let: \begin{align*} \chi_0(x) &= 1 \\[1ex] \chi_1(x) &= \begin{cases} 1, & \text{if } x = 0, \\ a, & \text{if } x = 1, \\ a^2, & \text{if } x = 2, \\ a^3, & \text{if } x = 3, \\ a^4, & \text{if } x = 4, \end{cases} & \chi_3(x) &= \begin{cases} 1, & \text{if } x = 0, \\ a^3, & \text{if } x = 1, \\ a, & \text{if } x = 2, \\ a^4, & \text{if } x = 3, \\ a^2, & \text{if } x = 4, \end{cases} \\[1ex] \chi_2(x) &= \begin{cases} 1, & \text{if } x = 0, \\ a^2, & \text{if } x = 1, \\ a^4, & \text{if } x = 2, \\ a, & \text{if } x = 3, \\ a^3, & \text{if } x = 4, \end{cases} & \chi_4(x) &= \begin{cases} 1, & \text{if } x = 0, \\ a^4, & \text{if } x = 1, \\ a^3, & \text{if } x = 2, \\ a^2, & \text{if } x = 3, \\ a, & \text{if } x = 4. \end{cases} \end{align*}

Also define:

$\hat f(\chi) = \sum {f}(a)\bar{\chi}(a)$ - The Fourier transform

$f(a)=\frac{1}{\rule{0pt}{0.65em} |G|}\sum \hat f(\chi)\chi_i(a)$ - The inverse Fourier transform

Then we can compute the transforms for each element of $G$.

Now my question is how would I do the same with the Klein group? I mean, can I proceed by doing this:

$$\chi(e)=1$$ $$\chi(a)=\alpha$$ $$\chi(b)=\alpha^2$$ $$\chi(c)=\alpha^3$$

Then proceed like the group $\mathbb{Z}_5$?

EDIT: I just checked and this is not possible as the Cayley table doesn't not match

EDIT: I have solved this problem now.

Answer 1

Since $K_4$ is abelian of order $4$, there are exactly four irreducible representations, all of dimension one. Since every element has order two, all values of the characters are $\chi(g)= \pm 1$. This implies the character table will take the form:

\begin{align*} \begin{array}{c | c c c c } & e & a & b & c\\ \hline \chi_1 & 1 & 1 & 1 & 1\\ \chi_2 & 1 & \pm 1 & \pm 1 & \pm 1\\ \chi_3 & 1 & \pm 1 & \pm 1 & \pm 1\\ \chi_4 & 1 & \pm 1 & \pm 1 & \pm 1\\ \end{array} \end{align*}

The only way to make all the rows of this table orthogonal, is by taking:

\begin{align*} \begin{array}{c | c c c c } & e & a & b & c\\ \hline \chi_1 & 1 & 1 & 1 & 1\\ \chi_2 & 1 & 1 & -1 & -1\\ \chi_3 & 1 & -1 & 1 & -1\\ \chi_4 & 1 & -1 & -1 & 1\\ \end{array} \end{align*}

Now we are in a position to construct the Fourier transform of the Klein group. Using the character table above and definition of Fourier transform, we have:

$$\hat{f}(\chi_1) = f(e)+f(a)+f(b)+f(c)$$

$$\hat{f}(\chi_2) = f(e)+f(a)-f(b)-f(c)$$

$$\hat{f}(\chi_3) = f(e)-f(a)+f(b)-f(c)$$

$$\hat{f}(\chi_4) = f(e)-f(a)-f(b)+f(c)$$

Computing the inverse Fourier transform using definition of inverse Fourier transform, we have:

$$f(e) = \frac{1}{4}[f(e)+f(e)+f(e)+f(e)] = f(e)$$

$$f(a) = \frac{1}{4}[f(a)+f(a)+f(a)+f(a)] = f(a)$$

$$f(b) = \frac{1}{4}[f(b)+f(b)+f(b)+f(b)] = f(b)$$

$$f(c) = \frac{1}{4}[f(c)+f(c)+f(c)+f(c)] = f(c)$$

as expected.