Exercise on saturated Tits system 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.  
 A: (a): First, assume that $b \in B \cap \hat{N}$.  We will show that $bnbn^{-1} \in B \cap N$ for all $n \in N$.  Let $n \in N$.  Since $b \in \hat{N} = \mathcal N_G(N)$, we have $bnb^{-1}  = n'$ for some $n' \in N$. Then $bn = n'b$, so $BnB = Bn'B$.  This implies that $n \equiv n' \pmod{T}$, so $n'n^{-1} = b_0$ for some $b_0 \in T$.  Then $$bnbn^{-1} = n'n^{-1} \in T = B \cap N$$ as required.
Next, let's show that if $b \in B \cap \hat{N}$, then $b \in \hat{T} = \bigcap\limits_{n \in N} nBn^{-1}$.  By what we just showed, we have for each $n \in N$, that $bnbn^{-1} = b_0$ for some $b_0 \in B \cap N$.  Then $$b = b_0nb^{-1}n^{-1} = n(n^{-1}b_0n)b^{-1}n^{-1}$$
which is in $nBn^{-1}$, since $N$ normalizes $B \cap N$.  
Finally, let's show that $\hat{T} \cap N = T$.  Since $\hat{T} \subseteq B$, we have $\hat{T} \cap N \subseteq B \cap N = T$.  Conversely, if $b_0 \in T = B \cap N$, then already $b_0 \in N$, so we just need to show that $b_0 \in nBn^{-1}$ for every $n \in N$.  But
$$b_0 = n(n^{-1}b_0n)n^{-1}$$
with $n^{-1}b_0n \in B$.
