# Subgroup of $\text{PSL}(2,\mathbb{Z})$ generated by $S$ and $T^2$

In the group $$PSL(2,\mathbb{Z})$$ (which acts on the upper half plane of $$\mathbb{C}$$), suppose $$S$$ is the inversion and $$T$$ is the translation by $$1$$, i.e. $$S= \left( {\begin{array}{cc} 0 & 1 \\ -1 & 0 \\ \end{array} } \right), T= \left( {\begin{array}{cc} 1 & 1 \\ 0 & 1 \\ \end{array} } \right).$$

From books on modular forms, the group $$PSL(2,\mathbb{Z})$$ is generated by $$S$$ and $$T$$. Let $$G$$ by the subgroup generated by $$S$$ and $$T^2$$, $$G=\langle S,T^2\rangle$$ Is $$G$$ a normal subgroup of $$PSL(2,\mathbb{Z})$$? Is $$[PSL(2,\mathbb{Z}):G]$$ finite? If so, are there studies on the modular forms with congruence group $$G$$?

Similarly one could ask the questions when $$G=\langle S,T^k\rangle, k\in \mathbb{Z}+$$!

It's of infinite index and generates $$\mathrm{PSL}_2(\mathbf{Z})$$ as normal subgroup (hence is not normal).
Let me first check that it generates $$\mathrm{PSL}_2(\mathbf{Z})$$ as normal subgroup. Write $$U=TS$$. (All equalities are meant in $$\mathrm{PSL}_2(\mathbf{Z})$$.) So $$U^3=1$$. Modding out $$\langle S,U\rangle$$ by the relators $$S,(US^{-1})^2$$ is equivalent to mod out by $$S,U^2$$, and hence to mod out by $$S,U$$ since $$U^3=1$$. Hence the quotient is trivial, which is precisely the given assertion.
Next, let me check that it has infinite index. I use that $$\mathrm{PSL}_2(\mathbf{Z})$$ is the free product of $$\langle S\rangle$$ and $$\langle U\rangle$$.
Every free product $$A\ast B$$ can be written as $$B^{\ast A}\rtimes A$$, where the free factors in the kernel are the $$aBa^{-1}$$ when $$a$$ ranges over $$A$$. Here we thus write $$\mathrm{PSL}_2(\mathbf{Z})= (\langle U\rangle\ast \langle SUS^{-1}\rangle)\rtimes \langle S\rangle$$. Write $$V=SUS^{-1}$$: then $$T^2=(US^{-1})^2=US^{-1}US=UV$$ and $$SUS^{-1}=VU$$. Thus, in the free subgroup $$F$$ of index two $$\langle U,V\rangle$$, the intersection with $$G$$ is generated by $$UV,VU$$.
Let us check that it has infinite index. Indeed, inside $$F$$, modding out by $$UV,VU$$ yields an infinite cyclic group. So the normal subgroup of $$F$$ generated by $$G\cap F$$ has infinite index, and hence $$G\cap F$$ has infinite index in $$F$$, which in turn implies that $$G$$ has infinite index in $$\mathrm{PSL}_2(\mathbf{Z})$$.
Maybe you can check what this strategy provides when replacing $$T^2$$ with $$T^k$$.