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.

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.

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.