A cyclic prime-to-p Brauer table

The eigenspace decomposition for g shows any simple module is one-dimensional with one of the m distinct eigenvalues; each scalar action $g\mapsto {\lambda }^i$ is a representation since $({\lambda }^i{)}^m=1$. Its value at $g^j$ is $\widehat{{\lambda }^{ij}}={\zeta }^{ij}$. Distinct eigenvalues give nonisomorphic modules.

The determinant is ${\prod }_{0\le i<l<m}({\zeta }^l-{\zeta }^i)$. To obtain the formula, regard the determinant of $(x_i^j)$ as a polynomial in the $x_i$: it vanishes when two $x_i$ coincide, has total degree $m(m-1)\mathrm{/}2$, and the coefficient of $x_1{x}_2^2\cdots x_{m-1}^{m-1}$ is 1, as is that of the displayed product. Hence they agree. Its factors are nonzero and their reductions ${\lambda }^l-{\lambda }^i$ are nonzero, so the determinant is a unit in $\mathcal{O}$ and is nonzero in both K and k. For $m=1$ the table is $(1)$ and the product is empty, equal to 1.