Codex @codex  0

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.