Let $P_0 =M_0N_0 \subset G$ be a minimal parabolic subgroup of a connected, reductive group with maximal split torus $A_0$. Let $\Phi^+ = \Phi(A_0,N_0)$ and $\Delta \subset \Phi^+$ be a set of simple roots. Then the standard parabolic subgroups $P = MN$ are in order preserving correspondence with the subsets of $\Delta$.
For $\theta, \Omega \subset \Delta$, it follows from the Bruhat decomposition that
$$P_{\theta} \cap P_{\Omega} = P_{\theta \cap \Omega}$$
It follows from the description of root subgroups that
$$N_{\theta}N_{\Omega} = N_{\theta \cap \Omega}$$
It is the case that $M_{\theta} \cap M_{\Omega} = M_{\theta \cap \Omega}$? The inclusion '$\supset$' is clear. A priori the left hand side is not connected or reductive, but it seems to be true for $G = \operatorname{GL}_n$.
Best Answer
Let $U_{\theta} = M_{\theta} \cap U_0$, which is the unipotent radical of the minimal parabolic $M_{\theta} \cap P_0$ of $M_{\theta}$. We have
$$M_{\theta \cap \Omega} \subset M_{\theta} \cap M_{\Omega} \subset P_{\theta} \cap P_{\Omega} = P_{\theta \cap \Omega} = M_{\theta \cap \Omega} \ltimes N_{\theta \cap \Omega}$$
Let $p \in M_{\theta} \cap M_{\Omega}$. We may write $p = mn$ for $m \in M_{\theta \cap \Omega}$ and $n \in N_{\theta \cap \Omega}$. The problem becomes to show that $M_{\theta} \cap M_{\Omega}$ has trivial intersection with $N_{\theta \cap \Omega}$. We have
$$M_{\theta} \cap M_{\Omega} \cap N_{\theta \cap \Omega} = U_{\theta} \cap U_{\Omega} \cap N_{\theta \cap \Omega}$$
This group is normalized by $A_0$, and this therefore connected and directly spanned by the root subgroups it contains. There are no roots in this intersection, so it must be trivial.