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I am interested in finding a reference for the following claim:

There exists a free subgroup $F_2$ of $\mathrm{SL}_2(\mathbb{Z})$ on two generators that does not contain any nontrivial unipotent elements.

Thanks in advance!

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    $\begingroup$ one can show that the commutator subgroup of the congruence subgroup $\Gamma (2)$ of level 2 in $SL_2(\mathbb Z)$ does not contain unipotent elements, but is free on infinitely many generators. You can then take any two elements in the generating set. $\endgroup$ Commented Jun 15, 2022 at 1:25
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    $\begingroup$ Much more is true: Every Zariski dense subgroup of $SL(n,R)$ contains a free subgroup without nontrivial unipotents. This is due to J.Tits $\endgroup$ Commented Jun 15, 2022 at 5:13
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    $\begingroup$ It seems to be the easiest instance of the ping-pong lemma, which implies that every Zariski-dense subgroup of $\mathrm{PSL}(2,\mathbf{R})$ contains a subgroup isomorphic to $F_2$ and QI-embedded in $\mathrm{PSL}(2,\mathbf{R})$. That it is QI-embedded prevents the existence of unipotents. $\endgroup$
    – YCor
    Commented Jun 15, 2022 at 8:44
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    $\begingroup$ It’s a fine question, but not appropriate for mathoverflow which is aimed at research-level questions. $\endgroup$
    – Ian Agol
    Commented Jun 15, 2022 at 11:59
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    $\begingroup$ @IanAgol - I agree that the math question is not research level, but what about the reference request? I’ve had a few times when I’ve tried (and failed) to find a perfect reference for something a bit easy… which I still did not want to prove myself… $\endgroup$
    – Sam Nead
    Commented Jun 15, 2022 at 13:16

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Consider the hyperbolic matrices $$ A = \begin{pmatrix} 2 & 1 \\ 1 & 1 \end{pmatrix} \quad \mbox{and} \quad b = \begin{pmatrix} 2 & -1 \\ -1 & 1 \end{pmatrix} $$ Working in the upper half plane model of $\mathbb{H}^2$, we take $a$ to be the (oriented!) geodesic from $0$ to $-1$ and $a'$ to be the geodesic from $1$ to $\infty$. Similarly, take $b$ to be the geodesic from $0$ to $1$ and $b'$ to be the geodesic from $-1$ to $\infty$. (It helps to draw a figure at this point.) Then $A$ takes $a$ to $a'$ and $B$ takes $b$ to $b'$, all preserving orientations. Also, the axis of $A$ is transverse to (but not perpendicular to) $a$ and $a'$; similarly the axis of $B$ is transverse to $b$ and $b'$.

We deduce that $A$ and $B$ generate a free rank two subgroup of $\mathrm{SL}(2, \mathbb{Z})$. However, their commutator $ABA^{-1}B^{-1}$ is parabolic. So, to answer the original question, we instead consider the subgroup generated by $A^2$ and $B^2$. A standard "ping-pong" argument shows that these generate a free group of rank two where all non-identity elements are hyperbolic.


I poked around in a few standard references, but did not find this exact statement. However it is "easily" deduced from material in various places. For example, you may enjoy reading Chapter 3 of Noneuclidean tesselations and their groups by Wilhelm Magnus. (Note the amazing collection of illustrations, mostly taken from the works of Fricke and Klein, starting on page 159.) In a somewhat different direction, you could use the "combination theorem" of Klein-Maskit. See Section VIII.A.1 of Kleinian groups by Bernard Maskit.

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    $\begingroup$ Schottky might be among the references? $\endgroup$
    – YCor
    Commented Jun 15, 2022 at 8:44
  • $\begingroup$ Unfortunately, I do not read German even nearly well enough. At a quick skim, Schottky's 1877 paper is mostly about uniformisation, and does not have many examples (and certainly no explicit two-by-two integer matrices...) $\endgroup$
    – Sam Nead
    Commented Jun 15, 2022 at 10:26

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