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:

Does there exist a finite-index subgroup $\Gamma \le \operatorname{SL}_2(\mathbf{Z})$ for which every cusp is irregular?

(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!)