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.



Existence of simultaneously normal finite index subgroups
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.


