How to associate a proper parabolic subgroup of a real s.s Lie group $G$ to a non-trivial unipotent element in a non uniform lattice in $G$? Let $G$ be a real semi-simple Lie group. Let $\Gamma$ be a non-uniform lattice in $G$.
Then it is known that $\Gamma$ contains a non-trivial unipotent element. When $\mathbb{R}$-rank of $G$ is 1, it is known that a unipotent element will be contained in a unique maximal unipotent subgroup of $G$, which then will be contained in a proper parabolic subgroup of $G$.
Question is , in higher rank case, can one find a proper parabolic subgroup of $G$ containing a unipotent element of $\Gamma$?  
 A: The question can be formulated in a better way: given a non-trivial unipotent element $u$ in $G$,how to associate a proper parabolic subgroup containing the unipotent element? This is always possible: over $\mathbb C$, this is just the Borel fixed point theorem for the unipotent group acting on $G/P$ for any parabolic subgroup. Over $\mathbb R$ a more "canonical way" is to take the unipotent one parameter group $U_1$ in which your unip lies, then take its normaliser $N_1$. Let $U_2$ be the unipotent radical of $N_1$ (it contains $U_1$). Let $N_2$ be the normaliser of $U_2$. Keep repeating this process, and at some time you get a unipotent group $U=U_n$ which is the unipotent radical of the normaliser $P=N_n$: $U_n=U_{n+1}$. Then a result of Borel-Tits says that $P$ is a parabolic proper subgroup containing $U\supset U_1\supset\{u\}$.  
[Edit]I will post the precise reference to Borel Tits tomorrow since I do not have access to it at home (this is not necessary, thanks to grghxy's link). If we assume (for the sake of simplicity) that $G$ is a real simple algebraic group of real rank at least two, then a lattice is necessarily arithmetic, and since it is non-uniform (contains a unipotent element) the lattice is contained in the $\mathbb Q$ points of a $\mathbb Q$ -structure on $G$. The above procedure actually gives us a parabolic subgroup defined over $\mathbb Q$ since all the groups $N_k$ and $U_k$ are defined over $\mathbb Q$. 
