I'm trying to solve exercise 6.2.11 (5) of Linear Algebraic group by Springer, but I'm a bit rusty in algebraic geometry, so I don't know how to proceed.
Let $P$ and $Q$ be parabolic subgroups in a linear algebraic group $G$ with $P\subseteq Q$.
Let $X$ be a closed subgroup of $G$ such that $XP=X$.
Then $XQ$ is closed.
My thought was that the hypothesis imply that both $\pi(X)$ and $\pi(Q)$ are closed and complete in $G/P$, where $\pi$ is the projection.
Moreover, $\pi^{-1}\left(\pi(XQ)\right)=XQ$, hence $XQ$ is closed if and only if its image is closed.
So, if I can prove that $\pi(XQ)\cong\pi(X)\times\pi(Q)$ it would be complete and closed and the claim would follow.
Unfortunately, I can't define a map from $\pi(X)\times\pi(Q)$ to $\pi(XQ)$, so this idea seems moot. Am I on the right track?
Consider the following maps $\pi_P : G \rightarrow G/P$ and $\pi_Q : G \rightarrow G/Q$ and $\pi_{P,Q} : G/P \rightarrow G/Q$. Note the last morphism is proper and hence is a closed map in particular.
The hypothesis $XP = X$ implies that $\pi_P^{-1}(\pi_P(X)) = XP = X$. Hence, $\pi_P^{-1}(\pi_P(X))$ is closed. Thus $\pi_P(X)$ is a closed subset of $G/P$. It follows that $\overline{X} := \pi_{P,Q}(\pi_P(X))$ is a closed subset in $G/Q$, since $\pi_{P,Q}$ is a closed map.
Thus, $\pi_Q^{-1}(\overline{X})$ is a closed set in $G$. But, $\pi_Q^{-1}(\overline{X}) = XQ$. We are done!
Note that we did not need the fact that $X$ is a subgroup.