Let a finite group $G$ $p$ a prime and $s \in G_{reg}$ the sub-group of $p$-regular elements of $G$ (i.e. the elements whose orders are prime to $p$).
The exponent of $G$, $m=p^am', (p,m')=1$ wich is the least commun multiple of the orders of elements of $G$.
The order of $G$, $o=p^rq, (p,q)=1$.
We have a representation $\rho: \: G \longrightarrow GL_n(K)$ with $char(K)=p$. I would like to write the Brauer character of $\rho$. But something is not quite clear in my mind.
For $g \in G_{reg}$ the order of $g$ is by definition not divisible by $p$, the characteristic of $K$ and $\rho(g)$ is diagonalizable.
Then I would like to know what are exactly the eigenvalues of $\rho(g) \in GL_n(K)$ ?
I know that those eigenvalues are roots of unity in $K$. But are they: $\mathscr{l}=|\langle g \rangle|$-th roots of unity ? Or $m'$-th roots of unity (where $m'$ is the $p$-regular part of the exponent of $G$) ? Or $m$-th roots of unity ?
There is confusion for me on that particular point. Many thanks in advance for any suggestions.
The eigenvalues of $\rho(g)$ are $|\langle g \rangle|$-th roots of unity in the algebraic closure of $K$ ! And every element of $G$ are solutions of $X^m-1=0$ where $m$ is the exponent of $G$.