 Solution

The degree-sum formula for irreducible characters givesThe number of linear characters is the order of the abelianization, so $|G/G^{\prime }|=3$ and $|G^{\prime }|=7$. The intersection of the kernels of all linear characters is $G^{\prime }$. Consequently $g_4,g_5$ are nonidentity elements of this order-seven subgroup and have order seven. The images of $g_2,g_3$ under a displayed linear character have order three, so their orders are divisible by three. Lagrange's theorem leaves three or twenty-one; an element of order twenty-one would make $G$ cyclic, contradicting its degree-three irreducible characters. ThereforeThe center is trivial. A nontrivial proper center would have order three or seven, giving a cyclic quotient and hence an abelian group, again a contradiction. Thus a centralizer of a nonidentity element is proper. It has order three for an order-three element, and seven for an order-seven element. The class sizes are therefore $1,7,7,3,3$.

Column orthogonality says ${\sum }_i|{\chi }_i(g){|}^2=|C_G(g)|$. For $g_2,g_3$, the three linear characters already contribute three, so the remaining two entries must both vanish. Orthogonality with the identity column says ${\sum }_i{\chi }_i(1)\overline{{\chi }_i(g)}=0$ for $g\ne 1$. At $g_4$ this gives $1+\alpha +{\chi }_5(g_4)=0$, so ${\chi }_5(g_4)=\beta$, using $\alpha +\beta =-1$. Similarly ${\chi }_5(g_5)=\alpha$. Thus the completed table isHere $\beta =\overline{\alpha }$, $|\alpha {|}^2=|\beta {|}^2=2$ and $\alpha \beta =2$. These relations, together with $1+\omega +{\omega }^2=0$, verify all weighted row orthogonality relations and the centralizer column norms.