= Solution
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 <1/a/solution> give
$$
\boxed{\chi_{S^2\rho}=(3,3,1,1,-1),\qquad
\chi_{\wedge^2\rho}=(1,1,-1,-1,1).}
$$
Comparison with the <character table> gives the <Clebsch-Gordan decomposition of group representations>
$$
\boxed{S^2\rho=L_{++}\oplus L_{+-}\oplus L_{-+},\qquad
\wedge^2\rho=L_{--}.}
$$
For 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 $\rho\otimes\rho=\bigoplus_{s,t}L_{st}$.
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