Cyclic subgroup of finite index in $\text{PSL}_2(\mathbb{Z})$

Question

As the title suggests, I am considering the question of whether $\text{PSL}_2(\mathbb{Z})$ contains any cyclic subgroups of finite index. For this recall that $\text{PSL}_2(\mathbb{Z})$ is the quotient of $\text{SL}_2(\mathbb{Z})$ (the group of all 2x2 matrices with integer coefficients and determinant one) by $\pm I$.

My intuition says such a subgroup should not exist, however, neither do I know how to prove this nor could I find an example of such a group.

Any help is very much appreciated.

Answer 1

$\left( \begin{array}{cc}1&1\\0&1\end{array} \right)^n = \left( \begin{array}{cc}1&n\\0&1\end{array} \right)$ does not commute with $\left( \begin{array}{cc}1&0\\1&1\end{array} \right)^n = \left( \begin{array}{cc}1&0\\n&1\end{array} \right)$ for any $n > 0$, so ${\rm PSL}(2,{\mathbb Z})$ has no abelian subgroup of finite index.

Answer 2

$\newcommand{\Z}{\mathbf{Z}}\newcommand{\PSL}{\text{PSL}_2}\newcommand{\SL}{\text{SL}_2}$I'll be borrowing some tools from geometric group theory. With that being said, let us show that there is no cyclic subgroup of $\PSL (\Z)$ of finite index. Let us assume towards a contradiction that such a subgroup exists and denote it by $G$. Since $[\PSL(\Z):G] < \infty$ it follows that they are quasi-isometric. In particular, both of them will have the same growth order.

On the one hand, we have that $G \cong \Z$. Thus, one has that $G$ has linear growth order.

On the other hand, it can be proved $\SL(\Z)$ has finite index subgroup isomorphic to a free group of rank 2 (you may read this thread). Therefore, $\SL(\Z)$ is quasi-isometric to a free group of rank 2. This conclusion tells us that $\SL(\Z)$ has exponential growth.

Finally, since $\PSL(\Z)$ is just a quotient of $\SL(\Z)$ by a finite subgroup, it follows that they are quasi-isometric and have the same growth order. Nevertheless, the first one has linear growth (because of the assumption we made) and the later has exponential growth.