Why is the image of $G$ under a $B-$adapted homomorphism normal? (Question from Bruhat Tits paper 1)

Question

I posted this question on MSE earlier, however could not elicit a reply. I am not sure if this belongs here, if not, please flag it. Moreover, I am not sure what tags to put on it, so if this question remains, please add appropriate tags.

I was reading Bruhat Tits' "Groupes réductifs sur un corps local : I. Données radicielles valuées", available here.

Let $(G,B,N,S)$ be a Tits system and $\phi : G \to G^{'}$ be a group homomorphism. It is called $B-$adapted (resp. $BN-$ adapted) if

1. $\ker (\phi) \subset B$, and
2. for every $g \in G^{'}$, there exists a $h\in G$ such that $\phi(hBh^{-1}) = g\phi(B)g^{-1}$ (resp. $\phi(hBh^{-1}) = g\phi(B)g^{-1}$ and $\phi(hNh^{-1}) = g\phi(N)g^{-1})$

Now let $\phi : G \to G^{'}$ be $B-$adapted.

In (1.2.14) Page 18 they claim that $\phi(G)$ is a normal subgroup of $G^{'}$ since we have the Bruhat decomposition. However, if I compute for some $w \in W$ (or equivalently $w \in N$)

$$ g\phi(BwB)g^{-1} = g\phi(B)\phi(w)\phi(B)g^{-1} = g\phi(B)g^{-1}g\phi(w)g^{-1}g\phi(B)g^{-1} = \phi(hBh^{-1})g\phi(w)g^{-1}\phi(hBh^{-1}), $$

I cannot take $g$ inside $\phi$ for $w$ since $\phi$ is only assumed to be $B-$adapted. I think we also need $BN-$adapted.

Am I missing something?

Answer 1

Bruhat-Tits are not saying that $\phi(G)$ is normal "since we have the Bruhat decomposition", they are saying it is normal since the conjugates of $B$ generate $G$.

(Proof. Let $a\in G$ and $g\in G'$. To see that $g\phi(a)g^{-1}\in\phi(G)$, pick $(b_i ,k_i)\in B\times G$ such that $a=\prod_i k_ib_ik_i^{-1}$. Put $g_i=g\phi(k_i)$. By hypothesis, we have $g_i\phi(b_i)g_i^{-1}=\phi(h_ic_ih_i^{-1})$ for some $(c_i,h_i)\in B\times G$. Therefore, $g\phi(a)g^{-1}=\prod_ig_i\phi(b_i)g_i^{-1}=\prod_i\phi(h_ic_ih_i^{-1})\in\phi(G)$.)