Constructing a homomorphism such that a given set is the kernel. Given the matrices with coefficients in $\mathbb Z_5$, I am asked to consider:
$$G= \left\{\begin{pmatrix}
a & 0 \\ 0 &d 
\end{pmatrix} \mid ad\neq 0 \bmod 5 \right\}$$
and:
$$ H= \left\langle \begin{pmatrix}
1 & 0 \\ 0 &-1 
\end{pmatrix}, \begin{pmatrix}
-1 & 0 \\ 0 &1 
\end{pmatrix}\right \rangle$$
And then show that $G/H$ is a group. The standard approach would be to show that $H$ is the kernel of a homomorphism. The question would be, how would I construct such a  group map/homomorphism?
I know $H$ consists of the elements:
$\begin{pmatrix}
1 & 0 \\ 0 &-1 
\end{pmatrix} ^2 =\begin{pmatrix}
-1 & 0 \\ 0 &1 
\end{pmatrix} ^2 = \begin{pmatrix}
1 & 0 \\ 0 &1 
\end{pmatrix} $. 
$ \begin{pmatrix}
  1 & 0 \\ 0 &-1 
 \end{pmatrix} = \begin{pmatrix}
 1 & 0 \\ 0 &4 
 \end{pmatrix}$
$ \begin{pmatrix}
-1 & 0 \\ 0 &1 
\end{pmatrix}=\begin{pmatrix}
4 & 0 \\ 0 &1 
\end{pmatrix} $ 
$\begin{pmatrix}
-1 & 0 \\ 0 &1 
\end{pmatrix}\begin{pmatrix}
1 & 0 \\ 0 &-1 
\end{pmatrix}= \begin{pmatrix}
-1 & 0 \\ 0 &-1 
\end{pmatrix} = \begin{pmatrix}
4 & 0 \\ 0 &4 
\end{pmatrix}   $
I would like to map all these elements to the matrix 
$\begin{pmatrix}  1 & 0 \\ 0 & 1\end{pmatrix}$
But I have trouble coming up with a good map of sorts.

Consider the homomorphism $\phi:G\to H$ given by $\phi(g)=g^2\ \forall g\in G$. $\ker(\phi)=H$ since $H$'s elements have only $\pm1$ on the diagonal, which square to 1, yielding the identity matrix (in contrast, $(\pm2)^2=-1$ so the other elements of $G$ are not in the kernel). Thus $G/H$ is a group.
$G$ happens to be abelian, so all its subgroups are normal, including $H$.
