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.

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.

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.

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.