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.

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
