TITLE: How to find a onto homomorphism between two groups?
QUESTION [7 upvotes]: Consider the following subgroups of $\text{SL}(2,\mathbb{Z})$ :

$A$ 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.

REPLY [5 votes]: 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.