Let $G$ be the group of all real matrices of the form $\displaystyle\left( \begin{smallmatrix} a & b \\ 0 & c \end{smallmatrix} \right)$ with $ac \neq 0$ under matrix multiplication. Let $H$ be the subgroup consisting of all the elements in which $a=c=1$. Use the first isomorphism theorem to show that $G/H$ is isomorphic to $(\mathbb{R}-\{0\})\times(\mathbb{R}-\{0\})$.
I'm not sure how to invoke the first isomorphism theorem.
The first isomorphism theorem tells us that if we have a map between groups
$$f \colon L \to K$$
then the image is isomorphic to the quotient by the kernel $$\operatorname{im}(f) \cong L/\ker f.$$
So to apply it to prove that $G/H \cong (\mathbf R \smallsetminus 0) \times (\mathbf R \smallsetminus 0)$ we can first construct a map
$$ f\colon G \to(\mathbf R \smallsetminus 0) \times (\mathbf R \smallsetminus 0)$$
that has kernel $H$ and is surjective.
So the question is if I give you an element of $G$ can you find a pair of non-zero elements of $\mathbf R$ to construct the map? Then try and prove that its a homomorphism and has kernel exactly $H$.