 Solution

The symmetric group $S_3$ has three conjugacy classes, determined by cycle type: the identity, three transpositions, and two three-cycles. Hence it has three complex irreducible representations. Two are the trivial representation $1$ and the sign representation $\epsilon$. Their dimensions are one, so the sum of squares of irreducible representation dimensions gives $1+1+d^2=6$, whence $d=2$ for the remaining representation $D$.

Write ${\chi }_D=(2,a,b)$ on these classes. Orthogonality to $1$ and $\epsilon$ gives $2+3a+2b=0$ and $2-3a+2b=0$. Thus $a=0,b=-1$, and the character table isThe character of a tensor product representation is the product of the factor characters. Thus ${\epsilon }^2=1$, $\epsilon \otimes D=D$, and ${\chi }_{D\otimes D}=(4,0,1)={\chi }_1+{\chi }_{\epsilon }+{\chi }_D$. These determine all the Clebsch-Gordan decomposition of group representations: