Question about the coset in $GL_2(\mathbb{R}) $ Considering $G=GL_2(\mathbb{R})$ the group of  invertible matrixes $2\times2$ with real entries and its following subgroup
$$H=\left\{\left(\begin{matrix}  c & 0\\0&c\end{matrix}\right)|c\in\mathbb{R}_+^* \right\} .$$
In order to show that there are infinite cosets of $H$ in $G$, I take $A, B\in G$ such that they are in the same coset, so $AB^{-1}\in H$.
If $A=\left(\begin{matrix}  a & b\\c&d\end{matrix}\right) $ and $b=\left(\begin{matrix}  r & s\\t&u\end{matrix}\right)$, then
$$AB^{-1} =\frac{1}{\det B} \left(\begin{matrix}  au-bt & - as+br\\cu-dt&-cs+dt\end{matrix}\right)\in H $$
if, and only if, $au-bt=-cs+dt$ and $cu-dt=-as+br=0$.
Another approach using the Theo Benedit's observation, if $A\in GL_2(\mathbb{R})$, a coset has the form
$$AH=\{cA; c \in \mathbb{R}^*\}. $$
Let's consider $d:= \det(A)\neq0$. If $ B\in H$, such that $\det(B) = b$ and then $ \det(AB)= \det(A)\cdot\det(B)=db \neq 0$. Then every element in the coset has non-zero determinant.
Conversely,if $M\in GL_2(\mathbb{R})$, with $\det(M)= m\neq 0$. Let $N= A^{-1}M$. Thus $M= AN$ and
$$\det(N)=\det(A^{-1}M)=\det(A^{-1})\cdot\det(M)=\frac{m}{d}\neq 0,$$
which means that $N\in H$ and $M =AN \in AH$. Hence, I proved that
$$AH=\{M\in GL_2(\mathbb{R});\det(M)= m\neq0\} $$
and since there are infinite possible choices for $ m $, we have infinite cosets.
 A: first of all, lets notice that a coset of $H$ is of the form $$D\cdot H=\left\{D\cdot \begin{pmatrix}
c & 0\\
0 & c
\end{pmatrix}\middle|\;c\in \mathbb{R}_+\right\}=
\left\{c\cdot D\cdot \begin{pmatrix}
1 & 0\\
0 & 1
\end{pmatrix}\middle|\;c\in \mathbb{R}_+\right\}=
\left\{c\cdot D \middle|\;c\in \mathbb{R}_+\right\}
$$ for some matrix $D\in GL_2(\mathbb{R})$.
Now, consider the following family of matrices:
$\left\{\begin{pmatrix}
1 & n\\
0 & 1 
\end{pmatrix}\,\middle|\; n\in\mathbb{N}\right\}\subset GL_2(\mathbb{R})$.
Now, lets assume that there exists $m,n\in \mathbb{N}$ s.t the matrices $A=\begin{pmatrix}
1 & n\\
0 & 1 
\end{pmatrix},\;
B=\begin{pmatrix}
1 & m\\
0 & 1
\end{pmatrix} 
 $
are in the same coset. than there exists matrix $D\in GL_2(\mathbb{R})$ s.t
$A=a\cdot D,\; B=b\cdot D$, for $\mathbb{R}\ni a,b>0$. so we can conclude that $\frac{1}{a}\cdot A=\frac{1}{b}\cdot B$ which implies $a=b$. it follows that $A=B$, so $m=n$.
So, each matrix from the family is in a different coset.
