 Solution

In a one-dimensional representation, the relation $srs=r^{-1}$ implies $\rho (r)=\rho (r{)}^{-1}$, so $\rho (r)=\delta \in \{\pm 1\}$. Independently $\rho (s)=ϵ\in \{\pm 1\}$. These give four irreducible representationsA two-dimensional representation $E$ is obtained by restricting the given matrices to the $yz$ plane:The eigenvalues of $E(r)$ are $i$ and $-i$. Its two eigenlines are interchanged by $E(s)$, so neither is invariant under the whole group. Thus $E$ is a complex irreducible representation.

Taking traces gives the character table, with class sizes included for the character inner product:For example $⟨{\chi }_E,{\chi }_E⟩=(4+4)/8=1$, and its inner product with each one-dimensional row is $(2-2)/8=0$. The four one-dimensional rows are mutually orthonormal by direct weighted summation. Their squared dimensions sum to $4\cdot 1^2+2^2=8=|G|$, and there are five rows for five conjugacy classes. This proves completeness of the list.