Finite groups of isometries of hyperbolic manifolds

 I have been blogging about things which one can do with all groups recently because I think it is especially nice when something can be done for all groups. Sometimes, there are restrictions on what can be done: e.g. the inverse Galois problem asks about finite groups because infinite field extensions are insane, or Mostow rigidity implies that for $n \geq 3$ a hyperbolic $n-$manifold can only have finitely many isometries. It is of course still extremely interesting to be able to show that all finite groups are realised in either of these problems.

For $n=2,3$ and any finite group $G$ previous work had shown that for any finite group there is a compact hyperbolic manifold with $G$ as its group of isometries, but this was highly specific to the low-dimensional setting. I would like to describe and sketch the proof of a fantastic result of Belilopetsky and Lubotzky that gives a complete solution to the problem in all dimensions:

Theorem 1: For any $n \geq 2$ and $G$ a finite group there exist infinitely many compact hyperbolic $n$-manifolds $M$ such that $\mathrm{Isom}(M) \cong \mathrm{Isom}^+(M) \cong G$.

Compact hyperbolic manifolds are quotients $\mathbb{H}^n/\Gamma$ for $\Gamma$ a cocompact torsion-free discrete subgroup of $H := \rm{Isom}(\mathbb{H}^n)$. It is know that the group of isometries of $M$ is isomorphic to $N_H(\Gamma)/\Gamma$. In addition to the strength and completeness of the result, the proof is an ingenious mix of group theory, number theory, and a sprinkling of hyperbolic geometry. A very similar proof to what we sketch below shows that one can replace compact with non-compact finite volume. 

The paper is relatively short but I found it extremely dense, so I'll try to unpack it to the best of my ability. I should warn the reader that there is really quite a lot of content in the paper, hence in this post as well, so some things which aren't familiar to the reader may have to be taken on faith. Very recently, Lubotzky--Stover managed to get the analogous results for complex hyperbolic manifolds of complex dimension 2, and remark that they would get the same in any dimension if they could find a non-arithmetic lattice of that dimension, which remains a big open problem. I hope to get to blogging about this at some point...

Group theory

Let \(\Gamma\) be a finitely generated group, \(\Delta\) a finite index normal subgroup of \(\Gamma\), and \(M\) is a normal subgroup of \(\Delta\) with \(\Delta/M\) being isomorphic to a free group \(F = F_r\) on \(r \geq 2\) generators. We denote by \(N = N_\Gamma(M)\) the normalizer of \(M\) in \(\Gamma\), so \(\Delta \leq N \leq \Gamma\). The group \(N\) acts by conjugation on \(F = \Delta/M\). Denote by \(C = C_N(\Delta/M)\) the kernel of this action and by \(D\) the subgroup of all elements of \(N\) which induce inner automorphisms of \(F\). Both \(C\) and \(D\) are normal in \(N\) and clearly
\[D = \Delta C.\]
Moreover, \(M\) is normal in \(D\), \(\Delta\) and in \(C\). As \(F = \Delta/M\) has a trivial center and hence intersects \(C/M\) trivially, \(\Delta \cap C = M\). So, taking mod \(M\) we get
\[
D/M = \Delta/M \times C/M.
\]
Moreover, \(\Delta/M\) is of finite index in \(D/M\) and hence \(C/M\) is a finite group.

In case that was confusing, here is a diagram:



The main theorem of this section is

Theorem 2: Let \(\Gamma\), \(\Delta\) and \(D\) be as above. For every finite group \(G\) there exist infinitely many finite index subgroups \(B\) of \(D\) with \(N_\Gamma(B)/B\) isomorphic to \(G\).


Recall the following:
 For a prime \( p \) and a group $G$ denote by \( \mathcal{L}_p(G) \) the family of all normal subgroups of \( G \) of \( p \)-power index. A group $G$ is said to be residually-$p$ if for any non-trvial element $g \in G$ there is a finite $p$-group $Q$ and a homomorphism $\phi: G \to Q$ such that $g \notin \ker(\phi)$. It is an exercise to show that this is equivalent to \(\bigcap_{N \in \mathcal{L}_p(G)} N = \{e\}\)

Theorem 3: Let \( F = F_r \) be a free group on \( r \geq 2 \) generators.

