Fourier transform of the Klein group 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 $G$?

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.
 A: 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.
