Clebsch-Gordan coefficients for the Kronecker product of 2-dimensional irreducible representations of $S_3$ How to calculate the Clebsch-Gordan coefficients.
I have computed the 2 dimensional irreducible representations of $S_3$:
$$ D^{(2)}(e)=I_{2},D^{(2)}(a)= \left(\begin{array}{cc}
 1 & 0 \\
 -1 & -1 \\
\end{array}\right), D^{(2)}(\alpha)= \left( \begin{array}{cc}
 0 & 1 \\
 -1 & -1 \\
\end{array}
 \right)
 $$ 
Now, I would like to calculate the Clebsch-Gordan decomposition of the Kronecker product of this 2 dimensional representation with itself, that is:
$$ D^{2\otimes 2}=\bigoplus_{l=1}^{3}a_{l} D^{(l)} $$ 
I calculated the coefficients using orthogonalization relations :
$$ D^{2\otimes 2}=D^{(1)}\oplus D^{(1')} \oplus D^{(2)} $$ 
Where 1 and 1' represent the trivial and alternating representations respectively.
This is what I've tried:
Take, for example, $D^{2\otimes 2}(a) $ :
$$ \left( \begin{array}{cccc}
 1 & 0 & 0 & 0 \\
 -1 & -1 & 0 & 0 \\
 -1 & 0 & -1 & 0  \\
 1 & 1 & 1 & 1  \\
\end{array} \right) $$ 
It can be seen that this matrix is not block diagonal. So, the task is to find a basis for which this matrix becomes block diagonal, because of Clebsch-Gordan . How can I find such basis?.
I don't know if this is the best method to tackle this problem. So, it would be nice if you can provide alternative methods to compute the Clebschs.
Any help is truly appreciated.
 A: Here is how I learn Clebsch-Gordan coefficients using character theory. For any representation $V$ of group $G$,
you can define its character, i.e. a group homomorphism $\chi_V:G\to \mathbf{C}$
such that $\chi_V(g)=\text{trace}(g)$. 
Class function. A function $f:G\to \mathbf{C}$ is a class function if 
it is constant under conjugacy class, i.e. $f(g)=f(hgh^{-1})$ for all $h,g\in G$.
Denote $C(G)$ be the vector space of all class functions on $G$. 
Note that $\chi_V$ is also a class function (why?). 
One can define a Hermitian inner product on $C(G)$ by (check this) 
$$ \langle f,g\rangle = \frac{1}{|G|}\sum_{s\in G}f(s)\overline{g(s)} $$
The whole point is that characters of all irreducible reprsentations of $G$
(irreducible characters for short) form an orthonormal basis of $C(G)$.
This implies that two representation $V,W$ are isomorphic iff their 
characters $\chi_V=\chi_W$.
From properties of characters, we find if $D^{2\otimes 2}
=a D^1\oplus bD^{1'}\oplus cD^{2}$ then its corresponding 
character $\chi_{2\otimes 2}=a\chi_1+b\chi_{1'}+c\chi_2$. Due to 
orthogonal relation, we find $a= \langle \chi_{2\otimes 2},\chi_1\rangle,
b=\langle  \chi_{2\otimes 2},\chi_1'\rangle, c= \langle \chi_{2\otimes 2},\chi_2\rangle$. Also from property of characters, we know 
$\chi_{2\otimes 2}=(\chi_2)^2$. Hence, to find $a,b,c$, it suffices to 
find irreducible characters $\chi_1,\chi_1',\chi_2$.
Note these characters are class functions, i.e. they are constant under 
conjugacy classes so we only need to find their value under each
conjugacy class of $S_3$. I think you can continue from here
knowing the definition of characters and you also know all $D^1,D^{1'},D^2$.
The final answer is $D^{2\otimes 2}=D^1\oplus D^{1'}\oplus D^2$. 