1. \( F \) is residually-\( p \) for any prime $p$.
2.  If \( \alpha \) is an automorphism of \( F \) such that \( \alpha(N) = N \) for any \( N \in \mathcal{L}_p(F) \) then \( \alpha \) is inner.

Proof sketch of Theorem 2: Denote the order of \(G\) by \(g=|G|\), and \(g'=|\operatorname{Aut}(G)|\). Also let \(d+1=[\Gamma:D]\) and \(e,\gamma_{1},\ldots,\gamma_{d}\) be representatives of the right cosets of \(D\) in \(\Gamma\), i.e. \(\Gamma\smallsetminus D=\bigcup_{i=1}^{d}D\gamma_{i}\).

Let \(x>\max\{g,\ g^{\prime}\}\) be a very large integer, to be determined later. Choose \(d+1\) primes \(p_{0}<p_{1}<\ldots<p_{d}\) with \(p_{0}\geq x^{2}\).

Now, if \(d=0\), choose a normal subgroup of \(\Delta\) of index \(p_{0}\) containing \(M\) and call it \(K\). If \(d>0\) the definition of \(K\) will be more delicate: We claim that for every \(i=1,\ldots,d\) there exists a normal subgroup \(K_{i}\supset M\) of index \(p_{i}^{\alpha_{i}}\) in \(\Delta\) for some \(\alpha_{i}\in\mathbb{N}\), such that \(K_{i}^{\gamma_{i}}\neq K_{i}\) (where \(K_{i}^{\gamma_{i}}=\gamma_{i}^{-1}K_{i}\gamma_{i}\)). Indeed, if not then conjugation by \(\gamma_{i}\) stabilizes all normal subgroups of \(\Delta\) containing \(M\) and of index \(p_{i}\)-power in \(\Delta\). As \(\Delta/M\cong F\) is residually-\(p_{i}\) (Theorem 3 item1), \(M\) is the intersection of these normal subgroups and hence \(\gamma_{i}\) normalizes \(M\) and so \(\gamma_{i}\in N_{\Gamma}(M)\). Moreover, \(\gamma_{i}\) acting on \(F=\Delta/M\) is now an automorphism of \(F\) which preserves any normal subgroup of \(F\) of \(p_{i}\)-power index. By Theorem 3 item 2, \(\gamma_{i}\) induces on \(F\) an inner automorphism and hence \(\gamma_{i}\in D\) in contradiction to the way \(\gamma_{i}\) was chosen. We define \(K:=\bigcap_{i=1}^{d}K_{i}\).

In both cases denote the index of \(K\) in \(\Delta\) by \(k\). Observe that \(k\geq x^{2}\) and \(K/M\) is a subgroup of \(\Delta/M=F=F_{r}\) of index \(k\) so by Schreier's theorem, \(K/M\) is a free group on \(1+k(r-1)\) generators.

Claim There is a subgroup \(A\) of finite index in \(\Delta\) such that the following hold:

\[N_{\Delta}(A)/A\simeq G,\ N_{\Gamma}(A)=N_{D}(A)\ \text{and}\ N_{\Gamma}(A) \simeq G\times C/M.\]

This claim is proved by a counting argument that we omit. The proof uses some other previous counting results proved by Lubotzky on the number of various types of subgroups of the free group. The fact that the various counts can be done to subgroups of any index is how one obtains infinitely many $B$.

Let us replace \(A\) by \(B=AC\). Since \(A\) and \(C\) are both in \(D\) and \(C\lhd D\), \(B\) is indeed a subgroup and it is contained in \(D\). It is also clear that \(B\cap\Delta=A\) (look at everything mod \(M!\)). We will show that \(N_{\Gamma}(B)/B\cong G\). This will finish the proof of the theorem.

