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Let $S=s=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$. Entrywise complex conjugation interchanges the diagonal entries of $r$ and fixes $s$, soSince $r,s$ generate the group, these identities hold for every element. Thus $S$ is an invertible intertwiner, provingThe character of a tensor product representation is the product of the factor characters. Hence ${\chi }_{R_2\otimes R_2}=(4,1,0)$. Its character inner products with the three rows of solution give multiplicities $(1,1,1)$, soFor explicit Clebsch-Gordan decomposition of group representations bases, let $e_{+},e_{-}$ be the coordinate basis. The symmetric tensor $e_{+}\otimes e_{-}+e_{-}\otimes e_{+}$ spans $1$; the antisymmetric tensor $e_{+}\otimes e_{-}-e_{-}\otimes e_{+}$ spans $\epsilon$; and the ordered pair $e_{-}\otimes e_{-},e_{+}\otimes e_{+}$ carries $R_2$, since $r$ acts by $\mathrm{diag}(\alpha ,{\alpha }^2)$ and $s$ exchanges them. In particular $S^2{R}_2=1\oplus R_2$ and ${\Lambda }^2{R}_2=\epsilon$.

Multiplying by the sign representation preserves the character of $R_2$, because its character is zero on all three reflections. Thus $\epsilon \otimes R_2\cong R_2$, and associativity givesThereforeThe dimensions give $8=1+1+3\cdot 2$. Its two one-dimensional summands can also be seen directly: $e_{+}^{\otimes 3}+e_{-}^{\otimes 3}$ and $e_{+}^{\otimes 3}-e_{-}^{\otimes 3}$ transform as $1$ and $\epsilon$, respectively.

The squares of $I,-I,\pm A,\pm B$ are $I$, whereas $(\pm C{)}^2=-I$. Therefore ${\chi }_{\rho }(g^2)=(2,2,2,2,-2)$. The symmetric square and exterior square formulas from solution giveComparison with the character table gives the Clebsch-Gordan decomposition of group representationsFor explicit bases, put $e_1=(1,0{)}^T,e_2=(0,1{)}^T$. The symmetric tensors $e_1\otimes e_1+e_2\otimes e_2$, $e_1\otimes e_2+e_2\otimes e_1$, and $e_1\otimes e_1-e_2\otimes e_2$ transform as $L_{++},L_{+-},L_{-+}$ respectively. The antisymmetric tensor $e_1\otimes e_2-e_2\otimes e_1$ transforms as $L_{--}$, also the determinant representation of $\rho$. Adding these direct sums recovers $\rho \otimes \rho ={⨁}_{s,t}{L}_{st}$.

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:

A tensor product of group representations has character equal to the product of its factor characters. Multiplying the rows of the character table in solution, using $s^2=2+t$, $t^2=2+s$ and $st=-1$, gives all the Clebsch-Gordan decomposition of group representations:For example, the character of $R_2\otimes R_2$ is $(4,s^2,t^2,0)=(4,2+t,2+s,0)$, the sum of the characters of $R_0$, $R_1$ and $R_3$. Likewise $R_2\otimes R_3$ has character $(4,-1,-1,0)={\chi }_2+{\chi }_3$. The trivial representation and $R_1\otimes R_1=R_0$ account for the remaining entries.

There is one complex irreducible representation for each conjugacy class, hence three. The trivial representation $1$ and sign representation $\epsilon$ are distinct and one-dimensional. The sum of squares of irreducible representation dimensions leavesCall the remaining irreducible representation $2$. The permutation representation on ${\mathbb{C}}^3$ has character equal to the number of fixed objects, namely $(3,1,0)$ on the ordered conjugacy classes $(e,\text{transpositions},\text{three-cycles})$. Removing its constant-vector trivial representation gives $(2,0,-1)$, whose character inner product with itself is one. This constructs $2$ and gives the character tableTo obtain every Clebsch-Gordan decomposition of group representations, multiply the characters pointwise. In particular, ${\chi }_{\epsilon }^2={\chi }_1$, ${\chi }_{\epsilon }{\chi }_2={\chi }_2$ andThus all ordered pairs of irreducible representations are covered by