 Solution

The $x$ axis is an invariant subspace on which $r$ acts as $+1$ and $s$ as $-1$. The $yz$ plane carries the irreducible representation $E$ from b. ThusThe character of a tensor product is the product of the characters, so ${\chi }_{R\otimes R}=(9,1,1,1,1)$. Weighted character inner products giveFor the trivial character, the multiplicity is $(9+1+2+2+2)/8=2$. For any other one-dimensional character, the weighted sum over the nonidentity classes is $-1$, giving $(9-1)/8=1$. For $E$, it is $(9\cdot 2-2)/8=2$.

Equivalently, $E\otimes E$ is the sum of all four one-dimensional representations, and tensoring $E$ by ${\chi }_{-,+}$ gives the same two-dimensional irreducible representation. The dimensions in the boxed decomposition sum to nine, as required.