First, note that if \(\gamma\in N_{\Gamma}(B)\) then \(\gamma\) also normalizes \(\Delta\) (since \(\Delta\lhd\Gamma\)) and hence:

\[A^{\gamma}=(B\cap\Delta)^{\gamma}=B^{\gamma}\cap\Delta^{\gamma}=B\cap\Delta=A,\]

so \(\gamma\in N_{\Gamma}(A)\). On the other hand every \(\gamma\in N_{\Gamma}(A)\) also normalizes \(C\), since \(N_{\Gamma}(A)\leq D\) and \(C\) is normal in \(D\). This shows that \(N_{\Gamma}(B)=N_{\Gamma}(A)\)

Some Bass-Serre theory

The standard reference for this section is Serre's book Trees. Let \( Q \) be a finite group of order \(\geq 3\) and \( T \) be a subgroup of \( Q \) satisfying the following:  
\((\ast)\) If \( N \leq T \leq Q \) and \( N \lhd Q \) then \( N = \{ e \} \).  

In particular, \( (\ast) \) implies that \( [Q : T] > 2 \).  

Let \( R = Q \ast_T Q \) be the free product of \( Q \) with itself amalgamated along \( T \) or \( R = Q \ast_T - \) the HNN-construction. There is a natural projection \( \pi : R \rightarrow Q \) whose kernel will be denoted by \( F = \text{Ker}(\pi) \). As \( Q \) is also a subgroup of \( R \) we have \( R = F \rtimes Q - \) a semi-direct product.  

Proposition 4 

1. \( F \) is a non-abelian free group.  
2.  If \( \alpha \) is an automorphism of \( R \) satisfying \( \alpha(F) = F \) and \( \alpha|_F = \text{id} \) then \( \alpha = \text{id} \).  

Proof: 1: By definition \( F \setminus \{ e \} \) does not meet any conjugate of \( Q \), so it acts freely on the tree associated to \( R \) by Bass-Serre theory. This implies that \( F \) is a free group. The rank \( r \) of \( F \) can be computed using a known formula. In particular, the condition \( [Q : T] > 2 \) implies that \( r \geq 2 \) and so \( F \) is a non-abelian free group.
The proof of 2 is a nice exercise. $\blacksquare$

Arithmetic groups

I am not really sure how much detail to give in this section. I thought about writing some posts on arithmetic groups, but I think Witte-Morris' book
https://arxiv.org/abs/math/0106063

is excellent and I won't do a better job so I'll mostly refer the reader to that text, which is where all of the material in this section comes from. 

Recall that on a linear, semisimple Lie group $G$ there is a Haar measure. If a discrete subgroup $\Gamma$ is such that the quotient $G/\Gamma$ has finite volume for the pushforward of the Haar measure then $\Gamma$ is said to be a lattice. 

One can treat the entries of $\mathrm{SL}_n(\mathbb{R)}$ as polynomial variables and hence talk about subgroups which are defined over $\mathbb{Q}$ by polynomial equations with rational coefficients. It is an important foundational theorem that when $G$ is defined over $\mathbb{Q}$ the set of its integer points $G_{\mathbb{Z}}$, i.e. the elements of the group where all the entries are integers, is a lattice, and is said to be an \emph{arithmetic lattice}. Anything commensurable to an arithmetic lattice is also defined to be an arithmetic lattice, and this is essentially the definition of arithmetic lattice, up to some further mild fiddling that can be ignored on a first reading. 

We will be interested in lattices in $H := \rm{Isom}(\mathbb{H}^n)$. There is an astonishing criterion for arithmeticity (well multiple, but we will discuss just this one). Recall that $\mathrm{Comm}_G (\Gamma) = \{ g \in G | g \Gamma g^{-1} \text{is commensurable to} \Gamma \}$.

Margulis' commensurability criterion: Let $G$ be a connected linear semi-simple Lie group with no compact factors. Then a lattice $\Gamma$ is arithmetic iff $\mathrm{Comm}_G (\Gamma)$ is dense in $G$.

