It is well known that if $K$ is a finite index subgroup of a group $H$, then there is a finite index subgroup $N$ of $K$ which is normal in $H$. Indeed, one can observe that there are only finitely many distinct conjugates $hKh^{-1}$ of $K$ with $h \in H$, and their intersection $N := \bigcap_{h \in H} h K h^{-1}$ will be a finite index normal subgroup of $H$. Alternatively, one can look at the action of $H$ on the finite quotient space $H/K$, and observe that the stabiliser of this action is a finite index normal subgroup of $H$.
But now suppose that $K$ is a finite index subgroup of two groups $H_1$, $H_2$ (which are in turn contained in some ambient group $G$, thus $K \leq H_1 \leq G$ and $K \leq H_2 \leq G$ with $[H_1:K], [H_2:K] < \infty$). Is it possible to find a finite index subgroup $N$ of $K$ which is simultaneously normal in both $H_1$ and in $H_2$ (or equivalently, is normal in the group $\langle H_1 H_2 \rangle$ generated by $H_1$ and $H_2$)?
The observation in the first paragraph means that we can find a finite index subgroup $N$ which is normal in $H_1$, or normal in $H_2$, but it does not seem possible to ensure normality in both $H_1$ and $H_2$ simultaneously. However, I was not able to find a counterexample (though it has been suggested to me that the automorphism groups of trees might eventually provide one).
By abstract nonsense one can assume that the ambient group $G$ is the amalgamated free product of $H_1$ and $H_2$ over $K$, but this does not seem to be of too much help.
I'm ultimately interested in the situation in which one has finitely many groups $H_1,\ldots,H_m$ rather than just two, but presumably the case of two groups is already typical.
Best Answer
I think the answer to your question is no. Take $G=PSL_d(\mathbb{Q}_p)$. It is a simple group. Take $H_1=PSL_d(\mathbb{Z}_p)$ and take $H_2=H_1^g$ for some $g \in G$ so that $H_1 \ne H_2$. Now, if I am not mistaken $H_1$ and $H_2$ are maximal in $G$ so together they generate $G$. Also, $G$ commensurates $H_1$ since $H_1$ is open in $G$ and profinite. So $K=H_1 \cap H_2$ is open and of finite index in both $H_1$ and $H_2$. But as $G$ is simple, $K$ contains no non-trivial normal subgroup of $G$.
I am sure you can do something similar with Lie groups and lattices.