Recall that if $\Gamma$ is a finite-index subgroup of $\operatorname{SL}\_2(\mathbf{Z})$, then a *cusp* of $\Gamma$ is an orbit of $\Gamma$ on the set $\mathbf{P}^1\_{\mathbf{Q}}$. If $-1\notin \Gamma$, then for any cusp $c$, the stabilizer of $c$ in $\Gamma$ (well-defined up to conjugacy) is an infinite cyclic group generated by an element conjugate in $\operatorname{SL}\_2(\mathbf{Z})$ to either $\begin{pmatrix} 1 & h \\ 0 & 1\end{pmatrix}$ or $\begin{pmatrix} -1 & h \\ 0 & -1 \end{pmatrix}$ for some $h \in \mathbf{N}$ (the width of the cusp $c$), and we say that $c$ is **regular** in the first case and **irregular** in the second.


My question, as in the title, is:



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> Does there exist a finite-index subgroup $\Gamma \le \operatorname{SL}\_2(\mathbf{Z})$ for which every cusp is irregular?
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(It would suffice to find a finite-index *normal* subgroup with at least one irregular cusp, since the cusps of a normal subgroup are either all regular or all irregular; but I don't know if that helps!)



Does there exist a finite-index subgroup of SL2Z with all cusps irregular?
The thrice-punctured sphere can be represented as a quotient of the upper half-plane by $\Gamma(2)$. One may take as a fundamental domain the region contained in the geodesics between $i\infty, 0, 1, -1$, where $1$ and $-1$ are identified by the "translation by 2" operator. The fundamental group of the punctured sphere is free on 2 generators, and we may choose these to be $\begin{pmatrix} -1 & 2 \\ 0 & -1 \end{pmatrix}$ (which induces translation by -2) and $\begin{pmatrix} -1 & 0 \\ 2 & -1 \end{pmatrix}$ (which rotates around the cusp at zero - you can check this by noting where the endpoints of the geodesics go). One usually sees the negatives of these matrices in descriptions of generators of $\Gamma(2)$, so they may be unfamiliar.


At any rate, this group has index 12 in $SL\_2(\mathbb{Z})$, since $\Gamma(2)$ is a direct product of $\pm 1$ with this group (see e.g., [Andy Putman's answer here](https://mathoverflow.net/questions/29700/generators-for-congruence-group-gamma2/29709#29709)). To check that it's normal, I conjugated the above matrices against the standard S and T generators by hand and looked at the results mod 4, where this group has only four representatives out of 48 invertible matrices. I'd be interested to hear a computation-free method.


