Suppose that we have a closed embedding $G\_1\hookrightarrow G\_2$ of reductive groups (say over $\mathbb{Q}$), and suppose that we have a *maximal* parabolic sub-group $P\_2\subset G\_2$, and a *minimal* parabolic $P\_1\subset G\_1$. Is it possible to have two different maximal parabolic sub-groups of $G\_1$ contained in $P\_2$ and containing $P\_1$? 


Actually, is it even possible that there are two different maximal parabolics of $G\_1$ contained in $P\_2$? 


Even more optimistically, if there is a maximal parabolic sub-group of $G\_1$ contained in $G\_1\cap P\_2$, does that make $G\_1\cap P\_2$ a parabolic sub-group and thus equal to the maximal parabolic it contains?


EDIT: As Jim Humphreys points out below, there is one pathological case, where $G\_1\cap P\_2$ might be *all* of $G\_1$. In which case, *all* parabolic sub-groups of $G\_1$ will be contained in $P\_2$! This is the case, for example, when $G\_1$ is a Levi sub-group of $G\_2$. But Angelo's answer below shows that, *excluding this possibility*, the answer to my third question is 'yes'.



Can the intersection of a maximal parabolic with a closed sub-group contain more than one maximal parabolic?
If $P$ is a parabolic subgroup of a reductive group $G$ and $H$ is a closed subgroup of $G$ containing $P$, then $G/H$ is a quotient of $G/P$, so it is projective, and $H$ is parabolic. It follows that a maximal parabolic subgroup is maximal among all proper closed subgroups, which would seem to imply that the answer to all of your questions is positive.


