Consider the group $GL_n(\mathbb{F}_p)$. We have the following subgroups :
-$\Sigma_n$ the symmetric group (permutation matrices)
-$B_n$ the Borel subgroup (upper triangular matrices)
-$U_n$ the unipotent subgroup (upper triangular matrices with all diagonal entries equal to $1$).
Consider the ring $R=\mathbb{F}_p[GL_n(\mathbb{F}_p)]$. If $H$ is a subgroup of $GL_n(\mathbb{F}_p)$, we denote by $\overline{H}:=\sum_{h\in H}h\in R$ and if $H\subseteq \Sigma_n$, we denote by $\tilde{H}:=\sum_{h\in H} sign(h) h\in R$, where $sign:\Sigma_n\rightarrow\{\pm 1\}$ is the usual map.
One defines the Steinberg idempotent as : $e_n=\overline{B}_n\tilde{\Sigma}_n/[GL_n(\mathbb{F}_p):U_n]\in R$. Here are my questions :
(1) Why $e_n$ is idempotent in $R$ ? Is it possible to give a "bear hand" proof of this in this case? In the original paper of Steinberg, it isn't done explicitly. If an explanation is given by using Chevalley's group (which I have no knowledge of), can someone please state the properties which are used?
(2) [Solved, it was an easy question] Why if $p=2$ we have $e_n=\overline{B}_n \overline{\Sigma}_n$ in $R$ ?