## 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.

-

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.

-
More generally, if $k$ is non-arch. local field and $G$ is conn'd semisimple $k$-gp which is $k$-split and simply conn'd (e.g., ${\rm{SL}}_d$, ${\rm{Sp}}_ {2g}$, Spin gps, etc.) then Tits proved proper open subgps of $G(k)$ are compact (since $G(k)$ is gen'td by subgps $U(k)$ for unip. radicals $U$ of Borel $k$-subgps, due to the hypotheses on $G$). So max'l compact open subgps are max'l subgps. Same then holds for $G(k)/Z$, where $Z$ is finite center of $G(k)$, but $G(k)/Z$ is simple. So distinct max'l compact subgps $H_1$ and $H_2$ of $G(k)/Z$ (which always exist) are always a counterex. – BCnrd Aug 5 2010 at 10:47
Just a comment that when $d=2$, $G$ acts on the Bass-Serre tree of similarity classes of rank 2-$\mathbb{Z}_p$ lattices in $\mathbb{Q}_p^2$, and $H_i$ are stabilizers of distinct vertices of this tree. So this shows that there is an example which comes from automorphism groups of trees, as suggested to Tao. Also, this demonstrates that $G$ is an amalgamated product of conjugates of $H_1$ fixing adjacent vertices, amalgamated over the edge stabilizer. – Agol Aug 5 2010 at 17:03
The group of Burger and Mozes: Lattices in product of trees, Inst. Hautes Études Sci. Publ. Math. No. 92 (2000), 151--194, gives a counterexample too. It is an amalgamated product $G$ of two free groups over a subgroup $K$ which has finite indexes in both of the free groups, but the group $G$ itself if simple. – Mark Sapir Sep 3 2010 at 20:07
If $H_1,\dots H_n$ are profinite and open in the topological group G, they have arbitrarily small open normal subgroups in common if and only if they generate a residually discrete subgroup of $G$. See Corollary 4.1 of: P-E. Caprace, N. Monod, Decomposing locally compact groups into simple pieces, Math. Proc. Cambridge Philos. Soc. 150 Nr. 1 (2011), 97--128. – Colin Reid Jul 29 2011 at 13:56
As $H_1$ and $H_2$ are topologically finitely generated in Yiftach's example, there are pairs of elements $P = \{x,y\}$ such that $x$ and $y$ individually are contained in open compact subgroups of $G$, but $P$ is not contained in the normaliser of any open compact subgroup. – Colin Reid Jul 29 2011 at 14:13

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.

-

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 $b^2_1=0$.

-

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 $\Lambda$.

-

Is there any qualitative version of the well known fact mentioned in the first paragraph? Namely if $[H:K]=n$ then what is the minimum index in $H$ for a proper $N$?

-
 You should ask this as a separate question. Hint: the 'obvious' upper bound (which I believe is optimal) is n!. – HW Jul 28 2011 at 23:09 @HW: How about the 'almost as obvious' upper bound $n!/2$ for $n>2$ ;-)? – Someone Jul 29 2011 at 8:21 @Someone: Think about the 'obvious' copy of $S_{n-1}$ inside $S_n$ (the symmetric group on $n$ elements). This shows that $n!$ is optimal. – HW Jul 29 2011 at 11:42 @HW: You're right. I wonder what I was thinking... – Someone Aug 2 2011 at 8:26

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 $L$.

-
 Claim 1 is false as it would imply that for any group $G$ and normal subgroup $N\leq G$, any subgroup $H\leq N$ of of finite index in $N$ has a subgroup of finite index which is normal in $G$. I can't find the mistake in the proof though. – Drike Dec 1 2011 at 17:40