Brauer character

Remarks

The eigenvalues lie in $\mu_{p'}(k)$ because $\rho(g)^m = \id$ with $m$ the order of $g$, prime to $p$. Such an isomorphism $\phi$ exists; it is usually fixed through a $p$-modular system, and other choices change $\beta_\rho$ by a Galois automorphism. Example: for the trivial representation of degree $d$, all eigenvalues are $1$, so $\beta(g) = d$ for every $g \in G_{p'}$; for $d = p$ this is $p$, while the trace in $k$ is $0$. Non-example: let $G = \Z/p$ and let $V = k^2$, with the generator acting by $\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$. The generator is not $p$-regular, so the Brauer character has no value there. It is defined only at the identity, with value $2$. Brauer characters are constant on conjugacy classes, additive on short exact sequences, and $\beta_\rho(1) = \dim V$. The irreducible Brauer characters are linearly independent, and their number is the number of conjugacy classes of $p$-regular elements.