Abstract algebra question: abelian group. $H=\left\{\begin{pmatrix}1 & b \\ 0 & 1\end{pmatrix} : b \in\mathbb{R}\right\}$
$G=\left\{\begin{pmatrix}a & b \\ 0 & d\end{pmatrix}: a, b, d \in\mathbb{R}, ad\ne0\right\}$
$H$ is normal subgroup of $G$. Show that $G/H$ is abelian.
 A: Verifying that $H$ is a subgroup of $G$ only requires showing that the relevant properties hold:


*

*$1\in H$ (what's $1$ in this case?)

*if $x,y\in H$, then $xy\in H$ (do the multiplication)

*if $x\in H$, then $x^{-1}\in H$ (note that $\left(\begin{smallmatrix}1 & b\\0 & 1\end{smallmatrix}\right)^{-1}=\left(\begin{smallmatrix}1 & -b\\0 & 1\end{smallmatrix}\right)$)


In order to show that $H$ is normal in $G$, what's the condition to verify? Note that
$$
\begin{pmatrix}a & b \\ 0 & d\end{pmatrix}^{\!-1}=
\begin{pmatrix}d^{-1} & -b \\ 0 & a^{-1}\end{pmatrix}
$$
so that you can easily do the required computation.
Hint for $G/H$ abelian: the map $\varphi\colon G\to (\mathbb{R}\setminus\{0\})\times(\mathbb{R}\setminus\{0\})$ defined by
$$
\varphi\colon\begin{pmatrix}a & b \\ 0 & d\end{pmatrix}
\mapsto(a,d)
$$
is a homomorphism, when the codomain is considered as a group under coordinatewise multiplication.
A: Let $x,x'\in G$ with
$$
x=\begin{bmatrix}
a & b \\
0 & d
\end{bmatrix},\:\:\:
x'=\begin{bmatrix}
a' & b' \\
0 & d'
\end{bmatrix}$$
Note that
$\overline{x}\overline{x'}=\overline{xx'}=\overline{\begin{bmatrix}
aa' & ab'+bd' \\
0 & dd'
\end{bmatrix}}$ and 
$\overline{x'}\overline{x}=\overline{x'x}=\overline{\begin{bmatrix}
aa' & a'b+b'd \\
0 & dd'
\end{bmatrix}}$ so it remains to find some element $h\in H$ such that $$\begin{bmatrix}
aa' & ab'+bd' \\
0 & dd'
\end{bmatrix}h=\begin{bmatrix}
aa' & ab'+bd' \\
0 & dd'
\end{bmatrix}\begin{bmatrix}1 & s \\
0 & 1
\end{bmatrix}=\begin{bmatrix}
aa' & a'b+b'd \\
0 & dd'
\end{bmatrix}.$$
Can you find such an $s\in\mathbb{R}$. Remember that $ad\neq 0$ and $a'd'\neq 0$ so when dividing through by something, check that it's non-zero using these criteria.
