# Non-residually finite matrix groups

By Malcev's theorem, every finitely generated linear group is residually finite (RF). On the other hand, say, the group of rational numbers is linear, but is not residually finite. Thus, one has to impose some restrictions on linear groups to avoid this example. Discreteness sounds like a reasonable hypothesis. It is a nice and easy exercise to show that every discrete subgroup of $PSL(2, {\mathbb R})$ is residually finite.

Question 1. (This question came from Fanny Kassel) Are there non-RF discrete subgroups of $PSL(2, {\mathbb C})$?

On the other hand, almost surely, there are no simple discrete infinite subgroups $\Gamma$ of rank 1 Lie groups. (Take a high power of a hyperbolic element in $\Gamma$, then its normal closure in $\Gamma$ should have infinite index in $\Gamma$.) This argument fails however in the higher rank case because of the Margulis' normal subgroups theorem for higher rank lattices.

Question 2. Are there infinite simple discrete subgroups of $SL(n, {\mathbb R})$?

It is hard to imagine that such things could exist, but I see no way to rule them out...

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In Q2 you probably want to add some restrictions. Otherwise, pick any faithful complex representation of a finite simple group, and restrict scalars to $\mathbb R$ to get a simple discrete subgroup of some $\mathrm{GL}(n,\mathbb R)$ :) – Mariano Suárez-Alvarez Feb 24 '12 at 19:48
Yes, of course: I now added "infinite" to Question 2. Thanks. – Misha Feb 24 '12 at 20:16
Just for the record: Both $SL(2, {\mathbb R})$ and $SO(4,1)$ contain discrete non-RF subgroups. These discrete subgroups are central ${\mathbb Z}_2$-extensions of infinite free products of finite cyclic groups. – Misha Feb 24 '12 at 20:48
Welcome to MO, Misha! – Alain Valette Feb 25 '12 at 8:18
Maybe you should create a different question for question 2. The answer may be of quite a different nature from question 1. – Ian Agol Feb 25 '12 at 17:39

The answer to Question 1 is yes.

For each prime $p>2$, take the von Dyck group $D(p,p,\infty)$, generated by two rotations of order $p$ whose product is a parabolic $q_p$. These groups lie in $PSL(2,\mathbb{R})< PSL(2,\mathbb{C})$. We may normalize the parabolic element in each of these to be $q_p: z\mapsto z+1$. For each $p$, we may conjugate $D(p,p,\infty)$ by a parabolic element $\alpha_p: z\mapsto z+ir(p)$, where $r(p)>0$ is some sequence chosen to grow fast enough so that the resulting group $\Gamma=\langle \alpha_p^{-1} D(p,p,\infty) \alpha_p, p\ prime \rangle$ is isomorphic to the abstract infinite amalgamated product $D(3,3,\infty) \ast_{q_3=q_5} D(5,5,\infty) \ast_{q_5=q_7} D(7,7,\infty) \ast \cdots$. This is proved using the Klein combination theorem.

To see that $\Gamma$ is not residually finite, we see that $q_3$ is contained in the kernel for any homomorphism $\varphi:\Gamma \to K$, where $K$ is finite. Choose $p> |K|$, then $\varphi_{|D(p,p,\infty)}=Id$, since any element of order $p$ must be sent to the identity. So $q_p$ must lie in the kernel, but $q_p=q_3$ from the amalgamated product structure, so $\varphi(q_3)=1$.

With a bit more care (such as choosing $D(p_i,q_i,\infty)$ where $p_i,q_i$ are coprime primes $\to \infty$ as $i\to \infty$), one may guarantee that the group has no finite quotients.

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Very nice! Thanks, Ian! – Misha Feb 25 '12 at 6:53

The answer to question 2 is no [edit :] when the group is finitely generated. It follows from Malceev's result. In the case of finitely presented, here is a simple proof.

Proof : Let $G$ be a group defined by a finite number $n$ of generators and a finite number of relations.

Let $k$ be a field. Pick an element $g$ in $G$.

The set of morphisms $\varphi:G\to GL_m(k)$ such that $\varphi(g)\ne 1$ can be seen as an algebraic subvariety of $M_m(k)^n$. In fact this variety is defined over $\mathbf Z$.

So if this variety admits a point over a field $k$ of characteristic $0$, then it has to have points over infinitely many finite fields.

Thus if the group is simple and infinite, the variety is empty.

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In the infinitely presented finitely generated case use the Hilbert basis theorem. – Mark Sapir Feb 25 '12 at 16:11
Yes, of course. The nontrivial case is of infinitely generated groups. The point is Question 2 is that we do not know how to construct infinite quotients because of Margulis and we do not know how to construct finite quotients because of counter-examples in Question 1. – Misha Feb 25 '12 at 17:26
Misha, Since the subgroup need not be algebraic, Margulis' theorem probably does not apply here. There are lots of infinitely simple linear groups. The problem is whether one of them can be discrete. One can probably use some model theory here (Alexander Borovik and Oleg Belegradek). – Mark Sapir Feb 25 '12 at 18:26