Gromov - Piatetski-Shapiro construction

It turns out that for many Lie groups, basically all lattices are arithmetic, which is an amazing result not discussed here. For hyperbolic manifolds, however, very many of them aren't arithmetic. Here is an important construction of cocompact non-arithmetic lattices in \(\mathbb{H}^n\):

 One starts with two non-commensurable torsion-free arithmetic lattices \(L_{1}\) and \(L_{2}\) in \(H\), such that each of the corresponding factor manifolds \(W_{i}=L_{i}\backslash\mathcal{H}^{n}\) admits a totally geodesic hypersurface \(Z_{i}\) (\(i=1,2\)) and \(Z_{1}\) is isometric to \(Z_{2}\). Assume that \(Z_{i}\) (\(i=1,2\)) separates \(W_{i}\) into two pieces \(X_{i}\cup Y_{i}\) (the non-separating case can be treated in a similar way). Then a new manifold \(W\) is defined by gluing \(X_{1}\) with \(Y_{2}\) along \(Z_{1}\) (which is isometric to \(Z_{2}\)). In particular, \(W\) itself has a properly embedded totally geodesic hypersurface \(Z\) (isometric to \(Z_{1}\) and \(Z_{2}\)) and so \(\pi_{1}(W)=\pi_{1}(X_{1})\ast_{\pi_{1}(Z)}\pi_{1}(Y_{2})\). 
 \(\Gamma_{0}=\pi_{1}(W)\) is a non-arithmetic lattice in \(H\cong{\rm O}_{0}(n,1)\) which can be supposed to be contained in \({\rm SO}_{0}(n,1)\) (so \(W\) is orientable) and \(\pi_{1}(Z)\) is a subgroup (in fact, a lattice) in a conjugate of \({\rm SO}_{0}(n-1,1)\).

 Showing that everything glues up nicely requires attention, but all of this is done in detail in Witte-Morris' book.

Let \(\Gamma={\rm Comm}_{\,H}(\Gamma_{0})=\{g\in H\mid|\Gamma_{0}:\Gamma_{0} \cap g^{-1}\Gamma_{0}g|<\infty\}\) be the commensurability group of \(\Gamma_{0}\). Since \(\Gamma_{0}\) is non-arithmetic, the commensurability criterion implies that \(\Gamma\) is also a lattice, a maximal lattice in \({\rm Isom}(\mathcal{H}^{n})\).

Let now \(\Lambda\) be a finite index normal subgroup of \(\Gamma\) which is contained in \(\Gamma_{0}\). So \(\Lambda\) is the fundamental group of a finite sheeted cover \(W^{\prime}\) of \(W\). Pulling back the hypersurface \(Z\) to \(W^{\prime}\), we deduce that \(W^{\prime}\) also admits a properly embedded totally geodesic hypersurface and hence \(\Lambda=\Lambda_{1}\ast_{\Lambda_{3}}\Lambda_{2}\) or \(\Lambda=\Lambda_{1}\ast_{\Lambda_{3}}\) is a non-trivial free product with amalgam or an HNN-construction.

It turns out that the groups \(\Lambda_{1}\) and \(\Lambda_{2}\) are Zariski dense in \({\rm SO}(n,1)\). This implies that they satisfy something called strong approximation, which is basically that if the defining equations have solutions mod $p$ for a prime then there is a lift to a solution with integer coefficients (think Chinese remainder theorem). By the construction these groups are contained in \(\Lambda\leq\Gamma\). 

We now make a short digression on the definition of ring of definition due to Vinberg:

