Let $H<G$ be a closed Lie subgroup of the connected Lie group $G$. Let $x\in G/\Gamma$ for $\Gamma<G$ discrete subgroup.
Assume $g_i\in G$ such that $\lim g_i=e$, and also $g_ix\in Hx$ for all $i$.
A "basic exercise" (one of them, at least) my advisor gave me is to prove that $g_i\in H$ for large enough $i$.
I'd love some tips here.
What I figured alone is the following (some sketch, but missing one part): We have $g_i \in H\Gamma$ and so $g_i=h_i\gamma_i$. I would love to show that $\lim h_i=h$ and $\lim\gamma_i=\gamma$ for some $h,\gamma$ (I don't know if it's true). In that case, since $\Gamma$ is discrete subgroup it is also closed and so $\forall i\geq i_0\::\gamma_i=\gamma_{i_0}$ and so $h=\gamma_{i_0}^{-1}\in H$ so $\forall i\geq i_0\:: \gamma_i\in H$ and indeed $g_i=h_i\gamma_i \in H$.
How can I prove that $h_i$ and $\gamma_i$ indeed converge? Is there a way around it? Do I know that the $h_i$ or $\gamma_i$ must be bounded (and therefore have a converging subsequence)?
What you are asked to prove is simply false. Consider $G$, the additive group of real numbers and $H, \Gamma$ noncommensurable infinite cyclic subgroups of $G$, i.e. subgroups which generate a dense subgroup.