Proving that $G/N$ is an abelian group 
Let $G$ be the group  of all $2 \times 2$ matrices of the form $\begin{pmatrix} a & b \\ 0 & d\end{pmatrix}$ where $ad \neq 0$ under matrix multiplication. Let  $N=\left\{A \in G \; \colon \; A = \begin{pmatrix}1 & b \\ 0 & 1\end{pmatrix} \right\}$ be  a subset  of the group $G$. Prove that  $N$ is  a normal subgroup of $G$ and prove that  $G/N$ is abelian group.

Here is my attempt!
To prove $N$ is normal I consider  the group homomorphism $f \colon G \to \mathbb R^*$ given by $f(B) = \det(B)$ for all $B$ in $G$. And I see that $f(N)$ is all the singleton $\{1\}$ since $\{1\}$ as a subgroup of $\mathbb R^*$  is normal, it follows that $N$ is also normal. Is this proof helpful  here? Then how to prove that $G/N$ is Abelian? I know $G/N$ is a collection of left cosets. 
Thank you.
 A: I would like to add one more way to solve the problem. If $A= \begin{pmatrix} a && b \\ 0 &&c \end{pmatrix}$ and $B= \begin{pmatrix} d && e \\ 0 &&f \end{pmatrix}$ then $[A,B]=ABA^{-1}B^{-1}=\begin{pmatrix} 1 & -\frac{e}{f}+\frac{\frac{-db}{c}+\frac{ae+bf}{c}}{f} \\ 0 & 1 \end{pmatrix}$. Substituting $e=0$, $f=1$, $c=1$, $b=1$ and $d=1-k$ for any given k we see that $N=G'$ is in fact the (first) derived subgroup of $G$ and therefore $G/N$ is necessarily Abelian (even the largest Abelian factor group of $G$).
Of couse this proof presumes a little more knowledge on the part of the reader than Martin's excellent solution.
A: One way is using first isomorphism theorem.
To do this you should find a group homomorphism such that $\operatorname{Ker} \varphi=N$.
Let us try $\varphi: G\to \mathbb R^*\times \mathbb R^*$ given by
$$\begin{pmatrix} a & b \\ 0 & d\end{pmatrix} \mapsto (a,d).$$
(By $\mathbb R^*$ I denote the group $\mathbb R^*=\mathbb R\setminus\{0\}$ with multiplication. By $G\times H$ I denote the direct product of two groups, maybe your book uses notation $G\oplus H$ for this.)
It is relatively easy to verify that $\varphi$ is a surjective homomorphism. It is clear that $\operatorname{Ker} \varphi=N$. Hence, by the first isomorphism theorem,
$$G/N \cong \mathbb R^*\times\mathbb R^*$$
This is a commutative group.

If you prefer, for any reason, not using the first isomorphism theorem, you could also try to verify one of equivalent definitions of normal subgroup and then describe the cosets and their multiplication.
In this case you have
$$\begin{pmatrix} a & b \\ 0 & d \end{pmatrix}
\begin{pmatrix} 1 & b' \\ 0 & 1 \end{pmatrix}
\frac1{ad}
\begin{pmatrix} d & -b \\ 0 & a \end{pmatrix}=
\begin{pmatrix} 1 & \frac{ab'}d \\ 0 & 1 \end{pmatrix}$$ 
(I have omitted the computations), which shows that $xNx^{-1}\subseteq N$ for any $x\in G$.
You can find out easily that cosets are the sets of the form
$$\{\begin{pmatrix} x & y \\ 0 & z \end{pmatrix}; y\in\mathbb R\}$$ 
for $x,z\in\mathbb R\setminus\{0\}$ and that the multiplication of cosets representatives $\begin{pmatrix} x & 0 \\ 0 & z \end{pmatrix}$ is coordinate-wise.
A: This can another solution.
Consider the relation of equivalence: 
$$a\equiv b\pmod H \iff a^{-1}b\in H$$ 
where $a$ and $b$ are elements of a group $G$ while $H$ is a subgroup of $G$ Note that the equivalent class $[a]=aH$ the left coset of $a$ (generic element of this group). Note that every triangular upper matrix is congruent to diagonal matrix. Indeed if $$A=\begin{pmatrix} a & b \\ 0 & d \end{pmatrix}$$ then $A$ is congruent to: $$
\begin{pmatrix} a & b' \\ 0 & d \end{pmatrix}$$ in the preceding equivalence relation. Therefore every element of $G$ stands for a diagonal matrix in the equivalence class : $$Nab=Nba$$ because $a$ and $b$ are diagonal matrix but diagonal matrix commute
A: We can apply the property of the determinant to show that the normal subgroup test is satisfied. 
For the second part, we can use the fact that G is abelian since it forms a diagonal matrix. Then we show that elements in the factor group commute.
A: We want to show that $G/N$ is abelian.
That is the product any 2 left cosets of N commute.
I.e $(Na)(Nb)=(Nb)(Na)$ for every $a,b\in G$.
Note that $(Na)(Nb)=Nab$ and $(Nb)(Na)=Nab$
In other words we need to show that $Nab=Nba$ for  every $a,b\in G$.
Thus It would be enough to prove that 
$aba^{-1}b^{-1}\in N$
(If $Nx=Ny \Rightarrow xy^{-1}\in N$.)
Try It Out...I Got :)