Let \( V \) be a finite-dimensional vector-space over a field \( F \) and \( A \) a subring of \( F \). A set \( L \subset V \) is called an \( A \)-lattice if it is a finitely generated \( A \)-submodule and if the natural linear mapping \( F \otimes_A L \to V \) is an isomorphism (apologies that lattice is used in two very different senses in this post). If \( A \) is a principal ideal ring then every \( A \)-lattice has a basis that is simultaneously a basis of the space \( V \) over \( F \). Let \( K \) be a subfield of \( F \); then the \( K \)-lattice coincides with the \( K \)-form of \( V \). \(\Delta\) denotes a family of linear transformations in \( V \). The integrally closed Noetherian ring \( A \subset F \) is said to determine the family \(\Delta\) if in \( V \) there is an \( A \)-lattice that is invariant with respect to \(\Delta\) or \(\Delta\) is determined over \( A \). If \( A \) is a principal ideal ring this determines of \(\Delta\) over \( A \) means that there is a basis in which the \(\Delta\)-transformations are represented by matrices with elements of \( A \). If \(\Delta\) is determined over \( A \) and \( B \supseteq A \), then also \(\Delta\) is determined over \( B \). Assuming that \( F \) is algebraically closed and of characteristic 0 and \( G \) is a semi-simple algebraic group, not necessarily connected, and \(\Gamma\) a Zariski-dense subgroup, Vinberg shows that the integrally closed Noetherian ring \( A \subset F \) determines the group \( \operatorname{Ad} \Gamma \) (the adjoint group of \(\Gamma\)) if and only if \( A \supseteq \operatorname{Tr} \operatorname{Ad} \gamma \) for all \(\gamma \in \Gamma\), there is a smallest field determining \( \operatorname{Ad} \Gamma \), and if this is a number field then there is a smallest ring determining the group.

Let \(\mathcal{O}\) (resp., \(\mathcal{O}_{i}\), for \(i=1,2\)) be the minimal ring of definition of \(\Gamma\) (resp., \(\Lambda_{i}\)). As \(\Lambda\) is finitely generated group, \(\mathcal{O}\) is a finitely generated ring. In fact, it is contained in some number field \(k\). This last claim is true for all lattices in \({\rm O}(n,1)\) if \(n\geq 3\) by the local rigidity of these lattices. But it also follows (for every \(n\), including \(n=2\)) for the Gromov-Piatetski-Shapiro lattices directly from their construction.

Thus \(\mathcal{O}\) is a ring of \(S\)-integers in some (real) number field \(k\). For \(i=1,2\), \(\mathcal{O}_{i}\) is a subring of \(\mathcal{O}\), so it is the ring of \(S_{i}\)-integers of some subfield \(k_{i}\) of \(k\) for a suitable finite set \(S_{i}\) of primes in \(k_{i}\). We can assume \(\Gamma\leq{\rm SO}(n,1)(\mathcal{O})\) and \(\Lambda_{i}\leq{\rm SO}(n,1)(\mathcal{O}_{i})\) for \(i=1,2\).

Geometric realisation

We will use the above setup to show

Theorem 5: For every \(n\geq 2\) there exist a maximal cocompact non-arithmetic lattice \(\Gamma\) in \(H\) with subgroups \(M\), \(\Delta\) and \(D\) satisfying the following:

1. \(\Delta\lhd\Gamma\) and \([\Gamma:\Delta]<\infty\).
2. \(M\lhd\Delta\), \(\Delta/M\) is a non-abelian free group.
3. \([\Gamma:D]<\infty\), \(D\leq N_{\Gamma}(M)\) and \(D=\{\delta\in N_{\Gamma}(M)\mid\delta\text{ induces an inner automorphism on }\Delta/M\}\).
4.  \(D\) is torsion-free and contained in some principal congruence subgroup $\Gamma(l) \leq \rm{SO}(n,1)$.

Proof: Strong approximation implies that for almost every maximal ideal \(\mathcal{P}\) of \(\mathcal{O}\) with finite quotient field \(\mathbb{F}_{q}=\mathcal{O}/\mathcal{P}\), \(q=|\mathcal{O}/\mathcal{P}|\), the image of \(\Gamma\) in \(\mathrm{PO}_{n+1}(\mathbb{F}_{q})\) contains \(\mathrm{PA}_{n+1}(\mathbb{F}_{q})\) which is of index at most two in \(\mathrm{PSO}_{n+1}(\mathbb{F}_{q})\). The same also applies to \(\Lambda\) since \(\Lambda\) is of finite index in \(\Gamma\). Moreover, \(\Lambda_{i}\) is also Zariski dense for \(i=1,2\), so a similar statement holds for \(\Lambda_{i}\) with respect to the ring \(\mathcal{O}_{i}\). 

By Chebotarev density theorem, there exist infinitely many primes \(l\) in \(\mathbb{Q}\) which split completely in \(k\) (and hence also in \(k_{i}\)). (Chebotarev was previously discussed on this blog but the reader is welcome to take this result on faith.)

