Show that the quotient group $T/N$ is abelian let $T= \begin{bmatrix}a&b\\0&d\end{bmatrix}, a, b, d \in \mathbb{R}, ad \neq 0$
And let $N = \begin{bmatrix}1&b\\0&1\end{bmatrix}, b\in \mathbb{R}$
I showed that N is a normal subgroup of T
Now I have to show that the quptient group $\dfrac{T}{N}$ is an abelian group (I am not sure what that quotient group represents).
I would need a hint to solve this problem
 A: Define a map $\phi:T\to\mathbb{R}^{\times}\times\mathbb{R}^{\times}$ by
$$ \phi\Big(\begin{bmatrix}a&b\\0&d\end{bmatrix}\Big)=(a,d) $$
This is a homomorphism because if $t_j=\begin{bmatrix}a_j&b_j\\0&d_j\end{bmatrix}\in T$, $j=1,2$, then
$$ \begin{bmatrix}a_1&b_1\\0&d_1\end{bmatrix}\begin{bmatrix}a_2&b_2\\0&d_2\end{bmatrix}=\begin{bmatrix}a_1a_2&a_1b_2+b_1d_2\\0&d_1d_2\end{bmatrix}$$
hence $\phi(t_1t_2)=(a_1a_2,d_1d_2)=(a_1,d_1)(a_2,d_2)=\phi(t_1)\phi(t_2)$.
Since $\mathbb{R}^{\times}\times\mathbb{R}^{\times}$ is an abelian group, all that's left to do is to show that $\phi$ is surjective, and that the kernel is $N$, then use the first isomorphism theorem.
A: Lemma. Let $H$ be a normal subgroup of a group $G$. Then $G/H$ is abelian if and only if $aba^{-1}b^{-1}\in H$ whenever $a,b\in G$.
A quick proof can be found here.
Now, to apply the lemma to our situation, let
\begin{align*}
A &=
\left[\begin{array}{rr}
a & b \\
0 & c
\end{array}\right] &
B &=
\left[\begin{array}{rr}
w & x \\
0 & y
\end{array}\right]
\end{align*}
be elements of $T$.
Then
$$
ABA^{-1}B^{-1}=
\left[\begin{array}{rr}
1 & -\frac{\frac{b w}{c} - \frac{a x + b y}{c}}{y} - \frac{x}{y} \\
0 & 1
\end{array}\right]
$$
so $ABA^{-1}B^{-1}\in T$. The lemma then implies that $G/H$ is abelian.
