Existence of simultaneously normal finite index subgroups 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.
 A: This is not an answer to the question but just a remark. In 0708.4327, Proposition 7.3 says:

Let $G$ be a countable discrete group and let $H_1, H_2 \subset G$ be finitely
  generated infinite subgroups. Assume that $[H_1 : H_1 \cap H_2]$ and $[H_2 : H_1 \cap H_2]$ are finite. If $\beta_1^{(2)}(\langle H_1, H_2 \rangle ) \neq 0$, then the inclusion $H_1 \cap H_2 \subset \langle H_1, H_2\rangle$ has finite index.

Hence, assuming $\beta_1^{(2)}(\langle H_1, H_2 \rangle ) \neq 0$ one gets that $K$ has finite index in $\langle H_1, H_2\rangle$ and one is back in the classical case.
For a group $G$, the quantity $\beta_1^{(2)}(G)$ is called first $\ell^2$-Betti number and takes values in $[0,\infty]$. It vanishes for amenable groups and is non-zero for free groups, more precisely: $\beta_1^{(2)}(F_n) = n-1$. Unfortunatelly, the first $\ell^2$-Betti number tends to vanish in many cases.
A: There are cases in which this does hold.  Greenberg proved that if $H_1$ and $H_2$ are Fuchsian groups with a common finite-index subgroup then each $H_i$ is of finite index in $\langle H_1,H_2\rangle$.  I've no doubt that this is known more generally for quasiconvex subgroups of word-hyperbolic groups, although a reference currently eludes me.
Further remark.  Of course, Greenberg's theorem follows from the $l_2$-Betti number result that Andreas mentioned.  But there are word-hyperbolic examples, such as fundamental groups of hyperbolic 3-manifolds, with $b^2_1=0$. 
A: 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.
A: I ran across this question and thought I would generalize Henry Wilton's positive example. If $G$ is a word hyperbolic group, and if $H_1,H_2$ are two quasiconvex subgroups of $G$ which have the same limit set $\Lambda$ in the Gromov boundary of $G$, then the answer is positive. This is because $H_1,H_2$ necessarily have finite index in the stabilizer subgroup of $\Lambda$.
A: The answer is positive if we add a bit more symmetry in the assuptions (and if I made no mistake !). There is a beautiful theorem by Schlichting :
Theorem (Schlichting). Let $G$ be a group
and $\mathfrak H$ a family of subgroups of $G$. Assume that the index $H/H\cap K$ remains
bounded for any members $H$ and $K$ of $\mathfrak H$. Then there is a subgroup $N$ of $G$ invariant under the group of
automorphisms of $G$ fixing $\mathfrak{H}$ setwise such that $N/N\cap H$ and $H/ N\cap H$ remain bounded for any $H$ in
$\mathfrak{H}$.
In every proof I know of Schlichting's Theorem (there are 3 or 4), the subgroup $N$ is obtained as a finite extension of a finite intersection of member of $\mathfrak H$. One can actually do better :
Claim 1. The subgroup $N$ obtained in Schlichting's Theorem is the intersection of finitely many members of $\mathfrak H$.
Corollary 1. $G$ is a group, $H_1,\dots,H_n$ are subgroups of $G$, and $H$ is a subgroup of every $H_i$ such that $H_i/H$ is finite. If every $H_i$ normalises $\bigcap_{i=1}^n H_i$, then $H$ has a subgroup of finite index wich is normal in every $H_i$.
Proof of Corollary 1. Let $\mathfrak H$ the set of $\langle H_1,\dots, H_n\rangle$-conjugates of $H$. By Schlichting's theorem applied to the family $\mathfrak H$ inside the group $I=\bigcap_{i=1}^n H_i$, there is a subgroup $N$ of finite index in $I$ which is normal in every $H_i$. By Claim 1, $N$ is a finite intersection of $\langle H_1,\dots, H_n\rangle$-conjugates of $H$ so it must be a subgroup of $H$.
Let us proove Claim 1 now.
Definition 1 (Wagner). Let $\mathcal L$ be a lattice. A rank on
$\mathcal L$ is a function from $\mathcal L^2$ to $\mathbf N\cup\{\infty\}$
satisfying the following properties:
1) There is some $k$ in $\mathbf N$ such that $\delta(a,a)$ equals $k$ for all $a$ in
$\mathcal L$.
2) $\delta$ is increasing in the first argument and decreasing in the second.
3) If $a'\geq a\geq a\geq b\geq b'$ are elements of $\mathcal L$, if $\delta(a,b)$
and $\delta(a',b')$ are equal and finite, then $a=a'$ and $b=b'$.
4) There is an increasing function $g$ from $\mathbf{N}^2$ to $\mathbf N$ such that
whenever $a\geq b\geq c$ and both $\delta(a,b)$ and $\delta(b,c)$ are finite, then
$\delta(a,c)\leq g(\delta(a,b),\delta(b,c))$.
5) There is an increasing function $f$ from $\mathbf N^2$ to $\mathbf N$ such that
whenever both $\delta(a,c)$ and $\delta(b,c)$ are finite, then $\delta(a\lor b,c)\leq
f(\delta(a,c),\delta(b,c))$.
