Show that $G/N$ is isomorphic to $\mathbb{R}^{*} \times \mathbb{R}^{*}$

Question

I want to solve the following:

Let $G$ be the multiplicative group of elements of the form $$ \begin{pmatrix} a & b \\ 0 & c \\ \end{pmatrix} $$ where $a\neq 0, c\neq0$, $a,b,c \in \mathbb{R}$. Let $N$ be a subgroup of $G$ that consists of elements of the form $$ \begin{pmatrix} 1 & b \\ 0 & 1 \\ \end{pmatrix}. $$

Show that $N$ is a normal subgroup of $G$ and prove that $G/N$ is isomorphic to $\mathbb{R}^{*} \times \mathbb{R}^{*}$.

I have managed to show the part about $N$ being a normal subgroup. But I'm stuck on the second part.

My initial idea was to use the first isomorphism theorem, and find a surjective homomorphism $f: G \rightarrow \mathbb{R}^{*} \times \mathbb{R}^{*} $ with kernel $N$. I believe we then have to consider $\mathbb{R}^{*} \times \mathbb{R}^{*}$ an additive group, otherwise we'll never have $f(A*B) = f(A)*f(B)$, since $f(A*B)$ will be an element of the form $(a,b)$ and $f(A)*f(B)$ will be a scalar.

But I haven't been able to "construct" such a homomorphic function $f$, so if anyone has any ideas I would be greatful.

Answer 1

Consider $f:G\rightarrow \mathbb{R}^*\times\mathbb{R}^*$ defined by $f\left(\matrix{a&b\cr 0&c}\right)=(a,c)$ show that $f$ is an epimorphism and its kernel is $N$.

$\mathbb{R}^*\times\mathbb{R}^*$ is the multiplicative group.

Answer 2

Well, I'd consider the mapping $G\rightarrow {\Bbb R}^*\times {\Bbb R}^*$ defined by $$\left(\begin{array}{cc} a & b\\ 0 &c\end{array}\right)\mapsto (a,c).$$