Show that $G/H\cong\mathbb{R}^*$. 
Let $G:=
\bigg\{\left( \begin{array}{ccc}
a & b  \\
0 & a  \\
\end{array} \right)\Bigg| a,b \in \mathbb{R},a\ne 0\bigg\}$.  Let $H:=
\bigg\{\left( \begin{array}{ccc}
1 & b  \\
0 & 1  \\
\end{array} \right) \Bigg| b \in \mathbb{R}\bigg\}$.
  I know that $H$ is normal subgroup of $G$.  I need to prove that $G/H\cong\mathbb{R}^*$.

My attempt:
$$G/H=\{gH\mid g\in G\}$$
$$
\left( \begin{array}{ccc}
a & b  \\
0 & a  \\
\end{array} \right)
$$
$$
\left( \begin{array}{ccc}
1 & b  \\
0 & 1
\end{array} \right)
=
\left( \begin{array}{ccc}
a & ab+b  \\
0 & a
\end{array} \right)
$$
And there are $\infty$  solutions from the form :$
\left( \begin{array}{ccc}
a & ab  \\
0 & a
\end{array} \right)
$.
$$
\begin{vmatrix}
\left( \begin{array}{ccc}
a & ab  \\
0 & a
\end{array} \right)\end{vmatrix}=\mathfrak{c}=|\mathbb{R}^*|
$$
$$\Rightarrow G/N\cong \mathbb{R}^*$$
Is it correct? is there another way to solve this?
 A: I don't quite understand your solution, but here is another way to do it: 
You can get the isomorphism from the First Isomorphism Theorem. For example, consider the map
$$
\phi: G \to \mathbb{R}^\times
$$
given by 
$$
\phi\pmatrix{a & b \\ 0 & a} = a.
$$
You have to show that this map


*

*is a homomorphism,

*is surjective,

*has $H$ as kernel.


Then the First Isomorhpism Theorem will tell you that $G / H \simeq \mathbb{R}^\times$.

If you don't know the First Isomorphism Theorem, you can define the map
$$
\psi: G/H \to \mathbb{R}^\times
$$
by
$$
\psi \pmatrix{a & b \\ 0 & a}H = a
$$
and show that this map is


*

*well defined,

*a homomorphism,
3, surjective,

*injective.


(Is you do this, then you pretty much prove the First Isomorphism Theorem in this special case.)
A: How about considering the map $\varphi:G\to\mathbb{R}^*$ sending $\begin{bmatrix}a&b\\0&a\end{bmatrix}\mapsto a$ for all $a\in\mathbb{R}^*$ and $b\in\mathbb{R}$?  Why is it a group homomorphism?  What are the kernel and the image of $\varphi$?
A: Here is the "backwards" proof:
Suppose $A = \begin{bmatrix}a&b\\0&a\end{bmatrix},A' = \begin{bmatrix}a'&b'\\0&a'\end{bmatrix}$.
Then $AH = A'H \iff a = a'$, since $AA'^{-1} = \begin{bmatrix}a/a'&(-ab')/a'^2+b/a'\\0&a/a'\end{bmatrix}$
(Note how we use the fact that $a' \neq 0$).
Note as well that $AH = (aI)H$.
Thus $a \mapsto (aI)H$ is the desired isomorphism.