Theorem 1 (Wagner). $\mathcal L$ is a lattice with rank $\delta$
and $\mathfrak F$ is a family of elements in $\mathcal L$ such that
$\delta(a_0,a_1\land\dots\land a_i)\leq n(i)$ for all $i$ in $\mathbf N$ and
$a_0,\dots,a_i$ in $\mathfrak F$. There is some $f$ in $\mathfrak F$ and $m$ in
$\mathbf N$ such that $f$ is fixed by all automorphisms of $\mathcal L$ fixing $\mathfrak
F$ setwise and leaving $\delta$ invariant, with in addition $\delta(a,f)\leq m$ and
$\delta(f,a)\leq f(n(1),f(k,n(1)))$. More precisely, $f$ is a finite intersection
$\bigwedge_{i=1}^ma_i\lor a_{m+1}$ where $a_0,\dots,a_m,a_{m+1}$ are finite intersections
of members of $\mathfrak F$.
Using Mark Sapir's idea here, one can say:
Lemma 1. $G$ is a group, $H$ and $K$ are subgroups of $G$ and $\sigma$ is an
group automorphism of $G$. If $\sigma$ stabilises $H\cup K$ setwise, then either $H\cup K$ is a group
or $\sigma$ stabilises $H\cap K$.
Proof of Lemma 1. The group $\sigma H$ is the union of the two
groups $\sigma H\cap H$ and  $\sigma H\cap K$. This can happen only if either $\sigma
H=\sigma H\cap H$ or $\sigma H=\sigma H\cap K$. In the first case, $\sigma H\subset H$,
and in the second $\sigma H\subset K$. Similarly, either $\sigma K\subset K$, or $\sigma
K\subset H$. They are four cases to deal with and in fact two by symmetry. Assume first that $\sigma
K\subset H$ and $\sigma H\subset H$. Then $K\subset\sigma^{-1}H$ and
$H\subset\sigma^{-1}H$. As $\sigma^{-1}$ stabilises $H\cup K$, we either have
$\sigma^{-1}H\subset H$ or $\sigma^{-1}H\subset K$, so the first case leads to $K\subset
H$ or $H\subset K$. Second case : $\sigma K\subset H$ and
$\sigma H\subset K$ which yield $\sigma (H\cap K)\subset (H\cap K)$. This also holds for
$\sigma^{-1}$, so $\sigma(H\cap K)=H\cap K$.
Proof of claim 1. We call a coset in $G$ a left coset $gH$ of any
subgroup $H$ of $G$ and consider the set $\mathcal C$ of finite unions of cosets in $G$. By
convention, an empty union is empty so the emptyset is in $\mathcal C$. The set $\mathcal
C$ is partially ordered by inclusion. If $gH$ and $gK$ are two cosets in $G$, the intersection $gH\cap gK$ is either empty or a coset of $H\cap K$. Equiped with intersection and union, $\mathcal C$ forms a distributive lattice.
For any $A$ in $\mathcal C$, we say that a subgroup $H$ of $G$ is represented in $A$ if $gH\subset A$ for some $g$ in $G$. If $B\subset A$ and $B$ is non empty, the group $\{1\}$ is represented in $B$, so it is always possible to find a familly $(H_i)_{i\in I}$ of subgroup represented in $B$ and elements $(g_i)_{i\in I}$ in $G$ such that$\bigcup_{i\in I} g_i H_i\cup B=A$. We call such a union a $B$-covering of $A$ of size $|I|+1$. If $B\subset A$, we define the rank $\delta(A,B)$ to be the minimal size of $B$-coverings of $A$. For arbitrary $A$ and $B$ in $\mathcal C$, we extend the definition by putting $\delta(A,B)$ equal to $\delta(A,A\cap B)$. Note that every automorphism of $G$ leaves $\delta$ invariant. If $H$ and $K$ are two subgroups of $G$, then $\delta(H,K)$ is the usual index $[H:H\cap K]$. It is not difficult to check that this is indeed a rank in the sense of Definition 1, taking addition and multiplication for $f$ and $g$.
Now consider $\mathcal L$ the sub-lattice of $\mathcal C$ generated by the elements in
$\mathfrak H$. The number $\delta(H,H')$ is bounded by $n$ for every $H,H'$ in $\mathfrak
H$. By Theorem 1, there is some $L$ in $\mathcal L$  and a natural number $p$ such
that $\delta(H,L)$ is at most $p$ and $\delta(L,H)$ is at most $2n+1$ for all $H$ in
$\mathfrak H$, and such that $L$ is invariant by any automorphism of $G$ fixing
$\mathfrak H$ setwise. But, as the lattice is distributive, $L$ is also the union of two
groups $A$ and $B$ commensurable with $\mathfrak H$. By Lemma 1 either $A\cup
B$ or $A\cap B$ do the job.The End
Corollary 2. Your question has a positive answer if and only if there is at least one subgroup $L$ of $G$ commensurable with $H_1$ and normalised by every $H_i$.
Proof. If $K$ is of finite index in every $H_i$, set $I=L\cap K$ and apply Schlichting to the set of $\langle H_1,\dots, H_n\rangle$-conjugates of $I$ inside $L$.
