How to find a onto homomorphism between two groups? Consider the following subgroups of $\text{SL}(2,\mathbb{Z})$ :


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*$A$ the subgroup of  matrices with determinant $1$ :


\begin{bmatrix}4\mathbb{Z}+1&8\mathbb{Z}\\4\mathbb{Z}&4\mathbb{Z}+1\end{bmatrix}


*

*$B$ the subgroup of  matrices with determinant $1$ :


\begin{bmatrix}2\mathbb{Z}+1&8\mathbb{Z}\\4\mathbb{Z}&2\mathbb{Z}+1\end{bmatrix}
I want some onto homomorphism from $B$ to $A$ whose kernel is \begin{bmatrix}1&0\\0&1\end{bmatrix}\begin{bmatrix}-1&0\\0&-1\end{bmatrix}
How to get this? I have no idea how to find the map.
 A: Let $$\beta=\begin{pmatrix}
2a+1 & 8 b \\
4c&2d+1
\end{pmatrix} \in B .$$ Then $4 ad+ 2a+2d+1 -32bc = 1$, that is $2ad+a+d = 16 bc$, which yields $a \equiv b \pmod2$. Thus, either the elements on the diagonal of $\beta$ are both congruent to $1$ or both congruent to $3$ modulo $4$. In the first case I set $\varphi (\beta) = \beta$ and I call $\beta$ even, in the second case $\varphi(\beta) = - \beta$ and call $\beta$ odd. Parity behaves as usual and thus $\varphi$ is a homomorphism. Its image is $A$ since $A$ is the subgroup of the even matrices of $B$ and its kernel is $\{\pm I_2\}$. Hence
$\varphi$ is the desired homomorphism.
A: This is not an answer, but it might help in some way:
Since $C=\left\{\begin{pmatrix}
1&0\\0&1
\end{pmatrix},\begin{pmatrix}
-1&0\\0&-1
\end{pmatrix}\right\}$ is a normal subgroup of $B$, there exists a surjective morphism $g:B\rightarrow G$ such that $\frac{B}{C}=\frac{B}{\ker(g)}\cong G$. If the desired map exists, we need to show that $\frac{B}{C}\cong A$ .
Now we want to find an isomorphism $\phi: \frac{B}{C}\rightarrow A:\begin{pmatrix}
2x+1&8a\\
4b & 2y+1
\end{pmatrix}\cdot C\mapsto ?$
It has to be well-defined, so if $X,Y\in B$ are such that $XY^{-1}\in C$, then they need to get mapped to the same thing. In particular $X$ and $-X$ need to get mapped to the same element (in fact this all you need to get a well-defined function). This might give some hint on how to construct this map (which would also solve the problem). Again, this is exactly the same as the original question, but now we see that under $\phi$, $X$ and $-X$ need to get mapped to the same thing which might help to see some formula's.
