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 $H$.


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.


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