The Hecke group of level two, $\Gamma\_{0}(2)$, is an index-$2$ subgroup of the Fricke group of level two, $\Gamma\_{0}^{+}(2)$, i.e. $\left[\Gamma\_{0}^{+}(2):\Gamma\_{0}(2)\right] = 2$. The index of $\Gamma\_{0}(2)$ in the modular group, $\text{SL}(2,\mathbb{Z})$ is $\left[\text{SL}(2,\mathbb{Z}):\Gamma\_{0}(2)\right] = 3$. Using the multiplicative property of indices, we find that the index of $\Gamma\_{0}^{+}(2)$ in the modular group turns out to be fractional, i.e.
$$\mu = \left[\text{SL}(2,\mathbb{Z}):\Gamma\_{0}^{+}(2)\right] = \frac{\left[\text{SL}(2,\mathbb{Z}):\Gamma\_{0}(2)\right]}{\left[\Gamma\_{0}^{+}(2):\Gamma\_{0}(2)\right]} = \frac{3}{2}.$$
How can this be given that the indices of all other Fricke groups (of prime levels) in the modular group are integers?



Fractional group index?
$\Gamma\_0^+(2)$ is not a subgroup of ${\rm SL}\_2(\mathbb{Z})$: the elements of $\Gamma\_0^+(2)$ not in $\Gamma\_0(2)$ are fractional linear transformations such as $z \mapsto -1/(2z)$ that are represented by integer matrices (such as $\left(0\;-1\atop2\;\phantom-0\right)$) of determinant $2$. [Likewise for $\Gamma\_0^+(p)$ with $p$ an odd prime, even though the "index" $(p+1)/2$ turns out to be an integer in that case.]


