Let $D = \text{diag}(1,d)\in M\_{2}(\mathbb{Z})$ be a $2\times 2$ matrix, where $d$ is an odd integer. We define the subgroup $\Gamma\_D\subset M\_{4}(\mathbb{Z})$ as:
$$\Gamma\_D := \left\lbrace R\in M\_{4}(\mathbb{Z}) \: | \: R
\left(\begin{matrix}
0 & D \\ 
-D & 0
\end{matrix}\right) R^t = 
\left(\begin{matrix}
0 & D \\ 
-D & 0
\end{matrix}\right)
\right\rbrace,$$
and the subgroup $\Gamma\_D(D)\subset\Gamma\_D$ as:
$$\Gamma\_2(1,d):= \left\lbrace \left(\begin{matrix}
a & b \\ 
c & d
\end{matrix}\right)\in \Gamma\_D \: | \: a-Id\_2 \equiv b \equiv c \equiv d-Id\_2 \equiv 0 \: mod(D)
\right\rbrace,$$ 
where $A\equiv 0\: mod(D)$ means that $A = D\cdot B$ for some $B\in M\_2(\mathbb{Z})$.


Does there exists an exact sequence


$$1 \to \Gamma\_2(2,2d) \to \Gamma\_2(1,d) \to Sp(4,\mathbb{Z}/2\mathbb{Z}) \to 1,$$


where the last map is the reduction $mod\ 2$ of the entries of the matrices? In order to prove this I wanted to check the index of the level subgroups inside $\Gamma\_2$. I know that there exists a formula by Igusa for the level $n$ subgoups, and I wonder if also these indexes are known, or at least easily computable.



Index of congruence modular subgroup of level (1,d)
Yes, there is a short exact sequence


$$ 1 \to \Gamma\_2(2,2d) \to \Gamma\_2(1,d) \to Sp(4,\mathbb{Z}/2\mathbb{Z}) \to 1. $$


One can obtain this sequence as follows.
Consider the group scheme $G := Sp(f)$ defined over $\mathbb{Z}$ where $f$ is the
symplectic form defined above, i.e. $f(u,v) = u^t\begin{pmatrix} 0 & D \\ -D & 0 \end{pmatrix}v$.


The group of $\mathbb{Z}$-points is 
$G(\mathbb{Z}) = \Gamma\_D$, this is, the subgroup of matrices in $GL\_4(\mathbb{Z})$ which preserve the symplectic form.
As a group scheme over $\mathbb{Q}$ the group $G$ is just the symplectic group. However, over $\mathbb{Z}$ they are not isomorphic. Fortunately, since $d$ is odd, we have an isomorphism
$G \times \mathbb{Z}\_2 \cong Sp(4)\times \mathbb{Z}\_2$ over the ring of 2-adic integers. In particular, $G(\mathbb{Z}\_2)$ surjects onto $Sp(4,\mathbb{Z}/2\mathbb{Z})$.


So it suffices to verify that $\Gamma\_2(1,d)$ is dense in $G(\mathbb{Z}\_2)$, which follows from strong approximation.
(For instance, the principal congruence subgroup $\Gamma\_D(d) \subset \Gamma\_2(1,d)$ is dense in $G(\mathbb{Z}\_2)$ -- since $d$ is odd we obtain this group by applying a congruence condition away from $2$).


The same argument applies to all primes away from $d$, so at these primes you may also compute the index of congruence subgroups just as in the symplectic group.


