$[\mathrm{SL}_2(\mathbb{Z}):\Gamma_0(2)]=3$

The index of $\Gamma_0(2)$ in the modular group.

For $N\ge 1$ let

The claim is the case $N=2$ of the classical index formula $[\mathrm{SL}_2(\mathbb Z):\Gamma_0(N)]=N\prod_{p\mid N}\left(1+\tfrac1p\right)$, namely

Why it is true. Reduction modulo 2 sends $\Gamma_0(2)$ onto the group of upper triangular matrices in $\mathrm{SL}_2(\mathbb F_2)$; since $\mathrm{SL}_2(\mathbb F_2)\cong S_3$ has order 6 and its upper triangular subgroup has order 2, the index is 3.

A proof that avoids surjectivity of the reduction map goes through the first column. For $\gamma\in \mathrm{SL}_2(\mathbb Z)$ let $v(\gamma)=(\gamma_{00},\gamma_{10})\bmod 2$. Since $\det\gamma=1$, the vector $v(\gamma)$ is never $(0,0)$, so it takes one of the three values $(1,0)$, $(0,1)$, $(1,1)$, all of which occur. If $h\in\Gamma_0(2)$ then $h_{10}$ is even and $\det h=1$ forces $h_{00}$ odd, so the first column of $\gamma h$ is congruent modulo 2 to $h_{00}$ times that of $\gamma$; hence $v(\gamma h)=v(\gamma)$ and $v$ is constant on left cosets. Conversely if $v(\gamma)=v(\gamma')$ then, writing $\gamma=\begin{pmatrix}a&b\\c&d\end{pmatrix}$ and $\gamma^{-1}=\begin{pmatrix}d&-b\\-c&a\end{pmatrix}$, the lower-left entry of $\gamma^{-1}\gamma'$ is $-ca'+ac'\equiv -ca+ac=0 \pmod 2$, so $\gamma^{-1}\gamma'\in\Gamma_0(2)$. Thus $v$ induces a bijection between $\mathrm{SL}_2(\mathbb Z)/\Gamma_0(2)$ and $\mathbb F_2^2\setminus\{0\}$, a set of size 3.

Formalization note. Subgroup.index is Nat.card (G ⧸ H), so the statement asserts exactly that this coset space has three elements.