Show that $K\triangleleft G$ and $G/K\cong\mathbb{R}^*\times\mathbb{R}$

Let $G=\left\{ \begin{bmatrix} a & b \\ 0 & c \\ \end{bmatrix} \middle| \, a,b,c\in\mathbb{R};\,a\ne0,\,c\ne0 \right\}$. If $K=\left\{ \begin{bmatrix} 1 & b \\ 0 & 1 \\ \end{bmatrix} \middle| \, b\in\mathbb{R} \right\}$, show that $K\triangleleft G$ and $G/K\cong\mathbb{R}^*\times\mathbb{R}$.

I proved that $G$ is a normal subgroup of $G$ relatively easy. But now I am trying to prove that $G/K\cong\mathbb{R}^*\times\mathbb{R}$. I am not sure exactly what $\mathbb{R}^*$ is. I am sure this will require the isomorphism theorem. So I'll need a homomorphism $G$ to some codomain and I think I'll need to show that $K$ is the kernel of that homomorphism. But I am not sure exactly what to do.

• $\mathbb{R}^* = \mathbb{R} - \{0\}$. That is, it is the set of real numbers without $0$. – Decaf-Math Jan 22 '16 at 5:23
• Add the operation 'multiplication' also in the comment@pyrazolam. – Error 404 Jan 22 '16 at 5:42
• @Burgundy What operation is associated with $G$? – Error 404 Jan 22 '16 at 5:49
• I assumed the operation of $G$ is matrix multiplication since $a\ne0$ and $c\ne 0$. So the inverse would exist. – Burgundy Jan 22 '16 at 6:05

Apart from looking at elements of order $2$ (which is a very elegant way to see the isomorphism cannot hold), you can also count elements. This is not about the reals, but should work in any field, so take a field $F$ with $q$ elements. We have $|G|=q(q-1)^2$ and $|K|=q$, hence $|G/K|=(q-1)^2$. Thus $G/K \cong F^* \times F$ cannot hold.
These heuristics suggest $G/K \cong F^* \times F^*$ and this turns out to be true, since $K$ is the kernel of the surjective map $$G \to F^* \times F^*, \begin{pmatrix}a&b\\0&c\end{pmatrix} \mapsto (a,c)$$ This map can easily checked to be a group homomorphism (basically this is saying that diagonal elements of a triangular matrix multiply if you multiply the matrices)
$\mathbb{R}^{\star} \times \mathbb{R}$ has only one element of order $2$, namely $(-1,0)$, whereas $G/K$ has at least two elements of order $2$, namely $$\begin{pmatrix} -1&0\\ 0&1 \end{pmatrix} \qquad \begin{pmatrix} 1&0\\ 0&-1 \end{pmatrix}$$