Thus for every prime ideal \(\mathcal{P}\) of \(\mathcal{O}\) which lies above such \(l\), \(\mathcal{O}/\mathcal{P}=\mathcal{O}_{i}/\mathcal{O}_{i}\cap\mathcal{P}\cong \mathbb{F}_{l}\). Moreover, if we replace \(\Lambda\) by the intersection of all the index 2 subgroups in it (note that this intersection is characteristic in \(\Lambda\) and so normal in \(\Gamma\)), we can assume that for infinitely many rational primes \(l\), the images of \(\Lambda\), \(\Lambda_{1}\) and \(\Lambda_{2}\) are exactly the groups \(\mathrm{PA}_{n+1}(\mathbb{F}_{l})\).

Choose such a prime \(l\). We obtain a homomorphism

\[\pi:\Lambda\to Q=\mathrm{PA}_{n+1}(\mathbb{F}_{l})\]

with \(\pi(\Lambda_{1})=\pi(\Lambda_{2})=Q\) while \(T=\pi(\Lambda_{3})\leq\mathrm{PSO}_{n}(\mathbb{F}_{l})\) is a proper subgroup of \(Q\) (and by choosing \(l\) sufficiently large we can assume that the index of \(T\) in \(Q\) is as large as we want). For later use we observe that if \(T\) contains a normal subgroup \(N\) of \(Q\) then \(N=\{e\}\). Indeed, if \(n\neq 3\) or \(n=3\) and \(Q=\mathrm{PA}_{4}^{-}\), \(Q\) is a finite simple group and there is nothing to prove. Suppose \(n=3\), \(Q=\mathrm{PA}_{4}^{+}\cong\mathrm{PA}_{5}\times\mathrm{PA}_{5}\). The only possibility for \(N\neq\{e\}\) is \(N=\mathrm{PA}_{5}\). The image \(T\) of \(\Lambda_{3}\) in \(\Omega_{4}\) is equal to \(\mathrm{Stab}_{\Omega_{4}}(U)\), the stabilizer of a 3-dimensional subspace \(U\) of \(V=\mathbb{F}_{4}^{+}\) and hence indeed it is isomorphic to \(\Omega_{3}\), but it cannot be a normal subgroup of \(\Omega_{4}\). For if this is the case, then for every \(g\in\Omega_{4}\), \(\mathrm{Stab}_{\Omega_{4}}(gU)=gTg^{-1}=T\). This implies \(gU=U\) (otherwise \(\Omega_{3}\) would preserve a 2-dimensional subspace). Now, this means that \(U\) is \(\Omega_{4}\) invariant, which is a contradiction. (We recall that \(\mathrm{PA}_{4}^{+}\cong\mathrm{PA}_{5}\times\mathrm{PA}_{5}\cong\mathrm{PSL}(2 )\times\mathrm{PSL}(2)\) via the action of \(\mathrm{SL}(2)\times\mathrm{SL}(2)\) on the \(2\times 2\) matrices by \((g,h)(A)=gAh^{-1}\), with \(\det(A)\) being the invariant quadratic form. But, the action on the 4-dimensional space is irreducible.)

The universal property of free products with amalgam and HNN-constructions implies that there exists a homomorphism

\[\tilde{\pi}:\Lambda\to R=Q*_{T}Q\text{ (or }=Q*_{T})\]

depending on \(\Lambda=\Lambda_{1}*_{\Lambda_{3}}\Lambda_{2}\) or \(\Lambda=\Lambda_{1}*_{\Lambda_{3}}\). The group \(R\) is mapped by \(\tilde{\pi}\) onto \(Q\) with a kernel \(\bar{F}\) which is a non-abelian free group by Proposition 4 item 1. Let \(M=\mathrm{Ker}(\tilde{\pi})\) and \(\Delta=\mathrm{Ker}(\tilde{\pi}\circ\tilde{\pi})\). We have:



It is easy to see that \(M \lhd \Delta\) and \(\Delta/M \cong F\). Also, \(\Delta = \Delta \cap \Gamma(l)\) where the congruence subgroup \(\Gamma(l)\) is the kernel of the projection of \(\Gamma\) to \({\rm PO}_{n+1}(\mathbb{F}_{l})\). Thus \(\Delta\) is a finite index normal subgroup of \(\Gamma\). We therefore have the properties (1) and (2) of the theorem.

