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.
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)