How the quotient group of following matrix group looks like? 
Consider the following subsets of the group of $2\times 2$ non-singular matrices over $\mathbb{R}$: 
  $G=\left\{ \begin{bmatrix}a&b\\ 0&d\end{bmatrix}: a,b,d\in \mathbb{R}, ad=1\right\}$
$H=\left\{ \begin{bmatrix}1&b\\ 0&1\end{bmatrix}: b\in \mathbb{R}\right\}$
Which of the following statements are correct? 
  
  
*
  
*$G$ forms a group under matrix multiplication.
  
*$H$ is a normal subgroup of $G$.
  
*The quotient group $G/H$ is well defined and is abelian. 
  
*The quotient group $G/H$ is well defined and is isomorphic to the group of 2x2 diagonal matrices (over reals) with determinant 1.
  

So I checked that option 1 is true. Also for all $g\in G$, $g^{-1}Hg=H$ so $H$ is a normal subgroup of $G$. Option 2 is also correct. But I am stuck with option 3 and 4. Since $H$ is normal, we can talk about the quotient group $G/H$, but I don't know how this quotient group is look like ? and also the well define part. Also I need help with the 4th option. How can I do this? Any help would be great. thanks.
 A: Hint:
Let $K$ be the group of $2\times 2$ diagonal matrices (over reals) with determinant $1$.  
Define $f:G\rightarrow K$ by $$\begin{pmatrix}a&b\\0&d\end{pmatrix}\mapsto \begin{pmatrix}a&0\\0&d\end{pmatrix}$$
Try to verify that this is a surjective homomorphism with kernel $H$.
Then apply the first isomorphism theorem.
A: A coset of $H$ looks like
$$ \begin{bmatrix} a & b \\ 0 & d \end{bmatrix} H= \left\{ \begin{bmatrix} a & b \\ 0 & d\end{bmatrix}\begin{bmatrix} 1 & h \\ 0 & 1 \end{bmatrix} : ad=1 \right\} $$
Multiply out
$$ \begin{bmatrix} a & b \\ 0 & d\end{bmatrix}\begin{bmatrix} 1 & h \\ 0 & 1 \end{bmatrix}=\begin{bmatrix} a & ah+b \\ 0 & d\end{bmatrix} $$
Can you think of an $h$ that makes this matrix particularly nice? If so, you could create a very nice set of coset representatives. In particular, a set of coset representatives which is itself a subgroup.
Proposition. If $G$ is a group and $H,K$ are subgroups such that $H$ is normal and $K$ is a set of coset representatives for $H$, then $G/H\cong K$.