Let now \(D = \{\delta \in N_{\Gamma}(M) \mid \delta\) induces an inner automorphism on \(F \cong \Delta/M\}\). Since \(D\) contains \(\Delta\), we are left only with proving that \(D\) is torsion-free. Denote \(C = \{\delta \in N_{\Gamma}(M) \mid \delta|_{\Delta/M} = \text{id}\}\). Then \(D = \Delta C\). We will show that \(C \leq \Gamma(l)\) which will prove that \(D \leq \Gamma(l)\). As \(\Gamma(l)\) is torsion-free and contained in $\rm{SO}(n,1)$ (when \(l\) is large enough), we will deduce that \(D\) has no torsion. 

Let \(c \in C\). The element \(c\) acts on \(\Delta/M \cong R\) with restriction to \(\Delta/M \cong F\) being trivial. Such an automorphism of \(R\) is trivial by Proposition 4 item 2, so \(c\) acts trivially on \(\Delta/M\), i.e. \([c, \Delta] \subseteq M\). Taking this mod \(\Gamma(l)\) we deduce that \(c\) centralizes \(Q = {\rm PO}_{n+1}(\mathbb{F}_{l})\) in \({\rm PO}_{n+1}(\mathbb{F}_{l})\).


Claim: $c$ is trivial in $Q$

Proof: Let \( f \) be an \( m \)-dimensional quadratic form over a finite field \( \mathbb{F} \) of characteristic \( p > 2 \). If \( m \) is odd, there is a unique, up to isomorphism, orthogonal group \( O(f) = O_m(\mathbb{F}) = O_m \). If \( m \) is even there are two groups \( O_m^+ \) and \( O_m^- \) corresponding to the cases when \( f \) splits and does not split over \( \mathbb{F} \), respectively. Let \( SO(f) \), \( PSO(f) \) denote the corresponding special orthogonal and projective special orthogonal groups, and let \( \Omega(f) = [O(f), O(f)] \) be the commutator subgroup of \( O(f) \). The projective group \( P\Omega(f) = P\Omega_m^+(\mathbb{F}) \) is generally simple and is contained in \( PSO(f) \) with index at most 2. More precisely, \( P\Omega_m \) is simple if \( m \geq 5 \) or \( m = 3 \) and \( p > 3 \); in case \( m = 4 \), \( P\Omega_4^- \) is simple but \( P\Omega_4^+ = P\Omega_3 \times P\Omega_3 \) is a direct product of two groups which are simple if \( p > 3 \). Note also that the centralizer of \( P\Omega(f) \) in \( PO(f) \) is always trivial. $\blacksquare$

This implies that \(c \in \Gamma(l)\) and we are finally done. $\blacksquare$

Proof of Theorem 1: Let \( n \geq 2 \), \( G \) is a finite group, \(\Gamma\), \(\Delta\) and \( D \) are lattices as in Theorem 5. By Theorem 2, there exist infinitely many finite index subgroups \( B \) of \( D \) with \( N_{\Gamma}(B)/B \cong G \). As \(\Gamma\) is non-arithmetic and maximal, by the commensurability criterion, \(\Gamma = \text{Comm}_{H}(\Gamma) = \{ g \in H \mid |\Gamma : \Gamma \cap g^{-1} \Gamma g| < \infty \}\). We deduce that \( N_{H}(B) \leq \text{Comm}_{H}(\Gamma) = \Gamma \), so \( N_{H}(B) = N_{\Gamma}(B) \) and hence \( N_{H}(B)/B \cong G \).  \( N_{H}(B)/B \cong \text{Isom}(B \setminus \mathcal{H}^{n}) \) and all isometries are contained in $\rm{SO}(n,1)$, hence orientation preserving, so we are done. $\blacksquare$