# 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

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

2. $N$ 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

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.

• I want to know how did find that there is such natural map from G to D. how do I think about these things? Commented Feb 3, 2015 at 6:23
• Take product of two elements $A,B$ in $G$; the non-diagonal entry in the product $AB$ will be quite complicated, but the diagonal entries in $AB$ are the products of diagonal entries in $A$ and $B$. Thus, sending any element of $G$ to a diagonal matrix will define a homomorphism. Commented Feb 3, 2015 at 6:41
• Thanks it makes sense atleast in this question Commented Feb 3, 2015 at 7:02

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.

• I see its not easy to guess what choice of $\phi$to make. There has to some system to find that right choice. In a exam u can't wait enough. Plz help little more Commented Feb 3, 2015 at 6:19
• @singularity The other answerer gave you the answer, so, look at that answer and then look at my hint again. The system is exactly to construct the map that sends the normal subgroup to the identity. If you review this case until you understand how this is done, then compare to another example, you will understand how to find other answers of this kind on a test. You don't have to guess the map at all: it is as simple as it can possibly be. You just change what needs to be changed to get what you want. Consider this deeply and you will understand. Commented Feb 3, 2015 at 10:14