quotient group G/N order and isomorphic group Consider group
$$G= \left\{ 
\begin{pmatrix}
a & b\\ 
0 & a^{-1} \end{pmatrix} \colon a,b\in \mathbb{R}, a> 0\right \}$$
with usual matrix multiplication.
Let 
$$N= \left\{ 
\begin{pmatrix}
1 & b\\ 
0 & 1 \end{pmatrix} \colon b\in \mathbb{R}\right \}.$$
Question is to verify following two statements


*

*$N$ is normal subgroup of $G$ and quotient group $G/N$ is of finite order

*$N$ is normal subgroup of $G$ and quotient group $G/N$ is isomorphic to $\mathbb{R}^{+}$
My attempt:
I have checked that $N$ is a subgroup and it is also a normal subgroup.
My doubt:
 I need suggestions to 
1. find  $G/N$.
2. Then how to see if it has finite order or not and also how to see if it is isomorphic to $\mathbb{R}^{+}$
thanks
 A: Statement (1) is wrong; $G/N$ is actually of infinite order (which is in Statement (2))
Consider the group 
$$D=\left\{ \begin{pmatrix} a & 0 \\ 0 & a^{-1} \end{pmatrix} \colon a\in\mathbb{R}^+\right\}.$$
Show that $D\cong  \mathbb{R}^+$. 
Consider the natural map from $G$ to $D$ 
$$\begin{pmatrix} a & b \\ 0 & a^{-1}\end{pmatrix} \mapsto 
\begin{pmatrix} a & 0 \\ 0 & a^{-1}\end{pmatrix}.$$
Check that this is a surjective homomorphism with kernel $N$, which will answer your question.
A: There is a shortcut way of doing this, and it's an important idea to remember.

A subgroup $N$ of a group $G$ is normal if and only if it is the kernel of some homomorphism $\varphi:G\rightarrow H$.

You already know this in the form of the first isomorphism theorem, with $H=G/N$ and $\varphi$ surjective, but sometimes there can be other, more obvious choices for $H$ if you take a step back and allow $\varphi$ to not be surjective. (Then, to use the isomorphism theorem, you instead consider $\varphi(G)$, which will be proper in $H$.)
But let's get down to brass tacks: in this case, you want to figure out a good choice for $H$ and $\varphi$ so that $N$ ends up being the kernel of $\varphi:G\rightarrow H$. What do you have to do to $\left(\begin{array}{ll}a&b\\0&a^{-1}\end{array}\right)$ so that it is sent to the identity if and only if $a=1$? What kind of map can you make to accomplish that?
After you figure the previous part out, my second hint is that you can find another homomoprhism (an isomorphism) from $G/H$ to $\mathbb{R}$.  I think this part will be obvious once you get the previous part.
