Prove $G/H \cong (\mathbb{R}-\{0\})\times(\mathbb{R}-\{0\})$

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.

• – Lord Shark the Unknown Nov 24 '18 at 4:57
• The homomorphism defined seems to be incorrect. – Yadati Kiran Nov 24 '18 at 4:59
• $H$ should be the kernel of the mapping. – Yadati Kiran Nov 24 '18 at 9:32

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 $$H$$.

HINT: Define $$f:G \to (\mathbb{R} - \{0\})\times(\mathbb{R}-\{0\})$$ by $$f\left(\begin{smallmatrix} a & b \\ 0 & c \end{smallmatrix}\right)=(a,c)$$. Show that it's an epimorphism and then use first isomorphism theorem.