I'm trying to work through the following exercise in Bourbaki, Lie Groups and Lie Algebras.
For part (a), we are assuming $b \in B$ normalizes $N$, and we are supposed to show that $bnb^{-1}n^{-1} \in T = B \cap N$ for all $n \in N$. The hint is to write $bn = (bnb^{-1})b$ use Theorem 1, which says that
$$w \mapsto BwB$$
is an injective on $N/T$. Already $bnb^{-1}n^{-1}$ lies in $N$. Thus, if we can show that $Bbnb^{-1}n^{-1}B = B$, it will follow that $bnb^{-1}n^{-1} \in B \cap N$. I'm probably just tired but I can't get anywhere on this problem. I would appreciate any further hints.