Suppose we have a $(B,N)$-pair for a group $G$. The group $W:=N/(B\cap N)$ is known to be generated by a set $S$ of involutions. Apparently $$ S=\{w\in W:B\cup BwB\text{ is a group}\}. $$
The reason for this is that for any $t\in N(w)$, $BtB\subset Bw^{-1}BwB$, where $N$ is the map from $W$ into the power set of the $W$-conjugates of $S$, sending $s$ to $\{s\}$ for $s\in S$ and for $x,y\in W$, $N(xy)=N(y)\Delta y^{-1}N(x)y$, where $\Delta$ denotes the symmetric difference.
How does this fact imply $B\cup BwB$ is a group only if $|N(w)|=1$?