Show that $G/H\cong\mathbb{R}^*$.

Question

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?

Answer 1

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

1. is a homomorphism,
2. is surjective,
3. 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

1. well defined,
2. a homomorphism, 3, surjective,
3. injective.

(Is you do this, then you pretty much prove the First Isomorphism Theorem in this special case.)

Answer 2

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 $\varphi$?

Answer 3

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.