Let $G$ be an algebraic group. Let $\mathcal{B}$ be the set of all Borel subgroups of $G$. Denote the identity component of subgroup $H\subset G$ by $H^\circ$.
I want to show that $$\mathscr{R}G=\left(\bigcap_{B\in \mathcal{B}} B\right)^\circ$$ is a normal subgroup of $G$.
I think I have the rough idea down, but it's not 'clicking'.
- So I consider $g\in G,b\in \mathscr{R}G$, then since $b$ is in every $B\in\mathcal{B}$,$$gbg^{-1}\in gBg^{-1}$$ for each $B\in \mathcal{B}$, but by the conjugacy theorem, each of these is some Borel subgroup $B'\in \mathcal{B}$.
- I haven't yet proved this, but I believe that by the conjugacy theorem $$\{gBg^{-1}\}_{B\in\mathcal{B}}=\mathcal{B}$$ for any $g\in G$. (Is this true?)
- So I can conclude $gbg^{-1}\in B$ for every $B$ and hence $gbg^{-1}\in \cap_{B\in \mathcal{B}} B$, now I just need to consider the identity component.
You can check that "if $B$ is a Borel subgroup, then $gBg^{-1}$ also". On the other hand, for each Borel subgroup $B$, $B=g(g^{-1}Bg)g^{-1}=gB'g^{-1}$ is also the form $gB'g^{-1}$ for some Borel subgroup $B'$. Hence $$\{gBg^{-1}\}_{B\in\mathcal{B}}=\mathcal{B}.$$
So \begin{align*} g(\mathscr{R}G)g^{-1}&=g\left(\bigcap_{B\in \mathcal{B}} B\right)^\circ g^{-1}\\ &=\left(g\left(\bigcap_{B\in \mathcal{B}} B \right)g^{-1}\right)^\circ\\ &=\left(\bigcap_{B\in \mathcal{B}}g B g^{-1}\right)^\circ\\ &=\left(\bigcap_{B\in \mathcal{B}} B \right)^\circ\\ &=\mathscr{R}G.\\ \end{align*} This follows by the map \begin{aligned} f: & G \to G \\ & x \mapsto gxg^{-1} \end{aligned} is a homeomorphism. Hence $\mathscr{R}G$ is a normal subgroup.